In a cross between a carrier female for red color blindness and a normal male, what proportion of their female progeny is expected to show the trait?

In the scenario described, color blindness is an X-linked recessive trait, which means that the gene responsible for the condition is located on the X chromosome. Males have one X and one Y chromosome (XY), while females have two X chromosomes (XX).

A carrier female for red color blindness, who has one normal vision allele and one color blindness allele, can be represented as X(C)X(c), where X(C) is the X chromosome with the normal vision allele and X(c) is the X chromosome with the color blindness allele. The normal male, representing his X chromosome with normal vision, can be represented as X(C)Y.

When these parents produce offspring, the combinations of their alleles result in the following possible genotypes:

• From the mother (carrier female): X(C) (normal vision) or X(c) (colorblind)

• From the father (normal male): X(C) (normal vision) or Y (which does not contribute to color blindness in females)

The potential combinations for female progeny would be:

1. X(C) from the mother and X(C) from the father = X(C)X(C) (normal vision)

2. X(c) from the mother and X(C) from