Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

2. In Drosophila, brittle (b), ebony body (e) and gaucho (g) are three recessive

ID: 196177 • Letter: 2

Question

2. In Drosophila, brittle (b), ebony body (e) and gaucho (g) are three recessive genes. If brittle, ebony females are crossed with homozygous gaucho body males, the resulting F1 progeny are all wild-type. If heterozygous F1 females are mated with brittle, ebony body, gaucho males, the following 1,000 progeny appear: mahea Ban brittle, ebony 390 ucho 420 britle Gaucho ss ony brittle 22 ebony, gaucho 28 ild-type 8 rittle, gaucho, ebony 12 a) Draw the chromosomal composition of the F1 females. b) Derive a map for the three (3) genes. Ann E. Reynolds Lee M. Silver

Explanation / Answer

Answer:

a). F1 genotype: bGe / BgE

b). b-------------14cM-----------g---7cM----e

Hint: Always recombinant genotypes are smaller than the non-recombinant genotypes.

Brittle, ebony / gaucho = bGe / BgE

Hence, the parental (non-recombinant) genotypes is bGe / BgE

Brittle, ebony—390---bGe

Gaucho—420---BgE

Brittle, gaucho—65---bgE

Ebony-55--BGe

Brittle—22--bGE

Ebony, gaucho—28--Bge

Wild-type—8--BGE

Brittle, gaucho, ebony—12--bge

Total = 1000

1).

If single crossover occurs between b & G..

Normal combination: bG / Bg

After crossover: bg/BG

bg progeny= 65+12=77

BG progeny = 55+8=63

Total this progeny = 140

The recombination frequency between b&G = (number of recombinants/Total progeny) 100

RF = (140/1000)100 = 14%

2).

If single crossover occurs between G & e..

Normal combination: Ge / gE

After crossover: GE / ge

GE progeny= 22+8=30

ge progeny = 28+12=40

Total this progeny = 70

The recombination frequency between G&e = (number of recombinants/Total progeny) 100

RF = (70/1000)100 = 7%

3).

If single crossover occurs between b & e..

Normal combination: be / BE

After crossover: bE/bE

bE progeny= 65+22=77

Be progeny = 55+28=83

Total this progeny = 160

The recombination frequency between b&e = (number of recombinants/Total progeny) 100

RF = (160/1000)100 = 16%

Recombination frequency (%) = Distance between the genes (cM)

b-------------14cM-----------g---7cM----e

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote