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
++ $\rightarrow$ 39
ab $\rightarrow$ 31
a+ $\rightarrow$ 17
+b $\rightarrow$ 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.
$Distance \, between \, a \, and \, b \, = \, \frac{N o . \, o f \, R e c o m b i n a n t s}{T o t a l \, n o . \, o f \, p r o g e n y} \, x \, 100 \, =\frac{30}{100} \, x \, 100 \, = \, 30 \, cM$