1. Tardigrade
  2. Question
  3. Biology
  4. An individual homozygous for gene a and b is crossed with wild type and F1 was back crossed with the double recessive. The appearance of the offsprings is as follows ++ arrow 39 ab arrow 31 a+ arrow 17 +b arrow 13 Find out the distance between genes a and b.

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Q. 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.

NTA AbhyasNTA Abhyas 2020Principles of Inheritance and Variation

Solution:

Solution
(++) = 39; (ab) = 31; (a+) = 17;(b+) = 13.
Total number of progeny = 39 + 31 + 17 + 13 = 100.
No. of recombinants = (a+) + (b+) = 13 + 17 = 30.
$Distance \, between \, a \, and \, b \, = \, \frac{N o . \, o f \, R e c o m b i n a n t s}{T o t a l \, n o . \, o f \, p r o g e n y} \, x \, 100 \, =\frac{30}{100} \, x \, 100 \, = \, 30 \, cM$