The "big five" assumptions are the ones listed in the main text. However, the basic formulation of Hardy-Weinberg equilibrium also relies on a few other assumptions;

Allele and genotype frequencies don't differ between males and females. That is, the basic form of Hardy-Weinberg does not cover sex-linked genes.

so, is there a modified formula?

  • $\begingroup$ What "main text" are you referring to? $\endgroup$ Mar 31, 2018 at 19:25
  • $\begingroup$ Also, exactly what is your question (modified formula of what?)? Is it on how to calculate genotypic frequencies in the presence of sex-linkage, or on how sex-linkage affects the HW-equilibrium? $\endgroup$ Mar 31, 2018 at 19:33

1 Answer 1


Solving Hardy-Weinberg problems

First of all, you might want to have a look at the post

and eventually at

After reading the first post, you should be able to answer your question yourself. I encourage you to try it before reading what follows.

Hardy-Weinberg for sex-linked loci

I will make an example. Let's consider a case where males are XY and females XX (like in mammals for example). Let A and B be two alleles of a bi-allelic present on the X chromosome. The frequency of these two alleles in the entire population are $p$ and $q$. Let's assume the locus is present on the X chromosome. We will assume that the allele frequency do not differ between males and females.

In males, there are two possible genotypes. A and B. Their frequency among males are noted $f_{m,A}$ and $f_{m,B}$ are

$$f_{m,A} = p$$

$$f_{m,B} = q$$

In females, there are three possible genotypes. AA, AB (or BA) and BB. Their frequency among females are noted $f_{f,AA}$, $f_{f,AB}$ and $f_{f,BB}$ and are

$$f_{f,AA} = p^2$$ $$f_{f,AB} = 2pq$$ $$f_{f,BB} = q^2$$

If the sex ratio (the ratio of the number of males over the number of females) is $r$, then the genotypes frequencies in the overall population are the ones from above multiplied by $r$ and $1-r$ for males and females, respectively.


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