SAY Write your answer in the space provided or on a separate

SAY. Write your answer in the space provided or on a separate sheet of pap In the fruit fly. Drarephila melanu raster, a spineless (no wing a male that is claret bristles) female ny mated to (dark eyes) and hairless (no th otypically wild-type F1 f nai progeny were mated to fully homozygous (mutant) males, and the following progeny (1000 total) were observed, claret, spineless 18 309 claret, hairless hairless, claret, spineless 140 hairless hairless spineless (a) Which gene is in the middle? (b) With respect to the three genes mentioned in the problem, what are the genotypes of the homozygous Parents used in making the wild F1 heteroaygote? (c What are the map distances between the three genes? (d) What is the coefficient of coincidence?

Solution

Answer:

(1) (a) Spineless = (s + +) [Parental]

Claret hairless = (+ c h) [Parental]

Claret = =(+ c +) [double cross over]

hairless,spineless = (s + h) [double cross over]

Therefore, Parentals = s + + , + c h

and Double crossovers = + c + , s + h

Analysing, we get: s+ s+, +c +c and h+ +h.

h is different, so, h (hairless) is in the middle.

(b) The genotypes of the homozygous parents are = s + + / s + + and + c h / + c h

(c) The map distances have been calculated as under:

(Spineless – hairless) = (Number of recombinants)/1000) * 100

=  (32 + 38 + 12 + 18)/(1000) * 100

= (100/1000) * 100 = 10% = 10 map units (mu)

(hairless-claret) = (140 + 130 + 12 + 18)/1000) * 100

= (300/1000) * 100

= 30%

= 30 map units (mu)

(d) Coefficient of coincidence = observed double crossovers/expected double crossovers

So, we need to calculate expected double cross overs:

Expected double cross overs = recombination frequency in region 1 (map units / 100) X recombination frequency in region 2 (map units / 100)

= .(10/100) * (30/100)

= 0.1 * 0.3

= 0.03

Therefore, Coefficient of coincidence = [(18 + 12)/1000] / 0.03

= (30/1000) / 0.03

= 0.03 / 0.03 = 1