Consider the following three autosomal recessive mutations in Drosophila: vestig
ID: 144610 • Letter: C
Question
Consider the following three autosomal recessive mutations in Drosophila: vestigial wings (v); wild type is long (v+) black body color (b); wildtype is gray (b+) plum eyes (p); wildtype is red (p+) A vestigal, gray, red female (homozygous for all three genes) is crossed with a long wing, black, plum male (homozygous for all three genes). The F1 female progeny are mated with triple homozygous recessive males. Here is the phenotypic data for the F2 progeny: vestigal; gray; red 580 long wings; black; plum 592 vestigal; black; red 45 long; gray; plum 40 vestigal; black; plum 89 long; gray; red 94 vestigal; gray; plum 3 long; black; red 5 A total of 1448 progeny were counted. Which one of the following values is the approximate distance between the plum eye color and black body color loci? Note: there will be some rounding error so your answer might not be exactly the same.
Explanation / Answer
Recessive character
Wild type character
VESTIGIAL WING (v)
Long wing (v+)
Black body (b)
Gray body (b+)
Plum eye (p)
Red eye (p+)
The F2 phenotypes were:
Genotype
Number of progeny
vb+p+
580
v+bp
592
vbp+
45
v+b+p
40
vbp
89
v+b+p+
94
vb+p
3
v+bp+
5
So the gene order will be: vbp
Arrange all the above genotype as per the order:
Genotype
Number of progeny
Vp+b+
580
v+bp
592
Vp+b
45
v+pb+
40
Vpb
89
v+p+b+
94
Vpb+
3
v+p+b
5
Linkage distance = (# of double crossover product) + (# of single crossover product) * 100
Total number of progeny
So map distance= (3+5)+(45+40)*100 /1448
=( 8+85 ) * 100 /1448
= 93*100/1448
=9300/1448
= 6.4 cM
cM: centimorgan ( unit of linkage map)
Recessive character
Wild type character
VESTIGIAL WING (v)
Long wing (v+)
Black body (b)
Gray body (b+)
Plum eye (p)
Red eye (p+)