Free Inheritance MCQs with Answers

61. Total number of linkage groups in a normal human is:

• A. 02
• B. 23
• C. 46
• D. 92

Explanation: All the genes on one chromosome form a single linkage group, and the number of linkage groups equals the haploid chromosome number, which for humans is 23. The figure 46 is the diploid number, counting both members of each homologous pair, but the two homologues carry the same set of loci and so form one group, not two. Ninety two would be the count of chromatids after replication.

62. Self-fertilization in plants through successive generations can lead to the development of:

• A. Hybrid breeds of plants
• B. Variations in coming generation
• C. True breeding plants
• D. Adaptation with their environment

Explanation: Repeated self pollination steadily removes heterozygotes, since each generation halves their proportion, until the line is homozygous and breeds true for every character. This is exactly how Mendel obtained the pure lines his experiments depended on. Hybrids and new variation arise from cross fertilisation, which is the opposite process.

63. A baby girl is born with hemophilia, which is an X-linked recessive disorder. What are the most likely genotypes of her parents?

• A. The mother is a carrier and the father is normal
• B. The mother is hemophiliac and the father is normal
• C. The mother is carrier and the father is hemophiliac
• D. Both are normal

Explanation: A girl has two X chromosomes and needs the recessive allele on both, one from each parent, so the father must be haemophiliac himself and the mother must carry at least one copy. A carrier mother with a normal father could only produce affected sons, which is why the first option fails. This is why affected females are so rare.

64. In a pea plant seed color is determined by two alleles: Y (Yellow, dominant) and y (green, recessive). Which parental cross would most likely result in offspring showing a 1:1 ratio of yellow to green seeds?

• A. Yy x YY
• B. Yy x yy
• C. Yy x Yy
• D. YY x yy

Explanation: Crossing a heterozygote with a homozygous recessive gives half Yy and half yy, that is one yellow to one green, which is the classic test cross. Yy by Yy gives three yellow to one green, and any cross involving YY gives all yellow because every offspring receives at least one dominant allele. Recognising the test cross by its 1:1 outcome is the quickest route.

65. Mendel crossed a plant with round yellow seeds (RRYY) and a plant with wrinkled green seeds (rryy), what was the phenotype of all F1 offspring?

• A. All round green
• B. All round yellow
• C. All wrinkled yellow
• D. All wrinkled green

Explanation: Every offspring receives R and Y from one parent and r and y from the other, so all are RrYy and both dominant characters show, giving round yellow seeds. The recessive wrinkled and green phenotypes reappear only in the F2, in the familiar 9:3:3:1 ratio. This uniform F1 is what Mendel's law of dominance describes.

66. If a normal person marries with colour blind female what will be the possibility of normal male child:

• A. 0%
• B. 25%
• C. 50%
• D. 75%

Explanation: A colour blind mother carries the defective allele on both her X chromosomes, and a son receives his only X from her, so every son must be colour blind. The daughters each receive a normal X from the father and are carriers with normal vision. This is the clearest illustration of why an X linked recessive trait passes from mother to son.