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

++ $\to$ 39

ab $\to$ 31

a+ $\to$ 17

+b $\to$ 13

Find out the distance between genes a and b.

• A. 31
• B. 30
• C. 42.8
• D. 3

Answer 1

Answer: (B)

(++) = 39; (ab) = 31; (a+) = 17;(b+) = 13.
Total number of progeny = 39 + 31 + 17 + 13 = 100.
No. of recombinants = (a+) + (b+) = 13 + 17 = 30.
$Distancebetweenaandb=\frac{No.ofRecombinants}{Totalno.ofprogeny}x100=\frac{30}{100}x100=30cM$