3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
Which statement best describes the role of checkpoints in the cell cycle?
Answer and reasoning
AThey are points at which a cell moves on automatically once it has grown to a certain size. A student who thinks size alone decides when a cell divides picks this. Size is one condition checked, but checkpoints also check DNA damage, DNA replication and chromosome attachment, and they can halt the cycle.
BThey are points at which the cell's progress is held until conditions for the next stage are met.Correct Checkpoints are internal controls: at each one, conditions such as DNA damage, the completion of DNA replication or the attachment of chromosomes to the spindle are checked, and the cycle is halted until they are met.
CThey are points at which any cell with a problem is destroyed at once, instead of being held there. A student who thinks stopping at a checkpoint means death picks this. A checkpoint halts the cycle while a problem is corrected; only a cell whose problem cannot be corrected may undergo apoptosis.
DThey are points in mitosis at which the cell's DNA is replicated, so each daughter cell gets a copy. A student who thinks DNA is replicated during mitosis picks this. DNA is replicated in S phase, before mitosis; checkpoints control whether the cell moves from one stage to the next.
Which statement best explains how cyclins and cyclin-dependent kinases (CDKs) control progression through the cell cycle?
Answer and reasoning
ACDK levels rise and fall during the cycle, while cyclin, present at a constant level, switches each CDK on. A student who swaps the roles of the two proteins picks this. It is cyclin concentration that rises and falls; CDK concentration stays fairly constant, and CDK activity changes because cyclin binds to it.
BCyclins are the kinases that phosphorylate target proteins, and CDKs carry the cyclins to these proteins. A student who thinks the cyclin is the enzyme picks this. The CDK, the cyclin-dependent kinase, phosphorylates target proteins; the cyclin is the regulatory protein that the CDK must bind to be active.
CA CDK starts the next stage once the cell has grown large enough, whether or not a cyclin is bound. A student who thinks size alone controls division picks this. A CDK is active only when bound to a cyclin; cell size is one of several conditions checked at checkpoints, not a switch for the CDK.
DEach CDK is switched on only while bound to its cyclin, then adds phosphate groups to target proteins.Correct A CDK is a kinase that is active only when bound to its cyclin. Cyclin concentration rises during the cycle; when enough cyclin–CDK complex has formed, the CDK phosphorylates target proteins, and the changes in these proteins move the cell past a checkpoint into the next stage.
In a hypothetical mammal, protein Q halts the cell cycle at the G1 checkpoint when a cell's DNA is damaged and triggers apoptosis if the damage cannot be repaired. A mutation in a skin cell inactivates protein Q. Which statement best explains how this mutation can lead to cancer?
Answer and reasoning
AThe cell then divides faster at every stage, and fast division is what makes a cell cancerous. A student who thinks cancer is defined by fast division picks this. Losing protein Q removes a control on division; it does not speed up every stage, and what marks cancer cells is division without the normal controls.
BThis single mutation is enough, on its own, to turn the cell into a cancer cell. A student who thinks one mutation causes cancer picks this. Cancer usually requires mutations in several genes; loss of Q makes further mutations more likely to persist.
CThe damaged cell divides more often because it needs to replace the cells that died. A student who explains division by what a cell needs picks this. Division is controlled by checkpoints and signals; losing Q removes a checkpoint control, and the cell does not divide to meet a need.
DCells with damaged DNA keep dividing, and further mutations build up in their descendants.Correct Without protein Q, cells with damaged DNA are neither halted for repair nor removed by apoptosis. They keep dividing, so damaged DNA is copied and further mutations accumulate in the cell line; cancer usually develops when mutations in several genes remove the controls on division.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
4.6.A.1 Cell cycle checkpoint Fix
Cell cycle checkpoint
An internal control point in the cell cycle at which the cell's progress can be halted until conditions for the next stage are met, for example until damaged DNA is repaired or until every chromosome is attached to the spindle.
G1 checkpoint
A checkpoint near the end of G1 at which a cell's size, nutrients, signals from other cells and DNA damage are checked before DNA replication begins. A cell that does not pass may remain in G1 or enter G0.
G2 checkpoint
A checkpoint at the end of G2 at which the completion of DNA replication and the condition of the DNA are checked before mitosis begins.
M checkpoint
A checkpoint during mitosis that delays anaphase until every chromosome is attached to spindle fibers from both poles, so that sister chromatids separate correctly.
Chi-square test
A statistical test that compares observed counts with the counts expected under a null hypothesis: χ² = Σ(o − e)²/e, with degrees of freedom equal to the number of categories minus one. If χ² exceeds the critical value at p = 0.05, the null hypothesis is rejected.
Students often think A cell divides automatically once it has grown large enough; size alone decides when it passes from one stage to the next. In fact No. Size is one of the conditions checked, but progression through the cell cycle is controlled at several checkpoints, which also check signals from other cells, DNA damage, DNA replication and the attachment of chromosomes to the spindle.
Students often think A cell that cannot pass a checkpoint is destroyed at once, so stopping at a checkpoint means death. In fact No. A checkpoint halts the cycle until conditions are met; the cell remains alive, for example while damaged DNA is repaired. Only if the problem cannot be corrected may the cell undergo apoptosis.
4.6.A.2 Cyclin Fix
Cyclin
A regulatory protein whose concentration rises and falls in a regular pattern during the cell cycle; it is made and then broken down in each cycle.
Cyclin-dependent kinase (CDK)
A protein kinase whose concentration stays fairly constant through the cell cycle but which is active only when bound to a cyclin. An active CDK phosphorylates target proteins, changing their activity.
Cyclin–CDK complex
A cyclin bound to its CDK. The complex is the active kinase: it phosphorylates proteins that move the cell past a checkpoint, and its activity falls when the cyclin is broken down.
Students often think The CDK's concentration rises and falls during the cell cycle, while the cyclin is present at a constant level. In fact The cyclin. Its concentration rises and falls in each cycle, while the concentration of its CDK stays fairly constant; the CDK's activity rises and falls because it depends on binding the cyclin.
Students often think A CDK is active whenever it is present, so CDK activity stays the same throughout the cell cycle. In fact No. A CDK is active only when it is bound to a cyclin. Because the cyclin's concentration rises and falls, CDK activity rises and falls even though the amount of CDK stays nearly constant.
4.6.B.1 Cancer Fix
Cancer
A disease in which cells divide without the normal controls on the cell cycle, usually as a result of mutations in several genes for proteins that regulate the cycle, such as checkpoint proteins.
Apoptosis
Programmed cell death: an orderly, regulated process in which a cell dismantles itself. A cell whose DNA damage cannot be repaired may be removed by apoptosis instead of dividing.
Students often think A single mutation in a cell-cycle gene is enough on its own to turn a normal cell into a cancer cell. In fact Usually not. Cancer usually develops when a cell line accumulates mutations in several genes that control the cell cycle, such as genes for checkpoint proteins and for proteins that promote division.
Students often think Cancer is caused by cells dividing faster than normal; fast division is what makes a cell cancerous. In fact No. What distinguishes cancer cells is that they divide without the normal controls, ignoring checkpoints and signals that would stop them. Many normal cells, such as some cells in the skin and bone marrow, divide rapidly, and some cancer cells divide slowly.
9 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 9
Two cultures of a hypothetical mammalian cell line were grown under the same conditions, except that one was treated at time 0 with a chemical that damages DNA. The graph shows the number of cells in each culture over the next 40 hours. Which statement is supported by the data?
Answer and reasoning
AThe number of treated cells fell, as cells that are stopped at a checkpoint die immediately. A student who thinks a cell stopped at a checkpoint dies picks this. The number of treated cells rose slightly, from 100 to 110 thousand; the cells were halted, not killed.
BBoth cultures grew at the same rate, since cells divide once they have grown large enough. A student who thinks size alone decides when a cell divides picks this. The control culture quadrupled while the treated culture barely grew; DNA damage halted division whatever the cells' size.
CThe number of treated cells stayed nearly constant, while the number of control cells kept rising.Correct The treated culture grew only from 100 to 110 thousand cells in 40 hours, while the control culture grew from 100 to 400 thousand. The treated cells survived but almost stopped dividing, as expected if checkpoints halt cells with damaged DNA.
DThe number of control cells rose by about the same amount in each 10-hour period. A student who expects a straight-line relationship picks this. The control culture gained 40, 60, 80 and then 120 thousand cells in successive 10-hour periods, far from the same amount each time, because each new cell can divide again.
In an untreated culture of a hypothetical cell line, 50% of cells are in G1, 30% in S, 15% in G2 and 5% in mitosis. A second culture was treated with a chemical that prevents spindle fibers from attaching to chromosomes, and 200 of its cells were classified by phase. The table shows the results. A student tests the null hypothesis that the treated culture has the same distribution of cells among the phases as the untreated culture. Which statement correctly gives the chi-square value and the conclusion at p = 0.05?
Answer and reasoning
Aχ² = 9.31, which is greater than 7.81, so reject the null hypothesisCorrect Expected numbers are 100, 60, 30 and 10. χ² = 0.49 + 0.42 + 0.30 + 8.10 = 9.31. With 4 − 1 = 3 degrees of freedom, the critical value at p = 0.05 is 7.81; 9.31 > 7.81, so the null hypothesis is rejected. The excess of cells in mitosis fits the M checkpoint holding cells whose chromosomes are not attached to spindle fibers.
Bχ² = 9.31, which is less than 9.49, so fail to reject the null hypothesis A student who takes the degrees of freedom to be the number of categories picks this. With four phases there are 3 degrees of freedom, not 4, so the critical value is 7.81, not 9.49.
Cχ² = 4.65, which is smaller than 7.81, so fail to reject the null hypothesis A student who uses percentages instead of counts picks this: (46.5 − 50)²/50 + (27.5 − 30)²/30 + (16.5 − 15)²/15 + (9.5 − 5)²/5 = 4.65. χ² must be calculated from the observed and expected numbers of cells.
Dχ² = 9.31, which is above 7.81, so fail to reject the null hypothesis A student who thinks a χ² above the critical value supports the null hypothesis picks this. χ² = 9.31 exceeds the critical value of 7.81 for 3 degrees of freedom, so the deviation from the expected numbers is unlikely to be due to chance and the null hypothesis is rejected.
Working Expected numbers for 200 cells under the null hypothesis: G1 0.50 × 200 = 100; S 0.30 × 200 = 60; G2 0.15 × 200 = 30; mitosis 0.05 × 200 = 10. χ² = (93 − 100)²/100 + (55 − 60)²/60 + (33 − 30)²/30 + (19 − 10)²/10 = 0.49 + 0.42 + 0.30 + 8.10 = 9.31. Degrees of freedom = 4 phases − 1 = 3; critical value at p = 0.05 = 7.81. Since 9.31 > 7.81, reject the null hypothesis: the treated culture's distribution differs, with more cells in mitosis, as expected if the M checkpoint holds cells whose chromosomes are not attached. Distractors: using 4 degrees of freedom gives a critical value of 9.49, and 9.31 < 9.49 leads to 'fail to reject'; using percentages (46.5, 27.5, 16.5, 9.5 observed vs 50, 30, 15, 5 expected) gives χ² = 0.245 + 0.208 + 0.150 + 4.050 = 4.65 < 7.81; reversing the decision rule gives fail to reject even though 9.31 > 7.81.
The diagram shows a dividing cell from an organism whose body cells each contain 4 chromosomes. In this cell, the M checkpoint is delaying the start of anaphase. Based on the diagram, which feature of the cell is the reason for the delay?
Answer and reasoning
AThe cell contains 8 chromosomes, twice the number in the organism's body cells. A student who counts each sister chromatid as a chromosome picks this. The cell has four centromeres, so it has 4 chromosomes, each made of two sister chromatids: the normal number for a cell in mitosis.
BThe DNA in the chromosomes is still being replicated, as replication happens during mitosis. A student who thinks DNA is replicated during mitosis picks this. DNA is replicated in S phase; each chromosome shown already consists of two sister chromatids, so replication is complete.
CThe sister chromatids are still joined, though they should separate during metaphase. A student who thinks sister chromatids separate in metaphase picks this. Sister chromatids stay joined until anaphase; joined chromatids are normal at this stage.
DThe cell has one chromosome attached to spindle fibers from only one pole.Correct The M checkpoint delays anaphase until every chromosome is attached to spindle fibers from both poles. Three chromosomes are attached to both poles and lined up in the middle, but the fourth is attached only to the left pole, so its sister chromatids could not be separated correctly.
The graph shows the concentration of a cyclin, the concentration of its cyclin-dependent kinase (CDK) and the activity of the CDK during two cell cycles in a hypothetical cell line. The phases of the cell cycle are shown above the graph. Which statement is supported by the data?
Answer and reasoning
ACDK activity is highest when cyclin concentration is highest, although CDK concentration does not change.Correct In each cycle, CDK activity and cyclin concentration both peak in M phase and then fall together, while CDK concentration stays at 60 throughout. CDK activity follows the cyclin, because the CDK is active only when bound to its cyclin.
BCDK concentration rises and falls in each cell cycle, while the concentration of the cyclin stays the same. A student who swaps the roles of the two proteins picks this. The graph shows the cyclin concentration rising and falling in each cycle, while the CDK concentration stays constant.
CCDK activity stays at one level throughout each cycle, as CDK molecules are present at all times. A student who thinks a CDK is active whenever it is present picks this. CDK activity rises to a peak in M phase and falls sharply, although the CDK is present throughout; activity depends on the cyclin.
DCDK concentration falls after each M phase, as the CDK is used up in phosphorylating proteins. A student who thinks enzymes are used up as they work picks this. CDK concentration stays at 60 throughout; it is the cyclin concentration that falls after M phase.
In a hypothetical cell line, a cyclin–CDK complex phosphorylates proteins that the cell needs to pass the G2 checkpoint and enter mitosis. The cells are treated with a chemical that binds to the CDK and blocks its kinase activity but does not affect the cyclin. Which prediction about the treated cells is best supported?
Answer and reasoning
AThey enter mitosis normally, as the cyclin itself phosphorylates the target proteins. A student who thinks the cyclin is the kinase picks this. The CDK is the enzyme that phosphorylates the target proteins; with its kinase activity blocked, the proteins are not phosphorylated, even though the cyclin is present.
BThey die at once, since a cell that fails to pass a checkpoint is destroyed immediately. A student who thinks a cell stopped at a checkpoint dies picks this. The cells are halted before mitosis; a halt at a checkpoint does not by itself kill a cell.
CThey stop before mitosis, and their cyclin still builds up, as the chemical does not act on it.Correct Without CDK activity, the target proteins are not phosphorylated, so the cells cannot pass the G2 checkpoint and are held before mitosis. The chemical does not affect the cyclin, so the cyclin is still made and builds up as it normally does.
DThey stop before mitosis, and their cyclin level stays constant, as only CDK levels cycle. A student who swaps the roles of cyclin and CDK picks this. It is the cyclin whose concentration rises and falls in each cycle; the chemical does not stop the cyclin from being made, so it still builds up.
The model shows how a cyclin and a cyclin-dependent kinase (CDK) control the passage of a cell through a checkpoint. A mutation makes the cyclin resistant to breakdown. Based on the model, which prediction about CDK activity after the cell passes the checkpoint, together with its explanation, is best supported?
Answer and reasoning
ACDK activity falls, as the CDK is broken down while the cyclin remains. A student who swaps the roles of cyclin and CDK picks this. The model shows the cyclin, not the CDK, being broken down after the checkpoint; with the cyclin resistant to breakdown, the CDK stays bound to it and active.
BCDK activity stays high, as the CDK stays bound to cyclin that is not broken down.Correct In the model, a CDK is active while bound to cyclin, and CDK activity normally falls after the checkpoint because the cyclin is broken down, leaving the CDK without cyclin and inactive. If the cyclin is not broken down, the CDK stays bound to it, so CDK activity stays high.
CCDK activity falls, as the CDK is used up in phosphorylating its targets. A student who thinks enzymes are used up as they work picks this. The CDK is a kinase, an enzyme, and is not consumed; in the model CDK activity falls only because the cyclin is broken down, which the mutation prevents.
DCDK activity stays high, as a CDK is active whenever it is present in a cell. A student who thinks a CDK is active whenever it is present picks this. The prediction is right but the explanation is not: the model shows that a CDK without cyclin is inactive, so CDK activity stays high only because the cyclin, which is not broken down, stays bound to it.
Protein Q normally halts the cell cycle when a cell's DNA is damaged. A researcher wants to test whether the loss of protein Q causes cells to keep dividing after their DNA is damaged by radiation. She has cultures of normal cells and of cells that lack protein Q. Which experimental design is most appropriate?
Answer and reasoning
ANormal cells given no radiation, and cells lacking Q that are given radiation A student who accepts a comparison in which two variables differ picks this. These groups differ both in protein Q and in radiation, so a difference could be caused by either.
BNormal cells and cells lacking Q, both exposed to the same dose of radiationCorrect The two groups differ only in whether they have protein Q; both have their DNA damaged in the same way. A difference in continued division can then be attributed to the loss of Q.
CCells lacking Q given radiation, with no second culture of cells needed A student who thinks a single group can show an effect picks this. Without irradiated normal cells for comparison, continued division could not be attributed to the loss of Q.
DCells lacking Q given radiation, and further cells lacking Q given no radiation A student who thinks the control is always the untreated group picks this. Both groups lack Q, so the comparison shows the effect of radiation, not the effect of losing Q.
A cell's DNA is severely damaged, and the damage cannot be repaired. The proteins that control the cell's checkpoints function normally. Which prediction about this cell is best supported?
Answer and reasoning
AIt becomes a cancer cell, as any damage to a cell's DNA leads to the development of cancer. A student who thinks any DNA damage causes cancer picks this. With normal checkpoints, a cell with irreparable damage is halted and may undergo apoptosis rather than dividing.
BIt may still divide on schedule, as cells divide once they have grown large enough. A student who thinks size alone decides when a cell divides picks this. Checkpoints check DNA damage as well as size, so a cell with damaged DNA is halted.
CIt is halted at a checkpoint and may undergo apoptosis, so the damage is not passed on.Correct Checkpoint proteins halt a cell with damaged DNA; if the damage cannot be repaired, the cell may undergo apoptosis. Either way, the cell does not divide and pass the damaged DNA to daughter cells.
DIt divides sooner than usual, as a damaged cell needs to be replaced by new ones. A student who thinks cells divide because they need to picks this. Division is controlled by checkpoints; with normal checkpoint proteins, a cell with damaged DNA is halted, not hurried through division.
A student claims that protein Q is needed for cells with damaged DNA to undergo apoptosis. Normal cells and cells lacking protein Q were either exposed to radiation that damages DNA or not. The graph shows the mean percentage of cells undergoing apoptosis 24 hours later (n = 4 cultures), with error bars representing ±2 SE of the mean. Which observation from the graph best supports the student's claim?
Answer and reasoning
ARadiation raised apoptosis significantly in normal cells but not in cells lacking Q.Correct In normal cells, the bars for no radiation (1.5–4.5) and radiation (24–32) do not overlap, so DNA damage greatly increased apoptosis. In cells lacking Q, the bars (0.5–3.5 and 3–7) overlap, so no significant increase is shown. Damaged cells underwent apoptosis only when Q was present, which supports the claim.
BUntreated cells lacking Q had a lower mean apoptosis than untreated normal cells. A student who treats any difference between means as real picks this. The bars for the two untreated groups overlap widely, so the 1-point difference could be due to chance; it also involves no DNA damage.
CIrradiated cells lacking Q had a low level of apoptosis, about 5% of all cells. A student who thinks a single group can show an effect picks this. Without comparing this value with irradiated normal cells, it cannot show that Q is needed for apoptosis.
DUntreated normal cells had less apoptosis than irradiated cells lacking Q. A student who accepts a comparison in which two variables differ picks this. These groups differ in both protein Q and radiation, so the comparison cannot show the role of Q, and their bars overlap.
Working No test statistic is calculated; the comparison rests on the ±2 SE error bars. Normal cells: no radiation 3 (1.5–4.5) vs radiation 28 (24–32), no overlap, so radiation significantly increases apoptosis. Cells lacking Q: no radiation 2 (0.5–3.5) vs radiation 5 (3–7), the bars overlap, so no significant increase is shown. This contrast supports the claim. Untreated normal (3) vs untreated lacking Q (2): bars overlap, so no difference is shown. A single group (lacking Q, radiation, 5%) gives no comparison. Untreated normal vs irradiated lacking Q differ in two variables.
Compiled from the AP Biology Course and Exam Description (effective Fall 2025) and our question bank · Specialist review in progress. How these pages are made · Free, no account
