Genetics Engineering help?

Question

The recessive mutations b (black body color), st (scarlet eye color), and hk (hooked bristles) identify three autosomal genes in D. melanogaster. The following progeny were obtained from a testcross of females heterozygous for all three genes.


black, scarlet 243
black 241
black, hooked 15
black, hooked, scarlet 10
hooked 235
hooked, scarlet 226
scarlet 12
wild-type 18

What conclusions are possible concerning the linkage relationship of these three genes? What was the genotype of the female heterozygote? Calculate any appropriate map distances.

Answer 1

Take each of the three traits seperately and compare to expected mendalian ratios. (Testcross means the male was completely recessive. If the female is a heterozygote then the expected ratio is 1:1, if homozygous dominant 1:0, if homozygous recessive 0:1).

For example
243+241+15+10=509 black (1000/509=2)
235+226+12+18=491 not black (2)
==>approximately 1:1
The female had to be Bb because both types are visable.

scarlet =491
not scarlet=509
==>1:1, STst

hooked=514
not hooked=486
==>1:1, HKhk

Female is B b, ST st, HK hk

For linkage/mapping distance, take the total progeny that do not look like either parent and had to come from a crossover event and divide by the total number of progeny. Assume that the combinations you see most often are parental (243+241+235+226). Work with two traits at a time. Significant deviation from mendalian ratios indicates linkage.

b and st
BST(parental)=235+18=253
bst(parental)=243+10=253
Bst=226+12=238
bST=241+15=256
parental=506, recombination=494...approximately1:1 no linkage

b and hk
BHK=12+18
bhk=15+10
Bhk(parental)=235+226
bHK(parental)=243+241
parental=945, recombination=55
55/1000=.055(100)=5.5cM between b and hk

Your map distance is cM.