Question

During research on cob length of maize, a plant biotechnologist found that this trait is inherited by the additive effects of multiple genes. He gave names to the genes as A, B and C. He started hybridisation with pure lines and finally he self-crossed two individuals having all the genes for this trait in heterozygous conditions. How many offspring can he expect, who will be exactly the same genotypically as the parents with which he started the experiments?

• A. 1/64
• B. 2/64
• C. 8/64
• D. 16/64

Answer 1

The correct option is B 2/64

This is a case of polygenic inheritance, so the expression of the phenotypes shows continuous variations which are contributed not only by the two alternating parents but also the intermediates.

In the cross,
AABBCC $\times$ aabbcc (Parents)
$\Downarrow$
AaBbCc $F_1$ progeny
$\Downarrow$
AaBbCc $\times$ AaBbCc (selfing)

This cross will give rise to 64 progenies, and among them, only 2 will have exactly similar composition of genotypes as well as phenotypes of their parents (first quadrant and last quadrant).