Take the NUMS Biology — Cell Structure and Function (Set 7): 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 cell is 30 µm long. In a micrograph it measures 6 cm. What is the magnification?
- A. ×200
- B. ×2000 ✓
- C. ×1800
- D. ×20 000
Answer: B. ×2000
Explanation: Convert 6 cm to 60 000 µm; magnification = image size ÷ actual size = 60 000 ÷ 30 = 2000.
Q2. A micrograph has a scale bar 2 cm long labelled 5 µm. A bacterium on the same micrograph measures 1.2 cm. What is its actual length?
- A. 6 µm
- B. 12 µm
- C. 2 µm
- D. 3 µm ✓
Answer: D. 3 µm
Explanation: Each cm represents 5 ÷ 2 = 2.5 µm, so 1.2 cm is 1.2 × 2.5 = 3 µm.
Q3. Two structures lie 0.1 µm apart. The resolving power of a good light microscope is about 0.2 µm. What will be seen?
- A. Two separate objects, because stains improve the resolving power
- B. One object, because light cannot pass through a thin specimen
- C. One blurred object, as the points are within the resolution limit ✓
- D. Two separate objects, because the magnification is high enough
Answer: C. One blurred object, as the points are within the resolution limit
Explanation: Resolution is the smallest distance at which two points are seen as separate. Points 0.1 µm apart are below 0.2 µm, so they merge regardless of magnification.
Q4. A researcher wants to study the three-dimensional surface of a pollen grain. Which instrument is the best choice?
- A. A light microscope with phase contrast, which shows living cells
- B. A scanning electron microscope, which scatters electrons from the surface ✓
- C. A light microscope with Gram stain, which shows wall thickness
- D. A transmission electron microscope, which passes electrons through thin slices
Answer: B. A scanning electron microscope, which scatters electrons from the surface
Explanation: A scanning electron microscope gives a 3-D view of surfaces, whereas a transmission electron microscope shows internal ultrastructure of thin sections.
Q5. Two cubical cells have sides of 10 µm and 20 µm. What is the surface area to volume ratio of the larger cell, compared with the smaller?
- A. 0.6 µm⁻¹, which is equal to that of the smaller cell
- B. 0.3 µm⁻¹, which is half that of the smaller cell ✓
- C. 0.6 µm⁻¹, which is double that of the smaller cell
- D. 0.3 µm⁻¹, which is a quarter of the smaller cell
Answer: B. 0.3 µm⁻¹, which is half that of the smaller cell
Explanation: Ratio = 6/side. For 10 µm it is 0.6 and for 20 µm it is 0.3 µm⁻¹, half as large. Larger cells exchange materials less efficiently per unit volume.
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Q6. A plant cell has a solute (osmotic) potential of −0.9 MPa and a pressure potential of +0.4 MPa. What is its water potential?
- A. +0.5 MPa
- B. −0.9 MPa
- C. −0.5 MPa
- D. −1.3 MPa
Q7. Cell X has a water potential of −0.5 MPa. Adjacent cell Y has a solute potential of −1.1 MPa and a pressure potential of +0.3 MPa. What happens?
- A. No net movement, because both cells are turgid
- B. Water moves from X to Y, because X has the higher pressure potential and so a stronger push
- C. Water moves from X to Y as Y has the lower (more negative) potential
- D. Water moves from Y to X, because Y has the higher solute concentration and the greater tendency to lose water
Q8. Potato cylinders of 2.0 g are placed in sucrose solutions for 3 hours. Mass after: 0.1 M gives 2.3 g, 0.3 M gives 2.0 g, 0.5 M gives 1.7 g. What was the percentage change in mass in 0.5 M sucrose, and what does 0.3 M indicate?
- A. −15%, and 0.3 M is roughly isotonic with the cell sap
- B. −15%, and 0.3 M is hypotonic to the cell sap
- C. −17.6%, and 0.3 M is roughly isotonic with the cell sap
- D. −0.3%, and 0.3 M is roughly isotonic with the cell sap
Q9. Onion epidermal cells in a sucrose solution show a clear gap between the shrunken protoplast and the cell wall. What fills the gap?
- A. The external sucrose solution, which passes through the wall
- B. Air, which has entered through the stomata of the epidermis
- C. Pure water, which has leaked out of the shrinking vacuole
- D. Cytoplasm, which has been squeezed out of the shrinking cell
Q10. A Paramecium is moved from pond water to a more dilute medium. What change is expected in its contractile vacuole?
- A. It discharges less frequently, because less water enters by osmosis
- B. It stops, because the cell wall resists water entry
- C. It discharges more frequently, because more water enters by osmosis
- D. It discharges at the same rate, because salts are pumped out
Q11. Each cycle of the sodium–potassium pump moves 3 Na⁺ out and 2 K⁺ in using one ATP. How many ions move in each direction when 30 ATP molecules are used?
- A. 90 Na⁺ out and 90 K⁺ in
- B. 90 Na⁺ out and 60 K⁺ in
- C. 30 Na⁺ out and 20 K⁺ in
- D. 60 Na⁺ out and 90 K⁺ in
Q12. The uptake of a sugar by a cell rises with external concentration, then levels off. Adding cyanide, which stops ATP production, does not change the uptake. The sugar enters by:
- A. facilitated diffusion, as the plateau shows carriers are saturated
- B. active transport, where the plateau shows that the ATP supply is saturated
- C. osmosis, where the plateau shows equal water potential
- D. simple diffusion, where the plateau shows the membrane is full
Q13. In familial hypercholesterolaemia, blood LDL cholesterol is very high. Which cellular defect explains this?
- A. Defective sodium pumps, so cholesterol cannot leave the cell at all
- B. Defective lysosomes, so cholesterol is made in excess of need
- C. Defective ribosomes, so LDL particles cannot be assembled in cells
- D. Defective LDL receptors, so receptor-mediated endocytosis of LDL fails
Q14. In Pompe disease a lysosomal enzyme that breaks down glycogen is missing, and glycogen builds up in muscle cells. Which statement best explains why lysosomes are involved?
- A. Lysosomes synthesise glycogen from free glucose in the cell
- B. Lysosomes transport glycogen out through the cell membrane
- C. Lysosomes release ATP from glycogen by aerobic oxidation
- D. Lysosomes contain hydrolytic enzymes that digest large molecules
Q15. Catalase in peroxisomes converts hydrogen peroxide as 2H₂O₂ → 2H₂O + O₂. What mass of oxygen forms from 34 g of H₂O₂? (H₂O₂ = 34 g/mol, O₂ = 32 g/mol)
- A. 32 g
- B. 16 g
- C. 17 g
- D. 8 g
Q16. Cells are given radioactive amino acids for a few minutes and then moved to non-radioactive medium. The radioactivity is found first in the rough ER, then in the Golgi, then in vesicles at the cell surface. This sequence shows the route of:
- A. a cytoskeleton protein being made on free ribosomes
- B. a lipid being made in the smooth ER and moved to the nucleus
- C. a secretory protein moving through the endomembrane system
- D. a mitochondrial protein moving from the Golgi to the matrix
Q17. Hepatocytes of a person taking a drug for several weeks show more smooth ER, and the person needs higher doses for the same effect. Which role of smooth ER does this show?
- A. Synthesis of proteins for the drug's receptors in the cell
- B. Packaging of drugs into secretory vesicles for export
- C. Storage of the drug as a lipid droplet in the cytosol
- D. Detoxification of drugs by enzymes in the smooth ER membrane
Q18. Chloramphenicol blocks protein synthesis on 70S ribosomes but not 80S ribosomes. In a eukaryotic cell treated with it, which protein synthesis would be affected most directly?
- A. Synthesis in the nucleolus, which has 70S ribosomes
- B. Synthesis inside mitochondria, which have 70S ribosomes
- C. Synthesis on the rough ER, which has 70S ribosomes
- D. Synthesis in the cytosol, which has 70S ribosomes
Q19. The 30S and 50S subunits join to form a bacterial ribosome. What is its sedimentation coefficient, and why?
- A. 20S, because the smaller value is subtracted from the larger one
- B. 70S, because S values reflect sedimentation rate, not simple mass
- C. 80S, because the two subunit values are simply added together
- D. 40S, because the two subunit values are averaged to one value
Q20. A cell loses its nucleolus. Which activity will eventually fail?
- A. Transcription of mRNA, so no genes are expressed in the cell
- B. Formation of the nuclear envelope after division
- C. Replication of DNA, so the cell can no longer divide at all
- D. Assembly of ribosomal subunits, so protein synthesis declines
Q21. Dividing cells are treated with colchicine, which blocks microtubule assembly. Other cells are treated with cytochalasin, which blocks actin filaments. What is expected?
- A. Both cell types stall in prophase with a broken nuclear envelope
- B. Colchicine cells stall in metaphase and cytochalasin cells become binucleate
- C. Colchicine cells become binucleate and cytochalasin cells stall in metaphase
- D. Both cell types complete division but make no spindle fibres
Q22. In Kartagener syndrome, the dynein arms of cilia are missing. Which set of problems is most likely?
- A. Muscle weakness and anaemia, because red cells cannot divide properly
- B. Poor vision and deafness, because neurons lack their myelin sheath
- C. Repeated chest infections and male infertility, as cilia cannot beat
- D. Fragile bones and slow clotting, because collagen is faulty in the body
Q23. A centriole is built of nine triplets of microtubules. How many microtubules are there in one centrosome containing two centrioles?
Q24. A plant tissue is treated with pectinase and the cells separate easily. Which structure was digested?
- A. The primary wall, which is made of cellulose microfibrils
- B. The plasmodesmata, which contain smooth ER
- C. The tonoplast, which surrounds the vacuole
- D. The middle lamella, which cements neighbouring cell walls
Q25. A fluorescent dye injected into one cardiac muscle cell soon appears in its neighbours. Which junction allows this?
- A. Tight junctions, which seal the space between the cells
- B. Gap junctions, which form channels joining adjacent cells
- C. Plasmodesmata, which link animal cells to the wall
- D. Desmosomes, which anchor cells with intermediate filaments
Q26. A potato tuber is exposed to light for several days and the surface turns green. Which change has occurred?
- A. Amyloplasts have developed into chloroplasts
- B. Amyloplasts have developed into leucoplasts
- C. Chromoplasts have developed into amyloplasts
- D. Chloroplasts have developed into chromoplasts
Q27. Mouse and human cells are fused. The membrane proteins labelled with different fluorescent dyes are at first separated, but they mix completely within about 40 minutes. This experiment supports:
- A. the presence of a protein layer on both sides of a lipid bilayer
- B. the rigid nature of the membrane, in which proteins are fixed
- C. the transfer of proteins across the membrane by flip-flop
- D. the fluid nature of the membrane, in which proteins move laterally
Q28. Which feature of the plasma membrane explains why the A, B and O antigens are found only on the outer surface of red blood cells?
- A. Phospholipids flip-flop rapidly between the two faces of the bilayer
- B. Cholesterol has its hydroxyl group facing the cytoplasm side only
- C. Glycolipids and glycoproteins carry their sugar chains on the outer face
- D. Integral proteins span the membrane with both ends on the outside
Q29. A protein with a signal peptide is made by ribosomes in the cytosol. The signal peptide directs the ribosome to attach to the rough ER. This protein is most likely to be:
- A. destined for the matrix of the mitochondrion
- B. destined for secretion or for the cell membrane
- C. destined for the nucleus through a nuclear pore
- D. destined to stay free in the cytosol
Q30. Epithelial cells lining the small intestine are joined near their upper (lumen) ends by tight junctions. What is their main function?
- A. They allow cytoplasm to flow freely from cell to cell
- B. They carry electrical signals from nerve cells to the lining
- C. They anchor the cells together against mechanical stress
- D. They prevent substances from leaking between cells into the body