Biochemistry End-of-Year 2011 (2) — Past Paper OmpathStudy

Biochemistry EOY 2011 - Clean Past Paper Complete, structured transcription of all 10 source pages. The original paper carries no printed answer key — the model answers below have been composed from standard biochemistry teaching to fill that gap. Section C statements are of the "identify which are true/false" type; each question gives one overall Answer line naming the true statement(s), then explains every option. Section A - Structured questions (Short/Long Answer) Question 1 Outline the criteria used to classify compounds as neurotransmitters; mention three ways in which neurotransmitter action is terminated; and indicate the possible effect of premature degradation of acetylcholine by acetylcholinesterase. Model answer: Criteria for classifying a substance as a neurotransmitter: (1) it is synthesised and stored in the presynaptic neuron; (2) it is released into the synaptic cleft on depolarisation/Ca²⁺ influx in amounts sufficient to affect the postsynaptic cell; (3) exogenous application reproduces the same effect as the endogenously released substance; (4) specific postsynaptic receptors exist for it; (5) a defined mechanism exists for removing it from the synapse. Three ways neurotransmitter action is terminated: enzymatic degradation in the cleft (e.g. acetylcholinesterase hydrolysing acetylcholine); re-uptake into the presynaptic terminal by specific transporters (e.g. dopamine, noradrenaline, serotonin transporters); simple diffusion away from the synaptic cleft. Effect of premature degradation of acetylcholine by acetylcholinesterase: the postsynaptic response is cut short/weakened because ACh is cleared before it can fully activate its receptors — this reduces cholinergic transmission (e.g. reduced muscle contraction at the neuromuscular junction). This is the opposite problem to AChE inhibition (organophosphate poisoning), where ACh accumulates and over-stimulates the synapse. --- Question 2 Derive the Henderson-Hasselbalch equation from a hypothetical weak acid HA; calculate the amount of sodium ethanoate required in 1,000 mL of 0.01 M acetic acid to make a buffer of pH 5; and briefly state the importance of buffers in the human body. Model answer: Derivation: for HA ⇌ H⁺ + A⁻, Ka = [H⁺][A⁻]/[HA]. Rearranging, [H⁺] = Ka·[HA]/[A⁻]. Taking negative logs: −log[H⁺] = −log Ka − log([HA]/[A⁻]), i.e. pH = pKa + log([A⁻]/[HA]) — the Henderson-Hasselbalch equation. Calculation (using pKa of acetic acid ≈ 4.76): 5 = 4.76 + log([A⁻]/[HA]) → log([A⁻]/[HA]) = 0.24 → [A⁻]/[HA] ≈ 1.74. With [HA] = 0.01 mol in 1 L, [A⁻] (sodium ethanoate) needed ≈ 0.0174 mol. Mass = 0.0174 mol × 82 g/mol (molar mass of sodium acetate) ≈ 1.4 g of sodium ethanoate. Importance of buffers in the body: they resist changes in pH caused by metabolic acid/base production, keeping blood and intracellular pH within the narrow physiological range (≈7.35–7.45) needed for normal enzyme activity and protein structure — e.g. the bicarbonate, phosphate and protein buffer systems. --- Question 3 Describe in full the four enzymatic steps leading to a two-carbon-shorter palmitoyl-CoA. Model answer: One round of fatty acid β-oxidation removes two carbons via four enzymatic steps: (1) Oxidation — acyl-CoA dehydrogenase (FAD-linked) introduces a trans-Δ² double bond, forming trans-Δ²-enoyl-CoA and FADH₂. (2) Hydration — enoyl-CoA hydratase adds water across the double bond to give L-3-hydroxyacyl-CoA. (3) Oxidation — 3-hydroxyacyl-CoA dehydrogenase (NAD⁺-linked) oxidises the hydroxyl to a keto group, forming 3-ketoacyl-CoA and NADH. (4) Thiolytic cleavage — thiolase (β-ketothiolase) uses a fresh CoA-SH to cleave off acetyl-CoA, leaving an acyl-CoA two carbons shorter than the starting molecule (e.g. palmitoyl-CoA → myristoyl-CoA + acetyl-CoA). --- Question 4 Derive the Michaelis-Menten equation and explain the significance of the Km of an enzyme. Model answer: For E + S ⇌ ES → E + P, applying the steady-state assumption (rate of ES formation = rate of ES breakdown) gives v₀ = Vmax[S] / (Km + [S]), where Km = (k₋₁ + k₂)/k₁. Significance of Km: it is the substrate concentration at which the reaction proceeds at half its maximum velocity (v₀ = Vmax/2). A low Km indicates high apparent affinity of the enzyme for its substrate (half-maximal rate reached at low [S]); a high Km indicates low affinity. Km is used to compare an enzyme's affinity for different substrates and is independent of enzyme concentration. --- Question 5 State four functions of nucleotides in humans with one example each; draw and label a cytosolic ribosome; explain why the genetic code is degenerate; state permitted wobble base-pairing; and complete the supplied transcription/translation sequence table. Model answer: Four functions of nucleotides: (1) energy currency — ATP powers cellular reactions; (2) intracellular signalling — cyclic AMP (cAMP) acts as a second messenger; (3) structural monomers of nucleic acids — dAMP/dGMP/dCMP/dTMP build DNA; (4) coenzyme components — NAD⁺, F