Question

You self an individual with the following genotype (locus C is on a separate chromosome from A and B): A b C tot a B с 10 m.u. What proportion of offspring will be: ab/ab: c/c ?

Answer 1

From the gene mapping positions, we can see that Ab is present in the same chromosome and aB also present in the same chromosome. So, the chromosome containing ab is only possible if a crossover event take place between these two genes. Now, the map distance between these two genes is 10 m.u. which signifies the recombination frequency is 10% or 0.1. So, for 10% of cases the ab gametes will be formed.

Now, for both the gamete should be ab, to form a individual ab/ab. So, now the probability is (0.1*0.1) = 0.01 = 1/100.

Now, for the gene C, a gamete with c will be produced in 50% cases or the frequency is 0.5. So, a individual with a genotype c/c is possible in a frequency of (0.5*0.5) = 0.25 = 25/100 = 1/4.