Biology

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?

2 Answer

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

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