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NEET UG BIOLOGY MCQ: PRINCIPLE OF INHERITANCE AND VARIATION

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An individual homozygous for genes cd\text{cd} is crossed with wild type and F1F_1 crossed back with
the double recessive. The appearance of the offspring is as follows.
++→903cd→897+d→98c+→102\begin{array}{cccccccccccccccccccc} { + + }& \to &{903} \\ {{\text{cd}}}& \to &{897} \\ { + {\text{d}}}& \to &{98} \\ {{\text{c}} + }& \to &{102} \end{array}
The distance between the genes c\text{c} and d\text{d} is
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Correct option: D
10 map units
Option D (10 map units) is correct because the recombination frequency is calculated as (number of recombinants / total offspring) * 100, which in this case is ((98 + 102) / (903 + 897 + 98 + 102)) * 100 = 10.

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