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APBIO-U04-L01 A molecule released into the blood

Topic 4.1 · Cell Communication · 84 steps

A black-and-white photograph of a startled young woman, eyes wide and mouth open, a hand raised to her face; beside it a schematic: the heart marked, the two adrenal glands marked above the kidneys, dots of the released molecule in a blood vessel, and the lens of the eye marked
A black-and-white photograph of a startled young woman, eyes wide and mouth open, a hand raised to her face; beside it a schematic: the heart marked, the two adrenal glands marked above the kidneys, dots of the released molecule in a blood vessel, and the lens of the eye marked

Photo: D. Sharon Pruitt, Wikimedia Commons, CC BY 2.0 (cropped and resized).

A door slams behind you in a dark hallway. Within half a minute your heart is hammering and your hands shake.

Nothing touched your heart. Two small glands on top of your kidneys released a molecule into your blood, and cells all over your body detected it. The lens of your eye is bathed in fluid filtered from that same blood, and does nothing.

What did the glands release? And why does the heart answer it when the lens does not?

Unit 4 · Cell Communication and Cell Cycle

1A molecule released into the blood

2

Video: Watch: The molecule the glands released

The cells of the two adrenal glands release epinephrine into the blood, and the blood carries the same molecule to every cell in the body.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L01a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L01a.mp4

3

How can cells a body’s length apart act together within a minute?

4

One cell releases a molecule. Other cells detect it.

5

The molecule carries no instructions.

6

The molecule counts as a message only because some cells carry a protein built to bind it. Those cells respond.

7

Cells with no such protein do nothing, however much of the molecule reaches them.

8

So the same molecule, from the same blood, makes the heart pound and leaves the lens alone.

9

Each of the four has a name: the molecule, the cell that releases it, the protein that binds it, and the cells that respond. Start with the molecule.

10

The two glands on top of your kidneys are the adrenal glands. When the door slams, their cells release a molecule into the blood passing through them.

Two kidneys with an adrenal gland on top of each, and a blood vessel between them into which the glands release molecules, drawn as scattered dots
Two kidneys with an adrenal gland on top of each, and a blood vessel between them into which the glands release molecules, drawn as scattered dots
11

That molecule is epinephrine, the hormone that readies your body to move fast. Its everyday name is adrenaline.

12
Check q1

Epinephrine is a hormone.

How does a hormone travel from the cell that releases it to the cells it acts on?

  1. A. ✓ The blood carries it
  2. B. It moves along a nerve
    A nerve carries an electrical signal along itself.
    A hormone is a molecule released into the blood, and the blood carries it.

Why: A hormone is a signal molecule carried in the blood to cells elsewhere in the body.

13

Within half a minute the blood has carried the epinephrine to every cell in your body: heart, liver, skin, and the fluid around the lens of your eye.

14

A molecule that one cell releases and that carries information to other cells is called a . Epinephrine is a chemical signal.

15

The epinephrine molecule carries no instructions inside it. It is the same small molecule wherever it goes.

16

The molecule counts as information only because some cells are built to detect it.

17

What you are expected to know Identify the chemical signal in a described case: the molecule one cell releases that carries information to other cells.

18
Check q2

A yeast cell releases a small molecule into the liquid around it. A second yeast cell, some distance away, detects the molecule and stops dividing.

Which of the following is the chemical signal?

  1. A. The liquid around the yeast cells
    The liquid carries the molecule from one yeast cell to the other.
    The chemical signal is the molecule the liquid carries.
  2. B. ✓ The molecule the first yeast cell released
  3. C. The second yeast cell’s stopping dividing
    Stopping dividing is what the second yeast cell does after the molecule reaches it.
    The chemical signal is the molecule that reached it.

Why: A chemical signal is a molecule one cell releases that carries information to other cells.
The first yeast cell released the molecule into the liquid.
The second yeast cell detected that molecule.
So the molecule is the chemical signal.

19
Check q3

A student says: “The epinephrine molecule carries instructions inside it that tell the heart to beat faster.”

Is the student correct?

  1. A. Yes — the instructions travel inside the molecule
    Epinephrine is the same small molecule wherever it goes.
    The heart cell is built to detect it, and the heart cell decides the response.
  2. B. ✓ No — the heart cell detects it and responds in its own way

Why: Epinephrine is the same small molecule in every part of the blood.
It carries no instructions inside it.
Heart cells are built to detect epinephrine, and they respond in their own way.
So the heart cell, not the molecule, decides the response.

20The cell that sends the signal

21

Video: Watch: The signaling cell

The adrenal cells released the epinephrine, so they are the signaling cells; the heart muscle cell only received the molecule.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L01b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L01b.mp4

22

Now look at the other end of the message: the cell that released the molecule.

Two kidneys with an adrenal gland on top of each, and a blood vessel between them into which the glands release molecules, drawn as scattered dots
Two kidneys with an adrenal gland on top of each, and a blood vessel between them into which the glands release molecules, drawn as scattered dots
23

The cells of the adrenal glands released the epinephrine into the blood.

24

The cell that releases the signal is called the . Here the adrenal cells are the signaling cells.

25

The heart muscle cell did not release the epinephrine. It only received the molecule.

26

So the heart muscle cell is not a signaling cell here.

27

What you are expected to know Identify the signaling cell in a described case: the cell that releases the chemical signal.

28
Check q4

A yeast cell releases a small molecule into the liquid around it. A second yeast cell, some distance away, detects the molecule and stops dividing.

Which of the two yeast cells is the signaling cell?

  1. A. ✓ The first yeast cell
  2. B. The second yeast cell
    The second yeast cell detects the molecule and responds.
    The signaling cell is the cell that releases the molecule, and that is the first yeast cell.

Why: The signaling cell is the cell that releases the molecule.
The first yeast cell released the molecule into the liquid.
So the first yeast cell is the signaling cell.
The second yeast cell detected the molecule and responded.

29Quick quiz: chemical signal and signaling cell mixed practice

30
Check q5

What is a chemical signal?

  1. A. The change a cell makes when a molecule reaches it
    The change is the cell’s response.
    The chemical signal is the molecule that reached the cell.
  2. B. The blood or liquid that carries a molecule between cells
    The blood carries the molecule.
    The chemical signal is the molecule itself.
  3. C. ✓ A molecule one cell releases that carries information to other cells

Why: A chemical signal is a molecule one cell releases.
The molecule carries information to other cells.
Epinephrine, released by the adrenal cells into the blood, is one.

31
Check q6

What is a signaling cell?

  1. A. ✓ The cell that releases the chemical signal
  2. B. The cell that detects the chemical signal
    The cell that detects the signal responds to it.
    The signaling cell is the cell that released it.
  3. C. Any cell the blood carries the chemical signal past
    The blood carries the signal past many cells that never released it.
    The signaling cell is the one cell type that released it.

Why: The signaling cell is the cell that releases the chemical signal.
The adrenal cells released the epinephrine, so the adrenal cells are the signaling cells.

32
Practice writing an answer

When food reaches the stomach, cells in the stomach lining release a molecule into the blood. Cells in the pancreas detect the molecule and release digestive juice.

(a) Identify the signaling cell and the chemical signal in this case. (1 pt)

Model answer The signaling cells are the cells in the stomach lining. They released the molecule.
The chemical signal is the molecule they released into the blood. It carried information to the pancreas cells.
Rubric
  • Award 1 point for: the stomach-lining cells as the signaling cells and the released molecule as the chemical signal (both needed).

Slip Naming the pancreas cells as the signaling cells. The pancreas cells detected the molecule; the signaling cell is the cell that released it.

33
Check q7

A leaf cell chewed by a caterpillar releases a molecule into the plant’s sap. Cells throughout the plant detect it and make a bitter compound.

Which of the following is the chemical signal?

  1. A. The sap that carries the molecule
    The sap moves the molecule through the plant.
    The chemical signal is the molecule the sap carries.
  2. B. The bitter compound the cells make
    The bitter compound is what the detecting cells make after the molecule reaches them.
    The chemical signal is the molecule that reached them.
  3. C. ✓ The molecule released into the sap

Why: The chemical signal is the molecule one cell released that carries information to other cells.
The chewed leaf cell released a molecule into the sap.
So that molecule is the chemical signal.

34
Check q8

An egg cell releases a molecule into the water around it. A sperm cell detects the molecule and swims toward the egg.

Which of the two cells is the signaling cell?

  1. A. The sperm cell
    The sperm cell detects the molecule and swims.
    The signaling cell is the cell that releases the molecule.
    The egg cell released the molecule.
  2. B. ✓ The egg cell

Why: The signaling cell is the cell that releases the molecule.
The egg cell released the molecule into the water.
So the egg cell is the signaling cell.
The sperm cell detected the molecule and responded.

35
Check q9

A nerve ending releases a molecule into the gap beside a muscle cell. The muscle cell contracts.

Which of the following is the chemical signal?

  1. A. ✓ The molecule released into the gap
  2. B. The fluid in the gap
    The fluid in the gap carries the molecule from the nerve ending to the muscle cell.
    The chemical signal is the molecule itself.
  3. C. The contraction of the muscle cell
    The contraction is what the muscle cell does after the molecule reaches it.
    The chemical signal is the molecule that reached it.

Why: The chemical signal is the molecule one cell released that carries information to another cell.
The nerve ending released a molecule into the gap.
So that molecule is the chemical signal.

36
Check q10

Cells of a gland in the neck release a molecule into the blood. Bone cells throughout the body grow faster.

Which of the two cell types is the signaling cell?

  1. A. The bone cell
    The bone cells grow faster after the molecule reaches them.
    The signaling cell is the cell that releases the molecule.
    The gland cells released the molecule.
  2. B. ✓ The gland cell

Why: The signaling cell is the cell that releases the molecule.
The gland cells released the molecule into the blood.
So the gland cell is the signaling cell.
The bone cells detected the molecule and responded.

37
Check q11

A damaged skin cell of a minnow releases a molecule into the water. Other minnows detect the molecule and flee.

Which of the following is the chemical signal?

  1. A. The damage to the skin
    The damage made the skin cell release the molecule.
    The chemical signal is the molecule.
  2. B. The fleeing of the other minnows
    The fleeing is what the other minnows do after the molecule reaches them.
    The chemical signal is the molecule that reached them.
  3. C. ✓ The molecule released into the water

Why: The chemical signal is the molecule one cell released that carries information to other cells.
The damaged skin cell released a molecule into the water.
So that molecule is the chemical signal.

38The protein that binds the signal

39

Video: Watch: A pocket that fits epinephrine

Heart muscle cells and liver cells carry a protein set in the membrane with a pocket that epinephrine fits; lens cells carry none, and the epinephrine washes past them.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L01c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L01c.mp4

40

Blood carries the same epinephrine past three kinds of cell: a heart muscle cell, a liver cell and a cell of the eye’s lens.

A blood vessel across the top carrying scattered molecules of epinephrine, and beneath it a heart muscle cell, a liver cell and a lens cell, each with molecules around it; the heart cell contracts harder, the liver cell releases glucose, the lens cell shows no change
A blood vessel across the top carrying scattered molecules of epinephrine, and beneath it a heart muscle cell, a liver cell and a lens cell, each with molecules around it; the heart cell contracts harder, the liver cell releases glucose, the lens cell shows no change
41

The heart muscle cell responds: it contracts harder and faster. The liver cell responds: it releases glucose into the blood.

42

The lens cell does nothing.

43

The same molecule, at the same concentration, washed past all three. The difference is in the cells.

44
Check q12

A channel protein is set through a cell’s plasma membrane.

Which of the following does the channel protein do?

  1. A. ✓ It lets a substance cross the membrane
  2. B. It stores the cell’s energy as glucose and fat
    A cell stores energy in glucose and fat molecules in the cytosol, not in a membrane protein.
    A channel protein lets a substance cross the membrane.

Why: A membrane protein is a protein set through the plasma membrane.
A channel or carrier protein lets a substance the bilayer blocks cross the membrane.

45

Heart muscle cells and liver cells carry a protein set in the plasma membrane that binds epinephrine.

The heart muscle cell and the liver cell each carry three small proteins set in the membrane with a molecule of epinephrine held on each; the lens cell's membrane carries none, and its molecules float free
The heart muscle cell and the liver cell each carry three small proteins set in the membrane with a molecule of epinephrine held on each; the lens cell's membrane carries none, and its molecules float free
46

Here is one small piece of the heart muscle cell’s membrane, enlarged. Epinephrine fits a pocket on that protein and is held there.

One small piece of a heart muscle cell's plasma membrane, enlarged: the fluid outside the cell is shaded one way above the membrane and the cytosol inside is shaded another below it; one receptor protein is set through the membrane, a molecule of epinephrine sits in a pocket on the receptor's outer face, and other molecules float free outside
One small piece of a heart muscle cell's plasma membrane, enlarged: the fluid outside the cell is shaded one way above the membrane and the cytosol inside is shaded another below it; one receptor protein is set through the membrane, a molecule of epinephrine sits in a pocket on the receptor's outer face, and other molecules float free outside
47

The pocket has the shape of epinephrine. Other molecules in the blood do not fit the pocket, so the protein binds epinephrine and lets the rest wash past.

48

Lens cells carry no such protein, so the epinephrine washes past them.

49

A protein on or in a cell that binds a particular chemical signal is called a for that signal.

50

What you are expected to know Describe a receptor: a protein on or in a cell that binds one particular chemical signal by fitting it in a pocket.

51
Check q13

A hormone in the blood reaches a kidney cell, a muscle cell and a skin cell at the same concentration. Only the kidney cell responds.

Which of the three cells carries a receptor for the hormone?

  1. A. The muscle cell
    The muscle cell showed no response.
    A cell responds only when a receptor binds the signal, so the muscle cell carries no receptor for this hormone.
  2. B. All three cells
    Only the kidney cell responded.
    The muscle cell and the skin cell carry no receptor for this hormone.
  3. C. ✓ The kidney cell

Why: A receptor is a protein that binds the signal.
A cell responds only when its receptor binds the signal.
Only the kidney cell responded.
So only the kidney cell carries a receptor for the hormone.

52Quick quiz: receptor mixed practice

53
Check q14

What is a receptor for a chemical signal?

  1. A. The molecule one cell releases to carry information
    The released molecule is the chemical signal.
    The receptor is the protein that binds it.
  2. B. The blood vessel that carries the signal to the cell
    The blood carries the signal to the cell.
    The receptor is the protein that binds the signal when it arrives.
  3. C. ✓ A protein on or in a cell that binds that signal

Why: A receptor is a protein on or in a cell.
It binds one particular chemical signal by fitting it in a pocket.

54
Practice writing an answer

Heart muscle cells carry a receptor for epinephrine.

(a) State what a receptor is. (1 pt)

Model answer A receptor is a protein on or in a cell that binds one particular chemical signal, fitting the signal in a pocket.
Rubric
  • Award 1 point for: a protein on or in the cell that binds a particular chemical signal.

55Why the heart and not the lens

56

Video: Watch: Target cells

The heart muscle cell and the liver cell carry a receptor for epinephrine and respond; the lens cell carries none and does nothing, however much epinephrine reaches it.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L01d.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L01d.mp4

57

The heart muscle cell and the liver cell each carry a receptor for epinephrine. Each responded when the epinephrine reached it.

A blood vessel across the top carrying scattered molecules of epinephrine, and beneath it a heart muscle cell, a liver cell and a lens cell, each with molecules around it; the heart cell contracts harder, the liver cell releases glucose, the lens cell shows no change
A blood vessel across the top carrying scattered molecules of epinephrine, and beneath it a heart muscle cell, a liver cell and a lens cell, each with molecules around it; the heart cell contracts harder, the liver cell releases glucose, the lens cell shows no change
58

A cell that carries a receptor for a signal, and so responds when the signal arrives, is called a for that signal. Heart muscle cells and liver cells are target cells for epinephrine.

59

Each target cell responds in its own way. The heart muscle cell contracts harder; the liver cell releases glucose.

60

The lens cell carries no receptor for epinephrine. So nothing in the lens cell binds the epinephrine, and nothing inside it changes.

61

A cell with no receptor for a signal does nothing, however much of the signal reaches it.

62

What you are expected to know Identify a target cell: a cell that carries the receptor for a signal and so responds when the signal arrives.

63
Check q15

Epinephrine reaches a liver cell, a heart muscle cell and a lens cell at the same concentration. The liver cell releases glucose and the heart muscle cell contracts harder. The lens cell shows no change.

Which of the following cells are target cells for epinephrine?

  1. A. The heart muscle cell only
    The liver cell released glucose when the epinephrine reached it.
    So the liver cell responded, and a cell that responds to the signal is a target cell.
  2. B. ✓ The liver cell and the heart muscle cell
  3. C. All three: the heart muscle cell, the liver cell and the lens cell
    A target cell is a cell that responds to the signal.
    The lens cell showed no change.
    So the lens cell is not a target cell.

Why: A target cell is a cell that responds to the signal.
The liver cell responded: it released glucose.
The heart muscle cell responded: it contracted harder.
So both are target cells for epinephrine.
The lens cell showed no change, so the lens cell is not a target cell.

64
Practice writing an answer

Epinephrine reaches a liver cell, a heart muscle cell and a lens cell at the same concentration. The liver cell and the heart muscle cell respond. The lens cell shows no change.

(a) Explain why the lens cell shows no change. (1 pt)

Model answer A cell responds to a signal through a receptor.
A receptor is a protein on or in the cell that binds the signal.
The liver cell and the heart muscle cell carry a receptor that binds epinephrine.
The lens cell carries no receptor for epinephrine.
So nothing in the lens cell binds the epinephrine.
The epinephrine washes past the lens cell.
So nothing inside the lens cell changes, however much epinephrine reaches it.
Rubric
  • Award 1 point for: the lens cell carries no receptor for epinephrine (no protein that binds it), so nothing in the lens cell binds the epinephrine and nothing inside the cell changes.

Slip Saying the lens cell ignores the epinephrine, or has no use for it. The lens cell has no protein that binds the epinephrine, so it cannot detect the epinephrine at all.

65
Check q16

A student looks at the same three cells and says: “The lens cell is ignoring the epinephrine, because the lens cell has no use for it.”

Is the student correct?

  1. A. Yes — a cell responds only to the signals it needs
    A cell cannot weigh what it needs.
    A cell responds only when a receptor binds the signal.
    The lens cell has no receptor for epinephrine, so it cannot detect it.
  2. B. ✓ No — nothing in the lens cell can detect the epinephrine

Why: A cell detects a signal only through a receptor, a protein that binds the signal.
The lens cell carries no receptor for epinephrine.
So nothing in the lens cell can detect the epinephrine; the epinephrine washes past it.

66
Check q17

A nerve ending releases a molecule into the gap beside a muscle cell, and the muscle cell responds. A researcher adds a blocker to the gap. The nerve ending still releases the molecule, the molecule still fills the gap at its usual concentration, and the muscle cell stops responding.

Which of the following has the blocker done?

  1. A. The blocker has broken the molecule down in the gap
    A molecule that is being broken down falls in concentration.
    The molecule still fills the gap at its usual concentration.
    So nothing broke the molecule down.
  2. B. The blocker has occupied the receptors on the nerve ending’s surface
    The nerve ending still releases the molecule, so nothing at the nerve ending has changed.
    The muscle cell is the cell that stopped responding, and its receptors bind the molecule.
  3. C. ✓ The blocker has occupied the receptors on the muscle cell’s surface

Why: The molecule still fills the gap at its usual concentration.
The nerve ending still releases the molecule, so nothing there has changed.
The muscle cell stopped responding, so the change is in the muscle cell.
The muscle cell detects the molecule through its surface receptors, so the blocker occupied them.

67

Half a minute ago a door slammed behind you in a dark hallway. Your heart is hammering and your hands shake.

68

The cells of the two adrenal glands above your kidneys released epinephrine into your blood. The adrenal cells were the signaling cells, and epinephrine was the chemical signal.

69

The blood carried the epinephrine to every part of your body, the heart muscle cells and the fluid around the lens cells among them.

70

Heart muscle cells carry a receptor for epinephrine. The epinephrine bound the receptor, so the heart muscle cells contracted harder and faster.

71

Lens cells carry no receptor for epinephrine. The epinephrine washed past them, so nothing in the lens changed.

72

The same molecule, from the same blood: the heart is a target cell for epinephrine, and the lens is not.

73Quick quiz: target cell mixed practice

74
Check q18

What is a target cell for a chemical signal?

  1. A. The cell that releases the signal into the blood
    The cell that releases the signal is the signaling cell.
  2. B. Any cell the blood carries the signal past
    The blood carries the signal past many cells that carry no receptor for it.
    Those cells do nothing, so they are not target cells.
  3. C. ✓ A cell that carries a receptor for the signal

Why: A target cell carries a receptor for the signal.
So when the signal arrives, the receptor binds it and the cell responds.

75
Practice writing an answer

Epinephrine reaches every cell in the body.

(a) State what makes a cell a target cell for epinephrine. (1 pt)

Model answer A target cell for epinephrine carries a receptor for epinephrine, so it responds when epinephrine arrives.
Rubric
  • Award 1 point for: it carries a receptor for epinephrine (and so responds when epinephrine arrives).

76Mixed practice mixed practice

77
Check q19

When food reaches the small intestine, cells lining the small intestine release a molecule into the blood. Cells in the pancreas detect it and release digestive enzymes. Liver cells, reached by the same blood, show no change.

Which of the following is the signaling cell?

  1. A. The pancreas cell
    The pancreas cell detects the molecule.
    The signaling cell is the cell that released the molecule.
  2. B. The liver cell
    The liver cell neither released nor detected the molecule.
  3. C. All three cells
    Only the cell that releases the molecule is the signaling cell.
    The blood just carries the molecule.
  4. D. ✓ The cell lining the small intestine

Why: The signaling cell is the cell that releases the chemical signal.
The cell lining the small intestine released the molecule into the blood.
So the cell lining the small intestine is the signaling cell.

78
Check q20

Cells in the pancreas release hormone G into the blood. Liver cells release glucose when G reaches them. Leg muscle cells, reached by the same blood, show no change.

Which of the following cells carry a receptor for G?

  1. A. ✓ The liver cells
  2. B. The leg muscle cells
    The leg muscle cells showed no change.
    A cell responds only when its receptor binds the signal, so the leg muscle cells carry no receptor for G.
  3. C. Both the liver cells and the leg muscle cells
    The leg muscle cells showed no change when G reached them.
    So the leg muscle cells carry no receptor for G.

Why: A cell responds to a signal only when it carries a receptor that binds the signal.
The liver cells responded: they released glucose.
The leg muscle cells showed no change.
So the liver cells carry a receptor for G, and the leg muscle cells do not.

79
Check q21

Two cell types respond to the same hormone in different ways: liver cells release glucose, and heart muscle cells contract harder.

Why do the two cell types respond differently to one molecule?

  1. A. The liver cells receive a different form of the hormone from the heart cells
    The blood carries one hormone to both.
    The difference lies in the cells.
  2. B. The heart cells receive the hormone first and use up its message
    The hormone reaches both cell types, and each responds in its own way.
  3. C. The hormone carries a different set of instructions to each cell type
    The hormone is the same small molecule everywhere.
    It carries no instructions inside it.
  4. D. ✓ Each cell type responds in its own way once its receptor binds the hormone

Why: The hormone carries no instructions inside it.
Each cell type carries the receptor and responds in its own way.
So the same molecule makes a liver cell release glucose and a heart muscle cell contract harder.

80
Check q22

When food reaches the small intestine, cells lining it release a molecule into the blood. The molecule reaches the pancreas cells and the liver cells. The pancreas cells release digestive enzymes; the liver cells show no change.

Which of the following cells are target cells for the molecule?

  1. A. The liver cells
    A target cell responds to the signal.
    The liver cells did not respond.
  2. B. Both the pancreas cells and the liver cells
    Reaching a cell is not enough.
    A target cell responds to the signal, and the liver cells showed no change.
  3. C. ✓ The pancreas cells

Why: A target cell is a cell that responds to the signal.
The pancreas cells responded: they released digestive enzymes.
So the pancreas cells are the target cells.
The liver cells showed no change, so the liver cells are not target cells.

81
Check q23

A skin cell carries no receptor for hormone K. A researcher gives the skin cell one hundred times the usual concentration of K.

Which of the following does the skin cell do?

  1. A. It responds weakly
    A cell responds only when a receptor binds the signal.
    The skin cell has no receptor for K, so nothing in it binds K at any concentration.
  2. B. It responds as strongly as a bone cell
    A cell responds only when a receptor binds the signal.
    The skin cell has no receptor for K, so nothing in it binds K at any concentration.
  3. C. ✓ It does nothing

Why: A cell detects a signal only through a receptor.
The skin cell carries no receptor for K.
So nothing in the skin cell binds K, however much K reaches it.
The skin cell does nothing.

82
Check q24

A root cell of a bean plant, attacked by a fungus, releases a molecule into the soil water. Bacteria nearby detect the molecule and move toward the root.

Which of the following is the chemical signal?

  1. A. The fungus attacking the root
    The fungus made the root cell release the molecule.
    The chemical signal is the molecule.
  2. B. ✓ The molecule released into the soil water
  3. C. The bacteria moving toward the root
    The movement is what the bacteria do after the molecule reaches them.
    The chemical signal is the molecule that reached them.

Why: A chemical signal is a molecule one cell releases that carries information to other cells.
The root cell released a molecule into the soil water.
The bacteria detected that molecule.
So the molecule is the chemical signal.

83
Practice writing an answer

Cells of a gland in the neck release hormone K into the blood. The blood carries K to bone cells and to skin cells at the same concentration. The bone cells grow faster. The skin cells show no change.

(a) Explain how this case demonstrates that a cell’s response to a chemical signal depends on the receptor the cell carries. (2 pt)

Model answer The blood carried the same hormone K to both cell types at the same concentration.
The bone cells responded, so the bone cells carry a receptor that binds K.
The skin cells showed no change, so the skin cells carry no receptor for K.
The molecule was the same for both cell types; only the receptor differed.
So the receptor a cell carries, not the signal, decides whether the cell responds.
Rubric
  • Award 1 point for: the bone cells respond because they carry a receptor that binds K.
  • Award 1 point for: the skin cells show no change because they carry no receptor for K, although the same K at the same concentration reached them.

Slip Saying the skin cells do not need K. The skin cells carry no protein that binds K, so nothing in them detects it.

Glossary

chemical signal
A molecule released by one cell that carries information to other cells. Epinephrine, released by the adrenal glands into the blood, is a chemical signal. Also called a signaling molecule.
signaling cell
The cell that releases a chemical signal. When a door slams, the cells of the adrenal glands are the signaling cells.
receptor
A protein on or in a cell that binds a particular chemical signal by fitting it in a pocket. Heart muscle cells carry a receptor for epinephrine; lens cells carry none.
target cell
A cell that carries a receptor for a signal and so responds when the signal arrives. A cell with no receptor for the signal does nothing, however much of it reaches the cell.

APBIO-U04-L01B Predict from the receptor

Topic 4.1 · Cell Communication · 53 steps

A photograph of a leaf surface under the microscope: pairs of pale, rounded guard cells, each pair with a slit-like pore between them, scattered among larger, darker, many-sided pavement cells; beside it a schematic of one pore between two curved guard cells with flat pavement cells around them, labelled
A photograph of a leaf surface under the microscope: pairs of pale, rounded guard cells, each pair with a slit-like pore between them, scattered among larger, darker, many-sided pavement cells; beside it a schematic of one pore between two curved guard cells with flat pavement cells around them, labelled

Photo: AioftheStorm, Wikimedia Commons, CC0 (resized).

Here is the surface of a leaf, seen through a microscope. Each pore in it sits between two guard cells, and flat pavement cells fill the space between the pores.

In a drought a plant hormone reaches every cell of that surface. Guard cells carry 9.2 units of the hormone’s receptor; pavement cells carry 0.4 units.

Thirty minutes after the hormone arrives, each pore has narrowed from 8.0 μm wide to 2.5 μm wide, and the pavement cells have not changed. Could you have predicted that from the receptor alone?

Unit 4 · Cell Communication and Cell Cycle

1Which cells respond?

2

Video: Watch: Reading the receptor bar chart

Guard cells carry 9.2 units of the hormone’s receptor and pavement cells 0.4 units; the cells that carry the receptor respond, and the cells that carry almost none do nothing, however much hormone arrives.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L01Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L01Ba.mp4

3

How do you predict which cells will answer a signal? You count the receptor each cell type carries.

4

The cells that carry the receptor respond. The cells that carry almost none do nothing.

5

More hormone changes how strongly the guard cells respond.

6

More hormone changes nothing in a pavement cell. A cell with almost no receptor stays as it is at any concentration.

7

So a bar chart of receptor per cell type is enough to say which cells move and which stay still.

8

And the receptor alone is the reason you give.

9
Check q1

A cell carries no receptor for a signal. The signal reaches the cell.

Which of the following does the cell do?

  1. A. It responds weakly
    A cell responds only when a receptor binds the signal.
    This cell has no receptor, so nothing in it binds the signal.
  2. B. ✓ It does nothing

Why: A cell detects a signal only through a receptor.
This cell carries no receptor for the signal.
So nothing in the cell binds the signal, and the cell does nothing.

10

Here is the surface of a leaf, drawn. Each pore in it is a stoma, and it sits between a pair of guard cells.

The surface of a leaf: flat pavement cells fill the surface, and a pair of curved guard cells surrounds a pore
The surface of a leaf: flat pavement cells fill the surface, and a pair of curved guard cells surrounds a pore
11

Flat pavement cells fill the surface between the pores.

12

Now suppose a drought comes. A plant hormone reaches every cell of the leaf surface.

13

The bar chart shows how much of the hormone’s receptor each kind of cell carries.

Bar graph of receptor amount in units: guard cells 9.2, pavement cells 0.4, with gridlines every 2 units
Bar graph of receptor amount in units: guard cells 9.2, pavement cells 0.4, with gridlines every 2 units
14

Guard cells carry 9.2 units of the receptor. Pavement cells carry 0.4 units.

15

In leaves whose cells lack the receptor altogether, every cell reads 0.3 units. So a reading near 0.3 units means almost no receptor.

16

The guard cells carry the receptor. So the guard cells will respond.

17

The pavement cells carry almost none. So the pavement cells will do nothing.

18

Thirty minutes after the hormone arrives, the guard cells have narrowed each pore from 8.0 μm wide to 2.5 μm wide. The pavement cells show no change.

19

More hormone changes how strongly the guard cells respond.

20

More hormone changes nothing in the pavement cells. A cell with almost no receptor stays as it is at any concentration.

21

In the leaves that lack the receptor, the hormone moves the pore width from 7.9 μm to 7.6 μm. A change that small does not count as a response.

22

What you are expected to know Predict, from a bar chart of how much receptor each cell type carries, which cells respond when the signal arrives and which do not.

23Quick quiz: does this cell respond? mixed practice

24
Check q2

Skin cells carry a receptor for signal Q. Bone cells carry none. Signal Q reaches both.

Which of the following cells respond?

  1. A. ✓ The skin cells
  2. B. The bone cells
    The cells that carry the receptor respond.
    The skin cells carry the receptor for Q. The bone cells carry none.
  3. C. Both the skin cells and the bone cells
    A cell responds only when it carries a receptor that binds the signal.
    The bone cells carry none.
    So the bone cells do nothing.

Why: Only cells that carry the receptor respond.
The skin cells carry the receptor for Q, so the skin cells respond.
The bone cells carry none, so the bone cells do nothing.

25
Check q3

The graph below shows how much of the receptor for hormone H three cell types carry. Cells that lack the receptor read 0.3 units. Hormone H reaches all three cell types.

Bar graph of receptor for hormone H in three cell types, in units, with gridlines every 2 units
Bar graph of receptor for hormone H in three cell types, in units, with gridlines every 2 units

Do the kidney cells respond to H?

  1. A. ✓ Yes
  2. B. No
    The kidney cells carry 7.5 units of the receptor for H, far above the 0.3 units of cells that lack it.
    So H binds the kidney cells and they respond.

Why: The kidney cells carry 7.5 units of the receptor for H. That is far above the 0.3 units of cells that lack the receptor.
So H binds the kidney cells’ receptors, and the kidney cells respond.

26
Check q4

The graph below shows how much of the receptor for hormone H three cell types carry; cells that lack the receptor read 0.3 units. Hormone H reaches all three cell types.

Bar graph of receptor for hormone H in three cell types, in units, with gridlines every 2 units
Bar graph of receptor for hormone H in three cell types, in units, with gridlines every 2 units

Do the skin cells respond to H?

  1. A. Yes
    The skin cells carry 0.2 units of the receptor, below the 0.3 units of cells that lack it.
    So the skin cells carry almost none: nothing in them binds H.
  2. B. ✓ No

Why: The skin cells carry 0.2 units of the receptor for H.
That is below the 0.3 units of cells that lack the receptor.
So the skin cells carry almost no receptor.
Nothing in them binds H.
So the skin cells do nothing.

27
Check q5

The graph below shows how much of the receptor for hormone H three cell types carry; cells that lack the receptor read 0.3 units. Hormone H reaches all three cell types. A researcher then gives the skin cells ten times as much H.

Bar graph of receptor for hormone H in three cell types, in units, with gridlines every 2 units
Bar graph of receptor for hormone H in three cell types, in units, with gridlines every 2 units

Do the skin cells respond now?

  1. A. Yes
    More hormone changes how strongly receptor-carrying cells respond, never which cells respond.
    The skin cells carry almost no receptor, so there is still almost nothing to bind H.
  2. B. ✓ No

Why: More hormone changes how strongly the cells that carry the receptor respond.
It never changes which cells respond.
The skin cells carry almost no receptor for H.
So at ten times the concentration there is still almost nothing in the skin cells to bind H.

28
Check q6

The graph below shows how much of the receptor for hormone M three cell types of a mouse carry. Cells that lack the receptor read 0.3 units. Hormone M reaches all three cell types.

Bar graph of receptor for hormone M in three cell types of a mouse, in units, with gridlines every 2 units
Bar graph of receptor for hormone M in three cell types of a mouse, in units, with gridlines every 2 units

Do the bone cells respond to M?

  1. A. ✓ Yes
  2. B. No
    The bone cells carry 4.4 units of the receptor for M, far above the 0.3 units of cells that lack it.
    So M binds the bone cells and they respond.

Why: The bone cells carry 4.4 units of the receptor for M. That is far above the 0.3 units of cells that lack the receptor.
So M binds the bone cells’ receptors, and the bone cells respond.

29
Check q7

The graph below shows how much of the receptor for hormone M three cell types of a mouse carry. At a first concentration of M the bone cells responded weakly. A researcher gives the bone cells ten times as much M.

Bar graph of receptor for hormone M in three cell types of a mouse, in units, with gridlines every 2 units
Bar graph of receptor for hormone M in three cell types of a mouse, in units, with gridlines every 2 units

Which of the following do the bone cells do now?

  1. A. The bone cells respond as before
    At the first concentration the bone cells responded only weakly.
    More M means more of their receptors bind M, so the bone cells respond more strongly.
  2. B. The bone cells stop responding
    The bone cells carry the receptor for M.
    More M means more of their receptors bind M, so the response grows.
  3. C. ✓ The bone cells respond more strongly

Why: The bone cells carry the receptor for M, and at the first concentration they responded weakly.
More hormone changes how strongly the cells that carry the receptor respond.
So at ten times the concentration more of the bone cells’ receptors bind M, and the bone cells respond more strongly.

30Why: the receptor alone

31

Video: Watch: The reason is the receptor

The guard cells respond because they carry 9.2 units of the receptor; the pavement cells stay as they are because they carry 0.4 units, almost none; the amount of hormone never changes which cells respond.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L01Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L01Bb.mp4

32

Now suppose someone asks you why the pores narrowed and the pavement cells did not move.

33

The reason is the receptor alone.

34

The guard cells carry 9.2 units of the receptor. So the hormone binds the guard cells, and the guard cells respond.

35

The pavement cells carry 0.4 units, almost none. So the hormone has almost nothing to bind in a pavement cell, and the pavement cell stays as it is.

36

The amount of hormone changes how strongly the guard cells respond. The amount of hormone never changes which cells respond.

37

So a justification names the receptor each cell type carries. It does not name what the cell is for, and it does not name how much hormone arrived.

38

What you are expected to know Justify, from the receptor each cell type carries, a prediction of which cells respond to a signal.

39
Practice writing an answer

Hormone P is released into the blood by cells of a gland in a rat’s neck. Researchers test whether P makes cells take up calcium. They measure how much of the receptor for P three cell types carry: bone cells, kidney cells and skin cells. The graph shows the results. In this measurement, cells that lack the receptor read 0.3 units. The researchers then give each cell type the same concentration of P for 30 minutes.

Bar graph of receptor for hormone P in three cell types of a rat, in units, with gridlines every 2 units
Bar graph of receptor for hormone P in three cell types of a rat, in units, with gridlines every 2 units

(a) Predict which of the three cell types take up calcium when P arrives, and justify your prediction from the graph. (2 pt)

Model answer The bone cells and the kidney cells take up calcium; the skin cells do not.
A cell responds only when it carries a receptor that binds the signal.
The bone cells carry 7.9 units of the receptor for P and the kidney cells 5.6 units, so P binds both and both respond.
The skin cells carry 0.2 units: almost no receptor.
So the skin cells stay as they are.
Rubric
  • Award 1 point for: the prediction that the bone cells and the kidney cells respond and the skin cells do not.
  • Award 1 point for: the justification that only cells carrying the receptor (7.9 and 5.6 units) can bind P and respond; the skin cells carry almost none (0.2 units).

Slip Predicting that the kidney cells do not respond because they carry less receptor than the bone cells. Both cell types carry the receptor far above 0.3 units, so both cell types respond.

(b) The researchers then give the skin cells ten times the concentration of P. Predict whether the skin cells now take up calcium, and justify your prediction. (1 pt)

Model answer The skin cells still take up no calcium.
More P changes how strongly the cells that carry the receptor respond.
More P changes nothing in a cell that carries almost none.
The skin cells carry 0.2 units of the receptor: almost none.
So at ten times the concentration P still has almost nothing to bind in a skin cell.
Rubric
  • Award 1 point for: the skin cells still do not respond, because the amount of signal changes how strongly receptor-carrying cells respond, never which cells respond; with almost no receptor (0.2 units) there is almost nothing for the extra P to bind.

Slip Predicting that the skin cells now respond a little. A cell with almost no receptor has almost nothing to bind the hormone at any concentration.

40
Check q8

A student looks at the guard cells (9.2 units of the receptor) and the pavement cells (0.4 units, near the 0.3 units read by cells that lack the receptor) and says: “If the plant made ten times as much hormone, the pavement cells would respond too.”

Is the student correct?

  1. A. Yes — more hormone would reach more of the pavement cells
    The hormone already reaches every pavement cell.
    A pavement cell carries almost no receptor, so more hormone has almost nothing more to bind.
  2. B. ✓ No — the pavement cells carry almost no receptor

Why: The amount of hormone changes how strongly the cells that carry the receptor respond.
It never changes which cells respond.
The pavement cells carry 0.4 units of the receptor, almost none.
So at ten times the concentration the hormone still has almost nothing to bind in a pavement cell.

41

Here is the leaf surface again, thirty minutes after the drought hormone reached every cell of it.

42

The guard cells carry 9.2 units of the hormone’s receptor. The hormone bound them, and each pore narrowed from 8.0 μm wide to 2.5 μm wide.

43

The pavement cells carry 0.4 units, almost none. The hormone had almost nothing to bind in them, and the pavement cells did not change.

44

You could have predicted both from the bar chart alone. The receptor each cell type carries is the whole reason.

45Mixed practice mixed practice

46
Check q9

A mouse’s liver cells carry 6.0 units of the receptor for signal S. At a low concentration of S the liver cells respond weakly. A researcher raises the concentration of S one hundred times.

Which of the following do the liver cells do at the higher concentration?

  1. A. ✓ The liver cells respond more strongly
  2. B. The liver cells respond as weakly as before
    The liver cells carry 6.0 units of the receptor.
    At one hundred times the concentration more of their receptors bind S. So the liver cells respond more strongly.
  3. C. The liver cells stop responding
    The liver cells carry the receptor for S.
    More S means more of their receptors bind S, so the response grows.

Why: The amount of signal changes how strongly the cells that carry the receptor respond.
The liver cells carry 6.0 units of the receptor.
So at one hundred times the concentration more of the liver cells’ receptors bind S, and the liver cells respond more strongly.

47
Check q10

The graph below shows how much of the receptor for hormone M three cell types of a mouse carry; cells that lack the receptor read 0.3 units. A student predicts that the gut cells respond to M.

Bar graph of receptor for hormone M in three cell types of a mouse, in units, with gridlines every 2 units
Bar graph of receptor for hormone M in three cell types of a mouse, in units, with gridlines every 2 units

Which of the following justifies the prediction?

  1. A. The gut cells need M to digest food
    A cell cannot weigh what it needs.
    A cell responds only when its receptor binds the signal.
  2. B. The blood carries M to the gut cells first
    Every cell the blood reaches gets M.
    Only a cell that carries the receptor responds.
  3. C. ✓ The gut cells read 5.8 units of the receptor

Why: A cell responds to a signal only when it carries a receptor that binds the signal.
The gut cells carry 5.8 units of the receptor for M, far above the 0.3 units of cells that lack it.
So the receptor is the justification.

48
Check q11

A mouse’s fat cells carry 0.1 units of the receptor for signal S, below the 0.3 units measured in cells that lack the receptor. At a low concentration of S the fat cells showed no change. A researcher raises the concentration of S one hundred times.

Which of the following do the fat cells do at the higher concentration?

  1. A. The fat cells now respond weakly
    The fat cells carry almost no receptor, so they have almost nothing to bind S at any concentration.
    They show no response at all, not a weak one.
  2. B. The fat cells now respond as strongly as the liver cells
    More signal changes how strongly receptor-carrying cells respond, never which cells respond.
    The fat cells carry almost no receptor, so at any concentration almost nothing in them binds S.
  3. C. ✓ The fat cells still stay as they are

Why: The amount of signal changes how strongly the cells that carry the receptor respond.
It never changes which cells respond.
The fat cells carry 0.1 units of the receptor: almost none.
So however much S arrives, almost nothing binds it, and the fat cells stay as they are.

49
Check q12

Cells lining a frog’s gut carry 6.3 units of the receptor for hormone R. The frog’s skin cells carry 0.2 units. Cells that lack the receptor read 0.3 units. R reaches both cell types.

Which of the following cells respond to R?

  1. A. ✓ The gut cells
  2. B. The skin cells
    The skin cells carry 0.2 units of the receptor: almost none.
    The gut cells carry 6.3 units, so R binds the gut cells.
  3. C. Both the gut cells and the skin cells
    The skin cells carry 0.2 units of the receptor, below the 0.3 units of cells that lack it.
    Nothing in the skin cells binds R.

Why: Only cells that carry the receptor respond.
The gut cells carry 6.3 units of the receptor for R, so R binds the gut cells and they respond.
The skin cells carry 0.2 units, almost none, so the skin cells do nothing.

50
Check q13

A frog’s skin cells carry 0.2 units of the receptor for hormone R; cells that lack the receptor read 0.3 units. A researcher doubles the concentration of R around the skin cells.

Which of the following do the skin cells do?

  1. A. The skin cells now respond weakly
    The skin cells carry almost no receptor for R.
    Doubling R leaves almost nothing in the skin cells for R to bind.
  2. B. ✓ The skin cells still do nothing

Why: The amount of hormone changes how strongly the cells that carry the receptor respond.
It never changes which cells respond.
The skin cells carry 0.2 units of the receptor: almost none.
So at twice the concentration R still has almost nothing to bind, and the skin cells do nothing.

51
Check q14

Kidney cells carry 7.5 units of the receptor for hormone H and respond to it. A student says: “The kidney cells respond because the kidney needs H to make urine.”

Is the student correct?

  1. A. Yes — a cell responds to the signals it needs
    A cell cannot weigh what it needs.
    A cell responds only when its receptor binds the signal.
  2. B. ✓ No — the kidney cells respond because they carry the receptor for H

Why: A cell responds to a signal only when it carries a receptor that binds the signal.
The kidney cells carry 7.5 units of the receptor for H.
So H binds the kidney cells’ receptors, and the kidney cells respond.
What the kidney needs plays no part.

52
Practice writing an answer

Hormone F is released into the blood by cells of a gland at the base of the brain. Researchers test whether F makes cells release stored fat. They measure how much of the receptor for F three cell types carry: fat cells, a mutant fat-cell line, and skin cells. The graph shows the results. The researchers then give each cell type the same concentration of F for 30 minutes and measure the fat released into the dish. In this measurement of receptor, cells that lack the receptor read 0.3 units.

Bar graph of receptor for hormone F in the three cell types tested, in units, with gridlines every 2 units
Bar graph of receptor for hormone F in the three cell types tested, in units, with gridlines every 2 units

(a) Identify the signaling cell and the chemical signal in this case. (1 pt)

Model answer The signaling cells are the cells of the gland at the base of the brain, which release hormone F into the blood.
The chemical signal is hormone F, the molecule that carries information to the other cells.
Rubric
  • Award 1 point for: the gland cells as the signaling cells and hormone F as the chemical signal (both needed).

Slip Naming the fat cells as the signaling cells. The fat cells respond; the signaling cell is the one that releases the molecule.

(b) Predict which of the three cell types release fat when the researchers add F, and justify your prediction using the graph. (2 pt)

Model answer The fat cells release fat; the mutant fat cells and the skin cells do not.
A cell responds to a signal only when it carries a receptor that binds the signal.
The fat cells carry 8.4 units of the receptor for F, so F binds them and they respond.
The mutant fat cells carry 0.2 units and the skin cells 0.3 units: almost no receptor.
So they stay as they are.
Rubric
  • Award 1 point for: the prediction that the fat cells respond and the mutant fat cells and skin cells do not.
  • Award 1 point for: the justification that only cells carrying the receptor (8.4 units) can bind F and respond; the other two carry almost none (0.2 and 0.3 units).

Slip Predicting that the mutant fat cells respond because they are fat cells. What matters is the receptor they carry, and they carry almost none.

(c) The researchers then raise the concentration of F ten times for all three cell types. Predict how the fat released by the mutant fat-cell line and by the skin cells changes, and justify your prediction. (1 pt)

Model answer The mutant fat cells and the skin cells still release no fat.
More F changes how strongly the cells that carry the receptor respond.
More F changes nothing in cells that carry almost none.
The mutant fat cells carry 0.2 units of the receptor and the skin cells 0.3 units: almost none.
So at ten times the concentration F still has almost nothing to bind in these cells, and they release no fat.
Rubric
  • Award 1 point for: the mutant fat cells and the skin cells still release no fat, because the amount of signal changes how strongly receptor-carrying cells respond, never which cells respond; with almost no receptor (0.2 and 0.3 units) there is almost nothing for the extra F to bind.

Slip Predicting that the mutant fat cells and skin cells now respond a little. A cell with almost no receptor has almost nothing to bind the hormone at any concentration.

APBIO-U04-L02 Cells that touch

Topic 4.1 · Cell Communication · 33 steps

Three pairs of cells: two cells touching; two cells with a tiny gap between them that scattered molecules cross; two cells far apart with a blood vessel between them
Three pairs of cells: two cells touching; two cells with a tiny gap between them that scattered molecules cross; two cells far apart with a blood vessel between them

Here are two immune cells. The cell on the left has captured part of a virus. It presses itself against the cell on the right.

Within minutes, the cell on the right changes what it is doing. Now put a fine mesh between the two cells. The mesh lets molecules through, but it keeps the cells apart. With the mesh in place, the cell on the right does nothing at all. How did the message pass?

Unit 4 · Cell Communication and Cell Cycle

1Cells that touch

2

Video: Watch: Cells that touch

Two immune cells pressed together: a protein on the surface of one binds a receptor on the surface of the other where they touch, nothing enters the fluid between them, and a fine mesh that keeps the cells apart stops the message.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02a.mp4

3

How does a message pass between two cells that touch?

4

A protein on the surface of one cell binds a receptor on the surface of the other. The two proteins bind where the cells touch.

5

Nothing is released into the fluid between the cells.

6

So the message travels no further than the touch. A mesh that keeps the cells apart stops it.

7

Touch is the shortest distance a message can travel between cells. The mesh test is how you recognize it.

8
Check q1

What is a receptor for a chemical signal?

  1. A. The molecule a cell releases to carry a message
    The molecule a cell releases is the chemical signal itself.
    The receptor is the protein that binds it.
  2. B. ✓ A protein on or in a cell that binds that signal
  3. C. A cell that changes what it does when the signal arrives
    A cell that responds is a target cell.
    The receptor is the protein on or in it that binds the signal.

Why: A receptor is a protein on or in a cell.
It binds one particular chemical signal by fitting it in a pocket.
A cell that carries the receptor responds when the signal arrives.

9

Here is a drawing of the two immune cells pressed together.

Two immune cells pressed together; a surface protein on the left cell binds a receptor on the surface of the right cell where they touch, both labeled, and the fluid around them carries no molecules
Two immune cells pressed together; a surface protein on the left cell binds a receptor on the surface of the right cell where they touch, both labeled, and the fluid around them carries no molecules
10

A protein on the surface of the cell holding the virus fragment binds a receptor on the surface of the second cell.

11

The message passes where the two cells touch. Nothing is released into the fluid between them.

12

The cell holding the virus fragment is an antigen-presenting cell, and the cell it presses against is a helper T cell. A killer T cell touches the infected cell it will destroy in the same way.

13

Now imagine a fine mesh between the two cells. The mesh lets molecules through, but it keeps the two surfaces apart.

The same two immune cells held apart by a fine mesh drawn as a dashed vertical band between them; the surface protein on the left cell and the receptor on the right cell face each other across the mesh without touching; small molecules are scattered on both sides of the mesh
The same two immune cells held apart by a fine mesh drawn as a dashed vertical band between them; the surface protein on the left cell and the receptor on the right cell face each other across the mesh without touching; small molecules are scattered on both sides of the mesh
14

The surface protein cannot reach the receptor. So the message does not pass.

15

A released molecule would cross the mesh and carry its message through. A message that passes by touch stops at the mesh.

16

Signaling in which a protein on the surface of one cell binds a receptor on the surface of a cell it touches is called .

17

Some touching cells are also joined by channels that let small molecules pass straight from one cell into the next: gap junctions in animals, plasmodesmata in plants.

18

What you are expected to know Describe signaling by cell-to-cell contact: a protein on the surface of one cell binds a receptor on the surface of the cell it touches, and nothing is released into the fluid between them.

19
Check q2

A cell lining the gut begins to make mucus when an immune cell presses against it. With a fine mesh between the two cells, one that lets molecules through but keeps the cells apart, the gut cell makes no mucus.

Which of the following is how the message passes from the immune cell to the gut cell?

  1. A. By a molecule the immune cell releases into the fluid around both cells
    A released molecule would cross the mesh and reach the gut cell.
    With the mesh in place, the gut cell made no mucus.
  2. B. By a molecule that the mesh stops from reaching the gut cell’s surface
    The mesh lets molecules through.
    Only the two cell surfaces are kept apart.
  3. C. ✓ By a protein on the immune cell’s surface binding a receptor on the gut cell

Why: The mesh lets molecules through, so a released molecule would still reach the gut cell.
The gut cell made mucus only when the immune cell pressed against it.
So the message passes where the cells touch: a surface protein binds a receptor.

20
Practice writing an answer

A cell lining the gut begins to make mucus when an immune cell presses against it. With a fine mesh between the two cells, one that lets molecules through but keeps the cells apart, the gut cell makes no mucus.

(a) Explain why the mesh stops the message. (1 pt)

Frame The message passes when …

Model answer The message passes when a protein on the immune cell’s surface binds a receptor on the gut cell’s surface.
The two proteins can bind only where the cells touch.
The mesh keeps the two cells apart.
So the surface protein cannot reach the receptor.
Nothing is released into the fluid, so nothing crosses the mesh.
So the message does not pass.
Rubric
  • Award 1 point for: the message passes by a surface protein binding a receptor where the cells touch, the mesh keeps the two surfaces apart, and nothing is released that could cross the mesh.

Slip Saying the mesh blocks a released molecule. The mesh lets molecules through; it is the two surfaces that cannot touch.

21

Back to the two immune cells: the cell on the left holding part of a virus, pressed against the cell on the right.

22

A protein on the surface of the left cell binds a receptor on the surface of the right cell where they touch. Within minutes, the right cell changes what it is doing.

23

With a fine mesh between them, the two surfaces cannot touch. Nothing is released into the fluid, so nothing crosses the mesh.

24

So the right cell does nothing.

25Quick quiz: cell-to-cell contact mixed practice

26
Check q3

What is cell-to-cell contact?

  1. A. A cell releases a molecule that spreads through the fluid to the cells beside it
    In cell-to-cell contact, nothing is released into the fluid.
    The two cell surfaces bind.
  2. B. ✓ A protein on one cell’s surface binds a receptor on the surface of a cell it touches
  3. C. The blood carries a molecule from one cell to cells far away
    In cell-to-cell contact, the two cells touch.
    No blood carries anything between them.

Why: In cell-to-cell contact, two cells touch.
A protein on the surface of one binds a receptor on the surface of the other.
Nothing is released into the fluid between them.

27
Check q4

Two cells communicate by cell-to-cell contact.

What is released into the fluid between them?

  1. A. A small molecule
    A released molecule would spread through the fluid.
    In cell-to-cell contact, the message passes only where the surfaces touch.
  2. B. The surface protein
    The surface protein stays fixed in the cell’s membrane.
    It binds the receptor on the cell it touches.
  3. C. ✓ Nothing

Why: The protein on one cell’s surface binds the receptor on the other cell’s surface.
Both proteins stay in their membranes.
So nothing is released into the fluid between the cells.

28
Practice writing an answer

Two cells in an embryo are pressed together, and one of them responds to the other.

(a) Describe signaling by cell-to-cell contact. (1 pt)

Model answer Two cells touch.
A protein on the surface of one cell binds a receptor on the surface of the other.
Nothing is released into the fluid between them.
Rubric
  • Award 1 point for: a protein on the surface of one cell binds a receptor on the surface of a cell it touches, with nothing released into the fluid.

29Mixed practice mixed practice

30
Check q5

In an embryo, a cell changes what it becomes only when a neighboring cell’s surface is pressed against it. Fluid the neighbor grew in has no effect on it.

Did the message need the two cells to touch?

  1. A. ✓ Yes
  2. B. No
    The neighbor’s fluid had no effect.
    So no released molecule carries the message; the two surfaces had to touch.

Why: The cell responds only when the neighbor’s surface presses against it.
The neighbor’s fluid does nothing.
So a protein on one surface binds a receptor on the other: cell-to-cell contact.

31
Check q6

Cells at the edge of a wound release a molecule into the fluid. Cells a few cell-widths away, with fluid between them and the wound cells, begin dividing within a day.

Did the message need the two cells to touch?

  1. A. Yes
    Fluid lies between the responding cells and the wound cells.
    The molecule reached them through that fluid.
  2. B. ✓ No

Why: The wound cells released a molecule into the fluid.
Cells a few cell-widths away, with fluid between, responded.
So the message crossed the fluid, and no touch was needed.

32
Check q7

A killer T cell presses against an infected cell, and within minutes the infected cell begins to die. Infected cells a few cell-widths away from the T cell are unaffected.

Did the message need the two cells to touch?

  1. A. ✓ Yes
  2. B. No
    A released molecule would spread through the fluid to the infected cells a few cell-widths away, and they would begin to die too.
    They are unaffected.

Why: Only the infected cell the T cell presses against dies.
The infected cells a few cell-widths away are unaffected.
So no released molecule carries the message through the fluid.
So a protein on the T cell’s surface binds a receptor on the infected cell where the two touch.

Glossary

cell-to-cell contact
Signaling in which two cells touch and a protein on the surface of one binds a receptor on the surface of the other, so the message passes with nothing released into the fluid between them. Also called direct contact.

APBIO-U04-L02D Nearby cells

Topic 4.1 · Cell Communication · 60 steps

A nerve ending on the left with a narrow gap between it and a muscle cell; dots of a released molecule sit in the gap; further right, after a break in the drawing, a second muscle cell with no dots near it
A nerve ending on the left with a narrow gap between it and a muscle cell; dots of a released molecule sit in the gap; further right, after a break in the drawing, a second muscle cell with no dots near it

Here is a nerve ending beside a muscle cell. A gap a fraction of a micrometer wide separates them.

Each time a signal reaches the nerve ending, it releases a molecule into that gap. The molecule crosses the gap. The muscle cell contracts within a millisecond. Muscle cells a millimeter further away never respond. Why does the message stop at the cell next door?

Unit 4 · Cell Communication and Cell Cycle

1A signal for the cells next door

2

Video: Watch: A signal for the cells next door

A nerve ending releases a molecule into the gap beside a muscle cell; the molecule diffuses across the gap to the receptors on the cell beside it and no further.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Da.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Da.mp4

3

How does a message reach the cells next door and no further?

4

The sending cell releases a molecule into the fluid around it.

5

The molecule spreads by diffusion to the cells beside it.

6

Within moments, an enzyme destroys the molecule, or the nearby cells take it up. So the molecule never travels further.

7

Knowing why the reach is short lets you predict what happens when the enzyme that destroys the signal is blocked.

8
Check q1

A cell releases a molecule into the fluid around it. Near the cell, the molecule is concentrated; further away, it is dilute.

Which way does the molecule spread?

  1. A. ✓ From where it is more concentrated to where it is less concentrated
  2. B. From where it is less concentrated to where it is more concentrated
    Particles move at random, so more leave the crowded region than arrive in it.
    The molecule spreads away from the cell, where it is concentrated.
  3. C. It stays where the cell released it
    Particles move constantly and at random.
    So the molecule spreads out from where it is concentrated.

Why: Particles move constantly and at random.
More leave the region where the molecule is concentrated than arrive there.
So the molecule spreads from where it is more concentrated to where it is less concentrated: diffusion.

9

Here is a drawing of the nerve ending, the gap and the muscle cell.

A nerve ending on the left and a muscle cell on the right with a narrow gap between them; molecules released by the nerve ending are scattered across the gap, and receptor proteins sit on the muscle cell's membrane facing the gap
A nerve ending on the left and a muscle cell on the right with a narrow gap between them; molecules released by the nerve ending are scattered across the gap, and receptor proteins sit on the muscle cell's membrane facing the gap
10

The nerve ending releases a small molecule into the gap.

11

The molecule diffuses across the gap. It binds receptors on the muscle cell within a millisecond.

12

The molecule spreads only to the cell beside the nerve ending. It never reaches the muscle cells a millimeter further away.

13

A signal that a cell releases into the fluid and that spreads by diffusion to the target cells in its immediate neighborhood is called a .

14

In an embryo, some cells release a signal into the fluid. The concentration of the signal falls with distance from them.

15

How much of the signal reaches a nearby cell decides which cell type that cell becomes. Signals like these are called morphogens.

16

An infected plant cell releases a signal that makes its neighbors strengthen their cell walls. This is called the plant immune response.

17

What you are expected to know Describe a local regulator: a signal a cell releases into the fluid that spreads by diffusion to the cells in its immediate neighborhood.

18
Check q2

When skin is cut, the damaged cells release a molecule. Cells within a fraction of a millimeter begin dividing faster within a day.

How does the molecule reach the nearby cells?

  1. A. ✓ It spreads through the fluid by diffusion
  2. B. The blood carries it to them
    Cells a fraction of a millimeter away are beside the damaged cells.
    The molecule diffused to them through the fluid; no blood carried it.
  3. C. The damaged cells press against them
    The damaged cells released the molecule into the fluid.
    It reached cells they do not touch.

Why: The damaged cells released the molecule into the fluid.
The molecule spread by diffusion to the cells a fraction of a millimeter away.
A signal that spreads by diffusion to nearby cells is a local regulator.

19Quick quiz: local regulator mixed practice

20
Check q3

What is a local regulator?

  1. A. A signal the blood carries through the whole body
    A local regulator reaches only the cells in its immediate neighborhood.
    No blood carries it.
  2. B. A protein on one cell’s surface that binds a receptor on a cell it touches
    A local regulator is released into the fluid.
    A surface protein that binds on touch is cell-to-cell contact.
  3. C. ✓ A signal a cell releases that spreads by diffusion to the cells beside it

Why: A local regulator is a signal a cell releases into the fluid.
It spreads by diffusion.
It reaches the target cells in its immediate neighborhood.

21
Check q4

A cell releases a local regulator.

Which cells does it reach?

  1. A. Only the cell it touches
    A local regulator is released into the fluid, so it reaches cells the sender does not touch.
  2. B. ✓ The cells in its immediate neighborhood
  3. C. Cells throughout the body
    A local regulator reaches only the cells in its immediate neighborhood.
    No blood carries it through the body.

Why: The cell releases the local regulator into the fluid.
The local regulator spreads by diffusion.
So it reaches the cells beside the sender.

22
Practice writing an answer

Cells lining a healing cut release a local regulator.

(a) Describe what a local regulator is. (1 pt)

Model answer A local regulator is a signal a cell releases into the fluid around it.
The signal spreads by diffusion.
It reaches the target cells in its immediate neighborhood.
Rubric
  • Award 1 point for: a signal a cell releases that spreads by diffusion to the cells in its immediate neighborhood.

23Why the message stops at the cell next door

24

Video: Watch: Why the message stops at the cell next door

The molecule in the gap is broken down by an enzyme within moments, or taken up by the nearby cells, so it is gone before it can spread further.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Db.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Db.mp4

25

Why does the molecule never reach the muscle cells a millimeter further away?

26

Here is the nerve ending, the gap and the muscle cell again.

A nerve ending on the left and a muscle cell on the right with a narrow gap between them; molecules released by the nerve ending are scattered across the gap, and receptor proteins sit on the muscle cell's membrane facing the gap
A nerve ending on the left and a muscle cell on the right with a narrow gap between them; molecules released by the nerve ending are scattered across the gap, and receptor proteins sit on the muscle cell's membrane facing the gap
27

Within moments, an enzyme in the gap breaks the molecule down.

28

Some local regulators are not broken down. The nearby cells take them up instead.

29

Either way, the molecule is gone within moments. It is destroyed or taken up before it can spread further.

30

A molecule spreading by diffusion moves in every direction. Enzymes destroy it before it has spread more than a few cell-widths.

31

So a local regulator reaches only nearby cells because enzymes destroy it, or cells take it up, within moments.

32

What you are expected to know Explain why a local regulator reaches only nearby cells: enzymes destroy it, or cells take it up, within moments.

33
Practice writing an answer

When skin is cut, the damaged cells release a local regulator. Cells within a fraction of a millimeter begin dividing faster within a day. Cells a centimeter away carry the same receptor and show no change. Enzymes break the molecule down within minutes of its release.

(a) Explain why the signal acts only on the nearby cells. (1 pt)

Model answer The local regulator spreads from the damaged cells by diffusion.
Enzymes break it down within minutes.
So it is gone before it has spread more than a fraction of a millimeter.
The cells a centimeter away carry the receptor, but the local regulator never reaches them.
So nothing in those cells changes.
Rubric
  • Award 1 point for: the molecule spreads by diffusion and is broken down within minutes, so it is gone before it reaches the cells a centimeter away; those cells carry the receptor, but the molecule never reaches them.

Slip Saying the distant cells lack the receptor. They carry it; the molecule is destroyed before it reaches them.

34
Check q5

A student says: “The signal from the cut acts only on the nearby cells because the signal knows to stop once it reaches them.”

Is the student correct?

  1. A. Yes — the signal stops spreading once it has reached its target cells
    A molecule spreading by diffusion has no way to stop.
    Enzymes break the molecule down within minutes.
    So the molecule is gone before it spreads further.
  2. B. ✓ No — enzymes destroy the signal before it can spread further

Why: A molecule has no way of knowing where to go.
The molecule spreads by diffusion in every direction.
Enzymes break the molecule down within minutes.
So the molecule is destroyed before it can spread further than the nearby cells.

35
Check q6

Cells at the edge of a wound release a local regulator. Cells within 0.2 mm respond. Cells 2 mm away carry the receptor and show no change. Now imagine a drug blocks the enzyme that breaks the local regulator down.

Which cells respond now?

  1. A. None of the cells, within 0.2 mm or beyond
    Blocking the enzyme does not stop the cells from releasing the local regulator.
    The cells within 0.2 mm still respond.
  2. B. Only the cells within 0.2 mm, as before
    With the enzyme blocked, the local regulator is no longer destroyed within moments.
    It keeps spreading.
  3. C. ✓ The cells within 0.2 mm and cells further away too

Why: The enzyme destroyed the local regulator within moments, so it reached only the cells within 0.2 mm.
With the enzyme blocked, the local regulator is not destroyed.
It keeps spreading by diffusion.
So cells further away, which carry the receptor, now respond too.

36The nerve cell’s local regulator

37

Video: Watch: The nerve cell’s local regulator

The molecule a nerve ending releases into the gap beside its target is a local regulator: a neurotransmitter.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Dc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Dc.mp4

38

Here is the nerve ending, the gap and the muscle cell once more.

A nerve ending on the left and a muscle cell on the right with a narrow gap between them; molecules released by the nerve ending are scattered across the gap, and receptor proteins sit on the muscle cell's membrane facing the gap
A nerve ending on the left and a muscle cell on the right with a narrow gap between them; molecules released by the nerve ending are scattered across the gap, and receptor proteins sit on the muscle cell's membrane facing the gap
39

The molecule the nerve ending releases into the gap is a local regulator.

40

It diffuses across the gap to the muscle cell beside it. An enzyme in the gap destroys it within a millisecond.

41

The local regulator a nerve cell releases into the gap beside its target is called a .

42

A nerve cell’s target can be a muscle cell, a gland cell or another nerve cell. Each one carries receptors for the neurotransmitter on the side facing the gap.

43

What you are expected to know Identify a neurotransmitter as the local regulator a nerve cell releases into the gap beside its target.

44
Check q7

A nerve ending sits a fraction of a micrometer from a gland cell, with a gap between them. Each time a signal reaches the nerve ending, a molecule crosses the gap. The gland cell responds within two milliseconds, and the molecule is gone from the gap within one millisecond.

Which of the following kinds of signal is the molecule?

  1. A. A hormone
    A hormone is carried in the blood through the whole body.
    This molecule crossed a gap a fraction of a micrometer wide by diffusion; no blood carried it.
  2. B. A local regulator released by the gland cell
    The gland cell is the target: it responds.
    The nerve ending released the molecule.
  3. C. ✓ A neurotransmitter
  4. D. A surface protein bound where the two cells touch
    A surface protein binds only where two cells touch, and nothing is released.
    Here a molecule crossed a gap.

Why: The nerve ending released the molecule into the gap.
The molecule diffused across the gap and was gone within a millisecond.
A signal that diffuses to the next cell and is destroyed at once is a local regulator.
A local regulator released by a nerve cell is a neurotransmitter.

45

Back to the nerve ending beside the muscle cell, with a gap a fraction of a micrometer wide between them.

46

Each time a signal reaches the nerve ending, it releases a neurotransmitter into the gap. The neurotransmitter diffuses across the gap, and the muscle cell contracts within a millisecond.

47

An enzyme in the gap destroys the neurotransmitter within a millisecond. So it never reaches the muscle cells a millimeter further away.

48Quick quiz: neurotransmitter mixed practice

49
Check q8

What is a neurotransmitter?

  1. A. A hormone a nerve cell releases into the blood to reach distant cells
    A neurotransmitter is released into a gap, not into the blood.
    It reaches only the cell beside the nerve ending.
  2. B. ✓ The local regulator a nerve cell releases into the gap beside its target
  3. C. The receptor on a muscle cell that binds a nerve cell’s signal
    The receptor is the protein that binds the neurotransmitter.
    The neurotransmitter is the molecule the nerve cell releases.

Why: A nerve cell releases a molecule into the gap beside its target.
The molecule diffuses across the gap and is destroyed within a millisecond, so it is a local regulator.
A local regulator a nerve cell releases is a neurotransmitter.

50
Check q9

Which kind of cell releases a neurotransmitter?

  1. A. A gland cell
    A gland cell can be a nerve cell’s target.
    It receives the neurotransmitter.
  2. B. A muscle cell
    A muscle cell can be a nerve cell’s target.
    It receives the neurotransmitter.
  3. C. ✓ A nerve cell

Why: A neurotransmitter is the local regulator a nerve cell releases.
Muscle cells and gland cells are targets that receive it.

51
Practice writing an answer

A nerve ending sits beside a heart muscle cell, with a narrow gap between them.

(a) Describe what a neurotransmitter is. (1 pt)

Model answer A neurotransmitter is a local regulator.
A nerve cell releases it into the gap beside its target cell.
It diffuses across the gap to that cell.
Rubric
  • Award 1 point for: the local regulator a nerve cell releases into the gap beside its target.

52Mixed practice mixed practice

53
Check q10

Cells in an inflamed joint release a molecule into the fluid. Within minutes, cells a few cell-widths away begin making a protective protein. The molecule is gone from the joint within ten minutes.

Which of the following kinds of signal is the molecule?

  1. A. A signal passed by cell-to-cell contact
    Cells a few cell-widths away responded.
    So the molecule crossed the fluid between cells.
  2. B. ✓ A local regulator
  3. C. A hormone
    The molecule reached only cells a few cell-widths away and was gone within ten minutes.
    No blood carried it.

Why: The molecule diffused to cells a few cell-widths away and was destroyed within ten minutes.
A signal that acts only in its immediate neighborhood and is quickly destroyed is a local regulator.

54
Check q11

A nerve ending releases a molecule into the gap beside a cell in the gut wall. The gut cell contracts within a millisecond, and the molecule is gone from the gap within a millisecond.

Which of the following kinds of signal is the molecule?

  1. A. A hormone
    A hormone is carried in the blood.
    This molecule diffused across a gap a fraction of a micrometer wide.
  2. B. A local regulator released by the gut cell
    The gut cell is the target: it responds.
    The nerve ending released the molecule.
  3. C. ✓ A neurotransmitter

Why: The nerve ending released the molecule into the gap beside its target.
The molecule diffused across the gap and was gone within a millisecond.
A local regulator a nerve cell releases into the gap beside its target is a neurotransmitter.

55
Check q12

A cell lining the lung releases a local regulator.

How does the local regulator reach the cells beside it?

  1. A. ✓ By diffusion through the fluid
  2. B. In the blood
    The blood carries hormones through the whole body.
    A local regulator spreads only through the fluid around the cell that released it.
  3. C. Through the two cells touching
    A local regulator is released into the fluid.
    It reaches cells the sender does not touch.

Why: The lung cell releases the local regulator into the fluid around it.
The local regulator spreads by diffusion to the cells beside the lung cell.

56
Check q13

A local regulator released by a kidney cell reaches only cells a few cell-widths away.

Why does it travel no further?

  1. A. It is too large to diffuse further
    A larger molecule diffuses more slowly, but it still spreads.
    Enzymes destroy the local regulator within moments.
  2. B. ✓ Enzymes destroy it, or cells take it up, within moments
  3. C. It stops spreading once it reaches a cell that carries its receptor
    A molecule spreading by diffusion has no way to stop.
    Enzymes destroy it before it spreads further.

Why: The local regulator spreads by diffusion in every direction.
Within moments, enzymes destroy it, or cells take it up.
So it is gone before it has spread more than a few cell-widths.

57
Check q14

A student says: “A neurotransmitter is a hormone, because a nerve cell releases it into the fluid.”

Is the student correct?

  1. A. Yes — a nerve cell releases it into the fluid, so it is a hormone
    A hormone is carried in the blood through the whole body.
    A neurotransmitter diffuses across a gap to one cell and is destroyed within a millisecond.
  2. B. ✓ No — a neurotransmitter is a local regulator, not a hormone

Why: A nerve cell releases the neurotransmitter into the gap beside its target.
The neurotransmitter diffuses across the gap and is destroyed within a millisecond.
A signal that reaches only the cell beside the sender is a local regulator, not a hormone.

58
Check q15

A nerve ending releases a neurotransmitter into the gap beside a muscle cell. Now imagine a drug blocks the enzyme in the gap that destroys the neurotransmitter.

What happens to the muscle cell’s response?

  1. A. ✓ It lasts longer
  2. B. It ends sooner
    The enzyme ended the response by destroying the neurotransmitter.
    With the enzyme blocked, the neurotransmitter stays in the gap.
  3. C. It does not change
    The enzyme was what removed the neurotransmitter from the gap.
    Blocking it lets the neurotransmitter keep binding the receptors.

Why: The enzyme destroys the neurotransmitter within a millisecond.
With the enzyme blocked, the neurotransmitter stays in the gap and keeps binding the muscle cell’s receptors.
So the muscle cell’s response lasts longer.

59
Practice writing an answer

In a healing wound, the damaged cells release a local regulator. Cells within 0.5 mm of the wound begin dividing. Cells 3 cm away carry the same receptor and do not divide.

(a) Explain why only the cells within 0.5 mm divide, even though the cells 3 cm away carry the receptor too. (1 pt)

Model answer The damaged cells release the local regulator into the fluid.
It spreads by diffusion, so it reaches the cells within 0.5 mm first.
Within moments, enzymes destroy it, or cells take it up.
So it is gone before it has spread 3 cm.
The cells 3 cm away carry the receptor, but the local regulator never reaches them.
So they do not divide.
Rubric
  • Award 1 point for: the local regulator is destroyed or taken up within moments of its release, so it reaches only the cells within 0.5 mm and never the cells 3 cm away, which carry the receptor but never bind it.

Slip Saying the cells 3 cm away lack the receptor. They carry it; the local regulator never reaches them.

Glossary

local regulator
A signal a cell releases into the fluid that spreads by diffusion to the target cells in its immediate neighborhood. Enzymes destroy it, or cells take it up, within moments, so it travels no further.
neurotransmitter
The local regulator a nerve cell releases into the gap beside its target cell, which responds within a millisecond.

APBIO-U04-L02E Far away

Topic 4.1 · Cell Communication · 55 steps

A pancreas cell on the left releases dots of insulin into a blood vessel that runs across the whole picture to a liver cell on the far right; dots are scattered along the vessel
A pancreas cell on the left releases dots of insulin into a blood vessel that runs across the whole picture to a liver cell on the far right; dots are scattered along the vessel

Here is a cell in the pancreas. It releases insulin into the blood.

A liver cell a body’s length away starts storing glucose about a minute later. A neurotransmitter crosses the gap beside a muscle cell in a millisecond. How does a message travel the length of a body, and why does it take so much longer to start?

Unit 4 · Cell Communication and Cell Cycle

1A signal in the blood

2

Video: Watch: A signal in the blood

Cells in the pancreas release insulin into the blood; the blood carries it through the whole body, and every liver cell and muscle cell that carries an insulin receptor responds, wherever it sits.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Ea.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Ea.mp4

3

How does one cell reach cells all over the body?

4

It releases a hormone into the blood. The blood carries the hormone everywhere.

5

Every cell that carries the hormone’s receptor responds, wherever it sits. So one hormone reaches many organs at once.

6

The response starts more slowly than a local regulator’s. The blood has to carry the molecule, and it dilutes the molecule on the way.

7

Touch, nearby and far away are the three distances a message can travel. A table of the three is how you place any new case.

8
Check q1

Testosterone is made in the testes. The blood carries it to muscle and bone cells, which respond by growing.

What kind of chemical signal is testosterone?

  1. A. ✓ A hormone
  2. B. A neurotransmitter
    A neurotransmitter is released into the gap beside one cell.
    Testosterone is carried in the blood to cells throughout the body.

Why: Testosterone is made in one part of the body.
The blood carries it to cells elsewhere, which respond.
A chemical signal made in one part of the body that changes what happens elsewhere is a hormone.

9

Here is a drawing of the pancreas cell, the blood vessel and the liver cell.

A pancreas cell on the left releases molecules into a blood vessel that runs across the picture to a liver cell on the far right; molecules are scattered along the vessel
A pancreas cell on the left releases molecules into a blood vessel that runs across the picture to a liver cell on the far right; molecules are scattered along the vessel
10

When the concentration of glucose in the blood rises, cells in the pancreas release insulin into the blood. Insulin is a hormone.

11

The blood carries insulin through the whole body.

12

Within a minute, every liver cell that carries an insulin receptor has begun storing glucose as glycogen, wherever in the liver it sits.

13

Muscle cells carry the insulin receptor too. They respond by taking up glucose from the blood.

14

A cell with no insulin receptor does nothing, however much insulin the blood brings it. So one hormone reaches many organs at once, but only the cells that carry its receptor respond.

15

Testosterone and estrogen are steroid hormones. They are made in the testes and ovaries and act on target cells throughout the body.

16

Growth hormone, from a gland at the base of the brain, reaches bone and muscle cells and makes them grow. Thyroid hormones, from a gland in the neck, set how fast most of the body’s cells use fuel.

17

What you are expected to know Describe how a hormone reaches its target cells: the blood carries it through the whole body to every cell that carries its receptor.

18
Check q2

Cells in the kidney release a molecule into the blood. Over the next few days, cells in the bone marrow, a body’s length away, make more red blood cells.

Which of the following kinds of signal is the molecule?

  1. A. A signal passed by cell-to-cell contact
    The bone marrow lies a body’s length from the kidney.
    The two kinds of cell never touch.
  2. B. A local regulator
    A local regulator is destroyed within moments and reaches only its neighbors.
    This molecule reached cells a body’s length away.
  3. C. ✓ A hormone

Why: The kidney cells released the molecule into the blood.
The blood carried it a body’s length to the bone marrow.
Target cells there responded.
A signal the blood carries to distant target cells is a hormone.

19Why a hormone takes longer

20

Video: Watch: Why a hormone takes longer

A neurotransmitter diffuses a fraction of a micrometer in a millisecond; insulin is carried a body’s length by the blood, and diluted on the way, so its response starts about a minute later. The three distances side by side in one table.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Eb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Eb.mp4

21
Check q3

A cell releases a local regulator.

How does the local regulator reach the cells beside it?

  1. A. The blood carries it to them
    The blood carries hormones.
    A local regulator spreads only through the fluid around the cell that released it.
  2. B. ✓ It spreads through the fluid by diffusion
  3. C. The two cells touch each other
    A local regulator is released into the fluid.
    It reaches cells the sender does not touch.

Why: The cell releases the local regulator into the fluid around it.
The local regulator spreads by diffusion to the cells in its immediate neighborhood.

22

Now compare two journeys. A neurotransmitter crosses the gap beside a muscle cell in a millisecond.

23

Insulin reaches the liver cells about a minute after the pancreas cells release it. Why does the hormone take so much longer?

24

The neurotransmitter diffuses across a gap a fraction of a micrometer wide. That takes a millisecond.

25

Insulin has to travel a body’s length, so the blood carries it all the way from the pancreas to the liver.

26

That journey takes far longer than diffusion across a gap. The blood also dilutes the insulin on the way, mixing it into all the blood in the body.

27

So a hormone’s response starts more slowly than a local regulator’s: the blood has to carry the hormone a long way, and it dilutes the hormone on the way.

28

Here is a table comparing the three distances a message can travel: touch, nearby and far away.

A table comparing the three distances a message can travel, touch, nearby and far away, on five rows: what carries the message (a surface protein binding a receptor on the cell it touches; a molecule diffusing through the fluid between cells; a molecule carried in the blood), how far it reaches (only the cell it touches; the cells in its immediate neighborhood; every cell in the body that carries its receptor), how fast the response starts (once the two surfaces bind; within moments, a millisecond across a nerve ending's gap; more slowly, about a minute after insulin is released), the name (cell-to-cell contact; a local regulator; a hormone), and one example (two immune cells pressed together, one holding part of a virus; a neurotransmitter crossing the gap to a muscle cell; insulin carried from the pancreas to the liver)
29

For example, an immune cell holding part of a virus presses against a second immune cell, and nothing is released into the fluid. This message travels by touch, because the two surfaces bind where the cells touch.

30

But a nerve ending releases a molecule into the gap beside a muscle cell, and an enzyme destroys it within a millisecond. This message travels to a nearby cell, because the molecule spreads only to the cell beside it.

31

But a pancreas cell releases insulin into the blood, and liver cells a body’s length away respond. This message travels far away, because the blood carries the molecule through the whole body.

32

What you are expected to know Explain why a hormone’s response starts more slowly than a local regulator’s: the blood has to carry the hormone, and dilutes it on the way.

33
Check q4

A neurotransmitter makes a muscle cell respond within a millisecond of its release. Insulin makes liver cells respond about a minute after its release.

Why does the hormone take so much longer to act?

  1. A. Insulin is a larger molecule and diffuses more slowly across the gap
    There is no gap to cross.
    Insulin rides the blood a body’s length, and the blood is slow.
  2. B. Liver cells carry fewer receptors than muscle cells
    The delay is in the journey, not the target.
    The blood carries the hormone far before it arrives.
  3. C. ✓ The blood has to carry insulin all the way to the liver cells
  4. D. Insulin has to be broken down before the liver cells can respond
    Breakdown ends a signal.
    The delay comes from the blood carrying insulin to distant cells.

Why: A neurotransmitter diffuses across a gap a fraction of a micrometer wide.
Insulin rides the blood a body’s length before it reaches the liver cells.
That is a far slower journey.
So the response starts later.

34
Practice writing an answer

The two glands above the kidneys release epinephrine into the blood. Muscle cells in the toes respond about a minute later. A nerve ending releases a neurotransmitter into the gap beside a toe muscle cell, and the muscle cell responds within a millisecond.

(a) Explain why the epinephrine’s response starts so much later than the neurotransmitter’s. (1 pt)

Frame The neurotransmitter only has to …

Model answer The neurotransmitter only has to diffuse across a gap a fraction of a micrometer wide.
So it reaches the muscle cell within a millisecond.
The epinephrine enters the blood above the kidneys.
The blood has to carry it a body’s length to the toes.
The blood also dilutes it on the way.
So the response in the toes starts about a minute later.
Rubric
  • Award 1 point for: the neurotransmitter diffuses across a gap a fraction of a micrometer wide, while the blood has to carry the epinephrine a body’s length (and dilutes it on the way), so the epinephrine’s response starts later.
35
Check q5

A student says: “A hormone acts on every cell the blood carries it to.”

Is the student correct?

  1. A. ✓ No — a hormone acts only on cells that carry its receptor
  2. B. Yes — the blood carries the hormone to every cell
    The blood does carry the hormone to every cell.
    But a cell responds only when it carries a receptor that binds the hormone.

Why: The blood carries a hormone to every cell.
But a cell responds only when it carries a receptor that binds the hormone.
Cells with no receptor for the hormone do nothing.
So a hormone acts only on its target cells.

36

Back to the cell in the pancreas, releasing insulin into the blood.

37

The blood carries the insulin through the whole body, a body’s length to the liver. About a minute later, every liver cell that carries an insulin receptor begins storing glucose as glycogen.

38

A neurotransmitter only has to diffuse across a gap a fraction of a micrometer wide. So it reaches the muscle cell in a millisecond.

39

The blood has to carry insulin a body’s length. So insulin reaches the liver cells about a minute later.

40Quick quiz: touch, nearby or far away? mixed practice

41
Check q6

Cells lining the gut release a molecule after a meal. Over the following hours, bone cells throughout the skeleton, from the skull to the toes, respond to it.

Which of the following kinds of signal is this?

  1. A. Cell-to-cell contact
    The bones lie far from the gut.
    Only the blood could carry a molecule from the gut to every bone in the skeleton.
  2. B. A local regulator
    A local regulator is broken down within moments and reaches only its neighbors.
  3. C. ✓ A hormone

Why: One cell type released a molecule.
Target cells of another type far away, throughout the skeleton, responded hours later.
Only the blood carries a signal that far to that many cells.
So the molecule is a hormone.

42
Check q7

A damaged cell in the gum releases a molecule. Within seconds the cells touching it and those a few cell-widths away begin making a protective protein. The molecule is gone from the tissue within a minute.

Which of the following kinds of signal is this?

  1. A. Cell-to-cell contact
    Cells a few cell-widths away also responded.
    So the molecule crossed the fluid between cells.
  2. B. ✓ A local regulator
  3. C. A hormone
    The molecule reached only cells a few cell-widths away and was gone within a minute.
    No blood carried it.

Why: The molecule diffused to cells a few cell-widths away and was destroyed within a minute.
A signal that acts only in its immediate neighborhood and is quickly destroyed is a local regulator.

43
Check q8

In an embryo, a cell changes what it becomes only when a neighboring cell’s surface is pressed against it. Fluid the neighbor grew in has no effect on it.

Which of the following kinds of signal is this?

  1. A. ✓ Cell-to-cell contact
  2. B. A local regulator
    A local regulator acts through the fluid.
    The neighbor’s fluid had no effect.
    So no released molecule carries the message.
  3. C. A hormone
    No blood carries a signal between two touching cells in an embryo.
    The response needed the two cell surfaces to touch.

Why: The cell responds only when the neighbor’s surface presses against it.
The neighbor’s fluid does nothing.
So a protein on one surface binds a receptor on the other: cell-to-cell contact.

44
Check q9

A nerve ending releases a molecule into the gap beside a heart muscle cell. The cell slows its beat within a millisecond, and the molecule is gone from the gap within a millisecond.

Which of the following kinds of signal is this?

  1. A. Cell-to-cell contact
    A molecule crossed the gap.
    In cell-to-cell contact, nothing is released.
  2. B. ✓ A local regulator
  3. C. A hormone
    The molecule crossed a gap beside one cell and was gone within a millisecond.
    No blood carried it.

Why: The nerve ending released a molecule into the gap beside its target.
The molecule diffused across the gap and was gone within a millisecond.
A signal that acts on the cell beside it and is destroyed at once is a local regulator: here, a neurotransmitter.

45
Check q10

Fat cells release a molecule into the blood. About an hour later, cells in the brain, a body’s length away, respond to it.

Which of the following kinds of signal is this?

  1. A. Cell-to-cell contact
    The brain lies a body’s length from the fat.
    Only the blood could carry the molecule there.
  2. B. A local regulator
    A local regulator is destroyed within moments and reaches only its neighbors.
  3. C. ✓ A hormone

Why: The fat cells released the molecule into the blood.
The blood carried it a body’s length to the brain.
Target cells there responded about an hour later.
A signal the blood carries to distant cells is a hormone.

46
Check q11

A killer T cell presses against an infected cell, and within minutes the infected cell begins to die. Infected cells a few cell-widths away from the T cell are unaffected.

Which of the following kinds of signal is this?

  1. A. ✓ Cell-to-cell contact
  2. B. A local regulator
    A released molecule would spread through the fluid to the infected cells a few cell-widths away, and they would begin to die too.
    They are unaffected.
  3. C. A hormone
    The blood would carry a molecule to every infected cell.
    Only the infected cell the T cell presses against dies.

Why: Only the infected cell the T cell presses against dies.
The infected cells a few cell-widths away are unaffected.
So no released molecule carries the message through the fluid.
So a protein on the T cell’s surface binds a receptor on the infected cell where the two touch: cell-to-cell contact.

47Mixed practice mixed practice

48
Check q12

Cells in the stomach wall release a molecule into the blood before a meal. About 20 minutes later, cells in the brain respond to it, and you feel hungry.

Which of the following kinds of signal is this?

  1. A. Cell-to-cell contact
    The brain lies far from the stomach.
    The two kinds of cell never touch.
  2. B. A local regulator
    A local regulator is destroyed within moments and reaches only its neighbors.
    This molecule reached the brain.
  3. C. ✓ A hormone

Why: The stomach cells released the molecule into the blood.
The blood carried it to the brain.
Target cells there responded.
A signal the blood carries to distant target cells is a hormone.

49
Check q13

An immune cell that has captured part of a bacterium presses against a second immune cell, which begins dividing within a day. With a fine mesh between them that lets molecules through but keeps the cells apart, the second cell stays as it was.

Which of the following kinds of signal is this?

  1. A. ✓ Cell-to-cell contact
  2. B. A local regulator
    A local regulator would cross the mesh through the fluid.
    With the mesh in place, the second cell stayed as it was.
  3. C. A hormone
    The blood plays no part between two cells pressed together.
    The response needed the two surfaces to touch.

Why: The mesh lets molecules through, so a released molecule would still reach the second cell.
The second cell divided only when the first pressed against it.
So a surface protein binds a receptor where the cells touch: cell-to-cell contact.

50
Check q14

The blood carries growth hormone to every cell in the body.

Which cells respond to it?

  1. A. All the cells in the body
    A cell responds only when it carries a receptor that binds the hormone.
    Cells with no receptor do nothing.
  2. B. ✓ Only the cells that carry its receptor

Why: The blood carries growth hormone to every cell.
A cell responds only when it carries a receptor that binds it.
So only the cells that carry the growth hormone receptor respond.

51
Check q15

A molecule released by one cell reaches only the cells beside it and is gone within a minute.

What carries the message to those cells?

  1. A. ✓ Diffusion through the fluid
  2. B. The blood
    The blood carries a molecule through the whole body.
    This molecule reached only the cells beside the sender.
  3. C. Surface proteins binding where the cells touch
    The molecule was released, so it left the cell’s surface.
    It reached cells the sender does not touch.

Why: The molecule was released into the fluid and reached only the cells beside the sender.
It spread to them by diffusion through the fluid.
It is a local regulator.

52
Check q16

A nerve ending releases a neurotransmitter into the gap beside a muscle cell. A gland releases a hormone into the blood at that very moment.

Which response starts first?

  1. A. The distant cells’ response to the hormone
    The blood has to carry the hormone a long way before it arrives.
    The neurotransmitter has only a gap to cross.
  2. B. ✓ The muscle cell’s response to the neurotransmitter
  3. C. The two responses start at the same moment
    The neurotransmitter crosses its gap in a millisecond.
    The hormone is still being carried in the blood.

Why: The neurotransmitter diffuses across a gap a fraction of a micrometer wide, so the muscle cell responds within a millisecond.
The blood has to carry the hormone to its distant target cells.
So the muscle cell’s response starts first.

53
Check q17

After a meal, cells lining the small intestine release a molecule into the blood. Within an hour, cells in the pancreas, several centimeters away, respond to it.

Which of the following kinds of signal is this?

  1. A. Cell-to-cell contact
    The intestine cells and the pancreas cells never touch.
    A molecule was released into the blood.
  2. B. A local regulator
    A local regulator spreads through the fluid only to the cells beside the sender.
    This molecule traveled several centimeters in the blood.
  3. C. ✓ A hormone

Why: The intestine cells released the molecule into the blood.
The blood carried it several centimeters to the pancreas.
A signal the blood carries to target cells elsewhere is a hormone.

54
Practice writing an answer

Cells in a bruised muscle release a molecule into the fluid. Cells within 0.3 mm begin repairing themselves within an hour. Muscle cells 2 cm away carry the same receptor and show no change. The molecule is gone from the muscle within five minutes.

(a) Determine which kind of signal the molecule is, and support your decision with the observations. (2 pt)

Model answer The molecule is a local regulator.
The bruised cells released it into the fluid, so the message did not pass by touch.
It reached only the cells within 0.3 mm, and it was gone within five minutes.
The cells 2 cm away carry the receptor but never received it.
A hormone in the blood would have reached them.
So the molecule spread by diffusion only to the cells nearby: a local regulator.
Rubric
  • Award 1 point for the decision: a local regulator.
  • Award 1 point for the support: the molecule was released into the fluid (so not cell-to-cell contact), reached only cells within 0.3 mm and was gone within five minutes, while cells 2 cm away with the receptor did not respond (so not carried by the blood).

Slip Calling the molecule a hormone because it was released. A hormone is carried in the blood to cells throughout the body; this molecule never reached the cells 2 cm away.

APBIO-U04-L02B Single cells counting their neighbors

Topic 4.1 · Cell Communication · 66 steps

A schematic conical flask holding three rod-shaped cells, dark, beside a photograph of a conical flask of a crowded bacterial culture glowing blue-green in a dark room
A schematic conical flask holding three rod-shaped cells, dark, beside a photograph of a conical flask of a crowded bacterial culture glowing blue-green in a dark room

Photo: Bathyctena, Wikimedia Commons, CC BY-SA 4.0 (cropped and resized).

Here are two flasks of the same light-producing bacterium. The sparse flask holds 100,000 cells per mL, and its water is almost dark.

The crowded flask holds 100,000,000 cells per mL, and the whole flask glows. Each bacterium is a single cell on its own in the water, and the cells make light only when they are crowded. How does one bacterium know how crowded it is?

Unit 4 · Cell Communication and Cell Cycle

1Crowded cells switch on

2

Video: Watch: Crowded cells switch on

Each bacterium releases a small signal molecule into the water. Few cells leave the molecule dilute; many cells build it up, and once it has built up enough every cell switches on its light.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Ba.mp4

3
Check q1

A cell releases a signal molecule into the fluid around it, and the molecule spreads by diffusion.

Which cells respond to the molecule?

  1. A. Every cell the molecule reaches
    A cell with no receptor for the molecule does nothing, however much of the molecule reaches it.
  2. B. ✓ Only the cells that carry a receptor for it

Why: A cell responds to a signal only through a receptor that binds it.
So only the cells that carry the receptor respond.

4

How can a single cell count its neighbors?

5

Each bacterium releases a small signal molecule into the water. Each bacterium also carries a receptor for that same molecule.

6

When cells are few, the molecule diffuses away and stays dilute.

7

When cells are dense, the molecule builds up. The molecule binds the receptors, and every cell switches on the same behavior: here, making light.

8

Here are the two flasks, with what is in the water drawn in.

Two flasks of a light-producing bacterium: the left holds a few rod-shaped cells and a few scattered signal molecules and is dark; the right is crowded with cells and signal molecules and glows
Two flasks of a light-producing bacterium: the left holds a few rod-shaped cells and a few scattered signal molecules and is dark; the right is crowded with cells and signal molecules and glows
9

Each rod is one bacterium, a single cell.

10

At 100,000 cells per mL the culture is almost dark. At 100,000,000 cells per mL the culture glows brightly.

11

Each cell releases a small signal molecule into the water around it.

12

When cells are few, the molecule diffuses away through the water and stays dilute. When cells are many, the molecule builds up in the water.

13

Once the molecule has built up enough, every cell in the flask switches on the same behavior: here, making light.

14

Bacteria detecting how crowded they are through a released signal that builds up as the cells become dense is called .

15

A quorum is the smallest number of members that must be present before a group can make a decision.

16

These bacteria make light only once enough cells are present.

17

What you are expected to know Describe quorum sensing: bacteria switch on one behavior only when they are crowded, sensing the crowd through a signal molecule they release that builds up as the cells become dense.

18
Check q2

Suppose a culture of the light-producing bacterium at 20,000 cells per mL is left to multiply for a day. By the next day it holds 200,000,000 cells per mL.

Which of the following happens to the concentration of the signal molecule in the water?

  1. A. ✓ The concentration rises
  2. B. The concentration stays the same
    Every cell releases the signal molecule.
    By the next day, 10,000 times as many cells release it into the same water.
  3. C. The concentration falls
    More cells release more of the molecule into the same water.

Why: Each cell releases the signal molecule into the water.
By the next day, 10,000 times as many cells release it into the same volume of water.
So the concentration of the signal molecule rises.

19
Check q3

Two flasks hold the same light-producing bacterium. One holds 30,000 cells per mL. The other holds 300,000,000 cells per mL.

Which flask glows?

  1. A. The flask at 30,000 cells per mL
    At 30,000 cells per mL the signal molecule diffuses away and stays dilute.
    The cells stay dark.
  2. B. ✓ The flask at 300,000,000 cells per mL

Why: The cells make light only when the signal molecule has built up enough.
At 300,000,000 cells per mL, many cells release the molecule into the same water, so it builds up.
So only the flask at 300,000,000 cells per mL glows.

20Quick quiz: quorum sensing mixed practice

21
Check q4

What is quorum sensing?

  1. A. ✓ Bacteria sensing the crowd through a molecule they all release
  2. B. Bacteria touching their neighbors to count how many there are
    The count works through a molecule released into the water, not through touch.
  3. C. Bacteria detecting the light their neighbors give off
    Each cell detects a released signal molecule, not light.

Why: Each bacterium releases a signal molecule into the water.
When cells are dense, the molecule builds up.
Bacteria detecting the crowd through that built-up signal is quorum sensing.

22
Practice writing an answer

Many kinds of bacteria change what they do once they become crowded.

(a) State what quorum sensing is. (1 pt)

Model answer Quorum sensing is bacteria detecting how crowded they are through a signal molecule that each cell releases and that builds up as the cells become dense.
Rubric
  • Award 1 point for: bacteria detecting how crowded they are through a released signal molecule that builds up as the cells become dense.
  • Accept: ‘bacteria sensing their own density through a molecule they all release’.

23How one cell counts the crowd

24

Video: Watch: How one cell counts the crowd

Each cell carries a receptor for the signal molecule. In the sparse flask the molecule stays below the threshold concentration and the receptors stay empty; in the dense flask it passes the threshold, binds the receptors, and every cell switches on.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Bb.mp4

25

Each cell also carries a receptor for the signal molecule it releases.

The two flasks again, sparse and dark on the left, crowded and glowing on the right; beneath each, one cell drawn enlarged with a receptor inside it: in the sparse flask's cell the receptor is empty and the signal molecule is floating away outside; in the crowded flask's cell one signal molecule is passing through the cell's surface and another has entered and is held on the receptor
The two flasks again, sparse and dark on the left, crowded and glowing on the right; beneath each, one cell drawn enlarged with a receptor inside it: in the sparse flask's cell the receptor is empty and the signal molecule is floating away outside; in the crowded flask's cell one signal molecule is passing through the cell's surface and another has entered and is held on the receptor
26

The receptor binds the signal molecule only when the molecule’s concentration in the water is high enough.

27

The concentration just high enough to bind the receptors is called the .

28

In the sparse flask, each cell’s molecule diffuses away through the water and stays dilute.

29

So the concentration stays below the threshold concentration. The receptors stay empty, and the flask stays dark.

30

In the dense flask, many cells release the molecule into the same water, and the concentration rises past the threshold concentration.

31

So the molecule binds the receptors. Every cell switches on its light.

32

Each cell switches on automatically once the signal molecule binds its receptor. No cell decides anything.

33

The concentration passes the threshold concentration only when many cells are releasing the molecule.

34

So a bacterium counts its neighbors by the concentration of the signal molecule they all release.

35

What you are expected to know Explain why a crowded culture glows and a sparse culture stays dark, using the signal molecule’s concentration, the threshold concentration and the receptor.

36
Practice writing an answer

A culture of a second light-producing bacterium holds 40,000 cells per mL. Each cell releases its own signal molecule and carries a receptor for it. The culture stays dark.

(a) Explain why the culture stays dark. (1 pt)

Model answer Each cell releases its signal molecule into the water.
At 40,000 cells per mL, few cells release the molecule into the water.
So the molecule diffuses away and stays dilute.
So its concentration stays below the threshold concentration.
So the molecule does not bind the receptors.
So no cell switches on its light.
Rubric
  • Award 1 point for: with few cells the signal molecule stays dilute, below the threshold concentration, so it does not bind the receptors and no cell switches on its light.

Slip Saying the cells have no receptors. Every cell carries the receptor; the molecule is too dilute to bind it.

37
Check q5

A student says: “The bacteria wait until enough of them are present, then they decide together to switch on the light.”

Is the student correct?

  1. A. ✓ No — each cell switches on by itself once the signal is concentrated enough
  2. B. Yes — the bacteria sense their neighbors and then decide together when to act
    A bacterium has no way to weigh its neighbors.
    Each cell switches on automatically once the signal molecule binds its receptor.

Why: Each cell carries a receptor for the signal molecule.
Each cell switches on its light automatically once the signal molecule’s concentration is high enough to bind that receptor.
The concentration is high enough only when many cells are releasing the molecule.

38
Check q6

A mutant strain of the bacterium carries the receptor but releases no signal molecule. A researcher adds fluid from a dense culture of the normal strain, with every cell removed, to a dense culture of the mutant.

Which of the following does the mutant culture do?

  1. A. The mutant culture glows only once its own cells are denser
    Density matters only because it raises the signal molecule’s concentration.
    The added fluid supplies the molecule directly.
  2. B. The mutant culture stays dark
    The mutant cannot make the signal molecule.
    But the mutant carries the receptor, and the added fluid supplies the molecule.
  3. C. ✓ The mutant culture glows

Why: The mutant carries the receptor and lacks only the signal molecule.
The fluid from the dense normal culture carries the molecule above its threshold concentration.
So the molecule binds the mutant’s receptors, and the culture glows.

39The fluid alone

40

Video: Watch: The fluid alone

Cell-free fluid from the dense culture makes a sparse culture glow; fresh fluid leaves one dark. The fresh-fluid culture is the control, and the result shows that the signal molecule, not the crowding, causes the light.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Bc.mp4

41
Check q7

What does the control in an experiment show?

  1. A. ✓ The result when the tested factor is absent
  2. B. The result when every factor is present
    The control lacks the one factor under test.
  3. C. The largest result the experiment can produce
    The control is a comparison, not a maximum.

Why: The control is given the same treatment as the other tubes or dishes but lacks the factor under test.
So it shows the result with that factor absent.

42

The two flasks leave one question open. Does the crowding itself switch on the light, or does the molecule in the water?

43

A researcher takes fluid from the dense culture and removes every cell from it.

44

She adds this cell-free fluid to a sparse culture. The sparse culture glows.

Two sparse cultures: the left has received fluid taken from a dense culture with its cells removed, and glows; the right has received fresh fluid and stays dark
Two sparse cultures: the left has received fluid taken from a dense culture with its cells removed, and glows; the right has received fresh fluid and stays dark
45

She adds fresh fluid, in which no cell has grown, to a second sparse culture. That culture stays dark.

46

She treats the fresh-fluid culture the same way but leaves out the factor under test, the fluid the dense cells grew in. So the fresh-fluid culture is the control.

47

The control shows that adding fluid on its own switches nothing on.

48

The dense culture’s fluid carried no cells, so no cell in the sparse culture touched a cell from the dense one.

49

So the light came from something in the fluid: the signal molecule, at the high concentration the dense cells had built up.

50

So the signal molecule, not the crowding itself, causes the light.

51

What you are expected to know Explain how the cell-free-fluid test shows that the signal molecule, not the crowding itself, causes the light.

52
Check q8

Cell-free fluid from a dense culture of the light-producing bacterium raises a sparse culture from 1 to 140 light units. Cell-free fluid from a sparse culture leaves it at 1 light unit, and so does fresh fluid with its nutrients removed.

Which of the following do the results show causes the light?

  1. A. Contact between bacteria packed closely together
    Fluid with every cell removed switched the light on.
    So no cell needed to touch another.
  2. B. ✓ A released molecule that has reached a high enough concentration
  3. C. Light given off by neighboring bacteria
    The sparse culture was dark until it received fluid from the dense culture.
    A molecule in that fluid, not light, set it glowing.
  4. D. Nutrients used up in the crowded culture
    Fresh fluid with its nutrients removed left the sparse culture dark.
    The dense culture’s fluid switched the light on: something it contained, not something the culture lacked.

Why: Fluid from the dense culture, with every cell removed, made a sparse culture glow.
Fluid from the sparse culture did nothing.
Fluid with its nutrients removed did nothing.
So a molecule the dense cells released, at a high enough concentration, causes the light.

53
Practice writing an answer

Cell-free fluid from a dense culture of the light-producing bacterium raises a sparse culture from 1 to 140 light units. Fresh fluid leaves a sparse culture at 1 light unit.

(a) Explain how these results demonstrate that a released signal molecule, rather than the crowding itself, causes the light. (1 pt)

Model answer The cell-free fluid carried no cells.
So no cell in the sparse culture touched a cell from the dense culture.
The sparse culture still glowed.
So the light came from something dissolved in the fluid.
Fresh fluid left a sparse culture dark.
So the fluid on its own switches nothing on.
So the light came from a molecule the dense cells had released into their fluid.
Rubric
  • Award 1 point for: the cell-free fluid carried no cells yet switched the light on, so a dissolved molecule the dense cells released causes the light; the fresh-fluid culture shows that fluid on its own does nothing.

Slip Saying the fluid carries light, or carries cells. The fluid carries only the signal molecule; every cell was removed.

54

Two flasks of the same light-producing bacterium, one sparse and one crowded.

55

At 100,000 cells per mL each cell’s signal molecule diffuses away and stays dilute. The receptors stay empty, and the flask is dark.

56

At 100,000,000 cells per mL the signal molecule has built up past its threshold concentration. It binds the receptors, and every cell switches on its light.

57

A bacterium counts its neighbors by the concentration of the signal molecule they all release.

58Mixed practice mixed practice

59
Check q9

A soil bacterium releases a signal molecule into the water around it and makes a slime coat only when it is crowded. One cell of this bacterium sits alone in a drop of water.

Which of the following does the cell do?

  1. A. ✓ The cell makes no slime coat
  2. B. The cell makes a thin slime coat
    One cell’s molecule diffuses away and stays dilute.
    Below the threshold concentration the receptor stays empty, and nothing is switched on.
  3. C. The cell makes a full slime coat
    One cell alone cannot build the molecule up to its threshold concentration.

Why: One cell releases the signal molecule into the drop.
The molecule diffuses away and stays dilute, below its threshold concentration.
So the molecule does not bind the cell’s receptor.
So the cell makes no slime coat.

60
Check q10

A researcher dilutes a dense, glowing culture of the light-producing bacterium a thousand-fold with fresh growth medium. Once the signal molecule stops binding their receptors, these cells stop making light within about an hour.

Which of the following happens to the light over the next hour?

  1. A. The light stays on
    The light depends on the signal molecule binding the receptors.
    Diluted a thousand-fold, the molecule is below its threshold concentration, so the receptors are empty.
  2. B. The light brightens
    The light depends on the signal molecule’s concentration.
    Dilution lowers the concentration a thousand-fold.
  3. C. ✓ The light goes out

Why: The light depends on the signal molecule’s concentration.
Diluting a thousand-fold drops the molecule below its threshold concentration.
So the molecule no longer binds the receptors.
So the cells stop making light, and within about an hour the light is out.

61
Check q11

A student says: “In the sparse flask the cells have no receptors. That is why it stays dark.”

Is the student correct?

  1. A. Yes — a cell in a sparse flask makes no receptor for the signal
    Every cell of the bacterium carries the receptor, crowded or not.
    In the sparse flask the signal molecule is too dilute to bind it.
  2. B. ✓ No — every cell carries the receptor; the signal is too dilute to bind it

Why: Every cell carries the receptor.
In the sparse flask, few cells release the signal molecule, so it stays dilute, below the threshold concentration.
So the molecule does not bind the receptors, and the flask stays dark.

62
Check q12

A mutant strain of the light-producing bacterium releases the signal molecule but carries no receptor for it. A researcher grows the mutant to 500,000,000 cells per mL.

Which of the following does the mutant culture do?

  1. A. The mutant culture glows
    The signal molecule builds up, but nothing in the cells binds it.
  2. B. The mutant culture glows dimly
    With no receptor, nothing in a cell binds the molecule.
    So nothing is switched on at all.
  3. C. ✓ The mutant culture stays dark

Why: The mutant cells release the signal molecule, so at 500,000,000 cells per mL it builds up past its threshold concentration.
But the mutant carries no receptor for the molecule.
So nothing in the cells binds it.
So no cell switches on its light, and the culture stays dark.

63
Check q13

A researcher adds fluid from a dense culture of the light-producing bacterium, with every cell removed, to one sparse culture. She adds fresh fluid to a second sparse culture.

Which of the following is the control?

  1. A. ✓ The sparse culture given fresh fluid
  2. B. The sparse culture given the cell-free fluid
    The cell-free fluid carries the factor under test.
    The control lacks that factor.
  3. C. The dense culture the fluid came from
    The dense culture is where the fluid came from, not a culture given a treatment.

Why: The factor under test is the fluid the dense cells grew in.
The control is the culture given the same treatment but lacking that factor.
The fresh-fluid culture receives fluid in which no cell grew.
So the sparse culture given fresh fluid is the control.

64
Practice writing an answer

A researcher dilutes a dense, glowing culture of the light-producing bacterium a thousand-fold with fresh growth medium. Once the signal molecule stops binding their receptors, these cells stop making light within about an hour. An hour after the dilution the light has gone out.

(a) Explain why the light goes out. (1 pt)

Model answer The light depends on the concentration of the signal molecule.
Diluting the culture a thousand-fold lowers the signal molecule’s concentration a thousand-fold.
So the concentration falls below the threshold concentration.
So the signal molecule no longer binds the receptors.
A cell makes light only while the signal molecule binds its receptor.
So every cell stops making light, and within the hour the light is out.
Rubric
  • Award 1 point for: dilution drops the signal molecule below its threshold concentration, so the receptors are no longer bound; a cell makes light only while its receptor is bound, so the cells stop making light.

Slip Saying the cells lost their receptors. The receptors are still there; the signal molecule is too dilute to bind them.

65
Check q14

The graph below shows the light produced by cultures of a second light-producing bacterium at four densities.

Bar graph of light produced, in light units, at four culture densities of a second light-producing bacterium, with gridlines every 50 light units
Bar graph of light produced, in light units, at four culture densities of a second light-producing bacterium, with gridlines every 50 light units

Between which two densities does the light first switch on?

  1. A. Between 20,000 and 200,000 cells per mL
    Light went from 2 to 3 light units there.
    The culture is still almost dark.
  2. B. ✓ Between 200,000 and 2,000,000 cells per mL
  3. C. Between 2,000,000 and 20,000,000 cells per mL
    The culture was already glowing at 2,000,000 cells per mL, at 40 light units.
    The switch from dark to glowing came one step earlier.
  4. D. At every step
    Below the threshold concentration the light barely changes.
    The switch happens where the signal molecule first binds the receptors.

Why: The culture is almost dark at 20,000 and 200,000 cells per mL.
It is glowing at 2,000,000 cells per mL, at 40 light units.
So the signal molecule passes its threshold concentration between 200,000 and 2,000,000 cells per mL.

Glossary

threshold concentration
The concentration of a signal molecule just high enough to bind its receptors. Below it the receptors stay empty; above it the molecule binds them and the cell responds.
quorum sensing
Bacteria detecting how crowded they are: each cell releases a small signal molecule and carries a receptor for it, so when cells are dense the signal builds up past a threshold concentration, binds the receptors, and the whole population switches on the same behavior.

APBIO-U04-L02C Which way did the message travel?

Topic 4.1 · Cell Communication · 49 steps

A dish in which plain donor immune cells and shaded recipient cells are mixed; some donors touch recipients, and small molecules are scattered in the fluid between the cells
A dish in which plain donor immune cells and shaded recipient cells are mixed; some donors touch recipients, and small molecules are scattered in the fluid between the cells

Here is a dish of immune cells. Some of the cells are donor cells that have been activated; the rest are recipient cells that have not. Mixed together, the recipients respond within hours.

The donors are touching some of the recipients, and they may also be releasing a molecule into the fluid. Either could carry the message. Which way did the message travel from the donors to the recipients, and how could you tell?

Unit 4 · Cell Communication and Cell Cycle

1The dish you compare against

2

Video: Watch: The dish you compare against

Recipients respond to fluid the donors grew in, with every donor cell removed, and stay as they were in fresh fluid. The fresh-fluid dish is the control: the same treatment minus the factor under test.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Ca.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Ca.mp4

3
Check q1

In an enzyme experiment, a tube of hydrogen peroxide with no tissue in it sits beside the tubes of peroxide with tissue.

Which of the following is the no-tissue tube?

  1. A. A condition kept the same in every tube
    A condition kept the same in every tube is a control variable, not a tube.
  2. B. ✓ The control
  3. C. The condition the experimenter changes
    The condition the experimenter changes is the independent variable: here, whether tissue is present.

Why: The no-tissue tube is given the same peroxide for the same time but lacks the factor under test, the tissue.
So the no-tissue tube is the control.

4

How do you tell touch from a released molecule? A membrane that stops cells, and two dishes of fluid, settle the question for any pair of cells.

5

Suppose a researcher tests whether the activated donor cells release a molecule that makes the recipients respond.

6

She takes fluid the donor cells grew in and removes every donor cell from it.

7

She adds this cell-free fluid to recipient cells in a fresh dish. The recipients respond.

Two dishes of recipient cells: the left has received fluid the donor cells grew in, with the donor cells removed, and its cells respond; the right has received fresh fluid and its cells stay as they were
Two dishes of recipient cells: the left has received fluid the donor cells grew in, with the donor cells removed, and its cells respond; the right has received fresh fluid and its cells stay as they were
8

She adds fresh fluid, in which no donor cell has grown, to a second dish of recipients. Those recipients stay as they were.

9

She treats the fresh-fluid dish the same way as the first dish but leaves out the factor under test, the fluid the donors grew in.

10

So the fresh-fluid dish is the control.

Two dishes of recipient cells: the left has received fluid the donor cells grew in, with the donor cells removed, and its cells respond; the right has received fresh fluid and its cells stay as they were
Two dishes of recipient cells: the left has received fluid the donor cells grew in, with the donor cells removed, and its cells respond; the right has received fresh fluid and its cells stay as they were
11

The control shows that adding fluid on its own produces no response.

12

So the response in the first dish can be credited to the fluid the donors grew in.

13

What you are expected to know Identify the control in a described signaling experiment.

14
Check q2

A fungus growing in a dish stops a second fungus 3 mm away from growing. A researcher sets up two more dishes of the second fungus: one given fluid the first fungus grew in, filtered to remove every cell; one given fresh fluid.

Which dish is the control?

  1. A. The dish with both fungi
    The dish with both fungi carries the factor under test, the first fungus.
    The control lacks that factor.
  2. B. The filtered-fluid dish
    The filtered fluid carries what the first fungus released.
    The control lacks that factor.
  3. C. ✓ The fresh-fluid dish

Why: The factor under test is the first fungus and what it releases.
The fresh-fluid dish is given the same treatment but has neither.
So the fresh-fluid dish is the control.

15
Check q3

A yeast cell releases a mating signal into the fluid around it. A researcher adds fluid that a mating yeast culture grew in, with every cell removed, to one dish of fresh yeast cells. She adds fresh fluid to a second dish of fresh yeast cells.

Which dish is the control?

  1. A. ✓ The dish given fresh fluid
  2. B. The dish given the cell-free fluid
    The cell-free fluid carries the factor under test, what the mating cells released.
  3. C. The mating culture the fluid came from
    The mating culture is where the fluid came from, not a dish given a treatment.

Why: The factor under test is the fluid the mating culture grew in.
The dish given fresh fluid receives the same treatment but lacks that fluid.
So the dish given fresh fluid is the control.

16
Check q4

A researcher tests whether wounded leaf cells release a molecule that makes healthy leaf cells produce a defense protein. One dish holds healthy cells given fluid the wounded cells sat in, filtered to remove every cell. A second dish holds healthy cells given fresh fluid. A third dish holds healthy cells touching wounded cells.

Which dish is the control?

  1. A. The filtered-fluid dish
    The filtered-fluid dish carries the factor under test, the fluid the wounded cells sat in.
  2. B. ✓ The fresh-fluid dish
  3. C. The dish where healthy cells touch wounded cells
    The dish where healthy cells touch wounded cells carries the wounded cells themselves.
    The control lacks the factor under test.

Why: The factor under test is the fluid the wounded cells sat in.
The fresh-fluid dish receives fluid in which no wounded cell sat.
So the fresh-fluid dish is the control.

17
Practice writing an answer

A fungus growing in a dish stops a second fungus 3 mm away from growing. The second fungus also stops growing in fluid the first fungus grew in, filtered to remove every cell. In fresh fluid, the second fungus keeps growing.

(a) Explain why the researcher needs the fresh-fluid dish before she can credit the second fungus’s response to a molecule from the first fungus. (1 pt)

Model answer Fresh fluid might change the second fungus on its own.
The fresh-fluid dish shows what fresh fluid does with no first fungus at all: the second fungus keeps growing.
The filtered-fluid dish differs from the fresh-fluid dish in one way: the first fungus grew in that fluid.
So the difference between the two dishes can be credited to a molecule the first fungus released into the fluid.
Rubric
  • Award 1 point for: the fresh-fluid dish shows the result with the first fungus’s fluid absent (the second fungus keeps growing), so the response to the filtered fluid can be credited to a molecule the first fungus released; without it, the fluid itself could not be ruled out.

Slip Saying the fresh-fluid dish shows the second fungus responds to any fluid. The second fungus kept growing in fresh fluid; it responded only to the fluid the first fungus grew in.

18Touch, or a released molecule?

19

Video: Watch: Touch, or a released molecule?

Recipients respond across a membrane that stops cells but lets molecules through, so the message is a released molecule; the fluid test shows the molecule came from the donors; only recipients that carry the receptor respond.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Cb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L02Cb.mp4

20
Check q5

A nerve ending releases a molecule into the narrow gap beside a muscle cell, and the muscle cell contracts.

Which of the following carried the message?

  1. A. A hormone carried in the blood
    The molecule crossed a narrow gap to the cell beside it, not the whole body in the blood.
  2. B. Touch between the two cells
    The nerve ending released a molecule into the gap; the two cells did not need to touch.
  3. C. ✓ A molecule released to the cells nearby

Why: The nerve ending released a molecule into the gap.
The molecule crossed the gap to the muscle cell beside it.
So a molecule released to the cells nearby carried the message.

21

Now consider the mixed dish: activated donor immune cells among recipient cells. The donors touch some recipients, and they may also release a molecule into the fluid.

22

Either touch or a released molecule could carry the message.

23

A researcher places the donor cells on one side of a membrane and the recipient cells on the other. The membrane stops cells but lets molecules through.

A dish with donor immune cells on the left and recipient cells on the right, separated by a membrane drawn as a dashed line that stops cells but lets molecules through; molecules are scattered on both sides
A dish with donor immune cells on the left and recipient cells on the right, separated by a membrane drawn as a dashed line that stops cells but lets molecules through; molecules are scattered on both sides
24

No donor cell can reach a recipient. The recipients still respond.

25

So the message crossed the membrane as a molecule dissolved in the fluid. Touch is ruled out.

26

The fluid test settles where the molecule came from. Cell-free fluid the donors grew in made recipients respond; fresh fluid did not.

27

So the molecule came from the donor cells.

28

The molecule reaches the recipients by diffusion through the fluid.

29

Only recipient cells that carry a receptor for the molecule respond to it. A recipient with no receptor does nothing, however much of the molecule reaches it.

30

For any pair of cells, the test has two steps.

31

1 Separate the two kinds of cell with a membrane that stops cells but lets molecules through. If the message still crosses, it is a released molecule.

32

2 Add fluid the sending cells grew in, with every cell removed, to fresh cells, and add fresh fluid to another dish. If only the grown-in fluid works, the molecule came from the sending cells.

33

What you are expected to know Decide, from a membrane-separation test or a cell-free-fluid test, whether a message traveled by touch or by a released molecule.

34
Check q6

A fungus growing on one side of a dish makes a second fungus on the other side stop growing, across a gap of 3 mm. No thread of either fungus touches the other.

Which of the following does the 3 mm gap show about the message?

  1. A. The message passes where the two fungi touch
    No thread of either fungus touches the other, and the second fungus still stopped growing.
    So the message crossed the gap without contact.
  2. B. ✓ The message crosses the gap as a released molecule

Why: No thread of either fungus touches the other.
The second fungus still stopped growing.
So the message crossed the 3 mm gap without contact.
A message that crosses a gap through fluid is a released molecule.

35
Check q7

Activated donor immune cells release a molecule into the fluid around them. Recipient immune cells that carry a receptor for the molecule respond within hours to fluid the donor cells grew in. A researcher adds the same fluid to recipient cells that carry no receptor for the molecule.

Which of the following do these recipients do?

  1. A. The recipients respond
    A cell responds only through a receptor, and these recipients carry none.
  2. B. The recipients respond weakly
    With no receptor, nothing in the cell binds the molecule.
    So nothing changes at all.
  3. C. ✓ The recipients stay as they were

Why: Only cells that carry the receptor respond.
These recipients have no receptor for the molecule.
So the fluid reaches them and nothing in them binds the molecule.
So they stay as they were.

36
Practice writing an answer

Cells of a wounded tomato leaf make nearby unwounded cells produce a defense protein within an hour. A researcher separates wounded cells from unwounded cells with a membrane that stops cells but lets molecules through. The unwounded cells still produce the defense protein.

(a) Explain how this result demonstrates that the message from the wounded cells is a released molecule rather than contact. (1 pt)

Model answer The membrane stops cells.
So no wounded cell can touch an unwounded cell.
The unwounded cells still produce the defense protein.
So the message crossed the membrane without contact.
The membrane lets molecules through.
So the message is a molecule the wounded cells released into the fluid.
Rubric
  • Award 1 point for: the membrane stops contact yet the unwounded cells still respond, so the message crossed as a molecule the wounded cells released into the fluid.

Slip Saying the result shows which molecule carried the message. The membrane test shows only that a released molecule, not contact, carried it.

37

Donor cells and recipient cells shared one dish, and the donors were touching some recipients.

38

Touch is not how the message traveled. Recipients responded across a membrane no cell could cross.

39

Fluid the donors grew in was enough on its own. The message is a molecule the donors released into the fluid.

40

The dish given fresh fluid stayed as it was. That dish, the control, shows that the fluid alone does nothing.

41Mixed practice mixed practice

42
Check q8

A gland in the neck releases a molecule into the blood, and bone cells all over the body respond to it.

Which of the following carried the message?

  1. A. Touch between the gland cells and the bone cells
    The bone cells sit all over the body, far from the gland.
    No gland cell touches them.
  2. B. A molecule released to the cells nearby
    A molecule released to nearby cells reaches only its own neighborhood.
    These bone cells sit all over the body.
  3. C. ✓ A hormone carried in the blood

Why: The gland released the molecule into the blood.
The blood carried the molecule through the whole body.
Bone cells all over the body responded.
So a hormone carried in the blood carried the message.

43
Check q9

Activated liver cells sit on one side of a membrane that stops cells but lets molecules through. Fat cells sit on the other side. The fat cells begin releasing their stored fat.

Which of the following does the result show about the message?

  1. A. ✓ The message crossed the membrane as a released molecule
  2. B. The message passed where liver cells touched fat cells
    The membrane stops cells.
    No liver cell could touch a fat cell.

Why: The membrane stops cells, so no liver cell touched a fat cell.
The fat cells still responded.
The membrane lets molecules through.
So the message crossed as a molecule the liver cells released.

44
Check q10

A strain of the second fungus lacks the receptor for the molecule the first fungus releases. A researcher adds filtered fluid the first fungus grew in to this strain.

Which of the following does this strain do?

  1. A. ✓ The strain keeps growing
  2. B. The strain stops growing
    With no receptor, nothing in the cells binds the molecule.
    So nothing in them changes.
  3. C. The strain grows more slowly
    A cell without the receptor does nothing at all.
    The molecule reaches it and nothing binds it.

Why: The fluid carries the first fungus’s molecule to the cells.
These cells have no receptor for the molecule.
So nothing in them binds it.
So the strain keeps growing.

45
Check q11

In the mixed dish, the activated donor cells were touching some of the recipient cells. A student says: “Because the donors were touching the recipients, the message must have passed by touch.”

Is the student correct?

  1. A. Yes — cells that touch pass their message by touch
    Touching cells can still send a released molecule.
    The membrane test showed the message crossing with no contact at all.
  2. B. ✓ No — the recipients also responded across a membrane no cell could cross

Why: Cells that touch may still release a molecule into the fluid.
Recipients responded across a membrane that stopped every donor cell.
So the message traveled as a released molecule, not by touch.

46
Check q12

A researcher sets up only one dish: recipient cells given fluid the donor cells grew in. The recipients respond.

Which of the following is still needed before the response can be credited to a molecule from the donors?

  1. A. A dish of donors given fresh fluid
    The question is whether the recipients’ response comes from the donors’ molecule.
    The control lacks that molecule and keeps everything else.
  2. B. A second dish of recipients given the donors’ fluid
    Repeating shows the result is steady.
    It does not show that the fluid alone would have done nothing.
  3. C. A dish of donors and recipients touching
    Contact is a different question.
    What is missing is the control that lacks the tested factor, the donors’ molecule.
  4. D. ✓ A dish of recipients given fresh fluid

Why: The control is given the same treatment but lacks the factor under test.
Recipients given fresh fluid show what fluid alone does.
Only with that dish can the response to the donors’ fluid be credited to a molecule the donors released.

47
Check q13

A researcher adds fluid the activated donor immune cells grew in, with every cell removed, to one dish of recipient cells. She adds fresh fluid to a second dish of recipients.

Which of the following does the fresh-fluid dish show?

  1. A. ✓ That fluid on its own produces no response
  2. B. That the donors release a molecule
    The fresh-fluid dish has no donor fluid in it.
    It shows what happens with the donors’ molecule absent.
  3. C. That the recipients carry the receptor
    The fresh-fluid dish carries no signal for a receptor to bind.
    It shows what fluid alone does.

Why: The fresh-fluid dish is the control: the same treatment minus the fluid the donors grew in.
Its recipients stay as they were.
So the fresh-fluid dish shows that fluid on its own produces no response.

48
Practice writing an answer

Researchers place skin cells from the edge of a wound (W cells) in a dish, with fresh skin cells (F cells) 2 mm away. The F cells begin dividing within a day. The researchers set up the three dishes drawn here. In dish 1, a mesh that lets molecules through but keeps cells apart separates W cells from F cells; the F cells divide. In dish 2, the researchers add fluid that W cells grew in, with every W cell removed, to F cells; the F cells divide. In dish 3, the researchers add fresh fluid to F cells; the F cells do not divide.

Three dishes drawn as rounded rectangles: dish 1 holds wound-edge cells on the left and fresh cells on the right with a vertical mesh line between them; dish 2 holds fresh cells in fluid the wound-edge cells grew in; dish 3 holds fresh cells in fresh fluid
Three dishes drawn as rounded rectangles: dish 1 holds wound-edge cells on the left and fresh cells on the right with a vertical mesh line between them; dish 2 holds fresh cells in fluid the wound-edge cells grew in; dish 3 holds fresh cells in fresh fluid

(a) Identify the control in this investigation. (1 pt)

Model answer Dish 3, the F cells given fresh fluid, is the control.
It receives the same treatment as dish 2 but lacks the factor under test, the fluid that W cells grew in.
Rubric
  • Award 1 point for: dish 3 (fresh fluid) as the control, because it lacks the factor under test.

Slip Naming dish 2 as the control. Dish 2 carries the tested factor; the control is the dish that lacks it.

(b) Explain how the results from dish 1 and dish 3 together show that the message from W cells to F cells is a released molecule rather than direct contact. (1 pt)

Model answer In dish 1 the mesh kept every W cell away from the F cells.
The F cells still divided.
So the message crossed the mesh as a molecule dissolved in the fluid, and contact is ruled out.
In dish 3 the F cells received fresh fluid and did not divide.
So fresh fluid on its own does nothing.
Therefore the molecule came from the W cells.
Rubric
  • Award 1 point for: the F cells responded with the mesh preventing contact (so a released molecule carried the message), and fresh fluid alone gave no response (so the molecule came from the W cells).

Slip Using dish 2 alone as the proof. Dish 2 shows the fluid works; dish 1 is what rules out contact, and dish 3 is what rules out the fluid itself.

(c) F cells carry a receptor for the W cells’ molecule, called R. The researchers add the fluid that W cells grew in to a line of F cells lacking R. Predict whether these cells divide, and justify your prediction. (2 pt)

Model answer The F cells lacking receptor R do not divide.
The molecule released by the W cells reaches them in the fluid.
A cell responds to a signal only through a receptor that binds the signal.
Receptor R is the receptor that binds the W cells’ molecule.
These cells have no receptor R.
So nothing in them binds the molecule.
So nothing inside them changes, however much of the molecule the fluid carries.
Rubric
  • Award 1 point for: the prediction that the cells lacking receptor R do not divide.
  • Award 1 point for: the justification that a cell responds only through a receptor that binds the signal, so the W cells’ molecule reaches the cells lacking R and nothing in them binds it, and nothing inside them changes.

Slip Predicting a weaker response. A cell without the receptor does nothing at all; the amount of signal changes only how strongly receptor-carrying cells respond.

APBIO-U04-P41 Practice questions: Topic 4.1

Topic 4.1 · Cell Communication · 10 MCQ · 2 FRQ · for APBIO-U04-T41

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. The first free-response question walks you through the reasoning one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Where a graph carries error bars, the caption says what the bars represent.

Video: Watch first: Topic 4.1 summary: how a message travels

A message passes by touch, to the cells next door, or through the blood; the same molecule reaches many cells; only the cells that carry a receptor for it respond.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-T41-summary.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-T41-summary.mp4

Q1 P41-q01

Before a caterpillar sheds its outer layer, cells of a gland in the front of its body release a molecule into its blood. Skin cells all over its body detect the molecule and begin building a new outer layer.

Which of the following gives the signaling cell and the chemical signal?

  1. A. The skin cell; the molecule it detects
    The skin cell responds: it detects the molecule.
    The signaling cell is the one that releases the molecule, the gland cell.
  2. B. The gland cell; the blood that carries the molecule
    The blood only carries the molecule from the gland to the skin.
    The chemical signal is the molecule itself.
  3. C. The skin cell; the new outer layer it builds
    The new outer layer is the response, what the skin cells do after they detect the molecule.
    The signal is the molecule.
  4. D. ✓ The gland cell; the molecule it releases

Why: The signaling cell is the cell that releases the molecule: the gland cell.
The chemical signal is the molecule it releases, which carries information to the skin cells that detect it.

Q2 P41-q02

The molting hormone in the caterpillar's blood reaches its muscle cells at the same concentration as its skin cells. The muscle cells show no change.

Why do the muscle cells show no change?

  1. A. Muscle cells receive less of the hormone than skin cells do
    The blood carries the hormone to both cell types at the same concentration.
    The difference is in the cells: the muscle cells have nothing that binds it.
  2. B. ✓ Muscle cells carry no receptor that binds the hormone
  3. C. Muscle cells detect the hormone but have no use for a new outer layer
    A cell with no receptor detects nothing.
    Muscle cells carry no protein that binds the hormone, so the hormone washes past them.
  4. D. Muscle cells destroy the hormone before it reaches their surface
    The hormone reaches the muscle cells as it reaches every cell in the blood.
    They show no change because nothing on or in them binds it.

Why: A cell responds to a signal only if it carries a receptor, a protein that binds that signal.
Muscle cells carry no receptor for the molting hormone, so however much reaches them, nothing binds it and nothing inside them changes.

Q3 P41-q03

The table below gives how much of the receptor for the molting hormone three cell types of the caterpillar carry. In this measurement, cells that lack the receptor read 0.7 units. The hormone reaches all three in the blood at the same concentration.

Amount of the receptor for the molting hormone carried by three cell types of a caterpillar, in units per cell.
Amount of the receptor for the molting hormone carried by three cell types of a caterpillar, in units per cell.

Which of the following predicts which cell types respond to the hormone, and explains why?

  1. A. ✓ Skin cells and gut-lining cells respond, because only they carry the receptor
  2. B. Skin cells respond only, because gut-lining cells carry too little receptor
    Gut-lining cells carry 4.7 units of the receptor, far above the 0.7 units of cells that lack it, so the hormone binds them and they respond.
  3. C. All three cell types respond, in proportion to the receptor each carries
    Muscle cells read 0.7 units: the no-receptor reading.
    Almost no receptor means almost nothing to bind the hormone, so the cell stays as it is rather than responding a little.
  4. D. All three cell types respond, because the hormone reaches all three
    A cell responds only through a receptor that binds the signal, and the muscle cells read 0.7 units, the reading of cells that lack the receptor.

Why: The cells that carry the receptor respond; the cells that carry almost none do nothing.
Skin cells (9.3 units) and gut-lining cells (4.7 units) carry the receptor, so the hormone binds them and they respond.
Muscle cells read 0.7 units, the no-receptor reading, so they stay as they are.

Q4 P41-q04

A researcher injects caterpillars with the molting hormone at the usual amount, ten times the usual amount and thirty times the usual amount, and measures the response of two cell types. The table below gives the results, in units. Muscle cells carry 0.7 units of the receptor, the level measured in cells that lack the receptor.

Response of two cell types to three amounts of the molting hormone.
Response of two cell types to three amounts of the molting hormone.

Which of the following predicts what the muscle cells do at one hundred times the usual amount, and explains why?

  1. A. The muscle cells respond weakly, in proportion to the extra hormone
    More hormone changes how strongly cells that carry the receptor respond; it changes nothing in cells that carry almost none.
  2. B. The muscle cells respond as strongly as cells that carry the receptor, because the extra hormone makes up for the missing receptor
    A response needs a receptor to bind the hormone, and muscle cells carry almost none.
  3. C. ✓ The muscle cells stay as they are, because they carry almost no receptor
  4. D. The muscle cells stay as they are, because the injected hormone never reaches them
    The blood carries the injected hormone to every cell, muscle cells included.
    They stay as they are because they carry almost no receptor to bind it.

Why: The muscle cells showed no response at the usual amount, ten times or thirty times.
A cell with no receptor binds nothing, so more hormone changes nothing in it.
The muscle cells carry 0.7 units: the no-receptor reading.
So at one hundred times they still stay as they are.

Q5 P41-q05

When a sperm cell reaches an egg, within seconds the egg's surface changes so that no second sperm can attach. Sperm held just out of reach of the egg by a mesh that lets molecules through, and fluid that sperm swam in, both leave the egg unchanged.

Which of the following gives the kind of signaling this is, and what shows it?

  1. A. A quorum-sensing signal; the egg changes only once enough sperm have gathered around it
    Quorum sensing needs a released molecule that builds up.
    One sperm reaching the egg is enough, and sperm held out of reach by the mesh have no effect.
  2. B. ✓ Direct contact; the egg changes only when a sperm touches it
  3. C. A local regulator; the sperm releases a molecule that diffuses through the mesh to the egg
    The mesh lets molecules through, yet sperm held behind it leave the egg unchanged.
    So no released molecule carries the message.
  4. D. A local regulator; the fluid the sperm swam in carried the sperm's molecule to the egg
    Fluid that sperm swam in leaves the egg unchanged, so the sperm release no molecule into fluid that could carry the message.
    The sperm itself must touch the egg.

Why: Only a sperm touching the egg makes it respond.
Sperm held out of reach by a mesh leave the egg unchanged, and so does fluid they swam in.
So no released molecule carries the message.
A protein on the sperm binds a receptor on the egg it touches: direct contact.

Q6 P41-q06

Cell fragments that plug a tear in a blood vessel release a burst of a molecule. A researcher measures the concentration of the molecule at distances from the plug, one minute and ten minutes after the release. The table below gives the results as a percent of the concentration at the plug at one minute. Muscle cells at every distance carry the same receptor for the molecule; only the muscle cells within 4 mm of the plug contract.

Concentration of the molecule at five distances from the plug, one minute and ten minutes after the release.
Concentration of the molecule at five distances from the plug, one minute and ten minutes after the release.

Why does the molecule act only on the muscle cells near the plug?

  1. A. The blood carries the molecule only as far as the nearest muscle cells
    A local regulator spreads by diffusion; the blood carries hormones, and blood flowing along the vessel would carry the molecule far past 4 mm.
  2. B. The muscle cells near the plug pull the molecule toward them, so none is left to reach the far cells
    Nothing pulls the molecule; it spreads by diffusion in every direction.
    Its concentration falls with distance because an enzyme breaks it down within minutes.
  3. C. The molecule is too large to diffuse further than 4 mm
    Given time, diffusion carries a molecule of any size further.
    By ten minutes the molecule has gone from every distance: it was destroyed, not held back by its size.
  4. D. ✓ The molecule is broken down within minutes, so it diffuses only a short way before it is gone

Why: At one minute the molecule reaches 4 mm and not 10 mm.
At ten minutes the molecule has gone from every distance, so it is broken down, not spreading slowly.
A molecule broken down that fast diffuses only a short way.
Only the muscle cells it reaches contract.

Q7 P41-q07

Two molecules act on the muscle cells in the wall of the bladder. A nerve ending less than 0.1 micrometers from a muscle cell releases molecule 1; the muscle cell contracts within a millisecond, and an enzyme in the gap destroys molecule 1 within five milliseconds. Molecule 2 arrives in the blood from a gland in the brain and stays in the blood for about an hour.

Which of the following classifies the two molecules?

  1. A. Molecule 1 is a hormone; molecule 2 is a local regulator
    Molecule 1 never enters the blood, so it is not a hormone.
    Molecule 2 rides the blood for an hour from a gland, so it is not a local regulator.
  2. B. Molecule 1 is a contact signal; molecule 2 is a hormone
    A contact signal stays in the membrane of the cell that carries it.
    Molecule 1 is released into the gap and crosses it.
  3. C. ✓ Molecule 1 is a neurotransmitter; molecule 2 is a hormone
  4. D. Both molecules are neurotransmitters
    Molecule 2 rides the blood for an hour from a gland in the brain to the bladder.
    A molecule carried in the blood to distant cells is a hormone.

Why: Molecule 1 crosses a gap under 0.1 micrometers and is destroyed within milliseconds.
A molecule a nerve ending releases beside its target is a neurotransmitter.
Molecule 2 travels in the blood from a gland to the bladder.
A molecule the blood carries to distant cells is a hormone.

Q8 P41-q08

Cells in a hen's ovary release a molecule into the blood. Within an hour, liver cells and cells lining the tube where the eggshell forms, both far from the ovary, have changed what they make.

Which of the following gives the kind of signaling this is, and what shows it?

  1. A. ✓ A hormone in the blood; one molecule reached distant organs of other cell types
  2. B. A local regulator; the molecule diffused from the ovary to the nearby tube
    A local regulator reaches only its immediate neighbors before it is destroyed.
    This molecule changed liver cells far from the ovary, which only the blood can reach.
  3. C. Direct contact; the ovary cells touched the cells that responded
    The ovary touches neither the liver nor the tube.
    One molecule reached both because the blood carried it through the body.
  4. D. A local regulator; the response took an hour to appear
    A local regulator acts within milliseconds to minutes on its neighbors.
    A signal that takes an hour and reaches distant organs rode the blood.

Why: A hormone released by one cell type is carried in the blood through the whole body and acts on target cells of other cell types wherever they are, so its response starts more slowly than a local signal's and reaches several organs at once.

Q9 P41-q09

A soil bacterium releases an antibiotic that kills rival bacteria. A sparse culture releases almost none; a dense culture releases a great deal. Each cell releases a small signal molecule and carries a receptor for it. When a researcher adds the purified signal molecule to a sparse culture at the concentration found in a dense culture, the sparse culture releases the antibiotic within an hour.

Which of the following makes the sparse culture given the signal release the antibiotic?

  1. A. Crowding, because the added molecules fill the space between the cells
    Nothing was added but the signal molecule; the cells are as far apart as before.
    What changed is the concentration of the signal around each cell.
  2. B. The antibiotic carried into the culture with the added molecule
    The signal molecule was purified; no antibiotic was added.
    The antibiotic measured was released by the sparse culture's own cells after the signal bound their receptors.
  3. C. The cells decide, from the signal, that enough neighbors are present, and choose to release the antibiotic
    A bacterium decides nothing.
    When the signal’s concentration around a cell is high enough, it binds the cell’s receptors and the cell releases the antibiotic automatically.
  4. D. ✓ The signal molecule, at a concentration high enough to bind the cells' receptors

Why: Each cell carries a receptor for the signal molecule.
In a sparse culture the signal diffuses away and stays below the threshold.
Adding the purified signal at a dense culture’s concentration binds the receptors, so every cell switches on.
The signal, not the crowding, causes the response: quorum sensing.

Q10 P41-q10

To test whether the antibiotic response comes from the added signal molecule, two sparse cultures are set up: one given the purified signal molecule in fresh medium, and one given fresh medium alone. After three hours the antibiotic each culture has released is measured.

Which of the following statements is the null hypothesis for this comparison?

  1. A. The purified signal molecule makes a sparse culture release more antibiotic
    A null hypothesis states that the factor changed has no effect on the quantity measured.
    The claim that the signal raises the release is the alternative, not the null.
  2. B. ✓ The purified signal molecule makes no difference to the amount of antibiotic a sparse culture releases
  3. C. Fresh medium alone makes a sparse culture release less antibiotic than the signal molecule does
    A null hypothesis says the factor changed makes no difference to the quantity measured.
    Saying fresh medium gives less antibiotic claims a difference: the alternative, not the null.
  4. D. Crowding makes no difference to the amount of antibiotic a sparse culture releases
    The two cultures differ in one thing: whether the purified signal molecule was added.
    Both cultures are sparse, so crowding is not what this comparison varies.

Why: A null hypothesis states that the factor changed makes no difference to the quantity measured.
Here the factor is whether the purified signal was added, and the quantity is the antibiotic released in three hours.
So: the signal molecule makes no difference to the antibiotic a sparse culture releases.

FRQ 1 P41-frq1 · Scientific Investigation scaffolded

Roots of many plants form a partnership with a soil fungus. Root cells release a molecule into the soil water, and the fungus's threads grow toward the root. Researchers test how the message travels. In every dish the researchers place a small piece of the fungus 10 mm from the source and measure the growth of its threads toward the source after 24 hours, in six dishes per treatment. Dish A: a living root tip. Dish B: a living root tip behind a mesh that lets molecules through but keeps root cells and fungal threads apart. Dish C: a drop of water that roots grew in for a day, with every root cell removed. Dish D: a drop of fresh water. The table below gives the mean growth toward the source for each dish, with ±2SE.

Mean growth of the fungus's threads toward the source in 24 hours, six dishes per treatment, with ±2SE for each mean.
Mean growth of the fungus's threads toward the source in 24 hours, six dishes per treatment, with ±2SE for each mean.

(a) Identify the signaling cells and the chemical signal in this partnership. (1 pt)

Frame The signaling cells are …, and the chemical signal is …

Hint The signaling cell is the one that sends the message, not the one that reacts to it; the signal is the thing that travels.

Model answer The signaling cells are the root cells, and the chemical signal is the molecule they release into the soil water.
Rubric
  • Award 1 point for both: the root cells are the signaling cells, and the molecule they release into the soil water is the chemical signal.
  • Do not award the point for naming the fungus as the signaling cell or the water as the signal.

Slip Naming the fungus as the signaling cell. The fungus responds; the signaling cell is the one that releases the molecule.

(b) Identify the control dish, and explain what it shows. (1 pt)

Frame The control is dish …, because it …; it shows that …

Hint Which dish is treated like the others but has the tested factor left out, and what would you expect to see in it if the factor were doing nothing?

Model answer The control is dish D, the drop of fresh water, because it is treated like the others but has the tested factor, anything from roots, left out.
It shows that water on its own gives almost no growth toward the source (0.4 mm), so growth toward the other sources can be credited to something from the roots.
Rubric
  • Award 1 point for: dish D (fresh water) is the control: it lacks the factor under test (anything from roots) and shows that water on its own gives almost no growth toward the source (0.4 mm), so growth in the other dishes can be credited to something from the roots.
  • Do not award the point for dish C as the control (it carries the tested factor) or for 'it shows the experiment worked'.

Slip Naming dish C as the control. Dish C carries the tested factor, water that roots grew in; the control is the dish that lacks it.

(c) State the null hypothesis for the comparison between dish C and dish D. (1 pt)

Frame The null hypothesis is that the water the roots grew in … the growth of the fungus toward the source.

Hint A null hypothesis names the factor changed and the quantity measured. What does it say about the effect of the one on the other?

Model answer The null hypothesis is that the water the roots grew in makes no difference to the growth of the fungus's threads toward the source, compared with fresh water.
Rubric
  • Award 1 point for: water that roots grew in makes no difference to the growth of the fungus's threads toward the source, compared with fresh water.
  • Do not award the point for a prediction of a difference in either direction, or for a statement that names neither the root water nor the growth.

Slip Writing the researchers' prediction ('root water makes the fungus grow toward it') as the null. The null predicts no difference.

(d) Explain how dish B and dish D together show that the message from the root is a released molecule rather than contact between root and fungus. (1 pt)

Frame In dish B the fungus grew toward the root although …, so …; dish D shows that …, so the molecule came from …

Hint What did the mesh stop, and what did it let through? And what does the fresh-water dish rule out?

Model answer In dish B the mesh kept every root cell and every fungal thread apart.
The fungus still grew 5.8 mm toward the root.
So the message crossed the mesh as a molecule dissolved in the water, and contact is ruled out.
In dish D the fungus was given fresh water and grew only 0.4 mm toward it.
So water on its own gives almost no growth.
Therefore the molecule that crossed the mesh came from the roots.
Rubric
  • Award 1 point for: in dish B the fungus grew toward the root (5.8 mm) although the mesh kept root cells and fungal threads apart, so the message crossed as a molecule in the water and contact was ruled out; dish D shows that water on its own gives almost no growth (0.4 mm), so the molecule came from the roots.
  • Do not award the point for using dish C alone (it shows the water works but does not by itself rule out contact in the other dishes) or for an answer with no reference to what the mesh does.

Slip Explaining from dish C alone. Dish C shows the root water works; dish B is what rules out contact, and dish D is what rules out the water itself.

(e) A strain of the fungus carries no receptor for the root's molecule. Determine how far its threads grow toward the source in dish C, and justify your decision. (1 pt)

Frame The strain's threads would grow about … mm toward the source, because …

Hint What has to happen at the fungus's surface before anything inside it can change?

Model answer The strain’s threads grow about 0.4 mm toward the source, the same as toward fresh water.
A cell responds to a signal only through a receptor that binds it.
The molecule reaches this strain in the water.
But nothing in the strain binds the molecule.
So nothing inside it changes, and its threads grow no more toward the source than toward fresh water.
Rubric
  • Award 1 point for: the decision (about 0.4 mm, the same as dish D: no growth toward the source) AND the reasoning it rests on (the molecule reaches the fungus, but nothing in it binds the molecule, so nothing inside it changes).
  • Do not award the point for the decision alone, or for 'less growth, because it takes up less of the molecule'.

Slip Deciding on slower growth toward the source. A cell without the receptor shows no response at all; the amount of signal changes only how strongly receptor-carrying cells respond.

FRQ 2 P41-frq2 · Scientific Investigation

A soil bacterium releases an antibiotic that kills rival bacteria. Each cell releases a small signal molecule and carries a receptor for it. Researchers grow four sets of six flasks: a sparse culture (30,000 cells per mL); a dense culture (300,000,000 cells per mL); a sparse culture given the purified signal molecule in fresh medium, at the concentration found in a dense culture; and a sparse culture given fresh medium alone. After three hours they measure the antibiotic released into the fluid, in units per mL. The graph below shows the means, and the error bars represent ±2SE.

Antibiotic released into the fluid in three hours by four cultures of the soil bacterium, six flasks each, in units per mL. Error bars represent ±2SE. Gridlines every 5 units per mL.
Antibiotic released into the fluid in three hours by four cultures of the soil bacterium, six flasks each, in units per mL. Error bars represent ±2SE. Gridlines every 5 units per mL.

(a) Explain why the sparse culture releases almost no antibiotic. (1 pt)

Model answer Every cell releases the signal molecule, but in a sparse culture it diffuses away into the fluid and stays dilute, below the threshold concentration that binds the receptors.
With nothing bound, the cells stay switched off and release almost no antibiotic.
Rubric
  • Award 1 point for: with few cells, the signal molecule each cell releases diffuses away and stays below the threshold concentration that binds the receptors, so the cells stay switched off.
  • Do not award the point for 'sparse cells make no signal' or for 'the cells do not need the antibiotic'.

Slip Saying sparse cells make no signal. They release it as dense cells do; the concentration stays too low because few cells are releasing it.

(b) Identify the control for the culture given the purified signal, and explain what it shows. (1 pt)

Model answer The control is the sparse culture given fresh medium alone.
It is treated the same way as the signal flasks but lacks the factor under test, the signal molecule.
It shows that adding medium by itself gives almost no antibiotic, 0.6 units per mL, so the release in the signal flasks is credited to the signal.
Rubric
  • Award 1 point for: the sparse culture given fresh medium alone is the control: it lacks the tested factor (the signal) and shows that adding medium by itself gives almost no antibiotic (0.6 units per mL), so the release in the signal flasks can be credited to the signal molecule.
  • Do not award the point for the dense culture as the control or for 'it shows the experiment worked'.

Slip Naming the dense culture as the control. The dense culture shows what a crowded population does; the control for the added signal is the same sparse culture given medium with the signal left out.

(c) Support the claim that the signal molecule, rather than crowding, switches on the antibiotic, using evidence from the error bars. (1 pt)

Model answer The bars represent ±2SE.
The signal flasks’ bar runs from 34 to 44 units per mL, and the fresh-medium bar from 0.3 to 0.9 units per mL.
The two bars do not overlap, so the difference is unlikely to be chance.
The signal flasks stayed sparse, yet the signal raised their antibiotic.
So the signal molecule, not crowding, switches on the antibiotic.
Rubric
  • Award 1 point for: the evidence (the signal flasks' bar, 34 to 44 units per mL, does not overlap the fresh-medium flasks' bar, 0.3 to 0.9) AND the reasoning (the signal flasks stayed sparse, so crowding cannot explain the rise; the signal alone raised the antibiotic, and the non-overlap shows the difference is unlikely to be chance).
  • Accept adding that the signal flasks' bar overlaps the dense culture's (38 to 46), so the signal alone gives a release these data cannot tell from a dense culture's. Accept: readings within half a gridline (2.5 units per mL) of those values. Do not award the point for a comparison of means with no use of the bars, or for the bars with no link to the claim.

Slip Quoting the bars and stopping. Supporting the claim needs the link: the flasks stayed sparse, so the rise came from the signal, not from crowding.

(d) The dense culture is diluted a hundred-fold with fresh medium, to 3,000,000 cells per mL. In this bacterium the signal molecule reaches the concentration that binds the receptors only when a culture is denser than about 10,000,000 cells per mL. Predict what happens to the diluted culture's release of antibiotic three hours after the dilution, and justify your prediction. (1 pt)

Model answer Three hours after the dilution, the release of antibiotic has fallen to almost nothing.
The diluted culture holds 3,000,000 cells per mL, below the 10,000,000 per mL threshold.
So the signal’s concentration falls below what binds the receptors, and each cell switches off.
No new antibiotic-making protein is made, and the proteins already built soon stop.
So the release fades to almost nothing.
Rubric
  • Award 1 point for: release falls to almost nothing, because 3,000,000 cells per mL is below the density (about 10,000,000 cells per mL) at which the signal reaches the concentration that binds the receptors, so the receptors are no longer bound and each cell switches off.
  • Accept 'falls to almost nothing' with or without the delay while proteins already built keep working. Do not award the point for 'release continues because the cells are already switched on' with no fading, for 'release rises because each cell has more room', or for a prediction that does not compare the new density with the threshold density.

Slip Predicting that release continues because the cells 'have already switched on'. Each cell responds only while the signal's concentration stays above the threshold; dilution removes it.

APBIO-U04-T41 End-of-topic test: Cell Communication

Topic 4.1 · Cell Communication · 19 MCQ · 2 FRQ

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the two free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. Then open the scoring guide and mark your own work against it. Where a graph carries error bars, the caption says what the bars represent.

Q1 T41-q01

When a tadpole is ready to become a frog, cells of its thyroid gland release a molecule into its blood. Cells of the tail detect the molecule and the tail begins to shrink; cells of the growing legs detect it and grow faster.

Which of the following gives the signaling cell and the chemical signal?

  1. A. The tail cell; the molecule it detects
    The tail cell responds: it detects the molecule.
    The signaling cell is the cell that releases the molecule, and that is the thyroid gland cell.
  2. B. ✓ The thyroid gland cell; the molecule it releases into the blood
  3. C. The thyroid gland cell; the blood that carries the molecule
    The blood only carries the molecule from the gland to the tail and the legs.
    The chemical signal is the molecule itself.
  4. D. The tail cell; the shrinking of the tail
    The shrinking is the response, what the tail cells do after they detect the molecule.
    The signal is the molecule that reached them.

Why: The signaling cell is the cell that releases the molecule: the thyroid gland cell.
The chemical signal is the molecule it releases, which the blood carries and which the tail cells and leg cells detect.

Q2 T41-q02

When a salmon swims from the sea into a river, cells of a gland near its brain release a hormone into its blood. Gill cells respond by pumping less salt out of the fish. Fin cells, bathed in the same blood at the same concentration, show no change.

Which of the following cells are target cells for the hormone?

  1. A. ✓ The gill cells only
  2. B. The gland cells and the gill cells
    The gland cells release the hormone: they are the signaling cells, not targets.
    A target cell responds to the signal, and only the gill cells respond.
  3. C. The gill cells and the fin cells
    A target cell responds to the signal.
    The fin cells are bathed in the hormone and show no change.
    So the fin cells are not target cells.
  4. D. Every cell the blood carries the hormone to
    The blood carries the hormone to every cell, but a target cell is one that responds, because it carries a receptor.
    Only the gill cells respond.

Why: A target cell responds to a signal because it carries a receptor for it.
The gill cells respond, so they are target cells.
The fin cells receive the same hormone and show no change, so they are not.
The gland cells release the hormone: they are signaling cells.

Q3 T41-q03

After a meal, cells of the gut wall release a hormone into the blood. The graph below shows how much of the receptor for this hormone three cell types carry, as the mean of six samples of each; the error bars represent ±2SE. In this measurement, cells that lack the receptor read 0.5 units. A researcher then gives each cell type the same concentration of the hormone.

Amount of the receptor for the gut hormone carried by three cell types, in units per cell, six samples of each. Error bars represent ±2SE. Gridlines every 1 unit.
Amount of the receptor for the gut hormone carried by three cell types, in units per cell, six samples of each. Error bars represent ±2SE. Gridlines every 1 unit.

Which of the following predicts which cell types respond to the hormone, and explains why?

  1. A. ✓ Pancreas cells and gallbladder muscle cells respond, because only they carry the receptor
  2. B. Pancreas cells respond only, because gallbladder muscle cells carry too little receptor
    Gallbladder muscle cells carry 5.4 units of the receptor, far above the 0.5 units of cells that lack it, so the hormone binds them and they respond.
  3. C. All three cell types respond, in proportion to the receptor each carries
    Lung cells read 0.5 units: the no-receptor reading.
    Almost no receptor means almost nothing to bind the hormone, so the cell stays as it is rather than responding a little.
  4. D. All three cell types respond, because the hormone reaches all three
    A cell responds only through a receptor that binds the signal.
    The lung cells read 0.5 units, the no-receptor reading, so nothing in them binds the hormone.

Why: The cells that carry the receptor respond; the cells that carry almost none do nothing.
Pancreas cells (8.1 units) and gallbladder muscle cells (5.4 units) carry the receptor, so the hormone binds them and they respond.
Lung cells read 0.5 units, the no-receptor reading, so they stay as they are.

Q4 T41-q04

In a developing embryo, R cells lie next to P cells. A protein set in the surface of an R cell binds a receptor on the surface of the P cell touching it, and that P cell becomes a nerve cell. Researchers grow R cells and P cells on opposite sides of a mesh that lets molecules through but keeps the cells apart.

Which of the following predicts what the P cells do, and explains why?

  1. A. The P cells become nerve cells, because the mesh lets R's surface protein through with other molecules
    R’s protein is set in R’s membrane, not released into the fluid.
    The mesh lets through dissolved molecules only, so the surface protein never reaches a P cell’s receptor.
  2. B. The P cells become nerve cells, but only once enough R cells have built up on their side of the mesh
    Quorum sensing needs a molecule released into the fluid.
    Here the message is a protein fixed in R’s surface, and however many R cells gather, none touches a P cell.
  3. C. ✓ The P cells do not change, because R's surface protein stays in R's membrane and cannot reach P's receptor
  4. D. The P cells do not change, because the mesh breaks down the molecules that R releases
    A mesh does not break molecules down; it lets them through.
    The P cells stay unchanged because the message passes only by touch, and no R cell touches them.

Why: R’s surface protein passes the message only where an R cell touches a P cell.
The mesh keeps every R cell away from the P cells.
R’s protein stays in R’s membrane, so it never reaches a P cell’s receptor.
So no P cell becomes a nerve cell.

Q5 T41-q05

A pollen grain lands on the stigma at the tip of a flower. Within minutes the stigma cell beneath it softens its surface so the pollen can grow into it. Pollen grains held 0.1 mm above the stigma, and fluid washed off pollen grains and dropped onto the stigma, both leave the stigma cells unchanged.

Which of the following gives the kind of signaling this is, and what shows it?

  1. A. ✓ Direct contact; the stigma cell changes only when a pollen grain touches it
  2. B. A local regulator; the pollen grain releases a molecule that diffuses across the short gap to the stigma cell
    Pollen held 0.1 mm above the stigma leaves the stigma cells unchanged.
    A released molecule would diffuse across so small a gap within minutes, and the cell would respond.
  3. C. A local regulator; the fluid washed off the pollen carried the pollen's molecule to the stigma cell
    Fluid washed off pollen grains leaves the stigma cells unchanged, so the pollen releases no molecule into fluid that could carry the message.
    The grain itself must touch.
  4. D. A quorum-sensing signal; the stigma cell changes only once enough pollen grains have landed on it
    Quorum sensing needs a released molecule that builds up.
    Here one grain landing is enough, and a grain held just above the stigma has no effect.

Why: Only a grain touching the stigma cell makes it respond.
A grain held 0.1 mm above it leaves it unchanged, and so does fluid washed off grains.
So no released molecule carries the message.
A molecule on the grain’s surface binds a receptor on the cell it touches: direct contact.

Q6 T41-q06

Cells lining the inside of a blood vessel release a small gas molecule. Muscle cells wrapped around the vessel within a few cell-widths of the lining relax within seconds, and the vessel widens there. Muscle cells further along the vessel carry the same receptor and stay contracted. An enzyme destroys the gas within a few seconds of its release. A researcher then adds a drug that blocks this enzyme.

Predict the effect of the drug on which muscle cells relax.

  1. A. Only the nearest muscle cells relax, as before
    The gas reached only the nearest cells because the enzyme destroyed it within seconds.
    With the enzyme blocked, the gas lasts longer and diffuses further before it is gone.
  2. B. The nearest muscle cells stop relaxing
    The enzyme only ends the signal by destroying the gas.
    Blocking it leaves the gas bound for longer, so the nearest cells relax as before, and for longer.
  3. C. Muscle cells throughout the body relax
    The gas spreads by diffusion through the vessel wall.
    Blocking the enzyme lets it diffuse further along the vessel; it does not put the gas into the blood.
  4. D. ✓ Muscle cells further along the vessel relax too

Why: The enzyme destroys the gas within seconds, so it reaches only the nearest cells.
The drug blocks the enzyme, so the gas lasts longer.
So it diffuses further along the vessel before it is gone.
Cells further along carry the receptor, so the gas now binds them and they relax.

Q7 T41-q07

Researchers study a molecule that a nerve ending releases beside a muscle cell in the wall of the gut. The table below gives what they measured.

Measurements of the molecule released at the gut nerve ending.
Measurements of the molecule released at the gut nerve ending.

Which of the following kinds of signal is the molecule?

  1. A. A hormone
    The table shows none of the molecule in the blood.
    The molecule crosses a gap of 0.02 micrometers and acts in under a millisecond.
  2. B. ✓ A neurotransmitter
  3. C. A contact signal
    The nerve ending releases the molecule, and the molecule crosses a gap.
    Contact signals never leave the cell that carries them.
  4. D. A quorum-sensing signal
    The table shows an enzyme destroying the molecule within milliseconds, so it never builds up.

Why: A nerve ending releases the molecule into a gap of 0.02 micrometers, beside one muscle cell.
The muscle cell responds in under a millisecond.
An enzyme destroys the molecule within milliseconds, and none reaches the blood.
A molecule a nerve cell releases beside its target is a neurotransmitter.

Q8 T41-q08

When extra blood stretches the upper chambers of a person's heart, cells lining those chambers release a molecule into the blood. About twenty minutes later, cells in the kidney, in the lining of blood vessels in the legs and in the adrenal glands have all changed their activity; these organs lie far apart.

Which of the following identifies the kind of signaling, and the evidence for it?

  1. A. A local regulator; the molecule diffused from the heart to the nearby blood vessels
    A local regulator reaches only nearby cells.
    This molecule changed cells in the kidney and in the legs, far from the heart; only the blood reaches that far.
  2. B. Direct contact; the heart pressed against each organ that responded
    The heart touches none of the kidney, the leg vessels or the adrenal glands.
    One molecule reached all three because the blood carried it through the whole body.
  3. C. ✓ A hormone in the blood; one molecule reached several distant organs of other cell types
  4. D. A local regulator; the response took twenty minutes to appear
    A local regulator acts within milliseconds to minutes on its neighbors.
    A signal that takes twenty minutes and reaches distant organs traveled in the blood.

Why: One cell type releases a hormone into the blood.
The blood carries the hormone through the whole body.
Every target cell that carries its receptor responds, wherever it is.
So one molecule from the heart changed cells in three distant organs.
The blood’s journey is why the response started slowly.

Q9 T41-q09

In a female mammal, a hormone released by the ovary reaches cells lining the uterus and milk-making cells in the breast, and both begin responding at about the same time, well after the release. A neurotransmitter released by a nerve ending acts on its target within a millisecond.

Which of the following explains why the hormone reaches both the uterus cells and the breast cells, and why its response starts more slowly?

  1. A. The hormone diffuses through the tissues from the ovary to the uterus and then on to the breast
    Diffusion carries a molecule only a short way before it is too dilute, and the breast lies far from the ovary.
    The blood carries the hormone to both.
  2. B. The ovary cells touch the cells of the uterus and of the breast
    The ovary lies far from the uterus lining and farther from the breast; no ovary cell touches either.
    The hormone reaches them in the blood.
  3. C. The uterus cells and breast cells take a long time to detect the hormone because they carry few receptors
    A cell with the receptor binds the hormone as soon as it arrives.
    The delay is the journey: the blood carries the hormone through the whole body.
  4. D. ✓ The blood carries the hormone through the whole body, a slower journey that also dilutes it

Why: The ovary releases the hormone into the blood.
The blood carries it through the whole body, reaching the uterus lining and the breast alike.
That journey takes far longer than crossing a tiny gap, and the blood dilutes the hormone.
So both organs respond well after the release.

Q10 T41-q10

A bacterium that rots vegetables releases enzymes that digest plant cell walls. Cultures were grown to four densities, and the enzyme released into the fluid was measured; the table below gives the results. Each cell releases a small signal molecule and carries a receptor for it.

Wall-digesting enzyme released into the fluid by cultures of the vegetable-rotting bacterium grown to four densities.
Wall-digesting enzyme released into the fluid by cultures of the vegetable-rotting bacterium grown to four densities.

Why do the two sparse cultures release almost no enzyme?

  1. A. A sparse culture has too few cells to make any of the signal molecule
    Every cell releases the signal molecule, sparse culture or dense.
    In a sparse culture the molecule diffuses away, and its concentration stays too low to bind the receptors.
  2. B. ✓ In a sparse culture the signal molecule diffuses away and stays below the concentration that binds the receptors
  3. C. Each cell in a sparse culture releases the enzyme, but the enzyme is too dilute in the fluid to measure
    If each cell released enzyme steadily, a hundred times the cells would give a hundred times the enzyme.
    Instead 2 units became 96: a threshold, not a steady trickle.
  4. D. The signal molecule in a sparse culture is a different molecule from the one in a dense culture
    The signal molecule is the same molecule at every density.
    What changes is its concentration: in a dense culture it accumulates past the threshold that binds the receptors.

Why: Each cell releases the signal molecule and carries its receptor.
When cells are few, the molecule diffuses away and stays below the threshold that binds the receptors, so almost no enzyme is released.
When cells are dense, the molecule passes the threshold, binds the receptors, and the population switches on.

Q11 T41-q11

A strain of the vegetable-rotting bacterium carries no receptor for the signal molecule but releases the signal normally. A dense culture of this strain releases almost no cell-wall-digesting enzyme. A researcher then adds the purified signal molecule at the concentration found in a dense culture.

Which of the following predicts what the receptor-lacking culture does, and explains why?

  1. A. The culture releases the enzyme, because the added signal reaches every cell
    The signal reaching a cell is not enough.
    A cell switches on only when the signal binds its receptor, and these cells carry no receptor for it.
  2. B. Almost none of the enzyme appears, because the added signal is broken down before it can bind
    The added signal surrounds the cells at a dense culture’s concentration, and nothing destroys it first.
    The culture stays off because its cells have no receptor to bind the signal.
  3. C. ✓ The culture still releases almost no enzyme, because its cells have nothing that binds the signal
  4. D. The culture releases the enzyme, because the added signal takes the place of the missing receptor
    A signal molecule and a receptor are different things: the signal is what binds, the receptor is the protein that binds it.
    Without the receptor, more signal changes nothing.

Why: Bacteria switch on the shared behavior when the signal molecule binds their receptors.
This strain releases the signal but carries no receptor for it, so however much signal surrounds the cells, nothing binds it and the culture stays switched off.

Q12 T41-q12

Cells taken from the back of a chick embryo (B cells) make nearby cells (N cells) start producing a cartilage protein. Researchers give N cells and skin cells from the same embryo fluid that B cells grew in, with every B cell removed. The N cells make the protein; the skin cells, in the same fluid, stay as they were.

Which of the following explains how the message reaches the N cells, and why only the N cells answer it?

  1. A. ✓ A molecule the B cells release spreads through the fluid; the N cells carry a receptor for it and the skin cells carry none
  2. B. A molecule the B cells release spreads through the fluid; the N cells use it up before it reaches the skin cells
    The N cells and the skin cells sit in the same fluid, so the molecule reaches both.
    The skin cells stay unchanged because nothing in them binds it: no receptor.
  3. C. The B cells pass the message by touching the N cells; the skin cells are never touched
    Every B cell was removed before the N cells received the fluid, yet the N cells make the protein.
    So a released molecule carries the message through the fluid.
  4. D. A molecule the B cells release spreads through the fluid; the skin cells detect it but have no use for a cartilage protein
    A cell detects a signal only through a receptor that binds it.
    The skin cells carry no receptor for the B cells’ molecule, so nothing inside them changes.

Why: The fluid carries the message with no B cell present, so the B cells released a molecule into it.
The molecule reaches N cells and skin cells alike.
The N cells carry a receptor that binds it, so they respond.
The skin cells carry none, so nothing in them changes.

Q13 T41-q13

Four behaviors of single-celled organisms are described.

Which of the following is an example of quorum sensing?

  1. A. A gut bacterium switches on the genes for digesting a sugar when the concentration of that sugar in the gut rises
    The sugar is food in the gut, not a molecule the cells release.
    Quorum sensing counts cells through a signal molecule the cells themselves release.
  2. B. A yeast cell stops dividing once neighboring cells press against its surface on every side
    Touch passes a message only between cells that press together.
    Quorum sensing works through a molecule released into the fluid, which builds up as the cells become dense.
  3. C. ✓ A soil bacterium forms a tough coat once the cells are dense enough for a molecule they each release to build up around them
  4. D. A pond bacterium swims toward the light once the pigment inside each cell has absorbed enough light to change shape
    Light comes from outside the population.
    In quorum sensing the signal is a molecule the bacteria release, and its concentration reports how crowded they are.

Why: Bacteria switch on a shared behavior only when they are crowded.
Each cell releases the same signal molecule into the fluid.
Dense cells push its concentration past the threshold concentration that binds the receptors.
Then every cell switches on the coat: quorum sensing.

Q14 T41-q14

When a sea star's arm is bitten, cells damaged at the wound release a molecule. Within an hour, cells up to 0.4 mm from the wound begin laying down new skeleton; cells further along the arm, which carry the same receptor, do nothing. The molecule is gone from the tissue within a few minutes of its release.

By which of the following routes did the message travel from the damaged cells to the cells that responded?

  1. A. ✓ By diffusion through the fluid between cells, to cells within a short distance
  2. B. By the body fluid that circulates through the sea star, to cells throughout the arm
    A molecule carried around the body would reach the receptor-carrying cells further along the arm, and they would respond.
    They do nothing, so no circulating fluid carried it.
  3. C. By surface proteins, to the cells touching the damaged cells
    Cells up to 0.4 mm from the wound responded, many cell-widths away from the damaged cells.
    A molecule diffusing through the fluid between cells reached them.
  4. D. By the fluid between cells, to every cell of the arm at once
    Cells further along the arm carry the receptor and do nothing, so the molecule never reached them.
    Gone within minutes, it diffuses only a short way.

Why: The damaged cells release the molecule into the fluid.
Cells up to 0.4 mm away respond: the message reaches beyond touching cells.
Cells further along carry the receptor and do nothing: the molecule is gone before it diffuses that far.
So the route is diffusion to nearby cells.

Q15 T41-q15

A tadpole’s thyroid gland releases one molecule into its blood. That one molecule makes the tail cells shrink and the leg cells grow.

Which of the following does this show about the molecule?

  1. A. The molecule carries one set of instructions for tail cells and another for leg cells
    The molecule is the same small molecule wherever the blood carries it, and it carries no instructions.
    Each cell type binds it and responds in its own way.
  2. B. The molecule changes into a different molecule on its way to the legs
    The blood carries one molecule to both the tail and the legs.
    The difference in response lies in the cells, not in the molecule.
  3. C. The tail cells receive the molecule first and pass a changed form of it to the legs
    The tail cells detect the molecule and respond; they pass nothing on.
    The legs receive the same molecule from the same blood.
  4. D. ✓ The molecule carries no instructions; each cell type responds in its own way once its receptor binds it

Why: A chemical signal carries no instructions inside it.
It counts as information only because some cells are built to detect it, and each kind of cell responds in its own way once its receptor binds the molecule: tail cells shrink, leg cells grow.

Q16 T41-q16

Lung cells show no change when the gut hormone reaches them. Researchers make a line of lung cells that carries the receptor for the gut hormone in its membrane. When the hormone reaches these cells, they respond.

Which of the following factors do the results show decides whether a cell responds to the hormone?

  1. A. Whether the hormone can reach the cell in the blood
    The hormone reached the ordinary lung cells too, and they showed no change.
    What changed the outcome was giving the cells the receptor.
  2. B. ✓ Whether the cell carries a receptor that binds the hormone
  3. C. Whether the cell is the kind of cell the hormone is meant for
    The cells are still lung cells; the only change is the receptor they now carry.
    A cell responds because it carries the receptor, not because of its kind.
  4. D. Whether the concentration of hormone around the cell is high enough
    The concentration was the same for the ordinary lung cells, which did nothing.
    The receptor, not the amount of hormone, decided the response.

Why: A target cell is a cell that responds to a signal because it carries a receptor, a protein that binds that signal.
The ordinary lung cells lack the receptor and show no change; the same cells given the receptor respond.
So carrying the receptor is what decides the response.

Q17 T41-q17

Four signals are described.

Which of them is a hormone?

  1. A. ✓ A molecule released by the testes into the blood that makes muscle cells all over the body grow larger
  2. B. A molecule released by a nerve ending into the gap beside a muscle cell, gone within milliseconds
    A molecule a nerve cell releases into the gap beside its target is a neurotransmitter, a local regulator.
    It never enters the blood and is gone within milliseconds.
  3. C. A molecule released by a cut cell that acts on cells a few cell-widths away and is gone within minutes
    A molecule that acts on its neighbors and is gone within minutes is a local regulator.
    It spreads by diffusion, and the blood carries none of it.
  4. D. A protein on the surface of one immune cell that binds a protein on the surface of another
    A surface protein binding a protein on a touching cell is direct contact.
    Nothing is released and nothing travels.

Why: A hormone is a signal carried in the blood to target cells of other cell types throughout the body.
The testes’ molecule enters the blood and reaches muscle cells all over the body, so it is a hormone.
The other three signals never enter the blood.

Q18 T41-q18

In a dense culture of the vegetable-rotting bacterium, every cell begins releasing the cell-wall-digesting enzyme within the same few minutes.

Why do the cells all switch on together?

  1. A. The cells signal each other by touch, and the message spreads from cell to cell across the culture
    The signal is a released molecule dissolved in the fluid; the cells are not touching.
    Every cell in the fluid is bathed in the same concentration at the same time.
  2. B. The first cell to switch on releases the enzyme, and the enzyme switches on the other cells
    The enzyme digests plant cell walls; it is the response, not the signal.
    What switches the cells on is the small signal molecule binding their receptors.
  3. C. The cells wait until they sense that the population is large enough, then agree to switch on
    Bacteria decide nothing and wait for nothing.
    Each cell responds automatically once the signal's concentration is high enough, and the concentration is high only when many cells are releasing it.
  4. D. ✓ The signal passes its threshold concentration for every cell at once, and each cell responds automatically

Why: Each cell releases the signal and carries a receptor for it.
In a dense culture the signal accumulates past its threshold concentration throughout the fluid, so it binds the receptors of every cell at about the same time, and each cell responds automatically.
That is quorum sensing.

Q19 T41-q19

The table below gives how much of the receptor for hormone Z four cell types of a frog carry. In this measurement, cells that lack the receptor read 0.5 units. A researcher then adds hormone Z to all four cell types at the same concentration.

Amount of the receptor for hormone Z carried by four cell types of a frog, in units per cell.
Amount of the receptor for hormone Z carried by four cell types of a frog, in units per cell.

Which of the following cell types respond to hormone Z?

  1. A. Skin cells only
    Muscle cells carry 5.5 units of the receptor, almost as much as skin cells, so they bind hormone Z and respond too.
  2. B. ✓ Skin cells and muscle cells
  3. C. Skin cells, muscle cells and lens cells
    Lens cells read 0.5 units, the reading of cells that lack the receptor: almost none.
    They have almost nothing to bind hormone Z, so they stay as they are.
  4. D. Skin cells, muscle cells, lens cells and kidney cells
    Kidney cells and lens cells both read 0.5 units, the reading of cells that lack the receptor.
    They carry almost none, so nothing in them binds hormone Z.

Why: Cells respond when they carry the receptor.
Skin cells (6.2 units) and muscle cells (5.5 units) carry it, so hormone Z binds them and they respond.
Kidney cells and lens cells both read 0.5 units, the no-receptor reading, so they carry almost none and do nothing.

FRQ 1 T41-frq1 · Scientific Investigation

Cells of a soil amoeba live as single cells while there are bacteria to eat. When the bacteria are gone, some cells release a small molecule, and nearby cells crawl toward the source. Well-fed cells carry a receptor for the molecule. Researchers set up four dishes, drawn below, each holding well-fed F cells. Dish 1: starved S cells beyond a mesh that lets molecules through but keeps cells apart. Dish 2: fluid S cells had lived in, every S cell removed. Dish 3: fresh fluid. Dish 4: the S-cell fluid, given to F cells from a line with no receptor for the molecule. After 20 minutes the researchers count the percent of F cells crawling toward the source (in dish 1, the mesh) in six fields of view per dish. The graph gives the means for dishes 1 to 3 with ±2SE error bars; dish 4 has still to be counted.

Top: the four dishes. Bottom: percent of F cells crawling toward the source after 20 minutes, mean of six fields of view per dish. Error bars represent ±2SE. Gridlines every 10%. Dish 4 has still to be counted.
Top: the four dishes. Bottom: percent of F cells crawling toward the source after 20 minutes, mean of six fields of view per dish. Error bars represent ±2SE. Gridlines every 10%. Dish 4 has still to be counted.

(a) Identify the independent variable and the dependent variable in this investigation. (1 pt)

Model answer Independent variable: what the F cells are given (starved cells behind a mesh, fluid the starved cells lived in, or fresh fluid; and, in dish 4, whether the F cells carry the receptor).
Dependent variable: the percent of F cells crawling toward the source after 20 minutes.
Rubric
  • Award 1 point for both: independent variable, what is added to the F cells (S cells behind a mesh, fluid the S cells lived in, or fresh fluid; accept 'the source of the fluid' or 'whether starved cells or their fluid are present'); dependent variable, the percent of F cells crawling toward the source after 20 minutes.
  • Accept 'the treatment condition of each dish (dish 4 varies the cell line as well as the fluid)' as the independent variable. Accept the F cells' receptor status (dish 4) as a further part of the independent variable. Do not award the point if the two variables are reversed, or if the amoeba species or the 20-minute count time is named as a variable.

Slip Naming the crawling as the independent variable. The crawling is what is measured; what the researchers change is what they add to each dish.

(b) Explain why dish 3 is included in the investigation. (1 pt)

Model answer Dish 3 is the control.
The F cells in dish 3 get fresh fluid: the same treatment as dish 2, but lacking the factor under test, the fluid the S cells lived in.
Dish 3 shows that fluid on its own makes only 4% of F cells crawl toward the source.
So the rise to 57% in dish 2 can be credited to a molecule the S cells released.
Rubric
  • Award 1 point for: dish 3 is the control: it shows how many F cells crawl toward fluid that no S cell has lived in (4%), so a rise above that in dish 2 can be credited to a molecule the S cells released into their fluid.
  • Accept 'it shows that fluid on its own gives no response'. Do not award the point for 'it shows the experiment worked' or for naming a condition kept the same (the 20 minutes, the temperature) as the reason.

Slip Saying the control 'shows the experiment worked'. The control gives the response with the tested factor absent; without it, the 57% in dish 2 could not be credited to anything the S cells released.

(c) Support the claim that a molecule released by the S cells, rather than contact with S cells, causes the crawling, using evidence from the error bars. (1 pt)

Model answer Dish 2 held no S cells, so no F cell could touch one.
The dish 2 bar runs from 51% to 63%.
The dish 3 bar runs from 2% to 6%.
The bars do not overlap, so the difference is unlikely to be chance.
So the fluid the S cells lived in carried a released molecule, and that molecule made the F cells crawl.
Rubric
  • Award 1 point for: the evidence (dish 2 held no S cells, and its bar, 51 to 63%, does not overlap the dish 3 bar, 2 to 6%) AND the reasoning that links it to the claim (no S cell was present to touch, so the fluid the S cells lived in carried a released molecule that made the F cells crawl, and the non-overlap shows the rise is unlikely to be chance).
  • Accept the dish 1 evidence instead (the mesh kept every S cell away, yet the F cells crawled; the dish 1 bar, 56 to 66%, overlaps the dish 2 bar), with the same link stated. Accept: readings within half a gridline (5 percentage points) of those values, that is 46–56 to 58–68 for dish 2 and 0–7 to 1–11 for dish 3. Do not award the point for evidence with no link to the claim, for reasoning with no data named, or for a comparison of the means alone.

Slip Naming the evidence and stopping. Supporting a claim needs the link: say why dish 2 against dish 3 rules out contact and leaves a released molecule as the only signal.

(d) Predict the result for dish 4, and justify your prediction. (1 pt)

Model answer About 4% of the F cells in dish 4 crawl toward the source, the same as in dish 3.
The molecule reaches these cells in the S-cell fluid.
A cell responds to a signal only through a receptor that binds the signal.
This line carries no receptor for the molecule.
So nothing in the cells binds the molecule, and nothing inside them changes.
They crawl no more than cells given fresh fluid.
Rubric
  • Award 1 point for: about 4% of the F cells crawl toward the source, the same as dish 3 (no rise), because these F cells carry no receptor for the molecule: the molecule reaches them in the fluid, nothing in them binds it, and nothing inside them changes.
  • Do not award the point for 'fewer crawl, because they take up less of the molecule' (a cell with no receptor shows no response at all) or for a prediction with no reason.

Slip Predicting a weaker response. A cell without the receptor does nothing at all; the amount of signal changes only how strongly receptor-carrying cells respond.

FRQ 2 T41-frq2 · Conceptual Analysis

When the blood carries too little oxygen, cells in the kidney release a hormone into the blood. Over the following days, cells in the bone marrow of every bone make more red blood cells. Liver cells and skin cells, bathed in the same blood, show no change. Researchers measured how much of the receptor for the hormone each cell type carries, in units per cell, in six samples of each; the graph below shows the means, and the error bars represent ±2SE. In this measurement, cells that lack the receptor read 0.5 units.

Amount of the receptor for the kidney hormone carried by three cell types, in units per cell, six samples of each. Error bars represent ±2SE. Gridlines every 1 unit.
Amount of the receptor for the kidney hormone carried by three cell types, in units per cell, six samples of each. Error bars represent ±2SE. Gridlines every 1 unit.

(a) Describe how the hormone gets from the kidney cells to the marrow cells of the leg bones. (1 pt)

Model answer The kidney cells release the hormone into the blood, and the blood carries it through the whole body.
It reaches the marrow cells of the leg bones, and of every other bone, in the same blood.
Rubric
  • Award 1 point for: the kidney cells release the hormone into the blood, and the blood carries it through the whole body, so it reaches the marrow cells wherever they are (including the leg bones).
  • Accept with or without a note that the response starts more slowly than a local signal's or that the hormone reaches many organs at once.
  • Do not award the point for 'it diffuses through the tissues to the bones' or for 'the kidney cells touch the marrow'.

Slip Saying the hormone diffuses through the tissues from the kidney to the bones. Diffusion carries a signal only a few cell-widths before it is destroyed or diluted away; a hormone rides the blood.

(b) Explain why the liver cells show no change although the hormone reaches them. (1 pt)

Model answer Liver cells carry almost none of the receptor for the hormone: 0.6 units per cell against 7.6 in marrow cells, and 0.5 units is the reading of cells that lack the receptor.
The hormone reaches the liver cells in the blood.
A cell responds to a signal only through a receptor that binds it.
With almost no receptor, nothing in a liver cell binds the hormone.
So nothing inside the liver cell changes.
Rubric
  • Award 1 point for: liver cells carry almost none of the receptor (0.6 units against 7.6 in marrow cells; 0.5 units is the no-receptor reading), so no protein on or in them binds the hormone and nothing inside them changes; only cells that carry the receptor are target cells.
  • Do not award the point for 'the liver does not need more red blood cells', 'the liver ignores the hormone', or 'less hormone reaches the liver'.

Slip Saying the liver cells 'have no need' for the hormone or 'ignore' it. A cell that does not respond has no protein that binds the signal; nothing is being weighed or chosen.

(c) The researchers then raise the concentration of the hormone in the blood twenty-fold and measure the hormone bound to each cell type, in units per cell, before and after: marrow cells 6.8 before and 7.1 after; liver cells 0.2 and 0.2; skin cells 0.2 and 0.2. In this measurement, cells with no receptor read 0.2 units. Using these data, predict how the liver cells respond at the higher concentration, and justify your prediction. (1 pt)

Model answer The liver cells still show no change.
At the normal concentration the liver cells bind 0.2 units of hormone, the no-receptor reading.
At twenty times the concentration they still bind 0.2 units, the no-receptor reading.
A cell responds only through a receptor that binds the hormone, and the liver cells bind none at either concentration.
So the liver cells show no change at twenty times the concentration.
Rubric
  • Award 1 point for: the liver cells still show no change, because the bound hormone stays at the no-receptor reading (0.2 units) at twenty times the concentration, so nothing in the liver cells binds the hormone at either concentration.
  • Do not award the point for 'the liver cells respond a little now' or for a prediction that does not use the bound-hormone data.

Slip Predicting a weak response in proportion to the extra hormone. More signal cannot make a receptor-less cell respond; it only strengthens the response of cells that already carry the receptor.

(d) Determine, using the error bars, whether the data show that liver cells and skin cells differ in how much of the receptor they carry. Justify your decision. (1 pt)

Model answer The bars represent ±2SE, the range each true mean is likely to lie in.
The liver bar runs from 0.4 to 0.8 units.
The skin bar runs from 0.3 to 0.7 units.
The bars overlap, so the difference between 0.6 and 0.5 could be chance.
So the data do not show that the two cell types differ.
Both cell types read close to the 0.5 units of cells that lack the receptor.
Rubric
  • Award 1 point for: the decision (the data do not show a difference) AND the reasoning it rests on (the liver bar, 0.4 to 0.8 units, and the skin bar, 0.3 to 0.7 units, overlap, so the difference between the means could be chance and the null hypothesis of no difference is not rejected).
  • Do not award the point for the decision alone, for a comparison of the means alone (0.6 against 0.5), or for deciding that liver cells carry more of the receptor than skin cells, or for deciding that the two cell types carry the same amount of receptor (overlap shows no difference, not equality).

Slip Deciding from the means alone, 0.6 against 0.5. Determining needs the decision and what it rests on: the two bars overlap.

APBIO-U04-L03 The ligand fits

Topic 4.2 · Introduction to Signal Transduction · 89 steps

A liver cell with two receptors set in its membrane: on the left the insulin receptor holds insulin, a compact folded molecule of 51 amino acids, in a large pocket; on the right the epinephrine receptor holds a single small molecule in a small pocket
A liver cell with two receptors set in its membrane: on the left the insulin receptor holds insulin, a compact folded molecule of 51 amino acids, in a large pocket; on the right the epinephrine receptor holds a single small molecule in a small pocket

Here is a liver cell. Its membrane carries receptors for insulin and receptors for epinephrine, side by side.

Insulin is 51 amino acids in two short chains that fit together. Epinephrine is a molecule of a few dozen atoms. Each one binds its own receptor and never the other’s.

A receptor is a protein with a pocket. Why does each pocket take one molecule and refuse the other?

Unit 4 · Cell Communication and Cell Cycle

1The ligand

2

Video: Watch: The ligand

Insulin fits the pocket of the insulin receptor by shape and charge, just as a substrate fits an enzyme’s active site; the signal molecule that binds a receptor is that receptor’s ligand.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03a.mp4

3

A receptor binds only the signal molecule whose shape and charge match its pocket.

4

An enzyme’s active site binds its substrate the same way.

5

Insulin is far too large for the epinephrine receptor’s pocket.

6

Epinephrine cannot fill the insulin receptor’s pocket.

7

The receptor holds its signal molecule, changes shape, and lets the signal molecule go unchanged.

8

So a cell with receptors for several signals answers each signal separately.

9

Start with the insulin receptor.

10
Check q1

A substrate fits into an enzyme’s active site.

What decides whether a molecule fits the active site?

  1. A. ✓ Its shape and charge
  2. B. Its size alone
    Two molecules of the same size can differ in shape and in where their charges sit; only the one whose shape and charge match the site fits it.

Why: A substrate fits an enzyme’s active site when its shape and its charges match the site.

11

Here is the insulin receptor of a liver cell, set in the plasma membrane.

A stretch of membrane, the fluid outside the cell shaded one way above it and the cytosol shaded another below it, with two receptors set through it: the insulin receptor on the left holds a compact folded molecule of insulin in a large pocket; the epinephrine receptor on the right holds a single small molecule in a small pocket; a free molecule of epinephrine floats beside the insulin receptor's pocket, too small to fill it
A stretch of membrane, the fluid outside the cell shaded one way above it and the cytosol shaded another below it, with two receptors set through it: the insulin receptor on the left holds a compact folded molecule of insulin in a large pocket; the epinephrine receptor on the right holds a single small molecule in a small pocket; a free molecule of epinephrine floats beside the insulin receptor's pocket, too small to fill it
12

On its outer face is a pocket, and insulin fits it.

13

A signal molecule fits its receptor’s pocket in the same way as a substrate fits an active site: by shape and charge.

14

The signal molecule that binds a receptor is called that receptor’s .

15

Insulin is the insulin receptor’s ligand.

16

Epinephrine is the epinephrine receptor’s ligand.

17

What you are expected to know Identify the ligand in a described case: the signal molecule that binds the receptor.

18
Check q2

A plant hormone reaches a root cell and binds a pocket on a protein set in the root cell’s membrane.

Which is the ligand?

  1. A. The pocket on the protein
    The pocket is where the ligand fits; the ligand is the molecule that fits it.
  2. B. The membrane protein
    The receptor holds the pocket; the ligand is the signal molecule it binds.
  3. C. The root cell
    The root cell is the target cell; the ligand is the signal molecule that binds the cell’s receptor.
  4. D. ✓ The plant hormone

Why: The ligand is the signal molecule that binds the receptor: the plant hormone.
The protein is the receptor and the pocket is where the hormone fits.

19Quick quiz: ligand mixed practice

20
Check q3

What is a ligand?

  1. A. The receptor’s pocket
    The pocket is part of the receptor; the ligand is the molecule that fits the pocket.
  2. B. The cell that carries the receptor
    The cell that carries the receptor is the target cell; the ligand is the molecule that binds the receptor.
  3. C. ✓ The signal molecule that binds a receptor

Why: A ligand is the signal molecule that binds a receptor.
Insulin is the ligand of the insulin receptor.

21
Practice writing an answer

Epinephrine binds a receptor on a heart muscle cell.

(a) Describe what a ligand is, naming the ligand in this case. (1 pt)

Model answer A ligand is the signal molecule that binds a receptor.
Epinephrine binds the receptor on the heart muscle cell.
So epinephrine is the ligand.
Rubric
  • Award 1 point for: a ligand is the signal molecule that binds a receptor, and here the ligand is epinephrine.

22The binding site and the ligand-binding domain

23

Video: Watch: The pocket and the region that holds it

The pocket a ligand fits is the receptor’s binding site; the region of the receptor that holds the pocket is its ligand-binding domain, a domain being a region of a protein with one job.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03b.mp4

24

Here is a receptor set in a membrane, with its ligand in the pocket.

A receptor set through a membrane band, the fluid outside the cell shaded one way above the band and the cytosol shaded another below it; a small round molecule sits in a cup at the top of the receptor; one pointer marks the cup and is labeled binding site, another marks the round molecule and is labeled ligand
A receptor set through a membrane band, the fluid outside the cell shaded one way above the band and the cytosol shaded another below it; a small round molecule sits in a cup at the top of the receptor; one pointer marks the cup and is labeled binding site, another marks the round molecule and is labeled ligand
25

The pocket a molecule fits is called a .

26

An enzyme’s active site is its binding site for its substrate.

27

A receptor’s pocket is its binding site for its ligand.

28

A region of a protein with one job is called a domain.

29

The region of the receptor that holds the pocket is called its .

The same receptor set through a membrane band with its ligand in the cup at the top; one pointer marks the cup and is labeled binding site; another marks the receptor's outer region above the band and is labeled ligand-binding domain
The same receptor set through a membrane band with its ligand in the cup at the top; one pointer marks the cup and is labeled binding site; another marks the receptor's outer region above the band and is labeled ligand-binding domain
30

The ligand-binding domain faces the outside of the cell, so the ligand reaches it from the fluid outside.

31

What you are expected to know Identify, on a drawing of a receptor, its binding site (the pocket) and its ligand-binding domain (the region of the receptor that holds the pocket).

32
Check q4

The model below shows a receptor set in a membrane with its ligand. Four positions are marked 1 to 4.

A receptor set through a membrane band, the fluid outside the cell shaded one way above the band and the cytosol shaded another below it; a small round shape sits in a cup at the top of the receptor; four positions are marked: 1 on the upper part of the round shape, 2 on the receptor's outer face beside the cup, above the band, 3 on the receptor within the band, 4 on the receptor below the band
A receptor set through a membrane band, the fluid outside the cell shaded one way above the band and the cytosol shaded another below it; a small round shape sits in a cup at the top of the receptor; four positions are marked: 1 on the upper part of the round shape, 2 on the receptor's outer face beside the cup, above the band, 3 on the receptor within the band, 4 on the receptor below the band

Which numbered position marks the receptor’s ligand-binding domain?

  1. A. Position 1
    Position 1 marks the ligand itself, which is not part of the receptor.
    The ligand-binding domain is the region of the receptor that holds the pocket: position 2.
  2. B. ✓ Position 2
  3. C. Position 3
    Position 3 marks the region of the receptor within the membrane.
    The ligand-binding domain faces outward and holds the pocket the ligand fits.
  4. D. Position 4
    Position 4 marks the region of the receptor facing the inside of the cell.
    The ligand never reaches that region; the ligand-binding domain is the outer region with the pocket.

Why: The ligand-binding domain is the region of the receptor that holds the pocket the ligand fits: the part above the membrane, position 2.
Position 1 is the ligand itself.

33Quick quiz: binding site, ligand-binding domain mixed practice

34
Check q5

What is a binding site?

  1. A. The signal molecule that fits a receptor’s pocket
    The signal molecule that fits a receptor’s pocket is its ligand; the binding site is the pocket the ligand fits.
  2. B. The whole folded receptor protein
    The binding site is one pocket on the protein, not the whole protein.
  3. C. ✓ The pocket on a protein that one molecule fits

Why: A binding site is the pocket on a protein that one molecule fits by shape and charge.
An enzyme’s active site is its binding site for its substrate.
A receptor’s pocket is its binding site for its ligand.

35
Check q6

What is a receptor’s ligand-binding domain?

  1. A. The ligand, the molecule that fits the pocket
    The ligand is the signal molecule; it is not part of the receptor.
  2. B. ✓ The region of the receptor that holds the pocket
  3. C. The middle part, set within the membrane
    The part within the membrane holds no pocket; the ligand-binding domain is the region that does.

Why: A domain is a region of a protein with one job.
The ligand-binding domain is the region of the receptor that holds the pocket its ligand fits.

36
Practice writing an answer

A receptor is one folded protein chain with several regions.

(a) Describe what a domain of a protein is, and name the domain of a receptor that holds the pocket its ligand fits. (1 pt)

Model answer A domain is a region of a protein.
That region does one job.
The region of a receptor that holds the pocket its ligand fits is the ligand-binding domain.
Rubric
  • Award 1 point for: a domain is a region of a protein with one job, and the region that holds the pocket is the ligand-binding domain.

37Why a receptor binds only its own ligand

38

Video: Watch: One pocket, one ligand

Epinephrine is too small to fill the insulin receptor’s pocket, and insulin is too large for the epinephrine receptor’s; a receptor makes no product from its ligand but holds it, changes shape, and lets it go unchanged.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03c.mp4

39

Here are the insulin receptor and the epinephrine receptor of the liver cell again.

A stretch of membrane, the fluid outside the cell shaded one way above it and the cytosol shaded another below it, with two receptors set through it: the insulin receptor on the left holds a compact folded molecule of insulin in a large pocket; the epinephrine receptor on the right holds a single small molecule in a small pocket; a free molecule of epinephrine floats beside the insulin receptor's pocket, too small to fill it
A stretch of membrane, the fluid outside the cell shaded one way above it and the cytosol shaded another below it, with two receptors set through it: the insulin receptor on the left holds a compact folded molecule of insulin in a large pocket; the epinephrine receptor on the right holds a single small molecule in a small pocket; a free molecule of epinephrine floats beside the insulin receptor's pocket, too small to fill it
40

Epinephrine is far smaller than insulin.

41

So epinephrine cannot fill the insulin receptor’s pocket.

42

Epinephrine fits the pocket of the epinephrine receptor.

43

Insulin is far too large for that pocket.

44

So each receptor binds its own ligand and passes other molecules by.

45

A ligand is not a substrate.

46

An enzyme makes a product from its substrate.

47

A receptor makes no product from its ligand.

48

The receptor holds the ligand in its pocket.

49

While the ligand is held, the receptor changes its own shape.

50

Then the receptor lets the ligand go, unchanged.

51

So a cell that carries receptors for several signals answers each signal separately.

52

What you are expected to know Explain why a receptor binds only its own ligand and lets it go unchanged.

53
Check q7

A molecule of insulin binds the pocket of an insulin receptor, stays for a few seconds, and is released again.

Which of the following has happened to the insulin molecule by the time it is released?

  1. A. ✓ The insulin molecule is unchanged
  2. B. The insulin molecule has gained a phosphate group from the receptor
    The receptor is the molecule that changes shape while the insulin is bound.
    The insulin leaves as it arrived.
  3. C. The insulin molecule has been converted into a different signal molecule
    A receptor makes no product.
    The receptor releases the same insulin molecule.
  4. D. The insulin molecule has been split into two smaller products
    An enzyme makes products from its substrate.
    A receptor is not an enzyme.
    The receptor holds the insulin, changes its own shape, and lets the insulin go as it arrived.

Why: Insulin is the receptor’s ligand.
A ligand is not a substrate.
The receptor makes no product from its ligand.
The receptor holds the insulin, changes its own shape while the insulin is bound, and releases the same insulin molecule unchanged.

54
Practice writing an answer

A molecule of insulin binds the pocket of an insulin receptor, stays for a few seconds, and is released again. The released insulin molecule is unchanged.

(a) Explain why the insulin molecule is unchanged when the receptor releases it. (1 pt)

Model answer Insulin is the receptor’s ligand.
A ligand is not a substrate.
An enzyme makes a product from its substrate.
A receptor is not an enzyme, so it makes no product from its ligand.
The receptor holds the insulin in its pocket.
While the insulin is held, the receptor changes its own shape.
Then the receptor lets the insulin go.
So the released insulin molecule is the same molecule that arrived.
Rubric
  • Award 1 point for: a receptor is not an enzyme and makes no product from its ligand; it holds the ligand, changes its own shape, and releases the ligand unchanged.
55
Check q8

Insulin binds the pocket of its receptor on a muscle cell and is released a few seconds later. A student says: “The receptor is an enzyme, and insulin is its substrate, so the insulin comes out changed.”

Is the student correct?

  1. A. Yes — the receptor changes the insulin into a product
    An enzyme makes a product from its substrate; a receptor makes no product from its ligand.
    The receptor holds the insulin, changes its shape, and releases the insulin unchanged.
  2. B. ✓ No — the receptor changes its own shape and releases the insulin unchanged

Why: A receptor is not an enzyme, and a ligand is not a substrate.
The receptor holds the insulin.
The receptor changes its own shape while the insulin is bound.
Then the receptor releases the same insulin molecule, unchanged.

56
Check q9

A liver cell carries two kinds of receptor. One kind has a small pocket, sized for epinephrine, a molecule of a few dozen atoms. A molecule of insulin, far larger than epinephrine, reaches the small pocket.

Predict what happens.

  1. A. ✓ The small pocket does not hold the insulin
  2. B. The small pocket holds the insulin
    A pocket holds only a molecule whose shape and charge match it.
    Insulin is far too large for a pocket sized for epinephrine.
  3. C. The small pocket widens and holds the insulin
    A receptor’s pocket has one shape.
    A molecule that does not fit that shape is not held.

Why: A receptor’s pocket holds only a molecule whose shape and charge match it.
Insulin is far too large for a pocket sized for epinephrine.
So the small pocket does not hold the insulin, and the insulin passes by.

57What a ligand can be

58

Video: Watch: Peptide or small molecule

Insulin and growth hormone are chains of amino acids, so they are peptides; epinephrine and testosterone have no amino acids and only a few dozen atoms, so they are small molecules.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03d.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03d.mp4

59
Check q10

Amino acids join one after another to make a chain.

Which bond joins each amino acid to the next?

  1. A. A hydrogen bond
    Hydrogen bonds hold a folded chain in shape; they do not join one amino acid to the next.
  2. B. An ionic bond
    An ionic bond joins two ions; amino acids in a chain are joined by a covalent bond.
  3. C. ✓ A peptide bond

Why: Two amino acids join by a peptide bond.
A chain of amino acids is held together by peptide bonds, one between each pair.

60

Insulin is 51 amino acids, joined by peptide bonds into two short chains that fit together.

Three ligands drawn as molecules: insulin, fifty-one amino acids in two short chains, one of 21 and one of 30, lying side by side so they fit together; epinephrine, a single ring with a short tail; testosterone, four fused carbon rings
Three ligands drawn as molecules: insulin, fifty-one amino acids in two short chains, one of 21 and one of 30, lying side by side so they fit together; epinephrine, a single ring with a short tail; testosterone, four fused carbon rings
61

A ligand that is a chain of amino acids is called a .

62

A ligand of a few dozen atoms with no amino acids is called a .

63

For example, insulin is a peptide, because insulin is built of chains of amino acids.

64

And growth hormone is a peptide, because growth hormone is a chain of amino acids: 191 of them, a whole protein.

65

But epinephrine is a small molecule, because epinephrine has no amino acids and only a few dozen atoms.

66

And testosterone is a small molecule, because testosterone has no amino acids: four carbon rings and a few attached atoms.

67

Here is a table sorting the four ligands into peptides and small molecules.

A two-column table sorting four ligands: peptides, insulin (51 amino acids in two chains) and growth hormone (a chain of 191 amino acids, a whole protein); small molecules, epinephrine (one carbon ring with a short tail, a few dozen atoms) and testosterone (four fused carbon rings, no amino acids)
68

So a ligand is one of two kinds: a peptide, which may be a whole protein, or a small molecule.

69

What you are expected to know Classify a named ligand as a peptide (a chain of amino acids, or a whole protein) or as a small molecule.

70
Check q11

A hormone of the adrenal gland is built of four fused carbon rings with a few attached atoms.

Which kind of ligand is this molecule?

  1. A. A peptide
    A peptide is a chain of amino acids.
    This hormone has no amino acids: four fused carbon rings with a few attached atoms, a few dozen atoms in all.
  2. B. ✓ A small molecule

Why: A peptide is a chain of amino acids.
This hormone is four fused carbon rings with a few attached atoms, and it has no amino acids.
A few dozen atoms make a small molecule.
So this hormone is a small molecule.

71
Check q12

A signal molecule released by gut cells is a folded chain of 120 amino acids.

Which kind of ligand is this molecule?

  1. A. ✓ A peptide
  2. B. A small molecule
    Folding packs the chain into a compact shape, but every one of the 120 amino acids is still there.
    A chain of amino acids is a peptide, whatever its length.

Why: A peptide is a chain of amino acids, short or long.
This signal molecule is a chain of 120 amino acids.
A chain long enough to be a whole protein is still a chain of amino acids.
So this signal molecule is a peptide.

72
Check q13

A signal molecule released by nerve cells is a chain of 5 amino acids.

Which kind of ligand is this molecule?

  1. A. ✓ A peptide
  2. B. A small molecule
    A small molecule is a few dozen atoms with no amino acids.
    A chain of 5 amino acids is still a chain of amino acids, so it is a peptide.

Why: A peptide is a chain of amino acids.
This signal molecule is a chain of 5 amino acids.
A short chain is still a chain of amino acids.
So this signal molecule is a peptide.

73
Check q14

A signal molecule made by the cells lining a blood vessel is a gas of two atoms.

Which kind of ligand is this molecule?

  1. A. A peptide
    A peptide is a chain of amino acids.
    This gas has no amino acids.
    Two atoms make a very small molecule.
    So this signal molecule is a small molecule.
  2. B. ✓ A small molecule

Why: A peptide is a chain of amino acids.
This signal molecule is two atoms, with no amino acids.
So this signal molecule is a small molecule.

74
Check q15

A plant signal molecule is a single carbon ring with a short tail, about 20 atoms in all.

Which kind of ligand is this molecule?

  1. A. A peptide
    A peptide is a chain of amino acids.
    This molecule has no amino acids.
    About 20 atoms make a small molecule.
    So this plant signal is a small molecule.
  2. B. ✓ A small molecule

Why: A peptide is a chain of amino acids.
This plant signal molecule is one carbon ring with a short tail, about 20 atoms, and it has no amino acids.
So this plant signal molecule is a small molecule.

75

Back to the liver cell with receptors for insulin and receptors for epinephrine side by side in its membrane.

76

Insulin’s shape and charges match the insulin receptor’s pocket, so that pocket holds insulin.

77

Epinephrine is far too small to fill that pocket.

78

Epinephrine’s shape and charges match the epinephrine receptor’s pocket, so that pocket holds epinephrine.

79

Insulin is far too large for that pocket.

80

So each pocket takes one molecule and refuses the other.

81Mixed practice mixed practice

82
Check q16

Epinephrine reaches a heart muscle cell and binds a pocket on a protein set in the cell’s membrane.

Which is the ligand?

  1. A. The heart muscle cell
    The heart muscle cell is the target cell; the ligand is the molecule that binds its receptor.
  2. B. The protein set in the membrane
    The protein is the receptor; the ligand is the signal molecule it binds.
  3. C. The pocket on the protein
    The pocket is the binding site; the ligand is the molecule that fits it.
  4. D. ✓ The epinephrine

Why: The ligand is the signal molecule that binds the receptor.
Epinephrine binds the protein’s pocket, so epinephrine is the ligand.

83
Check q17

A hormone molecule fits a pocket on a receptor protein.

What is the pocket called?

  1. A. ✓ The binding site
  2. B. The ligand
    The ligand is the hormone molecule that fits the pocket, not the pocket.
  3. C. The ligand-binding domain
    The ligand-binding domain is the region of the receptor that holds the pocket; the pocket itself is the binding site.

Why: The pocket a molecule fits is the binding site.
The region of the receptor that holds the pocket is the ligand-binding domain.

84
Check q18

A thyroid hormone is two carbon rings with a few attached atoms, about 35 atoms in all.

Which kind of ligand is this molecule?

  1. A. A peptide
    A peptide is a chain of amino acids.
    This hormone has no amino acids.
  2. B. ✓ A small molecule

Why: A peptide is a chain of amino acids.
This thyroid hormone is two carbon rings with a few attached atoms, and it has no amino acids.
About 35 atoms make a small molecule.
So this hormone is a small molecule.

85
Check q19

A cell carries receptors for a growth signal and no other receptors. Insulin reaches the cell.

Which of the following happens?

  1. A. Insulin binds the growth-signal receptors
    The growth-signal receptors’ pockets fit the growth signal by shape and charge; insulin does not fit them.
  2. B. The cell makes a new receptor for insulin
    A cell’s receptors are the proteins it already carries; a signal arriving does not make the cell build a receptor for it.
  3. C. ✓ Insulin passes by and binds nothing

Why: A receptor’s pocket holds only the ligand whose shape and charge match it.
The cell’s receptors fit the growth signal, not insulin.
So insulin passes by and binds nothing.

86
Check q20

A molecule of growth hormone binds its receptor on a bone cell and is released a few seconds later.

Which of the following describes the released molecule?

  1. A. A smaller molecule, with part cut off
    A receptor is not an enzyme; it cuts nothing off its ligand.
  2. B. A different signal molecule
    A receptor makes no product from its ligand.
  3. C. ✓ The same molecule, unchanged

Why: Growth hormone is the receptor’s ligand.
A receptor makes no product from its ligand.
The receptor holds the growth hormone, changes its own shape, and releases the same molecule unchanged.

87
Check q21

A receptor set in a membrane has a region that holds the pocket its ligand fits.

What is that region called?

  1. A. The binding site
    The binding site is the pocket itself; the region of the receptor that holds the pocket has its own name.
  2. B. ✓ The ligand-binding domain
  3. C. The ligand
    The ligand is the signal molecule that fits the pocket; it is not part of the receptor.

Why: A domain is a region of a protein with one job.
The region of the receptor that holds the pocket is the ligand-binding domain.

88
Practice writing an answer

Glucagon is a hormone: a chain of 29 amino acids. A liver cell carries glucagon receptors and epinephrine receptors in the same membrane. When glucagon reaches the cell, 90% of the glucagon receptors bind glucagon, and none of the epinephrine receptors bind it.

(a) Explain how these results demonstrate that a receptor binds its ligand by shape and charge. (1 pt)

Model answer Glucagon is the glucagon receptor’s ligand.
Glucagon’s shape and charges match the glucagon receptor’s pocket, so that pocket holds glucagon.
The epinephrine receptor’s pocket is sized for epinephrine, a molecule of a few dozen atoms.
Glucagon is far larger, and its charges do not match that pocket.
So the epinephrine receptor does not hold glucagon.
So each receptor binds only the ligand whose shape and charge fit its pocket.
Rubric
  • Award 1 point for: glucagon fits the glucagon receptor’s pocket by shape and charge and so is bound; it does not fit the epinephrine receptor’s pocket (far too large, charges not matching) and so is not bound; so the fit by shape and charge decides which receptor binds it.

Glossary

ligand
The signal molecule that binds a receptor. Insulin is the ligand of the insulin receptor.
binding site
The pocket on a protein that a particular molecule fits by shape and charge. An enzyme’s active site is its binding site for its substrate; a receptor’s pocket is its binding site for its ligand.
ligand-binding domain
The region of a receptor that holds the pocket its ligand fits. A domain is a region of a protein with one job.
peptide
A ligand that is a chain of amino acids, short or long. Insulin (51 amino acids in two chains) and growth hormone (191 amino acids, a whole protein) are peptides.
small molecule
A ligand of a few dozen atoms with no amino acids. Epinephrine and testosterone are small molecules.

APBIO-U04-L03B The message crosses, the molecule stays outside

Topic 4.2 · Introduction to Signal Transduction · 99 steps

A muscle cell with labeled insulin molecules, drawn as ringed dots, all outside its membrane: three are held on receptors set in the membrane and the rest float free; glucose, drawn as hexagons, passes in through two open transporters in the membrane; one more transporter rises from inside toward the surface; a legend names glucose, the receptor and the transporter
A muscle cell with labeled insulin molecules, drawn as ringed dots, all outside its membrane: three are held on receptors set in the membrane and the rest float free; glucose, drawn as hexagons, passes in through two open transporters in the membrane; one more transporter rises from inside toward the surface; a legend names glucose, the receptor and the transporter

Here are muscle cells in a dish, given insulin tagged with a label that shows where each molecule goes.

Ten minutes later the cells are taking up four times as much glucose as before. Yet none of the labeled insulin has crossed the membrane into the cytosol.

Something crossed the membrane, and it was not the insulin. What was it?

Unit 4 · Cell Communication and Cell Cycle

1The intracellular domain

2

Video: Watch: The change reaches the inner face

Labeled insulin stays outside the muscle cells, yet the cells take up glucose; with insulin bound, the receptor holds a new shape all the way to the region facing the cytosol, its intracellular domain.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03Ba.mp4

3

How does a signal that never enters the cell still change the cell?

4

The ligand fits the receptor’s binding site.

5

While the ligand is bound, the receptor holds a different shape all the way through to the region facing the cytosol.

6

That inner shape change is passed to the next molecule, and the next.

7

The information crosses the membrane; the signal molecule does not.

8

Start with the muscle cells in the dish.

9
Check q1

A molecule binds a protein at one site.

What can that binding do to the rest of the protein?

  1. A. ✓ Bend the protein’s shape elsewhere
  2. B. Nothing beyond the site it binds
    A protein is one folded chain.
    A molecule bound at one site pulls on the fold, so the shape changes elsewhere too.

Why: A molecule bound at one site can bend a protein’s shape elsewhere, and a protein’s shape is its job.

10

Here are the muscle cells in the dish again.

A muscle cell, its inside shaded, with labeled insulin molecules, drawn as ringed dots, all outside the membrane and three of them held by receptors, one receptor labeled; inside the cell glucose transporters move up to the membrane, and glucose molecules pass in through the transporters already there
A muscle cell, its inside shaded, with labeled insulin molecules, drawn as ringed dots, all outside the membrane and three of them held by receptors, one receptor labeled; inside the cell glucose transporters move up to the membrane, and glucose molecules pass in through the transporters already there
11

None of the labeled insulin has crossed the membrane into the cytosol.

12

Yet within ten minutes the cells are taking up four times as much glucose.

13

The cells have moved glucose transporters from inside the cell to the surface.

14

Something crossed the membrane, and it was not the insulin.

15

The ligand fits the receptor’s binding site by shape and charge.

Two panels of the same receptor set through a membrane, the fluid outside shaded above the membrane and the cytosol shaded below it: on the left the pocket is empty and the inner region has a straight base; on the right the ligand sits in the pocket and the inner region has changed shape, with a bulge on its inner face
Two panels of the same receptor set through a membrane, the fluid outside shaded above the membrane and the cytosol shaded below it: on the left the pocket is empty and the inner region has a straight base; on the right the ligand sits in the pocket and the inner region has changed shape, with a bulge on its inner face
16

While the ligand is bound, the receptor holds a different shape.

17

So another part of the receptor can now act on the next molecule.

18

For a receptor set in the membrane, the part that changes is the region facing the cytosol.

19

That region is called the receptor’s .

20

‘Intra’ means inside, and ‘cellular’ means of the cell: the intracellular domain is the region of the receptor inside the cell.

21

What you are expected to know Identify the intracellular domain: the region of a membrane receptor that faces the cytosol and changes shape while the ligand is bound.

22
Check q2

Insulin binds the pocket of its receptor on a liver cell, and the receptor changes shape.

Which region of the receptor acts on the next molecule inside the cell?

  1. A. The ligand-binding domain
    The ligand-binding domain faces the outside of the cell; it holds the insulin, and nothing inside the cell touches it.
  2. B. The binding site
    The binding site is the pocket the insulin fits, on the outer face of the receptor.
  3. C. ✓ The intracellular domain

Why: With insulin bound, the receptor holds a different shape all the way to its intracellular domain, the region facing the cytosol.
That region is the part inside the cell, so it is the part that acts on the next molecule.

23Quick quiz: intracellular domain mixed practice

24
Check q3

What is a receptor’s intracellular domain?

  1. A. The pocket that the ligand fits
    The pocket the ligand fits is the binding site, on the outer face of the receptor.
  2. B. ✓ The region of the receptor that faces the cytosol
  3. C. The middle part, set within the membrane
    The part within the membrane is surrounded by the phospholipid tails; the intracellular domain is the region below it, in the cytosol.

Why: A domain is a region of a protein with one job.
The intracellular domain is the region of the receptor that faces the cytosol.
While the ligand is bound, the receptor holds a different shape all the way to this region.

25
Practice writing an answer

A receptor is set through the plasma membrane of a muscle cell.

(a) Describe what the receptor’s intracellular domain is. (1 pt)

Model answer The intracellular domain is a region of the receptor.
It is the region that faces the cytosol.
While the ligand is bound, the receptor holds a different shape all the way to this region.
Rubric
  • Award 1 point for: the region of the receptor facing the cytosol (inside the cell), which changes shape while the ligand is bound.

26How the message crosses

27

Video: Watch: The message crosses, the molecule stays outside

The bound ligand holds the receptor in a new shape through to its intracellular domain; that change acts on the next molecule inside, and the next. The relay of the message through the inside of the cell is signal transduction.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03Bb.mp4

28

Here is the receptor with its ligand bound, its inner face changed.

Two panels of the same receptor set through a membrane, the fluid outside shaded above the membrane and the cytosol shaded below it: on the left the pocket is empty and the inner region has a straight base; on the right the ligand sits in the pocket and the inner region has changed shape, with a bulge on its inner face
Two panels of the same receptor set through a membrane, the fluid outside shaded above the membrane and the cytosol shaded below it: on the left the pocket is empty and the inner region has a straight base; on the right the ligand sits in the pocket and the inner region has changed shape, with a bulge on its inner face
29

That change of shape on the inner face is the first link in a relay through the inside of the cell.

30

The changed intracellular domain acts on the next molecule inside the cell.

31

That molecule changes, and acts on the next.

32

The relay of the message through the inside of the cell is called .

33

‘Transduction’ means leading across: the message is led across, from the receptor into the cell.

34

The information crosses the membrane.

35

The signal molecule does not.

36

What moves inside is a change of protein shape, passed from molecule to molecule.

37

What you are expected to know Explain how a signal that never enters the cell still changes the cell.

38
Check q4

A researcher gives fat cells a peptide hormone tagged with a label that shows where each molecule goes. Ten minutes later every labeled molecule is still outside the cells, and the cells have begun breaking down their stored fat.

Which of the following carried the message from the outside of the membrane to the inside of the cell?

  1. A. The hormone molecule, passed through the receptor’s pocket into the cytosol
    Every labeled molecule is still outside the cells after ten minutes, so no hormone molecule crossed the membrane.
    The pocket holds the hormone; it does not pass the hormone through.
  2. B. A piece of the hormone molecule, cut off in the pocket and passed into the cytosol
    A receptor is not an enzyme: it makes no product from the hormone.
    The receptor holds the hormone, changes shape, and releases it unchanged; no piece is cut off.
  3. C. ✓ A change in the shape of the receptor protein

Why: Every labeled molecule is still outside, so no hormone crossed the membrane.
Yet the cells changed, so something else carried the message.
The bound hormone holds the receptor in a different shape, right through the membrane.
So the receptor’s change of shape carried the message across and started the relay.

39
Practice writing an answer

A researcher gives fat cells a peptide hormone tagged with a label that shows where each molecule goes. Ten minutes later every labeled molecule is still outside the cells, yet the cells have begun breaking down their stored fat.

(a) Explain how the message crossed the membrane into the cell while every hormone molecule stayed outside. (1 pt)

Model answer The hormone is the receptor’s ligand and fits its binding site by shape and charge.
While the ligand is bound, the receptor holds a different shape.
The receptor is set through the membrane, so the new shape reaches its intracellular domain.
The changed intracellular domain acts on the next molecule inside the cell, and that molecule changes the next.
So the message travels inside as a change of protein shape while the hormone itself stays outside.
Rubric
  • Award 1 point for: the bound ligand changes the receptor’s shape all the way to its intracellular domain, and that inner change acts on the next molecule inside the cell (signal transduction), so the message crosses while the hormone does not.
40
Check q5

A researcher gives fat cells a labeled peptide hormone. After ten minutes every labeled molecule is still outside the cells, and the cells have begun breaking down their stored fat. A student says: “The cells changed, so some of the hormone must have entered the cells.”

Is the student correct?

  1. A. ✓ No — the hormone stays outside, and the receptor’s change of shape enters
  2. B. Yes — a cell changes only when the signal molecule enters it
    Every labeled hormone molecule is still outside, so no hormone entered.
    The receptor changes shape through to its intracellular domain; that change of shape is what enters the cell.

Why: Every labeled molecule is still outside, so no hormone entered.
The hormone binds its receptor at the surface.
The receptor changes shape through to its intracellular domain.
That change of shape is what enters the cell, and it starts the relay that breaks down the fat.

41
Check q6

Two cultures of muscle cells: one carries normal insulin receptors; the other carries receptors whose intracellular domain is altered but whose pocket is unchanged. A researcher adds insulin to both cultures.

Predict which culture moves glucose transporters to the cell surface.

  1. A. ✓ The normal culture only
  2. B. The altered culture only
    Insulin binds the altered receptor’s pocket, but its altered intracellular domain cannot change shape.
    So the relay never starts, and the transporters stay inside.
  3. C. Both cultures
    Both receptors bind insulin.
    The normal receptor changes shape through to its intracellular domain; the altered receptor passes nothing on.
    So the normal culture alone moves its transporters.
  4. D. Neither culture
    The normal receptor has a working pocket and a working intracellular domain.
    Insulin binds, the receptor changes shape through to its inner face, and the relay starts.

Why: Insulin binds both kinds of receptor.
A normal receptor then changes shape through to its intracellular domain.
That inner change starts the relay that moves the transporters.
The altered receptor cannot change shape inside, so its relay never starts and its transporters stay inside.

42Quick quiz: signal transduction mixed practice

43
Check q7

What is signal transduction?

  1. A. The ligand binding its receptor at the cell surface
    The ligand binding its receptor happens at the surface, before the relay inside begins.
  2. B. ✓ The relay of the message through the inside of the cell
  3. C. The signal molecule crossing the membrane into the cytosol
    The signal molecule stays outside; what moves inside is a change of protein shape.

Why: Signal transduction is the relay of the message through the inside of the cell.
It starts from the change in the receptor’s shape, and one molecule changes the next.

44
Practice writing an answer

A hormone binds its receptor on the surface of a liver cell.

(a) Describe what signal transduction is. (1 pt)

Model answer Signal transduction is the relay of the message through the inside of the cell.
It starts from the change in the receptor’s shape.
One molecule changes the next.
Rubric
  • Award 1 point for: the relay of the message through the inside of the cell, one molecule changing the next, starting from the receptor’s change of shape.

45Three stages, one name for the whole

46

Video: Watch: Reception, transduction, response

Epinephrine binds a heart-cell receptor at 1 second (reception); a relay molecule rises inside at 3 seconds (transduction); the cell contracts harder at 9 seconds (the cellular response). The whole ordered relay is a signal transduction pathway.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03Bc.mp4

47

Now consider epinephrine reaching a heart muscle cell.

Three boxes joined by arrows: epinephrine binds its receptor, labeled reception, at 1 second; a relay molecule rises inside the cell, labeled transduction, at 3 seconds; the cell contracts harder, labeled cellular response, at 9 seconds
Three boxes joined by arrows: epinephrine binds its receptor, labeled reception, at 1 second; a relay molecule rises inside the cell, labeled transduction, at 3 seconds; the cell contracts harder, labeled cellular response, at 9 seconds
48

At 1 second, epinephrine is bound to a receptor in the cell’s membrane.

49

At 3 seconds, a relay molecule inside the cell begins to rise.

50

At 9 seconds, the cell contracts more strongly.

51

The ligand binding its receptor is called .

52

The relay inside, one molecule changing the next, is signal transduction.

53

What the cell finally does differently is called the : here, contracting more strongly.

54

The response is what the cell does, not any step of the relay that gets there.

55

‘Response’ here means the cell’s response, the change in what the cell does; it is a narrower use of the everyday word.

56

The whole ordered relay, from the receptor binding its ligand to the response, is called a .

57

Here is a table of the three stages, what happens in each, and the heart muscle cell’s example with its time.

A table of the three stages of a signal transduction pathway: reception, the ligand binds its receptor, in the heart muscle cell epinephrine binds its receptor at 1 second; transduction, molecules inside the cell relay the message, one changing the next, a relay molecule rises at 3 seconds; cellular response, what the cell finally does differently, the cell contracts harder at 9 seconds
58

What you are expected to know Name the three stages of a signal transduction pathway in order, with what happens in each.

59
Check q8

A signal transduction pathway has three stages.

In which order do the three stages happen?

  1. A. ✓ Reception, transduction, cellular response
  2. B. Transduction, reception, cellular response
    The ligand must bind its receptor before any molecule inside can relay the message.
  3. C. Reception, cellular response, transduction
    The relay inside must reach the end before the cell does anything differently.

Why: The ligand binds its receptor first: reception.
Then molecules inside the cell relay the message: transduction.
Last, the cell does something differently: the cellular response.

60Quick quiz: reception, cellular response, signal transduction pathway mixed practice

61
Check q9

What is reception?

  1. A. ✓ The ligand binding its receptor
  2. B. One molecule inside the cell changing the next
    One molecule inside the cell changing the next is transduction, the stage after reception.
  3. C. What the cell finally does differently
    What the cell finally does differently is the cellular response, the last stage.

Why: Reception is the first stage of a signal transduction pathway.
Reception is the ligand binding its receptor.

62
Check q10

What is the cellular response?

  1. A. The ligand binding its receptor
    The ligand binding its receptor is reception, the first stage.
  2. B. One molecule inside the cell changing the next
    One molecule inside the cell changing the next is transduction, a step of the relay.
  3. C. ✓ What the cell finally does differently

Why: The cellular response is the last stage of a signal transduction pathway.
The cellular response is what the cell finally does differently, such as contracting harder or releasing glucose.

63
Check q11

What is a signal transduction pathway?

  1. A. The pocket on the receptor that a ligand fits
    The pocket a ligand fits is the binding site, one part of the receptor.
  2. B. ✓ The whole relay from the receptor to the cellular response
  3. C. The first stage only, the ligand binding its receptor
    The ligand binding its receptor is reception, the first stage; the pathway is the whole relay from receptor to response.

Why: A signal transduction pathway is the whole ordered relay.
It starts with the receptor binding its ligand and ends with the cellular response.

64
Practice writing an answer

A hormone binds its receptor on a kidney cell, and half a minute later the cell begins moving water channels to its surface.

(a) Describe what happens in the transduction stage of this pathway. (1 pt)

Model answer Transduction is the relay inside the cell.
The receptor’s intracellular domain changes shape and acts on the next molecule.
One molecule changes the next, through the inside of the kidney cell.
The relay carries the message from the receptor to the water channels.
Rubric
  • Award 1 point for: molecules inside the cell relay the message, one changing the next, from the changed receptor toward the response.

65Which stage is this event?

66

Video: Watch: Naming the stage

Five events in a liver cell, each placed in its stage: insulin binding and the receptor’s shape change are reception; relay proteins changing inside are transduction; the cell storing glucose is the cellular response.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03Bd.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L03Bd.mp4

67

One convention decides where reception ends.

68

Reception ends when the receptor has changed shape.

69

That shape change is the last event of reception, not the first event of transduction.

70

Transduction begins when the changed receptor acts on the next molecule inside the cell.

71

For example, insulin fits the pocket of its receptor on a liver cell. This event is reception, because the ligand is binding its receptor.

72

And the receptor changes shape, with insulin bound. This event is still reception, because reception ends when the receptor has changed shape.

73

But a relay protein inside the cell changes shape. This event is transduction, because a molecule inside the cell is passing the message on.

74

And a second relay protein is switched on. This event is transduction too, because a molecule inside the cell is passing the message on.

75

But the cell begins storing glucose. This event is the cellular response, because it is what the cell finally does differently.

76

Here is a table sorting the five events in the liver cell into their stages.

A two-column table of five events in a liver cell and the stage each belongs to: insulin fits the pocket of its receptor, reception; the receptor changes shape with insulin bound, reception; a relay protein inside the cell changes shape, transduction; a second relay protein is switched on, transduction; the cell begins storing glucose, cellular response
77

What you are expected to know Classify a described event as reception, transduction or the cellular response.

78
Check q12

A stomach cell begins releasing acid.

Which stage of the pathway is this event?

  1. A. Reception
    Reception is the ligand binding its receptor.
    Releasing acid is what the stomach cell finally does differently.
    So releasing acid is the cellular response.
  2. B. Transduction
    Transduction is the relay inside the cell, one molecule changing the next.
    Releasing acid is what the stomach cell finally does differently.
    So releasing acid is the cellular response.
  3. C. ✓ Cellular response

Why: The cellular response is what the cell finally does differently.
The stomach cell begins releasing acid.
So releasing acid is the cellular response.

79
Check q13

A hormone fits the pocket of a receptor on a bone cell.

Which stage of the pathway is this event?

  1. A. ✓ Reception
  2. B. Transduction
    Transduction is the relay inside the cell.
    The hormone fitting the receptor’s pocket is the ligand binding its receptor.
    So this event is reception.
  3. C. Cellular response
    The response is what the cell finally does differently.
    The hormone fitting the receptor’s pocket is the ligand binding its receptor.
    So this event is reception.

Why: Reception is the ligand binding its receptor.
The hormone fitting the pocket of the receptor is the ligand binding its receptor.
So this event is reception.

80
Check q14

With its ligand bound, a receptor in a kidney cell’s membrane changes shape.

Which stage of the pathway is this event?

  1. A. ✓ Reception
  2. B. Transduction
    The receptor changing shape with its ligand bound is the last event of reception.
    Transduction begins when the changed receptor acts on the next molecule inside the cell.
  3. C. Cellular response
    The response is what the cell finally does differently.
    The receptor changing shape with its ligand bound is the end of reception.

Why: Reception is the ligand binding its receptor, and reception ends when the receptor has changed shape.
The receptor changing shape with its ligand bound is the last event of reception.
So this event is reception.

81
Check q15

A guard cell narrows the pore beside it.

Which stage of the pathway is this event?

  1. A. Reception
    Reception is the ligand binding its receptor.
    Narrowing the pore is what the guard cell finally does differently.
    So narrowing the pore is the cellular response.
  2. B. Transduction
    Transduction is the relay of molecules inside the cell.
    Narrowing the pore is what the guard cell finally does differently.
    So narrowing the pore is the cellular response.
  3. C. ✓ Cellular response

Why: The cellular response is what the cell finally does differently.
The guard cell narrows the pore beside it.
So narrowing the pore is the cellular response.

82
Check q16

One relay protein inside a kidney cell changes the shape of a second relay protein.

Which stage of the pathway is this event?

  1. A. Reception
    Reception is the ligand binding its receptor at the surface.
    One relay protein changing the shape of the next is the relay inside the cell.
    So this event is transduction.
  2. B. ✓ Transduction
  3. C. Cellular response
    The response is what the cell finally does differently.
    One relay protein changing the next is a step of the relay inside the cell.
    So this event is transduction.

Why: Transduction is the relay inside the cell, one molecule changing the next.
One relay protein changing the shape of a second relay protein is exactly that.
So this event is transduction.

83
Check q17

A signal binds a receptor on a gland cell. A relay protein inside the cell changes shape. Then the cell releases a stored hormone into the blood. A student says: “The relay protein changing shape is the cell’s response to the signal.”

Is the student correct?

  1. A. Yes — the relay protein changing shape is the cellular response
    The relay protein changing shape is one molecule inside the cell changing: that is transduction.
    The cellular response is what the cell finally does differently: here, releasing the stored hormone.
  2. B. ✓ No — that change is transduction, and the response is releasing the hormone

Why: The response is what the cell finally does differently, not a step of the relay.
The relay protein changing shape is transduction.
The cell releasing the stored hormone is the cellular response.

84
Check q18

A hormone binds a receptor on a liver cell. The receptor changes shape. Then the receptor’s intracellular domain acts on a relay protein. A student says: “Transduction began when the receptor changed shape.”

Is the student correct?

  1. A. ✓ No — transduction began when the receptor acted on the relay protein
  2. B. Yes — transduction begins as soon as the receptor changes shape
    Reception ends when the receptor has changed shape.
    Transduction begins when the changed receptor acts on the next molecule inside the cell: here, the relay protein.

Why: Reception is the ligand binding its receptor, and reception ends when the receptor has changed shape.
Transduction begins when the changed receptor acts on the next molecule inside the cell.
Here that molecule is the relay protein.
So transduction began when the receptor acted on the relay protein.

85

Back to the muscle cells in the dish, given insulin tagged with a label that shows where each molecule goes.

86

Every labeled insulin molecule stayed outside the cells.

87

Yet within ten minutes the cells took up four times as much glucose.

88

What crossed the membrane was a change in the shape of the receptor.

89

That change started a relay inside, and the relay moved glucose transporters to the surface.

90

So the insulin stayed outside, and the cells still took up glucose.

91Mixed practice mixed practice

92
Check q19

In a yeast cell, a mating signal binds its receptor at 0 seconds; a relay protein inside changes shape at 5 seconds; the cell stops dividing at 60 seconds.

Which event is transduction?

  1. A. ✓ The relay protein changing shape at 5 seconds
  2. B. The mating signal binding its receptor at 0 seconds
    The ligand binding its receptor is the first stage; transduction is the relay inside that follows.
  3. C. The cell stopping its division at 60 seconds
    Stopping division is what the cell finally does differently; transduction is the relay that gets there.
  4. D. The mating signal being released by the other cell
    Transduction happens inside the target cell, after reception.

Why: Transduction is the relay inside the cell, one molecule changing the next.
The relay protein changing shape at 5 seconds is a step of that relay.

93
Check q20

In a liver cell, epinephrine binds a receptor; the receptor changes shape; a relay molecule rises; an enzyme is switched on; the cell releases glucose.

What is the name for this whole ordered relay, from receptor to response?

  1. A. Reception
    Reception is the first stage only: the ligand binding its receptor.
    The whole relay has a name of its own.
  2. B. ✓ A signal transduction pathway
  3. C. A cellular response
    The cellular response is the last stage only: what the cell finally does.
    The whole relay from receptor to response has a name of its own.

Why: The whole ordered relay, from the receptor binding its ligand to the cell’s response, is a signal transduction pathway.

94
Check q21

The model below shows three steps of a signaling pathway in a liver cell, numbered 1 to 3.

Three boxes joined by arrows and numbered: step 1, epinephrine bound to its receptor; step 2, a relay molecule rises inside the cell; step 3, the cell releases glucose
Three boxes joined by arrows and numbered: step 1, epinephrine bound to its receptor; step 2, a relay molecule rises inside the cell; step 3, the cell releases glucose

Which numbered step is the cellular response?

  1. A. Step 1
    Step 1, epinephrine bound to its receptor, is reception, the first stage, not what the cell finally does.
  2. B. Step 2
    Step 2, the relay molecule rising, is transduction; the response is the change in what the cell does.
  3. C. ✓ Step 3
  4. D. Steps 1 and 2 together
    Reception and transduction together are still not what the cell finally does differently; step 3, releasing glucose, is.

Why: The cellular response is what the cell finally does differently: step 3, the cell releasing glucose.
Step 1 is reception and step 2 is transduction.

95
Check q22

Insulin binds its receptor on the surface of a fat cell.

Which is the first change inside the cell?

  1. A. Insulin appears in the cytosol
    Insulin stays outside; the change that enters is a change of the receptor’s shape.
  2. B. Glucose transporters reach the surface at once
    The transporters move only after the relay inside has passed the message on, minutes later.
  3. C. The receptor’s pocket changes shape on the outer face
    The pocket holds the ligand; the change that starts the relay is on the receptor’s inner face.
  4. D. ✓ The receptor’s intracellular domain changes shape

Why: With insulin bound, the receptor holds a different shape all the way to its intracellular domain, the region facing the cytosol.
That change on the inner face is the first change inside the cell.

96
Check q23

A peptide hormone binds its receptor on the surface of a kidney cell, and the cell begins moving water channels to its surface.

Which of the following crossed the membrane?

  1. A. The hormone molecule
    The hormone binds the pocket on the outer face and stays outside.
  2. B. A piece of the hormone molecule
    A receptor makes no product from its ligand; no piece is cut off.
  3. C. ✓ A change in the receptor’s shape

Why: The hormone stays outside, held in the receptor’s pocket.
The receptor changes shape through to its intracellular domain.
That change of shape is what crosses the membrane and starts the relay inside.

97
Check q24

A plant signal binds a receptor on a root cell. A relay protein inside the root cell changes shape. Then the root cell begins to grow toward water.

Which event is the cellular response?

  1. A. The signal binding its receptor
    The signal binding its receptor is reception, the first stage.
  2. B. The relay protein changing shape
    The relay protein changing shape is transduction, a step of the relay inside.
  3. C. ✓ The root cell growing toward water

Why: The cellular response is what the cell finally does differently.
The root cell growing toward water is what it finally does differently.
So that event is the cellular response.

98
Practice writing an answer

Hormone G binds a receptor on the surface of skin cells, and within an hour the cells make more of a repair protein, measured as repair protein output in units. The researchers give three cell lines the same concentration of G. The results are in the table.

cell linereceptors binding G (%)repair protein output (units)
normal92rises from 10 to 80
line R (altered ligand-binding domain)3stays at 10
line K (altered intracellular domain)91stays at 10
Directly activating a relay protein inside the cell raises output to 80 units in all three lines. The model shows the receptor and where each line differs.

A receptor set through a membrane band, the fluid outside shaded above the band and the cytosol shaded below it, with its ligand in the outer pocket; the region above the membrane is marked as altered in cell line R, and the region below the membrane is marked as altered in cell line K
A receptor set through a membrane band, the fluid outside shaded above the band and the cytosol shaded below it, with its ligand in the outer pocket; the region above the membrane is marked as altered in cell line R, and the region below the membrane is marked as altered in cell line K

(a) Identify the stage of signaling that fails in line R, and justify your answer using the binding data. (1 pt)

Model answer Reception fails in line R.
In normal cells 92% of the receptors bind G.
In line R only 3% of the receptors bind G.
So the altered ligand-binding domain no longer fits G.
Reception is the ligand binding its receptor, and here G is barely bound, so the pathway never starts.
Rubric
  • Award 1 point for: reception, because G barely binds the altered ligand-binding domain (3% against 92%).

Slip Naming transduction. The relay inside is intact in line R (activating a relay protein directly restores the output); the failure is that G is not bound.

(b) Explain why line K shows no rise in output although 91% of its receptors bind G. (1 pt)

Model answer In line K 91% of the receptors bind G, so reception happens.
The intracellular domain of line K’s receptor is altered.
So the receptor cannot change shape on its inner face when G is bound.
The change of shape never reaches the inside of the cell.
So signal transduction never starts, and the output stays at 10 units.
Rubric
  • Award 1 point for: the altered intracellular domain cannot pass the change of shape to the inside of the cell, so transduction does not start even though G is bound.
  • Accept the bypass evidence as the justification: directly activating a relay protein raises line K’s output to 80 units, so every step after the receptor works and the break is at the receptor’s inner face.

Slip Saying that G does not bind line K’s receptors. It does (91%); the failure comes after binding, at the inner face.

(c) Predict the effect on line K’s output of raising the concentration of G one hundred-fold, and justify your prediction. (2 pt)

Model answer Line K’s output stays at 10 units.
At the ordinary concentration 91% of line K’s receptors already bind G, so more G can add almost no binding.
Binding is not the problem: the altered intracellular domain cannot change shape whatever sits in the pocket.
So the change of shape still cannot reach the inside of the cell.
Only a change inside the cell, such as activating the relay protein directly, raises the output.
Rubric
  • Award 1 point for: the prediction that the output stays at 10 units (no rise).
  • Award 1 point for: the justification that the receptors are already almost all bound (91%) and the block is at the intracellular domain, so more ligand cannot pass the change of shape inward.

Slip Predicting a rise because more G means a stronger signal. More ligand changes only how many receptors are bound, and line K’s are already bound; the break is inside.

Glossary

intracellular domain
The region of a receptor set in the membrane that faces the cytosol. While the ligand is bound, the receptor holds a different shape all the way to this inner region.
signal transduction
The relay of a message through the inside of a cell, one molecule changing the next, starting from the change in the receptor’s shape.
reception
The first stage of a signaling pathway: the ligand binds its receptor. Reception ends when the receptor has changed shape. That shape change is the last event of reception, not the first event of transduction.
cellular response
The last stage of a signaling pathway: what the cell finally does differently, such as contracting harder or releasing glucose. The response is what the cell does, not a step of the relay that gets there.
signal transduction pathway
The whole ordered relay from the receptor binding its ligand, through the relay of molecules inside the cell, to the cellular response.

APBIO-U04-L04 Stopped at the surface, or straight through

Topic 4.2 · Introduction to Signal Transduction · 38 steps

One cell with two hormones arriving: insulin, a folded chain, held at a receptor on the surface; testosterone, a small molecule, passing through the membrane, binding a receptor inside and moving with it into the nucleus
One cell with two hormones arriving: insulin, a folded chain, held at a receptor on the surface; testosterone, a small molecule, passing through the membrane, binding a receptor inside and moving with it into the nucleus

Here is one cell in the path of a drop of your blood, and the drop carries both insulin and testosterone.

Insulin is stopped at every cell’s surface. Testosterone slips straight through the membrane. Minutes later, testosterone is found in the nucleus, bound to a protein.

Why is one hormone stopped at the surface? Why does the other get inside?

Unit 4 · Cell Communication and Cell Cycle

1Stopped at the surface, or straight through

2

Video: Watch: Where a receptor waits

Insulin, large and polar, is held at the cell surface and binds a receptor set in the membrane. Testosterone, small and nonpolar, slips through the membrane and binds a receptor in the cytosol; the bound pair moves into the nucleus.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L04a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L04a.mp4

3
Check q1

A membrane’s oily middle lets some molecules through on their own and holds others back.

A stretch of phospholipid bilayer drawn as two rows of heads with their tails touching in the middle; the fluid outside the cell is shaded above it and the fluid inside below it; a small four-ringed steroid passes down through the bilayer, and a folded chain of insulin stays above it
A stretch of phospholipid bilayer drawn as two rows of heads with their tails touching in the middle; the fluid outside the cell is shaded above it and the fluid inside below it; a small four-ringed steroid passes down through the bilayer, and a folded chain of insulin stays above it

Which of these crosses the oily middle of a membrane on its own?

  1. A. A large polar molecule
    The oily middle holds a large polar molecule back.
  2. B. An ion
    The oily middle holds back a charged particle, however small.
  3. C. ✓ A small nonpolar molecule

Why: Only small nonpolar molecules cross the oily middle on their own.
Ions, large molecules and polar molecules are held back.

4

Where does a receptor wait for its ligand? The ligand’s size and polarity decide.

5

Read a ligand’s size and polarity, and you can say where its receptor waits before anyone shows you.

6

Insulin is large and polar. So insulin cannot cross the oily middle of the membrane.

7

Insulin binds a receptor at the cell surface. That receptor sits in the membrane with its binding site facing outward.

8

A receptor set in the plasma membrane with its binding site facing outward is called a .

9

Testosterone is small and nonpolar. So testosterone slips through the oily middle of the membrane.

10

Testosterone binds a receptor inside the cell. That receptor waits in the cytosol or the nucleus.

11

A receptor inside the cell, in the cytosol or the nucleus, is called an .

12

Now consider epinephrine, the hormone your body releases when you are frightened. Epinephrine is small, like testosterone, but polar.

13

The oily middle of the membrane holds back a polar molecule, even a small one. So epinephrine stays outside, and epinephrine binds a cell-surface receptor.

14

So a ligand’s size and polarity set where its receptor waits.

15

A polar or large ligand binds a cell-surface receptor. A small nonpolar ligand binds an intracellular receptor.

One cell, the fluid outside shaded above the membrane and the cytosol shaded below it: a receptor set in the membrane holds insulin in an outward-facing pocket; testosterone passes through the membrane to a receptor waiting in the cytosol, and an arrow shows the bound pair moving into the nucleus
One cell, the fluid outside shaded above the membrane and the cytosol shaded below it: a receptor set in the membrane holds insulin in an outward-facing pocket; testosterone passes through the membrane to a receptor waiting in the cytosol, and an arrow shows the bound pair moving into the nucleus
16

Here is a table comparing the three ligands: their size, their polarity, whether each crosses the oily middle, and the kind of receptor each binds.

Table comparing insulin, testosterone and epinephrine by size, polarity, whether each crosses the oily middle of the membrane, and the kind of receptor each binds
17

A steroid entering the cell is the exception to the rule that the ligand stays outside. Even then the hormone acts only through its receptor, never on its own.

18

What you are expected to know Predict, from a ligand’s size and polarity, whether its receptor is a cell-surface receptor or an intracellular receptor.

19
Check q2

A researcher gives cells two ligands, P and S. The model below shows where each ligand is found after ten minutes.

One cell with a membrane across the top and a nucleus below, the fluid outside shaded differently from the cytosol; filled ringed dots of ligand P sit only outside the membrane, and open ringed dots of ligand S sit inside the nucleus
One cell with a membrane across the top and a nucleus below, the fluid outside shaded differently from the cytosol; filled ringed dots of ligand P sit only outside the membrane, and open ringed dots of ligand S sit inside the nucleus

Which of the following describes ligand S?

  1. A. A peptide
    A peptide is a chain of amino acids, large and polar, and the membrane’s oily middle holds it back.
    S was found in the nucleus, so S crossed the membrane.
  2. B. A large polar molecule
    The membrane’s oily middle holds a large polar molecule back.
    S was found in the nucleus, so S crossed the membrane.
  3. C. ✓ A small nonpolar molecule

Why: S was found inside the nucleus, so S crossed the plasma membrane.
The membrane’s oily middle lets only small nonpolar molecules through.
So S is small and nonpolar, and S binds an intracellular receptor.
P stayed outside, so P cannot cross: a peptide or a polar molecule.

20
Practice writing an answer

Growth hormone is a peptide, a chain of 191 amino acids, that makes bone cells grow. Its receptor is a cell-surface receptor.

(a) Explain why the receptor for growth hormone sits in the plasma membrane with its binding site facing outward. (1 pt)

Model answer Growth hormone is a chain of 191 amino acids.
A chain of amino acids is large and polar.
The middle of the plasma membrane is oily.
An oily middle holds back a large polar molecule.
So growth hormone cannot cross the membrane.
So growth hormone can bind a receptor only at the surface of the cell.
Therefore its receptor sits in the membrane.
The binding site faces outward, because the hormone stays outside the cell.
Rubric
  • Award 1 point for: a large polar peptide cannot cross the membrane’s oily middle, so it binds a receptor at the cell surface, whose binding site faces outward.
21
Check q3

Testosterone is a small nonpolar hormone that slips through the plasma membrane. A student says: “Because testosterone gets inside the cell, it can act on the cell’s DNA by itself, with no receptor.”

Is the student correct?

  1. A. ✓ No — testosterone changes the cell only through its receptor
  2. B. Yes — a hormone that enters the cell needs no receptor
    Inside the cell, testosterone binds an intracellular receptor; the bound pair moves into the nucleus and acts on the DNA.
    A hormone never acts without its receptor.

Why: A hormone acts only through its receptor.
Testosterone does enter the cell.
Inside the cell, testosterone binds an intracellular receptor.
The bound pair moves into the nucleus and attaches to the DNA.
So testosterone changes the cell through its receptor, never on its own.

22

Back to the one cell in the path of a drop of your blood, with insulin and testosterone both arriving.

23

Insulin is large and polar. So the membrane stops insulin at the cell surface, where insulin binds a cell-surface receptor.

24

Testosterone is small and nonpolar. So testosterone slips through the membrane to an intracellular receptor, and minutes later the bound pair is in the nucleus.

25Quick quiz: cell-surface receptor, intracellular receptor mixed practice

26
Check q4

What is a cell-surface receptor?

  1. A. ✓ A receptor set in the plasma membrane with its binding site facing outward
  2. B. A receptor set in the plasma membrane with its binding site facing the cytosol
    A cell-surface receptor binds a ligand that stays outside the cell, so its binding site faces outward.
  3. C. A receptor waiting in the cytosol or the nucleus
    A receptor in the cytosol or the nucleus is an intracellular receptor.

Why: A cell-surface receptor is set in the plasma membrane.
Its ligand cannot cross the membrane, so its binding site faces outward.

27
Check q5

What is an intracellular receptor?

  1. A. A ligand that has crossed the membrane into the cell
    A ligand is the signal molecule; the receptor is the protein that binds it.
  2. B. ✓ A receptor inside the cell, in the cytosol or the nucleus
  3. C. A receptor set in the plasma membrane with its binding site facing outward
    A receptor set in the plasma membrane with its binding site facing outward is a cell-surface receptor.

Why: An intracellular receptor is a receptor inside the cell, in the cytosol or the nucleus.
It binds a small nonpolar ligand that has slipped through the membrane.

28
Practice writing an answer

A liver cell carries cell-surface receptors and intracellular receptors.

(a) State what a cell-surface receptor is. (1 pt)

Model answer A cell-surface receptor is a receptor set in the plasma membrane with its binding site facing outward.
Rubric
  • Award 1 point for: a receptor set in the plasma membrane with its binding site facing outward (or: facing the fluid outside the cell).

(b) State what an intracellular receptor is. (1 pt)

Model answer An intracellular receptor is a receptor inside the cell, in the cytosol or the nucleus.
Rubric
  • Award 1 point for: a receptor inside the cell (in the cytosol or the nucleus).
29
Check q6

A plant signal molecule is small and nonpolar, and it dissolves in oil.

Which kind of receptor does this signal molecule bind?

  1. A. A cell-surface receptor
    A small nonpolar molecule dissolves into the membrane’s oily middle and crosses into the cell.
    A ligand that crosses the membrane binds an intracellular receptor.
  2. B. ✓ An intracellular receptor

Why: A small nonpolar molecule dissolves into the membrane’s oily middle and crosses into the cell.
So this plant signal binds a receptor inside the cell.
Its receptor is an intracellular receptor.

30
Check q7

A hormone is a chain of 84 amino acids.

Which kind of receptor does this signal molecule bind?

  1. A. ✓ A cell-surface receptor
  2. B. An intracellular receptor
    A chain of amino acids is large and polar, and the membrane’s oily middle holds it back.
    So this hormone stays outside and binds a cell-surface receptor.

Why: A chain of 84 amino acids is a peptide, large and polar.
The membrane’s oily middle holds a large polar molecule back.
So this hormone binds a receptor at the cell surface.
Its receptor is a cell-surface receptor.

31
Check q8

A signal molecule is an ion carrying a full charge.

Which kind of receptor does this signal molecule bind?

  1. A. ✓ A cell-surface receptor
  2. B. An intracellular receptor
    The membrane’s oily middle holds back a charged particle, however small.
    So the ion stays outside and binds a cell-surface receptor.

Why: An ion carries a full charge.
The membrane’s oily middle holds back a charged particle, however small the ion is.
So this signal molecule binds a receptor at the cell surface.
Its receptor is a cell-surface receptor.

32
Check q9

A hormone of the ovary is a steroid, built of four fused carbon rings.

Which kind of receptor does this signal molecule bind?

  1. A. A cell-surface receptor
    A steroid is small and nonpolar, so it dissolves into the membrane’s oily middle and crosses into the cell.
    This hormone binds an intracellular receptor.
  2. B. ✓ An intracellular receptor

Why: A steroid is small and nonpolar.
A small nonpolar molecule dissolves into the membrane’s oily middle and crosses into the cell.
So this hormone binds a receptor inside the cell.
Its receptor is an intracellular receptor.

33
Check q10

A signal molecule is small and polar, like a sugar.

Which kind of receptor does this signal molecule bind?

  1. A. ✓ A cell-surface receptor
  2. B. An intracellular receptor
    The membrane’s oily middle holds a polar molecule back, even a small one.
    So this signal molecule stays outside and binds a cell-surface receptor.

Why: This signal molecule is polar.
The membrane’s oily middle holds a polar molecule back, even a small one.
So this signal molecule binds a receptor at the cell surface.
Its receptor is a cell-surface receptor.

34
Check q11

A signal molecule made by the cells lining a blood vessel is a gas of two atoms, small and nonpolar.

Which kind of receptor does this signal molecule bind?

  1. A. A cell-surface receptor
    A small nonpolar molecule dissolves into the membrane’s oily middle and crosses into the cell.
    So this gas binds an intracellular receptor.
  2. B. ✓ An intracellular receptor

Why: This gas is small and nonpolar.
A small nonpolar molecule dissolves into the membrane’s oily middle and crosses into the cell.
So this signal molecule binds a receptor inside the cell.
Its receptor is an intracellular receptor.

35Mixed practice: where the receptor waits mixed practice

36
Check q12

Signal molecule Z is a chain of 14 amino acids released into the blood by gut cells.

Predict where the receptor for Z sits on its target cells.

  1. A. In the nucleus, attached to the DNA
    A chain of amino acids is large and polar, so it cannot cross the membrane.
    A receptor at the cell surface binds it.
  2. B. In the cytosol, near the nucleus
    The oily middle of the membrane holds a chain of amino acids back.
  3. C. In the membrane, with the binding site facing the cytosol
    The ligand stays outside, so the binding site must face outward to bind it.
  4. D. ✓ In the membrane, with the binding site facing outward

Why: A chain of 14 amino acids is a peptide, large and polar, so it cannot cross the membrane.
Its receptor is a cell-surface receptor with the binding site facing outward.

37
Check q13

A researcher gives cells a small nonpolar hormone, and the cells respond.

Predict where the hormone molecules are found ten minutes later.

  1. A. Outside the cell, bound to a receptor in the membrane
    A small nonpolar molecule crosses the oily middle and binds a receptor inside the cell.
  2. B. ✓ Inside the cell, bound to their receptor in the nucleus
  3. C. Outside the cell, free in the fluid
    The hormone is not released after binding.
    Within ten minutes the bound pair has moved into the nucleus, and the hormone is still bound.
  4. D. Inside the cell, free in the cytosol, acting on the DNA alone
    A hormone acts only through its receptor; the pair attaches to the DNA together.

Why: A small nonpolar hormone crosses the membrane, binds its intracellular receptor, and the bound pair moves into the nucleus.
Ten minutes later the hormone molecules are there, bound to the receptor.

Glossary

cell-surface receptor
A receptor set in the plasma membrane with its binding site facing outward. It binds ligands that cannot cross the membrane, such as insulin and other peptides.
intracellular receptor
A receptor inside the cell, in the cytosol or the nucleus. It binds small nonpolar ligands, such as testosterone and cortisol, that slip through the membrane.

APBIO-U04-L04B The receptor goes to the DNA

Topic 4.2 · Introduction to Signal Transduction · 60 steps

Two drawings of the same liver cell: before cortisol, small receptor dots are scattered through the cytosol and the nucleus holds none; ten minutes after cortisol arrives, the receptor dots, each with a cortisol molecule (a small hexagon) attached, sit inside the nucleus; an arrow between the two drawings marks the ten minutes
Two drawings of the same liver cell: before cortisol, small receptor dots are scattered through the cytosol and the nucleus holds none; ten minutes after cortisol arrives, the receptor dots, each with a cortisol molecule (a small hexagon) attached, sit inside the nucleus; an arrow between the two drawings marks the ten minutes

Cortisol is a steroid hormone. The adrenal glands release cortisol when the body needs fuel.

Before cortisol arrives, a liver cell’s cortisol receptor sits mainly in the cytosol. Ten minutes after cortisol arrives, most of that receptor is in the nucleus.

Within an hour, the cell’s output of one cortisol-controlled protein has risen four-fold.

What did the receptor do in the nucleus? Why did the protein take an hour?

Unit 4 · Cell Communication and Cell Cycle

1A gene

2

Video: Watch: A gene

A long strand of DNA; one stretch of it is highlighted, and an arrow leads from that stretch to the insulin receptor protein. A stretch of DNA that carries the instructions for making one protein is a gene.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L04Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L04Ba.mp4

3
Check q1

Every cell of your body carries DNA.

Where does DNA hold its information?

  1. A. ✓ In the order of its bases
  2. B. In the length of its strands
    A longer strand holds more bases; the information is in which base comes where.
  3. C. In the number of its strands
    DNA has two strands in every cell; the information is in the order of the bases along them.

Why: DNA is a chain of bases.
The information is held in the order of those bases along the chain.

4

What does an intracellular receptor do once its ligand has bound? To answer that, you need two words first.

5

Every cell of your body carries the same DNA. The information in DNA is held in the order of its bases.

6

One stretch of that DNA carries the instructions for making the insulin receptor protein.

A long strand of DNA with one stretch highlighted as the gene for the insulin receptor; an arrow leads from the gene to the insulin receptor protein
A long strand of DNA with one stretch highlighted as the gene for the insulin receptor; an arrow leads from the gene to the insulin receptor protein
7

Ribosomes make the insulin receptor protein in the order set by a copy of those instructions.

8

A stretch of a cell’s DNA that carries the instructions for making one protein is called a .

9

The gene for the insulin receptor is one gene. The gene for keratin, the protein hair is built from, is another.

10

What you are expected to know Describe what a gene is: a stretch of a cell’s DNA that carries the instructions for making one protein.

11
Check q2

Keratin is the protein hair is built from. Cells at the base of a hair make keratin.

Which of the following is the gene for keratin?

  1. A. ✓ The stretch of DNA that carries the instructions for making keratin
  2. B. The keratin protein itself, folded and built into the hair
    Keratin is the protein the gene’s instructions make; the gene is the instructions.
  3. C. The ribosome that makes keratin from a copy of the instructions
    A ribosome makes the protein from a copy of the instructions; the gene is the instructions.

Why: A gene is a stretch of a cell’s DNA that carries the instructions for making one protein.
The stretch of DNA that carries the instructions for making keratin is the keratin gene.

12Quick quiz: the gene mixed practice

13
Check q3

What is a gene?

  1. A. A protein a cell has made from a copy of its instructions
    A protein is what the gene’s instructions make.
  2. B. ✓ A stretch of a cell’s DNA that carries the instructions for making one protein
  3. C. The whole of a cell’s DNA, every stretch of it together
    A cell’s DNA carries many genes; one gene is one stretch of it.

Why: A gene is one stretch of a cell’s DNA.
That stretch carries the instructions for making one protein.

14
Practice writing an answer

A liver cell’s DNA carries thousands of genes.

(a) State what a gene is. (1 pt)

Model answer A gene is a stretch of a cell’s DNA that carries the instructions for making one protein.
Rubric
  • Award 1 point for: a stretch of DNA that carries the instructions for making one protein.
15
Check q4

A stretch of a liver cell’s DNA carries the instructions for making the enzyme catalase.

Is this stretch of DNA a gene?

  1. A. ✓ Yes
  2. B. No
    This stretch of DNA carries the instructions for making one protein, catalase.

Why: This stretch of DNA carries the instructions for making one protein, catalase.
So this stretch of DNA is a gene: the catalase gene.

16
Check q5

A catalase molecule, a folded protein, sits in the cytosol of a liver cell.

Is this catalase molecule a gene?

  1. A. Yes
    A gene is a stretch of DNA; catalase is the protein made from the catalase gene’s instructions.
  2. B. ✓ No

Why: A gene is a stretch of a cell’s DNA.
A catalase molecule is a protein, not DNA.
So the catalase molecule is not a gene.

17
Check q6

A ribosome in a liver cell is making a catalase molecule.

Is this ribosome a gene?

  1. A. Yes
    A gene is a stretch of DNA; a ribosome makes the protein from a copy of the gene’s instructions.
  2. B. ✓ No

Why: A gene is a stretch of a cell’s DNA.
A ribosome is the structure that makes the protein.
So the ribosome is not a gene.

18Using a gene: gene expression

19

Video: Watch: Using a gene

A stomach cell carries the gene for pepsin and makes pepsin; a muscle cell carries the same gene and makes none. A cell using a gene’s instructions to make its protein is gene expression; a cell changes gene expression when it starts, stops, speeds or slows that making.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L04Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L04Bb.mp4

20

A stomach cell carries the gene for pepsin, the enzyme that digests protein in the stomach. The stomach cell uses that gene’s instructions and makes pepsin.

A long strand of DNA with the gene for pepsin highlighted and an arrow to the pepsin protein; below, a stomach cell holds three pepsin molecules and a muscle cell holds none
A long strand of DNA with the gene for pepsin highlighted and an arrow to the pepsin protein; below, a stomach cell holds three pepsin molecules and a muscle cell holds none
21

A muscle cell carries the same gene. The muscle cell does not use that gene, so the muscle cell makes no pepsin.

22

A cell is expressing a gene when it is using that gene’s instructions to make the protein. Making the protein from its gene is called .

23

So the stomach cell expresses the gene for pepsin. The muscle cell has the same gene and does not express it.

24

A cell changes gene expression when it starts, stops, speeds or slows the making of a protein.

25

What you are expected to know Describe gene expression: a cell using a gene’s instructions to make that gene’s protein.

26

What you are expected to know Identify a change in gene expression: the cell starts, stops, speeds or slows the making of a protein.

27
Check q7

After a signal arrives, a skin cell makes four times as much of a protective protein as before.

Which of the following changed in the skin cell?

  1. A. The cell gained a new gene for the protein
    The cell already had the gene; the cell changed how fast it used the gene.
  2. B. The signal molecule became the protein
    The cell built the protein from the gene’s instructions; the signal only changed how fast.
  3. C. The order of bases in the gene changed
    The order of its bases stayed the same; the cell sped up making the protein from it.
  4. D. ✓ The cell changed the expression of the gene for the protein

Why: The cell already carried the gene and sped up making the protein from it.
Speeding up the making of a protein is a change in gene expression.

28
Check q8

A pancreas cell makes the enzyme amylase. A skin cell from the same body makes no amylase at all.

Which statement explains the missing amylase?

  1. A. The skin cell’s copy of the gene has a different order of bases
    The skin cell carries the same gene as the pancreas cell; it is not using it.
  2. B. The skin cell makes amylase and then breaks it down
    The skin cell never makes amylase: the gene is present and not expressed.
  3. C. ✓ The skin cell has the amylase gene and does not express it

Why: Every cell of the body carries the same genes.
A cell carries a protein only if it expresses that protein’s gene.
The skin cell has the amylase gene and does not express it.

29Quick quiz: gene expression mixed practice

30
Check q9

What is gene expression?

  1. A. A cell copying its DNA before it divides
    Copying DNA is not making a protein from a gene.
  2. B. A cell carrying a gene in its DNA without using it
    Every cell carries the gene; expressing it is using its instructions to make the protein.
  3. C. ✓ A cell using a gene’s instructions to make that gene’s protein

Why: Gene expression is a cell using a gene’s instructions to make that gene’s protein.

31
Practice writing an answer

A liver cell makes the insulin receptor protein from its gene.

(a) State what gene expression is. (1 pt)

Model answer Gene expression is a cell using a gene’s instructions to make that gene’s protein.
Rubric
  • Award 1 point for: a cell using a gene’s instructions to make the protein (making the protein from its gene).
32
Check q10

A stomach cell makes the enzyme pepsin.

Is the stomach cell expressing the pepsin gene?

  1. A. ✓ Yes
  2. B. No
    The stomach cell is making pepsin, so it is using the pepsin gene’s instructions.

Why: The stomach cell is making pepsin.
So the stomach cell is using the pepsin gene’s instructions.
The stomach cell is expressing the pepsin gene.

33
Check q11

A muscle cell carries the pepsin gene and makes no pepsin.

Is the muscle cell expressing the pepsin gene?

  1. A. Yes
    Carrying a gene is not expressing it; the muscle cell is not making pepsin.
  2. B. ✓ No

Why: The muscle cell carries the pepsin gene.
The muscle cell is not making pepsin.
So the muscle cell is not expressing the pepsin gene.

34
Check q12

After a signal arrives, a bone cell makes twice as much of one protein as before.

Has the bone cell changed its gene expression?

  1. A. ✓ Yes
  2. B. No
    The bone cell sped up the making of a protein; speeding up is a change in gene expression.

Why: The bone cell sped up the making of one protein.
A cell changes gene expression when it starts, stops, speeds or slows the making of a protein.
So the bone cell has changed its gene expression.

35The receptor goes to the DNA

36

Video: Watch: The receptor goes to the DNA

Cortisol crosses the membrane and binds its intracellular receptor in the cytosol; the bound pair moves into the nucleus and attaches to the DNA; there it changes the expression of particular genes, and over the following hour the cell’s output of one protein rises four-fold. No relay is needed: the receptor itself reaches the DNA.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L04Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L04Bc.mp4

37
Check q13

Cortisol is a steroid: small and nonpolar.

Which kind of receptor does cortisol bind?

  1. A. A cell-surface receptor
    A small nonpolar molecule slips through the membrane’s oily middle and binds a receptor inside the cell.
  2. B. ✓ An intracellular receptor

Why: Cortisol is small and nonpolar.
So cortisol slips through the membrane into the cell.
Its receptor is an intracellular receptor.

38

Cortisol is a steroid hormone from the adrenal glands. Cortisol raises the glucose in the blood.

39

Before cortisol arrives, a liver cell’s cortisol receptor is found mainly in the cytosol. Ten minutes after cortisol arrives, most of the receptor is in the nucleus.

Two panels of the same liver cell: before cortisol, receptor molecules are scattered through the cytosol and the nucleus holds none; ten minutes after cortisol arrives, the receptor molecules, each with a cortisol molecule attached, sit inside the nucleus on the DNA
Two panels of the same liver cell: before cortisol, receptor molecules are scattered through the cytosol and the nucleus holds none; ten minutes after cortisol arrives, the receptor molecules, each with a cortisol molecule attached, sit inside the nucleus on the DNA
40

Cortisol crossed the membrane. In the cytosol, cortisol bound its intracellular receptor.

41

The bound pair moved into the nucleus. In the nucleus, the bound pair attached to the DNA.

42

Attached to the DNA, the bound pair changes the expression of particular genes. So over the following hours the cell makes more of some proteins and less of others.

43

Here is a bar graph of the cell’s output of one cortisol-controlled protein, before cortisol and one hour after. Within the hour, the output has risen four-fold, from 1.0 unit to 4.0 units.

Bar graph of the output of a cortisol-controlled protein in units: 1.0 before cortisol, 4.0 one hour after, with gridlines every unit
Bar graph of the output of a cortisol-controlled protein in units: 1.0 before cortisol, 4.0 one hour after, with gridlines every unit
44

Making a protein from its gene takes time. So the response to cortisol takes an hour, not seconds.

45

An intracellular receptor needs no relay of other molecules. The receptor itself is the molecule that reaches the DNA.

46

What you are expected to know Describe, for a steroid hormone such as cortisol, the steps from the hormone entering the cell to the cell changing its gene expression.

47

What you are expected to know Explain why a response that changes gene expression takes an hour or more.

48
Check q14

A steroid hormone has entered a cell and bound its intracellular receptor.

Which of the following describes how the bound receptor changes what the cell makes?

  1. A. The bound receptor switches on a relay of proteins that carry the message to the nucleus
    An intracellular receptor needs no relay.
    The receptor itself is the molecule that reaches the DNA.
  2. B. The hormone leaves the receptor and attaches to the DNA on its own
    The hormone acts only through its receptor.
    The bound pair goes to the DNA together.
  3. C. ✓ The bound receptor itself moves into the nucleus and attaches to the DNA
  4. D. The bound receptor moves to the cell surface and lets more hormone in
    The bound pair moves inward, into the nucleus, where the genes are.

Why: An intracellular receptor needs no relay.
With the hormone bound, the receptor itself moves into the nucleus.
In the nucleus the bound receptor attaches to the DNA.
Attached to the DNA, the bound receptor changes the expression of particular genes.

49
Practice writing an answer

A researcher gives liver cells cortisol. Ten minutes later most of the cortisol receptor has moved from the cytosol into the nucleus. After an hour the cell’s output of a cortisol-controlled protein has risen from 1.0 to 4.2 units.

(a) Explain how cortisol binding its receptor led to the rise in the protein’s output. (1 pt)

Model answer Cortisol is a small nonpolar molecule.
So cortisol crosses the plasma membrane into the cytosol.
In the cytosol, cortisol binds its intracellular receptor.
The bound pair moves into the nucleus.
In the nucleus, the bound pair attaches to the DNA.
Attached to the DNA, the bound pair raises the expression of the protein’s gene.
So the cell makes the protein faster.
Making protein takes time, so after an hour the output has risen four-fold.
Rubric
  • Award 1 point for: the bound pair (cortisol and its receptor) moves into the nucleus, attaches to the DNA and raises the expression of the protein’s gene, so the cell makes more of the protein.
50
Check q15

A cell line carries a cortisol receptor that binds cortisol normally but cannot move into the nucleus. A researcher gives the cells cortisol and measures the output of a cortisol-controlled protein after an hour.

Predict the protein output after the hour.

  1. A. More of the protein than before cortisol
    The receptor changes gene expression only by attaching to the DNA in the nucleus.
    This receptor cannot enter the nucleus, so the bound pair never reaches the DNA.
  2. B. ✓ The same amount of the protein as before cortisol
  3. C. Less of the protein than before cortisol
    A bound pair that cannot enter the nucleus never reaches the DNA.
    So nothing changes the gene’s expression, up or down.

Why: Cortisol enters the cells and binds the receptor.
The receptor changes gene expression only by entering the nucleus and attaching to the DNA.
This receptor cannot enter the nucleus.
So the bound pair never reaches the DNA, and the output stays at its level before cortisol.

51
Check q16

A researcher holds every relay protein in muscle cells switched off with a drug, then gives the cells testosterone. Ten minutes later the receptor, with testosterone bound, sits in the nucleus attached to the DNA. Over the following hours the cells make more of a testosterone-controlled protein, as untreated cells do.

What does this result show about how the message reaches the DNA?

  1. A. A relay of proteins carried the message before the drug had time to switch them off
    Every relay protein was already switched off when testosterone arrived, yet the message still reached the DNA and the protein output rose.
    So no relay protein carried the message.
  2. B. ✓ The bound receptor itself carried the message to the DNA
  3. C. Testosterone reached the DNA on its own, and the receptor followed it into the nucleus
    A hormone acts only through its receptor.
    The receptor was found in the nucleus with testosterone bound, so the bound pair is what attached to the DNA.
  4. D. The receptor released a small molecule, and that small molecule carried the message to the DNA
    The receptor itself was found in the nucleus, attached to the DNA.
    With the hormone bound, the receptor is the molecule that reaches the DNA: no messenger, no relay.

Why: Every relay protein was held off, yet the bound receptor reached the DNA and the protein output rose.
So no relay was needed.
Testosterone bound its receptor, and the bound pair moved into the nucleus and attached to the DNA.
The receptor itself carries the message to the DNA.

52

Back to the liver cell and the cortisol that reached it. Before cortisol arrived, the cortisol receptor sat in the cytosol.

53

Cortisol crossed the membrane and bound that receptor. The bound pair moved into the nucleus and attached to the DNA.

54

Attached to the DNA, the bound pair raised the expression of one gene. Making that gene’s protein took time, so within an hour the output had risen from 1.0 unit to 4.0 units.

55Mixed practice: the receptor goes to the DNA mixed practice

56
Check q17

After weeks of training, a muscle cell makes more of one contractile protein than before.

Which of the following changed in the muscle cell?

  1. A. ✓ The expression of the protein’s gene
  2. B. The order of bases in the protein’s gene
    The gene’s bases stayed in the same order; the cell made the protein from it faster.
  3. C. The number of copies of the protein’s gene
    The cell carries the same copy of the gene as before; the cell sped up making the protein from it.

Why: The muscle cell already carried the gene for the contractile protein.
The muscle cell sped up making the protein from that gene.
Speeding up the making of a protein is a change in gene expression.

57
Check q18

Estrogen is a steroid hormone. A researcher gives estrogen to cells of the womb lining.

Predict where the estrogen receptor is found ten minutes later.

  1. A. In the plasma membrane, with its binding site facing outward
    A steroid is small and nonpolar, so it crosses the membrane; its receptor waits inside the cell.
  2. B. In the cytosol, with no estrogen bound
    Estrogen binds its receptor inside the cell.
    Ten minutes later the bound pair is in the nucleus, on the DNA.
  3. C. ✓ In the nucleus, with estrogen bound, attached to the DNA

Why: Estrogen is a steroid, small and nonpolar, so it crosses the membrane and binds its intracellular receptor.
The bound pair sits in the nucleus, attached to the DNA.
Ten minutes later the receptor is there, with estrogen bound.

58
Check q19

Epinephrine makes a heart cell contract harder within seconds. Cortisol changes a liver cell’s output of a protein over an hour or more.

Why does the cortisol response take so much longer?

  1. A. Cortisol has further to travel in the blood than epinephrine
    Both hormones ride the blood.
    The difference is what the cell does after the hormone arrives.
  2. B. Cortisol is a larger molecule and crosses the membrane slowly
    Cortisol is small and nonpolar and crosses the membrane quickly.
    The slow part comes after.
  3. C. ✓ Cortisol changes gene expression, and making new protein takes time
  4. D. The liver cell carries fewer receptors than the heart cell does
    The delay is in the response.
    Building new protein from a gene takes far longer than making a cell contract.

Why: Epinephrine’s response is a quick change in what the cell is already doing.
Cortisol’s response is a change in gene expression.
The cell makes more or less of particular proteins, and building them takes an hour or more.

59
Practice writing an answer

Cortisol is a steroid hormone. Researchers grow normal liver cells and liver cells that lack the cortisol receptor. Each line is grown with or without cortisol for one hour, and the output of a cortisol-controlled protein is measured in units. The graph shows the results so far: normal cells produce 1.0 unit without cortisol and 4.2 units with cortisol; the receptor-lacking cells produce 1.0 unit without cortisol. The receptor-lacking cells with cortisol have yet to be measured.

Bar graph of protein output in units: normal cells without cortisol 1.0, normal cells with cortisol 4.2, receptor-lacking cells without cortisol 1.0, with gridlines every unit
Bar graph of protein output in units: normal cells without cortisol 1.0, normal cells with cortisol 4.2, receptor-lacking cells without cortisol 1.0, with gridlines every unit

(a) Identify the dependent variable in this investigation. (1 pt)

Model answer The dependent variable is the output of the cortisol-controlled protein, measured in units after one hour.
Rubric
  • Award 1 point for: the protein output (in units) as the dependent variable.

Slip Naming cortisol as the dependent variable. Cortisol, present or absent, is what the researchers change; the output is what they measure.

(b) Predict the protein output of the receptor-lacking cells given cortisol, and justify your prediction. (2 pt)

Model answer Cortisol is small and nonpolar, so cortisol crosses the membrane of these cells as it crosses any cell’s membrane.
But a hormone acts only through its receptor.
These cells have no cortisol receptor, so nothing binds the cortisol.
Therefore no bound pair moves into the nucleus and attaches to the DNA.
So the expression of the protein’s gene stays as it was, and the receptor-lacking cells produce about 1.0 unit, the same as without cortisol.
Rubric
  • Award 1 point for: the prediction that output stays at about 1.0 unit (no rise).
  • Award 1 point for: the justification that cortisol acts only through its receptor, so without the receptor no bound pair reaches the DNA and gene expression stays unchanged.

Slip Predicting a rise because cortisol can still enter the cell. Entering is not enough; the hormone changes gene expression only as a pair with its receptor.

(c) Explain how cortisol reaches its receptor in a normal liver cell. (1 pt)

Model answer Cortisol is a steroid, a small nonpolar molecule.
A small nonpolar molecule dissolves into the oily middle of the plasma membrane.
So cortisol passes through the membrane into the cytosol.
Its intracellular receptor waits in the cytosol, so cortisol binds the receptor inside the cell.
Rubric
  • Award 1 point for: cortisol is small and nonpolar, so it crosses the plasma membrane and binds its intracellular receptor in the cytosol.

Slip Placing the cortisol receptor on the cell surface. A steroid crosses the membrane and binds a receptor inside the cell; surface receptors are for ligands that cannot cross.

Glossary

gene
A stretch of a cell’s DNA that carries the instructions for making one protein.
gene expression
A cell using a gene’s instructions to make that protein. A cell changes gene expression when it starts, stops, speeds or slows the making of a protein.

APBIO-U04-L05 A phosphate as a switch

Topic 4.2 · Introduction to Signal Transduction · 91 steps

A muscle cell drawn as a long rounded shape; inside it a relay protein carrying one phosphate group marked P, with an arrow from the protein to two glucose transporters set in the cell membrane at the top; two more transporters wait inside the cell
A muscle cell drawn as a long rounded shape; inside it a relay protein carrying one phosphate group marked P, with an arrow from the protein to two glucose transporters set in the cell membrane at the top; two more transporters wait inside the cell

Inside a muscle cell, one relay protein moves glucose transporters to the cell surface. It does that only while it carries a phosphate group.

Add a chemical that stops the phosphate being added, and the transporters stay inside the cell. Take the phosphate away again, and the protein stops.

What is one phosphate group doing to a whole protein?

Unit 4 · Cell Communication and Cell Cycle

1A phosphate as a switch

2

Video: Watch: A phosphate as a switch

A kinase transfers a phosphate group from ATP onto a relay protein. The added phosphate changes the protein’s shape, and the new shape switches the protein’s activity on.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05a.mp4

3

How does a cell switch a relay protein on and off?

4

One enzyme transfers a phosphate group from ATP onto the relay protein.

5

The added phosphate changes the relay protein’s shape. The new shape switches the relay protein’s activity on, or for some proteins off.

6

A second enzyme removes the phosphate. The relay protein returns to its resting shape.

7

Inside a cell, the relay is often a chain of these switches. Each switched-on protein switches on the next.

8

Read the switch, and you can predict what any blocked enzyme does to the response.

9
Check q1

A protein inside a cell is phosphorylated.

What happens to the protein?

  1. A. ✓ It gains a phosphate group from ATP and changes shape
  2. B. It loses a phosphate group and is broken down
    Phosphorylation adds a phosphate group to the protein.
    Nothing is removed, and the protein is not broken down.

Why: In phosphorylation, ATP’s outer phosphate moves onto the protein.
The protein then changes shape.

10

Now consider the muscle cell’s relay protein. Here is the switch that turns it on.

A kinase transfers a phosphate from ATP onto a resting relay protein; the protein changes shape and is switched on; ADP is left
A kinase transfers a phosphate from ATP onto a resting relay protein; the protein changes shape and is switched on; ADP is left
11

An enzyme transfers a phosphate group from ATP onto the relay protein. ADP is left.

12

The added phosphate changes the relay protein’s shape.

13

The new shape switches the relay protein’s activity on.

14

The enzyme that transfers the phosphate from ATP onto the protein is called a .

15

Here is the kinase’s reaction written out as a word equation.

The kinase’s reaction as a word equation: relay protein plus ATP becomes relay protein–phosphate plus ADP, with kinase written above the arrow
The kinase’s reaction as a word equation: relay protein plus ATP becomes relay protein–phosphate plus ADP, with kinase written above the arrow
16

In the muscle cell, a kinase phosphorylates the relay protein. The phosphorylated relay protein moves glucose transporters to the cell surface.

17

Adding a phosphate does not always switch a protein on. For some relay proteins, removing the phosphate is what switches them on.

18

You read which way a protein’s switch works from what the cell does.

19

What you are expected to know Explain how a kinase switches a relay protein: the kinase transfers a phosphate group from ATP onto the protein, the added phosphate changes the protein’s shape, and the new shape switches the protein’s activity on or, for some proteins, off.

20
Check q2

Relay protein R moves glucose transporters to a muscle cell’s surface only while R carries a phosphate. A chemical blocks the kinase that phosphorylates R. Then the signal arrives.

Predict what happens to the glucose transporters.

  1. A. The transporters move to the surface as usual
    Only the kinase adds the phosphate that switches R on.
    The kinase is blocked, so R never gains a phosphate.
  2. B. ✓ The transporters stay inside the cell
  3. C. The transporters move to the surface faster than usual
    No kinase is adding phosphates, so R never gains one.
    R stays in its resting shape.

Why: R is switched on by a phosphate.
Only the kinase can add that phosphate.
The kinase is blocked.
So R never gains a phosphate, and R stays in its resting shape.
So the transporters stay inside the cell.

21
Practice writing an answer

Relay protein R moves glucose transporters to a muscle cell’s surface only while R carries a phosphate. A chemical blocks the kinase that phosphorylates R. Then the signal arrives, and the transporters stay inside the cell.

(a) Explain how this result demonstrates that a kinase switches R on. (1 pt)

Model answer R moves the transporters only while R carries a phosphate.
A kinase is the enzyme that transfers a phosphate from ATP onto R.
The chemical blocks that kinase.
So no phosphate is transferred onto R.
So R stays in its resting shape and does not move the transporters.
Blocking the kinase alone stopped the response, so the kinase’s phosphate is what switches R on.
Rubric
  • Award 1 point for: the blocked kinase cannot transfer a phosphate from ATP onto R, so R is never switched on and cannot move the transporters; removing only the kinase removed the response.
22
Check q3

A relay protein in a liver cell gains a phosphate from a kinase. A student says: “Adding a phosphate always switches a relay protein on.”

Is the student correct?

  1. A. ✓ No — for some relay proteins, removing the phosphate is what switches them on
  2. B. Yes — a phosphate added by a kinase always switches the protein on
    Adding a phosphate changes a protein’s shape.
    For some relay proteins the phosphate-free shape is the active one.

Why: The phosphate changes the protein’s shape.
For some relay proteins that new shape is active.
For others the phosphate-free shape is active, so removing the phosphate switches them on.
Which way the switch works is read from what the cell does.

23
Check q4

In a plant cell, relay protein S keeps a growth response switched on only while S carries no phosphate. A hormone normally switches on the kinase that phosphorylates S, and the growth stops. A chemical blocks that kinase. Then the hormone arrives.

Predict what happens to the growth response.

  1. A. The growth response stops
    The hormone stops growth only through the kinase.
    The kinase is blocked, so S keeps its phosphate-free, active shape.
  2. B. ✓ The growth response continues
  3. C. The growth response stops, then restarts
    No kinase phosphorylates S at any point.
    S never leaves its active shape.

Why: S is active while S carries no phosphate.
The hormone works by switching on the kinase that phosphorylates S.
The kinase is blocked.
So S never gains a phosphate and stays active.
So the growth response continues.

24Quick quiz: kinase mixed practice

25
Check q5

A kinase acts on a relay protein.

What does the kinase transfer onto the protein?

  1. A. ✓ A phosphate group
  2. B. A molecule of ATP
    The ATP gives up only its outer phosphate group; the rest leaves as ADP.
  3. C. An amino acid
    A kinase moves a phosphate group, not an amino acid.

Why: A kinase transfers a phosphate group from ATP onto a protein.

26
Check q6

A kinase phosphorylates a relay protein.

Where does the phosphate group come from?

  1. A. From the receptor
    The receptor gives up no phosphate.
  2. B. ✓ From ATP
  3. C. From the relay protein itself
    The relay protein gains the phosphate; the phosphate comes from elsewhere.

Why: A kinase takes the phosphate group from ATP and transfers it onto the protein.

27
Check q7

A kinase has just put a phosphate group onto a relay protein.

What happens to the relay protein’s shape?

  1. A. ✓ The shape changes
  2. B. The shape stays the same
    The added phosphate changes the protein’s shape.

Why: The added phosphate changes the protein’s shape.
The new shape is what switches the protein’s activity.

28
Check q8

A kinase takes a phosphate group from a molecule of ATP.

What is left of the ATP?

  1. A. Nothing
    Only one phosphate group leaves the ATP.
  2. B. ✓ ADP
  3. C. A second ATP
    No new ATP is made here.

Why: ATP loses its outer phosphate group.
ADP is left.

29
Check q9

Relay protein T is active only while T carries a phosphate. A kinase phosphorylates T.

Is T now switched on or off?

  1. A. ✓ Switched on
  2. B. Switched off
    T is active while T carries a phosphate, and T now carries one.

Why: T is active while T carries a phosphate.
The kinase put a phosphate onto T.
So T is switched on.

30
Check q10

Relay protein U is active only while U carries no phosphate. A kinase phosphorylates U.

Is U now switched on or off?

  1. A. Switched on
    U is active only without a phosphate, and U now carries one.
  2. B. ✓ Switched off

Why: U is active only while U carries no phosphate.
The kinase put a phosphate onto U.
So U is switched off.

31
Practice writing an answer

A relay protein inside a cell is switched on by a kinase.

(a) State what a kinase does. (1 pt)

Model answer A kinase transfers a phosphate group from ATP onto a protein, and the added phosphate changes the protein’s shape.
Rubric
  • Award 1 point for: a kinase transfers (adds) a phosphate group from ATP onto a protein.
  • Accept: a kinase phosphorylates a protein using ATP.

32Taking the phosphate off again

33

Video: Watch: Taking the phosphate off again

A phosphatase removes the phosphate a kinase added. The relay protein returns to its resting shape, and the switch ends.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05b.mp4

34

Now consider the muscle cell after the signal has passed. The relay protein still carries its phosphate.

Two panels: on the left a kinase transfers a phosphate from ATP onto a resting relay protein, which changes shape and is switched on, leaving ADP; on the right a phosphatase removes the phosphate and the protein returns to its resting shape
Two panels: on the left a kinase transfers a phosphate from ATP onto a resting relay protein, which changes shape and is switched on, leaving ADP; on the right a phosphatase removes the phosphate and the protein returns to its resting shape
35

So the relay protein keeps moving glucose transporters to the surface.

36

A second enzyme removes the phosphate from the relay protein.

37

The relay protein returns to its resting shape. The relay protein stops.

38

The enzyme that removes the phosphate again is called a .

39

Here is the phosphatase’s reaction written out as a word equation.

The phosphatase’s reaction as a word equation: relay protein–phosphate plus water becomes relay protein plus phosphate, with phosphatase written above the arrow
The phosphatase’s reaction as a word equation: relay protein–phosphate plus water becomes relay protein plus phosphate, with phosphatase written above the arrow
40

Together the two enzymes make a switch.

41

The kinase puts the phosphate on. The phosphatase takes the phosphate off.

42

For a protein that is active without its phosphate, the phosphatase is the enzyme that switches it on.

43

What you are expected to know Explain what a phosphatase does: it removes the phosphate a kinase added, so the relay protein returns to its resting shape and the switch ends.

44
Check q11

Relay protein Q in a liver cell carries a phosphate and is switched on.

Which enzyme returns Q to its resting shape?

  1. A. ✓ A phosphatase
  2. B. A kinase
    A kinase adds a phosphate; Q already carries one.

Why: Q carries a phosphate.
A phosphatase removes a phosphate from a protein.
So the phosphatase returns Q to its resting shape.

45
Check q12

In a relay, protein Q is switched off while it carries a phosphate and switched on when the phosphate is removed.

Which enzyme switches Q on?

  1. A. A kinase
    A kinase adds a phosphate from ATP, and Q is switched off while it carries a phosphate.
  2. B. ✓ A phosphatase

Why: A phosphatase removes a phosphate from a protein.
Q is active without its phosphate.
So the phosphatase is the enzyme that switches Q on.

46
Check q13

In a yeast cell, only the form of protein R without a phosphate switches on the mating response. A researcher adds a phosphatase inhibitor. Before the inhibitor, 20% of R carries a phosphate; ten minutes after, 80% of R carries a phosphate.

Predict what the inhibitor does to the mating response.

  1. A. The response rises
    Phosphate-free R is the active form here.
    The inhibitor blocks the phosphatase, so more R keeps its phosphate.
  2. B. The response stays the same
    The total amount of R is unchanged.
    But the share of R in the active, phosphate-free form fell from 80% to 20%.
  3. C. ✓ The response falls

Why: Only phosphate-free R is active here.
The phosphatase removes phosphates from R.
The inhibitor blocks the phosphatase, so the phosphates stay on.
The phosphate-free share of R falls from 80% to 20%.
So the mating response falls.

47Quick quiz: phosphatase mixed practice

48
Check q14

Which enzyme removes a phosphate group from a protein?

  1. A. A kinase
    A kinase adds a phosphate group.
  2. B. ✓ A phosphatase

Why: A phosphatase removes a phosphate group from a protein.

49
Check q15

Which enzyme transfers a phosphate group from ATP onto a protein?

  1. A. ✓ A kinase
  2. B. A phosphatase
    A phosphatase removes a phosphate group.

Why: A kinase transfers a phosphate group from ATP onto a protein.

50
Check q16

A relay protein has just lost its phosphate and returned to its resting shape.

Which enzyme acted on it?

  1. A. A kinase
    A kinase would have added a phosphate, not removed one.
  2. B. ✓ A phosphatase

Why: The protein lost a phosphate.
A phosphatase is the enzyme that removes a phosphate.
So a phosphatase acted on it.

51
Check q17

A relay protein has just gained a phosphate and changed shape.

Which enzyme acted on it?

  1. A. ✓ A kinase
  2. B. A phosphatase
    A phosphatase would have removed a phosphate, not added one.

Why: The protein gained a phosphate.
A kinase is the enzyme that adds a phosphate from ATP.
So a kinase acted on it.

52
Check q18

Which of the two enzymes uses ATP?

  1. A. ✓ A kinase
  2. B. A phosphatase
    A phosphatase uses water, not ATP; the phosphate leaves as inorganic phosphate.

Why: A kinase takes the phosphate it transfers from ATP.
A phosphatase needs no ATP.

53
Check q19

Relay protein V is active only while V carries a phosphate.

Which enzyme switches V off?

  1. A. A kinase
    A kinase adds the phosphate that keeps V active.
  2. B. ✓ A phosphatase

Why: V is active while V carries a phosphate.
A phosphatase removes the phosphate.
So the phosphatase switches V off.

54
Check q20

Relay protein W is active only while W carries no phosphate.

Which enzyme switches W off?

  1. A. ✓ A kinase
  2. B. A phosphatase
    A phosphatase removes phosphates, and W is active without one.

Why: W is active while W carries no phosphate.
A kinase adds a phosphate to W.
So the kinase switches W off.

55
Practice writing an answer

A relay protein inside a cell carries a phosphate.

(a) State what a phosphatase does. (1 pt)

Model answer A phosphatase removes a phosphate group from a protein, so the protein returns to its resting shape.
Rubric
  • Award 1 point for: a phosphatase removes (takes off) a phosphate group from a protein.

56Each kinase switches on the next

57

Video: Watch: Each kinase switches on the next

In a yeast cell, the activated receptor switches on kinase 1, kinase 1 phosphorylates kinase 2, and kinase 2 phosphorylates kinase 3 in the nucleus. Every kinase takes a fresh phosphate from a fresh ATP.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05c.mp4

58

Now consider a yeast cell. A mating signal binds its receptor.

A receptor in the membrane at the top left, the fluid outside the cell shaded above the membrane and the cytosol shaded below it, then three kinases in a chain leading from the membrane down toward the nucleus: kinase 1 at the membrane, kinase 2 in the cytosol, kinase 3 in the nucleus; at each kinase an ATP gives a phosphate and leaves as ADP; times 5 s, 20 s and 45 s are marked
A receptor in the membrane at the top left, the fluid outside the cell shaded above the membrane and the cytosol shaded below it, then three kinases in a chain leading from the membrane down toward the nucleus: kinase 1 at the membrane, kinase 2 in the cytosol, kinase 3 in the nucleus; at each kinase an ATP gives a phosphate and leaves as ADP; times 5 s, 20 s and 45 s are marked
59

Kinase 1, at the membrane, is switched on 5 seconds later. Kinase 2, in the cytosol, is switched on at 20 seconds.

60

Kinase 3, in the nucleus, is switched on at 45 seconds.

61

The activated receptor switches on kinase 1. Kinase 1 phosphorylates kinase 2 and switches it on. Kinase 2 phosphorylates kinase 3.

62

The last phosphorylated protein produces the response.

63

A chain of kinases in which each one phosphorylates the next and switches it on is called a .

64

Each step uses a fresh phosphate from a fresh ATP. What travels down the chain is activation, not a phosphate.

65

The same phosphate is never handed along the chain. Every kinase takes a new one from ATP.

66

What you are expected to know Describe a phosphorylation cascade: the activated receptor switches on the first kinase, each kinase phosphorylates the next and switches it on, the last phosphorylated protein produces the response, and each step uses a fresh phosphate from a fresh ATP.

67
Check q21

In the yeast cell, preventing the activation of kinase 1 stops kinase 2 and kinase 3 from being activated. Preventing the activation of kinase 2 stops kinase 3 and leaves kinase 1 active.

How does the mating signal reach the nucleus?

  1. A. ✓ Each activated kinase phosphorylates the next, from the receptor toward the nucleus
  2. B. The mating signal crosses the membrane and travels to the nucleus itself
    The mating signal binds a receptor at the surface.
    The kinases inside pass the message on, one to the next.
  3. C. The activated receptor leaves the membrane and phosphorylates all three kinases in turn
    The receptor stays put.
    Blocking kinase 2 stops only kinase 3, so kinase 2 is what switches kinase 3 on.
  4. D. Each kinase binds the mating signal at its own place in the cell
    Blocking kinase 1 stops the others, so they depend on kinase 1, not on the signal.

Why: Blocking kinase 1 stops both later kinases.
Blocking kinase 2 stops only kinase 3.
So each kinase activates the next in order.
The signal reaches the nucleus as a chain of phosphorylations, a phosphorylation cascade.

68
Practice writing an answer

In the yeast cell, preventing the activation of kinase 2 stops kinase 3 from being activated but leaves kinase 1 active.

(a) Explain how this result demonstrates that the kinases form a phosphorylation cascade. (1 pt)

Model answer The activated receptor switches on kinase 1.
So kinase 1 does not depend on kinase 2, and kinase 1 is activated as usual.
Kinase 2 is the kinase that phosphorylates kinase 3.
Kinase 2 cannot be activated.
So kinase 2 never phosphorylates kinase 3, and kinase 3 stays switched off.
A break stops everything after it and nothing before it, so each kinase switches on only the next one in the chain.
Rubric
  • Award 1 point for: the receptor activates kinase 1 directly, but only kinase 2 phosphorylates kinase 3, so kinase 1 is unaffected and kinase 3 is never switched on — each kinase activates only the next.
69
Check q22

In the yeast cell, kinase 3 in the nucleus has just gained a phosphate group. A student says: “Kinase 2 handed its own phosphate on to kinase 3.”

Is the student correct?

  1. A. Yes — the phosphate travels down the chain from kinase to kinase
    A kinase transfers a phosphate from ATP, not from itself.
    Kinase 2 keeps its own phosphate.
  2. B. ✓ No — kinase 2 transferred a fresh phosphate from an ATP molecule onto kinase 3

Why: A kinase transfers a phosphate from ATP, not from itself.
Kinase 2 keeps its own phosphate.
Kinase 2 takes a fresh phosphate from a fresh ATP and transfers it onto kinase 3.
What travels down the chain is activation.

70

Back to the muscle cell, its relay protein and the one phosphate group on it. A kinase transferred that phosphate from ATP onto the relay protein.

71

The phosphate changed the relay protein’s shape. The new shape moved glucose transporters to the cell surface.

72

The chemical that stopped the phosphate being added was a kinase blocker. With no phosphate, the relay protein kept its resting shape, so the transporters stayed inside.

73

A phosphatase took the phosphate off again. The relay protein returned to its resting shape and stopped.

74

One phosphate group changes a whole protein’s shape, and the shape is the switch.

75Quick quiz: phosphorylation cascade mixed practice

76
Check q23

A phosphorylation cascade has kinases 1, 2 and 3 in order.

What switches on kinase 2?

  1. A. ✓ Kinase 1
  2. B. The receptor
    The receptor switches on only the first kinase.
  3. C. Kinase 3
    Kinase 3 comes after kinase 2 in the chain.

Why: In a phosphorylation cascade, each kinase phosphorylates the next.
Kinase 1 phosphorylates kinase 2 and switches it on.

77
Check q24

In a phosphorylation cascade, kinase 2 phosphorylates kinase 3.

Where does the phosphate that goes onto kinase 3 come from?

  1. A. From kinase 2’s own phosphate
    Kinase 2 keeps its own phosphate.
  2. B. From kinase 1
    Kinase 1 is two steps back and passes on no phosphate.
  3. C. ✓ From a fresh ATP

Why: Every kinase takes a fresh phosphate from a fresh ATP.
Kinase 2 transfers that phosphate onto kinase 3.

78
Check q25

What switches on the first kinase of a phosphorylation cascade?

  1. A. The last kinase in the chain
    The last kinase produces the response; it comes at the end.
  2. B. ✓ The activated receptor
  3. C. The ligand, after it enters the cell
    The ligand stays outside; the receptor carries the message across.

Why: The ligand activates the receptor.
The activated receptor switches on the first kinase.

79
Check q26

What travels down a phosphorylation cascade from kinase to kinase?

  1. A. ✓ Activation: each kinase switches on the next
  2. B. One phosphate, handed from kinase to kinase
    Each kinase keeps its phosphate and takes a fresh one from ATP for the next.

Why: Each kinase phosphorylates the next with a fresh phosphate from ATP.
So what travels down the chain is activation, not a phosphate.

80
Check q27

A phosphorylation cascade has kinases 1, 2 and 3 in order. A chemical blocks kinase 3. The signal binds its receptor.

Which kinases are switched on?

  1. A. Kinase 1 only
    Kinase 1 switches on kinase 2 as usual; only kinase 3 is blocked.
  2. B. ✓ Kinases 1 and 2
  3. C. None of the three
    The receptor still switches on kinase 1, and kinase 1 still switches on kinase 2.

Why: The receptor switches on kinase 1.
Kinase 1 phosphorylates kinase 2 and switches it on.
Kinase 3 is blocked.
So kinases 1 and 2 are switched on.

81
Check q28

Kinase 2 has just phosphorylated kinase 3.

Does kinase 2 still carry its own phosphate?

  1. A. ✓ Yes
  2. B. No
    Kinase 2 took the phosphate for kinase 3 from ATP, not from itself.

Why: A kinase transfers a phosphate from ATP, not from itself.
So kinase 2 keeps its own phosphate.

82
Practice writing an answer

A yeast cell carries three kinases in a chain.

(a) State what a phosphorylation cascade is. (1 pt)

Model answer A phosphorylation cascade is a chain of kinases in which each kinase phosphorylates the next and switches it on.
Rubric
  • Award 1 point for: a chain of kinases, each phosphorylating (switching on) the next.

83Mixed practice mixed practice

84
Check q29

A relay protein in a plant root cell is switched on while it carries a phosphate.

Which enzyme switches it off?

  1. A. A kinase
    A kinase adds the phosphate that keeps the protein switched on.
  2. B. ✓ A phosphatase

Why: The protein is switched on while it carries a phosphate.
A phosphatase removes the phosphate.
So the phosphatase switches the protein off.

85
Check q30

A cell carries a cascade of three kinases, in which kinase 1 activates kinase 2 and kinase 2 activates kinase 3. A chemical blocks kinase 2. Then the signal binds its receptor.

Predict which kinases are active.

  1. A. ✓ Kinase 1 only
  2. B. Kinase 3 only
    The receptor activates kinase 1 as usual; the chain breaks at kinase 2, so kinase 3 stays off.
  3. C. Kinases 1 and 3
    Only kinase 2 phosphorylates kinase 3, and kinase 2 is blocked.
  4. D. All three kinases
    The receptor activates only kinase 1; each later kinase is switched on by the one before it.

Why: The receptor activates kinase 1 as usual.
Kinase 1 cannot switch on the blocked kinase 2.
Only kinase 2 can switch on kinase 3.
So kinase 1 alone is active.

86
Check q31

A kinase in a heart cell acts on a relay protein.

Which molecule supplies the phosphate group?

  1. A. ADP
    ADP is what is left after the phosphate has been taken.
  2. B. ✓ ATP
  3. C. Water
    Water is used by a phosphatase, not by a kinase.

Why: A kinase transfers a phosphate group from ATP onto the protein.
ADP is left.

87
Check q32

In a frog egg cell, relay protein M is active only while M carries a phosphate. A researcher adds a chemical that blocks every phosphatase in the cell, then the hormone that switches on M’s kinase. The hormone is washed away five minutes later.

Predict what M does after the hormone is gone.

  1. A. ✓ M stays active
  2. B. M switches off as usual
    Only a phosphatase can remove M’s phosphate, and every phosphatase is blocked.
  3. C. M switches off faster than usual
    No phosphatase is working, so nothing removes M’s phosphate.

Why: M is active while M carries a phosphate.
A phosphatase is the enzyme that removes the phosphate.
Every phosphatase is blocked.
So M keeps its phosphate and stays active.

88
Check q33

A phosphorylation cascade in a yeast cell ends at kinase 3 in the nucleus, which phosphorylates one last protein.

What does that last phosphorylated protein do?

  1. A. It switches kinase 1 back on
    Each kinase switches on the next, toward the nucleus; nothing feeds back to kinase 1.
  2. B. ✓ It produces the response
  3. C. It removes the phosphates from the kinases
    Phosphatases remove phosphates; the last protein is the one that acts.

Why: In a phosphorylation cascade, each kinase switches on the next.
The last phosphorylated protein produces the response.

89
Check q34

A student says: “A relay protein switched on by a phosphate stays switched on until the cell breaks the protein down.”

Is the student correct?

  1. A. ✓ No — a phosphatase removes the phosphate again
  2. B. Yes — the switch ends only when the cell breaks the protein down
    The protein is not broken down; a phosphatase removes the phosphate and the same protein returns to rest.

Why: A phosphatase removes the phosphate a kinase added.
The protein returns to its resting shape.
So the switch ends without the protein being broken down.

90
Practice writing an answer

In a yeast cell, relay protein N switches on the mating response only while N carries a phosphate. A researcher blocks every kinase in the cell and adds the mating signal. The mating response does not appear. In a second cell the researcher blocks every phosphatase instead, adds the mating signal, then washes it away. The mating response continues for an hour.

(a) Explain how the two results together demonstrate that a kinase and a phosphatase form a switch for N. (2 pt)

Model answer N is active only while N carries a phosphate.
A kinase transfers a phosphate from ATP onto N.
With every kinase blocked, N never gains a phosphate.
So N is never switched on, and no response appears.
A phosphatase removes the phosphate from N.
With every phosphatase blocked, N keeps its phosphate after the signal is gone.
So N stays switched on, and the response continues.
The kinase switches N on and the phosphatase switches N off.
Rubric
  • Award 1 point for: with the kinases blocked, no phosphate is transferred onto N, so N is never switched on and no response appears.
  • Award 1 point for: with the phosphatases blocked, N’s phosphate is never removed, so N stays switched on after the signal is gone — the kinase turns the switch on and the phosphatase turns it off.

Slip Saying the phosphatase inhibitor stopped the signal being removed. The signal was washed away; what kept the response going was the phosphate that no phosphatase could remove from N.

Glossary

kinase
An enzyme that transfers a phosphate group from ATP onto a protein. The added phosphate changes the protein’s shape and switches its activity on or off.
phosphatase
An enzyme that removes a phosphate group from a protein, returning the protein to the shape it had before a kinase phosphorylated it.
phosphorylation cascade
A chain of kinases in which each one phosphorylates the next and switches it on, from the activated receptor to the protein that produces the response. Each step uses a fresh phosphate from a fresh ATP.

APBIO-U04-L05B A small molecule spreads the message

Topic 4.2 · Introduction to Signal Transduction · 107 steps

A fan of five tiers widening downward: 40 bound receptors, 4,000 cAMP molecules, 4,000 kinases switched on, 400,000 enzyme molecules switched on, 40,000,000 glucose molecules released
A fan of five tiers widening downward: 40 bound receptors, 4,000 cAMP molecules, 4,000 kinases switched on, 400,000 enzyme molecules switched on, 40,000,000 glucose molecules released

Here is a liver cell one minute after epinephrine reached it: 40 of its receptors have a molecule of epinephrine bound.

Inside the cell, one small relay molecule has gone from about 200 copies to about 4,200. Glucose is leaving the cell six times faster than before. The counts here are scaled down so you can follow them.

How did 40 bound receptors become 4,000 new molecules inside?

Unit 4 · Cell Communication and Cell Cycle

1A small molecule spreads the message

2

Video: Watch: A small molecule spreads the message

The activated receptor switches on an enzyme in the membrane. The enzyme makes a small molecule inside the cell, and that molecule spreads through the cytosol and switches on kinases: a second messenger.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05Ba.mp4

3

How does a message from a few receptors reach the whole cytosol? And how does it grow on the way?

4

The activated receptor switches on an enzyme in the membrane. That enzyme makes a small molecule from ATP.

5

The small molecule spreads fast through the cytosol and switches on kinases.

6

One enzyme makes many of the small molecules before it is switched off. Each kinase acts on many proteins.

7

So every enzyme step multiplies the signal. A few bound receptors end in a very large response.

8
Check q1

An enzyme makes a small molecule at one spot in a cell’s cytosol. Nothing carries the molecule.

What happens to the molecule over the next few seconds?

  1. A. ✓ It spreads through the cytosol by diffusion
  2. B. It stays where it was made
    Molecules in a liquid never stay still; they move and spread out.

Why: Molecules in a liquid move at random.
So the small molecule spreads out from where it was made: it diffuses through the cytosol.

9

Now consider the liver cell before epinephrine arrives. About 200 copies of one small molecule sit in the cytosol.

A liver cell with a receptor and an enzyme set in its membrane, the fluid outside shaded above the membrane and the cytosol shaded below it; before epinephrine, a few copies of a small molecule are scattered in the cytosol
A liver cell with a receptor and an enzyme set in its membrane, the fluid outside shaded above the membrane and the cytosol shaded below it; before epinephrine, a few copies of a small molecule are scattered in the cytosol
10

One minute after epinephrine binds, about 4,200 copies of the small molecule are in the cytosol.

The same liver cell one minute after epinephrine: epinephrine sits in the receptor's pocket, the enzyme beside it is switched on, and many copies of the small molecule are scattered through the cytosol
The same liver cell one minute after epinephrine: epinephrine sits in the receptor's pocket, the enzyme beside it is switched on, and many copies of the small molecule are scattered through the cytosol
11

The activated receptor switched on an enzyme in the membrane. That enzyme made the small molecule.

12

The small molecule spreads quickly through the cytosol. It switches on every kinase it reaches.

13

When a receptor is activated, the cell makes or releases a small molecule inside itself.

14

It spreads through the cytosol and switches on target proteins.

15

A small molecule that carries the message inside the cell like this is called a .

16

The ligand outside the cell is the first messenger.

17

A second messenger is a small molecule, not a protein. A small molecule diffuses faster than a protein, so the second messenger spreads through the cytosol in seconds.

18

Calcium ions (Ca²⁺) are another second messenger. A signal opens a store inside the cell, and calcium ions flood out of the store into the cytosol.

19

What you are expected to know Describe a second messenger: a small molecule made or released inside the cell when a receptor is activated, which spreads through the cytosol and switches on target proteins such as kinases.

20
Check q2

Epinephrine binds a receptor on the surface of a liver cell. Within a minute, kinases deep in the cytosol are switched on. The epinephrine stays outside the cell.

Which molecule carries the message from the membrane to the kinases?

  1. A. The epinephrine itself
    Epinephrine stays outside the cell, so it cannot reach kinases deep in the cytosol.
  2. B. The receptor protein
    The receptor stays in the membrane.
    The receptor switches on an enzyme, and the enzyme makes a second messenger.
  3. C. ✓ A second messenger

Why: The activated receptor switches on an enzyme in the membrane.
That enzyme makes a second messenger inside the cell.
The second messenger spreads through the cytosol and switches on the kinases.

21
Check q3

In a muscle cell, a nerve’s signal molecule binds a receptor. Within a fraction of a second, calcium ions flood out of a store into the cytosol and switch on the proteins that make the muscle contract.

Which of these is the second messenger?

  1. A. The nerve’s signal molecule
    The signal molecule is the first messenger; it stays outside the cell.
  2. B. The receptor
    The receptor is a protein in the membrane; it does not spread through the cytosol.
  3. C. ✓ The calcium ions

Why: The calcium ions are released inside the cell when the receptor is activated.
They spread through the cytosol and switch on target proteins.
So the calcium ions are the second messenger.

22Quick quiz: second messenger mixed practice

23
Check q4

Where is a second messenger made or released?

  1. A. ✓ Inside the cell
  2. B. Outside the cell
    The ligand outside is the first messenger; the second messenger appears inside the cell.

Why: A second messenger is made or released inside the cell when a receptor is activated.

24
Check q5

Which is the first messenger?

  1. A. ✓ The ligand outside the cell
  2. B. The small molecule inside the cell
    The small molecule inside carries the message onward; it is the second messenger.

Why: The ligand outside the cell is the first messenger.
The small molecule made inside is the second.

25
Check q6

Is a second messenger a protein or a small molecule?

  1. A. A protein
    A protein would spread far more slowly through the cytosol.
  2. B. ✓ A small molecule

Why: A second messenger is a small molecule.
A small molecule diffuses quickly through the cytosol.

26
Check q7

What does a second messenger do when it reaches a kinase?

  1. A. ✓ It switches the kinase on
  2. B. It breaks the kinase down
    Nothing is broken down; the kinase is switched on.
  3. C. It carries the kinase to the nucleus
    The kinase stays where it is; it is switched on.

Why: A second messenger switches on the target proteins it reaches, such as kinases.

27
Check q8

Epinephrine binds its receptor and stays outside the liver cell.

Is epinephrine a second messenger?

  1. A. Yes
    A second messenger is made or released inside the cell; epinephrine stays outside.
  2. B. ✓ No

Why: Epinephrine stays outside the cell.
A second messenger appears inside the cell.
So epinephrine is the first messenger, not the second.

28
Check q9

After a receptor is activated, calcium ions flood from a store inside the cell into the cytosol.

Are the calcium ions a second messenger?

  1. A. ✓ Yes
  2. B. No
    The calcium ions are released inside the cell when the receptor is activated, and they spread through the cytosol.

Why: The calcium ions are released inside the cell when the receptor is activated.
They spread through the cytosol and switch on target proteins.
So they are a second messenger.

29
Practice writing an answer

A hormone binds a receptor on a liver cell, and a small molecule inside the cell rises.

(a) State what a second messenger is. (1 pt)

Model answer A second messenger is a small molecule made or released inside the cell when a receptor is activated; it spreads through the cytosol and switches on target proteins.
Rubric
  • Award 1 point for: a small molecule made or released inside the cell when a receptor is activated, which spreads through the cytosol and switches on target proteins (any two of: made inside, spreads, switches on targets).

30How the cell makes cyclic AMP

31

Video: Watch: How the cell makes cyclic AMP

The activated receptor switches on an enzyme in the membrane. The enzyme makes cyclic AMP from ATP: ATP → cyclic AMP + two linked phosphates.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05Bb.mp4

32
Check q10

ATP is a molecule the cell has in plenty.

How many phosphate groups does one ATP molecule carry?

  1. A. One
    ATP carries a chain of three phosphate groups.
  2. B. Two
    ADP carries two; ATP carries three.
  3. C. ✓ Three

Why: ATP is adenosine with a chain of three phosphate groups.
The T stands for tri, three.

33

Now consider the liver cell one minute after epinephrine binds. Epinephrine sits in the receptor’s pocket, and the receptor has changed shape through to its intracellular domain.

The same liver cell one minute after epinephrine: epinephrine sits in the receptor's pocket, the enzyme beside it is switched on, an arrow leads from ATP to cAMP, and many cAMP molecules are scattered through the cytosol
The same liver cell one minute after epinephrine: epinephrine sits in the receptor's pocket, the enzyme beside it is switched on, an arrow leads from ATP to cAMP, and many cAMP molecules are scattered through the cytosol
34

The changed receptor switches on an enzyme beside it in the membrane.

35

The switched-on enzyme makes the small molecule from ATP.

36

The enzyme cuts two of ATP’s three phosphate groups off, still joined to each other. The enzyme bends the one phosphate left into a ring.

37

So the new molecule keeps one of ATP’s three phosphates. The other two leave together, not as ADP.

38

The ring-shaped molecule the enzyme makes from ATP is called . ‘Cyclic’ means ring-shaped.

39

Here is the enzyme’s reaction written out as a word equation.

ATP becomes cyclic AMP plus two linked phosphates
40

cAMP is the second messenger in the liver cell. cAMP switches on the kinase that starts the breakdown of glycogen, the liver’s store of glucose.

41

The enzyme makes cAMP only while the receptor is activated.

42

What you are expected to know Describe how the cell makes cAMP: the activated receptor switches on an enzyme in the membrane, and that enzyme makes cyclic AMP from ATP.

43
Check q11

In a liver cell, the level of cAMP rises twenty-fold within a minute of epinephrine binding.

Which molecule is the cAMP made from?

  1. A. Epinephrine
    Epinephrine stays outside the cell; nothing inside is made from it.
  2. B. ✓ ATP
  3. C. Glycogen
    Glycogen is the glucose store that the pathway breaks down; cAMP is made from ATP.

Why: The switched-on enzyme in the membrane makes cAMP from ATP.
Two linked phosphates come off, and the one phosphate left is bent into a ring.

44
Practice writing an answer

Epinephrine binds a receptor on the surface of a liver cell. Soon afterwards the cAMP inside the cell has risen twenty-fold, and kinases deep in the cytosol are switched on.

(a) Explain how epinephrine binding at the surface made the cAMP inside the cell rise. (1 pt)

Model answer Epinephrine binds its receptor.
The receptor changes shape through to its intracellular domain.
The changed receptor switches on an enzyme set in the membrane.
The switched-on enzyme makes cAMP from ATP.
One enzyme molecule acts again and again, so one enzyme makes many cAMP molecules.
So the cAMP inside the cell rises twenty-fold.
Rubric
  • Award 1 point for: the activated receptor switches on a membrane enzyme that makes cAMP from ATP inside the cell, and the enzyme acts many times.

45Quick quiz: cyclic AMP mixed practice

46
Check q12

What is cAMP made from?

  1. A. Glucose
    Glucose is what the pathway releases at the end.
  2. B. ADP
    ADP has only two phosphates; the enzyme starts from ATP.
  3. C. ✓ ATP

Why: The enzyme makes cAMP from ATP.

47
Check q13

Where is the enzyme that makes cAMP?

  1. A. ✓ In the cell membrane
  2. B. In the nucleus
    The enzyme sits beside the receptor, in the membrane.
  3. C. Outside the cell
    The enzyme is part of the cell, set in its membrane.

Why: The enzyme that makes cAMP is set in the cell membrane, beside the receptor.

48
Check q14

What switches on the enzyme that makes cAMP?

  1. A. cAMP itself
    cAMP is the enzyme’s product; it comes after the enzyme is switched on.
  2. B. ✓ The activated receptor
  3. C. The kinase
    The kinase is switched on by cAMP, later in the pathway.

Why: The ligand activates the receptor.
The activated receptor switches on the enzyme that makes cAMP.

49
Check q15

In the liver cell, what does cAMP do?

  1. A. ✓ It switches on a kinase
  2. B. It binds the receptor from outside
    cAMP is made inside the cell and stays there.
  3. C. It carries glucose out of the cell
    cAMP carries a message, not glucose.

Why: cAMP is the second messenger.
It spreads through the cytosol and switches on the kinase that starts glycogen breakdown.

50
Check q16

What is left of the ATP after cAMP is made from it?

  1. A. ADP
    ADP is left when a kinase takes one phosphate; here two phosphates come off together.
  2. B. ✓ Two linked phosphates
  3. C. Nothing
    The two phosphates that were cut off are left.

Why: ATP → cyclic AMP + two linked phosphates.
The two phosphates cut off are left, still joined to each other.

51
Check q17

What does ‘cyclic’ in cyclic AMP mean?

  1. A. ✓ Ring-shaped
  2. B. Made every minute
    Cyclic here describes the molecule’s shape, not a timing.

Why: The one phosphate left on the molecule is bent into a ring.
‘Cyclic’ means ring-shaped.

52
Practice writing an answer

Epinephrine binds a receptor on a liver cell.

(a) State what cyclic AMP is. (1 pt)

Model answer Cyclic AMP is a ring-shaped small molecule that an enzyme in the membrane makes from ATP; it is the cell’s second messenger.
Rubric
  • Award 1 point for: a (ring-shaped) small molecule made from ATP by an enzyme in the membrane, which acts as a second messenger.

53Why a few receptors give a large response

54

Video: Watch: Why a few receptors give a large response

An activated enzyme acts on many molecules before it is switched off. So each enzyme step multiplies the number of activated molecules, and a few bound receptors end in a very large response: amplification.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05Bc.mp4

55
Check q18

One enzyme molecule is surrounded by a supply of its substrate.

How many substrate molecules can that one enzyme molecule act on?

  1. A. Exactly one, and then it is used up
    An enzyme is not used up by the reaction it speeds up.
    The moment the product leaves, the enzyme is free to act again.
  2. B. ✓ Many, one after another

Why: An enzyme is not used up by the reaction it speeds up.
So one enzyme molecule acts on substrate molecule after substrate molecule.

56

Now consider the liver cell’s counts again. Forty bound receptors switched on forty enzyme molecules.

Two tiers: 40 bound receptors, then 4,000 cAMP molecules
Two tiers: 40 bound receptors, then 4,000 cAMP molecules
57

Those forty enzymes made about 4,000 new molecules of cAMP.

58

Each enzyme molecule made many molecules of cAMP before it was switched off.

59

The cAMP switched on kinases. Each kinase phosphorylated many enzyme molecules before it was switched off.

Five tiers widening downward: 40 bound receptors, 4,000 cAMP molecules, 4,000 kinases switched on, 400,000 enzyme molecules switched on, 40,000,000 glucose molecules released
Five tiers widening downward: 40 bound receptors, 4,000 cAMP molecules, 4,000 kinases switched on, 400,000 enzyme molecules switched on, 40,000,000 glucose molecules released
60

An activated enzyme acts on many molecules before it is switched off. So each enzyme step multiplies the number of activated molecules.

61

A few bound receptors end in a very large response.

62

This multiplying of the signal at each enzyme step is called .

63

A phosphorylation cascade relays the message. Amplification is the multiplying.

64

Imagine a relay of one-to-one steps, in which each protein activates exactly one molecule of the next. That relay would carry the message without amplifying it.

65

What you are expected to know Explain why a few bound receptors give a very large response: an activated enzyme acts on many molecules before it is switched off, so each enzyme step multiplies the number of activated molecules.

66
Check q19

One pathway relays a signal through four proteins, each of which activates exactly one molecule of the next. A second pathway relays a signal through four enzyme steps.

Which pathway amplifies the signal?

  1. A. ✓ The pathway of four enzyme steps
  2. B. The pathway of one-to-one steps
    In the one-to-one pathway each protein activates exactly one molecule of the next.
    So the number of activated molecules never grows.
  3. C. Both pathways equally
    An enzyme acts on many molecules before it is switched off, so an enzyme step multiplies the signal.
    A one-to-one hand-off does not.
  4. D. Neither pathway
    Each enzyme in the enzyme pathway acts on many molecules before it is switched off.
    So each enzyme step multiplies the number of activated molecules.

Why: An activated enzyme acts on many molecules before it is switched off.
So each enzyme step in the enzyme pathway multiplies the number of activated molecules.
The one-to-one pathway relays the message without amplifying it.

67
Practice writing an answer

In a liver cell responding to a hormone, 2 bound receptors switch on 40 enzyme molecules, and those enzymes release 1,200 glucose molecules.

(a) Explain how this case demonstrates amplification. (1 pt)

Model answer An activated enzyme acts on many molecules before it is switched off.
So each enzyme step multiplies the number of activated molecules.
Here 2 bound receptors switched on 40 enzyme molecules: the receptor-to-enzyme step multiplied the signal.
Those 40 enzymes released 1,200 glucose molecules: the enzyme-to-glucose step multiplied it again.
So 2 bound receptors ended in 1,200 glucose molecules, a very large response from a few receptors.
Rubric
  • Award 1 point for: each enzyme acts on many molecules before it is switched off, so each step multiplies the count (2 → 40 → 1,200), which is amplification.
68
Check q20

A student says: “The liver cell’s response is so large because the phosphorylation cascade relays the signal from kinase to kinase.”

Is the student correct?

  1. A. Yes — relaying the signal down the chain is what makes it large
    A relay of one-to-one steps would carry the message without making it larger.
    The multiplying comes from each enzyme acting many times.
  2. B. ✓ No — each enzyme acts on many molecules before it is switched off

Why: A phosphorylation cascade relays the message.
A relay of one-to-one steps would carry the message without multiplying it.
Each enzyme acts on many molecules before it is switched off.
So the enzyme steps, not the relaying, make the response large.

69Quick quiz: amplification mixed practice

70
Check q21

One activated enzyme acts on many molecules before it is switched off.

Does that step amplify the signal?

  1. A. ✓ Yes
  2. B. No
    One enzyme activating many molecules multiplies the count.

Why: One activated molecule becomes many activated molecules.
So the step amplifies the signal.

71
Check q22

One relay protein activates exactly one molecule of the next protein.

Does that step amplify the signal?

  1. A. Yes
    One activated molecule becomes one activated molecule; the count does not grow.
  2. B. ✓ No

Why: One activated molecule becomes one activated molecule.
The count does not grow, so the step does not amplify the signal.

72
Check q23

What is meant by amplification?

  1. A. The signal is passed from one kinase to the next
    Passing the signal on is relaying; amplification is the multiplying.
  2. B. ✓ The signal is multiplied at each enzyme step
  3. C. The signal is switched off
    Switching off ends the signal; amplification makes it larger.

Why: Amplification is the multiplying of the signal at each enzyme step.
An activated enzyme acts on many molecules before it is switched off.

73
Check q24

At one step of a pathway, 3 activated molecules switch on 300 molecules of the next protein.

Does that step amplify the signal?

  1. A. ✓ Yes
  2. B. No
    Three activated molecules became 300; the count grew.

Why: Three activated molecules became 300 activated molecules.
The count grew, so the step amplifies the signal.

74
Check q25

In the liver cell, epinephrine binds one receptor, and that receptor switches on one enzyme molecule. The enzyme then makes many molecules of cAMP.

Which step multiplies the number of activated molecules?

  1. A. Epinephrine binding the receptor
    One epinephrine molecule activates one receptor; the count does not grow there.
  2. B. ✓ The enzyme making many molecules of cAMP

Why: One receptor switches on one enzyme.
One enzyme makes many molecules of cAMP.
So the enzyme step multiplies the count.

75
Check q26

A kinase switches on many enzyme molecules before a phosphatase switches the kinase off.

Does that step amplify the signal?

  1. A. ✓ Yes
  2. B. No
    One kinase switching on many enzymes multiplies the count.

Why: One activated kinase becomes many activated enzymes.
So the step amplifies the signal.

76
Practice writing an answer

A few bound receptors on a liver cell end in a very large release of glucose.

(a) State what amplification is. (1 pt)

Model answer Amplification is the multiplying of the signal at each enzyme step: an activated enzyme acts on many molecules before it is switched off.
Rubric
  • Award 1 point for: the multiplying of the signal at each enzyme step (each activated enzyme acts on many molecules).

77Calculate the amplification

78

Video: Watch: Calculate the amplification

The amplification at one step is the molecules activated or newly made at that step, divided by the molecules that activated them. Worked through for the liver cell’s 40 receptors and 4,000 new cAMP molecules.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05Bd.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05Bd.mp4

79

Now consider how much one step multiplies the signal. Here is a number for it.

80

The amplification at one step is the number of molecules activated, or newly made, at that step, divided by the number of molecules that activated them.

The amplification at one step of a pathway: the molecules activated (or newly made) at that step, divided by the molecules that activated them
81

The answer is a count per activating molecule, such as cAMP molecules per bound receptor.

82
Worked example

In a liver cell, 40 receptors have epinephrine bound. Over the next minute, the cell’s cAMP rises from 200 molecules to 4,200 molecules. Calculate the amplification at this step, as cAMP molecules made per bound receptor.

Write down the values in the question:
bound receptors = 40
cAMP before = 200 molecules
cAMP after = 4,200 molecules
Write down the equation:
amplification=activated moleculesactivating molecules
Substitute the values into the equation:
amplification=activated moleculesactivating molecules
new cAMP=4,200−200=4,000molecules
amplification=4,00040
amplification=100cAMP per receptor
83

The amplification at this step is 100 cAMP molecules per bound receptor. Each bound receptor switched on an enzyme that made a hundred molecules of cAMP before the pathway was switched off.

84
Check q27 numeric entry

In a liver cell, 30 receptors have epinephrine bound. Over the next minute, the cell’s cAMP rises from 250 molecules to 3,850 molecules.

Calculate the amplification at this step, as cAMP molecules made per bound receptor.

Part 1. Subtract the starting count from the final count. How many new cAMP molecules did the cell make?

Answer: 3600 molecules  (tolerance ±0.5)

Working
Subtract the starting count from the final count:
new cAMP=3,850−250=3,600molecules

Part 2. Divide the new cAMP molecules by the bound receptors. How many cAMP molecules did the cell make per bound receptor?

Answer: 120 cAMP per receptor  (tolerance ±0.5)

Working
Divide the activated molecules by the molecules that activated them:
amplification=3,60030=120cAMP per receptor

Answer: 120 cAMP per receptor  (tolerance ±0.5)

Working
Write down the values in the question:
bound receptors = 30
cAMP before = 250 molecules
cAMP after = 3,850 molecules
Write down the equation:
amplification=activated moleculesactivating molecules
Substitute the values into the equation:
amplification=activated moleculesactivating molecules
new cAMP=3,850−250=3,600molecules
amplification=3,60030
amplification=120cAMP per receptor
85

What you are expected to know Calculate the amplification at a step of a pathway as the number of activated (or newly made) molecules divided by the number of activating molecules.

86
Check q28 numeric entry

In a third liver cell, 50 receptors have epinephrine bound, and over the next minute the cell’s cAMP rises from 300 molecules to 5,800 molecules.

Calculate the amplification at this step, as cAMP molecules made per bound receptor.

Answer: 110 cAMP per receptor  (tolerance ±0.5)

Working
Write down the values in the question:
bound receptors = 50
cAMP before = 300 molecules
cAMP after = 5,800 molecules
Write down the equation:
amplification=activated moleculesactivating molecules
Substitute the values into the equation:
amplification=activated moleculesactivating molecules
new cAMP=5,800−300=5,500molecules
amplification=5,50050
amplification=110cAMP per receptor
87

Back to the liver cell one minute after epinephrine reached it: 40 bound receptors, and about 4,200 molecules of cyclic AMP in the cytosol where there were 200.

88

Each bound receptor switched on one enzyme molecule in the membrane. Each enzyme molecule made many molecules of cyclic AMP from ATP before it was switched off.

89

The cyclic AMP, a second messenger, spread through the cytosol and switched on kinases. Each kinase switched on many enzymes.

90

Every enzyme step multiplied the count. So 40 bound receptors became 4,000 new molecules inside, and the glucose poured out six times faster than before.

91Quick quiz: amplification calculations mixed practice

92
Check q29 numeric entry

In a kidney cell, 10 receptors have a hormone bound. Over the next minute, the cell’s cAMP rises from 100 molecules to 1,600 molecules.

Calculate the amplification at this step, as cAMP molecules made per bound receptor.

Answer: 150 cAMP per receptor  (tolerance ±0.5)

Working
Write down the values in the question:
bound receptors = 10
cAMP before = 100 molecules
cAMP after = 1,600 molecules
Write down the equation:
amplification=activated moleculesactivating molecules
Substitute the values into the equation:
amplification=activated moleculesactivating molecules
new cAMP=1,600−100=1,500molecules
amplification=1,50010
amplification=150cAMP per receptor
93
Check q30 numeric entry

In a yeast cell, a mating signal binds 4 receptors. Each activated receptor switches on kinase molecules, and 680 kinase molecules are found switched on.

Calculate the amplification at this step, as kinase molecules switched on per bound receptor.

Answer: 170 kinases per receptor  (tolerance ±0.5)

Working
Write down the values in the question:
bound receptors = 4
kinase molecules activated = 680
Write down the equation:
amplification=activated moleculesactivating molecules
Substitute the values into the equation:
amplification=activated moleculesactivating molecules
amplification=6804
amplification=170kinases per receptor
94
Check q31 numeric entry

In a heart cell, 20 activated kinase molecules phosphorylate enzyme molecules, and 3,800 enzyme molecules are found switched on.

Calculate the amplification at this step, as enzyme molecules switched on per activated kinase.

Answer: 190 enzymes per kinase  (tolerance ±0.5)

Working
Write down the values in the question:
activated kinases = 20
enzyme molecules activated = 3,800
Write down the equation:
amplification=activated moleculesactivating molecules
Substitute the values into the equation:
amplification=activated moleculesactivating molecules
amplification=3,80020
amplification=190enzymes per kinase
95
Check q32 numeric entry

In a plant root cell, 7 receptors have a hormone bound. Over the next minute, a relay molecule inside the cell rises from 60 molecules to 1,180 molecules.

Calculate the amplification at this step, as relay molecules made per bound receptor.

Answer: 160 relay molecules per receptor  (tolerance ±0.5)

Working
Write down the values in the question:
bound receptors = 7
relay molecules before = 60 molecules
relay molecules after = 1,180 molecules
Write down the equation:
amplification=activated moleculesactivating molecules
Substitute the values into the equation:
amplification=activated moleculesactivating molecules
new relay molecules=1,180−60=1,120molecules
amplification=1,1207
amplification=160relay molecules per receptor

96Mixed practice mixed practice

97
Check q33

In a kidney cell, a hormone binds receptors on the surface. Within a minute a small molecule made inside the cell has risen ten-fold, and proteins deep in the cytosol are switched on.

Which of these is the first messenger?

  1. A. The receptor
    The receptor receives the message; it is not a messenger.
  2. B. ✓ The hormone
  3. C. The small molecule
    The small molecule is made inside the cell after the receptor is activated: the second messenger.

Why: The hormone is the ligand outside the cell.
The ligand outside is the first messenger.
The small molecule made inside is the second.

98
Check q34

A student says: “cAMP is a small protein that the receptor releases into the cytosol.”

Is the student correct?

  1. A. Yes — the receptor releases cAMP into the cytosol
    The receptor releases nothing; it switches on an enzyme, and the enzyme makes cAMP from ATP.
  2. B. ✓ No — cAMP is a small molecule that an enzyme makes from ATP

Why: cAMP is not a protein.
An enzyme in the membrane, switched on by the activated receptor, makes cAMP from ATP.
cAMP is a small molecule, so it spreads quickly.

99
Check q35

A pathway relays a signal through five proteins. Each activated protein activates exactly one molecule of the next. A cell has 3 receptors bound.

How many molecules of the fifth protein are activated?

  1. A. ✓ 3
  2. B. 15
    Each step activates exactly one molecule of the next; the count never grows.
  3. C. Many more than 15
    No step here is an enzyme acting on many molecules, so nothing multiplies.

Why: Each activated protein activates exactly one molecule of the next.
So 3 bound receptors end in 3 activated molecules at every step, including the fifth.
A one-to-one relay does not amplify.

100
Check q36

In a liver cell, epinephrine has just bound its receptor.

What switches on the enzyme that makes cAMP?

  1. A. Epinephrine, after it enters the cell
    Epinephrine stays outside the cell.
  2. B. ✓ The activated receptor
  3. C. The kinase
    The kinase comes later; cAMP switches the kinase on.

Why: Epinephrine activates the receptor.
The activated receptor switches on the enzyme beside it in the membrane.
That enzyme makes cAMP.

101
Check q37

A cell makes a second messenger and a relay protein at its membrane.

Which spreads through the cytosol faster?

  1. A. ✓ The second messenger
  2. B. The relay protein
    A protein is far larger than a second messenger and diffuses more slowly.

Why: A second messenger is a small molecule.
A small molecule diffuses faster than a large protein.
So the second messenger spreads through the cytosol faster.

102
Check q38

In a cell, 5 activated receptors switch on 5 enzyme molecules, and those enzymes make 2,500 molecules of a second messenger.

Which step amplified the signal?

  1. A. The receptors switching on the enzymes
    Five receptors switched on five enzymes; the count did not grow there.
  2. B. ✓ The enzymes making the second messenger
  3. C. Both steps equally
    Only the enzyme step grew the count, from 5 molecules to 2,500 molecules.

Why: Five receptors switched on five enzymes: the count stayed at 5.
Five enzymes made 2,500 molecules of the second messenger: the count grew.
So the enzyme step amplified the signal.

103
Check q39 numeric entry

In a liver cell responding to a different hormone, 2 bound receptors activate 40 enzyme molecules, and those enzymes release 1,200 glucose molecules.

Calculate the amplification at each of the two steps. Enter the amplification at the enzyme-to-glucose step, as glucose molecules released per activated enzyme.

Part 1. Divide the activated enzyme molecules by the bound receptors. What is the amplification at the receptor-to-enzyme step, in enzymes per receptor?

Answer: 20 enzymes per receptor  (tolerance ±0.5)

Working
Divide the activated molecules by the molecules that activated them:
amplification=402=20enzymes per receptor

Answer: 30 glucose per enzyme  (tolerance ±0.5)

Working
Write down the values in the question:
activated enzymes = 40
glucose molecules activated = 1,200
Write down the equation:
amplification=activated moleculesactivating molecules
Substitute the values into the equation:
amplification=activated moleculesactivating molecules
amplification=1,20040
amplification=30glucose per enzyme
104
Check q40 numeric entry

In a frog egg cell, 3 receptors have a hormone bound. Over the next minute, a relay molecule inside the cell rises from 150 molecules to 2,550 molecules.

Calculate the amplification at this step, as relay molecules made per bound receptor.

Answer: 800 relay molecules per receptor  (tolerance ±0.5)

Working
Write down the values in the question:
bound receptors = 3
relay molecules before = 150 molecules
relay molecules after = 2,550 molecules
Write down the equation:
amplification=activated moleculesactivating molecules
Substitute the values into the equation:
amplification=activated moleculesactivating molecules
new relay molecules=2,550−150=2,400molecules
amplification=2,4003
amplification=800relay molecules per receptor
105
Check q41

A kinase has put a phosphate onto a relay protein, and the relay protein is switched on.

Which enzyme removes that phosphate again?

  1. A. Another kinase
    A kinase adds a phosphate from ATP; it removes none.
  2. B. ✓ A phosphatase

Why: A phosphatase removes a phosphate group from a protein.
So the phosphatase takes the phosphate off, and the relay protein returns to its resting shape.

106
Practice writing an answer

Researchers give liver cells epinephrine. One minute later, 25 receptors per cell have epinephrine bound, and the cAMP in each cell has risen from 450 molecules to 3,950 molecules. The cells release glucose at six times their resting rate. In a second dish, the researchers first treat the cells with a phosphatase inhibitor. The researchers add epinephrine to both dishes, and after five minutes they wash it away.

(a) Calculate the amplification at the receptor-to-cAMP step, as cAMP molecules made per bound receptor. (1 pt)

Answer: 140 cAMP per receptor  (tolerance ±0.5)

Model answer The amplification at the receptor-to-cAMP step is 140 cAMP molecules per bound receptor.
Working
Write down the values in the question:
bound receptors = 25
cAMP before = 450 molecules
cAMP after = 3,950 molecules
Write down the equation:
amplification=activated moleculesactivating molecules
Substitute the values into the equation:
amplification=activated moleculesactivating molecules
new cAMP=3,950−450=3,500molecules
amplification=3,50025
amplification=140cAMP per receptor
Rubric
  • Award 1 point for: 140 cAMP molecules per receptor. The resting 450 molecules are subtracted first: (3,950 − 450) ÷ 25 = 140. An answer of 158 (3,950 ÷ 25, the resting cAMP not subtracted) earns no point.

(b) Explain how the cells’ glucose release demonstrates amplification. (1 pt)

Model answer The steps that make the response large are enzyme steps: the enzyme that makes cAMP, and the kinases.
An activated enzyme acts on many molecules before it is switched off.
So each enzyme step multiplies the number of activated molecules.
The enzyme that makes cAMP made 3,500 cAMP molecules from 25 bound receptors.
Each kinase then switched on many enzymes, and each of those enzymes released many glucose molecules.
So 25 bound receptors ended in glucose leaving six times faster.
Rubric
  • Award 1 point for: each enzyme acts on many molecules before it is switched off, so every enzyme step multiplies the number of activated molecules, and a few bound receptors end in a very large response.

Slip Answering that the phosphorylation cascade relays the signal. A relay of one-to-one steps would carry the message without multiplying it; the multiplication comes from each enzyme acting many times.

(c) Predict how the glucose release of the inhibitor-treated cells compares with the untreated cells during the ten minutes after the epinephrine is washed away, and justify your prediction. (2 pt)

Model answer In untreated cells, once the epinephrine is gone, phosphatases remove the phosphates that the kinases added.
So the relay proteins and enzymes return to their resting shapes, and glucose release falls.
In the treated cells the phosphatases are inhibited.
So the relay proteins stay phosphorylated and switched on, although the ligand is gone.
Therefore the treated cells keep releasing glucose at a high rate while the untreated cells return toward their resting rate.
Rubric
  • Award 1 point for: the prediction that the treated cells keep releasing glucose at a high rate after the wash, while the untreated cells fall back.
  • Award 1 point for: the justification that phosphatases normally remove the phosphates and switch the relay off, so inhibiting them leaves the relay proteins phosphorylated and active.

Slip Predicting that the treated cells stop as soon as the epinephrine is washed away because the receptors are empty. The receptors are empty in both dishes; what keeps the treated cells going is the phosphates that are never removed.

Glossary

second messenger
A small molecule made or released inside the cell when a receptor is activated, which spreads through the cytosol and switches on target proteins such as kinases. The ligand outside the cell is the first messenger.
cyclic AMP (cAMP)
A very common second messenger: a ring-shaped small molecule made from ATP by an enzyme in the membrane that the activated receptor switches on. In a liver cell, cAMP switches on the kinase that starts the breakdown of glycogen.
amplification
The multiplying of a signal at each enzyme step of a pathway: an activated enzyme acts on many molecules before it is switched off, so a few bound receptors end in a very large response.

APBIO-U04-L05C Switching the signal off

Topic 4.2 · Introduction to Signal Transduction · 54 steps

A nerve ending drawn as a rounded bulb at the top, a muscle cell as a long band at the bottom, and between them a narrow gap holding a scatter of neurotransmitter molecules, three receptors set through the muscle cell's surface, labeled, and one enzyme molecule
A nerve ending drawn as a rounded bulb at the top, a muscle cell as a long band at the bottom, and between them a narrow gap holding a scatter of neurotransmitter molecules, three receptors set through the muscle cell's surface, labeled, and one enzyme molecule

A nerve ending releases a single pulse of its neurotransmitter onto a muscle cell. The muscle twitches once.

An enzyme in the gap destroys the neurotransmitter within a millisecond. Now suppose a drug blocks that enzyme. The nerve fires once, and the muscle goes on contracting long after the pulse.

Why does a response outlive its signal when one enzyme is missing?

Unit 4 · Cell Communication and Cell Cycle

1Three places to stop

2

Video: Watch: Three places to stop

A cell ends a response in three places: the ligand leaves the receptor and is removed, an enzyme destroys the second messenger, and phosphatases remove the phosphates the kinases added.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05Ca.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05Ca.mp4

3

How does a cell stop responding once the danger has passed? It switches the response off in three places.

4

The ligand leaves the receptor and is broken down or carried away. So the receptor returns to its resting shape.

5

An enzyme destroys the second messenger within seconds of its being made.

6

Phosphatases remove the phosphates the kinases added. So each relay protein returns to rest.

7

Block any one of the three, and the response outlives its signal.

8
Check q1

A kinase has put a phosphate onto a relay protein.

Which enzyme removes that phosphate again?

  1. A. ✓ A phosphatase
  2. B. Another kinase
    A kinase adds phosphates; only a phosphatase removes one.

Why: A phosphatase removes a phosphate from a protein.
The relay protein returns to its resting shape.

9
Check q2

Epinephrine binds a receptor on a liver cell, and cAMP rises inside.

What makes the cAMP?

  1. A. Epinephrine, after it enters the cell
    Epinephrine stays outside the cell.
  2. B. The receptor itself
    The receptor switches on the enzyme; the enzyme makes the cAMP.
  3. C. ✓ An enzyme in the cell membrane

Why: The activated receptor switches on an enzyme beside it in the membrane.
That enzyme makes cAMP from ATP.

10

Now consider the liver cell after the danger has passed. The cell must stop releasing glucose.

The fan of the epinephrine pathway with three off-switches marked: the ligand leaving the receptor at the top, an enzyme breaking cAMP down at the second tier, and phosphatases removing phosphates at the kinases-and-enzymes tier
The fan of the epinephrine pathway with three off-switches marked: the ligand leaving the receptor at the top, an enzyme breaking cAMP down at the second tier, and phosphatases removing phosphates at the kinases-and-enzymes tier
11

First, the epinephrine leaves the receptor and is broken down or carried away in the blood. So the receptor returns to its resting shape.

12

Second, an enzyme breaks down the second messenger. cAMP is destroyed within seconds of being made, so its level falls as soon as the enzyme that makes it is switched off.

13

Third, phosphatases remove the phosphates the kinases added. So each relay protein returns to its resting shape.

14

Here is a table comparing the three off-switches: what each one removes, how fast, and what happens when it is blocked.

A table of the three off-switches: what each removes, how fast, and what happens when it is blocked
15

What you are expected to know Describe how a cell ends a response at the receptor: the ligand leaves the receptor and is broken down or carried away.

16

What you are expected to know Describe how a cell ends a response in the cytosol: an enzyme destroys the second messenger.

17

What you are expected to know Describe how a cell ends a response at its relay proteins: phosphatases remove the phosphates the kinases added.

18
Check q3

cAMP is destroyed by an enzyme within seconds of being made. A signal has been switched off, and the enzyme that makes cAMP has stopped.

Predict what happens to the cAMP level over the next few seconds.

  1. A. The cAMP level stays high until the ligand outside is broken down
    An enzyme inside the cell destroys cAMP within seconds of its being made, whatever the ligand outside is doing.
  2. B. The cAMP level rises
    No new cAMP is made.
    The old cAMP is destroyed within seconds.
  3. C. The cAMP level stays high for hours
    An enzyme destroys cAMP within seconds of its being made, so with no new cAMP made the level cannot stay high.
  4. D. ✓ The cAMP level falls within seconds

Why: One enzyme makes cAMP, and another enzyme destroys cAMP within seconds.
The enzyme that makes cAMP has stopped.
The enzyme that destroys cAMP keeps working.
So the cAMP level falls within seconds.

19
Check q4

Epinephrine has left a liver cell’s receptor and been carried away in the blood.

What happens to the receptor?

  1. A. ✓ It returns to its resting shape
  2. B. It stays in its activated shape
    The receptor is activated only while the ligand sits in its pocket.
  3. C. It is broken down
    The receptor is not broken down; it returns to rest and can be used again.

Why: The receptor’s shape is changed by the ligand in its pocket.
The ligand has left.
So the receptor returns to its resting shape.

20When one off-switch fails

21

Video: Watch: When one off-switch fails

Block one off-switch and the response outlives its signal. The measurement that stays high after the signal is gone names the switch that failed.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05Cb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L05Cb.mp4

22

Now suppose one of the three off-switches is blocked. The response outlives its signal.

The fan of the epinephrine pathway with three off-switches marked: the ligand leaving the receptor at the top, an enzyme breaking cAMP down at the second tier, and phosphatases removing phosphates at the kinases-and-enzymes tier
The fan of the epinephrine pathway with three off-switches marked: the ligand leaving the receptor at the top, an enzyme breaking cAMP down at the second tier, and phosphatases removing phosphates at the kinases-and-enzymes tier
23

Imagine the enzyme that destroys the neurotransmitter at a muscle cell is blocked. The neurotransmitter keeps binding its receptors, so the muscle keeps responding.

24

Imagine a cell’s phosphatases are blocked. Its relay proteins keep their phosphates after the ligand is washed away, so the response continues.

25

The level that stays high after the signal is gone names the switch that failed.

26

Here is a table showing which level stays high and which switch has failed.

A table matching the level that stays high after the signal is gone to the switch that has failed
27

One more case is in the table.

28

Suppose cAMP stays high, and the cell destroys added cAMP at the normal rate.

29

Then the cell must still be making cAMP.

30

So the enzyme that makes cAMP has stayed switched on.

31

What you are expected to know Name the off-switch that has failed from the level that stays high after the signal is gone.

32
Check q5

In normal cells, after the ligand is washed away, cAMP falls from 100 units to 7 units in ten minutes. In treated cells it falls only from 100 units to 88 units. In both, no ligand is bound to the receptors one minute after the wash, and extracts of both break down added cAMP at the same rate.

Where has the off-switch failed in the treated cells?

  1. A. ✓ The enzyme that makes cAMP is still switched on
  2. B. cAMP is no longer being broken down
    Extracts of both groups break cAMP down at the same rate, so the enzyme that destroys cAMP works normally.
  3. C. The phosphatases have stopped removing phosphates
    Phosphatases act on the relay proteins.
    The measurement here is cAMP, and cAMP stays high only if the cell keeps making it.
  4. D. The ligand is still bound to the receptors
    One minute after the wash, no ligand is bound in either group.

Why: The ligand is gone.
The enzyme that destroys cAMP works normally.
Yet cAMP stays high.
So the cell must still be making new cAMP.
The enzyme that makes cAMP has stayed switched on after the ligand left.

33
Check q6

A researcher treats muscle cells with a phosphatase inhibitor, then adds insulin. The cells move glucose transporters to the surface. After five minutes the researcher washes the insulin away.

Predict what happens to the response over the next ten minutes, compared with untreated cells.

  1. A. The response stops when the insulin is washed away, as in untreated cells
    Normally phosphatases remove the phosphates the kinases added, so the relay switches off once the insulin goes.
    Here the phosphatases are blocked.
  2. B. The response reverses
    The inhibitor blocks the phosphatases.
    So the phosphates stay on the relay proteins, and the relay stays switched on.
  3. C. ✓ The response continues

Why: Phosphatases switch the relay off by removing the phosphates the kinases added.
The inhibitor blocks the phosphatases.
So the relay proteins keep their phosphates after the insulin is gone.
So the transporters stay at the surface, and the response continues.

34
Practice writing an answer

A researcher treats muscle cells with a phosphatase inhibitor, then adds insulin. The cells move glucose transporters to the surface. After five minutes the researcher washes the insulin away. Over the next ten minutes the response continues, while in untreated cells it stops.

(a) Explain how this result demonstrates that phosphatases are one of the cell’s off-switches. (1 pt)

Model answer Insulin binding its receptor switches on kinases inside the cell.
The kinases put phosphates onto relay proteins, and the phosphates hold the relay proteins switched on.
In untreated cells, phosphatases remove those phosphates once the insulin is gone, so the response stops.
In the treated cells the inhibitor blocks the phosphatases.
So the phosphates stay on, and the relay proteins stay switched on.
Blocking only the phosphatases let the response outlive its signal, so the phosphatases are an off-switch.
Rubric
  • Award 1 point for: phosphatases normally remove the phosphates and switch the relay off; with the phosphatases inhibited the relay proteins stay phosphorylated and active, so the response continues — removing that one enzyme removed the switch-off.
35

Back to the nerve ending, its single pulse of neurotransmitter, and the muscle cell that twitched once.

36

Normally the enzyme in the gap destroys the neurotransmitter within a millisecond. The ligand is removed, so the receptors return to rest and the muscle relaxes.

37

The drug blocked that enzyme. The neurotransmitter stayed in the gap and kept binding the receptors.

38

So the muscle went on contracting long after the pulse. The response outlived its signal because one off-switch, the removal of the ligand, was missing.

39Quick quiz: which off-switch failed? mixed practice

40
Check q7

After the wash, epinephrine is still bound to the liver cell’s receptors.

Which off-switch has failed?

  1. A. ✓ The ligand was not removed
  2. B. The second messenger was not removed
    The level that stays high is the amount of bound ligand, not cAMP.
  3. C. The phosphates were not removed
    The level that stays high is the amount of bound ligand, not the phosphates.

Why: The ligand is still bound after the wash.
So the ligand was not removed from the receptor.

41
Check q8

No ligand is bound. The level of cAMP stays high. The cell breaks down added cAMP very slowly.

Which off-switch has failed?

  1. A. The ligand was not removed
    No ligand is bound, so the ligand was removed.
  2. B. ✓ The second messenger was not removed
  3. C. The phosphates were not removed
    The level that stays high is the level of cAMP, the second messenger.

Why: cAMP is the second messenger.
The cell breaks down cAMP very slowly, so the level of cAMP stays high.
So the second messenger was not removed.

42
Check q9

No ligand is bound. cAMP has fallen to normal. The relay proteins still carry their phosphates.

Which off-switch has failed?

  1. A. The ligand was not removed
    No ligand is bound, so the ligand was removed.
  2. B. The second messenger was not removed
    cAMP has fallen, so the second messenger was removed.
  3. C. ✓ The phosphates were not removed

Why: The relay proteins still carry their phosphates.
Phosphatases remove those phosphates.
So the phosphates were not removed.

43
Check q10

A nerve fires once. The neurotransmitter stays in the gap for seconds instead of a millisecond.

Which off-switch has failed?

  1. A. ✓ The ligand was not removed
  2. B. The second messenger was not removed
    The neurotransmitter is outside the muscle cell; it is the ligand, not a second messenger.
  3. C. The phosphates were not removed
    The amount that stays high is the neurotransmitter in the gap, the ligand.

Why: The neurotransmitter is the ligand.
It stays in the gap instead of being destroyed.
So the ligand was not removed.

44
Check q11

Insulin is washed away from muscle cells whose phosphatases are blocked. The glucose transporters stay at the surface.

Which off-switch has failed?

  1. A. The ligand was not removed
    The insulin was washed away, so the ligand was removed.
  2. B. The second messenger was not removed
    No second messenger is measured here; the phosphatases are what is blocked.
  3. C. ✓ The phosphates were not removed

Why: The phosphatases are blocked.
Phosphatases remove the phosphates the kinases added.
So the phosphates were not removed.

45
Check q12

A receptor is empty, yet the concentration of calcium ions in the cytosol stays high. The pump that returns calcium ions to their store is blocked.

Which off-switch has failed?

  1. A. The ligand was not removed
    The receptor is empty, so the ligand was removed.
  2. B. ✓ The second messenger was not removed
  3. C. The phosphates were not removed
    The level that stays high is the concentration of calcium ions, a second messenger.

Why: Calcium ions are a second messenger.
The pump that removes them from the cytosol is blocked, so their concentration stays high.
So the second messenger was not removed.

46Mixed practice mixed practice

47
Check q13

A drug blocks the enzyme in the gap that destroys the neurotransmitter at a muscle cell. The nerve ending then releases a single pulse of neurotransmitter.

Predict what the muscle cell does.

  1. A. The muscle cell responds once, as usual, and then rests until the next pulse
    The enzyme that destroys the neurotransmitter is blocked.
    So the neurotransmitter is never cleared from the gap, and it keeps binding the receptors.
  2. B. The muscle cell shows no response
    The drug blocks the enzyme in the gap.
    The receptors are free and keep binding the neurotransmitter.
  3. C. ✓ The muscle cell keeps responding
  4. D. The muscle cell responds more weakly than usual
    The breakdown of the neurotransmitter is blocked, so the neurotransmitter stays in the gap at full concentration and keeps binding.

Why: The response normally ends when the enzyme in the gap destroys the neurotransmitter, so the neurotransmitter leaves the receptors.
The drug blocks that enzyme.
So the neurotransmitter stays in the gap and keeps binding the receptors.
So the muscle cell keeps responding.

48
Check q14

A liver cell’s three off-switches all work. Epinephrine leaves the blood.

Predict what happens to the cell’s glucose release over the next few minutes.

  1. A. ✓ It falls back toward its resting rate
  2. B. It continues at the high rate
    With the ligand gone, cAMP destroyed and the phosphates removed, nothing keeps the pathway switched on.
  3. C. It rises further
    No new signal arrives, and the three off-switches end the old one.

Why: The epinephrine leaves the receptors, so the receptors return to rest.
An enzyme destroys the cAMP within seconds.
Phosphatases remove the phosphates from the relay proteins.
So the pathway switches off, and glucose release falls back toward its resting rate.

49
Check q15

Which molecules do phosphatases act on to end a response?

  1. A. The ligand in the receptor’s pocket
    The ligand is removed by leaving the receptor and being broken down or carried away.
  2. B. The cAMP in the cytosol
    cAMP is destroyed by a different enzyme.
  3. C. ✓ The relay proteins carrying phosphates

Why: Phosphatases remove phosphate groups from proteins.
So they act on the relay proteins the kinases phosphorylated.

50
Check q16

A drug blocks a cell’s phosphatases. The ligand is then washed away.

Which level stays high?

  1. A. The amount of ligand bound to the receptors
    The ligand was washed away and leaves the receptors as usual.
  2. B. The level of cAMP in the cytosol
    The enzyme that destroys cAMP still works, so cAMP falls.
  3. C. ✓ The number of relay proteins carrying a phosphate

Why: Phosphatases remove the phosphates from the relay proteins.
The phosphatases are blocked.
So the relay proteins keep their phosphates, and that count stays high.

51
Check q17

In a patient, the blood carries epinephrine away more slowly than normal, so the receptors’ pockets keep being refilled.

Which off-switch has failed?

  1. A. ✓ The ligand was not removed
  2. B. The second messenger was not removed
    The level that stays high is the amount of epinephrine at the receptors, not cAMP.
  3. C. The phosphates were not removed
    The level that stays high is the amount of epinephrine at the receptors, not the phosphates.

Why: Epinephrine is the ligand.
It is not carried away, so it keeps binding the receptors.
So the ligand was not removed.

52
Check q18

A student says: “Once the ligand leaves the receptor, the response stops at once, whatever happens inside the cell.”

Is the student correct?

  1. A. Yes — the ligand leaving is the only off-switch a cell has
    A cell with its phosphatases blocked keeps responding after the ligand is gone.
  2. B. ✓ No — the second messenger and the phosphates inside must be removed too

Why: The ligand leaving switches off only the receptor.
cAMP already made keeps switching on kinases until an enzyme destroys it.
Phosphorylated relay proteins stay switched on until phosphatases remove their phosphates.
So the response stops only when all three are removed.

53
Practice writing an answer

A nerve ending releases a single pulse of neurotransmitter onto a muscle cell, and the muscle twitches once. Normally an enzyme in the gap destroys the neurotransmitter within a millisecond. A drug blocks that enzyme. Now the nerve fires once, and the muscle goes on contracting for several seconds.

(a) Explain how this case demonstrates that removing the ligand is one of a cell’s off-switches. (1 pt)

Model answer Normally the enzyme in the gap destroys the neurotransmitter, the ligand, within a millisecond, so the ligand is removed.
With the ligand gone, the receptors return to rest and the muscle relaxes.
The drug blocks that enzyme, so the neurotransmitter stays in the gap and keeps binding the receptors.
So the muscle keeps contracting after a single pulse.
Blocking only the removal of the ligand made the response outlive its signal, so removing the ligand is an off-switch.
Rubric
  • Award 1 point for: destroying the neurotransmitter removes the ligand and ends the response; with that enzyme blocked the ligand keeps binding and the response continues, so ligand removal is an off-switch.

APBIO-U04-L06 A door that opens

Topic 4.2 · Introduction to Signal Transduction · 39 steps

A nerve cell tip releasing neurotransmitter onto a muscle cell; ions flow into the muscle cell within a millisecond
A nerve cell tip releasing neurotransmitter onto a muscle cell; ions flow into the muscle cell within a millisecond

Here is a muscle cell under the tip of a nerve cell.

The nerve cell releases a neurotransmitter, and the neurotransmitter lands on the muscle cell. Within a thousandth of a second, a current of ions flows into the muscle cell.

Now suppose a researcher tears a patch of membrane off the muscle cell, with nothing from inside the cell attached, and adds the neurotransmitter. The same current flows, just as fast. How can a response happen with no relay inside the cell at all?

Unit 4 · Cell Communication and Cell Cycle

1A door that opens

2

Video: Watch: The receptor that is itself the door

Acetylcholine binds a channel protein in the muscle cell’s membrane. The channel opens, and positive ions flow in within a millisecond. In a torn patch of membrane the same current flows: the channel alone is the door.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L06a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L06a.mp4

3
Check q1

An ion sits outside a cell. The membrane’s oily middle holds charged particles back.

How can the ion cross the membrane?

  1. A. ✓ Only through a membrane protein: here a channel, while it is open
  2. B. By dissolving into the oily middle once its concentration is high enough
    The oily middle holds a charged particle back at any concentration.
    The ion needs a membrane protein to cross: here, a channel.

Why: The oily middle holds ions back.
An ion crosses only through a membrane protein.
Here that protein is a channel.
The channel lets ions through only while it is open.

4

What happens when the receptor is itself the door? Some receptors are channel proteins.

5

Here is one such receptor in a muscle cell’s membrane. The channel is closed, with many positive ions outside the cell and few inside.

A stretch of a muscle cell's membrane with a channel protein set in it, closed; acetylcholine approaches from outside, where positive ions are many
A stretch of a muscle cell's membrane with a channel protein set in it, closed; acetylcholine approaches from outside, where positive ions are many
6

Acetylcholine is the neurotransmitter a nerve cell releases onto a muscle cell. Acetylcholine is this channel’s ligand.

7

The ligand fits the receptor’s binding site by shape and charge.

8

While the ligand is bound, the receptor holds a different shape.

9

So another part of the receptor can now act on the next molecule.

10

Here the part that changes is the channel itself. The channel opens, and positive ions flow down their concentration gradient into the cell within a millisecond.

Acetylcholine sitting in the pocket on the channel's outer corner: the channel is open and positive ions flow through it into the cytosol
Acetylcholine sitting in the pocket on the channel's outer corner: the channel is open and positive ions flow through it into the cytosol
11

No second messenger and no kinase stand between the ligand and the ions. The ligand moves the door itself.

12

Some receptors are themselves channels. Their ligand opens or closes them.

13

A receptor like this is called a .

14

‘Gated’ means opened and closed, like a door. The ligand is what opens or closes it.

15

Most ligand-gated channels open when their ligand binds. Some ligand-gated channels close instead.

16

Now suppose a researcher tears a patch of membrane from the muscle cell. The patch holds the channel and nothing from inside the cell.

A patch of membrane torn from the muscle cell, holding only the channel: acetylcholine binds and the same ion current flows
A patch of membrane torn from the muscle cell, holding only the channel: acetylcholine binds and the same ion current flows
17

She adds acetylcholine to the patch. The same current flows, within a millisecond.

18

So the torn patch shows that a ligand-gated channel needs nothing from inside the cell.

19

When acetylcholine leaves the binding site, the channel returns to its closed shape. The ion flow stops.

20

What you are expected to know Predict what happens the moment a ligand binds a ligand-gated channel: the channel opens (or, for some channels, closes), and ions flow down their concentration gradient within a millisecond, with no relay inside the cell.

21
Check q2

A researcher gives a muscle cell a drug that holds every relay protein inside the cell switched off. The researcher then adds acetylcholine.

Which of the following happens within a millisecond?

  1. A. No positive ions flow in
    The receptor is itself the channel: acetylcholine binds it and its new shape is an open channel.
    No relay protein stands in between, so the ions flow in as usual.
  2. B. Positive ions flow in, but only after several seconds
    A relay of proteins inside the cell takes seconds.
    Here the receptor is itself the channel, so its new shape is an open channel within a millisecond.
  3. C. ✓ Positive ions flow in as usual

Why: The receptor is itself the channel.
Acetylcholine binds the receptor, and the receptor changes shape.
The new shape is an open channel, so positive ions flow in the moment acetylcholine binds.
No relay protein stands between acetylcholine and the ions, so switching every relay protein off changes nothing.

22
Practice writing an answer

A researcher gives a muscle cell a drug that holds every relay protein inside the cell switched off, then adds acetylcholine. Positive ions flow in as usual, within a millisecond.

(a) Explain why the current still flows with every relay protein switched off. (1 pt)

Model answer The receptor for acetylcholine is a ligand-gated channel.
So the receptor is itself the channel.
Acetylcholine binds the receptor, and the receptor changes shape.
The new shape is an open channel.
So positive ions flow in the moment acetylcholine binds.
No relay protein stands between acetylcholine and the ions.
So switching every relay protein off changes nothing, and the current flows as usual.
Rubric
  • Award 1 point for: the receptor is itself the channel (a ligand-gated channel), so acetylcholine’s binding opens the channel directly; no relay protein stands between the ligand and the ions, so switching the relay proteins off changes nothing.
23
Check q3

Maya says: “When acetylcholine binds, a second messenger made inside the cell must open the channel.”

Is Maya correct?

  1. A. Yes — a second messenger made inside the cell opens the channel
    The receptor is itself the channel.
    Acetylcholine binds the receptor, and the receptor’s new shape is an open channel.
    The cell makes no second messenger, and needs none.
  2. B. ✓ No — acetylcholine’s binding opens the channel itself

Why: The receptor is itself the channel.
Acetylcholine binds the receptor, and the receptor changes shape.
The new shape is an open channel.
So acetylcholine’s binding opens the channel itself.
No second messenger and no kinase stand between acetylcholine and the ions.

24
Check q4

In a gland cell’s membrane, a receptor is itself a channel. At rest the channel is open, and positive ions flow in. When the ligand binds, the receptor’s new shape is a closed channel.

Which of the following happens to the ion flow when the ligand binds?

  1. A. ✓ The ion flow stops within a millisecond
  2. B. The ion flow speeds up within a millisecond
    The ligand binds, the receptor changes shape, and the new shape is a closed channel.
    So the ion flow stops.
  3. C. The ion flow stops, but only after several seconds
    The receptor is itself the channel.
    The ligand binds, and the receptor’s own shape changes.
    So the door moves within a millisecond, with no relay inside the cell.

Why: The ligand binds the receptor, and the receptor changes shape.
The receptor is itself the channel.
For this channel the new shape is closed.
So the ion flow stops within a millisecond, with no relay in between.

25

Back to the muscle cell under the tip of the nerve cell, where a current of ions flowed in within a thousandth of a second. The nerve cell released acetylcholine, and acetylcholine bound a receptor that is itself a channel.

26

The receptor’s own shape change opened the channel, and positive ions flowed in within a millisecond. That receptor is a ligand-gated channel.

27

In the torn patch of membrane, nothing from inside the cell was present, and the same current flowed. The ligand moves the door itself, so no relay inside the cell is needed.

28Quick quiz: ligand-gated channel mixed practice

29
Check q5

A ligand binds a ligand-gated channel, and the channel opens.

Which of the following opens the channel?

  1. A. A kinase adding a phosphate to the channel
    No kinase stands between the ligand and the ions.
    The receptor’s own shape change is the open channel.
  2. B. A second messenger made inside the cell
    No second messenger stands between the ligand and the ions.
    The receptor’s own shape change is the open channel.
  3. C. ✓ The receptor’s own change of shape

Why: The ligand binds the receptor, and the receptor changes shape.
The receptor is itself the channel.
So the receptor’s own change of shape opens the channel.

30
Check q6

A ligand-gated channel is open.

Which of the following passes through it?

  1. A. Its ligand
    The ligand stays in the binding site on the outside of the channel.
    Ions pass through the open channel.
  2. B. ✓ Ions
  3. C. cAMP
    cAMP is made inside the cell by an enzyme.
    Ions pass through the open channel.

Why: A ligand-gated channel is a channel protein.
Ions cross a membrane through channel proteins.
So ions pass through the open channel.

31
Check q7

A ligand binds a ligand-gated channel.

How soon do the ions start to flow?

  1. A. ✓ Within a millisecond
  2. B. After several seconds
    Several seconds is the time a relay of proteins inside the cell takes.
    Here the receptor is itself the channel, so the door moves at once.
  3. C. After several minutes
    Minutes is far too slow.
    The receptor is itself the channel, so the door moves at once.

Why: The receptor is itself the channel.
Its own shape change opens the channel, with no relay inside the cell.
So the ions flow within a millisecond.

32
Check q8

The ligand leaves the binding site of a ligand-gated channel that opened when the ligand bound.

Which of the following happens to the channel?

  1. A. The channel stays open
    The receptor holds its open shape only while the ligand is bound.
    With the ligand gone, the receptor returns to its closed shape.
  2. B. ✓ The channel returns to its closed shape

Why: The receptor holds a different shape only while the ligand is bound.
The ligand has left.
So the receptor returns to its closed shape, and the ion flow stops.

33
Check q9

Ligand-gated channels come in more than one kind.

Does every ligand-gated channel open when its ligand binds?

  1. A. Yes — every ligand-gated channel opens
    Most ligand-gated channels open when their ligand binds.
    Some ligand-gated channels close instead; either way the ligand moves the door.
  2. B. ✓ No — some ligand-gated channels close instead

Why: The ligand’s binding changes the receptor’s shape.
For most ligand-gated channels the new shape is open.
For some the new shape is closed.
So not every ligand-gated channel opens.

34
Check q10

A receptor is itself a channel protein, and its ligand opens it.

Which of the following is this receptor called?

  1. A. ✓ A ligand-gated channel
  2. B. A second messenger
    A second messenger is a small molecule made inside the cell, not a receptor.
  3. C. A kinase
    A kinase is an enzyme that adds a phosphate to a protein, not a receptor.

Why: The receptor is itself a channel, and its ligand opens it.
A receptor like this is called a ligand-gated channel.

35
Practice writing an answer

A muscle cell carries receptors for acetylcholine.

(a) State what a ligand-gated channel is. (1 pt)

Model answer A ligand-gated channel is a receptor that is itself a channel protein; its ligand’s binding opens the channel, or for some channels closes it.
Rubric
  • Award 1 point for: a receptor that is itself a channel, opened (or closed) by its ligand binding.

36Mixed practice mixed practice

37
Check q11

Two cells each bind a ligand at their surface. In the first cell the receptor is itself a channel protein. In the second cell the receptor hands its message to a phosphorylation cascade of three kinases inside the cell.

Which cell responds within a millisecond?

  1. A. ✓ The cell with the ligand-gated channel
  2. B. The cell with the phosphorylation cascade
    Each kinase in the cascade must be switched on before the next.
    That relay takes seconds.
  3. C. The two cells respond equally fast
    Only the channel acts within a millisecond.
    The phosphorylation cascade takes seconds, so the two are not equally fast.

Why: The ligand-gated channel opens the moment its ligand binds, so ions flow within a millisecond.
The second cell’s receptor hands the message to one kinase, then the next, then the next.
That relay takes seconds.

38
Check q12

A receptor’s ligand binds, and within a millisecond negative ions flow through the receptor itself.

Which of the following is this receptor?

  1. A. An intracellular receptor
    An intracellular receptor sits inside the cell and has no channel through it.
    Ions flow through this receptor itself, so the receptor is the channel.
  2. B. ✓ A ligand-gated channel
  3. C. A second messenger
    A second messenger is a small molecule made inside the cell, not a receptor.
    Ions flow through this receptor itself, so the receptor is the channel.

Why: The ions flow through the receptor itself.
So the receptor is the channel.
A receptor that is itself a channel is a ligand-gated channel.

Glossary

ligand-gated channel
A receptor that is itself a channel protein. When its ligand binds, the receptor’s shape change opens the channel (or, for some channels, closes it), and ions flow down their concentration gradient within a millisecond, with no relay inside the cell.

APBIO-U04-L06B A door that hands the message on

Topic 4.2 · Introduction to Signal Transduction · 65 steps

Epinephrine from a blood vessel arriving at a liver cell; glucose leaves the cell seconds later
Epinephrine from a blood vessel arriving at a liver cell; glucose leaves the cell seconds later

Here is a liver cell beside a blood vessel.

Epinephrine arrives in the blood and lands on the liver cell. Several seconds later, the first glucose leaves the cell. The receptor epinephrine binds is a membrane protein with no channel through it.

Now suppose a drug holds one helper protein on the inner face of the membrane switched off. Epinephrine still binds, and the receptor still changes shape. Yet the cell makes no cAMP, and no glucose leaves. What is the helper protein, and what does it hand on?

Unit 4 · Cell Communication and Cell Cycle

1A helper protein on the inner face

2

Video: Watch: The helper protein beside the receptor

Epinephrine binds a receptor with no channel through it. The receptor’s intracellular domain changes shape and switches on the helper protein beside it on the inner face of the membrane: the G protein.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L06Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L06Ba.mp4

3
Check q1

A receptor set in a cell’s membrane binds its ligand on the outside of the cell.

Which region of the receptor changes shape on the inner face of the membrane?

  1. A. ✓ Its intracellular domain
  2. B. Its ligand-binding domain
    The ligand-binding domain holds the ligand on the outside of the cell.
    The region facing the cytosol is the part that changes shape on the inner face.

Why: The ligand binds the ligand-binding domain outside the cell.
The receptor changes shape.
The part that changes on the inner face is the region facing the cytosol: the intracellular domain.

4

How does a receptor with no channel pass its message on? Epinephrine’s receptor on a liver cell is a membrane protein with no channel through it.

A liver cell's epinephrine receptor set in the membrane, a helper protein switched off on the inner face, and an enzyme further along; epinephrine approaches
A liver cell's epinephrine receptor set in the membrane, a helper protein switched off on the inner face, and an enzyme further along; epinephrine approaches
5

On the inner face of the membrane, beside the receptor, sits a helper protein. The helper protein is switched off until the receptor changes shape.

6

The ligand fits the receptor’s binding site by shape and charge.

7

While the ligand is bound, the receptor holds a different shape.

8

So another part of the receptor can now act on the next molecule.

9

Here the part that changes is the receptor’s intracellular domain. Its new shape switches on the helper protein beside it.

Epinephrine bound: the receptor's intracellular domain has changed shape (drawn with a notch) and switched on the helper protein beside it, which now carries a notch too
Epinephrine bound: the receptor's intracellular domain has changed shape (drawn with a notch) and switched on the helper protein beside it, which now carries a notch too
10

The helper protein is called a .

11

A surface receptor that hands its message to a G protein like this is called a .

12

‘Coupled’ means joined to work together: the receptor is coupled with its G protein.

13

Receptors of this kind are found in eukaryotic cells.

14

The G protein is not the receptor: the receptor binds the ligand, and the G protein waits beside it. The G protein makes nothing itself; it carries the switch onward.

15

What you are expected to know Describe how a G protein-coupled receptor switches on its G protein: the ligand binds, the receptor’s intracellular domain changes shape, and the new shape switches on the G protein on the inner face of the membrane.

16
Check q2

A researcher takes three proteins, X, Y and Z, from the membrane of a liver cell and studies them one at a time. X binds epinephrine. X, once bound, switches on Y, and Y switches on Z.

Which protein is the G protein?

  1. A. X
    X binds epinephrine, so X is the receptor.
    The receptor switches on the G protein.
    X switches on Y, so Y is the G protein.
  2. B. ✓ Y
  3. C. Z
    X binds epinephrine, so X is the receptor.
    X switches on Y, so Y is the G protein.
    Z is the protein the G protein switches on.

Why: X binds epinephrine, so X is the receptor.
The receptor switches on the G protein beside it.
X switches on Y, so Y is the G protein.

17
Check q3

Leo says: “The G protein is the enzyme that makes cAMP.”

Is Leo correct?

  1. A. ✓ No — the G protein switches the cAMP-making enzyme on
  2. B. Yes — the G protein makes cAMP from ATP
    The receptor switches the G protein on.
    The G protein moves along the membrane’s inner face and switches on a separate enzyme; that enzyme makes the cAMP from ATP.

Why: The G protein makes no cAMP.
The receptor switches the G protein on.
The G protein moves along the membrane’s inner face and switches on a separate enzyme.
That enzyme makes cAMP from ATP.
So the G protein carries the switch from the receptor to the enzyme.

18Quick quiz: G protein, G protein-coupled receptor mixed practice

19
Check q4

A helper protein sits on the inner face of a cell’s membrane, beside a receptor. The receptor switches it on.

Which of the following is the helper protein called?

  1. A. A kinase
    A kinase is an enzyme that adds a phosphate to a protein.
    The helper protein beside the receptor is a G protein.
  2. B. A second messenger
    A second messenger is a small molecule, not a protein.
    The helper protein beside the receptor is a G protein.
  3. C. ✓ A G protein

Why: A helper protein on the inner face of the membrane, switched on by the receptor beside it, is called a G protein.

20
Check q5

A cell carries G proteins.

Where does a G protein sit?

  1. A. ✓ On the inner face of the plasma membrane
  2. B. Outside the cell, in the fluid around it
    The G protein is switched on by the receptor’s intracellular domain.
    So the G protein sits inside the cell, on the inner face of the membrane.
  3. C. Inside the nucleus
    The G protein is switched on by the receptor’s intracellular domain, which faces the cytosol.
    So the G protein sits on the inner face of the membrane.

Why: The receptor’s intracellular domain faces the cytosol.
Its new shape switches on the G protein beside it.
So the G protein sits on the inner face of the plasma membrane.

21
Check q6

A G protein is switched off, then switched on.

Which of the following switches the G protein on?

  1. A. ✓ The receptor’s intracellular domain changing shape
  2. B. cAMP binding the G protein
    cAMP is made later in the relay, after the G protein has acted.
    The receptor’s intracellular domain switches the G protein on.
  3. C. A kinase adding a phosphate to the G protein
    No kinase stands between the receptor and the G protein.
    The receptor’s intracellular domain switches the G protein on.

Why: The ligand binds the receptor.
The receptor’s intracellular domain changes shape.
The new shape switches on the G protein beside it.

22
Check q7

A surface receptor changes shape when its ligand binds, and its new shape switches on a G protein beside it.

Which of the following is this receptor called?

  1. A. A ligand-gated channel
    A ligand-gated channel is itself a channel and switches on nothing beside it.
    A receptor that switches on a G protein is a G protein-coupled receptor.
  2. B. ✓ A G protein-coupled receptor
  3. C. An intracellular receptor
    An intracellular receptor sits inside the cell.
    This receptor sits at the surface and switches on a G protein: a G protein-coupled receptor.

Why: The receptor hands its message to a G protein.
A surface receptor that works with a G protein is called a G protein-coupled receptor.

23
Check q8

A student says: “The G protein is the part of the receptor that binds the ligand.”

Is the student correct?

  1. A. Yes — the G protein binds the ligand
    The receptor binds the ligand.
    The G protein is a separate protein on the inner face of the membrane, switched on by the receptor.
  2. B. ✓ No — the G protein is a separate protein beside the receptor

Why: The receptor binds the ligand.
The G protein is a separate protein beside the receptor on the inner face of the membrane.
The receptor switches the G protein on.

24
Check q9

A cell carries G protein-coupled receptors.

Where does a G protein-coupled receptor sit?

  1. A. ✓ In the plasma membrane
  2. B. In the cytosol, away from the membrane
    The G protein-coupled receptor is a surface receptor.
    It sits in the plasma membrane, with its binding site outside the cell.
  3. C. In the nucleus, bound to the DNA
    The G protein-coupled receptor is a surface receptor, not an intracellular receptor.
    It sits in the plasma membrane.

Why: A G protein-coupled receptor is a surface receptor.
So it sits in the plasma membrane, with its binding site outside the cell and its intracellular domain facing the cytosol.

25
Practice writing an answer

Epinephrine binds a receptor on a liver cell.

(a) State what a G protein is. (1 pt)

Model answer A G protein is a helper protein on the inner face of the membrane that the receptor switches on; it carries the switch onward.
Rubric
  • Award 1 point for: a helper protein on the inner face of the membrane, switched on by the receptor (and passing the switch on).

(b) State what a G protein-coupled receptor is. (1 pt)

Model answer A G protein-coupled receptor is a surface receptor that hands its message to a G protein when its ligand binds.
Rubric
  • Award 1 point for: a surface receptor that works through (hands its message to) a G protein.

26The relay to the response

27

Video: Watch: From the G protein to the glucose

The switched-on G protein moves along the inner face and switches on an enzyme. The enzyme makes many cAMP from ATP. cAMP switches on kinases, and the kinases switch on the enzymes that break glycogen down. Three proteins, three jobs.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L06Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L06Bb.mp4

28
Check q10

In a liver cell, an enzyme set in the membrane makes the second messenger cAMP.

Which molecule does the enzyme make cAMP from?

  1. A. Glucose
    Glucose is what the liver cell releases at the end of the relay.
    The enzyme makes cAMP from ATP.
  2. B. ✓ ATP
  3. C. Glycogen
    Glycogen is the liver’s store of glucose, broken down at the end of the relay.
    The enzyme makes cAMP from ATP.

Why: cAMP is cyclic AMP.
The enzyme set in the membrane makes cAMP from ATP.

29

The switched-on G protein moves along the inner face of the membrane. It switches on an enzyme set in the membrane.

The switched-on G protein has moved along the inner face and switched on the enzyme set in the membrane
The switched-on G protein has moved along the inner face and switched on the enzyme set in the membrane
30

This enzyme makes cAMP from ATP. Here is the word equation.

The word equation for making cyclic AMP: ATP becomes cyclic AMP and two linked phosphates
The word equation for making cyclic AMP: ATP becomes cyclic AMP and two linked phosphates
31

The enzyme now makes many cAMP molecules, which spread through the cytosol.

The enzyme makes many cAMP molecules from ATP; they spread through the cytosol
The enzyme makes many cAMP molecules from ATP; they spread through the cytosol
32

cAMP switches on kinases, and the kinases switch on the enzymes that break glycogen down. The first glucose leaves the cell several seconds after epinephrine arrived.

cAMP switches on kinases, one drawn as a switched-on kinase with its phosphate; the kinases switch on the enzymes that break glycogen down, and glucose leaves seconds later
cAMP switches on kinases, one drawn as a switched-on kinase with its phosphate; the kinases switch on the enzymes that break glycogen down, and glucose leaves seconds later
33

Glycogen is the liver’s store of glucose. Here is the word equation for breaking it down.

glycogen becomes glucose, many molecules of glucose from one molecule of glycogen
34

Here is a table of the three proteins of the relay and the job each does.

A table of the three proteins of the relay and the job each does: the receptor binds the ligand; the G protein carries the switch across the inner face of the membrane; the enzyme makes cAMP from ATP
35

Now suppose a drug holds the G protein switched off. Epinephrine still binds its receptor, and the receptor still changes shape.

Epinephrine bound to its receptor while the G protein is blocked: the receptor changes shape, but the enzyme stays off and no cAMP is made
Epinephrine bound to its receptor while the G protein is blocked: the receptor changes shape, but the enzyme stays off and no cAMP is made
36

But the G protein is what switches on the enzyme. So the enzyme makes no cAMP, and no glucose leaves.

37

So the block-the-G-protein test shows that the message passes through the G protein on its way to the enzyme.

38

Here are the two doors side by side in one membrane, drawn to the same scale.

The two doors side by side in one membrane, drawn to the same scale: a ligand-gated channel with ions passing, and a G protein-coupled receptor with its G protein, enzyme and cAMP
The two doors side by side in one membrane, drawn to the same scale: a ligand-gated channel with ions passing, and a G protein-coupled receptor with its G protein, enzyme and cAMP
39

A ligand-gated channel is itself the door and opens within a millisecond. A G protein-coupled receptor hands the message to a G protein, an enzyme and cAMP, which takes seconds and multiplies the signal.

40

Some surface receptors are neither a channel nor a G protein-coupled receptor. Their inner part is itself a kinase, and the ligand’s binding switches that kinase on.

41

One simplification: we treat the G protein as switched on by its receptor and switched off again later. How the G protein switches itself off is left out here.

42

What you are expected to know Describe the relay after a G protein-coupled receptor: the G protein switches on an enzyme in the membrane, the enzyme makes many cAMP from ATP, cAMP switches on kinases, and the kinases switch on the enzymes that make the response.

43
Check q11

A researcher gives liver cells epinephrine while a drug holds their G proteins switched off. Epinephrine binds its receptors as usual.

Which of the following happens to the cAMP inside the cells?

  1. A. cAMP rises as usual
    The G protein switches on the enzyme that makes cAMP.
    The drug holds the G protein off, so the enzyme never switches on and no new cAMP is made.
  2. B. cAMP rises, but more slowly
    The G protein is the only link between the receptor and the enzyme.
    With the G protein held off, the enzyme never switches on, so no new cAMP is made.
  3. C. ✓ cAMP stays at its resting level

Why: Epinephrine binds the receptor, which changes shape.
The new shape normally switches on the G protein, but the drug holds the G protein off.
The G protein is what switches on the enzyme that makes cAMP.
So the enzyme stays off, and cAMP stays at its resting level.

44
Practice writing an answer

A researcher gives liver cells epinephrine while a drug holds their G proteins switched off. Epinephrine binds its receptors as usual, but cAMP inside the cells stays at its resting level.

(a) Explain why cAMP stays at its resting level although epinephrine binds its receptors. (1 pt)

Model answer Epinephrine binds the receptor, and the receptor’s intracellular domain changes shape.
The new shape normally switches on the G protein beside it.
But the drug holds the G protein switched off.
The G protein is what switches on the enzyme in the membrane.
So the enzyme stays off.
The enzyme is what makes cAMP from ATP.
So the enzyme makes no new cAMP, and cAMP stays at its resting level.
Rubric
  • Award 1 point for: the G protein carries the switch from the receptor to the enzyme that makes cAMP; with the G protein held off, the enzyme is never switched on, so no cAMP is made even though epinephrine binds and the receptor changes shape.
45
Check q12

A researcher studies three proteins, X, Y and Z, from the membrane of a liver cell. X binds epinephrine, X switches on Y, and Y switches on Z. One of the three makes cAMP from ATP.

Which protein makes the cAMP?

  1. A. X
    X binds epinephrine, so X is the receptor.
    The receptor switches on the G protein; the enzyme that makes cAMP is the last protein in the chain.
  2. B. Y
    Y is switched on by X, the receptor, so Y is the G protein.
    The G protein passes the switch on to the enzyme, Z.
  3. C. ✓ Z

Why: X binds epinephrine, so X is the receptor.
X switches on Y, so Y is the G protein.
Y switches on Z, so Z is the enzyme.
The enzyme is what makes cAMP from ATP.
So Z makes the cAMP.

46
Check q13

A researcher gives a liver cell epinephrine at time zero and records four events over the next seconds.

Which of the following is the order of the four events?

  1. A. cAMP rises → G protein switched on → receptor changes shape → glycogen broken down
    The enzyme makes cAMP.
    The G protein switches the enzyme on.
    The receptor switches the G protein on.
    So the receptor changes shape first, then the G protein, then cAMP.
  2. B. G protein switched on → receptor changes shape → cAMP rises → glycogen broken down
    The G protein waits, switched off, until the receptor’s intracellular domain changes shape.
    So the receptor changes shape first.
  3. C. Receptor changes shape → cAMP rises → G protein switched on → glycogen broken down
    The G protein switches on the enzyme, and the enzyme makes cAMP.
    So the G protein acts before cAMP rises.
  4. D. ✓ Receptor changes shape → G protein switched on → cAMP rises → glycogen broken down

Why: Epinephrine binds, and the receptor’s intracellular domain changes shape.
The new shape switches on the G protein.
The G protein switches on the enzyme, and the enzyme makes cAMP, so cAMP rises.
cAMP switches on kinases, and the kinases switch on the enzymes that break glycogen down.

47

Back to the liver cell beside the blood vessel, where the first glucose left several seconds after epinephrine arrived. Epinephrine bound a receptor with no channel through it, and the receptor’s intracellular domain changed shape.

48

The new shape switched on the helper protein beside it: the G protein. The G protein switched on the enzyme that makes cAMP from ATP.

49

cAMP switched on kinases, and the kinases switched on the enzymes that break glycogen down. Seconds later, and multiplied, the glucose left.

50

When the drug held the G protein switched off, epinephrine still bound and the receptor still changed shape.

51

But the message passes through the G protein on its way to the enzyme. So the cell made no cAMP, and no glucose left.

52Quick quiz: which kind of door? mixed practice

53
Check q14

A neurotransmitter binds a receptor on a nerve cell in the spinal cord. Negative ions flow into the cell within a millisecond.

Which kind of receptor is this?

  1. A. ✓ Ligand-gated channel
  2. B. G protein-coupled receptor
    A G protein-coupled receptor hands the message on to a G protein and an enzyme: a relay of seconds.
    Ions flowing within a millisecond means the receptor is the channel.

Why: Ions flow within a millisecond.
That is too fast for a relay of G protein, enzyme and cAMP.
So the receptor itself is the channel: a ligand-gated channel.

54
Check q15

A hormone binds a receptor on a kidney cell. Over the next few seconds, cAMP rises inside the cell.

Which kind of receptor is this?

  1. A. Ligand-gated channel
    A ligand-gated channel lets ions through and makes no cAMP.
    An enzyme switched on by a G protein makes cAMP, so a rise in cAMP means a G protein-coupled receptor.
  2. B. ✓ G protein-coupled receptor

Why: cAMP rises. cAMP is made by an enzyme, and the enzyme is switched on by a G protein.
So the receptor hands its message to a G protein: a G protein-coupled receptor.

55
Check q16

A ligand binds a receptor on a nerve cell. The receptor’s own shape change closes a channel through it.

Which kind of receptor is this?

  1. A. ✓ Ligand-gated channel
  2. B. G protein-coupled receptor
    The channel passes through the receptor itself.
    A receptor that is itself a channel, opened or closed by its ligand, is a ligand-gated channel.

Why: The channel passes through the receptor itself.
The receptor’s own shape change closes it.
Some ligand-gated channels close when their ligand binds.
So this is a ligand-gated channel.

56
Check q17

A drug holds a heart cell’s G proteins switched off. The heart cell’s response to its hormone stops.

Which kind of receptor does the hormone bind?

  1. A. Ligand-gated channel
    A ligand-gated channel is itself the door: no G protein stands between ligand and ions.
    A response that stops when G proteins are held off needs a G protein.
  2. B. ✓ G protein-coupled receptor

Why: The response stops when the G proteins are held off.
So the response needs a G protein.
A receptor that hands its message to a G protein is a G protein-coupled receptor.

57
Check q18

An odor molecule binds a receptor on a cell in the nose. The receptor switches on a helper protein on the inner face of the membrane.

Which kind of receptor is this?

  1. A. Ligand-gated channel
    A ligand-gated channel moves its own door and switches on nothing.
    A helper protein on the inner face of the membrane, switched on by the receptor, is a G protein.
  2. B. ✓ G protein-coupled receptor

Why: The receptor switches on a helper protein on the inner face of the membrane.
That helper protein is a G protein.
So the receptor is a G protein-coupled receptor.

58
Check q19

A researcher blocks every kinase and every enzyme inside a muscle cell. The cell’s receptor for a neurotransmitter still lets positive ions in within a millisecond.

Which kind of receptor is this?

  1. A. ✓ Ligand-gated channel
  2. B. G protein-coupled receptor
    A G protein-coupled receptor needs an enzyme to make cAMP, and every enzyme is blocked.
    Yet the ions still flow, within a millisecond.
    So the receptor itself is the channel.

Why: Every kinase and every enzyme is blocked, and the ions still flow within a millisecond.
So nothing inside the cell relays the message.
The receptor itself is the channel: a ligand-gated channel.

59Mixed practice mixed practice

60
Check q20

A hormone binds its receptor on a liver cell. A drug holds the cell’s G proteins switched off, and the cell’s cAMP stays at rest.

Which kind of receptor does the hormone bind?

  1. A. Ligand-gated channel
    A ligand-gated channel uses no G protein, so blocking G proteins would not change its response.
    This response needed the G protein.
  2. B. ✓ G protein-coupled receptor

Why: The G proteins were held off, and cAMP stayed at rest.
So the receptor’s message passes through a G protein.
A receptor that hands its message to a G protein is a G protein-coupled receptor.

61
Check q21

In a pathway that passes through a G protein, the receptor changes shape when its ligand binds.

Which component acts first after the receptor changes shape?

  1. A. ✓ The G protein
  2. B. The enzyme that makes cAMP
    The G protein switches the enzyme on, so the G protein acts before the enzyme.
  3. C. cAMP
    The enzyme makes cAMP, two steps after the receptor.

Why: The receptor’s new shape switches on the G protein.
The G protein switches on the enzyme.
The enzyme makes cAMP.
So the G protein acts first.

62
Check q22

Epinephrine binds its receptor on a liver cell, and cAMP rises inside the cell.

Which component makes the cAMP from ATP?

  1. A. The receptor
    The receptor binds epinephrine and changes shape.
    It makes no cAMP.
  2. B. The G protein
    The G protein carries the switch to the enzyme.
    It makes no cAMP.
  3. C. ✓ An enzyme set in the membrane

Why: The G protein switches on a separate enzyme in the membrane.
That enzyme makes cAMP from ATP.

63
Check q23

A drug destroys cAMP inside a heart cell as soon as the cell makes it. Epinephrine binds the cell’s G protein-coupled receptors as usual.

What happens to the cell’s response?

  1. A. It happens as usual
    The response needs cAMP to switch on the kinases.
    With cAMP destroyed at once, nothing switches them on.
  2. B. ✓ It is switched off
  3. C. It is faster than usual
    cAMP is the second messenger.
    Destroying it removes the message; it cannot speed the response.

Why: The receptor and the G protein still switch on the enzyme, so the enzyme makes cAMP.
The drug destroys the cAMP at once.
So no cAMP reaches the kinases, and the response is switched off.

64
Practice writing an answer

A gland cell carries two receptors. One receptor is a ligand-gated channel: when its ligand binds, positive ions flow in within a millisecond. The other receptor is a G protein-coupled receptor: when its ligand binds, cAMP rises over several seconds. A drug holds every G protein in the cell switched off. A researcher then gives the cell both ligands.

(a) Determine which of the two receptors still produces its response with the G proteins held off, and justify your decision. (1 pt)

Model answer The ligand-gated channel still produces its response.
The ligand-gated channel is itself the door, so its ligand’s binding opens the channel with no G protein in between.
So the ions still flow in.
The G protein-coupled receptor passes its message to a G protein.
The G protein is held off, so the enzyme never switches on and makes no cAMP.
So the G protein-coupled receptor produces no response.
Rubric
  • Award 1 point for: the decision (the ligand-gated channel) AND the reasoning it rests on (the channel opens with no G protein in between, while the G protein-coupled receptor needs the G protein to switch on the cAMP-making enzyme).

Slip Deciding without the reasoning, or saying neither receptor responds. The drug touches only the G protein, which the ligand-gated channel never uses.

Glossary

G protein
A helper protein on the inner face of the membrane. It is switched on by a receptor that has changed shape, moves along the inner face, and switches on an enzyme set in the membrane.
G protein-coupled receptor
A cell-surface receptor found in eukaryotic cells that works through a G protein: when its ligand binds, its intracellular domain changes shape and switches on the G protein, which switches on an enzyme in the membrane, which makes many cAMP from ATP.

APBIO-U04-L07 Three kinds of response

Topic 4.2 · Introduction to Signal Transduction · 52 steps

Three cells, each with a signal molecule arriving at a receptor: a gland cell whose vesicle is fusing with its membrane and releasing insulin outside; a liver cell with a nucleus and new enzyme molecules inside; a skin cell at a cut, drawn as one cell becoming two
Three cells, each with a signal molecule arriving at a receptor: a gland cell whose vesicle is fusing with its membrane and releasing insulin outside; a liver cell with a nucleus and new enzyme molecules inside; a skin cell at a cut, drawn as one cell becoming two

Here are three cells, each receiving a signal: a gland cell, a liver cell and a skin cell at a cut.

Within two seconds, the gland cell’s stored vesicles fuse with its membrane and release insulin. Over an hour, the liver cell makes an enzyme it did not have before. A day after damaged tissue signals it, the skin cell at the cut grows and splits in two.

A pathway ends in what the cell does differently. How many kinds of ending are there?

Unit 4 · Cell Communication and Cell Cycle

1Three kinds of response

2

Video: Watch: Three kinds of response

A gland cell releases insulin from vesicles; a liver cell makes a new enzyme; a skin cell at a cut splits in two. Three kinds of response come up again and again: secretion, a change in gene expression, and growth and division.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L07a.mp4

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3

What can a cell do at the end of a pathway?

4

A cell can do one of three common kinds of response.

5

The cell can send a molecule out. Vesicles fuse with the membrane and release the molecule outside the cell.

6

The cell can change which genes it expresses. So it starts or stops making particular proteins.

7

Or the cell can grow and divide.

8

Three kinds of response come up again and again. When a response is one of these three, name which kind it is.

9
Check q1

A vesicle inside a gland cell holds insulin. The vesicle moves to the plasma membrane.

In exocytosis, what happens next?

  1. A. ✓ The vesicle fuses with the plasma membrane and releases the insulin outside the cell
  2. B. The vesicle breaks open inside the cell and the insulin stays in the cytosol
    In exocytosis the vesicle’s membrane joins the plasma membrane.
    So the insulin inside the vesicle ends up outside the cell.

Why: The vesicle fuses with the plasma membrane.
Its membrane becomes part of the plasma membrane.
So its contents, the insulin, are released outside the cell.

10

A pathway ends in what the cell does differently. That last stage is the cellular response.

11

Here is a drawing of three common kinds of cellular response.

Three kinds of cellular response: vesicles releasing their contents, new proteins being made from genes, and one cell growing and splitting into two
Three kinds of cellular response: vesicles releasing their contents, new proteins being made from genes, and one cell growing and splitting into two
12

Suppose a gland cell’s vesicles fuse with its membrane and release insulin outside the cell. This response is , because the cell sent a molecule out.

13

Now suppose a liver cell makes an enzyme it did not have before. This response is a change in gene expression, because the cell started making a new protein.

14

Now suppose a skin cell at a cut grows and splits in two. This response is growth and division, because one cell became two.

15

Now suppose a gland cell makes more insulin than before and packs it into new vesicles, and nothing leaves the cell. This response is not secretion, because nothing has left the cell.

16

This response is a change in gene expression, because the cell is making more of a protein.

17

Here is a table comparing the three kinds of response: what the cell does in each, and one example.

A table of three common kinds of cellular response, what the cell does in each, and one example: secretion, vesicles fuse with the membrane and release a molecule outside the cell, a gland cell releasing insulin; a change in gene expression, the cell starts, stops, speeds or slows making a protein, a liver cell making a new enzyme; growth and division, the cell grows and splits into two, a skin cell at a cut dividing
18

What you are expected to know Classify a described cellular response as secretion, as a change in gene expression, or as growth and division.

19Quick quiz: which kind of response? mixed practice

20
Check q2

Three hours after a signal arrives, a kidney cell holds a transporter protein that was absent before.

Which kind of response is this?

  1. A. Secretion
    The transporter appeared inside the cell, and the cell never held it before.
    So the cell has started making it from its gene.
    That is a change in gene expression.
  2. B. ✓ A change in gene expression
  3. C. Growth and division
    The cell has not split.
    The cell holds a protein it did not hold before, so the cell has started expressing that protein’s gene.

Why: The cell holds a protein it never held before.
So the cell has started expressing that protein’s gene.
That is a change in gene expression.

21
Check q3

Within a second of its signal arriving, an immune cell releases a stored chemical from vesicles packed beside its membrane.

Which kind of response is this?

  1. A. ✓ Secretion
  2. B. A change in gene expression
    The chemical was already stored in vesicles.
    The vesicles fused with the membrane and sent the chemical out.
    That is secretion.
  3. C. Growth and division
    The cell did not grow or split.
    Vesicles fused with the membrane and sent a stored chemical out.
    That is secretion.

Why: Vesicles fused with the membrane and sent a stored chemical out of the cell.
A cell sending a molecule out is secretion.

22
Check q4

Two days after a signal reaches them, cells lining a healing gut have doubled in number.

Which kind of response is this?

  1. A. Secretion
    Nothing left the cells.
    The cells grew and split, so there are twice as many.
    That is growth and division.
  2. B. A change in gene expression
    The cells doubled in number, so each cell grew and split into two.
    That is growth and division.
  3. C. ✓ Growth and division

Why: The cells have doubled in number.
So each cell grew and split into two.
That is growth and division.

23
Check q5

Five hours after a hormone arrives, a muscle cell has tripled its output of a fuel-burning enzyme.

Which kind of response is this?

  1. A. Secretion
    The enzyme stays inside the muscle cell.
    The cell is making three times as much of it, so the cell has changed how fast it expresses the enzyme’s gene.
  2. B. ✓ A change in gene expression
  3. C. Growth and division
    The cell has not split.
    The cell is making three times as much of one enzyme, so the cell has changed how fast it expresses that enzyme’s gene.

Why: The cell now makes three times as much of one enzyme.
Making more of a protein means expressing its gene faster.
That is a change in gene expression.

24
Check q6

A pancreas cell receives a signal. Vesicles packed beside its membrane fuse with it and release digestive enzymes into a duct.

Which kind of response is this?

  1. A. ✓ Secretion
  2. B. A change in gene expression
    The enzymes were already made and packed in vesicles.
    The vesicles fused with the membrane and sent the enzymes out.
    That is secretion.
  3. C. Growth and division
    The cell did not grow or split.
    Vesicles fused with the membrane and sent the enzymes out of the cell.
    That is secretion.

Why: Vesicles fused with the membrane and released the enzymes outside the cell.
A cell sending a molecule out is secretion.

25
Check q7

Two days after damaged tissue signals them, bone cells near a fracture have begun dividing.

Which kind of response is this?

  1. A. Secretion
    Nothing left the bone cells.
    The bone cells began splitting into two.
    That is growth and division.
  2. B. A change in gene expression
    The bone cells began splitting into two.
    That is growth and division.
  3. C. ✓ Growth and division

Why: The bone cells began dividing.
A cell growing and splitting into two is growth and division.

26A signal to grow and divide

27

Video: Watch: A signal to grow and divide

Damaged tissue at a cut releases molecules; the molecules bind receptors on nearby skin cells; a day later those cells grow and divide. A signal whose message is grow and divide is called a growth factor.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L07b.mp4

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28

Now consider the skin cell at the cut again. Damaged tissue around the cut releases molecules.

Three kinds of cellular response: vesicles releasing their contents, new proteins being made from genes, and one cell growing and splitting into two
Three kinds of cellular response: vesicles releasing their contents, new proteins being made from genes, and one cell growing and splitting into two
29

The molecules bind receptors on the skin cells nearby. A day later, those skin cells grow and divide.

30

A signal that tells a cell to grow and divide is called a .

31

There are many kinds of growth factor. What they share is the message: grow and divide.

32

A growth factor is the signal, not the response. The response is the cell growing and dividing.

33

What you are expected to know Identify a growth factor: a signal whose message to the cell is grow and divide.

34
Check q8

A surgeon removes part of a person’s liver. Damaged tissue at the cut edge releases molecules, and two days later liver cells near the edge begin dividing.

Which of the following are the released molecules?

  1. A. ✓ Growth factors
  2. B. Neurotransmitters
    A neurotransmitter is released by a nerve cell onto the cell beside it.
    These molecules come from damaged tissue and tell nearby cells to divide: a growth factor’s job.
  3. C. Second messengers
    A second messenger is made inside a cell and stays inside it.
    These molecules leave the damaged tissue and bind receptors on other cells: a growth factor’s job.

Why: Damaged tissue released the molecules, and the molecules told nearby cells to grow and divide.
A signal that tells a cell to grow and divide is called a growth factor, whatever its exact kind.

35
Check q9

A researcher adds a growth factor to one dish of resting connective-tissue cells and gives a second dish plain medium.

Predict the number of cells in the first dish after two days, compared with the second dish.

  1. A. Fewer cells
    A growth factor tells a cell to grow and divide, so each cell becomes two.
    The count rises.
  2. B. The same number of cells
    A growth factor’s message is grow and divide.
    In two days the first dish’s cells have divided, so it holds more cells.
  3. C. ✓ More cells

Why: A growth factor is a signal to grow and divide.
The cells in the first dish received it, so they grew and divided.
The cells in the second dish received none, so they did not.
So the first dish holds more cells.

36

Here again are the three cells: a gland cell, a liver cell and a skin cell at a cut, each receiving a signal.

37

The gland cell’s vesicles fused with its membrane and released insulin. That response is secretion.

38

The liver cell made an enzyme it did not have before. That response is a change in gene expression.

39

The skin cell at the cut grew and split in two. That response is growth and division, and the signal from the damaged tissue was a growth factor.

40

Three cells, three endings, and each ending is one of the three common kinds.

41Quick quiz: growth factor mixed practice

42
Check q10

Which of the following is a growth factor?

  1. A. A signal that tells a cell to release a stored molecule
    A signal that makes a cell release a stored molecule ends in secretion.
    A growth factor’s message is grow and divide.
  2. B. ✓ A signal that tells a cell to grow and divide
  3. C. A protein a cell makes while it grows
    A growth factor is a signal from outside the cell.
    A protein the cell makes while growing is part of the response, not the signal.

Why: A growth factor is a signal.
Its message to the cell is grow and divide.

43
Practice writing an answer

A growth factor reaches a cell.

(a) State what a growth factor is. (1 pt)

Model answer A growth factor is a signal that tells a cell to grow and divide.
Rubric
  • Award 1 point for: a signal (molecule) that tells a cell to grow and divide.

44Mixed practice mixed practice

45
Check q11

Within a second of its signal arriving, a nerve ending’s vesicles fuse with its membrane and release a neurotransmitter onto the muscle cell beside it.

Which kind of response is this?

  1. A. ✓ Secretion
  2. B. A change in gene expression
    The neurotransmitter was already packed in vesicles.
    The vesicles fused with the membrane and sent it out.
    That is secretion.
  3. C. Growth and division
    The nerve ending did not grow or split.
    Its vesicles sent a molecule out.
    That is secretion.

Why: The vesicles fused with the membrane and released the neurotransmitter outside the nerve ending.
A cell sending a molecule out is secretion.

46
Check q12

Six hours after a signal arrives, a white blood cell holds a protein it did not hold before. The protein stays inside the cell.

Which kind of response is this?

  1. A. Secretion
    The protein stays inside the cell, so nothing was sent out.
    Starting to make a new protein is a change in gene expression.
  2. B. ✓ A change in gene expression
  3. C. Growth and division
    The cell has not split.
    The cell has started making a protein it did not make before.
    That is a change in gene expression.

Why: The cell is making a protein it did not make before.
So the cell has started expressing that protein’s gene.
That is a change in gene expression.

47
Check q13

A researcher gives a root cell a plant signal. Two days later there are four cells where there was one.

Which kind of response is this?

  1. A. Secretion
    Nothing left the root cell.
    One cell became four, so the cell grew and split, and its daughters split again.
  2. B. A change in gene expression
    One cell became four.
    A cell growing and splitting into two is growth and division.
  3. C. ✓ Growth and division

Why: One cell became four.
So the cell grew and split into two, and each of those cells grew and split again.
That is growth and division.

48
Check q14

A growth factor binds receptors on a connective-tissue cell.

Which of the following is the cell’s response?

  1. A. The cell releases the growth factor
    The growth factor came from outside the cell and bound the cell’s receptors.
    The cell’s response is what the cell does next: it grows and divides.
  2. B. ✓ The cell grows and divides
  3. C. The cell makes more growth factor
    The growth factor is the signal, not the response.
    The message it carries is grow and divide, so the response is the cell growing and dividing.

Why: A growth factor is a signal whose message is grow and divide.
The signal bound the cell’s receptors.
So the cell’s response is to grow and divide.

49
Check q15

Researchers find a molecule released by damaged kidney tissue. Kidney cells given the molecule double in number in two days.

Which of the following names the molecule?

  1. A. A second messenger
    A second messenger is made inside a cell and stays inside it.
    This molecule is released by tissue and acts on other cells.
  2. B. A neurotransmitter
    A neurotransmitter is released by a nerve cell onto the cell beside it.
    This molecule comes from damaged tissue and makes cells divide.
  3. C. ✓ A growth factor

Why: The molecule came from damaged tissue and made kidney cells grow and divide.
A signal whose message is grow and divide is a growth factor.

50
Check q16

Four hours after a hormone arrives, a gland cell is making twice as much of an enzyme it already made.

Which kind of response is this?

  1. A. Secretion
    The enzyme stays inside the gland cell.
    The cell is making twice as much of it, so the cell has sped up expressing the enzyme’s gene.
  2. B. ✓ A change in gene expression
  3. C. Growth and division
    The cell has not split.
    The cell is making twice as much of one enzyme, so the cell has sped up expressing that enzyme’s gene.

Why: The cell now makes twice as much of one enzyme.
Making more of a protein means expressing its gene faster.
That is a change in gene expression.

51
Practice writing an answer

A researcher adds a molecule released by damaged skin to a dish of resting skin cells. A second dish of the same cells gets plain medium. After two days the first dish holds 2,100 cells per square millimetre and the second dish holds 1,000 cells per square millimetre.

(a) Explain how this result demonstrates that the molecule is a growth factor. (2 pt)

Model answer A growth factor is a signal whose message to a cell is grow and divide.
The cells given the molecule doubled in number in two days, from 1,000 to 2,100 cells per square millimetre.
So each cell grew and split into two.
The cells given plain medium did not divide.
So the molecule is what made the cells grow and divide.
Therefore the molecule carries the message grow and divide, and that makes it a growth factor.
Rubric
  • Award 1 point for: the cells given the molecule grew and divided (doubled in number), while the cells given plain medium did not, so the molecule caused the growth and division.
  • Award 1 point for: a signal whose message to the cell is grow and divide is a growth factor, so the molecule is one.

Glossary

secretion
A cell sending a molecule out: vesicles fuse with the plasma membrane and release the molecule outside the cell. A gland cell releasing insulin is secretion.
growth factor
A signal that tells a cell to grow and divide. Damaged tissue at a wound releases growth factors, and the cells around it begin dividing about a day later. There are many kinds; what they share is the message.

APBIO-U04-L07B Same signal, different cells

Topic 4.2 · Introduction to Signal Transduction · 74 steps

A blood vessel carrying epinephrine, with a heart muscle cell below it on the left and a liver cell below it on the right
A blood vessel carrying epinephrine, with a heart muscle cell below it on the left and a liver cell below it on the right

Here are two of your cells in the same second: a heart muscle cell and a liver cell, with epinephrine arriving at both from the blood.

Your heart beats faster and harder. Your liver cells break stored glycogen down and release glucose. Both cell types used the same kind of receptor and the same rise in cAMP.

How can one molecule mean two different things?

Unit 4 · Cell Communication and Cell Cycle

1Same signal, different cells

2

Video: Watch: Same signal, different cells

Epinephrine reaches a heart muscle cell and a liver cell. Both carry the same kind of receptor, and cAMP rises in both. The heart cell’s kinases reach its contraction proteins; the liver cell’s kinases reach its glycogen-breaking enzymes. The cell decides what it does with the message.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L07Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L07Ba.mp4

3

Why does one signal do different things in different cells?

4

The difference is not in the signal, and not in the receptor.

5

The difference is in the proteins each cell holds after the cAMP rises.

6

The heart cell’s kinases reach its contraction proteins. The liver cell’s kinases reach its glycogen-breaking enzymes.

7

The cell decides what it does with a message.

8

You can draw a pathway as boxes and arrows.

9

Then you can read it link by link: where the message crosses the membrane, which two steps multiply it, and which box each measurement belongs to.

10
Check q1

A fat cell and a heart muscle cell carry the same DNA, yet they hold different proteins.

Why do the two cells hold different proteins?

  1. A. ✓ The two cells express different genes
  2. B. The two cells carry different genes
    Every cell of the body carries the same DNA and so the same genes.
    The two cells differ in which of those genes they express.

Why: The two cells hold different proteins because they express different genes.

11

Epinephrine in the blood reaches a heart muscle cell and a liver cell in the same second. The molecule is the same.

The epinephrine pathway drawn as boxes and arrows in a heart muscle cell and a liver cell: identical from epinephrine through receptor, G protein and enzyme to cAMP
The epinephrine pathway drawn as boxes and arrows in a heart muscle cell and a liver cell: identical from epinephrine through receptor, G protein and enzyme to cAMP
12

Both cells carry the same kind of receptor for epinephrine: a G protein-coupled receptor. In both, the G protein, the enzyme and the rise in cAMP follow.

13

Then the paths part. In the heart cell, the kinases that cAMP switches on reach the cell’s contraction proteins, and the cell contracts harder.

The two pathways continue: the same kinases, then contraction proteins and a harder contraction in the heart cell, glycogen-breaking enzymes and glucose leaving in the liver cell
The two pathways continue: the same kinases, then contraction proteins and a harder contraction in the heart cell, glycogen-breaking enzymes and glucose leaving in the liver cell
14

In the liver cell, the same kinases reach the enzymes that break glycogen down, and glucose leaves the cell.

15

The difference is not in the signal. The difference is in the proteins each cell holds.

16

The liver cell holds glycogen-breaking enzymes. The heart cell holds contraction proteins.

17

So the same ligand, through the same kind of receptor, ends at whatever proteins that cell carries. The cell decides what it does with the message.

18

A cell that carries no epinephrine receptor does nothing when epinephrine reaches it.

19

What you are expected to know Explain why one signal produces different responses in different cells: the response depends on which receptor, which relay proteins and which proteins for the kinases to act on each cell contains.

20
Check q2

Hormone H binds the same kind of G protein-coupled receptor on a fat cell and on a heart muscle cell, and cAMP rises in both. The fat cell releases fatty acids; the heart cell contracts harder.

Which of the following is the first point at which the two pathways differ?

  1. A. ✓ At the proteins the kinases act on
  2. B. At the G protein on the inner face
    The enzyme makes cAMP only after the G protein switches it on.
    So the G protein worked the same way in both cells.
  3. C. At the second messenger, cAMP
    The same second messenger, cAMP, rose in both cells.
    So the pathways still match at cAMP.
    The paths part after cAMP, at the proteins the kinases reach.
  4. D. At the receptor that binds hormone H
    Both cells carry the same kind of receptor, and hormone H binds both the same way.
    So the pathways match at the receptor.

Why: Both cells carry the same receptor, and cAMP rises in both.
In both cells cAMP switches on kinases.
The kinases reach the proteins each cell holds: fat-releasing enzymes in one, contraction proteins in the other.
So the pathways first differ at the proteins the kinases act on.

21
Practice writing an answer

Hormone H binds the same kind of G protein-coupled receptor on a fat cell and on a heart muscle cell, and cAMP rises in both. The fat cell releases fatty acids; the heart cell contracts harder.

(a) Explain why the same hormone makes the fat cell release fatty acids and the heart cell contract harder. (1 pt)

Model answer Both cells carry the same kind of receptor, so hormone H binds both the same way.
In both cells the receptor switches on a G protein, the enzyme makes cAMP, and cAMP switches on kinases.
The fat cell holds fat-releasing enzymes, so its kinases switch those on and fatty acids leave.
The heart cell holds contraction proteins, so its kinases act on those and it contracts harder.
The two cells hold different proteins because they express different genes.
Rubric
  • Award 1 point for: the pathway is the same as far as the kinases, and the kinases then act on whatever response proteins each cell holds (fatty-acid-releasing enzymes in the fat cell, contraction proteins in the heart cell), because the two cells express different genes.
22
Check q3

Priya says: “The heart cell must receive a different form of hormone H from the fat cell.”

Is Priya correct?

  1. A. Yes — each cell type receives its own form of hormone H
    The blood carries one hormone H to both cells, and hormone H stays outside both cells.
    The difference is inside the cells, in the proteins the kinases reach.
  2. B. ✓ No — the same hormone H reaches both cells

Why: The blood carries one molecule, hormone H, to both cells, and both cells bind it with the same kind of receptor.
The difference is inside the cells: the fat cell holds fatty-acid-releasing enzymes, and the heart cell holds contraction proteins.

23
Check q4

A researcher gives two cell types the same hormone. Both cell types carry its receptor, cAMP rises in both, and the kinase is switched on in both. Cell type P releases a stored product; cell type Q shows no change at all.

Which difference between P and Q explains the result?

  1. A. Q’s receptor binds the hormone more weakly
    Q’s cAMP rose, so the hormone bound Q’s receptor and the receptor worked.
    The pathway stops later than binding.
  2. B. Q’s G protein stays switched off
    The G protein switches on the enzyme that makes cAMP.
    Q’s cAMP rose.
    So Q’s G protein worked.
  3. C. Q breaks its cAMP down before it can act
    Q’s kinase was switched on, so Q’s cAMP acted before it was broken down.
    What Q lacks is a protein for the pathway to end at.
  4. D. ✓ Q holds no protein for the kinase to act on

Why: cAMP rose and the kinase was switched on in both cell types, so the receptor, the G protein, the enzyme and the kinase worked in both.
Q shows no response because Q holds no protein for the kinase to switch on.
The pathway has nowhere to end.

24Read the model, link by link

25

Video: Watch: Read the model, link by link

The epinephrine pathway of a liver cell drawn as boxes and arrows: epinephrine the ligand, the receptor across the membrane where the message crosses, the G protein and the enzyme along the inner face, cAMP the second messenger, the kinases, and the glycogen-breaking enzymes that end the pathway.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L07Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L07Bb.mp4

26

Here is the epinephrine pathway of a liver cell drawn as a model: each component a box, each step an arrow.

The epinephrine pathway of a liver cell drawn as a model: each component a box, each step an arrow, from epinephrine outside the cell to glucose leaving it
The epinephrine pathway of a liver cell drawn as a model: each component a box, each step an arrow, from epinephrine outside the cell to glucose leaving it
27

Read it link by link. Epinephrine is the ligand.

28

The receptor’s binding site holds the epinephrine. So the receptor’s intracellular domain changes shape.

The same model marked where the message crosses the membrane: at the receptor
The same model marked where the message crosses the membrane: at the receptor
29

That shape change is the message crossing the membrane. The epinephrine itself stays outside.

30

The G protein and the enzyme relay the message along the membrane.

31

cAMP is the second messenger. It spreads through the cytosol and switches on the kinases.

32

The kinases carry the message on to the glycogen-breaking enzymes. Those enzymes are the response proteins: the pathway ends at them.

33

What you are expected to know Name the role of each component on a drawn pathway model, in order: ligand, receptor and its domains, G protein or channel, second messenger, kinases, response protein.

34

What you are expected to know Say where the message crosses the membrane on a drawn pathway model.

35
Check q5

Here is a different pathway, for hormone P, drawn as a model with its boxes numbered 1 to 7.

A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface
A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface

At which numbered box does the message cross the membrane?

  1. A. Box 1
    Hormone P stays outside the cell.
    The message crosses when the receptor’s intracellular domain changes shape.
  2. B. ✓ Box 2
  3. C. Box 3
    The G protein is already on the inner face of the membrane.
    The message reached the G protein because the receptor changed shape.
  4. D. Box 5
    cAMP at box 5 is made inside the cell, after the message has already crossed.
    The crossing is the receptor’s intracellular domain changing shape.

Why: Hormone P binds the receptor’s outside, and the receptor’s intracellular domain changes shape.
That shape change is the message crossing the membrane: box 2.

36
Check q6

Here is the hormone P pathway drawn as a model with its boxes numbered 1 to 7.

A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface
A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface

Which box is the second messenger?

  1. A. Box 3
    The G protein is a protein on the inner face of the membrane.
    A second messenger is a small molecule made inside the cell.
  2. B. Box 4
    The enzyme makes the second messenger.
    The enzyme is not the messenger itself.
  3. C. ✓ Box 5
  4. D. Box 6
    The kinase is a protein that the second messenger switches on.
    The kinase is not the messenger.

Why: cAMP, at box 5, is the second messenger.
It is a small molecule made inside the cell when the receptor is activated.
cAMP spreads through the cytosol and switches on the kinase.

37Place a measurement on the model

38

Video: Watch: Place a measurement on the model

On the liver cell’s model two steps multiply the message, the enzyme making many cAMP and each kinase acting on many proteins; a measurement belongs to one box, so ‘cAMP rises’ is transduction and ‘glucose leaves’ is the response; and a drug that stops one box leaves every box before it working and every box after it quiet.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L07Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L07Bc.mp4

39
Check q7

An enzyme in a cell is switched on.

How many molecules does it act on before it is switched off?

  1. A. ✓ Many molecules, one after another
  2. B. One molecule only
    An enzyme is not used up when it acts.
    It acts on one molecule, lets it go, and acts on the next.

Why: An enzyme is not used up when it acts.
So one switched-on enzyme acts on many molecules, one after another, until it is switched off.

40

Two steps on the model multiply the message.

The same model with the two multiplying steps marked, the enzyme making many cAMP and each kinase acting on many proteins, and three measurements placed on it: epinephrine bound at the receptor is reception, cAMP rising is transduction, glucose leaves is the response
The same model with the two multiplying steps marked, the enzyme making many cAMP and each kinase acting on many proteins, and three measurements placed on it: epinephrine bound at the receptor is reception, cAMP rising is transduction, glucose leaves is the response
41

One enzyme makes many cAMP. Each kinase phosphorylates many proteins.

42

This multiplying is amplification. A few bound receptors end in a very large response.

43

A measurement belongs to one box. ‘Epinephrine bound at the surface’ is reception.

44

‘cAMP rises’ is transduction. ‘Glucose leaves the cell’ is the response.

45

Now suppose a drug stops one box working.

46

Every box before it still works. Every box after it goes quiet.

47

So the measurements tell you which box the drug acts at: the last box that still works is just before it.

48

What you are expected to know Name the two steps that amplify the message on a drawn pathway model.

49

What you are expected to know Place a described measurement on the model: the box it belongs to, and the stage that box is in.

50

What you are expected to know Work out which box a drug acts at from which measurements still change and which do not.

51
Check q8

In the hormone P model, one bound receptor ends in thousands of transporters reaching the surface.

A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface
A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface

Which two boxes hold the enzymes that multiply the message?

  1. A. Boxes 1 and 2
    One hormone molecule binds one receptor, one to one.
    Nothing is multiplied there.
  2. B. Boxes 2 and 3
    A G protein carries the message on to one enzyme.
    The big multiplications are the enzyme steps: one enzyme makes many cAMP, and one kinase phosphorylates many proteins.
  3. C. ✓ Boxes 4 and 6
  4. D. Boxes 5 and 7
    cAMP at box 5 is a small molecule, not an enzyme.
    The transporters at box 7 are the response.
    Neither acts on many molecules, so neither multiplies the message.

Why: Enzymes multiply.
The enzyme at box 4 makes many cAMP from ATP, and the kinase at box 6 phosphorylates many proteins.
So a few bound receptors end in thousands of transporters.

52
Check q9

A researcher gives cells with the hormone P pathway hormone P together with a drug. Hormone P binds its receptors as normal, and cAMP rises as normal. The proteins that the kinase normally phosphorylates gain no phosphate, and the transporters stay inside the cell.

A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface
A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface

At which box does the drug act?

  1. A. At box 3, the G protein
    cAMP rose as normal, and the enzyme makes cAMP only after the G protein switches it on.
    So the G protein worked.
  2. B. At box 4, the enzyme
    The enzyme makes cAMP, and cAMP rose as normal.
    So the enzyme worked.
  3. C. ✓ At box 6, the kinase
  4. D. At box 7, the transporters
    The kinase’s target proteins gained no phosphate, so the kinase never acted.
    The transporters stayed inside because the step before them failed, not because the drug hit them.

Why: Hormone P bound and cAMP rose, so boxes 1 to 5 worked.
The kinase’s target proteins gained no phosphate, so the kinase at box 6 never acted.
The transporters at box 7 stayed inside because the step before them failed.
So the drug acts at box 6.

53

Here again are the two cells: a heart muscle cell and a liver cell, with epinephrine arriving at both from the blood in the same second.

54

Both cells bound the same molecule with the same kind of receptor. In both, the same cAMP rose.

55

The heart cell’s kinases reached its contraction proteins, so the heart cell contracted harder.

56

The liver cell’s kinases reached its glycogen-breaking enzymes, so glucose left the liver cell.

57

One molecule meant two different things because the two cells hold different proteins.

58Quick quiz: read the model mixed practice

59
Check q10

Here is the pathway for hormone Q drawn as a model with its boxes numbered 1 to 7.

A model of the hormone Q pathway drawn as boxes numbered 1 to 7: hormone Q, receptor across the membrane, G protein, enzyme, cAMP, kinase, contraction proteins
A model of the hormone Q pathway drawn as boxes numbered 1 to 7: hormone Q, receptor across the membrane, G protein, enzyme, cAMP, kinase, contraction proteins

Which box is the ligand?

  1. A. ✓ Box 1
  2. B. Box 2
    The receptor at box 2 binds the ligand.
    The ligand is the molecule that arrives from outside: hormone Q.
  3. C. Box 5
    cAMP at box 5 is made inside the cell.
    The ligand is the molecule that arrives from outside: hormone Q.

Why: The ligand is the signal molecule that binds the receptor.
Hormone Q, at box 1, arrives from outside and binds the receptor.

60
Check q11

Here is the hormone Q pathway drawn as a model with its boxes numbered 1 to 7.

A model of the hormone Q pathway drawn as boxes numbered 1 to 7: hormone Q, receptor across the membrane, G protein, enzyme, cAMP, kinase, contraction proteins
A model of the hormone Q pathway drawn as boxes numbered 1 to 7: hormone Q, receptor across the membrane, G protein, enzyme, cAMP, kinase, contraction proteins

At which box does the message cross the membrane?

  1. A. Box 1
    Hormone Q stays outside the cell.
    The message crosses when the receptor’s intracellular domain changes shape.
  2. B. ✓ Box 2
  3. C. Box 3
    The G protein sits on the inner face of the membrane.
    The message had already crossed when the receptor changed shape.

Why: Hormone Q binds the receptor’s outside, and the receptor’s intracellular domain changes shape.
That shape change is the message crossing the membrane: box 2.

61
Check q12

Here is the hormone Q pathway drawn as a model with its boxes numbered 1 to 7.

A model of the hormone Q pathway drawn as boxes numbered 1 to 7: hormone Q, receptor across the membrane, G protein, enzyme, cAMP, kinase, contraction proteins
A model of the hormone Q pathway drawn as boxes numbered 1 to 7: hormone Q, receptor across the membrane, G protein, enzyme, cAMP, kinase, contraction proteins

Which box makes the second messenger?

  1. A. Box 3
    The G protein switches the enzyme on.
    The enzyme, not the G protein, makes cAMP.
  2. B. ✓ Box 4
  3. C. Box 5
    cAMP at box 5 is the second messenger itself.
    The enzyme at box 4 makes it.

Why: The enzyme at box 4 makes cAMP from ATP.
cAMP is the second messenger, so box 4 makes the second messenger.

62
Check q13

Here is the hormone Q pathway drawn as a model with its boxes numbered 1 to 7.

A model of the hormone Q pathway drawn as boxes numbered 1 to 7: hormone Q, receptor across the membrane, G protein, enzyme, cAMP, kinase, contraction proteins
A model of the hormone Q pathway drawn as boxes numbered 1 to 7: hormone Q, receptor across the membrane, G protein, enzyme, cAMP, kinase, contraction proteins

Which box is the response protein?

  1. A. Box 4
    The enzyme at box 4 makes cAMP.
    The pathway ends at the contraction proteins, box 7.
  2. B. Box 6
    The kinase at box 6 phosphorylates the contraction proteins.
    The pathway ends at the contraction proteins, box 7.
  3. C. ✓ Box 7

Why: The pathway ends at the proteins that make the cell respond.
The contraction proteins at box 7 make the cell contract, so box 7 is the response protein.

63
Check q14

In cells with the hormone Q pathway, a researcher measures a rise in cAMP.

A model of the hormone Q pathway drawn as boxes numbered 1 to 7: hormone Q, receptor across the membrane, G protein, enzyme, cAMP, kinase, contraction proteins
A model of the hormone Q pathway drawn as boxes numbered 1 to 7: hormone Q, receptor across the membrane, G protein, enzyme, cAMP, kinase, contraction proteins

Which stage of the pathway is this measurement?

  1. A. Reception
    Reception is hormone Q binding its receptor.
    cAMP is made inside the cell, after that.
  2. B. ✓ Transduction
  3. C. The cellular response
    The response is the cell contracting.
    cAMP rising is the message being relayed inside the cell.

Why: cAMP at box 5 is made inside the cell and carries the message on.
So a rise in cAMP is transduction.

64
Check q15

In cells with the hormone Q pathway, a researcher gives hormone Q and a drug together. Hormone Q binds as normal, and cAMP stays at its resting concentration.

A model of the hormone Q pathway drawn as boxes numbered 1 to 7: hormone Q, receptor across the membrane, G protein, enzyme, cAMP, kinase, contraction proteins
A model of the hormone Q pathway drawn as boxes numbered 1 to 7: hormone Q, receptor across the membrane, G protein, enzyme, cAMP, kinase, contraction proteins

Between which two boxes does the drug act?

  1. A. ✓ Between boxes 2 and 5
  2. B. Between boxes 5 and 6
    A block after cAMP would leave cAMP rising as normal.
    cAMP never rose, so the block is before cAMP.
  3. C. Between boxes 6 and 7
    A block after the kinase would leave cAMP rising as normal.
    cAMP never rose, so the block is before cAMP.

Why: Hormone Q bound as normal, so boxes 1 and 2 worked.
cAMP at box 5 never rose, so no cAMP was made.
So the drug acts after the receptor and before cAMP: at the G protein or the enzyme, between boxes 2 and 5.

65Mixed practice mixed practice

66
Check q16

A researcher adds hormone P to cells at time zero and follows two measurements: cAMP inside the cells, and transporters at the surface.

A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface
A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface

Which of the following changes first?

  1. A. The transporters at the surface
    The kinase waits, switched off, until cAMP switches it on.
    So cAMP has to be made first.
    Then the kinase acts.
    Then the transporters move.
  2. B. cAMP and the transporters change at the same instant
    The transporters move only after the kinase has phosphorylated its targets.
    Each box acts on the next, so cAMP rises first.
  3. C. ✓ The cAMP inside the cells

Why: Each component of the pathway switches on the next, in order.
The enzyme makes cAMP within seconds.
Then cAMP switches on the kinase.
Only after the kinase has acted do the transporters move.
So cAMP rises first.

67
Check q17

In cells with the hormone P pathway, a researcher adds hormone P and then counts the phosphate groups on the kinase’s target proteins.

A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface
A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface

Which stage of the pathway is this measurement?

  1. A. Reception
    Reception is the ligand binding its receptor.
    Phosphates added inside the cell come later, in the relay.
  2. B. ✓ Transduction
  3. C. The cellular response
    The response is what the cell finally does: the transporters reach the surface.
    The phosphorylations are the relay that gets the cell there.

Why: The kinase adds phosphates to its target proteins inside the cell.
That is the message being relayed: transduction.
The response is what those proteins then make the cell do.

68
Check q18

A cell type carries none of hormone P’s receptor. A researcher adds hormone P to a dish of these cells at ten times its normal concentration.

Predict the cells’ response.

  1. A. A response once enough hormone P is added
    With no receptor, nothing binds the hormone, however much arrives.
    So nothing inside the cell changes.
  2. B. ✓ No response at any concentration
  3. C. A slow response, through gene expression only
    Every response, fast or slow, begins with the ligand binding its receptor.
    These cells have no receptor for hormone P, so no response begins.
  4. D. A response through the G protein, bypassing the receptor
    The G protein waits, switched off, until a receptor changes shape.
    With no receptor, nothing switches the G protein on.

Why: A cell with no receptor for a signal does not respond, however much signal reaches it.
Nothing binds the hormone, so nothing inside the cell changes.

69
Check q19

Two cell types, a kidney cell and a gland cell, both bind hormone P and both raise cAMP. The kidney cell moves transporters to its surface; the gland cell releases stored granules. Both cell types use the same kind of kinase.

A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface
A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface

At which numbered box do the two pathways first differ?

  1. A. Box 2
    Both cell types bind hormone P, so both receptors work the same way.
  2. B. Box 3
    Both cell types raise cAMP, and the G protein is what leads to cAMP being made, so both G proteins worked the same way.
  3. C. Box 5
    Both cell types raised cAMP, so both use the same second messenger.
  4. D. ✓ Box 7

Why: Both cell types match through binding, cAMP and the same kinase.
They differ at the proteins the kinase acts on: transporters in the kidney cell, the granule-releasing proteins in the gland cell.
That is box 7, the response.

70
Check q20

In one cell type with the hormone P pathway, hormone P binds, cAMP rises, and the kinase’s target proteins gain phosphate as normal. Yet the transporters stay inside the cell.

A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface
A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface

Which box is faulty?

  1. A. Box 4, the enzyme
    cAMP rose as normal.
    So the enzyme that makes cAMP worked.
  2. B. Box 6, the kinase
    The kinase’s target proteins gained phosphate as normal.
    So the kinase worked.
  3. C. ✓ Box 7, the transporters

Why: Every measurement up to the kinase’s targets was normal, so boxes 1 to 6 worked.
The transporters still did not move.
So the fault is in the transporters themselves, box 7.

71
Check q21

Sam says: “In the hormone P pathway, the kinase at box 6 acts before the enzyme at box 4 makes any cAMP.”

A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface
A model of the hormone P pathway drawn as boxes numbered 1 to 7: hormone P, receptor across the membrane, G protein, enzyme, cAMP, kinase, transporters move to the surface

Is Sam correct?

  1. A. Yes — the kinase switches on first and then makes cAMP
    The kinase does not make cAMP; the enzyme at box 4 does.
    The kinase waits, switched off, until cAMP switches it on.
  2. B. ✓ No — cAMP is made first and then switches the kinase on

Why: Each box acts on the next in order.
The enzyme, box 4, makes cAMP.
cAMP, box 5, then switches on the kinase, box 6.
So the enzyme makes cAMP before the kinase acts.

72
Check q22

A bar graph’s caption says: ‘error bars represent ±2SE’.

What does each error bar show?

  1. A. ✓ The range the true mean is likely to lie in
  2. B. The largest and the smallest reading in that set of dishes
    The bar runs two standard errors either side of the mean.
    It shows where the true mean is likely to lie, not the spread of the readings.

Why: An error bar of ±2SE runs two standard errors above and below the mean.
So it shows the range the true mean is likely to lie in.

73
Practice writing an answer

Researchers study hormone P on gland cells that release granules, using the pathway model shown. Six dishes each receive one of three treatments: no hormone P; hormone P; or hormone P together with a drug that holds G proteins switched off. After 10 minutes the researchers count the granules released per cell. The graph shows the mean for each treatment; the error bars show ±2SE.

The hormone P pathway model beside a bar graph of granules released per cell in three treatments: no hormone, hormone P, and hormone P with a G protein blocker; error bars show ±2SE
The hormone P pathway model beside a bar graph of granules released per cell in three treatments: no hormone, hormone P, and hormone P with a G protein blocker; error bars show ±2SE

(a) Identify the independent variable and the dependent variable in this experiment. (1 pt)

Model answer The independent variable is the treatment each dish receives (no hormone P, hormone P, or hormone P with the G protein blocker).
The dependent variable is the number of granules released per cell.
Rubric
  • Award 1 point for: the treatment (hormone P present or absent, with or without the blocker) as the independent variable AND granules released per cell as the dependent variable.

Slip Naming cAMP or the G protein as a variable. Neither is measured here; the experimenters change the treatment and count granules.

(b) Explain why the dishes given no hormone P are included. (1 pt)

Model answer The no-hormone dishes are the control.
They show how many granules the cells release on their own, with hormone P absent.
Any rise above that level in the other dishes can then be credited to hormone P.
Rubric
  • Award 1 point for: the no-hormone dishes give the release with hormone P absent (a control), so the effect of hormone P can be measured against it.

Slip Saying the control ‘shows the experiment worked’. The control gives the baseline; without it, 42 granules per cell could not be credited to hormone P.

(c) Using the model, predict the level of cAMP in the dishes given hormone P with the blocker, compared with the dishes given hormone P alone, and justify your prediction. (1 pt)

Model answer cAMP stays at its resting level in the blocker dishes, far below the level in the hormone-P dishes.
In the model the G protein sits between the receptor and the enzyme that makes cAMP.
With the G protein held off, hormone P still binds, and the receptor still changes shape.
But the G protein cannot switch the enzyme on.
So the enzyme never makes cAMP.
Therefore the cAMP concentration stays low.
Rubric
  • Award 1 point for: cAMP stays low (at rest) with the blocker, because the G protein acts before the enzyme that makes cAMP, so the enzyme is never switched on.

Slip Predicting that cAMP rises but the granules are not released. That would need a block after cAMP; the blocker acts at the G protein, before the enzyme.

(d) The researchers claim the drug acts at the G protein rather than at the receptor. Describe one measurement that would test this claim, and state the result that would support it. (1 pt)

Model answer Measure how much hormone P is bound to the receptors in the blocker dishes.
If the binding is the same as in the hormone-P dishes, the receptor is working.
So the drug acts after the receptor, at the G protein.
Rubric
  • Award 1 point for: measuring hormone P bound at the receptor (or the receptor’s shape change) in the blocker dishes, with unchanged binding supporting the claim.
  • Accept: adding a G protein locked in its switched-on shape to blocker-treated cells and seeing granule release restored.

Slip Proposing to measure granules again with more hormone P. More hormone cannot tell a blocked receptor from a blocked G protein; a measurement at the receptor itself can.

APBIO-U04-P42 Practice questions: Topic 4.2

Topic 4.2 · Introduction to Signal Transduction · 10 MCQ · 2 FRQ · for APBIO-U04-T42

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. The first free-response question walks you through the reasoning one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Amplification at a step is the number of molecules made or activated at that step divided by the number of molecules that activated them; counts of molecules are given as measured. Where a graph carries error bars, the caption says what the bars represent.

Video: Watch first: Topic 4.2 summary: across the membrane and inside

The ligand stays outside and the receptor's shape change crosses the membrane; inside, a relay of switches and second messengers amplifies the message; the cell decides what it does with the message.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-T42-summary.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-T42-summary.mp4

Q1 P42-q01

A yeast cell carries a receptor on its surface that binds glucose. When glucose binds, the cell speeds up its growth. A sugar with the same atoms as glucose but a different arrangement of its parts reaches the receptor and is passed by.

Why does the receptor bind glucose and pass the other sugar by?

  1. A. ✓ The receptor's binding site fits glucose by shape and charge, and the other sugar fits it poorly
  2. B. The receptor binds whichever sugar arrives first, and glucose arrives first
    Binding is a fit, not a race.
    The binding site's shape and charges match glucose; a sugar with a different arrangement does not fit well and is passed by.
  3. C. The receptor is an enzyme that changes glucose into a product, and the other sugar is not a substrate it can change
    A receptor changes nothing: it holds glucose, changes its own shape, and lets glucose go unchanged.
    Glucose fits the site by shape and charge; the other sugar fits poorly.
  4. D. The receptor binds any sugar, and the cell decides afterwards which one to respond to
    A receptor binds only ligands that fit its binding site.
    Nothing about the cell 'decides'; the fit at the receptor is what selects glucose.

Why: A ligand fits its receptor's binding site by shape and charge, as a substrate fits an active site.
Glucose fits the site; a sugar with the same atoms arranged differently has a different shape, fits poorly, and is passed by, so each receptor binds its own ligand.

Q2 P42-q02

A peptide hormone reaches a mosquito’s salivary gland cells, and the cells begin releasing an enzyme. Researchers follow a labeled form of the hormone over the first minute: at 2 seconds it is bound at the cell surface; at 5 seconds the receptor's intracellular domain has changed shape; at 30 seconds a relay protein inside the cell carries a new phosphate. At every time, the amount of labeled hormone inside the cells is zero.

What crossed the membrane to start the change inside?

  1. A. The hormone itself, through a pore in the receptor
    Every hormone molecule stayed outside, and a peptide cannot cross the membrane.
    The receptor holds the hormone; it passes nothing through.
  2. B. A phosphate group, carried in by the hormone from outside
    The phosphate on the relay protein came from ATP inside the cell, added by a kinase; the hormone stayed outside and carried nothing in.
  3. C. ✓ A change in the receptor's shape, passed to its intracellular domain
  4. D. Sugar, carried in by the hormone
    The sugar is stored inside the cell and leaves it; nothing carried sugar in.
    What crossed was a change in the receptor's shape.

Why: None of the labeled hormone is ever inside the cells.
At 5 seconds the intracellular domain has changed shape; the relay protein gains its phosphate only at 30 seconds.
So the first change inside is the intracellular domain changing shape.
That change crosses the membrane; the hormone stays outside.

Q3 P42-q03

The drawing below shows a receptor set in a cell's membrane, with four positions numbered 1 to 4.

A receptor set in a cell's membrane; positions 1 to 4 are marked.
A receptor set in a cell's membrane; positions 1 to 4 are marked.

Which numbered position marks the site where the ligand binds?

  1. A. Position 1
    Position 1 sits on the ligand-binding domain, the outward region, but away from the pocket.
    The ligand fits only into the pocket cut into that region: the binding site.
  2. B. ✓ Position 2
  3. C. Position 3
    Position 3 is the part of the receptor set in the membrane.
    The ligand binds on the part that faces the fluid outside the cell, where the ligand arrives.
  4. D. Position 4
    Position 4 faces the cytosol: the intracellular domain, which changes shape while the ligand is bound.
    The ligand binds in the pocket of the outward region.

Why: The ligand-binding domain is the receptor's outward region.
The pocket cut into it, position 2, is the site where the ligand fits by shape and charge.
Position 1 is on that domain, away from the pocket.
Position 3 sits in the membrane; position 4 is the intracellular domain.

Q4 P42-q04

An insect's molting hormone is built of four carbon rings and dissolves in oil. Its flight hormone is a chain of ten amino acids. Both travel in the insect's blood.

Which of the following classifies each ligand and places its receptor?

  1. A. Molting hormone: a peptide with a receptor at the surface; flight hormone: a small molecule with a receptor inside the cell
    Four carbon rings are a small nonpolar molecule that crosses the membrane and binds a receptor inside the cell.
    A peptide is held back at the surface.
  2. B. Both hormones are small molecules with receptors at the surface
    Traveling in the blood does not show a molecule's make-up.
    The flight hormone is a chain of amino acids, a peptide, and only the molting hormone crosses the membrane.
  3. C. Both hormones are peptides with receptors inside the cell
    A chain of amino acids is a peptide that binds a receptor at the cell surface.
    Four rings that dissolve in oil pass through the membrane to a receptor inside.
  4. D. ✓ Molting hormone: a small molecule with a receptor inside the cell; flight hormone: a peptide with a receptor at the surface

Why: A ligand’s chemistry sets where its receptor sits.
The molting hormone, four carbon rings that dissolve in oil, is a small nonpolar molecule that crosses to an intracellular receptor.
The flight hormone, a chain of ten amino acids, is a peptide that binds a cell-surface receptor.

Q5 P42-q05

After the molting hormone arrives, an insect's skin cells make large amounts of the protein that hardens its new outer layer.

Which of the following is the gene for the hardening protein?

  1. A. The hardening protein itself, stored in the skin cells until the hormone arrives
    The protein is what the gene’s instructions build.
    The gene is the stretch of DNA that carries those instructions.
  2. B. The molting hormone, which carries the instructions for the protein into the skin cells
    A hormone carries no instructions inside it; it binds a receptor and is released unchanged.
    The instructions for the protein are in the cell’s DNA.
  3. C. ✓ The stretch of the skin cell's DNA that carries the instructions for making the hardening protein
  4. D. The chain of amino acids that the skin cell joins together to build the hardening protein
    The amino acid chain is the protein being built.
    The order to join them in is read from the gene, a stretch of the cell’s DNA.

Why: A gene is a stretch of a cell’s DNA that carries the instructions for making one protein.
The skin cell reads the hardening protein’s gene and joins amino acids in the order it gives.
So the gene is the DNA stretch, not the protein, the hormone or the amino acids.

Q6 P42-q06

In a kidney cell, relay protein Y moves water channels to the surface only while Y carries a phosphate. One cell line carries a changed form of Y: the amino acid that the kinase would phosphorylate has been replaced, so no kinase can add a phosphate to Y. The hormone arrives at these cells.

Which of the following best predicts what happens to the water channels, and why?

  1. A. ✓ The water channels stay inside, because Y is never switched on
  2. B. The water channels move to the surface as usual, because Y is still present
    Y moves the channels only while it carries a phosphate.
    No kinase can add one to the changed Y, so Y is never switched on and the channels stay inside.
  3. C. The water channels move to the surface as usual, because the kinase adds the phosphate to a different part of Y
    A kinase transfers a phosphate onto one particular amino acid of its target.
    That amino acid has been replaced, so the kinase has nowhere to put the phosphate.
  4. D. The water channels stay inside, because the hormone can no longer bind its receptor
    The receptor is unchanged and binds the hormone as usual.
    The change is in Y: it cannot be phosphorylated, so it is never switched on and the channels stay inside.

Why: A kinase transfers a phosphate from ATP onto one amino acid of its target, and the phosphate switches the protein on.
The changed Y has lost that amino acid, so no kinase can phosphorylate Y.
Y stays in its resting shape, and the water channels stay inside.

Q7 P42-q07

When a fish's fin is cut, cells at the wound release molecule N into the fluid around them. Fin cells that carry the receptor for molecule N begin dividing within a day, and the fin regrows.

Which of the following is the growth factor in this case?

  1. A. ✓ Molecule N
  2. B. The receptor on the fin cells
    The receptor is the protein on the fin cell that binds the signal.
    A growth factor is the signal itself, whose message is grow and divide.
  3. C. The dividing fin cells
    The dividing cells are the response.
    The growth factor is the signal that told them to grow and divide.
  4. D. The kinases that relay the message inside the fin cells
    The kinases relay the message inside the cell.
    The growth factor is the signal that arrived from outside the cell.

Why: A growth factor is a signal whose message to the cell is grow and divide.
Molecule N is the signal released at the wound, and the fin cells’ response is to divide.
So molecule N is the growth factor.

Q8 P42-q08

In a kidney cell, 6 receptors have the hormone bound, and over the next minute the cell's cAMP rises from 330 molecules to 2,430 molecules.

What is the amplification at this step, as cAMP molecules made per bound receptor?

  1. A. 55 cAMP per receptor
    The amplification counts the new cAMP: 2,430 minus 330 is 2,100 molecules, divided by 6.
  2. B. ✓ 350 cAMP per receptor
  3. C. 405 cAMP per receptor
    Subtract the starting count first: 2,430 minus 330 is 2,100, then divide by 6.
  4. D. 12,600 cAMP per receptor
    Amplification is the number of new molecules made per activating molecule, so the 2,100 new cAMP are divided by the 6 receptors.

Why: Amplification at a step is the number of new molecules made at that step divided by the number of molecules that activated them.
The working below gives 350 cAMP per receptor.

Q9 P42-q09

A gill cell of a snail carries a receptor for a neurotransmitter. When the neurotransmitter binds, negative ions flow into the cell within a millisecond. Throughout, the cell's cAMP stays at its resting level, and no relay protein inside the cell gains a phosphate.

What kind of receptor is this, and why does the current need no change inside the cell?

  1. A. ✓ A ligand-gated channel; the receptor's own shape change opens a channel through it, with no relay inside
  2. B. A G protein-coupled receptor; the G protein opens the channel without needing cAMP
    A G protein-coupled receptor works through a G protein and an enzyme, taking seconds.
    Here cAMP never changed and the current came within a millisecond: no relay carried the signal.
  3. C. An intracellular receptor; the neurotransmitter enters the cell and opens the channel itself
    A neurotransmitter binds a receptor at the cell surface and stays outside.
    What changes is the receptor's shape, and the receptor is itself the channel.
  4. D. A cell-surface receptor whose inner part is a kinase; it phosphorylates the channel, which then opens
    A kinase would add a phosphate to a relay protein, and no relay protein gained a phosphate.
    The receptor's own shape change opens the channel.

Why: A receptor that is itself a channel protein is a ligand-gated channel.
The ligand’s binding changes the receptor’s shape, the channel through it opens, and ions flow within a millisecond.
No second messenger or kinase stands between ligand and ions, so cAMP and the relay proteins stay as they were.

Q10 P42-q10

The insect's flight hormone binds the same kind of cell-surface receptor on fat cells and on flight-muscle cells. The fat cells release stored sugar; the muscle cells burn fuel faster. When cAMP itself is injected into fat cells with no hormone present, they release stored sugar; when cAMP is injected into flight-muscle cells, they burn fuel faster.

What does the injection result show about where the two pathways differ?

  1. A. Before cAMP, at the receptor: the two cell types bind the hormone differently
    A difference at the receptor would disappear when cAMP is injected past it, yet each cell type still gave its own response.
  2. B. At cAMP: the two cell types make different second messengers
    The same molecule, cAMP, was injected into both, and each cell type responded in its own way.
    So the difference lies after cAMP.
  3. C. Before cAMP, at the G protein: the two cell types' G proteins switch on different enzymes
    The G protein and its enzyme only make cAMP; injecting cAMP bypasses them, yet the responses still differed, so the split is downstream.
  4. D. ✓ After cAMP: the kinases switched on by cAMP reach different response proteins in the two cell types

Why: Injected cAMP alone gave each cell its own response, so the paths part after cAMP.
The same cAMP switches on kinases in both.
The kinases reach each cell’s own response proteins: sugar-releasing enzymes in fat cells, fuel-burning enzymes in muscle cells.
So the receiving cell’s proteins set the response.

FRQ 1 P42-frq1 · Scientific Investigation scaffolded

When the body is short of water, a peptide hormone from the brain reaches the cells lining the kidney's tubes and makes them save water. The hormone binds a G protein-coupled receptor on the cell surface; the G protein switches on an enzyme in the membrane; the enzyme makes cAMP from ATP (word equation: ATP → cyclic AMP + two linked phosphates); cAMP switches on a kinase; and the kinase phosphorylates the proteins that move water channels to the cell's surface. Researchers give kidney cells one of four treatments, six dishes each: no hormone; the hormone; the hormone together with a drug that blocks the enzyme that makes cAMP; or cells first treated with a phosphatase inhibitor, then given the hormone for five minutes, then washed free of hormone. Ten minutes after each treatment they count the water channels at the cell surface. The graph below shows the means for the first three treatments, and the error bars represent ±2SE; the fourth has still to be counted. In the dishes given the hormone alone, 10 receptors per cell have the hormone bound after one minute, and the cAMP in each cell has risen from 280 molecules to 2,180 molecules.

Water channels counted at the surface of kidney cells 10 minutes after each treatment, six dishes per treatment. Error bars represent ±2SE. Gridlines every 20 channels per cell. The fourth treatment has still to be counted.
Water channels counted at the surface of kidney cells 10 minutes after each treatment, six dishes per treatment. Error bars represent ±2SE. Gridlines every 20 channels per cell. The fourth treatment has still to be counted.

(a) Identify the independent variable and the dependent variable in this investigation. (1 pt)

Frame The independent variable is …, and the dependent variable is …

Hint Which thing did the researchers change between dishes, and which thing did they count to see the effect?

Model answer The independent variable is the treatment each dish receives (no hormone, the hormone, the hormone with the enzyme blocker, or the hormone then a wash in inhibitor-treated cells), and the dependent variable is the number of water channels at the cell surface after ten minutes.
Rubric
  • Award 1 point for both: independent variable, the treatment each dish receives (no hormone, hormone, hormone with the enzyme blocker, hormone then wash with the phosphatase inhibitor); dependent variable, the number of water channels at the cell surface after ten minutes.
  • Do not award the point if the two are reversed, or if cAMP or the receptor is named as a variable.

Slip Naming cAMP as a variable. cAMP is a molecule of the relay; the researchers change the treatment and count channels.

(b) Identify the stage of the pathway that the rise in cAMP belongs to, and justify your choice. (1 pt)

Frame The rise in cAMP is …, because …

Hint Where in the cell does the rise happen, and is it what the cell finally does differently?

Model answer The rise in cAMP is transduction, because cAMP is a second messenger made inside the cell that carries the message on from the receptor to the kinase.
Reception is the hormone binding its receptor, and the cellular response is the water channels reaching the surface.
Rubric
  • Award 1 point for: transduction, because cAMP is a molecule inside the cell relaying the message from the receptor toward the response (the ligand binding is reception, and the channels reaching the surface is the response).
  • Do not award the point for 'reception' or for 'the response'.

Slip Calling the rise in cAMP the response. The response is what the cell finally does differently, the channels moving; cAMP is a step of the relay that gets there.

(c) Explain why the dishes given the hormone with the enzyme blocker show about the same number of channels as the dishes given no hormone. (1 pt)

Frame With the enzyme blocked, no … is made, so … stays off and …

Hint Which molecule does the blocked enzyme make, and what does that molecule normally switch on next?

Model answer The blocked enzyme is the enzyme that makes cAMP, and cAMP is what switches on the kinase.
The kinase is what phosphorylates the proteins that move the channels.
With no cAMP made, the kinase stays off, so those proteins are never phosphorylated and the channels stay inside.
The hormone still binds and the G protein is still switched on, but the message stops at the enzyme.
So these cells look like cells given no hormone.
Rubric
  • Award 1 point for: the blocked enzyme is the one that makes cAMP, so no cAMP is made; with no cAMP the kinase stays off, the proteins that move the channels are never phosphorylated, and the channels stay inside, as in cells given no hormone (the hormone still binds and the G protein is still switched on).
  • Do not award the point for 'the hormone cannot bind' or for 'the drug destroys the channels'.

Slip Saying the drug stops the hormone binding. The block is inside, at the enzyme; binding and the G protein are unchanged, and everything after the enzyme stays off.

(d) Calculate the amplification at the cAMP step in the dishes given the hormone alone, as the increase in cAMP molecules per bound receptor. (1 pt)

Frame The amplification at this step is … cAMP molecules per bound receptor.

Hint Amplification is the number of new molecules made divided by the number of molecules that made them. Which of the two cAMP counts is the level before the hormone acted?

Answer: 190 cAMP per receptor  (tolerance ±0.5)

Model answer The amplification at this step is 190 cAMP molecules per bound receptor.
Working
Write down the values in the question:
bound receptors = 10
cAMP before = 280 molecules
cAMP after = 2,180 molecules
Write down the equation:
tex: \text{amplification} = \frac{\text{activated molecules}}{\text{activating molecules}}
Substitute the values into the equation:
tex: \text{amplification} = \frac{\text{activated molecules}}{\text{activating molecules}}
tex: \text{new cAMP} = 2,180 - 280 = 1,900\,\text{molecules}
tex: \text{amplification} = \frac{1,900}{10}
tex: \text{amplification} = 190\,\text{cAMP per receptor}
Rubric
  • Award 1 point for: 190 cAMP molecules per bound receptor (the increase, 2,180 − 280 = 1,900 molecules, divided by 10 bound receptors).
  • Do not award the point for 218 (2,180 divided by 10, with the resting 280 molecules not subtracted) or for 28 (the resting count divided by 10).

(e) Support the claim that the cells of the fourth treatment (hormone for five minutes, then washed away, in cells treated with the phosphatase inhibitor) still hold many channels at their surface when the researchers count, using the pathway. (1 pt)

Frame Phosphatases normally …, but in these cells …, so the proteins that move the channels …, so the channels …

Hint What do phosphatases normally do to the relay proteins once the hormone has gone, and what does the inhibitor do to that?

Model answer Phosphatases normally switch the relay off by removing the phosphates the kinase added.
In these cells the phosphatases are inhibited.
So the proteins that move the channels keep their phosphates.
So those proteins stay switched on although the hormone is gone.
So the channels stay at the surface, far more than the 12 per cell of the no-hormone dishes.
Rubric
  • Award 1 point for: the evidence (in these cells the phosphatases are inhibited; phosphatases normally switch the relay off by removing the phosphates the kinase added) AND the reasoning (so the proteins that move the channels keep their phosphates and stay switched on after the hormone is gone, so the channels stay at the surface, far more than the 12 per cell in the no-hormone dishes).
  • Do not award the point for 'the phosphatases are blocked' with no link to the channels, or for 'the receptors are empty, so the channels go back inside'.

Slip Naming the inhibitor and stopping. Supporting the claim needs the link: what the phosphatases would have done, and what their inhibition leaves switched on.

FRQ 2 P42-frq2 · Conceptual Analysis

An insect's molting hormone is a small nonpolar molecule built of four carbon rings. When it reaches the insect's skin cells, over the next day the cells make large amounts of the protein that hardens the new outer layer. The same insect's flight hormone, a chain of ten amino acids, reaches its fat cells in the blood, and within a minute the cells begin releasing stored sugar. The flight hormone binds a receptor on the fat cell's surface; inside the cell, cAMP rises and a kinase phosphorylates the enzymes that release the stored sugar.

(a) Determine where the receptor for the molting hormone sits in a skin cell, using the hormone's chemistry. (1 pt)

Model answer The receptor sits inside the cell, in the cytosol or the nucleus.
The molting hormone is small and nonpolar.
So it dissolves into the oily middle of the plasma membrane and passes through.
So it binds a receptor inside the cell: an intracellular receptor.
Rubric
  • Award 1 point for: the decision (inside the cell, in the cytosol or the nucleus) AND the reasoning it rests on (a small nonpolar molecule dissolves into the membrane's oily middle and passes through, so it binds a receptor inside the cell, an intracellular receptor).
  • Do not award the point for the location alone, or for 'at the surface, because it arrives in the blood'.

Slip Deciding on the surface because the hormone arrives in the blood. Arriving in the blood does not decide whether a hormone crosses the membrane; a small nonpolar molecule crosses and binds a receptor inside the cell.

(b) Describe what the bound receptor does to change what the skin cell makes. (1 pt)

Model answer With the hormone bound, the receptor itself moves into the nucleus and attaches to the DNA.
There the bound pair changes the cell's gene expression: the cell starts making large amounts of the hardening protein from that protein's gene.
An intracellular receptor needs no relay of other molecules; the receptor is the molecule that reaches the DNA.
Rubric
  • Award 1 point for: the hormone binds its receptor, the bound pair moves into the nucleus and attaches to the DNA, and the cell changes its gene expression, making more of the hardening protein; the receptor itself reaches the DNA, with no relay of other molecules.
  • Accept: the receptor already waits in the nucleus, binds the hormone there, and the bound receptor attaches to the DNA.
  • Do not award the point for 'the receptor switches on a phosphorylation cascade' or for 'the hormone attaches to the DNA on its own'.

Slip Adding a relay of kinases. An intracellular receptor carries the message to the DNA itself; the relay belongs to cell-surface receptors.

(c) Explain why the new hardening protein takes many hours to appear. (1 pt)

Model answer The hardening protein is a new protein: the cell had none of it before.
The response is a change in gene expression.
So the gene has to be expressed and the protein built from its instructions.
Building a protein takes hours.
Rubric
  • Award 1 point for: the response is a change in gene expression: the protein did not exist in the cell before, so its gene has to be expressed and the protein built, which takes hours.
  • Do not award the point for 'the hormone crosses the membrane slowly' or for 'the hormone has far to travel'.

Slip Blaming the hormone's journey or its crossing of the membrane. The hormone is inside the cell within minutes; the slow part is making a protein that did not exist before.

(d) Within a minute of the flight hormone leaving the blood, the fat cells stop releasing sugar. Describe two changes inside a fat cell that end the response. (1 pt)

Model answer An enzyme breaks the cAMP down, so the cAMP concentration falls and the kinase switches off.
Phosphatases remove the phosphates the kinase added, so the enzymes that release the sugar return to their resting shape and stop.
Rubric
  • Award 1 point for two of: the flight hormone leaves the receptor and is removed, so the receptor returns to its resting shape; an enzyme breaks cAMP down, so the kinase switches off; phosphatases remove the phosphates the kinase added, so the sugar-releasing enzymes stop.
  • Do not award the point for one change alone, or for 'the cell has no sugar left' or 'the receptor is destroyed'.

Slip Naming one change and stopping. The task asks for two: the cell ends a response by removing the ligand, destroying the second messenger and removing the phosphates.

APBIO-U04-T42 End-of-topic test: Introduction to Signal Transduction

Topic 4.2 · Introduction to Signal Transduction · 19 MCQ · 2 FRQ

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the two free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. Then open the scoring guide and mark your own work against it. Counts of molecules are given as measured. Where a graph carries error bars, the caption says what the bars represent.

Q1 T42-q01

The drawing below shows a receptor set in a cell's membrane with its binding site facing outward, and three signal molecules, 1, 2 and 3, found in the fluid around the cell. The binding site is lined with negative charges.

A receptor set in a cell's membrane, its binding site lined with negative charges and facing outward; the fluid outside the cell is shaded paler than the fluid inside. The three signal molecules 1, 2 and 3 found in the fluid around the cell are drawn above the membrane, each marked with the charge it carries.
A receptor set in a cell's membrane, its binding site lined with negative charges and facing outward; the fluid outside the cell is shaded paler than the fluid inside. The three signal molecules 1, 2 and 3 found in the fluid around the cell are drawn above the membrane, each marked with the charge it carries.

Which molecule is this receptor's ligand?

  1. A. Molecule 2
    Molecule 2 has the right shape, but it carries a negative charge and the lining is negative too.
    Like charges push apart, so molecule 2 is held only poorly.
  2. B. Molecule 3
    Molecule 3’s positive charge is attracted by the lining, but it is the wide oval: the wrong shape for the tall narrow site.
    Shape and charge both have to match.
  3. C. ✓ Molecule 1
  4. D. Molecules 1 and 2
    Molecule 2 has the shape but the wrong charge; like charges push apart.
    Only molecule 1 has both the shape and a charge the lining attracts.

Why: A ligand fits its binding site by shape and charge.
Molecule 1 has the site’s shape and a positive charge against the negative lining, so it is held.
Molecule 2 is pushed away by like charges.
Molecule 3 has the right charge but the wrong shape.

Q2 T42-q02

Acetylcholine released from nerve endings reaches heart muscle cells, which beat more slowly, and salivary gland cells, which release saliva. Both cell types carry a receptor that binds acetylcholine.

Why do the two cell types respond differently to the same molecule?

  1. A. ✓ Each cell type holds its own receptor, relay and response proteins, so the same ligand ends at different proteins
  2. B. Acetylcholine changes into a different molecule on its way from the nerve ending to the salivary gland cells
    Acetylcholine is the same molecule wherever a nerve ending releases it.
    The difference lies inside the two cells, in the proteins each one holds.
  3. C. The heart cells receive their acetylcholine from the blood, and the gland cells receive theirs from a nerve ending
    Both cell types receive acetylcholine from nerve endings beside them.
    How the ligand arrives does not decide the response; the proteins in the receiving cell do.
  4. D. The salivary cells receive more acetylcholine than the heart cells do, and a stronger signal gives a different kind of response
    More signal makes a response stronger; it cannot turn slower beating into saliva release.
    The kind of response is set by the response proteins the cell holds.

Why: The response depends on which relay and response proteins the cell holds.
The same acetylcholine binds both cell types’ receptors.
But each pathway ends at that cell’s own proteins: contraction proteins in the heart cell, saliva-releasing proteins in the gland cell.
So the cell, not the signal, sets the response.

Q3 T42-q03

Hormone C, which lowers the calcium in the blood, is a chain of 32 amino acids. Melatonin, the hormone that makes a person sleepy at night, is built of two fused carbon rings with two short chains attached, about 30 atoms in all.

Which of the following classifies the two ligands?

  1. A. Hormone C is a small molecule; melatonin is a peptide
    A chain of amino acids is a peptide, whatever its size; two rings with short chains attached is a small molecule.
  2. B. Both ligands are peptides
    Being a hormone says how a molecule travels, not what it is made of.
    A chain of 32 amino acids is a peptide; two rings is a small molecule.
  3. C. Both ligands are small molecules
    Traveling in the blood does not show what a molecule is made of.
    Hormone C, a chain of amino acids, is a peptide; only melatonin is a small molecule.
  4. D. ✓ Hormone C is a peptide; melatonin is a small molecule

Why: A ligand is one of two kinds: a peptide, a chain of amino acids that may be a whole protein, or a small molecule.
Hormone C, 32 amino acids joined in a chain, is a peptide.
Melatonin, two rings with short chains and about 30 atoms, is a small molecule.

Q4 T42-q04

Bone cells carry a cell-surface receptor for hormone C. One cell line carries a shortened receptor that lacks the intracellular domain but keeps the ligand-binding domain and the part set in the membrane. Hormone C binds the normal and the shortened receptors equally well. Within a minute of hormone C binding, a relay protein inside normal cells gains a phosphate; in the shortened-receptor cells the relay protein stays as it was.

What does the comparison show about how the message gets into the cell?

  1. A. Hormone C passes through the receptor into the cytosol, and the shortened receptor is too short to let it through
    A peptide cannot cross the membrane, and nothing passes through this receptor; what crosses is the receptor’s change of shape, which the shortened receptor cannot carry inside.
  2. B. ✓ The bound receptor changes shape all the way to its intracellular domain, and that change starts the relay inside
  3. C. The relay protein gains its phosphate directly from hormone C, which hands the phosphate over at the cell surface
    The relay protein’s phosphate comes from ATP, added by a kinase.
    Hormone C never reaches the relay protein; the message comes through the receptor’s intracellular domain.
  4. D. The shortened receptors hold hormone C less firmly than the normal receptors do, so the relay inside them is weaker
    Binding is equal in both lines, so the binding site works in both.
    The shortened receptor simply has no intracellular domain to change shape and start the relay.

Why: While the ligand is bound, the receptor holds a different shape.
For a membrane receptor, the part that acts inside is its intracellular domain.
Binding was equal in both lines, so reception happened in both.
Only the receptor with an intracellular domain could pass the shape change inside.

Q5 T42-q05

Hormone C arrives at a bone cell. At 3 seconds it is bound to receptors in the membrane. At 20 seconds a relay molecule inside the cell has risen ten-fold. At 5 minutes the cell has stopped breaking down bone.

Which of the following classifies the three events?

  1. A. 3 s: reception; 20 s: cellular response; 5 min: transduction
    A relay molecule rising inside the cell is a step of the relay: transduction.
    The cell stopping its breakdown of bone is what it finally does differently: the response.
  2. B. 3 s: reception; 20 s: reception; 5 min: cellular response
    Reception is finished when the receptor has changed shape.
    A relay molecule rising inside the cell at 20 seconds is the relay carrying the message on: transduction.
  3. C. 3 s: transduction; 20 s: reception; 5 min: cellular response
    Hormone C binding its receptor at 3 seconds is reception, the first stage.
    The relay molecule rising at 20 seconds is transduction, the stage that follows.
  4. D. ✓ 3 s: reception; 20 s: transduction; 5 min: cellular response

Why: Reception is the ligand binding its receptor: hormone C bound at 3 seconds.
Transduction is molecules inside the cell relaying the message: the relay molecule rising at 20 seconds.
The cellular response is what the cell finally does differently: stopping its breakdown of bone at 5 minutes.

Q6 T42-q06

Thyroid hormone is a small nonpolar molecule that dissolves in oil. Hormone C is a chain of 32 amino acids. Both hormones reach a bone cell in the blood.

Where does each hormone bind its receptor?

  1. A. Thyroid hormone at the cell surface; hormone C inside the cell
    A small nonpolar molecule crosses the membrane and binds a receptor inside the cell.
    A chain of amino acids cannot cross and binds a receptor at the cell surface.
  2. B. Both at the cell surface
    How a hormone arrives does not decide whether it can cross the membrane.
    Thyroid hormone is small and nonpolar, so it crosses and binds a receptor inside the cell.
  3. C. ✓ Thyroid hormone inside the cell; hormone C at the cell surface
  4. D. Both inside the cell
    Only a small nonpolar molecule crosses the membrane on its own.
    Hormone C, a peptide, is held back by the oily middle, so its receptor sits at the surface.

Why: A ligand’s chemistry sets where its receptor sits.
Thyroid hormone is small and nonpolar, so it slips through the membrane’s oily middle and binds an intracellular receptor.
Hormone C is a peptide, large and polar, so the membrane holds it back and it binds a cell-surface receptor.

Q7 T42-q07

Cells lining a person's small intestine make large amounts of lactase, the enzyme that digests milk sugar; that person's muscle cells make none. After a signal reaches the intestine-lining cells, they make three times as much lactase over the next few hours.

What is the difference between the intestine-lining cells and the muscle cells, and what changed in the intestine-lining cells after the signal?

  1. A. ✓ Intestine-lining cells express the lactase gene and muscle cells do not; the signal changed how fast the gene is expressed
  2. B. Intestine-lining cells carry the lactase gene and muscle cells lack it; the signal added more copies of the gene
    Every cell carries the same DNA, so the muscle cells have the lactase gene too and simply do not express it.
    The signal added no genes.
  3. C. Both cell types express the gene, but muscle cells destroy the lactase; the signal stopped that destruction
    Muscle cells make no lactase at all: the gene is present and not expressed.
    Nothing is being made and destroyed.
  4. D. Intestine-lining cells express the gene and muscle cells do not; the signal molecule itself was made into lactase
    The cell built the lactase from the gene's instructions, as always.
    The signal changed only how fast the gene was expressed; the signal molecule stayed outside the cell.

Why: A cell expresses a gene when it uses the gene’s instructions to make the protein.
Intestine-lining cells express the lactase gene; muscle cells have the gene and do not express it.
Speeding up the making of a protein is a change in gene expression: that is what the signal produced.

Q8 T42-q08

In a tadpole's tail cells, the receptor for thyroid hormone is found in the nucleus once the hormone has arrived. Over the next six hours the cells make far more of an enzyme that breaks down the tail's connective tissue.

How does the hormone change what the tail cells make?

  1. A. A relay of kinases carries the message from a cell-surface receptor to the nucleus
    Thyroid hormone is small and nonpolar, and its receptor is inside the cell, found attached to the DNA.
    An intracellular receptor needs no relay: the receptor itself reaches the DNA.
  2. B. ✓ The bound receptor moves to the DNA and changes which genes the cell expresses
  3. C. The hormone attaches to the DNA on its own, and the receptor follows it in
    A hormone acts only through its receptor.
    The hormone binds the receptor inside the cell, and the bound pair moves into the nucleus and attaches to the DNA.
  4. D. The receptor converts the hormone into the tissue-digesting enzyme
    The enzyme is a protein built from its gene's instructions.
    The hormone and its receptor change how much of it the cell makes; nothing is made from the hormone itself.

Why: Thyroid hormone crosses the membrane and binds its intracellular receptor.
The bound pair moves into the nucleus and attaches to the DNA.
The cell changes its gene expression, making more of the tissue-digesting enzyme over the following hours.
The receptor itself is the molecule that reaches the DNA.

Q9 T42-q09

In a fruit fly cell, relay protein W is switched on only while it carries a phosphate group. Two minutes after a growth factor binds the cell's receptor, W carries a phosphate and is active.

Which enzyme switched W on, and where did the phosphate come from?

  1. A. ✓ A kinase, which transferred a phosphate from ATP onto W
  2. B. A kinase, which transferred a phosphate from the growth factor onto W
    The growth factor stays outside the cell and gives up nothing to the relay.
    A kinase takes the phosphate it transfers from ATP.
  3. C. A kinase, which transferred a phosphate from the receptor onto W
    The receptor passes on a change of shape, not a phosphate.
    The phosphate a kinase adds comes from ATP.
  4. D. A phosphatase, which removed a phosphate from W
    A phosphatase removes phosphates.
    W is switched on by gaining a phosphate, so the enzyme that switched it on is a kinase.

Why: A kinase transfers a phosphate group from ATP onto the protein.
The added phosphate changes the protein’s shape, and the new shape switches the protein on or off.
W is active with its phosphate, so a kinase switched it on, and the phosphate came from ATP.

Q10 T42-q10

In a plant cell, relay protein V is switched off while it carries a phosphate and switched on when the phosphate is removed. A researcher adds a drug that blocks the cell's phosphatases, and then the signal arrives.

Predict what happens to V's activity compared with untreated cells.

  1. A. V is switched on more strongly
    For V it is removing the phosphate that switches it on.
    With the phosphatases blocked, more V keeps its phosphate, so more V stays off.
  2. B. V's activity is unchanged
    The total amount of V is unchanged, but the share in its active, phosphate-free form falls, because the phosphatases that would remove the phosphates are blocked.
  3. C. V is switched on faster
    The drug blocks the phosphatases, so phosphates that kinases add are no longer removed.
    Nothing here blocks a kinase.
  4. D. ✓ V stays mostly switched off

Why: Adding a phosphate does not always switch a protein on: for some relay proteins removing the phosphate activates them.
V is active without its phosphate.
With the phosphatases blocked, phosphates stay on V.
So less V is in its active form, and V stays mostly switched off.

Q11 T42-q11

In a clam's egg cell, the receptor for a maturation hormone switches on kinase A; kinase A phosphorylates kinase B; kinase B phosphorylates kinase C; and kinase C phosphorylates the proteins that start the egg's maturation. One egg carries a changed form of kinase B that no kinase can phosphorylate. The hormone binds this egg's receptors.

Predict which kinases are switched on in this egg.

  1. A. ✓ Kinase A only
  2. B. Kinase C only
    The receptor switches on kinase A as usual.
    The chain breaks at kinase B, which cannot be phosphorylated, so kinase C, which only kinase B can switch on, stays off.
  3. C. Kinases A and C
    Only kinase B phosphorylates kinase C.
    The changed kinase B cannot be phosphorylated, so kinase B is never switched on, and kinase C stays off.
  4. D. Kinases A, B and C
    Another kinase has to transfer the phosphate: A onto B, and B onto C.
    The changed kinase B cannot receive one, so kinase B and kinase C stay off.

Why: In a phosphorylation cascade each kinase phosphorylates the next.
The receptor switches on kinase A.
Kinase A cannot phosphorylate the changed kinase B, so B stays off.
Only kinase B can phosphorylate kinase C, so C stays off too.
So kinase A alone is on.

Q12 T42-q12

Hormone S reaches a fat cell and binds a receptor on the cell's surface. Within a minute the cAMP inside the cell has risen from 420 molecules to 3,540, and kinases throughout the cytosol are switched on.

Which statement describes the role of cAMP here?

  1. A. cAMP is the ligand: hormone S is turned into cAMP as it crosses the membrane into the cytosol
    Hormone S stays outside the cell; it is the first messenger.
    cAMP is made inside, from ATP, by an enzyme the activated receptor switches on.
  2. B. ✓ cAMP is a second messenger: a small molecule made from ATP inside the cell that switches on the kinases
  3. C. cAMP is a protein of the relay, phosphorylated by the receptor and passed from one kinase to the next along the cytosol
    cAMP is not a protein: it is a small molecule made from ATP, and it switches kinases on rather than being phosphorylated itself.
  4. D. cAMP is the cellular response: the fat cell releases it into the blood as the product that hormone S called for
    The response is what the fat cell finally does differently, after the kinases have acted.

Why: A second messenger is a small molecule made inside the cell when a receptor is activated; it spreads through the cytosol and switches on kinases.
cAMP, made from ATP by an enzyme the receptor switches on, is a common example.
The ligand outside, hormone S, is the first messenger.

Q13 T42-q13

In the fat cell, 12 receptors have hormone S bound, and over the next minute the cell's cAMP rises from 420 molecules to 3,540 molecules.

What is the amplification at this step, as cAMP molecules made per bound receptor?

  1. A. 35 cAMP per receptor
    The amplification counts the new cAMP the bound receptors caused: 3,540 minus 420 is 3,120 molecules, divided by 12.
  2. B. ✓ 260 cAMP per receptor
  3. C. 295 cAMP per receptor
    Subtract the starting count first: 3,540 minus 420 is 3,120 new molecules, then divide by 12.
  4. D. 37,440 cAMP per receptor
    Amplification is the number of new molecules made per activating molecule, so the 3,120 new cAMP are divided by the 12 receptors.

Why: Amplification at a step is the number of new molecules made at that step divided by the number of molecules that activated them.
The working below gives 260 cAMP per receptor.

Q14 T42-q14

In a salivary gland cell, 4 bound receptors lead to 180 molecules of the cAMP-making enzyme being switched on, and those enzyme molecules go on to make 10,800 molecules of cAMP.

Which of the following explains why 4 bound receptors end in 10,800 molecules of cAMP?

  1. A. Each bound receptor itself makes 2,700 molecules of cAMP from ATP before the hormone leaves it
    The receptor makes no cAMP; it switches on the G protein, which switches on the enzyme.
    The enzyme makes the cAMP from ATP.
  2. B. Once made, each cAMP molecule makes copies of itself, so the count doubles again and again
    The enzyme makes cAMP from ATP; a cAMP molecule cannot copy itself.
    The count grows because each activated enzyme keeps making cAMP.
  3. C. ✓ Each activated enzyme molecule makes cAMP molecule after molecule until it is switched off, so each step multiplies the count
  4. D. The 4 bound receptors let 10,800 hormone molecules into the cell, and the cell turns each one into a cAMP
    The hormone stays outside the cell, bound to the receptor.
    The enzyme inside makes cAMP from ATP, molecule after molecule.

Why: An activated enzyme acts on many molecules before it is switched off.
Each bound receptor switches on 45 enzyme molecules.
Each enzyme molecule makes 60 molecules of cAMP from ATP.
So each step multiplies the count, and 4 receptors end in 10,800 cAMP: amplification.

Q15 T42-q15

In fat cells, the response to hormone S ends within minutes of the hormone being washed away. In one dish a researcher first treats the cells with a drug that blocks the enzyme that breaks cAMP down. The researcher then adds hormone S to the dish and, after five minutes, washes the hormone away.

Predict what happens to fatty acid release from the treated cells over the next ten minutes, compared with untreated cells.

  1. A. The treated cells stop releasing fatty acids at once
    In untreated cells the response ends because cAMP is destroyed within seconds.
    Here that breakdown is blocked, so the cAMP already made keeps the kinases on.
  2. B. ✓ The treated cells keep releasing fatty acids
  3. C. The treated cells release fatty acids more slowly
    Making cAMP stops in both dishes once the hormone is gone.
    In the treated cells the cAMP is not broken down, so its concentration stays high and the response continues.
  4. D. The treated cells stop at the same time as the untreated cells
    Phosphatases act on the relay proteins downstream.
    While the cAMP concentration stays high, the kinases add phosphates faster than the phosphatases remove them, so the response continues.

Why: A cell ends a response by removing the ligand, breaking down the second messenger and reversing the phosphorylations.
Here the cAMP-destroying enzyme is blocked.
So the cAMP made during the five minutes stays after the hormone is washed away.
So the kinases stay on, and the fatty acids keep leaving.

Q16 T42-q16

Researchers grew two kinds of salivary gland cell: normal cells, and cells whose changed G protein is permanently switched on. Three treatments, six dishes each: normal cells with no signal; normal cells given signal D; changed cells with no signal. After one minute they measured cAMP; the graph shows the means with ±2SE error bars. In the dishes given no signal, no signal D was bound to any receptor.

cAMP inside salivary gland cells one minute after each treatment, in molecules per cell, six dishes per treatment. Error bars represent ±2SE. Gridlines every 1,000 molecules per cell.
cAMP inside salivary gland cells one minute after each treatment, in molecules per cell, six dishes per treatment. Error bars represent ±2SE. Gridlines every 1,000 molecules per cell.

Which of the following claims about the G protein's job in this pathway is best supported by the results?

  1. A. The G protein stops the cell making cAMP until signal D arrives
    The permanently switched-on G protein gave its cells as much cAMP as normal cells given signal D.
    So a switched-on G protein turns cAMP production on, not off.
  2. B. The G protein binds signal D at the surface and pulls it into the cell
    The changed cells made cAMP with nothing bound to any receptor.
    So the G protein acts after binding, and a G protein already switched on needs no ligand at all.
  3. C. ✓ The G protein carries the switch from the bound receptor to the cAMP-making enzyme
  4. D. The G protein breaks cAMP down once the signal is gone
    A permanently switched-on G protein made cAMP pile up, as much as signal D does in normal cells.
    So the G protein switches on the cAMP-making enzyme, not breakdown.

Why: cAMP rose with no signal bound, so the G protein acts after the receptor.
A switched-on G protein raised cAMP as far as signal D raises it.
So a switched-on G protein switches on the cAMP-making enzyme.
So the G protein carries the switch from receptor to enzyme.

Q17 T42-q17

Cells of the electric organ of a ray carry a receptor for acetylcholine. When acetylcholine binds, positive ions flow into the cell within a millisecond. Researchers purified the receptor protein and set it, on its own, into an artificial membrane that holds no other protein. When the researchers added acetylcholine, positive ions crossed this membrane within a millisecond.

What kind of receptor is this, and how does it produce the current?

  1. A. A G protein-coupled receptor; cAMP made inside the cell opens a separate channel
    The artificial membrane held no protein but the receptor, yet the ions crossed within a millisecond.
  2. B. An intracellular receptor; acetylcholine enters the cell and opens a channel from inside
    Acetylcholine binds a receptor at the cell surface; here there was no cell to enter.
    What changes is the receptor’s shape, and the receptor is itself the channel.
  3. C. A cell-surface receptor whose inner part is a kinase; it phosphorylates a channel, which then opens
    A kinase would have to phosphorylate a separate channel protein, and the artificial membrane held no protein but the receptor.
    The receptor's own shape change opens the channel.
  4. D. ✓ A ligand-gated channel; the receptor's own shape change opens a channel through it

Why: A receptor that is itself a channel is a ligand-gated channel.
The bound ligand changes the receptor’s shape: the channel through it opens, and ions flow within a millisecond.
Nothing else stands between ligand and ions.
So the receptor alone, in a membrane with no other protein, passes the ions.

Q18 T42-q18

Three responses to signals are recorded. A cell of the adrenal gland releases stored epinephrine within two seconds of its signal arriving. A liver cell holds twice as much of a new enzyme six hours after its signal arrives. Cells in a root tip begin dividing two days after a plant hormone arrives.

Which of the following classifies the three responses, in order?

  1. A. A change in gene expression; secretion; growth and division
    Releasing stored epinephrine in two seconds is secretion: packed vesicles fuse with the membrane.
    An enzyme that doubles over six hours is new protein: a change in gene expression.
  2. B. ✓ Secretion; a change in gene expression; growth and division
  3. C. Secretion; growth and division; a change in gene expression
    A liver cell holding twice as much of a new enzyme has expressed a gene; nothing has divided.
    The root tip cells dividing two days later are growth and division.
  4. D. Growth and division; a change in gene expression; secretion
    The adrenal cell released a molecule it had stored, so its response is secretion.
    The dividing root cells are the growth and division response.

Why: The adrenal cell’s vesicles fused with its membrane and released epinephrine: secretion.
The liver cell made new protein, a new enzyme: a change in gene expression.
The root tip cells began dividing: growth and division.

Q19 T42-q19

The model below shows the growth factor pathway of a fruit fly cell as boxes numbered 1 to 6. A researcher adds the growth factor and then counts the phosphate groups on kinase B (box 4).

The growth factor pathway of a fruit fly cell drawn as boxes 1 to 6 joined by arrows; the shaded band is the plasma membrane.
The growth factor pathway of a fruit fly cell drawn as boxes 1 to 6 joined by arrows; the shaded band is the plasma membrane.

Which stage of the pathway does this measurement belong to, and at which box does the message cross the membrane?

  1. A. ✓ Transduction; box 2
  2. B. Transduction; box 3
    Kinase A at box 3 is already inside the cell, and it is switched on because the receptor at box 2 changed shape.
    The crossing is at box 2.
  3. C. Transduction; box 4
    The message has already crossed by box 4.
    It crosses at the receptor, box 2, whose intracellular domain changes shape when the growth factor binds outside.
  4. D. Reception; box 2
    Reception is the growth factor, box 1, binding its receptor, box 2.
    Phosphates added to kinase B inside the cell are the relay carrying the message on: transduction.

Why: Phosphates added to a kinase inside the cell are the message being relayed, one molecule changing the next: transduction.
The message crosses the membrane at the receptor, box 2, where the growth factor binds outside and the receptor's intracellular domain changes shape inside.

FRQ 1 T42-frq1 · Scientific Investigation

Plant cells defend themselves against fungi. A short chain of sugars broken from a fungus's cell wall binds a receptor on the surface of a plant cell. Inside the cell the activated receptor phosphorylates kinase 1, kinase 1 phosphorylates kinase 2, kinase 2 phosphorylates kinase 3, and the cell then makes a defense protein over the following hours. Researchers grow plant cells in eighteen dishes and give six dishes each of three treatments: fresh medium; the sugar fragment; or the sugar fragment together with a drug that stops kinase 2 from being phosphorylated. After six hours they measure the defense protein in each dish, in units. The graph below shows the mean for each treatment, and the error bars represent ±2SE.

Defense protein made by cultured plant cells in six hours, in units, six dishes per treatment. Error bars represent ±2SE. Gridlines every 5 units.
Defense protein made by cultured plant cells in six hours, in units, six dishes per treatment. Error bars represent ±2SE. Gridlines every 5 units.

(a) Identify the independent variable and the dependent variable in this investigation. (1 pt)

Model answer Independent variable: the treatment each dish receives (fresh medium, the sugar fragment, or the sugar fragment with the kinase 2 blocker).
Dependent variable: the amount of defense protein the cells make in six hours, in units.
Rubric
  • Award 1 point for both: independent variable, the treatment the dishes receive (fresh medium, the sugar fragment, or the fragment with the kinase 2 blocker); dependent variable, the amount of defense protein made in six hours, in units.
  • Accept 'whether the fragment and the blocker are present' for the independent variable. Do not award the point if the two are reversed, or if kinase 2 or the receptor is named as a variable.

Slip Naming kinase 2 as a variable. Nothing about kinase 2 is measured or set as a level; the researchers change the treatment and measure the protein.

(b) State the null hypothesis for the comparison between the dishes given the sugar fragment and the dishes given fresh medium. (1 pt)

Model answer The sugar fragment makes no difference to the amount of defense protein the plant cells make in six hours, compared with cells given fresh medium.
Rubric
  • Award 1 point for a no-difference statement that names the tested factor and the measured result: the sugar fragment makes no difference to the amount of defense protein the cells make in six hours (compared with fresh medium).
  • Accept 'the fragment has no effect on defense protein output'. Do not award the point for a prediction of a difference in either direction, for a null about the blocker, or for a statement that names neither the fragment nor the protein.

Slip Writing the researchers' own prediction ('the fragment raises the defense protein') as the null. The null hypothesis predicts no difference, and it names both the factor changed and the quantity measured.

(c) Predict how the amount of phosphate carried by kinase 1 in the dishes given the fragment with the blocker compares with the dishes given the fragment alone, and justify your prediction. (1 pt)

Model answer Kinase 1 carries about the same amount of phosphate in both sets of dishes.
The receptor switches on kinase 1 first; kinase 1 then phosphorylates kinase 2.
The drug stops kinase 2 from being phosphorylated, a step after kinase 1.
So kinase 1 is switched on as usual, and only kinase 2, kinase 3 and the defense protein after them are affected.
Rubric
  • Award 1 point for: about the same amount of phosphate on kinase 1 in both, because kinase 1 is switched on by the receptor before kinase 2 in the chain, and the drug acts at kinase 2; each kinase is switched on by the one before it, so blocking a later step leaves the earlier steps as they were.
  • Do not award the point for 'less phosphate on kinase 1 because the pathway is blocked' or for 'more phosphate on kinase 1 because it cannot pass it on' (each kinase keeps its own phosphate and takes a fresh one from ATP for the next).

Slip Predicting less phosphate on kinase 1 because 'the pathway is blocked'. A block stops everything after it and nothing before it; kinase 1 is switched on by the receptor, which the drug leaves alone.

(d) Support the claim that the rise in defense protein produced by the sugar fragment needs kinase 2, using evidence from the error bars. (1 pt)

Model answer The bars represent ±2SE.
The fragment bar runs from 34 to 42 units and the fresh-medium bar from 3 to 5 units.
They do not overlap, so the rise from 4 to 38 units is unlikely to be chance.
So the fragment raised the defense protein.
The fragment-plus-blocker bar runs from 3 to 7 units and overlaps the fresh-medium bar.
So with kinase 2 blocked the fragment made no difference these data can show.
Therefore the rise needs kinase 2.
Rubric
  • Award 1 point for: the evidence (the fragment bar, 34 to 42 units, does not overlap the fresh-medium bar, 3 to 5 units; the fragment-plus-blocker bar, 3 to 7 units, overlaps the fresh-medium bar) AND the reasoning (the fragment raised the protein, and with kinase 2 blocked the data show no rise, so the rise needs kinase 2). Both bar readings and the link are needed.
  • Do not award the point for the readings with no link to the claim, for a comparison of means alone, or for 'the blocker lowered the protein' (the blocker dishes are compared with fresh medium, and their bars overlap).

Slip Quoting the bars and stopping, or reasoning with no readings. Supporting a claim needs the evidence and the link that ties it to the claim.

FRQ 2 T42-frq2 · Analyze Model

Some frogs darken their skin in dim light. Hormone J, a peptide released into the blood by a gland at the base of the brain, reaches the skin cells, and within minutes their pigment granules spread through the cell, so the skin darkens. The model below shows the pathway as boxes 1 to 7; hormone J binds a G protein-coupled receptor. Researchers treated skin cells with a drug: the X on the model marks the step the drug blocks, between box 5 and box 6.

The pathway by which hormone J darkens a frog skin cell, drawn as boxes 1 to 7 joined by arrows. The shaded band is the plasma membrane. The X marks the step the drug blocks: box 5 switching on kinase K.
The pathway by which hormone J darkens a frog skin cell, drawn as boxes 1 to 7 joined by arrows. The shaded band is the plasma membrane. The X marks the step the drug blocks: box 5 switching on kinase K.

(a) Describe what happens at the receptor (box 2) when hormone J binds, and how this passes the message to the G protein (box 3). (1 pt)

Model answer Hormone J fits the receptor's binding site by shape and charge, and while it is bound the receptor holds a different shape all the way to its intracellular domain, the part facing the cytosol.
That new shape switches on the G protein sitting beside it on the inner face of the membrane.
Hormone J itself stays outside the cell; what crosses the membrane is the change in the receptor's shape.
Rubric
  • Award 1 point for: hormone J fits the receptor's binding site by shape and charge, and while it is bound the receptor holds a different shape; the change reaches its intracellular domain, whose new shape switches on the G protein on the inner face of the membrane, while hormone J itself stays outside.
  • Accept 'the receptor changes shape on its inner side and that switches on the G protein'. Do not award the point for 'hormone J passes through the receptor to the G protein' or for 'the receptor phosphorylates the G protein'.

Slip Sending hormone J into the cell. A peptide cannot cross the membrane; the message crosses as a change of protein shape.

(b) Identify the second messenger in the model, and describe what it does. (1 pt)

Model answer The second messenger is cAMP, box 5.
Enzyme E, switched on by the G protein, makes many molecules of cAMP from ATP; cAMP is a small molecule, so it spreads quickly through the cytosol and switches on kinase K.
Rubric
  • Award 1 point for: cAMP (box 5), a small molecule made from ATP by enzyme E when the G protein switches the enzyme on; it spreads through the cytosol and switches on kinase K.
  • Do not award the point for naming the G protein or enzyme E as the second messenger, or for 'cAMP is a protein'.

Slip Naming enzyme E as the second messenger. The enzyme makes the second messenger; the small molecule it makes, cAMP, is the messenger.

(c) Make a claim about the level of the box 5 molecule and the darkening of the skin when hormone J reaches the drug-treated cells, compared with untreated cells given hormone J, and support your claim using the model. (2 pt)

Model answer cAMP at box 5 rises in the drug-treated cells as in untreated cells.
The skin of the drug-treated cells stays pale.
The X sits between box 5 and box 6, kinase K.
Hormone J still binds the receptor, so the G protein switches on enzyme E.
Enzyme E makes cAMP, so cAMP rises.
The X stops cAMP switching kinase K on, so kinase K stays off.
So the pigment granules stay clumped and the skin stays pale.
Rubric
  • Award 1 point for the claim: the box 5 molecule (cAMP; accept 'the box 5 molecule' or 'the second messenger' without the name, since the identification is scored in (b)) rises as it does in untreated cells, AND the skin stays pale. Both halves are needed. Do not award the point for 'cAMP stays low' or for 'the skin darkens as usual'.
  • Award 1 point for the support: the evidence from the model (the X sits between cAMP, box 5, and kinase K, box 6) AND the reasoning (everything before the X, boxes 1 to 5, works as before, so cAMP rises; cAMP cannot switch kinase K on, so kinase K stays off and the pigment granules stay clumped). Do not award the point for naming the position of the X with no link to the claim, or for reasoning that never refers to the model.

Slip Claiming that the cAMP concentration stays low, or giving the reasoning without pointing to where the X sits on the model. The block sits after cAMP, so cAMP is still made; supporting a claim means naming the evidence and linking it to the claim.

(d) Cells of the frog's adrenal gland carry the same kind of receptor for hormone J, the same G protein, enzyme E and kinase K, and respond to hormone J by releasing a steroid hormone. Explain why the two cell types respond differently to hormone J. (1 pt)

Model answer Hormone J is the same molecule at both cells, and in both cells the signal passes along the same pathway as far as kinase K.
The difference is in the response proteins each cell holds.
In a skin cell, kinase K reaches the proteins that move pigment granules; in an adrenal cell it reaches the proteins that release the steroid hormone.
So the same ligand ends at whatever proteins the cell has.
Rubric
  • Award 1 point for: the response depends on the proteins each cell holds; the pathway is identical to kinase K, and kinase K then reaches whatever response proteins that cell has: the proteins that move pigment granules in the skin cell, the proteins that release the steroid in the adrenal cell. Specificity lives in the receiving cell's proteins, not in hormone J.
  • Do not award the point for 'the adrenal cells receive more hormone J', 'hormone J is different in the adrenal gland', or 'the adrenal cells have a different receptor' (the stimulus says the receptor is the same kind).

Slip Putting the difference in the hormone or in the amount of it. More hormone makes a response stronger; it cannot turn pigment movement into steroid release. The proteins at the end of the pathway differ.

APBIO-U04-L08 Bananas in a paper bag

Topic 4.3 · Signal Transduction Pathways · 62 steps

A photograph of a pile of bananas, some yellow and ripe, some still green and hard; beside it a drawing of a closed paper bag labelled ripe banana on the left and green banana on the right, with small dots for gas crossing from the ripe one to the green one
A photograph of a pile of bananas, some yellow and ripe, some still green and hard; beside it a drawing of a closed paper bag labelled ripe banana on the left and green banana on the right, with small dots for gas crossing from the ripe one to the green one

Photos: Scott Webb, Wikimedia Commons, CC0 (resized; one banana cropped from it); Evan-Amos, Wikimedia Commons, CC BY-SA 3.0 (resized).

Here are two bananas sealed in a paper bag: one ripe and yellow, one hard and green.

Two days later the green banana is soft and sweet. The ripe fruit gave off a gas, and the green fruit’s cells detected it.

What did the gas make the cells do, and why did it take two days?

Unit 4 · Cell Communication and Cell Cycle

1Genes switched on by a gas

2

Video: Watch: Genes switched on by a gas

Ethylene crosses the bag, binds a receptor on a green banana cell, and the message reaches the DNA; hours later the cell holds new enzymes.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L08a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L08a.mp4

3
Check q1

A liver cell has the gene for the insulin receptor, and the liver cell makes the insulin receptor protein from that gene.

Which of the following is the liver cell doing with that gene?

  1. A. ✓ Expressing the gene
  2. B. Copying the gene into a neighboring cell
    A gene stays in the cell’s own DNA.
    Making the protein from the gene is expressing it.
  3. C. Removing the gene from its DNA
    The gene is still in the DNA: the cell is reading it, not removing it.

Why: A cell is expressing a gene when it is using that gene’s instructions to make the protein.
The liver cell is making the insulin receptor from its gene.
So the liver cell is expressing the gene.

4

What is the slowest response a signal can make a cell give, and why is it slow?

5

The gas is called ethylene. Ethylene binds receptors in the green banana’s cells, and the last activated protein of the pathway reaches the DNA.

6

There it changes gene expression. The cell starts making enzymes it was not making before: one softens the cell walls, and one turns starch into sugar.

7

Enzymes have to be built, so the response takes hours to days. Once built, the enzymes keep working after the signal is gone.

8

Compare epinephrine at a liver cell. Epinephrine switches on enzymes the liver cell already holds, so glucose leaves within seconds and stops when the signal stops.

9

Fast and gone, or slow and lasting, tells you whether the cell made a new protein.

10

Ethylene is a gas that ripening fruit gives off. The ripe banana’s cells release it into the air of the bag.

A closed paper bag on day 0: a photograph of a ripe yellow banana on the left and a hard green banana on the right, with small dots for molecules of ethylene gas spreading from the ripe one to the green one
A closed paper bag on day 0: a photograph of a ripe yellow banana on the left and a hard green banana on the right, with small dots for molecules of ethylene gas spreading from the ripe one to the green one
11

Ethylene diffuses through the air in the bag to the green banana’s cells. There it binds its receptors.

12

The receptor changes shape and passes the message to the relay inside the cell. The last activated protein of this pathway reaches the DNA.

A fruit cell with ethylene bound at a receptor in its membrane, the receptor, membrane and relay proteins labeled, the relay passing the message to the nucleus, and new enzymes appearing in the cytosol
A fruit cell with ethylene bound at a receptor in its membrane, the receptor, membrane and relay proteins labeled, the relay passing the message to the nucleus, and new enzymes appearing in the cytosol
13

There it changes gene expression. The cell starts making enzymes it was not making before: one softens the cell walls, another turns starch into sugar.

14

Here is the word equation for the reaction the second new enzyme speeds up.

starch gives sugar
15

Other signals change gene expression the same way. A yeast cell that detects a mating signal expresses its mating genes.

16

Dense bacteria, in quorum sensing, express the genes for making light.

17

Cytokines are signals that make immune cells express the genes for dividing.

18

What you are expected to know Describe how a signal changes gene expression: the last activated protein of the pathway reaches the DNA, and the cell starts or stops making particular proteins.

19
Check q2

A researcher gives yeast cells a mating signal, a peptide that stays outside the cell. Two hours later the cells hold mating proteins that were absent before.

Which of the following produced the new proteins?

  1. A. The receptor released proteins stored in vesicles
    The cells held no mating proteins before the signal.
    So there were no mating proteins in vesicles to release.
    The cells had to make the proteins from their genes.
  2. B. The signal entered the nucleus and was copied into protein
    The mating signal is a peptide, so it stays outside the cell.
    The last activated protein of the relay reaches the DNA, not the signal.
  3. C. ✓ The cells expressed their mating genes
  4. D. cAMP was converted into the proteins
    cAMP is a small messenger molecule made from ATP.
    It is never turned into protein; proteins are made from genes.

Why: The mating signal is a peptide, so it stays outside the cell.
The signal binds its receptor, and the receptor passes the message to the relay inside.
The last activated protein of the relay reaches the DNA.
So the cells express their mating genes and build the mating proteins.

20Slow to start, and lasting

21

Video: Watch: Slow to start, and lasting

The same green banana cell over two days: the enzymes appear slowly, and they keep working after the ethylene has gone from the bag.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L08b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L08b.mp4

22

A response that works through gene expression is slow to start. The enzymes have to be built, and the banana takes about two days.

The same bag on day 2: photographs of two yellow bananas, both now ripe and soft
The same bag on day 2: photographs of two yellow bananas, both now ripe and soft
23

The response also lasts. Once the cell has made the enzymes, they keep working after the ethylene is gone.

24

Ethylene binds a receptor and, in time, lets go again. The enzymes the cell built because of the ethylene keep working for days.

25

The signal has to stay long enough for the enzymes to be built. In one experiment, fruit given ethylene for less than six hours and then kept in ordinary air stayed unripe.

26

What you are expected to know Explain why a response through gene expression is slow to start and lasting: the enzymes have to be built, and once built they keep working after the signal is gone.

27
Practice writing an answer

A researcher gives yeast cells a mating signal, a peptide that stays outside the cell. Two hours later the cells hold mating proteins that were absent before.

(a) Explain why the mating proteins appear two hours after the signal, rather than within seconds. (1 pt)

Model answer The cells held no mating proteins before the signal arrived.
So the mating proteins had to be made.
To make a protein, the cell expresses the protein’s gene and builds the protein from its instructions.
Building a protein from its gene takes minutes to hours.
So the mating proteins appear only after two hours.
A response that only switches on a protein the cell already holds takes seconds, but here there was no such protein to switch on.
Rubric
  • Award 1 point for: the proteins did not exist before, so the cells had to express the mating genes and build the proteins, and building a protein from its gene takes hours (a protein already present could be switched on in seconds).
28
Check q3

Tomatoes picked green spend one day in air containing ethylene, then a week in ordinary air. In earlier experiments, tomatoes given ethylene for less than six hours stayed unripe once it was removed. By the end of the week the tomatoes are red and soft. A student says: “The ethylene must stay bound to the receptors all week, or the ripening would stop.”

Is the student correct?

  1. A. Yes — a ligand stays bound until the response is complete
    The enzymes the cells built during the day in ethylene keep working all week.
    An enzyme that already exists works whether or not ethylene is bound.
  2. B. ✓ No — the enzymes already built keep working whether or not ethylene is bound
  3. C. No — the ethylene entered the cells and is used up slowly over the week
    Ethylene is a ligand.
    The receptor binds ethylene and lets it go unchanged.
    Ethylene is not used up.

Why: During the day in ethylene, the cells built the ripening enzymes.
An enzyme keeps working for as long as it exists.
So the ripening enzymes keep working all week.
Ethylene does not have to stay bound for an enzyme that already exists to work.
So the student is wrong.

29
Practice writing an answer

Tomatoes picked green spend one day in air containing ethylene, then a week in ordinary air. By the end of the week they are red and soft.

(a) Explain why the tomatoes go on ripening for a week after the ethylene is gone. (1 pt)

Model answer During the day with ethylene, ethylene bound the receptors in the tomato cells.
The relay passed the message on, and the last activated protein reached the DNA.
So the cells expressed the genes for the ripening enzymes and built them.
An enzyme keeps working for as long as it exists.
So the ripening enzymes keep softening the fruit and turning its starch to sugar all week.
The ethylene was needed only to start the enzymes being made.
Rubric
  • Award 1 point for: the ethylene changed gene expression, so the cells built the ripening enzymes, and enzymes already built keep working after the signal is gone.
30
Check q4

A researcher dilutes dense light-producing bacteria into fresh medium, where the signal concentration is far below its threshold. The bacteria keep glowing for an hour before fading. By the next day the light is gone.

Why do the bacteria keep glowing for an hour after the dilution?

  1. A. ✓ The light-making proteins were already made, and they last for a while
  2. B. The signal molecules already bound to the receptors stay bound for about an hour
    A ligand binds and lets go within seconds; in fresh medium the receptors empty and stay empty.
    What lasts for the hour is the light-making protein already built.
  3. C. The receptors hold their changed shape for about an hour after the signal leaves
    A receptor returns to its resting shape as soon as its ligand leaves.
    What lasts for the hour is the light-making protein the cells had already built.

Why: Quorum sensing changed gene expression: the dense cells built the light-making proteins.
After the dilution, the signal is too dilute to bind, so no new light-making protein is made.
The light-making proteins already made keep working for a while, so the glow fades over an hour.

31Changing what the cell is doing right now

32

Video: Watch: Changing what the cell is doing right now

Epinephrine at a liver cell: cAMP rises, kinases add phosphates to enzymes already there, glucose leaves within seconds, and phosphatases switch it all off again.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L08c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L08c.mp4

33
Check q5

A kinase has added a phosphate to an enzyme in a muscle cell, and the phosphate has switched the enzyme on.

Which of the following removes that phosphate again?

  1. A. A second kinase
    A kinase adds a phosphate to a protein.
    Removing a phosphate is a different enzyme’s job.
  2. B. ✓ A phosphatase
  3. C. The receptor
    The receptor sits in the membrane and binds the ligand outside the cell.
    The phosphate is on an enzyme inside the cell.

Why: A kinase adds a phosphate to a protein.
The enzyme that removes the phosphate again is called a phosphatase.
So a phosphatase removes the phosphate, and the enzyme switches off.

34

Not every response needs a new protein. Epinephrine at a liver cell switches on enzymes the cell already holds, and glucose leaves within seconds.

35

Suppose a researcher gives liver cells a 2-second pulse of epinephrine and measures their cAMP and their glycogen breakdown. Here is a table of what she measures.

A table of the timing after a 2-second pulse of epinephrine to liver cells: at 5 seconds cAMP is four times its resting level while glycogen breakdown has barely changed; at 15 seconds breakdown is 3.5 times its resting rate; at 45 seconds cAMP is nearly back to rest
A table of the timing after a 2-second pulse of epinephrine to liver cells: at 5 seconds cAMP is four times its resting level while glycogen breakdown has barely changed; at 15 seconds breakdown is 3.5 times its resting rate; at 45 seconds cAMP is nearly back to rest
36

At 5 seconds, cAMP is four times its resting level, and glycogen breakdown has barely changed.

37

At 15 seconds, glycogen breakdown is 3.5 times its resting rate. By 45 seconds, cAMP is nearly back to its resting level.

38

No gene was expressed. cAMP switched on kinases, and the kinases added phosphates to enzymes that were there all along.

39

Insulin does the same at a muscle cell: it moves glucose transporters the cell already has to its surface, within minutes.

40

When the signal stops, the phosphatases remove the phosphates, and the response stops.

41

So a signal can change what a cell is doing this second, using proteins already there: fast to start, and fast to stop.

42

Here is a table comparing the two kinds of response: how fast each starts, and how long it lasts.

A two-row contrast: switching on a protein the cell already has takes seconds and stops when the signal stops; expressing a gene to make a new protein takes minutes to hours and lasts after the signal is gone
A two-row contrast: switching on a protein the cell already has takes seconds and stops when the signal stops; expressing a gene to make a new protein takes minutes to hours and lasts after the signal is gone
43

A response through gene expression takes minutes to hours. That response lasts after the signal is gone.

44

A response through existing proteins takes seconds. That response stops when the signal stops.

45

What you are expected to know Describe a response that uses proteins the cell already has: an existing enzyme, channel or transporter is switched on or off, so the change takes seconds and stops when the signal stops.

46
Check q6

A researcher gives fat cells a 3-second pulse of a hormone. At 4 seconds cAMP is up fivefold, and the breakdown of stored fat has barely changed. At 12 seconds fat breakdown is three times its resting rate. At 40 seconds cAMP is nearly back to rest.

Which of the following sequences fits the timing?

  1. A. Receptor binds → stored fat broken down → cAMP made from the released fat
    At 4 seconds cAMP had already risen fivefold, and fat breakdown had barely begun.
    So cAMP comes first.
  2. B. cAMP leaves the cell → binds the receptor → stored fat broken down
    The cell makes cAMP inside and uses it inside.
    The hormone, outside the cell, binds the receptor.
  3. C. Receptor binds → genes for new enzymes expressed → stored fat broken down
    Fat breakdown was up threefold within 12 seconds.
    Expressing a gene and building an enzyme takes minutes to hours.
    So the enzymes were already there.
  4. D. ✓ Receptor binds → cAMP rises → existing enzymes switched on → stored fat broken down

Why: The hormone bound its receptor, and cAMP rose first, within 4 seconds.
cAMP switched on kinases.
The kinases added phosphates to fat-breaking enzymes the cell already held, switching them on.
So fat breakdown rose by 12 seconds.
No new protein was made, which is why the response took seconds.

47

Back to the two bananas sealed in a paper bag, one ripe and yellow, one hard and green. The ripe banana’s ethylene bound receptors in the green banana’s cells.

48

Those cells expressed the genes for the enzymes that soften the flesh and turn starch into sugar. Building the enzymes took two days, and the enzymes kept working after the bag was opened.

49Quick quiz: a new protein, or a protein already there? mixed practice

50
Check q7

Within three seconds of a hormone arriving, a muscle cell’s stored glycogen begins to break down.

Which of the following did the cell do?

  1. A. The cell made a new protein
    Three seconds is far too short to express a gene and build a protein.
    So the glycogen-breaking enzymes were already in the cell, and the hormone switched them on.
  2. B. ✓ The cell switched on proteins it already had

Why: The response began within three seconds.
Expressing a gene and building a protein takes minutes to hours.
So no new protein was made in three seconds.
So the muscle cell switched on glycogen-breaking enzymes it already held.

51
Check q8

Six hours after a plant hormone arrives, a root cell holds an enzyme it lacked before.

Which of the following did the cell do?

  1. A. ✓ The cell made a new protein
  2. B. The cell switched on proteins it already had
    The cell lacked the enzyme before the hormone arrived, so there was nothing to switch on.
    The cell expressed the enzyme’s gene and built the enzyme, which takes hours.

Why: The root cell lacked the enzyme before the hormone arrived.
So the enzyme had to be made.
The cell expressed the enzyme’s gene and built the enzyme.
Building a protein from its gene takes hours, which fits the six hours.
So the cell made a new protein.

52
Check q9

A signal arrives, and within a minute a channel in the cell’s membrane opens and ions flow in.

Which of the following did the cell do?

  1. A. The cell made a new protein
    A minute is far too short to express a gene and build a protein.
    The channel was already in the membrane, and the signal opened it.
  2. B. ✓ The cell switched on proteins it already had

Why: The channel opened within a minute.
A minute is too short to make a new protein.
So the channel was already there, and the signal switched it on: the cell switched on a protein it already had.

53
Check q10

A day after a signal arrives, a gland cell holds a protein that was absent before, and the protein is still there after the signal is gone.

Which of the following did the cell do?

  1. A. ✓ The cell made a new protein
  2. B. The cell switched on proteins it already had
    The cell held none of this protein before, so there was nothing to switch on.
    It expressed the gene and made the protein, which took a day.

Why: The gland cell lacked the protein before.
So the cell had to make the protein by expressing its gene.
Making a protein takes hours, and a protein that is made lasts after the signal is gone.
So the cell made a new protein.

54
Check q11

Within a minute of a signal, a gland cell’s stored vesicles fuse with its membrane and release their contents; when the signal stops, the release stops.

Which of the following did the cell do?

  1. A. The cell made a new protein
    The vesicles were already in the cell; the signal made them fuse with the membrane.
    A minute is too short to make a protein, and release stopped with the signal.
  2. B. ✓ The cell switched on proteins it already had

Why: The vesicles were already stored in the cell.
The signal switched on the proteins that make vesicles fuse with the membrane, within a minute.
When the signal stopped, the release stopped.
Both observations show the same thing: the cell switched on proteins it already had.

55Mixed practice mixed practice

56
Check q12

Which of the following responses works through a change in gene expression?

  1. A. Epinephrine switching on a liver cell’s existing enzymes
    The enzymes were already in the liver cell.
    A phosphate switched them on in seconds, and no gene was expressed.
  2. B. Insulin moving existing transporters to a muscle cell’s surface
    The transporters already existed in the muscle cell.
    Insulin moved them to the surface, and no gene was expressed.
  3. C. ✓ Ethylene making a fruit cell produce new softening enzymes
  4. D. Acetylcholine opening a channel in a muscle cell
    The channel is the receptor itself, and the channel was already in the membrane.
    Acetylcholine opened it, and no gene was expressed.

Why: The softening enzymes did not exist in the fruit cell before the signal.
So the fruit cell had to express their genes and build them.
That is a change in gene expression, and it is why ripening takes days.

57
Check q13

Which of the following responses is complete within seconds?

  1. A. ✓ A liver cell’s existing enzymes breaking glycogen down
  2. B. A fruit cell making new enzymes that soften its cell walls
    The fruit cell must make the new enzymes from their genes, and making a protein takes hours.
  3. C. A skin cell dividing after a growth factor arrives
    A division takes about a day.
  4. D. A yeast cell making new mating proteins
    The yeast cell must make the new proteins from their genes, and making a protein takes hours.

Why: The liver cell already holds its glycogen-breaking enzymes.
A phosphate switches them on within seconds.
Every other response listed needs a new protein or a whole division, and those take hours to a day.

58
Check q14

Within 20 seconds of a hormone arriving, a fat cell’s existing lipid-breaking enzyme is active. If a researcher adds a phosphatase inhibitor, the enzyme stays active long after the hormone is washed away.

What switched the enzyme on?

  1. A. The gene for the enzyme was expressed
    20 seconds is far too short to express a gene and build a protein.
    The enzyme was already there.
  2. B. ✓ A kinase added a phosphate to the enzyme the cell already had
  3. C. The hormone bound the enzyme directly
    The hormone stays outside the cell and binds its receptor.
    The enzyme is inside the cell.
  4. D. cAMP was built into the enzyme’s structure
    cAMP switches on kinases; cAMP is not built into proteins.

Why: The enzyme already existed in the fat cell.
A kinase added a phosphate to the enzyme, which switched it on within 20 seconds.
Normally a phosphatase removes the phosphate when the hormone is gone, but the inhibitor stops that.
So the phosphate stays on and the enzyme stays active.

59
Check q15

Dense light-producing bacteria begin to glow about an hour after their quorum signal reaches them. A researcher adds a chemical that stops all gene expression, and then adds the signal.

Predict what the bacteria do over the next hour.

  1. A. They glow as normal
    A dark cell holds no light-making proteins; it makes them by expressing the genes for light.
    The chemical stops all gene expression, so no light-making protein is made.
  2. B. They glow, but later than normal
    The chemical stops gene expression for as long as it is present.
    With no gene expressed, no light-making protein is ever made, however long the cells wait.
  3. C. ✓ They stay dark

Why: The quorum signal binds its receptors, and the relay reaches the DNA.
Then the cells express the light-making genes and build the proteins, taking about an hour.
The chemical stops all gene expression, so the light-making proteins are never made and the bacteria stay dark.

60
Check q16

A researcher seals a hard green banana alone in a bag and releases a small amount of ethylene gas into the bag. No ripe banana is present.

Predict the green banana after two days.

  1. A. Still green and hard
    The ripe banana’s only part in ripening the green one is to release ethylene.
    Ethylene from any source binds the same receptors and starts the same pathway.
  2. B. ✓ Soft and sweet

Why: Ethylene in the bag binds receptors on the banana cells.
The receptors change shape and the relay passes the message on to the DNA.
The cells express the softening-enzyme genes and build the enzymes, taking about two days.
So after two days the banana is soft and sweet.

61
Practice writing an answer

A researcher seals a hard green banana alone in a bag and releases a small amount of ethylene gas into the bag. No ripe banana is present. After two days the banana is soft and sweet.

(a) Explain how this result demonstrates that ethylene alone is the signal that changes gene expression in the green banana’s cells. (1 pt)

Model answer The ripe banana is only ethylene’s usual source.
The released ethylene binds the receptors on the green banana’s cells.
The receptors change shape, and the relay passes the message on to the DNA.
So the cells express the genes for the softening enzymes and the starch-to-sugar enzyme, and build them.
Those enzymes soften the flesh and turn its starch to sugar.
So the banana ripens with no ripe banana present: the gas alone was the signal.
Rubric
  • Award 1 point for: ethylene is the signal (the ripe banana is only its source), so released ethylene binds the receptors, the pathway changes gene expression, the softening enzymes are made, and the banana ripens with no ripe banana present.

APBIO-U04-L08B From a changed cell to a changed organism

Topic 4.3 · Signal Transduction Pathways · 40 steps

A photograph of a ripe banana cut into pieces on a white plate, the yellow skin around soft pale flesh; beside it a small drawing of one banana cell holding three new enzyme molecules, labelled one cell, one enzyme at a time
A photograph of a ripe banana cut into pieces on a white plate, the yellow skin around soft pale flesh; beside it a small drawing of one banana cell holding three new enzyme molecules, labelled one cell, one enzyme at a time

Photo: Judgefloro, Wikimedia Commons, CC0 (resized).

Cut the ripe banana open: soft flesh, sweet taste, yellow skin.

Those are features you can see and taste on the whole fruit. Yet everything the gas did happened inside single cells, one enzyme at a time.

How does a change inside cells become a change you can see on the whole organism?

Unit 4 · Cell Communication and Cell Cycle

1What counts as phenotype

2

Video: Watch: What counts as phenotype

The cut banana: its softness, sweetness and yellow skin are features of the whole fruit, and together they are its phenotype; the base sequence of its DNA is not.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L08Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L08Ba.mp4

3
Check q1

Every cell of a banana plant carries the plant’s DNA, and one stretch of that DNA carries the instructions for the softening enzyme.

Which of the following is that stretch of DNA called?

  1. A. ✓ A gene
  2. B. A receptor
    A receptor is a protein that binds a signal.
    The instructions for making a protein are a stretch of DNA.
  3. C. A ligand
    A ligand is the signal molecule that binds a receptor.
    The instructions for making a protein are a stretch of DNA.

Why: A stretch of a cell’s DNA that carries the instructions for making one protein is called a gene.
The instructions for the softening enzyme are one such stretch.
So that stretch of DNA is a gene.

4

Cut the ripe banana open: soft flesh, sweet taste, yellow skin. Those are features you can see and taste on the whole fruit.

A photograph of a ripe banana cut into pieces: yellow skin around soft pale flesh; the caption names its soft flesh, sweet taste and yellow skin as features you can see and taste on the whole fruit
A photograph of a ripe banana cut into pieces: yellow skin around soft pale flesh; the caption names its soft flesh, sweet taste and yellow skin as features you can see and taste on the whole fruit
5

How does a molecular event show up on a whole organism?

6

An organism has a set of features you can see or measure: what its cells make, how its cells behave, how its body looks.

7

A signal that changes gene expression, or what the cell is doing, in enough cells changes the features you can see or measure. Enough banana cells made the softening enzyme, so the banana went soft.

8

An organism’s set of observable features is called its .

9

Phenotype means anything about the organism or its cells that can be observed or measured: what a cell makes, how it behaves, how the body looks.

10

The DNA sequence itself is not the phenotype. The DNA sequence is called the .

11

The banana’s sweetness is part of its phenotype, because you can taste it. The base sequence of its softening-enzyme gene is not, because that sequence is the genotype.

12

What you are expected to know Identify what counts as an organism’s phenotype: its observable features, including what its cells make and how they behave; the DNA sequence itself is the genotype.

13Quick quiz: phenotype mixed practice

14
Check q2

Which of the following is an organism’s phenotype?

  1. A. ✓ Its set of observable features
  2. B. The base sequence of its DNA
    The base sequence of the DNA is the genotype.
    The phenotype is what can be observed or measured about the organism.
  3. C. The signals its cells receive
    A signal is a molecule that binds a receptor.
    The phenotype is the set of features the organism shows.

Why: The phenotype is the organism’s set of observable features.
The DNA sequence is the genotype, and a signal is a molecule.
So the phenotype is the set of observable features.

15
Check q3

Which of the following is the name for an organism’s DNA sequence?

  1. A. The phenotype
    The phenotype is the set of features that can be observed or measured.
    The DNA sequence is the genotype.
  2. B. ✓ The genotype

Why: The DNA sequence is the set of instructions inside the cell.
That set of instructions is called the genotype.
So the DNA sequence is the genotype.

16
Practice writing an answer

A ripe banana is soft, sweet and yellow.

(a) State what an organism’s phenotype is. (1 pt)

Model answer An organism’s phenotype is its set of observable features: anything about the organism or its cells that can be observed or measured.
Rubric
  • Award 1 point for: the organism’s observable (or measurable) features.
17
Check q4

A startled person’s heart is beating at 130 beats per minute.

Is the heart rate part of the person’s phenotype?

  1. A. ✓ Yes
  2. B. No
    The phenotype is the organism’s set of observable features.
    A heart rate can be measured on the person.
    So the heart rate is part of the phenotype.

Why: The phenotype is the organism’s set of observable features.
A heart rate can be measured on the whole person.
So the heart rate is part of the phenotype.

18
Check q5

A petal cell of a sunflower carries a pigment gene with a particular DNA sequence.

Is the DNA sequence part of the sunflower’s phenotype?

  1. A. Yes
    A gene’s DNA sequence is the genotype: the instructions the cell reads.
    The phenotype is what those instructions lead to, such as the petal’s color.
  2. B. ✓ No

Why: A gene’s DNA sequence is the set of instructions inside the cell: the genotype.
The phenotype is what can be observed or measured about the organism or its cells.
The sequence leads to the petal color, and the petal color is part of the phenotype.
The sequence itself is not.

19
Check q6

A corn plant’s stem is 180 cm tall.

Is the stem height part of the plant’s phenotype?

  1. A. ✓ Yes
  2. B. No
    Height is measured on the plant itself with a ruler.
    So the stem height is part of the phenotype.

Why: The phenotype is the set of features you can see or measure on the whole organism.
A stem’s height can be measured on the plant.
So the stem height is part of the phenotype.

20
Check q7

One base in the receptor gene of a liver cell differs from the usual sequence.

Is the changed base sequence part of the organism’s phenotype?

  1. A. Yes
    The base sequence of the receptor gene is the genotype: the instructions the cell reads.
    The phenotype is what those instructions lead to, such as a differently shaped receptor.
  2. B. ✓ No

Why: The base sequence of the receptor gene is the genotype: the instructions the cell reads.
The phenotype is anything about the organism that can be observed or measured.
A changed sequence can lead to a changed receptor and response, and those are phenotype.
The sequence itself is not.

21
Check q8

A ripe banana’s flesh tastes sweet.

Is the sweetness part of the banana plant’s phenotype?

  1. A. ✓ Yes
  2. B. No
    Sweetness is a feature of the fruit itself.
    So the sweetness is part of the phenotype.

Why: The phenotype is the set of observable features of the organism.
The sweetness of the flesh can be tasted on the whole fruit.
So the sweetness is part of the phenotype.

22Enough cells change, and the organism changes

23

Video: Watch: Enough cells change, and the organism changes

One softened cell in a hard banana, then thousands: the fruit softens only when enough cells have changed.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L08Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L08Bb.mp4

24

The banana’s phenotype changed because a signal changed gene expression in enough of its cells. A gas in a bag turned a hard green fruit soft and yellow.

A photograph of a ripe banana cut into pieces: yellow skin around soft pale flesh; the caption names its soft flesh, sweet taste and yellow skin as features you can see and taste on the whole fruit
A photograph of a ripe banana cut into pieces: yellow skin around soft pale flesh; the caption names its soft flesh, sweet taste and yellow skin as features you can see and taste on the whole fruit
25

Suppose only one cell in the banana had made the softening enzyme. That one cell would have softened, and the fruit would have stayed hard.

26

Instead, enough cells changed at once. So the whole fruit softened, and the change showed on the whole organism.

27

A signal that changes what enough cells are doing changes the phenotype too. Epinephrine in your blood is why a startled person’s heart pounds.

28

In an embryo, signals set where legs and wings form. Genes called HOX genes control those signals, so a change in that signaling changes the animal’s body plan.

29

What you are expected to know Explain how a signal changes an organism’s phenotype: enough cells change what they make or do, so the change shows on the whole organism.

30
Check q9

A plant hormone makes the cells of a leaf stop making their green pigment, and over a week the whole leaf turns yellow.

Which of the following is the change in phenotype?

  1. A. The hormone binding its receptor
    Binding is an event inside each leaf cell that starts the change.
    The phenotype is the feature the change leads to, the one seen on the leaf.
  2. B. The gene for the pigment being switched off
    The gene being switched off is the cause inside each cell: the cell stops making green pigment.
    The phenotype is the feature that cause leads to: the leaf turning yellow.
  3. C. The relay proteins changing shape
    A relay protein changing shape is an event inside one cell, part of the pathway that causes the change.
    The phenotype is the feature the pathway leads to.
  4. D. ✓ The leaf turning yellow

Why: The phenotype is anything about the organism that can be observed or measured.
Binding, the relay and the switched-off gene are steps inside each cell: they cause the change.
The yellow leaf is the feature those steps lead to.
So the leaf turning yellow is the change in phenotype.

31
Check q10

A drug blocks the pathway by which epinephrine makes the heart beat faster and harder. A person taking the drug stands up to speak in front of a crowd, and epinephrine floods their blood.

Predict the change in the person’s phenotype, compared with a person taking no drug.

  1. A. Their heart cells stop expressing the receptor gene
    The drug blocks epinephrine’s pathway.
    The drug does not stop the heart cells expressing the receptor gene.
    The phenotype is what is measured on the person: the heart rate.
  2. B. Their liver cells stop releasing glucose
    The drug blocks the heart cells’ pathway.
    The liver cells’ epinephrine pathway is untouched.
  3. C. ✓ Their heart rate rises less than it otherwise would
  4. D. Their heart rate rises more than it otherwise would
    Epinephrine reaches the heart cells, but their pathway is blocked.
    So the epinephrine cannot make the heart speed up.
    So the heart rate rises less, not more.

Why: Epinephrine speeds the heart through a pathway in heart cells, and the drug blocks that pathway.
So epinephrine cannot make the heart beat faster.
The heart rate is a feature that can be measured on the person.
So the change in phenotype is a heart rate that rises less.

32
Practice writing an answer

A hard green banana sits in a room with almost no ethylene in the air. A researcher rests a ripe apple against one spot on its skin for an hour and then moves the apple away. The few banana cells under that spot bind ethylene and build the softening enzyme. Every other cell in the banana binds no ethylene. Two days later the banana is still hard.

(a) Explain how this case demonstrates that a signal changes an organism’s phenotype only when it changes enough cells. (1 pt)

Model answer The softening enzyme works inside the cell that made it.
So only the few cells under the spot softened.
Firmness is a feature of the whole fruit, so firmness is part of the banana’s phenotype.
Almost every cell in the banana stayed unchanged and hard.
So the whole fruit stayed hard, and the phenotype did not change.
The phenotype changes only when enough cells change.
Rubric
  • Award 1 point for: the enzyme changed only the few cells that made it, firmness is a feature of the whole fruit, and almost all the cells were unchanged, so the phenotype (the firmness) did not change; a phenotype changes only when enough cells change.
33

Back to the ripe banana cut open: soft flesh, sweet taste, yellow skin. Ethylene changed gene expression in enough of its cells, so the features of the whole fruit changed.

34

The softness and sweetness you can taste are the phenotype that a gas changed.

35Mixed practice mixed practice

36
Check q11

A hormone makes the pigment-holding cells in a fish’s skin spread their pigment across each cell within a few minutes, and the skin darkens.

Which of the following did the pigment cells do?

  1. A. The cells expressed a gene and made a new protein
    A few minutes is too short to express a gene and build a new protein.
    Proteins the pigment cells already held moved the pigment.
  2. B. ✓ The cells switched on proteins they already had
  3. C. The cells changed the base sequence of a gene
    The base sequence of the cells’ genes did not change.
    The darker skin came from proteins the cells already held, switched on within minutes.

Why: The pigment moved within a few minutes.
A few minutes is too short to make a new protein.
So proteins the cells already held moved it: the cells switched on proteins they already had.
Enough pigment cells changed at once for the whole fish to look darker: the phenotype.

37
Check q12

A hormone reaches the cells of a flower bud. Over a week the petal cells express a pigment gene, and the petals open red instead of white.

Which of the following is the change in the plant’s phenotype?

  1. A. ✓ The petals being red
  2. B. The hormone binding its receptors
    Binding is the event inside each petal cell that starts the change.
    The phenotype is the feature the change leads to, seen on the flower.
  3. C. The pigment gene being expressed in each petal cell
    Expressing the gene is the cause inside each cell.
    The phenotype is the feature that cause leads to: red petals.
  4. D. The relay proteins changing shape in each petal cell
    A relay protein changing shape is a step inside one cell.
    The phenotype is the feature all those steps lead to.

Why: The phenotype is anything about the organism that can be observed or measured.
Binding, the relay and the expressed gene are steps inside each petal cell.
Red petals are the feature those steps lead to, seen on the whole flower.
So the red petals are the change in phenotype.

38
Check q13

A mouse has brown fur. Each of its fur cells makes brown pigment.

Which of the following is part of the mouse’s phenotype?

  1. A. The base sequence of its fur-color gene
    The base sequence is the genotype: the instructions the fur cells read.
    The phenotype is what those instructions lead to.
  2. B. ✓ The brown color of its fur
  3. C. The signal molecule that reaches its fur cells
    A signal molecule is a ligand that binds a receptor.
    The phenotype is a feature of the mouse itself.

Why: The phenotype is the set of features that can be observed or measured on the organism.
The brown fur can be seen on the mouse.
The base sequence is the genotype, and the signal is a molecule.
So the brown fur is part of the phenotype.

39
Practice writing an answer

A person hears a loud bang. Within seconds epinephrine floods their blood and reaches every muscle cell in their heart. Each heart muscle cell contracts sooner and harder than before. A nurse measures the person’s heart rate: 130 beats per minute, up from 70.

(a) Explain how this case demonstrates that a signal which changes what enough cells are doing changes the organism’s phenotype. (1 pt)

Model answer Epinephrine binds receptors on the heart muscle cells.
Each heart muscle cell contracts sooner and harder: a change in what the cell is doing, using proteins the cell already has.
Epinephrine reached every heart muscle cell, so enough cells changed at once.
The heart rate is measured on the whole person, so it is part of the phenotype.
So the change in enough cells showed as a heart rate of 130 beats per minute: a changed phenotype.
Rubric
  • Award 1 point for: epinephrine changed what each heart muscle cell does, it reached enough cells at once, and the heart rate is a feature measured on the whole person, so the phenotype changed (70 to 130 beats per minute).

Glossary

genotype
An organism’s DNA sequence: the instructions its cells read. The genotype is not the phenotype; the phenotype is what those instructions lead to.
phenotype
An organism’s set of observable features: anything about the organism or its cells that can be observed or measured, such as a banana’s softness, sweetness and yellow skin. The DNA sequence itself is the genotype, not the phenotype. A signal that changes gene expression, or what the cell is doing, in enough cells changes the phenotype.

APBIO-U04-L08C A signal that says: dismantle

Topic 4.3 · Signal Transduction Pathways · 31 steps

Two photographs of a mouse embryo's foot: on the left a paddle-shaped foot with webbing joining the toes; on the right the same kind of foot a little later, its five toes separate; an arrow between them is labelled a signal reaches the webbing cells
Two photographs of a mouse embryo's foot: on the left a paddle-shaped foot with webbing joining the toes; on the right the same kind of foot a little later, its five toes separate; an arrow between them is labelled a signal reaches the webbing cells

Photos: Geyer et al. 2017, Wikimedia Commons, CC BY-SA 4.0 (cropped).

In a mouse embryo’s foot, webbing cells sit between the toes.

On day 13 a signal from the surrounding tissue binds receptors on the webbing cells. By day 15 the toes are separate, and the toe cells beside the webbing are unharmed.

The webbing cells did not burst and they were not cut away. What did the signal tell them to do?

Unit 4 · Cell Communication and Cell Cycle

1A cell dismantles itself

2

Video: Watch: A cell dismantles itself

A signal binds a webbing cell; the cell switches on its own enzymes, cuts up its DNA and proteins, and packages itself into pieces that its neighbors take up.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L08Ca.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L08Ca.mp4

3

Apoptosis is a cell dismantling itself in an orderly way. The webbing cells between an embryo’s fingers and toes die this way.

Two photographs of a mouse embryo's foot: before, a paddle-shaped foot with webbing joining the toes; after, the webbing is gone and the five toes are separate
Two photographs of a mouse embryo's foot: before, a paddle-shaped foot with webbing joining the toes; after, the webbing is gone and the five toes are separate
4
Check q1

In a developing embryo, the cells of the webbing between the fingers die as a normal part of development. Each cell dismantles itself in an orderly way.

Which of the following names this kind of cell death?

  1. A. ✓ Apoptosis
  2. B. Diffusion
    Diffusion is the spreading of molecules from where they are concentrated to where they are not.
    A cell dismantling itself in an orderly way is apoptosis.
  3. C. Phosphorylation
    Phosphorylation is a kinase adding a phosphate to a protein.
    A cell dismantling itself in an orderly way is apoptosis.

Why: A cell dismantling itself in an orderly way, as a normal part of development, is called apoptosis.
The webbing cells die this way.
So the kind of cell death is apoptosis.

5

Can a signal tell a cell to end itself? Yes.

6

The signaled cell takes itself apart from the inside, and its neighbors clear the pieces away. Nothing leaks out, so the organism loses only the cell it meant to lose.

7

Apoptosis can be a response to a signal. In a mouse embryo’s foot, a signal from the surrounding tissue binds receptors on the webbing cells.

8

The signaled cell activates its own enzymes. They cut up its DNA and its proteins, and the cell packages itself into tidy pieces.

A cell that has received a signal dismantles itself: its own enzymes, labeled, cut up its contents; it packages itself into pieces; two pieces are drawn inside a neighboring cell that has taken them up, two more still outside it
A cell that has received a signal dismantles itself: its own enzymes, labeled, cut up its contents; it packages itself into pieces; two pieces are drawn inside a neighboring cell that has taken them up, two more still outside it
9

Neighboring cells take the pieces up.

10

What you are expected to know Describe apoptosis as a response to a signal: the signaled cell activates its own enzymes, cuts up its DNA and proteins, and packages itself into pieces that neighboring cells clear away.

11
Check q2

In a developing mouse foot, the cells between the toes receive a signal on day 13. By day 15 the toes are separate.

Which of the following happened to the webbing cells?

  1. A. Each toe cell released enzymes onto the webbing cell beside it and digested it from outside
    The signal reached the webbing cells themselves, and each switched on its own enzymes; those enzymes did the cutting inside the cell.
    The toe cells only took up the pieces.
  2. B. ✓ Each webbing cell activated enzymes inside itself and cut up its DNA and proteins
  3. C. Each webbing cell swelled until its membrane burst and its contents spilled out
    A burst cell spills its contents and damages the cells around it.
    The toes formed with the tissue beside the webbing unharmed, so the webbing cells did not burst.
  4. D. Each webbing cell lost the blood supply that fed it and starved
    The webbing cells were fed by the same tissue as the toe cells.
    What they received was a signal, which switched on their own dismantling enzymes.

Why: The signal bound the webbing cells’ receptors, telling each cell to dismantle itself.
Each cell activated its own enzymes, which cut up its DNA and proteins.
The cell packaged itself into pieces, which neighboring cells took up.
So the webbing was removed and the toes were separate.

12
Check q3

A student says: “A cell that dies by apoptosis has been damaged from outside, like a cell that is burst by heat.”

Is the student correct?

  1. A. Yes — every death of a cell starts with damage from outside
    A cell burst by heat is damaged from outside, and its contents spill.
    A cell in apoptosis receives a signal, switches on its own enzymes and dismantles itself from inside.
  2. B. ✓ No — a cell that dies by apoptosis dismantles itself from inside, after a signal

Why: A cell burst by heat is damaged from outside, and its contents spill.
A cell that dies by apoptosis is not damaged from outside.
It receives a signal, switches on its own enzymes and dismantles itself from inside.
So the student is wrong: apoptosis is orderly and signaled.

13Why apoptosis protects the organism

14

Video: Watch: Why apoptosis protects the organism

The dismantling cell beside its neighbors: nothing spills, the pieces are cleared, and a virus inside is cleared with them.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L08Cb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L08Cb.mp4

15

Nothing leaks out of the dismantling cell. So its neighbors are unharmed, and a healthy organism removes cells it no longer needs.

16

Now consider a cell infected by a virus. A signal from an immune cell tells the infected cell to dismantle itself, so the virus is packaged up and cleared away with the cell.

17

The signal can also be an absence. A cell that loses contact with its neighbors stops receiving a survival signal, and it dismantles itself.

18

Here is a table comparing the four things a signal can make a cell do: how fast each happens, whether it lasts, and an example of each.

A table of the four things a signal can make a cell do, with how fast each happens, whether it lasts, and an example: change what it is doing this second, seconds, stops when the signal stops, epinephrine making a liver cell release glucose; change which genes it expresses, hours to days, lasts after the signal is gone, ethylene making a banana cell build softening enzymes; change the organism's phenotype when enough cells change, as fast as the cells change, as long as the cells stay changed, a green banana turning soft and yellow; dismantle itself, hours, the cell is gone for good, a webbing cell between a mouse embryo's toes
19

All four begin with a ligand and a receptor.

20

What you are expected to know Explain why apoptosis protects the organism: nothing leaks to harm the neighbors, unneeded or dangerous cells are removed, and a lost survival signal can trigger it.

21
Practice writing an answer

In a developing mouse foot, the cells between the toes receive a signal on day 13. Each webbing cell dismantles itself from inside. By day 15 the toes are separate, and the toe cells beside the old webbing are healthy.

(a) Explain why the toe cells beside the webbing were unharmed. (1 pt)

Model answer The signal switched on each webbing cell’s own enzymes.
Those enzymes worked inside the webbing cell.
They cut up the cell’s DNA and proteins.
The webbing cell packaged its contents, enzymes included, into tidy pieces wrapped in membrane.
So nothing spilled out of the webbing cell.
Neighboring cells took the pieces up.
So no enzyme and no spilled contents ever reached the toe cells, and the toe cells were unharmed.
Rubric
  • Award 1 point for: the webbing cell’s own enzymes worked inside it and the cell packaged its contents into pieces that neighbors took up, so nothing leaked out onto the toe cells.
22
Check q4

A researcher pulls a cell away from its neighbors in a tissue. The cell stops receiving a signal it had been receiving. Within hours the cell shrinks and breaks into pieces, which nearby cells take up.

Which of the following caused the cell’s death?

  1. A. A torn membrane
    A torn cell spills its contents and harms its neighbors.
    This cell packaged itself into pieces that nearby cells took up unharmed: an orderly dismantling from inside.
  2. B. Starvation
    The fluid around the cell still holds glucose.
    What the cell lost was a signal from its neighbors.
  3. C. The loss of a signal to divide
    A cell that receives no signal to divide simply does not divide.
    It does not break into pieces within hours.
  4. D. ✓ The loss of a survival signal

Why: The cell’s neighbors had been sending it a survival signal, which kept its dismantling enzymes switched off.
Pulling the cell away removed that signal.
With the signal gone, the cell switched its dismantling enzymes on.
So the cause of death was the loss of the survival signal.

23

Back to the mouse embryo’s foot, with webbing cells between the toes on day 13. A signal from the surrounding tissue bound receptors on the webbing cells.

24

Each webbing cell activated its own enzymes, cut itself up and packaged the pieces for its neighbors to clear. By day 15 the toes were separate, and the toe cells beside them were unharmed.

25Mixed practice mixed practice

26
Check q5

Two cells in a piece of tissue die. A researcher heats the first cell until its membrane bursts, and its contents spill onto the cells around it, which are damaged. The second cell receives a signal from a neighboring cell; it shrinks and breaks into small packages, which the cells around it take up unharmed.

Which cell died by apoptosis?

  1. A. The heated cell
    A signal is a molecule that binds a receptor.
    Heat destroyed the first cell from outside, and its contents spilled onto its neighbors.
    That is damage, not apoptosis.
  2. B. ✓ The signaled cell
  3. C. Both cells
    Apoptosis is one kind of death: the cell dismantles itself after a signal, and nothing leaks.
    The heated cell burst and its contents damaged its neighbors: damage, not apoptosis.

Why: The first cell burst from heat, and its spilled contents damaged its neighbors: damage, not apoptosis.
A signal switched on the second cell’s own enzymes.
Those enzymes cut up the cell, which packaged itself into pieces its neighbors took up unharmed.
So the signaled cell died by apoptosis.

27
Practice writing an answer

Two cells in a piece of tissue die. A researcher heats the first cell until its membrane bursts, and its contents spill onto the cells around it, which are damaged. The second cell receives a signal from a neighboring cell; it shrinks and breaks into small packages, which the cells around it take up unharmed.

(a) Explain how the two deaths demonstrate that apoptosis is an orderly, signaled dismantling of a cell from inside. (1 pt)

Model answer The second cell received a signal, which switched on its own enzymes.
Those enzymes cut up its DNA and proteins, and the cell packaged itself into small pieces.
Nothing leaked out, so the cells around it took the pieces up unharmed.
That is apoptosis: orderly, signaled, and from inside.
The heated cell received no signal: heat burst its membrane from outside, and its contents damaged its neighbors.
So only the signaled cell’s death was apoptosis.
Rubric
  • Award 1 point for: the signaled cell dismantled itself after a signal (its own enzymes, contents packaged, nothing leaked, neighbors unharmed), whereas the heated cell was destroyed from outside and its contents spilled, so only the signaled cell’s death is apoptosis.
28
Check q6

A cell infected by a virus receives a signal from a passing immune cell and, within hours, dismantles itself, virus and all.

Which of the following describes the outcome for the organism?

  1. A. ✓ The cell and its virus are packaged into pieces, neighbors clear them, and the infection stops there
  2. B. The infected cell’s contents spill onto the cells around it, and the virus spreads to those cells
    In apoptosis nothing spills: the cell packages its contents, virus included, into pieces, and its neighbors take the pieces up.
    So the virus is cleared, not spread.
  3. C. The infected cell’s dismantling enzymes leak out onto the cells around it and damage them
    The dismantling enzymes work inside the cell, and the cell packages them up with everything else.
    Nothing leaks onto the neighbors.
  4. D. The cells around the infected cell are told to dismantle themselves too, and a patch of tissue is lost
    The immune cell signaled the infected cell only.
    The neighbors received no signal to dismantle, so they carry on and take up the pieces.

Why: The immune cell’s signal tells the infected cell to dismantle itself.
The infected cell packages itself, virus included, into pieces, and neighboring cells take the pieces up; nothing leaks.
So the virus is cleared with the cell, and the infection stops there.

29
Check q7

A researcher tests whether a treatment changes a measured result.

What is the null hypothesis?

  1. A. ✓ The statement that the treatment makes no difference to the measured result
  2. B. The statement of what the researcher expects the treatment to do
    What the researcher expects is the prediction.
    The null hypothesis is the statement that the tested factor makes no difference.

Why: The null hypothesis is the statement that the tested factor makes no difference to the measured result.

30
Practice writing an answer

Researchers test whether the gas from a ripe banana ripens a green one. They prepare three kinds of sealed bag, six bags of each. Bag 1 holds a green banana alone. Bag 2 holds a green banana and a ripe one. Bag 3 holds a green banana, a ripe one and an ethylene absorber. The absorber removes ethylene from the air as fast as it is released. After two days the researchers measure the firmness of the green banana. Firmness is the force, in newtons, needed to press a blunt metal rod 5 mm into the fruit. The graph shows the results for bags 1 and 2; the error bars show ±2SE. Bag 3 has not yet been measured.

A bar graph of the firmness of the green banana after two days, in newtons, for bag 1 (green banana alone) and bag 2 (green banana with a ripe one), with ±2SE error bars; bag 3, with an ethylene absorber, is marked with a question mark
A bar graph of the firmness of the green banana after two days, in newtons, for bag 1 (green banana alone) and bag 2 (green banana with a ripe one), with ±2SE error bars; bag 3, with an ethylene absorber, is marked with a question mark

(a) Identify the independent variable and the dependent variable in this experiment. (1 pt)

Model answer The independent variable is what each bag holds with the green banana (nothing, a ripe banana, or a ripe banana with an ethylene absorber).
The dependent variable is the firmness of the green banana after two days, in newtons.
Rubric
  • Award 1 point for: the contents of the bag (ripe banana present or absent, absorber present or absent) as the independent variable AND the firmness of the green banana as the dependent variable.

Slip Naming ethylene concentration as the dependent variable. It is what the treatments change; what is measured is firmness.

(b) State the null hypothesis for the comparison of bag 2 with bag 3. (1 pt)

Model answer Null hypothesis: the ethylene absorber makes no difference to the firmness of the green banana; bags 2 and 3 will have the same mean firmness after two days.
Rubric
  • Award 1 point for: a null hypothesis stating that the absorber makes no difference to the green banana’s firmness (bag 2 and bag 3 the same).

Slip Writing the prediction (‘bag 3 stays firm’) as the null hypothesis. The null hypothesis is the statement of no difference.

(c) Explain why bag 1, the green banana alone, is included. (1 pt)

Model answer Bag 1 is the control.
Bag 1 shows how firm a green banana stays after two days with no ripe banana and no ethylene.
Bag 2 is compared against bag 1.
So the softening in bag 2 can be credited to what the ripe banana released, and not to the two days alone.
Rubric
  • Award 1 point for: bag 1 gives the firmness with the tested factor (the ripe banana’s gas) absent, so any difference in bag 2 can be credited to it.

Slip Saying bag 1 ‘tests whether bananas ripen’. It gives the baseline firmness that bag 2 is compared against.

(d) Predict the mean firmness of the green banana in bag 3 compared with bag 2, and justify your prediction using the pathway from ethylene to ripening. (1 pt)

Model answer Bag 3’s banana will be far firmer than bag 2’s, close to bag 1’s 38 N.
The ripe banana releases ethylene into the bag.
The absorber removes that ethylene from the air.
So no ethylene binds the receptors in the green banana’s cells.
So the receptors never change shape, and the relay never passes a message on.
So the genes for the cell-wall-softening enzymes are never expressed.
So the flesh stays hard.
Rubric
  • Award 1 point for: bag 3 firmer than bag 2 (near bag 1), because with the ethylene removed nothing binds the receptors, so the softening enzymes are never made.

Slip Predicting bag 3 softens because the ripe banana is still present. The banana releases the gas, but the absorber takes it out of the air before it can bind.

APBIO-U04-L09 A changed gene, a changed shape

Topic 4.3 · Signal Transduction Pathways · 84 steps

Two muscle cells, each with insulin (a dot, labeled) bound to a receptor (labeled) on its surface; glucose, drawn as hexagons and labeled, enters the left cell and does not enter the right one
Two muscle cells, each with insulin (a dot, labeled) bound to a receptor (labeled) on its surface; glucose, drawn as hexagons and labeled, enters the left cell and does not enter the right one

Here are two lines of muscle cells, both with insulin bound to their receptors.

A researcher gives both lines insulin. In both lines, insulin binds the receptors equally well.

In the first line the cells take up glucose. In the second line they do not, however much insulin the researcher adds. The second line’s receptors carry a change in the part facing the cytosol.

What is different about the second line’s receptor? Why can more insulin not mend it?

Unit 4 · Cell Communication and Cell Cycle

1A change in a gene

2

Video: Watch: A change in a gene

One base in the gene for a receptor is different; the receptor is built with one different amino acid at that spot; a change in the gene’s base sequence is a mutation.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L09a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L09a.mp4

3
Check q1

A liver cell’s DNA carries the instructions for making the insulin receptor protein.

What is the stretch of DNA that carries the instructions for one protein called?

  1. A. A ligand
    A ligand is the signal molecule that binds a receptor.
    The instructions for making the receptor are the gene.
  2. B. A ribosome
    A ribosome builds the protein.
    The instructions the ribosome follows are the gene.
  3. C. ✓ A gene

Why: A stretch of a cell’s DNA that carries the instructions for making one protein is called a gene.

4
Check q2

A ribosome builds a protein as a chain of amino acids, and the chain folds into its working shape.

What sets the protein’s final shape?

  1. A. ✓ The order of its amino acids
  2. B. The number of ribosomes that made it
    Ribosomes only build the chain.
    The chain folds by itself, according to the order of its amino acids.

Why: The order of amino acids sets how the chain folds, and so sets the protein’s shape.

5

What happens to a pathway when one of its proteins is built wrong?

6

A gene’s DNA sequence sets a protein’s amino acid order. The amino acid order sets the protein’s fold.

7

The fold is the protein’s job. So one changed base in a gene can change what the protein does.

8

In a receptor, the part of the receptor the change hits decides which stage of the pathway fails. One measurement, whether the ligand is found bound, tells you which part it was.

9

Suppose one base in the gene for a receptor is different. The receptor is usually built with one different amino acid at that spot.

A DNA band with one gene drawn as a box on it, the box zoomed into its row of bases, and the binding pocket of the receptor the gene codes for: on the left the normal gene and pocket; on the right one base is changed, one amino acid is different, and the pocket is misshapen
A DNA band with one gene drawn as a box on it, the box zoomed into its row of bases, and the binding pocket of the receptor the gene codes for: on the left the normal gene and pocket; on the right one base is changed, one amino acid is different, and the pocket is misshapen
10

When the sequence of a gene’s DNA changes, we call the change a , because the gene is no longer what it was.

11

Only a change in the base sequence itself is a mutation. A phosphate added to a protein, or a gene switched off, leaves the DNA sequence as it was.

12

What you are expected to know Describe a mutation: a change in the base sequence of a gene’s DNA.

13
Check q3

Which of the following changes to a cell is a mutation?

  1. A. A change in a receptor’s shape when its ligand binds
    A receptor’s shape change on binding is reception.
    The shape change reverses when the ligand leaves, and the gene is unchanged.
  2. B. A phosphate added to a protein by a kinase
    Phosphorylation changes a protein’s shape for a while.
    The gene that codes for the protein is unchanged.
  3. C. A change in which genes a cell expresses
    Expressing different genes uses the same DNA sequence.
    A mutation changes the sequence itself.
  4. D. ✓ A change in the sequence of a gene’s DNA

Why: A mutation is a change in the sequence of a gene’s DNA.
The other three change proteins or their use, and none of them changes any gene.

14Quick quiz: mutation mixed practice

15
Check q4

What is a mutation?

  1. A. A change in a protein’s shape while its ligand is bound
    A receptor changes shape while its ligand is bound, then changes back.
    Its gene’s sequence is unchanged.
  2. B. ✓ A change in the base sequence of a gene’s DNA
  3. C. A change in which proteins a cell is making
    A cell that makes different proteins is changing gene expression.
    Its DNA sequence is unchanged.

Why: A mutation is a change in the base sequence of a gene’s DNA.

16
Check q5

One base in the gene for an enzyme is changed from A to G.

Is this change a mutation?

  1. A. ✓ Yes
  2. B. No
    A base of the gene’s DNA sequence has changed.
    A change in the base sequence is a mutation.

Why: The gene’s base sequence has changed, so the change is a mutation.

17
Check q6

A kinase adds a phosphate to a receptor’s intracellular domain.

Is this change a mutation?

  1. A. Yes
    The phosphate changes the receptor protein, not its gene.
    The gene’s base sequence is as it was.
  2. B. ✓ No

Why: The phosphate is added to the protein.
The gene’s base sequence is unchanged, so the change is not a mutation.

18
Check q7

A lens cell has the gene for the insulin receptor and keeps it switched off.

Is this a mutation?

  1. A. Yes
    Not expressing a gene leaves its base sequence as it was.
    A mutation changes the sequence itself.
  2. B. ✓ No

Why: The lens cell’s gene has the normal base sequence.
The cell simply is not using it, so this is not a mutation.

19
Practice writing an answer

A change in a cell’s DNA is being described.

(a) State what a mutation is. (1 pt)

Model answer A mutation is a change in the base sequence of a gene’s DNA.
Rubric
  • Award 1 point for: a change in the base sequence (or the sequence) of a gene’s DNA.

20How a mutation changes what a protein does

21

Video: Watch: How a mutation changes what a protein does

One changed base, one changed amino acid, a different R group, a changed fold, a changed job; the three outcomes and what the data show for each.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L09b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L09b.mp4

22

Now consider the changed amino acid. One changed amino acid brings a different R group.

23

A different R group changes the fold at that spot. The fold is the job, so the protein’s job can change.

24

Here the changed amino acid sits in the ligand-binding domain. The pocket no longer fits the ligand.

25

So the receptor never changes shape when the ligand arrives. A misshapen pocket is one outcome; there are two others.

26

A mutation does not always remove a protein’s activity. Here is a table of the three things a mutation can do to a protein, and what the data show for each.

A table of the three things a mutation can do to a protein and what the data show for each: leaves the activity unchanged, the protein works as before; reduces or removes the activity, less response or none, however much ligand is added; locks the protein in its active shape, a response with no ligand present
27

The chain from gene to job has four links:
1 the DNA sequence
2 the amino acid order
3 the fold
4 the job.

28

A mutation changes link 1. So it can change every link after it.

29

What you are expected to know Explain how one changed base in a gene can change what its protein does.

30

What you are expected to know Identify from the data which of the three outcomes a mutation had: activity unchanged, activity reduced or removed, or the protein locked in its active shape.

31
Check q8

A mutation in the gene for an enzyme swaps one amino acid in the enzyme’s active site.

Predict the most likely effect on the enzyme.

  1. A. The enzyme is made in larger amounts
    A mutation in the gene changes the protein that is built.
    Here the change is one amino acid of the active site, so the active site’s shape changes.
  2. B. The substrate changes its own shape to fit
    A substrate has a fixed shape.
    The substrate binds only where the site’s shape and charge fit it.
  3. C. ✓ The substrate fits the active site less well, or not at all
  4. D. The enzyme gains a second active site
    One changed amino acid alters the pocket that exists.
    It does not build a new pocket.

Why: The amino acids that line the active site set its shape and charge.
The mutation swaps one of those amino acids.
So the pocket changes shape.
So the substrate fits the pocket less well, or not at all.
So the enzyme works less, or stops.

32
Check q9

Three different mutations in a receptor’s gene are studied. Mutant 1 binds its ligand and signals normally. Mutant 2 signals half as strongly. Mutant 3 signals with no ligand present at all. A student says: “Mutant 1 must have an unchanged gene, because a mutation always stops the protein working.”

Is the student correct?

  1. A. Yes — a receptor that works normally must come from an unchanged gene
    A mutation is a change in the gene’s DNA sequence.
    The changed amino acid can sit where it changes nothing about the fold, so the protein works as before.
  2. B. ✓ No — a mutation can leave the protein’s activity unchanged
  3. C. No — but a mutation can only lower activity, so mutant 3 carries no mutation
    Mutant 3 signals with no ligand, so its mutation locked the receptor in its active shape.
    A mutation can lock a protein on, not only slow or stop it.

Why: A mutation is a change in a gene’s DNA sequence.
One changed amino acid can sit where it changes nothing about the fold, so the protein works as before.
So mutant 1 can carry a mutation and still signal normally.
A mutation does not always stop the protein working.

33
Practice writing an answer

Three different mutations in a receptor’s gene are studied. Mutant 1 binds its ligand and signals normally. Mutant 2 signals half as strongly. Mutant 3 signals with no ligand present at all.

(a) Explain how a mutation can produce a receptor that signals with no ligand present, as mutant 3 does. (1 pt)

Model answer A mutation changes the sequence of the receptor’s gene, so one amino acid in the receptor is different.
One different amino acid changes the receptor’s fold at that spot.
In mutant 3 the changed fold holds the receptor’s inner part in its active shape.
Normally the inner part takes up its active shape only while a ligand is bound; here it is active all the time.
So the receptor passes the message on with no ligand bound.
Rubric
  • Award 1 point for: the changed amino acid locks the receptor’s inner part in its active shape, so the receptor signals whether or not a ligand is bound.

34Which stage fails

35

Video: Watch: Which stage fails

A mutation in the ligand-binding domain: the ligand never binds and reception fails; a mutation in the intracellular domain: the ligand binds, the inner part cannot take up its active shape, and transduction never starts.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L09c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L09c.mp4

36
Check q10

Insulin binds a receptor on a muscle cell’s surface.

Which stage of the signal transduction pathway is this?

  1. A. Transduction
    Transduction is the relay inside the cell, one molecule changing the next.
    The ligand binding its receptor comes before it.
  2. B. ✓ Reception
  3. C. The response
    The response is what the cell finally does differently.
    The ligand binding its receptor is the first stage.

Why: The ligand binding its receptor is called reception.
Insulin is the ligand, so insulin binding the receptor is reception.

37

Now consider the two lines of muscle cells from the opening. Both lines’ receptors bind insulin equally well.

38

The second line’s receptors carry a change in the part facing the cytosol. Here is a drawing of the two receptors side by side.

Two receptors side by side with insulin bound to both: the normal receptor's inner part has changed shape and the relay starts; the mutant's inner part is marked broken and unchanged
Two receptors side by side with insulin bound to both: the normal receptor's inner part has changed shape and the relay starts; the mutant's inner part is marked broken and unchanged
39

The ligand fits the receptor’s binding site by shape and charge.

40

While the ligand is bound, the receptor holds a different shape.

41

So another part of the receptor can now act on the next molecule.

42

In the second line the insulin fits and binds. The intracellular domain cannot take up its active shape.

43

Reception happens; transduction never starts.

44

Now imagine the mutation in the ligand-binding domain instead. The insulin no longer binds.

Two receptors side by side: insulin is bound to the normal one and its relay starts; the mutant's binding pocket is drawn as an open jagged notch of the wrong shape, and the insulin floats past unbound
Two receptors side by side: insulin is bound to the normal one and its relay starts; the mutant's binding pocket is drawn as an open jagged notch of the wrong shape, and the insulin floats past unbound
45

Reception itself fails. Nothing after it can start.

46

In both cases every protein after the receptor is intact. It sits idle, waiting for a message that never comes.

47

What you are expected to know Predict which stage of the pathway fails when a receptor’s ligand-binding domain carries a mutation.

48

What you are expected to know Predict which stage of the pathway fails when a receptor’s intracellular domain carries a mutation.

49
Check q11

A researcher gives two cultures of muscle cells the same insulin. Insulin binding is the same in both cultures. Only the first culture phosphorylates its receptors and takes up more glucose. The second culture’s receptors carry a mutation.

In the second culture, which stage of the pathway failed?

  1. A. ✓ Transduction
  2. B. Reception
    Reception is the ligand binding its receptor.
    Insulin binding was the same in both cultures, so reception succeeded; the message was lost after binding, in transduction.

Why: Reception is the ligand binding its receptor.
Insulin binding was the same in both cultures.
So insulin bound the second culture’s receptors, and reception succeeded.
The second culture’s receptors were never phosphorylated, and no extra glucose entered.
So the message was lost after binding.
The stage that failed is transduction.

50
Practice writing an answer

A researcher gives two cultures of muscle cells the same insulin. Insulin binding is the same in both cultures. Only the first culture phosphorylates its receptors and takes up more glucose. The second culture’s receptors carry a mutation in the intracellular domain.

(a) Explain why the second culture’s cells take up no extra glucose, even though insulin is bound to their receptors. (1 pt)

Model answer Insulin binds the second culture’s receptors, so reception succeeds.
The mutation is in the intracellular domain, so the inner part of the receptor cannot take up its active shape.
The inner part never acts on the next molecule inside the cell.
So transduction never starts.
The glucose transporters are never moved to the surface, so no extra glucose enters.
Rubric
  • Award 1 point for: the mutated intracellular domain cannot take up its active shape (or cannot act on the next molecule), so transduction never starts and the response (glucose uptake) never happens, even though insulin is bound.

51Quick quiz: which stage fails? mixed practice

52
Check q12

A gland cell’s receptor carries a mutation in its ligand-binding domain. The hormone arrives.

Which stage fails?

  1. A. ✓ Reception
  2. B. Transduction
    Transduction is the relay after binding.
    A mutated ligand-binding domain stops the binding itself, so the failure comes first, at reception.

Why: The ligand-binding domain is the pocket that holds the hormone.
The mutation leaves the pocket the wrong shape, so the hormone never binds.
Reception fails.

53
Check q13

A root cell’s receptor carries a mutation in its intracellular domain. The plant hormone binds the receptor as normal.

Which stage fails?

  1. A. Reception
    Reception is the ligand binding its receptor.
    The hormone binds as normal, so reception succeeds.
  2. B. ✓ Transduction

Why: The hormone binds, so reception succeeds.
The intracellular domain cannot take up its active shape, so it never acts on the next molecule.
Transduction fails.

54
Check q14

In a heart cell, a kinase two steps after the receptor carries a mutation and cannot be switched on. Epinephrine binds the receptor as normal.

Which stage fails?

  1. A. Reception
    Reception is epinephrine binding its receptor, and it binds as normal.
    The kinase is part of the relay inside the cell.
  2. B. ✓ Transduction

Why: Epinephrine binds, so reception succeeds.
The kinase is one molecule of the relay inside the cell.
The relay stops at the kinase, so transduction fails.

55
Check q15

A binding test on a fat cell’s mutant receptor finds no hormone bound, however much hormone the researcher adds.

Which stage fails?

  1. A. ✓ Reception
  2. B. Transduction
    Transduction is the relay after binding.
    Here nothing binds, so the failure is at reception.

Why: No hormone is found bound.
Reception is the ligand binding its receptor.
The hormone never binds, so reception fails.

56
Check q16

A kidney cell’s receptor holds its hormone as well as a normal receptor does, yet no relay protein inside the cell is switched on.

Which stage fails?

  1. A. Reception
    The receptor holds its hormone as normal, so reception succeeds.
    The failure comes after binding.
  2. B. ✓ Transduction

Why: The receptor holds its hormone, so reception succeeds.
No relay protein is switched on, so the message is lost after binding.
Transduction fails.

57Which domain broke

58

Video: Watch: Which domain broke

Ligand found bound and no response: the intracellular domain; no ligand found bound: the ligand-binding domain; more ligand cannot mend a receptor.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L09d.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L09d.mp4

59

How do you tell the two failures apart from outside the cell? One measurement does it: whether the ligand is found bound to the receptor.

60

Insulin found bound, and no response: the pocket works. The fault is in the intracellular domain.

61

No insulin found bound: the pocket is the wrong shape. The fault is in the ligand-binding domain.

62

In the second line no extra glucose enters, however much insulin the researcher adds. More ligand cannot mend a receptor.

63

What you are expected to know Identify which domain of a receptor carries the mutation from whether the ligand is found bound.

64
Check q17

A researcher gives two cultures of muscle cells insulin. The second culture’s receptors bind insulin as well as the first culture’s do, and the second culture’s message is lost after binding.

Which domain of the second culture’s receptor carries the mutation?

  1. A. ✓ The intracellular domain
  2. B. The ligand-binding domain
    The ligand-binding domain is the pocket that holds insulin.
    Insulin binds as well as ever, so the pocket is intact and the fault lies after binding.

Why: The ligand-binding domain is the pocket that holds the ligand.
Insulin binds as well as ever, so the pocket is intact.
The message is lost after binding.
The part of the receptor that acts after binding is the intracellular domain.
So the mutation is in the intracellular domain.

65

Back to the two lines of muscle cells, both with insulin bound to their receptors. The first line took up glucose; the second line did not, however much insulin the researcher added.

66

The second line’s receptors bound insulin, so reception happened. Their intracellular domains could not take up the active shape, so transduction never started.

67

Every protein after the receptor sat intact and idle. More insulin could only bind more receptors whose inner part cannot answer, so more insulin could not mend it.

68Quick quiz: which domain broke? mixed practice

69
Check q18

A bone cell’s receptor carries a mutation. Hormone is found bound to the receptor, and no relay protein is phosphorylated.

Which domain of the receptor carries the mutation?

  1. A. The ligand-binding domain
    Hormone is found bound to the receptor, so the pocket holds the hormone as it should.
    The message is lost after binding, at the inner part.
  2. B. ✓ The intracellular domain

Why: Hormone is found bound to the receptor.
So the ligand-binding domain works, and reception succeeded.
No relay protein is phosphorylated.
So the message was lost after binding, at the receptor’s inner part.
The mutation is in the intracellular domain.

70
Check q19

A root cell’s receptor carries a mutation. A plant hormone reaches the receptor, and a binding test finds the receptor empty.

Which domain of the receptor carries the mutation?

  1. A. ✓ The ligand-binding domain
  2. B. The intracellular domain
    A receptor with a broken intracellular domain still binds its ligand.
    Here the binding test finds the receptor empty, so the pocket itself is the wrong shape.

Why: The binding test finds the receptor empty.
So reception itself failed.
Reception fails when the pocket cannot hold the ligand.
The mutation is in the ligand-binding domain.

71
Check q20

A gill cell’s receptor carries a mutation. The receptor holds its hormone as well as a normal receptor does, yet cAMP stays at its resting level.

Which domain of the receptor carries the mutation?

  1. A. The ligand-binding domain
    The receptor holds its hormone as well as a normal receptor does, so the pocket is intact.
    The message is lost after binding.
  2. B. ✓ The intracellular domain

Why: The receptor holds its hormone as well as a normal receptor does.
So the ligand-binding domain works. cAMP stays at its resting level.
So the message was lost after binding, at the receptor’s inner part.
The mutation is in the intracellular domain.

72
Check q21

A fat cell’s receptor carries a mutation. The receptor holds its hormone as normal, and the cell shows no response.

Which domain of the receptor carries the mutation?

  1. A. The ligand-binding domain
    The receptor holds its hormone as normal, so the pocket is intact.
    The message is lost after binding.
  2. B. ✓ The intracellular domain

Why: The receptor holds its hormone as normal.
So reception succeeded.
The cell shows no response.
So the message was lost after binding, at the receptor’s inner part.
The mutation is in the intracellular domain.

73
Check q22

A skin cell’s receptor carries a mutation. At a hundred times the usual hormone concentration, still no hormone is found bound to the receptor.

Which domain of the receptor carries the mutation?

  1. A. ✓ The ligand-binding domain
  2. B. The intracellular domain
    A broken intracellular domain still binds its ligand.
    Here no hormone is found bound at any concentration, so the pocket is the wrong shape for the hormone.

Why: No hormone is found bound to the receptor, even at a hundred times the usual concentration.
More ligand cannot mend a pocket that is the wrong shape.
So reception itself fails.
The mutation is in the ligand-binding domain.

74
Check q23

A researcher gives a plant hormone to guard cells whose receptors carry a mutation. A binding test finds no hormone bound to the receptors, and the cells show no response. When the researcher activates a relay protein after the receptor directly, the cells respond normally.

Where is the fault?

  1. A. ✓ In the receptor’s ligand-binding domain
  2. B. In the relay protein after the receptor
    Activating the relay protein directly gave a normal response.
    So the relay protein and everything after it are intact.
  3. C. In the receptor’s intracellular domain
    A receptor with a broken intracellular domain still binds its ligand.
    Here nothing bound at all.
    So the fault is in the part that binds.
  4. D. In the response proteins
    The response proteins worked when the researcher activated the relay protein.
    So the response proteins are intact.

Why: The binding test found no hormone bound to the receptors.
So reception itself failed.
Reception fails when the ligand-binding domain cannot hold the ligand.
So the fault is in the ligand-binding domain.
Activating the relay protein directly gave a normal response.
So everything after the receptor is intact.

75Mixed practice mixed practice

76
Check q24

A researcher compares the gene for an enzyme in two lines of cells. In the second line, one base of the gene is different.

Which of the following describes the second line’s enzyme gene?

  1. A. It is expressed less
    How much a gene is expressed is how much of its protein the cell makes.
    One different base is a change in the sequence itself.
  2. B. It is phosphorylated
    A kinase phosphorylates proteins, not genes.
    One different base is a change in the gene’s sequence.
  3. C. ✓ It carries a mutation

Why: One base of the gene is different in the second line.
A change in the base sequence of a gene’s DNA is a mutation.

77
Check q25

A mutation in the gene for a G protein-coupled receptor leaves the receptor unable to bind its ligand. A researcher adds the ligand to cells carrying the mutant receptor.

Predict the cAMP inside the cells.

  1. A. cAMP rises as normal
    The G protein waits for the receptor to change shape.
    The receptor never binds its ligand, so the receptor never changes shape.
  2. B. cAMP rises less than normal
    The mutation removed binding.
    Nothing binds, so nothing follows.
  3. C. ✓ cAMP stays at rest

Why: The mutant receptor cannot bind its ligand.
So reception fails.
The receptor never changes shape, so the G protein is never switched on.
The G protein never switches on the enzyme that makes cAMP.
So cAMP stays at its resting level.

78
Check q26

A mutation changes one amino acid on the outer surface of an enzyme, far from the active site. The enzyme works exactly as before.

Which outcome did this mutation have?

  1. A. ✓ It left the activity unchanged
  2. B. It reduced the activity
    The enzyme works exactly as before.
    An enzyme with reduced activity would turn over less substrate each second.
  3. C. It locked the enzyme in its active shape
    An enzyme locked in its active shape would work all the time, whether or not it was being switched on.
    This enzyme works exactly as before.

Why: The changed amino acid sits far from the active site.
The fold at the active site is unchanged, so the enzyme works exactly as before.
The mutation left the activity unchanged.

79
Check q27

A hormone reaches a thyroid cell whose receptor carries a mutation. A binding test finds the receptor empty.

Which stage of the pathway fails?

  1. A. Transduction
    Transduction is the relay after binding.
    Here the hormone never binds, so the failure comes before the relay.
  2. B. ✓ Reception

Why: The binding test finds the receptor empty.
Reception is the ligand binding its receptor.
The hormone never binds, so reception fails.

80
Check q28

A receptor’s intracellular domain carries a mutation. The receptor’s hormone reaches the cell.

Which of the following is found?

  1. A. ✓ Hormone bound to the receptor, and no response
  2. B. No hormone bound to the receptor, and no response
    The ligand-binding domain is intact.
    The hormone still binds the receptor.
  3. C. Hormone bound to the receptor, and a normal response
    The intracellular domain cannot take up its active shape.
    Transduction never starts, so there is no response.

Why: The ligand-binding domain is intact, so the hormone binds.
The intracellular domain cannot take up its active shape.
So transduction never starts, and there is no response.

81
Check q29

A researcher gives a hundred times the usual hormone to cells whose receptor carries a mutation. A binding test finds no hormone bound to the receptors.

Predict the response.

  1. A. A normal response
    No hormone is found bound, even at a hundred times the usual concentration.
    Nothing binds, so no message starts.
  2. B. A weaker response than normal
    The hormone never binds, so no message ever starts.
    A weaker response would need some binding.
  3. C. ✓ No response

Why: No hormone is found bound to the receptors, even at a hundred times the usual concentration.
So reception fails.
Nothing after reception can start, so there is no response.

82
Check q30

A receptor’s ligand-binding domain carries a mutation that stops the ligand fitting its binding site. Its intracellular domain is normal.

Which of the following describes the receptor when its ligand arrives?

  1. A. Transduction fails, and the ligand still binds
    The ligand no longer fits its binding site, so it never binds.
    The failure is reception, not transduction.
  2. B. ✓ Reception fails, and the intracellular domain sits intact and idle
  3. C. Both stages happen, and the response is weaker
    The ligand no longer fits its binding site, so it never binds.
    A weaker response would need some binding; with none, there is no response at all.
  4. D. Neither stage happens, and the receptor is destroyed
    This mutation changes one amino acid.
    The receptor is built and sits in the membrane; only its pocket is the wrong shape.

Why: The mutation leaves the binding site the wrong shape, so the ligand never fits and never binds.
Reception fails.
The intracellular domain is normal, but no binding ever changes its shape, so it sits intact and idle.

83
Practice writing an answer

Researchers study two lines of fat cells. Line 1 carries the normal receptor for the hormone epinephrine. Line 2 carries a receptor with a mutation. Epinephrine binding its receptor normally raises cAMP inside the cell. The researchers give both lines the same epinephrine and measure the epinephrine bound per cell and the cAMP inside the cells. The table shows the results.

A results table for two lines of fat cells: line 1, normal receptor, 5200 epinephrine molecules bound per cell, cAMP 0.1 micromoles per litre with no epinephrine and 2.0 with epinephrine; line 2, mutant receptor, 5100 bound per cell, cAMP 0.1 and 0.1

(a) Identify the stage of the pathway that fails in line 2. (1 pt)

Model answer The stage that fails in line 2 is transduction.
Rubric
  • Award 1 point for: transduction.

Slip Naming reception. Line 2 binds as much epinephrine as line 1, so the ligand binds and reception succeeds.

(b) Identify which domain of the receptor carries the mutation in line 2, and support your answer using one measurement from the table. (2 pt)

Model answer The mutation is in the receptor’s intracellular domain.
Line 2 binds 5,100 epinephrine molecules per cell, about the same as line 1’s 5,200.
Binding is what the ligand-binding domain does, so that domain is intact in line 2.
Yet cAMP in line 2 stays at 0.1 µmol/L after epinephrine.
So the message is lost after binding.
The intracellular domain acts after binding, so the mutation is there.
Rubric
  • Award 1 point for: the intracellular domain.
  • Award 1 point for: the evidence (epinephrine bound per cell is the same in line 2 as in line 1) AND the reasoning (equal binding shows the ligand-binding domain works, so the loss of the cAMP rise must come after binding, in the intracellular domain).

Slip Citing the cAMP measurement alone. Unchanged cAMP shows the pathway failed somewhere; the binding measurement is what places the failure after reception.

(c) Predict the cAMP in line 2 when the researchers give ten times as much epinephrine, and justify your prediction. (1 pt)

Model answer cAMP in line 2 stays at 0.1 µmol/L.
More epinephrine can only bind more receptors.
Every receptor in line 2 has an intracellular domain that cannot take up its active shape.
So no bound receptor passes the message on, however many are bound.
The cAMP-making enzyme is never switched on, so cAMP does not rise.
Rubric
  • Award 1 point for: cAMP stays at rest (about 0.1 µmol/L), because more ligand only binds more receptors whose intracellular domain cannot pass the message on; more ligand cannot mend a receptor.

Slip Predicting a smaller rise than line 1. No bound receptor in line 2 passes the message on, so there is no rise at all.

Glossary

mutation
A change in the base sequence of a gene’s DNA. Because the sequence sets the protein’s amino acid order, a mutation can change the protein’s shape and so change or remove what it can bind and do; it can also lock the protein in its active shape.

APBIO-U04-L09B Find the break

Topic 4.3 · Signal Transduction Pathways · 70 steps

A photograph of five yeast cells under a microscope, oval and gray on a plain gray field, two of them with a small bud growing from one end; beside them the mating pathway drawn as five boxes, receptor, kinase 1, kinase 2, kinase 3, mating genes, with a small cross at kinase 2's corner, and two lines beneath: a kinase 3 active all the time switches the mating genes on; a kinase 1 active all the time changes nothing
A photograph of five yeast cells under a microscope, oval and gray on a plain gray field, two of them with a small bud growing from one end; beside them the mating pathway drawn as five boxes, receptor, kinase 1, kinase 2, kinase 3, mating genes, with a small cross at kinase 2's corner, and two lines beneath: a kinase 3 active all the time switches the mating genes on; a kinase 1 active all the time changes nothing

Here is a yeast cell. Its mating pathway passes its message receptor → kinase 1 → kinase 2 → kinase 3 → mating genes.

Suppose some yeast cells have a kinase 2 that does not work. Those cells show no response, even at 100 times the usual signal.

Give those cells a kinase 3 that is active all the time, and their mating genes switch on. Give them a kinase 1 that is active all the time instead, and nothing happens.

Where is the break? How do the two rescues show it?

Photo: Masur, Wikimedia Commons, public domain (resized).

Unit 4 · Cell Communication and Cell Cycle

1Upstream and downstream

2

Video: Watch: Upstream and downstream

A pathway passes its message one way, like a river; a component acting before a step is upstream of it, one acting after is downstream.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L09Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L09Ba.mp4

3
Check q1

In a liver cell’s epinephrine pathway, the receptor switches on a G protein, and the G protein switches on the enzyme that makes cAMP.

Which acts first after the receptor?

  1. A. ✓ The G protein
  2. B. The enzyme that makes cAMP
    The G protein switches the enzyme on.
    So the G protein acts before the enzyme.

Why: The receptor switches on the G protein.
The G protein then switches on the enzyme that makes cAMP.
So the G protein acts first after the receptor.

4

How do you find a break in a pathway you cannot see?

5

A pathway passes its message one way, like water down a river. Break one link, and everything before the break happens as normal while everything after it stops.

6

More signal cannot bypass the break. A component that works with no input restores the response only if it sits after the break.

7

That is how two rescues locate a break.

8

Here is a drawing of a pathway as four boxes. The message passes one way, from the receptor to the response, like water flowing down a river.

A four-box pathway, receptor to relay 1 to relay 2 to response, no box shaded, with upstream at the receptor end and downstream at the response end
A four-box pathway, receptor to relay 1 to relay 2 to response, no box shaded, with upstream at the receptor end and downstream at the response end
9

A component that acts before a step is called of that step, like a town upriver.

10

A component that acts after a step is called of that step, like a town downriver.

11

In the drawing, the receptor is upstream of relay 1. Relay 2 and the response are downstream of relay 1.

12

What you are expected to know Identify which components of a drawn pathway are upstream of a chosen step, and which are downstream.

13
Check q2

Here is a hormone pathway, boxes numbered 1 to 7. A researcher adds a drug that blocks the enzyme at box 4, then adds the hormone.

A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response
A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response

Which boxes are upstream of the blocked enzyme?

  1. A. Boxes 5, 6 and 7
    The message passes from the hormone at box 1 down to the response at box 7.
    Upstream means before the block.
    Boxes 1, 2 and 3 act before box 4.
  2. B. ✓ Boxes 1, 2 and 3

Why: The message passes one way, from the hormone at box 1 to the response at box 7.
A component is upstream of a step if it acts before that step.
Boxes 1, 2 and 3 act before the enzyme at box 4, so they are upstream of the block.

14
Check q3

In the hormone pathway shown, boxes numbered 1 to 7, the G protein is box 3 and cAMP is box 5.

A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response
A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response

Is cAMP upstream or downstream of the G protein?

  1. A. Upstream
    Upstream means acting before the step.
    cAMP at box 5 is made after the G protein at box 3 has acted.
  2. B. ✓ Downstream

Why: The G protein acts at box 3.
Two steps later the enzyme makes cAMP.
cAMP comes after the G protein, so cAMP is downstream of it.

15
Check q4

In the hormone pathway shown, boxes numbered 1 to 7, the hormone is box 1 and the enzyme is box 4.

A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response
A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response

Is the hormone upstream or downstream of the enzyme?

  1. A. ✓ Upstream
  2. B. Downstream
    Downstream means acting after the step.
    The hormone at box 1 acts before anything else in the pathway.

Why: The hormone at box 1 binds the receptor before the enzyme at box 4 acts.
The hormone comes before the enzyme, so the hormone is upstream of it.

16
Check q5

In the hormone pathway shown, boxes numbered 1 to 7, the G protein is box 3.

A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response
A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response

Which boxes are downstream of the G protein?

  1. A. ✓ Boxes 4, 5, 6 and 7
  2. B. Boxes 1 and 2
    Downstream means acting after the step.
    Boxes 1 and 2 act before the G protein at box 3.

Why: A component is downstream of a step if it acts after that step.
The enzyme, cAMP, the kinase and the response, boxes 4 to 7, all act after the G protein at box 3.
So boxes 4, 5, 6 and 7 are downstream of the G protein.

17
Check q6

In the hormone pathway shown, boxes numbered 1 to 7, the kinase is box 6 and the response is box 7.

A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response
A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response

Is the response upstream or downstream of the kinase?

  1. A. Upstream
    Upstream means acting before the step.
    The response at box 7 happens after the kinase at box 6 has acted.
  2. B. ✓ Downstream

Why: The kinase at box 6 acts, and the response at box 7 follows.
The response comes after the kinase, so the response is downstream of it.

18Quick quiz: upstream and downstream mixed practice

19
Check q7

A component acts before a chosen step in a pathway.

What is that component called?

  1. A. ✓ Upstream of the step
  2. B. Downstream of the step
    Downstream means acting after the step.
    A component that acts before the step is upstream of it.

Why: A component that acts before a step is called upstream of that step.

20
Check q8

What does it mean for a component to be downstream of a step?

  1. A. ✓ It acts after the step
  2. B. It acts before the step
    A component that acts before the step is upstream of it.
  3. C. It acts at the same moment as the step
    A pathway passes its message one component at a time.
    No two components act at the same moment.

Why: A component that acts after a step is called downstream of that step.

21
Practice writing an answer

A pathway passes its message from a receptor, through relay molecules, to a response.

(a) State what it means for a component to be upstream of a step in the pathway. (1 pt)

Model answer A component upstream of a step acts before that step; the message passes through the component before it reaches the step.
Rubric
  • Award 1 point for: the component acts before the step (or the message passes through it before reaching the step).

22Break one link

23

Video: Watch: Break one link

Kinase 2 does not work: the receptor and kinase 1 switch on as normal, kinase 3 and the mating genes stay off, and 100 times the signal changes nothing because the receptor is upstream of the break.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L09Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L09Bb.mp4

24
Check q9

A receptor’s intracellular domain carries a mutation. The ligand binds the receptor as normal.

Which stage of the pathway fails?

  1. A. Reception
    Reception is the ligand binding its receptor.
    The ligand binds as normal, so reception succeeds.
  2. B. ✓ Transduction

Why: The ligand binds, so reception succeeds.
The intracellular domain cannot take up its active shape, so it never acts on the next molecule.
Transduction never starts.

25

Now imagine one link of the pathway is broken. Here is the four-box pathway with relay 1 broken.

The four-box pathway broken at relay 1, marked with a small cross at the box's corner: the receptor upstream is shaded gray as activated, relay 2 and the response downstream are unshaded
The four-box pathway broken at relay 1, marked with a small cross at the box's corner: the receptor upstream is shaded gray as activated, relay 2 and the response downstream are unshaded
26

Everything upstream of the break happens as normal. Everything downstream of the break stops.

27

Now consider a yeast cell’s mating pathway: receptor → kinase 1 → kinase 2 → kinase 3 → mating genes.

A yeast cell's mating pathway drawn as five boxes: receptor, kinase 1, kinase 2, kinase 3, mating genes
A yeast cell's mating pathway drawn as five boxes: receptor, kinase 1, kinase 2, kinase 3, mating genes
28

Suppose kinase 2 is broken: it does not work. The receptor and kinase 1 are activated as normal.

The yeast pathway with kinase 2 not working, marked with a small cross at the box's corner: the receptor and kinase 1 are shaded gray as activated, kinase 3 and the mating genes are not
The yeast pathway with kinase 2 not working, marked with a small cross at the box's corner: the receptor and kinase 1 are shaded gray as activated, kinase 3 and the mating genes are not
29

Kinase 3 and the mating genes are downstream of kinase 2. They wait for a message that never comes.

30

These cells show no response, even at 100 times the usual signal. More signal cannot bypass a broken link.

31

More signal can only activate the receptor harder. The receptor is upstream of the break, so its message still stops at kinase 2.

32

The effect of a block is what stops downstream of it: kinase 3 is never activated, and the mating genes stay off. The broken kinase 2 itself is only where the break is.

33

What you are expected to know Predict which measurements still change when one component of a pathway is blocked.

34

What you are expected to know Explain why more signal cannot restore the response when a component of the pathway is blocked.

35
Check q10

In the hormone pathway shown, boxes numbered 1 to 7, a drug blocks the enzyme at box 4. The hormone arrives.

A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response
A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response

Does the G protein at box 3 switch on?

  1. A. ✓ Yes
  2. B. No
    The G protein at box 3 acts before the blocked enzyme at box 4.
    Everything upstream of a break still switches on.

Why: The G protein at box 3 is upstream of the blocked enzyme at box 4. Everything upstream of a break still switches on as normal.
So the hormone binds the receptor at box 2, and the receptor switches the G protein on.

36
Check q11

In the hormone pathway shown, boxes numbered 1 to 7, a drug blocks the enzyme at box 4. The hormone arrives.

A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response
A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response

Does cAMP at box 5 rise?

  1. A. Yes
    The enzyme at box 4 makes the cAMP.
    The drug blocks that enzyme.
    So no cAMP is made, and cAMP at box 5 does not rise.
  2. B. ✓ No

Why: The enzyme at box 4 makes the cAMP.
The drug blocks that enzyme.
cAMP at box 5 is downstream of the block, and everything downstream of a break stops.
So cAMP does not rise.

37
Check q12

In the hormone pathway shown, boxes numbered 1 to 7, a drug blocks the enzyme at box 4. The hormone arrives.

A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response
A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response

Does the response at box 7 happen?

  1. A. Yes
    The response at box 7 needs the kinase at box 6, and the kinase needs cAMP.
    No cAMP is made, so the response never starts.
  2. B. ✓ No

Why: The response at box 7 is downstream of the blocked enzyme at box 4.
Everything downstream of a break stops.
So the response does not happen.

38
Check q13

A bacterium responds to a rise in salt outside it through the pathway shown: protein S is phosphorylated, then protein R, then a reporter gene is expressed. In cells lacking R, S is still phosphorylated after a salt rise, and the reporter stays off.

A bacterium's salt pathway drawn as three boxes: protein S phosphorylated, then protein R phosphorylated, then a reporter gene expressed
A bacterium's salt pathway drawn as three boxes: protein S phosphorylated, then protein R phosphorylated, then a reporter gene expressed

After a salt rise in cells lacking S, which measurements change?

  1. A. Both R phosphorylation and reporter expression
    S is upstream of both R and the reporter, so both stop.
  2. B. R phosphorylation but not reporter expression
    S phosphorylates R.
    With S gone, R stays unphosphorylated.
  3. C. Reporter expression but not R phosphorylation
    The reporter is downstream of R, and R is downstream of S.
  4. D. ✓ Neither R phosphorylation nor reporter expression

Why: Protein S is the first link of the pathway.
Everything downstream of a missing link stops.
S phosphorylates R, so with S missing, R is never phosphorylated.
Phosphorylated R switches on the reporter gene, so the reporter is never expressed.
The salt rise changes neither measurement.

39
Practice writing an answer

A yeast cell’s mating pathway passes its message receptor → kinase 1 → kinase 2 → kinase 3 → mating genes. In a line of cells, kinase 2 is broken. A researcher gives these cells 200 times the usual mating signal, and the mating genes stay off.

(a) Explain how this result demonstrates that a break in a pathway blocks the message at any strength of signal. (1 pt)

Model answer The signal binds the receptor, and the receptor is upstream of kinase 2.
So 200 times the signal only activates the receptor and kinase 1 harder.
Kinase 1’s message has to pass through kinase 2, and kinase 2 is broken.
So the message stops at kinase 2, however strong it is.
Kinase 3 and the mating genes are downstream of the break, so they stay off at any concentration of signal.
Rubric
  • Award 1 point for: more signal acts only upstream of the break (the receptor and kinase 1), and every message must pass through the broken kinase 2, so nothing downstream is activated at any concentration.

40Locate the break

41

Video: Watch: Locate the break

An always-active component needs no input; an always-active kinase 3, downstream of the break, switches the mating genes on; an always-active kinase 1, upstream of it, changes nothing; so the break is located.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L09Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L09Bc.mp4

42

Suppose a component is locked in its active shape. It works with no input from the step before it.

43

When a component is active whether or not any signal arrives, we call it , because it needs no input to be switched on.

44

Give the cells with a broken kinase 2 an always-active kinase 3. The mating genes are expressed.

The yeast pathway with kinase 2 not working; kinase 3's own box is shaded and tagged always-active, and the mating genes are activated
The yeast pathway with kinase 2 not working; kinase 3's own box is shaded and tagged always-active, and the mating genes are activated
45

Kinase 3 sits downstream of the break. Everything after kinase 3 is intact, so the message reaches the genes.

46

Give the cells an always-active kinase 1 instead. Nothing happens.

The yeast pathway with kinase 2 not working; kinase 1's own box is shaded and tagged always-active, but kinase 3 and the mating genes stay off
The yeast pathway with kinase 2 not working; kinase 1's own box is shaded and tagged always-active, but kinase 3 and the mating genes stay off
47

Kinase 1 sits upstream of the break. Its message still has to pass through kinase 2.

48

So an always-active component restores the response only if it sits downstream of the break. That is how the two rescues locate the break: it lies between kinase 1 and kinase 3.

49

What you are expected to know Locate the break in a pathway from which always-active component restores the response.

50
Check q14

In the hormone pathway shown, boxes numbered 1 to 7, a drug blocks the enzyme at box 4. The researcher wants to restore the response with one addition.

A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response
A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response

Which single addition restores the response?

  1. A. Ten times more hormone
    More hormone activates the receptor harder.
    The receptor is upstream of the blocked enzyme, so its message still stops at box 4.
  2. B. An always-active receptor (box 2)
    The receptor at box 2 acts before the enzyme at box 4.
    Its message still has to pass through the blocked enzyme.
  3. C. ✓ An always-active kinase (box 6)

Why: Only a component downstream of the break can restore the response.
The kinase at box 6 sits downstream of the blocked enzyme at box 4.
An always-active kinase needs no cAMP to switch it on.
So it switches on the response at box 7 directly, and the response returns.

51
Check q15

A pathway’s components act in the order receptor → kinase A → kinase B → response. In a mutant line, kinase A is phosphorylated as normal after the signal, kinase B stays unphosphorylated, and there is no response.

Where is the break?

  1. A. At the receptor
    Kinase A was phosphorylated.
    So the receptor passed the message on as normal.
  2. B. ✓ Between kinase A and kinase B
  3. C. Between kinase B and the response
    Kinase B was never phosphorylated.
    So the message was lost before it reached kinase B, not after.
  4. D. In the signal molecule itself
    The signal activated the receptor and kinase A as normal.
    So the signal molecule is intact.

Why: Kinase A was phosphorylated as normal.
So the receptor and kinase A passed the message on as normal.
Kinase B stayed unphosphorylated.
So the message was lost between kinase A and kinase B.
So the break is that step, between kinase A and kinase B.

52

Back to the yeast cell whose mating pathway passes its message receptor → kinase 1 → kinase 2 → kinase 3 → mating genes. Its kinase 2 did not work, so 100 times the signal brought no response.

53

An always-active kinase 3, downstream of the break, switched the mating genes on. An always-active kinase 1, upstream of the break, changed nothing.

54

A changed protein changes everything downstream of it and nothing upstream. So the pattern of what still works tells you where the break is: here, at kinase 2.

55Quick quiz: always-active mixed practice

56
Check q16

What is an always-active kinase?

  1. A. ✓ A kinase that is active whether or not a signal arrives
  2. B. A kinase that is active only while a ligand is bound to the receptor
    A normal kinase is active only while the message reaches it.
    An always-active kinase needs no message.
  3. C. A kinase that a drug has blocked
    A blocked kinase passes nothing on.
    An always-active kinase passes the message on all the time.

Why: A component that is active whether or not any signal arrives is called always-active.

57
Check q17

A pathway passes its message receptor → protein P → protein Q → response. Protein P is broken. A researcher adds an always-active protein Q.

Does the response return?

  1. A. ✓ Yes
  2. B. No
    Protein Q sits downstream of the broken protein P.
    An always-active Q needs no message from P, so it switches on the response.

Why: Protein Q is downstream of the break at protein P.
An always-active protein Q needs no input from P.
So it acts on the response directly, and the response returns.

58
Check q18

A pathway passes its message receptor → protein P → protein Q → response. Protein Q is broken. A researcher adds an always-active protein P.

Does the response return?

  1. A. Yes
    Protein P sits upstream of the broken protein Q.
    Its message still has to pass through Q, and Q is broken.
  2. B. ✓ No

Why: Protein P is upstream of the break at protein Q.
An always-active protein P sends its message to Q.
Q is broken, so the message stops there and the response stays off.

59
Practice writing an answer

A component of a signal transduction pathway is described as always-active.

(a) State what always-active means for a component of a pathway. (1 pt)

Model answer An always-active component is active whether or not a signal arrives; it needs no input from the step before it.
Rubric
  • Award 1 point for: active whether or not a signal arrives (or needs no input from the step before it).

60Mixed practice mixed practice

61
Check q19

A researcher gives yeast cells whose kinase 2 is broken fifty times the normal concentration of mating signal.

Predict the response.

  1. A. A normal response
    Every message from the receptor must go through kinase 2, and kinase 2 is broken.
  2. B. A weaker response than normal
    The break is complete: kinase 2 passes nothing on.
  3. C. A response, but delayed by several hours
    Nothing downstream of kinase 2 is activated, however long the wait.
  4. D. ✓ No response at any concentration

Why: More signal cannot bypass a broken link.
Fifty times the signal activates the receptor and kinase 1 harder.
The receptor and kinase 1 are both upstream of the broken kinase 2, so the message still stops there.
So kinase 3 and the mating genes stay off at any concentration.

62
Check q20

In the hormone pathway shown, boxes numbered 1 to 7, a drug blocks the G protein at box 3. The hormone arrives.

A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response
A hormone pathway drawn as boxes numbered 1 to 7: hormone, receptor, G protein, enzyme, cAMP, kinase, response

Which boxes does the message still reach?

  1. A. ✓ Boxes 1 and 2
  2. B. Boxes 4 to 7
    Boxes 4 to 7 are downstream of the blocked G protein.
    Everything downstream of a break stops.
  3. C. All seven boxes
    The G protein at box 3 is blocked.
    The message reaches nothing downstream of it.
  4. D. No box
    The hormone at box 1 still binds the receptor at box 2.
    Everything upstream of a break happens as normal.

Why: The block is at the G protein, box 3.
Everything upstream of a break happens as normal: the hormone at box 1 binds the receptor at box 2.
Everything downstream of the break stops: the message never reaches boxes 4 to 7.

63
Check q21

A drug blocks the kinase in a heart muscle cell’s epinephrine pathway: receptor → G protein → cAMP-making enzyme → cAMP → kinase → the proteins that drive contraction. A researcher adds epinephrine.

Which of the following describes the effect of the block?

  1. A. cAMP stays at rest, so the cell does not contract harder
    The receptor, the G protein and the cAMP-making enzyme are all upstream of the kinase, so they still switch on and cAMP rises.
    The block stops the pathway after cAMP.
  2. B. ✓ cAMP rises as normal, but the cell does not contract harder
  3. C. cAMP rises higher than normal
    The kinase does not consume cAMP.
    So cAMP rises to its normal level and no higher.
  4. D. The kinase stays in its inactive shape, and nothing else in the cell changes
    Upstream of the kinase, epinephrine binds and cAMP rises, so something does change.
    Downstream, the contraction proteins are never switched on, so the cell does not contract harder.

Why: A block’s effect is what happens downstream of it.
The receptor, G protein and cAMP-making enzyme sit upstream of the kinase, so cAMP rises as normal.
The kinase is blocked, so it never adds phosphates to the proteins that drive contraction.
So the cell does not contract harder.

64
Check q22

A plant cell’s pathway passes its message receptor → protein A → protein B → protein C → the stomata close. Protein B is broken. The hormone arrives.

Which of the following is phosphorylated?

  1. A. Neither protein A nor protein C
    Protein A is upstream of the broken protein B.
    Everything upstream of a break happens as normal.
  2. B. Protein C only
    Protein C is downstream of the broken protein B.
    Nothing downstream of a break is activated.
  3. C. ✓ Protein A only
  4. D. Proteins A and C
    Protein C is downstream of the broken protein B.
    Only protein A, upstream of the break, is phosphorylated.

Why: Protein A is upstream of the broken protein B, so the receptor phosphorylates protein A as normal.
Protein B is broken, so it never phosphorylates protein C.
So protein A is phosphorylated and protein C is not.

65
Check q23

A researcher gives yeast cells whose kinase 3 is broken an always-active kinase 2 and twenty times the normal mating signal.

A yeast cell's mating pathway drawn as five boxes: receptor, kinase 1, kinase 2, kinase 3, mating genes
A yeast cell's mating pathway drawn as five boxes: receptor, kinase 1, kinase 2, kinase 3, mating genes

Predict the mating genes.

  1. A. Expressed as normal
    Kinase 2 acts on kinase 3, and kinase 3 is broken.
    So the message stops at kinase 3.
  2. B. Expressed, but weakly
    Kinase 3 is broken, so kinase 3 passes nothing on, however hard kinase 2 acts on it.
  3. C. ✓ Off

Why: The break is at kinase 3, and kinase 2 sits upstream of it.
An always-active kinase 2 sends its message to kinase 3, which is broken, so the message stops there.
So the mating genes stay off.
Twenty times the signal acts even further upstream, so it changes nothing.

66
Check q24

A plant cell’s pathway passes its message receptor → protein A → protein B → protein C → the stomata close. Protein B is broken. A researcher wants to restore the response with one addition.

Which single addition restores the closing of the stomata?

  1. A. Ten times more hormone
    More hormone activates the receptor harder.
    The receptor is upstream of the broken protein B, so the message still stops there.
  2. B. ✓ An always-active protein C
  3. C. An always-active protein A
    Protein A is upstream of the broken protein B.
    Its message still has to pass through B.

Why: Only a component downstream of the break can restore the response.
Protein C sits downstream of the broken protein B.
An always-active protein C needs no message from B.
So it closes the stomata directly.

67
Check q25

In the bacterium’s salt pathway, protein S is phosphorylated, then protein R, then a reporter gene is expressed. A researcher gives cells lacking protein R an always-active protein S, then raises the salt outside them.

A bacterium's salt pathway drawn as three boxes: protein S phosphorylated, then protein R phosphorylated, then a reporter gene expressed
A bacterium's salt pathway drawn as three boxes: protein S phosphorylated, then protein R phosphorylated, then a reporter gene expressed

Predict the reporter gene.

  1. A. The reporter is expressed as normal
    S acts on R, and R acts on the reporter.
    With R missing, S has nothing to pass its message to.
  2. B. The reporter is expressed only after the salt rise
    The salt rise acts at the top of the pathway.
    The reporter is expressed only through R, and R is missing.
  3. C. ✓ The reporter stays off

Why: Protein S sits upstream of protein R.
Protein R is missing, so R is the break.
An always-active S has nothing to pass its message to.
So the reporter stays off.
Only a component downstream of the break, such as an always-active R, would restore the reporter.

68
Check q26

A bar graph’s caption says: ‘error bars represent ±2SE’.

What does each error bar show?

  1. A. ✓ The range the true mean is likely to lie in
  2. B. The largest and the smallest reading in that set of dishes
    The bar runs two standard errors either side of the mean.
    It shows where the true mean is likely to lie, not the spread of the readings.

Why: An error bar of ±2SE runs two standard errors above and below the mean.
So it shows the range the true mean is likely to lie in.

69
Practice writing an answer

Researchers study the pathway shown in cultured muscle cells: hormone → receptor → kinase 1 → kinase 2 → glucose transporters move to the surface. They have normal cells and a mutant line whose receptor carries a change in its intracellular domain. Hormone binding per cell is the same in both lines. For each line, six dishes receive hormone and six receive none; the researchers measure glucose uptake after 30 minutes. The graph shows the means; the error bars show ±2SE.

The pathway model (hormone, receptor, kinase 1, kinase 2, transporters to the surface) beside a bar graph of glucose uptake for normal and mutant cells with and with no hormone; error bars show ±2SE
The pathway model (hormone, receptor, kinase 1, kinase 2, transporters to the surface) beside a bar graph of glucose uptake for normal and mutant cells with and with no hormone; error bars show ±2SE

(a) Identify the dependent variable in this experiment. (1 pt)

Model answer The dependent variable is the glucose uptake of the cells, in nanomoles per minute per million cells.
Rubric
  • Award 1 point for: glucose uptake (nmol per minute per million cells) as the dependent variable.

Slip Naming hormone binding as the dependent variable. Binding was checked and found equal; the measured outcome is glucose uptake.

(b) Explain why the researchers also gave each cell line dishes with no hormone. (1 pt)

Model answer The no-hormone dishes are the control for each line.
They show each line’s glucose uptake with the hormone absent.
So the rise seen when the researcher adds hormone can be credited to the hormone, and not to a difference between the lines.
Rubric
  • Award 1 point for: the no-hormone dishes give each line’s uptake with the tested factor absent (a control), so the effect of the hormone can be measured against it.

Slip Saying the no-hormone dishes ‘check the cells are alive’. They give the baseline uptake that the hormone dishes are compared against.

(c) Support the claim that the mutant receptor’s fault is in transduction rather than in reception, using one measurement from the description. (1 pt)

Model answer Hormone binding per cell is the same in the mutant line as in the normal line.
So the ligand-binding domain works, and reception succeeds.
The mutant cells still take up no extra glucose.
So the message is lost after binding.
The changed intracellular domain cannot take up its active shape.
So transduction never starts.
Rubric
  • Award 1 point for: the evidence (hormone binding per cell is the same in the mutant line as in the normal line) AND the reasoning (equal binding shows reception works, so the loss of the response must come after binding, in transduction: the intracellular domain cannot change shape).

Slip Citing the low rate of glucose uptake alone. A low rate of uptake shows the pathway failed; the binding measurement is what places the failure after reception.

(d) Predict the glucose uptake of mutant cells given an always-active kinase 2 with no hormone added, and justify your prediction. (1 pt)

Model answer Glucose uptake rises to about the level of normal cells given hormone, near 80 nmol per minute per million cells.
Kinase 2 sits downstream of the broken receptor.
Everything after kinase 2 is intact.
An always-active kinase 2 needs no message from the receptor.
So the always-active kinase 2 moves the transporters to the surface, whether or not the receptor ever changes shape.
Rubric
  • Award 1 point for: uptake rises (toward the hormone-treated normal level), because kinase 2 is downstream of the break and the transporters and other components after it are intact.

Slip Predicting no rise because the researcher added no hormone. The always-active kinase 2 needs no input from upstream; it acts on its own.

Glossary

upstream
A component of a pathway is upstream of a step if it acts before that step: the message passes through the component before it reaches the step, like a town upriver.
downstream
A component of a pathway is downstream of a step if it acts after that step: the message reaches the component only after it has passed the step, like a town downriver.
always-active
A component of a pathway that is active whether or not any signal arrives; it needs no input from the step before it. An always-active component restores a blocked pathway’s response only if it sits downstream of the break.

APBIO-U04-L10 Stuck on, switched off

Topic 4.3 · Signal Transduction Pathways · 103 steps

A culture dish seen from above holding cells, several of them in the middle of dividing, with no growth factor in the liquid around them
A culture dish seen from above holding cells, several of them in the middle of dividing, with no growth factor in the liquid around them

Here is a dish of cells taken from a tumor, washed free of every trace of growth factor.

The cells sit in the dish for two days. 79% of them divide anyway.

Add a drug that blocks their growth-factor receptor, and only 24% divide. That is the same as cells with no receptor at all.

What is switching division on with no signal present? And how did the drug find it?

Unit 4 · Cell Communication and Cell Cycle

1Broken on

2

Video: Watch: Broken on

A normal receptor at rest beside the mutant receptor whose inner part is locked on. The mutant passes the message on with nothing bound, so the cell divides without being told to.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L10a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L10a.mp4

3

A pathway can be broken off: one link cannot pass the message on. A pathway can also be broken on: one component passes the message on all the time, with no signal present.

4

Chemicals from outside the organism can do either. One kind of chemical switches a pathway on; another kind switches it off.

5

Three measurements each test one link of the pathway: 1 ligand bound, 2 second messenger raised, 3 response.

6

A change alters everything downstream of it and nothing upstream. Read the three measurements against that rule, and you can place the change and predict what the organism shows.

7
Check q1

A signal molecule reaches a cell and tells it to grow and divide.

What is such a signal called?

  1. A. A second messenger
    A second messenger is made inside a cell and stays inside it.
    This signal reaches the cell from outside.
  2. B. ✓ A growth factor
  3. C. A neurotransmitter
    A neurotransmitter is released by a nerve cell onto the cell beside it.
    Its message is not grow and divide.

Why: A signal that tells a cell to grow and divide is called a growth factor, whatever its exact kind.

8

Now consider the mutant cells in the dish. Here is a normal growth-factor receptor at rest, and beside it the mutant receptor from those cells.

Two receptors side by side with nothing bound to either: the normal receptor's inner part is inactive and passes no message on; the mutant receptor's inner part is locked in its active shape and passes the message on
Two receptors side by side with nothing bound to either: the normal receptor's inner part is inactive and passes no message on; the mutant receptor's inner part is locked in its active shape and passes the message on
9

The mutant receptor’s intracellular domain is locked in its active shape, with nothing in the binding site.

10

The locked receptor passes the message on all the time, with no ligand bound. So the cell keeps producing the response.

11

The response of this pathway is ‘grow and divide’. So the cell divides without being told to.

12

Here is a bar graph of the counts: the percentage of cells dividing in two days, in four groups of cells.

A bar graph of the percentage of cells dividing in two days: mutant cells with no growth factor 79%, mutant cells with growth factor 81%, mutant cells with a receptor-blocking drug 24%, and cells with no receptor 23%
A bar graph of the percentage of cells dividing in two days: mutant cells with no growth factor 79%, mutant cells with growth factor 81%, mutant cells with a receptor-blocking drug 24%, and cells with no receptor 23%
13

Of the mutant cells, 79% divide with no growth factor, and 81% divide with it. Of the cells with no receptor at all, 23% divide.

14

A drug that fits the mutant receptor holds it inactive. With the drug, division falls to 24%, the level of the cells with no receptor.

15

So the signal still starts at the receptor.

16

A response with no ligand present does not mean the pathway does nothing. The pathway is broken on.

17

One simplification: we treat the drug as holding the whole receptor in its inactive shape. Some real mutations lock the intracellular domain on; then a drug in the binding site changes nothing.

18
Check q2

A yeast kinase is locked in its active shape. It acts with no input from the step before it.

What is such a component called?

  1. A. Inactive
    An inactive component passes no message on.
    This kinase acts all the time.
  2. B. Blocked
    A blocked component is held in its inactive shape.
    This kinase is locked in its active shape.
  3. C. ✓ Always-active

Why: A component locked in its active shape works with no input from the step before it.
We call it always-active.

19

Here the always-active component is the receptor itself.

20

What you are expected to know Predict what a component locked in its active shape does to its pathway and to the cell.

21
Check q3

A line of gland cells carries a kinase locked in its active shape. The kinase sits in a pathway that ends in the secretion of a hormone.

Predict what the cells do.

  1. A. They secrete more hormone when the signal is present, and less without it
    The locked kinase needs no input.
    The locked kinase acts all the time.
  2. B. They secrete nothing
    This pathway is broken on: the kinase works with nothing upstream telling it to.
  3. C. ✓ They secrete the hormone with no signal present
  4. D. They secrete more hormone when more signal is added
    Downstream of a locked kinase, the cell produces its response whether or not any signal arrives.

Why: The kinase is locked in its active shape.
So the kinase phosphorylates its targets all the time.
So everything downstream of the kinase stays switched on all the time.
So the cells secrete the hormone with no signal at all.

22
Practice writing an answer

A line of gland cells carries a kinase locked in its active shape. The kinase sits in a pathway that ends in the secretion of a hormone. The cells secrete the hormone with no signal present.

(a) Explain why the gland cells secrete the hormone with no signal present. (1 pt)

Model answer A kinase is normally switched on only when the step above it passes it a message.
This kinase is locked in its active shape, so it is active whether or not any message reaches it.
An active kinase phosphorylates its target proteins, so the targets are switched on all the time.
Everything downstream of the kinase stays switched on all the time.
The pathway ends in secretion, so the cells secrete the hormone all the time, with no signal present.
Rubric
  • Award 1 point for: the locked kinase is active without any message from upstream, so it phosphorylates its targets continuously, everything downstream of it stays switched on, and the cell secretes with no signal.
23
Check q4

A student looks at the dish of mutant cells: 79% divide in two days with no growth factor present. The student says: “With no growth factor around, their growth-factor pathway must be doing nothing.”

Is the student correct?

  1. A. Yes — a pathway with no ligand present is switched off
    Dividing is the pathway’s response, and the mutant cells divide with no growth factor.
    So the components pass the message on with nothing starting them: broken on.
  2. B. ✓ No — the pathway produces its response all the time; it is broken on

Why: Dividing is the response at the end of the growth-factor pathway.
The mutant cells divide with no growth factor present.
So the pathway is producing its response with no ligand.
A pathway that produces its response with no ligand is broken on, not switched off.

24
Check q5

Here again are the mutant cells from the dish: the line that divides with no growth factor. The graph shows the percentage of cells dividing in two days in four groups of cells.

A bar graph of the percentage of cells dividing in two days: mutant cells with no growth factor 79%, mutant cells with growth factor 81%, mutant cells with a receptor-blocking drug 24%, and cells with no receptor 23%
A bar graph of the percentage of cells dividing in two days: mutant cells with no growth factor 79%, mutant cells with growth factor 81%, mutant cells with a receptor-blocking drug 24%, and cells with no receptor 23%

Which comparison shows that the mutant cells’ division still starts at the receptor?

  1. A. Mutant cells with no growth factor against mutant cells with growth factor
    Growth factor makes almost no difference to the mutant cells, 79% against 81%.
    So the comparison shows the cells divide with no ligand, but not where the signal starts.
  2. B. Mutant cells with growth factor against cells with no receptor
    Mutant cells and receptor-less cells are two different lines, so extra division in the mutant could start anywhere in its pathway.
  3. C. ✓ Mutant cells with no growth factor against mutant cells with the receptor-blocking drug
  4. D. Mutant cells with the drug against cells with no receptor
    The drug takes the mutant cells to the receptor-less level, 24% against 23%; it is the fall from 79% to 24% that shows the receptor drove the division.

Why: The drug holds the receptor inactive.
With the drug, division falls from 79% to 24%.
A drug at the receptor can only stop a signal that starts at the receptor.
So the fall from 79% to 24% shows that the mutant cells’ division starts at the receptor.

25Chemicals from outside

26

Video: Watch: Three molecules, one binding site

Epinephrine, a look-alike and a blocker in the same binding site. The look-alike triggers the shape change and switches the pathway on; the blocker fits, triggers nothing and keeps epinephrine out.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L10b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L10b.mp4

27

Molecules from outside the organism can bind a pathway’s components too. What such a molecule does depends on where it binds and on what happens when it binds.

28
Check q6

A competitive inhibitor enters an enzyme’s active site.

What does the inhibitor do there?

  1. A. ✓ It fills the site without being changed, so the substrate cannot enter
  2. B. It is changed into product, like the substrate
    An inhibitor is not a substrate: the enzyme changes nothing about it.
    The inhibitor just sits in the site and keeps the substrate out.

Why: A competitive inhibitor fills the active site without being changed, so the substrate cannot enter.

29

Here is the binding site of the epinephrine receptor, three times, with a different molecule in it each time.

Three receptors, each with a different molecule in its binding site: epinephrine and a look-alike both trigger the inner shape change and the message is passed on; a blocker fits but triggers nothing
Three receptors, each with a different molecule in its binding site: epinephrine and a look-alike both trigger the inner shape change and the message is passed on; a blocker fits but triggers nothing
30

The ligand fits the receptor’s binding site by shape and charge.

31

While the ligand is bound, the receptor holds a different shape.

32

So another part of the receptor can now act on the next molecule.

33

Here the ligand is epinephrine, and the pathway is switched on.

34

Now suppose a molecule is shaped enough like epinephrine to fit the binding site and produce the same shape change. We call such a molecule a .

35

A look-alike switches the pathway on, with no epinephrine present.

36

Now suppose a molecule fits the binding site but produces no shape change. The molecule just sits in the site.

37

While the molecule sits there, epinephrine cannot bind. So the pathway is switched off.

38

We call such a molecule a . A blocker fills the receptor’s binding site just like a competitive inhibitor fills an enzyme’s active site.

39

A beta blocker is a blocker that fits the epinephrine receptor on heart muscle cells.

40

Epinephrine cannot bind a site the beta blocker fills. So epinephrine cannot speed the heart.

41

Now suppose a chemical enters the cell and binds a relay protein, holding it in its inactive shape. The pathway stops at that protein.

42

For example, a kinase inhibitor added to liver cells with epinephrine lets cAMP rise as normal. But no glucose leaves the cells.

43

Here is a table of the three kinds of chemical: what each binds and how, and what the pathway then does.

A three-row table: a look-alike that fits and triggers the shape change switches the pathway on; a blocker that fits and triggers nothing switches it off; a chemical that holds a relay protein inactive switches it off from that protein on
A three-row table: a look-alike that fits and triggers the shape change switches the pathway on; a blocker that fits and triggers nothing switches it off; a chemical that holds a relay protein inactive switches it off from that protein on
44

What you are expected to know Predict what a chemical from outside does to a pathway, from where it binds and what happens when it binds.

45
Check q7

A person takes a beta blocker, which fits the epinephrine receptor on heart muscle cells without activating it. A door slams and epinephrine floods their blood.

Predict the person’s heart rate, compared with a person taking no drug.

  1. A. It rises as usual
    The blocker sits in many of the receptors’ binding sites.
    Epinephrine cannot bind a site the blocker fills.
    So fewer receptors are activated, and the heart rate rises less.
  2. B. It rises more
    The blocker fits the binding site and triggers no shape change.
    So the blocker adds nothing to the response, and it keeps epinephrine out.
  3. C. ✓ It rises less

Why: The beta blocker fits the receptor’s binding site and triggers no shape change, so it sits in the site and does nothing.
Epinephrine reaches the heart and finds many of its receptors occupied.
So fewer receptors are activated.
So the heart rate rises less than with no drug.

46
Practice writing an answer

A person takes a beta blocker, which fits the epinephrine receptor on heart muscle cells without activating it. A door slams and epinephrine floods their blood. The person’s heart rate rises less than it would with no drug.

(a) Explain why the beta blocker makes the heart rate rise less. (1 pt)

Model answer The beta blocker fits the epinephrine receptor’s binding site but produces no shape change, so it passes no message on.
One site holds one molecule at a time, so while the blocker sits in a site, epinephrine cannot bind it.
Epinephrine reaches the heart cells and finds many sites already filled by the blocker.
So fewer receptors bind epinephrine and change shape.
So fewer receptors pass the speed-up message on, and the heart rate rises less.
Rubric
  • Award 1 point for: the blocker fills binding sites without producing the shape change, so epinephrine binds fewer receptors, fewer receptors pass the speed-up message on, and the heart rate rises less.
47
Check q8

A researcher gives liver cells epinephrine alone, and the cells release glucose as normal. The researcher gives a second batch of those liver cells epinephrine together with a chemical; cAMP rises as normal, but their glucose release stays at its resting level.

Where does the chemical act?

  1. A. On the receptor’s binding site
    cAMP rose as normal, so epinephrine bound the receptor and the receptor passed the message on.
  2. B. ✓ On a kinase or an enzyme downstream of cAMP
  3. C. On the G protein or the enzyme that makes cAMP
    The G protein switches on the enzyme, and the enzyme makes the cAMP. cAMP rose as normal.
    So the G protein and the enzyme both worked at full speed.
  4. D. Nowhere in the pathway
    Cells from the same culture given epinephrine alone released glucose as normal.
    So the glycogen was there, and the pathway works.
    The chemical is what stopped the release.

Why: Cells given epinephrine alone released glucose: the pathway works.
With the chemical, cAMP rose as normal, so the receptor, G protein and cAMP-making enzyme all worked.
So the chemical acts after cAMP.
After cAMP come the kinases and then the glycogen-breaking enzymes, so the chemical acts on one of those.

48Quick quiz: look-alike, blocker, or a chemical holding a relay protein inactive? mixed practice

49
Check q9

A chemical enters a muscle cell and binds a kinase. Insulin still binds its receptor, and no glucose transporters move to the surface.

Which kind of chemical is this?

  1. A. A look-alike of the ligand
    A look-alike fits the receptor’s binding site and switches the pathway on.
    This chemical is inside the cell, on a kinase, and the pathway is off.
  2. B. A blocker in the binding site
    A blocker fits the receptor’s binding site and keeps the ligand out.
    Here insulin still binds its receptor, so the site is free; the chemical acts inside, on a kinase.
  3. C. ✓ A chemical holding a relay protein inactive

Why: Insulin still binds its receptor, so the binding site is free.
The chemical binds a kinase, a relay protein inside the cell, and the response stops.
So the chemical holds a relay protein inactive.

50
Check q10

A drug fits a receptor’s binding site and triggers no shape change. The hormone’s response is switched off.

Which kind of chemical is this?

  1. A. A look-alike of the ligand
    A look-alike triggers the receptor’s shape change and switches the pathway on.
    This drug triggers no shape change, and the pathway is off.
  2. B. ✓ A blocker in the binding site
  3. C. A chemical holding a relay protein inactive
    The drug sits in the receptor’s binding site, outside the cell.
    A relay protein is inside the cell.

Why: The drug fits the binding site and triggers no shape change.
So the drug just sits there, and the hormone cannot bind.
The pathway is switched off.
That is a blocker.

51
Check q11

A researcher sprays a plant with a chemical. The plant grows as if flooded with its own growth hormone, with no growth hormone present.

Which kind of chemical is this?

  1. A. ✓ A look-alike of the ligand
  2. B. A blocker in the binding site
    A blocker switches the pathway off.
    The plant is responding with no hormone present, so the chemical switched the pathway on.
  3. C. A chemical holding a relay protein inactive
    A chemical holding a relay protein inactive switches the pathway off from that protein on.
    The plant is responding, so the pathway is on.

Why: The plant responds as if its growth hormone were present, with no hormone around.
So the chemical switched the pathway on.
A chemical that fits the binding site and produces the receptor’s shape change does that.
That is a look-alike.

52
Check q12

A drug fills the epinephrine receptor’s binding site on heart cells. Epinephrine arrives, and the heart rate rises less than usual.

Which kind of chemical is this?

  1. A. A look-alike of the ligand
    A look-alike would speed the heart on its own.
    This drug makes the response smaller, so it keeps epinephrine out of the binding site.
  2. B. ✓ A blocker in the binding site
  3. C. A chemical holding a relay protein inactive
    The drug is in the receptor’s binding site, outside the cell.
    A relay protein is inside the cell.

Why: The drug fills the binding site, so epinephrine cannot bind those receptors.
The response is smaller.
A molecule that fills the binding site and keeps the ligand out is a blocker.

53
Check q13

A poison binds a relay protein inside a cell and holds it in its inactive shape. The hormone still binds its receptor, and the response stops.

Which kind of chemical is this?

  1. A. A look-alike of the ligand
    A look-alike binds the receptor’s site and switches the pathway on.
    This poison is inside the cell, and the pathway is off.
  2. B. A blocker in the binding site
    A blocker fills the receptor’s binding site.
    Here the hormone still binds its receptor, so the site is free; the poison is on a relay protein inside.
  3. C. ✓ A chemical holding a relay protein inactive

Why: The hormone still binds its receptor, so the binding site is free.
The poison holds a relay protein inside the cell in its inactive shape.
So the pathway is off from that protein on.
That is a chemical that holds a relay protein inactive.

54
Check q14

A chemical sits in a receptor’s binding site and produces the receptor’s shape change. The cells respond with no natural ligand around.

Which kind of chemical is this?

  1. A. ✓ A look-alike of the ligand
  2. B. A blocker in the binding site
    A blocker produces no shape change, and the pathway stays off.
    This chemical produces the shape change, and the cells respond.
  3. C. A chemical holding a relay protein inactive
    The chemical is in the receptor’s binding site, not on a relay protein inside the cell, and the pathway is on.

Why: The chemical fits the binding site and produces the receptor’s shape change.
So the receptor passes the message on as if the ligand were bound.
The cells respond with no natural ligand.
That is a look-alike.

55Read the fault off the table

56

Video: Watch: Three measurements, one rule

Ligand bound, second messenger raised, response: each tests one link. A change alters everything downstream of it and nothing upstream, so the pattern of yes and no places the change.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L10c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L10c.mp4

57

Now consider one set of cells measured three ways: normal cells given the ligand, receptor-mutant cells given the ligand, and receptor-mutant cells given an always-active kinase and no ligand.

58

Here is the table of the three measurements: ligand bound, second messenger raised, and response.

A table of three measurements, ligand bound, second messenger raised and response, for normal cells with the ligand, receptor-mutant cells with the ligand, and receptor-mutant cells given an always-active kinase with no ligand
A table of three measurements, ligand bound, second messenger raised and response, for normal cells with the ligand, receptor-mutant cells with the ligand, and receptor-mutant cells given an always-active kinase with no ligand
59

Each measurement tests one link:

  • Ligand bound: tests reception.
  • Second messenger: tests the relay up to that point.
  • Response: tests the whole chain.

60

Read the receptor-mutant row. No ligand is found bound, and nothing downstream follows.

61

Binding is the ligand-binding domain’s job. So the fault is at reception, in the ligand-binding domain.

62

Read the third row. The always-active kinase restores the response with no binding and no second messenger.

63

So the kinase sits downstream of the fault. Everything after the kinase is intact.

64

Both readings use one rule: a change alters everything downstream of it and nothing upstream.

65

To justify a claim about where a change acts, connect each measurement to that rule.

66

Now consider the dish of tumor cells again. The blocker held the receptor inactive, and division fell to 24%, the level of cells with no receptor.

67

A blocker at the receptor alters only what lies downstream of the receptor. If the always-active component sat downstream of the receptor, the blocker would change nothing.

68

Division fell. So the fault is at the receptor.

69

What you are expected to know Justify where in a pathway a change acts by connecting each of the three measurements to the rule that a change alters everything downstream of it and nothing upstream.

70
Check q15

A researcher gives cells their ligand and measures them three ways.

A one-row table for cells given the ligand: ligand bound no, second messenger raised no, response no
A one-row table for cells given the ligand: ligand bound no, second messenger raised no, response no

Where is the fault?

  1. A. ✓ At the ligand-binding domain
  2. B. Between ligand binding and the second messenger
    No ligand is found bound.
    The fault is at binding itself, before anything inside the cell.
  3. C. After the second messenger
    No ligand is found bound.
    The fault is at binding itself, before anything inside the cell.

Why: No ligand is found bound, so reception fails.
Binding is the ligand-binding domain’s job.
So the fault is at the ligand-binding domain.

71
Check q16

A researcher gives cells their ligand and measures them three ways.

A one-row table for cells given the ligand: ligand bound yes, second messenger raised no, response no
A one-row table for cells given the ligand: ligand bound yes, second messenger raised no, response no

Where is the fault?

  1. A. At the ligand-binding domain
    The ligand is found bound, so the ligand-binding domain works.
  2. B. ✓ Between ligand binding and the second messenger
  3. C. After the second messenger
    The second messenger is not raised.
    So the message is lost before the second messenger, not after it.

Why: The ligand is found bound, so reception works.
The second messenger is not raised, so the message is lost before it.
So the fault lies between ligand binding and the second messenger.

72
Check q17

A researcher gives cells their ligand and measures them three ways.

A one-row table for cells given the ligand: ligand bound yes, second messenger raised yes, response no
A one-row table for cells given the ligand: ligand bound yes, second messenger raised yes, response no

Where is the fault?

  1. A. At the ligand-binding domain
    The ligand is found bound, so the ligand-binding domain works.
  2. B. Between ligand binding and the second messenger
    The second messenger is raised.
    So every link up to the second messenger works.
  3. C. ✓ After the second messenger

Why: The ligand is found bound and the second messenger is raised, so every link up to the second messenger works.
The response does not follow.
So the fault lies after the second messenger.

73
Check q18

A researcher gives liver cells epinephrine and measures them three ways.

A one-row table for liver cells given epinephrine: epinephrine bound yes, cAMP raised no, glucose released no
A one-row table for liver cells given epinephrine: epinephrine bound yes, cAMP raised no, glucose released no

Where is the fault?

  1. A. At the ligand-binding domain
    Epinephrine is found bound, so the ligand-binding domain works.
  2. B. ✓ Between binding and cAMP
  3. C. Downstream of cAMP
    cAMP is not raised.
    So the message is lost before cAMP, not after it.

Why: Epinephrine is found bound, so reception works.
cAMP is not raised, so the message is lost before cAMP.
So the fault lies between binding and cAMP.

74
Check q19

A researcher gives cells no ligand and measures them three ways.

A one-row table for cells given no ligand: ligand bound no, second messenger raised yes, response yes
A one-row table for cells given no ligand: ligand bound no, second messenger raised yes, response yes

Where is the always-active component?

  1. A. ✓ Before the second messenger
  2. B. After the second messenger
    The second messenger is raised with no ligand bound.
    So something before the second messenger is passing the message on by itself.
  3. C. Nowhere: the pathway is switched off
    The response is produced with no ligand.
    The pathway is broken on, not switched off.

Why: No ligand is bound, yet the second messenger is raised.
So a component before the second messenger passes the message on with no input.
That component is the always-active one.

75
Practice writing an answer

Here again is the table of three measurements: normal cells with the ligand, receptor-mutant cells with the ligand, and receptor-mutant cells given an always-active kinase with no ligand.

A table of three measurements, ligand bound, second messenger raised and response, for normal cells with the ligand, receptor-mutant cells with the ligand, and receptor-mutant cells given an always-active kinase with no ligand
A table of three measurements, ligand bound, second messenger raised and response, for normal cells with the ligand, receptor-mutant cells with the ligand, and receptor-mutant cells given an always-active kinase with no ligand

(a) Justify the claim that the receptor-mutant cells’ change acts at the receptor’s ligand-binding domain, using two of the table’s measurements. (1 pt)

Model answer In the receptor-mutant cells given the ligand, no ligand is found bound.
Binding is what the ligand-binding domain does, so that domain has failed and reception fails.
In the receptor-mutant cells given an always-active kinase, the response returns with no binding and no second messenger.
So the kinase and everything after it are intact.
So the change acts before the kinase, at the receptor, and the missing binding places it in the ligand-binding domain.
Rubric
  • Award 1 point for: no ligand bound shows reception fails (the ligand-binding domain), AND the always-active kinase restoring the response shows everything downstream of the receptor is intact.
76
Check q20

Here is a different pathway: ligand → receptor → kinase A → kinase B → response genes expressed, with the table of measurements shown.

A table for a pathway from ligand to receptor to kinase A to kinase B to gene expression: normal cells with the ligand show binding, kinase A phosphorylated and the response; mutant X cells with the ligand show binding but no kinase A phosphorylation and no response; mutant X cells given an always-active kinase B with no ligand show no binding, no kinase A phosphorylation, and the response
A table for a pathway from ligand to receptor to kinase A to kinase B to gene expression: normal cells with the ligand show binding, kinase A phosphorylated and the response; mutant X cells with the ligand show binding but no kinase A phosphorylation and no response; mutant X cells given an always-active kinase B with no ligand show no binding, no kinase A phosphorylation, and the response

Where does mutant X’s change act?

  1. A. Before binding, at the ligand-binding domain
    Mutant X binds its ligand as normal, so the ligand-binding domain works.
  2. B. ✓ Between binding and kinase A
  3. C. Between kinase A and kinase B
    Kinase A is never phosphorylated in mutant X.
    So the message is lost before A, not after it.
  4. D. After kinase B, at the response genes
    An always-active kinase B restores the response.
    So the genes and everything after B are intact.

Why: Mutant X binds the ligand, so reception works.
Kinase A is never phosphorylated in mutant X, so the message is lost before kinase A.
An always-active kinase B restores the response, so everything from kinase B on is intact.
So the fault lies between binding and kinase A.

77What the organism shows

78

Video: Watch: From one cell to the whole organism

Every cell of a tissue carries the same fault, so every cell responds the same way. A pathway broken on gives tissue whose cells divide with or without growth factor; a pathway broken off gives tissue whose cells never divide.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L10d.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L10d.mp4

79
Check q21

A mouse’s skin cells divide about once a day.

Which of these is part of the mouse’s phenotype?

  1. A. The base sequence of its growth-factor receptor gene
    A gene’s base sequence is the gene itself.
    The phenotype is what can be seen or measured on the organism and its cells.
  2. B. ✓ How often its skin cells divide

Why: The phenotype is everything about an organism and its cells that can be seen or measured.
How often the skin cells divide can be measured.
So it is part of the phenotype.

80

A tissue is made of many cells. When every cell of the tissue carries the same fault, every cell responds the same way.

81

So the fault shows on the whole organism, as part of its phenotype.

82

Here is a table of how a growth-factor pathway can stand, no fault, broken off or broken on, and what the organism’s tissue shows for each.

A three-row table: a pathway with no fault, where cells divide only when growth factor is present; a pathway broken off, where cells never divide even with growth factor; a pathway broken on, where cells divide all the time with or without growth factor
A three-row table: a pathway with no fault, where cells divide only when growth factor is present; a pathway broken off, where cells never divide even with growth factor; a pathway broken on, where cells divide all the time with or without growth factor
83

What you are expected to know Predict the organism’s phenotype from a fault in one component of a pathway.

84
Check q22

An organism’s cells all carry an always-active kinase B in the pathway of the table shown.

A table for a pathway from ligand to receptor to kinase A to kinase B to gene expression: normal cells with the ligand show binding, kinase A phosphorylated and the response; mutant X cells with the ligand show binding but no kinase A phosphorylation and no response; mutant X cells given an always-active kinase B with no ligand show no binding, no kinase A phosphorylation, and the response
A table for a pathway from ligand to receptor to kinase A to kinase B to gene expression: normal cells with the ligand show binding, kinase A phosphorylated and the response; mutant X cells with the ligand show binding but no kinase A phosphorylation and no response; mutant X cells given an always-active kinase B with no ligand show no binding, no kinase A phosphorylation, and the response

Predict the organism’s phenotype.

  1. A. ✓ The response genes are expressed in every cell all the time
  2. B. The response genes are expressed only when the ligand is present
    An always-active kinase B needs no input from upstream.
    It acts whether or not the ligand arrives.
  3. C. The response genes are never expressed in any cell
    The kinase is locked on, so its targets are always switched on.
  4. D. The response genes are expressed only in cells that also carry mutant X
    The always-active kinase works on its own in any cell.
    Mutant X only showed that kinase B sits downstream of the receptor.

Why: An always-active kinase B phosphorylates its targets all the time.
So the response genes are expressed in every cell, with or without the ligand.
The pathway is broken on, and that is the phenotype the organism shows.

85
Check q23

An animal’s cells all carry a receptor locked in its active shape in the pathway whose response is ‘grow and divide’.

Predict the animal’s phenotype.

  1. A. Tissues whose cells divide only when growth factor is present
    A locked receptor needs no growth factor.
    The receptor passes the message on all the time.
  2. B. ✓ Tissues whose cells divide whether or not growth factor is present
  3. C. Tissues whose cells never divide, even when growth factor is present
    A locked receptor drives the pathway all the time, so the cells divide all the time.
    The pathway is broken on, not off.
  4. D. Tissues whose cells divide only when growth factor is absent
    Growth factor changes nothing for a locked receptor.
    The cells divide either way.

Why: A locked receptor drives the pathway continuously in every cell.
So the tissues’ cells divide whether or not any growth factor arrives.
That is the phenotype the organism shows.

86

A pathway can be broken on as well as off. Chemicals from outside can do either: a look-alike switches a pathway on; a blocker that fits without triggering switches it off.

87

The data on what still happens locate the change, and the change predicts what the organism shows.

88

Back to the dish of tumor cells, washed free of every trace of growth factor. 79% of the cells divided in two days, because their receptor was locked in its active shape.

89

The blocker fitted the receptor and held it inactive. Division fell to 24%, the level of cells with no receptor at all.

90

A blocker at the receptor alters only what lies downstream of the receptor. So the drug placed the fault at the receptor.

91Mixed practice mixed practice

92
Check q24

A dish of cells sits in a medium with nothing added to it. In these cells, a pathway that ends in secretion stays switched on all the time.

Which single change to one of the pathway’s proteins could explain this?

  1. A. A receptor whose binding site no longer fits the ligand
    A receptor that cannot bind its ligand never starts the pathway, so nothing would be secreted.
    These cells are secreting.
  2. B. ✓ A component locked in its active shape
  3. C. A receptor that binds its ligand more tightly than normal
    A receptor that binds its ligand tightly still needs a ligand to bind.
    The medium holds none, so the receptor sits at rest.
  4. D. A phosphatase made in far larger amounts than normal
    A phosphatase removes the phosphates that kinases add.
    More phosphatase removes them faster.
    So the pathway switches off sooner; it does not switch on.

Why: The medium holds no ligand, yet the cells secrete the hormone.
So the pathway is broken on.
An always-active component passes the message on with no input from the step before it.
So everything downstream of it stays switched on, and the cells secrete.

93
Check q25

A chemical enters a cell and holds one kinase of a hormone’s pathway in its inactive shape. A researcher then adds the hormone.

Predict the effect on the pathway’s response.

  1. A. The response is switched on as normal
    The hormone binds, and the receptor changes shape.
    But the kinase is held in its inactive shape, so it phosphorylates nothing.
    Nothing downstream of the kinase happens.
  2. B. The response is larger than normal
    A kinase held in its inactive shape phosphorylates nothing.
    So nothing downstream of the kinase is switched on, and the response cannot grow.
  3. C. The response is switched on, but later than normal
    The chemical holds the kinase in its inactive shape for as long as the chemical is bound.
    The message stops at the kinase, however long the cell waits.
  4. D. ✓ The response is switched off

Why: A chemical that holds a relay protein inactive breaks the pathway from that protein on.
The hormone binds its receptor as normal, and the receptor changes shape.
But the held kinase never passes the message on.
So nothing downstream of the kinase is switched on: the response is off.

94
Check q26

A researcher adds a blocker that fits a receptor’s binding site without triggering it, at a fixed dose. Then the researcher raises the natural ligand’s concentration tenfold.

Predict what happens to the response as the ligand rises.

  1. A. It stays switched off
    A blocker binds and leaves like any ligand.
    With more natural ligand around, the site is more often filled by the ligand, and the ligand triggers the shape change.
  2. B. It falls further
    The blocker and the natural ligand compete for one binding site; they cannot both sit in it at once.
    So more ligand means fewer sites held by the blocker.
  3. C. ✓ It returns toward normal
  4. D. It becomes far larger than normal
    The ligand wins a larger share of the sites, not more sites than exist.
    At most every site holds ligand, and that is the normal response.

Why: The blocker and the ligand compete for the same binding site, just like a competitive inhibitor competes with a substrate.
Raise the ligand tenfold and the ligand fills more of the sites.
So the response returns toward normal.

95
Check q27

A chemical enters a fat cell and binds the inner part of the epinephrine receptor, holding it in its inactive shape. The pathway’s components act in the order receptor → G protein → cAMP-making enzyme → cAMP → kinase → stored fat broken down. A researcher adds epinephrine.

Which measurements change when the researcher adds the epinephrine?

  1. A. ✓ Epinephrine bound at the receptor only
  2. B. None of the measurements
    Binding happens at the binding site outside the cell; the chemical sits on the inner part.
    So epinephrine still binds, and only the passing-on fails.
  3. C. Epinephrine bound and cAMP raised, but no fat broken down
    The receptor’s inner part is the first link after binding.
    Held inactive, it never switches on the G protein, so no cAMP is made.
  4. D. All three: epinephrine bound, cAMP raised, fat broken down
    Binding is upstream of the held inner part and still happens.
    Everything downstream of it stops.

Why: The chemical holds the receptor’s inner part inactive.
Binding happens upstream, outside the cell, so epinephrine binds as normal.
Downstream, the inner part never switches on the G protein: no cAMP, no active kinase, no fat broken down.
So epinephrine bound at the receptor is the one measurement that changes.

96
Check q28

The graph shows the percentage of cells dividing in two days for normal cells and for a second mutant line, mutant Y, including mutant Y cells whose receptor gene has been removed.

A bar graph of the percentage of cells dividing in two days: normal cells 22% with no growth factor and 80% with it; mutant Y cells 78% with no growth factor, 80% with it, 79% with a receptor blocker, and 78% with the receptor gene removed
A bar graph of the percentage of cells dividing in two days: normal cells 22% with no growth factor and 80% with it; mutant Y cells 78% with no growth factor, 80% with it, 79% with a receptor blocker, and 78% with the receptor gene removed

Where is mutant Y’s change?

  1. A. In the growth factor itself
    The cells divide with no growth factor at all.
    So the growth factor is not what drives them.
  2. B. In the receptor’s ligand-binding domain
    A broken binding domain would leave the pathway off.
    Mutant Y’s pathway is switched on.
  3. C. In the receptor’s intracellular domain
    A changed inner domain could ignore a blocker in the binding site.
    But mutant Y cells with no receptor gene still divide at 78%: the receptor is not the driver.
  4. D. ✓ Downstream of the receptor

Why: Mutant Y divides at 78% with no growth factor, so its pathway is broken on.
The receptor blocker leaves division at 79%.
Mutant Y cells with the receptor gene removed still divide at 78%.
So the receptor is not the driver: the always-active component sits downstream of it.

97
Check q29

A researcher gives a look-alike of a hormone to cells that carry no receptor for the hormone.

Predict the cells’ response.

  1. A. A normal response
    A look-alike works by producing the receptor’s shape change.
    With no receptor there is nothing for the look-alike to act on.
  2. B. A stronger response than the hormone gives
    With no receptor there is no response at all, strong or weak.
  3. C. A response through gene expression only
    Every response, fast or slow, begins at a receptor, and these cells have none.
  4. D. ✓ No response

Why: A look-alike switches a pathway on by fitting the receptor’s binding site and producing its shape change.
Cells with no receptor give the look-alike nothing to bind.
So the cells show no response.

98
Check q30

Two drugs, drug P and drug Q, each stop glucose release from liver cells given epinephrine. With drug P, cAMP still rises as normal. With drug Q, cAMP stays at its resting level.

Where does each drug act?

  1. A. Drug P upstream of cAMP; drug Q downstream of cAMP
    With drug P, cAMP still rises, so drug P acts after cAMP, not before.
    With drug Q, cAMP never rises, so drug Q acts before cAMP.
  2. B. Both drugs upstream of cAMP
    Drug P leaves the cAMP rise intact.
    So drug P acts after cAMP, not before.
  3. C. ✓ Drug P downstream of cAMP; drug Q upstream of cAMP
  4. D. Both drugs downstream of cAMP
    Drug Q stops cAMP from rising.
    So drug Q acts before cAMP is made.

Why: With drug P, cAMP rises.
So the message reaches cAMP, and drug P blocks a step downstream: a kinase or an enzyme.
With drug Q, cAMP never rises.
So drug Q blocks a step upstream: the receptor, the G protein or the cAMP-making enzyme.

99
Check q31

A researcher gives liver cells with an always-active G protein a receptor blocker, and no epinephrine.

Predict the cells’ cAMP.

  1. A. The cAMP concentration stays at rest
    The blocked receptor is upstream of the always-active G protein, and the always-active G protein needs no input from it.
  2. B. The cAMP concentration rises above its usual high level
    The blocker sits on the receptor and does nothing to the G protein.
    The G protein is active on its own, as before.
  3. C. ✓ The cAMP concentration stays high

Why: The G protein is always-active, so it switches on the enzyme that makes cAMP without any message from the receptor.
The receptor sits upstream of the G protein.
Blocking the receptor changes nothing downstream of an always-active component.
So the enzyme stays on, and the cAMP concentration stays high.

100
Check q32

A bar graph’s caption says: ‘error bars represent ±2SE’.

What does each error bar show?

  1. A. ✓ The range the true mean is likely to lie in
  2. B. The largest and the smallest reading in that set of dishes
    The bar runs two standard errors either side of the mean.
    It shows where the true mean is likely to lie, not the spread of the readings.

Why: An error bar of ±2SE runs two standard errors above and below the mean.
So it shows the range the true mean is likely to lie in.

101
Check q33

A researcher tests whether a treatment changes a measured result.

What is the null hypothesis?

  1. A. ✓ The statement that the treatment makes no difference to the measured result
  2. B. The statement of what the researcher expects the treatment to do
    What the researcher expects is the prediction.
    The null hypothesis is the statement that the tested factor makes no difference.

Why: The null hypothesis is the statement that the tested factor makes no difference to the measured result.

102
Practice writing an answer

Researchers investigate a line of cells taken from a skin tumor. These cells divide with no growth factor present. They use the pathway model shown: growth factor → receptor → kinase → the cell grows and divides. They set up six dishes of each of five treatments. Treatment 1 is normal cells with no growth factor. Treatment 2 is normal cells with growth factor. Treatment 3 is mutant cells with no growth factor. Treatment 4 is mutant cells with a receptor blocker and no growth factor. Treatment 5 is cells that carry no receptor. After two days they count the percentage of cells dividing in each dish. The graph shows the means; the error bars show ±2SE.

The pathway model (growth factor, receptor, kinase, the cell grows and divides) beside a bar graph of the percentage of cells dividing in five treatments, with ±2SE error bars
The pathway model (growth factor, receptor, kinase, the cell grows and divides) beside a bar graph of the percentage of cells dividing in five treatments, with ±2SE error bars

(a) Identify the dependent variable in this experiment. (1 pt)

Model answer The dependent variable is the percentage of cells dividing after two days.
Rubric
  • Award 1 point for: the percentage of cells dividing (in two days) as the dependent variable.

Slip Naming the receptor blocker or the cell line as the dependent variable. Those are what the researchers change; the count of dividing cells is what they measure.

(b) State the null hypothesis for the effect of the receptor blocker on the mutant cells. (1 pt)

Model answer Null hypothesis: the receptor blocker makes no difference to the percentage of mutant cells dividing; mutant cells with the blocker and mutant cells with no blocker divide at the same rate.
Rubric
  • Award 1 point for: a null hypothesis stating that the blocker makes no difference to the percentage of mutant cells dividing.

Slip Writing the prediction (‘the blocker lowers division’) as the null hypothesis. The null hypothesis is the statement of no difference.

(c) Explain why the cells that carry no receptor are included. (1 pt)

Model answer The receptor-less cells are a control: they show how much division happens with no signal from this receptor at all.
If the blocker brings the mutant cells down to that level, the blocker has removed all the division the receptor was driving.
Rubric
  • Award 1 point for: the receptor-less cells give the division level with the receptor’s signal absent (a control), against which the blocked mutant cells can be compared.

Slip Saying the receptor-less cells ‘show the receptor is needed for division’. They give the baseline level that the other treatments are compared against.

(d) Evaluate the claim that the mutant cells’ change acts at the receptor rather than downstream of it, using the data. (1 pt)

Model answer The claim is supported.
Mutant cells divide at 71% with no growth factor.
With the receptor blocker, division falls to 20%.
Its error bar, 17% to 23%, overlaps the no-receptor bar, 16% to 22%.
So the blocker brings the mutant cells down to the receptor-less level.
A blocker at the receptor can only stop a signal that starts at the receptor.
A change downstream would leave the percentage of cells dividing high.
So the change acts at the receptor.
Rubric
  • Award 1 point for: the judgement (the claim is supported) AND the ground (the receptor blocker lowers the mutant cells’ division to the receptor-less level, the bars overlapping, which can happen only if the signal starts at the receptor; a change downstream would leave division high with the blocker).

Slip Citing the 71% with no growth factor alone. That shows the cells divide with no ligand, but it cannot say where the change is; the blocker result does.

Glossary

look-alike
A molecule shaped enough like a receptor’s ligand to fit its binding site and produce the same shape change. A look-alike switches the pathway on with no natural ligand present.
blocker
A molecule that fits a receptor’s binding site but produces no shape change. While it sits in the site the ligand cannot bind, so the pathway is switched off. A beta blocker fits the epinephrine receptor on heart muscle cells.

APBIO-U04-P43 Practice questions: Topic 4.3

Topic 4.3 · Signal Transduction Pathways · 9 MCQ · 2 FRQ · for APBIO-U04-T43

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. The first free-response question walks you through the reasoning one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Where a graph carries error bars, they represent ±2SE, the range the true mean is likely to lie in.

Video: Watch first: Topic 4.3 summary: pathways broken on and off

A pathway can be broken on as well as off: a look-alike switches it on, a blocker switches it off; what still happens locates the change; the change predicts what the organism shows.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-T43-summary.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-T43-summary.mp4

Q1 P43-q01

A silk moth caterpillar's silk glands begin making silk protein about a day after a hormone reaches them. Before the hormone arrives, the gland cells contain no silk protein.

Why does the silk protein appear a day after the hormone, rather than within seconds?

  1. A. The hormone needs about a day to diffuse from the blood into the gland cells
    A hormone reaches its receptor, at the cell surface or inside the cell, within seconds of arriving; the day is spent expressing the silk genes and building the protein.
  2. B. ✓ The cells have to express the silk genes and build the protein, which takes hours
  3. C. The receptor takes a day to change shape once the hormone binds
    A receptor changes shape the moment its ligand binds.
    The slow part is expressing the genes and building the silk protein.
  4. D. cAMP has to build up for a day before the silk is released
    cAMP rises within seconds and switches on kinases; it does not store up.
    Silk protein is made from its genes, and the cells held none before.

Why: The gland cells held no silk protein, so it had to be made.
The hormone bound its receptor and the relay carried the signal.
The last activated protein reached the DNA, so the silk genes were expressed.
Expressing a gene and building its protein takes hours to a day.

Q2 P43-q02

When prey touches a sea anemone's tentacle, a signal reaches the tentacle cells, and within three seconds they release sticky mucus they had stored in vesicles. A researcher adds a chemical that stops all gene expression to the tentacle cells an hour before prey touches them.

Predict the mucus release when the prey touches.

  1. A. No release
    The mucus was made and packed into vesicles long before the prey arrived.
    Stopping gene expression stops new protein being made; it does not touch the vesicles already waiting.
  2. B. Release only after several hours
    The release does not wait on any gene.
    The signal makes vesicles that already exist fuse with the membrane, which takes seconds, with the chemical present or not.
  3. C. ✓ Release within three seconds, as before
  4. D. A smaller release than before
    None of the mucus is made fresh at the touch.
    Three seconds is far too short to build a protein, so the whole release comes from stored vesicles.

Why: The mucus already exists, packed in vesicles.
The signal makes those vesicles fuse with the membrane and release their contents within seconds.
No gene has to be expressed for that.
So a chemical that stops gene expression leaves the release unchanged: a change in cell function.

Q3 P43-q03

In a rabbit, a hormone released only in cold weather binds receptors on fur cells, and a pathway with one kinase makes the cells produce black pigment, so a normal rabbit's fur darkens in winter. In a normal rabbit the kinase carries a phosphate, and is switched on, only while the hormone is bound. A researcher measures the kinase in the fur cells of a mutant rabbit kept away from the hormone: the kinase carries its phosphate in summer and in winter.

Which of the following best predicts the mutant rabbit's phenotype?

  1. A. ✓ Black fur all year round
  2. B. Black fur only in warm weather
    The kinase carries its phosphate in summer and in winter, so the pigment pathway is on in both seasons; nothing switches it to the other season.
  3. C. White fur all year round
    The kinase is locked in its active shape, so the pathway is broken on: the kinase passes the signal on all the time, and the cells make pigment constantly.
  4. D. Darker fur than normal, but only in cold weather
    The kinase is already fully active; a message from the receptor cannot make it more active, so the hormone changes nothing.

Why: Normally the kinase gains its phosphate only when the cold-weather hormone binds.
In the mutant the kinase carries its phosphate all year, with no hormone.
A kinase carrying its phosphate is switched on, so the signal passes down the pathway all year.
So the fur is black all year round.

Q4 P43-q04

In a growing plant stem, certain cells receive a hormone. Over two days each of these cells empties, and what remains is a hollow tube of cell wall that carries water up the stem. The cells around the tubes stay healthy. A researcher applies a chemical that enters these cells and blocks protein-cutting enzymes; with the chemical present the cells stay full, and no tube forms.

Which of the following describes how these cells are emptied?

  1. A. Water pressure bursts the cells from outside, and the flow of water washes their contents away
    A burst cell spills its contents onto its neighbors, and the neighbors here stayed healthy.
    Blocking protein-cutting enzymes inside the cells also stopped the emptying; water pressure needs no enzymes.
  2. B. Enzymes released by the neighboring cells digest the contents from outside, and the flow clears them
    The chemical that stopped the emptying acted inside the emptying cells, on their enzymes.
    If the neighbors’ enzymes did the work from outside, blocking enzymes inside would have changed nothing.
  3. C. The cells dry out as water flows past them, and the dried contents are carried away
    Drying needs no enzymes, yet blocking the protein-cutting enzymes inside the cells kept the cells full.
    The emptying is done by enzymes, from inside, after a signal.
  4. D. ✓ Enzymes inside each cell, switched on by the hormone, break down its contents and clear them away

Why: The hormone binds the cell’s receptors, and the cell switches on its own enzymes.
Those enzymes cut up the cell’s contents, leaving only its cell wall.
Blocking those enzymes stops the emptying, so the work is done inside.
Nothing leaks, so the neighbors stay healthy: apoptosis.

Q5 P43-q05

Four changes in a liver cell are described.

Which of the following changes is a mutation?

  1. A. The cell makes twice as much of a receptor after a hormone arrives
    Making more of a protein is a change in gene expression.
    The gene’s base sequence is unchanged, so this is not a mutation.
  2. B. ✓ One base in the cell's gene for a receptor is swapped for a different base
  3. C. A receptor changes shape while its ligand is bound to it
    A receptor changes shape and changes back; its gene’s base sequence is untouched.
  4. D. A kinase adds a phosphate group to a relay protein
    A phosphate added to a protein changes the protein, not the DNA.
    A mutation is a change in a gene’s base sequence.

Why: A mutation is a change in the base sequence of a gene’s DNA.
Swapping one base for another changes that sequence, so it is a mutation.
Making more protein, a shape change and a phosphate added leave the DNA sequence as it was.

Q6 P43-q06

Root cells of a mutant tomato plant carry a hormone receptor whose intracellular domain has been changed by a mutation; the ligand-binding domain is normal. A researcher adds the hormone and measures three things: whether hormone is found bound to the receptors, whether the relay protein inside the cell is phosphorylated, and whether the roots respond.

Which pattern of results do you predict?

  1. A. Hormone bound: no; relay protein phosphorylated: no; response: none
    The ligand-binding domain is normal, so the hormone fits the pocket and is found bound.
    It is the inner part that cannot act.
  2. B. ✓ Hormone bound: yes; relay protein phosphorylated: no; response: none
  3. C. Hormone bound: yes; relay protein phosphorylated: yes; response: none
    The receptor's changed intracellular domain cannot take up its active shape, so it never phosphorylates the relay protein.
    The message stops at the receptor.
  4. D. Hormone bound: yes; relay protein phosphorylated: yes; response: weaker than normal
    The intracellular domain cannot pass the message on at all, so the relay protein is never phosphorylated and there is no response, not a weaker one.

Why: The ligand-binding domain is normal, so the hormone binds and is found bound to the receptors.
The intracellular domain is changed, so the inner part cannot take up its active shape.
So the relay protein is never phosphorylated.
So nothing downstream is switched on, and the roots show no response.

Q7 P43-q07

The model below shows a pathway in a snail's gill cells: a hormone binds a receptor, the receptor activates protein A, protein A phosphorylates kinase B, kinase B phosphorylates kinase C, and kinase C opens a channel. In a mutant line, kinase B is missing. A researcher adds the hormone.

The pathway in the snail's gill cells from the hormone to the channel. The dashed box is the component missing in the mutant line.
The pathway in the snail's gill cells from the hormone to the channel. The dashed box is the component missing in the mutant line.

Which pattern of measurements do you predict in the mutant cells?

  1. A. ✓ Hormone bound: yes; protein A active: yes; kinase C phosphorylated: no; channel: closed
  2. B. Hormone bound: yes; protein A active: yes; kinase C phosphorylated: yes; channel: closed
    Kinase C is phosphorylated by kinase B, and kinase B is missing.
    So kinase C stays unphosphorylated, and the channel it would open stays closed.
  3. C. Hormone bound: yes; protein A active: no; kinase C phosphorylated: no; channel: closed
    Protein A is upstream of the missing kinase B.
    The receptor activates protein A as normal; a break stops only what lies downstream of it.
  4. D. Hormone bound: yes; protein A active: yes; kinase C phosphorylated: yes; channel: open
    Binding is only the start of the pathway.
    Everything downstream of the missing kinase B receives no message, so kinase C stays unphosphorylated and the channel stays closed.

Why: Every component upstream of a missing component is activated as normal.
The hormone binds, and the receptor activates protein A.
Protein A has no kinase B to phosphorylate.
Everything downstream of the missing kinase B stops.
So kinase C is never phosphorylated, and the channel stays closed.

Q8 P43-q08

In a cell of a lizard's salt gland, a signal passes along this pathway: hormone → receptor → G protein → cAMP-making enzyme → cAMP → kinase → salt secreted. A researcher measures cAMP in normal salt-gland cells and in mutant salt-gland cells, with no hormone present: normal cells 2 units, mutant cells 60 units. With the hormone present, normal cells reach 58 units.

Which of the following best predicts what the mutant salt-gland cells do?

  1. A. They secrete salt only while the hormone is present
    A component before cAMP is locked on.
    So the cAMP concentration stays high with or without the hormone.
    So the kinase stays active with or without the hormone.
  2. B. They secrete no salt
    This pathway is broken on, not off.
    cAMP is made all the time.
    So every component downstream of cAMP is switched on.
  3. C. ✓ They secrete salt continuously, with no hormone present
  4. D. They secrete more salt only when extra hormone is added
    Extra hormone acts at the receptor, upstream of an enzyme that is already at full activity.
    It changes nothing downstream.

Why: Normal cells: 2 units of cAMP without hormone, 58 with it.
Mutant cells: 60 units without hormone, the hormone-bound level.
So in the mutant, a step before cAMP is always on.
cAMP switches on the kinase, so salt is secreted.
So the mutants secrete salt continuously, hormone or not.

Q9 P43-q09

A chemical from a soil fungus stops cells from responding to a hormone whose signal passes along this pathway: hormone → receptor → G protein → cAMP-making enzyme → cAMP → kinase → channel opens. Researchers know the chemical holds one component in its inactive shape. That component is either the G protein or the kinase.

Which single measurement, made with the hormone and the chemical both present, tells the two possibilities apart?

  1. A. Whether the hormone is found bound to the receptor
    The receptor is upstream of both the G protein and the kinase, so the hormone binds whichever component is held off: this measurement reads the same in both cases.
  2. B. Whether the channel opens
    The channel is downstream of both the G protein and the kinase.
    It stays closed whichever component is held off, so this measurement reads the same in both cases.
  3. C. Whether the kinase is present in the cells
    The chemical holds a component in its inactive shape; it does not remove it.
    The kinase is present in both cases.
  4. D. ✓ Whether cAMP rises

Why: cAMP sits between the G protein and the kinase.
G protein held off: the enzyme never switches on, so cAMP stays at rest.
Kinase held off: the enzyme works, so cAMP rises.
Binding is yes and the channel closed in both cases.
So cAMP is what differs.

FRQ 1 P43-frq1 · Scientific Investigation scaffolded

In wheat, a plant hormone, hormone B, makes the cells of the stem lengthen. Researchers propose that the signal passes along this pathway: hormone B → receptor → kinase K → protein F → genes for cell-wall-loosening enzymes expressed → cell lengthens. They compare a normal wheat line with a dwarf line whose stems stay short. The researchers spray six pots of each line with hormone B daily for ten days and six pots with water, then measure stem length. Two further tests on stem cells: a binding test finds as much hormone B bound to dwarf receptors as to normal ones; a kinase test finds kinase K present and normal in the dwarf line but never phosphorylated after hormone B. The graph below shows the mean stem lengths; the error bars represent ±2SE. The researchers plan a fifth group: dwarf seedlings carrying an always-active kinase K, sprayed with water.

Mean stem length of wheat seedlings after ten days of daily spraying, six pots per group. Error bars represent ±2SE. Gridlines every 5 cm.
Mean stem length of wheat seedlings after ten days of daily spraying, six pots per group. Error bars represent ±2SE. Gridlines every 5 cm.

(a) Identify the dependent variable in this experiment. (1 pt)

Frame The dependent variable is …, measured in ….

Hint Which quantity did the researchers measure at the end of the ten days, and what unit is on the graph's axis?

Model answer The dependent variable is the stem length of the seedlings after ten days, in centimeters.
Rubric
  • Award 1 point for: stem length (cm) after ten days as the dependent variable.
  • Do not award the point for the wheat line or the spray (independent variables), or for 'growth' with no measurement named.

Slip Naming hormone B or the wheat line as the dependent variable. Those are what the researchers changed; stem length is what they measured.

(b) Explain why seedlings of each line were also sprayed with water. (1 pt)

Frame The water-sprayed seedlings are the … for each line. They show …, so ….

Hint Think about what the hormone-sprayed pots are being compared against. What would you need to know about each line before you could credit any extra growth to hormone B?

Model answer The water-sprayed pots are the control for each line.
They show how long each line's stems grow in ten days with no added hormone, so any extra length in the hormone-sprayed pots of that line can be credited to hormone B rather than to a difference the lines have anyway.
Rubric
  • Award 1 point for: the water-sprayed pots give each line's stem length with hormone B absent (a control), so the effect of the hormone can be measured against it within each line.
  • Accept: 'a baseline for each line' with what is compared against it. Do not award the point for 'it is the control' alone, or for 'to check that water does nothing'.

Slip Writing 'they are the control' and stopping. Say what the control lets you compare: each line's growth with hormone B absent against its growth with hormone B present.

(c) State the null hypothesis for the effect of hormone B on the dwarf line's stem length. (1 pt)

Frame Null hypothesis: hormone B makes … to the … of the dwarf line; dwarf seedlings sprayed with hormone B and with water will ….

Hint A null hypothesis is a statement of no difference. Which two groups are being compared here, and which measurement?

Model answer Null hypothesis: hormone B makes no difference to the stem length of the dwarf line; dwarf seedlings sprayed with hormone B and dwarf seedlings sprayed with water will have the same mean stem length after ten days.
Rubric
  • Award 1 point for: a statement of no difference in stem length between dwarf seedlings sprayed with hormone B and dwarf seedlings sprayed with water.
  • Accept: 'hormone B has no effect on the dwarf line's stem length'. Do not award the point for a prediction of a difference in either direction, or for a null hypothesis about the normal line.

Slip Writing the expected result ('the dwarf plants stay short') as the null hypothesis. The null hypothesis is the statement of no difference between the two treatments.

(d) Evaluate the claim that the dwarf line's fault lies in the receptor's intracellular domain, using both tests. (1 pt)

Frame The claim is …. The binding test shows …, so reception …. The kinase test shows …, so the receptor's inner part ….

Hint Each test checks one link of the pathway. Which link does the binding test check, and which link does kinase K's phosphorylation check?

Model answer The claim is supported.
The binding test shows as much hormone B bound to the dwarf line’s receptors as to the normal line’s.
So the ligand-binding domain works, and reception succeeds.
The kinase test shows kinase K present and normal, yet never phosphorylated.
So the receptor’s inner part fails to take up its active shape and pass the message to kinase K.
Binding works and the very next step fails, so the fault is in the receptor’s intracellular domain.
Rubric
  • Award 1 point for: the judgement (the claim is supported) AND the ground (equal binding shows the ligand-binding domain works and reception succeeds, and kinase K, though present and normal, is never phosphorylated, so the receptor's inner part never passes the message on).
  • Accept 'supported: binding is normal but the receptor never activates kinase K'. Do not award the point for the judgement alone, for the short stems alone (that shows the pathway failed, not where), or for placing the fault in kinase K, which the kinase test rules out.

Slip Giving the judgement without the ground, or citing only the short stems. Short stems show the pathway failed somewhere; the two tests place the failure between binding and kinase K.

(e) Predict the stem length of the planned fifth group compared with the four groups on the graph, and justify your prediction. (1 pt)

Frame The fifth group's stems will be about … cm, close to the … group, because kinase K sits … of the fault and ….

Hint Find kinase K's place in the pathway relative to the fault you located in part (d). Are the components after it intact in the dwarf line?

Model answer The fifth group's stems will be long, about 32 cm, close to the normal line sprayed with hormone B.
Kinase K sits downstream of the faulty receptor, and an always-active kinase K phosphorylates protein F on its own, so the cell-wall-loosening genes are expressed and the cells lengthen whether or not the receptor ever passes a message on.
Rubric
  • Award 1 point for: long stems, about the length of the normal line sprayed with hormone B (about 32 cm), because kinase K is downstream of the faulty receptor and an always-active K drives protein F and the genes with no input from the receptor.
  • Accept: 'stems as long as the normal line's with hormone'. Do not award the point for 'short, because no hormone was sprayed' or 'short, because the receptor is faulty'.

Slip Predicting short stems because the pots get only water. An always-active kinase needs no message from the receptor; it acts on protein F by itself.

FRQ 2 P43-frq2 · Scientific Investigation

Fat cells release a hormone, hormone L, into the blood. Hormone L binds receptors on certain brain cells, and a pathway in those cells reduces appetite, so the animal eats less. Researchers study two strains of mice. Every day for two weeks, the researchers inject six mice of each strain with hormone L and six with saline (salt water with no hormone), and record the food each mouse eats per day. Blood tests show hormone L at 5 ng/mL in strain 1 mice and at 40 ng/mL in strain 2 mice. The graph below shows the mean food eaten per day; the error bars represent ±2SE.

Mean food eaten per day by mice of the two strains injected daily with saline or with hormone L, six mice per group. Error bars represent ±2SE. Gridlines every 1 g.
Mean food eaten per day by mice of the two strains injected daily with saline or with hormone L, six mice per group. Error bars represent ±2SE. Gridlines every 1 g.

(a) Explain why six mice of each strain were injected with saline. (1 pt)

Model answer The saline mice are the control for each strain.
They show how much each strain eats with no added hormone, and they receive the same daily injection, so any fall in food eaten by the hormone-injected mice of that strain can be credited to hormone L rather than to the injection or to a difference between the strains.
Rubric
  • Award 1 point for: the saline mice give each strain's food intake with the added hormone absent (a control), so the effect of injected hormone L can be measured against it within each strain, with the injection itself matched.
  • Accept: 'a baseline for each strain, so the hormone's effect can be seen'. Do not award the point for 'it is the control' alone, or for 'to see if saline changes appetite'.

Slip Writing 'they are the control' and stopping. Say what the control lets you compare: each strain's intake with hormone L absent against its intake with hormone L added.

(b) Using the graph, describe the effect of the hormone L injections on the food eaten by each strain. (1 pt)

Model answer In strain 1, hormone L cut the food eaten from 4.0 g per day to 2.6 g per day, and the two ±2SE bars (3.7 to 4.3 and 2.3 to 2.9) do not overlap, so the fall is a real effect.
In strain 2, food eaten was 7.8 g per day with saline and 7.7 g per day with hormone L; the bars overlap, so the injections made no difference that the data can show.
Rubric
  • Award 1 point for: strain 1 ate less with hormone L (4.0 g falling to 2.6 g per day, bars not overlapping), while strain 2 ate the same with or without it (7.8 g against 7.7 g, bars overlapping, so no difference is shown).
  • Accept: values read to within 0.1 g. Do not award the point for a description that gives no direction for strain 1, or that reads the overlapping strain 2 bars as a real difference.

Slip Reading the strain 2 means as a small fall. The bars overlap, so the data show no difference; overlap is read before any difference in means is claimed.

(c) Support the claim that strain 2's fault lies in the pathway inside its brain cells rather than in a shortage of hormone L, using evidence from the description and the graph. (1 pt)

Model answer Strain 2 mice have 40 ng/mL of hormone L in their blood, eight times strain 1’s: no shortage.
Injecting still more hormone L changed their food intake by nothing the bars can show.
Hormone present in plenty with no response shows the message is lost in the target cells.
So the fault is at the receptor or a component downstream of it.
Rubric
  • Award 1 point for: the evidence (strain 2's blood carries 40 ng/mL of hormone L, eight times strain 1's 5 ng/mL, and injecting more changed food intake by nothing the bars can show) AND the reasoning (hormone present in plenty with no response means the message is lost in the target cells, at the receptor or downstream of it).
  • Accept 'the hormone concentration is high yet the cells do not respond, so the fault is in the cells'. Do not award the point for the large food intake alone, for the 40 ng/mL with no link to the claim, or for 'strain 2 lacks the hormone'.

Slip Arguing from the large food intake alone. That shows the appetite pathway is failing; it is the high blood level of hormone L, with no response to injections, that rules out a shortage of the hormone.

(d) Predict the phenotype of strain 2 mice compared with strain 1 mice, and explain how a change in one protein produces it. (1 pt)

Model answer Strain 2 mice will be heavier and fatter than strain 1 mice.
One changed protein in their brain cells, the receptor or a relay component, stops hormone L’s message.
So the pathway that reduces appetite never passes the signal on.
So the mice eat about twice as much each day, 7.8 g against 4.0 g.
Repeated at every meal, that molecular fault shows on the whole animal as extra body fat: the phenotype.
Rubric
  • Award 1 point for: strain 2 mice are heavier (fatter), because a changed receptor or relay protein in the brain cells stops hormone L's message, so the pathway that reduces appetite never passes the signal on, the mice eat about twice as much every day, and the extra food is stored as fat.
  • Accept: 'heavier, because the cells never receive the message to eat less'. Do not award the point for 'they eat more' with no link to the failed pathway, or for a prediction that they are lighter.

Slip Stopping at 'strain 2 eats more'. The point needs the chain: a changed protein blocks the message, the appetite pathway never passes the signal on, the mice eat more, and the whole animal is heavier.

APBIO-U04-T43 End-of-topic test: Signal Transduction Pathways

Topic 4.3 · Signal Transduction Pathways · 18 MCQ · 2 FRQ

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the two free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. Then open the scoring guide and mark your own work against it. Where a graph carries error bars, they represent ±2SE, the range the true mean is likely to lie in.

Q1 T43-q01

In autumn a hormone reaches the cells at the base of a maple leaf's stalk. Over the next three days these cells build cell-wall-digesting enzymes that were absent from them before, the cell walls between them weaken, and the leaf drops.

Which of the following describes how the hormone produced the new enzymes?

  1. A. The hormone entered the cells and was itself rebuilt into the enzymes
    A hormone is a ligand: it binds a receptor and is released unchanged.
    The enzymes were built by the cells from their own genes over the three days.
  2. B. The receptor released enzymes the cells had stored in vesicles
    The cells held none of these enzymes before the hormone arrived, so there was nothing stored to release.
    New enzymes have to be made from their genes.
  3. C. ✓ The pathway's last activated protein reached the DNA and the cells expressed the enzymes' genes
  4. D. Kinases added phosphates to enzymes the cells already held, switching them on
    A phosphate switches on a protein that already exists, within seconds.
    These enzymes were absent before, and building them from their genes is what took three days.

Why: The hormone bound its receptor, and the relay proteins inside each cell passed the signal on.
The last activated protein of the pathway reached the DNA.
So the cells expressed the genes for the cell-wall-digesting enzymes and built them.
Building enzymes from their genes takes days: hence the three days.

Q2 T43-q02

A researcher gives a fish one 10-minute dose of a hormone. Two hours later its liver cells contain a protein they lacked before, and the protein is still present two days later, long after the hormone has gone from the blood.

Which of the following best explains both the two-hour delay and the protein outlasting the hormone?

  1. A. ✓ Building a protein from its gene takes hours, and the protein made keeps working after the signal has gone
  2. B. The hormone needs two hours to cross the membrane, and it stays bound inside the cell for days
    A hormone reaches its receptor, at the cell surface or inside the cell, within seconds.
    The hormone leaves again just as quickly.
    What lasts is the protein the cells made.
  3. C. cAMP takes two hours to build up, and it stays high for days after the hormone has gone
    cAMP rises within seconds of binding and is broken down within a minute of the signal stopping.
    The delay and the persistence both belong to the protein that was made.
  4. D. The receptor changes shape slowly, and it stays in its changed shape for days
    A receptor changes shape the moment its ligand binds and returns to rest when the ligand leaves.
    Nothing about the receptor takes two hours or lasts two days.

Why: The liver cells had to express the protein’s gene and build the protein, which takes hours: hence the two-hour delay.
A protein keeps working once it exists, so it outlasted the hormone.
Both features belong to a new protein, not to the hormone, the receptor or a second messenger.

Q3 T43-q03

A sea urchin egg releases a small peptide into the water. Within two seconds of the peptide reaching a sperm cell, the sperm's flagellum beats faster; when the peptide is washed away, the beat returns to normal within a minute.

Which of the following describes this response?

  1. A. The sperm expressed the genes for new flagellum proteins within the two seconds
    Expressing a gene and building a protein takes minutes to hours.
    A response that starts in two seconds and reverses in a minute uses proteins the cell already has.
  2. B. The peptide crossed into the sperm and bound the flagellum's motor proteins directly
    A peptide stays outside the cell; the message crosses the membrane through the receptor.
    The flagellum’s proteins were switched on from inside by the receptor’s relay.
  3. C. The peptide was taken up and used as fuel for the faster beat, and the beat slowed when the fuel was used up
    A signal is not a fuel: the receptor holds the peptide, and the sperm’s own stores power the beat.
    The beat slowed because the signal was gone.
  4. D. ✓ Proteins already in the flagellum were switched on, and switched off again when the signal stopped

Why: Two seconds on, a minute off: far too fast for new protein.
The peptide stayed outside and bound its receptor.
The receptor’s relay switched on proteins already in the flagellum.
When the peptide was gone, the phosphates were removed and the beat slowed.
A change in cell function.

Q4 T43-q04

In a cell, a signal passes along this pathway: hormone → receptor → G protein → cAMP-making enzyme → cAMP → kinase → channel opens.

Which of the following components is upstream of the cAMP-making enzyme?

  1. A. cAMP
    The enzyme makes cAMP, so cAMP comes after the enzyme in the pathway: downstream.
  2. B. ✓ The G protein
  3. C. The kinase
    cAMP switches the kinase on, so the kinase lies after the enzyme: downstream.
  4. D. The channel
    The channel opens last of all, so it is the most downstream component.

Why: The message passes one way, from receptor to response.
A component before a chosen step is upstream; one after it is downstream.
The G protein switches the enzyme on, so the G protein is upstream of the enzyme.
cAMP, the kinase and the channel come after it: downstream.

Q5 T43-q05

A hormone in the stem of a pea seedling makes the stem cells express genes for enzymes that loosen their cell walls, so each cell lengthens. Over a week the seedling grows 10 cm taller.

Which of the following is the change in the plant's phenotype?

  1. A. ✓ The seedling growing 10 cm taller
  2. B. The hormone binding its receptors in the stem cells
    Binding is the start of the pathway inside each cell, and it can be seen only with molecular tools.
    The phenotype is what shows on the whole plant.
  3. C. The base sequence of the cell-wall-loosening genes in the stem cells
    The base sequence of a gene is the instruction the cell reads, and the hormone did not change it.
    The phenotype is what shows on the plant: its height.
  4. D. The relay proteins in each cell changing shape
    A relay protein's shape change is a molecular event in one cell.
    The phenotype is the plant's set of observable features, and the height is one of them.

Why: An organism's phenotype is its set of observable features, what you can see or measure on the whole organism.
The signal changed gene expression in enough stem cells that the plant is 10 cm taller, and the height is the feature you can measure.

Q6 T43-q06

Growth hormone binds receptors on cells in a child's growing bones, and a pathway makes those cells divide, lengthening the bone. A child carries a mutation in the receptor's ligand-binding domain, so the binding site is misshapen and growth hormone floats past unbound. Blood tests show growth hormone at a normal level.

Which of the following predicts the child's height compared with a child whose receptors are normal, and gives the reason?

  1. A. Taller, because unbound growth hormone builds up in the blood
    Growth hormone acts only through its receptor.
    Hormone left in the blood does nothing to bone cells whose receptors it cannot bind, however much of it there is.
  2. B. The same height, because the hormone acts on bone cells directly
    A cell responds to a hormone only through a receptor that binds it.
    With no binding, the pathway that lengthens the bone never starts.
  3. C. ✓ Shorter, because the hormone's message never reaches the bone cells' pathway
  4. D. The same height, because more hormone makes up for the weaker fit
    The mutation removed the fit; it did not weaken it.
    More ligand cannot bind a site that is the wrong shape, and the blood level was normal in any case.

Why: The mutation changed the shape of the ligand-binding domain, so reception fails: growth hormone never binds, no signal passes down the pathway, and the bone cells divide less.
Repeated in every growing bone, that molecular fault shows on the whole child as a shorter height, the phenotype.

Q7 T43-q07

As a tadpole turns into a frog, a hormone reaches the cells of its tail, and over ten days the tail shrinks and disappears. Tail cells kept in a dish of nutrient medium, with no other tissue present, shrink and break up when a researcher adds the hormone and stay healthy with no hormone. In the tadpole, the tissue around the tail stays healthy throughout.

Which of the following describes how the tail cells were removed?

  1. A. Each cell beside the tail, on receiving the hormone, released enzymes that digested the tail cell next to it from outside
    Tail cells alone in a dish still broke up when the hormone was added.
    No neighbor was there to digest them, so the dismantling came from inside.
  2. B. ✓ Each tail cell, on receiving the hormone, cut up its own contents with its own enzymes, and nothing leaked out
  3. C. The tail cells swelled until their membranes gave way, and the blood carried their own spilled contents away
    A burst cell spills its contents and damages its neighbors, yet the tissue around the tail stayed healthy.
    The tail cells did not burst; each dismantled itself from inside.
  4. D. The hormone cut off the tail's blood supply, and the tail cells starved
    Tail cells in a dish, with no blood supply, stayed healthy until the hormone was added and broke up then.
    The hormone, not starvation, set the removal off.

Why: The hormone binds receptors in each tail cell, which activates its own enzymes, cuts up its contents and packages itself into pieces.
Tail cells alone in a dish do the same, so no other cell does the work.
Nothing leaks, so the tissue around the tail is unharmed.

Q8 T43-q08

In a chick embryo's wing, the cells between the future fingers dismantle themselves between day 7 and day 8, and the finger cells beside them are unharmed. A dismantling cell's own enzymes cut up its DNA and its proteins.

Which of the following explains why those enzymes do no harm to the finger cell beside the dismantling cell?

  1. A. The finger cells carry no receptor for the enzymes, so the enzymes cannot enter them
    The enzymes never leave the dismantling cell, so no receptor comes into it.
    The cell packages its contents into pieces that neighbors clear away.
  2. B. The finger cells destroy any enzyme that reaches them from a dismantling cell
    No enzyme reaches the finger cells.
    The dismantling cell keeps its enzymes inside and packages its contents into pieces that neighbors clear away.
  3. C. The dismantling cell bursts, and the blood carries its spilled contents away before they reach the finger cells
    A burst cell spills its contents onto its neighbors and harms them.
    A cell dismantling itself does not burst: it packages its contents into pieces.
  4. D. ✓ The dismantling cell packages its contents into pieces that neighboring cells clear away, so nothing leaks out

Why: The signaled cell activates its own enzymes and cuts up its DNA and proteins.
The cell packages itself into pieces.
Neighboring cells take the pieces in and clear them away.
So nothing leaks out, and the finger cells beside it are unharmed.

Q9 T43-q09

Two mutations in the gene for a G protein are studied in cells in which a signal passes along this pathway: hormone → receptor → G protein → cAMP-making enzyme → cAMP. Mutation 1 swaps one amino acid on the G protein's outer surface, far from where it contacts the receptor or the enzyme; these cells respond normally. Mutation 2 swaps one amino acid on the surface that contacts the enzyme; in these cells cAMP stays at its resting level.

Why do the two mutations have different effects?

  1. A. Every swapped amino acid destroys a protein's activity, so mutation 1 must have left the G protein's amino acids unchanged
    One swapped amino acid changes the shape only at that spot.
    A change where the G protein contacts nothing does nothing; a change where it contacts the enzyme stops it.
  2. B. ✓ Only mutation 2 changed the shape of a part the G protein uses to act on the enzyme
  3. C. Mutation 1 changed the hormone's shape, while mutation 2 changed the receptor's
    Both mutations are in the G protein's gene, so both change the G protein.
    The hormone and the receptor are built from other genes.
  4. D. Mutation 2 removed the G protein from the cell altogether, while mutation 1 made more of it
    One changed amino acid does not remove a protein.
    The G protein is present in both mutants; in mutant 2 the part that contacts the enzyme is misshapen.

Why: One changed amino acid changes the G protein’s shape at that spot.
Mutation 1 sits where the G protein contacts nothing, so it still works.
Mutation 2 sits where the G protein contacts the enzyme, so it can no longer switch the enzyme on.
So cAMP stays at rest.

Q10 T43-q10

A single base in the gene for a hormone receptor in a lizard's skin cells is changed. The cells' response to the hormone is weaker than normal. The cells make the normal number of receptor molecules.

Which chain of events explains how the changed base weakened the response?

  1. A. ✓ Changed DNA sequence → one changed amino acid → changed receptor shape → weaker fit for the hormone
  2. B. Changed DNA sequence → changed hormone shape → the hormone fits the receptor's binding site less well
    The mutation is in the receptor's gene, so the receptor is what changes.
    The hormone is built from a different gene and keeps its shape.
  3. C. Changed DNA sequence → the gene is read less often → the receptor is made in smaller numbers → weaker response
    The cells make the normal number of receptor molecules.
    So the fault is not in how many receptors are made.
    The changed base changes one amino acid of the receptor.
  4. D. Changed DNA sequence → the receptor is made as before → the receptor moves into the nucleus → the hormone cannot reach it
    One changed amino acid changes the receptor's shape; it does not move the receptor to a different place in the cell.
    A cell-surface receptor stays in the membrane.

Why: One changed base can change one amino acid of the receptor.
One changed amino acid changes the receptor’s fold at that spot.
The changed spot is in the ligand-binding domain, so the hormone fits less well and binds less often.
So the response is weaker.

Q11 T43-q11

Researchers give two lines of kidney cells the same hormone. Line 1 is normal; line 2 carries a mutation in the hormone's receptor. The researchers measure the amount of hormone bound to the receptors, and whether cAMP rises inside the cells after they add the hormone. The table below shows the results.

Hormone bound to the receptors, and whether cAMP rises inside the cells after the hormone is added, in the two lines of kidney cells given the same hormone.
Hormone bound to the receptors, and whether cAMP rises inside the cells after the hormone is added, in the two lines of kidney cells given the same hormone.

Which stage of the pathway failed in line 2's cells, and in which domain of the receptor is the mutation?

  1. A. Reception failed, so the mutation is in the ligand-binding domain
    The table shows as much hormone bound to line 2's receptors as to line 1's: 176 against 180 units per cell.
    Binding is reception, and it succeeded.
  2. B. Reception failed, so the mutation is in the intracellular domain
    Hormone was found bound to line 2’s receptors, so reception succeeded.
    The failure came after binding, in the receptor’s inner part: transduction, not reception.
  3. C. Transduction failed after binding succeeded, so the mutation is in the ligand-binding domain
    The ligand-binding domain did its job: 176 units bound, near the normal 180.
    The intracellular domain passes the message inward, and cAMP never rose, so that part failed.
  4. D. ✓ Transduction failed after binding succeeded, so the mutation is in the intracellular domain

Why: Hormone binds line 2’s receptors as well as line 1’s: 176 against 180 units.
So the ligand-binding domain works: reception succeeded.
cAMP never rises in line 2, so the receptor’s inner part cannot pass the message on.
So transduction never starts, and the mutation is in the intracellular domain.

Q12 T43-q12

A fish's gill cells carry a receptor for a salt-regulating hormone. Two mutant lines are found: in line P the mutation is in the receptor's ligand-binding domain, and in line Q it is in the intracellular domain. Cells of both lines show no response to the hormone.

Which single measurement tells the two lines apart?

  1. A. Whether the hormone crosses the plasma membrane into the cells
    A hormone that binds a cell-surface receptor stays outside the cell in both lines; the message, not the molecule, crosses.
    This measurement would read the same for P and Q.
  2. B. Whether the response proteins are present in the cells
    Every protein downstream of the receptor is intact in both lines; they are idle because no message reaches them.
    Both lines would show the response proteins present.
  3. C. ✓ Whether the hormone is found bound to the receptors
  4. D. Whether the cells respond when far more hormone is added
    More ligand cannot mend a receptor: line P’s site is the wrong shape for it, and line Q’s inner part cannot change shape once bound.
    Both lines stay unresponsive.

Why: A broken ligand-binding domain means the hormone never binds.
A broken intracellular domain means the hormone binds, but the inner part cannot change shape.
Downstream, both lines look the same: nothing happens.
So the one measurement that differs is whether hormone is found bound: no in P, yes in Q.

Q13 T43-q13

A mold's spores germinate when they detect sugar. The pathway is: sugar binds a receptor, which activates protein A; protein A phosphorylates kinase B; kinase B phosphorylates protein C; protein C reaches the DNA and the germination genes are expressed. A researcher gives sugar to normal spores and to two mutant strains and measures each step. The table below gives the results.

Measurements of the germination pathway in normal spores and two mutant strains, each given sugar.
Measurements of the germination pathway in normal spores and two mutant strains, each given sugar.

Where is the break in mutant strain 1's pathway?

  1. A. At the receptor's ligand-binding domain
    Protein A was activated, so the receptor bound the sugar and passed the message on.
    The receptor is upstream of the break and working.
  2. B. At the step from protein A to kinase B
    Kinase B was phosphorylated as normal, so protein A did its job.
    The message was lost after kinase B, not before it.
  3. C. ✓ At the step from kinase B to protein C
  4. D. At the germination genes themselves
    Protein C was never phosphorylated, so the message was lost before it reached the genes.
    The genes are downstream of a step that already failed; nothing shows them faulty.

Why: Every component upstream of a break is activated as normal and every component downstream stops.
The receptor, protein A and kinase B all did their jobs; protein C was never phosphorylated.
The break lies between kinase B and protein C: kinase B cannot phosphorylate C, or C cannot be phosphorylated.

Q14 T43-q14

Using the same table, a researcher makes one addition to mutant strain 2 to try to restore germination.

The same measurements of the germination pathway in normal spores and two mutant strains.
The same measurements of the germination pathway in normal spores and two mutant strains.

Which addition restores germination in mutant strain 2?

  1. A. ✓ An always-active kinase B
  2. B. An always-active protein A
    Protein A is upstream of the break; however active it is, its message has to pass to kinase B, and that step is the one that fails.
  3. C. Extra copies of the gene for protein C, so the cells hold more protein C
    More protein C changes nothing while nothing phosphorylates it.
    Protein C waits for a phosphate from kinase B, and kinase B is never switched on.
  4. D. One hundred times the sugar concentration
    More sugar only activates the receptor harder, at the top of the pathway; everything it triggers still stops at the break.

Why: Only an always-active component downstream of the break can restore the response.
The break is at the step from protein A to kinase B.
An always-active kinase B needs no input from protein A: it phosphorylates protein C, and the germination genes are expressed.

Q15 T43-q15

In a shrimp's skin, pigment cells make dark pigment when a hormone binds their receptors. A mutant line of pigment cells is studied. The graph below shows the pigment made in 24 hours by normal and mutant cells with and without the hormone, and by mutant cells whose receptor gene has been removed. No hormone is detectable in any culture's medium unless the researchers added it.

Pigment made in 24 hours by normal and mutant shrimp pigment cells, six cultures per treatment. Error bars represent ±2SE. Gridlines every 5 μg per million cells.
Pigment made in 24 hours by normal and mutant shrimp pigment cells, six cultures per treatment. Error bars represent ±2SE. Gridlines every 5 μg per million cells.

Which conclusion do the data support?

  1. A. The mutant cells' receptor cannot bind the hormone, so their pathway is switched off
    The mutant cells made as much pigment with no hormone as normal cells made with it.
    Their pathway is passing the signal on, not switched off.
  2. B. ✓ The mutant cells' receptor is locked in its active shape, so the signal passes down the pathway with no hormone
  3. C. A relay protein downstream of the receptor is stuck on and drives the pathway by itself
    Removing the receptor gene brought the mutant cells’ pigment down to the no-hormone level.
    A downstream component driving the pathway alone would have kept the pigment level high regardless.
  4. D. The mutant cells make pigment only when the hormone is present
    The mutant cells made 36 μg of pigment per million cells with no hormone at all, about the same as with it.
    The hormone made no difference to them.

Why: Mutant cells make full pigment with no hormone, so their pathway is broken on.
With the receptor gene removed, pigment falls to the normal resting level, so the signal that drives the pigment starts at the receptor: the mutant's receptor is locked in its active shape.

Q16 T43-q16

The body makes small peptides that bind receptors on certain brain cells; a pathway in those cells then lessens the sensation of pain. Morphine, a drug, fits the same binding site on the receptor and produces the same shape change that the peptides do.

Predict morphine's effect on these cells.

  1. A. The cells' response to the body's peptides is switched off while morphine is bound
    A molecule that fits the site and produces no shape change switches the pathway off.
    Morphine produces the same shape change as the peptide, so it switches the pathway on.
  2. B. Morphine is made into a second messenger inside the cells
    Second messengers are small molecules the pathway makes inside the cell.
    Morphine acts from outside, at the receptor's binding site.
  3. C. Morphine enters the cells and acts on the relay proteins directly
    Morphine binds the receptor's binding site, which faces outside the cell.
    The relay proteins are switched on by the receptor's shape change, as they are for the natural peptide.
  4. D. ✓ The cells respond as if the body's own peptide had bound, with no peptide present

Why: What matters at a receptor is the shape change.
A look-alike that fits the binding site and produces the ligand's shape change switches the pathway on exactly as the ligand would, so the cells respond with no natural peptide present.

Q17 T43-q17

Adenosine builds up in the brain through the day. It binds receptors on certain brain cells, and a pathway in those cells slows their activity, which is felt as drowsiness. Caffeine fits the adenosine receptor's binding site and produces no shape change in the receptor.

Which of the following explains why a person who drinks coffee feels less drowsy?

  1. A. ✓ Caffeine occupies binding sites, so fewer receptors bind adenosine and less signal passes down the pathway
  2. B. Caffeine activates the receptor more strongly than adenosine, speeding the cells up
    Caffeine produces no shape change, so it activates nothing.
    It works by sitting in the site so that adenosine cannot.
  3. C. Caffeine breaks adenosine down in the blood before it reaches the brain
    Caffeine is not an enzyme and does not change adenosine.
    It acts at the receptor, filling the binding site.
  4. D. Caffeine enters the cells and blocks a kinase in the slowing pathway
    Caffeine fits the receptor's binding site, on the outside of the cell.
    It blocks the pathway at reception, before any kinase is reached.

Why: A molecule that fits the binding site but produces no shape change blocks the natural ligand: while caffeine sits in the site, adenosine cannot bind, so less signal passes down the pathway that slows the cells, and the person feels less drowsy.

Q18 T43-q18

In a salamander's skin, gland cells secrete mucus when a hormone binds their receptors. The signal passes along this pathway: receptor → G protein → cAMP-making enzyme → cAMP → kinase → mucus secreted. The table below shows three measurements for normal cells given the hormone, for mutant V cells given the hormone, and for mutant V cells given a chemical that raises cAMP inside the cell directly, with no hormone.

Three measurements on the salamander's mucus-gland cells: whether hormone is found bound to the receptors, whether cAMP rises, and whether mucus is secreted.
Three measurements on the salamander's mucus-gland cells: whether hormone is found bound to the receptors, whether cAMP rises, and whether mucus is secreted.

Where does mutant V's change act?

  1. A. At the receptor's ligand-binding domain
    Mutant V cells bind the hormone as normal, so the ligand-binding domain works and reception succeeded.
  2. B. ✓ Between hormone binding and the making of cAMP
  3. C. Between cAMP and the kinase it switches on
    Raising cAMP directly restored secretion in mutant V cells, so the kinase and everything after it respond to cAMP as normal.
    The message was lost before cAMP.
  4. D. At the proteins that secrete the mucus
    Raising cAMP directly restored secretion in mutant V cells, so the proteins that secrete the mucus are intact.

Why: Mutant V cells bind the hormone, so reception works.
Given the hormone, mutant V cells make no cAMP, so the message is lost between binding and cAMP.
Raising cAMP directly restores secretion, so everything after cAMP is intact.
So the change acts between binding and the making of cAMP.

FRQ 1 T43-frq1 · Scientific Investigation

When skin is damaged, nearby cells release histamine. Histamine binds a receptor on the cells that line small blood vessels, and a pathway of relay proteins inside those cells, one of them protein K, loosens the junctions between the cells within minutes, so fluid leaks out of the vessel and the skin swells. Researchers grow sheets of vessel-lining cells and measure the volume of dye, in microliters, that leaks through each sheet in 10 minutes. There are six sheets in each of five treatments: (1) no histamine; (2) histamine; (3) histamine and drug X, a molecule that fits the histamine receptor's binding site and produces no shape change in the receptor; (4) cells carrying an always-active protein K, with no histamine; (5) cells carrying an always-active protein K, with drug X and no histamine. The graph below shows the mean volume leaked for each treatment; the error bars represent ±2SE.

Mean volume of dye leaking through a sheet of vessel-lining cells in 10 minutes, six sheets per treatment. Error bars represent ±2SE. Gridlines every 5 μL.
Mean volume of dye leaking through a sheet of vessel-lining cells in 10 minutes, six sheets per treatment. Error bars represent ±2SE. Gridlines every 5 μL.

(a) Identify the dependent variable in this experiment. (1 pt)

Model answer The dependent variable is the volume of dye that leaks through each sheet of cells in 10 minutes, in microliters.
Rubric
  • Award 1 point for: the volume of dye leaking through the sheet in 10 minutes (μL) as the dependent variable.
  • Accept: the leakiness of the cell sheet, measured as the dye leaked. Do not award the point for the treatment, the presence of histamine or of drug X (independent variables), or for 'swelling', which is not measured here.

Slip Naming histamine or drug X as the dependent variable. Those are what the researchers chose to add; the leaked volume is what they measured.

(b) Explain why treatment 1, cells given no histamine, is included. (1 pt)

Model answer Treatment 1 is the control.
It shows how much dye leaks through a sheet with the tested factor, histamine, absent, so the extra leak in treatment 2 can be credited to histamine and its pathway rather than to leakiness the sheets have anyway.
Rubric
  • Award 1 point for: treatment 1 gives the leak with histamine absent (a control), so the leak in treatment 2 can be compared against it and credited to histamine.
  • Accept: 'it is the baseline for comparison' with what is being compared. Do not award the point for 'it is the control' alone, or for 'it shows the cells are alive'.

Slip Writing 'it is the control' and stopping. The point is earned by saying what the control lets you compare: the leak with histamine absent against the leak with it present.

(c) Evaluate the claim that drug X acts at the receptor rather than downstream of protein K, using the data and the error bars. (1 pt)

Model answer The claim is supported.
With histamine, drug X cuts the leak from 38 μL to 6 μL, level with treatment 1’s 4 μL.
So drug X stops the message.
With an always-active protein K, drug X changes nothing: treatments 4 and 5 give 36 and 35 μL, and their bars overlap.
A drug acting downstream of protein K would have lowered treatment 5’s leak too.
So drug X acts upstream of protein K, at the receptor.
Rubric
  • Award 1 point for: the judgement (the claim is supported) AND the ground for it (treatments 4 and 5, cells with an always-active protein K without and with drug X, give the same leak, their ±2SE bars overlapping, so drug X changes nothing downstream of protein K and must act upstream, at the receptor; treatment 3 falling to treatment 1's level fits).
  • Accept a judgement grounded on treatments 4 and 5 with the overlap stated. Do not award the point for the judgement alone, for treatment 3 against treatment 2 alone (that shows the drug works, not where), or for a comparison of means with no reference to the error bars.

Slip Giving the judgement with no ground, or citing treatments 2 and 3 only. Only treatments 4 and 5, where protein K is locked on, show where the drug acts.

(d) A person takes drug X before being stung by a bee. Predict the swelling at the sting compared with a person who took no drug, and justify your prediction using the pathway. (1 pt)

Model answer The sting swells less.
Drug X sits in the binding sites of the histamine receptors on the vessel-lining cells, so the histamine released by the sting binds fewer receptors, less signal passes down the pathway that loosens the junctions, less fluid leaks out of the vessels, and the swelling, the feature you can see on the skin, is smaller.
Rubric
  • Award 1 point for: less swelling, because drug X occupies the receptors' binding sites so histamine binds fewer of them and less signal passes down the pathway that loosens the junctions (and lets fluid leak).
  • Accept: 'little or no swelling'. Do not award the point for 'less swelling' with no link to the receptor and the pathway, or for 'more swelling because histamine builds up'.

Slip Stopping at 'drug X blocks the receptor'. The point needs the chain to the visible feature: fewer receptors bound, less signal passes down the junction-loosening pathway, less fluid leaks, less swelling.

FRQ 2 T43-frq2 · Analyze Model

A barley seed stores starch. When the seed takes up water, its embryo releases a plant hormone, hormone B, which reaches the cells of the seed's outer layer. The model below shows how these cells respond: hormone B binds a receptor; the receptor activates kinase 1; kinase 1 phosphorylates kinase 2; kinase 2 phosphorylates protein F; protein F reaches the DNA and the cells express the gene for amylase, an enzyme that digests the stored starch into sugar for the growing embryo (word equation: starch → sugar, by amylase). In mutant strain M, the component marked in the model is absent: strain M seeds make no kinase 2.

The proposed pathway in the barley seed's outer-layer cells, from hormone B to starch digestion. The dashed box is the component missing in mutant strain M.
The proposed pathway in the barley seed's outer-layer cells, from hormone B to starch digestion. The dashed box is the component missing in mutant strain M.

(a) Using the model, describe the response of the outer-layer cells to hormone B, and identify the kind of response (a change in gene expression or a change in cell function). (1 pt)

Model answer The cells start making amylase, an enzyme they were not making before: protein F reaches the DNA and the amylase gene is expressed, so new enzyme is built and the stored starch is digested into sugar.
This is a change in gene expression, which is why it takes hours rather than seconds.
Rubric
  • Award 1 point for: the cells express the amylase gene and make new amylase, which digests the starch: a change in gene expression.
  • Accept: 'the cells switch on the amylase gene'. Do not award the point for 'the cells make sugar' with no mention of expressing the gene or making the enzyme, or for 'a change in cell function'.

Slip Calling the response a change in cell function. Existing proteins are switched on in seconds; here a new protein, amylase, has to be made from its gene.

(b) A researcher gives strain M seeds hormone B. Predict the amount of amylase these seeds make, and identify one component of the model that is still activated in them. (1 pt)

Model answer Strain M seeds make little or no amylase.
Hormone B still binds the receptor, because the receptor is upstream of the missing kinase 2.
The receptor still activates kinase 1, because kinase 1 is upstream of the missing kinase 2 too.
Kinase 1 has no kinase 2 to phosphorylate.
So protein F is never phosphorylated.
So the amylase gene is never expressed.
Rubric
  • Award 1 point for: little or no amylase, AND the receptor (or kinase 1) still activated, because it is upstream of the missing kinase 2.
  • Accept: either the receptor or kinase 1 as the component still activated. Do not award the point for 'kinase 2 is inactive' as the effect (the question asks what happens downstream), or for a prediction of normal amylase.

Slip Describing the missing kinase rather than the effect: 'kinase 2 does nothing'. The point is the measured outcome downstream (no amylase) and the upstream component that still works.

(c) A researcher gives strain M seeds an always-active protein F and no hormone. Make a claim about the amount of amylase these seeds make compared with normal seeds given hormone B, and support your claim using the model. (2 pt)

Model answer Strain M seeds given an always-active protein F make about as much amylase as normal seeds given hormone B.
On the model, protein F sits downstream of kinase 2.
Strain M lacks kinase 2, so normally protein F is never phosphorylated.
An always-active protein F needs no phosphate from kinase 2.
So protein F reaches the DNA on its own.
So the amylase gene is expressed and amylase is made, with no hormone B and no kinase 2.
Rubric
  • Award 1 point for the claim: about the same amount of amylase as normal seeds given hormone B (accept 'amylase is made even with no hormone'). Do not award the point for 'no amylase'.
  • Award 1 point for the support: the evidence from the model (protein F sits downstream of the missing kinase 2, and an always-active protein F needs no phosphate from kinase 2) AND the reasoning (so protein F reaches the DNA on its own, so the amylase gene is expressed with or without hormone B or kinase 2). Do not award the point for naming protein F as always-active with no link to the amylase gene being expressed, or for reasoning that never refers to the model.

Slip Claiming no amylase because no hormone was given, or reasoning without the model. The hormone sits above protein F in the pathway, and an always-active F acts without it; name where protein F sits relative to the missing kinase 2, then link that to the amylase gene.

(d) A chemical fits hormone B's binding site on the receptor and produces no shape change in the receptor. Predict the effect on starch digestion in normal seeds given both hormone B and the chemical, and explain how the chemical produces this effect. (1 pt)

Model answer Starch digestion falls: the seeds make little or no amylase.
The chemical sits in the receptor's binding site, so hormone B cannot bind; with no shape change in the receptor, kinase 1 is never activated, the message never reaches protein F, and the amylase gene stays off.
Rubric
  • Award 1 point for: less (or no) starch digestion, because the chemical occupies the binding site so hormone B binds fewer receptors and, producing no shape change, starts no transduction, so the amylase gene is not expressed.
  • Accept: 'the chemical blocks the receptor so the pathway is switched off' with the binding-site reasoning. Do not award the point for 'the chemical switches the pathway on' (that needs the shape change) or for 'the chemical breaks down the hormone'.

Slip Treating the chemical as a look-alike that switches the pathway on. A molecule that fits the site but produces no shape change blocks the natural ligand; only one that produces the shape change switches the pathway on.

APBIO-U04-L11 A set point, a stimulus, a response

Topic 4.4 · Feedback · 56 steps

A graph of blood glucose against time after a glass of orange juice: the line rises from the dashed set point at 90 mg/dL to a peak and comes back down to it
A graph of blood glucose against time after a glass of orange juice: the line rises from the dashed set point at 90 mg/dL to a peak and comes back down to it

Here is a graph of the glucose in one person’s blood over the two hours after a large glass of orange juice.

Fifteen minutes after the drink, the glucose has risen from about 90 to about 140 milligrams per deciliter. Two hours later it is back near 90, and the person did nothing to bring it down.

What pushed the glucose up? What brought it back? And what is special about 90?

Unit 4 · Cell Communication and Cell Cycle

1The set point

2
Check q1

A person walks from a warm room out into snow. Their core temperature stays near 37 °C.

What does homeostasis mean?

  1. A. ✓ Keeping the conditions inside the body steady while the conditions outside change
  2. B. Letting the conditions inside the body follow the conditions outside
    A body whose inside followed the outside would cool in snow and heat in sun.
    Homeostasis is the opposite: the inside is held steady.

Why: Homeostasis is keeping the conditions inside the body steady while the conditions outside change.

3

How do you describe a quantity the body holds steady? You need three words.

4

The first word names the value the quantity is held near. The second names the change that pushed the quantity away from that value.

5

The third names what the body did about that push. Start with the first word.

6

Between meals, the glucose in your blood sits near 90 milligrams per deciliter, written 90 mg/dL. The dashed line on the graph marks that value.

Blood glucose in mg/dL against time in minutes after the drink: the line starts at the dashed set point of 90, peaks at 140 after 15 minutes and returns to about 90 by 120 minutes; gridlines every 20 mg/dL
Blood glucose in mg/dL against time in minutes after the drink: the line starts at the dashed set point of 90, peaks at 140 after 15 minutes and returns to about 90 by 120 minutes; gridlines every 20 mg/dL
7

When a regulated quantity is held near one value, we call that value its . For blood glucose the set point is about 90 mg/dL.

8

The line never lies flat on the set point. The glucose keeps drifting a few mg/dL away from 90, and the body keeps bringing the glucose back.

9

So the glucose wobbles around its set point rather than sitting on it.

10

What you are expected to know Identify the set point in a described case: the value the regulated quantity is held near.

11

Video: Watch: The set point

Blood glucose between meals drifts a few mg/dL either side of 90 and is brought back each time. The value a regulated quantity is held near is its set point; the graph wobbles around it and never lies flat on it.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11a.mp4

12
Check q2

Between meals, a person’s blood glucose is measured every hour: 88, 93, 87, 91 and 90 mg/dL.

Which of the following is this person’s set point for glucose?

  1. A. 93 mg/dL
    The set point is the value the glucose is held near, not the highest value it reaches.
    Every reading is within a few mg/dL of 90.
  2. B. ✓ About 90 mg/dL
  3. C. 87 mg/dL
    The set point is the value the glucose is held near, not the lowest value it reaches.
    Every reading is within a few mg/dL of 90.

Why: Every reading is within a few mg/dL of 90 mg/dL.
The glucose is being held near 90 mg/dL.
The value a regulated quantity is held near is its set point.

13
Check q3

Between meals, a person’s blood glucose is measured every hour: 88, 93, 87, 91 and 90 mg/dL. A student says: “These readings show that homeostasis has failed, because the five readings differ.”

Is the student correct?

  1. A. Yes — homeostasis holds a quantity at exactly one value
    Homeostasis holds a quantity near its set point, never exactly on it.
    The glucose drifted a few mg/dL from its set point and was brought back each time: homeostasis working.
  2. B. ✓ No — the glucose is being held near its set point

Why: The glucose set point is about 90 mg/dL, and each reading is within a few mg/dL of it.
So the glucose drifted and was brought back each time.
Homeostasis holds a quantity near its set point, never exactly on it.
So the readings show homeostasis working.

14
Practice writing an answer

Between meals, a person’s blood glucose is measured every hour: 88, 93, 87, 91 and 90 mg/dL. The glucose is being held near its set point.

(a) Explain why readings that differ from one another by a few mg/dL show homeostasis working. (1 pt)

Model answer The set point of the glucose is about 90 mg/dL.
Homeostasis holds the glucose near the set point, never exactly on it.
Each reading is a few mg/dL away from the set point.
So each time the glucose drifted away from the set point, the body brought the glucose back.
So the readings scatter closely around the set point.
So readings that differ by a few mg/dL are homeostasis working.
Rubric
  • Award 1 point for: homeostasis holds the glucose near its set point, not exactly on it, so small drifts that are brought back each time (readings a few mg/dL either side of the set point) are the loop working.

15Quick quiz: set point mixed practice

16
Check q4

A quantity in the body is regulated.

What is the set point of that quantity?

  1. A. The change that moves the quantity away from its usual value
    A change that moves the quantity away is not the set point.
    The set point is the value the quantity is held near.
  2. B. What the body does about a change in the quantity
    What the body does is not a value.
    The set point is the value the quantity is held near.
  3. C. ✓ The value the quantity is held near

Why: The set point is the value a regulated quantity is held near, such as about 90 mg/dL for blood glucose.

17
Practice writing an answer

Blood glucose, core temperature and many other quantities in the body are regulated.

(a) State what the set point of a regulated quantity is. (1 pt)

Model answer The set point is the value the regulated quantity is held near.
Rubric
  • Award 1 point for: the value the quantity is held near (or kept close to).

18The stimulus

19

Now look at the first fifteen minutes of the graph. The juice pushed the glucose up from 90 mg/dL to 140 mg/dL.

The same graph with the rise labelled as the stimulus: glucose rises to 140 mg/dL
The same graph with the rise labelled as the stimulus: glucose rises to 140 mg/dL
20

When a change moves a regulated quantity away from its set point, we call the change a . The rise to 140 mg/dL is the stimulus here.

21

A fall away from the set point is a stimulus just as a rise is.

22

What you are expected to know Identify the stimulus in a described case: the change that moves the quantity away from its set point.

23

Video: Watch: The stimulus

The juice pushes the glucose from 90 to 140 mg/dL. A change that moves a regulated quantity away from its set point is a stimulus; a fall away from the set point is a stimulus just as a rise is.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11b.mp4

24
Check q5

After a bowl of rice, a person’s blood glucose rises from 85 mg/dL to 130 mg/dL in twenty minutes. By two hours it has fallen back to 85 mg/dL.

Which of the following is the stimulus?

  1. A. ✓ The rise from 85 to 130 mg/dL
  2. B. The 85 mg/dL the glucose returns to
    85 mg/dL is the value the glucose is held near: the set point.
    The stimulus is the change that moved the glucose away from that value.
  3. C. The fall from 130 back to 85 mg/dL
    The fall brings the glucose back toward its set point.
    The stimulus is the change that moved the glucose away from the set point: the rise.

Why: This person’s glucose is held near 85 mg/dL.
The rise to 130 mg/dL moved the glucose away from that value.
A change that moves the quantity away from its set point is the stimulus.

25Quick quiz: stimulus mixed practice

26
Check q6

A quantity in the body is regulated.

What is a stimulus?

  1. A. ✓ A change that moves the quantity away from its set point
  2. B. The value the quantity is held near
    The value the quantity is held near is the set point.
    A stimulus is a change that moves the quantity away from that value.
  3. C. What the body does about a change in the quantity
    What the body does about a change is not the change itself.
    A stimulus is a change that moves the quantity away from its set point.

Why: A stimulus is a change that moves a regulated quantity away from its set point, such as the rise in glucose after a sugary drink.

27
Practice writing an answer

Blood glucose, core temperature and many other quantities in the body are regulated.

(a) State what a stimulus is. (1 pt)

Model answer A stimulus is a change that moves a regulated quantity away from its set point.
Rubric
  • Award 1 point for: a change that moves the quantity away from its set point (or usual value).

28The response

29

Now look at what happened after the peak. Cells of the pancreas detected the rise and released insulin.

The same graph with the rise labelled as the stimulus and the fall labelled as the response: insulin released, glucose falls
The same graph with the rise labelled as the stimulus and the fall labelled as the response: insulin released, glucose falls
30

Over the next two hours the glucose fell back toward 90 mg/dL.

31

What the body does about a stimulus is called the . Here the response is the cells of the pancreas releasing insulin.

32

With these three words you can describe any quantity the body holds steady: its set point, the stimulus that moved it, and the response that brought it back.

33

What you are expected to know Identify the response in a described case: what the body does about the stimulus.

34

Video: Watch: The response

Cells of the pancreas detect the rise and release insulin, and the glucose falls back toward 90 mg/dL. What the body does about a stimulus is the response. Set point, stimulus, response: the three words for any quantity the body holds steady.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11c.mp4

35
Check q7

After a bowl of rice, a person’s blood glucose rises from 85 mg/dL to 130 mg/dL in twenty minutes. Cells of the pancreas release insulin, and by two hours the glucose has fallen back to 85 mg/dL.

Which of the following is the response?

  1. A. The person eating the rice
    Eating the rice is what caused the change.
    The response is what the body does about the change: releasing insulin.
  2. B. The rise to 130 mg/dL
    The rise is the change itself: the stimulus.
    The response is what the body does about the rise: releasing insulin.
  3. C. ✓ Cells of the pancreas releasing insulin

Why: The rise to 130 mg/dL moved the glucose away from its set point: the stimulus.
Cells of the pancreas released insulin because of the rise.
What the body does about a stimulus is the response.

36

Back to the large glass of orange juice. Fifteen minutes after the drink, the glucose in the blood had risen from about 90 mg/dL to about 140 mg/dL.

37

The 90 mg/dL is the set point: the value the glucose is held near. The rise to 140 mg/dL is the stimulus: the change that pushed the glucose away from it.

38

Cells of the pancreas releasing insulin is the response: what the body did about the rise. Two hours later the glucose was back near its set point of 90 mg/dL.

39Quick quiz: stimulus, set point or response? mixed practice

40
Check q8

A quantity in the body is regulated.

What is the response?

  1. A. The value the quantity is held near
    The value the quantity is held near is the set point.
    The response is what the body does about the change.
  2. B. ✓ What the body does about the change
  3. C. The change that moved the quantity away from its set point
    The change that moved the quantity away is the stimulus.
    The response is what the body does about that change.

Why: The response is what the body does about a stimulus, such as cells of the pancreas releasing insulin after blood glucose rises.

41
Practice writing an answer

A change moves a regulated quantity away from its set point, and the body acts.

(a) State what the response is. (1 pt)

Model answer The response is what the body does about the stimulus.
Rubric
  • Award 1 point for: what the body does about the stimulus (the change away from the set point).
42
Check q9

Ten minutes into a race, a runner’s core temperature has risen from 37 °C to 38.2 °C.

Which of the following is the rise to 38.2 °C?

  1. A. ✓ Stimulus
  2. B. Set point
    The set point is 37 °C.
    The rise to 38.2 °C moved her temperature away from it, and a move away from the set point is the stimulus.
  3. C. Response
    The runner’s body has not yet done anything about it.
    The rise is the change itself.
    A change that moves the quantity away from its set point is the stimulus.

Why: The runner’s core temperature is held near 37 °C.
The rise to 38.2 °C moved the temperature away from 37 °C.
A change that moves the quantity away from its set point is the stimulus.

43
Check q10

After a meal, cells of a dog’s pancreas release insulin.

Which of the following is the release of insulin?

  1. A. Stimulus
    The rise in glucose after the meal was the change.
    Releasing insulin is what the body does about that rise, and that is the response.
  2. B. Set point
    The set point is the value the glucose is held near.
    Releasing insulin is what the dog’s body does about the rise in glucose, and that is the response.
  3. C. ✓ Response

Why: The meal raised the dog’s glucose: that rise is the stimulus.
The pancreas cells release insulin because of the rise.
What the body does about a stimulus is the response.

44
Check q11

In cold water, a swimmer’s core temperature falls to 36.3 °C.

Which of the following is the fall to 36.3 °C?

  1. A. ✓ Stimulus
  2. B. Set point
    The set point is 37 °C.
    The fall to 36.3 °C moved his temperature away from it; a change that moves the quantity from its set point is the stimulus.
  3. C. Response
    The swimmer’s body has not yet done anything about the fall.
    A change that moves the quantity away from its set point is the stimulus.

Why: The swimmer’s core temperature is held near 37 °C.
The fall to 36.3 °C moved the temperature away from 37 °C.
A change that moves the quantity away from its set point is the stimulus.
A fall is a stimulus just as a rise is.

45
Check q12

A cold swimmer’s muscles shiver and release heat.

Which of the following is the shivering?

  1. A. Stimulus
    The change was the fall in his temperature.
    Shivering is what his body does about the fall.
    What the body does about a stimulus is the response.
  2. B. Set point
    The set point is the temperature his body holds near.
    Shivering is what his body does about the fall in temperature, and that is the response.
  3. C. ✓ Response

Why: The fall in the swimmer’s temperature is the stimulus.
The swimmer’s muscles shiver because of the fall.
What the body does about a stimulus is the response.

46
Check q13

A hen’s body holds its core temperature near 41 °C.

Which of the following is 41 °C?

  1. A. Stimulus
    Nothing has moved the hen’s temperature away from 41 °C.
    The value the body holds the temperature near is the set point.
  2. B. ✓ Set point
  3. C. Response
    41 °C is a value, not an action.
    The value the body holds the temperature near is the set point.

Why: 41 °C is the value the hen’s body holds its temperature near.
The value a regulated quantity is held near is the set point.
A hen’s set point is higher than a person’s 37 °C, but it is a set point all the same.

47
Check q14

Here is a record of a student’s core temperature and sweating as she exercises.

A table of a student's core temperature and sweating during exercise: 0 min 37.0 °C none; 2 min 37.2 none; 5 min 37.4 heavy; 10 min 37.2 heavy; 15 min 37.0 light
A table of a student's core temperature and sweating during exercise: 0 min 37.0 °C none; 2 min 37.2 none; 5 min 37.4 heavy; 10 min 37.2 heavy; 15 min 37.0 light

Which of the following is the stimulus?

  1. A. ✓ The rise to 37.4 °C
  2. B. The 37.0 °C at the start
    37.0 °C is the value her body holds the temperature near.
    The stimulus is the change that moved the temperature away from 37.0 °C: the rise to 37.4 °C.
  3. C. The heavy sweating
    The temperature rose first, and the heavy sweating began after the rise.
    The sweating is what her body did about the rise.
    The rise is the stimulus.

Why: Her temperature was 37.0 °C at the start: the set point.
By 5 minutes the temperature had risen to 37.4 °C.
That rise moved the temperature away from the set point, so the rise is the stimulus.
The heavy sweating that began at 5 minutes is the response.

48Mixed practice mixed practice

49
Check q15

After a run on a hot day, a dog’s core temperature has risen from 38.5 °C to 39.5 °C, and the dog pants.

Which of the following is the stimulus?

  1. A. The 38.5 °C the dog’s temperature is usually held near
    38.5 °C is the value the temperature is held near: the set point.
    The stimulus is the change that moved the temperature away from it.
  2. B. ✓ The rise to 39.5 °C
  3. C. The panting
    Panting is what the dog’s body does about the rise, not the change itself.
    The stimulus is the rise to 39.5 °C.

Why: The dog’s temperature is held near 38.5 °C.
The rise to 39.5 °C moved the temperature away from that value.
A change that moves the quantity away from its set point is the stimulus.

50
Check q16

After a run on a hot day, a dog’s core temperature has risen from 38.5 °C to 39.5 °C, and the dog pants.

Which of the following is the response?

  1. A. The 38.5 °C the dog’s temperature is usually held near
    38.5 °C is a value, not an action.
    The response is what the dog’s body does about the rise: panting.
  2. B. The rise to 39.5 °C
    The rise is the change itself: the stimulus.
    The response is what the dog’s body does about the rise: panting.
  3. C. ✓ The panting

Why: The rise to 39.5 °C is the stimulus.
The dog pants because of the rise.
What the body does about a stimulus is the response.

51
Check q17

In a warm room, a person’s core temperature has stayed near 37 °C for the whole afternoon.

Which of the following is 37 °C?

  1. A. Stimulus
    Nothing has moved the temperature away from 37 °C, so there is no change here.
    37 °C is the value the temperature is held near: the set point.
  2. B. ✓ Set point
  3. C. Response
    37 °C is a value, not something the body does.
    The value the temperature is held near is the set point.

Why: 37 °C is the value the person’s temperature is held near.
The value a regulated quantity is held near is the set point.

52
Check q18

During a long run, a person’s blood glucose falls from 90 mg/dL to 70 mg/dL, and cells of the pancreas release a hormone.

Which of the following is the stimulus?

  1. A. The release of the hormone
    Releasing the hormone is what the body does about the fall: the response.
    The stimulus is the fall itself.
  2. B. The 90 mg/dL the glucose is usually held near
    90 mg/dL is the value the glucose is held near: the set point.
    The stimulus is the change that moved the glucose away from it.
  3. C. ✓ The fall to 70 mg/dL

Why: The glucose is held near 90 mg/dL.
The fall to 70 mg/dL moved the glucose away from that value.
A change that moves the quantity away from its set point is the stimulus, whether the change is a fall or a rise.

53
Check q19

A cow eats a large meal of grain, and its blood glucose rises from 70 mg/dL to 110 mg/dL.

Which of the following is the rise to 110 mg/dL?

  1. A. ✓ Stimulus
  2. B. Set point
    The set point is the value the glucose is held near: about 70 mg/dL here.
    The rise moved the glucose away from that value: the stimulus.
  3. C. Response
    The cow’s body has not yet done anything about the rise.
    The rise is the change itself: the stimulus.

Why: The cow’s glucose is held near 70 mg/dL.
The rise to 110 mg/dL moved the glucose away from that value.
A change that moves the quantity away from its set point is the stimulus.

54
Check q20

A goldfish’s blood glucose is measured every hour: 62, 58, 60, 61 and 59 mg/dL.

Which of the following is the goldfish’s set point for glucose?

  1. A. 62 mg/dL
    The set point is the value the glucose is held near, not the highest value it reaches.
    Every reading is within a few mg/dL of 60.
  2. B. 58 mg/dL
    The set point is the value the glucose is held near, not the lowest value it reaches.
    Every reading is within a few mg/dL of 60.
  3. C. ✓ About 60 mg/dL

Why: Every reading is within a few mg/dL of 60 mg/dL.
The goldfish’s glucose is being held near 60 mg/dL.
The value a regulated quantity is held near is its set point.

55
Practice writing an answer

A rabbit’s core temperature is held near 39 °C. On a cold night the rabbit’s temperature falls to 38.2 °C. The rabbit’s muscles shiver and release heat, and its temperature returns to 39 °C.

(a) Identify the set point, the stimulus and the response in this case. (3 pt)

Model answer The set point is 39 °C, the value the rabbit’s temperature is held near.
The stimulus is the fall to 38.2 °C, the change that moved the temperature away from the set point.
The response is the shivering, what the rabbit’s body did about the fall.
Rubric
  • Award 1 point for: the set point is 39 °C (the value the temperature is held near).
  • Award 1 point for: the stimulus is the fall to 38.2 °C (the change away from the set point).
  • Award 1 point for: the response is the shivering (what the body does about the fall).

Glossary

set point
The value a regulated quantity is held near, such as about 90 mg/dL for blood glucose or 37 °C for core temperature.
stimulus
A change that moves a regulated quantity away from its set point, such as the rise in blood glucose after a sugary drink.
response
What the body does about a stimulus, such as cells of the pancreas releasing insulin after blood glucose rises.

APBIO-U04-L11B The response cancels the stimulus

Topic 4.4 · Feedback · 62 steps

A graph of blood glucose and blood insulin against time after a meal: the solid glucose line rises to 140 mg/dL at half an hour and falls back to 90 mg/dL by two hours; the dashed insulin line rises just after it and falls back to its resting level by two hours
A graph of blood glucose and blood insulin against time after a meal: the solid glucose line rises to 140 mg/dL at half an hour and falls back to 90 mg/dL by two hours; the dashed insulin line rises just after it and falls back to its resting level by two hours

Here is a graph of one person’s blood glucose and blood insulin over the two hours after a meal.

Half an hour after the meal, the glucose is about 140 milligrams per deciliter and the insulin is high. By two hours both are back down: the glucose is near 90 mg/dL, and the insulin is near its resting level. Nobody switched the insulin off.

What brought the glucose down? And why did the insulin fade with it?

Unit 4 · Cell Communication and Cell Cycle

1The insulin loop, step by step

2
Check q1

Cells of the pancreas release insulin into the blood, and liver cells a body’s length away respond to it.

What is a hormone?

  1. A. ✓ A signal molecule carried in the blood to cells across the body
  2. B. A signal molecule that passes only between two cells that touch
    A signal that passes only between touching cells cannot reach a liver cell a body’s length away.
    A hormone is carried in the blood to cells across the body.

Why: A hormone is a signal molecule released into the blood and carried to cells across the body.
Insulin is a hormone.

3

Why does a quantity come back to its set point on its own? You can see the answer by following the loop one step at a time.

4

The rise in glucose after the meal is the stimulus.

One box: glucose rises, labelled as the stimulus
One box: glucose rises, labelled as the stimulus
5

Cells of the pancreas detect the rise. They release insulin, a hormone carried in the blood to liver and muscle cells.

Three boxes joined by arrows: glucose rises, pancreas cells detect the rise, insulin released
Three boxes joined by arrows: glucose rises, pancreas cells detect the rise, insulin released
6

Liver and muscle cells take glucose out of the blood and store it as glycogen, the animal glucose store. The glucose in the blood falls.

Five boxes joined by arrows: glucose rises, pancreas cells detect the rise, insulin released, liver and muscle take glucose up, glucose falls
Five boxes joined by arrows: glucose rises, pancreas cells detect the rise, insulin released, liver and muscle take glucose up, glucose falls
7

Storing the glucose is a reaction. Inside the liver and muscle cells, many glucose molecules are joined into one glycogen molecule.

Word equation: glucose becomes glycogen. Inside liver and muscle cells, many glucose molecules are joined into one glycogen molecule.
8

As the glucose falls back toward 90 mg/dL, the pancreas cells detect a smaller rise. So they release less insulin.

9

The response fades as the stimulus shrinks.

10

What you are expected to know Describe the insulin loop step by step: glucose rises, pancreas cells release insulin, liver and muscle cells store glucose as glycogen, glucose falls, insulin release falls.

11

Video: Watch: The insulin loop, step by step

Glucose rises after a meal. Cells of the pancreas detect the rise and release insulin. Liver and muscle cells take glucose up and store it as glycogen, so the glucose falls. As the glucose falls, the pancreas cells release less insulin: the response fades as the stimulus shrinks.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11Ba.mp4

12
Check q2

Half an hour after a meal, a person’s blood glucose and insulin concentrations are both high. By two hours, both are back near their premeal values.

Which of the following removed the glucose from the blood?

  1. A. The pancreas cells releasing insulin into the blood
    Insulin is the message, not the thing that removes glucose from the blood.
    Insulin made the liver and muscle cells take glucose up, and that uptake removed the glucose.
  2. B. ✓ Liver and muscle cells taking glucose up
  3. C. The meal being digested
    Digestion put the glucose into the blood.
    Liver and muscle cells took the glucose out again, once insulin told them to.

Why: The rise in glucose was the stimulus.
The pancreas cells detected the rise and released insulin.
Insulin made the liver and muscle cells take glucose up.
So the glucose fell back toward its set point.

13
Practice writing an answer

Half an hour after a meal, a person’s blood glucose and insulin concentrations are both high. By two hours, the liver and muscle cells have taken the glucose up, and the glucose is back near its premeal value. The insulin is back near its premeal value too.

(a) Explain why the insulin came back down by two hours. (1 pt)

Model answer The rise in glucose was the stimulus.
The pancreas cells detected the rise and released insulin.
Insulin made the liver and muscle cells take glucose up.
So the glucose fell back toward its set point.
The pancreas cells then detected no rise.
So the pancreas cells released less insulin.
So the insulin fell as the glucose fell.
Rubric
  • Award 1 point for: the glucose came back toward its set point, so the pancreas cells detected no rise and released less insulin (the response fades as its stimulus shrinks).

14Feedback

15

Look at what the response did. The insulin brought the glucose down, and the rise in glucose was the very thing that triggered the insulin.

The five boxes with an arrow returning from the last box to the first, labelled: the response acts back on the change that triggered it
The five boxes with an arrow returning from the last box to the first, labelled: the response acts back on the change that triggered it
16

So the response acted back on the quantity that triggered it.

17

When the response acts back on its own trigger like this, we call the loop , because the result is fed back to its own cause.

18

For example, when your core temperature rises, you sweat, and the sweat cools you. This is feedback, because the response acts back on the change that triggered it.

19

But now consider a different case: you smell food, and your mouth waters. This is not feedback, because the saliva does not act back on the smell.

20

And when a swimmer’s core temperature falls, he shivers, and the shivering warms him. This is feedback again, because the response acts back on the change that triggered it.

21

What you are expected to know Identify feedback in a described case: a loop in which the response acts back on the very quantity that triggered it.

22

Video: Watch: Feedback

The insulin brought the glucose down, and the rise in glucose was what triggered the insulin: the response acted back on its own trigger. A loop like that is feedback. Sweating that cools a hot body is feedback; a mouth watering at a smell is not, because the saliva does nothing to the smell.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11Bb.mp4

23
Check q3

A bright light shines into a person’s eye. The pupil narrows, and less light reaches the back of the eye.

Is this loop feedback?

  1. A. ✓ Yes
  2. B. No
    The light reaching the back of the eye triggered the narrowing.
    The narrowing reduced that light.
    The response acted back on its own trigger: feedback.

Why: The bright light reaching the back of the eye is the change.
The pupil narrows because of that light.
The narrowing reduces the light reaching the back of the eye.
The response acts back on the very change that triggered it: feedback.

24
Check q4

A person hears a sudden loud bang and jumps.

Is this loop feedback?

  1. A. Yes
    The jump does nothing to the bang.
    The response does not act back on the change that triggered it, so this is not feedback.
  2. B. ✓ No

Why: The bang is the change.
The person jumps because of the bang.
The jump does nothing to the bang.
The response does not act back on the change that triggered it, so this is not feedback.

25Quick quiz: feedback mixed practice

26
Check q5

A change in the body triggers a response.

When is the loop called feedback?

  1. A. When the response happens within a few seconds of the change
    Speed does not make a loop feedback.
    Feedback is a loop in which the response acts back on the quantity that triggered it.
  2. B. When the response is carried by a hormone in the blood
    Many feedback loops use no hormone at all.
    Feedback is a loop in which the response acts back on the quantity that triggered it.
  3. C. ✓ When the response acts back on the very quantity that triggered it

Why: Feedback is a loop in which the response acts back on the very quantity that triggered it, as insulin acts back on the glucose that triggered its release.

27
Practice writing an answer

After a meal, cells of the pancreas release insulin, and the glucose that triggered the release comes back down.

(a) State what feedback is. (1 pt)

Model answer Feedback is a loop in which the response acts back on the very quantity that triggered it.
Rubric
  • Award 1 point for: a loop in which the response acts back on (affects) the quantity or change that triggered it.

28Negative feedback

29

In the glucose loop, the response reduces the very change that triggered it. Insulin brought the rise in glucose back down.

The five boxes with an arrow returning from the last box to the first, labelled: the response reduces the change that triggered it
The five boxes with an arrow returning from the last box to the first, labelled: the response reduces the change that triggered it
30

So the glucose moves back toward its set point, and the response fades away.

31

When the response reduces its own trigger, we call it , because the response opposes the change.

32

‘Negative’ here means opposing, not harmful. Negative feedback is what keeps your glucose in range between meals.

33

What you are expected to know Explain how negative feedback holds a quantity near its set point: the response reduces the very change that triggered it, so the quantity moves back toward its set point and the response fades away.

34

Video: Watch: Negative feedback

In the glucose loop the response reduces the very change that triggered it, so the glucose returns toward its set point and the response fades. That is negative feedback: negative means opposing, not harmful.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11Bc.mp4

35
Practice writing an answer

Ten minutes into a race, a runner’s core temperature has risen from 37 °C to 38.4 °C, and she is sweating heavily. The sweat evaporates and cools her. By the end of her cool-down her temperature is back at 37 °C, and the sweating has stopped.

(a) Explain how the runner’s sweating demonstrates negative feedback. (1 pt)

Model answer The rise in temperature to 38.4 °C was the stimulus.
The sweating was the response.
The sweat evaporated and cooled her, so her temperature fell.
The response reduced the very change that triggered it: negative feedback.
As her temperature returned to 37 °C, the stimulus shrank.
So the sweating faded and stopped.
Rubric
  • Award 1 point for: the sweating (the response) reduced the rise in temperature that triggered it, bringing the temperature back to its set point, and faded as the temperature returned.
36
Check q6

A student says: “Negative feedback must be bad for the body, because negative means harmful.”

Is the student correct?

  1. A. Yes — negative feedback works against the body
    In negative feedback the response opposes the change that triggered it.
    Insulin opposes the rise in glucose after a meal, bringing the glucose back to its set point.
  2. B. ✓ No — ‘negative’ does not mean harmful

Why: ‘Negative’ names what the response does to its trigger: the response opposes the change.
Insulin opposes the rise in glucose.
Opposing the rise brings the glucose back to its set point.
So negative feedback keeps the glucose in range between meals; it is not harmful.

37Quick quiz: negative feedback mixed practice

38
Check q7

A response acts back on the change that triggered it.

When is the feedback called negative feedback?

  1. A. ✓ When the response reduces the change that triggered it
  2. B. When the response harms the body
    ‘Negative’ means opposing, not harmful.
    Negative feedback is feedback in which the response reduces the change that triggered it.
  3. C. When the response makes the change larger
    A response that makes the change larger does not oppose it.
    Negative feedback is feedback in which the response reduces the change that triggered it.

Why: Negative feedback is feedback in which the response reduces the very change that triggered it, so the quantity returns toward its set point.

39
Practice writing an answer

After a meal, insulin brings the rise in glucose back down, and the insulin then fades.

(a) State what negative feedback is. (1 pt)

Model answer Negative feedback is feedback in which the response reduces the very change that triggered it.
Rubric
  • Award 1 point for: feedback in which the response reduces (opposes) the change that triggered it.

40The glucagon arm

41

Now consider the hours after the meal. The glucose drops below 90 mg/dL.

42

Other cells of the pancreas detect the drop and release glucagon. Glucagon is a hormone that tells liver cells to break glycogen down and release glucose into the blood.

The same loop correcting a drop instead of a rise: glucose falls below 90 mg/dL, other pancreas cells detect it, glucagon released, liver breaks glycogen down, glucose rises, with the return arrow labelled the same way
The same loop correcting a drop instead of a rise: glucose falls below 90 mg/dL, other pancreas cells detect it, glucagon released, liver breaks glycogen down, glucose rises, with the return arrow labelled the same way
43

Breaking glycogen down is a reaction too. Inside the liver cells, glycogen is broken down into glucose molecules, which leave the cell into the blood.

Word equation: glycogen becomes glucose. Inside liver cells, glycogen is broken down into glucose molecules, which leave the cell into the blood.
44

The glucose rises back toward 90 mg/dL. As it rises, the pancreas cells release less glucagon, so the response fades.

45

One loop, two arms. Insulin brings glucose down from above the set point.

46

Glucagon brings glucose up from below the set point. Each response fades as the glucose returns.

47

What you are expected to know Describe the glucagon arm step by step: glucose falls below the set point, other pancreas cells release glucagon, liver cells break glycogen down and release glucose, glucose rises, glucagon release falls.

48

Video: Watch: The glucagon arm

Hours after the meal the glucose drops below 90 mg/dL. Other pancreas cells detect the drop and release glucagon; liver cells break glycogen down and release glucose, so the glucose rises back. One loop, two arms: insulin brings glucose down from above the set point, glucagon brings it up from below.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11Bd.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11Bd.mp4

49
Check q8

Four hours after a meal, a person’s blood glucose has fallen to 70 mg/dL.

Predict what happens to the glucose over the next hour.

  1. A. The glucose stays at 70 mg/dL
    A fall below the set point is a stimulus.
    Other pancreas cells detect it and release glucagon, which makes the liver release glucose, so the glucose rises.
  2. B. The glucose falls further
    Insulin release falls when the glucose falls below 90 mg/dL.
    Below 90 mg/dL, other pancreas cells release glucagon, which makes the liver release glucose, so the glucose rises.
  3. C. ✓ The glucose rises back toward its set point

Why: A fall below the set point is a stimulus for the other arm of the loop.
Other pancreas cells detect the fall and release glucagon.
Glucagon makes the liver break glycogen down and release glucose into the blood.
So the glucose rises back toward its set point.

50
Check q9

A horse has not eaten since the morning. By the evening its blood glucose has fallen to 68 mg/dL, lower than its set point.

Which of the following happens next?

  1. A. Liver cells store glucose as glycogen
    Storing glucose as glycogen would lower the glucose further.
    Below the set point, the loop releases glucose instead: glucagon makes liver cells break glycogen down.
  2. B. ✓ Other cells of the pancreas release glucagon
  3. C. Cells of the pancreas release insulin
    Insulin is the response to a rise above the set point.
    Below the set point, other pancreas cells release glucagon.

Why: The horse’s glucose is below its set point.
A fall below the set point is the stimulus for the glucagon arm.
So other cells of the pancreas detect the fall and release glucagon.

51

Back to the meal: half an hour after eating, the person’s blood glucose was about 140 mg/dL, and the concentration of insulin in their blood was high.

52

The rise in glucose was the stimulus. Cells of the pancreas detected the rise and released insulin, and liver and muscle cells took the glucose up and stored it as glycogen.

53

The glucose fell back toward 90 mg/dL. As it fell, the pancreas cells released less insulin, so the insulin faded with the glucose: negative feedback.

54Mixed practice mixed practice

55
Check q10

Between meals, a person’s blood glucose drops to less than its set point, and glucagon is released.

Which cells release glucose into the blood?

  1. A. Cells of the pancreas
    Cells of the pancreas detect the fall and release glucagon; they release no glucose.
    Liver cells break glycogen down and release the glucose.
  2. B. ✓ Liver cells

Why: Glucagon is carried in the blood to liver cells.
Glucagon tells the liver cells to break glycogen down.
So liver cells release glucose into the blood, and the glucose rises back toward its set point.

56
Check q11

After a meal, insulin is released into the blood.

Which cells take glucose out of the blood?

  1. A. ✓ Liver and muscle cells
  2. B. Cells of the pancreas
    Cells of the pancreas detect the rise and release insulin; they do not store the glucose.
    Liver and muscle cells take glucose up in quantity and store it as glycogen.

Why: Insulin is carried in the blood to liver and muscle cells.
Those cells take glucose out of the blood and store it as glycogen.

57
Check q12

In negative feedback, a response follows a change.

What does the response do to the change that triggered it?

  1. A. ✓ Reduces it
  2. B. Leaves it unchanged
    A response that left the change unchanged would not bring the quantity back.
    In negative feedback the response reduces the change that triggered it.
  3. C. Makes it larger
    A response that made the change larger would not oppose it.
    In negative feedback the response reduces the change that triggered it.

Why: In negative feedback the response reduces the very change that triggered it, so the quantity moves back toward its set point.

58
Check q13

After a meal, a person’s blood glucose has come back down to 90 mg/dL.

What happens to the release of insulin as the glucose returns to 90 mg/dL?

  1. A. Insulin release stays at its peak
    The pancreas cells release insulin in answer to a rise in glucose.
    Once the glucose is back at 90 mg/dL there is no rise, so they release less insulin.
  2. B. ✓ Insulin release falls
  3. C. Insulin release rises
    Insulin rises when the glucose rises.
    As the glucose returns to 90 mg/dL, the pancreas cells detect no rise and release less insulin.

Why: The pancreas cells release insulin because they detect a rise in glucose.
As the glucose returns to 90 mg/dL, the rise shrinks.
So the pancreas cells release less insulin: the response fades as the stimulus shrinks.

59
Check q14

In type 1 diabetes, the pancreas cells that release insulin have been destroyed. After a meal, a person with type 1 diabetes has a blood glucose of about 190 mg/dL, and it stays there for hours. A student says: “The person’s set point for glucose has risen to 190 mg/dL.”

Is the student correct?

  1. A. Yes — the set point has risen
    The set point, the value the body defends, is still about 90 mg/dL.
    The insulin-releasing cells are destroyed, so nothing brings the glucose down.
  2. B. ✓ No — the set point is still about 90 mg/dL

Why: The set point is the value the loop defends: about 90 mg/dL, unchanged.
The insulin-releasing pancreas cells are destroyed, so the loop is missing a part.
So the glucose concentration stays high, far from the set point.
So the set point is still about 90 mg/dL.

60
Check q15

A person with type 1 diabetes, whose insulin-releasing pancreas cells have been destroyed, injects insulin before a meal. Their blood glucose comes back down after the meal.

Why does the injected insulin work when the person’s own insulin-releasing cells are destroyed?

  1. A. The injected insulin acts on liver and muscle cells without needing a receptor
    A hormone acts only by binding its receptor.
    Insulin stays outside the cell and binds the insulin receptor on the liver and muscle cells, and those receptors are still there.
  2. B. The injected insulin lowers the set point below the current glucose
    The set point stays near 90 mg/dL.
    The injection restores the response that brings the glucose back to the set point.
  3. C. The injected insulin stands in for the glucagon the person cannot make
    Type 1 diabetes destroys the cells that release insulin.
    Glucagon is still made, and glucagon raises glucose rather than lowering it.
  4. D. ✓ Liver and muscle cells still have insulin receptors

Why: Only one link of the loop is missing: the pancreas cells that release insulin.
The liver and muscle cells still carry their insulin receptors.
So the injected insulin binds those receptors.
So the liver and muscle cells take the glucose up, and the glucose comes back down.

61
Practice writing an answer

After a large meal, a dog’s blood glucose rises to 150 mg/dL. Two hours later it is 95 mg/dL, and the insulin in its blood is back near its resting level.

(a) Explain how the dog’s glucose returning to 95 mg/dL demonstrates negative feedback. (1 pt)

Model answer The rise in glucose to 150 mg/dL was the stimulus.
Cells of the pancreas detected the rise and released insulin: the response.
Insulin made liver and muscle cells take glucose up and store it as glycogen.
So the glucose fell back toward its set point.
The response reduced the very change that triggered it: negative feedback.
As the glucose fell, the pancreas cells released less insulin, so the response faded.
Rubric
  • Award 1 point for: the response (insulin, and the uptake of glucose it caused) reduced the rise in glucose that triggered it, bringing the glucose back toward its set point, and faded as the glucose returned.

Glossary

feedback
A loop in which the response to a change acts back on the very quantity that triggered it.
negative feedback
Feedback in which the response reduces the very change that triggered it, so the quantity moves back toward its set point and the response fades away. ‘Negative’ means opposing, not harmful.

APBIO-U04-L11C The same loop at three levels

Topic 4.4 · Feedback · 52 steps

Three panels: on the left a schematic pathway of three enzyme steps from A to the amino acid, with an arrow from the amino acid back to the first enzyme; in the middle a photograph of a single-celled pond organism, a Paramecium, one oval ciliated cell seen under a microscope; on the right three boxes, pancreas cells, insulin in the blood, liver and muscle cells, joined by arrows
Three panels: on the left a schematic pathway of three enzyme steps from A to the amino acid, with an arrow from the amino acid back to the first enzyme; in the middle a photograph of a single-celled pond organism, a Paramecium, one oval ciliated cell seen under a microscope; on the right three boxes, pancreas cells, insulin in the blood, liver and muscle cells, joined by arrows

Here are three loops of three sizes.

A bacterium makes an amino acid in three enzyme steps. When the amino acid piles up, it binds the first enzyme at a site away from its active site, and the enzyme slows. A Paramecium, a single-celled organism living in pond water, fills and empties its contractile vacuole faster as more water flows in. After a meal, your pancreas, blood and liver bring your glucose back to 90 mg/dL.

What do the three loops share? And how do you tell the sizes apart?

Unit 4 · Cell Communication and Cell Cycle

1One enzyme molecule is the whole loop

2
Check q1

After a meal, insulin brings the rise in blood glucose back down, and the insulin then fades.

In negative feedback, what does the response do to the change that triggered it?

  1. A. ✓ The response reduces the change
  2. B. The response makes the change larger
    A response that made the change larger would not bring the glucose back down.
    In negative feedback the response reduces the change that triggered it.

Why: Negative feedback is feedback in which the response reduces the very change that triggered it, so the quantity returns toward its set point.

3
Check q2

An inhibitor binds an enzyme at a site away from the active site and changes the enzyme’s shape.

What is the site where the inhibitor binds called?

  1. A. A ligand-binding domain
    A ligand-binding domain is the pocket on a receptor where its ligand binds.
    A site on an enzyme away from the active site is an allosteric site.
  2. B. ✓ An allosteric site

Why: A site on an enzyme other than the active site, where a molecule can bind, is called an allosteric site.

4

Does negative feedback need a whole body? It does not.

5

Negative feedback happens at three sizes: one enzyme molecule, one cell, one whole body. Start with the smallest.

6

A bacterium makes an amino acid in three enzyme steps: enzyme 1 turns A into B, enzyme 2 turns B into C, and enzyme 3 turns C into the amino acid.

A pathway of three enzymes, each drawn as a pill with a notch, its active site, facing the arrow it acts on: enzyme 1 turns A into B, enzyme 2 turns B into C, enzyme 3 turns C into the amino acid
A pathway of three enzymes, each drawn as a pill with a notch, its active site, facing the arrow it acts on: enzyme 1 turns A into B, enzyme 2 turns B into C, enzyme 3 turns C into the amino acid
7

When the amino acid piles up, it binds enzyme 1 at an allosteric site, a site away from the active site.

The same pathway with an arrow from the amino acid back to enzyme 1, where a dot on the far side of the pill from its notch marks the amino acid bound at a site away from the active site; caption: enzyme 1 slows
The same pathway with an arrow from the amino acid back to enzyme 1, where a dot on the far side of the pill from its notch marks the amino acid bound at a site away from the active site; caption: enzyme 1 slows
8

Bound there, the amino acid changes enzyme 1’s shape, and enzyme 1 slows.

9

With enzyme 1 slowed, less B forms. So less C forms, and the pathway makes less amino acid.

10

As the cell uses the amino acid up, the amino acid leaves enzyme 1, and the pathway speeds up again.

11

The end product reduced its own production. The response acted back on the change that triggered it and reduced it: negative feedback.

12

One enzyme molecule detected the change, and the same molecule responded. This is negative feedback at the molecular level.

13

What you are expected to know Explain how a pathway’s end product binding its first enzyme at an allosteric site is negative feedback at the molecular level: enzyme 1 slows, less product forms, and as the product is used up the pathway speeds again.

14

Video: Watch: One enzyme molecule is the whole loop

A bacterium makes an amino acid in three enzyme steps. When the amino acid piles up it binds enzyme 1 at an allosteric site, enzyme 1 slows, and less amino acid is made; as the cell uses the amino acid up, the pathway speeds again. One molecule detects and the same molecule responds: negative feedback at the molecular level.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11Ca.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11Ca.mp4

15
Check q3

In a pathway of three enzymes, the end product slows the pathway when it is plentiful.

Which of the following does the end product bind?

  1. A. ✓ The first enzyme, at a site away from its active site
  2. B. The last enzyme, in its active site
    The end product does not compete for the last enzyme’s active site.
    It binds the first enzyme at an allosteric site, so the whole pathway slows from enzyme 1 onward.
  3. C. The pathway’s starting substance
    The end product binds an enzyme, not a substance in the pathway.
    It binds the first enzyme at an allosteric site.

Why: The end product binds the pathway’s first enzyme at an allosteric site, a site away from the active site.
So the first enzyme slows, and every later step gets less to work on.

16
Check q4

In a yeast cell, a pathway of three enzymes makes a substance called T. When T is plentiful, T binds the pathway’s first enzyme, enzyme 1, at a site away from its active site.

What happens to the rate at which the pathway makes T?

  1. A. ✓ The pathway makes T more slowly
  2. B. The pathway makes T at the same rate
    T bound at the allosteric site changes enzyme 1’s shape, so enzyme 1 slows.
    Less B and C form, so the pathway makes less T.
  3. C. The pathway makes T faster
    T bound at the allosteric site slows enzyme 1, not speeds it.
    Less B and C form, so the pathway makes less T.

Why: T binds enzyme 1 at an allosteric site and changes its shape.
So enzyme 1 slows.
So less of each later substance forms, and the pathway makes T more slowly.

17
Practice writing an answer

In a yeast cell, a pathway of three enzymes makes a substance called T. When T is plentiful, T binds the pathway’s first enzyme, enzyme 1, at a site away from its active site, and enzyme 1 slows. As the cell uses T up, T leaves enzyme 1.

(a) Explain how T binding enzyme 1 demonstrates negative feedback. (1 pt)

Model answer The rise in T is the change.
T binds enzyme 1 at its allosteric site, so enzyme 1 slows: the response.
With enzyme 1 slowed, the pathway makes less T.
So the response reduced the very change that triggered it: negative feedback.
As the cell uses T up, T leaves enzyme 1 and the pathway speeds again, so the response fades as the change shrinks.
Rubric
  • Award 1 point for: T slowing enzyme 1 reduces the production of T, the very change that triggered the binding, so the response reduces its own trigger (and fades as T is used up).
18
Check q5

Suppose a mutation changes enzyme 1’s allosteric site in a bacterium, so that the pathway’s amino acid can no longer bind there. The bacterium has plenty of the amino acid.

Predict what happens to the pathway.

  1. A. The pathway slows, as before
    The amino acid slows enzyme 1 only by binding the allosteric site.
    With that site changed, the amino acid cannot bind, so nothing slows enzyme 1.
  2. B. ✓ The pathway keeps going at full speed
  3. C. The pathway stops completely
    The amino acid slows enzyme 1 only by binding the allosteric site.
    With that site changed, nothing slows enzyme 1, let alone stops the pathway.

Why: The amino acid slows enzyme 1 only while it is bound at the allosteric site.
The mutation means the amino acid can no longer bind there.
So nothing slows enzyme 1.
So the pathway keeps making the amino acid at full speed, however much has piled up.

19Molecular, cellular or organismal?

20
Check q6

Water enters a Paramecium, a single-celled pond organism, by osmosis all the time.

What does the Paramecium’s contractile vacuole do?

  1. A. ✓ Collects the water inside the cell and pushes it out
  2. B. Stops water from entering the cell across its membrane
    Nothing stops the water entering: it comes in by osmosis all the time.
    The contractile vacuole collects the water inside the cell and pushes it back outside.

Why: A contractile vacuole is a sac inside the cell that collects the incoming water and empties it outside through a pore.

21

Here is a photograph of a Paramecium, a single-celled organism that lives in pond water.

A photograph of a Paramecium under a microscope: one oval cell covered in fine hairs, with small clear vacuoles inside it, on a gray background
A photograph of a Paramecium under a microscope: one oval cell covered in fine hairs, with small clear vacuoles inside it, on a gray background
22

In pond water, water enters the Paramecium by osmosis all the time. Its contractile vacuole collects the water and empties it outside.

A Paramecium, labelled as one cell, drawn in cross-section in pond water, with arrows of water entering all round its membrane and a contractile vacuole pushing water back out through one point
A Paramecium, labelled as one cell, drawn in cross-section in pond water, with arrows of water entering all round its membrane and a contractile vacuole pushing water back out through one point
23

When more water flows in, the vacuole fills and empties faster. So one cell holds its own water content steady.

24

The size at which a loop detects and responds is called the of the loop.

25

To place a loop at its level, ask two questions. What detects the change, and what responds?

26

The amino acid binds one enzyme molecule, and that same molecule slows. This loop is molecular, because one molecule detects the change and the same molecule responds.

27

The Paramecium detects the extra water and pushes it out. This loop is cellular, because one cell detects the change and the same cell responds.

28

A Paramecium is a whole organism, but its loop is still cellular, because one cell does the detecting and the same cell does the responding.

29
Check q7

In a yeast cell, when the end product T of a pathway is plentiful, T binds the pathway’s first enzyme, enzyme 1, and slows it. A student says: “This loop is cellular, because it happens inside a single yeast cell.”

Is the student correct?

  1. A. Yes — the loop is cellular
    The level is set by what detects the change and what responds.
    Here one molecule, enzyme 1, both detects T and responds by slowing: molecular, even inside one cell.
  2. B. ✓ No — the loop is molecular

Why: The level of a loop is set by what detects the change and what responds.
T binds enzyme 1 and enzyme 1 slows, so one enzyme molecule both detects and responds.
One molecule detects and responds: the molecular level.
Where the loop happens does not set its level.

30

After a meal, pancreas cells detect the rise in glucose, and liver and muscle cells across the body respond. This loop is organismal, because organs across a body detect the change and respond.

Five boxes joined by arrows, glucose rises, pancreas cells detect the rise, insulin released, liver and muscle take glucose up, glucose falls, with an arrow returning from the last box to the first labelled: the response reduces the change that triggered it
Five boxes joined by arrows, glucose rises, pancreas cells detect the rise, insulin released, liver and muscle take glucose up, glucose falls, with an arrow returning from the last box to the first labelled: the response reduces the change that triggered it
31

Sweating when your core temperature rises above 37 °C is organismal too, because skin all over the body responds.

32

Here is a table comparing the three levels: what detects the change, what responds, and one example of each.

A table comparing the three levels of negative feedback on three rows: molecular, one enzyme molecule detects the change and the same molecule responds, example an amino acid binds enzyme 1 and slows it; cellular, one cell detects and the same cell responds, example a Paramecium empties its contractile vacuole faster; organismal, organs across a body detect and organs across the body respond, example pancreas cells release insulin and liver and muscle cells take glucose up
33

What you are expected to know Classify a described negative-feedback loop as molecular (one enzyme molecule detects and responds), cellular (one cell detects and responds; a single-celled organism’s loop is cellular) or organismal (organs across a body detect and respond).

34

Video: Watch: Molecular, cellular or organismal?

To place a loop, ask what detects the change and what responds. One enzyme molecule: molecular. One cell, as in a Paramecium emptying its vacuole faster: cellular, even though the Paramecium is a whole organism. Organs across a body, as in the glucose loop: organismal.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11Cb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11Cb.mp4

35
Check q8

Suppose one cell detects a change inside itself, and the same cell responds to it.

At which level is this loop?

  1. A. Molecular
    A molecular loop is one molecule detecting and responding.
    Here a whole cell detects and the same cell responds: the cellular level.
  2. B. ✓ Cellular
  3. C. Organismal
    An organismal loop needs organs across a body.
    Here one cell detects and the same cell responds: the cellular level.

Why: The level is set by what detects the change and what responds.
One cell detects and the same cell responds.
So the loop is at the cellular level.

36

Back to the three loops: the bacterium whose amino acid slows its own first enzyme, the Paramecium emptying its vacuole faster, and the pancreas, blood and liver bringing glucose back to 90 mg/dL.

37

All three are negative feedback: in each, the response reduces the change that triggered it.

38

They differ in what detects the change and what responds: one enzyme molecule is molecular, one cell is cellular, and organs across a body are organismal.

39Quick quiz: which level? mixed practice

40
Check q9

When a swimmer’s core temperature falls, muscles all over the body shiver and release heat.

At which level does this loop detect and respond?

  1. A. Molecular
    The level is set by what detects the fall and what responds.
    Here organs across the whole body detect the fall and respond: muscles everywhere shiver together.
  2. B. Cellular
    Muscles all over the body shiver together, told to by signals from elsewhere in the body.
    Organs across the whole body detect the fall and respond.
  3. C. ✓ Organismal

Why: The fall in temperature is detected and answered by organs across the whole body: muscles everywhere shiver together.
So the loop is at the organismal level.

41
Check q10

In a bacterium, the last product of a pathway that builds a nucleotide binds the pathway’s first enzyme and slows it.

At which level does this loop detect and respond?

  1. A. ✓ Molecular
  2. B. Cellular
    The pathway’s enzymes are inside one bacterial cell, but the level is set by what detects and responds.
    Here one enzyme molecule is bound by the product and slows.
  3. C. Organismal
    A bacterium is one cell, and the loop is smaller still: one enzyme molecule is bound by the pathway’s last product and slows.

Why: The pathway’s last product binds one enzyme molecule, and that enzyme molecule slows.
One molecule detects and one molecule responds.
So the loop is at the molecular level.

42
Check q11

When rain dilutes a pond, a single-celled alga’s contractile vacuole fills and empties more often.

At which level does this loop detect and respond?

  1. A. Molecular
    Water is a molecule, but the level is set by what detects and what responds.
    Here one whole cell detects the extra water and pushes it out with its vacuole.
  2. B. ✓ Cellular
  3. C. Organismal
    The alga is a whole organism, but a single-celled organism’s loop is cellular.
    The detecting and the responding both happen in one cell.

Why: One alga cell detects the extra water flowing in, and the same cell empties its vacuole faster.
One cell detects and one cell responds.
So the loop is at the cellular level, even though the alga is a whole organism.

43
Check q12

After a meal, cells of a horse’s pancreas release insulin, and liver and muscle cells across its body take glucose up.

At which level does this loop detect and respond?

  1. A. Molecular
    Insulin is a molecule, but the level is set by what detects and responds.
    Here pancreas cells detect the rise, and liver and muscle cells across the horse’s body respond.
  2. B. Cellular
    The pancreas cells detect the rise, the blood carries insulin, and liver and muscle cells across the horse’s body respond.
    Organs across a whole body are working together.
  3. C. ✓ Organismal

Why: Pancreas cells detect the rise in glucose.
Insulin travels in the blood.
Liver and muscle cells across the horse’s body respond.
Organs across a whole body detect and respond, so the loop is at the organismal level.

44
Check q13

When the calcium inside a nerve cell rises, that nerve cell detects the rise, and pumps in its membrane move calcium out until the calcium is back to its usual level.

At which level does this loop detect and respond?

  1. A. Molecular
    The level is set by what detects the change and what responds.
    Here one nerve cell detects its own calcium rising, and the same cell pumps the calcium out.
  2. B. ✓ Cellular
  3. C. Organismal
    Nothing outside the nerve cell takes part.
    The cell detects its own calcium and the cell pumps the calcium out.

Why: One nerve cell detects that its calcium has risen, and the same cell moves the calcium out through its own membrane.
One cell detects and one cell responds.
So the loop is at the cellular level.

45
Check q14

In a mold, the amino acid at the end of a pathway binds the pathway’s first enzyme away from its active site and slows it.

At which level does this loop detect and respond?

  1. A. ✓ Molecular
  2. B. Cellular
    The pathway sits inside a mold cell, but the level is set by what detects and responds.
    Here one enzyme molecule is bound by the amino acid and slows.
  3. C. Organismal
    The mold is the organism, but the loop is one enzyme molecule bound by the pathway’s own end product.

Why: The amino acid binds one enzyme molecule, and that enzyme molecule slows.
One molecule detects and one molecule responds.
So the loop is at the molecular level.

46Mixed practice mixed practice

47
Check q15

In a bacterium, a three-enzyme pathway makes an amino acid, and when the amino acid is plentiful it binds enzyme 1 at a site away from its active site. The cell suddenly uses up its whole store of the amino acid.

Predict what happens to the pathway.

  1. A. The pathway stops
    The bound amino acid is what slows enzyme 1.
    With the amino acid gone, enzyme 1 is free and works at full speed.
  2. B. The pathway keeps making the amino acid at the slowed rate
    The amino acid stays bound to enzyme 1 only while it is plentiful.
    Once the store is used up, the amino acid leaves enzyme 1, and enzyme 1 speeds up.
  3. C. ✓ The pathway speeds up

Why: The amino acid slows enzyme 1 only while the amino acid is bound to enzyme 1.
When the cell uses the amino acid up, the amino acid leaves the allosteric site.
So enzyme 1 speeds up, and the pathway makes more amino acid.
The response has faded with its stimulus.

48
Check q16

When a yeast cell holds plenty of ATP, ATP binds an enzyme of the pathway that makes ATP, at a site away from that enzyme’s active site, and the enzyme slows.

At which level does this loop detect and respond?

  1. A. ✓ Molecular
  2. B. Cellular
    The enzyme sits inside one yeast cell, but the level is set by what detects and responds.
    Here one enzyme molecule is bound by ATP and slows.
  3. C. Organismal
    A yeast cell is one cell, and the loop is smaller still: one enzyme molecule bound by ATP.

Why: ATP binds one enzyme molecule, and that enzyme molecule slows.
One molecule detects and one molecule responds.
So the loop is at the molecular level.

49
Check q17

When the salt around a single-celled alga rises, the alga detects the extra salt, and the same cell pumps salt out until the salt inside it is back to its usual level.

At which level does this loop detect and respond?

  1. A. Molecular
    Salt is made of ions, but the level is set by what detects and what responds.
    Here one whole cell detects the salt and the same cell pumps it out.
  2. B. ✓ Cellular
  3. C. Organismal
    The alga is a whole organism, but a single-celled organism’s loop is cellular.
    One cell detects and the same cell responds.

Why: One alga cell detects the extra salt, and the same cell pumps salt out.
One cell detects and one cell responds.
So the loop is at the cellular level.

50
Check q18

Between meals, a person’s blood glucose falls, other cells of the pancreas release glucagon, and liver cells, an organ away from the pancreas, release glucose into the blood.

At which level does this loop detect and respond?

  1. A. Molecular
    Glucagon is a molecule, but the level is set by what detects and responds.
    Pancreas cells detect the fall, and liver cells in an organ far from the pancreas respond.
  2. B. Cellular
    The pancreas cells detect the fall, the blood carries glucagon, and liver cells in an organ far from the pancreas respond.
    Organs across a whole body are working together.
  3. C. ✓ Organismal

Why: Pancreas cells detect the fall in glucose.
Glucagon travels in the blood.
Liver cells in an organ far from the pancreas respond.
Organs across a whole body detect and respond, so the loop is at the organismal level.

51
Practice writing an answer

When the calcium in a person’s blood falls, cells of a small gland in the neck detect the fall and release a hormone into the blood. Bone cells across the body respond by releasing calcium into the blood, and the blood calcium rises back toward its usual level.

(a) Explain why this loop is at the organismal level rather than the cellular level. (1 pt)

Model answer The level of a loop is set by what detects the change and what responds.
Cells of the gland in the neck detect the fall in calcium.
The hormone travels in the blood to bone cells across the body.
Bone cells across the body respond by releasing calcium.
Organs across a body detect and respond, so the loop is organismal.
In a cellular loop one cell would both detect and respond.
Rubric
  • Award 1 point for: the gland cells detect and bone cells across the body respond (organs across a body, joined by a hormone in the blood), so the loop is organismal; a cellular loop would have one cell doing both.

Glossary

the level of a loop
Set by what detects the change and what responds. Molecular: one enzyme molecule detects and the same molecule responds. Cellular: one cell detects and the same cell responds (a single-celled organism’s loop is cellular). Organismal: organs across a body detect and respond.

APBIO-U04-L11D Break one part of the loop

Topic 4.4 · Feedback · 39 steps

A graph of blood glucose against time after a meal for two people, with the set point at 90 mg/dL drawn as a line of dots: the solid line rises to 140 mg/dL and returns to the set point; the dashed line climbs past 200 mg/dL and stays high
A graph of blood glucose against time after a meal for two people, with the set point at 90 mg/dL drawn as a line of dots: the solid line rises to 140 mg/dL and returns to the set point; the dashed line climbs past 200 mg/dL and stays high

Here is a graph of blood glucose over the two hours after a meal, for a person without diabetes and for a person with type 1 diabetes.

In type 1 diabetes, the pancreas cells that release insulin have been destroyed. After the meal, the person’s blood glucose rises to 140 milligrams per deciliter, as it does in anyone. Then it keeps rising, past 200 mg/dL, and the concentration stays high for hours. Their set point has not changed: it is still about 90 mg/dL.

So what is broken?

Unit 4 · Cell Communication and Cell Cycle

1Name the part, follow the loop, say which way

2
Check q1

A mutation changes a receptor’s ligand-binding domain so that the pocket no longer fits its ligand. The ligand arrives at the cell.

What happens?

  1. A. ✓ The ligand does not bind, so reception fails
  2. B. The ligand binds, and the receptor responds as usual
    The changed pocket no longer fits the ligand.
    The ligand cannot bind, so the receptor never changes shape and reception fails.

Why: The ligand binds only a pocket that fits it.
The changed ligand-binding domain no longer fits the ligand.
So the ligand does not bind, and reception fails.

3
Check q2

Between meals, a person’s blood glucose drops to less than its set point.

Which hormone brings the glucose back up?

  1. A. Insulin
    Insulin brings glucose down from above the set point.
    Below the set point, other pancreas cells release glucagon, which makes liver cells release glucose.
  2. B. ✓ Glucagon

Why: A fall below the set point is the stimulus for the glucagon arm.
Other pancreas cells release glucagon.
Glucagon makes liver cells break glycogen down and release glucose, so the glucose rises.

4

What happens to a regulated quantity when one part of its loop is missing? The set point is not what breaks.

5

The loop that defends the set point is what breaks. To predict the effect, you follow the loop with the missing part left out.

6

In type 1 diabetes, the pancreas cells that release insulin have been destroyed. Here is the insulin loop with that box missing.

The insulin loop with the pancreas-cells box crossed out and a break drawn in the arrow after it; the three later boxes are faint and there is no return arrow
The insulin loop with the pancreas-cells box crossed out and a break drawn in the arrow after it; the three later boxes are faint and there is no return arrow
7

After a meal, glucose rises to 140 mg/dL as before. But no insulin is released.

8

So liver and muscle cells do not take the glucose up. The glucose concentration keeps rising, past 200 mg/dL, and stays high for hours.

Two glucose curves after the same drink, with the set point at 90 mg/dL drawn as a line of dots: the solid curve, labelled no diabetes, rises to 140 mg/dL and returns to 90; the dashed curve, labelled type 1 diabetes, climbs past 200 mg/dL and stays there; gridlines every 20 mg/dL
Two glucose curves after the same drink, with the set point at 90 mg/dL drawn as a line of dots: the solid curve, labelled no diabetes, rises to 140 mg/dL and returns to 90; the dashed curve, labelled type 1 diabetes, climbs past 200 mg/dL and stays there; gridlines every 20 mg/dL
9

The set point has not changed: it is still about 90 mg/dL. What is broken is the loop that defends it.

10

To predict what a missing or blocked part does, take three steps:

  1. Name the missing part.
  2. Follow the loop from that part to the quantity being measured.
  3. Say which way the quantity moves: to a higher value, to a lower value, or unchanged.

11

For type 1 diabetes, the missing part is the pancreas cells that release insulin. Without insulin, liver and muscle cells do not take glucose up.

12

So after a meal the glucose concentration ends higher than normal, and it stays high.

13

Now suppose a person’s liver cells have glucagon receptors that cannot bind glucagon. Between meals the glucose falls below 90 mg/dL, and glucagon is released as usual.

14

The missing part is the receptor on the liver cells. Glucagon cannot bind, so the liver cells never get the message.

15

So the liver cells do not break glycogen down, and this loop cannot bring the glucose back up. Between meals the glucose concentration ends lower than normal.

16

What you are expected to know Predict what happens to a regulated quantity when one part of its negative-feedback loop is missing or blocked: name the missing part, follow the loop to the measured quantity, and say which way it moves.

17

Video: Watch: Name the part, follow the loop, say which way

In type 1 diabetes the insulin-releasing cells are gone: glucose rises after a meal and nothing brings it down, so it stays high; the set point is unchanged. To predict any break: name the missing part, follow the loop to the measured quantity, and say which way it moves.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11Da.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L11Da.mp4

18
Check q3

A veterinarian gives a cat a drug that blocks the insulin receptors on its liver and muscle cells. The cat’s pancreas cells are normal. The cat then eats a meal, and its glucose rises.

Do the pancreas cells release insulin?

  1. A. ✓ Yes
  2. B. No
    The drug blocks the receptors on the liver and muscle cells.
    The pancreas cells are normal.
    So the pancreas cells detect the rise in glucose and release insulin as usual.

Why: The missing part is the receptor on the liver and muscle cells, not the pancreas cells.
The pancreas cells are normal.
So the pancreas cells detect the rise and release insulin as usual.
The break comes one step further on.

19
Check q4

A cat’s liver and muscle cells have their insulin receptors blocked by a drug. The cat eats a meal, its pancreas releases insulin, and the insulin reaches the liver and muscle cells.

Do the liver and muscle cells take the glucose up?

  1. A. Yes
    Insulin acts only by binding its receptors on the liver and muscle cells.
    The drug has blocked those receptors, so the cells are never told to take glucose up.
  2. B. ✓ No

Why: Insulin acts only by binding its receptor on the liver and muscle cells.
The drug has blocked those receptors.
So the insulin cannot bind.
So the liver and muscle cells are never told to take glucose up, and they do not.

20
Check q5

A cat’s liver and muscle cells have their insulin receptors blocked by a drug. Two hours after a meal, the veterinarian measures its blood glucose.

How does the cat’s blood glucose concentration compare with an untreated cat’s?

  1. A. Lower
    Insulin lowers glucose only by making the liver and muscle cells take glucose up.
    With the receptors blocked, they take no glucose up, so the glucose is not lowered.
  2. B. The same
    The set point has not changed.
    But the loop that defends the set point is broken at the receptor.
    So the glucose is not brought back to the set point.
  3. C. ✓ Higher

Why: The missing part is the receptor on the liver and muscle cells.
The pancreas cells release insulin, but the insulin cannot bind, so the liver and muscle cells do not take glucose up.
The response is missing.
So the glucose concentration stays high: higher than in an untreated cat.

21
Check q6

A drug stops the pancreas cells that make glucagon from releasing it. A person taking the drug then fasts for twelve hours.

Compared with a fast taken before the drug, predict the person’s blood glucose concentration at the end of the fast.

  1. A. Higher
    Liver cells break glycogen down when glucagon tells them to.
    With no glucagon released, the liver releases no glucose.
    So the glucagon arm cannot raise the glucose during the fast.
  2. B. The same
    The set point has not changed.
    But the loop that defends the set point from below is broken at the glucagon step.
    So the glucose is not brought back up.
  3. C. ✓ Lower

Why: During a fast the glucose falls below the set point.
The missing part is the glucagon release.
So the liver is never told to break glycogen down and release glucose.
This loop cannot bring the glucose back up.
So the glucose ends lower than in the fast before the drug.

22

Back to the person with type 1 diabetes, whose insulin-releasing pancreas cells have been destroyed. After the meal their glucose concentration rose to 140 mg/dL, kept rising past 200 mg/dL, and stayed high.

23

The missing part is the pancreas cells that release insulin. With no insulin, liver and muscle cells do not take the glucose up, so nothing brings the glucose down.

24

The set point is unchanged at about 90 mg/dL. What broke is the loop that defends it.

25Quick quiz: which way does the quantity move? mixed practice

26
Check q7

A drug stops a horse’s pancreas cells from releasing insulin. The horse eats a meal.

Compared with normal, predict the horse’s blood glucose concentration two hours after the meal.

  1. A. ✓ Higher
  2. B. The same
    With no insulin, liver and muscle cells are never told to take glucose up.
    So the rise after the meal is not brought down.
  3. C. Lower
    Insulin is what brings glucose down after a meal.
    With no insulin released, nothing lowers the glucose.

Why: The missing part is the insulin release.
Without insulin, liver and muscle cells do not take glucose up.
So the glucose concentration after the meal ends higher than normal.

27
Check q8

A mutation leaves a person’s liver cells with no glucagon receptors. The person fasts overnight.

Compared with normal, predict the person’s blood glucose concentration in the morning.

  1. A. Higher
    Glucagon is what raises glucose during a fast.
    With no receptors, the liver cells never get the message and release no glucose.
  2. B. The same
    Glucagon is released as usual, but the liver cells cannot bind it.
    So the liver cells release no glucose, and the fall is not brought back up.
  3. C. ✓ Lower

Why: The missing part is the glucagon receptor on the liver cells.
Glucagon cannot bind, so the liver cells do not break glycogen down.
This loop cannot bring the glucose back up, so the glucose concentration ends lower than normal.

28
Check q9

A person’s insulin-releasing pancreas cells have been destroyed, and their blood glucose concentration stays high after every meal.

Compared with before, predict the person’s set point for glucose.

  1. A. Higher
    The set point is the value the body defends, and it is still about 90 mg/dL.
    What is broken is the loop that defends it.
  2. B. ✓ Unchanged
  3. C. Lower
    The set point is the value the body defends, and it is still about 90 mg/dL.
    What is broken is the loop that defends it.

Why: The set point is the value the loop defends: about 90 mg/dL.
Destroying the insulin-releasing cells breaks the loop, not the set point.
So the set point is unchanged; the glucose sits far above it.

29
Check q10

A sheep’s pancreas cells that release glucagon are destroyed. The sheep then fasts for twelve hours.

Compared with normal, predict the sheep’s blood glucose concentration at the end of the twelve hours.

  1. A. Higher
    Glucagon is what raises glucose when the sheep has not eaten.
    With no glucagon, the liver releases no glucose.
  2. B. The same
    The glucose falls during the twelve hours as usual.
    With no glucagon, this loop cannot bring the glucose back up, so its concentration ends lower.
  3. C. ✓ Lower

Why: The missing part is the glucagon release.
Without glucagon, liver cells do not break glycogen down and release glucose.
This loop cannot bring the glucose back up, so the glucose concentration ends lower than normal.

30
Check q11

A drug blocks the insulin receptors on a dog’s liver and muscle cells. The dog eats a meal.

Compared with normal, predict the dog’s blood glucose concentration two hours after the meal.

  1. A. ✓ Higher
  2. B. The same
    Insulin is released, but it cannot bind the blocked receptors.
    So liver and muscle cells do not take glucose up, and the rise is not brought down.
  3. C. Lower
    Insulin lowers glucose only by binding its receptors on the liver and muscle cells.
    With those receptors blocked, nothing lowers the glucose.

Why: The missing part is the insulin receptor on the liver and muscle cells.
Insulin cannot bind, so those cells do not take glucose up.
So the glucose concentration after the meal ends higher than normal.

31
Check q12

A mutation leaves a person’s glucagon-releasing pancreas cells unable to detect a fall in glucose. The person fasts overnight.

Compared with normal, predict the person’s blood glucose concentration in the morning.

  1. A. Higher
    The glucose falls during the fast as usual.
    The pancreas cells do not detect the fall, so no glucagon is released and the glucagon arm cannot raise the glucose.
  2. B. The same
    The pancreas cells never detect the fall, so they release no glucagon.
    The liver releases no glucose, and the fall is not brought back up.
  3. C. ✓ Lower

Why: The missing part is the detection of the fall by the glucagon-releasing cells.
No glucagon is released, so liver cells do not release glucose.
This loop cannot bring the glucose back up, so the glucose concentration ends lower than normal.

32Mixed practice mixed practice

33
Check q13

A mutation leaves a person’s liver cells without the enzyme that breaks glycogen down. The person’s pancreas cells and glucagon receptors are normal. The person fasts overnight.

Compared with a person whose liver cells have the enzyme, predict the person’s blood glucose concentration by morning.

  1. A. ✓ Lower
  2. B. The same
    Glucagon is released and binds its receptors on the liver cells, but the step after reception, breaking glycogen down, needs the missing enzyme.
    So the liver cells release no glucose.
  3. C. Higher
    Glycogen releases glucose only when the enzyme breaks glycogen down.
    Without the enzyme, the glycogen stays put and the liver releases no glucose.

Why: The missing part is the enzyme that breaks glycogen down.
Glucagon is released and binds the liver cells.
But the liver cells cannot break glycogen down.
So the liver cells release no glucose.
This loop cannot bring the glucose back up during the fast, so the glucose ends lower.

34
Check q14

Ana and Ben drink the same glucose drink, and the graph shows their blood glucose over two hours.

Two glucose curves after the same drink, labelled Ana and Ben, with the set point at 90 mg/dL drawn as a line of dots: the solid curve rises and returns to the set point within two hours; the dashed curve climbs steadily and stays high; the axis runs from 0 to 200 mg/dL with gridlines every 20 mg/dL
Two glucose curves after the same drink, labelled Ana and Ben, with the set point at 90 mg/dL drawn as a line of dots: the solid curve rises and returns to the set point within two hours; the dashed curve climbs steadily and stays high; the axis runs from 0 to 200 mg/dL with gridlines every 20 mg/dL

Whose negative-feedback loop is missing a part?

  1. A. Ana’s
    Ana’s glucose rose and came back to the set point.
    A loop that brings the glucose back is working.
  2. B. ✓ Ben’s

Why: Ana’s glucose rises and comes back to the set point, so her loop is working.
Ben’s glucose rises and is never brought back down.
A loop that never brings the glucose back is missing a part.
So Ben’s loop is missing a part.

35
Check q15

After a glucose drink, a person’s blood glucose climbs and stays high for two hours.

Which arm of the person’s loop is missing a part?

  1. A. The glucagon arm
    The glucagon arm raises glucose from below the set point.
    The person’s glucose sits above the set point and is never brought down, which is the insulin arm’s job.
  2. B. ✓ The insulin arm

Why: The person’s glucose is above the set point.
The arm that brings glucose down from above the set point is the insulin arm.
The person’s glucose is never brought down.
So the insulin arm is missing a part.

36
Check q16

A person with type 1 diabetes, whose insulin-releasing pancreas cells have been destroyed, injects insulin before a meal, and their glucose comes back down afterward.

Which part of the loop does the injection stand in for?

  1. A. ✓ The pancreas cells releasing insulin
  2. B. The insulin receptors on liver and muscle cells
    The receptors on the liver and muscle cells are still there; the injected insulin binds them.
    What the person lacks is the release of insulin by pancreas cells.
  3. C. The liver cells breaking glycogen down
    Breaking glycogen down raises glucose; it belongs to the glucagon arm.
    What the person lacks is the release of insulin by pancreas cells.

Why: The missing part is the pancreas cells that release insulin.
The injection puts insulin into the blood in their place.
The insulin binds its receptors on liver and muscle cells, which take glucose up, so the glucose comes back down.

37
Check q17

In a dog, glucagon is released normally between meals, yet the concentration of glucose in the dog’s blood falls low and stays low. The dog’s liver cells hold plenty of glycogen.

Which of the following is the missing part?

  1. A. The pancreas cells that release glucagon
    Glucagon is released normally, so the pancreas cells are working.
    The break comes after the glucagon reaches the liver cells.
  2. B. ✓ The glucagon receptor on the liver cells
  3. C. The glycogen stored in the liver cells
    The liver cells hold plenty of glycogen.
    The glycogen is never broken down, because the liver cells never get glucagon’s message.

Why: Glucagon is released, so the pancreas cells are working.
The glycogen is there, yet the liver cells do not break it down.
So the liver cells never got glucagon’s message.
The missing part is the glucagon receptor on the liver cells.

38
Practice writing an answer

Researchers tested whether the insulin receptor on liver and muscle cells is needed for blood glucose to return toward its set point. They used two groups of ten mice: normal mice, and mice bred so that their liver and muscle cells lack the insulin receptor. Every mouse fasted overnight, then drank the same volume of glucose solution, and the researchers measured its blood glucose two hours later. The graph shows the two means with error bars that represent ±2SE.

Two bars of blood glucose two hours after a glucose drink, one for normal mice and one for mice whose liver and muscle cells lack the insulin receptor, each with a ±2SE error bar; the axis runs from 0 to 240 mg/dL with gridlines every 10 mg/dL and labels every 20; legend: error bars represent ±2SE
Two bars of blood glucose two hours after a glucose drink, one for normal mice and one for mice whose liver and muscle cells lack the insulin receptor, each with a ±2SE error bar; the axis runs from 0 to 240 mg/dL with gridlines every 10 mg/dL and labels every 20; legend: error bars represent ±2SE

(a) Identify the independent variable and the dependent variable in this experiment. (1 pt)

Model answer The independent variable is whether the mice’s liver and muscle cells carry the insulin receptor (normal mice or receptor-lacking mice).
The dependent variable is the blood glucose concentration two hours after the drink, in mg/dL.
Rubric
  • Award 1 point for: the independent variable is the presence or absence of the insulin receptor (normal versus receptor-lacking mice) and the dependent variable is the blood glucose two hours after the drink.

Slip Naming the glucose drink as the independent variable. Every mouse drank the same solution. The thing the researchers changed between the groups is the receptor.

(b) Justify the claim that the insulin receptor is needed for blood glucose to return toward its set point, using the error bars. (1 pt)

Model answer The normal mice’s bar runs from 90 to 100 mg/dL and the receptor-lacking mice’s bar from 200 to 220 mg/dL.
The two ±2SE bars do not overlap, so the difference is very unlikely to be chance.
The mice lacking the receptor stayed far above the set point, while the normal mice returned to the set point.
Rubric
  • Award 1 point for: the ±2SE bars (90–100 and 200–220 mg/dL, read against the gridlines) do not overlap, so the higher glucose in the receptor-lacking mice is unlikely to be chance.
  • Accept: ‘the bars do not overlap, so the difference is significant’, with the two ranges read from the graph to within half a gridline.

Slip Comparing the two means alone, about 95 against about 210 mg/dL. The claim rests on the ±2SE bars not overlapping.

(c) Explain the difference you justified in (b) by following the insulin loop in each group of mice. (1 pt)

Model answer In the normal mice, insulin bound its receptor on the liver and muscle cells, so they took glucose up and the glucose returned toward the set point.
In the mice lacking the receptor, the pancreas cells detected the rise and released insulin.
But the liver and muscle cells had no receptor to bind it, so reception failed.
So the liver and muscle cells did not take glucose up.
The response never happened, so the glucose concentration stayed high.
Rubric
  • Award 1 point for: the liver and muscle cells could not bind insulin, so they did not take glucose up and the rise was not reduced (the glucose stayed high).

Slip Saying the pancreas made no insulin. The pancreas cells are normal. The loop is broken one step further on, at the receptor.

(d) Predict how the insulin concentration in the blood of the receptor-lacking mice at the two-hour measurement compares with the normal mice’s, and justify your prediction. (1 pt)

Model answer The insulin concentration is higher in the receptor-lacking mice.
Their pancreas cells detect glucose, and at two hours the glucose concentration is still high, so the pancreas cells keep releasing insulin.
In the normal mice the glucose has returned near 90 mg/dL, so the stimulus is gone and insulin release has faded.
Rubric
  • Award 1 point for: insulin is higher in the receptor-lacking mice, because their glucose is still high so the pancreas cells keep releasing insulin (the response never fades).

Slip Predicting lower insulin because the receptor is missing. The receptor is on the liver and muscle cells. The pancreas cells that release insulin are normal, and they keep answering the high glucose concentration.

APBIO-U04-L12 Contractions that get stronger

Topic 4.4 · Feedback · 78 steps

A record of the contractions of one labor drawn as bars that grow taller hour by hour, then a dashed line marked birth, and no bars after it
A record of the contractions of one labor drawn as bars that grow taller hour by hour, then a dashed line marked birth, and no bars after it

Here is a record of the contractions during one labor: each one stronger than the last, until the baby is born.

Each contraction pushes the baby’s head harder against the cervix, the opening at the base of the uterus. Each push is followed by a stronger contraction than the one before. Nothing brings the contractions back down. They build until the baby is born, and then they stop.

Why does this loop drive the change further instead of cancelling it?

Unit 4 · Cell Communication and Cell Cycle

1The labor loop, step by step

2
Check q1

After a meal, blood glucose rises. The pancreas releases insulin, and liver and muscle cells take glucose up.

What does the response do to the rise that triggered it?

  1. A. ✓ The response reduces the rise
  2. B. The response increases the rise
    Taking glucose out of the blood brings the glucose concentration down.
    So the rise is reduced, not increased.

Why: In the glucose loop the response brings the glucose back down.
So the response reduces the change that triggered it.

3

What happens when the response feeds its own stimulus?

4

In labor, each contraction stretches the cervix more. The extra stretch triggers a stronger contraction.

5

So the change grows, round after round, until the birth removes the stimulus.

6

Here is that loop, one step at a time. The stimulus is a stretch: the baby’s head stretches the cervix, the opening at the base of the uterus.

7

Cells in the cervix detect the stretch. A gland at the base of the brain then releases a hormone into the blood, and that hormone is called oxytocin.

Three boxes joined by arrows: cervix stretched, labelled stimulus; cells there detect stretch; oxytocin released, labelled response
Three boxes joined by arrows: cervix stretched, labelled stimulus; cells there detect stretch; oxytocin released, labelled response
8

Oxytocin makes the muscle of the uterus contract harder.

9

The harder contraction pushes the baby’s head harder against the cervix. So the cervix is stretched more.

Five boxes joined by arrows: cervix stretched, cells there detect stretch, oxytocin released, harder contraction, cervix stretched more
Five boxes joined by arrows: cervix stretched, cells there detect stretch, oxytocin released, harder contraction, cervix stretched more
10

More stretch releases more oxytocin. More oxytocin makes a harder contraction, and the harder contraction stretches the cervix more again.

11

The response acted back on its own trigger, so this loop is feedback. But the response did not reduce the stretch: the response increased the stretch.

The five boxes with an arrow returning from the last box to the first, labelled: the response increases the change that triggered it
The five boxes with an arrow returning from the last box to the first, labelled: the response increases the change that triggered it
12

Birth removes the stretch. Once the baby is out, nothing stretches the cervix.

13

With no stretch, the gland stops releasing extra oxytocin. The loop that was driving the contractions ends.

14

Here is one labor recorded as the pressure inside the uterus during each contraction. Each contraction is stronger than the last, and after birth there are none.

Pressure in the uterus during each contraction, in mmHg, against hours since labor began: bars rising from 30 to 76 mmHg over seven hours, a dashed line marked birth, and none after it; gridlines every 10 mmHg
Pressure in the uterus during each contraction, in mmHg, against hours since labor began: bars rising from 30 to 76 mmHg over seven hours, a dashed line marked birth, and none after it; gridlines every 10 mmHg
15

What you are expected to know Describe the labor loop step by step: the baby’s head stretches the cervix, cells there detect the stretch, a gland releases oxytocin, oxytocin makes the uterus contract harder, the harder contraction stretches the cervix more, and birth removes the stretch.

16

Video: Watch: The labor loop, step by step

The baby’s head stretches the cervix. Cells there detect the stretch, and a gland at the base of the brain releases oxytocin. Oxytocin makes the uterus contract harder, and the harder contraction stretches the cervix more. Birth removes the stretch, and the loop ends.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L12a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L12a.mp4

17
Check q2

During labor, the stretch of the cervix triggers oxytocin release, oxytocin makes the uterus contract harder, and the harder contraction pushes the baby’s head against the cervix.

What does the response do to the stretch that triggered it?

  1. A. Reduces it
    Each contraction pushes the baby’s head harder against the cervix.
    So each contraction stretches the cervix more, not less.
  2. B. Leaves it unchanged
    The harder contraction pushes the baby’s head harder against the cervix.
    So the cervix is stretched more than before.
  3. C. ✓ Increases it

Why: The stimulus is the stretch of the cervix.
The response is a harder contraction of the uterus.
The harder contraction pushes the baby’s head harder against the cervix.
So the cervix is stretched more.
The response increases the very change that triggered it.

18
Practice writing an answer

During labor, the stretch of the cervix triggers oxytocin release, oxytocin makes the uterus contract harder, and the harder contraction pushes the baby’s head against the cervix. Each contraction is stronger than the one before.

(a) Explain why each contraction of labor is stronger than the one before. (1 pt)

Model answer The stretch of the cervix is the stimulus.
Cells in the cervix detect the stretch, so a gland at the base of the brain releases oxytocin.
Oxytocin makes the uterus contract harder.
The harder contraction pushes the baby’s head harder against the cervix, so the cervix is stretched more.
More stretch releases more oxytocin, so the next contraction is harder still.
So the response increases the very change that triggered it.
Rubric
  • Award 1 point for: each contraction stretches the cervix more, more stretch releases more oxytocin, and more oxytocin makes the next contraction harder (the response increases its own trigger, so each round is stronger).
19
Check q3

The contractions of labor grow stronger for hours. Then the baby is born, and the loop that was driving the contractions ends.

Why does the loop end?

  1. A. The gland at the base of the brain has used up its oxytocin
    The gland releases extra oxytocin for as long as the cervix is stretched.
    After the birth the stretch is gone, so the extra release stops.
  2. B. The contractions have brought the stretch back to its set point
    The stretch never came back down; each contraction increased it.
    The loop went round until an outside event, the birth, removed the stimulus.
  3. C. The uterus muscle is tired and can contract no harder
    The contractions were still growing stronger at the moment of birth, so the muscle had not tired.
    The loop ended because birth removed the stretch.
  4. D. ✓ Birth removes the stretch on the cervix

Why: The loop goes round again and again until an outside event removes the stimulus.
Birth removes the stretch on the cervix.
So nothing releases extra oxytocin, and the loop ends.

20The response increases its own trigger

21

In the glucose loop, the response reduced the change that triggered it. In the labor loop, the response increased the change that triggered it.

22

So in labor each round triggers a stronger response. The loop goes round again and again until an outside event removes the stimulus.

The five boxes with an arrow returning from the last box to the first, labelled: the response increases the change that triggered it
The five boxes with an arrow returning from the last box to the first, labelled: the response increases the change that triggered it
23

When the response increases its own trigger, we call the loop , because the response reinforces the change.

24

‘Positive’ here means reinforcing, not good, just as ‘negative’ meant opposing, not harmful.

25

Positive feedback holds no set point. It drives a change on purpose, further and further from where the change started.

26

Another loop of this kind seals a cut. Platelets, small cell fragments in the blood, stick to the break.

27

The platelets release signals that make more platelets stick, until the plug closes the break and the stimulus is gone.

28

What you are expected to know Explain positive feedback: the response increases the change that triggered it, so each round triggers a stronger response, and the loop goes round until an outside event removes the stimulus or the loop uses up what it feeds on.

29

Video: Watch: The response increases its own trigger

In the glucose loop the response reduced the change that triggered it. In the labor loop the response increased it, so each round triggered a stronger response until birth removed the stimulus. A loop like that is positive feedback: ‘positive’ means reinforcing, not good, and the loop holds no set point.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L12b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L12b.mp4

30
Check q4

A student says: “Labor is positive feedback because giving birth is good for the mother and the baby.”

Is the student correct?

  1. A. Yes — a loop with a good outcome is positive feedback
    The kind is decided by what the response does to its trigger.
    Each contraction increases the stretch that triggered it: positive feedback.
    ‘Positive’ means reinforcing, not good.
  2. B. ✓ No — ‘positive’ does not mean good

Why: ‘Positive’ names what the response does to its trigger: the response increases the change.
In labor each contraction increases the stretch that triggered it: positive feedback.
Whether the outcome is good has nothing to do with it.

31
Check q5

A mother nurses twins. The two infants suckle in turn, so she is suckled about twice as often as a mother nursing one infant. Suckling triggers milk release, and the milk keeps the infants suckling.

Compared with a mother nursing one infant, predict the amount of milk she produces.

  1. A. The same amount
    Positive feedback holds no set point.
    More suckling triggers more milk release, and more release keeps more suckling going.
    So milk production climbs with the suckling.
  2. B. Less
    Suckling triggers milk release, and the released milk keeps the infants suckling.
    Each round of suckling is a stimulus for more production.
    So more suckling brings more milk, not less.
  3. C. ✓ More

Why: Suckling is the stimulus.
Milk release is the response, and the released milk keeps the infants suckling.
Twice the suckling triggers more rounds of milk release.
So the mother of twins produces more milk.
Adding to the stimulus of a positive loop drives the loop harder.

32Quick quiz: positive feedback mixed practice

33
Check q6

What is positive feedback?

  1. A. ✓ Feedback in which the response increases the change that triggered it
  2. B. Feedback in which the response reduces the change that triggered it
    A response that reduces its trigger brings the quantity back toward its set point: that is negative feedback.
  3. C. Feedback in which the outcome is good for the organism
    The kind of loop is decided by what the response does to its trigger, not by whether the outcome is good.

Why: In positive feedback the response increases the very change that triggered it.
So each round triggers a stronger response.

34
Check q7

In the name positive feedback, what does ‘positive’ mean?

  1. A. Good
    ‘Positive’ says what the response does to its trigger.
    Whether the outcome is good has nothing to do with it.
  2. B. ✓ Reinforcing
  3. C. Rising
    The regulated quantity can be rising or falling in either kind of loop.
    ‘Positive’ says that the response reinforces the change.

Why: ‘Positive’ means reinforcing: the response increases the change that triggered it.
‘Negative’ meant opposing: the response reduces the change that triggered it.

35
Check q8

Does a positive-feedback loop hold a quantity at a set point?

  1. A. Yes
    Each round of a positive-feedback loop moves the quantity further from where it started.
    Nothing brings the quantity back to a value.
  2. B. ✓ No

Why: Positive feedback drives a change further and further from where it started.
So there is no value the loop holds the quantity near: no set point.

36
Check q9

What ends a positive-feedback loop?

  1. A. The response brings the change back to its set point
    A positive-feedback loop holds no set point.
    Its response increases the change; nothing in the loop brings the change back.
  2. B. The response reverses itself
    In labor each contraction was stronger than the last, right up to birth.
    The response never reversed; birth, an outside event, removed the stretch that drove it.
  3. C. ✓ An outside event removes the stimulus

Why: In positive feedback the response increases its own trigger, so no round brings the change back.
The loop ends when an outside event, such as birth, removes the stimulus.

37
Practice writing an answer

Two kinds of feedback loop: in one, the response reduces the change that triggered it; in the other, the response increases it.

(a) State what positive feedback is. (1 pt)

Model answer Positive feedback is a loop in which the response increases the very change that triggered it.
So each round triggers a stronger response, until an outside event removes the stimulus.
Rubric
  • Award 1 point for: the response increases (reinforces) the change that triggered it.

Slip Writing that positive feedback is feedback with a good outcome. ‘Positive’ names what the response does to its trigger.

38Which kind of loop? One question

39
Check q10

Which kind of feedback brings a quantity back toward its set point?

  1. A. ✓ Negative feedback
  2. B. Positive feedback
    Positive feedback drives a change further from where it started.
    The loop that brings a quantity back toward its set point is negative feedback.

Why: In negative feedback the response reduces the change that triggered it.
So the quantity comes back toward its set point.

40

Here are the two loops side by side. They differ in one arrow.

Two three-box loops side by side: the glucose loop, glucose rises, insulin released, glucose falls, with its return arrow labelled reduces; and the labor loop, cervix stretched, oxytocin released, cervix stretched more, with its return arrow labelled increases
Two three-box loops side by side: the glucose loop, glucose rises, insulin released, glucose falls, with its return arrow labelled reduces; and the labor loop, cervix stretched, oxytocin released, cervix stretched more, with its return arrow labelled increases
41

To decide which kind of loop you are looking at, ask one question: does the response reduce the change that triggered it, or increase it?

42

  • Reduces it: negative feedback. The quantity comes back toward its set point.
  • Increases it: positive feedback. The change grows until an outside event ends it.

43

For example, after a meal insulin brings the blood glucose back down. The response reduces the change that triggered it, so this is negative feedback.

44

But in labor each contraction stretches the cervix more. The response increases the change that triggered it, so this is positive feedback.

45

And when a newborn suckles, the milk released keeps the infant suckling, and the mother makes more milk. The response increases the change that triggered it, so this is positive feedback.

46

Ethylene is the gas a ripening fruit gives off, and ethylene makes nearby fruit ripen.

47

And when a ripening pear gives off ethylene, the pears beside it ripen and give off more ethylene. The response increases the change that triggered it, so this is positive feedback.

48

This loop can end without any outside event. Once every pear in the box is ripe, no pear is left to ripen, so no round can start: the loop has used up what it feeds on.

49

But when a hot runner sweats, the sweat cools the runner, and the falling temperature removes the trigger for sweating. The response reduces the change that triggered it, so this is negative feedback.

50

The runner’s temperature is falling. But the kind of loop is decided by what the response does to its trigger, not by which way the quantity moves.

51

And when a full infant stops suckling, the milk loop ends. The response still increased the change that triggered it, so this loop was still positive feedback.

52

Ask only that one question:

  • a falling quantity can be negative feedback (sweating);
  • a good outcome can come from positive feedback (a birth);
  • a loop that stops can still be positive feedback (a full infant).

53

A positive loop ends when an outside event removes the stimulus: a birth, a full infant, a sealed cut.

54

Or it ends when the loop uses up what it feeds on, as when every pear in the box is ripe. Neither ending is what makes a loop negative.

55

Here is a table of the loops you have seen, sorted by the one question.

A table of two columns: negative feedback, the response reduces the change: glucose after a meal, sweating when hot, a Paramecium's vacuole, an enzyme's end product; positive feedback, the response increases the change: labor, nursing and milk, ripening fruit and ethylene, platelets sealing a cut
A table of two columns: negative feedback, the response reduces the change: glucose after a meal, sweating when hot, a Paramecium's vacuole, an enzyme's end product; positive feedback, the response increases the change: labor, nursing and milk, ripening fruit and ethylene, platelets sealing a cut
56

What you are expected to know Classify a described loop as negative or positive feedback by one question, whether the response reduces or increases the change that triggered it, never by whether the quantity falls, whether the outcome is good, or whether the loop stops.

57

Video: Watch: Which kind of loop? One question

To decide which kind of loop you are looking at, ask one question: does the response reduce the change that triggered it, or increase it? Insulin after a meal reduces it: negative. Labor, nursing and ripening fruit increase it: positive. Sweating reduces it even though the temperature is falling: negative.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L12c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L12c.mp4

58

Back to the record of one labor: each contraction stronger than the last, until the birth, and then none. Where did that pattern come from?

59

The stretch of the cervix triggered release of oxytocin, and oxytocin made the uterus contract harder. The harder contraction stretched the cervix more, so the response increased its own stimulus, round after round.

60

That is positive feedback. Birth removed the stretch, and the loop had nothing left to feed on.

61Quick quiz: negative or positive feedback? mixed practice

62
Check q11

A cold swimmer shivers, the shivering warms him, and the warmer body shivers less.

Which kind of feedback is this?

  1. A. ✓ Negative feedback
  2. B. Positive feedback
    The trigger is the fall in his temperature.
    Shivering warms him, so the fall shrinks.
    The response reduces the change that triggered it.

Why: The trigger is the fall in the swimmer’s temperature.
The response is shivering.
Shivering warms him, so the fall shrinks.
The response reduces the change that triggered it: negative feedback.

63
Check q12

In an emptying bladder, urine flowing through the outlet triggers a stronger squeeze, and the stronger squeeze pushes more urine through.

Which kind of feedback is this?

  1. A. Negative feedback
    The trigger is urine flowing through the outlet.
    The stronger squeeze pushes more urine through.
    So the response increases the change that triggered it.
  2. B. ✓ Positive feedback

Why: The trigger is urine flowing through the outlet.
The response is a stronger squeeze.
The stronger squeeze pushes more urine through the outlet.
The response increases the change that triggered it: positive feedback.

64
Check q13

When a blood vessel is cut, each active molecule of a clotting enzyme activates more molecules of that enzyme.

Which kind of feedback is this?

  1. A. Negative feedback
    The trigger is active clotting enzyme appearing.
    Each active molecule activates more molecules.
    So the response increases the change that triggered it.
  2. B. ✓ Positive feedback

Why: The trigger is active clotting enzyme appearing.
The response is that each active molecule activates more molecules of the same enzyme.
So the number of active molecules climbs round after round.
The response increases the change that triggered it: positive feedback.

65
Check q14

When a dog gets hot, it pants, and the cooler dog pants less.

Which kind of feedback is this?

  1. A. ✓ Negative feedback
  2. B. Positive feedback
    The trigger is the rise in the dog’s temperature.
    Panting cools the dog, so the rise shrinks.
    The response reduces the change that triggered it.

Why: The trigger is the rise in the dog’s temperature.
The response is panting.
Panting cools the dog, so the rise shrinks.
The response reduces the change that triggered it: negative feedback.

66
Check q15

A ripe apple’s ethylene makes the apples beside it ripen, and the ripening apples give off more ethylene.

Which kind of feedback is this?

  1. A. Negative feedback
    Each apple that ripens gives off ethylene, adding to the very signal that made it ripen.
    The response increases the change that triggered it.
  2. B. ✓ Positive feedback

Why: The trigger is ethylene in the air.
The response is that the apples beside the ripe one ripen.
Each ripening apple gives off more ethylene.
So the response increases the change that triggered it: positive feedback.

67
Check q16

When amino acid X piles up in a bacterium, X slows the first enzyme of its own pathway, and the bacterium makes less X.

Which kind of feedback is this?

  1. A. ✓ Negative feedback
  2. B. Positive feedback
    The trigger is the rise in X.
    Slowing the first enzyme means less X is made, so the rise shrinks.
    The response reduces the change that triggered it.

Why: The trigger is the rise in X.
The response is that X slows the first enzyme of its own pathway.
So less X is made, and the rise shrinks.
The response reduces the change that triggered it: negative feedback.

68
Check q17

An immune cell’s signal makes nearby immune cells release more of the same signal, and the signal climbs for days. Once the virus is cleared, the signal release stops. A student says: “The loop stops, so it must be negative feedback.”

Is the student correct?

  1. A. Yes — a loop that stops is negative feedback
    Every positive loop ends when an outside event, here the virus being cleared, removes the stimulus.
    Each round makes more cells release the signal: the response increases its trigger.
  2. B. ✓ No — the loop is positive feedback

Why: The trigger is the signal; the response is more immune cells releasing the signal.
So the response increases the very change that triggered it: positive feedback.
The loop ends because an outside event, the virus being cleared, removes the stimulus.
So the ending does not make the loop negative.

69Mixed practice mixed practice

70
Check q18

An infant is nursing, and each bout of suckling brings another release of milk. A door slams. The startled infant turns its head away from the breast, and milk release stops.

What ended this positive-feedback loop?

  1. A. ✓ The startled infant stopped suckling
  2. B. The mother’s milk reached its set point
    Positive feedback holds no set point.
    A positive loop goes round again and again until something outside removes the stimulus.
    Here the slammed door removed it.
  3. C. The response reduced the suckling until none was left
    Milk release keeps the infant suckling; it does not reduce the suckling.
    The slammed door, not the milk, made the infant turn away.
  4. D. The loop turned into negative feedback
    The loop stayed positive to the end.
    A slammed door removed its stimulus.

Why: A positive loop goes round again and again until an outside event removes the stimulus.
The slammed door startled the infant.
The infant turned away and stopped suckling.
So there was no trigger left for milk release, and the loop ended.

71
Check q19

Here is the record of a second labor: the pressure in the uterus during each contraction.

Pressure in the uterus during each contraction, in mmHg, against hours since labor began, for a second labor: six bars, one at each hour from hour 1 to hour 6, then a dashed vertical line marked birth and no bars after it; the axis runs from 0 to 100 mmHg with gridlines every 10
Pressure in the uterus during each contraction, in mmHg, against hours since labor began, for a second labor: six bars, one at each hour from hour 1 to hour 6, then a dashed vertical line marked birth and no bars after it; the axis runs from 0 to 100 mmHg with gridlines every 10

Which of the following does the record show?

  1. A. Each contraction was weaker than the last
    The bars rise from 25 to 70 mmHg over six hours.
    So each contraction was stronger than the one before.
  2. B. ✓ Each contraction was stronger than the last
  3. C. The contractions held a steady pressure
    A set point would show as bars of about the same height.
    These bars climb steadily.
  4. D. The contractions weakened before they stopped
    The last bar before the dashed line marked birth is the tallest.
    The contractions stopped at birth, an outside event, while they were still growing.

Why: The bars rise from 25 mmHg to 70 mmHg, one contraction after another, until birth.
So each contraction was stronger than the last: the response increased its own trigger round after round.
An outside event, the birth, ended the loop.

72
Check q20

Sweating drives a hot runner’s core temperature down. A student says: “The temperature is falling, so sweating is positive feedback.”

Is the student correct?

  1. A. Yes — a falling quantity means positive feedback
    The kind is decided by what the response does to the change that triggered it.
    The trigger was the rise in temperature, and sweating reduces that rise: negative feedback.
  2. B. ✓ No — sweating is negative feedback

Why: The kind of loop is decided by what the response does to the change that triggered it.
The trigger was the rise in the runner’s temperature.
Sweating reduces that rise.
So sweating is negative feedback.

73
Check q21

A mother nurses one infant. Suckling triggers milk release, and the milk keeps the infant suckling. When the infant is a year old, the mother stops breastfeeding, so the suckling stops.

Predict what happens to her milk production over the following weeks.

  1. A. Her milk production rises
    Suckling is the stimulus of this loop.
    Removing the stimulus removes the response with it.
  2. B. Her milk production stays at the same level
    Milk release is triggered by suckling.
    With no suckling, nothing triggers the next release of milk.
  3. C. ✓ Her milk production falls until it stops

Why: Suckling is the stimulus for milk release.
Stopping breastfeeding removes the suckling.
With no stimulus, no new round of milk release is triggered.
So milk production falls until it stops.

74
Check q22

A person stands up quickly, and their blood pressure falls. Sensors in the arteries detect the fall, the heart beats faster, and the blood pressure rises back.

Which kind of feedback is this?

  1. A. ✓ Negative feedback
  2. B. Positive feedback
    The trigger is the fall in blood pressure.
    The faster heartbeat raises the pressure back, so the fall shrinks.
    The response reduces the change that triggered it.

Why: The trigger is the fall in blood pressure.
The response is a faster heartbeat.
The faster heartbeat raises the blood pressure back, so the fall shrinks.
The response reduces the change that triggered it: negative feedback.

75
Check q23

In labor, each contraction is stronger than the one before.

Which of the following explains why?

  1. A. The gland releases oxytocin at a steady rate throughout labor
    Oxytocin is released in response to stretch, not at a steady rate.
    More stretch releases more oxytocin, and more oxytocin makes a harder contraction.
  2. B. The muscle of the uterus warms up as it works, hour after hour
    The contractions grow because each one stretches the cervix more.
    More stretch releases more oxytocin, and more oxytocin makes the next contraction harder.
  3. C. ✓ Each contraction stretches the cervix more, releasing more oxytocin

Why: Each contraction pushes the baby’s head harder against the cervix, so the cervix is stretched more.
More stretch releases more oxytocin.
More oxytocin makes the next contraction harder.
So each contraction is stronger than the one before.

76
Check q24

Consider two loops: the glucose loop after a meal, and the labor loop.

Which of them holds a quantity at a set point?

  1. A. The labor loop only
    In labor each contraction stretches the cervix more.
    Nothing brings the stretch back to a value, so there is no set point.
  2. B. Both loops
    The glucose loop brings glucose back toward 90 mg/dL.
    The labor loop drives the stretch further and further, so only the glucose loop holds a set point.
  3. C. ✓ The glucose loop only

Why: The glucose loop brings blood glucose back toward 90 mg/dL, so it holds a set point.
In the labor loop each contraction stretches the cervix more, and nothing brings the stretch back to a value.
So only the glucose loop holds a set point.

77
Practice writing an answer

At the start of a nerve impulse, the inside of a nerve cell’s membrane becomes slightly less negative. That small change opens sodium channels in the membrane. Sodium ions (Na⁺) flow into the cell, and the inside becomes more positive still, which opens more sodium channels.

(a) Explain how the opening of the sodium channels demonstrates positive feedback. (1 pt)

Model answer The trigger is the inside of the membrane becoming more positive.
The response is sodium channels opening, so sodium ions flow in.
Each sodium ion carries a positive charge, so the inside becomes more positive still.
The more positive inside opens more sodium channels.
So the response increases the very change that triggered it, which is positive feedback.
Rubric
  • Award 1 point for: the inflow of sodium ions makes the inside more positive, which is the very change that opened the channels, so the response increases its own trigger (positive feedback).

Slip Describing the sodium flow without naming what triggered it. The point is that the response increases the change that triggered it.

(b) A student says: “The inside becomes positive for only a moment and then returns to negative, so the loop is negative feedback.” Evaluate the student’s claim. (1 pt)

Model answer The student is wrong.
The kind of loop is decided by what the response does to its trigger.
Opening sodium channels makes the inside more positive, the very change that opened them.
So the sodium-channel loop is positive feedback.
The return to negative is a separate event that ends the loop; the loop did not reduce its own trigger.
Rubric
  • Award 1 point for: the claim is wrong because the response increases the change that triggered it (positive feedback); a loop that ends is not thereby negative feedback.

Slip Calling the loop negative because the change is reversed later. Whether the loop stops is not the question; what the response does to its trigger is.

Glossary

positive feedback
Feedback in which the response increases the very change that triggered it, so each round triggers a stronger response, and the loop goes round again and again until an outside event removes the stimulus, or the loop uses up what it feeds on. ‘Positive’ means reinforcing, not good.

APBIO-U04-L12B Predict what a disruption does

Topic 4.4 · Feedback · 51 steps

A photograph of a bunch of green unripe bananas on the left and a photograph of a bunch of yellow ripe bananas on the right; between them a schematic sealed container drawn as a rounded box with a lid, holding six banana shapes, five open and one filled, with five dots of ethylene in the air above them
A photograph of a bunch of green unripe bananas on the left and a photograph of a bunch of yellow ripe bananas on the right; between them a schematic sealed container drawn as a rounded box with a lid, holding six banana shapes, five open and one filled, with five dots of ethylene in the air above them

Photos: Rosendahl, public-domain-image.com via Wikimedia Commons, public domain (cropped and resized); Augustus Binu, Wikimedia Commons, CC BY-SA 3.0 (resized).

Ripening fruit gives off a gas called ethylene, and ethylene makes nearby fruit ripen. Here, a student seals five unripe bananas and one ripe banana in a container at 20 °C.

After 24 hours the air in the container holds 11.4 microliters of ethylene per liter, and all five unripe bananas have ripened. The student sets up a second container the same way, but six hours in, the student puts an absorber inside it, and the absorber removes ethylene from the air.

How many bananas ripen in the second container, and how much ethylene is in its air at 24 hours?

Unit 4 · Cell Communication and Cell Cycle

1Predict what a disruption does

2
Check q1

Ripening fruit gives off ethylene, and ethylene makes nearby fruit ripen.

Which kind of feedback is this?

  1. A. Negative feedback
    Each fruit that ripens gives off more ethylene, adding to the very signal that made it ripen.
    The response increases the change that triggered it.
  2. B. ✓ Positive feedback

Why: The trigger is ethylene in the air.
The response is that nearby fruit ripen and give off more ethylene.
The response increases the change that triggered it: positive feedback.

3
Check q2

In type 1 diabetes the pancreas cells that release insulin have been destroyed.

After a meal, which way does the person’s blood glucose move over the next hours?

  1. A. ✓ The glucose rises and stays high
  2. B. The glucose rises, then falls back to 90 mg/dL
    Insulin is the response that brings glucose back down.
    With no insulin-releasing cells, nothing brings the glucose back.
  3. C. The glucose stays at 90 mg/dL
    The meal still raises the glucose.
    What is missing is the response that would bring it back down.

Why: The part missing is the insulin-releasing cells.
Insulin is what makes liver and muscle cells take glucose up.
With no insulin, nothing brings the glucose back down.
So the glucose rises and stays high.

4

How do you predict what removing one part of a loop does?

5

You take three steps: you name the loop’s kind, you name the part removed, and you follow the loop to the quantity being measured.

6

Suppose a student seals five unripe bananas and one ripe banana in a container at 20 °C. The ripening loop now goes round inside a box whose air the student can measure.

A sealed container holding five unripe bananas and one ripe banana at the start, with a few ethylene molecules scattered in the air; a legend beneath shows a photograph of unripe bananas beside an open oval and a photograph of ripe bananas beside a filled oval
A sealed container holding five unripe bananas and one ripe banana at the start, with a few ethylene molecules scattered in the air; a legend beneath shows a photograph of unripe bananas beside an open oval and a photograph of ripe bananas beside a filled oval
7

After 24 hours the air holds 11.4 microliters of ethylene per liter, written 11.4 μL/L. All five unripe bananas have ripened.

The same container at the start and after 24 hours: at the start a few ethylene molecules and one ripe banana; after 24 hours many ethylene molecules scattered through the air and all six bananas ripe; the photograph legend beneath
The same container at the start and after 24 hours: at the start a few ethylene molecules and one ripe banana; after 24 hours many ethylene molecules scattered through the air and all six bananas ripe; the photograph legend beneath
8

In this experiment, a banana that had ethylene around it for less than about six hours stayed unripe once the ethylene was removed.

9

A banana that had ethylene around it for longer went on ripening on its own, and it gave off ethylene of its own.

10

Now consider a second container set up the same way. Six hours in, the student places an absorber inside it, and the absorber removes ethylene from the air.

11

To predict what happens, take three steps:

  1. name the loop’s kind;
  2. name the part removed;
  3. follow the loop to the quantity being measured.

12

The part removed can be any one of these:

  • the starting stimulus
  • the detecting step: the cells that detect the change
  • the signal
  • the receptor for the signal
  • the responding step: the cells that respond

13

Step 1, the loop’s kind: each ripening banana gives off ethylene, and the ethylene makes the next banana ripen. The response increases the change that triggered it, so the loop is positive feedback.

14

Step 2, the part removed: the absorber removes the ethylene in the air. The ethylene is the signal that carries each round to the next banana.

15

Step 3, follow the loop to the quantity being measured: with the signal gone, a ripening banana no longer makes its neighbors ripen, so no new round starts.

16

Only the one or two bananas that already had about six hours of ethylene go on ripening. So fewer bananas ripen, and the concentration of ethylene in the air stays low.

Two containers after 24 hours: the one with no absorber, many ethylene molecules and all six bananas ripe; the one with an absorber block added at six hours, few ethylene molecules and only the original ripe banana plus one more ripe; the photograph legend beneath
Two containers after 24 hours: the one with no absorber, many ethylene molecules and all six bananas ripe; the one with an absorber block added at six hours, few ethylene molecules and only the original ripe banana plus one more ripe; the photograph legend beneath
17

At 24 hours the second container holds 1.6 μL/L of ethylene, and one more banana is ripe. The prediction was right.

18

The same three steps predict a disruption in labor.

19

In the labor loop, the muscle of the uterus contracts harder only when oxytocin binds its receptors on the muscle cells. With no new oxytocin, the contractions fade.

20

Any contraction that stretches the cervix restarts the loop: the stretch releases the woman’s own oxytocin, whatever caused that first contraction.

21

What you are expected to know Predict which way a regulated quantity moves when one part of its feedback loop is removed or added, by naming the loop’s kind, naming the part removed and following the loop to that quantity.

22

Video: Watch: Predict what a disruption does

To predict what removing one part of a loop does, take three steps: name the loop’s kind, name the part removed, and follow the loop to the quantity being measured. Ripening is positive feedback carried by ethylene; an absorber removes the ethylene; so no new round starts, fewer bananas ripen and the ethylene stays low.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L12Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L12Ba.mp4

23
Check q3

During labor, a doctor gives a drug that blocks the oxytocin receptors on the muscle cells of the uterus.

Which kind of loop is the drug acting on?

  1. A. Negative feedback
    In labor each contraction increases the stretch that triggered it.
    The response increases the change.
    So the labor loop is positive feedback.
  2. B. ✓ Positive feedback

Why: To predict a disruption, first name the loop’s kind.
In labor each contraction increases the stretch that triggered it.
The response increases the change that triggered it.
So the drug is acting on a positive-feedback loop.

24
Check q4

During labor, a drug blocks the oxytocin receptors on the muscle cells of the uterus. The cervix is stretched and oxytocin is released as before.

Does the uterus muscle contract harder?

  1. A. Yes
    Oxytocin acts only by binding its receptor on the uterus muscle cells.
    The drug has blocked those receptors, so the muscle is never told to contract harder.
  2. B. ✓ No

Why: The part removed is the receptor for oxytocin on the uterus muscle.
Oxytocin acts only by binding that receptor.
With the receptors blocked, the oxytocin cannot bind.
So the muscle is never told to contract harder, and it does not.

25
Check q5

A drug is blocking the oxytocin receptors on the muscle cells of a woman’s uterus during labor. The doctor follows her labor over the next hour.

Predict what happens to the contractions.

  1. A. The contractions grow stronger
    The muscle does not contract harder, so the cervix is stretched no further.
    With no extra stretch, no extra oxytocin is released.
    The loop has nothing to feed on.
  2. B. The contractions continue unchanged
    The stretch triggers oxytocin release, and oxytocin makes the muscle contract.
    The muscle does not feel the stretch itself.
    With the receptors blocked, the muscle receives no message.
  3. C. ✓ The contractions fade

Why: Follow the loop to the contractions.
The muscle does not contract harder, so the cervix is stretched no further.
So no extra oxytocin is released.
The loop has nothing to feed on.
So the contractions fade.

26
Check q6

A student seals five unripe bananas and one ripe banana in a container at 20 °C, with an ethylene absorber present from the very start.

Predict how many of the five unripe bananas are ripe after 24 hours.

  1. A. All five
    Each round needs ethylene in the air to reach the next fruit.
    The absorber removes the ethylene as fast as the ripe banana releases it, so no round starts.
  2. B. About half
    The absorber does not slow the ethylene down.
    The absorber removes the ethylene before it reaches the other bananas.
    So no round starts at all.
  3. C. ✓ Few or none

Why: The part removed is the ethylene in the air, the signal that carries each round.
The absorber removes the ethylene as soon as the ripe banana gives it off.
So the ripe banana cannot start the first round.
So few or none ripen, and the ethylene concentration stays low.

27
Check q7

A student seals six unripe bananas and no ripe banana in a container at 20 °C, with no absorber.

Predict how many of the six bananas are ripe after 24 hours.

  1. A. ✓ Few or none
  2. B. About half
    A round starts when a ripe fruit gives off ethylene.
    No banana in this container is ripe.
    So no banana gives off the ethylene that would start the first round.
  3. C. All six
    The ripe banana was the starting stimulus: the ripe banana gave off the ethylene that started the first round.
    With no ripe banana, nothing starts the loop.

Why: The loop has no starting stimulus.
A round starts when a ripe fruit gives off ethylene, and no banana in the container is ripe.
So nothing releases the ethylene that would start the first round.
So few or none ripen within 24 hours, and the ethylene concentration stays low.

28

Back to the two sealed containers, each holding five unripe bananas and one ripe banana at 20 °C. In the first, the ethylene climbed to 11.4 μL/L in 24 hours, and all five unripe bananas ripened.

29

In the second, the absorber went in at six hours and removed the ethylene from the air.

30

Ripening is positive feedback carried by ethylene, so with the signal gone no new round started. One more banana ripened, and the air held 1.6 μL/L.

31Quick quiz: the three steps mixed practice

32
Check q8

Suckling triggers milk release in a nursing mother, and the milk keeps the infant suckling. Suppose a drug stops the nerves in her nipple from detecting suckling.

Which kind of feedback is the nursing loop?

  1. A. Negative feedback
    The milk released keeps the infant suckling, which releases more milk.
    The response increases the change that triggered it.
  2. B. ✓ Positive feedback

Why: Suckling is the trigger, and milk release is the response.
The released milk keeps the infant suckling.
The response increases the change that triggered it: positive feedback.

33
Check q9

Suckling triggers milk release in a nursing mother. A drug stops the nerves in her nipple from detecting suckling.

Which part of the loop does the drug remove?

  1. A. The starting stimulus
    The infant still suckles, so the starting stimulus is there.
    The drug acts on the nerves, which no longer detect the suckling.
  2. B. ✓ The detecting step
  3. C. The responding step
    The breast can still release milk, so the responding step is intact.
    The drug acts on the nerves that detect the suckling.

Why: The drug acts on the nerves in the nipple.
Those nerves are the cells that detect the suckling.
So the part removed is the detecting step.

34
Check q10

Suckling triggers milk release in a nursing mother. A drug stops the nerves in her nipple from detecting suckling.

Predict what happens to the milk released at her next feeds.

  1. A. ✓ Less milk is released
  2. B. The same amount of milk is released
    Milk release is triggered only when suckling is detected.
    With the detecting step gone, no release is triggered.
  3. C. More milk is released
    The loop needs the detecting step to start each round.
    With that step gone, no round starts.

Why: The suckling is no longer detected.
So no milk release is triggered.
With no release, the loop has nothing to go round on.
So the milk released falls.

35
Check q11

When a runner gets hot, sweat glands release sweat, the sweat cools the runner, and the falling temperature removes the trigger for sweating. Suppose a drug stops the sweat glands from releasing sweat.

Which kind of feedback is the sweating loop?

  1. A. ✓ Negative feedback
  2. B. Positive feedback
    The trigger is the rise in temperature.
    Sweating cools the runner, so the rise shrinks.
    The response reduces the change that triggered it.

Why: The trigger is the rise in the runner’s temperature.
The response is sweating, which cools the runner.
The response reduces the change that triggered it: negative feedback.

36
Check q12

When a runner gets hot, sweat glands release sweat and the sweat cools the runner. A drug stops the sweat glands from releasing sweat.

Which part of the loop does the drug remove?

  1. A. The starting stimulus
    The runner still gets hot, so the starting stimulus is there.
    The drug acts on the sweat glands, which release no sweat.
  2. B. The detecting step
    The rise in temperature is still detected.
    The drug acts on the sweat glands, so the response never comes.
  3. C. ✓ The responding step

Why: The drug acts on the sweat glands.
The sweat glands are the cells that respond, by releasing sweat.
So the part removed is the responding step.

37
Check q13

When a runner gets hot, sweat glands release sweat and the sweat cools the runner. A drug stops the sweat glands from releasing sweat, and the runner keeps going.

Predict the runner’s core temperature over the next half hour.

  1. A. ✓ The runner’s core temperature keeps rising
  2. B. The runner’s core temperature stays where it was
    The runner’s muscles keep producing heat.
    With no sweat to carry the heat away, the temperature does not hold still.
  3. C. The runner’s core temperature falls back to 37 °C
    Sweat is what cools the runner.
    With no sweat, nothing brings the temperature back down.

Why: The runner’s muscles keep adding heat to the body.
Sweat is the response that carries the heat away.
With the response gone, nothing brings the temperature back down.
So the core temperature keeps rising.

38
Check q14

Ripening fruit gives off ethylene, and ethylene makes nearby fruit ripen. A grocer puts one ripe banana in a box of unripe avocados.

Which kind of feedback is the ripening loop?

  1. A. Negative feedback
    Each fruit that ripens gives off more ethylene, adding to the signal that made it ripen.
    The response increases the change that triggered it.
  2. B. ✓ Positive feedback

Why: The trigger is ethylene in the air.
The response is that nearby fruit ripen and give off more ethylene.
The response increases the change that triggered it: positive feedback.

39
Check q15

Ripening fruit gives off ethylene, and ethylene makes nearby fruit ripen. A grocer puts one ripe banana in a box of unripe avocados.

Which part of the loop does the ripe banana add?

  1. A. ✓ The starting stimulus
  2. B. The responding step
    The avocados themselves carry out the response.
    The ripe banana supplies the ethylene that starts the first round.
  3. C. The receptor for the signal
    The avocados already have their receptors for ethylene.
    The ripe banana supplies the ethylene that starts the first round.

Why: The ripe banana gives off ethylene into the air of the box.
Ethylene in the air is the signal that starts the first round of ripening.
So the ripe banana adds the starting stimulus.

40
Check q16

A grocer puts one ripe banana in a box of unripe avocados. A second box holds unripe avocados alone.

Compared with the second box, predict how many avocados in the first box are ripe after two days.

  1. A. Fewer
    The ripe banana gives off ethylene, which makes the avocados beside it ripen.
    It adds a starting stimulus; it removes nothing.
  2. B. The same number
    The box with no ripe fruit has no ethylene to start the first round.
    The ripe banana supplies that ethylene, so the loop starts.
  3. C. ✓ More

Why: The ripe banana gives off ethylene into the box.
The ethylene makes the avocados beside it ripen, and the ripening avocados give off more ethylene.
The box with no ripe fruit has nothing to start the first round.
So more avocados are ripe in the box with the banana.

41
Check q17

In pond water, water flows into a Paramecium, and the cell’s contractile vacuole fills and empties to push the water back out. Now imagine a Paramecium whose contractile vacuole has stopped emptying.

Which kind of feedback is the vacuole loop?

  1. A. ✓ Negative feedback
  2. B. Positive feedback
    The trigger is water building up in the cell.
    The vacuole pushes water out, so the build-up shrinks.
    The response reduces the change that triggered it.

Why: The trigger is water building up inside the cell.
The response is the vacuole emptying water out.
The response reduces the change that triggered it: negative feedback.

42
Check q18

In pond water, water flows into a Paramecium, and the cell’s contractile vacuole fills and empties to push the water back out. Imagine a Paramecium whose contractile vacuole has stopped emptying.

Which part of the loop is missing?

  1. A. The starting stimulus
    Water still flows into the cell from the pond, so the starting stimulus is there.
    What the cell has lost is the emptying that pushes the water out.
  2. B. ✓ The responding step
  3. C. The detecting step
    The cell still fills its vacuole as water builds up, so the water is detected.
    What is missing is the emptying that pushes the water out.

Why: The vacuole’s job in the loop is to empty water out of the cell.
That emptying is the loop’s response.
So the part missing is the responding step.

43
Check q19

In pond water, water flows into a Paramecium, and the cell’s contractile vacuole fills and empties to push the water back out. Imagine a Paramecium whose contractile vacuole has stopped emptying.

Predict the amount of water in the cell over the next minutes.

  1. A. ✓ The amount of water in the cell keeps rising
  2. B. The amount of water in the cell stays the same
    Pond water keeps flowing into the cell.
    The vacuole is what balanced that inflow, and the vacuole has stopped emptying.
  3. C. The amount of water in the cell falls
    Water keeps flowing in from the pond.
    With the vacuole no longer emptying, nothing pushes the water out.

Why: Water keeps flowing into the cell from the pond.
The vacuole is the response that pushes water out.
With the response gone, nothing removes the water.
So the amount of water in the cell keeps rising.

44Mixed practice mixed practice

45
Check q20

A woman’s labor has stalled: her contractions are weak and are growing no stronger. Her cervix detects stretch normally, and her own oxytocin release is normal, so her own loop works once contractions are strong enough to stretch the cervix. A doctor gives her oxytocin through a drip.

Predict what happens to the contractions.

  1. A. The contractions stay weak
    The receptors on the uterus muscle are intact.
    The loop stalled for lack of oxytocin, and the drip supplies oxytocin.
    So the muscle contracts harder.
  2. B. The contractions grow stronger only while the drip is connected
    The added oxytocin makes the uterus contract harder, which stretches the cervix more.
    The stretch triggers the woman’s own oxytocin release, so the loop feeds itself again.
  3. C. ✓ The contractions grow stronger and keep growing
  4. D. The contractions grow stronger once, then stop
    The stronger contraction stretches the cervix more.
    The extra stretch triggers the woman’s own oxytocin release.
    So the loop goes round again.

Why: The added oxytocin supplies the response.
The uterus contracts harder.
The harder contraction stretches the cervix more.
More stretch releases more of the woman’s own oxytocin.
So the positive loop feeds itself again, and the contractions keep growing.

46
Check q21

A student seals five unripe avocados and one ripe avocado in a container at 20 °C and puts an ethylene absorber inside after four hours. An unripe avocado needs about six hours with ethylene around it before it goes on ripening on its own.

Predict how many of the five unripe avocados are ripe after 24 hours.

  1. A. All five
    The absorber removes the ethylene after four hours.
    An avocado needs six hours with ethylene before it ripens on its own, so every unripe avocado stays unripe.
  2. B. About half
    Half would need round after round of ripening.
    The absorber went in at four hours, less than the six an avocado needs, so no unripe avocado goes on ripening.
  3. C. ✓ Few or none

Why: The absorber removes the ethylene that carries each round.
It went in at four hours, so each unripe avocado had ethylene around it for only four hours.
An avocado needs about six hours before it goes on ripening on its own.
So few or none of the five ripen.

47
Check q22

During a labor, a doctor gives a drug that stops the cells of the cervix from detecting stretch. Oxytocin receptors and the uterus muscle are unaffected.

Predict what happens to the contractions.

  1. A. The contractions grow stronger
    A stretch that is not detected triggers nothing.
    The loop needs the detecting step to start each round.
  2. B. The contractions continue unchanged
    Extra oxytocin is released when stretch is detected.
    With detection blocked, no extra oxytocin is released, and the oxytocin already in the blood is soon gone.
  3. C. The contractions become regular and steady
    The loop has been broken, not changed.
    Without its detecting step, no round can start.
  4. D. ✓ The contractions weaken and stop

Why: The loop is positive feedback.
The part removed is the detecting step.
With the stretch undetected, no extra oxytocin is released.
So the uterus does not contract harder.
So the contractions fade and stop.

48
Check q23

In an experiment, which of the following is the independent variable?

  1. A. The quantity measured to see the effect
    The quantity measured to see the effect is the dependent variable.
  2. B. ✓ The condition the experimenter deliberately changes
  3. C. The condition kept the same in every group
    A condition kept the same in every group is a controlled variable.

Why: The independent variable is the condition the experimenter deliberately changes.
The dependent variable is the quantity measured to see the effect of that change.

49
Check q24

Which of the following is a null hypothesis?

  1. A. ✓ The tested factor makes no difference to the measured result
  2. B. The tested factor increases the measured result
    A prediction of a difference in one direction is a hypothesis, not the null hypothesis.
  3. C. The result the experimenter hopes to see
    The null hypothesis is a statement about the tested factor, not a hope about the result.

Why: The null hypothesis states that the tested factor makes no difference to the measured result.
The experiment’s data are used to decide whether that statement is rejected.

50
Practice writing an answer

Ripening fruit gives off ethylene, a gas that makes nearby fruit ripen. A student set up three kinds of sealed container, five of each kind, and kept them at 20 °C for 24 hours. Container A held five unripe bananas and one ripe banana. Container B held the same, and the student added an ethylene absorber after six hours. Container C held five unripe bananas and no ripe banana. After 24 hours the student measured the ethylene in the air of each container and counted the ripe bananas. In each of the A containers all five unripe bananas had ripened; in the B containers one or two had; in the C containers none had. The graph shows the mean ethylene for each kind of container, with error bars that represent ±2SE.

Three bars of ethylene in the air after 24 hours, for container A (one ripe and five unripe bananas), container B (the same, with an absorber from six hours) and container C (five unripe bananas only), each with a ±2SE error bar; the axis runs from 0 to 14 μL/L with gridlines every 1 μL/L and labels every 2; legend: error bars represent ±2SE
Three bars of ethylene in the air after 24 hours, for container A (one ripe and five unripe bananas), container B (the same, with an absorber from six hours) and container C (five unripe bananas only), each with a ±2SE error bar; the axis runs from 0 to 14 μL/L with gridlines every 1 μL/L and labels every 2; legend: error bars represent ±2SE

(a) Identify the independent variable in this experiment. (1 pt)

Model answer The independent variable is what each container held: a ripe banana with no absorber (container A), a ripe banana with an ethylene absorber added at six hours (container B), or no ripe banana (container C).
Rubric
  • Award 1 point for: the independent variable is the container treatment, that is whether a ripe banana is present and whether an ethylene absorber is added at six hours.
  • Accept: ‘whether an ethylene absorber is present’ when the answer is framed on containers A and B alone.

Slip Naming the ethylene concentration as the independent variable. The ethylene is what was measured. The student changed what each container held.

(b) State the null hypothesis for the comparison between containers A and B. (1 pt)

Model answer The null hypothesis is that adding the ethylene absorber at six hours makes no difference to the ethylene in the air at 24 hours: the mean ethylene in the B containers equals the mean ethylene in the A containers.
Rubric
  • Award 1 point for: a no-difference statement that names the factor changed (the absorber) and the quantity measured (the ethylene in the air at 24 hours, or the number of ripe bananas).
  • Accept: ‘the absorber has no effect on how many bananas ripen’. Do not award the point for a prediction of a difference in either direction.

Slip Writing the expected result, ‘the absorber lowers the ethylene’, as the null. The null hypothesis predicts no difference.

(c) Explain why the student included container C. (1 pt)

Model answer Container C is the control.
Container C has everything the other containers have except the ripe banana, the factor that starts the loop.
Container C shows how much ethylene five unripe bananas give off on their own and how many ripen on their own.
So any extra ethylene or ripening in container A can be credited to the ripe banana.
Rubric
  • Award 1 point for: container C is the control, lacking the ripe banana, so it shows what happens with the starting stimulus absent and lets the ripening in A be credited to the ripe banana.

Slip Calling C a repeat of A. A control lacks the tested factor. Container C has no ripe banana at all, and that is its job.

(d) Justify the claim that the absorber lowered the ethylene in the air at 24 hours, using the error bars. (1 pt)

Model answer Container A’s bar runs from about 10.4 to 12.4 μL/L and container B’s from about 1.2 to 2.0 μL/L.
The two ±2SE bars do not overlap, so the difference is very unlikely to be chance, and the null hypothesis that the absorber makes no difference is rejected.
Rubric
  • Award 1 point for: the ±2SE bars for A (about 10.4–12.4 μL/L) and B (about 1.2–2.0 μL/L), read against the gridlines, do not overlap, so the lower ethylene with the absorber is unlikely to be chance.
  • Accept: readings within half a gridline of those values. Do not award the point for a comparison of the two means alone.

Slip Comparing the two means alone, about 11.4 against about 1.6 μL/L. The claim rests on the ±2SE bars not overlapping.

(e) Predict how many of the five unripe bananas in container B would be ripe at 24 hours if the absorber had been present from the start instead of from six hours, and justify your prediction by naming the kind of feedback loop involved. (1 pt)

Model answer None of the five: fewer than with the absorber added at six hours.
Ripening is a positive-feedback loop: each ripening fruit gives off more ethylene, which makes the next fruit ripen.
With the absorber present from the start, the ripe banana’s ethylene is removed before it reaches the others, so no round starts.
A banana needs about six hours with ethylene before it ripens on its own, and with the absorber present from the start none gets that time.
Rubric
  • Award 1 point for: fewer bananas ripen than with the absorber added at six hours (none of the five unripe bananas; accept at most one), because the loop is positive feedback carried by ethylene in the air and removing the ethylene from the start stops the first round.
  • Accept: six hours is about the time a banana needs with ethylene around it before it ripens on its own, so at six hours one or two had passed that time and from the start none would.

Slip Predicting the same result as before. Six hours of ethylene had already started the ripening of one or two bananas. With the absorber present from the start, there is no such head start.

APBIO-U04-P44 Practice questions: Topic 4.4

Topic 4.4 · Feedback · 9 MCQ · 2 FRQ · for APBIO-U04-T44

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. The first free-response question walks you through the reasoning one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Where a graph carries error bars, the caption says what the bars represent. Blood glucose is in milligrams per deciliter (mg/dL).

Video: Watch first: Topic 4.4 summary: feedback

Feedback is the response acting back on its own trigger; negative feedback reduces the change and holds a set point; positive feedback increases the change and holds none.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-T44-summary.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-T44-summary.mp4

Q1 P44-q01

Minutes after birth, a newborn's core temperature falls from 37.0 °C to 36.2 °C. Blood vessels in its skin narrow, and a special fat store in its back begins to burn, releasing heat. Its temperature rises back to 37.0 °C.

Which of these is the response in this loop?

  1. A. The fall in core temperature to 36.2 °C
    The fall is the stimulus: the change that moved the temperature away from its set point.
    The response is what the body did about it.
  2. B. ✓ The narrowed skin vessels and the burning fat store
  3. C. The 37.0 °C the newborn's body holds near
    37.0 °C is the set point, the value the temperature is held near.
    The response is an action, not a value.
  4. D. The cold air of the room
    The cold air caused the stimulus.
    The response is what the newborn's body did about the fall: narrowing its skin vessels and burning fat.

Why: The response is what the body does about a stimulus.
The fall to 36.2 °C was the stimulus, 37.0 °C is the set point, and the narrowing of the skin vessels and the burning of the fat store, which released heat, was the response.

Q2 P44-q02

A healthy adult's blood calcium is measured every six hours for two days. The readings are in the table. A student says: “These readings show that the calcium is out of control.”

Blood calcium of a healthy adult, measured every six hours for two days.
Blood calcium of a healthy adult, measured every six hours for two days.

Is the student correct, and why?

  1. A. Yes — a quantity under homeostatic control is held at exactly one value
    A quantity under homeostatic control is held near a value, never exactly at it.
    The calcium drifts a little from about 2.5 mmol/L and is brought back: the loop working.
  2. B. No — homeostasis is working, because the set point moves to match each new reading
    The set point stays near 2.5 mmol/L.
    Each reading drifts a little away from the set point and is brought back toward it; the set point itself does not move.
  3. C. Yes — each reading is a new stimulus, and the body never brings the calcium back
    Each drift is brought back: 2.6 was followed by 2.3, then 2.5.
    Uncorrected drifts would wander ever further from 2.5 mmol/L; these stay within 0.2 mmol/L.
  4. D. ✓ No — the calcium is under control, because each small drift away from about 2.5 mmol/L is brought back

Why: Homeostasis holds a quantity near its set point, never exactly on it.
The calcium drifts up to 2.6 mmol/L and down to 2.3 mmol/L and is brought back toward 2.5 mmol/L each time.
The scatter of readings is the loop working, so the student's claim is wrong.

Q3 P44-q03

After a hard sprint, the carbon dioxide in a runner’s blood has risen well above the value her body holds it near, and she is breathing fast and deep. Five minutes after she stops, her breathing is back to normal.

Why has her fast breathing stopped?

  1. A. ✓ The carbon dioxide in her blood is back near its set point, so her brain detects no excess
  2. B. Her lungs have breathed out all the carbon dioxide her body holds
    Her body makes carbon dioxide all the time; it is never all gone.
    Fast breathing stopped because the carbon dioxide was back near its set point.
  3. C. Her heart rate has fallen, and the slower heartbeat switched the fast breathing off
    The heartbeat does not switch breathing off.
    Fast breathing is triggered by excess carbon dioxide, and it faded because the excess was gone.
  4. D. Her body has raised the value it holds carbon dioxide near, so the excess no longer counts as a stimulus
    The set point stayed where it was.
    The carbon dioxide came back down to it, so the excess that triggered the fast breathing was gone.

Why: In negative feedback the response fades as the stimulus shrinks.
Fast, deep breathing carried carbon dioxide out of the blood.
The carbon dioxide fell back toward its set point.
Once the excess was gone, the brain triggered no more fast breathing.

Q4 P44-q04

Four negative-feedback loops are described in the table below.

Four negative-feedback loops.
Four negative-feedback loops.

Which of the following loops operates at the cellular level?

  1. A. The bacterium’s loop
    The loop is inside a cell, but what detects and responds is one enzyme molecule bound by its own product: the molecular level.
  2. B. The horse’s loop
    Insulin travels in the blood from the pancreas to liver and muscle cells: organs working together, the organismal level.
    Cellular responders do not make a loop cellular.
  3. C. ✓ The yeast cell’s loop
  4. D. The dog’s loop
    Panting uses the lungs, mouth and blood of a whole animal to shed heat: the organismal level.

Why: The level is set by what detects and what responds.
In the yeast cell’s loop, one cell detects its own acidity and pumps ions out: cellular.
The bacterium’s loop is one enzyme: molecular.
The dog’s and the horse’s loops use organs across a whole body: organismal.

Q5 P44-q05

In one breed of pig, a mutation changes the intracellular domain of the insulin receptor: insulin binds the receptor normally, but the intracellular domain no longer changes shape. The pigs’ pancreas cells and insulin are normal. A pig of this breed eats a meal of grain.

Which of the following predicts the pig’s blood glucose concentration three hours later, compared with a pig whose receptors are normal, with the correct reason?

  1. A. The same, because insulin still binds the receptor
    Binding alone passes nothing inside; the intracellular domain must change shape to pass the message inside.
    Here it does not, so the cells act as if no insulin arrived.
  2. B. ✓ Higher, because liver and muscle cells are never told to take glucose up
  3. C. Lower, because bound insulin stays on the receptor longer
    How long insulin stays bound changes nothing if the message never crosses the membrane.
    Glucose leaves the blood only when the cells are told to take it up.
  4. D. The same, because the set point has not changed
    A set point is a value the loop defends, not a mechanism.
    The loop is broken inside the target cells, so the glucose concentration is not brought back.

Why: Follow the loop: glucose rises after the meal, pancreas cells release insulin, insulin binds, but the intracellular domain never changes shape, so signal transduction fails and the liver and muscle cells do not take glucose up.
The response is missing, and the glucose concentration stays higher.

Q6 P44-q06

A cell has received the signal to dismantle itself. Inside it, a few molecules of a cutting enzyme are switched on. Each active cutting enzyme switches on more molecules of that enzyme, and within an hour thousands are active and the cell has taken itself apart.

Which of the following classifies this loop, with the correct reason?

  1. A. The loop ends once every enzyme molecule is active, so it is negative feedback
    Every positive-feedback loop ends when an outside limit removes the stimulus.
    Here, every molecule of the enzyme has been switched on.
    The ending does not make the loop negative feedback.
  2. B. Each active enzyme switches off the enzyme that activated it, so the number is held steady: negative feedback
    Nothing switches an active enzyme off, and nothing holds the number steady.
    Each active enzyme switches on more, so the number climbs from a few to thousands.
  3. C. Dismantling the cell is bad for it, so the loop is negative feedback
    The kind of loop is decided by what the response does to its trigger.
    Each active enzyme switches on more enzyme, so the response increases its trigger: positive feedback.
  4. D. ✓ Each active enzyme switches on more of its own kind, so the number of active enzymes climbs round after round: positive feedback

Why: The stimulus is active cutting enzyme, and the response is switching on more of the same enzyme.
The response increases the very change that triggered it, so each round is stronger: positive feedback.
The loop repeats until an outside limit, the cell having no more enzyme to activate, ends it.

Q7 P44-q07

When a person is under stress, a gland at the base of the brain releases a signal that makes the adrenal glands release the hormone cortisol. Cortisol in the blood acts on the cells of the brain gland and makes them release less of the signal. The cortisol level rises during the stress and then levels off.

Which of the following classifies this loop, with the correct reason?

  1. A. ✓ Cortisol acts back on the brain gland, so the gland releases less signal and the cortisol levels off: negative feedback
  2. B. Cortisol acts back on the brain gland, so the gland releases more signal and the cortisol climbs during stress: positive feedback
    Cortisol makes the brain gland release less of the signal, not more.
    So the response reduces its own trigger, and the cortisol levels off instead of climbing without limit.
  3. C. Cortisol helps the body cope with stress, so the loop is positive feedback
    The kind of loop is decided by what the response does to its trigger.
    Cortisol reduced the release of the signal that triggered it, so the loop is negative feedback.
  4. D. Stress is harmful, so the loop is negative feedback
    'negative' means opposing, not harmful.
    The loop is negative feedback because cortisol opposes the release of the signal that triggered it.

Why: Ask what the response does to its trigger.
The signal triggered cortisol release, and cortisol reduced the release of that signal.
The response opposed its trigger: negative feedback, which is why the cortisol levels off instead of climbing without limit.

Q8 P44-q08

During labor, each contraction of the uterus is stronger than the one before it.

Which of the following is the step by which one contraction leads to a stronger one?

  1. A. The contraction squeezes oxytocin out of the muscle of the uterus into the blood
    Oxytocin is released by a gland, not squeezed from the muscle.
    The stretch of the cervix by the baby’s head triggers the release.
  2. B. The contraction raises the mother’s blood pressure, and the higher pressure makes the heart pump oxytocin faster
    Blood pressure plays no part in the loop.
    The stretch of the cervix by the baby’s head triggers the release of more oxytocin.
  3. C. ✓ The contraction pushes the baby’s head against the cervix, and the stretch triggers the release of more oxytocin
  4. D. The contraction pushes the baby’s head back, which relieves the stretch on the cervix
    Each contraction stretches the cervix more, not less.
    Birth is what finally removes the stretch and ends the loop.

Why: The baby’s head stretches the cervix.
The stretch triggers the release of the hormone oxytocin.
Oxytocin makes the uterus contract harder.
The harder contraction stretches the cervix more, so more oxytocin is released and the next contraction is stronger still.

Q9 P44-q09

Some mice are bred so that the cells of their brain gland lack the receptor for cortisol; their adrenal glands and their cortisol are normal. A researcher keeps six normal mice and six receptor-lacking mice in a narrow tube for thirty minutes, a mild stress, and then measures the cortisol in their blood. The receptor-lacking mice’s mean cortisol concentration is far higher than the normal mice’s, and the two ±2SE ranges lie clear of each other.

Which of the following explains the higher cortisol concentration in the receptor-lacking mice?

  1. A. Their adrenal glands released cortisol on their own, without waiting for any signal from the brain gland
    The adrenal glands release cortisol only on the brain gland’s signal.
    The signal kept coming because the brain gland, with no cortisol receptor, never got cortisol’s message.
  2. B. ✓ Cortisol could not act on the brain gland, so the signal kept coming and the adrenal glands kept releasing cortisol
  3. C. Their set point for cortisol is higher, so their loop held the cortisol near that higher value
    The missing receptor broke the loop, not the set point.
    The brain gland never receives cortisol’s message and keeps releasing its signal, so cortisol climbs.
  4. D. Cortisol that cannot bind its receptor is not broken down, so the cortisol built up in the blood over the thirty minutes
    Binding a receptor is not how a hormone is removed; cortisol’s breakdown is unchanged in these mice.
    The cortisol climbed because the brain gland kept releasing its signal.

Why: The part removed is the brain gland’s cortisol receptor.
Normally cortisol binds it and the brain gland releases less signal, so cortisol levels off.
Here the message never arrives, so the signal keeps coming and cortisol keeps rising.
The ±2SE ranges lie clear, so the difference is not chance.

FRQ 1 P44-frq1 · Scientific Investigation scaffolded

Cells of a red alga make a red pigment in three enzyme steps: enzyme 1 turns A into B, enzyme 2 turns B into C, and enzyme 3 turns C into the pigment. When the pigment is plentiful it binds enzyme 1 at a site away from the active site. Students broke open alga cells to make an extract that contains all three enzymes, and measured how fast enzyme 1 made B. Six tubes received extract and A; six tubes received extract, A and a high concentration of the pigment. Everything else, including the temperature, was the same in every tube. The graph below shows the mean rate at which B formed in each set of tubes; the error bars represent ±2SE.

Mean rate at which enzyme 1 made B in alga extract, with and without added pigment; six tubes in each set, same temperature and same amount of A. Error bars represent ±2SE. Gridlines every 1 μmol/min.
Mean rate at which enzyme 1 made B in alga extract, with and without added pigment; six tubes in each set, same temperature and same amount of A. Error bars represent ±2SE. Gridlines every 1 μmol/min.

(a) Identify the independent variable and the dependent variable in this experiment. (1 pt)

Frame The independent variable is …, and the dependent variable is ….

Hint Which one thing did the students choose to make different between the two sets of tubes, and which quantity did they measure?

Model answer The independent variable is whether the pigment was added to the tube.
The dependent variable is the rate at which enzyme 1 made B, in μmol/min.
Rubric
  • Award 1 point for: independent variable, whether pigment was added to the tube (present or absent); dependent variable, the rate at which B formed, in μmol/min.
  • Do not award the point if the variables are reversed, or if the temperature, the extract or the substance A is named as the independent variable.

Slip Naming the extract or the substance A as the independent variable. Every tube had the same extract and the same A. The students changed only whether the pigment was present.

(b) State the null hypothesis for this experiment. (1 pt)

Frame There is no difference in … between tubes … and tubes ….

Hint A null hypothesis names the factor the students changed and the quantity they measured, and it predicts no difference.

Model answer There is no difference in the rate at which B forms between tubes with added pigment and tubes with none.
Rubric
  • Award 1 point for: there is no difference in the rate at which B forms between tubes with added pigment and tubes with none.
  • Do not award the point for a prediction of a difference in either direction.

Slip Writing the expected result, 'the pigment slows the pathway', as the null. The null predicts no difference.

(c) Support the claim that the pigment slows enzyme 1, using evidence from the error bars. (1 pt)

Frame The bar for extract and A runs from … to … μmol/min and the bar with added pigment from … to …, so the bars …, and ….

Hint The caption tells you what each bar represents. Compare the ends of the two bars before you compare the means, and recall what the overlap rule says about bars of that kind.

Model answer The bar for extract and A runs from 11.0 to 13.0 μmol/min.
The bar with added pigment runs from 2.2 to 3.8 μmol/min.
The two ±2SE bars do not overlap, so the difference is very unlikely to be chance.
So the added pigment slowed the rate at which enzyme 1 made B.
Rubric
  • Award 1 point for: the evidence (the ±2SE bars, 11.0 to 13.0 μmol/min with extract and A, 2.2 to 3.8 μmol/min with added pigment, do not overlap) AND the reasoning (so the lower rate with pigment is very unlikely to be chance, so the added pigment slowed enzyme 1).
  • Do not award the point for comparing the two means alone (12.0 against 3.0 μmol/min), or for the ranges with no link to the claim.

Slip Comparing the means alone, or quoting the bars with no link. Supporting the claim needs the non-overlap and what it shows about enzyme 1.

(d) Explain how the pigment slows its own production in a living alga cell. (1 pt)

Frame When the pigment is plentiful, it binds …, which …, so …, and ….

Hint What does a molecule bound at a site away from the active site do to an enzyme? Then follow the pathway from enzyme 1 through B and C to the pigment.

Model answer When the pigment is plentiful, it binds enzyme 1 at a site away from the active site.
Bound there, the pigment changes enzyme 1's shape and slows enzyme 1.
So less B is made, then less C, then less pigment.
The end product reduces its own production: negative feedback at the molecular level.
As the cell uses the pigment up, the pigment leaves enzyme 1 and the pathway speeds up again.
Rubric
  • Award 1 point for: the pigment binds enzyme 1 at a site away from its active site (an allosteric site), which changes the enzyme's shape and slows it, so less B is made, then less C and less pigment: the end product reduces its own production (negative feedback at the molecular level); when the cell uses pigment up, the pigment leaves enzyme 1 and the pathway speeds up again.
  • Accept 'binds an allosteric site on enzyme 1, changes its shape, slows it, so less pigment is made'. Do not award the point for 'the pigment competes with A for the active site', or for 'the pigment slows enzyme 1' with no shape change and no link to less pigment.

Slip Saying the pigment blocks the active site. The pigment binds a separate site and changes the enzyme's shape. Binding a separate site is what makes this feedback by the end product rather than competition with A.

(e) A mutant alga has an enzyme 1 that lacks the site the pigment binds. Predict how the amount of pigment in the mutant's cells compares with a normal alga's, and justify your prediction. (1 pt)

Frame The mutant's cells hold … pigment, because ….

Hint Name the part of the loop the mutant lacks, then follow the loop to the pigment and say which way the amount moves.

Model answer The mutant's cells hold more pigment.
The pigment has no site to bind on enzyme 1, so nothing slows the pathway as the pigment builds up.
Enzyme 1 keeps making B at full speed.
So more C and more pigment are made, and the pigment climbs well above the level a normal cell holds.
Rubric
  • Award 1 point for: more pigment (higher), because the site the pigment binds is missing, so the pigment can no longer slow enzyme 1 and the pathway makes pigment at full speed however much pigment has built up; the response that would reduce the pigment's own production is gone.
  • Do not award the point for 'less pigment', or for 'more pigment' with no reference to the missing binding site removing the slowing of enzyme 1.

Slip Predicting less pigment because 'the loop is broken'. Say which part is missing and follow the loop: the missing part is the brake on enzyme 1, so the pigment rises.

FRQ 2 P44-frq2 · Analyze Model

Cells lining the stomach release pepsin, an enzyme that digests protein, in an inactive form that has an extra stretch of amino acids folded over its active site. Stomach acid removes that stretch from a few molecules, making them active. Each active pepsin molecule can then cut the extra stretch off other inactive molecules, making them active too. The model below shows the loop. A drug that binds the active site of pepsin and stays there is swallowed with a meal; the step it blocks is marked X.

Model of the loop that activates pepsin in the stomach. The box marked X is the step the drug blocks.
Model of the loop that activates pepsin in the stomach. The box marked X is the step the drug blocks.

(a) Describe how the number of active pepsin molecules changes over the first minutes after the acid makes the first few active, using the model. (1 pt)

Model answer The number rises faster and faster.
The few molecules the acid activated each cut several inactive molecules into active ones.
Those new active molecules cut more.
So each round activates more than the round before, until nearly every pepsin molecule is active.
Rubric
  • Award 1 point for: it rises, and rises faster and faster, because each newly active molecule cuts and activates more, so every round activates more than the last.
  • Do not award the point for 'it rises' alone, or for 'it is held at a set level'.

Slip Describing a steady rise. The returning arrow means each round feeds the next, so the rise speeds up.

(b) Identify the kind of feedback in the model, and explain how the model shows it. (1 pt)

Model answer Positive feedback.
The stimulus is active pepsin appearing.
The response is active pepsin cutting inactive molecules into active ones.
The response increases the number of active pepsin molecules, the very change that triggered it.
The arrow returning from the last box to the cutting step shows the response feeding its own trigger.
Rubric
  • Award 1 point for: positive feedback, because the response, more pepsin molecules becoming active, increases the very change that triggered it, the number of active pepsin molecules; shown by the arrow returning from the 'number of active molecules rises' box to the cutting step.
  • Do not award the point for 'positive feedback' alone, or for 'positive because digestion is useful'.

Slip Naming the kind without saying what the response does to its trigger. 'Positive' is earned by the returning arrow.

(c) Determine the effect of the drug on the number of active pepsin molecules during the meal, and justify your decision. (1 pt)

Model answer The number of active pepsin molecules stays very low, near the few that the acid activates directly.
With its active site filled by the drug, an active pepsin molecule can cut nothing.
So the molecule activates no others.
Every round of the loop needs that cut.
So no round of the loop happens, and most of the pepsin stays inactive.
Rubric
  • Award 1 point for: the decision (the number stays very low, at most the few the acid activates) AND the reasoning it rests on (the drug fills the active site, so an active molecule can cut nothing, so pepsin activates no inactive molecules and no round of the loop happens).
  • Do not award the point for the decision alone, for 'the drug blocks pepsin' with no link to the number of active molecules, or for 'no pepsin is made'.

Slip Describing the blocked component and stopping. Follow the loop: no cutting, so no new active molecules, so the number stays low.

(d) In a healthy stomach the loop repeats for a while and then the number of active molecules stops rising. Explain what ends the loop. (1 pt)

Model answer The loop ends when something outside the loop removes the stimulus.
Once every inactive pepsin molecule has been cut, there is nothing left to activate.
As the food and enzyme leave the stomach, the active pepsin is gone too.
Positive feedback holds no set point.
A positive-feedback loop repeats until an outside limit ends it.
Rubric
  • Award 1 point for: the loop ends when an outside limit removes the stimulus, here the supply of inactive pepsin being used up (every molecule has been activated), or the food and enzyme leaving the stomach; a positive-feedback loop has no set point and ends only when something outside it removes the stimulus.
  • Do not award the point for 'the number reaches its set point' or for 'active pepsin switches itself off'.

Slip Saying the number 'reaches its set point'. Positive feedback holds no set point. An outside event or limit ends the loop.

APBIO-U04-T44 End-of-topic test: Feedback

Topic 4.4 · Feedback · 18 MCQ · 2 FRQ

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the two free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. Then open the scoring guide and mark your own work against it. Where a graph carries error bars, the caption says what the bars represent. Blood glucose is in milligrams per deciliter (mg/dL).

Q1 T44-q01

A saltwater fish holds the salt in its blood near 10 mg/mL. When the fish swims into saltier water, the salt in its blood rises to 12 mg/mL. Cells in its gills then pump salt out of the blood until it is back near 10 mg/mL.

Which of the following is the stimulus that the gill cells detect in this loop?

  1. A. ✓ The rise in blood salt to 12 mg/mL
  2. B. The 10 mg/mL the fish holds its blood salt near
    10 mg/mL is the set point: the value the salt is held near.
    The stimulus is the change that moved the salt away from it: the rise to 12 mg/mL.
  3. C. The pumping of salt out by the gill cells
    Pumping salt out is the response: what the fish does about the rise.
    The stimulus is the rise itself.
  4. D. The saltier water the fish swam into
    The saltier water caused the change, but the stimulus is the change in the regulated quantity that the gill cells detect: the blood salt rising to 12 mg/mL.

Why: A stimulus is a change that moves a regulated quantity away from its set point.
The regulated quantity is the salt in the blood, held near 10 mg/mL.
The rise to 12 mg/mL is the stimulus.
Pumping salt out is the response.

Q2 T44-q02

The graph below shows one healthy person's core temperature, measured every two hours for a day. The dashed line is at 37.0 °C.

One healthy person's core temperature, measured every two hours for a day. The dashed line is at 37.0 °C. Gridlines every 0.2 °C.
One healthy person's core temperature, measured every two hours for a day. The dashed line is at 37.0 °C. Gridlines every 0.2 °C.

Which of the following describes what the graph shows?

  1. A. The set point moved to each new reading in turn
    The set point stayed at 37.0 °C; the readings scattered around it.
    A set point is the value the quantity is held near, and it does not follow the readings.
  2. B. Homeostasis failed during the day, because the readings differ from one another
    Readings a few tenths of a degree apart are homeostasis working: a regulated quantity keeps drifting from its set point and being brought back, so its graph wobbles.
  3. C. ✓ The temperature wandered a few tenths of a degree from 37.0 °C and was brought back each time
  4. D. The temperature was held exactly at 37.0 °C all day
    The points, read against the gridlines, are 36.8, 37.2 and 36.9 °C.
    The line never lies flat on 37.0 °C.
    The line wobbles around 37.0 °C.

Why: Homeostasis holds a quantity near its set point, never exactly on it.
The temperature drifted to 37.2 °C and to 36.8 °C and was brought back toward 37.0 °C each time, so the line wobbles around the dashed set point.

Q3 T44-q03

A woman sits in a hot bath. Her core temperature rises to 37.8 °C and she begins to sweat heavily. She gets out, and within ten minutes the sweating has slowed and stopped.

Why did the sweating stop?

  1. A. Her sweat glands had used up their water and had to refill
    Sweat glands keep working as long as the body is too warm.
    The glands slowed because the temperature had come down, not because their water was used up.
  2. B. ✓ Sweating cooled her back toward 37 °C, so the rise that triggered it had shrunk away
  3. C. Her set point rose to 37.8 °C to match the new temperature
    The set point stayed at 37 °C; sweating brought the temperature back to it.
    A set point is a value the body defends; it did not move.
  4. D. Getting out of the bath switched the loop from negative feedback to positive feedback
    The loop was negative feedback throughout.
    Sweating reduced the rise that triggered it, and once the rise was gone there was nothing left to respond to.

Why: In negative feedback the response reduces the very change that triggered it.
Sweating carried heat away, the temperature fell back toward 37 °C, and as the rise shrank the response faded with it.

Q4 T44-q04

A healthy sheep has fasted overnight, and its blood glucose is 85 mg/dL. A veterinarian injects glucagon into the sheep's blood.

Predict the sheep's blood glucose concentration thirty minutes later.

  1. A. Lower than 85 mg/dL
    Taking glucose up is insulin's effect.
    Glucagon acts on liver cells and makes them release glucose into the blood, so the glucose rises.
  2. B. Still 85 mg/dL
    A set point is a value, not a mechanism.
    Injected glucagon acts on the liver whatever the set point is, and the liver releases glucose.
  3. C. Still 85 mg/dL until the sheep eats, then rising
    Glucagon acts on liver cells, not on food.
    The liver breaks stored glycogen down and releases the glucose, whether or not the sheep has eaten.
  4. D. ✓ Higher than 85 mg/dL

Why: Glucagon is the hormone of the loop's lower arm: it tells liver cells to break glycogen down and release glucose into the blood.
Injected glucagon does the same, so the glucose rises above 85 mg/dL.

Q5 T44-q05

A fungus makes a vitamin in three enzyme steps. The model below shows the pathway and a line that runs from the vitamin and ends in a bar under enzyme 1. The vitamin's shape is nothing like A's.

Model of the three-enzyme pathway that makes the vitamin in a fungus cell. Arrows run from each substance to the next; the line from the vitamin ends in a bar under enzyme 1.
Model of the three-enzyme pathway that makes the vitamin in a fungus cell. Arrows run from each substance to the next; the line from the vitamin ends in a bar under enzyme 1.

What does the bar-ended line represent?

  1. A. Enzyme 1 turns the vitamin back into A when the vitamin is plentiful
    The bar-ended line is a block, not an arrow: nothing is being turned into anything along it.
    The line shows the vitamin acting on enzyme 1 to slow it.
  2. B. ✓ The vitamin binds enzyme 1 at a site away from its active site and slows it
  3. C. The vitamin competes with A for the pocket on enzyme 1 where A binds
    A molecule that competed with A for its pocket would have to resemble A.
    The vitamin binds a separate site away from the active site and changes the enzyme’s shape.
  4. D. The vitamin speeds enzyme 1 up, so more vitamin is made
    A bar end marks inhibition.
    The end product slows the first enzyme, so the pathway makes less vitamin when the vitamin is already plentiful.

Why: In a pathway model an arrow means 'becomes' and a bar-ended line means 'inhibits'.
The end product, the vitamin, binds enzyme 1 at a site away from its active site, changes its shape and slows it: negative feedback at the molecular level, because the end product reduces its own production.

Q6 T44-q06

A liver cell in a mouse keeps the concentration of potassium ions inside itself near a set value. When potassium ions leak out faster than usual, pumps in the liver cell's membrane bring potassium ions back in faster, and the concentration inside returns to its usual value.

At which level is this negative feedback operating, and why?

  1. A. ✓ Cellular, because one cell detects the change and one cell responds to it
  2. B. Molecular, because potassium ions are the thing being moved
    The level is set by what detects the change and what responds, not by the size of what moves.
    Here a whole cell, with its membrane pumps, does both.
  3. C. Organismal, because the liver cell belongs to a whole animal
    Here one liver cell detects its own ion loss and its own pumps respond; no other organ is involved.
    The organismal level needs organs working across a body.
  4. D. Molecular, because the pumps are protein molecules
    Every response uses molecules.
    The molecular level means a single enzyme slowed by its own end product.
    Here a whole cell regulates the ion content of its interior.

Why: Negative feedback works at three levels, set by what detects and what responds: one enzyme molecule, one cell, or organs across a whole body.
One liver cell detects the loss of potassium ions and one liver cell responds, so the loop is cellular.

Q7 T44-q07

Three negative-feedback loops are described in the table below. A single-celled organism’s loop counts as cellular.

Three negative-feedback loops.
Three negative-feedback loops.

Which of the following gives the level of each loop?

  1. A. The plant cell’s loop cellular, the yeast’s loop cellular, the person’s loop organismal
    In the plant cell’s loop one enzyme molecule, bound by its own product, detects and responds: the molecular level.
    Being inside a cell does not make the loop cellular.
  2. B. The plant cell’s loop molecular, the yeast’s loop molecular, the person’s loop organismal
    The yeast cell detects its water loss, and the same cell makes the glycerol.
    Glycerol is a molecule, but glycerol does not detect and respond: the loop is cellular.
  3. C. ✓ The plant cell’s loop molecular, the yeast’s loop cellular, the person’s loop organismal
  4. D. The plant cell’s loop organismal, the yeast’s loop cellular, the person’s loop molecular
    The plant cell’s loop is one enzyme slowed by its own product: molecular.
    The person’s loop uses sensors, the heart and blood vessels across a whole body: organismal.

Why: The level is set by what detects and responds.
In the plant cell, one enzyme molecule is bound by the pathway’s product: molecular.
In the yeast’s loop, one cell detects its water loss and makes glycerol: cellular.
In the person’s loop, organs across the body detect and respond: organismal.

Q8 T44-q08

In a rare condition, a person's pancreas cells make insulin with one changed amino acid. The changed insulin has a different shape and does not fit the ligand-binding domain of the insulin receptor. The person's receptors, liver cells and muscle cells are all normal. The person eats a meal.

Which of the following predicts the person’s blood glucose concentration two hours after the meal, compared with a person whose insulin is normal, with the correct reason?

  1. A. Lower, because the pancreas releases extra insulin to make up for the poor fit
    More insulin of the wrong shape changes nothing: none of it can bind the receptor, so liver and muscle cells still leave the glucose in the blood.
  2. B. The same, because the set point of about 90 mg/dL is unchanged
    A set point is a value the loop defends, not a mechanism.
    The loop is broken at the binding step, so the glucose concentration is not brought back.
  3. C. Lower, because the changed insulin acts like glucagon
    A changed shape stops insulin fitting its own receptor.
    The changed shape does not make insulin fit glucagon's receptor.
    Nothing here raises the glucose faster than the meal already has.
  4. D. ✓ Higher, because the insulin is released but the cells never receive its message

Why: Follow the loop: glucose rises (the stimulus), pancreas cells release insulin, but the misshapen insulin cannot bind the receptor, so reception fails and liver and muscle cells do not take glucose up.
The response is missing, and the glucose concentration stays higher than normal.

Q9 T44-q09

A growth in a person's pancreas releases glucagon into the blood all the time, whether the blood glucose concentration is high or low. The person's insulin, receptors and liver are normal.

Predict the person's blood glucose concentration between meals, compared with a healthy person's.

  1. A. ✓ Higher than a healthy person’s
  2. B. The same as a healthy person’s between meals, raised only after meals
    Glucagon acts on the liver and is released all the time here.
    So the liver releases glucose between meals too, when a healthy person’s glucagon would have fallen.
  3. C. Lower than a healthy person’s
    Taking glucose up is insulin's effect.
    Glucagon makes liver cells break glycogen down and release glucose, which raises the blood glucose.
  4. D. The same as a healthy person’s, near 90 mg/dL
    In a healthy person glucagon release fades as glucose returns to 90 mg/dL.
    Here it never fades, so the glucose concentration sits higher than the set point.

Why: In the healthy loop, glucagon is released when glucose falls below the set point and its release fades as glucose returns.
A component that is always on removes that fading: the liver keeps releasing glucose, so between meals the glucose concentration is higher than 90 mg/dL.

Q10 T44-q10

In the days before an egg is released from a woman's ovary, cells in the ovary release the hormone estrogen. Estrogen makes a gland at the base of the brain release a second hormone, and that second hormone makes the ovary cells release even more estrogen. The estrogen level climbs faster and faster until the egg is released.

Which of the following classifies this loop, with the correct reason?

  1. A. More estrogen brings less of the second hormone, and less of the second hormone brings the estrogen back down: negative feedback
    The second hormone makes the ovary cells release even more estrogen, so each round raises the estrogen further.
    The response increased its trigger: positive feedback.
  2. B. ✓ More estrogen brings more of the brain gland's hormone, and that hormone brings more estrogen: positive feedback
  3. C. The estrogen is brought back to a set point each day: negative feedback
    The estrogen climbs faster and faster for days.
    Nothing brings the estrogen back to a set value.
    A set point would show as a level that stays steady.
  4. D. Estrogen release is good for reproduction: positive feedback
    The kind of loop is decided by what the response does to its trigger.
    Here more estrogen brings more estrogen, so the response increases its trigger: positive feedback.

Why: The stimulus is rising estrogen.
The response is more estrogen released, by way of the second hormone.
The response increases the very change that triggered it, so each round triggers a stronger response: positive feedback.
The loop repeats until an outside event, the egg’s release, removes the stimulus.

Q11 T44-q11

When a honeybee stings an animal near its hive, it releases an alarm chemical into the air. Nearby bees that detect the chemical fly at the animal and sting it, and each new sting releases more alarm chemical. The animal flees, and within a minute the stinging has stopped.

Why did the stinging stop?

  1. A. The alarm chemical reached its set point
    Positive feedback holds no set point.
    The stinging grew round after round and stopped only when the trigger, the animal, was gone.
  2. B. The bees had used up all their alarm chemical
    Each bee that stings releases alarm chemical.
    The release stops because there is nothing left to sting, not because the chemical is used up.
  3. C. ✓ The animal's leaving removed the stimulus, so no new stings released alarm chemical
  4. D. The loop switched to negative feedback once enough bees had stung
    The loop did not change kind.
    It was broken from outside: the animal left, so the stimulus was gone and no round could start.

Why: A positive-feedback loop repeats until an outside event removes the stimulus.
Here the animal fleeing is that event: with nothing to sting, no new alarm chemical is released and the loop has nothing to feed on.

Q12 T44-q12

Four events in a person’s body are described.

Which of the following is an example of feedback?

  1. A. A sudden loud noise makes the person’s heart beat faster, and the heartbeat stays fast for several minutes
    The faster heartbeat does nothing to the noise.
    Feedback needs the response to act back on what triggered it.
  2. B. The smell of fresh bread makes the person feel hungry, and the hunger grows until they eat
    Feeling hungry does nothing to the smell of bread.
    In feedback the response changes the very quantity that triggered it.
  3. C. Hearing their name called across a room makes the person turn toward the voice
    Turning the head does nothing to the voice.
    A response that leaves its trigger untouched is not feedback.
  4. D. ✓ Drinking two liters of water raises the water content of the person’s blood, and the kidneys pass more water out in the urine

Why: Feedback is a loop in which the response acts back on the very quantity that triggered it.
The rise in the blood’s water content triggered the kidneys.
The kidneys passed more water out, which lowered the blood’s water content.
So the response acted back on its own trigger: feedback.

Q13 T44-q13

On a hot afternoon a plant loses water through pores in its leaves, and its water content falls. As the water content falls, the pores close, and the loss of water slows. The plant's water content is still falling, more slowly, at the end of the afternoon.

Which of the following classifies this loop, with the correct reason?

  1. A. ✓ Closing the pores reduces the water loss that triggered it: negative feedback
  2. B. The water content keeps falling, so the loop is positive feedback
    Closing the pores slowed the water loss that triggered it: negative feedback.
    The response slowed the loss but did not stop it, so the water content still falls.
  3. C. Closing the pores adds to the fall in water content that triggered it: positive feedback
    Water loss triggered the pores to close, and closing them slowed the loss.
    A response that reduces the change that triggered it is negative feedback, not positive feedback.
  4. D. The water content is falling, and a falling quantity means negative feedback
    A falling quantity can sit in either kind of loop.
    This loop is negative feedback because closing the pores reduced the water loss that triggered it.

Why: Ask what the response does to its trigger.
Water loss triggered the pores to close, and closing them reduced the water loss.
The response opposed the change that triggered it: negative feedback, whichever way the quantity is moving.

Q14 T44-q14

A climber walks up a high mountain, where the air holds less oxygen. The oxygen in her blood falls lower than the value her body holds it near. Cells in her neck arteries detect the fall, and she breathes faster and deeper until the oxygen in her blood is back near that value.

Which of the following is the response in this loop?

  1. A. The fall in the oxygen in her blood
    The fall is the stimulus: the change that moved the oxygen away from its set point.
    The response is what her body did about the fall.
  2. B. The thin air of the mountain
    The thin air caused the stimulus.
    The response is what her body did about the fall in blood oxygen: breathing faster and deeper.
  3. C. ✓ The faster, deeper breathing
  4. D. The value her body holds the blood oxygen near
    The value the oxygen is held near is the set point.
    The response is an action, not a value.

Why: The response is what the body does about a stimulus.
The fall in blood oxygen was the stimulus.
The value the oxygen is held near is the set point.
Breathing faster and deeper, which brought the oxygen back, was the response.

Q15 T44-q15

The model below shows the loop that holds a saltwater fish's blood salt near 10 mg/mL. A poison blocks the salt pumps in the gill cells, the step marked X. The fish then swims into saltier water.

Model of the loop that holds the fish's blood salt near 10 mg/mL. The box marked X, with the broken arrow after it, is the step the poison blocks.
Model of the loop that holds the fish's blood salt near 10 mg/mL. The box marked X, with the broken arrow after it, is the step the poison blocks.

Predict what happens to the salt in the fish's blood.

  1. A. It rises to 12 mg/mL and returns to 10 mg/mL
    A set point is a value the loop defends, not a mechanism.
    With the pumps blocked, nothing moves salt out, so the salt is not brought back to 10 mg/mL.
  2. B. ✓ It rises and stays high
  3. C. It falls below 10 mg/mL
    The pumps are what move salt out of the blood.
    Blocking the pumps stops that movement.
    Salt keeps entering from the saltier water, and nothing removes the salt.
  4. D. It stays at 10 mg/mL
    Detecting the rise is not the response.
    The gill cells detect it, but the response, pumping salt out, is blocked, so detection alone changes nothing.

Why: Follow the loop with the marked step missing: salt enters from the saltier water and rises (the stimulus), the gill cells detect the rise, but the pumps are blocked, so no salt is moved out.
The response is missing and the salt rises and stays high.

Q16 T44-q16

Some mice carry a mutation in the cells of the brain that detect core temperature. Eight normal mice and eight mutant mice were kept in a room at 35 °C for two hours, and then their core temperatures were measured. The graph below shows the two means; the error bars represent ±2SE.

Mean core temperature of normal and mutant mice after two hours in a room at 35 °C, eight mice in each group. Each point is the mean; error bars represent ±2SE. Gridlines every 0.5 °C.
Mean core temperature of normal and mutant mice after two hours in a room at 35 °C, eight mice in each group. Each point is the mean; error bars represent ±2SE. Gridlines every 0.5 °C.

Which claim do the data support?

  1. A. Both groups regulated their temperature, because both means are above 37.0 °C
    The normal mice’s bar (36.8 to 37.6 °C) sits near the set point; the mutant bar (39.0 to 40.2 °C) sits far above it.
    The two bars do not overlap.
  2. B. The mutant mice have a set point of about 39.6 °C
    The mutation broke the loop that defends the set point, not the set point itself.
    The mutant mice could not detect the rise, so no response cooled them.
  3. C. The two groups differ only by chance, because eight mice in each group is too few to tell
    Each ±2SE bar is the range the true mean is likely to lie in.
    The bars do not overlap, so the difference is unlikely to be chance.
  4. D. ✓ The mutant mice failed to bring their temperature down: the bars do not overlap

Why: The normal bar runs 36.8 to 37.6 °C and the mutant bar 39.0 to 40.2 °C.
The ±2SE bars do not overlap, so the higher mean is unlikely to be chance.
The mutation removed the detecting step, so the rise triggered no response and the temperature was not brought back.

Q17 T44-q17

When a sperm enters a frog's egg, a little calcium is released into the egg's cytoplasm near the entry point. Released calcium triggers nearby stores to release more calcium, and a wave of calcium sweeps across the egg in about a minute. The wave stops once every store has emptied.

Which of the following classifies this loop, with the correct reason?

  1. A. The wave stops once the stores are empty, so the loop is negative feedback
    Every positive-feedback loop ends when something outside it removes the stimulus.
    Here the stimulus ends when every store has emptied.
    The ending does not make a loop negative feedback.
  2. B. Each release of calcium is followed by pumps that return the calcium to its set level: negative feedback
    Nothing returns the calcium while the wave spreads.
    Released calcium triggers the stores beside it to release more, so the release grows and spreads until the stores are empty.
  3. C. ✓ Each release of calcium sets off a larger release beside it, so the wave grows: positive feedback
  4. D. The wave helps the egg develop, so the loop is positive feedback
    The kind of loop is decided by what the response does to its trigger.
    Here each release causes a larger release, so the response increases its trigger: positive feedback.

Why: Released calcium triggers the stores beside it to release more calcium.
So the response increases the very change that triggered it.
Each release is larger than the last, so the wave spreads.
That is positive feedback.
The loop ends when an outside limit, the emptied stores, removes the stimulus.

Q18 T44-q18

A fungus makes a vitamin in three enzyme steps, and when the vitamin is plentiful it binds enzyme 1 at a site away from the active site and slows it. A researcher adds to the fungus a molecule that fits that same site on enzyme 1 and stays bound permanently. The cells then use their vitamin up.

Which of the following predicts what happens to the pathway once the vitamin is scarce, with the correct reason?

  1. A. It speeds up, because the vitamin has left enzyme 1
    The vitamin has left, but the added molecule has taken its place at the same site and stays.
    So enzyme 1 stays in its slowed shape: the brake stays on.
  2. B. ✓ It stays slowed, because the added molecule keeps enzyme 1 in its slowed shape
  3. C. It speeds up, because the added molecule competes with the vitamin and cancels it
    The two molecules do the same thing at that site: change enzyme 1's shape and slow it.
    The added one simply never leaves, so the slowing never fades.
  4. D. It stops completely, because the added molecule blocks the active site
    The added molecule fits the site away from the active site, where the vitamin binds.
    Enzyme 1 is slowed, as when the vitamin binds, but its active site is free.

Why: In the normal loop the brake fades when the vitamin leaves enzyme 1.
A molecule that fits the same site and stays bound holds enzyme 1 in its slowed shape whatever the vitamin level, so the response never fades and the pathway stays slowed even when the cell needs vitamin.

FRQ 1 T44-frq1 · Scientific Investigation

Researchers tested whether glucagon is needed for blood glucose to be brought back up when it falls. They bred rats whose pancreas cells make no glucagon; these rats’ insulin, insulin receptors and glucagon receptors are normal. Eight normal rats and eight glucagon-lacking rats, all the same age and kept at the same temperature, fasted for 18 hours with water available, and then each rat’s blood glucose was measured. The graph below shows the mean blood glucose of each group; the error bars represent ±2SE.

Mean blood glucose of normal rats and of rats whose pancreas cells make no glucagon, after an 18-hour fast; eight rats in each group. Error bars represent ±2SE. Gridlines every 10 mg/dL.
Mean blood glucose of normal rats and of rats whose pancreas cells make no glucagon, after an 18-hour fast; eight rats in each group. Error bars represent ±2SE. Gridlines every 10 mg/dL.

(a) Identify the independent variable and the dependent variable in this experiment. (1 pt)

Model answer Independent variable: whether the rats’ pancreas cells make glucagon.
Dependent variable: the blood glucose after the 18-hour fast, in mg/dL.
Rubric
  • Award 1 point for both: independent variable, whether the pancreas cells make glucagon (normal rats or glucagon-lacking rats); dependent variable, the blood glucose after the 18-hour fast, in mg/dL.
  • Accept 'the presence or absence of glucagon' for the independent variable. Do not award the point if the two variables are reversed, or if a condition kept the same for every rat (the 18-hour fast, the temperature, the age) is named as the independent variable.

Slip Naming the fast, the temperature or the age as a variable. Those were kept the same for every rat; the researchers changed whether the rats make glucagon and measured the blood glucose.

(b) Identify the control group in this experiment and justify your answer. State the null hypothesis for this experiment. (2 pt)

Model answer The control group is the normal rats.
The normal rats were the same age as the glucagon-lacking rats, were kept at the same temperature and fasted for the same 18 hours.
The normal rats differ from the glucagon-lacking rats only in making glucagon, the factor under test.
Null hypothesis: there is no difference in blood glucose after an 18-hour fast between normal rats and rats whose pancreas cells make no glucagon.
Rubric
  • Award 1 point for the control group WITH the reason: the normal rats, the group treated the same way (same age, same temperature, same 18-hour fast) but with glucagon present. Do not award the point for the normal rats named with no reason, for a condition kept the same for every rat (the fast, the temperature, the age) named as the control, or for 'the rats before the fast'.
  • Award 1 point for the null hypothesis as a no-difference statement naming the factor changed and the quantity measured: there is no difference in blood glucose after an 18-hour fast between normal rats and rats whose pancreas cells make no glucagon. Accept 'glucagon has no effect on fasting blood glucose'. Do not award the point for a prediction of a difference in either direction.

Slip Naming the 18-hour fast as the control. The fast is a condition every rat shared. The control is the group that has glucagon and is otherwise treated the same. Writing the null hypothesis as a prediction of a difference ('the glucagon-lacking rats have lower glucose'). The null hypothesis predicts no difference.

(c) Support the claim that glucagon is needed for blood glucose to be brought back up during a fast, using evidence from the error bars. (1 pt)

Model answer The normal rats’ bar runs from 80 to 92 mg/dL.
The glucagon-lacking rats’ bar runs from 50 to 62 mg/dL.
The two ±2SE bars do not overlap, so the difference between the two groups is very unlikely to be chance.
During the fast the glucagon-lacking rats’ glucose fell well below the set point and stayed there.
The normal rats’ glucose was held near 86 mg/dL.
So without glucagon the glucose is not brought back up: glucagon is needed.
Rubric
  • Award 1 point for: the evidence (the normal rats' ±2SE bar, 80 to 92 mg/dL, and the glucagon-lacking rats' bar, 50 to 62 mg/dL, do not overlap) AND the reasoning (the lower glucose in the glucagon-lacking rats is very unlikely to be chance; their glucose fell during the fast and was not brought back toward the set point, while the normal rats' was, so glucagon is needed).
  • Accept 'the bars do not overlap, so the difference is not chance', with the two ranges read from the graph and the link to glucagon stated. Do not award the point for a comparison of the two means alone (86 against 56 mg/dL), or for the ranges with no link to the claim.

Slip Comparing 86 with 56 mg/dL and stopping, or quoting the bars with no link. Supporting the claim needs the non-overlap and what it shows about glucagon.

(d) The researchers measured the rats after an 18-hour fast rather than one hour after a meal. Evaluate that decision as a test of whether glucagon is needed. (1 pt)

Model answer The decision was right.
Glucagon release rises when glucose falls below its set point, and falls after a meal.
So the glucagon arm of the loop is active during a fast.
One hour after a meal the glucose is above the set point, so the insulin arm is active instead.
Both groups have normal insulin and insulin receptors, so after a meal the two groups would look the same.
So only a fast tests the arm the rats lack.
Rubric
  • Award 1 point for: the judgement (the fast was the right choice; a measurement one hour after a meal would not show the effect) AND the ground (glucagon release rises when glucose falls below the set point and falls after a meal, so only during a fast is the glucagon arm active, and glucagon is what the rats lack; one hour after a meal the insulin arm is active, which both groups have intact, so the two groups would look alike).
  • Accept 'a good decision, because glucagon release rises during a fast and falls after a meal, so only a fast tests the glucagon arm'. Do not award the point for the judgement alone, or for 'fasting lowers the glucose' with no link to which arm of the loop is active.

Slip Judging the decision without naming which arm of the loop each measurement would test. After a meal the insulin arm is active, and both groups have it.

FRQ 2 T44-frq2 · Analyze Model

A species of bacterium that lives in seawater makes light, but only when its population is dense. Each cell releases a small signal molecule into the water. When the concentration of the signal molecule in the water is high enough, it enters the cells and binds a receptor protein inside each cell; the bound receptor switches on the genes for making light and the gene for making more of the signal molecule. The model below shows the loop. A researcher adds to the water an enzyme that destroys the signal molecule as fast as the cells release it; the point where the enzyme acts is marked X.

Model of the loop in the light-making bacterium. The break marked X on the returning arrow is where the added enzyme destroys the signal molecule.
Model of the loop in the light-making bacterium. The break marked X on the returning arrow is where the added enzyme destroys the signal molecule.

(a) Describe how the concentration of the signal molecule in the water changes over time once the cells begin to release it, using the model. (1 pt)

Model answer The concentration rises faster and faster.
A little signal binds some receptors.
Those cells release more signal.
The higher concentration binds more receptors.
So each round adds more signal to the water than the last, until every cell is releasing as much as it can.
Rubric
  • Award 1 point for: the concentration rises, and rises faster and faster (each rise makes the cells release more, so each round adds more signal than the last), until every cell is releasing at its maximum.
  • Accept 'it climbs faster and faster'. Do not award the point for 'it rises' alone with no reference to the rise growing, or for 'it is held at a set value'.

Slip Describing a steady rise, or a level that stays at a set point. The returning arrow means each round feeds the next, so the rise speeds up.

(b) Identify the kind of feedback shown in the model, and explain how the model shows it. (1 pt)

Model answer Positive feedback.
The stimulus is the rising concentration of signal molecule in the water, and the response, each cell releasing more signal molecule, increases that very concentration.
The arrow returning from the release box to the concentration box shows the response feeding its own trigger.
Rubric
  • Award 1 point for: positive feedback, because the response (a cell releasing more signal molecule) increases the very change that triggered it (the concentration of signal molecule in the water), shown by the arrow that returns from the release box to the signal-concentration box.
  • Do not award the point for 'positive feedback' alone, or for 'positive because light is useful to the bacteria'.

Slip Naming the kind without saying what the response does to its trigger. 'Positive' is earned by the returning arrow: the response increases the change that triggered it.

(c) Predict the effect of the added enzyme on light production by a dense population, and justify your prediction. (1 pt)

Model answer The cells stay dark.
The enzyme destroys the signal molecule as fast as it is released, so its concentration in the water never rises high enough to bind the receptors inside the cells.
With no bound receptor, the genes for light and for more signal are never switched on, no round of the loop starts, and the dense population makes no light.
Rubric
  • Award 1 point for: light production stays off (or stops), because the enzyme removes the signal molecule from the water before its concentration can rise, so receptors are not bound, the genes for light and for more signal are not switched on, and the loop cannot feed itself; the prediction must give the direction (no light, or less light) and the reason.
  • Do not award the point for 'less signal' with no link to light, or for 'the cells make light anyway because they are dense'.

Slip Describing the blocked component ('the signal is destroyed') and stopping. The point is downstream: no bound receptor, so no light, and the loop never starts.

(d) A mutant strain of the bacterium releases the signal molecule normally, but its receptor protein has a binding site of the wrong shape for the signal molecule. Determine whether a dense culture of the mutant strain, grown on its own, makes light, and justify your decision. (1 pt)

Model answer The dense mutant culture makes no light.
The mutant cells release the signal molecule, so the signal builds up in the water as the culture grows dense.
But the signal cannot fit the mutant receptor, so no receptor is bound.
An unbound receptor switches on no genes.
So the genes for light and for more signal stay off.
So no round of the loop starts, and the culture stays dark.
Rubric
  • Award 1 point for: the decision (no light) AND the reasoning it rests on (the signal molecule builds up in the water but cannot bind the mutant receptor, so no receptor is bound, the genes for light and for more signal are never switched on, and the loop never starts).
  • Do not award the point for the decision alone, for 'no light, because there is no signal' (the mutant makes the signal), or for 'light, because the culture is dense' with no reference to the receptor.

Slip Deciding 'light' because the culture is dense and the signal is present. Density matters only because the signal binds the receptor; here the receptor cannot be bound.

APBIO-U04-L13 New cells come from cells

Topic 4.5 · Cell Cycle · 64 steps

Two photographs of the same scraped knee: on the left, on the day of the fall, a raw red patch; in the middle, fifteen days later, a dry brown scab with pink new skin around its edge; on the right a drawing of the skin cells at the edge of the scrape enlarged, each with a nucleus, the edge cell just divided into two smaller cells
Two photographs of the same scraped knee: on the left, on the day of the fall, a raw red patch; in the middle, fifteen days later, a dry brown scab with pink new skin around its edge; on the right a drawing of the skin cells at the edge of the scrape enlarged, each with a nucleus, the edge cell just divided into two smaller cells

Photos: Bart Everson, Wikimedia Commons, CC BY 2.0 (both cropped and resized).

Here is a scraped knee on the day of the fall and fifteen days later, with the cells at the edge of the scrape drawn enlarged.

Scrape your knee on the pavement. Over the next couple of weeks, new skin closes in from the edge of the scrape until the raw patch is covered. Nothing floated in from outside to fill the gap. The skin cells at the edge of the scrape made more skin cells. Each new cell carries the same set of DNA as the cell it came from. Where do new cells come from, and what is the sequence one cell goes through to make two?

Unit 4 · Cell Communication and Cell Cycle

1New cells come from cells

2

Video: Watch: Two cells where there was one

The edge of the scrape a day after the fall: a skin cell at the edge divides into two daughter cells. Every new cell forms this way, from an existing cell dividing.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13a.mp4

3

Every new cell comes from an existing cell dividing in two.

4

Each division follows the same repeating sequence: the cell grows, copies its contents and divides. Start with the dividing.

5

Here is the edge of the scrape a day after the fall. The skin cells at the edge are dividing.

A row of skin cells on the left with a raw gap on the right; the cell at the edge of the row has just divided into two smaller cells, and an arrow points into the gap
A row of skin cells on the left with a raw gap on the right; the cell at the edge of the row has just divided into two smaller cells, and an arrow points into the gap
6

Each division makes two cells where there was one.

7

Every new cell forms this way: an existing cell divides into two. The two cells it makes are called its .

8
Check q1

A cell takes in what it needs, and gets rid of its waste, only across its surface.

Why does a growing cell stop growing and divide, instead of growing bigger and bigger?

  1. A. ✓ The cell’s volume grows faster than its surface area, so its surface cannot keep up with its needs
  2. B. The cell’s surface area grows faster than its volume, so it has more surface than it needs
    As a cell grows, its volume increases faster than its surface area.
    So the surface falls behind the needs of the inside.

Why: As a cell grows, its volume increases faster than its surface area.
The cell takes in what it needs only across its surface.
So past a certain size the surface cannot keep up, and the cell divides.

9

A cell keeps up with its needs only while it stays small. So a bigger organism has more cells, not bigger ones.

10

What you are expected to know Describe where new cells come from: an existing cell divides into two daughter cells, so a bigger organism has more cells, not bigger ones.

11
Check q2

A sea urchin embryo goes from one cell to sixteen cells in about four hours.

Where did the sixteen cells come from?

  1. A. The one cell swelled and then split into sixteen at once
    A cell divides into two, and it is the repeated doubling, one round after another, that reaches sixteen.
  2. B. New cells formed from the fluid around the embryo
    Every cell comes from an existing cell dividing.
    Nothing forms from the surroundings.
  3. C. ✓ Each cell divided into two, four times over
  4. D. Cells from the mother moved in to join the first cell
    The embryo’s cells are all descendants of the first cell, made by division.

Why: Every new cell comes from an existing cell dividing into two daughter cells.
One cell became two, two became four, four became eight, eight became sixteen: four rounds.

12
Check q3

A single yeast cell is sealed in a flask of sugar solution. By the next morning the flask holds thousands of yeast cells. A student says: “The new yeast cells formed from the sugar in the solution.”

Is the student correct?

  1. A. Yes — the new cells formed from the sugar
    Sugar is food: it supplies material for a cell to grow, but food never becomes a new cell on its own.
    Every new cell comes from an existing cell dividing.
  2. B. No — the new cells drifted in from the air
    The flask was sealed, so no yeast cell could enter it.
    Every cell in the flask came from the one cell that was put in.
  3. C. ✓ No — the new cells came from the one yeast cell dividing

Why: Sugar is food for the yeast cell, and food supplies material for growth.
Food does not become new cells.
Every new cell comes from an existing cell dividing.
So the thousands of cells came from the one yeast cell dividing again and again.

13
Practice writing an answer

A single yeast cell is sealed in a flask of sugar solution. By the next morning the flask holds thousands of yeast cells. The new cells did not form from the sugar.

(a) Explain how the flask came to hold thousands of yeast cells. (1 pt)

Model answer Every new cell comes from an existing cell dividing into two daughter cells.
The one yeast cell divided into two, and each daughter cell grew and divided into two more.
Each round of division doubled the number of cells, so after many rounds the flask held thousands.
The sugar was the cells’ food: it supplied the material they needed to grow between divisions.
So the sugar fed the dividing cells, and every cell came from the one cell dividing.
Rubric
  • Award 1 point for: the one yeast cell divided into two daughter cells, and repeated rounds of division doubled the number each time; the sugar was food for growth, not the source of the cells.

14Quick quiz: daughter cells mixed practice

15
Check q4

One skin cell divides into two.

What are the two new cells called?

  1. A. ✓ Daughter cells
  2. B. Parent cells
    The parent cell is the cell that divided.
    The two cells it made are its daughter cells.

Why: The two cells made when one cell divides are called its daughter cells.

16
Check q5

A yeast cell divides once.

How many daughter cells does one division make?

  1. A. 1
    A division splits one cell into two, not one.
  2. B. ✓ 2
  3. C. 4
    One division makes two cells.
    Four cells take two rounds of division.

Why: One cell divides into two daughter cells.
So one division makes 2.

17
Check q6

A liver cell divides into two cells.

Which cell is a daughter cell?

  1. A. The liver cell that divided
    The cell that divided is the parent cell.
    The cells it made are the daughter cells.
  2. B. ✓ One of the two cells made by the division

Why: A daughter cell is one of the two cells made when a cell divides.
So one of the two new cells is a daughter cell.

18
Practice writing an answer

One skin cell divides into two cells.

(a) State what a daughter cell is. (1 pt)

Model answer A daughter cell is one of the two cells made when one cell divides.
Rubric
  • Award 1 point for: one of the two cells made when a cell divides.
19
Check q7

A skin cell divides into two daughter cells.

Does each daughter cell carry the same set of DNA as the skin cell it came from?

  1. A. ✓ Yes
  2. B. No
    Every new cell carries the same set of DNA as the cell it came from.

Why: Every new cell carries the same set of DNA as the cell it came from.
So each daughter cell carries the same DNA as the skin cell.

20
Check q8

A cell divides. Then each of its two daughter cells divides.

How many cells are there now?

  1. A. 2
    Two cells is the count after the first division only.
    Each of those two then divided.
  2. B. 3
    Each daughter cell divided into two, not one.
    Two daughter cells made four.
  3. C. ✓ 4

Why: The first division made 2 daughter cells.
Each of the 2 divided into 2 more.
So there are 4 cells now.

21
Check q9

A new muscle cell forms in a growing child.

Where did the new muscle cell come from?

  1. A. ✓ An existing cell dividing
  2. B. Muscle protein hardening into a cell
    Protein is material a cell is built from, but material never becomes a cell on its own.
    Every new cell comes from an existing cell dividing.

Why: Every new cell comes from an existing cell dividing into two daughter cells.
So the muscle cell came from an existing cell dividing.

22What cell division is for

23

Video: Watch: Growth, repair, reproduction

A child grows taller, a scrape closes, a Hydra buds a new Hydra from its side: three things a cell divides for. Growth and repair add cells to the organism; in asexual reproduction the new cells become a new organism.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13b.mp4

24

Now consider a child who grows 5 cm taller in a year. Cells in the child’s bones divide, and the extra cells make the bones longer.

25

That is division for growth.

26

Now consider the scraped knee again. The edge cells divide again and again until the new cells fill the gap.

27

That is division for repair.

28

Here is a Hydra, a small animal about 1 cm long that lives in ponds. A new Hydra is budding from its side.

A photograph of a Hydra lit against a dark background: a slender orange-brown stalk with a crown of tentacles at the top, and a smaller Hydra with its own tentacles growing out of the stalk's side
A photograph of a Hydra lit against a dark background: a slender orange-brown stalk with a crown of tentacles at the top, and a smaller Hydra with its own tentacles growing out of the stalk's side
29

The bud drops off as a new animal, identical to its parent. One parent made a new organism by cell division alone: that is division for asexual reproduction.

30

Here is a lizard that has lost its tail. Cells at the stump divide again and again, and the new cells build a new tail: division for repair.

A photograph of a brown speckled lizard on a sunlit rock among green leaves; its tail ends in a short blunt stump where the tail was lost
A photograph of a brown speckled lizard on a sunlit rock among green leaves; its tail ends in a short blunt stump where the tail was lost
31

Here is a table of the three reasons a cell divides, with what the new cells do and one example of each.

A table of the three reasons a cell divides: growth, the new cells make the organism bigger, a child's bones lengthen; repair, the new cells replace cells that were lost or damaged, the scrape closes and the lizard's tail regrows; asexual reproduction, the new cells become a new organism, a Hydra buds
32

What you are expected to know Identify what a cell division is for: growth, repair or asexual reproduction.

33
Check q10

A lizard that has lost its tail regrows a new one over several weeks.

How does the new tail get longer?

  1. A. ✓ Cells at the stump divide again and again, adding more cells
  2. B. The cells at the stump swell to many times their size
    A cell keeps up with its needs only while it stays small.
    So growth means more cells, not bigger ones.
  3. C. Cells from the rest of the body shrink and move into the tail
    New tissue is made where it grows, by the cells there dividing.
  4. D. Material from the lizard’s food hardens into new tail cells
    Food supplies material, but every cell comes from an existing cell dividing.

Why: The cells at the stump divide again and again.
Each division adds two daughter cells where there was one.
The new cells build the tail: division for repair.

34Quick quiz: growth, repair or reproduction? mixed practice

35
Check q11

A cut on a finger closes over with new skin in a week.

Which of the following is this cell division for?

  1. A. Growth
    Growth makes the organism bigger.
    These new cells replace skin that was lost.
  2. B. ✓ Repair
  3. C. Asexual reproduction
    In asexual reproduction the new cells become a new organism.
    These new cells stay part of the finger.

Why: The new cells replace skin that was lost at the cut.
Replacing lost or damaged cells is repair.

36
Check q12

A bean seedling’s stem grows 3 cm longer in a week.

Which of the following is this cell division for?

  1. A. ✓ Growth
  2. B. Repair
    Repair replaces cells that were lost or damaged.
    Nothing was lost here: the new cells make the stem longer.
  3. C. Asexual reproduction
    In asexual reproduction the new cells become a new organism.
    These new cells stay part of the seedling.

Why: The new cells make the seedling bigger.
Making the organism bigger is growth.

37
Check q13

A strawberry plant grows a long thin stem along the ground. At the tip of the stem a new strawberry plant forms, identical to the parent plant.

Which of the following is this cell division for?

  1. A. Growth
    Growth makes the parent bigger.
    These new cells became a separate new plant.
  2. B. Repair
    Repair replaces cells that were lost or damaged.
    Nothing was lost: a new plant formed.
  3. C. ✓ Asexual reproduction

Why: The new cells became a new organism, identical to the one parent.
One parent making a new organism is asexual reproduction.

38
Check q14

A broken bone knits back together over six weeks.

Which of the following is this cell division for?

  1. A. Growth
    Growth makes the organism bigger.
    These new cells rebuild the bone that broke.
  2. B. ✓ Repair
  3. C. Asexual reproduction
    In asexual reproduction the new cells become a new organism.
    These new cells stay part of the bone.

Why: The new cells rebuild the bone at the break.
Replacing damaged cells is repair.

39
Check q15

A single-celled Paramecium in a pond divides into two Paramecium, each a separate living organism.

Which of the following is this cell division for?

  1. A. Growth
    Growth makes one organism bigger.
    Here there are two organisms where there was one.
  2. B. Repair
    Repair replaces lost or damaged cells within an organism.
    Here a new organism formed.
  3. C. ✓ Asexual reproduction

Why: One parent divided and the new cell is a separate new organism.
One parent making a new organism is asexual reproduction.

40The cell cycle

41

Video: Watch: Grow, copy, divide, again

An embryo goes from one cell to sixteen in four rounds of division. Each round is the same sequence: the cell grows, copies its contents and divides in two. That repeating sequence is the cell cycle.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13c.mp4

42

An embryo goes from one cell to two, then four, eight, sixteen: four rounds of division.

Five clusters in a row: one cell, two cells, four cells, eight cells and sixteen cells, with arrows between them
Five clusters in a row: one cell, two cells, four cells, eight cells and sixteen cells, with arrows between them
43

Each round is the same three events, in order:

  1. the cell grows;
  2. the cell copies its contents;
  3. the cell divides in two.

44

That repeating sequence is called the .

45

After one turn of the cell cycle, each daughter cell begins the same sequence again: grow, copy, divide.

46

What you are expected to know Describe the cell cycle: the repeating sequence in which a cell grows, copies its contents and divides in two.

47
Check q16

A skin cell goes through one turn of the cell cycle.

Which of the following is the order of events?

  1. A. The cell grows, then divides in two, then copies its contents
    A cell copies its contents before it divides, so that each daughter cell gets a copy.
  2. B. ✓ The cell grows, then copies its contents, then divides in two
  3. C. The cell divides in two, then grows, then divides in two again
    A cell copies its contents between growing and dividing.
    Two divisions with no copying between them leave the daughter cells short.

Why: One turn of the cell cycle is: the cell grows, the cell copies its contents, the cell divides in two.
The copying comes before the division.

48Quick quiz: cell cycle

49

What you are expected to know Say what the cell cycle is.

50

What you are expected to know Say which of its three events a cell does next.

51
Check q17

A cell grows, copies its contents, and divides in two. Each daughter cell then does the same.

What is this repeating sequence called?

  1. A. ✓ The cell cycle
  2. B. Asexual reproduction
    Asexual reproduction is one reason a cell divides.
    The repeating sequence of grow, copy, divide is the cell cycle.
  3. C. Repair
    Repair is one reason a cell divides.
    The repeating sequence of grow, copy, divide is the cell cycle.

Why: The repeating sequence in which a cell grows, copies its contents and divides in two is called the cell cycle.

52
Practice writing an answer

A cell grows, copies its contents, and divides in two. Each daughter cell then does the same.

(a) State what the cell cycle is. (1 pt)

Model answer The cell cycle is the repeating sequence a cell goes through: it grows, copies its contents, and divides into two daughter cells.
Rubric
  • Award 1 point for: the repeating sequence in which a cell grows, copies its contents and divides in two.
53
Check q18

A cell has just grown and copied its contents.

What does the cell do next in the cell cycle?

  1. A. Copies its contents again
    The contents are copied once per turn of the cycle.
    After copying, the cell divides.
  2. B. ✓ Divides in two

Why: One turn of the cell cycle is grow, copy, divide.
The cell has grown and copied.
So the cell divides in two next.

54
Check q19

A cell has just divided into two daughter cells.

What does each daughter cell do first in its own cell cycle?

  1. A. ✓ The daughter cell grows
  2. B. The daughter cell divides in two
    A daughter cell is small when it forms.
    It grows and copies its contents before it can divide.

Why: One turn of the cell cycle is grow, copy, divide.
A new daughter cell starts at the beginning.
So it grows first.

55
Check q20

An embryo goes from one cell to eight cells.

How many turns of the cell cycle separate the first cell from any one of the eight?

  1. A. ✓ 3
  2. B. 7
    Each turn doubles the count: 1, 2, 4, 8.
    That is three turns, not seven.
  3. C. 8
    Eight is the number of cells, not the number of turns.
    1 → 2 → 4 → 8 is three turns.

Why: Each turn of the cell cycle doubles the number of cells.
1 cell became 2, 2 became 4, 4 became 8.
So 3 turns separate the first cell from any one of the eight.

56
Check q21

A student says: “A skin cell can divide twice in a row, with no copying of its contents between the two divisions.”

Is the student correct?

  1. A. Yes — a cell can divide twice in a row without copying
    Each daughter cell must receive a copy of the contents.
    So a skin cell copies its contents before every division.
  2. B. ✓ No — a skin cell copies its contents before each division

Why: Each daughter cell must receive a copy of the contents.
So a skin cell copies its contents before each division.
Two divisions in a row would leave the second pair of daughter cells with no copy.

57

Now go back to the scraped knee. On the day of the fall the raw patch was open, and the skin cells at its edge were dividing.

A row of skin cells on the left with a raw gap on the right; the cell at the edge of the row has just divided into two smaller cells, and an arrow points into the gap
A row of skin cells on the left with a raw gap on the right; the cell at the edge of the row has just divided into two smaller cells, and an arrow points into the gap
58

Over the next couple of weeks, each edge cell went round the cell cycle again and again: it grew, copied its contents, and divided into two daughter cells.

59

The new daughter cells filled the gap from the edge inwards. That is why, fifteen days later, new skin had closed in over the scrape.

60Mixed practice mixed practice

61
Check q22

A bean seedling’s root grows 2 cm longer in one day.

Where do the new cells in the longer root come from?

  1. A. Existing root cells swelling to many times their size
    Root cells do lengthen after they form.
    But the new cells themselves come from cells near the tip dividing.
  2. B. Water from the soil hardening into new root cells
    Water is taken up by the root, but every cell comes from an existing cell dividing.
  3. C. ✓ Cells near the root tip dividing again and again

Why: Every new cell comes from an existing cell dividing.
Cells near the root tip divide again and again.
Their daughter cells are the new cells of the longer root.

62
Check q23

A gardener cuts a stem from a mint plant and puts it in water. Over two weeks the cut end grows new roots, and the cutting becomes a separate mint plant identical to the parent.

Which of the following is the new plant an example of?

  1. A. Growth
    Growth makes the parent plant bigger.
    Here the new cells became a separate plant.
  2. B. ✓ Asexual reproduction
  3. C. Repair
    Repair replaces cells that were lost or damaged.
    Here a whole new plant formed from one parent.

Why: The cutting became a new organism identical to its one parent.
One parent making a new organism by cell division is asexual reproduction.

63
Practice writing an answer

Researchers made a small wound in the skin of ten anesthetized mice. Twenty-four hours later they took a patch of skin one square millimeter in area from the edge of each wound, and a second patch from 5 mm away, and counted the cells caught in the act of dividing in each patch. The graph shows the two means with error bars that represent ±2SE.

A bar chart of dividing cells per square millimeter of skin, 24 hours after a wound, for two patches: at the wound edge, and 5 mm from the wound. The vertical axis runs from 0 to 60 with gridlines every 5 and labels every 10. The wound-edge bar is several times taller than the 5-mm bar. Each bar carries an error bar; the legend reads: error bars represent ±2SE
A bar chart of dividing cells per square millimeter of skin, 24 hours after a wound, for two patches: at the wound edge, and 5 mm from the wound. The vertical axis runs from 0 to 60 with gridlines every 5 and labels every 10. The wound-edge bar is several times taller than the 5-mm bar. Each bar carries an error bar; the legend reads: error bars represent ±2SE

(a) Identify the dependent variable in this experiment. (1 pt)

Model answer The dependent variable is the number of dividing cells per square millimeter of skin, counted 24 hours after the wound.
Rubric
  • Award 1 point for: the dependent variable is the number of dividing cells per mm² of skin (the quantity counted).

Slip Naming the distance from the wound as the dependent variable. The distance was set by the researchers. The count of dividing cells is what they measured.

(b) Support the claim that more cells are dividing at the wound edge than 5 mm away, using evidence from the error bars. (1 pt)

Model answer Read against the gridlines, the wound-edge bar runs from about 36 to about 48 dividing cells per mm².
The bar for 5 mm away runs from about 4 to about 8.
The two ±2SE bars do not overlap.
So the difference is very unlikely to be chance.
Therefore more cells are dividing at the edge.
Rubric
  • Award 1 point for: the evidence (the ±2SE bars, about 36–48 and about 4–8 cells per mm², read against the gridlines, do not overlap) AND the reasoning (so the higher count at the wound edge is unlikely to be chance, so more cells are dividing there).
  • Accept: readings within half a gridline (2.5 cells per mm²) of those values, that is 33.5–38.5 to 45.5–50.5 for the wound edge and 1.5–6.5 to 5.5–10.5 for 5 mm away. Do not award the point for a comparison of the two means alone.

Slip Comparing the two means alone, about 42 against about 6. The claim rests on the ±2SE bars not overlapping.

Glossary

daughter cells
The two cells made when one cell divides.
cell cycle
The repeating sequence a cell goes through: it grows, copies its contents, and divides into two daughter cells.

APBIO-U04-L13B What must be handed on

Topic 4.5 · Cell Cycle · 63 steps

Two cells side by side: on the left a cell whose nucleus is a circle filled with fine dots, labelled between divisions; on the right a cell holding six short dark rods and no nucleus outline, labelled about to divide, 6 of its 46 rods drawn; between them the words two meters of DNA in a nucleus 10 micrometers across
Two cells side by side: on the left a cell whose nucleus is a circle filled with fine dots, labelled between divisions; on the right a cell holding six short dark rods and no nucleus outline, labelled about to divide, 6 of its 46 rods drawn; between them the words two meters of DNA in a nucleus 10 micrometers across

Here is the DNA of one human body cell, drawn twice: on the left as it looks between divisions, on the right as it looks in a cell about to divide.

A human body cell holds 46 very long DNA molecules. Laid end to end they would stretch about two meters. They fit inside a nucleus about 10 μm across. Between divisions the nucleus looks grainy, with no separate threads to be seen. In a cell about to divide the same DNA shows as 46 short, dark rods. How does two meters of thread fit in, and why does it change shape before the cell divides?

Unit 4 · Cell Communication and Cell Cycle

1One DNA molecule, packaged: a chromosome

2

Video: Watch: The thread in the nucleus

One skin cell, its nucleus, and the 46 very long DNA molecules inside it, each wound on proteins. One DNA molecule packaged with its proteins is a chromosome. Two meters of DNA fit in a nucleus 10 μm across only because each molecule is wound.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13Ba.mp4

3
Check q1

A skin cell is about to divide. Its two daughter cells will each need its DNA.

Where in the skin cell is the DNA kept?

  1. A. ✓ In the nucleus
  2. B. In the cytosol, spread through the whole cell
    The cytosol is the fluid around the organelles.
    The DNA of a skin cell is kept inside one organelle, the nucleus.
  3. C. In the cell membrane
    The cell membrane is the cell’s outer boundary.
    The DNA is kept inside the nucleus, behind the nuclear envelope.

Why: A skin cell keeps its DNA inside the nucleus.
The nuclear envelope surrounds the DNA.

4

Every new skin cell carries the same DNA as the cell it came from. The DNA is inside the nucleus, behind the nuclear envelope.

One skin cell, labelled, with its nucleus drawn as a circle bounded by the nuclear envelope, and three long wavy threads inside it, each with small beads along it; a pointer ends on the envelope and another ends on one thread
One skin cell, labelled, with its nucleus drawn as a circle bounded by the nuclear envelope, and three long wavy threads inside it, each with small beads along it; a pointer ends on the envelope and another ends on one thread
5

A human body cell holds 46 very long DNA molecules. Each DNA molecule is wound on proteins.

6

One DNA molecule packaged with its proteins like this is called a .

One chromosome drawn as a single long thread of DNA wound around protein beads
One chromosome drawn as a single long thread of DNA wound around protein beads
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Laid end to end, the 46 DNA molecules of one cell would stretch about two meters. They fit inside a nucleus about 10 μm across only because each DNA molecule is wound around proteins.

8

What you are expected to know Identify a chromosome: one very long DNA molecule packaged with proteins.

9
Check q2

Here are three drawings of things found inside a cell, each with a label.

Three drawings in a row, lettered W, X and Y, each with a label beneath it. W is one small circle, labelled one protein molecule. X is one long wavy line with five small circles spaced along it, labelled one DNA molecule wound on protein beads. Y is a large bounded circle holding three separate wavy lines with small circles along them, labelled a nucleus holding all 46 such molecules
Three drawings in a row, lettered W, X and Y, each with a label beneath it. W is one small circle, labelled one protein molecule. X is one long wavy line with five small circles spaced along it, labelled one DNA molecule wound on protein beads. Y is a large bounded circle holding three separate wavy lines with small circles along them, labelled a nucleus holding all 46 such molecules

Which drawing shows one chromosome?

  1. A. W
    W is one protein molecule on its own.
    A chromosome is a whole DNA molecule wound on many proteins.
  2. B. ✓ X
  3. C. Y
    Y is the nucleus holding all 46 DNA molecules.
    A chromosome is one of those molecules with its proteins.

Why: A chromosome is one very long DNA molecule packaged with proteins.
X is one DNA molecule wound on protein beads.
So X is one chromosome.

10Quick quiz: chromosome mixed practice

11
Check q3

One very long DNA molecule is wound on proteins inside a nucleus.

What is the packaged molecule called?

  1. A. ✓ A chromosome
  2. B. A nucleus
    The nucleus is the organelle that holds all the DNA.
    One packaged DNA molecule is one chromosome.
  3. C. A daughter cell
    A daughter cell is a whole cell made by division.
    One packaged DNA molecule is one chromosome.

Why: One very long DNA molecule packaged with proteins is called a chromosome.

12
Practice writing an answer

A human body cell holds 46 chromosomes.

(a) State what a chromosome is. (1 pt)

Model answer A chromosome is one very long DNA molecule packaged with proteins.
Rubric
  • Award 1 point for: one very long DNA molecule packaged (wound) with proteins.
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Check q4

A human body cell holds 46 chromosomes. Its DNA is still uncopied.

How many very long DNA molecules does the cell hold?

  1. A. 1
    Each chromosome is its own DNA molecule.
    46 chromosomes are 46 DNA molecules.
  2. B. ✓ 46
  3. C. 92
    Each chromosome is one DNA molecule, not two.
    46 chromosomes are 46 DNA molecules.

Why: Each chromosome is one very long DNA molecule packaged with proteins.
So 46 chromosomes are 46 DNA molecules.

14
Check q5

A chromosome is made of two kinds of molecule.

Which two?

  1. A. ✓ DNA and protein
  2. B. DNA and sugar
    The DNA is wound on proteins, not on sugar.
    A chromosome is DNA and protein.
  3. C. Protein and fat
    A chromosome is built around one very long DNA molecule.
    The DNA is wound on proteins.

Why: A chromosome is one very long DNA molecule wound on proteins.
So a chromosome is DNA and protein.

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Check q6

The 46 chromosomes of one human cell would stretch about two meters if laid end to end.

How do they fit inside a nucleus 10 μm across?

  1. A. ✓ Each DNA molecule is wound around proteins
  2. B. Each DNA molecule is cut into short pieces
    Each chromosome stays one unbroken DNA molecule.
    The molecule fits because it is wound around proteins.
  3. C. The nucleus stretches to two meters
    The nucleus stays about 10 μm across.
    The DNA fits because each molecule is wound around proteins.

Why: Each DNA molecule is wound around proteins.
Winding packs two meters of thread into a space 10 μm across.

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Check q7

A body cell of a mouse, before its DNA is copied, holds 40 very long DNA molecules, each wound on proteins.

How many chromosomes does the mouse cell hold?

  1. A. 20
    Each DNA molecule with its proteins is one chromosome.
    40 DNA molecules are 40 chromosomes.
  2. B. ✓ 40
  3. C. 80
    One DNA molecule makes one chromosome, not two.
    40 DNA molecules are 40 chromosomes.

Why: One DNA molecule packaged with proteins is one chromosome.
The cell holds 40 such molecules.
So the cell holds 40 chromosomes.

17The whole set: the genome

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Video: Watch: All 46, to each daughter cell

A human skin cell’s 46 chromosomes are its complete set, its genome. A dividing cell must hand a whole genome, all 46 chromosomes, to each daughter cell.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13Bb.mp4

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A human skin cell’s 46 chromosomes together are its complete set. A cell’s complete set of chromosomes is called its .

One skin cell, labelled, with its nucleus drawn as a circle bounded by the nuclear envelope, and three long wavy threads inside it, each with small beads along it; a pointer ends on the envelope and another ends on one thread
One skin cell, labelled, with its nucleus drawn as a circle bounded by the nuclear envelope, and three long wavy threads inside it, each with small beads along it; a pointer ends on the envelope and another ends on one thread
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A dividing cell must hand a whole genome, all 46 chromosomes, to each daughter cell. That is the job the cell cycle is built around.

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What you are expected to know Identify the genome: a cell’s complete set of chromosomes, which a dividing cell must hand whole to each daughter cell.

22
Check q8

A human skin cell holds 46 chromosomes in its nucleus.

Which of these is the cell’s genome?

  1. A. Any one of its chromosomes
    A chromosome is one DNA molecule with its proteins.
    The genome is the whole set of chromosomes.
  2. B. The proteins its DNA is wound on
    The proteins only wind the DNA up.
    The genome is the DNA itself: all 46 chromosomes.
  3. C. The nucleus that holds the DNA
    The nucleus holds the genome.
    The nucleus is not the genome.
  4. D. ✓ The cell’s complete set of 46 chromosomes

Why: The genome is a cell’s complete set of chromosomes: for a human skin cell, all 46.

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Check q9

A student says: “When a skin cell with 46 chromosomes divides, each daughter cell gets 23 of them.”

Is the student correct?

  1. A. Yes — each daughter cell gets 23 chromosomes
    The genome is the cell’s complete set of chromosomes, and 23 is half of it.
    Each daughter cell needs the whole genome, so each gets all 46.
  2. B. No — each daughter cell gets 92 chromosomes
    A daughter cell needs one complete genome.
    One complete genome for a human skin cell is 46 chromosomes, the same number the parent cell had.
  3. C. ✓ No — each daughter cell gets all 46 chromosomes

Why: The genome is the cell’s complete set of chromosomes.
For a human skin cell the genome is 46 chromosomes.
Each daughter cell needs a complete genome.
So each daughter cell gets all 46 chromosomes.

24
Practice writing an answer

A skin cell with 46 chromosomes divides into two daughter cells. Each daughter cell receives all 46 chromosomes.

(a) Explain why each daughter cell must receive all 46 chromosomes. (1 pt)

Model answer Every new skin cell carries the same DNA as the cell it came from.
The genome is the cell’s complete set of chromosomes.
For a human skin cell the genome is 46 chromosomes.
A set of 23 chromosomes is half the genome.
A daughter cell with half the genome would not carry the same DNA as the parent cell.
So each daughter cell must receive the whole genome: all 46 chromosomes.
Rubric
  • Award 1 point for: the genome is the complete set of 46 chromosomes, and each daughter cell must carry the same DNA as the parent cell, so each daughter cell needs the whole set rather than half of it.
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Check q10

A body cell of a rhesus monkey holds 42 chromosomes. The cell divides into two daughter cells.

How many chromosomes does each daughter cell receive?

  1. A. 21
    Half a set is not a complete genome.
    Each daughter cell needs every one of the 42 chromosomes.
  2. B. ✓ 42
  3. C. 84
    Each daughter cell needs one complete genome.
    One complete genome for a rhesus monkey body cell is 42 chromosomes, the same number the parent cell had.

Why: A complete genome for a rhesus monkey body cell is 42 chromosomes.
Each daughter cell must receive a complete genome.
So each daughter cell receives all 42 chromosomes.

26Quick quiz: genome mixed practice

27
Check q11

A cell holds a complete set of chromosomes.

What is the complete set called?

  1. A. ✓ The cell’s genome
  2. B. The cell’s nucleus
    The nucleus is the organelle that holds the set.
    The set itself is the genome.
  3. C. One chromosome
    One chromosome is one member of the set.
    The whole set is the genome.

Why: A cell’s complete set of chromosomes is called its genome.

28
Practice writing an answer

A human skin cell holds 46 chromosomes in its nucleus.

(a) State what a cell’s genome is. (1 pt)

Model answer A cell’s genome is its complete set of chromosomes.
Rubric
  • Award 1 point for: the cell’s complete set of chromosomes.
29
Check q12

A body cell of a horse holds 64 chromosomes.

How many chromosomes make up the horse cell’s genome?

  1. A. 1
    One chromosome is one member of the set.
    The genome is the whole set: 64.
  2. B. 32
    Half the set is not the genome.
    The genome is the whole set: 64.
  3. C. ✓ 64

Why: The genome is the cell’s complete set of chromosomes.
The horse cell holds 64.
So its genome is 64 chromosomes.

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Check q13

A body cell of a horse holds 64 chromosomes and divides into two daughter cells.

How many chromosomes must each daughter cell receive?

  1. A. 32
    32 is half a genome.
    Each daughter cell needs the whole genome: 64.
  2. B. ✓ 64
  3. C. 128
    128 is two genomes.
    Each daughter cell needs one whole genome: 64.

Why: Each daughter cell must receive one complete genome.
The horse genome is 64 chromosomes.
So each daughter cell receives 64.

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Check q14

A student says: “Each chromosome is a genome.”

Is the student correct?

  1. A. ✓ No — the genome is the complete set of chromosomes
  2. B. Yes — each chromosome is a genome
    One chromosome is one DNA molecule with its proteins.
    The genome is the whole set of them.

Why: A chromosome is one DNA molecule packaged with proteins.
The genome is the cell’s complete set of chromosomes.
So one chromosome is one part of the genome, not the genome.

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Check q15

A daughter cell receives 22 of the 44 chromosomes its parent cell had.

Did the daughter cell receive a complete genome?

  1. A. Yes
    A complete genome is all 44 chromosomes.
    22 is half of them.
  2. B. ✓ No

Why: The parent cell’s genome is 44 chromosomes.
The daughter cell received 22.
22 is half of 44.
So the daughter cell did not receive a complete genome.

33Loose to work, tight to move

34

Video: Watch: Two forms of one chromosome

Between divisions each chromosome is spread out as a thin thread, chromatin, open to be read and copied. Before division each chromosome coils into a short thick condensed chromosome that can be moved without tangling or breaking.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13Bc.mp4

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Under a microscope, a cell between divisions shows a grainy nucleus. No separate chromosomes can be seen.

Two cells side by side: on the left a nucleus that looks grainy with no separate chromosomes; on the right a cell with six short dark rods and no nucleus drawn; the right caption says each rod is one chromosome, coiled
Two cells side by side: on the left a nucleus that looks grainy with no separate chromosomes; on the right a cell with six short dark rods and no nucleus drawn; the right caption says each rod is one chromosome, coiled
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Each chromosome is spread out as a long, thin thread of DNA wound on proteins. A chromosome spread out like this is called .

One chromosome spread out as a long thin thread of DNA wound on protein beads: chromatin
One chromosome spread out as a long thin thread of DNA wound on protein beads: chromatin
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Spread out, the DNA is open: the cell can read its genes and copy it. Loose is the form for working.

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A cell about to divide shows something different: short, thick, dark rods. Each rod is one chromosome coiled up tightly, and a chromosome coiled up like this is called a .

The same chromosome coiled into one short thick rod: a condensed chromosome
The same chromosome coiled into one short thick rod: a condensed chromosome
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Spread out, one chromosome is a thread a few centimeters long, in a cell about 10 μm across. A thread that long cannot be dragged across the cell without tangling or breaking.

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A short thick rod can be moved cleanly. Tight is the form for moving.

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Here is a table of the two forms of one chromosome: how each form looks, when the cell holds it, and what it is for.

A table of the two forms of one chromosome: chromatin, a long thin thread, between divisions, for reading genes and copying DNA; a condensed chromosome, a short thick rod, when the cell is about to divide, for moving without tangling or breaking
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What you are expected to know Explain why a cell’s DNA takes two forms: chromatin, spread out so it can be read and copied, and condensed chromosomes, coiled so they can be moved without tangling or breaking.

43Quick quiz: working or moving?

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What you are expected to know Tell which form a cell’s DNA is in, chromatin or condensed chromosomes, from what the microscope shows and from what the cell is doing.

45
Check q16

Between divisions, a chromosome is spread out as a long thin thread of DNA wound on proteins.

What is a chromosome in this form called?

  1. A. ✓ Chromatin
  2. B. A condensed chromosome
    A condensed chromosome is the coiled, rod-shaped form.
    The spread-out thread is chromatin.

Why: A chromosome spread out as a long thin thread is called chromatin.

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Check q17

Just before a cell divides, each chromosome coils up into a short thick rod.

What is a chromosome in this form called?

  1. A. Chromatin
    Chromatin is the spread-out thread.
    The coiled rod is a condensed chromosome.
  2. B. ✓ A condensed chromosome

Why: A chromosome coiled up tightly into a short thick rod is called a condensed chromosome.

47
Practice writing an answer

Between divisions, a cell’s nucleus looks grainy under the microscope.

(a) State what chromatin is. (1 pt)

Model answer Chromatin is a chromosome in its spread-out form: a long thin thread of DNA wound on proteins.
Rubric
  • Award 1 point for: a chromosome spread out as a long thin thread (the form between divisions).
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Check q18

A chromosome is in the form the cell can read genes from and copy.

Which form is that?

  1. A. ✓ Chromatin, spread out
  2. B. A condensed chromosome, coiled tight
    A coiled rod is closed up.
    The cell reads and copies its DNA in the spread-out form, chromatin.

Why: Spread-out DNA is open, so the cell can read it and copy it.
The spread-out form is chromatin.

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Check q19

A chromosome is in the form in which the cell can move it cleanly from one end of the cell to the other.

Which form is that?

  1. A. Chromatin, spread out
    A long thin thread tangles or breaks when dragged across a cell.
    The form for moving is the coiled rod, a condensed chromosome.
  2. B. ✓ A condensed chromosome, coiled tight

Why: A short thick rod can be moved without tangling or breaking.
The coiled form is a condensed chromosome.

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Check q20

Here is a cell seen under a microscope.

One cell drawn as a rounded rectangle with a large circle inside it; the circle is filled with small scattered dots and nothing else is drawn in the cell
One cell drawn as a rounded rectangle with a large circle inside it; the circle is filled with small scattered dots and nothing else is drawn in the cell

Which form is this cell’s DNA in?

  1. A. ✓ Chromatin, spread out
  2. B. Condensed chromosomes, coiled tight
    Condensed chromosomes show as separate dark rods.
    This nucleus shows no rods, only a grainy look.
    A grainy look means the DNA is spread out.
    So the DNA is chromatin.

Why: The nucleus looks grainy and shows no separate rods.
A grainy look means the DNA is spread out as long thin threads.
So the DNA is chromatin, spread out.

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Check q21

Here is a cell seen under a microscope.

One cell drawn as a rounded rectangle holding six short thick lines at slightly different angles, spread across the cell; no circle is drawn around them
One cell drawn as a rounded rectangle holding six short thick lines at slightly different angles, spread across the cell; no circle is drawn around them

Which form is this cell’s DNA in?

  1. A. Chromatin, spread out
    Chromatin shows as a grainy nucleus with no separate rods.
    This cell shows short thick rods.
    A rod is a chromosome coiled tight.
    So the DNA is in condensed chromosomes.
  2. B. ✓ Condensed chromosomes, coiled tight

Why: The cell shows short thick rods.
A rod is one chromosome coiled up tightly.
So the DNA is in condensed chromosomes, coiled tight.

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Check q22

A cell is reading its genes and copying its DNA.

Which form is its DNA in?

  1. A. ✓ Chromatin, spread out
  2. B. Condensed chromosomes, coiled tight
    A coiled rod is closed up, so the cell cannot read or copy it.
    The cell reads and copies its DNA while it is spread out: chromatin.

Why: Reading genes and copying DNA is the working job.
Spread-out DNA is open, so the cell can read it and copy it.
So the DNA is chromatin, spread out.

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Check q23

A cell is about to move its chromosomes to opposite ends of the cell.

Which form is its DNA in?

  1. A. Chromatin, spread out
    A long thin thread tangles or breaks when dragged across a cell; a short thick rod moves cleanly.
    So a cell coils each chromosome tight before moving them.
  2. B. ✓ Condensed chromosomes, coiled tight

Why: Moving chromosomes is the moving job.
A short thick rod can be moved without tangling or breaking.
So the DNA is in condensed chromosomes, coiled tight.

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Check q24

A liver cell is busy doing its job. Its next division is months away.

Which form is its DNA in?

  1. A. ✓ Chromatin, spread out
  2. B. Condensed chromosomes, coiled tight
    A working cell reads its genes while the DNA is spread out.
    Chromosomes coil only when a cell is about to divide, and this liver cell is not.

Why: A cell doing its job is reading its genes.
Reading genes is the working job, and the DNA is spread out for it.
The cell is not about to divide.
So the DNA is chromatin, spread out.

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Check q25

A root-tip cell has just begun to divide.

Which form is its DNA in?

  1. A. Chromatin, spread out
    A dividing cell is about to move its chromosomes.
    A long thin thread would tangle or break.
    So a dividing cell coils each chromosome into a short thick rod first.
  2. B. ✓ Condensed chromosomes, coiled tight

Why: A dividing cell is about to move its chromosomes.
Moving is the job of the tight form.
So the DNA is in condensed chromosomes, coiled tight.

56
Practice writing an answer

Before a cell divides, each of its chromosomes coils from a long thin thread into a short thick rod.

(a) Explain why each chromosome coils up before the cell divides. (1 pt)

Model answer A dividing cell must move its chromosomes to opposite ends of the cell.
Spread out, one chromosome is a thread a few centimeters long, in a cell about 10 μm across.
A long thin thread dragged across a cell would tangle or break.
A short thick rod can be moved cleanly.
So each chromosome coils up into a short thick rod before the cell moves it.
Tight is the form for moving.
Rubric
  • Award 1 point for: the chromosomes are about to be moved across the cell, and a long thin thread would tangle or break, whereas a short thick rod can be moved cleanly.
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Now go back to the two meters of DNA in one human body cell. Its 46 threads fit inside a nucleus 10 μm across because each thread is wound on proteins, and each wound thread is one chromosome.

Two cells side by side: on the left a nucleus that looks grainy with no separate chromosomes; on the right a cell with six short dark rods and no nucleus drawn; the right caption says each rod is one chromosome, coiled
Two cells side by side: on the left a nucleus that looks grainy with no separate chromosomes; on the right a cell with six short dark rods and no nucleus drawn; the right caption says each rod is one chromosome, coiled
58

Between divisions, the 46 chromosomes are spread out as chromatin, so the nucleus looks grainy. The cell reads and copies its DNA in this form.

59

In a cell about to divide, each of the 46 chromosomes has coiled into a condensed chromosome, so the same DNA shows as 46 short dark rods. The cell can move a rod without tangling or breaking it.

60Mixed practice mixed practice

61
Check q26

On a slide of an onion root tip, one cell shows short, thick, dark rods.

Which of the following is the cell about to do with its chromosomes?

  1. A. ✓ Move them across the cell
  2. B. Read the genes on them
    Genes are read while the DNA is spread out as chromatin.
    A coiled rod is closed up.
    A cell coils its chromosomes to move them.
  3. C. Copy them again
    Copying happens while the DNA is spread out as chromatin.
    The copying is already done when the rods form.
    The rods form for moving.

Why: Short thick dark rods are condensed chromosomes.
A chromosome coils into a rod so that it can be moved without tangling or breaking.
So the cell is about to move its chromosomes across the cell.

62
Practice writing an answer

A biologist looks at two cells on one slide of a growing onion root tip. In the first cell the nucleus looks grainy, with no separate threads to be seen. In the second cell the DNA shows as 16 short, thick, dark rods. An hour later the second cell has divided into two daughter cells, and the first cell has not.

(a) Identify the form of the DNA in the first cell. (1 pt)

Model answer The DNA in the first cell is chromatin: each chromosome is spread out as a long thin thread.
Rubric
  • Award 1 point for: chromatin (the spread-out form), from the grainy look with no separate rods.

(b) Explain how the difference between the two cells demonstrates why a chromosome coils up before a cell divides. (2 pt)

Model answer The first cell is not about to divide, so it holds its chromosomes as chromatin.
Spread out, the DNA is open, so the cell can read its genes and copy it.
The second cell divides an hour later, so it must move its chromosomes to opposite ends of the cell first.
A long thin thread dragged across a cell would tangle or break.
A short thick rod can be moved cleanly.
So the second cell has coiled each chromosome into a condensed chromosome.
Rubric
  • Award 1 point for: the first cell is not dividing and keeps its DNA spread out as chromatin, the open form for reading genes and copying DNA.
  • Award 1 point for: the second cell is about to divide and must move its chromosomes, and a coiled rod (a condensed chromosome) can be moved without tangling or breaking, whereas a long thin thread would tangle or break.

Glossary

chromosome
One very long DNA molecule packaged with proteins. A human body cell holds 46.
genome
A cell’s complete set of chromosomes: 46 in a human body cell.
chromatin
A chromosome in its spread-out form: a long thin thread of DNA wound on proteins, open to be read and copied. The form between divisions.
condensed chromosome
A chromosome coiled into a short thick rod that can be moved without tangling or breaking. The form in a cell about to divide.

APBIO-U04-L13C Copied and held together

Topic 4.5 · Cell Cycle · 71 steps

One fruit fly cell drawn three times: on the left eight short rods, each with a small dot part way along, labelled 8; in the middle eight X shapes, each two rods crossing at a small dot, labelled still 8; on the right a longer cell with eight V shapes at each end, the dot at each point facing the nearer end, labelled 16
One fruit fly cell drawn three times: on the left eight short rods, each with a small dot part way along, labelled 8; in the middle eight X shapes, each two rods crossing at a small dot, labelled still 8; on the right a longer cell with eight V shapes at each end, the dot at each point facing the nearer end, labelled 16

Here is one fruit fly cell drawn three times: before it copies its DNA, after it copies its DNA, and as it divides.

A fruit fly cell holds 8 chromosomes. Before it divides, it copies every one of its DNA molecules. So it now holds twice the DNA. Yet a biologist looking at it still counts 8 chromosomes. Later, as the cell divides, the same biologist counts 16. What is being counted, and when does the count change?

Unit 4 · Cell Communication and Cell Cycle

1Every DNA molecule copied: DNA replication

2

Video: Watch: The copy before the split

Before a cell divides it copies every one of its DNA molecules: one DNA molecule becomes two identical DNA molecules. The copying is DNA replication.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13Ca.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13Ca.mp4

3
Check q1

A body cell of a fruit fly holds 8 very long DNA molecules, each wound on proteins.

What is one DNA molecule with its proteins called?

  1. A. ✓ A chromosome
  2. B. A genome
    The genome is the cell’s complete set of chromosomes.
    One DNA molecule with its proteins is one chromosome.
  3. C. A nucleus
    The nucleus is the organelle that holds all the DNA.
    One DNA molecule with its proteins is one chromosome.

Why: One very long DNA molecule packaged with proteins is called a chromosome.
The fruit fly cell holds 8 of them.

4

Before it divides, the cell copies every one of its DNA molecules. One DNA molecule becomes two identical DNA molecules.

one DNA molecule becomes two identical DNA molecules
5

Copying every DNA molecule like this is called .

6

Each new molecule is a replica, an exact copy, of the original.

7

How the cell copies a DNA molecule is a later story. Here only the result matters: after DNA replication, the cell holds two identical copies of every DNA molecule.

8

What you are expected to know Describe DNA replication: before it divides, the cell copies every one of its DNA molecules.

9
Check q2

A cell is about to divide into two daughter cells.

Why did the cell copy its DNA first?

  1. A. So the chromosomes could coil up into short thick rods
    Coiling is for moving the chromosomes.
    Copying is what gives each daughter cell its own set.
  2. B. So the cell would have spare DNA in case some of it broke
    The copy is not held back.
    One copy of every chromosome goes to each daughter cell.
  3. C. So the cell could grow to twice its size before it divided
    Growth is separate from copying.
    Copying the DNA is what lets two cells each hold a complete genome.
  4. D. ✓ So each daughter cell can receive one copy of every chromosome

Why: DNA replication makes two identical copies of every DNA molecule.
When the cell divides, each daughter cell receives one copy of every chromosome.
So each daughter cell holds a complete genome.

10Quick quiz: DNA replication mixed practice

11
Check q3

Before a cell divides, it copies every one of its DNA molecules.

What is this copying called?

  1. A. ✓ DNA replication
  2. B. Cell division
    Cell division is the cell splitting into two daughter cells.
    The copying that comes before it is DNA replication.
  3. C. The genome
    The genome is the cell’s complete set of chromosomes.
    Copying every DNA molecule is DNA replication.

Why: Copying every DNA molecule before the cell divides is called DNA replication.

12
Practice writing an answer

A cell is about to divide.

(a) State what DNA replication is. (1 pt)

Model answer DNA replication is the copying of every DNA molecule in the cell before it divides.
Rubric
  • Award 1 point for: the copying of every DNA molecule (each into two identical copies) before the cell divides.
13
Check q4

One DNA molecule goes through DNA replication.

What does the cell hold afterwards?

  1. A. One molecule, twice as long
    Replication makes a second molecule.
    The original does not grow longer.
  2. B. ✓ Two identical copies of the molecule
  3. C. Two molecules carrying different genes
    Each new molecule is a replica, an exact copy, of the original.
    The two carry the same genes.

Why: DNA replication copies one DNA molecule into two.
Each new molecule is an exact copy of the original.
So the cell holds two identical DNA molecules.

14
Check q5

A cell holds 30 DNA molecules. It carries out DNA replication.

How many DNA molecules does the cell hold afterwards?

  1. A. 15
    Replication adds copies.
    The number of DNA molecules doubles from 30 molecules to 60 molecules.
  2. B. 30
    Every DNA molecule is copied, so the count doubles.
    30 molecules become 60.
  3. C. ✓ 60

Why: DNA replication copies every DNA molecule.
Each of the 30 molecules becomes two.
So the cell holds 60 DNA molecules.

15
Check q6

A cell is about to divide.

When does the cell carry out DNA replication?

  1. A. ✓ Before it divides
  2. B. While it divides
    The DNA is copied first, so that the cell has two copies to share out when it divides.
  3. C. After it divides
    A daughter cell needs a complete genome as soon as it forms.
    So the copying is done before the division.

Why: Each daughter cell must receive a complete genome.
So the cell copies its DNA before it divides.

16
Check q7

A student says: “DNA replication makes a copy of only the genes the cell is using.”

Is the student correct?

  1. A. ✓ No — every DNA molecule is copied in full
  2. B. Yes — only the genes in use are copied
    Each daughter cell needs a complete genome, not only the genes in use.
    So every DNA molecule is copied in full.

Why: Each daughter cell must receive a complete genome.
So DNA replication copies every DNA molecule in full, not only the genes in use.

17Two sister chromatids, one centromere

18

Video: Watch: The copied chromosome

After copying, each chromosome is two identical copies lying side by side, pinched together at one point: two sister chromatids joined at a centromere.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13Cb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13Cb.mp4

19

After copying, each chromosome is two identical copies of the same DNA molecule. The two copies lie side by side, pinched together at one point.

A condensed chromosome as one rod with a small white dot part way along it, an arrow labelled DNA replication, and then two oppositely tilted rods crossing at a small white dot; captions: one chromosome, and the same chromosome, copied
A condensed chromosome as one rod with a small white dot part way along it, an arrow labelled DNA replication, and then two oppositely tilted rods crossing at a small white dot; captions: one chromosome, and the same chromosome, copied
20

The two joined copies are called , because they are identical copies of one chromosome.

The same drawing: one rod, an arrow labelled DNA replication, two tilted rods crossing at a small white dot; a pointer from one of the two rods to the label sister chromatids
The same drawing: one rod, an arrow labelled DNA replication, two tilted rods crossing at a small white dot; a pointer from one of the two rods to the label sister chromatids
21

The point where the two sister chromatids are pinched together is called the , named for the center point where the pair is held.

The same drawing with both labels: the two rods labelled sister chromatids and the white dot where they cross labelled centromere
The same drawing with both labels: the two rods labelled sister chromatids and the white dot where they cross labelled centromere
22

What you are expected to know Identify the two sister chromatids and the centromere on a drawing of a copied chromosome.

23
Check q8

Here is one chromosome drawn after DNA replication, with three pointers.

Two tilted rods crossing at a small white dot. A pointer lettered W ends on the upper part of the left rod, a pointer lettered X ends on the dot where the rods cross, and a pointer lettered Y ends on the upper part of the right rod
Two tilted rods crossing at a small white dot. A pointer lettered W ends on the upper part of the left rod, a pointer lettered X ends on the dot where the rods cross, and a pointer lettered Y ends on the upper part of the right rod

Which letter marks the centromere?

  1. A. W
    W ends on one of the two rods.
    Each rod is one sister chromatid.
  2. B. ✓ X
  3. C. Y
    Y ends on one of the two rods.
    Each rod is one sister chromatid.

Why: The centromere is the point where the two sister chromatids are pinched together.
X ends on the dot where the two rods cross.
So X marks the centromere.

24
Check q9

Here is one chromosome drawn after DNA replication, with three pointers.

Two tilted rods crossing at a small white dot. A pointer lettered W ends on the upper part of the left rod, a pointer lettered X ends on the dot where the rods cross, and a pointer lettered Y ends on the upper part of the right rod
Two tilted rods crossing at a small white dot. A pointer lettered W ends on the upper part of the left rod, a pointer lettered X ends on the dot where the rods cross, and a pointer lettered Y ends on the upper part of the right rod

Which two letters mark the two sister chromatids?

  1. A. W and X
    X ends on the dot where the rods cross, the centromere.
    The sister chromatids are the two rods.
  2. B. ✓ W and Y
  3. C. X and Y
    X ends on the dot where the rods cross, the centromere.
    The sister chromatids are the two rods.

Why: Each sister chromatid is one of the two identical rods.
W ends on the left rod and Y ends on the right rod.
So W and Y mark the two sister chromatids.

25Quick quiz: sister chromatids and centromere mixed practice

26
Check q10

After DNA replication, one chromosome is two identical copies lying side by side.

What are the two copies called?

  1. A. ✓ Sister chromatids
  2. B. Daughter cells
    Daughter cells are whole cells made by division.
    The two joined copies of one chromosome are sister chromatids.
  3. C. Two genomes
    A genome is a cell’s whole set of chromosomes.
    The two joined copies of one chromosome are sister chromatids.

Why: The two identical copies of one chromosome, joined side by side, are called sister chromatids.

27
Check q11

Two sister chromatids are pinched together at one point.

What is that point called?

  1. A. The nuclear envelope
    The nuclear envelope surrounds the whole nucleus.
    The point where two sister chromatids are held is the centromere.
  2. B. ✓ The centromere
  3. C. The genome
    The genome is the cell’s whole set of chromosomes.
    The point where two sister chromatids are held is the centromere.

Why: The point where the two sister chromatids are pinched together is called the centromere.

28
Practice writing an answer

A chromosome has just been copied.

(a) State what sister chromatids are. (1 pt)

Model answer Sister chromatids are the two identical copies of one chromosome made by DNA replication, joined to each other at the centromere.
Rubric
  • Award 1 point for: the two identical copies of one chromosome, joined together.

(b) State what the centromere is. (1 pt)

Model answer The centromere is the point where the two sister chromatids are held together.
Rubric
  • Award 1 point for: the point where the two sister chromatids are held (pinched) together.
29
Check q12

A biologist compares the DNA in the two sister chromatids of one copied chromosome.

How do the two DNA molecules compare?

  1. A. The two DNA molecules carry different genes
    Sister chromatids are made by copying one DNA molecule.
    Each copy carries the same genes.
  2. B. ✓ The two DNA molecules are identical
  3. C. One DNA molecule is twice as long as the other
    Each sister chromatid is a full copy of the same DNA molecule.
    The two are the same length.

Why: DNA replication makes an exact copy of one DNA molecule.
The two sister chromatids are that molecule and its copy.
So the two DNA molecules are identical.

30
Check q13

One chromosome has been copied into two sister chromatids.

How many centromeres does the copied chromosome have?

  1. A. ✓ 1
  2. B. 2
    The two sister chromatids share one point where they are pinched together.
    That is one centromere.
  3. C. 4
    There are two sister chromatids, not four.
    The two share one centromere.

Why: The two sister chromatids are pinched together at one point.
That point is the centromere.
So one copied chromosome has one centromere.

31
Check q14

A student says: “The two sister chromatids of one chromosome are two different chromosomes.”

Is the student correct?

  1. A. Yes — they are two chromosomes
    The two copies are still joined at one centromere.
    Together they are one copied chromosome.
  2. B. ✓ No — they are two identical copies of one chromosome

Why: Sister chromatids are the two identical copies made when one chromosome is replicated.
They stay joined at one centromere.
So they are one chromosome, copied.

32Count the centromeres

33

Video: Watch: The counting rule

A joined pair of sister chromatids counts as one chromosome. To count chromosomes, count centromeres. Copying doubles the chromatids but not the chromosomes; the count changes only when the sisters are pulled apart.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13Cc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13Cc.mp4

34

A joined pair of sister chromatids still counts as one chromosome.

35

To count chromosomes, count centromeres. Each centromere marks one chromosome, copied or not.

Two cells: on the left a cell with 8 single rods, each with a small white dot part way along it, labelled 8 chromosomes, 8 chromatids; on the right the same cell after copying with 8 X shapes, each two rods crossing at a small white dot, labelled 8 chromosomes, 16 chromatids
Two cells: on the left a cell with 8 single rods, each with a small white dot part way along it, labelled 8 chromosomes, 8 chromatids; on the right the same cell after copying with 8 X shapes, each two rods crossing at a small white dot, labelled 8 chromosomes, 16 chromatids
36

So copying doubles the amount of DNA and the number of chromatids. The number of chromosomes stays the same.

37

Later, when the cell divides, the sister chromatids are pulled apart. Each separated chromatid then has a centromere of its own.

One long cell after its sister chromatids have been pulled apart: eight V shapes gathered toward the left end with the small white dot at each point facing left, and eight V shapes gathered toward the right end with the dot at each point facing right; labelled 16 centromeres: 16 chromosomes, 16 chromatids
One long cell after its sister chromatids have been pulled apart: eight V shapes gathered toward the left end with the small white dot at each point facing left, and eight V shapes gathered toward the right end with the dot at each point facing right; labelled 16 centromeres: 16 chromosomes, 16 chromatids
38

So each separated chromatid counts as one chromosome. The chromosome count changes only at that moment.

39

What you are expected to know State the counting rule: count centromeres, so a joined pair of sister chromatids is one chromosome and a separated chromatid is one chromosome.

40
Check q15

A cell copies its DNA, coils its chromosomes into rods, and later pulls the sister chromatids apart.

At which of these moments does the number of chromosomes in the cell change?

  1. A. When the DNA is copied
    Copying doubles the chromatids, but each joined pair still has one centromere and counts as one chromosome.
  2. B. When the chromosomes coil into rods
    Coiling changes the form of each chromosome, not how many there are.
  3. C. ✓ When the sister chromatids are pulled apart

Why: A pair of sister chromatids counts as one chromosome while they share a centromere.
Only when they are pulled apart does each become a chromosome of its own, and the count changes.

41
Check q16

A student says: “A cell with 10 chromosomes copies its DNA. Now it has 20 chromosomes.”

Is the student correct?

  1. A. Yes — the cell now has 20 chromosomes
    Copying turns each chromosome into two sister chromatids joined at one centromere.
    A joined pair counts as one chromosome, so the cell still has 10.
  2. B. ✓ No — the cell still has 10 chromosomes

Why: Copying doubles the chromatids to 20.
Each pair of sister chromatids is joined at one centromere and counts as one chromosome.
So the cell still has 10 chromosomes.

42
Practice writing an answer

A cell with 10 chromosomes copies its DNA. After the copying the cell still has 10 chromosomes.

(a) Explain why the cell still has 10 chromosomes after copying. (1 pt)

Model answer Copying turns each chromosome into two identical sister chromatids.
The two sister chromatids stay joined to each other at one centromere.
A joined pair of sister chromatids counts as one chromosome.
To count chromosomes, count centromeres.
The cell has 10 centromeres after copying.
So the cell still has 10 chromosomes, now made of 20 chromatids.
Rubric
  • Award 1 point for: each copied chromosome is two sister chromatids joined at one centromere and counts as one chromosome (count centromeres), so the cell has 10 chromosomes and 20 chromatids.

43Calculate the counts

44

Video: Watch: 8, 16, 8

The fruit fly cell worked through: 8 chromosomes and 8 chromatids before copying; 8 chromosomes and 16 chromatids after; 16 chromosomes when the sisters are pulled apart; 8 in each daughter cell.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13Cd.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L13Cd.mp4

45

Here is the fruit fly cell with 8 chromosomes before copying and after.

Two cells: on the left a cell with 8 single rods, each with a small white dot part way along it, labelled 8 chromosomes, 8 chromatids; on the right the same cell after copying with 8 X shapes, each two rods crossing at a small white dot, labelled 8 chromosomes, 16 chromatids
Two cells: on the left a cell with 8 single rods, each with a small white dot part way along it, labelled 8 chromosomes, 8 chromatids; on the right the same cell after copying with 8 X shapes, each two rods crossing at a small white dot, labelled 8 chromosomes, 16 chromatids
46

Before copying the cell holds 8 rods: 8 chromatids. After copying it holds 8 pairs: 16 chromatids, and still 8 centromeres.

47
Worked example

A fruit fly cell has 8 chromosomes. After its DNA is copied, how many chromatids does it hold?

Write down the values in the question:
chromosomes = 8
chromatids in each copied chromosome = 2
Write down the equation:
chromatids=2×chromosomes
Substitute the values into the equation:
chromatids=2×chromosomes
chromatids=2×8
chromatids=16
48

The count of chromosomes is unchanged: 8 before copying, 8 after. Only the chromatids doubled.

49

Now the cell divides, and the sister chromatids are pulled apart. Each separated chromatid has a centromere of its own, so each counts as a chromosome.

50
Worked example

The same fruit fly cell, with 8 chromosomes and 16 chromatids, pulls its sister chromatids apart. How many chromosomes does it hold at that moment, and how many does each daughter cell receive?

Write down the values in the question:
chromatids after copying = 16
daughter cells = 2
Write down the equation:
chromosomes when the sisters are apart=chromatids after copying
Substitute the values into the equation:
chromosomes when the sisters are apart=16
Write down the equation for each daughter cell:
chromosomes in each daughter cell=chromosomes when the sisters are apartdaughter cells
Substitute the values into the equation:
chromosomes in each daughter cell=162
chromosomes in each daughter cell=8
51

So each daughter cell holds 8 chromosomes again, the same number the parent cell started with.

52

What you are expected to know Calculate the numbers of chromosomes and chromatids in a cell before copying, after copying and after the sister chromatids separate.

53
Check q17 numeric entry

Here is a cell drawn after its DNA has been copied. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

One cell drawn after copying, holding six X shapes, each two rods crossing at a small white dot
One cell drawn after copying, holding six X shapes, each two rods crossing at a small white dot

Calculate the number of chromatids the cell holds.

Part 1. Count the centromeres. How many chromosomes does the cell hold?

Answer: 6  (tolerance ±0)

Working
Count the centromeres:
centromeres = 6
chromosomes = 6

Part 2. How many chromatids does each copied chromosome hold?

Answer: 2  (tolerance ±0)

Working
Count the chromatids in one copied chromosome:
chromatids in each copied chromosome = 2

Answer: 12  (tolerance ±0)

Working
Write down the values in the question:
chromosomes = 6
chromatids in each copied chromosome = 2
Write down the equation:
chromatids=2×chromosomes
Substitute the values into the equation:
chromatids=2×chromosomes
chromatids=2×6
chromatids=12
54
Check q18 numeric entry

A cell from a roundworm has 12 chromosomes. Its DNA is copied. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromatids in the cell after copying.

Answer: 24  (tolerance ±0)

Working
Write down the values in the question:
chromosomes = 12
chromatids in each copied chromosome = 2
Write down the equation:
chromatids=2×chromosomes
Substitute the values into the equation:
chromatids=2×chromosomes
chromatids=2×12
chromatids=24
55
Check q19 numeric entry

A human skin cell has 46 chromosomes. Its DNA is copied. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromatids in the cell after copying.

Answer: 92  (tolerance ±0)

Working
Write down the values in the question:
chromosomes = 46
chromatids in each copied chromosome = 2
Write down the equation:
chromatids=2×chromosomes
Substitute the values into the equation:
chromatids=2×chromosomes
chromatids=2×46
chromatids=92
56
Check q20 numeric entry

A dog cell has 78 chromosomes. Its DNA is copied, and later the sister chromatids are pulled apart. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in the cell at the moment the sister chromatids have been pulled apart.

Answer: 156  (tolerance ±0)

Working
Write down the values in the question:
chromosomes before copying = 78
chromatids in each copied chromosome = 2
Write down the equation:
chromatids after copying=2×chromosomes
Substitute the values into the equation:
chromatids after copying=2×chromosomes
chromatids after copying=2×78
chromatids after copying=156
Count the chromosomes once the sisters are apart, when each separated chromatid has a centromere of its own and counts as a chromosome:
chromosomes=chromatids after copying
chromosomes=156
57

Now go back to the fruit fly cell and the biologist counting its chromosomes. Before copying, the cell has 8 centromeres, so the biologist counts 8 chromosomes.

Two cells: on the left a cell with 8 single rods, each with a small white dot part way along it, labelled 8 chromosomes, 8 chromatids; on the right the same cell after copying with 8 X shapes, each two rods crossing at a small white dot, labelled 8 chromosomes, 16 chromatids
Two cells: on the left a cell with 8 single rods, each with a small white dot part way along it, labelled 8 chromosomes, 8 chromatids; on the right the same cell after copying with 8 X shapes, each two rods crossing at a small white dot, labelled 8 chromosomes, 16 chromatids
58

After DNA replication, each chromosome is two sister chromatids joined at one centromere. The cell still has 8 centromeres, so the biologist still counts 8 chromosomes, now made of 16 chromatids.

59

As the cell divides, the sister chromatids are pulled apart, and each separated chromatid has a centromere of its own. The cell now has 16 centromeres, so the biologist counts 16 chromosomes.

60Quick quiz: counts mixed practice

61
Check q21 numeric entry

A body cell of a horse has 64 chromosomes. Its DNA is copied. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromatids in the cell after copying.

Answer: 128  (tolerance ±0)

Working
Write down the values in the question:
chromosomes = 64
chromatids in each copied chromosome = 2
Write down the equation:
chromatids=2×chromosomes
Substitute the values into the equation:
chromatids=2×64
chromatids=128
62
Check q22 numeric entry

A cell of a garden pea has 14 chromosomes. Its DNA is copied. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in the cell after copying.

Answer: 14  (tolerance ±0)

Working
Count the centromeres:
centromeres before copying = 14
centromeres after copying = 14
Write down the equation:
chromosomes=centromeres
Substitute the values into the equation:
chromosomes=14
63
Check q23 numeric entry

A body cell of a chimpanzee has 48 chromosomes. Its DNA is copied. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromatids in the cell after copying.

Answer: 96  (tolerance ±0)

Working
Write down the values in the question:
chromosomes = 48
chromatids in each copied chromosome = 2
Write down the equation:
chromatids=2×chromosomes
Substitute the values into the equation:
chromatids=2×48
chromatids=96
64
Check q24 numeric entry

A body cell of a sheep has copied its DNA and now holds 108 chromatids. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in the cell.

Answer: 54  (tolerance ±0)

Working
Write down the values in the question:
chromatids = 108
chromatids in each copied chromosome = 2
Write down the equation:
chromosomes=chromatids2
Substitute the values into the equation:
chromosomes=1082
chromosomes=54
65
Check q25 numeric entry

A body cell of a rabbit has 44 chromosomes. Its DNA is copied, and later the sister chromatids are pulled apart. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in the cell at the moment the sister chromatids have been pulled apart.

Answer: 88  (tolerance ±0)

Working
Write down the values in the question:
chromosomes before copying = 44
chromatids in each copied chromosome = 2
Write down the equation:
chromosomes when the sisters are apart=2×chromosomes before copying
Substitute the values into the equation:
chromosomes when the sisters are apart=2×44
chromosomes when the sisters are apart=88
66
Check q26 numeric entry

A body cell of a cow has 60 chromosomes. It copies its DNA, pulls the sister chromatids apart, and divides into two daughter cells. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in each daughter cell.

Answer: 60  (tolerance ±0)

Working
Write down the values in the question:
chromosomes before copying = 60
chromatids in each copied chromosome = 2
daughter cells = 2
Write down the equations:
chromosomes when the sisters are apart=2×chromosomes before copying
chromosomes in each daughter cell=chromosomes when the sisters are apartdaughter cells
Substitute the values into the equations:
chromosomes when the sisters are apart=2×60=120
chromosomes in each daughter cell=1202
chromosomes in each daughter cell=60

67Mixed practice mixed practice

68
Check q27 numeric entry

A cell has copied its DNA and now holds 28 chromatids. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in the cell.

Answer: 14  (tolerance ±0)

Working
Write down the values in the question:
chromatids = 28
chromatids in each copied chromosome = 2
Write down the equation:
chromosomes=chromatids2
Substitute the values into the equation:
chromosomes=chromatids2
chromosomes=282
chromosomes=14
69
Check q28

After copying its DNA, a cell holds twice as much DNA as before.

Which of the following has happened to the number of chromosomes in the cell?

  1. A. ✓ The number of chromosomes is unchanged
  2. B. The number of chromosomes has doubled
    The amount of DNA doubled.
    But each copied chromosome is one joined pair of sister chromatids with one centromere, and a joined pair counts as one chromosome.
  3. C. The number of chromosomes has halved
    Copying adds DNA to the cell.
    Copying never removes a chromosome.

Why: Copying doubles the DNA and the chromatids.
Each pair of sister chromatids shares one centromere, so each pair counts as one chromosome.
So the chromosome count is unchanged.

70
Practice writing an answer

Researchers made a small wound in the skin of ten anesthetized mice. Twenty-four hours later, many of the skin cells at the edge of each wound were dividing. A mouse body cell holds 40 chromosomes.

(a) Explain how each new cell made at the wound edge comes to hold a complete genome of 40 chromosomes. (1 pt)

Model answer Before an edge cell divides, the cell copies every one of its DNA molecules.
So each of its 40 chromosomes becomes two identical sister chromatids joined at a centromere.
When the cell then divides, each daughter cell receives one chromatid of every pair.
So each new cell holds one copy of every chromosome: a complete genome of 40 chromosomes.
Rubric
  • Award 1 point for: the cell copies its DNA before dividing, so every chromosome is two identical sister chromatids, and each daughter cell receives one copy of every chromosome (all 40).

Slip Saying the parent cell splits its 40 chromosomes into two sets of 20. The chromosomes are copied first, so each daughter cell gets all 40.

(b) A mouse skin cell is examined just before it divides, when its chromosomes have coiled into rods. Calculate the number of chromatids it holds. (1 pt)

Answer: 80  (tolerance ±0)

Model answer The cell holds 80 chromatids: 40 chromosomes, each copied into two sister chromatids.
Working
Write down the values in the question:
chromosomes = 40
chromatids in each copied chromosome = 2
Write down the equation:
chromatids=2×chromosomes
Substitute the values into the equation:
chromatids=2×chromosomes
chromatids=2×40
chromatids=80
Rubric
  • Award 1 point for: 80 chromatids.

Glossary

DNA replication
The copying of every DNA molecule in the cell before it divides. It turns each chromosome into two identical sister chromatids.
sister chromatids
The two identical copies of one chromosome made by DNA replication, joined to each other at the centromere until the cell divides.
centromere
The point where two sister chromatids are held together. Counting centromeres counts chromosomes.

APBIO-U04-L14 The cycle in order

Topic 4.5 · Cell Cycle · 65 steps

One cell in a dish drawn at 0, 8, 16 and 23 hours, slightly larger each time with a grainy nucleus, and then at 24 hours as two smaller cells
One cell in a dish drawn at 0, 8, 16 and 23 hours, slightly larger each time with a grainy nucleus, and then at 24 hours as two smaller cells

Here is one cell in a dish, drawn at 0, 8, 16 and 23 hours, and then at the end of the day.

Suppose one cell in a dish is watched for a whole day. For about twenty-three of its twenty-four hours it looks as if nothing is happening: no visible chromosomes, no division, just a grainy nucleus in a slowly swelling cell. Then, in the last hour, it splits in two. What are the names for the quiet stretch and the busy hour, and how much of the day does each take?

Unit 4 · Cell Communication and Cell Cycle

1Interphase: the quiet stretch

2

For about twenty-three of its twenty-four hours, the cell in the dish showed a grainy nucleus and no division. That long stretch lasts from one division to the next.

A wheel with one large sector filling most of the circle, labelled interphase, about 23 h, and a thin sector left for the last hour, in which the cell divides
A wheel with one large sector filling most of the circle, labelled interphase, about 23 h, and a thin sector left for the last hour, in which the cell divides
3

The stretch between one division and the next is called , because inter means between.

4

Almost the whole day is interphase: about 23 of the 24 hours. The division that is so easy to see under the microscope takes only the last hour.

5

Video: Watch: The quiet stretch

One cell in a dish through a day: a grainy nucleus and no division for twenty-three hours, then the split in the last hour. The long stretch between one division and the next is interphase.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14a.mp4

6

What you are expected to know Identify interphase: the long stretch between one division and the next, about 23 hours of a 24-hour cycle.

7
Check q1

A mouse cell in a dish is watched for 20 hours. For 19 hours it keeps one grainy nucleus. In the last hour the cell splits in two.

Which stretch of the 20 hours is interphase?

  1. A. ✓ The first 19 hours
  2. B. The last hour
    A split into two cells is a division.
    Interphase is the stretch between one division and the next.
  3. C. The whole 20 hours
    The last hour is a division.
    Interphase ends when the division begins.

Why: Interphase is the stretch between one division and the next.
The cell divides only in the last hour.
So the first 19 hours are interphase.

8Quick quiz: interphase mixed practice

9
Check q2

A cell has kept one grainy nucleus for six hours, with no division.

Is the cell in interphase?

  1. A. ✓ Yes — the cell is in interphase
  2. B. No — the cell is dividing
    One grainy nucleus and no division is the stretch between divisions.

Why: The cell is between one division and the next.
That stretch is interphase.

10
Practice writing an answer

A cell in a dish is watched for a whole day.

(a) State what interphase is. (1 pt)

Model answer Interphase is the long stretch of the cell cycle between one division and the next.
Rubric
  • Award 1 point for: the stretch of the cycle between one division and the next.
11
Check q3

A cell’s nucleus is dividing into two nuclei.

Is the cell in interphase?

  1. A. Yes — the cell is in interphase
    A nucleus dividing into two is a division.
    Interphase is the stretch between divisions.
  2. B. ✓ No — the cell is dividing

Why: The nucleus is dividing.
A division is not the stretch between divisions.
So the cell is not in interphase.

12
Check q4

A cell’s cytoplasm is pinching in two.

Is the cell in interphase?

  1. A. Yes — the cell is in interphase
    The cytoplasm pinching in two is a division.
    Interphase is the stretch between divisions.
  2. B. ✓ No — the cell is dividing

Why: The cytoplasm is dividing.
A division is not the stretch between divisions.
So the cell is not in interphase.

13
Check q5

A cell finished dividing an hour ago. It has one nucleus and is slowly growing.

Is the cell in interphase?

  1. A. ✓ Yes — the cell is in interphase
  2. B. No — the cell is dividing
    The cell’s last division is over.
    The cell is growing with one nucleus, between one division and the next.

Why: The last division is over and the next has not begun.
The cell is between the two.
So the cell is in interphase.

14The three stages of interphase

15

Interphase has three stages in a fixed order:

  1. The first stage is a growth gap. It is called .
  2. The second stage copies the DNA. It is called .
  3. The third stage is a second growth gap. It is called .

The wheel with interphase split into three sectors in order: G1, 9 h; S phase, 10 h; G2, 4 h; and a thin sector labelled the last hour
The wheel with interphase split into three sectors in order: G1, 9 h; S phase, 10 h; G2, 4 h; and a thin sector labelled the last hour
16

The two gaps are called G1 and G2 because the G is short for gap, a growth gap. It is not the G of the G protein.

17

The copying stage is called S phase because the S is short for synthesis, the making of new DNA. It is not the s of the standard deviation.

18

Here is a table of the three stages of interphase. It compares what happens in each stage with the hours each stage takes in a rapidly dividing human cell, whose whole cycle takes 24 hours.

A table of the three stages of interphase in order, G1, S phase and G2, with what happens in each and the hours each takes in a 24-hour cycle: G1, the cell grows, 9 h; S phase, the DNA is copied, 10 h; G2, the cell grows again, ready to divide, 4 h
19

Video: Watch: G1, S phase, G2

The interphase sector of the wheel splits into three in order: G1, a growth gap of 9 hours; S phase, the DNA copied, 10 hours; G2, a second growth gap, 4 hours. G is short for gap, S for synthesis.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14b.mp4

20

What you are expected to know Name the three stages of interphase in order: G1, S phase, G2.

21

What you are expected to know State what happens in each stage of interphase: the cell grows in G1, the DNA is copied in S phase, the cell grows again in G2.

22

What you are expected to know State what the letters are short for: the G of G1 and G2 for gap, the S of S phase for synthesis.

23Quick quiz: G1, S phase or G2? mixed practice

24
Check q6

A cell is in interphase.

In which stage is its DNA copied?

  1. A. G1
    G1 is the first growth gap.
    The DNA is copied in the stage after G1.
  2. B. ✓ S phase
  3. C. G2
    G2 is the second growth gap.
    G2 begins after the copying is done.

Why: S phase is the copying stage.
The S is short for synthesis, the making of new DNA.

25
Practice writing an answer

A cell is in interphase.

(a) Name the three stages of interphase in order. (1 pt)

Model answer The three stages of interphase, in order, are G1, S phase and G2.
Rubric
  • Award 1 point for: G1, S phase, G2, in that order.
26
Check q7

A cell has just finished dividing.

Which stage of interphase does it enter first?

  1. A. ✓ G1
  2. B. S phase
    S phase is the second stage of interphase.
    The first stage is a growth gap.
  3. C. G2
    G2 is the third stage of interphase.
    The first stage is a growth gap.

Why: Interphase begins with the first growth gap.
The first growth gap is G1.

27
Check q8

A cell has just finished copying its DNA.

Which stage of interphase does it enter next?

  1. A. G1
    G1 comes before the copying.
    The stage after the copying is the second growth gap.
  2. B. S phase
    S phase is the copying stage.
    The copying is finished, so S phase is over.
  3. C. ✓ G2

Why: The DNA is copied in S phase.
The stage after S phase is the second growth gap, G2.

28
Check q9

The two gaps of interphase are written G1 and G2.

What is the G short for?

  1. A. ✓ Gap
  2. B. G protein
    A G protein is a switch protein in a signaling pathway.
    The G of G1 and G2 is the G of a growth gap.

Why: G1 and G2 are the two growth gaps of interphase.
The G is short for gap.

29
Check q10

The copying stage of interphase is written S phase.

What is the S short for?

  1. A. Standard deviation
    The standard deviation is a statistic that describes a set of measurements.
    The S of S phase is the S of making new DNA.
  2. B. ✓ Synthesis, the making of new DNA

Why: In S phase the cell makes new DNA.
The S is short for synthesis.

30Mitosis, cytokinesis and the whole turn

31

Then, in the last hour, the nucleus divides: the two sets of chromosomes are pulled apart into two nuclei. This stage is called .

The wheel with the last hour split into two thin sectors, labelled mitosis and cytokinesis, 1 h together
The wheel with the last hour split into two thin sectors, labelled mitosis and cytokinesis, 1 h together
32

Last, the cytoplasm divides. Now there are two cells.

33

The division of the cytoplasm is called , because cyto means cell and kinesis means movement: the cell’s contents are moved apart into two.

34

Mitosis is the division of the nucleus, not of the whole cell. The cell is not two cells until cytokinesis is done.

35

Each daughter cell then enters G1 of its own cycle. The wheel turns again.

The full wheel with an arrow around the rim from cytokinesis back into G1, labelled: daughter cells re-enter G1
The full wheel with an arrow around the rim from cytokinesis back into G1, labelled: daughter cells re-enter G1
36

In a rapidly dividing human cell in culture, one turn takes about 24 hours: 9 hours in G1, 10 in S phase, 4 in G2, and 1 hour for mitosis and cytokinesis together.

37
Worked example

A rapidly dividing human cell spends 9 hours in G1, 10 hours in S phase, 4 hours in G2 and 1 hour in mitosis and cytokinesis. How many hours of its 24-hour cycle are interphase?

Write down the values in the question:
G1 = 9 h
S phase = 10 h
G2 = 4 h
Write down the equation:
interphase=G1+S phase+G2
Substitute the values into the equation:
interphase=G1+S phase+G2
interphase=9+10+4
interphase=23h
38

So 23 of the 24 hours are interphase. The last hour is mitosis and then cytokinesis.

39

Video: Watch: The last hour, and round again

The last hour of the wheel splits in two: the nucleus divides, mitosis; then the cytoplasm divides, cytokinesis. An arrow from cytokinesis back into G1: each daughter cell begins its own turn.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14c.mp4

40

What you are expected to know State the five stages of the cell cycle in order: G1, S phase, G2, mitosis, cytokinesis, with each daughter cell entering G1 again.

41

What you are expected to know Distinguish mitosis, the division of the nucleus, from cytokinesis, the division of the cytoplasm.

42
Check q11

A cell holds two nuclei inside a single, undivided cytoplasm.

Which of the following describes where this cell is in its cycle?

  1. A. ✓ Mitosis is finished and cytokinesis has yet to happen
  2. B. Cytokinesis is finished and mitosis has yet to happen
    The nucleus divides first, in mitosis.
    The cytoplasm divides after, in cytokinesis.
    This cell has two nuclei and one cytoplasm.
    So mitosis has happened and cytokinesis has not.
  3. C. The cell is still in mitosis
    Mitosis is the division of the nucleus.
    This cell’s nucleus has divided into two nuclei, so mitosis is finished.
    The division of the cytoplasm is a separate stage, cytokinesis.
  4. D. The cell is in G2
    In G2 the cell still has one nucleus.
    The nucleus divides only in mitosis.
    So two nuclei mean mitosis has happened.

Why: Two nuclei in one cytoplasm mean the nucleus has divided.
The cytoplasm has not divided.
So mitosis is done, and cytokinesis is still to come.

43
Practice writing an answer

A cell holds two nuclei inside a single, undivided cytoplasm. Mitosis is finished and cytokinesis has yet to happen.

(a) Explain how the cell shows that mitosis is finished but cytokinesis has yet to happen. (1 pt)

Model answer Mitosis is the division of the nucleus.
This cell holds two nuclei.
So its nucleus has divided, and mitosis is finished.
Cytokinesis is the division of the cytoplasm.
This cell’s cytoplasm is still one undivided cytoplasm.
So cytokinesis has yet to happen.
The cell is not two cells until cytokinesis is done.
Rubric
  • Award 1 point for: two nuclei show the nucleus has divided (mitosis, finished), and one undivided cytoplasm shows the cytoplasm has not divided (cytokinesis, still to come).
44
Check q12

A cell in a growing embryo has just finished cytokinesis.

Which stage does each daughter cell enter next?

  1. A. G2 of the same cycle
    G2 belongs to the cycle that has just ended.
    Cytokinesis ends one cycle.
    Each daughter cell starts a new cycle at its first stage, G1.
  2. B. ✓ G1 of a new cycle
  3. C. S phase of the same cycle
    A new daughter cell grows in G1 first.
    Then the daughter cell copies its DNA in S phase.
  4. D. Mitosis of a new cycle
    A daughter cell cannot divide again straight away.
    The daughter cell must pass through G1, S phase and G2 before its next mitosis.

Why: Cytokinesis is the last stage of one cycle.
Each daughter cell then begins a new cycle at its first stage, G1.

45Quick quiz: which stage is the cell in? mixed practice

46
Check q13

In the last hour of a 24-hour cell cycle the cell divides, in two stages. One of the two stages divides the nucleus.

Which stage is it?

  1. A. ✓ Mitosis
  2. B. Cytokinesis
    Cytokinesis divides the cytoplasm.
    The division of the nucleus is mitosis.

Why: Mitosis is the division of the nucleus.

47
Check q14

In the last hour of a 24-hour cell cycle the cell divides, in two stages. One of the two stages divides the cytoplasm.

Which stage is it?

  1. A. Mitosis
    Mitosis divides the nucleus.
    The division of the cytoplasm is cytokinesis.
  2. B. ✓ Cytokinesis

Why: Cytokinesis is the division of the cytoplasm.

48
Practice writing an answer

In the last hour of its cycle a cell divides, in two stages.

(a) State what mitosis is. (1 pt)

Model answer Mitosis is the division of the nucleus: the two sets of chromosomes are pulled apart into two nuclei.
Rubric
  • Award 1 point for: the division of the nucleus (into two nuclei).

(b) State what cytokinesis is. (1 pt)

Model answer Cytokinesis is the division of the cytoplasm, which ends with two daughter cells.
Rubric
  • Award 1 point for: the division of the cytoplasm (into two cells).
49
Check q15

A cell’s DNA is being copied.

Which stage is the cell in?

  1. A. G1
    G1 is a growth gap.
    The DNA is copied in the stage after G1: S phase.
  2. B. ✓ S phase
  3. C. G2
    G2 is the second growth gap.
    G2 begins when the copying is done.
  4. D. Mitosis
    Mitosis divides the nucleus.
    The DNA was copied earlier, in S phase.

Why: S phase is the stage in which the DNA is copied.
The S is short for synthesis, the making of new DNA.

50
Check q16

The two sets of chromosomes in a cell are being pulled apart into two nuclei.

Which stage is the cell in?

  1. A. G1
    In G1 the cell grows with one nucleus.
    The nucleus divides in mitosis.
  2. B. S phase
    In S phase the DNA is copied inside one nucleus.
    The nucleus divides later, in mitosis.
  3. C. G2
    In G2 the cell prepares for division with one nucleus.
    The nucleus divides in the stage after G2: mitosis.
  4. D. ✓ Mitosis

Why: Mitosis is the division of the nucleus: the two sets of chromosomes are pulled apart into two nuclei.

51
Check q17

A daughter cell has just formed and is growing. Its DNA is still uncopied.

Which stage is the cell in?

  1. A. ✓ G1
  2. B. S phase
    In S phase the DNA is being copied.
    This cell’s DNA is still uncopied, so the copying has yet to begin.
  3. C. G2
    In G2 the DNA has already been copied.
    This cell’s DNA is still uncopied.
  4. D. Mitosis
    In mitosis the nucleus divides.
    This cell is growing with its DNA uncopied, which is the first stage of a new cycle.

Why: A new daughter cell enters G1 first.
In G1 the cell grows and its DNA is still uncopied.

52
Check q18

A cell has finished copying its DNA and is growing again before it divides.

Which stage is the cell in?

  1. A. G1
    In G1 the DNA is still uncopied.
    This cell has finished copying its DNA.
  2. B. S phase
    In S phase the copying is under way.
    This cell has finished copying, so S phase is over.
  3. C. ✓ G2
  4. D. Mitosis
    In mitosis the nucleus divides.
    This cell is still growing before its division, which is the second growth gap.

Why: G2 is the second growth gap.
It comes after the copying is done and before the division.

53
Check q19

The cytoplasm of a cell is pinching in two.

Which stage is the cell in?

  1. A. Mitosis
    Mitosis is the division of the nucleus.
    The division of the cytoplasm is a separate stage, cytokinesis.
  2. B. ✓ Cytokinesis

Why: Cytokinesis is the division of the cytoplasm.
The cytoplasm pinching in two is cytokinesis.

54
Check q20

A cell’s nucleus is dividing into two nuclei.

Which stage is the cell in?

  1. A. ✓ Mitosis
  2. B. Cytokinesis
    Cytokinesis divides the cytoplasm.
    The division of the nucleus is mitosis.

Why: Mitosis is the division of the nucleus.
A nucleus dividing into two nuclei is mitosis.

55
Check q21 numeric entry

A mouse cell line has a 22-hour cycle: 8 hours in G1, 9 hours in S phase, 4 hours in G2 and 1 hour for mitosis and cytokinesis together.

Calculate the number of hours these cells spend in interphase.

Answer: 21 h  (tolerance ±0.05)

Working
Write down the values in the question:
G1 = 8 h
S phase = 9 h
G2 = 4 h
Write down the equation:
interphase=G1+S phase+G2
Substitute the values into the equation:
interphase=G1+S phase+G2
interphase=8+9+4
interphase=21h

56The cell in the dish, again

57

Now go back to the cell in the dish, watched for a whole day. For twenty-three hours it showed a grainy nucleus and no division. Those twenty-three hours were interphase: G1, then S phase, then G2.

58

In the last hour the nucleus divided, which is mitosis. Then the cytoplasm divided, which is cytokinesis.

59

So the quiet stretch was interphase, 23 of the 24 hours. The busy hour was the division.

60Mixed practice mixed practice

61
Check q22

The five stages of the cell cycle are G1, G2, S phase, mitosis and cytokinesis.

Which list gives them in order, starting from G1?

  1. A. G1, G2, S phase, mitosis, cytokinesis
    S phase comes between the two gaps: G1, then S phase, then G2.
  2. B. ✓ G1, S phase, G2, mitosis, cytokinesis
  3. C. G1, S phase, mitosis, G2, cytokinesis
    G2 is the second growth gap and comes before mitosis.
  4. D. G1, S phase, G2, cytokinesis, mitosis
    The nucleus divides first, in mitosis.
    The cytoplasm divides after, in cytokinesis.

Why: The order is G1, S phase, G2 (together interphase), then mitosis and then cytokinesis.

62
Check q23 numeric entry

A hamster cell line has a 20-hour cycle: 7 hours in G1, 8 hours in S phase, 4 hours in G2 and 1 hour for mitosis and cytokinesis together.

Calculate the number of hours these cells spend in interphase.

Answer: 19 h  (tolerance ±0.05)

Working
Write down the values in the question:
G1 = 7 h
S phase = 8 h
G2 = 4 h
Write down the equation:
interphase=G1+S phase+G2
Substitute the values into the equation:
interphase=G1+S phase+G2
interphase=7+8+4
interphase=19h
63
Check q24 numeric entry

A cultured human cell line has a 24-hour cycle: 11 hours in G1, 8 hours in S phase and 4 hours in G2.

Calculate the time the cells spend in mitosis and cytokinesis together.

Answer: 1 h  (tolerance ±0.05)

Working
Write down the values in the question:
whole cycle = 24 h
G1 = 11 h
S phase = 8 h
G2 = 4 h
Write down the equation:
mitosis and cytokinesis=whole cycle−(G1+S phase+G2)
Substitute the values into the equation:
mitosis and cytokinesis=whole cycle−(G1+S phase+G2)
mitosis and cytokinesis=24−(11+8+4)
mitosis and cytokinesis=24−23
mitosis and cytokinesis=1h
64
Practice writing an answer

A researcher films one cell from a frog embryo for 30 hours. For the first 29 hours the cell keeps one grainy nucleus and slowly swells. In the next half hour its two sets of chromosomes are pulled apart into two nuclei. In the last half hour the cytoplasm pinches in two, and there are two cells.

(a) Identify the stage of the cell cycle the cell was in for the first 29 hours. (1 pt)

Model answer For the first 29 hours the cell was in interphase.
Rubric
  • Award 1 point for: interphase.

(b) Explain how the last hour of the film demonstrates that mitosis and cytokinesis are two different divisions. (1 pt)

Model answer Mitosis is the division of the nucleus.
In the first half hour the chromosomes were pulled apart into two nuclei: that was mitosis.
The cell still had one cytoplasm, so it was still one cell.
Cytokinesis is the division of the cytoplasm.
In the last half hour the cytoplasm pinched in two: that was cytokinesis.
The nucleus divided first and the cytoplasm after, so the two divisions are different events.
Rubric
  • Award 1 point for: the nucleus divided into two (mitosis) in the first half hour while the cytoplasm was still one, and the cytoplasm divided (cytokinesis) only in the last half hour, so the two are separate divisions in order.

Glossary

interphase
The long stretch of the cell cycle between one division and the next, made of G1, S phase and G2. About 23 hours of a 24-hour cycle.
G1
The first growth gap of interphase, straight after a division. The G is short for gap.
S phase
The stage of interphase in which the DNA is copied. The S is short for synthesis, the making of new DNA.
G2
The second growth gap of interphase, after the DNA has been copied and before the division.
mitosis
The division of the nucleus: the two sets of chromosomes are pulled apart into two nuclei. It is not the division of the whole cell.
cytokinesis
The division of the cytoplasm, after mitosis, which ends with two daughter cells.

APBIO-U04-L14B G1, S phase and G2

Topic 4.5 · Cell Cycle · 69 steps

One cell from a culture drawn at 0, 2, 4, 6 and 8 hours, slightly larger each time, with a grainy nucleus in each
One cell from a culture drawn at 0, 2, 4, 6 and 8 hours, slightly larger each time, with a grainy nucleus in each

Here is a culture of cells, sampled every two hours for eight hours. Each time, the mass of a cell and the DNA in it are measured.

Suppose a culture of cells is measured every two hours: the mass of each cell, and the DNA in it. From hour 0 to hour 2 the mass rises from 2.0 to 2.4 nanograms (ng) while the DNA stays at 6 picograms (pg) per cell. Later the DNA climbs from 6 pg to 12 pg. Later still the mass keeps rising and the DNA stays at 12 pg. What is the cell doing in each stretch?

Unit 4 · Cell Communication and Cell Cycle

1G1: the cell grows

2

In G1 the cell makes more of its organelles and more cytosol. So the cell grows.

3

In G1 the DNA has not yet been copied. So each chromosome is still one chromatid.

4

A cell in G1 has as many chromatids as it has chromosomes.

5

Here is a table of a culture of cells measured every two hours. The table compares the mass of each cell with the DNA in it.

A table of a cell culture measured every two hours: hour 0, mass 2.0 ng, DNA 6 pg; hour 2, 2.4 ng, 6 pg; hour 4, 2.8 ng, 9 pg; hour 6, 3.2 ng, 12 pg; hour 8, 3.6 ng, 12 pg
A table of a cell culture measured every two hours: hour 0, mass 2.0 ng, DNA 6 pg; hour 2, 2.4 ng, 6 pg; hour 4, 2.8 ng, 9 pg; hour 6, 3.2 ng, 12 pg; hour 8, 3.6 ng, 12 pg
6

From hour 0 to hour 2 the mass per cell rises from 2.0 ng to 2.4 ng. Over the same two hours the DNA per cell stays at 6 pg.

7

So from hour 0 to hour 2 the cells are growing with their DNA uncopied. Those cells are in G1.

8

Video: Watch: G1, the cell grows

A cell in G1 makes more organelles and more cytosol and swells; its DNA stays uncopied, one chromatid per chromosome. The table beside it: mass rising from 2.0 ng to 2.4 ng, DNA steady at 6 pg.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14Ba.mp4

9

What you are expected to know Describe G1: the cell makes more organelles and more cytosol and grows, while its DNA stays uncopied, one chromatid per chromosome.

10
Check q1

Here is a second culture of cells, measured every three hours.

A table of a second cell culture measured every three hours: hour 0, mass 3.0 ng, DNA 8 pg; hour 3, 3.3 ng, 8 pg; hour 6, 3.6 ng, 11 pg; hour 9, 3.9 ng, 14 pg; hour 12, 4.2 ng, 16 pg; hour 15, 4.5 ng, 16 pg
A table of a second cell culture measured every three hours: hour 0, mass 3.0 ng, DNA 8 pg; hour 3, 3.3 ng, 8 pg; hour 6, 3.6 ng, 11 pg; hour 9, 3.9 ng, 14 pg; hour 12, 4.2 ng, 16 pg; hour 15, 4.5 ng, 16 pg

Which interval shows cells in G1?

  1. A. ✓ Hours 0 to 3
  2. B. Hours 3 to 6
    From hour 3 to hour 6 the DNA goes from 8 to 11 pg.
    So the cells are copying their DNA, which is S phase.
  3. C. Hours 6 to 9
    From hour 6 to hour 9 the DNA goes from 11 to 14 pg.
    So the cells are still copying their DNA.
  4. D. Hours 12 to 15
    From hour 12 to hour 15 the DNA sits at 16 pg, the doubled amount.
    So these cells have finished S phase.

Why: In G1 the cell grows and its DNA stays uncopied.
From hour 0 to hour 3 the mass rises from 3.0 ng to 3.3 ng and the DNA stays at 8 pg.
So hours 0 to 3 are G1.

11
Practice writing an answer

In the second culture, measured every three hours, the cells were in G1 from hour 0 to hour 3.

A table of a second cell culture measured every three hours: hour 0, mass 3.0 ng, DNA 8 pg; hour 3, 3.3 ng, 8 pg; hour 6, 3.6 ng, 11 pg; hour 9, 3.9 ng, 14 pg; hour 12, 4.2 ng, 16 pg; hour 15, 4.5 ng, 16 pg
A table of a second cell culture measured every three hours: hour 0, mass 3.0 ng, DNA 8 pg; hour 3, 3.3 ng, 8 pg; hour 6, 3.6 ng, 11 pg; hour 9, 3.9 ng, 14 pg; hour 12, 4.2 ng, 16 pg; hour 15, 4.5 ng, 16 pg

(a) Explain how the table shows that the cells were in G1 from hour 0 to hour 3. (1 pt)

Model answer In G1 the cell grows, so its mass rises.
In G1 the DNA has not yet been copied, so the DNA per cell stays the same.
From hour 0 to hour 3 the mass per cell rises from 3.0 ng to 3.3 ng.
Over the same three hours the DNA per cell stays at 8 pg.
Rising mass with steady DNA is the pattern of G1.
So the cells were in G1 from hour 0 to hour 3.
Rubric
  • Award 1 point for: the mass per cell rises (the cell is growing) while the DNA per cell stays the same (the DNA has not yet been copied), which is the pattern of G1.
12
Check q2

A chromosome is still uncopied, before S phase.

How many chromatids is the chromosome made of?

  1. A. ✓ One chromatid
  2. B. Two sister chromatids
    The copying makes the second chromatid.
    An uncopied chromosome is still one chromatid.

Why: Before the copying, a chromosome is one chromatid.
After the copying, it is two sister chromatids joined at the centromere.

13
Check q3 numeric entry

A cell in G1 has 14 chromosomes. In G1, before its DNA is copied, each chromosome is one chromatid.

Calculate the number of chromatids the cell has.

Answer: 14  (tolerance ±0)

Working
Write down the values in the question:
chromosomes = 14
chromatids in each chromosome before S phase = 1
Write down the equation:
chromatids=1×chromosomes
Substitute the values into the equation:
chromatids=1×chromosomes
chromatids=1×14
chromatids=14

14S phase: the DNA is copied

15

During S phase every chromosome is copied into two sister chromatids. So the DNA per cell doubles.

16

Here is a graph of DNA per cell, in picograms on the y-axis, against time since the last division, in hours on the x-axis.

DNA per cell in picograms against hours since the last division: flat at 6 pg through G1 (0 to 9 h), rising from 6 to 12 pg through S phase (9 to 19 h), flat at 12 pg through G2 (19 to 23 h), and dropping to 6 pg at the division in the last hour; gridlines every 2 pg
DNA per cell in picograms against hours since the last division: flat at 6 pg through G1 (0 to 9 h), rising from 6 to 12 pg through S phase (9 to 19 h), flat at 12 pg through G2 (19 to 23 h), and dropping to 6 pg at the division in the last hour; gridlines every 2 pg
17

A skin cell has 6 pg of DNA in G1. Through S phase the DNA per cell climbs. By G2 the skin cell has 12 pg: the G1 amount, doubled.

18

The copied DNA stays in the cell through G2. It also stays through mitosis. The DNA per cell halves only when the cell splits in two at cytokinesis.

19

So a cell in G2 has 12 pg. A cell in mitosis also has 12 pg.

20

A reading of 12 pg cannot by itself tell G2 from mitosis.

21

Video: Watch: S phase, the DNA doubles

On the DNA-per-cell graph the line sits at 6 pg through G1, climbs to 12 pg through S phase, and stays at 12 pg through G2 and through mitosis, dropping to 6 pg only when the cell splits in two.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14Bb.mp4

22

What you are expected to know Describe S phase: every chromosome is copied into two sister chromatids, so the DNA per cell doubles.

23
Check q4

A cell of one type has 11 pg of DNA in G1. The cell passes through S phase.

What happens to the DNA per cell during S phase?

  1. A. The DNA per cell stays at 11 pg
    In S phase every chromosome is copied into two sister chromatids.
    The cell then has twice the DNA.
  2. B. ✓ The DNA per cell doubles, from 11 pg to 22 pg
  3. C. The DNA per cell halves, from 11 pg to 5.5 pg
    In S phase every chromosome is copied.
    Copying adds DNA to the cell.

Why: In S phase every chromosome is copied into two sister chromatids.
So the DNA per cell doubles, to twice the G1 amount.

24
Practice writing an answer

Cells of one type have 5 pg of DNA in G1. One of these cells has 10 pg. The cell is in G2 or in mitosis: the reading fits both stages.

(a) Explain why a cell in G2 and a cell in mitosis give the same reading of 10 pg. (1 pt)

Model answer A reading of 10 pg is double the G1 amount of 5 pg.
So the cell has finished S phase: every chromosome has been copied into two sister chromatids.
The copied DNA stays in the cell through G2.
The copied DNA also stays in the cell through mitosis.
The DNA per cell halves only when the cell splits in two at cytokinesis.
So a cell in G2 and a cell in mitosis both have 10 pg.
Rubric
  • Award 1 point for: 10 pg is the doubled amount, and the DNA stays doubled through G2 and through mitosis and halves only when the cell splits at cytokinesis, so both stages give the same reading.

25Place a cell from its DNA reading

26

Suppose a skin cell reads 9 pg of DNA. 9 pg is more than the G1 amount of 6 pg. 9 pg is less than the doubled amount of 12 pg.

The same DNA-content graph with a point marked at 9 pg, part-way through S phase
The same DNA-content graph with a point marked at 9 pg, part-way through S phase
27

So the copying has begun and has not finished. The skin cell is part-way through S phase.

28

Here is a table of the three kinds of DNA reading. It compares each reading with the stage the reading places the cell in.

A table of three DNA readings and the stage each places the cell in: the G1 amount, 6 pg, is G1; between the G1 amount and double, between 6 and 12 pg, is part-way through S phase; double the G1 amount, 12 pg, is past S phase, G2 or mitosis
29

Video: Watch: Place the cell from its DNA

Three readings on the DNA graph: 6 pg lands in G1; 9 pg lands part-way up the S-phase climb; 12 pg lands on the flat top, which is G2 or mitosis.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14Bc.mp4

30

What you are expected to know Place a cell in the cycle from its DNA per cell: the G1 amount is G1, between the G1 amount and double is S phase, double is G2 or mitosis.

31Quick quiz: place the cell from its DNA mixed practice

32
Check q5

The chart shows how much DNA a cell of this type has in G1, beside how much DNA this cell has.

A bar chart with two bars, labelled in G1 and this cell, on a vertical axis of DNA per cell in picograms from 0 to 20 with a gridline every 4 pg; the this-cell bar is twice the height of the in-G1 bar
A bar chart with two bars, labelled in G1 and this cell, on a vertical axis of DNA per cell in picograms from 0 to 20 with a gridline every 4 pg; the this-cell bar is twice the height of the in-G1 bar

Which stage is this cell in?

  1. A. G1
    The this-cell bar is twice the height of the in-G1 bar: 16 pg against 8 pg.
    Double the G1 amount means the DNA has been copied.
  2. B. Part-way through S phase
    During S phase the bar would stop between 8 and 16 pg.
    This cell’s bar reaches the full double, 16 pg, so the copying is complete.
  3. C. ✓ G2 or mitosis

Why: The in-G1 bar shows the uncopied amount, 8 pg.
The this-cell bar is twice as tall, at 16 pg.
Double the G1 amount is the copied amount.
So the cell is past S phase: in G2 or in mitosis.

33
Check q6

The chart shows how much DNA a cell of this type has in G1, beside how much DNA this cell has.

A bar chart with two bars, labelled in G1 and this cell, on a vertical axis of DNA per cell in picograms from 0 to 20 with a gridline every 4 pg; the this-cell bar is the same height as the in-G1 bar
A bar chart with two bars, labelled in G1 and this cell, on a vertical axis of DNA per cell in picograms from 0 to 20 with a gridline every 4 pg; the this-cell bar is the same height as the in-G1 bar

Which stage is this cell in?

  1. A. ✓ G1
  2. B. Part-way through S phase
    During S phase the bar would stand between 8 and 16 pg.
    This cell’s bar matches the in-G1 bar at 8 pg, so the copying has yet to begin.
  3. C. G2 or mitosis
    Past S phase the bar would be twice the height of the in-G1 bar, at 16 pg.
    This cell’s bar is the same height as the in-G1 bar, 8 pg.

Why: The in-G1 bar shows the uncopied amount, 8 pg.
The this-cell bar is the same height, 8 pg.
So the DNA has not been copied: the cell is in G1.

34
Check q7

The chart shows how much DNA a cell of this type has in G1, beside how much DNA this cell has.

A bar chart with two bars, labelled in G1 and this cell, on a vertical axis of DNA per cell in picograms from 0 to 10 with a gridline every 1 pg; the this-cell bar is taller than the in-G1 bar but less than twice its height
A bar chart with two bars, labelled in G1 and this cell, on a vertical axis of DNA per cell in picograms from 0 to 10 with a gridline every 1 pg; the this-cell bar is taller than the in-G1 bar but less than twice its height

Which stage is this cell in?

  1. A. G1
    The this-cell bar is taller than the in-G1 bar: 7 pg against 5 pg.
    More than the G1 amount means copying has begun.
  2. B. ✓ Part-way through S phase
  3. C. G2 or mitosis
    Past S phase the bar would be double the in-G1 bar, 10 pg.
    This bar stops at 7 pg, short of the double: the copying is part done.

Why: The in-G1 bar shows the uncopied amount, 5 pg.
Double that is 10 pg.
The this-cell bar stands between the two, at 7 pg.
So the cell is part-way through S phase.

35
Check q8

The chart shows how much DNA a cell of this type has in G1, beside how much DNA this cell has.

A bar chart with two bars, labelled in G1 and this cell, on a vertical axis of DNA per cell in picograms from 0 to 24 with a gridline every 4 pg; the this-cell bar is twice the height of the in-G1 bar
A bar chart with two bars, labelled in G1 and this cell, on a vertical axis of DNA per cell in picograms from 0 to 24 with a gridline every 4 pg; the this-cell bar is twice the height of the in-G1 bar

Which stage is this cell in?

  1. A. G1
    The this-cell bar is twice the height of the in-G1 bar: 20 pg against 10 pg.
    Double the G1 amount means the DNA has been copied.
  2. B. Part-way through S phase
    During S phase the bar would stop between 10 and 20 pg.
    This cell’s bar reaches the full double, 20 pg, so the copying is complete.
  3. C. ✓ G2 or mitosis

Why: The in-G1 bar shows the uncopied amount, 10 pg.
The this-cell bar is twice as tall, at 20 pg.
Double the G1 amount is the copied amount.
So the cell is past S phase: in G2 or in mitosis.

36
Check q9

The chart shows how much DNA a cell of this type has in G1, beside how much DNA this cell has.

A bar chart with two bars, labelled in G1 and this cell, on a vertical axis of DNA per cell in picograms from 0 to 10 with a gridline every 1 pg; the this-cell bar is the same height as the in-G1 bar
A bar chart with two bars, labelled in G1 and this cell, on a vertical axis of DNA per cell in picograms from 0 to 10 with a gridline every 1 pg; the this-cell bar is the same height as the in-G1 bar

Which stage is this cell in?

  1. A. ✓ G1
  2. B. Part-way through S phase
    During S phase the bar would stand between 4 and 8 pg.
    This cell’s bar matches the in-G1 bar at 4 pg, so the copying has yet to begin.
  3. C. G2 or mitosis
    Past S phase the bar would be twice the height of the in-G1 bar, at 8 pg.
    This cell’s bar is the same height as the in-G1 bar, 4 pg.

Why: The in-G1 bar shows the uncopied amount, 4 pg.
The this-cell bar is the same height, 4 pg.
So the DNA has not been copied: the cell is in G1.

37
Check q10

The chart shows how much DNA a cell of this type has in G1, beside how much DNA this cell has.

A bar chart with two bars, labelled in G1 and this cell, on a vertical axis of DNA per cell in picograms from 0 to 30 with a gridline every 5 pg; the this-cell bar is taller than the in-G1 bar but less than twice its height
A bar chart with two bars, labelled in G1 and this cell, on a vertical axis of DNA per cell in picograms from 0 to 30 with a gridline every 5 pg; the this-cell bar is taller than the in-G1 bar but less than twice its height

Which stage is this cell in?

  1. A. G1
    The this-cell bar is taller than the in-G1 bar: 21 pg against 14 pg.
    More than the G1 amount means copying has begun.
  2. B. ✓ Part-way through S phase
  3. C. G2 or mitosis
    Past S phase the bar would be double the in-G1 bar, 28 pg.
    This bar stops at 21 pg, short of the double: the copying is part done.

Why: The in-G1 bar shows the uncopied amount, 14 pg.
Double that is 28 pg.
The this-cell bar stands between the two, at 21 pg.
So the cell is part-way through S phase.

38G2: getting ready

39

In G2 the DNA has already been copied. Now the cell makes the proteins it will need for division.

A cell in G2 with a grainy nucleus labelled 12 pg of DNA, two small centrosomes drawn beside it, and labels for the proteins and ATP being made
A cell in G2 with a grainy nucleus labelled 12 pg of DNA, two small centrosomes drawn beside it, and labels for the proteins and ATP being made
40
Check q11

A cell has used some of its ATP.

What happens to the ATP the cell has used?

  1. A. ✓ The cell remakes it from ADP and Pi, using energy from food
  2. B. The used ATP is gone for good, and the cell makes no more
    Each hydrolysis of ATP leaves ADP and Pi.
    The cell remakes ATP from that ADP and Pi with energy from food.

Why: Each hydrolysis of ATP leaves ADP and Pi.
The cell remakes ATP from that ADP and Pi.
The energy comes from the food the cell breaks down.

41

Division will use a great deal of ATP. So in G2 the cell also produces ATP in large quantities.

42

The cell also finishes copying one more structure. In mitosis, the fibers that move the chromosomes will grow from this structure.

43

The structure the fibers grow from is called the . The cell enters mitosis with two centrosomes.

44

Now consider cells given a drug during S phase. The drug stops the copying of DNA.

45

The cells still grow to full size, but stretches of their DNA stay uncopied.

46

Video: Watch: G2, getting ready

A cell in G2 with 12 pg of DNA in its nucleus: proteins for the division being made, ATP being made in large amounts, and a second centrosome finished beside the first.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14Bd.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14Bd.mp4

47

What you are expected to know Describe G2: the cell makes the proteins it will need for division and produces ATP in large quantities.

48

What you are expected to know Describe how G2 ends: the cell finishes copying its centrosome, so it enters mitosis with two.

49
Check q12

A cell has finished S phase and is in G2.

Which of the following happens in G2?

  1. A. ✓ The cell makes proteins for division
  2. B. The nucleus divides
    The nucleus divides in mitosis, after G2.
  3. C. The cell leaves the cycle
    A cell leaves the cycle from G1.
    G2 is preparation for division.
  4. D. The DNA is copied
    The DNA is copied in S phase.
    G2 begins when the copying is done.

Why: G2 is the preparation stage.
The cell makes the proteins for division and produces ATP in quantity.

50
Check q13

A student says: “G2 is a rest gap. The cell does nothing in G2 until mitosis begins.”

Is the student correct?

  1. A. Yes — the cell rests in G2
    The G of G2 stands for gap, but it is a growth gap: the cell makes the proteins for division, produces ATP in large amounts and finishes copying its centrosome.
  2. B. ✓ No — the cell is busy in G2, preparing to divide

Why: G2 is a growth gap, not a rest.
In G2 the cell makes the proteins for division, produces ATP in large amounts and finishes copying its centrosome.

51
Practice writing an answer

A cell has finished S phase and is in G2. The cell is busy in G2, preparing to divide.

(a) Explain why the cell is busy in G2. (1 pt)

Model answer In G2 the cell makes the proteins it will need for division.
The cell also produces ATP in large quantities, because division will use a great deal of ATP.
The cell also finishes copying its centrosome, so it enters mitosis with two centrosomes.
Each of these jobs must be done before mitosis begins.
So the cell is busy in G2.
G2 is a growth gap, not a rest.
Rubric
  • Award 1 point for: in G2 the cell makes the proteins for division, produces ATP in large quantities and finishes copying its centrosome, all before mitosis, so G2 is busy preparation rather than rest.

52Quick quiz: the centrosome mixed practice

53
Check q14

In mitosis, fibers move the chromosomes.

From which structure do the fibers grow?

  1. A. ✓ The centrosome
  2. B. The nucleus
    The nucleus holds the chromosomes.
    The fibers grow from a separate structure beside it.
  3. C. The chromosome
    A chromosome is moved by the fibers.
    The fibers grow from a separate structure.

Why: The centrosome is the structure the fibers grow from.

54
Practice writing an answer

A cell enters mitosis with two centrosomes.

(a) State what the centrosome is. (1 pt)

Model answer The centrosome is the structure from which the fibers that move the chromosomes grow.
Rubric
  • Award 1 point for: the structure the fibers that move the chromosomes grow from.
55
Check q15

A cell copies its centrosome during interphase.

In which stage does it finish the copying?

  1. A. G1
    In G1 the cell grows with one centrosome.
    The copying of the centrosome is finished later, in the second growth gap.
  2. B. S phase
    In S phase the cell copies its DNA.
    The copying of the centrosome is finished in the stage after S phase.
  3. C. ✓ G2

Why: The cell finishes copying its centrosome in G2, so it enters mitosis with two.

56
Check q16

A cell is entering mitosis.

How many centrosomes does it have?

  1. A. One
    The cell finished copying its centrosome in G2.
    So it enters mitosis with two.
  2. B. ✓ Two

Why: In G2 the cell finishes copying its centrosome.
So it enters mitosis with two centrosomes.

57The culture, again

58

Now go back to the culture measured every two hours.

59

From hour 0 to hour 2 the mass per cell rose from 2.0 ng to 2.4 ng while the DNA per cell stayed at 6 pg. Those cells were in G1, growing with the DNA uncopied.

60

Later the DNA per cell climbed from 6 pg to 12 pg. Those cells were in S phase, copying every chromosome into two sister chromatids.

61

Later still the mass kept rising and the DNA stayed at 12 pg. Those cells were in G2, making the proteins, the ATP and the second centrosome for the division.

62Mixed practice mixed practice

63
Check q17

Here is the second culture again, measured every three hours.

A table of a second cell culture measured every three hours: hour 0, mass 3.0 ng, DNA 8 pg; hour 3, 3.3 ng, 8 pg; hour 6, 3.6 ng, 11 pg; hour 9, 3.9 ng, 14 pg; hour 12, 4.2 ng, 16 pg; hour 15, 4.5 ng, 16 pg
A table of a second cell culture measured every three hours: hour 0, mass 3.0 ng, DNA 8 pg; hour 3, 3.3 ng, 8 pg; hour 6, 3.6 ng, 11 pg; hour 9, 3.9 ng, 14 pg; hour 12, 4.2 ng, 16 pg; hour 15, 4.5 ng, 16 pg

During which interval were the cells copying their DNA?

  1. A. Hours 0 to 3
    From hour 0 to hour 3 the DNA stays at 8 pg.
    The cells are growing with their DNA uncopied, which is G1.
  2. B. Hours 0 to 15
    The DNA is steady at 8 pg up to hour 3 and steady at 16 pg from hour 12.
    It rises only between hour 3 and hour 12.
  3. C. ✓ Hours 3 to 12
  4. D. Hours 12 to 15
    From hour 12 to hour 15 the DNA sits at 16 pg, already doubled.

Why: The DNA rises from 8 pg to 16 pg between hour 3 and hour 12. So the cells were copying their DNA in that interval.
That interval is S phase.

64
Check q18

Here is a third culture of cells, measured every four hours.

A table of a third cell culture measured every four hours: hour 0, mass 2.0 ng, DNA 9 pg; hour 4, 2.3 ng, 9 pg; hour 8, 2.6 ng, 12 pg; hour 12, 2.9 ng, 15 pg; hour 16, 3.2 ng, 18 pg; hour 20, 3.5 ng, 18 pg
A table of a third cell culture measured every four hours: hour 0, mass 2.0 ng, DNA 9 pg; hour 4, 2.3 ng, 9 pg; hour 8, 2.6 ng, 12 pg; hour 12, 2.9 ng, 15 pg; hour 16, 3.2 ng, 18 pg; hour 20, 3.5 ng, 18 pg

In which interval had the cells finished copying their DNA?

  1. A. Hours 0 to 4
    From hour 0 to hour 4 the DNA stays at 9 pg.
    That is the G1 amount, so the copying has yet to start.
  2. B. Hours 4 to 8
    From hour 4 to hour 8 the DNA rises from 9 pg to 12 pg.
    So the copying is under way.
  3. C. Hours 8 to 12
    From hour 8 to hour 12 the DNA rises from 12 pg to 15 pg, between the G1 amount and the doubled amount, so the copying is still under way.
  4. D. ✓ Hours 16 to 20

Why: The G1 amount is 9 pg, so the copied amount is 18 pg.
The DNA sits at 18 pg from hour 16 to hour 20. So the copying is finished in that interval.

65
Check q19 numeric entry

A cell of one type has 7 pg of DNA in G1.

Calculate its DNA content in G2.

Answer: 14 pg  (tolerance ±0.05)

Working
Write down the values in the question:
DNA in G1 = 7 pg
Write down the equation:
DNA in G2=2×DNA in G1
Substitute the values into the equation:
DNA in G2=2×DNA in G1
DNA in G2=2×7
DNA in G2=14pg
66
Check q20

A cell has grown since its last division, but its DNA is still 6 pg, the uncopied amount.

Which stage is the cell in?

  1. A. S phase
    In S phase the DNA is being copied, so it would be above 6 pg.
  2. B. G2
    In G2 the DNA has been copied, so it would be 12 pg.
  3. C. ✓ G1
  4. D. Mitosis
    A cell in mitosis has copied its DNA and is dividing its nucleus.
    This cell is growing with its DNA uncopied.

Why: Growing with the DNA still uncopied is G1: the cell has made more organelles and cytosol, but S phase has not yet begun.

67
Check q21

A drug stops cells from copying their DNA.

In which stage of the cycle does the drug act?

  1. A. G1
    In G1 the cell grows.
    The DNA is copied in S phase, the stage that follows.
  2. B. G2
    G2 begins after the copying is complete.
  3. C. Mitosis
    Mitosis divides the nucleus.
    The DNA was copied earlier, in S phase.
  4. D. ✓ S phase

Why: The DNA is copied in S phase, so a drug that stops the copying acts there.

68
Practice writing an answer

Here is a table of a fourth culture of cells, measured every five hours: the mass of each cell and the DNA in it.

A table of a fourth cell culture measured every five hours: hour 0, mass 2.2 ng, DNA 7 pg; hour 5, 2.5 ng, 7 pg; hour 10, 2.8 ng, 10 pg; hour 15, 3.1 ng, 14 pg; hour 20, 3.4 ng, 14 pg

(a) Identify the interval in which the cells were in S phase. (1 pt)

Model answer The cells were in S phase from hour 5 to hour 15.
Rubric
  • Award 1 point for: hours 5 to 15.

(b) Explain how the table shows that the cells were in S phase during that interval. (1 pt)

Model answer In S phase every chromosome is copied, so the DNA per cell rises.
From hour 0 to hour 5 the DNA per cell stays at 7 pg, the G1 amount.
From hour 5 to hour 15 the DNA per cell rises from 7 pg to 14 pg.
14 pg is double the G1 amount, so by hour 15 every chromosome has been copied.
So the cells were in S phase from hour 5 to hour 15.
Rubric
  • Award 1 point for: the DNA per cell rises from the G1 amount, 7 pg, to double that amount, 14 pg, during the interval, and rising DNA per cell is the copying of every chromosome, which is S phase.

Glossary

centrosome
The structure from which the fibers that will move the chromosomes grow. The cell finishes copying it in G2 and enters mitosis with two.

APBIO-U04-L14C Stepping out of the cycle

Topic 4.5 · Cell Cycle · 35 steps

A rat’s liver drawn three times: whole before surgery; the small part left after 67% is cut away, with a straight cut edge; and the organ three days later, regrown to about half its first size
A rat’s liver drawn three times: whole before surgery; the small part left after 67% is cut away, with a straight cut edge; and the organ three days later, regrown to about half its first size

Here is a rat’s liver drawn three times: before surgery, just after 67% of it is removed, and three days later.

Almost all of an adult rat’s liver cells are not dividing; about 2% of them are copying their DNA. A surgeon removes 67% of the liver. Within a day, 17% of the remaining liver cells are copying their DNA, and within three days the number of liver cells has reached 146% of what was left. A mature nerve cell in the same rat has not divided since before birth and never will. Where were the liver cells before the surgery, and what called them back?

Unit 4 · Cell Communication and Cell Cycle

1G0: out of the cycle, still working

2
Check q1

A daughter cell has just formed after cytokinesis.

Which stage does it enter first?

  1. A. ✓ G1
  2. B. S phase
    S phase is the second stage of interphase.
    A new daughter cell grows first, in G1.
  3. C. G2
    G2 is the third stage of interphase.
    A new daughter cell grows first, in G1.

Why: Each daughter cell begins a new cycle at its first stage, G1.

3

Not every cell goes round the cycle. From G1 a cell can step out of the cycle.

The full cycle wheel with an arrow leaving G1 to a box labelled G0, out of the cycle but still working, and an arrow from G0 back into G1
The full cycle wheel with an arrow leaving G1 to a box labelled G0, out of the cycle but still working, and an arrow from G0 back into G1
4

Outside the cycle, the cell carries on its work without preparing to divide.

5

This state is called . The G is the gap of G1 and G2. The 0 says the cell is outside the cycle.

6

A mature nerve cell stays in G0 for life. It works for decades and never divides.

A mature nerve cell: a cell body with a grainy nucleus, short branching fibers on the left, and one long fiber running to the right; labelled in G0 for life
A mature nerve cell: a cell body with a grainy nucleus, short branching fibers on the left, and one long fiber running to the right; labelled in G0 for life
7

Now consider the rat’s liver cells. Almost all of them sit in G0 too.

8

Only about 2% of the liver cells are copying their DNA.

9

A surgeon removes 67% of the liver. Within a day, 17% of the remaining liver cells are copying their DNA.

A bar chart of the percent of the rat’s liver cells copying their DNA: before surgery, 2%; one day after 67% of the liver was removed, 17%; vertical axis 0 to 25% with a gridline every 5%
A bar chart of the percent of the rat’s liver cells copying their DNA: before surgery, 2%; one day after 67% of the liver was removed, 17%; vertical axis 0 to 25% with a gridline every 5%
10

Within three days the number of liver cells reaches 146% of the number left after the surgery.

11

A signal from the damaged liver reached the G0 liver cells. The signal called them back into the cycle at G1.

12

G0 is not death, and G0 is not a stall. A G0 cell is busy doing its job.

13

Many G0 cells re-enter the cycle when a signal arrives.

14

Video: Watch: Stepping out, and being called back

An arrow leaves the wheel at G1 into a box marked G0: the cell keeps working but does not prepare to divide. A nerve cell stays there for life. A liver cell steps back in when a signal from the damaged liver arrives.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14Ca.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L14Ca.mp4

15

What you are expected to know Describe G0: a cell leaves the cycle from G1 and carries on its work without preparing to divide.

16

What you are expected to know Name a cell that stays in G0 for life: a mature nerve cell.

17

What you are expected to know Name a cell that re-enters the cycle when a signal arrives: a liver cell after surgery.

18
Check q2

A mature nerve cell has worked normally for forty years, and its last division was before birth.

Which of the following is the nerve cell in?

  1. A. Mitosis
    A cell in mitosis has condensed chromosomes and does no normal work.
    This nerve cell works normally.
  2. B. S phase
    A cell in S phase is copying its DNA on the way to a division.
    This nerve cell is not preparing to divide.
  3. C. G1
    A cell in G1 is growing toward S phase.
    A cell that is out of the cycle altogether, and working, is in G0.
  4. D. ✓ G0

Why: A cell that works for decades without preparing to divide has left the cycle.
That cell is in G0. A mature nerve cell stays in G0 for life.

19
Check q3

A fat cell has done the same job, storing and releasing fat, for ten years. Its last division was ten years ago. A student says: “A cell that stays undivided for ten years must be stuck part-way through its cycle.”

Is the student correct?

  1. A. Yes — the fat cell is stuck at a stage inside its cycle
    A cell stuck inside the cycle is waiting for the next stage.
    Ten years with no preparation to divide means the fat cell has left the cycle for G0.
  2. B. ✓ No — the fat cell has left the cycle and is at no stage of it

Why: A cell that works for years and makes no preparation to divide has left the cycle from G1 into G0.
G0 is outside the cycle.
So the fat cell is at no stage of the cycle.

20
Practice writing an answer

A fat cell has done the same job, storing and releasing fat, for ten years. Its last division was ten years ago. The fat cell is in G0.

(a) Explain why the fat cell is in G0 rather than at a stage inside its cycle. (1 pt)

Model answer A cell stuck part-way through its cycle is waiting to move on to the next stage.
The fat cell has made no preparation to divide for ten years.
Instead the fat cell has carried on its work, storing and releasing fat.
A cell that carries on its work without preparing to divide has left the cycle from G1 into G0.
G0 is outside the cycle.
So the fat cell is at no stage of the cycle.
Rubric
  • Award 1 point for: the fat cell has left the cycle from G1 into G0, a state outside the cycle in which it carries on its work without preparing to divide, so it is at no stage of the cycle (and G0 is not death).
21
Check q4

Almost all of a mouse’s liver cells are in G0. Only 3% are copying their DNA. A surgeon removes 70% of the liver. A day later, 25% of the liver cells are copying their DNA. After three days the liver has 140% of the cells it had just after the surgery.

Which statement explains these results?

  1. A. The cells grew larger without dividing, and the larger cells were counted as more cells
    The count of liver cells rose to 140% of the start.
    A count of cells rises only when cells divide.
  2. B. The 3% of cells already cycling produced all the new cells while the G0 cells stayed out of the cycle
    The share of liver cells copying DNA rose from 3% to 25%.
    The 3% of cycling cells cannot raise that share eightfold, so G0 cells re-entered the cycle.
  3. C. ✓ G0 liver cells re-entered the cycle, copied their DNA and divided
  4. D. The liver cells copied their DNA after they had divided
    The DNA is copied in S phase, before mitosis.
    In the data, the copying at one day came before the extra cells at three days.

Why: The jump from 3% to 25% of liver cells copying DNA shows liver cells entering S phase.
The rise to 140% shows those cells dividing.
So G0 liver cells re-entered the cycle.

22Quick quiz: G0 mixed practice

23
Check q5

A muscle cell has done the same job for twenty years. Its last division was twenty years ago.

Which state is the muscle cell in?

  1. A. ✓ G0
  2. B. G1
    A cell in G1 is growing toward S phase and a division.
    This cell is not preparing to divide.

Why: A cell that works without preparing to divide has left the cycle.
That state is G0.

24
Practice writing an answer

A mature nerve cell has not divided since before birth.

(a) State what G0 is. (1 pt)

Model answer G0 is a state outside the cycle, entered from G1, in which a cell carries on its work but is not preparing to divide.
Rubric
  • Award 1 point for: a state outside the cycle (left from G1) in which the cell keeps working but does not prepare to divide.
25
Check q6

A cell steps out of the cycle into G0.

From which stage does it step out?

  1. A. ✓ G1
  2. B. S phase
    In S phase the cell is copying its DNA, on the way to a division.
    A cell leaves the cycle before that, from G1.
  3. C. G2
    In G2 the cell is preparing to divide.
    A cell leaves the cycle before that, from G1.

Why: A cell leaves the cycle from G1 into G0.

26
Check q7

A cell is in G0.

Is the cell dead?

  1. A. Yes — a cell outside the cycle is dead
    A G0 cell carries on its work.
    A nerve cell in G0 works for decades.
  2. B. ✓ No — the cell is busy doing its job

Why: G0 is not death.
A G0 cell is busy doing its job; it is only not preparing to divide.

27
Check q8

A cell is in G0.

Is the cell copying its DNA?

  1. A. Yes — every cell copies its DNA
    Copying the DNA is a preparation to divide, done in S phase.
    A G0 cell is outside the cycle and is not preparing to divide.
  2. B. ✓ No — a G0 cell is not preparing to divide

Why: The DNA is copied in S phase, inside the cycle.
A G0 cell is outside the cycle.
So it is not copying its DNA.

28The rat’s liver, again

29

Now go back to the rat’s liver. Before the surgery, almost all of its cells sat in G0: out of the cycle, working, with only about 2% copying their DNA.

30

After 67% of the liver was removed, a signal from the damaged liver called the G0 cells back into the cycle.

31

Within a day 17% of the remaining cells were copying their DNA. Within three days the liver had 146% of the cells that were left.

32

The rat’s mature nerve cell received no such call. It stays in G0 for life.

33Mixed practice mixed practice

34
Practice writing an answer

Researchers removed 67% of the liver from eight rats. Eight other rats were opened and closed by the same surgery with the liver left intact. Twenty-four hours later the researchers measured the percent of liver cells that were copying their DNA in each rat. Before surgery, almost all liver cells in both groups were in G0. A rat liver cell in G1 has 6.5 pg of DNA. The graph shows the two means with error bars that represent ±2SE.

A bar chart of the percent of liver cells copying DNA 24 hours after surgery for two groups of rats: 67% of the liver removed, and liver left intact. The vertical axis runs from 0 to 30% with gridlines every 2% and labels every 10%. The bar for the rats with 67% of the liver removed is many times taller than the bar for the rats with the liver left intact. Each bar carries an error bar; the legend reads: error bars represent ±2SE
A bar chart of the percent of liver cells copying DNA 24 hours after surgery for two groups of rats: 67% of the liver removed, and liver left intact. The vertical axis runs from 0 to 30% with gridlines every 2% and labels every 10%. The bar for the rats with 67% of the liver removed is many times taller than the bar for the rats with the liver left intact. Each bar carries an error bar; the legend reads: error bars represent ±2SE

(a) Identify the independent variable and the dependent variable in this experiment. (1 pt)

Model answer The independent variable is whether 67% of the liver was removed or left intact.
The dependent variable is the percent of liver cells copying their DNA 24 hours after surgery.
Rubric
  • Award 1 point for: the independent variable is the removal of 67% of the liver (removed versus left intact) and the dependent variable is the percent of liver cells copying DNA at 24 hours.

Slip Naming the surgery itself as the independent variable. The rats with 67% of the liver removed and the rats with the liver left intact both had the surgery. What differed between the two groups was whether 67% of the liver was cut away.

(b) Calculate the DNA content of a liver cell that has finished S phase and is now in G2. (1 pt)

Answer: 13 pg  (tolerance ±0.05)

Model answer The cell has 13 pg of DNA: the 6.5 pg of G1, copied.
Working
Write down the values in the question:
DNA in G1 = 6.5 pg
Write down the equation:
DNA after S phase=2×DNA in G1
Substitute the values into the equation:
DNA after S phase=2×DNA in G1
DNA after S phase=2×6.5
DNA after S phase=13pg
Rubric
  • Award 1 point for: 13 pg.

(c) Justify the claim that removing part of the liver made more liver cells copy their DNA, using the error bars. (1 pt)

Model answer Read against the gridlines, the ±2SE bar for the rats with 67% of the liver removed runs from about 14.5% to about 19.5%.
The ±2SE bar for the rats with the liver left intact runs from about 1% to about 2%.
The two bars do not overlap.
So the difference between the removed group and the intact group is very unlikely to be chance.
Therefore removing part of the liver made more liver cells copy their DNA.
Rubric
  • Award 1 point for: the ±2SE bars (about 14.5–19.5% and about 0.8–2.2%, read against the gridlines) do not overlap, so the higher percent copying DNA after the removal is unlikely to be chance.
  • Accept: readings within half a gridline (1 percentage point) of those values, that is 13.5–15.5 to 18.5–20.5 for the removed group and 0–1.8 to 1.2–3.2 for the intact group. Do not award the point for a comparison of the two means alone.

Slip Comparing the two means alone, about 17% against about 1.5%. The claim rests on the ±2SE bars not overlapping.

(d) Explain, in terms of G0 and the stages of the cycle, how liver cells that had left the cycle came to be copying their DNA a day later. (1 pt)

Model answer Before surgery the liver cells were in G0: out of the cycle, working but not preparing to divide.
After the surgery a signal from the damaged liver reached the liver cells.
The signal made the liver cells re-enter the cycle at G1, where they grew.
Then the liver cells moved into S phase, where the DNA is copied.
That is why 17% of the liver cells were copying DNA a day later.
Rubric
  • Award 1 point for: the cells were in G0 and re-entered the cycle (into G1) when a signal arrived, then moved into S phase, where DNA is copied.

Slip Saying the cells were held in S phase all along. Before surgery the liver cells were in G0, out of the cycle. The liver cells re-entered the cycle after the surgery.

Glossary

G0
A state outside the cycle, entered from G1, in which a cell carries on its work but is not preparing to divide. Some cells stay there for life; others re-enter the cycle when a signal arrives.

APBIO-U04-L15 Building the machine: prophase and metaphase

Topic 4.5 · Cell Cycle · 70 steps

A photograph of stained onion root-tip cells, one with tangled dark rods and others with dark grainy nuclei, beside five drawn cells in a row: three show a grainy nucleus, one shows scattered dark rods, one a single row of rods across its middle
A photograph of stained onion root-tip cells, one with tangled dark rods and others with dark grainy nuclei, beside five drawn cells in a row: three show a grainy nucleus, one shows scattered dark rods, one a single row of rods across its middle

Photo: staticd, Wikimedia Commons, CC BY-SA 3.0 (cropped and resized).

Here is a slide of an onion root tip: a photograph of its cells on the left, and the same kind of cells drawn on the right.

On a slide of an onion root tip, most cells look ordinary. Scattered among them are cells caught mid-division: in one, dark rods lie scattered; in another they stand in a single row across the middle. Nothing on the slide is moving, but these cells can be put in order.

What is happening in the cell with scattered rods, and what has happened by the time they stand in a row?

Unit 4 · Cell Communication and Cell Cycle

1The two poles and the mitotic spindle

2

Video: Watch: The two poles and the mitotic spindle

The two centrosomes move to opposite ends of the cell, and between them a framework of protein fibers grows that will move the chromosomes.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L15a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L15a.mp4

3
Check q1

In G2 a cell finished copying one structure, so it enters mitosis with two of them.

Which structure did the cell copy?

  1. A. ✓ The centrosome
  2. B. The centromere
    The centromere is the point where two sister chromatids are held together.
    The structure copied in G2 is the centrosome.
  3. C. The nucleus
    The nucleus divides in mitosis.
    The structure copied in G2 is the centrosome.

Why: In G2 the cell finished copying its centrosome.
So it enters mitosis with two centrosomes.

4

Here is an animal cell at the start of its division, drawn with four chromosomes.

A cell with four copied chromosomes near the middle, two centrosomes drawn as dots at the left and right ends of the cell, and short spindle fibers growing inward from each centrosome
A cell with four copied chromosomes near the middle, two centrosomes drawn as dots at the left and right ends of the cell, and short spindle fibers growing inward from each centrosome
5

The two centrosomes move to opposite ends of the cell.

6

The cell will divide toward these two ends. Each of the two ends is called a .

7

Between the two poles, a framework of protein fibers begins to grow.

8

This framework will move the chromosomes.

9

The framework is wide in the middle and narrow at the two poles, like a spindle for spinning thread. So the framework is called the .

10

What you are expected to know Identify the mitotic spindle and the poles: the two centrosomes move to opposite ends of the cell, the poles, and a framework of protein fibers that will move the chromosomes grows between them.

11
Check q2

As a cell starts to divide, a framework of protein fibers begins to grow between its two poles.

Which of the following is this framework?

  1. A. The connection that holds sister chromatids together
    The connection that holds sister chromatids together sits at the centromere.
    The spindle is the set of fibers that will pull on it.
  2. B. The nuclear envelope
    The nuclear envelope surrounds the nucleus.
    The fibers grow from the centrosomes at the poles.
  3. C. ✓ The mitotic spindle
  4. D. A condensed chromosome
    A condensed chromosome is DNA wound on proteins, coiled tight.
    The spindle is a separate set of protein fibers, and the spindle moves the chromosomes.

Why: The mitotic spindle is a framework of protein fibers growing between the two poles.
Its job is to move the chromosomes.

12Quick quiz: the poles and the mitotic spindle mixed practice

13
Check q3

A dividing cell has two ends that it will divide toward.

What is each of these ends called?

  1. A. A centrosome
    The centrosome is the structure that sits at each end.
    The end itself is a pole.
  2. B. ✓ A pole
  3. C. A centromere
    The centromere is the point where two sister chromatids are joined.
    The end of the cell is a pole.

Why: The cell divides toward its two ends.
Each end is called a pole.

14
Check q4

The mitotic spindle has grown between the two poles of a dividing cell.

What will the mitotic spindle do?

  1. A. Copy the chromosomes
    The chromosomes were copied in S phase, before the spindle grew.
    The spindle moves them.
  2. B. Coil the chromosomes into rods
    The chromosomes coil into rods on their own.
    The spindle moves them.
  3. C. ✓ Move the chromosomes

Why: The mitotic spindle is a framework of protein fibers.
Its job is to move the chromosomes.

15
Check q5

In a dividing cell, the two centrosomes have moved apart.

Where do the two centrosomes now sit?

  1. A. ✓ At the two poles
  2. B. In the middle of the cell
    The two centrosomes move to opposite ends of the cell, not to its middle.
  3. C. Inside the nucleus
    The centrosomes sit at the two ends of the cell, outside the nucleus.

Why: The two centrosomes move to opposite ends of the cell.
Each end is a pole, so each centrosome sits at a pole.

16
Check q6

The mitotic spindle is a framework of fibers.

What are the fibers made of?

  1. A. DNA
    DNA is what the chromosomes are made of.
    The spindle fibers are protein.
  2. B. ✓ Protein
  3. C. Phospholipid
    Phospholipids make up the cell's membranes.
    The spindle fibers are protein.

Why: The mitotic spindle is a framework of protein fibers.

17
Check q7

A dividing cell's mitotic spindle is wide in the middle and narrow at its two ends.

Where are the two poles?

  1. A. At the wide middle
    The fibers spread out in the middle of the cell.
    They come together at the two ends, the poles.
  2. B. ✓ At the narrow ends

Why: The fibers grow from a centrosome at each end of the cell.
Each end is a pole, so the spindle is narrow at the poles.

18
Practice writing an answer

A cell is dividing.

(a) State what a pole of a dividing cell is. (1 pt)

Model answer A pole is one of the two ends of a dividing cell, the ends the cell will divide toward.
Rubric
  • Award 1 point for: one of the two ends of the cell, the ends the cell divides toward (a centrosome sits at each).

(b) State what the mitotic spindle is. (1 pt)

Model answer The mitotic spindle is a framework of protein fibers that grows between the two poles of a dividing cell and moves the chromosomes.
Rubric
  • Award 1 point for: a framework of protein fibers between the two poles that moves the chromosomes.

19Prophase: the rods appear and the envelope breaks down

20

Video: Watch: Prophase

The copied chromosomes coil into visible rods, the spindle begins to grow between the poles, and the nuclear envelope breaks down.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L15b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L15b.mp4

21
Check q8

In S phase every chromosome was copied.

What does one chromosome consist of as the cell enters mitosis?

  1. A. ✓ Two sister chromatids joined at one centromere
  2. B. One long uncopied thread of DNA
    The single thread was the chromosome before S phase.
    Copying made two.
  3. C. Two different chromosomes stuck to each other
    The two are copies of one chromosome, not two different chromosomes.

Why: S phase copied each chromosome.
The two copies, sister chromatids, stay joined at one centromere.

22

Here is the cell again, drawn with its four chromosomes.

A cell with four copied chromosomes coiling into rods near the middle, a dashed nuclear envelope breaking up, two centrosomes at the left and right poles with short spindle fibers growing from each
A cell with four copied chromosomes coiling into rods near the middle, a dashed nuclear envelope breaking up, two centrosomes at the left and right poles with short spindle fibers growing from each
23

Each copied chromosome coils up into a short thick rod.

24

Each rod is still two sister chromatids joined at one centromere.

25

The spindle begins to grow between the poles.

26

Late in this stage, the nuclear envelope breaks down. Now the spindle fibers can reach the chromosomes.

27

When a cell shows rods appearing, the spindle starting and the nuclear envelope breaking down, it is in the first stage of mitosis.

28

This stage is called . Pro means before: prophase comes before every other stage of mitosis.

29

On the onion slide, a prophase cell shows dark rods scattered near its middle. No nuclear envelope is visible around them.

30

What you are expected to know Identify prophase by its events: the copied chromosomes condense into visible rods, the spindle begins to grow between the poles, and the nuclear envelope breaks down.

31
Check q9

Here is a dividing cell drawn with six chromosomes.

A cell drawn with six X shapes, each two rods crossing at a small white dot, lying at different angles and positions near the middle of the cell inside a dashed oval; a small circle at each end of the cell with three short lines reaching in from each
A cell drawn with six X shapes, each two rods crossing at a small white dot, lying at different angles and positions near the middle of the cell inside a dashed oval; a small circle at each end of the cell with three short lines reaching in from each

Which stage is the cell in?

  1. A. Interphase
    In interphase the chromatin is grainy inside an unbroken nuclear envelope.
    Here the chromosomes have coiled into rods and the envelope is breaking up.
  2. B. ✓ Prophase

Why: Scattered rods, each a joined pair, a breaking nuclear envelope, and fibers growing in from two poles: the cell is in prophase.

32Quick quiz: prophase mixed practice

33
Check q10

In a cell, the chromosomes are coiling into rods and the nuclear envelope is breaking down.

Is the cell in prophase?

  1. A. ✓ Yes
  2. B. No
    Rods appearing and the nuclear envelope breaking down are the events of prophase.

Why: The chromosomes coil into rods and the nuclear envelope breaks down in prophase.

34
Check q11

A cell's chromatin looks grainy inside an unbroken nuclear envelope.

Is the cell in prophase?

  1. A. Yes
    In prophase the chromatin has coiled into rods and the envelope is breaking down.
    Here neither has happened.
  2. B. ✓ No

Why: Grainy chromatin inside an unbroken nuclear envelope is interphase.
In prophase the chromatin coils into rods and the envelope breaks down.

35
Check q12

In a cell, spindle fibers from two poles reach in toward scattered rods.

Is the cell in prophase?

  1. A. ✓ Yes
  2. B. No
    Scattered rods with a spindle growing in from two poles are the picture of prophase.

Why: In prophase the chromosomes have coiled into scattered rods and the spindle is growing in from the poles.

36
Check q13

A cell is copying its DNA.

Is the cell in prophase?

  1. A. Yes
    DNA is copied in S phase, before mitosis begins.
    Prophase is the first stage of mitosis.
  2. B. ✓ No

Why: A cell copies its DNA in S phase.
Prophase comes later, as the first stage of mitosis.

37
Check q14

A cell has entered prophase.

Which of the following happens?

  1. A. ✓ The nuclear envelope breaks down
  2. B. The DNA is copied
    The DNA was copied in S phase, before mitosis began.
  3. C. The cytoplasm divides
    The cytoplasm divides in cytokinesis, after mitosis has ended.

Why: In prophase the chromosomes coil into rods, the spindle begins to grow, and the nuclear envelope breaks down.

38
Practice writing an answer

A cell has just entered mitosis.

(a) State what prophase is. (1 pt)

Model answer Prophase is the first stage of mitosis: the copied chromosomes coil into visible rods, the spindle begins to grow between the poles, and the nuclear envelope breaks down.
Rubric
  • Award 1 point for: the first stage of mitosis, with its events named: the chromosomes condense into rods, the spindle begins to grow, and the nuclear envelope breaks down (any two of the three).

39Metaphase: a single row across the middle

40

Video: Watch: Metaphase

Fibers from both poles attach to every chromosome at its centromere and pull the chromosomes into a single row along the equator, each still two joined sister chromatids.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L15c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L15c.mp4

41

Spindle fibers from both poles attach to every chromosome at its centromere.

A cell with four copied chromosomes in a single row along a dashed line across the middle, spindle fibers from both poles attached at each centromere
A cell with four copied chromosomes in a single row along a dashed line across the middle, spindle fibers from both poles attached at each centromere
42

The fibers pull. The chromosomes move into a single row.

43

The row lies halfway between the two poles, like the equator of a globe. This plane halfway between the poles is called the .

44

Each chromosome in the row is still two joined sister chromatids.

45

Fibers from the left pole hold one sister chromatid. Fibers from the right pole hold the other sister chromatid.

46

When the chromosomes stand in a single row along the equator, the stage is called .

47

One way to remember the name: in metaphase the chromosomes stand in the middle of the cell.

48

Scattered rods mean the cell has not yet reached metaphase.

49

What you are expected to know Identify metaphase by its one event: spindle fibers from both poles have attached to every chromosome at its centromere and pulled them into a single row along the equator, each chromosome still two joined sister chromatids.

50
Check q15

Here is a dividing cell drawn with six chromosomes.

A cell drawn with six X shapes, each two rods crossing at a small white dot, stacked in one line down the middle of the cell, with lines running from a small circle at each end of the cell to every dot
A cell drawn with six X shapes, each two rods crossing at a small white dot, stacked in one line down the middle of the cell, with lines running from a small circle at each end of the cell to every dot

Which stage is the cell in?

  1. A. Prophase
    In prophase the rods lie scattered and the envelope is breaking.
    Here the chromosomes stand in one row with fibers from both poles.
  2. B. ✓ Metaphase
  3. C. Interphase
    In interphase the chromatin is grainy inside an unbroken nuclear envelope.
    Here the chromosomes are rods in one row, and no envelope is drawn.

Why: One row of chromosomes along the equator, each still a joined pair, with fibers from both poles at every centromere: metaphase.

51
Check q16 numeric entry

A metaphase cell has 6 chromosomes lined up along its equator, each still two joined sister chromatids.

Calculate the number of chromatids in the row.

Answer: 12  (tolerance ±0)

Working
Write down the values in the question:
chromosomes = 6
chromatids in each copied chromosome = 2
Write down the equation:
chromatids=2×chromosomes
Substitute the values into the equation:
chromatids=2×chromosomes
chromatids=2×6
chromatids=12

52Quick quiz: metaphase and the equator mixed practice

53
Check q17

A cell's chromosomes stand in one row across its middle, with fibers from both poles at every centromere.

Which stage is the cell in?

  1. A. Prophase
    In prophase the rods lie scattered and the nuclear envelope is breaking down.
    Here the rods stand in one row.
  2. B. ✓ Metaphase

Why: One row of chromosomes along the equator, held by fibers from both poles, is metaphase.

54
Check q18

A cell's chromosomes lie scattered, and its nuclear envelope is breaking up.

Which stage is the cell in?

  1. A. ✓ Prophase
  2. B. Metaphase
    In metaphase the chromosomes stand in one row along the equator.
    Here the rods lie scattered.

Why: Scattered rods and a breaking nuclear envelope are the events of prophase.

55
Check q19

The equator of a dividing cell is a plane.

Where does the plane lie?

  1. A. ✓ Halfway between the two poles
  2. B. At one pole
    The poles are the two ends of the cell.
    The equator lies halfway between them.
  3. C. Just inside the nuclear envelope
    The nuclear envelope has broken down by metaphase.
    The equator lies halfway between the two poles.

Why: The equator is the plane halfway between the two poles, like the equator of a globe.

56
Check q20

In a metaphase cell, spindle fibers hold each chromosome at its centromere.

From how many poles do the fibers reach each chromosome?

  1. A. One
    Fibers from one pole hold one sister chromatid.
    Fibers from the other pole hold the other sister chromatid.
  2. B. ✓ Two

Why: Fibers from the left pole hold one sister chromatid and fibers from the right pole hold the other.
So fibers reach each chromosome from two poles.

57
Check q21

In a metaphase cell, every chromosome stands in the row along the equator.

Is each chromosome still two joined sister chromatids?

  1. A. ✓ Yes
  2. B. No
    At metaphase the connection at each centromere is still intact.
    Each chromosome in the row is a joined pair.

Why: At metaphase the sister chromatids are still joined at their centromere.
Fibers from both poles hold the joined pair in the row.

58
Practice writing an answer

A cell is in mitosis.

(a) State what the equator of a dividing cell is. (1 pt)

Model answer The equator is the plane halfway between the two poles of a dividing cell.
Rubric
  • Award 1 point for: the plane halfway between the two poles (where the chromosomes line up at metaphase).

(b) State what metaphase is. (1 pt)

Model answer Metaphase is the stage of mitosis in which spindle fibers from both poles have pulled every chromosome into a single row along the equator, each chromosome still two joined sister chromatids.
Rubric
  • Award 1 point for: the stage in which the chromosomes stand in a single row along the equator, held by fibers from both poles (each still two joined sister chromatids).

59The onion slide, again

60

Back to the slide of the onion root tip, with most of its cells looking ordinary and a few caught dividing.

61

The ordinary cells show grainy chromatin inside an unbroken nuclear envelope. They are in interphase.

62

The cell with dark rods scattered near its middle is in prophase. Its chromosomes have coiled into rods, its spindle is growing, and its nuclear envelope has broken down.

63

The cell with its rods in a single row across the middle is in metaphase. Fibers from both poles hold every chromosome at the equator.

64

What you are expected to know Read a dividing cell on the slide: scattered rods with a breaking envelope are prophase, and a single row along the equator is metaphase.

65Mixed practice mixed practice

66
Check q22

In a metaphase cell, every chromosome sits along the equator.

Which of the following holds each chromosome there?

  1. A. ✓ Spindle fibers from both poles
  2. B. The nuclear envelope
    The nuclear envelope broke down in prophase.
    So at metaphase there is no envelope around the row.
    The spindle fibers hold the row.
  3. C. Spindle fibers from one pole only
    Fibers from each pole hold one sister chromatid, so the two pulls balance and the chromosome stays at the equator.
    A one-sided pull would drag it toward that pole.
  4. D. The two centrosomes
    The two centrosomes sit at the poles.
    The fibers that grow from the centrosomes reach the chromosomes and hold them.

Why: At metaphase, spindle fibers from each pole are attached to every chromosome at its centromere.
Fibers from one pole hold one sister chromatid, and fibers from the other pole hold the other.
The two pulls balance.
So each chromosome is held in the row along the equator.

67
Check q23

Here is a dividing cell drawn with two chromosomes.

A cell drawn with two X shapes, each two rods crossing at a small white dot, lying at different angles near the middle of the cell inside a dashed oval; a small circle at each end of the cell with three short lines reaching in from each
A cell drawn with two X shapes, each two rods crossing at a small white dot, lying at different angles near the middle of the cell inside a dashed oval; a small circle at each end of the cell with three short lines reaching in from each

Which stage is the cell in?

  1. A. Interphase
    In interphase the chromatin is grainy inside an unbroken nuclear envelope.
    Here the chromosomes are rods and the envelope is breaking up.
  2. B. ✓ Prophase
  3. C. Metaphase
    In metaphase the chromosomes stand in one row along the equator.
    Here the two rods lie scattered inside a breaking envelope.

Why: Scattered rods, each a joined pair, a breaking nuclear envelope, and fibers growing in from two poles: prophase.

68
Check q24

Here is a dividing cell drawn with eight chromosomes.

A cell drawn with eight X shapes, each two rods crossing at a small white dot, stacked in one line down the middle of the cell, with lines running from a small circle at each end of the cell to every dot
A cell drawn with eight X shapes, each two rods crossing at a small white dot, stacked in one line down the middle of the cell, with lines running from a small circle at each end of the cell to every dot

Which stage is the cell in?

  1. A. Interphase
    In interphase the chromatin is grainy inside an unbroken nuclear envelope.
    Here the chromosomes are rods in one row, with no envelope.
  2. B. Prophase
    In prophase the rods lie scattered and the envelope is breaking.
    Here the rods stand in one row with fibers from both poles.
  3. C. ✓ Metaphase

Why: One row of joined pairs along the equator, with fibers from both poles at every centromere: metaphase.

69
Practice writing an answer

Here is a dividing cell drawn with four chromosomes.

A cell drawn with four X shapes, each two rods crossing at a small white dot, lying at different angles and positions near the middle of the cell inside a dashed oval; a small circle at each end of the cell with three short lines reaching in from each
A cell drawn with four X shapes, each two rods crossing at a small white dot, lying at different angles and positions near the middle of the cell inside a dashed oval; a small circle at each end of the cell with three short lines reaching in from each

(a) Identify the stage the cell is in. (1 pt)

Model answer The cell is in prophase.
Rubric
  • Award 1 point for: prophase.

(b) Explain how the drawing shows that the cell has entered mitosis and is still before metaphase. (1 pt)

Model answer The chromosomes have coiled into rods.
The nuclear envelope is breaking down.
Rods appearing and the envelope breaking down are the events of prophase, the first stage of mitosis.
So the cell has entered mitosis.
The rods lie scattered, not in one row along the equator.
In metaphase every chromosome stands in one row along the equator.
So the cell has not yet reached metaphase.
Rubric
  • Award 1 point for: rods (condensed chromosomes) and a breaking nuclear envelope show that mitosis has begun; the rods lie scattered rather than in one row along the equator, so metaphase has not been reached.

Slip Calling the cell metaphase because fibers reach the rods. Fibers reach the rods from prophase on. Metaphase needs the single row along the equator.

Glossary

pole
One of the two ends of a dividing cell, the ends the cell will divide toward. A centrosome sits at each pole, and the mitotic spindle grows between them.
mitotic spindle
A framework of protein fibers that grows between the two poles of a dividing cell, attaches to the chromosomes at their centromeres, and moves them.
prophase
The first stage of mitosis: the copied chromosomes condense into visible rods, the mitotic spindle begins to grow between the poles, and the nuclear envelope breaks down.
equator
The plane halfway between the two poles of a dividing cell, where the chromosomes line up at metaphase.
metaphase
The stage of mitosis in which spindle fibers from both poles have pulled every chromosome into a single row along the equator, the plane halfway between the poles, each chromosome still two joined sister chromatids.

APBIO-U04-L15B Two cells from one: anaphase, telophase and cytokinesis

Topic 4.5 · Cell Cycle · 104 steps

A photograph of stained onion root-tip cells, one with two groups of dark rods moving apart and one with two clumps of rods at its ends, beside four drawn cells: one with two clumps of rods at opposite ends, one with two nuclei, one with a flat plate across its middle, and one rounded cell pinched in at its waist
A photograph of stained onion root-tip cells, one with two groups of dark rods moving apart and one with two clumps of rods at its ends, beside four drawn cells: one with two clumps of rods at opposite ends, one with two nuclei, one with a flat plate across its middle, and one rounded cell pinched in at its waist

Photo: staticd, Wikimedia Commons, CC BY-SA 3.0 (cropped and resized).

Here is the same slide of an onion root tip, with cells caught after the single row: a photograph on the left, drawn cells on the right.

On the onion slide, one cell shows two groups of rods moving apart, with an emptying space between them. In the next cell each group sits inside a new envelope, the rods fading back to grainy. A third cell is building a flat plate across its center. On a second slide, of cells from an animal, a dividing cell is instead pinched in two around its middle.

What happens between the single row and the two finished cells?

Unit 4 · Cell Communication and Cell Cycle

1Anaphase: the sister chromatids separate, then move

2

Video: Watch: Anaphase

The connection at each centromere breaks, so the sister chromatids separate; then the spindle fibers pull them toward opposite poles, so each pole receives one copy of every chromosome.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L15Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L15Ba.mp4

3
Check q1

At metaphase, spindle fibers hold each chromosome in the row along the equator.

What is each chromosome in the row?

  1. A. ✓ Two sister chromatids joined at one centromere
  2. B. One single chromatid, not yet copied
    The chromosome was copied in S phase.
    At metaphase the two copies are still joined at one centromere.
  3. C. Two separate chromosomes, no longer joined
    The two copies are the sister chromatids of one chromosome.
    They stay joined at one centromere until anaphase.

Why: S phase copied each chromosome.
The two copies, the sister chromatids, stay joined at one centromere.
At metaphase they are still joined.

4

Now consider a metaphase cell drawn with four chromosomes, a moment after its row has formed. Two events happen, in order.

A cell with each of its four chromosome pairs split into two separate V-shaped chromatids, a white dot at each point, the two columns a clear gap apart either side of the equator, fibers from both poles attached at the dots
A cell with each of its four chromosome pairs split into two separate V-shaped chromatids, a white dot at each point, the two columns a clear gap apart either side of the equator, fibers from both poles attached at the dots
5

First, the connection at each centromere breaks. The sister chromatids separate.

6

Each separated chromatid now counts as a chromosome.

7

Second, the spindle fibers pull the separated chromatids toward opposite poles.

The same cell with two groups of V-shaped chromosomes moving apart toward the poles, fibers pulling, the middle emptying
The same cell with two groups of V-shaped chromosomes moving apart toward the poles, fibers pulling, the middle emptying
8

One group of chromatids moves toward each pole. The middle of the cell empties.

9

Separation comes before pulling. While the sister chromatids are still joined, fibers from both poles hold the pair at the equator, and nothing moves.

10

When the sister chromatids separate and are pulled to the poles, the stage is called , because ana means apart: the two groups move apart.

11

On the slide, an anaphase cell shows two V-shaped groups of rods moving apart.

12

Each pole receives one chromatid of every pair: one copy of every chromosome.

13

What you are expected to know Identify anaphase by its two events in order: the connection at each centromere breaks and the sister chromatids separate, each now a chromosome; then the spindle fibers pull them toward opposite poles.

14
Check q2

Here is a dividing cell drawn with six chromosomes.

A cell drawn with twelve V shapes, each with a small white dot at its point, in two columns of six: one column left of the middle with its points facing left, one column right of the middle with its points facing right; lines run from a small circle at each end of the cell to the dots of the nearer column
A cell drawn with twelve V shapes, each with a small white dot at its point, in two columns of six: one column left of the middle with its points facing left, one column right of the middle with its points facing right; lines run from a small circle at each end of the cell to the dots of the nearer column

Which stage is the cell in?

  1. A. Prophase
    In prophase the rods lie scattered inside a breaking envelope.
    Here the chromatids are in two ordered groups heading for the poles, with fibers pulling.
  2. B. Metaphase
    In metaphase every chromosome is a joined pair in one row.
    Here the row has split and the two halves are moving apart.
  3. C. ✓ Anaphase

Why: Two V-shaped groups moving toward opposite poles, pulled by the spindle, with no envelope yet: anaphase.

15
Check q3

A cell has its chromosomes in a single row along the equator, with fibers from both poles attached at every centromere.

Which of the following happens first as anaphase begins?

  1. A. ✓ The sister chromatids separate
  2. B. The spindle fibers pull the chromatids to the poles
    While the sister chromatids are joined, fibers from both poles hold each pair in place.
    The chromatids move only after they have separated.
  3. C. A nuclear envelope forms around each group of chromosomes
    Envelopes form after the chromosomes have reached the poles.
  4. D. The chromosomes line up in a row along the equator
    The row along the equator is already in place.
    Anaphase begins when the pairs split.

Why: Anaphase has two events in order.
First, the sister chromatids separate.
Only then do the fibers pull the separated chromatids toward the poles.

16
Check q4

A student says: “In anaphase the spindle fibers pull the sister chromatids apart, and that pulling is what breaks the connection at the centromere.”

Is the student correct?

  1. A. Yes — the pull from the two poles tears the sister chromatids apart
    Fibers from both poles pull on each joined pair, and nothing moves.
    The connection breaks first; then the fibers pull the separated chromatids toward opposite poles.
  2. B. ✓ No — the connection breaks first, and only then do the fibers pull

Why: Anaphase has two events in order.
First, the connection at each centromere breaks and the sister chromatids separate.
Second, the spindle fibers pull the separated chromatids toward opposite poles.
The pulling comes after the break, so the pulling is not what breaks the connection.

17
Practice writing an answer

At the start of anaphase the connection at each centromere breaks first. Only then do the spindle fibers pull the separated chromatids toward opposite poles.

(a) Explain why the pulling comes after the break rather than causing it. (1 pt)

Model answer At metaphase each chromosome is two sister chromatids joined at the centromere.
Fibers from each pole hold one sister chromatid, so the two pulls balance and the joined pair stays at the equator.
The cell cuts the connection at each centromere, as the first event of anaphase.
Only then is each chromatid held from one pole alone, so the fibers can pull it toward that pole.
So the pulling follows the break and does not cause it.
Rubric
  • Award 1 point for: while the sister chromatids are joined, fibers from both poles pull against each other and nothing moves; the connection at the centromere breaks first, and only then can the fibers pull each chromatid toward one pole.
18
Check q5

A spindle inhibitor is a drug that stops the spindle fibers pulling the separated chromatids toward opposite poles. A researcher treats dividing cells with a spindle inhibitor. In every treated cell the chromosomes are condensed and lie in a row along the equator, and each is still two sister chromatids joined at the centromere. No treated cell shows two groups moving apart.

Which stage of mitosis does the inhibitor stop from happening?

  1. A. Prophase
    The chromosomes have condensed into rods, so prophase has happened.
  2. B. Metaphase
    A row of chromosomes along the equator, each still two joined sister chromatids, is the picture of metaphase.
    So metaphase has happened; the stage after it fails.
  3. C. ✓ Anaphase

Why: Condensed chromosomes lie in a row along the equator, so prophase and metaphase have happened.
In anaphase the spindle fibers pull the separated chromatids toward opposite poles.
The inhibitor stops that pulling, so the sister chromatids stay joined and no two groups move apart.
Anaphase fails to happen.

19Quick quiz: anaphase mixed practice

20
Check q6

In a cell, the sister chromatids have just separated, and fibers are pulling them toward the poles.

Is the cell in anaphase?

  1. A. ✓ Yes
  2. B. No
    The sister chromatids have separated and the fibers are pulling: both events of anaphase are under way.

Why: Anaphase begins the moment the sister chromatids separate.
The fibers then pull them toward the poles.

21
Check q7

A cell's chromosomes stand in one row along the equator, each still a joined pair.

Is the cell in anaphase?

  1. A. Yes
    In anaphase the sister chromatids have separated and are moving apart.
    Here every pair is still joined in the row.
  2. B. ✓ No

Why: Joined pairs in one row along the equator is metaphase.
Anaphase begins only when the sister chromatids separate.

22
Check q8

In a cell, two groups of single chromatids move apart, and the middle of the cell is emptying.

Is the cell in anaphase?

  1. A. ✓ Yes
  2. B. No
    Two groups of single chromatids moving apart is the second event of anaphase.

Why: In anaphase the spindle fibers pull the separated chromatids toward opposite poles, so two groups move apart.

23
Check q9

A cell's rods lie scattered, and its nuclear envelope is breaking down.

Is the cell in anaphase?

  1. A. Yes
    In anaphase the sister chromatids have separated and two groups move apart.
    Here the rods are scattered and the envelope is still breaking.
  2. B. ✓ No

Why: Scattered rods and a breaking nuclear envelope are the events of prophase, not anaphase.

24
Check q10

In anaphase, the separated chromatids move toward the poles.

What pulls them?

  1. A. ✓ The spindle fibers
  2. B. The nuclear envelope
    The nuclear envelope broke down in prophase and has not yet re-formed.
    The spindle fibers pull the chromatids.
  3. C. The centromeres
    The centromere is the point the fibers hold.
    The spindle fibers pull the chromatids.

Why: The spindle fibers are attached at each centromere.
They pull the separated chromatids toward opposite poles.

25
Practice writing an answer

A cell is in mitosis.

(a) State what anaphase is. (1 pt)

Model answer Anaphase is the stage of mitosis in which the sister chromatids separate and the spindle fibers pull them toward opposite poles.
Rubric
  • Award 1 point for: the stage in which the sister chromatids separate and are pulled toward opposite poles (one copy of every chromosome to each pole).

26Telophase: two nuclei in one cell

27

Video: Watch: Telophase

The chromosomes arrive at the poles, the spindle breaks down, a new nuclear envelope forms around each set and the rods uncoil into chromatin: two nuclei in one cytoplasm.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L15Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L15Bb.mp4

28
Check q11

In a cell between divisions, one structure surrounds the chromatin and separates it from the cytoplasm.

Which structure?

  1. A. ✓ The nuclear envelope
  2. B. The cell membrane
    The cell membrane surrounds the whole cell.
    The nuclear envelope surrounds the chromatin.
  3. C. The cell wall
    The cell wall lies outside the cell membrane of a plant cell.
    The nuclear envelope surrounds the chromatin.

Why: The nuclear envelope surrounds the nucleus and separates the chromatin from the cytoplasm.

29

The chromosomes arrive at the two poles. The spindle that moved them breaks down.

The same cell with two clumps of V-shaped chromosomes at the poles, a white dot at each point, each clump inside a forming dashed nuclear envelope, and no spindle
The same cell with two clumps of V-shaped chromosomes at the poles, a white dot at each point, each clump inside a forming dashed nuclear envelope, and no spindle
30

A new nuclear envelope forms around each set of chromosomes.

31

Inside each envelope, the rods uncoil back into chromatin. The grainy look returns.

32

The cell now holds two nuclei in one shared cytoplasm. This stage is called , because telo means end: it is the last stage of mitosis.

33

On the slide, a telophase cell shows two clumps at opposite ends. Each clump sits inside a forming envelope, its rods fading.

34

What you are expected to know Identify telophase by its events: the chromosomes arrive at the poles, the spindle breaks down, a new nuclear envelope forms around each set and the chromosomes uncoil back into chromatin, so the cell holds two nuclei in one cytoplasm.

35
Check q12

Here is a dividing cell drawn with six chromosomes.

A cell drawn with two dashed ovals, one near each end of the cell, each holding six V shapes with a small white dot at each point; nothing is drawn between the two ovals
A cell drawn with two dashed ovals, one near each end of the cell, each holding six V shapes with a small white dot at each point; nothing is drawn between the two ovals

Which stage is the cell in?

  1. A. Prophase
    In prophase a single envelope breaks around scattered rods near the middle.
    Here two envelopes form round clumps at the poles.
  2. B. Metaphase
    In metaphase the chromosomes lie in one row along the equator.
    Here the chromosomes sit at the two poles.
  3. C. Anaphase
    In anaphase the fibers are pulling and no envelope has formed.
    Here the spindle is gone and envelopes are forming.
  4. D. ✓ Telophase

Why: Two clumps at the poles, each inside a forming nuclear envelope, the rods fading and no spindle: telophase.

36
Check q13

Two cells on a slide both show their chromosomes in two groups at opposite ends.

What tells a telophase cell from an anaphase cell?

  1. A. In telophase the chromosomes stand in a single row along the equator
    A single row is the stage before the sister chromatids separate.
  2. B. ✓ In telophase an envelope forms around each group and the spindle is gone
  3. C. In telophase the sister chromatids are still joined at their centromeres
    The sister chromatids separated at the start of anaphase.
    In telophase each group is a set of single chromosomes.
  4. D. In telophase the cytoplasm has already divided into two cells
    In telophase the two nuclei still share one cytoplasm.
    The split of the cytoplasm comes after telophase.

Why: In anaphase the groups are still moving and the fibers are pulling.
In telophase the groups have arrived at the poles.
The spindle has broken down.
An envelope is forming around each set.

37Quick quiz: telophase mixed practice

38
Check q14

Two clumps of rods sit at the poles of a cell, each inside a forming envelope, and no spindle is left.

Is the cell in telophase?

  1. A. ✓ Yes
  2. B. No
    Two clumps at the poles inside forming envelopes, with no spindle, are the events of telophase.

Why: In telophase the chromosomes have arrived at the poles, the spindle has broken down, and a new envelope forms around each set.

39
Check q15

In a cell, two groups of chromatids are still moving toward the poles, with fibers pulling them.

Is the cell in telophase?

  1. A. Yes
    In telophase the groups have arrived and the spindle has broken down.
    Here the fibers are still pulling.
  2. B. ✓ No

Why: Groups still moving, pulled by the fibers, is anaphase.
Telophase begins when the chromosomes arrive at the poles and the spindle breaks down.

40
Check q16

A cell holds two nuclei inside one shared cytoplasm.

Has mitosis ended?

  1. A. ✓ Yes
  2. B. No
    Mitosis is the division of the nucleus.
    Two nuclei means the nucleus has divided.

Why: Mitosis is the division of the nucleus.
The cell holds two nuclei, so the nucleus has divided and mitosis has ended.

41
Check q17

A cell is in telophase.

What happens to its rods?

  1. A. The rods are copied
    Copying happens in S phase, before mitosis.
    In telophase the rods uncoil.
  2. B. ✓ The rods uncoil back into chromatin
  3. C. The rods coil tighter
    The rods coiled up in prophase.
    In telophase they uncoil back into chromatin.

Why: In telophase a new envelope forms around each set of chromosomes, and inside it the rods uncoil back into chromatin.

42
Practice writing an answer

A cell is in mitosis.

(a) State what telophase is. (1 pt)

Model answer Telophase is the last stage of mitosis: the chromosomes arrive at the poles, the spindle breaks down, a new nuclear envelope forms around each set, and the chromosomes uncoil back into chromatin.
Rubric
  • Award 1 point for: the last stage of mitosis, in which a new nuclear envelope forms around each set of chromosomes at the poles (two nuclei in one cytoplasm).

43Quick quiz: which stage is this cell in? mixed practice

44
Check q18

Here is a dividing cell drawn with two chromosomes.

A cell drawn with two dashed ovals, one near each end of the cell, each holding two V shapes with a small white dot at each point; nothing is drawn between the two ovals
A cell drawn with two dashed ovals, one near each end of the cell, each holding two V shapes with a small white dot at each point; nothing is drawn between the two ovals

Which stage is the cell in?

  1. A. Prophase
    In prophase the joined pairs lie scattered and the envelope is breaking.
    Here two clumps of fading rods sit at the poles, each inside a forming envelope, with no spindle.
  2. B. Metaphase
    In metaphase every chromosome is a joined pair in one row along the equator.
    Here two clumps of fading rods sit at the poles inside forming envelopes, with no spindle.
  3. C. Anaphase
    In anaphase the fibers are still pulling single chromatids toward the poles.
    Here the rods have arrived and are fading inside forming envelopes, with no spindle left.
  4. D. ✓ Telophase

Why: Two clumps at the poles, each inside a forming nuclear envelope, the rods fading and no spindle: telophase.

45
Check q19

Here is a dividing cell drawn with six chromosomes.

A cell drawn with twelve V shapes, each with a small white dot at its point, in two columns of six with a clear gap between the columns, one column either side of the midline; the points of the left column face left and the points of the right column face right; lines run from a small circle at each end of the cell to the dots of the nearer column
A cell drawn with twelve V shapes, each with a small white dot at its point, in two columns of six with a clear gap between the columns, one column either side of the midline; the points of the left column face left and the points of the right column face right; lines run from a small circle at each end of the cell to the dots of the nearer column

Which stage is the cell in?

  1. A. Prophase
    In prophase the joined pairs lie scattered and the envelope is breaking.
    Here the sister chromatids have separated into single chromatids, and the fibers are pulling them toward the poles.
  2. B. Metaphase
    In metaphase every chromosome is a joined pair in one row along the equator.
    Here every pair has split into single V-shaped chromatids: anaphase has begun.
  3. C. ✓ Anaphase
  4. D. Telophase
    In telophase two clumps sit at the poles inside forming envelopes, with no spindle.
    Here single chromatids are still on the move, with the fibers pulling them toward the poles.

Why: The sister chromatids have separated into single chromatids and the fibers are pulling them toward the poles.
Anaphase begins the moment the sister chromatids separate, so this cell is in anaphase.

46
Check q20

Here is a dividing cell drawn with four chromosomes.

A cell drawn with four X shapes, each two rods crossing at a small white dot, stacked in one line down the middle of the cell, with lines running from a small circle at each end of the cell to every dot
A cell drawn with four X shapes, each two rods crossing at a small white dot, stacked in one line down the middle of the cell, with lines running from a small circle at each end of the cell to every dot

Which stage is the cell in?

  1. A. Prophase
    In prophase the joined pairs lie scattered and the envelope is breaking.
    Here the pairs stand in one row along the equator, with fibers at every centromere.
  2. B. ✓ Metaphase
  3. C. Anaphase
    In anaphase the sister chromatids have separated and the fibers pull them toward the poles.
    Here every chromosome is still a joined pair, standing in one row along the equator.
  4. D. Telophase
    In telophase two clumps sit at the poles inside forming envelopes, with no spindle.
    Here the pairs stand in one row along the equator, with a spindle at every centromere.

Why: One row of joined pairs along the equator, with fibers from both poles at every centromere: metaphase.

47
Check q21

Here is a dividing cell drawn with eight chromosomes.

A cell drawn with eight X shapes, each two rods crossing at a small white dot, lying at different angles and positions near the middle of the cell inside a dashed oval; a small circle at each end of the cell with three short lines reaching in from each
A cell drawn with eight X shapes, each two rods crossing at a small white dot, lying at different angles and positions near the middle of the cell inside a dashed oval; a small circle at each end of the cell with three short lines reaching in from each

Which stage is the cell in?

  1. A. ✓ Prophase
  2. B. Metaphase
    In metaphase every chromosome is a joined pair in one row along the equator.
    Here the rods are scattered, each still a joined pair, inside a breaking nuclear envelope.
  3. C. Anaphase
    In anaphase the sister chromatids have separated and the fibers pull them toward the poles.
    Here the rods are scattered, each still a joined pair, inside a breaking nuclear envelope.
  4. D. Telophase
    In telophase two clumps sit at the poles inside forming envelopes, with no spindle.
    Here the rods are scattered inside a breaking envelope, with fibers growing in from two poles.

Why: Scattered rods, each a joined pair, a breaking nuclear envelope, and fibers growing in from two poles: prophase.

48
Check q22

Here is a dividing cell drawn with four chromosomes.

A cell drawn with eight V shapes, each with a small white dot at its point, in two columns of four well apart on either side of the midline; the points of the left column face left and the points of the right column face right; lines run from a small circle at each end of the cell to the dots of the nearer column
A cell drawn with eight V shapes, each with a small white dot at its point, in two columns of four well apart on either side of the midline; the points of the left column face left and the points of the right column face right; lines run from a small circle at each end of the cell to the dots of the nearer column

Which stage is the cell in?

  1. A. Prophase
    In prophase the joined pairs lie scattered and the envelope is breaking.
    Here the sister chromatids have separated into single chromatids, and the fibers are pulling them toward the poles.
  2. B. Metaphase
    In metaphase every chromosome is a joined pair in one row along the equator.
    Here the sister chromatids have separated into single chromatids that the fibers pull toward the poles.
  3. C. ✓ Anaphase
  4. D. Telophase
    In telophase two clumps sit at the poles inside forming envelopes, with no spindle.
    Here single chromatids are still on the move, with the fibers pulling them toward the poles.

Why: The sister chromatids have separated into single chromatids and the fibers are pulling them toward the poles.
Anaphase begins the moment the sister chromatids separate, so this cell is in anaphase.

49
Check q23

Here is a dividing cell drawn with two chromosomes.

A cell drawn with two X shapes, each two rods crossing at a small white dot, one above the other on the midline of the cell, with lines running from a small circle at each end of the cell to each dot
A cell drawn with two X shapes, each two rods crossing at a small white dot, one above the other on the midline of the cell, with lines running from a small circle at each end of the cell to each dot

Which stage is the cell in?

  1. A. Prophase
    In prophase the joined pairs lie scattered and the envelope is breaking.
    Here the pairs stand in one row along the equator, with fibers at every centromere.
  2. B. ✓ Metaphase
  3. C. Anaphase
    In anaphase the sister chromatids have separated and the fibers pull them toward the poles.
    Here every chromosome is still a joined pair, standing in one row along the equator.
  4. D. Telophase
    In telophase two clumps sit at the poles inside forming envelopes, with no spindle.
    Here the pairs stand in one row along the equator, with a spindle at every centromere.

Why: One row of joined pairs along the equator, with fibers from both poles at every centromere: metaphase.

50Cytokinesis in an animal cell: the cleavage furrow

51

Video: Watch: The cleavage furrow

In an animal cell a ring of protein under the membrane tightens and pinches the cell in at a cleavage furrow until it splits into two daughter cells.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L15Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L15Bc.mp4

52
Check q24

A cell has finished mitosis.

Which process divides its cytoplasm into two cells?

  1. A. ✓ Cytokinesis
  2. B. Mitosis
    Mitosis is the division of the nucleus, and it has just ended.
    Cytokinesis divides the cytoplasm.
  3. C. Interphase
    Interphase is the long stretch between divisions.
    Cytokinesis divides the cytoplasm.

Why: Mitosis divides the nucleus.
Cytokinesis divides the cytoplasm, so that there are two cells.

53

Now consider an animal cell whose mitosis has just ended. A ring of protein just under the cell membrane tightens.

An animal cell pinched in at its middle by a deepening groove, a thick band at the top and bottom of the waist for the ring of protein, a nucleus in each lobe and arrows pointing inward at the groove
An animal cell pinched in at its middle by a deepening groove, a thick band at the top and bottom of the waist for the ring of protein, a nucleus in each lobe and arrows pointing inward at the groove
54

The ring pinches the cell in around its middle.

55

The groove it makes is called the , because the cell is cleaved, split, along it.

56

The furrow deepens until the cell splits in two.

57

Now there are two daughter cells, each with one nucleus.

58

What you are expected to know Describe cytokinesis in an animal cell: a ring of protein under the membrane tightens and pinches the cell in at a cleavage furrow until it splits into two daughter cells.

59
Check q25

Here are three animal cells drawn with four chromosomes, each at a different moment of its division.

Three drawn cells in a row, labelled A, B and C. A: an oval with two groups of four V shapes well apart on either side of its middle, lines from a small circle at each end to the nearer group. B: an outline narrowed to a waist at its middle, with a grainy circle in each half. C: an oval with two dashed ovals near its ends, each holding four V shapes with a dot at each point, nothing drawn between them
Three drawn cells in a row, labelled A, B and C. A: an oval with two groups of four V shapes well apart on either side of its middle, lines from a small circle at each end to the nearer group. B: an outline narrowed to a waist at its middle, with a grainy circle in each half. C: an oval with two dashed ovals near its ends, each holding four V shapes with a dot at each point, nothing drawn between them

Which cell is dividing its cytoplasm at a cleavage furrow?

  1. A. Cell A
    In cell A two groups of chromatids are still moving apart: anaphase.
    Its cytoplasm has not begun to divide.
  2. B. ✓ Cell B
  3. C. Cell C
    In cell C two envelopes are forming around the two sets of chromosomes: telophase.
    Its outline is not pinched in.

Why: Cell B is pinched in at its middle, with a nucleus in each half.
A ring of protein is tightening at the groove, the cleavage furrow.

60Quick quiz: cytokinesis and the cleavage furrow mixed practice

61
Check q26

A cell's nucleus has divided into two, but its cytoplasm is still one.

Has cytokinesis happened?

  1. A. Yes
    Cytokinesis is the division of the cytoplasm.
    Here the cytoplasm is still one.
  2. B. ✓ No

Why: Cytokinesis divides the cytoplasm.
The cytoplasm is still one, so cytokinesis has not happened.

62
Check q27

A groove is deepening all the way around the middle of a dividing animal cell.

What is the groove called?

  1. A. ✓ A cleavage furrow
  2. B. An equator
    The equator is the plane halfway between the poles, where the chromosomes lined up.
    The groove pinching the cell in is the cleavage furrow.
  3. C. A pole
    A pole is one of the two ends of the cell.
    The groove pinching the cell in is the cleavage furrow.

Why: A ring of protein pinches an animal cell in around its middle.
The groove it makes is the cleavage furrow.

63
Check q28

An animal cell is dividing its cytoplasm.

What tightens to pinch the cell in two?

  1. A. The spindle fibers
    The spindle broke down in telophase.
    A ring of protein under the membrane pinches the cell in.
  2. B. The nuclear envelope
    The nuclear envelope surrounds each nucleus.
    A ring of protein under the membrane pinches the cell in.
  3. C. ✓ A ring of protein under the membrane

Why: A ring of protein just under the cell membrane tightens.
It pinches the cell in at the cleavage furrow until the cell splits.

64
Check q29

Two daughter cells have just formed from one animal cell.

Which process finished last?

  1. A. Mitosis
    Mitosis divided the nucleus first.
    Cytokinesis then divided the cytoplasm, so the two cells formed last.
  2. B. ✓ Cytokinesis

Why: Mitosis divides the nucleus.
Cytokinesis divides the cytoplasm afterwards, and only then are there two cells.

65
Check q30

The cleavage furrow of an animal cell has just begun to form.

Has the nucleus already divided?

  1. A. ✓ Yes
  2. B. No
    Cytokinesis begins after mitosis has ended.
    So the nucleus has already divided into two.

Why: The cleavage furrow forms in cytokinesis.
Cytokinesis comes after mitosis, so the nucleus has already divided.

66
Practice writing an answer

An animal cell is dividing its cytoplasm.

(a) State what a cleavage furrow is. (1 pt)

Model answer A cleavage furrow is the groove that forms around the middle of a dividing animal cell as a ring of protein under the membrane tightens, and it deepens until the cell splits in two.
Rubric
  • Award 1 point for: the groove pinched into the middle of a dividing animal cell (by a ring of protein under the membrane) that deepens until the cell splits.

67Cytokinesis in a plant cell: the cell plate

68

Video: Watch: The cell plate

A plant cell's rigid cell wall cannot be pinched, so vesicles gather at the equator and fuse into a cell plate that grows outward into a new cell wall.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L15Bd.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L15Bd.mp4

69
Check q31

A plant cell has a rigid layer outside its cell membrane. An animal cell has no such layer.

What is the layer called?

  1. A. The nuclear envelope
    The nuclear envelope surrounds the nucleus, inside the cell.
    The rigid layer outside the membrane is the cell wall.
  2. B. ✓ The cell wall
  3. C. The cytoplasm
    The cytoplasm is the fluid inside the cell membrane.
    The rigid layer outside the membrane is the cell wall.

Why: A plant cell has a rigid cell wall outside its cell membrane.

70

Now consider a plant cell whose mitosis has just ended. A plant cell has a rigid cell wall, so the cell cannot be pinched in.

A plant cell drawn with a thick cell wall, a nucleus at each end, and a flat plate forming across the middle from small vesicles fusing at the equator
A plant cell drawn with a thick cell wall, a nucleus at each end, and a flat plate forming across the middle from small vesicles fusing at the equator
71

Instead the cell builds a new cell wall across its middle.

72

Vesicles gather at the equator.

73

The vesicles fuse together into a flat plate, just like vesicles fuse with the cell membrane in exocytosis. This flat plate is called the .

74

The cell plate grows outward until it reaches the old cell wall. Now there are two cells, each inside its own cell wall.

75

Here is an animal cell pinched in at its cleavage furrow, beside a plant cell building its cell plate.

Two panels side by side with a divider between them: on the left an animal cell pinched in at a waist with a nucleus in each lobe, labelled animal cell, cleavage furrow; on the right a plant cell drawn with a thick cell wall, a nucleus at each end and a flat plate across its middle, labelled plant cell, cell plate
Two panels side by side with a divider between them: on the left an animal cell pinched in at a waist with a nucleus in each lobe, labelled animal cell, cleavage furrow; on the right a plant cell drawn with a thick cell wall, a nucleus at each end and a flat plate across its middle, labelled plant cell, cell plate
76

What you are expected to know Describe cytokinesis in a plant cell: vesicles gather at the equator and fuse into a cell plate that grows outward into a new cell wall, because the rigid cell wall cannot be pinched.

77
Check q32

A yeast cell carries a faulty protein. Here is the yeast cell 40 minutes after its division began, when a normal yeast cell would have finished dividing.

A large rounded cell joined by a short open neck to a smaller rounded cell, the two drawn as one continuous outline that runs around the large cell, along the neck, around the small cell and back; a round grainy patch sits inside the large cell and another inside the small cell; no line crosses the neck between the two cells
A large rounded cell joined by a short open neck to a smaller rounded cell, the two drawn as one continuous outline that runs around the large cell, along the neck, around the small cell and back; a round grainy patch sits inside the large cell and another inside the small cell; no line crosses the neck between the two cells

Which process has the faulty protein stopped?

  1. A. ✓ Cytokinesis
  2. B. Mitosis
    The mother cell and the bud each hold a nucleus, so the nucleus has divided and mitosis is over.
    What failed is cytokinesis, the split of the cytoplasm.
  3. C. Telophase
    The mother cell and the bud each show a rounded grainy nucleus inside an envelope, so telophase is complete.
    What is missing is the split of the cytoplasm.

Why: The mother cell and the bud each hold a nucleus, so mitosis is complete.
The cytoplasm runs unbroken through the neck, so cytokinesis has not happened.
So the faulty protein stops cytokinesis.

78
Practice writing an answer

A yeast cell carries a faulty protein. Here is the yeast cell 40 minutes after its division began, when a normal yeast cell would have finished dividing. The faulty protein has stopped cytokinesis.

A large rounded cell joined by a short open neck to a smaller rounded cell, the two drawn as one continuous outline that runs around the large cell, along the neck, around the small cell and back; a round grainy patch sits inside the large cell and another inside the small cell; no line crosses the neck between the two cells
A large rounded cell joined by a short open neck to a smaller rounded cell, the two drawn as one continuous outline that runs around the large cell, along the neck, around the small cell and back; a round grainy patch sits inside the large cell and another inside the small cell; no line crosses the neck between the two cells

(a) Explain how the drawing shows that mitosis is complete and cytokinesis has failed. (1 pt)

Model answer Mitosis is the division of the nucleus.
The mother cell holds a rounded grainy nucleus.
The bud holds a second rounded grainy nucleus.
So the nucleus has divided into two nuclei, and mitosis is complete.
Cytokinesis is the division of the cytoplasm.
The outline runs unbroken from the mother cell through the open neck into the bud.
So the cytoplasm is still one cytoplasm.
So cytokinesis has failed.
Rubric
  • Award 1 point for: two nuclei (one in the mother cell, one in the bud) show that the nucleus has divided, so mitosis is complete; the cytoplasm runs unbroken through the neck, so the cytoplasm has not divided and cytokinesis has failed.

79Quick quiz: the cell plate mixed practice

80
Check q33

Vesicles have fused into a flat plate across the middle of a dividing plant cell.

What is the plate called?

  1. A. The cleavage furrow
    A cleavage furrow is the groove pinched into an animal cell.
    The flat plate of fused vesicles is the cell plate.
  2. B. ✓ The cell plate
  3. C. The equator
    The equator is the plane the plate forms along.
    The flat plate of fused vesicles is the cell plate.

Why: In a plant cell, vesicles fuse at the equator into a flat plate.
That plate is the cell plate.

81
Check q34

A dividing plant cell builds a cell plate rather than pinching in.

Why does the cell build a cell plate?

  1. A. The plant cell has no cell membrane
    A plant cell has a cell membrane, inside its cell wall.
    The cell wall is what stops the pinching.
  2. B. ✓ The plant cell’s rigid cell wall cannot be pinched
  3. C. The plant cell’s spindle is still in place
    The spindle broke down in telophase.
    The rigid cell wall is what stops the pinching.

Why: A plant cell has a rigid cell wall.
The cell wall cannot be pinched in, so the cell builds a cell plate instead.

82
Check q35

The cell plate of a dividing plant cell is growing outward.

What does the cell plate grow into?

  1. A. A new nuclear envelope
    The nuclear envelopes formed in telophase, around each set of chromosomes.
    The cell plate becomes the new cell wall.
  2. B. A new spindle
    The spindle broke down in telophase.
    The cell plate becomes the new cell wall.
  3. C. ✓ A new cell wall

Why: The cell plate grows outward until it reaches the old cell wall.
It becomes the new cell wall between the two daughter cells.

83
Practice writing an answer

A plant cell is dividing its cytoplasm.

(a) State what a cell plate is. (1 pt)

Model answer A cell plate is the flat structure that forms across the equator of a dividing plant cell as vesicles fuse, and it grows outward into the new cell wall between the two daughter cells.
Rubric
  • Award 1 point for: the flat plate of fused vesicles across the equator of a dividing plant cell that grows into the new cell wall.

84Quick quiz: furrow or plate? mixed practice

85
Check q36

On a slide, a dividing cell shows a flat plate forming across its center, between two nuclei.

Which of the following is forming?

  1. A. A cleavage furrow
    A furrow is the groove a protein ring pinches into an animal cell.
    Here vesicles fuse into a flat plate that grows into a new cell wall: the plant way.
  2. B. ✓ A cell plate

Why: A plant cell has a rigid cell wall, so it cannot be pinched.
Vesicles fuse at the equator into a flat plate, the cell plate, which grows into the new cell wall.

86
Check q37

On a slide, a dividing cell shows a deepening groove running all the way around its middle.

Which of the following is forming?

  1. A. ✓ A cleavage furrow
  2. B. A cell plate
    A cell plate is the new cell wall a plant cell builds from fused vesicles.
    Here a ring of protein pinches the cell in at a groove: the animal way.

Why: A ring of protein under the membrane pinches an animal cell in at a groove.
That groove is the cleavage furrow.

87
Check q38

A bean root cell is dividing its cytoplasm.

Which of the following is forming?

  1. A. A cleavage furrow
    A bean is a plant, and a plant cell’s rigid cell wall cannot be pinched, so it builds a plate from fused vesicles instead of a furrow.
  2. B. ✓ A cell plate

Why: A plant cell has a rigid cell wall, so it cannot be pinched.
Vesicles fuse at the equator into a flat plate, the cell plate, which grows into the new cell wall.

88
Check q39

A frog skin cell is dividing its cytoplasm.

Which of the following is forming?

  1. A. ✓ A cleavage furrow
  2. B. A cell plate
    A frog is an animal, and an animal cell is pinched in by a ring of protein; a cell plate is the cell wall a plant cell builds.

Why: A ring of protein under the membrane pinches an animal cell in at a groove.
That groove is the cleavage furrow.

89
Check q40

A ring of protein tightens just under the membrane of a dividing cell.

Which of the following is forming?

  1. A. ✓ A cleavage furrow
  2. B. A cell plate
    A cell plate is the new cell wall a plant cell builds from fused vesicles.
    Here a ring of protein pinches the cell in at a groove: the animal way.

Why: A ring of protein under the membrane pinches an animal cell in at a groove.
That groove is the cleavage furrow.

90
Check q41

Vesicles gather and fuse at the equator of a dividing cell.

Which of the following is forming?

  1. A. A cleavage furrow
    A furrow is the groove a protein ring pinches into an animal cell.
    Here vesicles fuse into a flat plate that grows into a new cell wall: the plant way.
  2. B. ✓ A cell plate

Why: A plant cell has a rigid cell wall, so it cannot be pinched.
Vesicles fuse at the equator into a flat plate, the cell plate, which grows into the new cell wall.

91The two slides, again

92

Back to the slide of the onion root tip, and its cells caught after the single row.

93

The cell with two groups of rods moving apart is in anaphase. Its sister chromatids have separated, and the fibers are pulling one copy of every chromosome toward each pole.

94

The cell with each group inside a new envelope is in telophase: two nuclei in one cytoplasm.

95

The onion cell building a flat plate across its center is in cytokinesis. Its vesicles are fusing into a cell plate, because its rigid cell wall cannot be pinched.

96

The animal cell on the second slide, pinched in around its middle, is in cytokinesis too. A ring of protein is tightening at its cleavage furrow.

97

Here is the whole division of one animal cell in one strip, for a cell with four chromosomes: prophase, metaphase, anaphase, telophase and cytokinesis.

Five panels in a row for an animal cell with four chromosomes: prophase, metaphase, anaphase, telophase and cytokinesis
Five panels in a row for an animal cell with four chromosomes: prophase, metaphase, anaphase, telophase and cytokinesis
98

What you are expected to know Read a dividing cell after the row: two groups moving apart are anaphase, two forming envelopes are telophase, and a cleavage furrow or a cell plate is cytokinesis.

99Mixed practice mixed practice

100
Check q42

A cell goes through the four stages of mitosis.

Which list gives the four stages in order?

  1. A. Prophase, anaphase, metaphase, telophase
    The row along the equator, metaphase, comes before the sister chromatids are pulled apart in anaphase.
  2. B. Metaphase, prophase, anaphase, telophase
    Prophase builds the spindle.
    Metaphase is the row the spindle pulls the chromosomes into.
  3. C. Prophase, metaphase, telophase, anaphase
    Telophase, with its two envelopes, comes after anaphase has pulled the groups to the poles.
  4. D. ✓ Prophase, metaphase, anaphase, telophase

Why: In prophase the rods appear and the spindle forms.
In metaphase the chromosomes stand in one row along the equator.
In anaphase the sister chromatids separate and move apart.
In telophase two nuclei form.

101
Check q43 numeric entry

A cell with 12 chromosomes lines them up at metaphase and enters anaphase. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes that arrive at each pole.

Answer: 12  (tolerance ±0)

Working
Write down the values in the question:
chromosomes at metaphase = 12
chromatids in each copied chromosome = 2
poles = 2
Write down the equations:
chromatids=2×chromosomes at metaphase
chromosomes at each pole=chromatidspoles
Substitute the values into the equations:
chromatids=2×12=24
chromosomes at each pole=242
chromosomes at each pole=12
102
Check q44

A student grows root tips with a spindle inhibitor for six hours. On a slide of these root tips, most dividing cells show condensed chromosomes, each still an X of two sister chromatids that have not separated. Cells with two clumps at the poles are rare.

Why are cells with two clumps at the poles rare?

  1. A. The inhibitor stops the chromosomes coiling into rods, so they are too tangled to be moved
    The slide shows condensed chromosomes, so the coiling has happened.
    The movement of the chromatids is what fails.
  2. B. ✓ The inhibitor stops the fibers pulling the chromatids apart, so the cells are held before anaphase
  3. C. The inhibitor speeds the cells through anaphase and telophase, so few cells are caught in those stages
    The inhibitor stops the fibers moving the chromosomes, so it cannot hurry the pulling.
    Most dividing cells are held with sister chromatids joined: a pile-up before anaphase.
  4. D. The inhibitor stops the DNA from being copied, so the cells never enter mitosis
    The chromosomes on the slide are copied: each is two sister chromatids.
    The block comes later, when the fibers should pull the separated chromatids toward opposite poles.

Why: The spindle fibers pull the separated chromatids to the poles in anaphase.
With the fibers stopped, the sister chromatids stay joined and never move, so the cells pile up before anaphase.
On the slide these held cells count as metaphase.
Few cells reach anaphase or telophase, the two-clump stages.

103
Practice writing an answer

Onion root tips were grown for six hours either in water or in water containing a spindle inhibitor, five root tips in each group. A slide was made of each root tip, and among its dividing cells the student counted how many were in metaphase, recording that number as a percent of all the dividing cells on the slide. The student counted a cell as in metaphase when its chromosomes were condensed and each was still two joined sister chromatids, whether or not they lay in a row. The graph shows the two means with error bars that represent ±2SE.

A bar chart of the percent of dividing onion root-tip cells that were in metaphase, for root tips grown in water only and for root tips grown with a spindle inhibitor. The vertical axis runs from 0 to 100% with gridlines every 10% and labels every 20%. The spindle-inhibitor bar is several times taller than the water-only bar. Each bar carries an error bar; the legend reads: error bars represent ±2SE
A bar chart of the percent of dividing onion root-tip cells that were in metaphase, for root tips grown in water only and for root tips grown with a spindle inhibitor. The vertical axis runs from 0 to 100% with gridlines every 10% and labels every 20%. The spindle-inhibitor bar is several times taller than the water-only bar. Each bar carries an error bar; the legend reads: error bars represent ±2SE

(a) Identify the independent variable and the dependent variable in this experiment. (1 pt)

Model answer The independent variable is whether the root tips were grown with the spindle inhibitor or in water only.
The dependent variable is the percent of dividing cells that were in metaphase.
Rubric
  • Award 1 point for: the independent variable is the presence or absence of the spindle inhibitor and the dependent variable is the percent of dividing cells in metaphase.

Slip Naming the number of dividing cells as the dependent variable. What was recorded is the percent of those cells that were in metaphase.

(b) Describe how the spindle holds each chromosome in place at the equator of an untreated metaphase cell. (1 pt)

Model answer Each chromosome at metaphase is still two joined sister chromatids.
Spindle fibers from both poles are attached to the chromosome at its centromere.
Fibers from one pole hold one sister chromatid.
Fibers from the other pole hold the other sister chromatid.
The two pulls balance.
So the chromosome is held in the single row along the equator.
Rubric
  • Award 1 point for: spindle fibers from both poles attached at the centromere, one sister chromatid held from each pole, so the chromosome is held in the row along the equator.

Slip Describing the sister chromatids being pulled apart. At metaphase the connection at the centromere is still intact. The fibers hold the joined pair in the row.

(c) Justify the claim that the inhibitor holds dividing cells at metaphase, using the error bars. (1 pt)

Model answer Read against the gridlines, the ±2SE bar for the water-only root tips runs from about 18% to about 26% of dividing cells in metaphase.
The ±2SE bar for the inhibitor root tips runs from about 79% to about 91%.
The two bars do not overlap.
So the difference between the water-only root tips and the inhibitor root tips is very unlikely to be chance.
Therefore the inhibitor holds dividing cells at metaphase.
Rubric
  • Award 1 point for: the ±2SE bars (about 18–26% and about 79–91%, read against the gridlines) do not overlap, so the higher percent in metaphase with the inhibitor is unlikely to be chance.
  • Accept: readings within half a gridline (5 percentage points) of those values, that is 13–23 to 21–31 for water only and 74–84 to 86–96 for the inhibitor. Do not award the point for a comparison of the two means alone.

Slip Comparing the two means alone, about 22% against about 85%. The claim rests on the ±2SE bars not overlapping.

(d) Predict, in order, the two events of the stage a treated cell enters next if the inhibitor is washed out and the cell continues its division. (1 pt)

Model answer First, the connection at each centromere breaks and the sister chromatids separate, each now a chromosome.
Second, the spindle fibers pull the separated chromatids toward opposite poles, so each pole receives one copy of every chromosome.
That is anaphase.
Rubric
  • Award 1 point for: the sister chromatids separate at the centromere first, and then the spindle fibers pull them toward opposite poles (the two events of anaphase, in that order).

Slip Putting the pulling first, or jumping to the nuclear envelopes re-forming. Separation comes before movement. The envelopes come later, in telophase.

Glossary

anaphase
The stage of mitosis in which the connection at each centromere breaks, the sister chromatids separate, and the spindle fibers pull them toward opposite poles, so each pole receives one copy of every chromosome.
telophase
The stage of mitosis in which the chromosomes arrive at the poles, the spindle breaks down, a new nuclear envelope forms around each set and the chromosomes uncoil back into chromatin: two nuclei in one cytoplasm.
cleavage furrow
The groove that forms around the middle of a dividing animal cell as a ring of protein under the membrane tightens, deepening until the cell splits in two.
cell plate
The flat structure that forms across the equator of a dividing plant cell as vesicles fuse, and grows outward into the new cell wall between the two daughter cells.

APBIO-U04-L16 Name the stage

Topic 4.5 · Cell Cycle · 66 steps

One whitefish embryo cell, fixed and stained: a row of chromosomes across the middle, with fibers reaching each one from both ends of the cell
One whitefish embryo cell, fixed and stained: a row of chromosomes across the middle, with fibers reaching each one from both ends of the cell

Here is one cell from a whitefish embryo, fixed and stained.

Its chromosomes stand in a straight line across the middle. Fibers reach each chromosome from both ends of the cell. The stain has frozen the cell at one instant. Which stage was that, and how would you know a drawing of it was right?

Unit 4 · Cell Communication and Cell Cycle

1Name the stage

2
Check q1

A stained dividing cell shows short dark rods.

What is each rod?

  1. A. ✓ A chromosome coiled tight for moving
  2. B. A protein fiber of the spindle
    Spindle fibers are protein.
    The stain binds DNA, so the rods are DNA.
  3. C. A vacuole full of stored water
    A vacuole is a sac of membrane holding water.
    The stain binds DNA.

Why: The stain binds DNA.
Before division each chromosome coils into a short thick rod so it can be moved.
So each rod is a coiled chromosome.

3

How do you name a stage from one frozen cell? You name it by the defining event in front of you, never by where the stage sits in the list.

4

The six stages happen in order. A slide shows them in no order.

5

Each cell on the slide was frozen at one instant, in whatever stage it had reached.

Six cells from one slide, each frozen at a different stage, in no particular order
Six cells from one slide, each frozen at a different stage, in no particular order
6

This cell shows a grainy tangle of chromatin inside an unbroken nuclear envelope, and no rods. This is interphase.

A cell in interphase: a grainy tangle of chromatin inside an unbroken nuclear envelope, and no rods
A cell in interphase: a grainy tangle of chromatin inside an unbroken nuclear envelope, and no rods
7

This cell shows separate rods scattered inside a breaking nuclear envelope, with fibers starting to grow from the two poles. This is prophase.

A cell in prophase: separate rods scattered inside a nuclear envelope that is breaking up, with short fibers growing in from the two poles
A cell in prophase: separate rods scattered inside a nuclear envelope that is breaking up, with short fibers growing in from the two poles
8

This cell shows the rods in one straight row across the equator, with fibers from both poles reaching every rod. This is metaphase.

A cell in metaphase: the rods in one straight row across the equator, with fibers from both poles reaching every rod, and no nuclear envelope
A cell in metaphase: the rods in one straight row across the equator, with fibers from both poles reaching every rod, and no nuclear envelope
9

This cell shows two groups of rods moving apart toward the poles, with an emptying space between them. This is anaphase.

A cell in anaphase: two groups of V-shaped rods moving apart toward the poles, with an emptying space between them
A cell in anaphase: two groups of V-shaped rods moving apart toward the poles, with an emptying space between them
10

This cell shows two clumps of rods at opposite ends, each inside a forming nuclear envelope. This is telophase.

A cell in telophase: two clumps of rods at opposite ends, each inside a forming nuclear envelope, drawn with a broken line
A cell in telophase: two clumps of rods at opposite ends, each inside a forming nuclear envelope, drawn with a broken line
11

This pair of cells shows cytokinesis: a furrow pinches the animal cell in two, and a plate grows across the plant cell.

Cytokinesis in an animal cell, pinched in at a furrow, and in a plant cell, where a plate is growing across the middle; each cell holds two nuclei
Cytokinesis in an animal cell, pinched in at a furrow, and in a plant cell, where a plate is growing across the middle; each cell holds two nuclei
12

Rods scattered through the cell, with a spindle growing, are prophase. The same rods in one row are metaphase.

Two cells side by side: scattered rods with a growing spindle, which is prophase, and the same rods in one row at the equator, which is metaphase
Two cells side by side: scattered rods with a growing spindle, which is prophase, and the same rods in one row at the equator, which is metaphase
13

The row is the one feature that separates prophase from metaphase.

14

Two groups already moving apart are anaphase, even when the groups are still close together. Movement apart names the stage, whatever the distance between them.

15

Here is a table of the six stages and the defining event that names each one.

A table of the six stages and the defining event that names each one
16

What you are expected to know Name the stage of one cell from the defining event it shows.

17

Video: Watch: Name the stage

Six cells from one slide, each frozen at a different stage. Grainy chromatin inside an unbroken envelope is interphase; scattered rods with a spindle starting are prophase; one row at the equator is metaphase; two groups moving apart are anaphase; two forming envelopes are telophase; a furrow or a plate is cytokinesis.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L16a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L16a.mp4

18
Check q2

Six cells from one root-tip slide are drawn below, numbered 1 to 6.

Six cells from one slide, numbered 1 to 6, each frozen at a different stage; the stages are in no particular order
Six cells from one slide, numbered 1 to 6, each frozen at a different stage; the stages are in no particular order

Which cell is in prophase?

  1. A. Cell 2
    Cell 2 shows grainy chromatin inside an unbroken envelope, which is interphase.
  2. B. ✓ Cell 4
  3. C. Cell 5
    Cell 5 is pinched in two by a furrow, with a nucleus on each side, which is cytokinesis.
  4. D. Cell 6
    Cell 6 has its rods in one straight row at the equator, which is metaphase.

Why: Cell 4 shows separate rods scattered inside an envelope that is breaking up, with fibers starting from the poles: prophase.

19Quick quiz: which stage is this cell in? mixed practice

20
Check q3

Here is one dividing cell, drawn as the stain shows it.

A cell drawn with two X shapes, one long and one short, stacked in one line down the middle of the cell; lines run from a small dot at each end of the cell to the crossing point of each X; no oval is drawn around them
A cell drawn with two X shapes, one long and one short, stacked in one line down the middle of the cell; lines run from a small dot at each end of the cell to the crossing point of each X; no oval is drawn around them

Which stage is the cell in?

  1. A. Prophase
    In prophase the rods lie scattered inside a breaking envelope.
    Here they stand in one row with fibers from both poles: metaphase.
  2. B. ✓ Metaphase
  3. C. Anaphase
    Here the X shapes have not separated.
    They stand in one row at the equator, held by fibers from both poles.
    That is metaphase.

Why: The two X shapes stand in one row across the middle of the cell.
Fibers from both poles reach each one.
One row at the equator is metaphase.

21
Check q4

Here is one dividing cell, drawn as the stain shows it.

A cell drawn with two X shapes, one long and one short, lying at different positions near the middle of the cell inside a dashed oval; a small dot at each end of the cell with two short lines reaching in from each
A cell drawn with two X shapes, one long and one short, lying at different positions near the middle of the cell inside a dashed oval; a small dot at each end of the cell with two short lines reaching in from each

Which stage is the cell in?

  1. A. ✓ Prophase
  2. B. Metaphase
    Here the X shapes lie scattered, not in a row, inside a breaking envelope, with the spindle only starting to grow: prophase.
  3. C. Anaphase
    Here the X shapes are still whole, so the sister chromatids have not separated.
    The rods lie scattered inside a breaking envelope.
    That is prophase.

Why: The two X shapes lie scattered inside an envelope that is breaking up.
The spindle is only starting to grow from the poles.
Scattered rods with a spindle starting are prophase.

22
Check q5

Here is one dividing cell from a plant, drawn as the stain shows it.

A cell drawn with a cell wall and four V shapes, each with a dot at its point, in two columns of two: one column left of the middle with its points facing left and one right of the middle with its points facing right; lines run from a small dot at each end of the cell to the points of the nearer column
A cell drawn with a cell wall and four V shapes, each with a dot at its point, in two columns of two: one column left of the middle with its points facing left and one right of the middle with its points facing right; lines run from a small dot at each end of the cell to the points of the nearer column

Which stage is the cell in?

  1. A. Metaphase
    Here the rods have left the equator.
    Two groups of V shapes are moving apart toward the poles.
    Two groups moving apart are anaphase.
  2. B. ✓ Anaphase
  3. C. Telophase
    No envelope is forming around either group yet.
    The fibers still reach the V shapes, and the groups are still moving apart.
    That is anaphase.

Why: Two groups of V shapes are moving apart toward the poles, with the fibers still pulling them.
Two groups moving apart are anaphase.
The cell wall shows a plant cell, and the cell wall changes nothing about the stage.

23
Check q6

Here is one dividing cell, drawn as the stain shows it.

A cell drawn with two dashed ovals, one near each end of the cell, each holding two V shapes with a dot at each point; nothing is drawn between the two ovals
A cell drawn with two dashed ovals, one near each end of the cell, each holding two V shapes with a dot at each point; nothing is drawn between the two ovals

Which stage is the cell in?

  1. A. Metaphase
    Here nothing is at the equator.
    Two clumps sit at opposite ends of the cell, each inside a forming envelope.
    Two forming envelopes are telophase.
  2. B. Anaphase
    Here the groups have arrived at the poles.
    An envelope is forming around each clump, and no fibers are drawn.
    That is telophase.
  3. C. ✓ Telophase

Why: Two clumps sit at opposite ends of the cell.
An envelope is forming around each clump, and the spindle is gone.
Two forming envelopes are telophase.

24
Check q7

Here is one cell, drawn as the stain shows it.

A cell drawn with one solid oval in its middle holding wavy threads; no rods and no lines from the ends of the cell
A cell drawn with one solid oval in its middle holding wavy threads; no rods and no lines from the ends of the cell

Which stage is the cell in?

  1. A. ✓ Interphase
  2. B. Prophase
    Here there are no rods.
    The cell holds one unbroken envelope with grainy chromatin inside it.
    Grainy chromatin inside an unbroken envelope is interphase.
  3. C. Telophase
    Here there is one nucleus, in the middle of the cell, with grainy chromatin inside it.
    One unbroken envelope with grainy chromatin is interphase.

Why: The cell holds one unbroken envelope with grainy chromatin inside it and no rods.
Grainy chromatin inside an unbroken envelope is interphase.

25
Check q8

Here is one cell from a plant, drawn as the stain shows it.

A cell drawn with a cell wall and a single straight line down its middle that stops short of the cell wall at both ends; on each side of the line a solid oval holding wavy threads
A cell drawn with a cell wall and a single straight line down its middle that stops short of the cell wall at both ends; on each side of the line a solid oval holding wavy threads

Which stage is the cell in?

  1. A. Anaphase
    Here there are no rods at all.
    Each side of the cell holds a finished nucleus, and a plate is growing across the middle.
    A plate is cytokinesis.
  2. B. Telophase
    The two nuclei are finished, and a plate is growing across the middle of the cell.
    The plate is the cytoplasm being divided, which is cytokinesis.
  3. C. ✓ Cytokinesis

Why: A plate is growing across the middle of a walled cell, with a finished nucleus on each side.
A plate dividing the cytoplasm is cytokinesis in a plant cell.

26Describe the stage

27
Check q9

A cell has finished S phase.

What is each of its chromosomes now?

  1. A. One single chromatid, not yet copied
    S phase copied every chromosome.
    So each chromosome is now two chromatids, not one.
  2. B. Two separate chromosomes, no longer joined
    The two copies stay joined at one centromere until anaphase.
    Joined, they count as one chromosome.
  3. C. ✓ Two sister chromatids joined at one centromere

Why: In S phase the cell copied each chromosome once.
The two copies are sister chromatids.
They stay joined at one centromere, so together they count as one chromosome.

28

Describing a correct drawing of a stage follows the same rules as naming it. The description starts with a count.

29

Here is a drawing of one cell’s four chromosomes after S phase.

Four copied chromosomes drawn as X shapes: two long and two short, each two sister chromatids joined at one centromere, eight chromatids in all
Four copied chromosomes drawn as X shapes: two long and two short, each two sister chromatids joined at one centromere, eight chromatids in all
30

Each chromosome is two sister chromatids joined at one centromere. So the drawing shows four chromosomes, and the number of chromatids is calculated from that count.

31
Worked example

A cell has four chromosomes after S phase. Calculate the number of chromatids in the cell.

Write down the values in the question:
chromosomes in the cell = 4
Write down the equation:
chromatids after S phase=2×chromosomes
Substitute the values into the equation:
chromatids after S phase=2×4=8
32

A correct drawing shows two long and two short chromosomes. That lets you check each daughter cell later: each must show two long and two short.

33

A correct drawing of metaphase shows:
1 the four chromosomes in one row at the equator
2 each still two joined chromatids
3 a fiber from each pole to every centromere
4 no nuclear envelope

A cell with four chromosomes at metaphase: four X shapes in one row at the equator, a fiber from each pole to every centromere, no nuclear envelope; the poles and the equator are labeled
A cell with four chromosomes at metaphase: four X shapes in one row at the equator, a fiber from each pole to every centromere, no nuclear envelope; the poles and the equator are labeled
34

A correct drawing of anaphase shows:
1 eight single chromatids
2 four moving toward each pole, with the centromere leading
3 fibers from the poles to each chromatid
4 still no nuclear envelope

A cell with four chromosomes at anaphase: eight single V-shaped chromatids, four moving toward each pole with the centromere leading, fibers from the poles to each, no nuclear envelope
A cell with four chromosomes at anaphase: eight single V-shaped chromatids, four moving toward each pole with the centromere leading, fibers from the poles to each, no nuclear envelope
35

A correct drawing of telophase shows:
1 four chromatids at each pole
2 an envelope forming around each set
3 no spindle

A cell with four chromosomes at telophase: four chromatids at each pole, a nuclear envelope forming around each set, and no spindle
A cell with four chromosomes at telophase: four chromatids at each pole, a nuclear envelope forming around each set, and no spindle
36

The spindle appears in prophase and is gone by telophase. The nuclear envelope is present in interphase, breaks up in prophase and re-forms in telophase.

37

What you are expected to know Describe a correct drawing of a cell with a stated number of chromosomes at a named stage: the number of chromatids, joined or single, where they are, and the spindle and the nuclear envelope only where they exist.

38

Video: Watch: Describe the stage

A cell with four chromosomes after S phase: four X shapes, eight chromatids. At metaphase the four stand in one row with a fiber from each pole and no envelope; at anaphase eight single chromatids move, four to each pole; at telophase four sit at each pole inside a forming envelope, with no spindle.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L16b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L16b.mp4

39
Check q10 numeric entry

A parent cell has eight chromosomes and is in anaphase. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of single chromatids moving toward the poles, in all.

Answer: 16  (tolerance ±0)

Working
Write down the values in the question:
chromosomes in the parent cell = 8
Write down the equation:
chromatids at anaphase=2×chromosomes
Substitute the values into the equation:
chromatids at anaphase=2×chromosomes
chromatids at anaphase=2×8=16
40
Check q11

A cell with eight chromosomes is in anaphase. Pieces of its chromosomes are moving toward the two poles.

Which of the following is each moving piece?

  1. A. ✓ A single chromatid
  2. B. Two sister chromatids joined at one centromere
    The sister chromatids separate at the start of anaphase.
    So each piece moving toward a pole is a single chromatid, with its centromere leading.

Why: The sister chromatids separate at the start of anaphase.
So each piece moving toward a pole is a single chromatid, with its centromere leading.
Eight single chromatids move toward each pole.

41
Check q12

A cell with eight chromosomes is in anaphase.

Is there a nuclear envelope around the moving chromatids?

  1. A. Yes
    The nuclear envelope broke up in prophase and forms again only in telophase, around each set at the poles.
    So at anaphase no envelope surrounds the moving chromatids.
  2. B. ✓ No

Why: The nuclear envelope broke up in prophase.
It forms again only in telophase, around each set of chromatids at the poles.
So at anaphase there is no nuclear envelope around the moving chromatids.

42Judge a drawing

43

Now suppose four students each draw the same cell at the same stage, and only one drawing is right. Judging the drawings uses the same features, checked in order.

44

Check five features:
1 the number of chromatids
2 joined in pairs or single
3 where the chromatids are
4 a spindle only from prophase to anaphase
5 a nuclear envelope only where it exists

45

A drawing that fails one check is wrong. The failed check says what should change.

46

What you are expected to know Judge a drawing of a cell at a named stage: pick the correct drawing from several, and say which feature of a wrong drawing should change.

47

Video: Watch: Judge a drawing

Four drawings of one cell at metaphase. Check the number of chromatids, joined or single, where they are, the spindle and the envelope, in order; the drawing that passes every check is correct, and the failed check says what should change in each wrong one.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L16c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L16c.mp4

48
Check q13

Four students drew a cell with four chromosomes at metaphase. Their drawings are numbered 1 to 4 below.

Four drawings of a cell with four chromosomes at metaphase, numbered 1 to 4: drawing 1 has the four X shapes in a row inside an unbroken nuclear envelope; drawing 2 has eight single rods in a row with fibers; drawing 3 has four X shapes in a row at the equator with fibers from both poles and no envelope; drawing 4 has four X shapes scattered with fibers
Four drawings of a cell with four chromosomes at metaphase, numbered 1 to 4: drawing 1 has the four X shapes in a row inside an unbroken nuclear envelope; drawing 2 has eight single rods in a row with fibers; drawing 3 has four X shapes in a row at the equator with fibers from both poles and no envelope; drawing 4 has four X shapes scattered with fibers

Which drawing shows metaphase correctly?

  1. A. Drawing 1
    The nuclear envelope broke up in prophase.
    So at metaphase the row of chromosomes stands in the open cell.
  2. B. Drawing 2
    At metaphase each chromosome is still two joined chromatids.
    Eight single rods would be the count for anaphase, and in anaphase the rods would be moving apart.
  3. C. ✓ Drawing 3
  4. D. Drawing 4
    Scattered rods with a growing spindle are prophase.
    Metaphase has one row at the equator.

Why: Drawing 3 has four X shapes in one row at the equator.
Each X is two joined chromatids.
A fiber runs from each pole to every centromere.
There is no nuclear envelope.
That is metaphase for a cell with four chromosomes.

49
Check q14

Four students drew a cell with two chromosomes, one long and one short, at anaphase. Their drawings are numbered 1 to 4 below.

Four drawings of a cell with two chromosomes at anaphase, numbered 1 to 4: drawing 1 has two X shapes moving to each pole; drawing 2 has two single V-shaped chromatids, one long and one short, moving to each pole with fibers and no envelope; drawing 3 has one chromatid moving to each pole; drawing 4 has two chromatids moving to each pole inside an unbroken nuclear envelope
Four drawings of a cell with two chromosomes at anaphase, numbered 1 to 4: drawing 1 has two X shapes moving to each pole; drawing 2 has two single V-shaped chromatids, one long and one short, moving to each pole with fibers and no envelope; drawing 3 has one chromatid moving to each pole; drawing 4 has two chromatids moving to each pole inside an unbroken nuclear envelope

Which drawing shows anaphase correctly?

  1. A. Drawing 1
    The sister chromatids separate at the start of anaphase, so each moving piece is a single chromatid.
    Two X shapes per pole would be eight chromatids: too many for two.
  2. B. ✓ Drawing 2
  3. C. Drawing 3
    Two chromosomes are four chromatids after S phase.
    So two chromatids travel to each pole, one long and one short.
  4. D. Drawing 4
    The nuclear envelope broke up in prophase.
    The chromatids move through the open cell.

Why: Drawing 2 has four single chromatids.
One long and one short chromatid move toward each pole with the centromere leading.
Fibers run from the poles, and there is no envelope.
That is anaphase for a cell with two chromosomes.

50
Check q15

A student’s drawing of a cell at telophase is shown below: two clusters at the poles, each inside a forming envelope, and each chromosome drawn as an X of two joined chromatids.

A student's drawing of a cell at telophase: two clusters at the poles, each inside a forming nuclear envelope drawn with a broken line, and every chromosome in each cluster drawn as an X of two joined chromatids
A student's drawing of a cell at telophase: two clusters at the poles, each inside a forming nuclear envelope drawn with a broken line, and every chromosome in each cluster drawn as an X of two joined chromatids

Which feature of the drawing should change?

  1. A. ✓ Each X should be a single chromatid
  2. B. The two clusters should be moved back into one row at the equator
    At telophase the chromatids have arrived at the poles.
    The row was two stages ago.
  3. C. A spindle should be drawn between the two clusters of chromosomes
    By telophase the spindle has broken down.
    The drawing is right to leave the spindle out.
  4. D. The forming envelopes should be replaced by one envelope around everything
    At telophase an envelope forms around each of the two sets, giving two nuclei in one cell.

Why: The sister chromatids separated at the start of anaphase.
So each cluster at telophase is made of single chromatids, one of every pair.
X shapes belong to prophase and metaphase, before the separation.

51
Practice writing an answer

A cell has six chromosomes: three long and three short.

(a) Describe this cell at anaphase: the number of chromatids, where the chromatids are moving, what the spindle fibers are doing, and what has happened to the nuclear envelope. (2 pt)

Model answer Twelve single chromatids are moving.
Six move toward each pole, three long and three short in each group.
The centromere of each chromatid leads.
Spindle fibers run from each pole to its six chromatids.
There is no nuclear envelope, because it broke up in prophase.The model drawing. The cell has six chromosomes and is at anaphase. Twelve single V-shaped chromatids are drawn, six moving toward each pole. Each group has three long and three short chromatids. Fibers run from each pole to its six chromatids. There is no nuclear envelope.anaphase: twelve single chromatids, six to each pole
Rubric
  • Award 1 point for: twelve single chromatids, six in each group (three long and three short), moving toward opposite poles with the centromeres leading.
  • Award 1 point for: spindle fibers from each pole to its chromatids, and no nuclear envelope (it broke up in prophase).

Slip Describing six X-shaped chromosomes moving to the poles. The sister chromatids separate at the start of anaphase. So each moving piece is a single chromatid, and there are twelve of them.

52

Back to the one cell from the whitefish embryo, fixed and stained. Its chromosomes stand in a straight line across the middle, and fibers reach each one from both ends of the cell.

53

One row at the equator, held by fibers from both poles, is metaphase.

54

A correct drawing of that cell shows every chromosome as two joined sister chromatids in one row, a fiber from each pole to every centromere, and no nuclear envelope.

55

A drawing that fails one of those checks is wrong, and the failed check says what should change.

56Quick quiz: which drawing is correct? mixed practice

57
Check q16

Four students drew a cell with three chromosomes at metaphase. Their drawings are numbered 1 to 4 below.

Four drawings of a cell with three chromosomes at metaphase, numbered 1 to 4: drawing 1 has three X shapes in a row inside an unbroken oval; drawing 2 has three X shapes in a row with lines from a dot at each end of the cell to every X; drawing 3 has six short single rods in a row with lines from both ends; drawing 4 has three X shapes scattered, with two short lines from a dot at each end
Four drawings of a cell with three chromosomes at metaphase, numbered 1 to 4: drawing 1 has three X shapes in a row inside an unbroken oval; drawing 2 has three X shapes in a row with lines from a dot at each end of the cell to every X; drawing 3 has six short single rods in a row with lines from both ends; drawing 4 has three X shapes scattered, with two short lines from a dot at each end

Which drawing shows metaphase correctly?

  1. A. Drawing 1
    The nuclear envelope broke up in prophase.
    At metaphase the row stands in the open cell, held by fibers.
  2. B. ✓ Drawing 2
  3. C. Drawing 3
    At metaphase each chromosome is still two joined chromatids.
    Three chromosomes are three X shapes, not six single rods.
  4. D. Drawing 4
    Scattered rods with a spindle starting are prophase.
    Metaphase has one row at the equator.

Why: Drawing 2 has three X shapes in one row at the equator.
Each X is two joined chromatids.
A fiber runs from each pole to every centromere.
There is no nuclear envelope.
That is metaphase for a cell with three chromosomes.

58
Check q17

Four students drew a plant cell with two chromosomes at anaphase. Their drawings are numbered 1 to 4 below.

Four drawings of a plant cell with two chromosomes at anaphase, numbered 1 to 4: drawing 1 has two X shapes moving toward each pole with lines from the poles; drawing 2 has one V shape moving toward each pole with a line from each pole; drawing 3 has two V shapes moving toward each pole inside one unbroken oval, with no lines; drawing 4 has two V shapes, one long and one short, moving toward each pole with lines from the poles and no oval
Four drawings of a plant cell with two chromosomes at anaphase, numbered 1 to 4: drawing 1 has two X shapes moving toward each pole with lines from the poles; drawing 2 has one V shape moving toward each pole with a line from each pole; drawing 3 has two V shapes moving toward each pole inside one unbroken oval, with no lines; drawing 4 has two V shapes, one long and one short, moving toward each pole with lines from the poles and no oval

Which drawing shows anaphase correctly?

  1. A. Drawing 1
    The sister chromatids separate at the start of anaphase.
    So each moving piece is a single chromatid, not an X.
  2. B. Drawing 2
    Two chromosomes are four chromatids after S phase.
    So two chromatids move toward each pole, not one.
  3. C. Drawing 3
    The nuclear envelope broke up in prophase.
    At anaphase the fibers pull the chromatids through the open cell.
  4. D. ✓ Drawing 4

Why: Drawing 4 has four single chromatids.
One long and one short chromatid move toward each pole with the centromere leading.
Fibers run from the poles, and there is no envelope.
That is anaphase for a cell with two chromosomes.

59
Check q18

Four students drew a cell with three chromosomes at prophase. Their drawings are numbered 1 to 4 below.

Four drawings of a cell with three chromosomes at prophase, numbered 1 to 4: drawing 1 has three X shapes in a row with lines from a dot at each end of the cell; drawing 2 has one solid oval holding wavy threads; drawing 3 has three single straight rods scattered inside a dashed oval, with two short lines from a dot at each end; drawing 4 has three X shapes scattered inside a dashed oval, with two short lines from a dot at each end
Four drawings of a cell with three chromosomes at prophase, numbered 1 to 4: drawing 1 has three X shapes in a row with lines from a dot at each end of the cell; drawing 2 has one solid oval holding wavy threads; drawing 3 has three single straight rods scattered inside a dashed oval, with two short lines from a dot at each end; drawing 4 has three X shapes scattered inside a dashed oval, with two short lines from a dot at each end

Which drawing shows prophase correctly?

  1. A. Drawing 1
    One row at the equator, held by fibers from both poles, is metaphase.
    In prophase the rods are still scattered.
  2. B. Drawing 2
    Grainy chromatin inside an unbroken envelope is interphase.
    In prophase the rods have formed and the envelope is breaking up.
  3. C. Drawing 3
    The cell copied its chromosomes in S phase, before prophase.
    So each rod in prophase is an X of two joined chromatids, not a single rod.
  4. D. ✓ Drawing 4

Why: Drawing 4 has three X shapes scattered inside a breaking envelope.
Each X is two joined chromatids.
Short fibers are starting to grow from the two poles.
That is prophase for a cell with three chromosomes.

60
Check q19

Four students drew a cell with two chromosomes at telophase. Their drawings are numbered 1 to 4 below.

Four drawings of a cell with two chromosomes at telophase, numbered 1 to 4: drawing 1 has two X shapes at each end of the cell, each pair inside a dashed oval; drawing 2 has two V shapes moving toward each pole with lines from the poles and no oval; drawing 3 has two V shapes, one long and one short, at each end of the cell, each pair inside a dashed oval, with no lines; drawing 4 has four V shapes together in the middle inside one unbroken oval
Four drawings of a cell with two chromosomes at telophase, numbered 1 to 4: drawing 1 has two X shapes at each end of the cell, each pair inside a dashed oval; drawing 2 has two V shapes moving toward each pole with lines from the poles and no oval; drawing 3 has two V shapes, one long and one short, at each end of the cell, each pair inside a dashed oval, with no lines; drawing 4 has four V shapes together in the middle inside one unbroken oval

Which drawing shows telophase correctly?

  1. A. Drawing 1
    The sister chromatids separated in anaphase.
    So each cluster at telophase holds single chromatids, not X shapes.
  2. B. Drawing 2
    Fibers still pulling two groups apart, with no envelope, is anaphase.
    In telophase the spindle has gone and an envelope forms around each set.
  3. C. ✓ Drawing 3
  4. D. Drawing 4
    In telophase an envelope forms around each of the two sets at the poles.
    One envelope around everything in the middle is interphase.

Why: Drawing 3 has two single chromatids at each pole, one long and one short.
An envelope is forming around each set.
There is no spindle.
That is telophase for a cell with two chromosomes.

61
Check q20

Four students drew an animal cell with three chromosomes at anaphase. Their drawings are numbered 1 to 4 below.

Four drawings of an animal cell with three chromosomes at anaphase, numbered 1 to 4: drawing 1 has three V shapes moving toward each pole with lines from the poles; drawing 2 has six V shapes moving toward each pole with lines from the poles; drawing 3 has three X shapes moving toward each pole with lines from the poles; drawing 4 has six V shapes all moving toward the left pole, with lines from that pole only
Four drawings of an animal cell with three chromosomes at anaphase, numbered 1 to 4: drawing 1 has three V shapes moving toward each pole with lines from the poles; drawing 2 has six V shapes moving toward each pole with lines from the poles; drawing 3 has three X shapes moving toward each pole with lines from the poles; drawing 4 has six V shapes all moving toward the left pole, with lines from that pole only

Which drawing shows anaphase correctly?

  1. A. ✓ Drawing 1
  2. B. Drawing 2
    Three chromosomes are six chromatids after S phase.
    So three chromatids move toward each pole, not six.
  3. C. Drawing 3
    The sister chromatids separate at the start of anaphase.
    So each moving piece is a single chromatid, not an X.
  4. D. Drawing 4
    Anaphase sends one chromatid of every pair to each pole.
    Six chromatids moving toward one pole would leave the other pole with nothing.

Why: Drawing 1 has six single chromatids.
Three move toward each pole with the centromere leading.
Fibers run from the poles, and there is no envelope.
That is anaphase for a cell with three chromosomes.

62Mixed practice mixed practice

63
Check q21

A cell on a slide shows condensed rods scattered through it before any row has formed, a nuclear envelope in pieces, and fibers growing in from both ends.

Which stage is the cell in?

  1. A. Interphase
    Interphase shows grainy chromatin inside an unbroken envelope.
    Here the rods have formed and the envelope is in pieces.
  2. B. ✓ Prophase
  3. C. Metaphase
    Metaphase has one row at the equator.
    These rods are scattered.
  4. D. Anaphase
    Anaphase has two groups heading for the poles.
    These rods are scattered with no groups.

Why: Separate rods scattered through the cell, an envelope breaking up and a spindle starting to grow are the defining events of prophase.

64
Check q22 numeric entry

The animal cell drawn below is in anaphase. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

An animal cell drawn with V shapes in two columns, one column left of the middle with the points facing left and one right of the middle with the points facing right, a dot at each point, and lines from a small dot at each end of the cell to the points of the nearer column
An animal cell drawn with V shapes in two columns, one column left of the middle with the points facing left and one right of the middle with the points facing right, a dot at each point, and lines from a small dot at each end of the cell to the points of the nearer column

Count the chromosomes the parent cell had before S phase.

Answer: 4  (tolerance ±0)

Working
Count the chromatids moving toward one pole:
chromatids moving toward the left pole = 4
Each chromatid is one copy of one chromosome:
chromosomes in the parent cell=4
65
Check q23

A cell from a frog embryo is pinched in around its middle, and a nucleus sits on each side of the pinch.

Which stage is the cell in?

  1. A. Anaphase
    The chromosomes are already inside two finished nuclei.
  2. B. Telophase
    The pinch around the middle is the cytoplasm being divided, which comes after telophase.
  3. C. Cytokinesis in a plant cell
    A plant cell, with its rigid cell wall, builds a plate.
    An animal cell is pinched in by a furrow.
  4. D. ✓ Cytokinesis in an animal cell

Why: A furrow pinching an animal cell in two, with a nucleus on each side, is cytokinesis in an animal cell.
A plant cell cannot be pinched, so it builds a plate instead.

APBIO-U04-L16B Why the two daughter cells match

Topic 4.5 · Cell Cycle · 41 steps

A photograph of a golden retriever standing among trees; to its right a drawn skin cell labeled 78 chromosomes, an arrow, and two smaller daughter cells each labeled 78 chromosomes
A photograph of a golden retriever standing among trees; to its right a drawn skin cell labeled 78 chromosomes, an arrow, and two smaller daughter cells each labeled 78 chromosomes

Photo: Dietmar Rabich, Wikimedia Commons, CC BY-SA 4.0 (resized).

A dog has 78 chromosomes in each body cell.

A skin cell of the dog copies its DNA, lines its chromosomes up, and divides. Each daughter cell holds 78 chromosomes, with DNA identical to the parent’s. The cell never counted. How does the sequence guarantee that both daughters get everything?

Unit 4 · Cell Communication and Cell Cycle

1Count through the sequence

2
Check q1

A cell is at anaphase.

What happens to the two sister chromatids of one chromosome?

  1. A. ✓ The two sister chromatids separate, and one moves toward each pole
  2. B. The two sister chromatids stay joined, and both move toward one pole
    The connection at the centromere breaks at the start of anaphase.
    Then the fibers pull the two chromatids toward opposite poles.
  3. C. The two sister chromatids separate, and both move toward the same pole
    The fibers from one pole are attached to one chromatid, and the fibers from the other pole to its sister.
    So the two chromatids move toward opposite poles.

Why: At the start of anaphase the connection at each centromere breaks.
So the sister chromatids separate.
Then the spindle fibers pull one chromatid toward each pole.

3

Why are the two daughter cells of mitosis identical to each other and to the parent cell? The answer comes from counting the chromatids through the whole sequence.

4

Now consider a cell with six chromosomes. In S phase the cell copies each of the six chromosomes once.

Six chromosomes before S phase, drawn as six single rods, and the same six after S phase, drawn as six X shapes: still six chromosomes, now twelve chromatids
Six chromosomes before S phase, drawn as six single rods, and the same six after S phase, drawn as six X shapes: still six chromosomes, now twelve chromatids
5

Here is a drawing of the same cell at four moments: the parent cell, metaphase, anaphase and the two daughter cells.

Four small panels in a row: a parent cell in interphase, its nucleus holding grainy chromatin, labeled 6 chromosomes; the cell at metaphase with six X shapes, twelve chromatids; the cell at anaphase with six chromatids moving to each pole; two daughter cells, each with a nucleus holding chromatin, labeled 6 chromosomes each
Four small panels in a row: a parent cell in interphase, its nucleus holding grainy chromatin, labeled 6 chromosomes; the cell at metaphase with six X shapes, twelve chromatids; the cell at anaphase with six chromatids moving to each pole; two daughter cells, each with a nucleus holding chromatin, labeled 6 chromosomes each
6
Worked example

A parent cell has six chromosomes. Calculate the number of chromatids at metaphase, the number of chromatids that reach each pole in anaphase, and the number of chromosomes in each daughter cell.

Write down the values in the question:
chromosomes in the parent cell = 6
Write down the equations:
chromatids at metaphase=2×chromosomes in the parent
chromatids reaching each pole=chromatids at metaphase2
chromosomes in each daughter=chromatids reaching each pole
Substitute the values into the equations:
chromatids at metaphase=2×6=12
chromatids reaching each pole=122=6
chromosomes in each daughter=6
7

Each single chromatid that reaches a pole now counts as one chromosome. So the number of chromosomes in each daughter cell equals the number in the parent cell.

8

What you are expected to know Calculate the number of chromosomes and chromatids at metaphase, at anaphase and in each daughter cell for a parent cell with a stated number of chromosomes.

9

Video: Watch: Count through the sequence

A parent cell with six chromosomes: S phase copies each one, so twelve chromatids stand in the row at metaphase; anaphase sends six to each pole; each single chromatid now counts as a chromosome, so each daughter cell holds six, the same as the parent.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L16Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L16Ba.mp4

10
Check q2 numeric entry

A parent cell with 20 chromosomes goes through S phase and reaches metaphase. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromatids in the cell at metaphase.

Answer: 40  (tolerance ±0)

Working
Write down the values in the question:
chromosomes in the parent cell = 20
Write down the equation:
chromatids at metaphase=2×chromosomes in the parent
Substitute the values into the equation:
chromatids at metaphase=2×20=40
11
Check q3 numeric entry

A parent cell with 20 chromosomes has 40 chromatids in its row at metaphase, and the cell goes into anaphase. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromatids that reach each pole.

Answer: 20  (tolerance ±0)

Working
Write down the values in the question:
chromatids at metaphase = 40
Write down the equation:
chromatids reaching each pole=chromatids at metaphase2
Substitute the values into the equation:
chromatids reaching each pole=402=20
12
Check q4 numeric entry

A parent cell with 20 chromosomes has sent 20 chromatids to each pole in anaphase, and the cell finishes cytokinesis. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in each daughter cell.

Answer: 20  (tolerance ±0)

Working
Write down the values in the question:
chromatids reaching each pole = 20
Each single chromatid now counts as one chromosome:
chromosomes in each daughter=chromatids reaching each pole=20
13
Check q5 numeric entry

A parent cell with 14 chromosomes goes through S phase, mitosis and cytokinesis. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in each daughter cell.

Answer: 14  (tolerance ±0)

Working
Write down the values in the question:
chromosomes in the parent cell = 14
Write down the equations:
chromatids at metaphase=2×chromosomes in the parent
chromosomes in each daughter=chromatids at metaphase2
Substitute the values into the equations:
chromatids at metaphase=2×14=28
chromosomes in each daughter=282=14
14
Check q6 numeric entry

A lily cell has 24 chromosomes. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromatids in the cell at metaphase.

Answer: 48  (tolerance ±0)

Working
Write down the values in the question:
chromosomes = 24
Write down the equation:
chromatids at metaphase=2×chromosomes
Substitute the values into the equation:
chromatids at metaphase=2×24=48

15Quick quiz: count through the sequence mixed practice

16
Check q7 numeric entry

A parent cell has 10 chromosomes. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromatids at metaphase.

Answer: 20  (tolerance ±0)

Working
Write down the values in the question:
chromosomes in the parent cell = 10
Each chromosome was copied once in S phase:
chromatids at metaphase=2×10=20
17
Check q8 numeric entry

A parent cell has 18 chromosomes. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in each daughter cell.

Answer: 18  (tolerance ±0)

Working
Write down the values in the question:
chromosomes in the parent cell = 18
Each chromosome was copied once, and anaphase sent one copy of each to each pole:
chromatids at metaphase=2×18=36
chromosomes in each daughter=362=18
18
Check q9 numeric entry

A parent cell has 32 chromosomes and is in anaphase. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromatids moving toward each pole.

Answer: 32  (tolerance ±0)

Working
Write down the values in the question:
chromosomes in the parent cell = 32
Each chromosome was copied once, and anaphase sends one copy of each to each pole:
chromatids at metaphase=2×32=64
chromatids reaching each pole=642=32
19
Check q10 numeric entry

A parent cell has 8 chromosomes and is in metaphase. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in the row at the equator.

Answer: 8  (tolerance ±0)

Working
Write down the values in the question:
chromosomes in the parent cell = 8
A joined pair of sister chromatids counts as one chromosome, so copying changes the chromatid count, not the chromosome count:
chromosomes at metaphase=8
20
Check q11 numeric entry

A cell in metaphase has 40 chromatids in its row. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes the parent cell had before S phase.

Answer: 20  (tolerance ±0)

Working
Write down the values in the question:
chromatids at metaphase = 40
Each chromosome became two chromatids in S phase:
chromosomes in the parent cell=402=20
21
Check q12 numeric entry

A parent cell has 12 chromosomes and is in anaphase. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of single chromatids moving in the cell, in all.

Answer: 24  (tolerance ±0)

Working
Write down the values in the question:
chromosomes in the parent cell = 12
Each chromosome was copied once in S phase, and every chromatid is moving:
chromatids at anaphase=2×12=24

22Why the two daughter cells match

23

In S phase every chromosome was copied once into two identical sister chromatids.

Four small panels in a row: a parent cell in interphase, its nucleus holding grainy chromatin, labeled 6 chromosomes; the cell at metaphase with six X shapes, twelve chromatids; the cell at anaphase with six chromatids moving to each pole; two daughter cells, each with a nucleus holding chromatin, labeled 6 chromosomes each
Four small panels in a row: a parent cell in interphase, its nucleus holding grainy chromatin, labeled 6 chromosomes; the cell at metaphase with six X shapes, twelve chromatids; the cell at anaphase with six chromatids moving to each pole; two daughter cells, each with a nucleus holding chromatin, labeled 6 chromosomes each
24

In anaphase one chromatid of every pair was sent to each pole.

25

So each daughter cell receives one complete copy of the genome.

26

Each chromatid was an exact copy of its sister. So the two daughter cells carry the same genome as each other and as the parent cell.

27

The match comes from one copy in S phase and one separation in anaphase.

28

A cell does not share its DNA out between the two daughter cells. Each daughter cell receives a copy of everything.

29

What you are expected to know Explain why each daughter cell of mitosis holds the parent’s full chromosome number with DNA identical to the parent’s.

30

Video: Watch: Why the two daughter cells match

In S phase every chromosome was copied once into two identical sister chromatids. In anaphase one chromatid of every pair was sent to each pole. So each daughter cell receives one complete copy of the genome. A cell does not share its DNA out; each daughter gets a copy of everything.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L16Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L16Bb.mp4

31
Practice writing an answer

A parent cell with 14 chromosomes goes through S phase, mitosis and cytokinesis. Each daughter cell holds 14 chromosomes, and the DNA of each daughter cell is identical to the parent’s.

(a) Explain why each daughter cell holds 14 chromosomes with DNA identical to the parent’s. (1 pt)

Model answer In S phase the parent cell copied each of its 14 chromosomes once.
So each chromosome became two identical sister chromatids.
So the cell entered mitosis with 28 chromatids, two copies of every chromosome.
In anaphase the sister chromatids separated.
The spindle sent one chromatid of every pair to each pole.
So each pole received one copy of every chromosome.
Therefore each daughter cell holds 14 chromosomes, and each chromosome is an exact copy of the parent’s.
Rubric
  • Award 1 point for: each chromosome was copied once in S phase into two identical sister chromatids (28 chromatids), and anaphase sent one chromatid of every pair to each pole, so each daughter cell receives one complete copy of the genome.
32
Check q13

A cell with ten chromosomes is about to divide. A student says: “The parent cell will share its ten chromosomes out, five to each daughter cell.”

Is the student correct?

  1. A. Yes — each daughter cell receives five chromosomes
    In S phase the parent cell copied every chromosome: twenty chromatids, two of each of the ten.
    Anaphase sends one of each pair to each pole.
  2. B. ✓ No — each daughter cell receives all ten chromosomes
  3. C. No — each daughter cell receives twenty chromosomes
    The twenty chromatids are two copies of each of the ten chromosomes.
    Anaphase sends one copy of each pair to each pole, so each daughter cell receives ten.

Why: The parent cell does not share out its ten chromosomes.
In S phase the parent cell copies every chromosome, so it enters mitosis with twenty chromatids.
Anaphase sends one copy of each pair to each pole.
So each daughter cell receives all ten chromosomes.

33
Check q14

A dog has 78 chromosomes in each body cell. A skin cell of the dog goes through mitosis and cytokinesis.

Which of the following describes the DNA of the two daughter cells?

  1. A. Half of the parent cell’s DNA in each daughter cell
    The parent cell copied all 78 chromosomes in S phase, and anaphase sent one copy of each to each pole.
    So each daughter cell holds all 78, not half.
  2. B. The same as each other, but different from the parent cell’s
    Each sister chromatid is an exact copy of the chromosome it came from, and each daughter cell receives one copy of every chromosome.
    So each daughter’s DNA matches the parent’s.
  3. C. ✓ The same as each other and as the parent cell’s

Why: The skin cell copied each of its 78 chromosomes in S phase into two identical sister chromatids.
Anaphase sent one chromatid of every pair to each pole.
So each daughter cell holds 78 chromosomes, and the DNA of each is the same as the other’s and as the parent cell’s.

34

Back to the dog with 78 chromosomes in each body cell, and its one skin cell that copied its DNA, lined its chromosomes up and divided into two daughter cells.

35

In S phase the skin cell copied each of its 78 chromosomes once into two identical sister chromatids. In anaphase the spindle sent one chromatid of every pair to each pole.

36

So each daughter cell holds 78 chromosomes, and the DNA of each is identical to the parent cell’s. The cell never counted: one copy and one separation did the work.

37Mixed practice mixed practice

38
Check q15 numeric entry

A human skin cell with 46 chromosomes goes through S phase, mitosis and cytokinesis. To count chromosomes, count centromeres. A joined pair of sister chromatids is one chromosome, and a separated chromatid is one chromosome.

Calculate the number of chromosomes in each daughter cell.

Answer: 46  (tolerance ±0)

Working
Write down the values in the question:
chromosomes in the parent cell = 46
Write down the equations:
chromatids at metaphase=2×chromosomes in the parent
chromosomes in each daughter=chromatids at metaphase2
Substitute the values into the equations:
chromatids at metaphase=2×46=92
chromosomes in each daughter=922=46
39
Practice writing an answer

A cell has six chromosomes: three long and three short. It goes through S phase, mitosis and cytokinesis.

(a) Calculate the number of chromosomes each daughter cell holds after cytokinesis. (1 pt)

Answer: 6  (tolerance ±0)

Model answer Each daughter cell holds six chromosomes, the same number as the parent.
Working
Write down the values in the question:
chromosomes in the parent cell = 6
Write down the equations:
chromatids at metaphase=2×chromosomes in the parent
chromosomes in each daughter=chromatids at metaphase2
Substitute the values into the equations:
chromatids at metaphase=2×6=12
chromosomes in each daughter=122=6
Rubric
  • Award 1 point for: six chromosomes in each daughter cell.
40
Practice writing an answer

A student grows onion roots in water and treats half of them with a spindle inhibitor, a chemical that stops the spindle fibers from moving the chromosomes. The other half stay in plain water. After six hours she fixes and stains root tips from both groups and examines the dividing cells. Each onion root-tip cell has 16 chromosomes.

(a) Identify the independent variable in this investigation. (1 pt)

Model answer The independent variable is the treatment the roots received: the spindle inhibitor or plain water.
Rubric
  • Award 1 point for: the presence or absence of the spindle inhibitor (the treatment) as the independent variable.

Slip Naming the number of dividing cells, or the stage the cells reach. Those are measured, so they are dependent variables. The independent variable is what the student changed.

(b) Predict what the chromosomes of a treated cell look like when the cell is frozen at the stage where the chemical stops it. Justify your prediction. (1 pt)

Model answer The treated cell shows condensed chromosomes.
Each chromosome is still two sister chromatids joined at the centromere.
In anaphase the spindle fibers pull the separated chromatids toward opposite poles.
With the fibers stopped, no chromatid moves toward a pole.
The sister chromatids stay joined and no two groups form, so anaphase never starts.
The chromosomes may lie in a row at the equator or lie scattered; either way the cell is held before anaphase.
Rubric
  • Award 1 point for: condensed chromosomes, each still two joined sister chromatids that never separate into two groups, because the stopped spindle fibers cannot pull separated chromatids toward opposite poles.
  • Accept either “held in a row at the equator” or “scattered, no row forms” for where the chromosomes lie. Do not award the point for “the DNA is not copied” or for “pulled to one pole”.

Slip Predicting that the chemical stops the DNA being copied. Copying happened in S phase, before the spindle existed. The inhibitor acts on the pulling, not the copying.

(c) Explain why each daughter cell from an untreated root-tip cell holds 16 chromosomes identical to the parent’s. (2 pt)

Model answer Each of the 16 chromosomes was copied once in S phase into two identical sister chromatids.
So the cell entered mitosis with 32 chromatids.
In anaphase the spindle sent one chromatid of every pair to each pole.
So each pole received 16 chromatids, one copy of every chromosome.
Therefore each daughter cell holds 16 chromosomes carrying the same genome as the parent.
Rubric
  • Award 1 point for: each chromosome copied once in S phase into two identical sister chromatids (32 chromatids in the cell).
  • Award 1 point for: anaphase sends one chromatid of each pair to each pole, so each daughter receives one complete copy of the genome (16 chromosomes).

Slip Writing that the cell divides its 16 chromosomes into two sets of eight. Every chromosome was doubled first, so each daughter receives 16, one copy of each.

APBIO-U04-L17 From counts to time

Topic 4.5 · Cell Cycle · 78 steps

Two tally tables: a root-tip slide of 100 cells with 88 in interphase, 6 in prophase, 3 in metaphase, 2 in anaphase and 1 in telophase; and a healing-skin slide of 200 cells with 36 in mitosis and 164 in interphase
Two tally tables: a root-tip slide of 100 cells with 88 in interphase, 6 in prophase, 3 in metaphase, 2 in anaphase and 1 in telophase; and a healing-skin slide of 200 cells with 36 in mitosis and 164 in interphase

Here is a class tally from a microscope session.

The class counted 100 cells on a root-tip slide: 88 in interphase, 6 in prophase, 3 in metaphase, 2 in anaphase, 1 in telophase. The whole cycle in these cells takes about 24 hours.

How long does metaphase take?

Unit 4 · Cell Communication and Cell Cycle

1

How can a slide of frozen cells tell you how long a stage lasts?

2

Every cell on the slide was frozen at a random moment of its own cycle.

3

A stage that takes a long time catches many cells in it at any instant. A stage that is over quickly catches few.

4

So the fraction of cells in a stage equals the fraction of the cycle that stage takes.

5

A count becomes a percent, and a percent becomes a time. Here is how.

6From counts to percentages

7
Check q1

Five students each count the cells in mitosis on one slide.

What is the mean of their five counts?

  1. A. ✓ The five counts added together, divided by five
  2. B. The biggest count minus the smallest count
    The biggest minus the smallest is the range of the counts.
  3. C. The count that sits in the middle when the five are put in order
    The middle count is the median.

Why: The mean of a set of values is the values added together, divided by how many values there are.
So the mean of five counts is the five counts added together, divided by five.

8

Here is a table of the class tally for the root-tip slide: 100 cells, each recorded in the stage it was frozen in.

The class tally for the root-tip slide: 100 cells in all; 88 in interphase, 6 in prophase, 3 in metaphase, 2 in anaphase, 1 in telophase
The class tally for the root-tip slide: 100 cells in all; 88 in interphase, 6 in prophase, 3 in metaphase, 2 in anaphase, 1 in telophase
9

88 of the 100 cells were in interphase and 3 were in metaphase.

10

Out of a hundred, a count reads straight off as a percent: 88% in interphase and 3% in metaphase.

11

Now suppose five students each count 200 cells on a slide from healing skin. Their counts of cells in mitosis are 34, 36, 38, 35 and 37.

12

Five people counting the same slide get slightly different counts. The class uses the mean of the five counts as the count for the slide.

13
Worked example

Five students counted 34, 36, 38, 35 and 37 cells in mitosis on the same slide. What count should the class use for the slide?

Write down the values in the question:
the five counts = 34, 36, 38, 35, 37
number of counts = 5
Write down the equation:
mean count=the counts added togethernumber of counts
Substitute the values into the equation:
mean count=34+36+38+35+375
mean count=1805
mean count=36cells
14

The healing-skin slide was counted out of 200, not out of 100. So its count of 36 has to be turned into a percent.

15

Here is the equation that turns a count into a percent.

The percent of cells in a stage: the count in that stage divided by the total counted, times 100
16
Worked example

Of 200 cells counted on the healing-skin slide, 36 were in mitosis. What percent of the cells were in mitosis?

Write down the values in the question:
cells in mitosis = 36
total cells counted = 200
Write down the equation:
percent in mitosis=cells in mitosistotal cells counted×100
Substitute the values into the equation:
percent in mitosis=36200×100
percent in mitosis=0.18×100
percent in mitosis=18%
17

A percent is a count out of a hundred. So two slides counted out of different totals can be compared once each count is a percent.

18

The percent of cells in mitosis, all four stages of mitosis taken together, is called the of the slide. The healing-skin slide’s mitotic index is 18%.

19

Video: Watch: From counts to percentages

Five students count the same slide and their counts are averaged. The count in a stage divided by the total counted, times 100, is the percent in that stage: 36 of 200 cells in mitosis is 18%. The percent in mitosis is the slide’s mitotic index.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L17a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L17a.mp4

20

What you are expected to know Calculate the percent of cells in a stage from a count: the cells in that stage divided by the total cells counted, times 100.

21
Check q2 numeric entry

Of 250 cells counted on a slide of uninjured skin, 10 were in mitosis.

Calculate the percent of the cells in mitosis.

Part 1. Divide the count by the total. What fraction of the cells were in mitosis, as a decimal?

Answer: 0.04  (tolerance ±0.0005)

Working
Divide the count in mitosis by the total counted:
cells in mitosistotal cells counted=10250=0.04

Answer: 4 %  (tolerance ±0.05)

Working
Write down the values in the question:
cells in mitosis = 10
total cells counted = 250
Write down the equation:
percent in mitosis=cells in mitosistotal cells counted×100
Substitute the values into the equation:
percent in mitosis=10250×100
percent in mitosis=0.04×100
percent in mitosis=4%
22
Check q3 numeric entry

On a slide of a whitefish embryo, where cells divide often, a student counted 120 cells and found 27 in mitosis.

Calculate the mitotic index of the slide as a percent.

Answer: 22.5 %  (tolerance ±0.05)

Working
Write down the values in the question:
cells in mitosis = 27
total cells counted = 120
Write down the equation:
mitotic index=cells in mitosistotal cells counted×100
Substitute the values into the equation:
mitotic index=27120×100
mitotic index=0.225×100
mitotic index=22.5%
23
Check q4 numeric entry

A class counted 50 cells on a root-tip slide and found 3 of them in prophase, so 6% of the cells on that slide were in prophase. A second class counts 400 cells on another slide from that root tip.

Predict the number of the 400 cells that are in prophase.

Answer: 24  (tolerance ±0)

Working
Write down the values in the question:
percent in prophase = 6%
total cells counted = 400
Write down the equation:
cells in prophase=percent in prophase100×total cells counted
Substitute the values into the equation:
cells in prophase=6100×400
cells in prophase=0.06×400
cells in prophase=24cells

24Quick quiz: mitotic index mixed practice

25
Check q5

What is the mitotic index of a slide?

  1. A. The percent of the cells that are in interphase
    Interphase is the part of the cycle outside mitosis.
  2. B. ✓ The percent of the cells that are in mitosis
  3. C. The number of cells counted on the slide
    The total counted is the bottom of the fraction, not the index.

Why: The mitotic index is the percent of the counted cells that are in mitosis, all four stages of mitosis taken together.

26
Practice writing an answer

A student counts the cells on a slide of an onion root tip and records the stage of each one.

(a) State what is meant by the mitotic index of the slide. (1 pt)

Model answer The mitotic index is the percent of the counted cells that are in mitosis: the cells in mitosis divided by the total cells counted, times 100.
Rubric
  • Award 1 point for: the percent (or fraction) of the counted cells that are in mitosis.
27
Check q6 numeric entry

Of 50 cells counted on a slide, 4 are in mitosis.

Calculate the mitotic index as a percent.

Answer: 8 %  (tolerance ±0.05)

Working
Write down the values in the question:
cells in mitosis = 4
total cells counted = 50
Write down the equation:
mitotic index=cells in mitosistotal cells counted×100
Substitute the values into the equation:
mitotic index=450×100
mitotic index=0.08×100
mitotic index=8%
28
Check q7 numeric entry

Of 300 cells counted on a slide, 18 are in prophase.

Calculate the percent of the cells in prophase.

Answer: 6 %  (tolerance ±0.05)

Working
Write down the values in the question:
cells in prophase = 18
total cells counted = 300
Write down the equation:
percent in prophase=cells in prophasetotal cells counted×100
Substitute the values into the equation:
percent in prophase=18300×100
percent in prophase=0.06×100
percent in prophase=6%
29
Check q8 numeric entry

Of 80 cells counted on a slide, 2 are in anaphase.

Calculate the percent of the cells in anaphase.

Answer: 2.5 %  (tolerance ±0.05)

Working
Write down the values in the question:
cells in anaphase = 2
total cells counted = 80
Write down the equation:
percent in anaphase=cells in anaphasetotal cells counted×100
Substitute the values into the equation:
percent in anaphase=280×100
percent in anaphase=0.025×100
percent in anaphase=2.5%
30
Check q9 numeric entry

Of 400 cells counted on a slide, 30 are in mitosis.

Calculate the mitotic index as a percent.

Answer: 7.5 %  (tolerance ±0.05)

Working
Write down the values in the question:
cells in mitosis = 30
total cells counted = 400
Write down the equation:
mitotic index=cells in mitosistotal cells counted×100
Substitute the values into the equation:
mitotic index=30400×100
mitotic index=0.075×100
mitotic index=7.5%

31Why a count gives a time

32

Video: Watch: Why a count gives a time

Every cell on the slide was frozen at a random moment of its own cycle. A long stage catches many cells at any instant; a short stage catches few. So a stage’s share of the cells equals its share of the cycle time. The cycle length itself has to be given.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L17b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L17b.mp4

33

Why does a count give a time? Every cell on the slide was frozen at a random moment of its own cycle.

34

Imagine two stages: a long one and a short one. At any instant, many cells are part-way through the long stage, and few are part-way through the short stage.

35

So a stage that takes a long time catches many cells in it. A stage that is over quickly catches few.

36

On the root-tip slide, interphase caught 88 of the 100 cells and metaphase caught 3. So interphase takes most of the cycle, and metaphase takes a small slice of it.

37

So the fraction of cells in a stage equals the fraction of the cycle that stage takes.

38

The count gives the fraction. The count cannot give the length of the whole cycle: a frozen slide shows no clock.

39

So the cycle length has to be given to you. In these root-tip cells it is about 24 hours.

40

Here is the rule as an equation.

The time a stage takes: its percent of the cells, as a fraction, times the length of the whole cycle, which the question gives
41

What you are expected to know Explain why the percent of cells found in a stage equals the percent of the cycle that stage takes.

42
Check q10

At the instant a slide was fixed, 5% of its cells were in prophase.

What does the 5% tell you about prophase?

  1. A. About 5% of the cells will ever go through prophase
    Every cycling cell passes through prophase.
    The 5% is how many cells were caught in prophase at one instant.
  2. B. Prophase takes about 5 hours
    The time is 5% of the cycle.
    Hours need the cycle length, which the count cannot give.
  3. C. ✓ Prophase takes about 5% of the whole cycle
  4. D. About 5% of each cell’s DNA had condensed
    In prophase every chromosome in the cell condenses.
    The 5% is a share of cells, each caught at a random moment.

Why: Each cell was frozen at a random moment.
So the fraction of cells caught in prophase equals the fraction of the cycle that prophase takes: about 5%.
The time in hours needs the cycle length as well.

43
Practice writing an answer

On a slide fixed at one instant, 2% of the cells are in telophase and 90% are in interphase.

(a) Explain how these two counts demonstrate that telophase is a much shorter stage than interphase. (2 pt)

Model answer Each cell on the slide was frozen at a random moment of its own cycle.
So a long stage catches many cells at any instant, and a short stage catches few.
Interphase caught 90% of the cells, so interphase takes about 90% of the cycle.
Telophase caught 2% of the cells, so telophase takes about 2% of the cycle.
So telophase is a much shorter stage than interphase.
Rubric
  • Award 1 point for: each cell was frozen at a random moment, so a stage’s share of the cells equals its share of the cycle time.
  • Award 1 point for: 90% of the cells in interphase against 2% in telophase, so interphase takes about 90% of the cycle and telophase about 2%.

Slip Writing that telophase is shorter because fewer cells reach it. Every cycling cell passes through telophase; the count is small because each cell spends little time there.

44
Check q11

On one slide, 12 cells are in prophase and 3 are in anaphase. A student says: “Prophase must be a faster stage than anaphase, because more cells got through it.”

Is the student correct?

  1. A. Yes — more cells in a stage means the stage is faster
    A fast stage catches few cells, because each cell is through it quickly.
    12 cells caught in prophase means prophase takes longer than anaphase.
  2. B. ✓ No — more cells in a stage means the stage takes longer
  3. C. No — the two stages take the same time
    12 cells against 3 is a four-times difference.
    So prophase takes about four times as long as anaphase.

Why: Every cell passes through both stages.
Each cell was frozen at a random moment.
A stage that takes longer catches more cells at that moment.
So 12 in prophase against 3 in anaphase means prophase takes longer, not less.

45From percentages to time

46

Video: Watch: From percentages to time

Three of a hundred cells in metaphase, on a 24-hour cycle: 3% of 24 hours is 0.72 hours, about 43 minutes. Hours to one decimal place; minutes to the nearest minute.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L17c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L17c.mp4

47

Back to the class tally: 3 of the 100 root-tip cells were in metaphase, so metaphase takes 3% of the cycle.

48

The whole cycle in these cells takes 24 hours. So metaphase takes 3% of 24 hours.

49

Here is the equation again. The percent, as a fraction, times the cycle length gives the time in the stage.

The time a stage takes: its percent of the cells, as a fraction, times the length of the whole cycle, which the question gives
50
Worked example

On the root-tip slide, 3% of the cells were in metaphase. The whole cycle takes 24 hours. How long does metaphase take, in hours and in minutes?

Write down the values in the question:
percent in metaphase = 3%
cycle length = 24 h
Write down the equation:
time in metaphase=percent in metaphase100×cycle length
Substitute the values into the equation:
time in metaphase=3100×24h
time in metaphase=0.03×24h
time in metaphase=0.72h
Convert the hours to minutes:
time in metaphase=0.72h×60min/h
time in metaphase=43.2min≈43min
51

3 cells in 100 is 3% of the cells. 43 minutes in 24 hours is 3% of the cycle. The count and the time are the same fraction of the whole.

52

Give a time in hours to one decimal place: 21.12 h is written 21.1 h. Never round hours to a whole hour.

53

Give a time in minutes to the nearest minute: 43.2 min is written 43 min.

54

What you are expected to know Calculate the time a cell spends in a stage from the percent of cells found in that stage and the stated length of the whole cycle: the percent as a fraction, times the cycle length.

55
Check q12 numeric entry

A class counted 50 cells on a root-tip slide and found 44 of them in interphase, so 88% of the cells were in interphase. The whole cycle takes 24 hours.

Calculate the time a cell spends in interphase, in hours to one decimal place.

Part 1. Write the percent as a decimal fraction. What fraction of the cycle is interphase?

Answer: 0.88  (tolerance ±0.0005)

Working
Divide the percent by 100:
percent in interphase100=88100=0.88

Part 2. Multiply that fraction by the cycle length. How many hours is interphase, before rounding?

Answer: 21.12 h  (tolerance ±0.005)

Working
Multiply the fraction by the cycle length:
0.88×24h=21.12h

Answer: 21.1 h  (tolerance ±0.05)

Working
Write down the values in the question:
percent in interphase = 88%
cycle length = 24 h
Write down the equation:
time in interphase=percent in interphase100×cycle length
Substitute the values into the equation:
time in interphase=88100×24h
time in interphase=0.88×24h
time in interphase=21.12h
Round the hours to one decimal place:
time in interphase≈21.1h
56
Check q13 numeric entry

In a culture of a human cell line, the whole cycle takes 20 hours. A DNA measurement shows that 45% of the cells are in S phase at any instant.

Calculate the time a cell spends in S phase, in hours.

Answer: 9 h  (tolerance ±0.05)

Working
Write down the values in the question:
percent in S phase = 45%
cycle length = 20 h
Write down the equation:
time in S phase=percent in S phase100×cycle length
Substitute the values into the equation:
time in S phase=45100×20h
time in S phase=0.45×20h
time in S phase=9.0h
57
Check q14 numeric entry

A class counted 50 cells on a root-tip slide and found 1 of them in anaphase, so 2% of the cells were in anaphase. The whole cycle takes 24 hours.

Calculate the time a cell spends in anaphase, in minutes to the nearest minute.

Answer: 29 min  (tolerance ±0.5)

Working
Write down the values in the question:
percent in anaphase = 2%
cycle length = 24 h
Write down the equation:
time in anaphase=percent in anaphase100×cycle length
Substitute the values into the equation:
time in anaphase=2100×24h
time in anaphase=0.48h
Convert the hours to minutes and round to the nearest minute:
time in anaphase=0.48h×60min/h
time in anaphase=28.8min≈29min
58

We made one simplification: every cell takes the same 24 hours round its cycle, and the slide caught each cell at a random moment. Real cycle lengths differ; a question gives you one cycle length to use.

59

Back to the class tally for the root-tip slide: 100 cells, with 88 in interphase, 6 in prophase, 3 in metaphase, 2 in anaphase and 1 in telophase, in cells whose whole cycle takes 24 hours.

60

Each cell was frozen at a random moment, so the 3 cells in metaphase mean that metaphase takes 3% of the cycle.

61

Metaphase takes 3% of the 24-hour cycle: 0.72 hours, about 43 minutes.

62Quick quiz: five quick times mixed practice

63
Check q15 numeric entry

In a culture whose cycle takes 20 hours, 10% of the cells are in G2.

Calculate the time a cell spends in G2, in hours.

Answer: 2 h  (tolerance ±0.05)

Working
Write down the values in the question:
percent in G2 = 10%
cycle length = 20 h
Write down the equation:
time in G2=percent in G2100×cycle length
Substitute the values into the equation:
time in G2=10100×20h
time in G2=0.10×20h
time in G2=2.0h
64
Check q16 numeric entry

In a culture whose cycle takes 16 hours, 25% of the cells are in S phase.

Calculate the time a cell spends in S phase, in hours.

Answer: 4 h  (tolerance ±0.05)

Working
Write down the values in the question:
percent in S phase = 25%
cycle length = 16 h
Write down the equation:
time in S phase=percent in S phase100×cycle length
Substitute the values into the equation:
time in S phase=25100×16h
time in S phase=0.25×16h
time in S phase=4.0h
65
Check q17 numeric entry

On a root-tip slide, 5% of the cells are in prophase. The whole cycle takes 24 hours.

Calculate the time a cell spends in prophase, in hours.

Answer: 1.2 h  (tolerance ±0.05)

Working
Write down the values in the question:
percent in prophase = 5%
cycle length = 24 h
Write down the equation:
time in prophase=percent in prophase100×cycle length
Substitute the values into the equation:
time in prophase=5100×24h
time in prophase=0.05×24h
time in prophase=1.2h
66
Check q18 numeric entry

In a culture whose cycle takes 30 hours, 50% of the cells are in G1.

Calculate the time a cell spends in G1, in hours.

Answer: 15 h  (tolerance ±0.05)

Working
Write down the values in the question:
percent in G1 = 50%
cycle length = 30 h
Write down the equation:
time in G1=percent in G1100×cycle length
Substitute the values into the equation:
time in G1=50100×30h
time in G1=0.50×30h
time in G1=15.0h
67
Check q19 numeric entry

In a culture whose cycle takes 20 hours, 1% of the cells are in telophase.

Calculate the time a cell spends in telophase, in minutes.

Answer: 12 min  (tolerance ±0.5)

Working
Write down the values in the question:
percent in telophase = 1%
cycle length = 20 h
Write down the equation:
time in telophase=percent in telophase100×cycle length
Substitute the values into the equation:
time in telophase=1100×20h
time in telophase=0.01×20h
time in telophase=0.2h
Convert the hours to minutes:
time in telophase=0.2h×60min/h=12min

68Mixed practice mixed practice

69
Check q20

A student counts 180 cells on a slide and records the stage of each one.

Which value is the slide’s mitotic index?

  1. A. The percent of the 180 cells that are in interphase
    Interphase is the part of the cycle outside mitosis.
  2. B. The number of the 180 cells that are in metaphase
    Metaphase is one stage of mitosis, and a count is not a percent.
  3. C. ✓ The percent of the 180 cells that are in mitosis

Why: The mitotic index is the percent of the counted cells that are in mitosis, all four stages taken together.

70
Check q21 numeric entry

A class counts 150 cells on a root-tip slide and finds 12 in prophase.

Calculate the percent of the cells in prophase.

Answer: 8 %  (tolerance ±0.05)

Working
Write down the values in the question:
cells in prophase = 12
total cells counted = 150
Write down the equation:
percent in prophase=cells in prophasetotal cells counted×100
Substitute the values into the equation:
percent in prophase=12150×100
percent in prophase=0.08×100
percent in prophase=8%
71
Check q22

On a slide from a plant’s shoot tip, 9 of 150 cells are in mitosis. On a slide from one of its leaves, 12 of 300 cells are in mitosis.

Which slide has the greater percent of cells in mitosis?

  1. A. ✓ The shoot-tip slide
  2. B. The leaf slide
    12 is the bigger count, but the leaf slide was counted out of twice as many cells.
    12 out of 300 is 4%, and 9 out of 150 is 6%.
  3. C. The two slides have the same percent
    9 out of 150 is 6%.
    12 out of 300 is 4%.
    The percents differ.

Why: A percent is a count out of a hundred.
9 out of 150 is 6%.
12 out of 300 is 4%.
So the shoot-tip slide has the greater percent of cells in mitosis.

72
Check q23

In one tissue, G1 takes ten hours and anaphase takes half an hour. A slide of that tissue is fixed at one instant.

Compared with anaphase, how many cells does G1 catch on the slide?

  1. A. ✓ More cells
  2. B. The same number of cells
    Each cell was frozen at a random moment.
    A longer stage catches more of the cells at that moment.
  3. C. Fewer cells
    A stage that takes longer catches more cells, not fewer.

Why: Each cell was frozen at a random moment of its own cycle.
G1 takes twenty times as long as anaphase.
So at any instant many more cells are part-way through G1 than through anaphase.
So G1 catches more cells.

73
Check q24

A student has worked out that metaphase takes 4% of the cycle and wants the time in hours.

Which value must the question give the student?

  1. A. The number of cells counted
    The count is already done; the 4% came from it.
  2. B. ✓ The length of the whole cycle
  3. C. The number of cells in interphase
    The interphase count is another share of the cells, not a time.

Why: The time in a stage is the stage’s fraction of the cycle times the cycle length.
A count of frozen cells cannot give the cycle length.
So the question must give the length of the whole cycle.

74
Check q25 numeric entry

On a root-tip slide, 4% of the cells are in anaphase. The whole cycle takes 24 hours.

Calculate the time a cell spends in anaphase, in minutes to the nearest minute.

Answer: 58 min  (tolerance ±0.5)

Working
Write down the values in the question:
percent in anaphase = 4%
cycle length = 24 h
Write down the equation:
time in anaphase=percent in anaphase100×cycle length
Substitute the values into the equation:
time in anaphase=4100×24h
time in anaphase=0.04×24h
time in anaphase=0.96h
Convert the hours to minutes and round to the nearest minute:
time in anaphase=0.96h×60min/h
time in anaphase=57.6min≈58min
75
Check q26

A student counts 500 cells and finds 25 in metaphase. She says: “So metaphase lasts 25 hours.”

Is the student correct?

  1. A. Yes — the count in a stage is its time in hours
    A count is a number of cells, not a number of hours.
    25 out of 500 is 5% of the cycle.
  2. B. ✓ No — the count gives a fraction of the cycle, not hours
  3. C. No — 25 cells means metaphase lasts 5 hours
    25 out of 500 is 5% of the cycle, and a percent is a share, not a number of hours.

Why: The count gives a fraction of the cycle: 25 out of 500 is 5%.
Hours need the cycle length as well, which the count cannot give.
So the student cannot say 25 hours from the count alone.

76
Check q27

In a culture of hamster cells the whole cycle takes 20 hours. In a culture of pig cells the whole cycle takes 40 hours. In both cultures, 5% of the cells are in prophase.

Compared with a hamster cell, how long does a pig cell spend in prophase?

  1. A. Half as long
    The pig cells’ cycle is the longer one.
    The same 5% of a longer cycle is a longer time.
  2. B. The same time
    The same percent of two different cycle lengths gives two different times.
  3. C. ✓ Twice as long

Why: The time in a stage is the stage’s fraction of the cycle times the cycle length.
Both cultures spend 5% of the cycle in prophase.
The pig cells’ cycle is twice as long as the hamster cells’.
So a pig cell spends twice as long in prophase.

77
Practice writing an answer

A student counts 300 cells on a slide from the growing tip of a bean root. She finds 45 cells in mitosis, of which 12 are in metaphase. In these cells the whole cycle takes 30 hours.

(a) Calculate the mitotic index of the slide. (1 pt)

Answer: 15 %  (tolerance ±0.05)

Model answer The mitotic index of the slide is 15%.
Working
Write down the values in the question:
cells in mitosis = 45
total cells counted = 300
Write down the equation:
mitotic index=cells in mitosistotal cells counted×100
Substitute the values into the equation:
mitotic index=45300×100
mitotic index=0.15×100
mitotic index=15%
Rubric
  • Award 1 point for: 15% of the cells in mitosis.

(b) Calculate the time a cell spends in metaphase, in hours to one decimal place. (1 pt)

Answer: 1.2 h  (tolerance ±0.05)

Model answer A cell spends 1.2 hours in metaphase.
Working
Write down the values in the question:
cells in metaphase = 12
total cells counted = 300
cycle length = 30 h
Write down the equations:
percent in metaphase=cells in metaphasetotal cells counted×100
time in metaphase=percent in metaphase100×cycle length
Substitute the values into the equations:
percent in metaphase=12300×100=4%
time in metaphase=4100×30h=1.2h
Rubric
  • Award 1 point for: 1.2 h (4% of a 30-hour cycle).

(c) The cycle length of 30 hours was measured on living cells. Explain why the length of the whole cycle has to be measured on living cells rather than worked out from the count on the slide. (2 pt)

Model answer Each cell on the slide was frozen at one instant of its own cycle.
So the count shows what share of the cells was in each stage at that instant.
A share of the cells is a share of the cycle, not a number of hours.
The same shares would be counted whether the cycle took 30 hours or 60 hours.
So the slide gives no cycle length, and the hours have to be measured on living cells.
Rubric
  • Award 1 point for: the slide shows each cell frozen at one instant, so a count gives only the share of cells (the fraction of the cycle) in each stage.
  • Award 1 point for: the shares would be the same for a 30-hour or a 60-hour cycle, so the count gives no clock or cycle length.

Slip Writing that the cycle length can be found by adding up the times of the stages. Each stage’s time was itself calculated from the given 30 hours; the slide supplied only the shares.

Glossary

mitotic index
The percent of the counted cells on a slide that are in mitosis, all four stages taken together: the cells in mitosis divided by the total cells counted, times 100. Thirty-six of 200 cells in mitosis is a mitotic index of 18%.

APBIO-U04-L17B Two tissues compared

Topic 4.5 · Cell Cycle · 80 steps

Two tally tables: a healing-skin slide of 200 cells with 36 in mitosis and 164 in interphase, and an uninjured-skin slide of 200 cells with 8 in mitosis and 192 in interphase
Two tally tables: a healing-skin slide of 200 cells with 36 in mitosis and 164 in interphase, and an uninjured-skin slide of 200 cells with 8 in mitosis and 192 in interphase

Here are two tally tables from the same microscope session.

Five students each counted 200 cells on a slide from the healing edge of a wound and 200 on a slide of uninjured skin. Their mean counts of cells in mitosis were 36 for the wound and 8 for the uninjured skin: 18% and 4%. Each mean has its ±2SE error bar, and the two error bars do not overlap.

Are more of the wound’s cells dividing? By how much? And what can the counts not tell you?

Unit 4 · Cell Communication and Cell Cycle

1

What can two tissues’ counts tell you, and what can they not?

2

A greater percent of cells in mitosis means more of the sampled cells are dividing, or that mitosis takes longer in that tissue.

3

Or the whole cycle is shorter in that tissue, so mitosis is a bigger share of it.

4

The count alone cannot say which. And the count never shows that every cell is cycling faster.

5

Percent change measures how much greater one percent is than the other.

6

A different pattern is a pile-up: one stage fills while the stages after it empty. The cells are being held at that stage.

7

Two tissues’ counts show one of these three patterns. Here is how to read each.

8More cells dividing, or a longer mitosis?

9

Video: Watch: More cells dividing, or a longer mitosis?

Healing skin at 18% of cells in mitosis, uninjured skin at 4%, ±2SE bars that do not overlap. More of the healing skin’s cells are dividing, or mitosis takes longer there, or the whole cycle is shorter there; the count cannot say which, and it does not show every cell cycling faster.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L17Ba.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L17Ba.mp4

10
Check q1

Two means each come with a ±2SE error bar, and the two error bars have a clear gap between them.

What does that tell you about the two means?

  1. A. ✓ The difference between the two means is very unlikely to be chance
  2. B. The difference between the two means could easily be chance
    Separate error bars show two means that differ by more than chance would explain.
    Overlapping bars are the ones that leave the question open.

Why: When two ±2SE error bars do not overlap, the difference between the means is very unlikely to be chance.

11

Here is a bar chart of the mean percent of cells in mitosis for the two slides, each mean with its ±2SE error bar.

Two bars: mean percent of cells in mitosis, 18.0 for healing skin and 4.0 for uninjured skin, with ±2SE error bars from 15.6% to 20.4% and from 2.8% to 5.2%; gridlines every 5 percent; the legend reads: error bars represent ±2SE
Two bars: mean percent of cells in mitosis, 18.0 for healing skin and 4.0 for uninjured skin, with ±2SE error bars from 15.6% to 20.4% and from 2.8% to 5.2%; gridlines every 5 percent; the legend reads: error bars represent ±2SE
12

The legend says the bars represent ±2SE, so that rule applies.

13

The healing-skin bar runs 15.6% to 20.4%. The uninjured-skin bar runs 2.8% to 5.2%.

14

The bars do not overlap. So the difference between 18% and 4% is very unlikely to be chance.

15

What does the greater percent mean? Each cell on a slide was frozen at one instant.

16

So the percent is how many of the sampled cells were caught in mitosis at that instant.

17

More cells are caught in mitosis when more of the sampled cells are dividing. More cells are also caught in mitosis when each dividing cell spends longer in mitosis.

18

More cells are also caught in mitosis when the whole cycle is shorter, so mitosis is a bigger share of the cycle.

19

So a greater percent of cells in mitosis has three possible causes: more of the sampled cells are dividing, or mitosis takes longer in that tissue, or the whole cycle is shorter there. The count alone cannot tell you which.

20

The count does not show that every cell in the healing skin is cycling faster.

21

More of the healing skin’s cells are caught in the act of dividing. That is all the count says.

22

What you are expected to know Interpret a greater percent of cells in mitosis: more of the sampled cells are dividing, or mitosis takes longer in that tissue, or the whole cycle is shorter there.

23
Check q2

Five students each counted 180 cells from a lizard’s regrowing tail and 180 from its old tail. The graph below shows the mean percent of cells in mitosis in each.

A bar chart of the mean percent of cells in mitosis for two tissues, the regrowing tail and the old tail of a lizard. The vertical axis runs from 0 to 20 percent with gridlines every 5 percent. The regrowing-tail bar is much taller than the old-tail bar. Each bar carries an error bar; the legend reads: error bars represent ±2SE
A bar chart of the mean percent of cells in mitosis for two tissues, the regrowing tail and the old tail of a lizard. The vertical axis runs from 0 to 20 percent with gridlines every 5 percent. The regrowing-tail bar is much taller than the old-tail bar. Each bar carries an error bar; the legend reads: error bars represent ±2SE

Compared with the old tail, how many of the regrowing tail’s cells are caught dividing?

  1. A. ✓ More of them are caught dividing
  2. B. Fewer of them are caught dividing
    The regrowing-tail bar sits well above the old-tail bar, and the two error bars do not overlap.
    So more of the regrowing-tail cells are caught dividing.
  3. C. The data do not show a difference
    The legend says the bars represent ±2SE.
    The two error bars do not overlap.
    So the difference between the means is very unlikely to be chance.

Why: The legend says the bars represent ±2SE.
The regrowing-tail bar sits well above the old-tail bar, and the two error bars do not overlap.
So more of the regrowing tail’s cells are caught dividing.

24
Check q3

Five students each counted 180 cells from a leaf and 180 from the shoot tip of the same tomato plant. The graph below shows the mean percent of cells in mitosis in each.

A bar chart of the mean percent of cells in mitosis for two tissues, a leaf and the shoot tip of the same tomato plant. The vertical axis runs from 0 to 20 percent with gridlines every 5 percent. The leaf bar is much shorter than the shoot-tip bar. Each bar carries an error bar; the legend reads: error bars represent ±2SE
A bar chart of the mean percent of cells in mitosis for two tissues, a leaf and the shoot tip of the same tomato plant. The vertical axis runs from 0 to 20 percent with gridlines every 5 percent. The leaf bar is much shorter than the shoot-tip bar. Each bar carries an error bar; the legend reads: error bars represent ±2SE

Compared with the shoot tip, how many of the leaf’s cells are caught dividing?

  1. A. More of them are caught dividing
    The leaf bar sits well below the shoot-tip bar, and the two error bars do not overlap.
    So fewer of the leaf cells are caught dividing.
  2. B. ✓ Fewer of them are caught dividing
  3. C. The data do not show a difference
    The legend says the bars represent ±2SE.
    The two error bars do not overlap.
    So the difference between the means is very unlikely to be chance.

Why: The legend says the bars represent ±2SE.
The leaf bar sits well below the shoot-tip bar, and the two error bars do not overlap.
So fewer of the leaf’s cells are caught dividing.

25
Check q4

Two classes each counted 180 cells on a slide of the same onion root tip, five students in each class. The graph below shows the mean percent of cells in mitosis on each slide.

A bar chart of the mean percent of cells in mitosis for two slides of the same onion root tip, one counted by class 1 and one by class 2. The vertical axis runs from 0 to 20 percent with gridlines every 5 percent. The two bars are of similar height, the class 2 bar a little taller. Each bar carries an error bar; the legend reads: error bars represent ±2SE
A bar chart of the mean percent of cells in mitosis for two slides of the same onion root tip, one counted by class 1 and one by class 2. The vertical axis runs from 0 to 20 percent with gridlines every 5 percent. The two bars are of similar height, the class 2 bar a little taller. Each bar carries an error bar; the legend reads: error bars represent ±2SE

Compared with the class 1 slide, how many of the class 2 slide’s cells are caught dividing?

  1. A. More of them are caught dividing
    The class 2 bar is a little taller, but the ±2SE bars overlap, so a higher mean could be chance.
  2. B. Fewer of them are caught dividing
    The class 2 bar is not lower, and the ±2SE bars overlap, so no lower share is shown.
  3. C. ✓ The data do not show a difference

Why: The legend says the bars represent ±2SE.
The two error bars overlap.
So the difference between the means could be chance, and the data do not show a difference between the two slides.

26
Check q5

Five students each counted 180 cells from a frog’s skin and 180 from its gut lining. The graph below shows the mean percent of cells in mitosis in each.

A bar chart of the mean percent of cells in mitosis for two tissues, the skin and the gut lining of a frog. The vertical axis runs from 0 to 20 percent with gridlines every 5 percent. The two bars are of similar height, the gut-lining bar a little taller. Each bar carries an error bar; the legend reads: error bars represent ±2SE
A bar chart of the mean percent of cells in mitosis for two tissues, the skin and the gut lining of a frog. The vertical axis runs from 0 to 20 percent with gridlines every 5 percent. The two bars are of similar height, the gut-lining bar a little taller. Each bar carries an error bar; the legend reads: error bars represent ±2SE

Compared with the gut lining, how many of the skin’s cells are caught dividing?

  1. A. More of them are caught dividing
    The skin bar is a little shorter, not taller, and the ±2SE bars overlap.
  2. B. Fewer of them are caught dividing
    The skin bar is a little shorter, but the ±2SE bars overlap, so a smaller share could be chance.
  3. C. ✓ The data do not show a difference

Why: The legend says the bars represent ±2SE.
The skin bar is a little shorter, but the two error bars overlap.
So the difference between the means could be chance, and the data do not show a difference.

27
Check q6

Five students each counted 180 cells from an adult mouse’s brain and 180 from its gut lining. The graph below shows the mean percent of cells in mitosis in each.

A bar chart of the mean percent of cells in mitosis for two tissues, the brain and the gut lining of an adult mouse. The vertical axis runs from 0 to 20 percent with gridlines every 5 percent. The brain bar is very short and the gut-lining bar is much taller. Each bar carries an error bar; the legend reads: error bars represent ±2SE
A bar chart of the mean percent of cells in mitosis for two tissues, the brain and the gut lining of an adult mouse. The vertical axis runs from 0 to 20 percent with gridlines every 5 percent. The brain bar is very short and the gut-lining bar is much taller. Each bar carries an error bar; the legend reads: error bars represent ±2SE

Compared with the gut lining, how many of the brain’s cells are caught dividing?

  1. A. More of them are caught dividing
    The brain bar sits well below the gut-lining bar, and the two error bars do not overlap.
    So fewer of the brain cells are caught dividing.
  2. B. ✓ Fewer of them are caught dividing
  3. C. The data do not show a difference
    The legend says the bars represent ±2SE.
    The two error bars do not overlap.
    So the difference between the means is very unlikely to be chance.

Why: The legend says the bars represent ±2SE.
The brain bar sits well below the gut-lining bar, and the two error bars do not overlap.
So fewer of the brain’s cells are caught dividing.

28
Check q7

Five students each counted 180 cells from a zebrafish’s regrowing fin and 180 from one of its uninjured fins. The graph below shows the mean percent of cells in mitosis in each.

A bar chart of the mean percent of cells in mitosis for two tissues, the regrowing fin and an uninjured fin of a zebrafish. The vertical axis runs from 0 to 20 percent with gridlines every 5 percent. The regrowing-fin bar is much taller than the uninjured-fin bar. Each bar carries an error bar; the legend reads: error bars represent ±2SE
A bar chart of the mean percent of cells in mitosis for two tissues, the regrowing fin and an uninjured fin of a zebrafish. The vertical axis runs from 0 to 20 percent with gridlines every 5 percent. The regrowing-fin bar is much taller than the uninjured-fin bar. Each bar carries an error bar; the legend reads: error bars represent ±2SE

Compared with the uninjured fin, how many of the regrowing fin’s cells are caught dividing?

  1. A. ✓ More of them are caught dividing
  2. B. Fewer of them are caught dividing
    The regrowing-fin bar sits well above the uninjured-fin bar, and the two error bars do not overlap.
    So more of the regrowing-fin cells are caught dividing.
  3. C. The data do not show a difference
    The legend says the bars represent ±2SE.
    The two error bars do not overlap.
    So the difference between the means is very unlikely to be chance.

Why: The legend says the bars represent ±2SE.
The regrowing-fin bar sits well above the uninjured-fin bar, and the two error bars do not overlap.
So more of the regrowing fin’s cells are caught dividing.

29
Check q8

After part of a rat’s liver is removed, students count 150 cells from the regrowing liver and 150 from a normal liver. They find 30 cells in mitosis in the regrowing liver and 3 in the normal liver.

Which conclusion do the counts support?

  1. A. Every cell in the regrowing liver divides faster
    The count shows that more of the sampled cells are in the act of dividing.
    The count does not show the speed of any one cell.
  2. B. ✓ More of the regrowing liver’s cells are caught dividing
  3. C. The regrowing liver holds more cells in total
    Both samples were 150 cells.
    The difference is in how many of those 150 cells were in mitosis.
  4. D. Each dividing cell in the regrowing liver spends less time in mitosis
    A stage that takes less time catches fewer cells.
    So a shorter mitosis would lower the count.
    The regrowing liver’s count is higher, not lower.

Why: Thirty of 150 against 3 of 150 is a far greater fraction of cells in mitosis.
So more of the regrowing liver’s cells are dividing, or its mitosis takes longer, or its whole cycle is shorter.
The count cannot say that every cell cycles faster.

30
Practice writing an answer

After part of a rat’s liver is removed, students count 150 cells from the regrowing liver and 150 from a normal liver. They find 30 cells in mitosis in the regrowing liver and 3 in the normal liver. More of the regrowing liver’s cells are caught dividing.

(a) Explain why a count of frozen cells leaves the speed of any single cell’s cycle unknown. (1 pt)

Model answer Each cell on a slide was frozen at one instant of its own cycle.
So a count of frozen cells shows how many of the sampled cells were caught in mitosis.
So more of the regrowing liver’s cells are caught dividing.
The speed of one cell’s cycle is its cycle length, and a count of frozen cells does not give the cycle length.
So the count leaves the speed of any single cell’s cycle unknown.
Rubric
  • Award 1 point for: a count of cells frozen at one instant shows how many of the sampled cells were caught dividing (more of them, or a longer mitosis, or a shorter cycle) and gives no cycle length, so the speed of any single cell’s cycle stays unknown.
31
Check q9

In a mouse’s healing gut, 12% of the cells are in mitosis. In that mouse’s uninjured gut, 2% are. A student says: “So every cell in the healing gut is cycling faster.”

Is the student correct?

  1. A. Yes — a bigger share means every cell cycles faster
    The count shows how many cells were caught dividing at one frozen instant.
    More caught dividing does not show that any one cell cycles faster.
  2. B. No — the healing gut holds more cells in total
    A percent is a count out of a hundred, so the samples are the same size.
    More of each hundred healing-gut cells were caught in mitosis.
  3. C. ✓ No — more of the healing gut’s cells are caught dividing

Why: Each cell on the slide was frozen at one instant.
So 12% against 2% shows that more healing-gut cells were caught dividing.
More of them are dividing, or mitosis takes longer, or the whole cycle is shorter.
The count does not show the speed of any one cell.

32A pile-up in one stage

33

Video: Watch: A pile-up in one stage

After a spindle inhibitor, metaphase rises from 50.0% to 85.0% of the cells in mitosis and anaphase falls from 20.0% to 2.5%. Cells reach metaphase and anaphase never starts. The stage that fills up is where the block acts.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L17Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L17Bb.mp4

34
Check q10

In anaphase, what pulls the separated chromatids toward opposite poles?

  1. A. The nuclear envelope
    The nuclear envelope broke down in prophase; it re-forms in telophase.
  2. B. ✓ The spindle fibers
  3. C. The cell membrane
    The cell membrane pinches in during cytokinesis, after the chromatids have separated.

Why: The spindle fibers attach to each chromosome at its centromere.
In anaphase the connection at each centromere breaks, so the sister chromatids separate.
Then the spindle fibers pull the separated chromatids toward opposite poles.

35

Now consider a different pattern. Suppose a student treats onion roots with a spindle inhibitor, a chemical that stops the spindle fibers from moving the chromosomes.

36

She sorts 120 cells in mitosis by stage, from untreated roots and from treated roots.

37

She counts a treated cell as metaphase when its chromosomes are condensed and each is still two joined sister chromatids, whether or not they lie in a row.

38

Here is a graph of her two counts, as percents of the 120 cells in mitosis.

Two panels of three bars each, percent of 120 cells in mitosis in metaphase, anaphase and telophase: untreated roots 50.0, 20.0 and 30.0; roots treated with a spindle inhibitor 85.0, 2.5 and 12.5; gridlines every 10 percent
Two panels of three bars each, percent of 120 cells in mitosis in metaphase, anaphase and telophase: untreated roots 50.0, 20.0 and 30.0; roots treated with a spindle inhibitor 85.0, 2.5 and 12.5; gridlines every 10 percent
39

Metaphase rose from 50.0% to 85.0% of the cells in mitosis, while anaphase fell from 20.0% to 2.5%.

40

In anaphase the spindle fibers pull the separated chromatids toward opposite poles. With the fibers stopped, the sister chromatids stay joined and nothing moves toward the poles.

41

So the treated cells reach metaphase, and anaphase never starts. The inhibitor holds the cells at metaphase.

42

A pile-up in one stage means the cells are being held at that stage. The stage that fills up tells you where the block acts.

43

What you are expected to know Interpret a pile-up in one stage: the cells are being held there, and the stage that fills up locates the block.

44
Check q11

A researcher adds a drug to dividing whitefish embryo cells. The graph below sorts 100 cells in mitosis by stage before and after the drug.

Two panels of bars, one above the other, on one scale of percent of 100 whitefish cells in mitosis from 0 to 100 with gridlines every 10 percent. Each panel has four bars, labelled prophase, metaphase, anaphase and telophase. The upper panel is labelled before the drug and the lower panel after the drug
Two panels of bars, one above the other, on one scale of percent of 100 whitefish cells in mitosis from 0 to 100 with gridlines every 10 percent. Each panel has four bars, labelled prophase, metaphase, anaphase and telophase. The upper panel is labelled before the drug and the lower panel after the drug

What does the drug do to the cells?

  1. A. The drug speeds the cells through anaphase
    Anaphase’s share fell because telophase’s share grew.
    The same 100 cells in mitosis now include many more cells held at telophase.
    So the pile-up is in telophase.
  2. B. The drug stops the chromosomes condensing in prophase
    Prophase did not fill up.
    Prophase fell from 42% to 30%.
    The big change is telophase climbing from 23% to 52%.
  3. C. ✓ The drug holds the cells at telophase
  4. D. The drug makes the cells skip metaphase
    Metaphase is still there at 12%.
    The change is a pile-up in telophase, where cells enter and cannot leave.

Why: Telophase rose from 23% to 52% of the cells in mitosis while the stages before it fell.
Cells reach telophase and cannot finish it.
So the drug holds the cells at telophase.
The stage that fills up is where the block acts.

45
Check q12

A researcher adds a drug to dividing animal cells. The graph below sorts 100 dividing cells by stage before and after the drug.

Two panels of bars, one above the other, on one scale of percent of 100 dividing cells from 0 to 60 with gridlines every 10 percent. Each panel has five bars, labelled prophase, metaphase, anaphase, telophase and cytokinesis. The upper panel is labelled before the drug and the lower panel after the drug
Two panels of bars, one above the other, on one scale of percent of 100 dividing cells from 0 to 60 with gridlines every 10 percent. Each panel has five bars, labelled prophase, metaphase, anaphase, telophase and cytokinesis. The upper panel is labelled before the drug and the lower panel after the drug

Where does the drug act?

  1. A. On the copying of DNA in S phase
    The cells all reached mitosis and went through its stages.
    The pile-up is at the very end.
  2. B. On the coiling of the chromosomes in prophase
    Prophase fell a little, from 25% to 20%.
    The stage that filled up is cytokinesis, from 15% to 42%.
  3. C. On the pulling apart of the sisters in anaphase
    Anaphase’s share fell rather than rose.
    The pile-up is later, in cells with two nuclei and no furrow.
  4. D. ✓ On the furrow that divides the cytoplasm in cytokinesis

Why: Cytokinesis rose from 15% to 42% of the dividing cells while every earlier stage fell.
Cells reach cytokinesis and cannot finish it, so the drug acts on the furrow that divides the cytoplasm.

46Quick quiz: which stage fills? mixed practice

47
Check q13

A researcher adds a drug to dividing cells. The graph below sorts 100 cells in mitosis by stage before and after the drug.

Two panels of bars, one above the other, on one scale of percent of 100 cells in mitosis from 0 to 100 with gridlines every 10 percent. Each panel has four bars, labelled prophase, metaphase, anaphase and telophase. The upper panel is labelled before the drug and the lower panel after the drug
Two panels of bars, one above the other, on one scale of percent of 100 cells in mitosis from 0 to 100 with gridlines every 10 percent. Each panel has four bars, labelled prophase, metaphase, anaphase and telophase. The upper panel is labelled before the drug and the lower panel after the drug

At which stage does the drug hold the cells?

  1. A. ✓ Prophase
  2. B. Metaphase
    Metaphase’s share fell.
    The stage that filled is prophase.
  3. C. Anaphase
    Anaphase’s share fell.
    The stage that filled is prophase.
  4. D. Telophase
    Telophase’s share fell.
    The stage that filled is prophase.

Why: Prophase rose from 40% to 70% of the cells in mitosis while every later stage fell.
Cells reach prophase and cannot leave it.
So the drug holds the cells at prophase.

48
Check q14

A researcher adds a drug to dividing cells. The graph below sorts 100 cells in mitosis by stage before and after the drug.

Two panels of bars, one above the other, on one scale of percent of 100 cells in mitosis from 0 to 100 with gridlines every 10 percent. Each panel has four bars, labelled prophase, metaphase, anaphase and telophase. The upper panel is labelled before the drug and the lower panel after the drug
Two panels of bars, one above the other, on one scale of percent of 100 cells in mitosis from 0 to 100 with gridlines every 10 percent. Each panel has four bars, labelled prophase, metaphase, anaphase and telophase. The upper panel is labelled before the drug and the lower panel after the drug

At which stage does the drug hold the cells?

  1. A. Prophase
    Prophase’s share fell.
    The stage that filled is metaphase.
  2. B. ✓ Metaphase
  3. C. Anaphase
    Anaphase’s share fell.
    The stage that filled is metaphase.
  4. D. Telophase
    Telophase’s share fell.
    The stage that filled is metaphase.

Why: Metaphase rose from 30% to 60% of the cells in mitosis while anaphase and telophase fell.
Cells reach metaphase and cannot leave it.
So the drug holds the cells at metaphase.

49
Check q15

A researcher adds a drug to dividing cells. The graph below sorts 100 cells in mitosis by stage before and after the drug.

Two panels of bars, one above the other, on one scale of percent of 100 cells in mitosis from 0 to 100 with gridlines every 10 percent. Each panel has four bars, labelled prophase, metaphase, anaphase and telophase. The upper panel is labelled before the drug and the lower panel after the drug
Two panels of bars, one above the other, on one scale of percent of 100 cells in mitosis from 0 to 100 with gridlines every 10 percent. Each panel has four bars, labelled prophase, metaphase, anaphase and telophase. The upper panel is labelled before the drug and the lower panel after the drug

At which stage does the drug hold the cells?

  1. A. Prophase
    Prophase’s share fell.
    The stage that filled is anaphase.
  2. B. Metaphase
    Metaphase’s share fell.
    The stage that filled is anaphase.
  3. C. ✓ Anaphase
  4. D. Telophase
    Telophase’s share did not change.
    The stage that filled is anaphase.

Why: Anaphase rose from 20% to 50% of the cells in mitosis while prophase and metaphase fell.
Cells reach anaphase and cannot finish it.
So the drug holds the cells at anaphase.

50
Check q16

A researcher adds a drug to dividing cells. The graph below sorts 100 cells in mitosis by stage before and after the drug.

Two panels of bars, one above the other, on one scale of percent of 100 cells in mitosis from 0 to 100 with gridlines every 10 percent. Each panel has four bars, labelled prophase, metaphase, anaphase and telophase. The upper panel is labelled before the drug and the lower panel after the drug
Two panels of bars, one above the other, on one scale of percent of 100 cells in mitosis from 0 to 100 with gridlines every 10 percent. Each panel has four bars, labelled prophase, metaphase, anaphase and telophase. The upper panel is labelled before the drug and the lower panel after the drug

At which stage does the drug hold the cells?

  1. A. Prophase
    Prophase’s share fell.
    The stage that filled is telophase.
  2. B. Metaphase
    Metaphase’s share fell.
    The stage that filled is telophase.
  3. C. Anaphase
    Anaphase’s share fell.
    The stage that filled is telophase.
  4. D. ✓ Telophase

Why: Telophase rose from 20% to 50% of the cells in mitosis while every earlier stage fell.
Cells reach telophase and cannot finish it.
So the drug holds the cells at telophase.

51
Check q17

A researcher adds a drug to dividing cells. The graph below sorts 100 cells in mitosis by stage before and after the drug.

Two panels of bars, one above the other, on one scale of percent of 100 cells in mitosis from 0 to 100 with gridlines every 10 percent. Each panel has four bars, labelled prophase, metaphase, anaphase and telophase. The upper panel is labelled before the drug and the lower panel after the drug
Two panels of bars, one above the other, on one scale of percent of 100 cells in mitosis from 0 to 100 with gridlines every 10 percent. Each panel has four bars, labelled prophase, metaphase, anaphase and telophase. The upper panel is labelled before the drug and the lower panel after the drug

At which stage does the drug hold the cells?

  1. A. Prophase
    Prophase’s share fell.
    The stage that filled is metaphase.
  2. B. ✓ Metaphase
  3. C. Anaphase
    Anaphase’s share fell.
    The stage that filled is metaphase.
  4. D. Telophase
    Telophase’s share fell.
    The stage that filled is metaphase.

Why: Metaphase rose from 25% to 55% of the cells in mitosis while anaphase and telophase fell.
Cells reach metaphase and cannot leave it.
So the drug holds the cells at metaphase.

52How much greater: percent change

53

Video: Watch: How much greater: percent change

From 4% of cells in mitosis in uninjured skin to 18% in healing skin: the change, 14 percentage points, divided by the initial 4%, times 100, is a 350% increase.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L17Bc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L17Bc.mp4

54
Check q18

A count rises from 20 cells to 25 cells.

How is the percent change worked out?

  1. A. The change divided by the final value, times 100
    Percent change is measured against where you started, so the change is divided by the initial value.
  2. B. The final value divided by the initial value, times 100
    That gives the final value as a percent of the start, not the change.
  3. C. ✓ The change divided by the initial value, times 100

Why: Percent change measures the change against the initial value.
So it is the change divided by the initial value, times 100.

55

How much greater is 18% than 4%? Percent change compares the two: from 4% in uninjured skin to 18% in healing skin.

56

Here is the equation, the same one you used for the change in mass of a potato core.

Percent change: the final value minus the initial value, divided by the initial value, times 100
57
Worked example

The percent of cells in mitosis is 4% in uninjured skin and 18% in healing skin. What is the percent change?

Write down the values in the question:
initial value = 4% (uninjured skin)
final value = 18% (healing skin)
Write down the equation:
percent change=final−initialinitial×100
Substitute the values into the equation:
percent change=18−44×100
percent change=144×100
percent change=+350%
58

A rise gets a positive answer and a fall a negative one. Keep the sign in front of your answer.

59

What you are expected to know Calculate the percent change between two populations’ percents of cells in mitosis: the final value minus the initial value, divided by the initial value, times 100.

60
Check q19 numeric entry

In a regrowing liver, 15% of the cells are in mitosis; in a normal liver, 6% are.

Calculate the percent change, from 6% to 15%.

Answer: 150 %  (tolerance ±0.5)

Working
Write down the values in the question:
initial value = 6% (normal liver)
final value = 15% (regrowing liver)
Write down the equation:
percent change=final−initialinitial×100
Substitute the values into the equation:
percent change=15−66×100
percent change=96×100
percent change=+150%
61
Check q20 numeric entry

In a fruit-fly embryo, 20% of the cells are in mitosis at one stage of development, against 5% in the adult fly’s gut.

Calculate the percent change, from 5% to 20%.

Answer: 300 %  (tolerance ±0.5)

Working
Write down the values in the question:
initial value = 5% (adult gut)
final value = 20% (embryo)
Write down the equation:
percent change=final−initialinitial×100
Substitute the values into the equation:
percent change=20−55×100
percent change=155×100
percent change=+300%
62
Check q21 numeric entry

Of 500 cells counted from a mouse’s bone marrow, 60 were in mitosis. Of 500 cells counted from that mouse’s skin, 25 were in mitosis.

Calculate the percent change, from the skin’s percent of cells in mitosis to the bone marrow’s percent of cells in mitosis.

Answer: 140 %  (tolerance ±0.5)

Working
Write down the values in the question:
cells in mitosis, bone marrow = 60 of 500
cells in mitosis, skin = 25 of 500
Write down the equations:
percent in mitosis=cells in mitosistotal cells counted×100
percent change=final−initialinitial×100
Substitute the values into the equations:
percent in mitosis, bone marrow=60500×100=12%
percent in mitosis, skin=25500×100=5%
percent change=12−55×100
percent change=75×100=+140%
63

Back to the two slides: five students each counted 200 cells from the healing edge of a wound and 200 from uninjured skin.

64

Their means were 18% of cells in mitosis in the wound and 4% in the uninjured skin, and the two ±2SE error bars do not overlap.

65

So more of the wound’s cells are dividing, or mitosis takes longer there, or the whole cycle is shorter there. The count cannot say which, and it does not show that every cell cycles faster.

66

How much greater? From 4% to 18% is a percent change of +350%.

67Quick quiz: four quick percent changes mixed practice

68
Check q22 numeric entry

The percent of cells in mitosis is 2% in one tissue and 5% in another.

Calculate the percent change, from 2% to 5%.

Answer: 150 %  (tolerance ±0.5)

Working
Write down the values in the question:
initial value = 2%
final value = 5%
Write down the equation:
percent change=final−initialinitial×100
Substitute the values into the equation:
percent change=5−22×100
percent change=+150%
69
Check q23 numeric entry

The percent of cells in mitosis is 10% in one tissue and 12% in another.

Calculate the percent change, from 10% to 12%.

Answer: 20 %  (tolerance ±0.5)

Working
Write down the values in the question:
initial value = 10%
final value = 12%
Write down the equation:
percent change=final−initialinitial×100
Substitute the values into the equation:
percent change=12−1010×100
percent change=+20%
70
Check q24 numeric entry

The percent of cells in mitosis is 8% in one tissue and 6% in another.

Calculate the percent change, from 8% to 6%. Keep the sign.

Answer: -25 %  (tolerance ±0.5)

Working
Write down the values in the question:
initial value = 8%
final value = 6%
Write down the equation:
percent change=final−initialinitial×100
Substitute the values into the equation:
percent change=6−88×100
percent change=−25%
71
Check q25 numeric entry

The percent of cells in mitosis is 4% in one tissue and 14% in another.

Calculate the percent change, from 4% to 14%.

Answer: 250 %  (tolerance ±0.5)

Working
Write down the values in the question:
initial value = 4%
final value = 14%
Write down the equation:
percent change=final−initialinitial×100
Substitute the values into the equation:
percent change=14−44×100
percent change=+250%

72Mixed practice mixed practice

73
Check q26

Five students each counted 250 cells from the lining of a rat’s cheek and 250 from the lining of its windpipe. The graph below shows the mean percent of cells in mitosis for each lining, with error bars.

A bar chart of the mean percent of cells in mitosis for two tissues, the lining of the cheek and the lining of the windpipe, each counted by five students. The vertical axis runs from 0 to 20 percent with gridlines every 5 percent. The two bars are of similar height, the cheek-lining bar a little taller. Each bar carries an error bar; the legend reads: error bars represent ±2SE
A bar chart of the mean percent of cells in mitosis for two tissues, the lining of the cheek and the lining of the windpipe, each counted by five students. The vertical axis runs from 0 to 20 percent with gridlines every 5 percent. The two bars are of similar height, the cheek-lining bar a little taller. Each bar carries an error bar; the legend reads: error bars represent ±2SE

What can the class conclude about the two linings?

  1. A. More of the cheek lining’s cells are dividing
    The ±2SE bars overlap: the cheek-lining bar runs about 9% to 15%, the windpipe-lining bar about 7% to 13%.
    So the higher mean could be chance.
  2. B. ✓ The data do not show a difference between the two linings
  3. C. The two linings have exactly the same share of dividing cells
    Overlapping ±2SE bars leave the question open.
    The true means may still differ.
  4. D. Cells of the windpipe lining spend longer in mitosis than cells of the cheek lining
    Overlapping bars support no claim either way about the two linings.

Why: The legend says the bars represent ±2SE.
The cheek-lining bar runs about 9% to 15%, the windpipe-lining bar about 7% to 13%.
The two error bars overlap.
So these counts do not show a difference between the two linings, which is not the same as showing them equal.

74
Check q27

In a salamander’s regrowing leg, 14% of the cells are in mitosis. In its uninjured leg, 3% are. One explanation is that more of the regrowing leg’s cells are dividing.

Which other explanation could also produce the greater percent in the regrowing leg?

  1. A. ✓ Mitosis takes longer in the regrowing leg
  2. B. The regrowing leg has more cells in total
    A percent is a count out of a hundred, so the size of the tissue does not change it.
  3. C. Mitosis takes less time in the regrowing leg
    A stage that takes less time catches fewer cells at any instant.
    So a shorter mitosis would lower the regrowing leg’s percent, not raise it.

Why: Each cell was frozen at one instant.
A dividing cell that spends longer in mitosis is more likely to be caught in mitosis.
So a longer mitosis in the regrowing leg would also raise its percent of cells in mitosis.

75
Check q28

A researcher adds a drug to dividing cells. Anaphase rises from 10% to 45% of the cells in mitosis, and telophase falls from 25% to 5%.

At which stage does the drug hold the cells?

  1. A. Prophase
    Prophase is not the stage whose share rose.
  2. B. Metaphase
    Metaphase is not the stage whose share rose.
  3. C. ✓ Anaphase
  4. D. Telophase
    Telophase’s share fell, from 25% to 5%.
    The cells are not reaching telophase.

Why: Anaphase’s share rose from 10% to 45% while telophase’s share fell.
The stage that fills up is where the block acts.
So the drug holds the cells at anaphase.

76
Check q29 numeric entry

In a plant’s root tip, 5% of the cells are in mitosis in dry soil and 6% in watered soil.

Calculate the percent change, from 5% to 6%.

Answer: 20 %  (tolerance ±0.5)

Working
Write down the values in the question:
initial value = 5% (dry soil)
final value = 6% (watered soil)
Write down the equation:
percent change=final−initialinitial×100
Substitute the values into the equation:
percent change=6−55×100
percent change=+20%
77
Check q30

A student sorts cells in mitosis by stage before and after a drug.

Which pattern shows that the drug holds the cells at one stage?

  1. A. ✓ One stage’s share rises while the stages after it fall
  2. B. Every stage’s share rises by the same amount
    The shares are percents of the cells in mitosis, so they cannot all rise.
  3. C. The mitotic index falls to zero
    A mitotic index of zero means no cells in mitosis at all, not cells held at one stage.

Why: Cells held at one stage keep arriving there and cannot leave.
So that stage’s share rises, and the stages after it receive fewer cells and fall.

78
Check q31

The percent of cells in mitosis rises from 4% in one tissue to 8% in another. A student says: “That is a 4% increase.”

Is the student correct?

  1. A. Yes — the two percents differ by 4
    The difference of 4 percentage points is the change, not the percent change.
    The change is divided by the initial 4%.
  2. B. ✓ No — the percent change is +100%
  3. C. No — the percent change is +200%
    The change, 4 percentage points, is as big as the initial value.
    A change as big as the initial value is +100%, not +200%.

Why: Percent change is the change divided by the initial value, times 100.
The change is 4 percentage points and the initial value is 4%.
So the percent change is +100%.

79
Practice writing an answer

A student treats onion root tips with a spindle inhibitor, a chemical that stops the spindle fibers from moving the chromosomes, for six hours. Untreated root tips grow alongside in plain water. She then fixes and stains both sets and sorts, from each set, 80 cells in mitosis that had reached metaphase or a later stage, recording each as metaphase, anaphase or telophase. She counted a cell as metaphase when its chromosomes were condensed and each was still two joined sister chromatids, whether or not they lay in a row. Her counts are in the table below.

A table of the student's counts of 80 cells in mitosis from untreated roots and from roots treated with a spindle inhibitor: metaphase 36 and 66, anaphase 20 and 4, telophase 24 and 10, total 80 and 80

(a) Calculate the percent of the treated cells in mitosis that were in metaphase. (1 pt)

Answer: 82.5 %  (tolerance ±0.05)

Model answer In the treated roots, 82.5% of the cells in mitosis were in metaphase.
Working
Write down the values in the question:
treated cells in metaphase = 66
treated cells in mitosis counted = 80
Write down the equation:
percent in metaphase=cells in metaphasetotal cells in mitosis×100
Substitute the values into the equation:
percent in metaphase=6680×100
percent in metaphase=0.825×100
percent in metaphase=82.5%
Rubric
  • Award 1 point for: 82.5% (accept 83%) of the treated cells in mitosis in metaphase.

(b) Identify the dependent variable in this investigation. (1 pt)

Model answer The dependent variable is the number of cells in mitosis in each stage (or the percent of cells in mitosis in each stage), which the student measured.
Rubric
  • Award 1 point for: the number or percent of cells in mitosis in each stage as the dependent variable.

Slip Naming the spindle inhibitor. The inhibitor is what the student changed, so it is the independent variable. The dependent variable is what she counted.

(c) Describe how the distribution of stages changed in the treated roots. Explain why the cells piled up at that stage. (2 pt)

Model answer In the treated roots, metaphase rose from 45.0% to 82.5% of the cells in mitosis.
Anaphase fell from 25.0% to 5.0%.
Telophase fell from 30.0% to 12.5%.
So the cells reached metaphase and were held there.
In anaphase the spindle fibers pull the separated chromatids toward opposite poles.
With the fibers stopped, the sister chromatids stayed joined and nothing moved toward the poles.
So anaphase never started, and the cells could not move on from metaphase.
Rubric
  • Award 1 point for: metaphase rose (from 45% to about 83%) while anaphase and telophase fell, so cells accumulated at metaphase.
  • Award 1 point for: in anaphase the spindle fibers pull the separated chromatids toward opposite poles, so with the fibers stopped the cells cannot leave metaphase.

Slip Writing that the inhibitor stopped the chromosomes condensing or the DNA copying. Both happened before the spindle forms. The cells reached metaphase and were held at the step the spindle carries out.

APBIO-U04-P45 Practice questions: Topic 4.5

Topic 4.5 · Cell Cycle · 10 MCQ · 2 FRQ · for APBIO-U04-T45

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. The first free-response question walks you through the reasoning one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Where a graph carries error bars, the caption says what the bars represent. Where a question gives the length of the whole cell cycle, use that length. Give times in hours to one decimal place and in minutes to the nearest minute.

Video: Watch first: Topic 4.5 summary, part 1: the cell cycle

Grow, copy, divide: G1, S phase, G2, mitosis, cytokinesis; count centromeres, and the DNA per cell halves only at cytokinesis; one copy and one separation give each daughter cell everything.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-T45-summary.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-T45-summary.mp4

Q1 P45-q01

Four cases of cells dividing are described.

Which of the following is cell division for asexual reproduction?

  1. A. An oat seedling’s first leaf lengthens by 3 cm in two days
    New cells that make the seedling bigger are growth.
    Asexual reproduction makes a new organism from one parent’s dividing cells.
  2. B. ✓ A potato plant grows a whole new plant from a bud on one of its tubers
  3. C. Bark grows back over a wound on a tree trunk
    New cells that replace lost or damaged cells are repair.
    Asexual reproduction makes a new organism, not a patch on the old one.
  4. D. An apple tree makes seeds after bees have pollinated its flowers
    A seed forms after pollen from one flower reaches another: two parents.
    Asexual reproduction makes a new organism from one parent’s cells, with the parent’s DNA.

Why: In asexual reproduction one parent’s cells divide and the new cells become a new organism carrying the parent’s DNA.
The new potato plant grew from the parent’s bud alone.
A lengthening leaf is growth; bark over a wound is repair; a seed comes from two parents.

Q2 P45-q02

Between divisions, a stained cell from a rice root shows an evenly grainy nucleus and no separate rods.

Why is the cell's DNA in this form between divisions?

  1. A. Grainy DNA is DNA that has been broken into short pieces ready for copying
    The DNA is intact: each chromosome is one very long molecule, spread out thinly as chromatin.
    Spread out is how it is read and copied.
  2. B. The rods have already been handed on to the daughter cells
    A cell between divisions holds its full set of chromosomes; they are spread out as chromatin, which is why no separate rods show.
  3. C. The chromosomes have coiled up tightly so they can be moved
    Coiled chromosomes show as distinct dark rods, and they appear only when the cell is about to divide.
    A grainy nucleus is the opposite: DNA spread out for working.
  4. D. ✓ Each chromosome is spread out as chromatin, open to be read and copied

Why: For most of the cycle each chromosome is spread out as chromatin, which shows as a grainy nucleus.
Loose is the form for working: the DNA is open to be read and copied.
Only when the cell is about to divide does each chromosome coil into a movable rod.

Q3 P45-q03

An elephant body cell has 56 chromosomes. In S phase it copies its DNA.

How many chromatids does the cell hold at the end of S phase?

  1. A. 28
    28 halves the set, as if copying shared the chromosomes out.
    Copying gives every chromosome a second, identical chromatid, so the chromatids double: 2 × 56 = 112.
  2. B. 56
    56 is the count before copying, one chromatid per chromosome.
    After S phase each of the 56 chromosomes is two sister chromatids joined at one centromere: 112 chromatids.
  3. C. ✓ 112
  4. D. 224
    224 doubles the count twice.
    Each chromosome becomes one pair of sister chromatids, so 56 chromosomes give 2 × 56 = 112 chromatids, and still 56 chromosomes.

Why: Copying turns each chromosome into two identical sister chromatids joined at one centromere, so the chromatids double while the chromosome count stays the same: chromatids=2×56=112, and the cell still holds 56 chromosomes.

Q4 P45-q04

A cell in a growing fern root has just finished cytokinesis and become two daughter cells.

Which three stages does each daughter cell pass through next, in order?

  1. A. ✓ G1, S phase, G2
  2. B. S phase, G1, G2
    A new daughter cell grows first, in G1, before it copies its DNA in S phase.
    The copying is followed by the second growth gap, G2.
  3. C. G2, S phase, G1
    After cytokinesis a daughter cell starts a new cycle at its first stage, G1, then S phase, then G2.
  4. D. Mitosis, G1, S phase
    A daughter cell has just come out of a division.
    It must grow (G1), copy its DNA (S phase) and prepare (G2) before its own mitosis.

Why: The stages of the cell cycle are G1, S phase, G2, then mitosis and cytokinesis, in that order.
Each daughter cell re-enters G1 of its own cycle.
So after cytokinesis the next three stages are G1, S phase and G2: the interphase that takes up most of the cycle.

Q5 P45-q05

Cells of one newt species hold about 40 pg of DNA in G1.

Which DNA reading shows a newt cell part-way through S phase?

  1. A. 20 pg
    20 pg is half the G1 amount.
    A cell never holds less DNA than the G1 amount.
    During S phase the amount rises from 40 pg toward 80 pg.
  2. B. 40 pg
    40 pg is the G1 amount, before copying has begun.
    A cell part-way through S phase holds more than 40 pg and less than double.
  3. C. ✓ 60 pg
  4. D. 80 pg
    80 pg is double the G1 amount, so the cell is in G2 or mitosis.
    Part-way through S phase the amount is between 40 and 80 pg.

Why: DNA content is steady at the G1 amount, doubles during S phase as every chromosome is copied, and stays doubled through G2.
For a cell with about 40 pg in G1, the copied amount is 80 pg.
A reading between them, 60 pg, places the cell part-way through S phase.

Q6 P45-q06

A cell from a growing lettuce root is watched under a microscope.

Which of the following describes the cell while it is in interphase?

  1. A. Its chromosomes are coiled into rods and lined up in one row across its middle
    A row of rods across the middle is metaphase, a stage of mitosis.
    In interphase the chromosomes are spread out and no rods are visible.
  2. B. Its nuclear envelope has broken down and spindle fibers are reaching its chromosomes
    A vanished envelope and a growing spindle are prophase, a stage of mitosis.
    In interphase the nuclear envelope is intact.
  3. C. Its cytoplasm is being divided in two by a cell plate
    A cell plate divides the cytoplasm in cytokinesis, after mitosis.
    Interphase is the long stretch before the next division begins.
  4. D. ✓ Its nucleus looks grainy with no separate chromosomes visible, and it is growing

Why: Interphase is the long stretch between one division and the next: about 23 hours of a 24-hour cycle.
The cell grows and copies its DNA, but no division is seen.
Its nucleus looks grainy, with no separate chromosomes visible.

Q7 P45-q07

Of the cells on a slide from the tip of a rye root grown at 20 °C, 15% are in mitosis. Of the cells on a slide from a root of the same plant grown at 5 °C, 6% are in mitosis.

What is the percent change in the percent of cells in mitosis, from the 20 °C root to the 5 °C root?

  1. A. −150%
    −150% divides the change, −9, by the final value, 6.
    Percent change divides the change by the starting value, 15.
  2. B. ✓ −60%
  3. C. −9%
    −9 is the difference, 6 − 15.
    Percent change divides that difference by the starting value, 15, and multiplies by 100.
  4. D. +60%
    The percent fell, so the change is negative.
    The difference, 6 − 15, divided by 15 and multiplied by 100 is −60%, and the sign is kept.

Why: Percent change is the change divided by the starting value, times 100.
The working below gives −60%: the percent of cells in mitosis fell by 60% in the cold root.

Q8 P45-q08

Two nuclei have formed in a dividing cell from a frog embryo, and the cytoplasm is now being divided.

Which structure divides the cytoplasm of this cell, and how is it made?

  1. A. ✓ A cleavage furrow, made by a ring of protein under the membrane tightening
  2. B. A cell plate, made by vesicles fusing at the equator into a new cell wall
    A cell plate is a plant cell’s way: its rigid cell wall cannot be pinched.
    A frog cell has no cell wall, so a ring of protein pinches it in.
  3. C. A nuclear envelope, made by the spindle fibers joining up across the middle
    Each nuclear envelope surrounds one set of chromosomes at a pole, and the spindle has broken down by this point.
    The cytoplasm is pinched in by a ring of protein.
  4. D. A centromere, made by the sister chromatids joining at the middle of the cell
    A centromere is the join between two sister chromatids, made when the DNA is copied in S phase.
    It has nothing to do with dividing the cytoplasm.

Why: A frog cell is an animal cell with no cell wall.
A ring of protein just under the membrane tightens.
The ring pinches the cell in at a cleavage furrow.
The cleavage furrow deepens until the cell splits into two daughter cells.

Q9 P45-q09

A skin cell at the edge of a healing scrape divides again and again.

Which of the following describes the cell cycle?

  1. A. The division of a cell’s nucleus into two nuclei, each with a full set of chromosomes
    The division of the nucleus is mitosis, one stage of the cycle.
    The cell cycle is the whole repeating sequence: grow, copy, divide.
  2. B. ✓ The repeating sequence in which a cell grows, copies its DNA and divides in two
  3. C. The movement of the separated chromatids from the equator to the two poles of the cell
    Chromatids moving to the poles is anaphase, one stage of mitosis.
    The cell cycle is the whole sequence from one division to the next.
  4. D. The long stretch between one division and the next, in which the cell rests
    The stretch between divisions is interphase, and the cell is not resting in it: the cell grows and copies its DNA.
    The cycle is the whole sequence, division included.

Why: Each edge cell grows, copies its DNA and divides in two.
Each daughter cell then starts the same sequence afresh.
That repeating sequence is the cell cycle.

Q10 P45-q10

A body cell of an alligator holds 32 chromosomes.

Which of the following is the cell’s genome?

  1. A. The 16 chromosomes that each daughter cell receives when the cell divides
    A dividing cell hands its whole genome to each daughter cell: all 32 chromosomes, not 16.
    Every chromosome is copied first, so each daughter cell receives a complete set.
  2. B. The set of proteins that the 32 DNA molecules are wound around
    The proteins are packaging.
    The genome is the DNA itself: the cell’s complete set of chromosomes.
  3. C. The single longest of the 32 DNA molecules
    One chromosome is one DNA molecule.
    The genome is the complete set of all 32.
  4. D. ✓ The complete set of 32 chromosomes, which each daughter cell receives in full

Why: A genome is a cell’s complete set of chromosomes: here, all 32.
Before the cell divides, every chromosome is copied.
So each daughter cell receives all 32, not 16.

FRQ 1 P45-frq1 · Scientific Investigation scaffolded

An earthworm that has lost the last segments of its tail grows them back over a few weeks. A student asks whether more of the cells in the regrowing tail are dividing than in the rest of the body. Five students each fix and stain one such earthworm, cut a thin slice through the regrowing tail and a second slice through the body halfway along the worm, and record the stage of 200 cells in each slice. The table gives one student’s counts for her regrowing-tail slice. The graph gives the mean percent of cells in mitosis for the five regrowing-tail slices and the five mid-body slices, with error bars that represent ±2SE. Take the whole cell cycle in an earthworm as 24 hours.

Top: the stage of each of the 200 cells the student counted on her slide of the regrowing tail. Bottom: the mean percent of cells in mitosis for the five students' regrowing-tail slides and five mid-body slides; error bars represent ±2SE. Gridlines every 5%.
Top: the stage of each of the 200 cells the student counted on her slide of the regrowing tail. Bottom: the mean percent of cells in mitosis for the five students' regrowing-tail slides and five mid-body slides; error bars represent ±2SE. Gridlines every 5%.

(a) Identify the dependent variable in this investigation. (1 pt)

Frame The dependent variable is …

Hint Which quantity did each student read off her slides and record, as against the thing the students chose to compare?

Model answer The dependent variable is the percent of the 200 cells in each slice that were in mitosis, the quantity the students measured.
Rubric
  • Award 1 point for: the dependent variable is the percent of cells in mitosis (or the stage each of the 200 cells was recorded in).
  • Accept "the number of cells in mitosis out of 200". Do not award the point for "the regrowing tail and the mid-body" (that is the independent variable) or for "200 cells" (a condition kept the same).

Slip Naming the part of the earthworm (regrowing tail or mid-body). That is what the students chose to compare, the independent variable; the dependent variable is what they counted.

(b) Calculate the percent of the cells on the regrowing-tail slide that were in mitosis. (1 pt)

Frame The percent of regrowing-tail cells in mitosis is … %.

Hint Which four stages in the table together make up mitosis, and what total does the percent equation divide their sum by?

Answer: 23 %  (tolerance ±0.05)

Model answer On the regrowing-tail slide 23% of the cells were in mitosis: 46 of the 200 cells were in prophase, metaphase, anaphase or telophase.
Working
Write down the values in the question:
cells in mitosis = 22 + 9 + 9 + 6 = 46
total cells counted = 200
Write down the equation:
tex: \text{percent in mitosis} = \frac{\text{cells in mitosis}}{\text{total cells counted}} \times 100
Substitute the values into the equation:
tex: \text{percent in mitosis} = \frac{\text{cells in mitosis}}{\text{total cells counted}} \times 100
tex: \text{percent in mitosis} = \frac{46}{200} \times 100
tex: \text{percent in mitosis} = 0.23 \times 100 = 23\,\%
Rubric
  • Award 1 point for: 23% (22 + 9 + 9 + 6 = 46 cells in mitosis out of 200).
  • Do not award the point for 11% (prophase alone), 77% (the interphase share), about 30% (46 divided by 154, the interphase count) or 22% (the five-slide mean read off the graph).

(c) Calculate the time, in hours to one decimal place, that a regrowing-tail cell spends in metaphase, taking the whole cycle as 24 hours. (1 pt)

Frame A regrowing-tail cell spends … hours in metaphase.

Hint Start from the share of the 200 cells that were in metaphase. What does the question give as the length of the whole cycle, and how does the time-in-stage equation use that length?

Answer: 1.1 h  (tolerance ±0.05)

Model answer A regrowing-tail cell spends 1.1 hours in metaphase (1.08 h to two decimal places), about 65 minutes.
Working
Write down the values in the question:
cells in metaphase = 9
total cells counted = 200
cycle length = 24 h
Write down the equations:
tex: \text{percent in metaphase} = \frac{\text{cells in metaphase}}{\text{total cells counted}} \times 100
tex: \text{time in metaphase} = \frac{\text{percent in metaphase}}{100} \times \text{cycle length}
Substitute the values into the equations:
tex: \text{percent in metaphase} = \frac{9}{200} \times 100 = 4.5\,\%
tex: \text{time in metaphase} = \frac{4.5}{100} \times 24\,\text{h}
tex: \text{time in metaphase} = 0.045 \times 24\,\text{h} = 1.08\,\text{h}
tex: 1.08\,\text{h} \approx 1.1\,\text{h} \text{ (to one decimal place)}
tex: 1.08\,\text{h} \times 60\,\text{min/h} = 64.8\,\text{min} \approx 65\,\text{min}
Rubric
  • Award 1 point for: 1.1 h (9 of 200 cells is 4.5%, and 4.5% of 24 h is 1.08 h, which is 1.1 h to one decimal place; about 65 minutes).
  • Accept 1.08 h. Do not award the point for 4.5 h (the percent read as hours), 0.045 h (the fraction with no cycle length) or 2.2 h (9 divided by 100 instead of 200).

(d) Support the claim that more of the regrowing tail's cells are dividing than of the mid-body's cells, using evidence from the error bars. (1 pt)

Frame The regrowing tail's bar runs from … to … and the mid-body's from … to …, so …

Hint What does the caption say each bar represents, and what does the overlap rule say about two bars of that kind?

Model answer The caption says the bars represent ±2SE, so the overlap rule applies.
The regrowing tail's bar runs from 19.0% to 25.0% of cells in mitosis.
The mid-body's bar runs from 2.4% to 5.6%.
The bars do not overlap.
So the difference between the two groups is very unlikely to be chance.
More of the regrowing tail's cells were caught in mitosis.
So more of the regrowing tail's cells are dividing.
Rubric
  • Award 1 point for: the evidence (the ±2SE bar for the regrowing tail runs from 19.0% to 25.0% and the bar for the mid-body from 2.4% to 5.6%, and they do not overlap) AND the reasoning (so the higher percent in mitosis in the regrowing tail is very unlikely to be chance, so more of the regrowing tail's cells were caught dividing).
  • Accept a comparison that quotes both ranges, says they are apart, and links that to the claim. Do not award the point for comparing the two means alone (22.0% against 4.0%), or for the ranges with no link.

Slip Comparing 22.0% with 4.0% and stopping, or quoting the bars with no link. Supporting the claim needs the non-overlap and what it shows.

(e) Evaluate the claim that a greater percent of cells in mitosis shows that every regrowing-tail cell cycles faster than a mid-body cell. (1 pt)

Frame The claim is …, because a count of frozen cells shows …; it cannot show …

Hint One slide freezes every cell at one instant. Think about what a count of frozen cells can tell you directly, and what about a single cell's cycle it cannot.

Model answer The claim is not supported.
A count of frozen cells shows how many were caught dividing at one instant.
So a greater percent in mitosis shows that more of the regrowing tail’s cells are dividing, or that mitosis takes longer in the regrowing tail.
The speed of a single cell’s cycle is its cycle length.
A count of frozen cells does not give the cycle length.
So the count cannot show that every regrowing-tail cell cycles faster.
Rubric
  • Award 1 point for: the judgement (the claim is not supported) AND the ground (a count of frozen cells shows how many were caught dividing, so a greater percent shows that more of the regrowing tail's cells are dividing, or that mitosis takes longer there; it cannot show the speed of any one cell's cycle, which would need the cycle length).
  • Accept any correct statement of what one frozen slide cannot show: whether any single cell cycles faster, or the length of the cycle. Do not award the point for the judgement alone, or for 'every regrowing-tail cell divides faster' stated as what the data show.

Slip Judging the claim without the ground, or accepting it. The count says how many of the sampled cells were dividing; it does not show the speed of any one cell.

FRQ 2 P45-frq2 · Conceptual Analysis

The drawing shows two cells from the root of Haplopappus, a small desert daisy whose body cells hold four chromosomes: two long and two short. The cells were fixed and stained at one instant and are labeled cell A and cell B.

Two cells from the root of Haplopappus, a small desert daisy whose body cells hold four chromosomes (two long and two short), fixed and stained at one instant.
Two cells from the root of Haplopappus, a small desert daisy whose body cells hold four chromosomes (two long and two short), fixed and stained at one instant.

(a) Identify the stage of mitosis shown by cell A. Describe the feature of the drawing that shows it. (1 pt)

Model answer Cell A is in anaphase.
The sister chromatids have separated.
Two groups of four single chromatids are moving apart toward opposite poles.
Their centromeres lead the way, pulled by fibers from the centrosomes.
No nuclear envelope surrounds them.
Rubric
  • Award 1 point for: anaphase, shown by two groups of single chromatids (V shapes with the centromere leading) moving apart toward opposite poles, pulled by the spindle fibers, with no envelope around them.
  • Accept "the sister chromatids have separated and are moving to the poles". Do not award the point for metaphase (there is no row at the equator) or for telophase (the fibers are still attached and no envelope is forming).

Slip Naming telophase because the chromosomes are in two groups. In telophase the fibers are gone and an envelope forms around each group. Here the fibers are still pulling and the groups are still on the move.

(b) Each G1 cell of this plant holds 5 pg of DNA. Calculate the amount of DNA in cell A. (1 pt)

Answer: 10 pg  (tolerance ±0.05)

Model answer Cell A holds 10 pg of DNA.
Every chromosome was copied in S phase, so the DNA per cell doubled from 5 pg to 10 pg.
The sister chromatids have separated, but the cell has not yet divided, so all 10 pg are still in cell A.
Working
Write down the values in the question:
DNA per G1 cell = 5 pg
copies of every chromosome after S phase = 2
Write down the equation:
tex: \text{DNA per cell after S phase} = 2 \times \text{DNA per G1 cell}
Substitute the values into the equation:
tex: \text{DNA in cell A} = 2 \times 5\,\text{pg} = 10\,\text{pg}
Rubric
  • Award 1 point for: 10 pg (the G1 amount doubled in S phase, and the cell has not yet divided).
  • Do not award the point for 5 pg (the G1 amount: the DNA has been copied and the cell has not yet split in two) or 20 pg (doubling twice).

(c) Explain why cell B is dividing its cytoplasm with a plate across its middle rather than with a groove pinching in. (1 pt)

Model answer Cell B is a plant cell with a rigid cell wall.
A cell wall cannot be pinched in the way an animal cell’s membrane can.
So vesicles gather at the equator and fuse into a cell plate, the structure across the middle of the drawing.
The cell plate grows outward until it reaches the old cell wall.
Then the cell plate becomes the new cell wall between the two daughter cells.
Rubric
  • Award 1 point for: a plant cell's rigid cell wall cannot be pinched in, so vesicles gather at the equator and fuse into a cell plate that grows outward into a new cell wall between the two daughter cells.
  • Accept "the cell wall cannot be pinched, so a new cell wall is built between the nuclei". Do not award the point for "plant cells have no ring of protein" with no reference to the cell wall, or for describing the cell plate as the spindle.

Slip Saying the cell plate is the spindle left behind. The spindle broke down in telophase. The cell plate is new cell wall, built from vesicles fusing at the equator.

(d) Justify the claim that the two cells formed from cell B carry the same genome as each other. (1 pt)

Model answer Before cell B divided, every one of its four chromosomes was copied once in S phase into two identical sister chromatids.
In anaphase the sister chromatids separated.
The spindle sent one chromatid of every pair to each pole.
So each of the two nuclei in cell B holds one copy of every chromosome.
Therefore the two daughter cells carry the same complete genome, four chromosomes each, identical to each other and to the parent.
Rubric
  • Award 1 point for: every chromosome was copied once in S phase into two identical sister chromatids, and anaphase sent one chromatid of every pair to each pole, so each daughter cell receives one complete copy of the genome: one of each of the four chromosomes, identical to the other daughter's.
  • Accept "each nucleus holds one copy of every chromosome because each was copied and the copies were separated". Do not award the point for "the parent shared its four chromosomes out, two to each side".

Slip Saying the four chromosomes were shared out two and two. The chromosomes were doubled first, in S phase, so each daughter cell receives all four.

APBIO-U04-T45 End-of-topic test: Cell Cycle

Topic 4.5 · Cell Cycle · 19 MCQ · 2 FRQ

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the two free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. Then open the scoring guide and mark your own work against it. Where a graph carries error bars, the caption says what the bars represent. Where a question gives the length of the whole cell cycle, use that length. Give times in hours to one decimal place and in minutes to the nearest minute.

Q1 T45-q01

A planarian is a flatworm about a centimeter long. Cut one in half across the middle and, within two weeks, each half has grown the parts it lost and is a complete worm.

Where do the cells that rebuild each half come from?

  1. A. ✓ Existing cells near the cut divide again and again into daughter cells
  2. B. Cells near the cut swell until they are large enough to fill the missing part
    A cell keeps up with its needs only while it stays small, so growth and repair add cells rather than enlarging them.
    The missing part is rebuilt by cell division.
  3. C. New cells form from the fluid that oozes from the cut surface
    Every new cell comes from an existing cell dividing into two; fluid holds no cells that could build a worm.
    The cells near the cut divide to make new ones.
  4. D. Cells drift in from the water and attach themselves to the cut surface
    The worm's new cells are all descendants of its own cells.
    Nothing floats in from outside.
    Cells near the cut divide, and their daughter cells build the missing part.

Why: Every new cell is made by an existing cell dividing into two daughter cells.
Repair, growth and, in an animal like this, reproduction by splitting all come from the same process.
Cells near the cut divide again and again.
Their daughter cells build the missing half.

Q2 T45-q02

A donkey body cell holds 62 chromosomes in its nucleus.

Which of the following describes one of these chromosomes?

  1. A. A short rod of DNA that exists only while the cell divides
    A chromosome is present through the whole cycle: between divisions it is spread out as chromatin, coiling into a short rod only when the cell is about to divide.
  2. B. One of the proteins the cell's DNA is wound around
    The proteins are the packaging.
    A chromosome is the DNA molecule itself, wound around those proteins.
  3. C. The whole set of DNA the cell hands to each daughter cell
    The whole set of chromosomes is the cell's genome.
    One chromosome is one of the 62 DNA molecules in that set.
  4. D. ✓ One very long DNA molecule packaged with proteins

Why: A chromosome is one very long DNA molecule packaged with proteins.
A donkey cell has 62 chromosomes, and together they make up its genome.
The proteins are packaging, and the rod shape is only the form a chromosome takes when the cell is about to divide.

Q3 T45-q03

A student adds a stain that binds DNA to two cells from a growing bean root. In cell X the nucleus looks evenly grainy. In cell Y there are short, dark rods and no nucleus outline.

Which cell is about to divide, and why does its DNA look as it does?

  1. A. Cell X; its DNA is spread out so it can be moved to the poles
    A grainy nucleus is chromatin, the spread-out form for reading and copying.
    A thread would tangle if moved, so a cell about to divide coils its chromosomes into rods.
  2. B. ✓ Cell Y; its chromosomes have coiled into rods that move without tangling
  3. C. Cell Y; its DNA has been copied, and copied DNA shows as rods
    Copying happens while the DNA is still spread out as chromatin, and it leaves the nucleus looking grainy.
    The rods appear later, when each chromosome coils up for the division.
  4. D. Cell X; a grainy nucleus means its chromosomes are already moving toward the poles
    A grainy nucleus is DNA spread out for reading and copying; nothing is moving in cell X.
    Chromosomes are moved only after they have coiled into rods.

Why: Chromatin, the spread-out form, shows as a grainy nucleus: the form for reading and copying.
A cell about to divide coils each chromosome into a short, thick rod.
A rod can be dragged across the cell without tangling: the form for moving.
Cell Y is the one about to divide.

Q4 T45-q04

A cat body cell has 38 chromosomes. During S phase it copies its DNA. Count chromosomes by their centromeres.

How many chromosomes and how many chromatids does the cell hold at the end of S phase?

  1. A. 19 chromosomes with 38 chromatids
    Copying adds a second chromatid to every chromosome: 38 chromosomes stay, and the chromatids double to 76.
  2. B. 38 chromosomes with 38 chromatids
    38 chromatids is the count before S phase, one chromatid per chromosome.
    Copying gives every chromosome a second, identical sister chromatid: 2 × 38 = 76 chromatids.
  3. C. ✓ 38 chromosomes with 76 chromatids
  4. D. 76 chromosomes with 76 chromatids
    76 chromosomes counts each chromatid as a chromosome.
    A joined pair of sister chromatids has one centromere and counts as one chromosome, so there are still 38.

Why: Copying doubles the chromatids, not the chromosomes: each of the 38 chromosomes becomes two identical sister chromatids joined at one centromere, and a joined pair still counts as one chromosome. chromatids=2×chromosomes=2×38=76, while the chromosome count stays at 38.

Q5 T45-q05

The model below shows the cell cycle of a cultured human cell as a wheel, one turn per cycle, with its five stages numbered 1 to 5 in order from a daughter cell's first stage.

The cell cycle of a cultured human cell as a wheel, one turn per cycle, with its five stages numbered 1 to 5 in order from a daughter cell's first stage. Stages 4 and 5 are drawn wider than their true share of the turn so that their numbers fit.
The cell cycle of a cultured human cell as a wheel, one turn per cycle, with its five stages numbered 1 to 5 in order from a daughter cell's first stage. Stages 4 and 5 are drawn wider than their true share of the turn so that their numbers fit.

In which numbered stage is the cell's DNA copied?

  1. A. Stage 1
    Stage 1 is G1, the first growth gap: the cell grows and makes more organelles, but its DNA is still uncopied.
  2. B. ✓ Stage 2
  3. C. Stage 3
    Stage 3 is G2, the second growth gap: the DNA has already been copied, and the cell is making the proteins, ATP and second centrosome it needs for the division.
  4. D. Stage 4
    Stage 4 is mitosis: the nucleus divides.
    The DNA was copied earlier, in the long stage before G2.

Why: From a daughter cell’s first stage the order is G1, S phase, G2, mitosis, cytokinesis.
So stage 2 is S phase, in which every chromosome is copied into two sister chromatids.
Stages 1 and 3 are the growth gaps G1 and G2; stages 4 and 5 are the division.

Q6 T45-q06

Cells of one type were measured every hour as they grew in a dish. Between hour 2 and hour 5 the mean mass of a cell rose from 1.6 to 2.2 ng while the DNA in each cell stayed at 3 pg, the amount a newly formed daughter cell holds.

Which stage were the cells in during these hours, and how can you tell?

  1. A. S phase; the mass rose, which shows the DNA was being copied
    Mass and DNA are separate measurements.
    In S phase the DNA content climbs above 3 pg; here it stayed at the daughter-cell amount, so copying had yet to begin.
  2. B. G2; a steady DNA content shows that the copying was finished
    In G2 the DNA content is steady at double the daughter-cell amount, 6 pg.
    A steady 3 pg is the uncopied amount, so these cells were before S phase.
  3. C. Mitosis; the mass rose as each cell got ready to split in two
    A cell in mitosis has copied its DNA, so it would hold 6 pg, and it divides its nucleus rather than growing.
    These cells were growing with uncopied DNA.
  4. D. ✓ G1; the cells were growing while their DNA was still uncopied

Why: The mass of each cell rose, so the cells were growing.
The DNA in each cell stayed at 3 pg, the daughter-cell amount, so the DNA had not been copied.
In G1 the cell grows while its DNA is still uncopied.
So the cells were in G1.

Q7 T45-q07

The graph shows the DNA per cell of a trout cell line through one 22-hour cycle. A new daughter cell holds 4 pg.

DNA per cell through one 22-hour cycle of a trout cell line; a new daughter cell holds 4 pg. Gridlines every 1 hour and every 1 pg.
DNA per cell through one 22-hour cycle of a trout cell line; a new daughter cell holds 4 pg. Gridlines every 1 hour and every 1 pg.

During which hours were the cells copying their DNA?

  1. A. Hours 0 to 7
    From hour 0 to hour 7 the line is flat at 4 pg, the uncopied amount: the cells were in G1, growing with their DNA still to be copied.
  2. B. ✓ Hours 7 to 15
  3. C. Hours 15 to 21
    From hour 15 to hour 21 the line is flat at 8 pg, double the daughter-cell amount: the copying was already finished and the cells were in G2.
  4. D. Hours 21 to 22
    At hour 22 the DNA per cell drops from 8 pg back to 4 pg: that is the division, when the doubled set is split between two daughter cells.

Why: DNA content is steady through G1, doubles during S phase as every chromosome is copied, and stays doubled through G2.
The line climbs from 4 pg to 8 pg between hour 7 and hour 15.
So that is when the cells were in S phase, copying their DNA.

Q8 T45-q08

In G2 a cell finishes copying one structure it will need for the division that follows.

Which structure is copied, and what will it do?

  1. A. ✓ The centrosome, from which the fibers that move the chromosomes will grow
  2. B. The nucleus, which will hold one set of chromosomes for each daughter cell
    The nucleus is not copied in G2; it divides during mitosis, after the chromosomes are pulled apart.
    What the cell finishes copying in G2 is what the spindle grows from.
  3. C. The centromere, which will hold the two sister chromatids together
    The centromere is the join between two sister chromatids, made in S phase.
    The similar-sounding structure copied in G2 is the centrosome, which the spindle fibers grow from.
  4. D. The cell wall, which will grow between the two daughter cells
    A new cell wall forms only in a plant cell, at cytokinesis, from a cell plate.
    G2 prepares the division: the structure copied is the one the spindle grows from.

Why: In G2 the cell makes the proteins for division, produces ATP in large quantities and finishes copying its centrosome.
The centrosome is the structure the chromosome-moving fibers grow from, so the cell enters mitosis with two.
The centromere, a different structure, is the join between two sister chromatids.

Q9 T45-q09

Cells lining the gut divide every few days. The muscle cells of the heart work for a whole lifetime; each one's last division was before birth.

Which statement describes the heart muscle cells?

  1. A. Heart muscle cells are held in G2, ready to divide as soon as the heart is damaged
    A cell held in G2 has copied its DNA and is preparing to divide.
    Heart muscle cells do ordinary work with no preparation to divide: they have left the cycle.
  2. B. Heart muscle cells stay in G1 and grow larger every year instead of dividing
    A cell in G1 is growing toward S phase and a division; a cell that never prepares to divide has left the cycle from G1 into G0.
  3. C. Heart muscle cells have died, which is why they never divide
    G0 is not death.
    A heart muscle cell in G0 is alive and busy, contracting many times a minute; it has left the cycle, which differs from dying.
  4. D. ✓ Heart muscle cells have left the cycle from G1 into G0 and work on there

Why: A cell can leave the cycle from G1 into G0 and carry on its work without preparing to divide.
Heart muscle cells and mature nerve cells stay there for life.
Gut-lining cells keep cycling.
G0 is neither death nor a stall: the cell is working.

Q10 T45-q10

Late in prophase the nuclear envelope breaks down.

Why does this breakdown matter for what happens next?

  1. A. ✓ The spindle fibers can now reach every chromosome and attach at its centromere
  2. B. The chromosomes can now uncoil back into chromatin so that their genes can be read
    Uncoiling comes at the other end of mitosis, in telophase.
    In prophase the chromosomes are coiling up, and the envelope goes so the fibers can reach them.
  3. C. The DNA can now be copied, because the copying enzymes can reach it
    The DNA was copied in S phase, before prophase, while the envelope was intact.
    The envelope breaks down so that the spindle can reach the copied chromosomes.
  4. D. The sister chromatids can now separate, because the envelope no longer holds them together
    The sister chromatids are held together at their centromere, not by the envelope, and that connection breaks at anaphase.
    The envelope goes so the fibers can reach the chromosomes.

Why: The spindle’s job is to move the chromosomes.
While the nuclear envelope surrounds the chromosomes, the fibers cannot get to them.
Once the envelope has broken down, fibers from both poles reach every chromosome and attach at its centromere.
That attachment pulls the chromosomes into a row at metaphase.

Q11 T45-q11

Two dividing cells are seen on a slide of a growing wheat root. In cell P, rods lie scattered through the cell. In cell Q, every rod stands in one line across the middle of the cell with fibers from both ends attached to it.

Which cell is in metaphase, and what marks the stage?

  1. A. Cell P; the spindle fibers have begun to reach some of the scattered chromosomes
    Fibers reaching some scattered rods is prophase still under way: the spindle is being built.
    Metaphase is reached only when every chromosome has been pulled into a single row.
  2. B. Cell P; the chromosomes have coiled into rods and the envelope has gone
    Rods and a vanished envelope are the marks of prophase, the stage before metaphase.
    What defines metaphase is the single row of chromosomes at the equator.
  3. C. ✓ Cell Q; fibers from both poles hold every chromosome in one row at the equator
  4. D. Cell Q; the sister chromatids have separated and started to move apart
    In cell Q each rod is still two joined sister chromatids, held at the equator by fibers from both poles.
    Separating the sister chromatids is anaphase’s first event.

Why: Metaphase has one defining event: fibers from both poles have attached to every chromosome at its centromere and pulled the chromosomes into a single row along the equator.
Each chromosome is still two joined sister chromatids.
Cell Q shows exactly that.
Cell P has yet to get there.

Q12 T45-q12

The drawing below shows one of the structures seen in a cell at prophase, with two parts marked J and K.

One of the structures seen in a cell at prophase, with two parts marked J and K.
One of the structures seen in a cell at prophase, with two parts marked J and K.

Which of the following identifies J and K?

  1. A. J is one of the two strands of a DNA double helix; K is a gene
    A chromatid is a whole DNA molecule wound on proteins, not one strand of the helix.
    K is the pinch where the two chromatids are held: the centromere.
  2. B. J is one chromosome; K is the point where it touches a second chromosome
    The X is one chromosome after copying: two identical sister chromatids.
    K is the centromere holding them together, not a contact between two chromosomes.
  3. C. J is the centromere; K is one of the two chromatids joined there
    The centromere is the one point at the pinch, K.
    Each arm of the X, like J, is a sister chromatid.
  4. D. ✓ J is one of the two sister chromatids; K is the centromere that holds the pair together

Why: After copying, a chromosome is two identical sister chromatids lying side by side.
Each diagonal of the X is one sister chromatid: J.
The one point that holds the pair together is the centromere: K.

Q13 T45-q13

At the end of mitosis, in telophase, the rods uncoil back into chromatin.

Why does the DNA return to the spread-out form at this point?

  1. A. ✓ Spread out, the DNA is open to be read, so each new nucleus can start its work
  2. B. Spread out, the DNA is easier for the spindle fibers to drag toward the two poles
    The moving is over: the chromosomes have reached the poles and the spindle has broken down.
    It was the coiled rod form that let them be moved without tangling.
  3. C. Spread out, the DNA takes up less room inside the new nuclear envelope
    Spread-out chromatin fills the nucleus.
    The coiled rod is the compact form.
    The DNA uncoils for a different reason: loose is the form for working, for reading and copying.
  4. D. Spread out, the sister chromatids can separate from each other
    The sister chromatids separated at the start of anaphase, before the chromosomes moved.
    By telophase each set is single chromatids, and the uncoiling is about reading the DNA again.

Why: The chromosomes were coiled into rods to be moved without tangling.
Once the chromosomes have arrived, each set uncoils back into chromatin, the spread-out form that leaves the DNA open to be read.
So the two new nuclei can get on with the cell’s work.

Q14 T45-q14

At the end of mitosis an animal cell pinches in around its middle until it splits. A plant cell at the same point does something different.

Why does a plant cell divide its cytoplasm by building a plate across its middle rather than pinching in?

  1. A. Its two nuclei sit too far apart for a ring of protein to pinch the cell between them
    The distance between the nuclei is similar in animal and plant cells.
    A plant cell cannot pinch in because of its cell wall, so it builds a new one.
  2. B. ✓ Its rigid cell wall cannot be pinched, so it builds a new cell wall between the nuclei
  3. C. Its spindle stays in place after mitosis and hardens into the plate
    The spindle breaks down in telophase, before the cytoplasm divides.
    The cell plate is built from vesicles that fuse at the equator and grows into a new cell wall.
  4. D. Its nuclear envelopes fuse across the middle and seal the two halves apart
    Each nuclear envelope surrounds one set of chromosomes at a pole; envelopes do not build cell walls.
    Vesicles fuse at the equator to form the cell plate.

Why: In an animal cell a protein ring pinches the cell in at a cleavage furrow.
A plant cell’s rigid cell wall cannot be pinched.
So instead vesicles fuse at the equator into a cell plate, which grows out to the old cell wall and becomes the new cell wall.

Q15 T45-q15

The drawing below shows one cell from a fish embryo, fixed and stained as it was dividing.

One cell from a fish embryo, fixed and stained; the cell is drawn with four chromosomes.
One cell from a fish embryo, fixed and stained; the cell is drawn with four chromosomes.

Which stage of mitosis does the drawing show?

  1. A. Metaphase
    Metaphase is one row of chromosomes across the equator with fibers from both poles.
    Here the chromosomes sit in two clusters at the ends, with no row and no fibers.
  2. B. Anaphase
    In anaphase two groups are on the move, pulled by fibers, with no envelope around them.
    Here the fibers are gone and an envelope is forming around each cluster.
  3. C. ✓ Telophase
  4. D. Cytokinesis
    Cytokinesis divides the cytoplasm: a cleavage furrow pinches the cell in.
    This cell has no cleavage furrow yet.
    It holds two forming nuclei in one undivided cytoplasm.

Why: Two clusters of chromosomes at opposite poles, each inside a forming nuclear envelope, with the spindle gone and the rods beginning to fade: those are the events of telophase.
The cleavage furrow of cytokinesis has yet to appear.

Q16 T45-q16

The four drawings below were made of one cell at prophase. The cell has two chromosomes, one long and one short.

Four drawings of the same cell, which has two chromosomes (one long and one short), each meant to show prophase.
Four drawings of the same cell, which has two chromosomes (one long and one short), each meant to show prophase.

Which drawing shows prophase correctly?

  1. A. Drawing 1
    Drawing 1 shows grainy chromatin inside an unbroken envelope: interphase, before the chromosomes have coiled into rods.
    In prophase the rods are visible and the envelope is breaking up.
  2. B. Drawing 2
    Drawing 2 has both chromosomes in one row at the equator with fibers from both poles: metaphase.
    In prophase the rods are scattered and the spindle just starting.
  3. C. Drawing 3
    Drawing 3 has four single rods.
    In prophase each chromosome is still two joined sister chromatids, drawn as an X, so a two-chromosome cell shows two X shapes.
  4. D. ✓ Drawing 4

Why: At prophase the copied chromosomes condense into rods, each still two sister chromatids joined at a centromere.
So a two-chromosome cell shows two X shapes, one long and one short, scattered.
The envelope is breaking up, and short fibers grow from the two poles.
Drawing 4 has all that.

Q17 T45-q17

A student counts 400 cells on a slide from the growing root tip of a leek seedling and finds 60 of them in one of the four stages of mitosis.

What is the mitotic index of the slide?

  1. A. 0.15%
    0.15 is the fraction of the cells in mitosis, 60 divided by 400.
    A percent is that fraction multiplied by 100: 15%.
  2. B. ✓ 15%
  3. C. 18%
    18% comes from dividing 60 by the 340 interphase cells instead of by all 400.
    The percent in a stage is its count divided by the total counted, times 100.
  4. D. 60%
    60 is the count, not the percent.
    The slide held 400 cells, not 100. Divide 60 by 400 and multiply by 100: 15%.

Why: The percent of cells in a stage is the count in that stage divided by the total counted, times 100, and the percent in mitosis is the slide's mitotic index: 60400×100=0.15×100=15%.

Q18 T45-q18

On a slide from a growing root tip, 12% of the cells are in one of the four stages of mitosis. Take the whole cell cycle in this root as 24 hours.

How long does mitosis take in these cells? Give the time in hours to one decimal place.

  1. A. 0.12 h
    0.12 is the fraction of the cycle that mitosis takes.
    The time is that fraction of the whole 24-hour cycle: 0.12 × 24 h = 2.88 h, about 2.9 hours.
  2. B. 2.0 h
    2.0 h comes from dividing the 24-hour cycle by 12.
    Mitosis takes 12% of the cycle, so multiply 24 h by the fraction 0.12: 2.88 h.
  3. C. ✓ 2.9 h
  4. D. 12 h
    12 h reads the percent as a number of hours.
    The 12% is a share of the cycle, and 12% of 24 h is 2.88 h, about 2.9 hours.

Why: At any instant the fraction of cells in a stage equals the fraction of the cycle that stage takes, so the time in a stage is the percent as a fraction times the cycle length: 12100×24h=2.88h, about 2.9 hours.

Q19 T45-q19

A student grew cells from a chick embryo for six hours in a chemical that stops the nuclear envelope from breaking down. She sorted 100 mitotic cells from the treated cells and 100 mitotic cells from untreated cells by stage, as shown in the table.

The stage of 100 mitotic cells from untreated chick-embryo cells and of 100 mitotic cells from cells grown for six hours in the chemical.
The stage of 100 mitotic cells from untreated chick-embryo cells and of 100 mitotic cells from cells grown for six hours in the chemical.

What does the chemical do to the dividing cells?

  1. A. ✓ It holds the cells in prophase
  2. B. It speeds the cells through anaphase
    Anaphase’s share fell from 10 to 6 of the 100 only because prophase’s share grew.
    The stage that filled up, prophase, is where the cells are held.
  3. C. It stops the DNA from being copied
    Every cell counted is in mitosis, so all of them copied their DNA.
    The change is which stage they are caught in: prophase rose from 45% to 82%.
  4. D. It holds the cells at the equator in metaphase
    Metaphase fell from 25% to 8%.
    The pile-up is one stage earlier, in prophase, which rose from 45% to 82%.
    Cells enter prophase and cannot leave it.

Why: A pile-up in one stage means the cells are held there.
Prophase rose from 45 to 82 of 100 mitotic cells while every later stage fell: the cells reach prophase and stop.
That fits: with the envelope intact, the fibers cannot reach the chromosomes.

FRQ 1 T45-frq1 · Scientific Investigation

A student tests the claim that root cells stop dividing as they get further from the tip of the root. She grows five garlic bulbs until each has roots about 30 mm long. From one root of each bulb she cuts a thin slice 1 mm from the tip and a second slice 10 mm from the tip, stains each slice and, under the microscope, records the stage of 320 cells in it. The table gives her counts for one slice cut 1 mm from the tip. The graph gives the mean percent of cells in mitosis for the five slices at each distance, with error bars that represent ±2SE. Take the whole cell cycle in this root as 24 hours.

Top: the stage of each of the 320 cells counted in one slice cut 1 mm from a garlic root tip. Bottom: the mean percent of cells in mitosis in the five slices at each distance; error bars represent ±2SE. Gridlines every 2%.
Top: the stage of each of the 320 cells counted in one slice cut 1 mm from a garlic root tip. Bottom: the mean percent of cells in mitosis in the five slices at each distance; error bars represent ±2SE. Gridlines every 2%.

(a) Identify the independent variable and the dependent variable in this investigation. (1 pt)

Model answer The independent variable is the distance of the slice from the root tip, 1 mm or 10 mm.
The dependent variable is the percent of the 320 cells that were in mitosis (the stage each cell was recorded in).
Rubric
  • Award 1 point for both: the independent variable is the distance of the slice from the root tip (1 mm or 10 mm) and the dependent variable is the percent of cells in mitosis (or the stage of each cell counted).
  • Accept "how far from the tip the cells were" for the independent variable and "the number of cells in mitosis out of 320" for the dependent variable. Do not award the point if the two are reversed, or if a condition kept the same (320 cells per slice, the stain, the 30 mm roots) is named as the independent variable.

Slip Naming the stain or the 320 cells per slice as a variable. Those were kept the same for every slice; the student changed the distance from the tip and measured the share of cells in mitosis.

(b) Calculate the time, in hours to one decimal place, that a cell 1 mm from the tip spends in prophase, taking the whole cycle as 24 hours. (1 pt)

Answer: 2.4 h  (tolerance ±0.005)

Model answer A cell 1 mm from the tip spends 2.4 hours in prophase, 144 minutes.
Working
Write down the values in the question:
cells in prophase = 32
total cells counted = 320
cycle length = 24 h
Write down the equations:
tex: \text{percent in prophase} = \frac{\text{cells in prophase}}{\text{total cells counted}} \times 100
tex: \text{time in prophase} = \frac{\text{percent in prophase}}{100} \times \text{cycle length}
Substitute the values into the equations:
tex: \text{percent in prophase} = \frac{32}{320} \times 100 = 10\,\%
tex: \text{time in prophase} = \frac{10}{100} \times 24\,\text{h}
tex: \text{time in prophase} = 0.10 \times 24\,\text{h} = 2.4\,\text{h}
tex: 2.4\,\text{h} \times 60\,\text{min/h} = 144\,\text{min}
Rubric
  • Award 1 point for: 2.4 h (32 of 320 cells is 10%, and 10% of 24 h is 2.4 h; 144 minutes).
  • Accept 2.40 h. Do not award the point for 10 h (the percent read as hours), 0.10 h (the fraction with no cycle length) or 4.8 h (the count divided by 160, the wrong total).

(c) Support the claim that a smaller share of the cells is dividing 10 mm from the tip than 1 mm from the tip, using evidence from the error bars. (1 pt)

Model answer The caption says the bars represent ±2SE, so the overlap rule applies.
The 1 mm bar runs from 13.6% to 18.4% of cells in mitosis.
The 10 mm bar runs from 0.4% to 2.0%.
The bars do not overlap.
So the difference between the two groups is very unlikely to be chance.
Far fewer of the cells 10 mm from the tip were caught in mitosis.
So fewer of those cells are dividing.
Rubric
  • Award 1 point for: the evidence (the ±2SE bar for 1 mm runs from 13.6% to 18.4% and the bar for 10 mm from 0.4% to 2.0%, and they do not overlap) AND the reasoning (so the lower percent in mitosis at 10 mm is very unlikely to be chance, so far fewer of those cells were caught dividing, so fewer of them are dividing).
  • Accept a comparison that quotes both ranges, says they are apart, and links that to the claim. Do not award the point for comparing the two means alone (16.0% against 1.2%), for the ranges with no link, or for a decision made from the table, which shows one slice only.

Slip Quoting 16.0% and 1.2% and stopping. Supporting the claim needs the bars and the link: no overlap, so the difference is not chance, so fewer cells are dividing.

(d) Explain, in terms of the cell cycle, why so few of the cells 10 mm from the tip are found in any stage of mitosis. (1 pt)

Model answer Near the tip the cells are cycling, so at any instant some are caught in each stage of mitosis.
Ten millimeters up the root the cells have left the cycle.
They stepped out of G1 into G0.
In G0 they carry on their work as root cells without preparing to divide.
A cell in G0 never enters mitosis, so almost none of these cells are found in any of its stages.
Rubric
  • Award 1 point for: the cells 10 mm from the tip have left the cycle, from G1 into G0, where they carry on their work as root cells without preparing to divide, so almost none are ever caught in a stage of mitosis.
  • Accept "they have stopped cycling and sit in G0". Accept "these cells have matured into working root cells and no longer divide" without the label G0: the biology is complete without the name. Do not award the point for "they divide faster, so they are caught less often" (a shorter mitosis would lower the count only a little, and these cells show almost none) or for "they are dead" (G0 cells are alive and working).
  • Do not award the point for naming G0 alone: the answer must say that these cells have left the cycle and are working, not preparing to divide.

Slip Saying the cells 10 mm from the tip are dividing quickly and so are rarely caught. A quick mitosis would still leave some cells in each stage. Almost none here means the cells are out of the cycle, in G0.

FRQ 2 T45-frq2 · Conceptual Analysis

The drawing shows three cells from the same insect embryo, fixed and stained at one instant and labeled cell 1, cell 2 and cell 3. Each body cell of this insect holds ten chromosomes: five long and five short. Every cell in the embryo is cycling.

Three cells from the same insect embryo, fixed and stained at one instant. Each body cell of this insect holds ten chromosomes: five long and five short.
Three cells from the same insect embryo, fixed and stained at one instant. Each body cell of this insect holds ten chromosomes: five long and five short.

(a) Identify the stage of mitosis shown by cell 3. Describe the feature of the drawing that shows it. (1 pt)

Model answer Cell 3 is in prophase.
Its chromosomes have coiled into visible rods.
Each rod is still two sister chromatids joined at a centromere.
The rods lie scattered rather than in a row.
The nuclear envelope around them is breaking into pieces.
Short spindle fibers are growing in from the centrosome at each pole.
Rubric
  • Award 1 point for: prophase, shown by the chromosomes condensed into rods (each still two joined sister chromatids, drawn as an X) lying scattered inside a nuclear envelope that is breaking up, with short spindle fibers starting from the two poles.
  • Accept any one of the three features (scattered rods with no row, the breaking envelope, the spindle starting to grow) with the stage named. Do not award the point for metaphase (the chromosomes are scattered, not in a row) or for interphase (the chromosomes are visible as rods).

Slip Naming metaphase because rods with fibers are visible. In metaphase every chromosome stands in one row at the equator. Scattered rods with the envelope still breaking up are prophase.

(b) Calculate the number of chromatids in cell 2. (1 pt)

Answer: 20  (tolerance ±0)

Model answer Cell 2 holds 20 chromatids.
Each of its ten chromosomes was copied in S phase into two sister chromatids.
At metaphase the sister chromatids are still joined.
Working
Write down the values in the question:
chromosomes in each body cell = 10
chromatids in each copied chromosome = 2
Write down the equation:
tex: \text{chromatids} = 2 \times \text{chromosomes}
Substitute the values into the equation:
tex: \text{chromatids} = 2 \times 10
tex: \text{chromatids} = 20
Rubric
  • Award 1 point for: 20 chromatids (ten chromosomes, each two sister chromatids at metaphase).
  • Do not award the point for 10 (the chromosome count) or 40 (doubling twice).

(c) Describe the two events, in order, that carry a cell's chromosomes from the arrangement in cell 2 to the arrangement in cell 1. (1 pt)

Model answer First, the connection at each centromere breaks, so the two sister chromatids of every chromosome separate; each one is now counted as a chromosome.
Second, the spindle fibers pull the separated chromatids toward opposite poles, so ten move toward each pole with their centromeres leading.
That is anaphase.
Cell 1 is caught in its second event.
Rubric
  • Award 1 point for both events in order: first the connection at each centromere breaks so the sister chromatids separate (each now counted as a chromosome), and then the spindle fibers pull the separated chromatids toward opposite poles.
  • Accept "the sister chromatids split, then the fibers pull one of each pair to each pole". Do not award the point if the pulling is put before the separation, or if only one of the two events is given.

Slip Writing that the fibers pull the sister chromatids apart. The pulling comes second: while the sister chromatids are joined, fibers from both poles hold each pair at the equator and nothing moves.

(d) Evaluate the claim that the two daughter cells formed from cell 1 will each hold ten chromosomes carrying the same genome as the parent cell. (1 pt)

Model answer The claim is correct.
Every chromosome was copied once in S phase into two identical sister chromatids.
So cell 1 held twenty chromatids, two copies of each of the ten chromosomes.
In anaphase the sister chromatids separated.
The spindle sent one chromatid of every pair to each pole.
So each pole received one copy of every chromosome.
Therefore each daughter cell holds ten chromosomes carrying the same genome as the parent.
Rubric
  • Award 1 point for: the judgement (the claim is correct) AND the ground (every chromosome was copied once in S phase into two identical sister chromatids, and anaphase sent one chromatid of every pair to each pole, so each daughter receives one complete copy of the genome: one of each of the ten chromosomes, identical to the parent's).
  • Accept 'correct, because each chromosome was copied and then the copies were separated, one to each daughter'. Do not award the point for the judgement alone, or for 'the parent shares its ten chromosomes out, five to each daughter'.

Slip Judging the claim without the ground, or saying the parent divides its ten chromosomes into two sets of five. The chromosomes were doubled first, in S phase.

APBIO-U04-L18 Division has to be switched on

Topic 4.6 · Regulation of Cell Cycle · 52 steps

Two photographs of the same thumb: on the left the cut is open, captioned wound open, 24 dividing cells in the sample; on the right the same cut has closed, captioned wound closed, 3 dividing cells in the sample
Two photographs of the same thumb: on the left the cut is open, captioned wound open, 24 dividing cells in the sample; on the right the same cut has closed, captioned wound closed, 3 dividing cells in the sample

Photos: Sintegrity, Wikimedia Commons, CC BY-SA 4.0 (both cropped and resized).

Here is the edge of a healing cut, with the count of dividing cells in one sample: 24 while the wound is open, 3 after it has closed.

The cells did not use up their nutrients. The same food reached them both times. Something switched division on, and something switched it off. What switches a cell’s division on, and what switches it off?

Unit 4 · Cell Communication and Cell Cycle

1What a cell needs before it divides

2

Video: Watch: What a cell needs before it divides

A cell enters the cycle only when five conditions are met: inside it, its size and its complete undamaged DNA; around it, nutrients, room to grow and a growth factor.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18a.mp4

3
Check q1

A growth factor reaches a receptor on a cell.

What does the growth factor tell the cell to do?

  1. A. ✓ Divide
  2. B. Stop dividing
    A growth factor is a signal from other cells.
    Its message is grow and divide.
  3. C. Leave the cycle
    A cell that leaves the cycle rests in G0.
    A growth factor tells a cell the opposite: grow and divide.

Why: A growth factor is a signal from other cells.
Its message is grow and divide.
So the cell divides.

4

Does a cell divide unless something stops it? It does not.

5

A cell enters the cycle only when the conditions inside it and around it are met.

6

Two conditions are inside the cell:

  • the cell is big enough;
  • its DNA is complete and undamaged.

7

Three conditions are around the cell:

  • nutrients;
  • room to grow;
  • a growth factor, a signal from other cells that tells the cell to divide.

8

Here is a table comparing the five conditions: where each one is, and what the cell detects.

A table of the five conditions a cell needs before it enters the cycle, with where each condition is and what the cell detects: the cell is big enough, inside, its own size; its DNA is complete and undamaged, inside, no break in its DNA; nutrients, around the cell, nutrients reaching it; room to grow, around the cell, no neighbor touching it on every side; a growth factor, around the cell, a growth factor bound to a receptor
9

Molecules reaching receptors start division. A tissue that is short of cells does not start it; only a signal does.

10

What you are expected to know State the five conditions a cell needs before it enters the cycle: inside it, its size and its complete undamaged DNA; around it, nutrients, room to grow and a growth factor.

11
Check q2

A student looks at the dividing cells along the edge of a cut. The student says: “These cells divide because the skin needs new cells there.”

Is the student correct?

  1. A. Yes — the skin’s need for new cells makes the cells at the edge divide
    A cell cannot detect what a tissue needs; it detects molecules.
    Signal molecules from the damaged tissue bind receptors on the edge cells and tell them to enter the cycle.
  2. B. ✓ No — signal molecules from the damaged tissue reach the cells’ receptors

Why: A cell cannot detect what a tissue needs; it detects molecules.
Signal molecules from the damaged tissue reach receptors on the edge cells.
The bound receptors tell the cells to enter the cycle.
So the cells divide because a signal reached them, not because of a need.

12The growth-factor experiment

13

Video: Watch: The growth-factor experiment

Two dishes of liver cells with the same nutrients; a growth factor added to one. Only that dish's cells divide often, so the signal, not the food, switched division on.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18b.mp4

14

Now consider liver cells in two dishes. A researcher gives both dishes the same nutrients.

15

The researcher adds a growth factor to one dish only.

16

Here is a graph of the divisions counted in one day: 27 per 100 cells in the dish with the growth factor, and 9 per 100 cells in the dish without it.

Two bars: divisions per 100 cells in one day for liver cells in a dish, 27 with a growth factor added and 9 without, with ±2SE error bars of 3 and 2; gridlines every 5; the legend reads: error bars represent ±2SE
Two bars: divisions per 100 cells in one day for liver cells in a dish, 27 with a growth factor added and 9 without, with ±2SE error bars of 3 and 2; gridlines every 5; the legend reads: error bars represent ±2SE
17

The dish without the growth factor is the control. It shows how often the cells divide with the same food and no signal.

18

The food was the same in both dishes. So the food did not cause the extra divisions.

19

The growth factor was the only difference. So the growth factor switched division on.

20

What you are expected to know Explain how the two dishes show that the growth factor, not the food, switched division on.

21
Check q3

A researcher grows liver cells in two dishes with the same nutrients and adds a growth factor to one dish only.

Which dish is the control?

  1. A. The dish given the nutrients and the growth factor
    The control leaves out the tested factor.
    The tested factor is the growth factor, so the control is the dish given only the nutrients.
  2. B. ✓ The dish given only the nutrients

Why: The control leaves out the one factor being tested.
The tested factor is the growth factor.
So the dish given only the nutrients is the control.

22
Check q4

Kidney cells in two dishes receive the same glucose, amino acids and oxygen. A researcher adds a growth factor to one dish only. Over two days the cells with the growth factor complete 58 divisions per 200 cells; the others complete 19. The cells that divided took about the same time to complete a cycle in both dishes.

Which of the following did the growth factor change?

  1. A. How fast each cell moved through mitosis
    The cells that divided took about the same time in both dishes.
    The growth factor acts before the cycle starts: the bound receptor tells the cell to enter the cycle.
  2. B. How much building material each cell had
    Both dishes had the same glucose, amino acids and oxygen.
    The growth factor added a signal, not material.
    The signal told more cells to enter the cycle.
  3. C. ✓ How many cells entered the cycle

Why: The nutrients were the same in both dishes, so the extra divisions came from the growth factor.
The growth factor binds its receptor on the cell.
The bound receptor tells the cell to enter the cycle.
So more cells entered the cycle, and more divisions were counted.

23
Practice writing an answer

Kidney cells in two dishes receive the same glucose, amino acids and oxygen. A researcher adds a growth factor to one dish only. Over two days the cells with the growth factor complete 58 divisions per 200 cells; the others complete 19. The cells that divided took about the same time to complete a cycle in both dishes.

(a) Explain why the growth factor raised the count of divisions although the nutrients were the same in both dishes. (1 pt)

Model answer The nutrients were the same in both dishes.
So the nutrients did not cause the extra divisions.
The growth factor is a signal.
The growth factor binds its receptor on a kidney cell.
The bound receptor tells the cell to enter the cycle.
So more cells entered the cycle in the dish with the growth factor.
Each cell that enters the cycle goes on to divide.
So more divisions were counted in that dish.
Rubric
  • Award 1 point for: the growth factor is a signal that, bound to its receptor, tells a cell to enter the cycle, so more cells entered the cycle and divided; the nutrients, the same in both dishes, supplied material but no signal.

24Room to grow

25

Video: Watch: Room to grow

Skin cells packed edge to edge stop dividing and start again when thinned out: a cell touched on every side receives a signal to stop, a molecule reaching a receptor.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18c.mp4

26

Room matters too. Now consider skin cells in a dish.

27

Cells spread thinly in the dish divide. The same cells packed edge to edge stop dividing.

28

When the researcher thins the packed cells out, they start dividing again.

29

Here is a graph of the cells that divided in one day: 42% of the thinly seeded cells, 7% of the packed cells, and 40% of the cells thinned out again.

Three bars: percent of skin cells that divided in one day, 42 in a thinly seeded dish, 7 in a packed dish, and 40 for cells moved from the packed dish to a thin one, with ±2SE error bars of 4, 2 and 4; gridlines every 10; the legend reads: error bars represent ±2SE
Three bars: percent of skin cells that divided in one day, 42 in a thinly seeded dish, 7 in a packed dish, and 40 for cells moved from the packed dish to a thin one, with ±2SE error bars of 4, 2 and 4; gridlines every 10; the legend reads: error bars represent ±2SE
30

A cell touched on every side by neighbors receives a signal to stop.

31

Contact, like a growth factor, is a molecule on one cell reaching a receptor on another.

32

So cells divide while the signals are present. They stop when the signals fade.

33

What you are expected to know Explain why packed cells stop dividing and thinned-out cells start again: a cell touched on every side receives a signal to stop.

34
Check q5

A researcher grows skin cells in three dishes of the same nutrient-rich medium, which the researcher replaces often, and counts the cells that divide in one day. The graph below shows the counts for a thinly seeded dish, a packed dish, and a packed dish the researcher has thinned out.

Three bars: percent of skin cells that divided in one day, 45 in a thinly seeded dish, 6 in a packed dish, and 43 for cells moved from the packed dish to a thin one, with ±2SE error bars of 4, 2 and 4; gridlines every 10; the legend reads: error bars represent ±2SE
Three bars: percent of skin cells that divided in one day, 45 in a thinly seeded dish, 6 in a packed dish, and 43 for cells moved from the packed dish to a thin one, with ±2SE error bars of 4, 2 and 4; gridlines every 10; the legend reads: error bars represent ±2SE

Which of the following held the packed cells’ cycle?

  1. A. ✓ Contact with neighbors on every side
  2. B. A shortage of nutrients
    The researcher replaced the medium often in every dish.
    So the nutrients stayed the same in every dish.
    Only the crowding differed.
  3. C. The age of the cells
    The same cells divided again at 43% as soon as the researcher thinned them out.
    So nothing about the cells had worn out.
  4. D. An inner timer in each cell
    The same cells divided or held depending only on how packed the dish was.
    So the signal came from outside the cell.

Why: The nutrients were the same in every dish.
The packed cells divided again as soon as they were thinned out.
So they had not worn out, and they were not short of food.
Only the crowding changed.
So contact with neighbors on every side held the cycle.

35
Check q6

A researcher keeps skin cells thinly spread in a dish, with nutrients but no growth factor.

Which of the following happens?

  1. A. Most of the cells enter the cycle
    No cell is touched on every side, so nothing signals stop.
    But no growth factor signals start.
    Division has to be switched on.
  2. B. ✓ Few of the cells enter the cycle

Why: No cell is touched on every side, so no stop signal reaches the cells.
No growth factor reaches the cells either, so no start signal reaches them.
Division has to be switched on by a signal.
So few of the cells enter the cycle.

36

Back to the healing cut on the thumb: 24 dividing cells in the sample while the wound was open, 3 after it had closed.

37

While the wound was open, the damaged tissue released growth factors. They reached receptors on the cells at the edge.

38

Every condition was met. So the edge cells entered the cycle and divided: 24 dividing cells in the sample.

39

When the gap closed, the signals faded. The growth-factor condition was no longer met.

40

So the edge cells stopped entering the cycle: 3 dividing cells in the same kind of sample.

41Quick quiz: more or fewer cells dividing? mixed practice

42
Check q7

A researcher adds a growth factor to a dish of liver cells that have plenty of nutrients.

What happens to the number of cells dividing?

  1. A. ✓ More cells divide
  2. B. Fewer cells divide
    A growth factor is a signal to divide.
    It binds its receptor, and the cell enters the cycle.
    So more cells divide.
  3. C. The same number divide
    The cells had nutrients but no signal.
    The growth factor is the signal.
    So more cells enter the cycle and divide.

Why: A growth factor is a signal to divide.
The growth factor binds its receptor on a liver cell.
The bound receptor tells the cell to enter the cycle.
So more cells divide.

43
Check q8

Skin cells in a dish have grown until every cell is touched on every side by neighbors.

What happens to the number of cells dividing?

  1. A. More cells divide
    A cell touched on every side receives a signal to stop.
    So fewer cells divide.
  2. B. ✓ Fewer cells divide
  3. C. The same number divide
    Contact on every side is a molecule reaching a receptor, and the message is stop.
    So fewer cells divide.

Why: A cell touched on every side by neighbors receives a signal to stop.
Contact is a molecule reaching a receptor.
So the packed cells stop entering the cycle, and fewer cells divide.

44
Check q9

A researcher moves cells from a packed dish into a dish where the cells lie far apart.

What happens to the number of cells dividing?

  1. A. ✓ More cells divide
  2. B. Fewer cells divide
    In the new dish no cell is touched on every side.
    So the stop signal is gone, and the cells enter the cycle again.
  3. C. The same number divide
    The crowding was the stop signal.
    The move removes the crowding.
    So the cells divide again.

Why: In the packed dish, contact on every side sent each cell a signal to stop.
In the new dish the cells lie far apart, so no cell is touched on every side.
The stop signal is gone.
So the cells enter the cycle again, and more cells divide.

45
Check q10

A researcher washes the growth factor out of a dish of dividing kidney cells. The nutrients stay.

What happens to the number of cells dividing?

  1. A. More cells divide
    The growth factor was the signal to enter the cycle.
    Without it, cells reaching G1 have no signal.
    So fewer cells divide.
  2. B. ✓ Fewer cells divide
  3. C. The same number divide
    Nutrients supply material, not the signal.
    The signal was the growth factor, and it is gone.
    So fewer cells divide.

Why: The growth factor was the signal to enter the cycle.
The researcher has washed it out.
So cells reaching G1 have no signal to go on.
Nutrients alone do not start the cycle.
So fewer cells divide.

46
Check q11

A researcher doubles the glucose in a dish of cells. No growth factor reaches the cells.

What happens to the number of cells dividing?

  1. A. More cells divide
    Glucose is a nutrient: it supplies material and is not a signal.
    With no growth factor there is no signal to enter the cycle, so the number dividing is unchanged.
  2. B. Fewer cells divide
    More glucose neither starts nor stops the cycle.
    The signal is missing before and after.
    So the number dividing stays the same.
  3. C. ✓ The same number divide

Why: Glucose is a nutrient.
A nutrient supplies material for a cell, but a nutrient is not a signal.
Division starts when a growth factor reaches a receptor.
No growth factor reaches these cells before or after the change.
So the number of cells dividing stays the same.

47
Check q12

A cut in the skin closes. The signal molecules from the damaged tissue fade away.

What happens to the number of skin cells dividing at the edge of the cut?

  1. A. More cells divide
    The signal molecules from the damaged tissue were what told the edge cells to enter the cycle.
    The signal is fading.
    So fewer cells divide.
  2. B. ✓ Fewer cells divide
  3. C. The same number divide
    Cells divide while the signals are present and stop when they fade.
    The signals are fading.
    So fewer cells divide.

Why: While the cut was open, signal molecules from the damaged tissue reached the receptors of the edge cells and told them to enter the cycle.
The cut has closed, and those signals are fading.
Cells divide while the signals are present and stop when they fade.
So fewer cells divide.

48Mixed practice mixed practice

49
Check q13

A dish of cells has nutrients and room to grow, but no growth factor. The cells stay in G1.

Why do the cells stay in G1?

  1. A. ✓ No signal to divide has reached the cells
  2. B. The cells have used up their building material
    The nutrients are present.
    What is missing is the signal to divide.

Why: Division has to be switched on by a growth factor binding a receptor.
No growth factor has arrived.
So the growth-factor condition is unmet, and the cells stay in G1.

50
Check q14

A skin cell at the edge of a healing cut divides. A week later the cut has closed, and the cell stops dividing.

Why does the cell stop dividing?

  1. A. The cell has no DNA left to copy
    A cell copies its DNA anew each cycle.
    DNA is never used up.
  2. B. ✓ The signal molecules from the damaged tissue have faded

Why: Signal molecules from the damaged tissue told the edge cells to enter the cycle.
The cut closed, and the signals faded.
Cells divide only while the signals are present, so division stops.

51
Practice writing an answer

A researcher grows skin cells in one dish until they are packed edge to edge, and grows skin cells thinly spread in a second dish. The researcher then gives both dishes fresh nutrients and the same growth factor. In one day, 8% of the packed cells divide and 44% of the thinly spread cells divide.

(a) Explain how the two dishes demonstrate that a cell enters the cycle only when every condition is met. (1 pt)

Model answer Both dishes have nutrients and a growth factor.
So both sets of cells have food and a signal to divide.
The packed cells are touched on every side by neighbors.
Contact on every side is a signal to stop.
So one condition is unmet in the packed dish: only 8% divide.
No thinly spread cell is touched on every side, so every condition is met, and 44% divide.
So a cell enters the cycle only when every condition is met.
Rubric
  • Award 1 point for: both dishes have nutrients and a growth factor, but the packed cells are touched on every side, a signal to stop, so one condition is unmet and few divide; the thinly spread cells have every condition met and divide, so a cell needs every condition met.

APBIO-U04-L18D Three checkpoints

Topic 4.6 · Regulation of Cell Cycle · 51 steps

Two cells in G2 drawn side by side. The left cell's copied DNA has a break in one stretch, labeled damaged stretch, and an arrow from it reads: waits 9 hours, then mitosis. The right cell's DNA is undamaged and its arrow reads: straight into mitosis
Two cells in G2 drawn side by side. The left cell's copied DNA has a break in one stretch, labeled damaged stretch, and an arrow from it reads: waits 9 hours, then mitosis. The right cell's DNA is undamaged and its arrow reads: straight into mitosis

A cell in G2 has finished copying its DNA, but one stretch of the copy is damaged. Repairing the damage takes nine hours.

The cell waits the whole nine hours before it starts mitosis. A cell beside it with undamaged DNA goes straight in. Nothing timed the wait. What held the first cell, and where on the cycle was it held?

Unit 4 · Cell Communication and Cell Cycle

1A checkpoint

2
Check q1

Here is the cell cycle drawn as a wheel.

The cell cycle drawn as a wheel of sectors sized by hours: G1, S phase, G2, then the thin slivers of mitosis and cytokinesis, with an arrow from cytokinesis back into G1
The cell cycle drawn as a wheel of sectors sized by hours: G1, S phase, G2, then the thin slivers of mitosis and cytokinesis, with an arrow from cytokinesis back into G1

Which stage comes straight after S phase?

  1. A. G1
    G1 comes before S phase.
    After the DNA is copied in S phase, the cell enters G2.
  2. B. ✓ G2
  3. C. Mitosis
    Mitosis comes after G2.
    The stage straight after S phase is G2.

Why: S phase copies the DNA.
The wheel turns G1, S phase, G2, then mitosis.
So G2 comes straight after S phase.

3

How does a cell wait for a condition instead of moving on by the clock?

4

At three places on the wheel, the cycle can be held until a condition is met.

The same wheel with three short bars drawn across its rim: one near the end of G1, one at the boundary between G2 and mitosis, and one inside the mitosis sliver
The same wheel with three short bars drawn across its rim: one near the end of G1, one at the boundary between G2 and mitosis, and one inside the mitosis sliver
5

A place like this is called a , because the cell is checked there before it goes on.

6

Video: Watch: A checkpoint

At three places on the cycle wheel the cycle can be held until a condition is met. A place like this is a checkpoint: a set of conditions, not a timer.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18Da.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18Da.mp4

7

A checkpoint is a set of conditions, not a timer.

8

The cell waits at a checkpoint for as long as a condition is unmet: a minute or a week.

9

The wait is not open-ended. A cell held for about a day with damage it cannot repair does not wait for ever: the cell is removed instead.

10

When the condition is met, the cell moves on.

11

What you are expected to know Describe a checkpoint: a place on the cycle where the cycle is held until a condition is met, for as long as that takes.

12
Check q2

A cell in G2 has finished copying its DNA, but a stretch of it is damaged. Repairing the damage takes nine hours.

When does the cell enter mitosis?

  1. A. After the usual length of G2
    A checkpoint is a set of conditions.
    The cell waits at the checkpoint at the end of G2 until its DNA is undamaged, however long that takes.
  2. B. As soon as the damage is detected
    Damage holds the cell at the checkpoint at the end of G2 until it is repaired.
  3. C. Never
    The checkpoint holds the cell until the damage has been repaired.
    Once the repair is complete, the condition is met, and the cell moves on.
  4. D. ✓ When the repair is complete

Why: A checkpoint holds the cell while its condition is unmet.
The condition at the end of G2 is undamaged DNA.
Until the repair is finished, the cell waits.
When the repair is complete, the condition is met, and the cell enters mitosis: nine hours late.

13
Check q3

A student says: “A cell waits at a checkpoint for a fixed time, and then it moves on.”

Is the student correct?

  1. A. ✓ No — the cell waits until the condition is met, however long that takes
  2. B. Yes — the wait at a checkpoint has one fixed length for every cell
    A checkpoint is a set of conditions: the cell waits at it for as long as a condition is unmet, a minute or a week.

Why: A checkpoint is a set of conditions, not a timer.
The cell waits at the checkpoint for as long as a condition is unmet.
When the condition is met, the cell moves on.
So the wait has no fixed length: it can be a minute or a week.

14
Practice writing an answer

A cell is still small. It sits in a nutrient-rich medium with a growth factor present, and its DNA is undamaged. A checkpoint holds the cell before it enters S phase.

(a) Explain why the checkpoint holds the cell although it has nutrients, a growth factor and undamaged DNA. (1 pt)

Model answer A cell enters the cycle only when every condition is met.
One condition inside the cell is its size.
This cell has nutrients, a growth factor and undamaged DNA.
This cell is still small.
So the size condition is unmet.
A checkpoint holds a cell for as long as any one condition is unmet.
So the checkpoint holds the cell until it has grown.
Rubric
  • Award 1 point for: size is one of the conditions a cell needs before it enters the cycle, and a checkpoint holds a cell while any one condition is unmet; this cell is too small, so it is held until it has grown.

15Quick quiz: checkpoint mixed practice

16
Check q4

A cell waits at a checkpoint.

What is the cell waiting for?

  1. A. ✓ A condition to be met
  2. B. A fixed time to pass
    A checkpoint is not a timer.
    The cell waits for its condition to be met.

Why: A checkpoint is a set of conditions.
The cell waits at it for as long as a condition is unmet.
So the cell is waiting for a condition to be met.

17
Practice writing an answer

A cell waits at one place on the cell cycle.

(a) State what a checkpoint is. (1 pt)

Model answer A checkpoint is a place on the cell cycle where the cell is held until a condition is met.
Rubric
  • Award 1 point for: a place on the cycle where the cell is held until a condition is met (a set of conditions, not a timer).
18
Check q5

A cell reaches a checkpoint. Its condition is met two minutes later.

What does the cell do?

  1. A. Waits the usual time anyway
    A checkpoint has no usual time.
    Once the condition is met, nothing holds the cell.
  2. B. ✓ Moves on at once

Why: A checkpoint holds a cell only while a condition is unmet.
The condition is met after two minutes.
So the cell moves on at once.

19
Check q6

A cell reaches the G1 checkpoint while it is still too small. Growing to full size takes the cell a week.

What does the cell do?

  1. A. ✓ Waits the whole week
  2. B. Moves on after the usual time
    A checkpoint is not a timer.
    While the condition is unmet, the cell stays where it is.

Why: A checkpoint holds a cell for as long as a condition is unmet.
Size is one of the G1 checkpoint’s conditions, and it stays unmet for a week.
So the cell waits the whole week.

20
Check q7

A cell reaches a checkpoint with every one of its conditions met.

Is the cell held?

  1. A. Yes
    A checkpoint holds a cell only while a condition is unmet.
    Every condition is met.
  2. B. ✓ No

Why: A checkpoint holds a cell only while a condition is unmet.
Every condition is met.
So nothing holds the cell, and it moves on.

21
Check q8

Two cells reach the same checkpoint at the same moment. One cell's condition is met at once. The other cell's condition is met a day later.

Which cell moves on first?

  1. A. ✓ The cell whose condition is met at once
  2. B. Both cells move on at the same time
    Each cell moves on when its own condition is met.
    One condition is met a day before the other.

Why: Each cell waits at the checkpoint until its own condition is met.
One cell's condition is met at once, so that cell moves on at once.
The other cell waits a day.
So the first cell moves on first.

22
Check q9

A cell has waited at a checkpoint for six hours.

What decides when the cell moves on?

  1. A. ✓ Whether its condition is met
  2. B. How long it has waited
    The length of the wait decides nothing.
    The cell moves on when its condition is met.

Why: A checkpoint is a set of conditions, not a timer.
The cell moves on when its condition is met.
So whether the condition is met decides when the cell moves on, not the six hours.

23Where the three checkpoints sit

24

Video: Watch: Where the three checkpoints sit

The G1 checkpoint near the end of G1, before the DNA is copied; the G2 checkpoint at the end of G2, before mitosis; the M checkpoint at metaphase, before the sister chromatids separate. A cell that fails a check stays where it is.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18Db.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18Db.mp4

25

The G1 checkpoint sits near the end of G1, before the DNA is copied.

The cell cycle wheel with three short bars across the rim, labeled: the G1 checkpoint near the end of G1, before S phase; the G2 checkpoint at the end of G2, before mitosis; the M checkpoint inside mitosis, at metaphase
The cell cycle wheel with three short bars across the rim, labeled: the G1 checkpoint near the end of G1, before S phase; the G2 checkpoint at the end of G2, before mitosis; the M checkpoint inside mitosis, at metaphase
26

The G2 checkpoint sits at the end of G2, before mitosis.

27

The M checkpoint sits at metaphase, before the sister chromatids separate.

28

Its M stands for mitosis. It has nothing to do with molar concentration.

29

A cell that fails a check stays where it is.

30

Here is a table comparing the three checkpoints: where each sits, which stage follows it, and where a cell that fails the check stays.

A table comparing the three checkpoints: the G1 checkpoint sits near the end of G1, before S phase, and a cell that fails it stays in G1 or moves out into G0; the G2 checkpoint sits at the end of G2, before mitosis, and a cell that fails it stays in G2; the M checkpoint sits at metaphase, before anaphase, in which the sister chromatids separate, and a cell that fails it stays at metaphase
31

What you are expected to know Place the three checkpoints on the cycle: the G1 checkpoint near the end of G1, the G2 checkpoint at the end of G2 and the M checkpoint at metaphase.

32
Check q10

The cycle wheel below has four positions marked 1 to 4.

The cell cycle wheel with four positions marked 1 to 4 on its rim: 1 near the end of G1, 2 in the middle of S phase, 3 at the boundary between G2 and mitosis, 4 inside mitosis
The cell cycle wheel with four positions marked 1 to 4 on its rim: 1 near the end of G1, 2 in the middle of S phase, 3 at the boundary between G2 and mitosis, 4 inside mitosis

Which position marks the G2 checkpoint?

  1. A. Position 1
    Position 1 is the G1 checkpoint, before the DNA is copied.
  2. B. Position 2
    No checkpoint sits at position 2.
    Position 2 is the middle of S phase, where the DNA is being copied.
  3. C. ✓ Position 3
  4. D. Position 4
    Position 4 is the M checkpoint, before the sister chromatids separate.

Why: The G2 checkpoint sits at the end of G2, just before mitosis.
Position 3 sits at the boundary between G2 and mitosis on the wheel.
So position 3 marks the G2 checkpoint.

33

Back to the two cells in G2, one with a damaged stretch in its copied DNA and one with its DNA undamaged.

34

Both cells reached the G2 checkpoint at the end of G2.

35

The undamaged cell’s condition was met. So it went straight into mitosis.

36

The damaged cell’s condition was unmet. So the G2 checkpoint held it in G2 while the damage was repaired: nine hours.

37

When the repair was complete, the condition was met. So the cell entered mitosis.

38Quick quiz: which checkpoint is next? mixed practice

39
Check q11

A cell has finished G2 and is about to begin mitosis.

Which checkpoint does the cell reach?

  1. A. G1 checkpoint
    The G1 checkpoint sits near the end of G1. This cell has finished G2, so its DNA is copied.
    The check at the end of G2 is the G2 checkpoint.
  2. B. ✓ G2 checkpoint
  3. C. M checkpoint
    The M checkpoint sits at metaphase.
    This cell has not begun mitosis yet.
    The check at the end of G2, before mitosis, is the G2 checkpoint.

Why: The G2 checkpoint sits at the end of G2, before mitosis.
This cell has finished G2 and is about to begin mitosis.
So the cell reaches the G2 checkpoint.

40
Check q12

A cell’s chromosomes are lined up in a single row across the middle of the cell.

Which checkpoint does the cell reach?

  1. A. G1 checkpoint
    A cell with its chromosomes in a row is at metaphase, inside mitosis.
    The check at metaphase is the M checkpoint.
  2. B. G2 checkpoint
    A cell with its chromosomes in a row across the middle is already in mitosis, at metaphase.
    The check there is the M checkpoint.
  3. C. ✓ M checkpoint

Why: Chromosomes lined up in a single row across the middle of the cell is metaphase.
The M checkpoint sits at metaphase, before the sister chromatids separate.
So the cell reaches the M checkpoint.

41
Check q13

A cell has grown through G1 and is about to start copying its DNA.

Which checkpoint does the cell reach?

  1. A. ✓ G1 checkpoint
  2. B. G2 checkpoint
    The G2 checkpoint sits after S phase, and this cell has not started S phase yet.
    The check before the DNA is copied is the G1 checkpoint.
  3. C. M checkpoint
    This cell is at the end of G1, long before mitosis.
    The check there is the G1 checkpoint.

Why: The G1 checkpoint sits near the end of G1, before the DNA is copied.
This cell has grown through G1 and is about to start copying its DNA.
So the cell reaches the G1 checkpoint.

42
Check q14

A cell is about to separate its sister chromatids.

Which checkpoint does the cell reach?

  1. A. G1 checkpoint
    A cell about to separate its sister chromatids is at metaphase, inside mitosis.
    The check there is the M checkpoint.
  2. B. G2 checkpoint
    A cell about to separate its sister chromatids is already in mitosis.
    The check just before the sister chromatids separate is the M checkpoint.
  3. C. ✓ M checkpoint

Why: The M checkpoint sits at metaphase, before the sister chromatids separate.
This cell is about to separate its sister chromatids.
So the cell reaches the M checkpoint.

43
Check q15

A new cell has just been formed by cytokinesis and starts to grow.

Which checkpoint does the cell reach first?

  1. A. ✓ G1 checkpoint
  2. B. G2 checkpoint
    A new cell begins in G1.
    The first checkpoint on the wheel after cytokinesis is the G1 checkpoint, near the end of G1.
  3. C. M checkpoint
    A new cell begins in G1, and it must grow and copy its DNA before mitosis.
    The first checkpoint it reaches is the G1 checkpoint.

Why: A new cell formed by cytokinesis begins in G1.
The G1 checkpoint sits near the end of G1.
So the G1 checkpoint is the first checkpoint the new cell reaches.

44
Check q16

A cell has just doubled its DNA.

Which checkpoint does the cell reach next?

  1. A. G1 checkpoint
    The G1 checkpoint sits before the DNA is copied.
    This cell has doubled its DNA, so it has finished S phase.
    The next checkpoint is the G2 checkpoint.
  2. B. ✓ G2 checkpoint
  3. C. M checkpoint
    A cell that has doubled its DNA has finished S phase and is in G2.
    The next checkpoint is the G2 checkpoint; the M checkpoint comes later, in mitosis.

Why: A cell that has doubled its DNA has finished S phase.
After S phase comes G2.
The G2 checkpoint sits at the end of G2.
So the G2 checkpoint is the next checkpoint the cell reaches.

45Mixed practice mixed practice

46
Check q17

A cell has copied all its DNA, and a stretch of the copy is damaged.

Which checkpoint holds the cell?

  1. A. G1 checkpoint
    The G1 checkpoint sits before the DNA is copied.
    This cell has finished copying.
  2. B. ✓ G2 checkpoint
  3. C. M checkpoint
    The M checkpoint sits at metaphase, inside mitosis, later than this.

Why: The DNA has been copied, so the cell is past S phase, in G2.
The G2 checkpoint sits at the end of G2, and its condition is undamaged DNA.
So the G2 checkpoint holds the cell.

47
Check q18

Before the sister chromatids separate, every chromosome must be attached to spindle fibers from both poles. At metaphase, one chromosome in a cell is attached to spindle fibers from one pole only.

What does the cell do?

  1. A. The cell separates the sister chromatids anyway
    The M checkpoint holds the sister chromatids together until every chromosome is attached from both poles.
  2. B. The cell goes back to G2 and rebuilds its spindle
    The cycle goes round one way only.
    A cell at metaphase does not return to G2.
  3. C. ✓ The cell waits at metaphase until both poles attach to the chromosome

Why: The M checkpoint sits at metaphase, and its condition is that every chromosome is attached from both poles.
One chromosome is not.
So the cell waits at metaphase until it is.

48
Check q19

A cell waits at a checkpoint.

Which statement describes a checkpoint?

  1. A. ✓ A set of conditions the cell waits to meet
  2. B. A timer that counts down a fixed time
    A cell waits at a checkpoint for as long as a condition is unmet: a minute or a week.
  3. C. The stage in which the DNA is copied
    The DNA is copied in S phase.
    A checkpoint is a hold between stages.

Why: A checkpoint is a set of conditions.
The cell waits for as long as a condition is unmet.
It moves on when the condition is met.

49
Check q20

A cell has grown to full size, has nutrients and a bound growth factor, and its DNA is undamaged.

What happens to the cell at the G1 checkpoint?

  1. A. The cell is held
    Every condition is met: size, nutrients, a growth signal and undamaged DNA.
    So nothing holds the cell.
  2. B. The cell moves out into G0
    A cell moves out into G0 when a condition stays unmet, not when every condition is met.
  3. C. ✓ The cell passes into S phase

Why: The G1 checkpoint sits near the end of G1, before S phase.
Every condition is met: size, nutrients, a growth signal and undamaged DNA.
So the cell passes into S phase.

50
Practice writing an answer

Researchers grow two lines of kidney cells in a nutrient-rich medium with a growth factor. In one line, a mutation breaks the G1 checkpoint’s check for damaged DNA; the other line is normal. The researchers damage the DNA of cells in both lines while the cells are in G1.

(a) Make a claim about which line’s cells go on into S phase after the damage. (1 pt)

Model answer The mutant cells go on into S phase.
The normal cells are held in G1.
Rubric
  • Award 1 point for the claim: the mutant cells go on into S phase and the normal cells are held. No reasoning is required for this point.

Slip Claiming that both lines are held because both have damaged DNA. Damage holds a cell only when the checkpoint detects it.

(b) Support your claim using the G1 checkpoint’s conditions. (1 pt)

Model answer One condition of the G1 checkpoint is undamaged DNA.
In the normal cells the check for damaged DNA works, so the damage is detected.
So the normal cells are held in G1.
In the mutant cells that check is broken, so the damage is never detected.
So the mutant cells pass the G1 checkpoint and copy their damaged DNA.
Rubric
  • Award 1 point for: the evidence (undamaged DNA is a G1 condition, and the mutant line’s check for it is broken) AND the reasoning (the normal cells detect the damage and are held; the mutant cells never detect it, so they pass into S phase).

Slip Giving the claim again with no support, or saying the mutant cells are held because their DNA is damaged. Damage holds a cell only when it is detected.

Glossary

checkpoint (G1 checkpoint, G2 checkpoint, M checkpoint)
A place in the cell cycle where the cycle is held until a condition is met. The G1 checkpoint sits near the end of G1, before S phase. The G2 checkpoint sits at the end of G2, before mitosis. The M checkpoint sits at metaphase, before the sister chromatids separate. A checkpoint is a set of conditions, not a timer.

APBIO-U04-L18B What each checkpoint checks

Topic 4.6 · Regulation of Cell Cycle · 77 steps

Two small bar charts of skin cells by stage, G1, S phase, and G2 with mitosis: untreated cells 46, 39 and 15 percent; six hours after ultraviolet light 74, 14 and 12 percent, the G1 bar much taller
Two small bar charts of skin cells by stage, G1, S phase, and G2 with mitosis: untreated cells 46, 39 and 15 percent; six hours after ultraviolet light 74, 14 and 12 percent, the G1 bar much taller

Here are skin cells sorted by stage, six hours after a dose of ultraviolet light. Ultraviolet light damages DNA.

Before the dose, 46% of the cells were in G1 and 39% were in S phase.

Six hours after the dose, 74% are in G1 and only 14% are in S phase.

Their food and their room did not change. Why have the cells piled up in G1?

Unit 4 · Cell Communication and Cell Cycle

1The G1 checkpoint: damaged DNA is not copied

2

Video: Watch: The G1 checkpoint

The G1 checkpoint checks that the cell is big enough, has nutrients, has received a growth signal and has undamaged DNA. Its hold means damaged DNA is not copied. A checkpoint holds; other proteins repair.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18BA.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18BA.mp4

3

What is each checkpoint looking for, and what goes wrong without it?

4

Each checkpoint checks for something. Its hold prevents a particular harm.

5

The G1 checkpoint checks size, nutrients, a growth signal and undamaged DNA. So damaged DNA is not copied.

6

The G2 checkpoint checks that the DNA is completely copied and undamaged. So no daughter cell receives an incomplete or damaged genome.

7

The M checkpoint checks that every chromosome is attached to spindle fibers from both poles. So no daughter cell ends up with an extra or a missing chromosome.

8

A checkpoint holds; other proteins repair.

9

That is why the ultraviolet-damaged skin cells piled up before S phase.

10
Check q1

The cycle wheel below has four positions marked 1 to 4.

The cell cycle wheel of sectors sized by hours, with four positions marked 1 to 4 on its rim: 1 near the end of G1, 2 in the middle of S phase, 3 at the boundary between G2 and mitosis, 4 inside mitosis
The cell cycle wheel of sectors sized by hours, with four positions marked 1 to 4 on its rim: 1 near the end of G1, 2 in the middle of S phase, 3 at the boundary between G2 and mitosis, 4 inside mitosis

At which position does the G1 checkpoint sit?

  1. A. ✓ Position 1
  2. B. Position 2
    Position 2 is the middle of S phase, where the DNA is being copied.
    The G1 checkpoint sits before S phase begins.
  3. C. Position 3
    Position 3 is the end of G2: the G2 checkpoint.
  4. D. Position 4
    Position 4 is metaphase: the M checkpoint.

Why: The G1 checkpoint sits near the end of G1, before the DNA is copied in S phase.
Position 1 is near the end of G1 on the wheel.
So the G1 checkpoint sits at position 1.

11

Here is the cycle wheel with the three checkpoints marked. At each checkpoint the cell waits until a condition is met.

The cell cycle wheel of sectors sized by hours, with three short bars across the rim, labeled: the G1 checkpoint near the end of G1, before S phase; the G2 checkpoint at the end of G2, before mitosis; the M checkpoint inside mitosis, at metaphase
The cell cycle wheel of sectors sized by hours, with three short bars across the rim, labeled: the G1 checkpoint near the end of G1, before S phase; the G2 checkpoint at the end of G2, before mitosis; the M checkpoint inside mitosis, at metaphase
12

The G1 checkpoint checks four conditions:

  • the cell is big enough
  • the cell has nutrients
  • a growth signal has reached the cell
  • the cell’s DNA is undamaged

13

If any one of the four conditions is unmet, the cell waits at the end of G1. It does not start S phase.

14

So damaged DNA is not copied.

15

The stop signal from contact on every side acts here too. A cell touched on every side is held in G1.

16

Here is a graph of the skin cells sorted by stage, before and six hours after the ultraviolet light.

Two stacked panels of three bars, percent of skin cells in G1, S phase, and G2 with mitosis: untreated 46, 39 and 15; six hours after ultraviolet exposure 74, 14 and 12; gridlines every 20 percent
Two stacked panels of three bars, percent of skin cells in G1, S phase, and G2 with mitosis: untreated 46, 39 and 15; six hours after ultraviolet exposure 74, 14 and 12; gridlines every 20 percent
17

Cells in G1 rose from 46% to 74%. Cells in S phase fell from 39% to 14%.

18

The ultraviolet light damaged the cells’ DNA. A checkpoint protein detected the damaged DNA.

19

So the checkpoint protein kept each cell from starting S phase.

20

So the damaged cells piled up at the G1 checkpoint.

21

A checkpoint holds; it does not repair. The hold gives other proteins time to repair the DNA.

22

While the cell waits, the damage is still there.

23

What you are expected to know Predict which cells the G1 checkpoint holds.

24

What you are expected to know State what the hold at the G1 checkpoint prevents.

25
Check q2

Researchers sorted untreated skin cells, and skin cells six hours after ultraviolet exposure, by stage; the graph below shows the percent in G1, S phase, and G2 with mitosis. DNA damage measured 3 units in the untreated cells and 19 units in the exposed cells.

Two stacked panels of three bars, percent of skin cells in G1, S phase, and G2 with mitosis: untreated 46, 39 and 15; six hours after ultraviolet exposure 74, 14 and 12; gridlines every 20 percent
Two stacked panels of three bars, percent of skin cells in G1, S phase, and G2 with mitosis: untreated 46, 39 and 15; six hours after ultraviolet exposure 74, 14 and 12; gridlines every 20 percent

At which checkpoint were the exposed cells held?

  1. A. ✓ The G1 checkpoint
  2. B. The G2 checkpoint
    The G1 share rose from 46% to 74% and the S-phase share fell from 39% to 14%: cells were held at the end of G1, the G1 checkpoint.
  3. C. The M checkpoint
    The share of cells in G2 with mitosis barely changed, from 15% to 12%.
    The stage that filled was G1. So the hold was at the G1 checkpoint.

Why: The exposed cells carried DNA damage.
The G1 share rose from 46% to 74%; the S-phase share fell from 39% to 14%.
So cells reached the end of G1 and did not start S phase.
The checkpoint before S phase is the G1 checkpoint.

26
Practice writing an answer

Researchers sorted untreated skin cells, and skin cells six hours after ultraviolet exposure, by stage. DNA damage measured 3 units in the untreated cells and 19 units in the exposed cells. In the exposed cells, the share in G1 rose from 46% to 74% and the share in S phase fell from 39% to 14%. The exposed cells were held at the G1 checkpoint.

(a) Explain why the damaged cells were held before S phase. (1 pt)

Model answer The ultraviolet light damaged the cells’ DNA: 19 units of damage against 3.
The G1 checkpoint checks that the DNA is undamaged before the DNA is copied.
A checkpoint protein detects the damaged DNA.
So the checkpoint protein keeps each damaged cell from starting S phase.
So the damaged cells wait at the end of G1 instead of copying their DNA.
The hold prevents a particular harm: damaged DNA is not copied into two daughter cells.
Rubric
  • Award 1 point for: the G1 checkpoint checks for undamaged DNA before S phase, a checkpoint protein detects the damage and holds the cell, so the damaged DNA is not copied.
27
Check q3

A student says: “While a damaged cell waits at a checkpoint, the checkpoint repairs the damaged DNA.”

Is the student correct?

  1. A. Yes — the checkpoint mends the DNA while the cell waits there
    A checkpoint holds; it does not repair.
    The hold gives other proteins time to repair the DNA.
    While the cell waits, the damage is still there.
  2. B. ✓ No — the checkpoint holds the cell; other proteins repair the DNA

Why: A checkpoint holds; it does not repair.
A checkpoint protein detects the damaged DNA and keeps the cell from starting S phase.
That pause gives other proteins time to repair the DNA.
While the cell waits, the damage is still there.
So other proteins repair, not the checkpoint.

28
Check q4

Skin cells reach the end of G1 with undamaged DNA and plenty of nutrients. Their receptors for growth factor are all empty.

Which of the following happens to the cells?

  1. A. ✓ The cells are held at the G1 checkpoint
  2. B. The cells pass into S phase
    The G1 checkpoint checks four conditions: size, nutrients, a growth signal and undamaged DNA.
    No growth signal has reached these cells, so one condition is unmet and they are held.
  3. C. The cells are held at the G2 checkpoint
    These cells are at the end of G1 and have not copied their DNA.
    The checkpoint they face is the G1 checkpoint.

Why: The G1 checkpoint checks four conditions: size, nutrients, a growth signal and undamaged DNA.
These cells have undamaged DNA and nutrients.
Their receptors are empty, so no growth signal has reached them.
So one condition is unmet.
So the cells are held at the G1 checkpoint.

29The G2 checkpoint: no daughter cell receives an incomplete or damaged genome

30

Video: Watch: The G2 checkpoint

The G2 checkpoint checks that the DNA is completely copied and undamaged before mitosis. Its hold means no daughter cell receives an incomplete or damaged genome.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18BB.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18BB.mp4

31

Now consider a cell that has reached the end of G2 with one stretch of its DNA still uncopied.

32

Suppose the cell entered mitosis anyway. The uncopied stretch has no second copy.

33

So one daughter cell would receive that stretch, and the other daughter cell would receive none of it.

34

That daughter cell’s genome would be incomplete.

35

The G2 checkpoint checks that the DNA is completely copied and undamaged.

36

So the cell waits in G2, before mitosis, until the copying is complete and any damage is repaired.

37

So no daughter cell receives an incomplete or damaged genome.

38

What you are expected to know Predict which cells the G2 checkpoint holds.

39

What you are expected to know State what the hold at the G2 checkpoint prevents.

40
Check q5

A researcher gives cells growing in a dish a chemical during S phase that slows DNA copying. The cells reach full size, but a test finds long stretches of DNA still uncopied.

Where are these cells held?

  1. A. In G1, before S phase
    These cells are already copying.
    The check that the copying is complete comes at the end of G2.
  2. B. ✓ In G2, before mitosis
  3. C. At metaphase, before the sister chromatids separate
    The G2 checkpoint holds a cell with uncopied DNA before mitosis.
    So the cells never reach metaphase.
  4. D. In telophase, before cytokinesis
    A cell with uncopied DNA is held at the G2 checkpoint.
    So the cells never reach telophase.

Why: The G2 checkpoint checks that the DNA is completely copied and undamaged before mitosis.
These cells still have long stretches of DNA uncopied.
So the G2 checkpoint’s condition is unmet.
So the cells are held in G2, before mitosis, even though they have reached full size.

41
Practice writing an answer

A researcher gives cells growing in a dish a chemical during S phase that slows DNA copying. The cells reach full size, but a test finds long stretches of DNA still uncopied. The cells are held in G2.

(a) Explain how the hold in G2 protects the daughter cells these cells would have produced. (1 pt)

Model answer Long stretches of the cells’ DNA are still uncopied.
The G2 checkpoint checks that the DNA is completely copied before mitosis.
So the G2 checkpoint holds the cells in G2.
If a cell divided with a stretch uncopied, one daughter cell would receive that stretch and the other would receive none of it.
So the hold prevents a daughter cell receiving an incomplete genome.
Rubric
  • Award 1 point for: the G2 checkpoint checks that the DNA is completely copied before mitosis, so the hold stops a cell dividing with uncopied DNA and prevents a daughter cell receiving an incomplete genome.

42The M checkpoint: no daughter cell ends up with an extra or a missing chromosome

43

Video: Watch: The M checkpoint

The M checkpoint checks that every chromosome is attached to spindle fibers from both poles. One unattached chromosome holds the whole cell at metaphase, so no daughter cell ends up with an extra or a missing chromosome.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18BC.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18BC.mp4

44
Check q6

A cell is at metaphase, with every chromosome lined up across the middle of the spindle.

At the start of anaphase, which chromosomes’ sister chromatids separate?

  1. A. One chromosome at a time, in turn
    Anaphase begins with one event: the sister chromatids of every chromosome separate together.
  2. B. Only the chromosomes that reached the row first
    Anaphase separates the sister chromatids of every chromosome at once; no chromosome goes first.
  3. C. ✓ Every chromosome, all at the same moment

Why: At the start of anaphase, the sister chromatids of every chromosome separate at the same moment.
The spindle fibers then pull one chromatid of each pair to each pole.

45

The M checkpoint checks that every chromosome is attached to spindle fibers from both poles.

46

Now consider a cell at metaphase with one chromosome attached to spindle fibers from one pole only.

47

Suppose anaphase began anyway. Both sister chromatids of that chromosome would be pulled to the same pole.

48

So one daughter cell would receive an extra chromosome, and the other daughter cell would be missing one.

49

All the sister chromatids separate together at the start of anaphase. So the cell cannot separate the other chromosomes and hold back one.

50

So the M checkpoint holds the whole cell at metaphase until that one chromosome is attached from both poles.

51

So no daughter cell ends up with an extra or a missing chromosome.

52

Here is a table comparing the three checkpoints: what each one checks, and the harm its hold prevents.

A table comparing the three checkpoints. Rows: the G1 checkpoint checks that the cell is big enough, has nutrients, has received a growth signal and has undamaged DNA, and its hold prevents damaged DNA being copied; the G2 checkpoint checks that the DNA is completely copied and undamaged, and its hold prevents a daughter cell receiving an incomplete or damaged genome; the M checkpoint checks that every chromosome is attached to spindle fibers from both poles, and its hold prevents a daughter cell ending up with an extra or a missing chromosome
53

What you are expected to know Predict which cells the M checkpoint holds.

54

What you are expected to know State what the hold at the M checkpoint prevents.

55
Check q7

In a dividing cell at metaphase, every chromosome but one is attached to spindle fibers from both poles. The last chromosome is attached from one pole only.

Which of the following happens next?

  1. A. Anaphase begins on time
    The M checkpoint’s condition is every chromosome attached from both poles.
    One chromosome is attached from one pole only.
    So the condition is unmet, and the cell waits.
  2. B. The attached chromosomes separate while the last one waits
    The sister chromatids of every chromosome separate together at the start of anaphase.
    So the whole cell waits for the last chromosome.
  3. C. ✓ The whole cell waits at metaphase
  4. D. The cell drops back into G2 and builds a new spindle
    A checkpoint holds the cell where it is.
    A checkpoint never returns a cell to an earlier stage.

Why: The M checkpoint checks that every chromosome is attached to spindle fibers from both poles.
One chromosome is attached from one pole only.
So the M checkpoint’s condition is unmet.
So the whole cell waits at metaphase until that chromosome is attached from both poles.

56
Practice writing an answer

In a dividing cell at metaphase, every chromosome but one is attached to spindle fibers from both poles. The last chromosome is attached from one pole only. The whole cell waits at metaphase.

(a) Explain why the whole cell waits for one chromosome, and what the wait prevents. (1 pt)

Model answer The M checkpoint’s condition is every chromosome attached to spindle fibers from both poles.
One chromosome is attached from one pole only, so the condition is unmet.
All the sister chromatids separate together.
So the cell cannot separate some chromosomes and hold back one; the whole cell waits.
If the cell went on, both sister chromatids of that chromosome could go to one pole.
So one daughter cell would get an extra chromosome and the other would be missing one.
Rubric
  • Award 1 point for: the M checkpoint’s condition is every chromosome attached from both poles, so one unattached chromosome leaves the condition unmet and the whole cell is held; the hold prevents a daughter cell receiving an extra or a missing chromosome.
57

Back to the skin cells six hours after their dose of ultraviolet light. Before the dose, 46% of the cells were in G1; six hours after it, 74% were.

58

Only 14% were in S phase, against 39% before. Their food and their room had not changed.

59

The ultraviolet light damaged their DNA. The G1 checkpoint checks that the DNA is undamaged before it is copied.

60

So a checkpoint protein held the damaged cells at the end of G1, and they piled up before S phase.

61

The hold gave other proteins time to repair the DNA before it was copied. Without the hold, every daughter cell would have received the damage.

62Quick quiz: which checkpoint holds this cell? mixed practice

63
Check q8

A cell finds a damaged stretch of DNA after S phase.

Which checkpoint holds the cell?

  1. A. G1 checkpoint
    The damage was found after S phase, so the cell has passed the G1 checkpoint.
    The next check for undamaged DNA is the G2 checkpoint.
  2. B. ✓ G2 checkpoint
  3. C. M checkpoint
    The M checkpoint checks how the chromosomes are attached, not whether the DNA is damaged.
    Damage found after S phase is caught at the G2 checkpoint.

Why: The G2 checkpoint checks that the DNA is completely copied and undamaged before mitosis.
This cell has finished S phase and carries damaged DNA.
So the G2 checkpoint holds the cell.

64
Check q9

One chromosome of a dividing cell is attached to spindle fibers from one pole only.

Which checkpoint holds the cell?

  1. A. G1 checkpoint
    A cell with a spindle is in mitosis.
    The attachment of chromosomes is checked at the M checkpoint.
  2. B. G2 checkpoint
    A cell with a spindle has already entered mitosis.
    The attachment of every chromosome from both poles is checked at the M checkpoint.
  3. C. ✓ M checkpoint

Why: The M checkpoint checks that every chromosome is attached to spindle fibers from both poles.
This cell has one chromosome attached from one pole only.
So the M checkpoint holds the cell.

65
Check q10

A cell reaches the end of G1 while it is still too small.

Which checkpoint holds the cell?

  1. A. ✓ G1 checkpoint
  2. B. G2 checkpoint
    Size is checked at the end of G1, before the DNA is copied.
  3. C. M checkpoint
    A cell at the end of G1 has not copied its DNA and has no spindle.
    Size is checked at the G1 checkpoint.

Why: The G1 checkpoint checks that the cell is big enough, has nutrients and a growth signal, and that its DNA is undamaged.
This cell is too small.
So the G1 checkpoint holds the cell.

66
Check q11

A cell has long stretches of DNA still uncopied when it reaches the end of G2.

Which checkpoint holds the cell?

  1. A. G1 checkpoint
    This cell has been copying its DNA, so it has passed the G1 checkpoint.
    Whether the copying is complete is checked at the G2 checkpoint.
  2. B. ✓ G2 checkpoint
  3. C. M checkpoint
    A cell with uncopied DNA is held before mitosis begins, at the G2 checkpoint.

Why: The G2 checkpoint checks that the DNA is completely copied and undamaged.
This cell has long stretches of DNA still uncopied.
So the G2 checkpoint holds the cell.

67
Check q12

The growth-factor receptors of a cell at the end of G1 are all empty.

Which checkpoint holds the cell?

  1. A. ✓ G1 checkpoint
  2. B. G2 checkpoint
    A growth signal is one of the G1 checkpoint’s conditions, checked before the DNA is copied.
  3. C. M checkpoint
    The M checkpoint checks the attachment of chromosomes.
    The growth signal is checked at the G1 checkpoint.

Why: The G1 checkpoint checks for a growth signal, as well as size, nutrients and undamaged DNA.
No growth factor is bound to this cell’s receptors, so no growth signal has reached it.
So the G1 checkpoint holds the cell.

68
Check q13

A cell’s DNA is damaged before the DNA has been copied.

Which checkpoint holds the cell?

  1. A. ✓ G1 checkpoint
  2. B. G2 checkpoint
    This cell has not copied its DNA yet.
    The earliest check for undamaged DNA is the G1 checkpoint, before S phase.
  3. C. M checkpoint
    The damage is found before S phase, long before any spindle forms.
    The G1 checkpoint holds the cell.

Why: The G1 checkpoint checks that the DNA is undamaged before the DNA is copied.
This cell’s DNA is damaged and has not been copied yet.
So the G1 checkpoint holds the cell, and the damage is not copied.

69Mixed practice mixed practice

70
Check q14

A cell is held at the M checkpoint.

Which harm does this hold prevent?

  1. A. Damaged DNA being copied into two daughter cells
    The G1 checkpoint stops damaged DNA being copied, before S phase.
  2. B. ✓ A daughter cell with an extra or a missing chromosome
  3. C. A daughter cell with an incomplete genome
    The G2 checkpoint prevents an incomplete genome; it checks that the copying is complete.

Why: The M checkpoint checks that every chromosome is attached to spindle fibers from both poles.
Its hold prevents a daughter cell ending up with an extra or a missing chromosome.

71
Check q15

A cell has finished S phase and is passing through G2. Its DNA is complete and undamaged.

Which checkpoint will check that the copying is complete?

  1. A. The G1 checkpoint
    The G1 checkpoint sits before S phase, before any copying has begun.
  2. B. The M checkpoint
    The M checkpoint checks how the chromosomes are attached to the spindle, not whether the DNA is copied.
  3. C. ✓ The G2 checkpoint

Why: The G2 checkpoint sits at the end of G2, after S phase.
The G2 checkpoint checks that the DNA is completely copied and undamaged.

72
Check q16

A cell at the end of G1 has undamaged DNA, plenty of nutrients and a bound growth factor, but it is still too small.

Which of the following happens to the cell?

  1. A. ✓ The cell is held at the G1 checkpoint
  2. B. The cell passes into S phase
    Size is one of the G1 checkpoint’s four conditions.
    The cell is too small, so one condition is unmet.
  3. C. The cell is held at the G2 checkpoint
    This cell is at the end of G1 and has not copied its DNA.
    The checkpoint it faces is the G1 checkpoint.

Why: The G1 checkpoint checks four conditions: size, nutrients, a growth signal and undamaged DNA.
This cell is too small, so one condition is unmet.
So the G1 checkpoint holds the cell.

73
Check q17

In a dividing lily root-tip cell, two chromosomes are attached to spindle fibers from one pole only. The cycle wheel below has four positions marked 1 to 4.

The cell cycle wheel of sectors sized by hours, with four positions marked 1 to 4 on its rim: 1 near the end of G1, 2 in the middle of S phase, 3 at the boundary between G2 and mitosis, 4 inside mitosis
The cell cycle wheel of sectors sized by hours, with four positions marked 1 to 4 on its rim: 1 near the end of G1, 2 in the middle of S phase, 3 at the boundary between G2 and mitosis, 4 inside mitosis

At which position is this cell held?

  1. A. Position 1
    A cell with a spindle has long passed G1; the attachment of chromosomes is checked at metaphase.
  2. B. Position 2
    No checkpoint sits at position 2.
    Position 2 is the middle of S phase, and a cell copying its DNA has no spindle yet.
  3. C. Position 3
    Position 3 is the G2 checkpoint.
    The G2 checkpoint checks that the DNA is completely copied and undamaged.
    The attachment of chromosomes to the spindle is checked later, at metaphase.
  4. D. ✓ Position 4

Why: The M checkpoint checks that every chromosome is attached to spindle fibers from both poles.
This cell has two chromosomes attached from one pole only, so the cell is held at the M checkpoint.
The M checkpoint sits at metaphase, inside the mitosis sliver on the wheel: position 4.

74
Check q18

A researcher gives a culture of cells a chemical that damages DNA. Two days later most of the cells hold 12 pg of DNA each, twice the 6 pg of a cell in G1, and every one of them is outside mitosis.

Where are the cells being held?

  1. A. At the G1 checkpoint
    Cells held in G1 would hold 6 pg of DNA.
    These cells hold 12 pg, so their DNA has been copied.
  2. B. In G0
    A cell in G0 has not copied its DNA and holds 6 pg.
    These cells have 12 pg, so they are inside the cycle, after S phase.
  3. C. ✓ At the G2 checkpoint
  4. D. At the M checkpoint
    A cell held at metaphase is in mitosis, and none of these cells is.

Why: A cell in G1 holds 6 pg; a cell that has finished S phase holds 12 pg.
These cells hold 12 pg, so they have finished S phase.
None is in mitosis, so they sit in G2.
The G2 checkpoint checks for undamaged DNA, so the damage holds them there.

75
Check q19

Two cells reach the G2 checkpoint at the same moment. One has a damaged stretch of DNA; the other’s DNA is complete and undamaged.

When does each cell enter mitosis?

  1. A. ✓ The undamaged cell now; the damaged cell after its repair
  2. B. Both cells enter after the same fixed wait
    A checkpoint is a set of conditions.
    Each cell is checked against the condition.
    Only the undamaged cell meets the condition now.
  3. C. Both cells enter at once, with no wait
    A checkpoint holds; it does not repair.
    Other proteins repair the DNA during the hold.
    Repair takes time, so the damaged cell is delayed.
  4. D. Neither cell enters until both are ready
    Each cell is held or released on its own conditions.
    One cell’s damage holds only that cell.

Why: A checkpoint is a set of conditions, checked in each cell on its own.
The undamaged cell meets the G2 checkpoint’s condition, so it enters mitosis now.
The damaged cell fails the checkpoint’s condition.
So it is held at the G2 checkpoint until its repair is complete, then enters mitosis.

76
Practice writing an answer

Researchers compared two kinds of cultured cell: cells with a working copy of a checkpoint protein, protein P, and cells whose gene for protein P carries a mutation that leaves the protein useless. The researchers exposed both kinds to the same DNA-damaging treatment, and both kinds had the same amount of damage right afterward. Six hours later the researchers counted the cells that had entered division and tested for DNA damage. Untreated cells of both kinds were counted alongside. The table below gives the results. The last row counts the cells present at six hours: for working P these were mostly cells that had not divided, for mutant P mostly daughter cells.

A table comparing cells with working checkpoint protein P and cells with a mutant, useless protein P: untreated cells that entered division in six hours, 92 percent and 90 percent; damaged cells that entered division, 14 percent and 78 percent; cells at six hours still carrying damage, 72 percent and 74 percent
A table comparing cells with working checkpoint protein P and cells with a mutant, useless protein P: untreated cells that entered division in six hours, 92 percent and 90 percent; damaged cells that entered division, 14 percent and 78 percent; cells at six hours still carrying damage, 72 percent and 74 percent

(a) Describe what the untreated cells show about the two kinds of cell. (1 pt)

Model answer With no DNA damage, the working-P cells and the mutant-P cells entered division at about the same rate: 92% and 90%.
So the mutation in protein P does not by itself stop cells dividing.
The untreated cells are the comparison group.
They show that any difference between the two kinds of cell appears only after DNA damage.
Rubric
  • Award 1 point for: untreated cells of both kinds divide at similar rates (92% and 90%), so protein P is not needed for division itself and the difference appears only after damage.

Slip Skipping the untreated rows. They are the control comparison: without them, a low division count in the damaged working-P cells could be blamed on the cells rather than on the checkpoint.

(b) Identify the null hypothesis for the effect of protein P on cells with damaged DNA. (1 pt)

Model answer The null hypothesis is that protein P makes no difference to the percent of damaged cells that enter division: working-P and mutant-P cells would divide at the same rate after damage.
Rubric
  • Award 1 point for: protein P makes no difference to the percent of damaged cells entering division (the two kinds divide at the same rate after damage).

Slip Writing the expected result (‘protein P holds damaged cells’) as the null hypothesis. The null hypothesis is the statement of no effect, which the data can then reject.

(c) Explain how the results support the claim that protein P holds the cycle after DNA damage rather than repairing the damage. (2 pt)

Model answer After damage, 14% of the cells with working protein P entered division, against 78% of the mutant-P cells.
So working protein P held the damaged cells at a checkpoint.
Damage was still detectable in 72% of the held cells at six hours, so protein P had not removed the damage.
So protein P holds the cycle and gives other proteins time to repair.
The mutant-P cells had no hold and divided with the damage: 74% of their daughters carried it.
Rubric
  • Award 1 point for: with working protein P far fewer damaged cells entered division (14% against 78%), so protein P holds the cycle after damage.
  • Award 1 point for: damage remained in most of the held cells (72%), so protein P does not itself repair the DNA; the hold gives time for repair, and mutant daughters inherit the damage.
  • Accept, for the second point: damage persisted about equally in both kinds of cell (72% against 74%), so the working protein had not removed it.

Slip Writing that protein P repairs the DNA. Damage was still present in 72% of the held cells six hours later; the protein’s job is the hold, and repair is done by other proteins in the time the hold buys.

APBIO-U04-L18C Cyclins rise and fall; CDKs wait for them

Topic 4.6 · Regulation of Cell Cycle · 93 steps

A curve of one protein's concentration through one cell cycle: low through G1 and S phase, climbing through G2, peaking as mitosis begins and collapsing as it ends; the stages G1, S phase, G2 and mitosis marked along the bottom
A curve of one protein's concentration through one cell cycle: low through G1 and S phase, climbing through G2, peaking as mitosis begins and collapsing as it ends; the stages G1, S phase, G2 and mitosis marked along the bottom

Here is a graph of how the concentration of one protein inside a dividing cell changes through one cycle.

The concentration is low through G1 and S phase. It climbs through G2. It peaks as mitosis begins and collapses as mitosis ends. The next cycle, it does the same again.

A cell waits at a checkpoint until a condition is met. Once the condition is met, what moves the cell on?

Unit 4 · Cell Communication and Cell Cycle

1A cyclin: a protein whose concentration rises and falls

2

Video: Watch: A cyclin

One protein’s concentration is low through G1 and S phase, climbs through G2, peaks as mitosis begins and collapses as mitosis ends, cycle after cycle. A protein like this is called a cyclin.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18CA.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18CA.mp4

3
Check q1

A cell has reached the G2 checkpoint with a damaged stretch of DNA.

How long does the cell wait there?

  1. A. ✓ Until the DNA is repaired, however long that takes
  2. B. A fixed time, and then it moves on
    A checkpoint is a set of conditions, not a timer.
    The cell moves on only when the condition is met.

Why: A checkpoint is a set of conditions.
The cell waits for as long as a condition is unmet.
The cell moves on when the condition is met.

4

What moves a cell past a checkpoint once its condition is met? Two kinds of protein move it on.

5

One is a protein whose concentration rises and falls in a repeating pattern through the cycle.

6

The other is a kinase that sits at a steady level and is active only while the first protein is bound to it.

7

Together, the pair switches on the proteins that start the next stage.

8

While a checkpoint’s condition is unmet, a damage signal keeps the pair switched off, however much of the first protein piles up.

9

It is the first protein, not the kinase, whose amount changes.

10

Here is the graph again: the concentration of the first protein through one cycle.

One protein's concentration through a cycle, low through G1 and S phase, climbing through G2, peaking as mitosis begins and collapsing as it ends
One protein's concentration through a cycle, low through G1 and S phase, climbing through G2, peaking as mitosis begins and collapsing as it ends
11

Through G1 and S phase, the protein’s concentration is low, about 10 units.

12

Through G2, its concentration climbs. As mitosis begins, it peaks at about 95 units.

13

As mitosis ends, the cell destroys the protein, and its concentration collapses to about 8 units.

14

A protein whose concentration rises and falls like this, in a repeating pattern through the cycle, is called a , because its concentration cycles.

15

The next cycle, the cyclin’s concentration climbs, peaks and collapses again.

16

A cell makes more than one cyclin. This graph follows one of them, the cyclin that peaks as mitosis begins.

17

What you are expected to know Say what a cyclin is.

18

What you are expected to know Read off a graph of a cyclin’s concentration where the concentration is highest and where it is lowest.

19
Check q2

On the graph below, one cyclin’s concentration is drawn through a cycle, with four moments marked 1 to 4.

One protein's concentration against time through one cycle, the stages G1, S phase, G2 and mitosis marked on the axis, gridlines every 20 units, four moments circled and numbered 1 to 4 on the curve
One protein's concentration against time through one cycle, the stages G1, S phase, G2 and mitosis marked on the axis, gridlines every 20 units, four moments circled and numbered 1 to 4 on the curve

At which marked moment is the cyclin’s concentration highest?

  1. A. Moment 1
    At moment 1, in G1, the curve sits at about 8 units, its lowest level.
  2. B. Moment 2
    At moment 2 the curve is still climbing through G2, at about 55 units.
  3. C. ✓ Moment 3
  4. D. Moment 4
    At moment 4 the curve has collapsed to about 12 units, at the end of mitosis.

Why: Read the curve at each marked moment.
Moment 3 sits at the peak, about 94 units, as mitosis begins.
So the cyclin’s concentration is highest at moment 3.

20Quick quiz: cyclin mixed practice

21
Check q3

What is a cyclin?

  1. A. A signal molecule released by other cells that tells a cell to divide
    A signal molecule from other cells that tells a cell to divide is a growth factor.
    A cyclin is a protein inside the cell.
  2. B. ✓ A protein whose concentration rises and falls in a repeating pattern through the cycle
  3. C. A protein present at the same level all the way round the whole cycle
    A cyclin’s concentration changes through the cycle: low through G1 and S phase, high as mitosis begins.

Why: A cyclin is a protein whose concentration rises and falls in a repeating pattern through the cycle.
Its concentration cycles; that is where the name comes from.

22
Practice writing an answer

One protein’s concentration in a dividing cell climbs through G2, peaks as mitosis begins and collapses as mitosis ends, cycle after cycle.

(a) State what a cyclin is. (1 pt)

Model answer A cyclin is a protein whose concentration rises and falls in a repeating pattern through the cell cycle.
Rubric
  • Award 1 point for: a protein whose concentration rises and falls (cycles) in a repeating pattern through the cell cycle.

23A CDK: a kinase that waits for a cyclin

24

Video: Watch: A cyclin-dependent kinase

A second protein sits at a steady level all the way round the cycle. It is a kinase, and it is active only while a cyclin is bound to it: a cyclin-dependent kinase (CDK). Active cyclin–CDK complexes appear when the cyclin is high and vanish when it is destroyed.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18CB.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18CB.mp4

25

A second protein moves the cell on together with the cyclin. This second protein is a kinase.

26
Check q4

A kinase is a kind of enzyme.

What does a kinase do?

  1. A. ✓ Transfers a phosphate group from ATP onto a protein
  2. B. Removes a phosphate group from a protein
    Removing a phosphate is a phosphatase’s job.
    A kinase adds one, taking it from ATP.

Why: A kinase transfers a phosphate group from ATP onto a protein, and the protein changes shape.

27

This kinase is present at about the same level all the way round the cycle.

28

This kinase is active only while a cyclin is bound to it.

29

A kinase like this is called a , because its activity depends on a bound cyclin.

Cyclin, an oval, and CDK, a rounded block, drawn apart and labeled inactive, and then bound together and labeled an active complex
Cyclin, an oval, and CDK, a rounded block, drawn apart and labeled inactive, and then bound together and labeled an active complex
30

Here is the graph again, with the CDK added as a flat broken line above the cyclin’s peak.

One cyclin's concentration through a cycle, low through G1 and S phase, climbing through G2, peaking as mitosis begins and collapsing as it ends, with CDK as a flat broken line at a steady level above the cyclin's peak
One cyclin's concentration through a cycle, low through G1 and S phase, climbing through G2, peaking as mitosis begins and collapsing as it ends, with CDK as a flat broken line at a steady level above the cyclin's peak
31

The CDK’s level does not change through the cycle. The cyclin’s concentration does.

32

So active cyclin–CDK complexes appear when the cyclin’s concentration is high.

33

When the cell destroys the cyclin, the complexes come apart and vanish.

34

It is the cyclin, not the CDK, whose amount changes. A cell full of CDK and empty of cyclin has no active complexes at all.

35

What you are expected to know Say what a cyclin-dependent kinase (CDK) is.

36

What you are expected to know Predict from a graph of a cyclin’s concentration when active cyclin–CDK complexes are present.

37
Check q5

On the graph below, one cyclin’s concentration and the CDK level are drawn through a cycle, with four moments marked 1 to 4.

One cyclin's concentration against time through one cycle, the stages G1, S phase, G2 and mitosis marked on the axis, gridlines every 20 units, CDK as a flat broken line, four moments circled and numbered 1 to 4 on the curve
One cyclin's concentration against time through one cycle, the stages G1, S phase, G2 and mitosis marked on the axis, gridlines every 20 units, CDK as a flat broken line, four moments circled and numbered 1 to 4 on the curve

At which marked moment are the most active cyclin–CDK complexes present?

  1. A. Moment 1
    At moment 1, in G1, the cyclin is scarce.
    A CDK is active only with a cyclin bound, so almost no active complexes exist at moment 1.
  2. B. Moment 2
    At moment 2 the cyclin is still rising.
    Complexes are forming, but the most exist when the cyclin’s concentration peaks, at moment 3.
  3. C. ✓ Moment 3
  4. D. Moment 4
    At moment 4 the cyclin has just been destroyed.
    With no cyclin bound, the complexes have come apart.

Why: A CDK is active only while a cyclin is bound to it.
The CDK’s own level is steady all the way round, so the number of active complexes follows the cyclin’s concentration.
The cyclin’s concentration peaks at moment 3, as mitosis begins.
So moment 3 has the most active complexes.

38
Check q6

A mutant cell makes the normal amount of CDK, but its gene for one cyclin is broken, so that cyclin is missing.

How much active complex does that cyclin’s CDK form?

  1. A. ✓ None
  2. B. Half the normal amount
    Activity needs the cyclin bound to the CDK.
    One partner alone gives no active complex.
  3. C. The normal amount
    With no cyclin bound, a cyclin-dependent kinase stays inactive, however much of it there is.

Why: A CDK is active only while a cyclin is bound to it.
This cell makes none of that cyclin.
So no cyclin–CDK complexes form.
So there is no active complex, whatever the amount of CDK the cell holds.

39
Practice writing an answer

A mutant cell makes the normal amount of CDK, but its gene for one cyclin is broken, so that cyclin is missing. That cyclin’s CDK forms no active complex.

(a) Explain why the CDK forms no active complex although the cell makes the normal amount of it. (1 pt)

Model answer A CDK is a cyclin-dependent kinase.
A CDK is active only while a cyclin is bound to it.
This cell’s gene for the cyclin is broken, so the cell makes none of that cyclin.
So there is no cyclin to bind the CDK.
So no cyclin–CDK complex forms.
The amount of CDK does not matter here: a cell full of CDK and empty of cyclin has no active complexes at all.
Rubric
  • Award 1 point for: the CDK is active only with a cyclin bound, and the cell makes none of that cyclin, so no active complex forms however much CDK the cell holds.
40
Check q7

A student says: “The amount of CDK in a cell rises through G2 and falls at the end of mitosis.”

Is the student correct?

  1. A. Yes — the CDK is made in G2 and destroyed at the end of mitosis
    The CDK is present at about the same level all the way round the cycle.
    The cyclin is what climbs through G2 and collapses at the end of mitosis.
  2. B. ✓ No — the CDK’s amount stays steady; the cyclin’s amount rises and falls

Why: The CDK’s amount stays steady all the way round the cycle.
It is the cyclin whose concentration rises through G2 and collapses at the end of mitosis.
So the student has given the CDK the cyclin’s pattern.
What rises and falls is the CDK’s activity, not its amount.

41Quick quiz: cyclin-dependent kinase (CDK) mixed practice

42
Check q8

What is a cyclin-dependent kinase (CDK)?

  1. A. A protein whose concentration rises and falls through the cycle
    A protein whose concentration rises and falls through the cycle is a cyclin.
    A CDK’s level stays steady.
  2. B. An enzyme that removes phosphate groups from the proteins that start the next stage
    An enzyme that removes a phosphate group is a phosphatase.
    A CDK is a kinase: it adds a phosphate group, from ATP.
  3. C. ✓ A kinase, present at a steady level, that is active only while a cyclin is bound to it

Why: A cyclin-dependent kinase (CDK) is a kinase present at a steady level through the cycle.
It is active only while a cyclin is bound to it; that is what cyclin-dependent means.

43
Check q9

One of the two proteins that move a cell past a checkpoint is present at a steady level all the way round the cycle, and is active only while the other is bound to it.

Which protein is it?

  1. A. A cyclin
    A cyclin’s concentration rises and falls through the cycle; it is not present at a steady level.
  2. B. ✓ A cyclin-dependent kinase (CDK)

Why: A CDK is present at a steady level through the cycle.
A CDK is active only while a cyclin is bound to it.
So the protein described is a cyclin-dependent kinase (CDK).

44
Check q10

One of the two proteins that move a cell past a checkpoint is destroyed at the end of mitosis, and its concentration collapses.

Which protein is it?

  1. A. ✓ A cyclin
  2. B. A cyclin-dependent kinase (CDK)
    The CDK’s amount stays steady all the way round the cycle; it is not destroyed at the end of mitosis.

Why: A cyclin’s concentration rises and falls through the cycle.
The cell destroys the cyclin at the end of mitosis, and its concentration collapses.
So the protein described is a cyclin.

45
Practice writing an answer

A cell holds the same amount of one kinase all the way round the cycle, yet that kinase is active only as mitosis begins.

(a) State what a cyclin-dependent kinase (CDK) is. (1 pt)

Model answer A cyclin-dependent kinase (CDK) is a kinase, present at a steady level through the cycle, that is active only while a cyclin is bound to it.
Rubric
  • Award 1 point for: a kinase (adds a phosphate group from ATP to a protein), present at a steady level, active only while a cyclin is bound to it.

46How the active cyclin–CDK complex moves the cell on

47

Video: Watch: How the active complex moves the cell on

The active cyclin–CDK complex transfers a phosphate group from ATP onto the proteins that start the next stage. The added phosphate changes their shape and switches them on. So rising cyclin turns the next stage on, and destroying the cyclin turns it off.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18CC.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18CC.mp4

48

Suppose a cell at the G2 checkpoint has repaired its DNA, and its cyclin concentration is high. What moves it on into mitosis?

49

A cell moves on from a checkpoint only when the proteins that start the next stage are switched on.

50

Entry into S phase and entry into mitosis are each driven by a cyclin–CDK complex.

51

A growth signal reaching the cell makes the cell start making the cyclin that drives entry into S phase.

52

With no growth signal, the cell makes none of that cyclin.

53

Here is a drawing of how the active complex switches the next stage on.

Cyclin and CDK drawn apart, labeled inactive, and bound together, labeled an active complex; an arrow from the complex to a box labeled next-stage protein, which gains a phosphate group, labeled P, and switches on
Cyclin and CDK drawn apart, labeled inactive, and bound together, labeled an active complex; an arrow from the complex to a box labeled next-stage protein, which gains a phosphate group, labeled P, and switches on
54

The CDK in the active complex transfers a phosphate group from ATP onto a protein that starts the next stage.

55

The added phosphate changes that protein’s shape.

56

The new shape switches the protein on, until a phosphatase removes the phosphate.

57

The proteins a CDK phosphorylates are the ones that start the next stage.

58

So rising cyclin turns the next stage on. Destroying the cyclin turns it off again.

59

What you are expected to know Explain how an active cyclin–CDK complex moves a cell into the next stage.

60
Check q11

A cell at the G2 checkpoint has an active cyclin–CDK complex.

What does the active complex do to the proteins that start mitosis?

  1. A. ✓ The complex adds a phosphate group to them
  2. B. The complex destroys them
    A kinase adds a phosphate.
    It is the cyclin that is destroyed, at the end of mitosis.
  3. C. The complex copies them
    Ribosomes make proteins.
    A kinase changes a protein’s shape by adding a phosphate.
  4. D. The complex removes a phosphate group from them
    A phosphatase removes the phosphate.
    The CDK adds it, and the added phosphate switches the protein on.

Why: A CDK is a kinase, so it transfers a phosphate group from ATP onto its target proteins.
The added phosphate changes each target protein’s shape, and the new shape switches the protein on.
This complex’s targets are the proteins that start mitosis.
So the active complex switches mitosis on.

61
Check q12

A drug stops a dividing cell from destroying its cyclin at the end of mitosis, so the cyclin’s concentration stays high.

What happens to the cyclin–CDK complex?

  1. A. The complex comes apart
    Destroying the cyclin is what normally pulls the complex apart.
    The drug prevents that.
    So the cyclin stays bound.
  2. B. The complex switches off as usual
    A CDK is active for as long as a cyclin is bound to it.
    The steady CDK level is not what ends mitosis; destroying the cyclin is.
  3. C. The complex cannot form
    The cyclin bound the CDK as its concentration rose, long before the end of mitosis.
    The complex has already formed.
  4. D. ✓ The complex stays active

Why: A CDK is active for as long as a cyclin is bound to it.
Destroying the cyclin is what normally switches the complex off at the end of mitosis.
The drug stops the cyclin being destroyed, so the cyclin stays bound and the complex stays active.

62How a checkpoint hold keeps the complex off

63

Video: Watch: How a checkpoint hold works

In cells with damaged DNA, cyclin climbs from 40 to 80 units while CDK activity stays near 10%. A damage signal keeps the complex switched off however much cyclin there is; when the repair is complete the signal ends and the cells enter mitosis.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18CD.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L18CD.mp4

64

Now consider cells whose DNA was damaged after S phase, so that they are held at the G2 checkpoint.

65

Here is a graph of the hold in numbers: one cyclin’s concentration, and the CDK’s activity, over the hour after the damage.

Two stacked graphs against time after a DNA-damaging treatment, 0 to 60 minutes: cyclin concentration rises from 40 to 60 to 80 units and sits at 75 at 60 minutes; CDK activity is 15, 10 and 12 percent at 0, 20 and 40 minutes and 85 percent at 60 minutes; a broken line marks repair complete at about 55 minutes
Two stacked graphs against time after a DNA-damaging treatment, 0 to 60 minutes: cyclin concentration rises from 40 to 60 to 80 units and sits at 75 at 60 minutes; CDK activity is 15, 10 and 12 percent at 0, 20 and 40 minutes and 85 percent at 60 minutes; a broken line marks repair complete at about 55 minutes
66

Cyclin climbs from 40 units to 80 units over the first 40 minutes. Yet CDK activity stays near 10%.

67

So the cells stay in G2.

68

A damage signal kept the cyclin–CDK complex switched off while the cyclin piled up.

69

At about 55 minutes the repair is complete, and the damage signal ends.

70

CDK activity jumps to 85%, and the cells enter mitosis within minutes.

71

Plenty of cyclin did not guarantee an active complex.

72

While a checkpoint’s condition is unmet, a damage signal keeps the complex switched off, however much cyclin there is.

73

One simplification: at the M checkpoint, the hold on the sister chromatids’ separation works through a different protein complex, not a cyclin–CDK complex. In this course a cyclin–CDK complex drives entry into S phase and into mitosis.

74

What you are expected to know Explain why a cell with plenty of cyclin can still wait at a checkpoint.

75
Check q13

A researcher exposes a culture of cells that have finished S phase to a DNA-damaging treatment at 0 minutes. The graph below shows one cyclin’s concentration and CDK activity over the next 45 minutes; repair was complete by 40 minutes, and the cells entered mitosis soon after 45 minutes.

Two stacked graphs against time after a DNA-damaging treatment, 0 to 45 minutes: cyclin concentration rises from 45 to 65 to 85 units and sits at 80 at 45 minutes; CDK activity is 12, 8 and 9 percent at 0, 15 and 30 minutes and 80 percent at 45 minutes; a broken line marks repair complete at about 40 minutes
Two stacked graphs against time after a DNA-damaging treatment, 0 to 45 minutes: cyclin concentration rises from 45 to 65 to 85 units and sits at 80 at 45 minutes; CDK activity is 12, 8 and 9 percent at 0, 15 and 30 minutes and 80 percent at 45 minutes; a broken line marks repair complete at about 40 minutes

During the first 30 minutes, which of the following describes the cyclin–CDK complex?

  1. A. Not yet formed, because cyclin is low
    Cyclin climbed from 45 to 85 units in the first 30 minutes, so plenty was there to bind the CDK.
    The complex formed; its activity was held down.
  2. B. ✓ Present but switched off
  3. C. Active
    CDK activity sat near 9% for the first 30 minutes while cyclin climbed.
    So the complex was present but switched off.

Why: Cyclin climbed from 45 to 85 units over the first 30 minutes.
So plenty of cyclin was present, and it bound the CDK: the complex was present.
Over the same 30 minutes, CDK activity stayed near 9%.
So the complex was switched off.
The complex was present but switched off.

76
Practice writing an answer

A researcher exposes a culture of cells that have finished S phase to a DNA-damaging treatment. The graph below shows one cyclin’s concentration and CDK activity over the next 45 minutes. For the first 30 minutes cyclin climbed while CDK activity stayed near 9%. Repair was complete by 40 minutes; CDK activity then jumped to 80%, and the cells entered mitosis soon after.

Two stacked graphs against time after a DNA-damaging treatment, 0 to 45 minutes: cyclin concentration rises from 45 to 65 to 85 units and sits at 80 at 45 minutes; CDK activity is 12, 8 and 9 percent at 0, 15 and 30 minutes and 80 percent at 45 minutes; a broken line marks repair complete at about 40 minutes
Two stacked graphs against time after a DNA-damaging treatment, 0 to 45 minutes: cyclin concentration rises from 45 to 65 to 85 units and sits at 80 at 45 minutes; CDK activity is 12, 8 and 9 percent at 0, 15 and 30 minutes and 80 percent at 45 minutes; a broken line marks repair complete at about 40 minutes

(a) Explain why CDK activity stayed low for 30 minutes although cyclin climbed, and why it then rose. (1 pt)

Model answer The treatment damaged the DNA, and the G2 checkpoint checks for undamaged DNA before mitosis.
While the DNA was damaged, a damage signal kept the cyclin–CDK complex switched off.
Cyclin climbed from 45 to 85 units and bound the CDK, but a switched-off complex transfers no phosphate groups.
So CDK activity stayed near 9%.
By 40 minutes other proteins had repaired the DNA, so the damage signal ended.
So the complex switched on, and CDK activity jumped to 80%.
Rubric
  • Award 1 point for: a damage signal kept the cyclin–CDK complex switched off while cyclin accumulated, so activity stayed low; repair ended the signal, the complex switched on, and the cells entered mitosis.
77
Check q14

A student says: “Once a cell has made enough cyclin, the cyclin–CDK complex is active and the cell enters mitosis.”

Is the student correct?

  1. A. ✓ No — a damage signal keeps the complex switched off however much cyclin there is
  2. B. Yes — enough cyclin bound to CDK always gives an active complex
    In the damaged cells, cyclin reached 85 units while CDK activity sat at 9%.
    A damage signal kept the complex switched off until the repair was complete.

Why: In damaged cells, cyclin climbed to 85 units while CDK activity stayed near 9%.
A damage signal kept the complex switched off.
Only when the repair was complete did the complex switch on.
So while a checkpoint’s condition is unmet, a signal keeps the complex off, whatever the cyclin level.

78

Back to the graph of one protein’s concentration through one cycle: low through G1 and S phase, climbing through G2, peaking as mitosis begins and collapsing as mitosis ends. That protein is a cyclin.

79

When the cyclin’s concentration is high and the checkpoint’s condition is met, the cyclin–CDK complex is active.

80

The active complex phosphorylates the proteins that start the next stage, and those proteins switch on. That is what moves the cell on.

81

Back, too, to the healing cut: 24 dividing cells at its edge while the wound was open, and 3 after it closed.

82

While the wound was open, signals from the damaged tissue kept the edge cells making the cyclin that drives them past the G1 checkpoint.

83

When the gap closed, the signals faded. So the cells stopped making that cyclin.

84

With no cyclin, no active complexes formed, and the cells stayed in G1.

85Mixed practice mixed practice

86
Check q15

On the graph below, one cyclin’s concentration and the CDK level are drawn through a cycle, with four moments marked 1 to 4.

One cyclin's concentration against time through one cycle, the stages G1, S phase, G2 and mitosis marked on the axis, gridlines every 20 units, CDK as a flat broken line, four moments circled and numbered 1 to 4 on the curve
One cyclin's concentration against time through one cycle, the stages G1, S phase, G2 and mitosis marked on the axis, gridlines every 20 units, CDK as a flat broken line, four moments circled and numbered 1 to 4 on the curve

At which marked moment has the cyclin just been destroyed?

  1. A. Moment 1
    At moment 1 the cyclin is low because it has not yet been made in quantity; the destruction happens at the end of mitosis.
  2. B. Moment 2
    Moment 2 is the climb through G2, where the cyclin is being made faster than it is destroyed.
  3. C. Moment 3
    At the start of mitosis the cyclin is at its highest, about to be destroyed.
  4. D. ✓ Moment 4

Why: The cyclin is destroyed at the end of mitosis.
So the curve collapses at the end of the M arc.
Moment 4 sits just after that collapse, where the cyclin’s concentration has fallen back to its low level.
So at moment 4 the cyclin has just been destroyed.

87
Check q16

A week after a cut has closed, the skin cells at its edge have stopped dividing. The signals from the damaged tissue that reached them while the wound was open have faded.

Which of the following describes the state of cyclin and CDK in these cells?

  1. A. A damage signal holds the complexes switched off
    A damage signal holds complexes off while cyclin is plentiful.
    Here the growth signals that kept the cells making cyclin have gone, so the cyclin concentration is low.
  2. B. The CDK has been destroyed
    The CDK stays at a steady level all along.
    It is the cyclin, not the CDK, whose amount changes.
  3. C. ✓ Cyclin has stopped accumulating
  4. D. The complexes are active
    Without cyclin accumulating there is no active complex to switch anything on.
    The cells wait in G1 with their next-stage proteins switched off.

Why: While the wound was open, signals from the damaged tissue kept the edge cells making cyclin.
Once the wound closed, the signals faded.
So the cells stopped making that cyclin, and cyclin stopped accumulating.
With no cyclin, no active cyclin–CDK complexes formed, so the cells stayed in G1.

88
Check q17

A cyclin drives a cell into mitosis.

At which point in the cycle does the cell destroy that cyclin?

  1. A. At the end of G1
    At the end of G1 the cyclin that drives entry into mitosis is still scarce; the cell has not yet made it in quantity.
  2. B. ✓ As mitosis ends
  3. C. At the start of G2
    At the start of G2 the cyclin’s concentration is beginning to climb, not collapsing.

Why: The cyclin that drives entry into mitosis peaks as mitosis begins.
The cell destroys it as mitosis ends, and its concentration collapses.

89
Check q18

An active cyclin–CDK complex adds a phosphate group to a protein that starts the next stage.

Where does that phosphate group come from?

  1. A. From the cyclin
    The cyclin binds the CDK and switches it on; the cyclin is not the source of the phosphate.
  2. B. From a phosphatase
    A phosphatase removes phosphate groups from proteins; it does not supply them.
  3. C. ✓ From ATP

Why: A CDK is a kinase.
A kinase transfers a phosphate group from ATP onto a protein.
So the phosphate group comes from ATP.

90
Check q19

A cell at the G2 checkpoint has repaired its DNA, the damage signal has ended, and its cyclin concentration is high.

Which of the following describes the cyclin–CDK complex?

  1. A. Not yet formed, because cyclin is low
    The cyclin concentration is high, so plenty of cyclin is bound to the CDK: the complex has formed.
  2. B. ✓ Active
  3. C. Present but switched off
    The damage signal has ended, so nothing holds the complex off; with cyclin bound, the complex is active.

Why: The cyclin concentration is high, so cyclin is bound to the CDK and the complex is present.
The DNA is repaired and the damage signal has ended, so nothing keeps the complex switched off.
So the complex is active, and the cell enters mitosis.

91
Check q20

In a culture of cells, one cyclin’s concentration is high and rising, but CDK activity is near 10%.

In which stage are these cells most likely to be?

  1. A. ✓ In G2, held by a damage signal
  2. B. In mitosis, dividing
    Cells in mitosis have an active complex; CDK activity would be high, not near 10%.
  3. C. In G1, with the cyclin destroyed
    In G1 the cyclin that drives entry into mitosis is low, not high and rising.

Why: High and rising cyclin with CDK activity near 10% means the complex is present but switched off.
A damage signal keeps a complex switched off while a checkpoint’s condition is unmet.
So the cells are held in G2 by a damage signal.

92
Practice writing an answer

Researchers measured one cyclin’s concentration and the CDK’s activity at four moments in undamaged cells. The table below gives the results. The amount of CDK was the same at all four moments.

A table of one cyclin's concentration and the CDK's activity at four moments in undamaged cells: start of G2, 30 units and 5 percent; late G2, 70 units and 40 percent; start of mitosis, 90 units and 80 percent; end of mitosis, 10 units and 6 percent

(a) Explain how these results demonstrate that the CDK’s activity depends on a cyclin. (2 pt)

Model answer The amount of CDK stayed the same, yet its activity rose from 5% to 80%.
So the amount of CDK did not set its activity.
The cyclin’s concentration rose from 30 to 90 units and fell to 10 units.
CDK activity was high only while the cyclin’s concentration was high: 80% at 90 units, 5% at 30 units.
So the CDK was active only while a cyclin was bound to it: its activity depended on the cyclin.
Rubric
  • Award 1 point for: the amount of CDK did not change while its activity changed (5% to 80% to 6%), so the CDK’s own amount did not set its activity.
  • Award 1 point for: CDK activity was high only when the cyclin’s concentration was high (80% at 90 units against 5% at 30 units and 6% at 10 units), so the CDK is active only with a cyclin bound; its activity depends on the cyclin.

Glossary

cyclin
A regulatory protein whose concentration rises and falls in a repeating pattern through the cell cycle. A cyclin-dependent kinase is active only while a cyclin is bound to it.
cyclin-dependent kinase (CDK)
A kinase, present at a roughly steady level through the cycle, that is active only while a cyclin is bound to it. The active complex transfers phosphate groups from ATP onto the proteins that start the next stage of the cycle, switching them on.

APBIO-U04-L19 Cancer: division without the controls

Topic 4.6 · Regulation of Cell Cycle · 60 steps

Two pairs of bars, percent of cells entering S phase without and with a growth factor: line N 8 and 61; line M 58 and 60
Two pairs of bars, percent of cells entering S phase without and with a growth factor: line N 8 and 61; line M 58 and 60

Here are two lines of lung cells, grown with and without a growth factor.

In line N, 8% of the cells enter S phase without the growth factor and 61% with it. In line M, 58% enter S phase without the growth factor and 60% with it. Line M’s cells enter the cycle on their own, signal or no signal.

Which control has line M lost, and what happens to a tissue made of cells like that?

Unit 4 · Cell Communication and Cell Cycle

1Three controls on division

2

Video: Watch: Three controls on division

A growth signal has to arrive, the checkpoints hold the cycle until each condition is met, and the cyclin–CDK complexes switch each stage on. Three controls, each a protein made from a gene.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19a.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19a.mp4

3
Check q1

Liver cells grow in two dishes with the same nutrients. A researcher adds a growth factor to one dish only.

In which dish do more cells divide?

  1. A. ✓ The dish that received the growth factor
  2. B. The dish that received nutrients only
    The food is the same in both dishes.
    The growth factor is the signal that switches division on, so the dish with it has more dividing cells.
  3. C. Both dishes, the same number in each
    The growth factor is the signal that switches division on.
    Only one dish has it, so only that dish’s cells divide often.

Why: The food is the same in both dishes.
The growth factor is the signal that tells a cell to enter the cycle.
Only one dish has the growth factor.
So more cells divide in that dish.

4
Check q2

A cell in G2 has one damaged stretch of DNA. Repairing the damage takes nine hours.

When does the cell enter mitosis?

  1. A. At the same time as a neighboring cell with undamaged DNA
    The G2 checkpoint holds a cell with damaged DNA.
    The neighboring cell has no damage, so nothing holds it.
  2. B. ✓ When the repair is complete, after the nine hours
  3. C. After a fixed wait, whether or not the repair is complete
    A checkpoint is a set of conditions, not a timer.
    The cell waits for as long as the damage is there.

Why: The G2 checkpoint checks that the DNA is undamaged.
The damage takes nine hours to repair.
The checkpoint holds the cell for as long as the condition is unmet.
So the cell enters mitosis when the repair is complete, after the nine hours.

5
Check q3

A cyclin is bound to its CDK, and the cyclin–CDK complex is active.

Which of the following does the active complex do?

  1. A. The complex destroys the cyclin at the end of the stage
    The cell destroys the cyclin as the stage ends, and the complex comes apart.
    While active, the complex adds a phosphate group to the proteins that start the next stage.
  2. B. The complex repairs the damaged stretches of the cell’s DNA
    Repair enzymes mend the DNA.
    The active complex adds a phosphate group to the proteins that start the next stage.
  3. C. ✓ The complex adds a phosphate group to the proteins that start the next stage

Why: The CDK in the active complex transfers a phosphate group from ATP onto a protein that starts the next stage.
The added phosphate switches that protein on.
So the active complex switches the next stage on.

6

What is cancer, in the terms of this unit? The answer starts with the controls on division.

7

A change in a gene can remove one of the controls on division.

8

Cells that have lost a control keep dividing when the tissue has no use for them. So the cells pile up.

9

Three controls act on division. The first is a growth signal: a growth factor has to reach the cell before the cell enters the cycle.

A pathway of six boxes joined by arrows: growth factor, receptor, relay proteins, cyclin made, cyclin–CDK active, cell passes the G1 checkpoint
A pathway of six boxes joined by arrows: growth factor, receptor, relay proteins, cyclin made, cyclin–CDK active, cell passes the G1 checkpoint
10

The second is the checkpoints: each one holds the cycle until its condition is met.

11

The third is the cyclin–CDK complexes: the active complex switches the next stage on.

12

Here is a table comparing the three controls: what each one does, and the case that showed it.

A table of the three controls on division, what each one does, and the case that showed it: a growth signal starts the cycle, the healing cut; the checkpoints hold the cycle until a condition is met, the cell that waited nine hours in G2; the cyclin–CDK complexes switch the next stage on, the protein whose concentration rose and fell
13

Each control is a protein. Each protein is made from a gene.

14

What you are expected to know Identify the three controls on division: a growth signal, the checkpoints and the cyclin–CDK complexes.

15
Check q4

A cell at the end of G2 waits until its DNA is completely copied.

Which control is this?

  1. A. A growth signal
    A growth signal starts the cycle.
    A hold at the end of G2 until a condition is met is a checkpoint.
  2. B. ✓ A checkpoint
  3. C. A cyclin–CDK complex
    A cyclin–CDK complex switches the next stage on.
    A hold at the end of G2 until a condition is met is a checkpoint.

Why: The cell waits at the end of G2 until a condition is met: the DNA is completely copied.
A place where the cycle is held until a condition is met is a checkpoint.

16
Check q5

A skin cell at the edge of a cut enters the cycle when a molecule from the damaged tissue reaches its receptor.

Which control is this?

  1. A. ✓ A growth signal
  2. B. A checkpoint
    A checkpoint holds a cell that is already in the cycle.
    A molecule from other cells that starts the cycle is a growth signal.
  3. C. A cyclin–CDK complex
    A cyclin–CDK complex acts inside the cell.
    A molecule from other cells that starts the cycle is a growth signal.

Why: A molecule released by the damaged tissue reaches a receptor on the skin cell.
The cell then enters the cycle.
A signal from other cells that starts the cycle is a growth signal.

17
Check q6

Inside a cell, a protein adds a phosphate group to the proteins that start mitosis.

Which control is this?

  1. A. A growth signal
    A growth signal arrives from outside the cell.
    The protein that adds a phosphate group to the proteins that start the next stage is a cyclin–CDK complex.
  2. B. A checkpoint
    A checkpoint holds the cycle.
    The protein that adds a phosphate group to the proteins that start the next stage is a cyclin–CDK complex.
  3. C. ✓ A cyclin–CDK complex

Why: The active cyclin–CDK complex adds a phosphate group to the proteins that start the next stage.
Here the next stage is mitosis.
So this protein is a cyclin–CDK complex.

18How a mutation removes a control

19

Video: Watch: One changed gene, one missing control

A mutation changes a gene, and the gene’s protein changes with it. A checkpoint protein that cannot hold lets damaged DNA be copied. A signaling protein locked in its active shape drives division with no growth factor. The daughters inherit the fault.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19b.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19b.mp4

20

A mutation changes a gene. So a mutation can change or remove what the gene’s protein does.

21

Now consider a mutation in the gene for a checkpoint protein. The changed protein cannot hold the cell.

22

So the cell passes a checkpoint it should have failed. It copies damaged DNA.

23

Now consider a mutation in the gene for a signaling protein. The changed protein is locked in its active shape.

24

A growth-factor receptor locked in its active shape sends the message with no growth factor bound. So the cell divides with no growth factor present.

The same pathway with the receptor locked in its active shape: the growth-factor box is empty and the step from it to the receptor is drawn as a break, yet the arrows from the receptor through the relay to cyclin and the G1 checkpoint are unbroken
The same pathway with the receptor locked in its active shape: the growth-factor box is empty and the step from it to the receptor is drawn as a break, yet the arrows from the receptor through the relay to cyclin and the G1 checkpoint are unbroken
25

The daughter cells inherit the same faulty gene. So the daughter cells carry the same missing control, and so do their daughters after them.

26

One simplification: we treat ‘a checkpoint protein’ as one protein doing the holding. In a real cell several proteins share the work of detecting damage and holding the cycle.

27

What you are expected to know Explain how a mutation in the gene for a control protein removes that control.

28

What you are expected to know Explain why the daughter cells inherit the missing control.

29
Check q7

Two lines of lung cells are grown with and without a growth factor. In line N, 8% of the cells enter S phase without the growth factor and 61% with it. In line M, 58% enter S phase without the growth factor and 60% with it. Line M carries a mutation.

Which control has line M lost?

  1. A. ✓ The requirement for a growth signal before the cycle starts
  2. B. The M checkpoint’s check that every chromosome is attached
    The measurement is entry into S phase, which the growth signal controls.
    The M checkpoint acts much later, at metaphase.
  3. C. The copying of the DNA during S phase
    Line M’s cells copy DNA well.
    The fault is that they start to without waiting for the signal.
  4. D. The destruction of cyclin at the end of mitosis
    Nothing here measures mitosis.
    Line M enters S phase whether or not the growth factor is present.

Why: Line N waits for the growth factor: 8% enter S phase without it, 61% with it.
Line M enters S phase either way: 58% and 60%.
So line M’s cells no longer wait for the growth factor.
So line M has lost the requirement for a growth signal.

30
Practice writing an answer

Two lines of kidney cells are grown with and without a growth factor. In line P, 6% of the cells enter S phase without the growth factor and 59% with it. In line Q, which carries a mutation, 55% enter S phase without the growth factor and 57% with it.

(a) Explain how the counts show that line Q has lost the requirement for a growth signal. (1 pt)

Model answer Line P waits for the growth factor: 6% of its cells enter S phase without it and 59% with it.
Line Q enters S phase either way: 55% without the growth factor and 57% with it.
So the growth factor makes almost no difference to line Q.
So line Q’s cells enter the cycle with no growth factor bound: the requirement for a growth signal is lost.
Rubric
  • Award 1 point for: line Q enters S phase at almost the same rate with and without the growth factor (55% and 57%) while line P needs it (6% against 59%), so line Q’s cells enter the cycle with no growth signal.

31A tumor, and cancer

32

Video: Watch: The lump, and the disease

Cells that have lost a control keep dividing where the tissue has no use for them, and pile up into a lump: a tumor. The disease of division without the controls is cancer. Its cells divide when they should not, not faster.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19c.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19c.mp4

33

Cells that have lost a control keep dividing where and when the tissue has no use for them. So the cells pile up into a lump.

A sheet of cells one cell thick, and on part of it a mound of extra cells three layers high, labeled: cells that kept dividing pile up into a lump
A sheet of cells one cell thick, and on part of it a mound of extra cells three layers high, labeled: cells that kept dividing pile up into a lump
34

A lump like this is called a .

35

The disease in which cells divide without the controls is called .

36

A cancer cell’s cycle can take as long as a normal cell’s cycle. Cancer cells do not divide faster because they are cancer.

37

Cancer cells divide when they should not, because a control that would have stopped them is missing.

38

What you are expected to know Describe a tumor as a lump of cells that kept dividing where the tissue has no use for them, and cancer as the disease in which cells divide without the controls.

39
Check q8

A student says: “Cancer cells are cells that go round the cell cycle faster than normal cells.”

Is the student correct?

  1. A. Yes — a cancer cell completes each cycle in less time than a normal cell
    A cancer cell’s cycle can take as long as a normal cell’s cycle.
    It divides when the tissue has no use for new cells, because a control is missing.
  2. B. ✓ No — cancer cells divide when they should not, because a control is missing

Why: Cancer is division without the controls.
A cancer cell’s cycle can take as long as a normal cell’s.
It divides when the tissue has no use for new cells, because a control that would have stopped it is missing.
So a cancer cell divides when it should not, not faster.

40
Practice writing an answer

A tumor in a patient’s skin doubles in size over a year. Its cells each complete a cell cycle in about 24 hours, the same time as the normal skin cells beside them. The normal skin stays the same thickness all year.

(a) Explain how the tumor grows although its cells cycle no faster than the normal skin cells. (1 pt)

Model answer A normal skin cell enters the cycle only when a growth signal reaches it and every checkpoint condition is met.
So normal skin cells divide only when the skin needs new cells.
The tumor’s cells have lost a control.
So they enter the cycle when the skin has no use for new cells.
So extra cells pile up.
So the tumor grows, although each cycle takes 24 hours, the same as a normal cell’s.
Rubric
  • Award 1 point for: the tumor’s cells enter the cycle when the tissue has no use for new cells because a control is missing, so extra cells pile up; the rate of growth comes from how many cells divide when they should not, not from a shorter cycle.
41

Back to the two lines of lung cells grown with and without a growth factor. In line N, 8% of the cells entered S phase without the growth factor and 61% with it.

42

Line N’s cells waited for the signal. So line N still has the growth-signal control.

43

In line M, 58% entered S phase without the growth factor and 60% with it. Line M’s cells did not wait for the signal.

44

So a mutation has removed line M’s requirement for a growth signal: a signaling protein in its cells sends the message with no growth factor bound.

45

Every daughter cell inherits the fault. So a tissue made of cells like line M’s keeps dividing where the tissue has no use for the cells, and the cells pile up into a tumor.

46Quick quiz: tumor, cancer mixed practice

47
Check q9

A lump of cells has formed in a tissue. The cells kept dividing where the tissue had no use for them.

Which of the following is the lump called?

  1. A. ✓ A tumor
  2. B. A tissue
    A tissue is a group of cells the body has a use for.
    A lump of cells the tissue has no use for is a tumor.
  3. C. A growth factor
    A growth factor is a signal molecule.
    A lump of cells the tissue has no use for is a tumor.

Why: The cells kept dividing where the tissue had no use for them, and piled up into a lump.
A lump like this is called a tumor.

48
Check q10

In a patient, a control on the cycle has been lost. Cells keep dividing where the tissue has no use for them.

Which of the following is this disease called?

  1. A. Apoptosis
    Apoptosis is a cell dismantling itself in an orderly way.
    The disease in which cells divide without the controls is cancer.
  2. B. ✓ Cancer

Why: The cells divide without the controls on the cycle.
The disease in which cells divide without the controls is called cancer.

49
Practice writing an answer

Division is controlled by a growth signal, the checkpoints and the cyclin–CDK complexes.

(a) State what a tumor is. (1 pt)

Model answer A tumor is a lump of cells that kept dividing where the tissue had no use for them.
Rubric
  • Award 1 point for: a lump of cells that kept dividing where (or when) the tissue had no use for them.

(b) State what cancer is. (1 pt)

Model answer Cancer is the disease in which cells divide without the controls on the cycle.
Rubric
  • Award 1 point for: the disease in which cells divide without the controls on the cell cycle (a control has been lost).
50
Check q11

A student says: “A tumor is a lump of cells that divide faster than normal cells.”

Is the student correct?

  1. A. Yes — each of a tumor’s cells completes its cycle in less time than a normal cell
    A tumor cell’s cycle can take as long as a normal cell’s cycle.
    The lump forms because the cells divide when the tissue has no use for them.
  2. B. ✓ No — a tumor’s cells divide when they should not, because a control is missing

Why: A tumor’s cells have lost a control.
So they divide when the tissue has no use for them, and pile up.
Their cycle can take as long as a normal cell’s cycle.

51
Check q12

A cell in a tissue is a cancer cell.

Which of the following makes it a cancer cell?

  1. A. ✓ The cell divides when it should not
  2. B. The cell completes each cycle in less time than a normal cell
    A cancer cell’s cycle can take as long as a normal cell’s cycle.
    What makes it a cancer cell is dividing when it should not.
  3. C. The cell is larger than the normal cells around it
    Size is checked at the G1 checkpoint and does not make a cell a cancer cell.
    What makes it a cancer cell is dividing when it should not.

Why: Cancer is division without the controls.
A cancer cell has lost a control that would have stopped it.
So it divides when it should not.

52
Check q13

After an injury, a tissue is short of cells. Its cells divide until the gap is closed, then stop.

Is this a tumor?

  1. A. Yes
    The tissue has a use for these cells, and they stop when the gap is closed.
    A tumor is cells the tissue has no use for, which keep coming.
  2. B. ✓ No

Why: The tissue is short of cells, so it has a use for the new cells.
The cells stop dividing when the gap is closed.
So the controls are working, and this is not a tumor.

53
Check q14

In a gut lining that is already complete, a group of cells keeps dividing and piles up into a lump.

Is this a tumor?

  1. A. ✓ Yes
  2. B. No
    The lining is complete, so the tissue has no use for these cells.
    Cells piling up where the tissue has no use for them are a tumor.

Why: The lining is complete, so the tissue has no use for new cells.
The cells keep dividing anyway and pile up into a lump.
So this is a tumor.

54Mixed practice mixed practice

55
Check q15

Cells are growing in a dish with a growth factor. A researcher washes the growth factor out.

Predict where the cells collect over the next day.

  1. A. At metaphase
    A missing growth factor acts long before any spindle exists.
    The cells never start the cycle.
  2. B. In G2
    Cells already past G1 finish their cycle.
    Cells that arrive in G1 have no signal to go on, so they rest there.
  3. C. In telophase
    The growth factor controls entry into the cycle, not the division of the cytoplasm.
  4. D. ✓ In G1 or G0

Why: The growth factor, the signal to enter the cycle, has been washed out.
So cells that reach G1 have no signal to pass the G1 checkpoint.
So those cells rest in G1 or drop into G0.
Nothing piles up inside the cycle: it is simply not entered.

56
Check q16

A cell carries a mutation in the checkpoint protein that detects damaged DNA. So no checkpoint holds this cell when its DNA is damaged. Its DNA is then damaged.

What happens to the damage?

  1. A. The damage is repaired at the G2 checkpoint instead
    No checkpoint holds this cell, so the G2 checkpoint does not hold it either.
    By then S phase has already copied the damaged DNA, so both daughters receive the damage.
  2. B. The damage stops the cell dividing for good
    Stopping is exactly what this cell’s checkpoints can no longer do: no checkpoint detects the damage.
  3. C. ✓ The damage is copied in S phase and passed to both daughter cells
  4. D. The damage is removed when the cell drops into G0
    Dropping into G0 is a response at the G1 checkpoint, and this cell’s G1 checkpoint never detects the damage.

Why: The G1 checkpoint is what keeps damaged DNA from being copied.
This cell’s checkpoint protein does not detect the damage, so no checkpoint holds the cell.
So the cell enters S phase carrying the damage.
S phase copies the DNA, damage included.
So both daughter cells receive the damage.

57
Check q17

A cell carries a mutation that locks a growth-factor receptor in its active shape. The cell sits in a dish with no growth factor in it.

Does the cell enter the cycle?

  1. A. ✓ Yes
  2. B. No
    The locked receptor sends the message with no growth factor bound.
    So the cell enters the cycle without the signal.

Why: A receptor locked in its active shape sends the message with no growth factor bound.
The message ends at the G1 checkpoint.
So the cell enters the cycle although the dish has no growth factor.

58
Check q18

Cells from a tumor and normal cells from the same tissue each complete one cell cycle in 22 hours.

Which of the following is the difference between the two kinds of cell?

  1. A. ✓ The tumor cells enter the cycle when the tissue has no use for them
  2. B. The tumor cells complete each cycle in less time
    Both kinds of cell take 22 hours for one cycle.
    The tumor cells differ in when they enter the cycle, not in how long it takes.
  3. C. The tumor cells copy their DNA twice in each cycle
    Every cell copies its DNA once per cycle, in S phase.
    The tumor cells differ in when they enter the cycle.

Why: Both kinds of cell take 22 hours for one cycle.
The tumor cells have lost a control.
So the tumor cells enter the cycle when the tissue has no use for new cells.

59
Practice writing an answer

Researchers grow cells from a patient’s tumor and normal cells from the same tissue in dishes with no growth factor. After two days, 3% of the normal cells and 49% of the tumor cells have divided. Every cell that divided took about 22 hours to complete its cycle.

(a) Explain how these results demonstrate that cancer is division with a control missing, rather than faster division. (1 pt)

Model answer Every cell that divided took about 22 hours, tumor cell or normal cell.
So the tumor cells did not cycle faster.
Only 3% of the normal cells divided, because a normal cell waits for a growth signal.
49% of the tumor cells divided with no growth factor present.
So the tumor cells entered the cycle without the signal: the growth-signal control is lost.
So the difference is how many cells divided, not how fast each cycled.
Rubric
  • Award 1 point for: the equal cycle time (about 22 hours) rules out faster division, AND the 49% against 3% with no growth factor shows the tumor cells enter the cycle without the growth signal, so a control is missing.

Glossary

tumor
A lump of cells that kept dividing where and when the tissue had no use for them, because a control on the cell cycle has been lost.
cancer
The disease in which cells divide without the controls on the cycle. A mutation has removed a control, and the daughter cells inherit the fault.

APBIO-U04-L19B The other outcome

Topic 4.6 · Regulation of Cell Cycle · 37 steps

Two bars, percent of leukemia cells undergoing apoptosis one day after the same DNA-damaging drug: drug-sensitive cells 64, drug-resistant cells 11
Two bars, percent of leukemia cells undergoing apoptosis one day after the same DNA-damaging drug: drug-sensitive cells 64, drug-resistant cells 11

Researchers gave two kinds of leukemia cell the same DNA-damaging drug.

After a day, 64% of the drug-sensitive cells had dismantled themselves by apoptosis. Only 11% of the resistant cells had. Both kinds carried the same damage, and both kinds had working checkpoints.

Yet 58% of the resistant cells had completed a division by the next day. What did the resistant cells lose, and why did they divide?

Unit 4 · Cell Communication and Cell Cycle

1A damaged cell removes itself

2

Video: Watch: The second outcome

A checkpoint holds the damaged cell while repair enzymes mend the DNA. When the damage cannot be repaired, the held cell is signaled to undergo apoptosis: it dismantles itself into packages, its neighbors clear them away, and the damage is never copied.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19Ba.mp4

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3
Check q1

In an embryo’s hand, the cells of the webbing between the fingers take themselves apart in an orderly way.

Which of the following is this called?

  1. A. ✓ Apoptosis
  2. B. Mitosis
    Mitosis is the division of a nucleus into two.
    A cell taking itself apart in an orderly way is apoptosis.
  3. C. Cytokinesis
    Cytokinesis is the division of the cytoplasm into two cells.
    A cell taking itself apart in an orderly way is apoptosis.

Why: The webbing cells take themselves apart in an orderly way as a normal part of development.
A cell doing this is undergoing apoptosis.

4

What happens to a cell whose DNA damage cannot be repaired? Start with what a checkpoint does.

5

A checkpoint holds the cell. The checkpoint does not repair the damage.

6

Repair enzymes do the mending while the hold lasts.

7

Suppose the repair succeeds. The condition is met, and the cell moves on.

8

Now suppose the damage cannot be repaired. The held cell is signaled to undergo apoptosis.

9

In apoptosis the cell dismantles itself in an orderly way. The cell packages up its own contents, and its neighbors clear the packages away.

A cell held at a checkpoint with one damaged stretch of DNA, and two arrows leading from it: upward, repair enzymes mend the DNA and the cell moves on; downward, the damage cannot be repaired, the cell dismantles itself into six small packages, and two neighboring cells clear the packages away, so the damage is never copied
A cell held at a checkpoint with one damaged stretch of DNA, and two arrows leading from it: upward, repair enzymes mend the DNA and the cell moves on; downward, the damage cannot be repaired, the cell dismantles itself into six small packages, and two neighboring cells clear the packages away, so the damage is never copied
10

So the cell never divides. The damage is never copied into a daughter cell.

11

A damaged cell has two possible outcomes. In the first, the cell goes on dividing, and a tumor can grow.

12

In this second outcome, the cell is removed, and the tissue is protected.

13

What you are expected to know Describe apoptosis as the outcome for damage that cannot be repaired: the held cell is signaled to dismantle itself, so the damage is never copied into a daughter cell.

14
Check q2

A skin cell exposed to ultraviolet light carries DNA damage beyond repair.

Which outcome protects the skin?

  1. A. The cell enters S phase and copies over the damaged stretch with a clean copy
    Copying damaged DNA makes two damaged copies.
    The checkpoint exists to stop exactly that.
  2. B. The cell divides quickly, so that healthy daughter cells can replace it in the skin
    Its daughters would inherit the damage.
    The protective outcome is that the cell never divides.
  3. C. ✓ Before it can divide, the cell packages up its contents for its neighbors to clear away
  4. D. Before it can divide, the cell bursts and spills its contents for its neighbors to clear away
    A burst cell spills its contents onto its neighbors and harms them.
    In apoptosis the cell packages its contents, and its neighbors clear the packages away.

Why: This skin cell carries damage beyond repair, so it is signaled to undergo apoptosis.
In apoptosis the cell packages up its contents, and its neighbors clear the packages away.
So the cell never divides.
So the damage is never copied into daughter cells, and the skin is protected.

15
Check q3

A student says: “When a damaged skin cell undergoes apoptosis, its contents spill onto its neighbors and harm them.”

Is the student correct?

  1. A. ✓ No — the cell packages its contents, and its neighbors clear the packages away
  2. B. Yes — the dying cell bursts and spills its contents onto its neighbors
    Apoptosis is orderly.
    The cell dismantles itself and packages up its own contents.
    Its neighbors clear the packages away.
    So nothing spills, and the neighbors are not harmed.

Why: Apoptosis is a cell dismantling itself in an orderly way.
The cell packages up its own contents, and its neighbors clear the packages away.
So nothing spills onto the neighbors, and they are not harmed.

16When the apoptosis response is lost

17

Video: Watch: The cell that was not removed

Two kinds of leukemia cell carry the same damage. The sensitive cells dismantle themselves; the resistant cells have lost the apoptosis response and survive. In the resistant cells, which were not removed, the hold lapsed after about a day, the cells divided, and both daughters carried the damage.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19Bb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19Bb.mp4

18

Now consider two kinds of leukemia cell given the same DNA-damaging drug. Here is a graph of the cells undergoing apoptosis after one day: 64% of the drug-sensitive cells, and 11% of the drug-resistant cells.

Two bars: percent of leukemia cells undergoing apoptosis one day after the same DNA-damaging drug, 64 for drug-sensitive cells and 11 for drug-resistant cells, with ±2SE error bars of 5 and 3; gridlines every 10; the legend reads: error bars represent ±2SE
Two bars: percent of leukemia cells undergoing apoptosis one day after the same DNA-damaging drug, 64 for drug-sensitive cells and 11 for drug-resistant cells, with ±2SE error bars of 5 and 3; gridlines every 10; the legend reads: error bars represent ±2SE
19

Both kinds of cell carried the same damage. Both kinds had working checkpoints, so both kinds were held.

20

The sensitive cells were signaled to undergo apoptosis. So 64% of them had dismantled themselves within a day.

21

The resistant cells had lost the apoptosis response. So the resistant cells survived damage they should not have survived.

22

In the resistant cells the hold did not last. After about a day the held cells resumed the cycle and divided.

23

Both daughter cells carried the damage. That is the route to a tumor.

24

What you are expected to know Explain what happened to the damaged cells that had lost the apoptosis response.

25
Check q4

A researcher gives drug-sensitive and drug-resistant leukemia cells the same DNA-damaging drug. Both kinds carry the same damage after 12 hours. After 24 hours, 64% of the sensitive cells had undergone apoptosis and 11% of the resistant cells had; 58% of the resistant cells had completed a division.

What have the resistant cells lost?

  1. A. The ability to copy their DNA before they divide
    The resistant cells completed divisions, so they copied their DNA.
    What they skipped was removing themselves.
  2. B. ✓ The response that removes a damaged cell before it divides
  3. C. The enzymes that repair the damage to their DNA
    Both kinds of cell carried the same damage at 12 hours.
    The difference came afterward, in whether the damaged cell dismantled itself.
  4. D. The receptor that lets the drug into the cell
    The resistant cells were damaged as much as the sensitive ones, so the drug reached them.
    The lost step is the response to the damage.

Why: Both kinds of cell carried the same damage at 12 hours.
The sensitive cells were signaled to undergo apoptosis and dismantled themselves.
The resistant cells had lost that response, so they survived.
So the resistant cells divided with the damage still in them: 58% had completed a division.

26
Practice writing an answer

A researcher gives drug-sensitive and drug-resistant leukemia cells the same DNA-damaging drug. Both kinds carry the same damage after 12 hours. After 24 hours, 64% of the sensitive cells had undergone apoptosis and 11% of the resistant cells had; 58% of the resistant cells had completed a division. The resistant cells have lost the response that removes a damaged cell before it divides. In a cell with no apoptosis response, the checkpoint hold on damage beyond repair lapses after about a day.

(a) Explain why 58% of the resistant cells had completed a division although they carried the damage. (1 pt)

Model answer A held cell with damage beyond repair is normally signaled to undergo apoptosis.
The sensitive cells received that signal and dismantled themselves: 64% did.
The resistant cells have lost the apoptosis response, so they were not removed.
In a cell that is not removed, the hold lapses after about a day and the cycle resumes.
So the resistant cells copied their damaged DNA and divided: 58% had completed a division.
Rubric
  • Award 1 point for: the resistant cells were not removed by apoptosis, so the hold did not last and they resumed the cycle and divided with the damage still in their DNA, passing it to both daughters.
27
Check q5

A mutation removes a cell’s apoptosis response but leaves its checkpoints working. The cell’s DNA is then damaged beyond repair. In a cell with no apoptosis response, a checkpoint hold lapses after about a day.

Predict what happens to the cell.

  1. A. ✓ The cell divides and passes the damage to both daughters
  2. B. The cell dismantles itself in an orderly way
    Dismantling itself in an orderly way is apoptosis, and this cell has no apoptosis response.
    Instead the hold lapses after a day and the cell divides, damage and all.
  3. C. The cell is held at a checkpoint and stays there, still carrying the damage
    The checkpoint holds the damaged cell at first.
    With no apoptosis response, the hold lapses after about a day.
    Then the cell copies its damaged DNA and divides.
  4. D. The cell repairs the damage while it waits at the checkpoint
    The hold gives repair enzymes time to work, but this damage cannot be repaired, so it stays in the cell.
    When the hold lapses, the cell divides, damage and all.

Why: The checkpoint holds the damaged cell at first, but a checkpoint does not repair.
With no apoptosis response, the held cell is not removed.
After about a day the hold lapses and the cell resumes the cycle.
So it copies its damaged DNA and divides: both daughters carry the damage.

28

Back to the two kinds of leukemia cell given the same DNA-damaging drug. Both kinds carried the same damage, and both kinds were held at a checkpoint.

29

The sensitive cells were signaled to undergo apoptosis. So 64% of them dismantled themselves within a day, and their damage was never copied.

30

The resistant cells had lost the apoptosis response. So only 11% of them dismantled themselves.

31

In the rest of the resistant cells, the hold lapsed after about a day. So 58% had completed a division by the next day, and both daughters of each carried the damage.

32Mixed practice mixed practice

33
Check q6

A skin cell’s DNA is damaged beyond repair. Its checkpoints and its apoptosis response both work.

Which of the following happens to the damage?

  1. A. The damage is copied into both daughter cells
    The held cell is signaled to undergo apoptosis before it divides.
    So the damage is never copied.
  2. B. The damage is repaired by the checkpoint that holds the cell
    A checkpoint holds; it does not repair.
    This damage is beyond repair, so the held cell dismantles itself.
  3. C. ✓ The damage is never copied

Why: The checkpoint holds the damaged cell.
The damage cannot be repaired, so the held cell is signaled to undergo apoptosis.
The cell dismantles itself before it divides.
So the damage is never copied.

34
Check q7

A checkpoint holds a cell whose DNA is damaged.

Which of the following repairs the damage?

  1. A. The checkpoint itself
    A checkpoint holds the cell; it does not repair.
    Repair enzymes mend the DNA while the hold lasts.
  2. B. ✓ Repair enzymes, while the hold lasts
  3. C. The daughter cells, after the division
    A cell that divided with the damage passes it to both daughters.
    Repair happens during the hold, before any division.

Why: A checkpoint holds the cell.
The hold gives repair enzymes time to work.
So repair enzymes mend the damage while the hold lasts.

35
Check q8

Researchers grew cells from a patient’s lump in two dishes and gave both dishes a drug that damages DNA. To one dish the researchers also added a second drug, which blocks the checkpoint protein that detects damaged DNA. The table below compares the two dishes after two days.

A table comparing the lump's cells after two days of the DNA-damaging drug alone and after the drug plus a checkpoint blocker: living cells in G2, 50 percent and 23 percent; living cells copying DNA, 17 percent and 37 percent; DNA damage per cell, high and just as high; cells that died, 31 percent and 9 percent

Why did fewer cells die in the dish given both drugs?

  1. A. The second drug repaired the DNA damage in the cells before they divided
    The damage stayed as high as before; only the response to it changed.
  2. B. ✓ Damaged cells were no longer held at G2; they divided instead of removing themselves
  3. C. The cells left the cycle into G0, where the damage does them no harm
    The cells kept copying DNA (37% in S phase) and dividing; they stayed in the cycle.
  4. D. The second drug stopped the first drug from entering the cells
    The damage measured the same, so the first drug reached the cells as before.

Why: The second drug blocked the checkpoint protein that detects damaged DNA.
So the damage was never detected: no hold in G2, and no signal to remove the cell.
The damaged cells carried on into mitosis.
So far fewer cells died, and the damage passed to daughter cells.

36
Practice writing an answer

Researchers give drug-resistant leukemia cells a second drug that restores the apoptosis response. They then give the cells the same DNA-damaging drug as before. Without the second drug, 11% of the resistant cells had undergone apoptosis after a day, and 58% had completed a division.

(a) Predict how the percent of resistant cells that complete a division changes with the second drug, and justify your prediction. (1 pt)

Model answer Far fewer than 58% complete a division.
The DNA-damaging drug damages the cells’ DNA as before, and the working checkpoints hold the cells.
With the apoptosis response restored, a held cell whose damage cannot be repaired is signaled to undergo apoptosis.
So the held cells dismantle themselves instead of surviving the hold.
A cell that has dismantled itself never divides.
So far more than 11% undergo apoptosis, and far fewer than 58% complete a division.
Rubric
  • Award 1 point for: the prediction (far fewer cells complete a division) AND the reason: with the apoptosis response restored, the held damaged cells are signaled to dismantle themselves, so they are removed before the hold lapses and cannot divide.

Slip Predicting no change because the DNA damage is the same. The damage is the same, but the response to it is not: the restored response removes the damaged cells before they divide.

APBIO-U04-L19C Break a control and predict

Topic 4.6 · Regulation of Cell Cycle · 70 steps

Three small bar charts of the lump's cells by stage, G1, S phase, G2 and mitosis: before the drug 30, 40, 20 and 10 percent with 5 percent dead; after two days of the drug 21, 17, 50 and 12 percent with 31 percent dead; drug plus a checkpoint blocker 28, 37, 23 and 12 percent with 9 percent dead
Three small bar charts of the lump's cells by stage, G1, S phase, G2 and mitosis: before the drug 30, 40, 20 and 10 percent with 5 percent dead; after two days of the drug 21, 17, 50 and 12 percent with 31 percent dead; drug plus a checkpoint blocker 28, 37, 23 and 12 percent with 9 percent dead

Here are three sets of counts from a patient’s lump: before a drug, after two days of the drug, and after the drug plus a second drug that blocks the checkpoint protein that detects damaged DNA.

Before treatment, 40% of the lump’s cells were copying DNA. After the drug, 17% were, half of the living cells sat in G2, and 31% had died. When the doctors added the checkpoint-blocking drug as well, the pile-up in G2 vanished and only 9% died.

What did each drug do, and how can you tell from where the cells piled up?

Unit 4 · Cell Communication and Cell Cycle

1Follow the control to the stage it governs

2

Video: Watch: Where the cells collect

Break one control and the cells collect at the step that control would have let them pass. A spindle inhibitor: metaphase. A cyclin that cannot be destroyed: mitosis never ends. An inhibitor of the complex that drives entry into mitosis: G2. A growth factor removed: the cycle is never entered.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19Ca.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19Ca.mp4

3
Check q1

Researchers sort 100 cycling cells by stage. 46 of them are in G2, against 20 in an untreated sample of 100.

Which of the following does the rise in G2 show?

  1. A. ✓ The cells spend longer in G2
  2. B. The cells spend less time in G2
    A stage that holds more of the cells at one moment is a stage the cells spend longer in.
  3. C. The cells copy their DNA in G2
    DNA is copied in S phase, before G2.
    More cells in G2 means the cells spend longer in G2.

Why: The share of cells found in a stage matches the share of the cycle spent in that stage.
More cells are in G2 than before.
So the cells spend longer in G2.

4

How do you predict what a disruption to the controls does? Follow the control to the stage it governs.

5

Cells collect at the step that control would have let them pass.

6

Now consider a spindle inhibitor: a drug that stops the spindle fibers from moving the chromosomes.

7

No chromosome is attached from both poles. So the M checkpoint’s condition is never met.

8

So cells collect at metaphase with their sisters still joined.

The cell-cycle wheel of sectors, G1, S phase, G2, mitosis and cytokinesis, with the three checkpoints drawn as bars across the rim, and a cluster of seven cells drawn just outside the rim before the M checkpoint; labeled: cells collect at metaphase
The cell-cycle wheel of sectors, G1, S phase, G2, mitosis and cytokinesis, with the three checkpoints drawn as bars across the rim, and a cluster of seven cells drawn just outside the rim before the M checkpoint; labeled: cells collect at metaphase
9

Now consider a cyclin that cannot be destroyed. The complex that drives entry into mitosis stays active, so cells cannot finish mitosis.

10

Here is a graph of the cells that had re-formed a nuclear envelope 30 minutes after mitosis began: 86% of normal cells, and 8% of the cells with the cyclin that cannot be destroyed.

Three bars: percent of cells that had re-formed a nuclear envelope 30 minutes after mitosis began, 86 for normal cells, 8 for cells with a cyclin that cannot be destroyed, and 82 for those cells when the cyclin is also stopped from binding CDK, with ±2SE error bars of 4, 3 and 4; gridlines every 20; the legend reads: error bars represent ±2SE
Three bars: percent of cells that had re-formed a nuclear envelope 30 minutes after mitosis began, 86 for normal cells, 8 for cells with a cyclin that cannot be destroyed, and 82 for those cells when the cyclin is also stopped from binding CDK, with ±2SE error bars of 4, 3 and 4; gridlines every 20; the legend reads: error bars represent ±2SE
11

Suppose the researcher also stops that cyclin from binding CDK. Then 82% of the cells re-form their envelopes.

12

So the active complex, not the leftover cyclin, held the cells in mitosis.

13

Now consider an inhibitor of a cyclin–CDK complex. Which stage fills depends on which complex the inhibitor blocks.

14

Here a researcher gave embryo cells an inhibitor of the complex that drives entry into mitosis. Here is a graph of the cells by stage: G2 rose from 20% to 46%, and mitosis fell from 30% to 6%.

Two stacked panels of four bars, percent of embryo cells in G1, S phase, G2 and mitosis: untreated 22, 28, 20 and 30; after an inhibitor of the complex that drives entry into mitosis 21, 27, 46 and 6; gridlines every 10 percent
Two stacked panels of four bars, percent of embryo cells in G1, S phase, G2 and mitosis: untreated 22, 28, 20 and 30; after an inhibitor of the complex that drives entry into mitosis 21, 27, 46 and 6; gridlines every 10 percent
15

That complex stays switched off. So cells collect in G2, and mitosis empties.

16

An inhibitor of the complex that drives entry into S phase does the same thing one stage earlier. That complex stays switched off, so cells collect in G1 and S phase empties.

17

Now consider a growth factor removed. No signal tells the cells to enter the cycle, so cells rest in G1 or drop into G0.

The cell-cycle wheel of sectors, G1, S phase, G2, mitosis and cytokinesis, with the three checkpoints drawn as bars across the rim, an arrow leaving G1 to a box labeled G0, out of the cycle but still working, and an arrow from G0 back into G1; labeled: no growth factor, cells rest in G1 or G0
The cell-cycle wheel of sectors, G1, S phase, G2, mitosis and cytokinesis, with the three checkpoints drawn as bars across the rim, an arrow leaving G1 to a box labeled G0, out of the cycle but still working, and an arrow from G0 back into G1; labeled: no growth factor, cells rest in G1 or G0
18

Nothing piles up inside the cycle. The cycle is simply not entered.

19

Here is a table comparing the five disruptions: the control each one breaks, the stage that fills, and whether division continues.

A table of five disruptions to the controls: the control each breaks, the stage that fills, and whether division continues. A spindle inhibitor breaks the M checkpoint's condition, cells collect at metaphase, division stops. A cyclin that cannot be destroyed keeps the complex that drives entry into mitosis active, cells stay in mitosis, division stops. An inhibitor of the complex that drives entry into mitosis, cells collect in G2, division stops. An inhibitor of the complex that drives entry into S phase, cells collect in G1, division stops. A growth factor removed breaks the growth signal, cells rest in G1 or G0 with no pile-up inside the cycle, division stops
20

What you are expected to know Predict, for a described disruption to a control, where the cells collect and whether division continues, by following the control to the stage it governs.

21
Check q2

A researcher gives dividing embryo cells a drug that blocks one cyclin–CDK complex: the complex whose cyclin peaks at the end of G1.

Which step does that complex normally switch on?

  1. A. ✓ Entry into S phase
  2. B. Entry into mitosis
    The complex that drives entry into mitosis has its cyclin peak at the end of G2.
    The blocked complex’s cyclin peaks at the end of G1, at S-phase entry.
  3. C. The separation of the sisters
    The sisters separate at metaphase, in the middle of mitosis.
    The blocked complex’s cyclin peaks at the end of G1, long before mitosis, where the cell enters S phase.

Why: A complex is active while its cyclin concentration is high.
The blocked complex’s cyclin peaks at the end of G1, so it acts there.
An active complex switches on the next stage, and the stage after G1 is S phase.
So it normally switches on entry into S phase.

22
Check q3

A researcher gives dividing embryo cells a drug that keeps one cyclin–CDK complex switched off: the complex whose cyclin peaks at the end of G1.

Which stage fills up?

  1. A. G2
    Cells collect at the step the blocked control would have let them pass.
    The blocked step is entry into S phase, at the end of G1, so G1 fills.
  2. B. Metaphase
    A cell that cannot enter S phase never copies its DNA and never reaches mitosis.
    So the cells collect in G1.
  3. C. ✓ G1

Why: Cells collect at the step the blocked control would have let them pass.
The blocked control is entry into S phase, at the end of G1.
So the cells reach the end of G1 and stop there.
So G1 fills up, and S phase empties.

23
Check q4

A researcher gives dividing embryo cells a drug that keeps the end-of-G1 cyclin–CDK complex switched off.

Does division continue?

  1. A. Yes
    A cell held in G1 never enters S phase, so it never copies its DNA and never reaches mitosis.
    So no new daughter cells form.
  2. B. ✓ No

Why: A cell held in G1 never enters S phase.
So it never copies its DNA, and it never reaches mitosis.
So no new daughter cells form, and division stops.

24
Check q5

A researcher sorts embryo cells by stage before and after adding a different drug: an inhibitor of the cyclin–CDK complex that drives entry into mitosis. The graph below shows the percent of cells in G1, S phase, G2 and mitosis.

Two stacked panels of four bars, percent of embryo cells in G1, S phase, G2 and mitosis: untreated 25, 25, 22 and 28; after the inhibitor 24, 24, 47 and 5; gridlines every 10 percent
Two stacked panels of four bars, percent of embryo cells in G1, S phase, G2 and mitosis: untreated 25, 25, 22 and 28; after the inhibitor 24, 24, 47 and 5; gridlines every 10 percent

In which stage does the inhibitor hold the cells?

  1. A. G1
    G1 stayed at about 24%.
    The stage that filled was G2, which rose from 22% to 47%.
  2. B. S phase
    S phase stayed at about 24%.
    The stage that filled was G2, which rose from 22% to 47%.
  3. C. ✓ G2
  4. D. Mitosis
    Mitosis fell from 28% to 5%.
    Cells were failing to enter mitosis, so they collected in the stage before it, G2.

Why: G2 rose from 22% to 47%.
Mitosis fell from 28% to 5%.
So cells reached the end of G2 and did not enter mitosis.
Entry into mitosis is driven by one cyclin–CDK complex.
This inhibitor keeps that complex switched off.
So the cells are held in G2, and mitosis empties.

25
Check q6

A researcher follows normal cells, and cells with a cyclin altered so that it cannot be destroyed, for 30 minutes after mitosis begins. The graph below shows the percent of cells that had re-formed a nuclear envelope.

Three bars: percent of cells that had re-formed a nuclear envelope 30 minutes after mitosis began, 86 for normal cells, 8 for cells with a cyclin that cannot be destroyed, and 82 for those cells when the cyclin is also stopped from binding CDK, with ±2SE error bars of 4, 3 and 4; gridlines every 20; the legend reads: error bars represent ±2SE
Three bars: percent of cells that had re-formed a nuclear envelope 30 minutes after mitosis began, 86 for normal cells, 8 for cells with a cyclin that cannot be destroyed, and 82 for those cells when the cyclin is also stopped from binding CDK, with ±2SE error bars of 4, 3 and 4; gridlines every 20; the legend reads: error bars represent ±2SE

What happens to the altered cells at the end of mitosis?

  1. A. The altered cells collect in G2
    These cells did begin mitosis.
    The fault shows at its end: only 8% re-formed a nuclear envelope.
  2. B. The altered cells collect in G1
    A cell reaches G1 only after it finishes mitosis, and only 8% of these cells re-formed a nuclear envelope.
  3. C. The altered cells finish mitosis and divide
    After 30 minutes only 8% of the altered cells had re-formed a nuclear envelope, against 86% of normal cells.
  4. D. ✓ The altered cells stay in mitosis

Why: After 30 minutes, 86% of normal cells had re-formed a nuclear envelope.
Only 8% of the cells with the cyclin that cannot be destroyed had.
Re-forming the nuclear envelope is the end of mitosis.
So the altered cells stay in mitosis.

26
Practice writing an answer

A researcher follows normal cells, and cells with a cyclin altered so that it cannot be destroyed, for 30 minutes after mitosis begins. After 30 minutes, 86% of the normal cells had re-formed a nuclear envelope; only 8% of the altered cells had. When the researcher also stops the altered cyclin from binding CDK, 82% of the cells re-form their envelopes. The altered cells stay in mitosis.

(a) Explain why the altered cells stay in mitosis, and what the 82% shows. (1 pt)

Model answer The complex that drives entry into mitosis switches mitosis on.
Destroying its cyclin at the end of mitosis switches mitosis off.
The altered cyclin cannot be destroyed, so it stays bound to its CDK.
So the complex stays active and mitosis stays on: the envelopes do not re-form.
Stopped from binding CDK, the altered cyclin forms no active complex, and 82% of cells re-form their envelopes.
So the active complex, not the leftover cyclin, held the cells in mitosis.
Rubric
  • Award 1 point for: the undestroyable cyclin keeps the cyclin–CDK complex active so mitosis stays switched on; stopping the cyclin binding CDK (82% finish) shows the active complex, not the cyclin alone, is what holds the cells.

27Quick quiz: which stage fills? mixed practice

28
Check q7

A researcher gives dividing cells a spindle inhibitor.

Which stage fills?

  1. A. G1
    The spindle acts in mitosis, long after G1.
    The M checkpoint holds the cells at metaphase, inside mitosis.
  2. B. G2
    The cells pass the G2 checkpoint and enter mitosis.
    The M checkpoint holds them at metaphase, inside mitosis.
  3. C. ✓ Mitosis

Why: The inhibitor stops the spindle fibers from moving the chromosomes.
So no chromosome is attached from both poles.
So the M checkpoint’s condition is never met.
So the cells collect at metaphase, which is inside mitosis.

29
Check q8

A researcher gives dividing cells an inhibitor of the cyclin–CDK complex that drives entry into mitosis.

Which stage fills?

  1. A. G1
    The blocked complex acts at the end of G2, not G1.
    The cells collect in G2.
  2. B. ✓ G2
  3. C. Mitosis
    The blocked complex is what lets cells enter mitosis.
    So mitosis empties, and G2 fills.

Why: The complex that drives entry into mitosis stays switched off.
So the cells reach the end of G2 and stop there.
So G2 fills.

30
Check q9

A researcher gives dividing cells an inhibitor of the cyclin–CDK complex that drives entry into S phase.

Which stage fills?

  1. A. ✓ G1
  2. B. G2
    A cell that cannot enter S phase never reaches G2.
    The cells collect in G1.
  3. C. Mitosis
    A cell that cannot enter S phase never copies its DNA and never reaches mitosis.
    The cells collect in G1.

Why: The complex that drives entry into S phase stays switched off.
So the cells reach the end of G1 and stop there.
So G1 fills.

31
Check q10

A researcher removes the growth factor from a dish of dividing cells.

Which stage fills?

  1. A. ✓ G1
  2. B. G2
    Cells already past G1 finish their cycle.
    Cells arriving in G1 have no signal to enter the cycle, so they rest in G1 or G0.
  3. C. Mitosis
    Cells already in mitosis finish it.
    Cells arriving in G1 have no signal to enter the cycle, so they rest in G1 or G0.

Why: With no growth factor, no signal tells a cell to enter the cycle.
So cells arriving in G1 rest there or drop into G0.
So G1 fills, and nothing piles up inside the cycle.

32
Check q11

A researcher gives dividing cells an altered cyclin that resists destruction at the end of mitosis.

Which stage fills?

  1. A. G1
    A cell reaches G1 only after it finishes mitosis, and these cells cannot finish mitosis.
  2. B. G2
    The cells enter mitosis as usual.
    The fault shows at the end of mitosis, which never comes.
  3. C. ✓ Mitosis

Why: Destroying the cyclin is what switches mitosis off.
The altered cyclin is never destroyed, so the complex stays active.
So the cells enter mitosis and stay there.

33
Check q12

A drug damages the DNA of dividing cells after S phase. The cells’ checkpoints all work.

Which stage fills?

  1. A. G1
    The damage happens after S phase, so the cells have already left G1.
    The G2 checkpoint holds them before mitosis.
  2. B. ✓ G2
  3. C. Mitosis
    The G2 checkpoint checks for undamaged DNA before mitosis.
    So the damaged cells never enter mitosis; they collect in G2.

Why: The G2 checkpoint checks that the DNA is undamaged before mitosis.
The damage happened after S phase, so the cells reach the G2 checkpoint with damaged DNA.
So the cells are held in G2, and G2 fills.

34Which control did the mutant lose?

35

Video: Watch: Read the lost control off the count

Clumps of tissue in dishes: normal clumps give 18 division events per 100 cells; clumps with an always-active signal protein give 52; clumps with a useless checkpoint protein give 47. The normal clumps are the control. Each rise names the control that was lost.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19Cb.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19Cb.mp4

36

Now consider a whole investigation, read the same way. Researchers grew clumps of tissue in dishes and counted the division events per 100 cells over 48 hours.

Four bars: division events per 100 cells in 48 hours for tissue clumps that are normal, 18; with an always-active signal protein, 52; with a useless checkpoint protein, 47; with both faults, 118; ±2SE error bars of 3, 5, 5 and 8; gridlines every 20; the legend reads: error bars represent ±2SE
Four bars: division events per 100 cells in 48 hours for tissue clumps that are normal, 18; with an always-active signal protein, 52; with a useless checkpoint protein, 47; with both faults, 118; ±2SE error bars of 3, 5, 5 and 8; gridlines every 20; the legend reads: error bars represent ±2SE
37

The normal clumps, with a working signal protein and a working checkpoint protein, gave 18 events per 100 cells.

38

The normal clumps are the control. Every other clump is compared with them.

39

Clumps with an always-active signal protein gave 52 events per 100 cells. The always-active protein sends the message with no growth factor present, so cells entered the cycle that should have waited.

40

So the signal control is lost in those clumps.

41

Clumps with a useless checkpoint protein gave 47 events per 100 cells. The useless protein cannot hold, so cells that should have waited passed the checkpoint.

42

So the hold is lost in those clumps.

43

To read which control a mutant lost: find the control group, then ask what each single change did on its own.

44

The independent variable is which controls the clumps carry. The dependent variable is the count of division events.

45

What you are expected to know Identify which control a mutant has lost from its division count against the control group: an always-active signal protein has lost the signal control, and a useless checkpoint protein has lost the hold.

46
Check q13

Clumps of kidney tissue were grown in dishes in a medium with no growth factor in it, and division events were counted per 200 cells in 48 hours. The normal clumps, with a working signal protein and a working checkpoint protein, gave 22. The clumps with an always-active signal protein gave 61. The clumps with a useless checkpoint protein gave 55.

What does the count of 61 show about the always-active signal protein?

  1. A. ✓ Cells enter the cycle with no growth factor to tell them to
  2. B. Cells copy their DNA twice in each cycle
    The count is of division events; a cell that divided more often still copied its DNA once per cycle.
  3. C. The checkpoint protein has been destroyed as well
    These clumps have a working checkpoint protein; only the signal protein is altered, and that alone raised the count.
  4. D. Mitosis takes less time in these cells
    More cells entered the cycle, because the signal protein sent the message without a growth factor.

Why: The signal protein sends the message to enter the cycle, normally only when a growth factor has bound.
With no growth factor in the medium, normal clumps gave 22 events and always-active clumps gave 61.
So the always-active protein sent cells into the cycle with no growth factor.

47
Practice writing an answer

Clumps of kidney tissue were grown in dishes in a medium with no growth factor in it, and division events were counted per 200 cells in 48 hours. The normal clumps, with a working signal protein and a working checkpoint protein, gave 22. The clumps with an always-active signal protein gave 61.

(a) Explain how the count of 22 from the normal clumps lets the count of 61 show what the always-active signal protein does. (1 pt)

Model answer The normal clumps have a working signal protein and grow in the same medium with no growth factor.
So their count of 22 events per 200 cells shows how often cells divide with no growth factor.
The always-active clumps differ from the normal clumps in one thing only: the signal protein.
So the extra 39 events, 61 against 22, came from the always-active signal protein.
So the always-active signal protein sends cells into the cycle with no growth factor.
Rubric
  • Award 1 point for: the normal clumps are the control, the same conditions with a working signal protein, so the rise from 22 to 61 can be attributed to the always-active signal protein alone.

48Do two faults add, or do more?

49

Video: Watch: More than the sum

The signal fault raised the count by 34 and the checkpoint fault by 29. If the two rises simply added, both faults would give 81. The clumps with both faults gave 118: each lost control lets more cells through the gap the other leaves.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19Cc.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-L19Cc.mp4

50

Now consider the clumps with both faults: an always-active signal protein and a useless checkpoint protein. They gave 118 events per 100 cells.

51

Would 118 be expected if the two faults simply added? Here is the working, as a table.

A table of the working: the rise from the signal fault, 52 minus 18 equals 34 events per 100 cells; the rise from the checkpoint fault, 47 minus 18 equals 29; expected if the rises add, 18 plus 34 plus 29 equals 81; measured with both faults, 118
52

118 events per 100 cells is far above 81. So together the two faults do more than their sum.

53

The always-active signal protein sends extra cells into the cycle. With the hold gone, none of those extra cells is held at the checkpoint either.

54

So each lost control lets more cells through the gap the other leaves.

55

To read two faults together: ask whether the two changes add up, or do more.

56

What you are expected to know Judge whether two faults together add or do more, by comparing the double mutant’s count with the count the two single rises would give if they simply added.

57
Check q14

Clumps of kidney tissue were grown in dishes, and division events were counted per 200 cells in 48 hours. Normal clumps gave 22, clumps with an always-active signal protein gave 61, and clumps with a useless checkpoint protein gave 55. Clumps carrying both faults gave 139.

Which statement about the two faults do the results support?

  1. A. The two faults act independently of each other
    The signal fault alone gave 61 events and the checkpoint fault alone 55.
    Both faults gave 139 events, far more than the two rises added together.
  2. B. Only the signal fault raises the count
    The checkpoint fault on its own raised the count from 22 to 55 events, so the checkpoint fault matters.
  3. C. ✓ Together the two faults do more than the sum of their effects
  4. D. The checkpoint fault cancels part of the signal fault’s effect
    The double mutant divided far more than either single mutant, so the faults reinforce each other.

Why: The signal fault raised the count from 22 to 61 events: a rise of 39.
The checkpoint fault raised it to 55 events: a rise of 33.
Added onto 22, both rises give 94.
Both faults gave 139, far above 94.
So the faults do more than their sum.

58
Check q15

Researchers counted the cells of a patient’s lump by stage before the patient’s doctors gave a drug, and again after two days of the drug. The graph below shows the counts, with the percent of cells that had died.

Two stacked panels of four bars, percent of the lump's living cells in G1, S phase, G2 and mitosis, with the percent of cells that died beside each title: before the drug 30, 40, 20 and 10 with 5 percent dead; after two days of the drug 21, 17, 50 and 12 with 31 percent dead
Two stacked panels of four bars, percent of the lump's living cells in G1, S phase, G2 and mitosis, with the percent of cells that died beside each title: before the drug 30, 40, 20 and 10 with 5 percent dead; after two days of the drug 21, 17, 50 and 12 with 31 percent dead

What did the drug do?

  1. A. The drug sped the cells through the cycle
    Half the cells sat in G2, which is a hold; cells that raced through the cycle would spread across the stages.
  2. B. ✓ The drug damaged the cells’ DNA
  3. C. The drug blocked the spindle
    The pile-up is in G2, before mitosis, and mitosis itself barely changed.
  4. D. The drug removed the growth signal
    G1 fell from 30% to 21%; the stage that filled was G2, the checkpoint that checks for damaged DNA.

Why: The drug damaged the cells’ DNA.
The G2 checkpoint checks for undamaged DNA, so the damaged cells were held in G2: half the living cells sat there.
A held cell with damage beyond repair is signaled to undergo apoptosis.
So many of the held cells removed themselves: 31% died.

59

Back to the patient’s lump: 40% of its cells copying DNA before the drug, and after two days of the drug 17% copying DNA, half the living cells in G2 and 31% dead.

60

The first drug damaged the cells’ DNA. The G2 checkpoint checks for undamaged DNA, so the cells that still had a working G2 checkpoint were held in G2: the pile-up.

61

A held cell with damage beyond repair is signaled to undergo apoptosis. So 31% of the cells died.

62

The second drug blocked the checkpoint protein that detects damaged DNA. So the damage was never detected, and the damaged cells were never held.

63

The pile-up in G2 vanished, and only 9% died.

64

Never held, the damaged cells carried on dividing. That is exactly the loss of control that made the lump in the first place.

65Mixed practice mixed practice

66
Check q16

A researcher adds a spindle inhibitor to dividing root-tip cells.

Predict what happens to the cells over the next few hours.

  1. A. The cells collect in G2 before mitosis begins
    Cells pass the G2 checkpoint and enter mitosis; they are held at metaphase, where the spindle should attach.
  2. B. The cells collect in G1 before copying their DNA
    The spindle acts in mitosis; cells reach it with their DNA already copied.
  3. C. The cells finish mitosis without a spindle
    The M checkpoint holds cells whose chromosomes are unattached, so the sisters never separate and no daughters form.
  4. D. ✓ The cells collect at metaphase with their sisters still joined

Why: The inhibitor stops the spindle fibers from moving the chromosomes.
So no chromosome is attached from both poles.
So the M checkpoint’s condition is never met.
So the cells collect at metaphase with their sisters still joined, and no new daughter cells form.

67
Check q17

Two drugs are tested on dividing cells. After drug X, cells collect in G2 and mitosis empties. After drug Y, cells enter mitosis and stay there, with their nuclear envelopes still broken down.

Which pair of actions fits the two drugs?

  1. A. X stops the cyclin being destroyed; Y keeps the complex that drives entry into mitosis switched off
    A cyclin never destroyed keeps mitosis on: cells stay in mitosis, drug Y’s pattern.
    A complex kept off never starts mitosis: cells collect in G2, drug X’s pattern.
  2. B. Both drugs block the spindle, at two different stages
    A spindle block holds cells at metaphase, inside mitosis.
    Drug X holds cells in G2; drug Y holds them at the end of mitosis.
    Neither is a metaphase block.
  3. C. ✓ X keeps the complex that drives entry into mitosis switched off; Y stops the cyclin being destroyed
  4. D. X removes the growth factor; Y blocks the G1 checkpoint
    A missing growth factor leaves cells resting in G1; a blocked G1 checkpoint lets cells into S phase.
    Neither is seen: X fills G2, Y keeps cells in mitosis.

Why: One cyclin–CDK complex drives entry into mitosis.
Drug X keeps that complex off, so cells cannot enter mitosis: G2 fills and mitosis empties.
Destroying that cyclin is what switches mitosis off.
Drug Y stops the cyclin being destroyed, so the complex stays active: cells enter mitosis and stay there.

68
Check q18

A cell line carries a mutation in the gene for a protein of the M checkpoint, so the M checkpoint cannot hold. A researcher then gives the cells a low dose of a spindle inhibitor, which leaves a few chromosomes attached from one pole only.

Predict what happens to the cells.

  1. A. The cells collect at metaphase with their sisters still joined
    Holding at metaphase is exactly what the lost M checkpoint can no longer do.
    The cells go on through mitosis with a few chromosomes attached from one pole only.
  2. B. ✓ The daughter cells receive extra or missing chromosomes
  3. C. The cells collect in G2, before mitosis begins
    The spindle acts in mitosis, after the G2 checkpoint.
    The cells enter mitosis and, with no hold at metaphase, go on through it.

Why: A few chromosomes are attached from one pole only.
The M checkpoint would hold the cell at metaphase.
This M checkpoint cannot hold.
So the cells go on through mitosis.
One pole pulls both sisters of those chromosomes.
So one daughter receives an extra chromosome and the other lacks it.

69
Practice writing an answer

Researchers grew clumps of gut tissue in dishes and counted division events per 50 cells over one day. Normal clumps carried a working signal protein A, which promotes entry into the cycle, and a working checkpoint protein I, which holds the cycle when conditions are unsuitable. Other clumps carried an always-active signal protein A, or a useless checkpoint protein I, or both. The graph below shows the counts; error bars represent ±2SE.

Four bars: division events per 50 cells in one day for gut-tissue clumps that are normal, 6; with an always-active signal protein A, 19; with a useless checkpoint protein I, 16; with both faults, 41; ±2SE error bars of 1, 2, 2 and 3; gridlines every 10; the legend reads: error bars represent ±2SE
Four bars: division events per 50 cells in one day for gut-tissue clumps that are normal, 6; with an always-active signal protein A, 19; with a useless checkpoint protein I, 16; with both faults, 41; ±2SE error bars of 1, 2, 2 and 3; gridlines every 10; the legend reads: error bars represent ±2SE

(a) Identify the dependent variable in this investigation. (1 pt)

Model answer The dependent variable is the number of division events per 50 cells in one day.
The researchers counted this number in every clump.
Rubric
  • Award 1 point for: the number of division events per 50 cells (in one day) as the dependent variable.

Slip Naming the mutations. Which proteins the clumps carry is what the researchers changed, so that is the independent variable; the dependent variable is what they counted.

(b) Explain why the clumps with the always-active signal protein A divided more often than the normal clumps. (1 pt)

Model answer Signal protein A carries the message to enter the cycle.
Normally protein A is active only when a growth factor has bound.
The always-active protein A sends the message with no growth factor present.
So cells that should have waited in G1 enter the cycle.
So the clumps with the always-active protein A divided more often: 19 events per 50 cells against 6.
Rubric
  • Award 1 point for: the always-active signal protein sends the enter-the-cycle message without a growth factor, so more cells enter the cycle and divide.

Slip Writing that the protein makes cells divide faster. The protein changes how many cells enter the cycle, not how quickly a cell moves round it.

(c) Evaluate the claim that the two faults together do more than the sum of their separate effects, using the counts. (1 pt)

Model answer The claim is supported.
The signal fault alone raised the count from 6 to 19 events per 50 cells, a rise of 13.
The checkpoint fault alone raised the count from 6 to 16 events, a rise of 10.
Adding the two rises onto the normal count gives 29 events per 50 cells.
The double mutant gave 41 events per 50 cells, far above 29.
So together the faults do more than the sum of their separate effects.
Working
Write down the rises:
rise from the signal fault = 19 − 6 = 13 events per 50 cells
rise from the checkpoint fault = 16 − 6 = 10 events per 50 cells
Add the rises to the normal count:
expected if the effects add = 6 + 13 + 10 = 29 events per 50 cells
Compare with the measured count:
measured for both faults = 41 events per 50 cells
41 is far above 29
Rubric
  • Award 1 point for: the judgement (the claim is supported) AND the ground: the double-mutant count (41) is well above the count the two single rises would give if added (about 29; accept 35 if the two single counts are added), so the faults reinforce each other.

Slip Comparing 41 with 19 or 16 alone. The claim is about the sum: the test is whether 41 exceeds what adding the two separate rises predicts.

(d) Predict the count for double-mutant clumps treated with a drug that returns signal protein A to its normal behavior, and justify your prediction. (1 pt)

Model answer About 16 division events per 50 cells.
The drug returns signal protein A to its normal behavior.
So the signal fault is removed.
The useless checkpoint protein I is still there.
So only the checkpoint fault is left.
The checkpoint fault alone gave 16 events per 50 cells.
So the treated double-mutant clumps should give about 16.
Rubric
  • Award 1 point for: a prediction near 16 events per 50 cells (accept 12 to 20) with the reason that the drug removes the signal fault and leaves only the checkpoint fault.

Slip Predicting 6, as if both faults were removed, or 41, as if the drug changed nothing. The drug removes one fault; the other stays and still shows in the count.

APBIO-U04-P46 Practice questions: Topic 4.6

Topic 4.6 · Regulation of Cell Cycle · 9 MCQ · 2 FRQ · for APBIO-U04-T46

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. The first free-response question walks you through the reasoning one part at a time, and you can open a hint for each part; the second is at the level of the test. When you finish a question, open the scoring guide and mark your own work against it. Where a graph carries error bars, the caption says what the bars represent. Every cell cycle in these questions is given with its length.

Video: Watch first: Topic 4.6 summary, part 1: switched on, held, moved on

Division is switched on by a growth signal, held at three checkpoints until each condition is met, and moved on by cyclin–CDK complexes.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-T46-summary.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U04-T46-summary.mp4

Q1 P46-q01

A cultured cell has reached full size, has nutrients around it and carries undamaged DNA. The cell's receptors for the growth factor are empty; the dish was set up with no growth factor added.

What does the cell do at the G1 checkpoint, and why?

  1. A. The cell passes, because its size, nutrients and DNA are all in order
    Size, nutrients and undamaged DNA are three of the G1 checkpoint’s conditions; a growth signal is the fourth.
    That condition is unmet, so the cell is held in G1.
  2. B. ✓ The cell waits in G1, or drops into G0, because the growth signal is missing
  3. C. The cell passes, because a growth factor is only needed after S phase
    The growth signal is checked at the G1 checkpoint, at entry into the cycle, before S phase.
    A cell with no signal does not start copying its DNA.
  4. D. The cell undergoes apoptosis, because a cell with no signal removes itself
    Apoptosis follows damage beyond repair, or a lost survival signal.
    A missing growth factor is neither: the cell waits in G1 or rests in G0 until a growth factor arrives.

Why: The G1 checkpoint is a set of conditions: size, nutrients, a growth signal and undamaged DNA.
Three of the four are met; no growth factor is bound, so the growth-signal condition is unmet.
So the cell waits in G1, or drops into G0, until the signal arrives.

Q2 P46-q02

In a culture of dividing cells, the time each cell spent at metaphase was measured. Most cells spent 4 to 8 minutes there. A few spent more than 2 hours, and each of those had one chromosome attached to spindle fibers from one pole only.

Which statement about the M checkpoint do these times support?

  1. A. The M checkpoint is a timer set to about six minutes, and damage makes the timer last longer
    A checkpoint is a set of conditions, not a clock.
    Most cells met the condition within minutes; the slow cells waited because one chromosome stayed attached from one pole only.
  2. B. The M checkpoint holds only cells whose DNA is damaged, and the slow cells had damaged DNA
    Damaged DNA is checked at the G1 and G2 checkpoints, before mitosis.
    At metaphase the condition is the attachment of every chromosome to spindle fibers from both poles.
  3. C. The M checkpoint attaches the last chromosome itself, and doing so takes about two hours
    A checkpoint holds; it does not do the work itself.
    The spindle attaches the chromosome, and the cell waits at metaphase until that has happened.
  4. D. ✓ The M checkpoint is a condition: each cell waits until its last chromosome is attached from both poles

Why: The M checkpoint checks that every chromosome is attached from both poles before the sister chromatids separate: a condition, not a timer.
Most cells met the condition within minutes.
A cell with one chromosome attached from one pole only did not, so it waited until that chromosome attached.

Q3 P46-q03

Four events in dividing tissues are described.

Which of the following describes a checkpoint holding a cell?

  1. A. ✓ A cell whose copied DNA carries a break waits at the end of G2 until the break is repaired
  2. B. A liver cell enters the cycle when a growth factor from a damaged neighbor binds its receptor
    A growth factor binding its receptor lets the cell enter the cycle.
    The cell goes on; nothing holds it.
  3. C. Rising cyclin forms active complexes that switch on the proteins that start S phase
    Active cyclin–CDK complexes switch the next stage on.
    The cell goes on; a checkpoint hold keeps a cell waiting.
  4. D. A cell that has grown to twice its starting size divides because it is now big enough
    A cell that divides has gone on to the next stage.
    A checkpoint holding a cell keeps it waiting until a condition, such as undamaged DNA, is met.

Why: A checkpoint holds the cycle until its condition is met.
Undamaged copied DNA is the G2 checkpoint’s condition.
So the cell with a break waits at the end of G2 until the break is repaired.
The other three events are a cell going on, not a cell held.

Q4 P46-q04

Cells of a fish embryo were sorted by stage, and the amounts of one cyclin and of its cyclin-dependent kinase (CDK) were measured in each group; the graph below gives each as a percent of that protein’s highest value.

Amounts of one cyclin (darker bars) and of its CDK (paler bars) in fish embryo cells sorted by stage, each as a percent of that protein's highest value. Gridlines every 20%.
Amounts of one cyclin (darker bars) and of its CDK (paler bars) in fish embryo cells sorted by stage, each as a percent of that protein's highest value. Gridlines every 20%.

Which of the following do the measurements show?

  1. A. Both proteins rise through G2 and fall at the end of mitosis
    The CDK’s bars are all within 94% to 100% of its highest value: its amount hardly changes.
    Only the cyclin climbs, from 8% to 100% at the start of mitosis.
  2. B. The CDK's amount rises and falls, and the cyclin's stays steady
    The cyclin’s bars run from 8% to 100% across the stages, while the CDK’s bars stay between 94% and 100%: the cyclin changes, the CDK stays steady.
  3. C. ✓ The cyclin's amount rises and falls, and the CDK's stays steady
  4. D. The two proteins are present in equal amounts at every stage
    Each protein is shown as a percent of its own highest value, and the cyclin’s bars change from stage to stage while the CDK’s do not.

Why: The cyclin’s amount changes: 8% in G1, 22% in S phase, 68% in G2, 100% at the start of mitosis.
The CDK stays between 94% and 100% throughout.
A CDK is active only while a cyclin is bound, so active complexes peak with the cyclin, as mitosis starts.

Q5 P46-q05

Cultured mouse cells short of one amino acid wait in G1. After an hour, a researcher adds a drug that blocks the checkpoint protein that detects the shortage. The table below gives measurements just before the drug and 15 minutes after it.

Cultured mouse cells short of one amino acid: measurements just before a drug that blocks the checkpoint protein detecting the shortage, and 15 minutes after it.
Cultured mouse cells short of one amino acid: measurements just before a drug that blocks the checkpoint protein detecting the shortage, and 15 minutes after it.

Which statement explains the rise in the activity of the cyclin-dependent kinase (CDK)?

  1. A. The cyclin passed the concentration that switches the CDK on only after the drug was added
    The cyclin stood at 82% of its peak before the drug, so plenty of complex had formed.
    A signal from the checkpoint protein was keeping it switched off.
  2. B. ✓ The shortage signal had been keeping the cyclin–CDK complex switched off, and the drug removed that signal
  3. C. The drug supplied the missing amino acid, so the G1 checkpoint's condition was met and the hold ended
    The shortage was there when the cells entered S phase, so the condition was never met.
    The drug blocked the protein that reports the shortage, so the signal stopped.
  4. D. The drug made the cells build more CDK, and the extra CDK bound the cyclin that was waiting
    The CDK level is steady; the cyclin was already at 82%, so the complex had formed.
    A signal from the checkpoint protein was holding it off.

Why: An unmet condition sends a signal that keeps the complex off, whatever the cyclin.
Here the shortage signal held the complex off: CDK activity 5%.
The drug blocked the checkpoint protein, ending the signal.
With nothing holding it off, the complex became active and the cells entered S phase.

Q6 P46-q06

Here are four cells from different tissues.

Which cell is a cancer cell?

  1. A. A cell in an embryo that completes a cycle every 12 hours while a growth factor is present
    A fast cycle with a growth factor present is normal division: the signal is there and the conditions are met.
    Cancer is division without the conditions, not fast division.
  2. B. A liver cell that has rested in G0 for three years with undamaged DNA
    A cell resting in G0 has left the cycle because a condition was unmet; that is a control working, not lost.
  3. C. A skin cell at the edge of a cut that divides for a week and then stops as the cut closes
    Dividing while the signals from a wound are present and stopping when they fade is exactly what a controlled cell does.
  4. D. ✓ A cell that passes the G1 checkpoint carrying damaged DNA, and whose daughters do the same

Why: Cancer is division without the controls.
A cancer cell passes a checkpoint it should fail: here, the G1 checkpoint with damaged DNA, and its daughters inherit the same missing hold.
Dividing quickly with a growth factor, resting in G0, and healing a cut then stopping are all normal.

Q7 P46-q07

Cells from a colon tumor have lost the apoptosis response: a held cell whose damage cannot be repaired no longer removes itself. Researchers give these tumor cells a DNA-damaging drug in a dish; the damage cannot be repaired. In these cells a checkpoint hold on damaged DNA lasts about a day.

Which of the following best predicts what most of the tumor cells have done by day 3?

  1. A. ✓ They have divided, and each daughter cell carries the damage
  2. B. They have removed themselves, so the count has fallen
    These tumor cells have lost the apoptosis response, so a held cell with damage beyond repair no longer removes itself.
    The damaged cells survive.
  3. C. They are still held at the checkpoint with their damage
    A checkpoint hold on damaged DNA lasts about a day.
    After that the hold lapses, and a cell that is still alive resumes the cycle.
  4. D. They have divided, and the daughter cells are free of the damage
    The damage was never repaired, so S phase copied the damaged DNA.
    Each daughter cell receives a copy of the damage.

Why: The tumor cells have lost the apoptosis response, so the damaged cells are not removed.
The damage cannot be repaired.
After about a day the hold lapses, and the surviving cells resume the cycle.
So by day 3 most have divided, and both daughter cells carry the damage.

Q8 P46-q08

The table below gives the percent of cultured mouse cells in each stage before and 24 hours after a researcher added a drug. The treated cells in G2 held twice the DNA of a G1 cell, their DNA was undamaged, and a growth factor was bound to their receptors.

Percent of cultured mouse cells in each stage before a drug was added and 24 hours after, each from one count of 200 cells.
Percent of cultured mouse cells in each stage before a drug was added and 24 hours after, each from one count of 200 cells.

Which action of the drug fits the counts?

  1. A. The drug blocks the spindle from attaching to the chromosomes
    A spindle block holds cells at metaphase, so mitosis would fill.
    Here mitosis fell to 4% and G2 filled: cells were not passing from G2 into mitosis.
  2. B. The drug damages DNA
    The G2 cells’ DNA was copied and undamaged, so every G2 condition was met.
    With no cyclin made, the mitosis-entry complex never forms.
  3. C. ✓ The drug stops the cell making the cyclin that drives entry into mitosis
  4. D. The drug speeds cells through S phase
    Faster S phase would not fill G2.
    G2 rose from 20% to 56% because cells reached the end of G2 and did not enter mitosis.

Why: G2 rose to 56%, mitosis fell to 4%: cells held at the end of G2.
Every G2 condition was met, so no condition caused the hold.
The drug stops the mitosis-entry cyclin being made, so no active complex forms.
So the mitosis proteins stay off, and cells collect in G2.

Q9 P46-q09

Normal cells from a mouse tissue show 20 division events per 100 cells in 48 hours in a dish with no growth factor added. Cells with an always-active relay protein in the growth-signal pathway show 56; cells with a useless checkpoint protein show 44; cells carrying both faults show 130.

Which of the following do the counts show about the two faults together?

  1. A. The faults cancel in part: the double mutant’s count is below the count for the always-active relay protein alone
    130 is above 56, the count for the always-active relay protein alone.
    The faults do not cancel.
  2. B. The faults add: the double mutant’s count equals the normal count plus the two single rises
    Adding the two single rises to the normal count gives 20 + 36 + 24 = 80.
    The double mutant’s count is 130, well above 80.
  3. C. ✓ The faults do more than add: the double mutant’s count is above the normal count plus the two single rises
  4. D. The second fault makes no difference: the double mutant’s count equals the count for the always-active relay protein alone
    The always-active relay protein alone gives 56; the double mutant gives 130.
    The second fault more than doubled the count.

Why: If the faults added, the double mutant’s count would be the normal count plus the two single rises.
The working below gives 80 expected against 130 measured.
130 is far above 80, so the two faults together do more than add.

FRQ 1 P46-frq1 · Scientific Investigation scaffolded

A drug used to treat cancer damages DNA. To see which cells it removes, researchers grew three kinds of human cell in dishes: cells from a tumor, cells from the lining of the gut (a tissue that replaces itself every few days), and muscle cells, 96% of which were resting in G0. Five dishes of each kind received the same dose of the drug, and 6 hours later all three kinds carried the same amount of DNA damage per cell. After 3 days the researchers counted the percent of cells in each dish that had undergone apoptosis. The graph below shows the mean of the five dishes for each kind of cell; error bars represent ±2SE. In these cells a checkpoint hold on damaged DNA lasts about a day; a held cell that has not been removed by then resumes the cycle.

Mean percent of cells that had undergone apoptosis 3 days after the same dose of a DNA-damaging drug, for three kinds of human cell; five dishes per kind. Error bars represent ±2SE. Gridlines every 10%.
Mean percent of cells that had undergone apoptosis 3 days after the same dose of a DNA-damaging drug, for three kinds of human cell; five dishes per kind. Error bars represent ±2SE. Gridlines every 10%.

(a) Identify the independent variable in this investigation. (1 pt)

Frame The independent variable is …

Hint Which thing did the researchers deliberately make different between the dishes, and which did they measure at the end?

Model answer The independent variable is the kind of cell in the dish: tumor, gut lining or muscle.
Rubric
  • Award 1 point for: the kind of cell (tumor, gut lining or muscle).
  • Do not award the point for the drug dose (the same in every dish) or for the percent of cells undergoing apoptosis (the dependent variable).

Slip Naming the drug as the independent variable. Every dish received the same dose; what differed between dishes was the kind of cell.

(b) Using the error bars, describe how the tumor cells compare with the gut lining cells. (1 pt)

Frame The tumor bar runs from … to …% and the gut lining bar from … to …%, so the bars …, and the data …

Hint The caption says what each bar represents; check whether the two bars share any part of their range before you decide what the data show.

Model answer The tumor bar runs from 57% to 69% and the gut lining bar from 40% to 50%, so the bars are clear of each other, and the data show that a larger percent of tumor cells than gut lining cells underwent apoptosis.
Rubric
  • Award 1 point for: the tumor bar (57% to 69%) and the gut lining bar (40% to 50%) are clear of each other, so the data show a real difference: a larger percent of tumor cells underwent apoptosis.
  • Accept the two ranges with "do not overlap, so the difference is real". Do not award the point for a comparison of the means alone (63% against 45%) with no use of the bars.

Slip Comparing 63% with 45% and stopping. The point is earned by using the bars: because they are clear of each other, the difference is more than chance.

(c) Explain why most of the muscle cells survived the drug. (1 pt)

Frame Most muscle cells were in …, outside the …, so they … and were never …

Hint Where in the cycle were the muscle cells at the start, and where do the three checkpoints sit?

Model answer Most muscle cells were in G0, outside the cycle.
A cell in G0 is not copying its DNA and never approaches a checkpoint.
So the damage in the muscle cells was never detected at a checkpoint.
A cell whose damage is never detected is neither held nor signaled to undergo apoptosis.
So the muscle cells survived.
Rubric
  • Award 1 point for: the muscle cells were resting in G0, outside the cycle, so they were neither copying DNA nor approaching any checkpoint; their damage was never detected at a checkpoint, so they were neither held nor signaled to undergo apoptosis, and they survived.
  • Accept "in G0, so they never reach a checkpoint". Do not award the point for "the drug did not enter muscle cells" (the damage was the same) or for "muscle cells repair DNA faster".

Slip Writing that the drug damaged muscle cells less. The stimulus says all three kinds carried the same damage; the muscle cells survived because, resting in G0, they reached no checkpoint.

(d) The gut lining cells are normal cells. Explain why the drug removed so many of them. (1 pt)

Frame Gut lining cells … every few days, so most were …; a cell held at a checkpoint with damage that … is signaled to …

Hint How often do gut lining cells pass through the checkpoints, and what does a checkpoint do with a damaged cell whose repair fails?

Model answer Gut lining cells divide constantly, because the lining replaces itself every few days.
So most gut lining cells were in the cycle.
A dividing cell reaches the G1 or G2 checkpoint, and these cells reached it carrying damaged DNA.
The checkpoint held the damaged cells.
The cells whose damage could not be repaired were signaled to undergo apoptosis.
Those cells dismantled themselves, so the damage was never copied.
Rubric
  • Award 1 point for: gut lining cells divide constantly (the tissue replaces itself every few days), so most were in the cycle and reached a checkpoint (G1 or G2) with damaged DNA; the checkpoint held them, and cells whose damage could not be repaired were signaled to undergo apoptosis.
  • Do not award the point for "the drug targets gut cells" or for "they divided too fast" with no checkpoint and no apoptosis signal.

Slip Stopping at "they divide a lot". The point needs the chain: dividing cells reach a checkpoint, the damaged cell is held, and damage beyond repair brings the apoptosis signal.

(e) The researchers propose adding a second drug that blocks the apoptosis response. Make a claim about how the tumor cells’ bar would change. (1 pt)

Frame The percent of tumor cells undergoing apoptosis would …

Hint The bar counts cells undergoing apoptosis. Which way does a drug that blocks that response move the count?

Model answer The percent of tumor cells undergoing apoptosis would fall sharply.
It would fall toward the few percent seen in muscle cells.
Rubric
  • Award 1 point for the claim: the percent of tumor cells undergoing apoptosis would fall sharply (toward the muscle cells' value). No reasoning is required for this point.
  • Do not award the point for 'it would rise' or 'it would stay the same'.

Slip Claiming that more cells die, as if two drugs must do more than one. The second drug blocks the very response that was killing the damaged cells.

(f) Support your claim in (e) using the chain from DNA damage to a dismantled cell. (1 pt)

Frame The second drug removes …, so the damaged cells …, so the percent …

Hint Which step in the chain from DNA damage to a dismantled cell does the second drug remove? Follow what happens to a damaged cell after that.

Model answer The second drug removes the last step of the chain from damage to a dismantled cell.
Damage is still detected at a checkpoint, so the damaged cells are still held.
But the detection no longer ends in the cell dismantling itself.
So the damaged tumor cells stay held, or divide with the damage still in them.
So fewer tumor cells undergo apoptosis, and the bar falls.
Rubric
  • Award 1 point for: the evidence (the second drug blocks the apoptosis response, the last step of the chain damage → hold at a checkpoint → apoptosis) AND the reasoning (damage is still detected and the cells are still held, but the detection no longer ends in the cell dismantling itself, so the damaged tumor cells stay held or divide carrying the damage instead of being removed, so fewer undergo apoptosis).
  • Do not award the point for 'the drug blocks apoptosis' with no link to what the damaged cells do instead, or for reasoning that never names the step removed.

Slip Naming the blocked step and stopping. Supporting the claim needs the link: what the damaged cells do instead of dismantling themselves, and what that does to the count.

FRQ 2 P46-frq2 · Scientific Investigation

A line of human cells carries a mutation in the gene for the cyclin-dependent kinase (CDK) that drives entry into mitosis: the altered CDK holds its active shape with or without a cyclin bound. The researchers gave normal cells and mutant cells a drug that stops DNA copying halfway through S phase. Six hours later the researchers counted the percent of cells that had entered mitosis, and 24 hours later the percent of daughter cells with pieces of chromosomes missing. The table below gives the results.

Normal cells and cells with an always-active CDK, after a drug stopped DNA copying halfway through S phase: the percent that entered mitosis within 6 hours, and the percent of daughter cells with pieces of chromosomes missing after 24 hours.
Normal cells and cells with an always-active CDK, after a drug stopped DNA copying halfway through S phase: the percent that entered mitosis within 6 hours, and the percent of daughter cells with pieces of chromosomes missing after 24 hours.

(a) Identify the null hypothesis for the effect of the mutation on entry into mitosis. (1 pt)

Model answer The null hypothesis is that the mutation makes no difference to the percent of cells entering mitosis after DNA copying is stopped: normal and always-active-CDK cells would enter mitosis at the same rate.
Rubric
  • Award 1 point for: the mutation makes no difference to the percent of cells that enter mitosis after DNA copying is stopped (normal and mutant cells enter mitosis at the same rate).
  • Do not award the point for a prediction of a difference in either direction ("mutant cells enter mitosis more often").

Slip Writing the expected result ("the mutant cells enter mitosis anyway") as the null hypothesis. The null hypothesis is the statement of no effect, which the data can then reject.

(b) Explain why the normal cells stayed out of mitosis. (1 pt)

Model answer The G2 checkpoint checks that the DNA is completely copied before mitosis.
With copying stopped halfway, that condition was unmet.
So a signal from the checkpoint kept the complex that drives entry into mitosis switched off.
An inactive complex transfers no phosphate groups, so the proteins that start mitosis were never switched on.
Therefore the normal cells were held before mitosis, and only 2% entered it.
Rubric
  • Award 1 point for: the G2 checkpoint checks that the DNA is completely copied; with copying stopped halfway the condition was unmet, so a signal kept the cyclin–CDK complex switched off, the proteins that start mitosis were never switched on, and the cells were held before mitosis (only 2% entered).
  • Accept "held at the G2 checkpoint because the DNA was incomplete" with the complex kept off. Do not award the point for "they had no cyclin" or for "the drug stopped mitosis directly".

Slip Writing that the drug itself blocked mitosis. The drug blocked copying; it was the cell's own G2 checkpoint that held the complex off and kept the cell out of mitosis.

(c) Explain how the always-active CDK let the mutant cells into mitosis with their DNA half copied. (1 pt)

Model answer The altered CDK holds its active shape whatever signals reach it.
So the G2 checkpoint's signal has nothing to switch off.
The altered CDK transferred phosphate groups from ATP onto the proteins that start mitosis.
Those proteins switched on although the DNA was only half copied.
So 44% of the mutant cells entered mitosis, and 61% of their daughters received a genome with pieces missing.
Rubric
  • Award 1 point for: the altered CDK is active with or without a cyclin bound and the checkpoint's signal cannot switch it off, so it transfers phosphate groups from ATP onto the proteins that start mitosis whatever the state of the DNA; mitosis begins with the DNA half copied, and the daughters receive incomplete genomes (61% with pieces missing).
  • Do not award the point for "the mutant cells have no checkpoint" (the checkpoint's signal is sent; the CDK ignores it) or for "the CDK copied the DNA faster".

Slip Writing that the mutant cells lack a G2 checkpoint. The checkpoint protein still sends its signal; the fault is that an always-active kinase takes no notice of it.

(d) A second drug blocks the kinase activity of this CDK. Determine the effect of adding it to the mutant cells that received the copying blocker, and justify your decision. (1 pt)

Model answer Far fewer mutant cells would enter mitosis, close to the normal 2%, and far fewer daughters would have pieces missing.
With its kinase activity blocked, the CDK can transfer no phosphate groups onto the proteins that start mitosis.
So those proteins stay off, whatever shape the CDK holds.
Therefore mitosis is never switched on.
Rubric
  • Award 1 point for: the decision (far fewer mutant cells enter mitosis, near the normal 2%, and far fewer daughters have pieces missing) AND the reasoning it rests on (with the kinase activity blocked, the proteins that start mitosis are never phosphorylated, so mitosis is never switched on, whatever shape the CDK holds).
  • Accept 'the mutant cells behave like normal cells: they stay out of mitosis', with the reasoning. Do not award the point for the decision alone, or for 'the cells enter mitosis anyway because the CDK is always active'.

Slip Deciding that the cells still enter mitosis because the CDK is always active. Being in the active shape only matters if the kinase can transfer phosphates; the second drug stops that.

APBIO-U04-T46 End-of-topic test: Regulation of Cell Cycle

Topic 4.6 · Regulation of Cell Cycle · 18 MCQ · 2 FRQ

Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the two free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. Then open the scoring guide and mark your own work against it. Where a graph carries error bars, the caption says what the bars represent. Every cell cycle in these questions is given with its length.

Q1 T46-q01

A researcher drags a fine tip across a sheet of mouse skin cells packed edge to edge in a dish, clearing a strip of cells; the medium, which the researcher replaces every day, is the same before and after. The table below gives the percent of cells entering S phase in a day.

Percent of mouse skin cells entering S phase in a day, before a strip was cleared from the packed sheet and over the two days after it, for cells along the edge of the strip and cells far from it.
Percent of mouse skin cells entering S phase in a day, before a strip was cleared from the packed sheet and over the two days after it, for cells along the edge of the strip and cells far from it.

Which explanation accounts for the difference?

  1. A. ✓ Cells at the edge of the strip lost contact on one side, so the signal that held them in G1 ended
  2. B. Clearing the strip released nutrients into the medium, and the edge cells took them up first
    The medium was unchanged, so nutrients were not the cause.
    What changed was a side free of neighbors, and contact on every side holds a cell in G1.
  3. C. The tip damaged the edge cells' DNA, and a damaged cell enters S phase to repair it
    Damaged DNA holds a cell at the G1 checkpoint; it never sends a cell into S phase.
    The edge cells entered S phase because their contact signal ended.
  4. D. Cells far from the strip had used up their share of the medium and could grow no further
    Cells far from the strip had the same medium as the edge cells.
    They stayed in G1 because they were still touched on every side: the hold signal.

Why: A cell touched on every side receives a signal that holds it in G1.
Clearing the strip freed one side of each edge cell, ending that signal.
So the edge cells entered the cycle: 38% reached S phase.
The medium was the same everywhere, so nutrients played no part.

Q2 T46-q02

Cultured bone cells receive the same nutrients in four dishes. A researcher adds a growth factor at a different concentration to each dish and counts the divisions per 100 cells in 24 hours. The results are in the table.

Divisions per 100 bone cells in 24 hours at four concentrations of the growth factor.
Divisions per 100 bone cells in 24 hours at four concentrations of the growth factor.

Which of the following do the counts at 0 and 100 ng/mL show about the growth factor?

  1. A. The growth factor is a nutrient
    A nutrient is building material, and the dishes already had the same nutrients.
    The growth factor is information: it binds a receptor and tells the cell to enter the cycle.
  2. B. ✓ The growth factor is a signal that starts the cycle
  3. C. The growth factor speeds up mitosis
    The growth factor acts before the cycle starts, at the decision to enter it.
    Thirty divisions per 100 cells means more cells entered, not that each mitosis finished faster.
  4. D. The growth factor holds cells at the G1 checkpoint
    A growth factor helps a cell past the G1 checkpoint rather than holding it.
    With none the cells were held in G1; with 100 ng/mL most went on.

Why: Division has to be switched on by a signal.
The growth factor binds a receptor, which tells the cell to enter the cycle.
With none, 3 cells per 100 divided; with 100 ng/mL, 30, on the same nutrients.
So the growth factor is the signal that starts the cycle.

Q3 T46-q03

The timeline below shows one 30-hour cycle of a cultured rat cell line, with four moments marked 1 to 4. A cell in this culture is held for eight hours because its copied DNA carries a break.

One 30-hour cycle of a cultured rat cell line drawn as a timeline, with four moments marked 1 to 4.
One 30-hour cycle of a cultured rat cell line drawn as a timeline, with four moments marked 1 to 4.

At which marked moment does the cell wait?

  1. A. Moment 1
    Moment 1, at 7 h, is the middle of G1, and no checkpoint sits there.
    This cell’s DNA is already copied, so it has passed G1 altogether.
  2. B. Moment 2
    Moment 2, at 13.5 h, is the end of G1: the G1 checkpoint, before the DNA is copied.
    This cell’s DNA is copied, so it has passed that point.
  3. C. ✓ Moment 3
  4. D. Moment 4
    Moment 4 is inside mitosis, where the M checkpoint checks chromosome attachment.
    A cell with broken copied DNA is held earlier, at the end of G2.

Why: A break in copied DNA is checked at the G2 checkpoint.
The G2 checkpoint sits at the end of G2, just before mitosis: moment 3, at 28 h.
The cell waits at the G2 checkpoint while the break is repaired.
The checkpoint holds the cell; other proteins repair the DNA.

Q4 T46-q04

In one culture, cells were timed from the start of G2 to the start of mitosis. Most took 4 hours. One cell that had been exposed to a burst of X-rays took 19 hours, and then entered mitosis with its DNA repaired.

Which statement about the G2 checkpoint explains the 19 hours?

  1. A. The G2 checkpoint is a timer that lasts longer in a cell that has been exposed to X-rays
    A checkpoint is a set of conditions, not a clock.
    The exposed cell waited 19 hours because its DNA stayed damaged that long.
  2. B. The G2 checkpoint repaired the DNA itself, and that repair takes about fifteen hours
    A checkpoint holds the cell; it does not repair the DNA.
    Other proteins repaired the DNA during the hold, and once it was undamaged the cell moved on into mitosis.
  3. C. The G2 checkpoint sent the cell back to S phase to copy its DNA a second time
    A checkpoint never returns a cell to an earlier stage.
    The cell stayed in G2, held at the G2 checkpoint, until the repair was complete.
  4. D. ✓ The G2 checkpoint is a set of conditions, and the cell waited there until its DNA was undamaged

Why: The G2 checkpoint checks that the DNA is completely copied and undamaged: a set of conditions, not a timer.
The X-rayed cell’s DNA stayed damaged for 19 hours, so the cell waited that long.
As soon as the repair was complete, the condition was met and the cell entered mitosis.

Q5 T46-q05

In a strain of yeast, a mutation removes the M checkpoint's hold. When these cells divide, 12% of the daughter cells carry one chromosome too many or one too few, against 0.1% of the daughters of normal yeast.

Which condition does the missing hold normally check before the sister chromatids separate?

  1. A. That the DNA has been completely copied and carries no damage
    Complete, undamaged DNA is the G2 checkpoint's condition, checked before mitosis begins.
    The M checkpoint, at metaphase, checks the chromosomes' attachment to the spindle.
  2. B. ✓ That every chromosome is attached to spindle fibers from both poles
  3. C. That the cell has grown large enough and has a growth signal bound
    Size and a growth signal are checked at the G1 checkpoint, before the DNA is copied.
    A missing chromosome in a daughter comes from a fault at metaphase.
  4. D. That the cleavage furrow has pinched the cytoplasm into two
    The cleavage furrow forms in cytokinesis, after the sister chromatids have separated.
    The M checkpoint acts earlier, at metaphase, before anaphase begins.

Why: The M checkpoint checks that every chromosome is attached to fibers from both poles before the sister chromatids separate.
With the hold removed, a cell enters anaphase with one chromosome unattached.
That chromosome is not pulled to its pole: one daughter cell gets an extra chromosome, the other lacks it.

Q6 T46-q06

A researcher exposes cells in the regrowing tail of a lizard to a chemical that cuts DNA strands after S phase is complete. The cells stop at the end of G2 and stay there while the cuts are repaired.

Which harm does this hold prevent?

  1. A. ✓ A daughter cell receiving a damaged genome
  2. B. Damaged DNA being copied in S phase
    Copying is finished: the chemical acted after S phase was complete.
    The hold keeps the damage out of the daughter cells, not out of S phase.
  3. C. A cell entering the cycle with no growth signal
    The growth signal is checked at the G1 checkpoint.
    These cells are at the end of G2, held by the check that their copied DNA is undamaged.
  4. D. A daughter cell receiving an extra chromosome
    An extra chromosome comes from a chromosome unattached at metaphase, the M checkpoint's condition.
    The G2 checkpoint's hold is about the state of the DNA, not the number of chromosomes.

Why: The G2 checkpoint checks that the DNA is completely copied and undamaged before mitosis.
Holding a cell with cut strands there means the cell does not divide while the damage is in it.
So no daughter cell receives a damaged genome.
The pause also gives repair proteins time to work.

Q7 T46-q07

The graph below shows one cyclin’s concentration and the activity of its cyclin-dependent kinase (CDK) through one 20-hour cycle of a cultured cell line.

One cyclin's concentration (solid) and the activity of its CDK (dashed) through one 20-hour cycle of a cultured cell line, each as a percent of its own maximum. Gridlines every 20%.
One cyclin's concentration (solid) and the activity of its CDK (dashed) through one 20-hour cycle of a cultured cell line, each as a percent of its own maximum. Gridlines every 20%.

Which statement do the two curves support?

  1. A. The cyclin is the kinase, and the CDK is the protein that the cyclin phosphorylates to begin mitosis
    The CDK is the kinase: a cyclin-dependent kinase.
    The cyclin is its partner, and the CDK is active only while a cyclin is bound to it.
  2. B. The CDK's activity climbs in G2 because the cell makes more CDK then and destroys it after mitosis
    The amount of CDK stays steady through the cycle.
    Its activity rises in G2 because the cyclin’s concentration rises and more complexes form, not because more CDK is made.
  3. C. ✓ CDK activity rises and falls with the cyclin because the CDK is active only while a cyclin is bound
  4. D. The cyclin climbs through G2 because the active CDK phosphorylates it, and the phosphate keeps it stable
    The cyclin rises through G2 because the cell makes it faster than it destroys it.
    The CDK phosphorylates the proteins that start mitosis, not the cyclin.

Why: The CDK level is steady, and the CDK is active only while a cyclin is bound.
So active complexes appear as the cyclin climbs through G2 and peak with it as mitosis begins.
The cyclin is destroyed at the end of mitosis, so the complexes vanish.
So activity follows concentration.

Q8 T46-q08

A researcher measures the activity of a cyclin-dependent kinase (CDK) in extracts made from cells in G1, adding or removing proteins as the table below shows.

Kinase activity of extracts made from cells in G1, with proteins added or removed.
Kinase activity of extracts made from cells in G1, with proteins added or removed.

Which explanation accounts for the results?

  1. A. The cyclin is itself the kinase, and the CDK plays no part in the activity that was measured
    The CDK is the kinase; the cyclin has no kinase activity of its own.
    Adding cyclin worked because it switched on the CDK that was already in the extract.
  2. B. The added CDK was destroyed as soon as it entered the extract, so it could add nothing to the activity
    Nothing in the extract destroys CDK.
    Adding more CDK changed nothing because CDK was never what was missing: with no cyclin bound, any amount of CDK stays inactive.
  3. C. The cyclin supplied the phosphate groups that the CDK transfers onto its target proteins
    The phosphate groups a kinase transfers come from ATP, not from cyclin.
    The cyclin's job is to bind the CDK and switch it on.
  4. D. ✓ The CDK is active only while a cyclin is bound to it, so the cyclin was the missing partner

Why: A cyclin-dependent kinase is active only while a cyclin is bound to it.
The G1 extract had CDK but almost no cyclin, so nothing switched the CDK on.
Adding cyclin formed cyclin–CDK complexes, and the complexes were active.
Adding more CDK added CDK with no cyclin to bind: zero activity.

Q9 T46-q09

At the end of mitosis a cell destroys the cyclin that binds its cyclin-dependent kinase (CDK). Within minutes the proteins that started mitosis switch off, and nuclear envelopes re-form.

Which of the following best explains why destroying the cyclin switches those proteins off?

  1. A. ✓ The CDK loses its cyclin and stops adding phosphate groups, so a phosphatase returns the proteins to their off shape
  2. B. The cyclin was the kinase that kept the proteins phosphorylated, so its loss ends the phosphorylation
    The cyclin has no kinase activity; it switches the CDK on by binding it.
    Once it is destroyed, the CDK adds no more phosphates and a phosphatase strips the rest.
  3. C. Destroying the cyclin destroys the CDK bound to it, so the cell must make new CDK before it can move on
    Destroying the cyclin leaves the CDK intact; the CDK is present at a steady level through the whole cycle.
    What ends is its activity, because no cyclin is bound.
  4. D. The fragments of the destroyed cyclin bind the mitosis proteins and hold them switched off
    Cyclin fragments play no part.
    The mitosis proteins switch off for two reasons: a phosphatase removes their phosphate groups, and the CDK, with no cyclin bound, adds no new ones.

Why: The mitosis proteins are on while they carry a phosphate from the cyclin–CDK complex.
Destroying the cyclin leaves the CDK with no partner, so it stops adding phosphates.
A phosphatase removes the phosphates already on the mitosis proteins.
So they return to their off shape, and mitosis switches off.

Q10 T46-q10

A cultured cell waits at the G1 checkpoint. The table below describes the cell, including the concentration of the cyclin that drives entry into S phase, the partner of the cell’s cyclin-dependent kinase (CDK).

A cultured cell waiting at the G1 checkpoint: the state of each G1 condition and of the cyclin that drives entry into S phase.
A cultured cell waiting at the G1 checkpoint: the state of each G1 condition and of the cyclin that drives entry into S phase.

Which of the following best explains the wait?

  1. A. The cyclin has to reach 100% of its peak before a single cyclin–CDK complex can form in the cell
    Complexes form as soon as cyclin binds CDK; there is no threshold at 100%.
    At 90% the cell holds plenty of complex, but a signal keeps it switched off.
  2. B. A small cell holds too little CDK to pair with the cyclin it has made, so no complex forms
    CDK is present at a roughly steady level in cells of every size, so a small cell has CDK to spare.
    The complex forms; it is held switched off.
  3. C. ✓ A signal from the unmet size condition keeps the cyclin–CDK complex switched off, so the S-phase proteins stay off
  4. D. The bound growth factor is a signal to stop, and it holds the cell in G1 until the cell has grown
    A growth factor is a signal to divide, one of the G1 checkpoint's conditions, and here it is met.
    The unmet condition is size.

Why: An unmet condition sends a signal that keeps the next stage’s proteins off.
Size is a G1 condition, and this cell is still small.
The S-phase complex is present, but the size signal keeps it off.
Only when the cell has grown does the complex switch S phase on.

Q11 T46-q11

In a tumor, the cell in which the gene for a checkpoint protein first mutated died years ago. The tumor has gone on growing ever since.

Why does the tumor keep growing after that first cell has died?

  1. A. The mutation spread from the first cell to its neighbors through the fluid around them
    A mutation is a change in a cell’s DNA and does not pass through fluid.
    It reaches other cells only when that DNA is copied into daughters.
  2. B. ✓ The first cell's daughters inherited the mutated gene, so they lack the same hold and pass it on to their daughters
  3. C. The tumor's cells divide faster than normal cells, so the loss of one cell makes no difference
    Cancer cells do not divide faster because they are cancer; they divide when they should not.
    The tumor grows because every daughter carries the same missing hold.
  4. D. The dying cell released a growth factor that has kept its neighbors dividing ever since
    A growth factor from a dying cell would fade within hours.
    The tumor has grown for years because the fault is in a gene, copied into every daughter cell.

Why: Each control is a protein made from a gene.
When the first cell divided, each daughter received the mutated gene and so lacks the hold.
Each daughter passes it on, so its daughters lack the hold too.
The fault is inherited down the line, so the cells keep dividing.

Q12 T46-q12

A line of cancer cells passes the G1 checkpoint and copies its DNA even when the DNA carries breaks. With no growth factor added, however, the cells stay in G1, as normal cells do.

In which gene is the line's mutation most likely to be?

  1. A. The gene for the growth-factor receptor, locking the receptor in its active shape
    A locked-on receptor would send the enter-the-cycle message with no growth factor bound, but this line still waits for it.
    The lost control is the check on damaged DNA.
  2. B. The gene for the cyclin that drives entry into mitosis, making the cyclin impossible to destroy
    A cyclin that cannot be destroyed keeps the mitosis-entry complex active, so cells cannot finish mitosis and collect there.
    This line’s fault shows at the G1 checkpoint.
  3. C. The gene for the phosphatase that removes phosphate groups from the proteins that start S phase
    Losing that phosphatase would leave the S-phase proteins on, driving cells into S phase whatever the conditions, even with no growth factor.
    This line still waits for the growth factor.
  4. D. ✓ The gene for a checkpoint protein that detects damaged DNA at the G1 checkpoint

Why: At the G1 checkpoint a checkpoint protein detects damaged DNA and keeps the cell out of S phase.
A mutation that leaves that protein useless removes that one hold, so the cells copy damaged DNA.
The cells still wait for a growth signal, so the rest of the checkpoint works.

Q13 T46-q13

Researchers gave bone marrow cells from two strains of mouse the same DNA-damaging drug in a dish and followed the cells for three days. In these cells a checkpoint hold on damaged DNA lasts about a day; a held cell that has not been removed by then resumes the cycle. The table below gives the results.

Bone marrow cells from two strains of mouse, given the same DNA-damaging drug in a dish and followed for three days.
Bone marrow cells from two strains of mouse, given the same DNA-damaging drug in a dish and followed for three days.

Which control has strain 2 lost?

  1. A. The enzymes that repair damaged DNA
    Strain 1’s damaged cells were not mended either: they were removed, and its count fell to 40%.
    Strain 2’s cells kept the same damage and divided with it.
  2. B. The requirement for a growth signal before the cycle starts
    Strain 2’s cells were held at the G1 checkpoint through the first day, so entry controls worked.
    What failed is the removal of a held cell whose damage stayed.
  3. C. ✓ The response that makes a cell with damage beyond repair remove itself
  4. D. The G1 checkpoint's hold on damaged DNA
    Strain 2’s cells were held at the G1 checkpoint through day one, so the checkpoint works.
    A held cell with damage beyond repair normally removes itself; strain 2’s did not.

Why: A held cell with damage beyond repair normally removes itself: strain 1’s count fell to 40%.
Strain 2’s cells were held through day one, so the checkpoint works, but were not removed.
The hold lapsed and they divided, damage and all: 150%.
So strain 2 lost the apoptosis response.

Q14 T46-q14

Cultured hamster cells were sorted by stage before and 18 hours after a drug was added; the graph below shows the percent in G1, S phase, G2 and mitosis. The cells had nutrients and a growth factor throughout, their DNA was undamaged, and the cells in G1 had reached full size.

Percent of cultured hamster cells in each stage before a drug was added and 18 hours after, each from one count of 200 cells. Gridlines every 20%.
Percent of cultured hamster cells in each stage before a drug was added and 18 hours after, each from one count of 200 cells. Gridlines every 20%.

Which action of the drug accounts for the change?

  1. A. ✓ The drug keeps the complex that drives entry into S phase switched off
  2. B. The drug removes the growth signal
    The growth factor was present throughout, so that condition was met.
    Cells held in G1 with the signal present means the block sits downstream of it.
  3. C. The drug damages DNA
    The DNA was undamaged, so that condition was met.
    Size, nutrients, signal and undamaged DNA were all present, so the hold came from the drug keeping the S-phase complex off.
  4. D. The drug stops the destruction of the cyclin that drives entry into S phase
    A cyclin that is not destroyed keeps its complex active, which would drive cells into S phase.
    Here cells failed to enter S phase, so the complex was off.

Why: G1 rose to 79% and S phase fell to 6%: cells were held at the end of G1.
Every G1 condition was met, so no condition caused the hold.
The drug keeps the S-phase-entry complex switched off.
So the S-phase proteins stay off, and cells collect in G1.

Q15 T46-q15

Cells in one dish receive a spindle inhibitor, a drug that stops the spindle fibers from moving the chromosomes; these cells are held at metaphase with their sister chromatids joined. Cells in a second dish receive the spindle inhibitor together with a drug that removes the M checkpoint’s hold.

Which of the following best predicts what the cells in the second dish do?

  1. A. The cells stay at metaphase
    Attachment was only a condition the M checkpoint checked.
    The second drug removes the hold, so the next stage’s proteins switch on and the cell leaves metaphase.
  2. B. ✓ The cells leave metaphase with their chromosomes unattached
  3. C. The cells drop back into G2 and build a new spindle
    A checkpoint hold never returns a cell to an earlier stage, and removing the hold does not either; the cells go forward from metaphase.
  4. D. The cells leave the cycle into G0
    G0 is entered from G1, at the start of the cycle.
    These cells are in mitosis.
    With no hold, the cells finish mitosis.

Why: The M checkpoint’s condition is that every chromosome is attached from both poles.
With a spindle inhibitor that is unmet, so normally the cell is held at metaphase.
The second drug removes the hold, so the next stage’s proteins switch on and the sister chromatids separate.

Q16 T46-q16

Cancer cell line A enters the cycle with no growth factor present, because a relay protein in its growth-signal pathway is always active; its checkpoints work. Cancer cell line B waits for the growth factor as normal cells do, but passes the G1 checkpoint carrying damaged DNA. A drug blocks line A's always-active relay protein.

Predict the effect of the drug on each line's division count in a dish with no growth factor added.

  1. A. Both lines' counts fall to normal, because both lines depend on the relay protein
    Line B’s fault is a checkpoint that lets damaged DNA through, downstream of the growth signal.
    Blocking the relay protein changes nothing for line B.
  2. B. Line B's count falls to normal; line A's stays as it was
    Line A is the line with the always-active relay protein; blocking that protein removes line A's fault, while line B's fault is in a checkpoint the drug does not touch.
  3. C. Both counts stay as they were, because a drug cannot restore a lost control
    A drug can block an always-active protein: the enter-the-cycle message then stops, and line A waits for a growth factor as normal cells do.
  4. D. ✓ Line A's count falls to normal; line B's stays as it was

Why: Line A divides with no growth factor: its relay protein sends the message on its own.
The drug blocks that protein, so line A waits in G1 and its count falls to normal.
Line B’s fault is a checkpoint protein the drug does not touch, so its count is unchanged.

Q17 T46-q17

Three lines of mouse cells were grown for 36 hours in a dish with no growth factor added. The table below gives their division counts, the percent of X-ray-damaged cells that entered S phase, and the percent of damaged cells that underwent apoptosis.

Three lines of mouse cells: division events per 100 cells in 36 hours in a dish with no growth factor added; the percent of X-ray-damaged cells that entered S phase within 6 hours; and the percent of damaged cells that had undergone apoptosis after 2 days.
Three lines of mouse cells: division events per 100 cells in 36 hours in a dish with no growth factor added; the percent of X-ray-damaged cells that entered S phase within 6 hours; and the percent of damaged cells that had undergone apoptosis after 2 days.

Which control has line 1 lost, and which has line 2 lost?

  1. A. ✓ Line 1 has lost the G1 checkpoint's hold on damaged DNA; line 2 has lost the requirement for a growth signal
  2. B. Line 1 has lost the requirement for a growth signal; line 2 has lost the G1 checkpoint's hold on damaged DNA
    Line 1’s division count with no growth factor is normal (3 per 100 cells), so it still waits for the signal; line 2, at 41, divides with no signal.
  3. C. Line 1 has lost the enzymes that repair DNA; line 2 has lost the M checkpoint's hold at metaphase
    Repair enzymes were never measured; what line 1 shows is damaged cells entering S phase (55%) instead of being held.
    Line 2’s damaged-cell counts are normal, so its checkpoints work.
  4. D. Line 1 has lost the apoptosis response only, with its checkpoints intact; line 2 has lost the destruction of cyclin
    With only apoptosis lost, damaged cells would still be held, and only about 3% would enter S phase; in line 1, 55% did, so the hold itself is gone.

Why: Line 1: 55% of damaged cells entered S phase against 3% normally, so its G1 hold on damaged DNA is lost.
Line 2: 41 divisions per 100 cells with no growth factor against 3, so it enters the cycle with no signal.
Line 2 still holds and removes damaged cells.

Q18 T46-q18

Skin cells at the edge of a fresh cut begin dividing within a day. A student says: “The cells divide because the skin needs new cells to close the gap.”

Is the student correct, and why?

  1. A. Yes — each edge cell senses the empty space beside it and divides to fill the need
    A cell has no sense of what the tissue needs.
    Division is switched on by molecules: growth factors released at the wound bind the cell’s receptors.
  2. B. Yes — the gap lets more nutrients reach the edge cells, and the extra food makes them divide
    Nutrients are building material; they do not tell a cell to divide.
    The signal is a growth factor binding a receptor.
  3. C. ✓ No — growth factors released at the wound bind the edge cells’ receptors and switch division on
  4. D. No — the cut damaged the edge cells’ DNA, and a damaged cell divides to replace itself
    Damaged DNA holds a cell at a checkpoint; it never starts a division.
    The edge cells divide because growth factors from the wound bind their receptors.

Why: Division is not a cell’s default; it has to be switched on.
Damaged tissue at the cut releases growth factors.
The growth factors bind receptors on the edge cells.
The bound receptors tell those cells to enter the cycle and divide.

FRQ 1 T46-frq1 · Scientific Investigation

Researchers took bladder cells from a healthy bladder and from a bladder tumor of the same patient and grew them as two cell lines. They seeded each line into five dishes at low density (cells far apart) and five dishes at high density (cells touching neighbors on every side), all in the same medium with the growth factor, and the medium was replaced every 6 hours. After 24 hours the researchers counted the percent of cells in S phase in each dish. The graph below shows the mean of the five dishes for each condition; error bars represent ±2SE.

Mean percent of cells in S phase after 24 hours for bladder cells from a healthy bladder (normal) and from a bladder tumor of the same patient, each seeded at low and at high density; five dishes per condition. Error bars represent ±2SE. Gridlines every 5%.
Mean percent of cells in S phase after 24 hours for bladder cells from a healthy bladder (normal) and from a bladder tumor of the same patient, each seeded at low and at high density; five dishes per condition. Error bars represent ±2SE. Gridlines every 5%.

(a) Identify the dependent variable in this investigation. (1 pt)

Model answer The dependent variable is the percent of cells in S phase in each dish after 24 hours, which the researchers counted.
Rubric
  • Award 1 point for: the percent of cells in S phase after 24 hours (the quantity the researchers counted in each dish).
  • Accept "the percent of cells copying their DNA". Do not award the point for the seeding density or the cell line, which the researchers set, or for "the number of cells".

Slip Naming the density or the cell line. The density and the cell line were what the researchers set; the dependent variable is what the researchers counted at the end.

(b) Support the claim that the fall in the normal cells' S-phase percent at high density is caused by contact with neighbors rather than by a shortage of nutrients, using the description of the experiment. (1 pt)

Model answer The medium was the same in every dish and was replaced every 6 hours.
So the high-density cells had the same nutrients and growth factor as the low-density cells.
The one thing that differed between the dishes was density.
At high density each cell was touched by neighbors on every side.
Contact on every side is a signal that holds a cell in G1.
Therefore contact, not a shortage of nutrients, cut the S-phase percent from 31% to 4%.
Rubric
  • Award 1 point for: the evidence (the medium, with its nutrients and growth factor, was the same in every dish and was replaced every 6 hours) AND the reasoning (so nutrients were equal at both densities, and the only difference was whether each cell was touched on every side, so the fall from 31% to 4% is credited to contact, a signal from neighbors that holds the cycle in G1).
  • Accept 'nutrients were kept the same, so the change in density is the only difference', with the link to contact stated. Do not award the point for 'crowded cells have less food', or for the medium fact with no link to the claim.

Slip Writing that crowded cells had too little food, or naming the medium with no link to the claim. The medium was the same in every dish; the change that came with density was contact with neighbors.

(c) Using the error bars, describe what the data show about the effect of density on the tumor cells, and identify the control on the cycle that these cells have lost. (1 pt)

Model answer The tumor cells' error bars overlap: 30% to 36% at low density and 27% to 33% at high density.
Overlapping bars mean the data show no difference between the two densities.
So density has no measurable effect on the tumor cells' S-phase percent.
The tumor cells have lost the contact control: the signal from neighbors on every side that holds normal cells in G1.
Rubric
  • Award 1 point for both parts together: the tumor cells' bars at low density (30% to 36%) and at high density (27% to 33%) overlap, so the data show no difference in S-phase percent between the two densities; the tumor cells have lost the contact signal from neighbors that holds normal cells in G1 when they are touched on every side.
  • Accept "the bars overlap, so no difference is shown; they ignore contact". Do not award the point for "the tumor cells divide faster", or for a conclusion drawn from the two means alone (33% against 30%).

Slip Reading the means alone (33% against 30%) as a small drop. The bars overlap, so the data have shown no difference; and the lost control is the contact signal, not the speed of the cycle.

(d) Tests on the tumor cells show that they enter the cycle only when a growth factor is bound to their receptors. The researchers plan to repeat the high-density tumor dishes with the growth factor left out of the medium. Predict the percent of tumor cells in S phase after 24 hours, and justify your prediction. (1 pt)

Model answer About 4%: few tumor cells will be in S phase.
The tumor cells enter the cycle only when a growth factor is bound to their receptors.
The control the tumor cells lost is the contact hold, not the growth-signal requirement.
So with the growth factor left out, the tumor cells wait in G1 or drop into G0, like normal cells with no signal.
Rubric
  • Award 1 point for: a low value, near the normal cells' high-density value (about 4%; accept 0% to 10%), with the reason that the tumor cells enter the cycle only when a growth factor is bound; the tumor cells lost the contact hold but still need the growth signal, so with the growth factor absent they wait in G1 or drop into G0.
  • Do not award the point for "about 30%, because cancer cells divide whatever the conditions", or for a prediction with no reason.

Slip Predicting about 30%, as if a cancer cell divides whatever the conditions. This line lost one control, contact; the task says the tumor cells still need the growth signal.

FRQ 2 T46-frq2 · Analyze Model

The graph below is a model of one 24-hour cycle in a cultured cell line. The graph shows the concentrations of two cyclins, cyclin X and cyclin Y, each as a percent of its own peak, and the amount of cyclin-dependent kinase (CDK) as a percent of its maximum. In this line, cyclin X bound to a CDK drives entry into S phase, and cyclin Y bound to a CDK drives entry into mitosis. The stages are marked along the top of the graph.

A model of one 24-hour cycle in a cultured cell line: the concentrations of cyclin X (solid) and cyclin Y (dashed), each as a percent of its own peak, and the amount of CDK (dotted) as a percent of its maximum. Gridlines every 20%.
A model of one 24-hour cycle in a cultured cell line: the concentrations of cyclin X (solid) and cyclin Y (dashed), each as a percent of its own peak, and the amount of CDK (dotted) as a percent of its maximum. Gridlines every 20%.

(a) Describe the pattern of cyclin Y's concentration through the cycle, using times from the graph. (1 pt)

Model answer Cyclin Y stays low, near 6% of its peak, through G1 and S phase, from 0 to about 17 h.
Cyclin Y rises through G2, from about 17 h, to its peak at 23 h as mitosis begins, and it collapses back to its low level by 24 h, the end of mitosis.
Rubric
  • Award 1 point for: cyclin Y stays low through G1 and S phase (about 0 to 17 h), rises through G2 (from about 17 h to 23 h), peaks at about 23 h as mitosis begins, and falls back to its low level by the end of mitosis (24 h).
  • Accept times within about an hour of these. Do not award the point for a description of cyclin X or of the CDK line, or for "it goes up and down" with no times.

Slip Describing cyclin X by mistake, or giving the shape with no times. The task asks for the pattern of cyclin Y with times read from the graph.

(b) Explain how the rise in cyclin Y moves the cell into mitosis, although the amount of CDK stays flat. (1 pt)

Model answer The CDK is present all the time, but the CDK is active only while a cyclin is bound to it.
As cyclin Y rises through G2, more cyclin Y–CDK complexes form.
Each active complex transfers a phosphate group from ATP onto the proteins that start mitosis.
The added phosphate changes those proteins' shape and switches them on, so mitosis begins.
The amount of CDK stays flat; what rises is the number of active complexes.
Rubric
  • Award 1 point for: the CDK is active only while a cyclin is bound to it, so as cyclin Y rises more active cyclin Y–CDK complexes form; the active complex transfers a phosphate group from ATP onto the proteins that start mitosis, the added phosphate changes their shape and switches them on, so mitosis begins; the amount of CDK stays the same, and it is the number of active complexes that rises.
  • Do not award the point for "cyclin Y is the kinase", for "more CDK is made", or for "the complex starts mitosis" with no phosphate transfer or shape change.

Slip Writing that cyclin Y switches on the mitosis proteins itself. Cyclin Y has no kinase activity; it binds the CDK, and the CDK transfers the phosphate groups.

(c) A mutation removes cyclin X from these cells. Make a claim about where the mutant cells collect and what happens to their division. (1 pt)

Model answer The mutant cells collect in G1, at the G1 checkpoint, or drop into G0.
The mutant cells never enter S phase, so division stops.
Rubric
  • Award 1 point for the claim: the cells collect in G1 (at the G1 checkpoint, or drop into G0) and division stops. No reasoning is required for this point.
  • Accept 'they stay in G1 and stop dividing'. Do not award the point for 'they collect in G2' or 'in mitosis', or for 'they divide with no DNA copied'.

Slip Predicting a pile-up in G2 or in mitosis. Cyclin X bound to a CDK drives entry into S phase. Without cyclin X, the cells never leave G1, so the pile-up is at the very first step of the cycle, in G1.

(d) Support your claim in (c) using the model. (1 pt)

Model answer On the model, cyclin X rises at the end of G1, and cyclin X bound to a CDK drives entry into S phase.
With no cyclin X, no cyclin X–CDK complex forms.
So the proteins that start S phase are never phosphorylated, and they stay off.
The cells never copy their DNA, so they never reach G2.
Cyclin Y's rise, which the model places in G2, therefore never comes.
So mitosis is never entered.
Rubric
  • Award 1 point for: the evidence from the model (cyclin X rises at the end of G1, and cyclin X bound to a CDK drives entry into S phase) AND the reasoning (with no cyclin X, no active cyclin X–CDK complex forms, so the proteins that start S phase are never switched on, so the cells never copy their DNA, never reach G2 and never enter mitosis).
  • Accept 'no cyclin X, no active complex, no entry into S phase', tied to the model. Do not award the point for 'the CDK works on its own', or for a justification built on cyclin Y alone.

Slip Arguing from cyclin Y or from the CDK line alone. The CDK is present, but with no cyclin X bound it stays inactive; cyclin Y rises only in cells that have reached G2.

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