Question

A heterozygous woman carrying the recessive gene for color blindness marries a man who is red/green color blind. Assume the dominant gene is XC (allelle for normal vision) and the recessive gene is Xc (determines clolr blindeness). The mother's genotype is XCXc and the father's is XcY. What percentage of children will be color blind? What percentage of children will be normal? What will be the genotype of the F1 generation?

Answer 1

Percentage of children to be color blind- 50%

Percentage of children to be normal- 50%

Genotypes of F1 generation: XCXc , XcXc, XCY, XcY

Step-by-step explanation:

If we cross multiply these genotypes XCXc and XcY, the resulting offspring would be:

XCXc – normal vision female- 25%

XcXc – colorblind female- 25%

XCY – normal male- 25%

XcY- color blind male- 25%