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: 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: 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.