Question

A cross in Drosophila involved the recessive, X-linked genes yellow (y), white (w), and cut (ch. A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal but whose wings were cut . The F, females were wild type for all three traits. • The F, males expressed the yellow-body and white-eye traits. The cross was carried to an F2 progeny, and only male offspring were tallied. On the basis of the data shown here, a genetic map was constructed Male Offspring Phenotype y + + A c + + + + + + + + + + *

Answer 1

%recombination = map unit

Map position can be calculated by first finding the distance between two genes and then adding or subtracting it from the position of given gene. Here, position of middle gene is given, using sign conventions, position of y gene can be calculated by subtracting the distance between genes y and w from 1.5. Position of ct gene can be calculated by adding the gene distance between ct and w to 1.5

Since the parental combinations are ++ct and yw+ , as these two combinations have maximum frequency, the combinations other than these two would be considered as recombinant.

% recombination between gene ct and w = (90+95)×100/1000

= 185/10 = 18.5 mu

The map position would be 18.5 + 1.5 = 20.0

% recombination between gene y and w = (9+6)×100/1000

15/10 = 1.5 mu

The map position would be 1.5 - 1.5 = 0.0