59 Year 2: Molecular Biology exam questions on BACHELOR OF MEDICINE AND BACHELOR OF SURGERY (MBChB) for medical students. Includes MCQs, answers, explanations a
This MCQ set contains 59 questions on BACHELOR OF MEDICINE AND BACHELOR OF SURGERY (MBChB) in the Year 2: Molecular Biology unit. Each question includes the correct answer and a detailed explanation for active recall and exam preparation.
Correct answer: E – DNA replication is catalyzed by RNA polymerase
DNA replication is catalyzed by DNA polymerase, not RNA polymerase. RNA polymerase is responsible for transcription (making RNA from DNA template), while DNA polymerase synthesizes new DNA strands during replication. ---
Correct answer: B – RNA polymerase
RNA polymerase is involved in transcription, not replication. All other enzymes are essential for DNA replication: DNA polymerase synthesizes new strands, ligase joins Okazaki fragments, helicase unwinds the DNA double helix, and primase synthesizes RNA primers. ---
Correct answer: E – C and D
Both DNA replication and PCR require RNA primers to initiate DNA synthesis. DNA polymerases cannot start synthesis de novo; they need a 3'-OH group provided by an RNA primer. ---
Correct answer: D – Nucleus and Cytosol respectively
In eukaryotes, transcription occurs in the nucleus where DNA is located, and translation occurs in the cytosol on ribosomes. This spatial separation allows for RNA processing before translation. ---
Correct answer: E – A and B
Both options A and B describe the same function using different wording. tRNA molecules pick up specific amino acids in the cytosol and transport them to the ribosome for protein synthesis. ---
Correct answer: A – It is a condition that occurs due to a missense mutation
Xeroderma pigmentosum is caused by mutations in DNA repair genes (often missense mutations), leading to defective nucleotide excision repair. This results in increased sensitivity to UV radiation and higher risk of skin cancer. ---
Correct answer: A – DNA polymerase 3'–5' proofreading enzyme
DNA polymerase has 3'-5' exonuclease activity (proofreading function) that removes incorrectly incorporated nucleotides, ensuring high fidelity during DNA replication. ---
Correct answer: B – Frameshift
Frameshift mutations involve insertion or deletion of nucleotides that change the reading frame, not single nucleotide changes. Point mutations are single nucleotide substitutions. ---
Correct answer: C – Taq DNA polymerase
PCR uses Taq DNA polymerase, a thermostable enzyme from Thermus aquaticus that can withstand the high temperatures required for PCR cycling. ---
Correct answer: B – It involves the combination of a target DNA to a host DNA
Recombinant DNA technology involves combining DNA from different sources, typically inserting target DNA into a vector (host DNA) to create recombinant molecules. ---
Correct answer: C – Phosphorylation
Phosphorylation is a post-translational modification of proteins, not a post-transcriptional modification of RNA. Post-transcriptional modifications include capping, splicing, editing, and polyadenylation. ---
Correct answer: D – mRNA is important for replication
mRNA is not involved in DNA replication; it's involved in protein synthesis (translation). DNA replication uses DNA polymerase and other replication machinery. ---
Correct answer: C – Br, Br, rr, rr
Cross: Br × rr produces 50% Br (black hair) and 50% rr (red hair) offspring. ---
Correct answer: C – Directional
The CCR5-Δ32 mutation provides resistance to HIV infection, representing directional selection favoring the mutant allele in populations exposed to HIV. ---
Correct answer: B – Change in gene frequency due to random effects in a small population
Genetic drift is the random change in allele frequencies that occurs in small populations due to sampling effects during reproduction. ---
Correct answer: A – Bottleneck event
A bottleneck event occurs when a population's size is significantly reduced, leading to loss of genetic diversity. The surviving 5% represent a genetic bottleneck. ---
Correct answer: E – 32%
If q² = 0.04, then q = 0.2 and p = 0.8. Heterozygous frequency = 2pq = 2(0.8)(0.2) = 0.32 or 32%. ---
Correct answer: A – Stabilizing
Sickle cell anemia represents balancing/stabilizing selection where heterozygotes have a survival advantage (malaria resistance) maintaining both alleles in the population. ---
Correct answer: B – Genetic drift
In small populations, rare traits can increase in frequency due to genetic drift, especially if they originate from a small number of founders. ---
Correct answer: A – The genes being far apart on the same chromosome
Crossover frequency is proportional to the physical distance between genes on the same chromosome. Genes far apart have higher recombination rates. ---
Correct answer: A – 48%
If q² = 0.16, then q = 0.4 and p = 0.6. Heterozygous frequency = 2pq = 2(0.6)(0.4) = 0.48 or 48%. ---
Correct answer: C – 42%
q² = 98/200 = 0.49, so q = 0.7 and p = 0.3. Heterozygous frequency = 2pq = 2(0.3)(0.7) = 0.42 or 42%. ---
Correct answer: C – 0.99
q² = 1/10,000 = 0.0001, so q = 0.01. Therefore, p = 1 - 0.01 = 0.99. ---
Correct answer: A – 0.02
Using p = 0.99 and q = 0.01 from previous question, carrier frequency = 2pq = 2(0.99)(0.01) = 0.0198 ≈ 0.02. ---
Correct answer: B – Substrate-level phosphorylation
Fermentation produces ATP through substrate-level phosphorylation without requiring electron transport or oxygen. ---