# Really having problem comprehending this Hardy-Weinberg example from biology textbook

Genetic equilibrium is a hypothetical state, but it is often used as a benchmark. Consider how the Hardy–Weinberg equations were used in early studies of an allele that causes hereditary hemochromatosis. Individuals affected by this disorder absorb too much iron from their food, and the excess accumulates in tissues and organs. Untreated, the iron overload results in fatigue and pain; in serious cases, cirrhosis of the liver, heart disease, diabetes, and other problems can occur.

The allele is inherited in an autosomal recessive pattern, so carriers (people heterozygous for the allele) have no symptoms. People of northern European descent have a very high risk of hereditary hemochromatosis compared with other populations; about 1 in 83 people of Irish descent have the disorder. Researchers discovered that about 28% of normal (unaffected) Irish individuals were carriers, which means the frequency of the allele among members of the Irish population (q) is 0.14. Thus, the frequency of Irish people who are homozygous for the allele (q2) is 0.0196, or 1 in 51 individuals. This calculated frequency is higher than the observed prevalence of the disorder, so the researchers concluded (correctly) that other factors probably contribute to the progression of symptoms.

I'm really having problem on understanding what the last paragraph is trying to say. It says if (1/83) have the disorder, that means q^2 = (1/83), and q = 0.11

That means p = 0.89, since p+q = 1. That also means that the heterozygous population is 2pq = 2(0.89)(0.11) = 0.20, or 20%.

I'm having trouble comprehending how they figured out q=0.14. I don't even get what "28% of normal individuals were carriers" mean. Does it mean 28% of homozygous dominant and heterozygous individuals combined? If so, how do you get q = 0.14 from that?

I just do not get what this is trying to say.

You get off track when you state:

(1/83) have the disorder, that means q^2 = (1/83), and q = 0.11

For that to be correct, each and every person who is homozygous for the problem allele must show symptoms of the disorder and no others can do so. When they stated "have the disorder" they meant have clinical symptoms of the disorder, and nothing about knowledge of the frequency of the problem allele in the population. Hence, you shouldn't have jumped to that conclusion and knew from that what q was.

The text says:

Researchers discovered that about 28% of normal (unaffected) Irish individuals were carriers, which means the frequency of the allele among members of the Irish population (q) is 0.14.

Here they are talking about results from genetic testing of normal individuals. The tests revealed that 28% were heterozygous, having both a normal allele and a problem allele. This is talking about the frequency of the alleles, and so this 0.28 should be the sum of the probabilities for the two heterozygous areas of the Punnett square, that is 2pq. When I solve that I get q ≈ 0.168 whereas the text shows 0.14. Whichever it is, the final sentence in the text excerpt would still apply.

Here the discussion is highlighting the subtlety between disease 'prevalence' and the 'penetrance' of the mutation causing the disease. The prevalence is an estimated value for the fraction of the population afflicted by the disease (which really should have statistical errors associated with it based on the sample size of their data). Even if the disease only occurs due to a genetic mutation, not all individuals with the mutation necessarily have symptoms. This is due to incomplete penetrance of the mutation. Think of penetrance like some probability that the genetic component wins over the environmental component, epistatic interactions based on the rest of the genome, and any other complicating factors you can think of that disrupt genetic determinism for Mendelian disorders. If the probability that having the mutation causes the disease is 1, and as mgkrebbs said no other mutations or environmental factors cause the disease, then the mutation is completely penetrant. In that case, and in that case only, you can (post-hoc) infer the allele frequency from the disease prevalence, which is what you tried to do. That said, measuring the penetrance virtually always requires knowing the allele frequency of the mutation, so this isn't how we figure out allele frequencies in human disease genetics. (As an aside, penetrance is still really hard to measure, and estimates in the current literature disagree depending on if they condition on having affected family members or if they use an unbiased sample.)

As for the allele frequency itself, the authors of your textbook made a small mistake in the calculation, as mgkrebbs pointed out. $$2p(1-p)=0.28$$ (again missing error bars), which can be solved to find $$p\approx0.17$$. There is another issue here, which is that there is very little information given about the allele itself with respect to the mode of inheritance (and other normal things to check). The author's claim it is autosomal recessive (which is not necessarily $$h=0$$, since the medical definition only requires unaffected homozygote individuals, and that assertion is based on the sample size of heterozygote carriers assessed; they may have other medical conditions that are not the same as the homozygote-induced medical conditions). I just looked up a few of the mutations responsible for hemochromatosis in OMIM (Online Mendelian Inheritance in Man, which is a database of genetic disease status, including mode of inheritance) and here are the results:

If you click on any of the entries, they mention that compound heterozygous mutations (one mutation from the mother, one from the father, in different locations but both in the same gene) can also cause the disease. This means that assuming it is a single mutant allele with a single frequency $$p$$ is incorrect (unless they included that in their estimates of the carrier frequencies via sequencing). If other mutations in the same gene can cause this disease, then one needs to compute both the carrier frequency over all possible mutations and then check the prevalence to see if the additional target size for compound hets is sufficient to explain the epidemiology. The allele frequency for compound hets if there are two alleles is $$p_{c.h.} = p_1^2 + 2p_1p_2 + p_2^2$$. If there are more than two alleles, this becomes a binomial expansion $$(\sum p_i)^2$$, which can grow quite quickly due to the cross terms $$2p_i p_j$$. This could be part of the issue between the discrepancy in disease prevalence and homozygote status, but it goes in the wrong direction! This likely means that there is a larger environmental component at work (back to this in a bit).

Looking more into this specific disease, here is an article about hemochromatosis in Ireland: https://www.liebertpub.com/doi/10.1089/109065701753145583, which cites the following statistic in the abstract: "Over 93% of Irish HH patients are homozygous for the $$HFE$$ gene $$C282Y$$ mutation, providing a reliable diagnostic marker of the disease in this population. However, the prevalence of the $$C282Y$$ mutation and that of the second $$HFE$$ gene mutation, $$H63D$$, have yet to be determined within the Irish population. ". Granted, this paper is from 2004, but I couldn't quickly find a more recent statistic on the allele composition of affected individuals. This means that there are not only multiple alleles involved, albeit with one responsible for the vast majority of Irish cases, but also that there are multiple genes involved (again increasing the target size).

So what's going on? The key is environment and one other factor: sex. "[A]pproximately 50% of women and 20% of men do not have an increase in serum ferritin level", cited here: https://www.sciencedirect.com/science/article/pii/S1542356505004131. Serum ferritin is responsible for the adverse component of the disease, so this paper tells us that women are a coin flip away from not having the disease even if they are homozygotes for the most common mutation causing the the disease in the Irish population. Additionally, they measure the 'heritability' (https://medlineplus.gov/genetics/understanding/inheritance/heritability/), which is the fraction of the variance in the population explained by genetics. They find that the heritability is $$0.35\pm0.1$$ (error bars, finally!!) out of a maximum 1. This means that 35% of the distribution of disease in the population can be explained by genetics alone (indicating something about how likely you are to get the disease if your parents are both carriers), and the remaining fraction is unexplained! Usually, we blame this on a number of environmental factors, but in this case, male/female status plays a major role (though not enough to explain 65% of the variance). Compound hets are likely involved in the remaining "missing" heritability (I honestly can't guess to what extent), and the rest could be from a number of things that are too far removed from the topic.

So there is your answer: compute the homozygote frequency from the carrier frequency (properly) and the difference between that and the disease prevalence is wrapped up in all of the human genetic considerations I outlined. So sorry this turned into a novel, but I wanted to address some of the details that go into the genetics of human disease!