Cell Communication and Cell Cycle: Unit Diagnostic (Biology Unit 4)

What this diagnostic is for

Use this page before you revise Unit 4, or just after finishing it. It has one short question per topic, and a second for the three largest topics (4.2, 4.3 and 4.5). Each answer names the study guide to read if you got it wrong.

These are original Marlbridge practice questions, not past exam questions. They are not calibrated against real exam results, so your total does not predict an exam score. Treat it as a map of what to revisit.

Calculator and data. A calculator is allowed on both sections of the Biology exam; here a basic one is enough. All data, cell lines and patients are fictional.

How to work. Answer everything before opening any answer, noting a reason for each choice.

Question 1 (multiple choice · 4.1)

Which example is matched with the correct way that cells communicate?

(A) Bacteria switch on biofilm genes only when crowded: long-distance signalling by a hormone.
(B) Cells in a limb bud form different digits depending on how much of a signal reaches them: a local regulator that forms a concentration gradient.
(C) A helper T cell is activated by an antigen-presenting cell: a local regulator that diffuses between them.
(D) Insulin makes muscle cells take up glucose: short-distance signalling, because muscle is close to the pancreas.

Answer and explanation

Answer: (B). A morphogen spreads from its source, so its concentration falls with distance, and cells respond to the level where they sit.

(A) is quorum sensing: the signal acts locally in the surrounding fluid, not through blood.
(C) is tempting, but the signal is held on the surface of the presenting cell: direct contact.
(D) is wrong: insulin travels in the blood to target cells all over the body.

If you missed this: read “Communication over short distances: local regulators” and “Comparing the routes” in the cell communication study guide.

Question 2 (multiple choice · 4.2)

Liver cells are treated with a chemical that stops their G proteins from exchanging GDP for GTP. Glucagon is then added. Which event still happens?

(A) Glucagon binds its receptor, and the receptor changes shape.
(B) Adenylyl cyclase converts ATP into cAMP.
(C) Protein kinase A is activated.
(D) Enzymes that break down glycogen are switched on.

Answer and explanation

Answer: (A). Reception comes before the G protein, so binding and the shape change still happen. A G protein holding GDP stays “off”, so the message stops there.

(B), (C) and (D) are all downstream of the G protein, so none happens. Adenylyl cyclase is a separate enzyme, not part of the receptor.

If you missed this: read “G protein-coupled receptors (GPCRs)” in the introduction to signal transduction study guide.

Question 3 (multiple choice · 4.2)

In a fictional pathway, kinase 1 phosphorylates kinase 2, which switches kinase 2 on. A drug blocks the protein phosphatase that removes this phosphate group. The signal is applied for 5 minutes and then washed away. Compared with untreated cells, what is the most likely result?

(A) There is no response, because kinase 2 cannot be phosphorylated while the phosphatase is blocked.
(B) The response stops sooner, because phosphate groups build up and block kinase 2.
(C) The response continues for longer after the signal is removed.
(D) There is no difference, because phosphatases act only on the receptor.

Answer and explanation

Answer: (C). Kinases switch proteins on by adding phosphate; phosphatases switch them off by removing it. With the phosphatase blocked, kinase 2 stays on after the signal has gone.

(A) mixes up the two enzymes: kinase 1, not the phosphatase, adds the phosphate.
(B) invents a mechanism. The phosphate is what keeps kinase 2 active.
(D) is wrong: phosphatases reverse phosphorylation anywhere in a cascade.

If you missed this: read “Protein phosphorylation” in the introduction to signal transduction study guide.

Question 4 (multiple choice · 4.3)

Liver cells are treated with a drug that blocks transcription. Epinephrine is added, and within 30 seconds the cells release glucose from glycogen, just as untreated cells do. Which statement best explains this?

(A) Epinephrine enters the nucleus and acts as a transcription factor that the drug cannot block.
(B) Epinephrine is itself an enzyme that breaks glycogen down.
(C) The response is a change in gene expression, which does not need transcription.
(D) The pathway switches on an enzyme the cell already has, by phosphorylation, so no new protein is needed.

Answer and explanation

Answer: (D). The pathway activates the glycogen-breaking enzyme already in the cell. The speed is a clue: making new proteins takes far longer.

(A) describes a steroid hormone, and a transcription factor would be useless while transcription is blocked.
(B) confuses the signal with the response. The cell’s own enzyme does the work.
(C) assumes every response involves genes; gene expression starts with transcription.

If you missed this: read “Changing what proteins already do” in the signal transduction pathways study guide.

Question 5 (multiple choice · 4.3)

In a fictional gland cell, ligand L binds receptor R, and R activates relay protein P. P activates two kinases: K1 leads to secretion of a stored hormone within minutes; K2 switches on genes for making more hormone over hours. A loss-of-function mutation stops K2 from being activated. What happens when L is added?

(A) Neither response happens, because the mutation is in the pathway.
(B) The genes are switched on, but the stored hormone is not secreted.
(C) The stored hormone is secreted normally, but the genes are not switched on.
(D) Both responses happen even without L.

Answer and explanation

Answer: (C). A mutation affects only the steps downstream of it. K2 is on one branch, so only the gene response is lost.

(A) treats the mutation as if it were before the branch point.
(B) reverses the two branches.
(D) describes a gain-of-function mutation; this one is loss of function.

If you missed this: read “When part of the pathway changes: mutations” in the signal transduction pathways study guide.

Question 6 (multiple choice · 4.4)

In a fictional patient, a small group of β cells releases insulin at a high rate all the time, whatever the blood glucose concentration. The rest of the loop works normally. Which outcome is most likely between meals?

(A) Blood glucose rises, because high insulin makes the liver release glucose.
(B) Blood glucose falls below the set point and stays low, because insulin release no longer falls when glucose falls.
(C) Blood glucose stays at the set point, because the β cells detect the fall and stop releasing insulin.
(D) Blood glucose rises above the set point, because positive feedback takes over.

Answer and explanation

Answer: (B). Normally, falling glucose shrinks the stimulus, so insulin release falls. Here insulin ignores glucose, so cells keep taking glucose up. Glucagon may partly oppose the fall, but cannot cancel a signal that never stops.

(A) gives insulin the job of glucagon.
(C) describes the normal loop; these cells ignore glucose.
(D) misuses “positive feedback”: nothing here strengthens the original change.

If you missed this: read “When feedback fails” in the feedback study guide.

Question 7 (multiple choice · 4.5)

Dividing animal cells are treated with a chemical that stops spindle microtubules from forming. DNA replication is not affected. After several hours, most of the dividing cells contain:

(A) loose chromatin in G1, because DNA cannot be replicated without a spindle.
(B) separated chromatids at opposite poles, but no new nuclear envelope.
(C) no chromosomes, after a cleavage furrow has divided the cell.
(D) condensed chromosomes, each of two sister chromatids, that cannot be lined up at the equator or pulled apart.

Answer and explanation

Answer: (D). The cells replicate their DNA and condense their chromosomes as normal. With no spindle fibres, nothing attaches at the centromeres, so chromosomes cannot be aligned or separated.

(A) puts the spindle in the wrong phase: replication does not use it.
(B) needs anaphase, and anaphase needs spindle fibres to pull the chromatids.
(C) is impossible: chromosomes cannot vanish.

If you missed this: read “Mitosis: sharing the copies out” in the cell cycle study guide.

Question 8 (short answer · 4.5)

A body cell of a fictional animal has 12 chromosomes. A student measures the DNA content of 1000 cells from a tissue in which every cell is dividing. The cycle takes 24 hours.

DNA per cell (G1 value = 1 unit)	1 unit	between 1 and 2 units	2 units
Number of cells	550	250	200

Under a microscope, 40 of the 1000 cells are in mitosis.

(a) State which phase or phases each group of cells is in. (b) Estimate how long a cell spends in S phase and in G2. (c) State the number of chromosomes and of DNA molecules in a cell in G1 and in a cell in G2.

Answer and explanation

(a) 1 unit: G1. Between 1 and 2 units: S phase (DNA being replicated). 2 units: G2, plus cells in mitosis that have not yet divided.

(b) S phase: 250 ÷ 1000 × 24 h = 6.0 h. The 2-unit group (G2 and mitosis): 200 ÷ 1000 × 24 = 4.8 h. Mitosis: 40 ÷ 1000 × 24 = 0.96 h. So G2 ≈ 4.8 − 0.96 = 3.84 h, about 3.8 h.

(c) G1: 12 chromosomes, 12 DNA molecules. G2: still 12 chromosomes, each of two sister chromatids, so 24 DNA molecules.

Watch for: “24 chromosomes in G2”. Replication doubles the DNA, not the chromosomes. Count centromeres.

If you missed this: read “Interphase: G1, S and G2” and “Worked example 2: chromosomes, chromatids and DNA content” in the cell cycle study guide.

Question 9 (short answer · 4.6)

Two fictional cell lines are grown with no growth signal. Mutant line C makes a G1 cyclin at a high level all the time. No cells die.

Day	0	1	2	3
Normal / 10⁵ cells	2.0	2.1	2.1	2.2
Mutant C / 10⁵ cells	2.0	4.0	8.0	16.0

(a) Calculate how many times the mutant population grew in 3 days, and its mean cycle length if every cell divides. (b) Explain the result for mutant C in terms of cyclins, CDKs and the G1 checkpoint. (c) Explain why this mutation is a step towards cancer, and name the process that could still remove the cell if its DNA became badly damaged.

Answer and explanation

(a) 16.0 ÷ 2.0 = 8 times = 2³, so 3 divisions in 72 h: about 24 h per cycle. Normal cells barely divide.

(b) A CDK is active only when bound to cyclin. Normally the G1 cyclin builds up only when conditions, including growth signals, favour division, so most cells stop at the G1 checkpoint. In mutant C the cyclin is always there, so the cyclin-CDK complex is always active, phosphorylates the proteins that start replication, and cells pass the checkpoint without a signal.

(c) The cells divide without the controlling signal, and each extra division is a chance for further mutations; cancer usually needs several. Apoptosis could still remove a cell with badly damaged DNA, if its damage-detecting checkpoint proteins still work.

Watch for: “the extra cyclin makes more CDK”. CDK amount stays roughly constant; cyclin raises CDK activity.

If you missed this: read “Cyclins and CDKs: the engine of the cycle” and “When control fails: cancer” in the regulation of the cell cycle study guide.

Your next step

Mark each question, then use this table. A short answer missed in one part counts as missed.

Topic	Question(s)	If you missed it, read
4.1 Cell communication	1	Study guide 4.1
4.2 Introduction to signal transduction	2, 3	Study guide 4.2
4.3 Signal transduction pathways	4, 5	Study guide 4.3
4.4 Feedback	6	Study guide 4.4
4.5 Cell cycle	7, 8	Study guide 4.5
4.6 Regulation of cell cycle	9	Study guide 4.6

How to use your result

Wrong answers first. Read the named section, then do that topic’s practice questions.
Lucky guesses count as misses. If your reason did not match the explanation, count it wrong.
Missed two of Questions 2, 3 and 5? All three ask what happens downstream of one changed step.
No misses? Go straight to the mixed unit review, which joins the topics in longer, exam-style questions.
Come back in a week and check your reasons again.