Biochemistry 2021 CAT — Past Paper Questions OmpathStudy

About this paper This is the University of Nairobi HBC100/UPC102/VBC100 Biochemistry 1 — End of Semester 1 Exam 2021 , transcribed from the scanned quiz-review screenshots. The scans show a Moodle quiz review with radio-button selections, but those on-screen markings turned out to be unreliable — cross-checking against chemistry/biochemistry first principles caught numerous wrong markings (including duplicated questions marked differently across different screenshots). Every answer below has been independently re-derived from first principles; the original marking is noted only where it differs from the derived correct answer. Questions and Answers Question 1 What is the concentration of hydroxide ions in an aqueous solution with a pH of 4.80? A. 4.2 x 10⁻⁹ M B. 1.6 x 10⁻⁵ M C. 2.0 x 10⁻⁸ M D. 6.3 x 10⁻¹⁰ M E. 3.6 x 10⁻¹² M ✅ Answer: D. 6.3 x 10⁻¹⁰ M Explanation: pOH = 14 − 4.80 = 9.20. [OH⁻] = 10⁻⁹·²⁰ = 6.3 × 10⁻¹⁰ M. Question 2 Chargaff's rules state that in typical DNA: A. A+T=G+C B. A=C C. A+G=T+C D. A=U E. A=G ✅ Answer: C. A+G=T+C Explanation: Chargaff's rules state A=T and G=C, so purines (A+G) equal pyrimidines (T+C). Question 3 Inadequate tryptophan in the diet can lead to low levels of: A. Vitamin A B. Nicotinic acid C. Pantothenic acid D. Pyridoxine E. Vitamin C ✅ Answer: B. Nicotinic acid Explanation: Tryptophan is a metabolic precursor for niacin (nicotinic acid, vitamin B3) synthesis, so a tryptophan-poor diet can cause niacin deficiency. Question 4 The following is the major role of vitamin E in the body: A. Participates in hydroxylation reactions B. Helps in blood clotting C. It is important in bone formation D. It causes male sterility E. It protects cells against oxidative damage by reactive oxygen species ✅ Answer: E. It protects cells against oxidative damage by reactive oxygen species Explanation: Vitamin E is a lipid-soluble antioxidant that protects cell membranes from peroxidation. Question 5 The products of basic hydrolysis of an ester are: A. Alcohol and carboxylic acid B. Another ester and water C. Carboxylate ion and alcohol D. Alcohol and water E. Carboxylic acid and water ✅ Answer: C. Carboxylate ion and alcohol Explanation: Base-catalyzed hydrolysis (saponification) deprotonates the carboxylic acid product, yielding a carboxylate ion plus an alcohol. Question 6 On keeping for a long time, fats undergo spontaneous hydrolysis in a process known as: A. Saponification B. Condensation C. Decomposition D. All of the above E. Hydrolytic rancidity ✅ Answer: E. Hydrolytic rancidity Explanation: Slow, spontaneous hydrolysis of stored fats releasing free fatty acids is termed hydrolytic rancidity. Question 7 The major role of vitamin D3 (cholecalciferol) in the body is: A. Control of calcium and phosphorous metabolism required in proper calcification of bones B. Transcription of specific genes that mediate growth and development C. Biological antioxidant D. Posttranslational modification of blood coagulation factors E. Is required in oxidation-reduction reactions ✅ Answer: A. Control of calcium and phosphorous metabolism required in proper calcification of bones Explanation: Vitamin D3's principal role is regulating calcium/phosphate homeostasis for bone mineralization. Question 8 On acids and bases: A. Weak acids form weak conjugate bases in water B. All Lewis acids are Bronsted acids C. All Bronsted acids are Lewis acids D. All Bronsted acids are Arrhenius acids E. Strong acids form strong conjugate bases in water ✅ Answer: C. All Bronsted acids are Lewis acids Explanation: A Bronsted acid donates H⁺; that proton is itself an electron-pair acceptor, so every Bronsted acid is also a Lewis acid. The reverse is not true (e.g., BF₃ is a Lewis acid but not a proton donor). Question 9 Calculate the pH of 0.025M of acetic acid, assuming that the dissociation does not significantly change the acid concentration, Ka = 8.5 x 10⁻⁵ A. 7.63 B. 4.12 C. 4.67 D. 2.84 E. 4.23 ✅ Answer: D. 2.84 Explanation: [H⁺] = √(Ka·C) = √(8.5×10⁻⁵ × 0.025) = √(2.125×10⁻⁶) = 1.46×10⁻³ M. pH = −log(1.46×10⁻³) = 2.84. Question 10 The pKa of a weak acid is equal to: A. The pKb of its conjugate base B. The pH of a solution containing equal amounts of the acid and its conjugate base C. The equilibrium concentration of its conjugate base D. Its relative molecular mass E. The base or acid concentration of the aqueous solution ✅ Answer: B. The pH of a solution containing equal amounts of the acid and its conjugate base Explanation: From Henderson–Hasselbalch, pH = pKa + log([A⁻]/[HA]); when [A⁻]=[HA] the log term is zero, so pH = pKa. Question 11 (i) 2, 0, 0, +½ and, (ii) 3, 1, -1, -½ A. They are both located in s orbitals B. They have the same spin C. They are at the same energy level D. They are located in different p-orbitals E. None of the other choices is true ✅ Answer: E. None of the other choices is true Explanation: Set (i) n=2, l=0 is a 2s electron; set (ii) n=3, l=1 is a 3p electron. Different subshells, different e