In mice, black coat colour (allele B) is dominant to brown coat colour (allele b). The offspring of a cross between a black mouse (BB) and a brown mouse (bb) were allowed to interbreed. What percentage of the progeny would have black coats ?

Options

A. `25 %`
B. `50 %`
C. `75 %`
D. `100 %`

Explanation

To solve the problem of determining the percentage of progeny with black coats from the interbreeding of mice, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Genotypes of the Parent Mice**:
- The black mouse has the genotype BB (homozygous dominant).
- The brown mouse has the genotype bb (homozygous recessive).

2. **Determine the Gametes Produced by Each Parent**:
- The black mouse (BB) can only produce gametes with the allele B.
- The brown mouse (bb) can only produce gametes with the allele b.

3. **Cross the Parent Mice**:
- When we cross the two parents (BB x bb), the possible combinations of alleles in the offspring are:
- All offspring will have the genotype Bb (heterozygous).

4. **Interbreeding the Offspring**:
- Now, we allow the Bb offspring to interbreed. The cross is Bb x Bb.

5. **Set Up a Punnett Square**:
- Create a Punnett square to visualize the possible genotypes of the offspring:
```
B b
-----------
B | BB | Bb |
-----------
b | Bb | bb |
-----------
```
- From the Punnett square, we can see the resulting genotypes:
- 1 BB (black)
- 2 Bb (black)
- 1 bb (brown)

6. **Calculate the Phenotypic Ratio**:
- The phenotypic ratio from the Punnett square is:
- 3 black (BB and Bb) : 1 brown (bb).

7. **Determine the Percentage of Black Coats**:
- Out of 4 total offspring, 3 will have black coats.
- Therefore, the percentage of progeny with black coats is:
\[ \frac{3}{4} \times 100 = 75\% \]

### Final Answer:
75% of the progeny will have black coats.