Heterozygote Advantage
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[edit] Introduction
If a gene has more than one allele, such that members of the population with two different alleles of that gene (heterozygotes) are favored by natural selection over members of the population with two identical alleles of that gene, then we say that the gene exhibits heterozygote advantage.
(The reader who has no idea what these words mean should consult our article on Mendelian Genetics before attempting to read further.)
[edit] Example: Warfarin resistance in rats
Rats have a gene that comes in two possible alleles, S and R (standing for Susceptible and Resistant). Rats with genotype SS are the original type, and are susceptible to the rat poison warfarin. Rats with genotype SR are resistant to warfarin poisoning, at the cost of requiring rather more vitamin K in their diets than an SS type rat. RR rats are also resistant to warfarin, but require twenty times more vitamin K in their diets than an SS type rat, and so are plagued by malnutrition.
When rats are targeted with warfarin, this gene exhibits heterozygote advantage: the SR form is favored by natural selection over both the SS and the RR genotypes: the slight extra dependence on vitamin K associated with the SR genotype is more than made up for by not dying of warfarin poisoning.
However, natural selection cannot produce a population consisting exclusively of rats with an SR genotype: for by Mendel's laws, if we had a population like that, the next generation (before the operation of natural selection) would be 25% SS, 25% RR, and only 50% SR.
What natural selection will do is maintain a mix of alleles in the gene pool. Consider the case where the gene pool is almost all S alleles. Then an S allele is likely to find itself in the same rat as another S allele, i.e. in a warfarin-susceptible rat. An R allele, on the other hand, is unlikely to find itself in company with another R allele, in a malnourished rat, but is much more likely to be found sharing the genotype with an S allele, conferring warfarin resistance without causing malnutrition. It follows that when R alleles are rare, natural selection favors R alleles over S alleles, and will increase their proportion in the gene pool.
Similarly, if S alleles are rare, natural selection will favor S alleles over R alleles, since in that case the typical S allele will be in an SR rat, whereas the typical R allele will be in an RR rat.
So natural selection will increase the proportion of S alleles if they are scarce, and increase the proportion of R alleles when they are scarce. Unless the population of rats in question is so small that one allele gets fixed in the gene pool by genetic drift, natural selection will maintain a balance of the alleles in the gene pool.
One interesting aspect of this is that the balance is dependent on the rats being targeted with warfarin. In the absence of warfarin, the R allele is nothing but bad news, and is quickly removed from the gene pool by natural selection. This situation is not uncommon in the genetics of pesticide resistance, and is a useful thing to know, since it has practical applications in the use of pesticides.
[edit] Example: sickle-cell anemia
The most famous example of heterozygote advantage is sickle-cell anemia in humans. This involves two forms of the gene for making β-globin: the alleles are usually referred to as A1 and A2.
A person with genotype A1A1 is susceptible to malaria; a person with genotype A1A2 is malaria-resistant, and a person with genotype A2A2 is malaria-resistant but will also suffer from sickle-cell anemia, a painful and dangerous illness.
One interesting aspect of this is that the beneficial aspect of the A2 allele was discovered on the grounds of population genetics, by the great J.B.S. Haldane, one of the founders of that discipline. He reasoned as follows.
He knew that sickle-cell anemia was caused by an allele that was recessive with respect to causing sickle-cell anemia; and he knew also that sickle-cell anemia was very bad news from the point of view of natural selection, with sufferers tending to die young. Now, if this was the only effect of the A2 allele, then natural selection would tend to remove that allele from the gene pool, and the proportion of the A2 allele in the population would be at that level where mutations are introducing new A2 alleles into the gene pool just as fast as natural selection can remove them.
But the selective disadvantage of having sickle cell anemia is so great, and the rate of mutation in humans so low, that this could not conceivably account for the high proportion of A2 alleles found in African populations. Therefore, concluded Haldane, in the light of evolution, the only explanation is that there must be heterozygote advantage --- there must be some great benefit, as yet undiscovered, to having the genotype A1A2. From the geographical distribution of the A2 allele, Haldane reasoned that this benefit must be resistance to some tropical disease, and he guessed, correctly as it turned out, that the disease was malaria.
[edit] Equilibrium
Suppose we have two alleles, P and Q, of a gene, exhibiting heterozygote advantage. As we mentioned in our discussion of warfarin resistance in rats, natural selection will maintain a mixture of the two alleles: for when the P allele is scarce, natural selection will favor it, and similarly when the Q allele is scarce, it will be favored by natural selection.
This means that somewhere in between the gene pool being 100% P and 100% Q, there is a point of equilibrium where natural selection favors P and Q equally. If the proportion of P alleles is greater than equilibrium, then natural selection will favor Q, and the proportion of P alleles in the gene pool will be driven down --- towards the equilibrium; and when the proportion of P alleles is less than equilibrium, then natural selection will favor P, and the proportion of P alleles in the gene pool will be driven up --- towards the equilibrium.
So while genetic drift may nudge the gene pool away from the equilibrium, natural selection will tend to drag it back, and so we should expect the proportions of P and Q in the gene pool to be more or less at the equilibrium point.
In fact, the system that we've just described has three equilibrium points. For if the gene pool was 100% P, then this would also be true of the next generation (ignoring, for now, the possibility of mutation). So if the P allele was at that level, it would stay at that level --- which is the definition of equilibrium. Similarly, a gene pool which was 100% Q would stay 100% Q.
However, these points of equilibrium are unstable. If you disturb the situation slightly --- if, for example, you add a scattering of Q alleles to a population which is otherwise 100% P --- then natural selection will favor the Q alleles and drive the gene pool away from being 100% P, and towards the equilibrium point that involves a mixture of P and Q.
By analogy, an unstable equilibrium is like a pencil balanced on its end: any slight nudge to this system will destroy the balance. By contrast, a stable equilibrium is like a pendulum hanging straight down: any slight nudge to this system will be corrected and the equilibrium restored.
[edit] Equilibrium: the mathematics
If we know the pressures that natural selection exerts on the genotypes PP, PQ, and QQ, then we can figure out what the equilibrium level of P and Q should be in the gene pool.
Suppose that the relative fitness of the genotypes PP, PQ, and QQ are expressed by the ratio 1-s : 1 : 1-t, where s and t are positive numbers less than 1.
We have discussed the concept of relative fitness in the main article on [[population Genetics|population genetics]], but to recapitulate: this means that the proportionate genetic contribution of an average PP zygote to the next generation, as compared to the genetic contribution of an average PQ zygote, are in the ratio 1-s : 1. Similarly, the proportionate genetic contribution of an average PQ zygote to the next generation, as compared to the genetic contribution of an average QQ zygote, are in the ratio 1 : 1-t.
Since 1 is greater than both 1-s and 1-t, this does indeed model heterozygote advantage. Notice that because it is the ratio of the three numbers that interests us, and not their absolute value, we can always describe the relative fitnesses of the genotypes in such a way that the fitness of the PQ genotype is assigned the value 1.
Now, let the proportion of P in the gene pool at the point where the population reproduces be denoted by p, and let the proportion of Q in the gene pool at this point be denoted by q. Since every copy of this gene is either P or Q, it follows that p+q = 1.
Then by Mendel's laws, the proportions of genotypes of the zygotes just after fertilization, and before any selection has taken place, will be p2 for the PP genotype, 2pq for the PQ genotype, and q2 for the QQ genotype.
These zygotes must grow up and undergo selection before reaching the point at which they are ready to reproduce (at which point exactly one generation will have passed.) So after selection, using our knowledge of the proportions of the zygotes and the information about the relative fitnesses of the genotypes, we find that the genotypes PP, PQ, and QQ will be in the ratio (1-s)p2 : 2pq : (1-t)q2.
To express the frequency of each allele as a proportion of the total gene pool, we need only divide each figure through by the sum of these three figures. That is, the proportion of PP genotypes in the gene pool will be (1-s)p2/((1-s)p2+2pq+(1-t)q2)
We can simplify the expression in the denominator by remembering that p+q = 1, and making use of the identity 1 = 12 = (p+q)2 = p2+2pq+q2.
Using this simplification, we find that (1-s)p2+2pq+(1-t)q2 = 1-sp2-tq2.
So, the proportion of PP genotypes in the gene pool after one generation will be (1-s)p2/(1-sp2-tq); the proportion of PQ genotypes will be 2pq/(1-sp2-tq), and the proportion of QQ genotypes will be (1-t)q2/(1-sp2-tq).
Given this information, it is easy to find the proportion of the P allele in the gene pool after one generation: it is the proportion of PP genotypes in the population plus half the proportion of PQ genotypes in the population. So we may write the following equation:
New value of p = ((1-s)p2+pq)/(1-sp2-tq).
At equilibrium, by definition, the value of p stays the same from generation to generation, so the equilibrium value of p is given by solving the equation:
p = ((1-s)p2+pq)/(1-sp2-tq2)
The reader may check that 0 and 1 are both solutions to this equation, corresponding to the two unstable points of equilibrium at which the gene pool is 100% Q or 100% P. We wish to find the third, stable, equilibrium point. We reason as follows.
If
- p = ((1-s)p2+pq)/(1-sp2-tq2)
then
- ((1-s)p2+pq)/(1-sp2-tq2)-p = 0
As we are looking for an equilibrium where p is not 0, we can assume that p is not 0 and divide both sides of this equation by p.
This gives us the equation
- ((1-s)p+q)/(1-sp2-tq2)-1 = 0
As we know that 1-sp2-tq2 is not 0 (the proof is left to the reader) we can use the identity 1 = (1-sp2-tq2)/(1-sp2-tq2) to rewrite this equation as:
- ((1-s)p+q)/(1-sp2-tq2) - (1-sp2-tq2)/(1-sp2-tq2)
Simplifying this, as we can because both the fractions have the same denominator, we get
- ((1-s)p+q-1+sp2+tq2)/(1-sp2-tq2) = 0
In order for the expression on the left to be equal to 0, it is necessary and sufficient for the numerator of the expression to be 0. So it's enough to solve the simpler equation
- (1-s)p+q-1+sp2+tq2 = 0
Using the identity p+q = 1, we get
- (1-s)p+q-p-q+sp2+tq2 = 0
Simplifying:
- -sp+sp2+tq2 = 0
So:
- -sp(1-p)+tq2 = 0
Recalling that q = 1-p, we can rewrite this as
- -spq+tq2 = 0
Now, the situation where q is 0 is one of the unstable points of equilibrium that we're not looking for. So we can assume that q is not 0, and divide both sides of the equation by q, giving the equation
- -sp + tq = 0
Recalling again that q = 1-p, we can rewrite this entirely in terms of p:
- -sp+t-tp = 0.
Finally, solving for p, we obtain the result:
- p = t / (s+t).
This gives the value of p at stable equilibrium.
