Question

"a man with normal vision, xb y, and a woman who is a carrier for color blindness, xb xb, mate. How many total phenotypes result from this cross?"

Answer 1

If Xb Y is males genotype (normal vision)

XB Xb i female genotype (XB is carrier for color blindnes)

If we cross them

Xb Y x XB Xb

F1 (offspring): XbXB XbXb XBY XbY

Female children might be carriers for color blindnes or normal vision (2 phenotypes)

Male children might have color blindnes or with normal vision (2 phenotypes)

Answer 2

The information in the question is incorrect, the correct information is:

A man with normal vision, XB Y, and a woman who is a carrier for color blindness, XB Xb, mate. How many total phenotypes result from this cross?

The answer is 2 phenotypes (offsprings with normal vision and color blindness)

Step-by-step explanation:

According to the information in the question, the allele B is for normal vision while the allele b is for color blindness. The allele B is dominant over allele b, since the color blind allele (b) is masked in an heterozygous state (Bb).

In a cross between a parent male that has normal vision (XB Y) and a female parent heterozygous for the trait i.e. possess different alleles (XB Xb). Note that, the Y chromosome of male is genetically inert I.e. contains no allele.

Four possible offsprings will be produced with distinct two (2) phenotypes: Normal vision and color blindness.

1 female offspring is homozygous for the normal vision allele (XBXB) while the other female offspring is heterozygous i.e. a carrier of the color blindness (XBXb), but phenotypically normal-visioned, since the normal vision allele is dominant.

One male offspring is color blind (XbY) while the other is normal-visioned (XBY).

Hence, in total, 2 phenotypes are produced.