B-blood group man maries A blood group woman. The first child has B-blood group. Genotype of the child is

Options

A. `I^(A)I^(O)`
B. `I^(B)I^(O)`
C. `I^(A)I^(B)`
D. `I^(B)I^(B)`

Answer

To determine the genotype of the child from a B-blood group man and an A-blood group woman, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Blood Groups and Possible Genotypes of Parents:**
- The man has a B blood group. The possible genotypes for a B blood group are:
- IB IB (homozygous)
- IB IO (heterozygous)
- The woman has an A blood group. The possible genotypes for an A blood group are:
- IA IA (homozygous)
- IA IO (heterozygous)

2. **Determine the Possible Genotypes of the Child:**
- The child has a B blood group. The possible genotypes for a B blood group are:
- IB IB
- IB IO

3. **Analyze the Possible Combinations:**
- If the father is IB IB, he can only pass on the IB allele. The mother, being A blood group, can pass either IA or IO. However, if she passes IA, the child would be AB (IA IB), which does not match the child's B blood group. Therefore, the father cannot be IB IB.
- If the father is IB IO, he can pass either IB or IO. The mother, being A blood group, can pass either IA or IO. If the mother passes IO, the child would inherit IB from the father and IO from the mother, resulting in the genotype IB IO, which corresponds to a B blood group.

4. **Conclusion:**
- Since the only possible genotype for the child that results in a B blood group is IB IO, we conclude that the genotype of the child is IB IO.

### Final Answer:
The genotype of the child is **IB IO**.