Take the NUMS Biology — Cell Structure and Function (Set 6): 30 multiple-choice questions (40 minutes) with instant scoring and a full answer review.…
The first 5 questions of this test, with the correct answer and explanation.
Q1. A photomicrograph shows a cell 45 mm long. The cell's real length is 15 µm. What is the magnification of the photograph?
- A. ×30,000
- B. ×300
- C. ×3,000 ✓
- D. ×675
Answer: C. ×3,000
Explanation: Magnification = image size ÷ real size = 45 mm ÷ 0.015 mm = 3,000.
Q2. In an electron micrograph, a scale bar labelled 5 µm measures 25 mm. A nucleus in the same picture measures 60 mm across. What is the real diameter of the nucleus?
- A. 24 µm
- B. 5 µm
- C. 12 µm ✓
- D. 30 µm
Answer: C. 12 µm
Explanation: 1 mm on the picture = 5/25 = 0.2 µm, so 60 mm = 12 µm.
Q3. A cube-shaped cell of side 2 µm grows into a cube of side 4 µm. How does its surface area to volume ratio change?
- A. It rises from 3 to 6 (doubled)
- B. It falls from 6 to 1.5 (quartered)
- C. It falls from 3 to 1.5 (halved) ✓
- D. It stays at 6 because shape is unchanged
Answer: C. It falls from 3 to 1.5 (halved)
Explanation: Ratio = 6/side. For side 2 it is 3; for side 4 it is 1.5. Doubling the side halves the ratio.
Q4. Onion epidermal strips were placed in sucrose solutions of different molarity. About half the cells were plasmolysed in 0.35 M sucrose, while 0.30 M gave none and 0.40 M gave nearly all. What does 0.35 M estimate?
- A. The solute potential of the cell sap at incipient plasmolysis ✓
- B. The wall pressure at which the cells burst in the solution
- C. The pressure potential of the cell wall at the point of full turgor
- D. The water potential of pure water at the same temperature
Answer: A. The solute potential of the cell sap at incipient plasmolysis
Explanation: At incipient plasmolysis the pressure potential is about zero, so cell water potential equals its solute potential, matching the solution in which 50% of cells plasmolyse.
Q5. A plant cell has solute potential −1.0 MPa and pressure potential +0.6 MPa. It is placed in a large volume of solution with water potential −0.7 MPa. What happens?
- A. Cell water potential is −1.6 MPa, so water enters the cell
- B. Cell water potential is −0.4 MPa, so water enters the cell
- C. Cell water potential is −0.4 MPa, so water leaves the cell ✓
- D. Cell water potential is −1.6 MPa, so water leaves the cell
Answer: C. Cell water potential is −0.4 MPa, so water leaves the cell
Explanation: Ψ = Ψs + Ψp = −1.0 + 0.6 = −0.4 MPa, which is higher than −0.7 MPa. Water moves towards the lower (more negative) value, so it leaves the cell.
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Q6. Mouse and human cells were fused, and their surface proteins, labelled with different fluorescent dyes, mixed completely within about 40 minutes at 37 °C but not when cooled. This supports the idea that:
- A. membrane proteins are fixed to the cytoskeleton at all temperatures
- B. membrane proteins can move sideways
- C. the membrane is a rigid protein–lipid–protein sandwich
- D. membrane lipids flip quickly between the two layers
Q7. Cholesterol within an animal cell membrane helps cells survive both warm and cool conditions because it:
- A. makes the bilayer fully rigid at every temperature, whether warm or cool
- B. restrains phospholipids when warm and disrupts their packing when cool
- C. forms the main transport channel for water and ions across the bilayer
- D. binds the membrane tightly to the cell wall and the cytoskeleton
Q8. In red blood cells, the sodium–potassium pump moves 3 Na⁺ out and 2 K⁺ in per ATP hydrolysed. If 600 ATP molecules are used, what is the net number of positive charges moved out of the cell?
- A. 300
- B. 1,800
- C. 1,200
- D. 600
Q9. The uptake rate of molecule X into a cell rises linearly with concentration and never plateaus, while that of molecule Z rises and levels off at high concentration. Which conclusion is best?
- A. X moves by active transport and Z by osmosis through the membrane
- B. X moves by simple diffusion; Z uses a carrier of limited number
- C. Both use active transport with an unlimited number of carriers
- D. X uses a carrier protein and Z moves by simple diffusion across the bilayer
Q10. An animal tissue is homogenised and centrifuged first at low speed, then the supernatant at medium speed. Which organelle is mostly found in the pellet from the medium-speed spin?
- A. Ribosomes
- B. Nuclei
- C. Mitochondria
- D. Smooth endoplasmic reticulum fragments
Q11. Tissue for cell fractionation is homogenised in a cold, buffered, isotonic solution. What is the purpose of the isotonic condition?
- A. To stop organelles bursting or shrinking by osmosis
- B. To slow down enzyme activity and prevent autolysis
- C. To keep the pH constant while the tissue is ground
- D. To increase organelle density so they sediment faster
Q12. Cells are fed radioactive amino acids for 3 minutes and then given normal amino acids. The radioactive secretory protein is traced first to the rough ER, later to the Golgi, then to vesicles near the plasma membrane. What does this show?
- A. That the Golgi apparatus makes the polypeptide chains on its own
- B. That lysosomes synthesise all the secretory proteins on site
- C. The route taken by a secretory protein through the cell
- D. That proteins travel from the plasma membrane back into the ER
Q13. A drug raises the pH inside lysosomes from about 5 to 7 without damaging the enzymes themselves. What is the likely consequence?
- A. Acid hydrolases work poorly
- B. Lysosomal enzymes become more active at neutral pH
- C. Lysosomes burst and digest the whole cell at once
- D. Mitochondria take over digestion of engulfed material
Q14. Which cell would you expect to have the most peroxisomes?
- A. A mature phloem sieve tube element, which transports sugars
- B. A red blood cell of an adult mammal, which carries oxygen
- C. A liver cell, active in detoxification and fat breakdown
- D. A resting sperm cell stored in the epididymis, awaiting release
Q15. Chloramphenicol blocks translation on 70S ribosomes. Why might it have side effects on human cells even though they have 80S ribosomes?
- A. Human nuclei contain 70S ribosomes
- B. Chloramphenicol binds to cytoplasmic 80S ribosomes with high affinity
- C. Human 80S ribosomes are made of 70S plus 10S
- D. Mitochondria have 70S-type ribosomes that can be affected
Q16. A protein made in the cytoplasm normally functions inside the nucleus. A mutation deletes its nuclear localisation signal. Where will the mutant protein be found?
- A. In the nucleus, because pores admit all molecules that diffuse to them
- B. In the cytoplasm, since it cannot be recognised for nuclear import
- C. Inside the nucleolus, since nuclear pores do not require any signal sequence
- D. In the lysosomes, since all unsorted proteins are sent there by default
Q17. Cells making large quantities of protein, such as plasma cells, usually have:
- A. no nucleolus and many peroxisomes
- B. many chloroplasts and large vacuoles
- C. a prominent nucleolus and abundant rough ER
- D. a very small nucleolus and mostly smooth ER
Q18. Colchicine is added to a dividing culture. Many cells accumulate at metaphase. The most likely reason is that colchicine:
- A. stops microtubules assembling
- B. stops DNA replication from starting during S phase
- C. dissolves the nuclear envelope and keeps it dissolved
- D. breaks sister chromatids apart far too early in mitosis
Q19. Cytochalasin, which blocks actin polymerisation, is added to dividing animal cells. After mitosis, many cells contain two nuclei. This is because:
- A. nuclear envelopes could not reform around chromatin
- B. spindle microtubules failed to separate chromosomes
- C. DNA replicated twice without mitosis
- D. the actin contractile ring could not form
Q20. Some men with Kartagener syndrome are infertile and suffer repeated lung infections, due to missing dynein arms in cilia and flagella. What is the cellular explanation?
- A. The nuclear pore complexes are blocked in these cells
- B. Ribosomes cannot attach to the rough ER in these cells
- C. The Golgi apparatus cannot package enzymes in these cells
- D. Microtubule doublets cannot slide
Q21. A 70S ribosome comprises a 50S and a 30S subunit. Why do the numbers not add up to 80?
- A. S values are only valid for eukaryotic ribosomes, not prokaryotic ones
- B. One of the subunits is always partly lost during centrifugation
- C. S measures sedimentation rate, which depends on shape and mass
- D. The 70S value was measured at a different temperature from the subunits
Q22. An animal cell and a plant cell are placed in distilled water. The animal cell bursts and the plant cell does not. The main reason is that the plant cell has:
- A. a larger vacuole that is filled only with pure water
- B. a rigid cellulose wall that resists swelling, giving turgor
- C. active pumps that keep removing water from the cell
- D. a cell membrane that is completely impermeable to water, so none can enter
Q23. Cardiac muscle cells contract in a coordinated way because ions pass quickly from one cell to the next through:
- A. plasmodesmata in a cellulose wall
- B. hemidesmosomes attached to basement membrane
- C. gap junctions in intercalated discs
- D. tight junctions that seal the cells
Q24. During ripening, green tomato fruit turns red as chloroplasts in the cells change to:
- A. chromoplasts that accumulate carotenoid pigments
- B. leucoplasts that store lipids and no pigments at all
- C. amyloplasts that store starch granules in the fruit
- D. etioplasts that lack pigments, as in dark-grown shoots
Q25. Paramecium in pond water has a contractile vacuole that pulses 12 times per minute. When placed in a more concentrated salt solution, the rate falls. The best explanation is that:
- A. less water enters by osmosis, so less needs to be expelled
- B. more water enters by osmosis, so less can be expelled at once
- C. the vacuole membrane has lost its ability to pump
- D. the vacuole is storing salts instead of water
Q26. A secretory protein is made by ribosomes on the rough ER. Which order describes its path before leaving the cell?
- A. Rough ER → vesicle → cis Golgi → trans Golgi → secretory vesicle
- B. Rough ER → peroxisome → trans Golgi → cis Golgi → secretory vesicle
- C. Rough ER → trans Golgi → cis Golgi → lysosome → plasma membrane
- D. Smooth ER → nucleus → cis Golgi → secretory vesicle
Q27. With a light microscope whose resolution limit is about 0.2 µm, which of these cannot be seen as a separate structure?
- A. A nucleus, about 10 µm across
- B. A chloroplast, about 5 µm long
- C. A mitochondrion, about 1 µm wide
- D. A ribosome, about 25 nm across
Q28. Hooke observed box-like compartments in thin slices of cork in 1665. What was he actually looking at?
- A. Bacterial colonies in the bark
- B. Chloroplasts in the cell sap
- C. Living cells with visible nuclei
- D. Empty walls of dead plant cells
Q29. In familial hypercholesterolaemia, cells lack functional LDL receptors. Why does blood cholesterol rise?
- A. LDL is not taken in by endocytosis
- B. Cholesterol cannot be synthesised in the liver at all
- C. LDL is destroyed in lysosomes much too quickly
- D. Cells overproduce the LDL receptors on their surface
Q30. A technician treats isolated membranes with a high-salt wash, and some proteins are released. Others are released only after adding detergent. What does this show?
- A. Salt-released proteins are integral; detergent-released ones are peripheral
- B. Salt dissolves the phospholipids, while detergent dissolves only the proteins
- C. Salt-released proteins are peripheral; detergent-released are integral
- D. Both groups are glycolipids