This question is background research for a science fiction / fantasy novel, my knowledge of genetic is very basic.

It seems obvious to me that for genetic disorders that are autosomal recessive, if those disorders prevent affected individuals from reproducing, the carrier and affected rates in a population will naturally be smaller. I'm not sure if this is the entire reason why say, blue eyes are more common that cystic fibrosis, but it definitely seems to be a contributing factor.

My question is this: say you have a (fictional) autosomal recessive trait that does not prevent reproduction. Let's call the initial carrier rate in the population for this trait C and the initial affected rate A.

In case it matters, the fictional trait I'm considering might reduce the chance of reproduction (as the people may be shunned, or considered less desirable as a mate) but there is nothing preventing reproduction. Something like "bright purple hair" where there is a stigma against affected individuals -- "your hair marks you as tainted" or something. I'm thinking A is around 0.1% (1 in 1000 births). (And if I understand the math right, C would then be around 6%, or 1 in 16 -- so that C * C / 4 gives A.)

Now let's say an oppressive government decides to prevent all affected individuals from reproducing (but carriers are left alone). Let's assume a 100% success rate for the government at rendering affected people infertile thus preventing their reproduction. What will happen to C and A over time? Is there some simple way to estimate how they will trend over several generations? I don't need exact numbers but I'm having trouble with even coming up with a guess.

  • $\begingroup$ Welcome to Biology.SE. I am not sure I understand your C and A correctly. Is it true that C+A=1? $\endgroup$
    – Remi.b
    May 18, 2016 at 1:47
  • $\begingroup$ Sorry @Remi.b, my terms come from common stuff like literature given to family members of babies born with a genetic disorder (I have some personal experience there). C is what they call the "carrier rate", the proportion of the population that is a carrier. I'm not sure if "affected rate" is a real term, but I mean the proportion of the population born with the disorder. $\endgroup$ May 18, 2016 at 1:55
  • $\begingroup$ Oh ok... so any individual that carry the recessive allele counts in the frequency C regardless of whether it has the disease, is that right? Btw, red hair is not a disease, as you said they can procreate! $\endgroup$
    – Remi.b
    May 18, 2016 at 1:58
  • $\begingroup$ Oh so maybe the terms I used like 'carrier' are more appropriate to disorders/diseases? I thought they were generic enough to apply to any trait!! $\endgroup$ May 18, 2016 at 2:01
  • $\begingroup$ Um, hmm yes I think individuals with the trait do count as part of frequency C. But if it made explanation/calculation easier it would be fine if you want to redefine the terms so that C does not include the "affected" individuals. $\endgroup$ May 18, 2016 at 2:05

1 Answer 1


In population genetics, with a simple one-locus two-allele model, we consider the frequency of the two alleles to be $p$ and $q$ where $p + q = 1$ (and thus $1-p=q$ and $1-q=p$). When an allele is recessive it is only expressed at the phenotypic level when the individual carries two copies of the allele.

Following from your example, gene $A$ might be responsible for hair colour, with alleles $A_1$ and $A_2$ where $A_2$ is a recessive allele responsible for purple hair. I will denote the frequency of $A_1$ with $p$ and frequency $A_2$ with $q$. This means that the frequency of individuals with purple hair ($f_{A_2A_2}$) is $q^2$ and the frequency of people with not-purple hair is $p^2 + 2pq$ (assuming an idealised population). As such, a population of 500 diploid individuals* where there is 100 copies of the $A_2$ allele would have the following genotypic (& phenotypic) frequencies:

$$f_{A_1 A_1} = p^2 = 0.9^2 = 0.81$$ $$f_{A_1 A_2} = 2pq = 2 \times 0.9 \times 0.1 = 0.18$$ $$f_{A_2 A_2} = q^2 = 0.1^2 = 0.01$$

So the allele for purple hair would be present in 19% of the population, but the trait would only manifest in 1% (only ~5.3%, $\frac{1}{19}$, of the people that carry the purple hair allele exhibit the trait). Therefore selection would be quite inefficient at removing this allele from the population, even if you could prevent all purple-haired individuals from reproducing. To find the frequency of $q$ required to give a genotypic/phenotypic frequency where 1 in $x$ individuals have purple hair just take the square root of $1/x$. E.g. to get 1 in 1000 individuals expressing purple hair

$$q = \sqrt{q^2} = \sqrt{(1/1000)}$$

What does this mean for evolution? Alleles that are recessive deleterious are very hard to remove from the population because, at low frequencies, they rarely produce the detrimental effect. Assuming complete dominance ($A_1 A_2$ = exactly like $A_1 A_1$), denoted by $h = 0$, selection only acts on $A_2 A_2$ homozygotes. The frequency of the $A_1$ allele increases at the rate $1/(1-sq)$ where $s$ is the relative fitness advantage. Mutation can also introduce more copies of the $A_2$ allele to the population such that there will be an equilibrium, balanced by the rate of loss due to selection, and rate of gain by mutation. Furthermore, to further build on your analogy, a neighbouring population, where purple-hair is not selected against, may bring copies of the allele in by migration.

Suggested further reading on evolution here and reading on eugenics would likely be of interest to you.

* 500 diploid individuals = 1000 gene copies. Thus $q = 100/(500 \times 2) = 0.1$

  • $\begingroup$ That helps immensely. And even better, one of your links led me to find the free PopG program which seems to provide a way of exploring these values over time, while doing the calculations and graphing for me! $\endgroup$ May 18, 2016 at 14:08
  • $\begingroup$ Just noted a couple minor correction -- in the third equation I think you mean q-squared not p-squared is 0.01. (sorry I don't know how to write equations) Also after you define the rate 1/(1-sq), I think you meant "where 1-sq is the relative fitness advantage" but I'm not sure. $\endgroup$ May 18, 2016 at 18:13

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