Question

In humans, the ABO blood type is under the control of autosomal multiple alleles. Color blindness is a recessive X-linked trait. Mom is type A and is color-blind. Dad is type B and has normal vision. They have a daughter who is type O and has normal vision. What is the probability that their next child will be a son who is type AB and is color-blind? Enter your answer as a fraction (for example: 1/5).

Answer 1

Answer:

1/8

Explanation:

The alleles that control the blood type are autosomal. The A and B alleles are codominant, and they are both dominant over the i allele.

The possible genotypes that determine the blood types are:

• Type A: $I^AI^A$, $I^Ai$

• Type B: $I^BI^B$, $I^Bi$

• Type AB: $I^AI^B$

• Type O: $ii$

Color blindness is a recessive X-linked trait. The possible genotypes and phenotypes are:

• Normal vision woman: $X^VX^V$, $X^VX^v$,

• Color-blind woman: $X^vX^v$

• Normal vision man: $X^VY$

• Color-blind man: $X^vY$

The mom is type A and color-blind. She is $X^vX^v$ and could be either $I^AI^A$ or $I^Ai$.

The dad is type B and has normal vision. He is $X^VY$ and could be either $I^BI^B$ or $I^Bi$.

Because the daughter is type O, her genotype for blood type is ii. That means that both parents must have the recessive i allele. The mom is $I^Ai$ and the dad is $I^Bi$.

To calculate the probability of having a type AB color-blind son, we need to do the Punnett Square (see attached image) and the rules of probability.

Given two independent events, the rules of probability state that the probability that both events occur is found by multiplying the probabilities of each event.

P(A and B) = P(A) x P(B)

Genes in different chromosomes assort independently, so the probability of having offspring with the genotype $X^vY$ , $I^AI^B$ is:

P= 1/4 x 1/2 = 1/8.