Unit 4 · 10–15% of the exam Beta

Unit 4 practice test: Cell Communication and Cell Cycle

12 exam-style questions from the unit, free, with an explanation for every option. Sets share one table, graph or experiment, as on the real exam.

Covers: Cell Communication; Introduction to Signal Transduction; Signal Transduction Pathways; Feedback; Cell Cycle; Regulation of Cell Cycle.

Data table

How far does an injected dye spread?

Researchers grew a single flat layer of liver cells in a dish. They injected one cell with a fluorescent dye and, 5 minutes later, counted how many neighboring cells had become fluorescent. They tested four dyes of different sizes, in untreated cells and in cells given a drug that closes gap junctions 10 minutes before the injection. Each value is the mean for 10 injected cells.

Number of neighboring cells fluorescent 5 minutes after one cell was injected
Dye	Molecular mass (daltons)	Untreated cells	Cells given gap-junction blocker
P	376	24	1
Q	457	22	1
R	3,000	0	0
S	10,000	0	0

1. Which statement best describes the results in untreated cells?

Dyes P and Q (457 daltons or less) reached over 20 neighbors; dyes R and S reached none.
Each dye reached about the same number of neighbors, whatever its size.
The larger the dye, the more neighboring cells it reached in 5 minutes.
Dye P reached fewer neighbors than dye S, because small dyes leak out of cells before they can spread.

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Untreated cells: dye P reached 24 neighbors and dye Q 22, while dyes R and S reached 0. There is a sharp cut-off between 457 and 3,000 daltons.

Correct: Dyes P and Q (457 daltons or less) reached over 20 neighbors; dyes R and S reached none.: These are the untreated values: about 22-24 cells for the two small dyes and zero for the two large ones.
Each dye reached about the same number of neighbors, whatever its size.: The counts range from 0 to 24, so size clearly matters.
The larger the dye, the more neighboring cells it reached in 5 minutes.: The trend runs the other way: the two largest dyes reached no neighbors.
Dye P reached fewer neighbors than dye S, because small dyes leak out of cells before they can spread.: Dye P reached 24 cells and dye S reached none.

2. Which claim is best supported by the data?

Molecules below a size limit between about 460 and 3,000 daltons pass from cell to cell through gap junctions within minutes.
Gap junctions pump small dyes into neighboring cells using ATP, and the blocker works by stopping ATP production.
The dyes leak out of the injected cell and are taken up from the medium by its neighbors.
Large dyes do reach the neighbors through gap junctions, but more slowly than small dyes do.

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Small dyes spread to about 22-24 neighbors and the blocker cuts that to 1, so they travel through gap junctions; dyes of 3,000 daltons or more do not spread at all.

Correct: Molecules below a size limit between about 460 and 3,000 daltons pass from cell to cell through gap junctions within minutes.: Both parts are supported: spread depends on open gap junctions, and only the small dyes spread.
Gap junctions pump small dyes into neighboring cells using ATP, and the blocker works by stopping ATP production.: Gap junctions are channels: molecules diffuse through them. Nothing in the data points to ATP or pumping.
The dyes leak out of the injected cell and are taken up from the medium by its neighbors.: Leaking and uptake from the medium would not be stopped by closing gap junctions, yet the blocker stopped nearly all spread.
Large dyes do reach the neighbors through gap junctions, but more slowly than small dyes do.: The large dyes reached zero cells in 5 minutes; the data give no sign that they pass at all.

3. What is the main reason for including the cells given the gap-junction blocker?

To test whether the dye spreads through gap junctions rather than by some other route
To make sure the injected cell was still alive and healthy after the injection was made
To keep the size of the dye the same from one trial to the next
To measure how quickly each dye diffuses through the cytoplasm

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If closing gap junctions stops the spread, the dye must travel through them. Without this comparison, leaking out and re-entering would be just as possible.

Correct: To test whether the dye spreads through gap junctions rather than by some other route: The blocker condition isolates one route: if spread stops when gap junctions close, they were the route.
To make sure the injected cell was still alive and healthy after the injection was made: The blocker says nothing about whether the cell survived; a living cell with closed junctions also stops spread.
To keep the size of the dye the same from one trial to the next: Dye size is the variable the researchers changed on purpose, set by which dye they chose, not by the blocker.
To measure how quickly each dye diffuses through the cytoplasm: The experiment counts neighbors reached, not the speed of movement inside one cell.

4. Heart muscle cells are joined by gap junctions that let ions flow between them. Predict the effect of the same blocker on a sheet of beating heart muscle cells.

Their contractions would fall out of step, because the electrical change could not pass from cell to cell.
Each cell would contract faster, because ions could no longer escape into neighboring cells.
The cells would stop making ATP, because glucose could no longer enter them.
There would be no change in the beat, because hormones carried in the blood set the timing of each contraction.

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Ions are small, like dyes P and Q, so they pass through gap junctions. The ion flow carries the electrical change from one heart cell to the next; blocking it leaves cells contracting out of step.

Correct: Their contractions would fall out of step, because the electrical change could not pass from cell to cell.: Ions are small enough to pass, so closing the junctions breaks the cell-to-cell spread of the electrical signal.
Each cell would contract faster, because ions could no longer escape into neighboring cells.: Ion flow to neighbors is what spreads the beat; blocking it does not make any one cell contract faster.
The cells would stop making ATP, because glucose could no longer enter them.: Glucose enters heart cells through membrane transporters from the tissue fluid, not through gap junctions.
There would be no change in the beat, because hormones carried in the blood set the timing of each contraction.: Hormones can change the heart rate, but each beat spreads from cell to cell through gap junctions within a fraction of a second.

Experimental setup

Fusing cells in different phases

In the 1970s, researchers fused pairs of mammalian cells that were in different phases of the cell cycle. Each fused cell had two nuclei sharing one cytoplasm. They then watched what the second nucleus did.

Results of fusing a cell in one phase with a cell in another
Cells fused	What the second nucleus did
S-phase cell + G1 cell	The G1 nucleus began copying its DNA at once, hours earlier than it would have alone.
S-phase cell + G2 cell	The G2 nucleus did not copy its DNA again; it waited until the S nucleus finished, then both entered mitosis.
M-phase cell + G1 cell	The G1 nucleus's envelope broke down and its uncopied chromosomes condensed at once, as single chromatids.
M-phase cell + G2 cell	The G2 nucleus's envelope broke down and its chromosomes condensed at once, each as two sister chromatids.

5. Which conclusion is best supported by the four results?

S-phase and M-phase cytoplasm contains factors that push another nucleus into that phase.
Each nucleus controls its own timing, and the cytoplasm it shares has no effect on it.
The G1 nucleus is the one that controls the timing of both nuclei in a fused cell.
Fusing any two cells makes both nuclei enter mitosis together, in whatever phase each one was in.

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Sharing cytoplasm with an S cell starts DNA copying in a G1 nucleus, and sharing with an M cell makes another nucleus condense its chromosomes: something in the cytoplasm drives each phase.

Correct: S-phase and M-phase cytoplasm contains factors that push another nucleus into that phase.: The second nucleus changes according to the other cell's phase, through the shared cytoplasm.
Each nucleus controls its own timing, and the cytoplasm it shares has no effect on it.: The G1 nuclei change their timing immediately after fusion, so the shared cytoplasm matters.
The G1 nucleus is the one that controls the timing of both nuclei in a fused cell.: The G1 nuclei follow the other cell's lead, not the reverse.
Fusing any two cells makes both nuclei enter mitosis together, in whatever phase each one was in.: S + G1 fusions lead to DNA copying, not to mitosis.

6. In the S + G2 fusion, the G2 nucleus did not copy its DNA again even though it shared cytoplasm with an S-phase nucleus. Which explanation best fits?

Something in a nucleus that has already copied its DNA blocks a second round of copying in the same cycle.
The G2 nucleus had too little cyclin of its own to respond to any signal coming from the S-phase cytoplasm.
The S-phase factor does not reach a G2 nucleus, because G2 nuclei have no pores in their nuclear envelope.
The G2 nucleus was damaged by the fusion, so it waited for repair at the G2 checkpoint before mitosis.

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The same S-phase cytoplasm starts copying in a G1 nucleus but not in a G2 nucleus, so the difference lies in the nucleus itself: once copied, DNA is not copied again until after mitosis.

Correct: Something in a nucleus that has already copied its DNA blocks a second round of copying in the same cycle.: The difference between the G1 and G2 nuclei shows a block in the already-copied nucleus.
The G2 nucleus had too little cyclin of its own to respond to any signal coming from the S-phase cytoplasm.: The G1 nucleus responded to the S signal without its own S-phase cyclin, so the G2 nucleus's cyclin level is not what decided this.
The S-phase factor does not reach a G2 nucleus, because G2 nuclei have no pores in their nuclear envelope.: Nuclei in every phase of interphase have pores; G2 nuclei exchange molecules with the cytoplasm.
The G2 nucleus was damaged by the fusion, so it waited for repair at the G2 checkpoint before mitosis.: Nothing suggests damage, and the G2 nucleus entered mitosis on schedule.

7. The M-phase factor is now known to be MPF, an active cyclin-CDK complex. Before fusion with a G2 cell, the M-phase cell is treated with a drug that blocks CDK kinase activity. Predict the result.

The G2 nucleus does not condense its chromosomes, because the factor that drives mitosis is inactive.
The G2 nucleus condenses its chromosomes as before, because the M-phase cell's cyclin is still present.
The G2 nucleus copies its DNA again, because blocking CDK sends the cell back to S phase.
The G2 nucleus condenses even faster, because the drug frees cyclin to act on the nucleus itself.

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MPF acts by phosphorylating target proteins. With its kinase activity blocked, the M-phase signal is gone, so the G2 nucleus is not pushed into mitosis.

Correct: The G2 nucleus does not condense its chromosomes, because the factor that drives mitosis is inactive.: No kinase activity, no phosphorylated targets, no condensation.
The G2 nucleus condenses its chromosomes as before, because the M-phase cell's cyclin is still present.: Cyclin works only by switching on the CDK's kinase activity, which is blocked.
The G2 nucleus copies its DNA again, because blocking CDK sends the cell back to S phase.: Blocking CDK does not reset a nucleus that has already copied its DNA.
The G2 nucleus condenses even faster, because the drug frees cyclin to act on the nucleus itself.: Cyclin acts through its CDK; on its own it does not condense chromosomes.

8. Some cancer drugs block the CDKs that work with G1 cyclins. Based on the fusion results and what you know of the cycle, predict their effect on cancer cells that depend on these CDKs.

The cells are held at the G1 checkpoint and do not start copying their DNA.
The cells are held in mitosis, because their chromosomes are unable to finish separating.
The cells copy their DNA twice before dividing, because G1 is skipped.
The cells divide faster, because they spend less time waiting in G1.

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G1 cyclin-CDK carries the cell past the G1 checkpoint into S phase. Blocked, the cancer cells stay in G1.

Correct: The cells are held at the G1 checkpoint and do not start copying their DNA.: The G1 cyclin-CDK acts at the G1 checkpoint, so the block holds cells there.
The cells are held in mitosis, because their chromosomes are unable to finish separating.: That would follow from keeping mitotic cyclin-CDK active, a different complex.
The cells copy their DNA twice before dividing, because G1 is skipped.: Blocking the G1 complex stops entry into S, so no copying happens at all.
The cells divide faster, because they spend less time waiting in G1.: Without the G1 push, cells do not leave G1, so they divide less.

9. What mainly decides whether a signaling molecule's receptor is on the cell surface or inside the cell?

Whether the molecule can cross the lipid bilayer, given its size and polarity
Whether the molecule arrives in the blood or diffuses over from a nearby cell
How high the molecule's concentration is in the fluid outside the cell
Whether the response that the cell gives to the signal is a fast or a slow one

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A ligand can reach an inside receptor only if it crosses the membrane, which small nonpolar molecules can and polar or large ones cannot.

Correct: Whether the molecule can cross the lipid bilayer, given its size and polarity: Membrane permeability, from size and polarity, sets where the receptor can be.
Whether the molecule arrives in the blood or diffuses over from a nearby cell: Hormones in the blood include both kinds, insulin (surface) and steroids (inside).
How high the molecule's concentration is in the fluid outside the cell: Concentration affects how many receptors are bound, not where they are.
Whether the response that the cell gives to the signal is a fast or a slow one: Speed is a consequence of the receptor type, not what decides it.

10. Cells use ATP to pump Ca²⁺ out of the cytoplasm, keeping its concentration there about 10,000 times lower than in the ER. How does this make Ca²⁺ an effective second messenger?

Opening a few channels lets Ca²⁺ diffuse in down its steep gradient, raising its level many-fold in milliseconds.
The pumps add a phosphate group to each Ca²⁺ ion as they move it, which switches the ion on so it can bind proteins.
Keeping Ca²⁺ inside the ER lets the ER release Ca²⁺ into the blood when a hormone signal is needed.
Low Ca²⁺ in the cytoplasm keeps most enzymes in the cell switched on until a signal raises the level.

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A steep concentration gradient stores potential energy. When channels open, Ca²⁺ rushes in by diffusion, producing a large, fast change that Ca²⁺-binding proteins detect.

Correct: Opening a few channels lets Ca²⁺ diffuse in down its steep gradient, raising its level many-fold in milliseconds.: The gradient lets a small opening make a big, fast change.
The pumps add a phosphate group to each Ca²⁺ ion as they move it, which switches the ion on so it can bind proteins.: Ions are not phosphorylated; pumps move them using ATP's energy.
Keeping Ca²⁺ inside the ER lets the ER release Ca²⁺ into the blood when a hormone signal is needed.: The ER releases Ca²⁺ into the cytoplasm, where it acts as a second messenger inside the cell.
Low Ca²⁺ in the cytoplasm keeps most enzymes in the cell switched on until a signal raises the level.: Most enzymes are not set by Ca²⁺; rising Ca²⁺ switches particular proteins on.

11. When a blood vessel is cut, platelets stick at the cut and release chemicals that make more platelets stick. Which statement best explains why this positive feedback loop does not keep growing through the bloodstream?

Once the clot seals the break, the exposed damaged wall that started the loop is covered, so the stimulus ends.
Platelets in the blood are used up after the first few seconds of clotting.
A negative feedback loop takes over as soon as clotting starts and reverses it.
Positive feedback loops weaken with each cycle until they stop by themselves.

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Positive feedback stops when an outside event removes the stimulus. Sealing the break covers the damaged wall that started the loop.

Correct: Once the clot seals the break, the exposed damaged wall that started the loop is covered, so the stimulus ends.: The loop ends when its trigger, the exposed damage, is gone.
Platelets in the blood are used up after the first few seconds of clotting.: Plenty of platelets remain; a clot uses a small fraction of them.
A negative feedback loop takes over as soon as clotting starts and reverses it.: Opposing systems exist, but they do not reverse clotting the moment it starts; the clot must form first.
Positive feedback loops weaken with each cycle until they stop by themselves.: Positive feedback grows stronger with each cycle; it needs an outside event to stop.

12. A drug stops the ring of actin filaments from tightening in animal cells but does not affect mitosis. Predict what dividing cells look like after the drug is added.

Single cells with two nuclei, because mitosis finishes but the cytoplasm is not divided
Cells stuck at metaphase, because the chromosomes fail to attach to the spindle
Two daughter cells with one nucleus each, because cytokinesis goes on using a cell plate
Cells with no nucleus, because the chromosomes are lost when the furrow fails

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The actin ring makes the cleavage furrow. Mitosis still separates the chromosomes into two nuclei, but without cytokinesis they stay in one cell.

Correct: Single cells with two nuclei, because mitosis finishes but the cytoplasm is not divided: Nuclear division without cytoplasm division gives one cell with two nuclei.
Cells stuck at metaphase, because the chromosomes fail to attach to the spindle: Spindle attachment uses microtubules, which the drug does not affect.
Two daughter cells with one nucleus each, because cytokinesis goes on using a cell plate: Animal cells divide with a furrow; the cell plate is how plant cells divide.
Cells with no nucleus, because the chromosomes are lost when the furrow fails: The chromosomes are already enclosed in two new nuclei; a failed furrow does not remove them.

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Practice has every question in the unit, with feedback after each one; the notes for every topic are free. Write a full answer with the justification trainer. Other units: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5 · Unit 6 · Unit 7 · Unit 8.