Question

1. The inheritance of color blindness in humans is due to a recessive gene located on the X chromosome (X linked). X⁺ (normal) > X^c (color blind)

a. If a color-blind boy is born to parents both of whom have normal vision, what are the genotypes of the three individuals?

b. What is the probability that the second child born to that couple will be a color-blind daughter?

Answer 1

Colorblindness is a condition which is inherited by the X-linked recessive gene present on X-chromosome. Here X+ is normal, dominant and Xc is colorblind, recessive.

Females have two X-linked alleles because females are XX, and male have one X-linked allele that is male are XY. So a male to be colorblind requires only a single Xc gene whereas a female to be colorblind requires two Xc genes, because for the expression of recessive character presence of two recessive alleles are necessary.

Qa) In the question it is given that both the parents are normal and a colorblind boy is born to them.

Answer-The genotype of father XY, mother XcX, boy XcY.

Explanation:-

Here the boy to be colorblind requires a single Xc gene from his mother, because it is a male the presence of a single recessive Xc gene is sufficient to express the recessive character.

Usually boys get the defective Xc gene from their mother, so here the genotype of mother will be XcX (she is a carrier of colorblindness), and genotype of father will be XY (as he is normal), because both the parents are normal, the genotype of the colorblind boy will be XcY.

Qb) Here it is asked the probability of the second child who is a girl born to that couple will be a colorblind.

Answer-

The chance of a girl being colorblind is 0%.

Explanation:-

There is no chance of the second child who is a daughter being colorblind, because a girl to become colorblind got to inherit two defective Xc genes one from father and one from mother, because the gene responsible for colorblndness is a recessive gene and for its expression, in females it requires two recessive genes. But here the father is normal, so daughters cannot be colorblind.

From the below punnette square it can be well studied.

When a women who is a carrier of color-blindness crossed with a male who is normal than the offspring will be

The genotypes of both the parents are female (XcX), male (XY)

When XcX crossed with XY

	X	Y
Xc	XcX (carrier female)	XcY (colorblind male)
X	XX(normal female)	XY(normal male)

So there is no chance of the second child being born to the couple is a colorblind daughter, because here their sons have 50% chance of being colorblind (XcY) and 50% chance of being normal male (XY) and their daughter have 50% chance of being a carrier (XcX) and 50% chance of being completely normal (XX).