Question

In maize coloured (C) and full endosperm (F) is dominant over colourless (c) and shrunken endosperm (f). \(F_1\) generation was subjected to test cross. It produced four phenotypes in the following percentages: \[ \begin{aligned} &\text{Coloured full }=48\% &\text{Coloured shrunken }=5\% &\text{Colourless full }=7\% &\text{Colourless shrunken }=40\% \end{aligned} \] Find out the distance between two non-allelic genes.

• A. 48 units
• B. 5 units
• C. 7 units
• D. 12 units

Hint:

Gene distance is calculated as \[ \boxed{ \text{Map distance (cM)} = \%\text{ recombinants}. } \] Here, \[ 5\%+7\%=12\%. \]

Answer 1

The Correct Option is D

Step 1: Identify parental and recombinant classes. The parental phenotypes are the most frequent: \[ 48\% \quad \text{and} \quad 40\%. \] The recombinant phenotypes are \[ 5\% \quad \text{and} \quad 7\%. \]

Step 2: Calculate recombination frequency. \[ \text{Recombination frequency} = 5+7 = 12\%. \] \[ \boxed{ \text{Map distance} = 12\ \text{map units (cM)}. } \]

Step 3: Choose the correct option. Hence, the correct option is \[ \boxed{(D).} \]