Two genes 'A' and 'B' are linked. In a dihybrid cross involving these two genes, the `F_(1)` heterozyogte is crossed with homozygous recessive parental type (aa bb). What would be the ratio of offspring in the next generation?

Options

A. `1:1:1:1`
B. `9:3:3:1`
C. `3:1`
D. `1:1`

Explanation

To solve the problem of the dihybrid cross involving two linked genes 'A' and 'B', we will follow a systematic approach.

### Step-by-Step Solution:

1. **Identify the Genotypes of the Parents**:
- The F1 heterozygote is represented as \( AaBb \).
- The homozygous recessive parent is represented as \( aabb \).

2. **Determine the Gametes Produced by Each Parent**:
- The F1 parent \( AaBb \) can produce the following gametes due to linkage:
- \( AB \) (dominant for both traits)
- \( Ab \) (dominant for A, recessive for B)
- \( aB \) (recessive for A, dominant for B)
- \( ab \) (recessive for both traits)
- The homozygous recessive parent \( aabb \) can only produce one type of gamete:
- \( ab \)

3. **Set Up the Punnett Square**:
- Since the homozygous recessive parent can only produce \( ab \), we will set up a Punnett square with the gametes from the F1 parent:

| | \( ab \) |
|-------|----------|
| \( AB \) | \( AaBb \) |
| \( Ab \) | \( Aabb \) |
| \( aB \) | \( aaBb \) |
| \( ab \) | \( aabb \) |

4. **Determine the Genotypes of the Offspring**:
- From the Punnett square, the genotypes of the offspring are:
- \( AaBb \) (1)
- \( Aabb \) (1)
- \( aaBb \) (1)
- \( aabb \) (1)

5. **Count the Offspring Ratios**:
- The offspring genotypes are:
- 1 \( AaBb \)
- 1 \( Aabb \)
- 1 \( aaBb \)
- 1 \( aabb \)
- Thus, the ratio of the offspring is:
- \( 1 : 1 : 1 : 1 \)

### Final Answer:
The ratio of the offspring in the next generation is \( 1 : 1 : 1 : 1 \).