A normal male (XY) married to another color blind (XcXc) female produces a color blind son (XcY) and a color blind carrier daughter (XcX) in the F1 lineage. Intermarriage between the F1 progeny results in F2 progeny with 1 normal son, 1 normal daughter, 1 color blind son and 1 color blind daughter.
here,
A normal male has a genotype of XY
Genotype XcXc of a color blind woman
The first progeny is F1 and the second progeny is F2
Parents : ♂ × ♀
Phenotype: Normal color blindness
Genotype : XY XcXc
Gamit : (X) (Y) (Xc) (Xc)
F1 Generation : XcX XcY
Self fertilization : F1×F1
Parents : ♂ × ♀
Phenotype: color blind carrier
Genotype : XcY XcX
Gamit : (Xc) (Y) (Xc) (X)
F2 Generatin (checker board)
Comment: From the checker board, out of 4 children, 1 normal son, 1 normal daughter, 1 color blind son and 1 color blind daughter. Color blindness requires one Xc gene in males and two Xc genes in females.
Phenotypic Ratio : Normal Son : Normal Daughter : Color Blind Son : Color Blind Daughter = 1:1:1:1