Question

1) A woman with blood type A marries a man with blood type B. Their three children have blood types A, AB, and B. Which trait, if any, is dominant? Explain. What are the genotypes of the parents? What is the probability of these parents having a child with blood type O? Show your Punnett square.

Answer 1

A woman with blood type A marries a man with blood type B. Blood types of 3 children are A, AB and B. We see that both A and B can be inherited and expressed together. Thus, neither A is dominant over B and vice versa. Blood group is a case of codominance. But if we look at O blood group and as explained later, O allele is recessive to both A and B. Thus, overall we that there are 3 alleles and out of that two (A and B) are codominant but dominant over O.

Coming to the parents genotype, we see that we have 3 different blood groups present in the children and each child gets one allele from one parent. Thus for one child to be AB, one parent needs to have the A allele whereas the other needs to have the B allele (already mentioned that woman is A type and man is B type).

Let us represent the alleles as IA, IB and i for A, B and O respectively.

Now, since the woman has blood group A, she can have any of the two combinations IAIA or IAi. Similarly, the man can have any of the two combinations IBIB or IBi. Since the children have all three blood groups thus any parent can't have 2 similar alleles because then there is no possibility of the other blood group.

Suppose one parent is IAIA and the other is IBi then no child can have B blood type. Thus, the genotype of the parents will be IAi and IBi. The punnet square is shown below.

	IA	i
IB	IAIB AB Blood group	IBi B Blood group
i	IAi A Blood group	ii O Blood group

We can see that the probability of having a child with blood group O is 25%.