Handout_-_average_excess_F08

Handout_-_average_excess_F08 - Natural Selection Let's talk...

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Natural Selection Let's talk about natural selection from the population genetics point of view. One of the easiest ways to start is to ignore all the stages at which selection can occur and to just lump them all together into a term that we will call Fitness . Fitness is defined as the (probability of survival) X (the probability of successful mating) X (fecundity). This is just the success of different genotypes at producing new individuals. Now we will assign a fitness value to genotypes that have been produced by random mating. At generation 0, before selection , we will have allele frequencies of p and q and genotype frequencies of p 2 , 2pq, and q 2 . Now let's introduce differential fitnesses of w BB , w Bb , w bb , which represent the average fitnesses of the individuals with each genotype It is convenient but not necessary that the w's are expressed as relative fitnesses , with one of the fitnesses (usually the highest) set equal to 1 and the others expressed as a proportion of that fitness. After selection , the genotypes will be represented in the following relative numbers: Genotype BB Bb bb Numbers p 2 w BB 2pqw Bb q 2 w bb These are just (the probability of getting the genotype with random mating) times (the genotypes' average viability, mating success, and fecundity). It is easier to think about fitnesses by considering only viability selection, but the math works with selection at any stage in the life cycle. Note that these are not frequencies anymore. [A good way to visualize this is to let p = .5 , q =.5, w BB = 1 and w Bb = w bb = 0. Now all of our individuals are BB, but p 2 w BB = p 2 , which not 1.] In order to make them frequencies, we need to divide these numbers by the total number. The total number is just the sum of the three genotype numbers: p 2 w BB + 2pqw Bb + q 2 w bb = W Note that W = the average fitness of the population (since the sum of frequencies of a class times the value of the class equals the average value. See, I told you that you Problem Set 1 would be useful! Note also that W is the average of all the individuals in the population, not just the unweighted average of the 3 genotype fitnesses.) So now our genotype frequencies after selection are: BB Bb bb p 2 w BB / W 2pqw Bb / W q 2 w bb / W We can calculate our allele frequencies after selection as the sum of homozygote frequency and 1/2 the heterozygote frequency, just as we did before: p after selection = [p 2 w BB / W] + [pqw Bb / W] = (p 2 w BB + pqw Bb ) / W if all w are the same (= no selection) then p after selection = p but if they differ then p after selection will not necessarily equal p. That is, if there is differential fitness among genotypes, allele frequencies may change due to selection . Since we will now mate randomly among this post-selection population, it also follows that
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Handout_-_average_excess_F08 - Natural Selection Let's talk...

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