Question

1. if a man with blood type b (one of whose parents had blood type o) marries a woman with blood type ab, what will be the theoretical percentage of their children with blood type b?

Explanation:

Step1: Determine Genotypes

The man has blood type B, and one parent is O (genotype \(ii\)). So his genotype must be \(I^B i\) (since to get an \(i\) allele from the O parent). The woman has blood type AB, so her genotype is \(I^A I^B\).

Step2: Set Up Punnett Square

The possible gametes from the man (\(I^B i\)) are \(I^B\) and \(i\). From the woman (\(I^A I^B\)) are \(I^A\) and \(I^B\). The Punnett square combinations are:

• \(I^B \times I^A\): \(I^A I^B\)
• \(I^B \times I^B\): \(I^B I^B\)
• \(i \times I^A\): \(I^A i\)
• \(i \times I^B\): \(I^B i\)

Step3: Calculate Blood Type B Proportion

Blood type B includes \(I^B I^B\) and \(I^B i\). From the Punnett square, 2 out of 4 (or 50%) of the offspring have blood type B ( \(I^B I^B\) and \(I^B i\) ).

Answer:

50%