When white eyed and miniature winged Drosophila meloaogaster is crossed with its wild type it produces following percent of recombinations

Options

A. `1.3%`
B. `37.2%`
C. `62.8%`
D. `98.7%`

Answer

To solve the problem of the recombination percentages when white-eyed and miniature-winged Drosophila melanogaster is crossed with its wild type, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Traits and Genotypes**:
- The traits in question are white eyes (w) and miniature wings (m).
- The wild type for Drosophila has normal eyes (W) and normal wings (M).
- The parental genotypes are:
- Female: white-eyed miniature (w m)
- Male: wild type (W M)

2. **Perform the Cross**:
- The cross can be represented as:
- Female (w m) x Male (W M)

3. **Determine the Expected Offspring**:
- The offspring will inherit one allele from each parent. Since we are dealing with two traits, we can use a Punnett square or consider the combinations:
- Possible gametes from the female: w m
- Possible gametes from the male: W M
- The resulting combinations in the offspring will be:
- W M (wild type)
- w m (white-eyed miniature)
- W m (normal wings, white eyes)
- w M (white eyes, normal wings)

4. **Calculate the Percentages**:
- According to the problem, the recombination percentages are given as:
- Parental types (W M and w m): 62.8%
- Recombinant types (W m and w M): 37.2%
- This indicates that the parental types are more frequent than the recombinant types.

5. **Interpret the Results**:
- The percentage of recombination (37.2%) indicates that there is loose linkage between the genes for eye color and wing size. This means that the genes are located on the same chromosome but are not tightly linked, allowing for some recombination to occur during meiosis.

### Final Answer:
The percentage of recombination when white-eyed and miniature-winged Drosophila melanogaster is crossed with its wild type is **37.2%**.

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