Determine the genotypes of the parents, if it is known that they have one healthy daughter

| March 1, 2021 | education

Determine the genotypes of the parents, if it is known that they have one healthy daughter, one daughter who is a carrier of the color blindness gene (d), one healthy and one sick son. The recessive gene for color blindness is localized on the X chromosome.

x (D) – normal vision, X (d) – color blindness. If the parents have a healthy daughter, a daughter who is a carrier of color blindness, as well as healthy and sick sons, then one of them is heterozygous (for example, the mother) and has the genotype X (D) X (d), and the father is X (D) Y.

Parents: (mother) X (D) X (d) × X (D) Y

Gametes: X (D) X (d) X (D) Y

Prospective offspring: X (D) X (D) daughter is healthy, X (D) X (d) daughter is a carrier of color blindness, X (D) Y is a healthy son, X (d) Y is a son of a color blind patient.



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