Below are the results of a threepoint mapping cross from Dro

Below are the results of a three-point mapping cross from Drosophila involving three recessive traits on the X-chromosome: sc = Scute bristles; s = Sable body; v = Vermillion eyes. Determine the combination of alleles on the chromosomes of the parent that produced the crossover gametes. Determine the correct sequence of the genes. 8e sure to show your work. Determine the map distances of the three genes.

Solution

(2) (a) The most abundant genotypes are the parental types. According to the given data, the number of offsprings with the genotype SC S V = 314 and the number of offsprings with the genotype + + + = 300. Therefore, the parental genotypes are :

SC S V and + + +.

(b) To determine the gene sequence, we need the parental genotypes as well as the double crossover geneotypes. The least frequent genotypes are the double-crossover geneotypes. According to the given data, the number of offsprings with the genotype SC S + = 10 and the number of offsprings with the genotype + + V = 11. Therefore, the double-crossover genotypes are SC S + and + + V.

We can see from the table that the V gene must be in the middle because the recessive + allele is now on the same chromosome as the SC and S alleles, and the dominant V allele is on the same chromosome as the recessive + and + alleles.

Therefore, the correct sequence of genes is SC V S.

(c) Total number of offsprings = 314 + 300 + 145 + 150 + 40 + 30 + 10 + 11 = 1,000

SC - V distance = [ 100 x { (145 +150+10+11) / 1,000 }] = 31.6 cM

V - C distance = [100 x {(40 + 30+ 10 + 11) / 1,000}] = 9.1 cM ( Ans)