An individual homozygous for gene a and b is crossed with wild type and F 1 was back crossed with the double recessive. The appearance of the offsprings is as follows ++ → 39 ab → 31 a+ → 17 +b → 13 Find out the distance between genes a and b.

Question

An individual homozygous for gene a and b is crossed with wild type and F 1 was back crossed with the double recessive. The appearance of the offsprings is as follows ++ → 39 ab → 31 a+ → 17 +b → 13 Find out the distance between genes a and b., 31, 30, 42.8, 3

MARKS App · Accepted Answer

(++) = 39; (ab) = 31; (a+) = 17;(b+) = 13. Total number of progeny = 39 + 31 + 17 + 13 = 100. No. of recombinants = (a+) + (b+) = 13 + 17 = 30. D i s tan c e b e t w e e n a a n d b = N o . o f R e c o m b i n a n t s T o t a l n o . o f p r o g e n y x 100 = 30 100 x 100 = 30 c M

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