Question

Question 5 [10 marks]

Gamete Frequency B1 B2 B1 B2

For the above gametes, the A locus is under selection and the B locus is neutral. Selection favours the A₁ allele with parameters h = 0.5 and s = 0.01, and the initial frequency of A₁ is p₁= 0.5 (the frequency of A₂ is q₁ = 0.5). What will be the frequency of allele B₁ after selection if its current frequency is p₂ = 0.5 and the coefficient of linkage disequilibrium between the loci is D = 0.25?

Answer 1

Assumptions of Hardy Weinberg: 1) diploid sexual population 2) infinite size, 3) random mating, 4) no selection, migration or mutation. This is a Null Model; obviously some of these assumptions will not hold in real biological situations. The theorem is useful for comparison to real-world situations where deviations from expectation may point to the action of certain evolutionary forces (e.g., mutation selection, genetic drift, nonrandom mating, etc.). Use a Punnet square to determine genotype frequencies: f(AA) = p², f(Aa) = 2pq, f(aa) = q²and p² + 2pq + q² = 1

One generation of random mating restores Hardy Weinberg equilibrium. H-W equilibrium is when the genotype frequencies are in the proportions expected based on the allele frequencies as determined by the relation p² + 2pq + q².

Example: consider a sample of 100 individuals with the following genotype frequencies:

	Observed Genotype Frequencies	Allele count	Allele frequency	Expected genotytpe frequencies under H-W
BB	0.5	1B	p = 25/200 = 0.125	p2 = (.125)2 = 0.015
Bb	0.5	1a, 1b		2pq = 2(.125)(.22) = 0.05
bb	0.5	1 b	p = 25/200 = 0.125	p2 = (.125)2 = 0.015

Observed are different from expected, thus some force must be at work to change frequencies.

NATURAL SELECTION

Selection occurs because different genotypes exhibit differential survivorship and/or reproduction. If we consider a continuously distributed trait (e.g., wing length, weight) with a strong genetic basis, the response to selection can be characterized by where in the distribution the "most fit" (greatest survivorship&reproduction) individuals lie. If after selection one extreme is most fit this is directional selection; if the intermediate phenotypes are the most fit this is stabilizing selection; if both extremes are the most fit this is disruptive selection.

R. A. Fisher proposed a simple bookkeeping, or population genetics, approach for one locus with two alleles: we have AA, Aa and aa in frequencies p², 2pq, q² . Define l_ii as the genotype-specific probability of survivorship, mii as the genotype-specific fecundity. We build a model that will predict the frequencies of alleles that will be put into the gamete pool given some starting frequencies at the preceding zygote stage;

Genotypes	Zygote	-----> ----->	Adult	-----> ----->	Gametes
AA	p2		lAA p2		mAA lAA p2
Aa	2pq		lAa 2pq		mAa lAa 2pq
aa	q2		laa q2		maa laa q2

The gamete column is what determines the frequencies of A and a that will be put into the gamete pool for mating to build the next generation's genotypes. We can simplify by referring to the fitness of a genotype as w_ii = m_ii l_ii . These fitness values will determine the contribution of that genotype to the next generation. Thus the frequency of A allele in the next generation p_t+1 (sometimes referred to as p') would be the contributions from those genotypes carrying the A allele divided by all alleles contributed by all genotypes:

p_t+1 = (w_AA p² + w_Aa pq)/(w_AA p² + w_Aa 2pq + w_aa q²). Or for the a allele,

q_t+1 = (w_aa q² + w_Aa pq)/(w_AA p² + w_Aa 2pq + w_aa q²). Note that the heterozygotes are not 2pq but pq because in each case they are only being considered for the one allele in question. If we scale all wii's such that the largest = 1.0 we refer to these as the relative fitnesses of the genotypes. A worked example where p = .4, q = .6 and w_AA = 1.0 w_Aa = 0.8 w_aa = 0.6:

Genotype frequencies are p² = 0.16, 2pq = 0.48, q² =0.36, thus:

p_t+1 = ((.16 x 1.0) + (.24 x .8))/((.16 x 1.0) + (.48 x .8) + (.36 x .6)) = .463; so q = .537 and thus f(AA)_t+1 = .215, f(Aa)_t+1 = .497 and f(aa)_t+1 = .288. Note both allele frequencies and genotype frequencies have changed (compare to what we saw with inbreeding). This can be continued with the new allele frequencies and so on. When will the selection process stop? when Dp = 0, i.e., when p_t+1 = p_t . In some situations this will stop only when one allele is selected out of the population (p = 1.0).

Now we can consider various regimes of selection (s = selection coefficient, (1-s) is fitness):

	AA	Aa	aa
I	1	1	1 - s	selection against recessive
II	1 - s	1 - s	1	selection against dominant
III	1	1 - hs	1 - s	incomplete dominance (0<h<1)
IV	1 - s	1	1 - t	selection for heterozygotes

Substitute the fitnesses (w_ii) in condition I above into the expression Dp = p_t+1 - p_t and prove for yourself that the equations on page 101 (eqn. 5.5) is related to the expression for p_t+1 shown above. First three are directional in that selection stops only when allele is eliminated. In I the elimination process slows down because as q becomes small the a alleles are usually in heterozygote state and there is no phenotypic variance. In II selection is slow at first because with q small most genotypes are AA so there is low phenotypic variance; as selection eliminates A alleles q increases and the frequency of the favored genotype (aa) increases so selection accelerates. III is like the worked example run to fixation/loss. IV is known as balancing selection due to overdominance (heterozygotes are "more" than either homozygote). Both alleles maintained in population by selection. This is an example of a polymorphic equilibrium (fixation/loss is also an equilibrium condition but it is not polymorphic). The frequencies of the alleles at equilibrium will be:

p_equil = t/(s + t); q_equil = s/(s+t).

Classic example = sickle cell anemia. A=normal allele; S=sickle allele. S should be eliminated because sickle cell anemia lowers fitness. S is maintained where malarial agent (Plasmodium falciparum) exists because AS heterozygotes are resistant to malaria. Note that S allele is very low frequency where there is no malaria (the selective coefficient of S is different because the environment is different). See figure 5.8, pg. 120; table 5.9, pg. 119.

Another way that genetic variation can be maintained is through multiple niche polymorphism (polymorphism maintained by environmental heterogeneity in selection coefficients). If different genotypes are favored in different niches, patches or habitats, both alleles can be maintained.

Yet another way to maintain variation by selection is through frequency dependent selection.Heterozygotes will have the highest average fitness although they are not the most fit in either habitat (see figure 5.12, pg. 124). The same dynamics would apply to temporal heterogeneity (spring and fall; winter and summer) assuming that selection did not eliminate one allele during the first period of selection. Classic example of temporal heterogeneity: third chromosome inversions of Drosophila pseudoobscura studied by T. Dobzhansky. Different chromosomal arrangements ("Standard" and "Chiricahua") show reciprocal frequency changes during the year.

If an allele's fitness is not constant but increases as it gets rare this will drive the allele back to higher frequency. See figure 5.9, pg. 121. Example: allele may give a new or distinct phenotype that predators ignore because they search for food using a "search image" (e.g., I like the green ones).

Most (by no means all) evolutionary biologists believe that selection plays a major role in shaping organic diversity, but it is often difficult to "see" selection. One reason is that selection coefficients can be quite small (1-s ~1) so the response to selection is small. When selection coefficients are large Dp can be large, but the problem here is that with directional selection fixation is reached in a few generations and we still can't "see" selection unless we are lucky enough to catch a population in the middle of the period of rapid change.

What affects the rate of change under selection? Recall that Dp = p_t+1 - p_t

Dp = [(w_AA p2 + w_Aa pq)/(w_AA p2 + w_Aa 2pq + w_aa q2)] - p . With some simple algebra we can rearrange this

equation to: Dp = (pq[p(w_AA - w_Aa) + q(w_Aa - w_aa)])/(w_AA p2 + w_Aa 2pq + w_aa q2)

Note that Dp will be proportional to the value of pq. This value (pq) will be largest when p=q=0.5 or, in English, when the variance in allele frequency is greatest. This is a simplified version of the main point of the fundamental theorem of natural selection modestly presented by R. A. Fisher.

It states that the rate of evolution is proportional to the genetic variance of the population. In the above example we have not explicitly defined the fitnesses wiis or the dominance relationships and these can have a major effect on Dp as written above.

Another important observation for looking at this Dp equation and plugging in some values is that selection always increases the mean fitness of the population. For example with p=0.4, q=0.6 and w_AA=1, w_Aa=0.8 and w_aa=0.6, the mean fitness (w'bar') = 0.76. After one generation of selection p' = 0.463 and q' = 0.537. Recalculating w'bar' we get wbar_t+1 = 0.78, which is greater than 0.76. When will this process stop? At fixation (or equilibrium with over dominance).

This treatment of the algebra of natural selection illustrates what selection alone can do to allele and genotype frequencies. In the next lectures we will consider other evolutionary forces (mutation gene flow, genetic drift), how they act alone, and eventually, how they interact with each of the other evolutionary forces.