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November 28, 2004
X chromosome evolution
A few weeks ago there were reports of a possible ’gene for homosexuality’ on the Xchromosome, which produced a higher incidence of homosexuality in males but increased fertility in females. In comments on the subject I suggested that a gene on the Xchromosome would tend to promote the reproductive fitness of females at the expense of males (up to a limit of twice the advantage to females) since an Xchromosome spends on average twice as long in females as it does in males. While making this comment I had qualms that the argument might be fallacious, so I have tried to think it through properly. My post on the Xchromosome waltz analysed what was meant by saying that an Xchromosome spends twice as long in females as it does in males. I also discovered while rereading Matt Ridley’s Genome that none other than Robert Trivers had already put forward essentially the same proposal: ‘Trivers argued that, because an X chromosome spends twice as much time in women as it does in men, a sexually antagonistic gene that benefited female fertility could survive even if it had twice as large a deleterious effect on male fertility’. [Note 1]. So the argument can’t be fallacious….. can it? ………… Well, not exactly, but it is tricky. Not for the first time, I wish I hadn’t got caught up in the problem! The complication is that in a sexually reproducing population, with equal numbers of males and females, the average reproductive fitness (number of offspring per individual) must be the same for males and females. A particular gene may promote greater fitness among females than in males (or vice versa), but any difference in fitness is constrained by the overall equality of fitness of males and females in the population. This is not a serious constraint when the gene is rare, but if the gene becomes common this factor needs to be taken into account. As the gene approaches fixation its fitness among both males and females must approach the population average (which assuming a stable population must be around two offspring per parent surviving to maturity). The exact pattern of evolution of the gene in males and females will depend on the details of relative fitness (including considerations of genetic dominance, etc). For further discussion see Note 2. (It’s tedious, but I include it on the offchance that someone may want to check the assumptions and workings.) On the assumption that the gene has constant fitness relative to its rival alleles in the same sex, the general outcome is that: a. a gene favouring fertility in females at the expense of males can survive if and only if the advantage in females is more than half the disadvantage in males b. provided the advantage in females is more than half the disadvantage in males, the gene will either go to fixation or settle at an equilibrium short of fixation, depending on the balance of advantage and disadvantage. No matter how great the advantage in females, there can still be an equilibrium frequency if the countervailing disadvantage in males is strong enough. Conversely, no matter how small the advantage, the gene could still go to fixation if the disadvantage in males is also sufficiently small. For more on the conditions for equilibrium, see Note 2. The theoretical possibility of a balanced equilibrium frequency is consistent with the observation of a persistent nonzero incidence of homosexuality in modern populations, though there may of course be other reasons for this, and homosexuality may not have a genetic basis at all (or there may be both genetic and environmental causes). It is also consistent with the quoted comment of Trivers, which refers to a gene ’surviving’, but not necessarily changing in frequency. On the face of it, a fitness advantage among females of at least half the disadvantage to males is a fairly modest requirement, so one might expect such genes on the Xchromosome to be common. One might also expect there to be cases where the disadvantage to males is so small (relative to the female advantage) that the gene would go to fixation. So it may seem a puzzle that such genes have not often been detected. One explanation may be that once the gene has gone to fixation its effects on male bearers would be ’invisible’ because there would be no male nonbearers to compare them with. As pointed out earlier, the average reproductive fitness of males in the population as a whole is necessarily the same as that of females, so the effects would not be evident in differential reproduction. The effect of the gene on males would only manifest itself in traits such as general health and longevity. Since males do tend to have shorter lives than females, this might be due in part to such ’invisible’ fitness disadvantages. (But note that in some organisms, e.g. birds and butterflies, males are the homogametic sex.) On the other hand, such genes may be difficult to find because in practice it is not so easy to increase female fitness at the expense of males. The factors which make females more fertile, such as efficient use of nutrition, are probably in general good for males too. Factors which make females more attractive to males may be an exception to this rule  which is why ‘genes for homosexuality’ are a plausible example  but attractiveness to males is not usually a major constraint on female fertility. I dare say that most of the above points have been made somewhere in the literature, but I don’t recall seeing them.
Note 2 It is reasonable to suppose that each genotype has a constant fitness relative to other genotypes in the same sex. (This would not always be the case, e.g. if there is frequencydependent selection, but we ignore that possibility.) In males there are only two genotypes, so the matter is relatively simple. We may assume that BY has fitness x (where x is the average number of offspring surviving to maturity per individual BY male), while AY has fitness (1c)x, with c representing the disadvantage or ’cost’ of the A gene in males (note that c cannot be greater than 1, since fitness cannot be less than zero). In a stable population with N males the total number of offspring surviving to maturity will be 2N. The frequency of AY is p, and of BY is 1p, so we have pN(1c)x + (1p)Nx = 2N, which gives x = 2/(1cp). The fitness of AY is therefore 2(1c)/(1cp), and there will be 2Np(1c)/(1cp) offspring of AY males in the next generation. (For simplicity I assume discrete generations.) Since each offspring of an AY male has a ½ probability of receiving an A gene, the total number of A genes in the offspring of males will be Np(1c)/(1cp). (All the offspring with an Xchromosome will of course be female, but that is not relevant for the present purpose.) The reduction in the number of A genes in the population due to the disadvantage of the A gene among males is therefore Np  Np(1c)/(1cp), which simplifies to Npc(1p)/(1cp). The position among females is more complicated, because there are three genotypes to consider. The simplest assumption is that gene effects are additive, so that the heterozygote AB has fitness halfway between that of AA and BB. We may therefore assume that BB has fitness x, AB has fitness (1+b)x, and AA has fitness (1+2b)x, with b representing the fitness ’benefit’ of the A gene among females. We assume that these relative fitnesses are constant. Unfortunately the size of x depends also on the proportions of the different genotypes in the population. If we assume random mating, resulting in HardyWeinberg equilibrium, the proportions are p^2, 2p(1p), and (1p)^2 for AA, AB, and BB respectively. Assuming a total of 2N offspring, this gives x = 2/(1+2bp), so the fitness of AA is 2(1+2b)/(1+2bp), and of AB is 2(1+b)/(1+2bp). The AA females have collectively 2Np^2(1+2b)/(1+2bp) offspring, while the AB females have 4Np(1p)(1+b)/(1+2bp) offspring. Since each AA female has two A genes, the expected number of A genes passed on by AA females is ½ x 2 x 2Np^2(1+2b)/(1+2bp), whereas the number of A genes passed on by AB females is ½ x 4Np(1p)(1+b)/(1+2bp). Adding and simplifying we get the total number of A genes passed on by females as 2Np(1+b+bp)/(1+2bp). Since there were 2Np copies of A among females in the first generation, the increase due to the advantage of the A gene among females is 2Np(1+b+bp)/(1+2bp)  2Np = 2Npb(1p)/(1+2bp). It should be noted that with increasing p, the fitness of the female genotypes AA and AB falls, as the numerator is constant while the denominator 1+2bp rises. Conversely, among males the fitness of the genotype AY rises, as the value of the denominator 1cp falls. An intuitive way of understanding this is that as A becomes more common, it is increasingly competing against itself rather than the alternative B, which is weaker than A among females but stronger than A among males. So as p increases, the competition gets stronger for females but weaker for males. We are finally in a position to compare the increase in A genes due to the advantage among females with the decrease due to the disadvantage among males: Increase: 2Npb(1p)/(1+2bp) One point that may be noted is that if b equals exactly ½ c, the decrease (for any nonzero p) is necessarily greater than the increase. No matter how small the frequency of the A gene in the population, it would to some extent be competing against itself, and this is sufficient in principle to tip the balance of advantage against A. With b = ½ c, the number of A genes must therefore gradually decline to zero. This may seem to conflict with the statement ascribed to Trivers, that such a gene could ‘survive even if it had twice as large a deleterious effect on male fertility’. This statement is true if the fitness disadvantage takes the form of a constant ratio in the number of offspring between the two sexes, rather than between competing alleles in the same sex. But this is not realistic unless the gene is rare; if it is common, the overall equality of fitness of males and females in the population must compress any fitness differential between male and female bearers of the A gene. For values of b greater than ½ c, the main point of interest is whether the A gene will go to fixation or settle at some equilibrium value of p. An equilibrium will be reached if the increase due to the advantage among females exactly equals the decrease due to the disadvantage among males, that is, if 2Npb(1p)/(1+2bp)  Npc(1p)/(1cp) = 0. A little algebraic manipulation shows that for values of p such that 1>p>0 this condition is satisfied if and only if 2b  c  4bcp = 0. If we fix any two of the variables we can solve the equation for the third: e.g. for p = ¾ and b = 2/3 we get c = 4/9. It is also useful to put the equation in the form p = (2bc)/4bc. It is evident that for any value of b, it is possible to find values of c so small that p would be greater than 1, indicating that for these values A would go to fixation. Perhaps less obviously, for any value of c it is possible to find values of b such that 1>p>0, i.e. an equilibrium short of fixation. If we take values of b converging from above towards ½ c, 2bc will converge to 0 while 4bc converges to 2c^2, therefore (2bc)/4bc must beyond some point in the convergence take on values between 0 and 1. So whatever the value of c, an equilibrium value of p can always be found with some value of b. This has all assumed that gene effects are additive. I have not worked out the effect of nonadditive fitness in detail, but I presume that if the fitness advantage of the homozygote AA (relative to BB) is less than twice that of AB, the equilibrium level of p would be lower than with additive gene effects, since AA homozygotes, which become more common as p increases, would add less to total female fitness than if the effects were additive, whereas the countervailing fitness disadvantage to AY males would be unaffected.
Posted by David B at
04:09 AM


