Free practice final
Cell Biology practice final (free, with answers)
An intro cell biology final usually asks how membranes and transporters work, how proteins reach the right compartment, how cells signal, and how the cell cycle, mitosis, meiosis, and cell death are controlled. These 15 questions test both the facts and the reasoning behind them. Work through each one before opening the answer, and read the explanation even when you got it right.
Topics
- Membrane Structure
- Membrane Transport
- Endomembrane System
- Cytoskeleton
- Cell Signaling
- Cell Cycle Control
- Mitosis and Meiosis
- Apoptosis and Cancer
15 questions
Try each one before you open its answer.
1. Which change in a membrane's phospholipids would make the bilayer more fluid at the same temperature?
- A Longer hydrocarbon tails
- B More fully saturated tails
- C More tails with cis double bonds
- D Converting cis double bonds in the tails to trans double bonds
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Answer: C. More tails with cis double bonds
A cis double bond puts a kink in the tail, so neighboring lipids pack less tightly and the bilayer stays more fluid. Longer tails have more contact with their neighbors, and saturated or trans tails are straighter and pack more tightly, so all three make the membrane less fluid.
2. Intestinal epithelial cells take up glucose through a Na⁺–glucose symporter, even when glucose is more concentrated inside the cell than in the gut. What directly supplies the energy for this uptake?
- A ATP hydrolysis by the symporter itself
- B Na⁺ moving down its electrochemical gradient into the cell
- C Glucose moving down its concentration gradient
- D A proton gradient made by the electron transport chain
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Answer: B. Na⁺ moving down its electrochemical gradient into the cell
This is secondary active transport: Na⁺ flowing into the cell down its gradient drives glucose in against its gradient. The Na⁺/K⁺ pump uses ATP to keep that Na⁺ gradient in place, but the symporter itself does not hydrolyze ATP.
3. At 37 °C, a cell has 140 mM K⁺ inside and 5 mM K⁺ outside. What is the K⁺ equilibrium (Nernst) potential, inside relative to outside? Use RT/F = 26.7 mV.
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Answer: About −89 mV
E = (RT/zF) ln([K⁺]out/[K⁺]in) = 26.7 mV × ln(5/140) ≈ 26.7 × (−3.33) ≈ −89 mV. The inside must be negative to hold back K⁺ that would otherwise leak out down its concentration gradient.
4. A soluble protein has an N-terminal ER signal sequence but no ER retention signal and no lysosomal sorting signal. Where does it end up?
- A Retained in the ER lumen
- B Released into the cytosol
- C Delivered to lysosomes
- D Secreted from the cell
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Answer: D. Secreted from the cell
The signal sequence sends the protein into the ER lumen, and with no other signals it follows the default route through the Golgi to the cell surface. Staying in the ER or going to lysosomes each requires an extra sorting signal.
5. In a neuron, a vesicle travels along an axonal microtubule from the cell body toward the axon terminal, where the microtubule plus ends point. Which motor protein most likely carries it?
- A Cytoplasmic dynein
- B Kinesin
- C Myosin II
- D Dynamin
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Answer: B. Kinesin
Kinesin walks toward microtubule plus ends, so it carries cargo outward to the axon terminal. Dynein moves toward minus ends (back to the cell body), myosins move on actin, and dynamin pinches off vesicles rather than moving them.
6. Soluble lysosomal hydrolases receive a sugar-phosphate tag in the Golgi that receptors in the trans Golgi network recognize, sending them toward lysosomes. What is this tag?
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Answer: Mannose 6-phosphate
Enzymes in the cis Golgi add a phosphate to mannose residues on these hydrolases, and mannose 6-phosphate receptors in the trans Golgi network package them into vesicles bound for endosomes and lysosomes. Without the tag, the hydrolases are secreted instead.
7. Cholera toxin chemically modifies Gαs so that it can no longer hydrolyze its bound GTP. What happens in an affected intestinal cell?
- A Adenylyl cyclase stays active and cAMP remains high
- B Gαs can no longer bind GTP, so cAMP falls
- C The receptor can no longer bind its ligand
- D Phospholipase C stays active and IP₃ remains high
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Answer: A. Adenylyl cyclase stays active and cAMP remains high
Gαs switches itself off by hydrolyzing GTP. Locked in the GTP-bound state, it keeps stimulating adenylyl cyclase, and the high cAMP drives the Cl⁻ and water loss behind cholera's diarrhea.
8. Which event directly allows sister chromatids to separate at the start of anaphase?
- A Cyclin B is synthesized, activating Cdk1
- B Separase cleaves cohesin after the APC/C triggers destruction of securin
- C Condensin compacts each chromatid
- D Kinetochore microtubules lengthen and push the sisters apart
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Answer: B. Separase cleaves cohesin after the APC/C triggers destruction of securin
Cohesin holds sisters together until the APC/C ubiquitylates securin, which is then degraded. Freed separase cuts cohesin, and the spindle pulls the sisters to opposite poles.
9. Activated Gαq stimulates an enzyme that cleaves the membrane lipid PIP₂ into IP₃ and diacylglycerol. Name the enzyme.
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Answer: Phospholipase C (phospholipase C-β)
Phospholipase C-β splits PIP₂ into two messengers. IP₃ opens Ca²⁺ channels in the ER, and diacylglycerol stays in the membrane, where together with Ca²⁺ it activates protein kinase C.
10. An organism has a diploid number of 2n = 8. How many chromosomes and chromatids does each cell have at metaphase II of meiosis?
- A 8 chromosomes, 16 chromatids
- B 4 chromosomes, 4 chromatids
- C 8 chromosomes, 8 chromatids
- D 4 chromosomes, 8 chromatids
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Answer: D. 4 chromosomes, 8 chromatids
Meiosis I separates homologs, so each cell keeps n = 4 chromosomes, but each chromosome still has two sister chromatids until anaphase II. That makes 4 chromosomes and 8 chromatids.
11. After DNA damage in G1, p53 accumulates and activates transcription of the gene for p21. How does p21 stop the cell cycle?
- A It activates the APC/C, which destroys all cyclins
- B It phosphorylates Rb, releasing E2F
- C It binds and inhibits G1/S and S-phase cyclin–Cdk complexes
- D It cleaves cohesin so that replication cannot begin
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Answer: C. It binds and inhibits G1/S and S-phase cyclin–Cdk complexes
p21 is a Cdk inhibitor protein: by blocking G1/S and S-phase cyclin–Cdk activity, it keeps the cell from entering S phase while the DNA is repaired. Phosphorylating Rb and releasing E2F would do the opposite and push the cell into S phase.
12. In one diploid cell, a pair of homologous chromosomes fails to separate in meiosis I, and meiosis II then proceeds normally. How many of the four gametes have an abnormal chromosome number?
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Answer: All four: two have an extra chromosome (n + 1) and two are missing one (n − 1).
Both homologs go to the same daughter cell in meiosis I, so neither cell has a normal set, and meiosis II passes that error to all four gametes. Nondisjunction in meiosis II, by contrast, makes only two of the four gametes abnormal.
13. In the intrinsic pathway of apoptosis, what does cytochrome c do once it is released from mitochondria into the cytosol?
- A It binds Apaf1, forming an apoptosome that activates caspase-9
- B It binds Fas death receptors at the plasma membrane
- C It enters the nucleus and directly cuts the DNA
- D It inhibits Bax and Bak to stop further release
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Answer: A. It binds Apaf1, forming an apoptosome that activates caspase-9
Cytosolic cytochrome c binds Apaf1, which assembles into the apoptosome and activates the initiator caspase-9. Caspase-9 then activates executioner caspases that dismantle the cell. Fas receptors belong to the separate extrinsic pathway.
14. Paclitaxel (taxol) binds microtubules and prevents them from depolymerizing. In which stage do dividing cells treated with it most often arrest?
- A G1, before DNA replication
- B S phase, during DNA replication
- C G2, before chromosomes condense
- D Mitosis, before anaphase
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Answer: D. Mitosis, before anaphase
A working spindle depends on microtubules that can grow and shrink. With microtubules frozen in place, kinetochores cannot attach and come under tension properly, so the spindle assembly checkpoint holds the cell in mitosis.
15. What is the term for a normal gene, such as RAS, that can be turned into a cancer-promoting gene by a gain-of-function mutation in just one copy?
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Answer: Proto-oncogene
A proto-oncogene normally promotes growth under tight control, and a mutation that makes its product overactive turns it into an oncogene. For example, a Ras that cannot hydrolyze GTP stays switched on. Tumor suppressor genes, by contrast, usually need both copies lost.