A female heterozygous drosophila melanogaster with the x linked genes of singed bristles (sn), crossveinlesss wings (cv), and vermillion eye colour (v) is testcrossed with a singed, crossveinless, vermillion male and the following progeny were obtained:
1 singed, crossveinless, vermillion 3
2 crossveinless, vermillion 392
3 vermillion 34
4 crossveinless 61
5 singed, crossveinless 32
6 singed, vermillion 65
7 singed 410
8 wildtype 3
a) what is the order of these genes on the X chromosome
b) what are the genetic map distances between sn & cv; sn & v; and cv & v
c) what is the coefficient of coincidence
I need a lot of detailed explanations.
In drosophila, sex is determined by the number of X chromosomes. In this question, since they didn't tell us how many X's each fly has, we can assume that the female has 2 X chromosomes, and the male has 1.
We know that the female is heterozygous, so she has different alleles for each gene on her 2 X chromosomes. The male only has one copy of each gene. I know that vermillion eyes and crossveinless wings are both recessive genes, and I can only assume that singed bristles are as well. So, a fly can only show these traits if they only have the recessive allele (symbolized below using a -) and not the dominant allele (symbolized below using a +).
The alleles can be symbolized as:
sn- = singed bristles
sn+ = non-singed bristles
cv- = crossveinless wings
cv+ = wings with crossveins
v- = vermillion eyes
v+ = non-vermillion eyes
The female has genotype sn+ sn- cv+ cv- v+ v- She is wild-type.
The male has genotype sn- cv- v- He is singed, crossveinless, and vermillion.
a) what is the order of these genes on the X chromosome?