APBIO-U06-L01 Where the instructions are kept
Here are two cells, drawn side by side to scale. One is a bacterium, about one thousandth of a millimeter long. The other is one of your skin cells, some twenty times wider. Each cell carries every instruction it needs in its DNA.
Where does each cell keep that DNA, and what does it look like?
Unit 6 · Gene Expression and Regulation
1The instructions are in the DNA
Where does a cell keep its DNA? A bacterium keeps its DNA on one closed circle, lying free in the cell.
Your cells keep theirs on 46 separate lines, inside the nucleus.
Each line is wound on beads of protein. So about two meters of DNA fit inside a nucleus about a hundredth of a millimeter across.
Where the DNA sits, and how it is packed, are the two facts you need to follow how a cell copies and reads it.
A strand of DNA carries information.
Where in the strand is the information held?
- A. In which bases are presentTwo strands can hold the same four bases and carry different information.
The information is in the order of the bases. - B. ✓ In the order of its bases
- C. In how many bases it hasTwo strands can hold the same number of bases and carry different information.
The information is in the order of the bases.
Why: The same bases in a different order carry a different message.
So the information is in the order of the bases along the strand.
A cell’s instructions are written in the order of the bases along its DNA. Biologists call a cell’s instructions its genetic information.
Suppose a bacterium is about to divide into two cells. Before it divides, it copies its DNA.
Then it hands one copy to each new cell. So each new cell carries the same instructions as the old cell.
A human body cell holds 46 chromosomes.
What is one chromosome?
- A. One protein that reads the DNA’s basesA chromosome is made of DNA, with proteins wound in it.
The proteins do not read the DNA. - B. One base pair of the DNAOne base pair is one rung of the DNA ladder.
A chromosome is a whole DNA molecule, millions of base pairs long, wound on proteins. - C. ✓ One very long DNA molecule wound on proteins
Why: A chromosome is one very long DNA molecule packaged with proteins.
A human body cell holds 46 of them.
Your body cells do the same. Before a skin cell divides, it copies all 46 of its chromosomes.
Each new skin cell then receives a full set of 46 chromosomes. With them it receives every instruction.
A human sperm fuses with a human egg at fertilization.
How many chromosomes does each of the two gametes carry?
- A. ✓ 23
- B. 4646 is the count in a body cell.
Each gamete carries one complete set, 23 chromosomes, so the zygote has 46. - C. 9292 would be two body cells’ worth.
Each gamete carries one complete set, 23 chromosomes.
Why: A gamete is haploid: one complete set.
One human set is 23 chromosomes.
The sperm’s 23 and the egg’s 23 make the zygote’s 46.
Video: Watch: The instructions are in the DNA
The two cells drawn to scale; a bacterium copying its DNA and handing one copy to each new cell; a sperm and an egg each bringing 23 chromosomes to the zygote.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L01a.mp4
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A parent hands instructions on the same way. A sperm carries 23 chromosomes copied from the father’s DNA, and an egg carries 23 copied from the mother’s DNA.
The two gametes fuse into the zygote. So the child’s first cell receives its instructions from both parents, in their DNA.
So a cell stores its genetic information in its DNA. Whenever the DNA is copied and handed on, the information is handed on with it.
What you are expected to know Name where a cell stores its genetic information: in its DNA.
What you are expected to know Describe how the genetic information is handed on: the DNA is copied and a copy goes to each new cell or into a gamete.
Suppose a bacterium divides into two new cells.
Which molecule carries the instructions the old cell hands to each new cell?
- A. The old cell’s proteinsProteins are built by following the instructions.
The instructions themselves are stored in the DNA, which the old cell copies and hands on. - B. ✓ The old cell’s DNA
- C. The old cell’s RNAThe instructions are stored in the DNA.
The old cell copies its DNA and hands one copy to each new cell.
Why: A cell stores its genetic information in its DNA.
Before dividing, the old cell copies its DNA.
Each new cell receives one copy.
A father’s instructions reach his child in a sperm.
In which molecule does the sperm carry them?
- A. ✓ The sperm’s DNA
- B. The sperm’s proteinsThe sperm’s DNA holds the father’s instructions.
The proteins were built from instructions; they do not carry them.
Why: A parent’s genetic information is in its DNA.
The sperm carries 23 chromosomes, each a DNA molecule copied from the father’s.
So the instructions travel in the sperm’s DNA.
22One circle or many lines
A bacterium is a prokaryotic cell.
Where does its DNA lie?
- A. Inside a nucleusA bacterium has no nucleus.
Its DNA lies in the nucleoid, a region of the cytosol with no membrane around it. - B. ✓ In the nucleoid
Why: A bacterium’s DNA lies in the nucleoid.
The nucleoid is the region of the cytosol where the DNA is, with no membrane around it.
An onion skin cell is a eukaryotic cell.
Which part of the cell holds its DNA?
- A. The cell wallThe cell wall is the stiff layer outside the cell.
The DNA sits inside the nucleus. - B. A ribosomeRibosomes build proteins.
The DNA sits inside the nucleus. - C. ✓ The nucleus
Why: In a eukaryotic cell the DNA sits inside the nucleus, wrapped in the nuclear envelope.
Video: Watch: One circle or many lines
The zoom from the two cells to the bacterium’s one closed circle, then to the skin cell’s 46 lines inside its nucleus; three cells sorted as prokaryotic or eukaryotic by their chromosomes.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L01b.mp4
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Here are the two cells again, with the DNA drawn inside each. Look at the bacterium first.
The bacterium’s DNA is one very long molecule. The two ends of that molecule are joined to each other.
So the molecule forms a closed circle. The circle lies in the nucleoid, and few proteins are attached to it.
In E. coli, the circle is about 4.6 million base pairs long.
When a chromosome’s DNA is one closed circle like this, it is called a , because the molecule has no free end.
Now look at the skin cell. Its DNA is in 46 separate molecules, and each molecule has two free ends.
Each molecule is wound on proteins along its whole length. All 46 sit inside the nucleus, wrapped in the nuclear envelope.
When a chromosome’s DNA is one long molecule with two free ends like this, it is called a , because the molecule lies in a line from one end to the other.
Here are the two cells with their chromosomes named. The bacterium has one circular chromosome, and the skin cell has 46 linear chromosomes.
A cell’s chromosomes tell you which kind of cell it is.
For example, here is a cell whose DNA is one circular chromosome lying in the nucleoid. This is a prokaryotic cell.
But here is a cell whose DNA is several linear chromosomes inside a nucleus. This is a eukaryotic cell.
And here is a much smaller cell with a cell wall, whose DNA is linear chromosomes inside a nucleus. This is still a eukaryotic cell.
The cell’s size and its cell wall change nothing. Its chromosomes decide it: one circle in the nucleoid, or lines inside a nucleus.
Here is a table of the two kinds of chromosome set, prokaryotic against eukaryotic.
What you are expected to know Classify a drawn or described chromosome set as prokaryotic (one circular chromosome in the nucleoid, few proteins attached) or eukaryotic (several linear chromosomes inside the nucleus, each wound on proteins).
Look at how the cell drawn below holds its DNA.
Which kind of cell is it?
- A. ✓ Prokaryotic
- B. EukaryoticThe DNA is one closed circle with no nucleus around it.
One circular chromosome lying free makes the cell prokaryotic.
Why: The cell’s DNA is one closed circle.
No nucleus wraps it.
One circular chromosome in the nucleoid is a prokaryotic cell’s chromosome set.
Look at where the DNA lies in the cell drawn below.
Which kind of cell is it?
- A. ProkaryoticThe DNA is in several lines inside a nucleus, not one circle lying free.
Linear chromosomes inside a nucleus make the cell eukaryotic. - B. ✓ Eukaryotic
Why: The compartment with a double edge is a nucleus.
Inside it lie several linear chromosomes.
Linear chromosomes inside a nucleus are a eukaryotic cell’s chromosome set.
The cell drawn below is small and has a thick cell wall. Look at how it holds its DNA.
Which kind of cell is it?
- A. ProkaryoticSmall size and a cell wall decide nothing.
The DNA lies as lines inside a nucleus, so the cell is eukaryotic. - B. ✓ Eukaryotic
Why: The compartment with a double edge is a nucleus, and the DNA inside it is linear.
Linear chromosomes inside a nucleus make the cell eukaryotic, whatever its size.
A cell’s DNA is one closed circle with a few proteins attached, lying in its cytosol.
Which kind of cell is it?
- A. ✓ Prokaryotic
- B. EukaryoticThe DNA is one circle lying in the cytosol, so there is no nucleus.
A eukaryotic cell keeps linear chromosomes inside a nucleus.
Why: One circular chromosome lying in the cytosol, with few proteins attached, is a prokaryotic cell’s chromosome set.
A cell about two thousandths of a millimeter long holds one DNA molecule whose two ends are joined to each other.
Which kind of cell is it?
- A. ✓ Prokaryotic
- B. EukaryoticA DNA molecule whose ends are joined is a closed circle: one circular chromosome.
That is a prokaryotic cell’s chromosome set.
Why: The two ends of the one DNA molecule are joined, so it is a circular chromosome.
One circular chromosome is a prokaryotic cell’s chromosome set.
A cell’s DNA is in 16 separate molecules, each with two free ends and each wound on proteins, inside a nucleus.
Which kind of cell is it?
- A. ProkaryoticA prokaryotic cell has one circular chromosome and no nucleus.
Sixteen linear chromosomes inside a nucleus make this cell eukaryotic. - B. ✓ Eukaryotic
Why: Each DNA molecule has two free ends, so each is a linear chromosome.
They sit inside a nucleus.
Linear chromosomes inside a nucleus are a eukaryotic cell’s chromosome set.
48Quick quiz: circular chromosome, linear chromosome mixed practice
A cell’s DNA can be joined into a circle or lie in a line.
What is a circular chromosome?
- A. One DNA molecule with two free ends, wound on proteinsA DNA molecule with two free ends is a linear chromosome.
- B. A closed ring of proteins with no DNA wound on itA chromosome is made of DNA.
A circular chromosome is a DNA molecule joined into a closed circle. - C. ✓ One DNA molecule whose two ends are joined into a closed circle
Why: A circular chromosome is one DNA molecule whose two ends are joined to each other, so the molecule forms a closed circle.
A cell’s DNA can be joined into a circle or lie in a line.
What is a linear chromosome?
- A. One DNA molecule whose two ends are joined into a closed circleA DNA molecule joined into a closed circle is a circular chromosome.
- B. ✓ One long DNA molecule with two free ends
- C. A straight protein fiber with no DNA in itA chromosome is made of DNA.
A linear chromosome is one long DNA molecule with two free ends.
Why: A linear chromosome is one long DNA molecule with two free ends, lying in a line from one end to the other.
A cell’s chromosome may be a circle or a line.
(a) State what a circular chromosome and a linear chromosome are. (1 pt)
A linear chromosome is one long DNA molecule with two free ends.
- Award 1 point for: a circular chromosome is a DNA molecule joined into a closed circle (no free ends), and a linear chromosome is a DNA molecule with two free ends.
E. coli’s one chromosome is a closed loop of DNA.
Which kind of chromosome is it?
- A. ✓ Circular
- B. LinearA closed loop has no free end.
A DNA molecule with no free end is a circular chromosome.
Why: The loop is closed, so the DNA molecule has no free end.
A DNA molecule joined into a closed circle is a circular chromosome.
One of a mouse’s chromosomes is a DNA molecule with two free ends.
Which kind of chromosome is it?
- A. CircularA circular chromosome has no free end.
A DNA molecule with two free ends is a linear chromosome. - B. ✓ Linear
Why: The molecule has two free ends, so it lies in a line from one end to the other.
That is a linear chromosome.
54Wound on beads
Go back to the skin cell. Its 46 linear chromosomes are inside a nucleus about a hundredth of a millimeter across.
One human body cell holds 46 DNA molecules.
Laid end to end, about how long would they be?
- A. About two millimetersTwo millimeters is far too short.
The 46 DNA molecules of one cell would stretch about two meters. - B. ✓ About two meters
Why: Laid end to end, the 46 DNA molecules of one human cell would stretch about two meters.
Two meters of DNA must fit inside a nucleus a hundredth of a millimeter across. So the cell packs the DNA very tightly.
Here is how the cell packs it, drawn three ways: one bead close up, a row of beads, and the row coiled.
Look at the bead close up. The DNA winds around a small cluster of proteins, like thread wound on a bead.
Look at the row. Bead follows bead along the DNA, so the chromosome looks like a string of beads.
Look at the coil. The string of beads coils on itself, and each coil packs the string shorter.
The small proteins that the DNA winds around, like thread on a bead, are called . A cluster of histones is one bead.
The histones pack the DNA. They are not the genes: the genes are in the DNA that is wound on them.
Between one division and the next, a cell reads and copies its DNA.
In which form are its chromosomes during this time?
- A. ✓ Spread out as chromatin
- B. Coiled into short thick rodsShort thick rods are the form for dividing.
Between divisions each chromosome is spread out as chromatin, so its DNA can be read.
Why: Between divisions each chromosome is spread out as chromatin, a long thin thread of DNA wound on proteins.
In that form the DNA can be read and copied.
Video: Watch: Wound on beads
One of the skin cell’s 46 lines pulled out of the nucleus and unwound: the DNA wrapped on bead after bead, the beaded string coiled, then coiled again into a short thick rod as the cell prepares to divide.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L01c.mp4
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Between divisions, the beaded string is coiled loosely. This loosely coiled form is chromatin: loosely wound, so the DNA can be read and copied.
When the cell is about to divide, the coiled string coils again, and again. Each chromosome becomes a short thick condensed chromosome.
A short thick chromosome can be moved to a new cell without tangling or breaking. In four words: loose to work, tight to move.
What you are expected to know Describe how a eukaryotic chromosome’s DNA is packed: wound on histone beads, the beaded string coiled, and coiled further into a condensed chromosome for division.
The drawing below shows part of one chromosome, with the DNA threading through a row of shapes.
What are the round shapes?
- A. GenesThe genes are in the DNA itself, the line that threads through.
Each round shape is a cluster of histone proteins that the DNA winds around. - B. ✓ Clusters of histone proteins
- C. Molecules of RNANo RNA is part of a chromosome’s packing.
Each round shape is a cluster of histone proteins that the DNA winds around.
Why: The line is the DNA.
It winds around each round shape like thread on a bead.
Each bead is a cluster of histone proteins.
A human skin cell’s nucleus is about a hundredth of a millimeter across. The DNA inside it would stretch about two meters if it were laid end to end. Just before the cell divides, each chromosome coils into a short thick rod.
(a) Explain how winding the DNA on histones lets two meters of it fit inside the nucleus. (1 pt)
Frame The DNA fits because …
The beaded string then coils on itself.
Each coil packs the same length of DNA into a shorter space.
So two meters of DNA take up a space far smaller than a hundredth of a millimeter across.
- Award 1 point for: the DNA winds on histone beads and the beaded string coils, so a long molecule occupies a much shorter (smaller) space.
(b) Explain why each chromosome coils tightest just before the cell divides. (1 pt)
Frame Each chromosome coils tightest before division because …
A loosely wound chromosome is a long thin thread.
A long thin thread would tangle or break when pulled across the cell.
A tightly coiled chromosome is a short thick rod, which the cell can move without tangling or breaking.
- Award 1 point for: the chromosome must be moved to a new cell, and a short thick (tightly coiled) chromosome can be moved without tangling or breaking, unlike a long thin thread.
A student looks at the string-of-beads drawing of a chromosome and says: “Each bead is one gene.”
Is the student correct?
- A. ✓ No: the beads pack the DNA, and the genes are in the DNA
- B. Yes: each bead is one gene, and the DNA links the genesA bead is a cluster of histones, proteins the DNA winds around.
The genes are in the order of the bases of the DNA itself.
Why: Each bead is a cluster of histone proteins.
The histones only pack the DNA.
The genes are in the DNA that is wound on the beads.
In a liver cell between divisions, a gene is being read.
How is that gene’s DNA wound?
- A. Tightly, as a condensed chromosomeTight coiling is for moving a chromosome during division.
A gene being read is on loosely wound chromatin. - B. ✓ Loosely, as chromatin
Why: Between divisions the DNA is loosely wound as chromatin.
Loosely wound DNA can be read.
So a gene being read is on loosely wound DNA.
Go back to the two cells drawn side by side to scale: the bacterium about one thousandth of a millimeter long, and the skin cell some twenty times wider.
In the bacterium, one closed circle of DNA lies in the nucleoid: one circular chromosome.
In the skin cell, 46 linear chromosomes sit inside the nucleus, each wound on histone beads. So about two meters of DNA fit in a nucleus about a hundredth of a millimeter across.
77Quick quiz: histone mixed practice
A eukaryotic chromosome is DNA wound on proteins.
What is a histone?
- A. A gene that the cell is currently readingA gene is a stretch of the DNA.
A histone is one of the proteins the DNA winds around. - B. The double membrane wrapping the nucleusThe membrane around the nucleus is the nuclear envelope.
A histone is one of the proteins the DNA winds around. - C. ✓ A protein the DNA winds around, like thread on a bead
Why: A histone is one of the small proteins that a eukaryotic chromosome’s DNA winds around, like thread on a bead.
A eukaryotic chromosome is DNA wound on proteins.
(a) State what a histone is. (1 pt)
- Award 1 point for: a histone is a protein that DNA winds around (packs the DNA into beads).
Between divisions, a chromosome’s DNA loops round small proteins along its whole length.
Which proteins does it loop round?
- A. RibosomesRibosomes build proteins and lie outside the nucleus.
The proteins a chromosome’s DNA loops round are histones. - B. ✓ Histones
Why: The proteins a chromosome’s DNA winds around, bead after bead, are histones.
A protein is part of a chromosome, and the DNA is wound around it.
Is that protein a histone?
- A. ✓ Yes
- B. NoA protein that DNA winds around inside a chromosome is a histone.
Why: Histones are the proteins DNA winds around in a chromosome.
So a protein with the DNA wound around it is a histone.
Other proteins attach to a chromosome, but the proteins the DNA winds around are the histones.
82Mixed practice mixed practice
A student says: “This cell is eukaryotic, so its DNA is one closed circle lying in its cytosol.”
Is the student correct?
- A. Yes: a eukaryotic cell’s DNA is one closed circle in its cytosolOne closed circle lying in the cytosol is a prokaryotic cell’s chromosome set.
A eukaryotic cell keeps linear chromosomes inside a nucleus. - B. ✓ No: a eukaryotic cell’s DNA is linear chromosomes inside a nucleus
Why: A eukaryotic cell keeps its DNA as linear chromosomes inside a nucleus.
One closed circle lying in the cytosol is a prokaryotic cell’s chromosome set.
So the student has described a prokaryotic cell, not a eukaryotic one.
A chromosome in a skin cell is coiled into a short thick rod.
What is the cell about to do?
- A. ✓ Divide into two cells
- B. Read a gene on that chromosomeA gene is read from loosely wound chromatin.
A chromosome coils into a short thick rod when the cell is about to divide.
Why: A chromosome coils tightly into a short thick rod so that it can be moved without tangling.
The cell moves its chromosomes when it divides.
So the cell is about to divide.
A skin cell is about to divide into two new cells.
Which step gives each new cell a full set of instructions?
- A. Building a full set of proteins before it dividesProteins are built by following the instructions; they do not carry them.
The cell copies its DNA, and each new cell receives one copy. - B. Making RNA before it dividesRNA is not where a cell stores its instructions.
The cell copies its DNA, and each new cell receives one copy. - C. ✓ Copying its DNA before it divides
Why: A cell stores its instructions in its DNA.
Before it divides, the cell copies its DNA.
Each new cell receives one complete copy, and with it every instruction.
A cell is about to divide. It must move each chromosome to a new cell in one piece.
What makes each chromosome easy to move?
- A. The DNA has been cut into short piecesThe DNA is not cut.
The whole molecule is coiled tightly, so it is short and thick. - B. The DNA has been taken off its histonesThe DNA stays wound on its histones.
The beaded string coils tightly, so the chromosome is short and thick. - C. ✓ The DNA is coiled tightly into a short thick rod
Why: The beaded string coils again and again.
So the chromosome becomes a short thick rod.
A short thick rod can be moved without tangling or breaking.
A cell about five thousandths of a millimeter across has a cell wall. Its DNA is linear chromosomes inside a nucleus.
Which kind of cell is it?
- A. ✓ Eukaryotic
- B. ProkaryoticSize and a cell wall decide nothing.
Linear chromosomes inside a nucleus make the cell eukaryotic.
Why: The DNA is linear chromosomes inside a nucleus.
That is a eukaryotic cell’s chromosome set, whatever the cell’s size.
Suppose a bacterium divides into two new cells.
What does each new cell receive?
- A. Half of the old cell’s DNA, uncopiedBefore it divides, the old cell copies its DNA.
Each new cell receives a complete copy, so each has every instruction. - B. The old cell’s proteins and none of its DNAThe instructions are in the DNA.
Each new cell receives a complete copy of the old cell’s DNA. - C. ✓ A complete copy of the old cell’s DNA
Why: Before dividing, the old cell copies its DNA.
Then it hands one complete copy to each new cell.
So each new cell carries every instruction the old cell had.
Here are the two cells from the start of the lesson, drawn to scale: a bacterium about one thousandth of a millimeter long, and a skin cell some twenty times wider.
(a) Describe the bacterium’s chromosome set: how many DNA molecules it has, their shape, and where in the cell they lie. (1 pt)
The molecule is joined into a closed circle: one circular chromosome.
It lies in the nucleoid, a region of the cytosol with no membrane around it.
- Award 1 point for: one circular DNA molecule (chromosome) lying in the nucleoid (in the cytosol, no membrane around it).
(b) Describe the skin cell’s chromosome set in the same way. (1 pt)
Each molecule is a line with two free ends: a linear chromosome.
All 46 lie inside the nucleus, wrapped in the nuclear envelope.
- Award 1 point for: many (46) linear DNA molecules (chromosomes) inside the nucleus. Accept ‘many’ for the count.
(c) Between divisions, the skin cell reads many of its genes. Describe how its chromosomes are wound at that time, and explain why they are wound that way. (1 pt)
The DNA is still wound on histone beads, but the beaded string is coiled loosely.
A gene can be read only where its DNA is loosely wound.
So the cell keeps its chromosomes loosely wound while it reads its genes.
- Award 1 point for: loosely wound (as chromatin, not a condensed chromosome), because the DNA can be read only where it is loosely wound.
Glossary
- circular chromosome
- One DNA molecule whose two ends are joined to each other, so it forms a closed circle with no free end. A prokaryotic cell's chromosome is a circular chromosome lying in the nucleoid.
- linear chromosome
- One long DNA molecule with two free ends, lying in a line from one end to the other and wound on proteins. A eukaryotic cell's chromosomes are linear chromosomes inside the nucleus.
- histone
- One of the small proteins that a eukaryotic chromosome's DNA winds around, like thread on a bead. A cluster of histones is one bead; the beaded string coils to pack the DNA.
APBIO-U06-L02 Extra circles
Look again inside the bacterium. Beside its one big circle of DNA sit three much smaller circles. They are not part of the chromosome.
What are they, and what do they carry?
Unit 6 · Gene Expression and Regulation
1Small circles beside the chromosome
What are the small circles of DNA in a bacterium? Each small circle is its own closed piece of DNA, separate from the chromosome.
Each small circle carries a few genes of its own.
Each small circle is copied on its own, apart from the chromosome. So a cell may hold several copies.
Small circles like these are found in bacteria and in some eukaryotes, such as yeast.
Bacteria pass these small circles from one cell to another. Biologists also use these circles to carry a human gene into a bacterium.
A bacterium’s one chromosome is a single DNA molecule.
Which shape is it?
- A. A line with two free endsA DNA molecule with two free ends is a linear chromosome, a eukaryotic cell’s kind.
A bacterium’s chromosome is one closed circle. - B. ✓ A closed circle
Why: The two ends of a bacterium’s DNA molecule are joined to each other.
So the molecule forms a closed circle: a circular chromosome.
Video: Watch: Small circles beside the chromosome
The bacterium magnified: its one chromosome, then the three small circles; one small circle close up with its genes; a plasmid copied on its own while the chromosome stays one; a yeast cell with a plasmid inside its nucleus.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L02a.mp4
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Here is the bacterium again, magnified. Look at the big circle first.
The big circle is the bacterium’s one chromosome. In E. coli, the chromosome is about 4.6 million base pairs long.
Along that circle lie about 4,000 genes. Almost every instruction the cell needs is on this one circle.
Now look at the three small circles. Each small circle is its own DNA molecule, far shorter than the chromosome.
The molecule’s two ends are joined to each other. So each small circle is closed, just like the chromosome.
Here is one small circle, close up. The thick stretches are its genes.
A small circle carries only a few genes of its own. For example, one gene may let the bacterium survive an antibiotic, a drug that kills bacteria.
Suppose that antibiotic reaches the bacterium. Following the gene, the bacterium builds a protein that breaks the antibiotic down.
So the bacterium lives, while bacteria without that small circle die.
When a small closed circle of DNA lies in a cell apart from the chromosome like this, it is called a .
Here is the bacterium again, with its chromosome and its three plasmids named.
The cell copies a plasmid on its own, apart from the chromosome.
Here is a bacterium with one plasmid, and the same bacterium later. The plasmid has been copied twice, while the chromosome is still one.
So a cell may hold several copies of one plasmid, while it holds only one chromosome.
Plasmids are not found only in bacteria. Some eukaryotes carry plasmids too.
For example, here is a yeast cell. Its linear chromosomes lie inside a nucleus, so the yeast cell is a eukaryote.
Beside those chromosomes, inside the nucleus, lies a plasmid.
The chromosome is the one big circle, carrying thousands of genes. A plasmid is a small extra circle, carrying a few.
So the rule is this: a plasmid is a small closed circle of DNA lying apart from the chromosome.
For example, look at the small closed circle at the left end of this bacterium. This is a plasmid, because it is a small closed circle of DNA lying apart from the chromosome.
But look at the big closed circle in the middle of the same bacterium. This is not a plasmid, because it is the chromosome itself.
And look at the small closed circle inside this yeast cell’s nucleus. This is still a plasmid, because it is a small closed circle of DNA lying apart from the chromosomes.
What you are expected to know Identify a plasmid on a drawing or in a description: a small closed circle of DNA lying apart from the chromosome.
What you are expected to know Describe what a plasmid carries: a few genes of its own, copied apart from the chromosome.
The cell drawn below holds two closed loops of DNA.
Which loop is the plasmid?
- A. ✓ The smaller loop
- B. The larger loopThe large loop is the cell’s chromosome.
The plasmid is the small closed loop lying apart from it.
Why: A plasmid is a small closed circle of DNA lying apart from the chromosome.
The larger loop is the chromosome.
So the smaller loop, lying apart from it, is the plasmid.
The cell drawn below holds three closed loops of DNA.
Which loops are the plasmids?
- A. The one larger loopThe one large loop is the cell’s chromosome.
The plasmids are the small closed loops lying apart from it. - B. ✓ The two smaller loops
Why: The larger loop is the chromosome.
Each smaller closed loop lying apart from it is a plasmid.
So the two smaller loops are the plasmids.
Look at the DNA in the round cell drawn below.
How many plasmids does the cell hold?
- A. 0The loop in the middle is the chromosome.
The much smaller closed loop lying apart from it is a plasmid, so the cell holds one. - B. ✓ 1
- C. 2The loop in the middle is the chromosome, not a plasmid.
Only the much smaller closed loop is a plasmid, so the cell holds one.
Why: The loop in the middle is the chromosome.
The much smaller closed loop lies apart from it, so it is a plasmid.
The cell holds one plasmid.
Look at the DNA in the cell drawn below.
How many plasmids does the cell hold?
- A. ✓ 0
- B. 1The one large loop is the chromosome, not a plasmid.
Counting it as a plasmid gives one; the cell holds none. - C. 2Two plasmids would be two small closed loops apart from the chromosome.
There is none, so the cell holds no plasmid.
Why: The one large loop is the chromosome.
A plasmid is a small closed circle lying apart from the chromosome, and there is none.
So the cell holds no plasmid.
Look at the DNA in the cell drawn below.
How many plasmids does the cell hold?
- A. 2Two small loops sit near the left end and two near the right end: four in all.
Each small loop is a plasmid. - B. 3Two small loops sit near each end, so there are four small loops, not three.
Each small loop is a plasmid. - C. ✓ 4
Why: The large loop is the chromosome.
Each small closed loop lying apart from it is a plasmid.
There are four small loops, so the cell holds four plasmids.
The cell drawn below has a cell wall, and one small closed loop of DNA lies inside it.
Is this cell a eukaryote?
- A. ✓ Yes
- B. NoThe DNA lies as linear chromosomes inside a nucleus, so the cell is a eukaryote.
The small closed loop is a plasmid, which some eukaryotes carry.
Why: The short lines are linear chromosomes, and they lie inside a nucleus.
Linear chromosomes inside a nucleus make the cell a eukaryote.
The small closed loop is a plasmid; some eukaryotes carry plasmids.
A student looks at a bacterium with its chromosome and two plasmids and says: “The plasmids carry all of the cell’s genes.”
Is the student correct?
- A. Yes: the plasmids carry every gene the cell hasThe chromosome carries thousands of genes.
Each plasmid carries only a few. - B. ✓ No: most of the cell’s genes are on the chromosome
Why: Each plasmid carries a few genes of its own.
The chromosome carries thousands of genes.
So most of the cell’s genes are on the chromosome, not on the plasmids.
Go back to the bacterium, magnified: beside its one big circle of DNA sit three much smaller circles.
Each small circle is a plasmid: a small closed circle of DNA, lying apart from the chromosome.
Each plasmid carries a few genes of its own.
Each plasmid is copied on its own, apart from the one big chromosome.
44Quick quiz: plasmid mixed practice
A bacterium’s DNA is a chromosome and, often, some extra pieces.
What is a plasmid?
- A. A protein that DNA winds around, like thread on a beadA protein that DNA winds around is a histone, part of a eukaryotic chromosome.
A plasmid is DNA: a small closed circle lying apart from the chromosome. - B. The one big circle of DNA that carries most of the cell’s genesThe one big circle carrying most of the genes is the chromosome.
A plasmid is a small closed circle of DNA lying apart from it. - C. ✓ A small closed circle of DNA lying apart from the chromosome
Why: A plasmid is a small closed circle of DNA that lies in a cell apart from the chromosome, carrying a few genes of its own.
A bacterium’s DNA is a chromosome and, often, some extra pieces.
(a) State what a plasmid is. (1 pt)
It carries a few genes of its own.
The cell copies it on its own, apart from the chromosome.
- Award 1 point for: a small circular (closed) DNA molecule separate from the chromosome. Accept with or without: carries a few genes; copied independently of the chromosome.
E. coli’s one big closed circle of DNA, millions of base pairs long, lies in its nucleoid.
Is that circle a plasmid?
- A. YesThe one big circle, millions of base pairs long, is the chromosome.
A plasmid is a small extra circle lying apart from it. - B. ✓ No
Why: The one big circle that carries most of the cell’s genes is the chromosome.
A plasmid is a small closed circle lying apart from the chromosome.
So the big circle is not a plasmid.
A small closed circle of DNA lies in a bacterium’s cytosol, apart from the chromosome, carrying a gene for surviving an antibiotic.
Is that circle a plasmid?
- A. ✓ Yes
- B. NoA small closed circle of DNA lying apart from the chromosome is a plasmid, whatever gene it carries.
Why: The circle is small, closed, and made of DNA.
It lies apart from the chromosome.
So it is a plasmid.
Glossary
- plasmid
- A small closed circle of DNA that lies in a cell apart from the chromosome. A plasmid carries a few genes of its own and is copied on its own; plasmids are found in bacteria and in some eukaryotes, such as yeast.
APBIO-U06-L02B Instructions written in RNA
Photo: Graham Beards, Wikimedia Commons, CC BY-SA 3.0 (cropped and resized).
Now consider a particle far smaller than the bacterium, sitting on its surface. The photograph shows many of them, each with a many-sided head and a short tail, fixed to the bacterium’s cell wall. It is not a cell. It has no ribosomes and makes nothing on its own. Yet it carries instructions, and once inside a cell it gets the cell to make copies of it. Some of these particles carry their instructions not as DNA but as RNA.
What is this particle, and how can RNA hold the instructions?
Unit 6 · Gene Expression and Regulation
1A particle that is not a cell
What is this particle, and what does it carry? It is genetic material inside a protein coat, and it can reproduce only inside a cell.
It is not a cell: it has no ribosomes and makes nothing on its own.
Many of these particles carry their instructions as DNA. Many others carry their instructions as RNA.
So the rule that genetic information is stored in DNA has one exception.
You tell DNA from RNA by one base: RNA carries uracil, and DNA carries thymine.
A bacterium builds its own proteins.
Which part of the bacterium builds them?
- A. ✓ The bacterium’s ribosomes
- B. The bacterium’s nucleoidThe nucleoid is the region where the bacterium’s DNA lies.
Its ribosomes build the proteins. - C. The bacterium’s cell wallThe cell wall is the stiff layer outside the cell.
Its ribosomes build the proteins.
Why: A ribosome joins amino acids into a protein.
A bacterium’s ribosomes float in its cytosol.
So the ribosomes build the bacterium’s proteins.
Video: Watch: A particle that is not a cell
The particle on the bacterium’s surface drawn to scale, then magnified: a protein coat with genetic material inside, no ribosomes; the bacterium building copies of it.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L02Ba.mp4
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Look at the particle drawn on the bacterium’s surface, and then drawn five times larger beside it.
The particle is about 20 % of the bacterium’s length: five of them laid end to end would span the bacterium.
Its outside is a coat made of protein. Inside the coat lies its genetic material.
Now look for ribosomes inside the coat. There are none.
So the particle cannot build a single protein on its own. Outside a cell, it makes nothing.
Now suppose the particle pushes its genetic material through the bacterium’s cell wall, into the cell.
The bacterium’s ribosomes read those instructions and build the particle’s coat proteins.
The bacterium copies the particle’s genetic material too.
The new coats and the new copies of the genetic material join into hundreds of new particles.
The new particles burst out of the bacterium and settle on other bacteria.
A particle like this, genetic material inside a protein coat that reproduces only inside a cell, is called a .
A virus is not a cell.
A cell has ribosomes of its own; a virus has none.
A cell reproduces on its own, by dividing; a virus reproduces only inside a cell.
Here is a table of a cell against a virus.
Two features decide whether something is a cell or a virus: does it have ribosomes of its own, and does it reproduce on its own?
For example, here is a rod with ribosomes in its cytosol, and it divides in two on its own. This is a cell.
But here is a particle with no ribosomes, and it reproduces only inside a cell. This is a virus.
And here is a particle as large as the rod, with no ribosomes, and it reproduces only inside a cell. This is still a virus.
The size decides nothing. The ribosomes and the way it reproduces decide it.
What you are expected to know Describe what a virus is: genetic material inside a protein coat, with no ribosomes, that reproduces only inside a cell.
A particle with a protein coat sits on a bacterium’s surface. It has no ribosomes, and it reproduces only inside the bacterium.
Is the particle a cell or a virus?
- A. A cellA cell has ribosomes of its own and reproduces on its own.
No ribosomes and reproducing only inside a cell make the particle a virus. - B. ✓ A virus
Why: The particle has no ribosomes.
It reproduces only inside a cell.
Genetic material in a protein coat, with no ribosomes and reproducing only inside a cell, is a virus.
A rod about one thousandth of a millimeter long has ribosomes in its cytosol, and it divides in two on its own.
Is the rod a cell or a virus?
- A. ✓ A cell
- B. A virusA virus has no ribosomes and reproduces only inside a cell.
Ribosomes of its own and dividing on its own make the rod a cell.
Why: The rod has ribosomes of its own.
It divides on its own.
Ribosomes of its own and dividing on its own make it a cell.
Suppose a round particle is about one fiftieth of a millimeter across. It holds a nucleus and ribosomes, and it divides on its own.
Is the particle a cell or a virus?
- A. ✓ A cell
- B. A virusA virus has no ribosomes and no nucleus.
Ribosomes of its own and dividing on its own make the particle a cell.
Why: The particle has ribosomes of its own.
It divides on its own.
Ribosomes of its own and dividing on its own make it a cell.
Suppose a particle is so large that a light microscope shows it. It has no ribosomes, and it reproduces only inside a cell.
Is the particle a cell or a virus?
- A. A cellSize decides nothing.
No ribosomes and reproducing only inside a cell make the particle a virus. - B. ✓ A virus
Why: The particle has no ribosomes.
It reproduces only inside a cell.
Its size changes neither feature, so it is a virus.
A particle with a protein coat gets a cell to build copies of it. On its own, it makes nothing.
Is the particle a cell or a virus?
- A. A cellA cell builds its own proteins and reproduces on its own.
A particle that reproduces only by getting a cell to copy it is a virus. - B. ✓ A virus
Why: The particle makes nothing on its own.
It reproduces only inside a cell.
Genetic material in a protein coat, with no ribosomes and reproducing only inside a cell, is a virus.
Suppose a particle is about 20 % of a bacterium’s length. It has ribosomes of its own, and it divides in two on its own.
Is the particle a cell or a virus?
- A. ✓ A cell
- B. A virusBeing small decides nothing.
Ribosomes of its own and dividing on its own make the particle a cell.
Why: The particle has ribosomes of its own.
It divides on its own.
Its small size changes neither feature, so it is a cell.
A student says: “A virus is just a very small cell.”
Is the student correct?
- A. ✓ No: a virus has no ribosomes and reproduces only inside a cell
- B. Yes: a virus is a cell too small to hold ribosomesEvery cell, however small, has ribosomes of its own and reproduces on its own.
A virus has no ribosomes and reproduces only inside a cell.
Why: A cell has ribosomes of its own and reproduces on its own.
A virus has no ribosomes.
A virus reproduces only inside a cell.
So a virus is not a cell, however small a cell can be.
37Quick quiz: virus mixed practice
Viruses and cells are both found in a drop of pond water.
What is a virus?
- A. A very small cell with ribosomes of its own but no nucleus around its DNAA small cell with ribosomes but no nucleus is a bacterium.
A virus has no ribosomes. - B. A protein that a cell builds inside itself and then sends outA virus carries genetic material, not only protein.
It is genetic material inside a protein coat. - C. ✓ Genetic material in a protein coat that reproduces only inside a cell
Why: A virus is genetic material inside a protein coat.
It has no ribosomes.
It reproduces only inside a cell.
Viruses and cells are both found in a drop of pond water.
(a) State what a virus is. (1 pt)
It has no ribosomes and makes nothing on its own.
It reproduces only inside a cell.
- Award 1 point for: genetic material (DNA or RNA) inside a protein coat, reproducing only inside a cell. ‘Not a cell’ or ‘no ribosomes’ completes it, but its absence does not lose the point.
A particle in the pond water has ribosomes of its own.
Could the particle be a virus?
- A. YesA virus has no ribosomes.
A particle with ribosomes of its own is a cell. - B. ✓ No
Why: A virus has no ribosomes.
So a particle with ribosomes of its own is not a virus.
A particle in the pond water has no ribosomes, and it reproduces only inside a cell.
Could the particle be a virus?
- A. ✓ Yes
- B. NoNo ribosomes and reproducing only inside a cell are a virus’s two features.
The particle could be a virus.
Why: A virus has no ribosomes and reproduces only inside a cell.
The particle has both features.
So the particle could be a virus.
42Instructions in RNA
Every cell stores its genetic information in its DNA. The bacterium’s one circle and the skin cell’s 46 lines are DNA.
Many viruses store their genetic information in DNA. The tailed virus in the photograph carries DNA.
Many other viruses store their genetic information in RNA.
The influenza virus, which causes flu, carries RNA. HIV, the virus that causes AIDS, carries RNA.
So a virus’s genetic material is DNA or RNA, and you need a way to tell which.
DNA and RNA each use four bases.
Which base does RNA carry in place of thymine?
- A. AdenineAdenine is in both DNA and RNA.
RNA carries uracil in place of thymine. - B. ✓ Uracil
- C. GuanineGuanine is in both DNA and RNA.
RNA carries uracil in place of thymine.
Why: DNA’s bases are adenine, thymine, guanine and cytosine.
RNA has no thymine.
In its place RNA carries uracil.
Video: Watch: Instructions in RNA
Two viruses drawn see-through: the tailed virus from the photograph with DNA inside its coat, and an influenza virus with RNA inside its coat; the bases of each read off to tell which is which.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L02Bb.mp4
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Here is a table of DNA against RNA: the three differences between them.
DNA holds its information in the order of its bases.
An RNA strand also has bases in an order. So an RNA strand can hold instructions in the same way.
Here are two viruses drawn see-through. The tailed virus from the photograph holds DNA; the influenza virus holds RNA.
Both viruses hold their instructions in the order of their bases. Only the kind of molecule differs.
To tell which a virus carries, look at its bases.
Thymine means DNA. Uracil means RNA.
For example, suppose a virus’s genetic material contains adenine, thymine, guanine and cytosine. This is DNA, because it contains thymine.
But suppose a virus’s genetic material contains adenine, uracil, guanine and cytosine. This is RNA, because it contains uracil.
Here is a table of the two cases with their verdicts.
The sugar tells you the same thing.
Deoxyribose means DNA. Ribose means RNA.
What you are expected to know Classify a virus’s genetic material as DNA or RNA from its bases: thymine means DNA, and uracil means RNA.
The influenza virus’s genetic material contains uracil and no thymine.
Which molecule is its genetic material?
- A. DNADNA contains thymine and never uracil.
Genetic material that contains uracil is RNA. - B. ✓ RNA
Why: The genetic material contains uracil.
Uracil is RNA’s base in place of thymine.
So the influenza virus’s genetic material is RNA.
The genetic material of a virus that infects bacteria contains thymine.
Which molecule is its genetic material?
- A. ✓ DNA
- B. RNARNA carries uracil in place of thymine and never contains thymine.
Genetic material that contains thymine is DNA.
Why: The genetic material contains thymine.
Thymine is a base of DNA and never of RNA.
So the virus’s genetic material is DNA.
Suppose a newly found virus carries genetic material whose bases are adenine, guanine, cytosine and thymine.
Which molecule is its genetic material?
- A. ✓ DNA
- B. RNARNA never contains thymine.
Genetic material with thymine among its bases is DNA.
Why: The bases include thymine.
Thymine is a base of DNA and never of RNA.
So the genetic material is DNA.
The sugar in HIV’s genetic material is ribose.
Which molecule is its genetic material?
- A. DNADNA’s sugar is deoxyribose.
Genetic material whose sugar is ribose is RNA. - B. ✓ RNA
Why: The sugar is ribose.
Ribose is RNA’s sugar; DNA’s sugar is deoxyribose.
So HIV’s genetic material is RNA.
Suppose a virus’s genetic material has the bases adenine, guanine, cytosine and uracil.
Which molecule is its genetic material?
- A. DNADNA never contains uracil.
Genetic material with uracil among its bases is RNA. - B. ✓ RNA
Why: The bases include uracil.
Uracil is a base of RNA and never of DNA.
So the genetic material is RNA.
Suppose a virus’s genetic material is built on the sugar deoxyribose.
Which molecule is its genetic material?
- A. ✓ DNA
- B. RNARNA’s sugar is ribose.
Genetic material built on deoxyribose is DNA.
Why: The sugar is deoxyribose.
Deoxyribose is DNA’s sugar; RNA’s sugar is ribose.
So the genetic material is DNA.
Go back to the particle on the bacterium’s surface: a particle about 20 % of the bacterium’s length, with a protein coat and no ribosomes.
It is a virus: instructions in a protein coat, no ribosomes, reproducing only inside a cell.
Its instructions may be written in RNA. So genetic material that contains uracil can still carry a full set of instructions.
Glossary
- virus
- Genetic material (DNA or RNA) inside a protein coat, that reproduces only inside a cell. A virus is not a cell: it has no ribosomes and makes nothing on its own.
APBIO-U06-L03 Two rings with one: the pairing rules
Here are the five bases, drawn as their rings. Adenine and guanine are each made of two rings. Cytosine, thymine and uracil are one ring each. In DNA, adenine always sits opposite thymine, and guanine always sits opposite cytosine.
Why is adenine never opposite guanine?
Unit 6 · Gene Expression and Regulation
1Two rings or one
Why do the bases pair the way they do? Every rung of the DNA ladder is one two-ring base joined to one one-ring base.
Two rings plus one ring is the same width at every rung. So the two backbones stay the same distance apart along the whole molecule.
DNA and RNA are built from five kinds of nitrogenous base.
Which base does RNA carry in place of thymine?
- A. AdenineAdenine is found in both DNA and RNA.
- B. CytosineCytosine is found in both DNA and RNA.
- C. ✓ Uracil
Why: RNA has no thymine.
In its place RNA carries uracil, so RNA’s four bases are adenine, uracil, guanine and cytosine.
Adenine pairs with thymine, or with uracil when the partner strand is RNA. Guanine pairs with cytosine.
Hydrogen bonds hold each pair together. The same four pairs hold in every living organism.
These rules let one strand fix the other. The whole unit rests on them.
Video: Watch: Two rings or one
The five bases drawn as their rings; the two-ring bases and the one-ring bases sorted into two groups and named.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L03a.mp4
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Until now every base was drawn as one plain ring. Here are the five bases drawn as their real rings, with their names.
Look at adenine. Its atoms form two rings that share one edge.
Look at guanine. Its atoms also form two rings that share one edge.
Now look at cytosine. Its atoms form one ring.
Thymine is one ring too. Uracil is one ring too.
When a base is made of two rings, like adenine and guanine, it is called a .
When a base is made of one ring, like cytosine, thymine and uracil, it is called a .
The two names are easy to mix up. Here is a hook: the longer name, pyrimidine, goes with the smaller base, the one ring.
For example, take adenine. Adenine is a purine, because it has two rings.
But take cytosine. Cytosine is a pyrimidine, because it has one ring.
And take thymine. Thymine is also a pyrimidine, because it has one ring.
But take guanine. Guanine is a purine, because it has two rings.
And take uracil. Uracil is a pyrimidine, because it has one ring.
So the five bases fall into two groups. The purines are adenine and guanine, and the pyrimidines are cytosine, thymine and uracil.
What you are expected to know Classify a nitrogenous base as a purine (two rings: adenine or guanine) or a pyrimidine (one ring: cytosine, thymine or uracil), from its drawn rings or from its name.
The base drawn below is one of the five bases.
Which kind of base is it?
- A. ✓ Purine
- B. PyrimidineThe drawn base has two rings that share one edge.
A two-ring base is a purine.
Why: The drawn base has two rings.
A base with two rings is a purine.
The base drawn below is one of the five bases, drawn turned round.
Which kind of base is it?
- A. ✓ Purine
- B. PyrimidineTurning the drawing round changes nothing.
The base has two rings, so it is a purine.
Why: The drawn base has two rings, whichever way round it is drawn.
A base with two rings is a purine.
The base drawn below is one of the five bases.
Which kind of base is it?
- A. PurineThe drawn base has one ring.
A one-ring base is a pyrimidine. - B. ✓ Pyrimidine
Why: The drawn base has one ring.
A base with one ring is a pyrimidine.
Thymine is one of the five bases.
Which kind of base is thymine?
- A. PurineThymine is made of one ring.
A one-ring base is a pyrimidine. - B. ✓ Pyrimidine
Why: Thymine is made of one ring.
A base with one ring is a pyrimidine.
Adenine is one of the five bases.
Which kind of base is adenine?
- A. ✓ Purine
- B. PyrimidineAdenine is made of two rings that share one edge.
A two-ring base is a purine.
Why: Adenine is made of two rings.
A base with two rings is a purine.
Uracil is one of the five bases.
Which kind of base is uracil?
- A. PurineUracil is made of one ring.
A one-ring base is a pyrimidine. - B. ✓ Pyrimidine
Why: Uracil is made of one ring.
A base with one ring is a pyrimidine.
Cytosine is one of the five bases.
Which kind of base is cytosine?
- A. PurineCytosine is made of one ring.
A one-ring base is a pyrimidine. - B. ✓ Pyrimidine
Why: Cytosine is made of one ring.
A base with one ring is a pyrimidine.
Guanine is one of the five bases.
Which kind of base is guanine?
- A. ✓ Purine
- B. PyrimidineGuanine is made of two rings that share one edge.
A two-ring base is a purine.
Why: Guanine is made of two rings.
A base with two rings is a purine.
An RNA strand from a virus carries four kinds of base: adenine, uracil, guanine and cytosine.
How many of the four are pyrimidines?
- A. OneUracil is one ring and cytosine is one ring.
Uracil and cytosine are both pyrimidines. - B. ✓ Two
- C. ThreeAdenine and guanine are two rings each, so they are purines.
Only uracil and cytosine are pyrimidines.
Why: Uracil is one ring and cytosine is one ring: two pyrimidines.
Adenine and guanine are two rings each: two purines.
33Quick quiz: purine, pyrimidine mixed practice
The five bases fall into two groups by their rings.
What is a purine?
- A. ✓ A base made of two rings
- B. A base made of one ringA base made of one ring is a pyrimidine.
- C. A sugar with five carbonsA purine is a base, not a sugar.
The five-carbon sugars are deoxyribose and ribose.
Why: A purine is a nitrogenous base made of two rings that share one edge: adenine or guanine.
The five bases fall into two groups by their rings.
What is a pyrimidine?
- A. A base made of two ringsA base made of two rings is a purine.
- B. A phosphate groupA pyrimidine is a base, not a phosphate group.
The phosphate group is the part of a nucleotide drawn as a circle marked P. - C. ✓ A base made of one ring
Why: A pyrimidine is a nitrogenous base made of one ring: cytosine, thymine or uracil.
The five bases fall into two groups by their rings.
(a) State what a purine and a pyrimidine are, and name the bases in each group. (1 pt)
A pyrimidine is a base made of one ring: cytosine, thymine or uracil.
- Award 1 point for: a purine is a two-ring base (adenine, guanine) and a pyrimidine is a one-ring base (cytosine, thymine, uracil).
A base drawn in a textbook has two rings that share one edge.
Which kind of base is it?
- A. ✓ Purine
- B. PyrimidineTwo rings that share one edge make a purine.
A pyrimidine is one ring.
Why: A base made of two rings is a purine.
A student says: “Guanine is a pyrimidine.”
Is the student correct?
- A. Yes: guanine is made of one ringGuanine is made of two rings that share one edge.
A two-ring base is a purine. - B. ✓ No: guanine is made of two rings, so it is a purine
Why: Guanine is made of two rings.
A base with two rings is a purine, not a pyrimidine.
39Why every rung is two rings with one
In a cell, the two strands of DNA twist around each other into a double helix.
Where do the two sugar-phosphate backbones lie?
- A. ✓ On the outside, with the paired bases inside
- B. On the inside, with the bases outsideThe bases point inward and pair across the middle.
The two backbones lie on the outside.
Why: In the double helix the two sugar-phosphate backbones lie on the outside.
The paired bases lie inside, between them.
Video: Watch: Why every rung is two rings with one
Three rung widths tried between the two backbones: two purines too wide, two pyrimidines too narrow, one purine with one pyrimidine filling the rung exactly.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L03b.mp4
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Now consider one rung of the DNA ladder. A rung is one base pair: a base on one backbone and its partner on the other backbone.
The two backbones lie the same distance apart along the whole molecule. So every rung has the same width to fill.
Here are three rungs tried between the two backbones: two purines, one purine with one pyrimidine, and two pyrimidines.
Two purines are two rings plus two rings: four rings across the rung. Four rings are too wide, so the two backbones would be pushed apart.
Two pyrimidines are one ring plus one ring: two rings across the rung. Two rings are too narrow, so the two bases cannot reach each other to bond.
One purine with one pyrimidine is two rings plus one ring: three rings across the rung. Three rings fill the rung exactly.
So every rung of DNA is one purine with one pyrimidine.
Adenine with thymine is two rings plus one ring. Guanine with cytosine is two rings plus one ring.
That is why adenine never sits opposite guanine. Two purines would make a rung too wide for the backbones.
What you are expected to know Explain why every base pair is one purine with one pyrimidine: two rings plus one ring is the same width at every rung, so the two backbones stay the same distance apart.
The two backbones of DNA lie a fixed distance apart.
Which pairing fills a rung between them exactly?
- A. A purine with a purineTwo purines are four rings across the rung.
Four rings are too wide for the space between the backbones. - B. ✓ A purine with a pyrimidine
- C. A pyrimidine with a pyrimidineTwo pyrimidines are two rings across the rung.
Two rings are too narrow to reach across the space.
Why: A purine is two rings and a pyrimidine is one ring.
Together they are three rings across the rung.
Three rings fill the space between the backbones exactly.
In every DNA molecule the two backbones lie the same distance apart along the whole molecule.
(a) Explain what would happen to the two backbones if two purines faced each other across one rung. (1 pt)
Frame If two purines faced each other across one rung, …
Each purine is two rings, so two purines are four rings.
Four rings are wider than the space between the two backbones.
So the two backbones would be pushed apart, and the rung could not form.
- Award 1 point for: two purines (four rings) are wider than the fixed space between the two backbones, so the backbones would be pushed apart (the rung cannot form).
A student says: “Adenine pairs with thymine because the two bases are the same size.”
Is the student correct?
- A. ✓ No: a two-ring base pairs with a one-ring base
- B. Yes: paired bases are always the same sizeAdenine is two rings and thymine is one ring.
A rung is one two-ring base with one one-ring base.
Why: Adenine is a purine with two rings.
Thymine is a pyrimidine with one ring.
Two rings plus one ring fill the rung, so the paired bases are not the same size.
Suppose a newly described base is made of two rings.
Which kind of partner must it have to fill a rung?
- A. ✓ A one-ring base
- B. A two-ring baseTwo rings plus two rings are four rings, too wide for the rung.
A two-ring base needs a one-ring partner.
Why: The new base is two rings.
A rung is three rings across.
So its partner must be one ring.
56The pairing table
One strand of a DNA molecule carries a guanine.
Which base lies opposite it on the other strand?
- A. AdenineAdenine pairs with thymine.
- B. ✓ Cytosine
- C. ThymineThymine pairs with adenine.
Why: In DNA guanine pairs with cytosine.
So opposite a guanine lies a cytosine.
Two water molecules sit side by side. The hydrogen on one carries a small positive charge, written δ+. The oxygen on the other carries a small negative charge, written δ−. The hydrogen is attracted to that oxygen.
What is this attraction called?
- A. ✓ A hydrogen bond
- B. A covalent bondA covalent bond is a shared pair of electrons inside a molecule.
The attraction between the two molecules is a hydrogen bond.
Why: A hydrogen with a small positive charge (δ+) is attracted to an oxygen or nitrogen with a small negative charge (δ−) on a neighboring molecule.
That attraction is a hydrogen bond.
Video: Watch: The pairing table
The four pairs settling into place with their hydrogen bonds drawn as dashed lines; an RNA strand bringing uracil opposite adenine; the pairing table filled in row by row.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L03c.mp4
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Now consider the rungs of a DNA molecule one by one. Each rung is one purine with one pyrimidine, held together by hydrogen bonds.
Adenine, two rings, sits opposite thymine, one ring. Two hydrogen bonds hold the pair, the same attraction that holds one water molecule to the next.
Guanine, two rings, sits opposite cytosine, one ring. Three hydrogen bonds hold the pair.
The hydrogen bonds decide which pyrimidine a purine pairs with. Adenine’s hydrogen bonds line up with thymine’s, and guanine’s line up with cytosine’s.
Adenine’s hydrogen bonds do not line up with cytosine’s. So adenine never pairs with cytosine, and guanine never pairs with thymine.
Now suppose the partner strand is RNA. RNA has no thymine; in its place it carries uracil, one ring.
So opposite an adenine on the DNA strand, the RNA strand carries uracil. Adenine pairs with uracil, held by two hydrogen bonds.
Guanine still pairs with cytosine, whether the partner strand is DNA or RNA.
A DNA strand never carries uracil, and an RNA strand never carries thymine. So a strand with thymine in it is DNA, and a strand with uracil in it is RNA.
Here is a table of the pairs: the base, its partner, the kind of partner strand, and the hydrogen bonds holding the pair.
What you are expected to know Name the base that pairs with a given base: adenine with thymine on a DNA strand, adenine with uracil on an RNA strand, guanine with cytosine on either; each pair held by hydrogen bonds.
The base on a DNA strand is adenine. The partner strand is DNA.
Which base sits opposite the adenine?
- A. CytosineCytosine pairs with guanine.
- B. ✓ Thymine
- C. UracilUracil is RNA’s base.
The partner strand here is DNA, so it carries thymine.
Why: Adenine pairs with thymine on a DNA strand.
So the DNA partner carries thymine opposite the adenine.
The base on a DNA strand is adenine. The partner strand is RNA.
Which base sits opposite the adenine?
- A. CytosineCytosine pairs with guanine.
- B. GuanineGuanine pairs with cytosine.
- C. ✓ Uracil
Why: The partner strand is RNA, which carries uracil in place of thymine.
Adenine pairs with uracil on an RNA strand.
The base on a DNA strand is guanine. The partner strand is RNA.
Which base sits opposite the guanine?
- A. AdenineAdenine pairs with thymine or with uracil.
- B. ✓ Cytosine
- C. UracilUracil pairs with adenine.
Why: Guanine pairs with cytosine on a DNA strand and on an RNA strand.
So the RNA partner carries cytosine opposite the guanine.
The base on a DNA strand is thymine. The partner strand is DNA.
Which base sits opposite the thymine?
- A. ✓ Adenine
- B. CytosineCytosine pairs with guanine.
- C. GuanineGuanine pairs with cytosine.
Why: Thymine pairs with adenine.
So the DNA partner carries adenine opposite the thymine.
The base on an RNA strand is uracil. The partner strand is DNA.
Which base sits opposite the uracil?
- A. ✓ Adenine
- B. GuanineGuanine pairs with cytosine.
- C. ThymineThymine and uracil never pair with each other.
Each of them pairs with adenine.
Why: Uracil pairs with adenine.
So the DNA partner carries adenine opposite the uracil.
The base on a DNA strand is cytosine. The partner strand is RNA.
Which base sits opposite the cytosine?
- A. AdenineAdenine pairs with thymine or with uracil.
- B. ✓ Guanine
- C. UracilUracil pairs with adenine.
Why: Cytosine pairs with guanine on a DNA strand and on an RNA strand.
So the RNA partner carries guanine opposite the cytosine.
A nucleic-acid strand carries thymine.
Could the strand be RNA?
- A. YesAn RNA strand carries uracil in place of thymine, never thymine.
- B. ✓ No
Why: RNA has no thymine.
A strand that carries thymine is DNA.
78The same rules in every living thing
Every known cell, from bacteria to humans, builds its ribosomes the same way.
What does a feature shared by all living things show?
- A. ✓ All living things descend from shared ancestors
- B. Bacteria descend from humansBacteria were on Earth billions of years before humans.
- C. Each kind of living thing arose on its ownA feature shared by every living thing was inherited from ancestors they share.
Why: A feature shared by all living things is evidence that all living things descend from shared ancestors: their common ancestry.
Video: Watch: The same rules in every living thing
The same four pairs in a bacterium, an oak tree and a human; a cell whose pairing changed, unable to copy its DNA and leaving no descendants.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L03d.mp4
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Now consider three living things: a bacterium, an oak tree and you.
In the bacterium, adenine pairs with thymine and guanine pairs with cytosine.
In the oak tree, adenine pairs with thymine and guanine pairs with cytosine.
In you, adenine pairs with thymine and guanine pairs with cytosine.
Here is a table of the pairs in the three living things.
Every organism ever examined pairs its bases this way. Every virus examined does too.
All living things descend from shared ancestors: their common ancestry. The first cells paired adenine with thymine and guanine with cytosine.
Every living thing since has inherited that pairing from the first cells.
Now imagine one cell in which the pairing changed, so that adenine no longer paired with thymine.
Each base would no longer have one partner. So one strand could no longer fix the other.
The cell could not copy its DNA. So it left no descendants, and the change died with it.
So the pairing never changed. It has been kept the same in every living thing since the first cells.
When a feature has been kept the same through evolution like this, in every descendant of the ancestor that first had it, it is called , because conserved means kept.
Base pairing is conserved in every living thing, because the first cells had it. So is the rule that every rung is two rings with one.
What you are expected to know Explain what it means that base pairing is conserved through evolution: every organism and virus pairs its bases the same way, because all inherited the pairing from a common ancestor and a change would stop the DNA being copied.
Base pairing is conserved through evolution.
Which of the following does conserved mean here?
- A. The pairing rules have changed a little in every generation since the first cellsA conserved feature has stayed the same in every descendant of the ancestor that first had it.
- B. ✓ The pairing rules have been kept the same in every living thing since the first cells
- C. Each kind of living thing has worked out its own pairing rulesEvery kind of living thing examined pairs its bases the same way.
Why: Conserved means kept the same.
Base pairing has been kept the same in every living thing since the first cells.
Every organism examined pairs adenine with thymine and guanine with cytosine.
(a) Explain what would happen to a cell in which adenine stopped pairing with thymine. (1 pt)
Frame The cell would …
One partner each is what lets one strand fix the other.
So one strand could no longer fix the other strand.
The cell could not copy its DNA.
So the cell would leave no descendants, and the change would die with it.
- Award 1 point for: with adenine’s partner changed, one strand no longer fixes the other, so the DNA cannot be copied and the cell leaves no descendants. Accept an answer that begins from rung width (adenine’s new partner would not fill the rung); accept ‘the cell could not copy its DNA, so it could not divide’ as the consequence.
A student says: “If one kind of organism changed its base-pairing rules, its descendants would simply inherit the new rules.”
Is the student correct?
- A. ✓ No: a cell whose pairing rules changed could not copy its DNA, so it would leave no descendants
- B. Yes: descendants inherit whatever pairing rules their ancestor had, changed or notOne partner for each base is what lets one strand fix the other.
A cell whose pairing changed could not copy its DNA, so it would leave no descendants.
Why: Base pairing is what lets one strand fix the other, so the DNA can be copied.
A cell whose pairing rules changed could not copy its DNA.
So it would leave no descendants, and the change would die with it.
A microbe is newly found in a hot spring.
Predict which base pairs with adenine in the microbe’s DNA.
- A. CytosineCytosine pairs with guanine in every organism.
- B. GuanineGuanine pairs with cytosine in every organism.
- C. ✓ Thymine
Why: Base pairing is conserved through evolution.
Every organism pairs adenine with thymine.
So the microbe’s adenine pairs with thymine.
Go back to the five bases drawn as their rings: adenine and guanine with two rings each, and cytosine, thymine and uracil with one ring each.
Two rings plus one ring make every rung the same width. Adenine pairs with thymine, or with uracil on an RNA strand, and guanine pairs with cytosine, each pair held by hydrogen bonds.
A bacterium, an oak tree and you pair them the same way.
103Quick quiz: conserved (through evolution) mixed practice
Biologists describe some features as conserved through evolution.
What does it mean that a feature is conserved through evolution?
- A. ✓ The feature has been kept the same in every descendant of the ancestor that first had it
- B. The feature is found only in the oldest living things, and the newest kinds have lost itA conserved feature is found in every descendant of the ancestor that first had it, the newest kinds as well as the oldest.
- C. The feature has changed a little with every generation since the ancestor that first had itA conserved feature has stayed the same in every descendant of the ancestor that first had it.
Why: Conserved means kept.
A feature conserved through evolution has been kept the same in every descendant of the ancestor that first had it.
Base pairing has been kept in every living thing since the first cells.
A feature can be conserved through evolution.
(a) State what it means for a feature to be conserved through evolution. (1 pt)
Each descendant inherited it.
Base pairing is conserved in every living thing, because the first cells had it.
- Award 1 point for: the feature has been kept the same (unchanged) through evolution because it was inherited from a common ancestor; ‘in every living thing since the first cells’ completes it for base pairing, but its absence does not lose the point.
Adenine pairs with thymine in a bacterium, in an oak tree and in a human.
Is base pairing conserved through evolution?
- A. ✓ Yes
- B. NoThe pairing is the same in every living thing, so it has been kept the same since the first cells.
That is what conserved through evolution means.
Why: The pairing is the same in a bacterium, an oak tree and a human.
A feature kept the same in every living thing is conserved through evolution.
Bird beaks differ in shape from one kind of bird to another.
Is beak shape conserved through evolution?
- A. YesA conserved feature is kept the same in every descendant of the ancestor that first had it.
Beak shape differs from one kind of bird to another. - B. ✓ No
Why: Beak shape differs from one kind of bird to another.
A feature that differs is not kept the same, so it is not conserved through evolution.
108Mixed practice mixed practice
The base on an RNA strand is cytosine. The partner strand is DNA.
Which base sits opposite the cytosine?
- A. AdenineAdenine pairs with thymine or with uracil.
- B. ✓ Guanine
- C. ThymineThymine pairs with adenine.
Why: Cytosine pairs with guanine on a DNA strand and on an RNA strand.
So the DNA partner carries guanine opposite the cytosine.
The base drawn below is one of the five bases, drawn on its side.
Which kind of base is it?
- A. ✓ Purine
- B. PyrimidineThe drawn base has two rings that share one edge.
A two-ring base is a purine.
Why: The drawn base has two rings.
A base with two rings is a purine.
A fungus is newly found in a cave.
Predict which base pairs with thymine in the fungus’s DNA.
- A. ✓ Adenine
- B. CytosineCytosine pairs with guanine in every organism.
- C. GuanineGuanine pairs with cytosine in every organism.
Why: Base pairing is conserved through evolution.
Every organism pairs thymine with adenine.
So the fungus’s thymine pairs with adenine.
Suppose two one-ring bases faced each other across one rung of DNA.
Which of the following would happen?
- A. The two bases would fill the rung exactlyTwo one-ring bases are two rings across the rung.
A rung needs three rings to be filled. - B. The two backbones would be pushed apartTwo rings across the rung are too narrow, not too wide.
Four rings would push the backbones apart. - C. ✓ The two bases could not reach each other to bond
Why: Two one-ring bases are two rings across the rung.
Two rings are narrower than the space between the backbones.
So the two bases could not reach each other to bond.
A DNA strand carries adenine, thymine, guanine and cytosine.
Which of its bases are purines?
- A. ✓ Adenine and guanine
- B. Thymine and cytosineThymine and cytosine are one ring each.
A one-ring base is a pyrimidine.
Why: Adenine and guanine are two rings each.
A base with two rings is a purine.
So the strand’s purines are adenine and guanine.
A nucleic-acid strand carries uracil.
Could the strand be DNA?
- A. YesA DNA strand carries thymine, never uracil.
- B. ✓ No
Why: DNA has no uracil.
A strand that carries uracil is RNA.
A student says: “Guanine pairs with cytosine because both bases have two rings.”
Is the student correct?
- A. ✓ No: cytosine is made of one ring
- B. Yes: paired bases have the same number of ringsCytosine is made of one ring.
A rung is one two-ring base with one one-ring base.
Why: Guanine is two rings and cytosine is one ring.
Two rings plus one ring fill the rung.
So the paired bases do not have the same number of rings.
As a reminder, here are the five bases again, drawn as their rings. A microbe newly found in a hot spring is examined. Its DNA pairs adenine with thymine and guanine with cytosine, and every rung is the same width.
(a) Explain how the microbe’s base pairing shows that base pairing is conserved through evolution. (1 pt)
Frame The microbe’s pairing shows this because …
A bacterium, an oak tree and a human pair their bases the same way.
All of them inherited the pairing from a common ancestor.
So the pairing has been kept the same in every living thing, which is what conserved through evolution means.
- Award 1 point for: the microbe pairs its bases exactly as every other organism does, so the pairing has been kept the same (inherited unchanged from a common ancestor).
(b) Explain how pairing one purine with one pyrimidine keeps every rung of the microbe’s DNA the same width. (1 pt)
Frame Every rung is the same width because …
A purine is two rings and a pyrimidine is one ring.
So every rung is three rings across.
Three rings fill the space between the two backbones exactly, at every rung.
- Award 1 point for: each rung is a two-ring base with a one-ring base, three rings across, so every rung has the same width.
Glossary
- purine
- A nitrogenous base made of two rings that share one edge: adenine or guanine. Every rung of DNA is one purine paired with one pyrimidine.
- pyrimidine
- A nitrogenous base made of one ring: cytosine, thymine or uracil. Every rung of DNA is one pyrimidine paired with one purine.
- conserved (through evolution)
- Kept the same through evolution, in every descendant of the ancestor that first had it, because each inherited it. Base pairing is conserved in every living thing: every organism and virus pairs adenine with thymine (or with uracil on an RNA strand) and guanine with cytosine.
APBIO-U06-L04 Use the rules
A laboratory reports one strand of a DNA sample: 5′-ATGCCTAAG-3′. Nothing else is reported. Can you write the other strand?
Now suppose the laboratory reported only “adenine 28 %”. Could you fill in the other three bases?
Unit 6 · Gene Expression and Regulation
1Write the partner strand
What can one strand tell you about the other? It tells you the whole of it.
Each base has one partner. So one strand fixes the other strand, base by base.
Each pair holds one A for one T and one G for one C. So one percentage fixes the other three.
A molecule whose two halves each carry the whole message can be pulled apart and rebuilt exactly. That is why DNA can be copied and inherited.
A strand of DNA has two different ends.
Which end carries the free phosphate group?
- A. ✓ The 5′ end
- B. The 3′ endThe 3′ end carries a free hydroxyl group on the last sugar’s carbon 3.
The free phosphate group sits at the 5′ end, on carbon 5.
Why: The last sugar’s carbon 5 carries a phosphate group joined to nothing further.
That free phosphate group marks the 5′ end.
Biologists write a strand of DNA as a row of letters.
From which end do they start?
- A. ✓ From its 5′ end
- B. From its 3′ endThe 3′ end is where a strand is finished, not started.
A strand is written from its 5′ end. - C. From either endThe two ends differ, so a strand read from the other end gives a different order of bases.
A strand is written from its 5′ end.
Why: The two ends of a strand differ, so the strand has a direction.
A strand is written from its 5′ end to its 3′ end.
The two ends of a strand differ, so a strand has a direction. A strand is written from its 5′ end to its 3′ end, and both ends are marked.
Here is the laboratory’s strand drawn flat, from its 5′ end at the left to its 3′ end at the right. A base hangs from every sugar.
Two paired strands of DNA lie side by side.
Where one strand has its 5′ end, which end does the other strand have?
- A. Its 5′ endPaired strands are antiparallel.
So where one strand has its 5′ end, the other has its 3′ end. - B. ✓ Its 3′ end
Why: Paired strands are antiparallel: they lie side by side and point in opposite directions.
So one strand’s 5′ end lies beside the other strand’s 3′ end.
The partner strand lies against the laboratory’s strand and runs the other way. So the partner’s 3′ end sits at the left, and its 5′ end sits at the right.
Here is the pairing table again. In DNA, adenine pairs with thymine, and guanine pairs with cytosine.
Take the laboratory’s strand one base at a time. Under each base, write its partner from the table.
Here is the partner written under the given strand. Its ends are marked the other way round: 3′ at the left and 5′ at the right.
Read the partner from its own 5′ end, at the right, and it is 5′-CTTAGGCAT-3′. Both rows name the same strand: one written under its partner, one written on its own.
Here are the two strands drawn flat. Hydrogen bonds join every base to its partner.
One strand of DNA carries an adenine. Its partner strand is RNA.
Which base sits opposite the adenine?
- A. ThymineThymine is a base of DNA.
An RNA strand carries uracil in its place, and uracil pairs with adenine. - B. GuanineGuanine pairs with cytosine.
Opposite adenine on an RNA strand sits uracil. - C. ✓ Uracil
Why: RNA carries uracil in place of thymine.
So on an RNA partner strand, uracil sits opposite adenine.
Video: Watch: Write the partner strand
The laboratory’s strand written out; under each base its partner appears, one at a time; the ends of the partner marked the other way round; then the same partner written from its own 5′ end, and the RNA partner with U in place of T.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L04a.mp4
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Now suppose the partner strand is RNA. RNA carries uracil in place of thymine, so under each adenine the partner carries uracil.
Here is the laboratory’s strand with an RNA partner written under it: 3′-UACGGAUUC-5′. Under G still sits C, and under C still sits G.
What you are expected to know Write the partner strand for a given DNA strand, antiparallel, with both ends marked, choosing T or U by the kind of partner strand asked for.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. One strand of DNA reads 5′-GCATTC-3′. Its DNA partner is written under it, with one base missing.
Which base is missing?
- A. AdenineThe gap sits under an adenine.
The partner strand carries adenine’s partner, thymine, not a copy of adenine. - B. ✓ Thymine
- C. UracilUracil belongs to RNA.
This partner strand is DNA, so under adenine sits thymine.
Why: The gap sits under the A of 5′-GCATTC-3′.
In DNA, adenine pairs with thymine.
So the missing base is thymine, and the partner reads 3′-CGTAAG-5′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. One strand of DNA reads 5′-TTGACG-3′.
Written under it, with its ends marked 3′ at the left and 5′ at the right, what does the DNA partner strand read?
- A. 3′-TTGACG-5′Copying the given strand gives the same bases, not each base’s partner.
Under T sits A, under G sits C. - B. 3′-CGTCAA-5′These letters are the partner strand reversed.
Under the given strand the partner reads 3′-AACTGC-5′. - C. ✓ 3′-AACTGC-5′
- D. 3′-AACUGC-5′Uracil belongs to RNA.
This partner strand is DNA, so under A sits T.
Why: Under each base write its partner: T gives A, T gives A, G gives C, A gives T, C gives G, G gives C.
The partner runs the other way, so its ends are 3′ at the left and 5′ at the right.
It reads 3′-AACTGC-5′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. One strand of DNA reads 5′-CAGTTA-3′.
Written from its own 5′ end, what does the DNA partner strand read?
- A. ✓ 5′-TAACTG-3′
- B. 5′-GTCAAT-3′These are the partners written left to right under the given strand.
That starts from the partner’s 3′ end, and a strand is written from its 5′ end. - C. 5′-CAGTTA-3′Copying the given strand gives the same bases, not each base’s partner.
Under C sits G, under A sits T. - D. 5′-ATTGAC-3′Reversing the given strand gives its own bases backward, not each base’s partner.
Under C sits G, under A sits T.
Why: Under each base write its partner: C gives G, A gives T, G gives C, T gives A, T gives A, A gives T.
Under the given strand that reads 3′-GTCAAT-5′.
The partner’s 5′ end is at the right, so from its own 5′ end it reads 5′-TAACTG-3′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. One strand of DNA reads 5′-GGATCA-3′.
Written from its own 5′ end, what does the DNA partner strand read?
- A. 5′-CCTAGT-3′These are the partners written left to right under the given strand.
That starts from the partner’s 3′ end, and a strand is written from its 5′ end. - B. 5′-GGATCA-3′Copying the given strand gives the same bases, not each base’s partner.
Under G sits C, under A sits T. - C. 5′-ACTAGG-3′Reversing the given strand gives its own bases backward, not each base’s partner.
Under G sits C, under A sits T. - D. ✓ 5′-TGATCC-3′
Why: Under each base write its partner: G gives C, G gives C, A gives T, T gives A, C gives G, A gives T.
Under the given strand that reads 3′-CCTAGT-5′.
The partner’s 5′ end is at the right, so from its own 5′ end it reads 5′-TGATCC-3′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. One strand of DNA reads 5′-TACGGA-3′. An RNA strand pairs with it along its whole length.
Written under the DNA strand, with its ends marked 3′ at the left and 5′ at the right, what does the RNA strand read?
- A. 3′-ATGCCT-5′Thymine belongs to DNA.
The partner strand is RNA, so under adenine sits uracil. - B. ✓ 3′-AUGCCU-5′
- C. 3′-UACGGA-5′Copying the DNA strand with U for T gives the same bases, not each base’s partner.
Under T sits A, under A sits U. - D. 3′-UCCGUA-5′These letters are the RNA strand reversed.
Under the DNA strand the RNA reads 3′-AUGCCU-5′.
Why: The partner is RNA, so under each adenine write uracil.
T gives A, A gives U, C gives G, G gives C, G gives C, A gives U.
Under the DNA strand the RNA reads 3′-AUGCCU-5′.
27One percentage fixes the other three
In every sample of double-stranded DNA, the amount of adenine equals the amount of thymine.
Why does the amount of adenine equal the amount of thymine?
- A. Adenine and thymine are the same sizeSize has nothing to do with the count.
Every adenine is paired one-to-one with a thymine, so the two counts match. - B. Each strand carries equal numbers of the four basesA single strand can carry any mix of bases.
Every adenine on one strand is paired with a thymine on the other, so across both strands the two counts match. - C. ✓ Every adenine is paired with a thymine on the other strand
Why: Every adenine on one strand is paired with a thymine on the other strand.
So there is one thymine for every adenine, and the two amounts are equal.
Video: Watch: One percentage fixes the other three
A bar for the four bases with only adenine 28 % filled; thymine fills to match; the rest of the bar splits equally between guanine and cytosine; the working written beside the bar.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L04b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L04b.mp4
Now suppose the laboratory reports one number only: its double-stranded sample is 28 % adenine. Every adenine is paired with a thymine, so thymine is 28 % too.
The four percentages add up to 100 %, because every base is one of the four. Here is that as an equation.
the four percentages add to 100 %
Every guanine is paired with a cytosine. So guanine and cytosine share what is left of the 100 %, equally.
Here is each pair’s rule as an equation. Thymine matches adenine, and guanine and cytosine share equally what adenine and thymine leave.
from one percentage, the other three
A double-stranded DNA sample is 28 % adenine. Calculate the percentage of each of the other three bases.
Here are the four percentages as one bar: adenine 28 %, thymine 28 %, guanine 22 % and cytosine 22 %.
What you are expected to know Calculate the other three base percentages of a double-stranded DNA sample from one given percentage, using A = T, G = C and a total of 100 %.
A double-stranded DNA sample is 31 % thymine.
Calculate the percentage of guanine.
Part 1. Every T is paired with an A. State the percentage of adenine, %A.
Answer: 31 % (tolerance ±0)
Part 2. Calculate the percentage left for guanine and cytosine together, %G + %C.
Answer: 38 % (tolerance ±0)
Answer: 19 % (tolerance ±0)
38Practice: one base’s percentage to its partner’s mixed practice
A double-stranded DNA sample is 21 % guanine.
Calculate the percentage of cytosine.
Answer: 21 % (tolerance ±0)
A double-stranded DNA sample is 34 % thymine.
Calculate the percentage of adenine.
Answer: 34 % (tolerance ±0)
A double-stranded DNA sample is 17 % cytosine.
Calculate the percentage of guanine.
Answer: 17 % (tolerance ±0)
A double-stranded DNA sample is 36 % adenine.
Calculate the percentage of thymine.
Answer: 36 % (tolerance ±0)
A double-stranded DNA sample is 23 % thymine.
Calculate the percentage of adenine.
Answer: 23 % (tolerance ±0)
A double-stranded DNA sample is 18 % cytosine.
Calculate the percentage of adenine.
Answer: 32 % (tolerance ±0)
45Two strands, one strand, or RNA?
A nucleic acid sample contains uracil.
Which nucleic acid is it?
- A. DNADNA’s four bases are adenine, thymine, guanine and cytosine.
Uracil is found in RNA and never in DNA. - B. ✓ RNA
Why: RNA carries uracil in place of thymine.
DNA never contains uracil, so a sample with uracil is RNA.
A nucleic acid is a single strand with no partner strand.
Must its amount of adenine equal its amount of thymine?
- A. YesEqual amounts come from one-to-one pairing across two strands.
A single strand has no partner, so its bases are paired with nothing. - B. ✓ No
Why: A single strand has no partner strand, so its bases are paired with nothing.
Nothing forces its amount of adenine to match its amount of thymine.
Video: Watch: Two strands, one strand, or RNA?
Three small tables of base percentages sorted one at a time: the uracil test first, then A against T, then G against C; each table lands in its row of the three-kinds table.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L04c.mp4
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A laboratory can measure the percentage of each base in a sample. Those four numbers tell you which kind of sample it is.
Here is a table of the three kinds of sample and the three tests that sort them: is uracil present, does A equal T, and does G equal C.
For example, here is a sample: adenine 30 %, thymine 30 %, guanine 20 %, cytosine 20 %. This is double-stranded DNA, because it has no uracil, A equals T and G equals C.
But here is a sample: adenine 31 %, thymine 19 %, guanine 25 %, cytosine 25 %. This is single-stranded DNA, because A does not equal T.
For a DNA sample, both tests must pass before it is double-stranded: A equals T and G equals C. If either pair is unequal, the sample is a single strand.
And here is a sample: adenine 30 %, uracil 20 %, guanine 25 %, cytosine 25 %. This is RNA, because it carries uracil.
Test for uracil first. A sample with uracil is RNA, whatever its percentages.
What you are expected to know Determine from a table of base percentages whether a sample is double-stranded DNA, single-stranded DNA or RNA.
The base percentages of a nucleic acid sample are in the table below.
Which kind of sample is it?
- A. ✓ Double-stranded DNA
- B. Single-stranded DNAAdenine matches thymine at 22 %.
Guanine matches cytosine at 28 %.
Matching amounts are what two paired strands give. - C. RNAThe sample has thymine and no uracil, so it is DNA.
Why: The sample has no uracil, so it is DNA.
Adenine matches thymine at 22 %.
Guanine matches cytosine at 28 %.
Matching amounts come from two paired strands: double-stranded DNA.
The base percentages of a nucleic acid sample are in the table below.
Which kind of sample is it?
- A. Double-stranded DNAAdenine is 27 % but thymine is 21 %.
In two paired strands every A has a T, so the two would be equal. - B. ✓ Single-stranded DNA
- C. RNAThe sample has thymine and no uracil, so it is DNA.
Why: The sample has no uracil, so it is DNA.
Adenine is 27 % and thymine is 21 %, so A does not equal T.
Unequal amounts can only come from a single strand: single-stranded DNA.
The base percentages of a nucleic acid sample are in the table below.
Which kind of sample is it?
- A. Double-stranded DNAThe sample contains uracil, and DNA never does.
- B. Single-stranded DNAThe sample contains uracil, and DNA never does.
- C. ✓ RNA
Why: The sample contains uracil.
Uracil is found in RNA and never in DNA, so the sample is RNA.
The base percentages of a nucleic acid sample are in the table below.
Which kind of sample is it?
- A. Double-stranded DNAA equals T, but guanine is 19 % and cytosine is 29 %.
In two paired strands every G has a C, so the two would be equal. - B. ✓ Single-stranded DNA
- C. RNAThe sample has thymine and no uracil, so it is DNA.
Why: The sample has no uracil, so it is DNA.
Guanine is 19 % and cytosine is 29 %, so G does not equal C.
Unequal amounts can only come from a single strand: single-stranded DNA.
The base percentages of a nucleic acid sample are in the table below.
Which kind of sample is it?
- A. Double-stranded DNAThe sample contains uracil, and DNA never does.
The uracil test comes before the matching test. - B. Single-stranded DNAThe sample contains uracil, and DNA never does.
- C. ✓ RNA
Why: The sample contains uracil, so it is RNA.
The uracil test comes first: a sample with uracil is RNA, whatever its percentages.
A laboratory measures the base percentages of a virus particle’s genetic material. They are in the table below.
Which kind of sample is it?
- A. Double-stranded DNAThe sample contains uracil, and DNA never does.
- B. Single-stranded DNAThe sample contains uracil, and DNA never does.
- C. ✓ RNA
Why: The virus’s genetic material contains uracil.
Uracil is found in RNA and never in DNA, so this virus carries its genetic information as RNA.
63Why DNA can be copied
The two strands of a DNA molecule are held together along their whole length.
Which bonds hold them together?
- A. Covalent bonds between the two backbonesThe two backbones never touch.
The strands hold together only where their bases pair, by hydrogen bonds. - B. ✓ Hydrogen bonds between the paired bases
Why: Between the two strands there are only hydrogen bonds, one set at every base pair.
The covalent bonds join the nucleotides along each backbone.
Video: Watch: Why DNA can be copied
The two strands of the laboratory’s sample part along their hydrogen bonds; loose nucleotides pair onto each strand one at a time; two identical molecules stand where one stood; a wrong base tried in one place and left sticking out as a mismatch.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L04d.mp4
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Now imagine the laboratory heats its DNA sample until the hydrogen bonds break and the two strands part.
Each strand comes away whole. Each strand still carries all of its bases in order.
Look at the first strand, 5′-ATGCCTAAG-3′. Each of its bases has one partner, so it fixes what a new partner strand must be: 3′-TACGGATTC-5′.
Look at the second strand, 3′-TACGGATTC-5′. It fixes a new partner in the same way: 5′-ATGCCTAAG-3′.
Suppose loose nucleotides pair onto each strand and a new partner grows against it. There are now two molecules, and each is identical to the one the laboratory started with.
So one DNA molecule can become two exact copies, because each strand carries everything needed to rebuild the other. A copy of the instructions can go to each new cell.
A molecule that carries the instructions from one cell to the next, and from parent to offspring, is called the . DNA is the hereditary material.
Now suppose a wrong base lands in a new strand: a guanine opposite an adenine.
Guanine does not pair with adenine, because their hydrogen bonds do not line up. So the two do not fit together.
The mistake shows as a pair that does not match.
What you are expected to know Explain why specific base pairing lets DNA serve as hereditary material: each strand carries the complete information needed to rebuild the other, so the molecule can be copied exactly.
Suppose heat parts the two strands of a DNA molecule. Loose nucleotides then pair onto each strand.
Which of the two strands can direct the building of a new partner strand?
- A. ✓ Both strands
- B. Only one of the two strandsEach base on either strand has one partner.
So each strand fixes a new partner, base by base. - C. Neither strandEach base has one partner, so a parted strand fixes what pairs onto it.
Both strands direct a new partner.
Why: Each base has one partner.
So each parted strand fixes, base by base, the new strand that pairs onto it.
Both strands direct a new partner.
Suppose heat parts the two strands of a DNA molecule. One strand reads 5′-TCGAAT-3′. Loose nucleotides then pair onto each strand, and a new partner strand grows against each.
(a) Explain how the base-pairing rules let each separated strand direct the building of a partner identical to the one it lost. (1 pt)
Frame Each strand can direct a new partner because …
Opposite each base on the separated strand, only that partner can pair.
So the order of bases on the separated strand fixes the order of bases on the new strand.
The old partner was fixed by the same rules, so the new partner has the same order of bases as the old one.
- Award 1 point for: each base pairs with only one partner, so the order of bases on the separated strand fixes the order on the new strand, which is therefore the same as the lost partner’s.
A student looks at the two separated strands and says: “Only one of the two strands holds the information; the other strand is just a copy of it.”
Is the student correct?
- A. ✓ No: each strand fixes the other, so each strand holds the whole message
- B. Yes: one strand holds the information and the other is a copyThe two strands are partners, not copies: each base faces its partner base.
Either strand fixes the other base by base, so each holds the whole message.
Why: Each base has one partner, so the order of bases on either strand fixes the order on the other.
Either strand alone can direct a new partner.
So each strand holds the whole message.
Suppose, as a new partner strand grows, a cytosine lands opposite an adenine.
How does the wrong base show?
- A. As a break in the sugar-phosphate backboneThe backbone is joined by covalent bonds and is not broken by a wrong base.
A wrong base shows as a pair that does not fit. - B. ✓ As a pair whose two bases do not fit each other
- C. No sign shows; any base pairs with any other baseAdenine pairs only with thymine, or with uracil on an RNA strand.
Adenine’s hydrogen bonds do not line up with cytosine’s, so the pair does not fit.
Why: Adenine’s hydrogen bonds line up with thymine’s, not with cytosine’s.
So cytosine opposite adenine does not fit, though the rung is the right width.
The wrong base shows as a mismatched pair.
Go back to the laboratory’s report of one strand, 5′-ATGCCTAAG-3′, with nothing else reported.
The other strand is 3′-TACGGATTC-5′. Each base on the reported strand has one partner, so the reported strand fixes the other.
From adenine 28 % come thymine 28 %, and guanine and cytosine 22 % each.
Each strand carries everything needed to rebuild its partner. A molecule like that can be copied.
85Mixed practice mixed practice
The base percentages of a nucleic acid sample are in the table below.
Which kind of sample is it?
- A. ✓ Double-stranded DNA
- B. Single-stranded DNAAdenine matches thymine at 19 %.
Guanine matches cytosine at 31 %.
Matching amounts are what two paired strands give. - C. RNAThe sample has thymine and no uracil, so it is DNA.
Why: The sample has no uracil, so it is DNA.
Adenine matches thymine at 19 %.
Guanine matches cytosine at 31 %.
Matching amounts come from two paired strands: double-stranded DNA.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. One strand of DNA reads 5′-GATTC-3′.
Written from its own 5′ end, what does the DNA partner strand read?
- A. 5′-CTAAG-3′These are the partners written left to right under the given strand.
That starts from the partner’s 3′ end, and a strand is written from its 5′ end. - B. ✓ 5′-GAATC-3′
- C. 5′-GATTC-3′Copying the given strand gives the same bases, not each base’s partner.
Under G sits C, under A sits T.
Why: Under each base write its partner: G gives C, A gives T, T gives A, T gives A, C gives G.
Under the given strand that reads 3′-CTAAG-5′.
The partner’s 5′ end is at the right, so from its own 5′ end it reads 5′-GAATC-3′.
A double-stranded DNA sample is 21 % thymine.
Calculate the percentage of cytosine.
Answer: 29 % (tolerance ±0)
A cell copies its DNA before it divides, and each new cell receives an exact copy.
What lets a DNA molecule be copied exactly?
- A. The two strands are identical to each otherThe two strands are partners, not copies: under A sits T, under G sits C.
Each strand fixes the other, and that is what lets the molecule be copied. - B. The sugar-phosphate backbone carries the informationThe backbone repeats unchanged along every strand.
The information is in the order of the bases, and each strand’s order fixes its partner’s. - C. ✓ Each strand carries everything needed to rebuild the other
Why: Each base has one partner, so each strand fixes the other base by base.
When the strands part, each directs a new partner identical to the one it lost.
So one molecule becomes two exact copies.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. One strand of DNA reads 5′-CCGAT-3′. An RNA strand pairs with it along its whole length.
Written under the DNA strand, with its ends marked 3′ at the left and 5′ at the right, what does the RNA strand read?
- A. 3′-GGCTA-5′Thymine belongs to DNA.
The partner strand is RNA, so under adenine sits uracil. - B. ✓ 3′-GGCUA-5′
- C. 3′-AUCGG-5′These letters are the RNA strand reversed.
Under the DNA strand the RNA reads 3′-GGCUA-5′.
Why: The partner is RNA, so under each adenine write uracil.
C gives G, C gives G, G gives C, A gives U, T gives A.
Under the DNA strand the RNA reads 3′-GGCUA-5′.
The base percentages of a nucleic acid sample are in the table below.
Which kind of sample is it?
- A. Double-stranded DNAThe sample contains uracil, and DNA never does.
- B. Single-stranded DNAThe sample contains uracil, and DNA never does.
- C. ✓ RNA
Why: The sample contains uracil.
Uracil is found in RNA and never in DNA, so the sample is RNA.
A double-stranded DNA sample is 33 % guanine.
Calculate the percentage of cytosine.
Answer: 33 % (tolerance ±0)
A laboratory heats a short piece of double-stranded DNA until its two strands part. It then adds loose nucleotides, and a new partner strand grows against each separated strand. The drawing shows one separated strand, 5′-ACGTTG-3′, with the new strand that grew against it. In the new strand, a thymine sits opposite a guanine.
(a) Explain how this experiment demonstrates that specific base pairing lets DNA serve as hereditary material. (1 pt)
The strand that grows against each separated strand is therefore identical to the partner it lost.
So the one molecule becomes two molecules with the same order of bases.
Exact copies like these can be handed to new cells, which is what hereditary material must allow.
- Award 1 point for: specific pairing means each separated strand fixes its new partner’s base order, so the two molecules produced are exact copies of the original, which is what hereditary material must allow.
(b) Explain how the thymine opposite the guanine shows itself as a mistake. (1 pt)
Guanine’s hydrogen bonds do not line up with thymine’s.
So the thymine and the guanine do not fit together, though the rung is the right width.
The pair does not match, and the mistake shows as a mismatched pair.
- Award 1 point for: thymine is not guanine’s partner (its hydrogen bonds do not line up with guanine’s), so the two do not pair properly and the error shows as a mismatched (non-fitting) pair. Accept with or without the hydrogen bonds; accept ‘the rung is the right width, but the bases do not pair’.
Glossary
- hereditary material
- The molecule that carries a cell's instructions from one cell to the next, and from parent to offspring. DNA is the hereditary material: each strand fixes the other, so the molecule can be copied exactly and handed on.
APBIO-U06-P61 Practice questions: Topic 6.1
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 one calculation 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: DNA and RNA structure, summed up
Where the DNA sits in the two kinds of cell; plasmids, and viruses whose genetic material is RNA; a purine with a pyrimidine keeps every rung the same width; A–T, A–U and G–C in every organism; the partner strand, the percentages, and why pairing is what makes DNA copyable.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-T61-summary.mp4
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Suppose a cell divides into two new cells before it has copied its DNA.
Which statement about the two new cells is correct?
- A. Each new cell holds a complete set, because the instructions are in the cell's proteins tooProteins are built from the instructions; they do not carry them.
Without a copy of the DNA, one set of instructions cannot serve two cells. - B. Each new cell holds a complete set, because RNA carries a second copy of the instructionsA cell's genetic information is stored in its DNA.
Without a copy of the DNA, one set cannot serve two cells. - C. ✓ Between them the two new cells hold one set of instructions, so one lacks part of it
- D. Neither new cell is harmed, because a cell needs its DNA only when it makes gametesEvery cell reads its DNA to build its proteins, all its life.
A cell lacking part of its instructions cannot build every protein it needs.
Why: A cell's instructions are in its DNA, one set per cell.
Copying the DNA before division gives each new cell a complete set.
Without the copy, one set is shared between two cells, so at least one lacks part of it.
A cell is half a millimeter long, hundreds of times the length of a typical bacterium. Its DNA is many copies of one closed circle lying in its cytosol, with few proteins attached.
Which of the following best identifies the cell type and the evidence for it?
- A. ✓ Prokaryotic: its DNA is a closed circle in the cytosol
- B. Eukaryotic: it is far larger than a bacteriumSize decides nothing.
A closed circle of DNA lying in the cytosol, with no nucleus, is a prokaryotic cell's chromosome set. - C. Eukaryotic: it holds many DNA moleculesThe many molecules are copies of one circular chromosome, not linear chromosomes in a nucleus.
A circle in the cytosol makes the cell prokaryotic. - D. Prokaryotic: it is a single cellBeing one cell decides nothing: a yeast is one cell and is eukaryotic.
The closed circle of DNA lying in the cytosol shows it.
Why: The cell's DNA is a closed circle: a circular chromosome.
It lies in the cytosol, so no nucleus wraps it.
A circular chromosome lying in the cytosol is a prokaryotic cell's chromosome set, whatever the cell's size.
The drawing shows a round cell with its DNA. One structure is labeled X.
Which of the following best identifies structure X?
- A. A piece broken off the chromosomeA plasmid is its own closed circle of DNA, separate from the chromosome and copied on its own.
Nothing has broken off: X is a closed circle. - B. A virus that has entered the cellA virus is genetic material in a protein coat.
X is a bare closed loop of DNA lying apart from the chromosome: a plasmid. - C. A cluster of histone proteinsHistones are the proteins a eukaryotic chromosome winds on.
This cell has one circular chromosome and no histone beads; X is a plasmid. - D. ✓ A plasmid
Why: The large loop is the cell's one circular chromosome.
X is a small closed loop of DNA lying apart from it.
A small closed circle of DNA lying apart from the chromosome is a plasmid.
A student looks at a particle with no nucleus and says: “It has no nucleus, so it is a virus.”
Which statement about the student's claim is correct?
- A. The student is right: every particle without a nucleus is a virusA bacterium has no nucleus and is a cell.
What shows a virus is that it reproduces only inside a cell. - B. ✓ The student is wrong: a bacterium has no nucleus and is a cell
- C. The student is right: a nucleus is what lets a cell reproduce on its ownA bacterium reproduces on its own with no nucleus.
Ribosomes and its own chemistry let a cell reproduce, not a nucleus. - D. The student is wrong: every virus has a nucleus of its ownA virus is genetic material in a protein coat, with no nucleus and no ribosomes.
Having no nucleus does not make a particle a virus, because bacteria have none either.
Why: A bacterium has no nucleus, and it is a cell: it has ribosomes and reproduces on its own.
So lacking a nucleus does not show that a particle is a virus.
A virus is shown by reproducing only inside a cell.
The student is wrong.
Which statement about the genetic material of viruses is correct?
- A. ✓ Many viruses carry DNA, and many others carry RNA
- B. Every virus carries RNAMany viruses carry their genetic information as DNA; a virus that infects bacteria is one.
Many others carry RNA. - C. Every virus carries DNA, as every cell doesEvery cell's genetic material is DNA, but many viruses carry RNA; influenza virus is one.
Viruses are the exception. - D. Every virus carries both DNA and RNA, one inside the otherA virus carries one kind of genetic material, DNA or RNA, inside its protein coat.
Many viruses carry DNA and many others carry RNA.
Why: Every cell's genetic material is DNA.
Among viruses, many carry DNA and many others carry RNA.
So the only correct statement is that many carry DNA and many others RNA.
The drawing shows two bases facing each other between the two backbones of a DNA molecule, at one rung.
Which statement about this rung is correct?
- A. Both bases are pyrimidines, so the rung is too narrowEach base is drawn as two fused rings, so each is a purine.
Two purines make a rung too wide. - B. The two bases have different numbers of ringsBoth bases are drawn with two rings.
Two two-ring bases make a rung too wide. - C. Any two bases can share a rung, so this rung is a normal oneA rung is one purine with one pyrimidine, three rings across.
Two purines are four rings, too wide: each base is pushed past its backbone rail. - D. ✓ Both bases are purines, so the rung is too wide
Why: Each base is two fused rings: a purine.
A rung is one purine with one pyrimidine, three rings across.
Two purines are four rings, so this rung is too wide: the bases push past the backbone rails.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. An RNA strand reads 5′-GAUC-3′, as drawn. It has paired with a DNA strand along its whole length.
Which of the following is the complementary DNA sequence, written 3′ to 5′?
- A. 3′-CUAG-5′A DNA strand never carries uracil.
Opposite adenine on a DNA strand sits thymine. - B. 3′-GAUC-5′These letters copy the RNA strand; the DNA strand carries each base's partner.
Under G sits C, and under A sits T. - C. ✓ 3′-CTAG-5′
- D. 3′-GATC-5′These letters copy the RNA strand with T for U, not its partners.
Under G sits C, under A sits T, under U sits A, under C sits G.
Why: Under each RNA base write its DNA partner: G takes C, A takes T, U takes A, C takes G.
The DNA strand runs the other way, so written 3′ to 5′ it follows the RNA strand's order.
It reads 3′-CTAG-5′.
Every organism ever examined pairs adenine with thymine and guanine with cytosine.
Which of the following explains why the pairing rules have stayed the same in every living thing?
- A. Every organism lives in conditions that favor these two pairs of bases over any other pairingOrganisms live in every kind of condition, from hot springs to ice.
All organisms share the rules because all inherited them. - B. ✓ A cell whose pairing rules changed could not copy its DNA and would leave no descendants
- C. Organisms exchange DNA with one another often enough to keep the rules the same everywhereExchanging DNA between organisms is rare, and it would spread a rule, not keep one.
The rules are the same because every descendant inherited them from the first cells. - D. Each organism works out the pairing rules afresh as it develops from a single cellThe rules are inherited in the chemistry of the bases, not worked out.
Every organism received them from its ancestors.
Why: Base pairing is what lets each strand fix its partner, so the DNA can be copied.
A cell whose pairing changed could not copy its DNA and would leave no descendants.
So every descendant of the first cells kept the same rules: base pairing is conserved through evolution.
Heat parts the two strands of a DNA molecule, and a new partner grows against each. A student says: “Each old strand needs its lost partner nearby, to show which bases to add.”
Which statement about the student's claim is correct?
- A. The student is right: without the lost partner nearby, the new bases are added by chanceNothing about the new bases is chance.
Each base of the old strand has one partner, so the old strand alone fixes every new base. - B. The student is right: the lost partner is copied first and then swapped inNothing is swapped in.
Loose nucleotides pair onto the old strand one by one, each fixed by the base it faces. - C. The student is wrong: the two old strands read each other to rebuild the partnersThe two old strands have parted and each works alone.
Each base has one partner, so each old strand fixes its own new partner. - D. ✓ The student is wrong: each base has one partner, so an old strand alone fixes its new partner
Why: Each base has one partner: A with T, G with C.
So each base of an old strand fixes the base that pairs onto it.
The old strand alone rebuilds a partner identical to the one it lost.
The student is wrong.
About two meters of DNA fit inside a nucleus about a hundredth of a millimeter across.
Which of the following is the first stage of the packing that makes this possible?
- A. ✓ The DNA winds on histone beads
- B. The chromosome coils into a short thick rodCoiling into a rod is the last stage, just before division.
The first stage is winding the DNA on histone beads. - C. The DNA joins its two ends into a circleA closed circle is a prokaryotic chromosome's shape, and it packs nothing.
A eukaryotic chromosome's DNA first winds on histone beads. - D. The DNA is cut into 46 short piecesThe 46 chromosomes are 46 separate molecules from the start; nothing is cut.
Each one's DNA first winds on histone beads.
Why: A eukaryotic chromosome's DNA first winds around clusters of histones, like thread on beads.
The beaded string then coils on itself.
Further coiling condenses the chromosome for division.
So winding on histone beads is the first stage.
A laboratory reports the base percentages of a double-stranded DNA sample from a soil bacterium. Cytosine is 14 %; no other base is reported.
(a) State the percentage of guanine in the sample. (1 pt)
Frame Guanine is … %, because …
Hint Look at the pairing rule for cytosine.
Answer: 14 % (tolerance ±0)
- Award 1 point for: 14 %.
(b) Calculate the percentage left for adenine and thymine together. (1 pt)
Frame Adenine and thymine together are … %.
Hint Use what the four percentages add up to.
Answer: 72 % (tolerance ±0)
- Award 1 point for: 72 %.
- Accept a value that follows correctly from the student's own answer to (a) (100 % minus twice that answer).
(c) Calculate the percentage of adenine. (1 pt)
Frame Adenine is … %.
Hint Compare the amounts of adenine and thymine in double-stranded DNA.
Answer: 36 % (tolerance ±0)
- Award 1 point for: 36 %.
- Accept a value that follows correctly from the student's own answer to (b) (half of that answer).
(d) Predict the percentage of uracil the laboratory would find in the sample, and justify your prediction. (1 pt)
Frame Uracil would be … %, because …
Hint Think about which of the two nucleic acids the sample is.
- Award 1 point for: 0 % AND the reason (DNA never carries uracil; uracil is RNA's base).
(e) A second sample from the same laboratory reads adenine 38 %, thymine 38 %, guanine 12 %, cytosine 12 %. Determine whether this sample is consistent with double-stranded DNA, and state what your decision rests on. (1 pt)
Frame The sample is consistent with …, because …
Hint Check each pair of bases, and check which base marks the sample as DNA.
- Award 1 point for: the decision (consistent with double-stranded DNA; accept "is double-stranded DNA") AND what it rests on (A = T and G = C, with no uracil).
Slip Deciding from the matching pairs alone. Matching amounts with uracil in place of thymine would be RNA; the thymine is part of the ground.
A soil bacterium holds two kinds of DNA molecule. The table shows what a laboratory finds for each.
(a) Identify which molecule is the plasmid. (1 pt)
- Award 1 point for: molecule 2.
(b) Describe two features in the table that identify it as the plasmid. (1 pt)
It is present in about 15 copies while the chromosome is present once, so it is copied on its own.
It carries one gene of its own, not the thousands the chromosome carries.
- Award 1 point for: two of — much smaller than the chromosome; many copies per cell (copied on its own, apart from the chromosome); carries a gene of its own (one, for resistance) while the chromosome carries the cell's genes.
Slip Naming the shape. Both molecules are closed circles, so shape does not separate them.
(c) The bacterium copies its chromosome and then divides, before it makes any more copies of molecule 2. Predict how many copies of molecule 1 and about how many copies of molecule 2 each daughter cell receives. (1 pt)
- Award 1 point for: one copy of molecule 1 AND about half of the plasmid copies (about 7 or 8; accept "about half of 15" or "several").
Slip Predicting about 15 copies of molecule 2 for each daughter. The stem says no more copies are made before the division, so the 15 copies are shared between the two daughters.
(d) Justify your prediction. (1 pt)
Molecule 2 is a plasmid, copied on its own, apart from the chromosome, to about 15 copies.
No more copies are made before the division, so the two daughters share the 15 copies.
Each daughter therefore receives about half of them.
- Award 1 point for: the chromosome is copied once and one copy goes to each new cell, while the 15 plasmid copies are not copied again before the division and are shared between the two daughters, so each gets about half.
APBIO-U06-T61 End-of-topic test: DNA and RNA Structure
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.
A mother's skin cell, one of her egg cells and a skin cell of her child all carry the same order of bases in one gene.
Which of the following explains why the three cells carry the same order?
- A. The child's cells rebuilt the gene's bases from the proteins the egg carriedProteins are built from the instructions; they do not carry them.
The egg carried the mother's DNA, copied. - B. ✓ The mother's cells copied their DNA, and the egg cell carried a copy to the child
- C. The egg cell carried the gene as RNA, and the child's cells turned the RNA back into DNAA cell's genetic material is DNA, and the egg cell carries DNA.
The order was handed on as DNA, not turned into DNA. - D. The same order of bases arises on its own in every human cellAn order of bases is not fixed by chance.
The child's gene matches because it was copied from the mother's DNA.
Why: A cell's genetic information is the order of bases in its DNA.
The mother's cells copied that DNA, and the egg cell received a copy.
The child's cells grew from that egg by copying the DNA again.
So the same order of bases is found in all three cells.
The drawing shows a cell with a thick cell wall.
Which of the following best identifies the cell type and the evidence for it?
- A. Eukaryotic: it has a cell wallA cell wall decides nothing: some eukaryotes have one too.
The DNA is one closed circle lying free, so the cell is prokaryotic. - B. Eukaryotic: it holds more than one chromosomeThe two small loops are plasmids, not chromosomes.
The cell's one chromosome is the closed circle in the cytosol, so the cell is prokaryotic. - C. Prokaryotic: it has a cell wallThe cell is prokaryotic, but the cell wall is no sign of it.
Its one closed circle of DNA lying free in the cytosol shows it. - D. ✓ Prokaryotic: its DNA is one closed circle in the cytosol
Why: The large loop is one DNA molecule joined into a closed circle.
No nucleus wraps it: it lies in the cytosol.
One circular chromosome in the cytosol is a prokaryotic cell's chromosome set.
The cell wall and the two small loops, which are plasmids, decide nothing.
A eukaryotic chromosome's DNA is packed in stages.
Which of the following lists the stages in order, from the bare double helix to the condensed chromosome?
- A. ✓ Double helix → wound on histone beads → beaded string coiled → condensed chromosome
- B. Double helix → beaded string coiled → wound on histone beads → condensed chromosomeThe DNA winds on the beads first.
Only a beaded string can then coil. - C. Double helix → wound on histone beads → condensed chromosome → beaded string coiledThe beaded string coils before the chromosome condenses.
The condensed chromosome is the last stage. - D. Wound on histone beads → double helix → condensed chromosome → beaded string coiledThe bare double helix is the least packed of all.
Winding on beads comes after it, and the condensed chromosome comes last.
Why: The bare double helix is the least packed.
It winds on histone beads, like thread on beads.
The beaded string coils on itself.
Further coiling makes the short thick condensed chromosome, the most packed.
A bacterium's plasmid carries a gene for resistance to an antibiotic. When the bacterium divides, one daughter cell receives no plasmid.
Which statement about that daughter cell is correct?
- A. It has lost most of its genesMost of a bacterium's genes are on its chromosome, which the daughter cell has.
The plasmid carried only a few genes. - B. It cannot divide againThe chromosome carries the genes the cell needs to live and divide.
Losing the plasmid loses only the plasmid's few genes. - C. ✓ It keeps every chromosome gene but is no longer resistant to the antibiotic
- D. It is still resistant, because a plasmid's genes are also on the chromosomeA plasmid's genes are its own, apart from the chromosome.
Without the plasmid, the gene for resistance is gone.
Why: A plasmid is a small closed circle of DNA apart from the chromosome, carrying a few genes of its own.
The chromosome carries most of the cell's genes.
So the daughter cell without the plasmid keeps every chromosome gene and loses only the resistance gene.
Pond water holds both cells and virus particles.
Which of the following observations would show that a particle from it is a virus rather than a cell?
- A. It is made partly of proteinEvery cell is made partly of protein too.
Protein alone shows nothing. - B. It is smaller than a bacteriumSize decides nothing.
A virus is known by what it cannot do: reproduce on its own. - C. ✓ It reproduces only inside a cell
- D. It contains a nucleic acidEvery cell contains nucleic acid too.
Nucleic acid is found in cells and viruses alike.
Why: A virus is genetic material in a protein coat, with no ribosomes.
It builds nothing on its own.
So it reproduces only inside a cell, which a cell never needs to do.
A protein coat, a small size and a nucleic acid are found in cells too.
Cells and viruses both carry genetic material.
Which of the following is the genetic material of some viruses and of no cell?
- A. ✓ A single strand built on ribose, with the bases adenine, uracil, guanine and cytosine
- B. A double strand built on deoxyribose, with the bases adenine, thymine, guanine and cytosineDouble-stranded DNA is every cell's genetic material.
Many viruses carry DNA too, so this sample could be either's. - C. One closed circle of DNA lying free in the cytosolA closed circle of DNA lying free is a prokaryotic cell's chromosome.
It could be a cell's. - D. Several linear molecules of DNA inside a nucleusLinear DNA molecules inside a nucleus are a eukaryotic cell's chromosomes.
They could be a cell's.
Why: Every cell's genetic material is DNA.
Some viruses carry their genetic information as RNA.
A strand built on ribose with uracil is RNA.
So only that sample could be a virus's and not a cell's.
A double-stranded DNA molecule is 3,000 base pairs long, so it holds 6,000 bases.
What percentage of its 6,000 bases are pyrimidines?
- A. 0 %Cytosine and thymine are one-ring bases, pyrimidines, and DNA carries both.
Every rung holds one pyrimidine. - B. 25 %One kind of base in four is not the count.
Every rung holds one purine and one pyrimidine, so half of all bases are pyrimidines. - C. ✓ 50 %
- D. It depends on the order of bases in the moleculeThe order sets which pyrimidine sits where, not how many there are.
Every rung holds exactly one pyrimidine, whatever the order.
Why: Every rung is one purine paired with one pyrimidine.
The molecule has 3,000 rungs, so it holds 3,000 pyrimidines.
3,000 of 6,000 bases is 50 %.
In every DNA molecule, each rung is one purine paired with one pyrimidine.
Which of the following follows from this rule?
- A. Each strand carries equal numbers of the four basesA single strand can carry any mix of bases.
The rule fixes the width of each rung, not the count of each base. - B. ✓ The two backbones stay the same distance apart along the molecule
- C. The two strands carry the same order of basesEach strand carries the partners of the other's bases, not the same bases.
The rule fixes the width of each rung. - D. Every rung is held by the same number of hydrogen bondsAn A–T rung has two hydrogen bonds and a G–C rung three.
The rule fixes the width, not the bond count.
Why: A purine is two rings and a pyrimidine one ring.
So every rung is three rings across, the same width.
The two backbones therefore stay the same distance apart along the whole molecule.
An RNA strand lies against a DNA strand, base to base.
Which of the following pairs could form between them?
- A. ✓ Adenine on the DNA strand with uracil on the RNA strand
- B. Thymine on the DNA strand with uracil on the RNA strandThymine and uracil are both one-ring bases, and a rung needs two rings with one.
Thymine on DNA pairs with adenine on RNA. - C. Guanine on the DNA strand with adenine on the RNA strandGuanine and adenine are both two-ring bases, too wide for a rung.
Guanine pairs with cytosine. - D. Cytosine on the DNA strand with uracil on the RNA strandCytosine pairs with guanine, on DNA or on RNA.
Uracil pairs with adenine.
Why: Adenine pairs with thymine on a DNA partner and with uracil on an RNA partner.
Guanine pairs with cytosine on either.
So adenine on the DNA strand takes uracil on the RNA strand.
Biologists describe base pairing as conserved through evolution.
Which of the following findings would contradict that description?
- A. A fungus whose DNA is 40 % adenine while an insect's is 25 %The share of each base differs from one organism to another.
The pairing rules are what is conserved, not the amounts. - B. ✓ A fungus whose DNA pairs adenine with guanine
- C. A bacterium whose chromosome is a closed circle while an insect's are linesThe shape of the chromosome differs between prokaryotes and eukaryotes.
The pairing rules are the same in both. - D. An insect whose DNA has more guanine–cytosine rungs than a bacterium'sHow many rungs are G–C differs from one organism to another.
Guanine still pairs with cytosine in both.
Why: Conserved through evolution means kept the same in every descendant of the ancestor that first had it.
Every organism examined pairs adenine with thymine and guanine with cytosine.
A fungus pairing adenine with guanine would break that rule.
Differing amounts, shapes and G–C counts leave the rule intact.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. One strand of DNA reads 5′-GTACCG-3′, as drawn. An RNA strand pairs with it along its whole length.
Written from its own 5′ end, what does the RNA strand read?
- A. 5′-CAUGGC-3′These letters are each base's partner in the given strand's order, so they run 3′ to 5′.
Written from its own 5′ end, the RNA strand reads 5′-CGGUAC-3′. - B. 5′-CGGTAC-3′An RNA strand never carries thymine.
Opposite each adenine of the DNA strand sits uracil. - C. ✓ 5′-CGGUAC-3′
- D. 5′-GCCAUG-3′These letters are the given strand reversed with U for T, not its partners.
Each base needs its partner: G takes C, T takes A, A takes U.
Why: Under 5′-GTACCG-3′ the partners are C, A, U, G, G, C, running 3′ to 5′.
So under the strand the RNA reads 3′-CAUGGC-5′.
Written from its own 5′ end, it reads 5′-CGGUAC-3′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. One strand of DNA reads 5′-AGGCTT-3′, as drawn.
Written under it, with its ends marked 3′ at the left and 5′ at the right, what does the DNA partner strand read?
- A. 3′-AGGCTT-5′These letters copy the given strand; the partner carries each base's partner.
Under A sits T, and under G sits C. - B. 3′-AAGCCT-5′These letters are the partners read from the given strand's 3′ end.
Written under the strand, the partners follow the given strand's order: 3′-TCCGAA-5′. - C. 3′-UCCGAA-5′A DNA strand never carries uracil.
Opposite adenine on a DNA partner sits thymine. - D. ✓ 3′-TCCGAA-5′
Why: Under each base write its DNA partner: A takes T, G takes C, C takes G, T takes A.
The partner strand runs the other way, so under the strand its ends read 3′ at the left and 5′ at the right.
It reads 3′-TCCGAA-5′.
A double-stranded DNA sample is 24 % cytosine.
Calculate the percentage of adenine in the sample.
- A. 24 %This is cytosine's share repeated.
Cytosine equals guanine; adenine comes from what is left. - B. ✓ 26 %
- C. 52 %This is the share left for adenine and thymine together.
They share it equally, so adenine is half of it. - D. 76 %This is everything except cytosine.
Guanine also equals 24 %, and the rest is shared between adenine and thymine.
Why: Cytosine equals guanine, so guanine is 24 %.
The four bases add to 100 %, so adenine and thymine together are %.
Adenine equals thymine, so adenine is %.
The table shows the base percentages of a sample from a leaf and a sample from a virus.
Which statement about the two samples is correct?
- A. Both samples are double-stranded DNA, because in each the paired amounts matchThe virus sample contains uracil, and DNA never does.
A sample with uracil is RNA, whatever its amounts. - B. The virus sample is single-stranded DNA, because it has no thymineA DNA strand always carries thymine and never uracil.
The uracil shows that the virus sample is RNA. - C. The leaf sample is single-stranded DNA, because guanine and cytosine are only 15 % eachA single strand shows as A unequal to T or G unequal to C.
In the leaf sample A = T and G = C. - D. ✓ The virus sample is RNA, because it contains uracil
Why: The leaf sample has thymine and no uracil, with A = T and G = C: double-stranded DNA.
The virus sample contains uracil.
Uracil is found in RNA and never in DNA.
So the virus sample is RNA, whatever its amounts.
A student measures the base percentages of a nucleic acid sample, shown in the table, and says: “The measurement must be wrong, because adenine always equals thymine.”
Which statement about the student's claim is correct?
- A. The student is right: every DNA sample has equal adenine and thymineAdenine equals thymine only when every adenine is paired with a thymine on a partner strand.
A single strand has no partner, so its amounts can differ. - B. The student is wrong: adenine equals thymine only in eukaryotic DNAThe pairing rules are the same in every organism.
What breaks A = T here is that the sample is a single strand. - C. ✓ The student is wrong: in a single strand of DNA the amounts need not match
- D. The student is right: guanine does not equal cytosine either, so the measurement is wrong twice overIn a single strand of DNA neither pair need match.
Two unequal pairs mean the sample is a single strand, not that the measurement is faulty.
Why: Adenine equals thymine when every adenine is paired with a thymine on the other strand.
A single strand has no partner, so its amounts need not match.
Adenine 37 % against thymine 13 %, guanine 28 % against cytosine 22 %: the sample is a single strand.
A bacterium's chromosome and a human chromosome are compared.
Which of the following correctly describes how the two chromosomes are packed?
- A. ✓ The human chromosome winds on histone beads, and the bacterium's chromosome does not
- B. Both chromosomes wind on beads of protein, as every chromosome doesHistone beads are a eukaryotic chromosome's packing.
A bacterium's circular chromosome has few proteins attached and no beads. - C. The bacterium's chromosome has no proteins of any kind attached, and the human chromosome has histonesA few proteins are attached to a bacterium's chromosome.
What it lacks is the histone beads that a human chromosome winds on. - D. Histones join the two ends of the bacterium's circle, and the human chromosome's two ends stay freeThe circle is closed because the DNA molecule's two ends are joined to each other.
No histones are involved; a human chromosome is linear, so its ends are free.
Why: A human chromosome's DNA winds on histone beads.
A bacterium's chromosome is one closed circle with few proteins attached and no histone beads.
So only the human chromosome winds on histone beads.
Researchers examining yeast cells find, inside the nucleus, a small closed circle of DNA that lies apart from the chromosomes and is copied on its own.
Is the molecule a plasmid?
- A. No: plasmids are found only in bacteriaPlasmids are found in prokaryotes and in some eukaryotes, yeast among them.
A small closed circle of DNA apart from the chromosomes is a plasmid. - B. Yes: any closed circle of DNA in a cell is a plasmidA bacterium's chromosome is a closed circle too, and it is not a plasmid.
This molecule lies apart from the chromosomes and is copied on its own. - C. No: a closed circle of DNA lying apart from the chromosomes is a virus that has entered the cellA virus is genetic material inside a protein coat.
A bare closed circle of DNA copied on its own inside the cell is a plasmid. - D. ✓ Yes: some eukaryotes, yeast among them, carry plasmids
Why: The molecule is a small closed circle of DNA apart from the chromosomes, copied on its own.
That is a plasmid.
Plasmids are found in prokaryotes and in some eukaryotes, such as yeast.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. A virus's genetic material is one single strand of DNA. A short stretch of it reads 5′-GTTACG-3′, as drawn. Inside a cell, loose DNA nucleotides pair onto the strand, and a partner strand made of DNA nucleotides grows against it along its whole length.
(a) Determine the sequence of the partner strand, written from its own 5′ end. (1 pt)
Written from its own 5′ end, the partner strand reads 5′-CGTAAC-3′.
- Award 1 point for: 5′-CGTAAC-3′.
- Accept 3′-CAATGC-5′ written under the given strand with both ends marked.
Slip Writing 5′-CAATGC-3′: the partners in the given strand's order, which run 3′ to 5′, not from the partner's own 5′ end.
(b) Explain why the double-stranded molecule can now be copied exactly, base for base. (1 pt)
So each strand fixes the whole order of bases on the other.
When the strands part, each directs a new partner identical to the one it lost.
So one molecule becomes two molecules with the same order of bases.
- Award 1 point for: each base has one partner, so each strand fixes its partner base by base and a separated strand rebuilds a partner identical to the one it lost (the copy carries the same order of bases).
Slip Saying the two strands are identical. They are partners, not copies; each carries what is needed to rebuild the other.
(c) Suppose instead that any base could pair with any other base. Predict what would happen to the order of bases when the molecule was copied. (1 pt)
The order of bases would change with every copy, so the information would be lost.
- Award 1 point for: the new strands would differ from the original (the order of bases would change at random / the information would be lost).
Slip Predicting that no partner strand would form. Bases would still pair; which base sat where would no longer be fixed.
(d) Justify your prediction. (1 pt)
With specific pairing, the old strand fixes each new base, so the instructions are copied exactly, and a wrong base shows as a pair that does not fit.
Without specific pairing, nothing fixes the new bases and nothing marks a wrong one.
So the instructions would change with every copy.
- Award 1 point for: without specific pairing the old strand no longer fixes the new strand's bases (and a wrong base no longer shows as a misfit), so the information cannot be copied exactly and handed on unchanged.
Slip Restating the prediction. The point is the mechanism: specific pairing is what makes the old strand fix the new one.
A laboratory measures the base percentages of double-stranded DNA from three distantly related organisms: a soil bacterium, a flowering plant and a fish. The table shows the results.
(a) Identify the relationship between the base percentages that holds in each organism. (1 pt)
- Award 1 point for: adenine equals thymine AND guanine equals cytosine in each organism.
(b) Explain how base pairing in double-stranded DNA accounts for this relationship. (1 pt)
So across the two strands there is one thymine for every adenine and one cytosine for every guanine.
So the two amounts in each pair are equal.
- Award 1 point for: each adenine is paired with a thymine and each guanine with a cytosine on the other strand, so the paired bases are present in equal amounts.
Slip Saying adenine and thymine are the same size. Size is not the reason; one-to-one pairing is.
(c) Support the claim that the base-pairing rules are conserved through evolution, using the data in the table. (1 pt)
Yet in all three, adenine equals thymine and guanine equals cytosine.
The same rules in organisms so distantly related show that each inherited the rules from their common ancestor and kept them.
So base pairing is conserved through evolution.
- Award 1 point for: evidence AND reasoning — the same A = T and G = C relationship holds in all three distantly related organisms although their overall amounts differ, so the pairing rules were inherited from a common ancestor and kept the same (conserved).
Slip Quoting the equal pairs without the reasoning. Support needs the evidence and what it shows: shared rules across distant lines show inheritance from a common ancestor.
(d) A fourth sample, a nucleic acid sample taken from the fish's cells, reads adenine 33 %, uracil 25 %, guanine 22 %, cytosine 20 %. Explain what this composition shows about the sample. (1 pt)
Adenine 33 % against uracil 25 %, and guanine 22 % against cytosine 20 %: neither pair matches.
In a double-stranded molecule every base is paired with its partner, so the amounts in each pair would match.
They do not, so the sample is a single strand of RNA.
- Award 1 point for: the sample is single-stranded RNA — RNA because it carries uracil (and no thymine), and a single strand because adenine does not equal uracil (or guanine does not equal cytosine): in a double strand every base is paired, so the amounts would match.
Slip Calling the sample DNA with a faulty measurement. Uracil marks RNA; and in a single strand the amounts need not match, so nothing is wrong with the measurement.
APBIO-U06-L05 Half old, half new
Here is a cell about to divide. Before it divides, every one of its DNA molecules is copied. Here is one of those molecules, both strands drawn dark. After copying there are two molecules.
Which parts of the two are the old dark strands, and which parts are new?
Unit 6 · Gene Expression and Regulation
1Copied once, before the cell divides
A cell is in interphase, the long stretch between one division and the next.
In which stage of interphase does the cell copy its DNA?
- A. G1G1 is the first growth gap.
The cell copies its DNA in the stage after G1. - B. ✓ S phase
- C. G2G2 is the second growth gap.
G2 begins after the copying is finished.
Why: S phase is the copying stage of interphase.
The S is short for synthesis, the making of new DNA.
A cell holds 12 DNA molecules. In S phase it carries out DNA replication.
How many DNA molecules does the cell hold afterwards?
- A. 6DNA replication adds copies; it does not remove molecules.
Each of the 12 molecules becomes two, so the cell holds 24. - B. 12Every one of the 12 molecules is copied.
Each molecule becomes two, so the cell holds 24. - C. ✓ 24
Why: DNA replication copies every DNA molecule.
Each of the 12 molecules becomes two identical molecules.
So the cell holds 24 DNA molecules.
A cell is about to go through meiosis: two divisions in a row.
How many times does it copy its DNA before the two divisions?
- A. Not at allThe four cells that meiosis makes each need a complete set of DNA.
So the cell copies its DNA once, before meiosis I. - B. ✓ Once, before the first division
- C. Twice, once before each divisionNothing is copied between meiosis I and meiosis II.
The one copying happens before meiosis I.
Why: A cell entering meiosis copies its DNA once, in S phase, before meiosis I.
Meiosis II follows with no copying between the two divisions.
When a DNA molecule is copied, what is old and what is new? Each strand serves as a pattern for a new partner strand.
Each base on a strand pairs with only one partner base. So the old strand fixes every base of its new partner.
So every daughter molecule is one old strand paired with one new strand: half old, half new.
The copying happens once, in S phase, before the cell divides. So each new cell gets a complete set.
Follow two rounds of copying in a drawing, and you can say, for any round, how many molecules there are and how many still carry an old strand.
Suppose a skin cell is about to divide. In S phase, before it divides, it copies the DNA molecule of every one of its 46 chromosomes.
Copying a DNA molecule like this is called DNA replication. Each DNA molecule becomes two identical molecules.
Then the cell divides. Each daughter cell receives one complete copy of every DNA molecule: a complete genome.
So no daughter cell is missing an instruction. That is why the copying comes before the division, every time.
Video: Watch: Copied once, before the cell divides
The cell about to divide; its DNA copied in S phase, every molecule becoming two; the division, each daughter cell receiving a complete set; a cell entering meiosis copying once and dividing twice.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L05a.mp4
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A cell entering meiosis copies its DNA once too, in S phase, and then divides twice. So there as well, the copying happens once and comes first.
Every new cell begins with a complete copy of the DNA. So the hereditary information continues unbroken from cell to cell, and from parent to offspring through the gametes.
What you are expected to know Name when a cell copies its DNA: once, in S phase, before it divides.
What you are expected to know Explain why the copying comes before the division: each daughter cell must receive a complete genome.
A cell divides at midnight.
When did the cell copy its DNA?
- A. ✓ Before midnight, in S phase
- B. At midnight, while it dividedDuring the division the cell shares out DNA that is already copied.
The copying happened earlier, in S phase. - C. After midnight, in each daughter cellEach daughter cell needs a complete genome as soon as it forms.
So the copying happened before the division.
Why: A cell copies its DNA in S phase, before it divides.
So the copying happened before midnight.
Now imagine a cell skips S phase and divides anyway.
Which of the following describes the DNA each daughter cell receives?
- A. ✓ Only part of one set
- B. One complete setThe DNA was not copied, so there is only one set to share between two cells.
Each daughter cell gets only part of it. - C. Two complete setsTwo complete sets would need the DNA copied twice.
This cell did not copy its DNA at all.
Why: The cell skipped S phase, so its DNA was not copied.
One set of DNA is shared between two daughter cells.
So each daughter cell receives only part of one set.
21Half old, half new
A cell copies one of its DNA molecules.
Why can one DNA molecule be copied exactly?
- A. The two strands are identical copiesThe two strands are partners, not copies: where one has adenine, the other has thymine.
Each strand can rebuild the other. - B. The backbone holds the informationThe information is in the order of the bases, not in the backbone.
Each strand’s bases can rebuild the other strand. - C. ✓ Each strand can rebuild the other
Why: Each base pairs with only one partner base.
So each strand carries everything needed to rebuild the other.
That is why one molecule can be copied exactly.
Two DNA strands pair base to base all the way along.
What are the two strands called?
- A. IdenticalIdentical strands would have the same bases at each position.
Paired strands have partner bases at each position. - B. ParallelPaired strands lie side by side, but the word for strands that pair all the way along is complementary.
- C. ✓ Complementary
Why: Two strands whose bases pair all the way along are complementary.
Video: Watch: Half old, half new
The all-dark molecule parting; nucleotides pairing onto each dark strand one at a time and a light strand growing against it; two molecules standing where one stood, each half dark and half light; the two results that do not happen fading out.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L05b.mp4
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Here is one DNA molecule with both strands drawn dark. Dark marks a strand that was there before the copying: an old strand.
In every drawing here, old strands are drawn dark and new strands are drawn light.
Now the copying begins. The two dark strands come apart, base pair after base pair.
Free nucleotides float in the nucleus. The cell builds a new strand against each dark strand, one nucleotide at a time, from those free nucleotides.
Each new strand is drawn light.
Which nucleotide goes in next? The dark strand decides.
Where the dark strand has adenine, only a thymine pairs with it. Where the dark strand has guanine, only a cytosine pairs with it.
Each base on a strand pairs with only one partner base, so the sequence of one strand fixes the sequence of the strand built against it.
Here is a stretch of one old strand written out, with the new strand built against it written beneath. A strand is written from its 5′ end to its 3′ end, and both ends are marked.
When a strand’s sequence fixes the sequence of the new strand built against it like this, the strand is called a , because a template is a pattern that a copy is shaped against.
Both dark strands are template strands. Each of them gets a new light partner.
So when the copying ends, there are two molecules. Each molecule is one dark strand paired with one light strand: half old, half new.
Copying that leaves each new molecule half old and half new is called , because half of each molecule is conserved, kept, from the old molecule.
Semiconservative means half of each molecule is old. It does not mean half of the molecules are old: neither molecule is all old, and neither is all new.
Here are three results drawn side by side, with the real one marked.
On the left, both dark strands stay together and both light strands pair with each other. On the right, dark and light pieces alternate along every strand.
Each dark strand takes its own light partner along its whole length. So the middle result is the one the copying gives.
What you are expected to know Describe semiconservative replication: the two strands separate, each serves as a template strand for a new complementary strand, so each daughter molecule is one old strand paired with one new strand.
One DNA molecule with both strands dark is copied once. In the drawings below, dark marks an old strand and light marks a new strand.
Which drawing shows the two molecules the copying gives?
- A. ✓ Drawing 1
- B. Drawing 2In drawing 2, dark and light pieces alternate along every strand.
A new strand is built against a whole old strand and stays paired with it. - C. Drawing 3In drawing 3, one molecule has both dark strands.
The two old strands come apart, and each one pairs with a new strand.
Why: The two old strands come apart.
A new strand is built against each old strand and stays paired with it.
So each molecule is one dark strand and one light strand: drawing 1.
One DNA molecule is copied once. Afterwards there are two molecules, and each of them is one old strand paired with one new strand.
(a) Explain why each of the two molecules must be half old. (1 pt)
Frame Each molecule is half old because …
Each old strand serves as a template strand.
Each base on a template strand pairs with only one partner base, so one new strand is built against each old strand.
Each old strand stays paired with the new strand built against it.
So each of the two molecules is one old strand and one new strand.
- Award 1 point for: the old strands separate and each serves as a template for a new strand that stays paired with it, so each molecule has one old strand and one new strand.
A student looks at the two molecules made by copying one DNA molecule and says: “One of the two molecules keeps both old strands.”
Is the student correct?
- A. ✓ No: the two old strands end up in different molecules
- B. Yes: one molecule keeps both old strandsThe two old strands come apart before the copying.
Each old strand ends up paired with a new strand, in a different molecule.
Why: The two old strands come apart.
Each old strand serves as a template for a new strand and stays paired with it.
So each molecule keeps one old strand, not both.
One DNA molecule is copied once. Then one of its two daughter molecules is heated until its strands come apart.
How many of its two strands are old?
- A. NoneA molecule with no old strand would be all new.
Each molecule made by the copying keeps one old strand. - B. ✓ One
- C. TwoTwo old strands in one molecule would mean the old strands stayed together.
They came apart, and each took a new partner.
Why: Each molecule made by the copying is one old strand paired with one new strand.
Heating parts the two strands.
So one of the two strands is old.
47Quick quiz: template strand, semiconservative replication mixed practice
A DNA molecule is being copied.
What is a template strand?
- A. The new strand built during the copyingThe new strand is built against the template strand; it is not the template.
- B. The two paired strands of one moleculeA template strand is one strand, the one a new strand is built against.
- C. ✓ A strand whose bases set the new strand’s bases
Why: A template strand is a strand whose sequence fixes the sequence of the new strand built against it.
A DNA molecule is copied.
What is semiconservative replication?
- A. ✓ Copying that leaves each new molecule half old and half new
- B. Copying that leaves one molecule all old and one all newOne molecule all old and one all new would keep the old strands together.
In semiconservative replication each old strand pairs with a new strand. - C. Copying that mixes old and new pieces along every strandOld and new pieces along one strand never happen.
In semiconservative replication each strand is whole: one old, one new.
Why: Semiconservative replication leaves each new molecule half old and half new: one old strand paired with one new strand.
A DNA molecule is being copied.
(a) State what a template strand is. (1 pt)
- Award 1 point for: a strand whose base sequence directs (fixes) the sequence of the new strand built against it.
During copying, a new strand is built against an old strand.
Which of the two is the template strand?
- A. ✓ The old strand
- B. The new strandThe new strand’s bases are set by the old strand, not the other way round.
The old strand is the template strand.
Why: The old strand’s sequence fixes the new strand’s sequence.
So the old strand is the template strand.
One DNA molecule goes through semiconservative replication once, giving two molecules.
How many of the two molecules contain an old strand?
- A. One of themEach old strand ends up in a different molecule, paired with a new strand.
So both molecules contain an old strand. - B. ✓ Both of them
Why: The two old strands come apart and each pairs with a new strand.
So both molecules contain one old strand.
After a copying, one molecule is found to be all new: both of its strands are new.
Could one round of semiconservative replication of one all-old molecule have made this molecule?
- A. YesOne round gives two molecules, each with one old strand.
A molecule with two new strands has no old strand. - B. ✓ No
Why: One round of semiconservative replication gives two molecules.
Each of them is one old strand paired with one new strand.
So neither molecule is all new.
54Counting the rounds
Video: Watch: Counting the rounds
The all-dark molecule copied once into two half-dark molecules, then each copied again into four; the two dark strands followed through both rounds; the counts 1, 2, 4 written beside the rows.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L05c.mp4
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Go back to the one molecule with both strands dark. Suppose it is copied, and then each of the two molecules is copied again.
Here is the copying drawn as a tree: one molecule, then two, then four. Old strands are dark and new strands are light.
Round one: the one molecule becomes two. Each of the two has one dark strand and one light strand, so both carry an old strand.
Round two: each of the two molecules is copied. In each molecule, the dark strand gets a new light partner, and the light strand gets a new light partner too.
So round two gives four molecules. Two of them still carry a dark strand, and two are all light: all new.
The dark strands are never made again. There are only ever two of them, so in every round exactly two molecules carry an old strand.
The count of molecules doubles every round: 1, 2, 4, 8. Here are the two counts as equations.
n is the number of rounds: the count of molecules doubles every round; the two dark strands are never made again
The molecules with no dark strand are all the rest: the count of molecules minus 2.
One all-dark DNA molecule is copied once. Calculate how many molecules there are, and how many of them carry a dark strand.
What you are expected to know Predict, for semiconservative replication, how many molecules there are after a stated number of rounds, how many still carry an old strand, and how many are all new.
One all-dark DNA molecule goes through two rounds of copying.
Calculate how many of the molecules are all new, with no dark strand.
Part 1. Calculate how many molecules there are after the two rounds.
Answer: 4 (tolerance ±0)
Part 2. Calculate how many of those molecules still carry a dark strand.
Answer: 2 (tolerance ±0)
Answer: 2 (tolerance ±0)
One all-dark DNA molecule goes through two rounds of copying. In the drawings below, dark marks an old strand and light marks a new strand.
Which row shows the four molecules after the second round?
- A. Row 1In row 1 every molecule carries a dark strand.
There are only two dark strands, so only two of the four molecules can carry one. - B. Row 2In row 2 one molecule has both dark strands.
The two dark strands came apart in round one and are in different molecules. - C. ✓ Row 3
Why: There are only ever two dark strands, in two different molecules.
The other two molecules are all light.
Row 3 shows two half-dark molecules and two all-light molecules.
Go back to the cell about to divide, and its one DNA molecule with both strands drawn dark.
After copying there are two molecules, and each has one dark strand and one light strand.
After a second round there are four, and only two of them still carry a dark strand.
71Practice: rounds of copying mixed practice
One all-dark DNA molecule goes through four rounds of copying.
Calculate how many molecules there are after the four rounds.
Answer: 16 (tolerance ±0)
One all-dark DNA molecule goes through eight rounds of copying, giving 256 molecules.
Calculate how many of the 256 molecules carry a dark strand.
Answer: 2 (tolerance ±0)
One all-dark DNA molecule goes through five rounds of copying.
Calculate how many molecules there are after the five rounds.
Answer: 32 (tolerance ±0)
One all-dark DNA molecule goes through seven rounds of copying, giving 128 molecules.
Calculate how many of the 128 molecules are all new, with no dark strand.
Answer: 126 (tolerance ±0)
One all-dark DNA molecule goes through six rounds of copying.
Calculate how many molecules there are after the six rounds.
Answer: 64 (tolerance ±0)
A bacterium’s one all-dark DNA molecule goes through three rounds of copying.
Calculate how many of the molecules are all new, with no dark strand.
Answer: 6 (tolerance ±0)
78Mixed practice mixed practice
One all-dark DNA molecule goes through three rounds of copying. A student says: “Only two of the molecules carry a dark strand.”
Is the student correct?
- A. ✓ Yes: only the two original strands are dark, one in each of two molecules
- B. No: each round doubles the dark strands, so more and more molecules carry oneEvery strand the cell builds is new, and new strands are light.
The two dark strands are the original pair, and each sits in one molecule.
Why: Every new strand is light, so the two dark strands are the only dark strands there are.
Each round, each dark strand is copied and stays in one molecule.
So after three rounds, only two molecules carry a dark strand.
A cell copies its DNA and then divides.
Why does the copying come before the division?
- A. So the cell has time to grow before it dividesThe cell grows in G1 and G2.
The copying is there so that each daughter cell can receive a complete set of DNA. - B. So the DNA coils tightly before the cell dividesThe DNA coils tightly for the division itself.
The copying is there so that each daughter cell can receive a complete set of DNA. - C. ✓ So each daughter cell receives a complete set of DNA
Why: Each daughter cell needs a complete genome.
Only a copied set can be shared as two complete sets.
So the copying comes before the division.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. A template strand reads 5′-GGTACA-3′.
Written under the template strand, with its ends marked 3′ at the left and 5′ at the right, what does the new strand built against it read?
- A. 3′-GGTACA-5′These are the template’s own letters, not their partners.
Under each base sits its partner: G gives C, T gives A, A gives T. - B. ✓ 3′-CCATGT-5′
- C. 3′-TGTACC-5′These are the partners read from the wrong end of the template.
Under the template’s first base, G, sits C.
Why: Each base on the template pairs with only one partner base.
Under each base write its partner: G gives C, G gives C, T gives A, A gives T, C gives G, A gives T.
Under the template that reads 3′-CCATGT-5′.
A cell is in G1, the first growth gap of interphase.
Has the cell copied its DNA yet in this cycle?
- A. YesG1 comes before S phase.
The cell copies its DNA in S phase, so in G1 the copying has not happened yet. - B. ✓ No
Why: The cell copies its DNA in S phase.
G1 comes before S phase.
So in G1 the cell has not yet copied its DNA.
One all-dark DNA molecule goes through ten rounds of copying, giving 1024 molecules.
How many of the 1024 molecules carry a dark strand?
- A. ✓ 2
- B. 10The count of dark strands does not grow by one each round.
The two dark strands are never made again, so two molecules carry one. - C. 1024Only two of the 1024 molecules contain a dark strand.
The rest are all light.
Why: The dark strands are never made again.
There are only ever two of them, in two different molecules.
So 2 molecules carry a dark strand, whatever the round.
A DNA molecule goes through semiconservative replication.
What does each old strand do during the copying?
- A. The old strand pairs again with the other old strandThe two old strands come apart and do not pair with each other again.
Each old strand pairs with a new strand. - B. ✓ The old strand serves as a template for a new strand
- C. The old strand breaks into nucleotides that the cell reusesThe old strand stays whole.
A new strand is built against it, base by base.
Why: The two old strands come apart.
Each base on an old strand pairs with only one partner base, so the old strand fixes the new strand built against it.
Each old strand is a template strand.
Here is one DNA molecule of a cell about to divide, both strands drawn dark. The cell copies the molecule once and divides. Each daughter cell receives one molecule, and each of those molecules has one dark strand and one light strand.
(a) Explain how the dark and light strands of the two molecules show that each old strand served as a template strand. (1 pt)
Frame The strands show this because …
The dark strand is old and the light strand is new.
The new strand is paired with the old strand along its whole length.
So the new strand was built against the old strand, base by base: the old strand served as a template strand.
- Award 1 point for: each molecule pairs one old (dark) strand with one new (light) strand, so each new strand was built against an old strand, which served as the template.
(b) Each daughter cell later copies its molecule again. Explain why only two of the four molecules that result still carry a dark strand. (1 pt)
Frame Only two carry a dark strand because …
There are two dark strands, one in each daughter cell’s molecule.
In the next copying each dark strand gets a new light partner, and each light strand gets a new light partner.
So two of the four molecules have a dark strand and two are all light.
- Award 1 point for: no new dark (old) strand is ever made, so the two original strands end up in two of the four molecules and the other two molecules are entirely new.
Glossary
- template strand
- A strand whose sequence fixes the sequence of the new strand built against it, because each base pairs with only one partner base. In DNA replication both old strands are template strands.
- semiconservative replication
- Copying that leaves each new DNA molecule half old and half new: one old strand paired with one new strand built against it. Half of each molecule is conserved from the old molecule; it does not mean half of the molecules are old.
APBIO-U06-L05B The bands in the tube
Suppose bacteria are grown for many generations in food containing a heavier form of nitrogen, so every base in their DNA is heavy. Then they are moved to ordinary, light nitrogen and allowed to copy their DNA once, then twice. Spun in a tube, heavy DNA settles lower than light DNA, and half-heavy DNA settles between them.
Before you look at the tube, work out the bands each way of copying would give. Then look, and decide which way of copying is real.
Unit 6 · Gene Expression and Regulation
1Predict the bands
One DNA molecule is copied once by semiconservative replication.
What is each of the two molecules made of?
- A. ✓ One old strand and one new strand
- B. Two old strands, or two new strandsThe two old strands come apart, and each pairs with a new strand.
So no molecule has two old strands. - C. Old and new pieces along each strandEach new strand is built whole against one old strand.
No strand is made of old and new pieces.
Why: The two old strands come apart.
Each old strand serves as a template for a new strand and stays paired with it.
So each molecule is one old strand and one new strand.
One all-old DNA molecule goes through two rounds of semiconservative replication, giving four molecules.
How many of the four molecules carry an old strand?
- A. NoneThe two old strands are never destroyed.
Each old strand sits in one of the four molecules, so two molecules carry an old strand. - B. ✓ Two
- C. FourThere are only two old strands.
So only two of the four molecules can carry one; the other two are all new.
Why: There are only ever two old strands.
After round two they sit in two different molecules.
So two of the four molecules carry an old strand, and two are all new.
A scientist builds glucose from a heavier form of carbon, ¹³C, and feeds it to cells.
Which of the following can tell the heavier carbon apart from ordinary carbon?
- A. ✓ An instrument that measures mass
- B. The cells’ enzymesEnzymes treat the heavier form of an atom exactly like the ordinary form.
Only an instrument tells the two apart, by mass. - C. NeitherAn instrument can tell the two forms apart by their mass.
That is what makes the heavier form a label.
Why: Enzymes treat ¹³C exactly like ordinary carbon.
An instrument can tell ¹³C apart by its greater mass.
So the heavier carbon is a label the scientist can trace.
How can bands in a tube prove that copying is half old, half new? Each way of copying predicts its own pattern of bands.
If the old strands stayed together, there would be one heavy band and one light band after the first round.
If each old strand took a new partner, every molecule would be half heavy after the first round: one band in the middle of the tube.
After the second round, half the molecules would be light: a middle band and a light band.
The tube showed one middle band after round one, and a middle band with a light band after round two. Only the half-old, half-new picture fits both rounds.
Reading a tube like this is how the exam asks about the copying.
Suppose bacteria are grown for many generations in food whose nitrogen is a heavier form, written ¹⁵N.
Every base in DNA contains nitrogen atoms. The bacteria build their bases from the food’s nitrogen, so every base in their DNA is heavy.
Enzymes treat heavy nitrogen exactly like ordinary nitrogen. So the bacteria grow and copy their DNA as usual.
A strand built from heavy bases is a heavy strand. After many generations, both strands of every DNA molecule are heavy.
Now the bacteria are moved to food with ordinary, light nitrogen. Every strand built after the move is built from light bases: a light strand.
So the old strands are heavy, and the new strands are light.
In every drawing here, old strands are drawn dark and new strands are drawn light. So here every dark strand is heavy and every light strand is light.
After a round of copying, biologists take the DNA out of the bacteria. They spin it very fast in a tube of salt solution for many hours, in a machine called a centrifuge.
Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them. Here is the tube with its three heights marked.
A molecule with two heavy strands is heavy DNA. It settles at the heavy mark.
A molecule with one heavy strand and one light strand is half-heavy DNA. It settles at the half-heavy mark, in the middle of the tube.
A molecule with two light strands is light DNA. It settles at the light mark.
All the molecules of one kind settle at the same height. Together they show up as one dark band across the tube.
The bigger a band’s share of the molecules in the tube, the thicker the band.
To predict the bands a way of copying gives, take three steps:
- Draw the molecules after the round, old strands dark and new strands light.
- For each molecule, count its heavy strands: two makes heavy DNA, one makes half-heavy DNA, none makes light DNA.
- Draw one band for each kind of DNA present, at its mark. A kind with more molecules gets a thicker band.
Take old strands kept together first. Round one gives two molecules, one with both strands heavy and one with both strands light.
Both strands heavy is heavy DNA, at the heavy mark. Both strands light is light DNA, at the light mark.
So round one gives one heavy band and one light band, and nothing in the middle.
Round two copies the heavy molecule and the light molecule. The heavy molecule gives one heavy molecule and one light molecule.
The light molecule gives two light molecules.
So round two gives four molecules: one heavy and three light. The tube shows a thin heavy band and a thick light band.
Now suppose each old strand takes a new partner instead. That is semiconservative replication. Work out its bands before the drawing shows them.
Bacteria whose DNA started fully heavy copy it once in light nitrogen by semiconservative replication, giving two molecules.
How many heavy strands does each of the two molecules carry?
- A. NoneEach new strand pairs with an old strand, and every old strand is heavy.
So each molecule carries one heavy strand. - B. ✓ One
- C. TwoThe two old strands come apart before the copying.
So no molecule keeps both heavy strands.
Why: The two heavy strands come apart.
Each heavy strand takes a new, light partner.
So each of the two molecules carries one heavy strand.
Bacteria whose DNA started fully heavy copy it once in light nitrogen by semiconservative replication, and a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them. Three tubes are drawn below, numbered 1, 2 and 3.
Which tube shows the bands after round one?
- A. Tube 1A heavy band needs molecules with two heavy strands, and a light band needs molecules with none.
After round one every molecule has one heavy strand and one light strand. - B. Tube 2A light band needs molecules with two light strands.
After round one every molecule still carries one heavy strand. - C. ✓ Tube 3
Why: After round one every molecule has one heavy strand and one light strand.
Every molecule is half-heavy DNA, so all of them settle at the half-heavy mark.
Tube 3 shows one half-heavy band.
Bacteria whose DNA started fully heavy copy it twice in light nitrogen by semiconservative replication, giving four molecules.
Which of the following describes the four molecules?
- A. One half-heavy and three lightEach of the two heavy strands ends up in its own molecule.
So two molecules are half-heavy, not one. - B. ✓ Two half-heavy and two light
- C. Four half-heavyThere are only two heavy strands, so only two molecules can carry one.
The other two molecules have two light strands each.
Why: There are only ever two heavy strands, and they sit in two different molecules.
Those two molecules are half-heavy.
The other two molecules have two light strands each: light DNA.
Bacteria whose DNA started fully heavy copy it twice in light nitrogen by semiconservative replication, and a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them. Three tubes are drawn below, numbered 1, 2 and 3.
Which tube shows the bands after round two?
- A. ✓ Tube 1
- B. Tube 2After round two, two of the four molecules have two light strands.
Those two molecules make a light band as well. - C. Tube 3A heavy band needs molecules with two heavy strands.
The two heavy strands came apart in round one and never pair again.
Why: After round two, two molecules are half-heavy and two are light.
Half-heavy DNA settles at the half-heavy mark and light DNA at the light mark.
Equal numbers give equally thick bands: tube 1.
Here are the two rounds drawn for each old strand taking a new partner: one half-heavy band after round one, then a half-heavy band and a light band, equally thick, after round two.
Take old and new mixed in pieces along every strand. After round one, half of every strand is old and half is new.
So every molecule is half heavy. Round one gives one half-heavy band, the same band that each old strand with a new partner gives.
In round two, every strand is mixed again with new light pieces. So 25 % of every strand is old, and every molecule is 25 % heavy.
A molecule that is 25 % heavy settles between the half-heavy mark and the light mark. So round two gives one band, a little above the middle, and never a light band.
Suppose old and new are mixed along every strand. Bacteria whose DNA started fully heavy copy it once in light nitrogen, and a centrifuge spins the DNA.
How many bands does the tube show?
- A. ✓ One
- B. TwoTwo bands need two kinds of molecule.
After round one every molecule is the same: half of every strand old, half new. - C. ThreeThree bands need three kinds of molecule.
After round one every molecule is the same: half heavy.
Why: After round one, half of every strand is old and half is new.
Every molecule is the same, half heavy.
So all the molecules settle at one height: one band.
Suppose old and new are mixed along every strand. Bacteria whose DNA started fully heavy copy it twice in light nitrogen, and a centrifuge spins the DNA.
Does the tube show a light band?
- A. YesA light band needs molecules with no heavy nitrogen at all.
Here every strand still carries some old, heavy pieces. - B. ✓ No
Why: Every strand is a mixture of old and new pieces.
So every molecule still carries some heavy nitrogen.
No molecule is light DNA, and the tube shows no light band.
Video: Watch: Predict the bands
The bacteria in heavy nitrogen, every strand dark; the move to light nitrogen; for each of the three ways of copying, the molecules after round one and round two drawn, then dropped into a tube and settling into their bands.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L05Ba.mp4
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The table below compares old strands kept together, each old strand with a new partner, and old and new mixed along every strand: the bands each gives after round one and after round two.
After round one, two ways give the same tube: each old strand with a new partner, and old and new mixed along every strand. After round two, every way gives its own tube.
What you are expected to know Predict the bands each way of copying gives after one and two rounds in light nitrogen, for DNA that started fully heavy.
Bacteria whose DNA started fully heavy copy it three times in light nitrogen by semiconservative replication, giving 8 molecules.
Calculate how many of the 8 molecules are light, with no heavy strand.
Answer: 6 (tolerance ±0)
Bacteria whose DNA started fully heavy copy it three times in light nitrogen by semiconservative replication, and a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them. A thicker band means a bigger share of the molecules. Three tubes are drawn below, numbered 1, 2 and 3.
Which tube shows the bands after round three?
- A. Tube 1A thick half-heavy band needs most molecules to carry a heavy strand.
Only two of the eight molecules carry one; the other six are light. - B. Tube 2Equally thick bands need equal numbers of half-heavy and light molecules.
After round three there are two half-heavy molecules and six light ones. - C. ✓ Tube 3
Why: After round three there are 8 molecules.
The two heavy strands sit in 2 half-heavy molecules; the other 6 are light.
The light band holds three times as many molecules, so it is the thicker band: tube 3.
50Read the tube
Video: Watch: Read the tube
The three predicted tubes for round one beside the tube seen, one prediction fading out; the three predicted tubes for round two beside the tube seen, a second prediction fading out; the one way of copying left standing.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L05Bb.mp4
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In 1958 two scientists, Matthew Meselson and Franklin Stahl, did this experiment with the bacterium E. coli.
Here are the tubes they saw: before the move to light nitrogen, after round one, and after round two.
Before the move, the tube shows one heavy band: every molecule had two heavy strands.
After round one, the tube shows one half-heavy band and nothing else.
After round two, the tube shows a half-heavy band and a light band, equally thick.
Here are the three predicted tubes for round one beside the tube they saw.
Old strands kept together predicted a heavy band and a light band. The tube shows neither of them.
The tube shows one half-heavy band.
So the old strands did not stay together. Round one rules that way of copying out.
Each old strand with a new partner predicted one half-heavy band. Old and new mixed along every strand predicted one half-heavy band too.
Round one fits both of those ways of copying. So round one cannot tell them apart.
Here are the three predicted tubes for round two beside the tube they saw.
Old and new mixed along every strand predicted one band between the half-heavy mark and the light mark, and never a light band.
The tube shows a light band: whole molecules with no heavy nitrogen at all. So the old pieces were not spread along every strand.
Round two rules that way of copying out.
Each old strand with a new partner predicted a half-heavy band and a light band, equally thick. That is exactly what the tube shows.
Only one way of copying fits both rounds: each old strand took a new partner. The copying is semiconservative replication, and half of each molecule is old.
Biologists call the first rejected way conservative copying, and the second rejected way dispersive copying. Neither is what a cell does.
What you are expected to know Judge which way of copying the bands in the tube support, and name the observation that rules out each of the other two.
Bacteria whose DNA started fully heavy copy it once in light nitrogen, and a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them. The tube is drawn below.
Which way of copying does this tube rule out?
- A. ✓ Old strands kept together
- B. Each old strand with a new partnerEach old strand with a new partner gives molecules with one heavy strand each.
Those settle at the half-heavy mark, the band the tube shows. - C. Old and new mixed along every strandOld and new mixed along every strand gives molecules that are half heavy after round one.
Those settle at the half-heavy mark, the band the tube shows.
Why: Old strands kept together gives one molecule with two heavy strands and one with two light strands.
That is a heavy band and a light band.
The tube shows one half-heavy band instead, so that way of copying is ruled out.
Bacteria are grown for many generations in heavy nitrogen, so every strand of their DNA is heavy. They are moved to light nitrogen and copy their DNA once, then twice. After each round a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them. Now suppose the tubes had shown a different result, drawn below: a heavy band and a light band after round one, and a thin heavy band and a thick light band after round two.
(a) Explain how the supposed bands after round one would rule out each old strand taking a new partner. (1 pt)
Frame The bands after round one would rule this out because …
Both molecules settle at the half-heavy mark, as one half-heavy band.
The supposed tube shows a heavy band and a light band, and no half-heavy band.
So each old strand did not take a new partner.
- Award 1 point for: each old strand with a new partner predicts one half-heavy band after round one, and the supposed tube shows no half-heavy band (a heavy band and a light band instead).
(b) Explain how the same supposed bands after round one would also rule out old and new mixed along every strand. (1 pt)
Frame The same bands would rule this out because …
Every molecule is then half heavy and settles at the half-heavy mark.
The supposed tube shows no half-heavy band.
So old and new were not mixed along every strand.
- Award 1 point for: old and new mixed along every strand also predicts one half-heavy band after round one, which the supposed tube does not show.
(c) Explain how the supposed bands after both rounds would support old strands kept together. (1 pt)
Frame The bands would support this because …
After round two it leaves one heavy molecule and three light molecules: a thin heavy band and a thick light band.
Both supposed tubes show exactly these bands.
So the bands would support old strands kept together.
- Award 1 point for: old strands kept together predicts a heavy band and a light band after round one and a thin heavy band with a thick light band after round two, matching both supposed tubes.
Bacteria whose DNA started fully heavy copy it once in light nitrogen, and the spun DNA shows one half-heavy band. A student says: “After this one round, two ways of copying still fit the tube. A second round is needed to tell them apart.”
Is the student correct?
- A. ✓ Yes: two ways of copying predict this round-one tube, and only the round-two tube separates them
- B. No: one half-heavy band after round one fits only one way of copying, so this tube settles itTwo ways of copying leave every molecule half heavy after round one.
Both ways predict this tube, so this tube alone cannot settle it.
Why: After round one, each old strand with a new partner gives one half-heavy band.
Old and new mixed along every strand gives one half-heavy band too.
So one tube after round one cannot tell the two apart; round two is needed.
Now imagine the experiment the other way round. Bacteria are grown for many generations in ordinary, light nitrogen, so every strand is light.
Then the bacteria are moved to heavy nitrogen.
Bacteria whose DNA started fully light copy it once in heavy nitrogen, and a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them. Three tubes are drawn below, numbered 1, 2 and 3.
Which tube shows the bands after round one?
- A. Tube 1A light band needs molecules with two light strands.
After round one every molecule has taken one new, heavy strand. - B. ✓ Tube 2
- C. Tube 3A heavy band needs molecules with two heavy strands.
After round one every molecule still carries its old, light strand.
Why: Each old, light strand takes a new partner built from heavy bases.
So every molecule has one light strand and one heavy strand: half-heavy DNA.
All of them settle at the half-heavy mark: tube 2.
Bacteria whose DNA started fully light copy it twice in heavy nitrogen, and a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them.
Which bands does the tube show after round two?
- A. One heavy bandOne heavy band needs every molecule to have two heavy strands.
Two of the four molecules still carry an old, light strand. - B. ✓ A half-heavy band and a heavy band
- C. A half-heavy band and a light bandA light band needs molecules with two light strands.
The two old, light strands came apart in round one and never pair again.
Why: There are only two old, light strands, and they sit in two different molecules.
Those two molecules are half-heavy.
The other two molecules have two new, heavy strands: heavy DNA.
So the tube shows a half-heavy band and a heavy band.
Go back to the tubes of DNA spun after each round of copying in light nitrogen, from bacteria whose every strand started heavy.
One half-heavy band after round one rules out old strands staying together.
A half-heavy band and a light band after round two rule out old and new mixed along every strand.
Each old strand took a new partner: half old, half new.
82Mixed practice mixed practice
Suppose the old strands stayed together whenever DNA is copied. Bacteria whose DNA started fully light copy it once in heavy nitrogen, and a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them.
Which bands would the tube show?
- A. One half-heavy bandOne half-heavy band needs every molecule to carry one light strand and one heavy strand.
With the old strands kept together, no molecule mixes the two. - B. A half-heavy band and a heavy bandA half-heavy band needs a molecule with one light strand and one heavy strand.
With the old strands kept together, the two old, light strands stay in one molecule. - C. ✓ A light band and a heavy band
Why: With the old strands kept together, the two old, light strands stay in one molecule: light DNA.
The two new strands, built from heavy bases, pair with each other: heavy DNA.
So the tube shows a light band and a heavy band.
Bacteria are grown for many generations in food whose nitrogen is the heavier form, ¹⁵N.
Which part of their DNA carries the heavy nitrogen?
- A. The sugarsA sugar is built from carbon, hydrogen and oxygen atoms.
The nitrogen atoms of DNA are in the bases. - B. ✓ The bases
- C. The phosphate groupsA phosphate group is built from phosphorus and oxygen atoms.
The nitrogen atoms of DNA are in the bases.
Why: Every base contains nitrogen atoms; the sugar and the phosphate group contain none.
The bacteria build their bases from the food’s nitrogen.
So the bases carry the heavy nitrogen.
Bacteria whose DNA started fully heavy copy it twice in light nitrogen, and the spun DNA shows one band, between the half-heavy mark and the light mark.
Which way of copying gives this band?
- A. Old strands kept togetherOld strands kept together gives a heavy band in every round.
This tube shows no heavy band. - B. Each old strand with a new partnerEach old strand with a new partner gives a half-heavy band and a light band after round two.
This tube shows one band, and none at the light mark. - C. ✓ Old and new mixed along every strand
Why: With old and new mixed along every strand, every molecule is the same after each round.
After round two every molecule is 25 % heavy.
A molecule that is 25 % heavy settles between the half-heavy mark and the light mark: one band there.
Suppose old and new were mixed along every strand whenever DNA is copied. Bacteria whose DNA started fully light copy it once in heavy nitrogen, and a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them.
At which mark do the molecules settle after round one?
- A. ✓ The half-heavy mark
- B. The heavy markA band at the heavy mark needs molecules with two heavy strands.
Here every strand is half old, light pieces and half new, heavy pieces. - C. The light markA band at the light mark needs molecules with no heavy nitrogen.
Here every strand has taken new, heavy pieces.
Why: With old and new mixed along every strand, half of every strand is old and light and half is new and heavy after round one.
So every molecule is half heavy.
All the molecules settle at the half-heavy mark: one band there.
Bacteria whose DNA started fully heavy copy it once in light nitrogen, and a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them. The tube drawn below shows the result a student predicted.
Which way of copying would give the bands the student predicted?
- A. ✓ Old strands kept together
- B. Each old strand with a new partnerEach old strand with a new partner gives molecules with one heavy strand each after round one.
Those settle as one half-heavy band only. - C. Old and new mixed along every strandOld and new mixed along every strand gives half-heavy molecules after round one.
Those settle as one half-heavy band only.
Why: A heavy band needs a molecule with two heavy strands, and a light band a molecule with two light strands.
Only old strands kept together gives both after round one.
So the student predicted old strands kept together.
Bacteria whose DNA started fully heavy copy it four times in light nitrogen by semiconservative replication, giving 16 molecules.
How many of the 16 molecules are half-heavy?
- A. 1One molecule keeping both heavy strands is old strands kept together.
Here the two heavy strands sit in two different molecules, each half-heavy. - B. ✓ 2
- C. 4The count of heavy strands does not grow with the rounds.
There are only two heavy strands, so two molecules are half-heavy.
Why: The heavy strands are never made again.
There are only ever two of them, in two different molecules.
So 2 of the 16 molecules are half-heavy, and the rest are light.
Bacteria are grown for many generations in heavy nitrogen, so every strand of their DNA is heavy. They are moved to light nitrogen and copy their DNA three times. A centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them. The tube is drawn below.
(a) Explain how the two bands show that each old strand took a new partner. (1 pt)
Frame The two bands show this because …
A molecule gets one heavy strand and one light strand only when an old, heavy strand takes a new, light partner.
A light band is made of molecules with two light strands, built after the move.
So the old strands came apart and each took a new partner.
- Award 1 point for: the half-heavy band is molecules with one old (heavy) strand paired with one new (light) strand, which only each old strand taking a new partner produces; the light band is molecules made of two new strands.
(b) Explain why the half-heavy band is thinner than the light band. (1 pt)
Frame The half-heavy band is thinner because …
Each heavy strand sits in its own molecule, so only two molecules are half-heavy.
After three rounds there are 8 molecules, so the other 6 are light.
A band with a bigger share of the molecules is thicker, so the light band is the thicker one.
- Award 1 point for: only two heavy strands exist and they are never remade, so only two of the eight molecules are half-heavy while six are light; a bigger share of the molecules makes a thicker band.
APBIO-U06-L06 The fork opens
The copying machine is drawn below, caught in the act. On the right, the two strands are still paired. On the left, they have been pulled apart, like a zipper half undone.
What is doing the unzipping? And why does the closed part just ahead of it not twist into a knot?
Unit 6 · Gene Expression and Regulation
1Helicase parts the strands
In a DNA molecule, two strands lie side by side with their bases facing each other.
What holds the two strands together along their length?
- A. Covalent bonds along each backboneThe covalent bonds along a backbone join the nucleotides of one strand.
Hydrogen bonds between paired bases hold the two strands together. - B. Covalent bonds between the paired basesPaired bases are held by hydrogen bonds, drawn as dashed lines.
Covalent bonds join the nucleotides of one strand, not the two strands. - C. ✓ Hydrogen bonds between the paired bases
Why: Where two bases face each other, hydrogen bonds hold them together.
There is a base pair at every step along the strands.
So hydrogen bonds hold the two strands together along their length.
One strand of DNA is a chain of nucleotides.
Which kind of bond joins one nucleotide to the next along the strand’s backbone?
- A. ✓ A covalent bond
- B. A hydrogen bondHydrogen bonds join paired bases across the molecule.
Along a backbone, sugar joins phosphate by covalent bonds.
Why: A backbone is sugar, phosphate, sugar, phosphate.
A covalent bond joins each sugar to the next phosphate group.
So covalent bonds join the nucleotides along the backbone.
A DNA molecule holds both hydrogen bonds and covalent bonds.
Which kind of bond takes less energy to break?
- A. ✓ The hydrogen bonds
- B. The covalent bondsA covalent bond is strong and takes much energy to break.
A hydrogen bond is weak and takes far less. - C. The two kinds take the same energyThe two kinds differ in strength.
A hydrogen bond is weak; a covalent bond is strong.
Why: A hydrogen bond is a weak attraction between two molecules or two parts of a molecule.
A covalent bond is a strong bond that joins atoms.
So the hydrogen bond takes less energy to break.
How does the double helix open for copying? An enzyme breaks the hydrogen bonds between the paired bases.
So the two strands come apart at a Y-shaped point.
Pulling two twisted strands apart over-twists the helix ahead of that point. Untwisting a rope from one end knots it further along in the same way.
A second enzyme cuts one strand ahead of the point, lets the twist unwind and seals the strand again. So the opening keeps moving.
Every drawing of copying in this unit shows this Y-shaped point. So learn its shape first, and the two enzymes that act at it.
The drawing below shows the copying caught in the act, with the parts named.
On the right, the two strands are still paired. Hydrogen bonds between the paired bases hold them together.
On the left, the two strands have come apart. Where they came apart, the hydrogen bonds between the paired bases have been broken.
Nothing else has been broken. Along each strand, every covalent bond of the backbone is still in place.
So each strand is still whole, from one end to the other.
Look at the point where the paired part turns into the parted part. There the drawing is shaped like the letter Y.
An enzyme sits at that point. It breaks the hydrogen bonds between the paired bases, one base pair after another.
Video: Watch: Helicase parts the strands
The paired molecule; an enzyme arriving and breaking the hydrogen bonds one base pair at a time; the two strands parting behind it into a Y; the Y moving along the molecule.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L06a.mp4
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A hydrogen bond is weak. So the enzyme breaks it.
A covalent bond is strong. So the enzyme leaves every covalent bond of the backbones in place.
So the two strands come apart. Each strand stays whole.
The enzyme that breaks the hydrogen bonds between the paired bases and parts the two strands is called , because it opens the double helix.
The Y-shaped point where the two strands come apart is called the , because the molecule forks into two arms there, the way a road forks.
Helicase moves along the molecule, into the paired part. So the replication fork moves with it, here to the right.
Behind the fork, each parted strand is on its own, with its bases exposed. Each parted strand is a template strand, ready for a new strand to be built against it.
What you are expected to know Describe what helicase does: it breaks the hydrogen bonds between paired bases, so the two strands separate at the replication fork.
What you are expected to know Name what each parted strand becomes: a template strand, exposed for copying.
A DNA molecule is being copied.
What does helicase do?
- A. Joins new nucleotides one at a time to a growing new strandJoining nucleotides builds the new strand; that is not helicase’s job.
Helicase parts the two old strands. - B. Pairs each new nucleotide with its partner base on the templatePairing nucleotides against the template builds the new strand.
Helicase parts the two old strands so that a template is exposed. - C. ✓ Breaks the hydrogen bonds between paired bases and parts the strands
Why: Helicase breaks the hydrogen bonds between paired bases.
So the two strands part, and each is exposed as a template strand.
A DNA molecule is being copied.
What is the replication fork?
- A. The far end of a template strandThe far end of a template is the end away from the fork.
The fork is the point where the two strands come apart. - B. ✓ The Y-shaped point where the two strands come apart
- C. The part of the molecule that is still pairedThe paired part lies ahead of the fork; the fork is at its edge.
The fork is the Y-shaped point where the strands come apart.
Why: At one point the paired strands come apart into two arms, and the drawing is shaped like a Y.
That point is the replication fork.
Helicase parts the two strands of a DNA molecule from one end to the other. A laboratory then measures the two single strands: each one is as long as the whole molecule was.
(a) Explain why each strand is still one whole piece. (1 pt)
Frame Each strand is still one whole piece because …
A hydrogen bond is weak, so helicase breaks it.
Along each backbone, covalent bonds join one nucleotide to the next.
A covalent bond is strong, and helicase leaves every covalent bond in place.
So each backbone stays joined from end to end, and each strand is one whole piece.
- Award 1 point for: helicase breaks the hydrogen bonds between paired bases and leaves the covalent bonds of each backbone in place, so each strand stays whole.
A student says: “Helicase cuts through each strand’s backbone to open the molecule.”
Is the student correct?
- A. Yes: helicase cuts through each strand’s backbone, so the molecule can openCutting a backbone would break a strand into two pieces.
Helicase breaks the hydrogen bonds between paired bases, and each strand stays whole. - B. ✓ No: helicase breaks only the hydrogen bonds between the paired bases; every backbone stays whole
Why: The two strands are held together by hydrogen bonds between the paired bases.
Helicase breaks those hydrogen bonds.
The covalent bonds of each backbone stay in place, so each strand stays whole.
Imagine a drug blocks helicase in a cell, and the cell reaches S phase.
What happens to the cell’s DNA?
- A. ✓ No fork opens: no template strand is exposed
- B. A fork opens but no new strand is built against the templatesOnly helicase parts the two strands.
With helicase blocked, the strands stay paired and no fork opens. - C. A fork opens; both strands are copied as usualHelicase is blocked, so no hydrogen bonds between paired bases are broken.
The strands stay paired and no fork opens.
Why: Helicase breaks the hydrogen bonds between the paired bases and parts the strands.
The drug blocks helicase, so the strands stay paired.
No fork opens, so no template strand is exposed.
32Quick quiz: helicase, replication fork mixed practice
A DNA molecule is being copied.
(a) State what helicase does. (1 pt)
So the two strands come apart at the replication fork, and each strand stands alone as a template strand.
- Award 1 point for: breaks the hydrogen bonds between paired bases so the two strands separate (unwinds the double helix).
(b) State what the replication fork is. (1 pt)
- Award 1 point for: the Y-shaped point (region) where the two strands separate during replication.
Helicase is at work on a DNA molecule with a replication fork.
Where on the molecule does helicase sit?
- A. ✓ At the point of the Y, where the strands part
- B. At the far end of a template strandThe far ends of the templates parted long ago.
Helicase sits where the strands are parting now: the point of the Y.
Why: Helicase breaks the hydrogen bonds where the strands are parting.
The strands part at the point of the Y, the replication fork.
So helicase sits at the point of the Y.
Behind the replication fork, each old strand is on its own.
What is each single strand behind the fork called?
- A. A new strandA new strand has yet to be built there.
Each parted old strand is a template strand, the pattern for a new strand. - B. ✓ A template strand
Why: Each parted strand is an old strand.
A new strand will be built against it, so it is a template strand.
Helicase is at work on a DNA molecule.
At the replication fork, what are the two strands doing?
- A. ✓ Coming apart
- B. Joining togetherHelicase at the fork breaks hydrogen bonds; it joins nothing.
At the fork the two strands are coming apart.
Why: Helicase sits at the fork and breaks the hydrogen bonds between paired bases.
So at the fork the two strands are coming apart.
37Reading the fork drawing
The fork drawing below has every part named.
The point of the Y is the replication fork. Helicase sits on it.
The two arms to the left of the fork are the two template strands. Each template strand is one old strand, drawn dark, with its bases exposed.
To the right of the fork, the two strands are still paired. That is the unopened part.
The fork moves into the unopened part. So in this drawing the fork moves to the right.
One strand of DNA has two ends that differ.
What are the two ends of a strand called?
- A. ✓ A 5′ end and a 3′ end
- B. A head end and a tail endHead and tail name the ends of an animal, not of a strand.
A strand’s ends are its 5′ end and its 3′ end. - C. A left end and a right endLeft and right depend on how the strand is drawn.
A strand’s own ends are its 5′ end and its 3′ end.
Why: A strand’s 5′ end carries a free phosphate group, and its 3′ end carries a free –OH.
So the two ends are the 5′ end and the 3′ end.
Each template strand has a 5′ end and a 3′ end. The drawing writes both ends: one at the far end, one at the fork.
Two paired DNA strands lie side by side.
How do the two paired strands lie?
- A. In the same directionPaired bases only fit together when the two strands run opposite ways.
So one strand’s 5′ end lies beside the other’s 3′ end. - B. ✓ In opposite directions
Why: The two strands of DNA are antiparallel.
One strand’s 5′ end lies beside the other strand’s 3′ end.
Video: Watch: Reading the fork drawing
The fork drawing named part by part: the point of the Y, the two template strands, the paired part, the arrow for the direction of travel; the ends read along each template; the drawing mirrored to open the other way and read again.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L06b.mp4
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The two template strands are antiparallel: they run in opposite directions. So their ends face the fork the opposite way round.
Follow the upper template strand from its far end to the fork.
Its 3′ end is at the far end, on the left. So the end that faces the fork is its 5′ end.
Follow the lower template strand from its far end to the fork.
Its 5′ end is at the far end, on the left. So the end that faces the fork is its 3′ end.
So at every fork, one template’s 5′ end faces the fork. The other template’s 3′ end faces the fork.
The same molecule is drawn the other way round below. The unopened part is on the left, so the fork moves to the left.
Find the point of the Y first.
Then find the paired part. The fork moves into it.
Then read the ends written on each template.
In this drawing the upper template’s 3′ end faces the fork. The lower template’s 5′ end faces the fork.
Which template is drawn on top changes from drawing to drawing. Only the written ends tell you which end faces the fork.
What you are expected to know Read a fork drawing: name the replication fork, the two template strands and the unopened part, say which way the fork moves, and read each template’s ends.
The drawing below shows a replication fork, with four parts lettered J to M.
Which letter marks the replication fork?
- A. ✓ J
- B. KK marks a single dark strand with its bases exposed: a template strand.
The fork is the point where the two strands come apart. - C. MM marks the part where the two strands are still paired.
The fork is the point where they come apart.
Why: The two strands come apart at one point, and the drawing forms a Y there.
Letter J marks that point: the replication fork.
The drawing below shows a replication fork, with four parts lettered J to M.
Which lettered parts are the template strands?
- A. K onlyL is also a single dark strand with its bases exposed.
K and L are both template strands. - B. ✓ K and L
- C. J and MJ marks the fork and M the paired part.
The template strands are the two parted arms, K and L.
Why: Behind the fork each old strand is on its own, its bases exposed.
K and L are those two single strands.
So K and L are the template strands.
The drawing below shows a replication fork, with four parts lettered J to M.
Which way does the fork move?
- A. To the leftThe paired part, M, lies to the right of the fork.
The fork moves into the paired part, so it moves to the right. - B. ✓ To the right
Why: The fork moves into the part that is still paired.
The paired part lies to the right of the fork.
So the fork moves to the right.
The drawing below shows a replication fork, with four parts lettered J to M. The 3′ and 5′ ends of each template are written at the far end and at the end nearest the fork.
Which end of the upper template strand, K, faces the fork?
- A. ✓ The upper template’s 5′ end
- B. The upper template’s 3′ endThe upper template’s 3′ end is written at its far end, at the left.
So its 5′ end faces the fork.
Why: Read the ends written on the upper template strand.
Its 3′ end is at the far end and its 5′ end is at the fork.
So its 5′ end faces the fork.
The drawing below shows a different replication fork, with four parts lettered N to Q.
Which letter marks the part where the two strands are still paired?
- A. NN is a single dark strand with its bases exposed: a template strand.
In the paired part, both strands lie together with their bases touching. - B. ✓ O
- C. QQ marks the point where the strands come apart.
The paired part is O.
Why: In the paired part the two strands lie together with their bases touching.
Letter O marks that part.
The drawing below shows a different replication fork, with four parts lettered N to Q.
Which way does the fork move?
- A. ✓ To the left
- B. To the rightThe paired part lies to the left of the fork.
The fork moves into the paired part, so it moves to the left.
Why: The fork moves into the part that is still paired.
Here the paired part lies to the left of the fork.
So the fork moves to the left.
The drawing below shows a different replication fork, with four parts lettered N to Q. The 3′ and 5′ ends of each template are written at the far end and at the end nearest the fork.
Which end of the upper template strand, P, faces the fork?
- A. The upper template’s 5′ endThe upper template’s 5′ end is written at its far end, at the right.
So its 3′ end faces the fork. - B. ✓ The upper template’s 3′ end
Why: Read the ends written on the upper template strand.
Its 5′ end is at the far end and its 3′ end is at the fork.
So its 3′ end faces the fork.
67Ahead of the fork: the twist
Go back to the replication fork, where helicase is parting the two strands.
A DNA molecule has two strands.
How are the two strands arranged along the molecule?
- A. ✓ Twisted round each other, a double helix
- B. Lying straight, side by sideTwo straight strands would form a flat ladder.
In DNA the two strands twist round each other: a double helix.
Why: The two strands of DNA twist round each other along the whole molecule.
That shape is the double helix.
Video: Watch: Ahead of the fork: the twist
A two-strand rope untwisted from one end, the twist piling up and kinking; the DNA ahead of the fork over-twisting as helicase parts the strands; topoisomerase cutting one strand, the twist unwinding, the strand sealed; the fork moving on.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L06c.mp4
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The two strands are twisted round each other. So to pull them apart, helicase must untwist them.
A rope is made of two strands twisted round each other. Hold the far end still, and untwist the rope from the near end.
The twist does not vanish. Untwisting the near end pushes the twist along the rope.
So the rope ahead twists tighter and tighter.
When the twist is tight enough, the rope kinks into a knot. Then the untwisting stops.
Let the far end turn freely instead.
The twist you push along unwinds at the far end. So the rope stays even.
The DNA ahead of the fork behaves like the rope.
Helicase untwists the strands at the fork. So the twist is pushed into the paired part ahead.
So the DNA ahead of the fork twists tighter than a double helix normally is. This extra twisting is called , because coils pile onto the coils already there.
Left alone, the over-twisted DNA would kink like the knotted rope.
Then helicase could not part the strands there. So the fork would stall.
A second enzyme acts on the paired part ahead of the fork. It cuts one strand.
The cut strand turns. So the extra twist unwinds.
Then the enzyme seals the strand again. The enzyme that cuts one strand ahead of the fork, lets the twist unwind and seals the strand is called .
Topoisomerase keeps the DNA ahead of the fork relaxed. So the fork keeps moving.
Topoisomerase does not separate the strands.
Helicase separates the strands at the fork. Topoisomerase relaxes the twist ahead of it.
What you are expected to know Explain what topoisomerase does and why it is needed: it cuts one strand ahead of the fork, lets the over-twist unwind and seals the strand, so the fork keeps moving.
A DNA molecule is being copied.
What does topoisomerase do?
- A. ✓ Cuts one strand of the DNA, lets the twist unwind, seals the strand
- B. Breaks the hydrogen bonds between the paired bases at the forkBreaking the hydrogen bonds between paired bases is helicase’s job.
Topoisomerase cuts and reseals one strand so the twist unwinds. - C. Builds a new strand against each exposed template strand, base by baseBuilding new strands comes after the fork has opened.
Topoisomerase cuts and reseals one strand so the twist unwinds.
Why: Topoisomerase cuts one strand of the over-twisted DNA.
The cut strand turns, so the extra twist unwinds.
Then topoisomerase seals the strand again.
Helicase is parting the strands at a replication fork.
What is supercoiling?
- A. The two strands of the molecule coming apart at the forkThe strands coming apart is the fork opening.
Supercoiling is the DNA twisted tighter than a double helix normally is. - B. ✓ The DNA twisted tighter than a double helix normally is
- C. A new strand pairing with its template, base by baseA new strand pairing with its template is the copying itself.
Supercoiling is the DNA twisted tighter than a double helix normally is.
Why: Helicase untwists the strands at the fork and pushes the twist along the DNA.
That DNA twists tighter than a double helix normally is.
The extra twisting is supercoiling.
Topoisomerase acts on a DNA molecule that is being copied.
Where does topoisomerase act?
- A. At the point of the forkHelicase acts at the point of the fork.
Topoisomerase acts ahead of it, where the twist piles up. - B. ✓ On the paired part ahead of the fork
- C. On the template strands behind the forkThe template strands behind the fork are already apart and untwisted.
The twist piles up ahead of the fork, and topoisomerase acts there.
Why: Helicase untwists the strands at the fork, and the twist is pushed into the paired part ahead.
Topoisomerase cuts one strand there, lets the twist unwind and seals the strand.
So topoisomerase acts on the paired part ahead of the fork.
Imagine a cell in which topoisomerase is blocked by a drug. Helicase begins to part the strands at a replication fork, and after a short distance the fork stalls.
(a) State what topoisomerase does. (1 pt)
So the fork keeps moving.
- Award 1 point for: relieves the over-twisting (supercoiling) ahead of the fork by cutting one strand, letting it unwind and rejoining it.
(b) Explain why the fork stalls in this cell. (2 pt)
Frame The fork stalls because …
Untwisting at the fork pushes the twist into the paired part ahead.
So the DNA ahead of the fork twists tighter and tighter: supercoiling.
Topoisomerase is blocked, so no strand is cut and the extra twist cannot unwind.
The over-twisted DNA kinks, and helicase cannot part the strands there.
So the fork stalls.
- Award 1 point for: untwisting at the fork over-twists (supercoils) the DNA ahead of it, and with topoisomerase blocked the extra twist is not relieved.
- Award 1 point for: the over-twisted DNA ahead cannot be parted, so the fork stops moving.
A student says: “Helicase parts the two strands, and topoisomerase relaxes the extra twist.”
Is the student correct?
- A. ✓ Yes: helicase separates the strands, and topoisomerase lets the extra twist unwind
- B. No: topoisomerase separates the strands at the fork, and helicase relaxes the twistHelicase breaks the hydrogen bonds between paired bases at the fork.
Topoisomerase cuts one strand ahead of the fork to let the twist unwind.
Why: Helicase separates the strands at the fork by breaking the hydrogen bonds between paired bases.
Topoisomerase acts ahead of the fork.
It cuts one strand, lets the extra twist unwind and seals the strand.
Suppose you untwist a two-strand rope from one end. You want the twist you push along to unwind rather than pile up.
Which end of the rope must be free to turn?
- A. ✓ The far end
- B. The end you are untwistingYou are already turning the end you untwist.
The twist piles up at the far end unless the far end can turn. - C. Either end will doThe twist travels to the far end.
Only the far end turning lets the twist unwind.
Why: Untwisting one end pushes the twist along the rope toward the far end.
A far end that is held still traps the twist, and the rope kinks.
A far end that is free to turn lets the twist unwind.
So the far end must be free to turn.
Go back to the half-open zipper: the copying machine caught in the act, the strands paired on the right and pulled apart on the left.
At the fork, helicase is breaking the hydrogen bonds between the paired bases.
Ahead of the fork, topoisomerase cuts one strand, lets the over-twist unwind and seals the strand. So the fork keeps moving.
99Quick quiz: topoisomerase, supercoiling mixed practice
As helicase parts the strands at a replication fork, extra twist piles up in the DNA. One enzyme relaxes that twist, so the fork keeps moving.
Which enzyme is it?
- A. HelicaseHelicase breaks hydrogen bonds between paired bases and cuts no strand.
Topoisomerase cuts one strand ahead of the fork and seals it. - B. ✓ Topoisomerase
Why: Topoisomerase cuts one strand ahead of the fork, lets the twist unwind and seals the strand.
So the enzyme is topoisomerase.
101Mixed practice mixed practice
At a replication fork, helicase is parting the two strands.
Where does the DNA twist tighter than usual?
- A. Behind the fork, on the template strandsThe template strands behind the fork are already apart and untwisted.
The twist is pushed ahead of the fork. - B. At the point of the forkAt the point of the fork helicase is untwisting the strands.
The twist it removes is pushed into the paired part ahead. - C. ✓ Ahead of the fork, in the paired part
Why: Helicase untwists the strands at the fork.
The twist is pushed into the paired part ahead of the fork.
So the DNA ahead of the fork twists tighter.
A drug blocks helicase after a fork has already moved part of the way along a DNA molecule.
What happens to the fork?
- A. ✓ The fork stops where it is
- B. The fork keeps moving as the paired strands ahead come apart on their ownPaired strands stay paired until their hydrogen bonds are broken.
Only helicase breaks them, so the fork goes no further. - C. The fork keeps moving as topoisomerase parts the strands aheadTopoisomerase cuts one strand and seals it; it breaks no hydrogen bonds between paired bases.
With helicase blocked, the fork goes no further.
Why: Helicase breaks the hydrogen bonds between the paired bases, and the fork moves with it.
The drug blocks helicase, so no more hydrogen bonds are broken.
So the fork stops where it is, and no more template is exposed.
The drawing below shows a replication fork. The 3′ and 5′ ends of each template are written at the far end and at the end nearest the fork.
Which end of the lower template strand faces the fork?
- A. The lower template’s 5′ endThe lower template’s 5′ end is written at its far end, at the right.
So its 3′ end faces the fork. - B. ✓ The lower template’s 3′ end
Why: Read the ends written on the lower template strand.
Its 5′ end is at the far end and its 3′ end is at the fork.
So its 3′ end faces the fork.
Topoisomerase cuts one strand of the DNA ahead of the fork.
What does the cut let happen?
- A. ✓ The extra twist unwinds
- B. The two strands come apartThe strands come apart at the fork, where helicase breaks the hydrogen bonds.
Topoisomerase’s cut lets the extra twist ahead unwind. - C. A new strand startsNew strands are built behind the fork, against the templates.
Topoisomerase’s cut lets the extra twist ahead unwind.
Why: Ahead of the fork the DNA is over-twisted.
Topoisomerase cuts one strand, so the cut strand can turn.
So the extra twist unwinds, and topoisomerase seals the strand.
Helicase has passed, and the two template strands behind the fork are on their own.
Which bonds are still in place in each template strand?
- A. The hydrogen bonds between paired basesHelicase broke the hydrogen bonds between the paired bases.
The covalent bonds along each backbone are still in place. - B. ✓ The covalent bonds along the backbone
Why: Helicase breaks only the hydrogen bonds between paired bases.
The covalent bonds along each backbone stay in place.
So each template strand is still held together by its backbone.
At a replication fork, the part of the molecule that is still paired lies to the left of the fork.
Which way does the fork move?
- A. ✓ To the left
- B. To the rightThe fork moves into the part that is still paired.
That part lies to the left, so the fork moves to the left.
Why: Helicase parts the strands where they are still paired.
So the fork moves into the paired part.
The paired part lies to the left, so the fork moves to the left.
The drawing below shows the copying machine caught in the act: the two strands paired on the right and pulled apart on the left. The solid-outlined oval at the point of the Y is helicase. The dashed oval further right, on the paired part, is topoisomerase.
(a) Explain how the drawing shows what helicase does. (1 pt)
Frame The drawing shows this because …
To the left of helicase the two strands are apart.
Helicase sits at the point where the paired part turns into the parted part.
So helicase is breaking the hydrogen bonds between the paired bases there, one pair after another.
Behind it each strand is on its own: a template strand.
- Award 1 point for: helicase sits at the boundary between paired and parted strands, so it is the enzyme breaking the hydrogen bonds between the paired bases and separating the strands.
(b) Explain how topoisomerase keeps the paired part ahead of the fork from knotting. (1 pt)
Frame Topoisomerase keeps it from knotting because …
The twist is pushed into the paired part ahead, so the DNA there twists tighter: supercoiling.
Topoisomerase cuts one strand ahead of the fork.
The cut strand turns, so the extra twist unwinds.
Topoisomerase then seals the strand.
So the DNA ahead stays relaxed, and the fork keeps moving.
- Award 1 point for: unwinding at the fork over-twists the DNA ahead (supercoiling); topoisomerase cuts one strand, lets the twist unwind and rejoins the strand, so the DNA ahead stays relaxed.
Glossary
- helicase
- The enzyme that breaks the hydrogen bonds between the paired bases and parts the two strands of a DNA molecule, opening the double helix at the replication fork so that each strand stands alone as a template strand.
- replication fork
- The Y-shaped point where the two strands of a DNA molecule come apart during copying. Helicase sits at the fork; the fork moves into the part of the molecule that is still paired.
- supercoiling
- Extra twisting of the DNA ahead of the replication fork: helicase untwists the strands at the fork and pushes the twist into the paired part ahead, so the DNA there twists tighter than a double helix normally is.
- topoisomerase
- The enzyme that cuts one strand of the DNA ahead of the replication fork, lets the extra twist unwind and seals the strand again, so the fork keeps moving. It relaxes the twist; it does not separate the strands.
APBIO-U06-L07 Only at the 3′ end
The fork is open and one template strand lies exposed. It reads 3′-ACGTTA-5′. Free nucleotides float nearby, and the enzyme that will join them, DNA polymerase, is ready.
Which end of the new strand will DNA polymerase build from? And why does nothing happen until a short piece of RNA is in place?
Unit 6 · Gene Expression and Regulation
1Nucleotides join at the 3′ end
A cell builds a new strand of DNA one nucleotide at a time.
At which end does the cell add every new nucleotide?
- A. At the 5′ endThe 5′ end never gains a nucleotide.
Every new nucleotide joins at the 3′ end. - B. ✓ At the 3′ end
Why: The strand’s last sugar carries a free hydroxyl group on its carbon 3.
The incoming nucleotide’s phosphate group bonds to that hydroxyl group.
So every new nucleotide joins at the 3′ end.
The fork drawn below moves to the right. Both ends of each template strand are marked.
Which end of the upper template faces the fork?
- A. ✓ The upper template’s 5′ end
- B. The upper template’s 3′ endRead the marks on the upper template: 3′ at its far left end, 5′ near the fork.
So its 5′ end faces the fork.
Why: The upper template is marked 3′ at its far left end and 5′ near the fork.
So the upper template’s 5′ end faces the fork.
Which way does a new strand grow? DNA polymerase can only join a nucleotide to the free 3′ end of a strand that already exists.
DNA polymerase chooses each nucleotide by pairing it with the template base.
So every new strand grows 5′ to 3′. DNA polymerase reads its template 3′ to 5′.
DNA polymerase cannot start a strand from nothing. So a short piece of RNA paired to the template gives it the first 3′ end.
That piece of RNA is later replaced with DNA. This one rule, join only at the 3′ end, decides which way every new strand is built.
Video: Watch: Nucleotides join at the 3′ end
The fork’s upper template exposed; a nucleotide arriving at the new strand’s 3′ end; the strand lengthening one nucleotide at a time, its growth arrow pointing toward the fork.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L07a.mp4
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The fork below moves to the right. The upper template lies exposed, with its 3′ end at the far left and its 5′ end at the fork.
Suppose a new strand has started against the upper template. The new strand lies antiparallel to the template, so its 5′ end is at the left and its 3′ end faces the fork.
An enzyme sits at the new strand’s 3′ end. It picks up a free nucleotide whose base pairs with the next template base.
The enzyme joins that nucleotide to the strand’s free 3′ end. The joined nucleotide brings a new free 3′ end, so the next nucleotide joins there.
The 5′ end never gains a nucleotide. Nucleotide by nucleotide, the strand lengthens toward its 3′ end.
The enzyme that joins nucleotides one at a time to the 3′ end of a growing strand is called .
A polymer is a long chain of repeating units. The ending -ase marks an enzyme.
So DNA polymerase is the enzyme that builds the DNA chain.
A polymerase can only join a new nucleotide to the free 3′ end of a strand, so every new strand grows in the 5′ to 3′ direction.
DNA polymerase reads the template the other way. It moves along the template from the template’s 3′ end toward its 5′ end.
On the upper template, the new strand’s 3′ end faces the fork. So this new strand grows toward the fork.
What you are expected to know Describe what DNA polymerase does: it joins nucleotides one at a time to the free 3′ end of a growing strand, choosing each by pairing it with the template base.
What you are expected to know State which way a new strand grows: from its 5′ end toward its 3′ end, 5′ to 3′.
A cell is copying its DNA.
What is DNA polymerase?
- A. ✓ The enzyme that joins nucleotides to a growing strand’s 3′ end
- B. The enzyme that parts the two template strands at the forkThe enzyme that parts the two strands at the fork is helicase.
DNA polymerase joins nucleotides to the growing strand. - C. The enzyme that joins each new nucleotide to a growing strand’s 5′ endA strand’s 5′ end never gains a nucleotide.
DNA polymerase joins each nucleotide at the free 3′ end.
Why: DNA polymerase is the enzyme that joins nucleotides one at a time to the free 3′ end of a growing strand, choosing each by pairing it with the template base.
In the drawing below, the dark strand is the template and the light strand is the new strand being built. Both ends of each strand are marked.
Which end of the new strand receives the next nucleotide?
- A. The left endThe left end of the new strand is its 5′ end.
DNA polymerase joins each new nucleotide at the 3′ end. - B. ✓ The right end
Why: Read the new strand’s ends: 5′ at the left, 3′ at the right.
DNA polymerase joins each new nucleotide at the free 3′ end.
So the next nucleotide joins at the right end.
In the drawing below, the dark strand is the template and the light strand is the new strand being built. Both ends of each strand are marked.
Which end of the new strand receives the next nucleotide?
- A. ✓ The left end
- B. The right endThe right end of the new strand is its 5′ end.
DNA polymerase joins each new nucleotide at the 3′ end.
Why: Read the new strand’s ends: 3′ at the left, 5′ at the right.
DNA polymerase joins each new nucleotide at the free 3′ end.
So the next nucleotide joins at the left end.
In the drawing below, a new strand is drawn on its own. Both of its ends are marked.
Which end of the new strand receives the next nucleotide?
- A. ✓ The left end
- B. The right endThe right end of this strand is its 5′ end.
DNA polymerase joins each new nucleotide at the 3′ end.
Why: Read the strand’s ends: 3′ at the left, 5′ at the right.
DNA polymerase joins each new nucleotide at the free 3′ end.
So the next nucleotide joins at the left end.
In the drawing below, a new strand is drawn on its own. Both of its ends are marked.
Which way does this new strand grow?
- A. To the leftThe strand’s 3′ end is at the right.
A strand grows toward its 3′ end. - B. ✓ To the right
Why: Read the strand’s ends: 5′ at the left, 3′ at the right.
Every new nucleotide joins at the 3′ end.
So the strand grows to the right.
In the drawing below, the dark strand is the template and the light strand is the new strand being built. Both ends of each strand are marked.
Which way does the new strand grow?
- A. ✓ To the left
- B. To the rightThe new strand’s 3′ end is at the left.
A strand grows toward its 3′ end.
Why: Read the new strand’s ends: 3′ at the left, 5′ at the right.
Every new nucleotide joins at the 3′ end.
So the new strand grows to the left.
A student watches a new DNA strand being built and says: “DNA polymerase adds each nucleotide at the 5′ end, so the new strand grows 3′ to 5′.”
Is the student correct?
- A. Yes: the new strand grows 3′ to 5′The 5′ end of a strand never gains a nucleotide.
DNA polymerase joins each nucleotide at the free 3′ end, so the strand grows 5′ to 3′. - B. ✓ No: the new strand grows 5′ to 3′
Why: DNA polymerase can only join a nucleotide to the free 3′ end of a strand.
So the strand lengthens at its 3′ end and its 5′ end stays where it began.
The new strand grows 5′ to 3′.
We say “DNA polymerase” as if it were one enzyme. A cell has several, each with its own job. Here they are all called DNA polymerase.
31Quick quiz: DNA polymerase mixed practice
DNA polymerase reaches a template base G.
Which nucleotide does DNA polymerase join to the growing strand?
- A. One carrying adenineAdenine pairs with thymine, not with guanine.
Against a template G, DNA polymerase adds a nucleotide carrying cytosine. - B. ✓ One carrying cytosine
- C. One carrying guanineGuanine does not pair with guanine; it pairs with cytosine.
DNA polymerase adds each nucleotide by pairing it with the template base.
Why: DNA polymerase chooses each nucleotide by pairing it with the template base.
Guanine pairs with cytosine.
So against a template G it joins a nucleotide carrying cytosine.
A cell is copying its DNA.
(a) State what DNA polymerase does. (1 pt)
- Award 1 point for: joins nucleotides to the 3′ end of the growing strand (new strand grows 5′ to 3′), each chosen by pairing with the template base.
A new strand being built so far reads 5′-GCA-3′.
Beside which base does DNA polymerase join the next nucleotide?
- A. ✓ Beside the A
- B. Beside the GThe G is at the strand’s 5′ end.
DNA polymerase joins each nucleotide at the 3′ end, beside the A.
Why: The strand is written 5′ to 3′, so its 3′ end is the A.
DNA polymerase joins each nucleotide at the free 3′ end.
So the next nucleotide joins beside the A.
35Why the polymerase needs a primer
Video: Watch: Why the polymerase needs a primer
A tube holding template strands, free nucleotides and DNA polymerase, and nothing happening in it; a short piece of RNA landing on a template; the first DNA nucleotide joining the RNA’s 3′ end and the new strand growing from there.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L07b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L07b.mp4
Suppose a test tube holds DNA template strands, free DNA nucleotides and DNA polymerase, and nothing else.
No new strand appears. DNA polymerase adds not one nucleotide.
DNA polymerase can only join a nucleotide to the free 3′ end of a strand that already exists. In the tube no new strand exists, so there is no free 3′ end to join to.
So DNA polymerase cannot start a strand from nothing. It can only lengthen a strand that has already begun.
In the cell, another enzyme gives DNA polymerase a start. It lays down a short piece of RNA paired to the template.
Look at the start of the new strand in the fork drawing: the small dark stub. That stub is the short piece of RNA.
A short piece of RNA laid down on the template to start a new strand is called an , because it gets the new strand started.
The RNA primer has a free 3′ end. DNA polymerase joins the first DNA nucleotide to that 3′ end.
Then the second nucleotide joins, and the third. The new strand grows from the RNA primer toward the fork.
Later, enzymes in the cell remove the RNA primer and replace it with DNA. So the finished strand is DNA from end to end.
Suppose the same test tube also holds RNA primers paired to its templates. DNA polymerase joins DNA nucleotides to each primer’s 3′ end, and new strands appear.
What you are expected to know Explain why DNA polymerase needs an RNA primer: DNA polymerase cannot start a strand from nothing, and the RNA primer supplies the first free 3′ end.
A cell is copying its DNA.
What is an RNA primer?
- A. The enzyme that joins DNA nucleotides to the growing strand’s 3′ endThe enzyme that joins nucleotides to the growing strand is DNA polymerase.
The RNA primer is the short piece of RNA it builds from. - B. A short piece of DNA that DNA polymerase copies before the templateThe primer is made of RNA, not DNA.
DNA polymerase does not copy it; it joins DNA nucleotides to the primer’s 3′ end. - C. ✓ A short piece of RNA paired to the template, giving the first free 3′ end
Why: An RNA primer is a short piece of RNA laid down paired to the template.
Its free 3′ end is where DNA polymerase joins the first DNA nucleotide.
Suppose a test tube holds DNA template strands, all four free DNA nucleotides and DNA polymerase. It holds no RNA at all.
Is any new DNA strand built in this tube?
- A. YesDNA polymerase can only lengthen a strand from a free 3′ end.
With no RNA primer there is no free 3′ end, so no nucleotide is added. - B. ✓ No
Why: DNA polymerase can only join a nucleotide to the free 3′ end of a strand that already exists.
The tube holds no RNA primer, so no strand has begun and there is no free 3′ end.
So no new DNA strand is built.
Imagine a cell in which the enzyme that lays down RNA primers is blocked by a drug. Helicase opens forks as usual, and the templates lie exposed, but no new strand starts.
(a) Explain why no new strand starts in this cell. (1 pt)
Frame No new strand starts because …
With the primer enzyme blocked, no short piece of RNA is laid down on the exposed templates.
So no strand has begun, and there is no free 3′ end.
So DNA polymerase adds no nucleotide.
- Award 1 point for: DNA polymerase can only add to an existing free 3′ end (cannot start a strand), and with no primer laid down there is no free 3′ end.
A student says: “DNA polymerase only lengthens a strand that has already begun, so the first 3′ end it adds to belongs to a primer.”
Is the student correct?
- A. ✓ Yes: the primer is laid down first, so its 3′ end is the first end DNA polymerase can add to
- B. No: DNA polymerase begins each strand itself, and the primer only marks where to startDNA polymerase joins nucleotides only to a free 3′ end that already exists.
The primer supplies that first 3′ end.
Why: DNA polymerase joins a nucleotide only to the free 3′ end of a strand that already exists.
An enzyme lays down a short RNA primer paired to the template.
So the first 3′ end DNA polymerase adds to is the primer’s.
Suppose a test tube holds DNA template strands, free DNA nucleotides, DNA polymerase, and RNA primers paired to the templates.
What does DNA polymerase do in this tube?
- A. ✓ DNA polymerase joins DNA nucleotides to each primer’s 3′ end
- B. DNA polymerase joins DNA nucleotides to each primer’s 5′ endA strand’s 5′ end never gains a nucleotide.
DNA polymerase joins each nucleotide at the primer’s free 3′ end. - C. DNA polymerase adds no nucleotideEach RNA primer supplies a free 3′ end paired to a template.
DNA polymerase joins DNA nucleotides to that 3′ end.
Why: Each RNA primer is paired to a template and has a free 3′ end.
DNA polymerase can join a nucleotide to a free 3′ end.
So DNA polymerase joins DNA nucleotides to each primer’s 3′ end, and new strands grow.
54Quick quiz: primer (RNA primer) mixed practice
A cell is copying its DNA.
(a) State what an RNA primer is. (1 pt)
- Award 1 point for: a short piece of RNA paired to the template, at the start of a new strand. Accept: the piece of RNA that supplies the first free 3′ end.
A new DNA strand has been finished in the cell.
What has happened to the RNA primer that started it?
- A. The RNA primer stays as RNA at the strand’s 5′ endThe RNA primer does begin the strand at its 5′ end, but it does not stay.
Enzymes remove the RNA primer and replace it with DNA. - B. ✓ Enzymes have replaced the RNA primer with DNA
- C. The RNA primer stays as RNA at the strand’s 3′ endThe primer sits at the strand’s 5′ end, where the strand began, not at its 3′ end.
Enzymes later replace the RNA primer with DNA.
Why: The RNA primer starts the strand.
Enzymes later remove the RNA primer and replace it with DNA.
So the finished strand is DNA from end to end.
57Write the new strand
One strand of DNA carries a C. A partner strand is being built against it.
Which base does the partner strand carry opposite the C?
- A. AdenineAdenine pairs with thymine.
Opposite a C sits a G. - B. CytosineCytosine does not pair with cytosine.
Opposite a C sits a G. - C. ✓ Guanine
Why: In DNA, cytosine pairs with guanine.
So the partner strand carries a G opposite the C.
Video: Watch: Write the new strand
The template 3′-ACGTTA-5′ written out with its RNA primer at the left; T, then G, C, A, A, T appearing one at a time to the right of the primer; the new strand 5′-TGCAAT-3′ and its growth arrow.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L07c.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L07c.mp4
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
A template strand is often written the other way round, with its 3′ end at the left. Then the new strand written beneath it starts from its 5′ end at the left.
The exposed template below reads 3′-ACGTTA-5′, with an RNA primer paired to it just before its first written base, at the left. The three blanks stand for template bases that are not written out.
DNA polymerase joins the first nucleotide to the RNA primer’s 3′ end. The first template base is A, so the first nucleotide added carries T.
DNA polymerase then moves to the right along the template, one base at a time.
It adds the nucleotides in this order:
- against A it adds T
- against C it adds G
- against G it adds C
- against T it adds A
- against T it adds A
- against A it adds T
The new strand reads 5′-TGCAAT-3′. Its 3′ end is at the right, so it grew to the right.
To write the new strand against a template written 3′ to 5′:
- Under each template base, write its partner: T under A, A under T, C under G, G under C.
- Mark the new strand 5′ at the left and 3′ at the right.
- The new strand grows toward its 3′ end, so its arrow points to the right.
What you are expected to know Write the new strand DNA polymerase builds against a template with its ends marked, and state which way the new strand grows.
Both ends of each strand are marked. The template is written 3′ to 5′, so that the new strand built against it reads 5′ to 3′ beneath it. A template strand reads 3′-TGCAGT-5′, and an RNA primer is paired to it just before the first written base, as drawn below.
Which base does the first nucleotide DNA polymerase adds carry?
- A. ✓ Adenine
- B. ThymineThe first template base is T, and thymine does not pair with thymine.
Opposite T, DNA polymerase adds a nucleotide carrying adenine. - C. UracilUracil belongs to RNA.
DNA polymerase adds DNA nucleotides, and opposite T it adds one carrying adenine.
Why: The first template base, beside the primer, is T.
Thymine pairs with adenine.
So the first nucleotide added carries adenine.
Both ends of each strand are marked. The template is written 3′ to 5′, so that the new strand built against it reads 5′ to 3′ beneath it. A template strand reads 3′-TGCAGT-5′, and an RNA primer is paired to it just before the first written base, as drawn below.
Which way does the new strand grow along this template?
- A. To the leftThe new strand’s 5′ end is at the primer, at the left.
A strand grows toward its 3′ end, so this one grows to the right. - B. ✓ To the right
Why: The RNA primer sits at the left, so the new strand’s 5′ end is at the left.
Every new nucleotide joins at the 3′ end.
So the new strand grows to the right.
Both ends of each strand are marked. The template is written 3′ to 5′, so that the new strand built against it reads 5′ to 3′ beneath it. A template strand reads 3′-GGATCA-5′, and an RNA primer is paired to it just before the first written base, as drawn below.
Written with both ends marked, what does the finished new strand read?
- A. 5′-GGATCA-3′Copying the template’s own letters gives the template again, not each base’s partner.
Under G sits C, under A sits T. - B. ✓ 5′-CCTAGT-3′
- C. 3′-CCTAGT-5′The new strand begins at the primer, at the left.
So its 5′ end is at the left and its 3′ end is at the right. - D. 5′-CCUAGU-3′Uracil belongs to RNA.
DNA polymerase adds DNA nucleotides, so under A sits T.
Why: Under each template base write its partner: G gives C, G gives C, A gives T, T gives A, C gives G, A gives T.
The new strand begins at the primer, so its 5′ end is at the left and its 3′ end at the right.
It reads 5′-CCTAGT-3′.
Both ends of each strand are marked. The template is written 3′ to 5′, so that the new strand built against it reads 5′ to 3′ beneath it. A template strand reads 3′-ATCCGA-5′, and an RNA primer is paired to it just before the first written base, as drawn below.
Written with both ends marked, what does the finished new strand read?
- A. ✓ 5′-TAGGCT-3′
- B. 5′-UAGGCU-3′Uracil belongs to RNA.
DNA polymerase adds DNA nucleotides, so under A sits T. - C. 3′-TAGGCT-5′The new strand begins at the primer, at the left.
So its 5′ end is at the left and its 3′ end is at the right. - D. 5′-ATCCGA-3′Copying the template’s own letters gives the template again, not each base’s partner.
Under A sits T, under T sits A.
Why: Under each template base write its partner: A gives T, T gives A, C gives G, C gives G, G gives C, A gives T.
The new strand begins at the primer, so its 5′ end is at the left and its 3′ end at the right.
It reads 5′-TAGGCT-3′.
In the drawing below, an RNA primer lies paired under the left part of a DNA template strand. The template’s 3′ end and 5′ end are marked. The four ends of the two strands are lettered J, K, L and M.
Which lettered end does DNA polymerase join nucleotides to?
- A. JEnd J is the template’s free 3′ end.
No template base lies opposite it, so DNA polymerase has nothing to pair the next nucleotide with. - B. KEnd K is the template’s 5′ end.
A 5′ end never gains a nucleotide. - C. LEnd L is the primer’s 5′ end.
A 5′ end never gains a nucleotide. - D. ✓ M
Why: The template’s 3′ end is J, so the primer, antiparallel to it, has its 3′ end at M.
End M is a free 3′ end with template bases beyond it to pair against.
So DNA polymerase joins nucleotides to end M.
Go back to the open fork, where the exposed upper template reads 3′-ACGTTA-5′, free nucleotides float nearby and DNA polymerase is ready.
An enzyme lays down a short RNA primer paired at the template’s 3′ end. DNA polymerase adds T, then G, then C, A, A, T to the primer’s 3′ end.
The new strand reads 5′-TGCAAT-3′. It grew to the right, toward the fork.
77Mixed practice mixed practice
Both ends of each strand are marked. The template is written 3′ to 5′, so that the new strand built against it reads 5′ to 3′ beneath it. A template strand reads 3′-GATTCG-5′, and an RNA primer is paired to it just before the first written base, as drawn below.
Written with both ends marked, what does the finished new strand read?
- A. ✓ 5′-CTAAGC-3′
- B. 3′-CTAAGC-5′The new strand begins at the primer, at the left.
So its 5′ end is at the left and its 3′ end is at the right. - C. 5′-GATTCG-3′Copying the template’s own letters gives the template again, not each base’s partner.
Under G sits C, under A sits T.
Why: Under each template base write its partner: G gives C, A gives T, T gives A, T gives A, C gives G, G gives C.
The new strand begins at the primer, so its 5′ end is at the left and its 3′ end at the right.
It reads 5′-CTAAGC-3′.
A test tube holds DNA template strands, DNA polymerase and RNA primers paired to the templates, but no free DNA nucleotides.
What happens in the tube?
- A. DNA polymerase builds a new DNA strand against each templateDNA polymerase joins free DNA nucleotides to a primer’s 3′ end.
The tube holds no free DNA nucleotides, so it has nothing to add. - B. ✓ DNA polymerase adds no nucleotide
- C. DNA polymerase lengthens each RNA primer with more RNADNA polymerase adds DNA nucleotides only.
The tube holds none, so nothing is added to the primers.
Why: Each RNA primer supplies a free 3′ end.
DNA polymerase joins free DNA nucleotides to that end.
The tube holds no free DNA nucleotides, so DNA polymerase has nothing to add.
DNA polymerase moves along a template strand as it builds the new strand.
Which way does DNA polymerase move along the template?
- A. ✓ Toward the template’s 5′ end
- B. Toward the template’s 3′ endThe new strand grows 5′ to 3′ and lies antiparallel to the template.
So DNA polymerase moves along the template toward the template’s 5′ end.
Why: The new strand grows 5′ to 3′.
The new strand lies antiparallel to the template.
So DNA polymerase moves along the template from its 3′ end toward its 5′ end.
A student says: “DNA polymerase lays down the RNA primer, then adds DNA nucleotides to it.”
Is the student correct?
- A. ✓ No: another enzyme lays down the primer; DNA polymerase only adds to it
- B. Yes: DNA polymerase itself lays down the RNA and then adds DNA to itDNA polymerase can only join nucleotides to a free 3′ end that already exists.
Another enzyme lays down the RNA primer.
Why: DNA polymerase cannot start a strand from nothing.
Another enzyme lays down the short RNA primer paired to the template.
DNA polymerase then joins DNA nucleotides to the primer’s 3′ end.
A new strand being built so far reads 5′-TTGC-3′. DNA polymerase joins a nucleotide carrying G to it.
What does the strand read now?
- A. 5′-GTTGC-3′DNA polymerase joins each nucleotide at the 3′ end, after the C.
A 5′ end gains no nucleotide. - B. ✓ 5′-TTGCG-3′
- C. 3′-TTGCG-5′The strand keeps its 5′ end at the left and its 3′ end at the right.
The new G joins at the 3′ end, on the right.
Why: The strand is written 5′ to 3′, so its 3′ end is the C.
DNA polymerase joins the new nucleotide at that 3′ end.
So the strand now reads 5′-TTGCG-3′.
An RNA primer is paired to a DNA template strand. The primer’s 3′ end points to the right.
Which way does the new DNA strand grow from the primer?
- A. To the leftDNA polymerase joins the first DNA nucleotide to the primer’s free 3′ end, on the right.
Each later nucleotide joins the new 3′ end, so the strand grows right. - B. ✓ To the right
Why: DNA polymerase joins the first DNA nucleotide to the primer’s free 3′ end.
That end points to the right.
Each new nucleotide joins the new 3′ end, so the strand grows to the right.
Two test tubes each hold DNA template strands, all four free DNA nucleotides and DNA polymerase. Tube 1 holds nothing else. Tube 2 also holds short RNA primers paired to the templates. After an hour, tube 1 holds no new DNA. Tube 2 holds new DNA strands, each grown from a primer in the 5′ to 3′ direction.
(a) Explain how the results of the two tubes demonstrate that DNA polymerase can only lengthen a strand that has already begun. (1 pt)
Frame The two tubes demonstrate this because …
DNA polymerase can only join a nucleotide to the free 3′ end of a strand that already exists.
In tube 1 no strand has begun, so there is no free 3′ end and DNA polymerase adds nothing.
In tube 2 each RNA primer supplies a free 3′ end, so DNA polymerase joins nucleotides to it.
So DNA polymerase lengthens a strand that has begun and starts none.
- Award 1 point for: the tubes differ only in the primers; DNA polymerase adds nucleotides only to an existing free 3′ end, which the primer supplies, so without a primer no strand is started.
(b) Explain why each new strand in tube 2 grew in the 5′ to 3′ direction. (1 pt)
Frame Each new strand grew 5′ to 3′ because …
Each joined nucleotide brings a new free 3′ end, so the strand lengthens at its 3′ end.
The 5′ end, at the primer, never gains a nucleotide.
So every new strand grows in the 5′ to 3′ direction.
- Award 1 point for: nucleotides are joined only at the free 3′ end (the 5′ end never gains one), so the strand lengthens 5′ to 3′.
Glossary
- DNA polymerase
- The enzyme that joins nucleotides one at a time to the free 3′ end of a growing strand, choosing each nucleotide by pairing it with the template base, so every new strand grows 5′ to 3′.
- primer (RNA primer)
- A short piece of RNA laid down paired to the template to start a new strand: it gives DNA polymerase the first free 3′ end to join nucleotides to. Enzymes later replace it with DNA.
APBIO-U06-L08 One strand in pieces
Look at the fork again, now with both new strands drawn. The upper new strand is one long piece pointing into the fork. The lower new strand is three short pieces pointing away from the fork, with gaps between them. The same DNA polymerase built both, by the same rule.
Why do the two strands differ?
Unit 6 · Gene Expression and Regulation
1Which new strand is which
DNA polymerase is building a new strand against a template.
To which end of the new strand does it join the next nucleotide?
- A. The 5′ endThe 5′ end carries a phosphate group and never gains a nucleotide.
DNA polymerase joins each nucleotide to the free 3′ end. - B. ✓ The 3′ end
Why: DNA polymerase can only join a nucleotide to the free 3′ end of a strand.
So every new strand grows in the 5′ to 3′ direction.
A replication fork is drawn below, moving to the right. Both templates’ ends are marked.
Which end of the upper template faces the fork?
- A. ✓ The upper template’s 5′ end
- B. The upper template’s 3′ endThe upper template’s 3′ end is at its far left, away from the fork.
Its 5′ end is at the fork.
Why: Read the marks on the upper template.
Its far left end is marked 3′ and the end at the fork is marked 5′.
So the upper template’s 5′ end faces the fork.
Why is one new strand made in pieces? The two template strands run in opposite directions, and every new strand can only grow 5′ to 3′.
On one template the new strand grows toward the opening fork. So it follows the fork in one long piece.
On the other template the new strand grows away from the fork. So DNA polymerase must start again each time the fork exposes more template. That new strand is made in short pieces.
A third enzyme joins the pieces into one strand. This part of the fork is the one the exam asks about most.
Video: Watch: Which new strand is which
The fork with both templates’ ends written; the upper new strand extending toward the fork as one arrow; the lower new strand appearing as three short arrows pointing away from the fork, each from its own primer; the two names arriving beside them.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L08a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L08a.mp4
The fork below has both new strands drawn. Both new strands are light, because both are new.
Read the ends of the two template strands first. The upper template has its 3′ end at the far left and its 5′ end at the fork.
The lower template has its 5′ end at the far left and its 3′ end at the fork.
Look at the upper new strand. It is one long piece, and its arrowhead faces the fork.
Look at the lower new strand. It is three short pieces, and each piece’s arrowhead faces away from the fork.
Each short piece starts from its own primer, drawn as a small dark stub at the piece’s fork-side end.
The new strand built in one long piece toward the fork is called the , because it leads the way: it follows the fork as the fork opens.
The new strand built in short pieces away from the fork is called the , because it lags behind: each piece can start only after the fork has exposed more template.
The ends decide. The leading strand is built against the template whose 5′ end faces the fork.
The lagging strand is built against the template whose 3′ end faces the fork.
For example, take the upper template of this fork: its 5′ end faces the fork. So the leading strand is built against it.
But take the lower template of the same fork: its 3′ end faces the fork. So the lagging strand is built against it.
And take the upper template of this fork, which moves to the left: its 5′ end faces the fork. So the leading strand is built against it.
But take the upper template of this fork, which also moves to the left: its 3′ end faces the fork. So the lagging strand is built against it.
The template may be drawn on top or underneath. The fork may move either way. Only the template’s end at the fork decides.
5′ at the fork gives the leading strand, and 3′ at the fork gives the lagging strand.
The table below compares the leading strand with the lagging strand: the way each grows, the pieces, the primers and the template’s end at the fork.
The leading strand needs one primer, at its start. The lagging strand needs a primer for every piece.
What you are expected to know Identify, on a drawn fork with the templates’ ends marked, which new strand is the leading strand and which is the lagging strand.
Two new strands are being built at a replication fork.
What is the leading strand?
- A. ✓ The new strand built in one long piece toward the fork
- B. The new strand that starts again each time the fork exposes more templateThe new strand that starts again each time the fork exposes more template is the lagging strand, made in short pieces.
- C. The template strand the one long piece is built againstThe leading strand is the new strand, not its template.
The template is the old strand it is built against.
Why: The leading strand is the new strand built in one long piece toward the fork.
Two new strands are being built at a replication fork.
What is the lagging strand?
- A. The template strand the short pieces are built againstThe lagging strand is the new strand, not its template.
The template is the old strand the pieces are built against. - B. The new strand that follows the fork from one primerThe new strand that follows the fork from one primer is the leading strand, built in one long piece.
- C. ✓ The new strand built in short pieces away from the fork
Why: The lagging strand is the new strand built in short pieces away from the fork.
A replication fork is drawn below, moving to the right. Both templates’ ends are marked, and no new strand has been built yet.
Against which template does DNA polymerase build the leading strand?
- A. ✓ The upper template
- B. The lower templateThe lower template has its 3′ end at the fork.
The leading strand is built against the template whose 5′ end faces the fork.
Why: The upper template has its 5′ end at the fork.
The leading strand is built against the template whose 5′ end faces the fork.
So DNA polymerase builds the leading strand against the upper template.
A replication fork is drawn below, moving to the right. Both templates’ ends are marked, and no new strand has been built yet.
Against which template does DNA polymerase build the leading strand?
- A. The upper templateThe upper template has its 3′ end at the fork.
The leading strand is built against the template whose 5′ end faces the fork. - B. ✓ The lower template
Why: The lower template has its 5′ end at the fork.
The leading strand is built against the template whose 5′ end faces the fork.
So DNA polymerase builds the leading strand against the lower template.
A replication fork is drawn below, moving to the left. Both templates’ ends are marked, and no new strand has been built yet.
Against which template does DNA polymerase build the leading strand?
- A. The upper templateThe upper template has its 3′ end at the fork.
The leading strand is built against the template whose 5′ end faces the fork. - B. ✓ The lower template
Why: The lower template has its 5′ end at the fork.
The leading strand is built against the template whose 5′ end faces the fork, whichever way the fork opens.
So DNA polymerase builds the leading strand against the lower template.
A replication fork is drawn below, moving to the left. Both templates’ ends are marked, and no new strand has been built yet.
Against which template does DNA polymerase build the leading strand?
- A. ✓ The upper template
- B. The lower templateThe lower template has its 3′ end at the fork.
The leading strand is built against the template whose 5′ end faces the fork.
Why: The upper template has its 5′ end at the fork.
The leading strand is built against the template whose 5′ end faces the fork, whichever way the fork opens.
So DNA polymerase builds the leading strand against the upper template.
A replication fork is drawn below, moving to the right. Both templates’ ends are marked, and no new strand has been built yet.
Against which template does DNA polymerase build the lagging strand?
- A. ✓ The upper template
- B. The lower templateThe lower template has its 5′ end at the fork, so it gives the leading strand.
The lagging strand is built against the template whose 3′ end faces the fork.
Why: The upper template has its 3′ end at the fork.
The lagging strand is built against the template whose 3′ end faces the fork.
So DNA polymerase builds the lagging strand against the upper template.
A replication fork is drawn below, moving to the left. Both templates’ ends are marked, and no new strand has been built yet.
Against which template does DNA polymerase build the lagging strand?
- A. The upper templateThe upper template has its 5′ end at the fork, so it gives the leading strand.
The lagging strand is built against the template whose 3′ end faces the fork. - B. ✓ The lower template
Why: The lower template has its 3′ end at the fork.
The lagging strand is built against the template whose 3′ end faces the fork.
So DNA polymerase builds the lagging strand against the lower template.
36Quick quiz: leading strand, lagging strand mixed practice
Two new strands are being built at a replication fork.
(a) State what the leading strand and the lagging strand are. (1 pt)
The lagging strand is the new strand built in short pieces away from the fork, against the template whose 3′ end faces the fork.
- Award 1 point for: the leading strand is built continuously toward the fork and the lagging strand in short pieces away from the fork.
At a replication fork, one new strand is being built in short pieces, each from its own primer.
Which new strand is it?
- A. The leading strandThe leading strand is built in one long piece from one primer.
- B. ✓ The lagging strand
Why: The new strand built in short pieces, each from its own primer, is the lagging strand.
At a replication fork, one new strand grows toward the fork in one long piece.
Which new strand is it?
- A. ✓ The leading strand
- B. The lagging strandThe lagging strand grows away from the fork in short pieces.
Why: The new strand that grows toward the fork in one long piece is the leading strand.
At a replication fork, one new strand grows away from the fork, and it starts again each time the fork exposes more template.
Which new strand is it?
- A. The leading strandThe leading strand grows toward the fork in one long piece and never starts again.
- B. ✓ The lagging strand
Why: The new strand that grows away from the fork and starts again each time more template is exposed is the lagging strand.
41Why one strand is in pieces
The two strands of a DNA molecule lie side by side, each with a 5′ end and a 3′ end.
In which directions do the two strands lie, 5′ to 3′?
- A. ✓ In opposite directions
- B. In the same directionThe two strands are antiparallel.
Where one strand has its 5′ end, the other has its 3′ end.
Why: The two strands of DNA are antiparallel.
Antiparallel means side by side, with the two strands in opposite 5′ to 3′ directions.
Video: Watch: Why one strand is in pieces
The two templates’ directions drawn as opposite arrows; the leading strand’s 3′ end following the fork; on the other template the 3′ end pointing away, a new primer landing each time the fork exposes more template, and the short pieces appearing one at a time.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L08b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L08b.mp4
Go back to the fork. The two template strands were the two paired strands of one molecule.
So the two template strands are antiparallel: they run in opposite directions. Where the upper template has its 5′ end, at the fork, the lower template has its 3′ end.
Every new strand pairs antiparallel with its template too. So where a template has its 5′ end, the new strand built against it has its 3′ end.
DNA polymerase can only add a nucleotide to a free 3′ end. So every new strand grows 5′ to 3′: its growing end is its 3′ end.
The upper template has its 5′ end at the fork. So the new strand built against it has its 3′ end at the fork.
That growing 3′ end points at the fork. As the fork opens, it exposes more of the upper template right at that growing end.
So DNA polymerase keeps adding nucleotides, and the leading strand follows the fork in one long piece.
The lower template has its 3′ end at the fork. So the new strand built against it has its 5′ end at the fork, and its growing 3′ end points away from the fork.
Below, a short stretch of each template is written out near the fork, with both ends marked, and the new strand built against it written beneath.
As the fork opens, it exposes more of the lower template at the fork. But the piece’s growing 3′ end is at its far end, pointing away from that newly exposed template.
DNA polymerase cannot add a nucleotide to the piece’s 5′ end. So that piece cannot grow into the newly exposed template.
Instead, an enzyme lays down a short RNA primer on the newly exposed template, near the fork. DNA polymerase starts a new piece from that primer, and the new piece grows away from the fork.
Each time the fork exposes more of the lower template, the same happens again: a new primer, then a new piece.
So the lagging strand is made in short pieces, each with its own primer, with a gap between one piece and the next.
The whole reason fits in one sentence.
The two template strands run in opposite directions and every new strand can only grow 5′ to 3′, so on one template the new strand grows toward the opening fork in one piece and on the other it must grow away from the fork in short pieces that start again each time more template is exposed.
The same DNA polymerase built both strands by the same rule. The two strands differ only because their templates run in opposite directions.
What you are expected to know Explain why the leading strand is made in one piece and the lagging strand in short pieces.
Two new strands are being built at a replication fork.
Which new strand needs several primers?
- A. The leading strandThe leading strand grows in one piece from one primer.
Each short piece of the lagging strand needs its own primer. - B. ✓ The lagging strand
Why: The lagging strand is made in short pieces.
Each piece starts from its own primer.
So the lagging strand needs several primers.
Suppose a fork moves to the left. The lower template has its 3′ end at the fork, and DNA polymerase is building a new strand against it.
(a) Explain why the new strand built against the lower template is made in short pieces. (2 pt)
Frame This new strand is made in short pieces because …
The new strand pairs antiparallel with its template, so its growing 3′ end points away from the fork.
As the fork moves left, it exposes new template at the piece’s 5′ end.
DNA polymerase cannot add to a 5′ end, so the piece cannot grow that way.
So an enzyme lays down a new primer there, and a new piece starts.
- Award 1 point for: the template’s 3′ end faces the fork, so the new strand (antiparallel, growing 5′ to 3′) grows away from the fork.
- Award 1 point for: DNA polymerase cannot extend the piece toward the fork, so each newly exposed stretch of template needs a new primer and a new piece.
A student looks at the short pieces being built at a fork and says: “These pieces grow 3′ to 5′, away from the fork.”
Is the student correct?
- A. ✓ No: every new strand grows 5′ to 3′, the short pieces included
- B. Yes: the pieces grow 3′ to 5′, away from the forkDNA polymerase adds each nucleotide to a free 3′ end.
So every new strand grows 5′ to 3′, the lagging strand included.
Why: DNA polymerase adds each nucleotide to a free 3′ end, so every new strand grows 5′ to 3′.
Each short piece of the lagging strand grows 5′ to 3′ too.
On its template that direction is away from the fork, so the pieces point away from the fork.
Suppose a fork moves to the left. On the upper template, the new strand grows away from the fork.
What happens each time the fork exposes more of the upper template?
- A. DNA polymerase extends the same piece toward the forkThe piece’s growing 3′ end points away from the fork.
DNA polymerase cannot add to its 5′ end, so it cannot extend into the new template. - B. ✓ An enzyme lays down a new primer and a new piece starts
- C. The new strand starts growing toward the fork insteadEvery new strand grows 5′ to 3′ only.
On this template that direction is away from the fork, and it does not change.
Why: The new strand’s growing 3′ end points away from the fork.
The newly exposed template is at the fork, at the piece’s 5′ end, where DNA polymerase cannot add.
So an enzyme lays down a new primer there and a new piece starts.
66Ligase seals the pieces
Along one DNA strand, each nucleotide is joined to the next.
Which bond joins them?
- A. A hydrogen bond between neighboring basesHydrogen bonds join a base to its partner on the other strand.
Along one strand the nucleotides are joined by covalent bonds in the backbone. - B. ✓ A covalent bond from one sugar to the next phosphate group
Why: Along a strand the nucleotides are joined through the sugar-phosphate backbone.
The bond is a covalent bond from one sugar to the next phosphate group.
Video: Watch: Ligase seals the pieces
The three short pieces of the lagging strand; each RNA primer removed and the space filled with DNA; ligase settling on each break in the backbone and closing it; the lagging strand becoming one continuous strand.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L08c.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L08c.mp4
Go back to the lagging strand: three short pieces of DNA, each begun from its own RNA primer, with a gap between one piece and the next.
Each primer is RNA, and RNA does not belong in a finished DNA strand. So the cell removes each RNA primer, and DNA polymerase fills the space it leaves with DNA.
Now every piece is DNA, and the pieces lie end to end along the template. But the sugar-phosphate backbone is still broken between one piece and the next.
At each break, the last sugar of one piece is not joined to the first phosphate group of the next piece. The bases on both sides are still paired with the template; only the backbone has a break in it.
An enzyme joins the backbone across the break. It forms the covalent bond from the sugar at the end of one piece to the phosphate group at the start of the next.
The enzyme that joins the sugar-phosphate backbone between two neighboring pieces is called , because to ligate means to tie together.
Ligase adds no nucleotides, and it pairs no bases. It only joins a backbone that is broken between two pieces.
After ligase has joined every break, the lagging strand is one continuous strand, just like the leading strand.
Suppose a cell’s ligase is inactive. The cell still copies its DNA, and its leading strand is still one piece.
But no enzyme joins the breaks between the short pieces. So its lagging strand stays in pieces when the copying ends.
What you are expected to know State what ligase does: it joins the sugar-phosphate backbone between neighboring pieces of the lagging strand, leaving one continuous strand.
Several enzymes work at a replication fork.
What is ligase?
- A. The enzyme that adds nucleotides to the 3′ end of a growing strandThe enzyme that adds nucleotides to a 3′ end is DNA polymerase.
- B. The enzyme that breaks the hydrogen bonds between paired basesThe enzyme that breaks the hydrogen bonds between paired bases is helicase.
- C. ✓ The enzyme that joins the sugar-phosphate backbone between two neighboring pieces
Why: Ligase is the enzyme that joins the sugar-phosphate backbone between two neighboring pieces of a new strand.
Imagine a cell that has no working ligase. The cell copies its DNA.
When the copying ends, which new strand is left in many pieces, apart from the one join where its first primer sat?
- A. The leading strandThe leading strand is built in one piece from one primer.
It has no breaks for ligase to join. - B. ✓ The lagging strand
- C. Both new strandsThe leading strand is built in one piece and needs no ligase.
Only the lagging strand has breaks between pieces.
Why: The lagging strand is built in short pieces with a break in the backbone between them.
Ligase joins those breaks.
With no working ligase, the lagging strand stays in pieces.
A student says: “Where each RNA primer was, DNA polymerase fills the space with DNA, and then ligase joins the backbone.”
Is the student correct?
- A. ✓ Yes: DNA polymerase adds the nucleotides, and ligase joins only the backbone
- B. No: ligase fills each primer’s space with DNA and then joins the backboneLigase adds no nucleotides.
DNA polymerase fills the space with DNA, and ligase then joins the backbone.
Why: DNA polymerase adds nucleotides, including the DNA that replaces each primer.
Ligase adds no nucleotides.
Ligase joins the sugar-phosphate backbone between two neighboring pieces.
Go back to the fork with both new strands drawn: the upper new strand one long piece, the lower new strand three short pieces.
The leading strand follows the fork in one piece.
The lagging strand grows away from the fork in short pieces, each from its own primer, and ligase seals the backbone between them.
86Quick quiz: ligase mixed practice
Several enzymes work at a replication fork.
(a) State what ligase does. (1 pt)
- Award 1 point for: ligase joins (forms the covalent bond in) the sugar-phosphate backbone between neighboring pieces of the lagging strand.
A new strand is growing at a replication fork.
Which enzyme adds each new nucleotide to its 3′ end?
- A. ✓ DNA polymerase
- B. LigaseLigase adds no nucleotides; it joins the backbone between two pieces.
Why: DNA polymerase adds each nucleotide to the free 3′ end of the growing strand.
Ligase joins two neighboring pieces of the lagging strand.
Which bond does ligase form?
- A. A hydrogen bond between two basesHydrogen bonds between bases hold a strand to its template; they form as the bases pair.
Ligase works on the backbone. - B. ✓ A covalent bond in the sugar-phosphate backbone
Why: Between two pieces the backbone is broken.
Ligase forms the covalent bond from one sugar to the next phosphate group.
90Mixed practice mixed practice
A cell copies its DNA with one enzyme missing. Afterwards its lagging strands are still in separate short pieces, but every piece is DNA and every base is paired.
Which enzyme was missing?
- A. HelicaseHelicase opens the fork.
Without helicase no fork opens and no new strand is made at all. - B. DNA polymeraseDNA polymerase adds the nucleotides of every piece.
Every piece was built, so DNA polymerase was working. - C. ✓ Ligase
Why: The pieces were built and paired, so DNA polymerase worked.
The breaks in the backbone between pieces were never joined.
Ligase joins those breaks, so ligase was the missing enzyme.
A replication fork is drawn below, moving to the left. Both templates’ ends are marked, and no new strand has been built yet.
Against which template does DNA polymerase build the lagging strand?
- A. ✓ The upper template
- B. The lower templateThe lower template has its 5′ end at the fork, so it gives the leading strand.
The lagging strand is built against the template whose 3′ end faces the fork.
Why: The upper template has its 3′ end at the fork.
The lagging strand is built against the template whose 3′ end faces the fork.
So DNA polymerase builds the lagging strand against the upper template.
On a lagging strand, each RNA primer has been replaced with DNA. One break remains between each pair of neighboring pieces.
What is broken there?
- A. The pairing between the bases and the templateEvery base is still paired with the template.
The break is in the backbone between two pieces. - B. ✓ The sugar-phosphate backbone
- C. The hydrogen bonds inside each pieceAlong one piece the nucleotides are joined by covalent bonds, not hydrogen bonds.
The break is in the backbone between two pieces.
Why: Two neighboring pieces lie end to end on the template.
The last sugar of one piece is not joined to the first phosphate group of the next.
So the sugar-phosphate backbone is broken there.
A student says: “The leading strand needs no primer, because it is made in one piece.”
Is the student correct?
- A. ✓ No: the leading strand starts from one primer
- B. Yes: only the lagging strand needs primersDNA polymerase cannot start a strand from nothing.
The leading strand starts from one primer and then grows in one piece.
Why: DNA polymerase can only add to an existing 3′ end.
So every new strand starts from a primer.
The leading strand starts from one primer; the lagging strand needs one per piece.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. A template strand near the fork reads 5′-CGTTAC-3′, and the fork is just past its left-hand end.
Which new strand does DNA polymerase build against this template?
- A. ✓ The leading strand
- B. The lagging strandThis template’s 5′ end is at the fork.
The template whose 3′ end faces the fork gives the lagging strand.
Why: The template’s left-hand end is marked 5′, and the fork is at that end.
The leading strand is built against the template whose 5′ end faces the fork.
So DNA polymerase builds the leading strand against this template.
A replication fork moves to the right. The lower template has its 5′ end at the fork.
Which way does the new strand built against the lower template grow?
- A. Away from the forkThe template’s 5′ end is at the fork, so the new strand’s 3′ end is at the fork.
A strand grows at its 3′ end: here, toward the fork. - B. ✓ Toward the fork
Why: The new strand pairs antiparallel with its template.
The template’s 5′ end is at the fork, so the new strand’s 3′ end is at the fork.
A strand grows at its 3′ end.
So this new strand grows toward the fork.
At a replication fork the leading strand is made in one long piece.
Why is the leading strand made in one long piece?
- A. ✓ The leading strand’s template has its 5′ end at the fork
- B. The leading strand’s template has its 3′ end at the forkA template whose 3′ end faces the fork gives a new strand growing away from the fork: the lagging strand.
- C. The leading strand’s template is exposed all at once when the fork opensThe fork exposes template a little at a time.
The leading strand keeps up because its growing end points at the fork.
Why: The leading strand’s template has its 5′ end at the fork.
So the new strand’s growing 3′ end is at the fork.
Each stretch the fork exposes is right at that growing end, so DNA polymerase keeps adding to one piece.
The replication fork below has both new strands drawn. The 3′ and 5′ ends of each template are written at the fork and at the far end.
(a) Explain how the ends of the two templates show that the upper new strand is the leading strand. (1 pt)
The upper template has its 5′ end at the fork.
So the upper new strand is the leading strand.
- Award 1 point for: the template whose 5′ end faces the fork gives the leading strand, and that is the upper template, so the upper new strand is the leading strand.
(b) Imagine that this cell has lost its ligase. Predict what the new strand drawn in short pieces looks like when the copying ends. (1 pt)
- Award 1 point for: the strand drawn in short pieces (the lagging strand) stays in separate pieces (the breaks between pieces are not joined).
(c) Explain why the other new strand is still one piece in that cell. (1 pt)
No breaks were ever made between pieces in it.
So it needs no ligase, and it is still one piece.
- Award 1 point for: the other new strand (the leading strand) is built continuously (one piece, one primer), so it has no breaks for ligase to join.
Glossary
- leading strand
- The new strand built in one long piece toward the replication fork, against the template whose 5′ end faces the fork. It follows the fork as the fork opens and needs one primer, at its start.
- lagging strand
- The new strand built in short pieces away from the replication fork, against the template whose 3′ end faces the fork. Each piece starts from its own RNA primer after the fork has exposed more template; ligase later joins the pieces.
- ligase
- The enzyme that joins the sugar-phosphate backbone between two neighboring pieces of a new strand, forming the covalent bond between them, after each RNA primer has been replaced with DNA. It adds no nucleotides.
APBIO-U06-L09 The whole fork, and what breaks it
The finished fork drawing is below, every part in place, but every label has been taken off. Nine parts need names.
Then suppose a test tube holds everything the copying needs except one of those nine. What would you find in the tube?
Unit 6 · Gene Expression and Regulation
1Nine parts to name
Helicase works at the point of the replication fork.
Which bonds does helicase break?
- A. ✓ The hydrogen bonds between paired bases
- B. The covalent bonds of the sugar-phosphate backboneThe backbone’s covalent bonds hold each strand together.
Helicase leaves both strands whole and parts them from each other.
Why: The two strands are held together by hydrogen bonds between paired bases.
Helicase breaks those hydrogen bonds.
So the two strands come apart at the fork.
Topoisomerase works near the replication fork.
Where does topoisomerase act?
- A. At the point of the forkHelicase acts at the point of the fork.
Topoisomerase acts ahead of it, where the over-twist builds. - B. ✓ Ahead of the fork, on the still-paired part
- C. Behind the fork, on the new strandsThe new strands behind the fork are not over-twisted.
The over-twist builds ahead of the fork, in the still-paired part.
Why: Pulling the two strands apart over-twists the DNA ahead of the fork.
Topoisomerase cuts one strand there, lets the twist unwind and seals the strand.
So topoisomerase acts ahead of the fork, on the still-paired part.
DNA polymerase needs an RNA primer before it adds its first nucleotide.
What does the RNA primer give DNA polymerase?
- A. ✓ A free 3′ end to add nucleotides to
- B. The first template base to readThe template’s bases are there before the primer lands.
What the polymerase lacks is a strand end to add to.
Why: DNA polymerase can only add a nucleotide to the free 3′ end of a strand.
A bare template has no strand paired to it, so no 3′ end to add to.
The primer pairs with the template and supplies the first 3′ end.
At a fork, one new strand is built toward the fork and the other away from it.
Which new strand is made in short pieces?
- A. The leading strandThe leading strand grows toward the fork in one long piece.
- B. ✓ The lagging strand
Why: The lagging strand grows away from the fork.
Each time more template is exposed it must start again from a new primer.
So the lagging strand is made in short pieces.
Every primer on the new strands has been replaced with DNA.
What does ligase join?
- A. Nucleotides to a free 3′ endAdding nucleotides to a 3′ end is DNA polymerase’s job.
- B. Two paired bases to each otherPaired bases are held by hydrogen bonds, which form on their own.
Ligase joins the backbone, not the bases. - C. ✓ The backbone between two neighboring pieces
Why: After the primers are replaced, the lagging strand is still separate pieces.
Ligase joins the sugar-phosphate backbone between neighboring pieces.
So the pieces become one continuous strand.
Video: Watch: Nine parts to name
The finished fork with every label removed; the labels returning one at a time, helicase first and ligase last; the table of the nine parts filling row by row.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L09a.mp4
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What does the whole copying machine look like, and what does each part do?
Helicase opens the fork. Topoisomerase relaxes the over-twist ahead of it.
A primer gives each new strand its start. DNA polymerase adds nucleotides at the 3′ end.
The leading strand follows the fork. The lagging strand is made in pieces that ligase joins.
Each part has one job. So taking one part away leaves one missing result you can predict, and the tube shows it.
That is how the exam asks about replication: a model to label, or a part removed.
The fork drawing below has every label taken off. Nine parts are lettered A to I.
In every drawing here, old strands are drawn dark and new strands are drawn light.
The 3′ and 5′ ends of each template are written at the fork and at the far end.
Read the drawing one part at a time. Each part gets its name back, and its one job.
Part A, the oval at the point of the Y, is helicase.
Helicase breaks the hydrogen bonds between paired bases. So the two strands come apart at the fork.
Part B, the oval on the still-paired part ahead of the fork, is topoisomerase.
Pulling the two strands apart over-twists the DNA ahead of the fork. Topoisomerase cuts one strand there, lets the over-twist unwind and seals the strand again.
So the fork keeps moving.
Parts C and D, the two dark strands of the Y’s arms, are the two template strands.
Each template is an old strand. Its bases decide the sequence of the new strand built against it.
The two templates run in opposite directions.
Part C, the upper template, has its 5′ end at the fork. Part D, the lower template, has its 3′ end at the fork.
Part E, the small dark block at the start of a new strand, is a primer.
An enzyme lays down a short RNA primer paired to the template. The primer gives DNA polymerase a free 3′ end to add to.
Part F, the rounded box at a growing end, is DNA polymerase.
DNA polymerase joins nucleotides one at a time to the free 3′ end of the new strand. So every new strand grows 5′ to 3′, from its primer toward the polymerase.
Part G, the one long light strand, is the leading strand. Its primer sits at the far end, and DNA polymerase sits at its fork end.
DNA polymerase reads its template 3′ to 5′. So against part C, the template whose 5′ end faces the fork, the new strand grows toward the fork in one piece.
Part H, the three short light pieces, is the lagging strand. Each piece starts from a primer at its fork-side end and grows away from the fork.
Against part D, the template whose 3′ end faces the fork, the new strand must grow away from the fork.
So the lagging strand is made in short pieces, each from its own primer. Each piece starts again where the fork has exposed more template.
Part I, the small thick-outlined oval on the join between two finished pieces, is ligase.
Ligase joins the sugar-phosphate backbone between neighboring pieces. So the short pieces become one continuous strand.
The table below lists the nine parts and the one job of each.
What you are expected to know Name each of the nine parts of the fork drawing, helicase to ligase, and state the one job of each.
The fork drawn below moves to the right, with its parts lettered J to R.
Which of the following is the part lettered M?
- A. ✓ Helicase
- B. TopoisomeraseTopoisomerase is drawn on the still-paired part ahead of the fork, not at the point.
- C. LigaseLigase is drawn on the join between two short pieces of the lagging strand.
Why: M is the oval at the point of the Y, where the two strands come apart.
The enzyme that parts the strands there is helicase.
The fork drawn below moves to the left, with its parts lettered J to R.
Which of the following is the part lettered K?
- A. HelicaseHelicase sits at the point of the Y.
K sits ahead of the fork, on the still-paired part. - B. LigaseLigase sits on the join between two short pieces of the lagging strand.
- C. ✓ Topoisomerase
Why: K is the dashed oval on the still-paired part ahead of the fork.
The enzyme that relaxes the over-twist there is topoisomerase.
The fork drawn below moves to the right, with its parts lettered J to R.
Which of the following is the part lettered L?
- A. DNA polymeraseDNA polymerase is drawn as a rounded box at a growing 3′ end.
- B. ✓ Ligase
- C. PrimerA primer is the small dark block at the start of a new strand.
Why: L is the small oval on the join between two short pieces of the lagging strand.
The enzyme that joins the backbone there is ligase.
The fork drawn below moves to the right, with its two templates lettered J and O.
Which letter marks the template whose 3′ end faces the fork?
- A. JJ is the upper template, and its 5′ end is written at the fork.
- B. ✓ O
Why: The templates are the two dark strands of the Y’s arms.
O is the lower template, and 3′ is written at its fork end.
So O is the template whose 3′ end faces the fork.
The fork drawn below moves to the left, with its parts lettered J to R.
Which of the following is the part lettered O?
- A. ✓ Primer
- B. DNA polymeraseDNA polymerase is the rounded box at a growing end, not a block at a strand’s start.
- C. LigaseLigase is the small oval on the join between two pieces.
Why: O is the small dark block at the far end of the long light strand, where that strand started.
The short RNA that starts each new strand is the primer.
The fork drawn below moves to the right, with its parts lettered J to R.
Which of the following is the part lettered Q?
- A. HelicaseHelicase is the oval at the point of the Y.
- B. LigaseLigase is the small oval on the join between two pieces.
- C. ✓ DNA polymerase
Why: Q is the rounded box at the growing 3′ end of a new strand.
The enzyme adding nucleotides there is DNA polymerase.
The fork drawn below moves to the right, with its parts lettered J to R.
Which of the following is the strand lettered N?
- A. A template strandA template strand is old and drawn dark.
N is drawn light: a new strand. - B. ✓ The leading strand
- C. The lagging strandThe lagging strand is in short pieces, their arrowheads facing away from the fork.
Why: N is one long light strand, so it is a new strand.
Its primer is at the far end and DNA polymerase at its fork end, so it grows toward the fork.
The new strand built toward the fork in one piece is the leading strand.
The fork drawn below moves to the left, with its parts lettered J to R.
Which of the following is the strand lettered R?
- A. A template strandA template strand is old, drawn dark and in one piece.
R is light and in pieces. - B. The leading strandThe leading strand is one long piece, from its primer at the far end to the polymerase at the fork.
- C. ✓ The lagging strand
Why: R is three short light pieces, so it is a new strand.
Each piece starts from a primer at the fork side and grows away from the fork.
The new strand built away from the fork in short pieces is the lagging strand.
The fork drawn below moves to the right, with its two templates lettered K and R.
Which letter marks the template whose 3′ end faces the fork?
- A. ✓ K
- B. RR is the lower template, and its 5′ end is written at the fork.
Why: The templates are the two dark strands of the Y’s arms.
K is the upper template, and 3′ is written at its fork end.
So K is the template whose 3′ end faces the fork.
DNA polymerase builds a new strand against a template.
Toward which end of the template does the new strand grow?
- A. ✓ Toward the template’s 5′ end
- B. Toward the template’s 3′ endDNA polymerase starts reading at the template’s 3′ end and moves away from it.
So the new strand grows toward the template’s 5′ end.
Why: DNA polymerase reads its template 3′ to 5′.
It moves toward the template’s 5′ end, adding a nucleotide at each step.
So the new strand grows toward its template’s 5′ end.
The fork drawn below moves to the left. The 3′ and 5′ ends of each template are written at the fork and at the far end. No new strand has been drawn yet.
Against which template will the leading strand be built?
- A. The upper templateThe upper template has its 3′ end at the fork.
A new strand built against it must grow away from the fork, in pieces. - B. ✓ The lower template
Why: DNA polymerase reads a template 3′ to 5′, so a new strand grows toward its template’s 5′ end.
The lower template has its 5′ end at the fork.
So the new strand built against it grows toward the fork in one piece: the leading strand.
The fork drawn below moves to the right, with its parts lettered J to R.
(a) Describe the job of the part lettered P at the position where it is drawn. (1 pt)
Ligase joins the sugar-phosphate backbone between the two neighboring pieces.
So the two pieces become one continuous strand.
- Award 1 point for: P (ligase) joins the backbone between two neighboring short pieces of the lagging strand, making one continuous strand.
(b) Explain how the ends of the template lettered J show which way the strand lettered N grows. (1 pt)
DNA polymerase reads a template 3′ to 5′, so a new strand grows toward its template’s 5′ end.
So N grows toward the fork.
- Award 1 point for: template J (the template N is built against) has its 5′ end at the fork, and a new strand grows toward its template’s 5′ end (DNA polymerase reads 3′ to 5′), so N grows toward the fork.
One simplification we made here. We drew one fork opening from one point.
In reality a chromosome of a cell with a nucleus begins copying at many points at once, and a bacterial chromosome at one point. Each point opens two forks that move in opposite directions. Your exam questions expect the one-fork picture.
54Take one part away
Suppose a test tube holds everything the copying needs:
- a DNA molecule
- free nucleotides
- helicase
- topoisomerase
- the enzyme that lays down primers
- DNA polymerase
- ligase
In the tube, forks open and new strands grow, just like in a cell.
Each part has one job. Take one part out of the tube, and that one job is not done.
Every job that needed that result stops too.
Take ligase out of the tube first. Helicase still opens the fork, and topoisomerase still relaxes the over-twist ahead of it.
Primers still land. DNA polymerase still builds the leading strand and the short pieces of the lagging strand.
Ligase’s one job is joining the backbone between neighboring pieces. No enzyme is left to do it.
So the lagging strand stays in pieces.
The leading strand was one piece from the start. So the leading strand is finished as normal.
Put ligase back and take helicase out. Helicase’s one job is breaking the hydrogen bonds between paired bases.
No enzyme is left to part the two strands. So no fork opens.
With no fork open, no template lies bare. So no primer lands, DNA polymerase adds no nucleotide, and no new strand grows.
The earlier a part’s job comes, the more goes missing without it.
Helicase works first, so without helicase everything is missing. Ligase works last, so without ligase only the joining is missing.
A student says: “With no enzyme to lay down primers, DNA polymerase starts each new strand itself as soon as a template lies exposed.”
Is the student correct?
- A. Yes: DNA polymerase pairs the first nucleotide to the bare template and starts the strand itselfDNA polymerase joins nucleotides only to a free 3′ end.
A bare template has no strand paired to it, so there is no 3′ end to start from. - B. ✓ No: DNA polymerase adds only to a free 3′ end, so with no primer no new strand starts
Why: DNA polymerase can only join a nucleotide to the free 3′ end of a strand.
The primer supplies the first 3′ end on a bare template.
With no primer laid down, no new strand starts.
Imagine a cell whose DNA polymerase is blocked by a drug. Forks open and primers land on the templates.
What happens at each primer?
- A. ✓ Nothing is added to the primer
- B. The primer grows from its 5′ end insteadNo enzyme adds nucleotides to a 5′ end.
Only DNA polymerase adds nucleotides, and only at a 3′ end. - C. Ligase extends the primer insteadLigase joins the backbone between two pieces already in place.
Ligase adds no nucleotides to a strand.
Why: The primer supplies a free 3′ end.
DNA polymerase’s one job is adding nucleotides to that end.
With DNA polymerase blocked, no nucleotides are added, and the primer stays bare.
Imagine a cell whose ligase is inactive. Forks open, both new strands are built and every primer is replaced with DNA.
Which new strand is left in many pieces, apart from the one join where its first primer sat?
- A. The leading strandThe leading strand grows toward the fork in one piece and needs no joining.
- B. ✓ The lagging strand
- C. Both new strandsThe leading strand is one piece already.
Only the lagging strand was built in pieces that need joining.
Why: Ligase’s one job is joining the backbone between neighboring pieces.
Only the lagging strand is built in pieces.
So with ligase inactive, the lagging strand stays in pieces.
Imagine a cell whose helicase is blocked by a drug. Every other part of the copying machine is present.
Is any template exposed for copying?
- A. YesHelicase is the enzyme that parts the two strands.
With helicase blocked, no fork opens and the bases stay paired. - B. ✓ No
Why: Helicase’s one job is breaking the hydrogen bonds between paired bases.
With helicase blocked, no fork opens.
So no template is exposed, and nothing else can start.
Video: Watch: Take one part away
The complete fork; one part fading out at a time, and the drawing showing what is missing: no fork opens, the fork stalls, no new strand starts, no nucleotides are added, the lagging strand stays in pieces.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L09b.mp4
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The table below lists the five parts and what you see when you take each one away.
What you are expected to know Predict what you see when you take one named part of the fork away, and explain it through that part’s one job.
Suppose a strain of bacteria has a ligase that stops working above 40 °C and works again below it. The bacteria are grown at 30 °C and at 42 °C, and the newly made DNA is collected. The table below shows what is found at each temperature.
(a) Explain how the short pieces found at 42 °C show what ligase does. (1 pt)
Frame The short pieces show this because …
Ligase’s job is joining the backbone between neighboring pieces.
Above 40 °C the ligase stops working, so no enzyme joins the pieces.
So the lagging strand stays in short pieces, and those are the pieces found at 42 °C.
- Award 1 point for: the lagging strand is made in short pieces that ligase joins; with the ligase not working above 40 °C the pieces are not joined, so the short pieces remain.
(b) The bacteria grown at 42 °C are cooled to 30 °C. Predict what the newly made DNA will look like after the bacteria are cooled, and justify your prediction. (2 pt)
At 30 °C the ligase works again.
Ligase’s one job is joining the sugar-phosphate backbone between neighboring pieces.
DNA polymerase has already built each short piece, paired with its template.
So ligase joins each break, and the short pieces become one continuous strand.
- Award 1 point for the prediction: the newly made DNA is in long (continuous) strands, the short pieces joined.
- Award 1 point for the justification: at 30 °C the ligase works again and joins the backbone between neighboring pieces.
Go back to the fork drawing with its labels returned: helicase, topoisomerase, the two templates with their ends, the primers, DNA polymerase, the leading strand, the lagging strand’s pieces, ligase.
Take ligase out of the tube and the lagging strand stays in pieces.
79Mixed practice mixed practice
A skin cell copies one of its DNA molecules, giving two molecules.
Where do the two old strands end up?
- A. ✓ One in each of the two molecules
- B. Both together in a single moleculeThe two old strands come apart before the copying.
Each old strand takes a new partner, in a different molecule. - C. In neither: both are broken down into nucleotidesThe old strands stay whole.
Each old strand serves as a template for the new strand built against it.
Why: The two old strands come apart.
Each old strand serves as a template for a new strand and stays paired with it.
So one old strand ends up in each of the two molecules.
A replication fork is drawn below, with helicase at the point of the Y.
In which direction does the fork move?
- A. ✓ To the left
- B. To the rightThe arms have already been parted.
The fork moves into the part still paired, which lies to the left.
Why: Helicase at the point of the Y parts the strands ahead of it.
The still-paired part lies to the left.
So the fork moves to the left.
A new strand is being built against a template. Its primer sits at its left end.
Toward which end does the new strand grow?
- A. Toward its 5′ end, to the leftThe primer sits at the new strand’s 5′ end.
Nothing is added there; nucleotides are added at the 3′ end. - B. ✓ Toward its 3′ end, to the right
Why: DNA polymerase adds each nucleotide to the free 3′ end of the new strand.
The primer is the strand’s 5′ end, at the left.
So the new strand grows toward its 3′ end, to the right.
Bacteria whose DNA started fully heavy copy it once in light nitrogen, and a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them. The tube is drawn below.
What does the one band show about the molecules?
- A. Each molecule has two heavy strandsMolecules with two heavy strands settle at the heavy mark, the lowest.
The band is at the half-heavy mark. - B. Each molecule has two light strandsMolecules with two light strands settle at the light mark, the highest.
The band is at the half-heavy mark. - C. ✓ Each molecule has one heavy strand and one light strand
Why: The band sits at the middle mark, where half-heavy DNA settles.
In semiconservative replication, half-heavy DNA is one heavy strand paired with one light strand.
So every molecule kept one old, heavy strand and gained one new, light strand.
The fork drawn below moves to the right. The 3′ and 5′ ends of each template are written at the fork and at the far end. No new strand has been drawn yet.
Against which template will the lagging strand be built?
- A. ✓ The upper template
- B. The lower templateThe lower template has its 5′ end at the fork.
The new strand built against it grows toward the fork in one piece: the leading strand.
Why: A new strand grows toward its template’s 5′ end.
The upper template has its 3′ end at the fork, so its new strand must grow away from the fork.
A strand growing away from the fork is made in pieces: the lagging strand.
A cell copies its DNA with one part of the copying machine blocked. Forks open along the DNA and the templates lie exposed, but no new strand starts.
Which part is blocked?
- A. The helicase that parts the strandsForks open, so helicase is at work.
The blocked part does its job after the fork has opened. - B. ✓ The enzyme that lays down primers
- C. The ligase that joins the piecesLigase joins pieces of a strand that has already been built.
Here no strand starts at all, so the missing job comes earlier.
Why: Forks open and templates are exposed, so helicase is working.
DNA polymerase needs a primer’s free 3′ end to add to.
No new strand starts, so no primer was laid down: the primer enzyme is blocked.
A student says: “The bases of each short piece are paired with the template before ligase arrives; ligase joins only the backbone.”
Is the student correct?
- A. ✓ Yes: each base is paired as the piece is built, and ligase joins the backbone
- B. No: ligase pairs each piece’s bases with the template as it joins the backboneLigase adds no nucleotide and pairs no base.
It joins the sugar-phosphate backbone between two pieces already in place.
Why: DNA polymerase pairs each nucleotide with its template base as it builds a piece.
Ligase adds no nucleotide and pairs no base.
Ligase joins the sugar-phosphate backbone between two neighboring pieces.
The fork drawn below moves to the left, with its parts lettered J to R.
Which letter marks the enzyme that parts the two strands?
- A. ✓ M
- B. PP is the small oval on the join between two short pieces: ligase, which joins the backbone there.
- C. QQ is the dashed oval on the still-paired part ahead of the fork: topoisomerase, which relaxes the over-twist there.
Why: The enzyme that parts the two strands is helicase.
Helicase works at the point of the Y, where the strands come apart.
M is the oval at the point, so M marks helicase.
The fork drawn below moves to the left. Its two templates are labeled upper template and lower template, and the 3′ and 5′ ends of each are written at the fork and at the far end. No new strand has been drawn yet.
(a) Determine which template the leading strand will be built against, and justify your answer using the ends written on the two templates. (2 pt)
DNA polymerase reads a template 3′ to 5′, so a new strand grows toward its template’s 5′ end.
The upper template has its 5′ end at the fork, so the new strand built against it grows toward the fork in one piece: the leading strand.
The lower template has its 3′ end at the fork, so its new strand grows away from the fork, in pieces.
- Award 1 point for the determination: the upper template.
- Award 1 point for the justification: a new strand grows toward its template’s 5′ end (DNA polymerase reads 3′ to 5′), and the upper template has its 5′ end at the fork, so the strand built against it grows toward the fork as the leading strand.
(b) Predict what happens to this fork’s movement if topoisomerase is blocked. Justify your prediction. (2 pt)
Helicase pulling the two strands apart over-twists the DNA ahead of the fork.
Topoisomerase’s one job is to cut one strand there, let the over-twist unwind and seal the strand.
With topoisomerase blocked, the over-twist builds up until the fork cannot move.
- Award 1 point for the prediction: the fork stalls (stops moving).
- Award 1 point for the justification: parting the strands over-twists the DNA ahead of the fork, and with topoisomerase blocked nothing cuts and unwinds that over-twist, so it builds up and the fork cannot move.
APBIO-U06-P62 Practice questions: Topic 6.2
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 one experiment 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: DNA replication, summed up
One copying before division; half old, half new; the bands in the tube; the fork, helicase and topoisomerase; the primer and DNA polymerase adding only at the 3′ end; leading and lagging; ligase; what breaks when one part is removed.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-T62-summary.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-T62-summary.mp4
A student says: “A cell copies its DNA while it divides, so each daughter cell receives half of the copies.”
Which statement about the student's claim is correct?
- A. The student is right: the cell copies its DNA and divides at the same time, sharing the copies outThe copying is finished before the division begins.
It happens in S phase of interphase. - B. ✓ The student is wrong: the copying comes first, in S phase, and each daughter cell gets a complete set
- C. The student is wrong: each daughter cell copies its own DNA after the division has finishedA daughter cell copies its DNA in its own S phase, but that is not how it receives a complete set.
The parent cell copied first, then divided. - D. The student is right: each daughter cell needs only half of the parent's DNA to workEach daughter cell needs a complete set.
The parent copied its DNA in S phase so that both cells could receive one.
Why: The cell copies its DNA in S phase, before it divides.
So when it divides, there are two complete sets.
Each daughter cell receives one complete set, and the student is wrong.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. One strand of a DNA molecule reads 5′-CTTGAC-3′, as drawn. During copying it serves as a template strand.
Written under it with its ends marked, what does the new strand built against it read?
- A. ✓ 3′-GAACTG-5′
- B. 3′-CTTGAC-5′These are the template's own letters.
The new strand carries each template base's partner: C takes G, T takes A, G takes C, A takes T. - C. 5′-GAACTG-3′The new strand runs the other way to its template.
Under a strand written 5′ to 3′, the new strand's 3′ end is at the left. - D. 3′-GAACUG-5′A DNA strand never carries uracil.
Opposite the template's A sits T.
Why: Under each template base write its partner: C gives G, T gives A, T gives A, G gives C, A gives T, C gives G.
The new strand runs the other way to its template, so its 3′ end is at the left.
It reads 3′-GAACTG-5′.
One all-old DNA molecule goes through 4 rounds of copying.
How many of the molecules made are built only from new strands?
- A. 2Two is the number of molecules that still carry an old strand.
The all-new molecules are all the others. - B. ✓ 14
- C. 16Sixteen is every molecule after four rounds.
Two of them carry an old strand, so 14 are all new. - D. 30Thirty is the all-new count after five rounds (32 − 2).
Four rounds give 16 molecules, 14 of them all new.
Why: Four rounds of doubling give 2⁴ = 16 molecules.
The two old strands sit in two molecules.
So 16 − 2 = 14 molecules are built only from new strands.
Suppose the old strands stayed together whenever DNA is copied. Bacteria whose DNA started fully heavy copy it three times in light nitrogen, and a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them. A thicker band means a bigger share of the molecules.
Which bands would the tube show?
- A. One half-heavy bandWith the old strands kept together, no molecule is ever half heavy.
One molecule is heavy and every other molecule is light. - B. A half-heavy band and a light band, equally thickA half-heavy molecule needs one heavy strand and one light strand paired together.
Old strands kept together never pair with a new one. - C. ✓ A thin heavy band and a thick light band
- D. A thick heavy band and a thin half-heavy bandHere the old strands stay together, so no molecule is half-heavy.
Seven of the eight molecules are light, so the light band is the thick one.
Why: The two old, heavy strands stay together as one heavy molecule.
Every other molecule is two light, new strands: seven of the eight after three rounds.
So a thin heavy band and a thick light band.
Bacteria whose DNA started fully heavy copy it once in light nitrogen, and the spun DNA shows one half-heavy band. A student claims that this tube alone proves each old strand took a new partner.
Which further observation would show whether each old strand took a new partner?
- A. ✓ The bands after a second round of copying in light nitrogen
- B. The thickness of the band after round oneEvery way of copying that gives one band after round one puts every molecule in it.
Its thickness tells the ways apart no better than its position. - C. The bands before any copying, in heavy nitrogenBefore copying, every strand is heavy, whatever way the copying will go.
One heavy band rules nothing out. - D. The number of molecules after round oneEvery way of copying doubles the molecules.
Two molecules fit all three ways.
Why: Old and new mixed along every strand also gives one half-heavy band after round one, so round one alone leaves two ways open.
After round two, each old strand with a new partner gives a light band; the mixed way never does.
The round-two tube settles it.
Helicase and topoisomerase both act on a DNA molecule that is being copied.
Which statement compares what the two enzymes do?
- A. Helicase cuts one strand ahead of the fork and seals it; topoisomerase breaks the hydrogen bonds at the forkThe jobs are the other way round.
Helicase parts the strands at the fork; topoisomerase works on the paired part ahead. - B. The two enzymes both break the hydrogen bonds between paired bases, helicase at the fork and topoisomerase ahead of itTopoisomerase breaks no hydrogen bonds.
It cuts one strand's backbone ahead of the fork, lets the twist unwind and seals it. - C. The two enzymes both cut one strand and seal it again, helicase at the fork and topoisomerase ahead of itHelicase cuts no strand.
It breaks the hydrogen bonds between paired bases, and each backbone stays whole. - D. ✓ Helicase breaks the hydrogen bonds at the fork; topoisomerase cuts one strand ahead of it and seals it
Why: Helicase breaks the hydrogen bonds between paired bases, so the strands part at the fork.
Parting them pushes a twist into the paired part ahead.
Topoisomerase cuts one strand there, lets the twist unwind and seals it.
A replication fork is drawn below, moving to the left. Both templates' ends are marked. DNA polymerase builds a new strand against the lower template.
Toward which marked end does DNA polymerase move?
- A. The lower template's 3′ end, at the forkDNA polymerase reads a template from its 3′ end toward its 5′ end.
The lower template's 3′ end is at the fork, so DNA polymerase moves away from the fork. - B. ✓ The lower template's 5′ end, at its far end
- C. The upper template's 5′ end, at the forkThis DNA polymerase builds against the lower template, so it moves along the lower template.
The upper template's ends are not the ones it reads. - D. The upper template's 3′ end, at its far endThe upper template is not the one this DNA polymerase reads.
Along the lower template it moves toward the 5′ end, at the far end.
Why: Every new strand grows 5′ to 3′, so DNA polymerase reads its template 3′ to 5′.
The lower template's 3′ end is at the fork and its 5′ end at the far end.
So DNA polymerase moves toward the lower template's 5′ end, at the far end.
Both ends of each strand are marked. The template is written 3′ to 5′, so that the new strand built against it reads 5′ to 3′ beneath it. A template strand reads 3′-TAGCCA-5′, as drawn. An RNA primer is paired to it just before its first written base, to the left.
Written with both ends marked, what does the finished new strand read?
- A. 3′-ATCGGT-5′The new strand's 5′ end sits under the template's 3′ end, at the left.
Its 3′ end is at the right, where it stopped growing. - B. 5′-TAGCCA-3′These are the template's own letters.
The new strand carries each template base's partner. - C. ✓ 5′-ATCGGT-3′
- D. 5′-AUCGGU-3′A DNA strand never carries uracil.
Opposite the template's A sits T.
Why: Under each template base write its partner: T gives A, A gives T, G gives C, C gives G, C gives G, A gives T.
The new strand grows from the primer at the left, so its 5′ end is at the left.
It reads 5′-ATCGGT-3′.
A student says: “The leading strand is built against the template whose 5′ end faces the fork.”
Which statement about the student's claim is correct?
- A. The student is right: the leading strand's own 3′ end is at the fork, so the template it is built against has its 3′ end there tooA new strand runs the other way to its template.
Where the leading strand's 3′ end is, its template's 5′ end is. - B. The student is wrong: the leading strand is built against the template whose 3′ end faces the fork, not the one whose 5′ end doesDNA polymerase reads a template from its 3′ end toward its 5′ end.
A template with its 3′ end at the fork is read away from the fork. - C. ✓ The student is right: DNA polymerase reads a template 3′ to 5′, so it reads toward the fork only when the template's 5′ end is there
- D. The student is wrong: the leading strand is built against whichever template is drawn on top of the fork, whatever ends are written on itWhich template is on top changes from drawing to drawing.
Only the template's end at the fork decides.
Why: A new strand grows toward its template's 5′ end.
To grow toward the fork in one piece, the leading strand needs a template whose 5′ end is at the fork.
So the student has the ends the right way round.
A cell copies its DNA with one part of the copying machine blocked. Forks open, but each stops after moving a short way, and the DNA ahead of each fork is twisted tighter than usual.
Which part is blocked?
- A. ✓ Topoisomerase
- B. HelicaseForks open, so helicase is at work.
What fails is the relaxing of the twist ahead of the fork. - C. DNA polymeraseWith DNA polymerase blocked, forks would open and move on with nothing added to the primers.
Here the forks themselves stop. - D. LigaseWithout ligase the lagging strand stays in pieces, but the forks keep moving.
Here the forks stall behind an over-twisted stretch.
Why: Helicase parts the strands and pushes the twist into the paired part ahead.
Topoisomerase relaxes that twist by cutting one strand, letting it unwind and sealing it.
With topoisomerase blocked, the DNA ahead over-twists and the fork stalls.
Bacteria from a yogurt culture are grown for many generations in ordinary, light nitrogen, so every strand of their DNA is light. They are moved to heavy nitrogen and copy their DNA once, then twice. After each round a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them. The two tubes are drawn below.
(a) Describe the bands in the tube after round one. (1 pt)
Frame After round one the tube shows …
Hint Read the drawn tube against the three marks. Say how many bands, and at which mark.
- Award 1 point for: one band at the half-heavy mark (no heavy band, no light band).
(b) State how many light strands each molecule carries after round one, using the tube. (1 pt)
Frame Each molecule carries … light strand(s).
Hint Read where the band sits against the three marks. Ask what a molecule at that mark is made of.
- Award 1 point for: one (the band is at the half-heavy mark, so each molecule is one light strand paired with one heavy strand).
(c) Determine the first round whose tube tells apart each old strand taking a new partner from old and new being mixed along every strand, and justify your answer. (1 pt)
Frame Round … is the first, because …
Hint Work out the bands each way of copying predicts after round one, then after round two. Find the first round where the two predictions differ.
After round two, each old strand taking a new partner gives a half-heavy band and a heavy band, while old and new mixed along every strand gives one band between the half-heavy and heavy marks.
The round-two tube shows a heavy band, which only each old strand taking a new partner gives.
- Award 1 point for: round two AND the ground — both ways predict one half-heavy band after round one, and only each old strand taking a new partner predicts a heavy band (two bands) after round two.
(d) For one starting molecule, calculate how many of the molecules after nine rounds carry a light strand. (1 pt)
Frame After nine rounds, … molecules carry a light strand.
Hint Only the original strands are light. Count how many original strands there are, and remember that each sits in its own molecule.
Answer: 2 molecules (tolerance ±0)
- Award 1 point for: 2.
A complete replication fork is drawn below, moving to the right, with nine parts lettered J to R. The 3′ and 5′ ends of each template are written at the fork and at the far end.
(a) Identify the part lettered R. (1 pt)
- Award 1 point for: topoisomerase.
(b) State how many RNA primers the strand lettered L and the strand lettered M have each needed to reach the stage drawn, one primer for each piece started, and explain why the two numbers differ. (2 pt)
M has needed three primers, one for each of its three pieces: two are drawn as blocks, and the first piece’s primer is already replaced with DNA.
L grows toward the fork in one piece, so it started once.
M grows away from the fork, so each new piece starts when the fork exposes more template.
DNA polymerase adds only to an existing 3′ end, so each piece needs its own primer.
- Award 1 point for: L one primer; M three (one for each of its three pieces, the first piece’s primer already replaced with DNA). Accept two for M if the student counts the two primer blocks drawn and says the first piece’s primer has been replaced.
- Award 1 point for: the strand growing away from the fork (M) starts a new piece from its own primer each time more template is exposed, because DNA polymerase adds only to an existing 3′ end; the strand growing toward the fork (L) is one piece started once.
(c) Predict what the drawing would show if the part lettered Q stopped working at the moment drawn. (1 pt)
With Q stopped, no more hydrogen bonds are broken, so the fork stops moving and the paired part stays paired.
No new template is exposed.
So the new strands grow only as far as the template already exposed, and then the copying stops.
- Award 1 point for: the fork stops moving (the paired part stays paired; no new template is exposed), so the new strands stop growing once the exposed template is used.
(d) A student looking at the drawing says: “The part lettered P adds the nucleotides to the new strands.” Evaluate the student’s claim. (1 pt)
P is ligase, the hollow oval spanning the gap between two pieces of M.
Ligase joins the backbone between two neighboring pieces; it adds no nucleotides.
DNA polymerase, the rounded box lettered O, adds the nucleotides to every new strand.
- Award 1 point for: the claim is rejected because P (ligase) joins the backbone between neighboring pieces and adds no nucleotides; DNA polymerase (O) adds them.
APBIO-U06-T62 End-of-topic test: DNA Replication
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.
Cells are grown in dishes in a laboratory. The scientists add a drug that blocks DNA polymerase to some dishes for one hour during interphase and to others for one hour during mitosis, then wash it away. The table shows the share of cells that went on to divide within the next day.
Which conclusion do the results support?
- A. The DNA is copied while the chromosomes separate, during mitosisThe drug given during mitosis left almost every cell dividing.
Blocking copying during mitosis changed nothing, so the copying was already done. - B. ✓ The DNA is copied during interphase, before the cell divides
- C. The drug given during mitosis stops the chromosomes from separatingAlmost every cell given the drug during mitosis still divided.
The chromosomes still separated. - D. Each daughter cell copies its DNA for itself after the division endsA cell that had not copied its DNA did not divide.
The copying comes before the division, in interphase.
Why: The drug blocks DNA polymerase, so no DNA is copied while it is present.
Given during interphase, it stopped most cells from dividing; given during mitosis, it changed almost nothing.
So the DNA is copied during interphase, before the cell divides.
The two strands of a short DNA molecule are separated. Each strand is placed in its own test tube with DNA polymerase, an RNA primer paired to it and marked free DNA nucleotides. Afterwards a marked strand is found in each tube, as shown.
Which conclusion do the two tubes support?
- A. The marked strands are copies of the old strands, with the same letters in the same orderRead base by base, each marked strand carries the partner of each old base, not the same letter.
G sits opposite C and A opposite T. - B. Each old strand was broken down and rebuilt from marked nucleotidesThe old strands are still there, unmarked.
The marked strand in each tube is a new strand built against the old one. - C. The two old strands came from two different DNA moleculesWritten from its own 5′ end, the strand placed in tube 2 is the partner of the strand placed in tube 1.
The two old strands were one molecule. - D. ✓ Each old strand served as a template for a new partner strand
Why: In each tube the marked strand pairs base for base with the old strand, G opposite C and A opposite T, and runs the other way.
The old strand set every base of the new one.
So each old strand served as a template for a new partner strand.
One all-old DNA molecule goes through several rounds of copying. Afterwards 62 of the molecules are built only from new strands.
How many rounds of copying were there?
- A. ✓ 6
- B. 7Seven rounds give 128 molecules, 126 of them all new.
Two molecules carry an old strand whatever the round, so 62 all-new means 64 in all. - C. 31The all-new molecules are not half of the molecules.
Every molecule but two is all new, so 62 all-new means 64 molecules, which is six doublings. - D. 62The molecules double each round, so the number of rounds is far smaller than the number of molecules.
Two carry an old strand; 62 + 2 = 64 = 2⁶.
Why: Only two molecules ever carry an old strand, so the molecules number 62 + 2 = 64.
Each round doubles the molecules: 1, 2, 4, 8, 16, 32, 64.
64 = 2⁶, so there were six rounds.
Bacteria whose DNA started fully light copy it three times in heavy nitrogen, and a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them. A thicker band means a bigger share of the molecules. Four tubes are drawn below, numbered 1 to 4.
Which tube shows the bands after round three?
- A. Tube 1Two equal bands at the half-heavy and heavy marks is round two: two half-heavy and two heavy molecules.
After round three the heavy molecules outnumber the half-heavy ones. - B. Tube 2A thick light band and a thin heavy band would need light molecules.
Every new strand is heavy here, so no molecule is light after round one. - C. ✓ Tube 3
- D. Tube 4One half-heavy band is round one.
By round three most molecules are two heavy strands.
Why: The two old, light strands sit in two molecules whatever the round, each paired with a heavy new strand: half-heavy.
After three rounds there are 8 molecules, so the other 6 are heavy.
A thin half-heavy band and a thick heavy band: tube 3.
Bacteria whose DNA started fully heavy copy it in light nitrogen, and after each round a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them.
Which observation rules out old and new being mixed along every strand?
- A. One half-heavy band after round oneOld and new mixed along every strand also gives one half-heavy band after round one.
Round one fits that way of copying. - B. ✓ A light band after round two
- C. No heavy band after round oneA heavy band after round one comes only from old strands kept together.
Its absence rules out that way, and leaves the mixed way possible. - D. More molecules after round two than after round oneEvery way of copying doubles the molecules each round.
The count is the same whichever way is real.
Why: With old and new mixed along every strand, every molecule carries some old, heavy DNA.
So no molecule is ever all light, and no light band ever appears.
A light band after round two rules that way out.
A bacterium carries a mutation that makes one of its copying enzymes stop working at high temperature. At that temperature its DNA stays paired from end to end, and no Y-shaped opening appears anywhere along it.
Which enzyme has stopped working?
- A. TopoisomeraseTopoisomerase relaxes the twist ahead of a fork that has already opened.
Without it, forks open and then stall. - B. DNA polymeraseDNA polymerase adds nucleotides to a strand on an exposed template.
Without it, forks still open and templates are exposed. - C. LigaseLigase joins the pieces of a new strand after they are built.
Without it, forks open and the lagging strand is left in pieces. - D. ✓ Helicase
Why: Helicase breaks the hydrogen bonds between paired bases, so the two strands come apart and a Y-shaped fork opens.
With helicase stopped, no hydrogen bonds are broken.
So the strands stay paired and no opening appears.
A replication fork is drawn below, with helicase at the point of the Y. Four parts are lettered J to M.
Which letter marks the part helicase opens next?
- A. JJ is a template strand, already parted and exposed.
Helicase opens the part where the strands are still paired. - B. KK is a template strand, already parted and exposed.
Helicase moves into the paired part, not back along an arm. - C. ✓ L
- D. MM is the point of the Y, where the strands are parting now.
The part helicase opens next is the paired part beside it.
Why: Helicase breaks the hydrogen bonds between paired bases.
It works on the part where the strands are still paired, moving into it.
L is the paired part, so L is opened next.
Plasmids are copied in test tubes with every part of the copying machine present. In some tubes one enzyme, here called enzyme X, is blocked. The table shows the share of plasmids that finished copying and the share whose forks stalled part-way.
Which statement about the tubes with enzyme X blocked do the results support?
- A. ✓ The twist pushed ahead of each fork built up and was not relaxed, so the fork stopped
- B. The two strands could not be parted anywhere, so no fork opened along the plasmidMost forks stalled part-way, so forks did open and move.
The strands were being parted; the block came later. - C. The short pieces of one new strand were never joined into one, so the fork stoppedUnjoined pieces lie behind the fork, on template already copied.
The fork ahead keeps moving whether or not they are joined. - D. No DNA nucleotide could be added to any primer on the exposed templates, so the fork stoppedWith no nucleotide added, the templates lie bare, but helicase still parts the strands and the fork moves on.
A missing polymerase does not stop the fork.
Why: With enzyme X blocked, most forks stalled part-way and most plasmids never finished copying.
A fork stalls when the twist pushed into the paired DNA ahead of it builds up and is not relaxed.
Topoisomerase relaxes that twist, so enzyme X is topoisomerase.
A template strand and the RNA primer paired to its first two bases are drawn. DNA polymerase is about to add the first DNA nucleotide.
Which nucleotide does DNA polymerase add, and to which end of the primer?
- A. A nucleotide carrying adenine, to the primer's 3′ endThe next template base is A, and A pairs with T.
The nucleotide added carries thymine. - B. ✓ A nucleotide carrying thymine, to the primer's 3′ end
- C. A nucleotide carrying adenine, to the primer's 5′ endA strand's 5′ end never gains a nucleotide, and A pairs with T.
Thymine joins the primer's 3′ end. - D. A nucleotide carrying thymine, to the primer's 5′ endDNA polymerase joins each nucleotide to a free 3′ end.
The primer's 5′ end gains nothing.
Why: DNA polymerase joins each nucleotide to the free 3′ end of a strand that already exists: the primer's 3′ end.
It chooses the nucleotide by pairing it with the next template base, A.
A pairs with T, so the nucleotide carries thymine.
In a cell, every new DNA strand begins with a short stretch of RNA, which is later replaced with DNA.
Which statement explains why the strand begins with RNA?
- A. RNA nucleotides pair with the template more firmly than DNA nucleotides do, so the start holdsPairing strength is not the point.
DNA polymerase needs an existing 3′ end to add to, and the RNA piece supplies it. - B. DNA polymerase copies the RNA piece first to learn which DNA nucleotides to add after itDNA polymerase reads the template strand to choose each nucleotide.
The RNA piece gives it a 3′ end to add to, not a pattern to copy. - C. Helicase parts the two strands only where a short piece of RNA is already paired to themHelicase breaks the hydrogen bonds between paired bases wherever it moves.
The RNA piece is laid down on a template that is already exposed. - D. ✓ DNA polymerase adds only to a 3′ end that already exists; the RNA piece gives the first one
Why: DNA polymerase can only join a nucleotide to the free 3′ end of a strand that already exists.
On a bare template no strand has begun.
Another enzyme lays down a short piece of RNA paired to the template, and its 3′ end is where DNA polymerase starts.
A template strand reads 3′-CCGATT-5′, as drawn. An RNA primer is paired to it just before its first written base, to the left.
Written with both ends marked, what does the finished new strand read?
- A. ✓ 5′-GGCTAA-3′
- B. 3′-GGCTAA-5′The new strand's 5′ end sits under the template's 3′ end, at the left.
Its 3′ end is at the right, where it stopped growing. - C. 5′-CCGATT-3′These are the template's own letters.
The new strand carries each template base's partner: C takes G, G takes C, A takes T, T takes A. - D. 5′-GGCUAA-3′A DNA strand never carries uracil.
Opposite the template's A sits T.
Why: Under each template base write its partner: C gives G, C gives G, G gives C, A gives T, T gives A, T gives A.
The new strand grows from the primer at the left, so its 5′ end is at the left.
It reads 5′-GGCTAA-3′.
A replication fork is drawn below, moving to the right. Both templates' ends are marked, and no new strand has been built yet.
Which statement describes the new strand DNA polymerase builds against the upper template?
- A. One long piece, growing toward the forkThe upper template has its 3′ end at the fork.
A new strand grows toward its template's 5′ end, so here it grows away from the fork, in pieces. - B. One long piece, growing away from the forkA strand growing away from the fork must start again each time the fork exposes more template.
So it is made in short pieces. - C. ✓ Short pieces, growing away from the fork
- D. Short pieces, growing toward the forkA strand growing toward the fork follows the fork in one long piece.
The pieces belong to the strand that grows away from it.
Why: The upper template has its 3′ end at the fork.
A new strand grows toward its template's 5′ end, so against the upper template it grows away from the fork.
Each time the fork exposes more template a new piece must start: short pieces.
Suppose a cell's DNA polymerase could join nucleotides to either end of a growing strand, the 5′ end as well as the 3′ end.
Which change at the replication fork would follow?
- A. Both new strands would grow away from the fork in short piecesA strand that can grow at either end can always grow toward the fork.
Following the fork, it has no reason to stop and start again. - B. The leading strand would be built in short piecesThe leading strand already grows toward the fork in one piece.
The change frees the other strand to do the same. - C. The new strand on the template with its 3′ end at the fork would grow away from the fork in one long pieceA strand that could grow at either end would grow toward the fork, where new template appears.
Growing away from the fork is forced only by the 3′-end rule. - D. ✓ Both new strands would grow toward the fork in one long piece
Why: One new strand is made in pieces because it grows only at its 3′ end, which points away from the fork.
A polymerase adding at either end could lengthen that strand at its fork-side end as the fork moves.
So both new strands would follow the fork in one piece.
A lagging strand is built from 12 short pieces. Every RNA primer on it has been replaced with DNA.
At how many places between neighboring pieces does ligase join the backbone?
- A. 6Ligase joins every pair of neighboring pieces, not every other pair.
Twelve pieces in a row have eleven joins between them. - B. ✓ 11
- C. 12Ligase joins between pieces, not once per piece.
Twelve pieces have eleven joins between them. - D. 24Ligase forms one covalent bond at each break in the backbone.
Eleven breaks, eleven joins.
Why: Between each pair of neighboring pieces there is one break in the backbone.
12 pieces in a row have 11 such places.
Ligase joins the backbone at each one.
A complete replication fork is drawn below, moving to the left, with nine parts lettered J to R.
Which lettered part joins nucleotides to the 3′ end of a growing new strand?
- A. KK is a primer, the dark block where a new strand began.
It gives DNA polymerase a 3′ end to add to; it adds nothing itself. - B. LL is topoisomerase, the dashed oval on the paired part ahead of the fork.
It cuts one strand and seals it; it adds no nucleotides. - C. ✓ O
- D. QQ is the leading strand, the new strand being built along its template.
Nucleotides are added to it; it adds none.
Why: DNA polymerase sits at the growing 3′ end of a new strand, drawn as a rounded box at the strand’s tip.
O is the box at the tip of the long light strand.
So O joins nucleotides to the 3′ end.
A cell copies its DNA with one part of the copying machine blocked. Forks open along the DNA, and RNA primers lie paired to the exposed templates, but no DNA nucleotide joins any primer.
Which part is blocked?
- A. HelicaseForks open, so helicase is at work.
The missing job comes after the primers are in place. - B. The enzyme that lays down primersPrimers lie paired to the templates, so that enzyme is at work.
What fails is the adding of DNA to them. - C. ✓ DNA polymerase
- D. LigaseLigase joins pieces that have already been built.
Here nothing is built at all, so the missing job comes earlier.
Why: Forks open and primers are laid down, so helicase and the primer enzyme are working.
DNA polymerase is the enzyme that joins DNA nucleotides to a primer's 3′ end.
No DNA is added, so DNA polymerase is blocked.
A student says: “A cell with no working ligase makes no new DNA at all.”
Which statement about the student's claim is correct?
- A. The student is right: ligase adds the nucleotides that build each new strandDNA polymerase adds the nucleotides of every new strand.
Ligase adds none; it joins the backbone between finished pieces. - B. ✓ The student is wrong: both new strands are built, but the lagging strand's pieces stay unjoined
- C. The student is wrong: only the leading strand is built, so the cell makes half of its new DNA and then stopsThe lagging strand's pieces are built too, each from its own primer.
Without ligase they are simply left unjoined. - D. The student is right: no primer can be laid down without ligasePrimers are laid down by another enzyme, before any piece is built.
Ligase acts last, on pieces that already exist.
Why: Ligase's one job is to join the backbone between neighboring pieces of the lagging strand.
Helicase, the primer enzyme and DNA polymerase build both new strands without it.
So the new DNA is made; only the joins are missing.
A replication fork is drawn below, moving to the left. Its two template strands are lettered L and M, and the 3′ and 5′ ends of each are written at the fork and at the far end. Both new strands have already been started.
Where is the next RNA primer laid down as the fork exposes more of each template?
- A. At the fork end of template LL has its 5′ end at the fork, so its new strand grows toward the fork in one piece.
It needs no new primer. - B. At the far end of template LL’s new strand started once, at the far end, and grows toward the fork.
Its one primer is already in place. - C. ✓ At the fork end of template M
- D. At the far end of template MEach new piece on M starts beside the fork and grows away from it.
Its primer is laid at the fork end.
Why: A new strand runs the other way to its template and grows only at its 3′ end.
M has 3′ at the fork, so its new strand grows away from it.
When the fork exposes more of M, a new piece starts at M’s fork end from a new primer.
Bacteria from a lake are grown for many generations in heavy nitrogen, so every strand of their DNA is heavy. They are moved to light nitrogen and copy their DNA once, twice and three times. After each round a centrifuge spins the DNA. Heavy DNA settles lower in the tube than light DNA, and half-heavy DNA settles between them. A thicker band means a bigger share of the molecules. The three tubes are drawn below.
(a) Identify the way of copying that the round-one tube alone rules out. (1 pt)
- Award 1 point for: old strands kept together (the conservative way).
Slip Naming the mixed way. Old and new mixed along every strand also gives one half-heavy band after round one, so round one does not rule it out.
(b) Describe how the bands change from round one to round two. (1 pt)
After round two there are two bands of equal thickness, at the half-heavy mark and the light mark.
- Award 1 point for: one half-heavy band → a half-heavy band and a light band of equal thickness (a light band appears beside the half-heavy band).
(c) Evaluate the claim that old and new are mixed along every strand, using the round-two tube. (1 pt)
With old and new mixed along every strand, every molecule keeps some heavy DNA, so no molecule is ever all light.
The round-two tube shows a light band, so some molecules are all light.
So old and new are not mixed along every strand.
- Award 1 point for: the claim is rejected because the mixed way gives no all-light molecule (one band between the half-heavy and light marks), and the round-two tube shows a light band.
Slip Accepting the claim because round two shows two bands. The mixed way gives one band, and never a light one.
(d) Explain why the half-heavy band is thinner after round three than after round two. (1 pt)
So the same two molecules are half-heavy in every round.
After round two those two are 2 of 4 molecules; after round three they are 2 of 8, a smaller share.
A band holding a smaller share of the molecules is thinner.
- Award 1 point for: the same two half-heavy molecules are 2 of 4 after round two and 2 of 8 after round three, a smaller share, so the band is thinner.
Slip Saying the half-heavy molecules are lost or that their heavy strands are diluted. The two original heavy strands stay whole, each paired with a light partner, in every round; only their share of the molecules falls.
A complete replication fork is drawn below, moving to the right, with nine parts lettered J to R. The 3′ and 5′ ends of each template are written at the fork and at the far end.
(a) Describe the job of the part lettered K at the position where it is drawn. (1 pt)
Helicase pushes the twist into that paired part, so it twists tighter than a double helix normally is.
Topoisomerase cuts one strand there, lets the extra twist unwind and seals the strand, so the fork keeps moving.
- Award 1 point for: K (topoisomerase) relaxes the over-twisting (supercoiling) in the paired DNA ahead of the fork by cutting a strand, letting it unwind and resealing it. Accept 'cuts the DNA' without 'one strand'.
(b) Explain, using the ends written on the two templates, why the strand lettered Q is made in short pieces. (1 pt)
A new strand pairs the other way round to its template, so Q's growing 3′ end points away from the fork.
DNA polymerase can add only at that 3′ end, so Q grows away from the fork.
Each time the fork exposes more of Q's template, a new piece must start from a new primer.
- Award 1 point for: Q's template has its 3′ end at the fork, so the new strand (which grows 5′ to 3′, away from the fork) cannot follow the fork and must start again from a new primer each time more template is exposed.
Slip Saying Q's template has its 5′ end at the fork. That template gives the one-piece strand, P.
(c) Identify the part lettered N. (1 pt)
- Award 1 point for: a primer (RNA primer).
(d) Explain what the drawing would show at the join between two pieces of Q if the part lettered R were removed and the copying ran to the end. (1 pt)
R is ligase, the enzyme that joins the backbone between two neighboring pieces.
DNA polymerase fills the space where each primer was but forms no join between pieces, so without R the breaks remain.
- Award 1 point for: the pieces stay unjoined (breaks in the backbone remain between them) because R is ligase, the only enzyme that joins the backbone between neighboring pieces; DNA polymerase adds nucleotides but makes no such join.
Slip Saying that Q is not built at all. Every piece is built by DNA polymerase; only the joining is lost.
APBIO-U06-L10 A message leaves the nucleus
Here is a pancreas cell. The gene for insulin sits on a chromosome inside the nucleus. The ribosomes that build insulin sit outside the nucleus, on the rough ER.
The DNA never leaves the nucleus. So how does the instruction get out?
Unit 6 · Gene Expression and Regulation
1The route from the gene to the ribosome
A pancreas cell carries the gene for insulin.
What is a gene?
- A. ✓ A stretch of DNA carrying the instructions for one RNA or protein
- B. A protein the cell makes from amino acidsInsulin is the protein.
The gene is the stretch of DNA that carries the instructions for making insulin. - C. The double membrane wrapped around the nucleusThe double membrane around the nucleus is the nuclear envelope.
A gene is a stretch of DNA.
Why: A gene is a stretch of a cell’s DNA that carries the instructions for making one RNA or protein.
A pancreas cell is a eukaryotic cell.
Where does the cell keep its DNA?
- A. In the cytosolA bacterium keeps its DNA in the cytosol.
A eukaryotic cell keeps its DNA inside the nucleus. - B. ✓ In the nucleus
- C. On the rough ERThe rough ER carries ribosomes on its surface.
The cell’s DNA is inside the nucleus.
Why: A eukaryotic cell keeps its DNA inside the nucleus, a body wrapped in the nuclear envelope.
The pancreas cell has a network of membranes just outside its nucleus, the ER.
Which part of the ER is the rough ER?
- A. ✓ The part with ribosomes on its surface
- B. The part with no ribosomes on itThe part of the ER with no ribosomes on it is the smooth ER.
Ribosomes on the surface make the ER look rough.
Why: The rough ER is the part of the ER with ribosomes on its surface.
The pancreas cell uses the insulin gene’s instructions and makes insulin.
Which of the following is this called?
- A. SecretionSecretion is the cell releasing a substance to the outside.
Using a gene’s instructions to make its protein is gene expression. - B. Cell divisionCell division is one cell becoming two.
Using a gene’s instructions to make its protein is gene expression. - C. ✓ Gene expression
Why: A cell using a gene’s instructions to make that gene’s protein is gene expression.
Video: Watch: The route from the gene to the ribosome
The pancreas cell drawn large; the insulin gene lighting up inside the nucleus; a copy in RNA made beside it, slipping out through a pore, reaching a ribosome on the rough ER; insulin appearing; the two names written at their steps.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L10a.mp4
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How does an instruction locked in the nucleus reach a ribosome?
The cell copies the gene into a message written in RNA. The copy leaves the nucleus through a pore in the nuclear envelope.
At the ribosome the message is read, and the protein is built.
In a cell with a nucleus, every gene for a protein takes this route.
Here is the pancreas cell drawn large. The insulin gene is a short stretch of one chromosome’s DNA, inside the nucleus.
The nuclear envelope has small openings in it, called pores. A pore is wide enough for a strand of RNA to pass through.
The ribosomes that build insulin sit on the rough ER, outside the nucleus.
The chromosome stays inside the nucleus. So no ribosome ever touches the insulin gene.
Step 1. Inside the nucleus, the cell makes a copy of the insulin gene. The copy is written in RNA, not DNA.
Copying a gene into RNA is called . To transcribe is to write out a copy.
Step 2. The RNA copy leaves the nucleus through a pore. The gene stays behind on its chromosome.
Step 3. Outside the nucleus, the RNA copy reaches a ribosome on the rough ER. The ribosome reads the copy and joins amino acids in the order the copy sets.
Reading the RNA copy and building the protein from it is called . The ribosome translates the order of bases into an order of amino acids.
Step 4. The ribosome builds insulin from the copy and passes it into the rough ER.
The gene never moved. A copy of the gene did the traveling.
So expressing a gene is this whole route. Transcription in the nucleus comes first, then translation at a ribosome.
What you are expected to know Describe the route from a gene to its protein: the cell copies the gene into RNA inside the nucleus (transcription), the copy leaves through a pore, and a ribosome builds the protein from the copy (translation).
A pancreas cell copies its insulin gene into RNA.
Where does this happen?
- A. ✓ In the nucleus
- B. At a ribosome, outside the nucleusThe gene is on a chromosome inside the nucleus.
The cell copies the gene where the gene is, in the nucleus.
Why: The insulin gene is inside the nucleus.
So the cell copies the gene into RNA inside the nucleus.
A ribosome builds insulin from an RNA copy of the gene.
Where does this happen?
- A. Inside the nucleusNo ribosome is inside the nucleus.
The ribosomes that build insulin sit on the rough ER, outside the nucleus. - B. ✓ Outside the nucleus, on the rough ER
Why: The ribosomes that build insulin sit on the rough ER.
The rough ER is outside the nucleus.
So the ribosome builds insulin outside the nucleus.
An instruction travels from the insulin gene to a ribosome.
What passes through the pore in the nuclear envelope?
- A. The gene’s DNAThe gene’s DNA stays on its chromosome inside the nucleus.
The RNA copy passes through the pore. - B. ✓ An RNA copy of the gene
Why: The cell copies the gene into RNA inside the nucleus.
The RNA copy passes through the pore.
The gene’s DNA stays inside.
A ribosome has finished building a molecule of insulin.
Where is the insulin gene’s DNA now?
- A. Inside the finished insulinInsulin is a protein, built from amino acids.
The gene’s DNA is not part of it. - B. At the ribosomeThe ribosome read an RNA copy of the gene, not the gene’s DNA.
The DNA never left the nucleus. - C. ✓ Still on its chromosome in the nucleus
Why: The gene’s DNA never leaves the nucleus.
A copy in RNA traveled to the ribosome.
So the gene’s DNA is still on its chromosome.
Suppose a muscle cell expresses the gene for a muscle protein.
Where does the muscle cell copy that gene into RNA?
- A. At a ribosome in the cytoplasmA ribosome reads an RNA copy; it does not make the copy.
The gene is copied where the gene is, inside the nucleus. - B. ✓ Inside the nucleus
Why: A muscle cell has a nucleus, and every gene for a protein takes the same route.
The gene is on a chromosome inside the nucleus.
So the muscle cell copies the gene into RNA inside the nucleus.
The route from the insulin gene to insulin is drawn below. Two of its steps carry a lettered marker, A and B.
In which order do the two lettered steps happen?
- A. ✓ B, then A
- B. A, then BStep A is at the ribosome, where the copy is read.
Step B is inside the nucleus, where the copy is made.
The copy is made before it is read.
Why: Step B is inside the nucleus: the cell copies the gene into RNA.
Step A is at the ribosome: the ribosome reads the copy.
The copy must exist before the ribosome can read it.
So B happens first, then A.
A student says: “The insulin gene’s DNA leaves the nucleus and travels to the ribosome, and the ribosome reads the DNA.”
Is the student correct?
- A. Yes: the gene’s DNA travels to the ribosome, and the ribosome reads the DNA thereThe gene’s DNA never leaves the nucleus.
The cell copies the gene into RNA, and the copy travels to the ribosome. - B. ✓ No: the ribosome reads an RNA copy of the gene, and the DNA stays in the nucleus
Why: The gene’s DNA stays on its chromosome inside the nucleus.
The cell copies the gene into RNA.
The RNA copy leaves through a pore and reaches the ribosome.
So the ribosome reads the RNA copy, not the DNA.
31Quick quiz: transcription, translation mixed practice
A cell expresses a gene in two steps.
What is transcription?
- A. ✓ Copying a gene into RNA
- B. Building a protein from an RNA copy of a geneBuilding a protein from an RNA copy is translation.
- C. Copying the cell’s whole DNA before the cell dividesCopying the whole DNA before division is DNA replication.
Transcription copies one gene into RNA.
Why: Transcription is copying a gene into RNA, inside the nucleus.
A cell expresses a gene in two steps.
What is translation?
- A. Copying a gene into RNACopying a gene into RNA is transcription.
- B. Moving an RNA copy out of the nucleus through a poreThe RNA copy leaves through a pore between the two steps.
Translation is the ribosome building the protein from the copy. - C. ✓ Building a protein from an RNA copy of a gene
Why: Translation is a ribosome reading an RNA copy of a gene and building the protein from it.
A cell expresses a gene in two steps, transcription and translation.
(a) State what transcription is and what translation is. (1 pt)
Translation is a ribosome reading the RNA copy and building the protein from it.
- Award 1 point for: transcription = copying a gene into RNA; translation = building the protein from the RNA copy (at a ribosome).
Inside a nucleus, a cell makes an RNA copy of a gene.
Which step is this?
- A. ✓ Transcription
- B. TranslationTranslation is the ribosome building the protein.
Making the RNA copy is transcription.
Why: Making an RNA copy of a gene is transcription.
A ribosome joins amino acids in the order an RNA copy sets.
Which step is this?
- A. TranscriptionTranscription is copying the gene into RNA.
A ribosome joining amino acids is translation. - B. ✓ Translation
Why: A ribosome building the protein from an RNA copy is translation.
A cell expresses a gene in two steps.
Which step happens outside the nucleus?
- A. TranscriptionTranscription happens where the gene is, inside the nucleus.
- B. ✓ Translation
Why: Translation happens at a ribosome.
Ribosomes are outside the nucleus.
So translation happens outside the nucleus.
A cell begins to express a gene.
Which step happens first?
- A. ✓ Transcription
- B. TranslationThe ribosome needs an RNA copy to read.
The copy is made first, by transcription.
Why: Transcription makes the RNA copy.
Translation reads the copy.
So transcription happens first.
39The messenger
In a cell, DNA is two paired strands.
How many strands does an RNA molecule have?
- A. TwoTwo paired strands is DNA.
RNA is a single strand with no partner strand beside it. - B. ✓ One
Why: RNA is a single strand: one backbone with bases along it, and no partner strand.
A ribosome makes a protein by joining amino acids one after another.
What sets the order of the amino acids?
- A. ✓ Messenger RNA (mRNA)
- B. Ribosomal RNA (rRNA)Ribosomal RNA is the RNA the ribosome is built from.
The mRNA passing through the ribosome sets the order. - C. The order the amino acids arrive inAmino acids arrive in no fixed order.
The mRNA passing through the ribosome sets the order.
Why: The ribosome joins amino acids in the order the mRNA sets.
Every strand of DNA or RNA has two different ends.
Which group does the 5′ end of a strand carry?
- A. A free –OH groupThe free –OH group is at the 3′ end.
The 5′ end carries a free phosphate group. - B. ✓ A free phosphate group
Why: The 5′ end of a strand carries a free phosphate group, and the 3′ end carries a free –OH group.
Video: Watch: The messenger
The RNA copy of the insulin gene drawn as one strand with its 5′ and 3′ ends marked; the ribosome starting at the 5′ end and moving toward the 3′ end; the same strand named messenger RNA; the pancreas cell shown again with the whole route.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L10b.mp4
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Go back to the RNA copy of the insulin gene. Here it is drawn on its own.
The copy is a single strand: one backbone with bases along it, and no partner strand.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
The order of the bases along the copy is the gene’s instruction, written in RNA. The copy carries the instructions of one gene only: the insulin gene.
The copy carries those instructions from the DNA in the nucleus to a ribosome in the cytoplasm. It is a messenger.
An RNA copy that carries one gene’s instructions from the DNA to a ribosome is called .
The strand a ribosome reads is this same molecule: the mRNA is the RNA copy of the gene.
A ribosome reads an mRNA in one direction only. It starts at the mRNA’s 5′ end and moves toward the 3′ end.
The insulin mRNA is a few hundred nucleotides long. It leaves the nucleus through a pore and reaches a ribosome on the rough ER.
What you are expected to know Describe messenger RNA (mRNA): a single-stranded RNA copy of one gene’s instructions that carries them from the DNA in the nucleus to a ribosome, which reads it from its 5′ end.
A cell makes a messenger RNA (mRNA) from its insulin gene.
Which instructions does that mRNA carry?
- A. ✓ The instructions of the insulin gene only
- B. The instructions of every gene on that chromosomeAn mRNA is a copy of one gene, not of the whole chromosome.
- C. The instructions of every gene in the cellAn mRNA is a copy of one gene, not of every gene in the cell.
Why: An mRNA is an RNA copy of one gene.
So the insulin mRNA carries the insulin gene’s instructions only.
A cell has just made an mRNA inside its nucleus.
Where does the mRNA go next?
- A. Deeper into the nucleus, to another chromosomeAn mRNA carries a gene’s instructions to a ribosome.
Ribosomes are outside the nucleus. - B. ✓ Out of the nucleus, to a ribosome
Why: An mRNA carries a gene’s instructions from the DNA to a ribosome.
Ribosomes are outside the nucleus.
So the mRNA leaves the nucleus through a pore and goes to a ribosome.
A student says: “An mRNA is a single strand, so its bases have no partner strand.”
Is the student correct?
- A. ✓ Yes: an mRNA is a single strand, and its bases have no partner strand
- B. No: an mRNA’s bases are paired with a partner strand, as a DNA molecule’s areTwo paired strands is DNA.
An mRNA is RNA: one strand, with no partner strand.
Why: An mRNA is RNA.
RNA is a single strand.
So an mRNA’s bases have no partner strand.
A ribosome begins to read an mRNA.
At which end of the mRNA does the ribosome start?
- A. At the 3′ endThe ribosome moves toward the 3′ end.
It starts at the 5′ end. - B. ✓ At the 5′ end
Why: A ribosome reads an mRNA in one direction: from the 5′ end toward the 3′ end.
So the ribosome starts at the 5′ end.
A ribosome on the rough ER is building a protein.
Which molecule is the ribosome reading?
- A. ✓ An mRNA
- B. The gene’s DNAThe gene’s DNA is inside the nucleus, where no ribosome is.
The ribosome reads the mRNA that came from the gene.
Why: The gene’s DNA stays inside the nucleus.
The mRNA carries the gene’s instructions to the ribosome.
So the ribosome reads the mRNA.
Go back to the pancreas cell.
Inside the nucleus, the cell copies the insulin gene into mRNA: transcription.
The mRNA leaves through a pore and reaches a ribosome on the rough ER. The ribosome builds insulin from it: translation.
62Quick quiz: messenger RNA (mRNA) mixed practice
A cell contains several kinds of RNA.
What is messenger RNA (mRNA)?
- A. The RNA a ribosome is built from, with its proteinsThe RNA a ribosome is built from is ribosomal RNA (rRNA).
- B. ✓ An RNA copy of one gene’s instructions, carried to a ribosome
- C. The two paired DNA strands that make up a geneTwo paired strands is DNA.
An mRNA is a single strand of RNA.
Why: Messenger RNA (mRNA) is a single-stranded RNA copy of one gene’s instructions, carried from the DNA to a ribosome.
A pancreas cell makes an RNA copy of its insulin gene.
(a) State what messenger RNA (mRNA) is and where it goes. (1 pt)
It leaves the nucleus through a pore and goes to a ribosome.
The ribosome reads it.
- Award 1 point for: mRNA = a (single-stranded) RNA copy of one gene’s instructions; it goes from the nucleus (through a pore) to a ribosome.
An RNA copy of the insulin gene leaves the nucleus, and a ribosome reads it.
Which kind of RNA is the copy?
- A. ✓ Messenger RNA (mRNA)
- B. Ribosomal RNA (rRNA)Ribosomal RNA is the RNA the ribosome is built from.
The copy the ribosome reads is messenger RNA.
Why: An RNA copy of a gene that a ribosome reads is messenger RNA (mRNA).
A ribosome has just read the 3′ end of an mRNA.
How much of the mRNA has the ribosome read?
- A. ✓ All of it
- B. Only its 3′ endA ribosome starts at the 5′ end and reads toward the 3′ end.
By the 3′ end it has read the whole mRNA. - C. About half of itA ribosome reads from the 5′ end to the 3′ end without stopping halfway.
At the 3′ end it has read all of it.
Why: A ribosome reads an mRNA in one direction, from its 5′ end toward its 3′ end.
The 3′ end is the last part it reads.
So by the 3′ end it has read all of the mRNA.
A student says: “One mRNA carries the instructions of every gene in the cell to the ribosome.”
Is the student correct?
- A. Yes: one mRNA copies every gene in the cell, so one mRNA carries all the cell’s instructionsAn mRNA is a copy of one gene.
So one mRNA carries one gene’s instructions only. - B. ✓ No: each mRNA is a copy of one gene, so the cell makes a different mRNA for each gene it expresses
Why: The cell copies one gene into one mRNA.
So one mRNA carries the instructions of one gene only.
The cell makes a different mRNA for each gene it expresses.
Glossary
- transcription
- Copying a gene into RNA, inside the nucleus. The RNA copy carries the gene’s instructions; the gene’s DNA stays on its chromosome.
- translation
- A ribosome reading an RNA copy of a gene and building the protein from it, joining amino acids in the order the copy sets. It happens outside the nucleus.
- messenger RNA (mRNA)
- An RNA copy of one gene’s instructions: a single strand that carries them from the DNA in the nucleus to a ribosome, which reads it from its 5′ end.
APBIO-U06-L11 Three RNAs, three jobs
Here are three RNA molecules from the same cell. One is a long strand with no partner. One is folded back on itself into a shape like a clover leaf, with an amino acid hanging from one end. One is folded up tightly with proteins into a ball.
The same four letters make all three. Why do they have three different shapes?
Unit 6 · Gene Expression and Regulation
1The folded strand that carries an amino acid
An RNA strand lies against another strand.
Which base pairs with adenine on an RNA strand?
- A. CytosineCytosine pairs with guanine.
On an RNA strand, adenine pairs with uracil. - B. GuanineGuanine pairs with cytosine.
On an RNA strand, adenine pairs with uracil. - C. ✓ Uracil
Why: On an RNA strand, adenine pairs with uracil.
Guanine pairs with cytosine, whether the strand is DNA or RNA.
A ribosome builds a protein from small units.
What are the units a protein is built from?
- A. NucleotidesNucleotides are the units of DNA and RNA.
A protein is built from amino acids. - B. ✓ Amino acids
- C. SugarsSugars are the units of a carbohydrate.
A protein is built from amino acids.
Why: A protein is a chain of amino acids.
There are twenty kinds of amino acid.
Video: Watch: The folded strand that carries an amino acid
A short RNA strand drawn straight; stretches of it pairing with other stretches; the strand folding back into an L; an enzyme fixing one amino acid to its 3′ end; the folded strand landing on an mRNA, its three free bases pairing with three bases of the message; the names written at their parts.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L11a.mp4
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Why does a cell make three kinds of RNA? Each kind has a different job, and the job comes from the shape.
The long strand is mRNA. It carries one gene’s instructions to a ribosome.
The clover leaf is the adaptor. It pairs three of its bases with three bases of the mRNA, and it carries the one amino acid those three bases mean.
The ball is RNA folded up with proteins: the ribosome itself, the builder.
As with proteins in Unit 1, the sequence sets the fold and the fold sets the job.
Suppose a cell makes a short RNA, about 80 nucleotides long. Here it is drawn straight, with fewer nucleotides than the real strand.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
Some stretches of this strand can pair with other stretches of the same strand. Adenine pairs with uracil, and guanine pairs with cytosine, just as between two strands.
So the strand folds back on itself. Hydrogen bonds hold each paired stretch together.
Flattened out on paper, the folded strand looks like a clover leaf. In the cell it is bent into an L shape, and that is how we draw it.
The fold brings the strand’s two ends together at one tip of the L. At the far tip, three bases stay unpaired.
An enzyme attaches one amino acid to the strand’s 3′ end. This strand is loaded with alanine, one of the twenty amino acids.
Each folded RNA of this kind is loaded with one particular amino acid. A different one carries a different amino acid.
Now bring the loaded strand to an mRNA. Here is part of an mRNA, 5′-AUGGCUAAG-3′, with its bases in groups of three.
Three bases on the mRNA read together are called a . This piece of mRNA has three codons: AUG, GCU and AAG.
The three unpaired bases at the far tip of the L pair with one codon. Here they pair with the codon GCU: C with G, G with C, A with U.
The three bases on the folded strand that pair with a codon are called the . This strand’s anticodon is 3′-CGA-5′.
Each codon means one amino acid. GCU means alanine, and the strand whose anticodon pairs with GCU carries alanine.
So this strand brings one amino acid to the codon that means it. A small RNA that pairs its anticodon with a codon and carries the amino acid that codon means is called .
What you are expected to know Describe transfer RNA (tRNA): a small RNA folded into an L, with an anticodon of three bases at one tip that pairs with a codon on the mRNA, and the one amino acid that codon means at its 3′ end.
A ribosome reads an mRNA three bases at a time.
What are three bases on the mRNA read together called?
- A. A geneA gene is a stretch of DNA.
Three bases on the mRNA read together are a codon. - B. An anticodonAn anticodon is the three bases on a tRNA.
Three bases on the mRNA read together are a codon. - C. ✓ A codon
Why: Three bases on the mRNA read together are called a codon.
A cell contains several kinds of RNA.
What is a transfer RNA (tRNA)?
- A. ✓ A small folded RNA carrying one amino acid to its codon
- B. A long RNA carrying one gene’s instructions to a ribosomeThe long RNA that carries one gene’s instructions to a ribosome is messenger RNA (mRNA).
- C. The RNA folded up with proteins into a ballThe RNA folded up with proteins into a ball is ribosomal RNA (rRNA).
Why: A transfer RNA (tRNA) is a small RNA folded into an L that carries one amino acid and pairs its anticodon with a codon on the mRNA.
A tRNA pairs with an mRNA.
What is an anticodon?
- A. The amino acid attached to the tRNAThe amino acid is attached at the tRNA’s 3′ end.
The anticodon is the three bases that pair with a codon. - B. ✓ Three bases on the tRNA that pair with a codon
- C. Three bases on the tRNA that are the same as the codon’sAn anticodon carries the partner of each codon base, not the same base.
A pairs with U, and G pairs with C.
Why: An anticodon is the three bases on a tRNA that pair with a codon on the mRNA.
A tRNA is drawn below with two lettered markers, P at one tip and Q at the other.
Which tip pairs with a codon on the mRNA?
- A. ✓ Tip P
- B. Tip QTip Q is where the strand’s two ends are.
The three unpaired bases at tip P pair with the codon.
Why: The anticodon is the three unpaired bases at the far tip of the L.
Tip P carries those three bases.
So tip P pairs with a codon.
A tRNA is drawn below with two lettered markers, P at one tip and Q at the other.
At which tip does the enzyme attach the amino acid?
- A. ✓ Tip P
- B. Tip QTip Q is the anticodon, which pairs with the codon.
The amino acid is attached at the 3′ end, at tip P.
Why: The enzyme attaches the amino acid at the strand’s 3′ end.
The strand’s two ends are at tip P.
So the amino acid is attached at tip P.
A tRNA pairs with an mRNA.
How many bases of the tRNA pair with the mRNA?
- A. OneA codon is three bases, so one base cannot pair with a whole codon.
The anticodon is three bases. - B. ✓ Three
- C. Every baseMost of the tRNA’s bases are paired within its own strand.
Only the three bases of the anticodon pair with the mRNA.
Why: The anticodon is three bases.
The anticodon pairs with one codon of three bases.
So three bases of the tRNA pair with the mRNA.
An enzyme loads a tRNA.
How many amino acids does one tRNA carry?
- A. ✓ One
- B. ThreeThree is the number of bases in the anticodon.
A tRNA carries one amino acid. - C. TwentyTwenty is the number of kinds of amino acid.
One tRNA carries one amino acid.
Why: The enzyme attaches one amino acid to the tRNA’s 3′ end.
So one tRNA carries one amino acid.
A tRNA’s anticodon pairs with the codon UUU of an mRNA. Two codons and the amino acids they mean are listed below.
Which amino acid does this tRNA carry?
- A. AlanineAlanine is the amino acid that the codon GCU means.
This tRNA pairs with the codon UUU, and the table says UUU means phenylalanine. - B. LysineThe table lists codons, and AAA is this tRNA’s anticodon, not its codon.
Its codon is UUU, which means phenylalanine. - C. ✓ Phenylalanine
Why: A tRNA carries the amino acid that its codon means.
This tRNA’s codon is UUU.
The table says UUU means phenylalanine.
So this tRNA carries phenylalanine.
33Quick quiz: transfer RNA (tRNA), anticodon, codon mixed practice
A strand is written from its 5′ end to its 3′ end, and both ends are marked. The codon 5′-AAG-3′ of an mRNA is drawn below. The anticodon is written from its 3′ end to its 5′ end, so each of its bases sits opposite its partner on the codon.
Which anticodon pairs with this codon?
- A. 3′-AAG-5′An anticodon carries the partner of each base, not the same base.
A pairs with U, and G pairs with C. - B. 3′-CUU-5′Written from its 3′ end, the anticodon’s first base sits opposite the codon’s first base, A.
The partner of A is U, not C. - C. ✓ 3′-UUC-5′
Why: Each base of the anticodon pairs with the codon base opposite it.
Opposite A is U.
Opposite the second A is U.
Opposite G is C.
So the anticodon is 3′-UUC-5′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. The codon 5′-GGC-3′ of an mRNA is drawn below. The anticodon is written from its 3′ end to its 5′ end, so each of its bases sits opposite its partner on the codon.
Which anticodon pairs with this codon?
- A. ✓ 3′-CCG-5′
- B. 3′-GCC-5′Written from its 3′ end, the anticodon’s first base sits opposite the codon’s first base, G.
The partner of G is C, not G. - C. 3′-GGC-5′An anticodon carries the partner of each base, not the same base.
G pairs with C, and C pairs with G.
Why: Each base of the anticodon pairs with the codon base opposite it.
Opposite G is C.
Opposite the second G is C.
Opposite C is G.
So the anticodon is 3′-CCG-5′.
A tRNA arrives at a ribosome.
Which of the following does its anticodon pair with?
- A. ✓ A codon on the mRNA
- B. A codon on the gene’s DNAThe gene’s DNA stays in the nucleus, where no ribosome is.
The anticodon pairs with a codon on the mRNA. - C. The anticodon of another tRNATwo tRNAs do not pair with each other.
Each anticodon pairs with a codon on the mRNA.
Why: An anticodon is three bases on a tRNA that pair with a codon on the mRNA.
Suppose a poison destroys every tRNA in a cell whose anticodon pairs with the codon GCU. The cell’s mRNAs, its other tRNAs and its ribosomes are unharmed. Two codons and the amino acids they mean are listed below.
(a) Identify the amino acid that the poison stops reaching the cell’s ribosomes. (1 pt)
- Award 1 point for: alanine.
(b) Explain why the ribosomes leave that amino acid out where they read the codon GCU. (1 pt)
Frame The ribosomes leave it out because …
At the ribosome, that anticodon pairs with the codon GCU, and the ribosome joins the alanine it brings to the protein.
With every such tRNA destroyed, no anticodon pairs with GCU.
So no alanine is brought to that codon, and the ribosome has none to add.
- Award 1 point for: the tRNA that pairs with GCU is the one carrying alanine, so with those tRNAs destroyed no alanine is brought to the codon GCU.
38The RNA the ribosome is built from
Every cell has ribosomes.
What does a ribosome do?
- A. Carries one gene’s instructions out of the nucleusAn mRNA carries a gene’s instructions to the ribosome.
The ribosome joins amino acids into a protein. - B. Copies a gene into RNACopying a gene into RNA is transcription, and it happens at the gene.
A ribosome joins amino acids into a protein. - C. ✓ Joins amino acids into a protein
Why: A ribosome makes a protein by joining amino acids in the order an mRNA sets.
A ribosome from a bacterium and a ribosome from a human cell have the same build.
Which of the following is that build?
- A. ✓ A smaller part and a larger part
- B. One part onlyEvery ribosome has two parts, a smaller one and a larger one.
- C. A hollow ball with a membraneA ribosome has no membrane.
It is two parts, a smaller one and a larger one.
Why: Every known cell’s ribosomes share one build: a smaller part and a larger part.
Video: Watch: The RNA the ribosome is built from
A long RNA strand pairing with itself in many places and folding into a compact ball; proteins settling onto the fold; two such parts, a smaller and a larger, coming together as a ribosome; an mRNA threading between them and a tRNA held in place; the name written.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L11b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L11b.mp4
Now consider the RNA folded up with proteins into a ball. This kind of RNA is long, up to thousands of nucleotides.
Its bases pair within the strand in many places. So the strand folds up into a compact shape.
Proteins bind to the folded RNA. The folded RNA and its proteins make one part of a ribosome.
The RNA a ribosome is built from is called : the same rRNA named beside the ribosome in Unit 2.
A ribosome is built in two parts, the smaller lump and the larger lump. Each part is called a subunit.
Each subunit is rRNA folded up with proteins bound to it. Here are the two subunits drawn with their insides showing.
About half of a ribosome’s mass, or more, is rRNA. The rest is protein.
Here is a ribosome from a human cell, as a model built from many electron-microscope pictures.

A ribosome from a human cell, as a model built from many electron-microscope pictures: a smaller lump on a larger lump. The threads winding through both are the folded rRNA; the solid surfaces are the proteins bound to it. Image: Ykust, Wikimedia Commons, CC BY 4.0 (cropped and resized).
The rRNA does the work. The proteins keep the rRNA in its folded shape.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
The small subunit’s rRNA binds the mRNA. The large subunit’s rRNA binds each tRNA and joins its amino acid to the protein being built.
What you are expected to know State ribosomal RNA’s job: rRNA molecules, folded and bound to proteins, build the ribosome’s two subunits, bind the mRNA and the tRNAs, and join the amino acids.
A long RNA folds up, proteins bind to it, and the RNA with its proteins makes one subunit of a ribosome.
What is this RNA called?
- A. Messenger RNA (mRNA)Messenger RNA is the copy of one gene that a ribosome reads.
It is not part of the ribosome. - B. ✓ Ribosomal RNA (rRNA)
- C. Transfer RNA (tRNA)Transfer RNA is a small RNA that carries one amino acid.
It is not part of the ribosome.
Why: The RNA a ribosome is built from is ribosomal RNA (rRNA).
This RNA, folded with its proteins, makes one subunit of a ribosome.
So it is ribosomal RNA.
A student says: “A ribosome is made of RNA as well as protein.”
Is the student correct?
- A. ✓ Yes: about half of a ribosome’s mass, or more, is rRNA, and the rest is protein
- B. No: a ribosome is protein, and the only RNA at a ribosome is the mRNA it readsEach subunit of a ribosome is rRNA folded up with proteins bound to it.
About half of a ribosome’s mass, or more, is rRNA.
Why: Each subunit of a ribosome is rRNA with proteins bound to it.
About half of the ribosome’s mass, or more, is rRNA.
So a ribosome is made of RNA and protein.
Suppose an antibiotic binds to the rRNA of a bacterium’s ribosomes and blocks the rRNA.
Which of the following stops in the bacterium?
- A. ✓ Building proteins
- B. Copying DNA before divisionDNA polymerase copies the DNA, not the ribosome.
Blocking the rRNA stops the ribosome building proteins. - C. Copying genes into RNACopying a gene into RNA happens at the gene, not at a ribosome.
Blocking the rRNA stops the ribosome building proteins.
Why: The rRNA binds the mRNA and the tRNAs and joins the amino acids.
With its rRNA blocked, the ribosome cannot join amino acids.
So the bacterium stops building proteins.
57Quick quiz: ribosomal RNA (rRNA) mixed practice
At a ribosome, one molecule joins each new amino acid to the protein being built.
Which molecule does the joining?
- A. The mRNAThe mRNA carries the instructions.
The large subunit’s rRNA joins the amino acids. - B. The rRNA of the small subunitThe small subunit’s rRNA binds the mRNA.
The large subunit’s rRNA joins the amino acids. - C. ✓ The rRNA of the large subunit
Why: The large subunit’s rRNA binds each tRNA and joins its amino acid to the protein being built.
A student says: “The proteins of a ribosome do no work; a ribosome would work just as well if they were removed.”
Is the student correct?
- A. Yes: the rRNA does all the work of the ribosome, so the proteins could be removed and the ribosome would still workThe rRNA does the work, but the proteins keep it in its folded shape.
Without its shape the rRNA cannot bind the mRNA or join the amino acids. - B. ✓ No: the proteins keep the rRNA in its folded shape, and the rRNA needs that shape to do its work
Why: The rRNA of a ribosome binds the mRNA and joins the amino acids.
The proteins keep the rRNA in its folded shape.
Without them the rRNA loses the shape it needs, so the ribosome cannot work.
Suppose the rRNA of a ribosome’s small subunit is damaged. The large subunit is unharmed.
(a) Identify the step of building a protein that fails first at this ribosome. (1 pt)
- Award 1 point for: binding (holding) the mRNA.
(b) Explain why the large subunit’s rRNA, though unharmed, joins no amino acids at this ribosome. (1 pt)
Frame The large subunit’s rRNA joins no amino acids because …
So no mRNA is held at the ribosome.
Each tRNA pairs its anticodon with a codon on the mRNA, so with no mRNA held, no tRNA is brought into place.
The large subunit’s rRNA joins the amino acid of a tRNA held in place, so with none in place it joins nothing.
- Award 1 point for: with no mRNA held, no tRNA pairs its anticodon with a codon, so the large subunit’s rRNA has no amino acid in place to join.
61Which RNA is this?
A cell copies one gene into RNA inside its nucleus, and a ribosome reads the copy.
What is the copy called?
- A. ✓ Messenger RNA (mRNA)
- B. Ribosomal RNA (rRNA)Ribosomal RNA is the RNA a ribosome is built from.
The copy of one gene that a ribosome reads is messenger RNA. - C. Transfer RNA (tRNA)Transfer RNA carries one amino acid.
The copy of one gene that a ribosome reads is messenger RNA.
Why: Messenger RNA (mRNA) is a single-stranded RNA copy of one gene’s instructions, carried from the DNA to a ribosome.
Video: Watch: Which RNA is this?
The three molecules side by side; each named in turn from its shape and its job; the table filling in row by row; six new descriptions placed in their rows.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L11c.mp4
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Go back to the three RNA molecules from the same cell.
The long open strand carries one gene’s instructions to a ribosome. This RNA is mRNA.
The folded L has an amino acid attached at its 3′ end. This RNA is tRNA.
The ball is RNA folded up with proteins bound to it, part of a ribosome. This RNA is rRNA.
What you are expected to know Classify an RNA as mRNA, tRNA or rRNA from its shape or its job.
An enzyme attaches lysine to the 3′ end of an RNA.
Which RNA is this?
- A. Messenger RNA (mRNA)An mRNA is a long open strand and carries no amino acid.
- B. Ribosomal RNA (rRNA)An rRNA is folded with proteins into a ribosome and carries no amino acid.
- C. ✓ Transfer RNA (tRNA)
Why: The loading enzyme attaches one amino acid to a tRNA’s 3′ end.
No other kind of RNA carries an amino acid.
So this RNA is a transfer RNA (tRNA).
An RNA’s two ends lie together at one tip of its fold, and three bases stay unpaired at the other tip.
Which RNA is this?
- A. Messenger RNA (mRNA)An mRNA stays an open strand.
It has no fold and no tips. - B. Ribosomal RNA (rRNA)An rRNA folds into a compact ball with proteins bound to it.
It has no tip with three unpaired bases. - C. ✓ Transfer RNA (tRNA)
Why: A tRNA folds into an L.
Its two ends lie together at one tip, and the three unpaired bases at the other tip are its anticodon.
So this RNA is a transfer RNA (tRNA).
A ribosome reads an RNA from its 5′ end toward its 3′ end and joins amino acids in the order that RNA sets.
Which RNA is this?
- A. ✓ Messenger RNA (mRNA)
- B. Ribosomal RNA (rRNA)An rRNA is part of the ribosome; the ribosome does not read it.
- C. Transfer RNA (tRNA)A tRNA brings one amino acid; the ribosome does not read it from end to end.
Why: The RNA a ribosome reads from its 5′ end, joining amino acids in the order it sets, is a messenger RNA (mRNA).
At a ribosome, an RNA holds the mRNA in place while tRNAs arrive and leave.
Which RNA is this?
- A. Messenger RNA (mRNA)The mRNA is the strand being held and read.
It does not hold itself. - B. ✓ Ribosomal RNA (rRNA)
- C. Transfer RNA (tRNA)A tRNA pairs its anticodon with one codon and then leaves.
It does not hold the mRNA.
Why: The small subunit’s rRNA binds the mRNA and holds it while it is read.
So the RNA that holds the mRNA in place is ribosomal RNA (rRNA).
Here is the whole comparison as a table: the shape of each RNA, its job, and where it does that job.
74The sequence sets the fold, and the fold sets the job
Hemoglobin is the protein in red blood cells that carries oxygen. Each of hemoglobin’s four pockets holds one oxygen molecule.
What lets hemoglobin do this job?
- A. ✓ Hemoglobin’s shape
- B. The number of amino acids in hemoglobinTwo proteins with the same number of amino acids can fold into different shapes.
The folded shape makes the pockets. - C. Hemoglobin’s chain of amino acids before it foldsAn unfolded chain has no pockets.
The chain folds, and the folded shape makes the pockets.
Why: A protein’s shape, built up through its levels of structure, is what lets it do its job.
Video: Watch: The sequence sets the fold, and the fold sets the job
Three straight strands with three different orders of bases; each folding in its own way, one staying open, one into an L, one into a ball with proteins; each set to its job; the three molecules side by side at the end.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L11d.mp4
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Go back to the tRNA. Why did its strand fold into an L?
Certain stretches of its bases can pair with other stretches of the same strand. Which stretches pair is set by the order of the bases.
So the order of the bases decides how the strand folds. The folded shape then decides what the molecule can do: hold one amino acid at one tip and pair with a codon at the other.
The mRNA’s order of bases does not fold it into a tight shape. It stays an open strand, so a ribosome can read it base by base.
The rRNA’s order of bases pairs it in many places. It folds into a compact ball that proteins bind to: a subunit of a ribosome.
So the same four letters give three shapes because the three orders of bases are different. The sequence sets the fold, and the fold sets the job, just as for the proteins of Unit 1.
In Unit 4 a gene was a stretch of DNA carrying the instructions for one protein.
Some genes carry the instructions for an RNA that is never translated: the genes for tRNAs and rRNAs. From here on, ‘gene’ includes those.
What you are expected to know Explain why the same four letters give three shapes: the order of the bases decides how an RNA strand folds on itself, and the fold decides the molecule’s job.
A cell makes an RNA strand, and the strand folds before any protein binds to it.
What decides how the strand folds?
- A. The strand’s lengthTwo strands of the same length fold differently if their orders of bases differ.
The order of the bases decides the fold. - B. The proteins that bind to itA tRNA folds with no protein bound to it.
Which stretches of a strand pair, and so how it folds, is set by the order of its bases. - C. ✓ The order of its bases
Why: Which stretches of a strand can pair with each other is set by the order of the bases.
So the order of the bases decides how the strand folds.
Suppose a cell makes two RNA strands of the same length, 90 nucleotides each, with different orders of bases. One strand folds into a tight shape. The other stays an open strand.
(a) Explain why the two strands fold differently. (1 pt)
Frame The two strands fold differently because …
A strand folds where one stretch of its bases can pair with another stretch of the same strand.
In the first strand, several stretches can pair, so the strand folds into a tight shape.
In the second strand, no stretch can pair with another, so the strand stays open.
- Award 1 point for: a strand folds where stretches of its own bases pair; the two orders of bases give different pairing (one pairs, one does not), so the folds differ.
A student says: “Every gene carries the instructions for a protein.”
Is the student correct?
- A. Yes: every gene carries the instructions for a protein, so every RNA copy of a gene is translatedThe genes for tRNAs and rRNAs carry the instructions for an RNA that is never translated into a protein.
- B. ✓ No: the genes for tRNAs and rRNAs carry the instructions for an RNA that is never translated
Why: A gene for a tRNA or an rRNA is copied into RNA.
That RNA does its job as RNA and is never translated.
So not every gene carries the instructions for a protein.
Suppose one base in a paired stretch of a tRNA is changed to a different base.
Can every base in that stretch still pair with its partner?
- A. YesA base pairs with one partner only.
The changed base does not match the base opposite it. - B. ✓ No
Why: In a paired stretch, each base pairs with the base opposite it.
The changed base does not pair with that partner.
So that base is left unpaired, and the fold changes there.
One exception: in RNA, G also pairs loosely with U, so a few changes still leave a pair.
Go back to the three RNA molecules from the same cell: the long strand, the folded L and the ball.
The long strand is mRNA, the messenger.
The clover leaf, drawn as an L, is tRNA: an anticodon at one tip and one amino acid at the other.
The ball is rRNA with proteins: the ribosome itself.
94Mixed practice mixed practice
A tRNA carries the amino acid glycine. Its anticodon pairs with one codon.
Which amino acid does that codon mean?
- A. ✓ Glycine
- B. AlanineAlanine is the amino acid the codon GCU means.
A tRNA carries the amino acid its own codon means. - C. Any of the twentyEach codon means one amino acid.
A tRNA carries the amino acid its codon means.
Why: A tRNA carries the amino acid that its codon means.
This tRNA carries glycine.
So its codon means glycine.
A student says: “Each time a tRNA arrives at the ribosome, the protein chain grows by one amino acid.”
Is the student correct?
- A. ✓ Yes: the enzyme loaded each tRNA with one amino acid, and the ribosome joins it to the chain
- B. No: the enzyme loaded each tRNA with three amino acids, one for each base of its anticodonThe three bases of the anticodon pair with one codon.
One codon means one amino acid, and the enzyme attaches one amino acid to the tRNA.
Why: The enzyme attaches one amino acid to a tRNA’s 3′ end.
At the ribosome, the tRNA’s anticodon pairs with one codon.
The ribosome joins the tRNA’s one amino acid to the chain.
So each arrival adds one amino acid.
Suppose a drug stops a cell making one kind of RNA. The cell still makes its proteins normally, but within a day it can build no new ribosomes.
Which RNA does the drug stop?
- A. Messenger RNA (mRNA)mRNAs carry genes’ instructions to ribosomes.
A ribosome is not built from them. - B. ✓ Ribosomal RNA (rRNA)
- C. Transfer RNA (tRNA)tRNAs bring amino acids to the ribosome.
A ribosome is not built from them.
Why: A ribosome is built from rRNA folded with proteins.
With no new rRNA, the cell can build no new ribosomes.
So the drug stops the cell making rRNA.
An RNA strand’s order of bases is such that no stretch of it can pair with another stretch.
What shape does the strand take?
- A. ✓ An open strand
- B. A folded LA strand folds into an L where stretches of its bases pair.
With no pairing, the strand stays open. - C. A ballA strand folds into a ball where many stretches of its bases pair.
With no pairing, the strand stays open.
Why: A strand folds only where one stretch of its bases pairs with another.
No stretch of this strand can pair.
So the strand stays an open strand.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. The codon 5′-CUG-3′ of an mRNA is drawn below. The anticodon is written from its 3′ end to its 5′ end, so each of its bases sits opposite its partner on the codon.
Which anticodon pairs with this codon?
- A. 3′-CAG-5′Written from its 3′ end, the anticodon’s first base sits opposite the codon’s first base, C.
The partner of C is G, not C. - B. 3′-CUG-5′An anticodon carries the partner of each base, not the same base.
C pairs with G, U with A, and G with C. - C. ✓ 3′-GAC-5′
Why: Each base of the anticodon pairs with the codon base opposite it.
Opposite C is G.
Opposite U is A.
Opposite G is C.
So the anticodon is 3′-GAC-5′.
A cell makes all three kinds of RNA.
Which kind is the shortest?
- A. Messenger RNA (mRNA)An mRNA carries a whole gene’s instructions: hundreds of nucleotides or more.
- B. Ribosomal RNA (rRNA)The largest rRNAs are thousands of nucleotides long.
- C. ✓ Transfer RNA (tRNA)
Why: A tRNA is under a hundred nucleotides long.
An mRNA is hundreds of nucleotides or more, and the largest rRNAs are thousands.
So the tRNA is the shortest.
Suppose a drug blocks the enzymes that attach amino acids to tRNAs.
Which of the following does the drug stop?
- A. ✓ tRNAs bringing amino acids to the ribosome
- B. The cell copying genes into mRNACopying a gene into mRNA does not use tRNAs.
- C. rRNA folding into its shapeAn rRNA folds by pairing within its own strand, without tRNAs.
Why: A tRNA brings the amino acid attached to its 3′ end to the ribosome.
With the enzymes blocked, no tRNA is loaded.
So no tRNA brings an amino acid to the ribosome.
Suppose a cell copies one of its genes into an RNA that is never translated. The RNA folds on itself, an enzyme attaches one amino acid to it, and at a ribosome it pairs three of its bases with an mRNA.
(a) Identify which kind of RNA this is. (1 pt)
- Award 1 point for: transfer RNA (tRNA).
(b) Explain what would happen to this RNA’s job if its strand stayed open instead of folding. (2 pt)
Frame If the strand stayed open, …
The fold is what holds three bases unpaired at one tip, as the anticodon, and the 3′ end with its amino acid at the other tip.
With no fold, there would be no anticodon to pair with a codon.
So the RNA could not bring an amino acid to its codon, and it would lose its job.
- Award 1 point for: without the fold there is no tip holding three bases unpaired as an anticodon (and no tip carrying the amino acid) — the two working parts depend on the fold.
- Award 1 point for: so the RNA could not carry an amino acid to its codon; the job is lost with the fold.
Glossary
- codon
- Three bases on an mRNA read together. Each codon means one amino acid.
- anticodon
- The three bases on a transfer RNA (tRNA) that pair with a codon on the mRNA.
- transfer RNA (tRNA)
- A small RNA folded into an L shape that pairs its anticodon with a codon on the mRNA and carries the one amino acid that codon means, attached at its 3′ end.
- ribosomal RNA (rRNA)
- The RNA a ribosome is built from: folded up with proteins bound to it, it makes the ribosome’s two subunits, binds the mRNA and the tRNAs, and joins the amino acids.
APBIO-U06-L12 One strand is read
Here is a gene, drawn as a stretch of the double helix. RNA polymerase has landed just before it. The gene has two strands, and the RNA copy will match only one of them.
Which strand does it match, and which way does the polymerase move?
Unit 6 · Gene Expression and Regulation
1RNA polymerase starts at the promoter
A cell expresses a gene in two steps.
Which step copies the gene into RNA?
- A. ✓ Transcription
- B. TranslationTranslation is a ribosome building the protein from the RNA copy.
Copying the gene into RNA is transcription.
Why: Transcription is copying a gene into RNA, inside the nucleus.
Video: Watch: RNA polymerase starts at the promoter
The gene drawn as a line with a filled box; the short open box just before it; RNA polymerase landing on the open box; the bent arrow appearing at the first copied base; the two strands parting there.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L12a.mp4
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How does RNA polymerase know where to start, which strand to read, and which way to go?
RNA polymerase binds a short stretch of DNA just before the gene. That stretch marks the start, and it picks out one of the two strands.
RNA polymerase reads that one strand from its 3′ end toward its 5′ end. It needs no primer and no helicase.
RNA polymerase builds the RNA from the 5′ end toward the 3′ end, adding at the 3′ end. DNA polymerase follows the same rule.
So the RNA carries the letters of the other strand, with U in place of T.
Here is the stretch of DNA drawn as a single line. The gene is the long filled box on the line.
Just before the gene sits a short DNA sequence, drawn as the small open box.
An enzyme copies the gene into RNA. The enzyme that joins RNA nucleotides into a strand is called , because it builds a polymer of RNA.
RNA polymerase cannot start copying just anywhere along the DNA. It must first bind that short sequence just before the gene.
The short sequence is a start sign. It marks where copying begins.
The start sign also faces one way along the DNA. So it picks out which of the two strands RNA polymerase will read.
A short DNA sequence just before a gene, where RNA polymerase binds to begin copying, is called the , because it promotes the copying of that gene.
The first base of the gene that RNA polymerase copies is drawn as a bent arrow on the line. The point where copying begins is called the .
RNA polymerase binds the promoter. Then it separates the two DNA strands at the transcription start site.
With the strands apart, RNA polymerase begins building the RNA.
What you are expected to know Describe the start of transcription: RNA polymerase binds the promoter, a short DNA sequence just before the gene, then separates the two strands at the transcription start site.
RNA polymerase is about to copy a gene.
Where does it bind first?
- A. The far end of the geneCopying begins at the start of the gene, not at its far end.
RNA polymerase binds the promoter, just before the gene. - B. ✓ The promoter, just before the gene
- C. A ribosomeA ribosome reads an RNA copy after it is made.
RNA polymerase binds DNA, at the promoter.
Why: RNA polymerase must bind the promoter before it can copy a gene.
The promoter sits just before the gene.
A gene sits on a stretch of DNA.
Where is the gene’s promoter?
- A. ✓ Just before the gene
- B. Inside the geneThe promoter is not part of the gene’s copied sequence.
It sits just before the gene. - C. On another chromosomeThe promoter is on the same DNA as its gene, just before it.
Why: A promoter is a short DNA sequence just before its gene.
RNA polymerase binds a gene’s promoter.
What does the promoter tell RNA polymerase?
- A. How long the gene isThe promoter marks the start of copying, not the end.
- B. ✓ Where copying begins
- C. Which amino acids to joinRNA polymerase joins RNA nucleotides, not amino acids.
A ribosome joins amino acids, later.
Why: The promoter is the start sign.
It marks where copying begins, and which strand is read.
RNA polymerase has bound the promoter.
What does it do next?
- A. Leaves the DNA and travels to a ribosomeRNA polymerase copies the gene where the gene is, on the DNA.
- B. Joins the two DNA strands more tightlyThe two strands must come apart for one of them to be read.
RNA polymerase separates them. - C. ✓ Separates the two DNA strands at the start site
Why: RNA polymerase separates the two DNA strands at the transcription start site.
Then it begins building RNA.
RNA polymerase has separated the two strands at the start site.
What does it build?
- A. A proteinA ribosome builds the protein, later, from the RNA copy.
- B. A DNA strandDNA polymerase builds DNA strands, during replication.
RNA polymerase builds RNA. - C. ✓ An RNA strand
Why: RNA polymerase joins RNA nucleotides into a strand.
So it builds an RNA strand.
Suppose the promoter of a gene is removed from the DNA.
Can RNA polymerase still copy that gene?
- A. YesRNA polymerase must bind the promoter to begin copying.
With no promoter, it never binds before the gene. - B. ✓ No
Why: RNA polymerase begins copying only after it binds the promoter.
The promoter is gone.
So RNA polymerase cannot copy that gene.
26Quick quiz: RNA polymerase, promoter, transcription start site mixed practice
A cell copies a gene into RNA.
What is RNA polymerase?
- A. ✓ The enzyme that joins RNA nucleotides into a strand
- B. The DNA sequence where copying beginsThe DNA sequence where copying begins is the transcription start site.
RNA polymerase is an enzyme. - C. The ribosome that reads the RNAA ribosome reads the RNA after it is made.
RNA polymerase is the enzyme that makes the RNA.
Why: RNA polymerase is the enzyme that joins RNA nucleotides into a strand, copying a gene into RNA.
A gene sits on a chromosome.
What is the gene’s promoter?
- A. The first base of the gene that RNA polymerase copiesThe first copied base is the transcription start site.
The promoter is the sequence just before it. - B. The enzyme that copies the gene into RNAThe enzyme is RNA polymerase.
The promoter is a DNA sequence. - C. ✓ A short DNA sequence just before the gene
Why: A promoter is a short DNA sequence just before a gene.
RNA polymerase binds it to begin copying the gene.
RNA polymerase copies a gene.
What is the transcription start site?
- A. ✓ The point on the gene where copying begins
- B. The place where the RNA copy leaves the nucleusThe RNA leaves the nucleus through a pore.
The transcription start site is on the DNA, where copying begins. - C. The enzyme that copies the gene into RNAThe enzyme that copies the gene into RNA is RNA polymerase.
The transcription start site is the point where copying begins.
Why: The transcription start site is the point on the gene where RNA polymerase begins copying.
RNA polymerase is about to copy a gene.
(a) State what the promoter is and what RNA polymerase does there. (1 pt)
RNA polymerase binds the promoter, separates the two DNA strands at the transcription start site and begins building RNA.
- Award 1 point for: the promoter is a (short) DNA sequence just before the gene; RNA polymerase binds it and begins copying (transcribing) the gene there.
In the gene drawing below, a large oval sits on the small open box just before the gene.
What is the oval?
- A. ✓ RNA polymerase
- B. The promoterThe promoter is the DNA sequence, drawn as the small open box.
The oval sitting on it is RNA polymerase.
Why: The open box is the promoter, a DNA sequence.
The oval bound to it is RNA polymerase.
In the gene drawing below, RNA polymerase is about to begin copying the gene.
Where does the copying begin?
- A. At the small open boxThe small open box is the promoter, where RNA polymerase binds before copying begins.
Copying begins at the bent arrow, the transcription start site. - B. ✓ At the bent arrow
- C. At the far end of the lineThe far end of the line is the gene’s end.
Copying begins at the bent arrow, the transcription start site.
Why: The bent arrow marks the transcription start site.
RNA polymerase binds the promoter, then begins copying at the start site, just after it.
33RNA polymerase against DNA polymerase
DNA polymerase lengthens a new DNA strand.
At which end of the new strand does it join each nucleotide?
- A. ✓ The 3′ end
- B. The 5′ endThe 5′ end never gains a nucleotide.
DNA polymerase joins each nucleotide to the free 3′ end.
Why: DNA polymerase can only join a nucleotide to the free 3′ end of a strand.
A test tube contains DNA template strands, free DNA nucleotides and DNA polymerase, and nothing else.
Does DNA polymerase build a new strand?
- A. YesDNA polymerase cannot start a strand from nothing.
It needs an RNA primer’s free 3′ end to join the first nucleotide to. - B. ✓ No
Why: DNA polymerase can only lengthen a strand that has already begun.
No primer is in the tube.
So no new strand is built.
At a replication fork the two DNA strands come apart.
Which enzyme breaks the hydrogen bonds and parts the two strands?
- A. DNA polymeraseDNA polymerase joins nucleotides to a growing strand.
Helicase parts the two strands. - B. LigaseLigase seals a gap between two pieces of a new strand.
Helicase parts the two strands. - C. ✓ Helicase
Why: Helicase breaks the hydrogen bonds between paired bases, so the two strands separate at the fork.
Video: Watch: RNA polymerase against DNA polymerase
A test tube with a gene, RNA nucleotides and RNA polymerase, and nothing else; RNA appearing; the table filling in row by row against DNA polymerase.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L12b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L12b.mp4
Now consider a test tube. In it are a gene with its promoter, free RNA nucleotides and RNA polymerase.
The tube contains no primer and no helicase.
RNA appears. RNA polymerase copied the gene with nothing else there.
So RNA polymerase needs no primer. It starts a new RNA strand from nothing, on the bare DNA.
And RNA polymerase needs no helicase. It separates the two DNA strands itself at the start site, and keeps them apart as it moves along the gene.
RNA polymerase joins RNA nucleotides. So the strand it builds is RNA.
DNA polymerase joins DNA nucleotides. So the strand it builds is DNA.
In one way the two enzymes are alike. Each enzyme joins every new nucleotide to the free 3′ end of the strand it is building.
What you are expected to know Compare RNA polymerase with DNA polymerase: both join each new nucleotide at the 3′ end of the strand they build.
What you are expected to know State how RNA polymerase differs from DNA polymerase: it needs no primer, it separates the two DNA strands itself, and it joins RNA nucleotides.
RNA polymerase begins copying a gene.
Does it need a primer to start the RNA strand?
- A. YesDNA polymerase needs a primer.
RNA polymerase starts a strand from nothing, on the bare DNA. - B. ✓ No
Why: RNA polymerase starts a new RNA strand from nothing.
So it needs no primer.
RNA polymerase lengthens the strand it is building.
Which nucleotides does it join into the strand?
- A. DNA nucleotidesDNA polymerase joins DNA nucleotides.
RNA polymerase joins RNA nucleotides. - B. ✓ RNA nucleotides
Why: RNA polymerase joins RNA nucleotides.
So the strand it builds is RNA.
RNA polymerase adds a nucleotide to the growing RNA.
At which end of the RNA does it add the nucleotide?
- A. ✓ The 3′ end
- B. The 5′ endThe 5′ end never gains a nucleotide.
RNA polymerase, like DNA polymerase, adds at the 3′ end.
Why: RNA polymerase joins each new nucleotide to the free 3′ end of the RNA, as DNA polymerase does on DNA.
Two enzymes build strands against DNA: RNA polymerase and DNA polymerase.
Which of the two separates the DNA strands itself?
- A. ✓ RNA polymerase
- B. DNA polymeraseAt a replication fork, helicase parts the strands for DNA polymerase.
Why: RNA polymerase separates the two DNA strands itself at the start site.
DNA polymerase relies on helicase.
A student says: “Like DNA polymerase, RNA polymerase needs an RNA primer and helicase before it can start.”
Is the student correct?
- A. Yes: both enzymes need a primer and helicaseRNA polymerase starts a strand from nothing and parts the two DNA strands itself.
- B. ✓ No: RNA polymerase needs neither a primer nor helicase
Why: RNA polymerase starts a new RNA strand from nothing, so it needs no primer.
RNA polymerase separates the two DNA strands itself, so it needs no helicase.
Here is the whole comparison as a table: RNA polymerase against DNA polymerase on four points.
54Only one strand is read
In DNA replication, a new strand is built against an old strand.
What is the old strand called?
- A. A primerA primer is the short RNA piece that starts the new strand.
The old strand copied against is the template strand. - B. ✓ A template strand
- C. A promoterA promoter is the DNA sequence where RNA polymerase binds.
The old strand copied against is the template strand.
Why: A strand whose sequence fixes the sequence of the new strand built against it is called a template strand.
An RNA strand is built against a DNA strand.
Which base does the RNA carry opposite an adenine (A) on the DNA?
- A. Cytosine (C)Cytosine pairs with guanine.
Opposite adenine, an RNA strand carries uracil. - B. Thymine (T)Thymine pairs with adenine on a DNA strand.
RNA has no thymine; it carries uracil in its place. - C. ✓ Uracil (U)
Why: RNA carries uracil in place of thymine.
So opposite an adenine, an RNA strand carries uracil.
Video: Watch: Only one strand is read
The gene opened at the polymerase, its bases lettered; RNA nucleotides pairing one at a time against the lower strand; the RNA’s letters appearing and matching the upper strand’s, with U for T.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L12c.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L12c.mp4
Go back to the gene. RNA polymerase has bound the promoter and parted the two strands at the transcription start site.
Here is the gene opened at the polymerase, with every base lettered. The two DNA strands are drawn dark, and the RNA is drawn light.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
RNA polymerase reads one strand only: here, the lower strand. It picks up a free RNA nucleotide and pairs its base with the next base on that strand.
Opposite a T on that strand it places an A. Opposite an A it places a U.
Opposite a G it places a C. Opposite a C it places a G.
So the strand RNA polymerase reads fixes the RNA, base by base. This is the same rule that made the copying of DNA work.
Each base on a strand pairs with only one partner base, so the sequence of one strand fixes the sequence of the strand built against it.
In replication, the old strand a new strand is built against is the template strand. Here the strand RNA polymerase reads does the same job, so it too is the template strand.
The other strand is not read. Its bases stay unpaired while RNA polymerase passes.
The strand of a gene that RNA polymerase does not read is called the .
The non-template strand is also called the coding strand, because its letters match the RNA’s.
Look at the letters again. The RNA carries the letters of the non-template strand, with U in place of every T.
Why does the RNA match the non-template strand? The RNA pairs with the template strand.
The non-template strand also pairs with the template strand. Two strands that pair with the same strand carry the same letters.
The one difference is the letter opposite A. DNA carries T there, and RNA carries U.
Each gene has its own template strand. The whole DNA molecule does not have one.
Each gene’s promoter picks out the strand RNA polymerase reads.
So for another gene further along the same chromosome, the other strand can be the template.
What you are expected to know Explain why the RNA copy of a gene has the letters of the non-template strand: RNA polymerase pairs each RNA nucleotide against the template strand, so the RNA matches the other strand, with U in place of T.
RNA polymerase copies a gene into RNA.
Which strand’s letters does the RNA carry?
- A. The template strand’sThe RNA pairs against the template strand, so its letters are the template’s partners.
Those are the non-template strand’s letters. - B. ✓ The non-template strand’s
Why: RNA polymerase pairs each RNA base against the template strand.
The non-template strand also pairs with the template strand.
So the RNA carries the non-template strand’s letters, with U for T.
Suppose RNA polymerase is copying a gene in a liver cell. At one position, the template strand carries a G. The RNA carries a C there, and the non-template strand also carries a C there.
(a) Explain why the RNA and the non-template strand carry the same base at that position. (1 pt)
Frame They carry the same base because …
Opposite the template’s G, the RNA gets a C.
The non-template strand also pairs with the template strand.
Opposite that same G, the non-template strand has a C.
So both carry a C at that position.
- Award 1 point for: both the RNA and the non-template strand are paired against the same template base (G), and G pairs only with C, so both carry C.
A student says: “RNA polymerase reads both strands of a gene and makes two RNAs, one from each strand.”
Is the student correct?
- A. Yes: each strand of the gene is read in turn, so RNA polymerase makes two RNAs from the geneRNA polymerase reads the template strand only.
So one gene gives one RNA, carrying the non-template strand’s letters. - B. ✓ No: RNA polymerase reads one strand of the gene only, the template strand, and makes one RNA
Why: The promoter picks out one strand of the gene, the template strand.
RNA polymerase pairs RNA nucleotides against the template strand only.
So RNA polymerase makes one RNA from the gene.
Suppose two genes sit on one chromosome. For gene 1, RNA polymerase reads one of the two DNA strands.
For gene 2, can RNA polymerase read the other strand?
- A. ✓ Yes
- B. NoWhich strand is read belongs to each gene, not to the whole molecule.
Gene 2’s promoter can point RNA polymerase along the other strand.
Why: Each gene’s promoter picks out the strand RNA polymerase reads.
Gene 2 has its own promoter.
So for gene 2 the other strand can be the template.
82Quick quiz: template strand, non-template (coding) strand mixed practice
RNA polymerase copies a gene.
Which strand of the gene is its template strand?
- A. ✓ The DNA strand RNA polymerase reads
- B. The DNA strand whose letters the RNA carriesThe strand whose letters the RNA carries is the non-template strand.
The template is the strand read. - C. The RNA strand that RNA polymerase buildsThe RNA is the copy.
The template strand is the DNA strand the copy is built against.
Why: The template strand of a gene is the DNA strand RNA polymerase reads, pairing each RNA nucleotide against its bases.
RNA polymerase copies a gene.
Which strand of the gene is its non-template strand?
- A. The strand RNA polymerase readsThe strand RNA polymerase reads is the template strand.
- B. ✓ The DNA strand RNA polymerase does not read
- C. The RNA strand that leaves the nucleusThe RNA that leaves the nucleus is the copy.
The non-template strand is a DNA strand of the gene.
Why: The non-template strand is the DNA strand RNA polymerase does not read.
Its letters match the RNA’s, with T for U.
A gene has two DNA strands.
(a) State what the template strand and the non-template strand of a gene are. (1 pt)
The non-template strand is the other DNA strand, which is not read and whose letters the RNA matches, with U for T.
- Award 1 point for: the template strand is the strand RNA polymerase reads (pairs against); the non-template strand is the strand not read (whose letters the RNA matches).
A gene has two DNA strands.
Which strand is also called the coding strand?
- A. The template strandThe template strand is the one read.
The coding strand is the second name of the non-template strand. - B. ✓ The non-template strand
Why: The non-template strand is also called the coding strand, because its letters match the RNA’s.
RNA polymerase pairs an RNA nucleotide against a base of the gene.
On which strand is that base?
- A. ✓ The template strand
- B. The non-template strandThe non-template strand is not read.
RNA polymerase pairs RNA nucleotides against the template strand.
Why: RNA polymerase reads the template strand.
So the base it pairs against is on the template strand.
88Read 3′ to 5′, build 5′ to 3′
DNA polymerase builds a new DNA strand.
In which direction does the new strand grow?
- A. From its 3′ end toward its 5′ endDNA polymerase joins each nucleotide at the 3′ end, so the strand lengthens toward its 3′ end.
- B. ✓ From its 5′ end toward its 3′ end
Why: DNA polymerase joins each nucleotide to the free 3′ end.
So every new strand grows from its 5′ end toward its 3′ end.
Video: Watch: Read 3′ to 5′, build 5′ to 3′
The opened gene with its ends written; RNA polymerase moving along the template toward its 5′ end; each nucleotide joining the RNA’s 3′ end; the same gene turned over, the template now on top, the polymerase moving the other way.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L12d.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L12d.mp4
Go back to the opened gene. Here it is again with the ends of every strand written, and with arrows for the directions.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
RNA polymerase joins each new RNA nucleotide to the free 3′ end of the RNA it is building. So the RNA grows from its 5′ end toward its 3′ end.
DNA polymerase follows the same rule. A polymerase can only join a new nucleotide to the free 3′ end of a strand, so every new strand grows in the 5′ to 3′ direction.
The RNA lies antiparallel to the template strand it pairs with. So as the RNA grows toward its own 3′ end, RNA polymerase moves toward the template’s 5′ end.
RNA polymerase reads the template strand from its 3′ end toward its 5′ end. It builds the RNA from its 5′ end toward its 3′ end.
In the drawing the template is the lower strand, written 3′ to 5′ from left to right. So RNA polymerase moves to the right.
The RNA’s 5′ end was built first, so it is the older end. The RNA’s 3′ end is the newest end, and it sits at RNA polymerase.
Now suppose a gene whose template is the upper strand, written 5′ to 3′ from left to right. RNA polymerase still reads the template from its 3′ end, at the right, toward its 5′ end, at the left.
So on this gene RNA polymerase moves to the left. The RNA’s 3′ end faces left, and the RNA grows leftward.
What you are expected to know State the directions of transcription: RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end, adding at the RNA’s 3′ end.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. A gene is drawn below, opened at RNA polymerase.
Which way does RNA polymerase move along the template strand?
- A. To the leftThe template strand’s 5′ end is at the right.
RNA polymerase moves toward the template’s 5′ end. - B. ✓ To the right
Why: RNA polymerase reads the template strand from its 3′ end toward its 5′ end.
Here the template’s 3′ end is at the left and its 5′ end at the right.
So RNA polymerase moves to the right.
RNA polymerase is part-way through building an RNA.
Which end of the RNA is the older end, built first?
- A. ✓ The 5′ end
- B. The 3′ endNew nucleotides join at the 3′ end, so the 3′ end is the newest end.
The 5′ end was built first.
Why: RNA polymerase adds each nucleotide at the RNA’s 3′ end.
So the 3′ end is always the newest.
The 5′ end was built first.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. A gene is drawn below, opened at RNA polymerase.
Which way does RNA polymerase move along the template strand?
- A. ✓ To the left
- B. To the rightThe template strand’s 5′ end is at the left.
RNA polymerase moves toward the template’s 5′ end.
Why: RNA polymerase reads the template strand from its 3′ end toward its 5′ end.
Here the template’s 3′ end is at the right and its 5′ end at the left.
So RNA polymerase moves to the left.
RNA polymerase begins reading a gene’s template strand.
Which end of the template strand does it read first?
- A. The 5′ endRNA polymerase moves toward the template’s 5′ end, so it reaches that end last.
- B. ✓ The 3′ end
Why: RNA polymerase reads the template strand from its 3′ end toward its 5′ end.
So it reads the 3′ end of the template first.
Go back to the gene with RNA polymerase landed just before it. Here it is again, opened at the polymerase.
RNA polymerase sits on the promoter, opens the two strands, and moves along the template strand from its 3′ end toward its 5′ end.
Behind it the RNA grows from its 5′ end toward its 3′ end. It carries the letters of the non-template strand, with U in place of T.
109Mixed practice mixed practice
Suppose the free 3′ end of a growing RNA is blocked, so nothing can join to it.
Can RNA polymerase add the next nucleotide?
- A. YesThe 5′ end never gains a nucleotide.
With the 3′ end blocked, there is no end for the next nucleotide to join. - B. ✓ No
Why: A polymerase joins a new nucleotide only to the free 3′ end of a strand.
The 3′ end is blocked, and the 5′ end never gains a nucleotide.
So RNA polymerase cannot add the next nucleotide.
Suppose a cell’s helicase is removed.
Can RNA polymerase still copy a gene into RNA?
- A. ✓ Yes
- B. NoHelicase parts the strands at a replication fork.
RNA polymerase parts the two strands itself and needs no helicase.
Why: RNA polymerase separates the two DNA strands itself at the transcription start site.
It never needed helicase.
So it can still copy the gene.
Suppose the template strand of a gene carries an A at one position.
Which base does the RNA carry at that position?
- A. Adenine (A)The RNA carries the partner of the template base, not the same letter.
Adenine’s partner on RNA is uracil. - B. Guanine (G)Guanine pairs with cytosine.
Opposite adenine, RNA carries uracil. - C. ✓ Uracil (U)
Why: RNA polymerase pairs each RNA base against the template base.
Opposite an A, an RNA strand carries U.
A student says: “RNA polymerase binds anywhere along the DNA and starts copying wherever it lands.”
Is the student correct?
- A. ✓ No: RNA polymerase binds a promoter and starts at the start site
- B. Yes: RNA polymerase starts copying wherever it happens to landRNA polymerase begins copying only after it binds a promoter, just before a gene.
Copying starts at the transcription start site.
Why: RNA polymerase binds a promoter, a short DNA sequence just before a gene.
Then it separates the strands at the transcription start site and begins copying there.
So it does not start wherever it lands.
Suppose the non-template strand of a gene carries a G at one position.
Which base does the RNA carry at that position?
- A. Cytosine (C)Cytosine is the partner of G.
The RNA matches the non-template strand’s letters, so it carries G. - B. ✓ Guanine (G)
- C. Uracil (U)Uracil sits opposite an A on the template, in place of T.
Here the non-template base is G, so the RNA carries G.
Why: The RNA carries the letters of the non-template strand, with U in place of T.
The non-template base is G.
So the RNA carries G there.
RNA polymerase copies a gene from its transcription start site to its far end.
Which end of the template strand does it reach last?
- A. The 3′ endRNA polymerase begins at the template’s 3′ end and moves away from it.
- B. ✓ The 5′ end
Why: RNA polymerase reads the template strand from its 3′ end toward its 5′ end.
So it reaches the template’s 5′ end last.
Two enzymes build new strands against DNA in a cell.
Which of them starts a strand with no primer?
- A. ✓ RNA polymerase
- B. DNA polymeraseDNA polymerase can only lengthen a strand that has already begun.
It needs an RNA primer. - C. BothDNA polymerase needs an RNA primer.
Only RNA polymerase starts a strand from nothing.
Why: RNA polymerase starts a new RNA strand from nothing, on the bare DNA.
DNA polymerase needs an RNA primer’s free 3′ end first.
RNA polymerase has just bound a gene’s DNA. Copying is about to begin.
Which sequence did it recognize and bind?
- A. The template strand’s 5′ endRNA polymerase reads the template from its 3′ end, and it binds the promoter before it reads anything.
- B. ✓ The promoter
- C. The transcription start siteThe transcription start site is where copying begins.
RNA polymerase binds the promoter, just before it, first.
Why: RNA polymerase binds the promoter before copying begins.
Copying then starts at the transcription start site, just after the promoter.
Suppose RNA polymerase is copying a gene in a skin cell. The RNA it has built so far is 40 nucleotides long. At the next position, the template strand carries a T.
(a) Explain why the 41st nucleotide of the RNA carries an A. (1 pt)
Frame The 41st nucleotide carries an A because …
The next template base is a T.
T pairs only with A.
So the nucleotide RNA polymerase adds carries an A.
- Award 1 point for: RNA polymerase pairs the new RNA base against the template base, and T pairs (only) with A.
(b) Explain why the 41st nucleotide joins at the RNA’s 3′ end. (1 pt)
Frame It joins at the 3′ end because …
The RNA’s 5′ end was built first and cannot take a nucleotide.
So the 41st nucleotide joins at the 3′ end, and the RNA grows 5′ to 3′.
- Award 1 point for: a polymerase joins nucleotides only at the free 3′ end (the 5′ end never gains one), so the strand grows 5′ to 3′.
Glossary
- RNA polymerase
- The enzyme that joins RNA nucleotides into a strand, copying a gene into RNA. It binds the promoter, separates the two DNA strands and builds the RNA against the template strand, adding at the RNA’s 3′ end. It needs no primer and no helicase.
- promoter
- A short DNA sequence just before a gene, where RNA polymerase binds to begin copying the gene. It marks where copying begins and which strand is read.
- transcription start site
- The point on a gene where RNA polymerase begins copying: the first base copied, drawn as a bent arrow just after the promoter.
- non-template strand
- The DNA strand of a gene that RNA polymerase does not read. Its letters match the RNA’s, with T where the RNA has U. It is also called the coding strand.
APBIO-U06-L13 Write the transcript
Here is a template strand: 3′-TAC GGA TTT-5′. RNA polymerase is about to copy it.
What RNA will it make, with its ends marked? Then: what comes off when the polymerase reaches the end of the gene, and can the gene be read again?
Unit 6 · Gene Expression and Regulation
1The RNA from a template strand
RNA polymerase is copying a gene into RNA.
Against which strand does RNA polymerase pair each RNA nucleotide?
- A. ✓ The template strand
- B. The non-template strandThe RNA ends up with the non-template strand’s letters, but it is not built against that strand.
RNA polymerase pairs each RNA nucleotide against the template strand. - C. Both strandsOnly one of the gene’s two strands is read.
RNA polymerase pairs each RNA nucleotide against the template strand.
Why: Only one of the gene’s two DNA strands is the template.
RNA polymerase pairs each RNA nucleotide against the template strand’s bases.
So the RNA has the same sequence as the other, non-template strand, with U in place of T.
RNA polymerase is building an RNA strand.
Which way does RNA polymerase build the RNA?
- A. ✓ From the RNA’s 5′ end toward its 3′ end
- B. From the RNA’s 3′ end toward its 5′ endA polymerase can only join a new nucleotide to the free 3′ end of a strand.
So RNA polymerase builds the RNA from its 5′ end toward its 3′ end.
Why: RNA polymerase adds each nucleotide at the RNA’s 3′ end, the same growth rule DNA polymerase follows.
So it builds the RNA from its 5′ end toward its 3′ end, and it moves along the template from the template’s 3′ end toward its 5′ end.
A template strand carries an A. An RNA strand is being built against it.
Which base does the RNA carry opposite the A?
- A. GuanineGuanine pairs with cytosine.
Opposite an A, an RNA strand carries uracil. - B. ✓ Uracil
Why: Adenine pairs with thymine in DNA and with uracil where the partner strand is RNA.
The strand being built is RNA.
So opposite the template’s A the RNA carries uracil.
Video: Watch: The RNA from a template strand
The template 3′-TAC GGA TTT-5′ written out; RNA polymerase moving along it from left to right; A, U, G, C, C, U, A, A, A appearing beneath one at a time; the finished RNA 5′-AUG CCU AAA-3′ with its ends marked.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L13a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L13a.mp4
How do you write the RNA a gene gives? Against a template strand, pair each base with its RNA partner, with U opposite A.
Write the RNA from its 5′ end to its 3′ end, beneath the template written from its 3′ end to its 5′ end.
For the template 3′-TAC GGA TTT-5′, the RNA is 5′-AUG CCU AAA-3′.
Given the non-template strand instead, copy its letters in the same direction and swap each T for U.
When RNA polymerase passes a stop sequence, it lets go of the RNA. The two DNA strands pair again, and another polymerase can start at the promoter.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
A template strand is often written the other way round, with its 3′ end at the left. Then the RNA written beneath it starts from its 5′ end at the left.
RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end.
Each base on a strand pairs with only one partner base, so the sequence of one strand fixes the sequence of the strand built against it. This is the same rule that made DNA copying work.
Here is the template strand from the opening, 3′-TAC GGA TTT-5′. Its 3′ end is at the left, so RNA polymerase starts at the left.
The template’s first base is T. Thymine pairs with adenine, so the first RNA nucleotide carries A.
RNA polymerase then moves to the right along the template, one base at a time.
Here is the order it adds the RNA nucleotides:
- against T it adds A
- against A it adds U
- against C it adds G
- against G it adds C
- against G it adds C
- against A it adds U
- against T it adds A
- against T it adds A
- against T it adds A
The RNA reads 5′-AUG CCU AAA-3′. Its 3′ end is at the right, so RNA polymerase built it toward the right.
Here are the steps for writing the RNA against a template written 3′ to 5′:
- Under each template base, write its RNA partner: U under A, A under T, C under G, G under C.
- Mark the RNA 5′ at the left and 3′ at the right.
- Check that no T appears in the RNA: RNA carries U, never T.
What you are expected to know Write the RNA that RNA polymerase makes from a given template strand, with the RNA’s 5′ and 3′ ends marked.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. A template strand reads 3′-AGC TTA CGG-5′, as drawn below. The RNA is written beneath it as 5′-___ AAU GCC-3′.
Which three letters fill the blank?
- A. ✓ UCG
- B. AGCThese are the template’s own first three letters, not their partners.
Under A sits U, under G sits C, under C sits G. - C. GCUThese are the right partners in the wrong order.
The RNA’s 5′ end is at the left, so the partner of the template’s first base comes first.
Why: The template’s first three bases are AGC.
Under A write U, under G write C, under C write G.
So the blank is UCG.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. A template strand reads 3′-AGC TTA CGG-5′, as drawn below. The RNA’s 5′ end is at the left.
What does the whole RNA read?
- A. 5′-AGC UUA CGG-3′Copying the template’s own letters gives the template again, not each base’s partner.
Under A sits U, under G sits C. - B. ✓ 5′-UCG AAU GCC-3′
- C. 5′-CCG UAA GCU-3′These are the partners written from the right end of the template.
The RNA’s 5′ end is at the left, so the partner of the template’s leftmost base comes first.
Why: Under each template base write its RNA partner: A gives U, G gives C, C gives G, T gives A, T gives A, A gives U, C gives G, G gives C, G gives C.
The RNA’s 5′ end is at the left.
So the RNA reads 5′-UCG AAU GCC-3′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. A template strand reads 3′-GTA CCC TAG-5′, as drawn below.
Written with both ends marked, what does the RNA read?
- A. 3′-CAU GGG AUC-5′RNA polymerase starts at the template’s 3′ end, at the left, and builds the RNA from its 5′ end.
So the RNA’s 5′ end is at the left. - B. 5′-CUA GGG UAC-3′These are the partners written from the right end of the template.
RNA polymerase starts at the template’s 3′ end, at the left. - C. ✓ 5′-CAU GGG AUC-3′
- D. 5′-GUA CCC UAG-3′Copying the template’s own letters gives the template again, not each base’s partner.
Under G sits C, under T sits A.
Why: Under each template base write its RNA partner: U under A, A under T, C under G, G under C.
RNA polymerase starts at the template’s 3′ end, at the left, so the RNA’s 5′ end is at the left.
The RNA reads 5′-CAU GGG AUC-3′.
25The RNA from the non-template strand
RNA polymerase copies a gene into RNA.
Which strand’s letters does the RNA match?
- A. The template strand’s lettersThe RNA is built against the template strand, so it carries the template’s partners, not its letters.
Those partners are the non-template strand’s letters. - B. ✓ The non-template strand’s letters
Why: RNA polymerase pairs each RNA nucleotide against the template strand’s bases.
The non-template strand is also paired against the template strand.
So the RNA has the same sequence as the non-template strand, with U in place of T.
Video: Watch: The RNA from the non-template strand
A gene’s two strands written out, one above the other; the RNA growing against the template and matching the non-template strand letter for letter; each T of the non-template strand shown beside the U the RNA carries in its place.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L13b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L13b.mp4
Now consider a gene whose non-template strand reads 5′-ATGCCT-3′. Its template strand, written beneath it, reads 3′-TACGGA-5′.
RNA polymerase pairs each RNA nucleotide against the template strand. So the RNA reads 5′-AUGCCU-3′.
Now compare the RNA with the non-template strand. The RNA has the non-template strand’s letters, in the same order, with U where the non-template strand has T.
So a non-template strand gives you a shortcut. You do not pair its bases; you copy its letters.
Here are the steps for writing the RNA from a non-template strand written 5′ to 3′:
- Copy its letters in the same order.
- Change every T to U.
- Keep the ends the same way round: 5′ at the left, 3′ at the right.
For 5′-ATGCCT-3′ the letters copied are ATGCCT, the T becomes U, and the RNA reads 5′-AUGCCU-3′.
What you are expected to know Write the RNA that RNA polymerase makes from a gene, starting from its non-template strand: the same letters, the same direction, with U in place of T.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. A gene’s non-template strand reads 5′-CCAGTT-3′, as drawn below. The RNA is written as 5′-CCAGU_-3′.
Which letter fills the blank?
- A. AA is the partner of the non-template strand’s last base, T.
The RNA copies the non-template strand’s letters; it does not pair with them, so the T becomes U. - B. ✓ U
Why: The RNA has the non-template strand’s letters in the same order, with U in place of T.
The non-template strand’s last letter is T.
So the RNA’s last letter is U.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. A gene’s non-template strand reads 5′-TCAGGT-3′, as drawn below.
Written with both ends marked, what does the RNA read?
- A. ✓ 5′-UCAGGU-3′
- B. 5′-AGUCCA-3′These are the partners of the non-template strand’s bases.
The RNA is built against the template strand, so it carries the non-template strand’s own letters, with U for T. - C. 3′-UCAGGU-5′The non-template strand and the RNA lie the same way round.
So the RNA’s 5′ end is at the left, where the non-template strand’s 5′ end is. - D. 5′-UGGACU-3′These are the right letters in the wrong order.
The RNA copies the non-template strand’s letters in the same order, from its 5′ end.
Why: The RNA has the non-template strand’s letters in the same order, with U in place of T.
TCAGGT with each T changed to U is UCAGGU.
The ends stay the same way round, so the RNA reads 5′-UCAGGU-3′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. A gene’s two strands are drawn below: the top strand reads 5′-GACTTC-3′ and the bottom strand reads 3′-CTGAAG-5′. RNA polymerase reads the bottom strand.
Which strand’s letters does the RNA match?
- A. ✓ The top strand’s letters
- B. The bottom strand’s lettersThe bottom strand is the template: RNA polymerase pairs each RNA nucleotide against it.
So the RNA carries the bottom strand’s partners, which are the top strand’s letters.
Why: RNA polymerase reads the bottom strand, so the bottom strand is the template strand.
The RNA is built against the template, so it has the other strand’s letters, with U for T.
The RNA matches the top strand’s letters.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. A gene’s two strands are drawn below: the top strand reads 5′-GATACC-3′ and the bottom strand reads 3′-CTATGG-5′. The bottom strand is the template strand.
Written with both ends marked, what does the RNA read?
- A. 5′-CUAUGG-3′These are the template strand’s own letters with U for T.
The RNA carries the template’s partners: the top strand’s letters, with U for T. - B. 3′-GAUACC-5′The RNA lies the same way round as the top strand, whose 5′ end is at the left.
So the RNA’s 5′ end is at the left. - C. ✓ 5′-GAUACC-3′
Why: The bottom strand is the template, so the RNA has the top strand’s letters with U in place of T.
GATACC with each T changed to U is GAUACC.
The RNA lies the same way round as the top strand, so it reads 5′-GAUACC-3′.
39The transcript comes off, and the gene is read again
RNA polymerase is about to start copying a gene.
What is the promoter?
- A. ✓ The DNA sequence just before the gene where RNA polymerase binds
- B. The first RNA nucleotide that RNA polymerase adds to the new RNAThe first RNA nucleotide is part of the RNA, not of the DNA.
The promoter is a DNA sequence just before the gene, where RNA polymerase binds. - C. The enzyme that separates the gene’s two strands before copyingRNA polymerase itself separates the two strands.
The promoter is a DNA sequence just before the gene, where RNA polymerase binds.
Why: The promoter is a DNA sequence just before the gene.
RNA polymerase binds the promoter, and the promoter marks where transcription begins and which strand is read.
Video: Watch: The transcript comes off, and the gene is read again
RNA polymerase reaching the stop sequence; the RNA coming off and the polymerase leaving the DNA; the two DNA strands pairing again; a second and a third polymerase starting at the promoter and moving along the gene, each trailing an RNA.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L13c.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L13c.mp4
Go back to RNA polymerase copying the gene. It moves along the template strand and adds each RNA nucleotide at the RNA’s 3′ end.
Near the end of the gene lies a short stretch of DNA that signals RNA polymerase to stop. We call it a stop sequence.
RNA polymerase passes the stop sequence. Then it lets go of the DNA, and the RNA comes off.
An RNA that RNA polymerase has made from a gene and released is called a , because it is a written-out copy of the gene.
Behind the polymerase, the two DNA strands pair again. The gene is exactly as it was before.
In a eukaryotic cell, enzymes in the nucleus change the transcript before it leaves the nucleus.
A transcript as it comes off the gene, before enzymes change it, is called , the primary transcript, because it comes before the mRNA.
The promoter is free again. Another RNA polymerase binds it and starts a new RNA.
Suppose one polymerase after another starts at the promoter. Then several polymerases sit along the gene at once, each trailing an RNA.
Each polymerase adds nucleotides as it moves. So the polymerase that has moved furthest along the gene trails the longest RNA.
Here is a real gene with many polymerases along it at once.

One gene photographed through an electron microscope, from an insect cell. The dark line is the DNA. Each thread sticking out from it is one RNA still being made, so many polymerases are copying this gene at once. The threads are short near the start and long near the end. Image: Hans-Heinrich Trepte, Wikimedia Commons, CC BY-SA 3.0 (cropped, rotated and resized).
So RNA polymerases make many transcripts from one gene. The gene is copied, and copied again, and it is never used up.
What you are expected to know Describe what happens when RNA polymerase reaches the end of a gene: it passes the stop sequence, releases the transcript, the two DNA strands pair again, and another polymerase can start at the promoter.
RNA polymerase passes the stop sequence at the end of a gene.
What comes off the DNA?
- A. The template strandThe template strand stays in the DNA and pairs again with the non-template strand.
Only the RNA comes off. - B. ✓ The RNA
- C. Both DNA strandsBoth DNA strands stay and pair again behind the polymerase.
Only the RNA comes off.
Why: RNA polymerase passes the stop sequence and lets go of the DNA.
The RNA comes off.
The two DNA strands pair again, so the DNA is unchanged.
Suppose a muscle cell needs a large amount of one protein. Within a day, RNA polymerases have made thousands of transcripts of that protein’s gene. The gene is still on its chromosome, unchanged.
(a) Explain why one gene can give thousands of transcripts. (1 pt)
Frame One gene can give thousands of transcripts because …
Behind it, the two DNA strands pair again, so the gene is unchanged.
The promoter is free again, so another RNA polymerase binds it and starts a new transcript.
So the same gene is copied again and again.
- Award 1 point for: the transcript is released and the DNA re-pairs unchanged, so another RNA polymerase can bind the promoter and copy the gene again.
A student says: “Each time a gene is transcribed, part of the gene’s DNA leaves with the RNA, so a gene is used up after a few transcripts.”
Is the student correct?
- A. ✓ No: the DNA stays and pairs again, and only the RNA leaves
- B. Yes: each transcript carries away part of the gene’s DNAThe RNA is built from RNA nucleotides paired against the template; no DNA nucleotide joins it.
The two DNA strands pair again behind the polymerase.
Why: RNA polymerase builds the transcript from RNA nucleotides, paired against the template strand.
At the stop sequence it releases the transcript and lets go of the DNA.
The two DNA strands pair again, so the gene is unchanged and can be read again.
Four RNA polymerases are copying one gene at once. They started one after another at the promoter, and the drawing below shows where each one is now.
Which polymerase has made the longest RNA so far?
- A. The polymerase nearest the promoterThe polymerase nearest the promoter started last and has moved least.
Each polymerase adds nucleotides as it moves, so this one has made the shortest RNA. - B. ✓ The polymerase furthest from the promoter
Why: Every polymerase started at the promoter and adds RNA nucleotides as it moves along the gene.
The polymerase furthest from the promoter has moved furthest.
So it has added most nucleotides and has made the longest RNA.
Go back to the template strand, 3′-TAC GGA TTT-5′. RNA polymerase pairs an RNA nucleotide against each base, U opposite A.
The RNA reads 5′-AUG CCU AAA-3′.
At the stop sequence the polymerase lets go of the transcript. The DNA pairs again, and the next polymerase starts at the promoter.
62Quick quiz: transcript, pre-mRNA (primary transcript) mixed practice
RNA polymerase has passed the stop sequence of a gene.
What is a transcript?
- A. The stretch of DNA between the promoter and the stop sequenceThat stretch of DNA is the gene itself.
The transcript is the RNA made from it. - B. ✓ The RNA that RNA polymerase has made from the gene and released
- C. The template strand after RNA polymerase has read itThe template strand stays in the DNA.
The transcript is the RNA built against it and released.
Why: A transcript is an RNA that RNA polymerase has made from a gene and released: a written-out copy of the gene.
In a eukaryotic cell, a transcript has just come off its gene.
What is this transcript called at this moment?
- A. ✓ Pre-mRNA, the primary transcript
- B. The template strandThe template strand is the DNA strand RNA polymerase read; it stays in the DNA.
The RNA just released is pre-mRNA. - C. The promoterThe promoter is the DNA sequence just before the gene where RNA polymerase binds.
The RNA just released is pre-mRNA.
Why: In a eukaryotic cell, enzymes in the nucleus change the transcript before it leaves the nucleus.
The transcript as it comes off the gene, before enzymes change it, is pre-mRNA, the primary transcript.
RNA polymerase has passed the stop sequence of a gene.
(a) State what a transcript is. (1 pt)
- Award 1 point for: the RNA made from a gene by RNA polymerase and released (a copy of the gene in RNA).
A transcript has just come off its gene.
Which kind of nucleotide is the transcript made of?
- A. DNA nucleotidesThe DNA nucleotides stay in the gene’s two strands.
RNA polymerase built the transcript from RNA nucleotides. - B. ✓ RNA nucleotides
Why: RNA polymerase pairs RNA nucleotides against the template strand.
So the transcript is made of RNA nucleotides.
A transcript reads 5′-GCAUUG-3′.
Which end of this transcript did RNA polymerase make first?
- A. ✓ The 5′ end
- B. The 3′ endRNA polymerase adds each nucleotide at the RNA’s 3′ end.
So the 5′ end was made first and the 3′ end last.
Why: RNA polymerase builds the RNA from its 5′ end toward its 3′ end.
So the 5′ end of the transcript was made first.
68Mixed practice mixed practice
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. A template strand reads 3′-CGG ATT ACA-5′, as drawn below.
Written with both ends marked, what does the RNA read?
- A. ✓ 5′-GCC UAA UGU-3′
- B. 3′-GCC UAA UGU-5′RNA polymerase starts at the template’s 3′ end, at the left, and builds the RNA from its 5′ end.
So the RNA’s 5′ end is at the left. - C. 5′-UGU AAU CCG-3′These are the partners written from the right end of the template.
RNA polymerase starts at the template’s 3′ end, at the left.
Why: Under each template base write its RNA partner: C gives G, G gives C, G gives C, A gives U, T gives A, T gives A, A gives U, C gives G, A gives U.
The RNA’s 5′ end is at the left.
The RNA reads 5′-GCC UAA UGU-3′.
After RNA polymerase passes the stop sequence at the end of a gene, it lets go of the DNA.
What happens to the gene’s two DNA strands?
- A. One strand leaves with the RNANo DNA strand leaves; the RNA comes off on its own.
The two DNA strands pair again. - B. Both strands stay apartThe strands were apart only where the polymerase held them apart.
Behind it they pair again. - C. ✓ The two strands pair again
Why: RNA polymerase held the two strands apart only where it sat.
When it lets go, the two DNA strands pair again.
So the gene is unchanged.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. A gene’s non-template strand reads 5′-TGGCAT-3′, as drawn below.
Written with both ends marked, what does the RNA read?
- A. 3′-UGGCAU-5′The non-template strand and the RNA lie the same way round.
So the RNA’s 5′ end is at the left. - B. 5′-ACCGUA-3′These are the partners of the non-template strand’s bases.
The RNA carries the non-template strand’s own letters, with U for T. - C. ✓ 5′-UGGCAU-3′
Why: The RNA has the non-template strand’s letters in the same order, with U in place of T.
TGGCAT with each T changed to U is UGGCAU.
The ends stay the same way round, so the RNA reads 5′-UGGCAU-3′.
A student says: “A gene can be transcribed only once, so a cell that needs more of a protein must first copy the gene’s DNA.”
Is the student correct?
- A. ✓ No: after each transcript is released, another RNA polymerase can start at the promoter
- B. Yes: each gene gives one transcript, so more protein needs more copies of the geneThe DNA is unchanged after a transcript is released.
Another RNA polymerase binds the free promoter and copies the same gene again.
Why: RNA polymerase releases the transcript at the stop sequence, and the DNA pairs again unchanged.
The promoter is free, so another RNA polymerase binds it.
So one gene gives many transcripts without the DNA being copied.
RNA polymerase is part way along a gene, building an RNA.
At which end of the RNA does RNA polymerase add the next nucleotide?
- A. At the RNA’s 5′ endA 5′ end never gains a nucleotide.
RNA polymerase adds each nucleotide at the RNA’s free 3′ end. - B. ✓ At the RNA’s 3′ end
Why: A polymerase can only join a new nucleotide to the free 3′ end of a strand.
So RNA polymerase adds the next nucleotide at the RNA’s 3′ end.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. A gene’s two strands are drawn below: the top strand reads 3′-TTGCAC-5′ and the bottom strand reads 5′-AACGTG-3′. The top strand is the template strand.
Written with both ends marked, what does the RNA read?
- A. 5′-UUGCAC-3′These are the template strand’s own letters with U for T.
The RNA is built against the template, so it carries the bottom strand’s letters, with U for T. - B. ✓ 5′-AACGUG-3′
- C. 3′-AACGUG-5′The RNA lies the same way round as the bottom strand, whose 5′ end is at the left.
So the RNA’s 5′ end is at the left.
Why: The top strand is the template, so the RNA has the bottom strand’s letters with U in place of T.
AACGTG with the T changed to U is AACGUG.
The RNA lies the same way round as the bottom strand, so it reads 5′-AACGUG-3′.
A pancreas cell is making insulin. The drawing below shows the insulin gene with three RNA polymerases along it at once. Each polymerase trails an RNA, and the three RNAs differ in length.
(a) Explain how the drawing demonstrates that one gene can be transcribed again and again. (1 pt)
Frame The drawing demonstrates this because …
Each one started at the promoter after the one before it had moved on.
So the gene was still there, unchanged, for each new polymerase to read.
The DNA stays, and RNA polymerase makes only RNA copies, so the gene can be transcribed again and again.
- Award 1 point for: several polymerases read the same gene one after another (the DNA stays unchanged; RNA polymerase makes only RNA copies), so the same gene is transcribed repeatedly.
(b) Explain why the three RNAs differ in length. (1 pt)
Frame The three RNAs differ in length because …
Each polymerase adds one RNA nucleotide at the RNA’s 3′ end for every template base it passes.
The polymerase furthest along the gene has passed the most template bases.
So its RNA is the longest, and the polymerase nearest the promoter trails the shortest.
- Award 1 point for: each polymerase adds one nucleotide per template base it passes, and the three started at the promoter at different times, so the one furthest along has added most (longest RNA) and the one nearest the promoter least.
Glossary
- transcript
- An RNA that RNA polymerase has made from a gene and released: a written-out copy of the gene.
- pre-mRNA (primary transcript)
- In a eukaryotic cell, a transcript as it comes off the gene, before enzymes in the nucleus change it.
APBIO-U06-L14 A cap and a tail
Suppose two mRNAs copied from the same gene are put into two cells. One mRNA has a cap at its 5′ end and a tail of about 200 adenine nucleotides at its 3′ end. The other has neither.
After an hour, one cell has made plenty of the protein. The other has made almost none, and its mRNA has already gone. Which is which, and why?
Unit 6 · Gene Expression and Regulation
1A cap on the 5′ end, a tail on the 3′ end
In a eukaryote, RNA polymerase passes the stop sequence at the end of a gene for a protein and releases the RNA it has built.
What is that released RNA called at this stage?
- A. A tRNAA tRNA is the small folded RNA that carries one amino acid.
- B. ✓ A pre-mRNA
- C. An rRNAAn rRNA is an RNA that builds the ribosome.
Why: At the end of the gene, RNA polymerase releases the RNA it built: the transcript.
In a eukaryote that transcript is not yet ready to be read, so it is called the pre-mRNA.
A single RNA strand has two different ends.
Which end carries the free phosphate group?
- A. ✓ The 5′ end
- B. The 3′ endThe 3′ end carries a free hydroxyl group on the last sugar’s carbon 3.
Why: The last sugar’s carbon 5 carries a phosphate group joined to nothing further.
That free phosphate group marks the 5′ end.
Video: Watch: A cap on the 5′ end, a tail on the 3′ end
The pre-mRNA as released; an enzyme joining one modified guanine nucleotide to its left, 5′, end; another enzyme adding adenine nucleotides one after another at its right, 3′, end until a long tail of A’s hangs there.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L14a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L14a.mp4
What are the cap and the tail for?
In a eukaryote, enzymes in the nucleus add two things to the new pre-mRNA before it leaves.
The enzymes join a cap, a changed guanine nucleotide, to its 5′ end. They add a tail of about 200 adenine nucleotides to its 3′ end.
The ribosome recognizes the cap and grips it. So an mRNA with no cap is barely read.
The tail protects the mRNA from the enzymes that break RNA down. So an mRNA with a short tail is destroyed sooner and gives less protein.
Go back to the end of transcription. RNA polymerase has passed the stop sequence and released the pre-mRNA into the nucleus.
The pre-mRNA is a single strand. Its 5′ end carries the free phosphate group, and its 3′ end carries the free hydroxyl group.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
In a eukaryote, this pre-mRNA is not sent to a ribosome as it is. Enzymes in the nucleus add something to each end first.
At the 5′ end, an enzyme joins one extra nucleotide to the strand. The extra nucleotide is a guanine nucleotide with a small chemical change: a modified guanine nucleotide.
A modified guanine nucleotide joined to the 5′ end of the pre-mRNA is called the , because it covers the strand’s 5′ end. Another name for it is the GTP cap.
At the 3′ end, another enzyme adds adenine nucleotides one after another, about 200 of them.
A stretch of about 200 adenine nucleotides at the 3′ end of the pre-mRNA is called the , because poly means many and A is adenine’s letter.
The 5′ cap always sits at the 5′ end. The poly-A tail always sits at the 3′ end.
Here are two drawings: the pre-mRNA as released, then the same pre-mRNA with its 5′ cap and poly-A tail.
In every drawing of this topic, the 5′ cap is drawn as a dot at the 5′ end and the poly-A tail as a row of A’s at the 3′ end.
Both enzymes work inside the nucleus. So the pre-mRNA already carries its 5′ cap and its poly-A tail when it leaves through a pore.
What you are expected to know Describe the two additions enzymes in the nucleus make to a eukaryotic pre-mRNA: a 5′ cap, a modified guanine nucleotide, joined to the 5′ end, and a poly-A tail of about 200 adenine nucleotides at the 3′ end.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. The pre-mRNA drawn below has an arrow at one end.
Which of the following do enzymes in the nucleus add at the marked end?
- A. ✓ One modified guanine nucleotide
- B. A stretch of adenine nucleotidesThe marked end is the 5′ end.
The adenine nucleotides of the poly-A tail are added at the 3′ end.
Why: The arrow marks the 5′ end of the strand.
The 5′ cap, one modified guanine nucleotide, is joined to the 5′ end.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. The pre-mRNA drawn below has an arrow at one end.
Which of the following do enzymes in the nucleus add at the marked end?
- A. One modified guanine nucleotideThe marked end is the 3′ end.
The modified guanine nucleotide of the 5′ cap is joined to the 5′ end. - B. ✓ A stretch of adenine nucleotides
Why: The arrow marks the 3′ end of the strand.
The poly-A tail, a stretch of adenine nucleotides, is added at the 3′ end.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. The pre-mRNA drawn below is written from right to left, and it has an arrow at one end.
Which of the following do enzymes in the nucleus add at the marked end?
- A. One modified guanine nucleotideThe marked end is the 3′ end, even though it is drawn on the left.
The 5′ cap is joined to the 5′ end. - B. ✓ A stretch of adenine nucleotides
Why: The arrow marks the end labeled 3′.
The poly-A tail, a stretch of adenine nucleotides, is added at the 3′ end, whichever side it is drawn on.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. The pre-mRNA drawn below is written from right to left, and it has an arrow at one end.
Which of the following do enzymes in the nucleus add at the marked end?
- A. ✓ One modified guanine nucleotide
- B. A stretch of adenine nucleotidesThe marked end is the 5′ end, even though it is drawn on the right.
The poly-A tail is added at the 3′ end.
Why: The arrow marks the end labeled 5′.
The 5′ cap, one modified guanine nucleotide, is joined to the 5′ end, whichever side it is drawn on.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. The pre-mRNA drawn below is written from right to left with its bases drawn upward, and it has an arrow at one end.
Which of the following do enzymes in the nucleus add at the marked end?
- A. One modified guanine nucleotideThe marked end is the 3′ end.
The 5′ cap is joined to the 5′ end. - B. ✓ A stretch of adenine nucleotides
Why: The arrow marks the end labeled 3′.
The poly-A tail, a stretch of adenine nucleotides, is added at the 3′ end.
Which way the bases are drawn changes nothing.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. In every drawing of this topic, the 5′ cap is drawn as a dot at the 5′ end and the poly-A tail as a row of A’s at the 3′ end. Three finished mRNAs are drawn below, numbered 1, 2 and 3.
In which drawing are the 5′ cap and the poly-A tail at the wrong ends?
- A. Drawing 1Drawing 1 has its dot at the end labeled 5′ and its A’s at the end labeled 3′.
Those are the right ends. - B. Drawing 2Drawing 2 is written from right to left, so its right end is the 5′ end.
Its dot sits at that 5′ end, where the cap belongs. - C. ✓ Drawing 3
Why: The 5′ cap always sits at the 5′ end and the poly-A tail at the 3′ end.
In drawing 3 the dot sits at the end labeled 3′ and the A’s at the end labeled 5′.
So drawing 3 has them at the wrong ends.
An enzyme in the nucleus adds a stretch of adenine nucleotides to a pre-mRNA.
To which end does it add them?
- A. The 5′ endThe 5′ end receives the cap, one modified guanine nucleotide.
- B. ✓ The 3′ end
Why: The adenine nucleotides are the poly-A tail.
The poly-A tail is always added at the 3′ end.
30Quick quiz: 5′ cap (GTP cap), poly-A tail mixed practice
Which of the following is the 5′ cap?
- A. The DNA sequence just before a gene, where RNA polymerase bindsThe DNA sequence just before a gene where RNA polymerase binds is the promoter.
- B. ✓ A modified guanine nucleotide joined to the 5′ end
- C. A stretch of adenine nucleotides at the 3′ endThe stretch of adenine nucleotides at the 3′ end is the poly-A tail.
Why: An enzyme in the nucleus joins one modified guanine nucleotide to the 5′ end of the pre-mRNA.
That nucleotide is the 5′ cap, also called the GTP cap.
Which of the following is the poly-A tail?
- A. ✓ A stretch of adenine nucleotides at the 3′ end
- B. A modified guanine nucleotide joined to the 5′ endA modified guanine nucleotide joined to the 5′ end is the 5′ cap.
- C. The three bases on an mRNA that a tRNA’s anticodon pairs withThree bases on an mRNA read together are a codon.
Why: An enzyme in the nucleus adds adenine nucleotides one after another at the 3′ end of the pre-mRNA.
That stretch of A’s is the poly-A tail.
State what the poly-A tail is made of and which end of the pre-mRNA carries it. (1 pt)
Enzymes in the nucleus add it at the 3′ end of the pre-mRNA, so the 3′ end carries it.
- Award 1 point for: adenine nucleotides AND the 3′ end.
In a eukaryotic cell, several kinds of nucleic acid are present at once.
Which of the following receives a 5′ cap and a poly-A tail?
- A. A tRNAA tRNA folds on itself and carries an amino acid; it receives neither.
- B. The gene’s template strandThe template strand is DNA and stays in the gene; it receives neither.
- C. ✓ A pre-mRNA
Why: The cap and the tail are added to the RNA that RNA polymerase released: the pre-mRNA.
Neither is added to a tRNA or to the gene’s DNA.
Where in the cell are the 5′ cap and the poly-A tail added?
- A. At a ribosome in the cytoplasmA ribosome reads the mRNA; it adds nothing to its ends.
- B. ✓ In the nucleus
Why: The enzymes that join the cap and add the tail work inside the nucleus.
So the pre-mRNA already carries both when it leaves through a pore.
36The ribosome grips the cap
A ribosome has two subunits, a smaller one and a larger one.
At the start of translation, which subunit binds the mRNA?
- A. ✓ The small subunit
- B. The large subunitThe large subunit’s rRNA binds the tRNAs and joins the amino acids.
The small subunit binds the mRNA.
Why: A ribosome has two subunits, a smaller one and a larger one.
The small subunit binds the mRNA at the start of translation.
A ribosome reads an mRNA in one direction.
At which end of the mRNA does it start?
- A. ✓ The 5′ end
- B. The 3′ endThe ribosome moves toward the 3′ end; it does not start there.
Why: A ribosome starts at the mRNA’s 5′ end and moves toward its 3′ end.
Video: Watch: The ribosome grips the cap
A capped, tailed mRNA reaching a ribosome; the ribosome settling on the cap at the 5′ end and then moving along the strand; a second mRNA with no cap drifting past ribosomes that do not settle on it.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L14b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L14b.mp4
Now follow the capped, tailed mRNA out of the nucleus. It reaches a ribosome in the cytoplasm.
The ribosome must begin reading at the mRNA’s 5′ end. So the ribosome has to find that end first.
The 5′ cap marks the 5′ end. The ribosome recognizes the 5′ cap and grips the mRNA there.
Only then does the ribosome move along the mRNA and read its bases.
Now imagine an mRNA that reaches the ribosome with no 5′ cap.
The ribosome finds nothing to recognize at the 5′ end. So few ribosomes grip the mRNA.
An mRNA that few ribosomes grip is barely translated. So the cell makes only a little of its protein.
The ribosome recognizes the 5′ cap and grips the mRNA there, so an mRNA with no cap is barely translated.
What you are expected to know Explain what the 5′ cap does when an mRNA reaches a ribosome.
What you are expected to know Predict how much protein a cell makes from an mRNA that has no cap.
Suppose a eukaryotic cell receives an mRNA that has a poly-A tail but no 5′ cap.
Compared with the same mRNA with its cap, how much protein does the cell make from it?
- A. More proteinWithout a cap, fewer ribosomes grip the mRNA, not more.
- B. About the same amountWithout a cap, few ribosomes grip the mRNA, so it is not read as often.
- C. ✓ Much less protein
Why: The ribosome recognizes the 5′ cap and grips the mRNA there.
An mRNA with no cap gives the ribosome nothing to recognize, so few ribosomes grip it.
So the mRNA is barely translated, and the cell makes much less protein.
Suppose the enzyme that joins the 5′ cap stops working in a cell. New pre-mRNAs still receive their poly-A tails and still leave the nucleus, but without a cap. The cell makes far less of their proteins.
Explain why the cell makes far less of these proteins. (1 pt)
Frame The cell makes far less because …
An mRNA with no cap gives the ribosome nothing to recognize at its 5′ end.
So few ribosomes grip these mRNAs, and they are barely translated.
So little protein is made from them.
- Award 1 point for: the ribosome recognizes and binds the 5′ cap, so an uncapped mRNA is bound by few ribosomes and is barely translated.
A student says: “The ribosome grips an mRNA at its poly-A tail, so an mRNA needs a tail to be read.”
Is the student correct?
- A. Yes: the ribosome grips the mRNA at its poly-A tailThe poly-A tail sits at the 3′ end, the far end from where reading starts.
- B. ✓ No: the ribosome grips the mRNA at its 5′ cap
Why: The ribosome starts reading at the mRNA’s 5′ end.
The 5′ cap marks that end, and the ribosome recognizes and grips the cap.
The poly-A tail is at the other end and is not where the ribosome grips.
Suppose a cell receives two copies of the same mRNA at the same moment. Both copies have a poly-A tail. One copy has a 5′ cap, and the other copy’s cap is missing.
Which copy gives the cell protein first?
- A. ✓ The copy with the 5′ cap
- B. The copy with no capWithout a cap, the ribosome has nothing to recognize, so this copy is barely read.
- C. Both at the same timeOnly the capped copy is gripped by ribosomes straight away.
Why: Ribosomes recognize the 5′ cap and grip the capped copy at once, so they read it first.
The uncapped copy gives them nothing to recognize, so it is barely read.
54The tail decides how long the mRNA lasts
Video: Watch: The tail decides how long the mRNA lasts
A capped, tailed mRNA in the cytoplasm being read by ribosomes; an enzyme at its 3′ end removing the A’s of the tail one by one while the rest of the strand stays whole; the tail gone, the enzyme breaking down the strand itself; beside it, a bare mRNA broken down almost at once.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L14c.mp4
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Now consider the other end of the same mRNA, out in the cytoplasm.
The cytoplasm contains enzymes that break RNA down. These enzymes remove nucleotides from an RNA strand’s 3′ end, one at a time.
On a tailed mRNA, the poly-A tail is the 3′ end. So the enzymes reach the tail first.
They remove the tail’s adenine nucleotides one by one. While some tail is left, the rest of the mRNA is untouched.
So the mRNA keeps being translated, and the cell keeps making its protein.
When the tail is gone, the enzymes break down the mRNA itself. Then no more protein is made from it.
An mRNA with a tail of about 200 adenine nucleotides may last for many hours. An mRNA with a short tail is destroyed sooner.
Now imagine the poly-A tail is removed from an mRNA the moment it leaves the nucleus.
The enzymes reach the mRNA itself at once. So the mRNA is destroyed within the hour, and the cell makes little protein from it.
The poly-A tail protects the mRNA from the enzymes that break RNA down, so an mRNA with a shorter tail is destroyed sooner and gives less protein.
How long an mRNA lasts is one way a cell controls how much protein it makes.
What you are expected to know Explain what the poly-A tail does.
What you are expected to know Predict how long an mRNA lasts, and how much protein it gives, when its tail is shortened or removed.
Suppose the poly-A tail is removed from an mRNA as soon as the mRNA leaves the nucleus.
Compared with the same mRNA with its tail, how long does it last in the cytoplasm?
- A. LongerThe tail is what the RNA-breaking enzymes remove first; without it nothing delays them.
- B. The same timeWithout the tail, the RNA-breaking enzymes reach the mRNA itself at once.
- C. ✓ A shorter time
Why: Enzymes in the cytoplasm remove nucleotides from an RNA’s 3′ end.
With no tail, they reach the mRNA itself at once and break it down.
So the mRNA lasts a shorter time.
Suppose the enzyme that adds the poly-A tail stops working in a cell. New pre-mRNAs still receive their 5′ caps and still leave the nucleus, but without a tail. Within an hour these mRNAs have gone from the cytoplasm, while the same mRNAs with tails last for many hours.
Explain why the new mRNAs are gone within the hour. (1 pt)
Frame They are gone within the hour because …
On a tailed mRNA those enzymes spend hours removing the tail’s adenine nucleotides before they reach the mRNA itself.
An mRNA with no tail has nothing at its 3′ end to protect it.
So the enzymes break down the mRNA itself at once.
- Award 1 point for: the poly-A tail protects the mRNA from the enzymes that break RNA down (they remove the tail first); without a tail the enzymes break down the mRNA itself straight away.
Suppose three copies of one gene’s mRNA, R, S and T, are made in a cell with poly-A tails of different lengths. Here is a table of the three tails. All three mRNAs have a 5′ cap.
Predict which of the three mRNAs is destroyed soonest. Justify your prediction. (2 pt)
Enzymes in the cytoplasm remove nucleotides from an RNA’s 3′ end, and the poly-A tail is what they remove first.
R and S each have a tail that the enzymes must remove before they reach the mRNA itself.
T has no tail, so the enzymes break down T itself at once.
- Award 1 point for the prediction: mRNA T (the one with no tail).
- Award 1 point for the justification: the poly-A tail protects the mRNA from the RNA-breaking enzymes (they remove the tail first), so T, with no tail, is broken down at once.
Here is the whole comparison as a table, the 5′ cap against the poly-A tail: which end each sits on, what it is, what adds it, what it does, and what the mRNA loses without it.
Go back to the two mRNAs copied from the same gene and put into two cells.
One mRNA has a 5′ cap and a poly-A tail. Ribosomes recognize its cap and grip it, and its tail keeps the RNA-breaking enzymes busy for hours.
So that mRNA is read many times and lasts, and its cell makes plenty of protein.
The other mRNA has neither. Few ribosomes grip it, and the enzymes break it down within the hour.
So that mRNA is barely read and is gone, and its cell makes almost none of the protein.
78Mixed practice mixed practice
Suppose an enzyme in a cell lengthens every poly-A tail to about 400 adenine nucleotides.
What happens to how long each mRNA lasts in the cytoplasm?
- A. Each mRNA lasts a shorter timeA longer tail protects the mRNA for more time, not less.
- B. Each mRNA lasts the same timeThe RNA-breaking enzymes take longer to remove a longer tail, so they reach the mRNA itself later.
- C. ✓ Each mRNA lasts longer
Why: The RNA-breaking enzymes remove the tail’s adenine nucleotides before they reach the mRNA itself.
A tail of 400 adenine nucleotides takes them longer to remove than a tail of about 200.
So each mRNA is broken down later and lasts longer.
An mRNA in a eukaryotic cell leaves the nucleus through a pore.
Does it already carry its 5′ cap and its poly-A tail as it passes through the pore?
- A. ✓ Yes
- B. NoBoth additions are made by enzymes inside the nucleus, before the mRNA leaves.
Why: The enzyme that joins the 5′ cap and the enzyme that adds the poly-A tail both work in the nucleus.
So the mRNA carries both before it reaches the pore.
A eukaryotic mRNA is 1,000 nucleotides long. A ribosome grips it and begins reading.
Which nucleotides of the mRNA does the ribosome read first?
- A. ✓ Those nearest the 5′ cap
- B. Those nearest the poly-A tailThe poly-A tail is at the 3′ end, the far end from where reading starts.
- C. Those in the middle of the strandReading starts at an end, not in the middle.
Why: The ribosome recognizes the 5′ cap and grips the mRNA there.
Then it moves along the mRNA from the 5′ end toward the 3′ end.
So it reads the nucleotides nearest the 5′ cap first.
Suppose two mRNAs from different genes are read by ribosomes at the same rate. One is destroyed after 2 hours and the other after 20 hours.
From which mRNA does the cell make more protein?
- A. The mRNA destroyed after 2 hoursAn mRNA destroyed after 2 hours is translated for only 2 hours.
- B. ✓ The mRNA destroyed after 20 hours
- C. The same amount from eachThe mRNA that lasts ten times longer is translated for ten times longer.
Why: Both mRNAs are read at the same rate, so the protein made depends on how long each lasts.
The mRNA that lasts 20 hours is translated for ten times longer.
So the cell makes more protein from it.
A student says: “The ribosome adds the 5′ cap to the mRNA when translation starts.”
Is the student correct?
- A. Yes: the ribosome joins the cap as it grips the mRNAThe ribosome recognizes a cap that is already there; it does not add one.
- B. ✓ No: an enzyme in the nucleus joins the cap before the mRNA leaves
Why: An enzyme in the nucleus joins the 5′ cap to the pre-mRNA.
The mRNA leaves the nucleus already capped.
The ribosome recognizes and grips the cap; it adds nothing to the mRNA.
Suppose an mRNA with a 5′ cap but no poly-A tail is put into a eukaryotic cell. The cell makes some of the protein in the first minutes, and then stops within the hour.
(a) Explain why the cell makes some of the protein at first. (1 pt)
Frame The cell makes some at first because …
Ribosomes recognize the 5′ cap and grip the mRNA there.
So ribosomes read the mRNA and build the protein while the mRNA is still there.
- Award 1 point for: the mRNA carries a 5′ cap, which ribosomes recognize and bind, so it is translated.
(b) Explain why the cell stops making the protein within the hour. (1 pt)
Frame The cell stops making the protein because …
The poly-A tail is what those enzymes remove first, and this mRNA has none.
So the enzymes break down the mRNA itself at once, and within the hour it is gone.
With the mRNA gone, the ribosomes have nothing to read, so the cell stops making the protein.
- Award 1 point for: with no poly-A tail to protect it, the RNA-breaking enzymes degrade the mRNA itself quickly, so translation stops.
Glossary
- 5′ cap (GTP cap)
- A modified guanine nucleotide that an enzyme in the nucleus joins to the 5′ end of a eukaryotic pre-mRNA. The ribosome recognizes the cap and grips the mRNA there, so an mRNA with no cap is barely translated.
- poly-A tail
- A stretch of about 200 adenine nucleotides that an enzyme in the nucleus adds to the 3′ end of a eukaryotic pre-mRNA. It protects the mRNA from the enzymes that break RNA down, so an mRNA with a shorter tail is destroyed sooner and gives less protein.
APBIO-U06-L15 Where did 1,500 nucleotides go?
A gene is transcribed in the nucleus into an RNA 2,400 nucleotides long. The mRNA found at the ribosomes for that gene is 900 nucleotides long. Nothing was cut from its ends.
Where did 1,500 nucleotides go?
Unit 6 · Gene Expression and Regulation
1What the enzymes cut out
In a eukaryote, RNA polymerase passes the stop sequence of a gene for a protein and releases its RNA. The RNA still needs work before it can leave the nucleus.
What is this RNA called?
- A. tRNAtRNA is the RNA that carries an amino acid to the ribosome.
- B. ✓ pre-mRNA
- C. The template strandThe template strand is the DNA strand RNA polymerase reads.
It stays in the gene.
Why: RNA polymerase releases its RNA when it passes a stop sequence.
In a eukaryote that RNA is not yet ready, so it is called pre-mRNA.
Enzymes in the nucleus add a cap to one end of a pre-mRNA and a tail of about 200 A's to the other end.
Which end receives the cap?
- A. ✓ The 5′ end
- B. The 3′ endThe 3′ end receives the tail of about 200 A's.
Why: The cap, a modified guanine nucleotide, joins the 5′ end.
The tail of about 200 A's joins the 3′ end.
Video: Watch: What the enzymes cut out
The pre-mRNA from the opening gene: its two introns loop out, the enzymes cut them away, and the three exons join end to end into the 900-nucleotide mature mRNA.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L15a.mp4
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Why is the mRNA shorter than the pre-mRNA it came from?
A eukaryotic pre-mRNA carries stretches the cell keeps, and stretches between them that the cell cuts out.
Enzymes cut out each unwanted stretch and join the kept stretches end to end.
In the opening gene, three kept stretches totaling 900 nucleotides stayed. Enzymes cut out two stretches totaling 1,500 nucleotides.
Keep a different set of stretches, and the same gene gives a different mRNA and a different protein.
Go back to the gene from the opening. Here it is drawn as a gene map: the promoter box at the left, the bent arrow where RNA polymerase starts, and the gene stretching to the right.
RNA polymerase reads the whole gene, from the start to the stop sequence. So the pre-mRNA is a copy of the whole gene: 2,400 nucleotides long.
In this gene, the instructions for the protein sit in three separate stretches. Two stretches between them carry none of the protein's instructions.
On the gene map, the three stretches that carry instructions are drawn as filled boxes. The two stretches between them are drawn as the line between the boxes.
A stretch of the pre-mRNA that is kept in the mRNA is called an , because it is expressed: a ribosome reads it.
A stretch of the pre-mRNA that is cut out before the mRNA leaves the nucleus is called an , because it sits in between the exons.
The opening gene has three exons: 300, 240 and 360 nucleotides long. It has two introns: 700 and 800 nucleotides long.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
Enzymes in the nucleus cut the pre-mRNA at both ends of each intron. Each intron loops out and is removed.
The enzymes then join the exons end to end: exon 1 to exon 2, and exon 2 to exon 3.
Cutting out the introns and joining the exons end to end is called , because to splice two ropes is to join their ends into one rope.
The spliced RNA, with its cap and its tail, is called the , because the cell has finished it: it is ready to leave the nucleus.
The two introns, 1,500 nucleotides in all, are gone. The three exons, 900 nucleotides in all, remain.
In the mature mRNA the exons sit in one continuous stretch, in the gene's order: exon 1, then exon 2, then exon 3.
The enzymes cut the introns out of the RNA copy only. The gene's DNA keeps its introns, and the next pre-mRNA from this gene carries them again.
Every length written on a gene map counts exons and introns only. Enzymes add the cap and the tail at the ends, and the lengths leave them out.
What you are expected to know Describe splicing: enzymes in the nucleus cut each intron out of the pre-mRNA and join the exons end to end, so the mature mRNA is shorter than the pre-mRNA and its exons sit in one continuous stretch.
A eukaryotic pre-mRNA has three exons and two introns. Enzymes in the nucleus splice it.
Which stretches are in the mature mRNA?
- A. The introns onlyThe introns are cut out.
Only the exons stay. - B. ✓ The exons only
- C. Both kinds of stretchThe introns are cut out and are gone from the mature mRNA.
Why: Splicing cuts each intron out of the pre-mRNA.
The exons are joined end to end.
So the mature mRNA carries the exons only.
Suppose a gene in a eukaryotic cell is transcribed into a pre-mRNA 3,000 nucleotides long. The mature mRNA that leaves the nucleus is 1,200 nucleotides long. Nothing was cut from its ends.
(a) Explain why the mature mRNA is shorter than the pre-mRNA. (1 pt)
Frame The mature mRNA is shorter because …
Enzymes in the nucleus cut each intron out of the pre-mRNA.
The enzymes joined the exons end to end.
So the mature mRNA carries the exons only, 1,200 nucleotides, and the 1,800 nucleotides of introns are gone.
- Award 1 point for: enzymes cut the introns out of the pre-mRNA and joined the exons, so the mature mRNA carries the exons only (the 1,800 nucleotides of introns are gone).
A student says: “Splicing cuts the introns out of the gene's DNA, so the gene loses its introns.”
Is the student correct?
- A. Yes: the gene loses its introns when its pre-mRNA is splicedThe enzymes cut the pre-mRNA, the RNA copy.
The gene's DNA is unchanged. - B. ✓ No: the enzymes cut the introns out of the pre-mRNA, and the gene's DNA keeps them
Why: Splicing acts on the pre-mRNA, the RNA copy of the gene.
The gene's DNA keeps every intron.
So the next pre-mRNA from this gene carries the introns again.
A eukaryotic gene has four exons.
How many introns lie between them?
- A. ✓ Three
- B. FourFour exons have three gaps between them.
Each gap is one intron. - C. FiveFive introns would need six exons.
Why: An intron sits between two neighboring exons.
Four exons have three gaps between them.
So the gene has three introns.
31Quick quiz: intron, exon, splicing, mature mRNA mixed practice
A pre-mRNA carries stretches the cell keeps and stretches the cell cuts out.
What is a stretch that is kept in the mature mRNA called?
- A. IntronAn intron is a stretch that is cut out.
- B. ✓ Exon
- C. TranscriptThe transcript is the whole RNA as it came off the gene.
A stretch kept within it is an exon.
Why: A stretch that is kept in the mature mRNA is an exon.
Ex- as in expressed: a ribosome reads it.
A pre-mRNA carries stretches the cell keeps and stretches the cell cuts out.
What is a stretch that is cut out before the mRNA leaves the nucleus called?
- A. ✓ Intron
- B. ExonAn exon is a stretch that is kept and read by a ribosome.
- C. PromoterThe promoter is a DNA sequence just before the gene, and nothing is cut out of the DNA.
A stretch cut out of the pre-mRNA is an intron.
Why: A stretch that is cut out before the mRNA leaves the nucleus is an intron.
It sits in between the exons.
Enzymes in the nucleus cut each intron out of a pre-mRNA and join the exons end to end.
What is this process called?
- A. TranscriptionTranscription is RNA polymerase copying the gene into RNA.
- B. TranslationTranslation is a ribosome reading the mRNA and joining amino acids.
- C. ✓ Splicing
Why: Cutting out the introns and joining the exons end to end is splicing.
A pre-mRNA has been capped, tailed and spliced. It is ready to leave the nucleus.
What is this RNA called now?
- A. pre-mRNApre-mRNA is the RNA before the cell has finished it.
- B. ✓ mature mRNA
- C. tRNAtRNA carries an amino acid to the ribosome; it is a different RNA.
Why: The spliced RNA with its cap and tail is the mature mRNA.
The cell has finished it, so it can leave the nucleus.
A eukaryotic pre-mRNA carries exons and introns.
(a) State what splicing is. (1 pt)
- Award 1 point for: enzymes cut the introns out of the pre-mRNA and join the exons end to end.
On a gene map, exons are drawn as filled boxes and introns as the line between them. A eukaryotic gene is drawn below.
How many exons does the gene have?
- A. ThreeCount the filled boxes: there are four.
- B. ✓ Four
- C. FiveCount the filled boxes: there are four, with three lines between them.
Why: Each filled box on the gene map is one exon.
The map has four filled boxes.
So the gene has four exons.
A eukaryotic gene is transcribed and its pre-mRNA is spliced.
Which is longer, the pre-mRNA or the mature mRNA?
- A. ✓ The pre-mRNA
- B. The mature mRNAThe mature mRNA has lost its introns.
- C. The two are the same lengthThe introns were cut out, so the two lengths differ.
Why: The pre-mRNA carries the exons and the introns.
The mature mRNA carries the exons only.
So the pre-mRNA is longer.
39How long is the mature mRNA?
Add the lengths of the exons, and you have the coding length of the mature mRNA.
The exons keep the order they had in the gene.
Video: Watch: How long is the mature mRNA?
The gene map with its lengths written on; the three exon lengths added one at a time to the coding length of the mature mRNA; the exons in the gene's order.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L15b.mp4
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Now consider a second gene. Here is its gene map with the length of every exon and every intron written on the drawing, in nucleotides.
The exons are 201, 120 and 282 nucleotides long. The introns are 450 and 640 nucleotides long.
Splicing cuts out both introns and joins the three exons. So the coding length of the mature mRNA is the three exon lengths added together.
the coding length of the mature mRNA: the exon lengths added; the introns are gone
Here is the calculation written out as working.
the worked gene: three exons added
The exons keep the order they had in the gene. So the mature mRNA reads exon 1, then exon 2, then exon 3.
The intron lengths play no part in the coding length. They tell you only how much longer the pre-mRNA was.
What you are expected to know Calculate the coding length of a mature mRNA from its gene map by adding the exon lengths; the exons keep the gene's order.
Now consider the gene drawn below. Its exon and intron lengths are written on the drawing, in nucleotides.
Calculate the coding length of its mature mRNA.
Part 1. How long are exons 1 and 2 together?
Answer: 510 nucleotides (tolerance ±0)
Part 2. Add exon 3. How long are all three exons together?
Answer: 690 nucleotides (tolerance ±0)
Answer: 690 nucleotides (tolerance ±0)
A eukaryotic gene has three exons, numbered 1, 2 and 3 from the promoter. Enzymes splice its pre-mRNA.
In which order do the exons sit in the mature mRNA?
- A. Exon 3, then exon 2, then exon 1Splicing joins the exons in the order they had in the gene.
Exon 1 is nearest the promoter, so it comes first. - B. ✓ Exon 1, then exon 2, then exon 3
Why: The enzymes cut out the introns and join each exon to the next.
The exons keep the gene's order.
So the mature mRNA reads exon 1, then exon 2, then exon 3.
52Quick quiz: the coding length mixed practice
The gene drawn below has its exon lengths written above the exons, in nucleotides.
Calculate the coding length of its mature mRNA.
Answer: 390 nucleotides (tolerance ±0)
The gene drawn below has its exon lengths written above the exons, in nucleotides.
Calculate the coding length of its mature mRNA.
Answer: 715 nucleotides (tolerance ±0)
The gene drawn below has its exon lengths written above the exons, in nucleotides.
Calculate the coding length of its mature mRNA.
Answer: 490 nucleotides (tolerance ±0)
The gene drawn below has its exon lengths written above the exons, in nucleotides.
Calculate the coding length of its mature mRNA.
Answer: 675 nucleotides (tolerance ±0)
The gene drawn below has four exons, with its exon lengths written above the exons, in nucleotides.
Calculate the coding length of its mature mRNA.
Answer: 855 nucleotides (tolerance ±0)
Suppose a cell splices the pre-mRNA of the gene drawn below. Its exon and intron lengths are written on the drawing, in nucleotides.
Calculate the coding length of the mature mRNA.
Answer: 785 nucleotides (tolerance ±0)
59Which exon was skipped?
Sometimes the enzymes skip an exon and cut it out along with the introns beside it.
The mature mRNA is then shorter than expected by exactly that exon's length.
Video: Watch: Which exon was skipped?
A mature mRNA drawn to scale beneath its gene map falls short of the expected length by one exon's length; the skipped exon is found by its length.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L15c.mp4
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Go back to the gene with exons of 201, 120 and 282 nucleotides. Its mature mRNA should have a coding length of 603 nucleotides.
Suppose the mature mRNA from this gene is drawn to scale beneath the gene map, and it is 483 nucleotides long.
The mature mRNA is 120 nucleotides shorter than expected. Exon 2 is 120 nucleotides long.
So the enzymes skipped exon 2: they cut it out along with the introns on either side of it. The mature mRNA reads exon 1, then exon 3.
To read which exon was skipped, take the measured length away from the expected length. Then find the exon with that length.
For example, this mature mRNA is 483 nucleotides long, 120 short. So exon 2 was skipped.
And this mature mRNA is 402 nucleotides long, 201 short. So exon 1 was skipped.
And this mature mRNA is 321 nucleotides long, 282 short. So exon 3 was skipped.
But this mature mRNA is 603 nucleotides long, nothing short. So no exon was skipped.
What you are expected to know Read which exon was skipped from a mature mRNA drawn to scale beside its gene map: the length missing from the expected coding length is the skipped exon's length.
A gene map with its exon lengths is drawn below, with its mature mRNA to scale and its length written in nucleotides.
Which exon was skipped?
- A. Exon 1Exon 1 is 350 nucleotides long.
The mature mRNA is 100 nucleotides short. - B. ✓ Exon 2
- C. Exon 3Exon 3 is 200 nucleotides long.
The mature mRNA is 100 nucleotides short. - D. No exon was skippedThe expected coding length is 650 nucleotides.
The mature mRNA is 550.
Why: Expected coding length: nucleotides.
Measured: 550 nucleotides.
Missing: nucleotides, the length of exon 2.
A gene map with its exon lengths is drawn below, with its mature mRNA to scale and its length written in nucleotides.
Which exon was skipped?
- A. Exon 1The expected coding length is 790 nucleotides.
The mature mRNA is 790: nothing is missing. - B. Exon 2The expected coding length is 790 nucleotides.
The mature mRNA is 790: nothing is missing. - C. Exon 3The expected coding length is 790 nucleotides.
The mature mRNA is 790: nothing is missing. - D. ✓ No exon was skipped
Why: Expected coding length: nucleotides.
Measured: 790 nucleotides.
Nothing is missing, so no exon was skipped.
A gene map with its exon lengths is drawn below, with its mature mRNA to scale and its length written in nucleotides.
Which exon was skipped?
- A. ✓ Exon 1
- B. Exon 2Exon 2 is 310 nucleotides long.
The mature mRNA is 230 nucleotides short. - C. Exon 3Exon 3 is 125 nucleotides long.
The mature mRNA is 230 nucleotides short. - D. No exon was skippedThe expected coding length is 665 nucleotides.
The mature mRNA is 435.
Why: Expected coding length: nucleotides.
Measured: 435 nucleotides.
Missing: nucleotides, the length of exon 1.
A gene map with its two exon lengths is drawn below, with its mature mRNA to scale and its length written in nucleotides.
Which exon was skipped?
- A. Exon 1Exon 1 is 420 nucleotides long.
The mature mRNA is 175 nucleotides short. - B. ✓ Exon 2
- C. No exon was skippedThe expected coding length is 595 nucleotides.
The mature mRNA is 420.
Why: Expected coding length: nucleotides.
Measured: 420 nucleotides.
Missing: nucleotides, the length of exon 2.
A gene map with its exon lengths is drawn below, with its mature mRNA to scale and its length written in nucleotides.
Which exon was skipped?
- A. Exon 1Exon 1 is 260 nucleotides long.
The mature mRNA is 440 nucleotides short. - B. Exon 2Exon 2 is 195 nucleotides long.
The mature mRNA is 440 nucleotides short. - C. ✓ Exon 3
- D. No exon was skippedThe expected coding length is 895 nucleotides.
The mature mRNA is 455.
Why: Expected coding length: nucleotides.
Measured: 455 nucleotides.
Missing: nucleotides, the length of exon 3.
78One gene, several mRNAs
The same pre-mRNA can be spliced keeping different exons in different cells.
So one gene gives several mRNAs and several related proteins.
The gene's DNA is the same in every cell. Only the choice of exons differs.
Video: Watch: One gene, several mRNAs
One pre-mRNA spliced two ways: the muscle cell keeps all four exons; the nerve cell cuts out exon 2 with the introns beside it; two mature mRNAs, two related proteins.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L15d.mp4
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Now consider one gene with four exons, in two cells of the same body: a muscle cell and a nerve cell.
Both cells transcribe the gene. Both pre-mRNAs are the same: exons 1, 2, 3 and 4, with three introns between them.
In the muscle cell, the enzymes cut out the three introns and keep all four exons. Its mature mRNA reads exon 1, 2, 3, 4.
In the nerve cell, the enzymes cut out exon 2 along with the introns on either side of it. Its mature mRNA reads exon 1, 3, 4.
Splicing the same pre-mRNA in more than one way, keeping different sets of exons, is called , because the cell has alternatives: which exons to keep.
Ribosomes read each mature mRNA into a protein. The muscle cell's protein has the part made from exon 2.
The nerve cell's protein lacks that part.
The two proteins are related: they share the parts made from exons 1, 3 and 4. They differ in one part.
So one gene gives two mature mRNAs and two related proteins.
The gene's DNA is the same in the muscle cell and the nerve cell. Only the choice of exons kept differs.
What you are expected to know Explain alternative splicing: the same pre-mRNA spliced keeping different sets of exons in different cells gives several mature mRNAs and several related proteins from one gene.
What you are expected to know Predict which part a protein lacks from the exons its mRNA kept.
Two cells in one body each make a mature mRNA from one gene. One mRNA carries exons 1, 2 and 3. The other carries exons 1 and 3.
How many genes are involved?
- A. TwoBoth mRNAs came from one gene.
The two cells spliced its pre-mRNA differently. - B. ✓ One
Why: Both cells transcribed the one gene into the same pre-mRNA.
One cell kept all three exons; the other cut out exon 2 with the introns beside it.
So one gene gave two mRNAs.
Suppose a gene has five exons. A muscle cell's mature mRNA from this gene carries exons 1, 2, 3, 4 and 5. A nerve cell's mature mRNA from the same gene carries exons 1, 2, 4 and 5.
(a) Explain how one gene gives the two cells two different proteins. (2 pt)
Frame One gene gives two different proteins because …
The muscle cell's enzymes cut out the introns only and keep all five exons.
The nerve cell's enzymes cut out exon 3 along with the introns beside it.
So the two mature mRNAs carry different sets of exons.
Ribosomes read each mRNA into a protein, so the nerve cell's protein lacks the part made from exon 3.
- Award 1 point for: the same pre-mRNA is spliced keeping different sets of exons in the two cells (alternative splicing), so the two mature mRNAs differ.
- Award 1 point for: ribosomes read the two different mRNAs into two different proteins; the nerve cell's protein lacks the part made from exon 3.
A student says: “A cell whose mature mRNA lacks exon 4 must have lost exon 4 from its DNA.”
Is the student correct?
- A. ✓ No: the gene's DNA keeps exon 4, and the cell's enzymes cut it out of the pre-mRNA
- B. Yes: a mature mRNA lacks an exon only when the cell's DNA has lost that exon from the geneThe DNA is the same in every cell of the body.
The enzymes chose which exons to keep in the pre-mRNA.
Why: The gene's DNA is the same in every cell.
The cell's enzymes cut exon 4 out of the pre-mRNA along with the introns beside it.
So the mature mRNA lacks exon 4 while the DNA keeps it.
Suppose a gene has six exons. A muscle cell's mature mRNA from this gene carries all six. A nerve cell's mature mRNA carries exons 1, 2, 3, 4 and 6.
Which part does the nerve cell's protein lack?
- A. The parts made from exons 1 to 4Exons 1 to 4 are in the nerve cell's mRNA, so its protein has those parts.
- B. The part made from exon 6Exon 6 is in the nerve cell's mRNA, so its protein has that part.
- C. ✓ The part made from exon 5
Why: The nerve cell's mRNA carries every exon but exon 5.
Ribosomes read the mRNA into the protein.
So the nerve cell's protein lacks the part made from exon 5.
Go back to the 2,400-nucleotide pre-mRNA and the 900-nucleotide mRNA at the ribosomes.
The cell kept three exons totaling 900 nucleotides. It cut out two introns totaling 1,500 nucleotides and joined the exons.
Keep a different set of exons, and the same gene gives a different protein.
102Quick quiz: alternative splicing mixed practice
One pre-mRNA is spliced keeping different sets of exons in different cells.
What is this called?
- A. ✓ Alternative splicing
- B. TranscriptionTranscription is RNA polymerase copying a gene into RNA.
- C. TranslationTranslation is a ribosome reading an mRNA into a protein.
Why: Splicing the same pre-mRNA in more than one way, keeping different sets of exons, is alternative splicing.
A eukaryotic gene is transcribed into a pre-mRNA with several exons.
(a) State what alternative splicing is. (1 pt)
- Award 1 point for: the same pre-mRNA spliced keeping different sets of exons, giving different mature mRNAs (and proteins) from one gene.
One gene's pre-mRNA is spliced two ways in two cells.
How many kinds of mature mRNA does the gene give?
- A. OneTwo ways of splicing keep two different sets of exons.
- B. ✓ Two
Why: Each way of splicing keeps a different set of exons.
Two ways give two different mature mRNAs.
Two proteins are made from one gene by alternative splicing.
How do the two proteins compare?
- A. Identical in every partThe two mRNAs carry different sets of exons, so the proteins differ.
- B. Unrelated in every partBoth proteins come from the exons of one gene, so they share most parts.
- C. ✓ Related, differing in some parts
Why: The two mRNAs share most exons and differ in the skipped ones.
So the proteins share most parts and differ where exons were skipped.
107Mixed practice mixed practice
A gene map with its exon lengths is drawn below, with its mature mRNA to scale and its length written in nucleotides.
Which exon was skipped?
- A. Exon 1Exon 1 is 275 nucleotides long.
The mature mRNA is 385 nucleotides short. - B. Exon 2Exon 2 is 155 nucleotides long.
The mature mRNA is 385 nucleotides short. - C. ✓ Exon 3
- D. No exon was skippedThe expected coding length is 815 nucleotides.
The mature mRNA is 430.
Why: Expected coding length: nucleotides.
Measured: 430 nucleotides.
Missing: nucleotides, the length of exon 3.
Enzymes splice a pre-mRNA that has five exons and four introns.
How many stretches, joined end to end, does the mature mRNA's coding sequence contain?
- A. FourFour is the count of introns, and the introns are cut out.
- B. ✓ Five
- C. NineNine counts the exons and the introns together; the introns are gone.
Why: Splicing cuts out the four introns.
The five exons are joined end to end.
So the coding sequence contains five stretches.
A student says: “Alternative splicing changes the gene's DNA, so each cell ends up with a different version of the gene.”
Is the student correct?
- A. Yes: each cell's DNA loses the exons its enzymes do not keepThe enzymes act on the pre-mRNA, the RNA copy.
The gene's DNA is unchanged in every cell. - B. ✓ No: the DNA is the same in every cell; the enzymes choose which exons to keep
Why: Alternative splicing acts on the pre-mRNA.
The gene's DNA is the same in every cell of the body.
So the cells differ in the exons kept, never in the gene.
Suppose a cell splices the pre-mRNA of the gene drawn below. Its exon and intron lengths are written on the drawing, in nucleotides.
Calculate the coding length of the mature mRNA.
Answer: 745 nucleotides (tolerance ±0)
In a eukaryotic cell, a gene has been transcribed and the RNA has been capped, tailed and spliced.
Which RNA leaves the nucleus?
- A. ✓ The mature mRNA
- B. The pre-mRNAThe pre-mRNA is not yet finished; it stays in the nucleus while the enzymes work on it.
Why: The spliced RNA with its cap and tail is the mature mRNA.
The cell has finished it.
So the mature mRNA leaves the nucleus through a pore.
A muscle cell and a nerve cell make two different mature mRNAs from one gene.
Which of the following is the same in the two cells?
- A. ✓ The gene's DNA
- B. The set of exons keptThe two cells keep different sets of exons; that is what differs.
- C. The protein madeThe two mRNAs differ, so the two proteins differ.
Why: Both cells carry the same gene in their DNA.
Their enzymes keep different sets of exons from the same pre-mRNA.
So the DNA is the same and the mRNAs and proteins differ.
Suppose a eukaryotic gene is transcribed into a pre-mRNA 3,450 nucleotides long. Its introns total 2,300 nucleotides.
Calculate the coding length of the mature mRNA.
Answer: 1150 nucleotides (tolerance ±0)
Suppose a eukaryotic gene has three exons of 410, 255 and 345 nucleotides, with introns of 950 and 620 nucleotides between them. A cell's mature mRNA from this gene has a coding length of 755 nucleotides.
(a) Determine which exon the cell's enzymes skipped. Justify your answer with the lengths. (2 pt)
Missing: nucleotides.
Exon 2 is 255 nucleotides long.
So the enzymes skipped exon 2.
- Award 1 point for: the expected coding length is 1,010 nucleotides, so 255 nucleotides are missing.
- Award 1 point for: exon 2 is 255 nucleotides long, so exon 2 was skipped.
(b) Predict how the protein made from this mature mRNA differs from the protein made when no exon is skipped. (1 pt)
Ribosomes read the mRNA into the protein.
So this protein lacks the part made from exon 2 and shares the rest.
- Award 1 point for: the protein lacks the part made from exon 2 (the rest is shared).
Glossary
- exon
- A stretch of a pre-mRNA that is kept in the mature mRNA and read by a ribosome (ex- as in expressed). On a gene map, exons are drawn as filled boxes.
- intron
- A stretch of a pre-mRNA that is cut out before the mRNA leaves the nucleus; it sits in between the exons. The gene's DNA keeps its introns; only the RNA copy loses them.
- splicing
- Enzymes in the nucleus cut each intron out of a pre-mRNA and join the exons end to end, so the mature mRNA is shorter than the pre-mRNA and its exons sit in one continuous stretch.
- mature mRNA
- The spliced RNA with its cap and its tail: the cell has finished it, and it is ready to leave the nucleus and be read by a ribosome.
- alternative splicing
- Splicing the same pre-mRNA in more than one way, keeping different sets of exons, so one gene gives several mature mRNAs and several related proteins. The gene's DNA is the same in every cell.
APBIO-U06-L16 From gene to mature message
Here is one drawing. At the top is a eukaryotic gene, at the bottom its mature mRNA, and between them every step in order.
Beside it is a gene in a bacterium, whose mRNA is ready the moment RNA polymerase makes it. Trace both routes.
Unit 6 · Gene Expression and Regulation
1One drawing, every step in order
RNA polymerase begins transcription at a fixed place on the DNA.
What is the promoter?
- A. ✓ A stretch of DNA just before the gene, marking its start
- B. The stretch of RNA at the very start of the transcriptThe promoter is DNA, not RNA.
RNA polymerase binds it before any RNA exists. - C. The enzyme that builds the RNA against the templateThe enzyme that builds the RNA is RNA polymerase.
The promoter is the DNA it binds.
Why: The promoter is a DNA sequence just before the gene.
RNA polymerase binds it, so it marks where transcription begins.
A gene has two DNA strands.
Which strand is the template strand?
- A. The strand whose letters match the RNA’s, with U for TThe strand whose letters the RNA matches is the non-template strand, also called the coding strand.
- B. ✓ The strand RNA polymerase reads and pairs RNA nucleotides against
Why: RNA polymerase reads one strand and pairs each RNA nucleotide against it.
That strand is the template strand.
In a eukaryote, RNA polymerase passes the stop sequence of a gene for a protein and releases the RNA.
What is the released RNA called?
- A. A mature mRNA, ready to be readThe released RNA still needs its cap and tail, and its introns must be cut out.
It is not yet mature. - B. ✓ A pre-mRNA, not yet ready to be read
- C. A template strandThe template strand is DNA.
It stays in the gene.
Why: The released RNA is the transcript.
In a eukaryote it is not yet ready to be read, so it is called the pre-mRNA.
Enzymes in the nucleus join a cap to a pre-mRNA.
Which end gets the cap?
- A. ✓ The 5′ end
- B. The 3′ endThe 3′ end gets the poly-A tail.
Why: The cap is a modified guanine nucleotide.
Enzymes join it to the pre-mRNA’s 5′ end.
Enzymes in the nucleus add a poly-A tail to a pre-mRNA.
What is the poly-A tail?
- A. A second RNA strand paired with the pre-mRNA’s 3′ endThe pre-mRNA stays a single strand.
The tail is a run of adenine nucleotides that enzymes add to its 3′ end. - B. ✓ About 200 adenine nucleotides added to the 3′ end
- C. The stretch of RNA copied from an intronAn intron’s RNA is cut out of the pre-mRNA, not added to it.
Why: The poly-A tail is a tail of about 200 adenine nucleotides.
Enzymes add it to the pre-mRNA’s 3′ end.
A eukaryotic pre-mRNA has stretches that enzymes keep and stretches that enzymes cut out.
Which stretches are the exons?
- A. The stretches that are cut outThe stretches that are cut out are the introns.
- B. ✓ The stretches that are kept and joined end to end
Why: Enzymes remove each intron and join the exons end to end.
So the exons are the stretches that are kept.
Video: Watch: One drawing, every step in order
The eukaryotic column filling in row by row: RNA polymerase on the promoter, the pre-mRNA released, the cap and tail added, the introns cut out, the mature mRNA leaving through the pore.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L16a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L16a.mp4
What happens between a gene and the message a ribosome can read?
In a eukaryote, RNA polymerase binds the promoter and reads the template strand from its 3′ end toward its 5′ end. It releases a transcript.
Enzymes join a cap to the transcript’s 5′ end and a tail to its 3′ end.
Enzymes cut out the introns and join the exons. The mature mRNA then leaves the nucleus through a pore.
In a prokaryote there is no nucleus, no cap, no tail and no intron. So ribosomes read the message while RNA polymerase makes it.
One labeled drawing and one table hold the whole topic.
Here is the whole drawing, with every label taken off. Every step from a eukaryotic gene to its mature mRNA is on it.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
The dark strand is DNA. The light strands are RNA, newly made.
Now take the drawing one row at a time, from the top.
The top row is the gene, drawn as before. The promoter box sits at the left and the bent arrow marks the transcription start site.
The filled boxes are the exons. The line between them is each intron.
Step 1. RNA polymerase binds the promoter.
The second row is the template strand, the one strand RNA polymerase reads. Its 3′ end is at the left, so RNA polymerase moves along it from left to right, 3′ to 5′.
Step 2. RNA polymerase pairs each RNA nucleotide against the template strand and builds the RNA from its 5′ end toward its 3′ end.
Step 3. RNA polymerase passes the stop sequence and releases the RNA. The released RNA is the pre-mRNA, the third row.
In the drawing, the pre-mRNA’s exons are the numbered boxes and its introns the thin lines between them, as on the gene. It carries both, copied straight from the gene.
Step 4. Enzymes join a cap to the pre-mRNA’s 5′ end and a tail of about 200 A’s to its 3′ end.
Step 5. Enzymes remove each intron and join the exons end to end. The RNA is now the mature mRNA.
Step 6. The mature mRNA leaves the nucleus through a pore and reaches a ribosome.
Here is the whole drawing again with every label on. Read it from the top: promoter, template strand, pre-mRNA, cap and tail, introns out, and export through the pore.
What you are expected to know Trace one eukaryotic gene to its mature mRNA on the labeled drawing: promoter, template strand and its direction, pre-mRNA, cap, tail, introns removed and exons joined, exit through a pore.
The drawing below shows a eukaryotic gene and its RNA at each step, with the parts lettered and the labels taken off.
Which of the following is the part lettered K?
- A. ✓ RNA polymerase
- B. The promoterThe promoter is the open box on the gene line.
K is the oval sitting over it. - C. The 5′ capThe cap is the dot at the 5′ end of an RNA.
K is the oval on the gene line.
Why: K is the large oval seated over the promoter box.
RNA polymerase is drawn as an oval on the promoter.
So K is RNA polymerase.
The drawing below shows a eukaryotic gene and its RNA at each step, with the parts lettered and the labels taken off.
Which of the following is the part lettered W?
- A. An intronAn intron is a stretch of plain line between two filled boxes on the gene line.
W is the open box at the left end, before the gene. - B. ✓ The promoter
- C. An exonAn exon is a filled box on the gene line.
W is the open box at the left end.
Why: W is the open box at the left end of the gene line, under the oval.
The promoter is drawn as a box at the left, just before the gene.
So W is the promoter.
The drawing below shows a eukaryotic gene and its RNA at each step, with the parts lettered and the labels taken off.
Which of the following is the part lettered L?
- A. ✓ The template strand
- B. The pre-mRNAThe pre-mRNA is a light row of boxes joined by thin lines.
L is the dark bar with tabs. - C. The mature mRNAThe mature mRNA is the short light bar at the bottom.
L is the dark bar with tabs.
Why: L is the dark bar under the gene, with its 3′ end at the left.
DNA is drawn dark, and this is the strand RNA polymerase reads.
So L is the template strand.
The drawing below shows a eukaryotic gene and its RNA at each step, with the parts lettered and the labels taken off.
Which of the following is the strand lettered S?
- A. The template strandThe template strand is the dark bar with tabs.
S is a light bar. - B. The mature mRNAThe mature mRNA has its introns removed, a cap and a tail.
S still has its thin intron lines and no cap or tail. - C. ✓ The pre-mRNA
Why: S marks the first light row under the template strand: its bracket spans the whole row.
The row’s boxes are still joined by thin intron lines, and it has no cap and no tail.
So S is the pre-mRNA, the RNA as it was released.
The drawing below shows a eukaryotic gene and its RNA at each step, with the parts lettered and the labels taken off.
Which of the following is the part lettered N?
- A. The poly-A tailThe tail is the row of A’s at the 3′ end.
N is the dot at the 5′ end. - B. An intronAn intron is a thin line between two boxes of the RNA.
N is the dot joined to the RNA’s end. - C. ✓ The 5′ cap
Why: N is the dot at the 5′ end of the RNA.
The cap is drawn as a dot on the 5′ end.
So N is the 5′ cap.
The drawing below shows a eukaryotic gene and its RNA at each step, with the parts lettered and the labels taken off.
Which of the following is the part lettered R?
- A. The 5′ capThe cap is the dot at the 5′ end.
R is the row of A’s at the 3′ end. - B. ✓ The poly-A tail
- C. An exonAn exon is a box of the RNA.
R is the row of A’s after the last box.
Why: R is the row of A’s at the RNA’s 3′ end.
The poly-A tail is drawn as a row of A’s on the 3′ end.
So R is the poly-A tail.
The drawing below shows a eukaryotic gene and its RNA at each step, with the parts lettered and the labels taken off.
Which of the following is happening at the part lettered P?
- A. ✓ An intron is being cut out
- B. The cap is being addedThe cap is added at the RNA’s 5′ end, and this RNA already has its cap.
P is a loop of thin line standing out of the RNA. - C. The tail is being addedThe tail is added at the 3′ end.
P is a loop between two boxes.
Why: P is a thin line looped out of the RNA between two boxes.
The thin lines are the introns, and a loop is one being removed.
So at P an intron is being cut out.
The drawing below shows a eukaryotic gene and its RNA at each step, with the parts lettered and the labels taken off.
Which letter marks the mature mRNA on its way to a ribosome?
- A. Part SPart S is the first light row, still with its thin intron lines.
The mature mRNA is the row with an arrow to the ribosome. - B. ✓ Part T
- C. Part VV is the loop being cut out of an RNA.
The mature mRNA is the bar with an arrow to the ribosome.
Why: T hangs from the bracket over the bottom bar, whose arrow leads to the ribosome outside the frame.
The bottom bar has its exons joined, a cap and a tail.
So T marks the mature mRNA on its way to a ribosome.
A eukaryotic gene is transcribed.
Which of the following is the last step before a ribosome can read the message?
- A. RNA polymerase releases the pre-mRNAReleasing the pre-mRNA comes before the cap, the tail and the splicing.
All of those still happen inside the nucleus. - B. Enzymes cut out the introns and join the exonsSplicing happens inside the nucleus, where no ribosome can reach the mRNA.
The mRNA still has to leave. - C. ✓ The mature mRNA leaves the nucleus through a pore
Why: The ribosomes sit outside the nucleus.
Enzymes in the nucleus finish the cap, the tail and the splicing first.
Then the mature mRNA leaves through a pore, and only then can a ribosome read it.
The drawing below shows a eukaryotic gene and its RNA at each step, with the parts lettered and the labels taken off.
(a) Describe the job of the part lettered X in the cell. (1 pt)
Frame The part lettered X is …
A ribosome recognizes the cap and binds there.
So the cap is what lets the mRNA be read.
- Award 1 point for: X is the 5′ cap, which the ribosome recognizes and binds, so the mRNA is translated.
(b) Explain how the ends written on the part lettered Y show which way RNA polymerase moves along it. (1 pt)
Frame The ends show this because …
On Y the 3′ end is written at the left and the 5′ end at the right.
So RNA polymerase moves along Y from left to right.
- Award 1 point for: RNA polymerase reads the template 3′ to 5′, and Y’s 3′ end is at the left, so RNA polymerase moves from left to right.
41A gene in a bacterium
A bacterium has one circular chromosome.
Where does the chromosome lie?
- A. Inside a nucleus, within the nuclear envelopeA bacterium has no nucleus.
Its chromosome lies free in the cytosol. - B. ✓ Free in the cytosol, with no envelope around it
Why: A bacterium is a prokaryotic cell.
It has no nucleus, so its one circular chromosome lies free in the cytosol.
Video: Watch: A gene in a bacterium
The bacterial column filling in beside the eukaryotic one: RNA polymerase on the gene in the cytosol, ribosomes settling on the mRNA while it is still being made, the bacterial column finishing first.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L16b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L16b.mp4
What happens to the message in a cell with no nucleus?
A bacterium’s gene lies in the cytosol, in the same fluid as its ribosomes.
Its mRNA gets no cap and no tail, and it has no introns to remove.
So ribosomes bind the mRNA and read it while RNA polymerase is still making it.
Now consider a gene in a bacterium.
A bacterium has no nucleus. So its gene lies in the cytosol, in the same fluid as its ribosomes.
RNA polymerase binds the promoter and reads the template strand, just like the eukaryote’s RNA polymerase. The mRNA grows from its 5′ end toward its 3′ end.
A bacterium’s genes have no introns. So there is nothing to cut out.
No enzyme joins a cap to the 5′ end, and no enzyme adds a tail to the 3′ end.
So the mRNA is ready the moment RNA polymerase makes it.
Ribosomes bind the mRNA near its 5′ end and begin reading it while RNA polymerase is still adding nucleotides at its 3′ end.
In a eukaryote the ribosomes are outside the nucleus. So they cannot reach the mRNA until it is mature and has left through a pore.
What you are expected to know Compare a prokaryote’s mRNA with a eukaryote’s: made in the cytosol beside the ribosomes, no cap, no tail, no introns, and read while it is still being made.
A bacterium transcribes one of its genes.
Where is the mRNA made?
- A. Inside a nucleusA bacterium has no nucleus.
Its gene lies in the cytosol, so the mRNA is made there. - B. ✓ In the cytosol, beside the ribosomes
Why: A bacterium’s gene lies in the cytosol, because the cell has no nucleus.
RNA polymerase makes the mRNA where the gene is.
So the mRNA is made in the cytosol, beside the ribosomes.
A bacterium and a eukaryotic cell each make an mRNA.
Which cell’s mRNA gets a poly-A tail?
- A. ✓ A eukaryote’s
- B. A bacterium’sNo enzyme adds a tail to a bacterium’s mRNA.
The tail is a eukaryote’s addition.
Why: Enzymes in a eukaryote’s nucleus add the poly-A tail.
A bacterium’s mRNA gets no tail.
So only the eukaryote’s mRNA has a poly-A tail.
A bacterium’s mRNA has just been made.
Does an enzyme join a cap to its 5′ end?
- A. YesA bacterium’s mRNA gets no cap.
Its ribosomes bind the mRNA without one. - B. ✓ No
Why: The cap is a eukaryote’s addition, made by enzymes in the nucleus.
A bacterium has neither the nucleus nor the cap.
So no enzyme joins a cap to its mRNA.
A bacterium’s gene is transcribed into mRNA.
Do enzymes cut introns out of the mRNA?
- A. YesA bacterium’s genes have no introns.
There is nothing to cut out. - B. ✓ No
Why: A bacterium’s genes have no introns.
So the mRNA has no introns, and no enzyme cuts anything out.
RNA polymerase is halfway along a bacterium’s gene.
Can a ribosome already be reading the mRNA?
- A. ✓ Yes
- B. NoThe gene and the ribosomes lie in the same cytosol.
A ribosome binds the mRNA’s 5′ end while the 3′ end is still being built.
Why: A bacterium has no nucleus, so the mRNA is made in the cytosol beside the ribosomes.
The mRNA needs no cap, no tail and no splicing.
So a ribosome binds its 5′ end and reads it while RNA polymerase is still building the 3′ end.
Suppose a bacterium and a liver cell each transcribe one of their own genes.
(a) State two differences between the bacterium’s mRNA and the liver cell’s mature mRNA. (1 pt)
The bacterium’s mRNA has no poly-A tail, and the liver cell’s mRNA has a tail on its 3′ end.
- Award 1 point for any two of: no cap against a 5′ cap; no tail against a poly-A tail; no introns to remove against introns cut out; made in the cytosol against made in the nucleus; read while being made against read only after it leaves the nucleus.
(b) Explain why a ribosome can read the bacterium’s mRNA the moment RNA polymerase makes it. (1 pt)
Frame A ribosome can read it at once because …
The bacterium has no nucleus, so the mRNA is already in the cytosol with the ribosomes.
So a ribosome binds the mRNA as soon as RNA polymerase makes its 5′ end.
- Award 1 point for: the mRNA needs no processing (no cap, tail or introns removed) AND there is no nucleus to leave, so ribosomes in the same cytosol bind it at once.
Here is the whole comparison as a table, a prokaryote’s message against a eukaryote’s, on five rows: where the cell makes it, the cap, the tail, the introns, and whether it is read while it is being made.
Go back to the two columns: the eukaryotic gene and its mature mRNA on the left, and a gene in a bacterium on the right.
In the eukaryote, RNA polymerase binds the promoter, reads the template strand 3′ to 5′, and releases the pre-mRNA.
Enzymes add the cap and the tail and cut out the introns. The mature mRNA then leaves through a pore and reaches a ribosome.
In the bacterium, the mRNA has no cap, no tail and no introns. Ribosomes reach it at once.
68Mixed practice mixed practice
Suppose an enzyme removes the poly-A tail from an mRNA in a eukaryotic cell. The mRNA keeps its cap.
Which of the following happens to the mRNA?
- A. ✓ Enzymes break the mRNA down sooner
- B. Ribosomes stop binding the mRNAThe ribosome recognizes the cap, and the cap is still there.
The tail protected the mRNA from the enzymes that break RNA down. - C. The mRNA lasts longerThe tail protected the mRNA from the enzymes that break RNA down.
Without it those enzymes reach the mRNA sooner.
Why: The cap is what the ribosome recognizes and binds, and the cap is still there.
The tail protected the mRNA from the enzymes that break RNA down.
Without the tail those enzymes break the mRNA down sooner.
A codon on an mRNA is drawn below. A strand is written from its 5′ end to its 3′ end, and both ends are marked. The anticodon is written from its 3′ end to its 5′ end, so each of its bases sits opposite its partner on the codon.
Which anticodon pairs with this codon?
- A. 3′-CCA-5′These are the codon’s own letters.
Each anticodon base is the partner of the codon base opposite it, not the same base. - B. 3′-UGG-5′These are the partner bases written from the wrong end.
The anticodon’s 3′ end lies opposite the codon’s 5′ end. - C. ✓ 3′-GGU-5′
Why: The anticodon pairs with the codon, so it is written 3′ to 5′ beneath the codon’s 5′ to 3′.
C pairs with G, C pairs with G, and A pairs with U.
So the anticodon is 3′-GGU-5′.
Suppose every pore in a eukaryotic cell’s nuclear envelope is closed just as the cell switches on a new gene.
Which step of expressing that gene stops?
- A. Transcription of the gene, inside the nucleusThe gene and RNA polymerase are inside the nucleus, so transcription continues.
- B. ✓ Translation of its mRNA, at the ribosomes
- C. Both steps at onceTranscription continues inside the nucleus.
Only the mRNA’s way out is blocked.
Why: Transcription happens inside the nucleus, so it continues.
The new mRNA cannot leave through a closed pore, so it never reaches a ribosome.
So translation of that gene’s mRNA stops.
A eukaryotic gene is transcribed from its start to its end into a transcript 3,000 nucleotides long. The mature mRNA’s coding part is 1,200 nucleotides long.
Which of the following explains the difference?
- A. ✓ Enzymes cut the introns out of the pre-mRNA
- B. RNA polymerase stopped before the end of the geneThe transcript covered the whole gene, so nothing was left uncopied.
- C. Enzymes removed the poly-A tailThe tail is added to the mRNA after transcription, not removed from it.
Why: The pre-mRNA carries introns between its exons.
Enzymes cut each intron out and join the exons.
So the mature mRNA’s coding part is shorter than the transcript.
A template strand is drawn below. A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end.
Which RNA does RNA polymerase build from this template?
- A. ✓ 5′-AAC GUG CUA-3′
- B. 5′-UUG CAC GAU-3′These are the template’s own letters with U for T.
The RNA pairs against the template: U opposite A, A opposite T, G opposite C, C opposite G. - C. 5′-AUC GUG CAA-3′These are the right partners written from the wrong end.
The RNA’s 5′ end lies opposite the template’s 3′ end, at the left.
Why: Each RNA base pairs with the template base opposite it: U opposite A, A opposite T, G opposite C, C opposite G.
The RNA’s 5′ end lies opposite the template’s 3′ end, at the left.
So the RNA is 5′-AAC GUG CUA-3′.
Suppose a gene’s stop sequence is deleted from the DNA.
What happens when RNA polymerase reaches the end of the gene?
- A. RNA polymerase releases the RNA at the gene’s end anywayRNA polymerase lets go of the DNA only after it passes a stop sequence.
With none there, nothing tells it to stop. - B. RNA polymerase turns round and reads the gene againRNA polymerase moves one way along the template.
It starts again only by binding the promoter. - C. ✓ RNA polymerase keeps adding nucleotides past the gene’s end
Why: RNA polymerase lets go of the DNA and releases the RNA only when it passes a stop sequence.
The stop sequence is gone.
So RNA polymerase keeps adding nucleotides past the gene’s end.
An mRNA is found with a cap on its 5′ end, a poly-A tail on its 3′ end and no introns.
In which kind of cell was the mRNA made?
- A. A prokaryotic cellA prokaryote’s mRNA gets no cap and no tail.
Only a eukaryote’s enzymes add them. - B. ✓ A eukaryotic cell
Why: Enzymes in a eukaryote’s nucleus add the cap and the tail and cut out the introns.
A prokaryote’s mRNA gets none of these.
So the mRNA was made in a eukaryotic cell.
RNA polymerase transcribes one gene on a chromosome 50 times in one hour.
What has happened to the gene’s DNA?
- A. The gene’s DNA has been used upTranscription makes RNA copies.
The DNA stays in the gene, and its two strands pair again behind the polymerase. - B. ✓ The gene’s DNA is unchanged
- C. There are now 50 copies of the gene’s DNAEach transcription makes an RNA copy, not a DNA copy.
The gene’s DNA is still the one copy it was.
Why: Transcription copies the gene into RNA.
The two DNA strands pair again behind RNA polymerase.
So the gene’s DNA is unchanged after 50 transcriptions.
Suppose a liver cell’s gene has two introns, and a bacterium’s gene is one exon with no intron. Each cell transcribes its gene, and RNA polymerase reads a stretch of template in each.
(a) A student says that enzymes must join a cap to the bacterium’s mRNA before a ribosome can bind it. Explain why the student is wrong. (1 pt)
Frame The student is wrong because …
A bacterium’s ribosomes bind the mRNA without a cap.
So a ribosome binds the mRNA as soon as RNA polymerase makes its 5′ end.
- Award 1 point for: a bacterium’s mRNA gets no cap, and its ribosomes bind the uncapped mRNA (as it is made).
(b) Explain why the coding part of the liver cell’s mature mRNA is shorter than the template stretch read, while the bacterium’s mRNA is as long as the stretch read. (2 pt)
Frame The coding part of the liver cell’s mature mRNA is shorter because …
Enzymes cut out each intron and join the exons end to end, so the coding part has lost the introns’ length.
The bacterium’s gene has no introns.
So nothing is cut out of its mRNA, and it stays as long as the stretch of template read.
- Award 1 point for: the liver cell’s pre-mRNA carries introns that enzymes cut out, joining the exons, so the mature mRNA’s coding part is shorter than the stretch read.
- Award 1 point for: the bacterium’s gene has no introns, so nothing is cut out of its mRNA and its length matches the stretch read.
APBIO-U06-P63 Practice questions: Topic 6.3
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 one gene 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: Transcription and RNA processing, summed up
Three RNAs and their jobs; RNA polymerase at the promoter; one template strand, read 3′ to 5′, the RNA built 5′ to 3′; the transcript released and the gene read again; a cap and a tail; introns out, exons joined; alternative splicing; a prokaryote's message against a eukaryote's.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-T63-summary.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-T63-summary.mp4
A student says: “The gene for a protein is in the nucleus, so the ribosomes that build the protein must be inside the nucleus too.”
Which statement about the student's claim is correct?
- A. The student is right: a ribosome must sit beside the gene it readsA ribosome never reads the gene's DNA.
It reads the mRNA copy, which leaves the nucleus through a pore. - B. ✓ The student is wrong: an mRNA copy of the gene travels out to ribosomes outside the nucleus
- C. The student is wrong: the gene's DNA travels out to the ribosomesThe DNA stays on its chromosome in the nucleus.
The cell copies the gene into mRNA, and the copy travels. - D. The student is right: every ribosome in a eukaryotic cell sits inside the nucleusRibosomes sit in the cytosol and on the rough ER, outside the nucleus.
The mRNA reaches them through the pores.
Why: The gene stays on its chromosome inside the nucleus.
The cell copies the gene into an mRNA.
The mRNA leaves through a pore and reaches a ribosome outside the nucleus.
So the ribosomes need not be inside the nucleus, and the student is wrong.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. The anticodon is written from its 3′ end to its 5′ end, so each of its bases sits opposite its partner on the codon. A tRNA's anticodon reads 3′-AAC-5′, as drawn.
Which codon on an mRNA does this tRNA pair with?
- A. 5′-AAC-3′These are the anticodon's own letters.
The codon carries each anticodon base's partner: A takes U, A takes U, C takes G. - B. 3′-UUG-5′The partners are right, but the codon's 5′ end sits opposite the anticodon's 3′ end.
Written to the convention, the codon reads 5′-UUG-3′. - C. 5′-CAA-3′This is the anticodon itself, written from its 5′ end.
The codon is the strand the anticodon pairs with, so its bases are the anticodon's partners. - D. ✓ 5′-UUG-3′
Why: Under each anticodon base write its partner: A takes U, A takes U, C takes G.
The codon runs the other way, so its 5′ end sits under the anticodon's 3′ end.
The codon reads 5′-UUG-3′.
Suppose a drug jams the rRNA of a ribosome's large subunit and leaves the small subunit untouched.
Which of the following can the ribosome still do?
- A. ✓ Bind an mRNA, but join no amino acids
- B. Join amino acids, but bind no mRNAThe large subunit's rRNA is what joins the amino acids.
With it jammed, joining stops; the small subunit still binds the mRNA. - C. Bind an mRNA and join amino acidsThe rRNA does the ribosome's work; the proteins keep it in shape.
With the large subunit's rRNA jammed, no amino acids are joined. - D. Neither bind an mRNA nor join amino acidsThe small subunit's rRNA binds the mRNA on its own.
Only the joining, the large subunit's job, stops.
Why: The small subunit's rRNA binds the mRNA.
The large subunit's rRNA joins each amino acid to the protein being built.
With the large subunit's rRNA jammed, the ribosome still binds an mRNA but joins no amino acids.
RNA polymerase is about to copy a gene into RNA.
Which of the following does it need to begin?
- A. An RNA primer with a free 3′ endDNA polymerase needs a primer; RNA polymerase starts a strand from nothing.
- B. Helicase to part the two strandsRNA polymerase parts the two strands itself at the start site.
- C. ✓ The gene's promoter
- D. A ribosome bound to the geneA ribosome reads mRNA in the cytoplasm.
Copying a gene needs RNA polymerase bound to the promoter.
Why: RNA polymerase binds the promoter, the DNA stretch just before the gene, and begins copying at the start site.
It needs no primer and no helicase.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Two genes, P and Q, overlap on one stretch of a chromosome. The two DNA strands of the stretch and the RNA each gene gives from it are shown.
Which strand is the template for each gene?
- A. The upper strand for gene P, and the lower strand for gene QGene P's RNA has the upper strand's letters with U for T, so the upper strand is P's NON-template strand.
The strand read is the other one. - B. ✓ The lower strand for gene P, and the upper strand for gene Q
- C. The lower strand for both genesGene Q's RNA has the lower strand's letters with U for T.
So the lower strand is Q's non-template strand, and Q's template is the upper strand. - D. The upper strand for both genesGene P's RNA matches the upper strand, so P's template is the lower strand.
Which strand is the template belongs to the gene, not the whole molecule.
Why: An RNA has the non-template strand's letters and pairs with the template strand.
Gene P's RNA matches the upper strand with U for T, so P's template is the lower strand.
Gene Q's RNA matches the lower strand, so Q's template is the upper strand.
Two RNA polymerases are copying one gene at the same time. The first started 20 seconds before the second.
Which statement is correct?
- A. ✓ The first trails the longer RNA now, and the two RNAs will be the same length when released
- B. The first trails the longer RNA now, and it will stay longer after releaseBoth polymerases copy the whole gene, from the promoter to the stop sequence.
Each RNA is the full length when released. - C. The second trails the longer RNA, because it moves faster to catch upThe two polymerases move at the same rate.
The one that started earlier has passed more template bases, so its RNA is longer. - D. The two RNAs are the same length at every momentEach polymerase adds one nucleotide per template base it passes.
The first has passed more bases, so its RNA is longer until both are released.
Why: Each polymerase started at the promoter and adds one nucleotide for every template base it passes.
The first has passed more bases, so its RNA is longer now.
Each polymerase lets go at the stop sequence, having copied the whole gene, so the two released RNAs are the same length.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. A template strand reads 3′-CGT ACA GGT-5′, as drawn.
Written with both ends marked, what does the RNA read?
- A. 5′-CGU ACA GGU-3′These are the template's own letters with U for T.
The RNA carries each template base's partner. - B. 3′-GCA UGU CCA-5′The partners are right, but the RNA's 5′ end sits under the template's 3′ end, at the left.
- C. 5′-ACC UGU ACG-3′These are the right partners in the wrong order.
The RNA's 5′ end is at the left, under the template's 3′ end. - D. ✓ 5′-GCA UGU CCA-3′
Why: Under each template base write its RNA partner: C gives G, G gives C, T gives A, A gives U.
The RNA's 5′ end sits under the template's 3′ end.
So the RNA reads 5′-GCA UGU CCA-3′.
A ribosome begins to read a eukaryotic mRNA.
Which statement names the end the ribosome recognizes first and what that end carries?
- A. The 3′ end, which carries a stretch of adenine nucleotidesThe stretch of adenine nucleotides is the poly-A tail, at the 3′ end.
The ribosome recognizes the cap at the 5′ end. - B. The 5′ end, which carries a stretch of adenine nucleotidesThe 5′ end carries the cap, a modified guanine nucleotide.
The adenine stretch is the tail at the 3′ end. - C. ✓ The 5′ end, which carries a modified guanine nucleotide
- D. The 3′ end, which carries a modified guanine nucleotideThe modified guanine nucleotide is the cap, at the 5′ end.
The ribosome recognizes it there.
Why: Enzymes join the 5′ cap, a modified guanine nucleotide, to the mRNA's 5′ end.
The ribosome recognizes the cap and grips the mRNA there.
So the ribosome recognizes the 5′ end first.
A eukaryotic gene is transcribed into a pre-mRNA 3,600 nucleotides long. Its introns total 2,750 nucleotides.
Calculate the coding length of the mature mRNA.
- A. ✓ 850 nucleotides
- B. 2,750 nucleotidesThis is the intron total, the stretch that is cut out.
The coding length is what remains. - C. 3,600 nucleotidesThis is the whole pre-mRNA.
The introns are cut out of it, so the mature mRNA is shorter. - D. 6,350 nucleotidesAdding the introns to the pre-mRNA counts them twice.
They are cut out, so they are taken away.
Why: The pre-mRNA carries the exons and the introns.
Enzymes cut out the introns, so the coding length is the pre-mRNA's length minus the intron total.
nucleotides.
A liver cell's mature mRNA and a bacterium's mRNA are compared.
Which of the following is carried by the liver cell's mature mRNA and by no bacterium's mRNA?
- A. Uracil in place of thymineEvery RNA carries uracil, in a bacterium as in a liver cell.
- B. A 5′ end and a 3′ endEvery strand has a 5′ end and a 3′ end.
What a bacterium's mRNA lacks is the cap on its 5′ end. - C. A copy of one gene's instructionsEvery mRNA carries a gene's instructions to a ribosome.
The cap is the eukaryote's addition. - D. ✓ A 5′ cap
Why: In a eukaryote, enzymes in the nucleus join a cap to the pre-mRNA's 5′ end.
A bacterium has no nucleus and no capping enzyme, so its mRNA has no cap.
So the cap is carried by the liver cell's mRNA and by no bacterium's.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. A eukaryotic gene is drawn as a gene map with every exon and intron length written on it, in nucleotides. Beneath it are the first nine bases of the template strand. A skin cell's mature mRNA from this gene keeps every exon. A gland cell's mature mRNA from the same gene has a coding length of 735 nucleotides.
(a) Identify the RNA that RNA polymerase makes from the nine template bases drawn, written with both ends marked. (1 pt)
Frame The RNA reads 5′-… … …-3′.
Hint Pair each template base with its RNA partner, U opposite A. The RNA's 5′ end sits under the template's 3′ end, at the left.
- Award 1 point for: 5′-ACU GGC AAG-3′.
- Accept the same nine letters without the spaces.
(b) Calculate the coding length of the skin cell's mature mRNA. (1 pt)
Frame The coding length is … nucleotides.
Hint Add the exon lengths only. The introns are cut out.
Answer: 885 nucleotides (tolerance ±0)
- Award 1 point for: 885 nucleotides.
(c) Determine which exon the gland cell's enzymes skipped. Justify your answer with the lengths. (1 pt)
Frame The gland cell skipped exon …, because …
Hint Subtract the gland cell's coding length, 735, from your answer to (b). Find the exon whose length equals the difference.
- Award 1 point for: exon 2 AND the ground — 885 − 735 = 150, the length of exon 2.
- Accept a decision that follows correctly from the student's own answer to (b).
(d) Predict how the gland cell's protein differs from the skin cell's protein. (1 pt)
Frame The gland cell's protein …
Hint Each exon carries the instructions for one part of the protein.
- Award 1 point for: the gland cell's protein lacks the part made from exon 2 (the rest is shared).
The drawing shows a eukaryotic gene with RNA polymerase on it, the RNA as released, and the mature mRNA. Five parts are lettered N to R.
(a) Identify the part lettered N, and describe what it does at the part lettered O. (1 pt)
It binds O, the promoter, separates the two DNA strands at the start site and begins building RNA.
- Award 1 point for: N is RNA polymerase AND it binds the promoter (O) to begin copying the gene.
(b) Explain what determines the length of the row lettered P. (1 pt)
So the length of P, the RNA it releases, is the length of the gene's exons and introns together.
- Award 1 point for: RNA polymerase transcribes the whole gene (exons and introns), so P is the gene's full length.
(c) Identify the part lettered Q, and state what it is made of. (1 pt)
The cap is one modified guanine nucleotide.
- Award 1 point for: the 5′ cap AND a modified guanine nucleotide. Accept 'GTP cap'.
(d) Explain why the row lettered R is shorter than the row lettered P. (1 pt)
Enzymes cut each intron out and join the exons end to end.
So R, the mature mRNA, carries the exons only and is shorter by the introns' length.
- Award 1 point for: enzymes cut the introns out of P and join the exons, so R carries the exons only.
- The point is for the mechanism (the introns cut out and the exons joined), not for the bare word 'introns'.
APBIO-U06-T63 End-of-topic test: Transcription and RNA Processing
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.
In a eukaryotic cell, one gene's newly made RNA is marked so that it can be followed. The table shows where the marked RNA is over the next thirty minutes.
Which statement explains the pattern in the table?
- A. ✓ The RNA leaves the nucleus through pores and reaches ribosomes in the cytoplasm
- B. The gene's DNA leaves the nucleus and is read at the ribosomesThe DNA stays on its chromosome inside the nucleus.
The RNA copy is what leaves. - C. The RNA is made in the cytoplasm and carried into the nucleus to be storedThe RNA starts inside the nucleus, where the gene is, and the share there falls.
It moves out, not in. - D. Ribosomes inside the nucleus read the RNA and send the finished protein outRibosomes sit outside the nucleus.
The RNA itself leaves through the pores to reach them.
Why: The gene's RNA is made in the nucleus, so at first every marked molecule is there.
The share in the nucleus falls as the share in the cytoplasm rises.
So the RNA leaves the nucleus, through the pores, and reaches the ribosomes in the cytoplasm.
An mRNA is drawn with its 3′ end at the left, as shown. A ribosome is about to read it.
At which end of the drawn mRNA does the ribosome start?
- A. At the left-hand end, marked 3′A ribosome reads from the 5′ end toward the 3′ end.
The 3′ end is where it finishes. - B. At the left-hand end, whichever end is marked thereA drawing can show a strand either way round.
The ribosome starts at the 5′ end, wherever it is drawn. - C. ✓ At the right-hand end, marked 5′
- D. At either end, whichever it reaches firstA ribosome starts at one end only, the 5′ end.
The written ends show which end that is.
Why: A ribosome reads an mRNA in one direction, from its 5′ end toward its 3′ end.
On this drawing the 5′ end is written at the right.
So the ribosome starts at the right-hand end.
The codon 5′-GAU-3′ is drawn.
Which anticodon pairs with this codon?
- A. 3′-GAU-5′These are the codon's own letters.
An anticodon carries each codon base's partner: G takes C, A takes U, U takes A. - B. ✓ 3′-CUA-5′
- C. 5′-CUA-3′These are the right partners, but the anticodon's 3′ end sits opposite the codon's 5′ end.
Written from its 3′ end, it reads 3′-CUA-5′. - D. 3′-CTA-5′An RNA never carries thymine.
Opposite the codon's A sits U.
Why: Under each codon base write its partner: G takes C, A takes U, U takes A.
The anticodon runs the other way, so its 3′ end sits under the codon's 5′ end.
It reads 3′-CUA-5′.
The table describes four RNAs found in one eukaryotic cell.
Which row describes ribosomal RNA (rRNA)?
- A. WAn open strand of thousands of nucleotides that a ribosome reads is a messenger RNA (mRNA).
- B. XA short strand folded into an L with an amino acid at its 3′ end is a transfer RNA (tRNA).
- C. YA short RNA paired to a DNA strand at a fork is the RNA primer of DNA replication.
- D. ✓ Z
Why: Ribosomal RNA is the RNA a ribosome is built from.
Long rRNA strands fold up with proteins into the two subunits.
The rRNA binds the mRNA and joins the amino acids.
Row Z describes that.
An RNA binds a small molecule only when it is folded. In the folded RNA, nucleotides 12–16 lie against nucleotides 31–35, the two stretches lying the opposite way round. A scientist made two changed versions of the RNA and measured the share of molecules that bound the small molecule. The table shows the results.
Which conclusion do the results support?
- A. The changed nucleotide in change 1 is the one that touches the small moleculeChange 2 carries the same changed nucleotide at position 14 and still binds.
So that nucleotide does not touch the small molecule; it pairs with position 33. - B. The number of guanine bases in the RNA sets how well it bindsAll three RNAs carry the same number of guanine bases.
What differs is whether positions 14 and 33 can pair. - C. ✓ The two regions pair with each other, and the pairing holds the fold
- D. The small molecule binds to the letters at positions 12–16, whatever the RNA's foldChange 1 and change 2 share the letters at positions 12–16, yet only change 2 binds well.
What differs is whether position 14 pairs with position 33: the fold decides.
Why: In the RNA as found, every base of one stretch has its partner in the other, so the two stretches pair and hold the fold.
Change 1 puts U opposite U at one position: that pair is lost, and binding falls.
Change 2 restores a partner there, and binding returns.
Genes P and Q sit side by side on one chromosome. A short stretch of DNA is deleted just before gene P. Afterward, RNA polymerase makes no RNA from gene P and still makes RNA from gene Q.
Which stretch was deleted?
- A. ✓ Gene P's promoter
- B. Gene P's template strandDeleting one strand alone is not possible; a deletion removes a stretch of both strands.
The stretch just before the gene where RNA polymerase binds is the promoter. - C. Gene P's stop sequenceThe stop sequence sits at the far end of the gene, where RNA polymerase lets go.
The deleted stretch sits just before the gene. - D. Gene Q's promoterGene Q is still copied, so its promoter is still there.
The stretch just before gene P is gene P's promoter.
Why: RNA polymerase binds a gene's promoter, a stretch of DNA just before the gene, before it copies the gene.
With that stretch deleted, RNA polymerase never binds before gene P, so it makes no RNA from it.
Gene Q keeps its own promoter, so it is still copied.
Four test tubes each hold a gene with its promoter and the enzyme and nucleotides listed in the table. The last column shows whether new RNA appeared.
Which statement do the four tubes support?
- A. RNA polymerase needs a primer to start, as DNA polymerase doesTube 1 holds no primer, and RNA still appears.
RNA polymerase starts a strand from nothing. - B. ✓ RNA polymerase makes RNA with no primer and no helicase
- C. DNA polymerase can build RNA when it is given RNA nucleotidesTube 4 gives DNA polymerase RNA nucleotides, and no RNA appears.
Only RNA polymerase joins RNA nucleotides. - D. RNA polymerase joins DNA nucleotides into an RNA strandTube 2 gives RNA polymerase DNA nucleotides, and no RNA appears.
RNA polymerase joins RNA nucleotides only.
Why: Tube 1 holds only RNA polymerase, RNA nucleotides and a gene with its promoter, and RNA appears.
So RNA polymerase needs no primer and no helicase.
Tubes 2 and 4 show that only RNA polymerase, given RNA nucleotides, makes RNA.
The two DNA strands of a short stretch of a gene, X and Y, and the RNA made from that stretch are shown.
Which strand did RNA polymerase read as the template, and how do the sequences show it?
- A. ✓ Strand X, because each RNA base is the partner of a base on strand X
- B. Strand Y, because the RNA has strand Y's letters with U in place of TThe RNA carries the letters of the strand that was NOT read.
Matching letters mark strand Y as the non-template strand. - C. Both strands, because each strand is the partner of the otherRNA polymerase reads one strand of a gene only.
The RNA pairs with strand X and matches strand Y. - D. Strand Y, because each RNA base is the partner of a base on strand YU pairs with A, never with T.
Read against strand Y, base for base, the RNA's letters are strand Y's own letters with U for T, not their partners.
Why: RNA polymerase pairs each RNA base against the template strand.
Read against strand X, base for base, the RNA's letters are strand X's partners.
The RNA matches strand Y letter for letter, with U for T, so strand Y is the non-template strand.
So strand X was the template.
RNA polymerase is part-way along a gene, with a growing RNA behind it.
Which statement describes the ends at that moment?
- A. The RNA's 5′ end is at the polymerase, and the polymerase is moving toward the template's 3′ endNew nucleotides join the RNA's 3′ end, so that end is at the polymerase.
The polymerase reads the template toward its 5′ end. - B. The RNA's 3′ end is at the polymerase, and the polymerase is moving toward the template's 3′ endRNA polymerase reads the template from its 3′ end toward its 5′ end.
It moves toward the template's 5′ end. - C. The RNA's 5′ end is at the polymerase, and the polymerase is moving toward the template's 5′ endThe RNA's 5′ end was made first and trails behind.
The 3′ end, where nucleotides join, is at the polymerase. - D. ✓ The RNA's 3′ end is at the polymerase, and the polymerase is moving toward the template's 5′ end
Why: RNA polymerase joins each new nucleotide to the RNA's free 3′ end, so the 3′ end is at the enzyme.
It reads the template strand from its 3′ end toward its 5′ end.
So it moves toward the template's 5′ end.
A template strand reads 3′-ACG TTA GCC-5′, as drawn.
Written with both ends marked, what does the RNA read?
- A. 5′-ACG UUA GCC-3′These are the template's own letters with U for T.
The RNA carries each template base's partner: A takes U, C takes G, G takes C, T takes A. - B. 3′-UGC AAU CGG-5′The partners are right, but the RNA's 5′ end sits opposite the template's 3′ end, at the left.
So the RNA reads 5′-UGC AAU CGG-3′. - C. ✓ 5′-UGC AAU CGG-3′
- D. 5′-UGC AAU GGG-3′Six of the nine partners are right, but the seventh base is G.
The template's seventh base is G, and G takes C, so the RNA's seventh base is C.
Why: Under each template base write its RNA partner: A gives U, C gives G, G gives C, T gives A.
The RNA's 5′ end sits under the template's 3′ end, at the left.
So the RNA reads 5′-UGC AAU CGG-3′.
One gene is drawn with five RNA polymerases along it at once. Each trails the RNA it has made so far.
At which end of the gene is the promoter?
- A. ✓ At the right-hand end, where the RNAs are shortest
- B. At the left-hand end, where the RNAs are longestA polymerase near the promoter has only just started, so its RNA is short.
The long RNAs belong to the polymerases furthest from the promoter. - C. In the middle of the gene, where the RNAs are of middling lengthEvery polymerase starts at the promoter and moves one way along the gene.
The promoter is at an end, where the RNAs are shortest. - D. At either end, because RNA polymerase can start from both ends of a geneA gene has one promoter, at one end.
Every polymerase on the gene started there.
Why: Every RNA polymerase on the gene bound the same promoter and started there.
Each polymerase adds one nucleotide for every template base it passes, so the further it has moved, the longer its RNA.
The shortest RNAs trail the polymerases that started last, nearest the promoter, at the right.
A gene's non-template strand reads 5′-GCTACA-3′, as drawn.
Written with both ends marked, what does the RNA read?
- A. 5′-CGAUGU-3′These are the partners of the non-template strand's bases.
The RNA carries the non-template strand's own letters, with U for T. - B. ✓ 5′-GCUACA-3′
- C. 3′-GCUACA-5′The non-template strand and the RNA lie the same way round.
So the RNA's 5′ end is at the left. - D. 5′-ACAUCG-3′These are the right letters in the wrong order.
The RNA copies the non-template strand's letters in the same order, from its 5′ end.
Why: The RNA has the non-template strand's letters in the same order, with U in place of T.
GCTACA with each T changed to U is GCUACA.
The ends stay the same way round, so the RNA reads 5′-GCUACA-3′.
In a eukaryotic cell, enzymes in the nucleus add two things to a pre-mRNA before it leaves.
Which of the following do the enzymes join to the pre-mRNA's 3′ end?
- A. One modified guanine nucleotideOne modified guanine nucleotide is the cap.
Enzymes join the cap to the 5′ end. - B. A stretch of about 200 thymine nucleotidesAn RNA never carries thymine.
The tail joined to the 3′ end is adenine nucleotides. - C. A second RNA strand, paired with the first along its lengthAn mRNA stays a single strand.
The enzymes add a tail of adenine nucleotides at its 3′ end. - D. ✓ A stretch of about 200 adenine nucleotides
Why: Enzymes in the nucleus join a cap, one modified guanine nucleotide, to the pre-mRNA's 5′ end.
Other enzymes add the poly-A tail, a stretch of about 200 adenine nucleotides, at its 3′ end.
So the 3′ end gets the adenine stretch.
Four copies of one gene's mRNA are made in a test tube. Each copy has or lacks a 5′ cap, and each copy has or lacks a poly-A tail. Each copy is put into a eukaryotic cell. The table shows how much protein each cell made in the first ten minutes and how long each mRNA lasted.
Which row is the mRNA with a 5′ cap and no poly-A tail?
- A. WRow W's mRNA made a lot of protein and lasted many hours.
Row W has a cap for the ribosomes and a tail against the RNA-breaking enzymes. - B. ✓ X
- C. YRow Y's mRNA made almost no protein, so few ribosomes gripped it: it has no cap.
It lasted many hours, so it has a tail. - D. ZRow Z's mRNA made almost no protein and was gone within the hour.
It has no cap and no tail.
Why: A ribosome grips an mRNA at its 5′ cap, so a capped mRNA gives a lot of protein at first.
The poly-A tail keeps the RNA-breaking enzymes busy for hours, so an mRNA with no tail is gone within the hour.
Row X: much protein, gone within an hour.
In a eukaryotic cell, a gene's pre-mRNA is 3,200 nucleotides long. Its mature mRNA has a coding length of 3,200 nucleotides.
Which statement about this gene is correct?
- A. Enzymes cut the introns out of the gene's DNA before it was transcribedIntrons are cut out of the RNA copy, never out of the DNA.
A pre-mRNA as long as its mature mRNA carried no introns to cut. - B. Enzymes cut the introns out after the mRNA left the nucleusSplicing happens in the nucleus, before the mRNA leaves.
Here nothing was cut, so the gene has no introns. - C. ✓ The gene has no introns
- D. The gene's exons were cut out and its introns keptExons are the stretches kept; introns are the stretches cut out.
Nothing was cut from this pre-mRNA.
Why: The mature mRNA's coding length is the exon total.
Introns are cut out of the pre-mRNA, so a pre-mRNA with introns is longer than its mature mRNA.
Here the two lengths are equal, so nothing was cut out.
So the gene has no introns.
The drawing shows a eukaryotic gene with RNA polymerase on it, the RNA as released, and the mature mRNA. Four parts are lettered.
Which lettered part is a stretch that enzymes cut out before the mRNA leaves the nucleus?
- A. JJ is the promoter, the DNA stretch RNA polymerase binds.
Nothing is cut out of the DNA. - B. KK is the 5′ cap, added to the RNA's 5′ end.
It is added, not cut out. - C. LL is the poly-A tail, added at the RNA's 3′ end.
It is added, not cut out. - D. ✓ M
Why: M marks a thin stretch of the released RNA between two filled boxes.
The filled boxes are exons and the thin stretches between them introns.
Enzymes cut each intron out and join the exons, so the mature mRNA has no thin stretches.
So M is an intron.
The table describes four mRNAs, one per row.
Which row describes an mRNA made by a bacterium?
- A. Row WRow W's mRNA carries a cap and a tail.
A bacterium's enzymes add neither, so its mRNA has no cap and no tail. - B. Row XRow X's mRNA carries a 5′ cap.
A bacterium has no capping enzyme, so its mRNA has no cap. - C. ✓ Row Y
- D. Row ZRow Z's mRNA carries a poly-A tail.
A bacterium adds no tail to its mRNA.
Why: In a eukaryote, enzymes in the nucleus add a 5′ cap and a poly-A tail to the pre-mRNA.
A bacterium has no nucleus and no such enzymes, so its mRNA carries no cap and no tail.
Row Y carries no cap and no tail.
A eukaryotic gene is drawn as a gene map with every exon and intron length written on it, in nucleotides. Beneath it are the first nine bases of the template strand. A liver cell and a kidney cell each transcribe this gene. The liver cell's mature mRNA keeps every exon. The kidney cell's mature mRNA has a coding length of 765 nucleotides.
(a) Identify the RNA that RNA polymerase makes from the nine template bases drawn, written with both ends marked. (1 pt)
The RNA's 5′ end sits under the template's 3′ end.
So the RNA reads 5′-CAG GUU ACG-3′.
- Award 1 point for: 5′-CAG GUU ACG-3′ (both ends marked).
- Accept the same nine letters without the spaces.
Slip Writing 5′-GUC CAA UGC-3′: the template's own letters with U for T, not their partners.
(b) Calculate the coding length of the liver cell's mature mRNA. (1 pt)
Answer: 960 nucleotides (tolerance ±0)
270 + 195 + 495 = 960 nucleotides.
- Award 1 point for: 960 nucleotides.
(c) Determine which exon the kidney cell's enzymes skipped. Justify your answer with the lengths. (1 pt)
Exon 2 is 195 nucleotides long, and no other exon is.
So the kidney cell skipped exon 2.
- Award 1 point for: exon 2, WITH the lengths — 960 − 765 = 195, the length of exon 2 (or 270 + 495 = 765).
- Accept consistent working from an incorrect (b) total: the exon whose length equals the student's own (b) total minus 765, with that subtraction shown.
Slip Subtracting an intron length. Introns are cut out in both cells; only exon lengths decide the coding length.
(d) Explain why the two cells' pre-mRNAs are the same length although the two mature mRNAs differ. (1 pt)
So both pre-mRNAs are the full length of the gene, 2,250 nucleotides.
The two mRNAs differ only afterward, when the kidney cell's enzymes cut out exon 2 along with the introns.
- Award 1 point for: both cells transcribe the same whole gene (exons and introns) into the pre-mRNA, and the difference arises at splicing, when the kidney cell cuts out exon 2 as well as the introns.
Slip Saying the kidney cell's gene is shorter. The DNA is the same in both cells; splicing, not the gene, differs.
A gene in a plant's leaf cells has four exons. In leaf cells in the light, the mature mRNA from this gene carries exons 1, 2, 3 and 4. In the same leaf cells in the dark, the mature mRNA carries exons 1, 2 and 4. The two mature mRNAs are drawn.
(a) Describe how the leaf cell's enzymes treat the pre-mRNA of this gene differently in the light and in the dark. (1 pt)
In the dark the enzymes cut out exon 3 along with the introns beside it and join exons 1, 2 and 4.
- Award 1 point for: the same pre-mRNA is spliced keeping different sets of exons — all four in the light, exons 1, 2 and 4 in the dark (exon 3 cut out with the introns). Accept 'alternative splicing' with the exon sets named.
(b) Explain why the protein made in the dark is related to the protein made in the light but differs from it. (1 pt)
Both mRNAs carry exons 1, 2 and 4, so both proteins share the parts made from those exons.
The dark mRNA lacks exon 3, so the dark protein lacks the part made from exon 3.
- Award 1 point for: the dark protein lacks the part made from exon 3 (the parts made from exons 1, 2 and 4 are shared).
Slip Saying the dark protein is unrelated. Three of the four exons are shared, so most of the protein is the same.
(c) Predict what a comparison of this gene's DNA in a leaf cell in the light and a leaf cell in the dark would show. (1 pt)
- Award 1 point for: the DNA is identical in the two cells (exon 3 is present in the gene in both).
Slip Predicting that exon 3 is missing from the gene in the dark. The exon is missing from the mRNA, not from the DNA.
(d) Justify your prediction. (1 pt)
Enzymes cut exon 3 out of the pre-mRNA, not out of the gene.
So the gene's DNA keeps exon 3 in the dark as in the light, and the same gene gives two different mRNAs.
- Award 1 point for: enzymes cut stretches out of the RNA copy (the pre-mRNA), never out of the DNA, so the gene is unchanged and one gene gives the two mRNAs.
APBIO-U06-L17 Where proteins are built
Suppose two cells each switch on a gene for the same kind of enzyme at the same moment. In the bacterium, the enzyme appears in about a minute. In the human cell it takes much longer.
Both cells have ribosomes. What is the bacterium skipping?
Unit 6 · Gene Expression and Regulation
1Free in the cytoplasm, or on the rough ER
Where are proteins built, and why is a bacterium so quick?
A pancreas cell has rough ER just outside its nucleus.
Where do the rough ER’s ribosomes sit?
- A. ✓ On the ER’s outer surface, facing the cytoplasm
- B. Inside the ER, in its interiorThe ER’s interior holds folded protein, not ribosomes.
The ribosomes sit on the outer surface, facing the cytoplasm.
Why: The rough ER is the part of the ER with ribosomes on its outer surface.
That surface faces the cytoplasm.
A ribosome on the rough ER is building insulin, a protein the pancreas cell exports.
Where does the ribosome pass the growing insulin chain?
- A. Into the cytoplasmA protein for the cytoplasm is released into the cytoplasm.
Insulin is exported, and its chain passes into the ER. - B. ✓ Into the inside of the ER
- C. Straight out of the cellInsulin leaves the cell only at the end of its route, in a vesicle.
The ribosome passes the chain into the ER first.
Why: The ribosome sits on the rough ER.
The ribosome pushes the insulin chain through the ER membrane into the inside of the ER.
There the chain folds.
Ribosomes build proteins in the cytoplasm of every cell.
In a eukaryote, ribosomes also sit on the outer face of the rough ER. Those ribosomes build the proteins the cell exports.
A bacterium has no nucleus. Its mRNA needs no cap, no tail and no splicing.
So a bacterium’s ribosomes attach to the mRNA while RNA polymerase is still making it.
A eukaryote’s mRNA must be finished, processed and carried out of the nucleus before a ribosome can touch it.
Two questions follow: where do the ribosomes sit, and when do they start?
Video: Watch: Free in the cytoplasm, or on the rough ER
The pancreas cell drawn large; the insulin mRNA reaching a ribosome on the rough ER and the insulin chain passing into the ER; a glycolysis enzyme built on a ribosome floating free and released into the cytoplasm; the bacterium with every ribosome free.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L17a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L17a.mp4
Here is the pancreas cell again, with the route from the insulin gene to insulin.
The ribosome that builds insulin sits on the rough ER, on the face that touches the cytoplasm.
As the ribosome joins the insulin chain together, it passes the chain through the ER membrane into the ER’s interior.
Insulin is a protein the pancreas cell exports. A cell builds every protein it exports on a ribosome on the rough ER.
Now consider an enzyme of glycolysis in the same pancreas cell. Glycolysis happens in the cytoplasm.
So this enzyme does its job in the cytoplasm. The cell never exports it.
The pancreas cell builds this enzyme on a ribosome that floats free in the cytoplasm, not attached to the ER. The free ribosome releases the finished enzyme into the cytoplasm.
So a eukaryotic cell builds proteins in two places: on ribosomes free in the cytoplasm, and on ribosomes attached to the rough ER.
Where a protein is built depends on where the protein is going. A protein for the cytoplasm is built on a free ribosome.
A protein for export is built on a rough-ER ribosome. Its growing chain passes into the ER, and from there vesicles carry it out of the cell.
Now consider a bacterium. A bacterium has no nucleus and no ER.
Every one of a bacterium’s ribosomes floats free in the cytoplasm. So a bacterium builds every protein, exported or not, on a free ribosome.
Translation happens on ribosomes in every cell, prokaryote or eukaryote. Only a eukaryote also has ribosomes on the rough ER.
What you are expected to know State where translation happens: on ribosomes free in the cytoplasm in every cell, and, in a eukaryote, also on ribosomes attached to the rough ER, whose growing proteins pass into the ER.
A cell in the stomach wall builds an enzyme it exports into the stomach.
Where is the enzyme built?
- A. On a free ribosome in the cytoplasmThe stomach cell exports this enzyme.
A cell builds every protein it exports on a rough-ER ribosome, and the chain passes into the ER. - B. ✓ On a ribosome on the rough ER
Why: The enzyme is exported.
A protein for export is built on a ribosome on the rough ER, and its chain passes into the ER.
So the enzyme is built on a rough-ER ribosome.
A liver cell builds an enzyme that works in its own cytoplasm.
Where is the enzyme built?
- A. ✓ On a free ribosome in the cytoplasm
- B. On a ribosome on the rough ERA rough-ER ribosome passes its chain into the ER, on the way out of the cell.
This enzyme stays in the cytoplasm, so a free ribosome builds it.
Why: The enzyme does its job in the cytoplasm and is never exported.
A protein for the cytoplasm is built on a free ribosome.
So the enzyme is built on a free ribosome.
A muscle cell builds a protein that breaks down the glycogen stored in its cytoplasm.
Where is the protein built?
- A. ✓ On a free ribosome in the cytoplasm
- B. On a ribosome on the rough ERA rough-ER ribosome passes its chain into the ER, on the way out of the cell.
This protein works in the cytoplasm, so a free ribosome builds it.
Why: The protein works in the cytoplasm and is never exported.
A protein for the cytoplasm is built on a free ribosome.
So the protein is built on a free ribosome.
A skin cell builds collagen, a protein it exports to the space outside the cell.
Where is the collagen built?
- A. On a free ribosome in the cytoplasmCollagen leaves the cell.
A protein for export is built on a rough-ER ribosome, and its chain passes into the ER. - B. ✓ On a ribosome on the rough ER
Why: Collagen is exported.
A protein for export is built on a ribosome on the rough ER.
So the collagen is built on a rough-ER ribosome.
A liver cell builds albumin, a protein it exports into the blood.
Where is the albumin built?
- A. On a free ribosome in the cytoplasmAlbumin leaves the cell.
A protein for export is built on a rough-ER ribosome, and its chain passes into the ER. - B. ✓ On a ribosome on the rough ER
Why: Albumin is exported.
A protein for export is built on a ribosome on the rough ER.
So the albumin is built on a rough-ER ribosome.
A bacterium builds an enzyme.
Where is the enzyme built?
- A. ✓ On a free ribosome in the cytoplasm
- B. On a ribosome on the rough ERA bacterium has no ER.
Every one of its ribosomes floats free in the cytoplasm.
Why: A bacterium has no ER.
Every one of its ribosomes floats free in the cytoplasm.
So the enzyme is built on a free ribosome.
32Read while it is still being made
Suppose a bacterium and a liver cell each carry a gene for the same kind of enzyme.
Which cell keeps its DNA inside a nucleus?
- A. The bacteriumA bacterium has no nucleus.
Its DNA lies free in the cytoplasm, in the nucleoid. - B. Both cellsA bacterium is a prokaryote and has no nucleus.
Only the liver cell, a eukaryotic cell, keeps its DNA inside a nucleus. - C. ✓ The liver cell
Why: A liver cell is a eukaryotic cell, so its DNA is inside a nucleus.
A bacterium is a prokaryote and has no nucleus.
In a eukaryote, a pre-mRNA gets a 5′ cap, a poly-A tail, and its introns cut out.
How many of these three does a bacterium’s mRNA get?
- A. ✓ None
- B. OneNo enzyme joins a cap to a bacterium’s mRNA, none adds a tail, and its genes have no introns.
It gets none of the three. - C. All threeThe cap, the tail and splicing are a eukaryote’s additions, made in the nucleus.
A bacterium’s mRNA gets none of the three.
Why: A bacterium’s mRNA gets no cap and no tail.
Its genes have no introns, so there is nothing to cut out.
So it gets none of the three.
Video: Watch: Read while it is still being made
A bacterial gene in the cytoplasm; RNA polymerase moving along it with the mRNA growing behind it; a ribosome binding near the free 5′ end and reading while the polymerase is still moving; more ribosomes strung along the mRNA like beads; then the human cell, where the mRNA is finished, capped, tailed, spliced and carried out through a pore before a ribosome touches it.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L17b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L17b.mp4
Suppose a bacterium switches on a gene for an enzyme. RNA polymerase binds the promoter and begins making the mRNA.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
A bacterium has no nucleus. So its gene lies in the cytoplasm, in the same fluid as its ribosomes.
RNA polymerase builds the mRNA from its 5′ end toward its 3′ end. So the mRNA’s 5′ end is finished first, while RNA polymerase is still adding nucleotides at the 3′ end.
No enzyme joins a cap to that 5′ end. The 5′ end is free, and a ribosome can bind the mRNA near it as it is.
A ribosome reads an mRNA from its 5′ end. So a ribosome binds near the free 5′ end and begins translating, while RNA polymerase is still transcribing the rest of the gene.
More ribosomes bind the mRNA behind the first one. Each ribosome builds its own copy of the enzyme as it moves along the mRNA.
Several RNA polymerases read the same gene, one behind another. Here is the gene with three of them, and every growing mRNA already carries ribosomes.
So in a bacterium, transcription and translation happen at the same time, on the same mRNA.
The first finished enzyme molecules appear within about a minute of the gene switching on.
Now consider a gene for the same kind of enzyme in a human liver cell.
The gene lies inside the nucleus. The ribosomes lie outside it, in the cytoplasm and on the rough ER.
The nuclear envelope stands between the gene and every ribosome. So no ribosome can reach the mRNA while RNA polymerase is making it.
RNA polymerase finishes the pre-mRNA. Then enzymes join the 5′ cap, add the poly-A tail and cut out the introns.
Only the finished, processed mRNA leaves the nucleus through a pore. Only then does a ribosome bind it.
So in a human cell, transcription finishes before translation begins. In a bacterium, translation begins while transcription is still going on.
What makes the bacterium quick is when its ribosomes start, not how fast they join amino acids.
What you are expected to know Explain why a bacterium’s ribosomes begin translating an mRNA while RNA polymerase is still making it: no nuclear envelope separates the gene from the ribosomes, and the mRNA needs no cap, tail or splicing.
Suppose a bacterium and a human cell each switch on a gene for the same kind of enzyme at the same moment.
In which cell does the first finished enzyme molecule appear sooner?
- A. In the human cellThe human cell’s mRNA is capped, tailed, spliced and carried out of the nucleus before a ribosome binds it.
The bacterium’s ribosomes bind the mRNA as it is made. - B. ✓ In the bacterium
- C. In both cells at the same momentThe bacterium’s ribosomes begin translating while RNA polymerase is still making the mRNA.
The human cell’s ribosomes wait for the processed mRNA to leave the nucleus.
Why: In the bacterium, ribosomes bind the mRNA while RNA polymerase is still making it.
In the human cell, the mRNA is processed and exported before any ribosome binds it.
So the bacterium’s enzyme appears sooner.
Suppose a gut bacterium and a cell lining the human gut each switch on a gene for an enzyme at the same moment. The bacterium holds finished enzyme long before the gut-lining cell holds any.
(a) Explain why the bacterium’s enzyme appears so much sooner. (2 pt)
Frame The bacterium’s enzyme appears sooner because …
Its mRNA needs no cap, no tail and no splicing.
So ribosomes bind the mRNA near its free 5′ end while RNA polymerase is still making the rest of it.
The gut-lining cell’s mRNA must be capped, tailed, spliced and carried out through a pore before any ribosome can bind it.
- Award 1 point for: in the bacterium there is no nucleus (no nuclear envelope) between the gene and the ribosomes AND the mRNA needs no cap, tail or splicing, so ribosomes bind it while RNA polymerase is still making it.
- Award 1 point for: the gut-lining cell’s mRNA must be processed (capped, tailed, spliced) and leave the nucleus through a pore before a ribosome can bind it, so its translation starts later.
Suppose a bacterium switches on a gene and builds the enzyme quickly. A student says: “The bacterium builds the enzyme quickly because its ribosomes join amino acids faster than a human cell’s ribosomes.”
Is the student correct?
- A. ✓ No: the bacterium’s ribosomes begin translating sooner, and that is what makes it quick
- B. Yes: a bacterium’s ribosomes join amino acids faster than a human cell’s ribosomesA bacterium’s ribosomes do join amino acids faster, but that saves less than a minute.
They begin translating minutes sooner, while RNA polymerase is still making the mRNA.
Why: A bacterium’s ribosomes bind the mRNA while RNA polymerase is still making it.
A human cell’s ribosomes wait until the mRNA is processed and has left the nucleus.
So the bacterium’s ribosomes start sooner.
Suppose a bacterium switches on a gene. One minute later, a scientist tests the cell for the gene’s mRNA and for the gene’s protein.
Which of the following does the scientist find?
- A. Neither the mRNA nor the proteinRNA polymerase begins making the mRNA as soon as the gene switches on.
Ribosomes bind that mRNA at once, so protein follows within about a minute. - B. The mRNA onlyRibosomes bind the bacterium’s mRNA while RNA polymerase is still making it.
So finished protein appears within about a minute, alongside the mRNA. - C. ✓ Both the mRNA and the protein
Why: RNA polymerase begins making the mRNA when the gene switches on.
Ribosomes bind the mRNA near its free 5′ end while it is still being made.
So within about a minute the cell holds both the mRNA and finished protein.
The drawing shows a bacterial gene being transcribed. Two shapes carry lettered markers, J and K.
Which marker sits on a ribosome?
- A. Marker JMarker J sits on the large oval seated on the DNA: RNA polymerase, making the mRNA.
The small paired ovals on the hanging strand are ribosomes. - B. ✓ Marker K
Why: The large oval on the DNA is RNA polymerase, and the strand hanging from it is the mRNA it is making.
The small paired ovals along that strand are ribosomes, already reading it.
Marker K sits on one of them.
Go back to the two cells that switched on a gene for the same kind of enzyme at the same moment.
The bacterium has no nucleus and nothing to process. So its ribosomes bind the mRNA while RNA polymerase is still making it.
The human cell caps, tails, splices and exports its mRNA first. Only then can a ribosome begin.
62Mixed practice mixed practice
A ribosome in a eukaryotic cell sits loose in the cytoplasm, away from the ER.
Which kind of protein is the ribosome most likely building?
- A. ✓ A protein for the cytoplasm
- B. A protein for exportA protein for export is built on a ribosome attached to the rough ER.
A ribosome loose in the cytoplasm, away from the ER, is a free ribosome. - C. Either kind, equally oftenWhere a ribosome sits matches where its protein is going.
A free ribosome builds proteins for the cytoplasm.
Why: The ribosome sits loose in the cytoplasm, away from the ER.
So the ribosome is a free ribosome.
A free ribosome builds a protein for the cytoplasm.
So the protein is most likely for the cytoplasm.
Suppose a human cell switches on a gene. Two minutes later, RNA copied from the gene exists in the cell.
Where is that RNA?
- A. In the cytoplasm, being read by ribosomesThe RNA must be capped, tailed, spliced and carried out through a pore before a ribosome can bind it.
Two minutes in, the RNA is still inside the nucleus. - B. ✓ Inside the nucleus, still being processed
- C. Inside the rough ERNo RNA enters the ER.
The RNA is made inside the nucleus and stays there until it is processed and exported.
Why: RNA polymerase copies the gene inside the nucleus.
Enzymes in the nucleus then add the cap and the tail and cut out the introns.
Only the processed mRNA leaves through a pore.
Two minutes in, the RNA is still inside the nucleus.
A bacterium builds an enzyme that it exports out of the cell.
Where is the ribosome that builds the enzyme?
- A. On the rough ERA bacterium has no ER.
Even an exported protein is built on a free ribosome in the cytoplasm. - B. Inside the nucleoid, on the DNAA ribosome reads an mRNA, not the DNA.
A bacterium’s ribosomes float free in the cytoplasm. - C. ✓ Free in the cytoplasm
Why: A bacterium has no ER and no nucleus.
Every one of its ribosomes floats free in the cytoplasm.
So the ribosome that builds the enzyme is free in the cytoplasm.
In a human cell, RNA polymerase is copying a gene into pre-mRNA.
Which happens first?
- A. ✓ RNA polymerase adds the last nucleotide of the pre-mRNA
- B. A ribosome joins the first two amino acids of the proteinNo ribosome can reach the pre-mRNA inside the nucleus.
The pre-mRNA is finished, processed and exported before a ribosome binds it. - C. The two events happen at the same momentThe nuclear envelope keeps every ribosome away from the pre-mRNA.
Translation begins only after the mRNA has left the nucleus.
Why: The gene is inside the nucleus and the ribosomes are outside it.
The pre-mRNA is finished, processed and carried out through a pore before a ribosome binds it.
So the last nucleotide is added first.
A human cell is transcribing a gene.
Which structure stands between that gene and the cell’s ribosomes?
- A. The plasma membraneThe plasma membrane is the cell’s outer boundary.
The gene and the ribosomes are both inside it. - B. The rough ERThe rough ER carries ribosomes on its surface.
The nuclear envelope, wrapped round the gene, separates the gene from those ribosomes. - C. ✓ The nuclear envelope
Why: The gene lies inside the nucleus.
The nuclear envelope wraps the nucleus, and every ribosome lies outside it.
So the nuclear envelope stands between the gene and the ribosomes.
A student says: “In a human cell, no ribosome can bind an mRNA until the mRNA has left the nucleus.”
Is the student correct?
- A. No: ribosomes bind the mRNA inside the nucleus, while RNA polymerase is still making itEvery ribosome in a human cell lies outside the nucleus, in the cytoplasm or on the rough ER.
The nuclear envelope stands between the gene and the ribosomes. - B. ✓ Yes: every ribosome lies outside the nuclear envelope, so the mRNA must leave the nucleus first
Why: The gene lies inside the nucleus.
Every ribosome lies outside the nuclear envelope.
So no ribosome can bind the mRNA until the mRNA has left through a pore.
Suppose a bacterium and a plant cell each switch on a gene for the same kind of enzyme. Now imagine a treatment that blocks every pore in the plant cell’s nuclear envelope. The same treatment reaches the bacterium too.
(a) Predict what happens to the plant cell’s production of the enzyme. (1 pt)
Its mRNA cannot leave the nucleus, so no ribosome can bind it, and translation never begins.
- Award 1 point for: no enzyme is made (or translation does not begin), because the mRNA cannot leave the nucleus to reach a ribosome.
(b) Explain why the bacterium’s production of the enzyme continues as before. (1 pt)
Its gene and its ribosomes lie in the same cytoplasm.
So ribosomes bind the mRNA as RNA polymerase makes it, and there is no pore to block.
- Award 1 point for: the bacterium has no nucleus (no nuclear envelope or pores), so its mRNA is already beside the ribosomes and is translated as it is made.
APBIO-U06-L18 Three letters, one amino acid
An mRNA arrives at a ribosome. It reads 5′-AUGGCUUACUAA-3′. The message uses four kinds of letter, and proteins use twenty kinds of amino acid.
One letter per amino acid would give only four amino acids. Two letters would give sixteen. How many letters must the ribosome read at a time?
Unit 6 · Gene Expression and Regulation
1Read in threes from the start
A ribosome reads an mRNA.
What are three bases on the mRNA read together called?
- A. A geneA gene is a stretch of DNA.
Three bases on the mRNA read together are a codon. - B. ✓ A codon
- C. An anticodonAn anticodon is the three bases on a tRNA that pair with a codon.
Three bases on the mRNA read together are a codon.
Why: Three bases on the mRNA read together are called a codon.
How many bases stand for one amino acid?
Three bases stand for one amino acid.
The ribosome reads the mRNA three bases at a time, one codon after another, beginning at the codon AUG.
Where the reading begins sets which bases group together.
Sixty-four codons are possible: sixty-one name an amino acid and three mean stop.
One chart lists all sixty-four codons and what each one means.
Video: Watch: Read in threes from the start
The message 5′-AUGGCUUACUAA-3′ arrives at a ribosome; its letters group into threes from the first A, AUG GCU UAC UAA; the same letters grouped from the second letter give different groups; the ribosome begins at AUG.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L18a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L18a.mp4
Here is the mRNA from the opening, 5′-AUGGCUUACUAA-3′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
An mRNA is written with four kinds of letter: A, U, G and C. A protein is built from twenty kinds of amino acid.
Suppose the ribosome read one letter at a time. Four kinds of letter could then stand for only four amino acids.
Now suppose the ribosome read two letters at a time. Four choices for the first letter and four for the second give sixteen different pairs, still too few for twenty amino acids.
Here is a table of the three choices: how many letters are read together, and how many different groups that gives.
Three letters at a time give sixty-four different groups. Sixty-four is more than enough for twenty amino acids.
So the ribosome reads the mRNA three bases at a time. Three bases on the mRNA read together are a codon.
So the ribosome reads one codon at a time.
Read from the first letter, the twelve letters of this mRNA group into four codons: AUG, GCU, UAC and UAA.
Now imagine the ribosome began one letter later, at the first U.
The same twelve letters would then group as UGG, CUU and ACU, with letters left over at each end. Every codon is now different.
Each codon means one amino acid. So a different set of codons means a different set of amino acids, and a different protein.
So where the reading begins decides which bases group together. The grouping of an mRNA’s bases into codons, set by where the reading begins, is called the .
The ribosome does not begin at just any letter. The ribosome begins at the codon AUG.
The codon where the ribosome begins reading, AUG, is called the .
Beginning at AUG sets the reading frame. From AUG onward, every three bases are one codon, with no gaps between codons and no overlaps.
Here is the ribosome with the mRNA threaded through it and the codons boxed in threes from AUG.
The codons are on the mRNA, and the ribosome reads them. The anticodons are on the tRNAs, which pair with the codons.
What you are expected to know Describe how the ribosome reads an mRNA: three bases at a time from the start codon AUG, which sets the reading frame.
The mRNA 5′-AUGGCUUACUAA-3′ is drawn three times, grouped in three different ways.
Which grouping shows the codons the ribosome reads?
- A. Grouping JGrouping J begins one letter after the start codon, and leaves an A on its own at the start.
The ribosome begins at AUG. - B. Grouping KGrouping K begins two letters after the start codon, and leaves AU on their own at the start.
The ribosome begins at AUG. - C. ✓ Grouping L
Why: The ribosome begins reading at the start codon AUG.
From AUG it reads three bases at a time, with no gaps.
Grouping L begins at AUG and groups every three letters from there.
Suppose an mRNA reads 5′-AUGCCUGGAUCG-3′. A ribosome that begins at the A of AUG builds one protein. Now imagine a second ribosome that began one base later, at the first U. It would build a different protein.
(a) Explain why the second ribosome would build a different protein. (2 pt)
Frame The second ribosome would build a different protein because …
Beginning at the A groups the bases as AUG CCU GGA UCG.
Beginning at the U groups them as A UGC CUG GAU CG, with letters left over.
Every codon is now a different three bases.
Each codon means one amino acid.
So the different codons mean different amino acids, and the protein is different.
- Award 1 point for: the ribosome reads three bases at a time from where it begins, so beginning one base later groups the bases into different codons (a different reading frame).
- Award 1 point for: each codon means one amino acid, so different codons mean a different order of amino acids, and so a different protein.
A student says: “The codons are the groups of three bases on the tRNA, and the mRNA carries the anticodons.”
Is the student correct?
- A. ✓ No: the ribosome reads the codons on the mRNA, and the anticodons are on the tRNA
- B. Yes: the codons are on the tRNA, and the anticodons are on the mRNAThe ribosome reads the mRNA, and the groups of three it reads are the codons.
Each tRNA carries an anticodon that pairs with one codon.
Why: The ribosome reads the mRNA three bases at a time.
Those groups of three on the mRNA are the codons.
A tRNA carries three bases that pair with a codon: its anticodon.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. Suppose an mRNA reads 5′-CAUGUCAGGUCC-3′, drawn below.
Which grouping shows the codons the ribosome reads?
- A. CAU GUC AGG UCCGrouping from the first letter, C, makes no group AUG.
The ribosome begins at the start codon AUG, one letter in. - B. ✓ C AUG UCA GGU CC
- C. CA UGU CAG GUC CGrouping from the third letter makes no group AUG.
The ribosome begins at the start codon AUG, one letter in.
Why: The ribosome begins reading at the start codon AUG.
In this mRNA, AUG is the second, third and fourth letters.
From there the ribosome reads three bases at a time: AUG, UCA, GGU, with CC left over.
34Quick quiz: reading frame, start codon mixed practice
Imagine a ribosome that began reading one base later than the start codon.
How many bases would each codon it reads have?
- A. 2A codon is always three bases.
Beginning one base later changes which bases group together, and every group is still three bases. - B. ✓ 3
- C. 4A codon is always three bases.
Beginning one base later changes which bases group together, and every group is still three bases.
Why: A ribosome reads three bases at a time wherever it begins.
Beginning one base later changes the reading frame: different bases group together.
Every codon it reads is still three bases.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. An mRNA reads 5′-AUGACUGGUCAC-3′, drawn below. The ribosome begins at AUG.
Which codon does the ribosome read second?
- A. ✓ ACU
- B. GACGAC overlaps the first codon by one base.
Codons do not overlap: the second codon begins right after the G of AUG. - C. UGAUGA overlaps the first codon by two bases.
Codons do not overlap: the second codon begins right after the G of AUG.
Why: The first codon is AUG, the first three bases.
Codons follow one another with no overlap and no gap.
So the second codon is the next three bases, ACU.
An mRNA is read at a ribosome.
(a) State what the reading frame of an mRNA is, and what sets it. (1 pt)
Where the ribosome begins reading, the start codon AUG, sets it.
- Award 1 point for: the grouping of the bases into codons (three at a time), set by where reading begins (the start codon AUG).
What is the start codon?
- A. ✓ The codon AUG, where the ribosome begins reading
- B. The first three bases at the mRNA’s 5′ end, whatever they areThe ribosome begins at AUG, and AUG need not be the first three bases of the mRNA.
The start codon is AUG. - C. The last three bases at the mRNA’s 3′ endThe ribosome reads toward the 3′ end, so the last three bases are read last.
The start codon is AUG, where reading begins.
Why: The codon where the ribosome begins reading, AUG, is called the start codon.
What is a reading frame?
- A. The codon the ribosome begins reading fromThe codon the ribosome begins from is the start codon.
The reading frame is the grouping of the bases into codons that the start sets. - B. ✓ The way an mRNA’s bases group into codons, set by the start
- C. The three bases on a tRNA that pair with one codonThree bases on a tRNA that pair with a codon are an anticodon.
The reading frame is the grouping of the mRNA’s bases into codons.
Why: The grouping of an mRNA’s bases into codons, set by where the reading begins, is called the reading frame.
40Sixty-four codons
A protein is a chain of amino acids.
How many kinds of amino acid do cells use to build proteins?
- A. 4Four is the number of kinds of base in an mRNA.
Cells build proteins from twenty kinds of amino acid. - B. ✓ 20
- C. 64Sixty-four is the number of different three-base groups.
Cells build proteins from twenty kinds of amino acid.
Why: Cells build proteins from twenty kinds of amino acid.
Video: Watch: Sixty-four codons
The sixty-four codons written out one by one and laid into the chart; the AUG line marked Met (start); the three lines marked stop; sixty-one lines left that each name an amino acid.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L18b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L18b.mp4
Suppose you write out every codon three bases can make. There are sixty-four of them.
Here is a table of all sixty-four codons and what each one means: four rows, four columns, four lines in every cell.
The set of rules saying which amino acid each codon means is called the . This chart is the genetic code written out as a table.
Sixty-one of the sixty-four codons each name one amino acid. Between them, those sixty-one codons name all twenty kinds of amino acid.
Three codons name no amino acid: UAA, UAG and UGA. Each of the three tells the ribosome that the message has ended.
A codon that names no amino acid and ends the message is called a . On the chart, the three stop codons read stop.
The codon AUG does two jobs. AUG is the start codon, and AUG also names an amino acid.
The amino acid that AUG names is called . On the chart, the AUG line reads Met (start).
The ribosome begins reading at AUG. AUG names methionine (Met).
So nearly every new polypeptide begins with methionine (Met).
What you are expected to know State the contents of the genetic code: sixty-four codons, of which sixty-one name an amino acid and three are stop codons.
What you are expected to know Identify the start codon, AUG, and the amino acid it names, methionine (Met).
How many different codons can three bases make?
- A. 12Twelve is four kinds of base times three positions.
Each position has four choices, so three bases make four times four times four codons: sixty-four. - B. 16Sixteen is the number of different two-base pairs.
Three bases make sixty-four different codons. - C. ✓ 64
Why: Three bases can make sixty-four different codons.
How many codons name an amino acid?
- A. 20Twenty is the number of kinds of amino acid.
Sixty-one codons name them, several codons to some amino acids. - B. ✓ 61
- C. 63Sixty-three would leave out only the start codon, and AUG does name an amino acid.
The three stop codons name none, so sixty-one codons name an amino acid.
Why: Three of the sixty-four codons are stop codons and name no amino acid.
The other sixty-one each name one amino acid.
How many codons are stop codons?
- A. 1One codon, AUG, is the start codon.
Three codons, UAA, UAG and UGA, are stop codons. - B. ✓ 3
- C. 4Four is the number of kinds of base.
Three codons, UAA, UAG and UGA, are stop codons.
Why: Three codons name no amino acid and end the message: UAA, UAG and UGA.
So there are three stop codons.
A ribosome reaches UGA, a stop codon.
Which of the following happens?
- A. ✓ The ribosome adds no amino acid: the message has ended
- B. The ribosome adds an amino acid and moves to the next codonUGA is one of the three stop codons.
A stop codon names no amino acid.
Why: UGA is a stop codon.
A stop codon names no amino acid.
A stop codon tells the ribosome that the message has ended.
Besides marking where reading begins, what does the codon AUG name?
- A. Any amino acidEach codon that names an amino acid names one particular amino acid.
AUG names methionine (Met). - B. No amino acidThe three stop codons name no amino acid.
AUG names methionine (Met). - C. ✓ Methionine (Met)
Why: AUG is the start codon, and AUG also names one amino acid.
The amino acid AUG names is methionine (Met).
With which amino acid does nearly every new polypeptide begin?
- A. ✓ Methionine (Met)
- B. Alanine (Ala)Alanine (Ala) is named by other codons, such as GCU.
The ribosome begins at AUG, which names methionine (Met). - C. A different one for each proteinThe ribosome begins reading every message at the same codon, AUG.
AUG names methionine (Met).
Why: The ribosome begins reading at the start codon AUG.
AUG names methionine (Met).
So nearly every new polypeptide begins with methionine (Met).
61Quick quiz: genetic code, stop codon, methionine (Met) mixed practice
A ribosome reaches a stop codon.
Which amino acid does the stop codon name?
- A. ✓ None
- B. Methionine (Met)Methionine (Met) is the amino acid the start codon AUG names.
A stop codon names no amino acid. - C. Whichever amino acid the previous codon namedEach codon means its own amino acid or means stop.
A stop codon names no amino acid.
Why: A stop codon names no amino acid.
The stop codon tells the ribosome that the message has ended.
A ribosome is reading an mRNA, codon by codon.
(a) State what a stop codon is, and what the ribosome does when it reaches one. (1 pt)
When the ribosome reaches a stop codon, it adds no amino acid, because the message has ended.
- Award 1 point for: a codon that names no amino acid (UAA, UAG or UGA); at it the ribosome adds no amino acid and the message ends.
What is the genetic code?
- A. The order of bases along one geneThe order of bases along one gene is that gene’s sequence.
The genetic code is the set of rules saying which amino acid each codon means. - B. ✓ The set of rules saying which amino acid each codon means
- C. The three bases on a tRNA that pair with a codonThree bases on a tRNA that pair with a codon are an anticodon.
The genetic code is the set of rules saying which amino acid each codon means.
Why: The set of rules saying which amino acid each codon means is called the genetic code.
What is a stop codon?
- A. ✓ A codon that names no amino acid and ends the message
- B. A codon that names one final amino acid before the message endsA stop codon names no amino acid.
The message ends at the stop codon, and nothing is added there. - C. A codon that names no amino acid, so the ribosome skips it and reads onThe ribosome does not skip a stop codon.
The message ends at the first stop codon the ribosome reaches.
Why: A codon that names no amino acid and ends the message is called a stop codon.
What is methionine (Met)?
- A. The three-letter codon AUG itselfAUG is a codon: three bases on the mRNA.
Methionine (Met) is the amino acid that AUG names. - B. The tRNA that pairs with the start codon AUGThe tRNA that pairs with AUG carries methionine (Met).
Methionine (Met) is the amino acid itself, not the tRNA. - C. ✓ The amino acid that the start codon AUG names
Why: The amino acid that the start codon AUG names is called methionine (Met).
67Read the chart
A strand is written from its 5′ end to its 3′ end, and both ends are marked. The codon 5′-GAG-3′ of an mRNA is drawn below.
Which base of this codon does the ribosome read first?
- A. ✓ The G at the 5′ end
- B. The A in the middleThe ribosome reads an mRNA from its 5′ end toward its 3′ end.
The first base of a codon is the one at its 5′ end. - C. The G at the 3′ endThe ribosome reads an mRNA from its 5′ end toward its 3′ end.
The base at the 3′ end is read last.
Why: The ribosome reads an mRNA from its 5′ end toward its 3′ end.
A codon is written the same way, 5′ to 3′.
So the ribosome reads the G at the 5′ end first.
Video: Watch: Read the chart
The codon GAG read on the chart: its first base picks the row, its second base picks the column, its third base picks the line in that cell, and the line reads Glu; then every line reading Leu found across the chart.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L18c.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L18c.mp4
Suppose a ribosome reaches the codon 5′-GAG-3′. The chart says which amino acid GAG names, in three steps.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon.
The first base, G, picks the row. Find the row marked G at the left of the chart.
The second base, A, picks the column. Find the column marked A along the top: the row and the column cross at one cell.
The third base, G, picks the line inside that cell. The cell’s four lines are for U, C, A and G in that order, so read the G line.
The G line of that cell reads GAG Glu. So the codon GAG names glutamic acid (Glu).
The chart writes every amino acid as a three-letter short name, such as Glu for glutamic acid.
The chart also works the other way, from an amino acid to its codons.
Suppose you want every codon that names tryptophan (Trp). Search the chart for Trp.
Trp appears on one line only, UGG. So tryptophan has one codon.
Now search the chart for leucine (Leu).
Leu appears on six lines: UUA, UUG, CUU, CUC, CUA and CUG. So leucine has six codons.
What you are expected to know Read the genetic code chart: find the amino acid a codon names, and find every codon that names a given amino acid.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The codon 5′-CAU-3′ is drawn above the chart.
Which amino acid does the codon CAU name?
- A. Threonine (Thr)ACU names threonine (Thr).
CAU is read with C for the row and A for the column. - B. ✓ Histidine (His)
- C. Glutamine (Gln)CAA names glutamine (Gln).
CAU ends in U, so its line is the U line of the cell.
Why: The first base, C, picks the row.
The second base, A, picks the column.
The third base, U, picks the line in that cell.
That line reads CAU His, so CAU names histidine (His).
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The codon 5′-AAC-3′ is drawn above the chart.
Which amino acid does the codon AAC name?
- A. Threonine (Thr)ACA names threonine (Thr).
AAC is read with A for the row and A for the column. - B. Lysine (Lys)AAA names lysine (Lys).
AAC ends in C, so its line is the C line of the cell. - C. ✓ Asparagine (Asn)
Why: The first base, A, picks the row.
The second base, A, picks the column.
The third base, C, picks the line in that cell.
That line reads AAC Asn, so AAC names asparagine (Asn).
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The codon 5′-GAC-3′ is drawn above the chart.
Which amino acid does the codon GAC name?
- A. ✓ Aspartic acid (Asp)
- B. Serine (Ser)AGC names serine (Ser).
GAC is read with G for the row and A for the column. - C. Glutamic acid (Glu)GAA names glutamic acid (Glu).
GAC ends in C, so its line is the C line of the cell.
Why: The first base, G, picks the row.
The second base, A, picks the column.
The third base, C, picks the line in that cell.
That line reads GAC Asp, so GAC names aspartic acid (Asp).
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The codon 5′-UGU-3′ is drawn above the chart.
Which amino acid does the codon UGU name?
- A. Valine (Val)GUU names valine (Val).
UGU is read with U for the row and G for the column. - B. Tryptophan (Trp)UGG names tryptophan (Trp).
UGU ends in U, so its line is the U line of the cell. - C. ✓ Cysteine (Cys)
Why: The first base, U, picks the row.
The second base, G, picks the column.
The third base, U, picks the line in that cell.
That line reads UGU Cys, so UGU names cysteine (Cys).
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The codon 5′-AGU-3′ is drawn above the chart.
Which amino acid does the codon AGU name?
- A. Aspartic acid (Asp)GAU names aspartic acid (Asp).
AGU is read with A for the row and G for the column. - B. ✓ Serine (Ser)
- C. Arginine (Arg)AGA names arginine (Arg).
AGU ends in U, so its line is the U line of the cell.
Why: The first base, A, picks the row.
The second base, G, picks the column.
The third base, U, picks the line in that cell.
That line reads AGU Ser, so AGU names serine (Ser).
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The codon 5′-AUC-3′ is drawn above the chart.
Which amino acid does the codon AUC name?
- A. ✓ Isoleucine (Ile)
- B. Tyrosine (Tyr)UAC names tyrosine (Tyr).
AUC is read with A for the row and U for the column. - C. Methionine (Met)AUG names methionine (Met).
AUC ends in C, so its line is the C line of the cell.
Why: The first base, A, picks the row.
The second base, U, picks the column.
The third base, C, picks the line in that cell.
That line reads AUC Ile, so AUC names isoleucine (Ile).
The chart is drawn below.
Which codon names tryptophan (Trp)?
- A. UGAUGA is the line above Trp in the same cell, and it reads stop.
The line that reads Trp is UGG. - B. UGCUGC reads Cys.
The line that reads Trp is UGG. - C. ✓ UGG
Why: Search the chart for Trp.
Trp appears on one line only.
That line is UGG.
The chart is drawn below.
How many codons name leucine (Leu)?
- A. 2Two lines in the top-left cell read Leu, UUA and UUG.
Four more lines read Leu in the cell below it. - B. 4Four lines read Leu in the cell below the top-left one.
Two more lines read Leu in the top-left cell. - C. ✓ 6
Why: Search the chart for Leu.
Two lines in the top-left cell read Leu: UUA and UUG.
All four lines of the cell below it read Leu: CUU, CUC, CUA and CUG.
So six codons name leucine (Leu).
Go back to the mRNA from the opening, 5′-AUGGCUUACUAA-3′.
Read in threes from the start codon AUG, its codons are AUG, GCU, UAC and UAA.
The chart gives Met, Ala and Tyr. UAA is a stop codon, so the message ends there.
Three letters give sixty-four codons: sixty-one name amino acids and three mean stop.
95Mixed practice mixed practice
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. An mRNA reads 5′-AUGCGU-3′, drawn above the chart. The ribosome begins at AUG.
Which amino acid does the second codon name?
- A. Methionine (Met)Methionine (Met) is named by the first codon, AUG.
The second codon is the next three bases, CGU. - B. ✓ Arginine (Arg)
- C. Cysteine (Cys)Cysteine (Cys) is named by UGC, which overlaps the first codon.
Codons do not overlap: the second codon is CGU.
Why: The first codon is AUG.
The second codon is the next three bases, CGU.
On the chart, the row C, the column G, the U line reads CGU Arg.
So the second codon names arginine (Arg).
A student says: “There are sixty-four codons, so cells must use sixty-four kinds of amino acid.”
Is the student correct?
- A. ✓ No: several codons name the same amino acid, so cells use only twenty kinds
- B. Yes: each codon names a different one, so there are sixty-four kindsCells use twenty kinds of amino acid.
Several codons can name the same amino acid, and three codons name none.
Why: Three of the sixty-four codons are stop codons and name no amino acid.
The other sixty-one name amino acids.
Several codons can name the same amino acid, so those sixty-one codons name only twenty kinds.
Imagine a ribosome that began reading an mRNA two bases after the A of its start codon.
Which of the following is true of the codons it reads?
- A. They are the same codons, read in a different orderBeginning two bases later groups different bases together.
Every codon it reads is a different three bases. - B. They are two bases longA codon is always three bases.
Beginning two bases later changes which three bases group together. - C. ✓ They are different codons
Why: A ribosome reads three bases at a time from where it begins.
Beginning two bases later groups different bases together.
So every codon it reads is a different codon.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. A ribosome reaches the codon 5′-UAA-3′, drawn above the chart.
What does the ribosome do at UAA?
- A. Adds one amino acid and moves to the next codonOn the chart, the UAA line reads stop.
A stop codon names no amino acid. - B. ✓ Adds no amino acid: the message has ended
Why: The row U, the column A, the A line reads UAA stop.
UAA is a stop codon, so it names no amino acid.
The ribosome adds no amino acid, because the message has ended.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. A ribosome reaches the codon 5′-CGA-3′, drawn above the chart.
Which base of the codon picks the row of the chart?
- A. ✓ The first base, C
- B. The second base, GThe second base picks the column.
The first base picks the row. - C. The third base, AThe third base picks the line inside the cell.
The first base picks the row.
Why: The first base of a codon picks the row of the chart.
The second base picks the column.
The third base picks the line in that cell.
Suppose an mRNA’s first AUG is its fourth, fifth and sixth bases. A student says: “The polypeptide this mRNA gives begins with methionine (Met).”
Is the student correct?
- A. No: the polypeptide begins with the amino acid the mRNA’s first three bases nameThe ribosome begins reading at the start codon AUG, not at the mRNA’s first three bases.
AUG names methionine (Met). - B. ✓ Yes: the ribosome begins at AUG wherever it sits, and AUG names methionine (Met)
Why: The ribosome begins reading at the first AUG, here the fourth, fifth and sixth bases.
AUG names methionine (Met).
So the polypeptide begins with methionine (Met).
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The codon 5′-AAU-3′ is drawn above the chart.
Which amino acid does the codon AAU name?
- A. Isoleucine (Ile)AUU names isoleucine (Ile).
AAU is read with A for the row and A for the column. - B. Lysine (Lys)AAA names lysine (Lys).
AAU ends in U, so its line is the U line of the cell. - C. ✓ Asparagine (Asn)
Why: The first base, A, picks the row.
The second base, A, picks the column.
The third base, U, picks the line in that cell.
That line reads AAU Asn, so AAU names asparagine (Asn).
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. Suppose an mRNA reads 5′-AUGUUCACCUGA-3′, drawn above the chart. A ribosome begins reading it at its start codon.
(a) Identify the codons the ribosome reads, in order. (1 pt)
- Award 1 point for: AUG UUC ACC UGA, in that order, beginning at AUG.
(b) Using the chart, identify the amino acid each of the first three codons names. (1 pt)
- Award 1 point for: methionine (Met), phenylalanine (Phe), threonine (Thr), in that order.
(c) Describe what the fourth codon tells the ribosome. (1 pt)
It names no amino acid, and it tells the ribosome that the message has ended.
- Award 1 point for: UGA is a stop codon — it names no amino acid and the message ends there.
(d) Explain why the ribosome reads this mRNA three bases at a time. (2 pt)
Read one base at a time, four bases could name only four amino acids.
Read two at a time, the sixteen possible pairs could name only sixteen.
Read three at a time, the sixty-four possible codons can name all twenty.
So the ribosome reads three bases at a time.
- Award 1 point for: four kinds of base read one or two at a time give only four or sixteen groups, fewer than the twenty kinds of amino acid.
- Award 1 point for: three bases at a time give sixty-four codons, enough to name all twenty amino acids (with codons to spare).
Glossary
- reading frame
- The grouping of an mRNA’s bases into codons, set by where the ribosome begins reading.
- start codon
- The codon where the ribosome begins reading an mRNA: AUG.
- genetic code
- The set of rules saying which amino acid each codon means: sixty-four codons, of which sixty-one name an amino acid and three are stop codons.
- stop codon
- A codon that names no amino acid and ends the message: UAA, UAG or UGA.
- methionine (Met)
- The amino acid that the start codon AUG names, so nearly every new polypeptide begins with it.
APBIO-U06-L19 Spare codons
Look at the code chart. Sixty-one codons name only twenty amino acids. GAA and GAG both mean glutamic acid, and leucine has six codons.
Why does the code have spares?
Unit 6 · Gene Expression and Regulation
1Several codons, one amino acid
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The codon 5′-AGC-3′ is drawn above the chart.
Which amino acid does the codon AGC name?
- A. Aspartic acid (Asp)GAC names aspartic acid (Asp).
AGC is read with A for the row and G for the column. - B. ✓ Serine (Ser)
- C. Arginine (Arg)AGA names arginine (Arg).
AGC ends in C, so its line is the C line of the cell.
Why: The first base, A, picks the row.
The second base, G, picks the column.
The third base, C, picks the line in that cell.
That line reads AGC Ser, so AGC names serine (Ser).
Why do most amino acids have more than one codon?
Three letters give sixty-four codons. Twenty kinds of amino acid need only twenty.
So most amino acids have several codons. The codons for one amino acid often differ only in their third letter.
Several codons can mean one amino acid. Each codon still means only one amino acid.
So a change to one base can change the codon and leave the amino acid the same.
Video: Watch: Several codons, one amino acid
The chart appears with its two Glu lines marked, then its six Leu lines. The count of amino acids with more than one codon rises to eighteen. One codon’s third letter changes, and its amino acid stays the same.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L19a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L19a.mp4
Here is the code chart.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon.
Sixty-one of the sixty-four codons each name one amino acid. Cells build proteins from twenty kinds of amino acid.
Sixty-one codons name only twenty amino acids. So some amino acids must have more than one codon.
Look at the cell in the row marked G and the column marked A.
Two of its four lines read Glu: GAA and GAG. So GAA and GAG both mean glutamic acid (Glu).
Now find every line that reads Leu.
Six lines read Leu: UUA, UUG, CUU, CUC, CUA and CUG. So leucine (Leu) has six codons.
Now find every line that reads Pro.
All four lines of one cell read Pro: CCU, CCC, CCA and CCG. The four codons share their first two letters and differ only in their third letter.
Count the amino acids that have more than one codon. Eighteen of the twenty amino acids have more than one codon.
Only two amino acids have a single codon: methionine (Met) has only AUG, and tryptophan (Trp) has only UGG.
Here is a table of five amino acids and their codons, from one codon up to six.
A code in which several codons mean the same amino acid is called a , because it has more codons than it needs. The genetic code is redundant.
Most codons that share an amino acid differ only in their third letter. Proline (Pro) has CCU, CCC, CCA and CCG; leucine (Leu) has CUU, CUC, CUA and CUG.
In eight of the chart’s sixteen cells, all four lines read one amino acid. In those cells the first two letters settle the amino acid, and the third letter makes no difference.
Now suppose one base of an mRNA changes. The codon 5′-CCA-3′ becomes 5′-CCG-3′.
On the chart, CCA reads Pro and CCG reads Pro. The codon has changed, and the amino acid has not.
The ribosome adds proline (Pro) at that position either way. So the protein it builds is the same as before.
So some changes to a base change the codon but not the amino acid. The protein stays the same.
Redundant does not mean unclear. Several codons can mean one amino acid, but no codon means two amino acids.
CCA means proline (Pro) and nothing else. Every codon means only one amino acid, so the code is never ambiguous.
What you are expected to know Explain what it means that the genetic code is redundant: most amino acids have more than one codon, often differing only in the third letter.
What you are expected to know Explain why a change to one base of a codon can leave the protein the same.
What you are expected to know Explain why the code is never ambiguous: each codon means only one amino acid.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The chart is drawn below.
Which pair of codons shows that the code is redundant?
- A. ✓ AAA and AAG
- B. UUU and UUAOn the chart, UUU reads Phe and UUA reads Leu.
Two codons that mean different amino acids show no spare. - C. UGG and UGAOn the chart, UGG reads Trp and UGA reads stop.
A stop codon names no amino acid.
Why: Redundant means several codons mean one amino acid.
On the chart, AAA reads Lys and AAG reads Lys.
So AAA and AAG are two codons for one amino acid.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. Suppose one base of an mRNA changes, so that the codon 5′-GGU-3′ becomes 5′-GGC-3′. The two codons are drawn above the chart. The ribosome builds the same protein as before.
(a) Explain why the protein is unchanged. (2 pt)
Frame The protein is unchanged because …
The code is redundant: several codons can mean one amino acid.
So the ribosome adds glycine (Gly) at that position whether the codon reads GGU or GGC.
The order of amino acids in the protein is the same as before, so the protein is unchanged.
- Award 1 point for: GGU and GGC both name the same amino acid, glycine (Gly) — the code is redundant (several codons for one amino acid).
- Award 1 point for: so the ribosome adds the same amino acid at that position, and the order of amino acids (the protein) is unchanged.
A student says: “The genetic code is redundant, so one codon can mean two different amino acids.”
Is the student correct?
- A. ✓ No: each codon means only one amino acid
- B. Yes: a redundant code lets one codon have two meaningsRedundant means several codons for one amino acid.
Each codon still means only one amino acid.
Why: Redundant means several codons can mean the same amino acid.
Every codon on the chart reads one amino acid or stop.
So no codon means two amino acids.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. Suppose one base of an mRNA changes, so that the codon 5′-CGU-3′ becomes 5′-CGC-3′. The two codons are drawn above the chart.
Does the amino acid at that position change?
- A. YesOn the chart, CGU reads Arg and CGC reads Arg.
Both codons mean arginine (Arg). - B. ✓ No
Why: On the chart, CGU reads Arg.
CGC also reads Arg.
Both codons mean arginine (Arg).
So the amino acid does not change.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. Suppose one base of an mRNA changes, so that the codon 5′-AAA-3′ becomes 5′-AAU-3′. The two codons are drawn above the chart.
Does the amino acid at that position change?
- A. ✓ Yes
- B. NoOn the chart, AAA reads Lys and AAU reads Asn.
A change in the third letter does not always leave the amino acid the same.
Why: On the chart, AAA reads Lys.
AAU reads Asn.
The two codons mean different amino acids, so the amino acid changes.
Go back to the chart, and to the cell in the row marked G and the column marked A.
GAA and GAG both mean glutamic acid (Glu). So a change from GAA to GAG changes the codon and not the protein.
Each codon still means only one amino acid.
42Quick quiz: redundant (degenerate) code mixed practice
How many codons name methionine (Met)?
- A. ✓ 1
- B. 2Two is the number of codons for glutamic acid (Glu).
Methionine (Met) has one codon, AUG. - C. 6Six is the number of codons for leucine (Leu).
Methionine (Met) has one codon, AUG.
Why: Only AUG reads Met on the chart.
So methionine (Met) has one codon.
A student says: “Two different codons can mean the same amino acid.”
Is the student correct?
- A. No: each amino acid has one codon of its ownEighteen of the twenty amino acids have more than one codon.
Two different codons can mean the same amino acid. - B. ✓ Yes: the code is redundant, so two codons can mean one amino acid
Why: The code is redundant.
Several codons can mean the same amino acid.
So two different codons can mean the same amino acid.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The codons 5′-CAU-3′ and 5′-CAA-3′ are drawn above the chart.
Do the two codons mean the same amino acid?
- A. YesOn the chart, CAU reads His and CAA reads Gln.
The two codons mean different amino acids. - B. ✓ No
Why: On the chart, CAU reads His.
CAA reads Gln.
So the two codons mean different amino acids.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The codons 5′-UCU-3′ and 5′-UCA-3′ are drawn above the chart.
Do the two codons mean the same amino acid?
- A. ✓ Yes
- B. NoOn the chart, UCU reads Ser and UCA reads Ser.
Both codons mean serine (Ser).
Why: On the chart, UCU reads Ser.
UCA also reads Ser.
So the two codons mean the same amino acid, serine (Ser).
The four codons GUU, GUC, GUA and GUG all mean valine (Val).
Which letter differs between these four codons?
- A. The first letterAll four codons begin with G.
Only the third letter differs. - B. The second letterAll four codons have U as their second letter.
Only the third letter differs. - C. ✓ The third letter
Why: The four codons all begin GU.
Their third letters are U, C, A and G.
So the third letter is the one that differs.
The genetic code is redundant.
(a) State what it means that the genetic code is redundant. (1 pt)
- Award 1 point for: several codons mean the same amino acid (most amino acids have more than one codon).
What does it mean that the genetic code is redundant?
- A. Each codon can mean several different amino acidsEach codon means only one amino acid.
Redundant means several codons can mean the same amino acid. - B. Every amino acid has exactly one codonOnly methionine (Met) and tryptophan (Trp) have one codon each.
Redundant means several codons can mean the same amino acid. - C. ✓ Several codons can mean the same amino acid
Why: A code in which several codons mean the same amino acid is called a redundant code.
Eighteen of the twenty amino acids have more than one codon.
Glossary
- redundant (degenerate) code
- A code in which several codons mean the same amino acid. The genetic code is redundant: eighteen of the twenty amino acids have more than one codon, often differing only in the third letter. Each codon still means only one amino acid.
APBIO-U06-L19B The same code everywhere
The code chart on your screen is the one a gut bacterium uses. It is the one a wheat plant uses. It is the one you use.
Why would three such different living organisms read the same code?
Unit 6 · Gene Expression and Regulation
1One chart for every living thing
Every known cell, from a bacterium’s to a human’s, contains ribosomes of the same basic build.
A feature shared by all living things is evidence of which of the following?
- A. All living things carry the same set of genesA bacterium and a human carry different sets of genes.
A feature shared by all living things is evidence of descent from shared ancestors. - B. All living things arose separately and later became alikeOrganisms that arose separately would not share one build of ribosome.
A feature shared by all living things is evidence of descent from shared ancestors. - C. ✓ All living things descend from shared ancestors
Why: Every known cell 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 common ancestry.
Base pairing is conserved through evolution.
What does conserved mean here?
- A. Changed a little in each kind of organism over timeEvery organism and virus pairs its bases the same way.
Conserved means kept the same. - B. ✓ Kept the same in every descendant of the first cells
- C. Found only in the oldest kinds of organism alive todayBase pairing is found in every living thing, the newest kinds as well as the oldest.
Conserved means kept the same in every descendant.
Why: Conserved means kept.
Base pairing has been kept the same in every descendant of the first cells that had it.
Why is one shared code evidence that all living organisms are related?
The same codon means the same amino acid in bacteria, plants and animals.
The simplest explanation is that all of them inherited the code from one ancestral population, the first cells. Descent from shared ancestors is common ancestry.
Video: Watch: One chart for every living thing
A gut bacterium, a wheat plant and a person each read the codon AUG. Each adds methionine (Met). One chart appears between them, and three arrows lead back from the three organisms to the first cells.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L19Ba.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L19Ba.mp4
Here is the code chart.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon.
Suppose a ribosome in a gut bacterium reaches the codon 5′-AUG-3′. The bacterium’s ribosome adds methionine (Met).
Now suppose a ribosome in a wheat plant reaches 5′-AUG-3′. The wheat plant’s ribosome adds methionine (Met).
Now suppose a ribosome in one of your own cells reaches 5′-AUG-3′. Your ribosome adds methionine (Met).
Three very different organisms read one codon the same way. The same is true of every codon on the chart.
A bacterium, a wheat plant and a human read each codon as the same amino acid.
A code that every living organism reads the same way is described as universal. The genetic code is universal.
Why would three such different organisms share one chart?
Nothing in the chemistry forces AUG to mean methionine (Met). Suppose AUG meant alanine (Ala) instead, in every cell of an organism.
That organism’s ribosomes would still build working proteins, because each codon would still mean one amino acid.
Here is a table of two codons: what each means in the real code, and what it could have meant in a code that could have worked.
So the pairings on the chart could have been different. Suppose each kind of living thing had arisen on its own.
Then each kind would have its own pairings. A bacterium’s chart, a wheat plant’s chart and your chart would all differ.
They do not differ. One chart serves all three.
The simplest explanation is that all three inherited the code from one ancestral population, the first cells. The first cells used this chart, and every descendant kept it.
Now suppose one cell changed the meaning of a codon. Every gene that used that codon would then build a different protein.
Most of the cell’s proteins would be built wrong. So the cell would die.
So the code was passed on unchanged, generation after generation.
The genetic code is conserved through evolution, just like base pairing.
A feature shared by all living things is evidence that all living things descend from shared ancestors. So one shared genetic code is evidence of common ancestry.
What you are expected to know Explain why a genetic code shared by nearly all living organisms is evidence that they share common ancestry.
The genetic code is described as universal.
What does universal mean here?
- A. Every living organism carries the same set of genesA bacterium and a human carry different sets of genes.
Universal describes the code, the meaning of each codon, not the genes. - B. Every living organism builds the same set of proteinsDifferent genes give different proteins.
Universal describes the code, the meaning of each codon, not the proteins. - C. ✓ Every living organism reads each codon as the same amino acid
Why: A gut bacterium, a wheat plant and a human each read AUG as methionine (Met).
Every codon means the same amino acid in each of them.
A code that every living organism reads the same way is universal.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. Suppose scientists find a microbe in glacier ice that no one has described before. They read three of its codons. Its ribosomes read 5′-AUG-3′ as methionine (Met), 5′-CAC-3′ as histidine (His) and 5′-ACG-3′ as threonine (Thr), exactly as the chart does. The three codons are drawn above the chart.
(a) Explain what the microbe’s code shows about the microbe’s ancestry. (2 pt)
Frame The microbe’s code shows that …
The microbe reads AUG, CAC and ACG as the same amino acids as a bacterium, a wheat plant and a human.
The pairings could have been different.
So one shared code is unlikely to have arisen twice.
The simplest explanation is that the microbe inherited the code from the first cells, as every other living organism did.
- Award 1 point for: the microbe reads its codons as the same amino acids as other organisms, and a feature shared by all living things is evidence of descent from shared ancestors (common ancestry).
- Award 1 point for: the pairings could have been different, so the shared code is most simply explained by inheritance from one ancestral population (the first cells), not by arising separately.
A student says: “The genetic code is the same in every living thing because it is the only code that could work.”
Is the student correct?
- A. ✓ No: a code with different pairings would work just as well
- B. Yes: only these pairings of codon to amino acid can build a protein that worksNothing in the chemistry forces AUG to mean methionine (Met).
A different set of pairings would build proteins just as well.
Why: Any set of pairings would build working proteins, as long as each codon meant one amino acid.
So the chart’s pairings could have been different.
Every living organism uses the same pairings because all of them inherited the code from the first cells.
Suppose one bacterium changes so that its ribosomes read one codon as a different amino acid from the one every other living thing reads. A student says: “Most of that bacterium’s proteins will now be built wrong, so the change dies with it.”
Is the student correct?
- A. No: the bacterium’s proteins still work, because each codon still means one amino acid, so the change is passed onThe bacterium’s genes were written for the usual meaning of that codon.
Every gene that uses the codon now gives a protein with a different amino acid at that spot. - B. ✓ Yes: every gene that uses that codon now gives a changed protein, so most proteins are built wrong and the bacterium dies
Why: The bacterium’s genes were written for the usual meaning of that codon.
Every gene that uses the codon now gives a protein with a different amino acid at that spot.
So most of the bacterium’s proteins are built wrong.
So the bacterium dies, and the change dies with it.
Suppose a human gene is placed in a bacterium. The bacterium’s ribosomes read the gene’s mRNA.
Which protein do the bacterium’s ribosomes build?
- A. A bacterial proteinThe mRNA carries the human gene’s codons.
The bacterium’s ribosomes read each codon as the same amino acid a human ribosome would. - B. ✓ The human protein
- C. No proteinThe bacterium’s ribosomes read the human mRNA’s codons with the same genetic code.
So they build a protein from it.
Why: The mRNA carries the human gene’s codons.
The bacterium’s ribosomes read each codon with the same genetic code as a human cell.
So each codon adds the same amino acid, in the same order.
So the protein is the human protein.
Suppose the mRNA for one of a horse’s blood proteins is placed in a tulip cell, and the tulip cell’s ribosomes read it.
Which of the following decides the protein the tulip cell’s ribosomes build from it?
- A. ✓ The codons on the horse’s mRNA, read with the same genetic code a horse uses
- B. The tulip cell’s own genes, which the ribosomes follow instead of the mRNAA ribosome reads the mRNA threaded through it, not the cell’s genes.
The horse’s codons decide the amino acids. - C. The ribosomes themselves: a tulip ribosome builds tulip proteins, whatever mRNA it readsA ribosome adds whatever amino acid each codon names.
The horse’s codons name the same amino acids in a tulip cell, so the protein is the horse’s.
Why: The mRNA carries the horse gene’s codons.
The tulip cell’s ribosomes read those codons with the same genetic code a horse uses.
So each codon adds the same amino acid, in the same order.
So the protein is the horse’s blood protein.
Go back to the chart on your screen, and to the three organisms reading it.
AUG means methionine (Met) in a gut bacterium, in wheat and in you. That is why a human gene put into a bacterium gives the human protein.
We say the code is the same in all living things. A handful of codons mean something slightly different in many mitochondria and in a few microbes, so the exact phrase is nearly universal.
APBIO-U06-L20 Start, build, stop
Freeze the ribosome three times. In the first frame a small subunit sits on the mRNA and one tRNA is arriving. In the second frame a chain of five amino acids hangs from the ribosome and a tRNA is leaving. In the third frame the chain floats free and the ribosome is in two pieces.
What happened between the frames?
Unit 6 · Gene Expression and Regulation
1Initiation: the ribosome finds the start
Suppose a eukaryotic mRNA has reached the cytoplasm.
What does the ribosome do at the mRNA’s 5′ cap?
- A. ✓ Recognizes the cap and grips the mRNA there
- B. Cuts the cap off before it reads the mRNAThe cap stays on the mRNA.
The ribosome recognizes the cap and grips the mRNA there. - C. Pairs a tRNA with the capA tRNA pairs with a codon, not with the cap.
The ribosome recognizes the cap and grips the mRNA there.
Why: The 5′ cap sits at the mRNA’s 5′ end.
The ribosome recognizes the cap and grips the mRNA there.
A ribosome begins reading an mRNA.
At which codon does the ribosome begin reading?
- A. The first three bases at the 5′ end, whatever they areAUG need not be the first three bases of the mRNA.
The ribosome begins at the start codon, AUG. - B. ✓ AUG
- C. UAAUAA is a stop codon, which ends the message.
The ribosome begins at the start codon, AUG.
Why: The codon where the ribosome begins reading is the start codon.
The start codon is AUG.
How does the ribosome turn an mRNA into a chain of amino acids?
It starts when the small subunit binds near the mRNA’s 5′ end and moves to the first AUG. The methionine (Met) tRNA pairs with that AUG, and the large subunit joins.
Then, codon by codon, a tRNA whose anticodon pairs with the next codon brings its amino acid. The ribosome joins that amino acid to the chain by a peptide bond and moves one codon toward the 3′ end.
At a stop codon no tRNA pairs. The chain is released and the ribosome comes apart.
The ribosome is drawn the same way in every frame, so you can read any frame a question shows.
Video: Watch: Initiation: the ribosome finds the start
The small subunit lands near the mRNA’s 5′ end and slides to the first AUG; the Met tRNA pairs its anticodon with AUG; the large subunit closes over them.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L20a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L20a.mp4
Suppose an mRNA reads 5′-GCAUGCCUGACUGGAAGUAA-3′ and has reached the cytoplasm of a eukaryotic cell.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
Its first AUG is the third, fourth and fifth bases. Read in threes from there, its codons are AUG, CCU, GAC, UGG, AAG and UAA.
The small subunit of a ribosome binds the mRNA near its 5′ end. In a eukaryote, the small subunit binds at the 5′ cap.
The small subunit then moves along the mRNA toward the 3′ end until it reaches the first AUG.
A tRNA loaded with methionine (Met) arrives. Its anticodon, 3′-UAC-5′, pairs with AUG.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon.
Now the large subunit joins. The two subunits close round the mRNA and the Met tRNA, and the ribosome is complete.
The ribosome finding the start codon and assembling on it, small subunit first and large subunit last, is called .
The ribosome assembles on an AUG. So nearly every new polypeptide begins with methionine (Met).
We say the ribosome begins at the first AUG along the mRNA. In reality the bases around an AUG matter too, and some mRNAs begin at a later AUG.
The working rule here is the first AUG.
What you are expected to know Describe how translation starts: the small subunit binds near the 5′ end and moves to the first AUG, the Met tRNA pairs with it, and the large subunit joins.
An mRNA has just reached the cytoplasm.
Which part of the ribosome binds the mRNA first?
- A. ✓ The small subunit
- B. The large subunitThe large subunit joins last, after the Met tRNA has paired.
The small subunit binds the mRNA first.
Why: The small subunit binds the mRNA near its 5′ end.
The large subunit joins only after the Met tRNA has paired with AUG.
A ribosome is about to assemble on an mRNA.
Where on the mRNA does its small subunit bind?
- A. Near the 3′ endThe ribosome reads toward the 3′ end, so the 3′ end is read last.
The small subunit binds near the 5′ end. - B. In the middleThe small subunit binds near one end and moves from there.
It binds near the 5′ end. - C. ✓ Near the 5′ end
Why: The ribosome reads an mRNA from its 5′ end toward its 3′ end.
So the small subunit binds near the 5′ end and moves to the first AUG.
The small subunit has reached the first AUG of an mRNA.
Which amino acid does the tRNA that pairs there carry?
- A. Alanine (Ala)Alanine (Ala) is named by other codons, such as GCU.
AUG names methionine (Met), so the tRNA that pairs with AUG carries Met. - B. ✓ Methionine (Met)
- C. No amino acidA tRNA carries the amino acid named by the codon it pairs with.
AUG names methionine (Met), so the tRNA that pairs with AUG carries Met.
Why: The codon at the start is AUG.
AUG names methionine (Met).
So the tRNA that pairs with AUG carries methionine (Met).
The Met tRNA has paired with the first AUG.
Which of the following happens next?
- A. ✓ The large subunit joins
- B. The small subunit leaves the mRNAThe small subunit stays on the mRNA: it is half of the ribosome that will read the mRNA.
The large subunit joins next. - C. The Met tRNA leavesThe Met tRNA stays paired with AUG; its Met will be the first amino acid of the chain.
The large subunit joins next.
Why: The small subunit and the Met tRNA sit on AUG.
The large subunit then joins them.
The ribosome is complete.
Suppose an mRNA has two bases before its first AUG.
Which codon does the ribosome assemble on?
- A. The first three bases at the 5′ endThe small subunit binds near the 5′ end and moves on until it reaches AUG.
The ribosome assembles on that AUG. - B. ✓ The AUG, which begins at the third base
Why: The small subunit binds near the 5′ end.
It moves along the mRNA to the first AUG.
The ribosome assembles on that AUG, wherever it sits.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Suppose an mRNA reads 5′-CAAUGGUCCAUUAA-3′. Three frames of a ribosome beginning to read it are drawn below, lettered J, K and L, in a scrambled order.
In which order do the frames happen?
- A. J, K, LFrame J already has the Met tRNA paired at AUG, so J is not first.
The small subunit binds near the 5′ end first: L. - B. ✓ L, J, K
- C. L, K, JFrame K has the large subunit joined, and the large subunit joins last.
The order is L, J, K.
Why: The small subunit binds near the 5′ end: L.
It moves to the first AUG and the Met tRNA pairs: J.
The large subunit joins: K.
29Quick quiz: initiation mixed practice
Suppose an mRNA’s first AUG is its tenth, eleventh and twelfth bases. A ribosome has just finished initiation on it.
How many of the mRNA’s bases lie before the codon under its first tRNA?
- A. NoneThe small subunit moves past the first nine bases to the first AUG.
The first tRNA pairs there, so nine bases lie before it. - B. ThreeThree is the length of one codon.
The first tRNA sits on the AUG that begins at the tenth base, so nine bases lie before it. - C. ✓ Nine
Why: The small subunit binds near the 5′ end and moves to the first AUG.
The first AUG begins at the tenth base.
The Met tRNA pairs with that AUG.
So nine bases lie before the codon under the first tRNA.
Imagine an mRNA with no AUG anywhere along it.
Which of the following happens?
- A. ✓ No ribosome assembles on it
- B. The ribosome assembles at the first codon insteadThe large subunit joins only after a Met tRNA has paired with an AUG.
With no AUG, no ribosome assembles.
Why: The small subunit binds and moves along the mRNA looking for AUG.
There is no AUG, so no Met tRNA pairs.
The large subunit never joins, so no ribosome assembles.
A eukaryotic mRNA has reached the cytoplasm.
(a) Describe the three events of initiation, in order. (1 pt)
The tRNA carrying methionine (Met) pairs its anticodon with that AUG.
The large subunit joins, and the ribosome is complete.
- Award 1 point for all three in order: the small subunit binds near the 5′ end and moves to the first AUG; the Met tRNA pairs with AUG; the large subunit joins.
What is initiation?
- A. The large subunit binding the mRNA before the small subunit joins itThe small subunit binds first and moves to the first AUG.
The large subunit joins last, once the Met tRNA has paired. - B. The ribosome assembling at the mRNA’s 3′ end and reading toward the 5′ endThe small subunit binds near the 5′ end and the ribosome reads toward the 3′ end.
The ribosome assembles on the first AUG, near the 5′ end. - C. ✓ The ribosome finding the start codon and assembling on it
Why: The ribosome finding the start codon and assembling on it, small subunit first and large subunit last, is called initiation.
34Elongation: one codon, one amino acid
A polypeptide is a chain of amino acids.
What is the bond that joins one amino acid to the next called?
- A. A hydrogen bondHydrogen bonds pair bases and hold folds; they do not join the chain.
A peptide bond joins one amino acid to the next. - B. A bond between a sugar and a phosphateA sugar joined to a phosphate is the backbone of a nucleic acid, not a protein.
A peptide bond joins one amino acid to the next. - C. ✓ A peptide bond
Why: The covalent bond between two amino acids in a chain is called a peptide bond.
A tRNA arrives at a ribosome.
What does the tRNA bring?
- A. ✓ The one amino acid named by the codon it pairs with
- B. The codon it will pair with, copied from the mRNAThe codon is already on the mRNA.
A tRNA carries an anticodon that pairs with it and brings the one amino acid that codon names. - C. A copy of the whole mRNA, read from the geneThe mRNA is already at the ribosome.
A tRNA brings the one amino acid named by the codon it pairs with.
Why: A tRNA pairs its anticodon with a codon.
It carries the one amino acid that codon means.
Now the ribosome is complete, with the Met tRNA on AUG. How does the chain grow?
Video: Watch: Elongation: one codon, one amino acid
One cycle slowed and repeated: a tRNA pairs with the next codon; the ribosome joins the chain to the new amino acid; the ribosome steps one codon toward the 3′ end; the emptied tRNA leaves.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L20b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L20b.mp4
Go back to the mRNA 5′-GC AUG CCU GAC UGG AAG UAA-3′. Suppose the chain so far is Met joined to Pro, and the Pro tRNA sits on the second codon, CCU.
The next codon is GAC. GAC names aspartic acid (Asp).
A tRNA loaded with Asp arrives. Its anticodon, 3′-CUG-5′, pairs with GAC.
Now two tRNAs stand side by side in the ribosome: the Pro tRNA holding the chain, and the Asp tRNA holding its Asp.
The large subunit’s rRNA joins the chain to the new amino acid by a peptide bond. The whole chain now hangs from the Asp tRNA, and the Pro tRNA is empty.
The ribosome then moves one codon along the mRNA, toward the 3′ end. The next codon, UGG, is now in place for the next tRNA.
The emptied Pro tRNA leaves the ribosome. An enzyme will load it with another Pro, ready for the next CCU.
That is one cycle:
- a tRNA pairs with the next codon,
- a peptide bond joins the chain to its amino acid,
- the ribosome moves one codon,
- the emptied tRNA leaves.
The cycle repeats at UGG and at AAG. Each codon adds one amino acid.
So after AAG the chain is Met-Pro-Asp-Trp-Lys.
The chain growing one amino acid per codon, cycle after cycle, is called .
Each step of elongation uses energy. The ribosome breaks down an energy carrier to drive each step, just as a pump breaks down ATP.
Notice where the chain grows. The ribosome joins each new amino acid to the amino acid added just before it.
So the chain grows at its newest end, never at Met.
Here is the whole mechanism in one sentence: the ribosome reads the mRNA three bases at a time, a tRNA whose anticodon pairs with each codon brings one specific amino acid, and the ribosome joins the amino acids in that order, so the base sequence sets the amino acid sequence.
What you are expected to know Describe one cycle of elongation: a tRNA pairs with the next codon, the ribosome joins its amino acid to the chain by a peptide bond, moves one codon toward the 3′ end, and the emptied tRNA leaves.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. Suppose a ribosome reads 5′-CAAUGGUCCAUUAA-3′. Three frames of one cycle are drawn below, lettered J, K and L, in a scrambled order.
In which order do the frames happen?
- A. J, K, LFrame K shows the ribosome already moved, but it moves only after the peptide bond has formed.
The bond forms in L, so L comes before K. - B. ✓ J, L, K
- C. L, J, KFrame L already has the chain on the Val tRNA, so L is not first.
The Val tRNA arrives first: J.
Why: The Val tRNA pairs with the next codon: J.
A peptide bond joins the chain to Val: L.
The ribosome moves one codon toward the 3′ end: K.
The emptied Met tRNA leaves at the same time, also in K.
A ribosome has moved to a new codon.
What pairs with that codon?
- A. An amino acid, directlyAn amino acid has no bases, so it cannot pair with a codon.
The anticodon of a tRNA pairs with the codon. - B. ✓ The anticodon of a tRNA
- C. The other strand of the mRNAAn mRNA is one strand; it has no partner strand.
The anticodon of a tRNA pairs with the codon.
Why: A codon is three bases on the mRNA.
Three bases pair with three partner bases.
The three partner bases are the anticodon of a tRNA.
A ribosome has just joined a new amino acid to a growing chain.
Where in the chain is the new amino acid?
- A. Next to methionine (Met), at the startMethionine (Met) was the first amino acid, and nothing is added beside it after the second.
Each new amino acid joins the newest end. - B. Somewhere in the middleThe ribosome joins each new amino acid to the amino acid added just before it.
So the new amino acid is at the newest end. - C. ✓ At the end that was added most recently
Why: The ribosome joins each new amino acid to the amino acid added just before it.
So the chain grows at its newest end.
The new amino acid is that end.
Suppose a ribosome has built the chain Met-Pro-Asp and is now joining tryptophan (Trp). A student says: “Trp is joined next to Met, at the start of the chain.”
Is the student correct?
- A. ✓ No: Trp is joined to Asp, the amino acid added just before it
- B. Yes: Trp is joined next to Met, at the start of the chainMet stays at the start of the chain.
The peptide bond joins Trp to Asp, the amino acid added just before it.
Why: The chain hangs from the Asp tRNA, with Asp at its newest end.
The peptide bond forms between Asp and the Trp on the arriving tRNA.
So Trp is joined to Asp, not to Met.
A peptide bond has just formed in a ribosome.
Which way does the ribosome move next?
- A. One codon toward the 5′ endThe ribosome reads from the 5′ end toward the 3′ end.
So it moves one codon toward the 3′ end. - B. ✓ One codon toward the 3′ end
Why: The ribosome reads the mRNA from its 5′ end toward its 3′ end.
After each peptide bond it moves one codon toward the 3′ end.
60Quick quiz: elongation mixed practice
A ribosome moves along an mRNA, one codon at a time.
How many amino acids does the chain gain each time the ribosome moves one codon?
- A. NoneA cycle ends with one more amino acid on the chain.
Each codon adds one amino acid. - B. ✓ One
- C. ThreeA codon is three bases, and it names one amino acid.
Each codon adds one amino acid.
Why: One tRNA pairs with each codon.
Each tRNA brings one amino acid.
So the chain gains one amino acid per codon.
An emptied tRNA has left a ribosome.
What happens to that tRNA?
- A. ✓ An enzyme loads it with its amino acid again
- B. The cell breaks it downThe tRNA is not broken down after one use.
An enzyme loads it with its amino acid again. - C. It stays empty and never returnsThe tRNA is used again and again.
An enzyme loads it with its amino acid again.
Why: A tRNA is emptied when its amino acid joins the chain.
An enzyme loads the tRNA with its amino acid again.
The loaded tRNA can pair with its codon again.
A ribosome is part way along an mRNA, with the growing chain hanging from the tRNA on its current codon.
(a) Describe what happens in one cycle of elongation, from the arrival of a tRNA to its departure. (2 pt)
The ribosome joins the chain to that amino acid by a peptide bond.
The ribosome moves one codon toward the 3′ end.
The emptied tRNA leaves.
- Award 1 point for: a tRNA whose anticodon pairs with the next codon brings its amino acid, and the ribosome joins it to the chain by a peptide bond.
- Award 1 point for: the ribosome moves one codon toward the 3′ end and the emptied tRNA leaves.
What is elongation?
- A. The ribosome adding one amino acid for every base it passesA codon is three bases, and each codon adds one amino acid.
Elongation adds one amino acid per codon, cycle after cycle. - B. The chain growing at methionine (Met), each new amino acid joined in front of itMethionine (Met) stays at the start of the chain.
Elongation adds each new amino acid at the chain’s newest end, one per codon. - C. ✓ The chain growing one amino acid per codon, cycle after cycle
Why: The chain growing one amino acid per codon, cycle after cycle, is called elongation.
65Termination: the chain is released
A ribosome reaches a stop codon.
Which of the following is true of a stop codon?
- A. It names methionine (Met)Methionine (Met) is named by the start codon, AUG.
A stop codon names no amino acid. - B. It marks where reading beginsReading begins at the start codon, AUG.
A stop codon names no amino acid and ends the message. - C. ✓ It names no amino acid
Why: A stop codon names no amino acid and ends the message.
A polypeptide has just been built.
What decides the shape it takes?
- A. The number of tRNAs in the cellThe tRNAs bring the amino acids; they do not fold the chain.
The order of amino acids decides the shape. - B. ✓ Its order of amino acids
- C. The ribosome that built itThe ribosome joins the amino acids in order and releases the chain.
The order of amino acids decides the shape.
Why: The order of amino acids in a polypeptide is its primary structure.
The primary structure decides the shape the protein takes.
The shape lets the protein do its job.
The chain has reached five amino acids and the ribosome sits on the last codon, UAA. What ends the reading?
Video: Watch: Termination: the chain is released
The ribosome arrives at UAA; no tRNA pairs with it; the finished chain slips free of its tRNA; the two subunits let go of the mRNA and part.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L20c.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L20c.mp4
Go back to the mRNA 5′-GC AUG CCU GAC UGG AAG UAA-3′. The chain is Met-Pro-Asp-Trp-Lys, hanging from the Lys tRNA on AAG.
The next codon is UAA. UAA is a stop codon.
No tRNA has an anticodon that pairs with a stop codon. So no tRNA arrives at UAA, and the ribosome adds no amino acid.
Instead, the ribosome releases the chain. The bond between the last amino acid, Lys, and its tRNA is broken, and the finished polypeptide floats free.
The two subunits then let go of the mRNA and separate from each other. The cell can use each subunit again on another mRNA.
A stop codon ending the reading, with the chain released and the subunits separated, is called .
The released polypeptide folds into its working shape. Its order of amino acids decides that shape, and the shape lets the protein do its job.
What you are expected to know Describe how translation ends: at a stop codon no tRNA pairs, the finished polypeptide is released, the subunits separate, and the polypeptide folds.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. A ribosome reaches the codon 5′-UGA-3′, drawn above the chart.
Which tRNA pairs with UGA?
- A. ✓ No tRNA
- B. A tRNA carrying methionine (Met)The Met tRNA pairs with AUG, the start codon.
No tRNA has an anticodon that pairs with a stop codon. - C. A tRNA carrying no amino acidEvery tRNA that pairs with a codon carries an amino acid.
No tRNA has an anticodon that pairs with a stop codon.
Why: UGA is a stop codon.
No tRNA has an anticodon that pairs with a stop codon.
So no tRNA pairs with UGA.
A ribosome reaches a stop codon.
How many amino acids does the stop codon add to the chain?
- A. ✓ None
- B. OneA stop codon names no amino acid, and no tRNA pairs with it.
It adds none.
Why: A stop codon names no amino acid.
No tRNA pairs with it, so no amino acid arrives.
The stop codon adds none.
A ribosome has just reached a stop codon.
What happens to the finished chain?
- A. The ribosome carries it to the next mRNAA ribosome takes no chain with it to another mRNA.
At the stop codon the ribosome releases the finished chain. - B. ✓ The ribosome releases it
- C. The ribosome keeps it until the mRNA’s 3′ endReading ends at the first stop codon, not at the 3′ end.
At the stop codon the ribosome releases the finished chain.
Why: No tRNA pairs with the stop codon.
The bond between the last amino acid and its tRNA is broken.
The ribosome releases the finished chain.
A ribosome has finished reading an mRNA at a stop codon.
What happens to the two subunits?
- A. They stay joined and read on to the 3′ endReading ends at the stop codon.
The two subunits let go of the mRNA and separate. - B. ✓ They let go of the mRNA and separate
Why: The chain has been released.
The two subunits let go of the mRNA and separate from each other.
The cell can use each subunit again.
A ribosome has just finished building a polypeptide.
What does the polypeptide do next?
- A. ✓ Folds into its working shape
- B. Stays as a straight chainThe released chain does not stay straight.
It folds into the shape its order of amino acids decides. - C. Pairs with another mRNAA polypeptide has no bases to pair with an mRNA.
It folds into its working shape.
Why: The released polypeptide folds.
Its order of amino acids decides its shape.
The shape lets the protein do its job.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Suppose a ribosome reads 5′-CAAUGGUCCAUUAA-3′. Three frames are drawn below, lettered J, K and L, in a scrambled order.
In which order do the frames happen?
- A. J, L, KFrame J has the chain floating free and the subunits apart, which happens last.
The order is K, L, J. - B. K, J, LFrame J has the chain floating free and the subunits apart, which happens last, not second.
The order is K, L, J. - C. ✓ K, L, J
Why: Initiation: the Met tRNA at AUG and the ribosome assembled, K.
Elongation: the chain growing on a tRNA, L.
Termination: the chain released and the subunits apart, J.
84Quick quiz: termination mixed practice
Suppose a ribosome has released its chain at a stop codon. A student says: “Its two subunits can now assemble on another mRNA and read it.”
Is the student correct?
- A. No: a ribosome reads one mRNA only, and the cell then breaks both of its subunits downThe subunits are not broken down after one mRNA.
They let go of the mRNA, separate, and the cell uses each again. - B. ✓ Yes: the cell uses each subunit again, on one mRNA after another
Why: At the stop codon the chain is released.
The two subunits let go of the mRNA and separate from each other.
The cell uses each subunit again on another mRNA.
A ribosome has reached a stop codon.
(a) Describe what happens at the stop codon, and what the released polypeptide does next. (1 pt)
The ribosome releases the finished polypeptide, and the two subunits separate from the mRNA.
The polypeptide folds into its working shape.
- Award 1 point for: no tRNA pairs and no amino acid is added; the polypeptide is released and the subunits separate; the polypeptide folds into its shape.
What is termination?
- A. The ribosome reaching the mRNA’s 3′ end and falling off thereReading ends at the first stop codon, before the 3′ end.
Termination is a stop codon ending the reading. - B. ✓ A stop codon ending the reading: the chain released, the subunits separated
- C. The ribosome adding a final amino acid at the stop codon before it lets the chain goA stop codon names no amino acid, so nothing is added there.
Termination is a stop codon ending the reading: the chain released, the subunits separated.
Why: A stop codon ending the reading, with the chain released and the subunits separated, is called termination.
88Count the amino acids
A ribosome reads along an mRNA.
Where does the ribosome stop reading?
- A. ✓ At the first stop codon it reaches
- B. At the mRNA’s 3′ endThe ribosome stops before the 3′ end if a stop codon comes first.
Reading ends at the first stop codon. - C. At the next AUG it reachesAUG is where reading begins.
Reading ends at the first stop codon.
Why: A stop codon ends the reading.
The ribosome stops at the first stop codon it reaches.
How long a polypeptide does an mRNA give? Count its codons, but count them right.
Video: Watch: Count the amino acids
The message 5′-AUGGCUUACUAA-3′ grouped in threes from AUG; Met, Ala and Tyr appear under the first three codons; the stop codon gets no name; the count stops at three.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L20d.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L20d.mp4
Suppose an mRNA reads 5′-AUGGCUUACUAA-3′.
Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid.
Read from the first AUG in threes, its codons are AUG, GCU, UAC and UAA.
The chart gives Met for AUG, Ala for GCU and Tyr for UAC. UAA is a stop codon and names no amino acid.
So this mRNA gives a polypeptide of three amino acids, not four. The stop codon is read, but it adds nothing to the chain.
Here is the rule as a word equation.
So the count has three moves:
- find the first AUG,
- count the codons from there to the first stop codon,
- leave the stop codon out of the count.
What you are expected to know Count the amino acids a short mRNA gives: read from the first AUG in threes to the first stop codon, which adds no amino acid.
Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. Suppose an mRNA reads 5′-AUGCACGGCACAGUUCGGUGA-3′, drawn above the chart.
Count the amino acids in the polypeptide this mRNA gives.
Part 1. Count the codons the ribosome reads, from the first AUG up to and including the first stop codon.
Answer: 7 (tolerance ±0)
Part 2. Count how many of those codons are stop codons.
Answer: 1 (tolerance ±0)
Answer: 6 (tolerance ±0)
Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. Suppose an mRNA reads 5′-GAAUGUCCAAGUUUGACCAACGUGCAUAGCC-3′, drawn above the chart.
Count the amino acids in the polypeptide this mRNA gives.
Answer: 8 (tolerance ±0)
Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. Suppose an mRNA reads 5′-AUGUGGUAA-3′, drawn above the chart. A student says: “It gives three amino acids: one for AUG, one for UGG and one for UAA.”
Is the student correct?
- A. ✓ No: UAA is a stop codon and adds no amino acid, so the polypeptide has two
- B. Yes: each of the three codons, AUG, UGG and UAA, adds one amino acid to the polypeptideUAA is a stop codon.
A stop codon names no amino acid, so it adds nothing to the chain.
Why: AUG names Met and UGG names Trp.
UAA is a stop codon and names no amino acid.
So the polypeptide has two amino acids, Met and Trp.
Go back to the three frozen frames.
Initiation: the small subunit finds the first AUG and the methionine (Met) tRNA pairs with it.
Elongation: codon by codon, a tRNA brings an amino acid, a peptide bond forms, and the ribosome moves on.
Termination: a stop codon arrives, no tRNA fits, and the chain is released.
107Mixed practice mixed practice
Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. Suppose an mRNA reads 5′-AUGACCUAGGCUAAA-3′, drawn above the chart.
Count the amino acids in the polypeptide this mRNA gives.
Answer: 2 (tolerance ±0)
A ribosome is reading an mRNA.
Which end of the mRNA does the ribosome reach last?
- A. The 5′ endThe small subunit binds near the 5′ end at the start.
The ribosome moves toward the 3′ end, so it reaches the 3′ end last. - B. ✓ The 3′ end
Why: The small subunit binds near the 5′ end.
The ribosome moves one codon at a time toward the 3′ end.
So the 3′ end is reached last.
A ribosome has just joined an amino acid to the chain.
Which of the following leaves the ribosome next?
- A. The mRNAThe mRNA stays threaded through the ribosome until a stop codon.
The emptied tRNA leaves. - B. The large subunitThe large subunit leaves only at termination.
The emptied tRNA leaves. - C. ✓ The emptied tRNA
Why: The peptide bond moved the chain onto the newest tRNA.
The tRNA that held the chain before is now empty.
The emptied tRNA leaves.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Suppose a ribosome reads 5′-GGAUGUUCAGCUGA-3′. Three frames are drawn below, lettered J, K and L, with the chart beneath them.
Which frame shows the ribosome at a stop codon?
- A. Frame JIn frame J the ribosome sits over UUC and AGC, and both name an amino acid.
The stop codon, UGA, is under the ribosome in frame K. - B. ✓ Frame K
- C. Frame LIn frame L the ribosome is assembling on AUG, the start codon.
The stop codon, UGA, is under the ribosome in frame K.
Why: Read the codons from AUG: AUG, UUC, AGC, UGA.
UGA is a stop codon.
In frame K the ribosome sits over AGC and UGA, and nothing pairs with UGA.
A student says: “Once methionine (Met) is joined to the second amino acid, the emptied Met tRNA leaves the ribosome.”
Is the student correct?
- A. No: the first tRNA holds the chain until the chain is released at terminationA tRNA leaves once its amino acid has been joined to the next one.
The Met tRNA is emptied by the first peptide bond and leaves. - B. ✓ Yes: the first peptide bond empties the Met tRNA, and the emptied tRNA leaves
Why: The first peptide bond joins Met to the second amino acid.
The chain now hangs from the second tRNA.
The emptied Met tRNA leaves the ribosome.
Two ribosomes read the same mRNA, one behind the other.
Which of the following is true of the two polypeptides they make?
- A. ✓ They have the same order of amino acids
- B. The second is shorterBoth ribosomes read from the first AUG to the first stop codon.
Both chains are the same length. - C. The second begins with a different amino acidBoth ribosomes assemble on the same AUG.
Both chains begin with methionine (Met).
Why: Each ribosome assembles on the first AUG and reads the same codons in the same order.
Each codon names the same amino acid.
So the two polypeptides have the same order of amino acids.
A ribosome is on the codon GCA of an mRNA.
Which of the following brings alanine (Ala) to the ribosome?
- A. The codon GCA itselfA codon is three bases on the mRNA. It carries no amino acid.
A tRNA whose anticodon pairs with GCA brings alanine (Ala). - B. ✓ A tRNA whose anticodon pairs with GCA
- C. The large subunitThe large subunit joins the amino acids; it carries none.
A tRNA whose anticodon pairs with GCA brings alanine (Ala).
Why: GCA names alanine (Ala).
The tRNA whose anticodon pairs with GCA is loaded with alanine (Ala).
That tRNA brings it to the ribosome.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. Suppose an mRNA reads 5′-AUGCACUGGGCAUAA-3′, drawn above the chart. Its codons, from AUG, are AUG, CAC, UGG, GCA, UAA. Now imagine a cell that has none of the tRNA whose anticodon pairs with UGG. A ribosome assembles on this mRNA’s AUG.
(a) Predict what happens when the ribosome reaches UGG. (1 pt)
The ribosome stays on UGG, and the chain stops growing at two amino acids, Met and His.
- Award 1 point for: no tRNA pairs with UGG, so no amino acid is added and the ribosome does not move on (the chain stops at Met and His).
(b) Explain your prediction in part (a). (2 pt)
Frame This happens because …
A peptide bond forms only when a tRNA has paired with the codon and brought its amino acid.
No tRNA pairs with UGG in this cell, so no peptide bond forms.
So the ribosome stays on UGG and never reaches GCA.
- Award 1 point for: the ribosome moves one codon only after a peptide bond has formed, and a peptide bond needs a paired tRNA with its amino acid.
- Award 1 point for: no tRNA pairs with UGG in this cell, so no peptide bond forms and the ribosome stays on UGG (it never reaches GCA).
Glossary
- initiation
- The ribosome finding the start codon and assembling on it: the small subunit binds near the mRNA’s 5′ end and moves to the first AUG, the Met tRNA pairs with that AUG, and the large subunit joins.
- elongation
- The chain growing one amino acid per codon, cycle after cycle: a tRNA pairs with the next codon, the ribosome joins its amino acid to the chain by a peptide bond, moves one codon toward the 3′ end, and the emptied tRNA leaves.
- termination
- A stop codon ending the reading: no tRNA pairs with it, the finished polypeptide is released, and the two subunits separate from the mRNA.
APBIO-U06-L21 Translate it
Here is a template strand of DNA: 3′-TAC CCG AAA ATT-5′. You have nothing else but the code chart.
What protein does it make?
Unit 6 · Gene Expression and Regulation
1Codon to anticodon
A tRNA arrives at a ribosome.
Which of the following does its anticodon pair with?
- A. A codon on the gene’s DNAThe gene’s DNA stays in the nucleus.
The anticodon pairs with a codon on the mRNA at the ribosome. - B. ✓ A codon on the mRNA
- C. The anticodon of another tRNATwo tRNAs do not pair with each other.
The anticodon pairs with a codon on the mRNA.
Why: An anticodon is the three bases on a tRNA that pair with a codon on the mRNA.
An RNA strand pairs with another RNA strand.
Which base pairs with adenine when the partner strand is RNA?
- A. ThymineThymine is adenine’s partner in DNA.
RNA has no thymine; adenine pairs with uracil. - B. ✓ Uracil
- C. GuanineGuanine pairs with cytosine.
Adenine pairs with uracil when the partner strand is RNA.
Why: Adenine pairs with thymine, or with uracil when the partner strand is RNA.
How do you get from a DNA sequence to the amino acids?
Write the mRNA against the template strand, U opposite A, from its 5′ end to its 3′ end: 5′-AUG GGC UUU UAA-3′.
Find the first AUG and read in threes to the first stop codon.
The chart gives Met, Gly, Phe, then stop.
The same steps, drawn as one chart from DNA to protein, make one model of the whole route.
Video: Watch: Codon to anticodon
A codon sits in the ribosome. A tRNA lowers its anticodon onto the codon, base against base. The anticodon runs the other way, 3′ to 5′, and each of its bases is the partner of the codon base beneath it.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L21a.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L21a.mp4
Suppose a ribosome sits on the codon 5′-GCU-3′. The tRNA that brings alanine (Ala) has paired with it.
Here is that tRNA over its codon, zoomed in. Its anticodon is 3′-CGA-5′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon.
Each base of the anticodon sits opposite one base of the codon. Opposite bases are partners.
Here is the pairing table again: guanine pairs with cytosine, and adenine pairs with uracil where the partner strand is RNA.
The anticodon runs the other way to the codon. So the codon is written 5′ to 3′, and the anticodon beneath it is written 3′ to 5′.
Here is the codon 5′-GCU-3′. Its first base, G, is at its 5′ end.
Under each base of the codon, write its partner in the same order: under G write C, under C write G, under U write A.
Mark the anticodon 3′ at the left and 5′ at the right: 3′-CGA-5′.
The same steps work the other way, from an anticodon to its codon: write the partner under each base, and swap the end marks.
Here are the steps for writing the anticodon that pairs with a codon:
- Under each base of the codon, write its partner: C under G, G under C, U under A, A under U.
- Mark the anticodon 3′ at the left and 5′ at the right.
- Check that no T appears: RNA carries U, never T.
What you are expected to know Write the anticodon that pairs with a given codon, or the codon a given anticodon pairs with, with both ends marked.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The codon 5′-GUA-3′ is drawn below. Its anticodon is written beneath it as 3′-CA_-5′, with the last base blank.
Which base fills the blank?
- A. ✓ U
- B. AThe blank sits opposite the codon’s last base, A.
An anticodon base is the partner of the codon base, and the partner of A is U.
Why: The blank sits opposite the codon’s third base, A.
Adenine pairs with uracil when the partner strand is RNA.
So the blank is U, and the anticodon is 3′-CAU-5′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The codon 5′-ACC-3′ is drawn below.
Which anticodon pairs with this codon?
- A. 3′-ACC-5′An anticodon carries the partner of each base, not the same base.
Opposite A is U, and opposite C is G. - B. 3′-GGU-5′Written from its 3′ end, the anticodon’s first base sits opposite the codon’s first base, A.
The partner of A is U, so the anticodon begins with U. - C. ✓ 3′-UGG-5′
Why: Each base of the anticodon pairs with the codon base opposite it.
Opposite A is U.
Opposite C is G.
Opposite the second C is G.
So the anticodon is 3′-UGG-5′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The codon 5′-CGG-3′ is drawn below.
Which anticodon pairs with this codon?
- A. 3′-CCG-5′Written from its 3′ end, the anticodon’s first base sits opposite the codon’s first base, C.
The partner of C is G, so the anticodon begins with G. - B. ✓ 3′-GCC-5′
- C. 3′-CGG-5′An anticodon carries the partner of each base, not the same base.
Opposite C is G, and opposite G is C.
Why: Each base of the anticodon pairs with the codon base opposite it.
Opposite C is G.
Opposite G is C.
Opposite the second G is C.
So the anticodon is 3′-GCC-5′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The tRNA drawn below has the anticodon 3′-GGA-5′.
Which codon does this anticodon pair with?
- A. ✓ 5′-CCU-3′
- B. 5′-GGA-3′A codon carries the partner of each anticodon base, not the same base.
Opposite G is C, and opposite A is U. - C. 5′-UCC-3′The anticodon’s first base, at its 3′ end, sits opposite the codon’s first base, at its 5′ end.
The partner of G is C, so the codon begins with C.
Why: Each base of the codon pairs with the anticodon base opposite it.
Opposite G is C.
Opposite the second G is C.
Opposite A is U.
So the codon is 5′-CCU-3′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The tRNA drawn below has the anticodon 3′-CGU-5′.
Which codon does this anticodon pair with?
- A. 5′-ACG-3′The anticodon’s first base, at its 3′ end, sits opposite the codon’s first base, at its 5′ end.
The partner of C is G, so the codon begins with G. - B. ✓ 5′-GCA-3′
- C. 5′-CGU-3′A codon carries the partner of each anticodon base, not the same base.
Opposite C is G, and opposite U is A.
Why: Each base of the codon pairs with the anticodon base opposite it.
Opposite C is G.
Opposite G is C.
Opposite U is A.
So the codon is 5′-GCA-3′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. Four tRNAs are drawn below, lettered J, K, L and M, each with its anticodon written 3′ to 5′. A ribosome sits on the codon 5′-AUG-3′.
Which tRNA can pair its anticodon with this codon?
- A. JThe anticodon of tRNA J copies the codon’s letters.
An anticodon carries each codon base’s partner: opposite A is U, not A. - B. KThe anticodon of tRNA K is written the other way round.
Read 3′ to 5′, its first base, C, sits opposite the codon’s A, and C is not A’s partner. - C. ✓ L
- D. MThe last base of tRNA M is U, opposite the codon’s G.
Guanine pairs with cytosine, not uracil.
Why: Each anticodon base must be the partner of the codon base opposite it.
Opposite A is U, opposite U is A, opposite G is C.
So the pairing anticodon is 3′-UAC-5′, tRNA L.
29From the mRNA to the amino acids
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The codon 5′-AGG-3′ is drawn above the chart.
Which amino acid does it name?
- A. ✓ Arg
- B. GluGlutamic acid (Glu) is the codon GAG.
AGG is read with A for the row and G for the column. - C. SerSerine (Ser) is the codon AGU.
AGG’s third base is G, so read the G line of the cell.
Why: The first base, A, picks the row.
The second base, G, picks the column.
The third base, G, picks the G line of that cell.
The line reads AGG Arg.
Suppose a ribosome reads an mRNA. The mRNA’s codons name Met, Thr, Glu and Trp, and then a stop codon follows.
How many amino acids does the finished chain have?
- A. ✓ Four
- B. FiveNo tRNA pairs with the stop codon.
The stop codon adds no amino acid.
The chain has four amino acids: Met, Thr, Glu and Trp.
Why: Met, Thr, Glu and Trp are four amino acids.
A stop codon names no amino acid.
No tRNA pairs with it.
So the stop codon adds nothing.
The finished chain has four amino acids.
Video: Watch: From the mRNA to the amino acids
The mRNA 5′-AUGGGCUUUUAA-3′ lies flat. The first AUG is boxed, and the letters after it are boxed in threes. Each codon lights its line on the chart, and Met, Gly and Phe appear beneath the codons. The last box reads stop, and nothing appears beneath it.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L21b.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L21b.mp4
Suppose a ribosome translates the mRNA 5′-AUGGGCUUUUAA-3′.
Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid.
Step 1. Find the first AUG. Here it is the first three letters.
Step 2. From that AUG, box the letters in threes: AUG GGC UUU UAA.
Step 3. Read each codon on the chart. AUG names methionine (Met).
GGC: the first base, G, picks the row; the second base, G, picks the column; the third base, C, picks the line. The line reads Gly, glycine.
UUU reads Phe, phenylalanine. UAA reads stop: the ribosome stops there and adds no amino acid.
So the mRNA gives the polypeptide Met, Gly, Phe.
Here are the steps for translating an mRNA with the chart:
- Find the first AUG.
- From that AUG, box the letters in threes.
- Read each codon on the chart: row, column, line.
- Stop at the first stop codon. It adds no amino acid.
What you are expected to know Translate an mRNA into its amino-acid sequence with the code chart: find the first AUG, read codon by codon, and stop at the first stop codon.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. The mRNA 5′-GAUGUUCCACUAG-3′ is drawn below.
Which codon does the ribosome read second?
- A. GAUGAU is the mRNA’s first three letters.
The ribosome begins at the first AUG, not at the first letter, and AUG is its first codon. - B. GUUGUU is the second group when the letters are boxed from the first letter.
The ribosome boxes in threes from the first AUG, one letter later. - C. ✓ UUC
Why: The first AUG begins at the second letter.
From that AUG the letters box as G AUG UUC CAC UAG.
So the second codon is UUC.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The mRNA 5′-C AUG UCC ACA UAG-3′ is drawn above the chart, boxed in threes from its AUG. Its second codon is UCC.
Which amino acid does the second codon name?
- A. ✓ Ser
- B. LeuLeucine (Leu) is the codon CUC.
UCC is read with U for the row and C for the column. - C. ProProline (Pro) sits in the row marked C.
UCC’s first base is U, so read the row marked U.
Why: The first base, U, picks the row.
The second base, C, picks the column.
The third base, C, picks the C line of that cell.
The line reads UCC Ser.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The mRNA 5′-C AUG UCC ACA UAG-3′ is drawn above the chart, boxed in threes from its AUG. Its third codon is ACA.
Which amino acid does the third codon name?
- A. GlnGlutamine (Gln) is the codon CAA.
ACA is read with A for the row and C for the column. - B. ✓ Thr
- C. AsnAsparagine (Asn) is the codon AAC.
ACA’s second base is C, so read the column marked C.
Why: The first base, A, picks the row.
The second base, C, picks the column.
The third base, A, picks the A line of that cell.
The line reads ACA Thr.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. A ribosome translates the mRNA 5′-AUGGUCCAGUAA-3′, drawn above the chart.
Which amino acids does the ribosome join, in order?
- A. Met, Cys, GlnCysteine (Cys) is the codon UGC.
The second codon is GUC: G picks the row and U picks the column, and the line reads Val. - B. Tyr, Gln, ValTyr, Gln, Val are the anticodons read as codons.
The chart is read with the mRNA’s codons: AUG, GUC, CAG. - C. ✓ Met, Val, Gln
Why: The first AUG is the first three letters, so the codons are AUG, GUC, CAG, UAA.
AUG reads Met, GUC reads Val, CAG reads Gln.
UAA is a stop codon and adds no amino acid.
So the ribosome joins Met, Val, Gln.
47From a template strand to the amino acids
RNA polymerase builds an RNA against a template strand. The template strand carries an A.
Which base does the RNA carry opposite that A?
- A. AdenineRNA polymerase pairs each RNA nucleotide against the template, so the RNA carries A’s partner, not A itself.
Opposite A the RNA carries uracil. - B. ✓ Uracil
Why: RNA polymerase pairs each RNA nucleotide with the template base opposite it.
Adenine pairs with uracil when the partner strand is RNA.
So opposite the template’s A the RNA carries uracil.
Video: Watch: From a template strand to the amino acids
The template 3′-TAC CCG AAA ATT-5′ is written out. The mRNA appears beneath it letter by letter, U opposite A. The codons are boxed from AUG, and Met, Gly and Phe appear beneath them from the chart.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L21c.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L21c.mp4
Go back to the template strand from the opening, 3′-TAC CCG AAA ATT-5′.
RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end.
Given a template strand, first write the mRNA against it. Then translate that mRNA: find the first AUG, read in threes, stop at the stop codon.
Step 1. Under each template base, write its RNA partner: under T write A, under A write U, under C write G, under G write C.
Mark the mRNA 5′ at the left and 3′ at the right: 5′-AUG GGC UUU UAA-3′.
The mRNA carries the letters of the gene’s other strand, the non-template strand, with U in place of T.
Step 2. Find the first AUG on the mRNA. Here it is the first three letters.
Step 3. Box the letters in threes from that AUG and read each codon on the chart: Met, Gly, Phe, then stop.
Here are the steps for translating a template strand:
- Under each template base, write its RNA partner: A under T, U under A, G under C, C under G. Mark the mRNA 5′ at the left and 3′ at the right.
- Find the first AUG on the mRNA.
- Box the letters in threes from that AUG and read each codon on the chart.
- Stop at the first stop codon. It adds no amino acid.
What you are expected to know Translate from a DNA template strand: write the mRNA first, then find the first AUG, read in threes and stop at the first stop codon.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. A template strand reads 3′-TAC TAA CGC ATC-5′, as drawn below.
Which mRNA does RNA polymerase build against it?
- A. ✓ 5′-AUGAUUGCGUAG-3′
- B. 5′-UACUAACGCAUC-3′This copies the template’s letters with U for T.
The RNA is built against the template: each base is the template base’s partner. - C. 3′-AUGAUUGCGUAG-5′The RNA is built from its 5′ end, beneath the template’s 3′ end.
So its 5′ end is at the left.
Why: Each RNA nucleotide pairs with the template base opposite it: A under T, U under A, G under C, C under G.
The RNA’s 5′ end sits beneath the template’s 3′ end, at the left.
So the mRNA is 5′-AUGAUUGCGUAG-3′.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The template strand is drawn above the chart. Write the mRNA built against the template strand. Read the mRNA from its first AUG in threes. The mRNA’s second codon is AUU.
Which amino acid does the second codon name?
- A. TyrTyrosine (Tyr) is the codon UAU.
AUU is read with A for the row and U for the column. - B. MetMethionine (Met) is the codon AUG, the G line of the cell.
AUU’s third base is U, so read the U line. - C. ✓ Ile
Why: The first base, A, picks the row.
The second base, U, picks the column.
The third base, U, picks the U line of that cell.
The line reads AUU Ile.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. The template strand is drawn above the chart. Write the mRNA built against the template strand. Read the mRNA from its first AUG in threes. The mRNA’s third codon is GCG.
Which amino acid does the third codon name?
- A. ✓ Ala
- B. ArgArginine (Arg) is the codon CGG.
GCG is read with G for the row and C for the column. - C. GlyGlycine (Gly) is the codon GGC.
GCG’s second base is C, so read the column marked C.
Why: The first base, G, picks the row.
The second base, C, picks the column.
The third base, G, picks the G line of that cell.
The line reads GCG Ala.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. A template strand reads 3′-TAC GAC TTC ACT-5′, as drawn above the chart.
Which amino acids does the ribosome join, in order, from the mRNA built against this template?
- A. Met, Ser, LysSerine (Ser) is the codon UCG.
The mRNA’s second codon is CUG: C picks the row and U picks the column, and the line reads Leu. - B. ✓ Met, Leu, Lys
- C. Tyr, Asp, Phe, ThrThis copies the template’s letters with U for T.
The mRNA is built against the template, base by partner base, so it begins AUG.
Why: Against 3′-TAC GAC TTC ACT-5′ the mRNA is 5′-AUG CUG AAG UGA-3′.
The first AUG is its first three letters.
AUG reads Met, CUG reads Leu, AAG reads Lys.
UGA is a stop codon and adds no amino acid.
So the ribosome joins Met, Leu, Lys.
64The whole route in one chart
A ribosome reads an mRNA and builds a protein from it.
What is this step called?
- A. TranscriptionTranscription is copying a gene into RNA.
Reading the RNA and building the protein from it is translation. - B. ReplicationReplication is copying the DNA before a cell divides.
Reading the RNA and building the protein from it is translation. - C. ✓ Translation
Why: Reading the RNA copy and building the protein from it is called translation.
A gene is copied into RNA in the nucleus.
Which enzyme builds the RNA?
- A. DNA polymeraseDNA polymerase builds new DNA during replication.
The enzyme that builds the RNA copy of a gene is RNA polymerase. - B. ✓ RNA polymerase
- C. A ribosomeA ribosome reads the mRNA and joins amino acids.
The enzyme that builds the RNA copy of a gene is RNA polymerase.
Why: RNA polymerase binds the promoter and reads the template strand.
It pairs RNA nucleotides against the template, so RNA polymerase builds the RNA.
In a eukaryote, a mature mRNA is ready in the nucleus.
How does the mature mRNA reach the cytoplasm?
- A. ✓ It passes through a pore in the nuclear envelope
- B. A ribosome pulls it out of the nucleusRibosomes work in the cytoplasm, outside the nucleus.
The mature mRNA leaves through a pore in the nuclear envelope.
Why: The nuclear envelope has openings called pores.
The mature mRNA leaves the nucleus through a pore and reaches the ribosomes in the cytoplasm.
Video: Watch: The whole route in one chart
The route map of the cell appears. Beneath it three boxes assemble themselves: DNA in the nucleus; an arrow named transcription, by RNA polymerase; mRNA, which leaves through a pore; an arrow named translation, by a ribosome with tRNAs; and the polypeptide.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L21d.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L21d.mp4
Go back to the route map of a eukaryotic cell: the gene in the nucleus, the mRNA leaving through a pore, the ribosome building the protein.
The same route can be drawn as a chart of three boxes. Here is the flow chart of the route from DNA to polypeptide.
Box 1 is the DNA, in the nucleus. RNA polymerase reads the template strand 3′ to 5′ and builds the mRNA 5′ to 3′.
That step is transcription. Its arrow names the step and the enzyme that acts.
Box 2 is the mRNA. In a eukaryote, enzymes in the nucleus add its cap and tail and cut out its introns.
Then the mRNA leaves through a pore.
Box 3 is the polypeptide. A ribosome in the cytoplasm reads the mRNA 5′ to 3′.
tRNAs pair their anticodons with the codons, and the ribosome joins the amino acids.
That step is translation. Its arrow names the step and what acts: a ribosome, with tRNAs.
A model of the route names four things at every step: the molecule, the place, what acts, and the direction of each strand.
Suppose a model draws the ribosome inside the nucleus. That model is wrong: the mRNA leaves the nucleus first, and ribosomes work in the cytoplasm.
To judge a model, read each step against the flow chart: the right molecule, in the right place, with the right thing acting.
What you are expected to know Represent the flow of information from DNA to mRNA to polypeptide as a labeled model, naming the molecules, the places, what acts at each step and the direction of each strand.
The flow chart of the route from gene to protein is drawn below with its words removed and six of its blanks marked by circled letters, J to P. Box 1 is marked J.
Which molecule belongs in box J?
- A. mRNAThe mRNA is made from the DNA, so it is the second box.
The route begins with the DNA in the nucleus. - B. ✓ DNA
- C. A polypeptideThe polypeptide is what the route ends with, at the ribosome.
The route begins with the DNA in the nucleus.
Why: The route begins in the nucleus with the gene’s DNA.
So box J, in the nucleus, is the DNA.
The flow chart of the route from gene to protein is drawn below with its words removed and six of its blanks marked by circled letters, J to P. The first arrow’s step is marked K.
Which step is K?
- A. TranslationTranslation is the second arrow, from the mRNA to the polypeptide.
The first arrow, from the DNA to the mRNA, is transcription. - B. ✓ Transcription
Why: The first arrow leads from the DNA to the mRNA.
Copying a gene into RNA is transcription.
So K is transcription.
The flow chart of the route from gene to protein is drawn below with its words removed and six of its blanks marked by circled letters, J to P. What acts at the first arrow is marked L.
What acts at L?
- A. ✓ RNA polymerase
- B. A ribosomeA ribosome acts at the second arrow, building the polypeptide.
At the first arrow, RNA polymerase builds the mRNA. - C. DNA polymeraseDNA polymerase copies DNA during replication.
At the first arrow, RNA polymerase builds the mRNA.
Why: The first arrow is transcription, from the DNA to the mRNA.
RNA polymerase reads the template strand and builds the mRNA.
So L is RNA polymerase.
The flow chart of the route from gene to protein is drawn below with its words removed and six of its blanks marked by circled letters, J to P. Box 2 is marked M.
Which molecule belongs in box M?
- A. DNAThe DNA stays in the nucleus and is the first box.
The molecule that leaves through a pore is the mRNA. - B. A polypeptideThe polypeptide is built at the ribosome and is the third box.
The molecule that leaves through a pore is the mRNA. - C. ✓ mRNA
Why: Box M is made in the nucleus and leaves through a pore.
The mRNA is made from the DNA in the nucleus and carries the gene’s message to a ribosome.
So M is the mRNA.
The flow chart of the route from gene to protein is drawn below with its words removed and six of its blanks marked by circled letters, J to P. What acts at the second arrow is marked N.
What acts at N?
- A. RNA polymeraseRNA polymerase builds the mRNA at the first arrow.
At translation, a ribosome reads the mRNA and tRNAs bring the amino acids. - B. ✓ A ribosome with tRNAs
Why: The second arrow is translation, from the mRNA to the polypeptide.
A ribosome reads the codons, and tRNAs pair with them and bring the amino acids.
So N is a ribosome with tRNAs.
The flow chart of the route from gene to protein is drawn below with its words removed and six of its blanks marked by circled letters, J to P. The direction the mRNA is read is marked P.
In which direction does the ribosome read the mRNA?
- A. ✓ From its 5′ end toward its 3′ end
- B. From its 3′ end toward its 5′ endThe ribosome finds the first AUG near the mRNA’s 5′ end and moves toward the 3′ end.
So the mRNA is read 5′ to 3′.
Why: The ribosome begins at the first AUG, near the mRNA’s 5′ end.
It moves one codon at a time toward the 3′ end.
So P reads 5′ to 3′.
A student drew the model below of the route from gene to protein in a eukaryote. One step is drawn in the wrong place.
Which step is drawn in the wrong place?
- A. ✓ Adding the cap and tail
- B. TranscriptionTranscription is drawn in the nucleus, where RNA polymerase reads the gene.
That place is right. - C. TranslationTranslation is drawn at a ribosome in the cytoplasm.
That place is right.
Why: The model adds the cap and tail after the mRNA has left through a pore.
Enzymes in the nucleus add the cap and tail before the mRNA leaves.
So the cap and tail are drawn in the wrong place.
Go back to the template, 3′-TAC CCG AAA ATT-5′. The mRNA built against it is 5′-AUG GGC UUU UAA-3′.
From the chart: Met, Gly, Phe, stop. Three amino acids.
And the whole route from DNA to protein is drawn in one chart.
92Mixed practice mixed practice
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. A student looks at the codon 5′-UCG-3′ and says: “The tRNA that pairs with it has the anticodon 3′-AGC-5′.”
Is the student correct?
- A. No: the anticodon is 3′-UCG-5′, the codon’s own three letters, copied and read the same wayAn anticodon carries the partner of each codon base, not the same letter.
Opposite U is A, opposite C is G, opposite G is C. - B. ✓ Yes: opposite U is A, opposite C is G and opposite G is C, so the anticodon is 3′-AGC-5′
Why: Each anticodon base is the partner of the codon base opposite it.
Opposite U is A.
Opposite C is G.
Opposite G is C.
So the anticodon is 3′-AGC-5′, and the student is correct.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. A tRNA is drawn below with its anticodon, 3′-AAC-5′.
Which codon on an mRNA does this tRNA pair with?
- A. 5′-GUU-3′The anticodon’s first base, at its 3′ end, sits opposite the codon’s first base, at its 5′ end.
The partner of A is U, so the codon begins with U. - B. 5′-AAC-3′A codon carries the partner of each anticodon base, not the same base.
Opposite A is U, and opposite C is G. - C. ✓ 5′-UUG-3′
Why: Each codon base is the partner of the anticodon base opposite it.
Opposite A is U.
Opposite the second A is U.
Opposite C is G.
So the codon is 5′-UUG-3′.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. Suppose a ribosome translates the mRNA 5′-AUGUGCAGAUAA-3′, drawn above the chart.
Which amino acids does it join, in order?
- A. ✓ Met, Cys, Arg
- B. Tyr, Thr, SerTyr, Thr, Ser are the anticodons read as codons.
The chart is read with the mRNA’s codons: AUG, UGC, AGA. - C. Met, Val, ArgValine (Val) is the codon GUC.
The second codon is UGC: U picks the row and G picks the column, and the line reads Cys.
Why: The first AUG is the first three letters, so the codons are AUG, UGC, AGA, UAA.
AUG reads Met, UGC reads Cys, AGA reads Arg.
UAA is a stop codon and adds no amino acid.
So the ribosome joins Met, Cys, Arg.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. A gene’s template strand reads 3′-TAC GAG CTA ATT-5′, as drawn above the chart.
Which amino acids does the ribosome join, in order?
- A. Met, Ser, AspSerine (Ser) is the codon UCC.
The mRNA’s second codon is CUC: C picks the row and U picks the column, and the line reads Leu. - B. ✓ Met, Leu, Asp
- C. Tyr, Glu, Leu, IleThis copies the template’s letters with U for T.
The mRNA is built against the template, base by partner base, so it begins AUG.
Why: Against 3′-TAC GAG CTA ATT-5′ the mRNA is 5′-AUG CUC GAU UAA-3′.
The first AUG is its first three letters.
AUG reads Met, CUC reads Leu, GAU reads Asp.
UAA is a stop codon and adds no amino acid.
So the ribosome joins Met, Leu, Asp.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. An mRNA reads 5′-CCAUGGCAUAA-3′, as drawn below. A student says: “The ribosome begins reading at the first letter, C, so its first codon is CCA.”
Is the student correct?
- A. ✓ No: the ribosome begins at the first AUG, so its first codon is AUG
- B. Yes: the ribosome begins at the mRNA’s first letterThe ribosome does not begin at just any letter.
It moves to the first AUG and reads in threes from there.
Why: The ribosome begins at the first AUG, not at the first letter.
Here the first AUG begins at the third letter.
So the first codon is AUG, and the letters before it are not read.
In a eukaryote, a gene is transcribed and its protein is built.
Which of the following happens in the cytoplasm?
- A. RNA polymerase copies the gene into pre-mRNARNA polymerase reads the gene, and the gene is inside the nucleus.
The mRNA is read in the cytoplasm. - B. Enzymes cut the introns out of the pre-mRNAThe introns are cut out inside the nucleus, before the mRNA leaves through a pore.
The mRNA is read in the cytoplasm. - C. ✓ A ribosome reads the mRNA and builds the polypeptide
Why: The gene is copied and its pre-mRNA processed inside the nucleus.
The mature mRNA leaves through a pore.
Ribosomes work in the cytoplasm.
So the mRNA is read and the polypeptide built in the cytoplasm.
A student describes the route from gene to protein in two steps: first transcription, then translation. The student says that what joins amino acids into a chain acts at transcription. The student says that what builds the mRNA acts at translation.
Which of the following is wrong with the student’s description?
- A. ✓ What acts at each step is written against the wrong step: the two actors are swapped
- B. The two steps are in the wrong order: translation comes before transcriptionTranscription comes first: the mRNA is copied from the DNA before a ribosome can read it.
The two actors are swapped. - C. The route should begin with replication, and DNA polymerase acts thereReplication copies DNA before a cell divides. It is not a step of this route.
The route begins with transcription, and RNA polymerase acts there.
Why: RNA polymerase reads the template strand and builds the mRNA: transcription.
A ribosome, with tRNAs, reads the mRNA and joins the amino acids: translation.
So the two actors are written against the wrong steps.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end. Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. A gene’s template strand reads 3′-TAC TAT CCC ATT-5′, drawn above the chart. RNA polymerase transcribes it, and a ribosome translates the mRNA.
(a) Identify the mRNA built against this template strand, with its ends marked. (1 pt)
The mRNA is 5′-AUG AUA GGG UAA-3′.
- Award 1 point for: 5′-AUGAUAGGGUAA-3′ (spacing optional), with the 5′ end at the left.
(b) Identify the amino acids the ribosome joins, in order. (1 pt)
AUG reads Met, AUA reads Ile, GGG reads Gly.
UAA is a stop codon and adds no amino acid.
So the ribosome joins Met, Ile, Gly.
- Award 1 point for: Met, Ile, Gly (three-letter or full names), with no amino acid for UAA.
(c) Explain how this example demonstrates that the order of bases on the template strand sets the order of amino acids in the polypeptide. (2 pt)
Frame The template’s bases set the amino acids because …
So the template’s order of bases fixes the mRNA’s order of bases.
The ribosome reads the mRNA three bases at a time, and each codon means one amino acid.
A tRNA whose anticodon pairs with each codon brings that amino acid.
So the ribosome joins the amino acids in codon order, and the template’s order sets that order.
- Award 1 point for: each template base pairs with one RNA partner, so the template’s base order fixes the mRNA’s base order.
- Award 1 point for: the ribosome reads the mRNA in codons, a tRNA whose anticodon pairs with each codon brings one specific amino acid, and the amino acids are joined in codon order.
APBIO-U06-L22 Change one thing
Suppose three test tubes each hold mRNA, ribosomes, amino acids and an energy supply.
Tube 1 also holds tRNA. Tube 2 holds no tRNA. Tube 3 holds tRNA, but its mRNA’s only AUG has been changed to AUC.
In which tubes do the ribosomes build a polypeptide?
Unit 6 · Gene Expression and Regulation
1Take one part away
A small ribosomal subunit binds an mRNA near its 5′ end.
Which codon does it move to before the large subunit joins?
- A. The first stop codonA stop codon ends translation; nothing begins there.
The small subunit moves to the first AUG. - B. ✓ The first AUG
- C. The codon nearest the 3′ endThe small subunit binds near the 5′ end and moves to the first AUG, not to the far end.
Why: The small subunit binds near the mRNA’s 5′ end.
The small subunit moves along to the first AUG.
The methionine (Met) tRNA pairs there, and the large subunit joins.
A ribosome has moved on to the next codon of an mRNA.
Which of the following brings the next amino acid to that codon?
- A. ✓ A tRNA paired with the codon
- B. RNA polymeraseRNA polymerase builds RNA in the nucleus; it carries no amino acid.
- C. The large subunit of the ribosomeThe large subunit joins the amino acids; it does not bring them.
Why: Each tRNA carries one amino acid.
Its anticodon pairs with the codon in the ribosome.
So a tRNA whose anticodon pairs with the codon brings the next amino acid.
A ribosome is reading along an mRNA.
Which of the following ends translation?
- A. The ribosome reaching the mRNA’s 3′ endTranslation ends at the first stop codon, which sits before the 3′ end.
- B. The mRNA’s poly-A tail entering the ribosomeThe tail is never read; translation ends at the first stop codon before it.
- C. ✓ A stop codon entering the ribosome
Why: No tRNA pairs with a stop codon.
So no amino acid is added there.
The polypeptide is released, and the two subunits separate: translation ends.
A eukaryotic mRNA reaches the ribosomes in the cytoplasm.
Where does a ribosome first grip it?
- A. Any codon along itThe ribosome grips the mRNA at one place only, the 5′ cap.
From the cap it moves along to the first AUG. - B. ✓ Its 5′ cap
- C. Its 3′ endThe ribosome recognizes the 5′ cap, at the other end of the mRNA.
Why: The ribosome recognizes the 5′ cap.
The ribosome grips the mRNA there and moves along to the first AUG.
What happens to translation when one part is missing or changed?
Name the part’s job, and the answer follows.
No tRNA: no amino acid is delivered, so no chain forms.
No start codon: the ribosome never begins.
No cap: the ribosome rarely binds.
A scarce tRNA: that codon is read slowly, so a message full of it is translated slowly.
Video: Watch: Take one part away
Two test tubes stand side by side. Both hold mRNA, ribosomes, amino acids and an energy supply. Only the left tube holds tRNA. In the left tube, tRNAs bring amino acids to a ribosome and a chain grows. In the right tube, the ribosome sits on the mRNA and nothing grows.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L22a.mp4
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Go back to tube 1. Tube 1 holds an mRNA, ribosomes, amino acids, tRNAs loaded with their amino acids, and an energy supply.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon.
Here is a ribosome in tube 1, part way along the mRNA 5′-AUG GCA GAU UGG UAA-3′.
A tRNA brings its amino acid to its codon. The ribosome joins that amino acid to the growing chain.
So the ribosomes in tube 1 build the polypeptide the mRNA codes for: Met, Ala, Asp, Trp.
To predict what happens when one part is missing, name that part’s job.
Then ask what happens when no part does that job.
Now consider tube 2. Tube 2 holds every part except tRNA.
A tRNA’s job is to bring one amino acid to its codon.
With no tRNA, no amino acid arrives at the ribosome.
The ribosome sits on the mRNA with nothing to join. So no chain forms, and no polypeptide is built in tube 2.
Now imagine a tube that holds mRNA, loaded tRNAs, amino acids and an energy supply, but no ribosomes.
The ribosome’s job is to hold the mRNA and a tRNA together and to join the amino acids.
With no ribosome, no tRNA is held against its codon, and no amino acid is joined to another.
So no chain forms. The tRNAs and amino acids float in the tube unused.
Now consider tube 3. Its mRNA’s only AUG has been changed to AUC.
The start codon’s job is to mark where the ribosome begins. The small subunit moves along the mRNA to the first AUG, and the methionine (Met) tRNA pairs there.
AUC is not AUG. The small subunit finds no AUG anywhere on this mRNA.
So the Met tRNA never pairs, and the large subunit never joins.
The ribosome never begins, so no polypeptide is built in tube 3.
AUC names isoleucine (Ile), but reading never starts. So no codon of this mRNA is read at all.
Now imagine a eukaryotic mRNA with no 5′ cap reaches the ribosomes in the cytoplasm.
The cap’s job is to be the place the ribosome grips.
With no cap, the ribosome has nothing to recognize at the 5′ end. So few ribosomes grip the mRNA.
So the mRNA is barely translated, and the cell makes far less polypeptide from it.
Now imagine a cell that holds very little of the tRNA that pairs with one codon.
A ribosome reaches that codon and waits. The ribosome moves on only after a tRNA has paired and its amino acid has been joined.
A scarce tRNA takes longer to arrive. So the ribosome waits longer at that codon.
With none of that tRNA, the ribosome waits for ever, and the chain stops there. With only a little, the ribosome waits longer than usual, and the chain still grows.
So a codon whose tRNA is scarce is read slowly. A message full of that codon is translated slowly.
Suppose a scientist measures how long a ribosome spends at each codon.
Here is a bar chart of the time spent at the four codons that name valine (Val). Each time is relative to the fastest of the four, and the gridlines are every 0.5.
The GUA bar reaches the 3.0 line, and the GUC bar reaches the 1.0 line. So a ribosome spends three times as long at GUA as at GUC.
A long time at a codon means its tRNA is scarce. So this cell holds far less GUA tRNA than GUC tRNA.
So a message that uses GUA for its valines is translated more slowly than a message that uses GUC.
What you are expected to know Predict what the ribosomes build when one part of translation is removed or changed.
What you are expected to know Explain each prediction from the job of the part that changed.
A tube holds mRNA, ribosomes, amino acids and an energy supply, but no tRNA, and it has made nothing. A scientist now adds a full supply of tRNAs loaded with their amino acids.
What do the ribosomes in the tube build now?
- A. No polypeptideThe tube now holds every part.
tRNAs bring the amino acids, and the ribosomes join them. - B. The polypeptide, more slowlyThe tube now holds a full supply of tRNAs.
Each codon’s tRNA arrives at the usual speed. - C. ✓ The polypeptide, at the usual rate
Why: The missing part was tRNA.
A tRNA’s job is to bring one amino acid to its codon.
Now tRNAs bring the amino acids, and the ribosomes join them.
So the ribosomes build the polypeptide at the usual rate.
Suppose a toxin stops every ribosome in a yeast cell from binding mRNA. The yeast cell’s mRNAs, tRNAs and amino acids are unchanged.
What do the yeast cell’s ribosomes build from its mRNAs?
- A. ✓ No polypeptide
- B. The polypeptide, more slowlyNo ribosome binds the mRNA, so no codon is read at all.
- C. The polypeptide, at the usual rateOnly a ribosome holds a tRNA against its codon and joins the amino acids.
Why: The ribosome’s job is to hold the mRNA and a tRNA together and to join the amino acids.
No ribosome binds the mRNA.
So no codon is read, and no amino acid is joined.
So the yeast cell’s ribosomes build no polypeptide.
Suppose an mRNA’s only AUG is changed to UUG. A tube holds that mRNA and every other part of translation.
What do the ribosomes in the tube build?
- A. ✓ No polypeptide
- B. The polypeptide, more slowlyUUG is not a start codon.
With no AUG, the small subunit never finds a start, and reading never begins. - C. The polypeptide, at the usual rateThe Met tRNA pairs only with AUG.
With no AUG, the ribosome never assembles.
Why: The start codon’s job is to mark where the ribosome begins.
This mRNA has no AUG.
So the Met tRNA never pairs, and the large subunit never joins.
Reading never starts.
So no polypeptide is built.
Suppose a eukaryotic cell’s capping enzyme works poorly, so most new copies of one of its mRNAs carry no 5′ cap. The enzyme is then repaired, so every new copy of that mRNA carries a cap.
How does the amount of polypeptide made from that mRNA change?
- A. The amount fallsThe cap is the place the ribosome grips.
More ribosomes grip a capped mRNA, not fewer. - B. The amount stays the sameWithout a cap, few ribosomes gripped the mRNA.
With the cap, many grip it. - C. ✓ The amount rises
Why: The cap’s job is to be the place the ribosome grips.
With no cap, few ribosomes gripped the mRNA.
With a cap, many ribosomes grip it and translate it.
So the amount of polypeptide made rises.
Suppose a cell holds very little of the tRNA that pairs with UCU. One of its mRNAs uses UCU at twenty of its codons.
What does a ribosome build from that mRNA?
- A. No polypeptideA little of that tRNA still arrives, so the ribosome waits and then moves on.
- B. ✓ The polypeptide, more slowly
- C. The polypeptide, at the usual rateA scarce tRNA takes longer to arrive, so the ribosome waits longer at every UCU.
Why: Only the UCU tRNA can bring the amino acid for UCU.
That tRNA is scarce.
A scarce tRNA takes longer to arrive.
The ribosome waits longer at every UCU, then moves on.
So the polypeptide is built, more slowly.
A student says: “In a tube with no tRNA, the ribosome joins the amino acids straight onto the codons, so the polypeptide is still made.”
Is the student correct?
- A. ✓ No: an amino acid reaches the ribosome only on a tRNA, so with no tRNA none is joined
- B. Yes: the ribosome pairs each codon directly with an amino acid, so the polypeptide is still madeAmino acids do not pair with codons.
A tRNA’s anticodon pairs with the codon, and the tRNA carries the amino acid.
Why: A tRNA’s anticodon pairs with the codon.
The tRNA carries the amino acid to the ribosome.
With no tRNA, no amino acid reaches the ribosome.
So nothing is joined, and no polypeptide is built.
Suppose a cell holds much more of the tRNA that pairs with UUC than of the tRNA that pairs with UUU. Both codons name phenylalanine (Phe). Two mRNAs code for the same polypeptide: one uses UUU at every phenylalanine, the other uses UUC. A student says: “The UUC message is translated faster, because a UUC tRNA arrives sooner at each phenylalanine codon.”
Is the student correct?
- A. No: the two messages give the same polypeptide, so a ribosome translates them at the same rateThe two messages give the same polypeptide, but the ribosome waits at each codon for its tRNA.
The cell holds more UUC tRNA, so the ribosome waits less at UUC. - B. ✓ Yes: the cell holds more UUC tRNA, so one arrives sooner at each UUC and the ribosome waits less
Why: Only the tRNA whose anticodon pairs with a codon can bring its amino acid.
The cell holds more UUC tRNA than UUU tRNA.
So a UUC tRNA arrives sooner, and the ribosome waits less at each UUC.
So the UUC message is translated faster.
Suppose a scientist measures how long a ribosome spends at each codon of a message. The bar chart below shows the time spent at the four codons that name glycine (Gly), relative to the fastest of them, with a gridline every 0.5. The scientist then writes two messages for the same polypeptide: the GGC message uses GGC at every glycine, and the GGG message uses GGG at every glycine.
(a) Identify the codon at which the ribosome spends the longest time. (1 pt)
So the ribosome spends the longest time at GGA.
- Award 1 point for: GGA.
(b) Predict which message is translated faster, the GGC message or the GGG message. (1 pt)
So the ribosome spends less time at each glycine codon of the GGC message.
The GGC message is translated faster.
- Award 1 point for: the GGC message (the ribosome spends less time at GGC than at GGG).
(c) Explain why the ribosome spends a different time at different codons for the same amino acid. (2 pt)
Frame The time differs because …
The four glycine codons are read by different tRNAs.
The cell holds different amounts of these tRNAs.
A scarce tRNA takes longer to arrive at the ribosome.
The ribosome joins the next amino acid and moves on only after a tRNA has paired.
So the ribosome waits longer at a codon whose tRNA is scarce.
- Award 1 point for: the tRNAs that read the different codons for one amino acid are present in different amounts (a scarce tRNA takes longer to arrive). Accept: different codons are read by different tRNAs.
- Award 1 point for: the ribosome moves on from a codon only after a tRNA has paired and its amino acid is joined, so it waits longer at a codon whose tRNA is scarce.
Here is a table of the five changes: the part removed or changed, its job, and what the ribosomes build.
Go back to the three tubes.
Tube 1 holds every part. Its ribosomes build the polypeptide.
Tube 2 holds no tRNA. No amino acid is delivered, so no chain forms.
Tube 3 has no start codon. The ribosome never begins, so no chain forms.
APBIO-U06-L22B From gene to trait, link by link
Suppose a plant carries two versions of one pigment gene. One version gives purple flowers. The other gives white flowers.
The two versions differ at one base of DNA. How does a change of one base end up as a change in a flower’s color?
Unit 6 · Gene Expression and Regulation
1Six links from allele to flower
A gene comes in more than one version.
What is each version of the gene called?
- A. A genotypeA genotype is the pair of alleles an organism carries for one gene.
One version of a gene is an allele. - B. ✓ An allele
- C. A chromosomeA chromosome carries many genes.
One version of one gene is an allele.
Why: When a gene comes in more than one version, each version is called an allele.
A plant carries the allele D on one homolog and the allele d on the other.
Which of the following is the plant’s genotype?
- A. ✓ Dd
- B. DD is one allele: the genotype of a gamete, which carries one homolog.
The plant carries two alleles, so its genotype is the pair, Dd.
Why: A body cell has two homologs, with one allele on each.
The plant’s genotype is the pair of alleles it carries: Dd.
Which of the following is part of a plant’s phenotype?
- A. The base sequence of its DNANo one can observe a base sequence by looking at the plant.
The phenotype is the plant’s set of observable features. - B. The alleles it carriesThe alleles a plant carries are its genotype.
The phenotype is the plant’s set of observable features, such as its flower color. - C. ✓ The color of its flowers
Why: An organism’s set of observable features is called its phenotype.
Flower color can be observed, so it is part of the phenotype.
Two polypeptides contain the same amino acids in a different order.
Which of the following is true of their shapes?
- A. They fold into the same shapeThe order of amino acids decides the shape a polypeptide folds into.
A different order gives a different shape. - B. ✓ They fold into different shapes
Why: The order of amino acids in a polypeptide is its primary structure.
The primary structure decides the shape the protein takes.
A different order gives a different shape.
Which of the following lets a protein do its job?
- A. ✓ Its shape
- B. Its lengthTwo proteins of the same length can have different shapes and different jobs.
A protein’s shape is what lets it do its job. - C. The number of copies the cell makesMore copies do more of the same job.
What lets each copy do its job is its shape.
Why: A protein folds into a particular shape.
That shape is what lets the protein do its job.
A ribosome translates an mRNA.
Which of the following sets the order of the amino acids it joins?
- A. The order in which tRNAs happen to reach the ribosomeA tRNA is used only when its anticodon pairs with the codon the ribosome is reading.
The order of the codons sets the order of the amino acids. - B. The size of the ribosomeThe ribosome’s size decides nothing about the order.
The order of the bases on the mRNA sets the order of the amino acids. - C. ✓ The order of the bases on the mRNA
Why: The ribosome reads the mRNA three bases at a time.
A tRNA whose anticodon pairs with each codon brings one specific amino acid.
The ribosome joins the amino acids in that order.
So the base sequence sets the amino acid sequence.
How does a genotype become a phenotype?
Follow the chain link by link.
The allele’s DNA sequence sets the mRNA sequence. The mRNA sequence sets the order of amino acids.
The order of amino acids sets how the protein folds. The fold sets what the protein does.
A working pigment enzyme makes purple pigment. An enzyme with a different shape makes none, so the flower is white.
Video: Watch: Six links from allele to flower
Six boxes fill one at a time: the purple allele’s DNA, its mRNA, its amino acid order, its fold, its job, and purple flowers. Then the white allele takes the first box, and the change travels down the chain box by box until the last box reads white flowers.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L22Ba.mp4
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Suppose one plant carries two copies of the purple version of the pigment gene. A second plant carries two copies of the white version.
The purple version and the white version are two alleles of one gene. The pair of alleles a plant carries is its genotype.
The color of a plant’s flowers is part of its phenotype.
Here is the flow chart of the route from a gene to its polypeptide: DNA, mRNA, polypeptide.
The chain from a genotype to a phenotype has six links: the allele’s DNA sequence, the mRNA, the order of amino acids, the fold, the job, and the trait.
Here are the six links as six boxes, one row for each allele.
A strand is written from its 5′ end to its 3′ end, and both ends are marked.
RNA polymerase reads the template strand from its 3′ end toward its 5′ end and builds the RNA from its 5′ end toward its 3′ end.
Here is the one position where the two alleles differ: three bases of the template strand.
The purple allele’s template strand reads 3′-CTA-5′ there. The white allele’s template strand reads 3′-CAA-5′.
The two alleles differ at the middle base: T in the purple allele, A in the white allele.
Under each template base, RNA polymerase places its partner: G opposite C, A opposite T, U opposite A.
So the purple allele’s mRNA reads 5′-GAU-3′ at that position, and the white allele’s mRNA reads 5′-GUU-3′.
One changed base in the DNA gives one changed base in the mRNA. The DNA sequence sets the mRNA sequence.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon.
Read the two codons on the chart.
GAU reads Asp, aspartic acid. GUU reads Val, valine.
So the purple allele’s enzyme carries aspartic acid (Asp) at that spot, and the white allele’s enzyme carries valine (Val).
Every other amino acid in the two enzymes is the same. One changed codon gives one changed amino acid.
So the mRNA sequence sets the order of amino acids.
Asp has a charged R group, and water is attracted to it. Val has a nonpolar R group, and water is not attracted to it.
So the chain folds differently around that spot. The order of amino acids sets the fold.
The enzyme’s job is to make purple pigment. The pigment’s building block is the enzyme’s substrate, and it binds in the enzyme’s active site.
In the purple allele’s enzyme, the active site has the working shape. The building block binds, and the enzyme makes pigment.
In the white allele’s enzyme, the different fold changes the active site. The building block does not bind, and the enzyme makes no pigment.
The fold sets what the enzyme does.
A petal cell with the working enzyme fills with purple pigment. So the flowers are purple.
A petal cell with the other enzyme makes no pigment. So the flowers are white.
What the enzyme does sets the trait.
One changed base, carried down six links, ends as a changed flower color: a genotype sets a phenotype link by link.
One more thing can change a flower’s look: a condition outside the plant can change how much of the enzyme its petal cells make, so the phenotype changes while the DNA does not.
What you are expected to know Explain how a plant’s genotype sets its phenotype, link by link: the allele’s DNA sequence sets the mRNA sequence, the mRNA sets the order of amino acids, the order sets the fold, and the fold sets what the enzyme does.
Suppose an allele’s template strand has one base swapped for a different base.
Which of the following changes in the mRNA built against it?
- A. ✓ One base
- B. Nothing: the mRNA stays the sameRNA polymerase places the partner of each template base.
A different template base gets a different partner, so one mRNA base changes. - C. Every base after the changed oneEach mRNA base is the partner of one template base.
Only the base opposite the changed one is different.
Why: RNA polymerase builds the mRNA by placing the partner of each template base.
One template base is different.
So one partner is different.
The mRNA changes at one base.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. Suppose one base of a codon changes, so 5′-UCA-3′ becomes 5′-UUA-3′. The two codons are drawn above the chart.
Which amino acid does the new codon, UUA, name?
- A. Serine (Ser)UCA, the old codon, names serine (Ser).
UUA is read with U for the row and U for the column, and its A line reads Leu. - B. ✓ Leucine (Leu)
- C. Phenylalanine (Phe)UUU and UUC, the U and C lines of that cell, name phenylalanine (Phe).
UUA ends in A, so its line is the A line, which reads Leu.
Why: The first base, U, picks the row.
The second base, U, picks the column.
The third base, A, picks the line in that cell.
That line reads UUA Leu, so UUA names leucine (Leu).
Suppose an enzyme carries an amino acid with a nonpolar R group at a spot where the working enzyme carries one with a charged R group.
Which of the following happens to the enzyme’s fold?
- A. The fold stays the sameWater is attracted to a charged R group and not to a nonpolar one.
The chain folds differently around the changed amino acid. - B. The whole enzyme unfoldsEvery other amino acid is the same, so most of the fold is the same.
The fold changes around the changed amino acid. - C. ✓ The fold changes around that spot
Why: The order of amino acids sets the fold.
Water is attracted to the charged R group and not to the nonpolar one.
So the chain folds differently around that spot, and the enzyme’s shape changes.
Suppose the pigment enzyme folds into a different shape from its working shape.
Which of the following does it make?
- A. More pigmentAn enzyme’s shape is what lets it do its job.
In a different shape the building block does not bind, so the enzyme makes no pigment. - B. ✓ No pigment
- C. Pigment of a different colorIn the different shape the pigment’s building block does not bind in the active site.
The enzyme makes no pigment of any color.
Why: A protein’s shape is what lets it do its job.
In the different shape the active site is changed.
The pigment’s building block does not bind.
So the enzyme makes no pigment.
A plant’s petal cells carry an enzyme that makes no pigment.
Which of the following do its flowers show?
- A. Purple flowersPurple flowers need purple pigment.
Cells whose enzyme makes no pigment have none, so the flowers are white. - B. ✓ White flowers
- C. Flowers of both colorsEvery petal cell carries the same enzyme, and it makes no pigment.
No cell is purple, so the flowers are white.
Why: The enzyme makes no pigment.
So no petal cell fills with purple pigment.
The flowers are white.
Suppose an animal carries two versions of one gene for an enzyme that breaks down a fat. The two versions differ at one base of DNA. An animal with two copies of the first version breaks the fat down. An animal with two copies of the second version cannot, so the fat builds up in its blood.
(a) Explain how the change in one base of the DNA leads to the change in the trait. (3 pt)
Frame The change in one base leads to the change in the trait because …
So one codon is different.
That codon names a different amino acid.
So the enzyme’s amino acid order is different.
The order of amino acids sets the fold.
So the enzyme folds into a different shape.
The shape lets the enzyme do its job.
So the changed enzyme no longer breaks the fat down.
So the fat builds up.
- Award 1 point for: the changed base changes one codon of the mRNA, and that codon names a different amino acid, so the enzyme’s order of amino acids changes.
- Award 1 point for: the order of amino acids sets the fold, so the enzyme folds into a different shape.
- Award 1 point for: the shape lets the enzyme do its job, so the changed enzyme no longer breaks the fat down and the fat builds up.
A student says: “Two plants that both carry two copies of the purple allele must have flowers of exactly the same shade, whatever conditions they grow in.”
Is the student correct?
- A. ✓ No: a condition outside the plant can change how much enzyme the petal cells make
- B. Yes: the same allele gives the same enzyme, so the two plants’ flowers must look exactly the sameThe two plants make the same enzyme.
A condition outside a plant can change how much of that enzyme its petal cells make.
Why: The same allele gives the same mRNA and the same enzyme.
A condition outside the plant, such as the temperature, can change how much of the enzyme the petal cells make.
Less enzyme makes less pigment.
So the shade can differ while the DNA does not.
The white allele of the pigment gene differs from the purple allele at one base. A student says: “A plant with two copies of this white allele still builds an mRNA and an enzyme from the gene. The enzyme just has a shape that makes no pigment.”
Is the student correct?
- A. No: a version that gives white flowers is never transcribed, so the plant makes no enzyme from itThe white allele differs from the purple allele at one base, and RNA polymerase still transcribes it.
The ribosome still translates the mRNA; the enzyme’s shape is what differs. - B. ✓ Yes: the white allele is transcribed and translated, and its enzyme has a shape that makes no pigment
Why: The white allele’s DNA sequence gives an mRNA that differs at one base.
The ribosome reads that mRNA and joins the amino acids, with Val in place of Asp.
The enzyme is built.
The enzyme folds into a different shape.
So the enzyme makes no pigment.
Suppose both of a plant’s alleles for one gene are a version with one changed base. The changed base gives one different amino acid in the enzyme that makes the flower’s scent, and that enzyme folds into a different shape.
Which of the following does the plant show?
- A. Flowers with the usual scentThe enzyme’s shape is what lets it make the scent.
In a different shape it makes none, so the flowers have no scent. - B. Flowers with a stronger scentA different shape stops the enzyme doing its job; it does not speed the job up.
The flowers have no scent. - C. ✓ Flowers with no scent
Why: The order of amino acids sets the fold, and the fold is different.
The shape lets the enzyme do its job.
So the enzyme with the different shape makes no scent.
What the enzyme does sets the trait: the flowers have no scent.
Go back to the plant with two versions of one pigment gene.
Sequence to mRNA, mRNA to amino acid order, order to fold, fold to job, job to trait: the purple allele gives a working enzyme, and the white allele gives an enzyme with a different shape.
A different allele changes the first link, and the change travels down every link after it.
APBIO-U06-L23 Backwards: RNA to DNA
Suppose a virus whose genetic material is RNA lands on a cell. Hours later, a copy of the virus’s genes sits inside one of the cell’s chromosomes, written in DNA.
The cell has no enzyme that writes DNA from RNA. The virus brought its own. What is that enzyme, and what happens next?
Unit 6 · Gene Expression and Regulation
1A virus that writes DNA from RNA
A particle sits on a cell’s surface. It has no ribosomes, and it reproduces only inside a cell.
What is the particle?
- A. ✓ A virus
- B. A bacteriumA bacterium is a cell: it has ribosomes of its own and divides on its own.
A particle with no ribosomes that reproduces only inside a cell is a virus. - C. A proteinA protein carries no genetic material and does not reproduce.
The particle reproduces inside a cell, so it is a virus.
Why: The particle has no ribosomes.
It reproduces only inside a cell.
Genetic material in a protein coat, with no ribosomes and reproducing only inside a cell, is a virus.
HIV’s genetic material contains uracil and no thymine.
Which molecule is HIV’s genetic material?
- A. DNADNA contains thymine and never uracil.
Genetic material that contains uracil is RNA. - B. ✓ RNA
Why: The genetic material contains uracil.
Uracil is RNA’s base in place of thymine.
So HIV’s genetic material is RNA.
How can information flow from RNA back to DNA?
Some RNA viruses carry an enzyme of their own that copies their RNA into DNA inside the cell they enter.
A second enzyme of the virus then inserts the DNA copy into one of the cell’s chromosomes.
The cell’s own RNA polymerase and ribosomes read that copy like one of the cell’s genes, and they build new virus particles.
Not every RNA virus does this: influenza has an RNA genome and no such enzyme.
Block the copying enzyme, and no DNA copy forms. So a drug that blocks the enzyme stops the virus from taking over new cells.
Video: Watch: A virus that writes DNA from RNA
The virus’s RNA enters the cell with the enzyme beside it. The enzyme moves along the RNA and builds a DNA strand against it. A second DNA strand forms against the first. The flow chart’s new arrow appears, pointing from RNA back to DNA.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L23a.mp4
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Suppose a virus lands on a cell. Its genetic material is RNA.
In Unit 4, a cell’s complete set of DNA was called its genome. A virus’s complete set of genetic material is its genome too.
This virus’s genome is RNA. Inside the coat, beside the RNA, sits an enzyme the virus brought with it.
The virus enters the cell. Its RNA and its enzyme come out of the coat into the cell.
The enzyme moves along the RNA and builds a DNA strand against it, one nucleotide at a time.
Each DNA base pairs with the RNA base opposite it: A opposite U, T opposite A, G opposite C, and C opposite G.
Then the enzyme builds a second DNA strand against the first. The result is a double-stranded DNA copy of the virus’s RNA.
No enzyme of the cell can do this. The cell’s enzymes copy DNA into DNA and DNA into RNA, never RNA into DNA.
Transcription copies DNA into RNA. This enzyme copies RNA into DNA, the reverse of transcription.
An enzyme that copies RNA into DNA is called , because it carries out transcription in reverse.
An RNA virus that carries reverse transcriptase and copies its RNA genome into DNA is called a , because retro means backwards.
HIV, the virus that causes AIDS, is a retrovirus. Inside its coat it carries an RNA genome and reverse transcriptase.

HIV particles, the small round bumps, on the surface of a white blood cell they are leaving. Image: C. Goldsmith, CDC Public Health Image Library #1197, public domain (resized).
Information usually flows from DNA to RNA to protein. In a retrovirus, information flows from RNA to DNA as well: the reverse of the usual direction.
What you are expected to know Describe a retrovirus: an RNA virus that carries its own enzyme, reverse transcriptase, which copies the virus’s RNA genome into DNA inside the cell it enters.
A retrovirus has entered a cell, and its reverse transcriptase is at work.
Which molecule does reverse transcriptase build?
- A. An RNA strandRNA polymerase builds RNA strands, against DNA.
Reverse transcriptase builds a DNA strand against the virus’s RNA. - B. A polypeptideA ribosome builds polypeptides.
Reverse transcriptase builds a DNA strand against the virus’s RNA. - C. ✓ A DNA strand
Why: Reverse transcriptase copies RNA into DNA.
Reverse transcriptase reads the virus’s RNA and builds a DNA strand against it.
Reverse transcriptase works inside a cell a retrovirus has entered.
Which molecule does reverse transcriptase read?
- A. The cell’s DNAThe cell’s DNA is read by the cell’s own polymerases.
Reverse transcriptase reads the virus’s RNA. - B. The virus’s coat proteinA protein carries no bases to copy.
Reverse transcriptase reads the virus’s RNA. - C. ✓ The virus’s RNA
Why: Reverse transcriptase copies RNA into DNA.
The RNA it reads is the virus’s own genome.
A retrovirus carries an RNA genome.
Besides its RNA genome, what does a retrovirus carry inside its coat?
- A. ✓ Reverse transcriptase
- B. A ribosomeA virus has no ribosomes; it uses the cell’s.
A retrovirus carries reverse transcriptase inside its coat. - C. A chromosomeA chromosome belongs to a cell.
A retrovirus carries reverse transcriptase inside its coat.
Why: A retrovirus carries its RNA genome and reverse transcriptase inside its coat.
The enzyme copies the RNA into DNA once both are inside the cell.
Reverse transcriptase is at work in a cell.
Which way does the information flow?
- A. From DNA to RNAFrom DNA to RNA is transcription, RNA polymerase’s job.
Reverse transcriptase copies RNA into DNA. - B. ✓ From RNA to DNA
- C. From RNA to proteinFrom RNA to protein is translation, the ribosome’s job.
Reverse transcriptase copies RNA into DNA.
Why: Reverse transcriptase reads RNA and builds DNA.
So the information flows from RNA to DNA.
Suppose a virus’s RNA alone, with no enzyme beside it, is put into a cell.
Can the cell’s own enzymes copy the virus’s RNA into DNA?
- A. YesA cell’s enzymes copy DNA into DNA and DNA into RNA.
Only a retrovirus’s reverse transcriptase copies a virus’s RNA into DNA. - B. ✓ No
Why: The cell’s own enzymes copy DNA into DNA and DNA into RNA.
No enzyme of the cell copies a virus’s RNA into DNA.
So the cell cannot copy the virus’s RNA into DNA.
30Quick quiz: retrovirus, reverse transcriptase mixed practice
Suppose a virus enters a cell. An enzyme soon builds a DNA copy of the virus’s RNA.
Which enzyme built the copy?
- A. ✓ The virus’s reverse transcriptase
- B. The cell’s RNA polymeraseRNA polymerase reads DNA and builds RNA.
Only reverse transcriptase reads RNA and builds DNA. - C. The cell’s DNA polymeraseDNA polymerase reads DNA and builds DNA.
Only reverse transcriptase reads RNA and builds DNA.
Why: The copy is DNA built against RNA.
No enzyme of the cell copies RNA into DNA.
So the virus’s reverse transcriptase built it.
What is a retrovirus?
- A. ✓ An RNA virus that carries reverse transcriptase and copies its RNA into DNA
- B. Any virus whose genetic material is RNA rather than DNAMany RNA viruses carry no reverse transcriptase and make no DNA.
A retrovirus is an RNA virus that copies its RNA into DNA with its own reverse transcriptase. - C. A virus that infects bacteria and carries DNA in its coatA virus that infects bacteria with a DNA genome makes no DNA copy of RNA.
A retrovirus copies its RNA into DNA with its own reverse transcriptase.
Why: An RNA virus that carries reverse transcriptase and copies its RNA genome into DNA is called a retrovirus.
What is reverse transcriptase?
- A. An enzyme of the cell that copies DNA into RNAThe enzyme that copies DNA into RNA is RNA polymerase.
Reverse transcriptase copies RNA into DNA. - B. An enzyme that joins amino acids into a chainThe ribosome joins amino acids into a chain.
Reverse transcriptase copies RNA into DNA. - C. ✓ A virus’s enzyme that copies RNA into DNA
Why: An enzyme that copies RNA into DNA is called reverse transcriptase.
A retrovirus carries it inside its coat.
A retrovirus carries an enzyme inside its coat.
(a) State what reverse transcriptase does. (1 pt)
It is the retrovirus’s own enzyme, carried inside its coat.
- Award 1 point for: copies RNA into DNA (builds a DNA copy of the virus’s RNA). ‘The virus’s own enzyme’ completes it, but its absence does not lose the point.
35Retrovirus or not?
Influenza, the virus that causes flu, carries RNA too. Is influenza a retrovirus?
An RNA genome alone does not make a retrovirus. The enzyme the virus carries decides it.
Video: Watch: Retrovirus or not?
Four viruses are opened one after another. In each, the enzyme inside the coat is drawn beside the genome, and a tick or a cross appears under the question: is it a retrovirus? Only the viruses that carry reverse transcriptase get the tick.
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A virus is a retrovirus when it carries reverse transcriptase, the enzyme that copies its RNA into DNA.
For example, suppose a virus carries an RNA genome and an enzyme that copies RNA into RNA. This virus is not a retrovirus, because it carries no reverse transcriptase.
But suppose a virus carries an RNA genome and an enzyme that copies RNA into DNA. This virus is a retrovirus, because it carries reverse transcriptase.
And suppose a virus carries an RNA genome and reverse transcriptase, and it infects a different kind of cell. This virus is still a retrovirus, because it carries reverse transcriptase.
But suppose a virus carries a DNA genome and no reverse transcriptase. This virus is not a retrovirus, because it carries no reverse transcriptase.
Influenza carries an RNA genome and an enzyme that copies RNA into RNA. Influenza carries no reverse transcriptase, so influenza is not a retrovirus.
HIV carries an RNA genome and reverse transcriptase. So HIV is a retrovirus.
What you are expected to know Classify a described virus as a retrovirus or as an ordinary RNA virus by the enzyme it carries: a retrovirus carries reverse transcriptase and copies its RNA into DNA.
Suppose a virus carries an RNA genome and reverse transcriptase.
Is this virus a retrovirus?
- A. ✓ Yes
- B. NoThe virus carries reverse transcriptase, so it copies its RNA into DNA.
An RNA virus that carries reverse transcriptase is a retrovirus.
Why: The virus carries reverse transcriptase.
Reverse transcriptase copies its RNA genome into DNA.
So the virus is a retrovirus.
Suppose a virus carries an RNA genome and an enzyme that copies RNA into RNA, and makes no DNA.
Is this virus a retrovirus?
- A. YesAn RNA genome alone does not make a retrovirus.
This virus carries no reverse transcriptase and makes no DNA, so it is not a retrovirus. - B. ✓ No
Why: The virus’s enzyme copies RNA into RNA, not into DNA.
The virus carries no reverse transcriptase.
So it is not a retrovirus.
Suppose a virus enters a cell, and inside the cell a DNA copy of the virus’s RNA genome appears.
Is this virus a retrovirus?
- A. ✓ Yes
- B. NoA DNA copy of the virus’s RNA can be built only by reverse transcriptase.
A virus whose RNA is copied into DNA by its own enzyme is a retrovirus.
Why: A DNA copy of a virus’s RNA can be built only by reverse transcriptase.
So this virus carries reverse transcriptase.
An RNA virus that carries reverse transcriptase is a retrovirus.
Suppose a virus is larger than most viruses. It carries an RNA genome and reverse transcriptase.
Is this virus a retrovirus?
- A. ✓ Yes
- B. NoSize decides nothing.
An RNA virus that carries reverse transcriptase is a retrovirus, whatever its size.
Why: The virus carries reverse transcriptase.
Its size changes nothing about the enzyme.
So the virus is a retrovirus.
Suppose a virus carries a DNA genome and no reverse transcriptase.
Is this virus a retrovirus?
- A. YesA retrovirus carries an RNA genome and reverse transcriptase.
This virus carries a DNA genome and no reverse transcriptase, so it is not a retrovirus. - B. ✓ No
Why: The virus carries no reverse transcriptase.
Its genome is DNA already, so nothing is copied from RNA into DNA.
So it is not a retrovirus.
Suppose a virus that infects birds carries an RNA genome and an enzyme that copies RNA into RNA.
Is this virus a retrovirus?
- A. YesThe kind of cell a virus infects decides nothing.
This virus carries no reverse transcriptase, so it is not a retrovirus. - B. ✓ No
Why: The virus’s enzyme copies RNA into RNA.
The virus carries no reverse transcriptase.
So it is not a retrovirus.
A student says: “Any virus with an RNA genome is a retrovirus.”
Is the student correct?
- A. ✓ No: only an RNA virus that carries reverse transcriptase is a retrovirus
- B. Yes: an RNA genome is what makes a virus a retrovirusMany RNA viruses carry no reverse transcriptase and make no DNA.
Reverse transcriptase decides it, not the RNA genome.
Why: Many viruses have RNA genomes and carry no reverse transcriptase.
An enzyme that copies RNA into RNA makes no DNA.
Only an RNA virus that carries reverse transcriptase is a retrovirus.
A student says: “Influenza has an RNA genome and no reverse transcriptase, so influenza is an ordinary RNA virus rather than a retrovirus.”
Is the student correct?
- A. No: influenza copies its RNA into DNA too, like HIVInfluenza’s enzyme copies RNA into RNA and builds no DNA.
With no reverse transcriptase, influenza is an ordinary RNA virus. - B. ✓ Yes: with no reverse transcriptase, influenza is an ordinary RNA virus
Why: Influenza’s enzyme copies RNA into RNA.
Influenza carries no reverse transcriptase.
So influenza has an RNA genome and is not a retrovirus.
Here is a table of an ordinary RNA virus against a retrovirus: the genome, the enzyme that copies the genome, whether the virus makes a DNA copy, and an example of each.
56The host builds the virus
RNA polymerase copies a gene.
Which molecule does RNA polymerase read as it works?
- A. A polypeptideA polypeptide is what a ribosome builds.
RNA polymerase reads the gene’s template strand, which is DNA. - B. An RNA strandRNA polymerase builds an RNA strand; it does not read one.
It reads the gene’s template strand, which is DNA. - C. ✓ A DNA strand
Why: RNA polymerase reads the template strand of a gene.
The template strand is DNA.
So RNA polymerase reads DNA and builds RNA.
A ribosome reads an mRNA.
Which molecule does the ribosome build?
- A. A DNA copy of the mRNAA ribosome builds no DNA.
It joins amino acids into a polypeptide. - B. ✓ A polypeptide
- C. A second mRNARNA polymerase builds mRNA, at the gene.
The ribosome joins amino acids into a polypeptide.
Why: The ribosome reads the mRNA three bases at a time.
A tRNA brings one amino acid for each codon.
The ribosome joins the amino acids into a polypeptide.
Now the DNA copy of the virus’s RNA lies inside the cell. What happens next?
The virus cannot copy its genes or build its coat on its own. The cell it has entered does that work.
Video: Watch: The host builds the virus
The DNA copy moves into the nucleus, and the virus’s second enzyme joins it into a chromosome. The cell’s RNA polymerase moves along the copy and builds RNA from it. The cell’s ribosomes read that RNA and build the virus’s proteins. New particles assemble and push out through the cell’s membrane.
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A cell that a virus has entered, and whose enzymes and ribosomes the virus uses, is called the , because the cell houses the virus and its genes.
The DNA copy moves into the host cell’s nucleus.
A second enzyme the virus carries cuts one of the host cell’s chromosomes and joins the DNA copy into the cut.
The copy now sits inside the chromosome’s DNA, like one of the cell’s own genes.
Joining the DNA copy into a host chromosome is called , because the copy becomes part of the host’s genome.
The host cell’s RNA polymerase reads the inserted copy and builds RNA from it, just as it does from the cell’s own genes.
Some of that RNA is the genome of new virus particles. Some of it is mRNA.
The host cell’s ribosomes read the mRNA and build the virus’s proteins: its coat proteins and its reverse transcriptase.
The new RNA genomes, the reverse transcriptase and the coat proteins assemble into new virus particles.
The new particles push out through the host cell’s membrane and leave the cell.
Here are the five moments as one strip, in the order they happen.
What you are expected to know Describe what happens after reverse transcriptase has made the DNA copy: the virus’s second enzyme inserts it into a host chromosome, the host’s RNA polymerase transcribes it, the host’s ribosomes translate it, and new particles assemble and leave the cell.
Frames J, K, L, M and N show five moments of a retrovirus inside a host cell, in a mixed order.
In which order do the five moments happen?
- A. ✓ K, N, M, J, L
- B. K, M, N, J, LThe DNA copy exists only after the enzyme has built it against the RNA.
The copy is built before it is joined into a chromosome. - C. N, K, M, L, JThe virus’s RNA must enter the cell before the enzyme can copy it.
The host’s RNA polymerase builds RNA from the copy before the ribosomes can read that RNA.
Why: The RNA and the enzyme enter the cell first.
The enzyme builds the DNA copy.
The copy is joined into a chromosome.
The host’s RNA polymerase reads it.
The host’s ribosomes build the virus, and the particles leave.
A retrovirus’s DNA copy has been integrated, and RNA is being built from it.
Whose RNA polymerase builds that RNA?
- A. ✓ The host cell’s
- B. The virus’sA retrovirus carries reverse transcriptase, which copies RNA into DNA.
It carries no RNA polymerase; the host cell’s RNA polymerase reads the copy.
Why: The copy sits in the host’s chromosome like one of the cell’s genes.
The host cell’s RNA polymerase reads the cell’s genes.
So the host cell’s RNA polymerase reads the copy too.
mRNA has been built from a retrovirus’s DNA copy.
Whose ribosomes build the virus’s proteins from it?
- A. The virus’sA virus has no ribosomes.
The host cell’s ribosomes build the virus’s proteins. - B. ✓ The host cell’s
Why: A virus has no ribosomes of its own.
The host cell’s ribosomes read the mRNA.
So the host cell’s ribosomes build the virus’s proteins.
RNA polymerase is reading a retrovirus’s DNA copy.
Where does the DNA copy sit while the polymerase reads it?
- A. Inside the virus’s coatThe RNA and the enzyme left the coat once they were inside the cell, and the coat plays no part after that.
The copy sits inside a host chromosome. - B. Floating free in the cytoplasmThe virus’s second enzyme joined the copy into a chromosome in the nucleus.
The copy sits inside a host chromosome. - C. ✓ Inside one of the host cell’s chromosomes
Why: The virus’s second enzyme joined the copy into a host chromosome.
The host’s RNA polymerase reads the chromosome’s genes there.
So the copy sits inside a host chromosome while it is read.
78Quick quiz: host cell, integration mixed practice
Suppose reverse transcriptase has just finished building the DNA copy of a retrovirus’s RNA, and the copy lies in the cytoplasm.
Has integration happened yet?
- A. YesIntegration is the joining of the copy into a host chromosome.
The chromosomes are in the nucleus, and this copy is still in the cytoplasm. - B. ✓ No
Why: Integration joins the DNA copy into a host chromosome.
The host’s chromosomes are in the nucleus.
This copy is still in the cytoplasm, so it is in no chromosome.
So integration has not happened yet.
What is a host cell?
- A. ✓ The cell a virus has entered and whose enzymes and ribosomes the virus uses
- B. A cell that the virus builds for itself out of its own coat proteinsA virus is built from genetic material and protein, never from a cell.
The host cell is the cell the virus has entered and uses. - C. A cell that copies a virus’s RNA into DNA with an enzyme of its ownNo enzyme of the cell copies a virus’s RNA into DNA; the retrovirus brings the enzyme that does.
The host cell is the cell the virus has entered and uses.
Why: A cell that a virus has entered, and whose enzymes and ribosomes the virus uses, is called the host cell.
What is integration?
- A. ✓ The joining of the DNA copy of the virus’s RNA into a host chromosome
- B. The virus’s RNA and enzyme entering the host cell through its membraneThe RNA entering the cell comes first, before any DNA copy exists.
Integration is the joining of the DNA copy into a host chromosome. - C. The host’s ribosomes building the virus’s proteins from its mRNAThe ribosomes build the proteins after the copy has been read.
Integration is the joining of the DNA copy into a host chromosome.
Why: Joining the DNA copy of the virus’s RNA into a host chromosome is called integration.
A retrovirus’s RNA has been copied into DNA inside a host cell.
(a) State what integration is. (1 pt)
The copy then sits in the host’s DNA like one of the cell’s own genes.
- Award 1 point for: the DNA copy is inserted (joined) into a host chromosome (the host’s genome).
83Block one step
Suppose a drug that blocks reverse transcriptase is inside a cell when a retrovirus enters it. What happens to the virus?
To predict what happens when one step is blocked, name that step’s job. Then follow the chain of steps after it.
Video: Watch: Block one step
The virus’s RNA enters the cell, but the blocked enzyme builds no DNA against it. The chromosome stays as it was, the cell’s RNA polymerase reads only the cell’s own genes, and no new particles form. Then a second run: the copy forms but is never joined into a chromosome, and the cell breaks it down.
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Reverse transcriptase’s job is to copy the virus’s RNA into DNA.
With the enzyme blocked, the virus’s RNA enters the cell, but no DNA copy forms.
With no DNA copy, there is nothing for the virus’s second enzyme to join into a chromosome.
The host’s RNA polymerase reads DNA. The virus’s genes are written in RNA only, so the host’s RNA polymerase never reads them.
So the host builds no viral RNA and no viral proteins, and no new particles form.
That is why a drug that blocks reverse transcriptase stops a retrovirus from taking over new cells.
Now imagine a change in a retrovirus’s genes that stops integration: reverse transcriptase builds the DNA copy, but the second enzyme cannot join it into a chromosome.
The copy floats free in the nucleus. The copy is part of no chromosome.
When the cell divides, it copies its chromosomes and passes them on. The cell does not copy the free DNA with them, so the daughter cells are left without it.
So the virus’s genes are lost from that cell, and the cell builds few or no new particles.
What you are expected to know Predict what happens to a retrovirus when one step is blocked: a drug that blocks reverse transcriptase, or a change that stops integration.
A drug that blocks reverse transcriptase is inside a cell that a retrovirus has just entered.
Does a DNA copy of the virus’s RNA form in the cell?
- A. YesReverse transcriptase is the only enzyme that copies RNA into DNA.
With it blocked, the RNA enters but no DNA copy forms. - B. ✓ No
Why: Reverse transcriptase copies the virus’s RNA into DNA.
The drug blocks reverse transcriptase.
So the virus’s RNA is in the cell, and no DNA copy of it forms.
Suppose a retrovirus’s reverse transcriptase has a changed shape and no longer works. The virus enters a cell.
(a) Explain why the cell builds no new virus particles. (3 pt)
Frame The cell builds no new virus particles because …
So there is no DNA copy to join into a host chromosome.
The host’s RNA polymerase reads DNA.
The virus’s genes are in RNA only.
So the host’s RNA polymerase builds no RNA from them.
With no viral mRNA, the host’s ribosomes build no viral proteins.
So no new particles assemble.
- Award 1 point for: reverse transcriptase copies the virus’s RNA into DNA, so with it not working no DNA copy forms (and none is joined into a chromosome).
- Award 1 point for: the host’s RNA polymerase reads DNA, so with no DNA copy it builds no viral RNA (the virus’s genes are never transcribed).
- Award 1 point for: with no viral mRNA the host’s ribosomes build no viral proteins, so no new particles assemble.
A student says: “A drug that blocks reverse transcriptase does not keep the virus’s RNA out of the cell. It acts only once the RNA and the enzyme are inside.”
Is the student correct?
- A. No: the drug works by keeping the virus’s RNA and enzyme from entering the cellThe virus enters the cell with its RNA and its enzyme whether or not the drug is there.
The drug blocks the enzyme’s copying step inside the cell. - B. ✓ Yes: entering the cell needs no reverse transcriptase, so the drug acts only on the copying step inside
Why: Entering the cell needs no reverse transcriptase.
So the virus’s RNA and its enzyme enter as usual.
The drug blocks reverse transcriptase, the copying step inside the cell.
So the drug acts only once the RNA and the enzyme are inside.
A student says: “The drug stops the virus because it blocks the host’s RNA polymerase from copying the virus’s RNA.”
Is the student correct?
- A. ✓ No: the host’s RNA polymerase reads DNA, and with the enzyme blocked no DNA copy exists
- B. Yes: the host’s RNA polymerase copies the virus’s RNA, and the drug blocks itThe drug blocks the virus’s reverse transcriptase, not the host’s RNA polymerase.
The host’s RNA polymerase reads DNA, and no DNA copy of the virus exists.
Why: The drug blocks reverse transcriptase.
So no DNA copy of the virus’s RNA forms.
The host’s RNA polymerase reads DNA only.
With no DNA copy, the host’s RNA polymerase never reads the virus’s genes.
Suppose a change in a retrovirus’s genes stops integration. Reverse transcriptase builds the DNA copy, but the virus’s second enzyme never joins it into a chromosome. The host cell then divides.
What happens to the virus’s genes?
- A. The host copies them before it next divides, like its own DNABefore it divides, the host copies its chromosomes.
The free copy is part of no chromosome, so it is left uncopied. - B. The host joins them into a chromosome itselfOnly the virus’s second enzyme joins the copy into a chromosome.
The host cell has no enzyme that does this. - C. ✓ They stay free and uncopied, and the host breaks them down
Why: The DNA copy is part of no chromosome.
The host copies its chromosomes only, before it divides.
The free copy is not copied, and the cell breaks it down.
So the virus’s genes are lost.
Here is a table of the step blocked against what follows: whether a DNA copy forms, what happens to the virus’s genes, and whether new particles form.
104A copy that stays
A cell is in interphase, the long stretch between one division and the next.
In which stage of interphase does the cell copy its DNA?
- A. G1G1 is the first growth gap.
The cell copies its DNA in the stage after G1. - B. ✓ S phase
- C. G2G2 is the second growth gap.
G2 begins after the copying is finished.
Why: S phase is the copying stage of interphase.
The S is short for synthesis, the making of new DNA.
Go back to the infected cell, with the DNA copy inside one of its chromosomes.
Suppose the cell divides. What happens to the copy?
Video: Watch: A copy that stays
The chromosome with the copy inside it is copied in S phase, so two chromosomes each carry the copy. The cell divides, and each daughter cell receives one. Years later the copy still sits in the chromosome, and the cell’s RNA polymerase still reads it.
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In S phase the cell copies every one of its chromosomes. The inserted copy is part of a chromosome, so the cell copies it with the rest.
The cell divides. Each daughter cell receives a complete set of chromosomes, with the copy in it.
In each daughter cell, the host’s RNA polymerase reads the copy like one of the cell’s own genes.
Now suppose a drug that blocks reverse transcriptase reaches these cells. Reverse transcriptase made the copy long ago, and its job there is done.
The drug blocks no other step. So the copies already in chromosomes stay where they are, and the cells go on reading them.
A DNA copy inside a chromosome stays there as long as the cell lives, and every cell that descends from it carries the copy too.
What you are expected to know Explain why a DNA copy inserted into a chromosome stays as long as the cell does: the cell copies it every S phase and reads it like its own genes, and blocking reverse transcriptase leaves it untouched.
A cell carries a retrovirus’s DNA copy inside one of its chromosomes. The cell divides.
Which of the following carries the copy afterwards?
- A. ✓ Both daughter cells
- B. One daughter cellThe cell copied the chromosome in S phase, so both new chromosomes carry the copy.
Each daughter cell receives one. - C. Neither daughter cellThe copy is part of a chromosome, and every chromosome is copied and passed on.
Both daughter cells carry it.
Why: The copy is part of a chromosome.
In S phase the cell copies every chromosome, the copy included.
Each daughter cell receives a complete set.
So both carry the copy.
Suppose one cell lining a person’s gut carries an integrated copy of a retrovirus’s DNA. Over a year, that cell’s descendants divide many times.
(a) Explain what happens to the copy each time one of these cells divides. (2 pt)
Frame Each time one of these cells divides, …
The copy is part of a chromosome.
In S phase the cell copies every chromosome.
So the cell copies the inserted DNA with the rest.
Each daughter cell receives one complete set of chromosomes, with the copy in it.
Every division repeats this.
So after a year every descendant carries the copy.
- Award 1 point for: the copy is part of a chromosome, so the cell copies it with the chromosome in S phase.
- Award 1 point for: each daughter cell receives a complete set of chromosomes, with the copy in it, and this repeats at every division.
A student says: “A drug that blocks reverse transcriptase removes the DNA copies already sitting in chromosomes.”
Is the student correct?
- A. ✓ No: reverse transcriptase only makes copies, so copies already in chromosomes stay
- B. Yes: with the enzyme blocked, the copies in the chromosomes break apartReverse transcriptase’s job was to make the copy, and that job is done.
Blocking it changes nothing about a copy already inside a chromosome.
Why: Reverse transcriptase makes the DNA copy.
A copy already inside a chromosome no longer needs the enzyme.
So the drug leaves those copies where they are.
Suppose a cell carrying an integrated copy of a retrovirus has finished dividing for good and lives on for years.
How long does the copy stay in the cell?
- A. Until the cell’s next S phaseA cell that divides no more has no next S phase.
The copy is part of a chromosome and stays as long as the cell lives. - B. For a few hoursThe copy is part of a chromosome, and the cell keeps its chromosomes for its whole life.
The copy stays as long as the cell lives. - C. ✓ As long as the cell lives
Why: The copy is part of one of the cell’s chromosomes.
A cell keeps its chromosomes for its whole life.
So the copy stays as long as the cell lives.
Go back to the infected cell.
Reverse transcriptase made a DNA copy of the virus’s RNA. The virus’s second enzyme joined the copy into a chromosome.
The host’s RNA polymerase read the copy, and the host’s ribosomes built new virus particles from it.
Block reverse transcriptase, and none of it starts. But a copy already inside a chromosome stays as long as the cell does.
124Mixed practice mixed practice
Suppose a eukaryotic cell builds a protein that it will send out of the cell.
Which ribosomes build that protein?
- A. ✓ Ribosomes attached to the rough ER
- B. Ribosomes free in the cytosolFree ribosomes build the proteins that stay in the cytosol.
A ribosome on the rough ER builds a protein the cell sends out.
Why: A protein the cell will send out passes into the rough ER as it is built.
So the ribosome building it sits on the rough ER.
A ribosome reaches the codon UAG on an mRNA.
Which of the following happens next?
- A. ✓ No tRNA pairs with it, and the ribosome releases the chain
- B. A tRNA brings the chain’s last amino acidUAG names no amino acid, so no tRNA brings one.
The ribosome releases the chain. - C. The ribosome moves on to the next AUG and keeps readingA stop codon ends the reading; the ribosome does not go looking for another AUG.
The ribosome releases the chain and comes apart.
Why: UAG is a stop codon.
No tRNA has an anticodon that pairs with it.
So the ribosome adds no amino acid, releases the chain and comes apart.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. A tRNA’s anticodon reads 3′-GGC-5′.
Which codon does this tRNA pair with?
- A. 5′-GGC-3′GGC is the anticodon’s own letters.
A codon and its anticodon pair base by base: G with C and C with G, so the codon is CCG. - B. ✓ 5′-CCG-3′
- C. 5′-GCC-3′GCC is the paired codon read from its 3′ end.
The anticodon’s 3′ end lies against the codon’s 5′ end, so the codon reads 5′-CCG-3′.
Why: Each anticodon base pairs with the codon base opposite it: G with C, C with G.
The anticodon 3′-GGC-5′ lies against the codon written 5′ to 3′.
So the codon is 5′-CCG-3′.
Suppose a test tube holds mRNA, ribosomes, amino acids and tRNAs loaded with their amino acids, but no energy supply.
What do the ribosomes in the tube build?
- A. ✓ No polypeptide
- B. The polypeptide, more slowlyEach step of translation uses energy, and the tube has none to give.
The ribosomes build no polypeptide. - C. The polypeptide, at the usual rateEach step of translation breaks down an energy carrier.
With no energy supply, the ribosomes build no polypeptide.
Why: Each step of translation uses energy from an energy carrier.
The tube has no energy supply.
So the ribosomes build no polypeptide.
The codons ACU and ACC both name threonine (Thr).
Which word describes the genetic code here?
- A. UniversalUniversal describes the code being the same in nearly all living things.
Several codons naming one amino acid is redundancy. - B. AmbiguousAmbiguous would mean one codon names two amino acids.
Here two codons name one amino acid: the code is redundant. - C. ✓ Redundant
Why: Two codons name the same amino acid.
A code in which several codons mean the same amino acid is called a redundant code.
A gene from a fish is placed in a bacterium, and the bacterium’s ribosomes read each codon as the same amino acid the fish’s ribosomes read.
Which idea does this show?
- A. Each species reads codons in its own wayThe bacterium read the fish’s codons as the fish does.
The code is nearly the same in all living things. - B. ✓ The genetic code is nearly universal
- C. A bacterium copies the tRNAs of any gene it receivesThe bacterium used its own tRNAs, and they paired with the fish gene’s codons.
The code is nearly the same in all living things.
Why: The bacterium’s ribosomes and tRNAs read the fish gene’s codons as the fish’s do.
So the two species share one code.
The genetic code is nearly universal.
Suppose one base of an allele changes, and the enzyme it codes for folds into a different shape.
Which link does the changed fold set next?
- A. The mRNA sequenceThe mRNA sequence comes before the fold in the chain: the DNA sets it.
The fold sets what the enzyme does. - B. The order of amino acidsThe order of amino acids comes before the fold: it sets the fold.
The fold sets what the enzyme does. - C. ✓ What the enzyme does
Why: The chain is DNA sequence → mRNA → amino acid order → fold → job → trait.
The fold is set by the amino acid order.
The fold sets what the enzyme does.
Suppose a retrovirus enters a cell during G1, many hours before the cell will copy its own DNA in S phase.
(a) Predict whether the virus’s RNA is copied into DNA before the cell reaches S phase. (1 pt)
- Award 1 point for: the DNA copy is made before S phase (reverse transcriptase does not wait for the cell’s own copying).
(b) Explain what sets the time at which the virus’s RNA is copied into DNA. (2 pt)
Frame The time of the copying is set by …
Reverse transcriptase is carried inside the virus’s coat and works as soon as it is inside the cell.
The cell’s DNA polymerase copies DNA into DNA, in S phase.
Copying RNA into DNA does not use the cell’s DNA polymerase.
So the copy is built while the cell is still in G1.
- Award 1 point for: reverse transcriptase is the virus’s own enzyme and it copies the virus’s RNA into DNA.
- Award 1 point for: the cell’s DNA polymerase copies DNA into DNA in S phase and is not the enzyme that copies RNA into DNA, so the virus’s step does not wait for it.
Glossary
- host cell
- The cell a virus has entered, and whose enzymes and ribosomes the virus uses to copy its genes and build new particles.
- integration
- The joining of the DNA copy of a retrovirus’s RNA into one of the host cell’s chromosomes, so the copy becomes part of the host’s genome.
- reverse transcriptase
- A retrovirus’s own enzyme, carried inside its coat, that copies the virus’s RNA into DNA inside the cell it enters: transcription in reverse.
- retrovirus
- An RNA virus that carries reverse transcriptase and copies its RNA genome into DNA inside the cell it enters. HIV is a retrovirus.
APBIO-U06-P64 Practice questions: Topic 6.4
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 one changed cell 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: Translation, summed up
Where translation happens; codons, the code chart, AUG and the three stops; redundancy and the near-universal code; initiation, elongation, termination; codon and anticodon; genotype to phenotype through the chain; the retrovirus that runs the flow backward.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-T64-summary.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-T64-summary.mp4
A cell lining a hen's oviduct builds ovalbumin, a protein it releases into the egg white.
Where is ovalbumin built?
- A. On a ribosome free in the cytoplasmA free ribosome releases its chain into the cytoplasm, where the protein stays.
Ovalbumin leaves the cell, so it is built on the rough ER. - B. On a ribosome inside the nucleusRibosomes work in the cytoplasm, outside the nucleus.
An exported protein is built on a ribosome on the rough ER. - C. Inside the ER, with no ribosomeOnly a ribosome joins amino acids into a chain.
A ribosome on the ER passes its growing chain into the ER. - D. ✓ On a ribosome attached to the rough ER
Why: Ovalbumin is released from the cell.
A protein the cell releases is built on a ribosome attached to the rough ER.
The ribosome passes the growing chain into the ER, and from there the protein is carried out of the cell.
Suppose a gene switches on in a bacterium and a gene of the same length switches on in a yeast cell. The table shows how long each cell takes to finish copying the gene into mRNA, and how long until the first ribosome starts translating that mRNA.
Which statement explains why translation starts so much sooner in the bacterium?
- A. ✓ Its ribosomes bind the mRNA's free 5′ end while RNA polymerase is still working
- B. Its RNA polymerase copies the gene into mRNA faster than the yeast cell's doesBoth cells finish copying the gene in 3 min.
The bacterium's ribosomes start at 0.5 min, before its RNA polymerase has finished. - C. Its mRNA leaves the nucleus sooner than the yeast cell's mRNA doesA bacterium has no nucleus.
Nothing stands between its DNA and its ribosomes. - D. Its ribosomes join amino acids to the chain faster than the yeast cell's doHow fast amino acids are joined sets how long a chain takes, not when reading starts.
The bacterium's ribosomes start reading before the mRNA is finished.
Why: In the bacterium nothing separates the DNA from the ribosomes, and the mRNA needs no processing.
So ribosomes bind its free 5′ end while RNA polymerase is still working.
In the yeast cell the mRNA is finished, processed and moved out of the nucleus first.
So the bacterium starts sooner.
Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. A ribosome reads the mRNA drawn.
Which codon does the ribosome read third?
- A. 5′-GGG-3′This is the second codon.
After AUG the ribosome reads GGG, then UUU. - B. 5′-GUU-3′These letters are grouped from the mRNA's first letter, one base before the AUG.
The codons run from the A of AUG in threes. - C. ✓ 5′-UUU-3′
- D. 5′-UUA-3′These letters begin one base after the AUG's first letter.
The codons run from the A of AUG without overlap: AUG, GGG, UUU.
Why: The ribosome begins at the first AUG and reads in threes from its A.
The codons are AUG, GGG, UUU, and so on.
So the third codon is UUU.
Three RNA bases make one codon, so there are 64 different codons.
Which statement about the 64 codons is correct?
- A. 44 name an amino acid and 20 are stop codonsThere are 20 kinds of amino acid, not 20 stop codons.
Only 3 codons are stops. - B. ✓ 61 name an amino acid and 3 are stop codons
- C. 20 name an amino acid and 44 are stop codonsEvery one of the 20 amino acids has at least one codon, and most have several.
Only 3 codons are stops. - D. 63 name an amino acid and 1 is a stop codonThree codons are stops, not one.
The other 61 name amino acids.
Why: Of the 64 codons, 3 are stop codons, which end the message and name no amino acid.
The other 61 each name one of the 20 amino acids.
The genetic code chart is drawn.
How many codons name histidine (His)?
- A. 1Only Met and Trp have a single codon each.
His has more than one. - B. ✓ 2
- C. 3Ile has three codons.
His has two, both in the C row, A column. - D. 4The C row, A column cell holds four codons, but its A and G lines read Gln.
Only two of the four lines read His.
Why: Find every line on the chart that reads His.
They are CAU and CAC, in the C row, A column.
The other two lines of that cell, CAA and CAG, read Gln.
So 2 codons name His.
A gene from a moss is put into a yeast cell. The yeast cell's ribosomes read the gene's mRNA and build the moss protein, with every amino acid in its place.
What does this result show?
- A. The yeast and the moss carry the same genesThe gene came from the moss; the yeast did not have it.
What the two share is the meaning of each codon. - B. The moss gene carries the instructions for building the ribosome that reads itThe yeast cell's own ribosomes read the mRNA.
A gene carries instructions for one protein, not for the ribosome. - C. A yeast cell can read any RNA, whatever code it is written inThe yeast's ribosomes give each codon the meaning the yeast's code gives it.
The moss protein comes out right because the moss's code gives each codon the same meaning. - D. ✓ The yeast and the moss use the same genetic code
Why: The yeast's ribosomes and tRNAs read each codon of the moss mRNA.
They put in the amino acid the yeast's code gives that codon.
Every amino acid comes out in place.
So each codon means the same in the yeast as in the moss: one shared code.
Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. The mRNA drawn above the genetic code chart is translated.
How many amino acids does the polypeptide contain?
- A. 4Two codons are missed.
Count every codon from AUG up to, not including, the stop codon. - B. 5Met, from the AUG, is counted.
Counting from the codon after AUG misses it. - C. ✓ 6
- D. 7The stop codon adds no amino acid.
Only the codons from AUG up to the stop codon count.
Why: Reading begins at the first AUG and continues in threes.
The codons up to and including the stop codon are 7.
The stop codon adds no amino acid.
So the polypeptide has 7 − 1 = 6 amino acids.
A strand is written from its 5′ end to its 3′ end, and both ends are marked. Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. A tRNA's anticodon reads 3′-GAA-5′, as drawn.
Which codon on an mRNA does this tRNA pair with?
- A. ✓ 5′-CUU-3′
- B. 5′-GAA-3′These are the anticodon's own letters.
The codon carries each anticodon base's partner: G takes C, A takes U, A takes U. - C. 3′-CUU-5′The partners are right, but the codon's 5′ end sits opposite the anticodon's 3′ end.
Written to the convention, the codon reads 5′-CUU-3′. - D. 5′-AAG-3′This is the anticodon itself, written from its 5′ end.
The codon is the strand the anticodon pairs with, so its bases are the anticodon's partners.
Why: Under each anticodon base write its partner: G takes C, A takes U, A takes U.
The codon runs the other way, so its 5′ end sits under the anticodon's 3′ end.
The codon reads 5′-CUU-3′.
A scientist measures how long a ribosome spends at each codon of an mRNA. The bar chart shows the time spent at the four codons that name alanine (Ala), relative to the fastest of them.
At which codon does the ribosome spend the longest time?
- A. GCUThe GCU bar reaches the 1.5 line.
The GCC bar is taller, at 3.5. - B. ✓ GCC
- C. GCAThe GCA bar reaches the 1.0 line.
The GCC bar is taller, at 3.5. - D. GCGThe GCG bar reaches the 2.0 line.
The GCC bar is taller, at 3.5.
Why: Each bar's top shows the time the ribosome spends at that codon.
The GCC bar is the tallest, reaching the 3.5 line.
So the ribosome spends the longest time at GCC.
The bar chart shows the time a ribosome spends at each of the four codons that name alanine (Ala), relative to the fastest. A cell makes two mRNAs for the same polypeptide: one uses 5′-GCA-3′ at every alanine, the other uses 5′-GCC-3′ at every alanine.
Which mRNA is translated faster, and why?
- A. The GCC mRNA, because the tRNA that reads GCC is the more plentifulA tall bar means a long wait, and a long wait means a scarce tRNA.
The GCC bar is the taller, so its tRNA is the scarcer. - B. Both at the same rate, because GCA and GCC name the same amino acidThe two codons are read by different tRNAs, present in different amounts.
The ribosome waits longer at the codon whose tRNA is scarcer. - C. The GCA mRNA, because its tRNA carries alanine more tightlyHow tightly the tRNA holds its amino acid is not measured here.
The bar shows how long the ribosome waits for the tRNA to arrive. - D. ✓ The GCA mRNA, because the ribosome spends less time at each GCA than at each GCC
Why: The GCA bar reaches the 1.0 line and the GCC bar the 3.5 line.
So the ribosome waits less at each GCA than at each GCC.
The polypeptide is the same, and the GCA mRNA is translated faster.
Codons are written as they read on the mRNA, 5′ to 3′; an anticodon is written 3′ to 5′ against its codon. Read from the first AUG in threes and stop at the first stop codon; the stop codon adds no amino acid. Suppose that in a yeast cell the enzyme that loads cysteine (Cys) onto its tRNAs stops working. The cell's tRNAs, ribosomes and mRNAs are unchanged, but no tRNA carries Cys. A ribosome moves on from a codon only after a loaded tRNA has paired with it and its amino acid is joined, so a ribosome never skips a codon. A ribosome begins reading the mRNA drawn above the genetic code chart.
(a) Identify the amino acid that the codon 5′-UGC-3′ names. (1 pt)
Frame 5′-UGC-3′ names …
Hint Find the row for the first base, U, and the column for the second base, G. Then read the line for the third base, C.
- Award 1 point for: Cys.
(b) Describe what the loaded tRNA that pairs with 5′-UGC-3′ does at that codon in a cell where the loading enzyme works. (1 pt)
Frame In a working cell, the tRNA …
Hint Say what its anticodon does and what it brings into the ribosome.
The ribosome joins that Cys to the growing chain.
- Award 1 point for: its anticodon pairs with UGC AND it brings the amino acid (Cys) that the ribosome joins to the chain.
(c) Predict how the polypeptide built in the changed yeast cell differs from the polypeptide a working cell builds from this mRNA. (1 pt)
Frame In the changed cell, the ribosome …
Hint Ask what arrives at the ribosome when it reaches the third codon.
The polypeptide is left unfinished, two amino acids long, instead of Met–Leu–Cys–Pro–Gly.
- Award 1 point for: the chain stops at the third codon (UGC), so the polypeptide is unfinished (Met–Leu only). Accept 'the ribosome stalls at UGC and no complete polypeptide is made'.
- A prediction that the ribosome skips UGC and builds Met–Leu–Pro–Gly (the chain with Cys left out) earns nothing.
(d) Justify your prediction. (1 pt)
Frame The ribosome stops there because …
Hint What must pair with the codon and deliver an amino acid before the ribosome can move on?
With the loading enzyme stopped, no tRNA carries Cys.
The ribosome joins the next amino acid and moves on only after a loaded tRNA has paired.
So it waits at UGC, and the chain grows no further.
- Award 1 point for: no tRNA carries Cys, so no amino acid is delivered at UGC, AND the ribosome moves on only after a loaded tRNA has paired and its amino acid is joined.
A retrovirus's DNA copy sits in one of a cell's chromosomes. Suppose a drug binds the promoter of that DNA copy and nothing else, so that the host cell's RNA polymerase cannot bind there. The drug reaches the cell before the host cell's RNA polymerase has read the copy even once.
(a) Identify the molecule that the host cell's RNA polymerase builds from the DNA copy in an untreated cell. (1 pt)
- Award 1 point for: RNA (the virus's RNA — its genome copies and its mRNA).
(b) Describe what the host cell's ribosomes do with the molecule you identified in part (a), in an untreated cell. (1 pt)
They build the virus's proteins: its coat proteins and its reverse transcriptase.
- Award 1 point for: they translate the virus's mRNA into the virus's proteins (coat proteins, reverse transcriptase).
(c) Make a claim about whether the treated cell builds new virus particles while the drug is present, and a second claim about whether the DNA copy is passed on when the treated cell divides. (2 pt)
The DNA copy is passed to both daughter cells when the cell divides.
- Award 1 point for: the treated cell builds no new virus particles while the drug is present.
- Accept for the first point: no new particles once any viral mRNA already present is used up.
- Award 1 point for: the DNA copy is passed on to both daughter cells when the cell divides.
Slip Claiming that the drug removes the copy. The drug blocks reading of the copy; the copy itself stays in the chromosome.
(d) Support both claims. (2 pt)
So the ribosomes get no viral mRNA and build no coat proteins or reverse transcriptase, and no new genome RNA is made.
So no particles can be assembled.
The copy is part of a chromosome, so the cell copies it in S phase with the rest and each daughter cell receives it.
- Award 1 point for: with the promoter blocked no viral RNA is made, so no viral proteins or genome copies are made, so no particles are assembled.
- Award 1 point for: the copy is part of a chromosome and is copied in S phase with the chromosome, so both daughter cells carry it.
APBIO-U06-T64 End-of-topic test: Translation
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.
The table describes four proteins made by eukaryotic cells and where each protein ends up.
Which protein is built on a ribosome attached to the rough ER?
- A. WThis protein stays inside the cell that made it.
A protein that stays in the cytoplasm is built on a free ribosome. - B. XThis enzyme works in the cytoplasm.
A free ribosome releases its chain into the cytoplasm, where the enzyme stays. - C. YThis protein stays inside the muscle cell.
A free ribosome builds it and releases it into the cytoplasm. - D. ✓ Z
Why: A ribosome on the rough ER passes its growing chain into the ER.
From the ER, the protein is packaged and carried out of the cell.
So a protein the cell releases is built on a rough-ER ribosome.
Only protein Z is released.
A gene in a soil bacterium switches on. Scientists can tell when three things first exist in the cell: the 5′ end of the gene's mRNA, the first amino acids of its polypeptide, and the 3′ end of the mRNA. The table shows which of the three are present at three times after the gene switches on.
Which conclusion do the observations support?
- A. ✓ Ribosomes began translating the mRNA before RNA polymerase had finished making it
- B. Ribosomes began translating the mRNA only after RNA polymerase had released itAt 40 s the first amino acids exist, and the mRNA's 3′ end does not.
So a ribosome was reading the mRNA while its 3′ end was still being made. - C. RNA polymerase built the mRNA from its 3′ end toward its 5′ endThe 5′ end exists at 15 s and the 3′ end only at 90 s.
RNA polymerase built the mRNA from its 5′ end toward its 3′ end. - D. The polypeptide was built before any of the mRNA existedAt 15 s the mRNA's 5′ end exists and no amino acids do.
The mRNA came first, and the ribosome read it.
Why: At 15 s the mRNA's 5′ end exists, so RNA polymerase has begun.
At 40 s the first amino acids exist, so a ribosome is reading the mRNA.
At 40 s the 3′ end does not yet exist, so RNA polymerase has not finished.
So translation began before transcription finished.
A ribosome reads the mRNA drawn.
Which codon does the ribosome read second?
- A. 5′-GCA-3′The ribosome does not begin at the mRNA's first letter.
It begins at the first AUG, so the two leading letters are never part of a codon. - B. ✓ 5′-UGC-3′
- C. 5′-UGU-3′These three letters begin one base after the AUG's first letter.
The codons run from the A of AUG in threes without overlap: AUG, then UGC. - D. 5′-UUU-3′This is the third codon.
After AUG the ribosome reads UGC, then UUU.
Why: The ribosome begins at the first AUG.
It reads in threes from the A of AUG, with no overlap and no gap.
The first codon is AUG and the second is UGC.
A scientist treats eukaryotic cells with each of four drugs. The table shows what is seen in the treated cells.
Which drug blocks termination?
- A. WIn cells treated with drug W the ribosome never assembles on the mRNA.
Drug W blocks initiation. - B. XIn cells treated with drug X the ribosome assembles but the chain stops growing.
Drug X blocks elongation. - C. ✓ Y
- D. ZIn cells treated with drug Z translation ends and the chain is released.
Drug Z acts after translation, on the folding.
Why: At a stop codon no tRNA pairs, the finished polypeptide is released and the subunits separate.
In cells treated with drug Y the ribosomes reach the stop codons and stay there with the chains attached.
So the release does not happen: drug Y blocks termination.
The codon 5′-UAU-3′ is drawn above the genetic code chart.
Which amino acid does this codon name?
- A. ✓ Tyr
- B. IleIle is the codon AUU: the second base read as the row and the first as the column.
The first base picks the row, the second the column. - C. HisHis is the codon CAU, in the C row.
The first base of UAU is U, so its row is the U row. - D. No amino acid: it is a stop codonThe stop codons in this cell are UAA and UAG.
The third base here is U, and the UAU line reads Tyr.
Why: The first base, U, picks the row; the second base, A, picks the column.
In that cell the line for the third base, U, reads Tyr.
So UAU names Tyr.
Three short mRNAs were translated in a cell. The table shows each mRNA and the polypeptide it gave. A fourth mRNA reads 5′-AUG CUC UUU UAA-3′.
Which polypeptide does the fourth mRNA give?
- A. Met–Leu–ProThe fourth mRNA's third codon is UUU, not CCC.
The third mRNA shows that UUU names Phe. - B. Met–Phe–LeuThe codons are read in order from AUG: CUC, then UUU.
The second amino acid comes from CUC and the third from UUU. - C. Met–LeuUAA is the stop codon, and it adds no amino acid.
UUU comes before it, and the third mRNA shows that UUU names Phe. - D. ✓ Met–Leu–Phe
Why: The first two mRNAs differ only at their second codon, CUU against CUC, and give the same polypeptide.
So CUU and CUC both name Leu.
The third mRNA shows that UUU names Phe.
So AUG CUC UUU UAA gives Met–Leu–Phe.
A soil bacterium, an octopus and an owl each read the codon 5′-UUU-3′ as phenylalanine (Phe), and so does every other organism tested.
Which of the following best explains why the shared code is evidence of common ancestry?
- A. The code is the only set of pairings that works, so every organism had to arrive at the same code on its ownNothing in the chemistry forces UUU to mean Phe; other pairings would work.
So the shared code was inherited, not forced. - B. ✓ A code with other pairings would have worked, so one shared code is most simply explained by inheritance from one ancestral population
- C. The three organisms exchanged genes with one another recently, and the shared genes carried one code to all of themA gene carries instructions written in the code; it does not carry the code itself.
The shared code is older than any recent exchange. - D. Every organism needs phenylalanine, so every organism must use UUU to name itNeeding an amino acid does not fix which codon names it.
Another codon could have named Phe just as well.
Why: Nothing in the chemistry forces UUU to mean Phe; a code with other pairings would work.
Yet every organism tested uses the same code.
The simplest explanation is that all of them inherited the code from one ancestral population.
So the shared code is evidence for common ancestry.
The small subunit of a ribosome has bound a eukaryotic mRNA near its 5′ end and moved along it to the first AUG.
Which of the following happens next?
- A. The large subunit joins, and then the small subunit moves on to the second codonThe Met tRNA pairs with the AUG before the large subunit joins.
The small subunit stays on the AUG until the ribosome is complete. - B. A peptide bond forms between methionine (Met) and the second amino acidNo amino acid is in place yet.
The Met tRNA must pair with the AUG, and the large subunit must join, before any bond forms. - C. ✓ A tRNA carrying methionine (Met) pairs its anticodon with the AUG, and then the large subunit joins
- D. The small subunit moves on to the next AUG along the mRNAThe ribosome assembles on the first AUG it reaches.
The small subunit stops there.
Why: The small subunit stops at the first AUG.
The tRNA carrying methionine (Met) pairs its anticodon with that AUG.
Then the large subunit joins, and the ribosome is complete.
A ribosome has joined the first two amino acids of a polypeptide. The next codon in the ribosome is 5′-UGC-3′, drawn above the genetic code chart.
Which of the following arrives next?
- A. ✓ Cys, on the tRNA whose anticodon reads 3′-ACG-5′
- B. Cys, on the tRNA whose anticodon reads 3′-UGC-5′These are the codon's own letters.
An anticodon carries each codon base's partner: U takes A, G takes C, C takes G. - C. Thr, on the tRNA whose anticodon reads 3′-ACG-5′The amino acid is the one the CODON names on the chart, Cys.
The anticodon's letters are not read on the chart. - D. Cys, on the tRNA whose anticodon reads 5′-ACG-3′The partners are right, but an anticodon's 3′ end sits opposite the codon's 5′ end.
Written from its 3′ end it reads 3′-ACG-5′.
Why: The codon UGC names Cys on the chart.
Under each codon base write its partner: U takes A, G takes C, C takes G.
The anticodon runs the other way, so it reads 3′-ACG-5′.
That tRNA arrives carrying Cys.
The mRNA drawn above the genetic code chart is translated.
How many amino acids does the polypeptide contain?
- A. 6Met, from the AUG, is the first amino acid and is counted.
Counting from the codon after AUG misses it. - B. ✓ 7
- C. 8The stop codon adds no amino acid.
Only the codons from AUG up to the stop codon count. - D. 9The leading letters before AUG and the letters after the stop codon are not read.
Only the codons from the first AUG to the first stop codon count.
Why: Reading begins at the first AUG and continues in threes.
The codons up to and including the stop codon are 8.
The stop codon adds no amino acid.
So the polypeptide has 8 − 1 = 7 amino acids.
The mRNA drawn above the genetic code chart is translated.
Which amino acids does the polypeptide contain, in order?
- A. Cys, Gln, IleThese codons begin one base after the A of AUG.
The ribosome begins at the AUG itself and reads in threes from there. - B. Tyr, Gly, LeuThese are the anticodons read on the chart.
The chart is read with the mRNA's codons, not with the tRNAs' anticodons. - C. ✓ Met, Pro, Asp
- D. Met, Pro, SerThe third codon is GAU: row G, column A.
Its U line reads Asp, not Ser.
Why: The message begins with AUG, so the codons are AUG, CCA, GAU, UGA.
On the chart AUG names Met, CCA names Pro and GAU names Asp.
UGA is a stop codon, so reading ends there.
The polypeptide is Met, Pro, Asp.
A gene's template strand reads 3′-TAC GAG TAT ATC-5′, drawn above the genetic code chart. RNA polymerase copies it, and a ribosome translates the mRNA.
Which amino acids does the polypeptide contain, in order?
- A. Tyr, Glu, Tyr, IleThese are the template's letters with U for T.
The mRNA carries each base's partner: A gives U, C gives G, G gives C, T gives A. - B. Met, Ser, IleThe second codon is CUC: row C, column U.
Its C line reads Leu, not Ser. - C. Asp, Ile, Leu, ValThe mRNA's 5′ end sits under the template's 3′ end, at the left.
The mRNA reads 5′-AUG CUC AUA UAG-3′ and begins with AUG. - D. ✓ Met, Leu, Ile
Why: Under each template base write its mRNA partner: T gives A, A gives U, C gives G, G gives C.
The mRNA reads 5′-AUG CUC AUA UAG-3′.
AUG names Met, CUC names Leu, AUA names Ile, and UAG is a stop codon.
So the polypeptide is Met, Leu, Ile.
A student describes the flow of information in a eukaryotic cell with four labels: ‘in the nucleus’, ‘transcription by RNA polymerase’, ‘read 5′ to 3′’ and ‘begins with Met’. The student now describes the same flow in a bacterium.
Which label must change?
- A. ‘transcription by RNA polymerase’RNA polymerase builds the mRNA in a bacterium too.
That label stays. - B. ✓ ‘in the nucleus’
- C. ‘read 5′ to 3′’A ribosome reads an mRNA from its 5′ end toward its 3′ end in every cell.
That label stays. - D. ‘begins with Met’The code is the same in a bacterium: AUG names Met, and the ribosome assembles on AUG.
That label stays.
Why: A bacterium has no nucleus.
Its DNA lies in the cytoplasm, in the same fluid as its ribosomes.
The other three labels hold in every cell: RNA polymerase transcribes, the ribosome reads 5′ to 3′, and AUG names Met.
So ‘in the nucleus’ is the label that must change.
A scientist measures how long a ribosome spends at each codon of an mRNA. The bar chart shows the time spent at the four codons that name threonine (Thr), relative to the fastest of them. The scientist can rewrite the mRNA, keeping the same polypeptide.
Which one change would make the mRNA's translation faster?
- A. ✓ Replacing every ACA with ACU
- B. Replacing every ACU with ACAThe ACA bar is taller than the ACU bar.
The ribosome would spend longer at each threonine codon. - C. Replacing every ACC with ACGThe ACG bar is taller than the ACC bar.
The ribosome would spend longer at each of those codons. - D. Replacing every ACG with ACAThe ACA bar is the tallest of the four.
The ribosome would spend the longest time at each of those codons.
Why: The ribosome spends the longest time at ACA, the tallest bar, and the shortest at ACU.
A scarce tRNA takes longer to arrive, so the ribosome waits longer at its codon.
Replacing every ACA with ACU keeps the threonines and cuts the waiting.
So the translation is faster.
An animal's two alleles of one gene for an enzyme differ at one base of DNA. One codon of the mRNA differs as a result: 5′-AUA-3′ in allele 1 is 5′-AGA-3′ in allele 2. The two codons are drawn above the genetic code chart.
How does the enzyme built from allele 2 differ from the enzyme built from allele 1?
- A. It is one amino acid shorterAGA names an amino acid, Arg, so the codon is still read.
A codon that ends a polypeptide is a stop codon, and AGA is not one. - B. Every amino acid after that point differsOnly one base changed, and the codon it sits in is still three bases long.
The codons after it are grouped and read as before. - C. ✓ One amino acid differs: Arg in place of Ile
- D. It is not built at allThe ribosome still finds the AUG and reads codon by codon.
At the changed codon it joins Arg instead of Ile and carries on.
Why: On the chart AUA names Ile and AGA names Arg.
The ribosome reads the changed codon as Arg.
Every other codon is the same, so every other amino acid is the same.
So the enzyme differs by one amino acid: Arg in place of Ile.
The table describes four viruses.
Which virus is a retrovirus?
- A. WVirus W carries a DNA genome and no enzyme.
A retrovirus carries RNA and reverse transcriptase, the enzyme that copies RNA into DNA. - B. XVirus X carries RNA, but its enzyme copies RNA into RNA.
An RNA genome alone does not make a retrovirus; reverse transcriptase does. - C. ✓ Y
- D. ZVirus Z's enzyme copies DNA into RNA, the usual direction.
A retrovirus's enzyme copies RNA into DNA.
Why: A retrovirus is an RNA virus that carries reverse transcriptase, the enzyme that copies its RNA into DNA.
Virus Y carries an RNA genome and an enzyme that copies RNA into DNA.
So virus Y is the retrovirus.
A retrovirus's DNA copy has joined one of the host cell's chromosomes.
Which enzyme reads that DNA copy to make the virus's RNA?
- A. The virus's reverse transcriptaseReverse transcriptase copies RNA into DNA, and that job is done.
The DNA copy is read by the enzyme that reads the cell's own genes. - B. The host cell's DNA polymeraseDNA polymerase copies DNA into DNA, in S phase.
Making RNA from DNA is RNA polymerase's job. - C. The virus's second enzyme, the one that inserted the copyThe virus's second enzyme cuts a chromosome and joins the copy in, and that job is done.
Reading a gene into RNA is the host cell's own enzyme's job. - D. ✓ The host cell's RNA polymerase
Why: Once the copy sits in a chromosome, it is read like one of the cell's own genes.
The enzyme that reads a gene into RNA is RNA polymerase, the host cell's own.
The host's ribosomes then read the virus's mRNA and build its proteins.
A drug that blocks reverse transcriptase reaches two cells of the same kind. A retrovirus enters cell W one hour after the drug arrives. A retrovirus entered cell X a week before the drug arrived, and one of cell X's chromosomes already carries the DNA copy.
Which cell builds new virus particles while the drug is present?
- A. Cell W onlyIn cell W the enzyme is blocked before it can copy the RNA into DNA.
With no DNA copy, no viral genes are read and no particles form. - B. ✓ Cell X only
- C. Both cellsIn cell W no DNA copy is made, so nothing is read and no particles form.
Only cell X, whose copy already exists, builds particles. - D. Neither cellCell X's copy is already in a chromosome, where reverse transcriptase has no job left to do.
The host's RNA polymerase reads the copy, and particles form.
Why: In cell W the drug stops the RNA being copied, so no DNA copy forms and no particles form.
In cell X the copy already sits in a chromosome; reverse transcriptase has no job left there.
The host's enzyme and ribosomes read the copy.
So only cell X builds particles.
The model drawn shows the flow of genetic information in a eukaryotic cell. Four of its labels are missing, and the blanks are lettered Q, R, S and T.
(a) Represent the flow of information by stating what belongs at each blank: the molecule at Q, the place at R, the step at S and, at T, the direction in which the template strand is read. (2 pt)
R is at a ribosome in the cytoplasm.
S is translation.
T is the template strand read 3′ to 5′.
The completed model is drawn.
- Award 1 point for: Q polypeptide AND S translation. Accept 'protein' for Q.
- Award 1 point for: R at a ribosome in the cytoplasm (accept 'in the cytoplasm' or 'at a ribosome') AND T the template strand is read 3′ to 5′ (accept 'read 3′ to 5′').
- Accept at T 'the mRNA is built 5′ to 3′' only where the student names it as the mRNA's direction; a bare '5′ to 3′' at T earns nothing.
Slip Writing 'read 5′ to 3′' at T: that is the direction the mRNA is read, written beneath the middle box. RNA polymerase reads the template strand the other way.
(b) Describe what happens at the step lettered S. (1 pt)
For each codon a tRNA brings the amino acid the codon names.
The ribosome joins the amino acids by peptide bonds into the polypeptide.
- Award 1 point for: a ribosome reads the mRNA codon by codon AND tRNAs bring the amino acids, which the ribosome joins into the polypeptide.
(c) A student claims that, in this cell, transcription and the step lettered S must happen in two different places. Using the model, support the claim. (1 pt)
The ribosomes work in the cytoplasm, outside the nuclear envelope.
So the mRNA is finished in the nucleus and leaves through a pore before a ribosome can read it.
- Award 1 point for: the DNA stays in the nucleus, so transcription happens there, AND the ribosomes work in the cytoplasm, so the mRNA must leave the nucleus (through a pore) before translation. Accept: the nuclear envelope separates the DNA from the ribosomes.
- The label 'then leaves through a pore' copied on its own earns nothing: the point wants where the DNA stays AND where the ribosomes work.
Slip Saying the ribosomes enter the nucleus to read the mRNA there. Ribosomes work in the cytoplasm; the mRNA travels to them.
A plant carries two alleles of one gene for an enzyme that builds the hard material of the plant's seed coats. The two alleles differ at one base of DNA, so one codon of the mRNA differs: 5′-GGG-3′ in the first allele's mRNA is 5′-GAG-3′ in the second allele's mRNA. The two codons are drawn above the genetic code chart. A plant with two copies of the first allele makes seeds with hard coats. A plant with two copies of the second allele makes seeds with soft coats.
(a) Describe how the difference between the two alleles changes the enzyme's order of amino acids. (1 pt)
So the second allele's enzyme carries Glu at one position where the first allele's enzyme carries Gly.
Every other amino acid is the same.
- Award 1 point for: one amino acid differs — Glu (from GAG) in place of Gly (from GGG) — and the rest of the order is unchanged.
Slip Saying the second allele's enzyme is shorter or is not made. GAG names an amino acid, so the codon is read and the chain goes on.
(b) Explain how that difference leads to seeds with soft coats. (1 pt)
The order of amino acids sets the fold, so the enzyme folds into a different shape around that spot.
The shape lets the enzyme do its job, and the changed shape no longer fits the coat's building block at its active site.
So the hard material is not built, and the seed coat stays soft.
- Award 1 point for: the changed amino acid changes the enzyme's fold (shape), so the enzyme no longer does its job (its active site no longer fits its building block), so the hard material is not built. Accept 'a different R group at that position changes the fold'.
Slip Stopping at 'the enzyme is different'. The point is the chain: a different amino acid → a different fold → the job is lost → the trait.
(c) A third allele differs from the first allele at the third base of the same codon: its mRNA carries 5′-GGA-3′ there. Predict the seed coats of a plant with two copies of the third allele. (1 pt)
- Award 1 point for: hard seed coats (the same as the first allele's).
(d) Justify your prediction. (1 pt)
So the third allele's enzyme has the same order of amino acids as the first allele's.
The same order gives the same fold and the same job.
So the enzyme builds the hard material, and the seed coats are hard.
- Award 1 point for: GGA and GGG both name Gly (the code is redundant), so the enzyme's amino acid order, fold and job are unchanged.
Slip Claiming soft coats because a base changed. A base change that keeps the amino acid changes nothing downstream.
APBIO-U06-L24 Not every gene, not all the time
Suppose E. coli is growing in a broth of glucose. A technician moves the cells into a broth of lactose instead. No lactose has ever reached these cells before. Within about ten minutes they are digesting it.
The genes for digesting lactose were in every cell all along. What was stopping those genes, and what let them go?
Unit 6 · Gene Expression and Regulation
1Not every gene, not all the time
A ripening banana gives off the gas ethylene. Ethylene binds receptors on a fruit cell, and a pathway inside the cell carries the message on. The cell then starts making the enzymes that soften the fruit.
Where does the pathway’s last activated protein act?
- A. On the cell membraneThe last activated protein of the pathway reaches the DNA.
There it changes which genes the cell expresses. - B. ✓ On the DNA
- C. On the ribosomesThe last activated protein does not act on the ribosomes.
It reaches the DNA and changes which genes the cell expresses.
Why: The last activated protein of the pathway reaches the DNA.
There it changes gene expression.
So the cell starts making the enzymes that soften the fruit.
UV light makes a person’s skin cells make more of the pigment melanin, and the skin tans.
What did the UV light change in the skin cells?
- A. ✓ How much the cells use their pigment genes
- B. The base sequence of their DNAA tan is not a change in any allele.
The UV light changed how much the skin cells use their pigment genes.
Why: A condition outside the organism changes which genes its cells use and how much.
So the phenotype changes while the DNA does not.
The UV light made the skin cells use their pigment genes more.
Does a cell use all of its genes all of the time?
No. A cell transcribes only some of its genes at any moment.
How often a cell transcribes each gene is controlled.
A gene that is present but not transcribed makes no protein.
Video: Watch: Not every gene, not all the time
Two E. coli cells with the same genome are drawn side by side. In the cell in glucose broth, the lactose genes sit as open boxes and no RNA comes off them. In the cell in lactose broth, the same genes fill in and RNA strands hang from them.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L24a.mp4
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Go back to the E. coli growing in glucose broth. Every one of its cells carries the genes for digesting lactose.
In glucose broth, RNA polymerase is not transcribing those genes. So no mRNA of them exists in the cell.
With no mRNA, the ribosomes build none of the enzymes. The genes are present, and the cell makes no enzyme from them.
Now the technician moves the cells into lactose broth. Within a few minutes, RNA polymerase begins transcribing the lactose genes.
The ribosomes read the new mRNA and build the enzymes. Within about ten minutes, the cells are digesting lactose.
The two cells carry the same genome. One cell transcribes the lactose genes, and the other does not.
E. coli cells look like this through an electron microscope. Each rod is one cell, and every cell carries the same genome.

E. coli cells photographed through an electron microscope. Each rod is one cell, and every cell carries the same genome. Image: NIAID, Wikimedia Commons, public domain (resized).
A gene a cell carries is not always a gene the cell uses. A gene is expressed only when the cell transcribes it and the ribosomes build its protein.
So a cell does not transcribe all of its genes all of the time. Which genes it transcribes, and how often, is controlled.
What you are expected to know Explain why a gene that a cell carries can make no protein: the cell is not transcribing the gene, so no mRNA of it exists and the ribosomes build nothing from it.
E. coli is growing in glucose broth, with no lactose anywhere near it.
Does each cell carry the genes for digesting lactose?
- A. ✓ Yes
- B. NoThe lactose genes are part of the cell’s genome, and every cell carries the whole genome.
The cell carries the lactose genes whether or not it uses them.
Why: Every cell carries its whole genome.
The lactose genes are part of that genome.
So each cell carries them, in glucose broth as in lactose broth.
E. coli is growing in glucose broth, with no lactose in it.
Is each cell making the enzymes that digest lactose?
- A. YesIn glucose broth, RNA polymerase is not transcribing the lactose genes.
With no mRNA of them, the ribosomes build none of the enzymes. - B. ✓ No
Why: In glucose broth, RNA polymerase is not transcribing the lactose genes.
So no mRNA of them exists in the cell.
So the ribosomes build none of the enzymes.
Suppose a bacterium carries the gene for an enzyme that breaks down lignin, the tough material in wood. In a broth with no lignin, the enzyme is absent from the bacterium.
(a) Explain why the enzyme is absent, though the bacterium carries its gene. (2 pt)
Frame The enzyme is absent because …
So no mRNA of the gene exists in the cell.
A ribosome builds a protein only from an mRNA.
So the ribosomes build none of the enzyme.
- Award 1 point for: the gene is not being transcribed (RNA polymerase is not copying it), so no mRNA of it exists in the cell.
- Award 1 point for: the ribosomes build a protein only from an mRNA, so with no mRNA they build none of the enzyme.
A student says: “A gene a cell has is a gene a cell uses. If the gene is in the DNA, the cell is making its protein.”
Is the student correct?
- A. ✓ No: a cell makes a gene’s protein only while it transcribes the gene, and it transcribes only some of its genes
- B. Yes: every gene in a cell’s DNA is transcribed, so the cell makes every protein its genes code forA cell transcribes only some of its genes at any moment.
A gene that is present but not transcribed makes no protein.
Why: A cell makes a gene’s protein only while RNA polymerase transcribes the gene.
A cell transcribes only some of its genes at any moment.
So a gene in the DNA is not always a gene the cell is using.
Suppose a tree’s bark is cut. Cells beside the cut make an enzyme that produces a sticky resin. In cells far from the cut, the enzyme is absent.
Which of the following is true of the cells far from the cut?
- A. They have lost the gene for the resin enzymeEvery cell of the tree carries the same genome, the gene for the resin enzyme with it.
The cells far from the cut are not transcribing that gene. - B. They carry a different version of the geneAll the tree’s cells grew from one fertilized egg, so they carry the same version of the gene.
The cells far from the cut are not transcribing it. - C. ✓ They carry the gene and are not transcribing it
Why: Every cell of the tree carries the same genome.
So the cells far from the cut carry the gene for the resin enzyme.
The enzyme is absent there, so those cells are not transcribing the gene.
24Which genes, and how much
Which of the following counts as part of a cell’s phenotype?
- A. ✓ The proteins the cell makes
- B. The base sequence of its DNAThe DNA sequence itself is the genotype.
What a cell makes can be observed or measured, so it is part of the phenotype.
Why: An organism’s set of observable features is its phenotype.
What a cell makes can be observed or measured.
So the proteins a cell makes are part of its phenotype.
Which genes a cell expresses, and how much of each protein it makes, is its phenotype.
Video: Watch: Which genes, and how much
The two cells again, one in glucose broth and one in lactose broth. Under each cell a bar rises for the amount of the lactose-digesting enzyme it holds: a sliver under the glucose-broth cell, a tall bar under the lactose-broth cell.
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Suppose a biologist measures the amount of the lactose-digesting enzyme in each of the two cells: the cell in glucose broth and the cell in lactose broth.
The cell in glucose broth holds about 3 units of the enzyme. The cell in lactose broth holds about 300 units.
The units are arbitrary: they compare the two cells, and the axis says so.
The cell in lactose broth transcribes the lactose genes far more often. So its ribosomes build far more of the enzyme.
How often a cell transcribes a gene sets how much of that gene’s protein the cell holds.
One cell digests lactose and the other does not. What a cell makes is part of its phenotype.
So that difference is a difference in phenotype.
The two cells carry the same DNA. They express different genes, and at different levels.
So their phenotypes differ.
So a cell’s phenotype is set by the genes it expresses and by how much of each protein it makes.
What you are expected to know Explain what sets a cell’s phenotype: which genes the cell expresses, and how much of each protein it makes.
Two cells of one organism carry the same genome, and their phenotypes differ.
Which of the following differs between the two cells?
- A. The genes their DNA carriesThe two cells carry the same genome, so they carry the same genes.
They differ in which of those genes they express, and how much. - B. ✓ Which genes they express, and how much
- C. The order of the genes along their DNAThe two cells carry the same genome, so their genes lie in the same order.
They differ in which genes they express, and how much.
Why: The two cells carry the same genome.
A cell’s phenotype is set by which genes it expresses and how much of each protein it makes.
So the two cells express different genes, or the same genes at different levels.
Two cells with identical DNA sit side by side. One makes an enzyme that the other does not. A student says: “The two cells have different phenotypes, even though their DNA is the same.”
Is the student correct?
- A. No: two cells with the same DNA always have exactly the same phenotypeWhat a cell makes is part of its phenotype.
One cell makes the enzyme and the other does not, so their phenotypes differ. - B. ✓ Yes: what a cell makes is part of its phenotype, so their phenotypes differ
Why: What a cell makes is part of its phenotype.
One cell makes the enzyme and the other does not.
So the two cells have different phenotypes with the same DNA.
Two cells of one person carry the same genome. The drawing gives the amount of hemoglobin mRNA in each cell, in arbitrary units. One of the two cells is a red blood cell precursor, a cell that will fill with hemoglobin.
Which cell is the red blood cell precursor?
- A. ✓ Cell J
- B. Cell KA cell that will fill with hemoglobin transcribes the hemoglobin genes at a high level.
Cell K holds no hemoglobin mRNA, so cell K is building no hemoglobin.
Why: A cell that will fill with hemoglobin transcribes the hemoglobin genes often.
So that cell holds a large amount of hemoglobin mRNA.
Cell J holds the hemoglobin mRNA, so cell J is the red blood cell precursor.
42Two parts to a switch
Insulin binds the insulin receptor and leaves every other receptor alone.
Why does insulin bind only its own receptor?
- A. Insulin is the only signal molecule in the bloodThe blood carries many signal molecules at once.
Insulin binds the one receptor whose pocket matches its shape and charges. - B. The insulin receptor pulls insulin toward itself from a distanceA receptor pulls nothing toward itself.
Insulin binds the one receptor whose pocket matches its shape and charges. - C. ✓ Insulin’s shape and charges match only that receptor’s pocket
Why: A receptor binds only the molecule whose shape and charges match its pocket.
Insulin’s shape and charges match the insulin receptor’s pocket and no other.
So insulin binds only its own receptor.
RNA polymerase is about to copy a gene into RNA.
Which stretch of DNA does RNA polymerase bind to begin?
- A. ✓ The promoter, just before the gene
- B. The middle of the gene, halfway along itRNA polymerase cannot start copying just anywhere along the DNA.
It binds the promoter, just before the gene, and copies from there. - C. The last base of the gene, at its far endCopying begins at the start of the gene, not at its end.
RNA polymerase binds the promoter, just before the gene.
Why: RNA polymerase must bind the promoter before it can copy a gene.
The promoter is a short DNA sequence just before the gene.
A switch on a gene has two parts: a stretch of DNA beside the gene, and a protein whose shape fits that stretch.
The protein raises or lowers how often RNA polymerase transcribes the gene.
In the bacterium, one such protein was sitting on the DNA beside the lactose genes, and lactose moved it.
Video: Watch: Two parts to a switch
The gene drawn as a line with its promoter box and RNA polymerase on it. A second open box appears beside the promoter, and a protein settles onto it. The RNA strands coming off the gene thin out. The protein lifts off, and the strands come thick again.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L24c.mp4
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What controls how often RNA polymerase transcribes a gene?
Suppose a gene is drawn as a line, with its promoter just before it. RNA polymerase binds the promoter to begin copying the gene.
Beside the promoter sits another short stretch of DNA. That stretch is not part of the gene, and RNA polymerase does not copy it into RNA.
A protein binds only the molecule whose shape and charges fit its binding site.
One protein in the cell has a binding site that fits this stretch of DNA. So that protein sits on this stretch, and on no other stretch.
While the protein sits there, RNA polymerase transcribes the gene less often than before.
Some proteins of this kind do the opposite. While they sit on their stretch, RNA polymerase transcribes the gene more often.
A stretch of DNA beside a gene, where a protein binds to control how often the gene is transcribed, is called a , because to regulate is to control.
The protein that binds a regulatory sequence and raises or lowers how often RNA polymerase transcribes the gene is called a .
The promoter and the regulatory sequence are both DNA beside the gene. RNA polymerase binds the promoter.
The regulatory protein binds the regulatory sequence. The regulatory protein is itself a protein.
So the cell builds the regulatory protein from one of its own genes.
The regulatory sequence is never copied into RNA. So a change in a regulatory sequence can change how much of the protein the cell makes, but not which amino acids the protein has.
What you are expected to know Describe the two parts of a control on transcription: a regulatory sequence, DNA beside the gene, and a regulatory protein that binds it because its shape fits and raises or lowers how often the gene is transcribed.
Suppose a protein sits on a short stretch of DNA next to a gene’s promoter. While the protein sits there, RNA polymerase transcribes the gene less often.
Which part of the switch is that stretch of DNA?
- A. The promoterThe promoter is where RNA polymerase binds.
The stretch the protein sits on, beside the gene, is the regulatory sequence. - B. ✓ The regulatory sequence
- C. The gene itselfThe gene is the stretch RNA polymerase copies into RNA.
The stretch the protein sits on is beside the gene: the regulatory sequence.
Why: The protein binds a stretch of DNA beside the gene and changes how often the gene is transcribed.
A stretch of DNA where a protein binds to control transcription is the regulatory sequence.
A protein sits on a stretch of DNA beside a gene and lowers how often the gene is transcribed.
What is that protein called?
- A. RNA polymeraseRNA polymerase copies the gene into RNA; it is the enzyme whose work is being lowered.
The protein sitting beside the gene is a regulatory protein. - B. ✓ A regulatory protein
- C. A promoterA promoter is a stretch of DNA, not a protein.
The protein sitting on the DNA beside the gene is a regulatory protein.
Why: The protein binds the DNA beside the gene and lowers how often RNA polymerase transcribes it.
A protein that does this is a regulatory protein.
Suppose a bacterium carries a regulatory protein for its genes for digesting citrate. Along the whole chromosome, the protein binds only one short stretch of DNA, beside those genes.
(a) Explain why the protein binds only that one stretch. (2 pt)
Frame The protein binds only that one stretch because …
The bases of that stretch, in that order, have a shape and charges that fit the protein’s binding site.
Every other stretch of the chromosome has a different order of bases.
So every other stretch has a different shape.
So the protein’s binding site fits nowhere else, and the protein binds only there.
- Award 1 point for: the protein binds only DNA whose shape and charges fit its binding site, and that stretch’s bases fit it.
- Award 1 point for: every other stretch has a different order of bases, so a different shape, so the binding site fits nowhere else.
Suppose the protein that controls a gene in a plant’s cells is bound to the regulatory sequence beside the gene. While it is bound, RNA polymerase transcribes the gene rarely. Now imagine that protein is removed from the cells.
What happens to how often RNA polymerase transcribes the gene?
- A. ✓ The gene is transcribed more often
- B. The gene is transcribed as rarely as beforeThe protein was lowering how often the gene is transcribed.
With the protein gone, RNA polymerase transcribes the gene more often. - C. The gene is no longer transcribedThe protein was lowering transcription, and RNA polymerase still binds the promoter.
With the protein gone, RNA polymerase transcribes the gene more often.
Why: While the regulatory protein was bound, RNA polymerase transcribed the gene rarely.
The protein was lowering how often the gene is transcribed.
With the protein gone, RNA polymerase transcribes the gene more often.
This table compares the two parts of a switch: what each is made of, where it sits, and what it does.
Go back to the E. coli moved from glucose broth into lactose broth.
Its lactose genes were present in every cell but not transcribed.
A regulatory protein sat on the regulatory sequence beside those genes, and RNA polymerase rarely transcribed them.
Lactose changed that. Within about ten minutes, the cells were digesting lactose.
Which genes a cell transcribes, and how much of each protein it makes, is its phenotype.
73Quick quiz: regulatory sequence, regulatory protein mixed practice
The stretch beside a gene where a regulatory protein binds.
Is this part DNA or protein?
- A. ✓ DNA
- B. ProteinA regulatory protein binds a stretch of DNA.
That stretch, the regulatory sequence, is DNA.
Why: The regulatory protein binds a stretch of DNA beside the gene.
That stretch is the regulatory sequence, and it is DNA.
The part whose binding site fits a stretch of DNA beside a gene.
Is this part DNA or protein?
- A. DNAA binding site is part of a protein.
The part that fits the stretch of DNA is the regulatory protein. - B. ✓ Protein
Why: A binding site belongs to a protein.
The part whose binding site fits the DNA is the regulatory protein.
The part that raises or lowers how often RNA polymerase transcribes a gene by sitting on the DNA beside it.
Is this part DNA or protein?
- A. DNAA stretch of DNA sits still; it cannot settle onto another stretch.
The part that sits on the DNA and changes transcription is the regulatory protein. - B. ✓ Protein
Why: The part that sits on the DNA beside the gene is the regulatory protein.
A regulatory protein raises or lowers how often the gene is transcribed.
The promoter, where RNA polymerase binds to begin copying.
Is this part DNA or protein?
- A. ✓ DNA
- B. ProteinRNA polymerase is a protein, and it binds the promoter.
The promoter it binds is a short sequence of DNA.
Why: The promoter is a short DNA sequence just before the gene.
RNA polymerase binds it, so the promoter is DNA.
A short run of bases beside a gene, left out of the RNA when the gene is transcribed.
Is this part DNA or protein?
- A. ✓ DNA
- B. ProteinA run of bases is a stretch of DNA.
The regulatory sequence is DNA beside the gene, and RNA polymerase leaves it out of the RNA.
Why: Bases in a row make up a stretch of DNA.
The regulatory sequence is such a stretch, and RNA polymerase leaves it out of the RNA.
The part of the switch that the cell builds from one of its own genes.
Is this part DNA or protein?
- A. DNAA cell does not build its DNA from a gene; the DNA carries the genes.
The part built from a gene is the regulatory protein. - B. ✓ Protein
Why: A cell builds proteins from its genes.
The regulatory protein is a protein, so the cell builds it from one of its genes.
What is a regulatory sequence?
- A. ✓ A stretch of DNA beside a gene where a regulatory protein binds
- B. The stretch of DNA that RNA polymerase copies into RNA to make the gene’s mRNAThe stretch RNA polymerase copies into RNA is the gene.
The regulatory sequence sits beside the gene and is never copied. - C. The first bases of a gene, which code for the protein’s first amino acidsThe first bases of a gene are part of the gene and are copied into RNA.
The regulatory sequence sits beside the gene and is never copied.
Why: A stretch of DNA beside a gene, where a protein binds to control how often the gene is transcribed, is called a regulatory sequence.
What is a regulatory protein?
- A. A protein that copies a gene into RNA, beginning at the promoterThe protein that copies a gene into RNA is RNA polymerase.
A regulatory protein binds the regulatory sequence and changes how often that copying happens. - B. A protein that RNA polymerase builds by copying the regulatory sequenceRNA polymerase builds RNA, not protein, and it never copies the regulatory sequence.
A regulatory protein binds the regulatory sequence and changes how often the gene is transcribed. - C. ✓ A protein that binds a regulatory sequence and changes how often the gene is transcribed
Why: The protein that binds a regulatory sequence and raises or lowers how often RNA polymerase transcribes the gene is called a regulatory protein.
A cell controls how often RNA polymerase transcribes one of its genes.
(a) State what a regulatory sequence is. (1 pt)
- Award 1 point for: a stretch of DNA beside (near) a gene where a regulatory protein binds. ‘Controls how often the gene is transcribed’ completes it, but its absence does not lose the point.
(b) State what a regulatory protein does. (1 pt)
- Award 1 point for: binds the regulatory sequence (DNA beside the gene) and raises or lowers (changes) how often the gene is transcribed.
83Mixed practice mixed practice
Now imagine the regulatory sequence beside a gene is removed from the DNA, and the promoter stays.
Which of the following can RNA polymerase still do?
- A. Bind the regulatory protein in place of the missing sequenceRNA polymerase binds DNA at the promoter, never a regulatory protein.
With the promoter still there, it binds the promoter and transcribes the gene. - B. Nothing: the gene can no longer be transcribed at allRNA polymerase binds the promoter, and the promoter is still there.
So RNA polymerase still binds it and transcribes the gene. - C. ✓ Bind the promoter and transcribe the gene
Why: RNA polymerase binds the promoter to begin copying a gene.
The promoter is still there.
So RNA polymerase binds the promoter and transcribes the gene; only the control by the regulatory protein is gone.
A cell carries a gene, and a biologist finds no mRNA of that gene anywhere in the cell.
Which of the following is true?
- A. The cell has lost the geneThe cell carries the gene.
A gene the cell is not transcribing has no mRNA, and the gene is still in the DNA. - B. ✓ The cell is not transcribing the gene
- C. The cell is translating the gene without any mRNARibosomes build a protein only from an mRNA.
With no mRNA of the gene, the cell is not transcribing it and builds none of its protein.
Why: RNA polymerase makes a gene’s mRNA by transcribing the gene.
No mRNA of the gene exists.
So the cell is not transcribing the gene.
Two cells of one animal carry the same genome. The table gives how many times each cell transcribes gene F in an hour.
Which cell holds more of protein F?
- A. ✓ Cell N
- B. Cell PCell N transcribes gene F 40 times in an hour, cell P only 2.
More transcription gives more mRNA, so cell N builds more of protein F.
Why: How often a cell transcribes a gene sets how much of that gene’s protein it holds.
Cell N transcribes gene F 40 times in an hour, cell P 2 times.
So cell N holds more of protein F.
Suppose a regulatory protein folds into a different shape. Its binding site now fits a stretch of DNA with a different order of bases from the regulatory sequence beside its gene.
Which of the following happens?
- A. The protein binds the promoter insteadA protein binds only DNA whose shape and charges fit its binding site.
The changed site fits the promoter no better, so the protein binds neither stretch. - B. ✓ The protein stops binding the regulatory sequence
- C. RNA polymerase stops binding the promoterRNA polymerase’s own binding site is unchanged.
RNA polymerase still binds the promoter; only the regulatory protein has lost its fit.
Why: A protein binds only the molecule whose shape and charges fit its binding site.
The protein’s binding site fits a different order of bases from the regulatory sequence.
So the protein stops binding the regulatory sequence, and it stops changing how often the gene is transcribed.
A student says: “A cell can carry a gene for years while the amount of that gene’s protein in it stays at zero.”
Is the student correct?
- A. No: a cell makes at least a little of every protein its genes code for, all the timeA cell transcribes only some of its genes.
A gene the cell never transcribes gives no mRNA, so the cell makes none of its protein. - B. ✓ Yes: the cell makes the protein only while it transcribes the gene, and it may never transcribe it
Why: A cell makes a gene’s protein only while RNA polymerase transcribes the gene.
A cell may leave some of its genes untranscribed for years.
So the cell carries those genes and makes none of their proteins.
Two cells of one organism hold different amounts of one protein. A student says: “Their genes for that protein must differ.”
Is the student correct?
- A. ✓ No: the two cells can carry the same gene and transcribe it at different rates
- B. Yes: the amount of a protein a cell holds is set by which version of the gene the cell carriesTwo cells of one organism carry the same genome.
The cell that transcribes the gene more often holds more of the protein, with the same gene.
Why: Two cells of one organism carry the same genome, so they carry the same gene.
How often a cell transcribes a gene sets how much of its protein the cell holds.
So the amounts can differ while the gene is the same.
Suppose a cell begins transcribing a gene ten times as often as before.
What happens to the amount of that gene’s protein in the cell?
- A. ✓ The amount of the protein rises
- B. The amount of the protein stays the sameMore transcription gives more mRNA of the gene.
The ribosomes build more of the protein from the extra mRNA. - C. The amount of the protein fallsMore transcription gives more mRNA, not less.
The ribosomes build more of the protein from the extra mRNA.
Why: Transcribing the gene more often gives more mRNA of the gene.
The ribosomes build the protein from that mRNA.
So the amount of the protein rises.
Suppose a biologist compares two kinds of cell from one limpet, a small sea snail. The two kinds of cell carry identical DNA. Cells of the first kind make a protein that builds the shell, and cells of the second kind make none of it.
(a) Predict what the biologist finds when looking for the shell-building protein’s mRNA in each kind of cell. (1 pt)
The second kind holds none, or far less, because it is not transcribing the gene.
- Award 1 point for: mRNA of the gene in the first kind of cell and none (or far less) in the second kind.
(b) Explain how these two kinds of cell show that a cell’s phenotype depends on which genes it expresses. (2 pt)
Frame The two kinds of cell show this because …
Only the first kind transcribes that gene.
So only the first kind holds its mRNA and builds the protein.
Making the protein is part of the first kind’s phenotype.
So the two kinds have different phenotypes with the same genes: they express different genes.
- Award 1 point for: both kinds of cell carry the gene (identical DNA), but only the first kind transcribes it and builds the protein.
- Award 1 point for: what a cell makes is part of its phenotype, so the two kinds have different phenotypes while carrying the same genes — the phenotype depends on which genes are expressed.
Glossary
- regulatory sequence
- A stretch of DNA beside a gene where a regulatory protein binds to control how often the gene is transcribed. It is not part of the gene, and RNA polymerase does not copy it into RNA.
- regulatory protein
- A protein that binds a regulatory sequence because its binding site fits that stretch of DNA, and raises or lowers how often RNA polymerase transcribes the gene.
APBIO-U06-L24B Always on, or on demand
The table above shows four genes in E. coli, each measured in three broths: glucose, lactose, and no sugar. The units are arbitrary.
Gene 1 is transcribed at about 100 units in all three broths. Gene 2 is at 2 units in glucose, 200 units in lactose, and 2 units with no sugar. Which of these genes is switched on and off, and which is simply always on?
Unit 6 · Gene Expression and Regulation
1Reading a gene’s level across conditions
Suppose a cell carries a gene, and RNA polymerase builds no RNA from it.
Is the cell building that gene’s protein?
- A. YesA ribosome builds a protein from an mRNA copy of its gene.
With no RNA built, the ribosomes have nothing to read, so no protein is built. - B. ✓ No
Why: A protein is built by a ribosome reading an mRNA copy of its gene.
RNA polymerase builds no RNA from this gene.
So the ribosomes have no copy to read.
So the cell builds none of that gene’s protein.
How do you tell a gene that is always on from one that waits for a signal?
Read its level across conditions.
A gene transcribed at about the same level in every condition is always on.
The genes for ribosomal proteins and for the enzymes of glycolysis are like this.
A gene transcribed only when a particular signal is present is on demand.
The lactose genes are like this, and the signal is lactose.
Sorting genes this way is the first move in reading any expression table.
Video: Watch: Reading a gene’s level across conditions
The four-gene table appears one row at a time. A ring moves along Gene 1’s row, and its three levels are read out. A ring moves along Gene 2’s row, and its three levels are read out. A sorting column fills in beside each gene.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L24Ba.mp4
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The table below lists the four E. coli genes, with each gene’s level in the three broths. The levels are in arbitrary units.
Read Gene 1 across the three broths: 103, 99 and 101 units.
The three levels differ by a few units only. So Gene 1 is at about the same level in every broth.
A gene at about the same level in every condition is always on. Gene 1 is always on.
Read Gene 2 across the three broths: 2, 200 and 2 units.
Gene 2’s level in lactose broth is far above its level in the other two broths. So Gene 2 is on in lactose broth only.
A gene transcribed only when a particular signal is present is on demand. Gene 2 is on demand, and its signal is lactose.
To sort any gene from its table, follow three steps.
- Read the gene’s level in every condition.
- If the levels differ by a few units, the gene is always on.
- If one level is far above the rest, the gene is on demand, and that condition holds its signal.
The level itself never sorts a gene.
A gene held low in every condition is always on, just like a gene held high in every condition. Only the change across conditions sorts it.
A gene transcribed at about the same level in every condition is called . ‘Constitutive’ means built in: the cell transcribes the gene as part of its basic make-up.
Its everyday name is a housekeeping gene, because the cell needs its product for everyday upkeep.
The genes for ribosomal proteins and for the enzymes of glycolysis are constitutively expressed. A cell needs ribosomes and glycolysis in every condition.
A gene transcribed only when a particular signal is present is called an , because the signal induces, or brings on, its transcription.
So Gene 1 is constitutively expressed, and Gene 2 is an inducible gene with lactose as its signal.
What you are expected to know Classify a gene from its levels across conditions: constitutively expressed when its level is about the same in every condition, inducible when its level is high only while its signal is present.
Now suppose you sort the table’s other two genes, Gene 3 and Gene 4, the same way.
The E. coli table is drawn below with Gene 3’s row ringed: its level in glucose broth, lactose broth and the broth with no sugar.
Is Gene 3’s level about the same in all three broths?
- A. ✓ Yes
- B. NoGene 3 reads 33, 36 and 33 units.
Those levels differ by three units at most, so Gene 3’s level is about the same in every broth.
Why: Gene 3 reads 33, 36 and 33 units across the three broths.
The three levels differ by three units at most.
So Gene 3’s level is about the same in every broth.
The E. coli table is drawn below with Gene 3’s row ringed: its level in each of the three broths.
Which kind of gene is Gene 3?
- A. ✓ Constitutively expressed
- B. InducibleAn inducible gene’s level is high in one condition only.
Gene 3 reads 33, 36 and 33 units: its levels differ by a few units only.
Why: Gene 3 reads 33, 36 and 33 units.
The levels differ by a few units only.
So Gene 3 is at about the same level in every broth, low as that level is.
A gene at about the same level in every condition is constitutively expressed.
The E. coli table is drawn below with Gene 4’s row ringed: its level in each of the three broths.
Which kind of gene is Gene 4?
- A. Constitutively expressedGene 4 reads 145, 14 and 14 units.
Its level in glucose broth is far above its level in the other two broths. - B. ✓ Inducible
Why: Gene 4 reads 145, 14 and 14 units.
Its level in glucose broth is far above its level in the other two broths.
So Gene 4 is transcribed only while glucose is present.
A gene transcribed only when a particular signal is present is inducible.
The E. coli table is drawn below with Gene 4’s row ringed: its level in each of the three broths.
Which broth holds Gene 4’s signal?
- A. ✓ Glucose broth
- B. Lactose brothIn lactose broth Gene 4 reads 14 units, its low level.
An inducible gene’s signal is present in the condition where its level is high: glucose broth. - C. The broth with no sugarWith no sugar Gene 4 reads 14 units, its low level.
An inducible gene’s signal is present in the condition where its level is high: glucose broth.
Why: Gene 4 reads 145 units in glucose broth and 14 units in the other two broths.
An inducible gene is transcribed only while its signal is present.
Gene 4 is high in glucose broth only.
So glucose is its signal.
Now suppose you sort six genes from other cells, one table each.
Suppose a bacterium’s gene is measured in a broth without nickel and in the same broth with nickel added. Its levels are drawn below.
Which kind of gene is it?
- A. ✓ Constitutively expressed
- B. InducibleThe gene reads 46 and 47 units across the conditions.
The levels differ by one unit only, so the gene is at about the same level in every condition.
Why: The gene reads 46 and 47 units.
The levels differ by a few units only.
So the gene is at about the same level in every condition.
A gene at about the same level in every condition is constitutively expressed.
Suppose a bacterium’s gene is measured in glucose broth, in galactose broth and in a broth with no sugar. Its levels are drawn below.
Which kind of gene is it?
- A. Constitutively expressedThe gene reads 190 units in galactose broth and 17 units in the other conditions.
Its level is high in one condition only. - B. ✓ Inducible
Why: The gene reads 17, 190 and 17 units.
Its level in galactose broth is far above its level in the other conditions.
So the gene is transcribed only while that condition is present.
A gene transcribed only when a particular signal is present is inducible.
Suppose a bacterium’s gene is measured in a broth without cobalt and in the same broth with cobalt added. Its levels are drawn below.
Which kind of gene is it?
- A. Constitutively expressedThe gene reads 220 units with cobalt and 21 units in the other condition.
Its level is high in one condition only. - B. ✓ Inducible
Why: The gene reads 21 and 220 units.
Its level with cobalt is far above its level in the other condition.
So the gene is transcribed only while that condition is present.
A gene transcribed only when a particular signal is present is inducible.
Suppose a gene in a plant’s leaf cells is measured in the dark, in dim light and in bright light. Its levels are drawn below.
Which kind of gene is it?
- A. ✓ Constitutively expressed
- B. InducibleThe gene reads 235, 240 and 230 units across the conditions.
The levels differ by ten units only, so the gene is at about the same level in every condition.
Why: The gene reads 235, 240 and 230 units.
The levels differ by a few units only.
So the gene is at about the same level in every condition.
A gene at about the same level in every condition is constitutively expressed.
Suppose a bacterium’s gene is measured in fresh water and in salty water. Its levels are drawn below.
Which kind of gene is it?
- A. ✓ Constitutively expressed
- B. InducibleThe gene reads 71 and 73 units across the conditions.
The levels differ by two units only, so the gene is at about the same level in every condition.
Why: The gene reads 71 and 73 units.
The levels differ by a few units only.
So the gene is at about the same level in every condition.
A gene at about the same level in every condition is constitutively expressed.
Suppose a gene in an animal’s cells is measured with no hormone present and with a hormone added. Its levels are drawn below.
Which kind of gene is it?
- A. Constitutively expressedThe gene reads 270 units with the hormone and 23 units in the other condition.
Its level is high in one condition only. - B. ✓ Inducible
Why: The gene reads 23 and 270 units.
Its level with the hormone is far above its level in the other condition.
So the gene is transcribed only while that condition is present.
A gene transcribed only when a particular signal is present is inducible.
Suppose a team measures three genes in an insect’s cells at three stages of its life: larva, pupa and adult. The table below shows each gene’s level at each stage.
Which gene is constitutively expressed?
- A. Gene 5Gene 5 reads 22, 165 and 27 units.
Its level in the pupa is far above its level at the other two stages. - B. Gene 6Gene 6 reads 205, 28 and 28 units.
Its level in the larva is far above its level at the other two stages. - C. ✓ Gene 7
Why: Gene 7 reads 52, 53 and 52 units.
The levels differ by one unit only.
So Gene 7 is at about the same level at every stage.
A gene at about the same level in every condition is constitutively expressed.
The table below sets out the whole comparison: a constitutively expressed gene beside an inducible gene.
Back to the table of four genes in E. coli, measured in glucose broth, lactose broth and a broth with no sugar.
Gene 1 sits at about 100 units in every broth, so it is constitutively expressed.
Gene 2 sits at 2, 200 and 2 units, so it is inducible, and lactose is its signal.
Gene 3 sits at about the same level in every broth, so it is constitutively expressed too.
Gene 4 sits at a high level in glucose broth only, so it is inducible, and glucose is its signal.
47Quick quiz: constitutively expressed, inducible gene mixed practice
Suppose a bacterium transcribes the gene for a protein that pumps a poison out of the cell only while that poison is in its broth.
Which kind of gene is it?
- A. Constitutively expressedA constitutively expressed gene is transcribed at about the same level in every condition.
This gene is transcribed only while the poison is present. - B. ✓ Inducible
Why: The gene is transcribed only while the poison is in the broth.
The poison is its signal.
A gene transcribed only when a particular signal is present is inducible.
What is a constitutively expressed gene?
- A. A gene transcribed at a high level, whatever the level of the other genesA constitutively expressed gene may sit at a high level or a low level.
What marks it is a level that is about the same in every condition. - B. ✓ A gene transcribed at about the same level in every condition
- C. A gene copied into both daughter cells when the cell dividesEvery gene is copied into both daughter cells at division.
A constitutively expressed gene is one transcribed at about the same level in every condition.
Why: A gene transcribed at about the same level in every condition is called constitutively expressed.
Its everyday name is a housekeeping gene.
What is an inducible gene?
- A. A gene transcribed at a low level in every conditionA gene at a low level in every condition is constitutively expressed.
An inducible gene sits at a high level only while its signal is present. - B. A gene whose base sequence changes when a signal arrivesA signal changes how often a gene is transcribed, never its base sequence.
An inducible gene is transcribed only when a particular signal is present. - C. ✓ A gene transcribed only when a particular signal is present
Why: A gene transcribed only when a particular signal is present is called an inducible gene.
The signal induces, or brings on, its transcription.
A cell transcribes some of its genes at a steady level whatever its conditions.
(a) State what it means for a gene to be constitutively expressed. (1 pt)
- Award 1 point for: transcribed (expressed) at about the same level in every condition (steadily, whatever the conditions). ‘Housekeeping gene’ completes it, but its absence does not lose the point.
A cell transcribes some of its genes only at certain times.
(a) State what an inducible gene is. (1 pt)
- Award 1 point for: transcribed (switched on) only when a particular signal (molecule, condition) is present. Naming an example signal, such as lactose, completes it, but its absence does not lose the point.
Glossary
- constitutively expressed
- Transcribed at about the same level in every condition, such as the genes for ribosomal proteins and for the enzymes of glycolysis. The everyday name for such a gene is a housekeeping gene.
- inducible gene
- A gene transcribed only when a particular signal is present, such as the lactose genes, whose signal is lactose.
APBIO-U06-L25 Switching on the lactose genes
Zoom in on the chromosome of E. coli, at its three genes for lactose. The three genes sit side by side, and one promoter sits in front of all three.
Between the promoter and the genes is a short stretch of DNA with a protein clamped onto it. RNA polymerase is making no mRNA from the three genes. What is the clamped protein doing, and what could move it?
Unit 6 · Gene Expression and Regulation
1Three genes, one switch
RNA polymerase is about to copy a gene into RNA.
Where on the DNA does RNA polymerase bind first?
- A. The first base of the geneRNA polymerase reaches the first base of the gene only after it has bound.
It binds the promoter, the short sequence just before the gene. - B. ✓ The promoter
- C. The end of the geneThe end of the gene is where copying stops.
RNA polymerase binds the promoter, the short sequence just before the gene.
Why: The promoter is the short DNA sequence just before a gene.
RNA polymerase binds the promoter, then begins copying at the gene.
E. coli carries its genes on one molecule of DNA.
Which shape is that molecule?
- A. A line with two free endsA line with two free ends is a eukaryote’s chromosome.
A bacterium’s chromosome is one closed circle. - B. ✓ A closed circle
Why: A bacterium such as E. coli carries one circular chromosome.
The molecule is one closed circle with no free end.
Why do three genes switch on together, and only when lactose is there?
The three lactose genes and their one promoter form one unit.
One mRNA carries all three genes, so the cell switches them on and off as a group.
The clamped protein grips the short stretch of DNA between the promoter and the genes.
The protein sits in RNA polymerase’s path, so the polymerase cannot move into the genes.
Lactose binds the protein and changes its shape, so the protein no longer fits the DNA and lets go. Then RNA polymerase transcribes the genes.
When the lactose is used up, the protein fits the DNA again and the genes fall silent.
Video: Watch: Three genes, one switch
RNA polymerase binds the one promoter and moves through the first gene, the second and the third. One long mRNA grows behind it. Ribosomes build all three enzymes from that one mRNA.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L25a.mp4
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Look at the three lactose genes on the E. coli chromosome. They sit side by side, and one promoter sits in front of all three.
RNA polymerase binds that one promoter. From there it moves through the first gene, then the second, then the third.
RNA polymerase builds one long mRNA as it moves. That one mRNA carries all three genes.
Ribosomes read that one mRNA and build all three enzymes from it: the enzymes that take up and break down lactose.
So the cell cannot transcribe one of the three genes on its own. The cell switches all three on together, and all three off together.
A group of genes with related jobs that sits under one promoter and is transcribed as one mRNA is called an , because its genes work as one unit.
Switching a group of genes on or off together, with one control, is called , because the genes are regulated in step with each other.
The operon for lactose is called the lac operon. Lac is short for lactose.
The lac operon holds three genes, one promoter and the short stretch of DNA between them. There is no single lactose gene.
What you are expected to know Describe an operon: a group of genes with related jobs under one promoter, transcribed as one mRNA, so that the whole group is switched on or off together.
RNA polymerase transcribes the lac operon.
How many mRNAs carry the operon’s genes?
- A. ✓ One
- B. ThreeRNA polymerase moves from the one promoter through all the genes without stopping.
It builds one mRNA that carries every gene of the operon.
Why: The operon has one promoter.
RNA polymerase binds it once and moves through every gene in turn.
So it builds one mRNA that carries all the genes.
Suppose a bacterium carries four genes for breaking down xylose, a sugar. The four genes sit side by side under one promoter.
(a) Explain why the cell switches all four genes on together. (2 pt)
Frame The cell switches all four genes on together because …
RNA polymerase binds that one promoter.
From there it moves through all four genes in turn.
So it builds one mRNA that carries all four genes.
Ribosomes build all four enzymes from that one mRNA.
- Award 1 point for: the four genes share one promoter, so RNA polymerase binds once and transcribes all four.
- Award 1 point for: the four genes are carried on one mRNA (so all four enzymes are made from it together).
A student says: “The lac operon is really one long gene.”
Is the student correct?
- A. ✓ No: the lac operon holds three genes, transcribed together
- B. Yes: the three enzymes are made from one stretch of DNA, so it is one geneOne mRNA can carry several genes.
The lac operon’s one mRNA carries three genes, and ribosomes build three enzymes from it.
Why: The lac operon holds three genes.
RNA polymerase transcribes the three genes onto one mRNA.
One mRNA carrying three genes is still three genes.
25Quick quiz: operon, coordinate regulation mixed practice
Suppose a bacterium’s three genes for taking up melibiose, a sugar, sit side by side under one promoter and are transcribed as one mRNA.
Do these three genes form an operon?
- A. NoGenes with related jobs under one promoter, transcribed as one mRNA, form an operon.
These three genes fit that description. - B. ✓ Yes
Why: The three genes have related jobs.
They sit under one promoter.
RNA polymerase transcribes them as one mRNA.
So the three genes form an operon.
What is an operon?
- A. A single gene that is transcribed many timesAn operon is a group of genes, not one gene.
Its genes share one promoter and are transcribed as one mRNA. - B. ✓ A group of genes under one promoter, transcribed as one mRNA
- C. A group of genes that each have their own promoterGenes with their own promoters are switched on and off one by one.
An operon’s genes share one promoter and one mRNA.
Why: A group of genes with related jobs that sits under one promoter and is transcribed as one mRNA is called an operon.
What is coordinate regulation?
- A. Switching one gene on at a fixed time each dayCoordinate regulation is about a group of genes, not the time of day.
One control switches the whole group on or off together. - B. Transcribing a gene at a steady rate whatever the conditionsA steady rate whatever the conditions is no switching at all.
Coordinate regulation switches a group of genes on or off together. - C. ✓ Switching a group of genes on or off together, with one control
Why: Switching a group of genes on or off together, with one control, is called coordinate regulation.
A bacterium switches several genes on and off as a group.
(a) State what an operon is. (1 pt)
- Award 1 point for: a group of genes under one promoter, transcribed as one mRNA. ‘With related jobs’ completes it, but its absence does not lose the point.
(b) State what coordinate regulation is. (1 pt)
- Award 1 point for: a group of genes switched on or off together (by one control).
30No lactose: the block
A protein sits on a short stretch of DNA next to a gene’s promoter. While the protein sits there, RNA polymerase transcribes the gene less often.
What is that stretch of DNA called?
- A. The promoterRNA polymerase binds the promoter.
The stretch a regulatory protein binds beside the gene is a regulatory sequence. - B. An mRNAAn mRNA is RNA copied from the gene.
The stretch the protein sits on is DNA: a regulatory sequence. - C. ✓ A regulatory sequence
Why: A protein that lowers how often a gene is transcribed binds a stretch of DNA beside the gene.
That stretch is a regulatory sequence.
Go back to the lac operon. Suppose there is no lactose in the cell.
Video: Watch: No lactose: the block
With no lactose in the cell, the repressor sits on the operator. RNA polymerase comes to the promoter and finds the repressor covering the DNA it needs. RNA polymerase does not start, and no mRNA forms.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L25b.mp4
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The clamped protein has a gene of its own, with its own promoter, right beside the operon. The cell transcribes that gene all the time, lactose or no lactose.
So the cell always holds some of the protein.
With no lactose in the cell, the protein binds the short stretch of DNA between the promoter and the genes.
The short stretch of DNA beside the promoter that this protein binds is called the , because the genes’ switch is operated from there.
A protein that binds the operator and stops transcription of the genes is called a , because it represses transcription: it holds transcription down.
The operator is a regulatory sequence of the lac operon, and the repressor is its regulatory protein: the two parts of a switch, with their own names here.
A repressor protein bound to the operator sits in RNA polymerase’s path on the DNA, so the polymerase cannot move into the genes and they are not transcribed.
We say the repressor blocks RNA polymerase’s path. In reality the bound repressor covers DNA that RNA polymerase needs to start, so the polymerase does not start on the genes.
The repressor binds the operator, which is DNA. It does not bind the promoter, and it does not bind RNA polymerase.
So with no lactose, RNA polymerase makes no mRNA from the three genes, and the cell makes almost none of the lactose enzymes.
How little is almost none? The cell makes about a thousandth of the amount it makes when lactose is present.
What you are expected to know Describe the lac operon switched off: with no lactose, the repressor is bound to the operator beside the promoter, so RNA polymerase cannot move into the genes and they are not transcribed.
An E. coli cell has no lactose in it.
Where does the repressor sit?
- A. ✓ On the operator
- B. On the promoterThe promoter is where RNA polymerase binds.
The repressor binds the operator, the short stretch of DNA beside the promoter. - C. On the genesThe genes are what RNA polymerase would transcribe.
The repressor binds the operator, the short stretch of DNA before the genes.
Why: With no lactose, the repressor keeps its shape that fits the operator.
So the repressor sits on the operator, between the promoter and the genes.
Suppose a bacterium carries an operon for breaking down rhamnose, a sugar. The cell has no rhamnose in it, and a repressor is bound to the operon’s operator.
(a) Explain why the operon’s genes stay silent. (2 pt)
Frame The operon’s genes stay silent because …
The operator lies between the promoter and the genes.
The bound repressor sits in RNA polymerase’s path on the DNA.
So RNA polymerase cannot move into the genes.
RNA polymerase makes no mRNA from the genes.
With no mRNA, ribosomes build none of the enzymes.
- Award 1 point for: the repressor bound to the operator sits in RNA polymerase’s path, so the polymerase cannot move into (transcribe) the genes.
- Award 1 point for: no mRNA is made from the genes, so ribosomes build none of the enzymes.
A student says: “The repressor switches the lactose genes off by binding RNA polymerase and holding it still.”
Is the student correct?
- A. ✓ No: the repressor binds DNA, not RNA polymerase
- B. Yes: the repressor holds RNA polymerase stillThe repressor binds DNA, not the polymerase.
Bound to the operator, it sits in the polymerase’s path and blocks it.
Why: The repressor binds the operator, which is DNA.
It does not bind RNA polymerase.
Sitting on the operator, it blocks the polymerase’s path into the genes.
Suppose an E. coli cell has no lactose in it. A student says: “The three lactose genes are still in this cell’s DNA. The repressor is only stopping RNA polymerase from transcribing them.”
Is the student correct?
- A. No: with no lactose, the cell has no copy of the lactose genes to transcribeEvery E. coli cell carries the whole genome, the lactose genes with it.
With no lactose, the repressor sits on the operator and the genes are not transcribed. - B. ✓ Yes: the genes are present in the DNA, and the repressor on the operator stops their transcription
Why: The lactose genes are part of the cell’s genome, so they are in the cell’s DNA whatever the broth.
With no lactose, the repressor is bound to the operator.
So RNA polymerase cannot move into the genes, and they are not transcribed.
Suppose a change in the operator’s DNA sequence means the repressor’s shape no longer fits it. The cell has no lactose.
Are the three lactose genes transcribed?
- A. NoThe repressor can only block the polymerase from the operator.
With an operator it cannot bind, RNA polymerase moves into the genes. - B. ✓ Yes
Why: The repressor no longer fits the changed operator, so it cannot bind there.
Nothing sits in RNA polymerase’s path.
So the polymerase moves into the genes and transcribes them, lactose or no lactose.
51Quick quiz: operator, repressor mixed practice
The lac operon has an operator and a repressor.
Which of the two is a stretch of DNA?
- A. ✓ The operator
- B. The repressorThe repressor is a protein.
The operator is the stretch of DNA the repressor binds.
Why: The operator is a short stretch of DNA beside the promoter.
The repressor is the protein that binds it.
What is the operator?
- A. A DNA sequence inside the first lactose geneThe operator lies before the genes, not inside one.
It is the short stretch of DNA beside the promoter that the repressor binds. - B. The DNA sequence where RNA polymerase first bindsThe DNA sequence where RNA polymerase first binds is the promoter.
The operator is the stretch beside it that the repressor binds. - C. ✓ The short stretch of DNA beside the promoter that the repressor binds
Why: The short stretch of DNA beside the promoter that the repressor binds is called the operator.
What is a repressor?
- A. ✓ A protein that binds the operator and stops transcription of the genes
- B. A protein that binds the promoter and helps RNA polymerase startA repressor stops transcription; it does not help it start.
It binds the operator, not the promoter. - C. A sugar that binds the operator and switches the genes offA repressor is a protein, not a sugar.
It binds the operator and stops transcription.
Why: A protein that binds the operator and stops transcription of the genes is called a repressor.
The lac operon is switched off when the cell has no lactose.
(a) State what the operator is. (1 pt)
- Award 1 point for: a stretch of DNA (beside the promoter, before the genes) that the repressor binds.
(b) State what a repressor is. (1 pt)
- Award 1 point for: a protein that binds the operator and stops (blocks) transcription.
56Lactose arrives
A signal molecule binds its receptor, a protein, in a cell’s membrane.
What happens to the receptor’s shape?
- A. The receptor’s shape stays the sameA receptor holds its signal molecule and changes shape.
That shape change is how the receptor passes the signal on. - B. ✓ The receptor’s shape changes
Why: A receptor holds its signal molecule.
The bound signal molecule changes the receptor’s shape.
A small molecule binds an enzyme at a site away from the active site, and the active site changes shape.
What is the site the small molecule binds called?
- A. A second active siteThe site the small molecule binds is a site other than the active site.
A site on an enzyme other than the active site is an allosteric site. - B. ✓ An allosteric site
Why: The small molecule binds a site other than the active site.
A site on an enzyme other than the active site is called an allosteric site.
Binding there changes the active site’s shape.
Now suppose lactose enters the cell.
Video: Watch: Lactose arrives
Lactose docks onto the repressor. The repressor’s shape shifts, and it no longer fits the operator. It lifts off the DNA, and RNA polymerase moves through all three genes, building one mRNA behind it.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L25c.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L25c.mp4
Lactose enters the cell and binds the repressor.
The bound lactose changes the repressor’s shape.
The repressor with its new shape no longer fits the operator. So the repressor lets go of the DNA.
Nothing now sits in RNA polymerase’s path. The polymerase moves from the promoter through the three genes and builds the one mRNA.
Ribosomes build the three lactose enzymes from that mRNA. The enzymes take lactose into the cell and break it down.
The switch is fast. The lac mRNA appears within a few minutes of lactose arriving, and the enzymes within about ten minutes.
The enzymes break the lactose down until it is used up. With no lactose left in the cell, no lactose is bound to the repressor.
The repressor’s shape returns to the one that fits the operator. The repressor binds the operator again, and transcription stops.
A small molecule that binds a repressor and so switches genes on is called an , because it induces transcription: it brings transcription about. Lactose is the lac operon’s inducer.
The inducer binds the repressor and changes its shape so it no longer fits the operator, the repressor leaves the DNA, and RNA polymerase transcribes the genes.
A switch that is off until a signal turns it on is called an . The lac operon is an inducible system, and lactose is its signal.
The lac operon’s three genes are inducible genes: transcribed only while lactose is present.
Lactose binds the repressor at one site, and the shape of the part that grips the operator changes.
An allosteric site is at work here: a small molecule binds one site on a protein and changes the shape of another part.
What you are expected to know Explain how lactose switches the lac operon on, and how the operon switches off again once the lactose is used up.
The lac operon is drawn three times below, in a scrambled order, with its words removed and a letter beside each drawing. In the drawings, the small circle is lactose. A full rectangle is the repressor in the shape that fits the operator. A block with its lower corners cut away is the repressor in its changed shape. An oval above the promoter is RNA polymerase off the DNA.
In which order do the three drawings happen, starting with no lactose in the cell?
- A. ✓ K, L, J
- B. K, J, LThe repressor leaves the operator only after lactose has bound it.
The drawing with lactose on the repressor comes before the one with the repressor off the DNA. - C. L, K, JWith no lactose in the cell, the repressor sits on the operator with nothing bound to it.
That drawing comes first; lactose binds next; the repressor leaves last.
Why: With no lactose, the repressor sits on the operator with nothing bound to it.
Then lactose binds the repressor and changes its shape.
Then the repressor leaves the DNA, and RNA polymerase moves into the genes.
So the order is K, L, J.
Suppose a bacterium carries an operon for breaking down sorbitol, a sugar. With no sorbitol in the cell, a repressor sits on the operon’s operator. Now sorbitol enters the cell.
(a) Explain how sorbitol switches the operon on. (3 pt)
Frame Sorbitol switches the operon on by …
The bound sorbitol changes the repressor’s shape.
The repressor with its new shape no longer fits the operator.
So the repressor leaves the DNA.
Nothing now sits in RNA polymerase’s path.
So RNA polymerase moves into the genes and transcribes them.
- Award 1 point for: sorbitol binds the repressor (not the DNA).
- Award 1 point for: the bound sorbitol changes the repressor’s shape, so the repressor no longer fits the operator and leaves the DNA.
- Award 1 point for: with the repressor gone, RNA polymerase moves into the genes and transcribes them.
A student says: “Lactose switches the lactose genes on by binding the operator and pushing the repressor off it.”
Is the student correct?
- A. ✓ No: lactose binds the repressor, not the DNA
- B. Yes: lactose binds the operator and clears the way for RNA polymeraseLactose binds the repressor, a protein.
The changed repressor then leaves the operator on its own; lactose touches no DNA.
Why: Lactose binds the repressor.
The bound lactose changes the repressor’s shape.
The repressor no longer fits the operator and leaves it.
Lactose itself binds no DNA.
A student says: “Once the cell has used up the lactose, the repressor fits the operator again, and transcription of the three genes stops.”
Is the student correct?
- A. No: once switched on, the operon stays on for the rest of the cell’s lifeThe switch is not one-way.
With the lactose gone, the repressor returns to its fitting shape and binds the operator again. - B. ✓ Yes: with no lactose bound to it, the repressor takes back the shape that fits the operator
Why: The enzymes break the lactose down until it is used up.
With no lactose bound, the repressor takes back the shape that fits the operator.
The repressor binds the operator again, and transcription stops.
Suppose a change in the repressor’s gene gives a repressor that lactose no longer binds. The repressor still fits the operator. Lactose enters the cell.
Are the three lactose genes transcribed?
- A. YesLactose can only lift the repressor by binding it and changing its shape.
A repressor that lactose does not bind keeps its shape and stays on the operator. - B. ✓ No
Why: Lactose does not bind this repressor.
So the repressor’s shape stays the one that fits the operator.
The repressor stays on the operator and blocks RNA polymerase, lactose or no lactose.
Go back to the operon with the protein clamped beside its promoter.
With no lactose, the repressor sits on the operator, and the polymerase cannot pass.
Lactose arrives and binds the repressor. The repressor changes shape and lets go.
RNA polymerase transcribes the three genes as one mRNA.
When the lactose is gone, the repressor returns to the operator.
We made one simplification here. We say lactose binds the repressor.
In reality the cell first turns a little lactose into a close relative, allolactose, and that is what binds. The exam accepts ‘lactose’.
88Quick quiz: inducer, inducible system mixed practice
A bacterium’s genes for breaking down a sugar are off until that sugar arrives in the cell. Then the cell transcribes them.
Is this an inducible system?
- A. ✓ Yes
- B. NoA switch that is off until a signal turns it on is an inducible system.
These genes are off until the sugar arrives, so the switch is inducible.
Why: The genes are off until a signal arrives.
The signal, the sugar, turns them on.
A switch that is off until a signal turns it on is an inducible system.
What is an inducer?
- A. A small molecule that binds the operator and lifts the repressor off itAn inducer binds the repressor, a protein, not the operator.
The changed repressor then leaves the operator on its own. - B. A small molecule that binds RNA polymerase and speeds it upAn inducer binds the repressor, not RNA polymerase.
It changes the repressor’s shape so the repressor leaves the DNA. - C. ✓ A small molecule that binds a repressor and so switches genes on
Why: A small molecule that binds a repressor and so switches genes on is called an inducer.
What is an inducible system?
- A. A switch that is on until a signal turns it offA switch that is on until a signal turns it off is the other way round.
An inducible system is off until a signal turns it on. - B. ✓ A switch that is off until a signal turns it on
- C. A gene that is transcribed at the same rate all the timeA gene transcribed at the same rate all the time has no switch.
An inducible system is off until a signal turns it on.
Why: A switch that is off until a signal turns it on is called an inducible system.
The lac operon is switched on when lactose enters the cell.
(a) State what an inducer is. (1 pt)
- Award 1 point for: a small molecule that binds the repressor and switches the genes on (lifts the block).
(b) State what an inducible system is. (1 pt)
- Award 1 point for: off until a signal turns it on.
93Mixed practice mixed practice
Suppose a change in the repressor’s gene means an E. coli cell makes no repressor at all. The cell has no lactose in it.
Are the lactose genes transcribed?
- A. NoOnly a repressor on the operator blocks RNA polymerase.
With no repressor, nothing sits in the polymerase’s path, lactose or no lactose. - B. ✓ Yes
Why: The cell has no repressor.
So nothing binds the operator.
Nothing sits in RNA polymerase’s path, so it transcribes the genes, lactose or no lactose.
A bacterium’s operon holds five genes.
How many promoters does the operon have?
- A. ✓ One
- B. FiveAn operon’s genes share one promoter.
RNA polymerase binds it once and transcribes all five genes as one mRNA.
Why: An operon is a group of genes under one promoter.
Five genes in one operon share one promoter.
Suppose a molecule shaped like lactose binds the lac repressor, but the lactose enzymes cannot break it down. The molecule enters an E. coli cell.
Are the lactose genes switched on?
- A. ✓ Yes
- B. NoThe switch turns on the moment the repressor changes shape.
A molecule that binds the repressor changes its shape whether or not the enzymes can break the molecule down.
Why: The molecule binds the repressor.
The bound molecule changes the repressor’s shape, so the repressor leaves the operator.
RNA polymerase transcribes the genes.
Whether the enzymes can break the molecule down changes none of this.
The cell makes the lac repressor from a gene of its own, right beside the operon.
Is that gene transcribed when the cell has no lactose?
- A. NoThe repressor must already be in the cell when there is no lactose, ready to sit on the operator.
The cell transcribes the repressor’s gene all the time. - B. ✓ Yes
Why: The repressor blocks the operon when the cell has no lactose.
To do that, the repressor must already be there.
So the cell transcribes the repressor’s gene all the time, lactose or no lactose.
An E. coli cell has lactose in it.
Where is the repressor?
- A. On the operatorWith lactose bound, the repressor no longer fits the operator.
It has left the DNA. - B. On the promoterThe repressor never binds the promoter.
With lactose bound, it fits no DNA and floats free. - C. ✓ Off the DNA, with lactose bound to it
Why: Lactose binds the repressor and changes its shape.
The changed repressor fits no DNA.
So the repressor is off the DNA, with lactose bound to it.
RNA polymerase has just transcribed the lac operon into one mRNA.
How many kinds of enzyme do ribosomes build from that one mRNA?
- A. OneThe one mRNA carries all three genes of the operon.
Ribosomes build a different enzyme from each gene. - B. ✓ Three
Why: The lac operon holds three genes.
Its one mRNA carries all three.
Ribosomes build three kinds of enzyme from it, one from each gene.
Suppose a change in the lac promoter’s DNA means RNA polymerase can no longer bind it. The cell has lactose in it.
Are the three lactose genes transcribed?
- A. YesLactose lifts the repressor, but transcription still starts at the promoter.
With no promoter to bind, RNA polymerase never starts. - B. ✓ No
Why: Lactose lifts the repressor off the operator.
But RNA polymerase must first bind the promoter.
It cannot, so it never moves into the genes, and all three stay silent.
Suppose a bacterium carries an operon for breaking down fucose, a sugar. The operon has one promoter, an operator and a repressor, and it works like the lac operon. Fucose enters the cell.
(a) Predict whether the cell transcribes the fucose genes once fucose is inside it. (1 pt)
- Award 1 point for: the genes are transcribed (switched on) once fucose is inside.
(b) Explain what happens to transcription of the fucose genes once the cell has used up all the fucose. (2 pt)
Frame Once the fucose is used up, …
The repressor returns to the shape that fits the operator.
So the repressor binds the operator again.
The bound repressor covers DNA that RNA polymerase needs to start.
So RNA polymerase makes no more mRNA from the fucose genes, and transcription stops.
- Award 1 point for: with no fucose bound, the repressor returns to the shape that fits the operator and binds it again.
- Award 1 point for: the bound repressor blocks RNA polymerase, so the fucose genes are no longer transcribed.
Glossary
- operon
- A group of genes with related jobs that sits under one promoter and is transcribed as one mRNA, so the whole group is switched on or off together. E. coli’s lac operon holds the three genes for taking up and breaking down lactose.
- coordinate regulation
- Switching a group of genes on or off together, with one control. An operon’s genes are regulated this way: one promoter, one mRNA, all the genes at once.
- operator
- The short stretch of DNA beside an operon’s promoter that the repressor binds. With the repressor on it, RNA polymerase cannot move into the genes.
- repressor
- A protein that binds the operator and stops transcription of the operon’s genes. The cell makes the lac repressor all the time from a gene of its own, and the repressor binds the operator when the cell has no lactose.
- inducer
- A small molecule that binds a repressor and so switches genes on: the bound inducer changes the repressor’s shape, the repressor leaves the operator, and RNA polymerase transcribes the genes. Lactose is the lac operon’s inducer.
- inducible system
- A switch on transcription that is off until a signal turns it on. The lac operon is an inducible system: its genes are off until lactose arrives.
APBIO-U06-L26 Switching off the tryptophan genes
Suppose a second operon sits in the same E. coli cell as the lac operon. Its five genes make the enzymes that build the amino acid tryptophan.
A technician adds tryptophan to the broth. Within minutes the cell stops transcribing those five genes. The switch has the same parts as the lactose switch: a promoter, an operator, a repressor. So what is different?
Unit 6 · Gene Expression and Regulation
1Tryptophan switches off its own genes
In the lac operon, lactose has just bound the repressor.
What does the repressor do next?
- A. ✓ Leaves the operator
- B. Grips the DNA more tightlyLactose changes the repressor’s shape so it no longer fits the operator.
The repressor lets go of the DNA instead of gripping it. - C. Binds RNA polymeraseThe lactose repressor binds DNA at the operator, never RNA polymerase.
With lactose bound, the repressor leaves the operator.
Why: Lactose binds the repressor.
The repressor’s shape changes, and it no longer fits the operator.
So the repressor leaves the operator.
A bacterium builds an amino acid in three enzyme steps. When the amino acid piles up, it binds the first enzyme and slows it.
Which kind of feedback is this?
- A. Positive feedbackIn positive feedback the response increases the change that triggered it.
Here the response reduces the pile-up. - B. ✓ Negative feedback
Why: The rising amino acid is the change.
Slowing the first enzyme makes less amino acid.
So the response reduces the change that triggered it.
A response that reduces the change that triggered it is negative feedback.
Why does one small molecule switch genes on, and another switch them off?
The parts are the same, and the logic is reversed.
On its own, the tryptophan operon’s repressor cannot bind the operator. So the five genes are transcribed while tryptophan is scarce.
When tryptophan is plentiful, tryptophan binds the repressor and changes its shape.
The repressor now fits the operator. The repressor binds the operator, and transcription of the five genes stops.
The product switches off the genes that make it. This switch-off is negative feedback.
The lactose switch is off until lactose turns it on. The tryptophan switch is on until tryptophan turns it off.
Video: Watch: Tryptophan switches off its own genes
The repressor drifts past the operator without binding. RNA polymerase moves through the five genes. Tryptophan arrives and binds the repressor. The repressor’s shape changes, and the repressor snaps onto the operator. RNA polymerase is blocked.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L26a.mp4
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Suppose E. coli grows in a broth with no tryptophan in it. The cell needs tryptophan to build its proteins.
So the cell must build tryptophan itself.
Five genes sit side by side on the chromosome. The five genes code for the enzymes that build tryptophan.
The operon for tryptophan is called the trp operon. Trp is short for tryptophan.
One promoter sits in front of all five genes. An operator sits between the promoter and the genes.
This operon has a repressor of its own, made all the time from its own gene.
On its own, the repressor has a shape that does not fit the operator. So the repressor drifts past the operator and never binds it.
RNA polymerase binds the promoter and moves through all five genes. One mRNA carries all five, and the cell builds the enzymes.
Now suppose the technician adds tryptophan to the broth. Tryptophan enters the cell.
Tryptophan binds the repressor. Bound tryptophan changes the repressor’s shape.
The repressor now fits the operator. So the repressor binds the operator.
A repressor protein bound to the operator sits in RNA polymerase’s path on the DNA, so the polymerase cannot move into the genes and they are not transcribed.
A small molecule that binds a repressor and makes it fit the operator is called a , because it works with the repressor to switch the genes off. Here, tryptophan is the corepressor.
The corepressor binds the repressor and changes its shape so it now fits the operator, the repressor binds, and transcription of the genes stops.
Within minutes of tryptophan arriving, the five genes fall silent.
The cell now has plenty of tryptophan. So the cell stops building more.
The product switches off the genes that make it.
In negative feedback, the response reduces the change that triggered it. Rising tryptophan is the change, and silencing the tryptophan genes reduces it.
In the lac operon, lactose binds the repressor, and the repressor leaves the operator. The genes switch on.
In the trp operon, tryptophan binds the repressor, and the repressor binds the operator. The genes switch off.
In both operons, a small molecule changes the repressor’s shape.
In the lac operon, the new shape no longer fits the operator. In the trp operon, the new shape fits the operator.
A switch that is on until a signal turns it off is called a , because the signal represses the genes. The trp operon is a repressible system, and the lac operon is an inducible system.
Take the tryptophan away, and the repressor loses its corepressor.
The repressor no longer fits the operator. The repressor leaves the operator, and the five genes are transcribed again.
What you are expected to know Explain how plentiful tryptophan switches off the trp operon, a repressible system.
A technician has just added tryptophan to a broth of E. coli. Three moments in the trp operon are drawn below in a mixed order, under the letters J, K and L. In the drawings, the small circle is tryptophan. An oval above the promoter is RNA polymerase off the DNA. A full rectangle is the repressor in the shape that fits the operator. A block with its lower corners cut away is the repressor in the shape that drifts past the operator.
In which order do the three moments happen?
- A. ✓ K, L, J
- B. K, J, LThe repressor binds the operator only after tryptophan has bound the repressor.
Tryptophan binds first. - C. J, L, KBefore tryptophan arrives, the repressor is free and the genes are transcribed.
The blocked operon is the last moment.
Why: With no tryptophan, the repressor is free and RNA polymerase moves through the genes.
Tryptophan binds the repressor.
The repressor then fits and binds the operator, and the polymerase is blocked.
Suppose a bacterium has an operon whose genes code for the enzymes that build a vitamin the cell needs. A technician adds that vitamin to the broth, and within minutes the operon’s genes fall silent.
(a) Explain why the cell gains by switching these genes off while the vitamin is plentiful. (2 pt)
Frame The cell gains because …
With the vitamin plentiful, the cell has no need to build more.
So the enzymes that build the vitamin are not needed.
Switching the genes off saves the energy and materials that building those enzymes would use.
- Award 1 point for: the cell already has enough of the vitamin, so the enzymes that build it are not needed.
- Award 1 point for: transcribing the genes and building the enzymes uses energy and materials, which the cell saves by switching the genes off.
A student says: “Tryptophan lifts the repressor off the operator, just like lactose does.”
Is the student correct?
- A. ✓ No: tryptophan gives the repressor the shape that fits the operator, so the repressor binds
- B. Yes: tryptophan and lactose both lift the repressor off the operatorLactose makes the repressor leave the operator.
Tryptophan does the opposite: it gives the repressor the shape that binds the operator.
Why: Tryptophan binds the repressor and changes its shape.
The new shape fits the operator.
So the repressor binds the operator instead of leaving it.
Suppose a bacterium’s operon codes for the enzymes that build a fat molecule for its membrane. The operon is a repressible system, and that fat molecule is its corepressor. The fat molecule becomes plentiful in the cell.
What does the operon’s repressor do?
- A. Binds the promoterA repressor binds the operator, never the promoter.
With its corepressor bound, this repressor fits and binds the operator. - B. Drifts free of the DNADrifting free of the DNA is what the lac repressor does when lactose binds it.
A corepressor makes its repressor bind the operator. - C. ✓ Binds the operator
Why: The fat molecule is the corepressor.
The fat molecule binds the repressor and changes its shape so the repressor fits the operator.
So the repressor binds the operator.
42Quick quiz: repressible system, corepressor mixed practice
Suppose a bacterium’s genes for building the amino acid arginine are transcribed while arginine is scarce. When arginine is plentiful, arginine binds the operon’s repressor and the genes fall silent.
Which molecule is the corepressor?
- A. The repressorThe repressor is the protein the corepressor binds.
Arginine is the small molecule that binds the repressor and makes it fit the operator. - B. ✓ Arginine
- C. RNA polymeraseRNA polymerase transcribes the genes and binds nothing that switches them off.
Arginine binds the repressor and makes it fit the operator.
Why: Arginine binds the repressor.
The repressor then fits and binds the operator, and the genes fall silent.
So arginine is the corepressor.
What is a corepressor?
- A. The protein that blocks RNA polymerase at the operator when no signal is presentA protein that binds the operator and blocks RNA polymerase is the repressor itself.
The corepressor is the small molecule that makes the repressor fit the operator. - B. The molecule that binds a repressor and makes it leave the operator, so the genes switch onA small molecule that binds a repressor and makes it leave the operator is an inducer.
A corepressor makes the repressor fit and bind the operator. - C. ✓ A small molecule that binds a repressor and gives it the shape that fits the operator
Why: A small molecule that binds a repressor and gives it the shape that fits the operator is called a corepressor.
What is a repressible system?
- A. ✓ A switch that is on until a signal turns it off
- B. A switch that is off until a signal turns it onA switch that is off until a signal turns it on is an inducible system.
A repressible system is on until a signal turns it off.
Why: A switch that is on until a signal turns it off is called a repressible system.
The trp operon is one.
The trp operon of E. coli is a repressible system.
(a) State what tryptophan does to the operon’s repressor. (1 pt)
- Award 1 point for: tryptophan binds the repressor and gives it the shape that binds the operator (makes it fit the operator).
(b) State what makes a system repressible rather than inducible. (1 pt)
- Award 1 point for: the genes are transcribed until a signal switches them off (the signal turns the system off, where an inducible system’s signal turns it on).
47Inducible or repressible?
The lac operon is off until lactose arrives and switches it on.
Which kind of system is the lac operon?
- A. A repressible systemA repressible system is on until a signal turns it off.
The lac operon is off until lactose turns it on. - B. ✓ An inducible system
Why: The lac operon is off until lactose turns it on.
A switch that is off until a signal turns it on is an inducible system.
How do you tell which kind of switch a described operon has?
An inducible system has a repressor, and a repressible system has a repressor too. In each system a small molecule binds the repressor.
So look past the repressor.
Ask one question: what does the signal do to the genes?
A signal that switches the genes on marks an inducible system. A signal that switches the genes off marks a repressible system.
The signal of an inducible system is usually a nutrient the genes break down. The signal of a repressible system is usually the product the genes build.
Video: Watch: Inducible or repressible?
Four operons appear one after another. In each, the signal arrives and the genes switch on or off. Under the written question, is this operon repressible?, a tick or a cross appears. Only the operons whose signal switches the genes off get the tick.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L26b.mp4
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One rule sorts every operon: an operon is inducible when its signal switches the genes on, and repressible when its signal switches the genes off.
For example, take the lac operon. Lactose arrives, and the genes switch on.
This operon is inducible, because its signal switches the genes on.
But take the trp operon. Tryptophan arrives, and the genes switch off.
This operon is repressible, because its signal switches the genes off.
Now suppose an operon’s genes build one of the building blocks of DNA. That building block arrives, and the genes switch off.
This operon is still repressible, because its signal switches the genes off.
But now suppose an operon’s genes break down a sugar from fruit. That sugar arrives, and the genes switch on.
This operon is inducible, because its signal switches the genes on.
The number of genes did not matter, and neither did the presence of a repressor. Only what the signal does to the genes sorted the four operons.
What you are expected to know Classify a described operon as inducible or repressible from what its signal does to the genes.
E. coli’s genes for breaking down the sugar arabinose are transcribed only when arabinose is present.
Is this operon an inducible system or a repressible system?
- A. ✓ An inducible system
- B. A repressible systemArabinose switches the genes on.
A signal that switches the genes on marks an inducible system.
Why: Arabinose is the signal, and the genes are transcribed only when arabinose is present.
The signal switches the genes on.
So the operon is an inducible system.
Suppose an operon’s genes break down a protein found in the broth. The genes are transcribed only while that protein is present.
Is this operon an inducible system or a repressible system?
- A. ✓ An inducible system
- B. A repressible systemThe protein in the broth switches the genes on.
A signal that switches the genes on marks an inducible system.
Why: The protein in the broth is the signal, and the genes are transcribed only while it is present.
The signal switches the genes on.
So the operon is an inducible system.
Suppose a bacterium’s genes for building the amino acid histidine fall silent when histidine is plentiful.
Is this operon an inducible system or a repressible system?
- A. An inducible systemHistidine switches the genes off.
A signal that switches the genes off marks a repressible system. - B. ✓ A repressible system
Why: Histidine is the signal, and the genes fall silent when histidine is plentiful.
The signal switches the genes off.
So the operon is a repressible system.
Suppose an operon’s genes build a pigment. When the pigment is plentiful, transcription of the genes stops.
Is this operon an inducible system or a repressible system?
- A. An inducible systemThe pigment switches the genes off.
A signal that switches the genes off marks a repressible system. - B. ✓ A repressible system
Why: The pigment is the signal, and transcription stops when the pigment is plentiful.
The signal switches the genes off.
So the operon is a repressible system.
Suppose an operon’s genes build a glue that fastens the cell to a surface. A signal molecule from neighboring cells switches the genes on.
Is this operon an inducible system or a repressible system?
- A. ✓ An inducible system
- B. A repressible systemThe genes build a product, yet their signal switches them on.
A signal that switches the genes on marks an inducible system, whatever the genes make.
Why: The signal molecule from neighboring cells switches the genes on.
A signal that switches the genes on marks an inducible system.
So the operon is an inducible system, though its genes build a product.
Suppose an operon’s genes are silent until a certain alcohol enters the cell. Then the genes are transcribed.
Is this operon an inducible system or a repressible system?
- A. ✓ An inducible system
- B. A repressible systemThe alcohol switches the genes on.
A signal that switches the genes on marks an inducible system.
Why: The alcohol is the signal, and the genes are transcribed once the alcohol enters.
The signal switches the genes on.
So the operon is an inducible system.
Suppose an operon’s genes build a building block of the cell wall. Plenty of that building block stops transcription of the genes.
Is this operon an inducible system or a repressible system?
- A. An inducible systemThe building block switches the genes off.
A signal that switches the genes off marks a repressible system. - B. ✓ A repressible system
Why: The building block is the signal, and plenty of it stops transcription.
The signal switches the genes off.
So the operon is a repressible system.
A student reads that the lac operon and the trp operon both use a repressor, and says: “Both operons have a repressor, but of the two only the trp operon is the repressible system, because only tryptophan switches its genes off.”
Is the student correct?
- A. No: having a repressor makes the lac operon repressible as well as the trp operonBoth operons have a repressor, but lactose switches the lac genes on.
Only a signal that switches the genes off marks a repressible system. - B. ✓ Yes: what the signal does to the genes sets the kind, and only tryptophan switches genes off
Why: Lactose switches the lac operon’s genes on, so the lac operon is an inducible system.
Tryptophan switches the trp operon’s genes off, so the trp operon is a repressible system.
The shared repressor does not sort them; what the signal does to the genes does.
This table compares the lac operon with the trp operon: the signal, what the signal makes the repressor do, the genes with and without the signal, what the genes are for, and the kind of system.
76On or off right now?
One of a bacterium’s genes is transcribed only while a particular sugar is in the broth.
What is such a gene called?
- A. A constitutively expressed geneA constitutively expressed gene is transcribed at about the same level in every condition.
A gene transcribed only when its signal is present is an inducible gene. - B. A regulatory sequenceA regulatory sequence is a stretch of DNA beside a gene, not a gene itself.
A gene transcribed only when its signal is present is an inducible gene. - C. ✓ An inducible gene
Why: The sugar is the gene’s signal, and the gene is transcribed only while the sugar is present.
A gene transcribed only when a particular signal is present is called an inducible gene.
Is a described operon being transcribed right now? Two decisions, taken in order, settle it.
- 1 Is the operon inducible or repressible?
- 2 Is the signal present or absent right now?
An inducible operon is on when its signal is present and off when its signal is absent. A repressible operon is off when its signal is present and on when its signal is absent.
The genes of an inducible operon are inducible genes: transcribed only while the signal is present. The genes of a repressible operon are transcribed until the signal arrives.
Video: Watch: On or off right now?
A described operon appears. First, what its signal does to the genes is written: on, so inducible. Then the broth is checked: the signal is present. The two answers pick out one row of the table, and the verdict appears: the genes are transcribed.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L26c.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L26c.mp4
Suppose E. coli grows in a broth that holds lactose.
1 Inducible or repressible? Lactose switches the lac genes on, so the lac operon is inducible.
2 Signal present or absent? The broth holds lactose, so the signal is present.
An inducible operon with its signal present is on. So right now the lac genes are transcribed.
What you are expected to know Predict whether the genes of a described operon are transcribed in a stated condition, by two decisions in order: inducible or repressible, then signal present or absent.
Suppose a bacterium’s operon codes for enzymes that break down a solvent that has leaked into its broth. The genes are transcribed only while the solvent is present.
Is this operon inducible or repressible?
- A. ✓ Inducible
- B. RepressibleThe solvent switches the genes on.
A signal that switches the genes on marks an inducible system.
Why: The solvent is the signal, and the genes are transcribed only while the solvent is present.
The signal switches the genes on.
So the operon is inducible.
A bacterium’s operon codes for enzymes that break down a solvent, and its genes are transcribed only while the solvent is present. Right now the broth holds no solvent.
Is the operon’s signal absent or present right now?
- A. ✓ Absent
- B. PresentThe signal is the solvent.
The broth holds no solvent, so the signal is absent.
Why: The solvent is the signal.
Right now the broth holds no solvent.
So the signal is absent.
A bacterium’s operon codes for enzymes that break down a solvent, and its genes are transcribed only while the solvent is present. Right now the broth holds no solvent.
Are the operon’s genes being transcribed right now?
- A. YesAn inducible operon is on only while its signal is present.
The solvent is absent, so the genes are off. - B. ✓ No
Why: The genes switch on only while the solvent is present, so the operon is inducible.
The broth holds no solvent, so the signal is absent.
An inducible operon with its signal absent is off.
So the genes are not being transcribed.
Suppose a bacterium’s operon codes for the enzymes that build an amino acid the cell needs. When that amino acid is plentiful, the genes fall silent. Right now the amino acid is plentiful in the cell.
Are the operon’s genes being transcribed right now?
- A. YesThe amino acid switches the genes off, so the operon is repressible.
The amino acid is present, so the genes are off. - B. ✓ No
Why: The amino acid switches the genes off, so the operon is repressible.
The amino acid is plentiful, so the signal is present.
A repressible operon with its signal present is off.
So the genes are not being transcribed.
Suppose a bacterium’s operon codes for enzymes that break down a sugar found in plant roots. The genes switch on only while that sugar is present. Right now the broth holds plenty of that sugar.
Are the operon’s genes being transcribed right now?
- A. ✓ Yes
- B. NoThe genes switch on only while the sugar is present, so the operon is inducible.
The sugar is present, so the genes are on.
Why: The sugar switches the genes on, so the operon is inducible.
The broth holds plenty of the sugar, so the signal is present.
An inducible operon with its signal present is on.
So the genes are being transcribed.
Suppose a bacterium’s operon builds a sugar that the cell stores for later. The operon is a repressible system, and the stored sugar is its corepressor. Right now the cell holds very little of that sugar. A student says: “Right now the operon’s genes are silent, because the cell is still waiting for the signal.”
Is the student correct?
- A. ✓ No: a repressible operon is on while its signal is absent, so its genes are transcribed
- B. Yes: every operon stays silent until its signal has arrived in the cellAn inducible operon waits for its signal; a repressible operon is on until its signal turns it off.
With the sugar scarce, the genes are transcribed.
Why: The operon is repressible.
The stored sugar, its signal, is scarce, so the signal is absent.
A repressible operon with its signal absent is on.
So the genes are being transcribed.
This table lists the four combinations of kind of system and signal, and whether the genes are transcribed in each.
Back to the trp operon and its five genes. Tryptophan binds the repressor and gives it the shape that fits the operator.
The repressor binds the operator, and transcription stops. The product switches off its own genes.
Take the tryptophan away, and the five genes are transcribed again.
98Mixed practice mixed practice
E. coli grows in a broth with plenty of tryptophan.
Where is the trp operon’s repressor?
- A. On the promoterA repressor binds the operator, never the promoter.
With tryptophan bound, the trp repressor sits on the operator. - B. ✓ On the operator
- C. Off the DNAOff the DNA is where the trp repressor sits when tryptophan is scarce.
With tryptophan bound, the repressor fits and binds the operator.
Why: Tryptophan is plentiful, so tryptophan binds the repressor.
The bound repressor fits the operator.
So the repressor sits on the operator.
Suppose an operon’s genes break down an amino acid from the broth for energy. The genes are transcribed only while that amino acid is present.
Is this operon an inducible system or a repressible system?
- A. ✓ An inducible system
- B. A repressible systemThe amino acid switches the genes on.
A signal that switches the genes on marks an inducible system, whatever the signal is made of.
Why: The amino acid is the signal, and the genes are transcribed only while it is present.
The signal switches the genes on.
So the operon is an inducible system.
Tryptophan is plentiful in an E. coli cell.
What gives the trp repressor the shape that fits the operator?
- A. RNA polymerase binding the promoterRNA polymerase binding the promoter changes nothing about the repressor.
Tryptophan binding the repressor changes the repressor’s shape. - B. Contact with the operatorThe repressor can bind the operator only once its shape fits.
Tryptophan binding gives it that shape. - C. ✓ Tryptophan binding the repressor
Why: Tryptophan binds the repressor.
Bound tryptophan changes the repressor’s shape.
The new shape fits the operator.
Suppose a bacterium’s operon is a repressible system whose corepressor is the amino acid its enzymes build. Right now that amino acid is scarce in the cell.
Are the operon’s genes being transcribed right now?
- A. ✓ Yes
- B. NoA repressible operon is on until its signal turns it off.
The amino acid is scarce, so the signal is absent and the genes are on.
Why: The operon is repressible.
The amino acid, its signal, is scarce, so the signal is absent.
A repressible operon with its signal absent is on.
So the genes are being transcribed.
In both the lac operon and the trp operon, a small molecule binds the repressor.
In which operon does the bound small molecule make the repressor leave the operator?
- A. The trp operonIn the trp operon, bound tryptophan makes the repressor fit and bind the operator.
In the lac operon, bound lactose makes the repressor leave. - B. ✓ The lac operon
Why: Lactose binds the lac repressor, and the repressor no longer fits the operator, so it leaves.
Tryptophan binds the trp repressor, and the repressor now fits the operator, so it binds.
E. coli has been growing with plenty of tryptophan. A technician moves the cells into a broth with no tryptophan.
What happens to the trp operon’s repressor?
- A. ✓ Loses its corepressor and leaves the DNA
- B. Stays where it is, with tryptophan still boundWith no tryptophan in the cell, tryptophan leaves the repressor.
Without its corepressor the repressor no longer fits the operator and lets go. - C. Binds RNA polymeraseA repressor binds DNA at the operator, never RNA polymerase.
Without tryptophan the repressor loses its fit and leaves the operator.
Why: With no tryptophan, the repressor loses its corepressor.
Without the corepressor the repressor no longer fits the operator.
So the repressor leaves the operator.
Tryptophan is plentiful, and the trp repressor sits on the operator.
What does RNA polymerase do?
- A. Moves through the five genes as usual and builds one mRNAThe repressor on the operator blocks RNA polymerase.
The polymerase transcribes none of the five genes. - B. Binds the operator in place of the repressor and moves into the genesOnly the repressor binds the operator.
RNA polymerase is blocked by the bound repressor and transcribes none of the genes. - C. ✓ Is blocked by the bound repressor and transcribes none of the five genes
Why: The repressor sits on the operator, on DNA that RNA polymerase needs to start.
So RNA polymerase cannot start on the genes.
The five genes are not transcribed.
Suppose a change in the gene for the trp repressor gives E. coli a repressor that tryptophan can no longer bind. The repressor keeps the shape that does not fit the operator. The cell grows in a broth with plenty of tryptophan.
(a) Predict whether the five tryptophan genes are transcribed. (1 pt)
- Award 1 point for: the genes are transcribed (the operon stays on).
(b) Explain your prediction. (2 pt)
Frame The repressor …
So the repressor never gains the shape that fits the operator, and it stays off the operator even though tryptophan is plentiful.
Nothing sits in RNA polymerase’s path.
So the polymerase moves through the five genes, and the five genes are transcribed.
- Award 1 point for: with tryptophan unable to bind, the repressor never takes the shape that fits the operator, so it does not bind the operator.
- Award 1 point for: with the operator free, RNA polymerase moves into the genes, so they are transcribed even with tryptophan plentiful.
Glossary
- corepressor
- A small molecule that binds a repressor and gives it the shape that fits the operator, so the repressor binds and the genes are switched off. In the trp operon, tryptophan is the corepressor.
- repressible system
- A switch that is on until a signal turns it off: the genes are transcribed until the signal binds the repressor as a corepressor and the repressor binds the operator. The trp operon is a repressible system.
APBIO-U06-L26B Two sugars at once
Suppose E. coli grows in a broth that holds both glucose and lactose. Lactose is present, so the repressor has left the operator.
Yet the cells make very little of the lactose enzymes. Move the same cells into lactose alone, and they make a great deal.
The repressor is off the DNA in both broths. So what else does the operon need before it is transcribed at full rate?
Unit 6 · Gene Expression and Regulation
1Lifting the block is not enough
In Unit 4, epinephrine bound a receptor on a liver cell, and an enzyme in the cell’s membrane then made cyclic AMP from ATP.
What did the cyclic AMP do inside the liver cell?
- A. Cyclic AMP left the cell and bound a receptor on a neighboring cellA second messenger works inside the cell that made it.
Cyclic AMP spread through the liver cell’s cytosol and switched on kinases. - B. ✓ Cyclic AMP spread through the cytosol and switched on kinases
- C. The cell broke the cyclic AMP down to release energyCyclic AMP carries a message; the cell does not use it as an energy source.
Cyclic AMP spread through the cytosol and switched on kinases.
Why: Cyclic AMP is a second messenger.
A second messenger is a small molecule that carries the receptor’s message inside the cell.
So cyclic AMP spread through the cytosol and switched on kinases.
Lactose enters an E. coli cell whose lac operon is switched off.
Which molecule does lactose bind?
- A. The operatorThe operator is DNA, and lactose binds a protein.
Lactose binds the repressor, which then leaves the operator. - B. RNA polymeraseRNA polymerase binds the promoter; lactose does not bind it.
Lactose binds the repressor, which then leaves the operator. - C. ✓ The repressor
Why: Lactose is the inducer.
The inducer binds the repressor and changes its shape so it no longer fits the operator.
So lactose binds the repressor.
Why is lifting the block not enough?
Transcription of the lac operon at full rate needs a second protein, bound to a stretch of DNA on the far side of the promoter from the operator.
That second protein binds the DNA only while cyclic AMP is bound to it.
The cell makes cyclic AMP only when glucose is scarce.
So the cells make the lactose enzymes in quantity only when lactose is present and glucose is not.
The repressor lowers how often the genes are transcribed, and the second protein raises it.
With the repressor bound, the second protein cannot help.
Video: Watch: Lifting the block is not enough
Lactose lifts the repressor off the operator in both broths. In lactose alone, cyclic AMP binds the second protein, the protein docks beside the promoter, and RNA polymerase transcribes the three genes again and again. In glucose plus lactose, the cell holds little cyclic AMP, the protein floats free, and RNA polymerase transcribes the genes only now and then. The four broths appear as four frames of the same drawing.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L26Ba.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L26Ba.mp4
Go back to E. coli in the broth of glucose plus lactose.
Lactose has bound the repressor. So the repressor has left the operator.
Nothing blocks RNA polymerase’s path into the genes.
Yet RNA polymerase transcribes the three genes only now and then. The cells make very little of the lactose enzymes.
Now suppose the same cells are moved into a broth of lactose alone.
On the far side of the promoter from the operator sits a second short stretch of DNA. In lactose alone, a protein is bound to that stretch.
With that protein bound, RNA polymerase binds the promoter far more often.
So RNA polymerase transcribes the three genes far more often, and the cells make a great deal of the lactose enzymes.
A regulatory protein that binds DNA beside a gene and raises how often RNA polymerase transcribes the gene is called an , because it turns transcription up.
The repressor and the activator are both regulatory proteins, and each binds its own regulatory sequence beside the promoter.
The repressor binds the operator and lowers how often the genes are transcribed. The activator binds its own stretch of DNA and raises how often the genes are transcribed.
Why was the activator bound in lactose alone, and not in glucose plus lactose?
The activator binds the DNA only while a small signal molecule is bound to it. That molecule is cyclic AMP.
In Unit 4, cyclic AMP carried epinephrine’s message inside a liver cell. In E. coli, cyclic AMP carries a message about glucose.
Cyclic AMP binds the activator and changes its shape. In that shape, the activator fits its stretch of DNA beside the promoter.
E. coli makes cyclic AMP only when glucose is scarce. With plenty of glucose in the cell, the cell holds very little cyclic AMP.
In glucose plus lactose, glucose is plentiful. So the cell holds very little cyclic AMP.
With no cyclic AMP bound, the activator does not fit its stretch of DNA. So the activator floats free in the cell.
Without the activator, RNA polymerase transcribes the genes only now and then, and the cells make very little of the lactose enzymes.
In lactose alone, glucose is scarce. So the cell makes cyclic AMP.
Cyclic AMP binds the activator, the activator binds beside the promoter, and RNA polymerase transcribes the genes again and again.
Lactose sets the repressor: with lactose present, the repressor is off the operator.
Glucose sets the activator: with glucose scarce, the cell makes cyclic AMP and the activator is bound beside the promoter.
Now imagine a broth with no sugar at all.
Glucose is scarce. So the cell makes cyclic AMP.
So the activator binds beside the promoter.
But no lactose is present. So the repressor sits on the operator and blocks RNA polymerase’s path.
The activator cannot help: RNA polymerase cannot pass the repressor, and the cells make very little of the lactose enzymes.
The repressor lowers how often the genes are transcribed, and the activator raises it. Full-rate transcription needs both: the repressor off the operator and the activator bound beside the promoter.
What you are expected to know Describe how the activator switches the lac operon up: bound beside the promoter it raises how often RNA polymerase transcribes the genes, and it binds only with cyclic AMP bound, which the cell makes only when glucose is scarce.
The activator of the lac operon floats free in an E. coli cell.
Which molecule must bind the activator before the activator can bind the DNA?
- A. LactoseLactose binds the repressor, not the activator.
Cyclic AMP binds the activator and gives it its binding shape. - B. ✓ Cyclic AMP
- C. GlucoseGlucose binds neither regulatory protein; glucose sets how much cyclic AMP the cell makes.
Cyclic AMP binds the activator and gives it its binding shape.
Why: The activator fits its stretch of DNA in one shape only.
Cyclic AMP binds the activator and changes it to that shape.
So cyclic AMP must bind the activator first.
Suppose E. coli grows in a broth with plenty of glucose.
How much cyclic AMP does the cell hold?
- A. ✓ Very little
- B. A great dealThe cell makes cyclic AMP only when glucose is scarce.
With plenty of glucose, the cell holds very little cyclic AMP.
Why: E. coli makes cyclic AMP only when glucose is scarce.
Here glucose is plentiful.
So the cell holds very little cyclic AMP.
Cyclic AMP has bound the activator in an E. coli cell.
Where on the lac operon does the activator bind?
- A. The operator, just before the first lactose geneThe operator is the repressor’s site.
The activator binds its own stretch of DNA, on the far side of the promoter. - B. The first of the three lactose genesThe genes are what RNA polymerase transcribes; no regulatory protein sits inside them.
The activator binds its own stretch of DNA, on the far side of the promoter. - C. ✓ The DNA on the far side of the promoter from the genes
Why: The activator’s job is to help RNA polymerase bind the promoter.
Its own stretch of DNA sits beside the promoter, on the far side from the operator.
So the activator binds the DNA on the far side of the promoter.
Suppose E. coli cells are growing in a broth of lactose alone. A technician adds glucose to the broth. Within an hour the cells are making far less of the lactose enzymes than before, though lactose is still present.
(a) Explain why the cells make less of the lactose enzymes after the technician adds glucose. (3 pt)
Frame The cells make less of the lactose enzymes because …
E. coli makes cyclic AMP only when glucose is scarce.
So the cell now holds very little cyclic AMP.
With no cyclic AMP bound, the activator leaves its stretch of DNA beside the promoter.
Lactose still holds the repressor off the operator.
So RNA polymerase can still pass, but without the activator it transcribes the three genes only now and then.
- Award 1 point for: glucose is now plentiful, so the cell makes little cyclic AMP (holds very little cyclic AMP).
- Award 1 point for: with no cyclic AMP bound, the activator does not bind beside the promoter (leaves the DNA).
- Award 1 point for: without the activator bound, RNA polymerase transcribes the lac genes less often, so less enzyme is made. Accept with or without: the repressor stays off the operator because lactose is still present.
A student says: “Glucose switches the lactose genes down by binding the repressor and holding it on the operator.”
Is the student correct?
- A. ✓ No: glucose acts through cyclic AMP and the activator, not through the repressor
- B. Yes: glucose binds the repressor and holds it on the operator, just as lactose lifts it offGlucose binds neither the repressor nor the operator.
Plenty of glucose means little cyclic AMP, so the activator leaves the DNA and transcription falls.
Why: Only lactose changes the repressor’s shape.
Glucose sets how much cyclic AMP the cell makes.
With plenty of glucose the cell makes little cyclic AMP.
So the activator leaves the DNA, and RNA polymerase transcribes the genes less often.
Suppose E. coli grows in a broth of glucose only.
How much of the lactose enzymes do the cells make?
- A. ✓ Very little
- B. A great dealWith no lactose, the repressor sits on the operator and blocks RNA polymerase.
The cells make very little of the lactose enzymes.
Why: No lactose is present, so the repressor sits on the operator.
RNA polymerase cannot pass it.
Glucose is plentiful too, so the activator is off the DNA.
So the cells make very little of the lactose enzymes.
Suppose E. coli grows in a broth of lactose only.
How much of the lactose enzymes do the cells make?
- A. Very littleLactose lifts the repressor off the operator, and with glucose scarce the activator is bound.
RNA polymerase transcribes the genes again and again. - B. ✓ A great deal
Why: Lactose binds the repressor, so the repressor is off the operator.
Glucose is scarce, so the cell makes cyclic AMP and the activator binds beside the promoter.
RNA polymerase transcribes the genes again and again.
So the cells make a great deal of the lactose enzymes.
Suppose E. coli grows in a broth of glucose and lactose.
How much of the lactose enzymes do the cells make?
- A. ✓ Very little
- B. A great dealThe repressor is off the operator, but with glucose plentiful the activator is off the DNA.
RNA polymerase transcribes the genes only now and then.
Why: Lactose is present, so the repressor is off the operator.
Glucose is plentiful, so the cell holds little cyclic AMP and the activator is off the DNA.
Without the activator, RNA polymerase transcribes the genes only now and then.
So the cells make very little of the lactose enzymes.
Suppose E. coli grows in a broth with no sugar at all.
How much of the lactose enzymes do the cells make?
- A. ✓ Very little
- B. A great dealThe activator is bound, but with no lactose the repressor sits on the operator.
RNA polymerase cannot pass, so the activator cannot help.
Why: Glucose is scarce, so the cell makes cyclic AMP and the activator binds beside the promoter.
No lactose is present, so the repressor sits on the operator.
RNA polymerase cannot pass the repressor, so the activator cannot help.
So the cells make very little of the lactose enzymes.
The table below lists the four broths: whether the repressor is on the operator, whether the activator is bound beside the promoter, and how much of the lactose enzymes the cells make.
In glucose plus lactose the cells make a little more than in glucose alone, and far less than in lactose alone.
Now imagine an E. coli cell that makes no cyclic AMP at all, growing in a broth of lactose only. A student says: “This cell makes far less of the lactose enzymes than a normal cell in the same broth, even though lactose is present.”
Is the student correct?
- A. No: with lactose present the repressor is off the operator, so the cells make the enzymes in quantity whatever else changesLifting the repressor is not enough for full-rate transcription.
With no cyclic AMP the activator stays off the DNA, and RNA polymerase transcribes the genes only now and then. - B. ✓ Yes: with no cyclic AMP the activator stays off the DNA, so RNA polymerase transcribes the genes only now and then
Why: Lactose lifts the repressor off the operator in this cell as in a normal one.
But this cell makes no cyclic AMP.
With no cyclic AMP bound, the activator stays off the DNA.
Without the activator, RNA polymerase transcribes the genes only now and then.
Go back to the two broths of E. coli.
In glucose plus lactose, the repressor is off the operator, but the activator is not bound.
So the cells make very little of the lactose enzymes.
In lactose alone, cyclic AMP binds the activator, the activator binds beside the promoter, and RNA polymerase transcribes the operon at full rate.
58Quick quiz: activator mixed practice
Suppose RNA polymerase transcribes one of a bacterium’s genes often only while a certain protein is bound to the DNA beside the gene’s promoter. Remove the protein, and RNA polymerase transcribes the gene far less often.
Which of the following is that protein?
- A. A repressorA bound repressor lowers how often a gene is transcribed, so removing it would raise transcription.
This protein raises transcription while bound, so it is an activator. - B. ✓ An activator
Why: The gene is transcribed often only while the protein is bound.
So the bound protein raises how often RNA polymerase transcribes the gene.
A regulatory protein that raises transcription is an activator.
What is an activator?
- A. ✓ A regulatory protein bound beside a gene that raises how often RNA polymerase transcribes the gene
- B. A small signal molecule that binds the repressor so that it no longer fits the operatorThe small molecule that lifts the repressor is the inducer, lactose.
An activator is a protein that raises how often RNA polymerase transcribes a gene. - C. The protein on the operator that blocks RNA polymerase’s path into the genesThe protein on the operator that blocks RNA polymerase is the repressor.
An activator is a protein that raises how often RNA polymerase transcribes a gene.
Why: A regulatory protein that binds DNA beside a gene and raises how often RNA polymerase transcribes the gene is called an activator.
E. coli grows in a broth of lactose only, and the activator is bound beside the lac promoter.
(a) State what the activator does at the lac operon. (1 pt)
With the activator bound, RNA polymerase transcribes the three lac genes far more often.
- Award 1 point for: bound beside the promoter, the activator raises how often RNA polymerase transcribes the lac genes (raises transcription). Accept with or without: it binds only with cyclic AMP bound to it.
Glossary
- activator
- A regulatory protein that binds DNA beside a gene and raises how often RNA polymerase transcribes the gene. At the lac operon it binds beside the promoter only while cyclic AMP is bound to it.
APBIO-U06-L27 Tags on the DNA
Suppose two cells from one person each carry a copy of the same gene. The two copies match letter for letter. In the first cell the gene is transcribed steadily.
In the second cell the gene is silent: its DNA is wound tight around its histones, and small chemical groups sit along it. Not one base differs between the two copies. So what does differ?
Unit 6 · Gene Expression and Regulation
1Two identical genes, two different fates
A eukaryotic chromosome’s DNA winds around small clusters of protein, like thread on beads.
What are those proteins called?
- A. ✓ Histones
- B. RibosomesRibosomes build polypeptides in the cytosol.
The proteins a chromosome’s DNA winds around are histones. - C. Regulatory proteinsA regulatory protein binds a stretch of DNA beside a gene.
The proteins the DNA winds around, bead after bead, are histones.
Why: A chromosome’s DNA winds around clusters of small proteins, like thread on beads.
Those proteins are histones.
RNA polymerase is transcribing a gene in one of a person’s cells.
How is that gene’s DNA wound at that moment?
- A. Tightly, in a short thick coilA short thick coil is the form for moving a chromosome during division.
A gene being read lies on loosely wound DNA. - B. ✓ Loosely, as a long thin thread
Why: Between divisions the DNA is loosely wound.
Only loosely wound DNA can be read.
So a gene being transcribed is on loosely wound DNA.
Suppose a copying error changes one base of a gene into a different base.
What is that change called?
- A. ✓ A mutation
- B. A regulatory sequenceA regulatory sequence is a stretch of DNA beside a gene where a regulatory protein binds.
A change in a gene’s base sequence is a mutation. - C. A phenotypeA phenotype is an organism’s observable features.
A change in a gene’s base sequence is a mutation.
Why: A change in a gene’s base sequence is a mutation.
One base replaced by another is one kind of mutation.
A regulatory protein raises how often a gene is transcribed.
What does the regulatory protein bind?
- A. An mRNAA regulatory protein acts before any mRNA exists.
It binds a stretch of DNA beside the gene, its regulatory sequence. - B. A ribosomeA ribosome reads mRNA in the cytosol.
A regulatory protein binds a stretch of DNA beside the gene, its regulatory sequence. - C. ✓ A stretch of DNA beside the gene
Why: A regulatory protein binds a regulatory sequence, a stretch of DNA beside the gene.
From there it raises or lowers how often RNA polymerase transcribes the gene.
How can two identical genes behave differently?
Small chemical tags on the DNA or on its histones change how tightly the DNA is wound.
RNA polymerase and regulatory proteins can only reach a gene that is loosely wound.
So the tags change how often the gene is transcribed, and the gene’s sequence stays the same.
One kind of tag tightens the winding and silences the gene. Another kind loosens the winding and lets the gene be read.
Enzymes can remove the tags. So the change can be undone.
Yet a dividing cell copies its tags to its daughter cells. So the change can also be passed on.
Video: Watch: Two identical genes, two different fates
Two copies of one gene lie on their histones. Small filled dots appear on one copy’s promoter, and its coil tightens. Small triangles appear on the other copy’s histones, and its coil loosens. RNA polymerase moves in and transcribes the loose copy.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L27a.mp4
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Look at the two copies of the gene. Each copy is wound on histones, like thread on beads.
In cell 1 the beads sit apart. The DNA between them lies open, and RNA polymerase reaches the gene.
In cell 2 the beads are packed together. The DNA is coiled tight, and RNA polymerase cannot reach the gene.
Small chemical groups sit along the copy in cell 2. The cell’s own enzymes put them there.
A small chemical group that a cell’s enzymes add to its DNA or to its histones, and that changes how tightly the DNA is wound, is called a tag.
There are two tags. One sits on the DNA itself, and the other sits on the histones.
The tag on the DNA is a methyl group: one carbon atom with three hydrogen atoms.
Enzymes add methyl groups to some of the cytosine bases of the DNA. The methyl groups that silence a gene sit on its promoter.
A promoter that carries many methyl groups winds tight on its histones.
RNA polymerase cannot reach the gene. So the gene is silent.
Adding methyl groups to the DNA is called . A gene whose promoter is heavily methylated is usually silent.
The tag on the histones is an acetyl group: two carbon atoms with an oxygen atom and three hydrogen atoms.
A histone carries a positive charge, and the DNA’s phosphate groups carry a negative charge. So the DNA clings to the histones.
An acetyl group cancels part of a histone’s positive charge. So the histone grips the DNA less tightly, and the DNA loosens.
Adding acetyl groups to histones is called .
A gene whose histones carry acetyl groups is loosely wound. So RNA polymerase can transcribe the gene.
Chemical tags on the DNA or on its histones change how tightly the DNA is wound.
RNA polymerase and regulatory proteins can only reach a gene that is loosely wound.
So the tags change how often the gene is transcribed without changing its sequence.
A change in a gene’s transcription that comes from tags, not from a change in its base sequence, is called an , because epi- means on top of.
The change sits on top of the gene’s sequence. The sequence itself is untouched.
A mutation changes the base sequence. An epigenetic change leaves every base as it was, and changes only the tags.
So the two copies of the gene differ in their tags, and match in every base. The copy in cell 2 carries methyl groups on its promoter, and its histones have lost their acetyl groups.
What you are expected to know Describe an epigenetic change: tags on the DNA or on its histones change how tightly the DNA is wound, and so how often RNA polymerase transcribes the gene, while the base sequence stays the same.
Two copies of one gene are drawn, marked J and K. In the drawings, beads set apart with the DNA open between them are a loosely wound gene. Beads packed along a wave are a tightly wound gene.
Which copy does RNA polymerase transcribe?
- A. Copy JCopy J is coiled tight on its histones.
RNA polymerase cannot reach a tightly wound gene. - B. ✓ Copy K
Why: Copy K lies open on beads set apart.
RNA polymerase can reach a loosely wound gene.
So RNA polymerase transcribes copy K.
Enzymes add many methyl groups to a gene’s promoter.
What happens to the gene’s base sequence?
- A. ✓ The sequence stays the same
- B. Some bases change into other basesA methyl group sits on a cytosine base.
The cytosine is still a cytosine, so the sequence is the same.
Why: A methyl group is a tag added on top of a base.
The base itself stays what it was.
So the base sequence stays the same.
Suppose a person’s bone-forming cells and retina cells both carry the same gene. In the bone-forming cells, acetyl groups sit on the gene’s histones and no methyl groups sit on its promoter. In the retina cells, many methyl groups sit on the promoter and the histones carry no acetyl groups.
(a) Explain why only the bone-forming cells transcribe the gene. (2 pt)
Frame Only the bone-forming cells transcribe the gene because …
Acetyl groups on the histones loosen the histones’ grip on the DNA.
So RNA polymerase can reach the promoter and transcribe the gene.
In the retina cells, methyl groups on the promoter and bare histones make the DNA wind tight.
So RNA polymerase cannot reach the gene there.
- Award 1 point for: acetyl groups on the histones loosen the winding, so RNA polymerase reaches the gene in the bone-forming cells.
- Award 1 point for: methyl groups on the promoter (with no acetyl groups on the histones) tighten the winding, so RNA polymerase cannot reach the gene in the retina cells.
Two cells from one person each carry a copy of the same gene. In one cell the copy is silent, and methyl groups sit on its promoter. A student says: “The silent copy must carry a mutation: the methyl groups changed its bases.”
Is the student correct?
- A. ✓ No: the methyl groups sit on the bases and change none of them, so the sequence is the same
- B. Yes: a methyl group added to a base turns that base into a different baseA methyl group sits on top of a cytosine, and the cytosine stays a cytosine.
The base sequence is unchanged, so there is no mutation.
Why: A mutation is a change in the base sequence.
A methyl group sits on a base and changes no base.
So the silent copy carries no mutation: the sequence is the same, and only the tags differ.
Suppose the histones around a gene in the cells lining a person’s esophagus lose their acetyl groups. A student says: “RNA polymerase now reaches this gene less easily, so these cells transcribe it less.”
Is the student correct?
- A. No: acetyl groups do not change the winding, and only methyl groups on the promoter doAcetyl groups on the histones loosen the winding.
When the histones lose them, the DNA winds tighter, and RNA polymerase reaches the gene less often. - B. ✓ Yes: with fewer acetyl groups the DNA winds tighter, so RNA polymerase reaches the gene less often
Why: Acetyl groups on histones loosen the DNA.
Histones that lose their acetyl groups grip the DNA more tightly.
Tightly wound DNA is harder for RNA polymerase to reach.
So these cells transcribe the gene less.
Suppose enzymes add acetyl groups to the histones around a gene in an otter’s liver cells.
How is that gene’s DNA wound afterwards?
- A. ✓ More loosely
- B. More tightlyAcetyl groups weaken the histones’ grip on the DNA.
A weaker grip loosens the winding; it does not tighten it. - C. The same as beforeAcetyl groups cancel part of a histone’s positive charge.
The histone then grips the DNA less tightly, so the winding changes.
Why: An acetyl group cancels part of a histone’s positive charge.
So the histone grips the DNA less tightly.
So the gene’s DNA is wound more loosely.
The table below compares the two tags: where each sits, what each does to the winding, what happens to transcription, and what removes each.
45Quick quiz: epigenetic change, DNA methylation, histone acetylation mixed practice
Suppose a cell adds methyl groups to the promoter of one of its genes, and the gene falls silent.
Is this an epigenetic change?
- A. ✓ Yes
- B. NoThe gene’s transcription changed because of tags on its DNA, with no change in the base sequence.
That is an epigenetic change.
Why: The gene’s transcription changed.
The cause was tags on the DNA, and the base sequence stayed the same.
So the change is an epigenetic change.
Enzymes add acetyl groups to the histones around a gene.
Which name fits this change?
- A. DNA methylationDNA methylation adds methyl groups to the DNA’s cytosine bases.
Adding acetyl groups to histones is histone acetylation. - B. ✓ Histone acetylation
Why: The tag added is an acetyl group.
The enzymes add it to the histones.
Adding acetyl groups to histones is histone acetylation.
What is an epigenetic change?
- A. A change in a gene’s base sequence that alters the protein the gene codes for, passed to every daughter cellA change in the base sequence is a mutation.
An epigenetic change leaves the base sequence as it was and changes the tags. - B. ✓ A change in how often a gene is transcribed, made by tags on the DNA or its histones, with the base sequence unchanged
- C. A change in how often a gene is transcribed that comes from a change in the base sequence of its promoterA change in the promoter’s base sequence is a mutation.
An epigenetic change comes from tags on the DNA or its histones, with no base changed.
Why: A change in a gene’s transcription that comes from tags, not from a change in its base sequence, is called an epigenetic change.
What is DNA methylation?
- A. Enzymes removing methyl groups from the histones around a geneMethyl groups go onto the DNA’s cytosine bases, not onto histones.
DNA methylation is enzymes adding them, not removing them. - B. RNA polymerase copying a methylated stretch of DNA into RNADNA methylation is a change to the DNA, made by enzymes.
RNA polymerase transcribes; it adds no methyl groups. - C. ✓ Enzymes adding methyl groups to some of the DNA’s cytosine bases
Why: Adding methyl groups to the DNA is called DNA methylation.
Enzymes add the methyl groups to cytosine bases.
The methyl groups on a gene’s promoter silence the gene.
What is histone acetylation?
- A. ✓ Enzymes adding acetyl groups to the histones around a gene
- B. Enzymes adding acetyl groups to the bases of the DNAAcetyl groups go onto the histones, not onto the DNA’s bases.
Histone acetylation is enzymes adding acetyl groups to histones. - C. Histones winding the DNA tighter so that a gene falls silentAcetyl groups loosen the winding; they do not tighten it.
Histone acetylation is enzymes adding acetyl groups to histones.
Why: Adding acetyl groups to histones is called histone acetylation.
A cell changes how often it transcribes a gene without changing the gene’s base sequence.
(a) State what an epigenetic change is. (1 pt)
- Award 1 point for: a change in a gene’s transcription (expression) caused by tags on the DNA or histones, with the base sequence unchanged.
(b) State what DNA methylation is. (1 pt)
On a gene’s promoter, the methyl groups silence the gene.
- Award 1 point for: adding methyl groups to the DNA (its cytosine bases). ‘On the promoter’ completes it, but its absence does not lose the point.
(c) State what histone acetylation is. (1 pt)
- Award 1 point for: adding acetyl groups to histones.
52Which way does transcription change?
RNA polymerase is about to transcribe a gene.
Which stretch of DNA does RNA polymerase bind to begin?
- A. The regulatory sequence beside the geneThe regulatory sequence is where a regulatory protein binds.
RNA polymerase binds the promoter, the stretch of DNA just before the gene. - B. The gene’s first codonA codon is read by the ribosome on the mRNA, later.
RNA polymerase binds the promoter, the stretch of DNA just before the gene. - C. ✓ The promoter
Why: RNA polymerase binds a stretch of DNA just before the gene.
That stretch is the promoter.
Now suppose a cell adds methyl groups to the promoter of a gene it has been transcribing. Which way does the gene’s transcription change: up, or down?
Video: Watch: Which way does transcription change?
One gene on its beads, four times over. Dots appear on the promoter, and the coil tightens. Triangles appear on the beads, and the coil loosens. The dots leave, and the coil loosens. The triangles leave, and the coil tightens. Under each, an arrow points down or up.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L27b.mp4
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To predict the direction, ask what the change in tags does to the winding.
A change that tightens the winding lowers the gene’s transcription. A change that loosens the winding raises it.
For example, suppose the cell adds methyl groups to the gene’s promoter. Transcription of the gene falls, because the DNA winds tighter.
But suppose the cell adds acetyl groups to the gene’s histones. Transcription of the gene rises, because the DNA winds looser.
And suppose the cell removes the methyl groups from the gene’s promoter. Transcription of the gene rises, because the DNA winds looser.
But suppose the cell removes the acetyl groups from the gene’s histones. Transcription of the gene falls, because the DNA winds tighter.
More methyl groups on the promoter, or fewer acetyl groups on the histones: transcription falls. Fewer methyl groups, or more acetyl groups: transcription rises.
What you are expected to know Predict which way a gene’s transcription changes from a described change in its tags.
A cell adds methyl groups to a gene’s promoter.
What happens to the gene’s transcription?
- A. RisesMethyl groups on the promoter tighten the winding.
Tighter winding keeps RNA polymerase out, so transcription falls. - B. ✓ Falls
- C. Stays the sameMethyl groups on the promoter tighten the winding.
RNA polymerase then reaches the gene less often, so transcription falls.
Why: Methyl groups on the promoter tighten the winding.
RNA polymerase reaches a tightly wound gene less often.
So transcription falls.
Enzymes remove the acetyl groups from a gene’s histones.
What happens to the gene’s transcription?
- A. RisesHistones that lose their acetyl groups grip the DNA more tightly.
Tighter winding lowers transcription. - B. ✓ Falls
- C. Stays the sameHistones that lose their acetyl groups grip the DNA more tightly.
Tighter winding lowers transcription, so it falls.
Why: Acetyl groups loosen the histones’ grip.
With the acetyl groups gone, the DNA winds tighter.
So RNA polymerase reaches the gene less often, and transcription falls.
Enzymes add acetyl groups to a gene’s histones.
What happens to the gene’s transcription?
- A. ✓ Rises
- B. FallsAcetyl groups loosen the histones’ grip on the DNA.
Looser winding lets RNA polymerase in more often, so transcription rises. - C. Stays the sameAcetyl groups loosen the histones’ grip on the DNA.
A looser winding lets RNA polymerase in more often, so transcription rises.
Why: Acetyl groups loosen the histones’ grip on the DNA.
RNA polymerase reaches a loosely wound gene more often.
So transcription rises.
Enzymes add methyl groups to the promoter of a neighboring gene only. This gene’s promoter stays as it was.
What happens to this gene’s transcription?
- A. RisesNo tag on this gene changed, so its winding is as it was.
Its transcription stays the same. - B. FallsThe methyl groups went onto the neighboring gene’s promoter.
This gene’s winding is unchanged, so its transcription stays the same. - C. ✓ Stays the same
Why: Methyl groups act on the promoter they sit on.
This gene’s promoter gained none.
So this gene’s winding is unchanged, and its transcription stays the same.
Enzymes remove the methyl groups from a gene’s promoter.
What happens to the gene’s transcription?
- A. ✓ Rises
- B. FallsMethyl groups tighten the winding.
With them gone, the DNA winds looser and transcription rises. - C. Stays the sameMethyl groups tighten the winding.
Removing them loosens the DNA, so transcription rises.
Why: Methyl groups on the promoter tighten the winding.
With the methyl groups gone, the DNA winds looser.
So RNA polymerase reaches the gene more often, and transcription rises.
A gene’s histones lose their acetyl groups, and its promoter gains methyl groups.
What happens to the gene’s transcription?
- A. RisesBoth changes tighten the winding.
Tighter winding lowers transcription. - B. ✓ Falls
- C. Stays the sameBoth changes tighten the winding.
Tighter winding keeps RNA polymerase out more often, so transcription falls.
Why: Losing acetyl groups tightens the winding.
Gaining methyl groups on the promoter tightens it further.
So RNA polymerase reaches the gene less often, and transcription falls.
Suppose a drug stops the enzymes that remove acetyl groups from histones. A cell takes up the drug, and acetyl groups build up on one gene’s histones.
What happens to that gene’s transcription?
- A. ✓ Rises
- B. FallsWith the removing enzymes stopped, acetyl groups build up on the gene’s histones.
More acetyl groups loosen the winding, so transcription rises. - C. Stays the sameAcetyl groups keep being added to the gene’s histones, and none are removed.
The extra acetyl groups loosen the winding, so transcription rises.
Why: Enzymes keep adding acetyl groups to the histones.
The drug stops the enzymes that remove them, so acetyl groups build up.
More acetyl groups loosen the winding.
So transcription rises.
71Undone, yet passed on
A cell is about to divide.
When does the cell copy its DNA?
- A. While it is dividingDuring the division the cell shares out DNA that is already copied.
The copying happened earlier, in S phase. - B. After it has dividedEach daughter cell receives a complete set of DNA at the division.
The copying happened before, in S phase. - C. ✓ In S phase, before it divides
Why: A cell copies its DNA once, in S phase, before it divides.
So each daughter cell receives a complete set.
A cell has just copied a DNA molecule in S phase.
What does each of the two new DNA molecules hold?
- A. ✓ One old strand and one new strand
- B. Two new strandsEach new molecule keeps one of the old strands.
DNA polymerase builds one new strand against it. - C. Two old strandsThe two old strands part, one into each new molecule.
Each old strand is paired with a new strand.
Why: The two old strands separate.
DNA polymerase builds a new strand against each old one.
So each new molecule holds one old strand and one new strand.
A cell divides by mitosis.
How does each daughter cell’s DNA sequence compare with the parent cell’s?
- A. Half of the parent’sEach daughter cell receives one complete copy of the genome, not half.
Its DNA sequence is identical to the parent’s. - B. ✓ Identical to the parent’s
- C. Different from the parent’s at many basesEvery chromosome was copied once in S phase into two identical chromatids, and each daughter received one of each pair.
Its DNA sequence is identical to the parent’s.
Why: In S phase every chromosome was copied once into two identical sister chromatids.
Anaphase sent one chromatid of every pair to each pole.
So each daughter cell’s DNA sequence is identical to the parent’s.
Now suppose a drug stops the enzymes that remove acetyl groups from histones. In cells that take up the drug, acetyl groups build up, and many genes are transcribed more.
Then a technician washes the drug out of those cells. Does their transcription stay at the raised rate, or fall back?
Video: Watch: Undone, yet passed on
Triangles crowd a gene’s beads and the coil lies open. The drug leaves, enzymes pull the triangles off, and the coil tightens again. Then a gene with dots on its promoter is copied in S phase, the dots reappear on the new DNA, and both daughter cells carry them.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L27c.mp4
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A tag is not part of the DNA’s base sequence. One set of enzymes adds the tag, and another set of enzymes removes it.
Once the drug is gone, the removing enzymes take the acetyl groups off the histones again.
The DNA winds tighter, and RNA polymerase reaches the gene less often. Transcription falls back to its old rate.
So an epigenetic change can be undone. Enzymes remove the tags, and the winding goes back to what it was.
Yet an epigenetic change can also be passed on. Suppose a cell carries a silent gene, its promoter covered in methyl groups, and the cell divides.
In S phase the cell copies its DNA. Each new DNA molecule holds one old strand and one new strand.
The old strand still carries its methyl groups. Enzymes then add methyl groups to the new strand to match the old strand’s pattern.
So each copied DNA molecule carries the same methyl groups as before.
The cell divides, and each daughter cell receives a DNA molecule with the same methyl pattern. The gene stays silent in both daughter cells.
So a tissue keeps its character through many divisions: the daughter cells inherit the parent cell’s tags.
Enzymes can remove the tags. So an epigenetic change can be undone.
A dividing cell copies the tags. So an epigenetic change can be passed on.
What you are expected to know Explain why an epigenetic change can be undone and can also be passed on: enzymes remove the tags, and a dividing cell copies its tag pattern to its daughter cells.
Suppose a drug stops the enzymes that remove acetyl groups from histones. Cells that take up the drug transcribe one gene more often than before. Then a technician washes the drug out of the cells.
Over the following days, does transcription of that gene return to its untreated rate?
- A. NoThe removing enzymes work again once the drug is gone.
They take the extra acetyl groups off, and transcription falls back. - B. ✓ Yes
Why: With the drug gone, the enzymes that remove acetyl groups work again.
They take the extra acetyl groups off the histones.
The DNA winds tighter, so transcription returns to its untreated rate.
Suppose a gene in a person’s scalp cells is silent, and its promoter carries many methyl groups. A month later, the same cells are transcribing the gene.
(a) Explain how the gene could have been switched back on. (2 pt)
Frame The gene could have been switched back on because …
Enzymes in the cell can remove methyl groups from the promoter.
With the methyl groups gone, the DNA winds more loosely.
So RNA polymerase can reach the promoter and transcribe the gene again.
- Award 1 point for: enzymes can remove the methyl groups (the tags are not part of the base sequence, so the change is reversible).
- Award 1 point for: with the methyl groups gone the DNA winds more loosely, so RNA polymerase can reach the gene (and transcribe it).
A student says: “Once methyl groups have silenced a gene, the gene stays silent in that cell for good.”
Is the student correct?
- A. ✓ No: enzymes can remove the methyl groups, and the cell can transcribe the gene again
- B. Yes: methyl groups become a permanent part of the DNA, so the gene stays silentMethyl groups are tags on the DNA, not part of its sequence.
Enzymes can remove them, and the gene can be transcribed again.
Why: Methyl groups are tags added by enzymes.
Other enzymes can remove them.
With the methyl groups gone, the DNA loosens and RNA polymerase reaches the gene.
So the gene can be transcribed again.
Suppose a cell in the lining of a person’s gut carries a silent gene, its promoter covered in methyl groups, and the cell divides. A student says: “Both daughter cells will keep the gene silent.”
Is the student correct?
- A. No: copying the DNA in S phase strips the methyl groups off, so both daughter cells transcribe the geneThe old strand keeps its methyl groups through S phase, and enzymes add matching methyl groups to the new strand.
So the copied DNA carries the same tags. - B. ✓ Yes: the daughters inherit the parent’s methyl pattern, so the gene stays silent in both
Why: In S phase the old strand keeps its methyl groups.
Enzymes add methyl groups to the new strand to match.
Each daughter cell receives DNA with the same methyl pattern.
So the gene stays silent in both.
In the cells of a person’s cornea, a gene’s promoter carries many methyl groups and the gene is silent. Those cells divide many times over a year.
Which of the following describes the gene in the cornea cells present a year later?
- A. Half of the cells carry the methyl groups but the other half do notEach division copies the methyl pattern onto both daughter cells’ DNA.
So every descendant carries the methyl groups. - B. The promoter has lost its methyl groups: the gene is transcribedDivision does not strip the methyl groups off.
Enzymes copy the pattern onto the new strand at each S phase, so the promoter stays methylated. - C. ✓ The promoter still carries the methyl groups: the gene is silent
Why: At each S phase, enzymes copy the parent strand’s methyl pattern onto the new strand.
Each division hands both daughters the same pattern.
So a year later the promoter is still methylated, and the gene is still silent.
Go back to the two copies of the same gene, one in each of two cells from one person. The two copies match letter for letter.
The copy in cell 2 carries methyl groups on its promoter, and its histones have lost their acetyl groups. So it is wound tight, and RNA polymerase cannot reach it.
Remove the tags, and the gene opens again.
99Mixed practice mixed practice
Suppose two cells from one plant each carry a copy of the same gene. One cell transcribes the gene, and the other leaves it silent.
Which of the following can explain the difference?
- A. ✓ Different tags sit on the two copies
- B. The two copies differ in base sequenceThe two cells grew from one fertilized egg, so both copies have the same base sequence.
Tags on the DNA or histones can differ between two copies of one gene. - C. One of the cells has lost the geneEvery cell of the plant carries the gene.
Tags on the DNA or histones can silence one copy and leave the other read.
Why: The two cells carry the same genome, so the two copies match base for base.
Tags on the DNA or its histones set how tightly each copy is wound.
A tightly wound copy is silent, and a loosely wound copy is transcribed.
A cell transcribes a gene about 12 times an hour. Enzymes then add acetyl groups to the gene’s histones.
Which rate is now most likely?
- A. About 3 times an hourAcetyl groups loosen the winding, so RNA polymerase reaches the gene more often.
The rate rises above 12 times an hour. - B. About 12 times an hourAcetyl groups loosen the winding, so RNA polymerase reaches the gene more often.
Looser winding raises the rate above 12 times an hour. - C. ✓ About 36 times an hour
Why: Acetyl groups loosen the histones’ grip on the DNA.
RNA polymerase reaches a loosely wound gene more often.
So the rate rises, to about 36 times an hour.
Which of the following does DNA methylation change?
- A. The order of the bases in the geneA methyl group sits on a cytosine and leaves the base order as it was.
DNA methylation changes how tightly the DNA is wound. - B. ✓ How tightly the DNA is wound
- C. The amino acids the gene codes forThe amino acids depend on the base sequence, which methyl groups leave unchanged.
DNA methylation changes how tightly the DNA is wound.
Why: Methyl groups on the promoter tighten the winding.
The base sequence stays the same, so the protein it codes for is the same.
DNA methylation changes only how tightly the DNA is wound.
Suppose a drug stops the enzymes that add methyl groups to DNA. Cells with several methylated, silent genes take up the drug and divide many times.
What happens to those genes in the later daughter cells?
- A. The genes stay silent in every daughter cellAt each S phase the new strand needs methyl groups added to match the old strand.
With the adding enzymes stopped, the pattern is lost over the divisions. - B. ✓ In later daughter cells the genes are transcribed again
- C. The genes’ base sequence changesThe drug stops methyl groups being added; it changes no base.
The methyl pattern is lost over the divisions, and the genes are transcribed again.
Why: At each S phase, enzymes normally add methyl groups to the new strand to match the old one.
The drug stops those enzymes.
So each division carries fewer methyl groups forward, the DNA loosens, and the genes are transcribed again.
A gene is silent, and its promoter carries many methyl groups.
Which of the following could switch the gene back on?
- A. RNA polymerase pushing the histones apartRNA polymerase cannot reach a tightly wound gene, let alone unwind it.
Enzymes removing the methyl groups loosen the DNA. - B. The cell copying its DNA in S phaseCopying the DNA copies the methyl pattern too, so the gene stays silent.
Enzymes removing the methyl groups loosen the DNA. - C. ✓ Enzymes removing the methyl groups from the promoter
Why: Methyl groups on the promoter hold the DNA tightly wound.
Enzymes can remove them.
With the methyl groups gone, the DNA loosens and RNA polymerase reaches the gene.
Suppose a skin cell in a person’s eyelid carries a methylated, silent gene and divides. Each daughter cell keeps the gene silent. A student says: “The daughters keep it silent because the methyl groups changed the gene’s base sequence, and they inherited that changed sequence.”
Is the student correct?
- A. ✓ No: the base sequence is unchanged, and the daughters inherited the parent cell’s methyl pattern
- B. Yes: a methyl group is a change to a base, and daughter cells inherit their parent’s base sequenceA methyl group sits on a base and changes no base.
The daughters inherited the parent’s methyl pattern, copied at S phase.
Why: A methyl group changes no base, so the sequence is unchanged.
At S phase the cell copied the methyl pattern onto the new DNA.
So the daughters inherited the parent cell’s tags, and the gene stays silent.
A gene in one cell is wound loosely on its histones. The same gene in a second cell is wound tightly.
In which cell can a regulatory protein reach the gene’s regulatory sequence?
- A. ✓ Only the first cell
- B. Only the second cellA regulatory sequence on tightly wound DNA is buried in the coil.
A regulatory protein can reach it only where the DNA is loosely wound. - C. Both cellsA regulatory protein cannot reach DNA that is coiled tight.
Only the loosely wound gene, in the first cell, is open to it.
Why: RNA polymerase and regulatory proteins can only reach a gene that is loosely wound.
The gene is loosely wound in the first cell only.
So only in the first cell can a regulatory protein reach the regulatory sequence.
Suppose cells in a laboratory dish transcribe a gene 15 times an hour. A technician adds a drug that stops the enzymes that remove acetyl groups from histones. Two days later, the treated cells transcribe the gene 45 times an hour.
(a) Explain why the transcription rate rose. (2 pt)
Frame The rate rose because …
Enzymes kept adding acetyl groups, so acetyl groups built up on the gene’s histones.
The histones gripped the DNA less tightly, and the DNA loosened.
So RNA polymerase reached the gene more often, and the rate rose to 45 times an hour.
- Award 1 point for: acetyl groups build up on the histones because the drug stops the enzymes that remove them.
- Award 1 point for: more acetyl groups loosen the winding, so RNA polymerase reaches the gene more often.
(b) Explain what happens to the transcription rate in the weeks after the technician washes the drug out. (2 pt)
Frame Once the drug is gone, …
They take the extra acetyl groups off the histones.
The histones grip the DNA more tightly, and RNA polymerase reaches the gene less often.
So the rate falls back toward 15 times an hour.
- Award 1 point for: the removing enzymes work again and take the acetyl groups off (the change is reversible).
- Award 1 point for: the DNA winds tighter, so RNA polymerase reaches the gene less often and the rate falls back toward its untreated value.
Glossary
- epigenetic change
- A change in how often a gene is transcribed that comes from tags on the DNA or its histones, not from a change in the base sequence. Epi- means on top of: the change sits on top of the gene’s sequence, which is untouched.
- DNA methylation
- Enzymes adding methyl groups to some of the DNA’s cytosine bases. A promoter carrying many methyl groups winds tight on its histones, and the gene is usually silent.
- histone acetylation
- Enzymes adding acetyl groups to the histones that a gene’s DNA is wound on. The acetyl groups weaken the histones’ grip, the DNA loosens, and RNA polymerase can transcribe the gene.
APBIO-U06-L28 One genome, many cell types
Suppose a biologist takes a muscle cell and a pancreas cell from one person. She looks for the insulin gene in both cells: both cells carry it. She looks for insulin mRNA in both cells: only the pancreas cell holds any.
The two cells carry the same DNA and do different jobs. How does a cell become one kind and not another?
Unit 6 · Gene Expression and Regulation
1Two cells, one genome
A skin cell divides by mitosis into two daughter cells.
How does each daughter cell’s genome compare with the parent cell’s?
- A. ✓ Each daughter cell carries the parent’s complete genome
- B. Each daughter cell carries half of the parent’s genomeIn S phase every chromosome was copied once, and anaphase sent one copy of every chromosome to each pole.
Each daughter cell receives the whole genome. - C. Each daughter cell carries only the genes it will useA dividing cell does not share its genes out by job.
Anaphase sends one copy of every chromosome to each pole, so each daughter cell receives the whole genome.
Why: In S phase every chromosome was copied once into two identical sister chromatids.
Anaphase sent one chromatid of every pair to each pole.
So each daughter cell carries one complete copy of the parent’s genome.
How does one genome make about two hundred kinds of cell?
Every body cell of a person carries the same genes.
Each kind of cell transcribes a different set of those genes.
The proteins one set makes, such as insulin in a pancreas cell, do that kind of cell’s job.
Video: Watch: Two cells, one genome
A muscle cell and a pancreas cell are drawn side by side with the same row of genes in each nucleus. In the pancreas cell the insulin gene fills in and RNA hangs from it. In the muscle cell a muscle-protein gene fills in and RNA hangs from it, and the insulin gene stays open.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L28a.mp4
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Go back to the muscle cell and the pancreas cell.
Both grew from one fertilized egg by mitosis. So both carry the same genome.
The insulin gene sits in the DNA of both cells.
In the pancreas cell, RNA polymerase transcribes the insulin gene. So the pancreas cell holds insulin mRNA, and its ribosomes build insulin.
In the muscle cell, RNA polymerase does not transcribe the insulin gene. So the muscle cell holds no insulin mRNA and builds no insulin.
The muscle cell transcribes other genes instead. One of them is the gene for a protein that shortens the cell when the muscle contracts.
The pancreas cell carries that muscle-protein gene too, and does not transcribe it.
So the two cells carry the same genes and transcribe different sets of them. The mRNAs in the two cells differ, and so do the proteins.
A person’s body has about two hundred kinds of cell. Every kind carries the same genome, and every kind transcribes its own set of genes.
A protein that one kind of cell makes and other kinds do not, such as insulin, is called a , because a tissue is a group of cells of one kind.
A cell becoming one kind of cell and not another, by expressing one set of its genes, is called , because the cell becomes different from the other kinds.
Different kinds of cell expressing different sets of the same genes is called .
Differentiation removes no genes. The muscle cell still carries the insulin gene, and RNA polymerase is not transcribing it there.
So what makes a muscle cell a muscle cell is the set of genes it expresses, and the proteins that set makes.
What you are expected to know Explain differentiation as differential gene expression: every body cell carries the same genome, and each kind of cell transcribes its own set of genes and makes its own tissue-specific proteins.
A biologist looks for the insulin gene and for insulin mRNA in a muscle cell and a pancreas cell from one person. The table gives what she finds.
Which line of the table differs between the two cells?
- A. Whether the cell carries the insulin geneBoth cells carry the insulin gene, because every body cell carries the whole genome.
The cells differ on the mRNA line. - B. ✓ Whether the cell holds insulin mRNA
- C. Neither line: the two cells match on bothThe two cells match on the gene line alone.
The pancreas cell holds insulin mRNA and the muscle cell holds none.
Why: Both cells carry the insulin gene, because every body cell carries the whole genome.
Only the pancreas cell transcribes the insulin gene.
So only the pancreas cell holds insulin mRNA.
Suppose a biologist compares two kinds of cell from one young animal: the cells that build the enamel of its teeth, and the cells of its skin. The enamel-building cells are packed with an enamel protein. The skin cells hold none of it, and they carry its gene.
(a) Explain how the two kinds of cell can make different proteins while carrying the same DNA. (2 pt)
Frame The two kinds of cell make different proteins because …
The enamel-building cells transcribe the enamel protein’s gene.
So the enamel-building cells hold its mRNA, and their ribosomes build the protein.
The skin cells carry the same gene and do not transcribe it.
So the skin cells hold no mRNA of it and build none of the protein.
Their DNA is the same.
The genes they transcribe differ.
- Award 1 point for: both kinds of cell carry the enamel protein’s gene, and only the enamel-building cells transcribe it, so only they hold its mRNA and build the protein.
- Award 1 point for: the two kinds of cell differ in which genes they express (differential gene expression), not in which genes they carry.
A student looks at a bone cell, which makes no insulin, and says: “During differentiation this cell lost the insulin gene.”
Is the student correct?
- A. ✓ No: the bone cell still carries the insulin gene and is not transcribing it
- B. Yes: a cell that makes none of a protein has removed that protein’s geneA differentiated cell keeps its whole genome.
The bone cell carries the insulin gene and does not transcribe it.
Why: Differentiation removes no genes.
The bone cell carries the insulin gene, as every body cell does.
RNA polymerase does not transcribe it there, so the bone cell makes no insulin.
A liver cell of one person makes a blood-clotting protein. A nerve cell of the same person makes none of it. A biologist looks for the protein’s gene and for its mRNA in both cells.
Which of the following does she find?
- A. The gene in the liver cell alone, with its mRNA in the liver cell aloneEvery body cell carries the whole genome, the clotting protein’s gene with it.
The nerve cell carries the gene and does not transcribe it. - B. The gene in both cells, with its mRNA in both cellsOnly the liver cell makes the protein, so only the liver cell is transcribing the gene.
The nerve cell holds no mRNA of it. - C. ✓ The gene in both cells, with its mRNA in the liver cell alone
Why: Every body cell carries the whole genome, so both cells carry the gene.
Only the liver cell makes the protein, so only the liver cell transcribes the gene.
So the liver cell alone holds the mRNA.
27Quick quiz: differentiation, tissue-specific protein, differential gene expression mixed practice
The light-sensing protein of the eye, made by the eye’s rod cells and by no other kind of cell.
Is this protein a tissue-specific protein?
- A. ✓ Yes
- B. NoRod cells make this protein, and no other kind of cell does.
A protein that one kind of cell makes and other kinds do not is a tissue-specific protein.
Why: Rod cells make the light-sensing protein, and no other kind of cell does.
So it is a tissue-specific protein.
RNA polymerase, made by every cell in the body.
Is this protein a tissue-specific protein?
- A. YesEvery kind of cell makes RNA polymerase.
A tissue-specific protein is made by one kind of cell and by no other kind. - B. ✓ No
Why: Every kind of cell makes RNA polymerase.
A protein that every kind of cell makes is not a tissue-specific protein.
An antibody, made by one kind of white blood cell and by no other kind of cell.
Is this protein a tissue-specific protein?
- A. ✓ Yes
- B. NoOne kind of white blood cell makes the antibody, and no other kind does.
A protein one kind of cell makes and other kinds do not is a tissue-specific protein.
Why: One kind of white blood cell makes the antibody, and no other kind of cell does.
So the antibody is a tissue-specific protein.
The protein of the sheath that wraps a nerve fiber, made by the sheath cells and by no other kind of cell.
Is this protein a tissue-specific protein?
- A. ✓ Yes
- B. NoThe sheath cells make this protein, and no other kind of cell does.
A protein that one kind of cell makes and other kinds do not is a tissue-specific protein.
Why: The sheath cells make the protein of the sheath, and no other kind of cell does.
So it is a tissue-specific protein.
The enzyme that copies the DNA before a cell divides, made by every dividing cell.
Is this protein a tissue-specific protein?
- A. YesEvery dividing cell, of every kind, makes this enzyme.
A tissue-specific protein is made by one kind of cell and by no other kind. - B. ✓ No
Why: Every dividing cell makes the enzyme that copies its DNA, whatever kind of cell it is.
So the enzyme is not a tissue-specific protein.
What is differentiation?
- A. A cell losing the genes it does not use, so that one set is leftA differentiating cell loses no genes; it keeps its whole genome.
It becomes one kind of cell by expressing one set of its genes. - B. ✓ A cell becoming one kind of cell and not another, by expressing one set of its genes
- C. A cell copying its genome once, so that each daughter cell receives the whole of itCopying the genome once before division is S phase; every dividing cell does it.
Differentiation is a cell becoming one kind by expressing one set of its genes.
Why: A cell becoming one kind of cell and not another, by expressing one set of its genes, is called differentiation.
What is a tissue-specific protein?
- A. ✓ A protein that one kind of cell makes and other kinds do not
- B. A protein that every kind of cell makes at about the same levelA protein every kind of cell makes belongs to no one tissue.
A tissue-specific protein is made by one kind of cell and by no other kind. - C. A protein that a cell makes from a gene that only that kind of cell carriesEvery kind of cell carries the same genes.
A tissue-specific protein is made by one kind of cell from a gene that every kind carries.
Why: A protein that one kind of cell makes and other kinds do not is called a tissue-specific protein.
What is differential gene expression?
- A. Different kinds of cell carrying different genes in their DNAEvery kind of cell carries the same genes.
Differential gene expression is different kinds of cell expressing different sets of those same genes. - B. ✓ Different kinds of cell expressing different sets of the same genes
- C. One kind of cell expressing a gene at a different level every hourA level changing hour by hour is one cell changing how much it expresses one gene.
Differential gene expression is different kinds of cell expressing different sets of genes.
Why: Different kinds of cell expressing different sets of the same genes is called differential gene expression.
A person’s body has about two hundred kinds of cell, and every kind carries the same genome.
(a) State what differentiation is. (1 pt)
- Award 1 point for: a cell becoming one kind of cell (taking on one job) by expressing one set of its genes. ‘And not another’ completes it, but its absence does not lose the point.
(b) State what a tissue-specific protein is. (1 pt)
- Award 1 point for: a protein made by one kind of cell (one tissue) and not by other kinds.
(c) State what differential gene expression is. (1 pt)
- Award 1 point for: different kinds of cell expressing different sets of genes from the same genome.
37The protein beside the gene
In a bacterium, a regulatory protein controls how often a gene is transcribed.
Where does the regulatory protein bind?
- A. On the mRNA that RNA polymerase copied from the geneA regulatory protein binds DNA, not RNA.
It binds the regulatory sequence, a stretch of DNA beside the gene. - B. On the ribosome that translates the gene’s mRNAA regulatory protein acts before any mRNA exists.
It binds the regulatory sequence, a stretch of DNA beside the gene. - C. ✓ On the regulatory sequence, a stretch of DNA beside the gene
Why: A regulatory protein binds a regulatory sequence.
The regulatory sequence is a stretch of DNA beside the gene.
Bound there, the protein raises or lowers how often RNA polymerase transcribes the gene.
A kinase acts on a relay protein inside a cell.
What does the kinase do to the relay protein?
- A. The kinase cuts the relay protein into short piecesA kinase cuts nothing.
It transfers a phosphate group from ATP onto the relay protein, and the added phosphate changes the protein’s shape. - B. The kinase carries the relay protein through the nuclear poreA kinase carries nothing into the nucleus.
It transfers a phosphate group from ATP onto the relay protein, and the added phosphate changes the protein’s shape. - C. ✓ The kinase adds a phosphate group from ATP to the relay protein
Why: A kinase transfers a phosphate group from ATP onto the relay protein.
The added phosphate changes the protein’s shape.
The new shape switches the protein’s activity on or, for some proteins, off.
A steroid hormone enters a cell and binds its receptor in the cytosol.
What does the bound receptor do next?
- A. ✓ The bound receptor moves into the nucleus and attaches to the DNA
- B. The bound receptor switches on a relay of proteins in the cytosolAn intracellular receptor needs no relay.
With the hormone bound, the receptor itself moves into the nucleus and attaches to the DNA. - C. The bound receptor leaves the cell together with the hormoneThe bound pair moves inward, into the nucleus, where the genes are.
There the bound receptor attaches to the DNA.
Why: With the hormone bound, the receptor itself moves into the nucleus.
In the nucleus the bound receptor attaches to the DNA.
Attached to the DNA, it changes the expression of particular genes.
In a eukaryotic cell, too, a regulatory protein binds the DNA beside a gene.
The protein helps RNA polymerase bind the promoter, or blocks it from binding there.
A signal switches such a protein on: a kinase adds a phosphate group to it, or a hormone binds it.
Video: Watch: The protein beside the gene
The gene map is drawn with its promoter and a regulatory sequence beside it. A rounded protein settles onto the regulatory sequence, RNA polymerase docks on the promoter, and RNA strands come off the gene. A kinase adds a phosphate group to a second protein floating above the DNA, and that protein settles onto the DNA.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L28b.mp4
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How does a pancreas cell come to transcribe the insulin gene, while a muscle cell leaves it alone?
Suppose a gene in a human cell is drawn as a line, with its promoter just before it. Beside the promoter sits a regulatory sequence, a short stretch of DNA.
In a eukaryotic cell, RNA polymerase on its own rarely binds a promoter. So with nothing on the regulatory sequence, the gene is rarely transcribed.
One protein in the cell has a binding site that fits the regulatory sequence. So that protein sits on the regulatory sequence, beside the promoter.
While the protein sits there, RNA polymerase binds the promoter far more often. So RNA polymerase transcribes the gene often.
Other proteins of this kind do the opposite.
While such a protein sits beside the promoter, it blocks RNA polymerase from binding there. So the gene is rarely transcribed.
In a eukaryotic cell, a regulatory protein that binds the DNA beside a gene and helps or blocks RNA polymerase binding there is called a , because it is one factor deciding how often the gene is transcribed.
A transcription factor is a protein. The regulatory sequence it binds is DNA.
The cell builds each transcription factor from one of its own genes, as it builds every protein.
A signal from outside the cell often switches a transcription factor on.
Suppose a growth signal reaches the cell. A kinase adds a phosphate group from ATP to a transcription factor.
The added phosphate changes the factor’s shape, and the new shape fits the regulatory sequence. So the factor binds the DNA.
Now suppose a steroid hormone enters the cell and binds its receptor. The bound receptor moves into the nucleus and binds the DNA beside a gene.
The bound receptor is itself a transcription factor.
Go back to the pancreas cell. The pancreas cell holds a transcription factor whose binding site fits the regulatory sequence beside the insulin gene.
So RNA polymerase transcribes the insulin gene there.
The muscle cell holds no such factor. So RNA polymerase rarely transcribes its insulin gene.
Each kind of cell holds its own set of transcription factors, and so transcribes its own set of genes.
What you are expected to know Describe a transcription factor: a eukaryotic regulatory protein that binds the DNA beside a gene and helps or blocks RNA polymerase binding there, switched on by a kinase adding a phosphate group to it or by a hormone binding it.
A transcription factor sits on the DNA beside a gene in a liver cell.
Which kind of molecule is the transcription factor?
- A. A stretch of DNAThe stretch of DNA beside the gene is the regulatory sequence.
The transcription factor is the protein bound to it. - B. ✓ A protein
- C. An mRNAAn mRNA is a copy of a gene, built by RNA polymerase.
A transcription factor is a protein, built by a ribosome from its own gene’s mRNA.
Why: A transcription factor is a regulatory protein.
It binds a stretch of DNA beside the gene, the regulatory sequence.
So the factor is a protein, and what it binds is DNA.
A transcription factor binds the DNA beside a gene.
Which of the following does the transcription factor change?
- A. The amino acids the gene’s protein is built fromThe gene’s base sequence sets its protein’s amino acids, and a transcription factor changes no base.
It changes how often RNA polymerase binds the promoter. - B. The bases the gene carries along its whole lengthA transcription factor binds the DNA and changes no base in it.
It helps or blocks RNA polymerase binding the promoter. - C. ✓ How often RNA polymerase binds the promoter and transcribes the gene
Why: A transcription factor binds the DNA beside the gene.
Bound there, it helps or blocks RNA polymerase binding the promoter.
So it changes how often the gene is transcribed, and nothing in the gene itself.
Suppose a growth signal reaches a cell. Normally a kinase then adds a phosphate group to a transcription factor, and within an hour RNA polymerase transcribes gene G often. Now imagine a biologist adds a chemical that blocks the kinase before the signal arrives. Gene G stays rarely transcribed.
(a) Explain why gene G stays rarely transcribed. (2 pt)
Frame Gene G stays rarely transcribed because …
Without the phosphate group, the factor keeps its resting shape.
The resting shape does not fit the regulatory sequence beside gene G.
So the factor stays off the DNA there.
With no factor bound beside the promoter, RNA polymerase rarely binds it.
So RNA polymerase rarely transcribes gene G.
- Award 1 point for: the blocked kinase adds no phosphate group, so the transcription factor keeps its resting shape and does not bind the DNA beside gene G.
- Award 1 point for: with no transcription factor bound, RNA polymerase rarely binds the promoter, so the gene is rarely transcribed.
A biologist finds a transcription factor bound beside a gene in a liver cell. A student says: “Somewhere in this liver cell’s DNA there is a gene for this factor.”
Is the student correct?
- A. No: a transcription factor is a stretch of DNA beside the gene, so no gene codes for itA transcription factor is a protein, and the stretch of DNA it binds is the regulatory sequence.
The cell builds the factor from one of its own genes. - B. ✓ Yes: the factor is a protein, and a cell builds every protein it holds from one of its own genes
Why: A transcription factor is a protein.
A cell builds each of its proteins from one of its own genes.
So the liver cell’s DNA carries a gene for this factor.
Suppose a protein in a plant cell binds the DNA beside a gene only after a kinase has added a phosphate group to the protein. Bound there, the protein blocks RNA polymerase from binding the promoter.
Is the bound protein a transcription factor?
- A. ✓ Yes
- B. NoThe bound protein sits beside a gene and blocks RNA polymerase from binding the promoter.
A protein that helps or blocks RNA polymerase binding there is a transcription factor.
Why: The bound protein binds the DNA beside a gene.
Bound there, the protein blocks RNA polymerase from binding the promoter.
A protein that binds the DNA beside a gene and helps or blocks RNA polymerase binding there is a transcription factor.
Go back to the muscle cell and the pancreas cell, which carry the same DNA.
Both cells carry the insulin gene, and only the pancreas cell transcribes it.
In the pancreas cell, a transcription factor sits on the DNA beside the insulin gene and helps RNA polymerase bind the promoter.
The muscle cell holds no such factor. So RNA polymerase rarely transcribes its insulin gene.
Each kind of cell expresses its own set of genes, and the proteins that set makes give the cell its job.
75Quick quiz: transcription factor mixed practice
RNA polymerase, bound to the promoter and copying the gene into RNA.
Is this molecule a transcription factor?
- A. YesRNA polymerase is the enzyme that copies the gene.
A transcription factor is the protein beside the gene that helps or blocks RNA polymerase binding the promoter. - B. ✓ No
Why: RNA polymerase copies the gene into RNA.
A transcription factor helps or blocks RNA polymerase binding the promoter; it is a different protein.
The regulatory sequence, a stretch of DNA beside the gene.
Is this molecule a transcription factor?
- A. YesThe regulatory sequence is DNA.
A transcription factor is the protein that binds it. - B. ✓ No
Why: A transcription factor is a protein.
The regulatory sequence is the stretch of DNA that the factor binds, so it is not the factor.
A protein bound to the DNA beside a gene in a nerve cell, helping RNA polymerase bind the promoter.
Is this molecule a transcription factor?
- A. ✓ Yes
- B. NoThis protein binds the DNA beside a gene and helps RNA polymerase bind the promoter.
A protein that does this is a transcription factor.
Why: The protein binds the DNA beside a gene.
Bound there, it helps RNA polymerase bind the promoter.
A protein that does this is a transcription factor.
A protein that sits on the DNA beside a gene in a skin cell and blocks RNA polymerase from binding the promoter.
Is this molecule a transcription factor?
- A. ✓ Yes
- B. NoA transcription factor helps or blocks RNA polymerase binding the promoter.
This protein blocks it, so it is a transcription factor.
Why: The protein binds the DNA beside a gene.
Bound there, it blocks RNA polymerase from binding the promoter.
A protein that helps or blocks RNA polymerase binding there is a transcription factor.
A kinase in the cytosol that adds a phosphate group to a regulatory protein.
Is this molecule a transcription factor?
- A. YesThe kinase switches the regulatory protein on; it binds no DNA itself.
A transcription factor binds the DNA beside a gene. - B. ✓ No
Why: A transcription factor binds the DNA beside a gene.
The kinase adds a phosphate group to a protein in the cytosol and binds no DNA.
So the kinase is not a transcription factor.
What is a transcription factor?
- A. ✓ A protein that binds the DNA beside a gene and helps or blocks RNA polymerase binding there
- B. A protein that copies a gene into RNA once it has bound the promoterThe protein that copies a gene into RNA is RNA polymerase.
A transcription factor binds the DNA beside the gene and helps or blocks RNA polymerase binding the promoter. - C. A protein that binds a gene’s mRNA and blocks the ribosome from reading itA transcription factor acts on the DNA, before any mRNA exists.
It binds beside the gene and helps or blocks RNA polymerase binding the promoter.
Why: In a eukaryotic cell, a regulatory protein that binds the DNA beside a gene and helps or blocks RNA polymerase binding there is called a transcription factor.
A eukaryotic cell controls how often RNA polymerase transcribes one of its genes.
(a) State what a transcription factor is. (1 pt)
- Award 1 point for: a protein that binds DNA near (beside) a gene and helps or blocks (raises or lowers) RNA polymerase binding there. Either of ‘helps’ or ‘blocks’ alone completes it.
83Mixed practice mixed practice
A kidney cell and a bone cell from one person are tested for the mRNA of three genes. The table gives the amount of each mRNA in arbitrary units.
Which gene codes for a tissue-specific protein?
- A. Gene JBoth cells hold 36 units of gene J’s mRNA, so both kinds of cell make its protein.
A tissue-specific protein is made by one kind of cell alone. - B. Gene KBoth cells hold about the same amount of gene K’s mRNA, so both kinds make its protein.
A tissue-specific protein is made by one kind of cell alone. - C. ✓ Gene M
Why: The bone cell holds 140 units of gene M’s mRNA and the kidney cell holds none.
So the bone cell transcribes gene M and the kidney cell does not.
A protein that one kind of cell makes and other kinds do not is a tissue-specific protein.
Suppose a growth signal reaches a cell. Within an hour, a kinase has acted, a transcription factor sits on the DNA beside one of the cell’s genes, and RNA polymerase transcribes that gene often.
Which of the following happens first after the signal arrives?
- A. The factor binds the DNA beside the geneThe factor binds the DNA only in its new shape.
First a kinase adds a phosphate group to the factor, and the added phosphate changes its shape. - B. RNA polymerase binds the promoter oftenRNA polymerase binds the promoter often only once a transcription factor sits beside it.
First a kinase adds a phosphate group to the factor. - C. ✓ A kinase adds a phosphate group to the factor
Why: The signal switches on a kinase.
The kinase adds a phosphate group to a transcription factor, and the factor changes shape.
The factor then binds the DNA beside the gene.
Then RNA polymerase binds the promoter often.
A skin cell and a bone cell from one person transcribe different sets of genes. A student says: “The two cells hold different sets of transcription factors.”
Is the student correct?
- A. No: the two cells hold the same proteins beside their genes; the genes they carry differThe two cells carry the same genes.
They transcribe different sets of them because they hold different transcription factors. - B. ✓ Yes: each kind of cell holds its own set of transcription factors
Why: A transcription factor helps or blocks RNA polymerase binding beside particular genes.
The two cells carry the same genes and transcribe different sets of them.
So the two cells hold different sets of transcription factors.
A steroid hormone has bound its receptor in a liver cell. The bound receptor sits on the DNA beside gene H, and RNA polymerase transcribes gene H often. Now imagine the hormone is removed from the cell.
What happens to how often RNA polymerase transcribes gene H?
- A. ✓ Transcription of gene H falls
- B. Transcription of gene H stays the sameThe receptor binds the DNA only with the hormone bound.
With the hormone gone, the receptor leaves the DNA, and RNA polymerase rarely binds the promoter. - C. Transcription of gene H risesThe bound receptor was helping RNA polymerase bind the promoter.
With the hormone gone, that help is gone, and RNA polymerase rarely binds the promoter.
Why: The receptor binds the DNA beside gene H only with the hormone bound.
With the hormone gone, the receptor leaves the DNA.
With no factor bound beside the promoter, RNA polymerase rarely binds it.
So transcription of gene H falls.
Suppose a biologist counts the genes each of two cells from one person is transcribing: about 9,000 in a skin cell and about 12,000 in a nerve cell.
Which cell carries more genes in its DNA?
- A. The skin cell carries more genesTranscribing fewer genes is not carrying fewer genes.
Both cells carry the whole genome; the skin cell transcribes a smaller set of it. - B. The nerve cell carries more genesTranscribing more genes is not carrying more genes.
Both cells carry the whole genome; the nerve cell transcribes a larger set of it. - C. ✓ Neither: both cells carry the same genes
Why: Every body cell carries the whole genome.
The two counts are the genes each cell is transcribing, not the genes it carries.
So the two cells carry the same genes, and the nerve cell transcribes more of them.
Suppose a biologist gives a skin cell the gene for the transcription factor that switches on a muscle cell’s muscle-protein genes, and the skin cell builds that factor.
What happens to the skin cell’s copies of those genes?
- A. ✓ RNA polymerase begins transcribing them
- B. Nothing: a skin cell carries no muscle-protein genesEvery body cell carries the whole genome, the muscle-protein genes with it.
The factor binds beside those genes and helps RNA polymerase bind their promoters. - C. The genes are removed from the skin cell’s DNAA transcription factor removes no gene.
It binds the DNA beside the muscle-protein genes and helps RNA polymerase bind their promoters.
Why: Every body cell carries the whole genome, so the skin cell carries the muscle-protein genes.
The factor binds the DNA beside those genes.
Bound there, it helps RNA polymerase bind their promoters.
So RNA polymerase begins transcribing them.
Suppose a biologist compares two kinds of cell from one bean plant: cells of a developing seed and cells of a leaf. The seed cells are packed with a storage protein, and the leaf cells hold none of it.
(a) Predict what a test for the storage protein’s gene finds in a leaf cell. (1 pt)
Every cell of the plant carries the same genome, whether or not it transcribes the gene.
- Award 1 point for: the gene is present in the leaf cell (every cell carries the whole genome).
(b) Explain why RNA polymerase transcribes this gene in the seed cells and why the leaf cells lack the storage protein. Name the kind of protein that decides which cells transcribe the gene. (2 pt)
Frame RNA polymerase transcribes the gene in the seed cells because …
The bound factor helps RNA polymerase bind the gene’s promoter.
So RNA polymerase transcribes the gene often.
The leaf cells carry the gene but hold no such factor.
With nothing bound beside the promoter, RNA polymerase rarely binds it.
So the leaf cells rarely transcribe the gene and build none of the storage protein.
- Award 1 point for: the seed cells hold a transcription factor that binds the DNA beside the gene and helps RNA polymerase bind the promoter, so the gene is transcribed there.
- Award 1 point for: the leaf cells carry the gene but lack that factor, so RNA polymerase rarely binds the promoter and the gene is rarely transcribed.
Glossary
- tissue-specific protein
- A protein that one kind of cell makes and other kinds do not, such as insulin in a pancreas cell. A tissue is a group of cells of one kind.
- differentiation
- A cell becoming one kind of cell and not another, by expressing one set of its genes. Differentiation removes no genes: the cell keeps its whole genome.
- differential gene expression
- Different kinds of cell expressing different sets of the same genes. Every body cell carries the same genome, and each kind transcribes its own set of it.
- transcription factor
- In a eukaryotic cell, a regulatory protein that binds the DNA beside a gene and helps or blocks RNA polymerase binding there. It is a protein; the regulatory sequence it binds is DNA.
APBIO-U06-L28B Genes that answer the same call
Suppose a human cell is warmed to 42 °C for a few minutes. Within minutes it switches on dozens of genes at once. They are the genes for the proteins that protect other proteins from heat.
These genes sit on different chromosomes, and each has its own promoter. In a bacterium, genes that work together often sit in one operon behind one promoter. How does a eukaryotic cell switch on dozens of scattered genes with one signal?
Unit 6 · Gene Expression and Regulation
1One sequence, many genes
A eukaryotic cell controls one of its genes with a protein.
Which of the following is a transcription factor?
- A. The stretch of DNA beside the gene where a regulatory protein bindsThe stretch of DNA beside a gene is the regulatory sequence.
The transcription factor is the protein that binds it. - B. The enzyme that binds the promoter and copies the gene into RNAThe enzyme that copies a gene into RNA is RNA polymerase.
A transcription factor is the protein that helps or blocks RNA polymerase binding beside the gene. - C. ✓ A protein that binds the DNA beside the gene and helps or blocks RNA polymerase
Why: A transcription factor is a protein.
It binds the DNA beside a gene and helps or blocks RNA polymerase binding there.
The three genes of E. coli’s lac operon switch on together.
What do the three genes share?
- A. ✓ One promoter for all three genes
- B. A promoter in front of each of the three genesAn operon’s genes sit under one promoter, not one each.
RNA polymerase binds that one promoter and transcribes all three genes as one mRNA. - C. A ribosome that builds all three enzymesMany ribosomes read the operon’s mRNA.
What the three genes share is one promoter and the one mRNA that carries them.
Why: An operon’s genes sit under one promoter.
RNA polymerase binds that promoter once and moves through all the genes.
So one mRNA carries all three genes.
How do scattered genes switch on together?
Each of the heat-shock genes carries the same short regulatory sequence.
One transcription factor, activated by heat, binds that sequence wherever it occurs.
One signal activates one factor, and that one factor switches on dozens of genes. A eukaryotic cell does coordinate regulation this way.
The genes can sit on different chromosomes. What they share is the sequence, not the location.
An operon gives one mRNA for the group. A shared factor gives one mRNA per gene.
Video: Watch: One sequence, many genes
Two heat-shock genes are drawn on two chromosomes, each with its own promoter. The same short box appears beside each promoter. A rounded factor floats free above the DNA. The cell is warmed, the factor settles onto both boxes at once, and RNA strands come off both genes.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L28Ba.mp4
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Go back to the human cell warmed to 42 °C. Look at two of its heat-shock genes, one on each of two chromosomes.
Each gene has its own promoter.
No promoter is shared. So RNA polymerase must bind each gene separately.
Beside each promoter sits a short regulatory sequence. The sequence beside the first gene has the same order of bases as the sequence beside the second gene.
Every one of the dozens of heat-shock genes carries a copy of this same short sequence beside its promoter.
The cell holds one transcription factor whose binding site fits this sequence. At 37 °C, the factor floats free in the nucleus and does not bind the DNA.
So at 37 °C, RNA polymerase rarely binds the promoters of the heat-shock genes, and the genes are nearly off.
Now the cell is warmed to 42 °C.
The heat changes the factor’s shape. The factor now binds its sequence.
The same sequence occurs beside every heat-shock gene. So copies of the factor bind beside every one of them.
A bound factor helps RNA polymerase bind the promoter next to it. So RNA polymerase binds the promoter of every heat-shock gene.
RNA polymerase transcribes each gene from its own promoter. So each gene gives its own mRNA, and the ribosomes build a protective protein from each.
One signal switched on dozens of genes at once.
Switching a group of genes on together with one control is coordinate regulation. One shared factor is how a eukaryotic cell does it.
The heat-shock genes sit on different chromosomes. What they share is the short sequence beside each of them, not a place on the DNA.
An operon’s genes share one promoter and come out as one mRNA. Genes with a shared factor each keep their own promoter and come out as one mRNA each.
What you are expected to know Explain how a eukaryotic cell switches on scattered genes together: each carries the same regulatory sequence, one transcription factor binds that sequence wherever it occurs, and RNA polymerase transcribes each gene from its own promoter.
The heat-shock genes of a human cell switch on together when the cell is warmed.
Which of the following do the heat-shock genes share?
- A. The same chromosomeThe heat-shock genes sit on different chromosomes.
What they share is the short regulatory sequence beside each of them. - B. ✓ The same short regulatory sequence
- C. The same promoterEach heat-shock gene has its own promoter.
What they share is the short regulatory sequence beside each promoter.
Why: Each heat-shock gene carries the same short regulatory sequence beside its promoter.
One heat-activated transcription factor binds that sequence wherever it occurs.
So the genes switch on together.
A human cell warmed to 42 °C transcribes its dozens of heat-shock genes.
How many mRNAs carry them?
- A. One mRNA for the whole groupOne mRNA for a whole group comes from an operon’s one promoter.
Each heat-shock gene has its own promoter, so RNA polymerase transcribes each gene into its own mRNA. - B. ✓ One mRNA per gene
Why: Each heat-shock gene has its own promoter.
RNA polymerase transcribes each gene from its own promoter.
So each gene gives its own mRNA.
Suppose cadmium, a poisonous metal, reaches a cell. Within an hour the cell switches on a set of genes that sit on several chromosomes. The protein of each gene binds cadmium and makes it harmless.
(a) Explain how one signal switches on genes that sit on several chromosomes. (2 pt)
Frame One signal switches on all of them because …
The cadmium activates one transcription factor.
That factor binds the sequence wherever it occurs, on whichever chromosome.
So the factor binds beside every one of the genes.
A bound factor helps RNA polymerase bind the promoter next to it.
So RNA polymerase transcribes every gene in the set.
- Award 1 point for: each gene in the set carries the same regulatory sequence, and one transcription factor activated by the signal binds that sequence wherever it occurs.
- Award 1 point for: so the factor binds beside every gene in the set, whichever chromosome it sits on, and RNA polymerase transcribes each of them (from its own promoter).
A student says: “Genes that switch on together must sit side by side on one chromosome.”
Is the student correct?
- A. ✓ No: the genes share a regulatory sequence and can sit on different chromosomes
- B. Yes: one signal reaches only genes that sit together on one chromosomeA transcription factor binds its sequence wherever it occurs.
Genes carrying that sequence switch on together from different chromosomes.
Why: One transcription factor binds its regulatory sequence wherever it occurs.
Genes that carry that sequence switch on together.
So the genes can sit on different chromosomes.
Suppose a hormone enters a cell and binds a transcription factor. The bound factor binds the same short sequence beside eighteen genes, and the factor switches on every one of them. Now imagine one of the eighteen genes carries a changed sequence, one that the factor’s binding site does not fit. The hormone arrives.
What happens to the eighteen genes?
- A. ✓ Only the gene with the changed sequence stays off
- B. All eighteen genes stay off, changed or notThe factor still binds its sequence beside the other seventeen genes.
Only the gene with the changed sequence stays off. - C. All eighteen genes switch on, changed or notThe factor’s binding site does not fit the changed sequence.
So the factor does not bind beside that gene, and that gene stays off.
Why: A transcription factor binds only a sequence its binding site fits.
The factor still binds beside the seventeen genes with the unchanged sequence, so they switch on.
It does not bind beside the gene with the changed sequence, so that gene stays off.
This table compares an operon with a group of genes that share a factor: what the genes share, how many mRNAs, where the genes sit, and which cells use each.
34One factor switches on the next
A signal pathway passes its message from a receptor to relay 1, then to relay 2, then to the response. Relay 1 is broken.
Which parts of the pathway are still activated when the signal arrives?
- A. Relay 2 and the responseRelay 2 and the response are downstream of the break.
The message never reaches them, so they are not activated. - B. ✓ The receptor
- C. None of themThe receptor is upstream of the break.
The signal still activates the receptor as normal.
Only the parts downstream of the break are not activated.
Why: The message passes one way, from the receptor to the response.
The receptor is upstream of the break, so the signal still activates the receptor.
Relay 2 and the response are downstream of the break.
The message never reaches them, so they are not activated.
How does a cell in an embryo become one kind of cell, step by step?
During development a signal switches on one transcription factor.
That factor switches on the genes for the next factors, and so on. So the cell commits to its fate step by step.
Remove one factor from the chain and every gene downstream of it stays off, while everything upstream is unchanged.
Video: Watch: One factor switches on the next
A signal arrives at a cell in an embryo. The first factor settles beside the genes for the second factor, and RNA comes off them. The second factor appears and settles beside the genes for the third. The third settles beside the cell’s final genes. Then the second factor is lifted out: the first row keeps going, and every row below it goes dark.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L28Bb.mp4
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Suppose a cell sits in an early embryo. A signal from the cells beside it reaches the cell.
The signal switches on one transcription factor: the first factor in a chain.
The first factor binds the regulatory sequence beside a small group of genes. Those genes code for a second transcription factor.
RNA polymerase transcribes those genes, and the ribosomes build the second factor.
The second factor binds the regulatory sequence beside another group of genes. Those genes code for a third factor, and the ribosomes build it.
The third factor binds beside the genes for the proteins that do the finished cell’s job. Those genes switch on, and the cell has become one kind of cell.
Each factor is a protein. So the cell builds each factor from one of its own genes.
Each factor switches on the genes for the next factor. So the genes switch on in a fixed order, one group after another.
The chain passes the message one way, from the signal to the final genes. Just as in a signal pathway, the signal is upstream and the final genes are downstream.
Now imagine the second factor is removed from the cell. The signal still arrives, and the first factor still binds beside the genes for the second factor.
But no second factor binds beside the genes for the third factor. So RNA polymerase does not transcribe them, and the cell makes no third factor.
With no third factor, the final genes stay off too. Everything downstream of the missing factor stays off.
Everything upstream of the missing factor is unchanged. The signal still switches on the first factor, and the first factor still binds its sequence.
Some steps of a chain switch on the genes for apoptosis. The webbing cells between an embryo’s fingers and toes dismantle themselves this way.
What you are expected to know Explain how a cell in an embryo switches its genes on in order: a signal switches on one transcription factor, and that factor switches on the genes for the next factor.
What you are expected to know Predict a break in the chain: remove one factor, and every gene downstream of it stays off while everything upstream is unchanged.
A chain of three transcription factors is drawn below with its words removed. The seated factors are marked J, K and L, and an arrow leads from each row’s genes to the factor those genes make.
Which factor switches on the genes for factor L?
- A. Factor JFactor J sits beside the genes whose arrow leads to factor K.
So factor J switches on the genes for factor K, not for factor L. - B. ✓ Factor K
- C. Factor LFactor L is made from the middle row’s genes.
The factor seated beside those genes is factor K.
Why: The arrow from the middle row’s genes leads to factor L, so those are the genes for factor L.
Factor K is the factor seated beside the middle row’s genes.
So factor K switches on the genes for factor L.
Suppose that in an embryo a signal switches on a first transcription factor. The first factor switches on the genes for a second factor, and the second factor switches on the genes for a skin pigment. In one embryo cell, the gene for the second factor is deleted. The signal still arrives.
(a) Explain why the pigment genes stay off in that cell. (2 pt)
Frame The pigment genes stay off because …
The signal still switches on the first factor.
The first factor still binds beside the genes for the second factor.
But the gene for the second factor is deleted.
So the ribosomes build no second factor.
The second factor is the protein that binds beside the pigment genes.
With no second factor bound there, RNA polymerase does not transcribe the pigment genes.
- Award 1 point for: the gene for the second factor is deleted, so the cell makes no second factor (the first factor still works, upstream of the break).
- Award 1 point for: the second factor is what binds beside the pigment genes, so with none made, RNA polymerase does not transcribe the pigment genes — they are downstream of the missing factor.
A student says: “Remove the second factor from the chain, and the first factor switches off too, because the chain is broken.”
Is the student correct?
- A. ✓ No: the first factor is upstream of the break, so it is unchanged; only the genes downstream stay off
- B. Yes: a break anywhere in a chain switches off every factor in it, upstream and downstream alikeThe chain passes its message one way, from the signal to the final genes.
A missing factor stops everything downstream of it and changes nothing upstream.
Why: The signal switches on the first factor, and the second factor plays no part in that.
So the first factor is unchanged when the second is removed.
Only the genes downstream of the second factor stay off.
Suppose a cell in an embryo lacks the first muscle factor: the transcription factor that a signal switches on first on the way to becoming a muscle cell.
What happens to the later muscle genes in that cell?
- A. The later muscle genes switch on as normalThe later muscle genes are downstream of the first muscle factor.
With no first factor, the chain never starts, so they stay off. - B. The later muscle genes switch on lateA chain does not start late on its own.
The first factor switches on the genes for the next, and with no first factor those genes stay off. - C. ✓ The later muscle genes stay off
Why: The first muscle factor switches on the genes for the next factor.
The cell lacks the first factor, so those genes stay off, and so does every gene downstream.
So the later muscle genes stay off, and the cell does not become a muscle cell.
Go back to the human cell warmed to 42 °C.
Dozens of its genes sit on different chromosomes, and each carries the same short sequence.
One heat-activated factor binds that sequence everywhere. So all of them switch on within minutes.
64Mixed practice mixed practice
Suppose a biologist finds that forty of a cell’s genes switch on together within an hour of a hormone arriving, and that the forty genes lie on several chromosomes. A student says: “Each of the forty genes must carry the regulatory sequence that the hormone’s transcription factor binds.”
Is the student correct?
- A. No: one signal switches on genes together only when they sit side by side on one chromosomeGenes on several chromosomes switched on together, so sitting side by side is not what they share.
What they share is the sequence the factor binds. - B. ✓ Yes: a factor switches on a gene only where its sequence occurs, so every one of the forty genes carries it
Why: A transcription factor binds only its regulatory sequence.
It switched on all forty genes, on several chromosomes.
So every one of the forty genes carries that sequence.
The chain of three transcription factors is drawn below with its words removed; the seated factors are marked J, K and L. In one cell where this chain should be at work, a biologist finds factor J bound to the DNA and factor K present, but no factor L, and the genes at the end of the chain are off.
Where is the break?
- A. At the genes for factor KFactor K is present, so the genes for factor K were transcribed and translated.
The break lies further down the chain. - B. At the genes at the end of the chainThe genes at the end of the chain are off because factor L is missing.
They are downstream of the break, not the break itself. - C. ✓ At the genes for factor L
Why: Factor K is present, so everything up to and including its genes worked.
Factor L is absent although factor K is there to switch its genes on.
So the break is at the genes for factor L, and the genes downstream of it are off.
Imagine the heat-activated transcription factor is removed from a human cell. The cell is then warmed to 42 °C.
What happens to the heat-shock genes?
- A. ✓ All of them stay off
- B. Some of them switch onEvery heat-shock gene relies on the same factor.
With the factor gone, none of them is switched on. - C. All of them switch onThe heat switches the genes on only through the factor.
With the factor gone, no protein binds beside the heat-shock genes.
Why: Heat switches on the heat-shock genes only by activating the one factor.
The factor is gone, so nothing binds the sequence beside the genes.
So all the heat-shock genes stay off.
In a chain of three transcription factors, imagine the gene for the third factor is deleted.
Which factors does the cell still make?
- A. None of the threeThe signal still switches on the first factor, and the first still switches on the genes for the second.
Both factors are upstream of the missing factor. - B. The first onlyThe second factor is made from genes that the first factor switches on.
The third factor plays no part in that, so the cell still makes the second. - C. ✓ The first and the second
Why: The first and second factors are upstream of the third.
The signal still switches on the first, and the first still switches on the genes for the second.
So the cell still makes the first and the second.
A transcription factor binds the sequence 5′-GAAGCTTC-3′ wherever it occurs. The table gives the short sequence beside the promoter of three genes.
Which genes switch on when the factor is activated?
- A. Gene R onlyGene S carries 5′-GAAGCTTC-3′ beside its promoter too.
The factor binds that sequence wherever it occurs, so gene S switches on as well. - B. ✓ Genes R and S
- C. Genes R, S and TGene T carries 5′-CCATGGAT-3′, a different sequence.
The factor’s binding site does not fit it, so gene T stays off.
Why: The factor binds 5′-GAAGCTTC-3′ wherever it occurs.
Genes R and S carry that sequence beside their promoters; gene T carries a different one.
So genes R and S switch on.
In a chain of three transcription factors, imagine the gene for the first factor is deleted.
Which factors does the cell still make?
- A. ✓ None of the three
- B. The third onlyThe third factor is made from genes that the second factor switches on.
With no first factor there is no second, and so no third. - C. The second and the thirdThe genes for the second factor are switched on by the first factor.
With no first factor, the cell makes no second factor, and so no third.
Why: The first factor switches on the genes for the second.
With no first factor, the cell makes no second factor.
With no second factor, it makes no third.
So the cell makes none of the three.
Suppose that in an embryo a signal switches on a first transcription factor. The first factor switches on the genes for a second factor. The second factor switches on thirty genes on several chromosomes. The cell then becomes a cell that builds the tough fibers of a tendon.
(a) Explain how the second factor switches on thirty genes on several chromosomes at once. (2 pt)
Frame The second factor switches on all thirty at once because …
The second factor binds that sequence wherever it occurs, on whichever chromosome.
So the second factor binds beside every one of the thirty genes.
A bound factor helps RNA polymerase bind the promoter next to it.
So RNA polymerase transcribes each of the thirty genes from its own promoter.
- Award 1 point for: each of the thirty genes carries the same regulatory sequence, and the second factor binds that sequence wherever it occurs.
- Award 1 point for: so the factor binds beside every one of the genes, on whichever chromosome, and RNA polymerase transcribes each (from its own promoter).
(b) Explain why the thirty genes switch on only after the second factor has been built, one step after the signal arrives. (2 pt)
Frame The thirty genes switch on only after the second factor has been built because …
The first factor binds beside the genes for the second factor.
RNA polymerase transcribes those genes, and the ribosomes build the second factor.
Only the second factor fits the sequence beside the thirty genes.
So the thirty genes switch on once the second factor exists, one step after the signal.
- Award 1 point for: the signal switches on the first factor, and the first factor switches on the genes for the second factor, which the ribosomes then build.
- Award 1 point for: only the second factor binds the sequence beside the thirty genes, so they switch on one step later, once the second factor has been built.
APBIO-U06-L29 Make the claim
The table above comes from an experiment on two genes, X and Y. A technician measured cells in a dish before and after adding a chemical to the dish. That chemical is the treatment.
The table gives the level of each gene’s mRNA and the level of its protein, in arbitrary units. Gene X: 5 units of mRNA before the treatment, 80 units after, and its protein rises with it. Gene Y: 40 units of mRNA before and 40 units after, and its protein falls by half. What is being controlled in each gene, and where?
Unit 6 · Gene Expression and Regulation
1From the table to the claim
Suppose the poly-A tail is removed from an mRNA the moment the mRNA leaves the nucleus.
Compared with the same mRNA with its tail, how much of its protein does the cell make?
- A. MoreThe poly-A tail protects the mRNA from the enzymes that break RNA down.
Without its tail, the mRNA is destroyed sooner, so the ribosomes build less of its protein. - B. The same amountAn mRNA with no tail is destroyed sooner than the same mRNA with its tail.
A shorter-lived mRNA gives less protein. - C. ✓ Less
Why: The poly-A tail protects the mRNA from the enzymes that break RNA down.
Without the tail, the enzymes destroy the mRNA sooner.
The ribosomes read the mRNA for less time.
So the cell makes less of its protein.
Suppose a bacterium’s gene is measured in a broth without a certain sugar and in the same broth with the sugar added. Its levels are drawn below.
Which kind of gene is it?
- A. ✓ Constitutively expressed
- B. InducibleAn inducible gene’s level is high in one condition only.
This gene reads 62 units without the sugar and 63 units with it: about the same level in both.
Why: The gene reads 62 units without the sugar and 63 units with it.
The two levels differ by one unit only.
So the gene is at about the same level in both conditions.
A gene at about the same level in every condition is constitutively expressed.
Suppose a cell begins transcribing one of its genes less often than before.
What happens to the level of that gene’s protein in the cell?
- A. The protein level risesLess transcription gives less mRNA of the gene.
The ribosomes build less of the protein from less mRNA. - B. ✓ The protein level falls
- C. The protein level stays the sameThe level of a protein follows how often its gene is transcribed.
Less transcription gives less mRNA, so the ribosomes build less of the protein.
Why: How often a cell transcribes a gene sets how much of that gene’s protein the cell holds.
The cell transcribes the gene less often, so it holds less of the mRNA.
The ribosomes build the protein from that mRNA.
So the level of the protein falls.
Suppose enzymes remove the methyl groups from a gene’s promoter in a plant cell.
What happens to the gene’s transcription?
- A. ✓ Transcription rises
- B. Transcription fallsFewer methyl groups on the promoter loosen the DNA’s winding.
RNA polymerase reaches the gene more easily, so transcription rises. - C. Transcription stays the sameRemoving methyl groups changes how tightly the DNA is wound.
Looser winding lets RNA polymerase reach the gene more often, so transcription rises.
Why: Methyl groups on a promoter wind the DNA tighter.
Enzymes remove them, so the DNA winds looser.
RNA polymerase reaches a loosely wound gene more easily.
So transcription of the gene rises.
What can a table of mRNA and protein levels tell you about control?
A claim about regulation names four parts: which gene, which direction, where the control acts, and the evidence line that shows it.
If the mRNA changes and the protein follows, the control acts at transcription.
If the mRNA is unchanged and the protein changes, the control acts after transcription: on how often ribosomes translate the mRNA, or on how long the protein lasts.
Gene X: its mRNA rose from 5 to 80 units and its protein rose too. So transcription of X was induced by the treatment.
Video: Watch: From the table to the claim
The table of genes X and Y appears one row at a time. A ring moves along gene X’s mRNA row, then along gene X’s protein row. A fourth column fills in beside gene X: where the control acts. The ring then moves along gene Y’s two rows, and the fourth column beside gene Y waits to be filled in.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L29a.mp4
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The table below lists genes X and Y, each measured before and after the treatment: the level of the gene’s mRNA, and the level of its protein. The units are arbitrary, and the footer says so.
Read gene X’s mRNA row: 5 units before the treatment, 80 units after. Gene X’s mRNA rose.
Read gene X’s protein row: 11 units before, 140 units after. Gene X’s protein rose too.
The protein followed the mRNA. More mRNA gave more protein.
More mRNA of a gene means RNA polymerase transcribed the gene more often. So the treatment acted at transcription: it made RNA polymerase transcribe gene X more often.
Now write the claim. A claim about regulation has four parts.
- Which gene: gene X.
- Which direction: up.
- Where the control acts: at transcription.
- The evidence line: gene X’s mRNA rose from 5 to 80 units, and its protein rose from 11 to 140 units.
Put together, the claim about gene X reads like this.
The treatment raised transcription of gene X. The evidence is that gene X’s mRNA rose from 5 to 80 units and its protein rose with it, from 11 to 140 units.
Two rows decide where the control acts: the mRNA row and the protein row.
If the mRNA changes and the protein follows, the control acts at transcription. RNA polymerase transcribed the gene more often, or less often.
If the mRNA is unchanged and the protein changes, the control acts after transcription.
After transcription, a protein’s level can still change: ribosomes can translate each mRNA more or less often, and the protein itself can last longer or be broken down sooner.
If neither the mRNA nor the protein changes, the treatment does not control that gene.
The direction is whatever the rows show: up when the levels rise, down when they fall.
What you are expected to know Make a claim from a table of mRNA and protein levels, naming its four parts.
Now suppose you make the claim about gene Y the same way, one part at a time.
The table is drawn below with gene Y’s mRNA row ringed.
Is gene Y’s mRNA level after the treatment the same as before?
- A. ✓ Yes
- B. NoGene Y’s mRNA reads 40 units before the treatment and 40 units after.
The two levels are equal, so the mRNA is unchanged.
Why: Gene Y’s mRNA reads 40 units before the treatment and 40 units after.
The two levels are equal.
So the treatment left gene Y’s mRNA unchanged.
The table is drawn below with gene Y’s protein row ringed.
How did gene Y’s protein level change?
- A. Gene Y’s protein roseGene Y’s protein reads 64 units before the treatment and 32 units after.
The level after is lower, so the protein fell. - B. ✓ Gene Y’s protein fell
- C. Gene Y’s protein stayed the sameGene Y’s protein reads 64 units before and 32 units after.
The two levels differ, so the protein changed: it fell by half.
Why: Gene Y’s protein reads 64 units before the treatment and 32 units after.
The level after is half the level before.
So gene Y’s protein fell.
The table is drawn below with gene Y’s two rows ringed. Its mRNA is unchanged, and its protein fell.
Where does the control on gene Y act?
- A. At transcriptionUnchanged mRNA means RNA polymerase transcribed gene Y as often as before.
Transcription did not change, so the control acts after it. - B. ✓ After transcription
Why: Gene Y’s mRNA is unchanged, so RNA polymerase transcribed gene Y as often as before.
Gene Y’s protein fell all the same.
So the protein level changed after transcription: ribosomes translated each mRNA less often, or the protein was broken down sooner.
The control acts after transcription.
The treatment lowered gene Y’s protein level, and the control acts after transcription. The table is drawn below with gene Y’s two rows ringed.
Which sentence is the evidence line for that claim?
- A. ✓ Gene Y’s mRNA stayed at 40 units while its protein fell from 64 to 32 units
- B. Gene Y’s protein fell from 64 to 32 units, so its mRNA must have fallen tooGene Y’s mRNA reads 40 units before and after: it did not fall.
An evidence line states the levels the table shows. - C. Gene Y’s mRNA stayed at 40 units, so the treatment did not control gene YGene Y’s protein fell from 64 to 32 units, so the treatment did control gene Y.
The evidence line names both rows: the unchanged mRNA and the fallen protein.
Why: An evidence line states the levels that show the claim.
The claim rests on two rows: the mRNA unchanged at 40 units, and the protein fallen from 64 to 32 units.
Only the first sentence states both rows as the table shows them.
So the claim about gene Y reads like this.
The treatment lowered gene Y’s protein level, and the control acts after transcription. The evidence is the unchanged mRNA beside the halved protein.
Now suppose a second experiment adds a different chemical to cells in a dish and measures two other genes, R and W, before and after.
The second table is drawn below with gene R’s two rows ringed.
Which claim do the data support about gene R?
- A. The treatment raised transcription of gene RGene R’s mRNA reads 110 units before the treatment and 15 units after.
The mRNA fell, so RNA polymerase transcribed gene R less often. - B. The treatment controls gene R after transcriptionGene R’s mRNA changed, from 110 to 15 units.
A change in the mRNA means the control acts at transcription. - C. ✓ The treatment lowered transcription of gene R
Why: Gene R’s mRNA fell from 110 to 15 units.
Less mRNA means RNA polymerase transcribed gene R less often.
Gene R’s protein fell with it, from 175 to 37 units.
The protein followed the mRNA, so the control acts at transcription, and the direction is down.
The second table is drawn below: genes R and W in cells before and after the technician added the second chemical, in arbitrary units. Gene W’s two rows are ringed.
(a) Make a claim about how the treatment controls gene W: the direction of the change, and where the control acts. (1 pt)
- Award 1 point for a claim that names BOTH the direction (gene W’s protein rose / the treatment raised the protein level) AND where the control acts (after transcription; not at transcription). A claim that the treatment raised transcription of gene W earns no point.
(b) Support your claim with evidence from the table and reasoning. (2 pt)
So RNA polymerase transcribed gene W as often as before.
Gene W’s protein rose from 39 to 135 units all the same.
More protein from the same amount of mRNA means ribosomes translated each mRNA more often, or the protein lasted longer.
So the control acts after transcription, and it raised gene W’s protein level.
- Award 1 point for the evidence: gene W’s mRNA is unchanged (56 units before and after) while its protein rose (39 to 135 units) — both rows cited with their levels.
- Award 1 point for the reasoning: unchanged mRNA means transcription did not change, and more protein from the same mRNA means the mRNA was translated more often (or the protein lasted longer), so the control acts after transcription.
- Accept consistent reasoning from a wrong (a): if the claim in (a) was wrong, award the reasoning point for a reasoning line that follows correctly from the rows the student cites.
The table below sets out the rule: what the mRNA row shows, what the protein row shows, and where the control acts.
An exam question may ask at what level a gene is regulated. That question asks where the control acts: at transcription, or after transcription.
Back to the table of genes X and Y, measured before and after the treatment.
Gene X: more mRNA and more protein. So transcription of X was induced by the treatment.
Gene Y: the same mRNA and less protein. So Y is controlled after transcription.
The claim names the gene, the direction and where the control acts, and it states the levels, in units, that show it.
44Mixed practice mixed practice
Suppose a bacterium’s operon codes for enzymes that break down starch. The table below gives the mRNA of the operon’s genes in plain broth and in the same broth with starch added.
Is this operon an inducible system or a repressible system?
- A. ✓ An inducible system
- B. A repressible systemStarch switches the genes on: 41 units of mRNA in plain broth, 225 units with starch.
A signal that switches the genes on marks an inducible system.
Why: The mRNA reads 41 units in plain broth and 225 units with starch added.
So starch switches the operon’s genes on.
An operon is inducible when its signal switches the genes on.
Suppose a change in the gene for the lac operon’s activator gives an activator that holds its DNA-binding shape at all times, even in a cell with plenty of glucose. The cells grow in a broth of glucose plus lactose.
How much of the lactose enzymes do the cells make?
- A. Very littleLactose is present, so the repressor is off the operator.
This activator binds beside the promoter even with plenty of glucose, so RNA polymerase transcribes the operon at full rate. - B. ✓ A great deal
Why: Lactose binds the repressor, so the repressor leaves the operator.
The changed activator binds beside the promoter even with plenty of glucose.
With the repressor off and the activator bound, RNA polymerase transcribes the three genes at full rate.
So the cells make a great deal of the lactose enzymes.
A gene is measured in two kinds of cell from one person: the cells lining the bladder and the cells of the tongue. The table below gives the level of the gene’s mRNA in each kind of cell.
Which of the following can explain the low level in the tongue cells?
- A. The gene’s promoter has been removed from the tongue cells’ DNAEvery body cell carries the same genome, the gene’s promoter with it.
The tongue cells hold the gene and rarely transcribe it. - B. The gene’s promoter carries a different base sequence in the tongue cellsBoth kinds of cell grew from one fertilized egg by mitosis, so the promoter’s base sequence is the same in both.
What differs is the tags on it. - C. ✓ The gene’s promoter carries many methyl groups in the tongue cells
Why: Both kinds of cell carry the same genome, so both carry the gene and its promoter.
The tongue cells hold 9 units of the mRNA against 245 units, so they rarely transcribe the gene.
A promoter with many methyl groups winds tight, so RNA polymerase rarely reaches it.
A technician measures the lac operon’s mRNA in a normal E. coli strain and in a strain with a changed repressor gene, each in a broth with no lactose and in a broth with lactose. The table is drawn below.
Which change to the changed strain’s repressor explains its row?
- A. The cell makes none of itWith no repressor, nothing blocks RNA polymerase, so the operon would be on in both broths.
The changed strain reads 8 units with no lactose: the operon is off. - B. Its shape no longer fits the operatorA repressor that no longer fits the operator leaves the operon on in both broths.
The changed strain reads 8 units with no lactose: the operon is off. - C. ✓ Lactose no longer binds it
Why: With no lactose, the changed strain reads 8 units, so its repressor still blocks the operon.
With lactose, the changed strain reads 9 units: lactose failed to lift the repressor.
Lactose lifts a repressor by binding it and changing its shape.
So lactose no longer binds this repressor.
Suppose a shortage of oxygen switches on one transcription factor in a cell, and that factor binds one short regulatory sequence wherever it occurs. Four genes on three of the cell’s chromosomes are measured before the shortage and 30 minutes into it; the table is drawn below.
Which genes carry the sequence that the factor binds?
- A. Gene N onlyGene N reads 76 units before and 78 units into the shortage: the factor left it as it was.
The factor’s sequence sits beside the genes that switched on. - B. ✓ Genes G, V and Z
- C. All four: genes G, N, V and ZGene N’s mRNA hardly changed, so the factor did not switch gene N on.
Only the three genes that switched on carry the factor’s sequence.
Why: The factor binds its sequence wherever it occurs.
Genes G, V and Z rose many-fold, so the factor switched them on.
Gene N stayed near 76 units, so the factor did not bind beside it.
So genes G, V and Z carry the sequence.
A gene is measured in cells before and after a treatment; the table is drawn below. A student says: “The protein rose, so the control on this gene acts after transcription.”
Is the student correct?
- A. ✓ No: the mRNA rose with the protein, so the treatment raised transcription of the gene
- B. Yes: a rise in protein shows a control acting after transcriptionThe gene’s mRNA rose from 35 to 105 units, so RNA polymerase transcribed it more often.
The protein followed the mRNA, so the control acts at transcription.
Why: The mRNA rose from 35 to 105 units, so RNA polymerase transcribed the gene more often.
The protein rose with it, from 59 to 177 units.
The protein followed the mRNA, so the control acts at transcription.
So the treatment raised transcription of the gene.
A gene is measured in a bacterium in three broths; the table is drawn below. A student says: “This gene is constitutively expressed, because its level is about the same in every broth.”
Is the student correct?
- A. No: its level is high in every broth, so the gene is switched on, and a switched-on gene is inducibleThe level itself never sorts a gene.
This gene reads 91, 93 and 92 units: about the same in every broth, so it is constitutively expressed. - B. ✓ Yes: its level is about the same in every broth, so it is constitutively expressed
Why: The gene reads 91, 93 and 92 units across the three broths.
The levels differ by two units at most, so the gene is at about the same level in every broth.
A gene at about the same level in every condition is constitutively expressed.
So the student is correct.
Suppose a biologist measures one gene in three kinds of cell from one person: spleen cells, hair-follicle cells and cartilage cells. The table below gives the level of the gene’s mRNA and the level of its protein in each kind of cell, in arbitrary units.
(a) Identify the kind of cell in which the gene is controlled after transcription. (1 pt)
- Award 1 point for: the cartilage cells.
(b) Justify your answer with evidence from the table. (2 pt)
So RNA polymerase transcribes the gene in both kinds of cell.
The cartilage cells hold none of the protein, while the spleen cells hold 170 units.
An mRNA that is present while its protein is absent is not being translated, or enzymes break the protein down as fast as ribosomes build it.
So in the cartilage cells the control acts after transcription.
- Award 1 point for the evidence: the cartilage cells hold the gene’s mRNA (195 units, more than the spleen cells’ 160) and none of its protein (0 units).
- Award 1 point for the reasoning: mRNA present means the gene is transcribed; protein absent from a transcribed gene means the control acts after transcription (the mRNA is not translated, or the protein is broken down). The hair-follicle cells, with no mRNA, do not transcribe the gene at all.
- Accept consistent reasoning from a wrong (a): if the cell named in (a) was wrong, award the reasoning point for a reasoning line that follows correctly from that cell's two rows.
APBIO-U06-P65 Practice questions: Topic 6.5
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 one silent gene 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: Regulation of gene expression, summed up
Which genes and how much decides the phenotype; constitutive and inducible genes; regulatory sequences and proteins; the lac and trp operons; epigenetic tags; one genome, many cell types; transcription factors in sequence; genes regulated together.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-T65-summary.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-T65-summary.mp4
A cone snail's venom-gland cells make an enzyme in its venom. A biologist tests one of the snail's foot cells for the enzyme's gene, for the gene's mRNA and for the enzyme. The table gives the results.
Which of the following describes the enzyme's gene in the foot cell?
- A. The foot cell does not carry the geneThe biologist found the gene in the foot cell.
Every cell of the snail carries the same genome, the enzyme's gene with it. - B. ✓ The foot cell carries the gene and is not transcribing it
- C. The foot cell carries the gene and transcribes it, and no enzyme is built from the mRNAThe biologist found no mRNA of the gene in the foot cell.
Ribosomes translate only an mRNA that exists, and RNA polymerase has made none. - D. The foot cell carries the gene and expresses it: the enzyme is builtThe biologist found no mRNA and no enzyme in the foot cell.
An expressed gene gives both: RNA polymerase makes the mRNA, and ribosomes build the enzyme from it.
Why: The biologist found the gene, so the foot cell carries it.
RNA polymerase makes mRNA only while it transcribes a gene, and the biologist found no mRNA, so RNA polymerase is not transcribing the gene.
With no mRNA, the ribosomes build no enzyme.
Read a gene's level across conditions: about the same in every condition means constitutively expressed; high in one condition only means inducible, and that condition holds its signal. Suppose a biologist measures a bacterium's gene in a broth with no zinc and in the same broth with zinc added. Its levels are drawn.
Which kind of gene is it?
- A. ✓ Constitutively expressed
- B. Inducible, with zinc as its signalThe gene reads 15 units with no zinc and 16 with zinc added: about the same.
An inducible gene's level is high in one condition only. - C. Inducible, with the absence of zinc as its signalThe gene reads 15 and 16 units: its level is about the same whether zinc is there or not.
An inducible gene's level is high in one condition only. - D. Neither: its level is too low to sortThe level itself never sorts a gene.
A gene held low in every condition is constitutively expressed, just like a gene held high in every condition.
Why: The gene reads 15 units with no zinc and 16 with zinc added.
The two levels differ by one unit, so the gene is at about the same level in both conditions.
A gene at about the same level in every condition is constitutively expressed, low as this one is.
A soil bacterium's gene codes for an enzyme. A biologist studies two strains, each with one change in its DNA. The first strain makes the normal amount of the enzyme's mRNA, 100 units, and its enzyme does not work. The second strain makes 5 units of the mRNA, and the enzyme it does make works normally.
Which stretch of DNA did the second strain's change most likely alter?
- A. A codon in the middle of the geneA changed codon changes an amino acid of the enzyme, so the enzyme may stop working.
The second strain's enzyme works normally; only its amount fell. - B. The gene's start codonWith no start codon, the ribosome never assembles on the mRNA and no enzyme is built.
The second strain builds a working enzyme, just less of it. - C. The gene's last codon, before the stop codonA change inside the gene changes the enzyme's amino acids, not how often the gene is transcribed.
The second strain's mRNA fell to 5 units. - D. ✓ The regulatory sequence beside the gene
Why: The regulatory sequence lies beside the gene, and RNA polymerase never copies it into the mRNA.
So a change in it changes how often the gene is transcribed, not the enzyme's amino acids.
The second strain makes 5 units of mRNA instead of 100, and its enzyme works normally.
In the drawing the block is the repressor, and the oval is RNA polymerase, off the DNA. The drawing shows a bacterium's operon for breaking down mannose, a sugar, in a cell with no mannose. The operon works like the lac operon.
Which lettered part is the operator?
- A. QQ is the box that RNA polymerase floats above: the promoter.
The operator is the box the repressor sits on. - B. ✓ U
- C. EE marks the three genes that RNA polymerase transcribes once the repressor has gone.
The operator is the short stretch the repressor sits on. - D. None of Q, U and EThe operator is a stretch of DNA, not the block.
The block is the repressor, and the box it sits on, U, is the operator.
Why: With no mannose, the repressor sits on the operator.
The block, the repressor, sits on the box lettered U.
So U is the operator, the short stretch of DNA beside the promoter that the repressor binds.
A biologist measures the mRNA of the mannose operon's genes in four cultures of the bacterium: with and without mannose in the broth, in a normal strain and in a strain that makes no repressor. The table gives the results.
Which of the following is mannose, for this operon?
- A. An activatorAn activator is a regulatory protein that binds DNA beside the promoter and raises transcription.
Mannose is a small molecule, and it acts through the repressor. - B. A corepressorA corepressor binds a repressor and makes it fit the operator, so the genes switch off.
With mannose present the genes switch on. - C. ✓ An inducer
- D. A repressorThe repressor is the protein that sits on the operator and blocks RNA polymerase.
Mannose is the small molecule that binds the repressor and lifts it off.
Why: With the repressor working, the mRNA level is very low without mannose and high with it: mannose switches the genes on.
With no repressor the level is high either way: mannose acts through the repressor.
A small molecule that binds a repressor and so switches genes on is an inducer.
Suppose a bacterium's operon holds the genes for building the amino acid lysine. Right now the cell holds very little lysine.
Which of the following describes the operon's genes and its repressor right now?
- A. Not transcribed, with the repressor on the operatorThe operon builds lysine, so lysine is its corepressor.
With little lysine the repressor has no corepressor, so it does not fit the operator. - B. Not transcribed, with the repressor off the DNAWith the repressor off the DNA, nothing sits in RNA polymerase's path.
RNA polymerase moves into the genes and transcribes them. - C. Transcribed, with the repressor on the operatorA repressor on the operator blocks RNA polymerase's path, so the genes would be silent.
With little lysine the repressor does not fit the operator and is off the DNA. - D. ✓ Transcribed, with the repressor off the DNA
Why: The operon builds lysine, so it is a repressible system with lysine as its corepressor.
Lysine is scarce, so the signal is absent.
Without its corepressor the repressor does not fit the operator, so it is off the DNA.
Nothing sits in RNA polymerase's path, so the genes are transcribed.
E. coli cells grow in a broth of glucose plus lactose. A technician removes the glucose from the broth and leaves the lactose.
An hour later, how much of the lactose enzymes are the cells making, compared with before?
- A. ✓ Far more
- B. About the sameWith glucose gone the cell makes cyclic AMP, and cyclic AMP binds the activator.
The bound activator binds beside the promoter, and RNA polymerase transcribes the operon far more often. - C. Far lessLactose is still present, so the repressor stays off the operator.
With glucose gone the activator binds too, and transcription rises. - D. None at allLactose keeps the repressor off the operator, so RNA polymerase's path stays clear.
Removing glucose lets the activator bind, and the cells make far more of the enzymes.
Why: Lactose is still present, so the repressor stays off the operator.
With the glucose gone, the cell makes cyclic AMP.
Cyclic AMP binds the activator, which then binds beside the promoter.
So RNA polymerase transcribes the operon far more often, and the cells make far more of the enzymes.
Cells from an eel grow in a dish. Two chemicals each silence the same gene in the cells. A technician exposes one dish to the first chemical and another dish to the second for two days, washes each chemical out, and grows the cells on for a month, through many divisions. After the month, the first dish's cells transcribe the gene again; the second dish's gene stays silent. The gene's base sequence is unchanged in both dishes.
Which of the following explains the two results?
- A. In both dishes, enzymes removed the tags the chemical addedTags that enzymes remove let the gene be transcribed again.
The second dish's gene is still silent a month later. - B. In both dishes, the cells copied the tags to their daughter cells at each divisionTags copied at every division keep the gene silent through the month.
The first dish's gene is transcribed again, so its tags were removed. - C. ✓ In the first dish, enzymes removed the tags; in the second, the cells copied the tags to their daughter cells at each division
- D. In the first dish, the cells copied the tags to their daughter cells at each division; in the second, enzymes removed the tagsA gene whose tags are removed is transcribed again.
The first dish's gene came back and the second dish's stayed silent, so the removal happened in the first dish.
Why: The base sequence is the same in both dishes, so the silencing was a change in tags.
The first dish's gene is transcribed again: enzymes removed its tags.
The second dish's gene stays silent through many divisions: at each S phase the cells copied its tags onto the new DNA.
A cell in a scorpion's claw controls how often RNA polymerase transcribes one of its genes.
Which of the following is a transcription factor?
- A. The enzyme bound to the promoter, copying the gene into RNAThe enzyme that copies a gene into RNA is RNA polymerase.
A transcription factor is the protein beside the gene that helps or blocks RNA polymerase binding. - B. The stretch of DNA beside the gene where a regulatory protein bindsThe stretch of DNA beside the gene is the regulatory sequence.
A transcription factor is the protein that binds it. - C. A kinase in the cytosol adding a phosphate group to a regulatory proteinThe kinase switches the regulatory protein on and binds no DNA itself.
A transcription factor binds the DNA beside a gene. - D. ✓ A protein bound to the DNA beside the gene, helping RNA polymerase bind the promoter
Why: A transcription factor is a protein.
It binds the DNA beside a gene and helps or blocks RNA polymerase binding the promoter there.
The protein bound beside the gene, helping RNA polymerase bind, is the transcription factor.
A biologist measures a gene in the cells of a sea slug's skin. The cells hold 210 units of the gene's mRNA and 0 units of its protein.
Which of the following can explain the absence of the protein?
- A. RNA polymerase is not transcribing the geneThe cells hold 210 units of the gene's mRNA.
An mRNA exists only because RNA polymerase transcribed the gene. - B. The gene's promoter carries many methyl groupsA heavily methylated promoter winds tight, and the gene is rarely transcribed.
This gene's mRNA is at 210 units, so it is transcribed. - C. ✓ Ribosomes are not translating the mRNA
- D. The cells have lost the geneA cell with no copy of the gene makes no mRNA of it.
These cells hold 210 units of the mRNA, so they carry the gene.
Why: The cells hold 210 units of the gene's mRNA, so RNA polymerase is transcribing the gene.
The protein is absent all the same.
A protein is built only when ribosomes translate its mRNA.
So the control acts after transcription: the ribosomes are not translating the mRNA.
Two cells of the iris of one person's eye each carry a copy of the same gene. In the first cell RNA polymerase transcribes the gene steadily. In the second cell the gene is silent, and its promoter carries many methyl groups.
(a) Describe what the methyl groups on the promoter do to the winding of the second copy's DNA. (1 pt)
Frame The methyl groups on the promoter make the DNA …
Hint Think of the DNA as thread wound on histone beads. Do the methyl groups pack the beads together or set them apart?
- Award 1 point for: the DNA winds more tightly (the histones pack together; the DNA coils tight).
(b) Describe how the base sequences of the two copies compare. (1 pt)
Frame The two copies …
Hint A methyl group sits on a cytosine base. Is the cytosine still a cytosine?
- Award 1 point for: the two copies have the same base sequence (the methyl groups change no base).
(c) Explain why the second copy stays silent. (1 pt)
Frame The second copy stays silent because …
Hint Which kind of winding can RNA polymerase reach: loose or tight?
RNA polymerase can only reach a gene that is loosely wound.
So RNA polymerase cannot reach the second copy's promoter.
- Award 1 point for: the tightly wound DNA keeps RNA polymerase from reaching the gene (its promoter), so it is not transcribed.
- Restating part (a), that the DNA is tightly wound, earns nothing on its own: the point wants RNA polymerase (or a transcription factor) kept off the promoter.
(d) Predict what happens to the silent gene in the two daughter cells when the second cell divides. (1 pt)
Frame When the second cell divides, …
Hint At S phase the old strand keeps its methyl groups. What do enzymes do to the new strand?
At S phase the old strand keeps its methyl groups, and enzymes add methyl groups to the new strand to match.
So each daughter cell receives a promoter with the same methyl pattern.
- Award 1 point for: the gene stays silent in both daughter cells, because the methyl pattern is copied onto the new DNA at S phase (the daughters inherit the parent's tags).
(e) Explain how the silent gene could be switched on again in the second cell. (1 pt)
Frame The gene could switch on again if …
Hint Which enzymes act on the methyl groups, and what does their loss do to the winding?
With the methyl groups gone, the DNA winds more loosely.
So RNA polymerase can reach the promoter and transcribe the gene.
- Award 1 point for: enzymes remove the methyl groups, so the DNA winds more loosely and RNA polymerase can reach (transcribe) the gene.
A biologist plans to measure how a pesticide changes gene expression in the gill cells of a fish. To compare a gene's level between treated and untreated fish, she needs a reference gene: a gene whose level stays the same whatever the fish's conditions, so that every other gene can be measured against it. She measures three candidate genes in gill cells of fish kept in clean water, in water with arsenic and in water with a weedkiller. The table gives the results.
(a) Identify the gene with the lowest level in the gill cells of fish kept in clean water. (1 pt)
- Award 1 point for: gene Q.
(b) Identify the gene whose level varies most across the three waters. (1 pt)
- Award 1 point for: gene U.
(c) The biologist chooses gene E as her reference gene for the pesticide experiment. Evaluate her choice using the data. (1 pt)
Gene E reads 88, 90 and 87 units across the three waters: about the same level in every condition, so it is constitutively expressed.
A gene whose level does not change with the fish's conditions is a fair yardstick for the genes that do.
- Award 1 point for: supported, because gene E's level is about the same in every condition (88 / 90 / 87 units; constitutively expressed), so it can serve as the unchanging reference.
Slip Choosing by level: a reference gene is chosen for a level that does not change, not for a high or a low level.
(d) Gene Q's level rises in water with arsenic. Explain how arsenic could raise transcription of gene Q. (1 pt)
The switched-on factor binds the regulatory sequence beside gene Q.
Bound there, the factor helps RNA polymerase bind gene Q's promoter, so RNA polymerase transcribes gene Q more often.
- Award 1 point for: arsenic switches on a transcription factor (regulatory protein) that binds the DNA beside gene Q and helps RNA polymerase bind the promoter, so gene Q is transcribed more often. Accept: arsenic acts through a signal pathway that switches the factor on.
APBIO-U06-T65 End-of-topic test: Regulation of Gene Expression
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.
A soil bacterium carries a gene for an enzyme that breaks down nylon. A biologist grows the bacterium in a broth with no nylon and tests one cell for three things: the gene, the gene's mRNA, and the enzyme.
Which set of results shows that the cell carries the gene and the gene is silent?
- A. Gene not found; mRNA not found; enzyme not foundA cell that carries the gene has the gene in its DNA whatever the broth.
A gene not found is a gene the cell does not carry. - B. Gene found; mRNA found; enzyme not foundAn mRNA of the gene exists only while RNA polymerase transcribes the gene.
mRNA found means the cell is transcribing it. - C. ✓ Gene found; mRNA not found; enzyme not found
- D. Gene found; mRNA found; enzyme foundThe enzyme is built by ribosomes from the gene's mRNA.
Gene, mRNA and enzyme all found means the gene is expressed.
Why: The gene sits in the cell's DNA whether or not the cell uses it, so the gene is found.
RNA polymerase makes mRNA only while it transcribes the gene, so no mRNA means no transcription.
Ribosomes build the enzyme only from that mRNA, so no enzyme either.
Four plants of one species carry the same version of a gene for an enzyme that builds the waxy layer on a leaf. The table gives, for each plant, the mRNA of that gene in its leaf cells and the thickness of the leaf wax.
Which of the following best explains the relationship between the gene and the wax?
- A. ✓ How often the leaf cells transcribe the gene sets how much of the enzyme they hold, and so how much wax they build
- B. The enzyme works faster in the plants whose leaf cells hold more of its mRNAThe mRNA level tells how much enzyme the ribosomes build, not how fast each enzyme molecule works.
More mRNA gives more enzyme. - C. The plants with thicker wax transcribe the gene more often because the wax signals the leaf cells to do soThe wax is the product of the enzyme the gene codes for.
More transcription gives more enzyme, and more enzyme gives more wax, not the other way round. - D. Each plant's enzyme has a different order of amino acids, so each builds wax at a different rateThe four plants carry the same version of the gene.
The same gene gives the same order of amino acids in every plant.
Why: The four plants carry the same version of the gene, so each builds the same enzyme.
More mRNA of the gene means the leaf cells transcribe it more often.
Ribosomes build more of the enzyme from more mRNA.
More enzyme builds more wax, so the wax is thicker.
A biologist grows a fungus's cells in a broth with iron scarce and in the same broth with iron plentiful. The table gives the mRNA of three of the fungus's genes in each broth.
Which of the genes are constitutively expressed?
- A. Gene Q onlyGene U reads 14 and 15 units: about the same level in both broths, low as that level is.
The level itself never sorts a gene. - B. Gene U onlyGene Q reads 205 and 198 units: about the same level in both broths.
A gene at about the same level in every condition is constitutively expressed, high or low. - C. ✓ Genes Q and U
- D. Genes Q, U and EGene E reads 12 units with iron scarce and 310 units with iron plentiful.
Its level is high in one condition only, so gene E is inducible.
Why: Gene Q reads 205 and 198 units, gene U 14 and 15: each stays at about the same level.
A gene at about the same level in every condition is constitutively expressed.
Gene E reads 12 and 310: its level is high in one condition only, so inducible.
In a yeast cell, a regulatory protein binds a 16-base stretch of DNA beside a gene. A biologist measures the gene's mRNA in cells that have the original stretch and in cells whose stretch is changed at four of its sixteen bases, each with and without the regulatory protein. The table gives the results.
Which of the following explains the pattern in the table?
- A. The regulatory protein binds the gene's mRNA and protects it, and the changed stretch codes for an mRNA it cannot bindA regulatory protein binds DNA beside the gene, before any mRNA exists.
The 16-base stretch is never copied into the mRNA. - B. ✓ The protein's binding site fits the original stretch only, so it raises transcription only there
- C. The four changed bases change four amino acids of the gene's protein, so less of it is madeThe stretch lies beside the gene and RNA polymerase does not copy it.
A change in it changes how much protein is made, not which amino acids the protein has. - D. RNA polymerase binds the changed stretch in place of the promoter, so transcription barely startsRNA polymerase binds the promoter, and the promoter is unchanged.
The stretch beside the gene is where the regulatory protein binds.
Why: A protein binds only DNA whose shape fits its binding site.
On the original stretch the protein binds: the mRNA rises from 10 to 52 units.
Four changed bases give the stretch a shape the binding site does not fit.
The protein stays off; the mRNA stays at 11 units.
A bacterium has three genes for taking up and breaking down trehalose, a sugar. They sit side by side on the chromosome. A biologist measures the mRNA of each gene in the normal strain, without and with trehalose, and in a changed strain, in which a change removed the DNA just before the first gene. The table gives the results.
Which of the following conclusions do the results best support?
- A. Each gene has its own promoter, and trehalose switches on the three promoters one after anotherRemoving the DNA before the first gene alone silenced all three genes.
Genes with their own promoters would still be transcribed from those promoters. - B. The three genes are one long gene, translated into one enzyme that does all three jobsOne mRNA can carry several genes, and ribosomes build a separate enzyme from each.
Three genes on one mRNA are still three genes. - C. Trehalose binds the DNA just before the first gene and switches the three genes onA sugar binds a protein, the repressor, and binds no DNA.
Trehalose switches the genes on by changing the repressor's shape. - D. ✓ They share one promoter and are transcribed as one mRNA, so all three switch on and off together
Why: In the normal strain all three genes rise to about 190 units when trehalose is added: they switch together.
Removing the DNA just before the first gene silences all three at once.
So one promoter there serves all three, and RNA polymerase transcribes them as one mRNA.
In the drawing, the small circle is a small molecule bound to the repressor. A full rectangle is the repressor in the shape that fits the operator. A block with its lower corners cut away is the repressor in its changed shape. An oval above the promoter is RNA polymerase off the DNA. A bacterium's operon holds three cellobiose-breaking genes. The drawing shows it just after cellobiose has entered the cell.
Which of the following happens next?
- A. ✓ The repressor leaves the operator, and RNA polymerase transcribes the three genes
- B. Cellobiose moves from the repressor onto the operator and pushes the repressor offCellobiose binds the repressor, a protein, and binds no DNA.
The repressor leaves because its new shape no longer fits the operator. - C. The repressor grips the operator more tightly, and the three genes stay offBound cellobiose has changed the repressor's shape to the one that does not fit the operator.
A repressor that does not fit lets go of the DNA. - D. Cellobiose leaves the repressor, and the repressor stays on the operator until the cellobiose is used upCellobiose has just entered the cell and is bound to the repressor.
With cellobiose bound, the repressor no longer fits the operator and leaves it.
Why: Cellobiose has bound the repressor and changed its shape: the corners are cut away.
The new shape no longer fits the operator, so the repressor lets go of the DNA.
Nothing then sits in RNA polymerase's path, so it moves into the three genes and transcribes them.
In the drawing, the small circle is a small molecule bound to the repressor. A full rectangle is the repressor in the shape that fits the operator. A block with its lower corners cut away is the repressor in its changed shape. An oval above the promoter is RNA polymerase off the DNA. A bacterium's operon holds four genes for building the amino acid serine. The drawing shows the operon at one moment.
Which of the following describes the cell at this moment?
- A. It holds little serine, and the four genes are transcribedA small molecule bound has given the repressor the shape that fits the operator.
That molecule is the operon's product, serine, so serine is plentiful. - B. ✓ It holds plenty of serine, and the four genes are not transcribed
- C. It holds plenty of serine, and the four genes are transcribedThe repressor sits on the operator, in RNA polymerase's path.
With the repressor on the operator, RNA polymerase transcribes none of the four genes. - D. It holds little serine, and the four genes are not transcribedThis operon builds serine, so serine is its corepressor.
The repressor fits the operator only with serine bound, and serine is bound here, so serine is plentiful.
Why: The operon builds serine, so serine is its corepressor.
The repressor is a full rectangle with a small molecule bound: serine has given it the shape that fits the operator.
So serine is plentiful.
With the repressor on the operator, RNA polymerase transcribes none of the four genes.
A bacterium can build the amino acid proline. A biologist measures the mRNA of four of the bacterium's operons in a broth with no proline added and in the same broth with proline added. The table gives the results.
Which operon is a repressible system with proline as its corepressor?
- A. Operon 1Operon 1 reads 8 units with no proline and 176 with proline: proline switches its genes on.
A signal that switches the genes on marks an inducible system. - B. Operon 2Operon 2 reads 61 and 58 units: about the same with and without proline.
Proline does not switch its genes at all. - C. Operon 3Operon 3 reads 4 and 5 units: its mRNA level is low in both broths.
A repressible operon is on while its corepressor is absent. - D. ✓ Operon 4
Why: A repressible operon is on until its corepressor switches it off.
Operon 4 reads 143 units with no proline and 12 with proline added: proline switches its genes off.
So operon 4 is a repressible system, and proline is its corepressor.
Suppose a change in an E. coli cell's DNA removes the stretch of DNA that the lac operon's activator binds, and changes nothing else. The cell grows in a broth of lactose alone, beside a normal cell in the same broth.
How much of the lactose enzymes does the changed cell make, compared with the normal cell?
- A. ✓ Far less, but some
- B. About the sameThe normal cell has the activator bound beside the promoter, so RNA polymerase transcribes often.
The changed cell has no stretch for the activator to bind. - C. Far moreThe activator raises how often RNA polymerase transcribes the operon.
With nowhere to bind, the activator cannot raise it. - D. None at allLactose has lifted the repressor off the operator, so nothing blocks RNA polymerase's path.
Without the activator, RNA polymerase still transcribes the genes now and then.
Why: In lactose alone both cells make cyclic AMP, and the activator takes its binding shape.
The normal cell's activator binds beside the promoter, and RNA polymerase transcribes often.
The changed cell has no stretch for its activator.
The repressor is off, so RNA polymerase still transcribes now and then.
In the drawings, beads set apart with the DNA open between them are a loosely wound gene. Beads packed along a wave are a tightly wound gene. Two cells of one goat each carry a copy of the same gene; the two copies match base for base. The drawing shows how each copy is wound on its histones: copy Q in the first cell and copy U in the second.
Which of the following could make copy Q wind the way copy U does?
- A. Enzymes adding methyl groups to copy Q's promoterMethyl groups on a promoter wind the DNA tighter.
Copy Q is already coiled tight; methyl groups would keep it so. - B. RNA polymerase binding copy Q's promoter and pushing the histones apartRNA polymerase cannot reach a tightly wound gene, let alone unwind it.
The winding changes when enzymes change the tags. - C. A change in one base of copy QA change in a base changes the gene's sequence, not how tightly its DNA is wound.
The winding is set by the tags on the DNA and the histones. - D. ✓ Enzymes adding acetyl groups to copy Q's histones
Why: Copy Q is coiled tight on its histones and copy U lies open.
An acetyl group cancels part of a histone's positive charge, so the histone grips the DNA less tightly.
So acetyl groups added to copy Q's histones loosen its winding, and copy Q winds like copy U.
Cells of a green alga grow in a dish. A signal molecule added to the dish makes the cells add methyl groups to one gene's promoter. A biologist treats four dishes of the cells as the table describes; the drug used in the last dish blocks the enzymes that remove methyl groups from DNA. The table gives the methyl groups on the promoter and the gene's mRNA in each dish.
Which of the following explains the difference between the third and fourth dishes?
- A. The signal changed the promoter's base sequence, and the drug undid that changeMethyl groups sit on top of bases and change none of them.
The base sequence is the same in every dish. - B. ✓ Once the signal is gone, enzymes remove the methyl groups from the promoter, and the drug stops those enzymes
- C. The drug binds RNA polymerase and keeps it off the promoterThe two dishes differ in the methyl groups on the promoter as well as in the mRNA.
A drug on RNA polymerase would leave the methyl groups as they were. - D. The cells copy the methyl groups onto new DNA when they divide, so the methyl groups are never lostThe third dish's cells lost their methyl groups within 12 hours of the signal's withdrawal.
Enzymes remove the tags, whatever the cells copy at division.
Why: The signal adds methyl groups to the promoter, and the mRNA falls to 19 units.
With the signal gone, enzymes remove the methyl groups and the mRNA returns to 78 units.
The drug blocks those enzymes, so the methyl groups stay and the mRNA stays at 21 units.
A biologist measures the mRNA of three genes in two kinds of cell from one person: the cells of a sweat gland and the sound-sensing cells of the inner ear. Both kinds of cell carry all three genes. The table gives the results.
Which of the following best explains why the sweat-gland cells transcribe gene Q often and the inner-ear cells transcribe it rarely?
- A. The inner-ear cells have lost gene Q during differentiationBoth kinds of cell carry gene Q: differentiation removes no genes.
The inner-ear cells hold the gene and rarely transcribe it. - B. The sweat-gland cells make far more RNA polymerase than the inner-ear cellsThe inner-ear cells transcribe gene U often, at 620 units, so they have RNA polymerase in plenty.
The factors bound beside a gene set how often it is transcribed. - C. ✓ Only the sweat-gland cells hold a transcription factor that binds the DNA beside gene Q
- D. Gene Q's promoter has a different base sequence in the two kinds of cellThe two kinds of cell carry the same DNA, promoter and all.
What differs is the proteins bound beside the gene.
Why: Both kinds of cell carry gene Q.
The sweat-gland cells read 480 units of gene Q's mRNA and the inner-ear cells 6.
A factor bound beside a gene helps RNA polymerase bind its promoter.
Each kind of cell holds its own factors, so only the sweat-gland cells hold gene Q's.
Suppose a change in an E. coli cell's DNA means the cell makes no lac repressor at all. The changed cell grows in a broth of glucose only. A normal E. coli cell grows in a broth of lactose only.
How much of the lactose enzymes does the changed cell make?
- A. ✓ Far less than the normal cell, but some
- B. About as much as the normal cellThe normal cell has the activator bound beside the promoter.
The changed cell has plenty of glucose, so little cyclic AMP, so its activator is off the DNA. - C. Far more than the normal cellNothing blocks RNA polymerase's path in the changed cell.
Without the activator bound, it still transcribes the operon only now and then. - D. None at allNo repressor sits on the changed cell's operator, so nothing blocks RNA polymerase's path.
RNA polymerase transcribes the genes now and then even without the activator.
Why: The normal cell has its repressor off and its activator bound: RNA polymerase transcribes often.
The changed cell has no repressor, so nothing blocks the path.
But glucose is plentiful, so little cyclic AMP, so the activator stays off the DNA.
RNA polymerase transcribes only now and then.
Copper ions enter a fungus's cell. Within an hour, six of the cell's genes switch on together. The six genes sit on four different chromosomes. Copper switches on one transcription factor in the cell.
Which of the following does each of the six genes have beside its promoter?
- A. One promoter shared by all six genesGenes on four chromosomes cannot share one promoter.
Each of the six has its own promoter, and RNA polymerase transcribes each gene from its own. - B. ✓ The regulatory sequence that the factor's binding site fits
- C. A copy of the gene for the transcription factorThe factor is a protein built from one gene somewhere in the genome.
What sits beside each of the six genes is the DNA sequence the factor binds. - D. A stretch of DNA that copper bindsCopper binds the transcription factor, a protein, and binds no DNA.
The factor then binds its sequence beside each gene.
Why: One transcription factor binds only the regulatory sequence its binding site fits, wherever that sequence occurs.
The six genes switched on together when that one factor was switched on.
So each carries the same short sequence beside its promoter, and the bound factor helps RNA polymerase bind each promoter.
A bacterium grows at 37 °C. A technician moves it into a broth at 12 °C. The table gives the mRNA and the protein of two of the bacterium's genes at 37 °C and after 30 minutes at 12 °C.
Which claim do the data support about gene Q?
- A. The cold raised transcription of gene Q, and the control acts at transcriptionGene Q's mRNA reads 30 units at 37 °C and 31 units at 12 °C: unchanged.
Unchanged mRNA means RNA polymerase transcribed gene Q as often as before. - B. The cold lowered transcription of gene Q, and the control acts at transcriptionGene Q's mRNA is unchanged and its protein rose from 20 to 95 units.
The change is in the protein, not in transcription. - C. The cold does not control gene Q, and neither of its rows changedGene Q's protein rose from 20 to 95 units.
A protein level that changes means the cold does control the gene. - D. ✓ The cold raised gene Q's protein level, and the control acts after transcription
Why: Gene Q's mRNA reads 30 units at 37 °C and 31 after 30 minutes at 12 °C: transcription did not change.
Gene Q's protein rose from 20 to 95 units all the same.
More protein from the same mRNA means the control acts after transcription, and the direction is up.
Cells from a sea sponge grow in a dish and transcribe a gene steadily. A technician adds a drug that blocks the enzymes that add acetyl groups to histones. After two days, the gene's histones carry half as many acetyl groups as before.
What happens to the gene's transcription over the two days?
- A. It risesAcetyl groups loosen the histones' grip on the DNA.
With fewer acetyl groups the DNA winds tighter, and RNA polymerase reaches the gene less often. - B. It stays the sameFewer acetyl groups on the histones change the winding: the histones grip the DNA more tightly.
Tighter winding lowers transcription. - C. ✓ It falls, but continues
- D. It stops altogetherHalf the acetyl groups remain on the histones, so the DNA winds tighter but not shut.
RNA polymerase still reaches the gene now and then.
Why: Enzymes keep removing acetyl groups, and the drug stops the enzymes that add them, so the acetyl groups fall to half.
Fewer acetyl groups let the histones grip the DNA more tightly, so RNA polymerase reaches the gene less often.
Transcription falls, yet continues now and then.
In a kale plant's leaf cells, a signal from a neighboring cell switches on a kinase. The kinase adds a phosphate group to a transcription factor, and the factor then binds the DNA beside gene Q. Suppose a change in the factor's gene gives a factor with no site for the kinase to add a phosphate group to.
What happens to gene Q in a leaf cell with the changed factor after the signal arrives?
- A. ✓ Gene Q stays rarely transcribed
- B. Gene Q is transcribed often at all times, signal or no signalWithout a phosphate group the factor keeps its resting shape, which does not fit the DNA beside gene Q.
A factor off the DNA cannot help RNA polymerase bind. - C. Gene Q is transcribed often only after the signal, as in a normal cellThe added phosphate group gives a normal factor the shape that fits the DNA.
This factor has no site for it, so the signal changes nothing. - D. Gene Q's protein is built with a changed order of amino acidsThe change is in the factor's gene, not in gene Q.
Gene Q's sequence, and so its protein's amino acids, are as before.
Why: The kinase switches the factor on by adding a phosphate group.
The changed factor has no site for it and keeps its resting shape.
That shape does not fit the DNA beside gene Q, so the factor stays off.
With no factor bound, RNA polymerase rarely binds the promoter.
Cells taken from a toad embryo grow in a dish. At 0 minutes a signal reaches them. Gene Q codes for a transcription factor. A cell that lacks gene Q never transcribes gene U after the signal. The table gives the mRNA of the two genes at four times after the signal.
Which of the following best explains why gene U's mRNA rises only after 15 minutes?
- A. The signal switches on gene U directly, and gene U's promoter is slower than gene Q'sA cell that lacks gene Q never transcribes gene U after the signal.
So gene U is switched on through gene Q's protein, not by the signal directly. - B. ✓ Ribosomes must first build gene Q's protein from its mRNA, and that protein then switches on gene U
- C. Gene Q's protein binds gene U's mRNA and protects it from being broken downA transcription factor binds DNA beside a gene, not an mRNA.
Gene U's mRNA rises because RNA polymerase transcribes gene U more often. - D. Gene Q's mRNA is translated only after gene U's mRNA has been madeGene Q's mRNA rises at 15 minutes and gene U's only at 60 minutes.
Gene Q's protein is built first, and gene U follows.
Why: At 15 minutes gene Q's mRNA has risen from 2 to 60 units: the signal switched gene Q on.
Ribosomes then build gene Q's protein, a transcription factor.
The factor binds beside gene U and helps RNA polymerase bind its promoter.
So gene U's mRNA rises later, at 60 minutes.
A biologist measures four genes in the root cells of one kind of plant growing in wet soil, in moist soil and in dry soil. For each gene she measures the level of its mRNA and the level of its protein. The results are drawn as a grid: each square is shaded by its value, darker for a larger value, and the value is written in the square, in arbitrary units.
(a) Identify the gene that is transcribed more often as the soil gets drier. (1 pt)
- Award 1 point for: gene Q.
(b) Describe the trend in gene E's mRNA and protein levels across the three soils. (1 pt)
Neither level changes as the soil gets drier.
- Award 1 point for: gene E's mRNA AND protein stay at about the same level across the three soils (no change with the soil). Accept: about 40 units and about 43 units in every soil.
Slip Reading a change of one or two units as a trend. Levels within a unit or two of each other are the same level.
(c) A student claims that drier soil lowers gene U's protein level by lowering transcription of gene U. Evaluate the student's claim using the data. (1 pt)
Gene U's mRNA reads 57, 58 and 57 units across the three soils: unchanged, so RNA polymerase transcribed gene U as often as before.
Gene U's protein fell from 72 to 30 to 7 units all the same.
So drier soil lowers gene U's protein level after transcription, not by lowering transcription.
- Award 1 point for: the claim is not supported, because gene U's mRNA is about the same in every soil (57 / 58 / 57 units) while its protein falls (72 → 30 → 7 units), so transcription of gene U did not fall.
Slip Reading the falling protein as falling transcription. The mRNA row is the measure of transcription, and it is flat.
(d) Explain how the root cells can hold a high level of gene O's mRNA and no gene O protein. (1 pt)
Ribosomes do not translate that mRNA, so they build no protein from it.
So the root cells hold gene O's mRNA and none of its protein.
- Award 1 point for: the mRNA is not translated (ribosomes do not read it), so no protein is built from it although the gene is transcribed. Accept: the protein is broken down as fast as it is built.
- Saying that the mRNA is broken down before it is read earns nothing: the mRNA is present at a high level.
In a trout embryo, a cell becomes a scale-forming cell through a chain of three transcription factors. The drawing shows the chain in one embryo cell, with its words removed. In the drawing, a rounded block on a small box is a transcription factor bound to its regulatory sequence, an oval is RNA polymerase, and a light strand hanging from a gene is its mRNA. The site each factor binds is lettered Q, U or E, an arrow at the top left marks the signal, and an arrow leads from each row's genes down to the factor those genes code for. In this cell, one factor's genes are transcribed but the factor itself is absent. The bottom row's genes code for the proteins that build a scale.
(a) Identify the transcription factor that this cell lacks. (1 pt)
- Award 1 point for: factor E (the third factor).
(b) Describe how, in a normal embryo cell, factor Q leads to factor U being made. (1 pt)
Bound there, factor Q helps RNA polymerase bind their promoter, so RNA polymerase transcribes those genes.
Ribosomes build factor U from the mRNA.
- Award 1 point for: factor Q binds the DNA (the regulatory sequence) beside the genes for factor U, so RNA polymerase transcribes those genes and factor U is built from their mRNA. Accept an answer that names the binding and the transcription without the ribosomes.
(c) Explain why the genes for the scale proteins stay off in this cell. (1 pt)
This cell has no factor E, so nothing binds the DNA beside those genes.
With no factor bound, RNA polymerase rarely binds their promoter, so the genes stay off.
- Award 1 point for: the scale-protein genes are downstream of the missing factor: no factor E binds beside them, so RNA polymerase does not transcribe them.
- Accept reasoning consistent with the factor named in part (a).
Slip Saying the scale genes are missing or damaged. The genes are present; the factor that switches them on is missing.
(d) Explain why the genes for factor U are still transcribed in this cell. (1 pt)
The signal still switches on factor Q, and factor Q still binds beside the genes for factor U.
So RNA polymerase still transcribes them: the missing factor E plays no part in that step.
- Award 1 point for: the genes for factor U are upstream of the missing factor: factor Q, switched on by the signal, still binds beside them, so a break later in the chain changes nothing before it.
- Accept reasoning consistent with the factor named in part (a).
Slip Saying the whole chain switches off. A break stops everything after it, not before it.
APBIO-U06-L30 Where the switch sits
Suppose a human gene is drawn as a line: a promoter, a start site, then the exons. RNA polymerase floats nearby, off the promoter. Then three proteins land on the promoter, and the polymerase docks against them.
Far to the left, 30,000 base pairs away, two more proteins sit on another short stretch of DNA. Those far proteins matter too. How can a protein 30,000 base pairs away change what happens at the promoter?
Unit 6 · Gene Expression and Regulation
1Factors first, then the polymerase
In a eukaryotic cell, a protein binds the DNA beside a gene and helps RNA polymerase bind the promoter there.
What is this protein called?
- A. A repressorA repressor blocks transcription; this protein helps RNA polymerase bind.
A regulatory protein that helps or blocks RNA polymerase binding beside a gene is a transcription factor. - B. A regulatory sequenceA regulatory sequence is the stretch of DNA the protein binds.
The protein that binds it and helps RNA polymerase bind is a transcription factor. - C. ✓ A transcription factor
Why: In a eukaryotic cell, a regulatory protein that binds the DNA beside a gene and helps or blocks RNA polymerase binding there is called a transcription factor.
RNA polymerase is about to copy a gene into RNA.
Where does RNA polymerase bind to begin copying?
- A. On the first exon of the gene, where the RNA beginsRNA polymerase copies the exons; it does not begin on them.
RNA polymerase binds the promoter, the short DNA sequence just before the gene. - B. On the gene’s mRNA, once the first copy existsNo mRNA exists before RNA polymerase begins copying.
RNA polymerase binds the promoter, the short DNA sequence just before the gene. - C. ✓ On the promoter, the short sequence just before the gene
Why: The promoter is a short DNA sequence just before a gene.
RNA polymerase binds the promoter to begin copying the gene.
The point where copying begins is the transcription start site.
Where are the switches on a human gene, and how do far ones reach the promoter?
In a eukaryotic cell, RNA polymerase cannot bind the promoter on its own.
Transcription factors bind the promoter first. Then RNA polymerase binds to them.
Video: Watch: Factors first, then the polymerase
A human gene is drawn as a line with its promoter and start site. RNA polymerase floats above the promoter and does not bind it. Three rounded factors land on the promoter side by side. RNA polymerase then docks against them over the start site, and RNA comes off the gene.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L30a.mp4
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Go back to the human gene drawn as a line: its promoter, its transcription start site, then its exons.
RNA polymerase floats near the promoter. In a human cell, RNA polymerase does not bind the promoter on its own.
In the drawings, an oval floating above the DNA is RNA polymerase off the DNA.
In a bacterium, RNA polymerase binds its promoter directly. A human cell’s RNA polymerase needs other proteins on the promoter first.
Three proteins in the nucleus have binding sites that fit the promoter. So the three proteins land on the promoter, side by side.
The three proteins help RNA polymerase bind beside the gene. So the three proteins are transcription factors.
The same set of factors binds the promoter of every human gene that codes for a protein. The drawings show three of them.
RNA polymerase has a binding site that fits the bound factors. So RNA polymerase docks against the factors, over the transcription start site.
In the drawings, an oval touching the shapes on the promoter is RNA polymerase docked against them.
Docked there, RNA polymerase separates the two strands and begins building RNA. So the gene is transcribed.
Now imagine the three transcription factors are removed from the nucleus. RNA polymerase floats near the promoter and does not bind it.
So no RNA is built from the gene. A gene that no RNA polymerase transcribes is silent.
RNA polymerase binds a human promoter through the factors. So the switch at the promoter has two steps: the factors bind the promoter, and RNA polymerase binds the factors.
What you are expected to know Describe how a eukaryotic gene is switched on at its promoter: transcription factors bind the promoter first and RNA polymerase then binds to them.
In a human liver cell, one gene is about to be transcribed.
Which binds the promoter first?
- A. RNA polymeraseA human cell’s RNA polymerase cannot bind the promoter on its own.
The transcription factors bind the promoter first, and RNA polymerase binds to them. - B. ✓ The transcription factors
Why: In a human cell, RNA polymerase cannot bind the promoter on its own.
The transcription factors bind the promoter first.
RNA polymerase then binds to the bound factors.
Suppose a biologist puts the DNA of a gene into a tube with RNA polymerase and a supply of RNA nucleotides. With a bacterium’s gene and the bacterium’s RNA polymerase, RNA polymerase builds RNA. With an oyster’s gene and the oyster’s RNA polymerase, RNA polymerase builds no RNA. She then adds the oyster’s transcription factors to the oyster’s tube. Now RNA polymerase builds RNA from the oyster gene.
(a) Explain why the oyster’s RNA polymerase built no RNA before she added the factors. (2 pt)
Frame The oyster’s RNA polymerase built no RNA before she added the factors because …
The promoter’s transcription factors were not in the tube.
So nothing sat on the promoter for RNA polymerase to bind to.
So RNA polymerase stayed off the DNA and built no RNA.
Once she added the factors, they bound the promoter.
RNA polymerase then bound to the factors and began building RNA.
- Award 1 point for: a eukaryotic RNA polymerase cannot bind the promoter on its own; it binds to transcription factors bound there.
- Award 1 point for: with no factors in the tube, nothing was bound to the promoter, so RNA polymerase did not bind and no transcription began (adding the factors let RNA polymerase bind).
A student says: “In a human cell, RNA polymerase binds the promoter directly, just as it does in a bacterium.”
Is the student correct?
- A. ✓ No: in a human cell, transcription factors bind the promoter, and RNA polymerase binds to them
- B. Yes: RNA polymerase binds every promoter directly, in a bacterium or a human cellA bacterium’s RNA polymerase binds its promoter directly.
A human cell’s RNA polymerase cannot; it binds the transcription factors bound to the promoter.
Why: In a bacterium, RNA polymerase binds the promoter directly.
In a human cell, RNA polymerase cannot bind the promoter on its own.
Transcription factors bind the promoter first, and RNA polymerase binds to them.
In a human skin cell, a set of transcription factors binds every promoter and RNA polymerase docks against them. Now imagine every copy of one of those factors is destroyed.
What happens to transcription of the cell’s genes?
- A. Transcription of the genes continues as beforeRNA polymerase binds the promoter through the full set of factors.
With one factor gone, the set is incomplete, and RNA polymerase does not dock. - B. ✓ Transcription of the genes stops
- C. Transcription of the genes speeds upLosing a factor removes RNA polymerase’s foothold on the promoter.
With one factor gone, RNA polymerase does not dock, and transcription stops.
Why: RNA polymerase binds a human promoter only through the bound factors.
With one factor destroyed, the set on the promoter is incomplete.
So RNA polymerase does not dock.
So transcription of the genes stops.
26The switch far away
At E. coli’s lac operon, a regulatory protein binds the DNA beside the promoter while cyclic AMP is bound to it. This protein is an activator.
What does the activator do to the lactose genes?
- A. Blocks RNA polymerase’s path into the genesBlocking RNA polymerase’s path is the repressor’s job, at the operator.
An activator raises how often RNA polymerase transcribes the genes. - B. Makes the repressor let go of the operatorMaking the repressor let go is the inducer’s job: lactose binds the repressor.
An activator binds DNA beside the promoter and raises how often RNA polymerase transcribes the genes. - C. ✓ Raises how often RNA polymerase transcribes the genes
Why: An activator is a regulatory protein that binds DNA beside a gene.
Bound there, it raises how often RNA polymerase transcribes the gene.
Some switches on a human gene lie far from the gene.
A far switch is a short stretch of DNA that proteins bind.
The DNA loops. So the proteins on the far stretch touch the factors at the promoter.
Video: Watch: The switch far away
The human gene is drawn with its factors and docked RNA polymerase. Far to the left, two rounded proteins sit on a short open stretch of DNA. The DNA between the stretch and the promoter rises into a loop, and the far proteins come down to touch the factors on the promoter. RNA polymerase molecules then come off the promoter one after another, each trailing RNA.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L30b.mp4
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Go back to the far proteins, 30,000 base pairs to the left of the promoter.
In the drawings, an open box on the DNA is a short stretch of DNA that proteins can bind. The open box far to the left is the far stretch.
The far stretch is a regulatory sequence, like the regulatory sequence beside a bacterial gene.
The far proteins’ binding sites fit the far stretch. So the far proteins bind the far stretch.
The drawing is not to scale. The real gap between the far stretch and the promoter is 30,000 base pairs of DNA.
A protein 30,000 base pairs away cannot reach the promoter along a straight line of DNA.
Inside the nucleus, the DNA is not a straight line. The DNA between the far stretch and the promoter bends into a loop.
In the drawings, the DNA drawn as a loop is the same DNA, bent back on itself.
The loop brings the far stretch to rest against the promoter. So the far proteins touch the transcription factors on the promoter.
Held by the far proteins, the factors hold RNA polymerase at the start site. So RNA polymerase begins transcribing the gene more often.
A regulatory sequence that proteins bind so that the DNA loops and the gene is transcribed more often is called an , because it enhances transcription: it turns the rate up.
The proteins bound to an enhancer raise how often the gene is transcribed. So these proteins are activators, the same kind of protein as the activator beside the lac promoter.
An activator in a human cell is a transcription factor: the activator binds the DNA beside a gene and helps RNA polymerase transcribe that gene.
A bacterium’s repressor and activator, and a eukaryotic cell’s transcription factors, are all regulatory proteins. Each of these proteins binds a regulatory sequence and changes how often a gene is transcribed.
Transcription factors bound to the promoter and to an enhancer contact RNA polymerase and hold it at the start site, so the gene is transcribed more often.
The enhancer is DNA. The activators bound to it are proteins.
An enhancer may lie before its gene, after its gene, or inside one of its introns, near or far. The loop can usually form from either side of the gene.
What you are expected to know Describe an enhancer: a regulatory sequence, near or far from its gene, that activators bind so the DNA loops and the gene is transcribed more often.
In a human cell, an enhancer lies 11,000 bp from a gene’s promoter.
Which molecules bind the enhancer?
- A. RNA polymeraseRNA polymerase binds the factors at the promoter, never the enhancer.
Activators bind the enhancer. - B. ✓ Activators
- C. The promoterThe promoter is DNA, and DNA does not bind DNA.
Activators, which are proteins, bind the enhancer; the loop brings them to the promoter.
Why: An enhancer is a regulatory sequence.
Activators have binding sites that fit it.
So activators bind the enhancer, and the DNA loops to bring them to the promoter.
Suppose an enhancer lies 33,000 bp before an axolotl gene’s promoter. While activators are bound to the enhancer, RNA polymerase transcribes the gene far more often than while the enhancer is bare.
(a) Explain how activators bound 33,000 bp from the promoter raise how often the gene is transcribed. (2 pt)
Frame Activators bound 33,000 bp from the promoter raise transcription because …
The loop brings the enhancer to rest against the promoter.
So the activators touch the transcription factors bound to the promoter.
Together, the activators and the factors hold RNA polymerase at the start site.
So RNA polymerase begins transcribing the gene more often.
- Award 1 point for: the DNA loops, bringing the enhancer (and its bound activators) into contact with the transcription factors at the promoter.
- Award 1 point for: the activators and factors together hold RNA polymerase at the start site, so the gene is transcribed more often.
A student says: “RNA polymerase binds the enhancer, and from there it slides along the DNA to the promoter.”
Is the student correct?
- A. ✓ No: activators bind the enhancer, and RNA polymerase binds the factors at the promoter
- B. Yes: RNA polymerase binds that far stretch first and then moves along to the promoterRNA polymerase binds no enhancer.
Activators bind the enhancer, and the loop brings them to the factors at the promoter, where RNA polymerase docks.
Why: Activators bind the enhancer.
The DNA loops, so the activators touch the factors at the promoter.
RNA polymerase binds those factors at the promoter; it never binds the enhancer.
Suppose an enhancer lies 17,000 bp before a sturgeon gene. A biologist cuts the enhancer out and joins it back into the DNA 6,300 bp after the gene’s end.
Does the moved enhancer still raise transcription of the gene?
- A. ✓ Yes
- B. NoActivators bind the enhancer wherever it sits.
The loop can usually form from either side of the gene, so the activators still reach the factors at the promoter.
Why: Activators bind the enhancer wherever it sits.
The DNA loops so the bound activators touch the factors at the promoter.
The loop can usually form from either side of the gene.
So the moved enhancer still raises transcription.
54Quick quiz: enhancer mixed practice
The stretch of DNA just before a gene where three transcription factors bind and RNA polymerase docks against them.
Is this part an enhancer?
- A. YesThe stretch just before the gene where the factors bind and RNA polymerase docks is the promoter.
An enhancer is a different stretch, which activators bind. - B. ✓ No
Why: The stretch just before the gene where the factors bind and RNA polymerase docks is the promoter.
An enhancer is a regulatory sequence that activators bind so the gene is transcribed more often.
So this part is not an enhancer.
A stretch of DNA 14,000 bp before a gene. While proteins sit on it, the DNA loops to the promoter and the gene is transcribed more often.
Is this part an enhancer?
- A. ✓ Yes
- B. NoThis stretch of DNA lies far from the gene, proteins bind it, and the gene is transcribed more often while they do.
A regulatory sequence like this is an enhancer.
Why: The part is a stretch of DNA far from the gene.
Proteins bind it, the DNA loops, and the gene is transcribed more often.
So the stretch is an enhancer.
A stretch of DNA inside a gene’s second intron. Proteins bound to it loop the DNA back to the promoter, and the gene is transcribed more often.
Is this part an enhancer?
- A. ✓ Yes
- B. NoAn enhancer can lie inside its gene, not only before or after it.
A stretch of DNA that bound proteins loop to the promoter to raise transcription is an enhancer.
Why: The part is a stretch of DNA that proteins bind.
The DNA loops and the gene is transcribed more often.
An enhancer may lie inside its gene, so this stretch is an enhancer.
A protein is bound to the DNA 21,000 bp from a gene. While the protein is bound, the gene is transcribed more often.
Is this protein an enhancer?
- A. YesAn enhancer is DNA.
A protein that binds far from a gene and raises transcription is an activator. - B. ✓ No
Why: An enhancer is a stretch of DNA.
This part is a protein bound to the DNA.
A protein that raises transcription is an activator, not an enhancer.
The stretch of DNA beside a bacterial promoter that the repressor binds, blocking RNA polymerase’s path into the genes.
Is this part an enhancer?
- A. YesThe stretch beside a bacterial promoter that the repressor binds is the operator.
A repressor bound there lowers transcription; an enhancer’s activators raise it. - B. ✓ No
Why: The stretch beside a bacterial promoter that the repressor binds is the operator.
The repressor bound there blocks transcription.
An enhancer is bound by activators and raises transcription, so this part is not an enhancer.
What is an enhancer?
- A. ✓ A regulatory sequence, near or far from its gene, that activators bind so the gene is transcribed more often
- B. A protein that binds DNA far from a gene and loops the DNA back to the gene’s promoterA protein that binds far from a gene and raises transcription is an activator.
An enhancer is the stretch of DNA the activators bind. - C. The stretch of DNA at the transcription start site where RNA polymerase docks against the factorsThe stretch at the start site where RNA polymerase docks is the promoter.
An enhancer is a regulatory sequence, near or far, that activators bind.
Why: A regulatory sequence that can lie far from its gene, which activators bind so that the DNA loops and the gene is transcribed more often, is called an enhancer.
A eukaryotic gene has switches near it and switches far from it.
(a) State what an enhancer is. (1 pt)
- Award 1 point for: a DNA sequence (regulatory sequence), which may lie far from the gene, that activator proteins bind to raise the gene’s transcription. ‘The DNA loops’ completes it, but its absence does not lose the point.
62Before the start site, or after it
In a signaling pathway, the message passes from the receptor to relay 1, then to relay 2, then to the response.
Which component is upstream of relay 1?
- A. Relay 2Relay 2 acts after relay 1, so relay 2 is downstream of it.
A component that acts before a step is upstream of that step: the receptor. - B. The responseThe response comes last, after relay 1, so the response is downstream of it.
The receptor acts before relay 1, so the receptor is upstream. - C. ✓ The receptor
Why: In a pathway, a component that acts before a step is called upstream of that step, like a town upriver.
The receptor acts before relay 1.
So the receptor is upstream of relay 1.
A gene map has a ruler of its own: the transcription start site.
A position on the DNA is named by which side of the start site it lies on.
Video: Watch: Before the start site, or after it
The human gene is drawn as a straight line with its promoter, its bent start arrow and its exons. RNA polymerase moves from the start arrow to the right along the gene. A bracket appears under the DNA to the left of the arrow, and a second bracket under the DNA to the right of it, and positions on each side are pointed to in turn.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L30c.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L30c.mp4
In Unit 4, upstream and downstream named the order of steps in a signaling pathway.
The receptor acts first. So the receptor is upstream of the relays.
On a gene map, the same two words name positions along the DNA. The ruler is different: here it is the transcription start site, not the order of steps in a pathway.
Go back to the human gene. RNA polymerase starts at the transcription start site and moves along the gene, to the right in the drawing.
A position on the DNA before the transcription start site is called of the start site. RNA polymerase moves away from an upstream position, like a river flowing away from a town upriver.
A position on the DNA after the transcription start site, inside the gene or beyond it, is called of the start site. RNA polymerase moves toward a downstream position, like a river flowing toward a town downriver.
The promoter sits at or just upstream of the start site. RNA polymerase docks there and moves away from it into the gene.
RNA polymerase copies through the gene’s exons and introns after it starts. So the exons and introns all lie downstream of the start site.
An enhancer may sit anywhere: far upstream of the start site, downstream beyond the gene’s last exon, or inside an intron.
What you are expected to know Classify a position on a gene map as upstream or downstream of the transcription start site, and state where a promoter sits and where an enhancer may sit.
Now suppose circled letters mark six positions on the gene map. In the map, the open box is the promoter and the filled boxes are the exons.
The bent arrow marks the transcription start site and points the way RNA polymerase moves. The map is not to scale.
Position J is marked on the gene map below, 2,000 bp from the transcription start site. In the drawing, the bent arrow marks the transcription start site.
Is position J upstream or downstream of the start site?
- A. ✓ Upstream
- B. DownstreamPosition J lies to the left of the bent arrow, before the start site.
A position before the start site is upstream of it.
Why: The bent arrow marks the transcription start site.
Position J lies before the arrow, on the side RNA polymerase moves away from.
So position J is upstream of the start site.
Position L is marked on the gene map below, 1,800 bp from the transcription start site. In the drawing, the bent arrow marks the transcription start site.
Is position L upstream or downstream of the start site?
- A. UpstreamPosition L lies to the right of the bent arrow, between the first and second exons.
A position after the start site is downstream of it. - B. ✓ Downstream
Why: The bent arrow marks the transcription start site.
Position L lies after the arrow, inside the gene, where RNA polymerase moves toward.
So position L is downstream of the start site.
Position N is marked on the gene map below, 9,500 bp from the transcription start site. In the drawing, the bent arrow marks the transcription start site.
Is position N upstream or downstream of the start site?
- A. UpstreamPosition N lies to the right of the bent arrow, past the gene’s last exon.
A position after the start site, even beyond the gene, is downstream of it. - B. ✓ Downstream
Why: The bent arrow marks the transcription start site.
Position N lies after the arrow, beyond the gene’s end.
A position after the start site is downstream of it, inside the gene or beyond it.
Position K is marked on the gene map below, 350 bp from the transcription start site. In the drawing, the bent arrow marks the transcription start site.
Is position K upstream or downstream of the start site?
- A. ✓ Upstream
- B. DownstreamPosition K lies to the left of the bent arrow, just before the promoter.
A position before the start site is upstream of it, however close.
Why: The bent arrow marks the transcription start site.
Position K lies before the arrow, just left of the promoter box.
So position K is upstream of the start site.
Position P is marked on the gene map below, 70 bp from the transcription start site. In the drawing, the bent arrow marks the transcription start site.
Is position P upstream or downstream of the start site?
- A. ✓ Upstream
- B. DownstreamPosition P lies on the promoter, to the left of the bent arrow.
The promoter sits at or just upstream of the start site.
Why: The bent arrow marks the transcription start site.
Position P lies on the promoter, before the arrow.
The promoter sits at or just upstream of the start site, so position P is upstream.
Position M is marked on the gene map below, 3,900 bp from the transcription start site. In the drawing, the bent arrow marks the transcription start site.
Is position M upstream or downstream of the start site?
- A. UpstreamPosition M lies to the right of the bent arrow, between the second and third exons.
A position inside the gene is after the start site, so it is downstream. - B. ✓ Downstream
Why: The bent arrow marks the transcription start site.
Position M lies after the arrow, inside the gene’s second intron.
So position M is downstream of the start site.
Where on a gene map can a promoter sit?
- A. Anywhere: upstream, downstream or inside the geneA sequence that can sit anywhere near or far is an enhancer.
The promoter sits at or just upstream of the start site, where RNA polymerase docks. - B. Downstream of the gene’s last exonDownstream of the last exon is beyond the gene, where RNA polymerase has already passed.
The promoter sits at or just upstream of the start site. - C. ✓ At or just upstream of the transcription start site
Why: RNA polymerase docks at the promoter and moves away from it into the gene.
So the promoter sits at or just upstream of the transcription start site.
This table compares the promoter with the enhancer: where each sits, how near, what binds it, and what happens when it is bound.
Go back to the human gene drawn as a line.
The three transcription factors bind the promoter, and RNA polymerase docks against them.
The far stretch 30,000 base pairs away is an enhancer. The activators on it loop the DNA back to the promoter.
The enhancer lies upstream of the start site, before it. The exons lie downstream, after it.
91Quick quiz: upstream, downstream mixed practice
A regulatory sequence sits inside a gene’s first intron.
Is this position upstream or downstream of the transcription start site?
- A. UpstreamThe introns lie inside the gene, which RNA polymerase copies after it starts.
A position inside the gene is downstream of the start site. - B. ✓ Downstream
Why: RNA polymerase starts at the transcription start site and copies through the gene’s exons and introns.
The first intron lies after the start site.
So the sequence is downstream of the start site.
A stretch of DNA sits 4,700 bp beyond a gene’s last exon.
Is this position upstream or downstream of the transcription start site?
- A. UpstreamThe last exon lies after the start site, and this stretch lies beyond it.
A position after the start site, inside the gene or beyond it, is downstream. - B. ✓ Downstream
Why: The gene’s last exon lies after the transcription start site.
The stretch lies beyond the last exon, further from the start site on the same side.
So the stretch is downstream of the start site.
A gene’s promoter.
Is this position upstream or downstream of the transcription start site?
- A. ✓ Upstream
- B. DownstreamRNA polymerase docks at the promoter and moves away from it into the gene.
The promoter sits at or just upstream of the start site.
Why: RNA polymerase docks at the promoter and moves away from it into the gene.
So the promoter sits at or just upstream of the transcription start site.
An enhancer sits on the far side of a gene’s promoter from the gene.
Is this position upstream or downstream of the transcription start site?
- A. ✓ Upstream
- B. DownstreamThe promoter sits just upstream of the start site, and this enhancer lies beyond the promoter, further from the gene.
So the enhancer is upstream too.
Why: The promoter sits at or just upstream of the start site.
The enhancer lies on the far side of the promoter from the gene, further from the start site on the same side.
So the enhancer is upstream of the start site.
On a gene map, what does upstream of the transcription start site mean?
- A. Acting before a step in a signaling pathwayActing before a step is Unit 4’s pathway sense of the word.
On a gene map, upstream means a position before the transcription start site. - B. ✓ Before the start site, on the side RNA polymerase moves away from
- C. After the start site, inside the gene or beyond itAfter the start site is downstream.
Upstream means before the start site, on the side RNA polymerase moves away from.
Why: On a gene map, a position on the DNA before the transcription start site is called upstream of the start site.
RNA polymerase moves away from an upstream position.
On a gene map, what does downstream of the transcription start site mean?
- A. Before the start site, where the promoter sitsBefore the start site is upstream, where the promoter sits.
Downstream means after the start site, inside the gene or beyond it. - B. Acting after a step in a signaling pathwayActing after a step is Unit 4’s pathway sense of the word.
On a gene map, downstream means a position after the transcription start site. - C. ✓ After the start site, inside the gene or beyond it
Why: On a gene map, a position on the DNA after the transcription start site, inside the gene or beyond it, is called downstream of the start site.
RNA polymerase moves toward a downstream position.
A gene map marks the transcription start site with a bent arrow.
(a) State what upstream of the transcription start site means on a gene map. (1 pt)
- Award 1 point for: before the transcription start site (on the side RNA polymerase does not move into). Not the pathway sense.
(b) State what downstream of the transcription start site means on a gene map. (1 pt)
- Award 1 point for: after the transcription start site (within the gene or beyond it). Not the pathway sense.
99Mixed practice mixed practice
In a human cell, a gene’s promoter has just been bound by its three transcription factors.
Which of the following happens next?
- A. ✓ RNA polymerase binds to the factors and docks over the start site
- B. The factors leave the promoter so that RNA polymerase can bind itRNA polymerase cannot bind a bare human promoter.
The factors stay, and RNA polymerase binds to them. - C. RNA polymerase binds an enhancer and slides to the promoterRNA polymerase binds no enhancer.
RNA polymerase binds the factors bound to the promoter and docks there.
Why: In a human cell, RNA polymerase cannot bind the promoter on its own.
The factors are now bound to the promoter.
So RNA polymerase binds to the factors and docks over the start site.
In a human cell, activators sit on an enhancer 36,000 bp upstream of a gene for a liver enzyme, the DNA is looped, and the gene is transcribed often. The promoter’s three factors and RNA polymerase are bound. Now imagine every one of those activators is removed from the cell.
What happens to transcription of the gene?
- A. Transcription of the gene stopsThe factors and RNA polymerase are still bound at the promoter.
Without the activators, RNA polymerase still docks, but less often. - B. Transcription of the gene stays as oftenThe activators were holding RNA polymerase at the start site more often.
With the activators gone, the loop is gone, and the gene is transcribed less often. - C. ✓ Transcription of the gene falls to its low rate
Why: The activators on the enhancer were holding RNA polymerase at the start site more often.
With the activators gone, the loop no longer forms.
The factors still bind the promoter, so RNA polymerase still docks there sometimes.
So transcription falls to its low rate.
A biologist finds a stretch of DNA 48,000 bp from a gene in a human cell. Proteins bind the stretch, and while they are bound the gene is transcribed more often. A student says: “That stretch is an enhancer of the gene.”
Is the student correct?
- A. No: an enhancer must lie within a few hundred base pairs of the promoter, and this stretch lies far beyond thatAn enhancer can lie far from its gene; the DNA loops it back to the promoter.
A stretch that proteins bind to raise transcription is an enhancer at any distance. - B. ✓ Yes: a stretch of DNA that proteins bind to raise a gene’s transcription is an enhancer, at any distance
Why: An enhancer is a regulatory sequence that can lie far from its gene.
Proteins bind this stretch, and the gene is transcribed more often while they do.
So the stretch is an enhancer of the gene.
RNA polymerase is copying a gene. RNA polymerase has just moved past a regulatory sequence.
Is that regulatory sequence upstream or downstream of the transcription start site?
- A. UpstreamRNA polymerase moves away from every upstream position and never reaches one.
A sequence it has moved past lies after the start site: downstream. - B. ✓ Downstream
Why: RNA polymerase starts at the transcription start site and moves toward downstream positions.
RNA polymerase has moved past this sequence.
So the sequence lies after the start site: downstream of it.
A lung cell and a gut cell from one person carry the same gene. In the lung cell, the promoter’s factors and RNA polymerase are bound, and no activator sits on the gene’s enhancer. In the gut cell, the factors and RNA polymerase are bound, and activators sit on the enhancer with the DNA looped.
Which cell transcribes the gene more often?
- A. ✓ The gut cell
- B. The lung cellIn the lung cell the enhancer is bare, so nothing holds RNA polymerase at the start site beyond the factors.
The gut cell’s bound activators hold it there more often. - C. The two cells, equally oftenBoth cells have RNA polymerase docked, but only the gut cell has activators on the enhancer.
Bound activators hold RNA polymerase at the start site more often.
Why: In both cells the factors and RNA polymerase are bound at the promoter.
Only in the gut cell do activators sit on the enhancer with the DNA looped.
The activators hold RNA polymerase at the start site more often.
So the gut cell transcribes the gene more often.
A hormone binds its receptor in a human cell. The pathway ends at a transcription factor, which binds an enhancer 52,000 bp upstream of a gene’s start site.
In the pathway, is the transcription factor upstream or downstream of the receptor?
- A. UpstreamUpstream on the gene map names the enhancer’s position on the DNA.
In the pathway, the factor acts after the receptor, so it is downstream of the receptor. - B. ✓ Downstream
Why: In a pathway, a component that acts after a step is downstream of that step.
The transcription factor acts after the receptor, at the end of the pathway.
So the factor is downstream of the receptor, even though the enhancer it binds lies upstream of the gene’s start site.
Now imagine a human cell has lost every copy of the promoter’s transcription factors, while activators still sit on one gene’s enhancer.
How much RNA is made from that gene?
- A. A great dealActivators raise transcription only by holding RNA polymerase at a promoter its factors have bound.
With no factors, RNA polymerase never docks. - B. A littleRNA polymerase binds a human promoter only through the promoter’s factors.
With no factors in the cell, RNA polymerase never docks.
So no RNA is made. - C. ✓ None
Why: In a human cell, RNA polymerase binds the promoter only through the bound transcription factors.
The cell has no factors.
So RNA polymerase never docks, and activators alone cannot dock it.
So no RNA is made.
Suppose a stretch of DNA lies 23,000 bp upstream of a quail gene’s transcription start site. While the quail’s regulatory proteins are bound to that stretch, RNA polymerase transcribes the gene often. While no such protein is bound, RNA polymerase transcribes the gene rarely.
(a) Identify the kind of regulatory sequence the stretch is. (1 pt)
- Award 1 point for: an enhancer.
(b) Explain how this case demonstrates that a switch far from a gene can act at the gene’s promoter. (2 pt)
Frame The far stretch acts at the promoter because …
The loop brings the bound activators to the promoter, where they touch the transcription factors.
The activators and the factors together hold RNA polymerase at the start site.
So RNA polymerase transcribes the gene often while the activators are bound.
With no activators bound, the loop does not form.
So RNA polymerase transcribes the gene rarely.
- Award 1 point for: the DNA loops, so the activators bound 23,000 bp away come into contact with the transcription factors at the promoter.
- Award 1 point for: the activators and factors together hold RNA polymerase at the start site, so the gene is transcribed often while they are bound and rarely when they are not (the case’s two rates).
Glossary
- enhancer
- A regulatory sequence that can lie far from its gene, upstream, downstream or inside it, which activators bind so that the DNA loops, the activators touch the transcription factors at the promoter, and the gene is transcribed more often. The enhancer is DNA; the activators bound to it are proteins.
- upstream
- On a gene map, a position on the DNA before the transcription start site, on the side RNA polymerase moves away from. A promoter sits at or just upstream of the start site. (In Unit 4, upstream named a component that acts before a step in a signaling pathway: the same word, a different ruler.)
- downstream
- On a gene map, a position on the DNA after the transcription start site, inside the gene or beyond it, on the side RNA polymerase moves toward. (In Unit 4, downstream named a component that acts after a step in a signaling pathway: the same word, a different ruler.)
APBIO-U06-L31 Reporter constructs
The table above shows five pieces of DNA, each joined to a gene whose protein glows green. Each joined piece is called a construct.
Construct 1 carries the promoter only. Construct 2 carries the promoter plus a candidate stretch of DNA from near a wing gene. Construct 3 carries the same stretch, flipped end to end. Constructs 4 and 5 repeat 2 and 3 in a leg cell instead of a wing cell. Which constructs glow, and what does each glow show about the stretch?
Unit 6 · Gene Expression and Regulation
1Reading a reporter-construct table
In a eukaryotic cell, a stretch of DNA far from a gene raises how often that gene is transcribed, and activator proteins bind that stretch.
What is the stretch of DNA called?
- A. A promoterThe promoter is where the transcription factors bind and RNA polymerase docks, at the start of the gene.
A far regulatory sequence that raises transcription is an enhancer. - B. An activatorAn activator is a protein.
The stretch of DNA that activators bind, far from the gene, is an enhancer. - C. ✓ An enhancer
Why: A regulatory sequence that can lie far from the gene, upstream, downstream or inside it, is an enhancer.
Activator proteins bind it.
The DNA loops so the bound activators touch the factors at the promoter, and the gene is transcribed more often.
Suppose an enhancer is cut out and joined back in on the other side of its gene.
Does the enhancer still raise transcription of the gene?
- A. ✓ Yes
- B. NoThe activators still bind the enhancer.
The DNA usually loops from either side of the gene, so the activators still touch the factors at the promoter.
Why: Activators bind the enhancer wherever it sits.
The DNA loops so the bound activators touch the transcription factors at the promoter.
The loop can usually form from either side of the gene.
So the enhancer still raises transcription.
A wing cell and a leg cell of one animal carry the same genes.
Why do the two cells transcribe different sets of genes?
- A. Each kind of cell has lost the genes it does not useDifferentiation removes no genes; both cells carry the whole genome.
Each kind of cell holds its own set of transcription factors, and so transcribes its own set of genes. - B. ✓ Each kind of cell holds its own set of transcription factors
- C. Each kind of cell carries different promoters on its genesThe two cells carry the same DNA, so every gene has the same promoter in both.
What differs is the set of transcription factors each kind of cell holds.
Why: Both cells carry the same genes.
Each kind of cell holds its own set of transcription factors.
A gene is transcribed often only where its factors are present.
So each kind of cell transcribes its own set of genes.
How do you test whether a stretch of DNA is a switch?
Join the candidate stretch to a gene whose protein is easy to see, and read the glow.
Promoter alone: a faint glow everywhere, so the promoter is enough for a little transcription.
Promoter plus the stretch: a bright glow in wing cells only, so the stretch is a wing-specific enhancer.
The stretch flipped: still bright, so this enhancer works either way round.
The glow reports where and how much the switch drives transcription; it reports nothing about the protein the real gene makes.
The drawing below shows one construct: pieces of DNA joined end to end.
An open box is a piece of DNA the team joined in. The filled box is the glow gene.
The arrow inside the stretch’s box shows which way round the stretch was joined.
The glow gene’s protein is easy to see: cells that transcribe the gene glow green under blue light.
The table below lists the five constructs, the cells each was put into, and the glow of each.
Construct 1 carries the promoter alone.
In wing cells and in leg cells, construct 1 glows faint.
So the promoter on its own gives a little transcription of the glow gene.
That faint glow is the level every other construct is compared with.
Construct 2 carries the stretch joined upstream of the promoter.
In wing cells, construct 2 glows bright.
Bright is far above the promoter’s own faint glow. So the stretch raised transcription in wing cells.
A stretch of DNA that bound proteins use to raise a gene’s transcription is an enhancer. In wing cells, the stretch is an enhancer.
Construct 3 carries the same stretch, flipped end to end.
In wing cells, construct 3 glows bright too.
So the stretch raises transcription either way round.
To read any row of a reporter-construct table, follow three steps.
- Read the glow of the promoter alone: the promoter’s own level.
- Read the glow of the row with the stretch, in the same cells.
- A brighter glow means the stretch raises transcription there; the same glow means it raises nothing.
Now suppose you read constructs 4 and 5, the stretch in leg cells, the same way.
The table below shows construct 1, the promoter alone, and construct 4, the stretch joined to the promoter and put into leg cells. In the drawings, an open box is a piece of DNA joined in, and the filled box is the glow gene. The arrow inside the stretch’s box shows which way round the stretch was joined.
What does construct 4’s glow show about the stretch?
- A. ✓ The stretch does not raise transcription in leg cells
- B. The stretch raises transcription in leg cellsConstruct 4 glows faint in leg cells, the same as the promoter alone.
A glow no brighter than the promoter’s own shows the stretch raised nothing.
Why: Construct 1, the promoter alone, glows faint in leg cells.
Construct 4, the stretch with the promoter, glows faint in leg cells too.
The glow is no brighter than the promoter’s own.
So the stretch does not raise transcription in leg cells.
The whole table of five constructs is drawn below. In the drawings, an open box is a piece of DNA joined in, and the filled box is the glow gene. The arrow inside the stretch’s box shows which way round the stretch was joined.
Which two constructs, with the stretch joined the original way round, together show that the stretch raises transcription in wing cells alone?
- A. ✓ Constructs 2 and 4
- B. Constructs 2 and 3Constructs 2 and 3 were both put into wing cells.
Comparing them shows only that flipping the stretch changed nothing. - C. Constructs 1 and 5Construct 1 carries no stretch, and construct 5 carries the stretch flipped, in leg cells.
Two constructs that differ in two ways cannot show what one difference does.
Why: Constructs 2 and 4 carry the same DNA: the stretch joined to the promoter.
Construct 2 glows bright in wing cells.
Construct 4 glows faint in leg cells.
The kind of cell is the only difference.
So the stretch raises transcription in wing cells and not in leg cells.
Wing cells hold activators whose binding sites fit the stretch.
The activators bind the stretch. The DNA loops.
The bound activators touch the transcription factors on the promoter.
So RNA polymerase transcribes the glow gene often. The glow is bright.
Leg cells hold no activator that fits the stretch.
So the stretch sits empty. The DNA does not loop.
The promoter’s own factors still hold RNA polymerase at the start site now and then.
So the glow gene is transcribed a little. The glow is faint.
The stretch is the same DNA in both kinds of cell. Only the activators each kind of cell holds differ.
A gene whose protein is easy to see, joined to a stretch of DNA to test what the stretch does, is called a , because its glow reports what the stretch does.
In these five constructs, the glow gene is the reporter gene.
A reporter gene shows where a stretch drives transcription, and how much.
The construct carries the reporter gene, not the wing gene. So the glow shows how much transcription the stretch drives, never what the wing gene’s protein does.
What you are expected to know Read a reporter-construct table: for each construct and kind of cell, read the glow, compare it with the promoter alone, and say whether the stretch raises transcription in those cells.
Now suppose you read six rows from other teams’ tables, one table each.
Suppose a team tests a stretch of DNA from near a root gene of a plant. The table below shows the promoter alone and the stretch with the promoter, both put into root cells. In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined.
What does the glow of the row with the stretch show?
- A. The stretch does not raise transcription in these cellsThe row with the stretch glows bright in root cells, and the promoter alone glows faint.
A glow brighter than the promoter’s own shows the stretch raised transcription. - B. ✓ The stretch raises transcription in these cells
Why: The promoter alone glows faint in root cells.
The row with the stretch glows bright in root cells.
Bright is far above the promoter’s own glow.
So the stretch raises transcription in root cells.
Suppose a team joins a stretch of DNA from near a root gene of a plant to the promoter, and puts the construct into the plant’s flower cells. The table below shows the promoter alone and the stretch with the promoter, both in flower cells. In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined.
What does the glow of the row with the stretch show?
- A. ✓ The stretch does not raise transcription in these cells
- B. The stretch raises transcription in these cellsThe row with the stretch glows faint in flower cells, the same as the promoter alone.
A glow no brighter than the promoter’s own shows the stretch raised nothing.
Why: The promoter alone glows faint in flower cells.
The row with the stretch glows faint in flower cells too.
The glow is no brighter than the promoter’s own.
So the stretch does not raise transcription in flower cells.
Suppose a team joins a stretch of DNA from near a seahorse’s snout gene to the promoter, flipped end to end, and puts it into the seahorse’s eye cells. The table below shows both rows in eye cells. In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined.
What does the glow of the row with the stretch flipped show?
- A. ✓ The stretch does not raise transcription in these cells
- B. The stretch raises transcription in these cellsThe row with the stretch flipped glows faint in eye cells, the same as the promoter alone.
A glow no brighter than the promoter’s own shows the stretch raised nothing.
Why: The promoter alone glows faint in eye cells.
The row with the stretch flipped glows faint in eye cells too.
The glow is no brighter than the promoter’s own.
So the stretch does not raise transcription in eye cells.
Suppose a team joins a stretch of DNA from near a snout gene of a seahorse to the promoter, flipped end to end, and puts it into the seahorse’s snout cells. The table below shows both rows in snout cells. In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined.
What does the glow of the row with the stretch flipped show?
- A. The stretch does not raise transcription in these cellsThe row with the stretch flipped glows bright in snout cells, and the promoter alone glows faint.
A glow brighter than the promoter’s own shows the stretch raised transcription. - B. ✓ The stretch raises transcription in these cells
Why: The promoter alone glows faint in snout cells.
The row with the stretch flipped glows bright in snout cells.
Bright is far above the promoter’s own glow.
So the stretch raises transcription in snout cells, flipped or not.
Suppose a team tests a stretch of DNA from near a gene of a bird in the bird’s skin cells. The table below shows the promoter alone and the stretch with the promoter, both in skin cells. In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined.
What does the glow of the row with the stretch show?
- A. The stretch does not raise transcription in these cellsThe row with the stretch glows bright in skin cells, and the promoter alone glows faint.
A glow brighter than the promoter’s own shows the stretch raised transcription. - B. ✓ The stretch raises transcription in these cells
Why: The promoter alone glows faint in skin cells.
The row with the stretch glows bright in skin cells.
Bright is far above the promoter’s own glow.
So the stretch raises transcription in skin cells.
Suppose a team joins a stretch of DNA to the promoter and the reporter gene. In wing cells that construct glows bright. Then the team joins the stretch straight to the reporter gene, with no promoter, in wing cells. The table below shows both rows. In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined.
Why does the no-promoter row show no glow?
- A. ✓ RNA polymerase had no promoter to bind
- B. The stretch blocked RNA polymeraseNothing here blocks RNA polymerase: with a promoter, the same stretch gives a bright glow in wing cells.
This construct has no promoter.
So RNA polymerase has nowhere to bind. - C. Wing cells hold no activator that fits the stretchA missing activator leaves the promoter’s own faint glow, as in leg cells.
No glow at all means no transcription: this construct has no promoter for RNA polymerase to bind.
Why: RNA polymerase binds a gene’s promoter, through the transcription factors.
This construct carries no promoter.
So RNA polymerase never binds, and the reporter gene is never transcribed.
No transcription gives no glow.
A student reads construct 2’s bright glow and says: “The wing gene’s protein glows green in wing cells.”
Is the student correct?
- A. ✓ No: the glow is the reporter gene’s protein, and the construct carries no wing gene
- B. Yes: the stretch came from near the wing gene, so the glow shows that gene’s proteinThe construct carries the stretch and the reporter gene only.
The glowing protein is the reporter gene’s; the wing gene’s own protein is never made from a construct.
Why: The construct carries the stretch joined to the reporter gene.
The wing gene itself is not in the construct.
So the glowing protein is the reporter gene’s.
The glow shows how much transcription the stretch drives, never what the wing gene’s protein does.
The table below shows construct 1 and construct 4. A student reads construct 4’s faint glow in leg cells and says: “No activator is bound to this stretch in leg cells.” In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined.
Is the student correct?
- A. No: the faint glow shows that the promoter does not work in leg cellsThe promoter alone gives a faint glow in leg cells too.
A faint glow shows the promoter working and the stretch raising nothing. - B. ✓ Yes: the stretch raised nothing in leg cells, so no activator was bound to it there
Why: Construct 4 glows faint in leg cells, the same as the promoter alone.
So the stretch raised no transcription there.
A stretch raises transcription only while an activator is bound to it.
So no activator is bound to this stretch in leg cells.
The table below sets out the five constructs, the glow of each, and what each glow shows.
56Quick quiz: reporter gene mixed practice
Suppose a team joins a stretch of DNA to the gene for an enzyme that turns a clear broth blue, and reads the color of the broth.
Which part of the construct is the reporter gene?
- A. The stretch of DNA being testedThe stretch is the DNA under test.
The reporter gene is the gene whose product is easy to see: the enzyme’s gene. - B. ✓ The enzyme’s gene
- C. The promoterThe promoter is where RNA polymerase binds.
The gene whose product is easy to see, the enzyme’s gene, is the reporter gene.
Why: The enzyme turns the broth blue, so its product is easy to see.
A gene whose product is easy to see, joined to a stretch to test it, is a reporter gene.
So the enzyme’s gene is the reporter gene.
What is a reporter gene?
- A. A gene that codes for the activator protein that binds an enhancer far from the gene it controlsAn activator’s gene codes for a DNA-binding protein that is not easy to see.
A reporter gene’s protein is easy to see and reports what a stretch does. - B. The gene whose protein is being studied, joined to a promoter so that its protein glows in every cellThe gene under study is never in the construct; only a stretch of DNA from near it is.
The reporter gene is the easy-to-see gene joined to that stretch. - C. ✓ An easy-to-see gene joined to a stretch of DNA to show what the stretch does to transcription
Why: A gene whose protein is easy to see, joined to a stretch of DNA to test what the stretch does, is called a reporter gene.
Its glow reports what the stretch does to transcription.
A team wants to find out whether a stretch of DNA is a switch on transcription.
(a) State what a reporter gene is. (1 pt)
- Award 1 point for: a gene whose protein (product) is easy to see (glows, colors), joined to a stretch of DNA to show whether or how much the stretch drives transcription. ‘Easy to see’ and ‘joined to the stretch under test’ are both needed; naming a glowing protein completes it but its absence does not lose the point.
60Predicting a new construct
On a gene map, a regulatory sequence sits beyond the end of a gene’s last exon.
Is that sequence upstream or downstream of the start site?
- A. UpstreamUpstream means before the transcription start site on the DNA.
The last exon lies after the start site, and this sequence lies beyond it: downstream. - B. ✓ Downstream
Why: Upstream means before the transcription start site on the DNA.
Downstream means after it.
The gene’s last exon lies after the start site, and the sequence lies beyond that exon.
So the sequence is downstream.
An enhancer is a regulatory sequence that activator proteins bind.
Where can an enhancer sit?
- A. ✓ Upstream of the gene, downstream of it, or inside it
- B. Only upstream of the promoter, near the start siteAn enhancer can lie far from the gene, on either side of it or inside it.
The DNA loops so the bound activators reach the promoter wherever the enhancer sits.
Why: An enhancer is a regulatory sequence that can lie far from the gene.
The DNA loops so the bound activators touch the factors at the promoter.
The loop can usually form from either side.
So an enhancer can sit upstream of the gene, downstream of it, or inside it.
Go back to construct 3: the stretch flipped end to end, put into wing cells.
On the exam, a claim about a stretch is supported with the row of the table that shows it, and the reasoning that connects that row to the claim.
Claim: the stretch raises transcription either way round.
Evidence: construct 3, the stretch flipped, glows bright in wing cells, just like construct 2, the stretch the original way round.
Reasoning: flipping changed only which way round the stretch was joined.
The glow did not fall. So the way round does not change what the stretch does.
Now suppose the team builds a sixth construct: the stretch joined downstream of the reporter gene, after the gene’s end.
Construct 6 carries the same stretch, the same promoter and the same reporter gene as construct 2. Only the stretch’s position has changed.
What you are expected to know Predict the glow of a new construct from the activator each kind of cell holds and from where an enhancer can sit, and support a claim about a stretch with the row of the table that shows it.
The table below shows constructs 1 to 6; construct 6 carries the stretch joined downstream of the reporter gene, and it is put into leg cells. In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined.
Which glow do you predict for construct 6 in leg cells?
- A. No glowConstruct 6 carries the promoter.
The promoter alone gives a faint glow in leg cells, whatever the stretch does. - B. ✓ A faint glow
- C. A bright glowLeg cells hold no activator that binds this stretch.
Constructs 4 and 5 glow faint in leg cells, and moving the stretch downstream adds no activator.
Why: Leg cells hold no activator that binds the stretch.
So the stretch raises nothing in leg cells, wherever it sits.
The promoter alone gives a faint glow.
So construct 6 glows faint in leg cells.
The table below shows constructs 1 to 6; construct 6 carries the stretch joined downstream of the reporter gene, and it is put into wing cells. In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined.
Which glow do you predict for construct 6 in wing cells?
- A. No glowConstruct 6 carries the promoter.
The promoter alone gives at least a faint glow, in wing cells as in leg cells. - B. A faint glowWing cells hold activators that bind the stretch.
An enhancer works from downstream of a gene as well as upstream: the DNA loops to the promoter from either side. - C. ✓ A bright glow
Why: Wing cells hold activators whose binding sites fit the stretch.
An enhancer can sit upstream of its gene or downstream of it.
The DNA loops so the bound activators touch the factors at the promoter.
So the reporter gene is transcribed often, and construct 6 glows bright.
The table below shows constructs 1 to 6. Imagine construct 6, the stretch joined downstream of the reporter gene, had glowed faint in wing cells instead. In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined.
What would that faint glow have shown?
- A. ✓ The stretch does not raise transcription from downstream of the reporter gene
- B. Wing cells have lost the activators that bind the stretchConstruct 2, the stretch upstream of the promoter, glows bright in wing cells.
So wing cells do hold the activators that bind the stretch. - C. The promoter has stopped working in wing cells, whatever the stretch doesA faint glow is the promoter’s own level.
A promoter that had stopped working would give no glow at all.
Why: The promoter alone gives a faint glow.
A faint glow from construct 6 would be no brighter than the promoter alone.
So the stretch would have raised nothing from downstream of the reporter gene.
Suppose a team tests a stretch of DNA from near a petal-color gene of a plant. The team joins the stretch to the promoter and the reporter gene, and reads the glow in petal cells and in leaf cells. The table below shows the three constructs and the glow of each. In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined.
(a) Make a claim about what the stretch does in petal cells. (1 pt)
- Award 1 point for: the stretch raises transcription (drives more transcription; is an enhancer, a switch that turns transcription up) in petal cells. The direction and the cells are both needed.
(b) Support your claim with evidence from the table and the reasoning that connects that evidence to the claim. (2 pt)
The promoter alone glows faint in petal cells.
The stretch is the only difference between those two constructs.
So the stretch raised transcription of the reporter gene in petal cells.
- Award 1 point for evidence: the row with the stretch in petal cells glows bright, against faint for the promoter alone.
- Award 1 point for reasoning: the stretch is the only difference between the two constructs, so the stretch raised transcription. Accept consistent reasoning from a wrong (a): award the reasoning point for a reasoning line that follows correctly from the rows the student cites.
Back to the five constructs: the reporter gene behind the promoter alone, behind the stretch, and behind the stretch flipped, put into wing cells and leg cells.
The promoter alone gives a faint glow.
With the stretch, the glow is bright, but only in wing cells.
Flipped, it is still bright.
So the stretch is a wing-specific enhancer that works either way round.
Glossary
- reporter gene
- A gene whose protein is easy to see, joined to a stretch of DNA to test what the stretch does; its glow shows where the stretch drives transcription, and how much.
APBIO-U06-L32 Blocking the way
Suppose the gene for a bacterium’s lac repressor is deleted. The cell now makes its lactose enzymes whether or not lactose is present.
In a second experiment, a liver cell lacks one of its proteins. That cell makes half the usual amount of a certain enzyme. One protein was removed in each case. In the first case transcription rose. In the second case transcription fell. What was each protein doing before it was removed?
Unit 6 · Gene Expression and Regulation
1A repressor blocks the way
In E. coli, a protein binds the operator of the lac operon and stops transcription of the lactose genes.
What is this protein called?
- A. An activatorAn activator raises how often genes are transcribed.
A protein that binds the operator and stops transcription is a repressor. - B. An inducerAn inducer is the small molecule that binds the repressor and makes it let go.
The protein bound to the operator, stopping transcription, is a repressor. - C. ✓ A repressor
Why: A protein that binds the operator and stops transcription of the genes is called a repressor.
It represses transcription: it holds transcription down.
In a human cell, one gene is wound tight on its histones, and a second gene is loosely wound.
Which gene can RNA polymerase reach?
- A. The tightly wound geneTightly wound DNA is buried in its coil.
RNA polymerase and regulatory proteins can only reach a gene that is loosely wound. - B. ✓ The loosely wound gene
- C. Both genes, equallyThe two genes differ in how tightly they are wound.
RNA polymerase can only reach the loosely wound gene.
Why: RNA polymerase and regulatory proteins can only reach a gene that is loosely wound.
The tightly wound gene is buried in its coil.
So RNA polymerase reaches the loosely wound gene.
How do you tell an activator from a repressor?
Remove it and watch which way the gene goes.
A repressor binds a regulatory sequence and blocks transcription.
At an operator, the bound repressor sits in RNA polymerase’s path.
In a eukaryotic cell, a bound repressor can also bring in proteins that wind the DNA tighter.
Video: Watch: A repressor blocks the way
The lac operon is drawn with its repressor on the operator and RNA polymerase floating above the promoter. Then a human gene is drawn as a line with a short open stretch of DNA far to its left. A blocky repressor lands on the stretch, and the DNA around the promoter and the gene winds into a tight coil. RNA polymerase floats above the coil and does not dock.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L32a.mp4
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Go back to the lac operon with no lactose in the cell.
In the drawings, a block sitting on an open box is a repressor bound to a short stretch of DNA.
In the drawings, an oval floating above the promoter is RNA polymerase off the DNA.
With no lactose, the repressor is bound to the operator.
A repressor protein bound to the operator sits in RNA polymerase’s path on the DNA, so the polymerase cannot move into the genes and they are not transcribed.
The operator is DNA. The repressor blocks transcription by binding DNA, not by binding RNA polymerase.
Bound to the DNA, the repressor lowers how often the genes are transcribed. A regulatory protein that lowers transcription is a negative regulator of its gene.
Now suppose a human gene is drawn as a line. Three transcription factors sit on its promoter, and RNA polymerase is docked against them.
Far to the left sits a short stretch of DNA with nothing on it.
One protein in the nucleus has a binding site that fits the short stretch. So that protein binds the stretch.
The bound protein brings in enzymes. The enzymes take acetyl groups off the histones around the gene.
With the acetyl groups gone, the histones grip the DNA more tightly. So the DNA around the promoter and the gene winds tight.
In the drawings, DNA drawn as a coil of beads is DNA wound tight on its histones.
RNA polymerase and the transcription factors can only reach a gene that is loosely wound. So nothing binds the promoter, and the gene is not transcribed.
The bound protein lowered transcription of the gene. So the bound protein is a repressor, the same kind of protein as the repressor at the operator.
This repressor never sat in RNA polymerase’s path. This repressor blocked transcription by winding the DNA tight.
At an operator, the bound repressor sits in RNA polymerase’s path.
In a eukaryotic cell, a bound repressor can also bring in proteins that wind the DNA tighter.
Either way, the repressor binds DNA. Either way, transcription of the gene is blocked.
A short stretch of DNA that a eukaryotic repressor binds is a regulatory sequence, just like the operator.
What you are expected to know Explain how a bound repressor blocks transcription: at an operator it sits in RNA polymerase’s path, and in a eukaryotic cell it can also bring in proteins that wind the DNA tighter.
In a eukaryotic cell, a repressor lowers transcription of one gene.
Which molecule does the repressor bind?
- A. The gene’s mRNAA repressor acts before any mRNA is made.
It binds a regulatory sequence of DNA beside the gene. - B. ✓ A regulatory sequence of DNA
- C. RNA polymeraseA repressor does not bind RNA polymerase.
It binds a regulatory sequence of DNA, and from there it blocks transcription.
Why: A repressor is a regulatory protein.
Its binding site fits a regulatory sequence of DNA beside the gene.
So the repressor binds that DNA, and transcription of the gene is blocked.
Suppose a biologist studies two silent genes. In a bacterium, a repressor is bound to the operator beside gene 1. In an octopus cell, a repressor is bound to a short stretch of DNA 7,500 bp from gene 2’s promoter.
(a) Explain how the bound repressor stops transcription of gene 1. (1 pt)
Frame The repressor bound to the operator …
So RNA polymerase cannot move into gene 1.
So gene 1 is not transcribed.
- Award 1 point for: the bound repressor sits in RNA polymerase’s path (covers DNA the polymerase needs), so the polymerase cannot move into gene 1 and gene 1 is not transcribed.
(b) Explain how the bound repressor stops transcription of gene 2, 7,500 bp from the promoter. (2 pt)
Frame The repressor bound far from the promoter …
So the DNA around gene 2’s promoter winds tight.
RNA polymerase and the transcription factors can only reach a gene that is loosely wound.
So nothing binds gene 2’s promoter, and gene 2 is not transcribed.
- Award 1 point for: the bound repressor brings in proteins (enzymes that remove acetyl groups) that wind the DNA around gene 2 tighter.
- Award 1 point for: tightly wound DNA cannot be reached by RNA polymerase or the transcription factors, so gene 2 is not transcribed.
- Accept for point 1: the bound repressor covers a sequence the gene’s activators need (its enhancer), so no activator binds and the DNA does not loop.
A student says: “In a eukaryotic cell, a repressor can block transcription only by sitting in RNA polymerase’s path, just like the repressor at an operator.”
Is the student correct?
- A. ✓ No: a eukaryotic repressor can also wind the DNA tight, so RNA polymerase never reaches the gene
- B. Yes: every repressor, in a bacterium or a eukaryotic cell, blocks transcription by sitting where RNA polymerase must passAt an operator the repressor does sit in the path.
A eukaryotic repressor can instead wind the DNA tight, and RNA polymerase never reaches the promoter.
Why: At an operator, the bound repressor sits in RNA polymerase’s path.
In a eukaryotic cell, a bound repressor can also bring in proteins that wind the DNA tighter.
Tightly wound DNA is out of RNA polymerase’s reach.
So the student is not correct.
In a eukaryotic cell, a repressor is bound beside a gene, and the DNA around the gene is wound tight. An activator whose binding site fits the gene’s enhancer enters the nucleus.
Does the activator raise transcription of the gene?
- A. YesA regulatory protein can only reach a sequence that is loosely wound.
The enhancer is buried in the tight coil, so the activator never binds it. - B. ✓ No
Why: The DNA around the gene is wound tight.
RNA polymerase and regulatory proteins can only reach a gene that is loosely wound.
The enhancer is buried in the coil.
So the activator cannot bind it, and transcription stays blocked.
35Lose the repressor
Lactose enters an E. coli cell and binds the lac repressor. The bound lactose changes the repressor’s shape.
What does the repressor do next?
- A. Grips the DNA more tightlyThe changed shape no longer fits the operator.
So the repressor lets go of the DNA. - B. ✓ Lets go of the operator
- C. Binds RNA polymeraseA repressor never binds RNA polymerase.
With lactose bound, the repressor no longer fits the operator and lets go.
Why: The bound lactose changes the repressor’s shape.
The repressor with its new shape no longer fits the operator.
So the repressor lets go of the DNA, and RNA polymerase transcribes the genes.
Delete the repressor’s gene, or change its binding site, and the gene is on when it should be off.
Video: Watch: Lose the repressor
The lac operon is drawn with its repressor on the operator and no lactose in the cell. The repressor vanishes, and RNA polymerase moves through the three genes trailing mRNA. Then the repressor is drawn floating beside a changed operator it cannot fit, and RNA polymerase moves through the genes again. Last, the human gene’s repressor vanishes, its coil loosens, and RNA polymerase docks.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L32b.mp4
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Go back to the lac operon with no lactose in the cell. The repressor sits on the operator, and the genes are not transcribed.
Now imagine the repressor’s gene is deleted. The cell makes no repressor at all.
Nothing sits on the operator. So nothing sits in RNA polymerase’s path, lactose or no lactose.
RNA polymerase moves from the promoter through the three genes. So the cell makes the lactose enzymes all the time.
The genes are on when they should be off.
Now imagine instead that the cell makes the repressor as usual, but the operator’s sequence has changed. The repressor’s binding site no longer fits the changed operator.
The repressor floats free and never binds. Again nothing sits in RNA polymerase’s path, and the genes are transcribed all the time.
Delete the repressor’s gene, change its binding site, or do both: the result is the same. The gene is on when it should be off.
Go back to the human gene with its repressor bound and its DNA wound tight. Now imagine the repressor’s gene is deleted.
No repressor binds the short stretch. So no enzymes are brought in to take the acetyl groups off the histones.
Other enzymes add acetyl groups to the histones. So the DNA around the gene loosens.
The transcription factors bind the promoter, and RNA polymerase docks against them. So the gene is transcribed.
A repressor that is never made, or a binding site it cannot fit, both leave the gene on when it should be off.
Suppose a signal molecule has bound a repressor and lifted it off the DNA. Deleting that repressor’s gene changes nothing: the gene was already on.
Adding a repressor does the reverse. Give a cell a repressor that fits a stretch of DNA beside a gene being transcribed, and that gene switches off.
What you are expected to know Predict what happens to a gene when its repressor is lost: its gene deleted, its binding site changed, or both, so the gene is on when it should be off.
A bacterium grows in a broth with none of the sugar its operon breaks down, and the operon’s repressor is bound to the operator. Now imagine the gene for the repressor is deleted.
What happens to transcription of the operon’s genes?
- A. ✓ Transcription switches on
- B. Transcription switches offWith no repressor made, nothing sits in RNA polymerase’s path.
The operon’s genes are transcribed, sugar or no sugar. - C. Transcription stays as it wasBefore the deletion the bound repressor blocked the genes.
With no repressor made, RNA polymerase moves into the genes.
Why: The repressor’s gene is deleted, so the cell makes no repressor.
Nothing sits on the operator, in RNA polymerase’s path.
So RNA polymerase transcribes the operon’s genes, with none of the sugar present.
E. coli grows in a broth with lactose. Lactose has bound the lac repressor, and the repressor has left the operator. Now imagine the gene for the repressor is deleted.
What happens to transcription of the lactose genes?
- A. Transcription switches onThe genes were already switched on: lactose had lifted the repressor off.
Deleting a repressor that was not bound changes nothing. - B. Transcription switches offDeleting a repressor cannot switch genes off.
The repressor was already off the operator, so transcription continues as before. - C. ✓ Transcription stays as it was
Why: With lactose present, the repressor was already off the operator.
Deleting its gene removes a protein that was not bound.
So nothing changes: the genes stay switched on.
In a parrot’s cell, a repressor keeps one gene silent by winding its DNA tight. Now imagine the repressor’s binding site changes. The repressor’s shape fits only the old sequence.
What happens to transcription of the gene?
- A. ✓ Transcription switches on
- B. Transcription switches offThe gene was already silent.
With the repressor unable to bind, the DNA loosens and the gene is transcribed. - C. Transcription stays as it wasThe repressor no longer binds, so no enzymes tighten the DNA.
The DNA loosens, and the gene is transcribed.
Why: The repressor no longer fits its binding site, so it never binds.
No enzymes are brought in to tighten the DNA, and the DNA around the gene loosens.
So RNA polymerase reaches the gene, and transcription switches on.
A bacterium’s genes for taking up one sugar are transcribed all the time, in a broth with none of that sugar. Now imagine a biologist gives the bacterium a gene for a repressor that fits the operator of those genes.
What happens to transcription of the sugar genes?
- A. Transcription switches onThe genes were already transcribed all the time.
A repressor that fits their operator binds it and blocks RNA polymerase’s path. - B. ✓ Transcription switches off
- C. Transcription stays as it wasThe new repressor fits the operator, so it binds there.
Bound, it sits in RNA polymerase’s path, and transcription stops.
Why: The new repressor’s binding site fits the operator.
So the repressor binds the operator and sits in RNA polymerase’s path.
So transcription of the sugar genes switches off.
In an earthworm’s cell, a hormone has bound a repressor and lifted it off the DNA, and the gene it controls is being transcribed. Now imagine the repressor’s gene is deleted.
What happens to transcription of the gene?
- A. Transcription switches onThe gene was already being transcribed: the hormone had lifted the repressor off.
Deleting a repressor that was not bound changes nothing. - B. Transcription switches offDeleting a repressor never switches a gene off.
The repressor was already off the DNA, so transcription continues as before. - C. ✓ Transcription stays as it was
Why: The hormone had already lifted the repressor off the DNA.
Deleting the repressor’s gene removes a protein that was not bound.
So nothing changes: the gene stays transcribed.
In a lobster’s cell, a repressor keeps one gene silent. Now imagine the repressor’s gene is deleted and its binding site is changed as well.
What happens to transcription of the gene?
- A. ✓ Transcription switches on
- B. Transcription switches offThe gene was already silent.
With no repressor made and no site for one to fit, the DNA loosens and the gene is transcribed. - C. Transcription stays as it wasBoth changes remove the block on the gene.
No repressor binds, the DNA loosens, and the gene is transcribed.
Why: No repressor is made, and no repressor could fit the changed site anyway.
Nothing brings in the enzymes that tighten the DNA, so the DNA loosens.
So RNA polymerase reaches the gene, and transcription switches on.
61Activator or repressor? Remove it and watch
At the lac operon, cyclic AMP binds a protein, and that protein then binds beside the promoter. RNA polymerase then transcribes the genes far more often.
What is this protein called?
- A. An inducerAn inducer is a small molecule that binds a repressor.
A regulatory protein that binds beside a gene and raises how often it is transcribed is an activator. - B. ✓ An activator
- C. A repressorA repressor lowers how often the genes are transcribed.
A regulatory protein that raises it is an activator.
Why: A regulatory protein that binds DNA beside a gene and raises how often RNA polymerase transcribes the gene is called an activator.
It turns transcription up.
A steroid hormone enters a human cell and binds its receptor.
What does the bound receptor do next?
- A. Opens a channel in the plasma membraneA steroid hormone’s receptor sits inside the cell, not in the membrane.
The bound receptor moves into the nucleus and binds the DNA beside a gene. - B. Binds a ribosome and starts translationThe bound receptor acts on DNA, not on a ribosome.
It moves into the nucleus and binds the DNA beside a gene. - C. ✓ Moves into the nucleus and binds the DNA beside a gene
Why: A steroid hormone’s receptor sits inside the cell.
The bound receptor moves into the nucleus and binds the DNA beside a gene.
The bound receptor is itself a transcription factor.
Remove an activator, and the gene’s transcription falls.
Removal that raises transcription names a repressor. Removal that lowers transcription names an activator.
A hormone receptor that enters the nucleus and raises transcription is an activator too.
Video: Watch: Activator or repressor? Remove it and watch
A gene is drawn with a regulatory protein bound beside it and RNA polymerase docked. The protein vanishes, and the mRNA count beside the gene falls. A second gene is drawn with a blocky repressor bound beside it and no transcript. The repressor vanishes, and RNA polymerase moves through the gene trailing mRNA. Beside each, the word activator or repressor appears.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L32c.mp4
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Go back to the liver cell that lacks one of its proteins.
A normal liver cell makes the enzyme’s mRNA at 130 units. The cell that lacks the protein makes it at 65 units.
The units are arbitrary: a bigger number means more mRNA.
Removing the protein halved transcription of the enzyme’s gene. So the protein had been raising transcription.
A regulatory protein that raises how often a gene is transcribed is an activator. So the liver cell’s missing protein is an activator.
The table below lists four removal experiments: the cell, what was removed, the gene’s mRNA before and after, and what the removed protein was.
For example, the liver cell lost one protein. The enzyme’s mRNA fell from 130 units to 65 units.
Removing this protein lowered transcription. So this protein had been raising transcription: an activator.
But the bacterium lost its lac repressor. Its lactose genes went from on only with lactose to on all the time.
Removing this protein raised transcription. So this protein had been lowering transcription: a repressor.
And suppose a flounder’s cell loses one protein. A gene’s mRNA rises from 26 units to 330 units.
Removing this protein raised transcription. So this protein had been lowering transcription: a repressor.
But suppose a maple cell loses one protein. A gene’s mRNA reads 72 units before the loss and 73 units after it.
Removing this protein left transcription unchanged. So this protein had been controlling nothing here: not a regulator of this gene.
Removal that raises transcription names a repressor. Removal that lowers transcription names an activator.
Removal that changes nothing names a protein the gene’s transcription did not depend on. So that protein is not a regulator of that gene here.
Each kind of protein binds its own kind of sequence. The liver cell’s activator binds an enhancer of the enzyme’s gene, or a stretch beside its promoter.
The bacterium’s repressor binds the operator. The flounder cell’s repressor binds a short stretch of DNA beside the gene: a sequence a repressor binds.
In Unit 4, a growth factor’s message ran from its receptor to the nucleus. That pathway ends at transcription factors that switch on the genes for cell division: activators of those genes.
A mutation that locks one such transcription factor on keeps the division genes transcribed with no growth factor. A locked-on factor is one route to a tumor.
Go back to the steroid hormone’s receptor. Once the hormone binds it, the receptor moves into the nucleus and binds beside a gene.
So the gene is transcribed more often.
Remove the receptor, and transcription of that gene falls, hormone or no hormone. So the hormone-bound receptor is an activator.
What you are expected to know Classify a regulatory protein as an activator or a repressor from what removing it does to transcription, and name the kind of sequence it binds.
Suppose a biologist deletes the gene for one regulatory protein in a carp’s cell. Transcription of a nearby gene rises to 20 times its level.
Is the protein an activator or a repressor?
- A. An activatorRemoving an activator lowers transcription.
Here removal raised transcription, so the protein had been lowering it: a repressor. - B. ✓ A repressor
Why: Removing the protein raised transcription of the gene.
So the protein had been lowering transcription.
A regulatory protein that lowers transcription is a repressor.
Suppose a hornet’s cell lacks one regulatory protein. Transcription of a gene it controls falls to 10 % of its level in a normal cell.
Is the protein an activator or a repressor?
- A. ✓ An activator
- B. A repressorRemoving a repressor raises transcription.
Here transcription fell without the protein, so the protein had been raising it: an activator.
Why: Without the protein, transcription of the gene fell.
So the protein had been raising transcription.
A regulatory protein that raises transcription is an activator.
Suppose an iguana’s cell loses a protein that had been bound to a short stretch of DNA right beside a gene’s promoter. Transcription of the gene falls by half.
Is the protein an activator or a repressor?
- A. ✓ An activator
- B. A repressorActivators bind beside promoters too, not only far away.
Removal lowered transcription, so the protein had been raising it: an activator.
Why: Removing the protein lowered transcription of the gene.
So the protein had been raising transcription.
A regulatory protein that raises transcription is an activator, wherever its stretch of DNA sits.
Suppose the gene for one regulatory protein in a cuttlefish’s cell is deleted. A gene that is silent in a normal cell is transcribed all the time in the cell that lacks the protein.
Is the protein an activator or a repressor?
- A. An activatorRemoving an activator lowers transcription: a silent gene would stay silent.
Here the gene switched on without the protein, so the protein had been lowering transcription: a repressor. - B. ✓ A repressor
Why: Without the protein, the silent gene is transcribed all the time.
So the protein had been lowering its transcription.
A regulatory protein that lowers transcription is a repressor.
Suppose a regulatory protein in a gull’s cell is destroyed. Transcription of a gene rises to five times its level. The protein’s binding stretch lies far from the gene’s promoter.
Is the protein an activator or a repressor?
- A. An activatorRemoving an activator lowers transcription.
Here removal raised it fivefold, so the protein had been lowering it: a repressor, even one bound far away. - B. ✓ A repressor
Why: Without the protein, transcription of the gene rose.
So the protein had been lowering transcription.
A regulatory protein that lowers transcription is a repressor, near the promoter or far from it.
Suppose a tuna’s cell loses one regulatory protein. The mRNA of a gene it controls falls to 5 % of its level in a normal cell.
Is the protein an activator or a repressor?
- A. ✓ An activator
- B. A repressorRemoving a repressor raises a gene’s mRNA.
Here the mRNA fell without the protein, so the protein had been raising transcription: an activator.
Why: Without the protein, the gene’s mRNA fell.
So the protein had been raising transcription of the gene.
A regulatory protein that raises transcription is an activator.
Suppose a steroid hormone enters a pike’s cell and binds its receptor. The bound receptor moves into the nucleus, binds beside a gene, and the gene’s transcription rises. Now imagine the receptor’s gene is deleted: with the hormone present, transcription of the gene falls.
Which kind of protein is the hormone-bound receptor?
- A. ✓ An activator
- B. A repressorRemoving a repressor raises transcription.
Removing the receptor lowered it, so the bound receptor had been raising transcription: an activator. - C. Not a regulatory proteinA receptor that enters the nucleus and binds DNA beside a gene is a transcription factor.
Removing it lowered transcription, so the bound receptor is an activator.
Why: The bound receptor binds the DNA beside the gene, so it is a transcription factor.
Removing the receptor lowered transcription of the gene.
So the bound receptor had been raising transcription: an activator.
The table below compares the two results of a removal experiment: what was lost, what it had been bound to, and one example of each.
Go back to the two experiments: the bacterium whose gene for its lac repressor was deleted, and the liver cell that lacks one of its proteins.
The deleted repressor had been blocking transcription. So without it, the lactose genes are transcribed all the time.
The missing liver protein had been raising transcription. So it was an activator.
Removal that raises transcription means a repressor was lost. Removal that lowers transcription means an activator was lost.
104Mixed practice mixed practice
In a crow’s cell, a repressor is bound 6,200 bp upstream of a gene’s promoter, and the gene is silent.
How does the bound repressor keep the gene silent?
- A. By binding RNA polymerase and holding it off the promoterA repressor binds DNA, not RNA polymerase.
Bound far from the promoter, it brings in enzymes that wind the DNA tight. - B. ✓ By bringing in enzymes that wind the DNA around the gene tight
- C. By sitting in RNA polymerase’s path, as at an operatorThe repressor is bound 6,200 bp upstream of the promoter, far from RNA polymerase’s path.
From there the repressor brings in enzymes that wind the DNA tight.
Why: The repressor is bound to DNA 6,200 bp from the promoter, not in RNA polymerase’s path.
It brings in enzymes that take acetyl groups off the histones.
So the DNA around the gene winds tight, and RNA polymerase cannot reach it.
Suppose a starfish’s cell loses one protein. The mRNA of a gene beside the protein’s binding stretch measures the same before the loss and after it.
What had the protein been doing to that gene?
- A. Raising its transcriptionRemoving a protein that raises transcription lowers the mRNA.
The mRNA did not change, so the protein was not regulating this gene. - B. Lowering its transcriptionRemoving a protein that lowers transcription raises the mRNA.
The mRNA did not change, so the protein was not regulating this gene. - C. ✓ Neither raising nor lowering it
Why: Removing the protein left the gene’s mRNA unchanged.
A protein that regulated the gene would have moved the mRNA one way or the other.
So the protein did not regulate that gene.
In a kestrel, a repressor keeps one gene silent. In one of the kestrel’s cells the repressor’s gene is deleted. A second cell still makes the repressor, but the repressor’s binding site has changed to a sequence the repressor does not fit. A student says: “Both cells now transcribe the gene.”
Is the student correct?
- A. No: only the cell with the deleted gene transcribes it; in the other the repressor still bindsA repressor cannot bind a site it no longer fits.
In both cells no repressor is bound, and the gene is transcribed. - B. ✓ Yes: in both cells no repressor is bound, so the gene is on when it should be off
Why: With the gene deleted, no repressor is made.
With the binding site changed, the repressor is made but cannot bind.
Either way nothing blocks the gene, so both cells transcribe it.
In a bacterium, removing one regulatory protein raises transcription of an operon’s genes.
Which sequence had the protein been bound to?
- A. ✓ The operator
- B. The promoterThe promoter is where RNA polymerase binds.
A protein whose removal raises transcription is a repressor, and a bacterial repressor binds the operator. - C. The genes’ mRNAA regulatory protein binds DNA, not mRNA.
A protein whose removal raises transcription is a repressor, bound to the operator.
Why: Removal raised transcription, so the protein had been lowering it: a repressor.
In a bacterium, a repressor binds the operator beside the promoter.
A koala’s cell transcribes one gene often. Now imagine a biologist gives the cell a gene for a repressor whose binding site fits a stretch of DNA beside that gene.
What happens to transcription of the gene?
- A. Transcription switches onThe gene was already transcribed often.
The new repressor binds beside it and winds the DNA tight, so transcription stops. - B. ✓ Transcription switches off
- C. Transcription stays as it wasThe new repressor fits a stretch beside the gene, so it binds there.
Bound, it brings in enzymes that wind the DNA tight, and transcription stops.
Why: The new repressor’s binding site fits the stretch beside the gene.
So the repressor binds there and brings in enzymes that wind the DNA tight.
So RNA polymerase can no longer reach the gene, and transcription switches off.
Suppose a biologist gives a sea urchin’s cell extra copies of the gene for one regulatory protein, so the cell makes far more of that protein. Transcription of a gene the protein controls falls to 20 % of its level.
Is the protein an activator or a repressor?
- A. An activatorMore of an activator would raise transcription.
Here more of the protein lowered it, so the protein lowers transcription: a repressor. - B. ✓ A repressor
Why: The cell makes far more of the protein.
Transcription of the gene falls.
So the protein lowers transcription.
A regulatory protein that lowers transcription is a repressor.
Suppose a moss cell lacks one regulatory protein. A gene that protein normally controls makes 38 units of mRNA in a normal cell and 114 units in the cell that lacks the protein. The units are arbitrary.
(a) Identify the kind of regulatory protein the cell lacks. (1 pt)
- Award 1 point for: a repressor.
(b) Justify your identification in (a), using the data. (2 pt)
A protein whose removal raises transcription had been lowering it.
A regulatory protein that lowers transcription is a repressor.
Had the protein been an activator, its removal would have lowered transcription instead.
- Award 1 point for: removal raised the gene’s mRNA (38 → 114 units), so the protein had been lowering transcription: a repressor.
- Award 1 point for: the contrast — removing an activator would have lowered transcription, so the direction of the change after removal names the kind of protein.
- Accept reasoning consistent with a wrong (a): if (a) named an activator, award the second point only for a line that correctly states that removing an activator lowers transcription and removing a repressor raises it.
APBIO-U06-L33 Different products, different jobs
Suppose two cells from one person lie side by side. The first is a red blood cell precursor, a young cell in the bone marrow that is filling with hemoglobin. The second is a cell from the lens of the eye, filling with a transparent protein and holding no hemoglobin at all.
Both grew from the same fertilized egg. A student claims the lens cell must have lost the hemoglobin gene. What would you measure to test that claim?
Unit 6 · Gene Expression and Regulation
1Same genes, different reading
A muscle cell and a pancreas cell from one person both carry the insulin gene. Only the pancreas cell transcribes it.
During differentiation, what happened to the insulin gene in the muscle cell?
- A. The muscle cell removed the insulin gene from its DNADifferentiation removes no genes.
The muscle cell still carries the insulin gene, and RNA polymerase is not transcribing it there. - B. ✓ The muscle cell kept the insulin gene in its DNA and does not transcribe it
Why: Every body cell carries the same genome.
Differentiation removes no genes.
So the muscle cell still carries the insulin gene, and RNA polymerase is not transcribing it there.
Cortisol, a steroid hormone, crosses a liver cell’s membrane and binds its receptor in the cytosol. The bound pair moves into the nucleus and attaches to the DNA.
What does the bound pair change?
- A. ✓ Which genes the cell expresses
- B. The base sequence of the cell’s DNAA hormone changes no base in the DNA.
Attached to the DNA, the bound pair changes the expression of particular genes. - C. The shape of the cell’s ribosomesThe bound pair attaches to the DNA, not to the ribosomes.
Attached there, it changes the expression of particular genes.
Why: The bound pair attaches to the DNA.
Attached to the DNA, the bound pair changes the expression of particular genes.
So over the following hours the cell makes more of some proteins and less of others.
In Chinese primroses, one plant opens red flowers at 20 °C and white flowers at 30 °C.
What did the warmth change in the petal cells?
- A. The base sequence of the pigment genesA condition outside the organism changes no base in the DNA.
The warmth changed which pigment genes the petal cells use and how much. - B. ✓ Which pigment genes the cells use, and how much
Why: A condition outside the organism changes which genes its cells use and how much.
So the phenotype changes while the DNA does not.
The warmth changed which pigment genes the petal cells use.
Do two very different cells differ in their genes, or in how they use them?
Measure two things: the DNA and the mRNA.
Sequence the DNA of both cells: both carry the hemoglobin gene and the lens-protein gene.
Measure the mRNAs: only the red blood cell precursor holds hemoglobin mRNA, and only the lens cell holds lens-protein mRNA.
The genes are the same, and the reading of them is not. The proteins each cell is packed with follow from that reading.
A signal or a hormone can change which genes a cell reads: weeks of endurance training raise the transcription of genes for mitochondrial proteins in muscle.
Video: Watch: Same genes, different reading
A red blood cell precursor and a lens cell are drawn side by side with the same row of genes in each nucleus. In the precursor the hemoglobin gene fills in and RNA hangs from it. In the lens cell the lens-protein gene fills in and RNA hangs from it, and the hemoglobin gene stays open.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L33a.mp4
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Go back to the red blood cell precursor and the lens cell.
Both grew from one fertilized egg by mitosis. So both carry the same genome, the hemoglobin gene with it.
In the drawings, a filled box with a light strand hanging from it is a gene being transcribed. An open box is a gene that is present and not transcribed.
Sequence the DNA of both cells. The hemoglobin gene is in both, with the same bases in the same order.
In the red blood cell precursor, RNA polymerase transcribes the hemoglobin gene. So that cell holds hemoglobin mRNA, and its ribosomes build hemoglobin.
In the lens cell, RNA polymerase does not transcribe the hemoglobin gene. So the lens cell holds no hemoglobin mRNA and builds no hemoglobin.
The lens cell transcribes a different gene instead: the gene for a transparent protein that fills the lens.
The red blood cell precursor carries that lens-protein gene too, and does not transcribe it.
So the two cells hold different mRNAs.
Ribosomes build a different protein from each mRNA. So the two cells hold different proteins.
The proteins a cell builds are its products.
Hemoglobin binds oxygen. So a cell packed with hemoglobin can carry oxygen: that is the red blood cell’s job.
The transparent protein lets light through. So a cell packed with it can pass light to the back of the eye: that is the lens cell’s job.
The two cells read the same genes differently. So the two cells build different products.
A cell’s products decide its job.
A cell’s reading of its genes is not fixed for life. A signal from outside the cell can change which genes RNA polymerase transcribes, and how often.
Suppose a person starts endurance training: long sessions of steady exercise, several times a week.
Each session sends signals through the muscle cells. After each session, RNA polymerase transcribes the genes for mitochondrial proteins more often than before.
The genes in the muscle cells’ DNA did not change. Their reading did.
So the muscle cells build more mitochondrial proteins. Over weeks of training, each muscle cell holds more mitochondria.
A hormone can change the reading in the same way. Cortisol bound to its receptor attaches to the DNA and changes which genes a liver cell reads.
What you are expected to know Explain how the same DNA, read differently, gives two cells different mRNAs, different proteins and so different jobs.
What you are expected to know Explain how a signal or a hormone changes which genes a cell reads.
A red blood cell precursor and a lens cell from one person grew from one fertilized egg. The precursor is packed with hemoglobin, and the lens cell is packed with a transparent protein instead.
Which of the following differs between the two cells?
- A. The genes their DNA carriesBoth cells grew from one fertilized egg by mitosis, so their DNA carries the same genes.
What differs is which genes each cell transcribes, so the mRNAs each holds. - B. ✓ The mRNAs they hold
- C. Both the genes their DNA carries and the mRNAs they holdThe two cells carry the same genes: both grew from one fertilized egg by mitosis.
Only the mRNAs differ, because each cell transcribes its own set of genes.
Why: Both cells grew from one fertilized egg by mitosis, so both carry the same genes.
The precursor transcribes the hemoglobin gene, and the lens cell does not.
So the precursor holds hemoglobin mRNA, and the lens cell holds none.
The mRNAs differ; the genes do not.
Suppose a biologist compares two cells of one barnacle: a cell of its cement gland and a cell of one of its feeding legs. Both cells carry the gene for the cement protein, which hardens and fixes the barnacle to a rock. Only the cement-gland cell holds the cement protein.
(a) Explain why fixing the barnacle to a rock is the cement-gland cell’s job rather than the feeding-leg cell’s. (1 pt)
Only the cement-gland cell holds the cement protein, which fixes the barnacle to its rock.
So fixing the barnacle to the rock is the cement-gland cell’s job.
The feeding-leg cell holds no cement protein.
So the feeding-leg cell does a different job: it sweeps food from the water.
- Award 1 point for: a cell’s products decide its job, and only the cement-gland cell holds the cement protein, so fixing the barnacle to the rock is that cell’s job (the feeding-leg cell, with none of the protein, has another job).
(b) Explain how the two cells come to hold different proteins, though they carry the same DNA. (2 pt)
Frame The two cells hold different proteins because …
So only the cement-gland cell holds cement-protein mRNA.
Ribosomes build a protein only from an mRNA.
So only the cement-gland cell builds the cement protein.
- Award 1 point for: the cement-protein gene is transcribed in the cement-gland cell only, so only that cell holds the cement-protein mRNA.
- Award 1 point for: ribosomes build a protein only from its mRNA, so only the cell with the mRNA builds the cement protein.
A student looks at a cell lining a person’s windpipe, which makes no hemoglobin, and says: “This cell lost the hemoglobin gene when it differentiated.”
Is the student correct?
- A. ✓ No: the cell carries the hemoglobin gene and is not transcribing it
- B. Yes: a cell that makes no hemoglobin has no hemoglobin geneEvery body cell carries the whole genome, and differentiation removes no genes.
The windpipe cell carries the hemoglobin gene, and RNA polymerase is not transcribing it.
Why: Every body cell of the person grew from one fertilized egg by mitosis.
So the windpipe cell carries the same genome, the hemoglobin gene with it.
A gene present and not transcribed gives no mRNA and no protein.
So the cell makes no hemoglobin, with the gene still there.
Suppose a person trains for endurance for several weeks. Afterwards, each muscle cell holds more mitochondria than before. A student says: “Training changed which genes the muscle cells read, and left the genes themselves unchanged.”
Is the student correct?
- A. No: training changed the base sequence of the genes for mitochondrial proteinsA signal changes no base in a cell’s DNA.
Each session made RNA polymerase transcribe the genes for mitochondrial proteins more often, with the genes unchanged. - B. ✓ Yes: signals from each session made RNA polymerase transcribe those genes more often
Why: Each training session sends signals through the muscle cells.
The signals make RNA polymerase transcribe the genes for mitochondrial proteins more often.
The genes themselves do not change.
So the muscle cells build more mitochondrial proteins and hold more mitochondria.
39Two lines of evidence
After a treatment, a gene’s mRNA level rose, and its protein level rose with it.
Where does that control act?
- A. After transcriptionA control after transcription leaves the mRNA unchanged and changes the protein.
Here the mRNA changed and the protein followed, so the control acts at transcription. - B. ✓ At transcription
Why: If the mRNA changes and the protein follows, the control acts at transcription.
The gene’s mRNA rose and its protein rose with it.
So the treatment acts at transcription.
A cell carries a gene. RNA polymerase leaves that gene untranscribed.
How much of that gene’s protein does the cell build?
- A. ✓ None of the protein
- B. A little of the proteinRibosomes build a protein only from an mRNA, never from the DNA.
A gene that is not transcribed gives no mRNA, so the ribosomes build none of its protein.
Why: RNA polymerase leaves the gene untranscribed.
So no mRNA of the gene exists in the cell.
Ribosomes build a protein only from an mRNA.
So the cell builds none of that gene’s protein.
How do you show that two cells carry the same genes and read them differently?
Give two lines of evidence from a table: the row that shows the gene in both cells, and the row that shows it read in one cell only.
Each row you cite is one line of evidence.
Video: Watch: Two lines of evidence
The table of the two cells appears one row at a time. A ring moves along the hemoglobin gene’s DNA row, and present is read out for both cells. The ring moves to the mRNA row, and the two levels are read out. A fourth column fills in beside each row with what the row shows.
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The table below gives three rows for the hemoglobin gene and three for the lens-protein gene, in the red blood cell precursor and in the lens cell.
Each gene has three rows: whether the gene is in the DNA, its mRNA level and its protein level. The levels are in arbitrary units.
Claim: the two cells carry the same genes and read them differently.
Read the hemoglobin gene’s DNA row: present in the red blood cell precursor, present in the lens cell.
So the two cells carry the same hemoglobin gene. That is the first line of evidence.
Read the hemoglobin mRNA row: 260 units in the red blood cell precursor, 0 units in the lens cell.
Only the red blood cell precursor holds hemoglobin mRNA. So only that cell transcribes the gene: the two cells read the same gene differently.
That is the second line of evidence.
Read the hemoglobin protein row: 340 units in the precursor, 0 units in the lens cell. The protein follows the mRNA.
The protein row shows that the two cells differ. On its own, it does not show whether the gene is in both cells.
A cell with no protein might lack the gene, or might carry the gene and not transcribe it.
So the two rows that support the claim are the DNA row and the mRNA row.
To support the claim from any such table, follow three steps.
- Find the DNA row: present in both cells shows the same gene in both.
- Find the mRNA row: mRNA in one cell only shows the gene read in one cell only.
- Write each row as one line of evidence.
What you are expected to know Support the claim that two cells carry the same genes and read them differently, with two lines of evidence from a table of DNA, mRNA and protein: the DNA row and the mRNA row, each with its values.
Now suppose you support the same claim with the lens-protein gene’s three rows.
The lens-protein gene’s three rows are drawn below for the red blood cell precursor and the lens cell. In the table, present means the gene is in the cell’s DNA, and the mRNA and protein levels are in arbitrary units.
Which row shows that the two cells carry the same lens-protein gene?
- A. ✓ The DNA row
- B. The mRNA rowThe mRNA row, 0 units against 290, shows the two cells reading the gene differently.
The DNA row, present in both cells, shows the gene in both. - C. The protein rowThe protein row, 0 units against 350, shows that the cells differ.
Only the DNA row, present in both cells, shows the gene in both.
Why: The DNA row reads present in the red blood cell precursor and present in the lens cell.
Present means the gene is in the cell’s DNA.
So both cells carry the lens-protein gene.
The lens-protein gene’s three rows are drawn below for the red blood cell precursor and the lens cell. In the table, present means the gene is in the cell’s DNA, and the mRNA and protein levels are in arbitrary units.
Which row shows that the two cells transcribe the lens-protein gene differently?
- A. The DNA rowThe DNA row reads present in both cells.
Present in both shows the gene in both, not how each cell reads it. - B. ✓ The mRNA row
- C. The protein rowThe protein row shows different amounts of the protein.
Reading a gene is transcribing it, and transcription shows in the mRNA row: 0 units against 290.
Why: Reading a gene is transcribing it, and transcription makes the gene’s mRNA.
The mRNA row reads 0 units in the precursor and 290 units in the lens cell.
So only the lens cell transcribes the lens-protein gene.
The table below sets out all six rows, with what each row shows beside it.
Suppose a biologist compares two cells of one cow: a cell of its udder, which makes casein, the main protein of milk, and a cell of its hoof. The table below gives, for each cell, whether the casein gene and a hoof-protein gene are in the DNA, and the mRNA and protein levels of each. In the table, present means the gene is in the cell’s DNA, and the mRNA and protein levels are in arbitrary units.
(a) A student cites only the casein protein row, 420 units in the udder cell and 0 units in the hoof cell, to show that both cells carry the casein gene. Evaluate whether that row alone shows it. (1 pt)
A cell with no casein might lack the casein gene, or might carry it and not transcribe it.
The protein row cannot tell those two apart.
Only a row that reads the DNA itself shows the gene in both cells.
- Award 1 point for: the row alone does not show it, because a cell with none of the protein might lack the gene or might carry it untranscribed (the protein row cannot tell those apart).
(b) Support the claim that the two cells carry the same genes and read them differently, with two lines of evidence from the table. (2 pt)
Casein mRNA is at 370 units in the udder cell and 0 units in the hoof cell, so only the udder cell transcribes the casein gene.
- Award 1 point for the DNA line: the casein gene (or the hoof-protein gene) is present in both cells, so the two cells carry the same gene.
- Award 1 point for the mRNA line: casein mRNA at 370 units in the udder cell and 0 in the hoof cell (or hoof-protein mRNA at 0 and 360 units), so the gene is transcribed in one cell only. The values, or ‘in one cell only’, are needed; the protein row earns no point here.
Back to the red blood cell precursor and the lens cell, which grew from one fertilized egg.
Sequence their DNA and both carry the hemoglobin gene.
Measure their mRNAs and only the red blood cell precursor holds hemoglobin mRNA.
The genes are the same, and the reading of them is not.
APBIO-U06-L34 Small RNAs that silence
Suppose a biologist gives a cell a short RNA, about twenty bases long, whose bases match part of gene L’s mRNA. Gene L is transcribed exactly as before. Yet the amount of L protein falls to almost nothing.
The gene is on. RNA polymerase is still copying it into mRNA. The protein is missing. Where did the message go?
Unit 6 · Gene Expression and Regulation
1A short RNA finds its match
A short RNA strand pairs with a stretch of an mRNA. One base on the short RNA is adenine.
Which base on the mRNA does that adenine pair with?
- A. ✓ Uracil
- B. ThymineThymine is a DNA base.
When the partner strand is RNA, adenine pairs with uracil. - C. GuanineGuanine pairs with cytosine.
When the partner strand is RNA, adenine pairs with uracil.
Why: Each base pairs with one partner base.
In DNA, adenine pairs with thymine.
When the partner strand is RNA, adenine pairs with uracil.
So the adenine on the short RNA pairs with a uracil on the mRNA.
An mRNA’s poly-A tail is shortened, and enzymes in the cytoplasm destroy that mRNA sooner than usual.
What happens to the amount of that mRNA’s protein the cell makes?
- A. ✓ The cell makes less of the protein
- B. The cell makes the same amount of the proteinRibosomes read an mRNA only while it lasts.
An mRNA destroyed sooner is read fewer times, so the cell makes less of its protein. - C. The cell makes more of the proteinAn mRNA destroyed sooner is read by ribosomes fewer times.
Fewer readings give less protein, not more.
Why: Ribosomes build the protein from the mRNA while the mRNA lasts.
The shortened tail lets the RNA-breaking enzymes reach the mRNA sooner.
So the mRNA is destroyed sooner and is read fewer times.
So the cell makes less of the protein.
A muscle cell and a nerve cell from one person carry the same genome.
Which of the following describes the genes the two cells transcribe?
- A. The muscle cell and the nerve cell transcribe the same set of genesThe muscle cell transcribes the genes for its contractile proteins, and the nerve cell does not.
Each kind of cell transcribes its own set of genes. - B. ✓ The muscle cell and the nerve cell each transcribe their own set of genes
Why: Every body cell carries the same genome.
Differentiation removes no genes.
Each kind of cell transcribes its own set of genes and makes its own tissue-specific proteins.
So the muscle cell and the nerve cell transcribe different sets of genes.
After a treatment, a gene’s mRNA level is unchanged and its protein level has fallen.
Where does the control on that gene act?
- A. At transcriptionA control at transcription changes the mRNA level, and the protein follows.
Here the mRNA is unchanged and the protein fell, so the control acts after transcription. - B. ✓ After transcription
Why: If the mRNA is unchanged and the protein changes, the control acts after transcription.
The mRNA is unchanged and the protein fell.
So the control acts after transcription.
How can a gene be transcribed and still make no protein?
A short RNA pairs with the matching stretch of an mRNA.
The paired mRNA is either cut up or blocked from the ribosome.
So the cell makes less of that protein, while the gene is transcribed as before.
A cell makes short RNAs of this kind itself. A laboratory can also make one and supply it to the cell.
The short RNA acts on the message, not on the DNA and not on the protein. So the short RNA is a control after transcription.
A short RNA made to match a stretch found only in one gene’s mRNA silences that gene and no other.
Video: Watch: A short RNA finds its match
Gene L’s mRNA is drawn as a long rail with one tab per base. A short RNA of about twenty bases drifts to it and pairs with the matching stretch, base to base. In one kind of cell an enzyme cuts the paired stretch and the pieces break down. In another kind of cell the message stays whole and a ribosome cannot load onto it. Above them, RNA polymerase keeps copying gene L as before.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L34a.mp4
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Go back to the cell that received the short RNA. Before the short RNA arrived, ribosomes read gene L’s mRNA and built the L protein.
In the drawings, an RNA is a rail with one short tab per base. The mRNA is the long rail, with its 5′ end at the left and its 3′ end at the right.
A ribosome grips the mRNA near its 5′ end and reads along it toward the 3′ end. So the cell builds the L protein.
Now suppose the short RNA enters the cell. The short RNA is about twenty bases long.
The short RNA’s bases are the partners of a stretch of the mRNA’s bases.
Where the short RNA carries adenine, that stretch of the mRNA carries uracil. Where the short RNA carries guanine, the stretch carries cytosine.
So the short RNA pairs with that stretch of the mRNA, base to base, just as two DNA strands pair.
In the drawing, the dashed lines are the hydrogen bonds between the paired bases.
The two strands are antiparallel. So the short RNA’s 3′ end lies toward the mRNA’s 5′ end.
Proteins in the cell recognize the paired stretch. One of two events follows.
In many cases, an enzyme cuts the paired mRNA into pieces. Other enzymes in the cell break the pieces down.
So the amount of L mRNA in the cell falls. Ribosomes build no protein from a cut message.
In other cases, the paired mRNA is not cut. The pairing stops ribosomes from loading onto the message.
So the L mRNA stays in the cell, whole and unread. Ribosomes build no protein from an unread message either.
Either way, ribosomes build far less L protein.
The short RNA pairs with the mRNA. The short RNA never binds the DNA of gene L, and the short RNA never touches the L protein.
So gene L itself is untouched. RNA polymerase transcribes gene L as often as before.
So this control acts after transcription. RNA polymerase copies the gene as before.
Enzymes cut the copies up, or the pairing blocks them.
When the copies are cut up, the amount of L mRNA falls, though transcription is unchanged. So the amount of mRNA alone does not show whether transcription changed.
To know that transcription is unchanged, measure transcription itself: how often RNA polymerase copies the gene.
A short RNA that pairs with a matching stretch of an mRNA, so that the mRNA is cut up or blocked from the ribosome, is called a , because it is only about twenty bases long.
A small RNA that a cell transcribes from one of its own genes is called a . Micro means very small.
A small RNA that a laboratory makes and supplies to a cell is called an , short for small interfering RNA, because it interferes with the reading of the message.
Cells make siRNAs of their own too. A small RNA that a laboratory supplies is always an siRNA.
The silencing of a gene by a small RNA is called , because the small RNA interferes with the gene’s message.
A cell uses its own microRNAs to turn down many of its genes. Each kind of cell transcribes its own set of microRNA genes.
So each kind of cell silences its own set of messages.
A small RNA pairs with the matching sequence on an mRNA, and the paired mRNA is either cut up or blocked from the ribosome, so less of that protein is made while the gene is transcribed as before.
A small RNA acts on the mRNA, not on the DNA and not on the protein. So a small RNA is a control after transcription.
A small RNA made to match a stretch found only in one gene’s mRNA silences that gene and no other.
What you are expected to know Describe how a small RNA silences a gene: it pairs with the matching stretch of the gene’s mRNA, and the paired mRNA is cut up or blocked from the ribosome, so the cell makes less protein while transcription is unchanged.
A biologist gives a cell a short RNA made to silence gene L.
Which molecule does the short RNA pair with?
- A. The DNA of gene LThe short RNA never binds the DNA.
Its bases are the partners of a stretch of gene L’s mRNA, and the two pair. - B. The L proteinA protein carries no bases for an RNA to pair with.
The short RNA pairs with the matching stretch of gene L’s mRNA. - C. ✓ The mRNA copied from gene L
Why: The short RNA’s bases are the partners of a stretch of gene L’s mRNA.
So the short RNA pairs with that stretch of the mRNA.
The short RNA never binds the DNA of gene L, and the short RNA never touches the L protein.
In a cell given the short RNA against gene L, the paired mRNA is not cut. The pairing stops ribosomes from loading onto the message.
What happens to the amount of L mRNA in the cell?
- A. The amount of L mRNA fallsAn mRNA that is not cut stays whole in the cell.
The amount of L mRNA stays the same; only its reading stops. - B. ✓ The amount of L mRNA stays the same
- C. The amount of L mRNA risesRNA polymerase copies gene L as often as before, so no extra mRNA is made.
The copies stay whole and unread, so the amount stays the same.
Why: The paired mRNA is not cut, so every copy stays whole.
RNA polymerase makes new copies as often as before.
So the amount of L mRNA in the cell stays the same.
In a second cell given the short RNA against gene L, an enzyme cuts each paired mRNA into pieces.
What happens to the amount of L mRNA in the cell?
- A. ✓ The amount of L mRNA falls
- B. The amount of L mRNA stays the sameOther enzymes break the pieces of a cut mRNA down.
So fewer whole copies of L mRNA remain, and the amount falls. - C. The amount of L mRNA risesRNA polymerase makes new copies no faster than before.
The cut copies are broken down, so the amount falls.
Why: An enzyme cuts each paired mRNA into pieces.
Other enzymes break the pieces down.
RNA polymerase makes new copies no faster than before.
So the amount of L mRNA in the cell falls.
Suppose a biologist gives the cells of a peanut seed a short RNA made to silence a gene for one of the proteins the seed stores as food. Over the following days the treated cells make far less of that stored protein than untreated cells do.
(a) Predict what the biologist finds on measuring how often RNA polymerase transcribes the stored-protein gene in the treated cells, compared with the untreated cells. (1 pt)
- Award 1 point for: transcription of the stored-protein gene is unchanged (the same in treated and untreated cells).
(b) Explain why the treated cells make less of the stored protein. (2 pt)
Frame The treated cells make less of the stored protein because …
The paired mRNA is cut up, or ribosomes cannot load onto it.
So ribosomes build less of the stored protein, while the gene is transcribed as before.
- Award 1 point for: the short RNA pairs with (is complementary to) the matching stretch of the stored-protein mRNA.
- Award 1 point for: the paired mRNA is cut up (destroyed) or blocked from the ribosome (not translated), so less of the protein is made.
In a cell given the short RNA against gene L, the L protein has almost gone. A student says: “The short RNA stopped RNA polymerase from transcribing gene L.”
Is the student correct?
- A. ✓ No: RNA polymerase transcribes gene L as before; the short RNA acts on the mRNA
- B. Yes: the short RNA stopped transcription, so the cell made no mRNA and no proteinThe short RNA never binds the DNA of gene L.
RNA polymerase transcribes gene L as before, and the short RNA pairs with the mRNA.
Why: The short RNA acts on the mRNA, not on the DNA.
RNA polymerase transcribes gene L as often as before.
The paired mRNA is cut up or blocked from ribosomes.
So the L protein falls while transcription is unchanged.
In a cell given the short RNA against gene L, the paired mRNA is blocked and not cut, and the L mRNA level is unchanged. A student says: “The cell still makes less L protein than before.”
Is the student correct?
- A. No: an mRNA that is still in the cell is read by ribosomes as beforeThe paired stretch stops ribosomes from loading onto the message.
Ribosomes build no protein from a whole mRNA that none of them reads. - B. ✓ Yes: ribosomes cannot load onto the paired mRNA, so they build less L protein
Why: The paired mRNA is not cut, so the L mRNA level is unchanged.
The pairing stops ribosomes from loading onto the message.
So ribosomes build less L protein.
So the cell makes less L protein than before.
Two kinds of cell from one person, the cells of the voice box and the cells of the appendix, transcribe the gene for a certain protein equally often. Only the voice-box cells also transcribe a microRNA gene whose short RNA pairs with that protein’s mRNA.
Which kind of cell holds more of the protein?
- A. The voice-box cellsIn the voice-box cells the microRNA pairs with the protein’s mRNA, and the paired mRNA is cut up or blocked.
So the voice-box cells build less of the protein. - B. ✓ The appendix cells
Why: The two kinds of cell transcribe the gene equally often.
Only the voice-box cells make the microRNA.
The microRNA pairs with the protein’s mRNA, and the paired mRNA is cut up or blocked from ribosomes.
So the voice-box cells build less of the protein, and the appendix cells hold more.
The table below compares the two outcomes for the paired mRNA, cut and blocked: what happens to how often the gene is transcribed, to the amount of that mRNA in the cell, and to the amount of that protein.
Back to the cell that received the short RNA, about twenty bases long, matching part of gene L’s mRNA.
The small RNA paired with gene L’s mRNA.
Enzymes cut the paired message up, or the pairing held ribosomes off it. So the L protein fell.
RNA polymerase transcribed gene L as often as before. Transcription never changed.
58Quick quiz: small RNA (microRNA, siRNA), RNA interference mixed practice
A cell of a person’s earlobe transcribes one of its own genes into a short RNA, which pairs with an mRNA in that cell.
Which kind of small RNA is this?
- A. ✓ microRNA
- B. siRNAA supplied small RNA is an siRNA.
A small RNA a cell transcribes from one of its own genes is a microRNA.
Why: The cell transcribed the short RNA from one of its own genes.
A small RNA a cell makes itself is called a microRNA.
A plant’s cells transcribe many of their own genes into short RNAs that pair with mRNAs and turn down those genes’ expression.
Which kind of small RNA is this?
- A. ✓ microRNA
- B. siRNANobody supplied these short RNAs: the plant’s cells transcribed them from their own genes.
Such small RNAs are microRNAs.
Why: The plant’s cells transcribed the short RNAs from their own genes.
A small RNA a cell makes itself is called a microRNA.
A biologist makes a short RNA and adds it to a dish of cells. In the cells, the short RNA pairs with one gene’s mRNA.
Which kind of small RNA is this?
- A. microRNAA cell transcribes its microRNAs from its own genes.
A small RNA made outside the cell and supplied to it is an siRNA. - B. ✓ siRNA
Why: The biologist made the short RNA and supplied it to the cells.
A small RNA supplied to a cell is an siRNA.
A researcher injects a short RNA into the cells of an animal embryo. The short RNA pairs with one of the embryo’s mRNAs.
Which kind of small RNA is this?
- A. microRNAThe embryo’s cells did not transcribe this short RNA; the researcher supplied it.
A supplied small RNA is an siRNA. - B. ✓ siRNA
Why: The researcher made the short RNA and injected it into the cells.
A small RNA supplied to a cell is an siRNA.
A cell transcribes one of its own genes into a short RNA. Where the short RNA pairs with an mRNA, that mRNA is cut up.
Is this molecule a small RNA?
- A. ✓ Yes
- B. NoA short RNA that pairs with an mRNA so that the mRNA is cut up is a small RNA.
A cell’s own small RNA is a microRNA.
Why: The molecule is a short RNA.
The short RNA pairs with an mRNA, and the paired mRNA is cut up.
So the molecule is a small RNA.
A folded RNA carries one amino acid to the ribosome and pairs its anticodon with a codon.
Is this molecule a small RNA?
- A. YesAn RNA that carries an amino acid to the ribosome is a tRNA.
A small RNA pairs with a stretch of an mRNA and silences it. - B. ✓ No
Why: The molecule carries an amino acid to the ribosome, so the molecule is a tRNA.
A tRNA pairs one anticodon with one codon and brings an amino acid.
A small RNA silences an mRNA, so this molecule is not a small RNA.
An enzyme lays down a short RNA at a replication fork, paired to the DNA template, to start a new DNA strand.
Is this molecule a small RNA?
- A. YesThat short RNA is a primer: it pairs with the DNA template and starts a new DNA strand.
A small RNA pairs with an mRNA and silences it. - B. ✓ No
Why: The molecule pairs with DNA, not with an mRNA.
Its job is to start a new DNA strand.
A small RNA pairs with an mRNA so that the mRNA is cut up or blocked, so this molecule is not a small RNA.
A biologist supplies a short RNA to cells in a dish. In the cells the short RNA pairs with one gene’s mRNA, and ribosomes stay off that mRNA.
Is this molecule a small RNA?
- A. ✓ Yes
- B. NoA short RNA that pairs with an mRNA so that the mRNA is blocked from the ribosome is a small RNA.
A supplied one is an siRNA.
Why: The molecule is a short RNA.
The short RNA pairs with an mRNA, and ribosomes stay off the paired mRNA.
So the molecule is a small RNA.
A protein cuts an mRNA into pieces.
Is this molecule a small RNA?
- A. YesA protein is not an RNA.
A small RNA is a short RNA that pairs with an mRNA; the cutting protein acts after the pairing. - B. ✓ No
Why: The molecule is a protein, not an RNA.
A small RNA is a short RNA that pairs with an mRNA.
So this molecule is not a small RNA.
What is a small RNA?
- A. A short RNA that pairs with a stretch of a gene’s DNA so that RNA polymerase cannot bind the promoter and transcribe the geneA small RNA pairs with the matching stretch of an mRNA, never with the DNA.
RNA polymerase transcribes the gene as before. - B. ✓ A short RNA that pairs with a matching stretch of an mRNA so that the mRNA is cut up or blocked from the ribosome
- C. A short RNA folded into an L shape that carries an amino acid to the ribosome and pairs with a codonAn RNA that carries an amino acid to the ribosome is a tRNA.
A small RNA pairs with an mRNA so that the mRNA is cut up or blocked.
Why: A short RNA of about twenty bases pairs with the matching stretch of an mRNA.
The paired mRNA is cut up or blocked from the ribosome.
Such a short RNA is called a small RNA.
What is RNA interference?
- A. ✓ The silencing of a gene by a small RNA that pairs with the gene’s mRNA
- B. The blocking of a gene’s promoter by a repressor protein bound to the DNAA repressor bound to the DNA blocks transcription.
RNA interference is silencing by a small RNA that pairs with the mRNA. - C. The removal of introns from a pre-mRNA before it leaves the nucleusThe removal of introns is splicing.
RNA interference is the silencing of a gene by a small RNA paired with its mRNA.
Why: A small RNA pairs with a gene’s mRNA.
The paired mRNA is cut up or blocked, so the gene is silenced.
That silencing is called RNA interference.
Suppose a cell has just taken up a short RNA.
(a) State what a small RNA is. (1 pt)
- Award 1 point for: a short RNA that pairs with (is complementary to) a stretch of an mRNA, after which the mRNA is cut up or its translation is blocked. ‘About twenty bases’ completes it, but its absence does not lose the point.
(b) State what RNA interference names. (1 pt)
- Award 1 point for: the silencing (lowered expression) of a gene by a small RNA paired with the gene’s mRNA.
Glossary
- small RNA
- A short RNA, about twenty bases long, that pairs with a matching stretch of an mRNA. The paired mRNA is then cut up or blocked from the ribosome, so the cell makes less of that protein while the gene is transcribed as before. A small RNA acts on the mRNA, not on the DNA and not on the protein.
- microRNA
- A small RNA that a cell transcribes from one of its own genes. Each kind of cell transcribes its own set of microRNA genes and so silences its own set of messages.
- siRNA
- A small RNA that a laboratory makes and supplies to a cell; short for small interfering RNA. Cells make siRNAs of their own too. In the cell it pairs with the matching stretch of one gene’s mRNA, and that mRNA is cut up or blocked from the ribosome.
- RNA interference
- The silencing of a gene by a small RNA that pairs with the gene’s mRNA, so that the mRNA is cut up or blocked from the ribosome and the cell makes less of the protein, while the gene is transcribed as before.
APBIO-U06-L34B Knock a gene down and read what it did
Suppose a biologist grows three lines of cells from one animal. Every cell carries the gene for a protein that cuts one particular mRNA: the cutting gene. The cells of the first line hold two working copies of the cutting gene, the cells of the second line hold one, and the cells of the third line hold none.
The biologist measures the mRNA that the protein cuts, the target mRNA, in each line: 10 units in the first line, 55 units in the second, and 100 units in the third. Nobody changed the target gene itself. What does the pattern say the cutting protein does, and how sure can you be?
Unit 6 · Gene Expression and Regulation
1Turn one gene down
A biologist gives a cell a short RNA made to silence gene L.
Which molecule does the short RNA pair with?
- A. The DNA of gene LThe short RNA never binds the DNA.
Its bases are the partners of a stretch of gene L’s mRNA, and the two pair. - B. The L proteinA protein carries no bases for an RNA to pair with.
The short RNA pairs with the matching stretch of gene L’s mRNA. - C. ✓ The mRNA copied from gene L
Why: The short RNA’s bases are the partners of a stretch of gene L’s mRNA.
So the short RNA pairs with that stretch of the mRNA.
The short RNA never binds the DNA of gene L, and the short RNA never touches the L protein.
In a cell given the short RNA against gene L, an enzyme cuts each paired mRNA into pieces, and the amount of L mRNA falls.
How often does RNA polymerase transcribe gene L in that cell?
- A. ✓ As often as before
- B. Less often than beforeThe short RNA acts on the mRNA, not on the DNA.
RNA polymerase transcribes gene L as often as before; the copies are cut up after they are made.
Why: The short RNA pairs with the mRNA and never binds the DNA of gene L.
So gene L itself is untouched.
RNA polymerase transcribes gene L as often as before, and the copies are cut up after they are made.
A biologist adds a chemical to a dish of cells. Gene Y’s mRNA stays at 40 units, and gene Y’s protein falls from 64 to 32 units.
Where does the chemical’s control on gene Y act?
- A. At transcriptionA control at transcription changes the mRNA level, and the protein follows.
Here the mRNA is unchanged and the protein fell, so the control acts after transcription. - B. ✓ After transcription
Why: If the mRNA is unchanged and the protein changes, the control acts after transcription.
Gene Y’s mRNA is unchanged and its protein fell.
So the control acts after transcription.
How do you find out what a gene does when you cannot see it work?
Take the gene away, or turn it down, and watch what the cell can no longer do.
An siRNA against a gene leaves the gene’s transcription as it was and lowers the gene’s mRNA and protein. What the cells can no longer do is the gene’s job.
Video: Watch: Turn one gene down
Cells crawl slowly across a dish. An siRNA against gene S enters half of them, pairs with gene S’s mRNA, and the paired copies are cut up while RNA polymerase keeps transcribing gene S above. A four-line table fills in beside the cells: transcription the same, mRNA and protein far lower, and the siRNA cells sitting still.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L34Ba.mp4
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Suppose a biologist grows cells from one animal in a laboratory dish. The cells crawl slowly across the dish.
The cells transcribe a gene, gene S, whose job nobody knows.
The biologist makes a short RNA whose bases match part of gene S’s mRNA: an siRNA against gene S.
The biologist gives half of the cells the siRNA. The other half get none: those cells are the control cells.
In the siRNA cells, the siRNA pairs with gene S’s mRNA, and enzymes cut the paired copies up. RNA polymerase transcribes gene S as before.
Two days later the biologist measures four things in each half of the cells. The table below sets them out for the control cells and the siRNA cells.
The four lines are how often RNA polymerase transcribes gene S, the level of S mRNA, the level of S protein, and what the cells do.
In the table, transcription is how often RNA polymerase transcribes the gene in an hour. The mRNA and protein levels are in arbitrary units.
Read the transcription line: 67 times an hour in the control cells, 67 times an hour in the siRNA cells.
Transcription is unchanged. The siRNA never touched the DNA of gene S.
Read the S mRNA line: 148 units in the control cells, 13 units in the siRNA cells.
The siRNA paired with gene S’s mRNA, and enzymes cut the paired copies up. So far less S mRNA is left.
Read the S protein line: 191 units in the control cells, 29 units in the siRNA cells.
Ribosomes had far fewer messages to read. So the siRNA cells built far less S protein.
Read the last line: the control cells crawl, and the siRNA cells stay still.
The siRNA cells hold almost no S protein, and they no longer crawl. So the cells need S’s protein to crawl.
So gene S’s job is to make the protein the cells need to crawl.
Nobody saw the S protein at work. Turning the gene down showed its job by what the cells could no longer do.
The siRNA knocks gene S down: the gene stays in place, and its protein almost goes.
When RNA polymerase transcribes a gene less often, the cell also holds less of its mRNA and builds less of its protein.
So the mRNA line and the protein line alone cannot tell a cut-up message from a gene transcribed less often.
The transcription line settles the question. If transcription is unchanged, enzymes cut the message up after RNA polymerase made it.
What you are expected to know Predict what an siRNA against a gene does to the gene’s transcription, its mRNA, its protein and the cells.
What you are expected to know Read what the siRNA cells can no longer do as the job of the gene.
Suppose cells in a dish take up a red dye through a channel protein in their membrane, so the cells turn red. A biologist gives half the cells an siRNA against the channel gene. The siRNA pairs with the channel mRNA, and the paired copies are cut up.
How often does RNA polymerase transcribe the channel gene in the siRNA cells, compared with the control cells?
- A. Less often than in the control cellsThe siRNA pairs with the channel mRNA and never binds the DNA of the channel gene.
RNA polymerase transcribes the gene as often as before. - B. ✓ As often as in the control cells
- C. More often than in the control cellsThe siRNA acts on the mRNA after it is made.
Nothing about the gene’s DNA or its promoter changed, so transcription is as often as before.
Why: The siRNA pairs with the channel mRNA.
The siRNA never binds the DNA of the channel gene.
So RNA polymerase transcribes the channel gene as often as in the control cells.
Suppose cells in a dish take up a red dye through a channel protein. A biologist gives half the cells an siRNA against the channel gene, and in those cells the paired channel mRNA is cut up. The other half are the control cells.
How does the level of channel mRNA in the siRNA cells compare with the control cells?
- A. ✓ Lower than in the control cells
- B. The same as in the control cellsEnzymes break the cut copies down.
RNA polymerase makes new copies no faster than before, so the level of channel mRNA falls. - C. Higher than in the control cellsRNA polymerase makes new copies no faster than before.
The cut copies are broken down, so the level falls rather than rises.
Why: The siRNA pairs with each copy of the channel mRNA, and enzymes cut the paired copies up.
RNA polymerase makes new copies no faster than before.
So the level of channel mRNA in the siRNA cells is lower.
Suppose cells in a dish take up a red dye through a channel protein. A biologist gives half the cells an siRNA against the channel gene. In those cells the paired channel mRNA is cut up, and they hold far fewer copies of the channel mRNA than the control cells.
How does the level of channel protein in the siRNA cells compare with the control cells?
- A. ✓ Lower than in the control cells
- B. The same as in the control cellsRibosomes build the channel protein only from channel mRNA.
With far fewer copies of the mRNA, the ribosomes build far less of the protein. - C. Higher than in the control cellsRibosomes build a protein only from its mRNA.
Fewer copies of the mRNA give less of the protein, not more.
Why: Ribosomes build the channel protein from the channel mRNA.
The siRNA cells hold far fewer copies of the channel mRNA.
So the ribosomes build far less channel protein, and its level is lower.
Suppose cells in a dish take up a red dye through a channel protein. A day after a biologist gave half the cells an siRNA against the channel gene, those siRNA cells hold far less channel protein than the control cells. Both halves sit in the red dye.
Which cells stay pale?
- A. The control cellsThe dye enters a cell through the channel protein.
The control cells hold the channel protein, so the dye enters them and they turn red. - B. ✓ The siRNA cells
Why: The dye enters a cell through the channel protein.
The siRNA cells hold far less channel protein.
So little dye enters the siRNA cells, and they stay pale.
Suppose a biologist feeds an insect’s larvae a short RNA made to silence a gene whose job is unknown. Untreated larvae shed their outer covering as they grow. The table below gives, for untreated larvae and treated larvae, how often the gene is transcribed, the levels of its mRNA and its protein, and what the larvae do. In the table, transcription is how often RNA polymerase transcribes the gene in an hour; mRNA and protein levels are in arbitrary units.
(a) Explain how the results show what the gene does. (2 pt)
Frame The results show what the gene does because …
The treated larvae never shed their outer covering, while the untreated larvae do.
So the larvae need the gene’s protein to shed their outer covering.
So the gene’s job is to make that protein.
- Award 1 point for: the short RNA lowered the gene’s protein (172 → 41 units), so the treated larvae lack that protein.
- Award 1 point for: the treated larvae can no longer shed their outer covering, so the gene’s protein is needed for shedding — that is what the gene does.
(b) Explain what the transcription line shows about how the short RNA acted. (1 pt)
So the short RNA left the gene itself untouched.
The short RNA paired with the gene’s mRNA, and enzymes cut the paired copies up after RNA polymerase made them.
- Award 1 point for: transcription is unchanged (83 times an hour in both), so the short RNA acted on the mRNA (paired with it, and the copies were cut up), not on the gene’s DNA.
Suppose cells given an siRNA against the gene for a sugar-transport protein hold far less of that protein than untreated cells. A student says: “The siRNA cut the sugar-transport gene out of the cells’ DNA.”
Is the student correct?
- A. ✓ No: the siRNA pairs with the gene’s mRNA, and the gene stays in the cells’ DNA
- B. Yes: the cells hold far less of the protein because they no longer carry its geneAn siRNA never binds a gene’s DNA; it pairs with the gene’s mRNA.
The gene stays in the DNA, and RNA polymerase transcribes it as before.
Why: The siRNA pairs with the gene’s mRNA, not with the gene’s DNA.
The gene stays in the cells’ DNA, and RNA polymerase transcribes it as before.
The paired mRNA copies are cut up, so the ribosomes build less protein.
The protein fell with the gene still in place.
Suppose the cells lining a leech’s mouth build a protein that stops the blood the leech drinks from clotting, so that blood stays liquid. A biologist gives those cells an siRNA against the protein’s gene.
Compared with the blood an untreated leech drinks, how fast does the blood a treated leech drinks clot?
- A. More slowly than the untreated leech’sThe siRNA lowers the clot-stopping protein, so less of it reaches the blood.
The blood clots faster, not more slowly. - B. As fast as the untreated leech’sThe siRNA pairs with the protein’s mRNA, so the mouth cells build less of the protein.
Less protein reaches the blood, so the blood clots faster. - C. ✓ Faster than the untreated leech’s
Why: The siRNA pairs with the clot-stopping protein’s mRNA, and the paired copies are cut up.
So the mouth cells build far less of the protein.
Less of the protein reaches the blood the leech drinks.
So that blood clots faster.
40Two copies, one copy, none
Two cells of one kind each carry a gene. The first cell holds two working copies of the gene. In the second cell one copy is broken and the other works.
Which cell makes more of the gene’s product?
- A. The cell with one working copyA cell makes a gene’s product from each working copy of the gene it holds.
Two working copies make more of the product than one. - B. ✓ The cell with two working copies
Why: A cell makes a gene’s product from each working copy of the gene it holds.
The first cell holds two working copies, the second holds one.
So the cell with two working copies makes more of the gene’s product.
On a bar chart, a short line with a cap at each end passes through the top of each bar. The caption reads: Error bars represent ±2SE.
What does each line show?
- A. ✓ The range the true mean is likely to lie in
- B. How spread out the individual readings wereA bar whose caption says standard deviation shows how spread out the readings were.
A ±2SE error bar shows the range the true mean is likely to lie in.
Why: The caption says the bars represent ±2SE.
Two standard errors either side of the mean is the range the true mean is likely to lie in.
So each line shows the range the true mean is likely to lie in.
A dose series turns a gene down in steps, with copies of the gene: two copies, one copy, none. Like an siRNA, the dose series shows what the gene’s protein does.
Read the three lines together. Each working copy adds its share of the protein.
So the three lines hold three doses of the protein.
The pattern across all three doses says what the protein does. The ±2SE ranges say how sure you can be.
Video: Watch: Two copies, one copy, none
Three cells sit side by side, each with two rods, and the cutting gene’s boxes fill in: two, one, none. Under each cell a cutting-protein box grows to its dose: tall, half, none. Then a target-mRNA box rises under each: short, taller, tallest. The three ±2SE ranges appear beside them with clear gaps between.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-L34Bb.mp4
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Go back to the three lines of cells. Every cell carries the cutting gene on a pair of chromosomes, one from each parent.
In the drawings, the two chromosomes carrying the cutting gene are two rods. A filled box on a rod is a working copy of the cutting gene, and an open box is a broken copy.
The cells with two working copies are the normal cells. Every other line is compared with them.
So the two-copy line is the control.
A cell makes the cutting protein from each working copy of the cutting gene it holds.
So the one-copy cells make about half the cutting protein that the two-copy cells make. The zero-copy cells make none.
The three lines hold three doses of the cutting protein: the usual amount, about half of it, and none.
The table below gives the target mRNA in each line as a mean with its standard error, SE, and the range the true mean is likely to lie in.
In the table, the target mRNA is in arbitrary units. The ±2SE range is two SE below the mean to two SE above.
With no working copies, the cells make no cutting protein: the target mRNA is at its full amount, 100 units.
With one working copy, the cells make about half the cutting protein: 55 units are left. About half of the full amount, 45 units, is gone.
With two working copies, the cells make the usual amount of cutting protein: 10 units are left. Most of the full amount, 90 units, is gone.
The more cutting protein a cell makes, the less target mRNA it holds. So the cutting protein destroys the target mRNA.
One copy destroys about half as much target mRNA as two copies. Half the dose gives about half the effect.
Nobody changed the target gene. The target mRNA changed because the amount of cutting protein changed.
The three ±2SE ranges are 6 to 14, 47 to 63 and 88 to 112 units. None of them overlaps another.
So the three true means are very unlikely to be the same, and the pattern across the three doses is real.
To read any dose series of this kind, follow four steps.
- Find the control: the two-copy line.
- Read the dose: one copy makes about half the protein of two.
- Read the measured amount across the lines.
- Say what the pattern shows; check that the ±2SE ranges do not overlap.
What you are expected to know Read a table of cells with two, one and no working copies of a gene, and say what the pattern shows the gene’s protein does.
Now suppose you read the three lines of cells yourself, one step at a time.
Three lines of cells from one animal hold two, one and no working copies of the cutting gene. The biologist broke copies of the gene in two of the lines and left the third line unchanged.
Which line is the control?
- A. ✓ The line with two working copies
- B. The line with one working copyThe one-copy cells hold a broken copy, so they are changed cells.
The normal cells, with two working copies, are the control. - C. The line with no working copiesThe zero-copy cells hold two broken copies, so they are changed cells.
The normal cells, with two working copies, are the control.
Why: The normal cells of the animal hold two working copies of the cutting gene.
The other lines are compared with the normal cells.
So the line with two working copies is the control.
Three lines of cells from one animal hold two, one and no working copies of the cutting gene. Compare the one-copy cells with the two-copy cells.
Which amount is about half as large in the one-copy cells?
- A. The number of chromosomes in each cellEvery cell holds the same pair of chromosomes; one copy of the gene on them is broken.
The cutting protein, made from each working copy, is about halved. - B. ✓ The amount of cutting protein the cells make
- C. The amount of target mRNA that RNA polymerase makesRNA polymerase transcribes the target gene as before in every line; nobody changed the target gene.
The cutting protein, made from each working copy, is about halved.
Why: A cell makes the cutting protein from each working copy of the cutting gene.
The one-copy cells hold half as many working copies as the two-copy cells.
So the one-copy cells make about half as much cutting protein.
Three lines of cells from one animal hold two, one and no working copies of the cutting gene. The table below gives the target mRNA in each line; the zero-copy line holds the full amount.
Going from one working copy to two, what happens to the amount of target mRNA that is gone?
- A. About halvesCompared with the zero-copy line, 45 units are gone in the one-copy cells and 90 units in the two-copy cells.
The amount gone about doubles. - B. Stays about the sameCompared with the zero-copy line, 45 units are gone in the one-copy cells and 90 units in the two-copy cells.
Twice the dose removes about twice as much. - C. ✓ About doubles
Why: The zero-copy line holds the full amount, 100 units.
The one-copy line holds 55 units, so 45 units are gone.
The two-copy line holds 10 units, so 90 units are gone.
So the amount gone about doubles.
Suppose a biologist grows three lines of cells from one plant. Every cell carries the gene for a protein that cuts one particular mRNA. The lines hold two, one and no working copies of that gene. The table below gives the target mRNA in each line. In the table, each target mRNA value is a mean in arbitrary units with its standard error (SE); the ±2SE range is two SE below the mean to two SE above.
(a) Describe the relationship between the number of working copies of the gene and the level of the target mRNA. (1 pt)
- Award 1 point for: the target mRNA rises as the working copies fall (an inverse relationship), with the direction stated, with or without the values.
(b) Explain what the pattern shows the protein does to the target mRNA, using the one-copy line. (2 pt)
The zero-copy cells hold the full amount, 243 units.
The one-copy cells hold 146 units, so 97 units are gone.
The two-copy cells hold 49 units, so 194 units are gone.
Half the protein removes half as much target mRNA, so the protein destroys the target mRNA.
- Award 1 point for: the more of the protein a line makes, the less target mRNA it holds, so the protein destroys (cuts up, breaks down) the target mRNA.
- Award 1 point for: the one-copy cells make about half the protein of the two-copy cells and lose about half as much target mRNA (97 against 194 units), so half the dose gives about half the effect.
(c) Explain what the ±2SE ranges of the one-copy line and the two-copy line show about the difference in target mRNA between them. (1 pt)
The two-copy line’s mean is 49 units with SE 4, so its ±2SE range is 41 to 57 units.
The two ranges do not overlap, so the two true means are very unlikely to be the same.
So the two lines almost certainly differ.
- Award 1 point for: the two ±2SE ranges, worked from the means and SEs (132 to 160 and 41 to 57 units), do not overlap, so the true means differ — the difference is very unlikely to be chance.
Back to the three lines of cells from one animal, with two, one and no working copies of the cutting gene.
Two working copies of the cutting gene gave 10 units of the target mRNA, one copy gave 55, and none gave 100.
So the cutting protein destroys the target mRNA, and half the dose gives about half the effect.
Nobody touched the target gene.
77Mixed practice mixed practice
A cell of the skin of a person’s heel transcribes one of its own genes into a short RNA, which pairs with an mRNA in that cell.
Which kind of small RNA is this?
- A. ✓ microRNA
- B. siRNAA supplied small RNA is an siRNA.
A small RNA a cell transcribes from one of its own genes is a microRNA.
Why: The cell transcribed the short RNA from one of its own genes.
A small RNA a cell makes itself is called a microRNA.
Suppose cells given an siRNA against a gene hold far less of the gene’s protein than untreated cells, and they can no longer take up magnesium from their broth. Untreated cells take up magnesium.
What does the result show about the gene?
- A. The siRNA cells transcribe the gene less often than the untreated cellsAn siRNA pairs with the gene’s mRNA and leaves transcription as it was.
The cells that lack the protein cannot take up magnesium, so the protein is needed for that. - B. ✓ The gene’s protein is needed for taking up magnesium
- C. The gene is transcribed only while magnesium is presentThe siRNA, not magnesium, changed these cells.
The cells that lack the gene’s protein cannot take up magnesium, so the protein is needed for that.
Why: The siRNA lowered the gene’s protein.
The cells that lack the protein can no longer take up magnesium.
So the gene’s protein is needed for taking up magnesium.
Suppose three lines of cells hold two, one and no working copies of the gene for a protein that cuts one particular mRNA. The target mRNA is lowest in the two-copy line, higher in the one-copy line, and highest in the zero-copy line. Nobody changed the target gene.
What does the pattern show the protein does?
- A. The protein makes the target mRNAA protein that made the target mRNA would give the most mRNA with two copies.
Two copies give the least, so the protein destroys the target mRNA. - B. The protein has no effect on the target mRNAThe three lines differ, and only the dose of the protein differs between them.
More protein, less target mRNA: the protein destroys it. - C. ✓ The protein destroys the target mRNA
Why: Two working copies make the most cutting protein, and that line holds the least target mRNA.
No working copies make none, and that line holds the most.
The more of the protein, the less target mRNA.
So the protein destroys the target mRNA.
In a dose series for a plant gene whose protein builds starch, cells with two working copies of the gene hold 212 units of the starch-building protein, cells with one working copy hold 106 units, and cells with no working copies hold 0 units. A student reads the series and says: “A cell with one working copy of the starch gene makes about half as much of the starch-building protein as a cell with two working copies.”
Is the student correct?
- A. No: the one-copy cells make the usual amount of the protein, and only the zero-copy cells make lessThe one-copy cells hold 106 units of the starch-building protein, and the two-copy cells hold 212 units.
106 units is about half of 212 units. - B. ✓ Yes: a cell makes the protein from each working copy, so one copy makes about half
Why: The two-copy cells hold 212 units of the starch-building protein.
The one-copy cells hold 106 units, about half of 212 units.
Each working copy of the starch gene adds its share of the protein.
So one working copy makes about half as much of the protein as two.
A biologist wants to silence one chosen gene in a dish of cells and leave every other gene as it is.
Which molecule does the biologist supply to the cells?
- A. ✓ A short RNA whose bases match part of that gene’s mRNA
- B. A repressor protein that binds beside that gene’s promoterA repressor binds DNA and acts at transcription.
A short RNA matching part of the gene’s mRNA pairs with that mRNA and no other. - C. A short RNA that pairs with that gene’s DNA at its promoter and blocks RNA polymeraseA small RNA never pairs with a gene’s DNA or blocks its promoter.
A short RNA matching part of the gene’s mRNA pairs with that mRNA alone.
Why: A short RNA made to match a stretch found only in one gene’s mRNA silences that gene and no other.
The short RNA pairs with the matching stretch of that gene’s mRNA.
The paired mRNA is cut up or blocked, so the cell makes less of that gene’s protein.
In a dose series, the one-copy line’s ±2SE range for the target mRNA is 141 to 157 units and the two-copy line’s is 149 to 167 units. A student says: “The one-copy line and the two-copy line differ in target mRNA.”
Is the student correct?
- A. ✓ No: the two ranges overlap, so the true means could be the same
- B. Yes: the two sample means are different numbers, so the lines differTwo different sample means can come from one true mean.
The ±2SE ranges overlap, so the data cannot show that the lines differ.
Why: Each ±2SE range is where its true mean is likely to lie.
The range 141 to 157 units overlaps the range 149 to 167 units.
So both true means could lie in the shared range, at the same value.
The data do not show that the two lines differ.
A biologist supplies a short RNA to a worm’s cells. The short RNA pairs with one gene’s mRNA, and the cells make far less of that gene’s protein.
Which term names what happened to the gene?
- A. SplicingSplicing removes introns from a pre-mRNA in the nucleus.
A short RNA pairing with a gene’s mRNA and silencing the gene is RNA interference. - B. TranslationTranslation is a ribosome building a protein from an mRNA.
Here a short RNA silenced the gene.
The silencing of a gene by a small RNA is RNA interference. - C. ✓ RNA interference
Why: The supplied short RNA paired with the gene’s mRNA.
The paired mRNA was cut up or blocked, so the cells made far less of the protein.
The silencing of a gene by a small RNA is called RNA interference.
Suppose cells grown in a dish join to their neighbors in a flat sheet. A biologist gives the cells a short RNA made to silence a gene whose job is unknown. In the treated cells the level of the gene’s mRNA is unchanged, the level of its protein is far lower, and the cells come loose from one another.
(a) Describe what the result shows about the job of the gene. (1 pt)
So the gene’s protein is needed to hold the cells together in the sheet.
- Award 1 point for: the gene’s protein is needed to hold the cells to their neighbors (to keep the sheet together).
(b) Explain how the short RNA lowered the protein while the level of the gene’s mRNA stayed unchanged. (2 pt)
The paired copies were not cut up, so the level of the mRNA stayed the same.
The pairing stopped ribosomes from loading onto the mRNA.
Ribosomes build no protein from an unread message.
So the level of the protein fell.
- Award 1 point for: the short RNA paired with (is complementary to) the gene’s mRNA.
- Award 1 point for: the paired mRNA was blocked from the ribosome (not translated) rather than cut up, so the mRNA level stayed while the protein fell.
APBIO-U06-P66 Practice questions: Topic 6.6
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 one dose series 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: Gene expression and cell specialization, summed up
Transcription factors and RNA polymerase at the promoter; enhancers near or far; upstream and downstream of the start site; repressors block; activator or repressor from the data; different reading makes a cell’s products; small RNAs silence a message; a knockdown and a dose series show what a gene does.
File: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-T66-summary.mp4
Video file not found on this machine: /Users/jamesmoore/Documents/AP Biology/course_preview/media/APBIO-U06-T66-summary.mp4
A biologist compares transcription in a bacterium and in a yeast cell, a eukaryote. In each cell, RNA polymerase is about to transcribe a gene.
In which of the two cells does RNA polymerase bind the promoter directly, with no other protein bound there first?
- A. ✓ The bacterium only
- B. The yeast cell onlyA yeast cell’s RNA polymerase cannot bind a promoter on its own.
Its transcription factors bind first, and RNA polymerase binds to them. - C. Both cellsIn a eukaryotic cell, RNA polymerase docks against the factors bound to the promoter.
Only a bacterium’s RNA polymerase binds its promoter directly. - D. Neither cellA bacterium’s RNA polymerase binds its promoter directly.
The two-step switch, factors then polymerase, is the eukaryote’s.
Why: In a bacterium, RNA polymerase binds the promoter directly.
In a yeast cell, a eukaryote, RNA polymerase cannot bind a promoter on its own.
Transcription factors bind the promoter first, and RNA polymerase binds to them.
A student lists four stretches of DNA and RNA found in cells.
Which of the following is an enhancer?
- A. The stretch just before a gene where the transcription factors bind and RNA polymerase docksThe stretch where the factors bind and RNA polymerase docks is the promoter.
An enhancer is a different stretch, which activators bind. - B. ✓ A stretch inside a gene’s second intron that bound proteins loop to the promoter to raise transcription
- C. A stretch of DNA beside a gene that a repressor binds, so the DNA around the gene winds tightA stretch a repressor binds lowers transcription.
An enhancer is bound by activators and raises transcription. - D. The stretch of an mRNA that a short RNA pairs withAn enhancer is DNA.
The stretch of an mRNA that a short RNA pairs with is RNA, and the pairing silences the message.
Why: An enhancer is a regulatory sequence that activators bind so the DNA loops and the gene is transcribed more often.
It may lie inside one of the gene’s introns.
So the stretch inside the second intron that bound proteins loop to the promoter is an enhancer.
On the gene map of a eukaryotic gene, a first regulatory sequence lies 5,200 bp before the transcription start site, beyond the promoter. A second lies 1,200 bp beyond the gene’s last exon.
Which of the following describes the two positions?
- A. Both sequences lie downstream of the start siteThe first sequence lies before the start site, on the side RNA polymerase moves away from.
A position before the start site is upstream. - B. The first lies downstream and the second upstreamRNA polymerase moves from the start site into the gene and on past its last exon.
The first sequence lies behind it: upstream; the second lies ahead: downstream. - C. Both sequences lie upstream of the start siteThe second sequence lies beyond the last exon, after the start site.
A position after the start site, inside the gene or beyond it, is downstream. - D. ✓ The first lies upstream and the second downstream
Why: RNA polymerase starts at the transcription start site and moves into the gene.
The first sequence lies 5,200 bp before the start site, on the side RNA polymerase moves away from: upstream.
The second lies beyond the last exon, after the start site: downstream.
In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined. A team tests a stretch of DNA from near a rhinoceros beetle’s horn gene. The table gives four constructs and the glow of each.
In which kind of cell does the stretch raise transcription?
- A. In antenna cells onlyIn horn cells the stretch gives a bright glow and the promoter alone a faint one.
The bright row is in horn cells, and antenna cells stay faint. - B. ✓ In horn cells only
- C. In both kinds of cellIn antenna cells the stretch gives a faint glow, the promoter’s own.
A glow no brighter than the promoter alone shows the stretch raised nothing there. - D. In neither kind of cellConstruct 3 glows bright in horn cells, far above the promoter’s faint glow.
So the stretch raised transcription there.
Why: The promoter alone glows faint in both kinds of cell.
With the stretch, horn cells glow bright: far above the promoter’s own, so the stretch raises transcription there.
With the stretch, antenna cells glow faint: no brighter than the promoter alone, so the stretch raises nothing there.
In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined. A team tests a stretch of DNA from near a rhinoceros beetle’s horn gene, in the four constructs the table gives. A student claims that the stretch raises transcription in horn cells.
Which construct, compared with the promoter alone in horn cells, is the evidence for the claim?
- A. Construct 2: the promoter alone, in antenna cellsConstruct 2 carries the promoter alone, in antenna cells.
Evidence for the stretch’s effect in horn cells needs the stretch, in horn cells. - B. ✓ Construct 3: the stretch with the promoter, in horn cells
- C. Construct 4: the stretch with the promoter, in antenna cellsConstruct 4 glows faint in antenna cells.
It shows what the stretch does in antenna cells, and the claim is about horn cells. - D. Constructs 3 and 4 togetherConstruct 4 sits in antenna cells, so it adds nothing about horn cells.
The evidence is the horn-cell row that differs from the promoter alone by the stretch only.
Why: Construct 3 carries the stretch with the promoter in horn cells and glows bright.
Construct 1, the promoter alone in horn cells, glows faint.
The stretch is the only difference between them.
So construct 3, read against construct 1, is the evidence for the claim.
In a cell of a hazel tree, a repressor binds a short stretch of DNA far from a gene’s promoter and brings in enzymes. The gene falls silent.
What do the enzymes the repressor brings in do?
- A. Adding acetyl groups to the histonesAcetyl groups on the histones loosen the DNA.
A repressor brings in enzymes that do the opposite, so the DNA around the gene winds tight. - B. Cutting the gene’s mRNA into piecesA repressor acts before any mRNA is made.
The enzymes it brings in change the histones, so RNA polymerase never reaches the gene. - C. ✓ Taking acetyl groups off the gene’s histones
- D. Taking the transcription factors off the promoterThe enzymes act on the histones, not on the factors.
With the acetyl groups gone the DNA winds tight, and the factors reach the promoter no more.
Why: The bound repressor brings in enzymes.
The enzymes take acetyl groups off the histones around the gene.
With the acetyl groups gone, the histones grip the DNA more tightly, so the DNA winds tight.
RNA polymerase and the factors reach only loosely wound DNA, so the gene is silent.
In a stoat’s cell, a repressor keeps one gene silent. Now imagine a change in the repressor’s gene gives the repressor a binding site that fits only a sequence found nowhere in the cell’s DNA.
What happens to transcription of the gene?
- A. ✓ Transcription switches on
- B. Transcription switches offThe gene was already silent.
Losing a repressor never lowers transcription. - C. Transcription stays as it wasThe changed repressor fits its old stretch no more, so it never binds there.
With no repressor bound, no enzymes are brought in, and the DNA loosens. - D. Transcription of every gene in the cell switches onOne repressor controls the gene, or genes, whose stretch it fits.
The other genes keep their own regulators, so only this gene changes.
Why: A repressor blocks transcription only while its binding site fits a stretch beside the gene.
The changed repressor fits its old stretch no more, so it never binds there.
No enzymes are brought in, the DNA loosens, and RNA polymerase reaches the gene.
So transcription switches on.
A biologist gives a tench’s cell extra copies of the gene for one regulatory protein, so the cell makes far more of that protein. Transcription of a gene the protein controls rises to six times its level in a normal cell.
Which kind of protein is it, for that gene?
- A. ✓ An activator
- B. A repressorMore of a repressor would lower transcription.
Here more of the protein raised it sixfold. - C. Not a regulatory proteinA protein whose amount changes a gene’s transcription regulates that gene.
Sixfold more transcription is a regulator’s work. - D. An enhancerAn enhancer is a stretch of DNA.
The protein that binds it and raises transcription is an activator.
Why: The cell makes far more of the protein.
Transcription of the gene rises to six times its level.
So the protein raises transcription.
A regulatory protein that raises transcription is an activator.
A student lists four short RNAs found in cells.
Which of the following is a microRNA?
- A. A short RNA an enzyme lays down at a replication fork to start a new DNA strandA short RNA that starts a new DNA strand is a primer.
A microRNA pairs with an mRNA and silences it. - B. A folded RNA that carries an amino acid to the ribosomeAn RNA that carries an amino acid is a tRNA.
A microRNA pairs with a stretch of an mRNA. - C. A short RNA a biologist supplies to cells, which pairs with one gene’s mRNAA small RNA supplied to a cell is an siRNA.
A microRNA is a small RNA the cell transcribes from one of its own genes. - D. ✓ A short RNA a cell transcribes from one of its own genes, which pairs with an mRNA
Why: A small RNA pairs with a matching stretch of an mRNA, and the paired mRNA is cut up or blocked.
A small RNA a cell transcribes from one of its own genes is a microRNA.
A supplied one is an siRNA.
Single cells of a pond alga glide across the bottom of a dish. A biologist gives half the cells an siRNA against one gene, whose job is unknown. The table gives, for control cells and siRNA cells, transcription of the gene, its mRNA, its protein and what the cells do.
Which of the following does the result show about the gene?
- A. The siRNA cells transcribe the gene less often than the control cellsTranscription reads 45 times an hour in both.
The siRNA acts on the mRNA and leaves transcription as it was. - B. The gene is transcribed only while the cells glideTranscription reads 45 times an hour in the siRNA cells, which stay in place.
The gene is transcribed whether the cells glide or not. - C. ✓ The gene’s protein is needed for gliding
- D. The gene’s protein stops the cells glidingThe cells that lack the protein stay in place.
A protein that stopped gliding would leave those cells gliding.
Why: The siRNA pairs with the gene’s mRNA, and the paired copies are cut up.
The siRNA cells hold 19 units of the protein against 210.
Those cells stay in place instead of gliding.
So the cells need the gene’s protein to glide.
A biologist grows three lines of cells from one kind of sea cucumber. Every cell carries the gene for a protein that cuts one particular mRNA: the cutting gene. The lines hold two, one and no working copies of the cutting gene; the biologist broke copies of the gene in two of the lines and left the third line as it was. The table gives the target mRNA in each line as a mean in arbitrary units with its standard error (SE); a ±2SE range runs from two SE below the mean to two SE above.
(a) Identify the control line. (1 pt)
Frame The control line is …
Hint Which line holds the cells nobody changed?
- Award 1 point for: the two-copy line (the normal, unchanged cells).
(b) Describe the relationship between the number of working copies and the target mRNA. (1 pt)
Frame As the working copies fall from two to none, the target mRNA …
Hint Read the three means in the order of the table.
- Award 1 point for: the target mRNA rises as the working copies fall (the inverse relationship), with the direction stated.
(c) Explain how much cutting protein the one-copy cells make, compared with the two-copy cells. (1 pt)
Frame The one-copy cells make …
Hint A cell makes the protein from every working copy it holds. How many does each of the two lines hold?
A cell makes the cutting protein from each working copy of the cutting gene.
One working copy is half as many as two.
- Award 1 point for: about half as much cutting protein, because the protein is made from each working copy and the line holds one copy instead of two.
(d) Explain what the pattern shows the cutting protein does to the target mRNA. (1 pt)
Frame The more cutting protein a line makes, …
Hint Put the three lines side by side: the dose of protein in each, and the target mRNA left in each.
So the cutting protein destroys the target mRNA.
- Award 1 point for: more cutting protein, less target mRNA, so the protein destroys (cuts up) the target mRNA; nobody changed the target gene.
(e) Explain what the ±2SE ranges of the one-copy line and the zero-copy line show about the difference between them. (1 pt)
Frame The two ranges …
Hint Work each range: two SE below the mean to two SE above. Do the two ranges share any value?
The one-copy line’s mean is 48 units with SE 3, so its ±2SE range is 42 to 54 units.
The zero-copy line’s mean is 84 with SE 4, so its range is 76 to 92 units.
So the two true means are very unlikely to be the same: the lines differ.
- Award 1 point for: the two ±2SE ranges, 42 to 54 and 76 to 92 units, do not overlap, so the true means differ (the difference is very unlikely to be chance).
Two strains of one kind of bacterium live in oily soil. A biologist tests a cell of each strain for two genes: a gene for an enzyme that breaks down oil, and a gene for an enzyme that builds the cell wall. The table gives, for each cell, whether each gene is in the DNA and the level of its mRNA and its protein. In the table, present or absent says whether the gene is in the cell’s DNA; the levels are in arbitrary units.
(a) Identify the gene that both strains carry and transcribe. (1 pt)
- Award 1 point for: the cell-wall enzyme gene.
(b) A student claims that the two strains carry the same genes and differ only in which genes they express. Evaluate the claim using the table. (1 pt)
The oil-enzyme gene is present in the first strain’s DNA and absent from the second’s.
So the two strains differ in the genes they carry, and the second strain’s missing mRNA and enzyme follow from the missing gene.
- Award 1 point for: not supported, because the DNA row shows the oil-enzyme gene present in the first strain and absent from the second, so the strains differ in their genes (the mRNA and protein rows follow from that).
Slip Reading the oil-enzyme mRNA row, 240 against 0 units, as a difference in expression alone. Check the DNA row first: a gene that is absent cannot be expressed at all.
(c) The biologist gives a cell of the second strain a copy of the oil-enzyme gene, joined to a promoter that the cell’s RNA polymerase binds. Predict what the cell holds a day later. (1 pt)
- Award 1 point for: the cell holds the oil-enzyme mRNA and the oil enzyme (it now expresses the gene).
(d) Justify your prediction in part (c). (1 pt)
So RNA polymerase transcribes the gene into mRNA.
Ribosomes build the enzyme from that mRNA.
So the cell holds both the mRNA and the enzyme.
- Award 1 point for: with the gene and its promoter in the cell, RNA polymerase transcribes the gene into mRNA and ribosomes build the enzyme from it. Accept reasoning consistent with the prediction in part (c).
APBIO-U06-T66 End-of-topic test: Gene Expression and Cell Specialization
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.
In a eukaryotic cell of a mackerel, a change in RNA polymerase gives it a binding site that fits none of the transcription factors bound to a promoter. The factors still bind every promoter as before.
Which of the following describes transcription of the cell’s protein-coding genes after the change?
- A. RNA polymerase transcribes the genes as often as beforeIn a eukaryotic cell, RNA polymerase binds a promoter only through the factors bound there.
A polymerase that fits none of the factors never docks. - B. ✓ RNA polymerase transcribes none of the genes
- C. RNA polymerase transcribes the genes more often than beforeRNA polymerase docks by fitting the bound factors.
Bound factors help nothing when RNA polymerase can no longer bind to them. - D. RNA polymerase transcribes only the genes with activators bound to an enhancerActivators raise transcription only by holding RNA polymerase at a promoter it has docked at.
A polymerase that fits no factor never docks, enhancer or not.
Why: In a eukaryotic cell, RNA polymerase cannot bind a promoter on its own.
RNA polymerase docks by fitting the transcription factors bound there.
The changed polymerase fits no factor, so it never docks.
So it transcribes none of the genes.
In cells from a lamprey, a team joins an enhancer and a promoter to a reporter gene in four constructs, with and without one nuclear protein. The table gives the reporter mRNA in each case.
Which of the following conclusions do the results best support?
- A. The protein raises transcription by binding the promoterThe promoter-only construct holds the protein and makes 9 units of mRNA.
A protein acting at the promoter would raise transcription with no enhancer present. - B. The protein raises transcription by binding the reporter mRNAA regulatory protein binds DNA beside the gene, before any mRNA exists.
With the enhancer changed, the protein raises nothing, though the mRNA is still made. - C. The joined DNA raises transcription by itself, with or without the proteinWith the enhancer and promoter and no protein, the cells make 8 units of mRNA.
The enhancer raises transcription only while the protein is present. - D. ✓ The protein raises transcription by binding the enhancer
Why: The mRNA reaches 100 units only with the intact enhancer and the protein together.
Without the protein it stays at 8 units, so the protein is needed.
With the enhancer changed, the protein raises nothing.
So the protein binds the enhancer, and from there raises transcription.
A biologist notes three regulatory sequences on the gene map of a eukaryotic gene. The first lies 850 bp before the transcription start site. The second lies inside the gene’s third intron. The third lies 1,500 bp beyond the gene’s last exon.
Which of the three sequences lie downstream of the transcription start site?
- A. The first onlyDownstream means after the start site, on the side RNA polymerase moves toward.
The first sequence lies before the start site: upstream. - B. The second onlyA sequence beyond the last exon lies after the start site too.
Downstream covers positions inside the gene and beyond it. - C. ✓ The second and the third
- D. The first, the second and the thirdThe first sequence lies 850 bp before the start site.
A position before the start site is upstream, however near.
Why: RNA polymerase starts at the transcription start site and moves into the gene.
The third intron and the stretch beyond the last exon both lie after the start site, where RNA polymerase moves toward.
So the second and third sequences are downstream.
The first lies before the start site: upstream.
In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined. A team tests a stretch of DNA from near a squid’s ink-pigment gene. The table gives the glow of three constructs, in units.
In which kind of cell does the stretch raise transcription?
- A. ✓ In ink-gland cells only
- B. In arm cells onlyIn arm cells the stretch gives 13 units, about the promoter’s own 12.
The bright row, 95 units, is in ink-gland cells. - C. In both kinds of cellIn arm cells the stretch gives 13 units, no brighter than the promoter alone.
A glow no brighter than the promoter’s own shows the stretch raised nothing there. - D. In neither kind of cellIn ink-gland cells the stretch raises the glow from 12 units to 95.
A glow far above the promoter’s own shows the stretch raised transcription there.
Why: The promoter alone glows 12 units in ink-gland cells.
With the stretch, ink-gland cells glow 95 units: far above the promoter’s own, so the stretch raises transcription there.
In arm cells the stretch gives 13 units, about the promoter’s own.
So the stretch raises nothing in arm cells.
In the drawings, an open box is a piece of DNA joined in, and the filled box is the reporter gene. The arrow inside the stretch’s box shows which way round the stretch was joined. A team tests a stretch of DNA from near a squid’s ink-pigment gene; the table gives three constructs and their glow. The fourth construct joins the same stretch after the reporter gene’s end, the original way round, with the promoter, in ink-gland cells.
Which glow should the team predict for the fourth construct?
- A. About 0 unitsThe fourth construct carries the promoter, so RNA polymerase can bind and transcribe the reporter gene.
The promoter alone gives at least 12 units. - B. About 12 unitsInk-gland cells hold activators that bind the stretch.
The DNA loops to the promoter from downstream of the gene as from upstream, so the stretch still raises transcription. - C. ✓ About 95 units
- D. About 190 unitsThe fourth construct carries one stretch, as the second did.
Its activators loop to the promoter once, so the glow is about the second construct’s, 95 units.
Why: Ink-gland cells hold activators whose binding sites fit the stretch.
An enhancer works from downstream of its gene as well as upstream: the DNA loops so the bound activators touch the factors at the promoter.
So the reporter gene is transcribed often, and the glow is about 95 units.
In a walrus’s cell, a repressor is bound to a short stretch of DNA 9,000 bp upstream of a gene’s transcription start site, and the gene is silent. The drawing shows the gene. A technician adds a drug that blocks the enzymes that take acetyl groups off histones.
An hour after the technician adds the drug, which of the following describes the gene and the repressor?
- A. The gene stays silent, and the repressor stays boundThe repressor blocks transcription by bringing in the enzymes that take acetyl groups off.
With those enzymes blocked, other enzymes keep adding acetyl groups, and the DNA loosens. - B. ✓ The gene is transcribed, and the repressor stays bound
- C. The gene stays silent, and the repressor leaves the DNAThe drug acts on the acetyl-removing enzymes and binds no repressor.
The repressor still fits its stretch and stays bound. - D. The gene is transcribed, and the repressor leaves the DNAThe drug binds no repressor, so the repressor stays on its stretch.
The enzymes it brings in are blocked, so the DNA around the gene loosens.
Why: The bound repressor brings in enzymes that take acetyl groups off the histones, and the DNA winds tight.
The drug blocks those enzymes.
Other enzymes keep adding acetyl groups, so the DNA around the gene loosens.
RNA polymerase reaches the promoter, so the gene is transcribed with the repressor bound.
In a scallop’s cell, a repressor keeps one gene silent by winding its DNA tight. The cell has also lost every copy of one of the transcription factors that bind every promoter. Now imagine the repressor’s gene is deleted.
What happens to transcription of the silent gene?
- A. Transcription switches onWith no repressor, the DNA loosens, and RNA polymerase could reach the promoter.
RNA polymerase docks only on the promoter’s full set of factors, and one factor is gone. - B. Transcription switches on at a low rateThe low rate of a bare promoter needs its factors bound.
With one factor gone, RNA polymerase never docks, so no RNA is made. - C. Transcription switches offThe gene was already silent.
Deleting a repressor never lowers transcription. - D. ✓ Transcription stays as it was
Why: Deleting the repressor’s gene removes the block: no enzymes are brought in, and the DNA loosens.
But RNA polymerase binds a promoter only through its full set of transcription factors.
One factor is missing from the cell, so RNA polymerase never docks.
The gene stays silent.
Four biologists each delete the gene for one regulatory protein in one cell and measure the mRNA of a gene that protein had been bound beside. The table gives the mRNA before and after the deletion, in each cell.
In which cell was the deleted protein a repressor of the gene?
- A. The haddock’s cellRemoving the protein lowered the mRNA from 84 units to 21.
A protein whose removal lowers transcription had been raising it: an activator. - B. ✓ The plaice’s cell
- C. The tench’s cellThe mRNA reads 62 units before the deletion and 60 after: unchanged.
A protein whose removal changes nothing was not regulating this gene. - D. The bream’s cellRemoving the protein lowered the mRNA from 160 units to 7.
Removal that lowers transcription names an activator.
Why: A repressor lowers transcription, so removing one raises the gene’s mRNA.
In the plaice’s cell the mRNA rose from 15 units to 240 after the deletion.
So that protein had been lowering transcription: a repressor.
The haddock’s and the bream’s proteins had been raising transcription.
A steroid hormone enters a herring’s cell and binds its receptor. A biologist gives the cell extra copies of the receptor’s gene, so the cell makes far more receptor. With the hormone absent, transcription of one gene stays at its normal level. With the hormone present, it rises to three times its level in a normal cell.
Which kind of protein is the hormone-bound receptor, for that gene?
- A. ✓ An activator
- B. A repressorMore of a repressor would lower transcription.
More receptor, with the hormone present, raised it. - C. Not a regulatory proteinThe bound receptor moves into the nucleus and binds beside the gene.
More of it changed transcription, so it regulates the gene. - D. A carrier that takes the hormone to the gene’s promoterA hormone binds no DNA.
The bound receptor itself binds beside the gene and raises transcription: the receptor is the transcription factor.
Why: The bound receptor binds the DNA beside the gene, so it is a transcription factor.
More receptor, with the hormone present, raised transcription to three times its level.
A regulatory protein that raises transcription is an activator.
Without the hormone the receptor is not switched on, so nothing changes.
A hagfish’s slime-gland cells are packed with the slime protein, which thickens the water around the fish when a predator bites. Its heart cells hold none of the protein. Sequencing shows the slime-protein gene in the DNA of both kinds of cell, base for base the same. A biologist finds the gene’s mRNA in the slime-gland cells and no mRNA of it in the heart cells.
Which of the following explains why only the slime-gland cells hold the slime protein?
- A. The heart cells lost the slime-protein gene when they differentiatedSequencing found the gene in the heart cells’ DNA.
Differentiation removes no genes. - B. The heart cells carry a changed copy of the gene, whose mRNA codes for a protein that does no jobThe gene reads the same, base for base, in both kinds of cell.
The heart cells hold no mRNA of it, so they build no protein from it at all. - C. Both kinds of cell transcribe the gene, and the heart cells’ ribosomes leave the mRNA unreadThe heart cells hold none of the gene’s mRNA.
A gene transcribed in both cells would give mRNA in both. - D. ✓ Only the slime-gland cells transcribe the gene, so only they hold mRNA for ribosomes to read
Why: Both cells carry the same slime-protein gene.
RNA polymerase transcribes it in the slime-gland cells only, so only they hold its mRNA.
Ribosomes build a protein only from an mRNA.
So only the slime-gland cells build the slime protein, and making slime is those cells’ job.
Suppose a chard plant is moved from shade into strong sunlight. Over several days its leaves redden: the leaf cells build far more of an enzyme that makes a red, sun-shielding pigment. The leaf cells’ DNA is unchanged.
Which of the following describes what the strong sunlight changed in the leaf cells?
- A. ✓ How often the leaf cells transcribe the enzyme’s gene
- B. The base sequence of the enzyme’s geneThe leaf cells’ DNA is unchanged, so the gene reads as before.
What changed is how often RNA polymerase transcribes it. - C. The number of copies of the enzyme’s geneThe DNA is unchanged, so the cells hold the same copies of the gene as before.
The reading of the gene changed, not its number. - D. The order of amino acids in the enzymeThe gene’s sequence is unchanged, so the enzyme’s amino acids are as before.
The cells build more of the same enzyme.
Why: A condition outside the plant changes which genes its cells read, and how often.
Signals set off by the strong light make RNA polymerase transcribe the enzyme’s gene more often.
Ribosomes build more of the enzyme from more mRNA.
The gene itself is unchanged.
A buckwheat plant transcribes one of its own genes into a short RNA that pairs with the mRNA of a seed-protein gene. A second buckwheat plant lacks the short RNA’s gene, so it has no short RNA. The table gives the seed-protein mRNA and the seed protein in the seeds of both plants.
Which of the following does the short RNA do in the normal plant?
- A. The short RNA pairs with the seed-protein mRNA, and enzymes cut the paired mRNA upA cut-up message leaves less mRNA.
The mRNA reads 48 units in the normal plant and 50 without the short RNA: the mRNA stays whole. - B. The short RNA binds the seed-protein gene’s promoter and lowers transcriptionA short RNA of this kind never binds a gene’s DNA.
The mRNA level is about the same with and without it, so transcription is unchanged. - C. ✓ The short RNA pairs with the seed-protein mRNA and keeps ribosomes off it
- D. The short RNA binds the seed protein and breaks it downA small RNA acts on the mRNA, never on the protein.
Its pairing with the mRNA is what keeps the protein low.
Why: The seed-protein mRNA reads about the same in both plants, so transcription is unchanged and the mRNA is whole.
The protein rises from 28 units to 118 without the short RNA.
So the short RNA pairs with the mRNA and stops ribosomes loading onto it: a control after transcription.
A biologist supplies a short RNA to cells from a lungfish. The short RNA pairs with the mRNA of one gene. The table gives, for untreated cells and the treated cells, transcription of the gene, the gene’s mRNA and its protein.
At which step does the short RNA act on the gene’s expression?
- A. At transcription, on the gene’s DNARNA polymerase transcribes the gene 36 times an hour in both kinds of cell.
Transcription is unchanged, so the short RNA left the DNA alone. - B. ✓ After transcription, on the mRNA
- C. After translation, on the proteinThe mRNA fell from 120 units to 9 before the protein fell.
A control on the protein alone would leave the mRNA at 120. - D. At transcription and after it, on both the DNA and the mRNATranscription reads 36 times an hour in both.
The DNA and its transcription are untouched; only the mRNA and the protein fell.
Why: RNA polymerase transcribes the gene as often as before: 36 times an hour.
The mRNA fell from 120 units to 9, so the copies were cut up after they were made.
Ribosomes had fewer messages, so the protein fell too.
The short RNA acts on the mRNA, after transcription.
Cells taken from a jellyfish flash blue light when they are shaken. A biologist gives half the cells an siRNA against one gene, whose job is unknown. The table gives, for control cells and siRNA cells, transcription of the gene, its mRNA, its protein and what the cells do.
Which of the following does the result show about the gene?
- A. The siRNA lowered how often the gene is transcribedTranscription reads 52 times an hour in both.
The siRNA acts on the mRNA and leaves transcription as it was. - B. The gene’s protein stops the cells flashingThe cells that lack the protein stay dark.
A protein that stopped the flash would leave those cells flashing. - C. The siRNA cells have lost the geneRNA polymerase still transcribes the gene 52 times an hour in the siRNA cells.
The gene is still in their DNA; only its mRNA and protein are low. - D. ✓ The gene’s protein is needed for the flash
Why: The siRNA pairs with the gene’s mRNA, and the paired copies are cut up.
The siRNA cells hold far less of the protein: 15 units against 176.
Those cells stay dark when shaken.
So the cells need the gene’s protein to flash.
A biologist gives cells an siRNA against one gene. The treated cells hold far less of the gene’s mRNA and far less of its protein than untreated cells. A student says the siRNA might have acted on the gene’s DNA instead of on its mRNA.
Which line of a results table would settle where the siRNA acted?
- A. ✓ How often RNA polymerase transcribes the gene
- B. The level of the gene’s mRNAA cut-up message and a gene transcribed less often both lower the mRNA.
The mRNA line alone cannot tell them apart. - C. The level of the gene’s proteinLess mRNA gives less protein either way.
The protein line follows the mRNA and settles nothing about the DNA. - D. What the cells can no longer doWhat the cells can no longer do shows the protein’s job.
It leaves the step the siRNA acted at open.
Why: If the siRNA had acted on the DNA, RNA polymerase would transcribe the gene less often.
An siRNA acts on the mRNA, so transcription is as often as before.
Only the transcription line shows how often RNA polymerase transcribes the gene.
So that line settles where the siRNA acted.
A pheasant’s feather cells transcribe one gene often; its beak cells transcribe the same gene rarely. Both kinds of cell carry the gene, its promoter and its enhancer, base for base the same, and both transcribe most of their other genes.
Which of the following explains the difference?
- A. The beak cells lack the transcription factors that bind every promoterThe beak cells transcribe most of their genes, so their promoters’ factors are present.
Those factors bind every promoter. - B. The enhancer lies farther from the promoter in the beak cellsThe two cells carry the same DNA, so the enhancer sits at the same distance in both.
What differs is the proteins bound to it. - C. The beak cells have lost the gene’s enhancerBoth kinds of cell carry the enhancer, base for base.
Differentiation removes no DNA. - D. ✓ Only the feather cells hold the activators that bind the enhancer
Why: Both cells carry the same gene, promoter and enhancer.
Each kind of cell holds its own transcription factors.
The feather cells hold activators that fit the enhancer, so the DNA loops and the gene is transcribed often.
The beak cells hold none, so the promoter alone gives rare transcription.
Three lines of cells hold two, one and no working copies of the gene for a protein that cuts one particular mRNA. The three lines hold different amounts of that target mRNA. Nobody changed the target gene.
Which of the following explains why the target mRNA differs between the lines?
- A. The target gene’s promoter differs between the linesNobody changed the target gene, promoter and all.
RNA polymerase transcribes it as often in every line. - B. The lines hold different numbers of chromosomesEvery cell carries the same pair of chromosomes; only copies of the cutting gene are broken.
The chromosome number is the same in every line. - C. ✓ The lines make different amounts of the cutting protein
- D. The target gene’s base sequence differs between the linesThe target gene is unchanged in every line.
The amount of cutting protein is what differs.
Why: A cell makes the cutting protein from each working copy of the cutting gene.
So the three lines make three amounts of the cutting protein.
The cutting protein destroys the target mRNA.
More cutting protein leaves less target mRNA, with the target gene untouched.
Eukaryotic cells use siRNAs to regulate certain genes. Once an siRNA has paired with an mRNA, a cutting protein cuts that mRNA up, and the mRNA is not translated. A biologist grows three lines of cells from one kind of sea squirt. The lines hold two, one and no working copies of the gene for the cutting protein: the cutting gene. She measures the mRNA of two other genes in each line, the pigment gene and the pump gene. The table gives each mean in arbitrary units with its standard error (SE); a ±2SE range runs from two SE below the mean to two SE above.
(a) Describe how an siRNA finds the one mRNA it silences. (1 pt)
So the siRNA pairs with that stretch, base to base, and with no other mRNA.
- Award 1 point for: the siRNA pairs (is complementary) with a matching stretch of the target mRNA’s bases, so it binds that mRNA only.
(b)(i) Identify the kind of graph that best shows the mean pigment mRNA of the three lines. (1 pt)
- Award 1 point for: a bar graph (a bar for each line’s mean).
Slip Choosing a line graph. The three lines are three categories, so their means are bars, and a line between them would claim values in between.
(b)(ii) Describe what would be plotted on each axis of that graph, with its unit. (1 pt)
The y-axis carries the mean pigment mRNA, in arbitrary units.
- Award 1 point for: the lines (working copies of the cutting gene) on the x-axis and mean pigment mRNA in units on the y-axis.
(b)(iii) Describe how the error bar for the one-copy line’s pigment mRNA would be drawn on that graph, giving its two ends. (1 pt)
The mean is 75 units and the SE 4, so the bar runs from 67 units to 83 units.
- Award 1 point for: a line through the bar’s top from 67 units to 83 units (the mean 75, two SE of 4 either side).
Slip Drawing one SE either side. The graph’s legend would read ±2SE, so each end lies two SE from the mean.
(c)(i) Determine all the lines whose pump mRNA is statistically the same as the pump mRNA of the two-copy line. (1 pt)
The two-copy line’s ±2SE range for pump mRNA is 56 to 72 units.
The one-copy line’s range is 48 to 72 units and the zero-copy line’s 62 to 78 units.
Both ranges overlap the two-copy line’s range, so the true means could be the same.
- Award 1 point for: both the one-copy and the zero-copy lines, because their ±2SE ranges (48 to 72 and 62 to 78 units) overlap the two-copy line’s (56 to 72 units).
Slip Naming the one-copy line alone. The zero-copy line’s mean is higher, but its range still overlaps the two-copy line’s, so the two are statistically the same.
(c)(ii) Describe the relationship between the number of working copies of the cutting gene and the pigment mRNA. (1 pt)
- Award 1 point for: the pigment mRNA rises as the working copies fall (an inverse relationship), with the direction stated, with or without the values.
(c)(iii) Calculate the percent by which the pigment mRNA of the zero-copy line is higher than the pigment mRNA of the one-copy line. (1 pt)
Answer: 60 % (tolerance ±0.5)
- Award 1 point for: 60 % (the rise of 45 units over the one-copy line’s 75 units, times 100).
(d)(i) The biologist claims that siRNA cutting plays a greater part in controlling the pigment gene’s expression than the pump gene’s. Support the claim using the table. (1 pt)
The pump mRNA stayed at about 60 to 70 units, and every line’s range overlaps the two-copy line’s.
So the cutting protein removes much of the pigment mRNA and barely touches the pump mRNA.
- Award 1 point for the comparison: the pigment mRNA changes greatly across the lines (30 → 75 → 120 units) while the pump mRNA barely changes (its ranges overlap), so cutting controls the pigment gene’s expression far more. Both genes must be compared.
Slip Citing the pigment rows alone. Support for a comparison needs the pump rows too.
(d)(ii) Sea squirts with no working copy of the cutting gene grow far darker than normal, and the pigment gene’s protein builds the dark pigment. Explain how the loss of the cutting protein could make these animals darker. (1 pt)
Ribosomes build more of the pigment-building protein from more mRNA.
More of the protein builds more dark pigment, so the animals grow darker.
- Award 1 point for: without the cutting protein the pigment mRNA is no longer cut up, so more of it persists, ribosomes build more pigment protein, and more pigment is made.
A pitcher plant catches insects in its pitchers, leaf cups filled with fluid. A biologist compares two cells of one pitcher plant: a cell lining the pitcher and a cell of the plant’s stalk. She tests each cell for two genes: the gene for a digestive enzyme released into the pitcher’s fluid, and a gene for one of the ribosome’s proteins. The table gives, for each cell, whether each gene is in the DNA and the level of its mRNA and its protein. In the table, present means the gene is in the cell’s DNA, and the levels are in arbitrary units.
(a) Identify the gene that both cells transcribe. (1 pt)
- Award 1 point for: the ribosome-protein gene.
(b) The biologist claims that the two cells carry the same genes and differ in which genes they express. Support the claim with two lines of evidence from the table. (2 pt)
Digestive-enzyme mRNA is at 410 units in the pitcher-lining cell and 0 units in the stalk cell, so only the pitcher-lining cell transcribes that gene.
- Award 1 point for the DNA line: the digestive-enzyme gene (or the ribosome-protein gene) is present in both cells, so the two cells carry the same genes.
- Award 1 point for the mRNA line: digestive-enzyme mRNA at 410 units in the pitcher-lining cell and 0 in the stalk cell, so the gene is expressed (transcribed) in one cell only. The values, or ‘in one cell only’, are needed; the protein row earns no point on either half of the claim.
Slip Citing the protein row. A cell with none of the protein might lack the gene or might carry it untranscribed, so the protein row settles neither half of the claim.
(c) When an insect falls into the pitcher, the pitcher-lining cells transcribe the digestive-enzyme gene more often within a few hours. Explain how a signal from the insect could raise transcription of the gene. (1 pt)
The switched-on factor binds the DNA beside the digestive-enzyme gene.
Bound there, the factor helps RNA polymerase bind the promoter, so RNA polymerase transcribes the gene more often.
- Award 1 point for: the signal switches on a transcription factor (an activator) that binds beside the gene and helps RNA polymerase bind the promoter, so the gene is transcribed more often. Accept: the signal lifts a repressor off the DNA beside the gene, so the DNA loosens and RNA polymerase reaches the promoter.
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