We have learned how selection can change the frequencies of alleles and genotypes in populations. Selection typically eliminates variation from within populations. (The general exception to this claim is with the class selection models we have called "balancing" selection where alleles are maintained in the population by overdominance, habitat-specific selection, or frequency dependent selection). If selection removes variation, soon there will be no more variation for selection to act on, and evolution will grind to a halt, right? This would be true if it were not for the reality of Mutation which will restore genetic variation eliminated by selection. Thus, mutations are the fundamental raw material of evolution.

We will be gin by considering what mutation will do as an evolutionary force acting by itself. Simply, mutation will change allele frequencies, and hence, genotype frequencies. Lets consider a "fight" between forward and backward mutation. Forward mutation changes the A allele to the a allele at a rate (u); backward mutation changes a to A at a rate (v). We can express the frequency (p) of the A allele in the next generation (pt+1) in terms of these opposing forward and reverse mutations, much like forward and reverse chemical equations: (pt+1) = pt(1-u) + qt(v). The first part on the right is accounts for alleles not mutated (1-u), and the second part accounts for the increase in p due to mutation from a to A (the frequency of a times the mutation rate to A). We can also describe the change in allele frequency between generations (Dp) as: Dp = (pt+1) - (pt). This is useful because it lets us calculate a theoretical equilibrium frequency which is defined as the point at which there is no more change in allele frequencies, i.e. when Dp = 0 which is when (pt+1) = (pt); from above: pt(1-u) + (1-p)t(v) = pt [remember, q=(1-p)]. Now solve for p and convince yourself that the equilibrium frequency = p = v/(u+v). Similarly the equilibrium frequency of q = u/(u+v).


In the real world we will generally not find specific evolutionary forces acting alone; there will always be some other force that might counteract a specific force of interest. Our ability to detect these opposing evolutionary forces depends, of course, on the relative strengths of the two (or more) forces. However, it is instructive to examine the conditions where evolutionary forces oppose one another to give us a feel for the complexity of evolutionary processes. Here we will consider a simple case where mutation introduces a deleterious allele into the population and selection tries to eliminate it.

As above we define the mutation rate (u) as the mutation rate to the "a" allele. This will tend to increase the frequency of a (i.e., q will increase). In fact, q increases at a rate of u(1-q); remember, (1-q) = p, or the frequency of the A allele. This mutation pressure will increase the number of alleles which selection can act against.

To select against the a allele, we first will assume complete dominance, i.e., that the deleterious effects of the a allele are only observed in the aa homozygote. Under these conditions, the frequency of "a" (q) decreases by selection at a rate of -sq2(1-q), where s is the selection coefficient. We won't derive this for you, but note that the amount of change generated by this selection is a function of the frequency of the aa homozygote (q2) and the frequency of the A allele (1-q). In other words, the amount of change is proportional to the amount of genetic variation in the population, as we showed last lecture.

If we put these terms for mutation and selection together, the amount of change in the a allele is :

Dq = u(1-q) - sq2(1-q)

Now, if the "fight" between selection and mutation is a "draw", we have a condition where there is no change in the frequency of the a allele since mutation is increasing q just as fast as selection is reducing it. Under these conditions, Dq = 0, and we say that the equilibrium allele frequency, q-hat, has been reached (in formal notation q-hat is q with a circumflex over it). We simply rearrange the above formula so that is becomes : u(1-q) = sq2(1-q). We solve this for q to give the equilibrium allele frequency , q-hat: q = sqrt(u/s) (sqrt stands for square root).

Most mutation rates are fairly small numbers (about 10-6), so this equation suggests that deleterious alleles will be maintained in mutation selection balance at fairly low frequencies. However, this statement is exactly what this mode is intended to illustrate: we cannot say what that frequency is until we know BOTH mutation rate and selection coefficient. However, we will refer back to mutation selection balance several times during the course, so it is essential that you have a feel for dynamics of this evolutionary interaction.


In population genetics, the term "migration" is really meant to describe Gene flow, defined as the movement of alleles from one area (deme, population, region) to another. Gene flow assumes some form of dispersal or migration (wind pollination, seed dispersal, birds flying, etc.) but dispersal is not gene flow (genes must be transferred, not just their carriers)

We can describe gene flow (migration) in a manner similar to mutation. Consider two populations, x and y with frequencies of the A allele of px and py . Now consider that some individuals from population y migrate into population x. The proportion of these y individuals that become parents in population x in the next generation = m. After the migration event, population x can be considered to consist of migrant individuals (proportion m) and non-migrant individuals (proportion [1-m]). Thus the frequency of the A allele in population x in the next generation (px t+1) is just the frequency in the non-migrant portion (= px [1-m]) plus the frequency in the migrant portion (py m). Thus: px t+1 = px t [1-m] + py m.

The change in allele frequency due to gene flow is Dp = (px t+1) - px t which is just;

[px t [1-m] + py m] - px t Multiplying through and canceling terms leaves us with:

Dp = -m(px t - py t). This makes intuitive sense: the change in p depends on the migration rate and the difference in p between the two populations. If we considered a grid or array of populations and focus on one of those populations as the recipient population with all other populations contributing equally to it, then py would be replaced by the average p for all the other populations. Many scenarios are possible.


To address the combined effects of gene flow and selection, we will invoke a "fight" similar to what we described for mutation selection balance above. Consider that some weak allele is wafting over to the other side of the tracks, so to speak, where they do not survive (e.g., fish swimming into New York harbor). There is an evolutionary pressure changing allele frequencies in one direction ( into the harbor), and an opposing evolutionary force eliminating those alleles (sewage killing off genetically intolerant fish). Depending on the relative strengths of these two opposing forces, an equilibrium condition can arise.

Lets consider the movement of the "a" allele, and assume that it is completely recessive in its phenotype of death-by-sewage. The change in allele frequency from the migration into the harbor can be defined as above: Dq = -m(qx t - qy t). (Note that we have changed p to q since we are considering the a allele; x and y refer to the two populations). The change in allele frequency due to selection against this allele is -sq2(1-q) (note that this is the same expression we used in the mutation selection balance above). Putting these two pieces together, we can write the expression for the change in allele frequency that is due to BOTH gene flow and selection: Dq = -m(qx t - qy t) - sqx2(1-qx). When the "fight" between gene flow and selection is a "draw", the system will be in equilibrium and there will be no change in q,

and -m(qx t - qy t) = sq2(1-q).

So, back to the fish. Lets say that aa homozygotes drop dead the minute they enter the East river (but that AA and Aa fish are fine). Outside New York Harbor, the frequencies of the two alleles are equal (p = q= 0.5). It is discovered that in the East river, q = 0.1 over many generations, and the Mayor wants to know what proportion of the fish in the East river come from the outside each generation. This information gives us all we need: it's in equilibrium, qx = 0.1 (East River), qy = 0.5 (outside), and s = 1.0 since the aa's are dead. Plug in the values and you get m = 0.023. This says that 2.3% of the fish in the East river must come from outside each generation to maintain the allele frequency at q = 0.1