Red-green color blindness is inherited as a recessive X-linked trait

Red-green color blindness is inherited as a recessive X-linked trait. What is the probability that a woman with phenotypically normal parents and a color-blind brother will have a color-blind son? Assume that she has no previous children. Short Answer. Please provide your final answer as a fraction

The correct answer and explanation is :

Final Answer:

The probability that the woman will have a color-blind son is 1/4.

Explanation:

Red-green color blindness is a recessive X-linked trait, which means that the gene responsible for this condition is located on the X chromosome. In this scenario, we need to consider the inheritance pattern and the genetic make-up of the woman in question.

Step-by-Step Analysis:

1. Color-Blind Brother’s Genotype:

• The woman has a color-blind brother, which means he must have inherited the X-linked color blindness allele from his mother (since males have only one X chromosome).
• Therefore, the brother’s genotype is XᴄY, where Xᴄ represents the X chromosome with the color-blind allele and Y represents the Y chromosome.

1. Mother’s Genotype:

• The mother of the woman and her color-blind brother is phenotypically normal but is a carrier of the color blindness allele because she has a color-blind son.
• Hence, the mother’s genotype must be XᴄX, where one of her X chromosomes carries the color-blind allele, and the other does not.

1. Father’s Genotype:

• The father of the woman is phenotypically normal and does not have color blindness. As males have only one X chromosome, he must have the normal X chromosome, so his genotype is XY.

1. Woman’s Genotype:

• The woman inherits one X chromosome from her father and one X chromosome from her mother.
• From her father, she must inherit the normal X (since the father’s genotype is XY).
• From her mother, she could inherit either the Xᴄ chromosome (which carries the color-blind allele) or the normal X chromosome.
• Therefore, the woman’s genotype is either XᴄX (carrier) or XX (normal). Since her brother is color-blind, it indicates that her mother carries the color-blind allele, so the woman must be a carrier, meaning her genotype is XᴄX.

1. Probability of Having a Color-Blind Son:

• The woman is a carrier (XᴄX), and if she has a son, he will inherit one of her X chromosomes (which is either Xᴄ or X) and the Y chromosome from the father.
• If the son inherits the Xᴄ chromosome, he will be color-blind (because males only have one X chromosome).
• The probability of the son inheriting the Xᴄ chromosome is 1/2, as there is a 50% chance he will inherit either Xᴄ or X from his mother.

Thus, the probability that the woman will have a color-blind son is 1/2.

Final Calculation:

The woman’s genotype is XᴄX. Her son will inherit either Xᴄ or X from her, and the probability of him inheriting the Xᴄ chromosome is 1/2. Therefore, the probability of having a color-blind son is:

1/2 (probability of son inheriting Xᴄ).

Hence, the final answer is 1/4.

Here is the diagram showing the inheritance of red-green color blindness, an X-linked recessive trait. It illustrates the possible genotypes of the woman and her children, highlighting the 1/4 probability of having a color-blind son.