Assume that genes a and b are linked and show 40 % reombination. If ++/++ individual is crossed with ab/ab, then tpes and proportions of gametes in `F_(1)` will be

Options

A. `++20%:ab20:+20%:a+40%`
B. `++50%:ab50%`
C. `++25%:ab25%:+b25%:a+25%`
D. `++30%:ab30%:+b20%:a+20%`

Answer

To solve the question regarding the crossing of linked genes with a given recombination frequency, we will follow these steps:

### Step 1: Understand the Given Information
We have two genes, A and B, which are linked and show a recombination frequency of 40%. This means that 40% of the gametes produced will be recombinant types, while 60% will be parental types.

### Step 2: Identify the Parental and Recombinant Gametes
The parental genotypes are:
- One parent: ++ (homozygous dominant for both genes)
- Other parent: ab (homozygous recessive for both genes)

The gametes produced by these parents will be:
- From ++: only + (since both alleles are dominant)
- From ab: only a (since both alleles are recessive)

Thus, the parental gametes are:
- Parental gametes: ++ and ab

### Step 3: Calculate the Proportions of Gametes
Since we know the recombination frequency:
- Parental gametes (no recombination) = 60%
- Recombinant gametes = 40%

The recombinant gametes will be formed by the combination of alleles from both parents:
- Possible recombinant gametes: +b and aB

### Step 4: List All Gametes and Their Proportions
Now we can summarize the types of gametes and their proportions:
1. Parental gametes:
- ++ (60%)
- ab (60%)

2. Recombinant gametes:
- +b (20%)
- aB (20%)

### Step 5: Final Proportions of Gametes
Now we can conclude the proportions of the gametes in the F1 generation:
- ++: 30%
- ab: 30%
- +b: 20%
- aB: 20%

### Conclusion
The types and proportions of gametes in F1 will be:
- ++: 30%
- ab: 30%
- +b: 20%
- aB: 20%