A woman is phenotypically normal, but her father had the sex-linked recessive condition of red-green color blindness. If she marries a man with normal vision, a. What is the probability that they will have a color blind son? b. What is the probability that they will have a color blind daughter? c. what is the probability that their two children will both have normal vision? d. what is the probability that their first child will have normal vision and their second child will be color blind?
The correct answer and explanation is:
a. Probability that they will have a color blind son
To solve this, we first need to understand the genetic inheritance pattern for color blindness, which is a sex-linked recessive trait located on the X chromosome. The father of the woman is color blind, which means his genotype is XᴄY (where Xᴄ represents the X chromosome carrying the color blind allele, and Y is the Y chromosome). The woman has normal vision but carries the allele for color blindness, so her genotype is XᴄX (heterozygous).
When she marries a man with normal vision, his genotype is XᴺY (where Xᴺ is the normal X chromosome and Y is the Y chromosome). The possible combinations for their children are:
- Sons inherit their Y chromosome from their father and one of the X chromosomes from their mother. The son could inherit either Xᴄ or Xᴺ from his mother.
- Daughters inherit one X chromosome from each parent. Since the father only contributes Xᴺ, the daughter will inherit one Xᴺ from him and one Xᴄ or Xᴺ from her mother.
For a son to be color blind, he must inherit the Xᴄ chromosome from his mother (because males have only one X chromosome). So, the probability of having a color blind son is 50% (since there is a 50% chance the son will inherit Xᴄ).
b. Probability that they will have a color blind daughter
For a daughter to be color blind, she would need to inherit the Xᴄ chromosome from both parents. The father contributes only a normal Xᴺ chromosome to his daughters. Therefore, it is impossible for the daughter to be color blind because she cannot inherit a color blind X from her father.
Thus, the probability of having a color blind daughter is 0%.
c. Probability that their two children will both have normal vision
The probability of having a child with normal vision depends on the combination of X chromosomes they inherit. For a son to have normal vision, he must inherit Xᴺ from his mother, and for a daughter to have normal vision, she must inherit at least one Xᴺ from each parent.
For each child, the probability of normal vision is:
- Sons: 50% chance of inheriting Xᴺ from their mother (since there are two options: Xᴄ or Xᴺ).
- Daughters: 100% chance of having normal vision, because they will inherit Xᴺ from their father and either Xᴄ or Xᴺ from their mother.
So, the probability of both children having normal vision (whether one is a son or daughter) is:
- 50% (normal son) × 100% (normal daughter) = 50%.
d. Probability that their first child will have normal vision and their second child will be color blind
For the first child to have normal vision, the probabilities are:
- Sons: 50% chance of normal vision (inherits Xᴺ from mother).
- Daughters: 100% chance of normal vision.
For the second child to be color blind, the probabilities are:
- Sons: 50% chance of color blindness (inherits Xᴄ from mother).
- Daughters: 0% chance of color blindness (since they inherit only normal X chromosomes from father).
Thus, the probability that the first child will have normal vision and the second will be color blind is:
- 50% (first son with normal vision) × 50% (second son with color blindness) = 25%.
In conclusion, the correct answers are:
a. 50%
b. 0%
c. 50%
d. 25%