If a plant heterozygous for a trait AaBb (AB/ab) is test crossed to obtain following result AaBb=520, aabb=380, Aab=46 and aaBb=54, then two genes are how many moprgan units apart from each other?

Options

A. 20 units
B. 10 units
C. 50 units
D. all are wrong

Answer

To determine how many Morgan units apart the two genes are from each other based on the given test cross results, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Progeny Counts**:
The progeny counts from the test cross are:
- AaBb = 520 (Parental type)
- aabb = 380 (Parental type)
- Aabb = 46 (Recombinant type)
- aaBb = 54 (Recombinant type)

2. **Calculate Total Progeny**:
Total progeny = AaBb + aabb + Aabb + aaBb
\[ \text{Total} = 520 + 380 + 46 + 54 = 1000 \]

3. **Identify Parental and Recombinant Types**:
- Parental types = AaBb + aabb = 520 + 380 = 900
- Recombinant types = Aabb + aaBb = 46 + 54 = 100

4. **Calculate Recombination Frequency**:
Recombination frequency (RF) is calculated as:
\[ RF = \frac{\text{Number of Recombinants}}{\text{Total Progeny}} = \frac{100}{1000} = 0.1 \]

5. **Convert Recombination Frequency to Morgan Units**:
Since 1 Morgan unit corresponds to a 1% recombination frequency, we can convert the recombination frequency to Morgan units:
\[ \text{Distance in Morgan units} = RF \times 100 = 0.1 \times 100 = 10 \text{ Morgan units} \]

### Final Answer:
The two genes are **10 Morgan units apart** from each other.

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