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A woman heterozygous for color blindness marries a color blind man. What are the rations of carrier daughters, color bind daughters, normal sons and color blind sons in ${F_1}$ generation?
(A) 1:2:2:1
(B) 2:1:1:2
(C) 1:1:1:1
(D) 1:1:2:2

Last updated date: 24th Jun 2024
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Hint: The color blindness is the eye defect in which the person cannot see different colors. These people struggle for a vision in the bright environment. It is the inherited disorder that might pass from one generation to the other. In color blindness, there will be damage in the cone cells of the retina that provide vision during day time.

Complete Answer:
Let x is considered as the recessive gene carrying the color blindness character in it. X and Y are the dominant gene of the female and the male that do not carry the genes of the color blindness. It is known that the chromosome of the male is resembled by XY and that of female is considered as the XX. In the ${F_1}$ generation, the carrier woman and the color blindness man is crossed to produce the offspring. Hence the heterozygous female is represented as Xx and the color blind ness man is represented as the xY.
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The obtained result in the ${F_1}$ generation is one carrier female, one color blindness daughter, one color blind son and the one normal son. Hence the ratio of the ${F_1}$ generation is $1:1:1:1$.
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Thus, the option (C) is correct.

Note: Take the given male and female chromosomes, inter cross it. From the obtained result it is known that the XY is the normal son without any color blindness, xY is the color blind son. The gene Xx is carrier female that carries the color blind gene to the next generation. xx is the color blind daughter obtained from the ${F_1}$ generation.