 The reproductive success of an individual over its lifetime is known as its fitness.

 When individuals differ in their fitness selection takes place.

 In practice, fitness can be difficult to measure over an organisms lifetime.

 Instead other measures that correlate well with lifetime fitness are used to estimate fitness: e.g. survival to reproductive age or reproductive success in a single season.

 The goal in studying selection is to relate variation in fitness to variation in phenotype.

 E.g. we can try to compare variation in fitness to an animal’s size or camouflage color or some other phenotypic measure.

 Remember, fitness is a result of the organisms entire phenotype.

 Population genetics, however, looks at the evolution of alleles at a single locus.

 Population geneticists condense all the components of fitness (survival, mating success, etc.) into one value of fitness called w.

 Evolution depends on changes in the gene pool so we need to consider how alleles affect fitness rather than how genotypes affect fitness.

 The general selection model (next slide) enables us to assess how individual alleles contribute to fitness.

Genotype Initial freq Fitness Abundance In gen t+1 A p w 1 2 A 11 1 p 2 X w 11 A 1 A 2pq w 12 2 2pq X w 12 A q w q 2 2 A 22 2 2 X w 22 Weighted freq. gen t+1 (p 2 X w 11 )/w (2pq X w 12 )/w (q 2 X w 22 )/w

The term “Abundance in gen t+1” tells us for each genotype its abundance relative to other genotypes in the next generation Abund. gen t+1 p 2 X w 11 2pq X w 12 q 2 X w 22 To convert these to true frequencies we standardize them by dividing them by the average fitness of the population w.

 w = p 2 X w 11 + 2pq X w 12 + q 2 X w 22  Note that the formula is the sum of the fitness values for each genotype multiplied by (i.e. weighted by) the genotype frequencies.

Normalized weighted freq. gen t+1 (p 2 X w 11 )/w (2pq X w 12 )/w (q 2 X w 22 )/w These are the frequencies of each genotype in generation t +1.

 Using these weighted genotype frequencies we can calculate the allele frequencies in generation t+1.

 Need to sum alleles across genotypes.

 For the allele A of the A 1 A 1 1 it will be the frequency homozygotes plus half the frequency of the heterozygotes.

 Frequency of allele A 1 [p (t+1) ]  P (t+1) = ( p 2 X w 11 + pq X w 12 )/w  Frequency of allele A 2 [q (t+1) ]  q (t+1) = ( q 2 X w 22 + pq X w 12 )/w

Starting allele frequencies: A 1 Fitness w 11 0.9

w 1.0

12 = 0.8, A 2 w 0.2

= 0.2

22 w = p 2 X w 11 + 2pq X w 12 + q 2 X w 22 = (0.64 X 0.9) + (0.32 X 1) + (0.04 X 0.2) = 0.576 + 0.32 + 0.008 = 0.904

P (t+1) = (p 2 X w 11 + pq X w 12 )/w P (t+1) = 0.64 X 0.9 + 0.16 X 1)/0.904

= 0.576 + 0.16/0.904

= 0.814

Allele A 1 has increased in abundance slightly. In this example the success of the alleles A 1 and A 2 is very sensitive to the frequency of A 2.

 In this example, heterozygotes have the highest fitness, but if A 2 becomes too common A 2 A 2 homozygotes begin to appear and these have very low fitness.

  At lower frequencies of A 2 then A 2 A 2 homozygotes will be rarer and the A 2 allele will increase.

In next slide we lower frequency of A 2 to 0.1.

Allele frequencies: A 1 Fitness w 11 0.9

= 0.9, A 2 w 12 1.0

= 0.1

w 22 0.2

w = p 2 X w 11 + 2pq X w 12 + q 2 X w 22 = (0.81 X 0.9) + (0.18 X 1) + (0.01 X 0.2) = 0.729 + 0.18 + 0.002 = 0.911

P (t+1) = (p 2 X w 11 + pq X w 12 )/w P (t+1) = (0.81 X 0.9 + 0.09 X 1)/0.911

= (0.729 + 0.09)/0.911

= 0.899 (allele A allele A 2 1 has declined very slightly from frequency of 0.9 and has increased to a frequency of 0.101

 There are other ways of computing the effects of selection on allele frequency.

 One approach uses something called the average excess of fitness.

 A relatively simple formula allows us to calculate the net fitness contribution of an allele, which is called the average excess of fitness.

 This is the difference between the average fitness of individuals with that allele and the average fitness of the entire population.

  For example, for the allele A 1 the average excess of fitness is a A1 = [p X (w 11 – w)] + [q X (w 12 – w)]     Where w 11 – w is the difference in fitness between A 1 A 1 individuals and the mean fitness of the population w. W 12 is fitness of A population 1 A 2 heterozygotes. W is mean fitness of P and q are allele frequencies See Box 6.5 in your text page 168 for derivation of this formula.

 The average excess of fitness can be used to calculate how much an allele frequency will change between generations  Δp = p X (a A1 /w)     Δp is change in allele frequency from one generation to the next p is the frequency of the A1 a A1 allele is the average excess of fitness Average fitness of the population is w

 If the average excess of fitness is positive then an allele will increase in frequency.

 If average excess of fitness is negative then the allele will decrease in frequency.

 Δp = p X (a A1 /w)   The equation tells us that how fast an allele increases or decreases depends on both the strength of selection (value of a (p).

A1 ) AND how common an allele is in the population Note that for rare alleles even strong selection will not necessarily result in a rapid increase in an allele’s frequency.

 Alleles can differ greatly in their fitness. E.g. some alleles cause severe diseases and are strongly selected against.

 Many alleles however differ only slightly in their average excess of fitness, but because the effect of the fitness difference compounds over time (just like interest on money) even small differences can result in big changes.

 The compounding effect of natural selection is more effective in large populations than small ones.

 In small populations drift can easily eliminate beneficial mutations. In larger populations drift has less of an effect.

  Effects of drift strong in small populations but weaker in large populations Small advantages in fitness can lead to large changes over the long term in large popultions.

 Relative fitness can be expressed in different ways but often the genotype with the highest fitness is designated as having a relative fitness of w = 1.  Genotypes with lower relative fitness then have values for w of between 0 and 1.

 Another way differences in relative fitness are sometimes expressed by using a parameter (s) called the selection coefficient to describe the reduction in fitness of one genotype vs the other.

 A genotype that has a 20% lower fitness than a competing one would have an s value of 0.2.

 Strength of selection has a strong influence on how fast an allele spreads.

     In pocket mice coat color is affected by a gene with two alleles D and d. D allele is dominant.

DD: dark phenotype Dd: dark phenotype Dd: light phenotype On dark backgrounds light phenotype will be selected against.

 The higher the value of s the more strongly natural selection acts.

 The mouse coat color example is an example of frequency-independent selection . The fitness of a trait is not associated with how common the trait is.

   The commonest form of frequency independent selection is directional selection.

Under directional selection one allele is consistently favored over the other allele so selection drives allele frequencies in only one direction towards a higher frequency of the favored allele.

Eventually favored allele may replace other alleles and become fixed.

 Whether an allele is dominant, recessive or has additive effects (is codominant) will have a strong influence on how fast it spreads in a population.

   Additive: allele yields twice the phenotypic effect when two copies present Dominance: dominant allele masks presence of recessive in heterozygote Recessive: two copies of recessive allele need to be present for alleles effect to be felt.

 Clavener and Clegg’s work on Drosophila.

 Two alleles for ADH (alcohol dehydrogenase breaks down ethanol) ADH F and ADH S

 Two Drosophila populations maintained: one fed food spiked with ethanol, control fed unspiked food.  Populations maintained for multiple generations.

 Experimental population showed consistent long-term increase in frequency of ADH F   Flies with ADH F allele have higher fitness when ethanol present. ADH F enzyme breaks down ethanol twice as fast as ADH S enzyme.

Fig 5.13

 Jaeken syndrome: patients severely disabled with skeletal deformities and inadequate liver function.

 Autosomal recessive condition caused by loss-of-function mutation of gene PMM2 codes for enzyme phosphomannomutase.  Patients unable to join carbohydrates and proteins to make glycoproteins at a high enough rate.  Glycoproteins involved in movement of substances across cell membranes.

 Many different loss-of-function mutations can cause Jaeken Syndrome.

 Team of researchers led by Jaak Jaeken investigated whether different mutations differed in their severity. Used Hardy Weinberg equilibrium to do so.

 People with Jaeken syndrome are homozygous for the disease, but may be either homozygous or heterozygous for a given disease allele.

 Different disease alleles should be in Hardy-Weinberg equilibrium.

 Researchers studied 54 patients and identified most common mutation as R141H.

 Dividing population into “other” R141H and alleles. Allele frequencies are: Other: 0.6 and R141H: 0.4.

    If disease alleles are in H-W equilibrium then we would predict genotype frequencies of Other/other: 0.36

Other/R141H: 0.48

R141H/R141H: 0.16

 Observed frequencies are: Other/Other: 0.2

Other/R141H: 0.8

R141H/R141H : 0 Clearly population not in H-W equilibrium.

  Researchers concluded that R141H is an especially severe mutation and homozygotes die before or just after birth.

Thus, there is selection so H-W assumption is violated.

 If an allele has a positive average excess of fitness then the frequency of that allele should increase from one generation to the next.

 Obviously, the converse should be true and an allele with a negative average excess of fitness should decrease in frequency.

 Dawson (1970). Flour beetles. Two alleles at locus: + and l.

 +/+ and +/l phenotypically normal.

 l/l lethal.

 Dawson founded two populations with heterozygotes (frequency of + and l alleles thus 0.5).

 Expected + allele to increase in frequency and l allele to decline over time.

 Predicted frequencies based on average excess if fitness estimates and observed allele frequencies matched very closely.

 l allele declined rapidly at first, but rate of decline slowed.

Fig 5.16a

 Dawson’s results show that when the recessive allele is common, evolution by natural selection is rapid, but slows as the recessive allele becomes rarer.

 Hardy-Weinberg explains why.

    When recessive allele (a) common e.g. 0.95 genotype frequencies are: AA (0.05) 2 0.0025AA Aa (2 (0.05)(0.95) aa (0.95) 0.095Aa 0.9025aa

With more than 90% of phenotypes being recessive, if aa is selected against expect rapid population change.

2

    When recessive allele (a) rare [e.g. 0.05] genotype frequencies are: AA (0.95) (0.05) 2 2 Aa 2(0.95)(0.05) aa 0.9025AA 0.095Aa 0.0025aa

Fewer than 0.25% of phenotypes are aa recessive. Most a alleles are hidden from selection as heterozygotes. Expect only slow change in frequency of a.

 What is the predicted allele frequency after one generation for the + allele in Dawson’s beetle experiment?

 We can calculate the average excess of fitness and use our formula for Δp (change in p) to find out.

Fitness w ++ 1.0

w +l 1.0

Allele frequencies + = 0.5, l = 0.5

Genotype frequencies in initial generation ++ = 0.25 (p 2 ) +l = 0.5 (2pq) ll = 0.25 (q 2 ) w ll 0.0

 w = p 2 X w ++ + 2pq X w +l + q 2 X w ll  = (0.25 X 1) + (0.5 X1) + (0.25 X 0)  = 0.75

       For the + allele the average excess of fitness is a + = [p X (w 11 – w)] + [q X (w 12 – w)] a+ = [0.5 (1 - 0.75 ) + [0.5 X (1 - 0.75)] = 0.25 Δp = p (a + / w) = 0.5 (0.25/0.75) = 0.167

P t+1 = P + Δp = 0.5 + 0.167 = 0.667

Fig 5.16a

 Graph shows allele frequency was exactly as predicted in beetle population.