Practice questions · Topic 4.6

Unit 4 · Practice for the Topic 4.6 end-of-topic test

You’ve gone through everything in this topic. The summary video below recaps it all, so you’re ready for the questions.

Watch first: Topic 4.6 summary, part 1: switched on, held, moved on

Video coming soon

Division is switched on by a growth signal, held at three checkpoints until each condition is met, and moved on by cyclin–CDK complexes.

Watch next: Topic 4.6 summary, part 2: when a control is lost

Video coming soon

A mutation removes a control and the daughter cells inherit the fault; cancer is division without the controls, not faster division; apoptosis removes a damaged cell before it divides.

These are practice questions in the shape of the topic test. Work through them before you take the test; every question tells you what it wanted.

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

Question 1

A cultured cell has reached full size, has nutrients around it and carries undamaged DNA. The cell's receptors for the growth factor are empty; the dish was set up with no growth factor added.

What does the cell do at the G1 checkpoint, and why?

A. The cell passes, because its size, nutrients and DNA are all in order B. The cell waits in G1, or drops into G0, because the growth signal is missing C. The cell passes, because a growth factor is only needed after S phase D. The cell undergoes apoptosis, because a cell with no signal removes itself

Question 2

In a culture of dividing cells, the time each cell spent at metaphase was measured. Most cells spent 4 to 8 minutes there. A few spent more than 2 hours, and each of those had one chromosome attached to spindle fibers from one pole only.

Which statement about the M checkpoint do these times support?

A. The M checkpoint is a timer set to about six minutes, and damage makes the timer last longer B. The M checkpoint holds only cells whose DNA is damaged, and the slow cells had damaged DNA C. The M checkpoint attaches the last chromosome itself, and doing so takes about two hours D. The M checkpoint is a condition: each cell waits until its last chromosome is attached from both poles

Question 3

Four events in dividing tissues are described.

Which of the following describes a checkpoint holding a cell?

A. A cell whose copied DNA carries a break waits at the end of G2 until the break is repaired B. A liver cell enters the cycle when a growth factor from a damaged neighbor binds its receptor C. Rising cyclin forms active complexes that switch on the proteins that start S phase D. A cell that has grown to twice its starting size divides because it is now big enough

Question 4

Amounts of one cyclin (darker bars) and of its CDK (paler bars) in fish embryo cells sorted by stage, each as a percent of that protein's highest value. Gridlines every 20%.

Cells of a fish embryo were sorted by stage, and the amounts of one cyclin and of its cyclin-dependent kinase (CDK) were measured in each group; the graph below gives each as a percent of that protein’s highest value.

Which of the following do the measurements show?

A. Both proteins rise through G2 and fall at the end of mitosis B. The CDK's amount rises and falls, and the cyclin's stays steady C. The cyclin's amount rises and falls, and the CDK's stays steady D. The two proteins are present in equal amounts at every stage

Question 5

Cultured mouse cells short of one amino acid: measurements just before a drug that blocks the checkpoint protein detecting the shortage, and 15 minutes after it.

Cultured mouse cells short of one amino acid wait in G1. After an hour, a researcher adds a drug that blocks the checkpoint protein that detects the shortage. The table below gives measurements just before the drug and 15 minutes after it.

Which statement explains the rise in the activity of the cyclin-dependent kinase (CDK)?

A. The cyclin passed the concentration that switches the CDK on only after the drug was added B. The shortage signal had been keeping the cyclin–CDK complex switched off, and the drug removed that signal C. The drug supplied the missing amino acid, so the G1 checkpoint's condition was met and the hold ended D. The drug made the cells build more CDK, and the extra CDK bound the cyclin that was waiting

Question 6

Here are four cells from different tissues.

Which cell is a cancer cell?

A. A cell in an embryo that completes a cycle every 12 hours while a growth factor is present B. A liver cell that has rested in G0 for three years with undamaged DNA C. A skin cell at the edge of a cut that divides for a week and then stops as the cut closes D. A cell that passes the G1 checkpoint carrying damaged DNA, and whose daughters do the same

Question 7

Cells from a colon tumor have lost the apoptosis response: a held cell whose damage cannot be repaired no longer removes itself. Researchers give these tumor cells a DNA-damaging drug in a dish; the damage cannot be repaired. In these cells a checkpoint hold on damaged DNA lasts about a day.

Which of the following best predicts what most of the tumor cells have done by day 3?

A. They have divided, and each daughter cell carries the damage B. They have removed themselves, so the count has fallen C. They are still held at the checkpoint with their damage D. They have divided, and the daughter cells are free of the damage

Question 8

Percent of cultured mouse cells in each stage before a drug was added and 24 hours after, each from one count of 200 cells.

The table below gives the percent of cultured mouse cells in each stage before and 24 hours after a researcher added a drug. The treated cells in G2 held twice the DNA of a G1 cell, their DNA was undamaged, and a growth factor was bound to their receptors.

Which action of the drug fits the counts?

A. The drug blocks the spindle from attaching to the chromosomes B. The drug damages DNA C. The drug stops the cell making the cyclin that drives entry into mitosis D. The drug speeds cells through S phase

Question 9

Normal cells from a mouse tissue show 20 division events per 100 cells in 48 hours in a dish with no growth factor added. Cells with an always-active relay protein in the growth-signal pathway show 56; cells with a useless checkpoint protein show 44; cells carrying both faults show 130.

Which of the following do the counts show about the two faults together?

A. The faults cancel in part: the double mutant’s count is below the count for the always-active relay protein alone B. The faults add: the double mutant’s count equals the normal count plus the two single rises C. The faults do more than add: the double mutant’s count is above the normal count plus the two single rises D. The second fault makes no difference: the double mutant’s count equals the count for the always-active relay protein alone

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 · 6 points

A drug used to treat cancer damages DNA. To see which cells it removes, researchers grew three kinds of human cell in dishes: cells from a tumor, cells from the lining of the gut (a tissue that replaces itself every few days), and muscle cells, 96% of which were resting in G0. Five dishes of each kind received the same dose of the drug, and 6 hours later all three kinds carried the same amount of DNA damage per cell. After 3 days the researchers counted the percent of cells in each dish that had undergone apoptosis. The graph below shows the mean of the five dishes for each kind of cell; error bars represent ±2SE. In these cells a checkpoint hold on damaged DNA lasts about a day; a held cell that has not been removed by then resumes the cycle.

Mean percent of cells that had undergone apoptosis 3 days after the same dose of a DNA-damaging drug, for three kinds of human cell; five dishes per kind. Error bars represent ±2SE. Gridlines every 10%.

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

Hint: Which thing did the researchers deliberately make different between the dishes, and which did they measure at the end?

A full-credit answer: The independent variable is the kind of cell in the dish: tumor, gut lining or muscle.

Check the box for each point your answer earns

the kind of cell (tumor, gut lining or muscle).

Do not award the point for the drug dose (the same in every dish) or for the percent of cells undergoing apoptosis (the dependent variable).

Common slip: Naming the drug as the independent variable. Every dish received the same dose; what differed between dishes was the kind of cell.

(b) Using the error bars, describe how the tumor cells compare with the gut lining cells. (1 point)

Hint: The caption says what each bar represents; check whether the two bars share any part of their range before you decide what the data show.

A full-credit answer: The tumor bar runs from 57% to 69% and the gut lining bar from 40% to 50%, so the bars are clear of each other, and the data show that a larger percent of tumor cells than gut lining cells underwent apoptosis.

Check the box for each point your answer earns

the tumor bar (57% to 69%) and the gut lining bar (40% to 50%) are clear of each other, so the data show a real difference: a larger percent of tumor cells underwent apoptosis.

Accept the two ranges with "do not overlap, so the difference is real". Do not award the point for a comparison of the means alone (63% against 45%) with no use of the bars.

Common slip: Comparing 63% with 45% and stopping. The point is earned by using the bars: because they are clear of each other, the difference is more than chance.

(c) Explain why most of the muscle cells survived the drug. (1 point)

Hint: Where in the cycle were the muscle cells at the start, and where do the three checkpoints sit?

A full-credit answer: Most muscle cells were in G0, outside the cycle.
A cell in G0 is not copying its DNA and never approaches a checkpoint.
So the damage in the muscle cells was never detected at a checkpoint.
A cell whose damage is never detected is neither held nor signaled to undergo apoptosis.
So the muscle cells survived.

Check the box for each point your answer earns

the muscle cells were resting in G0, outside the cycle, so they were neither copying DNA nor approaching any checkpoint; their damage was never detected at a checkpoint, so they were neither held nor signaled to undergo apoptosis, and they survived.

Accept "in G0, so they never reach a checkpoint". Do not award the point for "the drug did not enter muscle cells" (the damage was the same) or for "muscle cells repair DNA faster".

Common slip: Writing that the drug damaged muscle cells less. The stimulus says all three kinds carried the same damage; the muscle cells survived because, resting in G0, they reached no checkpoint.

(d) The gut lining cells are normal cells. Explain why the drug removed so many of them. (1 point)

Hint: How often do gut lining cells pass through the checkpoints, and what does a checkpoint do with a damaged cell whose repair fails?

A full-credit answer: Gut lining cells divide constantly, because the lining replaces itself every few days.
So most gut lining cells were in the cycle.
A dividing cell reaches the G1 or G2 checkpoint, and these cells reached it carrying damaged DNA.
The checkpoint held the damaged cells.
The cells whose damage could not be repaired were signaled to undergo apoptosis.
Those cells dismantled themselves, so the damage was never copied.

Check the box for each point your answer earns

gut lining cells divide constantly (the tissue replaces itself every few days), so most were in the cycle and reached a checkpoint (G1 or G2) with damaged DNA; the checkpoint held them, and cells whose damage could not be repaired were signaled to undergo apoptosis.

Do not award the point for "the drug targets gut cells" or for "they divided too fast" with no checkpoint and no apoptosis signal.

Common slip: Stopping at "they divide a lot". The point needs the chain: dividing cells reach a checkpoint, the damaged cell is held, and damage beyond repair brings the apoptosis signal.

(e) The researchers propose adding a second drug that blocks the apoptosis response. Make a claim about how the tumor cells’ bar would change. (1 point)

Hint: The bar counts cells undergoing apoptosis. Which way does a drug that blocks that response move the count?

A full-credit answer: The percent of tumor cells undergoing apoptosis would fall sharply.
It would fall toward the few percent seen in muscle cells.

Check the box for each point your answer earns

the claim: the percent of tumor cells undergoing apoptosis would fall sharply (toward the muscle cells' value). No reasoning is required for this point.

Do not award the point for 'it would rise' or 'it would stay the same'.

Common slip: Claiming that more cells die, as if two drugs must do more than one. The second drug blocks the very response that was killing the damaged cells.

(f) Support your claim in (e) using the chain from DNA damage to a dismantled cell. (1 point)

Hint: Which step in the chain from DNA damage to a dismantled cell does the second drug remove? Follow what happens to a damaged cell after that.

A full-credit answer: The second drug removes the last step of the chain from damage to a dismantled cell.
Damage is still detected at a checkpoint, so the damaged cells are still held.
But the detection no longer ends in the cell dismantling itself.
So the damaged tumor cells stay held, or divide with the damage still in them.
So fewer tumor cells undergo apoptosis, and the bar falls.

Check the box for each point your answer earns

the evidence (the second drug blocks the apoptosis response, the last step of the chain damage → hold at a checkpoint → apoptosis) AND the reasoning (damage is still detected and the cells are still held, but the detection no longer ends in the cell dismantling itself, so the damaged tumor cells stay held or divide carrying the damage instead of being removed, so fewer undergo apoptosis).

Do not award the point for 'the drug blocks apoptosis' with no link to what the damaged cells do instead, or for reasoning that never names the step removed.

Common slip: Naming the blocked step and stopping. Supporting the claim needs the link: what the damaged cells do instead of dismantling themselves, and what that does to the count.

Free-response score: 0 of 6

Free response 2 · Scientific Investigation · 4 points

A line of human cells carries a mutation in the gene for the cyclin-dependent kinase (CDK) that drives entry into mitosis: the altered CDK holds its active shape with or without a cyclin bound. The researchers gave normal cells and mutant cells a drug that stops DNA copying halfway through S phase. Six hours later the researchers counted the percent of cells that had entered mitosis, and 24 hours later the percent of daughter cells with pieces of chromosomes missing. The table below gives the results.

Normal cells and cells with an always-active CDK, after a drug stopped DNA copying halfway through S phase: the percent that entered mitosis within 6 hours, and the percent of daughter cells with pieces of chromosomes missing after 24 hours.

(a) Identify the null hypothesis for the effect of the mutation on entry into mitosis. (1 point)

A full-credit answer: The null hypothesis is that the mutation makes no difference to the percent of cells entering mitosis after DNA copying is stopped: normal and always-active-CDK cells would enter mitosis at the same rate.

Check the box for each point your answer earns

the mutation makes no difference to the percent of cells that enter mitosis after DNA copying is stopped (normal and mutant cells enter mitosis at the same rate).

Do not award the point for a prediction of a difference in either direction ("mutant cells enter mitosis more often").

Common slip: Writing the expected result ("the mutant cells enter mitosis anyway") as the null hypothesis. The null hypothesis is the statement of no effect, which the data can then reject.

(b) Explain why the normal cells stayed out of mitosis. (1 point)

A full-credit answer: The G2 checkpoint checks that the DNA is completely copied before mitosis.
With copying stopped halfway, that condition was unmet.
So a signal from the checkpoint kept the complex that drives entry into mitosis switched off.
An inactive complex transfers no phosphate groups, so the proteins that start mitosis were never switched on.
Therefore the normal cells were held before mitosis, and only 2% entered it.

Check the box for each point your answer earns

the G2 checkpoint checks that the DNA is completely copied; with copying stopped halfway the condition was unmet, so a signal kept the cyclin–CDK complex switched off, the proteins that start mitosis were never switched on, and the cells were held before mitosis (only 2% entered).

Accept "held at the G2 checkpoint because the DNA was incomplete" with the complex kept off. Do not award the point for "they had no cyclin" or for "the drug stopped mitosis directly".

Common slip: Writing that the drug itself blocked mitosis. The drug blocked copying; it was the cell's own G2 checkpoint that held the complex off and kept the cell out of mitosis.

(c) Explain how the always-active CDK let the mutant cells into mitosis with their DNA half copied. (1 point)

A full-credit answer: The altered CDK holds its active shape whatever signals reach it.
So the G2 checkpoint's signal has nothing to switch off.
The altered CDK transferred phosphate groups from ATP onto the proteins that start mitosis.
Those proteins switched on although the DNA was only half copied.
So 44% of the mutant cells entered mitosis, and 61% of their daughters received a genome with pieces missing.

Check the box for each point your answer earns

the altered CDK is active with or without a cyclin bound and the checkpoint's signal cannot switch it off, so it transfers phosphate groups from ATP onto the proteins that start mitosis whatever the state of the DNA; mitosis begins with the DNA half copied, and the daughters receive incomplete genomes (61% with pieces missing).

Do not award the point for "the mutant cells have no checkpoint" (the checkpoint's signal is sent; the CDK ignores it) or for "the CDK copied the DNA faster".

Common slip: Writing that the mutant cells lack a G2 checkpoint. The checkpoint protein still sends its signal; the fault is that an always-active kinase takes no notice of it.

(d) A second drug blocks the kinase activity of this CDK. Determine the effect of adding it to the mutant cells that received the copying blocker, and justify your decision. (1 point)

A full-credit answer: Far fewer mutant cells would enter mitosis, close to the normal 2%, and far fewer daughters would have pieces missing.
With its kinase activity blocked, the CDK can transfer no phosphate groups onto the proteins that start mitosis.
So those proteins stay off, whatever shape the CDK holds.
Therefore mitosis is never switched on.

Check the box for each point your answer earns

the decision (far fewer mutant cells enter mitosis, near the normal 2%, and far fewer daughters have pieces missing) AND the reasoning it rests on (with the kinase activity blocked, the proteins that start mitosis are never phosphorylated, so mitosis is never switched on, whatever shape the CDK holds).

Accept 'the mutant cells behave like normal cells: they stay out of mitosis', with the reasoning. Do not award the point for the decision alone, or for 'the cells enter mitosis anyway because the CDK is always active'.

Common slip: Deciding that the cells still enter mitosis because the CDK is always active. Being in the active shape only matters if the kinase can transfer phosphates; the second drug stops that.

Free-response score: 0 of 4

Multiple choice checked: 0 of 9 correct.

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