4.2 Examples using X-linked traits

1. First an example just of Mendelian segregation.
Red-green color blindness is controlled by an X-linked gene. The allele causing color-blindness is recessive to the normal allele. Let's say that the normal allele for the gene is Xb+, and that the allele which causes color blindness is Xb. Now, a mother who is Xb+/Xb marries a man who is Xb+/Y. If the percentages for Mendelian segregation hold, what proportion of the sons and what proportion of the daughters will be color blind?

Well, first of all, we know that the sons receive their Y chromosomes from the father, therefore they receive the X from their mother. Since the mother has two X's, and since the father doesn't contribute any, each son has 1/2 chance of receiving the Xb+ allele, and 1/2 chance of receiving the Xb allele. This means that half of the sons, on average, will be color blind.

The daughters receive an X from each parent. The father only has one X to contribute, while the mother can contribute one of two. Therefore, the chance that a daughter will be Xb+/Xb+ is 1*1/2, which is 1/2. The chance that a daughter will be Xb+/Xb is 1*1/2, wich is also 1/2. Since both of these genotypes lead to the normal phenotype, none of the daughters will be color blind.

2. Now an examples using population allele frequencies.
Suppose the Xg+ allele for the Xg-blood-group system has frequency 0.3, and the recessive Xg allele has frequency 0.7. (I have no idea whether these are realistic for any populations.) A woman who is Xg-positive married a man who is Xg-negative. What are the probabilities of the Xg-blood types in their children.

The man has genotype Xg/Y. He hands on the Xg allele to each daughter, and the Y-chromosome to each son. So both sons and daughters will be Xg-positive if they receive an Xg+ allele from their mother, and Xg-negative if they receive the recessive Xg allele from their mother.
So we need to find the probability that the Xg-positive woman has genotype Xg+/Xg, and the probability she has genotype Xg+/Xg+.
Now the population frequency, among women, of Xg+/Xg+ is 0.3*0.3 = 0.09
and of Xg+/Xg is 2*0.3*0.7 = 0.42.
So given she is Xg-positive, the probability she is Xg+/Xg+ is 0.09/(0.09+0.42) = 0.176, and in this case she must pass on an Xg+ allele to each child. The probability the child is Xg-positive is 1.
The probability she is Xg+/Xg is 0.42/(0.09+0.42) = 0.824, and in this case the child is Xg-positive with probabilty 1/2 and Xg-negative with probability 1/2.
So overall, the probability the child is Xg-positive is 0.176+(1/2)*(0.824) = 0.588, while the probability the child is Xg-negative is 0.412.