Question

At one gene locus in humans there are 2 alleles: one for curly hair and another for straight hair. Neither allele is dominant, and heterozygous individuals have wavy hair. At a second gene locus, 3 alleles control ABO blood type. The alleles at these 2 gene loci assort independently. A man with straight hair and A blood has children with a woman with wavy hair and B blood. Their first child has straight hair and O blood. What is the probability that their next child will have wavy hair and blood type AB?

Answer 1

Answer: 1/8 (approximately 12.5%)

Step-by-step explanation:

The key to figuring out this question is contained in the statement “The alleles at these 2 gene loci assort independently”. This means that the analysis of each locus can be undertaken in isolation first.

1) Let S = straight hair; C = curly hair. Since both alleles are codominant, it can be assumed that the straight-haired man has ‘SS’ genotype and the wavy-haired woman is heterozygous (“SC”). The Punnett chart is as follows:

S C

S SS SC

S SS SC

Thus, there is a 1/2 (50%) chance that any one of their offspring will have wavy hair.

2) We know that the man and the woman have blood types A and B respectively, but we also know that they have already had one child with type 0 blood. Since the only possible genotype for type 0 blood is “00”, and each “0” allele must have been inherited from one of the parents, we can assume that the parents have ”A0” and “B0” genotypes. See chart:

B 0

A AB A0

0 B0 00

There is a 1/4 (25%) chance that any one of their offspring will have type AB blood.

3) Finally: Since the alleles are segregated independently, we may apply the rule of multiplication for probabilities of independent events, as follows:

Probability of wavy hair x probability of type AB blood = 1/2 x 1/4 = 1/8