Here, He could stand for homozygote fitness or homosexual fitness, and He is hetro fitness:
(hetrozygotes only have one gene to pass on, so this term is divided by two)
Sanity check: He=Ho=1 ⇒ x=x i.e. neutrality implies stability at all levels of prevalence (exc. stochasisity)
He=(1-x * Ho)/(1-x)
So, if 5% of people are gay (supported by e.g. number of people signed up to ok cupid) x=0.22, and Ho=0 (oversimplfication) then He=1/0.78=1.28
If only 20% of gayness (to use the scientific term) is explained by genetics, then x=0.1 and He=1.01.
I don’t know enough to even guess whether He=1.28 is plausible, but He=1.01 certainly is a modest fitness increase.
As far as being supported by random mutations, well, the mutation rate is around 10^-8 (I think there are more sublties to this, but its accurate to a first approximation), and since homozygotes have prevalence of x^2, this is enough to support a prevalence [EDIT: a prevalence of hetrozygote carriers] of 10^-4.
the mutation rate is around 10^-8 [...] and since homozygotes have prevalence of x^2, this is enough to support a prevalence of 10^-4.
Surely that can’t be right. If an allele is present with probability 10^-8, the probability of its being present in one of two places is 1-(1-10^-8)^2, which is not anything like 10^-4; it’s almost exactly 2.10^-8.
(This doesn’t change the point I think you’re making, namely that there is no possible way that every instance of non-heterosexuality is the result of an independent mutation at the same site.)
What I meant to say was that if 10^-4 of the population are hetrozygote carriers, then 10^-8 will be homozygotes and their genes will be lost to the next generation (assuming zero fitness), so if new mutations are created at 10^-8, then a prevalence of 10^-4 of the population being hetrozygote carriers is the steady state. In this case, the proportion of homosexuals (or heamophiliacs or whatever) would be 10^-8 times the number of nucleotides that will cause the condition if they mutate.
If the gene in question is dominant, then it’s still 10^-8. But yes, homosexuality cannot be primary caused by random mutations.
Oh, I’m very sorry: I completely misunderstood what you were doing, and failed to read your last paragraph as an application of the calculation you’d just done.
Just to say it again in a different way for the benefit of anyone else who misunderstood in the same way as I did, the point is this:
In equilibrium, the rate at which a mutation enters the population has to equal the rate at which it leaves. If it’s rare and recessive, the main way it leaves will be by homozygotes being less fit. So, taking the simplest possible approximations everywhere: if the mutation rate is m and the prevalence of this thing in (the genes of) the population is f, then every new generation will gain a fraction m and lose a fraction f^2 from dead/infertile homozygotes, so we should have m=f^2. So, e.g., if m=10^-8 then we will eventually get f=10^-4.
All of this needs adjusting if having the thing heterozygously makes a difference to fitness, or if having it homozygously doesn’t reduce your fitness to zero, and that adjustment is what the more complicated formula in skeptical_lurker’s earlier comment is for.
I’ve studied population genetics.
Let x be the prevalence of the genes.
N hetrozygotes=2x(1-x)
N homozygotes=x^2
For stability:
x(1-x) He+Ho x^2=x
Here, He could stand for homozygote fitness or homosexual fitness, and He is hetro fitness:
(hetrozygotes only have one gene to pass on, so this term is divided by two)
Sanity check: He=Ho=1 ⇒ x=x i.e. neutrality implies stability at all levels of prevalence (exc. stochasisity)
He=(1-x * Ho)/(1-x)
So, if 5% of people are gay (supported by e.g. number of people signed up to ok cupid) x=0.22, and Ho=0 (oversimplfication) then He=1/0.78=1.28
If only 20% of gayness (to use the scientific term) is explained by genetics, then x=0.1 and He=1.01.
I don’t know enough to even guess whether He=1.28 is plausible, but He=1.01 certainly is a modest fitness increase.
As far as being supported by random mutations, well, the mutation rate is around 10^-8 (I think there are more sublties to this, but its accurate to a first approximation), and since homozygotes have prevalence of x^2, this is enough to support a prevalence [EDIT: a prevalence of hetrozygote carriers] of 10^-4.
Surely that can’t be right. If an allele is present with probability 10^-8, the probability of its being present in one of two places is 1-(1-10^-8)^2, which is not anything like 10^-4; it’s almost exactly 2.10^-8.
(This doesn’t change the point I think you’re making, namely that there is no possible way that every instance of non-heterosexuality is the result of an independent mutation at the same site.)
What I meant to say was that if 10^-4 of the population are hetrozygote carriers, then 10^-8 will be homozygotes and their genes will be lost to the next generation (assuming zero fitness), so if new mutations are created at 10^-8, then a prevalence of 10^-4 of the population being hetrozygote carriers is the steady state. In this case, the proportion of homosexuals (or heamophiliacs or whatever) would be 10^-8 times the number of nucleotides that will cause the condition if they mutate.
If the gene in question is dominant, then it’s still 10^-8. But yes, homosexuality cannot be primary caused by random mutations.
Oh, I’m very sorry: I completely misunderstood what you were doing, and failed to read your last paragraph as an application of the calculation you’d just done.
Just to say it again in a different way for the benefit of anyone else who misunderstood in the same way as I did, the point is this:
In equilibrium, the rate at which a mutation enters the population has to equal the rate at which it leaves. If it’s rare and recessive, the main way it leaves will be by homozygotes being less fit. So, taking the simplest possible approximations everywhere: if the mutation rate is m and the prevalence of this thing in (the genes of) the population is f, then every new generation will gain a fraction m and lose a fraction f^2 from dead/infertile homozygotes, so we should have m=f^2. So, e.g., if m=10^-8 then we will eventually get f=10^-4.
All of this needs adjusting if having the thing heterozygously makes a difference to fitness, or if having it homozygously doesn’t reduce your fitness to zero, and that adjustment is what the more complicated formula in skeptical_lurker’s earlier comment is for.