Question

Red-green colorblindness is an X-linked recessive mutation. A colorblind female has children with a normal male.

a. What are the parental genotypes?
b. What F1 phenotypic and genotypic ratios are produced from this cross?

Answer 1

Parental genotypes will be X(b)X(b) of the female parent and X(B)Y of male parent.
Female parent will produce only one kind of gamete that is [X(b)] and male gametes will be either [X(B)] or [Y]...
All the females in the F1 progeny will be carriers with genotype X(B)X(b) and all the males will be diseased with genotype X(b)Y. Phenotypes will be in accordance with the genotypes. :-)
Did you find my answer satisfactory? :-)

Answer 2

No im sorry im still confused on the phenotype and genotype part of it

Answer 3

Genotypic ratio is 1:1 for F1 progeny and in phenotype all females are healthy but carriers and all males are diseased. All males are red-green colour blind but females aren't. :-)