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

++ $\to$ 39

ab $\to$ 31

a+ $\to$ 17

+b $\to$ 13

Find out the distance between genes a and b.

Question

TestHub · 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.

$D i s \tan c e b e t w e e n a a n d b = \frac{N o . o f R e c o m b i n a n t s}{T o t a l n o . o f p r o g e n y} x 100 = \frac{30}{100} x 100 = 30 c M$

Zoology - Principles of Inheritance and Variation Question with Solution | TestHub

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