NUMS — Medical & Dental Entry Test

NUMS Biology — Bioenergetics (Set 6)

Take the NUMS Biology — Bioenergetics (Set 6): 30 multiple-choice questions (40 minutes) with instant scoring and a full answer review. Covers 30 NUMS…

What this test covers

30 NUMS Biology

Sample questions with answers

The first 5 questions of this test, with the correct answer and explanation.

Q1. Counting one glucose molecule fully oxidised through glycolysis, the link reaction and the Krebs cycle (NADH and FADH₂ before the electron transport chain), how many NADH and FADH₂ are formed in total?
- A. 6 NADH and 2 FADH₂
- B. 8 NADH and 4 FADH₂
- C. 10 NADH and 2 FADH₂ ✓
- D. 10 NADH and 4 FADH₂
Answer: C. 10 NADH and 2 FADH₂
Explanation: Glycolysis 2 NADH; link reaction 2 NADH; Krebs cycle (two turns) 6 NADH and 2 FADH₂. Total 10 NADH and 2 FADH₂.
Q2. Germinating seeds in a closed respirometer consume 20 mL of oxygen and release 14 mL of carbon dioxide in an hour. What is the RQ and the likely substrate?
- A. RQ 1.4, suggesting mostly carbohydrate
- B. RQ 0.7, suggesting mostly lipid ✓
- C. RQ 0.7, suggesting mostly carbohydrate
- D. RQ 1.4, suggesting mostly lipid
Answer: B. RQ 0.7, suggesting mostly lipid
Explanation: RQ = CO₂ produced ÷ O₂ used = 14/20 = 0.7, typical of fat oxidation (carbohydrate gives 1.0).
Q3. Isolated chloroplasts are mixed with blue DCPIP and illuminated. Which tube turns colourless?
- A. Chloroplasts in the dark, because DCPIP is reduced by the Calvin cycle enzymes
- B. Chloroplasts in the light, because DCPIP is reduced by electrons from water ✓
- C. Boiled chloroplasts in the light, because boiling releases electrons to DCPIP
- D. Chloroplasts in the light with DCMU added, because DCMU adds electrons
Answer: B. Chloroplasts in the light, because DCPIP is reduced by electrons from water
Explanation: DCPIP replaces NADP⁺ as an electron acceptor in the Hill reaction. It decolourises only when functional chloroplasts are illuminated.
Q4. Engelmann placed a filamentous alga in a drop containing aerobic bacteria and lit it with a spectrum of colours. Bacteria clustered mostly in the blue and red regions. What does this show?
- A. Chloroplasts fluoresce red under blue light only, which attracts bacteria
- B. Green light provides the most energy for photosynthesis in algae
- C. Bacteria prefer green light for their own growth and movement
- D. Photosynthesis is greatest in wavelengths that chlorophylls absorb best ✓
Answer: D. Photosynthesis is greatest in wavelengths that chlorophylls absorb best
Explanation: Oxygen released (which attracts aerobic bacteria) is highest where photosynthetic pigments absorb most, in the blue and red parts.
Q5. A leaf extract is run by paper chromatography in a non-polar solvent. The pigment travelling furthest is carotene. This is because carotene is:
- A. the most polar pigment, so it sticks to the paper least firmly
- B. the only pigment that does not absorb light
- C. the most non-polar pigment, so it dissolves best in the solvent ✓
- D. the heaviest pigment, so it is carried upwards fastest by the solvent
Answer: C. the most non-polar pigment, so it dissolves best in the solvent
Explanation: Pigments that are least polar are most soluble in the non-polar solvent and least attracted to the polar paper, so they move furthest.

The other 25 questions

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Q6. A steadily illuminated algal culture with plenty of CO₂ is suddenly deprived of CO₂. What happens to the levels of RuBP and PGA shortly afterwards?
- A. RuBP falls and PGA rises
- B. Both RuBP and PGA rise
- C. Both RuBP and PGA fall
- D. RuBP rises and PGA falls
Q7. In the same culture, steady light is switched off while CO₂ is still supplied. What happens to RuBP and PGA?
- A. PGA rises and RuBP falls
- B. PGA falls and RuBP rises
- C. Both PGA and RuBP rise
- D. Both PGA and RuBP fall
Q8. A leaf in darkness releases 2 units of CO₂ per hour. At a certain light intensity the net uptake of CO₂ is 6 units per hour. What is the gross photosynthetic rate in CO₂ units per hour, assuming respiration is unchanged?
- A. 6
- B. 8
- C. 12
- D. 4
Q9. The rate of photosynthesis plateaus as light intensity is increased. When the temperature is raised by 10 °C (still below denaturation) the rate rises again. Which factor was limiting at the plateau?
- A. Temperature, because enzyme-catalysed steps limited the rate
- B. Chlorophyll content, which fell as temperature rose in the leaf
- C. Light intensity, because it was still too low at the plateau
- D. Oxygen concentration, because it was too high for the cell
Q10. Maize (C4) and wheat (C3) are sealed together in a transparent chamber in sunlight. As CO₂ in the chamber falls, which statement is correct?
- A. Wheat continues to fix CO₂ at a lower concentration than maize can
- B. Maize continues to fix CO₂ at a lower concentration than wheat can
- C. Neither plant is able to photosynthesise at low CO₂
- D. Both stop at the same CO₂ concentration
Q11. A CAM plant such as a cactus is tested at dawn and at dusk. Which result is expected for the acidity (malic acid content) of its leaf cell sap?
- A. Highest at dusk, because malate is made during the day
- B. Highest at dawn, because malate was stored overnight
- C. Equal at dawn and dusk, because stomata stay shut
- D. Lowest at dawn, because CO₂ is released at night
Q12. Which conditions in a C3 leaf most favour photorespiration?
- A. Cool and bright, with high CO₂ and low internal O₂
- B. Hot, bright and dry, with stomata closed and high O₂
- C. Cool, dim and humid, with stomata wide open all day
- D. Dark and warm, with high CO₂ in the leaf air spaces
Q13. The uncoupler 2,4-dinitrophenol makes the inner mitochondrial membrane leaky to H⁺. In a respiring mitochondrial suspension, what is expected?
- A. Oxygen use stops completely while ATP synthesis rises sharply
- B. The Krebs cycle runs in reverse and glucose is formed
- C. Oxygen use continues, ATP synthesis falls, heat is released
- D. Oxygen use and ATP synthesis both rise to higher levels
Q14. In respiring mitochondria, cyanide blocks cytochrome oxidase. Which statement is correct about the carriers of the electron transport chain upstream of the block?
- A. They become more active, and the Krebs cycle speeds up
- B. They remain mostly oxidised, and NAD⁺ builds up
- C. They are unaffected, and ATP continues to be made
- D. They remain mostly reduced, and NADH builds up
Q15. Oligomycin blocks ATP synthase but not electron carriers. In isolated mitochondria with a substrate and oxygen, what is the immediate effect?
- A. Proton gradient collapses and O₂ use rises to a higher rate
- B. Proton gradient rises; electron flow and O₂ use slow down
- C. NADH is oxidised faster than before, with more O₂ used
- D. ATP production rises with no change in O₂ use or heat
Q16. In an acid-bath experiment, isolated thylakoids were soaked at pH 4 until the lumen was acidic, then moved to pH 8 with ADP and phosphate in darkness. ATP formed. This supports:
- A. the idea that a proton gradient alone drives ATP synthesis
- B. the idea that ATP is made by the Calvin cycle in the dark
- C. the idea that NADPH is needed to drive ATP synthesis
- D. the idea that light directly phosphorylates ADP in the stroma
Q17. Plants are given water labelled with ¹⁸O, and carbon dioxide containing only ordinary oxygen. Which product is expected to contain the heavy isotope first?
- A. Carbon dioxide released
- B. Glucose formed
- C. Starch stored
- D. Oxygen gas released
Q18. The herbicide DCMU blocks electron flow out of photosystem II. Which process can still occur in the illuminated chloroplast?
- A. Photolysis of water with release of O₂
- B. Reduction of NADP⁺ to NADPH using water-derived electrons
- C. Fixation of CO₂ at normal rate
- D. Cyclic photophosphorylation around photosystem I
Q19. Yeast growing in an airtight flask with no oxygen converts pyruvate into ethanol. What is the main metabolic purpose of this step?
- A. To regenerate NAD⁺
- B. To make acetyl CoA for the Krebs cycle
- C. To remove oxygen from the medium
- D. To produce more ATP through oxidative phosphorylation
Q20. 90 g of glucose (M = 180) is fermented completely by yeast to ethanol and CO₂. Using masses of 44 g per mole of CO₂ and 46 g per mole of ethanol, what masses of CO₂ and ethanol are formed?
- A. 44 g CO₂ and 92 g ethanol
- B. 88 g CO₂ and 92 g ethanol
- C. 22 g CO₂ and 23 g ethanol
- D. 44 g CO₂ and 46 g ethanol
Q21. Malonate, a competitive inhibitor of succinate dehydrogenase, is added to respiring mitochondria. What is the expected result?
- A. Succinate accumulates and oxygen use falls
- B. Fumarate accumulates and oxygen use rises sharply
- C. Citrate falls and oxygen use is unchanged
- D. Malate accumulates and ATP use rises
Q22. A patient with severe thiamine (vitamin B₁) deficiency has high blood pyruvate. Which step is most likely impaired?
- A. Glucose → glucose-6-phosphate, a step needing ATP and hexokinase
- B. Oxaloacetate → malate, a step needing a niacin coenzyme
- C. Pyruvate → acetyl CoA, a step needing a thiamine coenzyme
- D. Acetyl CoA → citrate, a step needing oxaloacetate and water
Q23. In autumn the green of a deciduous leaf fades and yellow and orange appear. This is because:
- A. xanthophylls are changed back into chlorophyll
- B. carotenoids are newly made in very large amounts in autumn
- C. chlorophyll breaks down, unmasking the carotenoids present
- D. chloroplasts all convert into red amyloplasts in the cell
Q24. Thylakoids from illuminated chloroplasts are transferred in the dark to a mixture of stroma enzymes, CO₂ and RuBP. Sugar is formed. This shows that:
- A. Calvin cycle enzymes use ATP and NADPH made by thylakoids
- B. the Calvin cycle can make its own ATP and NADPH
- C. light is needed to fix carbon dioxide directly in the stroma
- D. thylakoids themselves contain the enzyme Rubisco
Q25. Using the standard Calvin cycle stoichiometry (18 ATP and 12 NADPH per glucose made), how many ATP are needed to make 3 glucose molecules?
- A. 36
- B. 18
- C. 54
- D. 72
Q26. A 16-carbon saturated fatty acid is broken down completely to acetyl CoA by β-oxidation. How many acetyl CoA molecules are formed?
- A. 7
- B. 16
- C. 8
- D. 4
Q27. A shade plant and a sun plant are compared. The shade plant reaches its light compensation point at a lower light intensity. This means:
- A. it has a lower respiration rate than a sun plant at all times
- B. it can achieve net CO₂ gain at lower light than the sun plant
- C. it has a higher maximum rate of photosynthesis
- D. it needs more light than the sun plant to break even
Q28. An aquatic plant is 10 cm from a lamp and gives 20 bubbles per minute. Assuming only light limits the rate and intensity ∝ 1/distance², how many bubbles per minute are expected at 20 cm?
- A. 5
- B. 2.5
- C. 10
- D. 40
Q29. In a respirometer, a tube of boiled, killed peas is set up beside the tube with living peas. The purpose of this control is:
- A. to correct for changes in temperature and air pressure
- B. to remove oxygen from the apparatus before the start of the test
- C. to show that dead peas respire faster than live ones
- D. to supply additional carbon dioxide to the apparatus
Q30. In an actively respiring mitochondrion, in which compartment is the proton concentration highest, and in which direction do protons flow through ATP synthase?
- A. Matrix; from the cytosol across into the intermembrane space
- B. Intermembrane space; flowing from there into the matrix
- C. Matrix; from the matrix out into the intermembrane space
- D. Intermembrane space; from the matrix out into the cytosol