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AP Biology Unit 4 Practice Questions: Cell Communication and Cell Cycle

10 original exam-style questions on Cell Communication and Cell Cycle. Answer each one to see the explanation — no account needed.

Question 1 of 10 · Signal Transduction

A researcher investigates a signaling pathway in liver cells. When epinephrine binds to its receptor on the cell surface, a series of molecular events leads to the breakdown of glycogen. The figure below summarizes key steps.

Effect of epinephrine concentration on cAMP production and glycogen phosphorylase activity
Epinephrine (nM)Intracellular cAMP (pmol/mg protein)Glycogen Phosphorylase Activity (nmol/min/mg)
01.20.8
0.14.73.1
118.311.9
1041.527.2
10043.127.8
A mutation causes the G protein in the epinephrine pathway to lose its GTPase activity. Which of the following best predicts the effect of this mutation?
  1. A. The G protein will be permanently inactive because it cannot hydrolyze GTP to GDP.
  2. B. Glycogen phosphorylase will be inactivated due to lack of phosphorylation.
  3. C. Epinephrine will no longer bind to its receptor on the cell surface.
  4. D. Adenylyl cyclase will be constitutively active, leading to continuously elevated cAMP levels.
Show answer and explanation

Correct answer: D

The G protein is active when bound to GTP and inactive after hydrolyzing GTP to GDP. Loss of GTPase activity means the G protein stays GTP-bound (active) indefinitely, keeping adenylyl cyclase continuously stimulated and cAMP elevated. This is a constitutive gain-of-function mutation, not a loss of function.

Question 2 of 10 · Cell Cycle Checkpoints

Which of the following best describes a key difference between the spindle assembly checkpoint (SAC) and the G1/S checkpoint?
  1. A. The SAC monitors kinetochore attachment during mitosis, while the G1/S checkpoint monitors DNA integrity and cell size before S phase.
  2. B. The SAC is active only in cancer cells, while the G1/S checkpoint functions only in normal cells.
  3. C. Both checkpoints perform identical functions but at different points in the cell cycle.
  4. D. The SAC initiates DNA repair, while the G1/S checkpoint initiates chromosome condensation.
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Correct answer: A

The spindle assembly checkpoint (SAC) ensures all kinetochores are properly attached to spindle fibers before anaphase proceeds, preventing chromosomal mis-segregation. The G1/S (restriction) checkpoint assesses DNA damage, growth factor signaling, and cell size to determine whether the cell commits to another round of division. These are distinct sensors operating at distinct cell cycle transitions.

Question 3 of 10 · Cell Cycle

Students examined cells from a root tip meristem under a microscope and counted the number of cells in each phase of the cell cycle. Their results are summarized below.

Cell cycle phase distribution in onion root tip cells (n = 500 cells observed)
PhaseNumber of CellsPercentage of Cells
Interphase (G1, S, G2)44088%
Prophase326.4%
Metaphase112.2%
Anaphase81.6%
Telophase/Cytokinesis91.8%
If the total cell cycle duration in these cells is 24 hours, approximately how long does metaphase last?
  1. A. 0.5 hours
  2. B. 1.2 hours
  3. C. 2.2 hours
  4. D. 6.0 hours
Show answer and explanation

Correct answer: A

Time in a phase = (fraction of cells in that phase) × total cycle time. Metaphase fraction = 2.2% = 0.022; time = 0.022 × 24 h ≈ 0.53 hours, closest to 0.5 hours. Choosing 2.2 hours confuses the percentage value with the time in hours.

Question 4 of 10 · Types of Cell Signaling: Endocrine vs. Paracrine vs. Autocrine

In the process of quorum sensing, bacteria release small signaling molecules called autoinducers. As bacterial population density increases, autoinducer concentration rises. When autoinducer concentration crosses a threshold, a coordinated change in gene expression occurs across the population. Which of the following represents the best analogy to quorum sensing in multicellular organisms?
  1. A. Endocrine signaling, where a diffusible molecule travels through a shared medium to coordinate a response in cells expressing the appropriate receptor
  2. B. Gap junction communication, where signals pass directly between adjacent cells through protein-lined pores without entering extracellular space
  3. C. Autocrine signaling in a single isolated cell, where the signal molecule binds only to the cell that produced it
  4. D. Synaptic signaling between two specific neurons, where the signal is immediately degraded after receptor binding
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Correct answer: A

Quorum sensing resembles endocrine signaling in that diffusible molecules accumulate in a shared extracellular medium and trigger a coordinated response when their concentration reaches a threshold in cells bearing the appropriate receptor — analogous to how blood-borne hormones coordinate responses across distant tissues. Choice B (gap junctions) requires direct cell contact, unlike the diffusible autoinducer mechanism.

Question 5 of 10 · Types of Natural Selection

Researchers studied beak depth in a population of medium ground finches (Geospiza fortis) on Daphne Major Island across three years that included a severe drought. During the drought, small soft seeds were depleted and only large, hard seeds remained. Population size and mean beak measurements are shown below.

G. fortis Beak Depth: Pre-drought, Drought Peak, and Recovery
Survey YearPopulation Size (N)Mean Beak Depth (mm)Std. Deviation (mm)
1976 (pre-drought)12009.420.68
1977 (drought peak)1809.960.61
1978 (post-drought recovery)6209.840.65
Based on the data, which of the following best describes the pattern of natural selection on beak depth between 1976 and 1977, and the evidence that supports this conclusion?
  1. A. Stabilizing selection, because the standard deviation decreased from 0.68 to 0.61 mm, indicating that extreme beak sizes at both ends of the distribution were eliminated
  2. B. Sexual selection, because the 85% decline in population size created a bottleneck in which only large-beaked males could successfully compete for mates
  3. C. Disruptive selection, because the population split into two distinct beak-size classes — one feeding on soft seeds and one on hard seeds — as shown by the bimodal data
  4. D. Directional selection, because the mean beak depth shifted from 9.42 mm to 9.96 mm, indicating that deeper-beaked individuals had higher survival as hard seeds became the primary food source
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Correct answer: D

The mean beak depth increased by ~0.54 mm (a statistically large shift relative to the standard deviation) because birds with deeper beaks could crack the hard seeds that remained during the drought, conferring higher survival and shifting the population mean toward larger values — the hallmark of directional selection. Choice A is incorrect because stabilizing selection eliminates extremes without substantially shifting the mean, yet the mean here shifted markedly while the standard deviation changed only slightly.

Question 6 of 10 · Tumor Suppressors and Cell Cycle Control

A tumor suppressor gene normally encodes a protein that inhibits cell division. If a cell acquires a loss-of-function mutation in both copies of this gene (homozygous), which of the following outcomes would most likely result?
  1. A. The cell would divide more slowly because the tumor suppressor protein would now inhibit DNA replication
  2. B. The cell would undergo apoptosis because the absence of the tumor suppressor triggers programmed cell death
  3. C. The cell would lose the inhibitory check on the cell cycle, potentially leading to uncontrolled division
  4. D. The cell would be protected from cancer because loss of the inhibitor would activate competing checkpoints
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Correct answer: C

Tumor suppressor proteins provide inhibitory signals that keep the cell cycle in check — homozygous loss of function removes this brake, allowing cells to progress through the cycle without normal regulation, a necessary step toward tumorigenesis (as described by Knudson's two-hit hypothesis). Choice B is a common misconception; loss of specific tumor suppressors does not automatically trigger apoptosis.

Question 7 of 10 · Cell Cycle Checkpoints and Spindle Assembly

A student prepares onion root tip slides and observes cells in different stages of mitosis under a microscope. The student counts 200 cells and records the stage of each cell in the table below.

Distribution of Onion Root Tip Cells by Mitotic Stage
StageNumber of Cells ObservedPercentage of Cells
Interphase14271%
Prophase2814%
Metaphase147%
Anaphase63%
Telophase/Cytokinesis105%
A researcher treats the onion root cells with colchicine, a drug that depolymerizes microtubules. After treatment, which of the following changes to the data table would most likely be observed?
  1. A. The percentage of cells in anaphase would increase as chromosomes are pulled to the poles more rapidly
  2. B. The percentage of cells in telophase would increase as the spindle is needed only for cytokinesis
  3. C. The percentage of cells in interphase would increase as cells are prevented from entering mitosis
  4. D. The percentage of cells in metaphase would increase as cells become arrested with chromosomes unable to segregate
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Correct answer: D

Colchicine depolymerizes spindle microtubules, preventing chromosome movement during anaphase — cells align chromosomes at the metaphase plate but cannot separate them, causing metaphase arrest and accumulation. Choice A is incorrect because destroying microtubules blocks, not accelerates, chromosome segregation.

Question 8 of 10 · Stages of Mitosis

During which phase of mitosis do sister chromatids separate and move to opposite poles?
  1. A. Prophase, when chromosomes first condense
  2. B. Metaphase, when chromosomes align at the cell plate
  3. C. Anaphase, when cohesin is cleaved and chromatids are pulled apart by spindle fibers
  4. D. Telophase, when nuclear envelopes re-form around each chromosome set
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Correct answer: C

Sister chromatid separation defines anaphase: the protease separase cleaves cohesin proteins holding chromatids together, and motor proteins move chromatids toward opposite poles along spindle fibers. In telophase, chromatids have already arrived at the poles and nuclear envelopes are re-forming.

Question 9 of 10 · Types of Cell Signaling

Local signaling between adjacent cells using molecules that travel only short distances is called:
  1. A. Endocrine signaling, where hormones travel through the bloodstream to distant target cells.
  2. B. Autocrine signaling, where a cell secretes molecules that bind to its own receptors.
  3. C. Paracrine signaling, where signaling molecules diffuse locally to affect nearby cells.
  4. D. Synaptic signaling, where electrical impulses travel along axons to distant neurons.
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Correct answer: C

Paracrine signaling uses local chemical mediators (e.g., growth factors, prostaglandins) that act on cells in the immediate vicinity. Endocrine signaling is long-distance via the bloodstream; autocrine signaling acts on the same cell that secreted the signal; synaptic signaling is neuron-specific.

Question 10 of 10 · Signal Amplification in Enzyme Cascades

Epinephrine (adrenaline) triggers a signaling cascade in liver cells that leads to glycogen breakdown. The pathway involves a GPCR → adenylyl cyclase → cAMP → protein kinase A (PKA) → glycogen phosphorylase activation. Which feature of this pathway allows a small amount of epinephrine to produce a large metabolic response?
  1. A. Epinephrine is a steroid hormone that directly enters cells and activates multiple nuclear receptors simultaneously
  2. B. Signal amplification occurs because each activated enzyme in the cascade can activate many molecules of the next component, multiplying the signal at each step
  3. C. Epinephrine binds irreversibly to its GPCR, ensuring prolonged activation of the pathway even after the hormone is removed
  4. D. cAMP directly phosphorylates glycogen phosphorylase without requiring any intermediate kinase steps, enabling rapid response
Show answer and explanation

Correct answer: B

Signal amplification (cascade amplification) is the defining feature of enzyme-coupled signaling: one activated receptor activates many G proteins → each G protein activates adenylyl cyclase → each cyclase produces many cAMP molecules → each PKA molecule phosphorylates many glycogen phosphorylase kinase molecules → each activates many phosphorylase molecules, amplifying the original signal millions-fold. Choice A is incorrect because epinephrine is a hydrophilic amine that cannot cross the membrane and acts through cell-surface GPCRs.

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