Selected MCQ

Current Question
Phenylketonuria (PKU) is inherited as an autosomal recessive trait. If two carriers have a child, what is the probability the child will be unaffected (no mutant alleles)?
  • A. 100%
  • B. 75%
  • C. 50%
  • D. 25%
Correct Answer: B
Explanation:
For two carriers (Aa × Aa), the offspring genotype ratio is 1 AA: 2 Aa: 1 aa. 75% (AA + Aa) are unaffected (though 50% are carriers).
Related Question 1
If a couple consists of one carrier for an autosomal recessive allele and one homozygous normal, what fraction of their children will be carriers (heterozygotes)?
  • A. 0%
  • B. 25%
  • C. 50%
  • D. 100%
Correct Answer: C
Explanation:
A cross Aa (carrier) × aa (normal) yields 50% Aa (carriers) and 50% aa (unaffected, non-carriers).
Related Question 2
A mother carries a mutated X-linked recessive gene but is unaffected. What is the probability that a son of hers will express the trait?
  • A. 0%
  • B. 25%
  • C. 50%
  • D. 100%
Correct Answer: C
Explanation:
A carrier mother (X^mX) has a 50% chance of passing the mutant X to a son. If he receives it, he will express the recessive trait (since sons inherit only one X chromosome).
Related Question 3
In a large, randomly mating population under Hardy–Weinberg equilibrium, 9% of individuals display a recessive trait. What percentage are heterozygous carriers?
  • A. 18%
  • B. 42%
  • C. 9%
  • D. 81%
Correct Answer: B
Explanation:
If q²=0.09, then q=0.30 and p=0.70. Heterozygous frequency = 2pq = 2(0.70)(0.30) = 0.42 or 42%.
Related Question 4
In a pedigree, if two unaffected parents have an affected child, the trait is most likely:
  • A. Autosomal dominant
  • B. Autosomal recessive
  • C. X-linked dominant
  • D. X-linked recessive
Correct Answer: B
Explanation:
Autosomal recessive traits can skip generations. Two carriers (phenotypically normal) can produce an affected child (25% chance).
Related Question 5
A female exhibits an X-linked recessive trait. What must be true of her parents’ genotypes?
  • A. Her father must have the trait, and her mother must be at least a carrier
  • B. Both parents must be affected
  • C. Only her mother is affected
  • D. Either parent could have the trait independently
Correct Answer: A
Explanation:
For a daughter to be homozygous for an X-linked recessive allele, her father must have the allele (so he is affected) and her mother must carry at least one mutant X.