In: Biology
Black body (b) and purple eye (pr) are recessive autosomal mutations in Drosophila. Bridges crossed b/b females and pr/pr males. F2 cross produced 684 wild-type, 371 black-bodied, and 300 purple-eyes flies. Do these result indicate that the b and pr genes are closely linked? Explain. Remember that there is no crossing over in male Drosophila
b - allele for black body - recessive
pr - allele for purple eye - recessive
B - wild type for body colour
Pr - wild type for eye colour.
BB PrPr, Bb PrPr, Bb Prpr and BB Prpr are the wild types.
Bridges crossed b/b females and pr/pr males
In this cross, since nothing is said about eye colour genotype of female and body colour genotype of male, let's consider them as homozygous dominant for those respective traits.
Cross By Bridges:
PrPr bb (female) X prpr BB (male)
'X' represents cross or mating
This cross produces only one type of progeny in the F1 i.e., Prpr Bb
To produce F2, we need to cross amoung the F1 individuals.
Thr cross: (Considering the genes are not linked)
Prpr Bb (female) X Prpr Bb (male)
Each of the parents will produce 4 types of gametes viz. Pr B, Pr b, pr B and pr b
This is a typical dihybrid situation, which produces the typical ratio of 9:3:3:1.
That is :
9 wild : 9Black : 9 Purple : 1 black and purple
But, the data givem has a very huge departure from this ratio. Hence the genes are linked, i.e. are found on the same chromosome.
To find if both the genes are tightly linked, we have to find out the phenotypic ratio when genes are tightly linked, i.e. 0% recombination for both the parents. A scheme to find out that ratio is attached with this answer. Please refer that.
The ratio of Wild : Black : Purple if genes are tightly linked was found to be 2 : 1 : 1.
As per the data given, there are 684 wild, 371 black bodied and 300 purple eyed flies.
The phenotype is in the ratio of 2.28 : 1.23 : 1 (Wild : Black : Purple).
This is approximately equal to 2:1:1.
Hence, we can conclude that there is tight linkage between the two genes.