Genetics: Sex-Linked Traits

Biology Tutorial

Genetics: Sex-Linked Traits


In the study of genetics, one mode of non-Mendelian inheritance is sex-linked. This occurs when the gene or genes in question are on the X chromosome (and sometimes on the y chromosome). Remember, we humans have 46 chromosomes existing in pairs. The 23rd pair are sex chromosomes. In general, women have XX, and men have XY. This means our phenotypic ratios for the offspring are different than we would expect for a Mendelian trait or even for non-Mendelian traits that involve incomplete dominance or co-dominance.

Sample Problem

Robert has hemophilia, a sex-linked blood disease. His wife Emily does not and does not have any family history of it. His father’s family does not have any record of anyone with hemophilia. His mother’s grandfather had hemophilia. Robert has a daughter, who does not have hemophilia. Is there a chance she could have children with the disease? Why or why not?


We know that hemophilia is genetic and sex-linked. Based on the information that the trait does not appear every generation and ‘skips’ generations, we know it must be recessive, an allele that is only expressed if it is the only allele present. Notice, Robert has the trait, but his mother did not and her parents did not, but her grandfather did. It skipped two generations. It must be recessive.

Let’s use XH for an X chromosome with the normal, healthy allele, and let’s use Xh for an X chromosome with the allele for hemophila.

Robert’s genotype must be XhY. Remember, men have an X and a Y chromosome. Since he has hemophilia, his X chromosome must have the allele for hemophilia, thus Xh.

Since Emily does not have any family history of hemophilia, her genotype is probably XHXH.

Draw a Punnett square to show the cross between Robert and Emily. Remember, write the dominant allele before the recessive and the X before the Y.


Xh XHXh XHXh Notice, these are the possible genotypes of their daughter.

Y XHY XHY Notice, these are possible genotypes for a son.

You may notice that, if they have a son, he cannot have hemophilia and cannot pass it on. After all, he wouldn’t have the Xh to pass on.

The question is about their daughter. There are two genotypes written in the square for a girl, but they are identical. Both are XHXh, a heterozygous genotype. This matches what we know about their daughter; she does not have hemophila. After all, she has a XH, the dominant allele, that would hide the recessive Xh.

However, she can pass that Xh on to the next generation. In fact, this must be the same genotype as Robert’s mother. Remember, she also does not have hemophilia, but she passed the allele on to Robert. Robert’s daughter has the same chance as her grandmother.

Let’s look at some possible Punnett squares.
If she married a man with hemophilia, the cross would be XHXh x XhY


Xh XHXh XhXh They have a 50% chance of having a daughter with hemophilia.

Y XHY XhY They have a 50% chance of having a son with hemophilia.

If she married a man without hemophilia, the cross would be XHXh x XHY


XH XHXH XHXh Their daughter could be a carrier, but she cannot have the disease.

Y XHY XhY They have a 50% chance of having a son with hemophilia.

In short, YES, Robert’s daughter could have children with hemophilia, even if she marries a man without the disease. She has the recessive allele for hemophilia and may pass it on to a son, who only has one X chromosome.

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