Black body (b/b), curved wing(c/c), and purple eye color (pr/pr) are recessive t

Question

Black body (b/b), curved wing(c/c), and purple eye color (pr/pr) are recessive traits known to be on chromosome 2 in Drosophila. A testcross of a heterozygous female for all 3 genes was crossed to a black bodied, curved wing, purple eyed male fly and yielded the following progeny: Give the genotype of the parents in correct gene order and Drosophila notation. Fill in the 'Type column indicating which phenotypic classes are parental types (P), region 1 crossover types (r1), region 2 crossover types (r2), or double crossover types (dco). Calculate the map distances between the genes(give answer in map units): Calculate the value for interference for this cross.

Explanation / Answer

a and b. The highest number of progeny always indicates the parent genotypes.

Now, (b pr +) and (+ + c) are in highest number, so they are parents.

Gene order is (b c pr) that based on the least numbered double crossovers (DCO) (b pr c). It seems that comparison between the non-crossover (NCO) parent b pr + and DCO b pr c the c loci will fall between b and pr and + and + of two different parents.

Single crossovers (SCO): b + c, + pr +, + pr c, b + +

c. Order of the genes is b c pr

Now, map distance between b------>c

Map distance = % recombination = (# in SCO phenotypes + # in DCO phenotypes x 100)/(total # progeny)

= (49+45+5+2)x100/777= 12.998%= 12.998 m.u

distance between c----->pr = (27+23+5+2)x100/777=7.335%=7.335 m.u

distance between b----->pr = 12.998+7.335= 20.333 m.u

d. Interference =1- Coefficient of Coincidence (C.C)

C.C = Obs DCO/ Exp DCO

Obs DCO = 5+2/777=0.009=0.9%

Exp DCO = (% recomb. b---c) (% recomb. c--pr)

= 0.12998x0.07335= 0.95%

Now, C.C = 0.9/0.95= 0.9473

Now, Interference (I) = 1-C.C = 1-0.9473 =0.0527

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