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, we will follow these steps:

### Step 1: Understand the Genetic Cross
We are given that black coat color (allele B) is dominant over brown coat color (allele b). We start with a black mouse (BB) and a brown mouse (bb).

### Step 2: Determine the F1 Generation
When we cross the black mouse (BB) with the brown mouse (bb), all offspring will inherit one allele from each parent. The black mouse can only pass on the B allele, while the brown mouse can only pass on the b allele. Therefore, all offspring in the F1 generation will be heterozygous (Bb).

**F1 Generation:**
- Genotype: Bb
- Phenotype: All black coats

### Step 3: Interbreeding the F1 Generation
Next, we allow the F1 generation (Bb) to interbreed. We will perform a Punnett square to determine the genotypes of the F2 generation.

**Punnett Square for Bb x Bb:**

| | B | b |
|-----|-----|-----|
| **B** | BB | Bb |
| **b** | Bb | bb |

### Step 4: Determine the Genotypic and Phenotypic Ratios
From the Punnett square, we can see the following genotypes:
- 1 BB (homozygous black)
- 2 Bb (heterozygous black)
- 1 bb (homozygous brown)

**Genotypic Ratio:**
- 1 BB : 2 Bb : 1 bb

**Phenotypic Ratio:**
- 3 black (BB + Bb) : 1 brown (bb)

### Step 5: Calculate the Percentage of Black-Coated Progeny
To find the percentage of progeny with black coats, we take the phenotypic ratio:
- Total parts = 3 (black) + 1 (brown) = 4 parts
- Percentage of black-coated mice = (3 black / 4 total) × 100 = 75%

### Final Answer
**75% of the progeny would have black coats.**

---