The genotypes of husband and wife are `I^(A)I^(B)` and `I^(A)i`. Among the blood groups of their children how many different genotypes and phenotypes are possible

Options

A. 3 genotypes , 3 phenotypes
B. 3 genotypes , 4 phenotypes
C. 4 genotypes , 3 phenotypes
D. 4 genotypes , 4 phenotypes

Answer

To solve the problem of determining how many different genotypes and phenotypes are possible among the children of a couple with genotypes `I^(A)I^(B)` (husband) and `I^(A)i` (wife), we can follow these steps:

### Step 1: Identify the Genotypes of Parents
The genotypes of the husband and wife are:
- Husband: `I^(A)I^(B)`
- Wife: `I^(A)i`

### Step 2: Determine the Gametes
Next, we need to determine the gametes that each parent can produce:
- The husband can produce two types of gametes: `I^(A)` and `I^(B)`.
- The wife can produce two types of gametes: `I^(A)` and `i`.

### Step 3: Set Up a Punnett Square
We will create a Punnett square to visualize the combinations of gametes from both parents:
```
| I^(A) | I^(B) |
------------------------------
I^(A) | I^(A)I^(A) | I^(A)I^(B) |
------------------------------
i | I^(A)i | I^(B)i |
```

### Step 4: List the Offspring Genotypes
From the Punnett square, we can list the possible genotypes of the offspring:
1. `I^(A)I^(A)`
2. `I^(A)I^(B)`
3. `I^(A)i`
4. `I^(B)i`

### Step 5: Identify the Phenotypes
Now, we will determine the corresponding phenotypes for each genotype:
1. `I^(A)I^(A)` → Blood Group A
2. `I^(A)I^(B)` → Blood Group AB
3. `I^(A)i` → Blood Group A
4. `I^(B)i` → Blood Group B

### Step 6: Count Unique Genotypes and Phenotypes
- **Genotypes**: `I^(A)I^(A)`, `I^(A)I^(B)`, `I^(A)i`, `I^(B)i` → **4 unique genotypes**
- **Phenotypes**: Blood Group A (from `I^(A)I^(A)` and `I^(A)i`), Blood Group AB, Blood Group B → **3 unique phenotypes**

### Final Answer
Thus, the total number of different genotypes is **4** and the total number of different phenotypes is **3**.

### Conclusion
The correct answer is **4 genotypes and 3 phenotypes**.

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