Biochemistry End-of-Year 2014 (2) — Past Paper OmpathStudy

Bchem EOY 2014 (2) — Past Paper Questions & Answers University of Nairobi, Year One Examinations 2013/2014 (August 5, 2014) — HBC 100/UPC 100/VBC 100: Biochemistry. This paper carries no printed answer key; model answers below are written from standard biochemistry references, not copied from a marking scheme. --- Section A: Essay Questions (Attempt any THREE) — 30 Marks Question 1 The Henderson-Hasselbalch equation is important in the calculation of a pH of a buffer solution. (a) With a specific, physiological example, explain the term "buffer solution". [2] (b) Derive the Henderson-Hasselbalch equation using a hypothetical weak acid HA. [3] (c) How many grams of acetic acid (CH₃CO₂H) and sodium acetate are needed to make up to one litre of a 50 mM acetate buffer with a pH of 5.0? Ka = 1.76×10⁻⁵. [4] Model answer: (a) A buffer solution resists changes in pH upon addition of small amounts of acid or base. Physiological example: the bicarbonate buffer system (H₂CO₃/HCO₃⁻) keeps blood pH stable near 7.4 despite ongoing metabolic acid production. (b) For HA ⇌ H⁺ + A⁻: Ka = [H⁺][A⁻]/[HA] → [H⁺] = Ka·[HA]/[A⁻] → taking -log of both sides: pH = pKa + log([A⁻]/[HA]) . (c) pKa = -log(1.76×10⁻⁵) ≈ 4.75. Using pH = pKa + log([A⁻]/[HA]): 5.0 = 4.75 + log([A⁻]/[HA]) → [A⁻]/[HA] ≈ 1.78. With total [A⁻]+[HA] = 50 mM: [HA] ≈ 18.0 mM, [A⁻] ≈ 32.0 mM. Mass of acetic acid (MW 60): 0.018 mol × 60 ≈ 1.08 g . Mass of sodium acetate (MW 82): 0.032 mol × 82 ≈ 2.62 g . Question 2 (a) State 4 functions of carbohydrates. [4] (b) Explain the basis of reduction tests as a means of identifying sugars in unknown solutions. [2] (c) Convert the given Fischer projection into two possible cyclic structures and state their full stereoisomeric names. [4] (i) What are enzymes? [3] (ii) Explain how enzymes work. [3] (iii) Where do enzymes come from and where do they go? [4] Model answer: (a) Functions of carbohydrates: energy source/storage (glucose, glycogen, starch); structural role (cellulose, chitin, peptidoglycan); cell-cell recognition/signalling (glycoproteins, glycolipids, e.g. blood group antigens); components of nucleic acids and several coenzymes. (b) Reduction tests (Benedict's/Fehling's) rely on a reducing sugar's free/potential aldehyde or ketone group; in alkaline solution the sugar's carbonyl is oxidized while Cu²⁺ is reduced to Cu⁺, precipitating as brick-red copper(I) oxide. Non-reducing sugars (e.g. sucrose, lacking a free anomeric carbon) give a negative test. (c) The Fischer projection given (OH left, left, right, right at C2–C5, with a D-configuration at C5) is D-mannose . Cyclizing the C1 aldehyde with the C5-OH creates a new anomeric centre at C1, giving two possible pyranose ring forms: α-D-mannopyranose and β-D-mannopyranose (differing only in the orientation of the new C1-OH relative to the reference C5 substituent). (i) Enzymes are biological catalysts — almost all proteins (a few are catalytic RNAs) — that increase reaction rate by lowering activation energy without being consumed. (ii) Enzymes work by binding substrate at a specific active site, stabilizing the transition state (via precise geometric/electrostatic complementarity, often with induced fit), thereby lowering the activation energy needed for the reaction to proceed. (iii) Enzymes are synthesized by ribosomes (as directed by the genome) and are eventually degraded/turned over by cellular proteolytic systems (e.g. the ubiquitin-proteasome system, lysosomal proteases) once no longer needed or damaged. --- Question (unnumbered, continues Section A) With full details, describe the enzymatic reactions that form a reversible link between the pentose phosphate pathway and the glycolytic pathway. (a) Suppose skeletal muscles were devoid of lactate dehydrogenase, could it carry out strenuous activity? Explain in detail. (b) People with beriberi, a disease caused by thiamine deficiency, have elevated levels of blood pyruvate and α-ketoglutarate especially after consuming a meal rich in glucose. Write a detailed account on the biochemical basis of this elevation. Model answer: The non-oxidative phase of the PPP reversibly interconverts sugar phosphates with glycolytic intermediates via transketolase (transfers 2-carbon units, TPP-dependent) and transaldolase (transfers 3-carbon units): Xylulose-5-P + Ribose-5-P ⇌ Sedoheptulose-7-P + Glyceraldehyde-3-P (transketolase); Sedoheptulose-7-P + Glyceraldehyde-3-P ⇌ Fructose-6-P + Erythrose-4-P (transaldolase); Xylulose-5-P + Erythrose-4-P ⇌ Fructose-6-P + Glyceraldehyde-3-P (transketolase again). The net effect converts 3 pentose phosphates into 2 fructose-6-phosphate + 1 glyceraldehyde-3-phosphate — both direct glycolytic intermediates — making the link fully reversible depending on the cell's relative need for NADPH vs. ATP/ribose. (a) Without lactate dehydrogenase, muscle could not regenerate NAD⁺ from the NADH produced by glycolysis under anaerobic conditions. NAD⁺ would rapidly become limiting, glyceraldehyde-3-phos