Tngned yellow male? 9)Two recessive genes in Drosophilia (b and vg) produce blac

Question

Tngned yellow male? 9)Two recessive genes in Drosophilia (b and vg) produce black body and vestigial wings respectively. When wildtype flies are testcrossed the FI all have the dominant genes on one chromosome and the recessive genes on the homologous chromosome. Testcrossing the female F1 produced 1930 wildtype 412 black 1888 black and vestigial 370 vestigial (a) Calculate the distance between b and vg (b) Another recessive gene cn lies between the loci of b and vg producing cinnabar eye color. When wildtype flies are testcrossed the Fl are all trihybrid. Testcrossing the FI females produced 72 black, cinnabar 664 wildtype 68 vestigial 4 black, vestigial 652 black, cinnabar, vestigial 70 black 61 cinnabar, vestigial 8 cinnabar (c) Calculate the map distances (d) do the b-vg distances coincide between the two experiments? Explain of blood arRis being

Explanation / Answer

9. a) According to the data available in the question:-

B VG – 1930 (Wild Type parental progeny)

b   vg – 1888 (Recessive parental progeny)

B   vg – 412 (cross over)

b   VG – 370 (cross over)

The total no of progeny – (1930 + 1888 + 412+ 370) = 4600

Total no of cross over progeny – (412 + 370) = 782

Frequency of crossing over between b and vg is – (782/4600)*100 = 17%

So, the map distance – b....................17mu......................vg.

c) By comparing the wild type progeny and double cross over we can found than b and vg are same or B and VG same. So, they are apart from each other.

By arranging these genotypes –

+ + + - 664

b cn vg – 652

+ cn vg – 61 (Single cross over or S.C.O I)

b + + - 70 (S.C.O I)

+ + vg – 68 (S.C.O II)

b cn + - 72 (S.C.O II)

b + vg – 4 (D.C.O)

+ cn + -8

Total no. of progeny – 1599

Frequency of S.C.O I between b and cn is – (131/1599)*100 = 8.19

Frequency of S.C.O II between cn and vg is – (140/1599)*100 = 8.76

Frequency of D.C.O = 0.75

Therefore, map distance between b and cn is – (8.19 + 0.75) = 8.94 mu

Map distance between cn and vg is – (8.76 + 0.75) = 9.51 mu

Genetic map –

b--------------------8.94mu-----------------cn-----------------------------------9.51mu-------------------------------vg

d) Co-efficient of coincidence – (frequency of observed D.C.O)/(frequency of expected D.C.O)

Observed D.C.O = 0.75/100 = 0.0075

Expected D.C.O = (8.94/100)*(9.51/100) = 0.0085

Co-efficient of coincidence – 0.0075/0.0085 = 0.88.

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