AP Biology Unit 2 — reviewer copy

Reviewer copy generated from the built lesson pages and the test JSON · 36 lessons · 2140 steps · 403 figures · 463 lesson MCQ · 65 numeric · 55 lesson FRQ · 14 worked examples · 11 practice sets · 11 tests · 292 test/practice MCQ · 44 test/practice FRQ · 130 videos

APBIO-U02-L01 Two watery worlds and the film between them

Topic 2.3 · Plasma Membrane · 60 steps

A red blood cell drifting in a drop of blood, with a cross-section of its edge drawn beside it

Here is a red blood cell in a drop of blood.

Salty water fills it and salty water surrounds it, and everything that separates the two is a film two molecules thick: the plasma membrane. The salt inside stays inside.

Unit 2 · Cell Structure and Function

Video: Watch first: Plasma membrane

A red blood cell in a drop of blood, and a zoom to the film around it.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T23-intro.mp4

2Salty water on both sides

Video: Watch: The film between two watery worlds

Cytosol inside, extracellular fluid outside, and a film two molecules thick between them, arranged by water.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L01.mp4

Every cell has a boundary that controls what enters and leaves it: the plasma membrane, built on a lipid bilayer.

Here is the red blood cell cut through. The small box on its edge is enlarged on the right, and there is the plasma membrane: two rows of phospholipids, with water on both sides.

A red blood cell cut through, its fluid outside shaded one way and its fluid inside another, with one piece of its edge zoomed in: the plasma membrane, two rows of phospholipids with water on both sides

Inside the membrane, a watery solution fills the cell: water, with salts, sugar and proteins dissolved in it.

Outside the membrane is another watery solution, the liquid part of the blood: water again, with salts and sugar dissolved in it.

The watery solution filling a cell inside its plasma membrane is called the .

The watery solution outside a cell is called the . It is whatever fluid is outside the cell.

For a red blood cell, the extracellular fluid is the blood around it. For a cell lining the gut, the extracellular fluid is the gut fluid beside it.

For a muscle cell, the extracellular fluid is the fluid between the muscle fibers.

The cytosol contains water, and the extracellular fluid contains water. So the plasma membrane has water against both of its faces: cytosol on one face, extracellular fluid on the other.

What you are expected to know Identify the cytosol and the extracellular fluid on a drawing of a cell in its surroundings, and say that the plasma membrane has water against both of its faces.

Check q1

A cell is drawn cut through, with three positions marked A, B and C.

A cell cut through, with three positions marked A, B and C

Which position marks the cytosol (the fluid inside the cell)?

A. A
Position A is in the fluid outside the cell, the extracellular fluid.
B. B
Position B is on the plasma membrane itself, the film that separates the two solutions.
C. ✓ C

Why: The cytosol is the watery solution filling the cell inside its plasma membrane, and position C is in it.
Position A is in the extracellular fluid, and position B is on the membrane.

Check q2

A muscle cell sits in the fluid between the muscle fibers. That fluid is the cell’s extracellular fluid (the fluid outside the cell).

Which statement about its plasma membrane is right?

A. ✓ Water touches both of its faces
B. Water touches only its outer face
The cytosol filling the cell contains water, so the inner face of the membrane is in water too.
C. Water touches only its inner face
The fluid between the muscle fibers, the extracellular fluid, contains water, so the outer face of the membrane is in water too.
D. Water touches neither face
Both solutions the membrane touches contain water.

Why: Every cell is filled with cytosol and surrounded by extracellular fluid.
The cytosol contains water, and the extracellular fluid contains water.
So the plasma membrane has water against both of its faces.

16The parts of the membrane, named

Video: Watch: The parts of the membrane, named

Heads toward the water on both faces, the hydrophobic interior where the tails of the two layers touch, and a wrong drawing.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L01b.mp4

Here is the membrane, drawn large: the extracellular fluid (outside the cell) above it, the cytosol (inside the cell) below it.

A lipid bilayer with the extracellular fluid above it and the cytosol below it, the two regions shaded differently

Phospholipids in water build a bilayer. Water pulls on the heads and not on the tails, so two layers lie tail to tail.

The extracellular fluid contains water, and the cytosol contains water. Water pulls on the heads.

So the heads of one layer face the extracellular fluid, and the heads of the other layer face the cytosol.

The tails of the two layers touch in the middle, away from both watery sides. They make a band of hydrocarbon with no water in it.

This oily middle of the membrane, the band where the tails touch and water does not enter, is called the .

Before the next drawing, decide for yourself where two labels go on this membrane: “heads toward water” and “hydrophobic interior”.

Here is the labeled cross-section. “Heads toward water” goes on both faces, because both faces touch a watery solution. “Hydrophobic interior” goes across the middle, where the tails touch.

The labeled cross-section: heads toward water on both faces, and the hydrophobic interior across the middle

A drawing with tails facing the cytosol is wrong. The cytosol contains water, and water pulls on heads, not tails, so the face toward the cytosol shows heads.

What you are expected to know Label a drawn cross-section of a plasma membrane: heads toward the water on both faces, and the hydrophobic interior where the tails touch. Spot a drawing whose tails face a watery side as wrong.

Check q3

Three membranes are drawn, P, Q and R, each with water above it and water below it.

Three drawn membranes, P, Q and R, each between water above and water below

Which drawing shows a plasma membrane correctly?

A. Q only
Q has the tails facing the water and the heads in the middle, the reverse of a real membrane.
B. ✓ P only
C. R only
R’s lower layer is upside down, with its tails toward the water below.
D. P and R
R’s upper layer is right, but its lower layer shows tails to the watery side below, so R is not a plasma membrane.

Why: In P the heads face the water on both faces and the tails touch in the middle.
Q has the tails facing the water, and R has its lower layer upside down with tails toward the water below.

Check q4

A student labels the cross-section of a plasma membrane shown. She has written “hydrophobic interior” beside the face of the membrane that touches the cytosol.

A student's labeled cross-section of a plasma membrane: extracellular fluid above, cytosol below, and the words hydrophobic interior written beside the face that touches the cytosol

What is wrong with her label?

A. Nothing: the face toward the cytosol is the membrane’s dry face
The cytosol contains water, so the face toward it is not dry; the heads on that face sit in water.
B. ✓ The label belongs in the middle, where the tails of both layers touch
C. The label belongs on the other face, the one toward the extracellular fluid
Both faces of the membrane touch water, cytosol on one and extracellular fluid on the other.
D. The label belongs on the heads, since the heads keep water out of the membrane
The heads are the parts water pulls on; they sit in the water, not away from it.

Why: The hydrophobic interior is the band in the middle of the membrane where the hydrocarbon tails of the two layers touch.
Both faces of the membrane are in water; only the middle is oily.

30Where a membrane protein’s parts sit

Check q5

A protein folds up in water.

Where do its nonpolar R groups end up?

A. ✓ Tucked inside, away from the water
B. On the outside, in the water
Water has nothing to pull on in a nonpolar R group, so the nonpolar R groups gather together away from the water.
C. Spread evenly through the protein
Water pulls on polar and charged R groups and not on nonpolar ones, so the two kinds sort: polar outside, nonpolar inside.

Why: Water pulls on polar and charged R groups and has nothing to pull on in nonpolar ones, so the nonpolar R groups gather inside, away from the water.

Video: Watch: Where a membrane protein’s parts sit

The nonpolar region against the tails, the polar regions in the water on each side, and a water-filled passage.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L01c.mp4

There are three kinds of R group: nonpolar, polar and charged. Water pulls on polar and charged R groups, and not on nonpolar ones.

You’ve also seen a protein fold in water with its nonpolar R groups tucked inside, away from the water. The same pull decides where a protein sits in a membrane.

Here is a protein set into the membrane. It is one folded chain of amino acids, drawn as beads along its backbone, and it reaches from the extracellular fluid to the cytosol.

A membrane protein spanning the bilayer, drawn as a folded chain of beads: a polar region in the extracellular fluid, a nonpolar region shaded across the middle against the tails, and a polar region in the cytosol

A protein set into a membrane like this is called a .

This membrane protein has three regions. Its middle region has nonpolar R groups. Its two end regions have polar and charged R groups.

Water does not pull on nonpolar R groups, so nothing holds the middle region in the water. The nonpolar region sits against the hydrocarbon tails, in the hydrophobic interior.

Water does pull on polar and charged R groups. So one end region is held in the cytosol, and the other end region is held in the extracellular fluid.

So the membrane protein spans the membrane. Its nonpolar region is in the hydrophobic interior, and its two polar regions are in the water, one on each side.

Some membrane proteins have a water-filled passage through them. Here is one cut through: the passage reaches from the extracellular fluid to the cytosol, and water fills it.

A membrane protein cut through: a water-filled passage down its middle joins the extracellular fluid to the cytosol, water molecules sit in the passage, polar groups line it, and the protein's nonpolar regions face the tails

Water fills the passage, and water pulls on polar and charged R groups. So polar and charged R groups line the passage. The protein’s nonpolar regions face outward, against the tails.

What you are expected to know Predict, for a protein drawn in a membrane with its R groups marked, which regions lie against the tails (the nonpolar ones) and which face the cytosol or the extracellular fluid, or line a water-filled passage (the polar and charged ones).

Check q6

A membrane protein has a stretch of twenty nonpolar R groups in its middle and charged R groups at both ends.

Where does each part sit?

A. ✓ Nonpolar stretch against the tails; charged ends in the water on each side
B. Charged ends against the tails; nonpolar stretch in the water
Water pulls on charged R groups and not on nonpolar ones, so the two regions are the other way round.
C. The whole protein in the hydrophobic interior, since it is set into the membrane
Water is attracted to charged R groups, so the charged ends cannot sit in the oily middle.
D. The whole protein in the water, since proteins fold up in water
Water does not pull on nonpolar R groups, so a stretch of twenty of them does not stay in the water.

Why: Water pulls on charged R groups and not on nonpolar ones.
So the charged ends are attracted into the water on each side of the membrane, and the nonpolar stretch, which nothing attracts into the water, sits against the tails in the hydrophobic interior.

Check q7

Here is a plasma membrane split open between its two layers of phospholipids. Small particles stand out from both exposed faces.

A plasma membrane split open between its two layers: the upper layer lifted away from the lower one, and small particles standing out from both exposed faces

What could the particles be?

A. Phospholipid heads that had sunk into the middle of the membrane
Water pulls the heads to the two watery faces of the membrane; they do not sit among the tails.
B. Salt from the extracellular fluid, dissolved in the middle of the membrane
Salt is charged, water is attracted to it, and it stays in the watery solutions on either side.
C. ✓ Membrane proteins, set into the bilayer among the tails
D. Drops of water trapped between the two layers of the membrane
Water does not enter the layer of tails, so no drops are trapped there.

Why: Splitting the membrane between its layers opens the hydrophobic interior.
Membrane proteins are set into the hydrophobic interior by their nonpolar regions, so particles standing out from the exposed faces can be membrane proteins.

46Cholesterol, glycoproteins and glycolipids

Check q8

Cholesterol is a steroid. Its four carbon rings and its short tail are made of carbon and hydrogen only; one small –OH group sits at one end.

Is the ring-and-tail part of cholesterol polar or nonpolar?

A. Polar
A polar part carries partial charges, and carbon–hydrogen rings carry none.
B. ✓ Nonpolar

Why: Carbon and hydrogen share electrons almost equally.
So the rings and tail carry no partial charges, and that part of cholesterol is nonpolar.

Video: Watch: Cholesterol, glycoproteins and glycolipids

Cholesterol’s rings sit among the tails; glycoproteins and glycolipids carry their carbohydrate chains on the outer face only.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L01d.mp4

Cholesterol is a steroid that is not a hormone, sitting among the phospholipid tails of an animal cell membrane and steadying it.

A lipid bilayer between the extracellular fluid above and the cytosol below, the two regions shaded differently, with cholesterol molecules, drawn as four small rings, sitting among the tails

The red blood cell is an animal cell, so cholesterol sits in its membrane too: nonpolar, among the nonpolar tails, in the hydrophobic interior.

Some membrane proteins carry a short chain of sugar units. A protein with a carbohydrate chain attached is called a .

A glycoprotein and a glycolipid in the membrane, the extracellular fluid above shaded one way and the cytosol below another; both carbohydrate chains, drawn as beads, reach into the extracellular fluid

Some phospholipids carry a chain of sugar units too. A lipid with a carbohydrate chain attached is called a .

Every one of these carbohydrate chains faces the outside of the cell. On the red blood cell they reach into the blood, never into the cytosol.

What you are expected to know Point out, on a drawn membrane, cholesterol among the tails, a glycoprotein and a glycolipid, and say that their carbohydrate chains face the outside of the cell.

Check q9

The drawing shows part of an animal cell membrane with two molecules numbered 1 and 2. Molecules 1 and 2 each carry a carbohydrate chain.

A drawn animal cell membrane with two molecules numbered 1 and 2, each with a short chain of beads reaching into the extracellular fluid; the lead for 1 ends inside a rounded block that spans both layers, the lead for 2 ends on a round head in the outer layer

Which names fit the two molecules?

A. 1 is a glycolipid; 2 is a glycoprotein
The name follows what the chain is attached to, and molecule 1 is a protein while molecule 2 is a lipid.
B. ✓ 1 is a glycoprotein; 2 is a glycolipid
C. Molecules 1 and 2 are both glycoproteins, since both carry a chain of sugar units
The chain is the same kind on each molecule; what differs is what it is attached to, and molecule 2 is a phospholipid, not a protein.
D. Molecules 1 and 2 are both glycolipids, since both sit in the lipid bilayer
Sitting in the lipid bilayer does not make a molecule a lipid; molecule 1 is a protein.

Why: A protein with a carbohydrate chain attached is a glycoprotein, and a lipid with a carbohydrate chain attached is called a glycolipid.
Molecule 1 is a protein, so it is a glycoprotein; molecule 2 is a phospholipid, so it is called a glycolipid.

Check q10

A red blood cell carries glycoproteins and glycolipids in its membrane.

Which way do their carbohydrate chains face?

A. Toward the cytosol
No carbohydrate chain reaches into the cytosol; all of them face the outside of the cell.
B. Into the hydrophobic interior
Sugar chains are polar and water is attracted to them, so they do not sit among the tails.
C. Half toward each side
The chains are not shared between the two faces; all of them face the outside of the cell.
D. ✓ Toward the blood, outside the cell

Why: The carbohydrate chains on glycoproteins and glycolipids face the outside of the cell.
On a red blood cell that is the blood, the extracellular fluid, never the cytosol.

Check q11

The drawing shows an animal cell’s membrane in cross-section, with four positions marked 1 to 4.

An animal cell membrane in cross-section with four marked positions: 1 among the heads on the outer face, 2 in the cytosol just inside the membrane, 3 among the tails in the middle, 4 at the carbohydrate chain of a glycoprotein in the extracellular fluid; each lead ends on a marker dot at the position

Where does cholesterol sit in it?

A. At 1, among the heads on the outer face
Cholesterol is nonpolar, and water is not attracted to it among the heads.
B. At 2, in the cytosol, just inside the membrane
Cholesterol is nonpolar and does not stay in a watery solution such as the cytosol.
C. ✓ At 3, among the tails, in the hydrophobic interior
D. At 4, attached to the carbohydrate chains outside
Cholesterol carries no carbohydrate chain and does not attach to one.

Why: Cholesterol is nonpolar, so water is not attracted to it in either watery solution.
It sits where the membrane is dry: among the phospholipid tails, in the hydrophobic interior, where it steadies the membrane.

Every piece of the membrane sits where water puts it:
• Phospholipid heads and the polar parts of proteins: in the water.
• Phospholipid tails and the nonpolar parts of proteins: in the hydrophobic interior.
• Cholesterol: among the tails.
• Carbohydrate chains: on the outside of the cell.

The film around the red blood cell is phospholipids lined up by the salty water on both sides, with proteins, cholesterol, glycoproteins and glycolipids set into it, each held in place by water’s pull on its polar parts and not on its nonpolar parts.

Glossary

cytosol: The watery solution that fills a cell inside its plasma membrane.
extracellular fluid: The watery solution outside a cell: whatever fluid is outside it. For a red blood cell it is the blood; for a cell lining the gut it is the gut fluid.
hydrophobic interior: The oily middle of a membrane, where the hydrocarbon tails of the two phospholipid layers touch and water does not enter.
membrane protein: A protein set into a membrane: one folded chain whose nonpolar region sits against the tails and whose polar and charged regions sit in the water on each side, or line a water-filled passage.
glycoprotein: A protein with a carbohydrate chain attached; in a plasma membrane the chain faces the outside of the cell.
glycolipid: A lipid with a carbohydrate chain attached; in a plasma membrane the chain faces the outside of the cell.

APBIO-U02-L02 A mosaic that flows

Topic 2.3 · Plasma Membrane · 35 steps

A fused cell with red-tagged proteins on one half and green-tagged proteins on the other, and the same cell forty minutes later with the colors mixed

Here is a cell made by fusing two cells: a mouse cell and a human cell, their membrane proteins tagged red and green.

Just after fusing, the red proteins cover one half and the green proteins the other. Forty minutes later the two colors are spread evenly over the whole surface.

Unit 2 · Cell Structure and Function

1The pieces drift sideways

Video: Watch: A mosaic that flows

The red and green proteins of a fused cell mix because the membrane’s pieces drift sideways, staying heads-out the whole time.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L02.mp4

Here is the fused cell just after fusing. Every red-tagged protein is on one half of its surface, every green-tagged protein on the other.

The fused cell just after fusing: red-tagged proteins on the left half, green-tagged proteins on the right half

Forty minutes later, at 37 °C, red and green proteins are mixed evenly over the whole surface.

The same fused cell forty minutes later: red and green proteins mixed evenly over the whole surface

No new proteins were made in those forty minutes. The tagged proteins moved sideways through the membrane, into each other’s half.

The tags stayed reachable from the solution outside the cell the whole time. So the proteins kept their orientation as they moved: outside end out, inside end in.

Phospholipids drift sideways the same way, and so do cholesterol, glycoproteins and glycolipids. Every piece of the membrane drifts within its layer.

The pieces drift, but they do not flip. Heads stay toward the water and tails stay in the hydrophobic interior while everything slides past everything else.

A membrane is many different pieces that move, so the picture biologists use for it is called the : mosaic because it is built of many different pieces, fluid because they move.

So a membrane is not a rigid structure with its parts fixed in place. It is a sheet two molecules thick whose parts are always on the move.

What you are expected to know Explain why two sets of tagged proteins mix over the surface of a fused cell, and say what fluid and mosaic each mean in the fluid mosaic model.

Check q1

The phospholipids in one small patch of a cell’s membrane are tagged green and the proteins in that patch red. Where each color is, and how bright it is, is recorded at 0 and at 20 minutes.

Which result at 20 minutes shows that both phospholipids and proteins moved sideways within the membrane?

A. Both colors stay in the patch, no brighter than before
Tags that stayed in the patch would show that the molecules had not moved.
B. ✓ Both colors spread beyond the patch, no brighter overall
C. Both colors disappear from the patch and appear nowhere else
Drifting pieces leave the patch but stay in the membrane, so their colors appear elsewhere on the surface rather than vanishing.
D. Both colors stay in the patch and grow brighter
Brighter color in the same place would mean more tagged molecules there, not molecules that have moved.

Why: Pieces that drift sideways carry their tags out of the patch to other parts of the membrane, so both colors spread beyond the patch.
The total brightness does not change, because no tagged molecules were added or lost.

Check q2

A plasma membrane is described by the fluid mosaic model.

What do the two words mean?

A. ✓ Mosaic: many different pieces; fluid: the pieces drift sideways
B. Mosaic: many identical phospholipids; fluid: water flows through it
The pieces of a membrane are of many kinds, and fluid describes the pieces moving, not water flowing through.
C. Mosaic: pieces fixed like tiles in cement; fluid: the cytosol inside is liquid
The pieces of a membrane are not cemented in place, and fluid describes the membrane itself, not the cytosol.
D. Mosaic: many pieces; fluid: the pieces flip from one face to the other
The pieces do not flip from one face to the other; they keep their orientation, heads toward water and tails inside.

Why: Mosaic means the membrane is built of many different pieces: phospholipids, proteins, cholesterol, glycoproteins and glycolipids.
Fluid means those pieces drift sideways within their layer.

Check q3

A mouse cell and a human cell were fused into one cell. Only the membrane proteins were tagged, red on the mouse half and green on the human half, and within an hour the tagged proteins had spread over the whole surface.

Which pieces of a plasma membrane drift sideways within their layer?

A. Only the proteins, since only the proteins were seen to move
The tags showed the proteins moving, but the phospholipids, cholesterol, glycoproteins and glycolipids drift sideways too.
B. Only the phospholipids; each protein stays where it was set
The proteins drift too: in the fused cell, the tagged proteins of the two halves mixed within forty minutes.
C. None of them; each piece is cemented in its place
Nothing cements the pieces of a membrane in place: the tagged proteins moved, and so does every other piece.
D. ✓ All of them, phospholipids and proteins alike

Why: Every piece of a membrane drifts sideways within its layer: the phospholipids, the proteins, cholesterol, the glycoproteins and the glycolipids.
The tags showed the proteins moving, and the other pieces move the same way.
That is why the membrane is called fluid.

15Read a model of the membrane

Video: Watch: Read a model of the membrane

Five numbered parts, two questions each: what is it, and where does water put it.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L02b.mp4

Here is a model of a plasma membrane with five numbered parts: the extracellular fluid (outside the cell) above, the cytosol (inside the cell) below.

A model of a plasma membrane with five numbered parts: 1 a phospholipid, 2 a membrane protein, 3 cholesterol, 4 a glycoprotein, 5 a glycolipid; the extracellular fluid above is shaded one way and the cytosol below another

Read it one number at a time, asking two things of each part: what is it, and where does water put it?

Number 1 is a phospholipid: head to the water, tails inside. Many of them make the two-layer barrier that keeps the cytosol separate from the extracellular fluid.

Number 2 is a membrane protein: nonpolar regions against the tails, polar and charged regions in the water on each side.

Number 3 is cholesterol: among the tails, steadying the layer.

Number 4 is a glycoprotein and number 5 a glycolipid: their carbohydrate chains face the extracellular fluid.

And every one of the five drifts sideways within its layer.

Here is a second model, its parts lettered A to E. Name each lettered part to yourself, and say where water puts it.

A second model of a plasma membrane with five parts lettered A to E, the two watery regions shaded differently

Here is the labeled model. A is a glycolipid: chain in the extracellular fluid. B is cholesterol: among the tails. C is a phospholipid: head to the water, tails inside.

The second model labeled: A glycolipid, B cholesterol, C phospholipid, D glycoprotein, E membrane protein

D is a glycoprotein: chain outside, nonpolar middle against the tails. E is a membrane protein: nonpolar middle against the tails, polar and charged ends in the water.

What you are expected to know Name each part of a labeled model of a plasma membrane, say where water puts it and what it does to keep the inside separate from the outside, and say that all of them drift within the layer.

Check q4

Here is a third model of a plasma membrane, its parts lettered J to N.

A third model of a plasma membrane with five parts lettered J to N

Which part is a glycolipid?

A. J
J spans the membrane and carries no carbohydrate chain: it is a membrane protein.
B. ✓ K
C. L
L has a head and two tails but no carbohydrate chain: a plain phospholipid.
D. M
M is the four-ring steroid among the tails, cholesterol, and carries no carbohydrate chain.

Why: A glycolipid is a lipid with a carbohydrate chain attached.
K is a phospholipid with a chain of sugar units rising from its head into the extracellular fluid.

Check q5

In this model of a plasma membrane, with its parts lettered J to N, one part is cholesterol.

Which part is cholesterol?

A. J
J is a protein spanning the bilayer, not a steroid.
B. L
L has a head in the water and two tails: a phospholipid, not cholesterol.
C. ✓ M
D. N
N is a protein with a carbohydrate chain, a glycoprotein.

Why: M is the four-ring steroid, cholesterol.
It is nonpolar, so water puts it among the tails in the hydrophobic interior, where it steadies the layer.

Check q6

Part J of the same model spans the membrane from the extracellular fluid to the cytosol.

Which of its regions sit against the tails?

A. ✓ Its nonpolar regions
B. Its polar regions
Water pulls on polar R groups, so those regions sit in the water on either face.
C. Its charged regions
Water is attracted to charged R groups, so those regions stay in the cytosol or the extracellular fluid.
D. Its carbohydrate chain
J carries no carbohydrate chain, and a chain, being sugar, would sit in the extracellular fluid in any case.

Why: Water does not pull on nonpolar R groups, so a membrane protein’s nonpolar regions sit against the tails in the hydrophobic interior.
Its polar and charged regions sit in the water on each side.

Check q7

In the model of a membrane shown, one glycoprotein sits at the spot marked X, its carbohydrate chain in the extracellular fluid.

A model of a plasma membrane, the extracellular fluid above shaded one way and the cytosol below another, with one glycoprotein at the spot marked X; its carbohydrate chain, labeled, reaches into the extracellular fluid; the lead from X ends on the protein

In the living membrane, where is that glycoprotein a minute later?

A. At exactly the same spot, since a glycoprotein is held in place by the tails
Nothing holds a piece of the membrane still; every piece drifts sideways within its layer.
B. ✓ Elsewhere in the membrane, its chain still in the extracellular fluid
C. Elsewhere in the membrane, its chain now flipped into the cytosol
Pieces keep their orientation as they drift; a carbohydrate chain never crosses to the cytosol.
D. Out of the membrane and dissolved in the extracellular fluid
The glycoprotein’s nonpolar middle keeps it in the membrane; it does not dissolve into the water.

Why: The glycoprotein drifts sideways within its layer, so a minute later it is somewhere else in the membrane.
It keeps its orientation as it drifts, so its carbohydrate chain is still in the extracellular fluid.

Practice writing an answer

A student draws a model of a liver cell’s plasma membrane, shown below, with the extracellular fluid above and the cytosol below. In the drawing, a cholesterol molecule floats in the extracellular fluid, and a glycolipid has its carbohydrate chain pointing down into the cytosol. Under the drawing the student writes: once a protein is set into the membrane, it stays at that spot.

A student's drawing of a liver cell's plasma membrane: extracellular fluid above, cytosol below; a four-ring cholesterol molecule floating in the extracellular fluid; a glycolipid in the lower layer with its carbohydrate chain pointing down into the cytosol; a protein spanning the membrane; and the student's note written under the drawing

(a) Two pieces are drawn in the wrong place: the cholesterol molecule and the glycolipid’s carbohydrate chain. State where each belongs, and explain why cholesterol belongs where it does. (1 pt)

Frame Cholesterol belongs …, because it is …; the glycolipid’s carbohydrate chain belongs …

Model answer Cholesterol is a steroid: four carbon rings with no charged groups.
The rings are nonpolar, so water has nothing on them to hold.
Water pushes the rings out of both watery solutions.
The only place free of water is among the tails.
So cholesterol belongs among the tails, in the hydrophobic interior.
The glycolipid’s carbohydrate chain is a chain of sugar units.
Carbohydrate chains sit on the outer face of the membrane only, so the chain belongs in the extracellular fluid, not in the cytosol.

Rubric

Award 1 point for: cholesterol placed among the tails, in the hydrophobic interior, AND the glycolipid’s carbohydrate chain placed in the extracellular fluid, on the outer face.
Accept: either placement described in words; both placements are needed for the point. The reason for cholesterol (water has nothing on its nonpolar rings to hold) completes a full answer but is not needed for the point; no reason is asked for the chain.

Slip Putting cholesterol with the heads or out in the water. Cholesterol’s rings are nonpolar, so water has nothing on them to hold. Water pushes the rings in among the tails.

(b) Explain what holds each phospholipid with its head toward the water and its tails in the hydrophobic interior. (1 pt)

Model answer The extracellular fluid and the cytosol both contain water.
Water’s partial charges pull on the polar head, so the head is attracted into the water beside it.
The hydrocarbon tails carry no charges or partial charges, so water is not attracted to anything them.
Water pushes the tails out of the water, and they gather with the other tails in the hydrophobic interior.

Rubric

Award 1 point for: the polar head is attracted into the water by water’s partial charges, and the nonpolar tails, which water is not attracted to, are pushed together into the hydrophobic interior.
Accept: hydrophilic head and hydrophobic tails, provided the answer says what water does to each.

Slip Saying the pieces are glued or cemented in place. Nothing cements a membrane. Water pulls each polar head into the water, and water pushes each pair of nonpolar tails into the hydrophobic interior. Those two actions hold every phospholipid where it is.

(c) Make a claim about where the protein is a minute later in a living membrane, and support your claim. (1 pt)

Model answer A minute later the protein is somewhere else in the membrane, still spanning it the same way up.
Every piece of a membrane drifts sideways within its layer, because the membrane is fluid.
So the student’s note is mistaken: a protein set into the membrane does not stay at one spot.

Rubric

Award 1 point for: the claim, supported by evidence and reasoning: the protein has drifted sideways to another place in the membrane, keeping its orientation, because the pieces of a membrane drift within their layer (the membrane is fluid).
Accept: any wording that has the protein move along the membrane and stay the same way up.

Slip Having the protein flip over or float out into the water. Pieces drift sideways and keep their orientation; the protein’s nonpolar middle keeps it in the membrane.

You can now read any model of a plasma membrane: name each piece, say where water puts it, and say why the whole thing is called a fluid mosaic.

The red and green proteins mixed because the membrane is fluid: its pieces drift sideways, staying heads-out the whole time.

Glossary

fluid mosaic model: The description of a membrane as many different pieces (phospholipids, proteins, cholesterol, glycoproteins and glycolipids) that drift sideways within their layer: mosaic because of the many pieces, fluid because they move.

APBIO-U02-P23 Practice questions: Topic 2.3

Topic 2.3 · Plasma Membrane · 9 MCQ · 2 FRQ · for APBIO-U02-T23

These practice questions have the shape of the Topic 2.3 test. Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the first free-response question, you work through one scenario in small steps, and each step offers a hint if you want one. For the second, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it.

Video: Watch first: the plasma membrane, summed up

From the red blood cell to the film around it: two watery fluids, a bilayer arranged by water, proteins, cholesterol and carbohydrate chains, all drifting.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T23-summary.mp4

Q1 P23-q01

The drawing shows a white blood cell drifting in lymph, the watery fluid that fills the spaces between the cells of the body. A student labels the fluid inside its plasma membrane X and the fluid outside it Y.

A white blood cell in lymph, with a label inside the cell (X) and one in the fluid outside it (Y). The fluid outside is shaded darker than the fluid inside.

What does the plasma membrane have against each of its two faces, and why?

A. ✓ Water against both faces: X, the cytosol, and Y, the lymph, are both watery solutions
B. Water against its inner face only: Y, the lymph, is mostly fat
The lymph is a watery solution, not fat, so the outer face of the membrane is in water too.
C. Water against its outer face only: X, the cytosol, is oily like the membrane
The cytosol is a watery solution, not an oily one, so the inner face of the membrane is in water too.
D. Oil against both faces: the membrane is a film of lipid, so lipid lies on both sides
The membrane is a film of lipid, but the lipid is inside the film, where the tails of the two layers touch; both faces of the film sit in water.

Why: X is the cytosol, the watery solution filling the cell.
Y is the lymph, the watery fluid outside the cell: its extracellular fluid.
The cytosol and the lymph are both watery solutions.
So the membrane has water against both of its faces.

Q2 P23-q02

The drawing shows how a plasma membrane cut across appears under an electron microscope: two dark lines with a pale band between them. The dark lines are where the phospholipid heads lie, one line against the cytosol and one against the extracellular fluid.

A plasma membrane cut across, as it appears under an electron microscope.

What fills the pale band in the middle?

A. A layer of water trapped between two separate membranes
The pale band is the inside of one membrane, two molecules thick, with no water in it.
B. ✓ The hydrocarbon tails of both layers of phospholipids, touching in the middle
C. The phospholipid heads of both layers, packed together
The heads are the dark lines; they face the water on each side.
D. Membrane proteins lying flat between the two lines
Membrane proteins sit across the membrane and are scattered, not laid flat as a continuous band.

Why: A plasma membrane is two layers of phospholipids.
The heads face the water on both faces and show as the two dark lines; the hydrocarbon tails of both layers touch in the middle and form the pale band, the hydrophobic interior, which holds no water.

Q3 P23-q03

A cell lining the gut has the gut fluid, a watery solution, against one face of its plasma membrane and the cytosol against the other. Its membrane is two layers of phospholipids.

Which parts of the phospholipids touch the gut fluid, and which touch the cytosol?

A. Tails touch the gut fluid, and heads touch the cytosol
Tails carry no charges or partial charges, so water is not attracted to them and they touch neither solution.
B. Heads touch the gut fluid, and tails touch the cytosol
The layer against the cytosol has its heads in the cytosol, not its tails.
C. Tails touch the gut fluid and the cytosol; the heads sit in the middle
Water pulls on the heads, so the heads face the gut fluid and the cytosol; the tails, which water does not pull on, touch in the middle.
D. ✓ Heads touch the gut fluid and the cytosol; the tails touch neither

Why: The gut fluid contains water, and the cytosol contains water.
Water pulls on the polar heads, so in each layer the heads face the watery solution beside them: gut fluid on one face, cytosol on the other.
Water does not attract the tails, so the tails touch in the middle.

Q4 P23-q04

A membrane protein in a kidney cell has a water-filled passage through it, reaching from the extracellular fluid to the cytosol. Some of the protein’s R groups line the passage. Others lie on the protein’s outer surface, against the phospholipid tails.

Which R groups line the passage, and which face the tails?

A. Nonpolar R groups line the passage; polar and charged R groups face the tails
Water fills the passage and attracts polar and charged R groups; nonpolar R groups sit against the tails, where there is no water.
B. ✓ Polar and charged R groups line the passage; nonpolar R groups face the tails
C. Nonpolar R groups in both places, since the protein sits in an oily membrane
A membrane protein has a polar region and a nonpolar region, and the passage is filled with water, which attracts polar and charged R groups.
D. Polar and charged R groups in both places, since the protein is dissolved in water
The protein spans the membrane, so part of its surface faces the hydrocarbon tails, and the tails attract nothing polar or charged.

Why: Water decides where each part of a membrane protein sits.
Water fills the passage and attracts polar and charged R groups, so polar and charged R groups line the passage.
Water is not attracted to nonpolar R groups, so nonpolar R groups lie against the hydrocarbon tails.

Q5 P23-q05

A protein found in a plasma membrane has, at one end, a short stretch of six nonpolar R groups, a stretch far shorter than the thickness of the membrane's interior. Every other R group in the protein is polar or charged.

Where does this protein sit?

A. ✓ Its nonpolar end is tucked among the tails and the rest of the protein is in the water on one side
B. It spans the whole membrane, with an end in the water on each side
A protein spans the membrane only if a nonpolar stretch reaches all the way across the interior, and six nonpolar R groups reach only partway.
C. It floats free in the extracellular fluid with the nonpolar end exposed to the water
A nonpolar end exposed to water is pushed out of the water and into the tails.
D. It lies entirely within the hydrophobic interior, among the tails
Almost all of this protein is polar or charged, and water holds those regions.

Why: Each region of a membrane protein sits where water puts it.
Water is not attracted to the short nonpolar stretch, so it is pushed into the tails and anchors the protein, but it is too short to reach the other face.
The polar and charged regions stay in the water.

Q6 P23-q06

Glycoproteins and glycolipids are membrane proteins and membrane lipids with a carbohydrate chain attached.

Where in a plasma membrane are their carbohydrate chains?

A. On the inner face of the membrane, in the cytosol
No carbohydrate chain faces the cytosol.
B. Inside the membrane, among the tails
A carbohydrate chain is polar, water is attracted to it, and nothing polar sits among the tails.
C. ✓ On the outer face of the membrane, in the extracellular fluid
D. Mixed: some on the inner face and some on the outer face
The chains are not shared between the two faces.

Why: The carbohydrate chains of glycoproteins and glycolipids all face the outside of the cell.
So they sit on the outer face of the plasma membrane, in the extracellular fluid, and never on the inner face or among the tails.

Q7 P23-q07

The model shows a section of a gut cell’s plasma membrane, with the extracellular fluid above and the cytosol below. One part is lettered Y.

A section of a gut cell’s plasma membrane: extracellular fluid above, cytosol below, one part lettered Y.

What is Y, and why does it sit where it does?

A. A phospholipid: water attracts its head, and its two tails hang down into the interior
Y is a compact four-ring molecule with no head and no tails, not a phospholipid.
B. A membrane protein: its middle is nonpolar, so the whole protein is pulled into the tails
Y is a small four-ring molecule sitting wholly among the tails, not a protein spanning the membrane.
C. ✓ Cholesterol: its rings are nonpolar, so water pushes them out of both solutions into the tails
D. Cholesterol: its rings are polar, so water reaches among the tails and holds the rings there
The name is right but the reason is backwards: the rings are nonpolar, and water does not reach among the tails, so nothing there is attracted by water.

Why: Y is a compact molecule of four carbon rings: cholesterol, a steroid.
Cholesterol has no charged groups, and its rings are nonpolar, so water has nothing to attract.
Water pushes the rings out of both watery solutions.
The only place free of water is among the phospholipid tails.

Q8 P23-q08

The membrane proteins of a frog cell are tagged so that they glow. A laser is flashed on one small patch of the membrane; the tagged proteins in that patch stop glowing for good. Over the next few minutes the dark patch brightens again, while the total glow from the whole cell stays the same as it was just after the flash.

Why does the dark patch brighten again?

A. The cell made new glowing proteins and put them into the patch
New glowing proteins would raise the total glow from the cell, and the total stayed the same.
B. The proteins that stopped glowing recovered and glow again
The flash stopped those proteins glowing for good.
C. The patch of membrane was replaced by a new patch of membrane
Membrane is not swapped out in patches.
D. ✓ Glowing proteins from around the patch drifted sideways into it

Why: The pieces of a membrane drift sideways within their layer, which is the 'fluid' of the fluid mosaic model.
Glowing proteins around the patch drifted into it, and dark ones drifted out, so the patch brightened while the total glow from the cell stayed the same.

Q9 P23-q09

The lipids of a fish cell's plasma membrane are analyzed. One in five of the lipid molecules is a compact molecule of four carbon rings with no charged groups.

What is this molecule, and where in the membrane does it sit?

A. Cholesterol; among the phospholipid heads, facing the water
Cholesterol has no charged groups, so water is not attracted to it among the heads.
B. Cholesterol; dissolved in the cytosol just inside the membrane
Cholesterol is nonpolar and the cytosol is a watery solution.
C. ✓ Cholesterol; among the phospholipid tails in the interior
D. A glycolipid; on the outer face, carrying a carbohydrate chain
A glycolipid carries a carbohydrate chain, and this molecule has none.

Why: A compact four-ring molecule with no charged groups is cholesterol, a steroid.
Water is not attracted to it, so it sits among the phospholipid tails in the hydrophobic interior of an animal cell's membrane, where it steadies the layer.

FRQ 1 P23-frq1 · Analyze Model or Visual Representation scaffolded

The model shows a section of the plasma membrane of a cell lining the small intestine, at one instant. The gut fluid, a watery solution, is above; the cytosol, another watery solution, is below. Three parts are lettered P, Q and S.

A section of a gut cell’s plasma membrane: gut fluid above, cytosol below, three parts lettered P, Q and S.

(a) Identify parts P and S, and state whether a molecule like S could also be found in the lower layer with its sugar chain in the cytosol. (1 pt)

Frame P is a …; S is a …; a molecule like S … be found with its chain in the cytosol, because …

Hint S is the same kind of molecule as P, with a carbohydrate chain attached to its head. Think about which face of a plasma membrane carries carbohydrate chains.

Model answer P is a phospholipid, one of the molecules that make up the two layers.
S is a glycolipid: a phospholipid with a carbohydrate chain attached to its head.
A molecule like S would not be found with its chain in the cytosol.
The carbohydrate chains of glycolipids and glycoproteins face the outside of the cell only.
Here the outside of the cell is the gut fluid, this cell’s extracellular fluid, so every chain faces the gut fluid.

Rubric

Award 1 point for: P is a phospholipid and S is a glycolipid, and no: the carbohydrate chains of glycolipids (and glycoproteins) face the outside of the cell only, here the gut fluid, never the cytosol.
Accept: "a lipid with a carbohydrate chain" for S. Do not award the point if S is called a glycoprotein or if sugar chains are allowed on the cytosol side.

Slip Calling S a glycoprotein, or allowing sugar chains on both faces. S is a two-tailed lipid with a sugar chain, so it is a glycolipid, and such chains face the outside of the cell only.

(b) Describe where water puts the two parts of P: its head and its two tails. (1 pt)

Frame The head of P is …, so water …; the tails of P are …, so water …

Hint Think about which part of the molecule water is attracted to, and what happens to the part water is not attracted to.

Model answer The gut fluid and the cytosol both contain water.
The head of P is polar, so water attracts it.
So the head faces the watery solution beside it: the gut fluid for the top layer, the cytosol for the bottom layer.
The two tails of P are hydrocarbon chains with no charges or partial charges, so water is not attracted to them.
So the tails point away from the water and lie against the other layer’s tails in the middle.

Rubric

Award 1 point for: the head is polar (charged), so water attracts it and it faces the watery solution (gut fluid or cytosol); the tails are nonpolar hydrocarbon, so water is not attracted to them and they are pushed together into the middle, the hydrophobic interior.
Accept: "hydrophilic head toward the water, hydrophobic tails inside" provided the answer says what water does to each part.

Slip Saying the tails are 'attracted to each other'. The tails end up together because water pushes them out of the water. Water's pull on the heads holds each phospholipid at the face of the membrane.

(c) Part Q spans both layers. Its middle stretch has only nonpolar R groups, and its two ends have charged R groups. Explain why Q sits with its middle stretch inside the membrane and its two ends in the water. (1 pt)

Frame The middle stretch of Q has … R groups, so …; the two ends have … R groups, so …

Hint Use the same rule you used for P, applied to the R groups of the amino acids in each region of the protein.

Model answer Q is a protein: one folded chain of amino acids that spans both layers.
The middle stretch of Q has only nonpolar R groups; water is not attracted to them, so that stretch is pushed out of the water.
So the middle stretch lies among the hydrocarbon tails, in the hydrophobic interior.
The two ends of Q have charged R groups; water is attracted to them.
So one end stays in the gut fluid and the other in the cytosol.

Rubric

Award 1 point for: the middle stretch has only nonpolar R groups, which water is not attracted to, so it lies among the hydrocarbon tails; the two ends have charged R groups, which water holds, so each end stays in the watery solution on its side (gut fluid or cytosol).
Accept: "the nonpolar middle is hydrophobic and the charged ends are hydrophilic" with where each ends up stated.

Slip Saying a membrane protein must be hydrophobic all over. Most membrane proteins have a nonpolar region and a charged or polar region. Water holds the charged ends of Q, and that is what keeps Q spanning the membrane instead of sinking into it.

(d) Predict where Q is one hour later in the living membrane, and state one thing about Q that has stayed the same. (1 pt)

Frame An hour later, Q is …, because …; what has stayed the same is …

Hint What does the fluid mosaic model say happens to the parts of a membrane over time, and what did your answer to (c) say fixes the way Q sits?

Model answer An hour later Q is somewhere else along the membrane.
Every piece of a membrane drifts sideways within its layer, because the membrane is fluid.
What has stayed the same is how Q sits.
It still spans the membrane, with its nonpolar middle among the tails and its charged ends in the water, the same end in the cytosol as before.

Rubric

Award 1 point for: Q has drifted sideways to a different place within the membrane (the pieces of a membrane move sideways within their layer: the membrane is fluid), and it still spans the membrane the same way up, with its nonpolar middle in the interior and its charged ends in the water.
Accept: any statement that Q has moved along the membrane and any one preserved feature (still spans the membrane; still the same end in the cytosol; still has its middle among the tails).

Slip Predicting that Q stays put, or that it flips over. It drifts sideways, and it keeps its orientation, because water attracts its charged ends in place as it moves.

(e) Explain how the arrangement shown in the model, with heads to the water on both faces and the tails touching in the middle, lets the membrane keep the gut fluid and the cytosol as two separate solutions. (1 pt)

Frame Between the two solutions lies …, which …, so …

Hint Think about what fills the middle of the membrane from one side to the other, and what a dissolved, polar substance would have to pass through if it tried to cross.

Model answer The phospholipids form a continuous two-layer sheet.
From one face to the other, its interior is hydrocarbon tails with no charges or partial charges.
The polar and charged substances dissolved in the gut fluid and the cytosol are attracted by water.
Nothing in the hydrocarbon interior attracts them, so they stay on their own side of the sheet.
So the two solutions stay separate.

Rubric

Award 1 point for: the two layers of phospholipids form a continuous sheet whose interior is hydrocarbon tails with no charges or partial charges; the water and the polar and charged substances dissolved in each solution are attracted by water and are attracted by nothing in that interior, so the interior is a barrier between the two solutions (proteins, cholesterol and the sugar chains sit in it without breaking the sheet).
Accept: "the oily middle keeps the two watery solutions apart" provided the answer says what the middle is made of and why dissolved substances stay on their own side.

Slip Saying only 'the membrane is a barrier'. The point is earned by naming what the barrier is made of, the hydrocarbon interior, and why dissolved substances stay on their side of it.

FRQ 2 P23-frq2 · Conceptual Analysis

The phospholipids are extracted from red blood cell membranes. Spread on the surface of a dish of water, the phospholipids form a film one molecule thick, with every head down in the water and every tail up in the air. Shaken into the water instead, the same phospholipids form closed bubbles whose skin is two molecules thick, with water inside and outside each bubble. The phospholipids are then shaken into water a second time, together with a membrane protein that has a long stretch of nonpolar R groups in its middle and charged R groups at both ends, and the same bubbles form.

(a) Describe the two parts of a phospholipid and what water does to each. (1 pt)

Frame A phospholipid has a …, which water …, and two …, which water …

Model answer A phospholipid has a polar head, which water is attracted to and keeps in the water, and two hydrocarbon tails, which carry no charges or partial charges, so water is not attracted to them and pushes them out of the water.

Rubric

Award 1 point for: a polar (charged) head, which water is attracted to, and two nonpolar hydrocarbon tails, which water is not attracted to and pushes out of the water.
Accept: hydrophilic head and hydrophobic tails, provided the answer says what water does to each.

Slip Naming the parts as hydrophilic and hydrophobic with no account of what water does. The point needs the pull on the head and the push on the tails.

(b) Explain why the phospholipids on the surface form a film one molecule thick with the tails in the air, while the phospholipids shaken into the water form a skin two molecules thick. (1 pt)

Model answer On the surface of the dish, water is below each phospholipid and air is above it.
The head is polar, so water attracts it into the water.
The tails are hydrocarbon, so water is not attracted to them, and they point up into the air.
One layer satisfies both parts: one molecule thick.
Shaken into the water, the phospholipids have water on every side.
The only place the tails can hide from water is against other tails, so two layers form tail to tail: two molecules thick.

Rubric

Award 1 point for: on the surface, the tails can escape the water into the air while the heads stay in the water, so one layer is enough; surrounded by water on all sides, the tails have no air to escape into, so two layers lie tail to tail with the tails hidden between them and a layer of heads facing the water on each side.
Accept: "in water the tails can only hide from water by touching the tails of a second layer".

Slip Saying the phospholipids 'prefer' two layers. The two-layer skin forms because, with water on both sides, the tails can hide from water only by touching the tails of a second layer.

(c) Make a claim about where the protein ends up when the bubbles form. (1 pt)

Model answer The protein ends up in the skin of the bubbles, spanning it.
Its nonpolar middle lies among the hydrocarbon tails in the interior.
Its two charged ends sit in the water, one inside the bubble and one outside.

Rubric

Award 1 point for: the claim that the protein inserts into the skin of the bubbles and spans it, with its nonpolar middle among the tails and one charged end in the water inside the bubble and the other in the water outside. No reasoning is required for this point.
Accept: "it sits in the bilayer with its middle in the interior and its ends in the water". Do not award the point for the protein staying dissolved in the water or lying on the surface of the skin.

Slip Leaving the protein dissolved in the water. Its nonpolar middle is pushed out of the water into the tails, and its charged ends keep it spanning the skin.

(d) Support your claim, using what holds each region of the protein where it sits. (1 pt)

Model answer The middle stretch of the protein has only nonpolar R groups.
Water is not attracted to them.
So the water molecules stay attracted to one another and exclude that stretch, which settles among the hydrocarbon tails of the skin.
The two ends have charged R groups.
Water is attracted to charged R groups, so each end stays in the water on its side of the skin.
Therefore the protein is held at both ends with its middle in the tails, so it spans the skin.

Rubric

Award 1 point for: the evidence AND the reasoning: the middle stretch has only nonpolar R groups, so water is not attracted to it and excludes it into the hydrocarbon tails; the two ends have charged R groups, so water holds each end in the water on its side; with a stretch long enough to cross the interior, the protein spans the skin.
Accept: the same reasoning in terms of hydrophobic and hydrophilic regions, provided both regions are placed and the reason for each placement is given. Do not award the point for evidence about the R groups with no link to where water puts them.

Slip Justifying with 'the protein is hydrophobic'. Only its middle is; the charged ends held by water are what keep it spanning the skin rather than sinking into it.

APBIO-U02-T23 End-of-topic test: Plasma Membrane

Topic 2.3 · Plasma Membrane · 17 MCQ · 2 FRQ

Q1 T23-q01

The drawing shows a single animal cell floating in a drop of liquid, with three labels: X, Y and Z.

A cell floating in a drop of liquid, with three labels.

What are X, Y and Z?

A. ✓ X cytosol; Y plasma membrane; Z extracellular fluid
B. X extracellular fluid; Y plasma membrane; Z cytosol
X sits inside the membrane, where the cytosol is; Z is the liquid outside the cell, the extracellular fluid.
C. X cytosol; Y cell wall; Z extracellular fluid
Y marks the thin boundary of the cell itself, and a cell wall, where a cell has one, is a stiff layer outside that boundary.
D. X plasma membrane; Y cytosol; Z extracellular fluid
The plasma membrane is the thin boundary, not the region inside it.

Why: The cytosol is the watery solution filling the cell inside the plasma membrane (X); the plasma membrane is the thin film that bounds the cell (Y); the extracellular fluid is the watery solution outside it (Z).

Q2 T23-q02

A red blood cell drifts in blood. A student says: "The plasma membrane is a film of lipid, so the cell's surface must be oily on the outside, with water only on the inside."

Which statement about the two faces of the membrane is correct?

A. Only the inner face touches water; the outer face is oily
Blood is a watery solution, so the outer face has water against it too.
B. ✓ Both faces touch water: cytosol inside, blood outside
C. Only the outer face touches water; the cytosol is oily
The cytosol is a watery solution, not an oily one.
D. Neither face touches water; the film keeps water off
The membrane sits between two watery solutions.

Why: The cytosol inside the cell and the blood outside it are both watery solutions, so the plasma membrane has water against both of its faces.
The oily part is the middle of the film, where the tails of the two layers touch.

Q3 T23-q03

Three students drew a cross-section of a plasma membrane, with the extracellular fluid above and the cytosol below. Each phospholipid is drawn as a circle (the head) with two wavy tails.

Three drawings of a membrane cross-section, the extracellular fluid above and the cytosol below in each. Circles are phospholipid heads; wavy lines are tails.

Which drawing is correct, and why?

A. ✓ I: heads face the water on both sides, tails touch in the middle
B. II: tails face the water on both sides, heads touch in the middle
Drawing II puts the hydrocarbon tails against the water on both sides.
C. III: two layers, both with their heads toward the extracellular fluid
Drawing III stacks two layers the same way up, so the lower layer’s tails point into the cytosol, which is water.
D. I or II: either works, as long as there are two layers
Two layers are not enough on their own; in drawing II the tails face the water.

Why: Water is against both faces of the membrane, so the heads, which water pulls on, face outward on both sides, and the hydrocarbon tails, which water does not pull on, touch in the middle.
Only drawing I shows this.

Q4 T23-q04

On the labeled cross-section of a plasma membrane shown, a student has written “oily interior” beside the middle band, where the two layers of phospholipids touch.

A student’s labeled cross-section of a plasma membrane. Circles are phospholipid heads; wavy lines are tails.

Why is that middle band oily?

A. Cholesterol fills the middle and makes it oily
A bilayer made of phospholipids alone has the same oily middle, so cholesterol is not what makes it oily.
B. The heads of both layers touch there and exclude water
The heads are the charged and polar parts that water pulls on, and they face the water on both sides.
C. ✓ The hydrocarbon tails of both layers touch there
D. A layer of fat is trapped between two membranes
A membrane is one bilayer, not two membranes with fat between them.

Why: In each layer the hydrocarbon tails point away from the water, so the tails of the two layers touch in the middle.
Hydrocarbon tails carry no charges or partial charges, so this band is the hydrophobic interior, the oily middle of the membrane.

Q5 T23-q05

The drawing shows a student's model of a plasma membrane: a single layer of phospholipids, with every head facing the extracellular fluid and every tail pointing into the cytosol.

A student’s drawing of a plasma membrane. Circles are phospholipid heads; wavy lines are tails.

What is wrong with the drawing?

A. Nothing; a membrane is one layer of phospholipids
A single layer would leave one set of tails against water.
B. ✓ The tails are touching the watery cytosol
C. The heads should point into the cytosol instead
Turning the layer over only moves the problem, putting the tails against the extracellular fluid.
D. The tails should stick out into the extracellular fluid
Tails never face water on either side.

Why: The cytosol contains water, and water pushes hydrocarbon tails away.
In this drawing the tails touch the cytosol, so the drawing cannot be right.
A real membrane is two layers laid tail to tail: heads face the water on both sides, and the tails touch in the middle.

Q6 T23-q06

A student says a protein can sit in a membrane only if the whole protein is hydrophobic, because the membrane is oily.

Which statement corrects the student?

A. The student is right: membrane proteins are nonpolar all over
A protein that was nonpolar all over would have no part that water holds.
B. Membrane proteins are all hydrophilic and rest on the surface
Many membrane proteins span the whole membrane, which needs a nonpolar region.
C. Charge does not matter; a protein sits wherever it is made
Charge is exactly what decides where each part of the protein ends up.
D. ✓ Most membrane proteins have both polar and nonpolar regions

Why: Most membrane proteins have both polar and nonpolar regions.
Water attracts the polar and charged regions into the watery solutions on either side of the membrane.
Water is not attracted to the nonpolar region, so it sits against the hydrocarbon tails.

Q7 T23-q07

The figure shows a membrane protein drawn as a chain of amino acids with three regions, marked 1, 2 and 3. The R groups in 1 are charged, the R groups in 2 are all nonpolar, and the R groups in 3 are polar and charged.

A membrane protein drawn as a chain with three regions.

Where does each region sit when the protein is in a plasma membrane?

A. ✓ 1 and 3 in the water on each side; 2 against the tails
B. 2 in the water; 1 and 3 against the tails
Water holds the charged and polar regions, not the nonpolar one.
C. All three against the tails, since it is a membrane protein
1 and 3 carry charged and polar R groups, which water pulls into the solutions on each side; only 2, with nonpolar R groups, can sit against the tails.
D. 1 in the cytosol; 2 and 3 in the extracellular fluid
The R groups in 2 are nonpolar, so 2 cannot stay in either watery solution.

Why: Water pulls on charged and polar R groups, so 1 and 3 stay in the watery solutions on either side of the membrane.
Water is not attracted to nonpolar R groups, so 2 is pushed out of the water and lies against the hydrocarbon tails, spanning the membrane.

Q8 T23-q08

The drawing shows a plasma membrane split open between its two layers. Many bumps stand out from the two exposed faces.

A plasma membrane split open between its two layers, the upper layer lifted away; bumps stand out from both exposed faces.

What could the bumps be?

A. Phospholipid heads that had sunk into the middle
Heads are charged or polar, so water attracts them at the two faces; they do not sink into the oily middle.
B. ✓ Membrane proteins that reach into the hydrophobic interior
C. Sugar chains, which coat both faces of the membrane
Carbohydrate chains face the water outside the cell, not the middle of the membrane.
D. Drops of water trapped between the two layers
The middle of the membrane is where the tails of the two layers touch, and it holds no water.

Why: The split passes through the hydrophobic interior, where the tails of the two layers touch.
The only pieces that reach into that region are membrane proteins, whose nonpolar regions sit among the tails, so they show up as bumps on the split faces.

Q9 T23-q09

The model shows a plasma membrane with the extracellular fluid above and the cytosol below. Five parts are numbered.

A model of a plasma membrane with five numbered parts.

Which number marks a glycolipid?

A. 2
Number 2 is a protein spanning the membrane with no carbohydrate chain attached: a membrane protein.
B. 3
Number 3 is the small compact molecule tucked among the tails, cholesterol, with no carbohydrate chain.
C. 4
Number 4 has a carbohydrate chain attached to a protein, so it is called a glycoprotein.
D. ✓ 5

Why: A glycolipid is a lipid with a carbohydrate chain attached.
Number 5 is a phospholipid in the outer layer with a chain of sugar units on its head, facing the extracellular fluid.

Q10 T23-q10

A red blood cell drifts in blood. The blood is the cell’s extracellular fluid, outside the cell; the cytosol is inside the cell. A fragment of the cell’s plasma membrane is examined: one face of the fragment touched the blood and the other face touched the cytosol. Carbohydrate chains are found attached to proteins and lipids on one face of the fragment only.

A red blood cell in blood, and a fragment of its plasma membrane enlarged: one face touched the blood, the other touched the cytosol.

Which face carries the chains, and why?

A. The face that touched the cytosol: carbohydrate chains anchor the membrane from inside the cell
Carbohydrate chains face the outside of the cell, never the cytosol.
B. ✓ The face that touched the blood: carbohydrate chains face the outside of the cell
C. Either face: chains are attached on both faces in equal numbers
Carbohydrate chains sit on one face of the membrane only, the face toward the extracellular fluid, never on both.
D. Neither face: carbohydrate chains sit in the middle, among the tails
A chain of sugar units is polar and cannot sit among the hydrocarbon tails.

Why: The carbohydrate chains of glycoproteins and glycolipids face the extracellular fluid, outside the cell.
For a red blood cell, the extracellular fluid is the blood around it.
So the face carrying the chains is the face that touched the blood.

Q11 T23-q11

Cholesterol is a steroid with four carbon rings and no charged groups. In an animal cell's plasma membrane it is present in large amounts.

Where in the membrane does cholesterol sit?

A. ✓ Among the phospholipid tails
B. Among the heads, facing the water
Cholesterol has no charged groups and its rings are nonpolar, so water has nothing on cholesterol to pull toward the heads.
C. Dissolved in the cytosol just inside the membrane
Cholesterol's rings are nonpolar, so water pushes cholesterol out of the cytosol just as it pushes out oil.
D. Attached to the carbohydrate chains outside
The carbohydrate chains are polar and sit in the water outside the cell, where a nonpolar molecule cannot stay.

Why: Cholesterol has no charged groups, and its four carbon rings are nonpolar, so water has nothing on the rings to pull on.
Water pushes the rings out of both watery solutions.
The only place free of water is among the hydrocarbon tails, so cholesterol settles there.

Q12 T23-q12

A mouse cell and a human cell are fused into one cell. Before fusing, the mouse cell's membrane proteins were tagged to glow red and the human cell's to glow green. At 0 minutes, red is on one half of the fused cell and green on the other half. At 40 minutes at 37 °C, red and green are mixed evenly over the whole surface. The total brightness of each color is the same as at 0 minutes.

What does the result show about the membrane?

A. New proteins were made and spread over the surface
New proteins would add brightness, and the brightness did not change.
B. The two halves swapped phospholipids but kept their proteins
The tags are on the proteins, not the phospholipids, and it is the tagged proteins that spread.
C. ✓ The membrane proteins drifted sideways within the membrane
D. The proteins left the membrane, crossed the cytosol and re-entered
A membrane protein has a nonpolar region that water pushes out of the cytosol, so it cannot cross the cytosol.

Why: The same tagged proteins, with no new ones made, ended up spread evenly over the whole surface.
They got there by drifting sideways within the membrane, which is what the fluid mosaic model describes: the pieces of the membrane are not fixed in place.

Q13 T23-q13

A textbook calls the plasma membrane a fluid mosaic. A student reads "mosaic" as saying the membrane is a rigid sheet of tiles cemented in place.

What does each word describe?

A. ✓ Mosaic: many different pieces; fluid: the pieces drift sideways
B. Mosaic: the pieces are fixed in place; fluid: water fills the gaps
The pieces of a membrane are not fixed, and there are no water-filled gaps in the oily middle.
C. Mosaic: a pattern of identical phospholipids; fluid: it is liquid inside
The pieces are of many different kinds, not identical phospholipids, and fluid is about the pieces moving, not liquid inside the cell.
D. Mosaic: the proteins only; fluid: the phospholipids can leave the membrane
Mosaic covers every kind of piece, and the phospholipids do not leave the membrane.

Why: In the fluid mosaic model, "mosaic" says the membrane is built of many different pieces, and "fluid" says those pieces are not fixed in place but drift sideways within their layer.

Q14 T23-q14

In a membrane, a phospholipid drifts sideways past its neighbors many times a second.

As it drifts, how does the phospholipid stay oriented?

A. It tumbles freely; its head and tails point any way
Water pulls on the head and pushes the tails away the whole time, so the phospholipid cannot tumble.
B. It flips over on each move so that both faces get a turn
Flipping over would drag the charged head through the oily middle and put the tails in the water.
C. It stops moving once it has found its place
The pieces of a membrane never settle into fixed places.
D. ✓ Head toward the water, tails in the interior, the whole time

Why: The pieces of a membrane drift sideways within their layer while staying heads-out: water keeps pulling on the head and pushing the tails away, so the phospholipid moves along the membrane without turning over.

Q15 T23-q15

The model shows a plasma membrane with five numbered parts. The two watery sides are labeled side A and side B.

A model of a plasma membrane with five numbered parts, between side A and side B.

Which side is the extracellular fluid, and what tells you?

A. Side B: the phospholipid heads face it
Both faces of the bilayer are made of heads, so the heads cannot tell the two sides apart.
B. ✓ Side B: the carbohydrate chains face it
C. Side A: the protein’s charged ends face it
A membrane protein has charged ends in the water on both sides, so its ends do not mark the outside.
D. Cannot tell: both faces look the same to water
Both faces are watery but they are not the same: the carbohydrate chains sit on one face only.

Why: The carbohydrate chains of glycoproteins and glycolipids face the outside of the cell.
So the side the chains face is the extracellular fluid.
In this model the chains of the glycoprotein and the glycolipid face side B.

Q16 T23-q16

A student describes the location of each part of a plasma membrane. Four of the descriptions are listed.

Which description is wrong?

A. Phospholipids: two layers, heads to the water, tails inside
That description is right: the phospholipids form the two-layer barrier with heads to the water and tails inside.
B. Cholesterol: sits among the tails and steadies the layer
That description is right: cholesterol sits among the tails of an animal cell membrane and steadies the layer.
C. ✓ Membrane proteins: charged regions against the tails
D. Glycolipids: carry carbohydrate chains on the outer face
That description is right: a glycolipid carries its carbohydrate chain on the outer face.

Why: A membrane protein sits with its nonpolar regions against the tails and its polar or charged regions in the water on either side.
Charged regions are attracted by water and cannot sit in the hydrophobic interior, so the third description has it backwards.

Q17 T23-q17

Over an hour, the pieces of a plasma membrane move about.

Which of the following movements is impossible in a living plasma membrane?

A. A phospholipid drifting sideways past its neighbors, staying in its own layer
Drifting sideways within a layer is what every piece of a membrane does; that is the fluid in fluid mosaic.
B. A glycoprotein drifting sideways while its carbohydrate chain stays in the extracellular fluid
A glycoprotein does drift sideways, and its carbohydrate chain stays in the extracellular fluid as it drifts: the pieces of a membrane move sideways but keep the same way up.
C. ✓ A glycoprotein turning over so its carbohydrate chain faces the cytosol
D. Cholesterol shifting its position among the hydrocarbon tails
Cholesterol is nonpolar and moves about among the tails, which is where water keeps it.

Why: The pieces of a membrane drift sideways, but they do not turn over.
To flip, the glycoprotein’s polar carbohydrate chain would have to pass through the hydrocarbon tails, where nothing holds it.
Water keeps every piece the same way up.
The other three movements stay in one layer.

FRQ 1 T23-frq1 · Analyze Model or Visual Representation

The model shows a cross-section of an animal cell’s plasma membrane at one instant. The extracellular fluid is above and the cytosol below; both are watery solutions. Five kinds of piece are numbered 1 to 5.

(a) Identify the pieces numbered 1 to 5. (1 pt)

Model answer 1 is a phospholipid, one of the two layers of the bilayer; 2 is a membrane protein; 3 is cholesterol; 4 is a glycoprotein, a protein with a carbohydrate chain attached; 5 is a glycolipid, a lipid with a carbohydrate chain attached.

Rubric

Award 1 point for naming at least four of the five correctly: 1 = phospholipid (one of the two layers of the bilayer); 2 = membrane protein; 3 = cholesterol; 4 = glycoprotein (a protein with a carbohydrate chain attached); 5 = glycolipid (a lipid with a carbohydrate chain attached).
Accept: "lipid bilayer" or "phospholipid bilayer" for 1; "protein that spans the membrane" for 2. Do not award the point for naming only "phospholipids" and "proteins" without telling the membrane protein, the glycoprotein and the glycolipid apart.

Slip Naming only phospholipids and proteins. The glycoprotein and the glycolipid each carry a carbohydrate chain; the plain membrane protein does not. The glycoprotein is a protein; the glycolipid is a lipid.

(b) The part marked 2 spans both layers. Explain, using how water treats each of its parts, why it sits with its middle stretch inside the membrane and its two ends in the water. (1 pt)

Model answer The extracellular fluid and the cytosol both contain water.
The two ends of the membrane protein have charged and polar R groups, which water’s partial charges pull on, so one end is held in the extracellular fluid and the other in the cytosol.
The middle stretch of the protein has nonpolar R groups, and water is not attracted to it, so the water molecules stay with one another and exclude that stretch.
It lies against the hydrocarbon tails, in the hydrophobic interior.

Rubric

Award 1 point for: water's partial charges pull on the charged and polar R groups at the two ends, so those ends stay in the cytosol and the extracellular fluid; water is not attracted to the nonpolar R groups in the middle, so that stretch is excluded from the water and lies against the hydrocarbon tails (the hydrophobic interior).
Accept: "the nonpolar middle is hydrophobic so it hides among the tails; the charged ends are hydrophilic so they stay in the water" provided the answer says why (water pulls on charged or polar groups and not on nonpolar ones). Do not award the point for "like sits with like" with no mention of charges or water.

Slip Saying like sits with like, with no mention of water or charges. The point needs what water does: it is attracted to charged and polar groups and is not attracted to nonpolar ones.

(c) The model shows one instant. Describe how the part marked 2 moves over the next hour, and state one thing about it that stays the same as it moves. (1 pt)

Model answer Over the hour, the protein drifts sideways along the membrane, staying within the two layers.
It moves because the membrane is fluid: every piece drifts sideways within its layer.
What stays the same is the way the protein sits.
It still spans the membrane the same way up, with its nonpolar middle in the hydrophobic interior and its charged ends in the water on the same sides as before.

Rubric

Award 1 point for: the membrane protein moves sideways along the membrane, within the layer (not across the membrane or out of it), together with one unchanged feature: it still spans the membrane with its nonpolar middle in the interior and its charged ends in the water (or: it stays the same way up; its ends stay on the same sides).
Accept: any wording that has the protein move along the membrane and stay the same way up. Do not award the point for a movement that carries the protein out into the cytosol or the extracellular fluid, or for “nothing changes” without naming the unchanged feature.

Slip Moving the protein out into the cytosol or the extracellular fluid, or saying nothing changes without naming the unchanged feature. The protein moves along the membrane, and it keeps spanning it the same way up.

(d) Explain how the arrangement shown in the model, with heads to the water on both faces and the tails touching in the middle, lets the membrane keep the cytosol and the extracellular fluid as two separate solutions. (1 pt)

Model answer The extracellular fluid and the cytosol both contain water.
Water pulls the phospholipid heads and the polar parts of the proteins into the water on each face, and pushes the hydrocarbon tails and the nonpolar parts into the middle.
So the tails of the two layers lie together in a continuous oily band, the hydrophobic interior.
The polar and charged substances dissolved in each solution are attracted by water, and nothing in the oily band can hold them, so they stay on their own side.

Rubric

Award 1 point for: the two layers of hydrocarbon tails form a continuous oily band (the hydrophobic interior) between the two watery solutions; water pulls the heads and the polar parts of the proteins into the water and pushes the tails and the nonpolar parts into the interior, so the band stays continuous while the pieces drift; the dissolved polar and charged substances of each solution are attracted by water and are attracted by nothing in the band, so they stay on their own side.
Accept: “the oily middle has no water in it, so the two watery solutions do not mix through it” together with a statement of what water does to the pieces (pulls the polar parts into the water, pushes the nonpolar parts into the interior). Do not award the point for “the proteins hold the membrane together” or for “nothing can cross the membrane”.

Slip Saying the proteins hold the membrane together, or that nothing can cross. Water pulls each polar part into the water and pushes each nonpolar part into the interior, so the oily band stays continuous. That oily band keeps the dissolved polar substances of the two watery solutions from mixing through it.

FRQ 2 T23-frq2 · Conceptual Analysis

A membrane protein in a red blood cell spans the plasma membrane, reaching from the cytosol to the blood. Along its backbone, a stretch of about twenty amino acids in the middle has nonpolar R groups, and the stretches at both ends have charged and polar R groups. The cytosol, a watery salt solution, is on one side of the membrane and the blood, another watery solution, is on the other. A variant of the protein is made in which six of the amino acids in the middle stretch are replaced by amino acids with charged R groups.

(a) Describe the two solutions the plasma membrane sits between and what the interior of the membrane is made of. (1 pt)

Model answer The membrane sits between the cytosol on one side and the blood, this cell’s extracellular fluid, on the other.
The cytosol and the blood are both watery solutions, so the membrane has water against both faces.
Its interior is the hydrocarbon tails of the two phospholipid layers, lying tail to tail.
The tails carry no charges or partial charges, so this interior is the hydrophobic interior.

Rubric

Award 1 point for: the cytosol inside and the extracellular fluid (here the blood) outside are both watery solutions, so the membrane has water against both faces; its interior is the hydrocarbon tails of the two phospholipid layers, lying tail to tail, with no charges or partial charges (the hydrophobic interior).
Accept: "oily middle made of the phospholipid tails" for the interior. Both the two watery solutions and the tail interior are needed for the point.

Slip Describing only the solutions or only the interior. The answer needs both: two watery solutions, and an oily middle made of the tails.

(b) Explain why the normal protein sits spanning the membrane, with one end in the cytosol and the other in the blood. (1 pt)

Model answer The cytosol and the blood both contain water.
Water’s partial charges pull on the charged and polar R groups at the two ends of the protein, so each end is attracted into the water on its side.
Water is not attracted to the nonpolar middle stretch, so water pushes it out of the water and into the hydrocarbon tails.
So the protein settles with that stretch across the interior and one end in the water on each side.

Rubric

Award 1 point for: water's partial charges pull on the charged and polar R groups at the two ends, so those ends stay in the water on each side; water is not attracted to the nonpolar middle stretch, so it is pushed out of the water into the hydrocarbon tails, and the protein settles with that stretch across the interior and an end in the water on each side.
Accept: "the middle is hydrophobic and the ends are hydrophilic" provided the answer says what water does to each. Do not award the point for "the protein is made in the membrane" or "the protein is glued in by the phospholipids".

Slip Saying the protein is glued in by the phospholipids or made in the membrane. Water’s pull on the ends and push on the middle is what sets where it sits.

(c) Make a claim about what happens to the position of the variant protein. (1 pt)

Model answer The variant is no longer held spanning the membrane.
Its middle stretch now carries charged R groups.
Water pulls on charged R groups, so the middle stretch is pulled toward the water.
Therefore the protein does not settle across the membrane: it stays in the water, or it is pulled out into the cytosol or the blood, or it sits against one face without crossing it.

Rubric

Award 1 point for: the claim that the variant is no longer held spanning the membrane. Any one of these fates earns the point: it never settles into the membrane and stays in the water; it is pulled out of the membrane into the cytosol or the blood; it sits against one face of the membrane without crossing it. No reasoning is required for this point.
Do not award the point for ‘nothing changes’ or for ‘the membrane opens a hole to let the charged part through’.

Slip Saying nothing changes, or that the membrane opens a hole for the charged part. Charged groups in the middle stretch change where water puts the protein.

(d) Support your claim using what holds each part of a protein where it sits. (1 pt)

Model answer Where each part of a membrane protein sits is decided by which parts water is attracted to.
In the normal protein the middle stretch is all nonpolar, so it stays among the uncharged tails and anchors the protein in the hydrophobic interior.
In the variant the middle stretch carries charged R groups.
Water’s partial charges pull on them, and the tails cannot hold them.
Therefore that stretch pulls toward the water, and the protein loses its anchor.

Rubric

Award 1 point for: the evidence AND the reasoning: where each part of a membrane protein sits is decided by which parts water is attracted to; the charged R groups now in the middle stretch are pulled on by water's partial charges and are attracted by nothing among the uncharged tails, so the stretch that used to anchor the protein in the hydrophobic interior now pulls toward the water, and the protein loses the nonpolar stretch that kept it spanning.
Accept: a comparison with the normal protein, whose all-nonpolar middle is excluded from water and so stays in the tails. Do not award the point for "charged things cannot enter a membrane" with no reference to water holding them or to the uncharged tails.

Slip Saying charged things cannot enter a membrane, with no reference to water holding them or to the uncharged tails.

APBIO-U02-L03 What gets through the hydrophobic interior

Topic 2.4 · Membrane Permeability · 74 steps

A sealed bubble of pure membrane in a beaker of solution: oxygen, sodium ions, chloride ions and glucose outside it at the start; ten minutes later oxygen is inside and the rest still outside

Here is a bubble of pure membrane, no proteins in it, floating in a solution of oxygen, salt and sugar.

Ten minutes later the oxygen is inside. The salt and the sugar are still outside.

Unit 2 · Cell Structure and Function

1Who gets into a bubble of pure membrane

Video: Watch first: Membrane permeability

A bubble of pure membrane in a solution of oxygen, salt and sugar: ten minutes later the oxygen is inside.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T24-intro.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T24-intro.mp4

Video: Watch: Who gets into a bubble of pure membrane

Methane, oxygen, water, ammonia, ions and glucose arrive one at a time: three groups, sorted by size, polarity and charge.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L03.mp4

Here is the bubble again. The bubble is a sealed sheet of phospholipid bilayer with no proteins in it. Inside the bubble there is only water. Outside the bubble is the solution.

A sealed bubble of phospholipid bilayer, with no proteins in it, floating in a beaker of solution; the bilayer is labelled and the inside of the bubble holds only water

Put methane, CH₄, a small nonpolar molecule, in the solution. Within minutes there is methane inside the bubble.

Methane in the solution outside the bubble, and methane inside it within minutes; the phospholipid bilayer and the water inside are labelled

Oxygen, O₂, is also small and nonpolar: inside within minutes. Carbon dioxide, CO₂, and nitrogen, N₂: the same.

Methane, oxygen, carbon dioxide and nitrogen all inside the bubble within minutes; the phospholipid bilayer and the water inside are labelled

Small nonpolar molecules pass freely through a phospholipid bilayer.

Now water, H₂O: small and polar, with no full charge. A little water gets in, slowly. Ammonia, NH₃, is small, polar and uncharged too, and a little ammonia trickles in the same way.

Water and ammonia outside the bubble, with a little of each inside; the phospholipid bilayer and the water inside are labelled

Small polar molecules with no charge pass through a phospholipid bilayer in small amounts.

Now sodium ions, Na⁺, and chloride ions, Cl⁻. A sodium ion is smaller than an oxygen molecule, but each of the two ions carries a full charge. Neither ion gets in, even after hours.

Sodium ions, chloride ions and glucose outside the bubble, and none of them inside even after hours; the phospholipid bilayer and the water inside are labelled

Glucose: polar, no charge, but large. None gets in either.

Ions and large polar molecules do not cross a phospholipid bilayer on their own.

So size alone does not decide who gets through. A sodium ion is smaller than an oxygen molecule and is blocked. Charge and polarity decide.

Three groups, then:
1. Small nonpolar molecules: pass freely.
2. Small polar molecules with no charge: pass in small amounts.
3. Ions and large polar molecules: do not cross on their own.

Summary table with three columns: small nonpolar molecules pass freely; small polar molecules with no charge pass in small amounts; ions and large polar molecules do not cross on their own

A membrane that lets some substances through and holds others back is called .

To say how a substance crosses, use the table in three steps.

First, remember the table: small nonpolar molecules pass freely; small polar molecules with no charge pass in small amounts; ions and large polar molecules do not cross on their own.

Second, find the group the substance belongs to, from its size, its polarity and its charge.

Third, read off how that group crosses.

What you are expected to know Sort a substance into one of the three groups by how it crosses a phospholipid bilayer, using its size, polarity and charge, and say that a membrane which lets some substances through and holds others back is selectively permeable.

21Fluency quiz: will it cross? mixed practice

Check q1

Nitrogen, N₂, is a small nonpolar molecule.

Does N₂ cross a phospholipid bilayer on its own?

A. ✓ Yes
B. No
N₂ is small and nonpolar, and small nonpolar molecules pass freely through the bilayer.

Why: N₂ is small and nonpolar.
So N₂ is in the first group, small nonpolar molecules.
The first group passes freely.
So N₂ crosses a phospholipid bilayer on its own.

Check q2

A calcium ion, Ca²⁺, is small and carries a full charge.

Does Ca²⁺ cross a phospholipid bilayer on its own?

A. Yes
Ions are in the third group.
B. ✓ No

Why: Ca²⁺ carries a full charge, so Ca²⁺ is an ion.
Ions are in the third group, with the large polar molecules.
The third group does not cross on its own.
So Ca²⁺ does not cross a phospholipid bilayer on its own.

Check q3

Urea is a small polar molecule with no charge.

How does urea cross a phospholipid bilayer?

A. Urea passes freely
Urea is polar.
A small polar molecule with no charge is in the second group.
B. ✓ Urea passes in small amounts
C. Urea does not cross on its own
Urea is small, and urea has no charge.

Why: Urea is small and polar, with no charge.
So urea is in the second group, small polar molecules with no charge.
The second group passes in small amounts.
So urea passes through a phospholipid bilayer in small amounts.

Check q4

Sucrose, table sugar, is a large polar molecule.

How does sucrose cross a phospholipid bilayer?

A. Sucrose passes freely
Large polar molecules are in the third group.
B. Sucrose passes in small amounts
Sucrose is large.
Large polar molecules are in the third group.
C. ✓ Sucrose does not cross on its own

Why: Sucrose is large and polar.
So sucrose is in the third group, ions and large polar molecules.
The third group does not cross on its own.
So sucrose does not cross a phospholipid bilayer on its own.

Check q5

Ethane is a small nonpolar molecule.

How does ethane cross a phospholipid bilayer?

A. ✓ Ethane passes freely
B. Ethane passes in small amounts
Ethane is nonpolar.
Small nonpolar molecules are in the first group.
C. Ethane does not cross on its own
Ethane is small, nonpolar and uncharged.
Small nonpolar molecules are in the first group.

Why: Ethane is small and nonpolar.
So ethane is in the first group, small nonpolar molecules.
The first group passes freely.
So ethane passes freely through a phospholipid bilayer.

Check q6

Hydrogen peroxide, H₂O₂, is a small polar molecule with no charge.

How does hydrogen peroxide cross a phospholipid bilayer?

A. Hydrogen peroxide passes freely
Hydrogen peroxide is polar.
A small polar molecule with no charge is in the second group.
B. ✓ Hydrogen peroxide passes in small amounts
C. Hydrogen peroxide does not cross on its own
Hydrogen peroxide is small, and hydrogen peroxide has no charge.

Why: Hydrogen peroxide is small and polar, with no charge.
So hydrogen peroxide is in the second group, small polar molecules with no charge.
The second group passes in small amounts.
So hydrogen peroxide passes through a phospholipid bilayer in small amounts.

Check q7

A chloride ion, Cl⁻, is small and carries a full charge.

How does Cl⁻ cross a phospholipid bilayer on its own?

A. Cl⁻ passes freely
Ions do not cross on their own.
B. Cl⁻ passes in small amounts
Cl⁻ carries a full charge, not a partial one.
C. ✓ Cl⁻ does not cross on its own

Why: Cl⁻ carries a full charge.
Water is attracted to that charge.
Nothing in the hydrophobic interior attracts a charge.
So Cl⁻ stays in the water on its own side of the membrane.
Ions do not cross on their own, however small they are.

29Why the oily middle sorts them

Video: Watch: Why the oily middle sorts them

The hydrophobic interior carries no charges; water holds an ion or a polar molecule on its own side.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L03b.mp4

Water pushes a hydrocarbon out. Water’s partial charges are attracted to any charge or partial charge. A hydrocarbon has no charges and no partial charges. So nothing in a hydrocarbon attracts water.

Look into the middle of the bilayer. The hydrocarbon tails of the two layers touch there. The middle band of the membrane, made only of tails, is called the hydrophobic interior.

A membrane cross-section with the solution outside the bubble shaded above and the solution inside shaded below; the middle band of hydrocarbon tails is outlined with a dashed line and labelled hydrophobic interior; glucose sits in the water above the membrane, and the tails in the middle carry no charges or partial charges

The tails in the hydrophobic interior carry no charges and no partial charges.

Glucose has an –OH on almost every carbon. Each –OH carries partial charges. Water is attracted to those partial charges. So the water on glucose’s own side of the membrane attracts glucose and keeps it there.

Nothing in the hydrophobic interior attracts glucose. So glucose stays in the water on its own side.

A sodium ion carries a full charge. Water is attracted to a full charge even more strongly than to a partial charge. Nothing in the hydrophobic interior attracts a charge. So the sodium ion stays in the water.

So water attracts an ion or a polar molecule. Nothing in the hydrophobic interior attracts that ion or molecule. So the ion or molecule stays in the water and cannot cross on its own.

O₂ has no charges and no partial charges for water to be attracted to. So nothing keeps O₂ in the water. A small nonpolar molecule dissolves into the tails and passes out the other side.

A membrane cross-section with the solution outside the bubble shaded above and the solution inside shaded below; an oxygen molecule above the membrane, in the middle among the tails, and below it: it dissolves into the tails and passes out the other side

The membrane does not block ions because ions are too big. The membrane blocks ions because ions are charged, and nothing in the hydrophobic interior attracts a charge. Size matters only among polar molecules.

What you are expected to know Explain why a phospholipid bilayer sorts substances this way. The hydrocarbon tails in the middle of the membrane, the hydrophobic interior, carry no charges or partial charges. Water attracts an ion or a polar molecule, and nothing in the hydrophobic interior attracts it, so the ion or polar molecule cannot cross on its own. A small nonpolar molecule dissolves into the tails and passes out the other side.

Check q8

X and Y are two substances of the same small size. X is nonpolar. Y carries a full charge. X gets into protein-free bubbles of phospholipid bilayer. Y stays outside.

Why does X get in while Y stays out?

A. X and Y both slip between the phospholipids; Y moves more slowly and will catch up
Y does not cross on its own at all.
B. The heads keep Y in the water outside and push X through the middle
The heads face the water on both sides and push nothing through.
C. X and Y are the same size, so the difference is chance
X and Y are the same size and yet cross very differently, so charge decides.
D. ✓ X dissolves into the tails; water attracts Y and keeps Y outside

Why: The tails in the middle of the membrane carry no charges or partial charges.
X is nonpolar, so nothing in X attracts water; X dissolves into the tails and passes through.
Y carries a full charge, so water attracts Y and the tails do not.
So Y stays outside.

Check q9

A student says: “The bilayer blocks sodium ions because they are too big to fit between the tails.”

Which statement corrects the student?

A. Nothing needs correcting: size decides whether a particle crosses
Na⁺ is smaller than O₂, and O₂ passes.
B. ✓ Na⁺ is blocked because of its charge, not its size
C. Na⁺ is blocked because it is too small for the tails to catch
Size is not the reason either way.
Na⁺ carries a full charge.
D. Na⁺ is blocked by the charges on the heads, not by the interior
The heads sit in the water, where ions are at home.

Why: A sodium ion is smaller than an oxygen molecule, and oxygen passes.
So size is not what blocks the sodium ion.
The sodium ion carries a full charge.
Water is attracted to that charge.
Nothing in the hydrophobic interior attracts a charge.
So the sodium ion stays in the water.

43Channels and carriers: the protein doors

Video: Watch: Channels and carriers: the protein doors

A water-lined tunnel for chloride ions, and a pocket that changes shape for glucose.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L03c.mp4

A nerve cell takes in glucose and lets chloride ions in and out. Neither glucose nor chloride can cross the hydrophobic interior. Glucose and chloride cross through proteins set into the membrane.

One kind of membrane protein has a water-lined tunnel running through it. When the tunnel is open, one kind of ion or small molecule passes through it.

A channel protein labelled in the membrane, the solution outside shaded above and the cytosol shaded below: a water-lined tunnel through the protein, open, with a chloride ion passing through it

A protein like this is called a . Chloride ions cross a nerve cell’s membrane through a chloride channel.

Another kind of membrane protein has a pocket in it. The pocket opens to one side of the membrane. Glucose fits into the pocket.

A carrier protein in the membrane drawn at three moments, the solution outside shaded above and the cytosol shaded below. Moment 1: a pocket in the carrier opens to the outside, and a glucose molecule sits in it. Moment 2: the carrier has changed shape, and the same pocket now opens to the cytosol, with the glucose still in it. Moment 3: the glucose has left the pocket into the cytosol

When glucose sits in the pocket, the protein shifts into a different shape. In the new shape, the pocket is closed on the outside and open to the cytosol. So the glucose can leave the pocket, into the cytosol.

Then the empty protein shifts back. The pocket opens to the outside again, ready for the next glucose.

A protein like this is called a , also called a transport protein. Glucose enters a nerve cell through a glucose carrier.

Ions and large polar molecules cross a membrane only through channel proteins or carrier proteins. The bilayer itself never lets them through.

What you are expected to know Say that ions and large polar molecules cross a membrane only through membrane proteins. A channel protein is a water-lined tunnel that lets one kind of ion or small molecule through when it is open. A carrier protein binds the substance in a pocket and changes shape to move it across.

Check q10

Glucose and Z, a small nonpolar molecule, both enter a yeast cell. A drug blocks the cell’s glucose carriers.

Which substance stops entering the cell?

A. ✓ Glucose only
B. Z only
Z is small and nonpolar, so Z dissolves through the tails and needs no protein.
C. Glucose and Z
Z is small and nonpolar, so Z dissolves through the tails and needs no protein.
D. Neither glucose nor Z
Glucose is large and polar, so glucose cannot cross the bilayer itself.

Why: Glucose is large and polar.
So glucose crosses only through its carrier.
The drug blocks the carrier.
So glucose stops entering.
Z is small and nonpolar.
So Z dissolves through the tails and needs no protein.
So Z keeps entering.

Check q11

The drug is still blocking the yeast cell’s glucose carriers, and Z keeps entering.

Why does Z keep entering?

A. Z uses the glucose carrier, and the drug only slows the carrier down
The drug blocks that carrier, and Z is not glucose.
B. ✓ Z dissolves through the tails, so Z needs no protein
C. Z enters through a channel that the drug opens
A small nonpolar molecule needs no protein at all.

Why: Z is small and nonpolar.
Nothing in Z attracts water.
So Z dissolves into the tails and passes out the other side.
Z needs no protein.
So blocking a protein changes nothing for Z.

Check q12

Which of the following describes a channel protein?

A. A protein that binds its substance in a pocket and changes shape to move it across
Binding the substance in a pocket and changing shape is what a carrier does.
B. ✓ A water-lined tunnel that one kind of ion passes through when it is open
C. A protein that flips across the bilayer, ferrying an ion with it
A channel protein stays in place in the membrane.

Why: A channel protein is a water-lined tunnel through the membrane.
When the channel is open, one kind of ion or small molecule passes through it.
A carrier protein, in contrast, binds its substance in a pocket and changes shape to move it across.

57Reading uptake data

Video: Watch: Reading uptake data

Read the last column first, explain each row from the membrane, judge the prediction.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L03d.mp4

Here are uptake data for protein-free bubbles of phospholipid bilayer in a solution of three substances, A, B and C.

Uptake data for protein-free bubbles of phospholipid bilayer: substance A, small, nonpolar, no charge, 38% of the outside amount inside after ten minutes; B, small, polar, full charge, none inside; C, large, polar, no charge, none inside

Read the last column first. A crossed: after ten minutes the amount inside is 38% of the amount outside. B and C did not cross: none inside.

Then explain each result from the membrane. A is small and nonpolar. So A dissolved into the tails and passed through.

B carries a full charge. C is large and polar. Water attracts both B and C. Nothing in the hydrophobic interior attracts either of them. So neither B nor C crossed.

Before the test, a student predicted that C would cross freely, because the bubbles have no proteins to block it. Write down your answer before continuing: did C cross? Do the data support or contradict the prediction? Say why.

A model answer: C did not cross; none of C was inside after ten minutes. So the data contradict the prediction. C is polar, so water attracts C. Nothing in the hydrophobic interior attracts C. So the bilayer itself stops C, not a protein.

What you are expected to know Decide from uptake data whether each substance crossed a protein-free phospholipid bilayer. Explain each result from the membrane’s structure. Predict whether a new substance will cross on its own or needs a protein. Say whether the data support or contradict a prediction.

Check q13

A new substance is a large polar molecule.

Will the substance cross a phospholipid bilayer on its own?

A. ✓ No
B. Yes
Water is attracted to the partial charges on a polar molecule, and nothing in the hydrophobic interior attracts it, so a large polar molecule stays on its own side.

Why: The substance is large and polar.
Large polar molecules do not cross on their own.
Water is attracted to the partial charges on a polar molecule.
Nothing in the hydrophobic interior attracts the molecule.
So the molecule stays in the water on its own side.

A large polar molecule, or a charged one, will need a channel or a carrier to get across.

Check q14

Kidney cells sit in a solution of inulin, a large polar molecule. No inulin gets into the cells, and no membrane protein binds inulin.

Why does inulin stay outside the cells?

A. Inulin follows water, and water is not moving into the cells
Water trickles through the bilayer on its own, and a large polar molecule does not follow it.
B. The carbohydrate chains on the cell surface hold inulin out
The chains sit in the water outside the cell and are not the barrier.
C. ✓ Water attracts inulin, and nothing in the hydrophobic interior does
D. Inulin needs a higher concentration outside to push it through
Nothing in the interior attracts a polar molecule at any concentration.

Why: Inulin is large and polar.
Water is attracted to its partial charges.
The tails of the hydrophobic interior carry no charges or partial charges.
So nothing in the interior attracts inulin.
No membrane protein binds inulin, so inulin has no other route.
So inulin stays outside the cells.

Check q15

A new anesthetic gas is described: a small molecule, nonpolar, with no charge.

Will the gas cross a protein-free phospholipid bilayer?

A. ✓ Yes, on its own
B. No, not at all
Small nonpolar molecules cross a phospholipid bilayer freely, faster than water does.
C. Only through a carrier protein
Ions and large polar molecules need carriers; a small nonpolar molecule dissolves through the tails without one.
D. Only after it picks up a charge
A charge would stop the gas, not help it.

Why: The gas molecule is small, nonpolar and uncharged.
So nothing in the gas molecule attracts water.
So the gas molecule dissolves into the tails and passes through.
Like O₂ and CH₄, the gas crosses a phospholipid bilayer on its own.

Check q16

Ethylene glycol is a small polar molecule with no charge. A student predicts that it will cross a phospholipid bilayer in small amounts. In a ten-minute test with protein-free bubbles, a small nonpolar molecule reaches 100% inside; ethylene glycol reaches 15%.

What do the data show about the prediction?

A. They contradict it: 15% counts as no crossing
The prediction was a small amount.
15% is above zero, so some got in.
B. They say nothing: ten minutes is too short a test
The test did measure ethylene glycol, and its value, 15%, can be compared with the prediction.
C. ✓ They support it: 15% is a small amount, as predicted
D. They contradict it: a molecule this small should reach 100%
Only small nonpolar molecules reach 100%.
Water is attracted to a polar molecule, so only a small amount slips through.

Why: The prediction was that a little ethylene glycol would get in.
After ten minutes, 15% of the outside amount was inside.
15% is above zero but far below 100%, so only a little got in.
So the data support the prediction.

Practice writing an answer

Protein-free bubbles of phospholipid bilayer float in a solution of three substances. At the start each substance is at the same concentration outside the bubbles, and none is inside. Carbon dioxide (CO₂) is a small nonpolar molecule. Glycerol is a small polar molecule with no charge. A magnesium ion (Mg²⁺) is smaller than glycerol and carries a full +2 charge. After ten minutes the amount inside, as a percentage of the amount outside, is: CO₂ 100%; glycerol 4%; Mg²⁺ 0%. Before the test a student predicted that Mg²⁺, the smallest of the three, would get in fastest.

(a) Describe what the data show about how readily each of the three substances crossed the bilayer. (1 pt)

Model answer CO₂ crossed freely: after ten minutes CO₂ was as concentrated inside as outside.
Glycerol crossed only a little: 4% of the outside amount was inside after ten minutes.
Mg²⁺ stayed outside: none of the Mg²⁺ crossed.

Rubric

Award 1 point for: CO₂ crossed freely (inside equal to outside), glycerol crossed in small amounts (4%), and Mg²⁺ stayed outside (0%).
Accept: any wording that ranks the three correctly with the 4% read as a small crossing rather than none.

Slip Reading 4% as no crossing. Small polar molecules with no charge trickle through a phospholipid bilayer. 4% is a small crossing. 0% is the value that says none.

(b) Explain your answer to (a), using the structure of the membrane’s interior. (1 pt)

Model answer The hydrocarbon tails in the middle of the membrane carry no charges or partial charges.
CO₂ is nonpolar, so nothing in CO₂ attracts water.
So CO₂ dissolves into the tails and passes through freely.
Glycerol is polar, so water is attracted to its partial charges and nothing in the tails attracts it.
So only a little glycerol slips into the interior.
Mg²⁺ carries a full +2 charge, which water attracts far more strongly than glycerol’s partial charges.
So Mg²⁺ cannot cross on its own.

Rubric

Award 1 point for: the tails carry no charges or partial charges, so a nonpolar molecule dissolves through them, a polar molecule that water attracts trickles through in small amounts, and an ion’s full charge is attracted by water and by nothing in the interior, so the ion stays out.
Accept: the oily middle described as having nothing that attracts a charge, with the difference between polar and charged stated.

Slip Blaming size. Mg²⁺ is the smallest of the three and crossed least. Water is attracted to the charge on Mg²⁺, and that is what keeps Mg²⁺ out.

(c) Evaluate the student’s prediction against the data. Then predict how a living cell that needs Mg²⁺ takes it in. (2 pt)

Model answer The data contradict the prediction.
Mg²⁺ was the smallest of the three, and Mg²⁺ crossed least of all, 0%.
CO₂ is larger than Mg²⁺, and CO₂ crossed completely.
So charge decided, not size.
Mg²⁺ cannot cross the bilayer on its own.
So a living cell takes Mg²⁺ in through a membrane protein: a channel protein that lets that ion pass.

Rubric

Award 1 point for: the judgement (the data contradict the prediction) AND the ground for it: the data contradict the prediction, because the smallest substance, Mg²⁺, crossed least (0%) while the larger CO₂ crossed fully, so charge rather than size decided.
Award 1 point for: a living cell takes Mg²⁺ in through a membrane protein, a channel protein (or a carrier).
Accept: a protein door for the second point.

Slip Letting CO₂ speak for Mg²⁺: a small molecule crossed, so the smaller ion should too. The bilayer sorts by charge and polarity, not by size. An ion needs a protein door.

What crosses a phospholipid bilayer is decided by the hydrophobic interior: small nonpolar molecules pass, and ions and large polar molecules need a protein door.

The oxygen dissolved into the hydrophobic interior and came out the other side. Water attracted the salt and the sugar. Nothing in the tails attracted the salt or the sugar, so the salt and the sugar stayed outside. A real cell has channels and carriers that let salt and sugar through.

Glossary

selectively permeable: Letting some substances through and holding others back, as a membrane does: small nonpolar molecules pass, ions and large polar molecules do not cross on their own.
channel protein: A membrane protein with a water-lined tunnel through it; when the tunnel is open, one kind of ion or small molecule passes through.
carrier protein: A membrane protein (also called a transport protein) that binds a substance in a pocket on one side of the membrane, changes shape, and releases it on the other side.

APBIO-U02-L04 A drop of dye spreads

Topic 2.5 · Membrane Transport · 75 steps

A drop of red dye falling into a glass of still water, and the same glass an hour later, evenly pink

Here is a drop of dye falling into a glass of still water, and the same glass an hour later.

Nobody stirs it, and an hour later the whole glass is pink.

Unit 2 · Cell Structure and Function

1Counting molecules: the mole

Video: Watch first: Membrane transport

A drop of dye in a glass of still water, and the whole glass pink an hour later.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T25-intro.mp4

Video: Watch: Counting molecules: the mole

A mole is a count of about 6 × 10²³ particles: half a mole of glucose and of table sugar, same count, different mass.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L04a.mp4

How many dye molecules are in that glass? Far too many to count one by one. A single drop of the solution holds more molecules than there are grains of sand on every beach on Earth.

A substance dissolved in a liquid is called a solute. The liquid it is dissolved in is called the solvent. Here the dye is the solute, and water is the solvent.

Chemists count molecules in a fixed bundle, the way eggs are counted in dozens. The bundle is about 6 × 10²³ particles: a six followed by twenty-three zeros.

A count of about 6 × 10²³ particles of anything is called one .

Half a mole of glucose is about 3 × 10²³ glucose molecules. Half a mole of table sugar is about 3 × 10²³ table-sugar molecules. Same count.

Two beakers: half a mole of glucose with a mass of 90 g, and half a mole of table sugar with a mass of 171 g, each holding the same number of molecules

The two half-moles do not have the same mass. A table-sugar molecule has almost twice the mass of a glucose molecule. So half a mole of table sugar has a mass of 171 g, and half a mole of glucose has a mass of 90 g.

A mole is a count, not a mass. A mole of a heavier substance has more mass, but it holds the same number of particles.

So one mole of any dissolved substance in each liter of solution means the same number of dissolved particles per liter, whatever the substance is.

What you are expected to know Say that a mole is a fixed count of about 6 × 10²³ particles, so equal numbers of moles of two substances hold the same number of particles even when their masses differ.

Check q1

Half a mole of glucose (90 g) is dissolved in one liter of water. Half a mole of table sugar (171 g) is dissolved in another liter.

How do the numbers of dissolved molecules compare?

A. About twice as many glucose molecules
Half a mole is the same count for either substance.
B. About twice as many table-sugar molecules
A mole is a count, not a mass.
C. Cannot say from the masses
The masses are not needed.
Half a mole is half a mole, whatever the substance.
D. ✓ The same number in each

Why: A mole is a fixed count of about 6 × 10²³ particles.
So half a mole of anything is about 3 × 10²³ particles.
So the two solutions hold the same number of molecules.
Table sugar has more mass because each table-sugar molecule is heavier.

Check q2

A mole of water has a mass of 18 g. A mole of glucose has a mass of 180 g.

How many molecules are in each mole?

A. Ten times as many in the glucose
Each glucose molecule is simply ten times heavier than a water molecule.
B. Ten times as many in the water
A mole is the same count whatever the substance.
C. ✓ About 6 × 10²³ in each
D. It depends on the volume each takes up
A mole is a count of particles, fixed however much room the particles take up.

Why: One mole of any substance is about 6 × 10²³ particles.
Each glucose molecule is ten times heavier than a water molecule.
So a mole of glucose has ten times the mass of a mole of water.
The two moles hold the same number of molecules.

15Down or against the gradient

Video: Watch: Down or against the gradient

Molar concentration in mol/L, a concentration gradient, and which way is down.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L04b.mp4

To compare two solutions you need to say how concentrated each one is.

Here is a dialysis tube, a bag of thin membrane, in a beaker. Inside it, each liter of solution holds 0.80 mol of glucose. Outside, each liter holds 0.20 mol.

A dialysis tube in a beaker: glucose at 0.80 mol/L inside the tube and 0.20 mol/L outside

The number of moles of solute in each liter of solution is called the , written mol/L. Inside the tube glucose is at 0.80 mol/L. Outside it is at 0.20 mol/L.

Glucose is four times as concentrated inside as outside. A difference in concentration between two regions like this is called a .

Moving from where glucose is more concentrated to where it is less, here from inside the tube to outside, is moving down its concentration gradient.

The same tube with two arrows: down the gradient runs from 0.80 mol/L inside to 0.20 mol/L outside; against it runs the other way

Moving the other way, from 0.20 mol/L outside to 0.80 mol/L inside, is moving against its concentration gradient.

To find which way is down, compare the two concentrations. Down runs from the larger concentration to the smaller one.

What you are expected to know Given a substance’s concentration in mol/L in two connected regions, say which direction is down its concentration gradient and which is against it.

25Fluency quiz: which way is down? mixed practice

Check q3

The bar chart shows the concentration of glucose outside a yeast cell and inside it.

A bar chart with two bars, labelled outside the cell and inside the cell, showing the concentration of glucose in mol/L; each bar carries its value above it

Which direction is down the concentration gradient of glucose?

A. ✓ Into the cell
B. Out of the cell
Glucose is 0.60 mol/L outside and 0.10 mol/L inside; down a gradient runs from the larger concentration to the smaller, so into the cell.
C. Neither: there is no gradient
The two bars are different heights, so there is a gradient.

Why: Down a gradient runs from the larger concentration to the smaller.
Glucose is 0.60 mol/L outside and 0.10 mol/L inside.
The outside bar is taller.
So down the gradient is into the cell.

Check q4

The bar chart shows the concentration of urea outside a kidney cell and inside it.

A bar chart with two bars, labelled outside the cell and inside the cell, showing the concentration of urea in mol/L; each bar carries its value above it

Which direction is down the concentration gradient of urea?

A. Into the cell
Urea is 0.30 mol/L inside and 0.05 mol/L outside; down a gradient runs from the larger concentration to the smaller, so out of the cell.
B. ✓ Out of the cell
C. Neither: there is no gradient
The two bars are different heights, so there is a gradient.

Why: Down a gradient runs from the larger concentration to the smaller.
Urea is 0.30 mol/L inside and 0.05 mol/L outside.
The inside bar is taller.
So down the gradient is out of the cell.

Check q5

The bar chart shows the concentration of ethanol outside a yeast cell and inside it.

A bar chart with two bars, labelled outside the cell and inside the cell, showing the concentration of ethanol in mol/L; each bar carries its value above it

Which direction is down the concentration gradient of ethanol?

A. Into the cell
Ethanol is 0.20 mol/L on both sides, so the two bars are the same height and there is no gradient.
B. Out of the cell
Ethanol is 0.20 mol/L on both sides, so the two bars are the same height and there is no gradient.
C. ✓ Neither: there is no gradient

Why: A gradient is a difference in concentration.
Ethanol is 0.20 mol/L outside and 0.20 mol/L inside.
The two bars are the same height.
So there is no gradient, and no direction is down.

Check q6

The bar chart shows the concentration of carbon dioxide outside a muscle cell and inside it.

A bar chart with two bars, labelled outside the cell and inside the cell, showing the concentration of carbon dioxide in mol/L; each bar carries its value above it

Which direction is down the concentration gradient of carbon dioxide?

A. Into the cell
Carbon dioxide is 0.06 mol/L inside and 0.02 mol/L outside; down a gradient runs from the larger concentration to the smaller, so out of the cell.
B. ✓ Out of the cell
C. Neither: there is no gradient
The two bars are different heights, so there is a gradient.

Why: Down a gradient runs from the larger concentration to the smaller.
Carbon dioxide is 0.06 mol/L inside and 0.02 mol/L outside.
The inside bar is taller.
So down the gradient is out of the cell.

Check q7

The bar chart shows the concentration of an amino acid outside a liver cell and inside it.

A bar chart with two bars, labelled outside the cell and inside the cell, showing the concentration of the amino acid in mol/L; each bar carries its value above it

Which direction is down the concentration gradient of the amino acid?

A. ✓ Into the cell
B. Out of the cell
The amino acid is 0.008 mol/L outside and 0.002 mol/L inside; down a gradient runs from the larger concentration to the smaller, so into the cell.
C. Neither: there is no gradient
The two bars are different heights, so there is a gradient.

Why: Down a gradient runs from the larger concentration to the smaller.
The amino acid is 0.008 mol/L outside and 0.002 mol/L inside.
The outside bar is taller.
So down the gradient is into the cell.

Check q8

Inside a nerve cell, potassium ions, K⁺, are at 0.14 mol/L. Outside the cell, K⁺ is at 0.005 mol/L. A K⁺ moves from outside the cell to inside.

Which way is the K⁺ moving?

A. Down its gradient
K⁺ is 0.14 mol/L inside and 0.005 mol/L outside, so down its gradient means out of the cell; a K⁺ moving in goes the other way.
B. ✓ Against its gradient
C. Neither: there is no gradient
The two concentrations differ, so there is a gradient.

Why: K⁺ is 0.14 mol/L inside and 0.005 mol/L outside.
So K⁺ is more concentrated inside.
Down the gradient runs from inside to outside.
A K⁺ moving in is moving the other way.
So the K⁺ is moving against its concentration gradient.

32Why the dye spreads

Video: Watch: Why the dye spreads

Every dye molecule moves at random; more leave the crowded side than arrive: a net movement, diffusion.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L04c.mp4

Back to the dye. Nobody stirred the glass, yet the dye spread through all of it. Why?

In liquid water the molecules are always moving. The same motion is why the smell of cooking spreads across a room on its own.

Here is the edge of the pink patch, drawn large. Every dye molecule is moving, constantly and at random, in no particular direction.

Dye molecules at the edge of the pink patch, each moving in its own random direction: many on the left, few on the right

Particles move constantly and at random, so more of them leave a region where they are more concentrated than arrive from a region where they are less concentrated.

Look at the edge of the patch. Many dye molecules are inside the patch. Each one moves in its own random direction. So many dye molecules happen to wander out of the patch.

Few dye molecules are outside the patch. Each one moves in its own random direction too. So few dye molecules happen to wander into the patch.

More dye molecules wander out than wander in. The difference between the number leaving and the number arriving is called the . Here the net movement is outward.

That gives a net movement from the more concentrated region to the less concentrated one. This net movement down a concentration gradient is called .

No dye molecule knows where the clear water is. Each one moves at random. The net movement comes only from more molecules leaving the concentrated side than arriving.

What you are expected to know Explain why a substance spreads from where it is concentrated to where it is not. Particles move constantly and at random. So more of them leave a region where they are more concentrated than arrive from a region where they are less concentrated. That gives a net movement from the more concentrated region to the less concentrated one. This net movement down a concentration gradient is diffusion.

Check q9

A drop of dye sits in still water. In the next second, more dye molecules leave the pink patch than enter it.

Why do more dye molecules leave the patch than enter it?

A. ✓ Dye molecules move at random; more start inside the patch than outside
B. Each dye molecule is pushed toward the clear water
No dye molecule is pushed in any direction; each dye molecule moves at random.
C. Dye molecules are attracted to regions with less dye
A dye molecule is not drawn toward clear water.
D. Currents in the water carry the dye molecules outward from the patch
The water is still.
The spreading comes from the random motion of the dye molecules themselves.

Why: Every dye molecule moves constantly and at random, in no particular direction.
More dye molecules start inside the patch than outside.
So more dye molecules wander out than wander in.
So the net movement is outward.

Check q10

Oxygen from the air dissolves at the surface of a still pond. Over the day, the dissolved oxygen spreads downward through the water.

Why does the dissolved oxygen spread downward?

A. ✓ Oxygen molecules move at random; more start near the surface than deeper down
B. Oxygen molecules sink, because oxygen is heavier than water
Dissolved oxygen molecules do not sink; each one moves at random.
C. The deeper water pulls oxygen molecules down toward where there is less oxygen
Nothing pulls an oxygen molecule anywhere; each one moves at random.

Why: Every dissolved oxygen molecule moves constantly and at random.
More oxygen molecules start near the surface, where the oxygen dissolved, than deeper down.
So more oxygen molecules wander downward than wander back up.
So the net movement is downward.

46Equal on both sides: dynamic equilibrium

Video: Watch: Equal on both sides: dynamic equilibrium

Equal concentrations, molecules still crossing both ways in equal numbers.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L04d.mp4

An hour on, the whole glass is the same pink. Has the dye stopped moving?

Look at a dialysis bag with glucose at 0.40 mol/L both inside and outside. The bag lets glucose through.

In ten minutes, about 3 × 10²⁰ glucose molecules cross out of the bag. That is three hundred billion billion molecules.

A dialysis bag in a beaker, glucose at 0.40 mol/L on both sides, with about three hundred billion billion molecules crossing out and the same number crossing in over ten minutes

In the same ten minutes, about 3 × 10²⁰ glucose molecules cross into the bag. The same number cross in as cross out.

So the concentration stays at 0.40 mol/L on both sides. The glucose has not stopped moving. Molecules are moving in, and molecules are moving out, in equal numbers.

Movement both ways with no net movement is called a .

Molecules do not stop moving when the concentrations are equal. They keep crossing. Only the net movement is zero.

What you are expected to know Say that when a substance’s concentration is equal on both sides of a membrane its particles keep crossing both ways at equal rates, so there is no net movement: a dynamic equilibrium.

Check q11

An hour after a drop of dye fell into a glass of still water, the water in the glass is evenly pink everywhere.

What are the dye molecules doing now?

A. No longer moving
The dye molecules never stop moving; the net movement is what has fallen to zero.
B. ✓ Still moving at random, with no net change in where the dye is
C. Moving only toward regions with less dye
With the dye evenly spread there is no direction left to favor; as many dye molecules move one way as the other.
D. Moving faster than before
Nothing was slowing the dye molecules before; each dye molecule was moving at random the whole time and still is.

Why: The dye molecules never stop moving.
The dye is evenly spread.
So as many dye molecules move into any region as move out of it.
So there is no net change.
That is a dynamic equilibrium.

Check q12

A membrane separates two solutions. The concentration of a solute that can cross the membrane is the same on both sides. A student says: “That means the particles have stopped moving through the membrane.”

Which statement about the student’s claim is correct?

A. The student is wrong: particles keep crossing, but only from the side that was more concentrated at the start
A particle does not know which side was more concentrated at the start; each particle moves at random.
B. The student is right: with no concentration gradient, no particle crosses the membrane
With equal concentrations, as many particles cross one way as cross the other.
C. ✓ The student is wrong: particles cross in both directions, and the same number cross each way

Why: The particles never stop moving.
Particles cross the membrane in both directions.
With equal concentrations on the two sides, the same number cross each way.
So there is no net movement, and the concentrations stay equal.
That is a dynamic equilibrium.

58No energy used: passive transport

Video: Watch: No energy used: passive transport

Oxygen in, carbon dioxide out, glucose out of a bag: down the gradient, the cell uses no energy.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L04e.mp4

Oxygen is more concentrated in the blood than inside a working muscle cell. Oxygen diffuses straight through the bilayer into the cell, down its gradient. The cell uses no energy to make this happen.

A muscle cell, shaded apart from the blood around it: oxygen diffuses in down its gradient and carbon dioxide diffuses out down its gradient

Carbon dioxide is more concentrated inside the working cell than in the blood. Carbon dioxide diffuses out, down its gradient. Again the cell uses no energy.

Some glucose solution is placed inside a dialysis bag. The bag sits in a beaker of water. The glucose concentration is greater inside the bag than outside.

So glucose leaves the bag, down its concentration gradient. Nothing uses energy to make this happen: the glucose molecules move on their own.

Whichever way a substance goes, moving down its gradient uses no energy.

A crossing of a membrane in which the substance moves down its concentration gradient, with no energy used by the cell, is called .

When the substance passes straight through the bilayer, as O₂ and CO₂ do, the passive transport is called . Other routes across a membrane can be passive too.

Moving into a cell does not always use the cell’s energy. Down a gradient, the cell uses no energy, whichever way the substance is going.

A cell can also move a substance the other way, against its gradient, from less concentrated to more. That is not passive transport: the cell has to use energy. This kind of crossing is called active transport. A later lesson teaches active transport.

What you are expected to know Classify a crossing as passive transport when the substance moves down its concentration gradient with no energy used by the cell, and as simple diffusion when it passes straight through the bilayer.

Check q13

Oxygen, small and nonpolar, is more concentrated outside a muscle cell than inside. Oxygen enters the cell.

Which of the following describes this crossing?

A. Passive transport, but only through a carrier protein
Small nonpolar molecules pass straight through the bilayer with no protein.
B. A crossing that uses the cell’s energy
Oxygen molecules move at random, and more of them cross from the concentrated side outside than cross back.
C. ✓ Simple diffusion, a form of passive transport

Why: Oxygen is more concentrated outside the cell than inside.
So oxygen entering the cell moves down its concentration gradient.
Down a gradient, the cell uses no energy.
So the crossing is passive transport.
Oxygen passes straight through the bilayer.
So the crossing is simple diffusion.

Check q14

The muscle cell uses no energy as the oxygen enters.

Why does the cell use no energy on this crossing?

A. Oxygen brings its own energy into the cell, so the cell has none to supply
An oxygen molecule carries no energy for the crossing; the crossing needs no energy from anything.
B. ✓ Oxygen moves down its gradient, and movement down a gradient happens by itself
C. The blood pushes the oxygen molecules across the membrane and into the cell
The blood pushes nothing.
Oxygen molecules move at random, and more of them are outside the cell than inside, so more wander in than out with no push.

Why: Oxygen molecules move at random.
More oxygen molecules are outside the cell than inside.
So more oxygen molecules cross in than cross out.
That is movement down the gradient, and it happens by itself.
Nothing has to push the molecules.
So the cell uses no energy.

Check q15

Carbon dioxide made inside a cell leaves down its gradient through the bilayer. Glucose enters a red blood cell down its gradient through a carrier protein.

Which crossing is passive transport?

A. Only the carbon dioxide leaving
Passive transport is any crossing down the gradient with no energy used, whatever the route, so the glucose crossing through a carrier counts too.
B. Only the glucose entering
Down a gradient uses no energy in either direction, so the carbon dioxide leaving counts too.
C. ✓ Both crossings
D. Neither crossing
Each crossing moves down its own gradient, so the cell uses no energy on either.

Why: Passive transport is a crossing down the concentration gradient with no energy used by the cell.
The carbon dioxide moves down its gradient through the bilayer: no energy used.
The glucose moves down its gradient through a carrier: no energy used.
So both crossings are passive transport.

Substances spread down their gradients on their own, with no energy used, because their particles move at random. Count the particles in moles and any two solutions can be compared.

The dye spread because more of its molecules left the concentrated spot than came back. The dye went on spreading until the whole glass held the same number of dye molecules per liter.

Glossary

mole: A fixed count of particles, about 6 × 10²³. A mole of any substance holds this many particles, whatever the mass of the substance.
molar concentration: A concentration counted in moles of solute per liter of solution, written mol/L.
concentration gradient: A difference in the concentration of a substance between two regions. Down the gradient runs from where the substance is more concentrated to where it is less; against runs the other way.
net movement: The overall movement of a substance in one direction: the number of particles crossing one way minus the number crossing the other way.
diffusion: The net movement of a substance down its concentration gradient, which happens because its particles move constantly and at random, so more leave the concentrated region than arrive.
dynamic equilibrium: The state in which a substance’s particles keep crossing a membrane in both directions at equal rates, so there is no net movement and the concentrations stay equal.
passive transport: A crossing of a membrane in which the substance moves down its concentration gradient with no energy used by the cell.
simple diffusion: Passive transport in which the substance passes straight through the lipid bilayer, as oxygen and carbon dioxide do.

APBIO-U02-L04B Millimoles: counting in thousandths

Topic 2.5 · Membrane Transport · 22 steps

A beaker of glucose solution with its concentration written two ways: 0.005 mol/L, and 5 mmol/L

Glucose in your blood is at about 0.005 mol/L. Potassium ions inside a nerve cell are at about 0.14 mol/L, and outside the cell at about 0.005 mol/L.

Biologists rarely write these concentrations with all their zeros. They use a smaller unit.

Unit 2 · Cell Structure and Function

1A thousandth of a mole

Video: Watch: A thousandth of a mole

A millimole is one thousandth of a mole; converting both ways, one worked example each.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L04Ba.mp4

A mole is a count of about 6 × 10²³ particles.

One thousandth of a mole is called a , written mmol.

A long bar labelled 1 mol, divided into thin slices; the first slice is shaded and labelled 1 mmol, one thousandth of a mole

So a thousand millimoles make one mole.

To change moles into millimoles, multiply by a thousand.

Two boxes, mol on the left and mmol on the right, joined by two arrows: the upper arrow from mol to mmol is labelled multiply by 1000, the lower arrow from mmol to mol is labelled divide by 1000

Worked example

Write 0.005 mol in millimoles.

Write down the value in the question:

0.005 mol

Write down the rule:

1 mol is 1000 mmol

Multiply by 1000:

0.005 \times 1000 = 5 mmol

To change millimoles into moles, divide by a thousand.

Worked example

Write 250 mmol in moles.

Write down the value in the question:

250 mmol

Write down the rule:

1000 mmol is 1 mol

Divide by 1000:

250 \div 1000 = 0.25 mol

Concentrations work the same way. A concentration of 0.005 mol/L is 5 mmol/L: five millimoles of solute in each liter.

Most concentrations inside and around cells are written in mmol/L. Potassium ions inside a nerve cell are at 140 mmol/L. That is 0.14 mol/L.

What you are expected to know Convert between moles and millimoles, and between mol/L and mmol/L, using the rule that a millimole is one thousandth of a mole.

Now practice the conversion until it is quick: six short questions, each one a single step.

Check q1 numeric entry

A solution holds 0.012 mol of glucose.

Write this amount in millimoles.

Answer: 12 mmol (tolerance ±0.01)

Working

Write down the value in the question:

0.012 mol

Write down the rule:

1 mol is 1000 mmol

Multiply by 1000:

0.012 \times 1000 = 12 mmol

Check q2 numeric entry

A beaker holds 0.5 mol of sodium chloride.

Write this amount in millimoles.

Answer: 500 mmol (tolerance ±0.5)

Working

Write down the value in the question:

0.5 mol

Write down the rule:

1 mol is 1000 mmol

Multiply by 1000:

0.5 \times 1000 = 500 mmol

Check q3 numeric entry

A cell contains 3 mmol of potassium ions.

Write this amount in moles.

Answer: 0.003 mol (tolerance ±0.0001)

Working

Write down the value in the question:

3 mmol

Write down the rule:

1000 mmol is 1 mol

Divide by 1000:

3 \div 1000 = 0.003 mol

Check q4 numeric entry

The fluid inside a muscle cell holds 0.15 mol of potassium ions in each liter.

Write this concentration in mmol/L.

Answer: 150 mmol/L (tolerance ±0.5)

Working

Write down the value in the question:

0.15 mol/L

Write down the rule:

1 mol is 1000 mmol

Multiply by 1000:

0.15 \times 1000 = 150 mmol/L

Check q5 numeric entry

Glucose in a solution is at 60 mmol/L.

Write this concentration in mol/L.

Answer: 0.06 mol/L (tolerance ±0.0005)

Working

Write down the value in the question:

60 mmol/L

Write down the rule:

1000 mmol is 1 mol

Divide by 1000:

60 \div 1000 = 0.06 mol/L

Check q6 numeric entry

A drop of solution holds 0.0004 mol of a dye.

Write this amount in millimoles.

Answer: 0.4 mmol (tolerance ±0.005)

Working

Write down the value in the question:

0.0004 mol

Write down the rule:

1 mol is 1000 mmol

Multiply by 1000:

0.0004 \times 1000 = 0.4 mmol

From here on, concentrations in the course are written in mmol/L. A millimole is one thousandth of a mole, so 140 mmol/L of K⁺ inside a nerve cell is 0.14 mol/L.

The glucose in your blood, 0.005 mol/L, is 5 mmol/L.

Glossary

millimole: One thousandth of a mole, written mmol; a thousand millimoles make one mole. Concentrations in and around cells are usually written in mmol/L, millimoles of solute per liter of solution.

APBIO-U02-L05 Doors for ions and sugars

Topic 2.6 · Facilitated Diffusion · 88 steps

A red blood cell, and a strip of its membrane drawn large with a glucose carrier and a potassium channel set into it

Here is a red blood cell, with two of the proteins in its membrane drawn large: a glucose carrier and a potassium channel.

A red blood cell gets its energy from glucose, and glucose cannot cross its hydrophobic interior on its own. Yet glucose gets in, and the cell uses no energy to let it in.

Unit 2 · Cell Structure and Function

1A blocked solute keeps its gradient

Video: Watch first: Facilitated diffusion

A red blood cell with a glucose carrier and a potassium channel drawn large: glucose gets in, and the cell uses no energy.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T26-intro.mp4

Video: Watch: A blocked solute keeps its gradient

Iodine crosses into the starch bag and its gradient vanishes; starch cannot cross and its gradient stays.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L05.mp4

A dialysis bag of starch solution sits in a beaker of iodine solution. Iodine turns blue-black wherever it mixes with starch.

A dialysis bag of starch in iodine solution at the start, amber outside and white inside; twenty minutes later the inside is blue-black and the outside still amber

Twenty minutes later the inside of the bag is blue-black. The beaker is still amber.

The iodine, a small molecule, crossed into the bag. The iodine spread until it was as concentrated inside as outside. So the iodine gradient is gone.

The starch, a huge polymer, could not cross. The starch stayed inside, still far more concentrated in the bag than outside it. So the starch gradient stays.

If starch could cross freely, diffusion would even the starch out too, as it did the iodine. Then the starch gradient would vanish.

A solute can stay more concentrated on one side of a membrane only because the membrane does not let that solute diffuse freely across.

So a gradient can exist across a membrane only for a solute the membrane blocks. A blocked solute keeps its gradient. A free solute loses its gradient.

What you are expected to know Explain that a solute stays more concentrated on one side of a membrane only because the membrane does not let it diffuse freely across. A gradient exists only for a solute the membrane blocks.

12Fluency quiz: does the gradient change? mixed practice

Check q1

Glucose is more concentrated inside a dialysis bag than outside. The bag lets glucose through.

Does the glucose gradient change over the next hour?

A. Yes, the gradient grows
Diffusion evens a free solute out; it never piles it up.
B. ✓ Yes, the gradient shrinks
C. No, the gradient stays the same
The bag lets glucose through, so glucose diffuses out down its gradient and the two concentrations move toward each other.

Why: The bag lets glucose through.
So glucose diffuses out of the bag, down its gradient.
So the two concentrations move toward each other.
So the glucose gradient shrinks until it is gone.
A free solute loses its gradient.

Check q2

Starch is more concentrated inside a dialysis bag than outside. The bag blocks starch.

Does the starch gradient change over the next hour?

A. Yes, the gradient grows
The bag blocks starch.
B. Yes, the gradient shrinks
The bag blocks starch.
C. ✓ No, the gradient stays the same

Why: The bag does not let starch through.
So no starch crosses.
So the two concentrations stay as they are.
So the starch gradient stays.
A blocked solute keeps its gradient.

Check q3

Na⁺ is more concentrated outside a nerve cell than inside. Every sodium channel in the membrane is closed.

Does the Na⁺ gradient change while the channels stay closed?

A. Yes, the gradient grows
Every sodium channel is closed, so Na⁺ has no route across and none crosses.
B. Yes, the gradient shrinks
Every sodium channel is closed, so Na⁺ has no route across and none crosses.
C. ✓ No, the gradient stays the same

Why: Na⁺ is an ion, so Na⁺ cannot cross the phospholipid bilayer on its own.
Every sodium channel is closed.
So Na⁺ has no route across.
So no Na⁺ crosses.
So the Na⁺ gradient stays.
A blocked solute keeps its gradient.

Check q4

Iodine is more concentrated outside a dialysis bag than inside. The bag lets iodine through.

Does the iodine gradient change over the next hour?

A. Yes, the gradient grows
Diffusion evens a free solute out; it never piles it up.
B. ✓ Yes, the gradient shrinks
C. No, the gradient stays the same
The bag lets iodine through, so iodine diffuses in down its gradient and the two concentrations move toward each other.

Why: The bag lets iodine through.
So iodine diffuses into the bag, down its gradient.
So the two concentrations move toward each other.
So the iodine gradient shrinks until it is gone.
A free solute loses its gradient.

Check q5

K⁺ is more concentrated inside a muscle cell than outside. Potassium channels in the membrane open and stay open.

Does the K⁺ gradient change while the channels are open?

A. Yes, the gradient grows
Diffusion through an open route evens an ion out; it never piles it up.
B. ✓ Yes, the gradient shrinks
C. No, the gradient stays the same
The open channels give K⁺ a route, so K⁺ diffuses out down its gradient and the two concentrations move toward each other.

Why: The open potassium channels give K⁺ a route across the membrane.
So K⁺ diffuses out of the cell, down its gradient.
So the two concentrations move toward each other.
So the K⁺ gradient shrinks.
A solute with a route loses its gradient.

Check q6

A dialysis bag that lets glucose cross holds glucose at 0.6 mol/L. The bag sits in a beaker of glucose at 0.1 mol/L.

What is the situation an hour later?

A. Glucose is still at 0.6 mol/L inside the bag and 0.1 mol/L outside it
The bag lets glucose cross, so glucose leaves the bag until the two concentrations match.
B. Glucose is more concentrated inside the bag than it was at the start
Glucose crosses this membrane freely, so diffusion evens glucose out.
C. ✓ Glucose is at the same concentration inside the bag and outside it
D. All of the glucose has left the bag, and none is left inside it
Diffusion stops when the two concentrations match, so glucose stays in the bag at the outside concentration; it does not all leave.

Why: This bag lets glucose cross.
Glucose is more concentrated inside the bag than outside.
So more glucose leaves the bag than enters it.
That goes on until the concentration is the same inside and out.
A free solute loses its gradient.
Only a blocked solute keeps its gradient.

19Facilitated diffusion

Video: Watch: Facilitated diffusion

Glucose through a carrier, potassium through a channel, down the gradient, no energy: a protein only provides the route.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L05b.mp4

A millimole is one thousandth of a mole. Concentrations in cells are written in mmol/L: 8 mmol/L is 0.008 mol/L.

Oxygen is more concentrated outside a muscle cell than inside. Oxygen diffuses in straight through the bilayer. That is simple diffusion: no protein, and the cell uses no energy.

Glucose is at 8 mmol/L outside the muscle cell and 2 mmol/L inside. Glucose cannot cross the bilayer. Glucose enters through a glucose carrier, down its gradient. The cell uses no energy.

A membrane with the fluid outside the cell shaded above it and the cytosol shaded below: glucose at 8 mmol/L outside and 2 mmol/L inside, entering through a labelled glucose carrier down its gradient

Potassium ions, K⁺, are at 140 mmol/L inside a nerve cell and 5 mmol/L outside. K⁺ leaves through an open potassium channel, down its gradient. The cell uses no energy.

A membrane with the fluid outside the cell shaded above it and the cytosol shaded below: potassium ions at 140 mmol/L inside and 5 mmol/L outside, leaving through a labelled open potassium channel down their gradient

Now the other way. A cell takes glucose in from 2 mmol/L outside to 8 mmol/L inside, against its gradient. This crossing needs energy from the cell.

A membrane with the fluid outside the cell shaded above it and the cytosol shaded below: glucose at 2 mmol/L outside and 8 mmol/L inside, being moved in through a labelled membrane protein against its gradient

A crossing against the gradient is a different kind of crossing. It is called active transport. A later lesson teaches active transport.

When an ion or a large polar molecule moves down its concentration gradient through a channel or carrier protein, with no energy from the cell, that crossing is called : a diffusion the protein makes easier.

Facilitated diffusion is a form of passive transport: down the gradient, no energy used. Oxygen’s crossing is passive too, but it is simple diffusion, because no protein is involved.

A protein in the route does not mean the cell is using energy. A channel or carrier only provides the route. Down the gradient, no energy is used.

What you are expected to know Classify a crossing as facilitated diffusion when an ion or a large polar molecule moves down its concentration gradient through a channel or carrier protein with no energy from the cell, and say that it is a form of passive transport.

Check q7

Two artificial membranes sit in the same solution. Oxygen and glucose are more concentrated above each membrane than below it. Oxygen crosses both membranes. Glucose crosses only membrane 2.

Two strips of artificial membrane side by side, each with the solution above shaded darker than the solution below. Membrane 1 is a phospholipid bilayer with no proteins; membrane 2 is the same bilayer with two proteins set into it, labelled protein. Oxygen and glucose are drawn above both membranes; an oxygen arrow passes down through each membrane; a glucose arrow passes down through one of the proteins in membrane 2

Why does glucose cross only membrane 2?

A. ✓ Glucose needs a carrier, and only membrane 2 has carriers
B. The carriers in membrane 2 use energy to push glucose down through the bilayer
Glucose is more concentrated above membrane 2 than below it, so glucose moves down its gradient, and movement down a gradient needs no energy.
C. Oxygen crossing membrane 2 opens a path that glucose then follows through
Oxygen dissolves through the tails and leaves no path behind.

Why: Glucose is large and polar, so glucose cannot cross a phospholipid bilayer on its own.
Membrane 1 has no proteins, so glucose has no route through it.
Membrane 2 has glucose carriers, so glucose has a route.
Glucose moves down its gradient through the carriers: facilitated diffusion.

Check q8

K⁺ is at 150 mmol/L inside a muscle cell and 4 mmol/L outside. K⁺ leaves the cell through an open potassium channel. A student says: “This must use the cell’s energy, because a protein is doing the work.”

Is the student correct?

A. Yes
The channel only gives K⁺ a route.
B. ✓ No

Why: K⁺ is at 150 mmol/L inside and 4 mmol/L outside.
So K⁺ leaving the cell is moving down its gradient.
Movement down a gradient happens by itself.
The channel only gives K⁺ a route through the membrane.
So the cell uses no energy.

Practice writing an answer

Cl⁻ is at 110 mmol/L outside a nerve cell and 10 mmol/L inside. Cl⁻ enters the cell through an open chloride channel.

(a) Explain why Cl⁻ entering through the open channel uses no energy from the cell. (1 pt)

Frame Cl⁻ entering through the open channel uses no energy because …

Model answer Cl⁻ entering through the open channel uses no energy because Cl⁻ is moving down its concentration gradient, from 110 mmol/L outside to 10 mmol/L inside.
Cl⁻ ions move at random.
More Cl⁻ ions enter from the crowded outside than leave from the inside.
So the movement down the gradient happens by itself.
The channel only gives Cl⁻ a route through the membrane.
The channel does not push Cl⁻.
So the cell uses no energy.

Rubric

Award 1 point for: Cl⁻ moves down its concentration gradient (110 to 10 mmol/L), and movement down a gradient happens by itself, so the channel only provides the route and the cell uses no energy.
Accept: any wording that says the direction is down the gradient AND that a channel gives a route without pushing. Do not award the point for “it is passive transport” with no reason.

Slip Saying only that the crossing is passive, or facilitated diffusion, without saying why no energy is needed. The reason is the direction: down the gradient, the ions move by themselves, and the channel is only the route.

34Ions cross only through channels

Video: Watch: Ions cross only through channels

A sodium channel closed, open, closed: an ion moves only while a channel for it is open.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L05c.mp4

An ion is an atom carrying a full charge, such as a sodium ion, Na⁺, or a potassium ion, K⁺.

An ion carries a full charge. Water is attracted to that charge. Nothing in the hydrophobic interior attracts a charge. So an ion stays in the water on its own side of the membrane.

So, on their own, Na⁺ and K⁺ cross a membrane only through channel proteins.

A channel can be open or closed. A part of the channel protein can swing across the tunnel and block it, the way a door closes across a doorway. When that part swings back, the tunnel is open again.

Two panels of a membrane with the fluid outside the cell shaded above it and the cytosol shaded below. Left: a labelled sodium channel is closed; a part of the channel protein has swung across the tunnel and blocks it, and the sodium ions stay outside. Right: the same channel is open, and sodium ions flow in down their gradient

Na⁺ is at 145 mmol/L outside a nerve cell and 15 mmol/L inside. While every sodium channel is closed, no Na⁺ enters, gradient or not.

Open a sodium channel and Na⁺ flows in, down its gradient, through the channel.

Close the channel again and the flow stops. An ion moves only while a channel for it is open.

So which ions move, and when, depends on which channels are open. Open the potassium channels and K⁺ moves. Open the sodium channels and Na⁺ moves.

What you are expected to know Explain that because ions such as Na⁺ and K⁺ cross a membrane only through channel proteins, an ion moves only while a channel for it is open, so which ions move, and when, depends on which channels are open.

45Fluency quiz: does the ion move, and which way? mixed practice

Check q9

The bar chart shows Na⁺ outside a nerve cell and inside it. Every sodium channel in the membrane is closed.

A bar chart with two bars, labelled outside the cell and inside the cell, showing the concentration of Na⁺ in mmol/L; each bar carries its value above it

Does Na⁺ cross the membrane?

A. Yes, into the cell
Every sodium channel is closed, so Na⁺ has no route across, steep gradient or not.
B. Yes, out of the cell
Every sodium channel is closed, so Na⁺ has no route across, steep gradient or not.
C. ✓ No

Why: Na⁺ is an ion.
An ion crosses a membrane only through an open channel.
Every sodium channel is closed.
So Na⁺ has no route.
So no Na⁺ crosses, however tall the outside bar is.

Check q10

The bar chart shows the concentration of Na⁺ outside a nerve cell and inside it. Now the sodium channels are open.

Which way does Na⁺ move?

A. ✓ Into the cell
B. Out of the cell
Na⁺ is 145 mmol/L outside and 15 mmol/L inside, so down its gradient is into the cell.
C. Neither: there is no gradient
The two bars are different heights, so there is a gradient.

Why: The sodium channels are open, so Na⁺ has a route.
Na⁺ is 145 mmol/L outside and 15 mmol/L inside.
The outside bar is taller.
Down the gradient runs from the larger concentration to the smaller.
So Na⁺ moves into the cell.

Check q11

The bar chart shows K⁺ outside a nerve cell and inside it. The potassium channels are open.

A bar chart with two bars, labelled outside the cell and inside the cell, showing the concentration of K⁺ in mmol/L; each bar carries its value above it

Which way does K⁺ move?

A. Into the cell
K⁺ is 140 mmol/L inside and 5 mmol/L outside, so down its gradient is out of the cell.
B. ✓ Out of the cell
C. Neither: there is no gradient
The two bars are different heights, so there is a gradient.

Why: The potassium channels are open, so K⁺ has a route.
K⁺ is 140 mmol/L inside and 5 mmol/L outside.
The inside bar is taller.
Down the gradient runs from the larger concentration to the smaller.
So K⁺ moves out of the cell.

Check q12

The bar chart shows the concentration of calcium ions (Ca²⁺) outside a muscle cell and inside it. Every calcium channel in the membrane is closed.

A bar chart with two bars, labelled outside the cell and inside the cell, showing the concentration of Ca²⁺ in mmol/L; each bar carries its value above it

Does Ca²⁺ cross the membrane?

A. Yes, into the cell
Every calcium channel is closed, so Ca²⁺ has no route across, however steep the gradient.
B. Yes, out of the cell
Every calcium channel is closed, so Ca²⁺ has no route across, however steep the gradient.
C. ✓ No

Why: Ca²⁺ is an ion.
An ion crosses a membrane only through an open channel.
Every calcium channel is closed.
So Ca²⁺ has no route.
So no Ca²⁺ crosses, however much taller the outside bar is.

Check q13

The bar chart shows the concentration of chloride ions (Cl⁻) outside an airway cell and inside it. The chloride channels are open.

A bar chart with two bars, labelled outside the cell and inside the cell, showing the concentration of Cl⁻ in mmol/L; each bar carries its value above it

Which way does Cl⁻ move?

A. ✓ Into the cell
B. Out of the cell
Cl⁻ is 110 mmol/L outside and 30 mmol/L inside, so down its gradient is into the cell.
C. Neither: there is no gradient
The two bars are different heights, so there is a gradient.

Why: The chloride channels are open, so Cl⁻ has a route.
Cl⁻ is 110 mmol/L outside and 30 mmol/L inside.
The outside bar is taller.
Down the gradient runs from the larger concentration to the smaller.
So Cl⁻ moves into the cell.

Check q14

Nerve cells hold Cl⁻ at 20 mmol/L; the fluid outside holds 120 mmol/L. With the chloride channels open, Cl⁻ inside rises to 30 mmol/L in five minutes. With the channels blocked, Cl⁻ inside stays at 20 mmol/L.

What do the results show?

A. Cl⁻ crossed the phospholipid bilayer itself
If Cl⁻ crossed the bilayer itself, the cells with blocked channels would have changed too, and they did not.
B. Cl⁻ left the cells through the open channels
The inside concentration rose, so Cl⁻ went in, not out.
C. ✓ Cl⁻ entered the cells through the open channels

Why: Cl⁻ is an ion, and an ion crosses a membrane only through an open channel.
With the channels open, the inside rose from 20 to 30 mmol/L, so Cl⁻ moved in, down its gradient.
With the channels blocked, nothing changed, because Cl⁻ had no route.

Check q15

A kidney cell has a K⁺ gradient and a Na⁺ gradient across its membrane. Only its sodium channels are open.

Which ion moves?

A. K⁺
The potassium channels are closed.
B. Both K⁺ and Na⁺
The potassium channels are closed.
C. ✓ Na⁺

Why: An ion moves only while a channel for it is open.
Only the sodium channels are open.
So Na⁺ has a route through the membrane and K⁺ does not.
So Na⁺ moves and K⁺ stays where it is.

Check q16

A kidney cell holds sodium ions (Na⁺) at 12 mmol/L inside and 142 mmol/L outside. Its sodium channels are open.

Which way does Na⁺ move through the open sodium channels?

A. ✓ Into the cell
B. Out of the cell
Na⁺ is 142 mmol/L outside and 12 mmol/L inside, so down its gradient is into the cell.
C. Neither: there is no gradient
The two concentrations differ, so there is a gradient.

Why: Na⁺ is 142 mmol/L outside and 12 mmol/L inside.
Down its gradient runs from the larger concentration to the smaller.
The sodium channels are open, so Na⁺ has a route.
So Na⁺ moves into the cell.

54Aquaporins: channels for water

Video: Watch: Aquaporins: channels for water

Water channels open, 12% in five minutes; blocked, 3%: aquaporins carry water fast, in the direction it would go anyway.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L05d.mp4

Water, small and polar, crosses a phospholipid bilayer only slowly, in small amounts.

Kidney cells move a great deal of water across their membranes every day. One thing that helps them is a set of channel proteins in the membrane that let water through. These channels can be open or blocked.

Kidney cells are placed in a dilute solution, one in which water moves into the cells. With the water channels open, the cells gain 12% in mass in five minutes. That extra mass is the water that entered.

Two panels of a kidney cell's membrane with the fluid outside the cell shaded above it and the cytosol shaded below: with the labelled aquaporin open, water molecules pass through it and the cells gain 12% in five minutes; with the aquaporin blocked, water trickles through the phospholipid bilayer and they gain 3%

The same cells are tested with the water channels blocked. They gain only 3% in mass in the same five minutes. That 3% is the water that trickled through the phospholipid bilayer itself.

A channel protein for water is called an . Through aquaporins, large quantities of water cross a membrane quickly.

Aquaporins do not push water. Aquaporins are open channels. Water moves through them only in the direction it would move anyway, just far faster.

Cells whose energy supply is cut still gain 11%. So the cell uses no energy on the crossing. The aquaporins only provide the route.

What you are expected to know Say that water crosses a phospholipid bilayer only slowly, and that aquaporins, channel proteins for water, let large quantities of water cross a membrane quickly, in the direction it would move anyway, with no energy used by the cell.

Check q17

Two membrane sacs come from kidney cells. Sac 1 has working aquaporins; sac 2 has none. Both sacs sit in the same dilute solution. In five minutes, one sac gains 15% in mass and the other gains 4%.

Which sac gained 15%?

A. Sac 2, the sac with no aquaporins
Aquaporins let water cross a membrane far faster than the phospholipid bilayer does, so the sac with aquaporins gains water faster.
B. ✓ Sac 1, the sac with working aquaporins
C. Cannot tell: both sacs gained water
Both sacs gained water, but 15% is far more than 4%.

Why: Aquaporins let water cross a membrane far faster than the phospholipid bilayer does.
Sac 1 has working aquaporins.
So water enters sac 1 quickly.
So sac 1 gained the most, 15%.

Check q18

Sac 2 has no aquaporins, yet it gained 4% in mass.

Why did sac 2 gain water at all?

A. Sac 2 must have a tear in its membrane
A phospholipid bilayer lets water through slowly on its own, so a sac with no aquaporins gains a little without any tear.
B. The dilute solution pushed water in between the heads
Nothing pushes water; water molecules move at random, and a few pass between the tails.
C. ✓ Water trickles slowly through the phospholipid bilayer itself

Why: Water is small and polar, with no charge.
Small polar molecules with no charge pass through a phospholipid bilayer in small amounts, slowly.
So even with no aquaporins, a little water enters sac 2.
So sac 2 gains a little, 4%.

66What a change to a door does

Video: Watch: What a change to a door does

Block a door, add doors, reverse the gradient; lidocaine and the gliflozins.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L05e.mp4

Every crossing through a channel or carrier depends on two things: the route (is a door open?) and the gradient (which way is down?).

Block the door, and the substance that used it stops crossing. Its gradient stays.

Add doors, and the crossing speeds up in the same direction. A cell with twice as many open potassium channels loses K⁺ twice as fast: still leaving, still down its gradient.

Two strips of a cell's membrane side by side, the fluid outside the cell shaded above each and the cytosol below, K⁺ at 140 mmol/L inside and 5 mmol/L outside both. Cell 1's strip has one open potassium channel with an arrow leaving through it; cell 2's strip has two open potassium channels with a thicker arrow leaving through each. Each drawn channel stands for a hundred channels

Reverse the gradient, and the net direction through a carrier reverses. Put more glucose inside a cell than outside, and its glucose carrier lets glucose leave.

Cells have these doors in different numbers. Substances in the body can also block these doors. Some medicines work exactly this way.

Lidocaine, the anesthetic a dentist injects, blocks the sodium channels of nerve cells. With the sodium channels blocked, no Na⁺ enters those nerve cells, and the nerve cannot send its pain signal.

The gliflozin medicines for diabetes block a glucose carrier in kidney cells. With that carrier blocked, glucose that the kidney would have taken back into the blood stays in the urine and leaves the body.

What you are expected to know Predict the effect on a named substance’s movement when a channel or carrier is blocked, made more abundant, or its concentration gradient is reversed, reasoning from the route and the gradient.

Now predict what happens in three cases: block a door, add doors, reverse a gradient.

Check q19

Glucose is more concentrated outside a muscle cell than inside. The cell doubles the number of glucose carriers in its membrane.

What happens to the movement of glucose?

A. ✓ Glucose enters faster, in the same direction
B. Glucose enters more slowly
Each carrier is one route, and twice as many routes let more glucose cross each second, so glucose enters faster, not more slowly.
C. Glucose enters at the same speed as before
Each carrier is one route, and twice as many routes let more glucose cross each second.

Why: Each carrier is one route.
Twice as many carriers is twice as many routes.
So glucose enters faster.
The gradient is unchanged.
So the direction is unchanged: glucose still enters.

Check q20

Liver cells carry a carrier for a small molecule, X. The bar chart shows X outside a liver cell and inside it.

A bar chart with two bars, labelled outside the cell and inside the cell, showing the concentration of X in mmol/L; each bar carries its value above it

Which way does X move through the carrier?

A. Into the cell
X is 6 mmol/L inside and 1 mmol/L outside, so down its gradient is out of the cell; a carrier lets X move down its gradient in either direction.
B. ✓ Out of the cell
C. Neither: there is no gradient
The two bars are different heights, so there is a gradient.

Why: A carrier gives X a route in either direction.
The gradient sets the direction.
X is 6 mmol/L inside and 1 mmol/L outside.
The inside bar is taller.
So X moves out of the cell.

Check q21

The outside solution around the liver cells is changed. The bar chart shows X outside a liver cell and inside it now.

Which way does X move through the carrier now?

A. ✓ Into the cell
B. Out of the cell
The gradient has reversed.
A carrier lets X move down its gradient in either direction.
C. Neither: there is no gradient
The two bars are different heights, so there is a gradient.

Why: A carrier gives X a route in either direction.
The gradient sets the direction.
X is now 8 mmol/L outside and 2 mmol/L inside.
The outside bar is taller.
So X now moves into the cell.
Reverse the gradient, and the net movement through a carrier reverses.

Check q22

Two groups of muscle cells differ only in their glucose carriers: group L has 1,000 per cell, group H has 10,000. Glucose starts at 2 mmol/L inside every cell. The solution outside stays at 10 mmol/L.

Which group takes glucose in faster?

A. ✓ Group H
B. Group L
Each carrier is one route, and group H has ten times as many carriers as group L.
C. Both groups take glucose in at the same rate
Each carrier is one route, and group H has ten times as many carriers as group L.

Why: Each carrier is one route.
Group H has ten times as many carriers as group L.
So group H has ten times as many routes for glucose.
So group H takes glucose in faster.

Check q23

Two groups of muscle cells differ only in their glucose carriers: group L has 1,000 per cell, group H has 10,000. Once inside, the glucose stays as glucose: the cells leave it unchanged. The solution outside stays at 10 mmol/L.

Where does the glucose concentration inside the cells level off?

A. Group H above 10 mmol/L; group L below 10 mmol/L
Once the inside reaches 10 mmol/L, the outside’s value, there is no gradient left, in either group.
B. ✓ Both groups near 10 mmol/L
C. Only group H reaches 10 mmol/L; group L never changes
Group L still has 1,000 working carriers, so glucose enters group L too, just more slowly.

Why: A carrier only lets glucose move down its gradient.
Glucose moves from 10 mmol/L outside toward the inside until the inside reaches 10 mmol/L and no gradient is left.
That is true for both groups.
So both level off near 10 mmol/L; more carriers change the speed only.

Check q24

Placental cells carry glucose carriers between a mother’s blood and her baby’s blood. Mother A’s blood has glucose at 7 mmol/L; her baby’s blood has glucose at 4 mmol/L.

Which way does glucose move?

A. ✓ From Mother A to her baby
B. From the baby to Mother A
Mother A’s blood is at 7 mmol/L and her baby’s at 4 mmol/L.
C. Neither: there is no net movement
The two concentrations differ, so there is a gradient.

Why: A carrier lets glucose move down its gradient.
Mother A’s blood is at 7 mmol/L.
Her baby’s blood is at 4 mmol/L.
Down the gradient runs from 7 mmol/L to 4 mmol/L.
So glucose moves from Mother A to her baby.

Check q25

Between meals, a liver cell holds glucose at 6 mmol/L and the blood around it holds glucose at 4 mmol/L. Glucose carriers sit in the liver cell’s membrane.

Which way does glucose move?

A. From the blood into the liver cell
The gradient sets the direction.
The liver cell holds glucose at 6 mmol/L and the blood at 4 mmol/L.
B. Neither: there is no net movement
The two concentrations differ, so there is a gradient.
C. ✓ From the liver cell into the blood

Why: A carrier lets glucose move down its gradient in either direction.
The liver cell holds glucose at 6 mmol/L and the blood 4 mmol/L.
Down the gradient runs from 6 to 4 mmol/L.
So glucose moves from the liver cell into the blood.

Check q26

A red blood cell sits in a solution with glucose at 5 mmol/L outside and 1 mmol/L inside. A drug then blocks all of its glucose carriers.

What happens to the glucose?

A. Glucose enters through the bilayer instead
Glucose is large and polar and never crosses the bilayer itself.
B. Glucose leaves the cell
Glucose is still more concentrated outside than inside.
With no route, glucose does not move in either direction.
C. ✓ Glucose stops entering, and its gradient stays

Why: The carriers were glucose’s only route across the membrane.
The drug blocks them.
So glucose has no route and stops entering.
A solute that cannot cross keeps its gradient.
So glucose stays at 5 mmol/L outside and 1 mmol/L inside.

Practice writing an answer

Cells lining the small intestine carry a carrier protein for fructose, a sugar of about the same size and polarity as glucose. After a meal, fructose is at 5 mmol/L in the fluid of the gut and 1 mmol/L inside the cells, and fructose enters the cells.

(a) Describe how fructose enters the cells, and name the kind of transport. (1 pt)

Frame Fructose enters the cells by …; this is …

Model answer Fructose enters the cells by passing through the carrier protein, down its concentration gradient from 5 mmol/L in the gut fluid to 1 mmol/L inside the cells; this is facilitated diffusion, a form of passive transport.
Movement down a gradient happens by itself.
So the cell uses no energy on the crossing.

Rubric

Award 1 point for: fructose passes through the carrier down its concentration gradient (5 to 1 mmol/L) with no energy used by the cell: facilitated diffusion.
Accept: passive transport through a carrier, provided down the gradient and no energy used are both stated.

Slip Saying the cell uses energy because a protein is involved. The carrier only gives fructose a route. Down its gradient, fructose moves on its own.

(b) A drug blocks half of the fructose carriers. Predict what happens to the movement of fructose. (1 pt)

Model answer Fructose still enters the cells, in the same direction as before.
Each carrier is one route, and half the routes are now blocked.
So fructose enters at about half the rate.
The gradient is unchanged, so the direction is unchanged.

Rubric

Award 1 point for: fructose still enters (same direction) but more slowly, about half as fast, because fewer carriers give fewer routes.
Accept: slower, still inward.

Slip Reversing the direction or stopping the movement altogether. Fewer doors slow a crossing. Only the gradient sets its direction.

(c) Hours later, fructose in the gut fluid has fallen to 0.5 mmol/L while inside the cells it is still 1 mmol/L, and the drug has worn off. Predict the net movement of fructose now, and justify your prediction. (2 pt)

Model answer Fructose now leaves the cells.
The net movement is outward, into the gut fluid.
The carrier lets fructose pass in either direction.
The gradient now runs from 1 mmol/L inside to 0.5 mmol/L outside.
So the net movement follows the gradient and reverses.

Rubric

Award 1 point for: the net movement of fructose is now out of the cells into the gut fluid (the direction reverses).
Award 1 point for: the carrier gives fructose a route in either direction and fructose is now more concentrated inside (1 mmol/L) than outside (0.5 mmol/L), so the net movement follows the gradient.
Accept: the second point in any wording that names both the two-way route and the reversed gradient.

Slip Letting the cell’s need set the direction: the cell wants fructose, so it keeps coming in. A carrier follows the gradient. Reverse the gradient and the net movement reverses.

A membrane that blocks a solute keeps that solute’s gradient. A channel or carrier then lets an ion, a sugar or water cross down that gradient, and the cell uses no energy. Block the door, add more doors or reverse the gradient, and you can say what happens.

Glucose reaches the red blood cell’s inside through a carrier, down its gradient, with no energy used. Its potassium can leave only while a potassium channel is open. Water crosses fast through aquaporins.

Glossary

facilitated diffusion: Passive transport in which an ion or a large polar molecule moves down its concentration gradient through a channel or carrier protein, with no energy from the cell.
aquaporin: A channel protein for water. Through aquaporins, large quantities of water cross a membrane quickly, in the direction water would move anyway.

APBIO-U02-P26 Practice questions: Topic 2.6

Topic 2.6 · Facilitated Diffusion · 9 MCQ · 2 FRQ · for APBIO-U02-T26

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 for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. 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.

Video: Watch first: doors for ions and sugars, summed up

A blocked solute keeps its gradient; facilitated diffusion; channels open and closed; aquaporins; what a change to a door does.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T26-summary.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T26-summary.mp4

Q1 P26-q01

Root cells of a bean plant take in sucrose (table sugar) from the fluid around them, where it is 12 mmol/L; inside the cells it is 3 mmol/L. The sucrose enters through a carrier protein.

Which of the following describes this crossing?

A. Simple diffusion
Sucrose is a large polar molecule and cannot cross the bilayer on its own.
B. ✓ Facilitated diffusion
C. Active transport
Sucrose moves from 12 mmol/L to 3 mmol/L, down its gradient, so no energy is needed; the carrier only provides the route.
D. Osmosis
Osmosis is the net movement of water across a membrane.

Why: Sucrose is a large polar molecule, so it cannot cross the bilayer on its own.
Sucrose moves from 12 mmol/L to 3 mmol/L, down its gradient, through a carrier.
Movement down a gradient happens by itself, so the cell uses no energy.
That is facilitated diffusion.

Q2 P26-q02

Root cells of a barley plant hold K⁺ at 100 mmol/L; the soil water around them holds 1 mmol/L. A toxin locks every potassium channel in their membranes shut.

What is the net movement of K⁺ while the channels are locked?

A. K⁺ leaves the cells
K⁺ is a charged ion and cannot cross the phospholipid bilayer, however steep its gradient.
B. K⁺ enters the cells
K⁺ is an ion, and an ion crosses only through an open channel; every potassium channel is locked shut, so K⁺ has no route.
C. ✓ There is no net movement of K⁺
D. Not enough information to say
The stem gives everything needed: the ion, its concentrations and the state of its channels.

Why: K⁺ is a charged ion.
An ion crosses a membrane only through an open channel protein.
Every potassium channel is locked shut.
So K⁺ has no route.
So there is no net movement of K⁺, despite the 100-to-1 mmol/L gradient.

Q3 P26-q03

Two frogs lay eggs. The eggs of frog 1 have many aquaporins in their membranes; the eggs of frog 2 have very few. Both sets of eggs sit in a dilute solution, in which water moves into the eggs. After 30 minutes the eggs of frog 1 have swelled and burst; the eggs of frog 2 have gained about 2% in volume.

What does the comparison show about aquaporins?

A. Aquaporins push water into a cell using the cell's energy
An aquaporin is an open channel, not a powered protein.
B. Aquaporins are the only route by which water can cross a membrane
The eggs of frog 2 still gained 2% in volume, so water crossed their phospholipid bilayer slowly.
C. Aquaporins reverse the direction in which water crosses a membrane
Water entered both sets of eggs; the aquaporins changed how fast, not which way.
D. ✓ Aquaporins let large quantities of water cross a membrane quickly

Why: The eggs of frog 2 gained only 2%.
So water crosses a phospholipid bilayer only slowly.
The eggs of frog 1, with many aquaporins, filled far faster and burst.
An aquaporin is a channel protein for water.
So aquaporins let large quantities of water cross a membrane quickly.

Q4 P26-q04

A cell’s energy supply has been cut off. At its membrane, O₂, more concentrated outside than inside, crosses the bilayer inward. An amino acid, 6 mmol/L outside and 2 mmol/L inside, crosses inward, but only through a membrane protein. Na⁺, 145 mmol/L outside and 15 mmol/L inside, stays put while every sodium channel is shut.

Which of the following crossings is facilitated diffusion?

A. ✓ The amino acid's crossing
B. The O₂'s crossing
O₂ crosses with no protein at all, which is simple diffusion.
C. The Na⁺'s crossing
Na⁺ makes no crossing; its channels are shut.
D. All three: the O₂, the amino acid and the Na⁺
Only one of the three uses a protein and moves.

Why: Facilitated diffusion runs down the gradient, through a channel or carrier, with no energy from the cell.
The amino acid moves from 6 toward 2 mmol/L, down its gradient, through a membrane protein.
The energy supply has been cut, so the cell uses no energy.
So this is facilitated diffusion.

Q5 P26-q05

The figure shows lactate, a small molecule with a full charge that muscle makes during hard exercise, at two cells during a sprint. In a leg muscle cell lactate is 15 mmol/L inside and 3 mmol/L in the blood; in a heart muscle cell it is 3 mmol/L in the blood and 1 mmol/L inside. Both cells carry the same kind of lactate carrier.

Lactate concentrations at a leg muscle cell and a heart muscle cell during a sprint. Each cell has the same kind of lactate carrier in its membrane.

What is the net movement of lactate at each cell?

A. Out of both cells
A carrier has no direction of its own; in the heart cell lactate is higher in the blood than inside, so it moves in.
B. Into both cells
In the leg muscle cell lactate is 15 mmol/L inside and 3 mmol/L in the blood, so down its gradient is out.
C. ✓ Out of the leg muscle cell and into the heart muscle cell
D. Into the leg muscle cell and out of the heart muscle cell
Lactate is 15 mmol/L inside the leg cell and 3 outside, so it leaves; 3 outside the heart cell and 1 inside, so it enters.

Why: A carrier gives lactate a route in either direction; the gradient decides the direction.
At the leg muscle cell, lactate is 15 mmol/L inside and 3 mmol/L outside, so lactate leaves.
At the heart muscle cell, lactate is 3 mmol/L outside and 1 mmol/L inside, so it enters.

Q6 P26-q06

Sweat flows along a sweat duct. The cells lining the duct take Cl⁻ out of the sweat and into themselves as the sweat flows past: Cl⁻ is 60 mmol/L in the sweat and 15 mmol/L inside the cells, and it enters the cells through chloride channels. In one person those chloride channels are faulty and stay shut.

Predict what happens to the Cl⁻ in this person's sweat.

A. Cl⁻ enters the cells through the phospholipid bilayer
Cl⁻ is a charged ion; water attracts the charge, and nothing in the hydrophobic interior does, so Cl⁻ cannot cross the bilayer at all.
B. ✓ Cl⁻ stays in the sweat
C. Cl⁻ enters the cells through their sodium channels
Each kind of ion crosses only through its own channels; a sodium channel is a tunnel for Na⁺.
D. The cells pull Cl⁻ in through the shut channels
A shut channel is a shut door, and a cell cannot pull an ion through a tunnel that is blocked.

Why: Cl⁻ is an ion.
An ion crosses a membrane only through a channel protein.
In this person the chloride channels stay shut.
So Cl⁻ has no route into the cells, however steep the 60-to-15 mmol/L gradient.
So Cl⁻ stays in the sweat, and the sweat stays salty.

Q7 P26-q07

During each heartbeat, the sodium channels of a heart muscle cell open for about one millisecond and then shut again. Na⁺ is 140 mmol/L outside the cell and 10 mmol/L inside.

Which of the following describes the movement of Na⁺ over one heartbeat?

A. ✓ Na⁺ flows in only during the millisecond the channels are open
B. Na⁺ flows in steadily through the beat, channels open or shut
An ion crosses only through an open channel, so while the channels are shut Na⁺ has no route.
C. Na⁺ flows out during the millisecond the channels are open
Na⁺ is 140 mmol/L outside and 10 mmol/L inside, so down its gradient is inward.
D. Na⁺ flows in only if the cell uses energy while the channels are open
Movement down a gradient through an open channel needs no energy from the cell.

Why: Na⁺ is a charged ion, so it crosses only through a channel protein, and only while a channel is open.
For the millisecond the sodium channels are open, Na⁺ flows in, from 140 toward 10 mmol/L, down its gradient, with no energy used.
When the channels shut, the flow stops.

Q8 P26-q08

Root cells take up water from soil water through their aquaporins. A gardener over-fertilizes. The soil water becomes far more concentrated in solute than the root cells are. So water now moves out of the root cells into the soil water.

What part do the aquaporins play in this water movement?

A. The aquaporins close, so the water stays inside the cells
An aquaporin is an open channel that does not shut to protect the cell.
B. The aquaporins push water back in using the cells' energy
An aquaporin is an open channel, not a powered protein.
C. The aquaporins let solute into the cells along with the water
An aquaporin passes water and nothing else.
D. ✓ The aquaporins let water leave faster, following its own gradient

Why: An aquaporin is a channel protein for water.
Large quantities of water cross through it quickly.
Water moves through it in the direction water would move anyway.
In the over-fertilized soil, water’s direction is outward.
So the aquaporins let water leave the cells faster than the phospholipid bilayer would.

Q9 P26-q09

Each of the following describes a substance crossing a cell’s plasma membrane.

Which of the following four crossings is an example of facilitated diffusion?

A. CO₂ leaving a cell straight through the bilayer, down its gradient
CO₂ is small and nonpolar and crosses with no protein, which is simple diffusion.
B. ✓ An amino acid entering a cell through a carrier, from 9 mmol/L outside to 3 mmol/L inside, with no energy used
C. A sugar entering a cell through a protein, from 1 mmol/L outside to 6 mmol/L inside, stopping when the energy supply is cut
This sugar moves from 1 mmol/L outside to 6 mmol/L inside, against its gradient, and stops when the energy supply is cut, so the cell is using energy for it.
D. Water crossing a phospholipid bilayer slowly, on its own
Water crossing the phospholipid bilayer uses no protein, so this is simple diffusion, just slow.

Why: Facilitated diffusion is a crossing down the gradient, through a channel or carrier, with no energy from the cell.
The amino acid moves from 9 mmol/L outside toward 3 mmol/L inside: down its gradient.
It moves through a carrier and needs no energy.
So its crossing is facilitated diffusion.

FRQ 1 P26-frq1 · Conceptual Analysis scaffolded

Cells lining a kidney tubule pass glucose from the tubule fluid back to the blood, so that the body keeps the glucose rather than losing it in urine. On the side of each cell that faces the blood there are glucose carrier proteins. Between meals, glucose is 8 mmol/L inside these cells and 5 mmol/L in the blood.

(a) Identify the kind of membrane protein glucose uses to cross the membrane facing the blood, and explain why glucose needs a protein at all. (1 pt)

Frame Glucose crosses through a … protein. Glucose needs a protein because …

Hint What fills the middle of the membrane, and what kind of molecule is glucose?

Model answer Glucose crosses through a carrier protein.
Glucose needs a protein because glucose is a large polar molecule.
Water is attracted to glucose.
The hydrocarbon tails in the middle of the membrane carry no charges or partial charges.
So nothing in the tails attracts glucose.
So glucose cannot cross the bilayer on its own.

Rubric

Award 1 point for: a carrier protein (accept: a channel or carrier, or a membrane protein that binds glucose), AND glucose is a large polar molecule that water attracts, and nothing in the hydrophobic interior attracts it, because the hydrocarbon tails carry no charges or partial charges, and it cannot cross the bilayer on its own.
Accept: 'the oily middle of the membrane has nothing that attracts a polar molecule'. Both the protein and the reason are needed for the point.

Slip Naming the carrier without saying why glucose needs it, or saying glucose is 'too big'. It is the polarity of glucose against the uncharged interior that keeps it out.

(b) Describe the direction of the net movement of glucose, using the two concentrations, and name this kind of transport. (1 pt)

Frame Glucose moves from … (… mmol/L) to … (… mmol/L), down its …; this kind of transport is …

Hint Which side has the higher glucose concentration, and what is the name for a crossing of this kind?

Model answer Glucose moves from inside the cell (8 mmol/L) to the blood (5 mmol/L), down its concentration gradient; this kind of transport is facilitated diffusion, a form of passive transport.

Rubric

Award 1 point for: net movement from inside the cell (8 mmol/L) into the blood (5 mmol/L), down glucose's concentration gradient, AND the name facilitated diffusion (a form of passive transport).
Accept: 'from high to low, through the carrier, with no energy used: facilitated diffusion'. Both the direction with the concentrations and the name are needed.

Slip Giving the name without the direction, or the direction without the name. The name and the direction are both needed.

(c) State whether the cells use energy on this crossing, and explain why. (1 pt)

Frame The cells use no energy because …

Hint Think about what makes glucose molecules move down a gradient in the first place, and what the carrier adds to that.

Model answer The cells use no energy because glucose is moving down its concentration gradient, and movement down a gradient happens by itself.
Glucose molecules move at random.
More of them leave the crowded inside than arrive from the less crowded blood.
The carrier only provides the route.
So the cells use no energy.

Rubric

Award 1 point for: glucose is moving down its concentration gradient, and movement down a gradient happens by itself, because molecules move at random so more of them leave the more concentrated side than arrive from the less concentrated one; the carrier only provides the route.
Accept: 'movement down a gradient needs no energy; the carrier is just the door'. Do not award the point for 'because it is passive' with no reason.

Slip Saying 'because it is passive transport' and stopping. The point wants the reason downhill movement needs no energy.

(d) After a meal, blood glucose rises to 8 mmol/L, equal to the concentration inside the cells. Predict the net movement of glucose through the carriers. (1 pt)

Frame When the two concentrations are equal, the net movement of glucose is …, because …

Hint Think about what a carrier does to a glucose molecule that binds to it, from either face of the membrane. Then ask what is different about the two faces once the two concentrations are equal.

Model answer When the two concentrations are equal, the net movement of glucose is zero, because glucose molecules cross the carriers in both directions in equal numbers.
Molecules still cross the carriers.
As many go one way as the other.
That is a dynamic equilibrium.

Rubric

Award 1 point for: no net movement, because glucose molecules cross the carriers in both directions at equal rates when the concentrations are equal (a dynamic equilibrium).
Accept: 'glucose still crosses but the two flows cancel'. Do not award the point for 'glucose stops moving'.

Slip Saying glucose stops moving. It keeps crossing both ways; only the net movement is zero.

(e) A drug halves the number of glucose carriers in these cells. Predict the effect on the movement of glucose between meals, and justify your prediction. (1 pt)

Frame With half the carriers, glucose … at about … the rate, in the … direction, because …

Hint What does the number of carriers control, and what sets the direction of net movement?

Model answer With half the carriers, glucose still leaves the cells at about half the rate, in the same direction, because each carrier is one route, and halving the routes halves how many molecules cross per second, while the gradient from 8 mmol/L to 5 mmol/L is unchanged and still sets the direction.

Rubric

Award 1 point for: glucose still leaves the cells into the blood, in the same direction, but at about half the rate, because each carrier is one route and halving the routes halves how many molecules cross per second, while the gradient (8 to 5 mmol/L) is unchanged and still sets the direction.
Accept: 'slower, same direction, because fewer doors'. Do not award the point for a change of direction or for 'the cell now needs energy'.

Slip Changing the direction, or saying the cell now needs energy. Fewer doors slow a crossing; only the gradient sets its direction.

FRQ 2 P26-frq2 · Scientific Investigation

A scientist studies how an amino acid enters mouse cells. She sets up two dishes. Dish 1 holds untreated cells. Dish 2 holds cells she has treated with a substance that blocks their amino acid carriers. Each dish holds 50,000 cells in a solution with the amino acid at 10 mmol/L, and inside the cells it starts at 0 mmol/L. She measures the concentration inside the cells at 0, 5, 10 and 20 minutes; the table shows the results.

Concentration of the amino acid inside the cells over 20 minutes. The solution around the cells holds 10 mmol/L throughout.

(a) Describe how the amino acid enters the cells in dish 1, and name the kind of transport. (1 pt)

Frame In dish 1 the amino acid enters through …, moving …, using … energy from the cell; this is …

Model answer In dish 1 the amino acid enters through carrier proteins, moving down its concentration gradient from 10 mmol/L outside toward 0 mmol/L inside, using no energy from the cell; this is facilitated diffusion, a form of passive transport.
Movement down a gradient happens by itself, so the cell uses no energy on it.

Rubric

Award 1 point for: the amino acid enters through carrier proteins, moving down its concentration gradient from 10 mmol/L outside toward 0 mmol/L inside, with no energy from the cell: facilitated diffusion (a form of passive transport).
Accept: 'through the carriers, from high to low, with no energy used'. The route and the name are both needed.

Slip Saying the amino acid 'diffuses in' with no mention of the carriers. Dish 2 shows that with the carriers blocked almost none gets in.

(b) Explain why the investigation uses the same number of cells in each dish. (1 pt)

Model answer More cells would give more membrane and more carriers.
So more cells take up more amino acid.
With the same number of cells in each dish, that cause is ruled out.
So any difference between the dishes can be put down to the blocked carriers.

Rubric

Award 1 point for: with equal numbers of cells, a difference in the amino acid taken up between the dishes can be put down to the blocked carriers rather than to one dish having more cells, and so more membrane and more carriers, taking up more.
Do not award the point for 'it was a control' or 'to make it fair' with no statement of what an unequal count would have muddled.

Slip Writing 'it was a control' or 'to keep it fair' and stopping. Say what an unequal count would have muddled: more cells take up more, whatever their carriers are doing.

(c) Determine the concentration of the amino acid inside the dish-1 cells at 60 minutes. (1 pt)

Model answer The concentration inside is about 10 mmol/L, the same as the solution around the cells.
The amino acid moves down its gradient through the carriers.
Once the inside reaches 10 mmol/L there is no gradient left.
So the rise levels off there.

Rubric

Award 1 point for: the decision that the concentration is about 10 mmol/L, the same as the solution outside (accept 9.5 to 10 mmol/L), AND the reasoning it rests on: the rise levels off once the inside matches the outside, because there is then no gradient.
Do not award the point for a value above 10 mmol/L, for a fall back toward 0 mmol/L, or for the value alone with no reasoning.

Slip Continuing the rise past 10 mmol/L because it rose steadily before. The carriers cannot take the inside above the outside.

(d) Support the claim you made in (c), using what a carrier does for the amino acid. (1 pt)

Model answer The carriers only let the amino acid run down its gradient.
Once the inside reaches 10 mmol/L, the outside’s value, there is no gradient left.
So molecules cross in both directions at equal rates.
That is a dynamic equilibrium, so there is no net movement.
The cells use no energy on this amino acid.
Therefore the carriers cannot push the inside concentration any higher.

Rubric

Award 1 point for: the evidence AND the reasoning: the carriers only let the amino acid run down its gradient; once the inside reaches the outside's 10 mmol/L there is no gradient, so molecules cross in both directions at equal rates (a dynamic equilibrium) and there is no net movement; the cells use no energy on this crossing, so the carriers cannot push the inside any higher.
Accept: 'at 10 mmol/L the gradient is gone, so the net movement is zero; carriers give a route, not a push'. Do not award the point for 'the carriers get full' or 'the cells have no room left'.

Slip Saying the carriers 'get full' or the cell 'has no room left'. The rise stops because the gradient is gone, not because anything is full.

APBIO-U02-T26 End-of-topic test: Facilitated Diffusion

Topic 2.6 · Facilitated Diffusion · 16 MCQ · 2 FRQ

Q1 T26-q01

Glucose is at 7 mmol/L in the blood around a liver cell and 3 mmol/L inside the cell. Glucose enters the cell through a carrier protein.

Which of the following describes the movement of glucose into the cell?

A. Simple diffusion
Glucose is a large polar molecule and cannot cross the bilayer on its own; the stem says it uses a carrier.
B. ✓ Facilitated diffusion
C. Active transport
Glucose moves from 7 mmol/L to 3 mmol/L, toward the lower concentration; a carrier only provides the route.
D. Osmosis
Osmosis is the net movement of water across a membrane.

Why: Glucose is a large polar molecule, so it cannot cross the bilayer on its own.
Glucose moves from 7 mmol/L to 3 mmol/L, down its gradient, through a carrier protein.
Movement down a gradient happens by itself, so the cell uses no energy.
That is facilitated diffusion.

Q2 T26-q02

A guard cell, one of the pair of cells that open and close a pore in a leaf, holds K⁺ at 300 mmol/L inside; the fluid in the cell wall around it holds K⁺ at 10 mmol/L. As the pore closes, one of the cell’s potassium channels opens.

Which way does K⁺ move, and what is the crossing called?

A. In, through the channel; facilitated diffusion
K⁺ is at 300 mmol/L inside and 10 mmol/L outside, so down its gradient is outward.
B. Out, through the channel; simple diffusion
Simple diffusion is passage straight through the bilayer, which an ion cannot manage.
C. Out, through the bilayer; facilitated diffusion
K⁺ is an ion and cannot pass the hydrophobic interior.
D. ✓ Out, through the channel; facilitated diffusion

Why: K⁺ is 300 mmol/L inside and 10 mmol/L outside.
So down its gradient is outward.
K⁺ is an ion, so it crosses only through a channel protein.
Movement down a gradient happens by itself, so the cell uses no energy on it.
So the crossing is facilitated diffusion.

Q3 T26-q03

Four substances, W, X, Y and Z, are tested on protein-free phospholipid bilayers and on living cells. W crosses the protein-free bilayer and enters the cells; blocking the cells’ ATP supply leaves its entry unchanged. X is held back by the bilayer but enters the cells; blocking ATP leaves its entry unchanged. Y is held back by the bilayer but enters the cells; blocking ATP stops its entry. Z is held back by the bilayer and stays outside the cells.

Which substance enters the cells by facilitated diffusion?

A. Substance W
W crosses the protein-free bilayer, so W needs no protein; W enters by simple diffusion.
B. Substance Y
Y’s entry stops when ATP is blocked, so the cell is using energy to move Y; that is active transport.
C. ✓ Substance X
D. Substance Z
Z stays outside the cells, so Z has no crossing to classify.

Why: Facilitated diffusion is a crossing through a channel or carrier, down the gradient, with no energy from the cell.
X is held back by the protein-free bilayer, so X needs a membrane protein.
X’s entry is unchanged when ATP is blocked, so the cell uses no energy on X.

Q4 T26-q04

Cells lining a kidney tubule take in glucose even when it is 0.5 mmol/L in the tubule fluid and 6 mmol/L inside the cells, and the uptake almost stops when the cells' energy supply is cut off. A membrane protein is involved.

Is this crossing facilitated diffusion?

A. Yes: a membrane protein is involved
A protein is needed for facilitated diffusion, but so are a direction down the gradient and no energy used by the cell, and this crossing has neither.
B. Yes: glucose is a large polar molecule
Large polar molecules do need proteins, but that alone does not make a crossing facilitated diffusion.
C. ✓ No: glucose moves against its gradient and the cell uses energy
D. No: here glucose crosses the bilayer directly
Glucose cannot cross a bilayer directly at all.

Why: Facilitated diffusion moves a substance down its gradient, through a channel or carrier, with no energy from the cell.
Here glucose moves from 0.5 mmol/L to 6 mmol/L: against its gradient.
The uptake almost stops when the energy supply is cut, so the cell is using energy.

Q5 T26-q05

A squid nerve cell has Na⁺ at 440 mmol/L outside and 50 mmol/L inside. At first every sodium channel is shut. Then the sodium channels open. The cell's energy supply is blocked throughout, and no other protein moves Na⁺.

What is the net movement of Na⁺ while the channels are shut, and then once they open?

A. Inward both times: through the bilayer while shut, then through the channels once open
Na⁺ is charged and cannot cross the bilayer at all.
B. ✓ Almost none while shut; inward through the channels once open
C. None either time: charged ions cannot cross a membrane
Ions cannot cross the bilayer, but they can cross through a channel protein.
D. Almost none while shut; outward through the channels once open
Na⁺ is at 440 mmol/L outside and 50 mmol/L inside, so down its gradient is inward.

Why: Na⁺ is a charged ion, and an ion crosses a membrane only through a channel protein.
While every sodium channel is shut, Na⁺ has no route and almost none moves.
Once the channels open, Na⁺ flows in, from 440 toward 50 mmol/L, down its gradient, using no energy.

Q6 T26-q06

Two kinds of animal cell sit in a dilute solution, in which water enters cells. Cells of type 1 have many aquaporins in their membranes; cells of type 2 have very few.

Which cells gain water faster, and why?

A. ✓ Type 1: more aquaporins give water more routes, so it enters faster
B. Type 2: with fewer channels, water is forced through the bilayer faster
Nothing forces water through the bilayer; fewer channels is fewer routes.
C. Both equally: water crosses the bilayer itself, so channels make no difference
Water crosses a phospholipid bilayer only slowly, so channels make a large difference.
D. Type 1: aquaporins pump water in, and more pumps move more water
Aquaporins are open channels, not pumps.

Why: Water crosses the phospholipid bilayer only slowly.
An aquaporin is a channel protein for water, and each open aquaporin is one more route.
Type 1 cells have many aquaporins, so water has many routes into them.
So type 1 cells gain water faster.

Q7 T26-q07

Cells lining a sweat duct hold Cl⁻ at 20 mmol/L; the fluid outside holds 100 mmol/L. At 0 minutes the cells’ chloride channels are open. At 2 minutes a blocker shuts every chloride channel. Cl⁻ inside the cells is measured at 0, 2 and 4 minutes.

Which of the following gives the Cl⁻ concentration inside the cells, in mmol/L, at 0, 2 and 4 minutes?

A. 20, 20, 20
Cl⁻ does enter while the channels are open; an ion crosses through an open channel, down its gradient.
B. 20, 12, 12
Cl⁻ is 100 mmol/L outside and 20 mmol/L inside, so down its gradient is inward; the inside rises, it does not fall.
C. 20, 28, 36
The blocker shuts every channel at 2 minutes, so Cl⁻ has no route after that and the inside stops rising.
D. ✓ 20, 28, 28

Why: Cl⁻ is an ion, so it crosses only through an open channel.
From 0 to 2 minutes the channels are open, so Cl⁻ enters down its gradient: 20 to 28 mmol/L.
At 2 minutes the blocker shuts the channels, so Cl⁻ has no route and the inside stays at 28.

Q8 T26-q08

A muscle cell in the stomach wall of a frog holds Na⁺ at 15 mmol/L inside against 110 mmol/L outside, and K⁺ at 120 mmol/L inside against 3 mmol/L outside. A drug shuts every sodium channel in its membrane; its potassium channels stay open. A student predicts: “With the sodium channels shut, K⁺ will stop moving too, because the membrane is now closed to ions.”

Which of the following statements about the prediction is correct?

A. ✓ The prediction is wrong: K⁺ still leaves the cell through its own open potassium channels
B. The prediction is wrong: K⁺ still enters the cell through the open potassium channels
K⁺ is 120 mmol/L inside and 3 mmol/L outside, so down its gradient is out of the cell, not in.
C. The prediction is right: shutting one kind of channel closes the membrane to every ion
Each kind of ion has its own channels; shutting the sodium channels leaves the potassium channels open.
D. The prediction is right: K⁺ can move only while Na⁺ moves the other way to balance it
An ion moves down its own gradient through its own channel; K⁺ does not need Na⁺ to move at all.

Why: Each kind of ion crosses only through its own channels.
The drug shuts the sodium channels, so Na⁺ has no route and stays put.
The potassium channels stay open; K⁺ is 120 mmol/L inside and 3 mmol/L outside, so K⁺ leaves.
The membrane is closed to Na⁺ only.

Q9 T26-q09

Cells from a salivary gland sit in a dilute solution, in which water enters them, for five minutes. Cells with working aquaporins and a normal energy supply gain 18%. Cells with their aquaporins blocked gain 5%. Cells with working aquaporins but an energy supply cut to under 5% of normal gain 17%.

What do the three results show about aquaporins?

A. They pump water in, using the cell's energy
Cutting the energy supply to under 5% barely changed the gain, from 18% to 17%, so aquaporins are not powered.
B. They are the only route water can take
Cells with blocked aquaporins still gained 5%, so water still got in slowly through the bilayer itself.
C. ✓ They speed water's crossing and use no energy from the cell
D. Blocking them made water leave the cells
The blocked cells still gained mass, so water did not leave.

Why: Cells with their aquaporins blocked gained only 5%, so water crosses the bilayer only slowly.
Cells with working aquaporins gained 18%, so aquaporins let large quantities of water cross quickly.
Cells with almost no energy supply still gained 17%, so aquaporins use no energy: they are open channels.

Q10 T26-q10

A membrane protein lets water molecules stream through it, several hundred million every second, but lets no ions and no glucose through.

What is the protein?

A. A carrier that binds each water molecule and changes shape
A carrier binds one molecule at a time and changes shape for each, far too slow for hundreds of millions a second.
B. ✓ An aquaporin: a channel protein for water
C. A protein that pushes water in using the cell's energy
Nothing powers the flow; water moves through the protein only down its own gradient.
D. A glycoprotein on the outer face of the membrane
A glycoprotein is a protein with a carbohydrate chain on the outer face, not a tunnel through the membrane.

Why: An aquaporin is a channel protein for water.
It is a water-lined tunnel through the membrane.
Large quantities of water cross through it quickly.
Ions and larger molecules such as glucose do not fit through it.
It is open, not powered, so it needs no energy from the cell.

Q11 T26-q11

A potassium channel in a kidney-tubule cell lets K⁺ stream through it. Na⁺ is also present on both sides of the membrane, and a Na⁺ ion is smaller than a K⁺ ion, yet almost no Na⁺ passes through this channel.

What does this show about the channel?

A. The channel is a sieve that holds back every ion larger than its pore, whatever the ion’s charge
A sieve lets the smaller particle through, and Na⁺ is the smaller ion, yet Na⁺ is the one held back.
B. The channel passes ions in one direction only, from the outside of the cell to the inside
A channel is an open tunnel with no direction of its own, and Na⁺ was present on both sides of the membrane.
C. The channel uses the cell’s energy on each K⁺ it passes and uses none on Na⁺
A channel uses no energy on any ion; ions move through it down their gradients on their own.
D. ✓ The channel is built to pass one kind of ion, K⁺, and size alone does not decide which ion gets through

Why: A Na⁺ ion is smaller than a K⁺ ion, so if size decided, Na⁺ would pass at least as easily.
Almost no Na⁺ passes, so the channel is not a sieve.
A channel is a protein shaped to pass one kind of ion; this one is shaped for K⁺.

Q12 T26-q12

Two cells lining a kidney tubule hold Cl⁻ at 25 mmol/L inside, with 115 mmol/L in the tubule fluid outside. Cell 1 has 100 open chloride channels; cell 2 has 200.

Which of the following describes the net movement of Cl⁻ in the two cells at the start?

A. Out of both cells; faster in cell 2
Cl⁻ is at 115 outside and 25 inside, so down its gradient is inward.
B. Into cell 1, out of cell 2
Both cells have the same gradient, so both take Cl⁻ the same way.
C. ✓ Into both cells; faster in cell 2
D. Into both cells; equally fast
Each open channel is a route, and cell 2 has twice as many.

Why: Cl⁻ is 115 mmol/L outside both cells and 25 mmol/L inside, so Cl⁻ enters both, down its gradient.
Each open channel is one route.
Cell 2 has twice as many open channels as cell 1, so Cl⁻ enters cell 2 twice as fast.
Neither cell uses energy.

Q13 T26-q13

A red blood cell with glucose carriers sits in a solution; at 10 minutes it is moved into a second solution. The table gives the glucose concentration outside and inside the cell.

Glucose outside and inside the red blood cell.

What is the net movement of glucose at 20 minutes?

A. Into the cell, through the carriers
At 20 minutes glucose is 5 mmol/L inside and 2 mmol/L outside, so down its gradient is now outward.
B. ✓ Out of the cell, through the carriers
C. Out of the cell, through the phospholipid bilayer
Glucose is a large polar molecule and never crosses the phospholipid bilayer itself; its only route is the carriers.
D. No net movement of glucose
The new solution made a new gradient, 5 mmol/L inside against 2 mmol/L outside, so net movement starts again.

Why: A carrier gives glucose a route in either direction; the gradient decides the direction.
At 10 minutes both sides were 6 mmol/L: no net movement.
At 20 minutes glucose is 5 mmol/L inside and 2 mmol/L outside, so the gradient runs outward.
So glucose leaves through the carriers.

Q14 T26-q14

Cells lining a duct of the pancreas, with open chloride channels, sit in a solution with Cl⁻ at 100 mmol/L; inside the cells it is 40 mmol/L, and Cl⁻ is entering. The solution is then replaced by one with Cl⁻ at 15 mmol/L. The channels stay open.

What happens to the movement of Cl⁻ after the change?

A. ✓ Cl⁻ now leaves the cells through the same channels
B. Cl⁻ keeps entering the cells through the channels
A channel is an open tunnel with no direction of its own; after the change the gradient runs outward, so the ion passes the other way.
C. Cl⁻ stops crossing the membrane
The channels stay open and the two concentrations differ, so Cl⁻ still has a route and a gradient to follow.
D. Cl⁻ now leaves the cells through the phospholipid bilayer
Cl⁻ is a charged ion; water attracts the charge, and nothing in the hydrophobic interior does, so Cl⁻ never crosses the bilayer itself.

Why: The gradient sets the direction through an open channel.
Before the change Cl⁻ was 100 mmol/L outside and 40 mmol/L inside, so Cl⁻ entered.
After the change Cl⁻ is 15 mmol/L outside and 40 mmol/L inside, so the gradient runs outward.
So Cl⁻ leaves, using no energy.

Q15 T26-q15

Fat cells sit in a solution held at 6 mmol/L glucose; inside the cells glucose is 1 mmol/L. The hormone insulin makes a fat cell move extra glucose carriers into its membrane: about 500 carriers per cell without insulin, about 5,000 with it. One dish of cells gets insulin and one does not. Inside the cells, glucose stays as glucose.

Which of the following describes the two dishes?

A. The dish without insulin fills faster: with fewer carriers there is less crowding at each one
Fewer carriers is fewer routes, so the cells without insulin take glucose in more slowly, not faster.
B. Both dishes fill equally fast; both end near 6 mmol/L inside
Ten times the carriers is ten times the routes, so the cells with insulin take in glucose much faster at first.
C. The dish with insulin fills faster; the dish without insulin never changes
The cells without insulin still have 500 working carriers, so glucose enters them too, just more slowly.
D. ✓ The dish with insulin fills faster; both end near 6 mmol/L inside

Why: Each carrier is one route, so with insulin a cell has ten times as many routes and takes glucose in faster.
A carrier only lets glucose move down its gradient.
Once the inside reaches 6 mmol/L, there is no gradient left.
So both dishes stop rising near 6 mmol/L.

Q16 T26-q16

The figure shows part of a fat cell’s plasma membrane, with two membrane proteins and the concentration of K⁺ and of glucose on each side. A toxin binds the protein marked X.

Part of a fat cell’s plasma membrane with a potassium channel and a glucose carrier, the concentration of K⁺ and of glucose on each side, and a cross drawn over the protein marked X. The extracellular fluid is shaded darker than the cytosol.

What happens to K⁺ and to glucose while the toxin is bound?

A. ✓ K⁺ stops crossing; glucose keeps entering through its carrier, down its gradient
B. K⁺ stops crossing; glucose still enters, but the cell must now use energy to bring it in
Glucose still moves down its gradient, from 7 mmol/L to 2 mmol/L, and movement down a gradient needs no energy from the cell whether or not another protein is blocked.
C. K⁺ stops crossing; glucose enters faster, because its carrier now has the membrane’s traffic to itself
Each protein carries its own substance at its own rate; the glucose carriers were never sharing a route with K⁺, so blocking the channel gives them nothing extra.
D. K⁺ and glucose both stop crossing: one blocked protein shuts the whole membrane
Each protein serves its own substance; the glucose carrier is untouched, so glucose keeps crossing.

Why: K⁺ is an ion, so K⁺ crosses only through a potassium channel; the toxin blocks that channel, so K⁺ stops crossing.
Glucose uses a different protein, the glucose carrier, which the toxin does not touch.
Glucose is 7 mmol/L outside and 2 mmol/L inside, so it keeps entering.

FRQ 1 T26-frq1 · Conceptual Analysis

A muscle cell of a crab, bathed in seawater, holds K⁺ at 180 mmol/L inside and 12 mmol/L outside, and Na⁺ at 50 mmol/L inside and 500 mmol/L outside. Its membrane has potassium channels and sodium channels, most of them shut while the cell is at rest. A drug is applied that doubles the number of potassium channels that open.

(a) Describe how K⁺ crosses the membrane when a potassium channel opens, and name the kind of transport. (1 pt)

Model answer When a potassium channel opens, K⁺ passes through the open channel protein, a water-lined tunnel.
K⁺ moves down its concentration gradient, from 180 mmol/L inside to 12 mmol/L outside, so K⁺ leaves the cell.
Movement down a gradient happens by itself.
So the cell uses no energy on this crossing.
This kind of transport is facilitated diffusion, a form of passive transport.

Rubric

Award 1 point for: K⁺ passes through the open channel protein, a water-lined tunnel, moving down its concentration gradient from 180 toward 12 mmol/L, out of the cell, with no energy used by the cell; this is facilitated diffusion (a form of passive transport).
Accept: "through the channel, from high to low, out of the cell, with no energy used: facilitated diffusion". The route, the direction and the name are all needed for the point.

Slip Giving the route without the name, or the name without the route, or sending K⁺ inward. The route, the direction from the two concentrations and the name are all needed.

(b) Explain what keeps Na⁺ out of the cell while its channels are shut, even though its concentration is much higher outside. (1 pt)

Model answer Na⁺ is an ion, so it carries a full charge.
The hydrocarbon tails in the middle of the membrane carry no charges or partial charges.
So nothing in the tails attracts a charged ion.
Water is attracted to Na⁺.
So Na⁺ stays in the water and cannot cross the hydrophobic interior on its own.
Na⁺ can cross only through a channel.
With its channels shut, Na⁺ has no route.

Rubric

Award 1 point for: Na⁺ is a charged ion, and the hydrocarbon tails in the middle of the membrane carry no charges or partial charges, so water attracts Na⁺ and nothing in the hydrophobic interior does, and Na⁺ cannot cross on its own; it can cross only through a channel, so with its channels shut it has no route.
Accept: "ions cross only through channels, and the sodium channels are closed", provided the reason (the charge on the ion and the uncharged interior) is given.

Slip Saying the sodium channels are closed, with no reason why Na⁺ needs a channel at all. The charge on the ion and the uncharged interior are the reason.

(c) Make a claim about how the drug changes the movement of K⁺ when the potassium channels open, and support your claim. (1 pt)

Model answer K⁺ still leaves the cell, in the same direction as before.
The gradient sets the direction, and the drug does not change the gradient.
Each open channel is one route.
The drug doubles the number of open channels.
Therefore K⁺ leaves faster.

Rubric

Award 1 point for: the claim that K⁺ still leaves the cell (the direction is unchanged) but leaves faster, supported by the reasoning: each open channel is one route and the drug doubles the number open, while the gradient, which sets the direction, is unchanged.
Accept: "more K⁺ leaves per second" or "K⁺ leaves about twice as fast" with the reason. Do not award the claim alone, a change of direction, or "the cell now needs energy".

Slip Changing the direction, or saying the cell now needs energy. More doors speed a crossing; the gradient sets its direction.

(d) The cell is then bathed in a solution holding K⁺ at 180 mmol/L, equal to the inside, with the drug still present and the channels open. Considering only the concentrations (set aside any charge across the membrane), predict the net movement of K⁺ and justify your prediction. (1 pt)

Model answer There is no net movement of K⁺.
K⁺ is 180 mmol/L on both sides.
So K⁺ ions cross the channels in both directions in equal numbers.
That is a dynamic equilibrium.
A channel provides only the route.
With no gradient there is no net direction, however many channels are open.

Rubric

Award 1 point for: no net movement of K⁺, because with equal concentrations on both sides K⁺ ions cross the channels in both directions in equal numbers (a dynamic equilibrium); channels provide only the route, and with no gradient there is no net direction, however many channels are open.
Accept: "K⁺ still crosses both ways but the two flows cancel". Do not award the point for "K⁺ stops moving" or for "K⁺ still leaves because the drug opened more channels".
Accept also: an answer that gives 'no net movement by concentration' and adds that a negative inside would pull a little K⁺ inward; the concentration argument earns the point and the charge remark is not penalized.

Slip Saying K⁺ stops moving, or that it still leaves because the drug opened more channels. Ions still cross both ways; the two flows cancel.

FRQ 2 T26-frq2 · Analyze Model or Visual Representation

The model shows a cross-section of a muscle cell's plasma membrane with the extracellular fluid above and the cytosol below, and three routes across it. Route 1 is the bilayer itself. Route 2 is a channel protein: a water-lined tunnel through the membrane. Route 3 is a carrier protein, which binds one kind of molecule and changes shape. The table beside the model gives the concentration of each of three substances in the extracellular fluid and in the cytosol. The cell uses no energy as O₂, K⁺ or glucose crosses this membrane.

Three routes across a plasma membrane, numbered 1 to 3, with a table of the concentration of each substance in the extracellular fluid and in the cytosol. The extracellular fluid is shaded darker than the cytosol.

(a) For each of O₂, K⁺ and glucose, state which route it takes across this membrane. (1 pt)

Model answer O₂ takes route 1, straight through the bilayer, because O₂ is a small nonpolar molecule; K⁺ takes route 2, the channel, because K⁺ is an ion; glucose takes route 3, the carrier, because glucose is a large polar molecule that a carrier binds.

Rubric

Award 1 point for: O₂ takes route 1, straight through the bilayer; K⁺ takes route 2, a channel; glucose takes route 3, a carrier.
All three are needed for the point. No reason is required for the point, but a reason given must not contradict the route.

Slip Giving glucose the channel or K⁺ the carrier. A channel is a tunnel for one kind of ion; a carrier binds a molecule and changes shape.

(b) State the net direction of K⁺ and of glucose across this membrane, using the concentrations in the table. (1 pt)

Model answer K⁺ is 150 mmol/L in the cytosol and 4 mmol/L in the extracellular fluid.
So the net direction of K⁺ is out of the cell.
Glucose is 6 mmol/L in the extracellular fluid and 1 mmol/L in the cytosol.
So the net direction of glucose is into the cell.

Rubric

Award 1 point for: K⁺ moves from the cytosol to the extracellular fluid (out), because 150 mmol/L is higher than 4 mmol/L; glucose moves from the extracellular fluid to the cytosol (in), because 6 mmol/L is higher than 1 mmol/L.
Both directions must be correct for the point; K⁺ given as moving inward earns nothing.

Slip Sending K⁺ inward because most things seem to come in. Read the table: K⁺ is higher in the cytosol, so its net movement is out.

(c) Explain what decides whether a substance can use route 1, and why a crossing through route 2 or route 3 still needs no energy from the cell. (1 pt)

Model answer The hydrocarbon tails in the middle of the membrane carry no charges or partial charges.
K⁺ is an ion and glucose is a polar molecule; water is attracted to both, and nothing in the tails attracts either.
So neither can cross the hydrophobic interior on its own, and neither can use route 1.
Through a channel or a carrier, each moves down its own concentration gradient.
Movement down a gradient happens by itself, so the cell uses no energy: facilitated diffusion.

Rubric

Award 1 point for: the hydrocarbon tails in the middle of the membrane carry no charges or partial charges, so an ion (K⁺) or a polar molecule (glucose), which water attracts, has nothing in the hydrophobic interior that attracts it, so it cannot cross on its own; through a channel or carrier each moves down its own concentration gradient, and movement down a gradient happens by itself, because more particles leave the more concentrated side than arrive from the less concentrated one, so no energy is needed (facilitated diffusion).
Accept: "charged or polar things cannot enter the oily middle; the proteins give a route, and movement down a gradient needs no energy". Both halves are needed for the point.

Slip Explaining why they need a protein but not why the crossing needs no energy, or the other way around. Both halves are needed.

(d) Kidney cells have many aquaporins in their membranes. Describe what an aquaporin is in terms of this model, and what it does for the cell. (1 pt)

Model answer An aquaporin is a route-2 protein: a channel for water.
Water crosses the bilayer, route 1, only slowly on its own.
Through aquaporins, large quantities of water cross the membrane quickly.
Water moves down its own gradient, with no energy used by the cell.

Rubric

Award 1 point for: an aquaporin is a route-2 protein, a channel, for water; water crosses the bilayer (route 1) only slowly on its own, and aquaporins let large quantities of water cross the membrane quickly, down water's own gradient and with no energy used by the cell.
Accept: "a water channel; it speeds water's crossing greatly". Do not award the point for "it pumps water" or for calling it a carrier.

Slip Saying an aquaporin pumps water, or calling it a carrier. It is an open channel; water moves through it only the way it would move anyway.

APBIO-U02-L06 Pumping uphill

Topic 2.8 · Mechanisms of Transport · 79 steps

A fish in salty seawater, a marine fish, with one of its gill cells drawn large: chloride at 40 mmol/L inside the cell, 550 mmol/L in the seawater, and an arrow pointing out of the cell

Here is a gill cell of a fish that lives in salty seawater: a marine fish. The gill cell holds chloride at 40 mmol/L while the seawater around it holds 550 mmol/L, and all day the cell pushes chloride out into that seawater.

Every crossing you have seen so far moved a substance downhill, from where it was more concentrated to where it was less. This chloride is going the other way, and it keeps going as long as the fish is alive.

Unit 2 · Cell Structure and Function

1Uphill needs energy

Video: Watch first: Mechanisms of transport

A gill cell of a marine fish pushing chloride out into saltier seawater, all day.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T28-intro.mp4

A substance left to itself moves down its concentration gradient: oxygen into a muscle cell, glucose out of a dialysis bag.

Here is the gill cell’s membrane. Chloride is 40 mmol/L inside the cell and 550 mmol/L in the seawater outside.

A gill cell's plasma membrane drawn as a vertical band, the cytosol shaded to its left and the seawater shaded to its right: chloride at 40 mmol/L inside and 550 mmol/L outside, and chloride being moved outward through a labelled membrane protein

Down the gradient would be inward, from 550 mmol/L to 40 mmol/L. The gill cell moves chloride outward, from 40 mmol/L to 550 mmol/L.

Chloride is an ion. Quick recall before we go on.

Check q1

A chloride ion (Cl⁻) carries a full negative charge.

How does a chloride ion cross a membrane?

A. Straight through the phospholipid bilayer
Water attracts a full charge, and nothing in the hydrophobic interior does, so an ion cannot cross the bilayer on its own.
B. ✓ Only through a membrane protein

Why: An ion carries a full charge, so it crosses a membrane only through a channel or carrier protein.

Video: Watch: Uphill needs energy

A gill cell, a root cell and a muscle cell: which crossings stop when the cell’s energy is cut, and why.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L06.mp4

So the chloride is going out through a membrane protein, against its gradient.

Now imagine the cell’s energy source is removed. The pushing almost stops: for every ten chloride ions the cell was pushing out, it now pushes out one.

Two panels of the gill cell's membrane, the cytosol shaded to the left and the seawater to the right: with the cell unpoisoned the labelled membrane protein pushes chloride out strongly, ten ions in one second; with the cell poisoned so it has no energy to use, one ion in one second

Give the cell its energy source back, and the pushing comes back. So the gill cell uses energy to move chloride uphill.

A cell’s energy for work like this comes from a small molecule. This molecule is called .

When the gill cell moves chloride uphill, the gill cell uses ATP.

Here is a root cell. Nitrate is 0.30 mmol/L in the soil water and 5.4 mmol/L inside the cell.

A root cell's plasma membrane drawn as a vertical band, the cytosol shaded to its left and the soil water shaded to its right: nitrate at 5.4 mmol/L inside and 0.30 mmol/L outside, and nitrate being moved inward through a labelled membrane protein

The root cell keeps taking nitrate in, from 0.30 mmol/L to 5.4 mmol/L: against its gradient.

Lower the cell’s ATP and the uptake of nitrate falls by 90%. So the root cell uses ATP to take nitrate in.

The root cell shows the same pattern in the other direction: the gill cell pushed chloride outward, and the root cell pulls nitrate inward. In both cells the substance moves against its gradient, and the cell uses ATP.

Now a muscle cell. Glucose is 8 mmol/L in the blood and 2 mmol/L inside the cell, and glucose enters the cell through a carrier.

A muscle cell's plasma membrane drawn as a vertical band, the cytosol shaded to its left and the blood shaded to its right: glucose at 2 mmol/L inside and 8 mmol/L outside, glucose moving inward through a labelled glucose carrier, with the words no ATP used

Block the cell’s ATP and the glucose keeps coming in just as fast. So the muscle cell uses no energy on this crossing.

This glucose moves down its gradient, from 8 mmol/L to 2 mmol/L. A substance moving down its gradient needs no energy from the cell.

Check q2

Glucose enters the muscle cell down its gradient, through a carrier protein, and the cell uses no energy.

What is this kind of crossing called?

A. ✓ Facilitated diffusion
B. Active transport
Active transport is a crossing on which the cell uses energy; here the cell uses no energy.

Why: A crossing down the gradient, through a protein, with no energy used by the cell is facilitated diffusion.

A protein is involved, but a protein alone does not make a crossing use energy.

So the direction of the crossing, relative to the gradient, decides whether the cell uses energy.

Down its gradient, a substance moves on its own, and the cell uses no energy. Against its gradient, a substance moves only if the cell uses energy.

To say whether a cell uses energy on a crossing, use the rule in three steps.

First, compare the two concentrations.

Second, see which way the substance is moving: from the larger concentration to the smaller is down its gradient; from the smaller to the larger is against its gradient.

Third, read off the answer: down, the cell uses no energy; against, the cell uses energy.

What you are expected to know Decide whether a cell uses energy on a crossing from the direction of the crossing relative to the substance’s concentration gradient: down, the cell uses no energy; against, the cell uses energy.

30Fluency quiz: does the cell use energy? mixed practice

Check q3

The bar chart shows the concentration of glucose outside a red blood cell and inside it. Glucose moves into the cell.

A bar chart with two bars, labelled outside the cell and inside the cell, showing the concentration of glucose in mmol/L; each bar carries its value above it

Does the cell use energy on this crossing?

A. Yes
Glucose moves from 5 mmol/L to 1 mmol/L, down its gradient; down a gradient, the cell uses no energy.
B. ✓ No

Why: Down a gradient, the cell uses no energy; against a gradient, the cell uses energy.
Glucose is 5 mmol/L outside and 1 mmol/L inside.
Glucose moves in, from the larger concentration to the smaller.
So glucose moves down its gradient.
So the cell uses no energy on this crossing.

Check q4

The bar chart shows the concentration of sodium ions (Na⁺) outside a kidney cell and inside it. Na⁺ moves out of the cell.

Does the cell use energy on this crossing?

A. ✓ Yes
B. No
Na⁺ moves from 12 mmol/L to 130 mmol/L, against its gradient; against a gradient, the cell uses energy.

Why: Down a gradient, the cell uses no energy; against a gradient, the cell uses energy.
Na⁺ is 12 mmol/L inside and 130 mmol/L outside.
Na⁺ moves out, from the smaller concentration to the larger.
So Na⁺ moves against its gradient.
So the cell uses energy on this crossing.

Check q5

The bar chart shows the concentration of sucrose outside a plant cell and inside it. Sucrose moves into the cell.

A bar chart with two bars, labelled outside the cell and inside the cell, showing the concentration of sucrose in mmol/L; each bar carries its value above it

Does the cell use energy on this crossing?

A. ✓ Yes
B. No
Sucrose moves from 50 mmol/L to 400 mmol/L, against its gradient; against a gradient, the cell uses energy.

Why: Down a gradient, the cell uses no energy; against a gradient, the cell uses energy.
Sucrose is 50 mmol/L outside and 400 mmol/L inside.
Sucrose moves in, from the smaller concentration to the larger.
So sucrose moves against its gradient.
So the cell uses energy on this crossing.

Check q6

The bar chart shows the concentration of chloride ions (Cl⁻) outside a muscle cell and inside it. Cl⁻ moves into the cell.

Does the cell use energy on this crossing?

A. Yes
Cl⁻ moves from 120 mmol/L to 12 mmol/L, down its gradient; down a gradient, the cell uses no energy.
B. ✓ No

Why: Down a gradient, the cell uses no energy; against a gradient, the cell uses energy.
Cl⁻ is 120 mmol/L outside and 12 mmol/L inside.
Cl⁻ moves in, from the larger concentration to the smaller.
So Cl⁻ moves down its gradient.
So the cell uses no energy on this crossing.

Check q7

The bar chart shows the concentration of O₂ outside a muscle cell and inside it. O₂ moves into the cell.

A bar chart with two bars, labelled outside the cell and inside the cell, showing the concentration of O₂ in mmol/L; each bar carries its value above it

Does the cell use energy on this crossing?

A. Yes
O₂ moves from 0.10 mmol/L to 0.03 mmol/L, down its gradient; down a gradient, the cell uses no energy.
B. ✓ No

Why: Down a gradient, the cell uses no energy; against a gradient, the cell uses energy.
O₂ is 0.10 mmol/L outside and 0.03 mmol/L inside.
O₂ moves in, from the larger concentration to the smaller.
So O₂ moves down its gradient.
So the cell uses no energy on this crossing.

Check q8

The bar chart shows the concentration of an amino acid outside a gut cell and inside it. The amino acid moves into the cell.

A bar chart with two bars, labelled outside the cell and inside the cell, showing the concentration of the amino acid in mmol/L; each bar carries its value above it

Does the cell use energy on this crossing?

A. ✓ Yes
B. No
The amino acid moves from 3 mmol/L to 12 mmol/L, against its gradient; against a gradient, the cell uses energy.

Why: The amino acid is 3 mmol/L outside and 12 mmol/L inside.
The amino acid moves in, from the smaller concentration to the larger: against its gradient.
Against a gradient, the cell uses energy.
So the cell uses energy on this crossing.

37Active transport and pumps

Video: Watch: Active transport and pumps

The uphill crossings named, and the protein that does them.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L06b.mp4

A crossing in which the cell uses energy, usually from ATP, to move a substance through a membrane protein against its concentration gradient is called .

The gill cell moving chloride out is active transport. The root cell taking nitrate in is active transport.

A membrane protein that uses the cell’s energy to move a substance against its gradient is called a .

Here are two pumps. A cell lining the stomach has an acid pump: it moves hydrogen ions (H⁺) out into the stomach, against their concentration gradient, and uses ATP to do it.

A root cell has a pump that moves hydrogen ions out into the soil water, against their concentration gradient, using ATP. A hydrogen ion is a single proton, so this pump is called a proton pump.

What makes a crossing active transport is that the cell uses energy. A crossing can be active transport in either direction: the gill cell moves chloride out, and the root cell takes nitrate in.

A protein on its own is not enough: facilitated diffusion uses a protein, and the cell uses no energy, so facilitated diffusion is not active transport.

What you are expected to know Decide whether a crossing is active transport from whether the cell uses energy to move the substance against its gradient, whichever way it is going and whether or not a protein is involved.

Practice writing an answer

A thyroid cell takes iodide ions, I⁻, in from the blood. Iodide is thirty times as concentrated inside the cell as in the blood, and the cell keeps taking iodide in.

(a) Explain why the thyroid cell must use energy to take iodide in. (1 pt)

Model answer Iodide is more concentrated inside the cell than in the blood.
So down the gradient is out of the cell.
The cell moves iodide in, from the smaller concentration to the larger.
So iodide moves against its gradient.
A substance left to itself moves down its gradient, never against it.
So something must move iodide the other way, and that takes energy.
So the cell uses energy, from ATP, to take iodide in.

Rubric

Award 1 point for: iodide moves from the less concentrated blood into the more concentrated cytosol, against its gradient; a substance left to itself moves only down its gradient, so moving iodide the other way needs energy from the cell (ATP).
Accept: “uphill” for against its gradient. Do not award the point for “because a protein is involved” or “because iodide is an ion” with no reference to the direction relative to the gradient.

Slip Saying the cell uses energy because a protein is involved. A protein alone does not make a crossing use energy; glucose enters a muscle cell through a carrier and the cell uses no energy. The direction relative to the gradient is the reason.

Check q9

Glucose is 6 mmol/L in the blood and 2 mmol/L inside a liver cell. Glucose enters the cell through a carrier protein. A student says: “A membrane protein is doing the work, so the cell must be using energy.”

Which statement about the student’s claim is correct?

A. The student is right: the cell uses energy on this crossing
Glucose moves from 6 mmol/L to 2 mmol/L, down its gradient; the carrier only gives it a route, so the cell uses no energy.
B. ✓ The student is wrong: the cell uses no energy on this crossing

Why: Glucose is 6 mmol/L in the blood and 2 mmol/L inside the cell.
Glucose moves in, from the larger concentration to the smaller: down its gradient.
A substance moving down its gradient moves on its own; the carrier only gives glucose a route.
So the cell uses no energy.

Check q10

A muscle cell holds calcium ions (Ca²⁺) at a far lower concentration inside than outside. All day, Ca²⁺ leaves the cell through a membrane protein.

Which of the following results would show that this crossing is active transport?

A. Blocking the cell’s ATP supply leaves the movement unchanged
If cutting off the ATP changes nothing, the cell was using no energy on the crossing.
B. ✓ Blocking the cell’s ATP supply stops the movement
C. Ca²⁺ crosses through a membrane protein
Facilitated diffusion also uses a membrane protein, and the cell uses no energy on it, so a protein alone is not a sign of active transport.

Why: Active transport is a crossing on which the cell uses energy from ATP.
Ca²⁺ moves against its gradient, toward the side where it is already more concentrated.
If the movement stops when the ATP is cut off, the movement needed the ATP.
So the crossing is active transport.

50The sodium–potassium pump

Here is a nerve cell. Sodium ions (Na⁺) are at 15 mmol/L inside and 145 mmol/L in the fluid outside.

A nerve cell's plasma membrane drawn as a vertical band, the cytosol shaded to its left and the fluid outside shaded to its right: Na⁺ at 15 mmol/L inside and 145 outside, and K⁺ at 140 mmol/L inside and 5 outside

Potassium ions (K⁺) are at 140 mmol/L inside and 5 mmol/L outside.

Check q11

Look at the nerve cell in the figure.

Which ion has the higher concentration inside the cell?

A. ✓ Potassium ions
B. Sodium ions
The figure shows sodium ions at the lower concentration inside and the higher concentration outside.

Why: The figure shows potassium ions at the higher concentration inside the cell and the lower concentration outside.

Video: Watch: The sodium–potassium pump

For each ATP, three sodium ions out and two potassium ions in, cycle after cycle.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L06c.mp4

The cell keeps the concentration of sodium ions low inside, against its gradient, and the concentration of potassium ions high inside, against its gradient. So the cell must be using energy to keep them there.

In the membrane sits a pump. For each ATP the pump uses, it moves three sodium ions out of the cell and two potassium ions into the cell.

A nerve cell's plasma membrane drawn as a vertical band, the cytosol shaded to its left and the fluid outside shaded to its right, with a labelled pump protein set in it: three Na⁺ arrows leaving the cell through the pump and two K⁺ arrows entering, with one ATP used per cycle

This pump is called the .

The pump repeats its cycle without stopping: three sodium ions out and two potassium ions in. Cycle after cycle, the pump keeps the concentration of sodium ions low and of potassium ions high inside the cell.

The pump uses a great deal of ATP. A nerve cell uses most of its ATP on this one pump.

What you are expected to know Describe the sodium–potassium pump: for each ATP it uses it moves three sodium ions (Na⁺) out of the cell and two potassium ions (K⁺) in, keeping the concentration of Na⁺ low and of K⁺ high inside.

61Fluency quiz: the sodium–potassium pump mixed practice

Check q12

A nerve cell has sodium–potassium pumps in its membrane.

Which ion does the pump move out of the cell?

A. ✓ Na⁺
B. K⁺
The pump keeps the concentration of K⁺ high inside the cell, so the pump brings K⁺ in.

Why: The pump keeps the concentration of Na⁺ low inside the cell, so the pump moves Na⁺ out.
It keeps the concentration of K⁺ high inside, so it moves K⁺ in.

Check q13

A sodium–potassium pump sits in the membrane of a nerve cell.

Which way does the pump move K⁺?

A. Out of the cell
The pump keeps the concentration of K⁺ high inside the cell.
B. ✓ Into the cell

Why: The concentration of K⁺ is high inside the cell and low outside.
The pump keeps that concentration high inside.
So the pump moves K⁺ into the cell.

Check q14

A sodium–potassium pump in a nerve cell completes one cycle.

How many Na⁺ does the pump move out of the cell in that cycle?

A. One
Each cycle moves three Na⁺ out of the cell, not one.
B. Two
Each cycle moves three Na⁺ out of the cell; two is the number of K⁺ it moves in.
C. ✓ Three

Why: In each cycle the pump moves three Na⁺ out of the cell and two K⁺ in.
So the pump moves three Na⁺ out in that cycle.

Check q15

A sodium–potassium pump in a nerve cell completes one cycle.

How many K⁺ does the pump move into the cell in that cycle?

A. One
Each cycle moves two K⁺ into the cell, not one.
B. ✓ Two
C. Three
Each cycle moves two K⁺ into the cell; three is the number of Na⁺ it moves out.

Why: In each cycle the pump moves three Na⁺ out of the cell and two K⁺ in.
So the pump moves two K⁺ in during that cycle.

Check q16

The bar chart shows the concentration of K⁺ outside a kidney cell and inside it. The sodium–potassium pump moves K⁺ into the cell.

Is the pump moving K⁺ down its gradient or against it?

A. Down its gradient
K⁺ is 4 mmol/L outside and 135 mmol/L inside; the pump moves K⁺ in, from the smaller concentration to the larger, so against its gradient.
B. ✓ Against its gradient

Why: K⁺ is 4 mmol/L outside and 135 mmol/L inside.
The inside bar is taller.
The pump moves K⁺ in, from the smaller concentration to the larger.
So the pump moves K⁺ against its gradient.

Check q17

The bar chart shows the concentration of Na⁺ outside a heart muscle cell and inside it. The sodium–potassium pump moves Na⁺ out of the cell.

Is the pump moving Na⁺ down its gradient or against it?

A. ✓ Against its gradient
B. Down its gradient
Na⁺ is 8 mmol/L inside and 138 mmol/L outside; the pump moves Na⁺ out, from the smaller concentration to the larger, so against its gradient.

Why: Na⁺ is 8 mmol/L inside and 138 mmol/L outside.
The outside bar is taller.
The pump moves Na⁺ out, from the smaller concentration to the larger.
So the pump moves Na⁺ against its gradient.

68Why the pump needs ATP

Video: Watch: Why the pump needs ATP

Both ions moved against their gradients, so the pump needs ATP: an ATPase.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L06d.mp4

Look again at the nerve cell’s two gradients. Na⁺ is 15 mmol/L inside and 145 mmol/L outside.

The pump moves Na⁺ out, from 15 mmol/L to 145 mmol/L. So the pump moves Na⁺ against its gradient.

K⁺ is 140 mmol/L inside and 5 mmol/L outside. The pump moves K⁺ in, from 5 mmol/L to 140 mmol/L. So the pump moves K⁺ against its gradient too.

Both ions are moved against their gradients.

A substance moves against its gradient only if the cell uses energy. So the pump needs ATP, and the pump’s work is active transport.

A protein that uses ATP to do its work is called an . The sodium–potassium pump is one: the Na⁺/K⁺ ATPase.

What you are expected to know Explain why the sodium–potassium pump needs ATP: it moves both Na⁺ and K⁺ against their gradients, and a substance moves against its gradient only if the cell uses energy.

Check q18

A liver cell’s sodium–potassium pump uses one ATP on every cycle.

Why does the pump need ATP?

A. The pump is a protein, and every protein in a membrane uses ATP
A channel is a membrane protein too, and an ion moving down its gradient through a channel uses no energy from the cell.
B. The pump moves ions, and every ion carries a charge
A charged ion moving down its gradient through a channel uses no energy from the cell.
C. ✓ The pump moves both ions against their gradients

Why: A substance moving down its gradient moves on its own.
A substance moves against its gradient only if the cell uses energy.
The pump moves Na⁺ and K⁺ against their gradients.
So the pump needs energy, and the energy comes from ATP.

The gill cell pushes chloride ions uphill. The sodium–potassium pump pushes sodium ions uphill. In both, the substance moves against its gradient, and the cell uses ATP. When the cell has no ATP left, the pushing stops.

Glossary

ATP: A small molecule that a cell uses as its energy for work, such as moving a substance against its gradient.
active transport: A crossing in which the cell uses energy, usually from ATP, to move a substance through a membrane protein against its concentration gradient.
pump: A membrane protein that uses the cell’s energy to move a substance against its concentration gradient.
sodium–potassium pump: The pump that, for each ATP it uses, moves three Na⁺ out of the cell and two K⁺ in, both against their gradients, keeping the concentration of Na⁺ low and of K⁺ high inside.
ATPase: A protein that uses ATP to do its work. The sodium–potassium pump is one, the Na⁺/K⁺ ATPase.

APBIO-U02-L07 Why the inside of a cell is negative

Topic 2.8 · Mechanisms of Transport · 61 steps

A nerve cell with one electrode inside it and one in the fluid outside, both wired to a meter whose needle points to the negative side

Here is a nerve cell with two fine electrodes touching it, one inside and one in the fluid outside, wired to a meter. The meter reads negative: the inside carries less positive charge than the outside, and the inside stays that way for as long as the cell is alive and fed.

Unit 2 · Cell Structure and Function

1A charge difference across the membrane

Sodium ions (Na⁺) and potassium ions (K⁺) are ions. Quick recall before we count.

Check q1

Sodium ions (Na⁺) and potassium ions (K⁺) are ions.

What charge does each Na⁺ or K⁺ carry?

A. ✓ One full positive charge
B. One full negative charge
The plus sign on Na⁺ and K⁺ marks a positive charge.
C. No charge
An ion always carries a full charge; the plus sign marks it.

Why: Each sodium ion and each potassium ion carries one full positive charge, marked by the plus sign.

Video: Watch: A charge difference across the membrane

Three positive charges out for every two in, counted cycle by cycle to a negative meter reading.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L07.mp4

The sodium–potassium pump moves, for each ATP, three sodium ions out of the cell and two potassium ions in. Count that in charges.

A plasma membrane drawn as a vertical band, the cytosol shaded to its left and the fluid outside shaded to its right, with the labelled sodium–potassium pump set in it: one cycle counted in charges, three positive charges leaving the cell and two entering

Three positive charges leave and two arrive. Every cycle, one more positive charge ends up outside the cell than inside.

Cycle after cycle, the pump sends a little more positive charge out than in. This helps make the inside of the cell slightly negative relative to the outside.

A plasma membrane drawn as a vertical band, the cytosol shaded to its left and the fluid outside shaded to its right, with a row of positive charges along its outer face and a row of negative signs along its inner face

That charge difference is what the electrodes measure. The meter reads the inside relative to the outside: the inside carries less positive charge than the outside, so the meter reads negative.

When ions cross a membrane so that one face ends up with more positive charge than the other, the charge difference across the membrane is called a .

A membrane that has a membrane potential is called a . Its two faces carry different charges.

In a typical animal cell the inside is the negative face.

What you are expected to know Say what a membrane potential is: a difference in charge between the two faces of a membrane, and, in a typical animal cell, which face is negative.

Check q2

A membrane separates two solutions. Ions cross it so that the left side ends up with more positive charge than the right.

Which of the following does the membrane now have?

A. ✓ A membrane potential, with the right side negative relative to the left
B. A membrane potential, with the left side negative relative to the right
The left side carries more positive charge, so the left side is the positive side; the side with less positive charge, the right, is the negative side.
C. No membrane potential
The two faces now carry different charges, and a charge difference across a membrane is a membrane potential.

Why: The left side now carries more positive charge than the right.
So the two faces of the membrane carry different charges.
A charge difference across a membrane is a membrane potential.
The side with less positive charge is the negative side.
That is the right side.

Check q3

Two electrodes touch a living muscle cell, one inside and one in the fluid outside. The meter reads the inside relative to the outside, and it reads negative.

Which face of the membrane carries more positive charge?

A. The inner face
A negative reading means the inside carries less positive charge, not more.
B. ✓ The outer face
C. Neither; the two faces carry equal charge
Equal charge on both faces would read zero.
D. It cannot be told from a negative reading
A negative reading does tell you which face carries more positive charge: the outside, the side the meter is reading relative to.

Why: The meter reads the inside relative to the outside.
A negative reading means the inside carries less positive charge than the outside.
So the outer face is the face with more positive charge.

15What helps keep the inside negative

Video: Watch: What helps keep the inside negative

The pump cycles constantly; take the ATP away and the gradients and the charge difference drain.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L07b.mp4

The pump does not cycle once and stop. The pump cycles constantly, and every cycle uses one ATP.

While the pump cycles, three out and two in keeps sending a little more positive charge out than in. That is how the pump helps keep the inside slightly negative.

Two panels of a plasma membrane, the cytosol shaded to the left of the band and the fluid outside to the right, each with the labelled pump set in it: with the pump running the inner face is negative and the outer face positive; with ATP blocked the pump is idle and the charges are evening out

Now imagine the cell’s ATP supply is removed. The pump stops.

Nothing is sending more charge out than in any longer.

Check q4

The pump has stopped. A few channels for sodium ions and for potassium ions are open.

What does each ion do through its open channels?

A. ✓ It drifts down its own concentration gradient
B. It stays where it is
An open channel is a route, and an ion with a route and a gradient drifts down the gradient.
C. It is pushed against its gradient
Only a pump using ATP moves an ion against its gradient; a channel is an open route.

Why: An ion drifts down its gradient whenever a channel for it is open.

So, over hours, Na⁺ drifts in and K⁺ drifts out, and the two gradients drain away.

As they drain, the charge difference drifts toward zero: the inside becomes less negative. Gradients and membrane potential last only while the cell keeps using ATP.

What you are expected to know Explain how the pump builds and maintains the Na⁺ and K⁺ gradients and, with three positive charges out for every two in, helps keep the inside negative, and predict that when the cell has no ATP left the ion gradients and the membrane potential fade.

Check q5

An animal cell is alive and well fed.

Which of the following helps make the inside of the cell slightly negative?

A. The hydrophobic interior holding every ion back
Holding ions back keeps things as they are; the bilayer cannot make one face more positive than the other.
B. The pump moving three Na⁺ in for every two K⁺ out
The pump sends Na⁺ out and brings K⁺ in.
C. The ATP itself carrying a charge into the cell each cycle
The pump uses ATP to do its work; the ions the pump moves are what carry the charge, not the ATP.
D. ✓ The pump moving three Na⁺ out for every two K⁺ in

Why: Three sodium ions leave the cell for every two potassium ions that enter.
Each ion carries one positive charge, so three positive charges leave for every two that come in.
Cycle after cycle, that helps make the inside slightly negative.

Check q6

A heart muscle cell’s supply of ATP is cut off. Within a minute, every sodium–potassium pump in its membrane has stopped.

What happens to the cell’s membrane potential over the next hours?

A. The potential stays as it was
With the pump stopped, nothing moves Na⁺ out or K⁺ in; the ions drift through open channels and the charge difference drains.
B. ✓ The potential fades toward zero
C. The potential grows
An ion drifting down its gradient moves toward the side where it is less concentrated, and that evens the two faces out.

Why: With no ATP the pump stops.
Nothing now moves Na⁺ out of the cell or K⁺ in against their gradients.
Na⁺ drifts in and K⁺ drifts out through open channels.
As the gradients drain, the charge on the two faces evens out.
So the charge difference fades toward zero.

Practice writing an answer

A chemical stops a kidney cell from making ATP. Within a minute, every sodium–potassium pump in its membrane has stopped. Only a few of the cell’s ion channels are open.

(a) Describe what happens to the cell’s membrane potential over the next hours, and explain why the change takes hours rather than happening the moment the pumps stop. (1 pt)

Frame Over the next hours the membrane potential …, because …

Model answer Over the next hours the membrane potential fades toward zero, because the gradients the pumps built drain only slowly.
When the pumps stop, the Na⁺ and K⁺ gradients are still there.
Ions cross the membrane only through open channels, and only a few are open.
So Na⁺ drifts in and K⁺ drifts out slowly.
As the gradients drain, the charge difference fades: over hours, not at once.

Rubric

Award 1 point for: the membrane potential fades toward zero, AND the gradients the pumps built remain at first and drain only slowly, because ions cross only through the few open channels, so the charge difference fades as the gradients drain rather than all at once.
Accept: “the ions leak out slowly through channels” for the slow drain. Do not award the point for “the pump stops” alone, with no account of what happens to the ions afterwards.

Slip Saying the potential is the pump’s work from moment to moment, so it vanishes when the pump stops. The pump built the gradients, and the gradients hold the charge difference while they last. The gradients take hours to drain through the few open channels.

28Two pulls on an ion

Check q7

Two chloride ions, each carrying a full negative charge, drift close to each other in water.

What do the two ions do to each other?

A. ✓ They push each other apart
B. They pull each other together
Like charges repel, and both ions carry a negative charge.
C. Nothing: a charge acts only on water
A charge acts on any other charge nearby, and both ions are charged.

Why: Both ions carry a negative charge, and like charges repel, so the two ions push each other apart.

Video: Watch: Two pulls on an ion

The concentration pull and the charge pull, found one at a time for sodium and for potassium.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L07c.mp4

A substance’s net movement is down its concentration gradient. Opposite charges attract, and like charges repel.

Two things act on an ion at once. Its concentration difference pulls it one way, and the charge across the membrane pulls on it as well.

The charge across the membrane pulls a positive ion toward the negative inside. It pushes a negative ion away from the negative inside.

Take sodium ions (Na⁺): 145 mmol/L outside, 15 inside. Its concentration pulls it in. It is positive, and the inside is negative, so the charge pulls it in as well.

A plasma membrane drawn as a vertical band, the cytosol shaded to its left and the fluid outside shaded to its right: Na⁺ at 145 mmol/L outside and 15 inside, the inside marked negative, and two arrows both pointing into the cell, labelled concentration and charge

Now potassium ions (K⁺): 140 mmol/L inside, 5 outside. Its concentration pulls it out. It is positive, and the inside is negative, so the charge pulls it in. The two pulls point opposite ways.

A plasma membrane drawn as a vertical band, the cytosol shaded to its left and the fluid outside shaded to its right: K⁺ at 140 mmol/L inside and 5 outside, the inside marked negative, an arrow pointing out labelled concentration and an arrow pointing in labelled charge

An ion’s concentration difference and the charge difference across the membrane, taken together, are called its .

For an ion, concentration alone does not decide where it is pulled. The charge across the membrane pulls too, and it can pull the other way.

To find the two pulls on an ion, take them one at a time. First, the concentration pull: it pulls the ion toward the side where the ion is less concentrated.

Second, the charge pull: a positive ion is pulled toward the negative face; a negative ion is pushed away from it.

Third, put the two together: the same way, or opposite ways.

What you are expected to know Name the two pulls on an ion, its concentration difference and the charge across the membrane, and say which way each one pulls a given ion.

Check q8

The bar chart shows the concentration of chloride ions (Cl⁻) outside a nerve cell and inside it. The inside of the cell is negative.

Which way does the concentration pull act on Cl⁻?

A. ✓ Into the cell
B. Out of the cell
The concentration pull acts toward the side where the ion is less concentrated: Cl⁻ is 10 mmol/L inside and 110 mmol/L outside, so inward.

Why: The concentration pull acts toward the side where the ion is less concentrated.
Cl⁻ is 110 mmol/L outside and 10 mmol/L inside.
The inside bar is shorter.
So the concentration pull on Cl⁻ is into the cell.

Check q9

The bar chart shows the concentration of chloride ions (Cl⁻) outside a nerve cell and inside it. The inside of the cell is negative.

Which way does the charge pull act on Cl⁻?

A. Into the cell
Cl⁻ is a negative ion, and the inside of the cell is negative.
B. ✓ Out of the cell

Why: Cl⁻ is a negative ion.
The inside of the cell is negative.
Like charges repel.
So the charge pushes Cl⁻ away from the inside.
So the charge pull acts on Cl⁻ out of the cell.

Check q10

The bar chart shows the concentration of chloride ions (Cl⁻) outside a nerve cell and inside it. The inside of the cell is negative.

Do the two pulls on Cl⁻ point the same way or opposite ways?

A. The same way
Into the cell and out of the cell are opposite ways.
B. ✓ Opposite ways

Why: The concentration pull on Cl⁻ is into the cell.
The charge pull on Cl⁻ is out of the cell.
So the two pulls on Cl⁻ point opposite ways, as they do for K⁺.

Check q11

Calcium ions (Ca²⁺) are far more concentrated outside a cell than inside, and the inside of the cell is negative.

Which way does each pull act on Ca²⁺?

A. Concentration pulls it in; charge pulls it out
Opposite charges attract, so the negative inside pulls a positive ion in.
B. Concentration pulls it out; charge pulls it in
Calcium is more concentrated outside, so its concentration pulls it in, toward where there is less.
C. ✓ Concentration pulls it in; charge pulls it in
D. Concentration pulls it out; charge pulls it out
Calcium is more concentrated outside, and it is positive while the inside is negative.

Why: Calcium is more concentrated outside the cell than inside.
So its concentration pulls it in.
Ca²⁺ is a positive ion, and the inside of the cell is negative.
Opposite charges attract.
So the charge pulls it in as well, as with Na⁺.

Check q12

Bicarbonate ions (HCO₃⁻) are at 24 mmol/L outside a red blood cell and 12 mmol/L inside, and the inside of the cell is negative.

Which way does each pull act on HCO₃⁻?

A. ✓ Concentration pulls it in; charge pulls it out
B. Concentration pulls it out; charge pulls it in
Bicarbonate is more concentrated outside, so its concentration pulls it in, and it is a negative ion, so the negative inside pushes it out.
C. Concentration pulls it in; charge pulls it in
Like charges repel, so the negative inside pushes a negative ion out.
D. Concentration pulls it out; charge pulls it out
Bicarbonate is more concentrated outside, so its concentration pulls it in, toward where there is less.

Why: Bicarbonate is 24 mmol/L outside and 12 mmol/L inside.
So its concentration pulls it in.
HCO₃⁻ is a negative ion, and the inside of the cell is negative.
Like charges repel.
So the charge pushes it out.
The two pulls point opposite ways.

47The whole chain, and what breaks it

Video: Watch: The whole chain, and what breaks it

Four links from ATP to a negative inside, then a poison undoes them one after another.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L07d.mp4

Put the pieces in order:
1. ATP lets the pump cycle.
2. The pump moves three Na⁺ out of the cell and two K⁺ in.
3. That keeps the concentration of Na⁺ low and of K⁺ high inside.
4. Three positive charges out for every two in helps make the inside slightly negative.

Five linked steps: ATP; the pump runs; three Na⁺ out and two K⁺ in; Na⁺ low and K⁺ high inside; inside slightly negative

One simplification: in a real cell, potassium ions leaking out through open channels also carry positive charge out, and that leak does most of the work of keeping the inside negative. Exam questions expect the pump’s three out, two in as the answer.

Every step to the right depends on the one before it, so stopping the ATP undoes them all, one after another.

Now, a quick example. A poison stops a cell from making ATP.

Explain, step by step, what happens over the next hours to its sodium ions, its potassium ions and the charge across its membrane. Write one short sentence for each step, each on its own line.

A model answer:
With no ATP the sodium–potassium pump stops.
So nothing moves Na⁺ out or K⁺ in against their gradients.
Na⁺ drifts in and K⁺ drifts out through whatever channels are open, so both gradients drain away.
No pump is sending three positive charges out for every two in.
So the charge difference fades, and the inside becomes less negative.

What you are expected to know Explain how a cell sets up and keeps its Na⁺ and K⁺ gradients and its membrane potential, from ATP through the pump’s three-out, two-in to the negative inside, and predict what happens to each when ATP is blocked.

Check q13

A drug blocks the sodium–potassium pump directly. The cell still has plenty of ATP.

What happens to Na⁺ and K⁺ over the next hours?

A. Na⁺ drifts out of the cell and K⁺ drifts in
Na⁺ is more concentrated outside the cell and K⁺ inside, so each drifts down its gradient: Na⁺ in, K⁺ out.
B. Neither ion drifts
ATP does nothing to the ions on its own; only the pump uses ATP, and the pump is blocked.
C. ✓ Na⁺ drifts into the cell and K⁺ drifts out

Why: A blocked pump cannot use the cell’s ATP, so the pump is stopped.
Nothing now moves Na⁺ out or K⁺ in against their gradients.
Na⁺ is more concentrated outside, so Na⁺ drifts in through open channels, down its gradient.
K⁺ is more concentrated inside, so K⁺ drifts out.

Check q14

A drug has blocked a cell’s sodium–potassium pump; the cell still has plenty of ATP.

What happens to the inside of the cell over the next hours?

A. ✓ The inside becomes less negative
B. The inside stays as negative as before
No pump is sending three positive charges out for every two in, so the charge difference drains and the inside becomes less negative.
C. The inside becomes more negative
An ion drifting down its gradient moves toward the side where it is less concentrated, and that evens the two faces out.

Why: No pump is sending three positive charges out for every two in.
Na⁺ drifts in and K⁺ drifts out through open channels.
As the two gradients drain, the charge on the two faces evens out.
So the charge difference fades, and the inside becomes less negative.

Check q15

A nerve cell holds Na⁺ at 20 mmol/L inside and 140 outside, and K⁺ at 130 mmol/L inside and 6 outside.

Which of the following lets the cell keep these differences up?

A. The pump working, with no ATP available
Each pump cycle uses one ATP; with none available the pump stops and the gradients drain.
B. Ion channels open, with ATP available
Open channels let the ions drift down their gradients, evening the two sides out, and ATP does not change that.
C. ✓ The pump working, with ATP available
D. Ion channels open, with no ATP available
Open channels let the gradients drain and, with no ATP, nothing pushes the ions back up.

Why: Both ions must be moved against their gradients: Na⁺ out and K⁺ in.
Only the pump does that, and each cycle uses one ATP.
So the cell keeps these differences up only while the pump works and ATP is available.
Open channels only let the ions run back down.

Practice writing an answer

A scientist keeps frog muscle cells for several hours in a poison that stops them from making ATP. Their membranes stay intact and their channels behave as before. Then the scientist washes the poison away, and the cells make ATP again. Before the poison, the cells held Na⁺ at 10 mmol/L inside and 120 mmol/L outside, and K⁺ at 125 mmol/L inside and 2.5 mmol/L outside, and a meter reading the inside relative to the outside read negative.

(a) Describe the state of the cells’ Na⁺, K⁺ and membrane potential at the end of the hours in poison, compared with before. (1 pt)

Model answer With no ATP the pump has been stopped for hours.
Na⁺ inside has risen above 10 mmol/L, because Na⁺ drifted in through open channels.
K⁺ inside has fallen below 125 mmol/L, because K⁺ drifted out.
The inside is less negative than before.

Rubric

Award 1 point for: Na⁺ inside higher, K⁺ inside lower, and the inside less negative (a smaller membrane potential) than before; all three.
Accept: the gradients have faded and the charge difference has shrunk, with the direction of each change given.

Slip Saying nothing changed because the membrane was intact. An intact membrane still has open channels. Once the pump stops, ions drift down their gradients through them.

(b) Explain what the sodium–potassium pump does once ATP is available again, and what this does to the Na⁺ and K⁺ concentrations inside the cells. (1 pt)

Model answer With ATP available the pump cycles again.
Each cycle uses one ATP to move three Na⁺ out and two K⁺ in.
Both ions are moved against their concentration gradients.
So, cycle after cycle, Na⁺ inside falls back toward 10 mmol/L and K⁺ inside rises back toward 125 mmol/L.

Rubric

Award 1 point for: the pump uses ATP to move three Na⁺ out and two K⁺ in against their gradients, so Na⁺ inside falls and K⁺ inside rises back toward their former values.
Accept: against their gradients in place of the concentration figures; the counts three and two and the directions are needed.

Slip Having the ions return on their own through channels. Channels only let ions move down their gradients. Only the pump, using ATP, moves them back up.

(c) Predict what the meter reading does over the next hours, and justify your prediction. (2 pt)

Model answer The reading becomes more negative again.
Each Na⁺ and each K⁺ carries one positive charge.
Every cycle sends three positive charges out and brings only two in.
So more positive charge ends up outside than inside.
So the inside becomes slightly negative relative to the outside once more.

Rubric

Award 1 point for: the inside becomes more negative again (the reading returns toward its former negative value).
Award 1 point for: each pump cycle sends three positive charges out and brings two in, so less positive charge is left inside than outside.
Accept: the second point in any wording that counts three charges out for two in.
Also accept for the second point: K⁺ drifting out through open channels, down the gradient the pump rebuilds, carries positive charge out of the cell and leaves the inside negative.

Slip Having the pump make the inside positive because it brings K⁺ in. Count the charges: three out, two in. So the inside ends up with less positive charge and reads negative.

The meter reads negative partly because the pump has been sending out three positive charges for every two it brings in; stop the ATP and the reading, like the sodium and potassium gradients, drifts toward zero.

Glossary

membrane potential: A difference in charge between the two faces of a membrane. In a typical animal cell the inside is slightly negative relative to the outside.
polarized membrane: A membrane whose two faces carry different charges, that is, one that has a membrane potential.
electrochemical gradient: The two pulls on an ion taken together: its concentration difference across the membrane and the charge difference across the membrane.

APBIO-U02-P28 Practice questions: Topic 2.8

Topic 2.8 · Mechanisms of Transport · 10 MCQ · 2 FRQ · for APBIO-U02-T28

Video: Watch first: mechanisms of transport, summed up

From the gill cell’s chloride pump to the sodium–potassium pump, the charge it leaves across the membrane, and the two pulls on an ion.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T28-summary.mp4

Q1 P28-q01

In one second, a single sodium–potassium pump in a kidney cell completes about 100 cycles.

Which statement describes what the pump has done in that second?

A. Moved Na⁺ into the cell and K⁺ out of it, using 100 ATP in all
The sodium–potassium pump moves Na⁺ out of the cell and K⁺ in.
B. Moved Na⁺ out of the cell and K⁺ into it, using one ATP for all 100 cycles
Each cycle uses one ATP of its own.
C. ✓ Moved Na⁺ out of the cell and K⁺ into it, using 100 ATP in all
D. Moved Na⁺ and K⁺ both out of the cell, using 100 ATP in all
The pump carries K⁺ into the cell.
The pump completed 100 cycles.

Why: Each cycle moves Na⁺ out of the cell and K⁺ in, and uses one ATP.
The pump completed 100 cycles.
So the pump used 100 ATP.

Q2 P28-q02

A squid's nerve fiber holds K⁺ at 400 mmol/L inside and 20 mmol/L outside, and Na⁺ at 50 mmol/L inside and 440 mmol/L outside. Its sodium–potassium pumps move Na⁺ out and K⁺ in. A student says the pump needs ATP only for moving the Na⁺, because K⁺ enters the cell anyway.

Which statement corrects the student?

A. ✓ K⁺ is moved from 20 mmol/L to 400 mmol/L, against its gradient, so it needs ATP too
B. Na⁺ is moved down its gradient, so it needs no ATP; only the K⁺ needs ATP
Na⁺ goes from 50 mmol/L to 440 mmol/L, toward where it is already more concentrated, which is against its gradient.
C. Neither ion needs ATP; the pump is a channel that ATP holds open
A channel lets ions run down their gradients with no energy used, and this pump moves both ions uphill.
D. The student is right: an ion entering a cell is always moving down its gradient
Which way is down depends on the concentrations, not on in or out, and K⁺ is 20 mmol/L outside and 400 mmol/L inside.

Why: Na⁺ is 50 mmol/L inside and 440 mmol/L outside, and the pump moves Na⁺ out: against its gradient.
K⁺ is 400 mmol/L inside and 20 mmol/L outside, and the pump moves K⁺ in: against its gradient too.
Moving a substance against its gradient takes ATP, for K⁺ as well.

Q3 P28-q03

The inside face of a mouse muscle cell’s plasma membrane carries slightly more negative charge than the outside face. A scientist then gives the cell a drug, and two hours later the two faces carry equal charge.

How should the change be described?

A. The cell had a concentration gradient across its membrane and has now lost that gradient
A concentration gradient is a difference in how much of a substance sits on each side, and what was lost here is a difference in charge.
B. ✓ The membrane was polarized and has now lost its membrane potential
C. The cell was in dynamic equilibrium and the drug has now pushed the cell out of it
Dynamic equilibrium describes a substance equal on both sides and crossing equally each way, and this is a charge difference and its loss.
D. The membrane was carrying out active transport and the drug has now stopped that transport
Active transport is a process, moving a substance against its gradient using energy, and this is a state, a charge difference across the membrane.

Why: A charge difference across a membrane is a membrane potential.
A membrane that has one is polarized.
At first the inside face was slightly negative, so the membrane was polarized.
Two hours later the two faces carried equal charge.
So the membrane potential was gone.

Q4 P28-q04

A pump in the cells lining the stomach moves H⁺ out of the cell and K⁺ into it, one H⁺ out for every one K⁺ in. H⁺ and K⁺ each carry one positive charge.

Considered on its own, does this pump’s cycling leave the inside of the cell negative, and why?

A. Yes: H⁺ piling up outside the cell makes the outside face positive
K⁺ piles up inside just as H⁺ piles up outside, and each ion carries the same single positive charge, so the two piles balance.
B. Yes: any pump that uses energy polarizes the membrane
Using energy moves the ions against their gradients; whether a charge difference builds depends on the count of charges out and in, not on the energy used.
C. No: this pump moves both ions down their gradients
A pump moves each ion to the side where it is already more concentrated, against its gradient; that is what the energy is used for.
D. ✓ No: one positive charge out for every one in leaves the charge on the two faces unchanged

Why: H⁺ and K⁺ each carry one positive charge.
Each cycle of this pump sends one positive charge out and brings one in, so the charge on each face is unchanged.
So this pump builds two gradients but no charge difference.
The sodium–potassium pump sends three out for every two in.

Q5 P28-q05

Red blood cells for transfusion are stored in a refrigerator at 4 °C. The cold slows their sodium–potassium pumps to almost nothing. Over two weeks in the bag, the fluid around the cells gains K⁺ and the cells gain Na⁺; the membranes stay intact.

Explain the change.

A. The cold dissolves part of each membrane, so ions leak through the gaps
The membranes stay intact, and an intact membrane still has open channels.
B. The pumps work in reverse in the cold, pushing K⁺ out and pulling Na⁺ in
A slowed pump does less; it does not reverse.
C. ✓ With the pumps nearly stopped, Na⁺ drifts in and K⁺ drifts out down their gradients
D. The cells use their ATP to push K⁺ out, since a cold cell has no use for it
No cell uses ATP to throw away its K⁺.

Why: Channels only let ions run down their gradients.
Only the pump, using ATP, moves them back up.
The cold slows the pumps to almost nothing.
So Na⁺ drifts in and K⁺ drifts out through open channels, down their gradients, and over two weeks the gradients fade.

Q6 P28-q06

In an experiment a scientist makes the inside of a nerve cell slightly positive relative to the outside, while K⁺ stays at 140 mmol/L inside and 5 mmol/L outside. A potassium channel then opens.

How do the two pulls on K⁺ compare?

A. ✓ Both favor leaving the cell
B. Concentration favors leaving; the positive inside favors staying
K⁺ is a positive ion, and like charges repel, so a positive inside pushes K⁺ toward the outside.
C. Both favor staying in the cell
K⁺ is 140 mmol/L inside and 5 mmol/L outside, so its concentration gradient favors leaving.
D. Concentration favors staying; the positive inside favors leaving
K⁺ is 140 mmol/L inside and 5 outside, so concentration pushes it out, and a positive inside repels the positive K⁺ outward too.

Why: Two things pull on an ion: its concentration difference and the charge across the membrane.
K⁺ is 140 mmol/L inside and 5 mmol/L outside, so its concentration pulls it out.
The inside is positive, K⁺ is positive, and like charges repel, so the charge pushes it out too.

Q7 P28-q07

The figure shows two cells holding Na⁺ at 145 mmol/L outside and 15 mmol/L inside. Cell 1 has its inside slightly negative relative to the outside. Cell 2 has no charge difference across its membrane. A sodium channel opens in each cell at the same moment.

Two cells with the same Na⁺ concentrations. Cell 1 has a charge difference across its membrane (inside slightly negative); cell 2 has none. Each cell has one open sodium channel.

Compare the net movement of Na⁺ into the two cells at that moment, and explain the difference.

A. Equal in both: only the concentration difference moves an ion
For an ion, concentration alone does not settle the pull; the charge across the membrane pulls too.
B. Faster into cell 2: a charge difference slows every ion down
A charge difference does not slow ions in general; it pulls a positive ion toward the negative side.
C. Into cell 1 only: with no charge difference, Na⁺ has no reason to move
Na⁺ is 145 mmol/L outside and 15 mmol/L inside in cell 2 as well, so its concentration difference alone sends it in.
D. ✓ Faster into cell 1: its negative inside pulls the positive ion in as well

Why: Na⁺ is 145 mmol/L outside and 15 mmol/L inside in both cells, so its concentration pulls it into both.
In cell 1 the inside is negative; Na⁺ is positive, and opposite charges attract, so the charge pulls Na⁺ in as well.
So Na⁺ enters cell 1 faster than cell 2.

Q8 P28-q08

A newly formed animal cell builds its Na⁺ and K⁺ gradients and its membrane potential from scratch. Four events are listed out of order: (1) the concentration of Na⁺ is low and of K⁺ high inside, and the inside is slightly negative; (2) the cell makes ATP; (3) the pump moves Na⁺ out of the cell and K⁺ in; (4) ATP powers the pump.

Which order runs from the first cause to the final result?

A. 2, 4, 1, 3
The gradients and the charge difference (1) exist only after the pump has moved ions (3), so 1 cannot come before 3.
B. ✓ 2, 4, 3, 1
C. 4, 2, 3, 1
The pump does nothing without ATP, so the cell must make ATP (2) before ATP can power the pump (4).
D. 1, 3, 4, 2
A newly formed cell has no Na⁺ or K⁺ gradients and no charge difference yet; those (1) are the result the pump produces, not the start.

Why: The chain is ATP → ATP powers the pump → three Na⁺ out and two K⁺ in per cycle → the concentration of Na⁺ low and of K⁺ high inside, with the inside slightly negative.
That is 2, 4, 3, 1. Each step depends on the one before it.

Q9 P28-q09

A membrane protein found in the cells lining the stomach moves K⁺ into the cell against its concentration gradient. It is named an ATPase.

What does the name tell you about how the protein works?

A. ✓ The protein uses ATP to power the moving of K⁺
B. The protein makes ATP as K⁺ passes through it
An ATPase uses ATP; it does not make it.
C. The protein carries ATP across the membrane along with K⁺
ATP stays inside the cell and is used there; the protein moves only K⁺.
D. The protein is a channel for K⁺ that stays open while ATP is present
A channel lets an ion run down its gradient with no energy used, and this protein moves K⁺ against its gradient.

Why: “ATPase” names a protein that uses ATP to do its work.
This protein moves K⁺ against its gradient.
Moving a substance against its gradient is active transport, and the protein uses ATP to do it.
So the protein uses ATP to power the moving of K⁺.

Q10 P28-q10

A student measures Na⁺ at 145 mmol/L outside a cell and 15 mmol/L inside, and writes: 'This difference is the cell's membrane potential.'

Which statement corrects the student?

A. The student is right: any difference across a membrane is a membrane potential
A difference in the amount of a substance on each side is a concentration gradient, not a membrane potential.
B. That is a membrane potential only once the sodium channels open and Na⁺ flows in
Opening channels lets Na⁺ flow down its concentration gradient.
The difference in concentration stays a concentration gradient, open channels or not.
A membrane potential is a difference in charge.
C. ✓ That is a concentration gradient; a membrane potential is a difference in charge across the membrane
D. That is a membrane potential only for K⁺, the ion the cell holds high inside
Which ion is involved makes no difference; both differences are concentration gradients.

Why: Na⁺ at 145 mmol/L outside and 15 mmol/L inside is a difference in amount: a concentration gradient.
A membrane potential is a difference in charge between the two faces of the membrane.
It arises when ions cross so that one face carries more positive charge.

FRQ 1 P28-frq1 · Conceptual Analysis scaffolded

Scientists build artificial cells: small membrane sacs with sodium–potassium pumps set into their membranes. At the start, the fluid inside and outside every sac holds Na⁺ at 100 mmol/L and K⁺ at 100 mmol/L, and there is no charge difference across the membrane. A few ion channels in each membrane are open throughout. The scientists then add ATP to the inside of the sacs and follow the sacs for an hour, until the ATP has all been used.

(a) Identify what one cycle of the pump moves, and the energy source it uses. (1 pt)

Frame One cycle moves … Na⁺ … of the sac and … K⁺ …, using …

Hint How many Na⁺ and how many K⁺ cross in one cycle, and which way does each go? What does the pump use up each cycle?

Model answer One cycle moves three Na⁺ out of the sac and two K⁺ into it, using one ATP.

Rubric

Award 1 point for: three Na⁺ out of the sac and two K⁺ into it, using one ATP.
The counts (three and two), the directions (Na⁺ out, K⁺ in) and ATP are all needed for the point.

Slip Swapping the counts, or sending both ions the same way. Three Na⁺ out, two K⁺ in, one ATP.

(b) Predict how the Na⁺ and K⁺ concentrations inside the sacs change over the hour. (1 pt)

Frame Inside the sacs, Na⁺ … and K⁺ …, because the pump …

Hint Both ions start at 100 mmol/L on each side. Which way does the pump send each one?

Model answer Inside the sacs, Na⁺ falls below 100 mmol/L and K⁺ rises above 100 mmol/L, because the pump sends Na⁺ out and brings K⁺ in, cycle after cycle.
So the pump builds the two gradients from nothing.

Rubric

Award 1 point for: Na⁺ inside falls below 100 mmol/L and K⁺ inside rises above 100 mmol/L, because the pump keeps sending Na⁺ out and bringing K⁺ in, cycle after cycle, faster than the open channels let them drift back.
Accept: 'Na⁺ ends up at a low concentration inside and K⁺ at a high concentration inside'. Do not award the point for both rising or both falling, or for 'nothing changes because the concentrations start equal'.

Slip Saying nothing changes because the pump needs a gradient to work on. The pump makes the gradients; it does not need them.

(c) Explain how the running pump produces a charge difference across the membrane, and state which face ends up negative. (1 pt)

Frame Each cycle moves … positive charges out and … positive charges in, so …; the … face ends up negative.

Hint Think about the charge each ion carries. Then count the charges that cross in one cycle, out and in.

Model answer Each cycle moves three positive charges out and two positive charges in, so more positive charge leaves the sac than enters it; the inside face ends up negative.
Each Na⁺ and each K⁺ carries one positive charge, so three Na⁺ out is three positive charges out and two K⁺ in is two positive charges in.
So the membrane has a membrane potential, and the membrane is polarized.

Rubric

Award 1 point for: each cycle moves three positive charges out and only two in, so more positive charge leaves than enters and the inside face ends up slightly negative relative to the outside (a membrane potential; the membrane is polarized).
Do not award the point for 'because Na⁺ is outside' with no count of the charges, or for the outside face named as negative.

Slip Saying the inside is negative because Na⁺ is outside, with no count. The three-for-two count is the whole point.

(d) Make a claim about what happens to the two gradients and to the charge difference after the ATP has all been used. (1 pt)

Frame With the ATP gone, the pump …; Na⁺ inside …, K⁺ inside …, and the charge difference …

Hint What happens to the pump without ATP? A few channels are open: what can the ions do through them once nothing opposes it?

Model answer With the ATP gone, the pump stops.
Na⁺ drifts back in through the open channels, so Na⁺ inside rises.
K⁺ drifts back out, so K⁺ inside falls.
No pump is sending three positive charges out for every two in.
So the charge difference fades, and the inside becomes less negative.

Rubric

Award 1 point for the claim, all three parts: Na⁺ inside rises (drifts back in), K⁺ inside falls (drifts back out), and the charge difference fades (the inside becomes less negative, toward zero). No reasoning is required for this point.
Accept: 'the gradients fade and the membrane potential fades'. Do not award the point for 'nothing changes because the membrane is intact', or for the inside becoming more negative.

Slip Saying nothing changes because the membrane is intact. An intact membrane still has open channels, and the ions run down their gradients through them.

(e) Support your claim in (d) using the chain of events that runs from ATP to the membrane potential. (1 pt)

Frame The chain is ATP → … → … → … → …; every step depends on the one before it, so …

Hint Write the chain out, from ATP to the negative inside. Then find the link that the end of the hour takes away.

Model answer The chain is ATP → ATP powers the pump → three Na⁺ out and two K⁺ in → Na⁺ concentration low and K⁺ concentration high inside → inside slightly negative.
Every step depends on the one before it.
With no ATP the pump stops, so nothing moves the ions against their gradients.
So they drift back down their gradients through the open channels.
No pump is sending three positive charges out for every two in, so the charge difference fades as well.

Rubric

Award 1 point for: the evidence AND the reasoning: the chain is ATP → ATP powers the pump → three Na⁺ out and two K⁺ in → Na⁺ concentration low and K⁺ concentration high inside → inside slightly negative; every step depends on the one before it, so with no ATP the pump stops, nothing moves the ions against their gradients, they drift back down them through the open channels, and with no pump sending three positive charges out for two in, the charge difference fades too.
Accept: support that names the pump's need for ATP and the ions drifting down their gradients through open channels. Do not award the point for the chain alone with no link to the claim, or for 'the pump works in reverse'.

Slip Saying the pump works in reverse. It simply stops; the ions do the rest by drifting down their gradients.

FRQ 2 P28-frq2 · Conceptual Analysis

A drug used to treat some heart conditions binds to the sodium–potassium pump and slows it. Scientists give the drug to heart muscle cells in a dish and measure the Na⁺ and K⁺ concentrations inside the cells over four hours (table). Before the drug, the cells held Na⁺ at 10 mmol/L and K⁺ at 140 mmol/L inside, against 145 mmol/L Na⁺ and 5 mmol/L K⁺ outside, and their inside was slightly negative relative to the outside. The membranes stay intact, and the cells keep making ATP.

Concentrations of Na⁺ and K⁺ inside heart muscle cells after a pump-slowing drug is given. Outside the cells: Na⁺ 145 mmol/L, K⁺ 5 mmol/L throughout.

(a) Describe what the sodium–potassium pump does in an untreated heart muscle cell. (1 pt)

Frame In an untreated cell, each cycle of the pump moves … and …, both …, using …

Model answer In an untreated cell, each cycle of the pump moves three Na⁺ out of the cell and two K⁺ into it, both against their concentration gradients, using one ATP.
That is what keeps Na⁺ at 10 mmol/L and K⁺ at 140 mmol/L inside.

Rubric

Award 1 point for: each cycle uses one ATP to move three Na⁺ out of the cell and two K⁺ in, both against their concentration gradients, keeping Na⁺ low and K⁺ high inside.
The counts, the directions and 'against the gradient' (or 'uphill') are needed; the numbers 10/145 and 140/5 are not required.

Slip Moving both ions the same way, or down their gradients. Three Na⁺ out, two K⁺ in, both uphill.

(b) Explain how the changes in the table show that the drug has slowed the pumps. (1 pt)

Model answer Na⁺ inside rises from 10 mmol/L to 30 mmol/L.
K⁺ inside falls from 140 mmol/L to 124 mmol/L.
Each ion is drifting down its gradient through open channels.
A working pump undoes that drift as fast as it happens.
Here the drift shows through.
So the pumps are moving fewer ions back than before.
So the drug has slowed them.

Rubric

Award 1 point for: Na⁺ inside rises (10 to 30 mmol/L) and K⁺ inside falls (140 to 124 mmol/L), which is each ion drifting down its gradient through open channels; a working pump undoes that drift, so the drift showing through means the pump is moving fewer ions back than before.
Accept: 'the gradients are fading, and only a slowed pump lets that happen'. Do not award the point for reading the numbers with no link to the pump, or for 'the drug lets ions through the membrane' (the membrane is intact).

Slip Reading off the numbers with no link to the pump, or blaming the drug for 'letting ions through'. The membrane is intact; the ions use channels, and the pump is what no longer keeps up.

(c) Predict how the charge across the membrane changes over the four hours. (1 pt)

Model answer The inside becomes less negative relative to the outside over the four hours.
The membrane potential fades toward zero.

Rubric

Award 1 point for: the inside becomes less negative relative to the outside (the membrane potential fades toward zero).
Do not award the point for the inside becoming more negative, or for the charge reversing to positive; 'less negative' or 'closer to zero' earns the point.

Slip Predicting the inside becomes more negative because Na⁺ is entering. The charge difference depends on the pump's three-out, two-in count, and the pump is doing less of it.

(d) Justify your prediction, using the pump's three-out, two-in count and the changes in the table. (1 pt)

Model answer The pump keeps the inside negative by sending three positive charges out for every two it brings in.
The drug slows the pump, so the pump exports less positive charge each second.
Meanwhile Na⁺ keeps drifting in and K⁺ keeps drifting out through the open channels.
The table shows Na⁺ inside rose by 20 mmol/L while K⁺ inside fell by only 16 mmol/L, so more positive charge entered than left.
So the charge difference shrinks, and the inside becomes less negative.

Rubric

Award 1 point for: the pump keeps the inside negative by sending three positive charges out for every two it brings in; slowed, it exports less positive charge each second, while Na⁺ drifting in and K⁺ drifting out through open channels carry charge that evens the two faces, so the charge difference the pump was maintaining shrinks.
Accept a count from the table: Na⁺ inside rose by 20 mmol/L while K⁺ inside fell by 16 mmol/L, so more positive charge entered than left and the inside became less negative. Accept: 'less pumping, so less of the three-for-two excess that made the inside negative'. Do not award the point for 'the drug makes the membrane leaky'.

Slip Saying the drug makes the membrane leaky. The membrane is intact; the change is that the pump no longer keeps up with the drift through the channels.

APBIO-U02-T28 End-of-topic test: Mechanisms of Transport

Topic 2.8 · Mechanisms of Transport · 16 MCQ · 2 FRQ

Q1 T28-q01

Over a short time, a single sodium–potassium pump in a liver cell moves 150 Na⁺ out of the cell.

Over the same cycles, how many K⁺ did the pump move into the cell, and how many ATP did it use?

A. 150 K⁺ in, 150 ATP
The pump does not swap one Na⁺ for one K⁺, and it does not use one ATP per ion; it moves three Na⁺ and two K⁺ for each ATP.
B. 225 K⁺ in, 75 ATP
Three Na⁺ leave for every two K⁺ that enter, so fewer K⁺ enter than Na⁺ leave: 100, not 225.
C. ✓ 100 K⁺ in, 50 ATP
D. 100 K⁺ in, 150 ATP
The pump uses one ATP for a whole cycle, three Na⁺ and two K⁺ together, so 150 Na⁺ out used 50 ATP, not 150.

Why: One cycle of the pump moves three Na⁺ out, two K⁺ in, and uses one ATP.
So the Na⁺ count gives the number of cycles, and the cycles give the K⁺ and the ATP; the working is below.

Q2 T28-q02

In a nerve cell the cytosol holds 15 mmol/L Na⁺ and 140 mmol/L K⁺; the extracellular fluid holds 145 mmol/L Na⁺ and 5 mmol/L K⁺. The sodium–potassium pump moves Na⁺ out and K⁺ in.

Why does the pump need ATP to do this?

A. ✓ Both ions are moved from lower to higher concentration
B. Both ions are moved from higher to lower concentration
Na⁺ goes from 15 mmol/L to 145 mmol/L and K⁺ from 5 mmol/L to 140 mmol/L, each toward where it is already more concentrated.
C. Na⁺ goes against its gradient but K⁺ goes down its gradient
K⁺ is at 5 mmol/L outside and 140 inside, so moving it in is against its gradient too.
D. Any crossing through a membrane protein uses ATP
Channels and carriers let substances cross with no ATP at all when they move down their gradients.

Why: Na⁺ is 15 mmol/L in the cytosol and 145 mmol/L outside, and the pump moves Na⁺ out: toward where it is already more concentrated.
K⁺ is 140 mmol/L inside and 5 mmol/L outside, and the pump moves K⁺ in: the same.
Moving both against their gradients needs energy, from ATP.

Q3 T28-q03

A membrane protein in a muscle cell moves Ca²⁺ out of the cell, from 0.0001 mmol/L inside to 1.2 mmol/L outside. When the cell’s ATP supply is blocked, the movement of Ca²⁺ stops.

Which name fits this protein, and why?

A. A calcium channel: Ca²⁺ runs through it down its gradient, so the movement needs a route only
Ca²⁺ is moved toward the side where it is already more concentrated, which is against its gradient, and a channel only lets an ion run down its gradient.
B. A calcium carrier for facilitated diffusion: the protein changes shape but the cell uses no energy
The movement stops when ATP is blocked, so the cell is using energy on it, and facilitated diffusion uses no energy from the cell.
C. A calcium synthase: the protein makes ATP as Ca²⁺ passes through it
The protein uses ATP; blocking the ATP supply stopped it, which a protein that made ATP would not do.
D. ✓ A Ca²⁺ ATPase: the protein uses ATP to power the moving of Ca²⁺ against its gradient

Why: Ca²⁺ is 0.0001 mmol/L inside and 1.2 mmol/L outside, and the protein moves Ca²⁺ out: against its gradient.
When ATP is blocked the movement stops, so the protein uses ATP.
A protein that uses ATP to do its work is called an ATPase.
So this is a Ca²⁺ ATPase.

Q4 T28-q04

Inside a liver cell the concentration of Na⁺ is low and of K⁺ is high. Outside the cell the concentration of Na⁺ is high and of K⁺ is low.

Which of the following is the work of the sodium–potassium pump?

A. Na⁺ entering and K⁺ leaving the cell, both down their gradients, through open channels
Na⁺ entering and K⁺ leaving are each moving down their gradients, which is the drift through open channels that the pump undoes.
B. ✓ Na⁺ leaving and K⁺ entering the cell, both against their gradients, while ATP is present
C. K⁺ leaving the cell down its gradient, through a channel, whether or not ATP is present
K⁺ leaving the cell is moving down its gradient through a channel, on which the cell uses no energy, and it is not the pump’s work.
D. Glucose entering the cell through a carrier, down its gradient, with no ATP used
Glucose entering down its gradient through a carrier is facilitated diffusion, on which the cell uses no energy.

Why: The sodium–potassium pump moves three Na⁺ out and two K⁺ in for each ATP it uses.
The concentration of Na⁺ is low inside and high outside, so Na⁺ leaving is against its gradient; K⁺ entering is too.
Moving an ion against its gradient needs ATP.

Q5 T28-q05

Four membranes are described.

Which of the following membranes is polarized?

A. ✓ A membrane whose inner face carries slightly less positive charge than its outer face
B. A membrane with Na⁺ at 145 mmol/L on one side and 15 mmol/L on the other, and equal charge on its two faces
A difference in concentration is a concentration gradient, and the two faces of this membrane carry equal charge, so it has no membrane potential.
C. A membrane that ions cross in both directions at equal rates, with equal charge on its two faces
Equal crossing in both directions is a dynamic equilibrium, and this membrane’s two faces carry equal charge.
D. A membrane whose pump has stopped, leaving both faces with the same charge
A stopped pump leaves the two faces with the same charge, and a membrane potential is a difference in charge.

Why: A charge difference across a membrane is called a membrane potential, and a membrane that has one is polarized.
The first membrane’s inner face carries less positive charge than its outer face, so its two faces differ in charge.
The other three have equal charge on their two faces.

Q6 T28-q06

A thin artificial membrane separates two salt solutions, A and B, that start out identical, each with equal positive and negative charge. A protein in the membrane then moves positive ions from side A to side B; the negative ions stay where they are.

What is the result?

A. A is now positive relative to B: the membrane is polarized
Positive ions left A and arrived in B, so B has gained positive charge and A has lost it.
B. Both sides stay neutral: the negative ions follow
The negative ions cannot cross, so they stay behind on A, which is exactly why a charge difference builds up.
C. ✓ B is now positive relative to A: the membrane is polarized
D. No charge difference: only concentrations changed
The ions that moved carry charge, so moving them moved charge.

Why: Each positive ion carries charge.
Positive ions moved from side A to side B, while the negative ions stayed on side A.
So side B now carries more positive charge than side A.
A charge difference across a membrane is a membrane potential, so the membrane is polarized.

Q7 T28-q07

In a resting nerve cell, one sodium–potassium pump completes 1,000 cycles.

Compare the positive charge that left the cell through the pump over the 1,000 cycles with the positive charge that entered.

A. 2,000 more positive charges left than entered
Each cycle sends three positive charges out and brings two in, a difference of one.
B. 1,000 more positive charges entered than left
The pump sends three Na⁺ out for every two K⁺ it brings in.
C. No difference: each cycle carries positive ions both ways, so the charges cancel
The two directions do not carry equal charge: three positive charges go out and only two come in each cycle.
D. ✓ 1,000 more positive charges left than entered

Why: Each Na⁺ and each K⁺ carries one positive charge.
Each cycle sends three positive charges out and brings two in, one more out than in.
So over 1,000 cycles, 1,000 more positive charges left than entered; the working is below.

Q8 T28-q08

A newly formed animal cell starts with Na⁺ and K⁺ at the same concentrations inside as outside, and no charge difference across its membrane. Its sodium–potassium pumps then start pumping, with plenty of ATP. Its channels stay shut.

Predict how the cell changes over the next hour.

A. ✓ Na⁺ falls and K⁺ rises inside, and the inside becomes slightly negative
B. Na⁺ and K⁺ both rise inside, and the inside becomes positive
The pump moves the two ions opposite ways, Na⁺ out and K⁺ in, so Na⁺ inside falls while K⁺ inside rises.
C. The two gradients build, but the charge stays equal, since both ions are positive
Both ions are positive but the count is unequal, three out and two in, so a charge difference builds along with the gradients.
D. Nothing changes until a channel opens; pumps need an existing gradient to work on
The pump does not wait for a gradient; it makes one.

Why: ATP powers the pump.
Each cycle sends three Na⁺ out and brings two K⁺ in, so Na⁺ falls inside and K⁺ rises inside.
Each cycle also sends three positive charges out and brings two in, so more positive charge leaves than enters.
So the inside becomes slightly negative.

Q9 T28-q09

The figure shows a resting animal cell: the concentration of sodium ions (Na⁺) is high outside and low inside, the concentration of potassium ions (K⁺) is high inside and low outside, and the inside is slightly negative. The channel drawn lets only Na⁺ through, and it opens.

A resting animal cell membrane: which ion is high on each side, and the charge on each face. One channel is drawn. The extracellular fluid is shaded darker than the cytosol.

Which way do the two pulls on Na⁺ act, and what happens?

A. Concentration pulls Na⁺ in, charge pulls it out; little moves
Na⁺ is a positive ion and the inside of the cell is negative, so the charge pulls Na⁺ in, not out.
B. Concentration pulls Na⁺ out, charge pulls it in; little moves
The concentration of Na⁺ is high outside and low inside, so its concentration gradient runs inward.
C. ✓ The gradient and the charge both pull Na⁺ in; Na⁺ enters the cell
D. The gradient and the charge both pull Na⁺ out; Na⁺ leaves the cell
Na⁺ is more concentrated outside and the negative inside attracts a positive ion, so nothing pulls it out.

Why: Two things pull on an ion: its concentration difference and the charge across the membrane.
Na⁺ is more concentrated outside, so its concentration pulls it in.
Na⁺ is positive and the inside is negative; opposite charges attract, so the charge pulls it in too.
So Na⁺ enters.

Q10 T28-q10

The resting animal cell in the figure: the concentration of K⁺ is high inside and low outside, and the inside is slightly negative. Now suppose the channel drawn lets only K⁺ through, and it opens.

How do the two pulls on K⁺ compare?

A. Both favor leaving the cell
K⁺ is a positive ion and the inside is negative, and that attraction favors K⁺ staying in.
B. ✓ Concentration favors leaving; the negative inside favors staying
C. Both favor staying in the cell
K⁺ is at a higher concentration inside than out, so its concentration gradient favors leaving.
D. Concentration favors staying; the negative inside favors leaving
K⁺ is more concentrated inside, so concentration favors leaving; K⁺ is positive and the inside is negative, so the charge favors staying.

Why: K⁺ is more concentrated inside, so its concentration gradient favors leaving.
K⁺ is a positive ion and the inside is negative.
Opposite charges attract, so the charge favors staying.
So the two parts of its electrochemical gradient pull opposite ways, unlike Na⁺, where both pull in.

Q11 T28-q11

A student says the direction an ion moves through an open channel depends only on its concentration on the two sides of the membrane.

What is missing from the student's account?

A. Ions move through the bilayer itself, not through channels
Charged ions cannot cross the hydrophobic interior; they cross only through channel proteins, and the route is not what is missing.
B. Nothing: for an ion only concentration matters
An ion carries charge, so the charge difference across the membrane pulls on it too.
C. The ion must also be pumped using ATP
Through an open channel an ion moves by itself, with no ATP.
D. ✓ The charge across the membrane also pulls on an ion

Why: An ion carries a charge.
Two things pull on it at once: its concentration difference and the charge difference across the membrane.
Together they are its electrochemical gradient.
For K⁺ in a resting cell the two pull opposite ways.
So concentration alone does not settle which way an ion moves.

Q12 T28-q12

In an experiment a scientist makes the inside of a muscle cell slightly positive relative to the outside, while Ca²⁺ stays far more concentrated outside the cell than inside.

How do the two pulls on Ca²⁺ now compare?

A. ✓ Concentration favors entering; the positive inside favors staying out
B. Both favor entering, as in a resting cell
The charge part has changed: a positive inside repels a positive ion, so the charge no longer pulls Ca²⁺ in.
C. Both favor staying out
Ca²⁺ is still far more concentrated outside, so its concentration still favors entering.
D. Concentration favors staying out; the positive inside favors entering
Ca²⁺ is more concentrated outside, so concentration favors entering; Ca²⁺ is positive and the inside is now positive, so the charge favors staying out.

Why: Ca²⁺ is far more concentrated outside than inside.
So its concentration favors entering.
The inside is now positive.
Ca²⁺ is a positive ion, and like charges repel.
So the charge now favors staying out.
The two pulls oppose each other, as they do for K⁺ in a normal resting cell.

Q13 T28-q13

A student reads that a resting nerve cell's membrane is polarized, with the inside slightly negative, and concludes: 'So the cytosol must be full of negative ions and the fluid outside full of positive ions.'

Which statement corrects the student?

A. The student is right: polarizing a membrane means sorting positive ions to one side and negative ions to the other
Sorting every ion to one side would take an enormous amount of energy and would kill the cell.
B. The inside is negative because K⁺, the ion piled up inside, carries a negative charge
K⁺ is a positive ion.
C. ✓ Both fluids still hold both kinds of ion; the inside face just carries slightly more negative charge
D. The extra charge sits inside the membrane's oily middle, not in either fluid
The hydrocarbon tails in the middle carry no charges at all; the excess charge lies in the fluid right against each face.

Why: A membrane potential is a charge difference between the two faces of a membrane.
Both fluids still hold positive and negative ions in nearly equal numbers.
The difference is a small excess of charge right against each face.
That excess leaves the inside face slightly negative.

Q14 T28-q14

A plant toxin binds the sodium–potassium pumps of a kidney cell and stops them. The cell keeps making ATP, and its membrane is undamaged.

What happens to the cell’s Na⁺ and K⁺ gradients and to its membrane potential over the next hours, and why?

A. The gradients and the potential both hold: the cell’s ATP keeps the ions where they are without the pump
ATP does not act on the ions directly; ATP powers the pump, and the pumps are stopped.
B. ✓ The gradients and the potential both fade: ATP does nothing without a working pump, so the ions drift downhill
C. The gradients and the potential both hold: once the pump has set them up, the gradients keep themselves
Ions keep drifting back down their gradients whenever channels are open, so the gradients last only as long as the pump keeps undoing that drift.
D. The gradients fade but the membrane potential holds: the charge difference was set once and stays
The charge difference is not permanent; the ions drifting down their gradients even the two faces out as the gradients fade.

Why: The chain is ATP → pump → three Na⁺ out and two K⁺ in → gradients → inside negative.
The pumps are blocked, so the chain breaks at the pump.
The ions drift down their gradients through open channels, so the gradients and the charge difference fade.

Q15 T28-q15

Nerve cells kept at 4 °C for several hours lose much of their Na⁺ and K⁺ gradients, and their insides become less negative, because the cold slows their sodium–potassium pumps almost to a stop. The cells are then warmed back to body temperature, with plenty of ATP.

What happens over the next hours?

A. The pumps restart, but a gradient cannot be rebuilt once it is lost
The gradients were built by the pump in the first place, and a restarted pump rebuilds them the same way.
B. The gradients return, but the inside stays less negative
Once the K⁺ concentration gradient is rebuilt, K⁺ leaking out through open channels carries positive charge out again, and the pump’s three-out, two-in helps, so the inside turns negative again.
C. Na⁺ is moved in and K⁺ out until the two are equal
The pump moves Na⁺ out and K⁺ in, against their gradients, making the two sides unequal.
D. ✓ The pumps restart: Na⁺ out, K⁺ in, and the inside turns negative again

Why: The pumps restart.
Each cycle uses one ATP to move three Na⁺ out and two K⁺ in, rebuilding both concentration gradients.
K⁺, high inside again, leaks out through open channels and carries positive charge out; the pump’s three-out, two-in helps.
So the inside turns negative again.

Q16 T28-q16

A student explains why the inside of a resting cell is negative: "The sodium–potassium pump carries negative ions into the cell."

Which statement corrects the student?

A. ✓ Both ions the pump moves are positive; each cycle carries more positive charge out of the cell than it brings in, which helps make the inside negative
B. Both ions the pump moves are positive, so the pump adds as much positive charge to one face as it removes from the other; the inside is negative for some other reason
The pump does not move equal charge each way: three positive charges leave for every two that enter, so the pump does change the charge on the two faces.
C. The inside is negative because the pump moves Na⁺ out and K⁺ in in equal numbers, and a Na⁺ ion carries more positive charge than a K⁺ ion
Na⁺ and K⁺ carry the same single positive charge, and the pump does not move them in equal numbers: three Na⁺ leave for every two K⁺ that enter.
D. The inside is negative because Na⁺ leaving carries positive charge out of the cell, while K⁺ entering carries no charge in with it
Each K⁺ entering brings one positive charge in, so Na⁺ leaving is only half the story; three charges leave for every two that enter, which helps make the inside negative.

Why: Na⁺ and K⁺ each carry one positive charge.
Each cycle sends three Na⁺ out and two K⁺ in.
So each cycle leaves one more positive charge outside, helping make the inside negative.
K⁺ leaking out through open channels does most of the work.

FRQ 1 T28-frq1 · Conceptual Analysis

Nerve cells keep the concentration of Na⁺ low and of K⁺ high in their cytosol (about 15 mmol/L Na⁺ and 140 mmol/L K⁺ inside, against 145 and 5 mmol/L outside) and keep their inside slightly negative relative to the outside. Ions cross the membrane only through channels, and only when those channels are open; a few channels are open even in a resting cell. A scientist adds a poison that stops the cells from making ATP; the membrane is not damaged.

(a) Describe how the sodium–potassium pump keeps the concentration of Na⁺ low and of K⁺ high inside a resting cell. (1 pt)

Model answer Each cycle of the pump moves Na⁺ out of the cell, three Na⁺ per cycle, and K⁺ into the cell, two K⁺ per cycle; the pump can do this because it uses one ATP each cycle.
Both ions are moved against their concentration gradients.

Rubric

Award 1 point for: the pump uses ATP (one per cycle) to move three Na⁺ out of the cell and two K⁺ in, both against their concentration gradients.
Accept "against their gradients" without the numbers 15/145 and 5/140, but the counts three and two and the direction (Na⁺ out, K⁺ in) must be given. Do not award the point for a pump that moves both ions the same way, or that moves them down their gradients.

Slip Moving both ions the same way, or down their gradients. Three Na⁺ out, two K⁺ in, both uphill.

(b) Explain how the pump helps make the inside of the cell slightly negative. (1 pt)

Model answer Na⁺ and K⁺ each carry one positive charge.
Each cycle moves three positive charges out and only two in.
So more positive charge leaves than enters.
So the inside ends up slightly negative relative to the outside.

Rubric

Award 1 point for: each cycle moves three positive charges out and only two in, so more positive charge leaves than enters and the inside ends up slightly negative relative to the outside (a membrane potential; the membrane is polarized).
Do not award the point for an explanation in which the pump moves negative charge, or for "because Na⁺ is outside" with no count of the charges moved.
Also accept, in addition or in place: the pump keeps K⁺ high inside, and K⁺ drifting out through open channels down that gradient carries positive charge out, leaving the inside negative.

Slip Saying the inside is negative because Na⁺ is outside, with no count of the charges moved. The three-for-two count is the point.

(c) Make a claim about what happens to the Na⁺ and K⁺ concentrations inside the cells, and to the charge across the membrane, over the hours after the scientist adds the poison. (1 pt)

Model answer Na⁺ inside rises, because Na⁺ drifts in.
K⁺ inside falls, because K⁺ drifts out.
The inside becomes less negative.

Rubric

Award 1 point for the claim, all three parts: Na⁺ inside rises (Na⁺ drifts in), K⁺ inside falls (K⁺ drifts out), and the inside becomes less negative. No reasoning is required for this point.
Accept 'the gradients fade' for the first two if the direction of each drift is stated, and 'less negative' or 'closer to zero' for the charge. Do not require the charge difference to reach zero. Do not award the point for 'nothing changes because the membrane is intact', or for the inside becoming more negative.

Slip Saying nothing changes because the membrane is intact, or that the inside becomes more negative. Open channels let the ions run down their gradients once the pump stops.

(d) Support your claim with the chain of events from ATP to the charge across the membrane. (1 pt)

Model answer Without ATP the pump stops.
So nothing moves the ions against their gradients.
A few channels are open.
So each ion drifts down its concentration gradient: Na⁺ in and K⁺ out.
So the gradients fade.
No pump is sending three positive charges out for every two in.
So the charge difference the pump was maintaining fades too, and the inside becomes less negative.

Rubric

Award 1 point for: the evidence AND the reasoning: without ATP the pump stops, so nothing moves the ions against their gradients; each ion drifts down its concentration gradient whenever a channel for it is open (Na⁺ in, K⁺ out), so the gradients fade; and with no pump sending three positive charges out for every two in, the charge difference it was maintaining fades too (the inside becomes less negative; it need not reach zero).
Accept support that names the pump's need for ATP and the ions drifting down their gradients. Do not award the point for "the poison lets ions through the membrane" (the membrane is not damaged) or for "the pump works in reverse".

Slip Saying the poison lets ions through the membrane, or that the pump works in reverse. The membrane is undamaged; the ions use channels, and the pump simply stops.

FRQ 2 T28-frq2 · Analyze Model or Visual Representation

The model shows one sodium–potassium pump in the plasma membrane of a resting animal cell, with the concentration of Na⁺ and K⁺ on each side. The direction each ion is carried, the number of ions moved in one cycle, the pump’s energy source and the charges on the two faces of the membrane are not drawn.

One sodium–potassium pump in a resting animal cell membrane, with the concentration of each ion on each side. The direction each ion is carried, the number of ions moved per cycle, the pump’s energy source and the charges on the two faces are not drawn. The extracellular fluid is shaded darker than the cytosol.

(a) Describe one cycle of the pump: how many of each ion it moves, in which direction each ion is carried, and what the pump uses as its energy source. (1 pt)

Model answer In one cycle, three Na⁺ move from the cytosol, at 15 mmol/L, to the outside, at 145 mmol/L, and two K⁺ move from the outside, at 5 mmol/L, into the cytosol, at 140 mmol/L; each ion is moved from where it is less concentrated to where it is more concentrated, against its gradient.
The pump uses one ATP for the cycle.

Rubric

Award 1 point for: three Na⁺ are moved from the cytosol to the outside and two K⁺ from the outside into the cytosol, using one ATP; each ion is moved from where it is less concentrated to where it is more concentrated (against its gradient).
Accept directions given as concentrations (Na⁺ from 15 mmol/L to 145 mmol/L; K⁺ from 5 mmol/L to 140 mmol/L). Do not award the point if the direction of either ion is reversed, the 3:2 count is missing, or ATP is not named.

Slip Reversing the direction of either ion, leaving out the three-to-two count, or leaving out the ATP. The model gives only the concentrations; the directions, the numbers of ions and the ATP are yours to supply.

(b) State the charge on each face of the membrane that results from many cycles of the pump, and the direction in which the charge difference pulls a Na⁺ ion. (1 pt)

Model answer The inside face is slightly negative and the outside face is positive.
Na⁺ is a positive ion.
Opposite charges attract.
So the charge difference pulls Na⁺ into the cell.

Rubric

Award 1 point for: the inside face slightly negative and the outside face positive, AND Na⁺ pulled into the cell.
Accept: "inside negative, outside positive; the charge pulls Na⁺ in". Do not award the point for the inside given as positive, or for Na⁺ pulled out of the cell.

Slip Calling the inside positive, or sending Na⁺ out. Opposite charges attract: the negative inside pulls a positive ion in.

(c) Explain, using the concentrations in the model, why this pump needs an energy source while a channel uses no energy. (1 pt)

Model answer Both ions are moved toward the side where they are already more concentrated.
So both are moved against their concentration gradients.
Moving a substance against its gradient takes energy.
That energy comes from ATP, so this is active transport.
A channel only lets an ion move down its gradient.
Movement down a gradient needs no energy, so a channel uses no energy.

Rubric

Award 1 point for: both ions are moved toward the side where they are already more concentrated, so the pump moves them against their concentration gradients, which takes energy; that energy comes from ATP (active transport, defined by the energy the cell uses). A channel only lets an ion move down its gradient, which needs no energy.
Accept "uphill" for against the gradient. Do not award the point for "because it is a pump" or "because the ions are charged" with no reference to the direction relative to the gradients.

Slip Saying because it is a pump, or because the ions are charged. The reason is the direction relative to the gradients.

(d) Explain how the model relates to the larger idea that a cell must use energy to keep its inside different from its surroundings. (1 pt)

Model answer Ions drift down their gradients whenever channels are open.
So the differences the model shows, a low concentration of Na⁺ and a high concentration of K⁺ inside and a negative inside, would fade on their own.
The cell keeps them only by using ATP in the pump, cycle after cycle.
So the cell holds a steady internal state by using energy.

Rubric

Award 1 point for: ions drift down their gradients whenever channels are open, so the differences the model shows (a low concentration of Na⁺ and a high concentration of K⁺ inside, inside negative) would fade on their own; the cell keeps them only by continually using ATP in the pump, a steady internal state (homeostasis) held by using energy.
Accept "without ATP the gradients and the charge difference fade" as the link. Do not award the point for "the membrane keeps the ions in" with no mention of energy.

Slip Saying the membrane keeps the ions in, with no mention of energy. The membrane has channels; the pump, using ATP, is what holds the differences.

APBIO-U02-L08 Too big for any door

Topic 2.5 · Membrane Transport · 60 steps

A white blood cell with its membrane curving out on one side to wrap around a rod-shaped bacterium

Here is a white blood cell wrapping its membrane around a bacterium. The bacterium is a thousand times too big for any channel, and minutes later it is inside the cell.

Unit 2 · Cell Structure and Function

1Wrapping something in

Check q1

A magnesium ion, Mg²⁺, carries a full charge. Methane, CH₄, is small and nonpolar.

Which of the two crosses a membrane only through a channel or carrier protein?

A. ✓ The magnesium ion
B. Methane
Methane is small and nonpolar, so methane dissolves through the tails and needs no protein.
C. Both of them
Methane is small and nonpolar, so methane crosses the bilayer on its own.

Why: Water attracts a full charge and nothing in the hydrophobic interior does, so the magnesium ion crosses only through a channel or carrier protein.

Video: Watch: Wrapping something in

A bacterium a thousand times too big for any door: the membrane folds around it and pinches off a vesicle.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L08.mp4

A membrane has doors: channels and carriers, each letting through one kind of ion or small molecule.

Here is a bacterium beside a channel. The bacterium is about a thousand times wider than the channel, and wider than any pump or carrier too.

A channel protein in a membrane with a single ion passing through it, beside a bacterium drawn far larger than the channel

The white blood cell takes the bacterium in anyway. Watch the membrane. The membrane folds inward around the bacterium, making a deep pocket.

A cell membrane drawn as a line, the fluid outside shaded above it and the cytosol shaded below it, curving down into a deep round pocket that wraps almost all the way around a rod-shaped bacterium, leaving only a narrow opening at the top; the pocket holds the outside fluid

The pocket closes and pinches off. The bacterium is now inside the cell, wrapped in a small sac of membrane.

A cell membrane drawn as a flat line, the fluid outside shaded above it and the cytosol shaded below it, and beneath the line a closed circle of membrane with a bacterium inside; the fluid inside the circle carries the outside shade

A small sac of membrane like this is called a .

Taking something in by folding the membrane around it and pinching the pocket off is called .

Cells take in large molecules the same way. A liver cell wraps membrane around a protein particle too big for any carrier and pinches it off inside.

Bending membrane and pinching it off uses energy from ATP. When a white blood cell cannot make ATP, its membrane stays flat. So the bacterium stays stuck to the outside.

What you are expected to know Describe endocytosis: the membrane folds inward around a large molecule or particle and pinches off, leaving it inside the cell in a vesicle, and say that this uses energy from the cell.

Check q2

An amoeba, a single-celled pond organism, touches a smaller cell. The amoeba’s membrane spreads around the smaller cell, closes behind it, and pinches off inside.

Which of the following has taken place?

A. Exocytosis
In exocytosis a vesicle moves to the plasma membrane and releases its contents outside.
B. ✓ Endocytosis
C. Facilitated diffusion
A whole cell is far too large for any channel or carrier.

Why: The amoeba’s membrane folded around the smaller cell.
The membrane closed behind the smaller cell and pinched off.
So the smaller cell is now inside the amoeba, wrapped in a vesicle.
Taking something in by folding the membrane around it and pinching off is endocytosis.

Check q3

A white blood cell can no longer make ATP. Large protein particles that the cell normally takes in touch its membrane.

What happens to the particles?

A. The particles enter as before
Bending membrane around a particle and pinching it off uses energy from ATP, and the cell has none.
B. The particles enter and lie loose in the cytosol
With no ATP the membrane cannot fold in at all.
C. ✓ The particles stay outside

Why: Bending membrane and pinching it off uses energy from ATP.
The cell has no ATP.
So the membrane cannot bend around a particle.
So the membrane stays flat, and the particles stay outside.

Check q4

The white blood cell has no ATP. The particles stay outside.

Why do the particles stay outside?

A. The particles need ATP to swim to the membrane
The particles already touch the membrane, and they got there without ATP; particles do not swim, they move at random.
B. Without ATP the cell cannot pump the particles in through a carrier
A particle is far too large for any carrier or channel protein, with or without ATP; endocytosis is the only way in.
C. ✓ Bending and pinching off membrane uses energy from ATP

Why: Endocytosis bends the membrane around the particle and pinches the pocket off.
Bending and pinching off membrane uses energy from ATP.
The cell has no ATP.
So the membrane cannot bend.
So the particles stay outside.

16Releasing from a vesicle

Video: Watch: Releasing from a vesicle

A vesicle of insulin moves to the plasma membrane, fuses with it and spills its contents outside.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L08b.mp4

Now the reverse. A cell in the pancreas has made insulin, a protein that acts as a hormone, and packed it into a vesicle. The vesicle moves to the plasma membrane.

A cell membrane drawn as a line, the fluid outside shaded above it and the cytosol shaded below it, and beneath the line a vesicle filled with small molecules moving up toward the membrane

The vesicle’s membrane fuses with the plasma membrane and opens to the outside. The insulin spills out into the fluid outside the cell.

A cell membrane drawn as a line, the fluid outside shaded above it and the cytosol shaded below it, dipping into an open cup that holds the outside fluid, with small molecules leaving the cup into the space above

The vesicle is gone: its membrane has become part of the plasma membrane.

Releasing something by moving a vesicle to the plasma membrane and fusing the two is called .

Exocytosis uses energy from ATP too. When the cell cannot make ATP, the vesicles stay where they are. So almost no insulin reaches the outside.

Cells release waste this way, and useful products too: insulin from the pancreas, and the digestive proteins the pancreas sends to the gut.

What you are expected to know Describe exocytosis: a vesicle moves to the plasma membrane, fuses with it and releases its contents outside, its membrane becoming part of the plasma membrane, and say that this uses energy from the cell.

Check q5

After a signal, most of the vesicles near a pancreas cell’s membrane vanish, and digestive protein appears in the fluid outside the cell. In cells with no ATP, the vesicles stay and almost no protein appears outside.

Which of the following explains both results?

A. The protein crossed the bilayer one molecule at a time
A protein is a large polar molecule and cannot cross the bilayer on its own.
B. The membrane folded inward and took protein in from outside
Protein appeared outside the cells while the vesicles inside emptied, so material left the cell.
C. The protein left through carrier proteins that ATP powered
A carrier binds one small molecule at a time, and a protein is far too large for it, with or without ATP.
D. ✓ Vesicles fused with the membrane and released the protein, using ATP

Why: The vesicles vanished.
At the same time protein appeared outside the cell.
So the vesicles fused with the plasma membrane and released their protein outside.
In cells with no ATP, neither change happened.
So the fusing needed energy from ATP.
That is exocytosis.

Check q6

A student says: “Exocytosis is only how a cell gets rid of waste.”

Is the student correct?

A. Yes, the student is correct
Cells also release useful products by exocytosis.
The pancreas releases insulin this way.
B. ✓ No: useful products leave this way too

Why: Cells do release waste by exocytosis.
Cells also release useful products by exocytosis: the pancreas releases insulin this way, and the pancreas sends its digestive proteins to the gut this way.

Check q7

A drug stops vesicle membranes from fusing with the plasma membrane in a pancreas cell. The cell goes on making insulin and packing it into vesicles.

Where does the insulin end up?

A. In the blood, as before
The drug blocks the fusing.
B. Loose in the cytosol
Blocking the fusing leaves each vesicle whole, so the insulin stays sealed inside it.
C. ✓ Inside vesicles in the cell
D. In the plasma membrane
Insulin is a protein dissolved in the fluid inside the vesicle; it never sits in the bilayer.

Why: Insulin leaves the cell only when its vesicle fuses with the plasma membrane.
The drug blocks that fusing.
So the vesicles cannot open to the outside.
So the vesicles pile up inside the cell, full of insulin.
So less insulin reaches the blood.

28Name the way it crossed

Video: Watch: Name the way it crossed

Five ways across in one table, told apart by direction, protein, energy and size.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L08c.mp4

You’ve now seen five ways across a membrane:
1. Simple diffusion: a small nonpolar molecule moves down its gradient straight through the bilayer; no protein; the cell uses no energy.
2. Facilitated diffusion: an ion or a large polar molecule moves down its gradient through a channel or carrier; a protein; the cell uses no energy.
3. Active transport: a pump moves a substance against its gradient; a protein; the cell uses energy from ATP.
4. Endocytosis: the membrane wraps something too big for any door and takes it in as a vesicle; the cell uses energy.
5. Exocytosis: a vesicle fuses with the plasma membrane and releases its contents outside; the cell uses energy.

A summary table of five ways across a membrane: simple diffusion, facilitated diffusion, active transport, endocytosis and exocytosis, with columns for gradient, protein, energy and size

To tell them apart, ask about the substance:
1. Which way is it moving relative to its gradient?
2. Is a protein needed?
3. Does the cell use energy?
4. How big is it?

To name the way across, answer those questions for the substance, then find the row of the table that matches.

Glucose entering a cell through a carrier, from more concentrated to less, with no ATP used: down the gradient, protein, no energy, small. Facilitated diffusion.

Chloride leaving a gill cell from 40 mmol/L to 550 mmol/L, using ATP: against the gradient, protein, energy, small. Active transport.

Anything far too big for a door moves in a vesicle: in by endocytosis, out by exocytosis. The cell uses energy either way.

Together these five ways across are how a cell keeps its solutes and water where they need to be. Pumps set up the differences. Channels and carriers let the right things through, down their gradients. Vesicles move what no door can.

Now, a quick example. A kidney cell takes up an ion from a fluid where it is 2 mmol/L into a cytosol where it is 30 mmol/L, through a membrane protein. A poison that stops ATP production stops the uptake.

Name the way the ion crosses, and give the two observations that decide it. Write one short sentence for each step, each on its own line.

A model answer: active transport. The ion moves from 2 mmol/L to 30 mmol/L, which is against its gradient. The uptake stops when the cell cannot make ATP. So the cell uses energy from ATP to move the ion.

What you are expected to know Name the way a described substance crosses a membrane, simple diffusion, facilitated diffusion, active transport, endocytosis or exocytosis, from its direction relative to the gradient, whether a protein is needed, whether the cell uses energy and how big it is.

41Name the way from the evidence

Video: Watch: Name the way from the evidence

Which observation answers each of the four questions, worked on one kidney cell.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L08d.mp4

In a real experiment nobody tells you the direction, whether a protein is needed, whether the cell uses energy, or the size of what moves. You read each one from an observation.

Two concentrations tell you the direction relative to the gradient. A protein-free bubble result tells you whether a protein is needed. An ATP block tells you whether the cell uses energy. The size of what moves tells you whether it crossed through the bilayer, through a protein or in a vesicle.

What you are expected to know From the evidence given, decide whether a substance crossed through the bilayer, through a protein or in a vesicle.

What you are expected to know From the evidence given, decide whether the substance moved with or against its gradient.

What you are expected to know From the evidence given, decide whether the cell used energy.

What you are expected to know Name the way across: simple diffusion, facilitated diffusion, active transport, endocytosis or exocytosis.

Check q8

Solute X, a charged ion, enters a cell down its gradient. Blocking one membrane protein stops the movement. The cell’s ATP use stays the same.

Which of the following is the way X crosses?

A. ✓ Facilitated diffusion
B. Simple diffusion
Blocking a membrane protein stopped the movement, so X needs a protein.
C. Active transport
The cell’s ATP use stays the same, so the cell uses no energy on this movement.
D. Endocytosis
X is small enough to pass through a protein door, and blocking that door stopped X.

Why: X enters down its gradient.
Blocking a protein stops the movement, so X needs a protein.
The cell’s ATP use stays the same, so the cell uses no energy.
Down the gradient, through a protein, with no energy used is facilitated diffusion.

Check q9

Oxygen enters a muscle cell from 0.15 mmol/L in the blood to 0.03 mmol/L in the cytosol. Oxygen enters protein-free bubbles of phospholipid bilayer just as readily. Blocking the cell’s ATP leaves the rate unchanged.

Which of the following is the way oxygen crosses?

A. Endocytosis
Oxygen is a small molecule, and oxygen enters bubbles that have no membrane machinery at all.
B. Active transport
Blocking the cell’s ATP leaves the rate unchanged, so the cell uses no energy on this crossing.
C. Facilitated diffusion
Oxygen enters protein-free bubbles of phospholipid bilayer just as readily as it enters the cell, so oxygen needs no channel or carrier.
D. ✓ Simple diffusion

Why: Oxygen moves from 0.15 mmol/L to 0.03 mmol/L: down its gradient.
Oxygen enters protein-free bubbles just as readily, so oxygen needs no protein.
Blocking ATP leaves the rate unchanged, so the cell uses no energy.
Down the gradient, straight through the bilayer, no energy used: simple diffusion.

Practice writing an answer

A root cell of a plant sits in soil water. Three substances cross its plasma membrane. Nitrate ions (NO₃⁻) enter from the soil water, where they are at 0.2 mmol/L, into the cytosol, where they are already at 5 mmol/L, through a membrane protein; when the cell cannot make ATP, the uptake stops. Carbon dioxide is more concentrated inside the cell than in the soil water, and it leaves the cell. Carbon dioxide crosses protein-free bubbles of phospholipid bilayer just as readily. When the cell cannot make ATP, carbon dioxide leaves at the same rate as before. The cell releases a slimy polysaccharide, a very large molecule. Small sacs of membrane inside the cell move to the plasma membrane. The polysaccharide appears outside the cell just after each sac reaches the membrane.

(a) Identify the mechanism by which nitrate enters the cell, the mechanism by which carbon dioxide leaves, and the mechanism by which the polysaccharide leaves. (1 pt)

Model answer Nitrate enters by active transport.
Carbon dioxide leaves by simple diffusion.
The polysaccharide leaves by exocytosis.

Rubric

Award 1 point for: nitrate by active transport, carbon dioxide by simple diffusion, the polysaccharide by exocytosis; all three.
Accept: the names in any order, each tied to its substance.

Slip Calling the nitrate movement facilitated diffusion because a protein is involved. Facilitated diffusion and active transport both use a protein. Movement against the gradient that uses ATP is what marks active transport.

(b) Justify your classification of the nitrate movement using two observations from the description. (1 pt)

Model answer Nitrate moves from 0.2 mmol/L in the soil water to 5 mmol/L in the cytosol.
That is against its concentration gradient.
The uptake stops when the cell cannot make ATP.
So the cell uses energy from ATP to move nitrate.
Movement against the gradient that uses ATP is active transport.

Rubric

Award 1 point for: nitrate moves against its gradient (0.2 to 5 mmol/L) AND the uptake stops when ATP is blocked, so the cell uses energy from ATP.
Accept: uphill for against the gradient. Both observations are needed for the point.

Slip Using the protein as the deciding observation. Channels and carriers are proteins too. The direction relative to the gradient and the need for ATP are what decide.

(c) Explain how the nitrate movement helps the root cell keep its solute concentrations different from the soil water, and why the polysaccharide leaves the cell the way it does. (2 pt)

Model answer Nitrate is 5 mmol/L inside the cell and 0.2 mmol/L in the soil water, and diffusion alone would even the two sides out.
The cell uses ATP to move nitrate uphill, into the cell, so nitrate stays far more concentrated inside.
A channel or a carrier lets through one ion or small molecule at a time.
The polysaccharide is far too large for any channel or carrier.
So it travels in a vesicle, which fuses with the plasma membrane and opens outward: exocytosis.

Rubric

Award 1 point for: by using ATP to move nitrate against its gradient, the cell holds nitrate far more concentrated inside than in the soil water, a difference that passive crossings on their own would even out.
Award 1 point for: a channel or carrier passes one ion or small molecule at a time, so the polysaccharide, a very large molecule, is far too large for any channel or carrier, and it leaves in a vesicle that fuses with the plasma membrane and opens outward (exocytosis).
Accept: for the first point, a gradient exists only for a solute the cell pumps or the membrane blocks, tied to nitrate.

Slip Sending the polysaccharide through a carrier a piece at a time. A carrier binds one small molecule. Anything too big for a door travels in a vesicle.

The bacterium was wrapped in membrane and taken in whole: endocytosis, the cell’s answer to anything too big for a door.

53Fluency quiz: name the way across mixed practice

Check q10

Glycerol enters a cell down its gradient through a channel protein. The cell uses no energy.

Which way across is this?

A. Simple diffusion
Glycerol goes through a channel protein.
B. ✓ Facilitated diffusion
C. Active transport
Glycerol moves down its gradient and the cell uses no energy.

Why: Glycerol moves down its gradient.
Glycerol needs a protein.
The cell uses no energy.
Down the gradient, through a protein, with no energy used is the facilitated diffusion row.
So glycerol crosses by facilitated diffusion.

Check q11

Carbon dioxide leaves a cell down its gradient, straight through the bilayer. The cell uses no energy.

Which way across is this?

A. ✓ Simple diffusion
B. Facilitated diffusion
Carbon dioxide goes straight through the bilayer.
C. Active transport
Carbon dioxide moves down its gradient and the cell uses no energy.

Why: Carbon dioxide moves down its gradient.
Carbon dioxide needs no protein.
The cell uses no energy.
Down the gradient, no protein, with no energy used is the simple diffusion row.
So carbon dioxide crosses by simple diffusion.

Check q12

A liver cell moves bile salt out, from 0.05 mmol/L inside to 2 mmol/L in the bile, through a membrane protein.

Which way across is this?

A. Facilitated diffusion
Bile salt moves from 0.05 mmol/L to 2 mmol/L, against its gradient; movement against a gradient never happens on its own.
B. ✓ Active transport
C. Endocytosis
A bile salt molecule is small enough for a protein door, and it goes through a membrane protein.

Why: Bile salt moves from 0.05 mmol/L to 2 mmol/L: against its gradient.
Movement against a gradient never happens on its own, so the cell uses energy.
Bile salt goes through a membrane protein.
Against the gradient, through a protein, using energy is active transport.

Check q13

A cell takes in a bacterium by wrapping membrane around it and pinching the pocket off. The cell uses ATP.

Which way across is this?

A. ✓ Endocytosis
B. Exocytosis
The cell takes the bacterium in.
C. Active transport
A bacterium is far too large for any protein door.

Why: A bacterium is far too big for any door.
The membrane wraps around the bacterium and pinches off, taking the bacterium in.
Something too big for any door, wrapped in membrane and taken in, is the endocytosis row.
So the bacterium enters by endocytosis.

Check q14

A vesicle full of mucus moves to the plasma membrane, fuses with it and releases the mucus outside. The cell uses ATP.

Which way across is this?

A. Endocytosis
The vesicle releases the mucus outside.
A vesicle fusing with the plasma membrane and releasing its contents outside is the exocytosis row.
B. Active transport
Mucus is a very large molecule, far too large for any protein door; it leaves in a vesicle.
C. ✓ Exocytosis

Why: The mucus leaves in a vesicle.
The vesicle fuses with the plasma membrane and releases its contents outside.
That is the exocytosis row.
So the mucus leaves by exocytosis.

Check q15

A muscle cell takes up an amino acid from 1 mmol/L outside to 9 mmol/L inside through a membrane protein. The cell uses ATP.

Which way across is this?

A. Facilitated diffusion
The amino acid moves from less concentrated to more concentrated, which is against its gradient, and the cell uses ATP.
B. ✓ Active transport
C. Exocytosis
The amino acid comes in, through a membrane protein.

Why: The amino acid moves from 1 mmol/L to 9 mmol/L: against its gradient.
The amino acid goes through a membrane protein.
The cell uses ATP.
Against the gradient, through a protein, using ATP is active transport.

Glossary

vesicle: A small sac of membrane inside a cell.
endocytosis: Taking a large molecule or particle into a cell by folding the plasma membrane inward around it and pinching the pocket off as a vesicle. It uses energy from the cell.
exocytosis: Releasing the contents of a vesicle outside a cell: the vesicle moves to the plasma membrane, fuses with it and opens, and its membrane becomes part of the plasma membrane. It uses energy from the cell.

APBIO-U02-P25 Practice questions: Topic 2.5

Topic 2.5 · Membrane Transport · 10 MCQ · 2 FRQ · for APBIO-U02-T25

These practice questions have the shape of the Topic 2.5 test. Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the first free-response question, you work through one scenario in small steps, and each step offers a hint if you want one. For the second, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it.

Video: Watch first: diffusion and gradients, summed up

The mole and the millimole, the concentration gradient, random motion and diffusion, dynamic equilibrium, passive transport.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T25-summary.mp4

Q1 P25-q01

Two solutions are prepared, each 0.20 mol/L and each one liter: one of glycine, an amino acid of which one mole has a mass of 75 g, and one of tryptophan, an amino acid of which one mole has a mass of 204 g.

How do the numbers of dissolved amino acid molecules in the two liters compare?

A. The tryptophan liter holds about 2.7 times as many molecules
More grams is not more molecules: each tryptophan molecule has 2.7 times the mass of a glycine molecule, so equal numbers of molecules have 2.7 times the mass.
B. The glycine liter holds about 2.7 times as many molecules
Lighter molecules do not pack in more tightly; 0.20 mol/L fixes the count of molecules in each liter whatever their mass.
C. ✓ The two liters hold the same number of molecules, about 1.2 × 10²³ each
D. The glycine liter holds 75 molecules and the tryptophan liter holds 204
75 and 204 are the masses of one mole of each amino acid in grams, not counts of molecules; a mole is about 6 × 10²³ molecules.

Why: A mole is a fixed count of particles, about 6 × 10²³.
So 0.20 mol/L of any dissolved substance gives the same number of particles per liter: about 1.2 × 10²³ here.
Each tryptophan molecule is heavier, so the tryptophan liter has more mass, not more molecules.

Q2 P25-q02

In a kidney, fluid inside a tube holds urea at 0.30 mol/L. Blood flowing beside the tube holds urea at 0.005 mol/L. The tube wall between them lets urea through.

Which way is down urea's concentration gradient?

A. ✓ From the tube fluid into the blood, from 0.30 mol/L toward 0.005 mol/L
B. From the blood into the tube fluid, from 0.005 mol/L toward 0.30 mol/L
From 0.005 mol/L toward 0.30 mol/L runs from the less concentrated side to the more concentrated one, which is up the gradient.
C. Along the tube, in the direction the fluid flows
A concentration gradient runs across the tube wall between the two regions, whichever way the fluid flows.
D. Neither way: urea is spread evenly once both sides hold some
The two sides differ, 0.30 mol/L against 0.005 mol/L, so there is a gradient.

Why: A concentration gradient is a difference in concentration between two connected regions.
Down the gradient is from the more concentrated region to the less concentrated one.
The tube fluid is at 0.30 mol/L and the blood at 0.005 mol/L.
So down the gradient is into the blood.

Q3 P25-q03

A crystal of purple dye dissolves at the bottom of a beaker of still water, and over an hour the color spreads through the whole beaker. A student says: “The dye molecules sense where the water is clear and head for it.”

Which observation shows that the student is wrong?

A. The color spreads until the whole beaker is evenly purple from top to bottom
An even spread is what the student expects too: molecules heading for clear water would also end up spread evenly.
B. The color spreads upward and sideways from the crystal at the same time
Spreading in every direction is what the student expects too: clear water lies upward and sideways alike.
C. The color spreads through the beaker faster when the water is stirred
Stirring makes currents, and currents carry dye whether or not the molecules head anywhere; the observation does not show how the molecules move on their own.
D. ✓ Under a microscope, some dye molecules move back into the purple region

Why: Dye molecules move constantly and at random.
A molecule that headed for clear water would never move back into the purple region, yet some do move back.
So the molecules do not head anywhere.
Near the crystal the dye molecules are crowded, so more leave that region than arrive: diffusion.

Q4 P25-q04

In a fish's gill, water flowing past holds oxygen at 8 mg/L, and the blood inside the gill holds oxygen at 3 mg/L. Oxygen molecules cross the gill's thin lining in both directions.

What is the net movement of oxygen at the gill?

A. None: oxygen crosses in both directions, so the two movements cancel
Molecules cross both ways, but not in equal numbers when one side is more crowded.
B. ✓ Into the blood: more oxygen molecules cross from water to blood than from blood to water
C. Into the blood: oxygen molecules cross only from the water into the blood
Some oxygen molecules do cross from the blood back into the water; there are simply more crossing the other way.
D. Into the water: the fish breathes out its oxygen through the gill
Oxygen is more concentrated in the water, so the net movement is into the blood.

Why: Oxygen molecules cross the gill lining at random in both directions.
The water holds more oxygen, 8 mg/L, than the blood, 3 mg/L, so more oxygen molecules cross from water to blood than the other way.
The difference is a net movement of oxygen into the blood: diffusion.

Q5 P25-q05

A membrane separates two urea solutions, both at 0.40 mol/L. Urea can cross the membrane. The number of urea molecules crossing the membrane from left to right each minute is counted for an hour.

How does that number change over the hour?

A. Molecules cross for a few minutes, then crossing stops
Nothing switches the molecules off: they keep moving at random, and any molecule that reaches the membrane crosses it.
B. Crossing slows gradually over the hour, toward zero
The two concentrations stay equal all hour, so the number of molecules reaching the membrane each minute does not fall.
C. ✓ Molecules cross at about the same rate all hour
D. No molecule crosses at any time in the hour
At equal concentrations molecules cross equally in both directions; they do not stop crossing; the molecules never stop moving.

Why: Urea molecules move constantly and at random, and a molecule that reaches the membrane crosses it.
Both solutions stay at 0.40 mol/L, so the same number reach the membrane from the left each minute.
So the number crossing left to right stays the same; as many cross right to left.

Q6 P25-q06

Molecule V is a small nonpolar molecule at 6 mmol/L outside a skin cell and 1 mmol/L inside, and it enters the cell. V enters protein-free bubbles of phospholipid bilayer just as fast, and blocking the cell’s ATP leaves the rate unchanged.

Which of the following is the type of transport shown?

A. ✓ Simple diffusion
B. Active transport
V moves down its gradient and blocking ATP changes nothing, so the cell uses no energy; entering a cell is not what makes a crossing active.
C. Facilitated diffusion
V enters protein-free bubbles just as fast as it enters the cell, so V needs no protein; a small nonpolar molecule dissolves into the tails.
D. Endocytosis
V is a small molecule; endocytosis wraps membrane around particles far too large for any door, and it uses energy, which this crossing does not.

Why: V moves from 6 mmol/L to 1 mmol/L: down its concentration gradient.
V enters protein-free bubbles just as fast, so V needs no protein.
Blocking ATP leaves the rate unchanged, so the cell uses no energy.
Down its gradient, through the bilayer itself, using no energy: simple diffusion.

Q7 P25-q07

Inside a leaf cell, sucrose is at 0.30 mol/L; in the watery spaces of the cell wall outside it, sucrose is at 0.02 mol/L. Sucrose is a large polar molecule, and no sucrose carrier in the cell’s membrane is working.

What keeps the sucrose more concentrated inside the cell than outside?

A. The sucrose molecules inside have stopped moving, so none reach the membrane
Sucrose molecules move constantly and at random and reach the membrane all the time.
B. The cell uses ATP holding each sucrose molecule in place
Holding a gradient across a membrane the solute cannot cross uses no energy.
C. The cell keeps making sucrose faster than sucrose leaks out through the bilayer
Sucrose is a large polar molecule and cannot pass through the bilayer at all, so there is no leak for the cell to make up.
D. ✓ The membrane blocks sucrose, so diffusion cannot even the two sides out

Why: Sucrose is a large polar molecule, so sucrose cannot cross the bilayer.
No sucrose carrier is working, so sucrose has no other route.
So diffusion cannot even the two sides out.
A solute stays more concentrated on one side only because the membrane does not let it diffuse freely across.

Q8 P25-q08

A cell lining the stomach pushes hydrogen ions (H⁺) out into the stomach fluid, where H⁺ is at 150 mmol/L. Inside the cell, H⁺ is at 0.0001 mmol/L.

What must this crossing involve?

A. A channel protein only
An open channel lets an ion move only down its gradient, and here H⁺ is moving up.
B. ✓ A pump protein and energy from ATP
C. No protein at all
Ions cannot cross the bilayer on their own, whatever their size.
D. A carrier protein that uses no energy
H⁺ is far more concentrated outside, so leaving the cell is against its gradient.

Why: H⁺ leaves the cell from 0.0001 mmol/L to 150 mmol/L.
That is against its concentration gradient.
Only active transport moves a substance against its gradient.
So the cell uses energy from ATP to drive a pump, a membrane protein that moves H⁺ uphill.

Q9 P25-q09

A cell lining a newborn’s gut takes in whole antibody proteins from milk. Under the microscope, the cell’s membrane where a cluster of antibodies touches it is seen to dent inward, and then to close over the cluster.

How do the antibodies get into the cell, and where are they once inside?

A. By simple diffusion through the bilayer; loose in the cytosol
An antibody is a protein, a large polar molecule, and cannot cross the bilayer on its own; the membrane was seen to fold around the antibodies and close over them.
B. Through a channel protein; loose in the cytosol
A channel lets one ion or small molecule through at a time, and an antibody is a protein, far too large for any channel.
C. By exocytosis; enclosed in a vesicle of membrane
The membrane closed over the antibodies and took them in; exocytosis releases material outward.
D. ✓ By endocytosis; enclosed in a vesicle of membrane

Why: The plasma membrane dents inward around the antibodies and closes over them: it folds inward and pinches off.
So the antibodies end up inside the cell enclosed in a vesicle: endocytosis.
An antibody is far too large for any channel or carrier.
Bending and pinching off membrane uses energy.

Q10 P25-q10

A cell in a mammary gland holds milk proteins inside vesicles. The vesicles move to the plasma membrane and fuse with it, and the proteins appear in the milk outside. When a drug blocks the cell's ATP production, the vesicles stay inside and protein stops appearing in the milk.

What do these observations show?

A. ✓ The proteins leave by exocytosis, and fusing a vesicle with the membrane uses energy from the cell
B. The proteins cross the bilayer on their own, using ATP to squeeze between the tails
A protein is far too large and polar to pass through the bilayer, with or without ATP.
C. The proteins leave through a carrier protein that ATP powers
No carrier moves a whole protein, and the vesicles were seen fusing with the membrane.
D. The vesicles are taken in from the milk, which uses ATP
The vesicles form inside the cell and move outward, so material is leaving, not entering.

Why: A vesicle inside the cell moves to the plasma membrane, fuses with it and releases its contents outside.
That is exocytosis.
The vesicle's membrane becomes part of the plasma membrane.
Blocking ATP stopped the release.
So exocytosis uses energy from the cell.

FRQ 1 P25-frq1 · Conceptual Analysis scaffolded

Four substances cross the plasma membrane of a cell lining the small intestine an hour after a meal. Fructose, a sugar, enters through a membrane protein from the gut fluid, at 12 mmol/L, into the cytosol, at 1 mmol/L; blocking the cell’s ATP supply leaves its entry unchanged. Amino acid L enters through a different membrane protein from the gut fluid, at 0.5 mmol/L, into the cytosol, already at 8 mmol/L; blocking ATP stops this uptake. Oxygen enters from the blood, at 0.15 mmol/L, into the cytosol, at 0.05 mmol/L, and crosses protein-free bubbles of phospholipid bilayer just as readily. The cell releases mucus, a very large polysaccharide: small sacs of membrane move to the plasma membrane, and the mucus appears outside just after each sac reaches it.

(a) Identify the direction fructose moves relative to its concentration gradient, and state the two concentrations that tell you. (1 pt)

Frame Fructose moves … its concentration gradient, from … mmol/L in the … to … mmol/L in the …

Hint Compare the concentration where fructose starts with the concentration where it ends up.

Model answer Fructose moves down its concentration gradient, from 12 mmol/L in the gut fluid to 1 mmol/L in the cytosol: from where it is more concentrated to where it is less.

Rubric

Award 1 point for: fructose moves down its concentration gradient, from 12 mmol/L in the gut fluid to 1 mmol/L in the cytosol.
Accept: "from high to low" with the two values quoted.

Slip Reading 'through a protein' as a sign of uphill movement. The direction is set by the two concentrations, and 12 to 1 mmol/L is downhill.

(b) Describe what the two observations about amino acid L show about how L enters. (1 pt)

Frame L moves from … mmol/L to … mmol/L, which is … its gradient; and when ATP is blocked …, which shows …

Hint Taking the two concentrations and the ATP result together, what does each one tell you about how L gets in?

Model answer L moves from 0.5 mmol/L in the gut fluid to 8 mmol/L in the cytosol, which is against its concentration gradient; and when ATP is blocked the uptake stops, which shows that the cell uses energy from ATP to move L uphill through a pump: active transport.

Rubric

Award 1 point for: L moves against its gradient (0.5 mmol/L outside up to 8 mmol/L inside), and the uptake stops when ATP is blocked, so the cell is using energy to move it: active transport.
Accept: 'uphill' for against the gradient; both observations are needed for the point.

Slip Naming active transport from the protein alone. Channels and carriers are proteins too; what marks active transport is movement against the gradient using ATP.

(c) Explain what decides whether a crossing through a membrane protein uses energy from the cell, using fructose and L as your examples. (1 pt)

Frame Fructose’s entry uses no energy because …; L’s entry must use energy because …

Hint Think about which way random movement alone would carry each substance, and which way each one is actually going.

Model answer Fructose moves down its gradient, from 12 mmol/L to 1 mmol/L.
Fructose molecules move constantly and at random, so more leave the crowded gut fluid than return to it.
So the net movement into the cell happens by itself; the protein only gives fructose a route: facilitated diffusion.
L moves against its gradient, from 0.5 mmol/L to 8 mmol/L.
Random movement alone gives a net flow the other way, so the cell must use energy from ATP to push L uphill.

Rubric

Award 1 point for: fructose is moving down its gradient, and particles moving at random give a net movement down a gradient by themselves, so the protein only provides a route (facilitated diffusion, a form of passive transport); L is moving against its gradient, which random movement never does on its own, so the cell must use energy to drive it uphill.
Accept: 'downhill uses no energy; uphill needs energy', with each substance's direction relative to its gradient named; the random-movement mechanism is welcome but not required.

Slip Saying fructose's entry uses no energy because fructose is a sugar the cell needs. What makes it use no energy is the direction: down the gradient, which random movement supplies on its own.

(d) Identify the mechanism by which oxygen enters and the mechanism by which the mucus leaves, giving the clue in the description that decides each. (1 pt)

Frame Oxygen enters by …, because …; the mucus leaves by …, because …

Hint For each one, ask: protein or no protein, and how big is the thing that moves?

Model answer Oxygen enters by simple diffusion: it moves from 0.15 mmol/L to 0.05 mmol/L, down its gradient, and it crosses protein-free bubbles just as readily, so it needs no protein.
Oxygen is a small nonpolar molecule, and the bilayer itself is its route.
The mucus leaves by exocytosis: mucus is far too large for any channel or carrier, and it appears outside just as each sac of membrane reaches the plasma membrane.
So the sacs fuse with the plasma membrane and open outward.

Rubric

Award 1 point for: oxygen enters by simple diffusion (oxygen moves down its gradient, 0.15 to 0.05 mmol/L, and crosses protein-free bilayer, so it needs no protein); the mucus leaves by exocytosis (mucus is far too large for any channel or carrier, and it appears outside as each membrane sac reaches the plasma membrane, so the sacs fuse with the membrane and release it).
Accept: 'passive transport through the bilayer' for oxygen. Both mechanisms with their clues are needed for the point.

Slip Calling the release of mucus active transport because it uses energy. Exocytosis does use energy, but it moves a whole vesicle's worth of material, not one molecule through a pump.

(e) Predict what happens to the entry of fructose if fructose builds up in the cytosol until it is also at 12 mmol/L, and justify your prediction. (1 pt)

Frame When the two concentrations are equal, fructose …; this is because …

Hint Once fructose is at 12 mmol/L on both sides, compare the number of fructose molecules crossing into the cell each second with the number crossing out.

Model answer When the two concentrations are equal, fructose still crosses through the protein in both directions, but the net movement of fructose into the cell stops; this is because as many fructose molecules leave the cell as enter it.
So the concentration inside stops rising.
That is a dynamic equilibrium: the molecules still move, and only the net movement is zero.

Rubric

Award 1 point for: net movement of fructose into the cell stops; fructose molecules keep crossing through the protein in both directions at equal rates (a dynamic equilibrium), so the concentration inside no longer rises.
Accept: "no net movement; equal traffic both ways". Do not award the point for "fructose stops moving" or "the protein closes".

Slip Saying the fructose molecules stop moving, or the carrier shuts. They keep crossing both ways; equal concentrations give equal traffic, so no net change.

FRQ 2 P25-frq2 · Scientific Investigation

Root cells of a salt-marsh plant keep sodium ions (Na⁺) at 10 mmol/L inside while the soil water around them holds Na⁺ at 200 mmol/L. Researchers add labeled Na⁺ to the soil water and follow it for one hour in three trials, keeping temperature and solution volumes the same. The results are in the table.

Labeled Na⁺ and total Na⁺ inside the root cells after one hour in each trial.

(a) Describe how an ion such as Na⁺ can cross a plasma membrane, and what stops it crossing the bilayer on its own. (1 pt)

Frame An ion such as Na⁺ crosses the membrane only through …; it cannot cross the bilayer on its own because …

Model answer An ion such as Na⁺ crosses the membrane only through a membrane protein: a channel when the ion moves down its gradient, or a pump when the ion is moved against it.
It cannot cross the bilayer on its own because the hydrocarbon tails in the middle of the membrane carry no charges or partial charges.
Water attracts the ion.
Nothing in the interior attracts the ion.
So Na⁺ stays in the water.

Rubric

Award 1 point for: Na⁺ crosses only through membrane proteins (a channel, or a pump for uphill movement); it cannot cross the bilayer on its own because the hydrocarbon tails in the interior carry no charges or partial charges, so water attracts an ion and nothing in the interior does.
Accept: "nothing in the oily interior attracts a charge, so ions need a protein".

Slip Saying Na⁺ is too big to cross. Na⁺ is one of the smallest particles present; water attracts its charge, and that attraction is what keeps Na⁺ out of the interior.

(b) Explain what trial 2 shows about how the cells normally keep their Na⁺ at 10 mmol/L. (1 pt)

Model answer In trial 2 the cells could make no ATP.
Their Na⁺ climbed from 10 mmol/L toward 200 mmol/L.
So the low concentration of Na⁺ in trial 1 was being maintained by work.
A pump using ATP moves Na⁺ out against its gradient as fast as Na⁺ enters.
With no ATP the pump stops.
So Na⁺ diffuses in down its gradient until the inside approaches the outside.

Rubric

Award 1 point for: with ATP blocked, Na⁺ inside rose toward the outside level, so the cells normally use energy from ATP to pump Na⁺ out against its gradient (active transport) as fast as it leaks in; without the pumping, Na⁺ diffuses in down its gradient (200 to 10 mmol/L) and the two sides move toward equal.
Accept: "a pump using ATP removes Na⁺ as fast as it enters; stop the pump and diffusion evens the sides out".

Slip Saying the drug let Na⁺ in. The drug only stopped the ATP supply; Na⁺ was entering all along, and what failed was the pumping out.

(c) Describe one further trial to test whether the blocked channel of trial 3 is the route by which Na⁺ entered in trial 2, and predict its result. (1 pt)

Model answer Treat cells with both the drug that blocks ATP and the compound that blocks the channel, then add labeled Na⁺ for an hour.
If the channel is the route Na⁺ used in trial 2, almost no labeled Na⁺ will appear inside and the total will stay near 10 mmol/L even though the pump is off.
If Na⁺ still rises toward 150 mmol/L, it is entering by some other route.

Rubric

Award 1 point for: treat cells with both the ATP-blocking drug and the channel-blocking compound; predicted result: almost no labeled Na⁺ enters and the total inside stays near 10 mmol/L (if the channel is the entry route), whereas Na⁺ rising as in trial 2 would show another route.
Accept: any design that combines the two treatments (or blocks the channel in ATP-blocked cells) with a prediction that matches the claim being tested.

Slip Proposing to repeat trial 3 with more cells. The question is whether the channel is the entry route when the pump is off, which needs both treatments in one trial.

(d) Evaluate the claim that two different membrane proteins, working together, keep the concentration of Na⁺ inside these cells low, using the results of all three trials. (1 pt)

Model answer The claim is supported.
Trial 3: with the channel blocked, almost no labeled Na⁺ reaches the cytosol, so the channel is the way in.
Trial 2: with ATP blocked, Na⁺ is not removed and the inside climbs to 150 mmol/L, so a pump powered by ATP is the way out.
Trial 1: labeled Na⁺ enters, yet the total stays at 10 mmol/L, so the pump removes Na⁺ as fast as it enters.
So a channel and a pump together keep the concentration of Na⁺ low.

Rubric

Award 1 point for: the judgement that the claim is supported AND the ground for it from the trials: trial 3 shows Na⁺ enters through the channel (block it and labeled Na⁺ stops appearing inside); trial 2 shows a pump using ATP removes it (block ATP and Na⁺ builds up inside); trial 1 shows the two working together, labeled Na⁺ entering yet the total holding at 10 mmol/L because the pump removes Na⁺ as fast as the channel lets it in.
Accept: the three trials tied to entry, removal and the steady state in any order. Do not award the judgement alone, or a judgement grounded on only one trial.

Slip Judging the claim from trial 2 alone. The claim has two parts, a channel for entry and a pump for removal, and each needs its own trial as the ground.

APBIO-U02-T25 End-of-topic test: Membrane Transport

Topic 2.5 · Membrane Transport · 17 MCQ · 2 FRQ

Q1 T25-q01

A 0.25 mol sample of ethanol, with a mass of 11.5 g, is dissolved in one liter of water. A 0.25 mol sample of glycerol, with a mass of 23.0 g, is dissolved in another liter of water.

How do the numbers of dissolved molecules in the two liters compare?

A. About twice as many in the glycerol liter
The glycerol has twice the mass only because each glycerol molecule has about twice the mass of each ethanol molecule, not because there are more of them.
B. ✓ The same number in each
C. About half as many in the glycerol liter
A mole of a heavier substance holds just as many molecules as a mole of a lighter one; a mole is the same count for any substance.
D. More in the ethanol liter: smaller molecules pack in
How tightly molecules pack has nothing to do with how many a mole holds; a mole is a fixed count.

Why: A mole is a fixed count of particles, about 6 × 10²³.
So 0.25 mol of ethanol and 0.25 mol of glycerol hold the same number of molecules, about 1.5 × 10²³ each.
Each glycerol molecule is heavier, so the glycerol has more mass, not more molecules.

Q2 T25-q02

A bag made of thin membrane holds a dye solution at 0.10 mol/L. It is lowered into a beaker of the same dye at 0.60 mol/L. The membrane lets the dye through.

Which way is down the dye's concentration gradient, and what does 0.60 mol/L tell you about the beaker solution?

A. ✓ Into the bag; 0.60 mol of dye in each liter
B. Out of the bag; 0.60 mol of dye in each liter
The beaker at 0.60 mol/L is the more concentrated side, so down the gradient runs into the bag, not out of it.
C. Into the bag; 0.60 g of dye in each liter
Mol/L counts moles, not grams.
Down a gradient runs from the more concentrated region to the less concentrated one.
D. Out of the bag; 0.60 g of dye in each liter
Down the gradient runs into the bag, from 0.60 mol/L to 0.10 mol/L, and mol/L counts moles, not grams.

Why: Down a concentration gradient runs from the more concentrated region to the less concentrated one.
The beaker is at 0.60 mol/L and the bag at 0.10 mol/L, so down the gradient is into the bag.
A concentration in mol/L is a molar concentration: moles of solute per liter of solution.

Q3 T25-q03

A spoonful of sugar dissolves at the bottom of a glass of still, cold tea. An hour later the tea tastes equally sweet at the top and at the bottom. A student explains: "The sugar molecules sense where the tea is unsweetened and swim toward it."

Which statement corrects the student?

A. The sugar is carried up through the tea by currents
The tea is still and cold, so no current carries the sugar; the sugar spreads even so.
B. Sugar molecules push one another apart until they are evenly spaced
Sugar molecules do not repel one another into an even pattern; each one moves on its own, at random.
C. ✓ Each sugar molecule moves at random; more leave the crowded bottom than come back
D. Water molecules pull the sugar toward the unsweetened regions
Water molecules do not pull sugar anywhere; water molecules collide with each sugar molecule at random from every side.

Why: No sugar molecule knows where the unsweetened tea is; every sugar molecule moves constantly and at random.
More sugar molecules are at the crowded bottom than in the tea above it.
So more sugar molecules leave the bottom than arrive from above.
That net movement is diffusion.

Q4 T25-q04

In a placenta, the mother’s blood and the baby’s blood flow past each other with only thin cell layers between them, and the two bloods never mix. Oxygen crosses these cell layers freely. Dissolved oxygen was measured at three points along one placental vessel; the results are in the table.

Dissolved oxygen in the two bloods at three points along the vessel.

What is the net movement of oxygen at each point?

A. ✓ Into the baby’s blood at 1 and 2; none at 3
B. Into the baby’s blood at all three points
At point 3 the two bloods hold the same 5.5 mg/L, so oxygen crosses equally both ways and there is no net movement.
C. Into the baby’s blood at 1; into the mother’s blood at 2; none at 3
At point 2 the mother’s blood still holds more oxygen, 6 mg/L against 5 mg/L, so the net movement is still into the baby’s blood, only smaller.
D. Into the baby’s blood at 1 only; none at 2 or 3
A smaller difference still gives a net movement; at point 2, 6 mg/L against 5 mg/L, more oxygen leaves the mother’s blood than returns to it.

Why: At point 1 the mother’s blood holds 8 mg/L of oxygen and the baby’s 3 mg/L; at point 2, 6 and 5 mg/L.
At both, the mother’s blood holds more, so more oxygen molecules leave it than arrive: net movement into the baby.
At point 3 both hold 5.5 mg/L.

Q5 T25-q05

Two chambers are separated by a membrane that a dye can cross. Both chambers hold the dye at 0.40 mol/L. Labeled dye molecules are followed for one minute: 85 cross from left to right and 85 cross from right to left. The concentrations stay at 0.40 mol/L throughout.

What do these counts show?

A. The dye molecules have stopped moving
Molecules never stop moving; 170 molecules crossed in one minute.
B. A gradient must still exist between the chambers
Both chambers are at 0.40 mol/L, so there is no gradient.
C. The counts must be faulty; nothing crosses at equal concentrations
Equal concentrations do not stop molecules crossing; they make the number crossing each way equal.
D. ✓ Molecules keep crossing both ways; the net movement is zero

Why: The two chambers hold the dye at the same concentration.
So the dye molecules keep crossing in both directions equally often: 85 each way in a minute.
So there is no net movement.
That is a dynamic equilibrium: constant crossing, zero net change.

Q6 T25-q06

Cells lining a fish’s gut hold amino acid Y at 3 mmol/L. Just after a meal the gut fluid holds Y at 12 mmol/L. Between meals the gut fluid holds Y at 0.5 mmol/L.

Which way is down the concentration gradient of Y at each time?

A. Into the cells at both times: a cell always has a gradient running into it
A gradient has no fixed direction toward a cell; it runs from the more concentrated region to the less, and between meals the cells are the more concentrated region.
B. ✓ Into the cells after the meal; out of the cells between meals
C. Out of the cells after the meal; into the cells between meals
After the meal the gut fluid, at 12 mmol/L, is the more concentrated region, so down the gradient runs from the fluid into the cells, not out.
D. Out of the cells at both times: the cells hold Y, so Y flows out
Holding Y does not fix the direction; after the meal the gut fluid holds four times as much Y as the cells, so down the gradient runs into the cells.

Why: Down a concentration gradient runs from more concentrated to less.
After the meal Y is 12 mmol/L in the gut fluid and 3 mmol/L in the cells, so down the gradient is into the cells.
Between meals Y is 0.5 mmol/L outside and 3 mmol/L inside, so down is out.

Q7 T25-q07

A leaf cell in bright light takes in carbon dioxide and gives off oxygen. Dissolved carbon dioxide is at 0.6 mg/L in the water film outside the cell and 0.2 mg/L inside, and it enters. Dissolved oxygen is at 12 mg/L inside the cell and 9 mg/L outside, and it leaves.

How should the movement of carbon dioxide and the movement of oxygen be classified?

A. ✓ Both crossings are passive transport
B. Carbon dioxide entering is active transport; oxygen leaving is passive transport
Carbon dioxide moves from 0.6 mg/L to 0.2 mg/L, down its gradient, and movement down a gradient uses no energy from the cell whether the substance is entering or leaving.
C. Carbon dioxide entering is passive transport; oxygen leaving is active transport
Oxygen moves from 12 mg/L inside to 9 mg/L outside, down its gradient; the cell made the oxygen, but it does not push the oxygen out.
D. Both crossings are active transport
Both substances move down their gradients, and a cell uses energy only to move a substance against its gradient; being in bright light changes nothing about that.

Why: Passive transport is movement down the gradient with no energy used by the cell.
Carbon dioxide moves from 0.6 mg/L outside to 0.2 mg/L inside: down its gradient.
Oxygen moves from 12 mg/L inside to 9 mg/L outside: down its gradient too.
So both are passive transport.

Q8 T25-q08

Substance A, a small nonpolar molecule, gets into protein-free bubbles of phospholipid bilayer within minutes; fructose, a sugar, never does. Living cells with fructose carriers take in both: A from 3 mmol/L outside to 1 mmol/L inside, and fructose from 6 mmol/L outside to 2 mmol/L inside. The cells’ ATP use stays the same while they do.

How do A and fructose enter the living cells?

A. Both by simple diffusion through the bilayer
Fructose never got into the protein-free bubbles, so fructose cannot cross the bilayer on its own.
B. ✓ Both passive; A straight through the bilayer, fructose through a carrier
C. A uses no energy; fructose needs energy, since it needs a protein
The cells’ ATP use stays the same while fructose enters, so the carrier uses no energy from the cell; fructose is moving down its gradient.
D. Fructose through the bilayer; A through a carrier
A got into bubbles that had no proteins at all, so A needs no carrier, and fructose did not get in, so fructose cannot use the bilayer.

Why: A moves from 3 mmol/L to 1 mmol/L and fructose from 6 mmol/L to 2 mmol/L: both down their gradients, with ATP use unchanged, so both are passive.
A got into the protein-free bubbles, so A crosses the bilayer itself: simple diffusion.
Fructose never did, so it uses a carrier: facilitated diffusion.

Q9 T25-q09

A bag of thin membrane is filled with a cloudy starch solution and lowered into a beaker of amber iodine solution. Iodine turns blue-black when it mixes with starch. After twenty minutes the inside of the bag is blue-black and the beaker is still amber.

Which statement explains the two colors?

A. Starch crossed out into the beaker; the amber color hides the reaction there
Starch reaching the beaker would have mixed with the iodine there and turned the beaker blue-black; the beaker stayed amber.
B. Iodine and starch both crossed; the starch simply moves more slowly
Starch crossing at any speed would have turned the beaker blue-black within twenty minutes; the beaker stayed amber, so no starch crossed.
C. Neither crossed; iodine changes color whenever starch is near it
Iodine changes color only when it mixes with starch, and the mixing happened inside the bag, so iodine must have crossed in.
D. ✓ Iodine crossed in; the membrane blocks starch, so diffusion could not even it out

Why: The inside of the bag turned blue-black, so iodine crossed into the bag and mixed with the starch.
The beaker stayed amber, so no starch crossed out.
The membrane passes the small iodine molecules and blocks the large starch molecules.
A solute keeps a gradient only where the membrane blocks it.

Q10 T25-q10

A muscle cell holds calcium ions (Ca²⁺) at 0.0001 mmol/L inside against 1.2 mmol/L outside. For a short interval every calcium channel in its membrane is shut, and no protein is moving Ca²⁺.

What happens to the Ca²⁺ gradient during the interval, and why?

A. It fades fast: Ca²⁺ is small enough to pass between the tails
Size is not what decides here; Ca²⁺ is charged, and water attracts a charged ion and keeps it out of the hydrophobic interior however small the ion is.
B. ✓ It holds: Ca²⁺ cannot cross the hydrophobic interior on its own
C. It holds: the Ca²⁺ ions have stopped moving
The Ca²⁺ ions keep moving constantly and at random; the gradient holds because they have no route across, not because they are still.
D. It reverses: Ca²⁺ floods in through the bilayer until the inside holds more than the outside
Ca²⁺ cannot cross the bilayer at all on its own, so nothing floods in while the channels are shut.

Why: Ca²⁺ is charged; water attracts a charged ion, and nothing in the hydrocarbon tails does.
So Ca²⁺ cannot pass the hydrophobic interior on its own.
Every calcium channel is shut, so Ca²⁺ has no other route.
So the gradient holds until a channel opens: the membrane is selectively permeable.

Q11 T25-q11

Cells of a kelp, a large seaweed, hold iodide ions (I⁻) at 30 mmol/L, while the seawater around them holds I⁻ at 0.0005 mmol/L, and the cells keep taking more in. When a drug lowers the cells’ ATP supply, iodide uptake falls by 90%.

How is iodide entering the kelp cells?

A. Simple diffusion
Iodide is an ion, so it cannot cross the phospholipid bilayer on its own, and it moves against its gradient, which no substance does by itself.
B. Facilitated diffusion
Facilitated diffusion moves a substance down its gradient, using no energy.
Iodide moves from 0.0005 to 30 mmol/L, against its gradient.
The uptake falls when the ATP supply is lowered.
C. Endocytosis
An iodide ion is small enough for a protein door; endocytosis wraps membrane around particles far too large for any channel or carrier.
D. ✓ Active transport

Why: Iodide moves from 0.0005 mmol/L outside to 30 mmol/L inside: against its concentration gradient.
When the drug lowers the ATP supply, the uptake falls by 90%.
So the cell uses energy from ATP to move the iodide in.
Using energy to move a substance against its gradient is active transport.

Q12 T25-q12

A nerve cell releases a chemical messenger at its tip by exocytosis, thousands of vesicles a minute. Measurements show that while this goes on, the area of the plasma membrane at the tip grows.

Why does the membrane area grow?

A. The cell makes new membrane at the tip to replace what the vesicles carry out of the cell
The vesicles do not leave the cell; each vesicle fuses with the plasma membrane and stays part of it.
B. ✓ Each vesicle's membrane fuses with the plasma membrane and stays there
C. Water follows the messenger out, and the tip swells and stretches
Exocytosis does not swell the cell with water.
D. The membrane thins and spreads as the vesicles push through it
A vesicle does not push through the membrane; it fuses with it.

Why: In exocytosis a vesicle inside the cell moves to the plasma membrane, fuses with it and releases its contents outside.
The vesicle’s own membrane becomes part of the plasma membrane.
So every fusion adds a little membrane at the tip.
Endocytosis, which pinches membrane off, does the reverse.

Q13 T25-q13

The figure shows a single-celled pond organism touching a food particle far too large for any channel or carrier, and the same cell a little later. When the organism’s ATP supply is blocked, particles stay stuck to the outside of the cell.

Panel 1: the cell touches a food particle. Panel 2: the same cell a little later. The extracellular fluid is shaded darker than the cytosol.

What is the process shown, and what does the ATP result show?

A. Facilitated diffusion; large particles just take longer
No protein door is anywhere near big enough for this particle, and facilitated diffusion would not stop when ATP is blocked.
B. Exocytosis; the ATP result shows fusion needs energy
Exocytosis runs the other way, releasing a vesicle’s contents outside.
C. ✓ Endocytosis; folding and pinching off membrane uses energy
D. Active transport through a pump; the pump stalls without ATP
A pump moves one ion or molecule at a time, and this particle is thousands of times too large.

Why: The plasma membrane folded inward around the food particle and pinched off.
So the particle is inside the cell, enclosed in a vesicle.
That is endocytosis.
The process stops when ATP is blocked.
So bending and pinching off membrane uses energy from the cell.

Q14 T25-q14

In a pancreatic cell, vesicles filled with insulin, a protein hormone, wait near the plasma membrane. After a signal, the vesicles move to the membrane, fuse with it, and the insulin appears in the blood. A student says: "That cannot be exocytosis. Exocytosis is how a cell gets rid of waste."

Which response is correct?

A. The student is right; insulin leaves through a carrier protein
Insulin is a protein, far too large for any carrier.
B. The student is right; insulin is small enough to cross the bilayer
Insulin is a large polar protein and cannot pass the hydrophobic interior.
C. It is endocytosis, because the vesicle joins the plasma membrane
Endocytosis takes material in; here a vesicle moves out to the membrane and releases its contents outside.
D. ✓ It is exocytosis; cells release useful products this way too

Why: Exocytosis is a vesicle inside the cell moving to the plasma membrane, fusing with it and releasing its contents outside.
The vesicle’s membrane becomes part of the plasma membrane.
Cells release waste this way, and useful products too, such as insulin and digestive proteins.
Exocytosis uses energy from the cell.

Q15 T25-q15

Liver cells hold glycerol, a small polar molecule, at 0.5 mmol/L. Cells are placed in solutions with glycerol at 0.1, 0.5 and 2.0 mmol/L, and the net movement of glycerol is recorded after one minute. Untreated cells: out, none, in. Cells treated with a drug that blocks one channel protein: almost none in every solution. The cells’ ATP use is the same in every trial.

By what mechanism does glycerol cross these membranes, and what in the results shows it?

A. ✓ Facilitated diffusion: down its gradient through a channel, using no energy
B. Simple diffusion: down its gradient straight through the bilayer
The channel blocker stopped the movement, so glycerol needs a channel; it is not crossing the bilayer on its own in these cells.
C. Active transport: a protein is needed, so the cell must be using ATP
Facilitated diffusion uses a protein with no energy used; the cells’ ATP use is the same in every trial, so the cell uses no energy.
D. Active transport: in the 0.1 mmol/L solution glycerol moves against its gradient
In the 0.1 mmol/L solution glycerol moves out, from 0.5 mmol/L inside to 0.1 mmol/L outside, which is down its gradient.

Why: Glycerol moves out at 0.1 mmol/L, in at 2.0 mmol/L, and not at 0.5 mmol/L: down its gradient every time.
The channel blocker stops the movement, so glycerol needs a channel protein.
ATP use is the same in every trial, so the cell uses no energy.
That is facilitated diffusion.

Q16 T25-q16

A single-celled organism living in a pond keeps its inside far saltier than the pond water. Its membrane has channels through which salt ions constantly leak out, down their gradient.

Which mechanism lets the cell keep its inside salty, and why?

A. Facilitated diffusion: channels carry the ions back in
Channels only let ions move down their gradient, and that is outward here.
B. Simple diffusion: the ions re-enter through the bilayer
Ions cannot cross the bilayer on their own, and a downhill crossing would run outward.
C. ✓ Active transport: pumps return the ions uphill, using ATP
D. No mechanism is needed: the leak stops once the inside is saltier than the outside
A leak down a gradient never stops while the gradient lasts; the saltier the inside, the faster the ions leak out.

Why: The leak runs down the gradient and uses no energy, so the leak can only lower the inside concentration.
To keep the inside salty, the cell must move ions back in against their gradient.
Only active transport does that: pumps, using ATP.

Q17 T25-q17

A dye that cannot cross any membrane is dissolved in the fluid around a white blood cell. The cell then takes in a bacterium by endocytosis.

Where is the dye found inside the cell afterward, and why?

A. Nowhere inside the cell: a dye that cannot cross a membrane stays outside
The dye did not need to cross a membrane: the membrane folded around a pocket of outside fluid and carried the pocket, dye and all, into the cell.
B. Throughout the cytosol: the membrane opened to let the bacterium in
The membrane never opens in endocytosis; it folds inward and pinches off, so nothing from outside reaches the cytosol directly.
C. ✓ Inside the vesicle with the bacterium: the closing pocket trapped some outside fluid
D. On the outer face of the plasma membrane only: the dye was left behind when the membrane folded in
The fluid touching the membrane where it folded in was carried into the pocket, not left behind.

Why: The plasma membrane folds inward around the bacterium, making a pocket.
The pocket holds fluid from outside the cell, and that fluid holds the dye.
The pocket pinches off as a vesicle, so the vesicle holds the bacterium and some outside fluid, dye included.
The dye never crosses a membrane.

FRQ 1 T25-frq1 · Conceptual Analysis

A cell in the gill of a marine fish was studied. The table gives results for three substances that cross its plasma membrane. The researchers also watched the cell’s membrane. When a bacterium far too large for any channel or carrier touches the cell, the membrane dents inward around the bacterium and then closes over it. Mucus made inside the cell appears in the seawater just after small sacs of membrane inside the cell reach the plasma membrane.

Three substances crossing the plasma membrane of a gill cell: the concentration of each in the seawater and in the cytosol, the direction each moves, whether each crosses protein-free bubbles of phospholipid bilayer, and what happens to each crossing when the cell’s ATP is blocked.

(a) Identify the mechanism by which oxygen enters the cell and the mechanism by which glucose enters the cell, and justify each identification with observations from the table. (1 pt)

Model answer Oxygen enters by simple diffusion, straight through the bilayer: it is at 0.20 mmol/L outside and 0.05 mmol/L inside, so it moves down its gradient, and it crosses protein-free bubbles just as readily, so it needs no protein.
Glucose enters by facilitated diffusion, through a carrier: it is at 8 mmol/L outside and 2 mmol/L inside, so it moves down its gradient; it does not enter protein-free bubbles, so it needs a protein; and blocking ATP leaves its rate unchanged, so no energy is used.

Rubric

Award 1 point for: oxygen enters by simple diffusion (it moves down its gradient, from 0.20 mmol/L to 0.05 mmol/L, and it crosses protein-free bubbles, so it needs no protein) AND glucose enters by facilitated diffusion (it moves down its gradient, from 8 mmol/L to 2 mmol/L; it does not cross protein-free bubbles, so it needs a protein; blocking ATP changes nothing, so the cell uses no energy). Both mechanisms must be named, each with at least one correct observation supporting it.
Accept: "passive transport through the bilayer" for oxygen and "passive transport through a carrier (or a membrane protein)" for glucose. Do not award the point if either substance is said to need ATP, or if glucose is said to cross the bilayer itself.

Slip Calling glucose's entry simple diffusion because glucose moves down its gradient, or calling it active transport because a protein is involved. The bubble result decides the route; the ATP result decides whether the cell uses energy.

(b) Identify the mechanism by which Cl⁻ leaves the cell, and justify your identification with two observations from the table. (1 pt)

Model answer Cl⁻ leaves by active transport, through a pump, because Cl⁻ is at 40 mmol/L inside and 550 mmol/L outside and the cell moves Cl⁻ out, toward the side where Cl⁻ is already more concentrated, so Cl⁻ is moved against its concentration gradient, which a substance left to itself never does; and because blocking the cell’s ATP supply stops the movement, so the cell is using energy from ATP to move Cl⁻.

Rubric

Award 1 point for: active transport (through a pump), justified by BOTH observations: Cl⁻ moves from 40 mmol/L to 550 mmol/L, against its concentration gradient, and the movement stops when the cell’s ATP supply is blocked, so the cell is using energy on it.
Accept: "the cell pumps Cl⁻ out" as the name. One observation alone does not earn the point; the direction against the gradient and the dependence on ATP are both needed.

Slip Naming active transport from the direction alone, or from the ATP result alone. A test item asks for both observations because either one on its own could be read another way.

(c) Identify the process that takes the bacterium into the cell and the process that releases the mucus, and describe what surrounds the bacterium once it is inside the cell. (1 pt)

Model answer The bacterium is taken in by endocytosis: the membrane dents inward around the bacterium and closes over it.
The mucus is released by exocytosis: each sac of membrane is a vesicle that fuses with the plasma membrane and opens outward, releasing its mucus into the seawater.
Inside the cell the bacterium is surrounded by a closed vesicle of membrane, pinched off from the plasma membrane.

Rubric

Award 1 point for: the bacterium is taken in by endocytosis, the mucus is released by exocytosis, and inside the cell the bacterium is enclosed in a closed vesicle of membrane pinched off from the plasma membrane (the plasma membrane is continuous again around the cell).
Accept: "phagocytosis" for the bacterium. Do not award the point if the bacterium is said to be loose in the cytosol, or if the two process names are swapped.

Slip Leaving the bacterium loose in the cytosol. The pocket that closed over it becomes a sealed vesicle, and the bacterium stays inside that vesicle.

(d) Explain why passive crossings alone would erase the difference between the concentration of Cl⁻ inside the cell and in the seawater, and how the cell keeps the difference. (1 pt)

Model answer Passive crossings can only run down a gradient.
Cl⁻ is far more concentrated in the seawater than in the cytosol, so on their own they would let Cl⁻ leak in, toward 550 mmol/L, and erase the difference.
By using energy the cell can move Cl⁻ against its gradient, back out into the seawater, so the inside stays at 40 mmol/L.
A difference between inside and outside is kept only by continual use of energy.

Rubric

Award 1 point for: passive crossings only run down a gradient, so on their own they would let Cl⁻ leak in toward 550 mmol/L and erase the difference between inside and outside; by using energy the cell moves Cl⁻ against its gradient, back out, and so keeps the inside at 40 mmol/L; a difference is kept only by continual use of energy.
Accept an answer framed in general terms (passive crossings even things out; active transport keeps them uneven) provided it says what the energy is used for: movement against the gradient.

Slip Saying the cell needs energy 'to move things across' without saying which way. Passive crossings run down the gradient on their own; the energy is used for movement against the gradient.

FRQ 2 T25-frq2 · Scientific Investigation

Liver cells are kept in a solution containing substance Z, a large polar molecule, at 8 mmol/L; inside the cells Z starts at 2 mmol/L. Temperature and the volumes of solution are the same in every trial. The concentration of Z inside the cells after twenty minutes is in the table.

Concentration of Z inside the liver cells after twenty minutes in each trial.

(a) Describe how a large polar molecule such as Z can cross a plasma membrane, and what prevents it from crossing the bilayer on its own. (1 pt)

Model answer A large polar molecule such as Z crosses the membrane by passing through a membrane protein: a channel or a carrier.
What keeps it from crossing the bilayer on its own is the hydrophobic interior: the hydrocarbon tails in the middle of the membrane carry no charges or partial charges.
Water attracts a polar molecule.
Nothing in the hydrophobic interior attracts a polar molecule.
So Z stays in the water and cannot cross the bilayer on its own.

Rubric

Award 1 point for: a large polar molecule crosses only through a membrane protein (a channel or a carrier), because the hydrocarbon tails in the middle of the membrane carry no charges or partial charges, so water attracts a polar molecule and nothing in the hydrophobic interior does.
Accept: "it needs a channel or carrier protein because it cannot dissolve into the oily middle", in any wording that names both the protein route and the reason.

Slip Saying Z is too big to fit between the phospholipids. Water attracts its polar groups, and that attraction is what keeps Z out of the oily middle.

(b) Identify the variable changed between trials, the variable measured, and the trial that serves as the control. (1 pt)

Model answer The variable changed is the treatment of the cells: none, ATP blocked, or one membrane protein blocked.
The variable measured is how much Z enters, the concentration of Z inside after twenty minutes.
The control is trial 1, the untreated cells.

Rubric

Award 1 point for: the changed (independent) variable is the treatment of the cells (none, ATP blocked, membrane protein blocked); the measured (dependent) variable is how much Z enters (the concentration of Z inside after twenty minutes); the control is trial 1, the untreated cells.
All three are needed for the point. Accept "the amount of Z inside" for the measured variable.

Slip Swapping the changed and measured variables, or naming trial 2 as the control. The control is the untreated cells the others are compared with.

(c) Describe one further trial that would test whether the blocked protein can move Z against its concentration gradient, and predict its result. (1 pt)

Model answer Load untreated cells with Z at 8 mmol/L and place them in a solution of Z at 2 mmol/L.
That reverses the gradient.
Prediction: Z leaves the cells through the protein until the two sides are equal.
That shows the protein cannot move Z uphill.
With equal Z inside and out, there would be no net movement.

Rubric

Award 1 point for: a trial that reverses the gradient, for example untreated cells loaded with Z at 8 mmol/L placed in a solution of Z at 2 mmol/L (or with equal Z inside and out), with the prediction that Z leaves the cells (net movement out) through the protein until the two sides are equal (or, with equal concentrations, that there is no net movement), showing the protein cannot move Z uphill.
Accept: any design that makes the gradient run the other way or removes it, with a prediction that net movement follows the gradient (down it, or none when the sides are equal). Also accept: untreated cells kept in the 8 mmol/L solution until the concentration inside stops changing, with the prediction that the inside levels off at 8 mmol/L and never rises above it. Do not award the point for repeating trial 2 or trial 3 unchanged.

Slip Repeating trial 2 or trial 3 unchanged. The new trial has to make the gradient run the other way, or remove it.

(d) Evaluate the claim that Z enters the cells by facilitated diffusion, using the results of all three trials. (1 pt)

Model answer The claim is supported.
Z moves from 8 mmol/L to 2 mmol/L, down its gradient, so no energy is needed.
Trial 2 confirms this: blocking ATP changes nothing, so the cell is not using energy.
Trial 3 shows that a membrane protein is required, so the crossing is not simple diffusion.
Down the gradient, through a protein, with no energy used by the cell is facilitated diffusion.
Therefore the claim fits all three trials.

Rubric

Award 1 point for: the judgement that the claim is supported AND the ground for it from the trials: Z moves down its gradient (8 to 2 mmol/L), so no energy is needed, and trial 2 confirms it (blocking ATP changes nothing); trial 3 shows a membrane protein is required, so this is not simple diffusion; down the gradient, through a protein, with no energy used by the cell is facilitated diffusion.
Accept: the three pieces (downhill, ATP not needed, protein needed) each tied to its trial. Do not award the judgement alone, or a judgement grounded on only one trial.

Slip Judging from one trial. Each of the three pieces, downhill, no ATP needed and protein needed, comes from its own trial.

APBIO-U02-L09 Red cells in three beakers

Topic 2.7a · Tonicity and Osmoregulation · 81 steps

Three beakers of salt water, each with the same red blood cells: swollen in the first, normal in the second, shriveled in the third

Here are three beakers of salt water, and three samples of the same red blood cells. In one beaker they swell, in one they keep their shape, in one they shrivel.

The student added nothing to the cells and took nothing from them. The only difference is the water they were dropped into.

Unit 2 · Cell Structure and Function

1Water crosses; the sugar cannot

Video: Watch first: Tonicity and osmoregulation

Three beakers of salt water, the same red blood cells, three different fates.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T27-intro.mp4

Video: Watch: Water crosses; the sugar cannot

Why a bag of sugar solution in pure water gets heavier: the concentration of water, and a net movement.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L09.mp4

A dialysis bag is a bag of thin membrane. Fill one with 0.40 mol/L sucrose, table sugar, and hang it in a beaker of pure water.

Two panels: a bag of sucrose solution hanging in pure water, and the same bag an hour later, swollen, with arrows showing water entering

An hour later the bag is heavier and fatter. Water has gone in. The membrane lets water through but not sucrose, so the sucrose is still all inside.

Look at the water itself. A liter of pure water is all water. A liter of sucrose solution has sucrose taking up some of the room, so the concentration of water in it is lower.

Two boxes of equal size: one full of water molecules only, one holding sucrose molecules with fewer water molecules around them

Where there is more solute, the concentration of water is lower. Outside the bag, the water is more concentrated. Inside the bag, the water is less concentrated.

Here is why a substance spreads. Particles move constantly and at random. So more of them leave a region where they are more concentrated than arrive from a region where they are less concentrated.

That gives a net movement from the more concentrated region to the less concentrated one.

Here the particles are water molecules. More of them cross into the bag than cross out of it: a net movement of water toward the sucrose.

A membrane between pure water on the left and a sucrose solution on the right; sucrose molecules are stopped at the membrane while a large arrow carries water from left to right

The sucrose cannot leave to even things out. Only the water moves. The water keeps moving until the concentration of water is the same on both sides, or until something stops the water. Here the stretched bag stops the water.

The bag’s membrane lets some particles through and holds others back: it is selectively permeable.

This net movement of water across a selectively permeable membrane, from the side with less solute to the side with more, is called .

Osmosis moves the water, not the solute. The solute is the thing that cannot cross.

What you are expected to know Explain osmosis: when a membrane lets water through but not a solute, more water crosses from the side with less solute than crosses back, a net movement of water toward the side with more solute.

Check q1

A bag of 0.20 mol/L sucrose hangs in a beaker of 0.60 mol/L sucrose. The bag’s membrane lets water through and holds sucrose back.

Which of the following is the net movement?

A. Water moves into the bag
Water moves toward the side with more solute.
B. Sucrose moves into the bag
The membrane holds sucrose back, so the sucrose stays where it is.
C. ✓ Water moves out of the bag
D. Nothing moves
The membrane lets water through, and water molecules cross it constantly.

Why: There is more sucrose outside the bag than inside.
So the concentration of water is lower outside the bag than inside.
Therefore more water molecules cross out of the bag than cross into it.
This means the net movement of water is out of the bag, toward the side with more solute.

Check q2

A bag of 0.20 mol/L sucrose hangs in a beaker of 0.60 mol/L sucrose. The bag’s membrane lets water through and holds sucrose back. The net movement of water is out of the bag.

Why do more water molecules cross out of the bag than into it?

A. The sucrose molecules outside the bag pull the water molecules toward them
Sucrose pulls nothing; each water molecule moves at random.
B. The stretched bag squeezes its water out through the membrane
A bag with water leaving is going slack, not squeezing; each water molecule moves at random.
C. ✓ Water is less concentrated outside the bag, so fewer water molecules cross in than out

Why: Water molecules move constantly and at random.
Inside the bag the water is more concentrated, so many water molecules reach the membrane and cross out.
Outside, sucrose takes up some of the room, so the water is less concentrated and fewer cross in.
So more cross out than in.

Check q3

Which of the following movements is osmosis?

A. ✓ Water crossing from where solute is 0.10 mol/L to where it is 0.40 mol/L
B. Glucose crossing through a carrier from 8 mmol/L to 2 mmol/L
Osmosis is a movement of water.
Glucose moving down its gradient through a carrier is facilitated diffusion.
C. Oxygen crossing a bilayer from where it is more concentrated to where it is less
Osmosis is a movement of water.
Oxygen crossing a bilayer is simple diffusion.
D. Water crossing both ways at equal rates between two equal solutions
Between two equal solutions, water crosses both ways at equal rates, so there is no net movement, and osmosis is a net movement.

Why: Osmosis is the net movement of water across a membrane.
The water moves toward the side with more solute.
Only water crossing from 0.10 mol/L to 0.40 mol/L of solute is a net direction toward more solute.

19Three beakers, three words

Video: Watch: Three beakers, three words

Hypotonic, hypertonic, isotonic: the solution named relative to the cell, and what the cell does in each.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L09b.mp4

Back to the red blood cells. All three samples start at the same volume, which we call 100%. The three beakers hold NaCl at 0.10, 0.15 and 0.30 mol/L.

Three beakers of NaCl solution at 0.10, 0.15 and 0.30 mol/L, with red cells at 120, 90 and 60 femtoliters, and arrows showing water entering, crossing both ways, or leaving

In the first beaker, 0.10 mol/L, there is less solute outside than inside the cells. Water moves in by osmosis and the cells swell, to about 133% of their starting volume.

In the third beaker, 0.30 mol/L, there is more solute outside than inside the cells. Water moves out and the cells shrink, to about 67% of their starting volume.

In the second beaker, 0.15 mol/L, the solute outside matches the cells’ own. Water crosses both ways at equal rates, so there is no net movement and the volume stays constant at 100%.

Why does 0.15 mol/L NaCl match the cells? Table salt, NaCl, splits into two ions when it dissolves: sodium ions (Na⁺) and chloride ions (Cl⁻).

So a 0.15 mol/L NaCl solution holds 0.30 mol/L of dissolved particles, the same particle concentration as the cell’s cytosol. So water crosses both ways at equal rates.

In that second beaker the water has not stopped. The water is crossing constantly, in and out at the same rate. This is dynamic equilibrium.

Now the words. When a solution has less total solute than the cell in it, we call the solution to the cell, because hypo- means less. Water enters and the cell swells.

When a solution has more total solute than the cell, we call it to the cell, because hyper- means more. Water leaves and the cell shrinks.

When a solution has the same total solute as the cell, we call it to the cell, because iso- means the same. Water crosses both ways at equal rates and the volume stays constant.

Which of the three a solution is, relative to a cell, is called its .

The total concentration of dissolved particles that decides tonicity is called the of the solution.

The words describe the solution relative to the cell: the 0.10 mol/L beaker is hypotonic to the red cells, because it holds less solute than the cells, not the other way around.

To name the tonicity, compare the solute concentration of the solution with the solute concentration of the cell. Less than the cell is hypotonic. More than the cell is hypertonic. The same as the cell is isotonic.

What you are expected to know Classify the solution around a cell as hypotonic, hypertonic or isotonic from its total solute concentration relative to the cell’s, and say whether the cell swells, shrinks or keeps its volume constant.

36Fluency quiz: hypotonic, hypertonic or isotonic? mixed practice

Check q4

The bar chart shows the concentration of solute in a solution and in the red blood cells placed in it.

A bar chart with two bars, labelled in the solution and in the cell, showing the concentration of solute in mol/L; each bar carries its value above it

Which word describes the solution relative to the cells?

A. ✓ Hypotonic
B. Hypertonic
The solution bar is shorter than the cell bar, so the solution holds less solute than the cells: hypotonic.
C. Isotonic
The solution bar is shorter than the cell bar, so the solution holds less solute than the cells: hypotonic.

Why: Hypo- means less, hyper- means more, iso- means the same.
The solution bar is shorter than the cell bar: 0.10 mol/L against 0.30 mol/L.
So the solution has less solute than the cells.
So the solution is hypotonic to the cells.

Check q5

The bar chart shows the concentration of solute in a solution and in the red blood cells placed in it.

Which word describes the solution relative to the cells?

A. Hypotonic
The solution bar is taller than the cell bar, so the solution holds more solute than the cells: hypertonic.
B. ✓ Hypertonic
C. Isotonic
The solution bar is taller than the cell bar, so the solution holds more solute than the cells: hypertonic.

Why: Hypo- means less, hyper- means more, iso- means the same.
The solution bar is taller than the cell bar: 0.50 mol/L against 0.30 mol/L.
So the solution has more solute than the cells.
So the solution is hypertonic to the cells.

Check q6

The bar chart shows the concentration of solute in a solution and in the red blood cells placed in it.

Which word describes the solution relative to the cells?

A. Hypotonic
The two bars are the same height, so the solution holds the same solute as the cells: isotonic.
B. Hypertonic
The two bars are the same height, so the solution holds the same solute as the cells: isotonic.
C. ✓ Isotonic

Why: Hypo- means less, hyper- means more, iso- means the same.
The two bars are the same height: 0.30 mol/L in each.
So the solution has the same solute concentration as the cells.
So the solution is isotonic to the cells.

Check q7

The bar chart shows the concentration of solute in a solution and in the red blood cells placed in it.

Which way does water move?

A. ✓ Into the cells
B. Out of the cells
The solution bar is shorter than the cell bar.
C. Neither: water crosses both ways at equal rates
The two bars are different heights, so the two sides differ and water has a net direction.

Why: Water moves by osmosis toward the side with more solute.
The solution bar is shorter than the cell bar: 0.20 mol/L against 0.30 mol/L.
So the cells hold more solute than the solution.
So water moves into the cells, and the cells swell.

Check q8

The bar chart shows the concentration of solute in a solution and in the red blood cells placed in it.

Which way does water move?

A. Into the cells
The solution bar is taller than the cell bar.
B. ✓ Out of the cells
C. Neither: water crosses both ways at equal rates
The two bars are different heights, so the two sides differ and water has a net direction.

Why: Water moves by osmosis toward the side with more solute.
The solution bar is taller than the cell bar: 0.45 mol/L against 0.30 mol/L.
So the solution holds more solute than the cells.
So water moves out of the cells, and the cells shrink.

Check q9

The fluid dripped into a patient’s vein holds the same total solute as the patient’s blood cells.

Which word describes the fluid relative to those cells?

A. Hypotonic
Hypo- means less solute; the fluid holds the same total solute as the cells.
B. Hypertonic
Hyper- means more solute; the fluid holds the same total solute as the cells.
C. ✓ Isotonic

Why: Hypo- means less, hyper- means more, iso- means the same.
The fluid holds the same total solute as the cells.
So the fluid is isotonic to the cells.
Water crosses both ways at equal rates and the cells keep their volume.

Check q10

Pond water holds less total solute than the cells of a frog’s skin.

Which statement is correct?

A. ✓ The pond water is hypotonic to the skin cells
B. The skin cells are hypotonic to the pond water
The skin cells hold more solute than the pond water, so the skin cells are hypertonic to the pond water, not hypotonic.
C. The pond water is hypertonic to the skin cells
Hyper- means more, and the pond water has less total solute than the skin cells.

Why: The pond water has less total solute than the skin cells.
Hypo- means less.
So the pond water is hypotonic to the skin cells.

44What the cell does in each

Once you have named the solution, the cell’s response follows. In a hypotonic solution water enters and the cell swells. In a hypertonic solution water leaves and the cell shrinks. In an isotonic solution the volume stays constant.

What you are expected to know Predict whether a cell swells, shrinks or keeps its volume constant from the tonicity of the solution around it, and say what the water is doing in each case.

Check q11

Red blood cells sit in a solution that is hypertonic to them.

What happens to the cells?

A. Water enters the cells and they swell
A hypertonic solution has more solute than the cells, so the net movement of water is out of the cells.
B. Water stops crossing and the cells keep their volume constant
Water never stops crossing the membrane; here more water leaves the cells than enters.
C. Solute enters the cells and they swell
The membrane holds the solute back, so the solute does not enter; water is what moves.
D. ✓ Water leaves the cells and they shrink

Why: Hyper- means more.
A hypertonic solution has more total solute than the cells.
Water moves by osmosis toward the side with more solute.
So water moves out of the cells.
Therefore the cells shrink.

Check q12

Cells sit in an isotonic solution and keep their volume for an hour.

What is the water doing during that hour?

A. ✓ Crossing in both directions at equal rates
B. Not crossing the membrane at all
Water molecules keep crossing the membrane in both directions; only the net movement is zero.
C. Crossing inward only, slowly
A net movement of water inward would make the cells swell, and the cells keep their volume.

Why: Isotonic means equal total solute on the two sides.
Water crosses the membrane constantly in both directions.
The two rates are equal.
So there is no net movement of water.
Therefore the volume of the cells stays constant.

49Only solutes that cannot cross decide the tonicity

Video: Watch: Only solutes that cannot cross decide the tonicity

Sucrose and urea at the same concentration, and the different things they do to the same cells.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L09c.mp4

Two more beakers, each with the same red cells. One holds 0.50 mol/L sucrose, the other 0.50 mol/L urea, a small molecule. Same total solute in both.

Two rows of red cells at 0, 2 and 12 minutes: in 0.50 mol/L sucrose the cell shrinks and stays shrunken; in 0.50 mol/L urea it shrinks and then swells again

After two minutes the cells in both beakers have shrunk, from 100% of their starting volume to 72%. So far the two beakers behave alike.

By twelve minutes the two beakers differ. In sucrose the cells are still at 72% of their starting volume. In urea they have swollen again, past their starting size, to 118%.

The difference is the solute. Sucrose cannot cross the membrane. Urea can cross the membrane, so urea diffuses into the cells until it is equal on both sides.

Once urea is equal on both sides, urea no longer drives osmosis. The solution outside now holds no solute that cannot cross, so the cells’ own solute makes the inside the side with more solute. So water moves in and the cells swell.

Only solutes that cannot cross the membrane decide the tonicity of a solution. A solute that crosses freely spreads until it is equal on both sides and then drives no net movement of water.

So two solutions with the same total solute concentration can do different things to a cell. Equal total solute does not mean isotonic.

What you are expected to know Say that only solutes that cannot cross the membrane decide the tonicity of a solution, and predict what a cell does in a solution whose solute crosses freely.

Check q13

Two bags each hold 0.40 mol/L sucrose. Bag X hangs in 0.40 mol/L urea; bag Y hangs in 0.40 mol/L sucrose. Water and urea cross the membrane; sucrose stays put.

After an hour, which bag has gained water?

A. Y only
Bag Y has the same sucrose inside and outside, so water crosses bag Y’s membrane at equal rates and bag Y gains nothing.
B. ✓ X only
C. Both bags
Bag Y’s sucrose inside is matched by sucrose outside, so there is no net movement of water into bag Y.
D. Neither bag
Only solutes that cannot cross the membrane decide the tonicity; urea crosses, so urea no longer matters, and the two bags are not in the same situation.

Why: Urea crosses bag X’s membrane and spreads until equal on both sides; equal urea drives no osmosis.
That leaves 0.40 mol/L sucrose inside bag X and none outside, so water moves into bag X.
Bag Y has 0.40 mol/L sucrose inside and outside, so bag Y gains no water.

Check q14

Two bags each hold 0.30 mol/L sucrose. Bag P hangs in 0.30 mol/L glycerol, a small molecule that crosses the membrane; bag Q hangs in pure water. Sucrose stays put.

After an hour, which bag has gained water?

A. P only
Bag Q hangs in pure water with 0.30 mol/L sucrose inside, so water moves into bag Q too.
B. Q only
Glycerol crosses into bag P until it is equal on both sides, and then bag P’s sucrose is unmatched outside, so water moves into bag P too.
C. ✓ Both bags
D. Neither bag
Both bags hold sucrose that nothing outside matches once the glycerol has spread, so water moves into both.

Why: Glycerol crosses bag P’s membrane and spreads until it is equal on both sides.
So the glycerol no longer drives osmosis.
Nothing outside then matches bag P’s sucrose, so water moves into bag P.
Bag Q holds sucrose and hangs in pure water, so water moves into bag Q.
Therefore both bags gain water.

61Which way, and how much

Video: Watch: Which way, and how much

One red cell in three solutions, and what pure water does to it: osmotic lysis.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L09d.mp4

Water moves by osmosis from the hypotonic side toward the hypertonic side: from lower solute concentration toward higher.

Water does not move toward the side with more water overall. Water moves toward the side with more solute per liter.

A red cell holds about 0.30 mol/L of solute that cannot cross. Put the cell in 0.10 mol/L and water moves in, so the cell swells. Move the cell to 0.40 mol/L and water moves out, so the cell shrinks.

Put the cell in pure water. Water moves in and keeps moving in, because pure water can never match the solute inside the cell. The cell swells until its membrane tears.

A red cell in pure water at three stages: normal, swollen, and burst open

The bursting of an animal cell in a strongly hypotonic solution is called .

What you are expected to know Predict the direction of net water movement and the change in a cell’s volume from the solute concentrations inside and out, including bursting in a strongly hypotonic solution.

Check q15

A red cell with 0.30 mol/L of non-crossing solute inside sits in 0.10 mol/L of the same solute. As the cell starts to change, it moves into 0.40 mol/L of the same solute.

What does the cell do, in order?

A. Shrinks, then swells
In 0.10 mol/L the solute outside is less than inside, so water enters; in 0.40 mol/L the solute outside is more than inside, so water leaves.
B. ✓ Swells, then shrinks
C. Holds its volume, then shrinks
At 0.10 mol/L outside against 0.30 mol/L inside, water moves in and the cell swells before the solution is changed.
D. Swells, then swells further
Once the outside is 0.40 mol/L, more than the 0.30 mol/L inside, the net movement of water reverses and the cell shrinks.

Why: First the outside is 0.10 mol/L and the inside 0.30 mol/L.
The inside has more solute, so water moves in and the cell swells.
Then the outside is 0.40 mol/L and the inside 0.30 mol/L.
The outside has more solute, so water moves out and the cell shrinks.

Check q16

A red cell with 0.30 mol/L of non-crossing solute inside falls into pure water.

What happens to the cell?

A. The cell shrinks as its water leaves
There is more solute inside the cell than outside, so water moves in.
B. The cell keeps its volume constant
Water moves toward the side with more solute, and that side is the inside of the cell, so water enters.
C. The cell swells until the two sides are equal, then stays constant
Pure water outside can never match the solute inside, so the two sides never become equal and water keeps entering.
D. ✓ The cell swells until its membrane tears and it bursts

Why: Pure water has no solute; the cell holds 0.30 mol/L.
So water keeps moving into the cell.
Pure water can never match the solute inside, so the water never stops entering.
The membrane can stretch only so far, so it tears and the cell bursts: osmotic lysis.

The first beaker had less salt than the cells, so water moved in. The third beaker had more salt than the cells, so water moved out. The second beaker matched the cells, so water crossed both ways at equal rates.

72Mastery quiz: osmosis and tonicity, mixed mixed practice

Check q17

A cell holds 0.30 mol/L of solute that its membrane holds back. It sits in a 0.50 mol/L solution of the same solute.

Which way does water move, and why?

A. Into the cell, because the cell holds less solute
The solution holds 0.50 mol/L and the cell 0.30 mol/L, so the side with more solute is outside.
B. ✓ Out of the cell, because the solution holds more solute
C. Neither way, because both hold solute
The two concentrations differ, so water has a net direction.

Why: Water moves by osmosis toward the side with more solute.
The solution holds 0.50 mol/L and the cell holds 0.30 mol/L.
So the side with more solute is outside the cell.
Therefore water moves out of the cell.

Check q18

A solution holds 0.10 mol/L of solute; the cells placed in it hold 0.30 mol/L.

Which word describes the solution relative to the cells?

A. Hypertonic
Hyper- means more solute, and this solution holds less solute than the cells.
B. Isotonic
Iso- means the same solute, and 0.10 mol/L is less than 0.30 mol/L.
C. ✓ Hypotonic

Why: The solution holds 0.10 mol/L and the cells hold 0.30 mol/L.
So the solution holds less solute than the cells.
A solution with less solute than the cell is hypotonic to the cell.

Check q19

A cell is placed in a solution that is isotonic to it.

What is the water doing at the membrane?

A. ✓ Crossing in both directions at equal rates
B. Not crossing at all
Water molecules never stop crossing a membrane; only the net movement is zero.
C. Crossing inward only
A net inward movement would make the cell swell, and in an isotonic solution the volume stays constant.

Why: Isotonic means the solution holds the same solute as the cell.
Water molecules cross the membrane constantly in both directions.
The two rates are equal, so there is no net movement.
Therefore the cell’s volume stays constant.

Check q20

A red blood cell holds 0.30 mol/L of solute that cannot cross its membrane. Its membrane can stretch to hold 50% more volume before it tears. It falls into 0.25 mol/L of the same solute.

What happens to the cell?

A. It shrinks until the two sides match, then keeps its volume constant
The solution holds less solute than the cell, so water moves into the cell, not out of it.
B. ✓ It swells until the two sides match, then keeps its volume constant
C. It swells until its membrane tears
Diluting 0.30 mol/L to 0.25 mol/L takes 20% more volume, well inside the 50% the membrane can take, so the swelling stops before the membrane tears.

Why: The cell holds 0.30 mol/L and the solution 0.25 mol/L, so water moves into the cell.
As water enters, the cell’s contents are diluted.
When the contents reach 0.25 mol/L, the two sides match.
So the net movement of water stops, and the cell keeps its new volume.

Check q21

Cells hold 0.30 mol/L of solute that the membrane holds back. They sit in 0.30 mol/L urea, a small molecule that crosses the membrane.

What do the cells do over the next half hour?

A. They keep their volume constant, because the totals match
Equal total solute is not isotonic when one solute crosses the membrane.
B. They shrink and stay shrunken for the whole half hour
Urea enters the cells until it is equal on both sides, so the early shrinking reverses.
C. ✓ They swell, because urea spreads until it is equal on both sides

Why: Urea crosses the membrane and spreads until it is equal on both sides.
Equal urea drives no osmosis.
The cells’ own solute cannot cross, and nothing outside matches it.
So the inside is the side with more solute that cannot cross.
Therefore water moves in and the cells swell.

Check q22

Red blood cells sit in a solution that holds more solute than they do, and the cells are shrinking.

What are the water molecules doing at the membrane?

A. Crossing out of the cells only
Water molecules cross the membrane in both directions all the time; the cells shrink because more cross out than in.
B. ✓ Crossing both ways, but more cross out than in
C. Crossing both ways at equal rates
Equal rates give no net movement, and the cells are shrinking, so more water is leaving than entering.

Why: Water molecules move constantly and at random, so some cross the membrane in each direction.
Outside the cells, solute takes up some of the room, so the water is less concentrated there.
So fewer water molecules cross in than cross out.
Therefore the net movement is out, and the cells shrink.

Check q23

Sea water holds far more solute than the cells of a swimmer’s skin.

Which statement is correct?

A. ✓ The sea water is hypertonic to the skin cells
B. The skin cells are hypertonic to the sea water
The skin cells hold less solute than the sea water, so the skin cells are hypotonic to the sea water, not hypertonic.
C. The sea water is isotonic to the skin cells
Iso- means the same solute, and the sea water holds far more.

Why: The sea water holds more solute than the skin cells.
A solution with more solute than the cell is hypertonic to the cell.
So the sea water is hypertonic to the skin cells.

Practice writing an answer

Sheep red blood cells hold about 0.30 mol/L of solute that cannot cross their membranes. A student puts some cells into each of three dishes: dish 1 holds 0.10 mol/L sucrose; dish 2 holds 0.30 mol/L sucrose; dish 3 holds 0.30 mol/L urea. Sucrose cannot cross the membrane; urea is a small molecule that crosses it.

(a) Identify the tonicity of the solution in dish 1 relative to the cells, and describe what happens to the cells there. (1 pt)

Frame Dish 1 is … to the cells, because …; so the cells …

Model answer Dish 1 is hypotonic to the cells, because it holds less solute than the cells.
Water moves toward the side with more solute, so water moves into the cells.
So the cells swell.

Rubric

Award 1 point for: hypotonic, because dish 1 holds less solute (0.10 mol/L) than the cells (0.30 mol/L), AND the cells swell as water moves in.
Do not award the point for the label alone with no reason, or for the cells shrinking.

Slip Naming the tonicity with no reason. The point needs the comparison of the two solute concentrations.

(b) Make a claim about what the cells in dish 3 look like after an hour, compared with the cells in dish 2. (1 pt)

Model answer After an hour the cells in dish 3 are swollen, larger than the cells in dish 2.
The cells in dish 2 keep their volume constant.

Rubric

Award 1 point for: the claim that the dish-3 cells are swollen (larger than at the start) while the dish-2 cells keep their volume constant. No reasoning is required for this point.
Do not award a claim that the two dishes look the same because both hold 0.30 mol/L.

Slip Treating dish 3 like dish 2 because the totals match. A solute that crosses the membrane does not decide the tonicity.

(c) Support your claim, using what each solute does at the membrane. (1 pt)

Model answer Sucrose cannot cross the membrane, so in dish 2 the 0.30 mol/L outside matches the 0.30 mol/L inside and water crosses both ways at equal rates.
Urea crosses the membrane, so in dish 3 urea enters the cells until it is equal on both sides.
Equal urea drives no osmosis.
The cells’ own solute cannot cross, and nothing in dish 3 matches it.
So water moves into the dish-3 cells, and they swell.

Rubric

Award 1 point for: the evidence AND the reasoning: sucrose cannot cross, so dish 2 matches the cells and water crosses at equal rates; urea crosses until equal on both sides and then drives no osmosis, leaving the cells’ own solute unmatched, so water enters the dish-3 cells.
Do not award the point for evidence about urea crossing with no link to why water then enters.

Slip Saying urea crosses without saying what that leaves unmatched. The point needs the link from urea spreading to water entering.

Glossary

osmosis: The net movement of water across a selectively permeable membrane, from the side with less solute to the side with more.
hypotonic: Having less solute that cannot cross the membrane than the cell it surrounds, so that water enters the cell and the cell swells.
hypertonic: Having more solute that cannot cross the membrane than the cell it surrounds, so that water leaves the cell and the cell shrinks.
isotonic: Having the same solute that cannot cross the membrane as the cell it surrounds, so that water crosses both ways at equal rates and the cell’s volume stays constant.
tonicity: Whether a solution is hypotonic, hypertonic or isotonic relative to a cell. Only solutes that cannot cross the membrane decide it.
osmolarity: The total concentration of dissolved particles in a solution.
osmotic lysis: The bursting of an animal cell in a strongly hypotonic solution, when water entering by osmosis swells it past what its membrane can hold.

APBIO-U02-L10 Why celery goes limp

Topic 2.4 · Membrane Permeability · 93 steps

A crisp celery stalk snapping in two, a limp celery stalk bending, and an onion skin cell drawn with its cell wall and its plasma membrane

Here is a fresh celery stalk snapping, and beside it a stalk left on the counter overnight, bending. Nothing has been added or taken away but water.

Next to them is one cell from the skin of an onion, drawn with its two boundaries: a cell wall on the outside and a membrane inside it.

Unit 2 · Cell Structure and Function

1Two boundaries, one selector

Video: Watch: Two boundaries, one selector

Which cells have a cell wall: bacteria, archaea, fungi and plants, and not animal cells.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L10.mp4

A plant cell has a stiff layer outside its membrane, the cell wall, that holds the cell’s shape. Here it is on the onion cell.

A plant cell drawn with a thick outer cell wall and, just inside it, the plasma membrane around the cytosol

Plants are not alone. Bacteria, single-celled organisms, have a cell wall too, and so do the cells of fungi, such as yeast.

Five cells in a row: a bacterium, an archaeon, a yeast cell and a plant cell each drawn with an outer cell wall, and an animal cell with a membrane only, labelled no cell wall

Many cells keep their DNA inside a body wrapped in a membrane: the nucleus. A bacterium has no nucleus; its DNA lies in the cytosol.

A second group of single-celled organisms with no nucleus is called the . Their cells have a cell wall as well. Four groups with cell walls: bacteria, archaea, fungi and plants.

Animal cells have no cell wall. Their membrane is their only boundary.

The cell wall is a rigid boundary. The cell wall gives the cell a fixed shape. The cell wall also holds back some large substances that would otherwise reach the membrane.

What you are expected to know Say which four groups of organisms have cells with a cell wall outside the plasma membrane, and what the cell wall does.

10Fluency quiz: does this cell have a cell wall? mixed practice

Check q1

A cell from a fern leaf.

Does this cell have a cell wall?

A. ✓ Yes
B. No
A fern is a plant, and plants are one of the four groups with cell walls: bacteria, archaea, fungi and plants.

Why: Bacteria, archaea, fungi and plants have cell walls.
A fern is a plant.
So a fern leaf cell has a cell wall.

Check q2

A cell from a cat’s liver.

Does this cell have a cell wall?

A. Yes
A cat is an animal, and animals are not among the four groups with cell walls: bacteria, archaea, fungi and plants.
B. ✓ No

Why: Bacteria, archaea, fungi and plants have cell walls.
A cat is an animal, and animals are not in the list.
So a cat’s liver cell has no cell wall; its membrane is its only boundary.

Check q3

A bacterium from a person’s gut.

Does this cell have a cell wall?

A. ✓ Yes
B. No
Bacteria are one of the four groups with cell walls: bacteria, archaea, fungi and plants.

Why: Bacteria, archaea, fungi and plants have cell walls.
A bacterium is in the list.
So the gut bacterium has a cell wall.

Check q4

An archaeon from a salt lake.

Does this cell have a cell wall?

A. ✓ Yes
B. No
An archaeon is one of the archaea, one of the four groups with cell walls: bacteria, archaea, fungi and plants.

Why: Bacteria, archaea, fungi and plants have cell walls.
An archaeon is one of the archaea, which are in the list.
So the salt-lake archaeon has a cell wall.

Check q5

A cell from a sparrow’s flight muscle.

Does this cell have a cell wall?

A. Yes
A sparrow is an animal, and animals are not among the four groups with cell walls: bacteria, archaea, fungi and plants.
B. ✓ No

Why: Bacteria, archaea, fungi and plants have cell walls.
A sparrow is an animal, and animals are not in the list.
So the muscle cell has no cell wall; its membrane is its only boundary.

Check q6

A cell of the fungus that causes athlete’s foot.

Does this cell have a cell wall?

A. ✓ Yes
B. No
Fungi are one of the four groups with cell walls: bacteria, archaea, fungi and plants.

Why: Bacteria, archaea, fungi and plants have cell walls.
A fungus is in the list.
So the athlete’s-foot fungus cell has a cell wall.

Check q7

A human red blood cell.

Does this cell have a cell wall?

A. Yes
A human is an animal, and animals are not among the four groups with cell walls: bacteria, archaea, fungi and plants.
B. ✓ No

Why: Bacteria, archaea, fungi and plants have cell walls.
A human is an animal, and animals are not in the list.
So a red blood cell has no cell wall; its membrane is its only boundary.

18Which boundary does the selecting

Video: Watch: Which boundary does the selecting

Strip the cell wall away and the same solutes get in: the plasma membrane does the selecting.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L10b.mp4

A cell wall and a membrane both stand between the outside and the cytosol. Only one of them chooses what enters. To find which, take the cell wall away and see whether the choice changes.

The cell wall does not choose what enters the cell. Root cells with their cell walls took in 72% of solute 1 and 4% of solute 2. The same cells with the cell wall stripped off and the membrane intact took in 70% and 3%.

Two root cells: one with its cell wall and membrane, one with the cell wall removed and the membrane intact, each labelled with how much of solute 1 and solute 2 got in

The numbers barely move. The membrane inside the cell wall is still doing the selecting, as it always has.

What you are expected to know Decide from uptake data with and without the cell wall that the plasma membrane, not the cell wall, decides which solutes enter.

Check q8

Leaf cells with their cell walls take in 65% of solute X and 6% of solute Y. The same cells with their cell walls removed and their membranes intact take in 63% and 5%.

Which boundary decides which solute gets in?

A. The cell wall
If the cell wall chose, removing the cell wall would have changed which solute got in, and both numbers barely moved.
B. Neither boundary
Far more X got in than Y in both cases, so one boundary is selecting, and that boundary is still there without the cell wall.
C. The cell wall for X and the membrane for Y
The pattern, much X and little Y, is the same with the cell wall and without the cell wall, so one boundary is doing all of the selecting.
D. ✓ The plasma membrane

Why: With the cell wall in place, much X got in and little Y.
With the cell wall gone, the pattern is the same.
So the cell wall was not doing the selecting.
The plasma membrane is the boundary that remains, so the plasma membrane decides which solute gets in.

Check q9

Leaf cells with their cell walls take in 65% of solute X and 6% of solute Y; the same cells with their cell walls removed take in 63% and 5%.

Which observation shows which boundary does the selecting?

A. The walled cells took in slightly more of both solutes
Selecting means letting one solute in and holding the other out; the walled cells took in slightly more of both, which is not a choice between them.
B. ✓ Removing the cell wall barely changed either number
C. Far more X than Y got into the walled cells
Much X and little Y got in with the cell wall and without it.

Why: Removing the cell wall barely changed either number.
With the cell wall gone, much X still got in and little Y still got in.
So the cell wall was not the boundary making the choice.
The plasma membrane, which remained, was.

26Pressed against the cell wall

Check q10

A solution has a lower concentration of solutes than the cell placed in it.

Which word describes the solution relative to the cell?

A. Hypertonic
Hyper- means more solute, and this solution holds less solute than the cell.
B. ✓ Hypotonic

Why: The solution holds less solute than the cell.
A solution with less solute than the cell is hypotonic to the cell.

Video: Watch: Pressed against the cell wall

Water in, the contents press on the cell wall, the cell wall presses back: turgid, and a stalk that snaps.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L10c.mp4

Put a walled cell in a hypotonic solution: one with a lower concentration of solutes than the cell. Water moves in by osmosis.

A walled cell in a dilute solution, its cell wall and plasma membrane labelled: water arrows entering, and small arrows inside pressing outward against the cell wall on every side

The contents swell and press outward on the cell wall. The cell wall does not give. The cell wall presses back.

A cell pressed hard against its own cell wall like this is firm. Such a cell is called .

The outward push of the cell’s contents against the cell wall is called .

A celery stalk is thousands of these cells side by side. Each one pressed against its cell wall makes the whole stalk stiff, and stiff enough to snap.

What you are expected to know Explain why a walled cell in a hypotonic solution becomes turgid. Say that turgor pressure is what holds a soft plant part stiff.

Check q11

A leaf’s cells sit in a hypotonic fluid, and each cell is firm.

Which part of the cell is pressing on which other part?

A. The cell wall presses inward on the contents of the cell
Water is entering the cell, and the push starts from the swelling contents, not from the cell wall.
B. Water outside the cell presses inward on the cell wall
The support comes from inside the cell, not from the water outside.
C. ✓ The contents press out on the cell wall, which presses back
D. Nothing in the cell presses on anything else
A cell wall alone does not make a leaf firm; the same cell wall around shrunken contents belongs to a limp leaf.

Why: The fluid is hypotonic to the cell.
So water enters the cell by osmosis.
The contents swell.
The swollen contents press outward on the cell wall.
The cell wall does not give, so the cell wall presses back.
That outward push is turgor pressure.
Turgor pressure is what makes the cell firm.

Check q12

A soft plant stem is limp.

Which change would make it stiff again?

A. ✓ Water entering its cells
B. Water leaving its cells
Losing water is what made the stem limp; stiffness comes from water entering.
C. The cell walls thickening as the cells dry out
A dry cell has the same cell wall and is limp; the water inside pressing outward is what stiffens the stem.
D. Solute leaving its cells
The cell’s solute stays where it is; what changes is water entering by osmosis.

Why: Water enters each cell by osmosis.
The contents of each cell swell.
The swollen contents press against the cell wall.
That push is turgor pressure.
Each cell in the stem becomes firm.
So the whole stem becomes stiff.

37Why a walled cell does not burst

Video: Watch: Why a walled cell does not burst

The same pure water bursts an animal cell and leaves a walled cell whole.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L10d.mp4

Think of an animal cell in pure water: water keeps entering, the membrane can stretch only so far, and the cell bursts. That bursting is osmotic lysis.

Two panels, both in pure water: an animal cell swollen and burst open, and a walled cell whose membrane has swollen out until it presses on the cell wall and stopped there

Put a walled cell in the same pure water. Water enters just the same. The membrane swells outward until it presses on the cell wall, and the cell wall stops the membrane there.

The cell becomes turgid instead of bursting. The cell wall protects the cell from osmotic lysis.

The cell wall does not keep the water out. Water still enters; what the cell wall stops is the swelling.

Take the cell wall away and the protection is gone. Treat bacteria with a drug that stops them building a cell wall, and in a dilute solution the bacteria burst.

What you are expected to know Explain why a walled cell in a hypotonic solution becomes turgid instead of bursting, while an animal cell, or a bacterium stripped of its cell wall, bursts in the same solution.

Check q13

Bacteria in a very dilute solution swell but do not burst. A compound in the solution now stops the bacteria building their cell wall.

What happens to the bacteria in the same dilute solution?

A. The bacteria shrink
The solution is dilute, so water enters the bacteria; water does not leave them.
B. ✓ The bacteria may burst
C. The bacteria stay as they were
Water was entering with the cell wall in place too; the cell wall, not the membrane, stopped the swelling.
D. The bacteria stop taking in water
Water enters through the membrane whether or not there is a cell wall; the cell wall only limits how far the cell can swell.

Why: The solution is dilute, so water enters the bacteria by osmosis.
With a cell wall, the swelling stops at the cell wall.
Without a cell wall, nothing stops the swelling.
The membrane can stretch only so far.
So the membrane can tear and the bacteria can burst.

Check q14

A plant cell sits in pure water. It becomes turgid and stays whole.

Why does it stay whole?

A. ✓ Water enters, and the cell wall stops the membrane swelling further
B. Pure water has no solute, so no water enters
The cell has solute inside and the pure water outside has none, so water does move in.
C. The cell wall keeps water from entering the cell
Water crosses the cell wall and enters through the membrane; the cell wall stops the swelling, not the water.
D. Plant cell membranes are stronger and do not tear
A plant cell’s membrane is a bilayer like any other; the cell wall around the membrane is what takes the strain.

Why: Pure water has no solute, so water enters the cell by osmosis.
The membrane swells until it presses on the cell wall.
The cell wall does not give, so the cell wall takes the push and the membrane swells no further.
So the membrane does not tear: the cell is turgid, not burst.

47Measuring a cell: micrometers

Video: Watch: Measuring a cell: micrometers

A micrometer is one thousandth of a millimeter; converting both ways, one worked example each.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L10e.mp4

A cell is far too small to measure in millimeters. A red blood cell is about 0.008 mm across.

Cells are measured in micrometers. A micrometer, written μm, is one thousandth of a millimeter, so 1 mm is 1,000 μm.

So the red blood cell is 8 μm across. An onion skin cell is about 80 μm across.

To turn millimeters into micrometers, multiply by 1,000. To turn micrometers into millimeters, divide by 1,000.

Worked example

An onion skin cell is 0.08 mm across. How many micrometers is that?

Write down the value in the question:

0.08 mm

Write down the rule:

1 mm is 1,000 μm

Multiply by 1,000:

0.08 × 1,000 = 80 μm

What you are expected to know Convert a length between millimeters and micrometers: a micrometer, μm, is one thousandth of a millimeter, so multiply millimeters by 1,000 to get micrometers and divide micrometers by 1,000 to get millimeters.

55Fluency quiz: millimeters and micrometers mixed practice

Check q15 numeric entry

A pollen grain is 0.05 mm across.

Give its width in micrometers.

Answer: 50 μm (tolerance ±0.5)

Working

Write down the value in the question:

0.05 mm

Write down the rule:

1 mm is 1,000 μm

Multiply by 1,000:

0.05 × 1,000 = 50 μm

Check q16 numeric entry

A yeast cell is 5 μm across.

Give its width in millimeters.

Answer: 0.005 mm (tolerance ±0.0005)

Working

Write down the value in the question:

5 μm

Write down the rule:

1,000 μm is 1 mm

Divide by 1,000:

5 ÷ 1,000 = 0.005 mm

Check q17 numeric entry

A cheek cell is 0.06 mm across.

Give its width in micrometers.

Answer: 60 μm (tolerance ±0.5)

Working

Write down the value in the question:

0.06 mm

Write down the rule:

1 mm is 1,000 μm

Multiply by 1,000:

0.06 × 1,000 = 60 μm

Check q18 numeric entry

A large plant cell is 250 μm long.

Give its length in millimeters.

Answer: 0.25 mm (tolerance ±0.005)

Working

Write down the value in the question:

250 μm

Write down the rule:

1,000 μm is 1 mm

Divide by 1,000:

250 ÷ 1,000 = 0.25 mm

60When the water leaves

Video: Watch: When the water leaves

Onion cells in salt water: the contents shrink from a cell wall that keeps its shape — plasmolysis.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L10f.mp4

Now the other direction. Put onion skin cells in concentrated salt water, hypertonic to them. Water moves out.

An onion cell in concentrated salt water, with arrows showing water leaving: the cell wall keeps its shape at 80 micrometers across while the membrane and contents have shrunk to 66 micrometers and pulled away from the cell wall

The cell wall keeps its shape: still 80 μm across. The membrane and the contents inside it shrink, to 66 μm, and pull away from the cell wall.

The shrinking of a walled cell’s contents away from its cell wall like this is called .

Here is a photograph of plasmolyzed red onion cells. In each cell the red contents have pulled away from the cell wall, which keeps its shape.

Red onion skin cells under the microscope after plasmolysis: the red contents of each cell have shrunk away from its cell wall — Photograph of red onion skin cells after plasmolysis: in each cell the red contents have shrunk into a smaller mass and pulled away from the cell wall, which keeps its shape

Nothing now presses on the cell wall. Turgor is gone, and a plant part made of such cells goes limp.

Put the cells back in fresh water and plasmolysis reverses. Water moves in, so the contents swell. The contents press against the cell wall again.

To say what state a cell is in, ask which way the water moved. Water in and the contents pressing on the cell wall: turgid. Water out and the contents shrunk away from the cell wall: plasmolyzed. Water in with no cell wall to stop the swelling: burst.

What you are expected to know Predict that a walled cell in a hypertonic solution loses water: its contents shrink away from a cell wall that keeps its shape, turgor is lost and the tissue goes limp.

What you are expected to know Predict that a plasmolyzed cell put back in fresh water gains water: its contents swell and press against the cell wall again, so the tissue goes stiff.

71Fluency quiz: turgid, plasmolyzed or burst? mixed practice

Check q19

A root cell sits in rain water, which is hypotonic to it.

Which word describes the cell?

A. ✓ Turgid
B. Plasmolyzed
Rain water is hypotonic to the cell, so water moves in.
C. Burst
Water moves in, but the cell wall stops the swelling.

Why: Rain water is hypotonic to the cell.
So water moves in.
The contents swell and press on the cell wall.
The cell wall stops the swelling.
A walled cell with its contents pressing on the cell wall is turgid.

Check q20

An onion cell sits in concentrated salt water.

Which word describes the cell?

A. Turgid
Concentrated salt water is hypertonic to the cell, so water moves out.
B. ✓ Plasmolyzed
C. Burst
Concentrated salt water is hypertonic to the cell, so water moves out and the contents shrink.

Why: Concentrated salt water is hypertonic to the cell.
So water moves out.
The contents shrink and pull away from the cell wall, which keeps its shape.
A walled cell with its contents shrunk away from the cell wall is plasmolyzed.

Check q21

A red blood cell sits in pure water.

Which word describes the cell?

A. Turgid
Only a walled cell can be turgid.
The membrane tears: burst.
B. Plasmolyzed
Pure water is hypotonic to the cell, so water moves in, not out.
C. ✓ Burst

Why: Pure water is hypotonic to the cell.
So water moves in.
A red blood cell is an animal cell with no cell wall, so nothing stops the swelling.
The membrane can stretch only so far.
So the membrane tears: burst.

Check q22

A yeast cell sits in fresh water.

Which word describes the cell?

A. ✓ Turgid
B. Plasmolyzed
Fresh water holds less solute than the yeast cell, so water enters.
C. Burst
A yeast cell is a fungus cell, and fungus cells have a cell wall.

Why: A yeast cell is a fungus cell, so it has a cell wall.
Fresh water is hypotonic to the cell.
So water enters the cell.
The contents swell and press on the cell wall.
The cell wall presses back.
So the cell is turgid.

Check q23

A bacterium sits in strong brine, which holds far more solute than the cell.

Which word describes the cell?

A. Turgid
The brine holds more solute than the cell, so water leaves.
B. ✓ Plasmolyzed
C. Burst
The brine holds more solute than the cell, so water leaves, and a cell that loses water cannot burst.

Why: A bacterium has a cell wall.
The brine is hypertonic to the cell.
So water leaves the cell.
The contents shrink and pull away from the cell wall.
The cell wall keeps its shape.
So the cell is plasmolyzed.

77Reading a plasmolyzed cell

A drawing or a measurement of a cell tells you the state it is in. From the state, you can say which way the water moved, and what the cell will do if the solution changes.

What you are expected to know Read the state of a walled cell from what is seen, say which way the water moved to bring it there, and predict the reverse when the solution changes.

Check q24

Cells from a pondweed leaf sit in a sucrose solution stronger than their contents. The membrane has pulled away from the cell wall, and the contents take up less room than before.

What has happened?

A. Water entered the cells
The solution has more solute than the cells, so water moves out of the cells, not in.
B. Sucrose entered the cells and pushed water out
Sucrose does not cross the membrane; water is what leaves.
C. The cell wall shrank with the contents
The cell wall is rigid and keeps its shape; the contents are what shrink.
D. ✓ Water left the cells

Why: The sucrose solution has more solute than the cells.
So the solution is hypertonic to the cells.
Water leaves the cells by osmosis.
The contents shrink.
The cell wall is rigid, so the cell wall keeps its shape.
Therefore the membrane pulls away from the cell wall.
This is plasmolysis.

Check q25

Plasmolyzed onion cells move from salt water into fresh water.

What happens?

A. Nothing changes and the contents stay shrunken
Fresh water has less solute than the cells, so water enters and the contents swell back.
B. ✓ Water enters, and the contents swell against the cell wall again
C. Water leaves, and the contents shrink further
Fresh water has less solute than the cells, so water enters the cells rather than leaving them.
D. The cell wall swells and the contents stay shrunken
The cell wall is rigid and keeps its shape; the contents are what swell as water enters.

Why: Fresh water has less solute than the cells.
So fresh water is hypotonic to the cells.
Water moves into the cells by osmosis.
The contents swell.
The swollen contents press against the cell wall again.
So the cells are turgid once more.

82Where the water is kept

Video: Watch: Where the water is kept

The central vacuole holds the water that presses the cytosol against the cell wall.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L10g.mp4

Inside many cells are small sacs of membrane that store water, food or waste. These sacs are called vacuoles.

A plant cell with one large central vacuole filling most of its interior, beside an animal cell with several small vacuoles

In a plant cell, most of the water that presses on the cell wall is in one large vacuole in the middle.

This single large vacuole is called the plant cell’s . The central vacuole stores water and nutrients. As the central vacuole fills, it presses the cytosol outward against the cell wall.

Put wilted plant cells in water. The central vacuole grows from 4,800 μm³ to 7,200 μm³. The cytosol barely changes. The cell’s firmness rises fourfold.

Animal cells have vacuoles too, but smaller and more numerous, storing materials the cell needs.

What you are expected to know Describe the vacuoles of cells: one large central vacuole in a plant cell that stores water and nutrients and presses the cytosol against the cell wall as it fills, and smaller, more numerous vacuoles in an animal cell.

Check q26

Wilted lettuce cells are put in water and become firm again. Each cell’s central vacuole grows by 2,300 μm³; its cytosol grows by only 40 μm³.

Which statement explains why the cells become firm?

A. The cytosol took in most of the water, and its swelling made the cell firm
The cytosol grew by only 40 μm³; almost all of the water went into the vacuole.
B. The firmness rose on its own; the vacuole only stores nutrients
The vacuole’s swelling is what pressed the cytosol against the cell wall; the firmness did not rise on its own.
C. ✓ The central vacuole took in most of the water, and its swelling pressed the cell firm
D. The vacuole emptied its water into the cytosol, which swelled the cell
The vacuole grew and the cytosol barely changed, so the water went into the vacuole, not out of it.

Why: The central vacuole grew by 2,300 μm³.
The cytosol grew by only 40 μm³.
So almost all of the water went into the central vacuole.
The swelling vacuole pressed the cytosol outward against the cell wall.
That push is turgor pressure.
Therefore the cell became firm.

Check q27

A plant cell and an animal cell are compared.

Which statement describes their vacuoles?

A. Only the plant cell has vacuoles
Animal cells do have vacuoles; an animal cell’s vacuoles are smaller and more numerous.
B. Both cells have one large central vacuole
One large central vacuole is the plant cell’s; an animal cell has smaller, more numerous vacuoles.
C. The plant cell has several small vacuoles; the animal cell has one large central vacuole
The plant cell is the one with the one large central vacuole.
D. ✓ The plant cell has one large central vacuole; the animal cell has several smaller ones

Why: A plant cell has one large central vacuole, which stores water and nutrients.
As the central vacuole fills, it presses the cytosol against the cell wall and helps keep the cell turgid.
An animal cell has several smaller vacuoles, which store materials the cell needs.

The fresh stalk’s cells were pressed hard against their cell walls by the water in their central vacuoles. Overnight that water left. The contents shrank from the cell walls, so the stalk bent.

Glossary

archaea: A second group of single-celled organisms with no nucleus, whose cells, like those of bacteria, have a cell wall.
turgid: Firm because the cell’s contents, swollen with water, are pressed hard against its cell wall.
turgor pressure: The outward push of a walled cell’s contents against its cell wall when water has entered. Turgor pressure is what holds a soft plant part stiff.
plasmolysis: What happens to a walled cell in a hypertonic solution: water leaves, and the membrane and contents shrink away from a cell wall that keeps its shape.
central vacuole: The one large vacuole in a plant cell that stores water and nutrients and, as it fills, presses the cytosol outward against the cell wall.

APBIO-U02-P24 Practice questions: Topic 2.4

Topic 2.4 · Membrane Permeability · 10 MCQ · 2 FRQ · for APBIO-U02-T24

These practice questions have the shape of the Topic 2.4 test. Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the first free-response question, you work through one scenario in small steps, and each step offers a hint if you want one. For the second, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it.

Video: Watch first: membrane permeability, summed up

From the bubble of pure membrane to its hydrophobic interior and the two protein doors: what crosses, and why.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T24-summary.mp4

Q1 P24-q01

A sheet of pure phospholipid bilayer, with no proteins in it, separates two chambers of water. One chamber receives each substance in turn, at 10 mmol/L. Nitric oxide (NO), a small nonpolar gas, appears in the other chamber within seconds. Calcium ions (Ca²⁺) and lactose, a large sugar with many –OH groups, have still to appear after an hour.

Which of these substances would also cross the bilayer freely on its own?

A. ✓ Ethylene (C₂H₄), a hydrocarbon with two carbon atoms
B. Potassium ions (K⁺), which are smaller than lactose
K⁺ carries a charge, and the phospholipid bilayer blocks ions of every size, as it blocked Ca²⁺.
C. Sucrose, a sugar with many –OH groups
Sucrose is large and polar, like lactose, with –OH groups that water attracts.
D. Phosphate ions (PO₄³⁻), which are smaller than sucrose
A phosphate ion carries a full −3 charge, and nothing in the interior of the bilayer attracts a charge.

Why: A phospholipid bilayer lets small nonpolar molecules pass freely: NO and ethylene dissolve into the hydrocarbon tails.
Ions (Ca²⁺, K⁺, phosphate) and large polar molecules (lactose, sucrose) do not cross on their own.
A membrane that lets some substances through and holds others back is selectively permeable.

Q2 P24-q02

Protein-free bubbles of bilayer are floated in a solution of three substances: nitric oxide (NO), a small nonpolar gas; urea, a small polar molecule with no charge; and potassium ions (K⁺). Each substance is at the same concentration outside the bubbles, and none is inside at the start. After ten minutes the amount of each inside, as a percentage of the amount outside, is measured: one substance reached 100%, one reached 9% and one stayed at 0%.

Which set of results fits the three substances?

A. Nitric oxide 0%; urea 9%; potassium ions 100%
A potassium ion carries a full charge, so water attracts it and nothing in the tails does: potassium ions stay at 0%, not 100%.
B. Nitric oxide 9%; urea 100%; potassium ions 0%
Nonpolar nitric oxide dissolves into the tails and passes freely, so nitric oxide reaches 100%; water attracts the partial charges of urea, so only a little gets through.
C. ✓ Nitric oxide 100%; urea 9%; potassium ions 0%
D. Nitric oxide 100%; urea 0%; potassium ions 9%
An ion gets none of itself through, while a small polar molecule with no charge gets a little through; so urea is at 9% and potassium ions at 0%.

Why: Small nonpolar molecules pass freely, so nitric oxide reaches 100%.
Small polar molecules with no charge pass in small amounts, so urea reaches 9%.
Ions carry a charge and do not cross on their own, so potassium ions stay at 0%.

Q3 P24-q03

Methane (CH₄) and methanol (CH₃OH) are molecules of nearly the same size. Methane is nonpolar. Methanol is polar, with an –OH group, but carries no charge. Outside protein-free bubbles of phospholipid bilayer, methane reaches 100% of its outside level inside within a minute, while methanol reaches 20%. A student explains: "Methanol crosses less because methanol is the bigger molecule."

Which statement corrects the student?

A. Nothing needs correcting: the bigger of two molecules always crosses less
The two molecules are nearly the same size, so size cannot be what separates them.
B. ✓ Methanol is no bigger than methane: water attracts methanol's –OH group, and nothing in the tails does
C. Methanol crosses less because the phospholipid heads repel its –OH group
The heads face the water on both sides and push nothing away; the sorting happens in the hydrophobic interior, which has nothing to attract a polar group.
D. Methanol crosses less because it dissolves into the tails and then moves through them more slowly
A molecule that dissolved into the tails as easily as methane would reach 100% as methane does; methanol stays mostly in the water because water attracts its –OH group.

Why: Methane and methanol are nearly the same size, so size does not decide between them.
Methanol’s –OH group carries partial charges that water is attracted to, so only a little methanol slips through.
Nothing in methane attracts water, so methane passes freely.
Polarity, not size, separates them.

Q4 P24-q04

Three substances are tested, one at a time, outside protein-free bubbles of phospholipid bilayer. Ethylene, a small nonpolar molecule, reaches 100% of its outside level inside within a minute. Propane, a nonpolar molecule larger than ethylene, also reaches 100%. The fluoride ion (F⁻), which is smaller than an ethylene molecule, stays at 0%.

Which conclusion do the three results support?

A. Size decides: the smallest particle crosses first, so the F⁻ reading of 0% must be a mistake in the measurement
The F⁻ reading is the datum, not a mistake; the larger propane crossing as freely as ethylene already shows that size is not deciding.
B. Size decides: F⁻ is too small for the tails to catch and carry across, while both nonpolar molecules are big enough to be carried
The tails do not catch and carry anything; a nonpolar molecule dissolves into them on its own, and F⁻ is kept out by its charge, not by its size.
C. The heads decide: the heads pull F⁻ into the water outside and push the two nonpolar molecules through the middle of the membrane
The heads face the water on both sides and push nothing through; the sorting happens in the hydrophobic interior.
D. ✓ Charge decides: water attracts F⁻ and nothing in the tails does, while both nonpolar molecules dissolve into the tails whatever their size

Why: Propane is larger than ethylene, and both reached 100%, so a larger nonpolar molecule crosses as freely.
F⁻ is the smallest particle, and none crossed, so size is not what keeps F⁻ out.
F⁻ carries a full charge, and water attracts that charge.
So F⁻ stays in the water.

Q5 P24-q05

Two proteins in a nerve cell's membrane let through substances that the phospholipid bilayer itself blocks. When protein A is open, potassium ions (K⁺) stream through it, thousands every millisecond, with no pause. Protein B binds one amino acid molecule at a time, changes shape, and releases the molecule on the other side.

What kinds of protein are A and B?

A. Both carrier proteins; A binds K⁺ more loosely
A carrier binds its substance and changes shape for each molecule it moves, which is slow, and A moves thousands of ions a millisecond with no binding.
B. ✓ A is a channel protein; B is a carrier protein
C. A is a carrier protein; B is a channel protein
When a channel is open, ions stream through it without pausing; a carrier binds one molecule at a time.
A streams K⁺, so A is the channel.
D. Both channel proteins; B is a narrower one
B binds each amino acid and changes shape to move it, which no tunnel does.

Why: Ions cross a membrane only through proteins.
A channel protein is a water-lined tunnel; when open, one kind of ion streams through it, so A is a channel.
A carrier protein binds its substance and changes shape to move it across, one at a time, so B is a carrier.

Q6 P24-q06

A student predicts that a new drug molecule, which is small and nonpolar, will need a carrier protein to get into cells. The class tests it on protein-free bubbles of bilayer: after five minutes the drug inside the bubbles is at 95% of its level outside.

Do the data support the student's prediction?

A. Yes: the drug reached only 95%, so a carrier would be needed for the last 5%
A substance at 95% of its outside level after five minutes is crossing freely.
B. Yes: small molecules always need a carrier, and the bubbles are a special case
Small nonpolar molecules cross a phospholipid bilayer on their own, and the bubbles show exactly that.
C. ✓ No: the drug crossed a phospholipid bilayer on its own, so a cell needs no carrier for it
D. The data cannot test the prediction, since the bubbles have no proteins
Having no proteins is what makes the bubbles the right test.

Why: Protein-free bubbles test what the bilayer alone lets through.
The drug reached 95% inside in five minutes.
So the drug crosses the bilayer on its own, as a small nonpolar molecule does.
So a cell would need no carrier to take the drug in.
The data contradict the prediction.

Q7 P24-q07

Five organisms are examined: a mushroom, which is a fungus; a frog; an oak tree; a bacterium from yogurt; and an archaean, a single-celled organism with no nucleus, from a hot spring. Every cell of each has a plasma membrane.

Which of these organisms have cells with a cell wall outside the plasma membrane?

A. The oak tree only
Plants have cell walls, but so do fungi, bacteria and archaea.
B. The oak tree and the mushroom
Bacteria and archaea have cell walls as well as plants and fungi.
C. All five, the frog included
Animal cells have no cell wall outside the plasma membrane.
D. ✓ All but the frog

Why: The cells of bacteria, archaea, fungi and plants have a cell wall outside the plasma membrane.
Animal cells, such as the frog’s, have no cell wall.
Nearly all archaea have a cell wall, and this hot-spring archaean has one.

Q8 P24-q08

A wilted lettuce leaf is put into a bowl of cold water for twenty minutes and comes out crisp. Only water has entered the leaf.

What made the leaf crisp again?

A. Water filled the spaces between the cells and glued the cells to one another
Stiffness comes from inside each cell, not from between cells.
B. ✓ Water entered the cells; the swollen contents pressed on the cell walls, which pressed back
C. The cold water stiffened the cell walls, which had gone soft in the warm air
Cold does not stiffen a cell wall, and the leaf would go crisp in warm water too.
D. Water dissolved the cell walls, which then set again in a firmer, stiffer form
Cell walls do not dissolve in water.
The swelling contents press outward on each cell wall.

Why: The water in the bowl is hypotonic to the leaf's cells.
So water enters the cells by osmosis.
The swelling contents press outward on each cell wall.
The cell wall presses back.
So each cell is firm, or turgid.
That outward push, turgor pressure, holds a soft plant part stiff.

Q9 P24-q09

Pond algae, whose cells have cell walls, live in nearly pure water without harm. Red blood cells, which have no cell wall, put into the same water swell and burst within minutes.

Why do the algae survive in the water that bursts the red blood cells?

A. The algae's cell walls keep water from entering their cells at all
Water still enters the algae’s cells; the cell wall lets water through.
B. The algae's membranes are far stronger than a red blood cell's
The two kinds of membrane are alike; the cell wall is the difference.
C. ✓ Water enters the algae's cells too, but their cell walls stop the swelling
D. The water is hypotonic to a red blood cell but isotonic to an alga
Nearly pure water is hypotonic to both kinds of cell.

Why: Nearly pure water is hypotonic to both kinds of cell, so water enters both.
The alga’s membrane can swell only until it presses on the cell wall, which presses back.
So the alga becomes turgid instead of bursting.
The red blood cell has no cell wall, so nothing stops its swelling: osmotic lysis.

Q10 P24-q10

Slices of eggplant are sprinkled with salt. Within half an hour the slices are limp and sitting in liquid.

What has happened to the cells of the slices?

A. Salt entered the cells and dissolved part of each cell wall
Salt cannot cross the membrane on its own, and cell walls do not dissolve in salt.
B. Water entered the cells, and the cell walls stretched away from the contents
Water left the cells rather than entering; the liquid around the slices is water drawn out of them.
C. The cell walls shrank around the contents, squeezing liquid out
A cell wall is rigid and keeps its shape and size; the cell wall does not shrink around the contents.
D. ✓ Water left the cells toward the salty surface, and the contents shrank from the cell walls

Why: The salt makes the surface of the slice hypertonic to the cells, so water leaves the cells by osmosis.
The membrane and contents shrink, but the cell wall keeps its shape.
So the contents pull away from the cell wall: plasmolysis.
With nothing pressing on the cell walls, turgor is lost.

FRQ 1 P24-frq1 · Scientific Investigation scaffolded

A class floats protein-free bubbles of bilayer in a solution of four substances, each at the same concentration outside the bubbles and none inside: ethylene (C₂H₄), a small nonpolar gas; hydrogen peroxide (H₂O₂), a small polar molecule with no charge; calcium ions (Ca²⁺), which are smaller than an ethylene molecule; and lactose, a large sugar with –OH groups all over it. After ten minutes the amount of each inside the bubbles, as a percentage of the amount outside, is: ethylene 100%; hydrogen peroxide 8%; Ca²⁺ 0%; lactose 0%.

(a) Identify the substance that crossed the bilayer freely and the substance that crossed in small amounts. (1 pt)

Frame The substance that crossed freely is …, at …%; the one that crossed in small amounts is …, at …%

Hint What percentage would a substance reach if it crossed until it was as concentrated inside as outside? Which value fits neither that nor zero?

Model answer The substance that crossed freely is ethylene, at 100%: after ten minutes ethylene was as concentrated inside as outside.
The one that crossed in small amounts is hydrogen peroxide, at 8%: hydrogen peroxide reached 8% of its outside level.

Rubric

Award 1 point for: ethylene crossed freely (100%); hydrogen peroxide crossed in small amounts (8%).
Accept: the two named with their percentages in either order. Do not award the point if 8% is read as no crossing.

Slip Reading 8% as none. Small polar molecules with no charge trickle through a phospholipid bilayer; 8% is a small crossing, and 0% is the value that says none.

(b) Describe the interior of the bilayer: what it is made of and what it lacks. (1 pt)

Frame The interior of the bilayer is made of …, which carry no …

Hint Think about which part of each phospholipid ends up in the middle, and what kind of chemical groups it has and has not got.

Model answer The interior of the bilayer is made of the hydrocarbon tails of the phospholipids, the tails of both layers touching in the middle, which carry no charges or partial charges.
So nothing in the interior attracts a polar or charged particle.
This band of tails is the hydrophobic interior.

Rubric

Award 1 point for: the interior is the hydrocarbon tails of the phospholipids of both layers, touching in the middle; the tails carry no charges or partial charges (nothing that attracts a polar or charged particle), so the interior is hydrophobic.
Accept: "the oily middle, made of tails with no charges".

Slip Describing the heads instead of the tails. The heads face the water on each side; the interior, which decides what crosses, is the tails.

(c) Explain why the calcium percentage is 0% although a calcium ion is smaller than an ethylene molecule. (1 pt)

Frame Ca²⁺ carries …, so water …; the interior has …, so …; ethylene, by contrast, …

Hint Compare Ca²⁺ with ethylene property by property: size, polarity, charge. Then use what you said in (b) the interior lacks.

Model answer Ca²⁺ carries a full +2 charge, so water attracts Ca²⁺ strongly.
The interior has hydrocarbon tails with no charges or partial charges, so nothing in the interior attracts Ca²⁺.
So Ca²⁺ cannot leave the water on its own side.
Ethylene, by contrast, has nothing that attracts water.
So ethylene dissolves into the tails and passes through.
Charge decides, not size: the smaller particle is the one that stayed out.

Rubric

Award 1 point for: Ca²⁺ carries a full +2 charge, so water attracts Ca²⁺, and the hydrocarbon interior has no charges or partial charges that attract Ca²⁺, so Ca²⁺ cannot leave the water to cross; nothing in ethylene attracts water, so ethylene dissolves into the tails. Charge, not size, decides.
Accept: "water keeps the ion; the oily middle gives it nothing" with the contrast to ethylene stated or implied.

Slip Saying Ca²⁺ is blocked because it is too big. Ca²⁺ is the smaller of the two; water attracts its charge, and that attraction is what keeps Ca²⁺ out.

(d) Explain why the lactose percentage is also 0%. (1 pt)

Frame Lactose is … and …; its –OH groups …, so …

Hint Hydrogen peroxide is polar too, and a little of it got through. List the ways lactose differs from hydrogen peroxide, and think about what water does to each polar group.

Model answer Lactose is large and polar; its –OH groups carry partial charges, so water attracts every one of them.
Nothing in the hydrocarbon interior attracts lactose.
So lactose stays in the water.
Hydrogen peroxide is polar too, but hydrogen peroxide is small, so a little of it slips through.
Lactose is both polar and large.
So none of the lactose gets through.

Rubric

Award 1 point for: lactose is a large polar molecule; water's partial charges attract every one of its –OH groups, and nothing in the interior attracts them, so lactose stays in the water; being large as well as polar, it does not even trickle through as the small polar hydrogen peroxide does.
Accept: "large and polar: water attracts it all over and the tails attract none of it".

Slip Saying lactose stayed out only because it is large. Size matters among polar molecules; what keeps lactose in the water is that water attracts its many –OH groups.

(e) Make a claim about how a living gut cell takes in Ca²⁺, and support your claim with the bubble result. (1 pt)

Frame A living gut cell takes in Ca²⁺ through …; the bubbles show …, so …

Hint Compare the bubbles with a living gut cell’s plasma membrane. Ask what the cell’s membrane has in it that the bubbles lack, and whether that could give Ca²⁺ a way in.

Model answer A living gut cell takes in Ca²⁺ through a membrane protein, a channel protein or a carrier protein.
Ca²⁺ reached 0% inside the protein-free bubbles.
So Ca²⁺ cannot cross the bilayer itself.
A channel protein is a water-lined tunnel that lets the ion through.
A carrier protein binds the ion and changes shape to move it across.
Therefore a protein set into the membrane is the only way in.

Rubric

Award 1 point for: the claim that Ca²⁺ enters through a membrane protein, a channel (a water-lined tunnel for the ion) or a carrier, supported by the evidence AND the reasoning: Ca²⁺ reached 0% inside the protein-free bubbles, so Ca²⁺ cannot cross the bilayer on its own, so a protein door is the only way in.
Accept: "through a channel protein" or "through a carrier protein" with the 0% result and the reason that the bilayer blocks ions. Do not award the claim alone.

Slip Making the claim with no route named, or naming the route with no evidence from the bubbles. The point needs the protein door, the 0% result and the link.

FRQ 2 P24-frq2 · Conceptual Analysis

A florist keeps cut tulip stems standing in tap water, and the stems stay stiff and upright. A student moves one stem into a strong sugar solution and leaves it there for an hour.

(a) Describe what the cell wall does for a plant cell and what the plasma membrane inside it does. (1 pt)

Frame The cell wall is … and gives the cell …; the plasma membrane inside it …

Model answer The cell wall is a rigid boundary outside the plasma membrane and gives the cell its shape; the cell wall also holds back some large substances.
The plasma membrane inside it does the selecting: the plasma membrane decides which substances enter and leave the cell.

Rubric

Award 1 point for: the cell wall, outside the plasma membrane, is a rigid boundary that gives the cell its shape and holds back some substances; the plasma membrane inside it is what decides which substances enter and leave (does the selecting).
Accept: "the cell wall holds the shape; the membrane selects".

Slip Giving the cell wall the job of deciding what enters. The cell wall holds back only some large substances; the membrane inside it is the selector.

(b) Explain why the stem in tap water is stiff. (1 pt)

Model answer Tap water is hypotonic to the stem's cells.
So water enters the cells by osmosis.
The swelling contents press outward on each cell wall.
The cell wall presses back.
So every cell is firm, or turgid.
That outward push against the cell wall is turgor pressure.
Turgor pressure is what holds the soft stem stiff.

Rubric

Award 1 point for: tap water is hypotonic to the stem's cells, so water enters them by osmosis; the swelling contents press outward on each cell wall and the cell wall presses back, making each cell turgid; this turgor pressure holds the soft stem stiff.
Accept: "water enters, the contents push on the cell wall, the cell wall pushes back: turgor".

Slip Saying the cell walls are what make the stem stiff. A cell wall with shrunken contents inside it gives a limp stem; it is the pressure of water-filled contents against the cell walls that stiffens it.

(c) Make a claim about what happens to the cells' contents and to the stem during the hour in the strong sugar solution. (1 pt)

Model answer Water leaves the cells.
So the membrane and the contents shrink away from each cell wall.
The cell wall keeps its shape.
This is plasmolysis.
Nothing presses on the cell walls, so turgor is lost.
Therefore the stem droops.

Rubric

Award 1 point for: the claim that water leaves the cells, the membrane and contents shrink away from the cell wall (plasmolysis) while the cell wall keeps its shape, turgor is lost and the stem droops. No reasoning is required for this point.
Accept: water moving out with the loss of turgor and the stem going limp. Do not award the point for sugar entering the cells or for the cells swelling.

Slip Having the sugar enter the cells, or the cells swell. Sugar is a large polar molecule the membrane blocks; only water moves, and it moves out toward the sugar.

(d) Support the claim you made in (c), using what the cell wall and the plasma membrane each let through. (1 pt)

Model answer The plasma membrane lets water through and blocks sugar, a large polar molecule.
The strong sugar solution is hypertonic to the cells.
So water crosses the membrane outward by osmosis.
The cell wall is rigid, so it keeps its shape whether or not the contents press on it.
The cell wall lets water pass freely.
Therefore the membrane and the contents shrink inward, away from a cell wall that stays where it was.

Rubric

Award 1 point for: the evidence AND the reasoning: the plasma membrane lets water through but blocks sugar, a large polar molecule, so with more dissolved sugar outside than inside (a hypertonic surrounding) water crosses the membrane outward by osmosis; the cell wall is a rigid boundary that holds its shape and lets water pass freely, so the membrane and contents shrink away from it.
Accept: "the membrane blocks sugar and passes water, so water leaves by osmosis; the cell wall is rigid and lets water through, so it keeps its shape while the contents shrink". Do not award the point for support that gives the cell wall the job of selecting.

Slip Giving the cell wall the job of keeping water in, or having it shrink with the cell. The cell wall is rigid and lets water pass; the membrane blocks the sugar and lets the water out.

APBIO-U02-T24 End-of-topic test: Membrane Permeability

Topic 2.4 · Membrane Permeability · 16 MCQ · 2 FRQ

Q1 T24-q01

Sealed bubbles of phospholipid bilayer with no proteins in them float in a solution of four substances. At the start each substance is at 10 mmol/L outside the bubbles and 0 mmol/L inside. After ten minutes the concentration inside the bubbles is: propane (C₃H₈), a hydrocarbon, 7 mmol/L; calcium ions (Ca²⁺) 0 mmol/L; nitrate ions (NO₃⁻) 0 mmol/L; fructose, a sugar, 0 mmol/L.

Which substance crossed the bilayer?

A. ✓ Propane: a small nonpolar molecule
B. Fructose: a molecule that dissolves in water
Fructose stayed at 0 mmol/L inside; dissolving in water is what keeps a large polar molecule out of the tails, not what carries it through.
C. Calcium ions: the smallest particles present
Calcium ions stayed at 0 mmol/L inside; being small is not enough when the particle carries a charge.
D. Nitrate ions: a small charged particle
Nitrate ions stayed at 0 mmol/L inside; a charge keeps an ion in the water.

Why: Only propane appeared inside the bubbles.
Propane is a hydrocarbon: a small nonpolar molecule, which dissolves into the tails and passes through.
Calcium ions and nitrate ions carry full charges, and fructose is a large polar molecule; water attracts all three, so they stay outside.
The bilayer is selectively permeable.

Q2 T24-q02

Carbon dioxide molecules (CO₂) and potassium ions (K⁺) are both present in the blood outside a liver cell, and both are tiny. A bare potassium ion is smaller than a carbon dioxide molecule.

Which of the two crosses the phospholipid bilayer on its own, and why?

A. Both: anything this small slips between the tails
K⁺ is even smaller than CO₂ and yet K⁺ does not cross; being small is not enough.
B. K⁺ only: it is the smaller of the two, so it slips through more easily
K⁺ is the smaller particle and still stays out, so size is not what decides.
C. ✓ CO₂ only: it has no charge for water to attract, and K⁺ does
D. Neither: the bilayer blocks every dissolved substance
CO₂ does cross a phospholipid bilayer, as every small nonpolar molecule does.

Why: CO₂ is a small nonpolar molecule, so nothing in CO₂ attracts water, and CO₂ dissolves into the tails and passes through.
K⁺ carries a full charge, so water attracts K⁺ and the uncharged tails do not.
So K⁺ does not cross on its own, though it is smaller.

Q3 T24-q03

Four substances are tested one at a time against a protein-free bilayer: nitrogen gas (N₂); ammonia (NH₃), a small polar molecule with no charge; potassium ions (K⁺); and sucrose, a large sugar with many –OH groups.

Which one passes through in small amounts only?

A. N₂
N₂ is a small nonpolar molecule and passes freely, not in small amounts.
B. ✓ NH₃
C. K⁺
K⁺ is an ion whose full charge keeps it in the water; it does not cross at all.
D. Sucrose
Sucrose is a large polar molecule and does not cross a phospholipid bilayer on its own.

Why: NH₃ is a small polar molecule with no charge, like H₂O, and such molecules pass through a phospholipid bilayer in small amounts.
N₂ is a small nonpolar molecule, so N₂ passes freely.
K⁺ is an ion, and sucrose is a large polar molecule, so neither crosses on its own.

Q4 T24-q04

Four substances are dissolved in the fluid around a liver cell.

Which of the following can cross the cell's plasma membrane only through a membrane protein?

A. ✓ Potassium ions (K⁺)
B. Oxygen (O₂)
Oxygen is a small nonpolar molecule, so oxygen dissolves through the bilayer itself and needs no protein.
C. Carbon dioxide (CO₂)
Carbon dioxide is a small nonpolar molecule, so carbon dioxide dissolves through the bilayer itself and needs no protein.
D. Ethanol, a small polar molecule with no charge
A small polar molecule with no charge passes through the bilayer in small amounts on its own, so ethanol needs no protein to get across.

Why: A potassium ion carries a full charge; water attracts that charge, and nothing in the hydrophobic interior does.
So K⁺ cannot cross the bilayer on its own and needs a channel or carrier.
Oxygen and carbon dioxide are small and nonpolar, and ethanol is small and polar: all cross alone.

Q5 T24-q05

A student explains: "The membrane blocks glucose because glucose carries a charge, and water attracts charged particles."

Which statement corrects the student?

A. Nothing needs correcting: any substance that water dissolves must carry a charge
Water dissolves many substances that carry no charge, such as glucose and glycerol; being polar is enough for water to attract them.
B. Glucose has no charge and is blocked because it is too big to fit between the tails
Size is not what decides on its own; glycerol is small and polar and only trickles through, while much larger nonpolar molecules pass.
C. Glucose is blocked because it is nonpolar and the tails repel nonpolar molecules
Glucose is polar, not nonpolar, and a nonpolar molecule would dissolve into the tails and pass through.
D. ✓ Glucose has no charge: it is a large polar molecule, and water attracts its many –OH groups

Why: Glucose carries no charge: it is a large polar molecule with many –OH groups.
Water attracts each polar –OH group.
The hydrocarbon tails carry no charges or partial charges, so nothing in the tails attracts glucose.
So glucose stays in the water.

Q6 T24-q06

Two proteins in a muscle cell's membrane let through substances that the phospholipid bilayer itself blocks. When protein 1 is open, calcium ions (Ca²⁺) stream through it, millions every second, without stopping. Protein 2 takes lactate, a substance a hard-working muscle makes, one molecule at a time, holding each molecule for a moment before that molecule appears on the other side.

What kinds of protein are protein 1 and protein 2?

A. Protein 1 a carrier protein; protein 2 a channel protein
Protein 1 lets millions of ions stream through without stopping, which is a channel; protein 2 binds one lactate molecule at a time, which is a carrier.
B. ✓ Protein 1 a channel protein; protein 2 a carrier protein
C. Both channel proteins; protein 2 is a slower one
Protein 2 does not offer an open passage; protein 2 holds one lactate molecule at a time.
D. Both carrier proteins; protein 1 binds calcium more loosely
Protein 1 holds nothing and moves nothing one at a time; calcium ions stream through protein 1 while it is open.

Why: A channel protein is a water-lined tunnel; when open, one kind of ion streams through it, so protein 1 is a channel.
A carrier protein binds its substance and changes shape to move it across, one molecule at a time, so protein 2 is a carrier.

Q7 T24-q07

Liver cells sit in a solution holding equal amounts of fructose, a sugar, and of molecule Z, a small nonpolar molecule. Fructose inside the cells rises by 3.0 mmol/L every five minutes. Then a drug that blocks one membrane protein reaches the cells; fructose now rises by only 0.5 mmol/L every five minutes. Molecule Z rises by 2.0 mmol/L every five minutes both before and after the drug.

What do the results show?

A. ✓ Fructose enters through a carrier protein; Z crosses the bilayer itself
B. Fructose and Z both cross the bilayer itself; the drug thinned the membrane
If fructose crossed the bilayer itself, blocking one protein would not cut its entry from 3.0 mmol/L to 0.5 mmol/L, and a thinned membrane would have let Z in faster.
C. Fructose and Z both need a protein; the protein for Z was not blocked
Z is a small nonpolar molecule, so Z dissolves through the bilayer itself and needs no protein; the drug could not change its entry.
D. The drug made the bilayer hold back polar molecules
Z is nonpolar and was unaffected, and fructose was the only substance whose entry changed.

Why: Blocking one membrane protein cut the rise in fructose from 3.0 mmol/L to 0.5 mmol/L, so fructose was entering through that protein.
Fructose is a large polar molecule, so the protein is a carrier.
Z is small and nonpolar, so Z dissolves through the bilayer itself; the drug left Z unchanged.

Q8 T24-q08

Protein-free bubbles of bilayer are placed in a solution of a new molecule. The molecule is about the size of glucose, and it is polar, with many –OH groups but no charge.

What should the data show after ten minutes, and would a real cell need a protein to take the molecule in?

A. Nearly 100% inside; no protein needed
Nearly 100% is what a small nonpolar molecule gives, and this molecule is polar all over.
B. A small amount inside, as for water; no protein needed
Water gets through in small amounts because water is tiny as well as polar, and this molecule is as large as glucose.
C. ✓ About 0% inside; a real cell would need a carrier or channel
D. About 0% inside; a real cell could not take it in at all
A phospholipid bilayer on its own is not a whole cell; a real cell has proteins that take in large polar molecules.

Why: The new molecule is large and polar, like glucose.
Water attracts its –OH groups, and nothing in the hydrocarbon tails attracts the molecule.
So it does not cross a phospholipid bilayer on its own, and about 0% appears inside.
A real cell takes it in through a carrier or channel.

Q9 T24-q09

Before an experiment, a student predicts: "Anything with no charge crosses a phospholipid bilayer freely, whatever its size." Protein-free bubbles of bilayer are placed in four solutions. After ten minutes, the amount inside as a fraction of the amount outside: carbon dioxide (CO₂) 100%; water (H₂O) 12%; glucose (no charge, six –OH groups) 0%; chloride ions (Cl⁻) 0%.

Do the data support the student's prediction?

A. Yes: CO₂ and water, both uncharged, got in
Water is uncharged and only 12% got in, and glucose is uncharged and none got in.
B. ✓ No: uncharged glucose did not cross; polarity and size matter too
C. No: Cl⁻ did not cross, which shows nothing uncharged crosses
Cl⁻ is charged, so its result does not test a prediction about uncharged substances.
D. Cannot tell: the bubbles had no proteins, so the test is unfair
Leaving out proteins is what keeps the test fair: anything inside must have crossed the bilayer itself.

Why: Glucose has no charge, yet none crossed; water is uncharged, yet only 12% crossed.
So the data contradict the prediction.
Among uncharged substances, only small nonpolar ones such as CO₂ pass freely.
Small polar molecules pass in small amounts; large polar molecules do not cross.

Q10 T24-q10

Four cells are examined: a cell of a mold (a fungus) growing on bread, a bacterium from soil, a cell from a pine needle, and a cell lining a dog's gut. Each cell has a plasma membrane.

Which of them also have a cell wall outside the membrane?

A. The pine cell only
The mold cell and the bacterium have cell walls too: bacteria, archaea, fungi and plants all have a cell wall outside the plasma membrane.
B. The pine cell and the mold cell
Bacteria have cell walls as well.
The cells of bacteria, archaea, fungi and plants have a cell wall outside the plasma membrane.
C. ✓ The pine cell, the mold cell and the bacterium
D. The pine cell, the mold cell, the bacterium and the dog’s gut cell
Animal cells have no cell wall.
The cells of bacteria, archaea, fungi and plants have a cell wall outside the plasma membrane.

Why: The cells of bacteria, archaea, fungi and plants have a cell wall outside the plasma membrane; animal cells do not.
So the mold cell, the bacterium and the pine cell are walled, and the cell lining the dog's gut is not.

Q11 T24-q11

Cell walls are separated from ground-up leaf tissue, with no membrane attached to them, and placed in a solution of two dissolved substances, glucose and a blue dye. Both substances pass through the isolated cell walls freely. Intact cells of the same leaf, in the same solution, take in the glucose and hold the dye out.

Which boundary decides what enters the intact cells?

A. The cell wall: it is the outer boundary and touches each substance first
Touching a substance first is not the same as deciding whether it gets in; the isolated cell walls let both substances through.
B. ✓ The plasma membrane: the isolated cell walls let both substances through
C. Both boundaries together, but only while the cell wall is attached to the membrane
The isolated cell wall let both substances through on its own, so attaching it to the membrane adds no selecting that the membrane does not do alone.
D. The cell wall: grinding tore holes in the isolated cell walls
A cell wall passes water and small solutes whether or not it has been handled; the intact cells, with undamaged cell walls, still held the dye out.

Why: Both substances passed freely through the isolated cell walls, so the cell wall does not hold the dye out.
In the intact cells, the dye was held out.
So the boundary holding the dye out is the one inside the cell wall: the plasma membrane.
The plasma membrane is the selectively permeable boundary.

Q12 T24-q12

A fresh asparagus spear snaps when it is bent. The same spear left in dry air overnight bends without snapping. Nothing has entered or left the spear but water.

Why was the fresh spear stiff?

A. ✓ Its swollen cells pressed out on their cell walls, and the cell walls pressed back
B. Its cell walls were thicker; they thinned as water left overnight
Cell walls do not thin when a plant loses water.
C. Water had glued the cells to one another
Water inside the cells, not between them, made the spear stiff.
D. Its cells had no cell walls yet; cell walls form as a stalk dries
Plant cells have cell walls from the start, wet or dry.

Why: In the fresh spear, water had entered the cells.
The swelling contents pressed outward on each cell wall.
The cell wall pressed back.
So the cells were turgid, and this turgor pressure held the spear stiff.
Overnight the water left.
So the push on the cell walls fell, and the spear bent.

Q13 T24-q13

A root hair cell of a radish seedling is placed in pure water, a hypotonic surrounding.

What happens at the cell wall, and what is the cell then called?

A. The cell wall pulls the contents inward; turgid
The cell wall does not pull; the cell wall is pushed on and pushes back.
B. The contents pull away from the cell wall; limp
The contents pull away from the cell wall only when water leaves a cell, and in pure water, a hypotonic surrounding, water enters.
C. ✓ The contents push out on the cell wall and it pushes back; turgid
D. The cell wall stretches and grows thinner as water enters; turgid
The cell wall is rigid; the cell wall does not stretch or thin.

Why: Pure water is hypotonic to the root hair cell.
So water enters the cell by osmosis.
The swelling contents press outward on the cell wall.
The cell wall presses back.
So the cell becomes firm: the cell is turgid.
That outward push of the contents against the cell wall is turgor pressure.

Q14 T24-q14

Leaf cells have their cell walls digested away by enzymes, leaving the plasma membrane intact, in a solution that matches the cells' contents. The cells, now with no cell wall, are then moved into pure water.

What happens to these cells with no cell wall, and why?

A. They swell, then stop swelling when the membrane is stretched tight
A plasma membrane is a thin, flexible film that tears when it is stretched; it does not stop the swelling the way a rigid cell wall does.
B. ✓ They burst: water enters, and no cell wall stops the swelling
C. They become turgid: the membrane presses back on the contents as a cell wall would
Only a rigid cell wall can press back hard enough to make a cell turgid; a plasma membrane stretched by the swelling contents tears instead.
D. They stay the same size: without a cell wall, water cannot enter
The cell wall is not what lets water in; water crosses the plasma membrane, and pure water is hypotonic to the cells, so water enters.

Why: Pure water is hypotonic to the cells, so water enters them by osmosis.
A walled cell swells only until its membrane presses on the cell wall, so it becomes turgid.
These cells have no cell wall, and a plasma membrane cannot press back like a rigid cell wall.
So they burst: osmotic lysis.

Q15 T24-q15

Onion skin cells were drawn in two different solutions, one cell in each drawing. The cell wall measures 80 μm across in both drawings.

Two drawings of an onion skin cell, 1 and 2. The outer rectangle is the cell wall; the inner line is the plasma membrane.

Which drawing shows a cell in concentrated salt water, and what happened to it?

A. ✓ Drawing 2: water left by osmosis, and the contents shrank away from the cell wall
B. Drawing 1: water left by osmosis, and the cell wall shrank with the cell
The cell wall does not shrink; the cell wall is 80 μm across in both drawings.
C. Drawing 2: salt entered by osmosis and pushed the membrane inward
Osmosis is the net movement of water, not of salt.
D. Drawing 1: water entered and pressed the membrane against the cell wall
Drawing 1 shows the cell in fresh water, with the membrane pressed against the cell wall.

Why: Concentrated salt water is hypertonic to the cell.
So water leaves the cell by osmosis.
The membrane and the contents shrink.
The cell wall keeps its shape.
So the contents pull away from the cell wall.
This is plasmolysis, and drawing 2 shows it.

Q16 T24-q16

Red cabbage leaf cells in concentrated salt water have their contents shrunk away from their cell walls, and the piece of leaf is limp. The cells are then moved into fresh water.

Predict what happens.

A. Nothing: once the contents have shrunk from the cell wall they cannot refill
The change is not permanent.
The contents swell back against the cell wall.
B. The cell wall swells first, and the contents follow
The cell wall does not swell or shrink; the cell wall keeps its shape throughout.
C. The contents lose more water, and the leaf stays limp
Fresh water has far less dissolved in it than the cells' contents, so water moves into the cells.
D. ✓ Water enters and the contents press on the cell wall again; the leaf firms

Why: Fresh water is hypotonic to the cells.
So water enters the cells by osmosis.
The contents swell until the membrane presses against the cell wall again.
So turgor pressure returns, and the piece of leaf becomes firm.
Plasmolysis is reversed.

FRQ 1 T24-frq1 · Analyze Model or Visual Representation

The model shows a cross-section of a plasma membrane with the extracellular fluid above and the cytosol below. Two membrane proteins are drawn, X and Y. Four substances are shown in the extracellular fluid: 1, oxygen (O₂); 2, water (H₂O); 3, sodium ions (Na⁺); 4, glucose.

A plasma membrane with two proteins, X and Y, and four numbered substances in the extracellular fluid, which is shaded darker than the cytosol.

(a) Describe how readily each of the four substances shown, oxygen, water, sodium ions and glucose, crosses the bilayer itself, on its own, away from proteins X and Y. (1 pt)

Model answer O₂ passes freely through the bilayer.
H₂O passes in small amounts.
Na⁺ and glucose do not cross the bilayer on their own.

Rubric

Award 1 point for: O₂ passes freely; H₂O passes in small amounts; Na⁺ and glucose do not cross the bilayer on their own.
Accept: "a little", "slowly" or "only some" for water. All four must be placed correctly for the point.

Slip Letting Na⁺ through because it is small, or blocking water entirely. Small polar molecules with no charge trickle through; ions and large polar molecules do not.

(b) Identify which of the four substances passes through protein X and which passes through protein Y, and name the kind of protein each is. (1 pt)

Model answer Glucose passes through Y.
Y is a carrier protein: Y binds glucose and changes shape to move glucose across.
Na⁺ passes through X.
X is a channel protein: an open water-lined tunnel.
Neither substance passes between the phospholipids.

Rubric

Award 1 point for: glucose identified as passing through Y, named as a carrier protein (it binds glucose and changes shape to move it across); Na⁺ identified as passing through X, named as a channel protein (an open water-lined tunnel). Both paths must go through the proteins, not between the phospholipids.
Accept: "transport protein" for Y. Do not award the point if either substance is said to cross the bilayer between the phospholipids, or if the two proteins are named the wrong way around.

Slip Sending either substance between the phospholipids, or naming the two proteins the wrong way around. A tunnel is a channel; a pocket that binds is a carrier.

(c) Explain, using the structure of the interior of the membrane, what decides whether a dissolved substance crosses the bilayer on its own. (1 pt)

Model answer The hydrocarbon tails in the middle of the membrane carry no charges or partial charges.
Na⁺ carries a full charge, so water attracts Na⁺.
Nothing in the interior attracts Na⁺.
So Na⁺ cannot cross on its own.
O₂ has no charge or partial charge, so nothing in O₂ attracts water.
So O₂ dissolves into the tails and passes out the other side.

Rubric

Award 1 point for: the hydrocarbon tails in the middle of the membrane carry no charges or partial charges; water attracts a charged particle (Na⁺) and nothing in the interior attracts it, so it cannot cross on its own; a molecule with no charge or partial charge (O₂) has nothing that attracts water, so it dissolves into the tails and passes through.
Accept: "the interior is hydrophobic (oily), so charged and polar particles are kept in the water while nonpolar molecules dissolve through" provided the answer names the uncharged tails and what water attracts. Do not award the point for "small things cross and big things do not".

Slip Saying size decides. Na⁺ is smaller than O₂ and still stays out; its charge is what keeps it in the water.

(d) Explain how what the model shows lets a cell keep the solution inside it different from the solution outside. (1 pt)

Model answer The bilayer itself lets only small nonpolar molecules through freely, and a little of small polar molecules.
The bilayer holds back ions and large polar molecules.
So ions and large polar molecules cross only through the proteins the cell has put in its membrane.
So the membrane is selectively permeable.
So the cell controls what enters and leaves.
So the cytosol can be kept different from the extracellular fluid, for example in its amounts of salt and sugar.

Rubric

Award 1 point for: the bilayer itself lets only small nonpolar molecules (and a little of small polar ones) through and holds back ions and large polar molecules, so those cross only through the proteins the cell has put in its membrane; the membrane is selectively permeable, so the cell controls what enters and leaves, and the cytosol can be kept different from the extracellular fluid (for example, its own amounts of salt and sugar).
Accept: "the membrane lets some substances through and holds others back" (the membrane is selectively permeable) together with a link to the cell controlling its inside. Do not award the point for "nothing crosses the membrane" or for "the proteins block everything".

Slip Saying nothing crosses, or that the proteins block everything. The membrane lets some substances through and holds others back; that selectivity is what the cell uses.

FRQ 2 T24-frq2 · Scientific Investigation

A company designs a drug to kill a parasite that lives inside liver cells. The nonpolar drug is a small nonpolar molecule. The charged drug is the same molecule with a charged group added, so that it dissolves better in water. Each drug is supplied at the same concentration to living liver cells and to protein-free bubbles of bilayer. After ten minutes the amount inside is measured as a percentage of the amount outside. The results are in the table.

Amount of each drug inside, as a percentage of the amount outside, after ten minutes.

(a) State what the protein-free bubble result shows about the route the nonpolar drug takes into liver cells, and explain why the result from the living cells alone leaves that route open. (1 pt)

Model answer The bubble result shows that the nonpolar drug crosses the bilayer itself, with no protein: the bubbles are pure bilayer, and the nonpolar drug reached 95% inside them, the same as in the cells.
A living cell’s membrane has proteins in it as well as the bilayer.
So a molecule found inside a living cell could have come through the bilayer or through a channel or carrier.
The bubble result rules the proteins out.

Rubric

Award 1 point for: the nonpolar drug crosses the bilayer itself, with no protein (simple diffusion), because it reached the same 95% in bubbles that have no proteins; the living-cell result alone could not separate that route from passage through a channel or carrier protein, since a living membrane has both the bilayer and proteins.
Accept: "the bubbles are a control for the proteins". The route and the reason the control was needed are both required.

Slip Saying the bubble result shows only that the nonpolar drug "can cross". The point is what it rules out: any protein route.

(b) Explain, using the structure of the hydrophobic interior of the bilayer, why the charged drug stays out of the liver cells while the nonpolar drug gets in. (1 pt)

Model answer The hydrocarbon tails in the interior carry no charges or partial charges.
The nonpolar drug is a small nonpolar molecule, so the nonpolar drug dissolves into the tails and passes through to the other side.
The charged drug carries a full charge, which water attracts, so water keeps the charged drug in the water outside the cell.
Nothing in the tails attracts a charged group.
So the charged drug stays out.

Rubric

Award 1 point for: the hydrocarbon tails in the middle of the membrane carry no charges or partial charges; the nonpolar drug is small and nonpolar, so it dissolves into the tails and passes through; the charged drug carries a full charge, which water attracts, so nothing in the interior attracts the charged drug and the charged drug stays in the water outside.
Accept: "the interior is hydrophobic, so the charged drug cannot enter it", provided the answer says what water does to the charged group. Both drugs must be explained for the point.

Slip Saying the charged drug is blocked because it is bigger. The added group makes it only slightly larger; the charge is what water attracts.

(c) The company then makes the charged drug half its original size, keeping the charged group. Make a claim about whether the smaller charged drug enters the liver cells, and support your claim. (1 pt)

Model answer The smaller charged drug still does not enter.
The charged drug still carries a full charge.
Water still attracts that charge.
The hydrocarbon tails still carry no charges or partial charges, so nothing in the tails attracts the molecule.
Size is not what kept the charged drug out.
Therefore shrinking it changes nothing.

Rubric

Award 1 point for: the claim that it still does not enter (stays near 1%), supported by evidence AND reasoning: the charge, not the size, is what keeps it out; water still attracts the charged group, and nothing in the uncharged tails attracts it, however small the molecule is.
Accept a claim of "a very small amount" if the support says the charge is what decides. Do not award the claim alone, or "it enters because it is now small enough".

Slip Predicting that a smaller molecule slips through. A bare Na⁺ ion is smaller than an O₂ molecule and still does not cross; the charge decides.

(d) The company then finds that the charged drug does enter kidney cells, reaching 60% in ten minutes. Make a claim about what the kidney cells’ membrane has that the liver cells’ membrane lacks, and describe one test of your claim. (1 pt)

Model answer The kidney cells’ membrane has a protein that the charged drug can pass through, a channel or a carrier that fits the charged molecule; the liver cells’ membrane has none.
Test: make protein-free bubbles from the kidney cells’ lipids and supply the charged drug.
If the charged drug stays out of the bubbles but enters the kidney cells, a protein is the route.
Or block the proposed protein with a drug: if uptake falls, that protein was the route.

Rubric

Award 1 point for: the claim that the kidney cells' membrane has a membrane protein (a channel or a carrier) that gives the charged drug a route through the membrane, AND one test: for example, compare uptake into protein-free bubbles made from kidney-cell lipids (the charged drug should stay out), or block or remove the proposed protein and show that uptake of the charged drug into kidney cells falls.
Accept any test that isolates the protein from the bilayer (bubbles with and without kidney-cell membrane proteins; a drug that blocks one protein). A claim without a test, or a test without a claim, does not earn the point.

Slip Claiming that kidney cells have a "weaker" or "thinner" bilayer. A bilayer is a bilayer; only a protein can give a charged molecule a way through.

APBIO-U02-L11 Keeping water in balance

Topic 2.7a · Tonicity and Osmoregulation · 48 steps

A Paramecium in pond water beside a fish in a river

Here is a single-celled pond organism, a Paramecium, in a drop of pond water, and beside it a fish in a river.

A Paramecium lives in pond water that is almost pure. Water pours into it through its membrane every second, and it does not burst.

Unit 2 · Cell Structure and Function

1The contractile vacuole

Video: Watch: The contractile vacuole

Water pours into a Paramecium all the time; a sac collects it and throws it out through a pore.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L11.mp4

A red blood cell in pure water: water enters by osmosis, the cell swells, and it bursts.

The Paramecium’s cytosol holds far more solute than the pond water around it. The pond water is hypotonic to the cell.

So water enters the Paramecium by osmosis, all the time. The Paramecium has no cell wall, and nothing outside the Paramecium stops the swelling.

Inside the cell, a small sac of membrane collects the incoming water. A sac that collects water inside a cell and empties it outside is called a .

Here the contractile vacuole is filling: the vacuole grows as water gathers in it.

A Paramecium with water entering across its whole membrane and a sac inside filling with water

A few seconds later the vacuole squeezes shut and pushes its water out through a pore in the membrane. Then the vacuole starts filling again.

The same Paramecium with the sac squeezed small and its water leaving through a pore in the membrane

In pond water, a Paramecium empties its vacuole 18 times a minute. In water with 0.10 mol/L of solute dissolved in it, the Paramecium empties its vacuole only four times a minute.

Eighteen drops for pond water and four drops for 0.10 mol/L solute: how often the vacuole empties in one minute

Less solute outside means more water entering each minute.

More water entering means more water to throw out. Therefore the vacuole empties more often.

The vacuole does not keep water out. Water enters first; the vacuole removes it afterward.

To say how the emptying rate changes, compare the solute outside before and after the move. In every case here, the water outside holds less solute than the cytosol.

Less solute outside: the vacuole empties more often. More solute outside: the vacuole empties less often.

What you are expected to know Explain how a Paramecium survives in fresh water: water enters it by osmosis all the time, its contractile vacuole collects that water and empties it outside, and the vacuole empties more often when there is less solute outside.

16Fluency quiz: how often does the vacuole empty? mixed practice

Check q1

The bar chart shows the solute in the water around a Paramecium before and after a student moves it to new water.

A bar chart with two bars, labelled before the move and after the move, showing the solute in the water around the Paramecium in mol/L; each bar carries its value above it

After the move, how often does its contractile vacuole empty?

A. More often than before
After the move the water holds more solute, 0.05 mol/L against 0.00 mol/L, so less water enters the cell each minute.
B. ✓ Less often than before
C. About as often as before
The two bars are different heights, so the solute outside has changed.

Why: Compare the two bars.
After the move the water holds more solute: 0.05 mol/L against 0.00 mol/L.
More solute outside means less water enters the cell each minute.
So there is less water to throw out.
So the vacuole empties less often than before.

Check q2

The bar chart shows the solute in the water around a Paramecium before and after a student moves it to new water.

After the move, how often does its contractile vacuole empty?

A. ✓ More often than before
B. Less often than before
After the move the water holds less solute, 0.02 mol/L against 0.10 mol/L, so more water enters the cell each minute.
C. About as often as before
The two bars are different heights, so the solute outside has changed.

Why: Compare the two bars.
After the move the water holds less solute: 0.02 mol/L against 0.10 mol/L.
Less solute outside means more water enters the cell each minute.
So there is more water to throw out.
So the vacuole empties more often than before.

Check q3

The bar chart shows the solute in the water around a Paramecium before and after a student moves it to new water.

After the move, how often does its contractile vacuole empty?

A. More often than before
The two bars are the same height, 0.04 mol/L before and after, so the same amount of water enters each minute.
B. Less often than before
The two bars are the same height, 0.04 mol/L before and after, so the same amount of water enters each minute.
C. ✓ About as often as before

Why: Compare the two bars.
The two bars are the same height: 0.04 mol/L before and 0.04 mol/L after.
The solute outside has not changed.
So the same amount of water enters each minute.
So the vacuole empties about as often as before.

Check q4

The bar chart shows the solute in the water around a Paramecium before and after a student moves it to new water.

After the move, how often does its contractile vacuole empty?

A. More often than before
After the move the water holds more solute, 0.09 mol/L against 0.03 mol/L, so less water enters the cell each minute.
B. ✓ Less often than before
C. About as often as before
The two bars are different heights, so the solute outside has changed.

Why: Compare the two bars.
After the move the water holds more solute: 0.09 mol/L against 0.03 mol/L.
More solute outside means less water enters the cell each minute.
So there is less water to throw out.
So the vacuole empties less often than before.

Check q5

The bar chart shows the solute in the water around a Paramecium before and after a student moves it to new water.

After the move, how often does its contractile vacuole empty?

A. ✓ More often than before
B. Less often than before
After the move the water holds almost no solute, 0.00 mol/L against 0.08 mol/L, so more water enters the cell each minute.
C. About as often as before
The two bars are different heights, so the solute outside has changed.

Why: Compare the two bars.
After the move the water holds almost no solute: 0.00 mol/L against 0.08 mol/L.
Less solute outside means more water enters the cell each minute.
So there is more water to throw out.
So the vacuole empties more often than before.

22Emptying the vacuole uses ATP

Video: Watch: Emptying the vacuole uses ATP

Water comes in for free; squeezing the vacuole shut is what the cell pays ATP for.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L11b.mp4

Water comes into the Paramecium on its own. Water entering by osmosis uses no energy from the cell.

Throwing the water back out does not happen on its own. Squeezing the vacuole shut uses energy.

The cell uses ATP to squeeze the vacuole shut and push its water out through the pore.

What you are expected to know Say that water enters a Paramecium by osmosis with no energy used, and that emptying the contractile vacuole uses energy: the cell uses ATP to squeeze the vacuole shut.

Check q6

A Paramecium in pond water loses its ATP supply. Its membrane is undamaged.

What happens to the cell’s size over the next few minutes?

A. ✓ The cell swells
B. The cell stays the same size
Water enters by osmosis, and osmosis uses no ATP.
C. The cell shrinks
The vacuole only ever moves water out, and without ATP the vacuole cannot do even that.

Why: Pond water is hypotonic to the cell, so water enters by osmosis.
Osmosis uses no ATP, so the water keeps entering.
Emptying the contractile vacuole uses ATP, so with no ATP the vacuole fills but cannot empty.
The water that enters stays in the cell, so the cell swells.

Practice writing an answer

An Amoeba, a single-celled organism with no cell wall, lives in pond water. It empties its contractile vacuole about ten times a minute. A scientist moves it into water that holds exactly as much solute as its cytosol. The cell keeps making ATP.

(a) Predict how often the vacuole empties in the new water, and explain why. (1 pt)

Frame In the new water the vacuole … because …

Model answer In the new water the vacuole stops emptying, or empties only very rarely, because no net water enters the cell.
The new water holds as much solute as the cytosol.
So the water is isotonic to the cell.
Water crosses the membrane both ways at equal rates.
So there is no net movement of water into the cell.
The contractile vacuole only removes water that has entered.
With no water entering, there is no water to throw out.
So the vacuole stops emptying.

Rubric

Award 1 point for: the vacuole stops emptying (or empties far less often), AND the new water is isotonic to the cell, so there is no net movement of water in, AND the vacuole only removes water that has entered.
Accept: “the vacuole empties much less often” with the isotonic reasoning. Do not award the point for “the cell has no ATP” (it does) or for the vacuole emptying more often.

Slip Saying the vacuole empties as often as before because the cell still has ATP. ATP lets the vacuole squeeze, but the vacuole has nothing to squeeze out when no water enters.

30Osmoregulation

Video: Watch: Osmoregulation

A river fish and a marine fish, opposite problems, both solved at the gills.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L11c.mp4

Now two fish. One lives in a river, and river water holds little salt. The other lives in seawater, which is salty: a fish that lives in seawater is called a marine fish.

First the river fish. The fish’s blood holds far more solute than river water does, so river water is hypotonic to the fish.

Water enters the fish across its gills all the time, and salt leaks out of the gills into the river.

A freshwater fish: water enters across the gills, salt leaks out, dilute urine leaves at the rear and salt is pumped in at the gills

The fish gets rid of the water by making large amounts of very dilute urine.

The fish replaces the salt by pumping salt in through its gills, by active transport.

The marine fish has the reverse problem. Seawater holds more solute than the marine fish’s blood, so water leaves the fish across its gills.

A marine fish: water leaves across the gills, it drinks seawater, salt is pumped out at the gills and little urine is made

So a marine fish drinks seawater, pumps the extra salt out through its gills, and makes very little urine.

Controlling the water and solute levels inside a body by constantly moving water and solutes across membranes is called .

Osmoregulation never stops, because the leaks never stop. Growth and homeostasis both depend on this constant movement of water and solutes across membranes.

What you are expected to know Say what osmoregulation is: controlling the water and solute levels inside a body by constantly moving water and solutes across membranes.

What you are expected to know Describe how a freshwater fish controls its water and solute levels.

What you are expected to know Describe how a marine fish controls its water and solute levels.

Check q7

A fish lives in seawater, which holds more solute than the fish’s blood.

Which pair of actions keeps the fish’s water and salt in balance?

A. Making large amounts of dilute urine and pumping salt in at the gills
In seawater the fish loses water, so dilute urine would waste more water, and the fish already has too much salt.
B. Pumping water out across its gills by active transport
Cells pump solutes, not water; water follows by osmosis.
C. Making very little urine and pumping salt in at the gills
Little urine is right for a fish that is losing water, but seawater brings salt into the fish with every drink, so the fish already has too much salt.
D. ✓ Drinking seawater and pumping salt out through its gills

Why: Seawater holds more solute than the fish’s blood, so water leaves the fish across its gills by osmosis.
The fish replaces that water by drinking seawater, which brings in salt.
So the fish pumps the extra salt out through its gills.
The fish also makes very little urine.

Check q8

A freshwater fish’s blood holds more solute than the river around it. A poison stops its gill cells from pumping salt.

What happens to the salt level in its blood?

A. The salt level rises
The fish’s blood holds more salt than the river, so salt leaks out of the fish, not in.
B. ✓ The salt level falls
C. The salt level stays steady
The pumps were replacing salt that leaks out all the time.

Why: The fish’s blood holds more salt than the river, so salt leaks out into the river all the time, down its gradient.
The gill pumps were replacing that salt.
The poison stops the pumps, but the leak continues.
So the blood loses salt and gains none: the salt level falls.

Check q9

The poisoned fish’s salt level falls.

Why does the salt level fall?

A. Salt now enters the fish through its gills by osmosis
Osmosis is the net movement of water, not of salt.
B. The pumps were pushing salt out of the fish, and now the salt stays in the blood
A freshwater fish’s gill pumps bring salt in, to replace the salt that leaks out.
C. ✓ Salt keeps leaking out into the river, and none is pumped back in

The Paramecium does not burst because its contractile vacuole throws the incoming water back out as fast as the water arrives. The fish in the river does the same job with dilute urine and salt pumped in at its gills.

Glossary

contractile vacuole: A sac inside a cell that collects water entering the cell and squeezes it out through a pore in the membrane; the cell uses ATP to squeeze it shut. A Paramecium in pond water empties its contractile vacuole many times a minute.
osmoregulation: Controlling the water and solute levels inside a body by constantly moving water and solutes across membranes, as a freshwater fish does with dilute urine and salt pumped in at its gills.

APBIO-U02-P27 Practice questions: Topic 2.7a

Topic 2.7a · Tonicity and Osmoregulation · 9 MCQ · 2 FRQ · for APBIO-U02-T27

Video: Watch first: osmosis and tonicity, summed up

From the sugar bag in pure water to the three beakers: osmosis, hypotonic, hypertonic and isotonic, and what pure water does to a red cell.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T27-summary.mp4

Q1 P27-q01

A hen's egg has its shell dissolved away, leaving the thin membrane beneath, which lets water through but not sugar. A student measures the egg's mass and leaves the egg overnight in corn syrup, a concentrated sugar solution. In the morning the egg has less mass and is shrunken.

What moved, and which way?

A. ✓ Water left the egg for the syrup, toward the side with more solute
B. Sugar entered the egg from the syrup, toward the side with less sugar
The membrane holds sugar back, and anything entering the egg would raise its mass, not lower it.
C. Water entered the egg from the syrup, toward the side with more water
The egg lost mass, so water left it; water moves toward the side with more solute.
D. Sugar left the egg for the syrup, and water followed it out
The egg holds far less sugar than corn syrup, and the membrane blocks sugar anyway.

Why: The membrane lets water through but holds sugar back, so the sugar stayed where it was.
The syrup holds far more solute than the egg’s contents.
Water moves toward the side with more solute, so water left the egg for the syrup.
Losing water lowered the egg’s mass.

Q2 P27-q02

A student puts sea urchin eggs, each 270 pL (picoliters) in volume, into three solutions of a solute that cannot cross their membranes. After ten minutes: in 0.60 mol/L the eggs were 350 pL; in 1.00 mol/L, 270 pL; in 1.40 mol/L, 210 pL.

Which term describes the 0.60 mol/L solution relative to the eggs, and what did water do?

A. ✓ Hypotonic: water entered the eggs
B. Hypertonic: water left the eggs
In a hypertonic solution water leaves and the cell shrinks.
C. Isotonic: water crossed both ways at equal rates
In an isotonic solution the volume stays constant, as it did in 1.00 mol/L.
D. Hypertonic: water entered the eggs
A solution that makes water enter has less solute than the cell, not more.

Why: The eggs swelled from 270 pL to 350 pL, so water moved into the eggs.
Water moves toward the side with more solute, so the eggs hold more solute than the solution.
A solution with less solute than the cell is hypotonic to it.
1.00 mol/L is isotonic; 1.40 hypertonic.

Q3 P27-q03

A 0.30 mol/L solution of a solute that cannot cross cell membranes is isotonic to human red blood cells. Cells from a crab that lives in seawater, a marine crab, hold about 1.0 mol/L of solute.

Which term describes the same 0.30 mol/L solution relative to the crab cells, and why?

A. Isotonic: a solution's tonicity is fixed by its own concentration
Tonicity describes a solution relative to a particular cell, and the crab cells hold far more than the solution.
B. Hypertonic: 0.30 mol/L is a strong solution for any cell
Hypertonic means more solute than the cell, and the crab cells hold more than three times the solution’s 0.30 mol/L.
C. ✓ Hypotonic: the crab cells hold more solute than the solution
D. Isotonic: both kinds of cell are animal cells
Being an animal cell does not set how much solute a cell holds.

Why: Hypotonic, hypertonic and isotonic describe a solution relative to a particular cell.
The crab cells hold 1.0 mol/L of solute; the solution holds 0.30 mol/L, less than the crab cells.
A solution with less solute than the cell is hypotonic to the cell.

Q4 P27-q04

An animal cell holds 0.30 mol/L of solute that cannot cross its membrane. It sits in 0.30 mol/L glycerol, a small molecule that crosses the membrane freely.

Predict what happens to the cell's volume over the next half hour, and why.

A. The volume stays constant: the total solute concentration is 0.30 mol/L on both sides
Equal total solute does not make two solutions act the same; only solute that cannot cross the membrane drives osmosis.
B. ✓ The volume grows: only the cell's own solute now drives osmosis, so water enters
C. The volume shrinks: the glycerol inside adds to the cell's solute and draws water out
Glycerol equal on both sides pulls water neither way, and the outside has none of the cell’s non-crossing solute.
D. The volume stays constant: a solute that crosses the membrane carries water along with it
Solutes do not carry water; water moves by osmosis toward the side with more solute that cannot cross.

Why: Only solutes that cannot cross the membrane set the tonicity of a solution.
Glycerol crosses, so after half an hour it is equal inside and outside and drives no osmosis.
What is left counting is the cell’s own 0.30 mol/L of solute, which nothing outside matches.
So water enters, and the volume grows.

Q5 P27-q05

Red blood cells from a fish that lives in seawater, a marine fish, hold about 0.35 mol/L of solute that cannot cross their membranes. A few drops of the fish's blood fall into a freshwater stream, whose water holds almost no solute.

Predict what happens to the red blood cells.

A. The cells shrink as water leaves them for the stream
Water moves toward the side with more solute, which is the inside of the cells, so water enters.
B. The cells hold their volume: the stream is isotonic to them
Isotonic means the same solute inside and out, and the stream has almost none against the cells’ 0.35 mol/L.
C. ✓ The cells swell until their membranes tear: osmotic lysis
D. The cells swell a little and then hold, once the stream water matches their inside
The stream holds almost no solute, so the two sides can never match, and water keeps entering until the membranes tear.

Why: The stream water holds almost no solute; the cells hold 0.35 mol/L that cannot cross.
Water moves toward the side with more solute, so water moves into the cells.
The stream can never match the inside, so water keeps entering.
With no cell wall, the cells swell until they tear.

Q6 P27-q06

An Amoeba, a single-celled organism with no cell wall, lives in pond water that holds far less solute than its cytosol. In pond water its contractile vacuole fills and empties every 30 seconds. Moved into water holding 0.08 mol/L of a solute its membrane blocks, it empties its vacuole every 4 minutes.

Explain the change in the emptying rate.

A. The vacuole now keeps water out, so it rarely needs to empty
The vacuole collects water that has already entered.
B. ✓ Less water enters in the new solution, so the vacuole fills more slowly
C. Solute enters the Amoeba, and the vacuole must remove the solute instead
The solute cannot cross the membrane.
D. More water enters in the new solution, and the vacuole stores it
With more solute outside the difference is smaller, so less water enters.

Why: The Amoeba’s cytosol holds more solute than the water around it, so water enters by osmosis.
The contractile vacuole empties as often as incoming water fills it.
Raising the solute outside shrinks the difference, so less water enters each second.
So the vacuole fills more slowly.

Q7 P27-q07

A single-celled pond organism, a Paramecium, sits in pond water, which holds far less solute than its cytosol. At 0 minutes a student adds a chemical that stops the cell making ATP, and records how often the contractile vacuole empties and how wide the cell is. The results are in the table.

Contractile vacuole emptying rate and cell width after the chemical is added.

Which statement explains both trends, the falling emptying rate and the growing width?

A. The vacuole was pumping water into the cell; without ATP it stopped, so water now leaks in through the membrane instead
The vacuole collects water that has already entered and throws it out; water never needed the vacuole to get in.
B. Osmosis stopped when the cell used up its ATP, so the vacuole had no water to empty, and pond solute swelled the cell
Osmosis uses no energy from the cell, so water keeps entering, and the pond’s solute cannot cross the membrane.
C. Without ATP the membrane lets the pond’s solute in, and the entering solute makes the cell swell and drowns the vacuole
The membrane is undamaged and the solute cannot cross it; the cell swells because water enters and is no longer thrown out.
D. ✓ Water still enters by osmosis, but without ATP the vacuole can no longer empty it outside, so the water stays in the cell

Why: The pond water holds far less solute than the cytosol, so water enters by osmosis, which uses no energy and continues after the chemical is added.
Squeezing water out of the contractile vacuole uses ATP.
As the cell uses up its ATP, the vacuole empties less often.
The water that enters stays.

Q8 P27-q08

A seabird drinks seawater, which holds far more solute than its blood. A gland above each eye pumps salt out of the blood into a very salty fluid that drips from the bird's beak, and the bird's blood stays near 0.30 mol/L solute all day.

Which term names what the bird is doing?

A. Osmotic lysis
Osmotic lysis is a cell bursting after too much water enters it, and nothing here is bursting.
B. Facilitated diffusion
Facilitated diffusion runs down a gradient and uses no energy from the cell, and the gland is pumping salt against its gradient.
C. Plasmolysis
Plasmolysis happens in walled cells, and the bird has no walled cells.
D. ✓ Osmoregulation

Why: Controlling the water and solute levels inside a body by constantly moving water and solutes across membranes is osmoregulation.
The bird takes in salt with every drink of seawater.
The gland pumps that salt out of the blood.
So the bird’s blood holds steady at about 0.30 mol/L.

Q9 P27-q09

A frog's skin lets water through, and pond water holds far less solute than the frog's blood. The frog makes large amounts of very dilute urine, and cells in its skin pump salt in from the pond by active transport.

Which problem is each of these two activities solving?

A. ✓ The urine removes water that enters through the skin; the pumps replace salt that leaks out
B. The urine removes salt that enters through the skin; the pumps bring in water
Salt leaks out of the frog into the dilute pond, and cells pump solutes, not water.
C. The urine removes water the frog drank; the pumps remove salt that entered with it
A frog in pond water has no need to drink; water enters through its skin on its own.
D. Both activities keep water out of the frog
Nothing keeps the water out; water enters through the skin by osmosis all the time.

Why: Pond water holds far less solute than the frog’s blood, so water enters through the skin by osmosis.
The frog removes that water as large amounts of dilute urine.
Salt leaks out of the frog into the pond, down its gradient.
The frog replaces that salt by pumping salt in.

FRQ 1 P27-frq1 · Analyze Model or Visual Representation scaffolded

Cells scraped from the inside of a person's cheek hold about 0.30 mol/L of solute that cannot cross their membranes. A student puts some of the cells into each of three dishes of a solute that cannot cross the membranes: dish A at 0.10 mol/L, dish B at 0.30 mol/L and dish C at 0.50 mol/L. The drawing shows one cell from each dish after ten minutes, numbered 1 to 3 in no particular order.

One cheek cell from each dish after ten minutes, numbered 1 to 3 in no particular order.

(a) Identify the dish whose solution is hypotonic to the cells, and identify which drawn cell came from that dish. (1 pt)

Frame Dish … is hypotonic to the cells, because …; the cell from that dish is cell …

Hint Hypo- means less. Compare each dish's solute with the 0.30 mol/L inside the cells. Then ask what a cell does in a solution like that, and look for the drawn cell that has done it.

Model answer Dish A is hypotonic to the cells.
Dish A holds 0.10 mol/L of solute.
The cells hold 0.30 mol/L.
So dish A holds less solute than the cells, and so dish A is hypotonic to the cells.
Water moves toward the side with more solute.
So water entered the cells in dish A, and those cells swelled.
Cell 3 is the swollen cell.
So cell 3 came from dish A.

Rubric

Award 1 point for: dish A (0.10 mol/L), because it holds less solute than the cells, AND cell 3 (the swollen cell) as the cell from dish A.
Do not award the point for dish C (that is the hypertonic dish), for dish A with the reason reversed (more solute), or for dish A matched to cell 1 or cell 2.

Slip Picking the dish where the cells shrank, or matching dish A to the shrunken cell. Hypo- means less solute than the cell, and in less solute the cells swell.

(b) Describe what water is doing at the membranes of the cells in dish B. (1 pt)

Frame In dish B, water is …

Hint The cells in dish B kept their volume. Ask whether water molecules have stopped hitting the membrane, and what a steady volume tells you about any water that does cross.

Model answer In dish B, water is crossing the membranes constantly in both directions.
The 0.30 mol/L solution holds the same solute as the cells, so the solution is isotonic to them.
The two rates of crossing are equal.
So there is no net movement of water.
Therefore the cells keep their volume.

Rubric

Award 1 point for: water is crossing the membranes in both directions at equal rates, so there is no net movement (the solution is isotonic to the cells).
Do not award the point for "water has stopped crossing" or for "nothing is happening".

Slip Saying the water has stopped. Water molecules keep crossing; only the net movement is zero.

(c) Identify which drawn cell came from dish C, and explain why its shape changed. (1 pt)

Frame Cell … came from dish C. Its shape changed because water …, from … toward …

Hint Compare the solute inside the cells with the solute in dish C. Which way does water move across a membrane that lets water through but holds the solute back? Then look for the drawn cell that shows the result.

Model answer Cell 2 came from dish C.
Cell 2 is the shrunken, crinkled cell.
The cells hold 0.30 mol/L of solute.
Dish C holds 0.50 mol/L.
Water moves toward the side with more solute.
So water moved from the cells into the solution.
The solute cannot cross the membrane, so only water moved.
Therefore the cells lost volume, and their outlines crinkled.

Rubric

Award 1 point for: cell 2, because dish C holds more solute (0.50 mol/L) than the cells (0.30 mol/L), so water moved out of the cells by osmosis, from the side with less solute toward the side with more, and the cells lost volume; the solute itself cannot cross.
Accept "the solution is hypertonic, so water left the cells" for the explanation. Do not award the point for cell 3 or cell 1, for "solute entered the cells", or for water leaving with no comparison of the two sides.

Slip Having the solute move into the cells, or matching dish C to the swollen cell. The membrane holds the solute back; it is water that moves, toward the side with more solute.

(d) Predict what would happen to the cells in a fourth dish of pure water, and explain your prediction. (1 pt)

Frame In pure water the cells would … because …

Hint Pure water holds no solute at all. Compare that with the 0.30 mol/L inside the cells, and ask whether the two sides could ever become equal.

Model answer In pure water the cells would swell until their membranes tear and the cells burst.
This bursting is osmotic lysis.
Pure water holds no solute.
So pure water can never match the 0.30 mol/L inside the cells.
Water moves toward the side with more solute.
So water keeps moving into the cells by osmosis.
A cheek cell has no cell wall to stop the swelling.
Therefore the membrane stretches until it tears.

Rubric

Award 1 point for: the cells would swell until their membranes tear and they burst (osmotic lysis), because pure water has less solute than the cells, and no solute at all, so water keeps entering by osmosis and the two sides can never match; an animal cell has no cell wall to hold the swelling.
Accept "burst" with the reason that water keeps entering. Do not award the point for "swell and then stop" or for the cells shrinking.

Slip Having the cells swell and then hold. With no solute outside the two sides never match, so water keeps entering until the membrane tears.

(e) A fifth dish, D, holds 0.50 mol/L of a small solute that crosses the membranes freely. Make a claim about how the cells in dish D compare with the cells in dish C after an hour, and support your claim. (1 pt)

Frame After an hour the cells in dish D are … than the cells in dish C, because …

Hint Dish D's solute can cross the membrane; dish C's cannot. Ask where dish D's solute is after an hour, and what solute is then left unmatched on one side of the membrane.

Model answer After an hour the cells in dish D are larger than the cells in dish C, swollen past their starting size.
Only solutes that cannot cross the membrane set the tonicity of a solution.
The small solute in dish D crosses into the cells until it is equal on both sides.
So it no longer drives osmosis.
The cells’ own 0.30 mol/L of solute is then unmatched outside, so water moves in and the cells swell.
In dish C the solute stays outside, so those cells stay shrunken.

Rubric

Award 1 point for: the claim that the cells in D are larger (swollen past their starting size, or burst) while the cells in C stay shrunken, supported by evidence AND reasoning: the small solute crosses into the cells until it is equal on both sides, so it no longer drives osmosis; only the cells' own 0.30 mol/L of solute that cannot cross is left counting, with none outside to match it, so water moves in.
Accept "D's cells shrink at first, then swell" with the reason. Do not award the claim alone, or "the same as C, because both dishes hold 0.50 mol/L".

Slip Treating dish D like dish C because the totals match. Equal total solute is not isotonic; a solute that crosses stops counting once it is equal on both sides.

FRQ 2 P27-frq2 · Conceptual Analysis

Brine shrimp live in salt lakes whose water holds about 4.0 mol/L of solute, far more than seawater. A brine shrimp's body fluid stays near 0.40 mol/L. Its body surface is thin and lets water through, and it has gills on its legs whose cells can pump salt.

(a) Describe the direction of net water movement across the brine shrimp's body surface in the lake, and explain why water moves that way. (1 pt)

Frame Water moves … across the shrimp's body surface, because …

Model answer Water moves out of the shrimp across its body surface into the lake.
The lake water holds about 4.0 mol/L of solute.
The shrimp’s body fluid holds 0.40 mol/L.
So the lake water holds far more solute than the body fluid: the lake is hypertonic to the shrimp.
Water moves by osmosis toward the side with more solute.
Therefore water leaves the shrimp.

Rubric

Award 1 point for: water moves out of the shrimp into the lake, because the lake water holds far more solute than the body fluid (it is hypertonic to the shrimp) and water moves by osmosis toward the side with more solute.
Accept "the shrimp loses water" with the solute comparison. Do not award the point for water entering, or for the direction with no comparison of the two sides.

Slip Sending water into the shrimp because it lives in water. The lake is far saltier than the shrimp, so water leaves it.

(b) Describe two things the brine shrimp must do to keep its body fluid near 0.40 mol/L. (1 pt)

Model answer The shrimp drinks the lake water.
Drinking replaces the water that leaves across its body surface.
The lake water the shrimp drinks brings in salt.
So the shrimp’s gill cells pump that salt back out into the lake by active transport.
The shrimp also makes very little urine, so it keeps the water it has.

Rubric

Award 1 point for two of: drink the lake water to replace the water it loses; pump the salt that comes in (with the drink and from the lake) out through its gills by active transport; make very little urine (keep the water it has).
Accept "pump salt out" without naming active transport. Do not award the point for one action only, or for making large amounts of dilute urine.

Slip Describing a freshwater animal: dilute urine and salt pumped in would lose water and gain salt, the opposite of what a brine shrimp needs.

(c) A single-celled organism with no cell wall is moved from a freshwater pond into the salt lake. Predict what happens to its volume and to how often its contractile vacuole empties, and justify your prediction. (1 pt)

Model answer In the pond, the cell’s cytosol held more solute than the water around it.
So water entered the cell constantly, and the contractile vacuole filled and emptied many times a minute.
In the lake, the outside holds far more solute than the cytosol.
So water leaves the cell by osmosis, and the cell shrinks.
The contractile vacuole only ever removes water that has entered.
Now no water enters.
So the vacuole stops emptying.

Rubric

Award 1 point for: the cell shrinks (loses water), because the lake holds far more solute than its cytosol and water moves out by osmosis, AND its contractile vacuole stops emptying (or empties very rarely), because the vacuole only removes water that has entered the cell and now little or none does.
Accept "the vacuole has nothing to remove". Do not award the point for the vacuole emptying more often, or for the cell swelling.

Slip Having the vacuole speed up to fight the salt. It removes incoming water; when water leaves instead, it has nothing to do.

(d) Evaluate the claim that keeping its body fluid at 0.40 mol/L uses energy for as long as the brine shrimp lives in the lake. (1 pt)

Model answer The claim is correct.
Salt enters the shrimp all the time: with every drink of lake water, and from the lake down its gradient.
The only way to get that salt out is to pump it against its gradient by active transport.
Active transport uses ATP.
The water loss and the salt gain never stop, so the pumping never stops either.
Therefore holding the body fluid at 0.40 mol/L uses energy for as long as the shrimp lives in the lake.

Rubric

Award 1 point for: the judgement that the claim is correct AND the ground for it: salt enters the shrimp constantly (with every drink and from the lake, down its gradient) and must be pumped out against its gradient by active transport, which uses ATP; the leaks never stop, so the pumping never stops.
Accept "salt keeps coming in, so it keeps having to be pumped out, and pumping uses ATP". Do not award the judgement alone, or "water is pumped out" (cells pump solutes, not water), or a one-off use of energy.

Slip Having the shrimp pump water out. Cells pump solutes; water follows by osmosis. The energy goes on moving salt against its gradient.

APBIO-U02-T27 End-of-topic test: Tonicity and Osmoregulation

Topic 2.7a · Tonicity and Osmoregulation · 19 MCQ · 2 FRQ

Q1 T27-q01

A student fills a bag made of a thin membrane that lets water through but not sucrose with 0.25 mol/L sucrose (table sugar) solution, ties it, measures its mass and lowers it into a beaker of 0.05 mol/L sucrose solution. After an hour the bag has more mass. The student explains: 'Sucrose spread out of the bag into the beaker until it was even on both sides, and that is why the bag has more mass.'

What is wrong with the student's explanation?

A. Nothing; solute spreading out of a bag is what raises its mass
The membrane holds sucrose back, and a bag that lost solute would have less mass, not more.
B. ✓ Losing sucrose would lower the bag's mass; the bag gained water by osmosis
C. Sucrose did spread out, but more water came in than sucrose went out
Sucrose cannot cross this membrane, so none of it spread out; only water moved.
D. Sucrose spread into the bag from the beaker, not out of it
The membrane holds sucrose back in both directions, so no sucrose entered the bag; water did.

Why: The membrane lets water through but not sucrose, so the sucrose stayed inside the bag.
The beaker holds 0.05 mol/L of sucrose; the bag holds 0.25 mol/L.
Water moves toward the side with more solute, so water moved into the bag.
Water entering raised the bag’s mass.

Q2 T27-q02

A tank is divided into two chambers by a membrane that lets water through but not glucose. Chamber A holds 0.05 mol/L glucose and chamber B holds 0.25 mol/L glucose, filled to the same level, as in the drawing.

Two chambers separated by a membrane that lets water through but not glucose, filled to the same level. Each dot is a glucose molecule.

What is the net movement across the membrane over the next hour?

A. Glucose moves from B into A until both chambers hold 0.15 mol/L
The membrane holds glucose back, so the glucose stays where it is; only water moves.
B. Water moves from B into A, toward the side with less solute
Water moves toward the side with more solute, where there is less water per liter, and that side is B.
C. Water crosses both ways at equal rates, so the levels hold
The rates are equal only when the two sides hold the same solute, and A holds 0.05 mol/L against B’s 0.25 mol/L.
D. ✓ Water moves from A into B, toward the side with more solute

Why: The membrane holds glucose back, so the glucose stays where it is.
Chamber A holds less glucose than chamber B, so chamber A holds more water per liter.
So more water molecules cross from A into B than cross back.
B’s level rises; A’s falls.

Q3 T27-q03

A cook slices strawberries, sprinkles sugar over them and leaves the bowl for an hour. The slices are now sitting in a pool of sweet liquid.

Where did the liquid come from?

A. ✓ Water left the strawberry cells by osmosis, toward the sugar on their surfaces
B. Sugar entered the cells and pushed the water out of them
The cell membranes hold sugar back; the sugar stays on the surface.
C. Water moved from the sugar film into the cells, and the cells overflowed into the bowl
Water moves toward the side with more solute, and the sugar film holds far more solute than the cells, so water moved out of the cells.
D. Water gathered from the air onto the sugar
Air supplies far too little water to make a pool of liquid in an hour.

Why: The sugar dissolves in the thin film of water on each cut surface.
That film now holds far more solute than the strawberry cells.
Water moves across the cell membranes toward the side with more solute.
So water leaves the cells and collects in the bowl; the sugar stays outside.

Q4 T27-q04

A student puts cells from a dog's kidney, each 300 μm³ in volume, into three sucrose solutions. Sucrose cannot cross their membranes. After ten minutes their volumes were: 0.20 mol/L sucrose, 400 μm³; 0.30 mol/L sucrose, 300 μm³; 0.50 mol/L sucrose, 215 μm³.

Which term describes the 0.50 mol/L solution relative to the cells, and what did water do?

A. Hypotonic: water left the cells
Hypotonic describes a solution with less solute than the cell, in which water enters and the cells swell.
B. Hypotonic: water entered the cells
The cells shrank from 300 μm³ to 215 μm³, so water left the cells.
C. ✓ Hypertonic: water left the cells
D. Hypertonic: water entered the cells
A hypertonic solution has more solute than the cell, so water moves out of the cell toward the solution.

Why: The cells shrank from 300 μm³ to 215 μm³, so water moved out of the cells.
Water moves toward the side with more solute, so the 0.50 mol/L solution holds more solute than the cells: hypertonic.
0.30 mol/L is isotonic; 0.20 mol/L is hypotonic.

Q5 T27-q05

A saline solution for rinsing contact lenses is isotonic to the cells on the surface of the eye. After an hour in the saline, the volume of those cells is unchanged.

What is water doing at the membranes of those cells during that hour?

A. ✓ Crossing in both directions at equal rates
B. Not crossing in either direction
Water molecules never stop crossing a membrane they can pass through.
C. Entering a little faster than it leaves
Then the cells would gain water and swell, and their volume is unchanged.
D. Leaving a little faster than it enters
Then the cells would shrink, and their volume is unchanged.

Why: Isotonic means the solute concentration is the same inside the cells and outside.
Water molecules keep crossing the membrane in both directions.
The two rates are equal.
So there is no net movement of water.
Therefore the volume of the cells holds.
This is a dynamic equilibrium.

Q6 T27-q06

A student puts cells from a mouse's liver into three solutions of a solute that cannot cross their membranes. The table gives each solution's concentration and the cells' volume at the start and after ten minutes.

Volume of mouse liver cells before and after ten minutes in three solutions of a solute that cannot cross their membranes.

Which solution is isotonic to the cells, and what is water doing at their membranes in it?

A. Solution X: water enters faster than it leaves
The cells in X swelled from 100 μm³ to 128 μm³, so X has less solute than the cells: hypotonic.
B. ✓ Solution Y: water crosses both ways at equal rates
C. Solution Y: water has stopped crossing the membranes
Water never stops crossing a membrane it can pass through.
D. Solution Z: water leaves faster than it enters
The cells in Z shrank from 100 μm³ to 74 μm³, so Z has more solute than the cells: hypertonic.

Why: In solution Y the volume stayed at 100 μm³, so there is no net movement of water.
So Y has the same solute concentration as the cells: isotonic.
In Y, water crosses the membranes constantly in both directions at equal rates.
X is hypotonic; Z is hypertonic.

Q7 T27-q07

A 0.15 mol/L solution of a solute that cannot cross cell membranes is hypotonic to mouse cells, which hold about 0.30 mol/L of solute. Cells from a freshwater mussel hold about 0.05 mol/L of solute.

Which term describes the same 0.15 mol/L solution relative to the mussel cells, and why?

A. Hypotonic: a solution's tonicity is fixed by its own concentration
Tonicity describes a solution relative to a cell, so the same solution can be hypotonic to one cell and hypertonic to another.
B. Isotonic: both kinds of cell are animal cells
Being an animal cell does not set how much solute the cell holds.
C. Hypotonic: 0.15 mol/L is a dilute solution
Dilute compared with the mouse cells is still three times the solute of the mussel cells.
D. ✓ Hypertonic: the mussel cells hold less solute than the solution

Why: Hypotonic, hypertonic and isotonic describe a solution relative to a particular cell.
The solution holds 0.15 mol/L of solute; the mussel cells hold 0.05 mol/L.
So the solution holds more solute than the mussel cells: it is hypertonic to them, and water would leave the cells.

Q8 T27-q08

A student puts sheep red blood cells, each 30 μm³ in volume, into 0.40 mol/L urea and, separately, into 0.40 mol/L sucrose. Urea is a small molecule that crosses the cell membrane; sucrose cannot. The cells’ volumes are in the table.

Volume of the red blood cells in the two solutions.

Why did the cells in urea shrink and then swell?

A. The cells pumped water in using ATP once the urea arrived
Water is never pumped; it crosses on its own toward the side with more solute that cannot cross.
B. Both solutions hold 0.40 mol/L, so both are isotonic to the cells
Equal total solute does not make two solutions act the same; only solute that cannot cross the membrane drives osmosis.
C. Urea left the cells, so water had to enter to replace it
Urea started outside the cells with none inside, so the urea diffused in.
D. ✓ Urea entered until equal on both sides, and water followed it in

Why: At first both solutions held more solute than the cells, so water left and the cells shrank.
Urea crosses the membrane, so urea diffused in until equal on both sides and stopped driving osmosis.
The cells’ own solute then pulled water in, so they swelled.
Sucrose cannot cross.

Q9 T27-q09

Animal cells hold 0.30 mol/L of solute that cannot cross their membranes. Solution X holds 0.30 mol/L sucrose, which cannot cross. Solution Y holds 0.15 mol/L sucrose together with 0.15 mol/L urea, a small molecule that crosses the membrane freely. Both solutions total 0.30 mol/L of solute.

Predict the cells' volume after an hour in each solution.

A. In both solutions the volume stays constant, because each totals 0.30 mol/L
Equal total solute does not make two solutions act alike; only solute that cannot cross the membrane drives osmosis.
B. In X the cells swell; in Y the volume stays constant
In X, 0.30 mol/L sucrose outside matches the cells’ 0.30 mol/L, so the volume stays constant; it is in Y that urea crosses and the cells swell.
C. ✓ In X the volume stays constant; in Y the cells swell
D. In both solutions the cells shrink, because each holds solute the cells lack
Water moves toward the side with more solute that cannot cross, and neither solution has more of that than the cells.

Why: Only solutes that cannot cross the membrane set the tonicity.
In X, 0.30 mol/L sucrose outside matches the cells’ 0.30 mol/L, so the volume stays constant.
In Y, the urea crosses in until equal and drives no osmosis.
That leaves 0.15 mol/L sucrose against 0.30 mol/L inside, so the cells swell.

Q10 T27-q10

Animal cells hold 0.25 mol/L of solute that cannot cross their membranes. Some of these cells are to be kept at their starting volume for a two-hour experiment.

Which solution keeps the cells at their starting volume for the whole two hours?

A. 0.25 mol/L urea, a small molecule that crosses the membrane
Urea spreads until equal on both sides, after which the urea no longer drives osmosis and the cells swell.
B. 0.10 mol/L sucrose, which cannot cross the membrane
0.10 mol/L sucrose is hypotonic to the cells, so the cells swell from the start.
C. 0.50 mol/L sucrose, which cannot cross the membrane
0.50 mol/L sucrose is hypertonic to the cells, so the cells shrink from the start.
D. ✓ 0.25 mol/L sucrose, which cannot cross the membrane

Why: A solution keeps the cells at their volume only if its solute that cannot cross matches the cells’ 0.25 mol/L, and stays matched.
Sucrose cannot cross, so 0.25 mol/L sucrose stays matched for two hours.
Urea stops counting once equal on both sides, so any urea solution ends up hypotonic.

Q11 T27-q11

A cell from the lining of a fish's gut holds 0.35 mol/L of solute that cannot cross its membrane. A student puts it into a solution of 0.15 mol/L of the same kind of solute. After a few minutes the student replaces that solution with one of 0.50 mol/L. The cell stays intact throughout.

How does the cell's volume change, in order?

A. ✓ Swells, then shrinks
B. Shrinks, then swells
In 0.15 mol/L the cell has more solute inside than out, so water moves in and the cell swells first.
C. Swells, then keeps swelling
In 0.50 mol/L there is more solute outside than inside, so water leaves the cell.
D. Holds, then shrinks
0.15 mol/L is less than the cell’s 0.35 mol/L, so the two are not equal and water moves in.

Why: Water moves toward the side with more solute that cannot cross.
First the outside is 0.15 mol/L and the cell 0.35 mol/L, so water enters and the cell swells.
Then the outside is 0.50 mol/L and the cell 0.35 mol/L, so water leaves and the cell shrinks.

Q12 T27-q12

A few drops of blood fall into a tube of distilled water. Within a minute the mixture, which was cloudy red, turns a clear red that light passes straight through.

What happened to the red blood cells?

A. Water left the cells and they shrank to specks too small to scatter light
Water moves toward the side with more solute, which is the inside of the cells, so the cells swelled.
B. The cells' solute leaked out, so they became transparent and hard to see
The membranes hold the cells’ solute in; water is what crosses.
C. ✓ Water entered the cells until their membranes tore, and their contents spilled out
D. The cells dissolved in the water, as salt does
A cell is not a crystal that dissolves; the cell’s membrane tore.

Why: Distilled water has no solute, so it can never match the solute inside a red blood cell.
So water keeps moving into the cells by osmosis.
An animal cell has no cell wall, so each cell swells until its membrane tears: osmotic lysis.
Burst cells leave a clear solution.

Q13 T27-q13

Animal cells hold 0.30 mol/L of solute that cannot cross their membranes. A student puts two samples of these cells into solutions of the same solute: one sample into 0.20 mol/L, the other into 0.05 mol/L.

Compare the two samples after five minutes.

A. Both samples swell by the same amount, since both solutions are hypotonic
How much water enters depends on how far apart the two sides are, and 0.05 mol/L is further below 0.30 mol/L than 0.20 mol/L is.
B. Only the cells in 0.05 mol/L swell; 0.20 mol/L is close enough to count as isotonic
Isotonic is equal solute, and 0.20 mol/L is below the cells’ 0.30 mol/L, so water enters those cells too.
C. The cells in 0.20 mol/L shrink and the cells in 0.05 mol/L swell
Cells shrink only in a solution with more solute than they hold, and both solutions hold less.
D. ✓ Both samples swell, and the cells in 0.05 mol/L swell more

Why: Both solutions hold less solute than the cells’ 0.30 mol/L, so water enters the cells in both samples.
How much water enters depends on how far apart the two sides are.
0.05 mol/L is further below 0.30 mol/L than 0.20 mol/L is.
So the cells in 0.05 mol/L swell more.

Q14 T27-q14

A crayfish lives in a river. The river water holds far less solute than the crayfish’s blood. All day, cells in the crayfish’s gills pump salt in from the river by active transport.

Which problem does the salt pumping solve?

A. The crayfish takes in too much salt when it drinks river water, and the pumps balance that
A river animal does not need to drink; water enters it by osmosis, and river water carries very little salt in with it.
B. ✓ The crayfish loses salt to the river across its gills and skin, and the pumps replace it
C. The crayfish loses water to the river across its gills, and the pumped salt draws water back in
The river holds less solute than the blood, so water enters the crayfish by osmosis; the crayfish has too much water, not too little.
D. The crayfish’s blood holds too much salt for its cells, and the pumps move salt into the gills to store it
The pumps move salt from the river into the blood, because the blood is losing salt, not because it has too much.

Why: The river water holds far less solute than the crayfish’s blood, so salt diffuses out into the river all the time.
To keep its blood salt steady, the crayfish must bring salt back in.
From the dilute river into the saltier blood is against the gradient: pumping.

Q15 T27-q15

A single-celled freshwater organism with no cell wall is moved from pond water into a solution whose solute concentration matches that of its cytosol.

Predict what its contractile vacuole does in the new solution, and why.

A. ✓ The vacuole fills far more slowly and rarely empties, because almost no net water enters the cell
B. The vacuole empties more often, to make room for the water the cell now takes in
With the outside matching the cytosol there is no net gain of water for the vacuole to remove.
C. The vacuole empties at the same rate as in pond water, because it squeezes at a fixed rhythm
The vacuole has no fixed rhythm: it empties as often as incoming water fills it.
D. The vacuole fills with solute instead of water, to bring the cell back to its old concentration
A contractile vacuole collects water, not solute.

Why: A contractile vacuole collects the water that enters by osmosis and empties as often as that water fills it.
In the new solution the solute outside matches the cytosol.
So water crosses in and out at equal rates, with no net gain.
So the vacuole has almost nothing to collect.

Q16 T27-q16

A single-celled organism from a freshwater stream has no cell wall, and the solute used in this experiment cannot cross its membrane. The graph gives how often its contractile vacuole empties in four solutions of that solute.

How often the contractile vacuole of a single-celled stream organism empties, in four solutions of a solute that cannot cross its membrane. Gridlines every 5 emptyings per minute.

The organism is moved from the 0.12 mol/L solution back into the 0.00 mol/L solution. Predict the change in its vacuole, and why.

A. The vacuole empties less often, because there is less solute outside for it to remove
The vacuole removes water, and with less solute outside more water enters.
B. The vacuole empties at the same rate, because it squeezes at a fixed rhythm
The graph shows the rate changing with the outside solute, from 3 to 30 emptyings a minute.
C. ✓ The vacuole empties about ten times more often, because more water enters each minute with less solute outside
D. The vacuole stops emptying, because in pure water the vacuole keeps water out of the cell
Water enters first and the vacuole removes it afterward.

Why: The cytosol holds more solute than the water around the cell, so water enters by osmosis; less solute outside means more water enters each minute.
The contractile vacuole empties as often as water fills it.
The graph shows 3 emptyings a minute at 0.12 mol/L and 30 at 0.00 mol/L.

Q17 T27-q17

A sea turtle lives in seawater that has far more solute than its blood. It drinks seawater, and a gland beside each eye uses active transport to push salt out of its blood into very salty tears.

What problem is the turtle solving by drinking and pushing out salt?

A. It gains water from the sea and must get rid of the excess
Seawater has more solute than the turtle’s blood, so water moves out of the turtle.
B. It loses salt to the sea and must drink to replace it
With more salt outside than in, salt tends to enter the turtle.
C. ✓ It loses water to the saltier sea and takes in salt as it drinks
D. It has no water problem; the drinking is only for feeding
The turtle loses water to the saltier sea by osmosis, so the drinking replaces water.

Why: Seawater holds more solute than the turtle’s blood, and water moves toward the side with more solute.
So the turtle constantly loses water to the sea across its thin surfaces.
The turtle replaces that water by drinking seawater, which brings salt in.
So the turtle must push the salt out.

Q18 T27-q18

A salmon hatches in a river and later swims out to sea. In the river its blood holds far more solute than the water around it; in the sea it holds far less. In the river the salmon makes large amounts of dilute urine and pumps salt in through its gills.

Which set of changes must happen when the salmon reaches the sea?

A. ✓ The salmon drinks seawater, pumps salt out at its gills and makes little urine
B. The salmon stops drinking, makes even more dilute urine and pumps salt in faster
In the sea they would dehydrate and over-salt the fish.
C. The salmon seals its gills so that water and salt stop crossing
Gills must stay thin and open for the fish to take in oxygen, so water keeps leaving across them in the sea.
D. The salmon stops osmoregulating, because seawater and blood are close enough to be isotonic
Seawater holds about three times the solute of the salmon’s blood, so the fish must osmoregulate constantly.

Why: In the river, water enters across the gills and salt leaks out, so the salmon makes dilute urine and pumps salt in.
In the sea, the water holds more solute than its blood, so water leaves and salt enters.
So the salmon drinks seawater, pumps salt out and saves water.

Q19 T27-q19

After drinking a liter of water, a healthy person's blood becomes slightly more dilute. Over the next two hours the kidneys make a large volume of very dilute urine, and the blood's solute concentration returns to where it was.

What is the body doing over those two hours?

A. Osmosis: water leaves the blood for the urine because the urine holds more solute
Osmosis moves water toward more solute, and this urine holds less solute than the blood.
B. ✓ Osmoregulation: moving the extra water out so the blood's solute concentration stays steady
C. Active transport of water into the kidney, using ATP
Cells pump solutes, not water; water follows by osmosis.
D. Removing solute from the blood to match the extra water
Removing solute would dilute the blood further; the urine is dilute because water is being removed while solute is kept.

Why: Controlling the water and solute levels inside a body by constantly moving water and solutes across membranes is osmoregulation.
The drink added water to the blood, so the blood became more dilute.
The kidneys removed that extra water as dilute urine and kept the solute.

FRQ 1 T27-frq1 · Analyze Model or Visual Representation

Frog red blood cells hold about 0.22 mol/L of solute that cannot cross their membranes. A student adds a drop of frog blood to each of four dishes: 0.05 mol/L, 0.22 mol/L and 0.40 mol/L of a solute that cannot cross the membranes, and pure water. The drawing shows a cell from three of the dishes after ten minutes; the cell from the 0.40 mol/L dish is not drawn. In the pure-water dish only torn fragments of membrane were found.

A frog red blood cell from three of the dishes after ten minutes. The cell from the 0.40 mol/L dish is not drawn. In the pure-water dish only torn fragments of membrane were found.

(a) Identify the tonicity of the 0.40 mol/L solution relative to the cells, and predict what the cells in that dish looked like after ten minutes. (1 pt)

Model answer The 0.40 mol/L solution holds more solute than the 0.22 mol/L inside the cells.
So the 0.40 mol/L solution is hypertonic to the cells.
Water moves toward the side with more solute.
So water left the cells.
After ten minutes the cells were smaller than the cell drawn from the 0.22 mol/L dish, and their outlines were crinkled.

Rubric

Award 1 point for: the 0.40 mol/L solution is hypertonic to the cells (more solute than the cells' 0.22 mol/L), and the cells shrank (lost water; smaller than the 0.22 mol/L cell, with a crinkled outline).
Accept "more concentrated than the cells" for hypertonic if the word is not used, and "shrivelled" or "crenated" for the shape. Do not award the point for hypotonic, or for shrinking with the tonicity unnamed.

Slip Swapping the prefixes. Hyper- means more: a solution with more solute than the cell is hypertonic, and the cell in it shrinks.

(b) Explain what water is doing at the membranes of the cells in the 0.22 mol/L dish, where the cells kept their volume. (1 pt)

Model answer The 0.22 mol/L solution holds the same solute as the cells.
So the solution is isotonic to the cells.
Water molecules keep crossing the membrane in both directions.
The two rates are equal.
So there is no net movement of water.
Therefore the cells’ volume stays constant.

Rubric

Award 1 point for: the solution is isotonic to the cells (same solute concentration inside and out), so water crosses the membrane in both directions at equal rates; there is no net movement of water, so the volume stays constant.
Accept "as much water enters as leaves". Do not award the point for "water stopped crossing" or for "the solution is isotonic" with no statement about the two flows being equal.

Slip Saying the water stopped crossing. In an isotonic solution water crosses constantly in both directions; only the net movement is zero.

(c) Explain why only torn fragments of membrane were found in the dish of pure water. (1 pt)

Model answer Pure water has no solute at all.
So the pure water can never match the 0.22 mol/L inside the cells.
Water moves toward the side with more solute.
So water keeps moving into the cells by osmosis.
A frog red blood cell has no cell wall to stop the swelling.
So each cell swells until its membrane tears, and its contents spill out.
This is osmotic lysis.
Only the torn membranes are left.

Rubric

Award 1 point for: pure water holds no solute, so it can never match the 0.22 mol/L inside the cells; water keeps entering by osmosis (net movement toward the side with more solute), and an animal cell has no cell wall to hold it, so each cell swells until its membrane tears: osmotic lysis.
Accept "the cells burst" with the reason that water kept entering because the outside has less solute (or no solute). Do not award the point for "the cells dissolved" or for water entering with no reason given.

Slip Saying the cells dissolved. Cells burst because water keeps entering; the membrane tears when it can stretch no further.

(d) A fifth dish holds 0.40 mol/L of a small solute that crosses the membranes freely. Make a claim about what the cells look like after an hour, and support your claim. (1 pt)

Model answer After an hour the cells have swollen past their starting size, and some may have burst.
At first the outside holds more solute than the cells, so water leaves and the cells shrink.
The small solute crosses the membrane and enters the cells until it is equal on both sides.
So it then drives no osmosis.
the cells’ own 0.22 mol/L of solute is still there, and nothing outside matches it.
Therefore water moves into the cells, and the cells swell.

Rubric

Award 1 point for: the claim that after an hour the cells have swollen past their starting size (or burst), supported by evidence AND reasoning: the small solute crosses into the cells until it is equal on both sides and then no longer drives osmosis, leaving the cells' own 0.22 mol/L of solute that cannot cross, with none outside to match it, so water moves in.
Accept an answer that adds an early shrinking before the swelling. Do not award the claim alone, or "the cells stay shrunken, like the 0.40 mol/L dish".

Slip Treating the fifth dish like the 0.40 mol/L dish because the totals match. Equal total solute is not isotonic; a solute that crosses stops counting once it is equal on both sides.

FRQ 2 T27-frq2 · Conceptual Analysis

A killifish lives in a tidal creek. At low tide the creek holds almost fresh water, about 0.01 mol/L of solute; at high tide seawater fills it, about 1.0 mol/L. The fish's blood stays near 0.35 mol/L of solute all day. Its gills are thin and let water through. A single-celled organism with no cell wall lives in the same creek and has a contractile vacuole.

(a) Describe the direction of net water movement across the fish's gills at low tide and at high tide. (1 pt)

Model answer At low tide the creek water holds far less solute than the fish’s blood.
So the creek is hypotonic to the fish.
Water moves toward the side with more solute.
So at low tide water moves into the fish across its gills.
At high tide the seawater holds far more solute than the blood.
So the seawater is hypertonic to the fish.
Therefore at high tide water moves out of the fish across its gills.

Rubric

Award 1 point for both: at low tide water moves into the fish (the creek holds less solute than the blood, so it is hypotonic to the fish), and at high tide water moves out of the fish (seawater holds more solute than the blood, so it is hypertonic).
Accept the directions with the solute comparison stated in either form (less solute outside; more water outside). Do not award the point for one tide only, or for the directions reversed.

Slip Giving one direction for the whole day. The creek changes from hypotonic to hypertonic with the tide, so the net movement of water reverses.

(b) Explain how the fish keeps its blood near 0.35 mol/L at low tide, including what it does with water and what it does with salt. (1 pt)

Model answer At low tide water enters the fish across its gills all the time.
Salt leaks out of the fish into the dilute creek all the time.
The fish removes the water by making large amounts of very dilute urine.
The fish replaces the lost salt by pumping salt in through its gills by active transport.
Active transport uses energy from the fish.
Together the urine and the pumping hold the blood near 0.35 mol/L.

Rubric

Award 1 point for both: it gets rid of the water that enters by making large amounts of very dilute urine, AND it replaces the salt that leaks out into the creek by pumping salt in through its gills by active transport (using energy).
Accept "pumps salt in at the gills" without naming active transport. Do not award the point for urine alone or salt alone, or for the fish drinking water at low tide.

Slip Handling only the water. Salt leaks out down its gradient all the time, so it has to be pumped back in as constantly as the water is removed.

(c) Predict two changes in what the fish does when the tide comes in and the creek fills with seawater, and justify one of them. (1 pt)

Model answer When the tide comes in, the fish starts drinking seawater.
The fish also makes very little urine.
Its gills pump salt out instead of in.
The fish drinks because water now leaves across its gills toward the saltier sea, so the fish must replace the water it loses.
Dilute urine would waste more water, so the fish makes little urine.
Salt comes in with every drink and from the sea, so the fish has to pump that salt back out.

Rubric

Award 1 point for two changes from: it starts drinking the seawater; it pumps salt out through its gills instead of in; it makes much less urine (small amounts, concentrated); AND a justification for one of them: water now leaves across the gills toward the saltier sea, so it must replace water (drinking) and stop wasting it (little urine), or salt now enters with the drink and from the sea, so it must be pumped out.
Accept any two of the three changes. Do not award the point for the changes with no justification, or for a justification in which water enters the fish at high tide.

Slip Keeping the low-tide habits. Dilute urine and salt pumped in would dehydrate and over-salt a fish in seawater.

(d) Predict how often the single-celled organism's contractile vacuole empties at high tide compared with low tide, and explain the difference. (1 pt)

Model answer At low tide the creek is hypotonic to the cell, so water enters constantly by osmosis, and the contractile vacuole fills and empties many times a minute to throw it back out.
At high tide the seawater holds far more solute than the cell, so little or no net water enters.
The contractile vacuole only ever removes water that has already entered.
So at high tide the vacuole has almost nothing to collect and empties rarely, or stops.

Rubric

Award 1 point for: at high tide it empties far less often (or stops), because with about 1.0 mol/L of solute outside little or no net water enters the cell (water may even leave), and the contractile vacuole only removes water that has already entered; at low tide water pours in by osmosis and the vacuole fills and empties often.
Accept "stops emptying" or "empties rarely" for high tide. Do not award the point for "empties more often to pump water in" or for the vacuole keeping water out.

Slip Having the vacuole work harder in seawater. It removes water that has entered; when water stops entering, it has nothing to do.

APBIO-U02-L12 Water potential: a number for which way water goes

Topic 2.7b · Water Potential · 72 steps

A U-tube with a membrane across the bottom: pure water on the left, sugar solution standing higher on the right

Here is a U-tube with a membrane across the bottom: pure water on the left, sugar solution on the right.

The right-hand level climbs, and keeps climbing for as long as you watch. In principle the level would stop climbing only when the weight of the raised column pushed water back out as fast as osmosis brought water in. To say where that balance lies, we need a number for how strongly water tends to move.

Unit 2 · Cell Structure and Function

1Water potential: which way water tends to go

Check q1

A dialysis bag of 0.50 mol/L sucrose hangs in a beaker of 0.10 mol/L sucrose. The membrane lets water through and holds sucrose back.

Which of the following is the net movement?

A. ✓ Water moves into the bag
B. Water moves out of the bag
There is more sucrose inside the bag than outside, so water moves toward the inside, the side with more solute.
C. Sucrose moves out of the bag
The membrane holds sucrose back, so the sucrose stays where it is.

Why: There is more sucrose inside the bag than outside, so the net movement of water is into the bag, toward the side with more solute.

Video: Watch first: A level that keeps climbing

A U-tube whose sugar side keeps rising, and the question that needs a number.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T27B-intro.mp4

Video: Watch: Water potential: which way water tends to go

Osmosis as a number: higher psi to lower psi, in bar, zero at pure water.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L12.mp4

Osmosis is the starting point: across a membrane that holds solute back, water moves from the side with less solute to the side with more.

Here is the same fact as a number. We give each side of the membrane a value, and water moves from the side with the higher value to the side with the lower value.

Two compartments of solution, shaded differently, separated by a membrane drawn as a dashed line and labelled; the left compartment is at −2 bar and the right at −5 bar, with an arrow for water from the −2 bar side to the −5 bar side

That value, a measure of how strongly water tends to move, is called the , written Ψ (the Greek letter psi).

Water moves by osmosis from a region of higher water potential to a region of lower water potential: from −2 bar toward −5 bar, never the reverse.

Water potential is measured in a unit of pressure called the . The air around you presses on you at about 1 bar; a car tire holds about 2 bar.

Every scale needs a zero. We give pure water in an open container Ψ = 0 bar. Everything else is measured against it.

What you are expected to know Water moves between two regions from the higher Ψ to the lower; pure water in an open container has Ψ = 0 bar.

Check q2

Which of these has a water potential of exactly 0 bar?

A. A sugar solution in an open beaker
A sugar solution holds dissolved solute, and dissolved solute lowers its water potential below zero.
B. ✓ Pure water in an open beaker
C. Any sample, before water starts to move
Water potential is a tendency to move, so a sample has its value whether or not water is moving yet.
D. The cytosol of a red blood cell
The cytosol holds dissolved solutes, and dissolved solute lowers its water potential below zero.

Why: Pure water in an open container is the zero of the scale, with a water potential of 0 bar.
A sugar solution holds dissolved solute.
The cytosol holds dissolved solute.
Dissolved solute lowers water potential below zero.
So only the pure water in the open beaker sits at exactly 0 bar.

13Fluency quiz: which way does water move? mixed practice

Check q3

The bar chart shows the water potential of a solution and of a cell sitting in it.

A bar chart of water potential in bars for the solution and the cell: the zero line is at the top, each bar hangs down from it, and each bar carries its value at its lower end

Which way does water move?

A. ✓ Into the cell
B. Out of the cell
Water moves toward the lower water potential; the cell, at −7 bar, is lower than the solution, at −3 bar.
C. Neither: there is no net movement
The two bars are different lengths, so the two water potentials differ, and there is a net movement of water.

Why: Water moves from the higher water potential to the lower.
The solution is at −3 bar.
The cell is at −7 bar.
The cell’s bar reaches further below zero, so −7 bar is the lower water potential.
So water moves from the solution into the cell.

Check q4

The bar chart shows the water potential of a solution and of a cell sitting in it.

Which way does water move?

A. Into the cell
Water moves toward the lower water potential; the solution, at −8 bar, is lower than the cell, at −4 bar.
B. ✓ Out of the cell
C. Neither: there is no net movement
The two bars are different lengths, so the two water potentials differ, and there is a net movement of water.

Why: Water moves from the higher water potential to the lower.
The solution is at −8 bar.
The cell is at −4 bar.
The solution’s bar reaches further below zero, so −8 bar is the lower water potential.
So water moves from the cell out into the solution.

Check q5

The bar chart shows the water potential of a solution and of a cell sitting in it.

Which way does water move?

A. Into the cell
The solution and the cell are both at −6 bar, so the two water potentials are equal and there is no net movement.
B. Out of the cell
The solution and the cell are both at −6 bar, so the two water potentials are equal and there is no net movement.
C. ✓ Neither: there is no net movement

Why: Water moves from the higher water potential to the lower.
The solution is at −6 bar.
The cell is at −6 bar.
The two bars are the same length, so the two water potentials are equal.
So water crosses both ways at equal rates, and there is no net movement.

Check q6

The bar chart shows the water potential of pure water in an open dish and of a cell sitting in it.

A bar chart of water potential in bars for the pure water and the cell: the zero line is at the top, each bar hangs down from it, and each bar carries its value at its lower end

Which way does water move?

A. ✓ Into the cell
B. Out of the cell
Water moves toward the lower water potential; the cell, at −5 bar, is lower than the pure water, at 0 bar.
C. Neither: there is no net movement
The two bars are different lengths, so the two water potentials differ, and there is a net movement of water.

Why: Water moves from the higher water potential to the lower.
The pure water is at 0 bar; the cell is at −5 bar.
The cell’s bar reaches below zero, so −5 bar is the lower water potential.
So water moves from the pure water into the cell.

Check q7

The bar chart shows the water potential of a solution and of a cell sitting in it.

Which way does water move?

A. Into the cell
Water moves toward the lower water potential; the solution, at −9 bar, is lower than the cell, at −1 bar.
B. ✓ Out of the cell
C. Neither: there is no net movement
The two bars are different lengths, so the two water potentials differ, and there is a net movement of water.

Why: Water moves from the higher water potential to the lower.
The solution is at −9 bar.
The cell is at −1 bar.
The solution’s bar reaches far below zero, so −9 bar is the lower water potential.
So water moves from the cell out into the solution.

Check q8

A cell has a water potential of −5 bar. It sits in a solution with a water potential of −2 bar.

Which way does water move?

A. ✓ Into the cell
B. Out of the cell
Water moves toward the lower water potential; the cell, at −5 bar, is lower than the solution, at −2 bar.
C. Neither: there is no net movement
−5 bar and −2 bar differ, so the two water potentials differ and there is a net movement of water.

Why: Water moves from the higher water potential to the lower.
The solution is at −2 bar.
The cell is at −5 bar.
−5 bar is further below zero than −2 bar, so −5 bar is the lower water potential.
So water moves from the solution into the cell.

20Solute lowers water potential

Check q9

A glass of pure water stands open on a bench.

What is the water potential of the water in the glass?

A. ✓ 0 bar
B. Below 0 bar
Pure water in an open container holds no solute and has nothing pressing on it, so it sits at the zero of the scale.
C. Above 0 bar
Nothing presses on the water in an open glass, so nothing raises its water potential above zero.

Why: Pure water in an open container is the zero of the scale, so its water potential is 0 bar.

Video: Watch: Solute lowers water potential

Stir in sucrose and the number drops below zero: the solute potential, Ψs.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L12b.mp4

Start with pure water in an open beaker: Ψ = 0 bar. Stir in sucrose. Put a membrane between this beaker and pure water, and water tends to move into the beaker: its water potential is now below zero.

Three open beakers: pure water, 0.1 mol/L sucrose and 0.3 mol/L sucrose, with the solute potential falling from zero

Stir in more sucrose, and water is drawn in more strongly. The water potential drops further below zero.

The part of the water potential that comes from dissolved solute is called the , Ψs.

Ψs is zero for pure water and negative for every solution. Adding solute only ever lowers it; a solute potential is never positive.

So 0.1 mol/L sucrose has a negative Ψs. 0.3 mol/L sucrose has a more negative Ψs. Water moves toward the lower water potential, so water moves from the 0.1 mol/L solution toward the 0.3 mol/L solution. That is the direction osmosis takes.

What you are expected to know Say what dissolving solute does to water potential: it lowers it below zero, and more solute lowers it further.

What you are expected to know Predict which way water moves between two solutions: toward the more concentrated solution, which has the lower water potential.

What you are expected to know Say what the solute potential, Ψs, is: the part of water potential due to dissolved solute, zero for pure water and negative for every solution.

Check q10

Two open beakers hold glucose solutions: 0.2 mol/L and 0.5 mol/L.

How do their solute potentials compare?

A. The 0.2 mol/L solution has the more negative Ψs
The 0.2 mol/L solution holds less glucose, so its solute potential is the less negative of the two.
B. Both beakers have Ψs = 0 bar
An open beaker sets the pressure potential to zero, not the solute potential.
C. The 0.5 mol/L solution has a positive Ψs
Adding solute only ever lowers water potential, so a solute potential is never positive.
D. ✓ The 0.5 mol/L solution has the more negative Ψs

Why: Dissolved solute lowers the solute potential below zero.
More solute lowers it further.
The 0.5 mol/L solution holds more glucose than the 0.2 mol/L solution.
So the 0.5 mol/L solution has the more negative solute potential.

32Pressure raises water potential

Check q11

Two open beakers hold sucrose solutions: 0.1 mol/L and 0.4 mol/L.

Which solution has the lower solute potential?

A. The 0.1 mol/L solution
The 0.1 mol/L solution holds less sucrose, so its solute potential is the less negative of the two.
B. ✓ The 0.4 mol/L solution
C. Neither: an open beaker sets both to 0 bar
An open beaker sets the pressure potential to zero, not the solute potential.

Why: Dissolved solute lowers the solute potential below zero, and the 0.4 mol/L solution holds more solute, so its solute potential is the lower.

Video: Watch: Pressure raises water potential

Hold the solute steady, let the cell wall press: the pressure potential, Ψp.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L12c.mp4

Now hold the solute steady and change one other thing: pressure.

In a turgid plant cell, water has entered, the contents press outward on the cell wall, and the cell wall presses back on them.

An open beaker of solution with Ψp = 0 bar beside a walled cell; the cell wall is labelled and presses inward on the cell’s contents at 3 bar

Squeezed water tends to leave. So a solution under pressure has a higher water potential than the same solution left alone: the cell wall’s push raises the water potential of the contents.

The part of the water potential that comes from pressure is called the , Ψp.

A solution in an open beaker has nothing pressing on it beyond the air, and the air’s push is counted as zero, so its Ψp = 0 bar.

The contents of a turgid cell are pressed on by the cell wall. If the cell wall pushes at 3 bar, the contents have Ψp = +3 bar. In a turgid cell Ψp is positive.

What you are expected to know Pressure on a solution raises its water potential. The part due to pressure is the pressure potential, Ψp: zero in an open container, positive inside a turgid cell.

Check q12

A turgid plant cell’s contents press on its cell wall, and the cell wall presses back on them at 5 bar.

What is the pressure potential of the cell’s contents?

A. −5 bar
Pressure raises water potential, so the pressure potential of contents pressed on by a cell wall is positive.
B. 0 bar
Zero is the pressure potential of a solution with nothing pressing on it, and here the cell wall presses on the contents at 5 bar.
C. ✓ +5 bar
D. It depends on how much solute the cell holds
Solute sets the solute potential; the pressure potential comes from the push alone.

Why: The cell wall presses on the contents at 5 bar.
Pressure raises water potential.
So the push gives the contents a positive pressure potential, Ψp = +5 bar.
Solute sets the solute potential, not the pressure potential.

43Add the two parts

Video: Watch: Add the two parts

Ψ = Ψp + Ψs, worked once: values, equation, substitution, answer with sign and unit.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L12d.mp4

You’ve now seen both parts on their own: solute pulls the water potential down, pressure pushes it up. Together they give the whole water potential.

The equation Ψ = Ψp + Ψs with each symbol named: water potential equals pressure potential plus solute potential, all in bars

Ψ = Ψp + Ψs, all three in bars. The three symbols:

Ψ: the water potential, the whole value, which sets which way water moves.

Ψp: the pressure potential, the part that comes from pressure on the water.

Ψs: the solute potential, the part that comes from dissolved solute.

Worked example

A plant cell has Ψp = +3 bar and Ψs = −8 bar. What is its water potential?

Write down the values in the question:

Ψp = +3 bar

Ψs = −8 bar

Write down the equation:

Ψ = Ψp + Ψs

Substitute the values into the equation:

Ψ = (+3) + (−8)

Calculate:

Ψ = −5 bar

Check q13 numeric entry

A sucrose solution in an open beaker has Ψs = −4 bar, with nothing pressing on it beyond the air.

Calculate the water potential of the solution.

Part 1. Nothing presses on the solution beyond the air. What is its pressure potential, Ψp?

Answer: 0 bar (tolerance ±0.5)

Working

Write down the pressure potential of a solution in an open beaker:

Ψp = 0 bar

Part 2. What is its solute potential, Ψs?

Answer: -4 bar (tolerance ±0.5)

Working

Read the solute potential from the question:

Ψs = −4 bar

Answer: -4 bar (tolerance ±0.5)

Working

Write down the values in the question:

Ψp = 0 bar

Ψs = −4 bar

Write down the equation:

Ψ = Ψp + Ψs

Substitute the values into the equation:

Ψ = 0 + (−4)

Calculate:

Ψ = −4 bar

In an open beaker the water potential is just the solute potential. Inside a turgid cell the cell wall’s push lifts it toward zero.

What you are expected to know Calculate the water potential of a cell or a solution from its two parts with Ψ = Ψp + Ψs, in bars, writing the values, the equation, the substitution and the result.

Check q14 numeric entry

A plant cell has Ψp = +2 bar and Ψs = −9 bar.

Calculate its water potential.

Answer: -7 bar (tolerance ±0.5)

Working

Write down the values in the question:

Ψp = +2 bar

Ψs = −9 bar

Write down the equation:

Ψ = Ψp + Ψs

Substitute the values into the equation:

Ψ = (+2) + (−9)

Calculate:

Ψ = −7 bar

Check q15 numeric entry

A plant cell has a pressure potential of +4 bar and a solute potential of −10 bar.

Calculate its water potential.

Answer: -6 bar (tolerance ±0.5)

Working

Write down the values in the question:

Ψp = +4 bar

Ψs = −10 bar

Write down the equation:

Ψ = Ψp + Ψs

Substitute the values into the equation:

Ψ = (+4) + (−10)

Calculate:

Ψ = −6 bar

What you are expected to know Water potential adds two parts: solute pulls it down, pressure pushes it up, and water moves toward the lower total.

Back to the U-tube. The sugar gives the right-hand solution a solute potential below zero, so water crosses into it and the column climbs.

The same U-tube as at the start, at balance: a membrane across the bottom, labelled; pure water on the left at Ψ = 0 bar; sugar solution standing higher on the right, its raised column pressing down on the solution so that the solution’s water potential is back at zero

As the column rises, its weight presses harder on the solution. That is a pressure potential above zero, growing as the column rises.

If nothing else interfered, the level would stop climbing when the pressure of the raised column had pushed the solution’s water potential back up to zero, matching the pure water on the left.

60Mixed practice mixed practice

Check q16

In a drought, the soil water around a root cell dries to a water potential of −8 bar. The root cell is at −6 bar.

Which way does water move?

A. Water moves from the soil water into the root cell
−8 bar is lower than −6 bar, so water does not move toward the root cell.
B. There is no net movement of water
The two values differ, so there is a net movement of water.
C. Solute moves from the root cell into the soil water
The membrane holds solute back, so it is water that moves, not solute.
D. ✓ Water moves from the root cell into the soil water

Why: Water moves from the higher water potential to the lower.
The root cell is at −6 bar.
The dry soil water is at −8 bar.
−8 bar is lower than −6 bar.
So water moves from the root cell into the soil water.

Check q17

A student records Ψs = +3 bar for a sugar solution in an open beaker.

Is the student’s record correct?

A. Yes
A solute potential is never positive: dissolved solute only ever lowers water potential.
B. ✓ No

Why: A solute potential is never positive.
Dissolved solute only ever lowers water potential.
So the solute potential is zero for pure water and negative for every solution.
A sugar solution has sugar dissolved in it.
So its solute potential must be below zero, and +3 bar cannot be right.

Check q18

A student records the solute potential of a sugar solution in an open beaker as Ψs = +3 bar.

Why must the solute potential of a sugar solution be below zero?

A. ✓ Dissolved solute only ever lowers water potential, so a solution’s solute potential is always negative
B. An open beaker sets the solute potential to 0 bar, so the value must be 0 bar
An open beaker sets the pressure potential, Ψp, to zero, not the solute potential.
C. Only a concentrated sugar solution has a solute potential, so a dilute one has none
Even a dilute solution holds some solute, and any dissolved solute lowers the water potential below zero.

Why: Dissolved solute only ever lowers water potential.
So the solute potential is zero for pure water and negative for every solution.
A sugar solution has sugar dissolved in it.
So its solute potential is below zero, and +3 bar cannot be right.

Check q19 numeric entry

A sucrose solution in an open beaker has a solute potential of −7 bar.

Calculate its water potential.

Answer: -7 bar (tolerance ±0.5)

Working

Write down the values in the question:

Ψp = 0 bar

Ψs = −7 bar

Write down the equation:

Ψ = Ψp + Ψs

Substitute the values into the equation:

Ψ = 0 + (−7)

Calculate:

Ψ = −7 bar

Check q20

A turgid plant cell’s cell wall presses on its contents at 4 bar. Beside it stands an open beaker of the same solution.

Which statement gives the two pressure potentials?

A. The cell’s contents have Ψp = −4 bar; the beaker’s solution has Ψp = 0 bar
A cell wall pressing on the contents raises their water potential, so their pressure potential is positive.
B. The cell’s contents have Ψp = 0 bar; the beaker’s solution has Ψp = +4 bar
It is the cell’s contents that the cell wall presses on; nothing presses on the solution in the open beaker.
C. ✓ The cell’s contents have Ψp = +4 bar; the beaker’s solution has Ψp = 0 bar
D. The cell’s contents and the beaker’s solution both have Ψp = +4 bar
Nothing presses on a solution in an open beaker beyond the air, so its pressure potential is 0 bar.

Why: The cell wall presses on the cell’s contents at 4 bar, so their Ψp is +4 bar.
Nothing presses on the beaker’s solution, so its Ψp is 0 bar.

Check q21 numeric entry

A plant cell has Ψp = +5 bar and Ψs = −11 bar.

Calculate its water potential.

Answer: -6 bar (tolerance ±0.5)

Working

Write down the values in the question:

Ψp = +5 bar

Ψs = −11 bar

Write down the equation:

Ψ = Ψp + Ψs

Substitute the values into the equation:

Ψ = (+5) + (−11)

Calculate:

Ψ = −6 bar

Check q22 numeric entry

A fully turgid plant cell has Ψp = +6 bar and Ψs = −6 bar.

Calculate its water potential.

Answer: 0 bar (tolerance ±0.5)

Working

Write down the values in the question:

Ψp = +6 bar

Ψs = −6 bar

Write down the equation:

Ψ = Ψp + Ψs

Substitute the values into the equation:

Ψ = (+6) + (−6)

Calculate:

Ψ = 0 bar

Check q23

Which change raises the water potential of a plant cell’s contents?

A. More solute dissolving in the cell’s contents
Dissolving more solute lowers Ψs, so the water potential falls.
B. ✓ The cell wall pressing harder on the cell’s contents
C. The cell wall pressing less hard on the cell’s contents
Less pressure means a smaller Ψp, so the water potential falls.
D. The cell’s contents being poured into an open beaker, with nothing pressing on them
In an open beaker Ψp is 0 bar, so the water potential falls to the solute potential.

Why: Pressure raises water potential.
A cell wall pressing harder gives the contents a larger Ψp. Ψ = Ψp + Ψs, so a larger Ψp gives a higher water potential.
More solute lowers Ψs, so the water potential falls.
Less pressure gives a smaller Ψp, so the water potential falls too.

Check q24

Cell A has Ψp = +3 bar and Ψs = −10 bar. Cell B, touching it, has Ψp = +1 bar and Ψs = −5 bar.

Which way does water move between the two cells?

A. ✓ Water moves from cell B into cell A
B. Water moves from cell A into cell B
Cell B, at −4 bar, has the higher water potential, so water does not move toward cell B.
C. There is no net movement of water
The direction follows the whole water potential, the sum of the two parts.
D. Solute moves from cell A into cell B
The membranes hold solute back, so it is water that moves between the cells.

Why: Ψ = Ψp + Ψs.
Cell A: Ψ = (+3) + (−10) = −7 bar.
Cell B: Ψ = (+1) + (−5) = −4 bar.
−7 bar is lower than −4 bar.
So water moves from cell B into cell A.

Check q25 numeric entry

A plant cell has Ψp = +1.5 bar and Ψs = −8.5 bar.

Calculate its water potential.

Answer: -7 bar (tolerance ±0.05)

Working

Write down the values in the question:

Ψp = +1.5 bar

Ψs = −8.5 bar

Write down the equation:

Ψ = Ψp + Ψs

Substitute the values into the equation:

Ψ = (+1.5) + (−8.5)

Calculate:

Ψ = −7.0 bar

Practice writing an answer

A plant cell has a solute potential of −9 bar, and its cell wall presses on its contents at 2 bar. The cell sits in a sucrose solution in an open beaker; the solution’s solute potential is −3 bar.

(a) Calculate the water potential of the cell. (1 pt)

Model answer The cell wall presses on the cell’s contents at 2 bar.
So the pressure potential, Ψp, is +2 bar.
The solute potential, Ψs, is −9 bar. Ψ = Ψp + Ψs.
Adding the two parts gives −7 bar.
So the cell’s water potential is −7 bar.

Working

Write down the values in the question:

Ψp = +2 bar

Ψs = −9 bar

Write down the equation:

Ψ = Ψp + Ψs

Substitute the values into the equation:

Ψ = (+2) + (−9)

Calculate:

Ψ = −7 bar

Rubric

Award 1 point for: Ψ = (+2) + (−9) = −7 bar, negative and in bars.

Slip Subtracting the pressure potential to get −11 bar. Add the two parts. The cell wall’s push raises the water potential toward zero.

(b) Calculate the water potential of the solution. (1 pt)

Model answer The solution is in an open beaker.
Nothing presses on the solution beyond the air.
So its pressure potential, Ψp, is 0 bar.
Its solute potential, Ψs, is −3 bar. Ψ = Ψp + Ψs.
With Ψp at zero, the water potential equals the solute potential.
So the solution’s water potential is −3 bar.

Working

Write down the values in the question:

Ψp = 0 bar

Ψs = −3 bar

Write down the equation:

Ψ = Ψp + Ψs

Substitute the values into the equation:

Ψ = 0 + (−3)

Calculate:

Ψ = −3 bar

Rubric

Award 1 point for: Ψ = 0 + (−3) = −3 bar, with Ψp = 0 bar for the open beaker stated or used.

Slip Leaving the solution’s water potential as unknown because the question gave no pressure potential. An open beaker has Ψp = 0 bar.

(c) Predict the net movement of water between the cell and the solution, and justify your prediction with the two water potentials. (2 pt)

Model answer Water moves from the higher water potential to the lower.
The solution’s water potential is −3 bar.
The cell’s water potential is −7 bar.
−7 bar is lower than −3 bar.
So water moves from the solution into the cell.

Rubric

Award 1 point for: water moves into the cell (the cell gains water).
Award 1 point for: −7 bar is lower than −3 bar, and water moves toward the lower water potential.
Accept: the two values compared in either order, provided the lower one is named as where water goes.

Slip Sending water away from the cell because 7 is a bigger digit than 3. −7 bar is further below zero than −3 bar. So −7 bar is the lower water potential. Water moves toward the lower water potential, so water moves into the cell.

Glossary

water potential (Ψ): A measure of how strongly water tends to move, in bars. Water moves by osmosis from a region of higher water potential to a region of lower water potential; pure water in an open container is given Ψ = 0 bar.
bar: The unit of pressure used for water potential. The air presses on you at about 1 bar.
solute potential (Ψs): The part of a water potential that comes from dissolved solute. It is zero for pure water and negative for every solution; more solute makes it more negative.
pressure potential (Ψp): The part of a water potential that comes from pressure on the water. It is zero for a solution in an open container and positive for the contents of a turgid cell pressed against its cell wall.

APBIO-U02-L13 Solute potential from a concentration

Topic 2.7b · Water Potential · 79 steps

Two beakers, one of salt solution and one of sugar solution twice as concentrated, with a labelled red blood cell in each

Here are two beakers, one of salt solution and one of sugar solution, with a red blood cell in each.

Two solutions: one of salt, one of sugar, the sugar twice as concentrated as the salt. Yet a red blood cell dropped into each behaves exactly the same.

Unit 2 · Cell Structure and Function

1How many particles a solute makes

Video: Watch: How many particles a solute makes

Sucrose stays whole, NaCl splits in two: the ionization constant, i.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L13.mp4

A mole is a fixed count of particles, so 0.15 mol/L of any dissolved substance means the same number of dissolved units in each liter.

Dissolve one unit of sucrose, table sugar, and it stays whole: one unit, one particle.

One sucrose unit dissolving into one particle, beside one NaCl unit dissolving into a sodium ion and a chloride ion

Dissolve one unit of NaCl, table salt, and it splits into two: a sodium ion, Na⁺, and a chloride ion, Cl⁻.

So 0.15 mol/L NaCl puts twice as many dissolved particles into the water as 0.15 mol/L sucrose does.

The number of separate particles each dissolved unit splits into is called the solute’s , i.

The word ionization is used because a salt splits into ions when it dissolves.

Sucrose and glucose do not split, so i = 1. We treat NaCl as fully split into Na⁺ and Cl⁻, so i = 2.

i is not 1 for every solute. A salt that splits into ions counts each ion.

What you are expected to know Give the ionization constant of sucrose, glucose and NaCl, and say how many dissolved particles one solution puts in the water compared with another at the same molar concentration.

12Fluency quiz: what is i? mixed practice

Check q1

Sucrose, table sugar, stays whole when it dissolves.

What is the ionization constant, i, of sucrose?

A. ✓ i = 1
B. i = 2
Sucrose stays whole, so each dissolved unit is one particle, not two.
C. i = 3
Sucrose stays whole, so each dissolved unit is one particle, not three.

Why: The ionization constant, i, counts the separate particles each dissolved unit splits into.
Sucrose stays whole: one unit, one particle.
So the ionization constant of sucrose is 1.

Check q2

NaCl, table salt, splits into Na⁺ and Cl⁻ when it dissolves.

What is the ionization constant, i, of NaCl?

A. i = 1
Each NaCl unit splits into two ions, Na⁺ and Cl⁻, so i counts two particles, not one.
B. ✓ i = 2
C. i = 3
Each NaCl unit splits into two ions, Na⁺ and Cl⁻: two particles, not three.

Why: The ionization constant, i, counts the separate particles each dissolved unit splits into.
Each NaCl unit splits into two ions, Na⁺ and Cl⁻: one unit, two particles.
So the ionization constant of NaCl is 2.

Check q3

Glucose stays whole when it dissolves.

What is the ionization constant, i, of glucose?

A. ✓ i = 1
B. i = 2
Glucose stays whole, so each dissolved unit is one particle, not two.
C. i = 3
Glucose stays whole, so each dissolved unit is one particle, not three.

Why: The ionization constant, i, counts the separate particles each dissolved unit splits into.
Glucose stays whole: one unit, one particle.
So the ionization constant of glucose is 1.

Check q4

Potassium bromide, KBr, splits into potassium ions (K⁺) and bromide ions (Br⁻) when it dissolves.

What is the ionization constant, i, of KBr?

A. i = 1
Each KBr unit splits into two ions, K⁺ and Br⁻, so i counts two particles, not one.
B. ✓ i = 2
C. i = 3
Each KBr unit splits into two ions, K⁺ and Br⁻: two particles, not three.

Why: The ionization constant, i, counts the separate particles each dissolved unit splits into.
Each KBr unit splits into two ions, K⁺ and Br⁻: one unit, two particles.
So the ionization constant of KBr is 2.

Check q5

Calcium chloride, CaCl₂, splits into one calcium ion (Ca²⁺) and two chloride ions (Cl⁻) when it dissolves.

What is the ionization constant, i, of CaCl₂?

A. i = 1
Each CaCl₂ unit splits into three ions, one Ca²⁺ and two Cl⁻, so i counts three particles, not one.
B. i = 2
Each CaCl₂ unit splits into three ions, one Ca²⁺ and two Cl⁻: three particles, not two.
C. ✓ i = 3

Why: The ionization constant, i, counts the separate particles each dissolved unit splits into.
Each CaCl₂ unit splits into three ions, one Ca²⁺ and two Cl⁻: one unit, three particles.
A salt that splits into ions counts each ion.
So the ionization constant of CaCl₂ is 3.

Check q6

A student makes up two solutions: 0.20 mol/L NaCl and 0.20 mol/L glucose.

How many times more dissolved particles does the NaCl solution hold than the glucose solution?

A. Half as many
It is NaCl that splits into two ions; glucose stays whole.
B. The same number
The two solutions hold the same number of dissolved units, but each NaCl unit becomes two ions.
C. ✓ Twice as many

Why: Each NaCl unit splits into two ions, Na⁺ and Cl⁻.
So the ionization constant of NaCl is 2.
Each glucose unit stays whole as one particle.
So the ionization constant of glucose is 1.
The two solutions have the same molar concentration.
So the NaCl solution holds twice as many dissolved particles.

19Solute potential from a concentration

Video: Watch: Solute potential from a concentration

Ψs = −iCRT worked for 0.15 M NaCl at 25 °C: −7.43 bar.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L13b.mp4

More solute makes the solute potential more negative. Here is the equation that turns a concentration into a number of bars.

Ψ_{s} = - i C R T

Ψs is the solute potential in bar; i the ionization constant; C the molar concentration in mol/L; R the pressure constant, 0.0831 L·bar/(mol·K); T the temperature in kelvin. The minus sign makes every solute potential negative.

$Ψ_{s} = - i C R T$ . i is the ionization constant. C is the molar concentration in mol/L, also written as M. R is the pressure constant, 0.0831 L·bar/(mol·K). T is the temperature in kelvin.

This R is a fixed number, not an amino acid’s R group. This C is a concentration, not carbon.

A temperature in kelvin, written K, counts up from the coldest anything can be. That is about 273 degrees below zero on the Celsius scale. The steps are the size of a Celsius degree, so add 273: 25 °C is 298 K.

The minus sign at the front makes every answer negative, as a solute potential must be.

Worked example

What is the solute potential of 0.15 M NaCl at 25 °C?

Write down the values in the question:

i = 2

C = 0.15 mol/L

R = 0.0831 L·bar/(mol·K)

T = 25 + 273 = 298 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (2) (0.15) (0.0831) (298)

Calculate:

Ψ_{s} = −7.43 bar

Look at the units that went in: 0.15 mol/L, then 0.0831 L·bar/(mol·K), then 298 K. The L on top of 0.15 mol/L cancels the L underneath in R. The mol cancels the mol. The K in 298 K cancels the K underneath in R. Only bar is left. So −7.43 is in bar.

In every calculation on this page, R = 0.0831 L·bar/(mol·K), the value the formula sheet gives.

Check q7 numeric entry

A beaker holds 0.30 M sucrose at 25 °C, open to the air. R = 0.0831 L·bar/(mol·K).

Calculate its solute potential.

Part 1. Sucrose stays whole when it dissolves. What is its ionization constant, i?

Answer: 1 (tolerance ±0)

Working

Count the particles each dissolved unit makes:

i = 1 (one unit, one particle)

Part 2. What is the molar concentration, C?

Answer: 0.3 mol/L (tolerance ±0.005)

Working

Read the concentration from the question:

C = 0.30 mol/L

Part 3. What is the temperature, T, in kelvin?

Answer: 298 K (tolerance ±0.5)

Working

Add 273 to the Celsius temperature:

T = 25 + 273 = 298 K

Answer: -7.43 bar (tolerance ±0.005)

Working

Write down the values in the question:

i = 1

C = 0.30 mol/L

R = 0.0831 L·bar/(mol·K)

T = 25 + 273 = 298 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (1) (0.30) (0.0831) (298)

Calculate:

Ψ_{s} = −7.43 bar

0.30 M sucrose puts 0.30 mol/L of particles into the water. 0.15 M NaCl also puts 0.30 mol/L of particles into the water. The same number of particles gives the same solute potential, −7.43 bar. So a red blood cell behaves the same in the two beakers.

Check q8 numeric entry

A 0.10 M NaCl solution is at 22 °C. R = 0.0831 L·bar/(mol·K).

Calculate its solute potential.

Part 1. NaCl splits into Na⁺ and Cl⁻. What is its ionization constant, i?

Answer: 2 (tolerance ±0)

Working

Count the particles each dissolved unit makes:

i = 2 (one Na⁺ and one Cl⁻)

Part 2. What is the temperature, T, in kelvin?

Answer: 295 K (tolerance ±0.5)

Working

Add 273 to the Celsius temperature:

T = 22 + 273 = 295 K

Answer: -4.9 bar (tolerance ±0.005)

Working

Write down the values in the question:

i = 2

C = 0.10 mol/L

R = 0.0831 L·bar/(mol·K)

T = 22 + 273 = 295 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (2) (0.10) (0.0831) (295)

Calculate:

Ψ_{s} = −4.90 bar

What you are expected to know Calculate the solute potential of a solution in bars with $Ψ_{s} = - i C R T$ , using the right i, the concentration in mol/L, the pressure constant R and the temperature in kelvin.

Check q9 numeric entry

A 0.40 M sucrose solution is at 20 °C. R = 0.0831 L·bar/(mol·K).

Calculate its solute potential.

Answer: -9.74 bar (tolerance ±0.005)

Working

Write down the values in the question:

i = 1

C = 0.40 mol/L

R = 0.0831 L·bar/(mol·K)

T = 20 + 273 = 293 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (1) (0.40) (0.0831) (293)

Calculate:

Ψ_{s} = −9.74 bar

Check q10 numeric entry

A 0.20 M glucose solution is at 25 °C. Glucose stays whole when it dissolves. R = 0.0831 L·bar/(mol·K).

Calculate its solute potential.

Answer: -4.95 bar (tolerance ±0.01)

Working

Write down the values in the question:

i = 1

C = 0.20 mol/L

R = 0.0831 L·bar/(mol·K)

T = 25 + 273 = 298 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (1) (0.20) (0.0831) (298)

Calculate:

Ψ_{s} = −4.95 bar

Check q11 numeric entry

A 0.12 M NaCl solution is at 24 °C. NaCl splits into Na⁺ and Cl⁻ when it dissolves. R = 0.0831 L·bar/(mol·K).

Calculate its solute potential.

Answer: -5.92 bar (tolerance ±0.01)

Working

Write down the values in the question:

i = 2

C = 0.12 mol/L

R = 0.0831 L·bar/(mol·K)

T = 24 + 273 = 297 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (2) (0.12) (0.0831) (297)

Calculate:

Ψ_{s} = −5.92 bar

36Which way the water goes

Check q12

A cell has a water potential of −3 bar. It sits in a solution with a water potential of −8 bar.

Which way does water move?

A. ✓ Out of the cell
B. Into the cell
Water moves toward the lower water potential, and the solution at −8 bar is lower than the cell at −3 bar.
C. Neither: there is no net movement
−3 bar and −8 bar differ, so there is a net movement of water.

Why: Water moves from the higher water potential to the lower, and −8 bar is lower than −3 bar, so water moves out of the cell.

Video: Watch: Which way the water goes

Two water potentials side by side: toward the lower one, further below zero.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L13c.mp4

Two water potentials side by side settle the direction. Water moves toward the lower one, and keeps moving until the two are equal.

A cell at Ψ = −6 bar. In a solution at −2 bar it gains water. In a solution at −9 bar it loses water. In a solution at −6 bar there is no net movement.

A number line of water potential from −10 bar to 0 bar, a cell marked at −6 bar, solutions marked at −9, −2 and −6 bar, with arrows for water toward the lower value

Watch the signs. −9 bar is lower than −6 bar, further below zero, so water moves toward −9 bar. The number further from zero is the lower water potential.

When the question gives only a concentration, calculate that side’s water potential first, as for the sucrose solution above.

What you are expected to know Predict which way water moves between a cell and its surroundings from their water potentials, given or calculated: toward the lower, more negative, value, until the two are equal.

Check q13

A cell at Ψ = −4 bar sits in a solution at Ψ = −7 bar.

What happens to the cell?

A. ✓ The cell loses water
B. The cell gains water
−4 bar is the higher of the two water potentials, so water does not move toward the cell.
C. The cell’s volume stays the same; there is no net movement of water
The two values differ, so there is a net movement of water.
D. The cell gains solute until it reaches −7 bar
The membrane holds solute back, so it is water that moves, not solute.

Why: Water moves from the higher water potential to the lower.
The cell is at −4 bar.
The solution is at −7 bar.
−7 bar is lower than −4 bar.
So water moves out of the cell into the solution.
The cell loses water.

Check q14

A cell has a water potential of −5 bar. A student puts it, in turn, into four solutions.

In which solution does the cell lose water?

A. 0 bar
0 bar is the highest value here, so water moves from the pure water into the cell.
B. −2 bar
−2 bar is closer to zero, so it is the higher value, and water moves from that solution into the cell.
C. −5 bar
At −5 bar the solution equals the cell, so there is no net movement of water.
D. ✓ −8 bar

Why: A cell loses water only to a solution with a lower water potential than its own.
The cell is at −5 bar.
Of the four values, only −8 bar is lower than −5 bar.
So the cell loses water in the −8 bar solution.

46How hard the cell wall has to push

A walled cell with Ψs = −8 bar is dropped into 0.2 M sucrose at 22 °C. Which way does water go, and where does it stop?

First the solution. It is in an open beaker, so Ψp = 0 bar and its water potential is its solute potential.

Check q15 numeric entry

A beaker of 0.2 M sucrose at 22 °C stands open to the air. R = 0.0831 L·bar/(mol·K).

Calculate the water potential of the solution.

Answer: -4.9 bar (tolerance ±0.05)

Working

Write down the values in the question:

i = 1

C = 0.2 mol/L

R = 0.0831 L·bar/(mol·K)

T = 22 + 273 = 295 K

Ψp = 0 bar

Write down the equations:

Ψ_{s} = - i C R T

Ψ = Ψ_{p} + Ψ_{s}

Substitute the values into the first equation:

Ψ_{s} = - (1) (0.2) (0.0831) (295)

Calculate:

Ψ_{s} = −4.9 bar

Substitute the values into the second equation:

Ψ = 0 + (−4.9)

Calculate:

Ψ = −4.9 bar

Video: Watch: How hard the cell wall has to push

Water enters until the two match; Ψp = Ψ − Ψs gives +3.1 bar.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L13d.mp4

The cell’s water potential is −8 bar. −8 bar is lower than −4.9 bar. So water enters the cell. The contents swell and press against the cell wall. The cell wall presses back on the contents. So the cell’s Ψp rises.

A walled cell in a beaker of 0.2 M sucrose, the cell’s contents shaded paler than the solution and the cell wall labelled: water enters, the wall presses back, and the pressure potential rises until the cell’s water potential matches the solution’s

In these calculations we treat the cell’s solute potential as unchanged while it gains or loses water. Only its pressure potential changes.

Water stops entering when the cell’s water potential has risen to the solution’s: −4.9 bar. Then $Ψ = Ψ_{p} + Ψ_{s}$ , rearranged, gives the pressure the cell wall has reached.

Worked example

The cell, Ψs = −8 bar, stops gaining water when its water potential has risen to the solution’s, −4.9 bar. What pressure potential has the cell wall reached?

Write down the values in the question:

Ψ = −4.9 bar (the solution’s)

Ψs = −8 bar

Write down the equation:

Ψ = Ψ_{p} + Ψ_{s}

Make Ψp the subject:

Ψ_{p} = Ψ - Ψ_{s}

Substitute the values into the equation:

Ψ_{p} = (−4.9) - (−8)

Calculate:

Ψ_{p} = +3.1 bar

Check q16 numeric entry

A walled cell with a solute potential of −8 bar is dropped into pure water in an open beaker and left until water stops entering.

Calculate the pressure potential it reaches.

Part 1. What is the water potential of the pure water around the cell, Ψ?

Answer: 0 bar (tolerance ±0.5)

Working

Write down the water potential of pure water:

Ψ = 0 bar

Part 2. What is the cell’s solute potential, Ψs?

Answer: -8 bar (tolerance ±0.5)

Working

Read the cell’s solute potential from the question:

Ψs = −8 bar

Answer: 8 bar (tolerance ±0.5)

Working

Write down the values in the question:

Ψ = 0 bar (pure water)

Ψs = −8 bar

Write down the equation:

Ψ = Ψ_{p} + Ψ_{s}

Make Ψp the subject:

Ψ_{p} = Ψ - Ψ_{s}

Substitute the values into the equation:

Ψ_{p} = 0 - (−8)

Calculate:

Ψ_{p} = +8 bar

What you are expected to know Calculate the pressure potential a walled cell reaches in a solution of known water potential, when its own water potential has risen to match: $Ψ_{p} = Ψ - Ψ_{s}$ , the solution’s Ψ minus the cell’s Ψs.

Check q17 numeric entry

A walled cell with Ψs = −7 bar sits in a solution at Ψ = −3 bar until water stops entering.

Calculate the pressure potential the cell has reached.

Answer: 4 bar (tolerance ±0.5)

Working

Write down the values in the question:

Ψ = −3 bar (the solution’s)

Ψs = −7 bar

Write down the equation:

Ψ = Ψ_{p} + Ψ_{s}

Make Ψp the subject:

Ψ_{p} = Ψ - Ψ_{s}

Substitute the values into the equation:

Ψ_{p} = (−3) - (−7)

Calculate:

Ψ_{p} = +4 bar

58The sucrose solution that matches

Video: Watch: The sucrose solution that matches

Make C the subject: the concentration that leaves a potato core unchanged.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L13e.mp4

A potato core is at Ψ = −7.35 bar at 22 °C. Which sucrose solution would leave it unchanged, neither gaining nor losing water?

A potato core at −7.35 bar in a beaker of 0.30 M sucrose at −7.35 bar: no net movement of water

A solution in an open beaker has Ψ = Ψs, so the solution needs a solute potential of −7.35 bar. This time the concentration is the unknown, so $Ψ_{s} = - i C R T$ has to be rearranged to make C the subject.

Worked example

Which sucrose concentration at 22 °C has a solute potential of −7.35 bar?

Write down the values in the question:

Ψs = −7.35 bar

i = 1

R = 0.0831 L·bar/(mol·K)

T = 22 + 273 = 295 K

Write down the equation:

Ψ_{s} = - i C R T

Make C the subject:

C = \frac{- Ψ_{s}}{i R T}

Substitute the values into the equation:

C = \frac{7.35}{(1) (0.0831) (295)}

Calculate:

C = 0.30 mol/L

The two minus signs cancel: −Ψs is +7.35 bar, so the concentration comes out positive, as a concentration must.

Check q18 numeric entry

A cell is at Ψ = −4.9 bar at 22 °C. R = 0.0831 L·bar/(mol·K).

Calculate the sucrose concentration that matches it.

Part 1. The solution needs Ψs = −4.9 bar. In $C = \frac{- Ψ_{s}}{i R T}$ , what is the top of the fraction, −Ψs?

Answer: 4.9 bar (tolerance ±0.05)

Working

Change the sign of the solute potential:

- Ψ_{s} = - (−4.9) = +4.9 bar

Part 2. What is the temperature, T, in kelvin?

Answer: 295 K (tolerance ±0.5)

Working

Add 273 to the Celsius temperature:

T = 22 + 273 = 295 K

Part 3. Calculate iRT, the bottom of the fraction in $C = \frac{- Ψ_{s}}{i R T}$ .

Answer: 24.5 L·bar/mol (tolerance ±0.05)

Working

Multiply i, R and T:

i R T = (1) (0.0831) (295)

i R T = 24.5 L·bar/mol

Answer: 0.2 mol/L (tolerance ±0.005)

Working

Write down the values in the question:

Ψs = −4.9 bar

i = 1

R = 0.0831 L·bar/(mol·K)

T = 22 + 273 = 295 K

Write down the equation:

Ψ_{s} = - i C R T

Make C the subject:

C = \frac{- Ψ_{s}}{i R T}

Substitute the values into the equation:

C = \frac{4.9}{(1) (0.0831) (295)}

Calculate:

C = 0.20 mol/L

What you are expected to know Calculate the molar concentration of the sucrose solution whose water potential matches a cell’s or a tissue’s, by making C the subject of $Ψ_{s} = - i C R T$ , with Ψp = 0 bar for the open solution.

Check q19 numeric entry

A piece of plant tissue is at Ψ = −6.1 bar at 22 °C. R = 0.0831 L·bar/(mol·K).

Calculate the sucrose concentration that would leave it unchanged.

Answer: 0.25 mol/L (tolerance ±0.005)

Working

Write down the values in the question:

Ψs = −6.1 bar

i = 1

R = 0.0831 L·bar/(mol·K)

T = 22 + 273 = 295 K

Write down the equation:

Ψ_{s} = - i C R T

Make C the subject:

C = \frac{- Ψ_{s}}{i R T}

Substitute the values into the equation:

C = \frac{6.1}{(1) (0.0831) (295)}

Calculate:

C = 0.25 mol/L

What you are expected to know One equation, $Ψ_{s} = - i C R T$ , turns a concentration into a solute potential in bars. Two water potentials side by side tell you which way water goes. The same equation, rearranged, gives how hard a cell wall has to push, or which solution matches a cell.

0.15 M NaCl puts twice as many particles into the water as 0.15 M sucrose would. That is the same number of particles as 0.30 M sucrose. So both solutions come to −7.43 bar. So a red blood cell behaves the same in each.

69Mixed practice mixed practice

Check q20 numeric entry

A solution is at 30 °C.

Calculate its temperature in kelvin.

Answer: 303 K (tolerance ±0.5)

Working

Write down the values in the question:

temperature = 30 °C

Write down the equation:

T = °C + 273

Substitute the values into the equation:

T = 30 + 273

Calculate:

T = 303 K

Check q21

A cell at Ψ = −3 bar sits in a solution at Ψ = −3 bar.

What happens to the cell?

A. The cell gains water from the solution
Water enters a cell only from a higher water potential, and the solution’s value equals the cell’s.
B. ✓ There is no net movement of water
C. The cell loses water to the solution
Water leaves a cell only toward a lower water potential, and the solution’s value equals the cell’s.
D. The cell gains solute until its value drops below −3 bar
The membrane holds solute back, so it is water that moves, not solute.

Why: Water moves toward the lower water potential.
The cell is at −3 bar.
The solution is at −3 bar.
The two values are equal.
So there is no net movement of water.
The cell’s volume stays the same.

Check q22 numeric entry

A 0.25 M NaCl solution is at 22 °C. R = 0.0831 L·bar/(mol·K).

Calculate its solute potential.

Answer: -12.3 bar (tolerance ±0.05)

Working

Write down the values in the question:

i = 2

C = 0.25 mol/L

R = 0.0831 L·bar/(mol·K)

T = 22 + 273 = 295 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (2) (0.25) (0.0831) (295)

Calculate:

Ψ_{s} = −12.3 bar

Check q23 numeric entry

A 0.50 M sucrose solution is at body temperature, 37 °C. R = 0.0831 L·bar/(mol·K).

Calculate its solute potential.

Answer: -12.9 bar (tolerance ±0.05)

Working

Write down the values in the question:

i = 1

C = 0.50 mol/L

R = 0.0831 L·bar/(mol·K)

T = 37 + 273 = 310 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (1) (0.50) (0.0831) (310)

Calculate:

Ψ_{s} = −12.9 bar

Check q24 numeric entry

A walled cell with Ψs = −10 bar sits in a solution at Ψ = −2.5 bar until water stops entering.

Calculate the pressure potential the cell has reached.

Answer: 7.5 bar (tolerance ±0.05)

Working

Write down the values in the question:

Ψ = −2.5 bar (the solution’s)

Ψs = −10 bar

Write down the equation:

Ψ = Ψ_{p} + Ψ_{s}

Make Ψp the subject:

Ψ_{p} = Ψ - Ψ_{s}

Substitute the values into the equation:

Ψ_{p} = (−2.5) - (−10)

Calculate:

Ψ_{p} = +7.5 bar

Check q25 numeric entry

A piece of plant tissue is at Ψ = −9.0 bar at 27 °C. R = 0.0831 L·bar/(mol·K).

Calculate the sucrose concentration that would leave it unchanged.

Answer: 0.36 mol/L (tolerance ±0.005)

Working

Write down the values in the question:

Ψs = −9.0 bar

i = 1

R = 0.0831 L·bar/(mol·K)

T = 27 + 273 = 300 K

Write down the equation:

Ψ_{s} = - i C R T

Make C the subject:

C = \frac{- Ψ_{s}}{i R T}

Substitute the values into the equation:

C = \frac{9.0}{(1) (0.0831) (300)}

Calculate:

C = 0.36 mol/L

Check q26 numeric entry

A walled cell with Ψs = −9 bar is dropped into 0.15 M sucrose at 22 °C and left until water stops entering. R = 0.0831 L·bar/(mol·K).

Calculate the pressure potential the cell reaches.

Part 1. First the solution, in its open beaker. Calculate its water potential.

Answer: -3.7 bar (tolerance ±0.05)

Working

Substitute the values into the solute-potential equation:

Ψ_{s} = - (1) (0.15) (0.0831) (295)

Calculate:

Ψ_{s} = −3.7 bar

Add the pressure potential, which is zero for an open beaker:

Ψ = Ψ_{p} + Ψ_{s} = 0 + (−3.7) = −3.7 bar

Answer: 5.3 bar (tolerance ±0.05)

Working

Write down the values in the question:

i = 1

C = 0.15 mol/L

R = 0.0831 L·bar/(mol·K)

T = 22 + 273 = 295 K

Ψp = 0 bar (open beaker)

Ψs of the cell = −9 bar

Write down the equations:

Ψ_{s} = - i C R T

Ψ = Ψ_{p} + Ψ_{s}

Make Ψp the subject:

Ψ_{p} = Ψ - Ψ_{s}

Substitute the values into the first equation:

Ψ_{solution} = - (1) (0.15) (0.0831) (295)

Calculate:

Ψ_{solution} = −3.7 bar

Substitute the values into the second equation:

Ψ_{p} = (−3.7) - (−9)

Calculate:

Ψ_{p} = +5.3 bar

Check q27

A cell at Ψ = −7.0 bar sits in 0.12 M NaCl at 25 °C, whose water potential is −5.94 bar.

What happens to the cell?

A. The cell loses water
−5.94 bar is closer to zero, so it is the higher value, and water moves from the solution into the cell.
B. The cell’s volume stays the same
Water moves whenever the two values differ, however small the gap.
C. ✓ The cell gains water
D. The cell gains solute from the salt solution
The membrane holds solute back, so it is water that moves, not solute.

Why: Water moves from the higher water potential to the lower.
The solution is at −5.94 bar.
The cell is at −7.0 bar.
−7.0 bar is lower than −5.94 bar.
So water moves from the solution into the cell.
The cell gains water.

Practice writing an answer

Seawater floods a marsh and leaves the soil water at about 0.20 M NaCl at 27 °C, open to the air. A root cell of a marsh plant has a water potential of −8.0 bar. R = 0.0831 L·bar/(mol·K).

(a) Calculate the water potential of the soil water. (1 pt)

Answer: -9.97 bar (tolerance ±0.08)

Model answer NaCl splits into two ions, so i = 2.
The temperature in kelvin is 300 K. Ψs = −iCRT gives the soil water’s solute potential: −9.97 bar.
The soil water is open to the air.
So its pressure potential is 0 bar.
So its water potential equals its solute potential.
The soil water is at −9.97 bar.

Working

Write down the values in the question:

i = 2

C = 0.20 mol/L

R = 0.0831 L·bar/(mol·K)

T = 27 + 273 = 300 K

Ψp = 0 bar

Write down the equations:

Ψ_{s} = - i C R T

Ψ = Ψ_{p} + Ψ_{s}

Substitute the values into the first equation:

Ψ_{s} = - (2) (0.20) (0.0831) (300)

Calculate:

Ψ_{s} = −9.97 bar

Substitute the values into the second equation:

Ψ = 0 + (−9.97)

Calculate:

Ψ = −9.97 bar

Rubric

Award 1 point for: Ψ = −2 × 0.20 × 0.0831 × 300 = −9.97 bar (accept −9.9 to −10.0 bar), with i = 2 for NaCl and T = 300 K.
Accept: the working shown with one intermediate value rounded differently, provided the answer lands in the range.

(b) Predict which way water moves between the root cell and the soil water, and justify your prediction with the two water potentials. (2 pt)

Model answer Water moves from the higher water potential to the lower.
The root cell is at −8.0 bar.
The soil water is at −9.97 bar.
−9.97 bar is lower than −8.0 bar.
So water moves out of the root cell into the soil water.
The root cell loses water.

Rubric

Award 1 point for: water moves out of the root cell into the soil water (the cell loses water).
Award 1 point for: −9.97 bar is lower than −8.0 bar, and water moves toward the lower water potential.

Slip Sending water into the root because roots usually take up water. Here the salty soil water has the lower water potential. So water leaves the root cell.

(c) A walled root cell with a solute potential of −12.0 bar sits in this soil water until the net movement of water stops. Calculate the pressure potential it reaches. (1 pt)

Answer: 2.03 bar (tolerance ±0.07)

Model answer The net movement of water stops when the cell’s water potential equals the soil water’s water potential, −9.97 bar.
The cell’s solute potential stays at −12.0 bar.
Rearranging Ψ = Ψp + Ψs gives Ψp = Ψ − Ψs.
That gives Ψp = +2.03 bar.
So the cell wall presses on the contents at 2.03 bar.

Working

Write down the values in the question:

Ψ = −9.97 bar (the soil water’s)

Ψs = −12.0 bar

Write down the equation:

Ψ = Ψ_{p} + Ψ_{s}

Make Ψp the subject:

Ψ_{p} = Ψ - Ψ_{s}

Substitute the values into the equation:

Ψ_{p} = (−9.97) - (−12.0)

Calculate:

Ψ_{p} = +2.03 bar

Rubric

Award 1 point for: Ψp = (−9.97) − (−12.0) = +2.03 bar (accept +2.0 to +2.1 bar), positive and in bars.
Accept: a value carried forward correctly from a wrong answer to an earlier part; the point is for this part’s step.

Glossary

ionization constant (i): The number of separate particles each dissolved unit of a solute splits into: 1 for sucrose and glucose, which stay whole; 2 for NaCl, which splits into Na⁺ and Cl⁻.

APBIO-U02-L14 The potato lab

Topic 2.7b · Water Potential · 57 steps

Six beakers of sucrose solution from pure water to 1.0 mol/L (1.0 M), a potato core in each, and a balance

Here are six potato cores in six beakers of sucrose solution, from pure water to 1.0 M (M is short for mol/L), and the balance used to measure their mass.

After an hour some have gained mass, some have lost it, and one has barely changed. At one concentration between those beakers, a core would neither gain nor lose mass.

Unit 2 · Cell Structure and Function

1Percent change in mass

Video: Watch: Percent change in mass

One core, before and after: the change compared with the starting mass, sign kept.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L14.mp4

Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L14.mp4

The class measured the mass of each core, dropped the core into its beaker, and measured its mass again an hour later.

This core had a mass of 12.5 g at the start and 14.0 g at the end: it gained 1.5 g.

A potato core on a balance reading 12.5 g at the start and 14.0 g an hour later

A gain of 1.5 g means more for a small core than for a large one, so the change is compared with what the core started at.

Divide the change in mass by the mass the core started at, then multiply by 100. That gives the percent change in mass.

Worked example

A potato core had a mass of 12.5 g before and 14.0 g after an hour in the solution. What is the percent change in its mass?

Write down the values in the question:

initial mass = 12.5 g

final mass = 14.0 g

Write down the equation:

percent change = \frac{final - initial}{initial} \times 100

Substitute the values into the equation:

percent change = \frac{14.0 - 12.5}{12.5} \times 100

Calculate:

percent change = +12 %

A core that lost mass gets a negative answer. Keep the sign, and write it in front of your answer: −12.5%, for example. The sign says which way the water went.

Check q1 numeric entry

A potato core sat in a sucrose solution for an hour. Its mass was 8.0 g at the start and 7.0 g at the end.

Calculate its percent change in mass.

Part 1. Subtract the starting mass from the final mass, keeping the sign. What is the change in mass?

Answer: -1 g (tolerance ±0.05)

Working

Subtract the starting mass from the final mass:

7.0 - 8.0 = −1.0 g

Answer: -12.5 % (tolerance ±0.05)

Working

Write down the values in the question:

initial mass = 8.0 g

final mass = 7.0 g

Write down the equation:

percent change = \frac{final - initial}{initial} \times 100

Substitute the values into the equation:

percent change = \frac{7.0 - 8.0}{8.0} \times 100

Calculate:

percent change = −12.5 %

The difference alone, −1.0 g, is not the percent change. Only when it is compared with the starting mass does it say how large the loss was for this core.

What you are expected to know Calculate the percent change in mass of a sample: the change in mass divided by the starting mass, multiplied by 100, with the + or − sign written in front of the answer.

Check q2 numeric entry

A potato core sat in a sucrose solution for an hour. Its mass was 6.4 g at the start and 5.8 g at the end.

Calculate its percent change in mass.

Answer: -9.4 % (tolerance ±0.05)

Working

Write down the values in the question:

initial mass = 6.4 g

final mass = 5.8 g

Write down the equation:

percent change = \frac{final - initial}{initial} \times 100

Substitute the values into the equation:

percent change = \frac{5.8 - 6.4}{6.4} \times 100

Calculate:

percent change = −9.4 %

13Graph the six cores

Video: Watch: Graph the six cores

Set on the x-axis, measured on the y-axis, six points, one best-fit line.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L14b.mp4

Here are all six cores: the sucrose concentration each sat in, in M (M is short for mol/L), and its percent change in mass.

A table of six sucrose concentrations, 0.0 to 1.0 M, and the percent change in mass of the core in each

The concentration is what was set; the percent change is what was measured. What was set goes on the x-axis; what was measured goes on the y-axis.

Each axis is labeled with its quantity and its unit: sucrose concentration (M) on the x-axis and change in mass (%) on the y-axis. M is short for mol/L.

The scale fits the data: 0 to 1.0 M on the x-axis, and −30% to +20% on the y-axis, so that every point lands on the paper and the zero line sits inside the graph.

Now get some graph paper. Draw the two axes with their labels, units and scales, and plot the six points from the table.

Then check your points against these.

The six points plotted: change in mass in percent against sucrose concentration in M

The points fall close to a straight line, but not exactly on one. The best-fit line is the one straight line that runs as close as possible to all six points, with about as many points above it as below.

Draw one best-fit straight line through your points, then check it against this one. A best-fit line is also called a trend line. It does not join the dots, and it need not pass through any of them.

The six points with one best-fit straight line running close to all of them

What you are expected to know Describe how a line graph of percent change in mass against sucrose concentration is built: axes labeled with units, a scale that fits the data, every point plotted, and one best-fit straight line.

Check q3

A student graphs the results for six cores: the sucrose concentration each core sat in, and each core’s percent change in mass.

Which labels belong on the axes?

A. ✓ Sucrose concentration (M) on the x-axis and change in mass (%) on the y-axis
B. Change in mass (%) on the x-axis and sucrose concentration (M) on the y-axis
The concentration was set, so it goes on the x-axis; the change in mass was measured, so it goes on the y-axis.
C. Beaker number on the x-axis and change in mass (%) on the y-axis
The beaker number is not the quantity that was set; the sucrose concentration was set.
D. Sucrose concentration (M) on the x-axis and final mass (g) on the y-axis
The cores started at different masses, so a final mass alone cannot be compared; the percent change in mass can.

Why: The quantity that was set goes on the x-axis: the sucrose concentration was set.
The quantity that was measured goes on the y-axis: the percent change in mass was measured.
Each axis carries its unit.

Check q4

Another group plotted their six cores and drew two lines through the points, labeled 1 and 2.

Another group’s six points with two lines drawn through them, labeled 1 and 2, each label joined to its line by a short leader: one zigzags through every point, one is straight

Which line is the best-fit straight line?

A. Line 1
Line 1 bends at every point, and a best-fit line is straight.
B. ✓ Line 2
C. Neither line
Line 2 is straight and runs close to all six points, with some above it and some below.

Why: A best-fit straight line is one straight line that runs as close as possible to all the points, with some above it and some below.
Line 2 does that.
Line 1 joins the dots, so line 1 is not a best-fit line.

26Read the isotonic point and turn it into a water potential

Video: Watch: Read the isotonic point

Where the line crosses the x-axis, 0.38 M, becomes the potato’s own water potential, −9.3 bar.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L14c.mp4

Where the best-fit line crosses zero, a core would neither gain nor lose mass. That solution is isotonic to the potato: it has the same water potential as the potato’s cells, so there is no net movement of water.

The potato best-fit line crossing zero at about 0.38 M, with a dashed line dropped to the concentration axis

Read down from the crossing to the concentration axis: about 0.38 M. None of the six beakers held that solution; the line says what a seventh beaker would have done.

A 0.38 M sucrose solution in an open beaker has Ψp = 0 bar, so its water potential is its solute potential.

Worked example

What is the water potential of 0.38 M sucrose at 22 °C?

Write down the values in the question:

i = 1

C = 0.38 mol/L

R = 0.0831 L·bar/(mol·K)

T = 22 + 273 = 295 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (1) (0.38) (0.0831) (295)

Calculate:

Ψ_{s} = −9.3 bar

In that solution water would neither enter nor leave the potato. So the potato has the same water potential as the solution. The potato itself is at −9.3 bar.

In 0.2 M sucrose at 22 °C the solution is at −4.9 bar. −4.9 bar is higher than −9.3 bar. So water moves into the potato, and the potato gains mass. The core in the 0.2 M beaker was recorded at +9%.

In every calculation on this page, R = 0.0831 L·bar/(mol·K), the value the formula sheet gives.

Check q5 numeric entry

Here is a graph for a second potato, with its best-fit line and a dashed line dropped from the zero crossing to the concentration axis. The solutions were at 22 °C. R = 0.0831 L·bar/(mol·K).

A graph with sucrose concentration in mol/L along the bottom axis and percent change in mass up the side, with a horizontal line at zero; six plotted points fall from a gain at the left to a loss at the right, with a straight best-fit line through them, and a dashed line drops from the point where the best-fit line crosses the zero line to the concentration axis

Calculate the second potato’s water potential.

Part 1. Sucrose stays whole when it dissolves. What is its ionization constant, i?

Answer: 1 (tolerance ±0)

Working

Count the particles each dissolved unit makes:

i = 1 (one unit, one particle)

Part 2. Read where the best-fit line crosses zero. What is the concentration, C, of the matching solution?

Answer: 0.25 mol/L (tolerance ±0.005)

Working

Read where the dashed line crosses the concentration axis, midway between 0.2 and 0.3:

C = 0.25 mol/L

Part 3. What is the temperature, T, in kelvin?

Answer: 295 K (tolerance ±0.5)

Working

Add 273 to the Celsius temperature:

T = 22 + 273 = 295 K

Answer: -6.1 bar (tolerance ±0.05)

Working

Write down the values in the question:

i = 1

C = 0.25 mol/L

R = 0.0831 L·bar/(mol·K)

T = 22 + 273 = 295 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (1) (0.25) (0.0831) (295)

Calculate:

Ψ_{s} = −6.1 bar

Check q6 numeric entry

A seventh core from the second potato is placed in an open beaker of 0.15 M sucrose at 22 °C. R = 0.0831 L·bar/(mol·K).

Calculate the water potential of the 0.15 M solution.

Answer: -3.7 bar (tolerance ±0.05)

Working

Write down the values in the question:

i = 1

C = 0.15 mol/L

R = 0.0831 L·bar/(mol·K)

T = 22 + 273 = 295 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (1) (0.15) (0.0831) (295)

Calculate:

Ψ_{s} = −3.7 bar

The second potato is at −6.1 bar. The 0.15 M solution is at −3.7 bar. −3.7 bar is higher than −6.1 bar. So water moves into the second potato, and the second potato gains water.

What you are expected to know Read from a best-fit line the sucrose concentration at which the change in mass is zero, turn it into the tissue’s water potential with $Ψ_{s} = - i C R T$ at the stated temperature, and predict what the tissue does in a named solution.

Check q7

Here is a graph for carrot cores, with its best-fit line.

Six points for carrot cores with their best-fit straight line

At what sucrose concentration would a carrot core’s mass stay the same?

A. 0.0 M
0.0 M is where the gain is biggest; the mass stays the same where the line crosses zero.
B. 0.40 M
The core at 0.40 M gained 4%, so the crossing lies between 0.40 M and 0.60 M.
C. ✓ 0.50 M
D. 0.60 M
The core at 0.60 M lost 3%, so the line crosses zero a little before 0.60 M.

Why: A core’s mass stays the same where the best-fit line crosses zero.
The 0.40 M core gained mass.
The 0.60 M core lost mass.
So the crossing lies between them.
On the graph the line crosses zero at 0.50 M.

Check q8 numeric entry

For radish tissue the best-fit line crosses zero at 0.45 M. The solutions were at 22 °C. R = 0.0831 L·bar/(mol·K).

Calculate the water potential of the radish tissue.

Answer: -11 bar (tolerance ±0.05)

Working

Write down the values in the question:

i = 1

C = 0.45 mol/L (where the best-fit line crosses the x-axis)

R = 0.0831 L·bar/(mol·K)

T = 22 + 273 = 295 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (1) (0.45) (0.0831) (295)

Calculate:

Ψ_{s} = −11.0 bar

Check q9

A piece of radish tissue with a water potential of −11.0 bar sits in 0.60 M sucrose at 22 °C, a solution whose water potential is −14.7 bar.

What happens to the tissue?

A. The tissue gains water and mass
−11.0 bar is the higher of the two, so water does not move into the tissue.
B. The tissue’s mass stays the same; there is no net movement
The two values differ, so there is a net movement of water.
C. The tissue swells until it bursts
Plant cells have cell walls and do not burst, and here water is leaving the tissue, not entering it.
D. ✓ The tissue loses water and mass

Why: Water moves toward the lower water potential.
The tissue is at −11.0 bar.
The solution is at −14.7 bar.
−14.7 bar is lower than −11.0 bar.
So water moves from the tissue into the solution.
The tissue loses water, so its mass falls.

What you are expected to know Percent change in mass, a best-fit line, its zero crossing and $Ψ_{s} = - i C R T$ give the water potential of the potato tissue from six masses.

The concentration at which a core neither gains nor loses mass is where the best-fit line crosses zero: about 0.38 M. $Ψ_{s} = - i C R T$ turns that concentration into the potato’s own water potential, −9.3 bar.

44Mixed practice mixed practice

Check q10

Two cores sat in the same solution for an hour. Core 1 started at 5.0 g and gained 0.5 g; core 2 started at 10.0 g and gained 0.5 g.

Which core changed by the larger percent?

A. The two cores changed by the same percent
The same 0.5 g is a larger share of a smaller core, so the two percent changes differ.
B. ✓ Core 1 changed by the larger percent
C. Core 2 changed by the larger percent
Percent change compares the gain with the starting mass, and 0.5 g is a smaller share of 10.0 g than of 5.0 g.
D. The two cores cannot be compared
Percent change is used precisely so that cores of different starting masses can be compared.

Why: Percent change divides the change by the starting mass.
For core 1, 0.5 g out of 5.0 g is a tenth, +10%.
For core 2, 0.5 g out of 10.0 g is a twentieth, +5%.
So core 1 changed by the larger percent.

Check q11 numeric entry

A potato core sat in a sucrose solution for an hour. Its mass was 10.4 g at the start and 9.0 g at the end.

Calculate its percent change in mass.

Part 1. Subtract the starting mass from the final mass, keeping the sign. What is the change in mass?

Answer: -1.4 g (tolerance ±0.05)

Working

Subtract the starting mass from the final mass:

9.0 - 10.4 = −1.4 g

Answer: -13.5 % (tolerance ±0.05)

Working

Write down the values in the question:

initial mass = 10.4 g

final mass = 9.0 g

Write down the equation:

percent change = \frac{final - initial}{initial} \times 100

Substitute the values into the equation:

percent change = \frac{9.0 - 10.4}{10.4} \times 100

Calculate:

percent change = −13.5 %

Check q12

A group’s percent changes run from −24% to +15%.

Which y-axis scale fits the data?

A. ✓ From −30% to +20%
B. From 0% to +20%
Four of the points lie below zero, so a scale that starts at 0% would leave them off the paper.
C. From −24% to +15%
That puts the end points on the frame, and marks in round tens are easier to mark and to read.
D. From −100% to +100%
A wide scale crowds the points into a thin band, so the pattern is hard to read.

Why: A scale fits the data when three things hold.
Every point lands on the paper.
The zero line sits inside the graph.
The marks are round numbers.
The scale from −30% to +20% does all three.

Check q13 numeric entry

A potato core sat in a sucrose solution for an hour. Its mass was 7.5 g at the start and 8.7 g at the end.

Calculate its percent change in mass.

Answer: 16 % (tolerance ±0.5)

Working

Write down the values in the question:

initial mass = 7.5 g

final mass = 8.7 g

Write down the equation:

percent change = \frac{final - initial}{initial} \times 100

Substitute the values into the equation:

percent change = \frac{8.7 - 7.5}{7.5} \times 100

Calculate:

percent change = +16 %

Check q14 numeric entry

A potato core sat in a sucrose solution for an hour. Its mass was 9.6 g at the start and 11.0 g at the end.

Calculate its percent change in mass.

Answer: 14.6 % (tolerance ±0.05)

Working

Write down the values in the question:

initial mass = 9.6 g

final mass = 11.0 g

Write down the equation:

percent change = \frac{final - initial}{initial} \times 100

Substitute the values into the equation:

percent change = \frac{11.0 - 9.6}{9.6} \times 100

Calculate:

percent change = +14.6 %

Check q15

For a group’s cores, the best-fit line crosses zero at 0.42 M sucrose.

Which statement says what the crossing means?

A. A core in 0.42 M sucrose would gain the most mass
The biggest gain is in pure water, at the top left of the line; at the crossing the change is zero.
B. A core in 0.42 M sucrose would lose all of its water
Zero on this axis means zero change in mass, with as much water entering as leaving.
C. The 0.42 M beaker held the core with the smallest change
The crossing is read from the line; 0.42 M lies between the tested concentrations, so no beaker need have held it.
D. ✓ A core in 0.42 M sucrose would neither gain nor lose mass

Why: The y-axis is change in mass.
Where the line crosses zero, the change in mass is zero.
So a core in 0.42 M sucrose would neither gain nor lose mass.
That solution is isotonic to the tissue.

Check q16 numeric entry

For a group’s cores, the best-fit line crosses zero at 0.35 M sucrose. Their solutions were at 25 °C. R = 0.0831 L·bar/(mol·K).

Calculate the water potential of their tissue.

Answer: -8.7 bar (tolerance ±0.05)

Working

Write down the values in the question:

i = 1

C = 0.35 mol/L

R = 0.0831 L·bar/(mol·K)

T = 25 + 273 = 298 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (1) (0.35) (0.0831) (298)

Calculate:

Ψ_{s} = −8.7 bar

Check q17 numeric entry

A beaker holds 0.80 M sucrose at 22 °C, open to the air. R = 0.0831 L·bar/(mol·K).

Calculate the water potential of the solution.

Answer: -19.6 bar (tolerance ±0.05)

Working

Write down the values in the question:

i = 1

C = 0.80 mol/L

R = 0.0831 L·bar/(mol·K)

T = 22 + 273 = 295 K

Ψp = 0 bar

Write down the equations:

Ψ_{s} = - i C R T

Ψ = Ψ_{p} + Ψ_{s}

Substitute the values into the first equation:

Ψ_{s} = - (1) (0.80) (0.0831) (295)

Calculate:

Ψ_{s} = −19.6 bar

Substitute the values into the second equation:

Ψ = 0 + (−19.6)

Calculate:

Ψ = −19.6 bar

Check q18 numeric entry

A beaker holds 0.70 M sucrose at 20 °C (293 K), open to the air. Sucrose stays as whole molecules when it dissolves. R = 0.0831 L·bar/(mol·K).

Calculate the water potential of the solution.

Answer: -17 bar (tolerance ±0.1)

Working

Write down the values in the question:

i = 1

C = 0.70 mol/L

R = 0.0831 L·bar/(mol·K)

T = 293 K

Ψp = 0 bar

Write down the equations:

Ψ_{s} = - i C R T

Ψ = Ψ_{p} + Ψ_{s}

Substitute the values into the first equation:

Ψ_{s} = - (1) (0.70) (0.0831) (293)

Calculate:

Ψ_{s} = −17.0 bar

Substitute the values into the second equation:

Ψ = 0 + (−17.0)

Calculate:

Ψ = −17.0 bar

Check q19

A piece of beetroot tissue with a water potential of −9.5 bar is placed in a 0.70 M sucrose solution whose water potential is −17.0 bar.

What happens to the tissue’s mass?

A. It rises
−17.0 bar is lower than −9.5 bar, and water moves toward the lower water potential, so water leaves the tissue and its mass falls.
B. It stays the same
Two negative values can still differ, and water moves toward the lower one.
C. ✓ It falls

Why: The tissue is at −9.5 bar.
The solution is at −17.0 bar.
−17.0 bar is lower than −9.5 bar.
Water moves toward the lower water potential.
So water moves from the tissue into the solution.
The tissue loses water, so its mass falls.

Practice writing an answer

A piece of beetroot tissue with a water potential of −9.5 bar sits in a 0.70 M sucrose solution whose water potential is −17.0 bar. Its mass falls.

(a) Explain why the tissue’s mass falls. (1 pt)

Frame Water moves from … into …, because …

Model answer Water moves from the tissue into the solution, because the solution has the lower water potential.
The tissue is at −9.5 bar.
The solution is at −17.0 bar.
−17.0 bar is lower than −9.5 bar.
Water moves toward the lower water potential.
So water leaves the tissue.
The tissue loses water.
So the tissue’s mass falls.

Rubric

Award 1 point for: water moves from the tissue into the solution because the solution’s water potential, −17.0 bar, is lower than the tissue’s, −9.5 bar, and water moves toward the lower water potential; so the tissue loses water and mass.
Accept: the two values compared in either order, provided the lower one is named as where water goes and the loss of water is tied to the loss of mass.

Slip Reading −17.0 bar as the higher value because 17 is the bigger number. A more negative water potential is a lower water potential, and water moves toward the lower one.

Practice writing an answer

A second class repeats the investigation with cores of zucchini at 20 °C (T = 293 K), in sucrose solutions from 0.0 M to 1.0 M. They plot percent change in mass against sucrose concentration and draw a best-fit straight line. The line crosses zero at 0.30 M. R = 0.0831 L·bar/(mol·K).

(a) Describe what the zero crossing at 0.30 M tells you about that solution and the zucchini tissue. (1 pt)

Model answer The y-axis is change in mass.
Where the line crosses zero, the change in mass is zero.
So in 0.30 M sucrose a core would neither gain nor lose mass.
So there is no net movement of water between the tissue and that solution.
That solution is isotonic to the zucchini tissue.
So the tissue has the same water potential as the 0.30 M solution.

Rubric

Award 1 point for: 0.30 M sucrose is isotonic to the tissue: no net water movement, so the tissue’s water potential equals the solution’s.
Accept: the solution in which the tissue neither gains nor loses mass, with no net water movement stated.

Slip Reading the crossing as the most concentrated solution the tissue survives in. The crossing is the solution whose water potential matches the tissue’s. In that solution water moves neither in nor out on balance.

(b) Calculate the water potential of the zucchini tissue at 20 °C. (1 pt)

Answer: -7.3 bar (tolerance ±0.1)

Model answer The 0.30 M sucrose solution is in an open beaker.
So its Ψp = 0 bar.
So its water potential equals its solute potential.
Sucrose does not split into ions, so i = 1. Ψs = −iCRT gives the solution’s solute potential: −7.3 bar.
So the solution is at −7.3 bar.
The tissue has the same water potential as that solution.
So the tissue is at −7.3 bar.

Working

Write down the values in the question:

i = 1

C = 0.30 mol/L

R = 0.0831 L·bar/(mol·K)

T = 293 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (1) (0.30) (0.0831) (293)

Calculate:

Ψ_{s} = −7.3 bar

Rubric

Award 1 point for: Ψ = −1 × 0.30 × 0.0831 × 293 = −7.3 bar (accept −7.2 to −7.4 bar), negative and in bars, given as the tissue’s water potential.
Accept: the working shown with one intermediate value rounded differently, provided the answer lands in the range.

(c) The class puts a seventh core into 0.50 M sucrose at 20 °C. Predict whether it gains or loses mass, and justify your prediction with the two water potentials. (2 pt)

Model answer Ψs = −iCRT gives the 0.50 M solution’s water potential: −12.2 bar.
The tissue is at −7.3 bar.
−12.2 bar is lower than −7.3 bar.
Water moves toward the lower water potential.
So water moves from the tissue into the solution.
So the core loses mass.

Working

Write down the values in the question:

i = 1

C = 0.50 mol/L

R = 0.0831 L·bar/(mol·K)

T = 293 K

Write down the equation:

Ψ_{s} = - i C R T

Substitute the values into the equation:

Ψ_{s} = - (1) (0.50) (0.0831) (293)

Calculate:

Ψ_{s} = −12.2 bar

Rubric

Award 1 point for: the core loses mass (water leaves the tissue).
Award 1 point for: 0.50 M sucrose has Ψ = −1 × 0.50 × 0.0831 × 293 = −12.2 bar, lower than the tissue’s −7.3 bar, and water moves toward the lower water potential.
Accept: −12.1 to −12.3 bar for the solution.
Accept: a value carried forward correctly from a wrong answer to an earlier part; the point is for this part’s step.

Slip Sending water toward the higher, less negative, value. Water moves toward the lower water potential. −12.2 bar is lower than −7.3 bar, so water moves into the solution.

APBIO-U02-P27B Practice questions: Topic 2.7b

Topic 2.7b · Water Potential · 10 MCQ · 2 FRQ · for APBIO-U02-T27B

Video: Watch first: Two parts and one equation

Water potential, its two parts, Ψs = −iCRT, which way water goes and how hard the cell wall pushes.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T27B-summary.mp4

Q1 P27B-q01

A dry bean seed has a water potential of about −100 bar. It sits in a dish of pure water.

Which way does water move, and why?

A. ✓ Into the seed, because 0 bar is a higher water potential than −100 bar
B. Out of the seed, because 100 is a bigger number than 0
−100 bar lies far below 0 bar, so it is the lower water potential.
C. There is no net movement, because a dry seed holds no water to exchange
Water potential is a tendency to move; a seed has a value whether or not much water is present.
D. Into the seed, because the water at 0 bar has the lower water potential
The direction is right, but 0 bar is the higher value.

Why: Water potential measures how strongly water tends to move.
Water moves from higher water potential to lower.
Pure water in an open dish is at 0 bar.
The dry seed is at −100 bar, far lower.
So water moves into the seed.
The seed swells.

Q2 P27B-q02

A syringe holds sucrose solution at a concentration of 0.3 mol/L, written 0.3 M. A membrane across its tip lets water through but not sucrose, and the tip stands in an open beaker of 0.3 M sucrose at the same temperature. A student pushes the plunger in and holds it, pressing the solution inside at 2 bar.

Which way does water move across the membrane while the plunger is held?

A. There is no net movement of water
Equal concentrations give equal solute potentials, but water potential has a pressure part too.
B. ✓ Out of the syringe into the beaker
C. Into the syringe from the beaker
Pressure pushing on water raises its water potential, so the pressed solution tends to lose water.
D. Sucrose moves out of the syringe through the membrane
The membrane lets water through but not sucrose, however hard the plunger pushes.

Why: The two solutions have the same concentration, so the same solute potential.
The plunger gives the syringe solution a pressure potential of +2 bar; the beaker’s has Ψp = 0 bar.
Pressure raises water potential, so the syringe solution has the higher water potential.
So water moves into the beaker.

Q3 P27B-q03

A leaf cell has a pressure potential of +7 bar and a solute potential of −15 bar.

What is its water potential?

A. −22 bar
The two parts add.
Water potential is the sum of its two parts, Ψ = Ψp + Ψs. Ψp = +7 bar and Ψs = −15 bar.
B. +8 bar
The solute part, −15 bar, is bigger in size than the +7 bar pressure part, so the total stays negative.
C. ✓ −8 bar
D. −15 bar
−15 bar is only the solute part; the pressure of +7 bar raises the total.

Why: Water potential is the sum of the pressure potential and the solute potential.
The pressure raises it and the solute lowers it; the working is below.

Q4 P27B-q04

A student has a beaker of 0.10 M NaCl at 25 °C. NaCl splits completely into Na⁺ and Cl⁻ when it dissolves; glucose stays as whole molecules.

Which glucose solution at 25 °C has the same solute potential as the 0.10 M NaCl?

A. ✓ 0.20 M glucose
B. 0.10 M glucose
0.10 M glucose puts half as many dissolved particles into the water as 0.10 M NaCl, because each NaCl unit splits into two ions and each glucose molecule stays whole.
C. 0.05 M glucose
NaCl makes more particles per unit than glucose, so the glucose solution must be more concentrated, not less.
D. 0.40 M glucose
0.40 M glucose puts in four times the particles of 0.10 M glucose, and 0.10 M NaCl puts in only twice as many.

Why: Ψs = −iCRT depends on the concentration of dissolved particles, i × C.
Each NaCl unit splits into two ions, so i = 2; glucose stays whole, so i = 1.
So the glucose solution needs twice the molar concentration; the working is below.

Q5 P27B-q05

A 0.12 M NaCl solution sits in an open beaker at 24 °C. R = 0.0831 L·bar/(mol·K).

What is its solute potential?

A. −2.96 bar
NaCl splits into Na⁺ and Cl⁻, so i = 2.
B. +5.92 bar
The minus sign in Ψs = −iCRT makes every solute potential zero or negative.
C. −0.48 bar
Temperature goes in as kelvin.
NaCl splits into two ions, so i = 2.
D. ✓ −5.92 bar

Why: Dissolved solute lowers water potential, so Ψs is negative.
NaCl gives two particles per unit, so i = 2 doubles the effect; the working is below.

Q6 P27B-q06

The figure shows three cells in a row inside a root. Cell A, nearest the soil, has Ψ = −3 bar; cell B has Ψ = −6 bar; cell C, nearest the center of the root, has Ψ = −10 bar.

Three cells in a row inside a root, from the soil side to the center, with the water potential of each.

In which direction does water move along the row?

A. From C to B to A
−3 bar is the highest of the three water potentials, so water leaves cell A, it does not arrive there.
B. From B outward, to A and to C
Cell B, at −6 bar, is lower than cell A and higher than cell C, so water comes into B from A and leaves B for C.
C. No net movement of water along the row
The three water potentials differ, by 3 bar and then 4 bar, so water does move along the row.
D. ✓ From A to B to C

Why: Water moves by osmosis from a region of higher water potential to one of lower.
Cell A is at −3 bar, cell B at −6 bar, cell C at −10 bar.
−3 bar is higher than −6, which is higher than −10.
So water moves A to B to C.

Q7 P27B-q07

A walled cell has a solute potential of −12.0 bar; assume this does not change as water moves. It sits in an open beaker of 0.10 M sucrose at 27 °C until water stops entering. R = 0.0831 L·bar/(mol·K).

What pressure potential has the cell reached?

A. +12.0 bar
+12.0 bar is what the cell would reach in pure water; the sucrose solution is at −2.49 bar.
B. −9.51 bar
Water entering presses the contents against the cell wall, and that push raises Ψ.
C. ✓ +9.51 bar
D. +14.5 bar
The cell’s water potential has to match −2.49 bar, not −14.5 bar.

Why: In an open beaker Ψp = 0 bar, so the solution’s water potential is its solute potential.
Water stops entering when the cell’s water potential reaches the solution’s.
The cell’s solute potential stays at −12.0 bar, so the pressure potential must make up the difference; the working is below.

Q8 P27B-q08

A piece of plant tissue has a water potential of −9.00 bar at 25 °C. R = 0.0831 L·bar/(mol·K).

Which sucrose concentration, in an open beaker at 25 °C, has the same water potential as the tissue?

A. ✓ 0.363 mol/L
B. 0.182 mol/L
Sucrose stays whole when it dissolves, so i = 1.
C. 4.33 mol/L
Temperature goes in as kelvin, 298 K.
In an open beaker Ψp = 0 bar, so the solution’s Ψ is its Ψs.
D. −0.363 mol/L
A concentration is never negative; the two minus signs in C = −Ψs ÷ (iRT) cancel.

Why: The tissue neither gains nor loses water when the solution has the same water potential, −9.00 bar.
In an open beaker Ψp = 0 bar, so the solution's water potential is its solute potential.
Rearranging Ψs = −iCRT for C gives the concentration; the working is below.

Q9 P27B-q09

A cucumber core has a mass of 7.0 g before it is placed in a solution and 8.4 g an hour later.

What is the percent change in mass?

A. +1.4%
+1.4 g is the change in mass, not the percent change.
B. ✓ +20%
C. +17%
Percent change = (final − initial) ÷ initial × 100.
D. −20%
The core gained mass, so the change is positive.

Why: $percent change = \frac{final - initial}{initial} \times 100 = \frac{8.4 - 7.0}{7.0} \times 100 = +20 %$ .
The sign shows the core gained mass.

Q10 P27B-q10

A class cuts six cores from one parsnip, measures the mass of each, leaves each in a different sucrose solution at 23 °C for an hour, and measures the mass again. The graph gives the percent change in mass of each core and the best-fit straight line. R = 0.0831 L·bar/(mol·K).

Percent change in mass of parsnip cores after an hour in six sucrose solutions at 23 °C, one core per solution, with the best-fit straight line.

What is the water potential of the parsnip tissue?

A. −14.8 bar
Sucrose stays whole when it dissolves, so i = 1.
B. ✓ −7.38 bar
C. −0.573 bar
In kelvin the temperature is 296 K.
In an open beaker Ψp = 0 bar, so Ψ = Ψs.
D. −4.92 bar
−4.92 bar is the water potential of 0.20 M sucrose, the tested solution whose core changed least; the isotonic concentration is read from where the line crosses zero.

Why: Where the best-fit line crosses zero, a core neither gains nor loses mass.
The line crosses zero at 0.30 M, so 0.30 M sucrose has the tissue’s water potential.
In an open beaker Ψp = 0 bar, so Ψ = Ψs; the working is below.

FRQ 1 P27B-frq1 · Scientific Investigation scaffolded

A class investigates the water potential of apple tissue. The class cuts six cores from one apple with the same cutter, blots them dry and measures the mass of each. The class puts each core into a beaker of sucrose solution at 25 °C, at one of six concentrations from 0.0 M to 1.0 M. After two hours the class blots each core and measures its mass again. The membranes of the apple cells let water through but not sucrose. R = 0.0831 L·bar/(mol·K). The table shows the results; one value, for the 0.2 M core, is left for you to calculate.

Mass of each apple core before and after two hours in sucrose solution at 25 °C, and the percent change in mass.

(a) Identify the independent variable and the dependent variable in this investigation. (1 pt)

Frame The independent variable is …; the dependent variable is …

Hint Which quantity did the class set before the experiment began, and which did they measure at the end?

Model answer The independent variable is the sucrose concentration, in M.
The dependent variable is the percent change in mass of the core.

Rubric

Award 1 point for: independent variable, sucrose concentration (M); dependent variable, percent change in mass (accept: change in mass, or mass of the core).
Do not award the point if the two are reversed. Naming a controlled variable (temperature, time, the apple) in place of either earns nothing.

Slip Reversing the two, or naming a controlled variable such as temperature. The concentration was set; the change in mass was measured.

(b) Calculate the missing value for the 0.2 M core. (1 pt)

Frame (… − …) ÷ … × 100 = …%

Hint A percent change compares the change to a starting value. Which of the two masses is the starting value here, and what sign should a gain carry?

Model answer The starting mass is 6.2 g.
The final mass is 6.8 g.
percent change = ((6.8) − (6.2)) ÷ (6.2) × 100 = +9.7%.
So the 0.2 M core changed by +9.7%.

Working

Write down the values in the question:

initial mass = 6.2 g

final mass = 6.8 g

Write down the equation:

tex: \text{percent change} = \frac{\text{final} - \text{initial}}{\text{initial}} \times 100

Substitute the values into the equation:

tex: \text{percent change} = \frac{(6.8) - (6.2)}{(6.2)} \times 100

Calculate:

tex: \text{percent change} = +9.7\%

Rubric

Award 1 point for: +9.7% (accept +9.6 to +9.8%, with the positive sign and the % sign).
Do not award the point for +0.6 (the change in grams), for +8.8% (divided by the final mass) or for a negative value.

Slip Giving +0.6, the change in grams, or dividing by the final mass 6.8 g. Divide the change by the starting mass.

(c) Describe the pattern in the results. Determine the sucrose concentration that is isotonic to the apple tissue. (1 pt)

Frame As the sucrose concentration rises, the change in mass …; the change would be zero at about … M, because …

Hint Where in the table does the change in mass switch from a gain to a loss?

Model answer As the sucrose concentration rises, the gain in mass shrinks.
Above some concentration the gain turns into a loss.
The change is +3.9% at 0.4 M and −1.3% at 0.6 M.
So zero change falls between 0.4 M and 0.6 M.
−1.3% is nearer to zero than +3.9%, so the crossing lies nearer to 0.6 M.
Therefore about 0.55 M sucrose is isotonic to the tissue.

Rubric

Award 1 point for: the pattern (as the sucrose concentration rises, the gain in mass shrinks and becomes a loss) AND the decision that the isotonic concentration lies between 0.4 M (+3.9%) and 0.6 M (−1.3%), with the reasoning it rests on: zero change falls between a gain and a loss, nearer the smaller change. Any estimate from 0.50 M to 0.60 M earns the point; interpolation gives about 0.55 M.
Do not award the point for 0.4 M or 0.6 M read straight from the table with no interpolation, or for the pattern alone. Carry the student’s estimate forward into parts (e) and (f).

Slip Picking 0.4 M or 0.6 M straight from the table. Zero lies between the two rows; judge where.

(d) Explain why the cores in the more concentrated solutions lost mass. (1 pt)

Frame In the concentrated solutions the cores lost mass because water …, from … toward …

Hint Which side has more solute per liter, and which way does water move across a membrane that lets water through but not sucrose?

Model answer In the concentrated solutions, the beaker held more solute per liter than the apple cells.
So the solution had the lower water potential.
Water moves by osmosis from higher water potential to lower.
So water left the apple cells for the solution.
Sucrose cannot cross the membranes, so only water moved.
The water lost is the mass lost.
So the cores lost mass.

Rubric

Award 1 point for: water left the apple cells by osmosis (net water movement), because those solutions held more solute than the cells (had a lower water potential) and sucrose cannot cross the membranes; the water lost is the mass lost.
Accept: 'water moves toward the side with more solute, which was the beaker'. Do not award the point for 'sucrose entered the cores' or for 'the cores dried out' with no reference to solute or water potential.

Slip Having sucrose move into the cores. Sucrose stays put; water moves toward the side with more solute.

(e) Calculate the water potential of the apple tissue at 25 °C, using your estimate from part (c). (1 pt)

Frame The tissue's Ψ equals the Ψs of the isotonic solution: Ψs = −iCRT = … bar

Hint The formula is Ψs = −iCRT. What is Ψp for a solution in an open beaker, what is i for sucrose, and which unit does T need?

Model answer The apple tissue has the same water potential as the isotonic solution, 0.55 M sucrose.
In an open beaker Ψp = 0 bar, so Ψ = Ψs.
Sucrose has i = 1, and T = 298 K.
Ψs = −(1)(0.55)(0.0831)(298) = −13.6 bar.
So the apple tissue is at about −13.6 bar.

Working

Write down the values in the question:

i = 1

C = 0.55 mol/L

R = 0.0831 L·bar/(mol·K)

T = 298 K (25 + 273)

Ψp = 0 bar (open beaker), so Ψ = Ψs

Write down the equation:

tex: \Psi_s = -iCRT

Substitute the values into the equation:

tex: \Psi_s = -(1)(0.55)(0.0831)(298)

Calculate:

tex: \Psi_s = -13.6\,\text{bar}

Rubric

Award 1 point for: Ψ = Ψs of the isotonic solution = −1 × C × 0.0831 × 298, using the student’s own estimate from part (c) with i = 1 and T = 298 K. For C = 0.55 M this is −13.6 bar. Any value from −12.4 bar (C = 0.50 M) to −14.9 bar (C = 0.60 M) earns the point, provided it matches the student’s part (c) estimate; negative, in bars.
Do not award the point for a positive value, for 25 used in place of 298 K, or for i = 2.

Slip Using 25 in place of 298 K, or dropping the minus sign. T must be in kelvin, and a solute potential is negative.

(f) A seventh core from the same apple is placed in 0.70 M sucrose at 25 °C. Make a claim about whether it gains or loses mass, and support your claim using water potentials. (1 pt)

Frame The 0.70 M solution has Ψ = … bar, which is … than the tissue's … bar, so water moves … and the core … mass.

Hint Find the water potential of the 0.70 M solution, then ask which way water moves between two water potentials.

Model answer The core loses mass.
In an open beaker Ψ = Ψs.
For 0.70 M sucrose at 298 K, Ψs = −(1)(0.70)(0.0831)(298) = −17.3 bar.
The tissue is at about −13.6 bar.
So −17.3 bar is lower than −13.6 bar.
Water moves from higher water potential to lower.
Therefore water moves from the cells into the solution, and the core loses mass.

Working

Write down the values in the question:

i = 1

C = 0.70 mol/L

R = 0.0831 L·bar/(mol·K)

T = 298 K (25 + 273)

Ψ of the tissue = −13.6 bar (from part e)

Write down the equation:

tex: \Psi_s = -iCRT

Substitute the values into the equation:

tex: \Psi_s = -(1)(0.70)(0.0831)(298)

Calculate:

tex: \Psi_s = -17.3\,\text{bar}

−17.3 bar is lower than −13.6 bar, so water leaves the core

Rubric

Award 1 point for: the claim that the core loses mass, supported by evidence AND reasoning: 0.70 M sucrose at 25 °C has Ψ = −(1)(0.70)(0.0831)(298) = −17.3 bar (accept −17.2 to −17.4 bar), which is lower (more negative) than the tissue’s value from the student’s own part (e) (anywhere from −12.4 to −14.9 bar; about −13.6 bar for 0.55 M), so net water movement is from the cells into the solution.
Accept support from the table: 0.70 M lies between 0.6 M (−1.3%) and 0.8 M (−6.3%), both losses, so the core loses. Do not award the claim with no comparison of water potentials or table rows.

Slip Making the claim with no comparison of the two water potentials. The point needs the two values side by side.

FRQ 2 P27B-frq2 · Conceptual Analysis

A cell from the leaf of a pondweed has a rigid cell wall. Its solute potential is −6.5 bar; assume this does not change as the cell gains or loses water. A student puts the cell into an open beaker of 0.10 M sucrose at 25 °C. At the moment the cell goes in, its contents rest against the cell wall without pushing on it. R = 0.0831 L·bar/(mol·K).

(a) Describe what the pressure potential of a cell is, and state the pressure potential of the sucrose solution in the open beaker. (1 pt)

Frame Pressure potential is the part of water potential that comes from …; for the solution in the open beaker Ψp = … bar, because …

Model answer Pressure potential is the part of water potential that comes from pressure on the water.
Pressure pushing on water raises its water potential.
So the contents of a turgid cell, pressed against the cell wall, have a positive Ψp.
Nothing presses on the solution in the open beaker.
So the solution has Ψp = 0 bar.

Rubric

Award 1 point for: pressure potential is the part of water potential due to pressure on the water (pressure pushing on water raises its water potential; it is positive for the contents of a turgid cell pressed against its cell wall), AND Ψp = 0 bar for the solution in the open beaker, because nothing presses on it.
Accept: 'the push of the cell wall on the contents' as the description. Both the description and the 0 bar are needed for the point.

Slip Giving Ψp = 0 bar with no statement of what pressure potential is, or the other way around. Both parts are needed.

(b) Calculate the water potential of the 0.10 M sucrose solution. (1 pt)

Answer: -2.48 bar (tolerance ±0.05)

Model answer In an open beaker Ψp = 0 bar, so Ψ = Ψs.
Sucrose has i = 1, and T = 25 + 273 = 298 K.
Ψs = −(1)(0.10)(0.0831)(298) = −2.48 bar.
So the solution’s water potential is −2.48 bar.

Working

Write down the values in the question:

i = 1

C = 0.10 mol/L

R = 0.0831 L·bar/(mol·K)

T = 298 K (25 + 273)

Ψp = 0 bar (open beaker), so Ψ = Ψs

Write down the equation:

tex: \Psi_s = -iCRT

Substitute the values into the equation:

tex: \Psi_s = -(1)(0.10)(0.0831)(298)

Calculate:

tex: \Psi_s = -2.48\,\text{bar}

tex: \Psi = 0 + (-2.48) = -2.48\,\text{bar}

Rubric

Award 1 point for: Ψ = Ψs = −(1)(0.10)(0.0831)(298) = −2.48 bar (accept −2.4 to −2.5 bar), with Ψp = 0 bar in the open beaker.
Do not award the point for a positive value, for 25 used in place of 298 K, or for i = 2.

Slip Using 25 in place of 298 K, which gives −0.21 bar. Convert to kelvin first.

(c) Predict which way water moves at first, and calculate the pressure potential the cell reaches once net water movement has stopped. (1 pt)

Model answer Water moves from higher water potential to lower.
The solution is at −2.48 bar.
The cell is at −6.5 bar.
So water moves into the cell.
As water enters, the contents press against the cell wall.
The cell wall presses back, so the pressure potential rises.
Net movement stops when the cell's water potential equals −2.48 bar. Ψp = Ψ − Ψs = (−2.48) − (−6.5) = +4.0 bar.

Working

Write down the values in the question:

Ψ of the cell at equilibrium = Ψ of the solution = −2.48 bar

Ψs of the cell = −6.5 bar

Write down the equation:

tex: \Psi = \Psi_p + \Psi_s

Make Ψp the subject:

tex: \Psi_p = \Psi - \Psi_s

Substitute the values into the equation:

tex: \Psi_p = (-2.48) - (-6.5)

Calculate:

tex: \Psi_p = +4.0\,\text{bar}

Rubric

Award 1 point for: water moves into the cell, because the solution's −2.48 bar is higher than the cell's −6.5 bar, AND the cell's Ψ rises to match −2.48 bar, so Ψp = Ψ − Ψs = (−2.48) − (−6.5) = +4.0 bar (accept +3.9 to +4.1 bar).
Accept 'the cell becomes turgid' for the effect. Do not award the point for water leaving the cell, or for Ψp = +6.5 bar (the pure-water value).

Slip Giving +6.5 bar, the pressure the cell would reach in pure water. This solution is at −2.48 bar, so the cell's Ψ only has to rise that far.

(d) Calculate the sucrose concentration at 25 °C in which the cell's contents would just stop pressing on the cell wall (Ψp = 0 bar), and predict what the cell looks like in a solution more concentrated than that. (1 pt)

Model answer At Ψp = 0 bar the cell's water potential is its solute potential, −6.5 bar.
So the matching solution must be at −6.5 bar.
C = −Ψs ÷ (iRT) = (6.5) ÷ ((1)(0.0831)(298)) = 0.26 M sucrose.
In a more concentrated solution, the solution's water potential is lower than the cell's.
So water leaves the cell.
The contents shrink away from the cell wall (plasmolysis).
The cell wall keeps its shape.
The tissue goes limp.

Working

Write down the values in the question:

Ψp = 0 bar, so Ψ of the cell = Ψs = −6.5 bar

the solution must have Ψs = −6.5 bar

i = 1

R = 0.0831 L·bar/(mol·K)

T = 298 K (25 + 273)

Write down the equation:

tex: \Psi_s = -iCRT

Make C the subject:

tex: C = \frac{-\Psi_s}{iRT}

Substitute the values into the equation:

tex: C = \frac{(6.5)}{(1)(0.0831)(298)}

Calculate:

tex: C = 0.26\,\text{mol/L}

Rubric

Award 1 point for: at Ψp = 0 bar the cell's Ψ is its Ψs, −6.5 bar, so the solution must be at −6.5 bar: C = (6.5) divided by ((1)(0.0831)(298)) = 0.26 M (accept 0.25 to 0.27 M), AND in a more concentrated solution water leaves, the contents shrink away from the cell wall (plasmolysis) while the cell wall keeps its shape, and the tissue goes limp.
Accept 'the cell loses turgor' or 'goes limp' for the effect. Do not award the point for a negative concentration, for 25 used in place of 298 K, or for a cell that bursts (walled cells do not).

Slip Answering −0.26 M. A concentration is never negative; the two minus signs cancel when Ψs is negative.

APBIO-U02-T27B End-of-topic test: Water Potential

Topic 2.7b · Water Potential · 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 free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it. R = 0.0831 L·bar/(mol·K); temperatures in kelvin are °C + 273.

Q1 T27B-q01

Water potential (Ψ) is measured in bars. Pure water in an open beaker has Ψ = 0 bar.

What does a water potential measure?

A. ✓ How strongly water tends to move
B. How much water a solution holds
Pressing on a solution raises its Ψ without adding any water.
C. How much solute is dissolved in a solution
Pressing on a solution changes its Ψ with no change in solute, so Ψ is not a measure of solute.
D. How fast the water is moving
Ψ is a tendency, not a speed: water standing still in two beakers can have different Ψ.

Why: Water potential (Ψ) measures how strongly water tends to move.
Water moves by osmosis from higher water potential to lower.
The bar is the unit of pressure used.
Pure water in an open container is given Ψ = 0 bar.

Q2 T27B-q02

The figure shows three open beakers at the same temperature, with concentrations given in M, short for mol/L: pure water, 0.1 M sucrose and 0.3 M sucrose.

Three open beakers at the same temperature, filled to the same level.

Rank their water potentials from highest to lowest.

A. Beaker 3, then 2, then 1
Dissolving solute lowers water potential, and more solute lowers it further.
B. Beaker 2, then 1, then 3
Pure water in an open container is at 0 bar, the highest any of these can be.
C. All equal: none is under pressure
Solute is the other part of water potential, and the three beakers differ in solute.
D. ✓ Beaker 1, then 2, then 3

Why: The solute potential, Ψs, is zero for pure water, negative for any solution, and more negative the more solute is dissolved.
No beaker is under pressure, so Ψ = Ψs for each.
So pure water (0 bar) ranks highest, then 0.1 M sucrose, then 0.3 M sucrose.

Q3 T27B-q03

A beaker of sucrose solution stands open on the bench.

Which change lowers the water potential of the solution?

A. Adding pure water to the solution
Adding pure water leaves less solute per liter, so the water potential rises.
B. ✓ Dissolving more sucrose in the solution
C. Pressing on the surface of the solution with a piston
Pressure pushing on water raises its water potential.
D. Pouring half of the solution away
Pouring some away leaves the rest at the same concentration, so its water potential is unchanged.

Why: In an open beaker Ψp is 0 bar, so Ψ = Ψs.
Dissolved solute lowers the solute potential below zero.
More solute lowers it further.
So dissolving more sucrose lowers the water potential.
Diluting raises it.
Pressing raises it.
Pouring some away leaves it unchanged.

Q4 T27B-q04

A guard cell in a leaf sits in pond water in an open dish. The cell’s contents push outward on its cell wall at 7 bar and the cell wall pushes back, so the cell is turgid. Nothing presses on the pond water in the dish.

What are the pressure potentials of the cell contents and of the pond water?

A. Cell −7 bar; pond water 0 bar
Pressure pushing on water raises its water potential, so a turgid cell’s pressure potential is positive.
B. ✓ Cell +7 bar; pond water 0 bar
C. Cell 0 bar; pond water +7 bar
The pond water in the open dish has nothing pressing on it; it is the cell’s contents that are pressed against the cell wall.
D. Cell +7 bar; pond water +7 bar
Only the cell’s contents are under pressure; water in an open container has Ψp = 0 bar.

Why: Pressure potential, Ψp, is the part of water potential due to pressure.
The cell wall presses on the turgid guard cell’s contents at 7 bar, so the contents have Ψp = +7 bar.
Nothing presses on water in an open container, so the pond water has Ψp = 0 bar.

Q5 T27B-q05

Two beakers hold the same 0.2 M sucrose solution at the same temperature. One is open; in the other a piston presses on the solution at 3 bar, as in the drawing.

Two beakers of the same 0.2 M sucrose solution at the same temperature. The left is open; in the right a piston presses on the solution at 3 bar. Each dot is a sucrose molecule.

Compare the water potentials of the two solutions, and explain the difference.

A. ✓ The pressed solution has the higher water potential, by 3 bar, because pressure raises water potential
B. The pressed solution has the lower water potential, because pressure squeezes water out of it
Pressure does make water tend to leave, and a stronger tendency to leave is exactly what a higher water potential means.
C. The two water potentials are equal, because the two concentrations are equal
Equal concentrations give equal solute potentials, but water potential has two parts, and the piston adds pressure to one solution only.
D. The open solution has the higher water potential, 0 bar, because nothing presses on it
Nothing pressing gives the open solution Ψp = 0 bar, but the dissolved sucrose gives it a negative Ψs, so its Ψ is below zero.

Why: Ψ = Ψp + Ψs.
The two solutions have the same sucrose concentration, so the same negative Ψs.
The open solution has Ψp = 0 bar; the piston gives the other Ψp = +3 bar.
Pressure raises water potential, so the pressed solution’s Ψ is higher by 3 bar.

Q6 T27B-q06

A leaf cell has a pressure potential of +2.5 bar and a solute potential of −12 bar.

What is its water potential?

A. −14.5 bar
Ψ = Ψp + Ψs, and Ψp is +2.5 bar: the pressure part is added to −12 bar, not taken away from it.
B. ✓ −9.5 bar
C. +9.5 bar
The solute part, −12 bar, is bigger in size than the +2.5 bar pressure part, so the total stays negative.
D. −12 bar
−12 bar is only the solute part; the pressure of +2.5 bar raises the total.

Why: Water potential is the sum of the pressure potential and the solute potential.
The pressure raises it and the solute lowers it; the working is below.

Q7 T27B-q07

Two plant cells touch. Cell X has a pressure potential of +4 bar and a solute potential of −9 bar. Cell Y has a pressure potential of +1 bar and a solute potential of −7 bar, as in the drawing.

Two plant cells that touch, with the pressure potential and solute potential of each.

Which way does water move between the two cells, and why?

A. From cell Y into cell X, because +4 bar is higher than +1 bar
The direction follows the whole water potential, Ψ = Ψp + Ψs: X is at −5 bar and Y at −6 bar.
B. From cell Y into cell X, because −9 bar is lower than −7 bar
Cell X does hold more solute, but its cell wall presses harder too; the two parts add.
C. ✓ From cell X into cell Y, because −5 bar is higher than −6 bar
D. There is no net movement, because −5 bar and −6 bar are both negative
Two negative values can still differ, and water moves down that 1 bar difference.

Why: Add the two parts of each cell’s water potential.
Cell X comes to −5 bar and cell Y to −6 bar; the working is below.
Water moves from higher water potential to lower, and −5 bar is higher than −6 bar, so water moves from X into Y.

Q8 T27B-q08

Two beakers at the same temperature hold 0.10 M solutions: one of potassium chloride, KCl, and one of fructose, a sugar. KCl splits completely into K⁺ and Cl⁻ when it dissolves; fructose stays as whole molecules.

Which solution has more dissolved particles, and what value of i does each solute take?

A. Fructose, twice as many; i = 2 for fructose, 1 for KCl
Fructose stays as whole molecules, one particle per unit, so i = 1; it is KCl that splits into two ions.
B. The same number; i = 1 for both
Each KCl unit splits into two ions, K⁺ and Cl⁻, so 0.10 M KCl puts twice as many particles into the water as 0.10 M fructose.
C. ✓ KCl, twice as many; i = 2 for KCl, 1 for fructose
D. KCl, twice as many; i = 1 for both
The ionization constant i counts the separate particles each dissolved unit splits into.

Why: The ionization constant i is the number of separate particles each dissolved unit splits into.
Fructose does not split, so i = 1.
KCl splits fully into K⁺ and Cl⁻, so i = 2.
At the same molar concentration, 0.10 M KCl has twice the dissolved particles.

Q9 T27B-q09

A 0.18 M NaCl solution is at 30 °C. R = 0.0831 L·bar/(mol·K).

What is its solute potential?

A. ✓ −9.06 bar
B. −4.53 bar
NaCl splits into Na⁺ and Cl⁻, so i = 2.
C. +9.06 bar
The minus sign in Ψs = −iCRT makes every solute potential zero or negative.
D. −0.90 bar
Temperature has to be in kelvin: 30 °C is 303 K, and −0.90 bar comes from putting 30 in place of 303.

Why: Dissolved solute lowers water potential, so Ψs is negative.
NaCl gives two particles per unit, so i = 2 doubles the effect; the working is below.

Q10 T27B-q10

A 0.25 M glucose solution is at 30 °C. Glucose stays as whole molecules when it dissolves. R = 0.0831 L·bar/(mol·K).

What is the solute potential of the solution?

A. −12.6 bar
Glucose stays whole when it dissolves, so i = 1.
B. −0.623 bar
Temperature goes into the equation in kelvin, 303 K.
C. +6.29 bar
The minus sign in Ψs = −iCRT makes every solute potential zero or negative.
D. ✓ −6.29 bar

Why: Dissolved solute lowers water potential, so Ψs is negative.
Glucose stays whole, so i = 1, and the temperature is 303 K; the working is below.

Q11 T27B-q11

A cell has a water potential of −6 bar. It sits in a solution whose water potential is −9 bar.

What happens to the cell's water?

A. Net movement of water into the cell
−9 lies further below zero than −6, so −9 bar is the lower water potential.
B. Net movement of solute into the cell
A water potential describes how strongly water tends to move; it does not describe the solute moving.
C. No net movement of water
−9 and −6 differ by 3 bar. −9 bar lies further below zero than −6 bar, so the solution has the lower water potential.
D. ✓ Net movement of water out of the cell

Why: Water moves from higher water potential to lower.
The cell is at −6 bar.
The solution is at −9 bar.
−9 bar is lower than −6 bar.
So water leaves the cell for the solution until the two values are equal.

Q12 T27B-q12

A cell with a water potential of −5.0 bar sits in an open beaker of 0.14 M NaCl at 22 °C. NaCl splits completely into Na⁺ and Cl⁻ when it dissolves. R = 0.0831 L·bar/(mol·K).

What happens to the cell's water, and why?

A. The cell gains water: the solution, at −0.512 bar, is higher than the cell's −5.0 bar
In kelvin the temperature is 295 K.
In an open beaker Ψp = 0 bar, so the solution’s Ψ is its Ψs.
B. The cell gains water: the solution, at −3.43 bar, is higher than the cell's −5.0 bar
NaCl splits into Na⁺ and Cl⁻, so i = 2; −3.43 bar is the value with i = 1.
C. ✓ The cell loses water: the solution, at −6.86 bar, is lower than the cell's −5.0 bar
D. There is no net movement: the two values are both negative and within 2 bar of each other
Two negative values can still differ, and water moves down any difference.

Why: In an open beaker Ψp = 0 bar, so Ψ = Ψs.
NaCl splits into two ions, so i = 2; T = 295 K.
The solution comes to −6.86 bar, lower than the cell’s −5.0 bar, so the cell loses water; the working is below.

Q13 T27B-q13

A root cell of a salt-marsh plant has a rigid cell wall and a solute potential of −8 bar; assume this does not change as water moves. It sits in an open beaker of solution at Ψ = −3 bar until net water movement stops.

What pressure potential has the cell reached?

A. ✓ +5 bar
B. +8 bar
+8 bar is what the cell would reach in pure water; this solution is at −3 bar.
C. −5 bar
Water entering presses the contents against the cell wall, and that push raises Ψ, so Ψp is positive.
D. +11 bar
The cell’s Ψ has to match −3 bar, not −11 bar.

Why: Net movement stops when the cell’s water potential equals the solution’s, −3 bar.
The cell’s solute potential stays at −8 bar.
Ψ = Ψp + Ψs, so Ψp = Ψ − Ψs = (−3) − (−8) = +5 bar.
Water entered until the cell wall’s push raised Ψ by 5 bar.

Q14 T27B-q14

A walled cell has a solute potential of −5.5 bar; assume this does not change as water moves. It sits in an open beaker of solution with a water potential of −5.5 bar until its water potential matches the solution’s.

What pressure potential has the cell reached?

A. +5.5 bar
+5.5 bar is what the cell would reach in pure water; this solution is at −5.5 bar.
B. −5.5 bar
A pressure potential inside a walled cell is never negative, and here the cell wall is not pressing on the contents at all.
C. Cannot tell: the equation gives no answer when the two values are equal
The equation does give an answer: two equal values subtract to zero.
D. ✓ 0 bar

Why: The cell’s water potential matches the solution’s, −5.5 bar, and its solute potential stays at −5.5 bar.
Ψ = Ψp + Ψs, so Ψp = (−5.5) − (−5.5) = 0 bar.
The solution already equals the cell’s solute potential, so no net water enters to press on the cell wall.

Q15 T27B-q15

A cell from a moss leaf has a water potential of −3.0 bar. R = 0.0831 L·bar/(mol·K).

Which sucrose concentration, in an open beaker at 20 °C, has the same water potential as the moss cell?

A. 0.06 M
Sucrose does not split into ions, so i = 1.
B. ✓ 0.12 M
C. 0.25 M
0.25 M sucrose at 20 °C has Ψs = −6.1 bar, twice as negative as the cell’s.
D. 1.8 M
Temperature has to be in kelvin, 293 K.

Why: In an open beaker nothing presses on the solution, so Ψp = 0 bar and Ψ = Ψs.
So the solution's Ψs must equal the cell's Ψ.
Rearranging Ψs = −iCRT gives the concentration; the working is below.

Q16 T27B-q16

A piece of plant tissue has a water potential of −7.50 bar at 27 °C. NaCl splits completely into Na⁺ and Cl⁻ when it dissolves. R = 0.0831 L·bar/(mol·K).

Which NaCl concentration, in an open beaker at 27 °C, has the same water potential as the tissue?

A. 0.301 mol/L
NaCl splits into Na⁺ and Cl⁻, so each dissolved unit makes two particles and i = 2.
B. 1.67 mol/L
Temperature goes in as kelvin, 300 K.
In an open beaker Ψp = 0 bar, so the solution’s Ψ is its Ψs.
C. ✓ 0.150 mol/L
D. 6.65 mol/L
C = −Ψs divided by iRT.
In an open beaker Ψp = 0 bar, so the solution’s Ψ is its Ψs.

Why: The tissue neither gains nor loses water when the solution is also at −7.50 bar.
In an open beaker Ψ = Ψs.
NaCl splits into two ions, so i = 2.
Rearranging Ψs = −iCRT for C gives the concentration; the working is below.

Q17 T27B-q17

A potato core has a mass of 12.5 g before it is placed in a salt solution and 11.0 g afterward.

What is the percent change in mass?

A. −1.5%
−1.5 g is the change in mass, not the percent change.
B. +12%
The core lost mass, so the change is negative.
C. ✓ −12%
D. −14%
Percent change = (final − initial) ÷ initial × 100.

Why: Percent change compares the change in mass with the starting mass.
The core lost mass, so the sign is negative; the working is below.

Q18 T27B-q18

The table shows a student's results: the percent change in mass of potato cores soaked in 0.0, 0.2, 0.4, 0.6, 0.8 and 1.0 M sucrose, one value per concentration. The values carry some measurement scatter.

Percent change in mass of a potato core after soaking in each sucrose solution.

Which way of drawing the graph lets the isotonic concentration be read from these results?

A. Percent change on the x-axis, concentration (M) on the y-axis, one straight best-fit line
The variable the student set, concentration, goes on the x-axis.
B. Concentration (M) on the x-axis, percent change on the y-axis, points joined dot to dot
Joining the dots reproduces every wobble of measurement scatter instead of smoothing it out.
C. Concentration (M) on the x-axis, final mass (g) on the y-axis, one straight best-fit line
Final mass depends on how big each core was to start with; percent change removes that.
D. ✓ Concentration (M) on the x-axis, percent change on the y-axis, one straight best-fit line

Why: The variable the student set, sucrose concentration in M, goes on the x-axis.
The variable measured, percent change in mass, goes on the y-axis, with a scale that includes the negative values.
One straight best-fit line smooths the scatter.
Where the line crosses 0% is the isotonic concentration.

Q19 T27B-q19

A class cuts six cores from one sweet potato, measures the mass of each, leaves each in a different sucrose solution at 25 °C for an hour, and measures the mass again. The graph gives the percent change in mass of each core and the best-fit straight line. R = 0.0831 L·bar/(mol·K).

Percent change in mass of sweet potato cores after an hour in six sucrose solutions at 25 °C, one core per solution, with the best-fit straight line.

What is the water potential of the sweet potato tissue?

A. −24.8 bar
Sucrose stays whole when it dissolves, so i = 1.
B. ✓ −12.4 bar
C. −1.04 bar
In kelvin the temperature is 298 K.
In an open beaker Ψp = 0 bar, so Ψ = Ψs.
D. −9.91 bar
−9.91 bar is the water potential of 0.40 M sucrose, the tested solution whose core changed least; the isotonic concentration is read from where the line crosses zero.

Why: Where the best-fit line crosses zero, a core neither gains nor loses mass.
The line crosses zero at 0.50 M, so 0.50 M sucrose has the tissue’s water potential.
In an open beaker Ψp = 0 bar, so Ψ = Ψs; the working is below.

FRQ 1 T27B-frq1 · Scientific Investigation

A student investigates the water potential of potato tissue. The student cuts six cores from one potato with the same cutter, blots them dry and measures the mass of each. The student puts each core into a beaker of sucrose solution at 22 °C, at one of six concentrations from 0.0 M to 1.0 M. After 24 hours the student blots each core and measures its mass again. The membranes of the potato cells are selectively permeable: they let water through but not sucrose. R = 0.0831 L·bar/(mol·K). The results are in the table.

Mass of each potato core before and after 24 hours in sucrose solution at 22 °C.

(a) Identify the independent variable, the dependent variable, and one variable the student kept the same. (1 pt)

Model answer Independent variable: sucrose concentration (M).
Dependent variable: percent change in mass.
Kept the same: temperature (22 °C), time (24 hours), the potato used, the size and shape of the cores (same cutter), and blotting before weighing.

Rubric

Award 1 point for all three: independent variable, sucrose concentration (M); dependent variable, percent change in mass (or change in mass); one controlled variable such as temperature (22 °C), time (24 hours), the potato used, the size and shape of the cores (same cutter), or blotting before weighing.
Accept "mass of the core" as the dependent variable. Do not award the point if the independent and dependent variables are reversed.

Slip Reversing the independent and dependent variables. The concentration was set; the change in mass was measured.

(b) Explain why the core in 0.0 M sucrose gained mass. (1 pt)

Model answer Pure water has no solute.
The potato cells have solute dissolved in their cytosol.
So the pure water has the higher water potential.
Water moves by osmosis from higher water potential to lower.
Sucrose cannot cross the membranes, so only water moves.
So water moved into the potato cells.
The extra water adds mass, so the core gained mass.

Rubric

Award 1 point for: water moved into the potato cells by osmosis (net water movement) because pure water has less solute than the cells (or a higher water potential), and sucrose cannot cross the membranes; the extra water adds mass.
Accept: "water moves toward the side with more solute, which is inside the cells". Do not award the point for "the potato absorbed water" with no reference to solute concentration or water potential, or for an answer in which sucrose moves.

Slip Saying the potato absorbed water, with no reference to solute or water potential, or having sucrose move. Sucrose stays put; water moves toward the side with more solute.

(c) Determine the sucrose concentration that is isotonic to the potato tissue, and calculate the water potential of the tissue at 22 °C. (1 pt)

Model answer The change is +1.0% at 0.4 M and −8.1% at 0.6 M.
So zero change falls just above 0.4 M: about 0.42 M sucrose is isotonic to the tissue.
The tissue has the same water potential as that solution.
In an open beaker Ψp = 0 bar, so Ψ = Ψs; sucrose has i = 1 and T = 22 + 273 = 295 K.
Ψs = −(1)(0.42)(0.0831)(295) = −10.3 bar.
Therefore the tissue is at about −10.3 bar.

Working

Write down the values in the question:

i = 1

C = 0.42 mol/L

R = 0.0831 L·bar/(mol·K)

T = 295 K (22 + 273)

Ψp = 0 bar (open beaker), so Ψ = Ψs

Write down the equation:

Ψs = −iCRT

Substitute the values into the equation:

Ψs = −(1)(0.42)(0.0831)(295)

Calculate:

Ψs = −10.3 bar

Rubric

Award 1 point for: the decision that the isotonic concentration lies between 0.40 and 0.45 M AND the reasoning it rests on (the change is +1.0% at 0.4 M and −8.1% at 0.6 M, so zero change falls just above 0.4 M), AND the tissue's water potential from Ψs = −iCRT with i = 1 and T = 295 K: for 0.42 M, Ψ = −(1)(0.42)(0.0831)(295) = −10.3 bar (accept −9.8 to −11.0 bar, negative, in bars).
Accept 0.40 M read straight from the table, giving −9.8 bar. Score the water potential against the student's own estimate: any estimate from 0.40 M to 0.45 M carried correctly into Ψs = −iCRT earns the point. Do not award the point for a positive value, for a value not in bars, or for 22 used in place of 295 K.

Slip Using 22 in place of 295 K, or dropping the minus sign. T must be in kelvin, and a solute potential is negative.

(d) The student puts a seventh core from the same potato into 0.3 M sucrose at 22 °C. Make a claim about whether it gains or loses mass, and support your claim using water potential. (1 pt)

Model answer The core gains mass.
In an open beaker Ψ = Ψs.
For 0.3 M sucrose at 295 K, Ψs = −(1)(0.3)(0.0831)(295) = −7.4 bar.
The tissue is at about −10 bar.
So −7.4 bar is higher than −10 bar.
Water moves from higher water potential to lower.
Therefore water moves from the solution into the cells, and the core gains mass.

Working

Write down the values in the question:

i = 1

C = 0.3 mol/L

R = 0.0831 L·bar/(mol·K)

T = 295 K (22 + 273)

Write down the equation:

Ψs = −iCRT

Substitute the values into the equation:

Ψs = −(1)(0.3)(0.0831)(295)

Calculate:

Ψs = −7.4 bar

Rubric

Award 1 point for: the claim that the core gains mass, supported by evidence AND reasoning: 0.3 M sucrose has Ψ = −(1)(0.3)(0.0831)(295) = −7.4 bar, which is higher (less negative) than the tissue's water potential of about −10 bar, so net water movement is from the solution into the cells.
Accept support from the table: 0.3 M lies between 0.2 M (+8.0%) and 0.4 M (+1.0%), both of which gained, so the core gains. Do not award the claim with no comparison of water potentials or concentrations.

Slip Making the claim with no comparison of water potentials. The point needs the two values, or the two table rows, side by side.

FRQ 2 T27B-frq2 · Conceptual Analysis

A cell from the stem of a plant has a rigid cell wall. Its solute potential is −8 bar; assume this does not change as the cell gains or loses water. A student puts the cell into an open beaker of dilute solution whose water potential is −2 bar. At the moment the cell goes in, its contents rest against the cell wall without pushing on it. Later, once net water movement has stopped, the student pours the solution away and replaces it with 0.5 M sucrose at 22 °C. R = 0.0831 L·bar/(mol·K).

(a) Describe what water potential measures, and calculate the cell's water potential at the moment it is placed in the beaker. (1 pt)

Model answer Water potential measures how strongly water tends to move.
Water moves from higher water potential to lower.
At the start the contents rest against the cell wall without pushing on it, so Ψp = 0 bar. Ψ = Ψp + Ψs = 0 + (−8) = −8 bar.
So the cell’s water potential at the start is −8 bar.

Working

Write down the values in the question:

Ψp = 0 bar (the contents rest against the cell wall without pushing)

Ψs = −8 bar

Write down the equation:

Ψ = Ψp + Ψs

Substitute the values into the equation:

Ψ = (0) + (−8)

Calculate:

Ψ = −8 bar

Rubric

Award 1 point for: water potential measures how strongly water tends to move (water moves from higher Ψ to lower Ψ), AND at the start Ψ = Ψp + Ψs = 0 + (−8) = −8 bar.
Accept "the tendency of water to leave a region" as the description. Do not award the point for Ψ = +8 bar, or for −8 bar with no statement of what Ψ measures.

Slip Giving Ψ = +8 bar, or −8 bar with no statement of what Ψ measures. Both parts are needed.

(b) State which way water moves at first, and explain why the net movement later stops. (1 pt)

Model answer Water moves from higher water potential to lower.
The beaker is at −2 bar and the cell at −8 bar, so water moves from the beaker into the cell.
As water enters, the contents press against the cell wall, the cell wall presses back, and the pressure potential rises, raising the cell’s water potential.
When the cell’s water potential reaches −2 bar, Ψp = +6 bar and no difference is left, so net movement stops.

Working

Write down the values in the question:

Ψ = −2 bar (the solution’s, once net movement has stopped)

Ψs = −8 bar

Write down the equation:

Ψ = Ψp + Ψs

Make Ψp the subject:

Ψp = Ψ − Ψs

Substitute the values into the equation:

Ψp = (−2) − (−8)

Calculate:

Ψp = +6 bar

Rubric

Award 1 point for: water moves from the beaker (Ψ = −2 bar, higher) into the cell (Ψ = −8 bar, lower); as water enters, the contents press against the cell wall and the pressure potential rises, raising the cell's water potential until it equals −2 bar (Ψp = +6 bar), when there is no longer a difference to drive net movement.
Accept "the cell becomes turgid and its Ψ rises to match the solution's" without the +6 bar figure. Do not award the point for "water moves in until the concentrations are equal": the cell's solute potential stays at −8 bar, and it is pressure that closes the gap.

Slip Saying water moves in until the concentrations are equal. The cell’s solute potential stays at −8 bar; it is pressure that closes the gap.

(c) Calculate the sucrose concentration at 22 °C that would leave the cell's contents just resting against the cell wall, with no pressure on it, and justify your answer. (1 pt)

Answer: 0.33 mol/L (tolerance ±0.01)

Model answer The contents just rest against the cell wall, so Ψp = 0 bar.
So the cell’s water potential equals its solute potential, −8 bar.
Net water movement stops when the solution has the same water potential, −8 bar.
In an open beaker Ψ = Ψs, so the solution needs Ψs = −8 bar.
C = −Ψs ÷ (iRT) = (8) ÷ ((1)(0.0831)(295)) = 0.33 mol/L.
So 0.33 mol/L sucrose leaves the contents just resting against the cell wall.

Working

Write down the values in the question:

Ψ of the solution needed = −8 bar

i = 1 (sucrose does not split)

R = 0.0831 L·bar/(mol·K)

T = 295 K (22 + 273)

Write down the equation:

tex:\Psi_s = -iCRT

Make C the subject:

tex:C = \frac{-\Psi_s}{iRT}

Substitute the values into the equation:

tex:C = \frac{(8)}{(1)(0.0831)(295)}

Calculate:

tex:C = 0.33 \text{ mol/L}

Rubric

Award 1 point for: 0.33 mol/L (accept 0.32 to 0.33 mol/L), with the reason that at zero pressure potential the cell's water potential equals its solute potential (−8 bar), so the solution must also be at −8 bar for net movement to stop.
Do not award the point for using −12.3 bar (the 0.5 M solution) or for a concentration found with T = 22.

Slip Using the 0.5 M solution's −12.3 bar as the target. The target is the cell's own solute potential, −8 bar, because the pressure potential is zero when nothing presses on the cell wall.

(d) Predict what happens to the cell after the solution is replaced with 0.5 M sucrose at 22 °C. Include the water potential of the new solution. (1 pt)

Model answer The new solution is 0.5 M sucrose at 22 °C in an open beaker, so Ψ = Ψs = −(1)(0.5)(0.0831)(295) = −12.3 bar.
−12.3 bar is lower than the cell’s water potential, so water leaves the cell.
As water leaves, the contents stop pressing on the cell wall, so the pressure potential falls to zero and the cell loses turgor.
The contents shrink away from the cell wall (plasmolysis); the cell wall keeps its shape.

Working

Write down the values in the question:

i = 1

C = 0.5 mol/L

R = 0.0831 L·bar/(mol·K)

T = 295 K (22 + 273)

Ψp = 0 bar (open beaker), so Ψ = Ψs

Write down the equation:

Ψs = −iCRT

Substitute the values into the equation:

Ψs = −(1)(0.5)(0.0831)(295)

Calculate:

Ψs = −12.3 bar

Rubric

Award 1 point for: the new solution has Ψ = Ψs = −(1)(0.5)(0.0831)(295) = −12.3 bar (accept −12.2 to −12.4 bar); water leaves the cell, the push against the cell wall falls to zero (the cell loses turgor), and the contents shrink away from the cell wall (plasmolysis) while the cell wall keeps its shape.
Accept: 'water leaves the cell and the cell plasmolyzes' reasoned from the new solution's water potential alone: −12.3 bar is below −8 bar, the lowest water potential this cell can hold (Ψs = −8 bar with Ψp no lower than 0 bar), so water must leave whatever the cell's state after part (b). The −2 bar step from part (b) is not required.
Accept "the cell goes limp" or "loses turgor" for the effect on the cell. Do not award the point for a positive Ψ or for water entering the cell. No final pressure potential is required: with Ψs fixed at −8 bar the cell cannot reach −12.3 bar, so net water movement out continues and no equilibrium is reached; an answer saying so is correct, and an answer that leaves the end point open is not penalized.
Accept either treatment of the cell's solute potential as water leaves (held at −8 bar as the stem assumes, or becoming more negative as the contents concentrate); the prediction, water leaving until the contents pull away from the cell wall, is the same.

Slip Giving a positive Ψ for the solution, or having water enter the cell. −12.3 bar is lower than anything the cell can reach.

APBIO-U02-L15 Why a cell can't just grow bigger

Topic 2.2 · Cell Size · 79 steps

A mouse and an elephant, with one cell from each drawn to the same scale and the same size

Here are a mouse and an elephant, and one cell from each, drawn to the same scale. The two cells are the same size.

An elephant is more than a hundred thousand times heavier than a mouse, but its cells are no bigger than the mouse’s. Living things get bigger by adding cells, not by growing them.

Why can a cell not simply grow bigger?

Unit 2 · Cell Structure and Function

1Surface area and volume of a cube

Video: Watch first: a mouse cell and an elephant cell

An elephant is a hundred thousand times heavier than a mouse, yet their cells are the same size.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T22-intro.mp4

Video: Watch: Surface area and volume of a cube

Six faces of s by s make 6s²; s by s by s makes s³; the 4 μm cube worked step by step.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L15.mp4

Here is a model cell shaped like a cube, with sides 2 μm long.

A cube-shaped model cell with sides 2 μm long

A micrometer, written μm, is one thousandth of a millimeter. Most cells are a few micrometers to a few tens of micrometers across.

The cell’s surface is its six faces. Each face is a square 2 μm by 2 μm, so each face has an area of 4 μm².

Six faces of 4 μm² make a surface area of 24 μm². All six faces count, not only the one facing you.

Now the space inside. Cut the cube into little cubes 1 μm on a side: two along, two across, two up.

The 2 μm cube cut into little cubes 1 μm on a side: two along, two across, two up, eight in all

Eight little cubes fill the 2 μm cube, so its volume is 8 μm³.

The AP formula sheet writes the same two results for any cube. It calls the length of one side s. Here are its two formulas.

surface area = 6 s^{2}

volume = s^{3}

Surface area and volume of a cube of side s, as the AP formula sheet writes them

Each face is a square s by s, so each face has an area of s². Six faces make 6s². The volume is s by s by s, which is s³.

Here is a cube-shaped cell with sides 4 μm long.

A 2 μm cube beside a 4 μm cube, each with its surface area and volume

Worked example

A cube-shaped cell has sides 4 μm long. What are its surface area and its volume?

Write down the values in the question:

s = 4 μm

Write down the equations:

surface area = 6 s^{2}

volume = s^{3}

Substitute the values into the equations:

surface area = 6 {(4)}^{2}

volume = {(4)}^{3}

Calculate:

surface area = 96 {μm}^{2}

volume = 64 {μm}^{3}

What you are expected to know Calculate the surface area ( $6 s^{2}$ ) and the volume ( $s^{3}$ ) of a cube-shaped cell from its side length, with the right unit on each.

Check q1 numeric entry

A cube-shaped cell has sides 3 μm long.

Calculate its surface area.

Answer: 54 μm² (tolerance ±0.5)

Working

Write down the values in the question:

s = 3 μm

Write down the equation:

surface area = 6 s^{2}

Substitute the values into the equation:

surface area = 6 {(3)}^{2}

Calculate:

surface area = 54 {μm}^{2}

Check q2 numeric entry

A cube-shaped cell has sides 3 μm long.

Calculate its volume.

Answer: 27 μm³ (tolerance ±0.5)

Working

Write down the values in the question:

s = 3 μm

Write down the equation:

volume = s^{3}

Substitute the values into the equation:

volume = {(3)}^{3}

Calculate:

volume = 27 {μm}^{3}

17The surface-area-to-volume ratio

Video: Watch: The surface-area-to-volume ratio

24 μm² over 8 μm³ is 3 per μm: surface area on top, volume underneath, always.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L15b.mp4

Put the two numbers for the 2 μm cube side by side: 24 μm² of surface, for 8 μm³ of inside.

Worked example

The 2 μm cube has a surface area of 24 μm² and a volume of 8 μm³. How much surface does it have for each μm³ inside it?

Write down the values in the question:

surface area = 24 μm²

volume = 8 μm³

Write down the equation:

ratio = \frac{surface area}{volume}

Substitute the values into the equation:

ratio = \frac{24 {μm}^{2}}{8 {μm}^{3}}

Calculate:

ratio = 3 per μm

This cube has 3 μm² of surface for every 1 μm³ inside it.

Surface area divided by volume is called the cell’s , written SA/V for short.

The 2 μm cube's surface area over its volume: 3 μm² of surface for each μm³ inside

The unit comes out of the division too: $\frac{{μm}^{2}}{{μm}^{3}}$ leaves $\frac{1}{μm}$ , read as “per μm”: so many μm² of surface for each μm³ inside.

It is always surface area divided by volume, never volume divided by surface area.

What you are expected to know Calculate a cell’s surface-area-to-volume ratio by dividing its surface area by its volume, and give it in per μm.

Check q3 numeric entry

A cube-shaped cell with sides 4 μm long has a surface area of 96 μm² and a volume of 64 μm³.

Calculate its surface-area-to-volume ratio, to three significant figures.

Answer: 1.5 per μm (tolerance ±0.05)

Working

Write down the values in the question:

surface area = 96 μm²

volume = 64 μm³

Write down the equation:

SA/V = \frac{surface area}{volume}

Substitute the values into the equation:

SA/V = \frac{96 {μm}^{2}}{64 {μm}^{3}}

Calculate:

SA/V = 1.50 per μm

Check q4 numeric entry

A cube-shaped cell has sides 3 μm long.

Now calculate its surface-area-to-volume ratio, to three significant figures.

Part 1. First, calculate its surface area.

Answer: 54 μm² (tolerance ±0.5)

Working

Substitute the side length into the surface-area equation:

surface area = 6 {(3)}^{2}

Calculate:

surface area = 54 {μm}^{2}

Part 2. Next, calculate its volume.

Answer: 27 μm³ (tolerance ±0.5)

Working

Substitute the side length into the volume equation:

volume = {(3)}^{3}

Calculate:

volume = 27 {μm}^{3}

Answer: 2 per μm (tolerance ±0.05)

Working

Write down the values in the question:

s = 3 μm

Write down the equations:

surface area = 6 s^{2}

volume = s^{3}

SA/V = \frac{surface area}{volume}

Substitute the values into the first two equations:

surface area = 6 {(3)}^{2}

volume = {(3)}^{3}

Calculate:

surface area = 54 {μm}^{2}

volume = 27 {μm}^{3}

Substitute the results into the ratio equation:

SA/V = \frac{54}{27}

Calculate:

SA/V = 2.00 per μm

28Fluency quiz: which cell has the greater ratio? mixed practice

Check q5

The table shows the surface area and the volume of cell A and cell B.

A table with two rows, cell A and cell B, and two columns, surface area in μm² and volume in μm³; each row carries that cell's two values

Which cell has the greater surface-area-to-volume ratio?

A. ✓ Cell A
B. Cell B
Cell B has more surface in total, but its ratio, 2.0 per μm, is smaller than cell A’s 3.0 per μm.
C. The same
The two ratios differ: cell A’s is 3.0 per μm and cell B’s is 2.0 per μm.

Why: Cell A’s ratio, 3.0 per μm, is greater than cell B’s, 2.0 per μm.
So cell A has the greater ratio, though cell B has more surface in total.

Check q6

The table shows the surface area and the volume of cell C and cell D.

A table with two rows, cell C and cell D, and two columns, surface area in μm² and volume in μm³; each row carries that cell's two values

Which cell has the greater surface-area-to-volume ratio?

A. Cell C
Cell C has 100 μm² of surface and cell D 150 μm², over the same 50 μm³ of volume, so cell D has the greater ratio.
B. ✓ Cell D
C. The same
The two cells have the same volume but not the same surface.

Why: The two cells have the same volume, so the one with more surface has the greater ratio.
Cell D’s ratio, 3.0 per μm, is greater than cell C’s, 2.0 per μm.
So cell D has the greater ratio.

Check q7

The table shows the surface area and the volume of cell E and cell F.

A table with two rows, cell E and cell F, and two columns, surface area in μm² and volume in μm³; each row carries that cell's two values

Which cell has the greater surface-area-to-volume ratio?

A. Cell E
Cell E’s ratio is 1.5 per μm, and cell F’s is 1.5 per μm too.
B. Cell F
Size alone does not set the ratio: cell F’s ratio is 1.5 per μm, the same as cell E’s.
C. ✓ The same

Why: Cell F is smaller, but its surface and its volume are both half of cell E’s.
Both ratios come to 1.5 per μm.
So the two cells have the same ratio.

Check q8

The table shows the surface area and the volume of cell G and cell H.

A table with two rows, cell G and cell H, and two columns, surface area in μm² and volume in μm³; each row carries that cell's two values

Which cell has the greater surface-area-to-volume ratio?

A. ✓ Cell G
B. Cell H
Cell H has more surface than cell G, and far more volume too.
C. The same
The two ratios differ: cell G’s is 1.2 per μm and cell H’s is 0.8 per μm.

Why: Cell G’s ratio, 1.2 per μm, is greater than cell H’s, 0.8 per μm.
So cell G has the greater ratio, though cell H has more surface in total.

Check q9

The table shows the surface area and the volume of cell J and cell K.

A table with two rows, cell J and cell K, and two columns, surface area in μm² and volume in μm³; each row carries that cell's two values

Which cell has the greater surface-area-to-volume ratio?

A. Cell J
Cell J spreads the same 72 μm² of surface over more volume, 48 μm³ against cell K’s 36 μm³, so cell J has the smaller ratio.
B. ✓ Cell K
C. The same
The two cells have the same surface but not the same volume.

Why: The two cells have the same surface area, so the one with less volume has the greater ratio.
Cell K’s ratio, 2.0 per μm, is greater than cell J’s, 1.5 per μm.
So cell K has the greater ratio.

Check q10

Cell L has a surface area of 300 μm² and a volume of 500 μm³. Cell M has a surface area of 140 μm² and a volume of 70 μm³.

Which cell has the greater surface-area-to-volume ratio?

A. Cell L
Cell M’s ratio is 2.0 per μm, against cell L’s 0.6 per μm, so cell M has the greater ratio.
B. ✓ Cell M
C. The same
The two ratios differ: cell L’s is 0.6 per μm and cell M’s is 2.0 per μm.

Why: Cell M’s ratio, 2.0 per μm, is greater than cell L’s, 0.6 per μm.
So cell M has the greater ratio, though cell L has more surface in total.

35Grow the cell and the ratio falls

Here are three cubes with sides of 2, 4 and 8 μm, each twice the one before. The first two ratios are 3 per μm and 1.5 per μm.

Cubes of side 2, 4 and 8 μm: 3 per μm, 1.5 per μm, and the third ratio still to find

Check q11 numeric entry

A cube-shaped cell has sides 8 μm long.

Now calculate its surface-area-to-volume ratio, to three significant figures.

Part 1. First, calculate its surface area.

Answer: 384 μm² (tolerance ±0.5)

Working

Substitute the side length into the surface-area equation:

surface area = 6 {(8)}^{2}

Calculate:

surface area = 384 {μm}^{2}

Part 2. Next, calculate its volume.

Answer: 512 μm³ (tolerance ±0.5)

Working

Substitute the side length into the volume equation:

volume = {(8)}^{3}

Calculate:

volume = 512 {μm}^{3}

Answer: 0.75 per μm (tolerance ±0.005)

Working

Write down the values in the question:

s = 8 μm

Write down the equations:

surface area = 6 s^{2}

volume = s^{3}

SA/V = \frac{surface area}{volume}

Substitute the values into the first two equations:

surface area = 6 {(8)}^{2}

volume = {(8)}^{3}

Calculate:

surface area = 384 {μm}^{2}

volume = 512 {μm}^{3}

Substitute the results into the ratio equation:

SA/V = \frac{384}{512}

Calculate:

SA/V = 0.750 per μm

Video: Watch: Grow the cell and the ratio falls

3, 1.5, 0.75 per μm: each doubling multiplies surface by four and volume by eight, so the ratio halves.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L15c.mp4

Side 2 μm: 3 per μm. Side 4 μm: 1.5 per μm. Side 8 μm: 0.75 per μm. Each doubling halves the ratio.

Cubes of side 2, 4 and 8 μm: the ratio halves each time the side doubles

Doubling every side multiplied the surface area by four (24 to 96 μm²) but the volume by eight (8 to 64 μm³).

The bigger cube has more surface. The bigger cube also has far more inside. So the bigger cube has less surface for each μm³ inside, and its ratio falls.

As a cell grows, its volume increases faster than its surface area, so its surface-area-to-volume ratio falls.

Any shape does the same when it is scaled up in every direction: a ball twice as wide has four times the surface and eight times the volume.

What you are expected to know Explain that as a cell grows its volume increases faster than its surface area, so its surface-area-to-volume ratio falls: doubling every dimension multiplies the surface area by four but the volume by eight.

Check q12

A yeast cell grows without changing shape. Its diameter goes from 4.0 μm to 8.0 μm.

What happens to its surface-area-to-volume ratio?

A. ✓ The ratio halves
B. The ratio doubles
Doubling every dimension multiplies the surface area by four and the volume by eight, so surface area divided by volume falls.
C. The ratio stays the same
Size changes the ratio even when the shape does not.

Why: Doubling the diameter doubles every dimension.
An area has two dimensions, so the surface area is multiplied by four.
A volume has three dimensions, so the volume is multiplied by eight.
The ratio is surface area divided by volume.
So the ratio halves.

Practice writing an answer

A yeast cell grows without changing shape. Its diameter doubles, from 4.0 μm to 8.0 μm, and its surface-area-to-volume ratio halves.

(a) Explain why the ratio halves. (1 pt)

Frame Doubling the diameter multiplies the surface area by … and the volume by …, so the ratio …

Model answer Doubling the diameter multiplies the surface area by four and the volume by eight, so the ratio halves.
An area has two dimensions.
Doubling both multiplies the area by four.
A volume has three dimensions.
Doubling all three multiplies the volume by eight.
The ratio is surface area divided by volume.
So four times the surface over eight times the volume is half the ratio.

Rubric

Award 1 point for: the surface area is multiplied by four and the volume by eight (two dimensions against three), so surface area divided by volume is halved.
Accept: the volume grows faster than the surface area, with both factors named.

Slip Giving the surface area and the volume the same factor. An area has two dimensions and a volume has three, so doubling a length does not multiply the two alike.

Check q13

A student says: “A bigger cell has a bigger surface-area-to-volume ratio, because it has more surface.”

Is the student correct?

A. Yes
A bigger cell has more surface but far more volume, so surface area divided by volume is smaller.
B. ✓ No

Why: A bigger cell has a smaller surface-area-to-volume ratio.
A bigger cell does have more surface.
A bigger cell also has far more volume.
The ratio is surface area divided by volume.
So the bigger cell’s ratio is smaller.

Check q14

A bigger cell has a smaller surface-area-to-volume ratio than a smaller cell of the same shape, even though it has more surface.

Why does a bigger cell have a smaller ratio?

A. A bigger cell has less surface than a smaller cell
A bigger cell does have more surface than a smaller cell.
B. A bigger cell changes shape as it grows
The shape need not change at all.
The three cubes kept their shape, and their ratios still fell as they grew.
C. ✓ A bigger cell’s volume grows faster than its surface area

Why: A bigger cell has more surface.
A bigger cell has far more volume.
When every side doubles, the surface area is multiplied by four and the volume by eight.
So the volume grows faster than the surface area.
The ratio is surface area divided by volume.
So the ratio falls.

49Why the falling ratio keeps cells small

Video: Watch: Why the falling ratio keeps cells small

Exchange happens only across the surface while need grows with volume; the 20 μm cube’s membrane supplies only 60%.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L15d.mp4

Everything a cell takes in or gets rid of crosses its plasma membrane, mostly by diffusion. That membrane is the cell’s surface.

Here is a cube-shaped cell 10 μm across. It needs 100 million oxygen molecules a minute. Its membrane can bring in 120 million a minute, so the membrane keeps up.

A 10 μm cube takes in 120 million oxygen molecules a minute and needs 100 million; a 20 μm cube takes in 480 million and needs 800 million

Double every side to 20 μm. Its volume is eight times bigger, so it needs eight times as many oxygen molecules: 800 million a minute.

Its surface is four times bigger, so its membrane can bring in four times as many: 480 million a minute. Now the membrane cannot keep up.

Two facts explain those numbers. If a cell’s volume doubles, the amount it needs doubles. If its surface area doubles, the amount its membrane can bring in doubles.

The rules hold for any factor. Triple a cell’s volume and it needs three times as much. Halve its surface area and its membrane brings in half as much.

A cell takes in what it needs, and gets rid of waste, only across its surface. The amount it needs grows with its volume. So as a cell grows, its surface cannot keep up with its interior.

Worked example

The 10 μm cube’s membrane brings in 120 million oxygen molecules a minute and the cell needs 100 million. Double it to a 20 μm cube. What share of its need can its membrane now supply?

Write down the values in the question:

supply of the 10 μm cell = 120 million molecules a minute

need of the 10 μm cell = 100 million molecules a minute

every side × 2

Write down the equations:

new supply = {(side factor)}^{2} \times old supply

new need = {(side factor)}^{3} \times old need

share supplied = \frac{new supply}{new need} \times 100

Substitute the values into the first two equations:

new supply = {(2)}^{2} (120)

new need = {(2)}^{3} (100)

Calculate:

new supply = 480 million a minute

new need = 800 million a minute

Substitute the results into the share equation:

share supplied = \frac{480}{800} \times 100

Calculate:

share supplied = 60 %

The 20 μm cube’s membrane can supply only 60% of what the cell needs, where the 10 μm cube’s membrane supplied all of it with some to spare. The smaller cell, with the greater surface-area-to-volume ratio, exchanges materials with its surroundings faster for each μm³ of its volume.

There is a second reason too: in a bigger cell, whatever comes in has farther to travel before it reaches the middle.

So cells do not grow without limit. Past a certain size a cell stays small, or divides into two smaller cells, each with a greater ratio.

What you are expected to know Explain why a cell cannot keep growing: it exchanges materials only across its surface, its needs grow with its volume, so the surface cannot keep up and cells stay small or divide.

Check q15 numeric entry

A 6 μm cube-shaped cell can bring in 60 million oxygen molecules a minute across its membrane and needs 50 million. An 18 μm cube of the same kind, every side three times as long, needs oxygen at the same rate per μm³ and brings it in at the same rate per μm² of membrane.

Now calculate the share of the 18 μm cell’s need that its membrane can supply.

Part 1. First, calculate the 18 μm cell’s supply.

Answer: 540 million molecules a minute (tolerance ±0.5)

Working

Multiply the old supply by the surface-area factor, (3)²:

new supply = {(3)}^{2} (60)

Calculate:

new supply = 540 million a minute

Part 2. Next, calculate its need.

Answer: 1350 million molecules a minute (tolerance ±0.5)

Working

Multiply the old need by the volume factor, (3)³:

new need = {(3)}^{3} (50)

Calculate:

new need = 1350 million a minute

Answer: 40 % (tolerance ±0.5)

Working

Write down the values in the question:

supply of the 6 μm cell = 60 million molecules a minute

need of the 6 μm cell = 50 million molecules a minute

every side × 3

Write down the equations:

new supply = {(side factor)}^{2} \times old supply

new need = {(side factor)}^{3} \times old need

share supplied = \frac{new supply}{new need} \times 100

Substitute the values into the first two equations:

new supply = {(3)}^{2} (60)

new need = {(3)}^{3} (50)

Calculate:

new supply = 540 million a minute

new need = 1350 million a minute

Substitute the results into the share equation:

share supplied = \frac{540}{1350} \times 100

Calculate:

share supplied = 40 %

Check q16

Two cells have the same shape. Cell P is 4 μm across. Cell Q is 16 μm across.

Which cell exchanges materials with its surroundings more efficiently, for each μm³ of its volume?

A. ✓ Cell P
B. Cell Q
Cell Q’s larger surface serves a far larger volume.
C. Both the same
The same shape at four times the size does not have the same ratio.

Why: Everything a cell takes in crosses its surface.
The amount a cell needs grows with its volume.
Cell P is the smaller cell, so cell P has the greater surface-area-to-volume ratio.
So cell P exchanges materials faster for each μm³ of its volume.

Check q17

Cell P exchanges materials more efficiently for each μm³ of its volume than cell Q does.

Why does cell P exchange more efficiently for each μm³?

A. Cell P has more surface in total than cell Q
Cell Q, the larger cell, has more surface in total.
B. Cell P has less volume, so it needs less from its surroundings
Needing less in total is not the same as being supplied better for each μm³.
C. ✓ Cell P has more surface for each μm³ inside it

Why: Everything a cell takes in crosses its surface, and the amount it needs grows with its volume.
Cell P has more surface for each μm³ of volume than cell Q.
So each μm³ of cell P is supplied across more surface.
So cell P exchanges more efficiently for each μm³.

Here are the mouse and the elephant again.

A house mouse, close up, on bare soil — Two photographs side by side: a house mouse on soil, seen close up, and a group of African elephants wading across a river

Two photographs side by side: a house mouse on soil, seen close up, and a group of African elephants wading across a river

The elephant’s cells are no bigger than the mouse’s because a cell twice as wide has eight times the volume to feed through only four times the surface.

68Mixed practice mixed practice

Check q18 numeric entry

A cube-shaped cell has sides 5 μm long.

Calculate its surface area.

Answer: 150 μm² (tolerance ±0.5)

Working

Write down the values in the question:

s = 5 μm

Write down the equation:

surface area = 6 s^{2}

Substitute the values into the equation:

surface area = 6 {(5)}^{2}

Calculate:

surface area = 150 {μm}^{2}

Check q19 numeric entry

A cube-shaped cell has sides 7 μm long.

Calculate its volume.

Answer: 343 μm³ (tolerance ±0.5)

Working

Write down the values in the question:

s = 7 μm

Write down the equation:

volume = s^{3}

Substitute the values into the equation:

volume = {(7)}^{3}

Calculate:

volume = 343 {μm}^{3}

Check q20 numeric entry

A cell has a surface area of 150 μm² and a volume of 125 μm³.

Calculate its surface-area-to-volume ratio, to three significant figures.

Answer: 1.2 per μm (tolerance ±0.05)

Working

Write down the values in the question:

surface area = 150 μm²

volume = 125 μm³

Write down the equation:

SA/V = \frac{surface area}{volume}

Substitute the values into the equation:

SA/V = \frac{150 {μm}^{2}}{125 {μm}^{3}}

Calculate:

SA/V = 1.20 per μm

Check q21

A cell’s surface-area-to-volume ratio is 2 per μm.

Which statement describes what this ratio means?

A. The cell’s surface area is 2 μm²
The ratio compares surface with volume; it is not a surface area on its own.
B. ✓ The cell has 2 μm² of surface for each μm³ inside it
C. The cell has 2 μm³ of volume for each μm² of surface
The ratio is surface area divided by volume, not volume divided by surface area.
D. The cell is 2 μm across
Per μm is the unit left when μm² is divided by μm³; it is not a distance across the cell.

Why: Surface area divided by volume gives μm² of surface for each μm³ of inside.
A ratio of 2 per μm means 2 μm² of surface for every 1 μm³ of interior.

Check q22 numeric entry

A cube-shaped cell has sides 9 μm long.

Calculate its surface-area-to-volume ratio, to three significant figures.

Answer: 0.667 per μm (tolerance ±0.005)

Working

Write down the values in the question:

s = 9 μm

Write down the equation:

SA/V = \frac{6 s^{2}}{s^{3}}

Substitute the values into the equation:

SA/V = \frac{6 {(9)}^{2}}{{(9)}^{3}}

Calculate:

SA/V = 0.667 per μm

Check q23

A cell grows and keeps its shape. Every side ends up three times as long.

How do its surface area, volume and surface-area-to-volume ratio change?

A. Surface area × 3, volume × 3, ratio unchanged
Only a length triples when the side triples; an area grows with the square of the size and a volume with the cube.
B. Surface area × 27, volume × 9, ratio three times
Tripling the sides multiplies the surface area (two dimensions) by nine and the volume (three dimensions) by twenty-seven, so the ratio falls.
C. ✓ Surface area × 9, volume × 27, ratio 33% of what it was
D. Surface area × 9, volume × 9, ratio unchanged
Volume has a third dimension, so it grows twenty-sevenfold, not ninefold, and the ratio falls.

Why: An area has two dimensions.
Tripling both multiplies the surface area by nine.
A volume has three dimensions.
Tripling all three multiplies the volume by twenty-seven.
The ratio is surface area divided by volume.
So the ratio falls to a third.

Check q24 numeric entry

A 5 μm cube-shaped cell can bring in 30 million oxygen molecules a minute across its membrane and needs 20 million. A 10 μm cube of the same kind, every side twice as long, needs oxygen at the same rate per μm³ and brings it in at the same rate per μm² of membrane.

Calculate the share of the 10 μm cell’s need that its membrane can supply.

Answer: 75 % (tolerance ±0.5)

Working

Write down the values in the question:

supply of the 5 μm cell = 30 million molecules a minute

need of the 5 μm cell = 20 million molecules a minute

every side × 2

Write down the equations:

new supply = {(side factor)}^{2} \times old supply

new need = {(side factor)}^{3} \times old need

share supplied = \frac{new supply}{new need} \times 100

Substitute the values into the first two equations:

new supply = {(2)}^{2} (30)

new need = {(2)}^{3} (20)

Calculate:

new supply = 120 million a minute

new need = 160 million a minute

Substitute the results into the share equation:

share supplied = \frac{120}{160} \times 100

Calculate:

share supplied = 75 %

Check q25

A cell takes in everything it needs across its plasma membrane. The cell grows until its membrane can only just supply what its interior needs.

What happens to the cell next?

A. The cell keeps growing without limit
Growing every side adds surface by the square but volume by the cube, so the membrane falls further behind rather than keeping up.
B. The cell’s membrane grows thicker
A thicker membrane adds no surface area.
The cell exchanges materials across the area of its membrane, and that area is what the interior has outgrown.
C. The cell stops taking in materials
A living cell keeps taking in what it needs; the limit is on how fast its surface can pass it in.
D. ✓ The cell stays this size, or divides in two

Why: Everything the cell needs crosses its surface, and the amount it needs grows with its volume.
Past this size the surface cannot keep up with the volume.
So the cell stays this size, or divides into two smaller cells.
Each smaller cell has a greater ratio, so its surface copes.

Check q26 numeric entry

Two cube-shaped cells have sides of 2 μm and 6 μm.

Calculate how many times the 6 μm cell’s surface-area-to-volume ratio the 2 μm cell’s ratio is.

Answer: 3 (tolerance ±0)

Working

Write down the values in the question:

2 μm cell: s = 2 μm

6 μm cell: s = 6 μm

Write down the equations:

SA/V = \frac{6 s^{2}}{s^{3}}

factor = \frac{SA/V of the 2 μm cell}{SA/V of the 6 μm cell}

Substitute the values into the ratio equation:

2 μm cell: SA/V = \frac{6 {(2)}^{2}}{{(2)}^{3}}

6 μm cell: SA/V = \frac{6 {(6)}^{2}}{{(6)}^{3}}

Calculate:

2 μm cell: SA/V = 3 per μm

6 μm cell: SA/V = 1 per μm

Substitute the results into the factor equation:

factor = \frac{3}{1}

Calculate:

factor = 3

Practice writing an answer

Two cube-shaped cells of the same kind take in oxygen across their plasma membranes. Cell A has sides 3 μm long; cell B has sides 6 μm long. Use the formulas for a cube from the AP formula sheet.

(a) Calculate the surface-area-to-volume ratio of each cell, with units. (1 pt)

Frame Cell A: … per μm. Cell B: … per μm.

Model answer Cell A: 2.0 per μm.
Cell B: 1.0 per μm.

Working

Write down the values in the question:

cell A: s = 3 μm

cell B: s = 6 μm

Write down the equation:

SA/V = \frac{6 s^{2}}{s^{3}}

Substitute the values into the equation:

cell A: SA/V = \frac{6 {(3)}^{2}}{{(3)}^{3}}

cell B: SA/V = \frac{6 {(6)}^{2}}{{(6)}^{3}}

Calculate:

cell A: SA/V = 2.0 per μm

cell B: SA/V = 1.0 per μm

Rubric

Award 1 point for: both ratios as surface area divided by volume, with units: cell A 2.0 per μm and cell B 1.0 per μm.
Accept: 2 per μm and 1 per μm; the surface areas and volumes need not be shown.

Slip Dividing volume by surface area gives 0.5 per μm and 1.0 per μm and turns the comparison upside down.

(b) Determine which cell exchanges oxygen with its surroundings more efficiently for its size, using the two ratios. (1 pt)

Model answer Cell A exchanges oxygen more efficiently for its size.
Oxygen enters a cell only across its surface.
The amount of oxygen a cell needs grows with its volume.
Cell A has 2.0 μm² of membrane for each μm³ of interior.
Cell B has 1.0 μm² of membrane for each μm³ of interior.
So each μm³ of cell A is supplied across twice as much surface as each μm³ of cell B.

Rubric

Award 1 point for: the decision AND the reasoning it rests on: cell A, because it has more surface for each unit of volume (2.0 against 1.0 per μm), and exchange happens across the surface while need grows with volume.
Accept: a higher surface-area-to-volume ratio means more membrane for each unit of interior to exchange across.

Slip Picking cell B because its total surface, 216 μm², is larger. Cell B also has far more volume, 216 μm³. What matters is surface for each μm³ of volume, and cell A has more.

(c) Cell B goes on growing as a single cell. Predict what happens to the share of its oxygen need that its membrane can supply, and explain why. (1 pt)

Model answer The share falls.
As cell B grows, its volume increases faster than its surface area.
Its need for oxygen grows with its volume.
Its supply of oxygen grows with its surface area.
So the need grows faster than the supply.
So the membrane supplies a smaller and smaller share of what the interior needs, until it can no longer keep up.

Rubric

Award 1 point for: the share falls, because volume (and so need) grows faster than surface area as the cell grows, so the surface supplies a smaller share of the need.
Accept: the ratio keeps falling, so less surface serves each unit of volume.

Slip Saying the share rises because the bigger cell has more membrane in total. The bigger cell does have more membrane. It has far more volume behind that membrane, so the share falls.

Glossary

surface-area-to-volume ratio (SA/V): A cell’s surface area divided by its volume, given in per μm. It says how much surface the cell has for each unit of its inside, and it falls as the cell grows.

APBIO-U02-L16 Folded surfaces

Topic 2.2 · Cell Size · 44 steps

The folded lining of the small intestine, one of its cells with a surface folded into tiny fingers, and a root hair

Here is the lining of your small intestine, and one cell from it, drawn large. The lining is folded, and the cell’s surface is folded again into thousands of tiny fingers.

Flattened out, that lining would carry about thirty square meters of membrane. Beside it is a root cell from a plant, with one fine outgrowth reaching into the soil. Why would a cell fold its surface, or grow an outgrowth like that?

Unit 2 · Cell Structure and Function

1Folds and shapes that add surface

Video: Watch: Folds and shapes that add surface

Microvilli, a root hair, a flattened shape and cilia: more membrane for the same volume.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L16a.mp4

As a cell grows, its surface cannot keep up with its interior. Surface per unit of volume is what limits exchange.

There is a way around the limit: fold the surface, stretch it out or flatten it. Folding, stretching and flattening each add surface without adding much volume.

Here is a cell from the lining of your gut. On the side facing the food, its plasma membrane is folded into thousands of thin fingers.

A gut lining cell whose top membrane is folded into many thin fingers, beside a cell of the same volume with a flat top

Such fingers of membrane on a cell’s surface are called .

Microvilli are folds of the plasma membrane. Microvilli do not move.

Gut cells with microvilli have about three times the membrane area of gut cells without microvilli, and the same volume. Gut cells with microvilli take up nutrients about 2.7 times as fast.

Here is a young root in soil. Each cell on its surface has grown one long thin outgrowth that reaches out between the soil particles, into the water held there.

A young root in soil, covered along its length in a fuzz of fine root hairs — Photograph of a young white root lying on soil, its surface covered in a dense fuzz of fine root hairs, with soil grains around it

An outgrowth like this is called a .

A root hair adds membrane in contact with the soil water. A root hair adds very little volume. So the root cell takes in water and minerals faster.

A root cell with one long thin outgrowth reaching between soil particles into the water there

Shape does the same job. A flattened or long thin cell has more surface than a rounded cell of the same volume.

A flattened cell and a rounded cell of the same volume: the flattened one has more surface

Short hair-like projections on the surface of some cells are called . Each cilium is more membrane. So a cell with cilia has more surface than the same cell without cilia.

A cell whose upper surface carries many short hair-like cilia

What you are expected to know Microvilli and root hairs add membrane without adding much volume; flattened and long thin shapes and cilia add surface.

15Fluency quiz: does it add surface? mixed practice

Check q1

A gut cell folds its top membrane into microvilli.

Does this give the cell more surface for each μm³ of its volume?

A. ✓ Yes
B. No
Each microvillus is a fold of membrane, and a fold of membrane is more membrane with almost no added volume.

Why: Each microvillus is a fold of the plasma membrane.
A fold of membrane is more membrane.
The folds add almost no volume.
So the cell has more surface for each μm³ of its volume.

Check q2

A round cell grows bigger and keeps its shape.

Does this give the cell more surface for each μm³ of its volume?

A. Yes
The bigger cell has more surface, but its volume grows faster than its surface.
B. ✓ No

Why: The bigger cell has more surface.
The bigger cell has far more volume.
Its volume grows faster than its surface.
So the cell has less surface for each μm³ of its volume.

Check q3

A root cell grows a root hair.

Does this give the cell more surface for each μm³ of its volume?

A. ✓ Yes
B. No
A root hair adds membrane along its whole length and very little volume.

Why: A root hair is a long thin outgrowth.
A root hair adds membrane along its whole length.
A root hair adds very little volume.
So the cell has more surface for each μm³ of its volume.

Check q4

A cell’s plasma membrane becomes thicker.

Does this give the cell more surface for each μm³ of its volume?

A. Yes
A thicker membrane covers the same surface: neither the surface area nor the volume changes.
B. ✓ No

Why: A thicker membrane covers the same surface.
The surface area does not change.
The volume does not change.
So the cell has the same surface for each μm³ of its volume.

Check q5

A flattened cell rounds up into a ball, keeping its volume.

Does this give the cell more surface for each μm³ of its volume?

A. Yes
A rounded shape has the least surface for its volume, so rounding up loses surface.
B. ✓ No

Why: A rounded shape has the least surface for its volume.
The volume stays the same.
The rounded cell has less surface than the flattened cell had.
So the cell has less surface for each μm³ of its volume.

Check q6

A round cell flattens into a disc, keeping its volume.

Does this give the cell more surface for each μm³ of its volume?

A. ✓ Yes
B. No
A rounded shape has the least surface for its volume, so flattening the same volume adds surface.

Why: A rounded shape has the least surface for its volume.
The volume stays the same.
Flattening the same volume adds surface.
So the cell has more surface for each μm³ of its volume.

22Why the folds and hairs speed exchange

Video: Watch: Why the folds and hairs speed exchange

Nutrients and water cross only the membrane; folds and hairs add membrane without adding volume.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L16b.mp4

Now the reason, for two of the cases: microvilli on a gut cell, and a root hair on a root cell.

What you are expected to know Explain why a cell with microvilli, or a root hair, exchanges materials with its surroundings faster than a cell of the same volume without them.

Check q7

Two gut-lining cells have the same volume. One has microvilli on its top face. The other has a flat top face. The cell with microvilli takes up nutrients faster.

Why does the cell with microvilli take up nutrients faster?

A. The microvilli beat back and forth and sweep nutrients into the cell
Microvilli are folds of membrane, and folds do not beat.
B. The microvilli make the membrane thinner, so nutrients cross it faster
The membrane of a microvillus is the same membrane, folded outward.
C. The microvilli shorten the distance nutrients travel into the cell
The membrane is folded outward, so the distance across it is unchanged.
D. ✓ The microvilli add membrane for uptake without adding volume

Why: The two cells have the same volume, so the folds add no volume.
Each fold is more membrane.
Nutrients cross into a cell only across its membrane.
So the cell with microvilli has more membrane for nutrients to cross.
This is why it takes up nutrients faster.

Check q8

Here are two roots in the same moist soil. The cells of root A are smooth. The cells of root B have root hairs.

Which root takes in water faster?

A. Root A
Root A’s cells are smooth, so they have less membrane in contact with the soil water than root B’s cells with their root hairs.
B. ✓ Root B
C. Both at the same rate
The soil is the same but the membrane in the soil water is not.

Why: A root hair adds membrane in contact with the soil water.
Water enters a root cell only across its membrane.
Root B’s cells have root hairs, so root B’s cells have more membrane in the soil water.
So root B takes in water faster.

Practice writing an answer

The cells of root A are smooth. The cells of root B have root hairs. Both roots sit in the same moist soil.

(a) Explain why the root you chose takes in water faster. (1 pt)

Frame A root hair adds … in contact with the soil water, while adding very little …, so …

Model answer A root hair adds membrane in contact with the soil water, while adding very little volume, so root B takes in water faster.
Water enters a root cell only across its membrane.
Root B’s cells have root hairs.
So root B’s cells have far more membrane in the soil water than root A’s cells have.
So more water crosses into root B’s cells each minute.

Rubric

Award 1 point for: root hairs add membrane in contact with the soil water (and very little volume), water enters only across the membrane, so root B takes in water faster.
Accept: more surface in the soil water for the same volume of cell, so faster uptake.

Slip Saying the root hair stores water. A root hair is long and thin and holds very little. A root hair is a surface for water to cross, not a store.

29A leaf’s inner surface, and the pores that open it

Check q9

A root cell sits in rain water, and its contents swell.

Where does the cell’s outward push come from?

A. ✓ The swollen contents pressing on the cell wall
B. The cell wall pressing inward on the contents
The push starts from the swelling contents; the cell wall presses back only because the contents press on it.
C. The rain water pressing on the cell
The rain water is entering the cell, so the push comes from inside the cell, not from the water outside.

Why: Water enters by osmosis and the swollen contents press outward on the cell wall, and that outward push is turgor pressure.

Video: Watch: A leaf’s inner surface, and the pores that open it

Stomata open the leaf’s inner surface to the air; swollen guard cells bow apart, shrunken ones straighten and press together.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L16c.mp4

A leaf is a different case. Most of a leaf’s surface is inside the leaf, around the air spaces between its cells.

Pores in the leaf’s skin, called , open that large inner surface to the outside air.

Leaf skin under the microscope: two pores, each a slit between a pair of green guard cells, among larger pale cells — Photograph through a microscope of the skin of a leaf: large pale cells, and among them two pores, each a slit between a pair of green sausage-shaped guard cells

Each pore sits between a pair of cells that open and close it. The two cells are called , because they guard the pore.

Turgor pressure is the outward push of a walled cell’s swollen contents against its wall.

When the two guard cells swell with water, they bow apart. So the pore opens.

A pore in a leaf's skin between two guard cells: swollen guard cells bow apart and the pore is open; shrunken ones press together and it is closed

When the two guard cells lose water and shrink, they straighten and press together. So the pore closes.

What you are expected to know Stomata open a leaf’s large inner surface to the outside air; the pair of guard cells around each pore opens it when they swell and closes it when they shrink.

Check q10

On a hot, dry afternoon, a leaf’s guard cells lose water and shrink.

What happens to the pores?

A. The pores open wider
Swollen guard cells bow apart.
B. The pores stay as they were
The pore is the gap between the two guard cells, so their shape sets whether it is open.
C. ✓ The pores close

Why: The guard cells lose water and shrink.
Shrunken guard cells straighten and press together.
So the pore between them closes.

Check q11

The pores close.

Why do the pores close when the guard cells shrink?

A. The shrunken guard cells bow further apart
Bowing apart is what swollen guard cells do.
B. The shrunken guard cells collapse inward and fill the gap
A guard cell does not fall into the pore.
C. ✓ The shrunken guard cells straighten and press together

Why: The pore is the gap between the two guard cells.
When the guard cells lose water, they shrink.
Shrunken guard cells straighten and press together.
So the gap between them closes.
This cuts the air spaces inside the leaf off from the outside air.

The gut lining is folded, and each cell’s surface is folded again into microvilli, because every fold adds membrane for nutrients to cross without adding much volume behind it.

A root hair, a root cell’s fine outgrowth, adds membrane in contact with the soil water in the same way.

Surface per unit of volume sets how fast anything can be exchanged, and folding, stretching and flattening are how a cell gets more of it.

Glossary

microvilli: Thousands of thin fingers of plasma membrane on the surface of a cell, such as a gut lining cell, that add membrane area without adding much volume. Microvilli are folds and do not move.
root hair: A long thin outgrowth of a root cell that reaches between soil particles, adding membrane in contact with the soil water and very little volume.
cilia: Short hair-like projections on the surface of some cells; each is more membrane, so a cell with cilia has more surface than the same cell without them.
stomata: Pores in the skin of a leaf that open the leaf’s large inner surface to the outside air.
guard cells: The pair of cells on either side of a stoma. When the guard cells swell with water, they bow apart and the pore opens. When the guard cells shrink, the pore closes.

APBIO-U02-L16B Mouse and elephant: heat and fuel per gram

Topic 2.2 · Cell Size · 37 steps

A mouse and an elephant standing on the same floor, each with the food it eats in a day heaped beside it: the mouse's heap is about a fifth of its own size, the elephant's a small heap beside a huge body

Here are a mouse and an elephant in the same cool room, each beside the food it eats in a day.

The mouse’s heap is about 20% of its own body mass. The elephant’s is a few percent of its. Gram for gram, the mouse eats several times as much. Why does the small animal need so much more food for each gram of its body?

Unit 2 · Cell Structure and Function

1Big bodies lose heat slowly, per gram

Video: Watch: Big bodies lose heat slowly, per gram

Heat leaves across the skin and is held in every gram: the mouse has far more skin per gram.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L16Ba.mp4

Stand a mouse and an elephant in the same cool room. Each hour, the mouse loses far more heat per gram of body than the elephant does.

Here is why. A warm body loses heat to cooler air across its surface, its skin. The heat it holds is in its whole volume, in every gram of it.

A small body and a large body, each losing heat across its whole surface: the small one has much more surface for each gram inside

A mouse has a lot of skin for each gram of body. An elephant has very little skin for each gram. So the mouse’s heat has more skin to leave through, gram for gram.

Scaling a cell up in every direction makes its surface-area-to-volume ratio fall. The same is true of a whole animal: scale a body up and there is less skin for each gram inside.

The elephant still loses more heat in total, because the elephant has far more skin in total. Per gram, the mouse loses heat faster.

As a body gets bigger, the rate of heat exchange per unit of mass falls, even though the total rises.

What you are expected to know Explain that a smaller animal exchanges heat with its surroundings faster per unit of body mass, because its surface-area-to-volume ratio is greater, while a larger animal exchanges more in total.

10Fluency quiz: which loses heat faster per gram? mixed practice

Check q1

A vole has a mass of 25 g. A badger has a mass of 10 kg. The two animals sit out on a cold night.

Which loses heat faster per gram of body?

A. ✓ The vole
B. The badger
The vole is the smaller animal, so each gram of the vole has more skin to lose heat through.
C. Both at the same rate
The two animals do not have the same skin for each gram.

Why: Heat leaves a body across its skin.
The vole is the smaller animal.
So each gram of the vole has more skin to lose heat through.
So each gram of the vole loses heat faster than each gram of the badger.

Check q2

A turkey has a mass of 8 kg. A wren has a mass of 10 g. The two birds stand in the same cold air.

Which loses heat faster per gram of body?

A. The turkey
The wren is the smaller animal, so each gram of the wren has more skin to lose heat through.
B. ✓ The wren
C. Both at the same rate
The two birds do not have the same skin for each gram.

Why: Heat leaves a body across its skin.
The wren is the smaller animal.
So each gram of the wren has more skin to lose heat through.
So each gram of the wren loses heat faster than each gram of the turkey.

Check q3

Rabbit 1 and rabbit 2 are two adult rabbits of the same size and shape. The two rabbits sit in the same cold field.

Which loses heat faster per gram of body?

A. Rabbit 1
The two rabbits are the same size and shape.
B. Rabbit 2
The two rabbits are the same size and shape.
C. ✓ Both at the same rate

Why: Heat leaves a body across its skin.
Two bodies of the same size and shape have the same skin for each gram.
So each gram of the two rabbits loses heat at the same rate.

Check q4

A rabbit has a mass of 2 kg. A deer has a mass of 90 kg. The two animals stand in the same frosty field.

Which loses heat faster per gram of body?

A. ✓ The rabbit
B. The deer
The rabbit is the smaller animal, so each gram of the rabbit has more skin to lose heat through.
C. Both at the same rate
The two animals do not have the same skin for each gram.

Why: Heat leaves a body across its skin.
The rabbit is the smaller animal.
So each gram of the rabbit has more skin to lose heat through.
So each gram of the rabbit loses heat faster than each gram of the deer.

Check q5

A hippo has a mass of 1,500 kg. An elephant has a mass of 4,000 kg. The two animals stand in the same cool night air.

Which loses heat faster per gram of body?

A. The elephant
The hippo is the smaller animal, large as it is, so each gram of the hippo has more skin to lose heat through.
B. ✓ The hippo
C. Both at the same rate
The two animals do not have the same skin for each gram.

Why: Heat leaves a body across its skin.
The hippo is the smaller animal, large as it is.
So each gram of the hippo has more skin to lose heat through.
So each gram of the hippo loses heat faster than each gram of the elephant.

16Why the smaller animal loses heat faster per gram

Video: Watch: Why the smaller animal loses heat faster per gram

Doubling every dimension multiplies skin by four and mass by eight: half the skin for each gram.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L16Bb.mp4

Now the reason, on one pair.

What you are expected to know Explain why each gram of a smaller animal loses heat faster: heat leaves across the skin, and a smaller body has more skin for each gram.

Check q6

On a cold night, a squirrel loses heat faster per gram of body than a deer does.

Why does each gram of the squirrel lose heat faster?

A. The squirrel has more skin in total than the deer
The deer, the larger animal, has more skin in total.
B. The squirrel is warmer inside than the deer
The two mammals hold about the same body temperature.
C. ✓ The squirrel has more skin for each gram of its body

Why: Heat is held in every gram of a body.
Heat leaves across the skin.
Each gram of the squirrel has more skin than each gram of the deer.
So each gram of the squirrel loses its heat faster.

Practice writing an answer

A 5 g hummingbird and a 5 kg goose sit in the same cold air. Suppose each bird suddenly stopped making new heat. The hummingbird’s body temperature would fall faster than the goose’s.

(a) Explain why the hummingbird’s body temperature would fall faster. (1 pt)

Frame Heat leaves a body across its …. Each gram of the hummingbird has far more … than each gram of the goose, so …

Model answer Heat leaves a body across its skin.
Heat is held in every gram of a body.
Each gram of the hummingbird has far more skin than each gram of the goose.
So each gram of the hummingbird loses its heat faster.
So the hummingbird’s body temperature falls faster.

Rubric

Award 1 point for: heat leaves across the skin, and each gram of the hummingbird has more skin (a greater surface-area-to-volume ratio) than each gram of the goose, so each gram loses its heat faster and the body temperature falls faster.
Accept: the smaller body has the greater surface-area-to-volume ratio, so it loses heat faster for each gram.

Slip Saying the goose cools faster because it has more skin in total. The goose does have more skin in total. Each gram of the goose has far less skin, and a body’s temperature falls by what each gram loses.

22Small organisms use energy faster, per gram

Video: Watch: Small organisms use energy faster, per gram

Metabolic rate is the rate of using energy; a mouse that loses heat fast must make heat fast.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L16Bc.mp4

Back to the food. A mouse eats about 20% of its body mass in food each day. An elephant eats a few percent of its.

Food eaten each day as a share of body mass: about a fifth for a mouse, a few percent for an elephant

Per gram of body, the mouse uses energy several times as fast as the elephant.

The rate at which an organism uses energy is called its .

Here is why the mouse’s metabolic rate is so high. A mouse that loses heat fast must make heat fast. Making heat uses energy.

Per unit of body mass, a smaller organism made of many cells typically uses energy faster than a larger one.

The elephant still uses far more energy in total each day. The pattern is about the rate per gram.

What you are expected to know Metabolic rate is the rate at which an organism uses energy; per unit of body mass it is typically higher in smaller organisms made of many cells than in larger ones.

Check q7

A hamster has a mass of 40 g. A horse has a mass of 500 kg.

Per gram of body, which has the higher metabolic rate?

A. ✓ The hamster
B. The horse
The horse’s larger total is spread over 500 kg of body.
C. About the same
Metabolic rate per gram falls as body size rises, even between two mammals.

Why: Metabolic rate per gram of body is typically higher in smaller organisms.
The 40 g hamster is far smaller than the 500 kg horse.
So each gram of the hamster uses energy faster than each gram of the horse.
The horse uses more energy in total.

Check q8

Each gram of the hamster uses energy faster than each gram of the horse.

Why does each gram of the hamster use energy faster?

A. A gram of hamster holds more cells than a gram of horse does
A gram of hamster and a gram of horse hold about the same number of cells.
B. The hamster moves about more than the horse does each day
The pattern holds for animals at rest.
C. ✓ The hamster loses heat faster per gram, so must make heat faster

Why: Heat leaves a body across its skin.
Each gram of the hamster has more skin than each gram of the horse, so each gram of the hamster loses heat faster.
To stay warm, the hamster must make heat faster, and making heat uses energy.
So each gram uses energy faster.

Check q9

What does an organism’s metabolic rate measure?

A. How much of its body is muscle
Metabolic rate is about energy use, not about what the body is made of.
B. How much food it can store
Metabolic rate is a rate of energy use, not an amount of food held in store.
C. How fast it grows
A fully grown mouse still has a high metabolic rate; the rate is about energy use, not about growth.
D. ✓ How fast it uses energy

Why: Metabolic rate is the rate at which an organism uses energy.

Check q10

Four mammals of different sizes: a shrew, a cat, a horse and an elephant.

Which mammal eats the largest share of its own body mass in food each day?

A. The elephant
The elephant eats the most in total but only a few percent of its mass a day; per gram it uses energy slowly.
B. The horse
Energy use per gram falls as size rises, so the horse eats a smaller share of its mass than a small mammal does.
C. ✓ The shrew
D. The cat
The cat is small, but the shrew is far smaller, and the rate per gram is highest in the smallest mammal.

Why: Metabolic rate per gram of body is highest in the smallest organism.
The shrew is the smallest of the four, so the shrew uses energy fastest per gram.
The food an animal eats each day supplies the energy it uses.
So the shrew eats the largest share.

A mouse has far more skin for each gram of its body than an elephant, so each gram of the mouse loses heat faster. To stay warm the mouse must make heat faster, and making heat uses energy, so its metabolic rate per gram is several times the elephant’s.

That is why the mouse eats about 20% of its mass each day and the elephant only a few percent: the small animal has to replace, gram for gram, far more heat.

Glossary

metabolic rate: The rate at which an organism uses energy. Per unit of body mass it is typically higher in smaller organisms made of many cells than in larger ones.

APBIO-U02-L16C Spheres, rods and slabs

Topic 2.2 · Cell Size · 27 steps

Three cells side by side with their dimensions: a round cell of radius 2 micrometers, a rod 4 micrometers long and 0.5 micrometers in radius, and a flat box-shaped cell 8 by 4 by 2 micrometers

Here are three cells, each with its dimensions: a round alga cell of radius 2 μm, a rod-shaped bacterium 4 μm long and 0.5 μm in radius, and a flat cell 8 μm long, 4 μm wide and 2 μm high.

Which of the three has the most surface for each cubic micrometer of its volume? Looking is not enough to tell you. You need a number for each.

Unit 2 · Cell Structure and Function

1Spheres, rods and slabs

Video: Watch: Spheres, rods and slabs

The formula-sheet equations for a sphere, a cylinder and a box; the round cell worked to 1.50 per μm.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L16Ca.mp4

A cube-shaped cell’s surface-area-to-volume ratio is its surface area divided by its volume. Most cells are not cubes. The same steps work for any shape.

The AP formula sheet gives the formulas for each shape. Here each formula is written beside its cell. Use the π key on your calculator, and round only the final answer.

Here is a round alga cell of radius 2 μm.

A round cell of radius 2 μm, its radius drawn from the center to the edge

The formula sheet gives a sphere’s surface area and its volume from its radius r. Here are the two formulas.

surface area = 4 π r^{2}

volume = \frac{4}{3} π r^{3}

Surface area and volume of a sphere of radius r, as the AP formula sheet writes them

Worked example

A round alga cell has a radius of 2 μm. What are its surface area, its volume and its surface-area-to-volume ratio?

Write down the values in the question:

r = 2 μm

Write down the equations:

surface area = 4 π r^{2}

volume = \frac{4}{3} π r^{3}

SA/V = \frac{surface area}{volume}

Substitute the values into the first two equations:

surface area = 4 π {(2)}^{2}

volume = \frac{4}{3} π {(2)}^{3}

Calculate:

surface area = 50.3 {μm}^{2}

volume = 33.5 {μm}^{3}

Substitute the unrounded results into the ratio equation:

SA/V = \frac{50.27}{33.51}

Calculate:

SA/V = 1.50 per μm

Here is a rod-shaped bacterium: radius 0.5 μm, length 4 μm. The formula sheet treats a rod as a cylinder of radius r and height h, so the rod’s length is its h.

A rod-shaped cell drawn as a cylinder lying on its side: a flat circular end face at the left showing its radius, 0.5 μm, and a straight body, 4 μm long

Here are the cylinder formulas. The surface area has two terms: the curved side, 2πrh, and the two flat ends, 2πr².

surface area = 2 π r h + 2 π r^{2}

volume = π r^{2} h

Surface area and volume of a cylinder of radius r and height h, as the AP formula sheet writes them

Check q1 numeric entry

A rod-shaped bacterium has a radius of 0.5 μm and a length of 4 μm. Model it as a cylinder.

Now calculate its surface-area-to-volume ratio, to three significant figures. Carry the unrounded surface area and volume forward.

Part 1. First, calculate its surface area, to three significant figures.

Answer: 14.1 μm² (tolerance ±0.05)

Working

Substitute the values into the surface-area equation:

surface area = 2 π (0.5) (4) + 2 π {(0.5)}^{2}

Calculate:

surface area = 14.1 {μm}^{2}

Part 2. Next, calculate its volume, to three significant figures.

Answer: 3.14 μm³ (tolerance ±0.005)

Working

Substitute the values into the volume equation:

volume = π {(0.5)}^{2} (4)

Calculate:

volume = 3.14 {μm}^{3}

Answer: 4.5 per μm (tolerance ±0.05)

Working

Write down the values in the question:

r = 0.5 μm

h = 4 μm

Write down the equations:

surface area = 2 π r h + 2 π r^{2}

volume = π r^{2} h

SA/V = \frac{surface area}{volume}

Substitute the values into the first two equations:

surface area = 2 π (0.5) (4) + 2 π {(0.5)}^{2}

volume = π {(0.5)}^{2} (4)

Calculate:

surface area = 14.1 {μm}^{2}

volume = 3.14 {μm}^{3}

Substitute the unrounded results into the ratio equation:

SA/V = \frac{14.137}{3.1416}

Calculate:

SA/V = 4.50 per μm

The rod’s ratio, 4.50 per μm, is three times the round cell’s 1.50 per μm.

A flat cell is modeled as a box of length l, width w and height h. Here are the box formulas.

A flat box-shaped cell of length l, width w and height h

The box has three pairs of faces, so its surface area has three terms. Its volume is length by width by height.

surface area = 2 l h + 2 l w + 2 w h

volume = l w h

Surface area and volume of a box of length l, width w and height h, as the AP formula sheet writes them

What you are expected to know Calculate the surface area, volume and surface-area-to-volume ratio of a cell modeled as a sphere, a cylinder or a box.

Check q2 numeric entry

A flattened cell is modeled as a box 8 μm long, 4 μm wide and 2 μm thick.

Now calculate its surface-area-to-volume ratio, to three significant figures.

Part 1. First, calculate its surface area.

Answer: 112 μm² (tolerance ±0.5)

Working

Substitute the values into the surface-area equation:

surface area = 2 (8) (2) + 2 (8) (4) + 2 (4) (2)

Calculate:

surface area = 112 {μm}^{2}

Part 2. Next, calculate its volume.

Answer: 64 μm³ (tolerance ±0.5)

Working

Substitute the values into the volume equation:

volume = (8) (4) (2)

Calculate:

volume = 64 {μm}^{3}

Answer: 1.75 per μm (tolerance ±0.005)

Working

Write down the values in the question:

l = 8 μm

w = 4 μm

h = 2 μm

Write down the equations:

surface area = 2 l h + 2 l w + 2 w h

volume = l w h

SA/V = \frac{surface area}{volume}

Substitute the values into the first two equations:

surface area = 2 (8) (2) + 2 (8) (4) + 2 (4) (2)

volume = (8) (4) (2)

Calculate:

surface area = 112 {μm}^{2}

volume = 64 {μm}^{3}

Substitute the results into the ratio equation:

SA/V = \frac{112}{64}

Calculate:

SA/V = 1.75 per μm

Check q3 numeric entry

A round cell has a radius of 5 μm.

Calculate its volume, to three significant figures.

Answer: 524 μm³ (tolerance ±1)

Working

Write down the values in the question:

r = 5 μm

Write down the equation:

volume = \frac{4}{3} π r^{3}

Substitute the values into the equation:

volume = \frac{4}{3} π {(5)}^{3}

Calculate:

volume = 524 {μm}^{3}

Three cells: a round alga cell of radius 2 μm, a rod-shaped bacterium 4 μm long and 0.5 μm in radius, and a flat cell 8 μm by 4 μm by 2 μm.

The rod has 4.50 μm² of surface for each μm³ of its volume, the flat cell 1.75 μm² and the round cell 1.50 μm².

So the rod has the most surface for its volume and the round cell the least, because a rounded shape has the least surface for its volume. The numbers decide it, not the look of the cell.

20Mixed practice mixed practice

Check q4 numeric entry

A round yeast cell has a radius of 3 μm.

Calculate its surface area, to three significant figures.

Answer: 113 μm² (tolerance ±0.5)

Working

Write down the values in the question:

r = 3 μm

Write down the equation:

surface area = 4 π r^{2}

Substitute the values into the equation:

surface area = 4 π {(3)}^{2}

Calculate:

surface area = 113 {μm}^{2}

Check q5 numeric entry

A rod-shaped bacterium has a radius of 1 μm and a length of 5 μm. Model it as a cylinder.

Calculate its volume, to three significant figures.

Answer: 15.7 μm³ (tolerance ±0.05)

Working

Write down the values in the question:

r = 1 μm

h = 5 μm

Write down the equation:

volume = π r^{2} h

Substitute the values into the equation:

volume = π {(1)}^{2} (5)

Calculate:

volume = 15.7 {μm}^{3}

Check q6 numeric entry

A flat cell is modeled as a box 12 μm long, 6 μm wide and 2 μm thick.

Calculate its surface-area-to-volume ratio, to three significant figures.

Answer: 1.5 per μm (tolerance ±0.005)

Working

Write down the values in the question:

l = 12 μm

w = 6 μm

h = 2 μm

Write down the equation:

SA/V = \frac{2 l h + 2 l w + 2 w h}{l w h}

Substitute the values into the equation:

SA/V = \frac{2 (12) (2) + 2 (12) (6) + 2 (6) (2)}{(12) (6) (2)}

Calculate:

SA/V = 1.50 per μm

Check q7 numeric entry

A round cell has a radius of 1.5 μm.

Calculate its surface-area-to-volume ratio, to three significant figures.

Answer: 2 per μm (tolerance ±0.005)

Working

Write down the values in the question:

r = 1.5 μm

Write down the equation:

SA/V = \frac{4 π r^{2}}{\frac{4}{3} π r^{3}}

Substitute the values into the equation:

SA/V = \frac{4 π {(1.5)}^{2}}{\frac{4}{3} π {(1.5)}^{3}}

Calculate:

SA/V = 2.00 per μm

Check q8 numeric entry

A rod-shaped cell has a radius of 1 μm and a length of 6 μm. Model it as a cylinder.

Calculate its surface-area-to-volume ratio, to three significant figures.

Answer: 2.33 per μm (tolerance ±0.005)

Working

Write down the values in the question:

r = 1 μm

h = 6 μm

Write down the equation:

SA/V = \frac{2 π r h + 2 π r^{2}}{π r^{2} h}

Substitute the values into the equation:

SA/V = \frac{2 π (1) (6) + 2 π {(1)}^{2}}{π {(1)}^{2} (6)}

Calculate:

SA/V = 2.33 per μm

Practice writing an answer

A yeast cell is modeled as a sphere of radius 4 μm. A cell from the skin of a fish is modeled as a flat box 6 μm long, 6 μm wide and 1 μm thick. Use the formulas from the AP formula sheet.

(a) Calculate the surface-area-to-volume ratio of each cell, and identify which cell has the greater ratio. (2 pt)

Frame Yeast cell: … per μm. Skin cell: … per μm. The … cell has the greater ratio.

Model answer Yeast cell: 0.750 per μm.
Skin cell: 2.67 per μm.
The skin cell has the greater ratio.

Working

Write down the values in the question:

yeast cell: r = 4 μm

skin cell: l = 6 μm, w = 6 μm, h = 1 μm

Write down the equations:

sphere: SA/V = \frac{4 π r^{2}}{\frac{4}{3} π r^{3}}

box: SA/V = \frac{2 l h + 2 l w + 2 w h}{l w h}

Substitute the values into the equations:

yeast: SA/V = \frac{4 π {(4)}^{2}}{\frac{4}{3} π {(4)}^{3}}

skin: SA/V = \frac{2 (6) (1) + 2 (6) (6) + 2 (6) (1)}{(6) (6) (1)}

Calculate:

yeast: SA/V = 0.750 per μm

skin: SA/V = 2.67 per μm

Rubric

Award 1 point for: both ratios as surface area divided by volume, with units: yeast cell 0.75 per μm and skin cell 2.67 per μm (accept yeast 0.7–0.8 and skin 2.6–2.8 per μm; the surface areas and volumes need not be shown).
Award 1 point for: the skin cell identified as having the greater ratio, from the two ratios. A student whose ratios are out of range still earns this point for naming whichever cell their own ratios make greater.

Slip Dividing volume by surface area, or comparing total surface alone. The yeast cell has more surface in total, 201 against 96 μm², but less for each μm³ of interior.

APBIO-U02-L16D Which cell exchanges faster?

Topic 2.2 · Cell Size · 31 steps

A round cell 5 micrometers across beside a cube-shaped cell with sides of 7 micrometers

Here are a round cell 5 μm across and a cube-shaped cell with sides of 7 μm.

The cube has more surface in total. Which of the two exchanges materials with its surroundings faster for its size? The total surface does not tell you. You need each cell’s surface-area-to-volume ratio.

Unit 2 · Cell Structure and Function

1Which exchanges faster, and why

Here are a round cell 5 μm across, so of radius 2.5 μm, and a cube-shaped cell with sides of 7 μm. To compare them, work out each cell’s surface-area-to-volume ratio.

A round cell 5 μm across beside a cube-shaped cell with sides of 7 μm

Worked example

The round cell has a radius of 2.5 μm. What is its surface-area-to-volume ratio?

Write down the values in the question:

r = 2.5 μm

Write down the equations:

surface area = 4 π r^{2}

volume = \frac{4}{3} π r^{3}

SA/V = \frac{surface area}{volume}

Substitute the values into the first two equations:

surface area = 4 π {(2.5)}^{2}

volume = \frac{4}{3} π {(2.5)}^{3}

Calculate:

surface area = 78.5 {μm}^{2}

volume = 65.4 {μm}^{3}

Substitute the unrounded results into the ratio equation:

SA/V = \frac{78.54}{65.45}

Calculate:

SA/V = 1.20 per μm

Check q1 numeric entry

A cube-shaped cell has sides 7 μm long.

Now calculate its surface-area-to-volume ratio, to three significant figures.

Part 1. First, calculate its surface area.

Answer: 294 μm² (tolerance ±0.5)

Working

Substitute the side length into the surface-area equation:

surface area = 6 {(7)}^{2}

Calculate:

surface area = 294 {μm}^{2}

Part 2. Next, calculate its volume.

Answer: 343 μm³ (tolerance ±0.5)

Working

Substitute the side length into the volume equation:

volume = {(7)}^{3}

Calculate:

volume = 343 {μm}^{3}

Answer: 0.857 per μm (tolerance ±0.005)

Working

Write down the values in the question:

s = 7 μm

Write down the equations:

surface area = 6 s^{2}

volume = s^{3}

SA/V = \frac{surface area}{volume}

Substitute the values into the first two equations:

surface area = 6 {(7)}^{2}

volume = {(7)}^{3}

Calculate:

surface area = 294 {μm}^{2}

volume = 343 {μm}^{3}

Substitute the results into the ratio equation:

SA/V = \frac{294}{343}

Calculate:

SA/V = 0.857 per μm

Video: Watch: Which exchanges faster, and why

Round cell 1.20 per μm against cube 0.857 per μm: compare by the ratio, not the total surface.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L16Da.mp4

The round cell has 1.20 μm² of surface for each μm³ of its volume. The cube has 0.857 μm² for each μm³. So the round cell exchanges materials more efficiently for its size, though the cube has more surface in total.

A bar chart with two bars, labelled round cell and cube, showing each cell's surface-area-to-volume ratio in per μm; each bar carries its value above it

The same comparison decides shapes and folds. Whichever cell has more surface per unit of volume exchanges faster. That is why exchange surfaces fold, and why a small animal uses energy so fast.

What you are expected to know Predict which of two cells exchanges materials faster for its size from their surface-area-to-volume ratios.

9Fluency quiz: which exchanges faster for its size? mixed practice

Check q2

The bar chart shows the surface-area-to-volume ratio of cell A and cell B.

A bar chart with two bars, labelled cell A and cell B, showing each cell's surface-area-to-volume ratio in per μm; each bar carries its value above it

Which cell exchanges materials with its surroundings faster for its size?

A. ✓ Cell A
B. Cell B
Cell B’s ratio is 0.9 per μm and cell A’s is 2.4 per μm, so cell A has more surface for each μm³ of its volume.
C. Both the same
The two bars are different heights, so the two ratios differ.

Why: Everything a cell exchanges crosses its surface, so the cell with the greater surface-area-to-volume ratio exchanges faster for its size.
Cell A: 2.4 per μm.
Cell B: 0.9 per μm.
So cell A exchanges faster for its size.

Check q3

Cell C has 300 μm² of surface and a surface-area-to-volume ratio of 0.6 per μm. Cell D has 60 μm² of surface and a ratio of 1.8 per μm.

Which cell exchanges materials with its surroundings faster for its size?

A. Cell C
Cell C has more surface in total, but its ratio, 0.6 per μm, is smaller than cell D’s 1.8 per μm.
B. ✓ Cell D
C. Both the same
The two ratios differ: 0.6 per μm against 1.8 per μm.

Why: Everything a cell exchanges crosses its surface, so the cell with the greater surface-area-to-volume ratio exchanges faster for its size.
Cell C: 0.6 per μm.
Cell D: 1.8 per μm.
So cell D exchanges faster for its size, though cell C has more surface in total.

Check q4

The bar chart shows the surface-area-to-volume ratio of cell E and cell F. Cell F is the larger cell.

A bar chart with two bars, labelled cell E and cell F, showing each cell's surface-area-to-volume ratio in per μm; each bar carries its value above it

Which cell exchanges materials with its surroundings faster for its size?

A. Cell E
Cell E is the smaller cell, but the two bars are the same height: each cell’s ratio is 2.0 per μm.
B. Cell F
The two bars are the same height: each cell’s ratio is 2.0 per μm.
C. ✓ Both the same

Why: Everything a cell exchanges crosses its surface, so the cell with the greater surface-area-to-volume ratio exchanges faster for its size.
Cell E: 2.0 per μm.
Cell F: 2.0 per μm.
So the two cells exchange at the same rate for their size.

Check q5

Cell G has a volume of 500 μm³ and a surface-area-to-volume ratio of 0.5 per μm. Cell H has a volume of 20 μm³ and a ratio of 3.0 per μm.

Which cell exchanges materials with its surroundings faster for its size?

A. Cell G
More volume is more interior to supply, not more exchange; cell G’s ratio, 0.5 per μm, is the smaller.
B. ✓ Cell H
C. Both the same
The two ratios differ: 0.5 per μm against 3.0 per μm.

Why: Everything a cell exchanges crosses its surface, so the cell with the greater surface-area-to-volume ratio exchanges faster for its size.
Cell G: 0.5 per μm.
Cell H: 3.0 per μm.
So cell H exchanges faster for its size.

Check q6

The bar chart shows the surface-area-to-volume ratio of a flat cell and a round cell of the same volume.

A bar chart with two bars, labelled flat cell and round cell, showing each cell's surface-area-to-volume ratio in per μm; each bar carries its value above it

Which cell exchanges materials with its surroundings faster for its size?

A. ✓ The flat cell
B. The round cell
The round cell’s ratio is 1.1 per μm and the flat cell’s is 2.8 per μm, so the flat cell has more surface for each μm³ of its volume.
C. Both the same
The two volumes are the same but the two ratios are not.

Why: Everything a cell exchanges crosses its surface, so the cell with the greater surface-area-to-volume ratio exchanges faster for its size.
The flat cell: 2.8 per μm.
The round cell: 1.1 per μm.
So the flat cell exchanges faster for its size.

Check q7

Cell 1 has a surface area of 90 μm² and a volume of 30 μm³. Cell 2 has a surface area of 140 μm² and a volume of 70 μm³.

Which cell exchanges materials with its surroundings faster for its size?

A. ✓ Cell 1
B. Cell 2
Cell 1’s ratio is 3.0 per μm and cell 2’s is 2.0 per μm, so cell 1 exchanges faster for its size.
C. Both the same
The two ratios differ: 3.0 per μm against 2.0 per μm.

Why: The cell with the greater surface-area-to-volume ratio exchanges faster for its size.
Cell 1’s ratio, 3.0 per μm, is greater than cell 2’s, 2.0 per μm.
So cell 1 exchanges faster for its size.

16Blood cells and fish eggs

Video: Watch: Blood cells and fish eggs

Same volume, more surface: the greater ratio; twice the radius: half the surface per μm³, so the center is supplied more slowly.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L16Db.mp4

Now two real comparisons, each with its numbers: two blood cells, then two fish eggs.

Here are the two blood cells, drawn with the same volume: one flattened, one rounder.

Two blood cells of the same volume, 90 μm³: a flattened disc labelled 120 μm² of membrane and a rounder cell labelled 100 μm² of membrane

What you are expected to know Explain why a lower surface-area-to-volume ratio means slower supply to the center of a cell.

Check q8

Two blood cells each have a volume of 90 μm³. The flattened cell has 120 μm² of membrane. The rounder cell has 100 μm² of membrane.

Which cell has the greater surface-area-to-volume ratio?

A. The rounder cell
The rounder cell has less membrane, 100 μm², around the same 90 μm³, so its ratio is the smaller.
B. ✓ The flattened cell
C. Both the same
Equal volumes do not give equal ratios: the flattened cell has 120 μm² of membrane around 90 μm³, the rounder cell only 100 μm².

Why: The two cells have the same volume, so the one with more membrane has the greater ratio.
The flattened cell’s ratio, 1.3 per μm, is greater than the rounder cell’s, 1.1 per μm.
So the flattened cell has the greater ratio.

Check q9

The flattened blood cell has 120 μm² of membrane and a volume of 90 μm³.

Which working gives its surface-area-to-volume ratio?

A. 90 μm³ divided by 120 μm², giving 0.75 per μm
The ratio is surface area divided by volume.
B. 120 μm² minus 90 μm³, giving 30 per μm
A ratio compares two quantities by dividing, not by subtracting.
C. ✓ 120 μm² divided by 90 μm³, giving 1.3 per μm

Why: The ratio is surface area divided by volume, never volume divided by surface area.
So the surface area, in μm², goes on top and the volume, in μm³, goes underneath.
Dividing μm² by μm³ leaves per μm.
So the flattened cell’s ratio is 1.3 per μm.

Here are two fish eggs in the same water. The larger egg has twice the radius of the smaller one.

Two round fish eggs side by side: a small one of radius 0.5 mm and a large one of radius 1.0 mm, each with its radius drawn

Check q10

Two round fish eggs sit in the same water. One has a radius of 0.5 mm. The other has a radius of 1.0 mm.

Compared with the smaller egg, how much surface does the larger egg have for each mm³ of its volume?

A. A quarter as much
Doubling the radius multiplies the surface by four, not two; against a volume eight times as large that leaves half the surface per mm³, not a quarter.
B. ✓ Half as much
C. The same
The same shape at twice the size does not keep the same ratio; the volume grows faster than the surface.
D. Twice as much
Doubling the radius multiplies the surface by four but the volume by eight, so surface per mm³ falls, not rises.

Why: Doubling the radius multiplies the surface area by four.
It multiplies the volume by eight.
So the larger egg has half as much surface for each mm³ of its volume.

Check q11

Both eggs use oxygen at the same rate per mm³. The concentration of oxygen at the center of the larger egg is lower than at the center of the smaller egg.

Why is the concentration of oxygen lower at the center of the larger egg?

A. ✓ Each mm³ of the larger egg has less surface for oxygen to enter across
B. The larger egg has less surface in total than the smaller egg
The larger egg has more surface in total, not less.
C. The larger egg’s membrane is thicker, so oxygen crosses it more slowly
The two eggs have the same kind of membrane.

Why: Oxygen enters an egg only across its surface.
Each mm³ of the larger egg has half as much surface as each mm³ of the smaller egg.
So each mm³ of the larger egg gets oxygen more slowly.
So the concentration of oxygen at the center of the larger egg is lower.

Here are the round cell and the cube again. The round cell has 1.20 μm² of surface for each μm³ of its volume; the cube has 0.857 μm².

So the round cell exchanges faster for its size, though the cube has more surface in total. Whichever cell has more surface per unit of volume exchanges faster, and the ratios decide it.

28Mixed practice mixed practice

Check q12

A cell lining the gut has its top surface folded into many tiny projections. A second cell of the same volume has a smooth top surface.

Which statement justifies the claim that the folded cell exchanges materials with its surroundings faster?

A. The folded cell has the same surface as the smooth cell, only arranged in folds
Folding a surface adds surface; the two cells do not have the same surface.
B. The folded cell is larger than the smooth cell, so it has more surface in total
The two cells have the same volume; the folds add surface, not size.
C. The folds make the membrane thinner, so substances cross it more quickly
The membrane is the same thickness in both cells; the folds add more of it.
D. ✓ The folds give the folded cell more surface for the same volume, so its ratio is greater

Why: Exchange happens across a cell’s surface.
The amount a cell needs grows with its volume.
The two cells have the same volume.
The folds give the folded cell more surface, so it has the greater surface-area-to-volume ratio.
So the folded cell exchanges faster.

Practice writing an answer

Two cells take up nutrients across their surfaces. A yeast cell, modeled as a sphere of radius 4 μm, has a surface area of 201 μm², a volume of 268 μm³ and a surface-area-to-volume ratio of 0.75 per μm. A cell from the skin of a fish, modeled as a flat box 6 μm long, 6 μm wide and 1 μm thick, has a surface area of 96 μm², a volume of 36 μm³ and a ratio of 2.67 per μm.

(a) Identify which cell exchanges nutrients with its surroundings more efficiently for its size, and explain why. (1 pt)

Model answer The skin cell exchanges nutrients more efficiently for its size.
Everything a cell takes in crosses its surface.
The amount a cell needs grows with its volume.
The skin cell has 2.67 μm² of membrane for each μm³ of interior.
The yeast cell has 0.75 μm² of membrane for each μm³ of interior.
So each μm³ of the skin cell is supplied across more than three times as much surface as each μm³ of the yeast cell.

Rubric

Award 1 point for: the skin cell, because it has more surface per unit of volume (2.67 against 0.75 per μm), and exchange happens across the surface while need grows with volume.
Accept: a higher surface-area-to-volume ratio means more membrane for each unit of interior to exchange across.

Slip Picking the yeast cell because its total surface, 201 μm², is larger. The yeast cell also has far more volume, 268 μm³. What matters is surface for each μm³ of volume, and the skin cell has more.

(b) A single-celled pond organism has the same volume as the yeast cell but is shaped as a flat ribbon with about 600 μm² of surface. Evaluate the claim that the ribbon shape is an aid to exchange. (1 pt)

Model answer The claim is supported.
The ribbon has the same volume as the yeast cell, 268 μm³.
The ribbon has about 600 μm² of membrane around that volume; the yeast cell has 201 μm².
So the ribbon’s surface-area-to-volume ratio is about 2.2 per μm, against the yeast cell’s 0.75 per μm.
Nutrients cross into a cell only across its surface.
The ribbon has about three times the surface for each μm³ of interior.
So nutrients cross into the ribbon faster for the same volume, and the flat shape aids exchange.

Working

Write down the values in the question:

surface area = 600 μm²

volume = 268 μm³

Write down the equation:

SA/V = \frac{surface area}{volume}

Substitute the values into the equation:

SA/V = \frac{600}{268}

Calculate:

SA/V = 2.24 per μm

Rubric

Award 1 point for: the judgement that the claim is supported AND the ground for it: at the same volume the ribbon has about three times the surface (600 against 201 μm²), so its ratio is about 2.2 per μm against 0.75, and more surface per unit of volume means faster exchange for the same interior.
Accept: the comparison stated as more surface for the same volume, so a higher ratio and faster exchange, without the value 2.2.

Slip Saying flat cells are smaller. The two volumes are the same. The flat shape adds surface at the same volume. More surface for the same volume raises the ratio.

APBIO-U02-P22 Practice questions: Topic 2.2

Topic 2.2 · Cell Size · 10 MCQ · 2 FRQ · for APBIO-U02-T22

These practice questions have the shape of the Topic 2.2 test. Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the first free-response question, you work through one scenario in small steps, and each step offers a hint if you want one. For the second, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it. Formulas are given wherever a calculation needs them; use the π key on your calculator and give answers to three significant figures.

Video: Watch first: Sum up: why a cell can’t just grow bigger

The cube, its ratio, the ratio falling as it grows, and why cells stay small or divide.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T22-summary.mp4

Q1 P22-q01

A cube-shaped plant cell has sides 12 μm long. For a cube of side s, surface area = 6s² and volume = s³.

What are its surface area and its volume?

A. 144 μm² and 1,728 μm³
144 μm² is the area of one face only.
B. ✓ 864 μm² and 1,728 μm³
C. 864 μm² and 144 μm³
144 μm³ is 12² with the wrong unit; volume needs the third dimension.
D. 576 μm² and 1,728 μm³
576 μm² counts four faces and leaves out the top and bottom.

Why: A cube has six square faces, so the surface area is six times the area of one face.
The volume is the side length multiplied by itself three times.
Area is in μm² and volume in μm³; the working is below.

Q2 P22-q02

A cell has a surface area of 220 μm² and a volume of 200 μm³.

What is its surface-area-to-volume ratio?

A. ✓ 1.1 per μm
B. 0.91 per μm
0.91 is the volume divided by the surface area, the ratio the wrong way up.
C. 20 μm
20 is 220 − 200, a difference, not a ratio.
D. 1.1 μm²
The number is right but the unit is wrong.

Why: The surface-area-to-volume ratio is the surface area divided by the volume.
Dividing μm² by μm³ leaves per μm.
So the cell has 1.1 μm² of surface for each μm³ of interior; the working is below.

Q3 P22-q03

A spherical cell grows without changing shape. Its radius increases from 1 μm to 4 μm.

Compared with the small cell, how have its surface area, volume and surface-area-to-volume ratio changed?

A. Surface ×4, volume ×4, ratio unchanged
Making every dimension 4 times as long multiplies an area by 16 and a volume by 64, not by 4.
B. Surface ×16, volume ×16, ratio unchanged
Volume grows with the cube of the radius, 64 times, not 16.
C. ✓ Surface ×16, volume ×64, ratio 25% of what it was
D. Surface ×64, volume ×16, ratio 4 times what it was
The two factors are swapped.
The ratio is surface area divided by volume.

Why: An area has two dimensions, so it grows by the length factor squared.
A volume has three dimensions, so it grows by the length factor cubed.
The volume grows faster, so the ratio falls; the working is below.

Q4 P22-q04

Cells of a kind of bacterium normally divide whenever they reach 2 μm long. A drug stops them dividing but leaves them growing. At 8 μm long the cells take in oxygen too slowly for their needs and stop growing.

Why does the oxygen intake fall behind the cell's needs as the cell grows?

A. The membrane grows thicker as the cell grows, so oxygen crosses it more slowly than before
The membrane stays the same thickness as the cell grows.
B. A larger cell has less membrane in total than a small one, so less oxygen can cross into it
A larger cell has more membrane in total, not less.
C. The cell has used up all the oxygen dissolved in the liquid around it, so none is left to enter
The liquid around the cell holds plenty of oxygen; the small cells in it are fine.
D. ✓ Oxygen enters only across the surface, while the need for it grows with the volume, which grows faster

Why: A cell takes in what it needs and gets rid of waste only across its surface.
The amount it needs grows with its volume.
As the cell grows, its volume grows faster than its surface.
So past a certain size the surface cannot keep up with the interior.

Q5 P22-q05

A cube-shaped cell with sides 6 μm long can bring in, across its membrane, 120% of the oxygen its interior uses each minute. A cell of the same kind grows to a cube with sides 12 μm long. Oxygen intake is proportional to membrane area; oxygen use is proportional to volume.

What percentage of its oxygen need can the 12 μm cell's membrane supply?

A. ✓ 60%
B. 120%
Doubling every side multiplies the membrane by 4 but the volume by 8, so the share falls to half.
C. 240%
Intake is multiplied by 4, but the need by 8 because volume depends on three lengths, so the share is 4/8 of 120%, which is 60%.
D. 30%
Doubling the side multiplies the intake by 4 and the need by 8, so the share is divided by 8 ÷ 4 = 2, not by 4.

Why: Intake grows with membrane area and need grows with volume.
Doubling the side multiplies the area by 4 and the volume by 8, so the need grows twice as fast as the intake.
So the share the membrane can supply halves; the working is below.

Q6 P22-q06

The top face of each cell lining a kidney tubule carries thousands of finger-like folds of its plasma membrane, called microvilli. The cell takes useful substances back from the tubule fluid across that face. A cell with microvilli has almost the same volume as a cell with a flat top face.

What do the microvilli do to the cell’s surface-area-to-volume ratio, and why?

A. They lower the ratio: each fold adds interior that the membrane must serve
A fold of membrane adds surface, not interior; the cell’s volume is almost unchanged.
B. ✓ They raise the ratio: the folds add membrane area while adding almost no volume
C. They leave the ratio unchanged: the cell’s volume is almost the same
The volume is almost the same but the surface is not; the folds add membrane area.
D. They raise the ratio: the folds add volume for the cell to hold what it takes back
The folds add almost no volume; what they add is membrane area.

Why: Each microvillus is a fold of the plasma membrane, so the folds add membrane area.
The folds add almost no volume.
More surface over the same volume is a greater ratio.
So the microvilli raise the ratio, and the cell exchanges faster for its size.

Q7 P22-q07

A hamster with a mass of 30 g and a dog with a mass of 30 kg pass a cold night in the same barn. Both animals keep their bodies at about 38 °C.

Which animal loses more heat per gram of body each hour, and why?

A. The dog: it has more surface in total, so more heat leaves it
The dog loses more heat in total, but far less for each gram.
B. The dog: a larger body is warmer inside, so heat leaves each gram faster
Both animals are at about 38 °C inside.
C. ✓ The hamster: it has more surface for each gram of body, so each gram loses heat faster
D. Both animals lose the same per gram: they are at the same temperature in the same barn
Sharing a temperature and a barn does not give the two animals the same surface per gram.

Why: Heat leaves a body across its surface; the mass of the body makes the heat.
As an organism gets larger, its surface-area-to-volume ratio falls.
So the hamster has far more surface for each gram than the dog.
So each gram of hamster loses heat faster.

Q8 P22-q08

A scientist measures how fast a resting bat and a resting cow use energy. The bat has a mass of 10 g and the cow a mass of 600 kg. Per gram of body, the bat uses energy about 20 times as fast as the cow.

What does this show?

A. The bat's metabolic rate is 20 times the cow's in total
In total the cow uses far more energy; it is 60,000 times heavier.
B. Both animals have the same metabolic rate per gram; the bat is simply more active
The animals were measured at rest, and per gram they differ 20-fold.
C. The cow's cells use energy faster than the bat's
Per gram the bat uses energy 20 times as fast, so the bat’s cells use energy faster, not the cow’s.
D. ✓ Metabolic rate per gram of body is higher in the smaller organism

Why: Metabolic rate is the rate at which an organism uses energy.
Per gram of body, metabolic rate is typically higher in smaller organisms.
The bat is far smaller than the cow, so each gram of bat uses energy about 20 times as fast as each gram of cow.

Q9 P22-q09

A spherical yeast cell has a radius of 1 μm. For a sphere, surface area = 4πr² and volume = 4/3 πr³.

What are its surface area, volume and surface-area-to-volume ratio?

A. ✓ 12.6 μm², 4.19 μm³, 3.0 per μm
B. 12.6 μm², 12.6 μm³, 1.0 per μm
The volume has been given the same number as the surface area.
C. 3.14 μm², 4.19 μm³, 0.75 per μm
3.14 μm² is πr² alone, the area of a flat circle.
D. 12.6 μm², 4.19 μm³, 0.33 per μm
0.33 is the volume divided by the surface area.

Why: The ratio is the surface area divided by the volume.
So the cell has 3.0 μm² of surface for each μm³ inside; the working is below.

Q10 P22-q10

Two single-celled fungi take up nutrients across their surfaces. Cell J has a surface area of 80 μm² and a volume of 40 μm³. Cell K has a surface area of 120 μm² and a volume of 100 μm³.

Which cell exchanges nutrients with its surroundings more efficiently, for its size, and why?

A. K: it has more surface in total, 120 μm² against J's 80 μm²
More surface in total does not make exchange more efficient when the interior is also far larger.
B. ✓ J: its ratio is 2.0 per μm against K's 1.2 per μm, more surface per μm³
C. K: it has more volume, 100 μm³ against 40 μm³, so it holds more nutrients
More volume is more interior to supply, which makes exchange harder.
D. Neither: both are wrapped in the same kind of membrane
The membranes are alike but the amount of membrane per μm³ is not.

Why: Nutrients enter across the surface, and the need for them grows with the volume.
J has more surface for each μm³ of its volume.
So J exchanges nutrients more efficiently for its size; the working is below.

FRQ 1 P22-frq1 · Analyze Model or Visual Representation scaffolded

The model shows two cube-shaped cells from an early embryo. Cell A has sides 5 μm long; its surface area is 150 μm², its volume 125 μm³ and its surface-area-to-volume ratio 1.2 per μm. Cell B has sides 10 μm long; its values are left blank. For a cube of side s, surface area = 6s² and volume = s³. Everything a cell takes in, and every waste it gets rid of, crosses its surface.

Two cube-shaped cells from an early embryo, drawn to the same scale; cell B's values are left blank.

(a) Calculate the surface area and the volume of cell B, with units. (1 pt)

Frame Surface area of B = 6(…)² = … μm²; volume of B = (…)³ = … μm³

Hint Six faces for the surface; side × side × side for the volume.

Model answer Surface area of B = 6(10)² = 600 μm².
Volume of B = (10)³ = 1,000 μm³.

Working

Write down the values in the question:

s = 10 μm

Write down the equation:

tex: \text{surface area} = 6s^2

tex: \text{volume} = s^3

Substitute the values into the equation:

tex: \text{surface area} = 6(10)^2

tex: \text{volume} = (10)^3

Calculate:

tex: \text{surface area} = 600\,\text{μm}^2

tex: \text{volume} = 1{,}000\,\text{μm}^3

Rubric

Award 1 point for: surface area 600 μm² and volume 1,000 μm³, both with units.
Accept: 600 μm² and 1000 μm³. Do not award the point for 100 μm² (one face) or for volume given in μm².

Slip Giving 100 μm² for the surface. That is one face; a cube has six, and all six are surface.

(b) Calculate the surface-area-to-volume ratio of cell B, with its unit. (1 pt)

Frame Ratio of B = … divided by … = …, with the unit …

Hint Which quantity goes on top and which underneath? Then ask what unit is left when μm² is divided by μm³.

Model answer Ratio of B = 600 divided by 1,000 = 0.6, with the unit per μm.
The ratio is surface area divided by volume, and dividing μm² by μm³ leaves per μm.

Working

Write down the values in the question:

surface area = 600 μm²

volume = 1,000 μm³

Write down the equation:

tex: SA/V = \frac{\text{surface area}}{\text{volume}}

Substitute the values into the equation:

tex: SA/V = \frac{(600)}{(1{,}000)}

Calculate:

tex: SA/V = 0.6\,\text{per μm}

Rubric

Award 1 point for: 600 divided by 1,000 = 0.6 per μm (surface area divided by volume, with the unit).
Accept: 0.6 μm⁻¹. Do not award the point for the inverted ratio, 1.67.

Slip Dividing the volume by the surface area to get 1.67. The ratio is surface over volume: it says how much surface serves each μm³.

(c) Describe how the surface area, the volume and the ratio changed when the side doubled from 5 μm to 10 μm, using the numbers in the model. (1 pt)

Frame When the side doubled, the surface area went from … to …, a factor of …; the volume went from … to …, a factor of …; so the ratio …

Hint Divide each of B's values by A's to find the factor; then compare the two factors.

Model answer When the side doubled, the surface area went from 150 μm² to 600 μm².
That is a factor of 4.
The volume went from 125 μm³ to 1,000 μm³.
That is a factor of 8.
So the volume grew faster than the surface area.
So the ratio halved, from 1.2 per μm to 0.6 per μm.

Working

Write down the values in the question:

cell A: surface area = 150 μm², volume = 125 μm³, SA/V = 1.2 per μm

cell B: surface area = 600 μm², volume = 1,000 μm³, SA/V = 0.6 per μm

Write down the equation:

tex: \text{factor} = \frac{\text{value for cell B}}{\text{value for cell A}}

Substitute the values into the equation:

tex: \text{surface area factor} = \frac{(600)}{(150)}

tex: \text{volume factor} = \frac{(1{,}000)}{(125)}

tex: \text{ratio factor} = \frac{(0.6)}{(1.2)}

Calculate:

tex: \text{surface area factor} = 4

tex: \text{volume factor} = 8

tex: \text{ratio factor} = 0.5

Rubric

Award 1 point for: surface area ×4 (150 → 600 μm²), volume ×8 (125 → 1,000 μm³), so the volume grew faster than the surface and the ratio halved (1.2 → 0.6 per μm).
Accept: the factors 4 and 8 with the ratio described as falling or halving. Do not award the point for "both got bigger" with no factors.

Slip Saying both got bigger. The point needs the two factors, 4 and 8, or the ratio halving, to show that volume outran surface.

(d) Explain why the change in the ratio you described in (c) limits how large a cell can grow. (1 pt)

Frame A cell takes in what it needs only across its …, while the amount it needs grows with its …, so as it grows …

Hint What does each quantity in the ratio stand for in a living cell? Which one is the route in and out, and which one is the amount of cell that has to be supplied?

Model answer A cell takes in what it needs only across its surface.
The cell gets rid of waste only across its surface.
The amount it needs grows with its volume.
As the cell grows, its volume grows faster than its surface.
So less and less surface serves each μm³ of interior.
Past a certain size the surface can no longer bring in enough for the interior, and the cell cannot keep growing.

Rubric

Award 1 point for: a cell takes in what it needs and gets rid of waste only across its surface, while the amount it needs grows with its volume; as the cell grows its volume outruns its surface, so past a certain size the surface cannot supply the interior fast enough.
Accept: "supply grows with surface, demand grows with volume, and volume grows faster".

Slip Saying 'bigger cells need more' without linking surface to supply and volume to demand. The point is that the two grow at different rates.

(e) Predict what the embryo's cells do as they grow toward the size of cell B, and justify your prediction using the ratio. (1 pt)

Frame As the cells approach B's size they …, because …

Hint Think about what a growing cell can do so that its surface keeps up with its interior, and which of those an embryo's cells do.

Model answer As the cells approach B’s size, they divide rather than growing without limit.
A 10 μm cell has only 0.6 μm² of surface for each μm³ of interior.
That is half what a 5 μm cell has.
Dividing a 10 μm cell into smaller cells brings the ratio back up to 1.2 per μm.
So each new cell’s surface can again keep up with its interior.

Rubric

Award 1 point for: the cells divide (or stop growing) rather than growing without limit, because dividing a large cell into smaller ones raises the surface-area-to-volume ratio again (each 5 μm cell has 1.2 per μm against the 10 μm cell's 0.6), so the surface can keep up with the interior.
Accept: "they divide, restoring more surface per μm³". Do not award the point for "they grow bigger to get more surface".

Slip Predicting that the cells grow larger to gain surface. They do gain surface in total, but they gain volume faster; dividing is what restores surface per μm³.

FRQ 2 P22-frq2 · Conceptual Analysis

Two kinds of bacterium live in the same soil water and take up dissolved nutrients across their surfaces. Species R is rod-shaped and is modeled as a cylinder of radius 0.5 μm and length 2 μm. Species S is round and is modeled as a sphere of radius 0.75 μm. Formulas: cylinder, surface area = 2πrh + 2πr², volume = πr²h; sphere, surface area = 4πr², volume = 4/3 πr³.

(a) Calculate the surface-area-to-volume ratio of each cell. (1 pt)

Frame R: SA/V = … per μm. S: SA/V = … per μm.

Model answer R: SA/V = (2π(0.5)(2) + 2π(0.5)²) ÷ (π(0.5)²(2)) = 5.00 per μm.
S: SA/V = 4π(0.75)² ÷ (4/3 π(0.75)³) = 4.00 per μm.

Working

Write down the values in the question:

R: r = 0.5 μm, h = 2 μm

S: r = 0.75 μm

Write down the equation:

tex: \text{cylinder: } SA/V = \frac{2\pi rh + 2\pi r^2}{\pi r^2 h}

tex: \text{sphere: } SA/V = \frac{4\pi r^2}{\frac{4}{3}\pi r^3}

Substitute the values into the equation:

tex: \text{R: } SA/V = \frac{2\pi(0.5)(2) + 2\pi(0.5)^2}{\pi(0.5)^2(2)}

tex: \text{S: } SA/V = \frac{4\pi(0.75)^2}{\frac{4}{3}\pi(0.75)^3}

Calculate:

tex: \text{R: } SA/V = 5.00\,\text{per μm}

tex: \text{S: } SA/V = 4.00\,\text{per μm}

Rubric

Award 1 point for: R 5.0 per μm and S 4.0 per μm (surface area divided by volume; accept R 4.8–5.2 and S 3.8–4.2 per μm).
Accept: the ratios alone, with or without the surface area and volume written. Do not award the point for inverted ratios or for one cell only.

Slip Forgetting the two end circles of the cylinder (2πr²), or dividing volume by surface. Surface over volume, for both cells, with every face counted.

(b) Determine which species takes up nutrients more efficiently for its size, using the two ratios from (a). (1 pt)

Model answer Species R takes up nutrients more efficiently for its size.
R’s ratio is 5.0 per μm and S’s is 4.0 per μm.
So each μm³ of R has more membrane serving it.
Nutrients enter a cell only across its membrane.
The need for nutrients grows with the cell’s volume.
Therefore more surface for each μm³ of volume gives a faster supply.

Rubric

Award 1 point for: the decision that species R takes up nutrients more efficiently AND the reasoning it rests on: R's surface-area-to-volume ratio is higher (5.0 against 4.0 per μm), so R has more membrane for each μm³ of interior; nutrients enter only across the membrane while the need for them grows with the volume.
Accept: "R: more surface per unit of volume, so faster exchange for the same interior". Do not award the decision alone.

Slip Choosing S because its volume is larger. More volume is more interior to supply; what matters for efficiency is surface for each unit of volume.

(c) Predict what happens to the ratio of species S, and to how well its surface can supply its interior, if a cell doubles its radius to 1.5 μm before dividing. (1 pt)

Model answer Doubling the radius multiplies the surface area by 4, to 28.3 μm².
It multiplies the volume by 8, to 14.1 μm³.
So the ratio halves, to 2.0 per μm.
The surface supplies the interior less well, because each μm³ now has half as much surface to be supplied across.
So exchange cannot keep pace with the larger volume.

Working

Write down the values in the question:

r = 1.5 μm

Write down the equation:

tex: \text{surface area} = 4\pi r^2

tex: \text{volume} = \frac{4}{3}\pi r^3

tex: SA/V = \frac{4\pi r^2}{\frac{4}{3}\pi r^3}

Substitute the values into the equation:

tex: \text{surface area} = 4\pi(1.5)^2

tex: \text{volume} = \frac{4}{3}\pi(1.5)^3

tex: SA/V = \frac{4\pi(1.5)^2}{\frac{4}{3}\pi(1.5)^3}

Calculate:

tex: \text{surface area} = 28.3\,\text{μm}^2

tex: \text{volume} = 14.1\,\text{μm}^3

tex: SA/V = 2.00\,\text{per μm}

Rubric

Award 1 point for: the ratio halves to 2.0 per μm (surface ×4 to 28.3 μm², volume ×8 to 14.1 μm³), so the surface supplies the interior less well; exchange cannot keep pace with the larger volume.
Accept: "the ratio falls (halves) and exchange becomes less efficient" with the direction of both changes stated.

Slip Saying the ratio rises because the cell has more surface. It has more surface in total but far more volume, so less surface for each μm³.

(d) The cells lining a mammal's gut have surfaces folded into thousands of tiny projections. Support the claim that these folds use the same principle as the shapes of the two bacteria. (1 pt)

Model answer The folds are microvilli.
Microvilli add membrane surface to the cell while adding almost no volume.
So they raise the cell’s surface-area-to-volume ratio.
That is the same quantity that makes the rod-shaped R more efficient than the round S.
Therefore, in both cases, more surface for each unit of volume lets the membrane take up nutrients faster for the interior it serves.

Rubric

Award 1 point for: the evidence AND the reasoning: the folds (microvilli) add membrane surface at almost the same volume (evidence), so they raise the cell's surface-area-to-volume ratio, just as the rod shape gives R more surface per μm³ than the sphere; in both cases more surface for each unit of volume means faster exchange across the membrane (reasoning).
Accept: "folds add surface without adding volume; higher ratio; faster exchange", with the link to the bacteria's shapes stated. Do not award the point for "folds make the cell bigger", or for the principle stated with no link to the folds.

Slip Saying the folds make the cell larger. The volume barely changes; the folds add surface, and surface per unit of volume is what speeds exchange.

APBIO-U02-T22 End-of-topic test: Cell Size

Topic 2.2 · Cell Size · 17 MCQ · 2 FRQ

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 for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it. Formulas are given wherever a calculation needs them; use the π key on your calculator and give answers to three significant figures.

Q1 T22-q01

A cube-shaped cell has sides of 20 μm. For a cube of side s, surface area = 6s² and volume = s³.

What are its surface area and its volume?

A. 400 μm² and 8,000 μm³
400 μm² is the area of one face only.
B. ✓ 2,400 μm² and 8,000 μm³
C. 2,400 μm² and 400 μm³
400 is one face’s area, not the volume; the volume is the side length multiplied by itself three times.
D. 120 μm² and 8,000 μm³
120 μm² is six times the side length, not six times a face’s area.

Why: Each face of the cube is a 20 μm square, and a cube has six faces.
The volume is the side multiplied by itself three times; the working is below.

Q2 T22-q02

A cube-shaped cell has a surface area of 180 μm² and a volume of 150 μm³.

What is its surface-area-to-volume ratio?

A. ✓ 1.2 per μm
B. 0.83 per μm
0.83 is the volume divided by the surface area, the ratio the wrong way up.
C. 30 μm
30 is 180 − 150, a difference; a ratio compares two quantities by dividing, not by subtracting.
D. 1.2 μm²
The number is right but μm² divided by μm³ does not leave μm².

Why: The surface-area-to-volume ratio is surface area divided by volume.
180 μm² divided by 150 μm³ is 1.2 per μm.
The ratio says the cell has 1.2 μm² of surface for each μm³ of interior.

Q3 T22-q03

For a cell with a surface area of 60 μm² and a volume of 40 μm³, a student writes: $S A / V = \frac{40}{60} = 0.67 μm$ .

What is wrong with the student's working?

A. Nothing; 0.67 μm is correct
The student divided the volume by the surface area.
B. ✓ Upside down; it should be 60 divided by 40 = 1.5 per μm
C. It should be 60 − 40 = 20 μm
A ratio compares two quantities by dividing, not by subtracting.
D. Right division; the unit should be μm³
The division is upside down as well as the unit.

Why: The surface-area-to-volume ratio is surface area divided by volume.
60 μm² divided by 40 μm³ is 1.5 per μm.
The student divided the other way round.
Dividing μm² by μm³ leaves a unit of per μm, not μm.

Q4 T22-q04

An amoeba grows before it divides. Its width increases from 50 μm to 100 μm while its shape stays the same.

Compared with the smaller cell, how have its surface area, volume and surface-area-to-volume ratio changed?

A. Surface ×2, volume ×2, ratio unchanged
Doubling a length does not simply double an area or a volume.
B. Surface ×2, volume ×4, ratio halved
Surface area depends on two lengths, not one, so it is multiplied by 4.
C. ✓ Surface ×4, volume ×8, ratio halved
D. Surface ×8, volume ×4, ratio doubled
Area depends on two lengths (×4) and volume on three (×8); the ratio is area divided by volume, so it halves.

Why: An area depends on two lengths and a volume on three.
So doubling every dimension multiplies the surface area by 4 and the volume by 8, whatever the shape.
So the ratio halves; the working is below.

Q5 T22-q05

Four cells, K, L, M and N, are measured. Their surface areas and volumes are in the table.

Surface area and volume of the four cells.

Which cell exchanges materials with its surroundings most efficiently, and why?

A. K: it has the most surface
K has the most total surface, but that surface has to serve 200 μm³ of interior, only 1.2 per μm.
B. K: it has the most volume to fill
More volume means more interior needing supplies, not more exchange.
C. N: its surface is bigger than its volume
L and M also have more surface than volume; N gives only 120 divided by 80 = 1.5 per μm, less than M.
D. ✓ M: it has the most surface per unit of volume

Why: Everything a cell exchanges crosses its surface, and the amount it needs grows with its volume.
So the cell with the most surface for each μm³ exchanges most efficiently.
The ratios are K 1.2, L 2.0, M 3.0 and N 1.5 per μm, so M wins.

Q6 T22-q06

Cells in a plant tissue are modeled as cubes. A 10 μm cube’s membrane can bring in at most 180 units of oxygen a minute, and the cell uses 100 units a minute. Oxygen intake is proportional to membrane area; oxygen use is proportional to volume.

What percentage of its oxygen need can the membrane supply in a 20 μm cube and in a 30 μm cube?

A. 180% and 180%
Intake grows with area (×4 for a doubled side) and use with volume (×8), so the two do not grow by the same factor and the percentage falls.
B. 720% and 1,620%
Intake does rise to 720 and 1,620 units, but oxygen use rises too, to 800 and 2,700 units, because volume grows by 8 and 27.
C. ✓ 90% and 60%
D. 45% and 20%
Oxygen use grows with volume, by 8 and 27, not with area; the intake grows with area at the same time, by 4 and 9.

Why: Doubling the side multiplies area by 4 and volume by 8: intake 720 units, use 800 units, 90%.
Tripling the side multiplies area by 9 and volume by 27: intake 1,620 units, use 2,700 units, 60%.
The surface falls further behind as the cell grows.

Q7 T22-q07

A student writes: "A larger cell exchanges materials with its surroundings more efficiently than a smaller one, because it has more membrane."

Which statement about the larger cell corrects the student?

A. The larger cell has less membrane in total than a smaller cell, so it exchanges less efficiently
A larger cell does have more membrane in total.
B. The student is right: more membrane in total means more exchange with the surroundings
More membrane in total does not help when the interior has grown faster.
C. Size makes no difference to efficiency, since the membrane is the same thickness in both cells
Size makes a large difference, and membrane thickness is not the point.
D. ✓ The larger cell has more membrane, but far more interior to serve, so it exchanges less efficiently

Why: A cell takes in what it needs only across its surface, and the amount it needs grows with its volume.
As a cell grows, its volume grows faster than its surface.
So a larger cell has less surface for each μm³ and exchanges less efficiently, despite more total membrane.

Q8 T22-q08

Two samples of cells come from the lining of a kidney tubule, and the two kinds of cell have the same volume. In one sample the top face of each cell carries about a thousand finger-like folds of its plasma membrane; in the other, the top face is flat. Under the same conditions, the cells with the folded face take up glucose 3.1 times as fast.

Why do the cells with the folded face take up glucose faster?

A. The folds beat and sweep glucose into the cell
Folds of membrane do not move; cilia are the structures that beat.
B. ✓ The folds add membrane at the same volume: more surface for each unit of interior
C. The folds add volume, so the cell has more room to hold glucose
The two kinds of cell have the same volume, so the folds added membrane, not interior.
D. The folds make glucose move from low to high concentration into the cell
Nothing about a fold changes which way a substance diffuses.

Why: Each fold is a fold of the plasma membrane, so the folded face carries far more membrane.
The two kinds of cell have the same volume, so the folds add membrane without adding interior.
Glucose enters only across membrane: three times the membrane, three times the uptake.

Q9 T22-q09

On the underside of a leaf are thousands of tiny pores, each set between a pair of cells. When the pores are open, air reaches the many cell surfaces inside the leaf; when they are shut, the air stays outside.

What are the pores and the paired cells, and how do the cells open the pore?

A. Stomata and guard cells; the cells beat like cilia and fan air in through the pore
Guard cells do not beat; cilia are hair-like structures on other kinds of cell.
B. Stomata and guard cells; the cells lose water and shrink, pulling the pore open
Guard cells that lose water go limp and the pore closes.
C. ✓ Stomata and guard cells; the cells take in water, swell and bow apart, opening the pore
D. Microvilli and gut lining cells; the folds spread apart as the cells fill up with water
Microvilli are folds of a single gut cell’s membrane, not pores on a leaf.

Why: The pores are stomata; the two cells around each pore are guard cells.
Water entering the guard cells raises their turgor pressure.
So the guard cells swell and bow away from each other, and the pore opens.
Through the open pore, air reaches the leaf’s inner surface.

Q10 T22-q10

Root cells of the same plant each grow a single thin outgrowth into the soil. Cells whose outgrowths are 0.2 mm, 0.6 mm and 1.0 mm long take up water at 0.8, 1.9 and 3.1 units a minute. The outgrowths are all about the same width.

What is the outgrowth, and why does a longer one take up water faster?

A. ✓ A root hair; a longer one adds membrane across which water can enter
B. A root hair; a longer one reaches wetter soil, and its cell wall pumps the water in
A cell wall cannot pump anything; it is a stiff layer that lets water through.
C. A cilium; a longer one beats harder and drives water into the cell
Cilia are hair-like structures on the surface of some cells, not outgrowths a root cell grows into soil.
D. A root hair; a longer one holds more cytosol, which soaks up more water
A thin outgrowth adds very little volume.

Why: A root hair is a long, thin outgrowth of a root cell.
Because it is thin, almost everything a longer root hair adds is surface.
Water enters only across that surface, so a longer root hair has more membrane in the soil water and takes up more water each minute.

Q11 T22-q11

A hummingbird has a mass of about 5 g; a swan has a mass of about 10 kg. Both birds keep their bodies at about 40 °C while they rest in the same cool air. Four statements compare the heat the two birds lose to the air.

Which statement is wrong?

A. Each gram of the hummingbird loses heat faster than each gram of the swan
This statement is true: each gram of the hummingbird has far more skin to lose its heat through, so each gram loses heat faster.
B. The hummingbird has more skin for each gram of body than the swan has
This statement is true: a small body has more skin for each gram of body than a large body of the same shape.
C. The swan’s total heat loss is the greater, because it has far more skin altogether
This statement is true: the swan has far more skin altogether, so it loses more heat altogether.
D. ✓ The hummingbird loses more heat altogether, because each gram of it loses heat so fast

Why: Heat leaves a body across its skin.
The swan has far more skin altogether, so the swan loses more heat altogether.
The hummingbird has far more skin for each gram of body, so each gram of hummingbird loses heat faster.
Per gram, not in total.

Q12 T22-q12

Three mammals rest in the same cool room. The table gives each animal’s mass and its heat loss per gram of body per hour.

Mass and heat loss per gram for the three mammals.

What explains this pattern?

A. A larger body loses less heat in total, so each gram loses less
A larger body loses more heat in total, not less.
B. ✓ A smaller body has more surface for each gram of mass, so each gram loses heat faster
C. A smaller body is warmer inside, so heat leaves it faster
All three hold about the same body temperature.
D. A larger body has thicker skin, which is what slows heat loss
The same rule holds for animals with similar skin.

Why: Heat leaves a body across its surface; the mass of the body makes the heat.
As a body gets larger, its surface-area-to-volume ratio falls.
So each gram of a larger animal has less surface to lose heat across.
The shrew, the smallest, loses heat fastest per gram.

Q13 T22-q13

A hummingbird eats about half its body mass in food every day. A horse eats about 2% of its body mass a day. The food an animal eats each day supplies the energy it uses.

What does this show about the metabolic rates of the two animals?

A. The horse has the higher metabolic rate per gram of body
The horse eats far more in total but a far smaller share of its own mass.
B. Both animals have the same metabolic rate per gram; the hummingbird just eats more often
Eating half your mass a day against 2% is not the same energy use per gram.
C. ✓ The hummingbird has the higher metabolic rate per gram of body
D. The hummingbird has the higher metabolic rate in total
The horse eats kilograms of food a day, so in total it uses far more energy.

Why: The food an animal eats each day supplies the energy it uses.
So the share of its own mass an animal eats each day shows its metabolic rate per gram.
The hummingbird eats half its mass a day; the horse only 2%.
So the hummingbird’s rate per gram is higher.

Q14 T22-q14

A spherical cell has a radius of 10 μm. For a sphere, surface area = 4πr² and volume = 4/3 πr³.

What are its surface area, volume and surface-area-to-volume ratio?

A. ✓ 1,260 μm², 4,190 μm³, 0.30 per μm
B. 1,260 μm², 419 μm³, 3.0 per μm
419 μm³ uses r² in place of r³ in the volume.
C. 126 μm², 4,190 μm³, 0.030 per μm
126 μm² uses r in place of r² in the surface area.
D. 314 μm², 4,190 μm³, 0.075 per μm
314 μm² is πr², the area of a flat circle, not a sphere’s surface.

Why: The ratio is the surface area divided by the volume.
So the cell has 0.30 μm² of surface for each μm³ inside; the working is below.

Q15 T22-q15

A flattened cell is modeled as a rectangular solid 10 μm long, 5 μm wide and 0.5 μm thick. For a rectangular solid, surface area = 2lh + 2lw + 2wh and volume = lwh.

What are its surface area, volume and surface-area-to-volume ratio?

A. 57.5 μm², 25 μm³, 2.3 per μm
57.5 μm² counts only one of each pair of faces.
B. 115 μm², 25 μm³, 0.22 per μm
0.22 is the volume divided by the surface area.
C. 100 μm², 25 μm³, 4.0 per μm
100 μm² is just the two large faces and leaves out the four thin edge faces.
D. ✓ 115 μm², 25 μm³, 4.6 per μm

Why: A thin, flat shape carries a great deal of surface for its volume.
So its ratio is high; the working is below.

Q16 T22-q16

Here are two cells: a spherical cell 7 μm across (radius 3.5 μm) and a cube-shaped cell with sides of 11 μm. Sphere: surface area = 4πr², volume = 4/3 πr³; cube: 6s² and s³.

Which cell exchanges materials with its surroundings more efficiently, and why?

A. The cube: it has the larger surface area in total
The cube has more surface in total but far more volume for that surface to serve.
B. The cube: it has the larger volume, so more interior takes part in exchange
More volume means more interior needing supplies, not more exchange.
C. ✓ The sphere: its ratio is 0.86 per μm against the cube’s 0.55 per μm
D. The sphere: a sphere always has more surface for its size than a cube has
A sphere has the least surface of any shape for its volume; the sphere wins here only because it is much smaller than the cube.

Why: Exchange crosses the surface, and the need grows with the volume, so the cell with more surface for each μm³ exchanges more efficiently.
The sphere has the higher ratio; the working is below.

Q17 T22-q17

Two cells from a frog’s skin each have a volume of 120 μm³. One is flattened and has 160 μm² of plasma membrane; the other is nearly spherical and has 140 μm² of plasma membrane.

Which cell exchanges materials more efficiently, and what does that say about its shape?

A. ✓ The flattened cell, 1.33 against 1.17 per μm: flattening adds surface without adding volume
B. The rounder cell, 0.86 against 0.75 per μm: a rounder shape holds more surface
0.86 and 0.75 are volume divided by surface, the ratio the wrong way up.
C. Neither cell: the two volumes are equal, so the two ratios must be equal too
Equal volume fixes only the bottom of the ratio; the membrane areas differ.
D. The flattened cell, because being thinner it holds less volume for its membrane to serve
The two cells have the same volume, 120 μm³.

Why: Both cells have the same volume to serve.
A flattened shape spreads that volume under more surface, so flattening raises the ratio; the working is below.

FRQ 1 T22-frq1 · Analyze Model or Visual Representation

The model shows three cube-shaped plant cells with sides of 15 μm, 30 μm and 60 μm, with the surface area and volume of the first two; the values for the 60 μm cell are left blank. For a cube of side s, surface area = 6s² and volume = s³. The 15 μm cell’s surface-area-to-volume ratio is 0.4 per μm and the 30 μm cell’s is 0.2 per μm. Everything a cell takes in, and every waste it gets rid of, crosses its surface.

Three cube-shaped model cells drawn to the same scale, with the surface area and volume of the first two.

(a) Describe how the surface area and the volume change each time the side of the cell doubles, using the numbers for the 15 μm and 30 μm cells. (1 pt)

Model answer When the side doubles from 15 μm to 30 μm, the surface area changes from 1,350 μm² to 5,400 μm².
That is a factor of 4.
The volume changes from 3,375 μm³ to 27,000 μm³.
That is a factor of 8.
So the volume grows faster than the surface area.
So the surface-area-to-volume ratio halves, from 0.4 per μm to 0.2 per μm.

Working

Write down the values in the question:

15 μm cell: surface area = 1,350 μm², volume = 3,375 μm³

30 μm cell: surface area = 5,400 μm², volume = 27,000 μm³

Write down the equation:

tex: \text{factor} = \frac{\text{value for the 30 μm cell}}{\text{value for the 15 μm cell}}

Substitute the values into the equation:

tex: \text{surface area factor} = \frac{(5{,}400)}{(1{,}350)}

tex: \text{volume factor} = \frac{(27{,}000)}{(3{,}375)}

Calculate:

tex: \text{surface area factor} = 4

tex: \text{volume factor} = 8

Rubric

Award 1 point for: the surface area is multiplied by 4 (1,350 → 5,400 μm²) while the volume is multiplied by 8 (3,375 → 27,000 μm³), so the volume grows faster than the surface area (or: the ratio halves, 0.4 → 0.2 per μm).
Accept: "surface ×4, volume ×8" with the numbers from the model cited. Do not award the point for "both get bigger" or "the volume gets bigger" with no factors or numbers.

Slip Saying both get bigger. The point needs the factors, 4 for surface and 8 for volume, or the ratio halving.

(b) Calculate, for the 60 μm cell, its surface area, its volume and its surface-area-to-volume ratio, each with its unit. (1 pt)

Model answer The 60 μm cell has a surface area of 21,600 μm², a volume of 216,000 μm³ and a surface-area-to-volume ratio of 0.1 per μm.

Working

Write down the values in the question:

s = 60 μm

Write down the equation:

tex: \text{surface area} = 6s^2

tex: \text{volume} = s^3

tex: SA/V = \frac{6s^2}{s^3}

Substitute the values into the equation:

tex: \text{surface area} = 6(60)^2

tex: \text{volume} = (60)^3

tex: SA/V = \frac{6(60)^2}{(60)^3}

Calculate:

tex: \text{surface area} = 21{,}600\,\text{μm}^2

tex: \text{volume} = 216{,}000\,\text{μm}^3

tex: SA/V = 0.1\,\text{per μm}

Rubric

Award 1 point for: surface area 21,600 μm², volume 216,000 μm³, ratio 0.1 per μm (all three, with units).
Accept: 0.1 μm⁻¹ or "1 : 10" for the ratio. Do not award the point if the ratio is inverted (10) or if the units are missing from all three values.

Slip Inverting the ratio to 10, or leaving the units off. The ratio is surface area divided by volume, in per μm.

(c) Explain why the pattern you described in (a) limits how large a cell can grow. (1 pt)

Model answer A cell takes in what it needs only across its surface.
The cell gets rid of waste only across its surface.
The amount it needs, and the waste it makes, grow with its volume.
As the cell grows, its volume grows faster than its surface.
So past a certain size the surface cannot supply the volume, and the cell cannot keep growing.

Rubric

Award 1 point for: a cell takes in what it needs and gets rid of waste only across its surface, while the amount it needs (or the waste it makes) grows with its volume; so as the cell grows its surface cannot keep up with (cannot supply) its interior.
Accept: "there is not enough membrane to supply the volume" or "exchange across the surface cannot keep pace with the demand of the interior". Do not award the point for "the cell gets too big" with no reference to both surface and volume.

Slip Saying the cell gets too big. The point needs both halves: exchange happens across the surface, and need grows with the volume.

(d) Explain how the pattern in the model relates to the way living cells are built. (1 pt)

Model answer The ratio falls as a cell grows.
So a cell stays small, or divides, rather than growing without limit.
A small cell has enough surface to supply its interior.
A cell built for exchange goes further.
The gut-lining cell folds its membrane into microvilli.
The folds add surface while adding almost no volume.
So the folds raise the cell’s ratio, and exchange keeps pace with the interior.

Rubric

Award 1 point for EITHER: because the ratio falls as a cell grows, a cell stays small or divides rather than growing without limit, so its surface keeps up with its interior; OR: folding the membrane (microvilli, root hairs, a flattened shape) adds surface while adding almost no volume, raising the surface-area-to-volume ratio so that exchange keeps pace with the interior.
Accept: either link made in full (the pattern in the model tied to the reason). Do not award the point for naming microvilli or division without saying what they do to surface relative to volume.

Slip Naming microvilli or division without saying what they do to surface relative to volume. Tie the structure to the ratio.

FRQ 2 T22-frq2 · Conceptual Analysis

In pond water live a single-celled alga, modeled as a sphere of radius 2 μm, and a rod-shaped bacterium, modeled as a cylinder of radius 0.4 μm and length 4 μm. Both cells take up dissolved nutrients across their surfaces. Formulas: sphere, surface area = 4πr², volume = 4/3 πr³; cylinder, surface area = 2πrh + 2πr², volume = πr²h.

(a) Calculate the surface-area-to-volume ratio of each cell. (1 pt)

Model answer Alga: SA/V = 4π(2)² ÷ (4/3 π(2)³) = 1.50 per μm.
Bacterium: SA/V = (2π(0.4)(4) + 2π(0.4)²) ÷ (π(0.4)²(4)) = 5.50 per μm.

Working

Write down the values in the question:

alga: r = 2 μm

bacterium: r = 0.4 μm, h = 4 μm

Write down the equation:

tex: \text{sphere: } SA/V = \frac{4\pi r^2}{\frac{4}{3}\pi r^3}

tex: \text{cylinder: } SA/V = \frac{2\pi rh + 2\pi r^2}{\pi r^2 h}

Substitute the values into the equation:

tex: \text{alga: } SA/V = \frac{4\pi(2)^2}{\frac{4}{3}\pi(2)^3}

tex: \text{bacterium: } SA/V = \frac{2\pi(0.4)(4) + 2\pi(0.4)^2}{\pi(0.4)^2(4)}

Calculate:

tex: \text{alga: } SA/V = 1.50\,\text{per μm}

tex: \text{bacterium: } SA/V = 5.50\,\text{per μm}

Rubric

Award 1 point for: alga 1.5 per μm and bacterium 5.5 per μm (surface area divided by volume; accept alga 1.4–1.6 and bacterium 5.3–5.7 per μm).
Accept: the ratios alone, with or without the surface area and volume written. Do not award the point for inverted ratios (volume divided by surface) or for one cell only.

Slip Dividing volume by surface area, or working out one cell only. Surface over volume, for both cells.

(b) Determine which cell exchanges nutrients with the water more efficiently, for its size, using the ratios from (a). (1 pt)

Model answer The bacterium exchanges nutrients more efficiently for its size.
Everything a cell takes in crosses its surface.
The amount it needs grows with its volume.
The bacterium has 5.5 μm² of surface for each μm³ of volume, and the alga has 1.5 μm².
So each μm³ of the bacterium is supplied across almost four times as much membrane.
Therefore the bacterium is the more efficient exchanger.

Rubric

Award 1 point for: the decision that the bacterium exchanges more efficiently AND the reasoning it rests on: it has almost four times as much surface for each unit of volume (5.5 against 1.5 per μm), and everything a cell takes in crosses its surface while what it needs grows with its volume.
Accept: "the bacterium; a higher surface-area-to-volume ratio means more membrane per unit of interior to exchange across". Do not award the point for "the bacterium because it is smaller" with no link to surface per unit of volume, or for the decision alone.

Slip Saying the bacterium because it is smaller, with no link to surface per unit of volume. Small helps only because it raises the ratio.

(c) Predict what happens to the alga's surface-area-to-volume ratio, and to how well its surface can supply its interior, if it grows to a radius of 6 μm before dividing. (1 pt)

Model answer Tripling the radius multiplies the surface area by 9, to about 452 μm².
It multiplies the volume by 27, to about 905 μm³.
So the ratio falls to a third, 0.50 per μm.
The surface supplies the interior less well, because each μm³ of the alga now has a third as much surface to be supplied across.
So exchange cannot keep pace with the larger volume.

Working

Write down the values in the question:

r = 6 μm

Write down the equation:

tex: \text{surface area} = 4\pi r^2

tex: \text{volume} = \frac{4}{3}\pi r^3

tex: SA/V = \frac{4\pi r^2}{\frac{4}{3}\pi r^3}

Substitute the values into the equation:

tex: \text{surface area} = 4\pi(6)^2

tex: \text{volume} = \frac{4}{3}\pi(6)^3

tex: SA/V = \frac{4\pi(6)^2}{\frac{4}{3}\pi(6)^3}

Calculate:

tex: \text{surface area} = 452\,\text{μm}^2

tex: \text{volume} = 905\,\text{μm}^3

tex: SA/V = 0.500\,\text{per μm}

Rubric

Award 1 point for: the ratio falls to a third, 0.50 per μm (surface ×9 to about 452 μm², volume ×27 to about 905 μm³), so the surface supplies the interior less well; exchange cannot keep pace with the larger volume.
Accept: "the ratio falls (to a third) and exchange becomes less efficient" with the direction of both changes stated. Do not award the point for "the ratio rises because the cell has more surface".

Slip Saying the ratio rises because the cell has more surface. It has more surface in total but far more volume, so less surface for each μm³.

(d) A related alga has the same volume as the 2 μm-radius sphere (33.5 μm³) but grows as a flat disk 0.5 μm thick, with about 150 μm² of surface. Evaluate the claim that this flat shape is an aid to exchange. (1 pt)

Model answer The claim is supported.
The disk has the same volume as the sphere, 33.5 μm³.
But it has about 150 μm² of surface against the sphere’s 50.3 μm².
So the disk has about three times the surface for the same volume: a ratio of about 4.5 per μm against 1.5 per μm.
Nutrients cross into a cell only across its surface.
Therefore the disk takes in nutrients faster for the same interior, and the flat shape aids exchange.

Working

Write down the values in the question:

surface area = 150 μm²

volume = 33.5 μm³

Write down the equation:

tex: SA/V = \frac{\text{surface area}}{\text{volume}}

Substitute the values into the equation:

tex: SA/V = \frac{(150)}{(33.5)}

Calculate:

tex: SA/V = 4.5\,\text{per μm}

Rubric

Award 1 point for: the judgement that the claim is supported AND the ground for it: at the same volume, the disk has about three times the surface (150 μm² against 50.3 μm²), so its ratio is about 4.5 per μm against 1.5 per μm; more surface per unit of volume means nutrients can be exchanged faster for the same interior.
Accept: the ground stated as "more surface for the same volume, so a higher ratio and faster exchange" without the 4.5 figure. Do not award the judgement alone, or "flat cells are smaller", or a judgement with no comparison of surface to volume.

Slip Saying flat cells are smaller. The volume is the same; the shape adds surface, and that is what raises the ratio.

APBIO-U02-L17 Two kinds of cell

Topic 2.1 · Cell Structure and Function · 95 steps

An onion skin cell 100 μm long with a dark round body inside it, and beside it a bacterium 2 μm long drawn at the same scale as a small speck; the same bacterium is drawn again, magnified, in a circle at the right

Here are two cells as they would look under one microscope: an onion skin cell 100 μm long with a dark round body inside it, and, beside it, a bacterium 2 μm long, the speck. The circle shows that same bacterium magnified, so that you can see its shape.

Every living thing is made of one or more cells. These two look nothing alike. What do they share, and what splits them?

Unit 2 · Cell Structure and Function

1Ribosomes

Video: Watch first: two cells under one microscope

One cell’s insides, seen whole before any part is studied alone; which body does which job?

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T21-intro.mp4

Check q1

Which of the following is a polypeptide?

A. A chain of sugar units
A chain of sugar units is a polysaccharide.
B. ✓ A chain of amino acids
C. A chain of nucleotides
A chain of nucleotides is a nucleic acid, such as DNA or RNA.

Why: A polypeptide is a chain of amino acids.
A protein is made of one or more polypeptides.

Check q2

Which of the following is RNA?

A. ✓ A chain of nucleotides
B. A chain of amino acids
A chain of amino acids is a polypeptide.
C. A chain of sugar units
A chain of sugar units is a polysaccharide.

Why: RNA is a nucleic acid.
A nucleic acid is a chain of nucleotides.

Video: Watch: Ribosomes

One particle drawn large: an mRNA strand runs between its two lumps and a chain of amino acids grows out, one amino acid at a time.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L17.mp4

Inside a cell are huge numbers of tiny particles, far smaller than the cell, each built from RNA and protein. The particles drift through the cytosol.

Here is one such particle from a human cell, as a scientist’s model built from thousands of electron-microscope pictures. An electron microscope shows things far smaller than a light microscope can.

A model of a human ribosome: a smaller orange lump on top of a larger blue lump, with thin colored threads winding through both — A model of a human ribosome built from electron-microscope images: a smaller orange lump sitting on a larger blue lump, with thin colored threads winding through both

Here is the same particle drawn simply. A strand of another RNA runs between its two lumps.

A ribosome drawn simply: a large lump above a small lump, both of rRNA and protein; an mRNA strand runs between the two lumps; a chain of amino acids, drawn as beads, grows out of the top of the large lump

A chain of amino acids grows out of the particle, one amino acid at a time.

A small particle of RNA and protein, not enclosed in any membrane, that makes proteins by joining amino acids is called a .

The RNA a ribosome is built from is called .

The strand that runs through the ribosome is called a . An mRNA is a copy of the instructions for one protein, carried from the cell’s DNA.

The ribosome joins the amino acids in the order the mRNA sets. So each mRNA gives one particular protein.

A pancreas cell that makes large amounts of digestive protein is packed with ribosomes.

What you are expected to know Describe a ribosome: a particle of rRNA and protein, not enclosed in a membrane, that makes a protein by joining amino acids in the order an mRNA sets.

Check q3

Which of the following does a ribosome do?

A. Stores the cell’s DNA until the cell needs it
The cell’s DNA stays where it is.
Only a copy of the instructions for one protein, an mRNA, comes to the ribosome.
B. ✓ Joins amino acids into a protein, in the order an mRNA sets
C. Wraps finished proteins in a membrane and ships them out
A ribosome has no membrane of its own.
D. Joins sugar units into starch for the cell to store
A ribosome joins amino acids, not sugar units.

Why: An mRNA runs through the ribosome.
The mRNA carries the order of amino acids for one protein.
The ribosome joins amino acids one by one in that order.
So a ribosome makes a protein.

Check q4

Which of the following describes how a ribosome is built?

A. ✓ rRNA and protein, with no membrane around it
B. Protein only, inside its own membrane
A ribosome contains rRNA as well as protein.
C. DNA and protein, inside its own membrane
The nucleic acid in a ribosome is rRNA, not DNA.
D. rRNA only, with no protein
A ribosome is built from rRNA and from protein.

Why: A ribosome is built from ribosomal RNA and protein.
No membrane encloses a ribosome.

18Ribosomes in every cell

Video: Watch: Ribosomes in every cell

A bacterium’s ribosome beside a human cell’s: the same build, and what a shared feature is evidence of.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L17b.mp4

Here is a ribosome from a bacterium beside one from a human cell, each drawn from its measured structure. Both ribosomes are built from rRNA and protein, and both have the same build: a smaller lump and a larger lump.

A ribosome from the bacterium E. coli: a large purple-blue lump above a smaller green lump, with a red strand of mRNA between them and yellow shapes reaching into the gap — Two pictures side by side: a ribosome from the bacterium E. coli, drawn from its measured structure as a large purple-blue lump over a smaller green lump with a red strand between them; and a model of a ribosome from a human cell, a smaller orange lump on a larger blue lump

A ribosome from a human cell: a smaller orange lump on top of a larger blue lump, with thin colored threads winding through both — Two pictures side by side: a ribosome from the bacterium E. coli, drawn from its measured structure as a large purple-blue lump over a smaller green lump with a red strand between them; and a model of a ribosome from a human cell, a smaller orange lump on a larger blue lump

Every known cell, from bacteria to humans, contains ribosomes of the same basic build.

A feature shared by all living things is evidence that all living things descend from shared ancestors. Descent from shared ancestors is called .

In plain words: every living thing alive today, from bacteria to humans, descended from the same first living cells, and evolved from that one starting point.

Ribosomes are one such shared feature. DNA as the store of inherited instructions is another, and so is the plasma membrane.

Every known cell has all three. So all living things descended from the same first cells.

What you are expected to know Describe the build every known cell’s ribosomes share: rRNA and protein, a smaller lump and a larger lump.

What you are expected to know Explain why a feature shared by all living things is evidence of common ancestry.

Check q5

The ribosomes of a bacterium and the ribosomes of a human cell have the same basic build.

Which of the following is this evidence of?

A. Bacteria are descended from humans
Bacteria were on Earth billions of years before humans, so a shared feature was inherited from an ancestor both descend from, not passed from one to the other.
B. Each kind of cell developed its own ribosomes separately
Two separate inventions would not be expected to match in build.
C. The two cells make the same proteins
The two cells make different proteins.
The ribosome itself is what is the same, not the proteins the ribosome makes.
D. ✓ All living things descend from shared ancestors

Why: Both ribosomes are built from rRNA and protein.
A feature shared by all living things is evidence that all living things descend from shared ancestors.
So ribosomes of one basic build in every cell are evidence of common ancestry.

Check q6

A newly discovered single-celled organism is brought up from a deep-sea vent.

Which of the following is the organism most likely to contain?

A. No ribosomes; it would make proteins another way
No known cell makes proteins without ribosomes.
Every known cell has ribosomes of one basic build.
B. Ribosomes only if it turns out to be a bacterium
Every known cell has ribosomes, from bacteria to humans.
C. ✓ Ribosomes built from rRNA and protein
D. Ribosomes built from DNA instead of rRNA
No known ribosome is built from DNA.
The shared rRNA-and-protein build is exactly the evidence for common ancestry.

Why: Every known cell, from bacteria to humans, contains ribosomes built from rRNA and protein.
A newly discovered cell is expected to match every other known cell.
So the new organism most likely contains ribosomes built from rRNA and protein.

30A prokaryotic cell

Video: Watch: A prokaryotic cell

The bacterium drawn large and labeled, its DNA lying in a nucleoid with no membrane around it.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L17c.mp4

Now the bacterium, drawn large. Everything inside the bacterium is in constant motion.

A bacterium about 2 μm long: cell wall, plasma membrane inside it, cytosol full of ribosomes drawn as small dots, and a nucleoid where the DNA, one closed loop, lies with no membrane around it

Outside is its cell wall. Just inside the cell wall is its plasma membrane.

Inside the membrane is the cytosol. In the cytosol are thousands of ribosomes, drawn here as small dots.

In the middle is a region where the bacterium’s DNA lies. The DNA is usually a single circular molecule: one closed loop, drawn here as the loop of thread.

Nothing wraps that DNA. There is no membrane around it; the DNA simply lies in the cytosol.

The region of a prokaryotic cell where its DNA lies, with no membrane around it, is called the .

A nucleoid is not a body with an edge. A nucleoid is only the part of the cytosol where the DNA is.

Here is a real bacterium, cut through and photographed through an electron microscope. The whole cell is about 2 μm long.

A rod-shaped bacterium cut lengthways under the electron microscope: a thin dark boundary, a dark speckled interior and pale patches inside — Electron-microscope photograph of a rod-shaped bacterium cut lengthways: a thin dark boundary around it, a dark speckled interior, and several pale patches inside

The dark speckles filling the real cell are its ribosomes. The pale patches are its nucleoid.

Now look again at the dark round body in the onion cell, drawn large here. That body is the cell’s nucleus: the part that holds the cell’s DNA, its inherited instructions, wrapped in a membrane.

A close-up of part of the onion skin cell: pale cytosol with a few small dots, and the dark round body drawn with two rings around it; labels name the body as the nucleus, say that the DNA is inside it, and point at the membrane around it

The bacterium has no such body. A cell whose DNA is not enclosed in a nucleus is called a , or prokaryote. The name means ‘before a nucleus’: this kind of cell has no nucleus.

Bacteria are prokaryotes. So are archaea, a second group of single-celled organisms with no nucleus.

A prokaryote has DNA, and a prokaryote has ribosomes. Its DNA simply has no membrane around it.

What you are expected to know Identify, in a drawing of a bacterium, its cell wall, plasma membrane, cytosol, ribosomes and nucleoid.

What you are expected to know Explain what a nucleoid is: the region of a prokaryotic cell where its DNA lies, with no membrane around it.

What you are expected to know Identify a cell whose DNA is not enclosed in a nucleus as prokaryotic: bacteria and archaea.

Check q7

Here is a bacterium with two positions marked.

A bacterium with two positions marked: X on the closed loop of thread in the middle of the cell, Y on one of the small dots

Which of the following is the structure at X?

A. ✓ The nucleoid
B. The nucleus
A nucleus is DNA enclosed in a membrane.
C. A ribosome
The ribosomes are the many small dots, such as Y.
D. The cell wall
The cell wall is the outer layer around the whole cell.

Why: X is the closed loop of thread in the middle of the bacterium.
That loop is the bacterium’s DNA.
No membrane surrounds the DNA.
So X is the nucleoid.

Check q8

Which of the following is a nucleoid?

A. A small nucleus in a bacterium, enclosed in a single membrane
A nucleoid has no membrane around it at all.
B. A particle of rRNA and protein in a bacterium that makes its proteins
A particle of rRNA and protein is a ribosome.
C. ✓ The region of a bacterium where its DNA lies, with no membrane around it
D. The rigid layer outside a bacterium’s plasma membrane, around the cell
The rigid layer outside the plasma membrane is the cell wall.

Why: A bacterium is a prokaryotic cell.
Its DNA lies in one region of the cytosol.
No membrane surrounds that region.
That region is the nucleoid.

Check q9

Which of the following statements about a bacterium is correct?

A. The bacterium’s DNA is enclosed in a nucleus too small to see
A bacterium’s DNA lies in a nucleoid, with no membrane around it.
A nucleus is DNA enclosed in a membrane.
B. ✓ The bacterium has DNA and ribosomes, but no nucleus
C. The bacterium has no ribosomes, since it has no nucleus
Every known cell has ribosomes, bacteria included.
Ribosomes have no membrane and need no nucleus.
D. The bacterium has a nucleus, since every cell has one
Not every cell has a nucleus.
A prokaryotic cell keeps its DNA in a nucleoid with no membrane around it.

Why: A bacterium has DNA.
Its DNA lies in a nucleoid, with no membrane around it.
So the bacterium has no nucleus.
Every known cell has ribosomes.
So the bacterium has DNA and ribosomes, but no nucleus.

51The nucleus, its envelope and the organelles

Video: Watch: The nucleus, its envelope and the organelles

The onion cell drawn large: the nucleus inside its double membrane, and the membrane-wrapped bodies around it.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L17d.mp4

Now the onion skin cell, drawn large. Everything inside the onion cell is in constant motion too.

An onion skin cell about 100 μm long: cell wall, plasma membrane, ribosomes, membrane-wrapped bodies, and a nucleus wrapped in a double membrane, the nuclear envelope

Its boundary is a cell wall with the plasma membrane just inside, as in the bacterium. Its cytosol has ribosomes too.

The dark round body is the nucleus. Look at its edge: two membranes, one inside the other.

That double membrane is called the .

The nucleus holds the cell’s DNA, its inherited instructions. The nucleus is not the cell’s brain: the nucleus holds the instructions, it does not think.

Around the nucleus are other bodies. Each body is wrapped in its own membrane, and each body does a particular job for the cell.

A small body inside a cell that does a particular job is called an .

An organelle wrapped in its own membrane is called a . The nucleus is one, and so is each of the other organelles around it.

Two other membrane-bound organelles appeared in earlier lessons: the vacuole of a plant cell, and the vesicles that carry material into and out of a cell.

A ribosome has no membrane around it. So a ribosome is not a membrane-bound organelle.

What you are expected to know Identify the nucleus and its nuclear envelope in a drawing of a plant or animal cell.

What you are expected to know Identify the other membrane-bound organelles in the drawing, such as the vacuole and the vesicles.

What you are expected to know Explain what an organelle is.

Check q10

Here is the onion skin cell with two positions marked, X and Y.

An onion skin cell with two positions marked: the lead for X ends on the outer of two concentric rings around the large round body at the right; the lead for Y ends on the thin line just inside the thick outer border, at the bottom

Which of the following is the structure at X?

A. The nucleus
The nucleus is the whole dark round body.
B. The plasma membrane
The plasma membrane is the cell’s outer boundary, at Y.
C. ✓ The nuclear envelope
D. The nucleoid
A nucleoid is DNA with no membrane around it.

Why: X is on the two membranes, one inside the other, around the nucleus.
That double membrane is the nuclear envelope.

Check q11

Which of the following is a membrane-bound organelle?

A. A ribosome
A ribosome has no membrane around it.
B. The cytosol
The cytosol is the fluid the organelles sit in.
C. The cell wall
The cell wall lies outside the plasma membrane.
D. ✓ The nucleus

Why: The nucleus is a body inside the cell.
The nucleus does a job: holding the DNA.
The nucleus is wrapped in its own membrane, the nuclear envelope.
So the nucleus is a membrane-bound organelle.

68Prokaryotic or eukaryotic?

Video: Watch: Prokaryotic or eukaryotic?

Four cells sorted live by two steps: find the DNA, look for a membrane around it.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L17e.mp4

A cell whose DNA sits inside a nucleus, with membrane-bound organelles around it, is called a , or eukaryote. The name means ‘true nucleus’.

The cells of plants and animals are eukaryotic. So are the cells of fungi, such as yeast and mushrooms.

So are protists, single-celled eukaryotes such as the Paramecium: one cell, with a nucleus.

A cell whose DNA lies in a nucleoid, and which typically has no membrane-bound organelles, is prokaryotic: bacteria and archaea.

Only where the DNA sits decides it. A yeast cell is small and lives alone, and a yeast cell is eukaryotic: its DNA is inside a nucleus.

What splits cells into prokaryotic and eukaryotic is whether the DNA sits inside a nucleus and the interior is divided by membranes.

Summary: a prokaryotic cell, with its DNA drawn as one closed loop in a nucleoid and usually no membrane-bound organelles, beside a eukaryotic cell, with its DNA in a nucleus and membrane-bound organelles

A cell with no nucleus still has DNA. The DNA lies in the nucleoid.

To classify a cell, use two steps. First, find the DNA. Second, look for a membrane around the DNA.

If a nucleus wraps the DNA, the cell is eukaryotic. If the DNA lies in a nucleoid, with no membrane around it, the cell is prokaryotic. Nothing else about the cell decides it.

What you are expected to know Classify a shown cell as eukaryotic (DNA inside a nucleus, membrane-bound organelles: plants, animals, fungi, protists) or prokaryotic (a nucleoid, typically no membrane-bound organelles: bacteria and archaea).

80Fluency quiz: prokaryotic or eukaryotic? mixed practice

Check q12

Here is a cell 8 μm long, drawn as it appears through a microscope.

A cell 8 μm long: a cell wall (lead ends on the cell wall), a plasma membrane just inside it (lead ends on the membrane line), many ribosomes drawn as small dots, and its DNA drawn as one closed loop of thread in the middle of the cell; a scale bar under the cell shows 8 μm

Is this cell prokaryotic or eukaryotic?

A. ✓ Prokaryotic
B. Eukaryotic
The DNA lies as a closed loop in the middle of the cell with no membrane around it: a nucleoid.

Why: First, find the DNA: the DNA lies as a closed loop in the middle of the cell.
Second, look for a membrane around the DNA: there is none.
So the DNA is in a nucleoid.
So the cell is prokaryotic.

Check q13

Here is a cell about 100 μm long, drawn as it appears through a microscope.

A cell about 100 μm long: a cell wall, a plasma membrane just inside it, many ribosomes, several oval bodies each drawn with its own outline, and its DNA inside a large round body at the right that is drawn with two concentric rings; a scale bar at the right shows 30 μm

Is this cell prokaryotic or eukaryotic?

A. Prokaryotic
The DNA is inside the large round body drawn with two rings, a double membrane: a nucleus.
B. ✓ Eukaryotic

Why: First, find the DNA: the DNA is inside the large round body at the right.
Second, look for a membrane around the DNA: the body is drawn with two rings, a double membrane.
So the DNA is inside a nucleus.
So the cell is eukaryotic.

Check q14

Here is a single-celled organism 7 μm across, drawn as it appears through a microscope.

A round cell 7 μm across: a cell wall, a plasma membrane just inside it, many ribosomes, two small oval bodies each with its own outline, and its DNA inside a round body drawn with two concentric rings; a scale bar under the cell shows 5 μm

Is this cell prokaryotic or eukaryotic?

A. Prokaryotic
The DNA sits inside a body drawn with two rings, a double membrane, so the DNA is in a nucleus.
B. ✓ Eukaryotic

Why: First, find the DNA: the DNA is inside the round body drawn with two rings.
Second, look for a membrane around the DNA: the two rings are a double membrane.
So the DNA is inside a nucleus.
So the cell is eukaryotic.

Check q15

An archaeon from a hot spring is 1 μm long. Its DNA lies in a nucleoid.

Is the archaeon prokaryotic or eukaryotic?

A. ✓ Prokaryotic
B. Eukaryotic
The DNA lies in a nucleoid, and a nucleoid has no membrane around it.

Why: First, find the DNA: the DNA lies in a nucleoid.
Second, look for a membrane around the DNA: a nucleoid has no membrane around it.
So the archaeon is prokaryotic.

Check q16

A Paramecium is a single cell 200 μm long. Its DNA is enclosed by a nuclear envelope.

Is the Paramecium prokaryotic or eukaryotic?

A. Prokaryotic
A nuclear envelope is the double membrane of a nucleus, so the DNA is inside a nucleus.
B. ✓ Eukaryotic

Why: First, find the DNA: the DNA is enclosed by a nuclear envelope.
Second, look for a membrane around the DNA: a nuclear envelope is the double membrane of a nucleus.
So the DNA is inside a nucleus.
So the Paramecium is eukaryotic.

Check q17

A cell 3 μm long from pond water has a cell wall and ribosomes. Its DNA lies in a region with no membrane around it.

Is this cell prokaryotic or eukaryotic?

A. ✓ Prokaryotic
B. Eukaryotic
The DNA lies in a region with no membrane around it: a nucleoid.

Why: First, find the DNA: the DNA lies in a region in the cell.
Second, look for a membrane around the DNA: there is none.
So the DNA is in a nucleoid.
So the cell is prokaryotic.

87Only where the DNA sits decides it

Video: Watch: Only where the DNA sits decides it

One yeast cell made small, alone and walled while the verdict stays: only where the DNA sits decides it.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L17f.mp4

Students often sort a cell by its size, or by whether it lives alone, or by whether it has a cell wall. None of these decides the kind of cell.

A yeast cell is about 5 μm across and lives alone. A yeast cell is eukaryotic, because its DNA is inside a nucleus.

A bacterium has a cell wall, and a plant cell has a cell wall too. So a cell wall tells you nothing about which kind of cell you have.

What you are expected to know Explain that only where the DNA sits decides whether a cell is prokaryotic or eukaryotic, not its size, whether it lives alone or whether it has a cell wall.

Check q18

A student says: “A yeast cell is only 5 μm across and lives on its own, so a yeast cell must be prokaryotic.”

Is the student correct?

A. Yes
Only where the DNA sits decides the kind of cell, and a yeast cell’s DNA is inside a nucleus.
B. ✓ No

Why: Only where the DNA sits decides the kind of cell.
A yeast cell’s DNA is inside a nucleus.
So a yeast cell is eukaryotic, whatever its size and however it lives.

Check q19

A yeast cell is eukaryotic.

Which of the following is the reason?

A. A yeast cell has a cell wall
Bacteria have cell walls too, and bacteria are prokaryotic.
B. A yeast cell has ribosomes
Every known cell has ribosomes, bacteria included.
C. ✓ A yeast cell’s DNA is inside a nucleus

Why: Only where the DNA sits decides the kind of cell.
A yeast cell’s DNA is inside a nucleus.
So a yeast cell is eukaryotic.

Glossary

ribosome: A small particle of ribosomal RNA and protein, not enclosed in any membrane, that makes proteins by joining amino acids in the order a messenger RNA sets. Every known cell has ribosomes.
rRNA and mRNA: Ribosomal RNA (rRNA) is the RNA a ribosome is built from. Messenger RNA (mRNA) is a copy of the instructions for one protein, carried from the cell’s DNA to a ribosome, which joins amino acids in the order the mRNA sets.
common ancestry: Descent from shared ancestors. A feature shared by all living things, such as ribosomes of one basic build, is evidence of it.
prokaryotic cell (prokaryote): A cell whose DNA is not enclosed in a nucleus; it lies in a nucleoid, and the cell typically has no membrane-bound organelles. Bacteria and archaea are prokaryotes.
nucleoid: The region of a prokaryotic cell where its DNA, usually a single circular molecule, lies with no membrane around it. It is not a body with an edge, only the part of the cytosol where the DNA is.
organelle: A small body inside a cell that does a particular job. A membrane-bound organelle is one wrapped in its own membrane, such as the nucleus.
nuclear envelope: The double membrane, two membranes one inside the other, that encloses the nucleus.
eukaryotic cell (eukaryote): A cell whose DNA sits inside a nucleus and whose interior holds membrane-bound organelles. The cells of plants, animals, fungi and protists are eukaryotic.

APBIO-U02-L17B Where a cell makes the proteins it sends out

Topic 2.1 · Cell Structure and Function · 19 steps

A pancreas cell cut open, packed with folded membrane and dark packets of protein, beside a duct; small dots leave the cell and travel along the duct toward the gut

After a meal, your pancreas pours digestive proteins into your gut. Those proteins cut the starch, fats and proteins in your food into small molecules that your gut can absorb.

A pancreas cell made every one of those digestive proteins. Where in the cell? And how does a protein made inside a cell get out of it?

Unit 2 · Cell Structure and Function

1Proteins that leave the cell

Video: Watch: Proteins that leave the cell

A pancreas cell pours digestive proteins into the gut; every one was built, folded and wrapped inside the cell, on one network of folded membrane.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L17Ba.mp4

Here is a pancreas cell photographed through an electron microscope. An electron microscope shows things far smaller than a light microscope can.

A pancreas cell under the electron microscope: a large gray oval body in the middle, stripes of folded membrane filling the cell around it, and dark round blobs at the right — Electron-microscope photograph of a pancreas cell: a large gray oval body in the middle, layered stripes of membrane filling the cell around it, and several dark round blobs at the right edge

The large gray body in the middle is the nucleus. The dark round blobs are packets of digestive protein, waiting to be released.

Between them, stripes fill the cell: membrane, folded layer upon layer. The cell makes its digestive proteins on those folded membranes.

A protein that leaves a cell needs three things done to it.

Three panels in a row: a chain of beads being joined; the same chain folded into a compact shape; the folded shape inside a small circle of membrane

First, the protein is built: amino acids are joined into a chain. Second, the chain is folded into its working shape.

Check q1

A vesicle carrying insulin fuses with a pancreas cell’s plasma membrane and releases the insulin outside the cell.

Which of the following is this called?

A. Endocytosis
Endocytosis brings material into the cell, wrapped in a vesicle.
B. ✓ Exocytosis
C. Diffusion
Diffusion is the net movement of particles down a concentration gradient; no vesicle takes part.

Why: The vesicle fuses with the plasma membrane.
The vesicle’s contents spill outside the cell.
Releasing material from a vesicle this way is exocytosis.

Third, the folded protein is wrapped in membrane. Only a protein wrapped in membrane can leave the cell by exocytosis.

All three steps happen on and inside one network of folded membrane. That network is what fills the pancreas cell in the photograph.

The network of folded membrane where a cell builds, folds and wraps the proteins it exports is called the , or ER.

So the ER’s purpose, in one sentence: the ER is where a cell makes the proteins it sends out, and gets them ready to leave.

A pancreas cell exports huge amounts of protein every day. So a pancreas cell is packed with ER, as the photograph shows.

A pancreas cell also makes insulin, the protein it releases into the blood, on the same network, in the same way.

What you are expected to know Explain why a cell that exports protein needs a place to build, fold and wrap that protein.

What you are expected to know State what the endoplasmic reticulum is for: making the proteins a cell exports and getting them ready to leave.

Check q2

A cell in a salivary gland releases a digestive protein into saliva.

Which of the following is the job of the cell’s endoplasmic reticulum?

A. Holding the cell’s DNA, its inherited instructions
The nucleus holds the cell’s DNA.
B. Pumping the finished protein across the plasma membrane
The protein leaves in a vesicle that fuses with the plasma membrane; no pump carries it across.
C. ✓ Making the protein and getting it ready to leave

Why: The digestive protein leaves the cell.
A protein that leaves a cell is built, folded and wrapped in membrane on the ER.
So the ER’s job is to make the protein and get it ready to leave.

Check q3

Two cells are the same size. One is a pancreas cell that exports digestive protein all day. The other is a cell that exports no protein.

Which cell has more endoplasmic reticulum?

A. ✓ The pancreas cell
B. The cell that exports no protein
A cell that exports no protein has little use for the network that builds, folds and wraps protein for export.
C. Both cells have the same amount
The two cells do different amounts of one job, so they carry different amounts of the network that does that job.

Why: The ER is where a cell makes and wraps the proteins it exports.
The pancreas cell exports protein all day.
So the pancreas cell has more ER.

Glossary

endoplasmic reticulum (ER): A network of folded membrane inside a cell, where the cell makes the proteins it exports and gets them ready to leave: built from amino acids, folded, and wrapped in membrane.

APBIO-U02-L18 A folded network with ribosomes on it

Topic 2.1 · Cell Structure and Function · 44 steps

A pancreas cell cut open, packed with folded membrane, with one stretch of the membrane enlarged to show ribosomes on its outer surface

A cell in your pancreas makes insulin all day and sends it out into the blood. Insulin is a protein, and the cell makes it, folds it and wraps it in membrane on one network of folded membrane: the endoplasmic reticulum.

Inside the cell, much of that network is studded with ribosomes on its outside. What does the network look like up close, and why do the ribosomes sit on it?

Unit 2 · Cell Structure and Function

1Made on the network, folded inside it

Video: Watch: Made on the network, folded inside it

The endoplasmic reticulum branches out from the nuclear envelope; ribosomes on its rough part pass new protein into its interior to be folded.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L18a.mp4

A pancreas cell makes insulin, a protein, and sends it out of the cell. Every protein a cell exports is built, folded and wrapped in membrane on one network of folded membrane: the endoplasmic reticulum, or ER.

That is the ER’s purpose. Now its structure.

Here is the ER in a pancreas cell, drawn large. It branches out from the nuclear envelope, the double membrane around the nucleus, as a network of connected membrane sacs and tubes.

The nucleus with its double-membrane envelope, and a branching network of connected membrane tubes branching out from it

The membrane of the network is continuous with the outer membrane of the nuclear envelope: one folded sheet.

Over much of the network, ribosomes sit on the outer surface, the side facing the cytosol. Under the microscope that surface looks rough.

A stretch of the network in cross-section: ribosomes sit on its outer surface, facing the cytosol; the ER interior is shaded

Here is rough ER as it really looks through an electron microscope: membranes studded with ribosomes.

Rough ER under the electron microscope: parallel wavy membranes, each studded along its surface with small dark dots — Electron-microscope photograph of rough ER: many roughly parallel dark wavy lines, each line studded along both sides with small dark dots

The part of the ER with ribosomes on it is called the .

A ribosome joins amino acids into a protein. A ribosome on the rough ER passes the protein it is making through the membrane into the ER’s interior.

A ribosome on the rough ER passing a new protein chain into the ER's interior (shaded), where it is folded

The ER’s interior is a compartment separate from the cytosol. In the ER’s interior the new protein is folded before the cell sends the protein on.

A small piece of the ER membrane bulges outward, then pinches off as a closed sac with folded protein inside. Pinching off a piece of membrane as a closed sac is called budding off.

A vesicle pinching off from the end of an ER sac and moving away, carrying a folded protein

A small sac of membrane like this is a vesicle. The vesicles that bud off the ER carry the folded proteins onward.

The network also runs through much of the cell and helps the cell keep its shape.

A pancreas cell that makes insulin for export is crowded with rough ER: many ribosomes, all passing protein into the interior.

What you are expected to know Identify the endoplasmic reticulum in a cell drawing.

What you are expected to know Describe the rough ER: the part of the ER with ribosomes on its outer surface.

What you are expected to know Explain where a protein made on the rough ER goes: into the ER’s interior, apart from the cytosol, to be folded.

Check q1

A cell in a salivary gland makes a digestive protein that it will release into saliva.

Where is the protein made?

A. On a ribosome loose in the cytosol
A loose ribosome leaves its protein in the cytosol, and a protein bound for saliva must enter the ER to be exported.
B. ✓ On a ribosome on the rough ER
C. In the nucleus
The nucleus holds the DNA.
The nucleus makes no proteins.

Why: Ribosomes make proteins.
The ribosomes that make protein for export sit on the rough ER.
So the digestive protein is made on a ribosome on the rough ER.

Check q2

The salivary gland cell’s ribosome has just made the digestive protein.

Where does the protein go the moment the ribosome finishes it?

A. Into the cytosol
The ribosome sits on the rough ER and passes the protein through the ER membrane as it makes it, into the ER’s interior.
B. Straight out of the cell
The ribosome passes the protein into the ER’s interior; leaving the cell comes much later, by exocytosis.
C. ✓ Into the ER’s interior

Why: The ribosome sits on the rough ER.
The ribosome passes the protein through the ER membrane as it makes it.
So the protein goes into the ER’s interior, a compartment separate from the cytosol.
In the ER’s interior the protein is folded.

Check q3

The drawing shows part of a liver cell: a network of membrane tubes branching out from the nuclear envelope, with small dark dots along the tubes.

Part of a liver cell: a round body wrapped in a double membrane at the left, labelled nuclear envelope (the lead ends on the outer membrane), a branching network of connected membrane tubes branching out from it to the right, and small dark dots along the outside of the tubes

Which of the following is the network?

A. The plasma membrane
The plasma membrane is the cell’s outer boundary, not a network inside the cell.
B. The nuclear envelope
The nuclear envelope is only the double membrane around the nucleus.
C. The smooth ER
Smooth ER has a bare surface; these tubes carry dots, which are ribosomes.
D. ✓ The rough ER

Why: A network of membrane tubes branching out from the nuclear envelope is the ER.
The dots on the outer surface of the tubes are ribosomes.
ER with ribosomes on it is rough ER.
So the network is the rough ER.

22Where the ribosomes stop

Video: Watch: Where the ribosomes stop

Follow the same tube past the last ribosome: smooth ER, which makes membrane lipids and breaks down harmful molecules.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L18b.mp4

Follow the network further and the ribosomes stop. Here the membrane is bare, and the tubes look smooth.

One continuous membrane tube, its interior shaded: ribosomes on its surface along the left half, a bare surface along the right half

The part of the ER with no ribosomes on it is called the .

With no ribosomes, the smooth ER makes no protein. The smooth ER has two other jobs.

The smooth ER makes new lipids for the cell’s membranes.

The smooth ER breaks down harmful molecules. A liver cell that breaks down alcohol has abundant smooth ER.

To tell rough ER from smooth ER, look at the surface. Ribosomes on the surface: rough ER. A bare surface: smooth ER.

What you are expected to know Identify smooth ER by its bare surface.

What you are expected to know State the two jobs of the smooth ER: making membrane lipids and breaking down harmful molecules.

32Fluency quiz: rough ER or smooth ER? mixed practice

Check q4

Here is one stretch of a cell’s ER.

One stretch of membrane tube seen in section, its interior shaded, with a row of small dots sitting on its upper surface

Is this stretch rough ER or smooth ER?

A. ✓ Rough ER
B. Smooth ER
Small dots sit on the upper surface of the tube; those dots are ribosomes, and ER with ribosomes is rough ER.

Why: Look at the surface.
Small dots sit on the upper surface of the tube.
Those dots are ribosomes.
ER with ribosomes on its surface is rough ER.

Check q5

Here is a stretch of a cell’s ER.

One stretch of membrane tube seen in section, its interior shaded, with nothing on its surface

Is this stretch rough ER or smooth ER?

A. Rough ER
Nothing sits on the surface of the tube; ER with a bare surface is smooth ER.
B. ✓ Smooth ER

Why: Look at the surface.
Nothing sits on the surface of the tube.
ER with a bare surface is smooth ER.

Check q6

Here are two stretches of the same membrane network, X and Y.

Two stretches of the ER network, marked X and Y, each a membrane tube seen in section with its interior shaded; one stretch carries a row of small dots along its upper surface, the other carries none

Which stretch is smooth ER?

A. ✓ X
B. Y
Y carries a row of dots on its surface.

Why: Look at the surfaces.
X has a bare surface.
Y carries a row of dots, which are ribosomes.
ER with a bare surface is smooth ER.
So X is smooth ER.

Check q7

A stretch of ER in a liver cell makes membrane lipids and breaks down harmful molecules.

Is this stretch rough ER or smooth ER?

A. Rough ER
Rough ER carries ribosomes and makes protein.
B. ✓ Smooth ER

Why: Making membrane lipids and breaking down harmful molecules are the smooth ER’s two jobs.
So this stretch is smooth ER.

Check q8

A stretch of ER in a pancreas cell passes new insulin into the ER’s interior.

Is this stretch rough ER or smooth ER?

A. ✓ Rough ER
B. Smooth ER
Insulin is a protein, only a ribosome makes a protein, and ribosomes sit on the rough ER.

Why: Insulin is a protein.
Only a ribosome makes a protein.
Ribosomes sit on the rough ER, and they pass the protein into the ER’s interior.
So this stretch is rough ER.

Check q9

A stretch of ER has no ribosomes on its surface.

Is this stretch rough ER or smooth ER?

A. Rough ER
Rough ER has ribosomes on its surface; this stretch has none.
B. ✓ Smooth ER

Why: Rough ER has ribosomes on its surface.
This stretch has none.
ER with no ribosomes on it is smooth ER.

39What the smooth ER does

Video: Watch: What the smooth ER does

A liver cell and a pancreas cell side by side: each is crowded with the part of the ER that does its job.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L18c.mp4

A liver cell that breaks down alcohol every day is crowded with smooth ER. A pancreas cell that exports insulin is crowded with rough ER. Each cell has more of the ER that does its job.

What you are expected to know Explain why a cell has more of the part of the ER that does its job: rough ER to make protein for export, smooth ER to make membrane lipids and break down harmful molecules.

Check q10

A stretch of ER has no ribosomes on it: it is smooth ER.

Which of the following does this stretch do?

A. Makes the proteins that the cell exports
Only ribosomes make proteins.
B. Holds the cell’s DNA
The nucleus holds the cell’s DNA.
C. ✓ Makes membrane lipids and breaks down harmful molecules

Why: Smooth ER has no ribosomes on it.
So smooth ER makes no protein.
The smooth ER has two jobs.
The smooth ER makes new membrane lipids.
The smooth ER also breaks down harmful molecules such as alcohol.

Glossary

rough ER: The part of the endoplasmic reticulum with ribosomes on its outer surface. Proteins made there pass into the ER’s interior to be folded.
smooth ER: The part of the endoplasmic reticulum with no ribosomes on it. It makes membrane lipids and breaks down harmful molecules.

APBIO-U02-L18B Where the bubbles from the ER go

Topic 2.1 · Cell Structure and Function · 70 steps

A pancreas cell cut open: a folded network of membrane at the left, small bubbles leaving it, a stack of flat pouches in the middle, and bubbles heading for the cell surface at the right

A cell in your pancreas makes insulin, a protein, and sends it out into the blood. The insulin is built and folded on a network of folded membrane inside the cell. Then small bubbles carrying it pop out of that network.

But a protein fresh from that network is not finished. And it has to end up at the cell surface, nowhere else. Between the network and the surface sits a stack of flat pouches. The network makes and folds the protein. The stack finishes it, sorts it and packs it for delivery.

Unit 2 · Cell Structure and Function

1Small bubbles leave the folded network

Inside a pancreas cell there is a network of folded membrane. A membrane is a thin sheet of lipid that fences off one space from another.

Here is a piece of that network photographed through an electron microscope.

Electron-microscope photograph of the folded network inside a cell: many dark wavy lines running side by side, each line with dots along its edges; two labels below point at one line and at a run of dots

Each dark wavy line is one membrane. The dots along the membranes are ribosomes.

This network is the endoplasmic reticulum, or ER.

Endo means inside. Plasmic means the cell’s cytoplasm, the jelly-like material inside the cell.

Reticulum means a little net. So endoplasmic reticulum means a little net inside the cell’s cytoplasm.

Ribosomes on the ER make proteins. The ER folds each protein into its working shape.

Then small bubbles pop out of the folded network.

Three wavy membranes of the folded network with ribosome dots on them; at the right end of the middle membrane a bubble is pinching off; further right a free bubble holds a folded protein

Each bubble is a piece of the ER’s membrane that has pinched off. Inside it is a folded protein.

A small bubble of membrane with something inside it is a vesicle.

So a folded protein leaves the ER inside a vesicle. The protein travels inside the vesicle, never loose in the cytosol.

What you are expected to know Describe how a protein leaves the ER: a small bubble of membrane, a vesicle, pinches off the ER with the protein inside.

Check q1

What is a vesicle?

A. A ribosome
A ribosome is the small round body that makes proteins.
B. ✓ A small bubble of membrane
C. A folded protein
A folded protein is what travels inside the bubble.

Why: A piece of membrane pinches off, closed round its contents.
That small bubble of membrane is a vesicle.

Video: Watch: Small bubbles leave the folded network

Ribosomes on the folded network make a protein. The network folds it. A small bubble of membrane pinches off around the protein and carries it away.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L18Ba-r7.mp4

Check q2

What is a membrane?

A. ✓ A thin sheet of lipid that fences off a space
B. A chain of amino acids joined in a row
A chain of amino acids joined in a row is a protein.
C. A small round body inside the cell
A membrane is a sheet, not a body.

Why: A membrane is a thin sheet of lipid.
The sheet fences off the space on one side from the space on the other.

Check q3

What is the endoplasmic reticulum?

A. The cell’s outer boundary
The plasma membrane is the cell’s outer boundary.
B. The store of the cell’s DNA
The nucleus stores the cell’s DNA.
C. ✓ A network of folded membrane inside the cell

Why: The endoplasmic reticulum is the network of folded membrane inside the cell.
Ribosomes on it make proteins.
The network folds them.

Check q4

What does the name endoplasmic reticulum mean?

A. ✓ A little net inside the cell’s cytoplasm
B. A rigid cell wall around the cell
Endo means inside, not around.
C. A stack of flat pouches
Reticulum means a little net, not a stack.

Why: Endo means inside.
Plasmic means the cell’s cytoplasm.
Reticulum means a little net.
So the name means a little net inside the cell’s cytoplasm.

Check q5

A ribosome on the ER of a salivary gland cell has just finished a protein, and the ER has folded the protein.

How does the protein leave the ER?

A. Loose through the cytosol
A protein bound for the cell surface is never loose in the cytosol.
It stays inside membrane the whole way.
B. ✓ Inside a vesicle that pinches off the ER
C. Straight through the ER membrane
A folded protein is far too large to pass through a membrane on its own.

Why: The folded protein sits inside the ER.
A piece of the ER membrane pinches off around it as a closed bubble: a vesicle.
So the protein leaves the ER inside a vesicle.

Practice writing an answer

A vesicle carrying a folded protein has just left the ER of a liver cell.

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

Model answer A vesicle is a small bubble of membrane with something inside it.

Rubric

Award 1 point for: a small bubble (or sac) of membrane with contents inside.
Accept: a piece of membrane pinched off as a closed bubble around its contents.

Slip Calling the vesicle a protein. The protein is the cargo; the vesicle is the bubble of membrane around it.

21The stack of pouches has a name: the Golgi complex

Here is a pancreas cell cut open. Between the folded network and the cell’s surface sits a stack of flat pouches, one on top of another.

A pancreas cell cut open: the nucleus at the left, the folded network beside it, small bubbles, a stack of flat pouches between the network and the cell surface, and the cell surface at the right

Each pouch is a flat bag of membrane, like a balloon with the air let out. Four to eight pouches sit in one stack.

Here is one such stack photographed through an electron microscope.

Electron-microscope photograph of a stack of long dark curved lines lying side by side, with small circles around them; two labels below point at one curved line and at one small circle

Each long dark curved line is one flat pouch seen edge-on. The small circles around the stack are vesicles.

The vesicles that leave the ER travel to this stack.

A stack of flat membrane pouches that receives the vesicles from the ER is called the .

Golgi is a person’s name: Camillo Golgi was the scientist who first saw the stack. Complex means several parts working as one.

What you are expected to know Identify the Golgi complex in a drawing of a cell: a stack of flat membrane pouches between the rough ER and the cell surface.

Check q6

What is the Golgi complex?

A. A network of folded membrane with ribosomes on it
A network of folded membrane with ribosomes on it is the rough ER.
B. A small bubble of membrane
A small bubble of membrane is a vesicle.
C. ✓ A stack of flat membrane pouches

Why: The Golgi complex is the stack of flat membrane pouches.
The vesicles from the ER travel to it.

Check q7

Where does the Golgi complex sit?

A. ✓ Between the rough ER and the cell surface
B. Inside the nucleus
The nucleus holds the cell’s DNA, not the stack.
C. Outside the plasma membrane
The plasma membrane is the cell’s boundary.
The Golgi complex sits inside it, in the cytoplasm.

Why: The vesicles from the ER travel to the Golgi complex.
From the Golgi complex, vesicles travel on to the cell surface.
So the Golgi complex sits between the rough ER and the cell surface.

Check q8

Here is the pancreas cell again, with four of its parts lettered.

The same pancreas cell cut open, with four of its parts marked by lettered circles: A on the folded network, B on the stack of flat pouches, C on a small bubble, D on the round body at the left

Which lettered part is the Golgi complex?

A. A
A network of folded membrane with ribosomes on it is the ER.
B. ✓ B
C. C
A small bubble of membrane is a vesicle.
D. D
The round body holding the cell’s DNA is the nucleus.

Why: The Golgi complex is the stack of flat membrane pouches.
It sits between the ER and the cell surface.

Practice writing an answer

A gland cell sends protein to its surface by way of its Golgi complex.

(a) State what the Golgi complex is. (1 pt)

Model answer The Golgi complex is a stack of flat membrane pouches that receives the vesicles from the ER.

Rubric

Award 1 point for: a stack of flat membrane pouches that receives the vesicles from the ER.
Accept: a stack of flat membrane pouches between the ER and the cell surface.

Slip Describing a network of folded membrane with ribosomes on it. That is the rough ER; the Golgi complex is a stack of separate flat pouches.

34The Golgi complex finishes each protein

A vesicle from the ER reaches the stack. The vesicle’s membrane joins the membrane of the first pouch: the vesicle fuses with the pouch.

A stack of five flat pouches; at the left a vesicle is merging into the first pouch, its protein inside; the label says the protein is folded but not yet finished

The protein inside the vesicle spills into the first pouch.

The protein that arrives is folded, but it is not finished.

The Golgi complex finishes the protein. The Golgi complex checks the protein’s folding.

The Golgi complex also changes the protein chemically: it attaches small chemical groups to the protein.

So the ER makes and folds the protein. The Golgi complex then finishes it.

What you are expected to know Describe what the Golgi complex does to a protein that arrives from the ER: it checks the folding and changes the protein chemically, so the protein is finished.

Check q9

In a tear gland cell, a vesicle from the ER has just fused with the first pouch of the stack, and its protein has spilled into the pouch.

Which of the following does the Golgi complex do to that protein?

A. Builds it from amino acids
Ribosomes build proteins from amino acids.
The protein was already built when it arrived.
B. ✓ Checks its folding and changes it chemically
C. Stores it for the rest of the cell’s life
The protein moves on from the Golgi complex within minutes.

Why: The protein arrives folded but not finished.
The Golgi complex checks the protein’s folding.
The Golgi complex also attaches small chemical groups to the protein.
So the Golgi complex finishes the protein.

Video: Watch: The Golgi complex finishes each protein

A vesicle from the ER fuses with the first pouch. The Golgi complex checks the protein’s folding and attaches small chemical groups. Now the protein is finished.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L18Bc.mp4

Check q10

A protein made in a salivary gland cell passes through the Golgi complex on its way to the cell surface.

Why does the protein pass through the Golgi complex?

A. ✓ The protein is folded but not finished when it leaves the ER
B. The protein is not yet folded when it leaves the ER
The ER folds the protein before the protein leaves.
C. The protein is too large to leave the ER any other way
The protein leaves the ER inside a vesicle; its size does not send it to the Golgi complex.

Why: The ER folds the protein.
A folded protein from the ER is still not finished.
The Golgi complex checks the folding and changes the protein chemically.
So the protein passes through the Golgi complex to be finished.

Check q11

A student says: “The Golgi complex makes the cell’s proteins.”

Is the student correct?

A. Yes: the Golgi complex joins amino acids into proteins
Ribosomes join amino acids into a protein.
The protein is already built when it reaches the Golgi complex.
B. ✓ No: ribosomes make the proteins, and the Golgi complex finishes them

Why: Only ribosomes join amino acids into a protein.
The protein reaches the Golgi complex already built and folded.
The Golgi complex checks the folding and changes the protein chemically.
So the Golgi complex finishes proteins; it does not make them.

46Packed into another vesicle, addressed

The finished protein reaches the far side of the stack: the pouch nearest the cell surface.

The same stack of five flat pouches; at the right end of the last pouch another bubble is pinching off with a finished protein inside; a free vesicle further right heads for the cell surface

There a piece of the last pouch’s membrane pinches off around the protein. Another vesicle forms.

This is not the vesicle the protein arrived in: that vesicle became part of the first pouch.

The protein leaves in another vesicle.

Each vesicle that leaves is addressed to one destination. For a salivary protein, the destination is the cell surface.

Proteins for different destinations go into different vesicles. So the Golgi complex sorts proteins as well as finishing them.

So the ER makes and folds a protein. The Golgi complex finishes it, sorts it, and packs it into another vesicle for delivery.

What you are expected to know Describe how a finished protein leaves the Golgi complex: packed into another vesicle that pinches off the far side of the stack, addressed to one destination.

Check q12

A finished protein leaves the Golgi complex of a tear gland cell.

How does the protein leave?

A. Loose through the cytosol
A protein bound for the cell surface is never loose in the cytosol.
B. In the same vesicle it arrived in
The vesicle the protein arrived in fused with the first pouch and became part of it.
C. ✓ In another vesicle that pinches off the far side

Why: The finished protein reaches the last pouch, on the far side of the stack.
A piece of that pouch’s membrane pinches off around the protein.
So the protein leaves in another vesicle.

Video: Watch: Packed into another vesicle, addressed

On the far side of the stack another vesicle pinches off around the finished protein. It is addressed to one destination. The ER makes and folds; the Golgi complex finishes, sorts and packs.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L18Bd.mp4

57A sugar chain is attached: glycosylation

The commonest change the Golgi complex makes is to attach a short chain of sugar units to the protein.

A folded protein with a short chain of three sugar units attached to its upper right side

Attaching a chain of sugar units to a protein is called .

Glyco means sugar; -ation means the doing of it. So glycosylation means the attaching of sugar.

Check q13

A protein with a carbohydrate chain attached is called which of the following?

A. ✓ A glycoprotein
B. A phospholipid
A phospholipid is a membrane lipid with a phosphate head and two fatty-acid tails.
C. A polysaccharide
A polysaccharide is a long chain of sugar units with no protein.

Why: Glyco means sugar.
A protein with a sugar chain attached is a glycoprotein.

The sugar chain has a job. The sugar chain helps decide where the protein goes, or how the protein works.

What you are expected to know Describe glycosylation: attaching a short chain of sugar units to a protein, which helps decide where the protein goes or how it works.

Check q14

What is glycosylation?

A. Folding a protein into its working shape
The ER folds the protein.
B. ✓ Attaching a chain of sugar units to a protein
C. Breaking a protein into amino acids
Breaking a protein into amino acids is hydrolysis, not glycosylation.

Why: Glyco means sugar.
Glycosylation is attaching a chain of sugar units to a protein.

Check q15

Which of the following does the sugar chain do for the protein?

A. Supplies the protein with energy
The sugar chain is not broken down for energy; it stays on the protein.
B. Holds the protein’s amino acids together
Bonds between the amino acids hold the protein together, not the sugar chain.
C. ✓ Helps decide where the protein goes or how it works

Why: The Golgi complex attaches the sugar chain.
The chain helps decide where the protein goes, or how the protein works.

Practice writing an answer

A protein at the surface of a liver cell carries a short chain of sugar units that it did not have when it left the ER.

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

Model answer Glycosylation is the attaching of a chain of sugar units to a protein.

Rubric

Award 1 point for: attaching a carbohydrate (sugar) chain to a protein.
Accept: adding a sugar chain to a protein, as the Golgi complex does.

Slip Saying glycosylation is making sugar. The sugar units already exist; glycosylation attaches them to the protein.

Check q16

In a patient’s cells, ribosomes on the rough ER make a protein bound for the cell surface as normal, and the protein leaves the ER in vesicles as normal. But none of the protein ever reaches the surface. Inside the cells the protein piles up in the Golgi complex, folded correctly and already carrying its sugar chain.

Which job of the Golgi complex has failed?

A. Fusing with the vesicles that arrive from the ER
The protein is inside the Golgi complex, so the vesicles from the ER did fuse with it.
B. Checking the protein’s folding
The protein in the Golgi complex is folded correctly, so the folding check did its job.
C. Attaching a sugar chain to the protein
The protein already carries its sugar chain, so that job was done.
D. ✓ Packing the protein into vesicles addressed to the cell surface

Why: The protein reached the Golgi complex, so the vesicles from the ER were received.
The protein is folded correctly and carries its sugar chain, so the Golgi complex finished it.
The one Golgi job left is packing the protein into a vesicle addressed to the cell surface.
That job failed.

The rough ER makes and folds insulin. A vesicle carries it to the Golgi complex.

A pancreas cell with insulin's route drawn through it: rough ER at the left, a vesicle, the Golgi complex, another vesicle, the plasma membrane at the right, each stop labelled with what happens there

The Golgi complex finishes it, sorts it and packs it into another vesicle. That vesicle carries it to the cell surface.

Glossary

Golgi complex: A stack of flat membrane pouches between the rough ER and the cell surface. It receives the vesicles from the ER, finishes and sorts the proteins inside, and packs them into new vesicles addressed to their destinations. Golgi is the name of Camillo Golgi, the scientist who first saw the stack; complex means several parts working as one.
glycosylation: Attaching a chain of sugar units to a protein, the commonest change the Golgi complex makes. Glyco means sugar; -ation means the doing of it. The chain helps decide where the protein goes or how it works.

APBIO-U02-L18C A sac that digests

Topic 2.1 · Cell Structure and Function · 63 steps

A white blood cell cut open: a rod-shaped bacterium sits inside a bubble of membrane near the cell's edge, and a small sac lies nearby inside the cell

A white blood cell has just swallowed a bacterium whole. The bacterium sits inside the cell, wrapped in a bubble of the cell’s own membrane.

An hour from now the bacterium will be gone, taken apart into small molecules. The white blood cell itself will be unharmed. How does a cell take a bacterium apart without taking itself apart? The cell keeps its digesting tools sealed inside a small sac, and brings the sac to the bacterium.

Unit 2 · Cell Structure and Function

1The bacterium must go, the cell must stay

Here is a white blood cell photographed through an electron microscope. The cell is wrapping itself around two rod-shaped bacteria.

A large rough-surfaced cell, colored yellow, wrapping itself around two long orange rods — Electron-microscope photograph, colors added: a large rough-surfaced cell in yellow wrapping itself around two long orange rods

The cell folds its membrane round a bacterium and pulls the bacterium inside. The bacterium ends up wrapped in a bubble of membrane: a vesicle.

Check q1

Taking a bacterium into the cell wrapped in a vesicle is called which of the following?

A. Exocytosis
Exocytosis releases material from a vesicle to the outside of the cell.
B. ✓ Endocytosis
C. Diffusion
Diffusion is the net movement of particles down a concentration gradient; no vesicle takes part.

Why: The membrane folds round the bacterium and closes into a vesicle inside the cell.
Taking material in wrapped in a vesicle is endocytosis.

A bacterium is made of large molecules: proteins, lipids, polysaccharides and DNA.

To get rid of the bacterium, the cell breaks those large molecules into small ones.

Check q2

Which reaction breaks a large molecule into small ones by adding water?

A. ✓ Hydrolysis
B. Dehydration synthesis
Dehydration synthesis joins small molecules into a large one and releases water.
C. Diffusion
Diffusion moves particles; it breaks no bonds.

Why: Hydrolysis adds water at a joint in a large molecule.
The joint splits.
So hydrolysis breaks a large molecule into small ones.

Video: Watch: The bacterium must go, the cell must stay

A white blood cell swallows a bacterium whole. To get rid of it, the cell must break the bacterium’s large molecules into small ones. It must not break its own.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L18Ca-r7.mp4

The cell breaks the bacterium’s large molecules into small ones by hydrolysis. Then the cell can use the small molecules, or push them out.

A chain of eight beads at the left, an arrow, and eight separate beads at the right

Breaking something’s large molecules into small ones like this is called it. So to digest a bacterium is to take it apart, molecule by molecule.

But the cell itself is made of the same kinds of large molecule. Whatever digests the bacterium could digest the cell.

So the cell must keep its digesting tools somewhere safe.

What you are expected to know Explain what it means for a cell to digest something: break its large molecules into small ones by hydrolysis.

Check q3

What does it mean for a cell to digest a bacterium?

A. Wrap the bacterium in a membrane
Wrapping the bacterium in membrane takes it in; it is still whole.
B. Push the bacterium out of the cell
Pushing the bacterium out leaves it whole.
C. ✓ Break the bacterium’s large molecules into small ones

Why: Digesting is breaking large molecules into small ones by hydrolysis.
So to digest a bacterium is to break its large molecules into small ones.

Practice writing an answer

A white blood cell has taken in a bacterium. An hour later the bacterium has been digested.

(a) State what it means for the cell to digest the bacterium. (1 pt)

Model answer The cell breaks the bacterium’s large molecules into small ones by hydrolysis.

Rubric

Award 1 point for: breaking the bacterium’s large molecules into small molecules (by hydrolysis).
Accept: taking the bacterium apart into small molecules the cell can use or remove.

Slip Saying the cell pushes the bacterium out or stores it. Digesting means breaking the large molecules into small ones.

16A protein that speeds one reaction up: enzyme

Hydrolysis on its own is slow. Left alone in water, a protein takes many years to break into its amino acids.

The cell cannot wait years. So the cell uses special proteins.

Two rows: in the top row a chain of beads breaks into separate beads with the arrow labelled many years; in the bottom row the same chain has a large gray blob sitting on it and breaks into separate beads with the arrow labelled seconds

Each special protein speeds up one reaction millions of times. So a protein that took years to break into amino acids now breaks apart in seconds.

A protein that speeds up one particular reaction is called an . One enzyme speeds up one reaction.

An enzyme that speeds up hydrolysis, the breaking of a large molecule into small ones, is a digestive enzyme.

A digestive enzyme is itself a protein.

What you are expected to know Describe an enzyme: a protein that speeds up one particular reaction; a digestive enzyme speeds up hydrolysis.

Check q4

What is an enzyme?

A. A sugar chain attached to a protein
A sugar chain attached to a protein is the result of glycosylation.
B. ✓ A protein that speeds up one particular reaction
C. A small bubble of membrane
A small bubble of membrane is a vesicle.

Why: An enzyme is a protein.
An enzyme speeds up one particular reaction.

Check q5

What does a digestive enzyme do?

A. Carries a protein to the cell surface
Vesicles carry proteins to the cell surface.
B. ✓ Speeds up the breaking of a large molecule into small ones
C. Speeds up the joining of amino acids into a protein
Ribosomes join amino acids into a protein.

Why: A digestive enzyme speeds up hydrolysis.
Hydrolysis breaks a large molecule into small ones.

Check q6

In a liver cell, one substance speeds up a single reaction. With the substance present, the reaction takes seconds instead of years.

Which of the following is the substance?

A. A vesicle
A vesicle is a small bubble of membrane that carries things; it speeds up no reaction.
B. A membrane
A membrane is a thin sheet of lipid that fences off a space; it speeds up no reaction.
C. ✓ An enzyme

Why: With this substance present, one reaction takes seconds instead of years.
So this substance speeds up one particular reaction.
A protein that speeds up one particular reaction is an enzyme.
So the substance is an enzyme.

Practice writing an answer

A digestive enzyme in a white blood cell breaks proteins into amino acids in seconds; without the enzyme the same breaking would take years.

(a) State what an enzyme is. (1 pt)

Model answer An enzyme is a protein that speeds up one particular reaction.

Rubric

Award 1 point for: a protein that speeds up a (particular) reaction.
Accept: a protein that speeds one reaction up many times over.

Slip Calling an enzyme a sugar or a lipid. Enzymes are proteins.

28The acidic sac of enzymes: lysosome

The cell keeps its digestive enzymes sealed inside small sacs of membrane.

One round sac in section: a single outline, gray fluid inside, and seven small dark blobs scattered in the fluid

The fluid inside each sac is acidic, far more acidic than the cytosol around it.

Acidic means sharp, like lemon juice. The cytosol is more like water.

A small sac of membrane, acidic inside and holding digestive enzymes, is called a .

Lyso means to break apart; some means a body. So a lysosome is a body that breaks things apart.

Every animal cell has lysosomes. A white blood cell has many.

What you are expected to know Describe a lysosome: a small membrane sac, acidic inside, holding digestive enzymes.

Check q7

What is a lysosome?

A. A stack of flat membrane pouches
A stack of flat membrane pouches is the Golgi complex.
B. ✓ A small acidic sac holding digestive enzymes
C. A network of folded membrane
A network of folded membrane is the ER.

Why: A lysosome is a small sac of membrane.
The fluid inside is acidic.
The sac holds digestive enzymes.

Check q8

What does the name lysosome mean?

A. A body that stores water
A vacuole stores water; lyso means to break apart.
B. A body that builds proteins
Ribosomes build proteins; lyso means to break apart.
C. ✓ A body that breaks things apart

Why: Lyso means to break apart.
Some means a body.
So a lysosome is a body that breaks things apart.

Check q9

What is the fluid inside a lysosome like?

A. ✓ Acidic, far more acidic than the cytosol
B. The same acidity as the cytosol
The fluid inside a lysosome is far more acidic than the cytosol.
C. Far less acidic than the cytosol
The fluid inside a lysosome is more acidic, not less.

Why: The fluid inside a lysosome is acidic.
It is far more acidic than the cytosol around the lysosome.

Practice writing an answer

A liver cell contains many lysosomes.

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

Model answer A lysosome is a small sac of membrane, acidic inside, that holds digestive enzymes.

Rubric

Award 1 point for: a membrane sac holding digestive enzymes (acidic inside).
Accept: a small sac of membrane holding digestive enzymes.

Slip Describing a stack of flat pouches. That is the Golgi complex; a lysosome is one round sac.

40The sac fuses and the bacterium is taken apart

A bacterium sits inside a vesicle in the white blood cell. A lysosome moves to the vesicle.

Two panels. Before: a bubble of membrane holding a rod-shaped bacterium sits beside a smaller gray sac with dark blobs inside. After: one larger sac holds separate small beads, the bacterium digested into small molecules, with the dark blobs among them

The two membranes join into one: the lysosome fuses with the vesicle.

Now the enzymes, the acidic fluid and the bacterium share one space.

The enzymes speed up the hydrolysis of the bacterium’s large molecules. The enzymes digest the bacterium into small molecules.

An hour later the bacterium is gone.

What you are expected to know Describe how a lysosome digests what the cell took in: it fuses with the vesicle, so the enzymes and the bacterium share one acidic space.

Check q10

A white blood cell engulfs a bacterium. The bacterium sits inside a vesicle. An hour later the fluid inside that vesicle is acidic and holds digestive enzymes.

How did the digestive enzymes get into the vesicle?

A. The vesicle budded off from a lysosome
The vesicle formed at the plasma membrane when the cell wrapped the bacterium.
B. The enzymes diffused through the cytosol into the vesicle
A digestive enzyme is a large protein.
A large protein cannot cross a membrane on its own.
C. ✓ A lysosome fused with the vesicle
D. The bacterium carried the enzymes in from outside the cell
The enzymes are the white blood cell’s own, held in its lysosomes.

Why: A lysosome fused with the vesicle holding the bacterium.
Fusion joins the two membranes into one.
So the lysosome’s acidic fluid and its enzymes are now inside the vesicle with the bacterium.

Video: Watch: The sac fuses and the bacterium is taken apart

A lysosome fuses with the vesicle holding the bacterium. Enzymes, acidic fluid and bacterium share one space. The enzymes digest the bacterium.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L18Cd.mp4

Check q11

White blood cells from a patient engulf bacteria at the normal rate. Hours later, the bacteria are still whole inside the cells. One protein normally found inside lysosomes is missing.

Which of the following do lysosomes normally do?

A. ✓ Use digestive enzymes to break down the bacteria they receive
B. Keep engulfed bacteria whole, stored away as waste
In healthy cells, engulfed bacteria are digested within hours.
C. Make the digestive enzymes themselves, from amino acids
Enzymes are proteins.
Ribosomes make proteins.

Why: The cells took the bacteria in as normal.
So the failure came after that.
The missing lysosome protein is one of the digestive enzymes.
Without that enzyme the lysosomes could not break the bacteria down.
So lysosomes normally digest what the cell takes in, using their digestive enzymes.

50Why the rest of the cell is safe

The cell’s own parts are made of the same kinds of large molecule as the bacterium. So why is the rest of the cell safe from the enzymes?

A round sac with gray fluid and dark blobs inside; outside the sac, in the pale surrounding fluid, one dark blob sits alone

First reason: the enzymes are sealed inside the lysosome’s membrane.

A protein cannot cross a membrane on its own. So the enzymes stay in.

Second reason: the enzymes work fast only in acidic fluid. The cytosol is far less acidic.

So an enzyme that leaks into the cytosol works only slowly there.

So the lysosome digests the bacterium, and the rest of the cell is safe.

What you are expected to know Explain why a lysosome’s enzymes do not digest the rest of the cell: they are sealed inside the membrane, and they work only slowly at the cytosol’s lower acidity.

Check q12

A white blood cell’s lysosomes hold enzymes that could break down the cell’s own proteins.

Why is the rest of the cell safe from those enzymes?

A. They are destroyed as soon as the bacterium is digested
The enzymes stay in the lysosome and are used again.
B. They can digest only bacteria, never the cell’s own molecules
A digestive enzyme breaks a kind of molecule, and the cell is made of the same kinds.
C. ✓ They are sealed in, and work only slowly in the cytosol

Why: The enzymes are sealed inside the lysosome’s membrane.
A protein cannot cross a membrane on its own.
The enzymes work fast only in acidic fluid.
The cytosol is far less acidic, so any enzyme that leaked out would work only slowly.

Check q13

In a patient, the fluid inside the lysosomes is much less acidic than normal. Engulfed bacteria pile up undigested inside the patient’s white blood cells.

Which of the following explains the pile-up?

A. ✓ The enzymes need acidic fluid to work fast, and this fluid is not acidic enough
B. The lysosomes cannot fuse with vesicles when the fluid is less acidic
Fusion joins two membranes; the acidity of the fluid inside plays no part in it.
C. The cells cannot wrap bacteria in a vesicle when the fluid is less acidic
Wrapping a bacterium happens at the plasma membrane, before any lysosome is involved.

Why: The digestive enzymes work fast only in acidic fluid.
The patient’s lysosome fluid is much less acidic.
So the enzymes work only slowly.
So the bacteria pile up undigested.

Practice writing an answer

A few digestive enzymes leak out of a lysosome into the cytosol of a liver cell.

(a) Explain why the leaked enzymes do little harm to the cell. (1 pt)

Model answer The enzymes work fast only in acidic fluid.
The cytosol is far less acidic than the inside of the lysosome.
So the leaked enzymes work only slowly in the cytosol.
So they do little harm.

Rubric

Award 1 point for: the enzymes work fast only in acidic fluid, and the cytosol is much less acidic, so the leaked enzymes work only slowly.
Accept: the cytosol’s lower acidity slows the enzymes.

Slip Saying the enzymes are still sealed in. These enzymes have leaked out; the second safeguard, the cytosol’s lower acidity, is what limits the harm.

A white blood cell takes a bacterium apart without taking itself apart. The digestive enzymes stay sealed in the lysosome.

The lysosome fuses with the vesicle holding the bacterium. The enzymes work fast only inside that acidic space.

Glossary

digest: To break something’s large molecules into small ones by hydrolysis, so that the cell can use them or push them out. To digest a bacterium is to take it apart, molecule by molecule.
enzyme: A protein that speeds up one particular reaction. A digestive enzyme speeds up hydrolysis, the breaking of a large molecule into small ones.
lysosome: A small sac of membrane, acidic inside, holding digestive enzymes. Lyso means to break apart; some means a body: a body that breaks things apart.

APBIO-U02-L18D Where lysosomes come from, and what else they take apart

Topic 2.1 · Cell Structure and Function · 47 steps

A cell cut open: a folded network at the left with dots on it, an arrow to a stack of flat pouches in the middle, and an arrow on to a small round sac at the right

A lysosome is a small sac of digestive enzymes. An enzyme is a protein. So where did those enzymes come from, and how did they get into the sac?

The enzymes travel the same route as every other protein: ribosome, ER, Golgi complex, vesicle. And bacteria are not the only thing a lysosome digests. Lysosomes also clear away the cell’s own worn-out parts. And they help a cell take itself apart when its time comes.

Unit 2 · Cell Structure and Function

1Where the lysosome’s enzymes come from

The digestive enzymes inside a lysosome are proteins.

Check q1

Which of the following makes proteins?

A. ✓ Ribosomes
B. The Golgi complex
The Golgi complex finishes and packs proteins that ribosomes have already made.
C. Lysosomes
Lysosomes break proteins down; they make none.

Why: Only ribosomes join amino acids into a protein.
So ribosomes make proteins.

Video: Watch: Where the lysosome’s enzymes come from

A ribosome on the rough ER makes a digestive enzyme. The Golgi complex finishes it and packs it into a vesicle addressed to stay inside the cell. That vesicle becomes a lysosome.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L18D.mp4

Here is the route a digestive enzyme takes, with what happens at each stop.

A cell cut open with a protein's route drawn through it: rough ER at the left, an arrow to a vesicle, an arrow to a stack of flat pouches, an arrow to a second vesicle, an arrow to a round sac at the right; each stop labelled with what happens there

A ribosome on the rough ER joins amino acids into the enzyme.

The ribosome passes the enzyme into the ER’s interior. The ER folds it.

A vesicle pinches off the ER carrying the enzyme. The vesicle fuses with the Golgi complex.

The Golgi complex finishes the enzyme. Then the Golgi complex packs the enzyme into another vesicle.

This vesicle is addressed to stay inside the cell, not to go to the cell surface.

The vesicle holds digestive enzymes. The cell makes the fluid inside it acidic.

So the vesicle is now a lysosome.

So the route that carries insulin to the blood also carries digestive enzymes to a lysosome. Only the address differs.

What you are expected to know Describe where a lysosome’s enzymes come from: ribosomes on the rough ER make them, the Golgi complex finishes them, and a vesicle from the Golgi complex becomes the lysosome.

Check q2

A liver cell is making new lysosomes.

Where are the lysosome’s digestive enzymes made?

A. Inside a lysosome
A lysosome holds enzymes; it makes none.
B. In the Golgi complex
The Golgi complex finishes proteins that are already made.
C. ✓ On a ribosome on the rough ER

Why: Only ribosomes make proteins.
The ribosomes that make proteins for the ER route sit on the rough ER.
So the enzyme is made on a ribosome on the rough ER.

Check q3

The ER has folded the enzyme.

How does the enzyme travel from the ER to the Golgi complex?

A. ✓ Inside a vesicle that pinches off the ER
B. Loose through the cytosol
A protein on this route is never loose in the cytosol.
C. Through a hole in the ER membrane
A protein is far too large to pass through a membrane on its own.

Why: A piece of the ER membrane pinches off around the enzyme: a vesicle.
The vesicle fuses with the Golgi complex.
So the enzyme travels inside a vesicle.

Check q4

A vesicle carrying a digestive enzyme has fused with the Golgi complex.

Which of the following does the Golgi complex do with the enzyme?

A. Builds it from amino acids
The ribosome on the rough ER built the enzyme before it arrived.
B. ✓ Finishes it and packs it into another vesicle
C. Digests it into amino acids
Digesting is a lysosome’s job; the Golgi complex finishes and packs.

Why: The enzyme arrives folded but not finished.
The Golgi complex finishes it.
Then the Golgi complex packs it into another vesicle.

Check q5

A vesicle carrying finished digestive enzymes leaves a liver cell’s Golgi complex, addressed to stay inside the cell.

What happens to the vesicle?

A. It fuses with the plasma membrane and releases the enzymes outside the cell
A vesicle addressed to stay inside the cell never fuses with the plasma membrane.
B. It fuses with the nucleus
No vesicle from the Golgi complex fuses with the nucleus.
C. ✓ It becomes a lysosome

Why: The vesicle is addressed to stay inside the cell.
It holds digestive enzymes, and its fluid becomes acidic.
A small acidic sac of digestive enzymes is a lysosome.
So the vesicle becomes a lysosome.

Check q6

A cell in the lining of the gut stocks a new lysosome with digestive enzymes.

In which order do the enzymes pass through the cell’s parts?

A. The Golgi complex, then a vesicle, then the rough ER, then another vesicle, then the lysosome
Ribosomes on the rough ER make the enzymes, so the enzymes are in the rough ER before they reach the Golgi complex.
B. ✓ Rough ER, then a vesicle, then the Golgi complex, then another vesicle, then the lysosome
C. Rough ER, then the cytosol, then the Golgi complex, then the lysosome
A protein on this route is never loose in the cytosol.

Why: Ribosomes on the rough ER make the enzymes.
A vesicle carries the enzymes to the Golgi complex.
The Golgi complex finishes them and packs them into another vesicle.
That vesicle becomes the lysosome.

Check q7

A student says: “Lysosomes make their own digestive enzymes.”

Is the student correct?

A. ✓ No: ribosomes on the rough ER make the enzymes
B. Yes: a lysosome has ribosomes of its own inside
A lysosome has no ribosomes inside it; ribosomes sit on the rough ER and loose in the cytosol.

Why: Enzymes are proteins.
Only ribosomes make proteins.
The enzymes reach the lysosome by the ER and Golgi route.
So lysosomes do not make their own enzymes.

21Worn-out parts go the same way

Bacteria are not the only thing a lysosome digests. The cell’s own parts wear out.

The cell wraps a worn-out part in membrane. A lysosome fuses with that wrapping.

A round bubble of membrane holding a crumpled wavy line, beside a smaller gray sac with dark blobs inside; the two are about to touch

The enzymes digest the worn-out part into small molecules. The cell uses those small molecules again.

Cells at the back of your eye shed worn-out fragments every day. Neighboring cells take those fragments in and digest them in lysosomes.

What you are expected to know State what else lysosomes digest: the cell’s own worn-out parts.

Check q8

A cell at the back of the eye takes in worn-out fragments shed by its neighbors.

Which organelle digests the fragments?

A. The Golgi complex
The Golgi complex finishes and packs proteins.
B. ✓ Lysosomes
C. Ribosomes
Ribosomes make proteins.

Why: The cell wraps the fragments in membrane.
A lysosome fuses with that wrapping.
The lysosome’s enzymes digest the fragments.
So lysosomes digest the fragments.

Check q9

Which of the following do lysosomes digest?

A. Bacteria the cell takes in, and nothing else
Lysosomes also digest the cell’s own worn-out parts.
B. The cell’s own worn-out parts, and nothing else
Lysosomes also digest what the cell takes in, such as a bacterium.
C. ✓ Bacteria the cell takes in, and the cell’s own worn-out parts

Why: A lysosome fuses with the vesicle holding a bacterium and digests it.
A lysosome also fuses with the wrapping round a worn-out part and digests it.
So lysosomes digest both.

29Cells that take themselves apart: apoptosis

Look at the hand of an embryo about six weeks after it forms: skin webbing joins the fingers. Two weeks later the webbing is gone.

Two hands seen palm-down, side by side: the left hand has gray skin filling the gaps between its four fingers almost to the tips; the right hand has four separate fingers

The webbing cells were not injured. The webbing cells took themselves apart in an orderly way, on schedule, as a normal part of development.

Biologists call this programmed cell death: the death is planned, not an accident.

A cell taking itself apart in an orderly way as a normal part of development is called .

Apo means away; ptosis means falling: the cells fall away, like leaves from a tree.

Lysosomes take part in apoptosis. The lysosomes’ enzymes help break down the dying cell’s contents.

Neighboring cells take in the pieces and digest them.

Sometimes the webbing cells do not die. Then a baby is born with two fingers still joined by skin.

A newborn baby's hand held between an adult's fingers; two of the baby's fingers are joined by skin along their whole length — Photograph of a newborn baby's hand held between an adult's fingers: two of the baby's fingers are joined together by skin along their whole length

What you are expected to know Describe apoptosis: programmed cell death, in which a cell takes itself apart in an orderly way as a normal part of development, with lysosomes taking part.

Check q10

What is apoptosis?

A. A cell dying suddenly after an injury
A cell that dies after an injury was not on schedule; apoptosis is orderly and planned.
B. ✓ A cell taking itself apart in an orderly way, on schedule
C. A cell dividing into two new cells
A cell dividing makes two cells; apoptosis removes one.

Why: In apoptosis a cell takes itself apart in an orderly way.
It does so on schedule, as a normal part of development.
So apoptosis is programmed cell death.

Video: Watch: Cells that take themselves apart

The webbing between an embryo’s fingers is gone two weeks later. The webbing cells took themselves apart on schedule: apoptosis, programmed cell death, with lysosomes taking part.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L18Dc.mp4

Practice writing an answer

As a tadpole becomes a frog, the cells of its tail die on schedule and are cleared away.

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

Model answer Apoptosis is programmed cell death: a cell takes itself apart in an orderly way as a normal part of development.

Rubric

Award 1 point for: programmed cell death, or a cell taking itself apart in an orderly way as a normal part of development.
Accept: a cell dying on schedule, in an orderly way, as part of normal development.

Slip Describing death by injury. Apoptosis is planned and orderly; the cell was healthy and dies on schedule.

Check q11

As a tadpole becomes a frog, its tail shrinks and disappears. The tail cells were healthy when the shrinking began. Nothing injured them.

Which of the following is happening to the tail cells?

A. Death by injury
The tail cells were healthy when the shrinking started.
B. ✓ Apoptosis

Why: The tail cells were healthy, and nothing injured them.
The tail cells died in an orderly way, as a normal part of development.
A cell taking itself apart in an orderly way as a normal part of development is apoptosis.

Check q12

The tadpole’s tail cells are undergoing apoptosis.

Which organelle takes part?

A. ✓ Lysosomes
B. The Golgi complex
The Golgi complex finishes and packs proteins.
The Golgi complex does not take a cell apart.
C. The smooth ER
The smooth ER makes membrane lipids and breaks down harmful molecules.

Why: Lysosomes hold digestive enzymes.
In apoptosis the lysosomes’ enzymes help break down the cell’s contents.
So lysosomes take part in apoptosis.

Ribosomes on the rough ER make a lysosome’s enzymes. The Golgi complex finishes them.

A vesicle from the Golgi complex becomes the lysosome.

Lysosomes digest what the cell takes in, the cell’s own worn-out parts and, in apoptosis, the cell itself.

Glossary

apoptosis: Programmed cell death: a cell takes itself apart in an orderly way as a normal part of development, such as the webbing between an embryo’s fingers. Apo means away; ptosis means falling. Lysosomes take part.

APBIO-U02-L18E The endomembrane system

Topic 2.1 · Cell Structure and Function · 64 steps

A cell cut open, with the nucleus, a folded network of membrane, a stack of sacs and small vesicles between them; to the right, a vesicle pinches off one membrane and fuses with another

A pancreas cell makes insulin on its ER. An hour later that insulin is in the blood. On the way the insulin passed through the Golgi complex and two vesicles. The insulin was never loose in the cytosol.

Every membrane the insulin passed through is joined to the next. A small bubble of membrane pinches off one membrane and fuses with the next. The membranes joined like this work as one system. Together they finish, pack and move what the cell sends out. Which membranes are joined like this, and what does the whole set do for the cell?

Unit 2 · Cell Structure and Function

1Membranes joined into one system

Check q1

A small bubble of membrane pinches off the ER of a pancreas cell, with a folded protein inside it.

Which of the following is this bubble called?

A. ✓ A vesicle
B. A ribosome
A ribosome is a tiny machine made of protein and RNA, with no membrane.
C. A lysosome
A lysosome is an acidic sac of digestive enzymes, formed from vesicles that leave the Golgi complex.

Why: A small sac of membrane that pinches off a larger membrane is a vesicle.

Video: Watch: Membranes joined into one system

A vesicle pinches off the ER and fuses with the Golgi complex. The ER runs straight into the nuclear envelope. The membranes joined either way are one system: the endomembrane system.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L18Ea.mp4

To pinch off means that a patch of membrane bulges out and closes up into a bubble.

Here is a pancreas cell cut open. A vesicle pinches off the ER and fuses with the Golgi complex.

A cell cut open: the nucleus with its envelope at the left, three flattened sacs of rough ER joined to the envelope, a small vesicle, a stack of four Golgi sacs, a second vesicle, and a vesicle fusing with the cell's outer membrane at the right with dots outside it

A second vesicle pinches off the Golgi complex and fuses with the plasma membrane.

So the ER, the Golgi complex and the plasma membrane are joined. Membrane passes from one to the next, riding in vesicles.

The ER is joined to the nuclear envelope in a second way. The ER membrane runs straight into the nuclear envelope, as one continuous sheet, with no vesicle needed.

Every membrane joined to the others in either of these two ways works as part of one system.

Membranes inside a cell that are joined in either of these two ways are called the .

The name says what it is: endo means inside. So the endomembrane system is the system of membranes inside the cell.

Having a membrane is not enough for a structure to belong.

A structure belongs when membrane flows to it or from it. A vesicle pinches off it or fuses with it. Or its membrane runs straight into a member’s.

Together, the joined membranes finish, pack and move the proteins and lipids the cell makes. Insulin’s trip from the ER to the blood is this system at work.

What you are expected to know Explain what the endomembrane system is: the membranes inside a cell that are joined into one system.

What you are expected to know State the rule for belonging: a structure is a member when a vesicle pinches off it or fuses with it, or when its membrane runs straight into a member’s.

Check q2

What is the endomembrane system?

A. Every membrane a cell has, whether joined to another or on its own
A structure belongs only if membrane flows to it or from it.
B. The double membrane that wraps the nucleus
The double membrane around the nucleus is the nuclear envelope, one member of the system.
C. ✓ The membranes inside a cell that are joined into one system

Why: Vesicles pinch off one membrane and fuse with another.
The ER runs straight into the nuclear envelope.
The membranes joined either way are the endomembrane system.

Check q3

Which of the following makes a structure a member of the endomembrane system?

A. It has a membrane of its own, whether or not that membrane is joined to any other membrane
Having a membrane is not enough; membrane must flow to the structure or from it.
B. ✓ Membrane flows to it or from it: a vesicle pinches off it or fuses with it, or its membrane runs into a member’s
C. It sits inside the cell, between the nucleus and the plasma membrane, and does a job the cell needs
Ribosomes sit inside the cell and do a job the cell needs, yet they have no membrane and are outside the system.

Why: A structure belongs when membrane flows to it or from it.
A vesicle pinches off it or fuses with it: the Golgi complex.
Or its membrane runs straight into a member’s: the nuclear envelope.
Either way, its membrane is joined to the system.

Check q4

What does endo mean in the word endomembrane?

A. ✓ Inside
B. Between
Endo means inside, as in endocytosis: bringing material inside the cell.
C. Outside
Endo means inside; exo, as in exocytosis, means outside.

Why: Endo means inside.
So the endomembrane system is the system of membranes inside the cell.

Practice writing an answer

A liver cell has many membranes inside it.

(a) State what the endomembrane system of this cell is. (1 pt)

Model answer The endomembrane system is the set of membranes inside the cell that are joined into one system.
A structure belongs when a vesicle pinches off it or fuses with it, or when its membrane runs straight into a member’s.

Rubric

Award 1 point for: the membranes inside the cell that are joined into one system (by vesicles, or by running straight into one another).
Accept: the membranes that vesicles pinch off and fuse with, working as one system.

Slip Writing ‘every membrane inside the cell’. Only the joined structures belong: a vesicle pinches off each one or fuses with it, or its membrane runs straight into a member’s.

21The seven members

Seven membranes in a cell are joined this way. Here they are, labelled on one cell.

A cell cut open with its parts labelled: nucleus and nuclear envelope, rough ER with ribosomes, a small vesicle, a Golgi stack, more vesicles, a lysosome, a vacuole, and a vesicle fusing with the outer membrane

The table lists the seven members, how each one is joined to the others, and its job in the system.

member	how it is joined	its job in the system
ER	vesicles pinch off it; its membrane runs straight into the nuclear envelope	makes and folds proteins; makes membrane lipids
Golgi complex	vesicles from the ER fuse with it; new vesicles pinch off it	finishes, sorts and packs proteins
transport vesicles	each pinches off one member and fuses with another: they are the connection	carry proteins and membrane from one member to the next
lysosomes	formed from vesicles that pinch off the Golgi complex; vesicles fuse with them	digest what the cell takes in, and the cell’s worn-out parts
vacuoles	vesicles fuse with them	store water, food and waste
nuclear envelope	the ER membrane runs straight into it	encloses the nucleus
plasma membrane	vesicles fuse with it (exocytosis) and pinch off it (endocytosis)	releases material from the cell and takes material in

The seven members of the endomembrane system.

Table of the seven members of the endomembrane system, with how each is joined to the others and its job in the system

The transport vesicles are the vesicles that carry cargo from one member to the next: the same vesicles that pinch off the ER and fuse with the Golgi complex.

The nuclear envelope belongs because the ER membrane runs straight into it.

Transport vesicles belong because they are the connection itself.

Ribosomes are missing from the table. A ribosome has no membrane, so membrane cannot flow to it or from it.

What you are expected to know Identify the seven members of the endomembrane system: the ER, the Golgi complex, transport vesicles, lysosomes, vacuoles, the nuclear envelope and the plasma membrane.

Check q5

Is the nuclear envelope a member of the endomembrane system?

A. ✓ Yes
B. No
The nuclear envelope makes no vesicles, but it is joined to the ER another way.

Why: The ER membrane runs straight into the nuclear envelope.
A membrane that runs straight into a member is joined to the system.
So the nuclear envelope is a member.

Check q6

Is a ribosome a member of the endomembrane system?

A. Yes
A ribosome has no membrane at all.
B. ✓ No

Why: The endomembrane system is a set of joined membranes.
A ribosome has no membrane.
So membrane cannot flow to a ribosome or from it.
So a ribosome is outside the system.

Check q7

Is a lysosome a member of the endomembrane system?

A. ✓ Yes
B. No
Digesting is a lysosome’s job; being a member is about how its membrane is joined.

Why: A lysosome is a membrane sac.
Vesicles from the Golgi complex form it, and vesicles fuse with it.
So membrane flows to the lysosome.
So a lysosome is a member.

Check q8

Is a vacuole a member of the endomembrane system?

A. ✓ Yes
B. No
Storing water and food is a vacuole’s job; vesicles still fuse with its membrane.

Why: A vacuole is a large membrane sac.
Vesicles fuse with it.
So membrane flows to the vacuole.
So a vacuole is a member.

Check q9

Is the cell wall a member of the endomembrane system?

A. Yes
The cell wall is made of cellulose, not of membrane.
B. ✓ No

Why: The cell wall is a rigid layer with no membrane.
The cell wall lies outside the plasma membrane.
So membrane cannot flow to the cell wall or from it.
So the cell wall is outside the system.

Check q10

Is the plasma membrane a member of the endomembrane system?

A. ✓ Yes
B. No
The plasma membrane is the cell’s outer boundary, but it takes in and gives out membrane all the time.

Why: Vesicles fuse with the plasma membrane in exocytosis.
Vesicles pinch off the plasma membrane in endocytosis.
So membrane flows to the plasma membrane and from it.
So the plasma membrane is a member.

35Why it is in, why it is out

The rule again: a structure is a member when a vesicle pinches off it or fuses with it, or when its membrane runs straight into a member’s. Now the rule is applied to six cases, on the labelled cell.

For example, the Golgi complex is a member, because vesicles pinch off it and fuse with it.

And the plasma membrane is a member, because vesicles fuse with it and pinch off it.

And the nuclear envelope is a member, because the ER membrane runs straight into it.

And a transport vesicle is a member, because it pinches off one member and fuses with another: it is the connection itself.

But a ribosome is not a member, because a ribosome has no membrane.

The edge of a plant cell: a thick gray band across the top, a line just below it, and below that a white region holding a small circle at the left, two dark blobs at the right, and a circle merging into the line at the upper right

And the cell wall is not a member, because the cell wall has no membrane. The cell wall is a rigid layer outside the plasma membrane.

So the test is never where a structure sits. The test is whether membrane flows to it or from it.

What you are expected to know Explain why a given structure is a member of the endomembrane system, or is outside it, from the rule for belonging.

Check q11

The nuclear envelope makes no vesicles of its own. It is a member of the endomembrane system.

Which of the following is the reason?

A. The nuclear envelope has two membranes
Having a membrane is not the test; membrane must flow to a structure or from it.
B. The nuclear envelope is wrapped around the DNA
Wrapping the DNA is the nuclear envelope’s job, but that job joins it to no other membrane.
C. ✓ The ER membrane runs straight into the nuclear envelope

Why: The members are the membranes joined to one another.
The ER membrane runs straight into the nuclear envelope.
So the nuclear envelope is a member.

Check q12

A student says: “A ribosome sits inside the cell, so a ribosome is a member of the endomembrane system.”

Is the student correct?

A. Yes
A ribosome has no membrane, so membrane cannot flow to it or from it.
B. ✓ No

Why: Sitting inside the cell is not the test.
The test is whether membrane flows to the structure or from it.
A ribosome has no membrane.
So a ribosome is outside the system.

Check q13

A plant cell has a cell wall. The cell wall is outside the endomembrane system.

Which of the following is the reason?

A. The cell wall is made of cellulose rather than protein
What the cell wall is made of is not the test.
The cell wall has no membrane, so membrane cannot flow to it.
B. ✓ The cell wall has no membrane, and it lies outside the plasma membrane
C. Vesicles pinch off the cell wall and fuse with it
Vesicles fuse with the plasma membrane, on the inner side of the cell wall, never with the cell wall itself.

Why: The cell wall is a rigid layer with no membrane.
It lies outside the plasma membrane.
So membrane cannot flow to it or from it.
So the cell wall is outside the system.

Practice writing an answer

A transport vesicle carries a protein from the ER to the Golgi complex. The transport vesicle is a member of the endomembrane system.

(a) Explain why the transport vesicle is a member. (1 pt)

Model answer A transport vesicle is a small bubble of membrane.
It pinches off one member, the ER.
It fuses with another member, the Golgi complex.
So membrane flows through the vesicle from one member to the next.
So the transport vesicle is a member.

Rubric

Award 1 point for: the vesicle is membrane that pinches off one member and fuses with another, so it is joined to the system (it is the connection).
Accept: membrane flows through the vesicle from the ER to the Golgi complex.

Slip Writing ‘because it carries a protein’. The cargo is not the test. The vesicle is a member because its own membrane pinches off one member and fuses with another.

49What the system does, in one cell

Check q14

A vesicle from the ER, carrying a protein, fuses with a stack of flattened membrane sacs in a pancreas cell.

Which member of the endomembrane system finishes the protein and packs it into another vesicle?

A. The ER
The ER makes and folds the protein before the vesicle leaves it.
B. A lysosome
A lysosome digests; it finishes and packs nothing.
C. ✓ The Golgi complex

Why: The stack of flattened membrane sacs is the Golgi complex.
The Golgi complex finishes each protein and packs it into another vesicle.

Video: Watch: What the system does, in one cell

Insulin’s route drawn on the seven-member cell. Every arrow runs between two members. The system modifies, packages and moves proteins within one cell.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L18Ed.mp4

Here is the labelled cell again. The arrows show the route insulin takes: the ER, a vesicle, the Golgi complex, a vesicle, the plasma membrane, then out of the cell.

The same labelled cell, with arrows from the rough ER to a vesicle, to the Golgi stack, to a second vesicle, to the outer membrane, and out of the cell, the stops numbered 1 to 5

Every arrow runs between two members of the system.

At each stop, one member does one job: the ER makes and folds the insulin, and the vesicles carry it.

The Golgi complex finishes the insulin and packs it. The plasma membrane releases it.

So the endomembrane system modifies proteins and lipids, packages them into vesicles, and moves them from one place to another.

The system handles lipids and carbohydrates as well as proteins. The ER makes membrane lipids, and the Golgi complex attaches carbohydrates (sugar chains) to proteins.

Everything the system moves, it moves within one cell.

When insulin leaves the pancreas cell, the system’s work on that insulin is done. The blood carries it from there.

So one set of joined membranes, working as one system, folds, finishes, packs and moves what the cell sends out. That set is the endomembrane system.

What you are expected to know State what the endomembrane system does: it modifies, packages and moves proteins, lipids and carbohydrates within one cell.

Check q15

Which of the following does the endomembrane system do?

A. ✓ Modifies, packages and moves proteins within one cell
B. Joins amino acids into protein chains
Ribosomes join amino acids into chains; ribosomes have no membrane and are outside the system.
C. Makes the ATP that the cell’s pumps use
The endomembrane system handles proteins, lipids and carbohydrates; it makes no ATP.

Why: A protein is made and folded in the ER.
Vesicles carry it to the Golgi complex, which finishes and packs it.
A vesicle carries it to the plasma membrane, which releases it.
So the system modifies, packages and moves proteins within one cell.

Check q16

A student says: “The endomembrane system carries insulin from the pancreas cell to a muscle cell.”

Is the student correct?

A. Yes
Once insulin has left the pancreas cell, no membrane of that cell touches it again.
B. ✓ No

Why: The endomembrane system moves insulin inside the pancreas cell, from the ER to the plasma membrane.
Exocytosis releases the insulin into the blood.
The blood carries it to the muscle cell.
So the system moves insulin within one cell only.

Glossary

endomembrane system: The membranes inside a cell that are joined into one system, because a vesicle pinches off one and fuses with another, or because one membrane runs straight into the next: the ER, the Golgi complex, transport vesicles, lysosomes, vacuoles, the nuclear envelope and the plasma membrane. Endo means inside: the system of membranes inside the cell. Together they modify, package and move proteins, lipids and carbohydrates within one cell.

APBIO-U02-L19 The cell that ships protein

Topic 2.1 · Cell Structure and Function · 48 steps

A pancreas cell cut open at the left, packed with folded membrane, beside a blood vessel drawn as two long lines; small dots leave the cell and travel along the vessel

After a meal, sugar floods your blood. Cells in your pancreas answer by releasing insulin into the blood. Insulin is the protein that tells the rest of your body’s cells to take that sugar in.

Insulin is far too big to cross a membrane. Yet a pancreas cell makes every insulin molecule inside itself, and the insulin ends up outside. How does one insulin molecule get from the ribosome that makes it to the blood?

Unit 2 · Cell Structure and Function

1One insulin molecule, from the ribosome to the blood

Check q1

Insulin is a protein.

Which of the following makes proteins?

A. ✓ Ribosomes
B. The Golgi complex
The Golgi complex finishes and packs proteins that ribosomes have already made.
C. Lysosomes
Lysosomes break proteins and other large molecules down.

Why: A ribosome joins amino acids into a chain.
That chain is a protein.
So ribosomes make proteins.

Video: Watch: One insulin molecule, from the ribosome to the blood

One insulin molecule, stop by stop. A ribosome on the rough ER builds it. The ER folds it. A vesicle carries it. The Golgi complex finishes and repacks it. Another vesicle carries it. The plasma membrane releases it into the blood.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L19a.mp4

Here is a pancreas cell cut open, beside a blood vessel. One insulin molecule’s route is numbered 1 to 7, and each stop says what happens there.

A pancreas cell cut open, beside a blood vessel drawn as two vertical lines: the nucleus at the left, three flattened sacs of rough ER with dots on them, a small vesicle, a stack of four Golgi sacs, a second vesicle, a vesicle merging with the cell's outer membrane, and dots in the vessel; seven numbered badges mark the stops

Stop 1: a ribosome sitting on the rough ER joins amino acids into a chain. That chain is the insulin molecule.

Stop 2: the ribosome pushes the chain into the inside of the ER. There, the chain folds into its working shape.

Stop 3: a patch of the ER membrane bulges out around the folded insulin and closes up into a small bubble. That bubble is a vesicle, and it carries the insulin to the Golgi complex.

Stop 4: the vesicle fuses with the Golgi complex. The Golgi complex finishes the insulin.

Then the Golgi complex packs the insulin into another vesicle. That vesicle pinches off the far side of the stack.

Stop 5: that vesicle carries the insulin to the plasma membrane.

Stop 6: the vesicle fuses with the plasma membrane, and the insulin spills out of the cell. Releasing material this way is exocytosis.

Stop 7: the insulin is in the blood, which carries it to the rest of the body.

From the ER to the outside, the insulin stayed inside membrane the whole way. It was inside the ER, then inside a vesicle, then inside the Golgi complex, then inside another vesicle.

So the insulin never crossed a membrane. When the last vesicle fused with the plasma membrane, the inside of the vesicle opened to the outside of the cell. So the insulin was outside.

Every protein a cell sends out takes this same route: a digestive protein from a pancreas cell, a tear protein from a tear gland cell, a milk protein from a breast cell.

What you are expected to know Describe the route of a protein made for export: a ribosome on the rough ER, inside the ER, a vesicle, the Golgi complex, another vesicle, the plasma membrane, where exocytosis releases it.

Check q2

A cell in a tear gland makes a protein and releases it into tears.

Where is the protein made?

A. In the Golgi complex
Only ribosomes join amino acids into a protein.
B. On ribosomes loose in the cytosol
A loose ribosome would leave the protein in the cytosol.
C. ✓ On ribosomes on the rough ER

Why: Only ribosomes make proteins.
The ribosomes that make protein for export sit on the rough ER.
So the tear protein is made on ribosomes on the rough ER.

Check q3

A tear gland cell’s protein has been folded inside the ER.

How does the protein travel from the ER to the Golgi complex?

A. ✓ Inside a vesicle that pinches off the ER
B. Loose through the cytosol
A protein for export is never loose in the cytosol.
C. Straight through the ER membrane into the Golgi complex
A protein is far too large and polar to pass through a membrane on its own.

Why: A protein for export travels inside membrane the whole way.
A vesicle pinches off the ER with the protein inside.
The vesicle fuses with the Golgi complex.
So the protein travels inside a vesicle.

Check q4

In a tear gland cell, a vesicle carrying a protein has fused with the Golgi complex.

Which of the following does the Golgi complex do to the protein?

A. Breaks the protein down into amino acids
Breaking proteins down is a lysosome’s job.
B. Builds the protein from amino acids
The ribosomes on the rough ER built the protein before it arrived.
C. ✓ Finishes the protein and packs it into another vesicle

Why: The ribosomes built the protein already.
The Golgi complex finishes the protein, for example by attaching a sugar chain.
The Golgi complex then packs the protein into another vesicle, which pinches off the far side of the stack.

Check q5

Another vesicle, carrying a finished tear protein, leaves the Golgi complex.

Which membrane does this vesicle fuse with so that the protein leaves the cell?

A. The ER membrane
Fusing with the ER would return the protein to the inside of the ER, inside the cell.
B. The Golgi membrane
Fusing back with the Golgi complex would return the protein to the inside of the Golgi complex.
C. The nuclear envelope
Fusing with the nuclear envelope would put the protein in the nucleus, inside the cell.
D. ✓ The plasma membrane

Why: The protein leaves the cell.
Only fusion with the plasma membrane opens a vesicle to the outside.
So the vesicle fuses with the plasma membrane, and exocytosis releases the protein outside the cell.

Check q6

A cell in a salivary gland makes a protein and releases it into saliva.

In which order does the protein pass through the cell’s parts?

A. ✓ Rough ER, then a vesicle, then the Golgi complex, then a vesicle, then outside the cell
B. The Golgi complex, then a vesicle, then the rough ER, then a vesicle, then outside the cell
Ribosomes on the rough ER make the protein, so the protein is in the rough ER before it reaches the Golgi complex.
C. Rough ER, then the cytosol, then the Golgi complex, then outside the cell
A protein for export is never loose in the cytosol.
D. The cytosol, then straight through the plasma membrane, then outside the cell
The protein never passes straight through the plasma membrane; exocytosis releases it.

Why: Ribosomes on the rough ER make the protein.
A vesicle carries the protein to the Golgi complex.
The Golgi complex finishes the protein.
Another vesicle carries the protein to the plasma membrane.
Exocytosis releases the protein outside the cell.

Check q7

A cell in a breast makes a milk protein and releases it into milk. A scientist finds one molecule of the protein inside the Golgi complex.

Which of the following is the protein’s next stop?

A. The rough ER, where the protein was made
The protein left the rough ER before it reached the Golgi complex; it never goes back.
B. ✓ A vesicle that pinches off the Golgi complex
C. The cytosol, loose among the ribosomes
A protein for export is never loose in the cytosol.

Why: The Golgi complex finishes the protein.
Then the Golgi complex packs it into another vesicle.
That vesicle pinches off the Golgi complex and carries the protein to the plasma membrane.

23The membrane travels too

Check q8

A vesicle fuses with the plasma membrane and releases its contents outside the cell.

What happens to the vesicle’s own membrane?

A. ✓ It becomes part of the plasma membrane
B. It breaks up into pieces in the cytosol
Fusion joins two membranes into one; nothing is broken up.
C. It travels back to the ER
The vesicle’s membrane has joined the plasma membrane and stays there.

Why: Fusion joins the vesicle’s membrane and the plasma membrane into one continuous membrane.
So the vesicle’s membrane becomes part of the plasma membrane.

Video: Watch: The membrane travels too

A marked patch of ER membrane pinches off as part of a vesicle’s membrane. It joins the Golgi membrane, rides another vesicle, and ends up in the plasma membrane. Vesicles move membrane as well as cargo.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L19b-r7.mp4

Here a patch of the ER membrane is marked dark. The dark patch lets you follow the membrane itself, not the insulin inside.

Five drawings in a row joined by arrows: three flattened sacs with a short dark stretch on the top one; a small circle with a dark arc; a stack of four sacs standing side by side with a dark stretch on the right-hand one; a small circle with a dark arc; a horizontal line with a dark stretch in it

When a vesicle pinches off the ER, it takes that patch of the ER membrane with it. The patch is now part of the vesicle’s own membrane.

When the vesicle fuses with the Golgi complex, the vesicle’s membrane joins the Golgi membrane. The patch is now part of the Golgi complex.

Another vesicle pinches off the Golgi complex, carrying the patch in its membrane. That vesicle fuses with the plasma membrane. The patch is now part of the plasma membrane.

So vesicles move membrane as well as cargo. The lipids and proteins of the ER membrane end up in the Golgi membrane, and then in the plasma membrane.

A cell adds new membrane to its surface this way: the ER makes membrane lipids, and vesicles carry that new membrane outward.

What you are expected to know Explain why a marked lipid in the ER membrane appears later in the Golgi membrane, then in vesicle membranes, then in the plasma membrane: a vesicle’s own membrane joins each membrane it fuses with.

Check q9

Scientists mark a phospholipid in the ER membrane of a liver cell and follow the mark for an hour.

Which membrane does the marked phospholipid reach first?

A. The plasma membrane
A vesicle from the ER fuses with the Golgi complex first; the plasma membrane comes later.
B. ✓ The Golgi membrane
C. A lysosome’s membrane
A lysosome forms from vesicles that pinch off the Golgi complex, so the mark reaches the Golgi membrane before any lysosome.

Why: A vesicle pinches off the ER, carrying the marked phospholipid in its membrane.
The vesicle fuses with the Golgi complex.
So the mark reaches the Golgi membrane first.

Check q10

An hour after scientists mark a phospholipid in a liver cell’s ER membrane, the mark is in the plasma membrane. A student says: “The phospholipid must have travelled through the cytosol on its own.”

Is the student correct?

A. Yes
The mark travelled in the membranes of vesicles.
B. ✓ No

Why: A phospholipid stays in membrane.
A vesicle carried the marked patch in its membrane from the ER to the Golgi complex.
Another vesicle carried it from the Golgi complex to the plasma membrane.
So the phospholipid never travelled through the cytosol on its own.

Practice writing an answer

Scientists mark a phospholipid in the Golgi membrane of a liver cell.

(a) Explain which membrane the mark reaches next, and how it gets there. (1 pt)

Model answer A vesicle pinches off the Golgi complex.
The marked phospholipid is part of that vesicle’s membrane.
The vesicle fuses with the plasma membrane.
So the vesicle’s membrane becomes part of the plasma membrane.
So the mark reaches the plasma membrane next.

Rubric

Award 1 point for: the plasma membrane, reached because a vesicle pinches off the Golgi complex with the marked phospholipid in its membrane and fuses with the plasma membrane.
Accept: the mark rides in a vesicle’s membrane from the Golgi complex to the plasma membrane.

Slip Writing that the vesicle carried the phospholipid inside it, like cargo. The phospholipid is part of the vesicle’s membrane, and that membrane joins the plasma membrane when the vesicle fuses with it.

36Why the pancreas cell is crowded

Here is a real pancreas cell photographed through an electron microscope.

The cell is crowded with rough ER: folded membrane, layer upon layer, with ribosomes on it. The dark round blobs are vesicles full of protein, waiting to leave the cell.

Here is the same kind of cell drawn, with each crowded part labelled.

A cell cut open: the nucleus at the left, many flattened sacs with dots on them filling the middle, a stack of sacs at the upper right, and many dark round blobs near the right-hand edge

Each part does one step of shipping the protein. Ribosomes make the protein, and the rough ER folds it.

The Golgi complex finishes the protein and packs it. Vesicles carry it to the plasma membrane and out of the cell.

A pancreas cell ships protein all day. A cell that does one job all day needs lots of the parts that do that job.

So a pancreas cell is crowded with ribosomes, rough ER, Golgi complex and vesicles.

The same is true of any cell that sends out lots of protein: a tear gland cell, a salivary gland cell, a cell that makes antibodies (proteins that stick to bacteria).

What you are expected to know Explain why a cell that exports protein is crowded with ribosomes, rough ER, Golgi complex and vesicles: each does one step of making and shipping the protein.

Check q11

A cell in the blood makes antibodies — proteins that stick to bacteria — and releases them into the blood all day.

Which of the following is this cell crowded with?

A. Lysosomes and vacuoles
Lysosomes digest what a cell takes in, and vacuoles store; this cell’s job is sending protein out.
B. Smooth ER and lysosomes
Smooth ER has no ribosomes on it, so it makes no protein.
C. ✓ Rough ER, Golgi complex and vesicles

Why: Antibodies are proteins, made on ribosomes on the rough ER.
The Golgi complex finishes and packs it.
Vesicles carry it out.
The cell does this all day, so it is crowded with rough ER, Golgi complex and vesicles.

Practice writing an answer

A cell in the stomach lining makes a digestive protein and releases it into the stomach, in large amounts, all day.

(a) Explain why this cell has lots of ribosomes, rough ER, Golgi complex and vesicles. Write one short sentence per part. (1 pt)

Model answer Ribosomes on the rough ER make the digestive protein.
The rough ER folds the protein and passes it on in vesicles.
The Golgi complex finishes the protein and packs it into vesicles.
Vesicles carry the protein to the plasma membrane, where it leaves the cell.
The cell makes and releases a lot of this protein, so it needs lots of each part.

Rubric

Award 1 point for: ribosomes make the protein, the rough ER folds it, the Golgi complex finishes and packs it, vesicles carry it out, and the cell exports a lot of protein, so it needs lots of each.
Accept: each part does one step of making and shipping the protein, and the cell does that all day.

Slip Naming the parts without saying what each does. Each part has one step: make, fold, finish and pack, carry out.

APBIO-U02-L19B Where a cell gets its energy

Topic 2.1 · Cell Structure and Function · 85 steps

A long muscle cell cut open, packed with small oval bodies between striped fibers

Your leg muscles contract with every step. Each contraction needs energy. A muscle cell that works all day needs a steady supply of it.

Where does a muscle cell get its energy from? And where inside the cell is that energy released? The answer is a fuel the blood brings, and a tiny part inside the cell that burns it: the mitochondrion. First the fuel.

Unit 2 · Cell Structure and Function

1Where the energy comes from: glucose and oxygen

Check q1

A muscle cell pumps ions across its plasma membrane.

Which molecule supplies the energy for that work directly?

A. ✓ ATP
B. DNA
DNA carries the cell’s inherited instructions; it supplies no energy.
C. Oxygen
Oxygen is used up in respiration; it supplies no energy to the pump itself.

Why: ATP is the small molecule a cell uses to supply the energy for its work.
Pumping ions is work.
So ATP supplies the energy for pumping ions.

Video: Watch: Where the energy comes from

A muscle cell takes in glucose and oxygen from the blood. Carbon dioxide and water leave. The energy released goes into ATP.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L19Ba.mp4

A muscle cell that works all day needs a steady supply of energy. Where does that energy come from?

The muscle cell’s fuel is glucose, a sugar carried to it in the blood.

The muscle cell breaks the glucose down using oxygen, which the blood also brings.

Carbon dioxide and water are left. As the glucose breaks down, it releases energy.

The word equation for aerobic cellular respiration: glucose plus oxygen gives carbon dioxide plus water, and energy is released

Written as a word equation, the change reads glucose + oxygen → carbon dioxide + water, and energy is released.

Breaking glucose down with oxygen to release energy is called .

The name says what happens: aer means air, so aerobic means using oxygen.

Cellular means inside cells.

Respiration is the biologist’s word for releasing energy from food. It is not breathing, even though the word sounds like it.

Breathing brings the oxygen in; respiration uses it.

The cell transfers the released energy into ATP. ATP then supplies the energy for the cell’s work, such as contracting.

So a muscle cell gets its energy by respiration: glucose and oxygen in, carbon dioxide and water out, and ATP made from the energy released.

What you are expected to know State the word equation for aerobic cellular respiration: glucose + oxygen → carbon dioxide + water, with energy released.

What you are expected to know Describe what the cell does with the energy released: it transfers the energy into ATP.

Check q2

What is aerobic cellular respiration?

A. ✓ Breaking glucose down with oxygen to release energy
B. Joining carbon dioxide and water to build glucose
Respiration takes glucose apart; it builds nothing.
C. Breathing air in and out of the lungs
Breathing moves air into and out of the lungs; respiration is chemistry inside every cell.

Why: Aerobic means using oxygen.
Cellular means inside cells.
Respiration here is the breakdown of glucose that releases energy.
So aerobic cellular respiration is breaking glucose down with oxygen, inside cells, to release energy.

Check q3

What does “aerobic” mean?

A. Inside a cell
Cellular is the word that means inside a cell.
B. ✓ Using oxygen
C. Using light
Respiration uses no light.

Why: Aer means air.
The part of the air that respiration uses is oxygen.
So aerobic means using oxygen.

Check q4

A student says: “Respiration is another word for breathing.”

Is the student correct?

A. Yes
Breathing is a movement of the chest and lungs, not a chemical change.
B. ✓ No

Why: Breathing moves air into and out of the lungs.
Aerobic cellular respiration is the breakdown of glucose with oxygen inside every cell.
Breathing brings the oxygen; respiration uses it.
So the two words name different things.

Check q5

A muscle cell is making ATP by aerobic cellular respiration.

Which two substances must the blood keep bringing to the cell?

A. ✓ Glucose and oxygen
B. Carbon dioxide and water
Carbon dioxide and water are what is left at the end; the cell gives them off.
C. Oxygen and water
Water is one of the two products; respiration makes it.

Why: The word equation reads glucose + oxygen → carbon dioxide + water.
The substances on the left are used up.
The blood must keep replacing what is used up.
So the blood keeps bringing glucose and oxygen.

Check q6

A muscle cell has broken a glucose molecule down by aerobic cellular respiration.

Which two substances are left?

A. Oxygen and water
Oxygen is used up; carbon dioxide takes its place among the products.
B. Glucose and ATP
ATP is made from the energy released; the glucose itself has been broken down.
C. ✓ Carbon dioxide and water

Why: The word equation reads glucose + oxygen → carbon dioxide + water.
The substances on the right are what is left.
So carbon dioxide and water are left.

Practice writing an answer

A muscle cell breaks glucose down with oxygen to release energy.

(a) State the word equation for aerobic cellular respiration. (1 pt)

Frame glucose + … → … + …

Model answer glucose + oxygen → carbon dioxide + water
Energy is released.

Rubric

Award 1 point for: glucose + oxygen → carbon dioxide + water (energy released).
Accept: the two reactants and the two products written in words, in either order on each side.

Slip Writing carbon dioxide and water on the left. Glucose and oxygen go in; carbon dioxide and water come out.

(b) State what the cell does with the energy released. (1 pt)

Model answer The cell transfers the energy released into ATP.

Rubric

Award 1 point for: the energy is transferred into ATP (the molecule that supplies the cell’s work).

Slip Saying the energy is used up or lost. The cell captures it: the energy goes into ATP.

24The oval organelle with two membranes

Here is a muscle cell photographed through an electron microscope. Between the striped fibers that do the contracting sit columns of small rounded bodies.

Part of a muscle cell under the electron microscope: broad striped bands, and between them columns of small rounded bodies — Electron-microscope photograph of part of a muscle cell: broad striped bands running across the picture, and between them columns of small rounded bodies

Here are two of the same kind of body, from a lung cell, at higher magnification. Each body is an oval with two membranes.

Two rounded bodies under the electron microscope, each with a dark outline and short dark lines folded across the inside — Electron-microscope photograph of two rounded bodies in a cell, each with a dark outline and short dark lines reaching in from the outline across the inside; a 200 nanometer scale bar

The outer membrane is smooth. The inner membrane folds in again and again.

Here is one drawn in section, cut open. The outer membrane is smooth; the inner membrane folds in and out again and again.

One mitochondrion cut open: a smooth outer membrane drawn as an oval, and inside it an inner membrane drawn as one continuous line with rounded folds reaching inward from the top and the bottom; labels with leader lines name the outer membrane and the inner membrane

An oval organelle with a smooth outer membrane and a folded inner membrane is called a . One mitochondrion, many mitochondria.

The name comes from how they looked in early microscopes: mitos means thread, and chondrion means little grain.

Every eukaryotic cell has mitochondria. A muscle cell has thousands of them.

A mitochondrion has membranes. But no vesicle buds from it, and none fuses with it.

So a mitochondrion is not a member of the endomembrane system.

What you are expected to know Identify a mitochondrion: an oval organelle with a smooth outer membrane and an inner membrane folded again and again.

Check q7

What is a mitochondrion?

A. A stack of flat membrane sacs that finishes and packs proteins
A stack of flat membrane sacs is the Golgi complex.
B. ✓ An oval organelle with a smooth outer membrane and a folded inner membrane
C. A round membrane sac, acidic inside, holding digestive enzymes
A round sac of digestive enzymes is a lysosome.

Why: A mitochondrion has two membranes.
The outer membrane is smooth.
The inner membrane is folded again and again.

Check q8

Under the electron microscope a muscle cell shows columns of oval bodies. Each body has a smooth outer membrane and an inner membrane folded in again and again.

Which organelle is each body?

A. ✓ A mitochondrion
B. A lysosome
A lysosome has one membrane: it is a single round sac.
C. The rough ER
The rough ER is a network of tubes with ribosomes on its surface.

Why: Two membranes, the outer one smooth and the inner one folded again and again, mark a mitochondrion.
So each body is a mitochondrion.

Check q9

Is a mitochondrion a member of the endomembrane system?

A. Yes
No vesicle buds from a mitochondrion or fuses with one, so its membranes are joined to none of the others.
B. ✓ No

Why: The members of the endomembrane system are joined by vesicles or are continuous with the ER.
No vesicle buds from a mitochondrion or fuses with one.
So a mitochondrion is not a member.

Practice writing an answer

A muscle cell is full of mitochondria.

(a) Describe the two membranes of a mitochondrion. (1 pt)

Model answer A mitochondrion has an outer membrane and an inner membrane.
The outer membrane is smooth.
The inner membrane is folded again and again.

Rubric

Award 1 point for: a smooth outer membrane and an inner membrane folded again and again (highly folded).

Slip Describing one membrane only. A mitochondrion has two: the smooth outer one and the folded inner one.

39What a mitochondrion is for

A muscle cell breaks glucose down with oxygen to release energy. Where in the cell does that happen?

In two places. The cell first splits glucose into smaller pieces in the cytosol, without using oxygen. Those pieces enter a mitochondrion. Inside the mitochondrion, the steps that use oxygen finish the breakdown, and these steps release most of the energy.

So a mitochondrion is where the cell makes most of its ATP.

That is the mitochondrion’s purpose: to release the energy in glucose and transfer it into ATP, as fast as the cell needs it.

A mitochondrion stores no energy. It transfers energy from glucose into ATP.

The cell uses that ATP straight away.

What you are expected to know State the purpose of a mitochondrion: it carries out the oxygen-using steps of aerobic cellular respiration, so it is where the cell makes most of its ATP.

Check q10

Which of the following happens inside a mitochondrion?

A. Amino acids are joined into a protein chain
Ribosomes join amino acids into a protein.
B. Proteins are finished and packed into vesicles
The Golgi complex finishes and packs proteins.
C. ✓ Pieces of glucose are broken down with oxygen to make ATP

Why: The cell splits glucose into pieces in the cytosol.
The pieces enter a mitochondrion.
There, the steps that use oxygen finish the breakdown and release most of the energy, which goes into ATP.
So inside a mitochondrion, pieces of glucose are broken down with oxygen to make ATP.

Check q11

A student says: “Mitochondria store the cell’s energy until it is needed.”

Is the student correct?

A. Yes
ATP is used within seconds of being made.
B. ✓ No

Why: A mitochondrion breaks glucose down as the cell needs energy.
The energy released is transferred into ATP straight away.
The cell uses that ATP for its work.
So mitochondria make ATP when it is needed; they store no energy.

Practice writing an answer

A heart muscle cell contracts about once a second, all day and all night.

(a) State the purpose of the cell’s mitochondria. (1 pt)

Model answer The mitochondria carry out the oxygen-using steps of aerobic cellular respiration.
So the mitochondria are where the cell makes most of its ATP.

Rubric

Award 1 point for: the mitochondria carry out respiration (the oxygen-using breakdown of glucose) and make the cell’s ATP (energy from glucose transferred into ATP).
Accept: ‘break glucose down with oxygen to make ATP’, with or without ‘most of’ the cell’s ATP.

Slip Saying the mitochondria make the cell contract. The fibers contract; the mitochondria make the ATP that the contracting uses.

50Two membranes, two compartments

Two membranes make two separate spaces inside a mitochondrion.

The same mitochondrion cut open, with two shaded spaces: a shaded space between the outer membrane and the inner membrane, and a white space inside the folded inner membrane; labels with leader lines name each space and each membrane

One space lies between the outer membrane and the inner membrane. The other space lies inside the inner membrane.

Respiration breaks glucose down in a series of steps. Some steps happen in one space, and the other steps happen in the other space.

Each space holds a different mix of substances, so each step gets the mix it needs.

A space closed off by a membrane, where its own mix of substances reacts, is called a . A mitochondrion has two.

What you are expected to know Explain what the two membranes of a mitochondrion do: they divide the inside into two compartments, and different steps of respiration happen in different compartments.

Check q12

What is a compartment?

A. ✓ A space closed off by a membrane where its own mix of substances reacts
B. A fold in the inner membrane of a mitochondrion
A fold adds surface to a membrane; it closes off no space.
C. A small vesicle carrying cargo between organelles
A vesicle carries something from one place to another.
A compartment is a closed-off space where its own mix of substances reacts.

Why: A membrane closes a space off.
Inside that space, its own mix of substances reacts.
So a compartment is a space closed off by a membrane where its own mix of substances reacts.

Check q13

What do the two membranes of a mitochondrion do?

A. ✓ Divide the inside into two compartments
B. Make the mitochondrion oval
Shape comes from the outer membrane’s size; the two membranes’ job is to divide the inside.
C. Make the cell’s protein chains
Ribosomes make protein chains.

Why: A mitochondrion has an outer membrane and an inner membrane.
Between them is one space; inside the inner membrane is a second space.
Different steps of respiration happen in the two spaces.
So the two membranes divide the inside into two compartments.

Check q14

A drug dissolves the inner membrane of a mitochondrion.

What happens to the two compartments?

A. They stay separate
Only the inner membrane kept the two spaces apart; with it gone, nothing separates them.
B. A third compartment forms
A new compartment needs a new membrane; dissolving one makes none.
C. ✓ They become one space, and their contents mix

Why: The inner membrane separated the two compartments.
The drug removed the inner membrane.
So nothing separates the two spaces any more.
So they become one space, and their contents mix.

Practice writing an answer

A mitochondrion in a heart muscle cell has an outer membrane and an inner membrane.

(a) State what the two membranes of a mitochondrion do. (1 pt)

Model answer They divide the inside into two compartments.
Different steps of respiration happen in the two compartments.

Rubric

Award 1 point for: they divide the inside into two compartments (spaces), in which different steps of respiration happen.

Slip Saying the membranes make ATP. The mitochondrion makes ATP on and inside the membranes; the membranes’ own job is to divide the space.

61Why the inner membrane is folded

Check q15

Two mitochondria are the same size. One has an inner membrane folded again and again; the other has a smooth inner membrane.

Which mitochondrion has more inner-membrane surface?

A. ✓ The one with the folded inner membrane
B. The one with the smooth inner membrane
A smooth membrane lining a space has the least surface that space allows.
C. The two have the same surface
Folds pack extra membrane into the same space, so the folded one has more surface.

Why: The two mitochondria are the same size.
Folds pack extra membrane into the same space.
So the folded inner membrane has more surface.

Video: Watch: Why the inner membrane is folded

Two mitochondria of the same size. The one with more folds has more inner-membrane surface. So it has more places to make ATP at once, so it makes ATP faster.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L19Be.mp4

Look at the two mitochondria. The right-hand one has three times the folds, so three times the inner-membrane surface.

Two mitochondria of the same size drawn side by side: the left one with three inner-membrane folds, the right one with nine; labels name the smooth outer membrane on the left one and the folded inner membrane on the right one

The mitochondrion makes its ATP on the inner membrane.

More inner-membrane surface means more places to make ATP at once.

Folding the inner membrane again and again packs a large surface into a small mitochondrion.

So a mitochondrion with more folds has more places to make ATP at once. So it makes ATP faster.

The folds are not there to save room. They are there to add surface, so that the mitochondrion makes ATP faster.

What you are expected to know Explain why the inner membrane of a mitochondrion is folded: more surface for the same volume, so more places to make ATP at once, so the mitochondrion makes ATP faster.

Check q16

Why is the inner membrane of a mitochondrion folded again and again?

A. To take up less room inside the cell
Folds add surface; they save no room, because more membrane fits in the same space.
B. ✓ To give more surface for the ATP-making reactions
C. To let glucose cross into the mitochondrion
Glucose crosses through transport proteins in the membrane, not through folds.

Why: The mitochondrion makes its ATP on the inner membrane.
Folds pack more membrane surface into the same space.
More surface means more places to make ATP at once.
So the mitochondrion makes ATP faster.

Check q17

Two mitochondria are the same size. One has an inner membrane folded about twice as much as the other.

Which mitochondrion makes ATP faster?

A. The one with the less folded inner membrane
Less folding means less inner-membrane surface.
So this mitochondrion has fewer places to make ATP at once.
B. The two make ATP at the same rate
The same size does not mean the same surface: the more folded membrane has about twice the surface.
C. ✓ The one with the more folded inner membrane

Why: The mitochondrion makes its ATP on the inner membrane.
The more folded inner membrane has about twice the surface.
So the more folded mitochondrion has about twice as many places to make ATP at once.
So the more folded mitochondrion makes ATP faster.

Practice writing an answer

An endurance athlete’s muscle mitochondria have more inner-membrane folding than an untrained person’s.

(a) Explain why the athlete’s mitochondria make ATP faster than the untrained person’s. (2 pt)

Frame Folding gives the inner membrane …, so …, so …

Model answer Folding gives the inner membrane more surface for the same volume.
The mitochondrion makes its ATP on the inner membrane.
So the folded membrane has more places to make ATP at once.
So the mitochondrion makes ATP faster.

Rubric

Award 1 point for: folding gives the inner membrane more surface (area) in the same space.
Award 1 point for: more surface gives more places to make ATP at once, so ATP is made faster.

Slip Stopping at ‘more surface’. Say what the surface does: more places to make ATP at once, so the mitochondrion makes ATP faster.

74More mitochondria, more work

An untrained person starts running most days. Over several months of training, the number of mitochondria in each muscle cell almost doubles.

	number of mitochondria in one muscle cell
before training	100% (the starting number)
after several months of training	about 190%

training almost doubles the number of mitochondria in each muscle cell

A table with two rows: before training, the number of mitochondria in a muscle cell is set as 100 percent; after several months of training it is about 190 percent

Each mitochondrion is a place where glucose is broken down to make ATP. Twice as many mitochondria means twice as many places making ATP.

So the trained muscle cell makes ATP about twice as fast.

It now makes ATP as fast as its contractions use it. So it can contract harder, for longer.

The rule applies in both directions. A cell that does a lot of work carries a lot of mitochondria; a cell that does little work carries few.

What you are expected to know Predict how a change in the number of mitochondria changes what a muscle cell can do: more mitochondria, ATP made faster, harder work for longer.

Check q18

After 16 weeks of training, an athlete’s muscle cells contain almost twice as many mitochondria as before.

Which of the following can the muscle cells now do better than before?

A. Make glucose from carbon dioxide and water
A mitochondrion breaks glucose down; it does not build glucose.
B. ✓ Make ATP faster from glucose and oxygen
C. Store a larger reserve of ATP for later
Mitochondria make ATP as the cell needs it; they store none.
D. Ship the proteins they make faster
Shipping proteins is done by the rough ER, the Golgi complex and vesicles.

Why: A mitochondrion finishes breaking glucose down, using oxygen.
That breakdown transfers energy from glucose into ATP.
Twice as many mitochondria means twice as many places making ATP.
So the muscle cells make ATP about twice as fast.

Check q19

A trained muscle cell has twice as many mitochondria as an untrained one.

Why does that let it make ATP faster?

A. Each mitochondrion now works twice as fast
Training changed the number of mitochondria, not the speed of each one.
B. ✓ There are twice as many places where ATP is made
C. Each mitochondrion stores twice as much ATP
Mitochondria store no ATP; they make it as the cell needs it.

Why: Each mitochondrion breaks glucose down to make ATP.
Twice as many mitochondria means twice as many places making ATP.
So the cell makes ATP about twice as fast.

Check q20

A bird’s flight muscle cells beat its wings for hours. A cell in the bird’s skin does little work.

Which cell has more mitochondria?

A. ✓ The flight muscle cell
B. The skin cell
A cell that does little work needs little ATP, so it carries few mitochondria.
C. The two cells have the same number
The two cells do very different amounts of work, so they carry different numbers of mitochondria.

Why: Beating wings for hours is a lot of work.
Work uses ATP, and mitochondria make ATP.
A cell that uses a lot of ATP carries a lot of mitochondria.
So the flight muscle cell has more mitochondria.

So a muscle cell gets its energy from glucose and oxygen, broken down inside its mitochondria. The harder the cell works, the more mitochondria it carries.

Glossary

aerobic cellular respiration: Breaking glucose down with oxygen to release energy, inside a cell: glucose + oxygen → carbon dioxide + water, energy released. Aer means air, so aerobic means using oxygen; cellular means inside cells. The cell transfers the energy released into ATP.
compartment: A space closed off by a membrane, where its own mix of substances reacts. A mitochondrion has two: the space between its two membranes, and the space inside the inner membrane.
mitochondrion (plural: mitochondria): An oval organelle with two membranes: a smooth outer membrane and an inner membrane folded again and again. It carries out the oxygen-using steps of aerobic cellular respiration, so it is where a cell makes most of its ATP. From mitos, thread, and chondrion, little grain.

APBIO-U02-L19C The cell that catches light

Topic 2.1 · Cell Structure and Function · 69 steps

A potted plant with three leaves under rays of light, water at its roots and air around it

An oak tree grows from an acorn into tonnes of wood. It never eats.

It takes in water from the soil and carbon dioxide from the air. It catches light. Where does all that wood come from? A plant makes its own food, in one kind of cell above all: the leaf cell.

Unit 2 · Cell Structure and Function

1Carbon dioxide and water, with light

Check q1

In aerobic cellular respiration, glucose and oxygen become which two substances?

A. ✓ Carbon dioxide and water
B. Amino acids and water
Amino acids are what proteins are built from; respiration breaks glucose down.
C. Starch and oxygen
Starch is a store of glucose; respiration breaks glucose down into carbon dioxide and water.

Why: Respiration breaks glucose down with oxygen.
The glucose becomes carbon dioxide and water.
Energy is released as it does.

Video: Watch: Carbon dioxide and water, with light

A leaf cell takes in carbon dioxide and water. Using the energy of light, it joins them into glucose. Oxygen is left over.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L19Ca.mp4

A leaf cell takes in carbon dioxide from the air. Water reaches it from the roots.

Using the energy of light, the leaf cell joins carbon dioxide and water into glucose. Oxygen is left over and leaves the leaf.

The word equation for photosynthesis: carbon dioxide plus water gives glucose plus oxygen, using the energy of light

Written as a word equation, the change reads carbon dioxide + water → glucose + oxygen, using the energy of light.

Making glucose from carbon dioxide and water, using the energy of light, is called .

The name says what happens: photo means light, and synthesis means building. Photosynthesis is building with light.

Compare the two word equations. What respiration uses, glucose and oxygen, photosynthesis makes. What respiration makes, carbon dioxide and water, photosynthesis uses. Inside the cell, the two are carried out by different steps.

process	word equation	the energy
photosynthesis	carbon dioxide + water → glucose + oxygen	the energy of light, stored in glucose
aerobic cellular respiration	glucose + oxygen → carbon dioxide + water	released, and transferred into ATP

photosynthesis builds glucose; respiration breaks it down

A table comparing the two processes: photosynthesis, carbon dioxide plus water gives glucose plus oxygen, the energy of light stored in glucose; aerobic cellular respiration, glucose plus oxygen gives carbon dioxide plus water, energy released and transferred into ATP

Respiration breaks glucose down and releases energy. Photosynthesis builds glucose and stores the light’s energy in it.

That is where a tree’s wood comes from. The plant builds that glucose into its cell walls.

Those cell walls are wood. So almost all of the tree began as carbon dioxide and water.

What you are expected to know State the word equation for photosynthesis: carbon dioxide + water → glucose + oxygen, using the energy of light.

What you are expected to know Describe what photosynthesis does with the energy of light: it stores the energy in glucose.

Check q2

What is photosynthesis?

A. Breaking glucose down with oxygen to release energy
Breaking glucose down with oxygen is aerobic cellular respiration.
B. ✓ Making glucose from carbon dioxide and water using the energy of light
C. Taking food in from the soil through the roots
Roots take in water and minerals; the plant makes its own glucose in its leaves.

Why: Photo means light and synthesis means building.
Photosynthesis builds glucose from carbon dioxide and water.
The energy to build it comes from light.

Check q3

Where does the energy for photosynthesis come from?

A. ✓ Light
B. Glucose
Glucose is what photosynthesis makes.
C. Oxygen
Oxygen is left over and leaves the leaf.

Why: Photosynthesis builds glucose from carbon dioxide and water.
Building takes energy.
The leaf cell catches that energy from light.

Check q4

Which gas does photosynthesis give off?

A. Carbon dioxide
Carbon dioxide is taken in and used; oxygen is what is left over.
B. ✓ Oxygen

Why: Photosynthesis joins carbon dioxide and water into glucose.
Oxygen is left over.
So the gas given off is oxygen.

Check q5

A student says: “A plant gets its food from the soil.”

Is the student correct?

A. Yes
Soil holds no glucose.
B. ✓ No

Why: Food for a plant is glucose.
The plant makes glucose from carbon dioxide and water, using the energy of light.
The soil supplies water and minerals only.
So the plant makes its own food.

Practice writing an answer

A leaf cell in bright light takes in carbon dioxide and water.

(a) State the word equation for photosynthesis. (1 pt)

Frame … + … → … + …, using the energy of light

Model answer carbon dioxide + water → glucose + oxygen
The energy comes from light.

Rubric

Award 1 point for: carbon dioxide + water → glucose + oxygen, using light (energy).
Accept: the two reactants and the two products written in words, in either order on each side.

Slip Writing glucose and oxygen on the left. Carbon dioxide and water go in; glucose and oxygen come out.

20The parts of a leaf cell, and the green ones

Here are cells from a moss leaf, photographed through a light microscope. Each cell is packed with small green bodies.

A row of moss leaf cells under the light microscope, each packed with small bright green oval bodies — Light-microscope photograph of a row of moss leaf cells, each a box with a clear cell wall, packed with dozens of small bright green oval bodies

Here is one leaf cell drawn in section. Around the outside is the rigid cell wall, with the plasma membrane just inside it.

A leaf cell cut open, drawn as a rounded box: a thick cell wall, the plasma membrane just inside it, a nucleus at the left, a large central vacuole in the middle, six green oval chloroplasts, three small mitochondria; labels with leader lines name every part

Inside are the cytosol, the nucleus, a large central vacuole full of watery fluid, and a few mitochondria.

Check q6

The same leaf cell cut open, with no labels: a thick outer layer, a large space in the middle, a round body at the left, six green ovals and three small ovals

In the drawing, which part is the large fluid-filled space in the middle?

A. The nucleus
The nucleus is the round body at the left, holding the cell’s DNA.
B. The cell wall
The cell wall is the rigid layer around the outside.
C. ✓ The central vacuole

Why: A plant cell has one large central vacuole.
The central vacuole is full of watery fluid.
So the large fluid-filled space in the middle is the central vacuole.

And there are the green bodies that fill the cells in the photograph. The green substance inside them catches the light.

A green oval body in which a plant cell makes glucose by photosynthesis is called a . Chloro means green, and plast means a formed body.

Chloroplasts are found in plant cells and in algae, the simple plant-like organisms of ponds and seas. Animal cells have none.

So a leaf cell is a plant cell with lots of chloroplasts: the cell that catches light and makes the plant’s glucose.

What you are expected to know Identify the parts of a leaf cell: cell wall, plasma membrane, cytosol, nucleus, central vacuole, mitochondria and chloroplasts.

What you are expected to know Identify a chloroplast: the green oval organelle in a plant cell that makes glucose by photosynthesis.

Check q7

What is a chloroplast?

A. ✓ The green oval organelle that makes glucose by photosynthesis
B. The large fluid-filled space in the middle of a plant cell
The large fluid-filled space is the central vacuole.
C. The rigid layer around the outside of a plant cell
The rigid outer layer is the cell wall.

Why: Chloro means green, and plast means a formed body.
The green ovals in a leaf cell make the plant’s glucose by photosynthesis.
So a chloroplast is the green organelle that makes glucose by photosynthesis.

Check q8

Which of these cells contains chloroplasts?

A. A muscle cell
A muscle cell is an animal cell, and animal cells have no chloroplasts.
B. ✓ A leaf cell
C. A pancreas cell
A pancreas cell is an animal cell, and animal cells have no chloroplasts.

Why: Chloroplasts are found in plant cells and in algae.
A leaf cell is a plant cell.
So the leaf cell contains chloroplasts.

Check q9

A student keeps a plant in the dark for two days, so that its leaves hold no starch, then gives it 12 hours of light. Iodine, which turns starch (the plant’s store of glucose) dark, then darkens only the leaf cells that contain chloroplasts.

Which of the following does this show?

A. Light alone makes starch in any leaf cell
Every leaf cell had light.
Only the cells with chloroplasts turned dark.
B. The roots took in starch from the soil and sent it to the cells with chloroplasts
Roots take in water and minerals from the soil, not starch.
The starch was made in the leaf, from the glucose photosynthesis makes.
C. ✓ Only the cells with chloroplasts made glucose, stored as starch
D. The cells without chloroplasts made starch and exported it
A cell without chloroplasts makes no glucose.

Why: Photosynthesis is the light-driven making of glucose from carbon dioxide and water.
A cell makes that glucose inside its chloroplasts.
So only the cells with chloroplasts made glucose.
Some of that glucose was stored as starch.
So iodine darkened only those cells.

Practice writing an answer

A leaf cell contains chloroplasts.

(a) Describe a chloroplast. (1 pt)

Model answer A chloroplast is a green oval organelle in a plant cell.
The chloroplast joins carbon dioxide and water into glucose, using the energy of light.

Rubric

Award 1 point for: a green (oval) organelle in plant cells (and algae) that makes glucose by photosynthesis.

Slip Saying only ‘it is green’. Say what it is and what it does: an organelle that makes glucose by photosynthesis.

35Inside a chloroplast

Cut a chloroplast open. It has two membranes, one just inside the other, and both are smooth.

One chloroplast cut open: an outer membrane and an inner membrane, one just inside the other, both smooth, and five stacks of flat discs inside; labels with leader lines name the outer membrane, the inner membrane, the stacks of flat discs and the fluid around the discs

Inside the inner membrane sit stacks of flat membrane discs, like piles of coins.

The green, light-catching substance sits in the membrane of those discs. So the discs are where the light is caught.

Photosynthesis has two parts, in two places inside the chloroplast. The discs catch the light.

The fluid around the discs is where the glucose is built.

A chloroplast and a mitochondrion both have two membranes. Inside they differ: a mitochondrion has a folded inner membrane; a chloroplast has a smooth inner membrane and stacks of discs.

What you are expected to know Describe the inside of a chloroplast: two smooth membranes, one inside the other, and stacks of flat membrane discs where the light is caught.

Check q10

Where inside a chloroplast is light caught?

A. In the outer membrane
The outer membrane is a smooth boundary with none of the green, light-catching substance in it.
B. ✓ On the stacks of flat membrane discs
C. In the space between the two membranes
The space between the membranes holds none of the green, light-catching substance.

Why: The green, light-catching substance sits in the membrane of the flat discs.
The discs are stacked inside the inner membrane.
So light is caught on the stacks of discs.

Check q11

In an electron micrograph of a leaf cell, an oval organelle shows two smooth membranes, one inside the other, and stacks of flat membrane discs inside the inner one.

Which organelle is it?

A. A mitochondrion
A mitochondrion’s inner membrane is folded again and again, and it has no stacks of discs.
B. A Golgi complex
A Golgi complex is a single stack of flat sacs with no outer membranes around it.
C. ✓ A chloroplast

Why: Two smooth membranes, one inside the other, with stacks of flat discs inside, mark a chloroplast.
So the organelle is a chloroplast.

Check q12

An organelle has a smooth outer membrane and an inner membrane folded again and again.

Which organelle is it?

A. ✓ A mitochondrion
B. A chloroplast
A chloroplast’s inner membrane is smooth, with stacks of discs inside it.

Why: A folded inner membrane marks a mitochondrion.
A chloroplast’s inner membrane is smooth.
So the organelle is a mitochondrion.

Practice writing an answer

A chloroplast from a leaf cell is cut open.

(a) Describe the inside of a chloroplast. (1 pt)

Model answer Two smooth membranes, one inside the other.
Inside the inner membrane, stacks of flat membrane discs.
The discs are where the light is caught.

Rubric

Award 1 point for: two smooth membranes (one inside the other) and stacks of flat membrane discs inside, where light is caught.

Slip Describing a folded inner membrane. That is a mitochondrion; a chloroplast’s two membranes are smooth, and the discs are inside.

47Why stacks of flat discs

Check q13

A flat slab and a sphere hold the same volume.

Which has the larger surface area?

A. The sphere
A sphere has the least surface possible for its volume.
B. The two have the same surface area
The same volume does not mean the same surface: the flatter the shape, the more surface it has.
C. ✓ The flat slab

Why: A sphere packs its volume behind the smallest possible surface.
A flat slab spreads the same volume thin, so it has far more surface.
So the flat slab has the larger surface area.

The green, light-catching substance sits in membrane. The more membrane surface a chloroplast has, the more of that substance it can carry.

A thin flat disc drawn at the left and a ball drawn at the right, of the same volume; the disc is wide and thin, the ball compact

A flat disc has a large surface for its volume. A ball of the same volume has far less.

Stacking many flat discs packs a huge membrane surface into one small chloroplast.

So the chloroplast catches more light at once. So the chloroplast builds more glucose at once.

The chloroplast makes glucose faster.

What you are expected to know Explain why the membrane inside a chloroplast is arranged as stacks of flat discs: more surface for the same volume, so more of the green, light-catching substance, so more photosynthesis at once.

Check q14

Why is the membrane inside a chloroplast arranged as stacks of flat discs?

A. ✓ Flat discs give more membrane surface for the green, light-catching substance
B. Flat discs let the chloroplast take up less room in the cell
Stacked discs pack surface in; saving room is no part of their purpose.
C. Flat discs let water flow through the chloroplast more easily
Water crosses the membranes wherever they are; the discs are flat to give more surface.

Why: The green, light-catching substance sits in membrane.
Flat discs have a large surface for their volume.
Stacked, they pack a huge surface into the chloroplast.
So more of that substance fits.
So the chloroplast builds more glucose at once.

Practice writing an answer

Inside a chloroplast, the membrane that catches light is arranged as stacks of flat discs rather than as one round sac.

(a) Explain how the stacks of flat discs let the chloroplast make glucose faster. (2 pt)

Frame Flat discs have …, so …, so …

Model answer Flat discs have a large surface for their volume.
So stacks of discs pack a large membrane surface into the chloroplast.
The green, light-catching substance sits in that membrane, so more of it fits.
So the chloroplast catches more light at once.
So the chloroplast makes glucose faster.

Rubric

Award 1 point for: flat discs (stacked) give a large membrane surface for the same volume.
Award 1 point for: more surface holds more of the green, light-catching substance, so the chloroplast catches more light at once and makes glucose faster.

Slip Stopping at ‘more surface’. Say what the surface does: more of the light-catching substance, so more photosynthesis at once.

57Why a leaf cell keeps its mitochondria

Check q15

Which organelle makes ATP by aerobic cellular respiration?

A. ✓ Mitochondria
B. Chloroplasts
Chloroplasts make glucose by photosynthesis.
C. The Golgi complex
The Golgi complex finishes and packs proteins.

Why: Aerobic cellular respiration is the breakdown of glucose with oxygen.
Mitochondria are the organelles that break glucose down with oxygen.
So mitochondria make the ATP.

Video: Watch: Why a leaf cell keeps its mitochondria

A leaf cell by day and by night. Chloroplasts make glucose only in the light. Mitochondria turn glucose into ATP all the time.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L19Ce.mp4

A leaf cell is full of chloroplasts. Cut it open and there are mitochondria too.

Chloroplasts make glucose, and only while light falls on them. At night they make nothing.

The leaf cell needs ATP all the time: to build proteins, to move materials, in the dark as well as in the light.

Its mitochondria break glucose down with oxygen and make ATP, by day and by night.

So a leaf cell needs both. Chloroplasts make the glucose; mitochondria turn that glucose into ATP.

What you are expected to know Explain why a leaf cell has mitochondria as well as chloroplasts: chloroplasts make glucose only in the light, and the cell needs ATP all the time.

Check q16

A student says: “A leaf cell has chloroplasts, so a leaf cell has no need of mitochondria.”

Is the student correct?

A. Yes
Chloroplasts make glucose only in the light; the cell needs ATP in the dark too.
B. ✓ No

Why: Chloroplasts make glucose, and only in the light.
The cell needs ATP all the time, dark or light.
Mitochondria make that ATP from the glucose.
So the leaf cell needs its mitochondria.

Practice writing an answer

A leaf cell contains chloroplasts and mitochondria.

(a) Explain why a leaf cell needs mitochondria as well as chloroplasts. (1 pt)

Frame Chloroplasts …; mitochondria …; so …

Model answer Chloroplasts make glucose from carbon dioxide and water, using light.
Mitochondria break that glucose down with oxygen and make ATP.
The leaf cell needs ATP for its work, in the light and in the dark.
So the leaf cell needs mitochondria as well as chloroplasts.

Rubric

Award 1 point for: chloroplasts make glucose (photosynthesis) and mitochondria break glucose down to make ATP (aerobic cellular respiration), and the cell needs ATP, so the cell needs both.
Accept: the leaf cell makes glucose in its chloroplasts and gets ATP from that glucose in its mitochondria.

Slip Saying chloroplasts take the place of mitochondria. Chloroplasts make the glucose. Mitochondria break the glucose down to make the ATP the cell uses.

So a tree’s wood began as carbon dioxide and water, joined into glucose by its chloroplasts using the energy of light. Its mitochondria turn some of that glucose into ATP, day and night.

Glossary

photosynthesis: Making glucose from carbon dioxide and water, using the energy of light: carbon dioxide + water → glucose + oxygen. Photo means light, synthesis means building. A plant cell makes this glucose inside its chloroplasts.
chloroplast: A green oval organelle in plant cells and algae that makes glucose by photosynthesis. Chloro means green, plast means a formed body. Two smooth membranes, one inside the other, with stacks of flat membrane discs inside where the light is caught.

APBIO-U02-L19D Read a cell’s job from its parts

Topic 2.1 · Cell Structure and Function · 45 steps

Four cells cut open, side by side: a pancreas cell full of folded membrane, a long muscle cell full of oval bodies, a boxy leaf cell full of green ovals, and a round white blood cell holding small sacs

Every one of your cells carries the same set of organelles. Yet cut open a muscle cell, a pancreas cell, a leaf cell and a white blood cell, and they look nothing alike inside.

The difference is how much of each organelle a cell has. Give a cell a job, and it fills up with the organelle that does that job.

Unit 2 · Cell Structure and Function

1From a cell’s job to the organelle it is full of

Check q1

Which organelle digests a bacterium that a white blood cell has engulfed?

A. The smooth ER
The smooth ER makes membrane lipids and breaks down harmful molecules such as alcohol.
B. ✓ Lysosomes
C. The Golgi complex
The Golgi complex finishes and packs proteins.

Why: A lysosome is an acidic sac of digestive enzymes.
A lysosome fuses with the vesicle holding the bacterium.
So lysosomes digest the bacterium.

Video: Watch: Four cells, one set of organelles

The pancreas cell, the muscle cell, the leaf cell and the white blood cell: the same organelles, in different amounts, for four different jobs.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L19Da.mp4

Here are four cells and what each one does. Each cell is full of the organelle that does its job.

A pancreas cell under the electron microscope: a large gray oval nucleus, stripes of folded membrane filling the cell around it, dark round packets at the right — A table with four columns, one per cell: the pancreas cell makes and ships digestive protein and is full of ribosomes, rough ER and Golgi complex; the muscle cell contracts all day and is full of mitochondria; the leaf cell catches light and makes glucose and is full of chloroplasts, with mitochondria as well; the white blood cell engulfs and digests bacteria and is full of lysosomes; the last row shows a photograph of each cell

Part of a muscle cell under the electron microscope: broad striped bands with columns of small rounded bodies between them — A table with four columns, one per cell: the pancreas cell makes and ships digestive protein and is full of ribosomes, rough ER and Golgi complex; the muscle cell contracts all day and is full of mitochondria; the leaf cell catches light and makes glucose and is full of chloroplasts, with mitochondria as well; the white blood cell engulfs and digests bacteria and is full of lysosomes; the last row shows a photograph of each cell

	pancreas cell	muscle cell	leaf cell	white blood cell
its job	makes and ships digestive protein	contracts, all day	catches light and makes glucose	engulfs and digests bacteria
what it is full of	ribosomes, rough ER and Golgi complex	mitochondria	chloroplasts (and mitochondria)	lysosomes
the cell

The pancreas cell ships protein. So it is full of ribosomes, rough ER and Golgi complex.

The muscle cell contracts all day. Every contraction uses ATP.

So the muscle cell is full of mitochondria.

The leaf cell catches light and makes glucose. So it is full of chloroplasts, with mitochondria as well.

The white blood cell engulfs bacteria and has a lot to digest. So it is full of lysosomes.

All four cells have ribosomes, ER, Golgi complex, lysosomes and mitochondria. Only the leaf cell has chloroplasts.

What differs most is how much of each organelle a cell has.

To predict what a cell is full of, ask which organelle does the cell’s job. The cell is full of that organelle.

What you are expected to know Predict from a cell’s job which organelle it has lots of.

What you are expected to know Name the organelle for each job: protein export, work that uses ATP, catching light, digesting what the cell takes in.

Check q2

A cell does a lot of work that uses ATP.

Which organelle does it have lots of?

A. ✓ Mitochondria
B. Chloroplasts
Chloroplasts make glucose from carbon dioxide and water, using light; they make no ATP for the cell’s work.
C. Lysosomes
Lysosomes digest what the cell takes in.

Why: Work uses ATP.
Mitochondria make ATP by aerobic cellular respiration.
So a cell that does a lot of work has lots of mitochondria.

Check q3

A cell makes and releases a lot of protein.

Which organelle does it have lots of?

A. Lysosomes
Lysosomes digest; they make no protein.
B. ✓ Rough ER
C. Mitochondria
Mitochondria make ATP; they make no protein.

Why: Ribosomes on the rough ER make protein for export, and the ER folds it.
So a cell that makes and releases a lot of protein has lots of rough ER.

Check q4

A cell catches light and makes glucose.

Which organelle does it have lots of?

A. Mitochondria
Mitochondria break glucose down; they catch no light.
B. Lysosomes
Lysosomes digest; they catch no light.
C. ✓ Chloroplasts

Why: Photosynthesis, the making of glucose using the energy of light, happens inside chloroplasts.
So a cell that catches light has lots of chloroplasts.

Check q5

A cell takes in and digests a lot of material.

Which organelle does it have lots of?

A. ✓ Lysosomes
B. Chloroplasts
Chloroplasts catch light; they digest nothing.
C. Rough ER
Rough ER makes protein; it digests nothing.

Why: Lysosomes hold the digestive enzymes that break down what a cell takes in.
So a cell that digests a lot of material has lots of lysosomes.

Check q6

A cell in the stomach lining makes and releases large amounts of a digestive protein.

Which organelles is the cell most likely to have lots of?

A. Lysosomes
Lysosomes digest what a cell takes in; this cell’s job is to make and send out protein.
B. ✓ Ribosomes, rough ER and Golgi complex
C. Chloroplasts and a large central vacuole
Chloroplasts are found in plant cells; a stomach cell is an animal cell.

Why: The cell makes and exports a protein.
Ribosomes make the protein, the rough ER folds it, and the Golgi complex finishes and packs it.
So the cell has lots of ribosomes, rough ER and Golgi complex.

20How that organelle serves the job

Naming the organelle is half the job. The other half is saying what that organelle does for the cell, one sentence per organelle.

	pancreas cell	muscle cell	leaf cell	white blood cell
its job	makes and ships digestive protein	contracts, all day	catches light and makes glucose	engulfs and digests bacteria
what it is full of	ribosomes, rough ER and Golgi complex	mitochondria	chloroplasts (and mitochondria)	lysosomes
the cell

Take the pancreas cell. Ribosomes make the protein, and the rough ER folds it.

The Golgi complex finishes and packs the protein. Vesicles carry it to the cell surface.

Take the muscle cell. Its mitochondria make the ATP that each contraction uses.

Take the leaf cell. Its chloroplasts catch light and make glucose; its mitochondria turn that glucose into ATP.

Take the white blood cell. Its lysosomes digest the bacteria it engulfs.

What you are expected to know Explain how the organelle a cell is full of serves the cell’s job, one sentence per organelle.

Check q7

A muscle cell is packed with mitochondria.

How do the mitochondria serve the muscle cell’s job?

A. ✓ They make the ATP that contracting uses
B. They build the protein fibers that contract
Ribosomes build protein fibers.
C. They store glucose for the cell
Mitochondria store no glucose; they break glucose down to make ATP.

Why: Contracting is work, and work uses ATP.
Mitochondria make ATP by aerobic cellular respiration.
So the mitochondria supply the ATP that contracting uses.

Check q8

A white blood cell is full of lysosomes.

How do the lysosomes serve the white blood cell’s job?

A. They carry the cell towards bacteria
Lysosomes digest; they do not move the cell.
B. They make the proteins the cell exports
Ribosomes make proteins.
C. ✓ They digest the bacteria the cell engulfs

Why: The white blood cell engulfs bacteria.
Lysosomes hold digestive enzymes.
A lysosome fuses with the vesicle holding the bacterium and digests it.
So the lysosomes digest what the cell engulfs.

Practice writing an answer

A cell in the stomach lining makes and releases large amounts of a digestive protein. It has lots of ribosomes, rough ER, Golgi complex and vesicles.

(a) Explain how each of these four organelles serves the cell’s job. Write one short sentence per organelle. (2 pt)

Model answer The ribosomes make the protein.
The rough ER folds the protein and passes it on in vesicles.
The Golgi complex finishes the protein and packs it into vesicles.
The vesicles carry the protein to the cell surface, where it is released.

Rubric

Award 1 point for: ribosomes make the protein, and the rough ER folds it (and passes it on).
Award 1 point for: the Golgi complex finishes and packs the protein into vesicles, and vesicles carry it to the surface for release.

Slip Saying ‘the ribosomes help make protein’ and stopping. Name what each part does: makes, folds, finishes and packs, carries out.

31Which organelle is this cell packed with?

You can reason in the other direction too. Look at what a cell is packed with.

Name that organelle from its shape and its membranes. Then you know the cell’s job.

Name the organelle from its shape and its membranes.

organelle	membranes	inside	the cell’s job it serves
mitochondrion	smooth outer, folded inner	—	makes ATP for work
chloroplast	two smooth	stacks of flat discs	makes glucose from light
Golgi complex	one stack of flat sacs	—	finishes and packs protein
rough ER	one connected network with ribosomes on it	—	makes protein for export
lysosome	one round sac	digestive enzymes	digests what the cell takes in

name the organelle from its shape and its membranes

A table with five rows, one per organelle, comparing its membranes, what is inside it and the cell job it serves: mitochondrion, chloroplast, Golgi complex, rough ER, lysosome

What you are expected to know Identify the organelle a cell is packed with from its shape and membranes.

Check q9

Here are two organelles of the same kind from a lung cell, photographed through an electron microscope.

Two rounded bodies under the electron microscope, each with a dark outline and short dark lines reaching in across the inside — Electron-microscope photograph of two rounded bodies in a cell, each with a dark outline and short dark lines reaching in from the outline across the inside

Which organelle are they?

A. Chloroplasts
A chloroplast has a smooth inner membrane with stacks of discs inside.
B. ✓ Mitochondria
C. Golgi complexes
A Golgi complex is a stack of separate flat sacs.

Why: A smooth outer membrane surrounds an inner membrane folded in again and again.
Two membranes with the inner one folded mark a mitochondrion.

Check q10

Here is a cell photographed through an electron microscope. Stripes of membrane fill it, layer upon layer.

A cell under the electron microscope: a large gray oval body in the middle, layered stripes filling the cell around it, dark round blobs at the right — Electron-microscope photograph of a cell: a large gray oval body in the middle, layered stripes filling the cell around it, and several dark round blobs at the right edge

Which organelle is this cell packed with?

A. ✓ Rough ER
B. Mitochondria
Mitochondria are separate oval bodies, each with its own two membranes.
C. Lysosomes
Lysosomes are separate round sacs.

Why: The membrane fills the cell in long stripes, layer upon layer, as one connected network.
One connected network of folded membrane filling a cell is the ER.
ER that makes protein for export carries ribosomes: rough ER.
So the cell is packed with rough ER.

Check q11

Here are cells from a moss leaf, photographed through a light microscope. Each cell is packed with small green bodies.

A row of boxy plant cells under the light microscope, each packed with small bright green oval bodies — Light-microscope photograph of a row of boxy plant cells, each packed with dozens of small bright green oval bodies

Which organelle are these cells packed with?

A. Mitochondria
Mitochondria have no green substance in them.
B. Golgi complexes
A Golgi complex has no green substance in it.
C. ✓ Chloroplasts

Why: The bodies are green.
The green organelle in a plant cell is the chloroplast.
So the cells are packed with chloroplasts.

Check q12

A cell lining a kidney tubule moves ions from the tubule fluid into the blood all day, against their concentration gradients. The cell is packed with oval organelles, each with a smooth outer membrane and a folded inner membrane.

Which organelle is the cell packed with?

A. Lysosomes
A lysosome is a sac with a single membrane.
B. Golgi complexes
A Golgi complex is a stack of flat sacs.
C. Rough ER
Rough ER is one connected network with ribosomes on it.
D. ✓ Mitochondria

Why: Each organelle has a smooth outer membrane around a folded inner membrane.
Two membranes, the inner one folded, mark a mitochondrion.
So the cell is packed with mitochondria.

Check q13

A cell lining a kidney tubule pumps ions into the blood all day, against their concentration gradients. The cell is packed with mitochondria.

Why does this cell need so many mitochondria?

A. Mitochondria carry the ions across the membrane
Pump proteins in the membrane carry the ions.
B. ✓ Mitochondria make the ATP that the pumps use
C. Mitochondria digest the ions
Ions are not digested.
The pumps use ATP for that work.

Why: Moving ions against their concentration gradients is work.
Pumps in the membrane do that work, and the pumps use ATP.
Mitochondria make ATP.
So a cell that pumps all day is packed with mitochondria.

Check q14

Which of the following cells contains the most lysosomes?

A. A gland cell that ships protein
A gland cell that ships protein is full of ribosomes, rough ER and Golgi complex, and takes in little to digest.
B. A leaf cell that catches light
A leaf cell is full of chloroplasts and takes in nothing to digest.
C. ✓ A white blood cell that engulfs bacteria
D. A muscle cell that contracts
A muscle cell is packed with mitochondria, and digesting is no part of its job.

Why: Lysosomes digest what a cell takes in.
A white blood cell that engulfs bacteria has the most to digest.
So the white blood cell is the cell full of lysosomes.

Check q15

A digestive enzyme is made by ribosomes on the rough ER.

Which route takes it to a lysosome?

A. ✓ ER → vesicle → Golgi complex → vesicle → lysosome
B. ER → cytosol → lysosome
A protein on this route is never loose in the cytosol.
C. Ribosome → plasma membrane → lysosome
The enzyme is addressed to stay inside the cell; it never reaches the plasma membrane.

Why: A vesicle carries the enzyme from the ER to the Golgi complex.
The Golgi complex finishes it and packs it into another vesicle.
That vesicle becomes the lysosome.

Practice writing an answer

A sperm cell has three jobs, in order. First, the sperm swims for hours, beating its tail. When the sperm reaches the egg, it releases digestive enzymes from a sac at its tip. The enzymes cut through the egg’s outer coat. Then the sperm carries its DNA into the egg. The sperm’s middle section, just behind the head, is packed with one kind of organelle.

(a) Identify the organelle the sac of digestive enzymes at the tip most resembles, and justify your answer. (1 pt)

Model answer The sac most resembles a lysosome.
A lysosome holds digestive enzymes inside a membrane and breaks down what the cell needs digested.
The sac at the sperm’s tip also holds digestive enzymes inside a membrane, and its enzymes cut through the egg’s coat.
So the sac at the tip most resembles a lysosome.

Rubric

Award 1 point for: a lysosome, because it is a membrane sac of digestive enzymes with a digesting job (breaking down the egg’s coat).
Accept: a lysosome-like sac of digestive enzymes.

Slip Calling the sac a vacuole or just a vesicle, with no job named. Its contents are digestive enzymes, and a sac of digestive enzymes is a lysosome.

(b) Describe the route the digestive enzymes took, when the cell was developing, from the ribosomes that made them to the sac. (1 pt)

Model answer Ribosomes on the rough ER made the enzymes.
The ribosomes passed the enzymes into the ER’s interior.
Vesicles budded from the ER and carried the enzymes to the Golgi complex.
The Golgi complex finished the enzymes and packaged them into new vesicles.
Those vesicles formed the sac at the tip.

Rubric

Award 1 point for: made by ribosomes on the rough ER and passed into the ER’s interior, then a vesicle to the Golgi complex (finished and packaged), then a vesicle that forms the sac.
Accept: rough ER, then Golgi, then vesicle, then the sac, in that order.

Slip Having the enzymes made loose in the cytosol and then wrapped up. Ribosomes on the rough ER make the digestive enzymes, which travel inside membrane the whole way, through the Golgi complex, to the sac.

(c) Predict which organelle fills the middle section, and explain why that organelle is needed there. (1 pt)

Model answer The middle section is packed with mitochondria.
Beating the tail for hours is work.
The sperm uses ATP for that work.
Mitochondria make ATP by aerobic cellular respiration, breaking glucose down with oxygen.
So the section that drives the tail is packed with mitochondria.

Rubric

Award 1 point for: mitochondria, which make ATP (by aerobic cellular respiration) to power the tail’s beating.
Accept: mitochondria release energy from food to power movement.

Slip Naming ribosomes or rough ER because the tail is made of protein. The tail is already built. Hours of movement need ATP, and mitochondria make ATP.

The pancreas cell is full of rough ER and Golgi complex, the muscle cell with mitochondria, the leaf cell with chloroplasts, the white blood cell with lysosomes. The same organelles, in different amounts, because each cell does a different job.

APBIO-U02-P21 Practice questions: Topic 2.1

Topic 2.1 · Cell Structure and Function · 10 MCQ · 2 FRQ · for APBIO-U02-T21

These practice questions have the shape of the Topic 2.1 test. Answer every question. For each multiple-choice question, pick one option and press Check; the feedback gives the reasoning. For the first free-response question, you work through one scenario in small steps, and each step offers a hint if you want one. For the second, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it.

Video: Watch first: Topic 2.1 summary, part 1: two kinds of cell, and the network inside one

A bacterium and an onion skin cell to one scale: what every cell has, where the DNA sits, and the ER running out from the nuclear envelope.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T21-summary.mp4

Q1 P21-q01

An antibiotic binds to the ribosomes of a bacterium. Within minutes the bacterium stops making new proteins, although its DNA and its mRNA molecules are intact and its supply of amino acids is normal.

Why does protein-making stop?

A. Ribosomes make the mRNA, so no instructions reach the rest of the cell
The mRNA is intact.
An mRNA is copied from the DNA, not made by ribosomes.
B. ✓ Ribosomes join amino acids into proteins in the order an mRNA sets, and that is blocked
C. Ribosomes store the amino acids, which are now locked away inside them
Amino acids are dissolved in the cytosol, and their supply is normal.
D. Ribosomes hold the cell's DNA, so the instructions for every protein are lost
A bacterium’s DNA lies in its nucleoid, not in its ribosomes.

Why: A ribosome makes a protein by joining amino acids in the order a messenger RNA sets.
The DNA, the mRNA and the amino acids are all present.
The antibiotic blocks the ribosomes, so nothing joins the amino acids together.
So protein-making stops.

Q2 P21-q02

Cells from a mushroom, a fern, a trout and a soil bacterium are examined. Every one of them contains ribosomes built from rRNA and protein, with the same basic shape and the same job.

What is this shared feature evidence of?

A. ✓ Common ancestry: the four kinds of organism descend from shared ancestors
B. Ribosomes spreading from one kind of organism to another by infection
Each cell makes its own ribosomes from instructions in its own DNA.
C. Each kind of organism building the same particle by chance
A ribosome is a complicated particle.
Four groups arriving at the same complicated particle by chance is far less likely than all four inheriting it.
D. The four kinds of organism having identical DNA
The four organisms differ in most of their DNA.

Why: Every known cell contains ribosomes of the same basic build.
A feature shared by all living things is evidence that they descend from shared ancestors: their common ancestry.
Each group inherited its ribosomes rather than inventing them.

Q3 P21-q03

The drawing shows a cell 3 μm long from a pond. Y is a region where a DNA stain collects, with no membrane around it. A student labels Y “the nucleus”.

A pond cell 3 μm long. W: stiff outer layer; X: thin layer inside it; Y: region where a DNA stain collects; Z: one of the small dots throughout the cell.

Which statement corrects the student's label?

A. Y is a vacuole, a membrane sac of stored water, and the cell is a plant cell
A vacuole is a membrane-enclosed sac.
Y has no membrane around it.
B. Y is the nucleus after all, a very small one with a very thin envelope
A nucleus, whatever its size, is enclosed by a double membrane.
C. Y is a nucleus whose envelope the stain has dissolved
A DNA stain does not dissolve membranes.
Y has no membrane around it.
D. ✓ Y is the nucleoid, so the cell is prokaryotic

Why: Y is where the DNA stain collects, and no membrane surrounds Y.
DNA lying in a region with no membrane around it is a nucleoid.
A cell whose DNA lies in a nucleoid is prokaryotic.
W is its cell wall, X its plasma membrane and Z its ribosomes.

Q4 P21-q04

The drawing shows a liver cell 25 μm across: many structures scattered through the cytosol, each wrapped in its own membrane. The largest, 6 μm across, is bounded by a double membrane and is shaded where a DNA stain collects.

A liver cell 25 μm across, as seen through a microscope; the shading in the largest structure is where a DNA stain collects.

What are the wrapped structures, and what is the largest one?

A. Prokaryotic cells living inside the liver cell; the largest is their host's nucleoid
A prokaryotic cell’s nucleoid has no membrane around it; here the largest structure has a double membrane around its DNA.
B. Ribosomes, the particles that make protein; the largest is the nucleoid
A ribosome has no membrane around it.
A ribosome is far too small to show as a wrapped structure at this magnification.
C. ✓ Membrane-bound organelles; the largest is the nucleus, holding the cell's DNA
D. Vacuoles storing water; the largest is a vacuole that has filled with DNA
A vacuole stores water and other materials.
A vacuole does not hold the cell’s DNA.

Why: A structure inside a cell that does a particular job and is wrapped in its own membrane is a membrane-bound organelle.
The largest is bounded by a double membrane and holds the DNA: the nucleus.
A cell with a nucleus and membrane-bound organelles is eukaryotic.

Q5 P21-q05

Two single-celled organisms come from the same pond water. Cell A is 60 μm long and has a nucleus, mitochondria and no cell wall. Cell B is 3 μm long and has a cell wall, ribosomes and its DNA in a region with no membrane around it.

Which of the following classifies the two cells, and why?

A. ✓ Cell A is eukaryotic and cell B is prokaryotic, since only cell A keeps its DNA inside a nucleus
B. Both cells are prokaryotic, since both are single cells
Many eukaryotes are single cells: cell A has a nucleus, so it is eukaryotic; cell B’s DNA has no membrane around it, so it is prokaryotic.
C. Both cells are eukaryotic, since both have DNA
Every cell has DNA, so having DNA decides nothing; cell B's DNA lies in a region with no membrane around it, a nucleoid, so cell B is prokaryotic.
D. Cell A is prokaryotic and cell B is eukaryotic, since only cell B has a cell wall
Plants and fungi have cell walls and are eukaryotes; a cell wall does not make cell B prokaryotic, its DNA with no membrane around it does.

Why: Where the DNA sits decides the kind of cell.
Cell A keeps its DNA inside a nucleus, so cell A is eukaryotic.
Cell B keeps its DNA in a nucleoid, with no membrane around it, so cell B is prokaryotic.
Plants and fungi are eukaryotes with cell walls.
So a cell wall proves nothing.

Q6 P21-q06

Scientists give rats a drug that the rats’ liver cells must break down. Within days, the part of the liver cells’ endoplasmic reticulum that carries no ribosomes has doubled in area, and the drug disappears from the rats’ blood twice as fast.

What is this part of the ER, and what does its growth do for the cell?

A. Rough ER; more of it makes more protein for export
Rough ER carries ribosomes on its surface.
This ER has none.
B. ✓ Smooth ER; more of it breaks down harmful molecules faster
C. Rough ER; more of it breaks down harmful molecules faster
Rough ER is the ER that carries ribosomes.
D. Smooth ER; more of it makes more protein for export
ER with no ribosomes makes no protein.
The smooth ER makes membrane lipids and breaks down harmful molecules.

Why: The part of the endoplasmic reticulum with no ribosomes on it is the smooth ER.
The smooth ER makes new membrane lipids and breaks down harmful molecules such as alcohol or a drug.
The liver cells doubled their smooth ER, so they break the drug down twice as fast.

Q7 P21-q07

A drug destroys the stack of flattened membrane sacs in a gland cell. The cell's ribosomes on the rough ER keep making a protein for export, and the protein still folds in the ER's interior. But the protein that reaches the outside now lacks sugar chains that are normally added to it after it leaves the ER, and much of it is delivered to the wrong places in the cell.

What does this show the stack of sacs normally does?

A. Makes the protein, joining its amino acids in order on the sacs' surfaces
Protein-making went on as normal after the stack was destroyed.
B. Folds the protein for the first time, before it reaches the ER
The protein still folded in the ER’s interior.
C. Digests the protein once the cell has finished using it
Digesting worn-out material is the job of lysosomes.
D. ✓ Attaches the sugar chains and packages the protein into addressed vesicles

Why: A stack of flattened membrane sacs that receives proteins from the ER is the Golgi complex.
The Golgi complex modifies the proteins, here by attaching carbohydrate chains.
It then packages them into vesicles addressed to their destinations.
Without it, the protein lacks its chains and goes to the wrong places.

Q8 P21-q08

A cell at the back of the eye takes in worn-out fragments shed by its neighbors every day and normally digests them. In a patient, the fragments pile up undigested inside membrane-enclosed sacs in these cells, and the interior of those sacs is found to be far less acidic than normal. The cell still takes the fragments in.

What has gone wrong?

A. The cell has stopped taking the fragments in from its neighbors
The fragments are inside the cell, in sacs.
B. The Golgi complex has stopped making the proteins the cell needs
The Golgi complex does not make proteins.
The fault found is in the acidity of the sacs.
C. ✓ The lysosomes are too weakly acidic, so their digestive enzymes work slowly
D. The mitochondria have stopped supplying the sacs with the ATP they need
Nothing in the findings involves mitochondria.
The sacs holding undigested material are lysosomes, and their interior is less acidic than normal.

Why: Lysosomes are membrane-enclosed sacs whose acidic interior holds digestive enzymes.
Those enzymes work far faster at the lysosome’s acidity than at the cytosol’s.
So a lysosome that is not acidic enough digests slowly.
So material taken in piles up.

Q9 P21-q09

As a caterpillar changes into a moth inside its case, most of its larval muscle cells shrink, dismantle themselves and are cleared away on a fixed schedule. The cells are healthy and uninfected when this begins.

What is happening to the muscle cells, and which organelle takes part?

A. Osmotic lysis; the cells burst as water enters
Bursting from too much water is neither orderly nor on a schedule.
B. ✓ Apoptosis, programmed cell death; lysosomes take part in dismantling the cells
C. Endocytosis; neighboring cells swallow the muscle cells whole
Endocytosis takes material into a cell.
Endocytosis is not how a cell dies.
D. Exocytosis; the cells release their contents and empty out
Exocytosis releases a vesicle’s contents outside a cell.

Why: A cell that dismantles itself in an orderly way, on schedule, as a normal part of development is undergoing apoptosis, programmed cell death.
Lysosomes take part in apoptosis.
The lysosomes’ enzymes break down the cell’s own parts.

Q10 P21-q10

Scientists label a phospholipid in the membrane of a cell’s ER and follow the label for an hour. The label appears in the membranes of the Golgi complex, then in the membranes of small sacs near the cell surface, and finally in the plasma membrane itself.

What carried the labeled lipid from one membrane to the next, and what does the result show about these membranes?

A. ✓ Vesicles that budded from one membrane and fused with the next; one endomembrane system
B. The cytosol, through which the lipid diffused on its own; separate, unconnected membranes
A phospholipid stays in membrane.
A phospholipid does not dissolve into the cytosol and reappear elsewhere.
C. Ribosomes, which carried the lipid along with them; the membranes are all rough ER
Ribosomes make proteins.
Ribosomes carry nothing between compartments.
D. Mitochondria, which absorbed and released the lipid; the membranes are all mitochondrial
No vesicles run between mitochondria and the ER or the Golgi complex.

Why: Material moves between the ER, the Golgi complex, transport vesicles and the plasma membrane inside vesicles.
A vesicle buds from one membrane and fuses with the next.
So a lipid in the ER’s membrane ends up in the plasma membrane.
Membranes connected this way are members of the endomembrane system.

FRQ 1 P21-frq1 · Analyze Model or Visual Representation scaffolded

The model shows a cell from a mammary gland that makes a milk protein and releases it into the milk duct. Five parts are numbered. The part marked 1 is a large round body enclosed by a double membrane. The part marked 2 is a network of membrane sacs continuous with the membrane of 1, with small dots on its outer surface. The part marked 3 is a stack of flattened membrane sacs. The part marked 4 is a small membrane sac near the plasma membrane. The part marked 5 is an oval body with a smooth outer membrane and a highly folded inner membrane.

A mammary gland cell that exports a milk protein, with five parts numbered.

(a) Identify the parts marked 1 and 2, and identify the dots on the part marked 2. (1 pt)

Frame 1 is the …; 2 is the …; the dots on 2 are …

Hint A double membrane around a large round body, and a network continuous with it carrying particles on its surface, are two of the parts every cell that exports protein is full of.

Model answer The part marked 1 is the nucleus, enclosed by its nuclear envelope, a double membrane.
The part marked 2 is the rough ER, a network of membrane sacs continuous with the nuclear envelope.
The dots on the rough ER are ribosomes.

Rubric

Award 1 point for: 1 is the nucleus (in its nuclear envelope); 2 is the rough ER (rough endoplasmic reticulum); the dots are ribosomes. All three identifications are needed for the point.
Accept: "nucleus" without naming the envelope. Do not award the point for two of the three, or if the dots are called vesicles or 2 is called the Golgi complex.

Slip Naming 1 and 2 but leaving the dots unnamed, or calling 2 the Golgi complex. All three identifications are needed; the Golgi is a separate stack of flattened sacs (3), and a network continuous with the nuclear envelope and studded with ribosomes is the rough ER.

(b) Describe what the dots on the part marked 2 do, and where the milk protein goes the moment a dot finishes making it. (1 pt)

Frame The dots … in the order set by …; as the protein is made it passes into …

Hint What do the dots build, and from which building blocks? Look at where the dots sit in the model: what lies directly beneath them?

Model answer The dots are ribosomes.
The ribosomes join amino acids into the milk protein in the order set by a messenger RNA.
The messenger RNA is a copy of instructions carried from the DNA in the nucleus.
As each protein is made, the ribosome passes it straight into the interior of the ER.
The ER’s interior is a compartment separate from the cytosol.
The protein is folded there.

Rubric

Award 1 point for: the ribosomes join amino acids into the protein in the order an mRNA sets, and the protein passes straight into the interior of the ER, a compartment separate from the cytosol, where it is folded.
Accept: "ribosomes make the protein and pass it into the ER". Do not award the point if the protein is placed in the cytosol.

Slip Sending the finished protein into the cytosol. A protein for export is never loose in the cytosol. The ribosome passes the protein into the ER’s interior.

(c) Describe what the part marked 3 does to the protein when it arrives from the part marked 2. (1 pt)

Frame 3 receives the protein in …; it then … and …; finally it …

Hint Three verbs: what 3 takes in, what it changes, and what it sends out, and in what.

Model answer The part marked 3 is the Golgi complex.
The Golgi complex receives the protein in vesicles that bud from the ER.
The Golgi complex then finishes the protein and changes it chemically, for example by attaching carbohydrate chains.
Finally the Golgi complex packages the protein into vesicles addressed to the plasma membrane.

Rubric

Award 1 point for: the Golgi complex receives the protein in vesicles from the ER, finishes and chemically modifies it (for example by glycosylation, attaching carbohydrate chains), and packages it into vesicles addressed to its destination, here the plasma membrane.
Accept: any two of receive / modify / package, stated for this protein. Do not award the point for "3 makes the protein".

Slip Saying the Golgi complex makes the protein. Ribosomes on the rough ER make it; the Golgi complex receives, modifies and packages it.

(d) Explain how the protein gets from the part marked 4 to the milk duct, naming the process. (1 pt)

Frame 4 moves to …, and …, so the protein …; this is …

Hint Can a large polar protein cross a membrane on its own? If not, what has to happen between the part marked 4 and the plasma membrane for the protein to end up outside the cell?

Model answer The part marked 4 is a vesicle.
The vesicle moves to the plasma membrane.
The vesicle fuses with the plasma membrane, and the vesicle’s membrane becomes part of the plasma membrane.
So the milk protein inside the vesicle is released outside the cell, into the duct.
This is exocytosis.
A protein is far too large and polar to cross the membrane on its own, so the protein can leave only this way.

Rubric

Award 1 point for: the part marked 4, a vesicle, moves to the plasma membrane and fuses with it, so its contents are released outside the cell into the duct; this is exocytosis (and the vesicle's membrane becomes part of the plasma membrane).
Accept: the fusion and release described without the word exocytosis. Do not award the point for the protein passing through the plasma membrane on its own.

Slip Having the protein pass through the plasma membrane by itself. It leaves only when the vesicle fuses with the membrane and opens outward.

(e) A cell lining a blood vessel exports very little protein but has just as many of the part marked 5. Predict which numbered parts it has far less of than the mammary cell, and justify your prediction. (1 pt)

Frame The blood-vessel cell has far less … and …, because …; it keeps 5 because …

Hint Ask what each part does for a cell whose job is exporting protein, and what every cell needs whatever its job.

Model answer The blood-vessel cell has far less rough ER and far less Golgi complex, and fewer vesicles.
Those parts make, finish, package and ship protein for export.
The blood-vessel cell exports very little protein, so it needs little of them.
It keeps its mitochondria, because every cell needs ATP whatever its job.
It keeps its nucleus, because every cell needs its DNA.

Rubric

Award 1 point for: far less rough ER (which makes and folds protein for export) and Golgi complex (which finishes and packages it), and fewer vesicles, because it exports little protein; it keeps its mitochondria because every cell needs ATP.
Accept: the numbers 2, 3 and 4 in place of the names; the rough ER and the Golgi complex with the link to protein export, with or without the vesicles. Do not award the point for parts named with no link to the cell's job.

Slip Naming parts with no link to the job, or dropping the mitochondria. The point is earned by tying the rough ER and the Golgi complex to protein export and the mitochondria to a need every cell has.

FRQ 2 P21-frq2 · Conceptual Analysis

Two cells from one plant are compared. A leaf cell contains chloroplasts, mitochondria and one large central vacuole. The leaf cell is fully grown and no longer divides. The leaf cell exports almost no protein. A cell from the growing tip of a root divides about once a day. The root-tip cell contains mitochondria but no chloroplasts, and holds many small vacuoles that later fuse into one large central vacuole as the cell matures.

(a) Describe the structure of a mitochondrion and what its folded inner membrane does for the cell. (1 pt)

Frame A mitochondrion has …; the folds give …, so …

Model answer A mitochondrion has two membranes.
The outer membrane is smooth.
The inner membrane is folded again and again.
The two membranes divide the inside into compartments.
Aerobic cellular respiration, the oxygen-using breakdown of glucose, happens in those compartments.
The folds give the inner membrane a large surface.
So more of the reactions that make ATP happen at once, and ATP is made faster.

Rubric

Award 1 point for: a double membrane, a smooth outer membrane and a highly folded inner membrane, dividing the inside into compartments where aerobic cellular respiration happens; the folds give the inner membrane a large surface, so ATP can be made more efficiently.
Accept: "two membranes, the inner one folded; more surface for the reactions that make ATP".

Slip Saying mitochondria 'make energy'. They transfer energy from glucose into ATP; the folds matter because those reactions happen on the inner membrane's surface.

(b) Explain why the leaf cell has both chloroplasts and mitochondria while the root cell has mitochondria only. (1 pt)

Model answer Chloroplasts are where photosynthesis happens: the light-driven making of sugar from carbon dioxide and water.
A leaf cell is in the light, so it makes sugar, so it has chloroplasts.
A root cell is underground with no light, so it cannot make sugar, so it has no chloroplasts.
The root cell lives on sugar sent from the leaves.
Both cells need ATP.
Aerobic cellular respiration in mitochondria makes ATP.
So both cells have mitochondria.

Rubric

Award 1 point for: chloroplasts are the site of photosynthesis, the light-driven making of sugar from carbon dioxide and water, which a leaf cell in the light can do and a root cell underground cannot; every cell, leaf or root, needs ATP from aerobic cellular respiration, which happens in mitochondria, so both cells have them.
Accept: "the leaf catches light and makes sugar; both cells break sugar down for ATP".

Slip Saying plant cells have chloroplasts instead of mitochondria. The leaf cell has both; chloroplasts make the sugar, mitochondria break it down for ATP.

(c) Make a claim about which of the two cells has more rough ER and Golgi complex, and support your claim with evidence from the descriptions of the two cells. (1 pt)

Hint Ask what rough ER and Golgi complex do for a cell. Then ask which of the two cells needs more of that done.

Model answer The root-tip cell has more rough ER and more Golgi complex.
The root-tip cell divides about once a day.
So it must build new proteins and new membrane for each daughter cell.
Rough ER makes and folds those proteins.
The Golgi complex finishes and packages them.
Therefore the cell that builds new parts every day is full of both.

Rubric

Award 1 point for: the claim that the root-tip cell has more rough ER and Golgi complex, supported by evidence AND reasoning: it divides about once a day (evidence), so it must build new proteins and new membrane for each daughter cell, and rough ER makes and folds protein while the Golgi complex finishes and packages it (reasoning).
Do not award the point for the claim alone, or for the leaf cell: the stimulus says it exports almost no protein and no longer divides.
Do not award the point for "the root cell is more active" or "the leaf cell does photosynthesis" with no link to protein for new cell parts.

Slip Making the claim without the evidence, or naming the evidence (daily division) without linking it to protein for new cell parts. The point needs the claim, the evidence and the link.

(d) The root cell's small vacuoles fuse into one large central vacuole as the cell matures. Explain what that central vacuole does for the mature cell. (1 pt)

Model answer The central vacuole stores water and nutrients.
As the vacuole fills, it presses the cytosol against the cell wall.
That push against the cell wall is turgor pressure.
Turgor pressure keeps the mature cell firm.
A plant cell has one large central vacuole; an animal cell has many small vacuoles.

Rubric

Award 1 point for: the central vacuole stores water and nutrients and, as it fills, presses the cytosol against the cell wall, maintaining turgor pressure so the cell is firm.
Accept: "stores water; its pressure against the cell wall keeps the cell firm".

Slip Describing the vacuole as a store only. Its filling is what presses the cytosol against the cell wall and gives the cell its firmness.

APBIO-U02-T21 End-of-topic test: Cell Structure and Function

Topic 2.1 · Cell Structure and Function · 18 MCQ · 2 FRQ

Q1 T21-q01

A cell’s particles of RNA and protein join amino acids into a protein, in the order an mRNA sets.

What are the particles?

A. Pieces of the Golgi complex
The Golgi complex finishes and packages proteins that have already been made.
B. Pieces of rough ER
The rough ER does not make proteins.
The ribosomes on its surface do.
C. ✓ Ribosomes
D. Vesicles
A vesicle is a small sac of membrane.

Why: Ribosomes are small particles of ribosomal RNA and protein, not enclosed in any membrane.
A ribosome makes a protein by joining amino acids in the order a messenger RNA sets.
These particles do exactly that, so they are ribosomes.

Q2 T21-q02

Every known kind of cell, from an archaeon in a hot spring to a cell in a whale, stores its inherited instructions in DNA. The DNA of every cell is built the same way: a double helix of the same four kinds of nucleotide.

What does a feature shared by every living thing suggest?

A. ✓ That all living things descend from shared ancestors
B. That each kind of cell invented its own DNA, which happens to match the others
Many kinds of cell inventing the same complicated molecule by chance is far less likely than all of them inheriting it.
C. That the environment forces every cell to build its DNA the same way
The environment does not build DNA.
A cell copies its DNA from the DNA it inherited.
D. That every living thing carries the same DNA
Living things differ in most of the order of their DNA.

Why: DNA of one build is found in every living thing.
A feature shared by all living things is evidence that living things descend from shared ancestors: their common ancestry.
Each kind of cell inherited its DNA rather than inventing it.

Q3 T21-q03

The drawing shows a cell 2 μm long. Region X is where a DNA stain collects; no membrane separates it from the rest of the cell. The small dots marked Y are found throughout the cell. A stiff layer lies outside the plasma membrane.

A cell 2 μm long. X marks the region where a DNA stain collects; Y marks one of the small dots found throughout the cell.

What kind of cell is this, and what are X and Y?

A. Eukaryotic; X is the nucleus and Y are ribosomes
A nucleus is enclosed by a double membrane.
B. ✓ Prokaryotic; X is the nucleoid and Y are ribosomes
C. Prokaryotic; X is the nucleus and Y are vesicles
A nucleus has a membrane around it, and X has none.
D. Eukaryotic; X is a vacuole and Y are ribosomes
A vacuole is a membrane-enclosed sac.
X has no membrane, and X is where the DNA lies.

Why: A cell whose DNA is not enclosed in a nucleus is a prokaryotic cell.
Its DNA lies in the nucleoid (X) with no membrane around it.
The dots (Y) are its ribosomes, which every cell has.
The stiff outer layer is its cell wall.
Bacteria and archaea are prokaryotes.

Q4 T21-q04

A student writes: "A bacterium has no nucleus, so it has no DNA and cannot make its own proteins."

Which statement corrects the student?

A. The bacterium has DNA in a nucleus after all; the nucleus is just very small
A bacterium has no nucleus of any size.
B. The bacterium has no DNA, but its ribosomes make proteins without instructions
Every cell has DNA.
A ribosome cannot make a protein without the instructions an mRNA copies from that DNA.
C. The bacterium has DNA, but with no nucleus it has no ribosomes either
Ribosomes are not enclosed in any membrane.
Ribosomes do not need a nucleus.
D. ✓ The bacterium has DNA in a nucleoid, and ribosomes that make its proteins

Why: Having no nucleus means only that the DNA has no membrane around it.
A prokaryotic cell keeps its DNA, a single circular molecule, in the nucleoid.
Its ribosomes need no membrane.
So its ribosomes make its proteins.

Q5 T21-q05

The drawing shows a plant cell about 100 μm long: one large round body that is stained dark by a dye that binds DNA, with several smaller bodies around it, each wrapped in its own membrane.

A plant cell about 100 μm long, as seen under a microscope.

What is the dark round body, and what does it hold?

A. ✓ The nucleus, in its nuclear envelope; the cell's DNA
B. The nucleoid; the cell's DNA, with no membrane around it
A nucleoid is DNA with no membrane around it.
C. The central vacuole; stored water
The central vacuole is a clear, watery sac filling most of the cell.
D. A ribosome; the cell's proteins
A ribosome is a tiny particle with no membrane.

Why: The dark round body is the nucleus, enclosed by its nuclear envelope, a double membrane.
The nucleus holds the cell’s DNA, its inherited instructions.
The membrane-wrapped bodies around the nucleus are the cell’s other membrane-bound organelles.
Each organelle does a particular job inside its own membrane.

Q6 T21-q06

Four cells: Cell P, 2 μm long, has a cell wall and a nucleoid. Cell Q, 100 μm long, has a cell wall, a nucleus and chloroplasts. Cell R, a single-celled pond organism 150 μm long, has a nucleus and no cell wall. Cell S, 1 μm long, has a cell wall and a nucleoid.

Which of the cells are eukaryotic?

A. Q only
Cell R also has a nucleus, so R is eukaryotic too.
B. R only
Cell Q has a nucleus, so Q is eukaryotic despite its cell wall.
C. ✓ Q and R
D. P, Q, R and S
DNA in a nucleoid is the mark of a prokaryotic cell.

Why: A eukaryotic cell keeps its DNA inside a nucleus.
Cell Q, a plant cell, and cell R, a protist, each have a nucleus, so Q and R are eukaryotic.
Cells P and S keep their DNA in a nucleoid with no membrane, so they are prokaryotic, whatever their size.

Q7 T21-q07

A cell that makes and exports a protein all day is crowded with the network of membrane tubes shown in the drawing. The network’s outer surface is studded with ribosomes, drawn as small dots.

Part of the network of branching membrane tubes that crowds the cell, one stretch of it joined to the nuclear envelope; the small dots sit on the tubes’ outer surfaces.

What is this network, and what happens to a protein made by the ribosomes on it?

A. Rough ER; the protein is folded in the cytosol, on the outer surface
A protein for export is never loose in the cytosol.
B. ✓ Rough ER; the protein enters its interior and is folded there
C. Golgi complex; the protein is made inside it and packaged
The Golgi complex is a stack of flattened sacs with no ribosomes on it.
D. Rough ER; the ER itself makes the protein, and the ribosomes store it
The ER does not make proteins.
Ribosomes make proteins, and ribosomes do not store them.

Why: Membrane folded into a network and studded with ribosomes is the rough endoplasmic reticulum.
The ribosomes on its surface pass the proteins they make into the ER’s interior.
The ER’s interior is a compartment separate from the cytosol.
The proteins are folded there before vesicles carry them on.

Q8 T21-q08

Part of a cell's endoplasmic reticulum has no ribosomes on its surface.

What does the cell do in this part of the ER?

A. Makes proteins for export
Ribosomes make proteins.
This part of the ER has no ribosomes.
B. Digests worn-out parts of the cell
Digesting worn-out parts of the cell is what lysosomes do.
C. Packages proteins into vesicles for delivery
Packaging finished proteins into vesicles is the Golgi complex’s job.
D. ✓ Makes new membrane lipids and breaks down harmful molecules

Why: ER with no ribosomes on it is the smooth ER.
In the smooth ER the cell makes new lipids for its membranes.
In the smooth ER the cell also breaks down harmful molecules such as alcohol.

Q9 T21-q09

In a gland cell, a protein for export is a plain chain of amino acids when it leaves the ribosomes on the rough ER. The same protein carries sugar chains when it leaves the cell. Between the ER and the plasma membrane, the protein passes through a stack of flattened membrane sacs.

What did the stack of flattened sacs do to the protein?

A. ✓ Added the sugar chains and packed it into vesicles
B. Joined its amino acids together in order
The protein was already a complete chain of amino acids when it left the ribosomes.
C. Stored it unchanged; the sugar chains were added at the plasma membrane
The sugar chains are attached inside the Golgi complex; the plasma membrane adds nothing to a protein.
D. Broke it into amino acids, which were rebuilt into a new protein
The protein that leaves the cell is the same chain, now with sugar chains added.

Why: A stack of flattened membrane sacs that receives proteins from the ER is the Golgi complex.
The Golgi complex chemically modifies the proteins, here by glycosylation, attaching carbohydrate chains.
The Golgi complex then packages the proteins into vesicles addressed to their destinations, in this case the cell surface.

Q10 T21-q10

In a patient, one digestive enzyme normally found inside lysosomes is missing. In the patient’s cells, worn-out parts of the cell that are normally broken down instead build up, whole, inside membrane-enclosed sacs.

What does this show about lysosomes?

A. They store the cell’s worn-out parts without breaking them down
In healthy cells the worn-out parts are broken down, not stored.
B. They make their own enzymes, on ribosomes inside them
Lysosomes have no ribosomes inside them.
Their enzymes are made by ribosomes and reach the lysosome through the ER and Golgi complex.
C. ✓ They digest the cell’s worn-out parts using the enzymes held inside them
D. They carry the worn-out parts out of the cell by exocytosis
Lysosomes break material down inside the cell.
Lysosomes do not carry material out.

Why: Lysosomes are membrane-enclosed sacs whose acidic interior holds digestive enzymes.
Digestive enzymes speed up hydrolysis, the breaking of large molecules into small ones.
Lysosomes digest the cell’s own worn-out parts as well as material the cell takes in.
With one enzyme missing, the worn-out parts stay whole and build up.

Q11 T21-q11

In a developing embryo, far more nerve cells form than the body will keep. The surplus nerve cells then die in an orderly, controlled way, on schedule, as a normal part of development.

What is this process, and which organelle takes part in it?

A. Osmotic lysis; the cells swell and burst
Bursting from taking in too much water is neither orderly nor programmed.
B. ✓ Apoptosis; lysosomes take part in dismantling the cells
C. Endocytosis; neighboring cells engulf the surplus cells
Endocytosis takes material into a cell.
Endocytosis is not a way a cell dies.
D. Exocytosis; the cells release their contents and shrink away
Exocytosis releases a vesicle’s contents outside a cell.

Why: A cell that dismantles itself in an orderly way as a normal part of development is undergoing apoptosis, programmed cell death.
Lysosomes take part in apoptosis.
The lysosomes’ enzymes break down the cell’s own parts.

Q12 T21-q12

A cell's endomembrane system is the set of membranes that work together to modify, package and transport proteins, lipids and polysaccharides within the cell.

Which list names only members of the endomembrane system?

A. ER, Golgi complex and mitochondria
Mitochondria have their own double membrane apart from the system.
B. Ribosomes, ER and Golgi complex
Ribosomes have no membrane at all.
C. Chloroplasts, mitochondria and the nuclear envelope
Chloroplasts and mitochondria have their own double membranes.
D. ✓ ER, Golgi complex, lysosomes and transport vesicles

Why: The endomembrane system’s members are the ER, the Golgi complex, lysosomes, vacuoles, transport vesicles, the nuclear envelope and the plasma membrane.
These are the membranes vesicles bud from and fuse with.
Mitochondria, chloroplasts and ribosomes are not members.

Q13 T21-q13

A cell in the stomach lining makes a digestive protein and exports it into the stomach.

In which order is the protein found, from where the ribosomes make it to where it leaves the cell?

A. ✓ Rough ER → vesicle → Golgi complex → vesicle → outside the cell
B. Golgi complex → vesicle → rough ER → vesicle → outside the cell
Ribosomes on the rough ER make the protein.
C. Rough ER → cytosol → plasma membrane → outside the cell
A protein for export never enters the cytosol.
D. Cytosol → straight through the plasma membrane → outside the cell
A protein is far too large and polar to pass through the plasma membrane on its own.

Why: Ribosomes on the rough ER make the protein and pass it into the ER’s interior.
The protein leaves in a vesicle that fuses with the Golgi complex.
The Golgi complex finishes it and packages it into another vesicle.
That vesicle fuses with the plasma membrane: exocytosis.

Q14 T21-q14

The mitochondria in heart muscle have about 1.5 times as much folding of their inner membrane as the mitochondria in skin cells. Given the same oxygen and glucose, equal numbers of heart-muscle mitochondria make 180 units of ATP a minute; skin-cell mitochondria make 120.

Why do the heart-muscle mitochondria make ATP faster?

A. The folds store energy that is released as ATP when needed
Mitochondria do not store energy.
Mitochondria transfer energy from glucose into ATP.
B. The folds enlarge the inner space where ATP builds up
Folding the inner membrane adds membrane, not inner space.
C. ✓ The folds give the inner membrane more surface on which ATP can be made
D. The folds let oxygen and fuel cross the inner membrane freely
The folds do not change what crosses the membrane.

Why: A mitochondrion’s inner membrane is highly folded.
The folds give the inner membrane a large surface.
The reactions of aerobic cellular respiration that make ATP happen on that membrane.
So more folding means more surface, and more ATP is made in the same time.

Q15 T21-q15

A leaf has green patches, whose cells contain chloroplasts, and white patches, whose cells contain none. A student keeps the plant in the dark for two days to clear starch from the leaf. The student then gives the leaf water, carbon dioxide and light for 12 hours, and covers one green patch so that it stays dark. Iodine, which turns starch dark, then darkens only the uncovered green patches. The white patches and the covered green patch stay pale.

What do the results show?

A. Light makes any plant cell produce starch
The white patches had light too.
The white patches made no starch.
B. Chloroplasts make starch in the dark as well as in the light
The covered green patch has chloroplasts.
The covered green patch made no starch.
C. The white-patch cells made sugar too, but stored it as something other than starch
Making sugar from carbon dioxide and water happens only in chloroplasts.
D. ✓ Starch was made only where there were both light and chloroplasts

Why: Chloroplasts are the site of photosynthesis, the light-driven making of sugar from carbon dioxide and water.
Only cells with chloroplasts made sugar, and only in the light.
Some of that sugar was stored as the starch that iodine stains.
Not every plant cell has chloroplasts.

Q16 T21-q16

A green alga is a single cell that lives in pond water. In daylight the cell gives off oxygen and builds up starch. Day and night, the cell also takes up some oxygen. The cell contains chloroplasts and mitochondria.

Which statement about this cell's organelles is right?

A. Its chloroplasts replace mitochondria, so it has no need of them
The alga has both organelles and needs both.
B. ✓ Its chloroplasts carry out photosynthesis while its mitochondria carry out respiration
C. Its chloroplasts carry out respiration and its mitochondria photosynthesis
Chloroplasts make sugar from carbon dioxide and water in the light, giving off oxygen.
D. Its chloroplasts only store starch; its mitochondria make the oxygen
The oxygen given off comes from photosynthesis in the chloroplasts.

Why: The alga’s cell, like a plant cell, has both organelles.
Chloroplasts carry out photosynthesis, making sugar from carbon dioxide and water and giving off oxygen.
Mitochondria carry out aerobic cellular respiration, the oxygen-using breakdown of glucose that makes ATP, so the cell takes up oxygen day and night.

Q17 T21-q17

A cook puts a limp celery stalk in pure water. An hour later the stalk is stiff again. In each of its cells the central vacuole has gained about 2,400 μm³ of water and the cytosol about 50 μm³.

What explains the stalk’s recovery?

A. The cytosol took in most of the water and swelled enough to firm the cell
The cytosol grew by only 50 μm³.
The vacuole grew by 2,400 μm³.
B. The firmness rose independently of the vacuole, which only stores materials
The stalk stiffened as the vacuoles filled.
C. ✓ The central vacuole took in most of the water and pressed the cytosol against the cell wall
D. The vacuole released water into the cytosol, so both swelled
The vacuole grew.

Why: A plant cell has one large central vacuole that stores water and nutrients.
Here each vacuole took in 2,400 μm³ of water; the cytosol gained only 50 μm³.
As each vacuole filled, it pressed the cytosol against the cell wall.
So turgor pressure returned, and the stalk stiffened.

Q18 T21-q18

Three cells are cut open. Cell 1 lines the gut and releases large amounts of a digestive protein into the gut. Cell 2 is a heart muscle cell that contracts about once a second, all day. Cell 3 is a white blood cell in the lung that engulfs dust particles and bacteria and destroys them.

Which organelles should each cell be full of?

A. ✓ 1: rough ER and Golgi complex; 2: mitochondria; 3: lysosomes
B. 1: lysosomes; 2: rough ER and Golgi complex; 3: mitochondria
Lysosomes digest; they do not make or export protein.
C. 1: mitochondria; 2: lysosomes; 3: rough ER and Golgi complex
Mitochondria make ATP; they do not export protein.
D. 1: rough ER and Golgi complex; 2: lysosomes; 3: mitochondria
Contracting all day uses ATP, made by mitochondria.

Why: A cell’s job shows in what it is full of.
Exporting protein needs ribosomes and rough ER to make and fold it, and Golgi complex to finish and package it.
Contracting all day uses ATP, made by mitochondria.
Destroying engulfed bacteria is digestion, done by lysosomes.

FRQ 1 T21-frq1 · Analyze Model or Visual Representation

The model shows a plasma cell, a white blood cell that makes an antibody protein and releases it into the blood. Six parts are numbered. The part marked 3 is a large round body enclosed by a double membrane. The part marked 4 is a network of membrane sacs, continuous with the membrane of 3, with small dots on its outer surface. The part marked 2 is a stack of flattened membrane sacs. The cell’s outer boundary is drawn as the outer line. The part marked 5 is a small membrane sac near that boundary. The part marked 6 is a membrane sac with an acidic interior. The part marked 1 is an oval body with a smooth outer membrane and a folded inner membrane.

A plasma cell, a white blood cell that exports an antibody protein into the blood, with six parts numbered; the outer line is the cell’s boundary.

(a) Identify the parts marked 3 and 4, and describe what the dots on the part marked 4 are and what they do. (1 pt)

Model answer The part marked 3 is the nucleus, enclosed by its nuclear envelope.
The part marked 4 is the rough ER.
The dots on the rough ER are ribosomes.
Ribosomes join amino acids into a protein in the order an mRNA sets.
The ribosomes on the rough ER pass each protein they make into the ER’s interior.

Rubric

Award 1 point for: 3 is the nucleus (enclosed by its nuclear envelope) and 4 is the rough ER; the dots are ribosomes, which make proteins (joining amino acids in the order an mRNA sets) and pass them into the ER's interior.
Accept: "rough endoplasmic reticulum" for 4, and "ribosomes, which make protein" for the dots. Do not award the point if the dots are called vesicles or if 4 is called the Golgi complex.

Slip Calling the dots vesicles, or 4 the Golgi complex. The Golgi is a stack of flattened sacs; a network continuous with the nuclear envelope and studded with ribosomes is the rough ER.

(b) Explain how the parts marked 4, 2 and 5 work together to get the antibody protein out of the cell. (1 pt)

Model answer Ribosomes on the rough ER make the antibody protein and pass it into the ER’s interior, where it is folded.
A vesicle buds from the ER carrying the protein and fuses with the Golgi complex.
The Golgi complex modifies the protein, for example by attaching carbohydrate chains, and packages it into a vesicle.
The vesicle moves to the plasma membrane and fuses with it.
So the protein is released outside the cell: exocytosis.

Rubric

Award 1 point for: the protein made on the rough ER is folded in the ER's interior and leaves in a vesicle that fuses with the Golgi complex, which modifies it (for example by attaching carbohydrate chains) and packages it into a vesicle that moves to the plasma membrane and fuses with it, releasing the protein outside by exocytosis.
Accept: the order ER → vesicle → Golgi → vesicle → plasma membrane with the Golgi's role given as finishing and packaging. Do not award the point if the protein is said to pass through the cytosol or straight through the plasma membrane.

Slip Having the protein pass through the cytosol between organelles, or straight through the plasma membrane. A protein for export travels inside membrane the whole way and leaves only when a vesicle fuses with the plasma membrane.

(c) Explain the role of the part marked 3 in making this protein. (1 pt)

Model answer The part marked 3 is the nucleus.
The nucleus holds the cell’s DNA.
The DNA carries the instructions for the antibody protein.
An mRNA copy of those instructions is made in the nucleus.
The mRNA is carried out of the nucleus to the ribosomes on the rough ER.
The ribosomes join amino acids in the order the mRNA sets.
So the nucleus supplies the instructions the ribosomes follow when they build the protein.

Rubric

Award 1 point for: the nucleus holds the cell's DNA, which carries the instructions for the protein, AND an mRNA copy of those instructions is carried out to the ribosomes on the rough ER, which build the protein in the order the mRNA sets.
Accept: "the nucleus holds the instructions the ribosomes follow" with the mRNA copy named. Do not award the point for "the nucleus controls the cell" or "the nucleus makes the protein" with no mention of the instructions or their copy.

Slip Saying the nucleus makes the protein. Ribosomes make the protein; the nucleus holds the DNA and sends out the mRNA copy that tells the ribosomes the order of amino acids.

(d) A skin cell has far less of the parts marked 4 and 2 than this plasma cell, but the same amount of the parts marked 3, 6 and 1. Explain how this difference relates to what each cell does. (1 pt)

Model answer The plasma cell makes and exports large amounts of antibody protein.
Rough ER makes and folds protein for export, and the Golgi complex finishes and packages it, so the plasma cell needs a great deal of both.
A skin cell exports little protein, so it needs little of either.
Every cell needs a nucleus for its DNA, lysosomes for digestion and mitochondria for ATP, so a skin cell and a plasma cell both keep their nucleus, lysosomes and mitochondria.

Rubric

Award 1 point for: the plasma cell's job is to make and export large amounts of protein, so it needs much rough ER (to make and fold the protein) and Golgi complex (to finish and package it); a skin cell exports little protein and so needs little of either; the parts every cell needs (a nucleus for its DNA, lysosomes for digestion, mitochondria for ATP) are the same in both.
Accept: "a cell has more of the organelles its job uses" provided the rough ER and Golgi are linked to protein export. Do not award the point for "the plasma cell is bigger" or for naming the parts with no link to the cell's job.

Slip Saying the plasma cell is bigger, or naming the parts with no link to the job. The point is earned by tying rough ER and Golgi to making and exporting protein.

FRQ 2 T21-frq2 · Conceptual Analysis

A cell in a salivary gland makes a digestive protein that breaks down starch in food, and exports it into saliva. The protein is made by ribosomes on the rough ER. The protein found in the saliva carries carbohydrate chains. The new protein at the ribosomes has none. A drug stops vesicles from budding off the ER in these cells. Everything else in the cell keeps working, including the ribosomes.

(a) Describe the route the digestive protein normally takes from the ribosomes that make it to the saliva. (1 pt)

Model answer From the ribosomes on the rough ER, the protein first enters the ER’s interior.
A vesicle buds from the ER carrying the protein.
The protein then travels in that vesicle to the Golgi complex, which the vesicle fuses with.
A second vesicle buds from the Golgi complex carrying the protein.
The protein leaves the cell when the second vesicle fuses with the plasma membrane and releases the protein into the saliva: exocytosis.

Rubric

Award 1 point for: ribosomes on the rough ER → the ER's interior → a vesicle → the Golgi complex → a vesicle → the plasma membrane, where exocytosis releases it outside the cell (into the saliva).
Accept: the sequence in words without the word exocytosis, provided it ends with a vesicle fusing with the plasma membrane. Do not award the point if the protein passes through the cytosol at any stage.

Slip Sending the protein through the cytosol or straight across the plasma membrane. It stays inside membrane compartments and vesicles the whole way.

(b) Describe where the cell adds the carbohydrate chains to the protein, and what else happens to the protein there. (1 pt)

Model answer The carbohydrate chains are added in the Golgi complex.
The Golgi complex receives the protein in vesicles from the ER.
The Golgi complex finishes the protein by attaching carbohydrate chains: glycosylation.
The Golgi complex then packages the protein into vesicles addressed to the cell surface.

Rubric

Award 1 point for: the carbohydrate chains are added in the Golgi complex, which receives the protein in vesicles from the ER, chemically modifies it (glycosylation, attaching carbohydrate chains), and packages it into vesicles addressed to its destination, here the cell surface.
Accept: the Golgi named with any two of receive / modify / package, stated for this protein. Do not award the point for the chains being added at the ribosomes or at the plasma membrane, or for "the Golgi makes the protein".

Slip Placing the change at the ribosomes or the plasma membrane, or saying the Golgi makes the protein. Ribosomes on the rough ER make it; the Golgi receives, modifies and packages it.

(c) Make a claim about what happens to the export of the digestive protein while the drug acts, and where the protein ends up inside the cell. (1 pt)

Model answer Export stops: no digestive protein reaches the saliva.
The ribosomes still make the protein, and the ER’s interior still folds it.
But no vesicle can bud from the ER, so the protein cannot leave the ER.
So the protein builds up in the interior of the rough ER.
Therefore none reaches the Golgi complex or the plasma membrane.

Rubric

Award 1 point for: the claim that export stops (no digestive protein reaches the saliva) and that the protein builds up in the interior of the rough ER, where it is still made and folded but from which it cannot leave. No reasoning is required for this point.
Accept: "it accumulates in the ER and none is released". Do not award the point for "it leaks into the cytosol and is released anyway" or for "the ribosomes stop making it".

Slip Having the protein leak into the cytosol and get out anyway, or the ribosomes stop. The ribosomes keep working; the protein piles up inside the ER.

(d) Support your claim, using how a protein for export moves from one compartment to the next inside a cell. (1 pt)

Model answer A protein for export travels only inside membrane compartments.
It moves from one compartment to the next only when a vesicle buds off one membrane and fuses with the next.
A protein is a large polar molecule, so it cannot cross a membrane on its own.
The drug stops vesicles budding from the ER.
Therefore the protein has no route out of the ER to the Golgi complex or the plasma membrane.

Rubric

Award 1 point for: the evidence AND the reasoning: the protein travels only inside membrane compartments (ER interior → vesicle → Golgi → vesicle), moving between them only when vesicles bud off one membrane and fuse with the next; a protein is a large polar molecule that cannot cross a membrane on its own, so with no vesicles budding from the ER it has no route out of the ER to the Golgi or the plasma membrane.
Accept: "it needs vesicles to move between compartments and cannot cross membranes by itself". Do not award the point for support based on the ribosomes stopping, which the scenario rules out.

Slip Justifying from the ribosomes stopping, which the scenario rules out. The block is in the vesicles, and vesicles are the only way a protein moves between compartments.

APBIO-U02-L20 Rooms with their own rules

Topic 2.9 · Cell Compartmentalization · 93 steps

A cell cut through, with one lysosome drawn inside it: the lysosome is a round sac with a dark membrane, shaded inside to show it is acidic; the cytosol around it is not shaded; the dots in the lysosome are digestive enzymes

Inside every one of your cells sit small sacs full of digestive enzymes. Those enzymes break proteins down.

So why do the enzymes not break the cell down? The answer is the membrane around each sac. The membrane keeps the inside of the sac acidic. And the enzymes work well only in acid.

We start with one sac, the lysosome. Then we look at the same idea across the whole cell. Then we look at two more jobs a membrane does. It keeps reactions apart. Folded, it gives reactions more surface. Last, we ask what happens when a membrane fails.

Unit 2 · Cell Structure and Function

1The same enzymes, three places

Video: Watch first: two places in one cell

The inside of a lysosome is acidic and the cytosol is not. The membrane between them keeps them different.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T29-intro.mp4

Video: Watch: The same enzymes, three places

The same enzymes, in the lysosome and in the cytosol: how fast do they break proteins down in each?

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L20.mp4

The lysosome contains enzymes. These enzymes break proteins down.

A lysosome drawn in section inside a cell: a round sac with a membrane, shaded inside, with dots for its digestive enzymes; labels with leader lines name the membrane, the enzymes and the acidic inside

These enzymes are built to work best in the conditions inside the lysosome. And the inside of the lysosome is acidic.

For example, drop some proteins into the lysosome. The enzymes break the proteins down quickly.

One frame: the enzymes and some proteins inside the lysosome, which is acidic; the proteins are cut into pieces

Because the enzymes work best in acidic conditions.

But now, drop the same proteins into the cytosol. The enzymes break the proteins down much more slowly.

One frame: the same enzymes and proteins in the cytosol, which is not acidic; the protein chains stay whole

Because the cytosol is not acidic.

Check q1

Some enzymes are taken out of a lysosome and dropped into water with some proteins.

One frame: the same enzymes and proteins in water, which is not acidic; the protein chains stay whole

How fast do the enzymes break the proteins down?

A. Quickly
Water is not acidic.
The enzymes work best in acidic conditions.
B. ✓ Slowly

Why: The enzymes work best in acidic conditions.
Water is not acidic.
So the enzymes break the proteins down slowly.

In water, the enzymes break the proteins down slowly. Water is not acidic. And the enzymes work best in acidic conditions.

where the proteins are	acidic?	how fast the enzymes break the proteins down
inside the lysosome	yes	quickly
in the cytosol	no	slowly
in water	no	slowly

the enzymes work best in acidic conditions

A table of the three places: inside the lysosome, acidic, the enzymes break the proteins down quickly; in the cytosol, not acidic, slowly; in water, not acidic, slowly

So this is why the lysosome has a membrane around it: to keep the inside acidic.

A lysosome inside a cell: its shaded inside is acidic, the cytosol around it is not, and the lysosome's membrane is the boundary between them

A pump puts the acid there: a pump in the lysosome’s membrane uses ATP to move hydrogen ions (H⁺) into the lysosome, and hydrogen ions make the inside acidic.

The acid cannot cross the membrane freely. So the acid stays inside the lysosome.

So the inside of the lysosome stays acidic. And so the enzymes inside keep breaking proteins down quickly.

What you are expected to know Predict how fast the lysosome’s enzymes break proteins down inside the lysosome, in the cytosol and in water: quickly, slowly, slowly.

What you are expected to know Explain why the lysosome has a membrane around it: the membrane keeps the inside acidic, and the enzymes work best in acidic conditions.

Check q2

A lysosome sits in the cytosol of a living cell.

Where do the lysosome’s enzymes break proteins down quickly?

A. ✓ Only inside the lysosome
B. Only in the cytosol
The cytosol is not acidic.
So the enzymes work slowly there.
C. In both places equally
The two places differ: the lysosome is acidic and the cytosol is not.

Why: The enzymes work best in acidic conditions.
The inside of the lysosome is acidic.
The cytosol is not.
So the enzymes break proteins down quickly only inside the lysosome.

Check q3

The lysosome’s enzymes break proteins down quickly inside the lysosome and slowly in the cytosol.

Why?

A. The lysosome holds many more protein molecules than the cytosol does
How many proteins there are does not change how fast each one is broken down.
B. The lysosome’s membrane speeds the enzymes up as they touch it
The membrane does not speed the enzymes up.
The membrane keeps the acid inside.
C. ✓ The lysosome is acidic, the cytosol is not, and the enzymes need acid

Why: The enzymes work best in acidic conditions.
The inside of the lysosome is acidic.
The cytosol is not acidic.
So the enzymes break proteins down quickly inside the lysosome and slowly in the cytosol.

Check q4

A student says: “The lysosome’s membrane speeds the digestive enzymes up, so the enzymes work only next to the membrane.”

Is the student correct that the membrane speeds the enzymes up?

A. Yes
The membrane does not speed the enzymes up.
It keeps the acid in.
B. ✓ No

Why: The membrane does not speed the enzymes up.
The membrane keeps the acid inside the lysosome.
So the whole inside of the lysosome is acidic.
The enzymes break proteins down quickly in acid.
So the enzymes break proteins down quickly everywhere inside the lysosome, not only next to the membrane.

Practice writing an answer

Yeast cells keep digestive enzymes inside their vacuoles. The inside of a yeast vacuole is acidic. The cytosol around it is not. The vacuole’s membrane does nothing to the enzymes themselves.

(a) Explain why the enzymes break proteins down quickly inside the vacuole even though the membrane does nothing to them. (1 pt)

Model answer The digestive enzymes work best in acidic conditions.
The acid cannot cross the vacuole’s membrane freely.
So the acid stays inside the vacuole.
So the inside of the vacuole stays acidic.
So the enzymes inside the vacuole break proteins down quickly.

Rubric

Award 1 point for: the enzymes work best in acid, and the membrane keeps the acid inside the vacuole, so the enzymes have the acidic conditions they need.
Accept: ‘the membrane keeps the inside acidic’ with the link to the enzymes needing acid.

Slip Saying the membrane activates the enzymes. The membrane keeps the acid in, and the acid is what the enzymes need.

22Compartments with their own conditions

Video: Watch: Compartments with their own conditions

The lysosome, the central vacuole and the ER each have a membrane that lets the inside be chemically different from the cytosol. This is called compartmentalization.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L20e.mp4

This does not just happen in the lysosome.

Lots of compartments in the cell have a membrane around them. The membrane lets the inside be chemically different from the cytosol.

For example, cut a lemon. The juice is sour.

The sourness comes from the central vacuoles of the lemon’s cells. The inside of each central vacuole is acidic, and the cytosol around it is not.

The vacuole’s membrane keeps the acid inside the vacuole.

Another example: the endoplasmic reticulum stores calcium ions. The concentration of calcium ions inside the ER is thousands of times higher than in the cytosol.

The ER’s membrane keeps the calcium ions inside the ER.

compartment	what is different inside	what its membrane keeps in
lysosome	acidic	the acid
central vacuole (lemon cell)	acidic	the acid
endoplasmic reticulum	a high concentration of calcium ions	the calcium ions

each compartment’s membrane lets the inside be chemically different from the cytosol

A table of three compartments: the lysosome, acidic inside, its membrane keeps the acid in; the central vacuole of a lemon cell, acidic inside, its membrane keeps the acid in; the endoplasmic reticulum, lots of calcium ions inside, its membrane keeps the calcium ions in

Dividing a cell’s inside into membrane-bound compartments, each with its own conditions, is called .

The name says what it means. A compartment is a separate room.

So compartmentalization means dividing the cell into separate rooms, each with its own conditions.

What you are expected to know State what a membrane around a compartment lets the compartment do: hold conditions different from the cytosol.

What you are expected to know State what compartmentalization is: dividing a cell’s inside into membrane-bound compartments, each with its own conditions.

Check q5

What is compartmentalization?

A. ✓ Dividing a cell’s inside into membrane-bound rooms, each with its own conditions
B. Folding a membrane again and again to give it far more surface area
Folding a membrane adds surface; compartmentalization is about dividing the inside into rooms.
C. Building a rigid cell wall around the outside of the whole cell
A cell wall sits outside the whole cell; compartmentalization is about the inside.

Why: A compartment is a separate room.
A membrane around a compartment lets the inside hold its own conditions.
So compartmentalization is dividing a cell’s inside into membrane-bound compartments, each with its own conditions.

Practice writing an answer

A eukaryotic cell contains lysosomes, a central vacuole and an endoplasmic reticulum, each wrapped in its own membrane.

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

Model answer Compartmentalization is dividing a cell’s inside into membrane-bound compartments.
Each compartment holds its own conditions.

Rubric

Award 1 point for: dividing the cell’s inside into membrane-bound compartments (rooms), each able to hold conditions different from the cytosol.

Slip Saying only ‘the cell has organelles’. Say what the membranes do: they let each compartment hold its own conditions.

Check q6

Cut a lemon and the juice is sour. The sourness comes from the central vacuoles of the lemon’s cells, which are far more acidic than the cytosol around them.

In the living lemon cell, which of the following keeps the cytosol from becoming as acidic as the vacuole?

A. The cell wall
The cell wall is outside the whole cell.
B. ✓ The vacuole’s membrane
C. The cytosol’s proteins
Nothing destroys the acid.

Why: A membrane around a compartment lets the inside hold conditions different from the cytosol.
The acid cannot cross the vacuole’s membrane freely.
So the acid stays inside the vacuole.
So the cytosol stays as it is: not acidic.

Check q7

The vacuole’s membrane keeps the cytosol of the lemon cell from becoming acidic.

Why does the membrane keep the cytosol from becoming acidic?

A. The acid is made inside the vacuole, so the acid never leaves the vacuole
Where the acid is made does not keep it in; the membrane is a barrier the acid cannot cross freely.
B. The membrane destroys any acid that touches it before the acid can leave
A membrane destroys nothing.
C. ✓ The acid cannot cross the membrane freely, so the acid stays inside the vacuole

Why: A dissolved substance spreads through any fluid it can reach.
The vacuole’s membrane is a barrier the acid cannot cross freely.
So the acid stays inside the vacuole.
So the cytosol stays as it is: not acidic.

Check q8

In a muscle cell, the concentration of calcium ions inside the ER is thousands of times higher than in the cytosol.

Which of the following keeps the ER’s calcium ions out of the cytosol?

A. ✓ The ER’s membrane
B. The ribosomes on the ER
Ribosomes build proteins; they do not hold calcium ions in.
C. The plasma membrane
The plasma membrane is around the whole cell; the calcium ions are inside the ER, inside the cell.

Why: A membrane around a compartment lets the inside hold conditions different from the cytosol.
The calcium ions cannot cross the ER’s membrane freely.
So the calcium ions stay inside the ER.
So the ER’s calcium ions stay out of the cytosol.

41Reactions that would spoil each other

Video: Watch: Reactions that would spoil each other

Internal membranes keep reactions that would interfere with one another in separate compartments.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L20b.mp4

In the cytosol, ribosomes build new proteins all the time. Inside the lysosomes, digestive enzymes break proteins down.

A cell in which a ribosome in the cytosol is building a protein chain while, inside a lysosome, digestive enzymes are breaking a protein down

Suppose the digestive enzymes floated in the same fluid as the ribosomes. The digestive enzymes would break down each new protein as fast as the ribosomes built it.

The lysosome’s membrane keeps the digestive enzymes inside the lysosome. So the digestive enzymes never reach the proteins being built.

Internal membranes keep reactions that would interfere with one another in separate compartments.

So each reaction gets the conditions it needs.

What you are expected to know Explain that internal membranes keep reactions that would interfere with one another in separate compartments, so each reaction gets the conditions it needs.

Check q9

In one cell, reaction 1 needs acidic conditions and reaction 2 stops in acid. Yet the cell keeps both reactions at full speed at the same time.

Which of the following makes this possible?

A. The two reactions take turns, one after the other
The case says both reactions are at full speed at the same time.
So neither reaction is waiting its turn.
B. Both reactions happen together in the cytosol
The cytosol has one acidity.
Reaction 1 needs acid, and reaction 2 stops in acid, so one acidity cannot suit both.
C. ✓ Each reaction has its own membrane-bound compartment

Why: Internal membranes keep reactions that would interfere with one another in separate compartments.
One compartment is acidic and holds reaction 1.
The other compartment is not acidic and holds reaction 2.
So each reaction gets the acidity it needs.
So both reactions are at full speed at the same time.

Check q10

Reaction 1 needs acidic conditions and reaction 2 stops in acid.

Why does the cytosol suit only one of the two reactions?

A. ✓ The cytosol can hold only one acidity at a time
B. The cytosol is too crowded for two reactions
The cytosol has room for thousands of reactions at once.
C. Reaction 2 would use up the products of reaction 1
Nothing in the case says reaction 2 uses reaction 1’s products.

Why: The cytosol is one fluid.
So the cytosol has one acidity at a time.
Reaction 1 needs acid.
Reaction 2 stops in acid.
One acidity cannot suit both.
So in the cytosol, at least one of the two reactions is slow.

51More membrane, more places to work

Video: Watch: More membrane, more places to work

Folding a membrane increases the surface area, so more reactions happen at once; the three jobs of internal membranes.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L20c.mp4

Many of a cell’s reactions happen on the surface of a membrane. Proteins in the membrane carry those reactions out.

The reactions that make ATP happen on the surface of a mitochondrion’s inner membrane.

Two mitochondria of the same size, cut open: one with few inner-membrane folds, one with many; a label with an arrow reads 'reactions happening on the membrane surface' and points at the surface of a fold

Folding a membrane increases the surface area. That increases the area for reactions to happen.

So more reactions happen at once. So the mitochondrion with more folds makes ATP faster.

The inner membrane does one more thing. It keeps the conditions on its two sides different. The reactions that make ATP need that difference.

That is the third thing internal membranes do for a cell. Here are all three:

Hold conditions. The inside of a compartment can differ from the cytosol, like the acidic lysosome.
Keep reactions apart. Reactions that would interfere with one another happen in separate compartments.
Add surface. More membrane means more reactions happening on the membrane surface at the same time.

What you are expected to know Explain that folding a membrane increases the surface area, so more reactions happen on the membrane at the same time, as in the folded inner membrane of a mitochondrion.

What you are expected to know Explain that the reactions that make ATP need the conditions on the two sides of the inner membrane to differ.

Check q11

Two mitochondria, X and Y, are the same size and get the same glucose and oxygen. The inner membrane of X is folded twice as much as the inner membrane of Y.

Two mitochondria of the same size, cut open, labelled X and Y; the inner membrane of X is drawn with about twice as many folds as the inner membrane of Y

Which mitochondrion makes ATP faster?

A. Y
The reactions that make ATP happen on the inner membrane, and Y has less inner-membrane surface.
B. ✓ X
C. Neither
X’s inner membrane is folded twice as much as Y’s.
So X has twice the inner-membrane surface.

Why: The reactions that make ATP happen on the inner membrane.
X’s inner membrane is folded twice as much as Y’s.
So X has twice the inner-membrane surface.
More surface means more of the ATP-making reactions happen at the same time.
So X makes ATP faster.

Check q12

Mitochondrion X makes ATP faster than mitochondrion Y.

Why does X make ATP faster?

A. X’s folds bring the glucose closer to the inner membrane, so the reactions start sooner
Folds do not bring fuel closer; the glucose reaches the membrane the same way in both mitochondria.
B. X’s folds leave more room inside, so more of the ATP-making reactions fit in the space
The reactions that make ATP happen on the inner membrane, not in the space inside.
C. ✓ X has more inner-membrane surface, so more of the ATP-making reactions happen at once

Why: The reactions that make ATP happen on the inner membrane.
Folding a membrane increases the surface area.
X’s inner membrane is folded twice as much as Y’s.
So X has twice the surface.
More surface means more of those reactions happen at the same time.
So X makes ATP faster.

63Break a compartment, predict what fails

Video: Watch: Break a compartment, predict what fails

Two synthetic cells, one room against two rooms; then the two questions to ask when a compartment breaks.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L20f.mp4

Researchers build synthetic cells: tiny bags of membrane filled with enzymes. Here is an experiment with two of them.

The experiment uses two enzymes, E1 and E2.

E1 works best in acid. E1 turns a starting substance into a middle substance.

E2 works best out of acid. E2 turns the middle substance into the product.

Cell A has one room. E1 and E2 are mixed together in it, and the room is mildly acidic.

Two synthetic cells drawn side by side. Cell A is one shaded room holding E1 and E2 mixed, mildly acidic, 29 units of the product. Cell B has two rooms: room 1, shaded acidic, holds E1; room 2, unshaded, holds E2; the dashed membrane 3 between them lets the middle substance cross; 84 units of the product

After 20 minutes, cell A has made 29 units of the product.

Cell B has two rooms. Room 1 is acidic and holds E1. Room 2 is not acidic and holds E2.

Membrane 3, between the two rooms, lets the middle substance cross from room 1 to room 2.

After 20 minutes, cell B has made 84 units of the product. Why so much more?

In cell B, E1 sits in acid, in room 1. E2 sits out of acid, in room 2. So each enzyme works at its best.

And the middle substance crosses membrane 3. So E2, in room 2, receives what E1 makes in room 1.

In cell A, both enzymes sit at the same mild acidity. That acidity is too weak for E1 and too strong for E2. So both enzymes work slowly.

So the membrane does two jobs. It keeps different conditions in each room. And it lets the middle substance cross into the next room.

To predict what fails when a compartment breaks, ask two questions:

What conditions did its membrane keep inside?
What did its membrane keep apart?

Then say what happens once those conditions are lost, or once the things kept apart reach each other.

What you are expected to know Predict what fails when a compartment is disrupted: its acid lost, its membrane made leaky, or its boundary blocked.

What you are expected to know Justify each prediction from the conditions the membrane kept inside and what the membrane kept apart.

Check q13

A drug neutralizes the acid inside a cell’s lysosomes, so their insides sit at the cytosol’s acidity. Their membranes stay intact and their enzymes stay inside them.

What happens to the digestion of bacteria the cell takes in?

A. Digestion speeds up
The digestive enzymes work best in acid, and the acid is gone.
B. Digestion goes on as before
Every part is in place, but the enzymes break proteins down quickly only in acid, and the acid is gone.
C. ✓ Digestion slows

Why: The drug has removed the acid from the lysosome.
The digestive enzymes work best in acid.
So digestion slows, even though every part is still in place.

Check q14

Digestion in the lysosomes slows after the drug neutralizes their acid, although their membranes and enzymes are intact.

Why does digestion slow?

A. The drug has broken the digestive enzymes apart
The drug neutralizes acid, and the enzymes are whole.
B. ✓ The enzymes need the acidic conditions the lysosome no longer holds
C. The enzymes have leaked out into the cytosol
The membranes are intact, so the enzymes are still inside.

Why: The digestive enzymes work best in acid.
The lysosome’s membrane used to keep acidic conditions inside.
The drug has neutralized the acid.
So the lysosome no longer holds the conditions its enzymes need.
So digestion slows.

Check q15

A poison makes the inner membrane of a mitochondrion leaky, so the conditions on its two sides become the same. The enzymes there are unharmed.

What happens to the mitochondrion’s ATP output?

A. The ATP output rises
A leaky membrane cannot keep the inside different from the outside, and the ATP-making reactions depend on that difference.
B. The ATP output stays the same
Intact enzymes are not enough: the ATP-making reactions need the difference in conditions across the inner membrane, and a leaky membrane cannot keep it.
C. ✓ The ATP output falls

Why: The reactions that make ATP need the conditions the inner membrane keeps on each side.
A leaky inner membrane cannot keep them different.
So the ATP output falls, even though every enzyme is intact.

Check q16

In synthetic cell B, E1 turns the starting substance into the middle substance in the acidic room. E2 turns the middle substance into the product in the room that is not acidic. The researchers replace the membrane between the two rooms with one that blocks the middle substance. Everything else stays the same.

Predict the amount of the product made in 20 minutes.

A. More than cell B made before
Blocking the middle substance stops E2 from receiving anything to work on.
B. About the same as cell B made before
The conditions in each room are the same as before.
But E2 no longer receives any of the middle substance.
C. ✓ Almost none, far less than cell B made before

Why: E2 makes the product only from the middle substance.
The new membrane blocks the middle substance in the acidic room.
So E2 receives none of it.
So almost none of the product is made, however well each enzyme’s room suits it.

Practice writing an answer

A drug makes the membranes of a cell’s lysosomes leaky. Acid leaks out of the lysosomes into the cytosol. The digestive enzymes leak out too.

(a) Predict two effects on the cell, and justify each. (2 pt)

Model answer First effect: the lysosomes stop digesting what the cell takes in.
The acid has leaked out, so the inside of each lysosome is no longer acidic.
The digestive enzymes work best in acid.
Second effect: the digestive enzymes reach the cell’s own proteins in the cytosol.
The membrane used to keep the enzymes apart from those proteins.
The enzymes work slowly at the cytosol’s acidity, but they now reach proteins they never reached before.

Rubric

Award 1 point for: digestion inside the lysosomes slows or stops, because the acid the enzymes need has leaked out (the conditions the membrane kept are lost).
Award 1 point for: the enzymes reach the cell’s own proteins in the cytosol, because the membrane no longer keeps them apart (the separation is lost). Accept a note that they work slowly there.

Slip Giving one effect twice in different words. One effect is about the conditions the membrane kept inside; the other is about what it kept apart.

Practice writing an answer

In an inherited disease, a person’s cells make normal digestive enzymes. But the Golgi complex packs the enzymes into vesicles that fuse with the plasma membrane.
So the enzymes are released into the extracellular fluid instead of being delivered to lysosomes. The lysosomes form, with acidic insides, but hold almost no enzymes.
The extracellular fluid, like the cytosol, is not acidic. The person’s white blood cells still take in bacteria by endocytosis.

(a) Describe what a lysosome’s membrane does for the digestive enzymes inside a healthy cell. (1 pt)

Model answer The membrane keeps the inside of the lysosome acidic.
The digestive enzymes work best in acid.
The membrane also keeps the enzymes apart from the cytosol.
In the cytosol, ribosomes are building the cell’s own proteins, which the enzymes would break down.

Rubric

Award 1 point for: the membrane keeps the acidic conditions the enzymes need inside the lysosome, and keeps the enzymes apart from the cytosol (and the proteins being built there).
Accept: either the conditions kept or the separation, provided the answer says what that does for the enzymes or the cell.

Slip Saying the membrane just holds the enzymes in place. What matters is the conditions it keeps, acid inside, and the separation from the cytosol.

(b) Predict what happens to the bacteria the white blood cells take in. (1 pt)

Model answer The lysosomes hold almost no enzymes.
So the bacteria the cells take in are not digested.
The bacteria build up inside the cells, still wrapped in membrane.

Rubric

Award 1 point for: the bacteria build up undigested inside the cells (in vesicles or lysosomes), because the lysosomes hold almost no enzymes.
Accept: digestion of what the cell takes in fails, with the reason given.

Slip Having the cytosol’s own proteins digest the bacteria. Digestion is the job of the lysosome’s enzymes, and they were sent out of the cell.

(c) Support the claim that the released enzymes do little damage to the tissue around the cells. (1 pt)

Model answer The released enzymes break proteins down quickly only in acidic conditions.
A lysosome’s membrane keeps those conditions inside the lysosome.
The extracellular fluid is not acidic.
So outside the cell the enzymes break proteins down slowly.
So they do little damage.

Rubric

Award 1 point for: the evidence AND the reasoning: the enzymes break proteins down quickly only in acid; the extracellular fluid is not acidic, so outside the cell they break proteins down slowly.
Accept: any wording that ties the enzymes’ slowness outside to the acidity they lack there.

Slip Expecting the enzymes to digest the tissue because they are digestive. Away from the acidic conditions a lysosome’s membrane keeps, they work slowly.

87Summary

The lysosome’s enzymes work best in acid. The lysosome’s membrane keeps the acid inside.

So the enzymes break proteins down quickly inside the lysosome. In the cytosol they break proteins down slowly.

Lots of compartments have a membrane that lets the inside be chemically different from the cytosol. Dividing the cell into such rooms is called compartmentalization.

Internal membranes do three things: they hold conditions, they keep reactions apart, and they add surface for reactions to happen on.

When a compartment breaks, ask what conditions its membrane kept inside and what its membrane kept apart. Then predict what fails.

Glossary

compartmentalization: Dividing a cell’s inside into membrane-bound compartments, each able to hold conditions different from the cytosol. A compartment is a separate room; compartmentalization is dividing the cell into separate rooms, each with its own conditions.

APBIO-U02-P29 Practice questions: Topic 2.9

Topic 2.9 · Cell Compartmentalization · 9 MCQ · 2 FRQ · for APBIO-U02-T29

Video: Watch first: Topic 2.9 in one picture

Hold conditions, keep reactions apart, add surface: what a cell’s internal membranes do, and how to predict what fails when a compartment breaks.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T29-summary.mp4

Q1 P29-q01

In the storage cells of a sprouting bean seed, digestive enzymes sit inside small membrane-bound bodies together with stored protein. The interior of each body is acidic; the cytosol around it is close to neutral. The enzymes cut stored protein quickly in acid and barely at all at the cytosol's acidity.

What lets the enzymes work inside the bodies and hardly at all in the cytosol around them?

A. The cytosol holds a substance that destroys the digestive enzymes
Nothing destroys the enzymes.
At the cytosol’s acidity they are still there, working slowly.
B. The membrane of each body speeds the digestive enzymes up
A membrane is a boundary, not a worker.
C. The digestive enzymes only cut protein that is stored inside a body
The enzymes cut the same stored protein in both tests.
D. ✓ The membrane of each body keeps its interior acidic while the cytosol is not

Why: The enzymes cut protein quickly only in acid.
A membrane around a compartment lets the inside hold conditions different from the cytosol.
The body’s membrane holds the acid inside.
So the enzymes work inside the bodies and hardly at all in the cytosol.

Q2 P29-q02

A scientist builds an artificial cell: a single outer membrane filled with fluid. The scientist considers four ways of arranging its contents.

Which arrangement gives the artificial cell compartmentalization?

A. Its proteins gathered into a dense cluster at the center, with no membrane around them
A cluster with no membrane around it evens out with the fluid around it.
B. ✓ Several membrane-bound sacs inside it, each holding fluid with its own conditions
C. A rigid cell wall added around the outside of the outer membrane
A cell wall sits outside the whole cell.
Compartmentalization is about membranes dividing the inside.
D. The fluid inside stirred so that every substance is spread evenly throughout
Even spreading is the opposite of compartmentalization.

Why: Compartmentalization is the dividing of a cell’s interior into membrane-bound compartments.
Each compartment can hold conditions different from the fluid around it.
Membrane-bound sacs inside the artificial cell are exactly that.

Q3 P29-q03

Cells of a mustard leaf make a bitter compound that damages a plant's protein-building machinery when the compound reaches it. The cells store the compound inside their central vacuoles, while ribosomes in the cytosol build new proteins all day.

Why does the cell benefit from storing the compound in the vacuole?

A. ✓ The compound never reaches the ribosomes, so protein building goes on undisturbed
B. The vacuole is where the compound is made, so it has to stay there
The benefit is that the vacuole’s membrane keeps the compound away from the ribosomes; where it was made gives the ribosomes no protection.
C. The compound would leak out of the cell if it were loose in the cytosol
The plasma membrane already keeps the cell’s contents inside.
D. The vacuole's membrane makes the compound more bitter to insects
A membrane does nothing to the compound itself.

Why: The compound damages the protein-building machinery when the compound reaches it.
The vacuole’s membrane holds the compound inside the vacuole, away from the ribosomes in the cytosol.
So the ribosomes build proteins undisturbed.

Q4 P29-q04

Cells of a mustard leaf store a bitter compound inside their central vacuoles. The compound damages ribosomes when it reaches them, and the ribosomes in the cytosol of these cells build new proteins all day. The vacuole membrane becomes leaky, so the bitter compound spreads into the cytosol.

Predict the effect on protein building in the cytosol, and explain why.

A. Protein building speeds up: the compound now reaches the ribosomes and helps them
The compound damages the protein-building machinery.
Reaching the ribosomes hurts them.
B. Protein building is unchanged: the ribosomes are still in the cytosol where they were
The ribosomes are unchanged, but the compound that used to be held in the vacuole now reaches them.
C. ✓ Protein building slows: the compound now reaches the ribosomes it damages
D. Protein building stops only inside the vacuole, where the ribosomes have moved
Nothing moved the ribosomes.
The compound is what has moved.

Why: The vacuole’s membrane was keeping the compound away from the ribosomes.
Once the membrane leaks, the compound spreads into the cytosol by diffusion.
The compound reaches the ribosomes and damages them.
So protein building slows.

Q5 P29-q05

The flight muscle cells of a hummingbird hold mitochondria whose inner membranes are folded far more than those in a chicken's breast muscle cells. The reactions that make ATP happen on the inner membrane. The two kinds of mitochondria are about the same size.

What do the extra folds do for the hummingbird's muscle cells?

A. The folds store fuel for the flight muscle to burn later
Folds are membrane, and membrane stores no fuel.
B. The folds keep the inside of each mitochondrion warmer, so the reactions speed up
Folds do not warm anything.
C. ✓ The folds give more membrane surface, so more ATP-making reactions happen at once
D. The folds let each mitochondrion hold more fluid than a chicken's
Folding a membrane inward takes up space.
It does not add fluid.

Why: The reactions that make ATP happen on the inner membrane, so they happen only where there is membrane.
Folding puts far more inner membrane inside a mitochondrion of the same size.
So more ATP-making reactions happen at the same time, which flight muscle needs.

Q6 P29-q06

Cells lining the gut that take in fat from food have far more smooth ER membrane than their neighbors. The reactions that rebuild absorbed fat into fats the body can use happen on the membranes of the smooth ER.

What does the extra ER membrane do for these cells?

A. ✓ The extra membrane gives more surface on which the fat-building reactions can happen at once
B. The extra membrane stores the absorbed fat until the cell needs it
The smooth ER does not store the fat.
C. The extra membrane lets the cell get by with fewer mitochondria than its neighbors
Mitochondria have a different job, making ATP.
D. The extra membrane thickens the plasma membrane so that fat can cross it more easily
The ER is inside the cell, separate from the plasma membrane.

Why: The fat-building reactions happen on the smooth ER membrane.
The surface of a membrane sets how many of its reactions can happen at the same time.
So more smooth ER lets these cells rebuild more fat at once.

Q7 P29-q07

In many plant cells the central vacuole is acidic while the cytosol is close to neutral, and the reactions that happen in the cytosol work best close to neutral. The vacuole’s membrane becomes leaky to acid.

Predict the effect on the reactions in the cytosol, and explain why.

A. The reactions are unaffected: the acid stays in the vacuole
A leaky membrane lets the acid diffuse out.
B. ✓ The reactions slow: the acid spreads into the cytosol, which is no longer close to neutral
C. The reactions speed up: acid speeds up every reaction
These reactions work best close to neutral.
There is no rule that acid speeds every reaction.
D. The reactions move into the vacuole, where the conditions are now the same as outside
Reactions happen where their proteins are, and those proteins stay in the cytosol.

Why: The vacuole’s membrane was holding the acid inside the vacuole.
Once the membrane leaks, the acid spreads into the cytosol by diffusion.
The cytosol is no longer close to neutral.
The reactions in the cytosol work best close to neutral, so they slow.

Q8 P29-q08

In a damaged cell, the membrane of one lysosome breaks. The lysosome’s acid and its digestive enzymes reach the mitochondria next to it. The reactions that make ATP happen on proteins in each mitochondrion’s inner membrane, and those reactions work well only in conditions close to neutral.

Predict the effect on how fast the cell makes ATP, and why.

A. Faster: the acid speeds up the ATP-making reactions on the inner membrane
The ATP-making reactions work well only close to neutral.
B. Unchanged: the inner membrane and its ATP-making proteins are still in place
The membrane and its proteins are in place, but the conditions they worked in are gone, and the enzymes are breaking the proteins down.
C. Faster: the digestive enzymes break down fuel right beside the inner membrane
Digestive enzymes take proteins apart.
Beside the inner membrane they attack the very proteins that make ATP.
D. ✓ Slower: the acid and the digestive enzymes damage the inner membrane’s proteins

Why: The lysosome’s membrane held its acid and digestive enzymes away from the mitochondria.
Once it breaks, the acid reaches the mitochondria, and the ATP-making reactions work only close to neutral.
The digestive enzymes break down the inner-membrane proteins that make ATP.
So ATP is made more slowly.

Q9 P29-q09

The figure shows one compartment inside a cell, before and after a change. A reaction that happens inside the compartment needs substance Y at 10 mmol/L or more. Before the change Y is at 30 mmol/L inside the compartment and 0.3 mmol/L in the cytosol; afterward it is at 1.2 mmol/L in both.

One compartment inside a cell, before and after a change to its membrane, with the concentration of substance Y inside the compartment and in the cytosol.

What most likely changed, and what follows for the reaction?

A. Most of the Y in the cell was destroyed, so the reaction stops for lack of Y
Nothing needed destroying.
The compartment’s 30 mmol/L spread through the much larger cytosol gives about 1.2 mmol/L everywhere.
B. Y was pumped out of the compartment, so the reaction now happens in the cytosol instead
Pumping Y out would use energy and would leave the cytosol above the compartment, not equal to it.
C. ✓ The membrane became leaky to Y, so Y spread out by diffusion and the reaction stops
D. Y was concentrated in the cytosol, so the reaction speeds up there
1.2 mmol/L in the cytosol is far below the 10 mmol/L the reaction needs.

Why: The membrane kept Y at 30 mmol/L in one place.
Once the membrane leaks, Y diffuses out until it is equal on both sides.
The cytosol is much larger, so both settle at about 1.2 mmol/L, below the 10 mmol/L the reaction needs.
So the reaction stops.

FRQ 1 P29-frq1 · Conceptual Analysis scaffolded

A company wants yeast cells to make a useful compound, Q, in two steps. Enzyme 1 turns a starting substance into an intermediate, J, and works best in acidic conditions; J damages ribosomes if it reaches them. Enzyme 2 turns J into Q and works best in conditions close to neutral, like the cytosol. Design A places both enzymes in the cytosol: after a day the cells hold 6 units of Q, and they grow slowly. Design B places enzyme 1 inside the yeast's vacuole, whose interior is acidic, and enzyme 2 in the cytosol; the vacuole's membrane lets J pass slowly out into the cytosol. After a day the cells hold 48 units of Q and grow normally.

A sketch of the two designs: the first move in a question like this is to draw the cell, the vacuole inside it, and where each enzyme sits. Dots are ribosomes.

(a) Explain why enzyme 1 makes more J in design B than in design A. (1 pt)

Frame Enzyme 1 works best in …; in design B it sits in …, so …, whereas in design A it sits in …, so …

Hint In which conditions does enzyme 1 work best, and which conditions does it sit in, in each design?

Model answer Enzyme 1 works best in acidic conditions.
In design B, enzyme 1 sits inside the vacuole.
The vacuole’s membrane keeps the interior acidic, so enzyme 1 works at its best.
In design A, enzyme 1 sits in the cytosol.
The cytosol is close to neutral, so enzyme 1 works slowly.
So enzyme 1 makes more J in design B.

Rubric

Award 1 point for: enzyme 1 works best in acid; in design B it sits inside the acidic vacuole, so it works at its best, whereas in design A it sits in the cytosol, close to neutral, where it works slowly.
Accept: 'the vacuole's membrane gives enzyme 1 the acid it needs; the cytosol does not'.

Slip Saying the vacuole 'protects' enzyme 1 with no mention of acidity. It is the acidic conditions the membrane holds that matter.

(b) Explain why the cells grow normally in design B but slowly in design A. (1 pt)

Frame In design A, J forms in …, where it …; in design B, J forms in …, and the J that leaks out …, so …

Hint In each design, where does J form, and what sits nearby that J could affect?

Model answer In design A, J forms in the cytosol.
The ribosomes are in the cytosol, so J reaches them and damages them.
Protein building slows, so the cells grow slowly.
In design B, J forms inside the vacuole, away from the ribosomes.
The J that leaks out slowly is turned into Q by enzyme 2 in the cytosol.
So little J ever reaches the ribosomes, and the cells grow normally.

Rubric

Award 1 point for: in design A, J forms in the cytosol, where the ribosomes are, and damages them, so protein building and growth slow; in design B, J forms inside the vacuole, away from the ribosomes, and the J that leaks out slowly is turned into Q by enzyme 2, so little of it reaches the ribosomes.
Accept: 'the vacuole membrane keeps J apart from the ribosomes'. Do not award the point for 'the cells have more Q' with no mention of J and the ribosomes.

Slip Explaining the growth from the amount of Q. Growth depends on the ribosomes, and the question is whether J reaches them.

(c) Predict the yield of Q from a design C: enzyme 1 in the vacuole and enzyme 2 in the cytosol, as in design B, but with a vacuole membrane that blocks J completely. (1 pt)

Frame Design C would give … Q, because …

Hint In design C, where is J made and where is enzyme 2? What lies between them?

Model answer Design C would give almost no Q, less even than design A’s 6 units.
The J that enzyme 1 makes stays trapped in the vacuole.
Enzyme 2 is in the cytosol, so it never receives any J.

Rubric

Award 1 point for: the prediction that design C gives very little Q, close to zero, lower than design A's 6 units. No reasoning is required for this point.
Do not award a prediction near design B's 48 units, or one equal to design A.

Slip Predicting a yield near 48 units because the acidity is right. Right conditions do nothing for enzyme 2 if J never reaches it.

(d) Justify your prediction in (c), using what the vacuole's membrane must hold in and what it must let through for this process to work. (1 pt)

Frame For the process to work, the membrane must hold in … and let through …; design C's membrane …, so …

Hint What must stay inside the vacuole for enzyme 1 to do its job, and what must be able to leave it for enzyme 2 to do its job? A quick sketch of the vacuole inside the cell, with each enzyme in its place, makes this easier to see.

Model answer For the process to work, the membrane must do two things.
It must hold in the acid, so that enzyme 1 works.
It must let J through to the cytosol, where enzyme 2 turns it into Q.
Design C’s membrane holds the acid but traps J.
So enzyme 2 receives no J, and no Q forms, however well enzyme 1 works.
Design B gave 48 units only because J could pass out of the vacuole.

Rubric

Award 1 point for: the evidence AND the reasoning: the membrane must hold in the acid, so that enzyme 1 works, AND let J through to the cytosol, where enzyme 2 turns it into Q; design C's membrane holds the acid but traps J, so enzyme 2 receives no J and no Q forms, however well enzyme 1 works.
Accept: reasoning that cites design B's yield as depending on J passing out of the vacuole. Do not award the point for support from acidity alone.

Slip Justifying from acidity alone. A compartment helps only if what the next step needs can still get through.

FRQ 2 P29-frq2 · Analyze Model or Visual Representation

The model shows part of a cell from a mustard leaf. The central vacuole is shaded; its shading marks an acidic interior that holds a bitter defensive compound at 80 mmol/L, while the cytosol holds the compound at 0.2 mmol/L. The compound stops ribosomes from working if it reaches them. The mitochondrion is drawn with its outer membrane and its folded inner membrane, where the reactions that make ATP happen. The dots in the cytosol are ribosomes, building new proteins.

Part of a mustard leaf cell: the central vacuole (shaded: acidic, holding the defensive compound at 80 mmol/L); a mitochondrion with a folded inner membrane; ribosomes in the cytosol. The cytosol holds the compound at 0.2 mmol/L.

(a) Describe what the vacuole's membrane does for the cell, in terms of the conditions it holds inside and what it keeps apart. (1 pt)

Frame The vacuole's membrane keeps the inside … and holds the compound at …, and it keeps the compound apart from …

Model answer The vacuole’s membrane keeps the inside acidic.
It holds the compound at 80 mmol/L inside while the cytosol has only 0.2 mmol/L.
It also keeps the compound apart from the ribosomes in the cytosol.
The compound would stop those ribosomes if it reached them.

Rubric

Award 1 point for: the membrane lets the vacuole's interior hold conditions different from the cytosol (acidic, and the compound at 80 mmol/L against 0.2 mmol/L outside), AND it keeps the compound away from the ribosomes it would stop.
Both halves, the conditions held and what is kept apart, are needed for the point.

Slip Saying the membrane 'holds the contents in place'. Name the conditions it holds and the thing it keeps the compound away from.

(b) Explain how the membrane makes possible a concentration of 80 mmol/L inside the vacuole beside 0.2 mmol/L in the cytosol. (1 pt)

Model answer Dissolved particles move at random and spread out.
Without a membrane, the compound would spread until it was evenly mixed.
The compound stays 400 times more concentrated inside the vacuole.
So the membrane must not let the compound diffuse freely across it.
The membrane is what makes the difference in concentration possible.

Rubric

Award 1 point for: dissolved particles move at random and spread, so the compound could stay 400 times more concentrated in the vacuole only because the membrane does not let it diffuse freely across; without the membrane, diffusion would even the two sides out.
Accept: 'a gradient can exist across a membrane only for a substance the membrane blocks'. Do not award the point for 'the vacuole is where it is made' with no reference to the membrane blocking diffusion.

Slip Saying the compound is concentrated because it is made in the vacuole. Whatever makes it, only a membrane that blocks it keeps it from spreading.

(c) Predict the effect on the ribosomes if the vacuole’s membrane is dissolved, and justify your prediction. (1 pt)

Model answer The ribosomes slow or stop building proteins.
With the membrane gone, the compound spreads out of the vacuole by diffusion into the cytosol.
It reaches the ribosomes and stops them.
The acid spreads out too, so the cytosol becomes acidic as well.

Rubric

Award 1 point for: the ribosomes slow or stop building proteins, because the compound spreads out of the vacuole by diffusion into the cytosol, reaches the ribosomes and stops them (the cytosol may also turn acidic).
Accept either the compound or the acid as the agent, provided it is said to reach the ribosomes. Do not award the point for 'nothing changes' or for a prediction with no justification.

Slip Predicting no change because the ribosomes themselves are untouched. Their surroundings change: what the membrane held in now reaches them.

(d) Support the claim that the mitochondrion in the model shows a third benefit of internal membranes, different from the two in part (a). (1 pt)

Model answer The mitochondrion’s inner membrane is folded.
Folding fits far more membrane into the same space.
The ATP-making reactions happen on that membrane, so more of them happen at the same time.
So the mitochondrion shows a third benefit of internal membranes: they add surface for reactions, as well as holding conditions and keeping things apart.

Rubric

Award 1 point for: the evidence AND the reasoning: the mitochondrion’s inner membrane is folded; folding fits more membrane into the same space, so more of the ATP-making reactions happen at the same time; that is a third benefit (more surface for reactions), different from holding conditions and keeping things apart.
Accept: 'more surface area for membrane-bound reactions' with the link to more reactions at once or faster ATP-making, named as a benefit different from the two in (a).

Slip Saying the folds store energy or let fuel in. They add surface for the reactions that make ATP.

APBIO-U02-T29 End-of-topic test: Cell Compartmentalization

Topic 2.9 · Cell Compartmentalization · 16 MCQ · 2 FRQ

Answer every question. For each multiple-choice question choose one answer and press Check; the feedback gives the reasoning. For the free-response questions, write one short sentence for each step of your reasoning, each on its own line, and make every link clear (so, because, therefore). That is what the exam’s ‘paragraph form’ means for you: linked sentences, not bullet points. Then open the scoring guide and mark your own work against it.

Q1 T29-q01

In a yeast cell, the fluid inside the vacuole is about 100 times more acidic than the cytosol around it. Digestive enzymes taken from the vacuole break protein down quickly in a test tube at the vacuole’s acidity, and barely at all at the cytosol’s acidity. In the living cell the enzymes break protein down quickly, and the cytosol stays close to neutral.

What lets the vacuole’s enzymes break protein down quickly in the living cell while the cytosol around them stays close to neutral?

A. The cytosol contains a substance that removes any acid that reaches it.
Nothing removes the acid.
The acid stays where it is, inside the vacuole.
B. ✓ The vacuole’s membrane keeps its interior acidic while the cytosol is not.
C. The vacuole’s membrane itself speeds the enzymes up.
A membrane is a boundary, not a worker.
D. The digestive enzymes only work on protein that is inside the vacuole.
The enzymes were given protein in both tests.

Why: The digestive enzymes need acidic conditions to work well.
A membrane around a compartment lets the inside hold a different acidity from the cytosol around it.
The vacuole’s membrane holds the acid inside the vacuole.
So the enzymes break protein down quickly inside the vacuole, and the cytosol stays close to neutral.

Q2 T29-q02

Inside a chloroplast, the reactions that capture light happen on the membranes of many flat discs stacked inside it. A plant grown in dim light builds chloroplasts that hold more stacked discs than the chloroplasts of the same plant grown in bright light.

What does the extra disc membrane do for the dim-light plant’s cells?

A. The extra discs store light for the chloroplast to use at night.
Light cannot be stored.
A chloroplast uses light as it arrives.
B. The extra discs hold more fluid, so each chloroplast swells and catches more light.
A flat membrane disc holds almost no fluid, and swelling would not change how much light reaches the chloroplast.
C. ✓ The extra discs let more of the light-capturing reactions take place at the same time.
D. The extra discs keep the chloroplast’s interior separate from the cytosol.
The chloroplast’s outer membranes already separate its interior from the cytosol; the discs are inside them.

Why: The light-capturing reactions happen on the membranes of the flat discs.
A reaction that happens on a membrane can happen in more places when there is more membrane.
More stacked discs means more disc membrane, so more light-capturing reactions happen at once.
So the plant captures more light.

Q3 T29-q03

In a cell, substance X is at 40 units of concentration inside a small membrane-bound organelle and at 1 unit in the cytosol. The reaction that uses X happens inside the organelle. A treatment dissolves the organelle’s membrane, while everything else in the cell stays in place.

What happens to the concentration of X at the site of the reaction, and why?

A. ✓ The concentration falls, because X spreads by diffusion through the whole cytosol.
B. The concentration rises, because the cytosol's X adds to the organelle's.
The cytosol holds only 1 unit of X.
C. The concentration stays the same, because X cannot move without a membrane to carry it.
A membrane does not carry X.
A membrane holds X in, and without one X moves by diffusion.
D. The concentration stays the same, because the organelle's proteins hold X in place.
The organelle’s proteins do not hold X in place.

Why: The membrane was what kept X concentrated in one place.
With the membrane gone, X moves at random out into the cytosol.
The cytosol is much larger than the organelle, so the 40 units spread thin.
So the concentration of X at the site of the reaction falls.

Q4 T29-q04

A cell breaks down its worn-out organelles inside lysosomes. At the same time, ribosomes in its cytosol are joining amino acids into new proteins.

Why does the cell keep the breaking-down inside lysosomes rather than letting it happen in the cytosol?

A. A lysosome is where the digestive enzymes are made, so digestion has to happen there.
Lysosomes do not make their enzymes.
Ribosomes make them, and they reach the lysosome through the ER and Golgi.
B. Worn-out organelles are too large to be taken apart in the open cytosol.
A worn-out organelle is wrapped in membrane and delivered whole to a lysosome, so its size is no obstacle; the lysosome exists to keep the enzymes away from everything else.
C. Loose in the cytosol, the digestive enzymes would drift away and be lost from the cell.
The plasma membrane already keeps everything inside the cell.
D. ✓ The digestive enzymes never reach the new proteins and so cannot break them down.

Why: Loose in the cytosol, the digestive enzymes would break down the new proteins as fast as the ribosomes built them.
Keeping digestion inside lysosomes keeps the enzymes and the new proteins apart.
So protein-building and digestion do not interfere with each other.

Q5 T29-q05

In a test tube, reaction 1 turns substance A into product P, and reaction 2 turns P into a waste. A student sets up two tubes, leaves the reactions for 20 minutes, then measures the P present. Tube 1, both reactions together in one solution: 9 units of P. Tube 2, reaction 1 and its supply of A inside a sealed bag whose membrane holds P in, with reaction 2 outside the bag: 61 units of P.

What does the comparison show?

A. The membrane speeds up reaction 1.
A membrane is a boundary.
It does not speed up a reaction.
B. ✓ Separating the reactions stops reaction 2 from using up P as it forms.
C. Reaction 2 cannot happen at all unless a membrane is present nearby.
Reaction 2 happened in tube 1, with no membrane anywhere.
D. The bag lets A spread more evenly, so more P forms.
How evenly A spreads does not change how much P is made.

Why: In one tube, reaction 2 destroys P as soon as reaction 1 makes it, so only 9 units are left.
In the bag, the membrane holds P in, so reaction 2 never reaches it, and P builds up to 61 units.
Internal membranes in a cell keep interfering reactions apart.

Q6 T29-q06

Inside a membrane-bound compartment of a cell, a reaction makes substance Q. In the cytosol outside it, a second reaction happens that Q slows down. The compartment’s membrane becomes leaky to Q, but nothing else changes.

Predict the effect on the second reaction, and explain why.

A. ✓ The second reaction slows, because Q now spreads into the cytosol where that reaction happens.
B. The second reaction speeds up, because Q leaves the compartment and stops competing with it.
Q spreading out of the compartment is the problem, not the cure.
C. The second reaction is unchanged, because its own proteins are still in the cytosol.
The proteins of the second reaction are unchanged, but their surroundings are not.
D. The second reaction speeds up, because a leaky membrane lets reactants reach it faster.
The second reaction happens in the cytosol, outside the compartment.

Why: The compartment’s membrane was keeping Q away from the second reaction.
Once the membrane leaks, Q diffuses out into the cytosol.
The second reaction happens in the cytosol, and Q slows it down.
So the second reaction slows.

Q7 T29-q07

The figure shows two membrane compartments of the same outer size. A reaction happens only on the surface of the inner membrane of each.

Two compartments of the same outer size. In design 1 the inner membrane is smooth; in design 2 it is folded.

What advantage does design 2 have?

A. Its folds hold the contents more tightly, so nothing leaks out.
Whether a membrane leaks depends on what it is made of, not on whether it is folded.
B. Its folds let the compartment hold more fluid than design 1.
Folding a membrane inward takes up space.
It does not add fluid.
C. ✓ More membrane fits inside it, so more of the reaction can happen at once.
D. Its folds keep the inside warmer, so the reaction speeds up.
Folds do not warm anything.

Why: The reaction happens only on the inner membrane.
Folding puts far more membrane inside the same outer size.
So design 2 has more surface on which the reaction can happen, and more of the reaction happens at the same time.

Q8 T29-q08

A single-celled pond organism, a Paramecium, digests the food it takes in inside small membrane-bound food vacuoles, whose interior is acidic. A researcher adds a compound that brings the interior of each food vacuole to the cytosol’s acidity. The vacuole membranes stay whole, and the digestive enzymes stay inside.

Predict the effect on the digestion of food inside the food vacuoles.

A. ✓ Digestion slows, because the digestive enzymes have lost the acidic conditions they need.
B. Digestion speeds up, because the enzymes now work at the same acidity as the enzymes of the cytosol.
The cytosol’s acidity is the wrong acidity for these enzymes.
C. Digestion is unchanged, because the digestive enzymes are still inside the food vacuoles.
The enzymes are still there, but the conditions they need are not.
D. Digestion stops, because the compound breaks the digestive enzymes apart.
The compound changed the acidity, not the enzymes.

Why: The food vacuole’s membrane was keeping its interior acidic.
The compound brings the interior to the cytosol’s acidity.
The digestive enzymes barely work without acidic conditions.
So digestion slows.

Q9 T29-q09

In a eukaryotic cell, one compartment holds reactions that work well only in acidic conditions, and a second compartment holds reactions that work well only where there is no acid. A defect makes the two membranes fuse into one compartment, whose interior settles at a mild acidity, halfway between the two.

Predict the effect on the two sets of reactions.

A. Both sets speed up, because their reactants are now in one place.
Bringing reactants together does not help reactions that need different conditions.
B. Only the acid-needing reactions slow; the others are unaffected.
The second set is affected too.
Those reactions worked well only where there was no acid, and the merged interior is mildly acidic throughout.
C. Neither set changes, because each set stays on its own side of the merged space.
Nothing keeps the two sides apart any more.
D. ✓ Both sets slow, because the merged compartment can have only one acidity.

Why: One compartment can hold only one acidity.
The merged compartment settles at a mild acidity, halfway between the two.
That is too weak for the reactions that need acid and too strong for the reactions that acid stops.
So both sets slow.

Q10 T29-q10

A mutant yeast strain has mitochondria whose inner membranes carry about 33% of the folds found in the normal strain’s mitochondria, so each mutant inner membrane has about 33% of the normal membrane area. The reactions that make ATP happen only on the inner membrane. The mutant’s outer membrane and the rest of the cell are unchanged.

Predict how fast the mutant strain’s mitochondria make ATP, compared with the normal strain’s.

A. At the same rate, because the outer membrane and the rest of the cell are unchanged.
The ATP-making reactions happen only on the inner membrane, not on the outer membrane or elsewhere in the cell, and the inner membrane now has 33% of its area.
B. ✓ More slowly, because fewer of the ATP-making reactions can take place at the same time.
C. Faster, because substances cross a less folded inner membrane more easily.
Fewer folds do not help substances cross the inner membrane; what has changed is the membrane area on which the reactions happen.
D. More slowly, because a less folded inner membrane holds less fuel for the reactions.
The inner membrane does not hold fuel; the folds add membrane area on which the reactions happen.

Why: The reactions that make ATP happen only on the inner membrane.
The mutant’s inner membrane has about 33% of the normal membrane area.
So the mutant has about 33% of the surface for those reactions, and fewer happen at the same time.
So ATP is made more slowly.

Q11 T29-q11

In a single eukaryotic cell, two sets of reactions happen at the same moment: one set works only where a substance, Z, is concentrated; the other set is shut down by Z.

How can both sets happen at the same time?

A. The cytosol has patches rich in Z and patches free of Z, with nothing between them.
A patch of Z with nothing around it would spread by diffusion until the cytosol was evenly mixed.
B. The Z-needing reactions happen only after the others have finished.
The two sets happen at the same moment.
C. ✓ Each set happens in its own membrane-bound compartment with its own conditions.
D. The cell raises and lowers the Z concentration of its whole cytosol back and forth.
The whole cytosol cannot be rich in Z and free of Z at once.

Why: A membrane around a compartment lets the inside hold conditions different from the cytosol.
One compartment can hold Z concentrated while another holds almost none.
So the Z-needing reactions can happen in one compartment and the Z-stopped reactions in another, at the same moment.

Q12 T29-q12

One way internal membranes help a cell is by keeping reactions that would interfere with each other apart.

Which of the following is an example of a membrane keeping interfering reactions apart?

A. ✓ Digestive enzymes held in lysosomes, apart from proteins being built in the cytosol.
B. A plant cell's central vacuole filling with water and pressing the cytosol against the cell wall.
The vacuole is storing water and keeping the cell firm.
C. A mitochondrion's inner membrane folded into many pleats.
Folding adds membrane surface for reactions.
That is a different benefit of internal membranes.
D. A vesicle fusing with the plasma membrane to release its contents outside.
A vesicle fusing with the plasma membrane is exocytosis, a way of releasing material from the cell.

Why: Digestive enzymes would break down the new proteins if the enzymes reached them.
The lysosome’s membrane keeps the enzymes and the new proteins apart.
So this is a case of a membrane keeping two interfering processes apart.

Q13 T29-q13

The reactions that break down alcohol happen on the membranes of the smooth ER. In the liver cells of someone who drinks alcohol regularly, the amount of smooth ER membrane grows.

What does the extra membrane do for the cell?

A. The extra membrane stores the alcohol until the cell is ready to deal with it.
The smooth ER does not store alcohol.
Its membranes are where the alcohol-breaking reactions happen.
B. The extra membrane dilutes the alcohol so that it does less harm to the cell.
A membrane does not dilute anything.
C. The extra membrane lets the cell get by with fewer mitochondria than before.
Mitochondria have a different job, making ATP.
D. ✓ The extra membrane lets the cell break down more alcohol at the same time.

Why: The reactions that break down alcohol happen on the smooth ER membrane.
A reaction that happens on a membrane can happen in more places when there is more membrane.
So more smooth ER lets the cell break down more alcohol at the same time.

Q14 T29-q14

The figure shows part of a cell. Measurements show that the shaded region has stayed acidic for hours while the cytosol around it has stayed close to neutral.

Part of a cell. The shaded region is acidic; the cytosol around it is not. The question mark points at the region's boundary.

What must the boundary of the shaded region be, and why?

A. A thick layer of protein that binds the acid and holds it in place.
Proteins do not soak up acid and hold a region acidic for hours.
B. ✓ A membrane; without one the acid would spread through the cytosol.
C. Nothing; the region makes new acid as fast as the old spreads away.
If acid were spreading out as fast as it was made, the cytosol next to the region would be acidic too.
D. A thicker, jelly-like patch of cytosol that slows the spread of acid.
Thicker cytosol slows diffusion a little.
It does not stop diffusion for hours.

Why: Particles of dissolved acid move constantly and at random.
An acidic region with nothing around it would even out with the cytosol.
The region has stayed acidic for hours while the cytosol has not.
So a membrane must surround the region, holding its conditions different from the cytosol.

Q15 T29-q15

A treatment dissolves every internal membrane in a eukaryotic cell, but its plasma membrane stays whole and all of its proteins stay in place.

Which result would you expect?

A. Nothing changes, because all the enzymes for the reactions are still present.
The proteins are present, but everything is now mixed into one cytosol at one acidity and one set of concentrations.
B. Every reaction speeds up, because reactants no longer have to cross membranes.
Reactants being together does not help when each reaction has lost the conditions it needs and interfering reactions are mixed together.
C. ✓ Reactions that needed acid slow down, and reactions once kept apart now happen in the same fluid.
D. Only the mitochondria's reactions stop; every other reaction continues as before.
Mitochondria are not the only compartments.
Lysosomes, the ER, the Golgi complex and the nucleus all lose their separate conditions too.

Why: Internal membranes let compartments hold their own conditions.
Internal membranes also keep interfering reactions apart.
Dissolve them, and every compartment’s contents mix into one cytosol at one acidity.
So reactions that needed acid or a concentrated reactant slow down, and the digestive enzymes reach the proteins they should never touch.

Q16 T29-q16

Four kinds of cell are compared.

Which of the following cells shows compartmentalization?

A. A bacterium with its DNA in a nucleoid and ribosomes spread through its cytosol.
A nucleoid is a region, not a membrane-bound compartment.
B. A mature red blood cell with no nucleus and no organelles.
A cell with no organelles has no membrane-bound compartments inside it.
C. An archaeon, a prokaryote, with its DNA in a nucleoid and no membrane-bound organelles.
An archaeon is a prokaryote: its nucleoid is a region with no membrane around it, and it has no membrane-bound compartments.
D. ✓ A yeast cell with a nucleus, mitochondria and a vacuole, each wrapped in membrane.

Why: Compartmentalization is the dividing of a cell’s interior into membrane-bound compartments.
Each compartment can hold its own conditions.
The yeast cell’s nucleus, mitochondria and vacuole are membrane-bound compartments.
So the yeast cell shows compartmentalization.

FRQ 1 T29-frq1 · Conceptual Analysis

Researchers build synthetic cells to test a two-step process. Enzyme E1 turns substance S into an intermediate, I, and works only where calcium ions are concentrated. Enzyme E2 turns I into the final product, P, and is shut down by concentrated calcium ions. Every synthetic cell holds the same amounts of S, E1 and E2 in the same total volume, and P is measured after 20 minutes. The figure shows three designs. Design 1: two chambers separated by an internal membrane that I can cross but calcium ions cannot, E1 in a calcium-rich chamber and E2 in a chamber with almost none: 96 units of P. Design 2: one mixed chamber at a middling calcium ion concentration: 31 units. Design 3: two chambers separated by the same kind of membrane, both at the middling concentration: 34 units.

Three synthetic-cell designs and the amount of P each made in 20 minutes. Shading marks the calcium ion concentration in each chamber; the dashed line is an internal membrane that I can cross but calcium ions cannot.

(a) Describe what a membrane around a compartment does for the reactions inside it. (1 pt)

Model answer The membrane around a compartment lets the inside hold conditions different from those outside it.
Those conditions include acidity and the concentration of a substance, such as calcium ions.
This matters for the reactions inside because a reaction that needs those conditions can then happen inside the compartment.

Rubric

Award 1 point for: the membrane lets the inside of the compartment hold conditions (such as acidity, or the concentration of a substance) different from those outside it, so a reaction that needs those conditions can happen there.
Accept: 'it keeps the inside acidic while the outside is not', or any equivalent stated in terms of conditions held different across the membrane.
Do not award: 'the membrane holds the contents in place' or 'protects the contents' with no mention of conditions differing.

Slip Saying the membrane holds the contents in place or protects them. The point is that conditions inside can differ from conditions outside.

(b) Explain why design 1 produced far more P than design 2. (1 pt)

Model answer In design 1 the membrane keeps calcium ions concentrated in E1’s chamber and almost absent from E2’s chamber, so each enzyme works in the conditions it needs.
I passes through the membrane from E1’s chamber to E2’s, so E2 receives the I it needs.
In design 2 both enzymes sit at a middling calcium ion concentration, which suits neither: E1 works slowly and E2 is partly shut down.
So both steps are slow, and far less P forms.

Rubric

Award 1 point for: in design 1 each enzyme works in the conditions it works in (E1 where calcium ions are concentrated, E2 where there are almost none) while I passes through the membrane from one chamber to the other; in design 2 both enzymes sit at a middling calcium ion concentration that suits neither, so both steps are slow.
Accept: an answer that names both elements, the conditions kept different by the membrane and I able to cross it.

Slip Naming the separate conditions without saying that I can cross, or the other way around. Both statements are needed.

(c) Predict the yield of P from a design 4: two chambers, calcium-rich and almost calcium-free as in design 1, but separated by an internal membrane that blocks I. (1 pt)

Model answer A very low yield of P, close to zero.
That is lower than in any of the three designs tested.

Rubric

Award 1 point for: the prediction of a very low yield, lower than in any of the three designs tested (close to zero P). No reasoning is required for this point.
Accept: 'almost no P', 'much less than 31 units'. Do not award a prediction equal to or greater than design 1's 96 units.

Slip Predicting a yield near design 1’s 96 units because the calcium ion concentrations are right. Right conditions do nothing for E2 if I never reaches it.

(d) Justify your prediction, using the results of designs 1 and 3. (1 pt)

Model answer E2 makes P only from I.
In design 4 the membrane blocks I, so the I that E1 makes stays in the calcium-rich chamber.
So E2 receives none, and little or no P forms, however well each chamber suits its enzyme.
Design 1 gave 96 units only because I could pass from E1’s chamber to E2’s.
Design 3 shows that passage alone, with the wrong conditions, still gave 34 units.
Therefore a compartment helps only if what the next step needs can still get through.

Rubric

Award 1 point for: E2 can make P only from I, so with I blocked in E1's chamber E2 receives none and little or no P forms, however good the conditions in each chamber; design 1's high yield depended on I passing between the chambers.
Accept: reasoning that also uses design 3 (I could cross but conditions were the same: 34 units, close to design 2's 31) to show that both separated conditions and passage of I are needed, and that design 4 has the first without the second. Support that rests on the logic (E2 needs I; I is blocked) with design 1 cited for the passage earns the point.

Slip Justifying from the calcium ion concentrations alone. Right conditions do nothing for E2 if I never reaches it; design 1 worked because I could cross, and design 4 removes the crossing.

FRQ 2 T29-frq2 · Analyze Model or Visual Representation

The model shows part of a eukaryotic cell. The lysosome, drawn shaded, holds digestive enzymes. The fluid inside the lysosome is acidic: its pH is 4.8. The cytosol is not acidic: its pH is 7.2. The lower the pH, the more acidic the fluid. The mitochondrion is drawn with its outer membrane and its inner membrane; the reactions that make ATP happen on the inner membrane. The dots in the cytosol are ribosomes, building new proteins.

Part of a eukaryotic cell: a lysosome (shaded), a mitochondrion with its two membranes, and ribosomes in the cytosol.

(a) Explain what keeps the fluid inside the lysosome at pH 4.8 while the cytosol around it stays at pH 7.2. (1 pt)

Model answer A pump in the lysosome’s membrane moves hydrogen ions into the lysosome.
The pump uses ATP to do this.
Hydrogen ions make the fluid acidic, so the inside is at pH 4.8.
The lysosome’s membrane is a barrier the hydrogen ions cannot cross freely.
So the hydrogen ions stay inside the lysosome and do not mix into the cytosol.
So the inside stays at pH 4.8 while the cytosol stays at pH 7.2.

Rubric

Award 1 point for: a pump in the lysosome’s membrane uses ATP to move hydrogen ions into the lysosome, against their concentration gradient, and those hydrogen ions are what keep the inside acidic (pH 4.8).
Accept with or without: the membrane as the barrier the hydrogen ions cannot cross freely, so they do not mix into the cytosol and the cytosol stays at pH 7.2.
Accept ‘a proton pump’ or ‘an H⁺ pump’ for the pump; accept ‘acid’ or ‘H⁺’ for hydrogen ions.
Do not award the point for the membrane barrier alone, with no pump moving hydrogen ions in: a barrier keeps a difference, it does not make one.

Slip Saying the membrane alone keeps the inside acidic. The membrane keeps the hydrogen ions in, but it is the pump, using ATP, that moves them in and makes the inside acidic.

(b) Explain how the folding of the mitochondrion’s inner membrane benefits the cell. (1 pt)

Model answer The ATP-making reactions happen on the inner membrane.
Folding fits more membrane into the same space.
So the mitochondrion has more surface for the ATP-making reactions.
More of those reactions happen at the same time, and ATP is made faster.

Rubric

Award 1 point for: folding fits more membrane into the same space, giving more surface for the ATP-making reactions, so more of them happen at the same time and ATP is made faster.
Accept: 'more surface area for the reactions' with the link to more reactions at once or to faster ATP making.

Slip Saying the folds store energy or let fuel in. They add surface for the reactions that make ATP.

(c) Predict what happens to the pH inside the lysosome if the cell has no ATP left, and justify your prediction. (1 pt)

Model answer The pH inside the lysosome rises toward the cytosol’s pH 7.2.
Hydrogen ions are more concentrated inside the lysosome than in the cytosol.
So hydrogen ions leak out of the lysosome, down their concentration gradient.
The pump that moves hydrogen ions back in uses ATP, so with no ATP the pump stops.
So nothing replaces the hydrogen ions that leak out.
The fluid inside has fewer hydrogen ions, so it is less acidic: the pH rises toward 7.2.

Rubric

Award 1 point for: the prediction that the pH inside the lysosome rises toward the cytosol’s pH 7.2, AND the reason: hydrogen ions leak out of the lysosome down their concentration gradient, and with no ATP nothing moves hydrogen ions back in.
Accept for the prediction: ‘the inside becomes less acidic’ or ‘the pH difference fades’. Accept for the reason: ‘the hydrogen ions diffuse out and are not replaced’.
Do not award the point for the prediction alone, or for ‘the pump stops’ with no account of where the hydrogen ions go.

Slip Predicting that the pH stays at 4.8 because the membrane holds the acid in. The membrane slows the leak but does not stop it; with no pump working, nothing replaces the hydrogen ions that leak out.

(d) Support the claim that internal membranes let many different processes happen at the same time in one cell, using the lysosome, the mitochondrion and the ribosomes in the model. (1 pt)

Model answer Internal membranes divide the cell into compartments.
The lysosome’s membrane holds acidic conditions inside the lysosome.
The lysosome’s membrane also keeps the digestive enzymes away from the ribosomes and the new proteins they build.
The mitochondrion’s folded inner membrane adds surface for the ATP-making reactions.
So digestion, protein-building and ATP-making all happen at the same time in one cell.

Rubric

Award 1 point for: the evidence AND the reasoning: the lysosome’s membrane holds its own conditions (acid) and keeps the digestive enzymes away from the ribosomes’ new proteins; the mitochondrion’s folded inner membrane adds surface for the ATP-making reactions; so digestion, protein-building and ATP-making happen at the same time in one cell.
Accept: any two of the three benefits (own conditions; interfering processes kept apart; more surface), each tied to a feature in the model, with the conclusion that many processes happen at once in one cell.

Slip Listing the organelles without saying what each membrane does. Tie at least two benefits to features in the model, then draw the conclusion.

APBIO-U02-L21 A cell inside a cell

Topic 2.10 · Origins of Cell Compartmentalization · 45 steps

A mitochondrion and a bacterium drawn side by side at the same scale: the mitochondrion an oval with a second membrane just inside its outer one and rounded folds reaching in; the bacterium a rod with a wall and a loop of DNA; a one-micrometer scale bar between them

Here is a mitochondrion. And here is a bacterium. They look pretty similar.

Biologists believe that is no accident. Mitochondria evolved from bacteria, over many thousands of generations and millions of years. Here is that story: how a bacterium ended up inside a cell, and why a mitochondrion today is not a bacterium.

Unit 2 · Cell Structure and Function

1A mitochondrion and a bacterium

Video: Watch first: a part of your cell that looks like a bacterium

Here is a mitochondrion, and a bacterium. They look alike. Biologists believe that is because mitochondria evolved from bacteria.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T210-intro.mp4

Here is a mitochondrion, photographed through an electron microscope. And here is a bacterium.

Two mitochondria under the electron microscope: oval bodies with a second membrane inside the outer one and many folds reaching in — Two electron-microscope photographs side by side: on the left two mitochondria, oval, each with a second membrane inside the outer one and many folds reaching in; on the right a bacterium cut lengthwise, a rounded shape with a thick boundary and a paler patch inside

A bacterium cut lengthwise under the electron microscope: a rounded shape with a thick boundary and a paler patch inside where its DNA lies — Two electron-microscope photographs side by side: on the left two mitochondria, oval, each with a second membrane inside the outer one and many folds reaching in; on the right a bacterium cut lengthwise, a rounded shape with a thick boundary and a paler patch inside

They look pretty similar. The mitochondrion is oval, and so is the bacterium.

A mitochondrion and a bacterium drawn side by side at the same scale beside a one-micrometer scale bar; labels with leader lines name the mitochondrion's outer membrane and folded inner membrane, and the bacterium's thick cell wall and its ribosomes

The mitochondrion is one to two micrometers long, and so is the bacterium.

Biologists believe that is because mitochondria evolved from bacteria.

Over many thousands of generations, and millions of years, bacteria became mitochondria.

That is the claim. Next, we look at how biologists think it happened.

What you are expected to know State the claim: mitochondria evolved from bacteria.

What you are expected to know Describe how a mitochondrion and a bacterium look alike: the same shape, and about the same size.

Check q1

What do biologists believe about where mitochondria came from?

A. ✓ Mitochondria evolved from bacteria
B. Mitochondria formed from folds of the cell’s own plasma membrane
A fold of the cell’s own membrane would look like the cell, not like a bacterium.
C. Bacteria evolved from mitochondria that escaped from cells
Bacteria lived free long before there were any cells with mitochondria.

Why: A mitochondrion looks like a bacterium: the same shape and about the same size.
Biologists believe that is because mitochondria evolved from bacteria.

Check q2

A mitochondrion and a bacterium are drawn side by side at the same scale.

How do the two look alike?

A. A cell wall around each of them
A bacterium has a cell wall.
A mitochondrion has none.
B. A nucleus inside each of them
A mitochondrion has two membranes; a bacterium has one membrane inside its cell wall.
C. ✓ The same shape, and about the same size

Why: The mitochondrion is oval, and so is the bacterium.
The mitochondrion is one to two micrometers long, and so is the bacterium.
So the two share their shape and their size.

13How it happened

Video: Watch: How it happened

A large cell took a bacterium inside and kept it. The bacterium lived and reproduced there, generation after generation. One organism living inside another: endosymbiosis.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L21.mp4

Long ago, a large cell took a bacterium inside itself.

Three panels: a large cell beside a small bacterium; the bacterium inside the large cell; generations later, the large cell has divided and each new cell carries some of the bacterium's descendants

The large cell did not digest the bacterium. The large cell kept it.

The bacterium lived on inside the large cell. And the bacterium reproduced there.

When the large cell divided, each new cell carried some of the bacteria. The bacteria were passed on like this for many thousands of generations.

Over those generations, the descendants of the bacterium became mitochondria.

One organism living inside another is called .

The name says what it means. Endo means inside.

Symbiosis means living together. So endosymbiosis means living together, one inside the other.

What you are expected to know Describe how biologists think it began: a large cell took a bacterium inside and kept it.

What you are expected to know Describe what the bacterium did inside the large cell: it lived and reproduced there for many generations.

What you are expected to know State what the bacterium’s descendants became: mitochondria.

What you are expected to know State what endosymbiosis is: one organism living inside another.

Check q3

What is endosymbiosis?

A. ✓ One organism living inside another
B. One cell digesting another
The large cell did not digest the bacterium; it kept the bacterium alive inside.
C. Two organisms living side by side
Endo means inside: one lives inside the other, not beside it.

Why: Endo means inside.
Symbiosis means living together.
So endosymbiosis is one organism living inside another.

Practice writing an answer

Long ago, a large cell took a bacterium inside itself and kept it alive there.

(a) State what endosymbiosis means. (1 pt)

Model answer Endosymbiosis means one organism living inside another.

Rubric

Award 1 point for: one organism living inside another (living together, one inside the other).

Slip Saying ‘two organisms living together’ with no ‘inside’. Endo means inside.

Check q4

A large cell took a bacterium inside itself.

According to the endosymbiosis account, what did the large cell do with the bacterium?

A. The large cell digested the bacterium
A digested bacterium leaves no descendants.
The bacterium’s descendants became mitochondria.
B. ✓ The large cell kept the bacterium alive inside itself
C. The large cell pushed the bacterium back out
The bacterium stayed inside and reproduced there.

Why: The large cell did not digest the bacterium.
The large cell kept the bacterium alive inside itself.
The bacterium lived and reproduced there for many generations.

Check q5

In the endosymbiosis account, which event came first?

A. The bacterium’s descendants became mitochondria
The descendants became mitochondria much later, after many generations inside the cell.
B. The bacterium reproduced inside the large cell
The bacterium could not reproduce inside the cell before the cell took it inside.
C. ✓ The large cell took the bacterium inside

Why: First, the large cell took the bacterium inside.
Then the bacterium lived and reproduced there, for many generations.
Over those generations its descendants became mitochondria.

31Not a bacterium any more

The bacteria that moved in changed over the generations.

The bacteria came to depend on the cell around them. They lost the ability to live on their own.

A mitochondrion today cannot live outside its cell. A mitochondrion is an organelle, not a bacterium.

So mitochondria descend from bacteria. But mitochondria are not bacteria.

What you are expected to know Explain why a mitochondrion today is not a bacterium: it descends from one, but over the generations it became an organelle that cannot live on its own.

Check q6

A student says: “Mitochondria are bacteria living inside our cells.”

Is the student correct that mitochondria are bacteria?

A. Yes
Mitochondria descend from bacteria, but a mitochondrion today cannot live on its own; it is an organelle.
B. ✓ No

Why: Mitochondria descend from bacteria.
Over the generations, the bacteria that moved in became organelles.
A mitochondrion today cannot live on its own.
So a mitochondrion is not a bacterium.

Check q7

A student says: “Mitochondria are bacteria living inside our cells.” The student is wrong.

Why is the student wrong?

A. Mitochondria are folds of the cell’s own membrane, so no bacterium was ever inside
A large cell took a bacterium inside and kept it.
The bacterium’s descendants became mitochondria.
B. ✓ Mitochondria became organelles that cannot live on their own
C. Mitochondria are bacteria the cell took in recently and will soon digest
The bacterium was taken inside long ago, not recently, and its descendants were never digested.

Why: The bacterium that moved in was a free-living cell.
Its descendants reproduced inside the cell, generation after generation.
Over those generations they became organelles that depend on the cell.
So a mitochondrion today cannot live on its own.
So a mitochondrion is not a bacterium, although it descends from one.

Practice writing an answer

A student says: “Mitochondria are bacteria living inside our cells.”

(a) Explain why the student is wrong, and state what a mitochondrion is instead. (1 pt)

Model answer Mitochondria descend from bacteria.
Over many generations inside the cell, the bacteria changed.
They lost the ability to live on their own.
So a mitochondrion today cannot live outside its cell.
A mitochondrion is an organelle, not a bacterium.

Rubric

Award 1 point for: mitochondria descend from bacteria but became organelles that depend on the cell and cannot live on their own, so a mitochondrion is not a bacterium.

Slip Saying the student is wrong because mitochondria have nothing to do with bacteria. They descend from bacteria; the error is ‘are bacteria’.

40Summary

A mitochondrion looks like a bacterium: the same shape, and about the same size. Biologists believe that is because mitochondria evolved from bacteria.

Long ago, a large cell took a bacterium inside and kept it. The bacterium lived and reproduced there for many thousands of generations.

One organism living inside another is called endosymbiosis.

The bacterium’s descendants became mitochondria: organelles that depend on the cell and cannot live on their own.

Glossary

endosymbiosis: One organism living inside another. Endo means inside; symbiosis means living together. Mitochondria descend from a bacterium that a large cell took inside and kept, and that lived and reproduced there for many generations.

APBIO-U02-L21B Five pieces of evidence that mitochondria came from bacteria

Topic 2.10 · Origins of Cell Compartmentalization · 72 steps

Five small panels in a row, numbered 1 to 5, each a small drawing: a loop of DNA, a cluster of dots, two ovals one inside the other, an oval pinching in two, and a mitochondrion beside a bacterium

Biologists claim that mitochondria evolved from bacteria. So, why do we believe it?

Well, let us examine the evidence. There are five pieces. We take them one at a time, then put them together and ask which facts count as evidence at all.

Unit 2 · Cell Structure and Function

1Evidence 1: its own DNA, in one loop

Video: Watch: The five pieces of evidence

Its own DNA in one loop, its own ribosomes, two membranes, splitting in two, a bacterium’s size: five pieces of evidence that mitochondria came from bacteria.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L21b.mp4

Here is the mitochondrion again, beside the bacterium.

A mitochondrion has its own DNA, separate from the DNA in the nucleus.

Three drawings: a mitochondrion with one closed loop of DNA inside it; a bacterium with one closed loop of DNA; a nucleus with long strands of DNA that have two ends

The mitochondrion’s DNA is one closed loop. A bacterium’s DNA is also one closed loop.

The DNA in the nucleus is different. The nucleus’s DNA is in long strands, each with two ends.

So the mitochondrion’s DNA looks like a bacterium’s DNA. It does not look like the nucleus’s DNA.

Evidence 1. A mitochondrion has its own DNA, in one closed loop, just like a bacterium’s.

What you are expected to know State evidence 1: a mitochondrion has its own DNA, in one closed loop, just like a bacterium’s and unlike the nucleus’s long strands.

Check q1

What form does a mitochondrion’s own DNA take?

A. ✓ One closed loop
B. Long strands with two ends
Long strands with two ends are the form of the DNA in the nucleus.
C. A mitochondrion has no DNA of its own
A mitochondrion does have its own DNA, separate from the nucleus’s.

Why: A mitochondrion has its own DNA.
That DNA is one closed loop.

Check q2

A mitochondrion has its own DNA.

Whose DNA does the mitochondrion’s DNA look like?

A. The DNA in the cell’s nucleus
The nucleus’s DNA is in long strands with two ends.
B. ✓ A bacterium’s DNA
C. Neither: no other DNA is a closed loop
A bacterium’s DNA is one closed loop too.

Why: A bacterium’s DNA is one closed loop.
The mitochondrion’s DNA is one closed loop.
The nucleus’s DNA is in long strands with two ends.
So the mitochondrion’s DNA looks like a bacterium’s.

12Evidence 2: its own ribosomes

A mitochondrion has its own ribosomes. Ribosomes build proteins.

A mitochondrion and a bacterium, each with small ribosome dots drawn inside it

Some antibiotics kill bacteria by stopping the bacteria’s ribosomes. Such an antibiotic stops only ribosomes built like a bacterium’s.

Researchers gave one such antibiotic to a bacterium, and to a eukaryotic cell with mitochondria.

the ribosomes	with the antibiotic, do they keep building proteins?
in a bacterium	no: they stop
inside a mitochondrion	no: they stop
in the cell’s cytosol	yes: they keep going

an antibiotic that stops a bacterium’s ribosomes

A table of the antibiotic experiment: the bacterium's ribosomes stop building proteins; the mitochondrion's ribosomes stop; the ribosomes in the cytosol keep building proteins

The antibiotic stopped the bacterium’s ribosomes. The antibiotic also stopped the mitochondria’s ribosomes.

But the antibiotic did not stop the ribosomes in the cell’s cytosol. Those ribosomes kept building proteins.

So the mitochondrion’s ribosomes are built like a bacterium’s ribosomes. They are not built like the cell’s own ribosomes.

Evidence 2. A mitochondrion has its own ribosomes, built like a bacterium’s.

What you are expected to know Explain evidence 2: an antibiotic that stops a bacterium’s ribosomes stops the mitochondrion’s ribosomes but not the cytosol’s, so the mitochondrion’s ribosomes are built like a bacterium’s.

Check q3

An antibiotic that stops a bacterium’s ribosomes also stops the ribosomes inside a cell’s mitochondria. The ribosomes in the cell’s cytosol keep building proteins.

What does this result show about the mitochondrion’s ribosomes?

A. ✓ They are built like a bacterium’s ribosomes
B. They are built like the cytosol’s ribosomes
The antibiotic stopped the mitochondrion’s ribosomes and left the cytosol’s working, so the two are built differently.
C. They are not real ribosomes
The mitochondrion’s ribosomes build the mitochondrion’s proteins; the antibiotic stopped that.

Why: The antibiotic stops ribosomes built like a bacterium’s.
The antibiotic stopped the mitochondrion’s ribosomes.
The antibiotic left the cytosol’s ribosomes working.
So the mitochondrion’s ribosomes are built like a bacterium’s, not like the cytosol’s.

Check q4

The claim: mitochondria evolved from bacteria. A mitochondrion’s ribosomes are built like a bacterium’s ribosomes.

Why does this support the claim?

A. ✓ A descendant of a bacterium would keep the bacterium’s kind of ribosomes
B. Ribosomes build proteins, and bacteria are made of proteins
Every cell’s ribosomes build proteins.
The evidence is the build of the mitochondrion’s ribosomes, which matches a bacterium’s.
C. Only bacteria and mitochondria have ribosomes of any kind
Every cell has ribosomes; the cell’s cytosol has ribosomes too, of a different build.

Why: A bacterium has ribosomes of one particular build.
A descendant of a bacterium would keep that build.
The mitochondrion’s ribosomes have a bacterium’s build.
So the ribosomes support the claim that mitochondria evolved from bacteria.

23Evidence 3: two membranes

A mitochondrion has two membranes: an outer membrane and an inner membrane.

A mitochondrion cut open, drawn large: an outer membrane, an inner membrane just inside it, and the inner membrane's rounded folds reaching into the middle; labels with leader lines name the outer membrane, the inner membrane and the folds

Now think about how a cell takes something in. The cell’s membrane wraps around the thing it takes in.

Three panels: a bacterium near the edge of a large cell; the cell's membrane folding around the bacterium; the bacterium inside the cell with two membranes around it

So a bacterium taken inside a cell could end up with two membranes around it.

A mitochondrion has two membranes. That is what we would expect if a bacterium had been taken inside a cell.

Two membranes are what we would expect if the claim is true. But two membranes alone do not prove it. Something else could have two membranes too.

Evidence 3. A mitochondrion has two membranes. That is what we would expect if a bacterium had been taken inside a cell.

What you are expected to know State evidence 3: a mitochondrion has two membranes, which is what we would expect if a bacterium had been taken inside a cell, and which does not prove it.

Check q5

A cell takes a bacterium inside by wrapping its membrane around it.

How many membranes would we expect around the bacterium once it is inside?

A. One
The bacterium keeps its own membrane, and the cell’s membrane wraps around that.
B. ✓ Two
C. None
The bacterium has a membrane of its own before it is taken in.

Why: The bacterium has its own membrane.
The cell’s membrane wraps around the bacterium.
So we would expect two membranes around it.

Check q6

The claim: mitochondria evolved from bacteria. A student says: “A mitochondrion has two membranes. That proves the claim.”

Is the student correct that two membranes prove the claim?

A. Yes
Two membranes are what we would expect.
But something else could have two membranes too.
B. ✓ No

Why: Two membranes are what we would expect if a bacterium had been taken inside a cell.
So two membranes are what the claim predicts.
But something else could have two membranes too.
So two membranes alone do not prove the claim.

33Evidence 4: splitting in two

How does a cell get new mitochondria? A new mitochondrion forms only when an existing one splits in two.

A mitochondrion lengthening, pinching in the middle, and separating into two

The mitochondrion lengthens. It pinches in the middle. It separates into two.

That is just like how a bacterium reproduces. A bacterium splits in two.

Evidence 4. New mitochondria form only when an existing one splits in two, just like bacteria do.

What you are expected to know State evidence 4: new mitochondria form only when an existing one splits in two, just like bacteria do.

Check q7

How does a bacterium reproduce?

A. ✓ It splits in two
B. It buds a new bacterium off the Golgi complex
A bacterium has no Golgi complex.
C. It is built by the cell around it
A bacterium lives free; no cell builds it.

Why: A bacterium lengthens and pinches in the middle.
It separates into two bacteria.
So a bacterium reproduces by splitting in two.

Check q8

The claim: mitochondria evolved from bacteria. A time-lapse film shows the mitochondria inside a cell lengthening, pinching in the middle and separating into two, while the cell itself stays a single cell.

Does the film support the claim that mitochondria evolved from bacteria?

A. ✓ Yes
B. No
Bacteria reproduce by splitting in two, and the mitochondria in the film split the same way.

Why: A bacterium reproduces by splitting in two.
The mitochondria in the film split in two the same way.
A descendant of a bacterium would keep that way of reproducing.
So the film supports the claim.

41Evidence 5: a bacterium’s size

The fifth piece of evidence is the size. A mitochondrion is about one to two micrometers long. So is a bacterium.

Evidence 5. A mitochondrion is about the size of a bacterium.

What you are expected to know State evidence 5: a mitochondrion is about the size of a bacterium.

Check q9

About how big is a mitochondrion?

A. About the size of a ribosome
A ribosome is far smaller: a mitochondrion holds many ribosomes.
B. About the size of a cell’s nucleus
A nucleus is about five times wider than a mitochondrion.
C. ✓ About the size of a bacterium

Why: A mitochondrion is about one to two micrometers long.
A bacterium is about one to two micrometers long too.
So a mitochondrion is about the size of a bacterium.

46The five together

Here are the five pieces of evidence that mitochondria came from bacteria:

Its own DNA, in one closed loop, just like a bacterium’s.
Its own ribosomes, built like a bacterium’s.
Two membranes, which is what we would expect if a bacterium had been taken inside a cell.
New mitochondria form only when an existing one splits in two, just like bacteria do.
A mitochondrion is about the size of a bacterium.

Summary: five small panels, numbered 1 to 5, one for each piece of evidence: its own DNA in one loop, its own ribosomes, two membranes, splits in two, the size of a bacterium

No one piece proves the claim on its own. Together, the five pieces are hard to explain any other way.

Not every fact about a mitochondrion is evidence for the claim. To sort a fact, ask what the fact is about.

For example: a mitochondrion has its own ribosomes. A bacterium has ribosomes too. So this fact supports the claim.

But: a muscle cell has many mitochondria. That fact is about how many there are. So this fact does not tell us where mitochondria came from.

And: a mitochondrion makes ATP. That fact is about what a mitochondrion does today. So this fact does not tell us where mitochondria came from.

So: a fact that names a feature a bacterium also has supports the claim.

Some facts are about what a mitochondrion does today, where it sits, or how many there are. Those facts do not tell us where mitochondria came from.

What you are expected to know Identify the five pieces of evidence that mitochondria came from bacteria: its own DNA in one loop, its own ribosomes, two membranes, splitting in two, a bacterium’s size.

What you are expected to know Sort a fact about mitochondria: a fact that names a feature a bacterium also has supports the claim.

What you are expected to know Sort a fact about mitochondria: a fact about what they do today, where they sit or how many there are does not tell us where they came from.

58Fluency quiz: does the fact support the claim? mixed practice

Check q10

The claim: mitochondria evolved from bacteria. The fact: a heart muscle cell has many more mitochondria than a skin cell.

Does this fact support the claim that mitochondria evolved from bacteria?

A. Supports the claim
How many mitochondria a cell has tells us what the cell uses them for.
It does not tell us where they came from.
B. ✓ Not about where mitochondria came from

Why: Ask what the fact is about.
How many mitochondria a cell has tells us what the cell uses them for.
A count like this does not tell us where mitochondria came from.

Check q11

The claim: mitochondria evolved from bacteria. The fact: mitochondria have their own ribosomes, built like a bacterium’s.

Does this fact support the claim that mitochondria evolved from bacteria?

A. ✓ Supports the claim
B. Not about where mitochondria came from
A bacterium has ribosomes of that build, and a descendant of a bacterium would keep them.

Why: Ask what the fact is about.
The fact names a feature of the mitochondrion: ribosomes built like a bacterium’s.
A bacterium has ribosomes of that build.
So the fact supports the claim.

Check q12

The claim: mitochondria evolved from bacteria. The fact: a mitochondrion is about the size of a bacterium.

Does this fact support the claim that mitochondria evolved from bacteria?

A. ✓ Supports the claim
B. Not about where mitochondria came from
A bacterium has that size, and a descendant of a bacterium would keep it.

Why: Ask what the fact is about.
The fact names a feature of the mitochondrion: its size.
A bacterium has that size.
So the fact supports the claim.

Check q13

The claim: mitochondria evolved from bacteria. The fact: the mitochondria in a leaf cell sit next to the chloroplasts.

Does this fact support the claim that mitochondria evolved from bacteria?

A. Supports the claim
Where an organelle sits in the cell does not tell us where the organelle came from.
B. ✓ Not about where mitochondria came from

Why: Ask what the fact is about.
The fact says where the mitochondria sit in the cell.
Where an organelle sits does not tell us where the organelle came from.

Check q14

The claim: mitochondria evolved from bacteria. The fact: a mitochondrion is wrapped in two membranes.

Does this fact support the claim that mitochondria evolved from bacteria?

A. ✓ Supports the claim
B. Not about where mitochondria came from
Two membranes are what we would expect if a bacterium had been taken inside a cell.

Why: Ask what the fact is about.
The fact names a feature of the mitochondrion: two membranes.
Two membranes are what we would expect if a bacterium had been taken inside a cell.
So the fact supports the claim.

Check q15

The claim: mitochondria evolved from bacteria. The fact: mitochondria make the ATP that the rest of the cell uses.

Does this fact support the claim that mitochondria evolved from bacteria?

A. Supports the claim
What an organelle does for its cell today describes the partnership now, not where the organelle came from.
B. ✓ Not about where mitochondria came from

Why: Ask what the fact is about.
The fact says what mitochondria do for the cell today: make ATP.
What an organelle does today does not tell us where the organelle came from.

Check q16

The claim: mitochondria evolved from bacteria. Watching a time-lapse film of mitochondria splitting in two, a student says: “This proves it: each mitochondrion is still a free-living bacterium inside the cell.”

Is the student correct that each mitochondrion is still a free-living bacterium?

A. Yes
Splitting in two is evidence that mitochondria came from bacteria; a mitochondrion today cannot live outside the cell.
B. ✓ No

Why: Bacteria split in two, and so do mitochondria.
So the film is evidence that mitochondria came from bacteria.
But a mitochondrion today cannot live outside the cell.
So a mitochondrion is not a free-living bacterium.

Practice writing an answer

A time-lapse film shows the mitochondria in a cell lengthening, pinching in the middle and splitting in two, while the cell itself does not divide. A student says: “This proves that each mitochondrion is still a free-living bacterium.”

(a) Explain why the student is wrong, and state what the film is actually evidence for. (1 pt)

Model answer A mitochondrion today cannot live outside the cell.
So a mitochondrion is not a free-living bacterium.
One piece of evidence proves nothing on its own.
Bacteria reproduce by splitting in two.
The mitochondria in the film split in two the same way.
So the film is evidence that mitochondria came from bacteria.

Rubric

Award 1 point for: a mitochondrion today cannot live on its own, so it is not a free-living bacterium (and one observation does not prove anything), AND the film is evidence that mitochondria came from bacteria, because bacteria also reproduce by splitting in two.

Slip Saying the student is wrong because mitochondria do not divide like bacteria. They do, and that likeness is the evidence. The student’s errors are ‘proves’ and ‘still a bacterium’.

Practice writing an answer

A student claims that the mitochondria of a yeast cell came from bacteria, and lists five observations:
1. An antibiotic that stops a bacterium’s ribosomes stops the mitochondria from making their own proteins, and leaves protein-making in the cytosol untouched.
2. The mitochondria lie close to the cell’s surface.
3. The mitochondria have their own DNA, in one closed loop.
4. The mitochondria move about the cell along protein tracks in the cytosol.
5. New mitochondria appear only when an existing one lengthens and splits in two.

(a) Identify the observations that support the claim. (1 pt)

Model answer Observations 1, 3 and 5 support the claim.
Observations 2 and 4 are not evidence about where the mitochondria came from.

Rubric

Award 1 point for: 1, 3 and 5 support the claim; 2 and 4 do not tell us where the mitochondria came from; all five sorted correctly.
The sorting alone earns the point; no reasons are required.

Slip Counting 2 or 4 as support because they are about mitochondria. Where an organelle sits and how it moves describe its life in the cell now, and do not tell us where it came from.

(b) Explain why observation 1 supports the claim. (1 pt)

Model answer Ribosomes make proteins.
The antibiotic stops a bacterium’s ribosomes.
The antibiotic stopped the mitochondria’s protein-making.
So the mitochondria’s ribosomes are built like a bacterium’s.
The antibiotic left the cytosol’s protein-making alone.
So the yeast’s own ribosomes are built differently.
A descendant of a bacterium would keep the bacterium’s kind of ribosomes.
So observation 1 supports the claim.

Rubric

Award 1 point for: the antibiotic acts on ribosomes of a bacterium’s build; it stopped the mitochondria’s ribosomes but not the cytosol’s, so the mitochondria’s ribosomes are built like a bacterium’s, which a descendant of a bacterium would have.

Slip Saying the antibiotic shows the mitochondria are bacteria. It shows their ribosomes are built like a bacterium’s: a trace of their ancestry.

68Summary

Five pieces of evidence say that mitochondria came from bacteria: its own DNA in one loop, its own ribosomes, two membranes, splitting in two, and a bacterium’s size.

No one piece proves the claim. Together they are hard to explain any other way.

A fact supports the claim if it names a feature a bacterium also has. A fact about what a mitochondrion does today, where it sits or how many there are does not tell us where mitochondria came from.

APBIO-U02-L21C The chloroplast’s story

Topic 2.10 · Origins of Cell Compartmentalization · 53 steps

A chloroplast drawn as an oval with stacks of flat discs inside, beside a cyanobacterium drawn as a rod with stacks of flat membrane inside, with a scale bar

Mitochondria are not the only organelle that came from a bacterium. A plant cell’s chloroplasts have a story of their own.

Here is the story. First, the claim biologists make about chloroplasts. Then the five features that support it. Then one more piece of evidence: a DNA comparison, worked through base by base.

Unit 2 · Cell Structure and Function

1A second cell that moved in

Video: Watch: A second cell that moved in

Some bacteria photosynthesize: cyanobacteria. A chloroplast looks like a cyanobacterium, and biologists believe that is because chloroplasts evolved from cyanobacteria.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L21Ca.mp4

Some bacteria photosynthesize. These bacteria make glucose from carbon dioxide and water.

These bacteria use the energy of light to do it, just like a leaf cell does.

A bacterium that photosynthesizes is called a . Cyano means blue-green, the color of these bacteria.

Here is a chloroplast beside a cyanobacterium. They look similar.

A chloroplast and a cyanobacterium drawn side by side: the chloroplast an oval with two membranes and stacks of flat discs inside; the cyanobacterium a rod with a cell wall, a loop of DNA and flat membranes inside; labels with leader lines; a scale bar

The chloroplast is oval, and so is the cyanobacterium. The chloroplast is about two micrometers long, and so is the cyanobacterium. The chloroplast catches light on flat membranes, and so does the cyanobacterium.

Biologists believe that is because chloroplasts evolved from cyanobacteria.

Long ago, one cell took a cyanobacterium inside itself and kept it. That cell was the ancestor of plants and of algae, the green, plant-like life of ponds and seas.

Over many generations, the cyanobacterium’s descendants became chloroplasts.

So this is a second case of endosymbiosis: a second cell that moved in and stayed.

What you are expected to know State what a cyanobacterium is: a bacterium that photosynthesizes.

What you are expected to know State the claim for the chloroplast: chloroplasts evolved from cyanobacteria taken inside the ancestor of plants and algae.

Check q1

What is a cyanobacterium?

A. ✓ A bacterium that photosynthesizes
B. A bacterium that lives inside a plant cell today
Cyanobacteria live free, in ponds, seas and soil.
C. A chloroplast that has left its cell
A chloroplast cannot live outside its cell.

Why: A cyanobacterium makes glucose from carbon dioxide and water, using the energy of light.
So a cyanobacterium is a bacterium that photosynthesizes.

Practice writing an answer

Cyanobacteria grow in ponds and seas.

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

Model answer A cyanobacterium is a bacterium that photosynthesizes.

Rubric

Award 1 point for: a bacterium that photosynthesizes (makes glucose from carbon dioxide and water using light).

Slip Saying ‘a blue-green organism’ with no mention of photosynthesis. The photosynthesis is what matters for the chloroplast’s story.

Check q2

What do biologists believe about where chloroplasts came from?

A. Chloroplasts formed from folds of the plant cell’s own plasma membrane
A fold of the cell’s own membrane would look like the cell, not like a cyanobacterium.
B. Chloroplasts evolved from mitochondria that turned green in the light
Mitochondria and chloroplasts came from different bacteria; neither came from the other.
C. ✓ Chloroplasts evolved from cyanobacteria taken inside the ancestor of plants

Why: A chloroplast looks like a cyanobacterium.
Biologists believe that is because chloroplasts evolved from cyanobacteria.
The ancestor of plants and algae took a cyanobacterium inside and kept it.

17The chloroplast’s own five pieces of evidence

The chloroplast has the same five features as the mitochondrion.

feature	a cyanobacterium	a mitochondrion	a chloroplast
its own DNA, in one closed loop	yes	yes	yes
its own ribosomes, built like a bacterium’s	yes	yes	yes
two membranes	one membrane, inside its cell wall	yes	yes
new ones form only by splitting in two	yes	yes	yes
about a bacterium’s size	yes	yes	yes

the chloroplast shares the mitochondrion’s five features

A table of five features across a cyanobacterium, a mitochondrion and a chloroplast: its own DNA in one closed loop (yes, yes, yes); its own ribosomes, built like a bacterium's (yes, yes, yes); two membranes (one membrane inside its cell wall; yes; yes); new ones form by splitting in two (yes, yes, yes); about a bacterium's size (yes, yes, yes)

A chloroplast has the same five features as a mitochondrion:

Its own DNA, in one closed loop.
Its own ribosomes, built like a bacterium’s.
Two membranes, which is what we would expect if a cyanobacterium had been taken inside a cell.
New chloroplasts form only when an existing one splits in two, just like bacteria do.
A chloroplast is about a bacterium’s size.

So the same five pieces of evidence support the claim that chloroplasts came from cyanobacteria.

What you are expected to know Identify the five features a chloroplast shares with a mitochondrion: its own DNA in one loop, its own ribosomes, two membranes, splitting in two, a bacterium’s size.

Check q3

Which of the following is a feature a chloroplast shares with a cyanobacterium?

A. A nucleus with a membrane around it
Neither a chloroplast nor a cyanobacterium has a nucleus.
B. ✓ Its own DNA, in one closed loop
C. Long strands of DNA with two ends
Long strands with two ends are the form of the DNA in the plant cell’s nucleus.

Why: A cyanobacterium keeps its DNA as one closed loop.
A chloroplast has its own DNA, also in one closed loop.
So the closed loop of DNA is a feature the two share.

Check q4

How do new chloroplasts form?

A. The Golgi complex makes them
The Golgi complex makes vesicles and lysosomes, not chloroplasts.
B. The nucleus buds them off
The nucleus makes no organelles.
C. ✓ An existing chloroplast splits in two

Why: New chloroplasts form only when an existing one splits in two.
That is just like how bacteria reproduce.

24Why DNA can tell us who is related to whom

Check q5

When a cell divides, what does each new cell receive?

A. ✓ A copy of the parent cell’s DNA
B. Half of the parent cell’s DNA, uncopied
The parent cell copies its DNA first, so each new cell gets a full copy.
C. New DNA made from scratch
DNA is copied from the parent’s DNA, not made from scratch.

Why: A cell copies its DNA before it divides.
Each new cell receives a copy of the parent cell’s DNA.

A cell copies its DNA and passes the copy to its offspring.

Three rows of the same short stretch of DNA: a parent cell's, its offspring's with one base changed and circled, and a later descendant's with two bases changed and circled

The copy is almost exact. Now and then, one base changes.

Over many generations, those small changes build up.

So two cells whose ancestors separated long ago have many differences in their DNA. Two cells whose ancestors separated recently have few.

Turn that around. DNA with more matching bases came from a closer relative.

What you are expected to know Explain why DNA with more matching bases came from a closer relative: DNA is copied from parent to offspring, and small changes build up over the generations.

Check q6

Cell X’s DNA matches cell Y’s DNA at 19 of 20 bases. Cell X’s DNA matches cell Z’s DNA at 8 of 20 bases.

Which cell is the closer relative of cell X?

A. Cell Z
Cell Z’s DNA matches at only 8 of 20 bases.
B. ✓ Cell Y
C. The two are equally close relatives
19 matching bases against 8 is a large difference.

Why: DNA with more matching bases came from a closer relative.
Cell Y’s DNA matches cell X’s at 19 of 20 bases.
Cell Z’s matches at only 8.
So cell Y is the closer relative of cell X.

Check q7

Cell Y’s DNA matches cell X’s DNA at more bases than cell Z’s DNA does. So cell Y is the closer relative of cell X.

Why does DNA with more matching bases come from a closer relative?

A. Related cells copy each other’s DNA while they live side by side
A cell copies only its own DNA, and passes the copy to its offspring.
B. ✓ DNA is copied from parent to offspring, and changes build up over generations
C. A cell with more matching bases simply has more DNA in total than the other
The count is of matching bases in the same stretch, not of how much DNA a cell has.

Why: DNA is copied from parent to offspring.
Now and then a base changes, and the changes build up over the generations.
Cells whose ancestors separated recently have had few generations for changes to build up.
So their DNA still matches at more bases.

34Comparing the DNA, step by step

Video: Watch: The DNA comparison

The chloroplast’s DNA compared, base by base, with three other sources.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L21d.mp4

Biologists took a short stretch of a chloroplast’s DNA: 20 bases in a row.

One row of 20 boxes, each holding one letter, A, T, G or C: a 20-base stretch of the chloroplast's DNA

Step 1. Compare it, base by base, with the same stretch from a cyanobacterium. 18 of the 20 bases match.

Two rows of 20 bases, the chloroplast's above the cyanobacterium's; 18 of the lower boxes are shaded because their base matches the one above; the count reads 18 of 20 bases match

Step 2. Compare it, base by base, with the same stretch from the plant’s own nucleus. 6 of the 20 bases match.

Two rows of 20 bases, the chloroplast's above the plant nucleus's; 6 of the lower boxes are shaded because their base matches the one above; the count reads 6 of 20 bases match

Step 3. Compare it, base by base, with the same stretch from another bacterium, one that does not photosynthesize. 7 of the 20 bases match.

Two rows of 20 bases, the chloroplast's above another bacterium's; 7 of the lower boxes are shaded because their base matches the one above; the count reads 7 of 20 bases match

Now put the three counts side by side.

the chloroplast’s DNA compared with	bases that match, out of 20
a cyanobacterium	18
the plant’s own nucleus	6
another bacterium (not a cyanobacterium)	7

the three counts, side by side

A table of the three counts: the chloroplast's DNA matches the cyanobacterium at 18 of 20 bases, the plant's own nucleus at 6 of 20, another bacterium at 7 of 20

DNA with more matching bases came from a closer relative. The cyanobacterium’s DNA matches the chloroplast’s at 18 of 20 bases. The nucleus’s DNA matches at only 6.

So the chloroplast’s closest relative is the cyanobacterium, not the plant’s own nucleus.

That is another piece of evidence that chloroplasts descended from a bacterium: their DNA shares many bases with a cyanobacterium’s.

What you are expected to know Determine an organelle’s closest relative from a DNA comparison: the source whose DNA matches at the most bases.

What you are expected to know State what the chloroplast’s DNA comparison supports: chloroplasts descended from a cyanobacterium.

Check q8

A 20-base stretch of a chloroplast’s DNA is compared with the same stretch from three sources. The cyanobacterium’s DNA matches at 18 bases, the plant’s nucleus at 6, and another bacterium at 7.

Which source is the chloroplast’s closest relative?

A. The plant’s nucleus
The nucleus’s DNA matches at only 6 of 20 bases.
B. The other bacterium
The other bacterium’s DNA matches at only 7 of 20 bases.
C. ✓ The cyanobacterium

Why: DNA with more matching bases came from a closer relative.
The cyanobacterium’s DNA matches at 18 of 20 bases, more than the nucleus’s 6 or the other bacterium’s 7.
So the cyanobacterium is the chloroplast’s closest relative.

Check q9

The chloroplast’s DNA matches a cyanobacterium’s at 18 of 20 bases and the plant’s own nucleus at 6 of 20.

Which conclusion does the DNA comparison support?

A. ✓ Chloroplasts descended from a cyanobacterium
B. Chloroplasts formed from a piece of the plant’s own nuclear DNA
DNA that came from the nucleus would still match the nucleus’s at most bases.
C. The DNA comparison does not tell us where chloroplasts came from
DNA with more matching bases came from a closer relative, so an 18-of-20 match does tell us about ancestry.

Why: DNA with more matching bases came from a closer relative.
The chloroplast’s DNA matches the cyanobacterium’s at 18 of 20 bases and the nucleus’s at only 6.
So the chloroplast’s closest relative is the cyanobacterium.
So the comparison supports the conclusion that chloroplasts descended from a cyanobacterium.

Practice writing an answer

Biologists compare a 20-base stretch of a chloroplast’s DNA with the same stretch from a cyanobacterium and from the plant’s own nucleus. The cyanobacterium’s DNA matches at 18 bases. The nucleus’s DNA matches at 6.

(a) Explain how the comparison supports the claim that chloroplasts descended from a cyanobacterium. (1 pt)

Model answer DNA with more matching bases came from a closer relative.
The chloroplast’s DNA matches the cyanobacterium’s at 18 of 20 bases.
It matches the plant’s own nucleus at only 6 of 20.
So the chloroplast’s closest relative is the cyanobacterium.
So the chloroplast descended from a cyanobacterium, not from the plant’s own DNA.

Rubric

Award 1 point for: DNA with more matching bases came from a closer relative; 18 of 20 against 6 of 20 makes the cyanobacterium the closer relative, so the chloroplast descended from a cyanobacterium rather than from the plant’s own DNA.

Slip Quoting the two numbers without saying what more matching bases means: a closer relative.

(b) Predict how the two counts would differ if chloroplasts had formed from the plant’s own nuclear DNA, and justify your prediction. (1 pt)

Model answer The nucleus’s DNA would match at most of the 20 bases.
The cyanobacterium’s DNA would match at only a few.
DNA with more matching bases came from a closer relative.
DNA that came from the nucleus would have the nucleus as its closest relative.
So the two counts would be the reverse of what was observed.

Rubric

Award 1 point for: the match to the nucleus would be high and the match to the cyanobacterium low (the reverse of the observed counts), because DNA copied from the nucleus would still match the nucleus at most bases.

Slip Predicting that the counts would stay the same. The counts follow ancestry: DNA that came from the nucleus would stay closest to the nucleus’s.

49Summary

Chloroplasts evolved from cyanobacteria, the bacteria that photosynthesize. The ancestor of plants and algae took a cyanobacterium inside and kept it.

A chloroplast has the mitochondrion’s five features: its own DNA in one loop, its own ribosomes, two membranes, splitting in two, and a bacterium’s size.

Its DNA adds one more piece of evidence. DNA with more matching bases came from a closer relative. And the chloroplast’s DNA matches a cyanobacterium’s at 18 of 20 bases.

Glossary

cyanobacterium: A bacterium that photosynthesizes: it makes glucose from carbon dioxide and water, using the energy of light. Cyano means blue-green, the color of these bacteria. Chloroplasts descend from cyanobacteria.

APBIO-U02-L21D Two ways to organize a cell’s inside

Topic 2.10 · Origins of Cell Compartmentalization · 26 steps

A bacterium, a rod with a thick wall and a loop of DNA inside it, beside a eukaryotic cell drawn with a nucleus and two smaller compartments inside it

Here is a bacterium. And here is a liver cell, a eukaryotic cell. The bacterium keeps its DNA in one place, and so does the liver cell. The bacterium keeps particular reactions in particular places, and so does the liver cell.

So both cells have an inside with regions. But they organize those regions in two different ways. We look first at the bacterium’s way, then at the eukaryotic cell’s way. Last, we ask which two of the eukaryotic cell’s compartments were once bacteria.

Unit 2 · Cell Structure and Function

1A bacterium’s inside

Video: Watch: Two ways to organize a cell’s inside

A bacterium keeps its DNA in a nucleoid with no membrane around it and has no membrane-bound organelles. A eukaryotic cell separates its regions with membranes.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L21Da.mp4

A bacterium’s inside has regions. Its DNA lies in one region, the nucleoid.

A bacterium drawn large: a rod with a thick cell wall, a cell membrane just inside it, a loop of DNA lying in a region with no membrane around it, and ribosome dots in the cytosol; labels with leader lines

No membrane surrounds the nucleoid.

Its ribosomes float in its cytosol. A bacterium has no membrane-bound organelles.

So a bacterium has regions with their own jobs. But no membrane separates one region from another.

What you are expected to know Describe how a bacterium organizes its inside: a nucleoid with no membrane around it, ribosomes, and no membrane-bound organelles.

Check q1

Where is a bacterium’s DNA?

A. ✓ In a nucleoid: a region with no membrane around it
B. In a nucleus: a compartment with a membrane around it
A nucleus with a membrane around it is a eukaryotic cell’s compartment.
A bacterium has none.
C. Spread evenly through the whole cell
A bacterium’s DNA lies in one region, the nucleoid.

Why: A bacterium’s DNA lies in one region.
That region is the nucleoid.
No membrane surrounds the nucleoid.

Practice writing an answer

A soil bacterium is cut open and viewed under an electron microscope.

(a) Describe how the bacterium organizes its inside. (1 pt)

Model answer Its DNA lies in one region, the nucleoid.
No membrane surrounds the nucleoid.
Its ribosomes float in its cytosol.
The bacterium has no membrane-bound organelles.

Rubric

Award 1 point for: the DNA lies in a nucleoid with no membrane around it, and the cell has no membrane-bound organelles (ribosomes float in the cytosol).

Slip Saying the bacterium has a nucleus. Its DNA lies in a region, the nucleoid, with no membrane around it.

10A eukaryotic cell’s inside

A eukaryotic cell’s inside has regions too.

A eukaryotic cell drawn large: a nucleus with a membrane around it, the endoplasmic reticulum, a mitochondrion and a chloroplast, each with its own membrane, and ribosome dots in the cytosol; labels with leader lines

But the eukaryotic cell separates its regions with membranes.

Its DNA lies inside the nucleus. A membrane surrounds the nucleus.

Its other regions have membranes around them too: the ER, the mitochondria, the chloroplasts.

Here are the two cells side by side.

A bacterium beside a eukaryotic cell: the bacterium a rod with a nucleoid region holding its DNA loop, ribosomes and no membrane-bound organelles; the eukaryotic cell with a nucleus, the ER, a mitochondrion and a chloroplast, each a region with a membrane around it, and ribosome dots; labels with leader lines

Both cells have regions with their own jobs. The bacterium does not separate its regions with membranes. The eukaryotic cell does.

	a bacterium	a eukaryotic cell
where its DNA is	a nucleoid: a region with no membrane around it	a nucleus: a compartment with a membrane around it
membrane-bound organelles	none	many: the ER, mitochondria, chloroplasts and more
how its regions are separated	not by membranes	by membranes

both cells have regions with their own jobs; the eukaryotic cell separates its regions with membranes

A table comparing the inside of a bacterium with the inside of a eukaryotic cell: where its DNA is (a nucleoid, no membrane around it; a nucleus, with a membrane around it); membrane-bound organelles (none; many: ER, mitochondria, chloroplasts); how its regions are separated (not by membranes; by membranes)

Two of the eukaryotic cell’s membrane-bound compartments were once bacteria themselves: the mitochondria and the chloroplasts.

What you are expected to know Describe how a eukaryotic cell organizes its inside: regions separated by membranes.

What you are expected to know State which two of those compartments were once bacteria: the mitochondria and the chloroplasts.

Check q2

A bacterium and a liver cell both keep their DNA in one region and both keep particular reactions in particular places.

Which statement about their insides is correct?

A. The bacterium separates its regions with membranes; the liver cell does not
The bacterium is the cell with no internal membranes.
B. Both cells separate their regions with membranes
A bacterium’s nucleoid is a region with no membrane around it.
C. ✓ The liver cell separates its regions with membranes; the bacterium does not

Why: Both cells have regions with their own jobs.
The liver cell is eukaryotic: it separates its regions with membranes.
The bacterium’s nucleoid has no membrane around it, and a bacterium has no membrane-bound organelles.
So the liver cell separates its regions with membranes and the bacterium does not.

Check q3

Which two compartments of a eukaryotic cell descend from bacteria?

A. The nucleus and the ER
The nucleus and the ER are the cell’s own membranes.
B. ✓ The mitochondria and the chloroplasts
C. The Golgi complex and the lysosomes
The Golgi complex and the lysosomes are made by the cell itself.

Why: Mitochondria descend from bacteria taken inside a large cell.
Chloroplasts descend from cyanobacteria taken inside the ancestor of plants and algae.
So the mitochondria and the chloroplasts are the two compartments that descend from bacteria.

22Summary

A bacterium has regions. Its DNA lies in the nucleoid, with no membrane around it. It has no membrane-bound organelles.

A eukaryotic cell has regions too. It separates its regions with membranes: the nucleus, the ER, the mitochondria, the chloroplasts.

Two of those compartments were once bacteria: the mitochondria and the chloroplasts.

APBIO-U02-P210 Practice questions: Topic 2.10

Topic 2.10 · Origins of Cell Compartmentalization · 9 MCQ · 2 FRQ · for APBIO-U02-T210

Video: Watch first: Topic 2.10 in one picture

Endosymbiosis, the five traces of the cell that moved in, and how to weigh an observation against the claim.

File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T210-summary.mp4

Q1 P210-q01

In the roots of a pea plant, some cells house nitrogen-fixing bacteria. Each bacterium lives inside a root cell, wrapped in a membrane the plant makes, and supplies the plant with nitrogen compounds while the plant supplies it with sugar.

Which term names this kind of relationship?

A. Compartmentalization
Compartmentalization is the dividing of one cell’s interior by membranes.
B. ✓ Endosymbiosis
C. Common ancestry
Common ancestry is what living things share when they descend from the same ancestors.
D. Active transport
Active transport is moving a substance across a membrane against its gradient.

Why: Endosymbiosis is one organism living inside another.
The bacteria live inside the pea root’s cells, and each partner supplies the other.
That is the same kind of relationship the ancestors of mitochondria and chloroplasts had with their host cells.

Q2 P210-q02

Biologists make a claim about where chloroplasts came from.

What is the claim?

A. Chloroplasts are cyanobacteria that live inside plant cells today and could live outside them if freed
Chloroplasts descend from cyanobacteria, but today they are organelles that cannot live outside the cell.
B. Chloroplasts formed when a plant cell folded its plasma membrane around some of its own DNA
Folds of the host’s membrane would carry the host’s own DNA.
C. Chloroplasts were built by the Golgi complex from vesicles filled with a green pigment
Vesicles from the Golgi complex carry no DNA or ribosomes.
D. ✓ Chloroplasts descend from free-living cyanobacteria taken inside a host cell and passed on with it

Why: The claim is that once free-living prokaryotic cells were taken inside a larger host.
They survived there and reproduced with the host over the generations.
This is endosymbiosis.

Q3 P210-q03

The mitochondria of a bread mold each hold their own DNA: one closed loop, separate from the DNA in the mold's nucleus, which is in long molecules with two free ends.

Which feature of a bacterium does this match, and what does the match suggest?

A. ✓ A bacterium's DNA is also one closed loop, as descent from a bacterium predicts
B. A bacterium has no DNA of its own, so the match is with the nucleus instead
Every bacterium has DNA, kept as a single closed loop in its nucleoid.
C. A bacterium's DNA is in long molecules with free ends, like the nucleus's
A bacterium keeps its DNA as one closed loop, not as long molecules with free ends.
D. The mitochondrion's DNA was copied from the nucleus and closed into a loop later
The nucleus’s DNA is in long molecules with free ends and is separate from the mitochondrion’s.

Why: A bacterium keeps its DNA as one closed loop in its nucleoid.
The mitochondrion’s own DNA has that same form.
The nucleus’s DNA is in long molecules with free ends.
So the mitochondrion’s DNA is like a bacterium’s, one of the features expected if mitochondria descend from a bacterium.

Q4 P210-q04

A researcher adds an antibiotic that stops bacterial ribosomes to a culture of single-celled green algae. Protein-making by the ribosomes in the algae's cytosol continues at 97% of its normal rate; protein-making by the ribosomes inside their chloroplasts falls to 4%. In a culture of free-living cyanobacteria, protein-making falls to 3%.

What do the three results show?

A. The antibiotic reaches the chloroplasts but never the cytosol
To reach a chloroplast the antibiotic has to pass through the cytosol first.
B. The chloroplasts are free-living cyanobacteria that happen to sit in the cytosol
Sharing a ribosome design with cyanobacteria shows where chloroplasts came from, not that they are free-living now.
C. ✓ Chloroplast ribosomes are built like bacterial ones, as descent from a bacterium predicts
D. The cytosol's ribosomes were damaged in a different way that the measurement missed
The cytosol’s ribosomes kept working at 97%.
The antibiotic does not act on ribosomes of the eukaryotic build.

Why: The antibiotic acts on ribosomes of the bacterial build.
It stopped the cyanobacteria’s ribosomes (3%) and the chloroplasts’ ribosomes (4%) alike.
It left the cytosol’s ribosomes (97%) alone.
So chloroplast ribosomes are built like a bacterium’s.

Q5 P210-q05

A student writes: 'A mitochondrion's two membranes prove that a cell was engulfed, and the inner membrane is the engulfed bacterium's own membrane.'

Which statement corrects the student?

A. ✓ Two membranes fit engulfment but do not prove it, and do not show which membrane came from which cell
B. The two membranes do prove engulfment, but the outer membrane, not the inner one, is the bacterium's own membrane
One observation on its own proves nothing.
The observation does not show which membrane came from which cell.
C. Two membranes count against engulfment, because a free-living bacterium is wrapped in one membrane, not two
A cell taken inside another could end up with two membranes around it.
D. The student is right on both counts: the two membranes are proof, and the inner membrane is the bacterium's
A double membrane is consistent with engulfment but does not prove it.

Why: A cell taken inside another could end up with two membranes around it.
So a double membrane is consistent with the endosymbiosis account.
One observation alone proves nothing.
The observation also does not show which membrane came from which cell, so biologists claim no origin for either membrane.

Q6 P210-q06

A biologist counts the chloroplasts in a growing leaf cell: 12 one morning and 24 the next morning, although the cell itself has not divided. Each new chloroplast appeared when an existing one pinched in the middle and split. Each chloroplast is about 5 μm long, the size of a large bacterium.

What do these observations add to the case that chloroplasts descend from bacteria?

A. The observations prove that chloroplasts are bacteria that divide inside the cell whenever they choose to
Splitting on their own timetable shows how chloroplasts reproduce, not that they are bacteria.
B. The observations add nothing, because how an organelle multiplies does not tell us where it came from
How the organelle multiplies is one of the five features that support the claim.
C. The observations show that chloroplasts are made by the cell's own membranes budding off and pinching free
The chloroplasts came from existing chloroplasts splitting, not from any other membrane budding off.
D. ✓ The observations show two features a descendant of a bacterium would have: splitting in two, and a bacterium’s size

Why: A bacterium reproduces by splitting in two.
A chloroplast multiplies the same way, on its own timetable rather than when the cell divides.
A bacterium is a few micrometers long, and so is a chloroplast.
These are two features a descendant of a bacterium would have.
Each supports the claim without proving it alone.

Q7 P210-q07

A student studies the mitochondria of a newly found protist and lists four observations.

Which observation supports the claim that these mitochondria descend from a bacterium?

A. The mitochondria are most numerous in the cells that swim fastest
Where the mitochondria are most numerous says what they do for the cell, not where they came from.
B. ✓ New mitochondria form only when an existing one lengthens and splits in two
C. The mitochondria are surrounded on all sides by cytosol
Every organelle sits in cytosol.
That does not tell us where an organelle came from.
D. The mitochondria turn brown with the stain the student used
The color an organelle takes with a stain does not tell us where it came from.

Why: Bacteria reproduce by splitting in two.
So this is one of the features expected of a descendant of a bacterium.
The other three observations say what the mitochondria do, where they sit, or how they stain.
None of those tells us where the mitochondria came from.

Q8 P210-q08

A soil bacterium and a single-celled alga, a eukaryote, both keep their DNA in one region of the cell, and in both, particular reactions happen in particular places.

Which statement compares their interiors correctly?

A. Both cells divide their interiors with internal membranes into membrane-bound organelles
A bacterium typically has no membrane-bound organelles.
Its DNA lies in a nucleoid with no membrane around it.
B. Neither has any membrane inside; in both, every reaction happens in one open space
The alga, a eukaryote, is full of internal membranes.
C. ✓ The alga separates its regions with membranes; the bacterium's nucleoid has no membrane around it
D. The bacterium has a nucleus; the alga keeps its DNA loose in the cytosol
The alga, a eukaryote, has a nucleus; the bacterium keeps its DNA in a nucleoid with no membrane around it.

Why: Both kinds of cell have regions with their own jobs.
A prokaryote such as the bacterium has a nucleoid and ribosomes, but no membrane-bound compartments.
A eukaryote such as the alga separates its regions with internal membranes into a nucleus and other organelles.

Q9 P210-q09

A protist holds an organelle that makes ATP without using oxygen. Biologists compare a 40-position stretch of the organelle’s DNA with the same stretch from three sources. Matching positions out of 40: with a bacterium of kind X, 37; with the protist’s own nucleus, 9; with a bacterium of kind Y, 14.

Which conclusion does the comparison support?

A. The organelle formed from the protist's own nuclear DNA, since 9 positions still match
Nine matches out of 40 is about what two unrelated stretches share by chance.
B. The organelle is a closer relative of bacterium Y than of bacterium X
Fourteen is far short of 37.
Bacterium Y is not the close relative.
C. The protist's nucleus descends from bacterium X
The 37-of-40 match belongs to the organelle, not the nucleus.
D. ✓ The organelle descends from a prokaryote related to bacterium X

Why: DNA with more matching bases came from a closer relative.
The organelle’s DNA matches bacterium X at 37 of 40 positions and its own cell’s nucleus at only 9.
So the organelle’s closest relative is bacterium X, which is what descent from a prokaryote related to X predicts.

FRQ 1 P210-frq1 · Conceptual Analysis scaffolded

A student studies a single-celled alga from a lake. An alga is a plant-like cell that photosynthesizes.
Inside each alga sits a rod-shaped compartment about 2 μm long, wrapped in two membranes.
The compartment holds its own DNA, one closed loop, and its own ribosomes. An antibiotic that stops bacterial ribosomes also stops the compartment’s ribosomes, while the alga’s cytosol ribosomes keep working.
New compartments form only when an existing one splits in two, and each daughter alga receives some when the alga divides.
A 50-position stretch of the compartment’s DNA matches a free-living bacterium at 46 positions and the alga’s own nucleus at 11.
The compartment supplies the alga with nitrogen compounds, and the alga is found mostly in lakes poor in nitrogen.

(a) Make a claim about the origin of the compartment. (1 pt)

Frame The compartment descends from a …, which was …, and which then …

Hint Use the account given for mitochondria and chloroplasts as your pattern: what kind of cell, what happened to it, and what followed over the generations?

Model answer The compartment descends from a once free-living bacterium.
That bacterium was taken inside an ancestor of the alga.
So it survived there and reproduced with the alga, generation after generation.
This is endosymbiosis.

Rubric

Award 1 point for: the claim that the compartment descends from a once free-living prokaryote (a bacterium) that was taken inside an ancestor of the alga and reproduced with it over the generations (endosymbiosis). No reasoning is required for this point.
The host cell and the descent over generations must both be present. Do not award 'it is a bacterium living in the alga' with no ancestor and no generations.

Slip Leaving out the host cell or the passing of generations. Both points are part of the claim.

(b) Identify two features of the compartment, other than its DNA, that support the claim, and state what each shows. (1 pt)

Frame First, the compartment …, which shows …; second, it …, which shows …

Hint Which of the compartment's features in the stimulus would you expect a free-living bacterium to have?

Model answer First, its ribosomes are stopped by an antibiotic that stops bacterial ribosomes, while the cytosol’s ribosomes keep working.
So its ribosomes are built like a bacterium’s, not supplied by the host.
Second, new compartments form only when an existing one splits in two.
That is just like how a bacterium reproduces.

Rubric

Award 1 point for any two, each with what it shows: its own ribosomes that the antibiotic stops (built like bacterial ribosomes, so not supplied by the host); two membranes (what we would expect if one cell was taken inside another); forming only by splitting in two (just like bacteria reproduce); about 2 μm long (a bacterium's size).
Both features need a reason. Do not accept 'a membrane around it' alone, because every organelle has one. Do not accept the DNA: part (c) scores it.

Slip Listing two features with no reasons, or giving 'a membrane around it'. Every organelle has a membrane; the two membranes are what we would expect if one cell was taken inside another.

(c) Explain how the DNA comparison supports the claim. (1 pt)

Frame DNA with more matching bases came from a …; the compartment's DNA matches … at … of 50 but … at only … of 50, so …

Hint What does a closer DNA match between two cells tell you about how closely they are related?

Model answer DNA with more matching bases came from a closer relative.
The compartment’s DNA matches the free-living bacterium at 46 of 50 positions.
It matches the alga’s own nucleus at only 11 of 50.
So the compartment came from a bacterium-like cell, not from the host’s own DNA.

Rubric

Award 1 point for: DNA with more matching bases came from a closer relative; the compartment's DNA matches the free-living bacterium at 46 of 50 positions but the alga's own nucleus at only 11, so the compartment came from a bacterium-like cell rather than from the host's own DNA.
Accept the comparison in words with the inference drawn. Do not award the point for the numbers alone, or for reading the 11 of 50 as evidence that the nucleus made the compartment.

Slip Quoting the numbers without drawing the inference. Say what more matching bases means: a closer relative.

(d) Identify one observation in the list that is irrelevant to where the compartment came from, and explain why it is irrelevant. (1 pt)

Frame The observation that … is irrelevant to where the compartment came from, because it tells you … rather than …

Hint Ask what each observation is about. An observation about where the alga lives, or about what the compartment does for the alga today, would be just as true whatever the compartment’s origin.

Model answer The observation that the alga lives mostly in nitrogen-poor lakes is irrelevant to where the compartment came from.
It tells you where the alga lives, not what the compartment is made of or where it came from.
Where the alga lives would be the same whatever the compartment’s origin.
(The observation that the compartment supplies nitrogen compounds is also a full answer: what it does today would be the same whatever its origin.)

Rubric

Award 1 point for: either observation, ‘the alga is found mostly in lakes poor in nitrogen’ or ‘the compartment supplies the alga with nitrogen compounds’, identified as irrelevant to where the compartment came from. A student who names both earns the point.
The reason must say that the observation describes the alga’s ecology or the partnership’s benefit, not what the compartment is made of or where it came from.
Do not award the point for any of the five features (DNA, ribosomes, membranes, division, size) named as irrelevant.

Slip Picking the size or the two membranes as irrelevant. The size and the two membranes are both features a bacterial ancestor would leave behind. Where the alga lives, and what the compartment does for the alga, are the observations that do not tell us where it came from.

(e) The student says: 'A compartment like this should grow on its own if we take it out of the alga.' Biologists find that the alga’s nucleus carries the instructions for many of the compartment’s proteins, and that the alga makes those proteins and moves them into the compartment. Predict what happens when a biologist places the compartments alone in lake water. Then use your prediction to evaluate the student’s claim. (1 pt)

Frame The claim is …, because placed alone in lake water the compartments …; the free-living cell was the …, and over the generations …

Hint Which cell does the claim say was free-living? What could change in a lineage that has lived inside a host for countless generations?

Model answer Placed alone in lake water, the compartments do not grow and divide.
The alga makes many of the compartment’s proteins and moves them in.
Alone, the compartment cannot get those proteins.
So it may work for a short time, then stop.
So the student’s claim is not supported.
The free-living cell was the ancestor.
Over the generations inside the alga, its descendants became organelles that depend on the alga.
So being unable to live alone is what the account expects, not evidence against it.

Rubric

Award 1 point for: the prediction that the compartments placed alone do not grow and divide (they may work for a short time, then stop), because the alga makes many of their proteins and moves them in, AND the judgement that the student’s claim is not supported: the free-living cell was the ancestor; over the generations inside the alga its descendants became organelles that depend on the alga, so being unable to live alone is what the account expects.
Accept: 'it is an organelle now, not a bacterium', with the prediction. Do not award the judgement alone, or a prediction that the compartments live and multiply like free-living bacteria.

Slip Judging the claim without the ground, or arguing that a free-living ancestor makes the compartment free-living now. The descendant is an organelle; dependence is what the account expects.

FRQ 2 P210-frq2 · Conceptual Analysis

Aphids, small insects that feed on plant sap, carry bacteria inside special cells in their bodies. Each bacterium sits inside a membrane pocket made by the host cell and has its own single membrane. The bacteria make amino acids that the aphid cannot get from sap. They pass from a mother aphid to her eggs, so every aphid is born with them. Their DNA is one closed loop, about a seventh the length of a free-living relative's, and nobody has managed to grow them outside an aphid.

(a) Describe how this relationship resembles the one biologists propose for the origin of mitochondria. (1 pt)

Frame Like the ancestor of mitochondria, these bacteria …, and they …

Model answer Like the ancestor of mitochondria, these bacteria live inside a larger host cell.
They are passed on with the host from one generation to the next.
They supply the host with something it needs.
This is one organism living inside another: endosymbiosis.

Rubric

Award 1 point for: bacteria living inside a larger host cell and passed on with the host from one generation to the next (endosymbiosis), each partner supplying the other.
Accept: 'one organism living inside another and reproducing with it'. Living inside the host and passing down the generations are both needed.

Slip Saying only that the bacteria are 'inside the aphid'. The passing down the generations is what matches the mitochondrion's story.

(b) Explain why the bacteria's failure to grow outside an aphid is what the biologists' account predicts, rather than evidence against it. (1 pt)

Model answer The account says the ancestor was free-living, not the descendant.
Over countless generations inside the host, the bacteria have come to depend on the aphid.
Their DNA has already shrunk to a seventh of a free-living relative’s.
Mitochondria show the same dependence.
So failing to grow alone is what the account expects of a long-term endosymbiont.

Rubric

Award 1 point for: over countless generations inside a host, a once free-living cell comes to depend on the host and loses the ability to live alone (its DNA is already much shorter than a free-living relative's); mitochondria show the same dependence, so being unable to live alone is what the account expects of a long-term endosymbiont.
Accept: 'the ancestor was free-living; the descendant depends on the host'. Do not award the point for 'they are not bacteria any more' with no reference to the change over generations.

Slip Treating the failure to grow alone as evidence against a free-living ancestor. Dependence is what long residence inside a host produces.

(c) Predict how a stretch of the bacteria's DNA would compare with the same stretch from a free-living relative and from the aphid's own nucleus. (1 pt)

Model answer The bacteria’s DNA would match the free-living relative far more closely than the aphid’s own nuclear DNA.
The aphid’s nuclear DNA would match at about the level two unrelated stretches share by chance.

Rubric

Award 1 point for: the bacteria's DNA would match the free-living relative far more closely than the aphid's nucleus (the nucleus would match at about the level of chance).
Do not award a prediction that the aphid's nucleus is the closer match.

Slip Predicting a close match with the aphid's nucleus because the two have lived together so long. Living together does not make DNA sequences converge; ancestry sets the match.

(d) Biologists say that the aphid arrangement, a host-made membrane pocket around a bacterium that keeps its own membrane, is consistent with a mitochondrion having two membranes, and they stop short of saying it proves how mitochondria arose. Justify the biologists' use of 'consistent with' rather than 'proves'. (1 pt)

Model answer A cell taken inside another would end up with the host’s pocket membrane around its own membrane.
So two membranes are what the account predicts.
That is why a mitochondrion’s two membranes are consistent with it.
But an observation the account predicts is evidence, not proof.
Nobody observed the ancient event, and two membranes could have arisen some other way.
So biologists say the evidence is consistent with the account and claim nothing more.

Rubric

Award 1 point for: a cell taken inside another would end up with the host's membrane around its own, so two membranes are what the account predicts; but an observation the account predicts is one piece of evidence, not proof, since the ancient event itself was never observed and two membranes could have arisen in some other way, so biologists claim only what the evidence supports.
Accept: 'consistent with means the evidence agrees with the claim; proof would need to rule out every other explanation'. Do not award 'the two membranes prove engulfment', or 'the outer membrane is the host's' stated as fact.

Slip Treating 'consistent with' as 'proves', or stating as fact which of a mitochondrion's membranes came from the host. The aphid case shows how two membranes could arise; it does not settle what happened to mitochondria.

APBIO-U02-T210 End-of-topic test: Origins of Cell Compartmentalization

Topic 2.10 · Origins of Cell Compartmentalization · 16 MCQ · 2 FRQ

Q1 T210-q01

Inside the cells of a reef coral live tiny single-celled algae, plant-like cells that photosynthesize. Each alga stays inside its coral cell for life, and when the coral cell divides the algae are passed on to both new cells.

Which term names this kind of relationship?

A. ✓ Endosymbiosis
B. Compartmentalization
Compartmentalization is the dividing of one cell’s interior by membranes.
C. Common ancestry
Common ancestry is what living things share when they descend from the same ancestors.
D. Osmosis
Osmosis is water crossing a membrane.
The question is about two organisms living one inside the other.

Why: Endosymbiosis is one organism living inside another.
The algae live and reproduce inside the coral’s cells.
That is the same kind of relationship the ancestors of mitochondria and chloroplasts had with their host cells.

Q2 T210-q02

A gardening book states: 'Every plant cell contains living cyanobacteria, and these bacteria make the plant's sugar.'

Which statement corrects the book?

A. Chloroplasts are unrelated to cyanobacteria; they formed from folds of the plant cell's own membranes.
A chloroplast carries its own circular DNA and bacterium-like ribosomes, and its DNA matches a cyanobacterium’s far more closely than the plant’s own.
B. Chloroplasts are cyanobacteria that can be taken out of a plant cell and grown on their own.
A chloroplast taken out of its cell does not grow or divide on its own.
C. ✓ Chloroplasts descend from cyanobacteria but are now organelles that cannot live on their own.
D. Plant cells took cyanobacteria in only recently, so the bacteria are still being digested.
The cell that moved in was taken inside long ago, not recently, and its descendants were never digested: they reproduced with the host, generation after generation.

Why: The ancestor was a free-living cyanobacterium taken inside a host cell long ago.
Over the generations, its descendants became organelles that depend on the plant cell.
So today’s chloroplasts are not bacteria, and they cannot live outside the cell.

Q3 T210-q03

The endosymbiosis account gives a sequence of events that led to today's mitochondria.

Which sequence is it?

A. A host cell folded its membrane around some of its own DNA, and the enclosed part became a mitochondrion.
A fold of the host’s membrane would enclose the host’s own DNA.
B. ✓ A free-living prokaryote was taken inside a larger host cell and reproduced with it over the generations.
C. A prokaryote grew a nucleus, then shrank until it fitted inside a larger cell and became a mitochondrion.
Prokaryotes have no nucleus.
The mitochondrion’s ancestor never had one.
D. A host cell built mitochondria from vesicles budded off its Golgi complex and filled them with proteins.
Vesicles from the Golgi complex carry no DNA or ribosomes.

Why: That is what endosymbiosis is: one organism living inside another.
A mitochondrion’s own bacterium-like DNA and ribosomes are the traces of that separate cell.
Its descendants are today’s mitochondria.

Q4 T210-q04

A plant cell has both mitochondria and chloroplasts. Each kind of organelle has its own DNA and its own ribosomes, and the two kinds do different jobs.

Under the endosymbiosis account, what is the smallest number of times a prokaryote was taken into an ancestor of this cell?

A. Once; a single prokaryote gave rise to both organelles.
The two organelles trace back to two different kinds of bacterium.
B. Never; both organelles formed from folds of the host's own membrane.
Folds of the host’s membrane would not carry their own DNA and ribosomes.
C. Four times, once for each membrane of the two organelles.
Membranes are not counted one event each.
A double membrane is what we would expect from one cell being taken inside another once.
D. ✓ Twice, one prokaryote for the mitochondria and another for the chloroplasts.

Why: Mitochondria descend from one kind of bacterium.
Chloroplasts descend from a different kind, a cyanobacterium.
One cell cannot be the ancestor of both.
So the cell’s ancestors must have taken in a prokaryote at least twice.

Q5 T210-q05

A mitochondrion holds its own DNA: a single circular molecule, separate from the DNA in the cell's nucleus, which is in long molecules with free ends.

What does this add to the case that mitochondria descend from bacteria?

A. ✓ A bacterium's DNA is also one circular molecule, as descent from a bacterium predicts.
B. Every organelle has its own DNA, so this tells nothing about ancestry.
Most organelles have no DNA at all.
Only mitochondria and chloroplasts do, and theirs is bacterium-like.
C. The DNA shows the mitochondrion is still a living bacterium.
Having its own DNA does not make it a free-living cell.
D. It shows the mitochondrion's DNA was copied from the cell's nucleus.
The nucleus’s DNA is in long molecules with free ends, not a circle.

Why: A bacterium keeps its DNA as one circle in its nucleoid.
The mitochondrion’s own DNA has that same form.
The nucleus’s DNA is in long molecules with free ends.
So the mitochondrion’s DNA is like a bacterium’s, not like the nucleus’s.

Q6 T210-q06

Biologists compare the same 30-position stretch of DNA from a moss’s chloroplast, the same moss’s nucleus, and two bacteria. Matching positions out of 30: chloroplast and cyanobacterium (a photosynthetic bacterium), 27; chloroplast and the moss’s nucleus, 8; chloroplast and a soil bacterium, 10.

Which conclusion does the comparison support?

A. The moss's nucleus, not the chloroplast, descends from a cyanobacterium.
The 27-of-30 match is between the chloroplast and the cyanobacterium, not the nucleus.
B. ✓ Chloroplasts descend from a cyanobacterium-like prokaryote.
C. Chloroplasts formed from the moss's own nuclear DNA, since 8 positions still match.
Eight matches out of 30 is about what two unrelated stretches of DNA give by chance.
D. Chloroplasts are closer relatives of the soil bacterium, since 10 is more than 8.
Ten against eight is a difference of two positions, no more than chance would give.

Why: DNA with more matching bases came from a closer relative.
The chloroplast’s DNA matches the cyanobacterium at 27 of 30 positions.
It matches its own moss’s nucleus at only 8.
So the chloroplast’s closest relative is the cyanobacterium, which is what descent from a cyanobacterium-like prokaryote taken into a host predicts.

Q7 T210-q07

A researcher gives cells from a trout’s heart muscle an antibiotic that stops bacterial ribosomes from building proteins. Protein-making inside the cells’ mitochondria stops. Protein-making in the cells’ cytosol carries on almost unaffected. A second drug, one that stops eukaryotic ribosomes, does the reverse: it stops protein-making in the cytosol and leaves the mitochondria working.

What do the two results show?

A. Mitochondria are free-living bacteria that happen to sit in the cytosol of the trout's cells.
Sharing a ribosome design with bacteria shows where mitochondria came from, not that they are free-living now.
B. The ribosomes in the cytosol are not true ribosomes.
The cytosol’s ribosomes build the cell’s proteins all day, and the second drug, which stops eukaryotic ribosomes, stopped them.
C. The antibiotic reaches the mitochondria but never the cytosol.
To reach a mitochondrion the antibiotic has to pass through the cytosol first.
D. ✓ Mitochondrial ribosomes are built like bacterial ones, as descent from a bacterium predicts.

Why: The antibiotic acts on ribosomes of the bacterial build.
It stopped the mitochondrial ribosomes.
It left the cytosol’s eukaryotic ribosomes working.
The second drug acts on ribosomes of the eukaryotic build, and it did the reverse.
So mitochondrial ribosomes are built like a bacterium’s, and the cytosol’s ribosomes are not.

Q8 T210-q08

The drawing shows a mitochondrion as it appears under the electron microscope: it is wrapped in two membranes, an outer one and an inner one.

A mitochondrion in section, with its outer membrane and its inner membrane labeled.

Which statement about this observation is correct?

A. It proves that a cell was taken inside another; no other explanation is possible.
One observation on its own rarely proves anything.
B. It shows the mitochondrion was built from folds of the nuclear envelope, which is also double.
A mitochondrion is not connected to the nuclear envelope.
C. ✓ It is consistent with one cell having been taken inside another, but does not prove it.
D. It shows the mitochondrion still lives as an independent cell inside the host.
How many membranes wrap an organelle does not tell us whether it can live alone.

Why: A cell taken inside another could end up with two membranes around it.
So the observation is what the account predicts.
But one observation on its own proves nothing.
The other features are needed to make the case strong.

Q9 T210-q09

A time-lapse film follows a moss leaf cell for 8 hours. Every new chloroplast appears when an existing one lengthens, pinches in the middle and splits into two. The cell itself stays a single cell throughout the film.

Which claim is best supported by this observation?

A. ✓ New chloroplasts arise only when an existing one splits in two.
B. Chloroplasts are made by the ER and Golgi complex like other membrane-bound organelles.
Every new chloroplast in the film came from an existing one, not from the ER or the Golgi complex.
C. Chloroplasts are independent organisms that could live outside the cell.
Splitting on their own timetable shows how chloroplasts reproduce, not that they could survive alone.
D. Chloroplasts split because they are wearing out and being recycled by the cell.
The film shows the number of chloroplasts rising, not falling.

Why: Every new chloroplast in the film came from an existing one splitting.
So none came from the ER or the Golgi complex.
A bacterium reproduces by splitting in two.
So this is one of the features expected of a descendant of a bacterium.

Q10 T210-q10

The figure shows a bacterium and a mitochondrion drawn at the same scale.

A bacterium (left) and a mitochondrion (right) drawn at the same scale. The bacterium shows its cell wall, a loop of DNA and ribosomes; the mitochondrion shows its outer membrane and its folded inner membrane.

What does the size comparison contribute to the case for endosymbiosis?

A. It proves the mitochondrion was once this very kind of bacterium.
Size alone cannot identify an ancestor.
Many things are about 2 μm long.
B. ✓ A mitochondrion is about the size of a bacterium, as expected if it descends from one.
C. Being so small shows that mitochondria came from the host's vesicles.
Vesicles carry no DNA or ribosomes.
A mitochondrion has both.
D. Their size shows that mitochondria are prokaryotic cells today.
A mitochondrion is an organelle inside a eukaryotic cell, not a cell of its own.

Why: If mitochondria descend from bacteria taken inside a larger cell, they should be about the size of bacteria.
They are: one to a few micrometers long.
Size is one of the five features the account predicts.
On its own it is weak, and it identifies no particular ancestor.

Q11 T210-q11

Suppose biologists made a new finding about mitochondria.

Which finding would count against the claim that mitochondria descend from bacteria?

A. Mitochondria turn out to be unable to live outside the cell.
Being unable to live alone today is what the account expects of a descendant that became an organelle.
B. Mitochondria turn out to be present in plants and fungi as well as animals.
A prokaryote taken in by an early eukaryotic ancestor would be passed on to all of its descendants.
C. An antibiotic that stops bacterial ribosomes turns out to stop their ribosomes too.
Ribosomes stopped by a bacterial antibiotic are built like bacterial ribosomes.
D. ✓ Their DNA turns out to match the nucleus more closely than any bacterium's.

Why: The claim predicts mitochondrial DNA closer to a bacterium’s than to the nucleus’s.
DNA matching the nucleus would make the organelle look like something the host built from its own DNA.
So that finding would count against the claim.

Q12 T210-q12

An archaeon is a prokaryote. A Paramecium is a single-celled eukaryote.

How does the archaeon's interior differ from the Paramecium's?

A. Both cells have membrane-bound organelles, but the archaeon's are much smaller.
An archaeon, like other prokaryotes, typically has no membrane-bound organelles.
B. The archaeon has no organized regions at all; its DNA and ribosomes drift anywhere.
An archaeon’s interior is organized.
Its DNA is gathered in a nucleoid, and its ribosomes are its protein-building sites.
C. ✓ The archaeon has a nucleoid and ribosomes but no membrane-bound compartments.
D. The archaeon has a nucleus but no other membrane-bound organelles.
A nucleus is a membrane-bound compartment, and an archaeon has none.

Why: A prokaryote has specialized regions: the nucleoid where its DNA lies, and ribosomes.
It typically has no internal membranes dividing its interior.
A eukaryote such as a Paramecium partitions its interior with internal membranes into a nucleus and other organelles.

Q13 T210-q13

A student lists observations about chloroplasts as evidence that they descend from a prokaryote:
1. They have their own circular DNA.
2. They have their own ribosomes.
3. They are found in leaf cells rather than root cells.
4. They have a double membrane.
5. They split in two.

Which observation is irrelevant to where chloroplasts came from?

A. Their own circular DNA
Circular DNA of their own is what a bacterial ancestor would leave behind.
B. ✓ Found in leaf cells rather than root cells
C. Their double membrane
A double membrane is consistent with one cell having been taken inside another.
D. They split in two
Splitting in two is how bacteria reproduce.

Why: Where in the plant a chloroplast is found tells you where the plant uses it.
It does not tell you where the chloroplast came from.
The other four observations are features a prokaryotic ancestor would leave behind, so they support the claim.

Q14 T210-q14

A newly found single-celled eukaryote holds a compartment that photosynthesizes. A biologist compares its features with those of a lysosome from the same cell. Compartment: two membranes; its own circular DNA; its own ribosomes; about 2 μm long; new ones form by splitting in two; its DNA matches a cyanobacterium at 17 positions out of 20 and the cell’s nucleus at 5. Lysosome: one membrane; no DNA; no ribosomes; new ones are made by the Golgi complex.

Which conclusion do these observations support?

A. ✓ The compartment descends from a prokaryote taken inside the cell's ancestor; the lysosome does not.
B. Both organelles descend from prokaryotes, because both are wrapped in membrane.
Being wrapped in membrane is true of every organelle.
C. Neither descends from a prokaryote, because both are organelles that cannot live alone.
Being unable to live alone is expected of any descendant that became an organelle.
D. The lysosome descends from a prokaryote; the compartment was built by the Golgi complex.
The compartment has the two membranes, its own circular DNA and its own ribosomes; the lysosome is made by the Golgi complex.

Why: The compartment shows all five expected features: two membranes, its own circular DNA matching a cyanobacterium, its own ribosomes, a bacterium’s size, and reproduction by splitting.
The lysosome shows none of them.
New lysosomes are made by the cell’s own Golgi complex.

Q15 T210-q15

A student argues: 'A chloroplast survives only inside a plant cell, so its ancestor must have lived inside a cell too, rather than as a free-living bacterium.'

What is wrong with this argument?

A. Nothing; a chloroplast that cannot live alone cannot have free-living ancestors.
A descendant that became an organelle loses the ability to live alone, so not living alone today is no evidence against a free-living ancestor.
B. The fact is wrong: chloroplasts kept in the light do live and multiply outside a cell.
A chloroplast removed from its cell does not live on.
C. Chloroplasts formed from folds of the host's own membranes, so living alone is not the issue.
Chloroplasts carry their own circular DNA and bacterium-like ribosomes.
D. ✓ Descending from a free-living ancestor does not require living on your own today.

Why: The account says the ancestor was free-living.
Over countless generations inside the host, its descendants became organelles that depend on the cell.
So a chloroplast that cannot live alone is what the account expects.

Q16 T210-q16

A plant cell has two kinds of ribosome. Those in its cytosol are of the eukaryotic build. Those inside its chloroplasts are smaller and built like a bacterium's.

What does the chloroplast's having its own, bacterium-like ribosomes add to the case for endosymbiosis?

A. It shows the host cell built the chloroplast's ribosomes to a smaller design of its own.
If the host had built them they would be of the host’s eukaryotic build, like the ribosomes in the cytosol.
B. Nothing new: every organelle has ribosomes of its own.
Most organelles have no ribosomes at all.
Only mitochondria and chloroplasts do.
C. ✓ It shows the ribosomes came in with a bacterium-like cell, not from the host.
D. It shows the chloroplast is still a bacterium that could live on its own.
Having bacterium-like ribosomes shows where the chloroplast came from, not what it is now.

Why: If the host had supplied the chloroplast’s ribosomes, they would be of the host’s eukaryotic build, like the ribosomes in the cytosol.
Instead they are built like a bacterium’s.
So the chloroplast descends from a bacterium-like cell that brought its own ribosomes with it.

FRQ 1 T210-frq1 · Analyze Model or Visual Representation

The model shows a bacterium and a mitochondrion from an animal cell, drawn at the same scale and labeled with features seen under the microscope. The bacterium is labeled: circular DNA; ribosomes; reproduces by splitting in two. The mitochondrion is labeled: two membranes; new ones form by splitting in two. A scale bar shows 1 μm, and each is about 1 to 2 μm long. Biologists also compared the same 20-position stretch of DNA: the mitochondrion’s DNA matched an oxygen-using bacterium’s at 17 positions and the animal’s own nuclear DNA at 6.

A bacterium (left) and a mitochondrion (right) at the same scale, labeled with the features seen in each.

(a) Describe two features in the model, other than its DNA, that the mitochondrion shares with the bacterium. (1 pt)

Model answer New mitochondria form by splitting in two, just like the bacterium reproduces.
The mitochondrion is also about the same size as the bacterium: both are about 1 to 2 μm long.

Rubric

Award 1 point for: any two of: new ones form by splitting in two (just like the bacterium reproduces); a size like the bacterium's (both about 1 to 2 μm); its own ribosomes (the mitochondrion has them, though the model labels only the bacterium's).
Accept: the two membranes named as a feature we would expect if one cell had been taken inside another. Do not accept 'a membrane around it' alone, because every organelle has one. Do not accept the DNA: part (c) scores it.

Slip Giving a membrane around it as a shared feature. Every organelle has one; the two membranes are what we would expect if one cell was taken inside another.

(b) Describe the event that the endosymbiosis account says connects a bacterium like this one to the mitochondrion. (1 pt)

Model answer A bacterium like this one was taken inside a larger host cell.
It survived there and reproduced with the host over the generations.
Its descendants became the mitochondrion.
This is endosymbiosis.

Rubric

Award 1 point for: the bacterium (or a cell like it) was taken inside a larger host cell and survived and reproduced there over the generations.
Accept: 'taken inside a host cell', 'engulfed by a host cell and kept alive', or 'endosymbiosis' with the host cell named. Do not award 'the bacterium turned into a mitochondrion' with no host cell mentioned.

Slip Saying the bacterium turned into a mitochondrion with no host cell. The host cell is part of the event.

(c) Explain how the DNA comparison supports the claim that the mitochondrion descends from a bacterium. (1 pt)

Model answer DNA with more matching bases came from a closer relative.
The mitochondrion’s DNA matches the bacterium at 17 of 20 positions.
It matches the nucleus of its own cell at only 6.
So the mitochondrion’s DNA came from a bacterium-like cell, not from the host’s own DNA.

Rubric

Award 1 point for: DNA with more matching bases came from a closer relative; the mitochondrion's DNA matches the bacterium (17 of 20) far more closely than the nucleus of its own cell (6 of 20), so it came from a bacterium-like cell rather than from the host's own DNA.
Accept: the comparison stated in words ('much closer to the bacterium than to the nucleus') with the inference drawn.

Slip Stating the numbers without drawing the inference, or reading the 6 of 20 as evidence that the nucleus made the mitochondrion.

(d) Explain how this model relates to the larger difference between how prokaryotic and eukaryotic cells organize their interiors. (1 pt)

Model answer A prokaryote such as the bacterium has specialized regions, a nucleoid and ribosomes.
It typically has no membrane-bound compartments.
A eukaryotic cell partitions its interior with internal membranes into organelles.
The endosymbiosis account explains how some of those compartments came to exist.
Mitochondria and chloroplasts are descendants of prokaryotes taken inside a host.

Rubric

Award 1 point for: a prokaryote such as the bacterium has specialized regions (nucleoid, ribosomes) but no membrane-bound compartments, while a eukaryotic cell partitions its interior with internal membranes into organelles; the endosymbiosis account explains how some of those compartments (mitochondria, chloroplasts) came to exist, as descendants of prokaryotes taken inside a host.
Accept: either the comparison plus the link to origin, or a statement that mitochondria and chloroplasts are membrane-bound compartments that a eukaryotic cell gained by taking in prokaryotes.

Slip Comparing the two kinds of cell without linking the model to where some compartments came from, or the other way around.

FRQ 2 T210-frq2 · Conceptual Analysis

A single-celled eukaryote from a pond holds a green compartment that photosynthesizes. Biologists record the following.
1. The compartment is wrapped in two membranes.
2. It holds its own DNA, one circular molecule.
3. It holds its own ribosomes. An antibiotic that stops bacterial ribosomes stops them too, while leaving the ribosomes in the cell’s cytosol working.
4. New compartments form only when an existing one splits in two.
5. The compartment is about 2 μm long.
6. It sits near the middle of the cell.

In an experiment, a biologist gently breaks the eukaryote open and places its intact compartments alone in pond water in the light, where free-living cyanobacteria grow well. The cell’s own DNA carries most of the instructions the compartment needs, and the cell makes many of the compartment’s proteins and moves them into it.

(a) Describe the claim that biologists make about the origin of chloroplasts and mitochondria. (1 pt)

Model answer The claim is that chloroplasts and mitochondria were once free-living prokaryotic cells.
A larger host cell took those cells inside itself.
They survived there and reproduced with the host over the generations.
This is endosymbiosis.

Rubric

Award 1 point for: they descend from once free-living prokaryotic cells that were taken inside a larger host cell, survived and reproduced with it over the generations (endosymbiosis).
Accept: 'a bacterium taken inside another cell that stayed and reproduced with it'. The host cell and descent over generations must both be present.

Slip Leaving out the host cell or the descent over generations. The host cell and the descent are both part of the claim.

(b) Support the claim that this compartment descends from a cyanobacterium, using two of the recorded observations and saying what each shows. (1 pt)

Model answer Its own circular DNA: a single circular DNA molecule is what a bacterium has.
So the compartment carries its own bacterium-like instructions.
Its own ribosomes: an antibiotic that stops bacterial ribosomes stops them, while the cell’s own ribosomes keep working.
So the compartment’s ribosomes are built like a bacterium’s, not supplied by the host.
Each feature is a feature of a free-living prokaryote, kept by a descendant that now lives inside the host.

Rubric

Award 1 point for: the evidence AND the reasoning: two of: its own circular DNA (like a bacterium's DNA); its own ribosomes that the antibiotic stops (built like bacterial ribosomes, so not supplied by the host); two membranes (what we would expect if one cell was taken inside another); forming only by splitting (just like bacteria reproduce); a size like a bacterium's; each with what it shows.
Accept: any two with a reason for each. Observation 6, that it sits near the middle of the cell, does not tell us where the compartment came from and earns nothing; naming it does not lose the point if two valid observations are also given with reasons.

Slip Listing two observations without saying what each shows, or using its position near the middle of the cell, which does not tell us where the compartment came from.

(c) Predict what happens to the compartments placed alone in pond water in the light. (1 pt)

Model answer The compartments do not survive and reproduce as free-living cells.
They may photosynthesize for a short time.
But they do not grow or divide, so they soon stop working.
Free-living cyanobacteria grow in the same water.

Rubric

Award 1 point for: the prediction that they do not survive and reproduce as free-living cells (they may photosynthesize for a short time, but they do not grow or divide and soon stop working), even though free-living cyanobacteria grow in the same water. No reasoning is required for this point.
Accept: 'they will not grow or divide the way the cyanobacteria do', or 'they photosynthesize briefly but cannot grow or divide', or 'they die'. Do not award a prediction that they live and multiply as free-living cells.

Slip Predicting that they live and multiply like the cyanobacteria. They descend from free-living cells but are organelles now.

(d) Justify your prediction, using the biologists' account of the origin of chloroplasts. (1 pt)

Model answer The account says the ancestor was free-living.
Over the generations inside the host, its descendants became organelles.
They reproduce with the host, so they depend on it.
Therefore they are no longer bacteria that can live on their own.
So descent from a free-living cell is consistent with the compartments failing to live free today.

Rubric

Award 1 point for: the account says the ancestor was free-living; over countless generations inside the host the descendants became organelles that depend on the host cell and cannot live on their own, so descent from a free-living cell does not make the compartment free-living today.
Accept: 'it is now an organelle, not a bacterium; being unable to live alone is what the account expects'.
The point needs only the dependence argument from the endosymbiosis account. Extra detail about instructions moved to the host's nucleus is not required and earns nothing extra.

Slip Arguing that because the ancestor was free-living the compartment should be too. The account expects dependence: it is an organelle now, and cannot live alone.

Could not render

video file missing: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T24-intro.mp4
video file missing: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-T26-summary.mp4
video file missing: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U02-L14.mp4