When a wheat variety of red kernels (homozygous for two nonallelic and independent dominant genes) is crossed with white kernelled wheat (homozygous for two recessive nonallelic independent genes), the phenotypic ratio in F2 generation would be

(a) 9:7 (b) 1:10:4:1 (c) 1:4:6:4:1 (d) 1:2:4:2:4:2:1

Please explain the answer.

The correct option is c) 1:4:6:4:1 

The phenotypic ratio in F2 generation would be 1:4:6:4:1 
1(dark red): 4(deep red): 6(medium red): 4(light red): 1(white).

This cross  involves characters( kernel colour in wheat) which show polygenic inheritance.  In polygenic inheritance, dihybrid ratio of 1:4:6:4:1 is obtained instead of 9:3:3:1 in F2 generation. In polygenic inheritance, traits are controlled by many genes.  As the number of copies of dominant genes increases than intensity of kernel colur also increases in wheat. 

 
In the given cross red kernel (R₁R₁R₂R₂ is crossed with a white kernel( r₁r₁r₂r₂ )

Parents : R₁R₁R₂R_{2        X}           r₁r₁r₂r₂ ​
Gametes: R₁R2                r₁r₂
F1 :  R₁r_1​R₂ r₂

F2: Selfing      R₁r_1​R₂ r_{2​​    X}      R₁r_1​R₂ r₂

F2 generation
 

	R1R2​	R1r2​​	r1​R2	r1​r2​​
R1R2​	R1R1R2​​R2​ ( Dark red)	R1R1R2​ r2​​(Deep red)	R1r1R2​ ​R2 ​​(Deep red)	R1r1R2​ ​r2​​ (medium)
R1r2​	R1R1R2r2​​ ​​(Deep red)	R1R1r2​​r2​​ (medium)	R1​​r1​R2r2 (medium)	R1​​r1​r2​​r2 (light red)
r1​R2	R1r1​R2  R2  ​​(Deep red)	R1r1​R2  r2​​ (medium)	r1​r1R2  ​R2 (medium)	r1​r1R2  ​r2​​ (light red)
r1​r2	R1r1​R2r2  (medium)	R1r1​r2r2​​ (light red)	r1​r1​R2r2 (light red)	r1​r1r2​​​r2 (White)

The phenotypic ratio in F2 generation would be 1:4:6:4:1