Question

In a plant species, the genotype \( DDEE \) is crossed with the genotype \( ddee \); and the F1 is test crossed. Considering that the two genes are linked and 20 map unit (cM) apart in the chromosome, the percentage (%) of the test cross progeny with the genotype \( ddee \) is ........................ (answer in integer)

Hint:

When calculating the percentage of a specific genotype in a test cross involving linked genes, consider the recombination frequency (given in cM) to determine the proportion of recombinant and parental gametes.

Answer 1

Step 1: Understanding the problem The genes are linked, which means the two loci are on the same chromosome and do not assort independently. The distance between the two genes is given as 20 cM, which corresponds to 20% recombinant offspring. This means that 20% of the gametes will be recombinant, and 80% will be parental.
Step 2: Calculating recombinant and parental gametes
- The parental gametes are \( DE \) and \( de \), which make up 80% of the gametes.
- The recombinant gametes are \( De \) and \( dE \), which make up 20% of the gametes.
Step 3: Genotype of F1 and test cross
The F1 plants are \( DdEe \). When test crossed with \( ddee \), the possible gametes from the F1 plant will be:
- Parental: \( DE \) and \( de \) (80%)
- Recombinant: \( De \) and \( dE \) (20%)
The test cross will produce the following combinations:
- \( ddee \) (from \( de \) and \( de \))
- \( ddeE \) (from \( de \) and \( dE \))
- \( Ddee \) (from \( DE \) and \( de \))
- \( DDEe \) (from \( DE \) and \( dE \))
Step 4: Percentage of \( ddee \) genotype
The genotype \( ddee \) is produced by the combination of two \( de \) gametes. The percentage of this genotype is: \[ \frac{1}{4} \times 80% = 20% \] Final Answer: \[ \boxed{20} \]