Question

2. In consideration of the following genetic cross: XBXb x XBY Color blindness is a recessive trait (b) and normal vision is dominant (B). If the parents above had children: a. What is the possibility of them having a son that is color blind? b. What is the possibility of them having a daughter that is color blind? c. What would the genotypes have to be in order for them to have a color blind daughter?

Answer 1

a. 25% possibility.
b. 0% possibility.
c. The genotype would need to be homozygous (XbXb).

Answer 2

• The possibility of them having a son that is color blind is 25%.

The possibility of them having a daughter that is color blind is 0%.

The genotype of the daughter in order to be color blind must be XbXb.

• For obtaining the answer consider the cross:

XBXb X. XBY-----> XBXB XBY XbXB XbY

From the results of the above cross and give that color blindness is a recessive trait, the phenotypes of the progenies with the obtained genotypes will be as follows:

XBXB ---> female with normal vision

XBY------> male with normal vision

XbXB----> female with normal vision

XbY------> colur blind male

• Therefore the possibility of having a color-blind son is: 1/4 X 100 = 25%

• The possibility of having a color-blind daughter is 0% since no female is color blind.

• In order for them to have a colour blind daughter, it is necessary for the father to carry a color-blind gene and pass it on to the daughter. Since this is a recessive trait the genotype of the colorblind daughter must be XbXb.