Huntington's Disease - Math and Biology

Huntington’s is autosomal dominant.

Therefore, the genotypes for someone affected by Huntington’s would be either **Hh **or hh.

So if I were to cross someone affected by Huntington’s with someone unaffected, the possibilities would be as follows (note that I did not specify whether the parent affected was homozygous or heterozygous).

Hh x hh


HH x hh

And the probability of yielding offspring affected by Huntington’s between an affected parent and a non-affected parent would be 75%, right, as 75% is the average between 100% and 50%.

No, it depends on the probability that the parent with Huntington’s is homozygous or heterozygous. This is unlikely to be 50-50.

This is theoretical. I’m not concerned with the real-world probability of finding someone homozygous for Huntington’s.

Yes, but only if HH and Hh are equally likely.

But since the H allele is so rare in the population, the frequency of HH is negligible compared to Hh. As a first approximation, the prevalence of the H allele is about 1/10,000 (going from wiki…). The frequency of HH parents will thus be 1/100,000,000 (assuming completely random assortment) and Hh parents will be ~ 1/10,000.

And in the real world, I would bet that other circumstances would further reduce the frequency of HH individuals. Namely, if both parents are carriers, they would probably avoid having children. And I wouldn’t be surprised if HH individuals had a much more severe form of the disease. While there’s probably some information about that in animal models, it would be too rare to study in patients. (In fact, I recall that in some animal models, and perhaps patients, there is a correlation between the quantity of mutant Huntington protein and severity of symptoms).

Thank you for verifying. So can we say that this is wrong:

The affected parent could just be HH. Therefore, 50% is wrong. The probability in theory is 75%.

But in the real world, finding someone with double H’s is rare. The real-world probability would be x, as defined by 50%<x<75%

No, the probability in theory depends on the relative frequency of homozygous and heterozygous Huntington’s. Only in a naive theory where you assumed them to be equally frequent would you get the 75% result. But why should we be so naive?

Yes, and we can easily have a theory which incorporates this. “Theory” does not have to mean “Stupid theory”.

For my purposes I am not concerned with the relative frequencies of Hh or HH.

I wanted to make sure that one could average probabilities across Punnett squares. I take that one can indeed do that.

True. And actually, in the hypothetical case where H is the only allele in the population, the probability could be 100%. Really, you want the average of 50% and 100%, weighted according to the frequency of parental genotypes. In my approximation above, that weighted average works out to be 50.01%. Which is a hell of a lot closer to 50% than 75% or 100%.

You “can” do anything. The question is what the result means.

If you want to determine the frequency of dominant property P among the children of someone with P and someone without P, you need to know the relative frequencies of the genes for and not for P. You’ll be taking an average, but it will be a WEIGHTED average, where the weights depend on these frequencies.

You also want to weigh the averages according to the penetrations. 5 percent of people who have the dominant allele will not express symptoms of the disease.

Missed edit:

Even in Punnet-square land (where I spend a lot of my time!) you will usually have more information about the parental frequencies. I.e. you’re doing a cross with H? x hh. But where did you get the H? individuals? Perhaps the grandparents were Hh x Hh, and you pick out H? to be parents in the next generation. In that case, you have a 1/3 chance of getting HH, and 2/3 of getting Hh. Or if the grandparents were H? x hh, any H? offspring will be Hh for sure.

The assumption that Hh and HH are equally likely is naive in just about any example I can think of.

Yeah, well, I’m doing back-of-the-envelope math where 95% = 100% :cool:

It’s not an unreasonable assumption on standardized tests such as the MCAT.

If you don’t have any actual information on the prevalence, then the only reasonable assumption you can make is “some unknown rate higher than 50%”. If anything, it’d be more reasonable to assume that there are no homozygotes than that they’re equal in number to the heterozygotes: Dominant-gene diseases are inherently guaranteed to be rare.

I don’t agree. Given that the MCAT is testing some degree of scientific knowledge, it is not unreasonable to assume that one should know that the disease frequency of Huntington’s disease is low, thus the allele frequency is very very low, thus the probability of having a homozygous pair is low. In order for homo and heterozygous to be equally likely you would have to have about 2/3 of the population affected by Huntingtons.

Having taken the MCAT twice, I can say that they would give you the background information on the prevalence of both, just so to avoid this kind of problem.


For an autosomal dominant disorder the defective form of the gene gets the capital letter. Therefore a patient with Huntington;s would be either Hh heterozygote or HH homozygote.


But the result of that average is meaningless. Your question is asking about the probabilities resulting from a single cross of a Huntington’s individual with a healthy individual. The Huntington’s individual is either HH or Hh. HH will pass the H gene 100% of the time. Hh will pass on the gene 50% of the time. There is no in between probability that is meaningful, unless you consider the relative frequencies of hetero- and homozygotes.

**Let’s assume I wasn’t talking about Huntington’s here. **I was just talking about some autosomal dominant disease. Because we do not know the specific autosomal dominant disease, we do not know the relative frequencies of the genotypes.

And for this autosomal disease, we are told to cross an “affected” parent with an “unaffected” parent. We want to chance that the offspring will be affected. The answer choices are below. Which would you pick?

a) 0%
b) 25%
c) 50%
d) 75%

Here’s my reasoning:
(G for generic autosomal dominant disease)

Gg x gg => 4 offspring; 2 of which will be affected.

GG x gg => 4 offspring; all of which will be affected.

(2 + 4)/8 = 75%. D.

I understand all the information about the relative frequencies of Huntington’s disease genotypes. I understand that this cross ignores a lot of relevant information. Please just entertain the question as presented.

Correct, I made a typo. I wish I could edit my original post, but the time window has passed.

As a grad student in biology, I would answer that the correct probability cannot be determined without more information. Implicit in your answer is the assumption that the two situations are equally likely. You simply cannot assume that. Your answer is not consistent with the idea that the frequencies are unknown; it is true ONLY if the frequencies are 50/50.