In a cross between individuals homozygous for (a, b) and wild type `(+ +).` In this cross 700 out of 1000 individuals were of parental type. Then the distance between a and b is :-

Options

A. 70 map unit
B. 35 map unit
C. 30 map unit
D. 15 map unit

Answer

To solve the problem, we need to determine the genetic distance between two genes (a and b) based on the number of recombinant progeny produced in a cross between homozygous individuals for these genes and wild-type individuals. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Cross
- We have two parents: one homozygous for alleles (a, b) and the other homozygous for wild-type alleles (+, +).
- The parental types are (a, b) and (+, +).

### Step 2: Identify the Progeny Types
- In this cross, we are told that out of 1000 progeny, 700 are of parental type.
- This means that the remaining progeny are recombinant types.

### Step 3: Calculate the Number of Recombinant Progeny
- Total progeny = 1000
- Parental type progeny = 700
- Therefore, the number of recombinant progeny = Total progeny - Parental type progeny = 1000 - 700 = 300.

### Step 4: Use the Formula for Genetic Distance
- The formula to calculate the distance between two genes (in map units) is:

\[ \text{Distance (in map units)} = \left( \frac{\text{Number of recombinant progeny}}{\text{Total progeny}} \right) \times 100 \]

### Step 5: Substitute the Values into the Formula
- Number of recombinant progeny = 300
- Total progeny = 1000

Substituting these values into the formula gives:

\[ \text{Distance} = \left( \frac{300}{1000} \right) \times 100 = 30 \text{ map units} \]

### Final Answer
- The distance between genes a and b is **30 map units**.

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