- #1
TytoAlba95
- 132
- 19
Homework Statement
In a linkage map, two genes A and B, are 70 cM apart. If individuals heterozygous for both the genes are test crossed number of progeny with parental phenotype will be:
1. equal to the number of progeny with recombinant phenotype
2. more than the number of progeny with recombinant phenotype
3. less than the number of progeny with recombinant phenotype
4. could be more or less than the number of progeny with recombinant phenotype depending on whether the genes are linked in cis or trans, respectively2. Solution
The answer is 1.
3. Attempt
Since the genes are 70 cM apart on a linkage map. The recombinant frequency is more than 50% (In case of true linkage the recombinant frequency should always be less than 50%, i.e. is less than parental frequency, I don't know the reason why.) So this is a case of independent assortment. So in a test cross, the frequency would be 1:1.