An individual homozygous for gene a and b is cross with wild type and `F_1` was back crossed with the double recessive. The appearance of the offspring is as follows
`++ to 39`
`ab to 31`
`a+ to 17`
`+b to 13`
Find out the distance between genes a and b .

Options

A. 31
B. 30
C. 42.8
D. 3

Answer

To solve the problem of finding the distance between genes A and B based on the given offspring phenotypes, we can follow these steps:

### Step 1: Identify the Parental and Recombinant Types
From the problem, we have the following offspring phenotypes and their counts:
- Parental types:
- ++ (wild type) = 39
- ab (double recessive) = 31
- Recombinant types:
- a+ (recombinant) = 17
- +b (recombinant) = 13

### Step 2: Calculate Total Offspring
To find the total number of offspring, we sum all the counts:
\[ \text{Total Offspring} = 39 (++) + 31 (ab) + 17 (a+) + 13 (+b) = 100 \]

### Step 3: Calculate the Number of Recombinants
Next, we need to determine the total number of recombinant offspring:
\[ \text{Total Recombinants} = 17 (a+) + 13 (+b) = 30 \]

### Step 4: Calculate Recombination Frequency
The recombination frequency (RF) is calculated using the formula:
\[ \text{RF} = \frac{\text{Number of Recombinants}}{\text{Total Offspring}} \times 100 \]
Substituting the values we found:
\[ \text{RF} = \frac{30}{100} \times 100 = 30\% \]

### Step 5: Convert Recombination Frequency to Centimorgans
In genetics, the distance between genes is often expressed in centimorgans (cM), where 1% recombination frequency is equivalent to 1 centimorgan. Thus, the distance between genes A and B is:
\[ \text{Distance} = 30 \text{ cM} \]

### Conclusion
The distance between genes A and B is **30 centimorgans**.

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