In Drosophila melanogaster, black body (b) is recessive to normal color (b\'), p
ID: 162526 • Letter: I
Question
In Drosophila melanogaster, black body (b) is recessive to normal color (b'), purple eyes (pr) are recessive to red eyes (pr^+), and vestigial wings (vg) is recessive to normal wings (vg^+). A female fly with a black body, purple eyes and vestigial wings is crossed to a wild type male. the F1 females produced by this cross were mated with black, purple and vestigial males in a tester. the progeny resulting from the testcross are listed below. Determine order of these genes on the chromosome. Calculate the map distances between the genes. Determine the coefficient of coincidence and the interference among these genes.Explanation / Answer
(a)order of genes: vg b pr
(b)
Distance between b and pr is 16.5 cM
Distance between vg and b is 4.4 cM
Distance between v and pr= 4.4 + 16.5= 20.9cM
(c)
Coefficient of coincidence=1.13/0.49 =2.30
inference=1-COC = 1-2.30=-1.3
Workout:
Highest number are parental type
Lowest numbers are recombinant (Single and double crossover)
Parental types are:
vg b pr
vg+ b+ pr+
Double crossover (DC):
vg+ b pr
vg b+pr+
Comparing parental type with double crossover types
Parental:vg b pr
DC: vg+ b pr
We find that the vg+ is different,
Next
Parental:vg+ b+ pr+
DC: vg b+pr+
We find that the vg+ is different,
Order of the uncommon gene is first, so comparing the position of v with other alleles we can find that the order of the genes is
vg b pr
Single point crossover between vg and b will produce
vg+bpr+=25
vgb+pr=28
RF= (single crossover+double crossover)/total offsprings x100
=(25+28+10+8)/1587 x100
=4.4%
Distance between vg and b is 4.4 cM
Single point crossover between b and pr will produce
vg+b+pr=120
vg b pr+=125
RF= (single crossover+double crossover)/total offsprings x100
=(120+125+10+8)/1587 x100
=16.5%
Distance between b vand pr is 16.5 cM
Distance between v and pr= 4.4 + 16.5= 20.9cM
Coefficient of coincidence(COC)= observed double crossover frequency/Expected frequency
observed double crossover frequency=(10+8)/1587 x 100=1.13
Expected frequency= frequency at first crossover x frequency at 2nd cross over
=[(25+28)/1587 x (120+125)/1587]x100
=0.033 x 0.15 x100
=0.49
Coefficient of coincidence=1.13/0.49 =2.30
inference=1-COC = 1-2.30=-1.3