Regulation of the Cell Cycle

Unit 4 · Topic 4.6 end-of-topic test

Suggested time: about 47 minutes. Answer everything, then press Submit the test to see the feedback and scoring guides.

Answer every question. For each multiple-choice question, pick one option. When you have answered every question, press Submit the test; the feedback then gives the reasoning for each. For the two free-response questions, write one short sentence per step of your reasoning, in order, and show any calculation. In the exam, write the steps as a paragraph: a bulleted list is not scored. Then open the scoring guide and mark your own work against it. Where a graph carries error bars, the caption says what the bars represent. Every cell cycle in these questions is given with its length.

Question 1

Percent of mouse skin cells entering S phase in a day, before a strip was cleared from the packed sheet and over the two days after it, for cells along the edge of the strip and cells far from it.

A researcher drags a fine tip across a sheet of mouse skin cells packed edge to edge in a dish, clearing a strip of cells; the medium, which the researcher replaces every day, is the same before and after. The table below gives the percent of cells entering S phase in a day.

Which explanation accounts for the difference?

A. Cells at the edge of the strip lost contact on one side, so the signal that held them in G1 ended B. Clearing the strip released nutrients into the medium, and the edge cells took them up first C. The tip damaged the edge cells' DNA, and a damaged cell enters S phase to repair it D. Cells far from the strip had used up their share of the medium and could grow no further

Question 2

growth factor (ng/mL)	0	1	10	100
divisions per 100 cells in 24 h	3	11	29	30

Divisions per 100 bone cells in 24 hours at four concentrations of the growth factor.

Cultured bone cells receive the same nutrients in four dishes. A researcher adds a growth factor at a different concentration to each dish and counts the divisions per 100 cells in 24 hours. The results are in the table.

Which of the following do the counts at 0 and 100 ng/mL show about the growth factor?

A. The growth factor is a nutrient B. The growth factor is a signal that starts the cycle C. The growth factor speeds up mitosis D. The growth factor holds cells at the G1 checkpoint

Question 3

One 30-hour cycle of a cultured rat cell line drawn as a timeline, with four moments marked 1 to 4.

The timeline below shows one 30-hour cycle of a cultured rat cell line, with four moments marked 1 to 4. A cell in this culture is held for eight hours because its copied DNA carries a break.

At which marked moment does the cell wait?

A. Moment 1 B. Moment 2 C. Moment 3 D. Moment 4

Question 4

In one culture, cells were timed from the start of G2 to the start of mitosis. Most took 4 hours. One cell that had been exposed to a burst of X-rays took 19 hours, and then entered mitosis with its DNA repaired.

Which statement about the G2 checkpoint explains the 19 hours?

A. The G2 checkpoint is a timer that lasts longer in a cell that has been exposed to X-rays B. The G2 checkpoint repaired the DNA itself, and that repair takes about fifteen hours C. The G2 checkpoint sent the cell back to S phase to copy its DNA a second time D. The G2 checkpoint is a set of conditions, and the cell waited there until its DNA was undamaged

Question 5

In a strain of yeast, a mutation removes the M checkpoint's hold. When these cells divide, 12% of the daughter cells carry one chromosome too many or one too few, against 0.1% of the daughters of normal yeast.

Which condition does the missing hold normally check before the sister chromatids separate?

A. That the DNA has been completely copied and carries no damage B. That every chromosome is attached to spindle fibers from both poles C. That the cell has grown large enough and has a growth signal bound D. That the cleavage furrow has pinched the cytoplasm into two

Question 6

A researcher exposes cells in the regrowing tail of a lizard to a chemical that cuts DNA strands after S phase is complete. The cells stop at the end of G2 and stay there while the cuts are repaired.

Which harm does this hold prevent?

A. A daughter cell receiving a damaged genome B. Damaged DNA being copied in S phase C. A cell entering the cycle with no growth signal D. A daughter cell receiving an extra chromosome

Question 7

One cyclin's concentration (solid) and the activity of its CDK (dashed) through one 20-hour cycle of a cultured cell line, each as a percent of its own maximum. Gridlines every 20%.

The graph below shows one cyclin’s concentration and the activity of its cyclin-dependent kinase (CDK) through one 20-hour cycle of a cultured cell line.

Which statement do the two curves support?

A. The cyclin is the kinase, and the CDK is the protein that the cyclin phosphorylates to begin mitosis B. The CDK's activity climbs in G2 because the cell makes more CDK then and destroys it after mitosis C. CDK activity rises and falls with the cyclin because the CDK is active only while a cyclin is bound D. The cyclin climbs through G2 because the active CDK phosphorylates it, and the phosphate keeps it stable

Question 8

Kinase activity of extracts made from cells in G1, with proteins added or removed.

A researcher measures the activity of a cyclin-dependent kinase (CDK) in extracts made from cells in G1, adding or removing proteins as the table below shows.

Which explanation accounts for the results?

A. The cyclin is itself the kinase, and the CDK plays no part in the activity that was measured B. The added CDK was destroyed as soon as it entered the extract, so it could add nothing to the activity C. The cyclin supplied the phosphate groups that the CDK transfers onto its target proteins D. The CDK is active only while a cyclin is bound to it, so the cyclin was the missing partner

Question 9

At the end of mitosis a cell destroys the cyclin that binds its cyclin-dependent kinase (CDK). Within minutes the proteins that started mitosis switch off, and nuclear envelopes re-form.

Which of the following best explains why destroying the cyclin switches those proteins off?

A. The CDK loses its cyclin and stops adding phosphate groups, so a phosphatase returns the proteins to their off shape B. The cyclin was the kinase that kept the proteins phosphorylated, so its loss ends the phosphorylation C. Destroying the cyclin destroys the CDK bound to it, so the cell must make new CDK before it can move on D. The fragments of the destroyed cyclin bind the mitosis proteins and hold them switched off

Question 10

A cultured cell waiting at the G1 checkpoint: the state of each G1 condition and of the cyclin that drives entry into S phase.

A cultured cell waits at the G1 checkpoint. The table below describes the cell, including the concentration of the cyclin that drives entry into S phase, the partner of the cell’s cyclin-dependent kinase (CDK).

Which of the following best explains the wait?

A. The cyclin has to reach 100% of its peak before a single cyclin–CDK complex can form in the cell B. A small cell holds too little CDK to pair with the cyclin it has made, so no complex forms C. A signal from the unmet size condition keeps the cyclin–CDK complex switched off, so the S-phase proteins stay off D. The bound growth factor is a signal to stop, and it holds the cell in G1 until the cell has grown

Question 11

In a tumor, the cell in which the gene for a checkpoint protein first mutated died years ago. The tumor has gone on growing ever since.

Why does the tumor keep growing after that first cell has died?

A. The mutation spread from the first cell to its neighbors through the fluid around them B. The first cell's daughters inherited the mutated gene, so they lack the same hold and pass it on to their daughters C. The tumor's cells divide faster than normal cells, so the loss of one cell makes no difference D. The dying cell released a growth factor that has kept its neighbors dividing ever since

Question 12

A line of cancer cells passes the G1 checkpoint and copies its DNA even when the DNA carries breaks. With no growth factor added, however, the cells stay in G1, as normal cells do.

In which gene is the line's mutation most likely to be?

A. The gene for the growth-factor receptor, locking the receptor in its active shape B. The gene for the cyclin that drives entry into mitosis, making the cyclin impossible to destroy C. The gene for the phosphatase that removes phosphate groups from the proteins that start S phase D. The gene for a checkpoint protein that detects damaged DNA at the G1 checkpoint

Question 13

Bone marrow cells from two strains of mouse, given the same DNA-damaging drug in a dish and followed for three days.

Researchers gave bone marrow cells from two strains of mouse the same DNA-damaging drug in a dish and followed the cells for three days. In these cells a checkpoint hold on damaged DNA lasts about a day; a held cell that has not been removed by then resumes the cycle. The table below gives the results.

Which control has strain 2 lost?

A. The enzymes that repair damaged DNA B. The requirement for a growth signal before the cycle starts C. The response that makes a cell with damage beyond repair remove itself D. The G1 checkpoint's hold on damaged DNA

Question 14

Percent of cultured hamster cells in each stage before a drug was added and 18 hours after, each from one count of 200 cells. Gridlines every 20%.

Cultured hamster cells were sorted by stage before and 18 hours after a drug was added; the graph below shows the percent in G1, S phase, G2 and mitosis. The cells had nutrients and a growth factor throughout, their DNA was undamaged, and the cells in G1 had reached full size.

Which action of the drug accounts for the change?

A. The drug keeps the complex that drives entry into S phase switched off B. The drug removes the growth signal C. The drug damages DNA D. The drug stops the destruction of the cyclin that drives entry into S phase

Question 15

Cells in one dish receive a spindle inhibitor, a drug that stops the spindle fibers from moving the chromosomes; these cells are held at metaphase with their sister chromatids joined. Cells in a second dish receive the spindle inhibitor together with a drug that removes the M checkpoint’s hold.

Which of the following best predicts what the cells in the second dish do?

A. The cells stay at metaphase B. The cells leave metaphase with their chromosomes unattached C. The cells drop back into G2 and build a new spindle D. The cells leave the cycle into G0

Question 16

Cancer cell line A enters the cycle with no growth factor present, because a relay protein in its growth-signal pathway is always active; its checkpoints work. Cancer cell line B waits for the growth factor as normal cells do, but passes the G1 checkpoint carrying damaged DNA. A drug blocks line A's always-active relay protein.

Predict the effect of the drug on each line's division count in a dish with no growth factor added.

A. Both lines' counts fall to normal, because both lines depend on the relay protein B. Line B's count falls to normal; line A's stays as it was C. Both counts stay as they were, because a drug cannot restore a lost control D. Line A's count falls to normal; line B's stays as it was

Question 17

Three lines of mouse cells: division events per 100 cells in 36 hours in a dish with no growth factor added; the percent of X-ray-damaged cells that entered S phase within 6 hours; and the percent of damaged cells that had undergone apoptosis after 2 days.

Three lines of mouse cells were grown for 36 hours in a dish with no growth factor added. The table below gives their division counts, the percent of X-ray-damaged cells that entered S phase, and the percent of damaged cells that underwent apoptosis.

Which control has line 1 lost, and which has line 2 lost?

A. Line 1 has lost the G1 checkpoint's hold on damaged DNA; line 2 has lost the requirement for a growth signal B. Line 1 has lost the requirement for a growth signal; line 2 has lost the G1 checkpoint's hold on damaged DNA C. Line 1 has lost the enzymes that repair DNA; line 2 has lost the M checkpoint's hold at metaphase D. Line 1 has lost the apoptosis response only, with its checkpoints intact; line 2 has lost the destruction of cyclin

Question 18

Skin cells at the edge of a fresh cut begin dividing within a day. A student says: “The cells divide because the skin needs new cells to close the gap.”

Is the student correct, and why?

A. Yes — each edge cell senses the empty space beside it and divides to fill the need B. Yes — the gap lets more nutrients reach the edge cells, and the extra food makes them divide C. No — growth factors released at the wound bind the edge cells’ receptors and switch division on D. No — the cut damaged the edge cells’ DNA, and a damaged cell divides to replace itself

How to tackle the free-response questions. Read the verb first: describe asks what you see or know; explain asks why or how, so name the mechanism; predict asks what will happen and why; justify asks for the evidence that supports a claim. Each point is earned by one idea, stated in a sentence that names the thing and the mechanism. Extra words earn nothing; a wrong extra can lose the point. If there is a figure or table, use what it shows. When you finish, check the box for each point your answer earns and compare your sentences with the full-credit answer.

Free response 1 · Scientific Investigation · 4 points

Researchers took bladder cells from a healthy bladder and from a bladder tumor of the same patient and grew them as two cell lines. They seeded each line into five dishes at low density (cells far apart) and five dishes at high density (cells touching neighbors on every side), all in the same medium with the growth factor, and the medium was replaced every 6 hours. After 24 hours the researchers counted the percent of cells in S phase in each dish. The graph below shows the mean of the five dishes for each condition; error bars represent ±2SE.

Mean percent of cells in S phase after 24 hours for bladder cells from a healthy bladder (normal) and from a bladder tumor of the same patient, each seeded at low and at high density; five dishes per condition. Error bars represent ±2SE. Gridlines every 5%.

(a) Identify the dependent variable in this investigation. (1 point)

A full-credit answer: The dependent variable is the percent of cells in S phase in each dish after 24 hours, which the researchers counted.

Check the box for each point your answer earns

the percent of cells in S phase after 24 hours (the quantity the researchers counted in each dish).

Accept "the percent of cells copying their DNA". Do not award the point for the seeding density or the cell line, which the researchers set, or for "the number of cells".

Common slip: Naming the density or the cell line. The density and the cell line were what the researchers set; the dependent variable is what the researchers counted at the end.

(b) Support the claim that the fall in the normal cells' S-phase percent at high density is caused by contact with neighbors rather than by a shortage of nutrients, using the description of the experiment. (1 point)

A full-credit answer: The medium was the same in every dish and was replaced every 6 hours.
So the high-density cells had the same nutrients and growth factor as the low-density cells.
The one thing that differed between the dishes was density.
At high density each cell was touched by neighbors on every side.
Contact on every side is a signal that holds a cell in G1.
Therefore contact, not a shortage of nutrients, cut the S-phase percent from 31% to 4%.

Check the box for each point your answer earns

the evidence (the medium, with its nutrients and growth factor, was the same in every dish and was replaced every 6 hours) AND the reasoning (so nutrients were equal at both densities, and the only difference was whether each cell was touched on every side, so the fall from 31% to 4% is credited to contact, a signal from neighbors that holds the cycle in G1).

Accept 'nutrients were kept the same, so the change in density is the only difference', with the link to contact stated. Do not award the point for 'crowded cells have less food', or for the medium fact with no link to the claim.

Common slip: Writing that crowded cells had too little food, or naming the medium with no link to the claim. The medium was the same in every dish; the change that came with density was contact with neighbors.

(c) Using the error bars, describe what the data show about the effect of density on the tumor cells, and identify the control on the cycle that these cells have lost. (1 point)

A full-credit answer: The tumor cells' error bars overlap: 30% to 36% at low density and 27% to 33% at high density.
Overlapping bars mean the data show no difference between the two densities.
So density has no measurable effect on the tumor cells' S-phase percent.
The tumor cells have lost the contact control: the signal from neighbors on every side that holds normal cells in G1.

Check the box for each point your answer earns

both parts together: the tumor cells' bars at low density (30% to 36%) and at high density (27% to 33%) overlap, so the data show no difference in S-phase percent between the two densities; the tumor cells have lost the contact signal from neighbors that holds normal cells in G1 when they are touched on every side.

Accept "the bars overlap, so no difference is shown; they ignore contact". Do not award the point for "the tumor cells divide faster", or for a conclusion drawn from the two means alone (33% against 30%).

Common slip: Reading the means alone (33% against 30%) as a small drop. The bars overlap, so the data have shown no difference; and the lost control is the contact signal, not the speed of the cycle.

(d) Tests on the tumor cells show that they enter the cycle only when a growth factor is bound to their receptors. The researchers plan to repeat the high-density tumor dishes with the growth factor left out of the medium. Predict the percent of tumor cells in S phase after 24 hours, and justify your prediction. (1 point)

A full-credit answer: About 4%: few tumor cells will be in S phase.
The tumor cells enter the cycle only when a growth factor is bound to their receptors.
The control the tumor cells lost is the contact hold, not the growth-signal requirement.
So with the growth factor left out, the tumor cells wait in G1 or drop into G0, like normal cells with no signal.

Check the box for each point your answer earns

a low value, near the normal cells' high-density value (about 4%; accept 0% to 10%), with the reason that the tumor cells enter the cycle only when a growth factor is bound; the tumor cells lost the contact hold but still need the growth signal, so with the growth factor absent they wait in G1 or drop into G0.

Do not award the point for "about 30%, because cancer cells divide whatever the conditions", or for a prediction with no reason.

Common slip: Predicting about 30%, as if a cancer cell divides whatever the conditions. This line lost one control, contact; the task says the tumor cells still need the growth signal.

Free-response score: 0 of 4

Free response 2 · Analyze Model · 4 points

The graph below is a model of one 24-hour cycle in a cultured cell line. The graph shows the concentrations of two cyclins, cyclin X and cyclin Y, each as a percent of its own peak, and the amount of cyclin-dependent kinase (CDK) as a percent of its maximum. In this line, cyclin X bound to a CDK drives entry into S phase, and cyclin Y bound to a CDK drives entry into mitosis. The stages are marked along the top of the graph.

A model of one 24-hour cycle in a cultured cell line: the concentrations of cyclin X (solid) and cyclin Y (dashed), each as a percent of its own peak, and the amount of CDK (dotted) as a percent of its maximum. Gridlines every 20%.

(a) Describe the pattern of cyclin Y's concentration through the cycle, using times from the graph. (1 point)

A full-credit answer: Cyclin Y stays low, near 6% of its peak, through G1 and S phase, from 0 to about 17 h.
Cyclin Y rises through G2, from about 17 h, to its peak at 23 h as mitosis begins, and it collapses back to its low level by 24 h, the end of mitosis.

Check the box for each point your answer earns

cyclin Y stays low through G1 and S phase (about 0 to 17 h), rises through G2 (from about 17 h to 23 h), peaks at about 23 h as mitosis begins, and falls back to its low level by the end of mitosis (24 h).

Accept times within about an hour of these. Do not award the point for a description of cyclin X or of the CDK line, or for "it goes up and down" with no times.

Common slip: Describing cyclin X by mistake, or giving the shape with no times. The task asks for the pattern of cyclin Y with times read from the graph.

(b) Explain how the rise in cyclin Y moves the cell into mitosis, although the amount of CDK stays flat. (1 point)

A full-credit answer: The CDK is present all the time, but the CDK is active only while a cyclin is bound to it.
As cyclin Y rises through G2, more cyclin Y–CDK complexes form.
Each active complex transfers a phosphate group from ATP onto the proteins that start mitosis.
The added phosphate changes those proteins' shape and switches them on, so mitosis begins.
The amount of CDK stays flat; what rises is the number of active complexes.

Check the box for each point your answer earns

the CDK is active only while a cyclin is bound to it, so as cyclin Y rises more active cyclin Y–CDK complexes form; the active complex transfers a phosphate group from ATP onto the proteins that start mitosis, the added phosphate changes their shape and switches them on, so mitosis begins; the amount of CDK stays the same, and it is the number of active complexes that rises.

Do not award the point for "cyclin Y is the kinase", for "more CDK is made", or for "the complex starts mitosis" with no phosphate transfer or shape change.

Common slip: Writing that cyclin Y switches on the mitosis proteins itself. Cyclin Y has no kinase activity; it binds the CDK, and the CDK transfers the phosphate groups.

(c) A mutation removes cyclin X from these cells. Make a claim about where the mutant cells collect and what happens to their division. (1 point)

A full-credit answer: The mutant cells collect in G1, at the G1 checkpoint, or drop into G0.
The mutant cells never enter S phase, so division stops.

Check the box for each point your answer earns

the claim: the cells collect in G1 (at the G1 checkpoint, or drop into G0) and division stops. No reasoning is required for this point.

Accept 'they stay in G1 and stop dividing'. Do not award the point for 'they collect in G2' or 'in mitosis', or for 'they divide with no DNA copied'.

Common slip: Predicting a pile-up in G2 or in mitosis. Cyclin X bound to a CDK drives entry into S phase. Without cyclin X, the cells never leave G1, so the pile-up is at the very first step of the cycle, in G1.

(d) Support your claim in (c) using the model. (1 point)

A full-credit answer: On the model, cyclin X rises at the end of G1, and cyclin X bound to a CDK drives entry into S phase.
With no cyclin X, no cyclin X–CDK complex forms.
So the proteins that start S phase are never phosphorylated, and they stay off.
The cells never copy their DNA, so they never reach G2.
Cyclin Y's rise, which the model places in G2, therefore never comes.
So mitosis is never entered.

Check the box for each point your answer earns

the evidence from the model (cyclin X rises at the end of G1, and cyclin X bound to a CDK drives entry into S phase) AND the reasoning (with no cyclin X, no active cyclin X–CDK complex forms, so the proteins that start S phase are never switched on, so the cells never copy their DNA, never reach G2 and never enter mitosis).

Accept 'no cyclin X, no active complex, no entry into S phase', tied to the model. Do not award the point for 'the CDK works on its own', or for a justification built on cyclin Y alone.

Common slip: Arguing from cyclin Y or from the CDK line alone. The CDK is present, but with no cyclin X bound it stays inactive; cyclin Y rises only in cells that have reached G2.

Free-response score: 0 of 4

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