Suppose a and b are recessive alleles of genes located 20 ma
Suppose a and b are recessive alleles of genes located 20 map units apart on chromosome 2 in Drosophila melanogaster. For the P generation, true-breeding a + males are mated to true-breeding + b females. In the F1 generation the heterozygous a +/+ b flies are crossed to each other. What proportion of F2 flies exhibit the recessive a phenotype and dominant + phenotype for b in the next generation (i.e., are like the a + males of the P generation)? You may wish to use a Punnet square again, but remember, there is no recombination in Drosophila males
Solution
Answer:
F1 X F1
a +/+b (male) X a +/+b (female)
Due to no recombination, only two types of gametes are produced. Those are a+ & +b.
Distance between the genes = % Recombination frequency.
Females produce the following gametes
Parental gametes (a + & + b )= 80% (each 40%)
Non parental gametes(+ + & a b) = 20% (each 10%)
| a + (0.4) | + b (0.4) | + + (0.1) | a b (0.1) | |
| a + (0.5) | a +/a + (0.2%) | a +/+ b(0.2%) | a +/+ + (0.05%) | a +/a b (0.05%) | 
| + b(0.5) | a +/ + b (0.2%) | + b/ + b (0.2%) | + b/+ + (0.05% | + b /a b (0.05%) | 

