IB Biology HL Theme C: Interaction and interdependence questions

Answer: A. Induced-fit binding involves a change in shape of both the enzyme and the substrate. For P, both shape-change scores are positive, so both molecules change shape during complex formation.
B: uses the enzyme’s positive score but overlooks that the substrate’s score is zero.
C: uses the substrate’s positive score but overlooks that the enzyme’s score is zero.
D: shows no shape change in either molecule and therefore is not induced-fit binding.
Remember: induced-fit binding involves shape changes in both interacting molecules.

Answer: D. Oxidative decarboxylation of 3C pyruvate leaves a 2C acetyl group. CoA transfers it to 4C oxaloacetate, forming 6C citrate; Krebs cycle oxidations and decarboxylations yield ATP and reduced NAD while regenerating oxaloacetate. A: gives the link reaction and citrate formation but stops before stating the cycle’s outputs and regeneration of oxaloacetate. B: reverses the redox direction; pyruvate is oxidized, not reduced. C: treats oxaloacetate as a net reactant, rather than a molecule regenerated by the cycle. Remember: acetyl groups enter the Krebs cycle by combining with oxaloacetate, which is then regenerated.

Answer: C. Only mixture 3 formed TP, and it contained both ATP and reduced NADP. This matches the conversion of GP into TP, which uses ATP and reduced NADP.

A: assumes ATP alone is sufficient; mixture 1 formed no TP. B: assumes reduced NADP alone is sufficient; mixture 2 formed no TP. D: selects the mixture without either compound, which formed no TP.

Remember: GP conversion to TP requires both ATP and reduced NADP.

Answer: D. Epinephrine binds its transmembrane receptor, which activates a G protein. The activated G protein stimulates adenylate cyclase to produce cAMP inside the cell; cAMP acts as the second messenger that relays the signal to intracellular targets.
A: omits the G protein, incorrectly making the receptor activate adenylate cyclase directly.
B: gives the pathway correctly but misidentifies cAMP as the first messenger; epinephrine is the first messenger.
C: reverses the order of G protein and adenylate cyclase activation.
Remember: the first messenger binds the receptor; the second messenger relays the signal inside the cell.

Answer: C. ATP supplies energy for sodium–potassium pumping, which establishes and maintains ion concentration gradients. With less ATP, pump activity decreases; over time, the gradients weaken and the resting potential becomes less negative.

A: predicts weaker gradients but reverses their effect on the resting potential. B: assumes the pump becomes more active despite reduced ATP availability. D: treats the pump as dissipating, rather than establishing and maintaining, concentration gradients.

Remember: ATP-driven pumping maintains the ion gradients that support the resting potential.

Answer: B. Increased blood CO₂ can lower pH. Brainstem chemoreceptors respond to the pH change by signalling the diaphragm and intercostal muscles to increase ventilation; greater CO₂ removal then helps pH recover and reduces the stimulus.

A: stops at the initial pH change and omits the feedback described in III.
C: reverses the response in II and omits the supported CO₂–pH relationship in I.
D: includes II, which incorrectly says ventilation decreases when pH falls.
Remember: increased ventilation removes CO₂, helping blood pH rise.

Answer: D. Antibiotic use selects for resistant bacteria because susceptible bacteria are more likely to be killed. With less unnecessary antibiotic use, the selective advantage of multiresistant bacteria is reduced, slowing their increase in frequency.

A: Treats antibiotics as directly causing resistance mutations rather than selecting bacteria that already have resistance.
B: Confuses reduced selection with blocked inheritance; resistant bacteria can still pass resistance to offspring.
C: Assumes resistance disappears without antibiotics; reduced use does not itself remove resistance.
Remember: antibiotics select among bacterial variants; they do not direct bacteria to become resistant.

Answer: A. The Fischer projection shows an aldehyde at C1 and the characteristic OH arrangement of D-glucose. As population density rises with a fixed supply of the only limiting resource, competition reduces the amount available per individual and slows population growth, tending to return the population towards carrying capacity.

B: reverses the effect of increasing density on the resource available per individual. C: misidentifies glucose as fructose; the displayed structure has a terminal aldehyde group. D: treats a fixed resource supply as density-independent, although its availability per individual changes with population density.

Remember: density-dependent factors tend to counter changes in population size relative to carrying capacity.

Answer: D. Heat is released at both stages shown, so add the heat outputs for each cell type: autotrophic cells, 40 + 25 = 65 kJ; heterotrophic cells, 30 + 20 = 50 kJ per 100 kJ input.

A: counts only the heat released during cell respiration. B: counts only the heat released when ATP is used. C: gives the totals but reverses the cell-type labels.

Remember: energy transfers are not 100% efficient, including when ATP is produced and when it is used.