Question

Female with stubble bristles crossed to a a male with stubble bristles

Info:

SB-Stubble Bristles

+ Wild Type

Female 422 SB

Male 406 SB

Female 192 +

Male 187 +

1. Draw the cross and the F1 generation

2. Predict the F1 geneotypes and phenotypic ratios

3. Use goodness of fit/apriori chi squared (X2) analysis to confirm the most likely F1 phenotypic ratio.

4. How is Stubble Bristles inherited ?

Answer 1

a)Since Stubble bristle is a lethal allele found on chromosome 3 of Drosophila melanogaster therefore any individual with a homozygous stubble bristle gene condition cannot survive.

Let the genotype for Homozygous dominant Stubble bristle(individual cannot survive)= SS

and the genotype for Homozygous recessive condition be(unaffected individual) = ss

Henceforth

The genotype of female fly =Ss

The genotype of male fly = Ss

Monohybrid Cross for F1 generation :-

	S	s
S	SS	Ss
s	Ss	ss

b) Genotypic ratio for F1 progeny=1:2:1(SS : 2 x Ss : ss)

Genotypic ratio for F1 progeny=3:1(3 stubble bristled : 1 normal)

c)chi square

	Female	Male	Total
Stubble bristle	422=a	406=b	828
Wild type	192=c	187=d	379
Total	614	593	1207

Now

=

=0.1815

d)The gene for Stubble bristle is present on 3rd chromosome linked with the gene of curly wings, due to which Stubble bristles are inherited via linkage if a fly has a curly wings then by gene's innate nature it will possese Stubble bristles.