BACHELOR OF MEDICINE AND BACHELOR OF SURGERY (MBChB) | MCQ Quiz | OmpathStudy Kenya

Question 1

Which of the following statements is NOT TRUE of replication in eukaryotes?

Accepted Answer

DNA replication occurs in the nucleus — 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. ---

Question 2

Which of the following enzymes is NOT involved in the process of replication?

Accepted Answer

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. ---

Question 3

Which of the following requires an RNA primer in order to proceed?

Accepted Answer

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. ---

Question 4

In eukaryotes, transcription and translation occur in...

Accepted Answer

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. ---

Question 5

The function of tRNA is to...

Accepted Answer

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. ---

Question 6

Which of the following statements is TRUE about Xeroderma pigmentosum?

Accepted Answer

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. ---

Question 7

Which of the following is important in maintaining DNA replication fidelity?

Accepted Answer

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. ---

Question 8

Which of the following is NOT a type of point mutation?

Accepted Answer

Frameshift — Frameshift mutations involve insertion or deletion of nucleotides that change the reading frame, not single nucleotide changes. Point mutations are single nucleotide substitutions. ---

Question 9

DNA replication in Polymerase Chain Reaction is catalyzed by

Accepted Answer

Taq DNA polymerase — PCR uses Taq DNA polymerase, a thermostable enzyme from Thermus aquaticus that can withstand the high temperatures required for PCR cycling. ---

Question 10

Which of the following is TRUE about recombinant DNA technology?

Accepted Answer

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. ---

Question 11

Post-transcriptional modification includes all the following EXCEPT

Accepted Answer

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. ---

Question 12

Which of the following statements is FALSE ?

Accepted Answer

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. ---

Question 13

If one parent has genotype Br (black hair) and the other rr (red hair), possible genotypes of offspring are

Accepted Answer

Br, Br, rr, rr — Cross: Br × rr produces 50% Br (black hair) and 50% rr (red hair) offspring. ---

Question 14

Which selection type fits the CCR5 mutant gene?

Accepted Answer

Directional — The CCR5-Δ32 mutation provides resistance to HIV infection, representing directional selection favoring the mutant allele in populations exposed to HIV. ---

Question 15

Genetic drift is defined as

Accepted Answer

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. ---

Question 16

A population loses 95% of fish due to an electrical fault. What type of genetic drift is this?

Accepted Answer

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. ---

Question 17

In a population, 4% have sickle cell anemia. What percentage is heterozygous (advantageous to malaria)?

Accepted Answer

16% — If q² = 0.04, then q = 0.2 and p = 0.8. Heterozygous frequency = 2pq = 2(0.8)(0.2) = 0.32 or 32%. ---

Question 18

What selection type fits sickle cell anemia?

Accepted Answer

Stabilizing — Sickle cell anemia represents balancing/stabilizing selection where heterozygotes have a survival advantage (malaria resistance) maintaining both alleles in the population. ---

Question 19

A small population exhibits high frequency of "blue-skin" from one couple. This is an example of

Accepted Answer

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. ---

Question 20

High crossover frequency between two genes is most likely due to

Accepted Answer

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. ---

Question 21

If 16% show recessive phenotype in Hardy-Weinberg equilibrium, percentage of heterozygotes is

Accepted Answer

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%. ---

Question 22

In a population of 200, 98 show recessive phenotype. Predicted heterozygous frequency is

Accepted Answer

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%. ---

Question 23

Albinism occurs in 1 in 10,000. What is the frequency of the dominant allele?

Accepted Answer

0.99 — q² = 1/10,000 = 0.0001, so q = 0.01. Therefore, p = 1 - 0.01 = 0.99. ---

Question 24

Carrier frequency for albinism (above question) is

Accepted Answer

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. ---

Question 25

Energy acquisition by glucose fermentation requires

Accepted Answer

Substrate-level phosphorylation — Fermentation produces ATP through substrate-level phosphorylation without requiring electron transport or oxygen. ---

Question 26

Embden-Meyerhof and Entner-Doudoroff pathways are also called

Accepted Answer

Glycolysis — Both pathways are variants of glycolysis - the metabolic pathway that breaks down glucose to produce energy. ---

Question 27

Electron acceptors in anaerobic respiration include

Accepted Answer

All of the above — In anaerobic respiration, various compounds serve as terminal electron acceptors including nitrate, sulfate, and fumarate. ---

Question 28

Most common CO₂ fixation pathway in chemolithotrophs

Accepted Answer

Calvin Cycle — The Calvin cycle (Calvin-Benson-Bassham cycle) is the most widespread CO₂ fixation pathway used by chemolithotrophs. ---

Question 29

Passive diffusion is driven by

Accepted Answer

Concentration difference — Passive diffusion is primarily driven by concentration gradients, with molecules moving from high to low concentration without energy input. ---

Question 30

Terminal electron acceptor in aerobic respiration is

Accepted Answer

Oxygen — In aerobic respiration, oxygen serves as the final electron acceptor in the electron transport chain. ---

Question 31

Energy acquisition by glucose fermentation requires

Accepted Answer

Substrate-level phosphorylation — Same as Question 25 - fermentation relies on substrate-level phosphorylation for ATP production. ---

Question 32

Role of bacteriochlorophyll in anoxygenic photosynthesis is to

Accepted Answer

Use light energy to energize electrons — Bacteriochlorophyll captures light energy and uses it to excite electrons, initiating the photosynthetic electron transport process. ---

Question 33

Anoxygenic photosynthetic bacteria oxidize

Accepted Answer

Sulfide — Anoxygenic photosynthetic bacteria typically oxidize sulfide compounds (like H₂S) rather than water, and don't produce oxygen. ---

Question 34

Major energy capture in aerobic glucose respiration occurs by

Accepted Answer

Electron transport from NADH — Most ATP in aerobic respiration is produced through oxidative phosphorylation via electron transport from NADH and FADH₂. ---

Question 35

Which of the following is NOT a plasma membrane function?

Accepted Answer

Chromosome segregation — Chromosome segregation occurs in the nucleus/cytoplasm during cell division, not at the plasma membrane level. # GENETICS MCQ ANSWERS

Question 36

For a recessive trait to appear, the individual must receive the variant genes from both parents.

Accepted Answer

True — For a recessive trait to be expressed, an individual must be homozygous recessive (have two copies of the recessive allele), meaning they must inherit one recessive allele from each parent.

Question 37

Assume B is a dominant allele for black hair and r is a recessive allele for red hair. If one parent has black hair with the genotype Br, and the other parent has red hair with the genotype rr, what are the potential genotypes for their children?

Accepted Answer

Br, Br, rr, rr — Cross: Br × rr. The Br parent can contribute B or r, while the rr parent can only contribute r. Possible offspring: Br (50%) and rr (50%).

Question 38

Which selection type fits CCR5 mutant gene?

Accepted Answer

Directional — The CCR5-Δ32 mutation provides resistance to HIV infection, creating a selective advantage in environments where HIV is present, leading to directional selection favoring the mutant allele.

Question 39

The characteristics an individual expresses due to their genetic makeup are called:

Accepted Answer

Phenotypes — Phenotype refers to the observable characteristics or traits of an organism that result from the interaction of its genotype with the environment.

Question 40

Genetic drift can be defined as:

Accepted Answer

Gene frequencies change over time because of random effects due to a small population size — Genetic drift is the random change in allele frequencies in a population, and its effects are more pronounced in smaller populations.

Question 41

Imagine you raise goldfish as a pet dealer. You have over 10,000 fish in one large tank but, due to an electrical problem, 95% of the fish perish one night. The remaining 5% are left to breed and repopulate, passing their genes and traits on to future generations. What type of genetic drift would this be considered?

Accepted Answer

Bottleneck event — A bottleneck event occurs when a population's size is significantly reduced for at least one generation, leading to reduced genetic diversity.

Question 42

In a certain group of African people, 4 percent are born with sickle cell anemia. What percentage of the group has the selective advantage of being more resistant to malaria than those individuals who are homozygous for normal hemoglobin or for sickle cell anemia?

Accepted Answer

16% — If 4% have sickle cell anemia (ss), then q² = 0.04, so q = 0.2 and p = 0.8. Heterozygotes (Ss) have malaria resistance: 2pq = 2(0.8)(0.2) = 0.32 = 32%.

Question 43

Which selection type fits sickle cell anemia?

Accepted Answer

Stabilizing — Sickle cell anemia shows balancing selection (a type of stabilizing selection) where heterozygotes have an advantage (malaria resistance) over both homozygotes.

Question 44

In a small group of people living in a remote area, there is a high incidence of "blue skin," a condition that results from a variation in the structure of hemoglobin. All of the "blue skinned" residents can trace their ancestry to one couple, who were among the original settlers of this region. The unusually high frequency of "blue skin" in the area is an example of:

Accepted Answer

Genetic drift — This is specifically the founder effect, a type of genetic drift where a small founding population leads to high frequency of certain alleles in the derived population.

Question 45

Which of the following is the most likely explanation for a high rate of crossing over between two genes?

Accepted Answer

The two genes are far apart on the same chromosome — Crossover frequency is directly proportional to the distance between genes on a chromosome. The farther apart genes are, the higher the probability of crossover events between them.

Question 46

A population is in Hardy-Weinberg equilibrium. The gene of interest has two alleles, with 16% of the population portraying the recessive phenotype. Which percentage of the population is heterozygous?

Accepted Answer

48% — If 16% show recessive phenotype, then q² = 0.16, so q = 0.4 and p = 0.6. Heterozygotes = 2pq = 2(0.6)(0.4) = 0.48 = 48%.

Question 47

If 98 out of 200 individuals in a population express the recessive phenotype, what percentage of the population would you predict to be heterozygous?

Accepted Answer

42% — Recessive frequency = 98/200 = 0.49, so q² = 0.49, q = 0.7, p = 0.3. Heterozygotes = 2pq = 2(0.3)(0.7) = 0.42 = 42%.

Question 48

Albinism is a hereditary disease due to absence of the skin pigment melanin. The allele for albinism is recessive to the allele for normal pigment. One person in 10,000 is albino. What is the frequency of the dominant allele?

Accepted Answer

0.99 — If 1/10,000 is albino, then q² = 0.0001, so q = 0.01. Therefore, p = 1 - 0.01 = 0.99.

Question 49

What is the frequency for carriers in Question 14 above?

Accepted Answer

0.02 — Carriers are heterozygotes: 2pq = 2(0.99)(0.01) = 0.0198 ≈ 0.02.

Question 50

This is true of populations which are included in Hardy-Weinberg equilibrium:

Accepted Answer

Mating is random — Hardy-Weinberg equilibrium requires random mating, large population size, no migration, no mutation, and no natural selection.

BACHELOR OF MEDICINE AND BACHELOR OF SURGERY (MBChB) | MCQ Quiz | OmpathStudy Kenya

Questions, Answers & Explanations