First some standard sociobiological assumptions:
There is a locus in the genome for sexual orientation.
There are 2 alleles at that locus, one 'for' homosexuality h, and one
'for'
heterosexuality H.
Heterosexuality is dominant to homosexuality, ie. heterozygotes Hh are
heterosexual (there is some fairly good evidence for this from Dean
Hamer's
group)
The frequency of H is p and the frequency of h is q.
p + q = 1 (ie. there are no other alleles at this locus)
Let us assume that we are in a society where there is no taboo against
homosexuality, and homosexuals generally tend to be non-reproductive.
Therefore, if we take a group of males sorted by genotypes and look at
their
progeny we might score them thus;
HH (heterosexual) fitness 1 Hh (het. carrier) fitness 1 hh (homosexual) fitness 1-s
s = the selection co-efficient against homosexuality. In this tabooless
society, we can assume that homosexuals reproduce at only 10% so the rate
of
heterosexuals, so s = 0.9
Now, remember p+ q = 1
so : (p+q)squared = 1
so: psquared + 2pq + qsquared = 1
There are psquared HH, 2pq Hh and qsquared hh
Thus the frequency of homosexuals in the population is the square of the frequency of the recessive gene that 'causes the behaviour' (excuse the sociobiology-speak)
If we have 1 in 10 men gay, then q = 0.316
The contribution of the 3 genotypes to the next generation is therefore:
HH psquared
Hh 2pq
hh qsquared (1-s)
The total population in the next generation is thus:
psquared + 2pq + qsquared (1-s)
Since psquared + 2pq + qsquared = 1
Let's normalise this to 1 by dividing everything by 1 - sqsquared
Thus the genotypes in generation 2 are:
HH psquared / (1 - sqsquared)
Hh 2pq / (1 - sqsquared)
hh qsquared (1-s) / (1- sqsquared)
What is the frequency of our homosexual gene in generation 2?
It is:
qsquared (1-s) / (1 - sqsqaured) + pq/ (1-sqsquared)
Why? because Hh carry one copy of the homosexuality gene and hh carry 2.
Because p = 1 - q we can simplify this to:
(q - sqsquared) / (1 - sqsquared)
delta q, the rate at which q changes over the generations is:
delta q = -spqsquared (1- sqsquared)
Remember we postulated that in our tabooless society, s = 0.9, so in generation 2 the frequency of the homosexual allele is:
0.316 - (0.9*0.684*0.316*0.316)/(1-[0.316*0.316*0.9])
So in generation 2 we have q down from 0.316 to 0.25
and the frequency of homosexual men is only: 0.25 squared = 6.25% of the population down from 10% in the generation before:
So, 1 thing is clear from this:
If homosexuality is genetic and recessive, and if homosexual men do not
breed as much as heterosexual men, then there will be a slow decrease in
the
population of homosexuals.
So far so good, I assume both Aaron and I agree thus far.
But here is where Aaron goes badly wrong. He assumes that if the taboo is
imposed and our homosexual men are forced to breed as heterosexuals do,
then
the frequency will start to climb back up. WRONG.
In generation 3 our model society is seized by a temendous taboo. All homosexual men are forced to go underground, get married and have families just like heterosexual men.
What is s?
Remember s was 0.9, but now s = 0
so what is delta q now???
delta q = -spqsquared (1-sqsquared)
What do we see on the top line of our equation? s, and s= 0
So delta q is now zero. THE IMPOSITION OF THE TABOO DOES NOT CHANGE THE FREQUENCY OF THE HOMOSEXUAL ALLELE. IN FACT IT WILL HOLD IT STEADY.
This is where Aaron has gone wrong,
The only thing which will change the frequency of q is mutation pressure,
as
I modeled in Friday's post. However ,as I demonstrated then, it will only
increase very slowly over millions of years.
to reiterate: (excuse capitals, used for emphasis only)
THE RELEASE OF A TABOO IS RELEVANT TO A DECREASE IN FREQUENCY OF
HOMOSEXUAL
ALLELES. BUT THE IMPOSITION OF A TABOO (contra Aaron) HAS NO EFFECT ON
INCREASING THE FREQUENCY OF SAID ALLELE.
Hardy-Weinberg equations have been standard in population genetics since 1913.