Lecture 7 Fitness and Selection - Biology 206

Biology 206: Evolutionary Genetics Lecture 7 Fitness and selection Readings: E&Z Ch.6: 6.6, 6.7  H-W Assumptions: 1) There is no selection (all individuals U have equal probabilities of...

Biology 206: Evolutionary Genetics Lecture 7 Fitness and selection Readings: E&Z Ch.6: 6.6, 6.7  H-W Assumptions: 1) There is no selection (all individuals U have equal probabilities of RE T survival and reproduction) S LEC I TH 2) There is no mutation (genes do not change from one allelic state to another) 3) There is no migration (genes are not added from outside the population) R E C T U E L is infinitely large) 4) There are no chance events (population A S T L 5) There is random mating  The concept of fitness Fitness: the expected reproductive success of an individual with a particular phenotype (or genotype) Components of fitness: – Survival to reproductive age – Mating success – Fecundity FITNESS of a genotype: (probability of SURVIVAL) x (average # offspring: REPRODUCTION)  Fitness and Selection 100 mice total Fitness and Selection hetero homo  Fitness and Selection changed Allele frequency evolution  Relative Fitness Relative fitness: fitness of a genotype standardized by comparison to other genotypes Calculate the relative fitness (w) of a genotype by dividing it by the fitness of the most fit genotype in the population  How does natural selection cause evolutionary change? Natural selection occurs when genotypes differ in average fitness Goal: predict change in allele frequency from one generation to next Information needed for prediction a. initial allele frequency b. relative fitness of the genotypes at the locus of interest  How does selection cause change in genotype frequencies? In a population of mice, the allele frequencies are: B1 (p) = 0.6 B2(q) = 0.4 Initially, start with population in HWE. Mice that are B1B1 have a survival rate of 100%, mice that are B1B2 survive at 75% and mice that are B2B2 survive at 50%. Genotype B1B1 B1B2 B2B2 Total re Befo tion Frequency at birth c sele Relative fitness (w) r Frequency at reproduction Afte tion c sele  How does selection cause change in genotype frequencies? Genotype B1B1 B1B2 B2B2 Total re Befo tion Frequency at birth 0.36 0.48 0.16 1 c sele Relative fitness (w) 1 0.75 0.5 r Frequency at reproduction 0.36(1) = 0.36 0.48(0.75) = 0.36 (0.16)(0.5) = 0.08 0.80 Afte tion c sele  How does selection cause change in genotype frequencies? Genotype B1B1 B1B2 B2B2 Total re Befo tion Frequency at birth 0.36 0.48 0.16 1 c sele Relative fitness (w) 1 0.75 0.5 r Frequency at reproduction 0.36(1) = 0.36 0.48(0.75) = 0.36 (0.16)(0.5) = 0.08 0.80 Afte tion c sele Problem: our frequencies no longer sum to one! We need to scale our numbers after selection by the overall proportion of the population that survives to reproduce!  How does selection cause change in genotype frequencies? Genotype B1B1 B1B2 B2B2 Total re Befo tion Frequency at birth 0.36 0.48 0.16 1 c sele Relative fitness (w) 1 0.75 0.5 r Frequency at reproduction 0.36(1) = 0.36 0.48(0.75) = 0.36 (0.16)(0.5) = 0.08 0.80 Afte tion c sele Relative freq. at 0.36/0.80 = 0.45 0.36/0.80 = 0.45 0.08/0.80 = 0.1 1 reproduction Problem: our frequencies no longer sum to one! We need to scale our frequency after selection by the overall proportion of the population that survives to reproduce!  How does selection cause change in genotype frequencies? Genotype B1B1 B1B2 B2B2 Total re Befo tion Frequency at birth 0.36 0.48 0.16 1 c sele Relative fitness (w) 1 0.75 0.5 r Frequency at reproduction 0.36(1) = 0.36 0.48(0.75) = 0.36 (0.16)(0.5) = 0.08 0.80 Afte tion c sele Relative freq. at 0.36/0.80 = 0.45 0.36/0.80 = 0.45 0.08/0.80 = 0.1 1 reproduction Genotype B1B1 B1B2 B2B2 Total efo re B c t ion Frequency at birth P (p2) H (2pq) Q (q2) 1 sele Relative fitness (w) w11 w12 w22 r Frequency at reproduction w11p2 w122pq w22q2 " ! Afte tion c sele Relative freq. at w11p / " 2 ! w122pq / " ! w22q2 / " ! 1 reproduction Fitness and Selection  How does selection cause change in genotype frequencies? Genotype B1B1 B1B2 B2B2 Total re Befo tion Frequency at birth 0.36 0.48 0.16 1 c sele Relative fitness (w) 1 0.75 0.5 r Frequency at reproduction 0.36 0.36 0.08 0.80 Afte tion c sele Relative freq. at 0.36/0.80 = 0.45 0.36/0.80 = 0.45 0.08/0.80 = 0.1 1 reproduction Genotype B1B1 B1B2 B2B2 Total efo re B c t ion Frequency at birth P (p2) H (2pq) Q (q2) 1 sele Relative fitness (w) w11 w12 w22 r Frequency at reproduction w11p2 w122pq w22q2 " ! Afte tion c sele Relative freq. at w11p / " 2 ! w122pq / " ! w22q2 / " ! 1 reproduction 2 B 42 Bz 811 ! is the mean fitness of the population " (calculated as the sum of the fitness weighted genotype frequencies): ! = w11p + w122pq + w22q2 " 2  How does selection cause change in allele frequencies? Genotype B1B1 B1B2 B2B2 Total efo re B c t ion Frequency at birth P (p2) H (2pq) Q (q2) 1 sele Relative fitness (w) w11 w12 w22 r Frequency at reproduction w11p2 w122pq w22q2 " ! Afte tion c sele Relative freq. at w11p / " 2 ! w122pq / " ! w22q2 / " ! 1 reproduction Allele frequencies in the zygotes forming the next generation can then be calculated from genotype frequencies: pt+1= w11p2 / " ! + 1/2(w122pq / ") ! qt+1= w22q2 / " ! + 1/2(w122pq / ") ! how do allele free charge not genotype freq E&Z BOX 6.5 Natural selection causes changes in allele frequencies between generations ! ### " ##$ ! =! + !& # $ # $ ! #$$ " ##$ & =& + !& # $ # $ Note: this is what we worked through on the previous slide.  PRACTICE PROBLEM A population of shrimp are susceptible to a harmful bacteria. Shrimp with genotype aa have 60% survival rate compared to shrimp with either of the other genotypes, which both have equally high survival. In a population of 500 shrimp, you find the following genotype frequencies: AA: 150, Aa: 230, aa: 120. What are the genotype frequencies after one round of selection?  PRACTICE PROBLEM 500 individuals, AA: 150, Aa: 230, aa: 120 aa have 60% survival What are the genotype frequencies after one round of selection? 150 230 120 500 Genotype AA Aa aa Total Frequency initial 150 230500 0.46 0.24 1 500 0.3 1201500 Relative fitness (w) 1 I 0.6 After selection 0.3 0.46 0.24 0.6 0.144 " ! 9990W 0.904 Freq. after selection 0.30.904 1 TO33 0.51 0.16 AA50.33 Ads0.51 28 0.16 " != allele A 30.33 4210.51 D s 0.16 4210.51 7  Some notes about calculating fitness If you are given only survival or only fecundity then you use that for absolute fitness If you are given both, then you multiply survival x fecundity to get absolute fitness You then calculate relative fitness, by dividing the absolute fitness of each genotype by the value of the most fit genotype (ie. how fit is each genotype relative to the most fit). I will post additional practice problems in OnQ REMINDER: Relationships among alleles at a locus Additive: allele yields twice the phenotypic effect when two copies present Probability of survival to reproduction 1.0 for AA 0.5 for Aa 0 for aa Dominance: dominant allele masks presence of recessive in heterozygote Probability of survival to reproduction 1.0 for AA 1.0 for Aa 0 for aa REMINDER: Relationships among alleles at a locus Additive: allele yields twice the phenotypic effect when two copies present Additive / codominant Dominance: dominant allele masks presence of doesn't always recessive in heterozygote bad pmeon 13 Why is it difficult for new dominant alleles to go to fixation in a population, even if they increase in frequency very rapidly through selection? A. Additive alleles prevent them from going to fixation B. Residual recessive alleles end up “hiding” in the remaining heterozygous individuals C. Because deleterious mutations are always recessive D. Populations with dominant alleles always suffer from high rates of mutation  Demonstration of directional selection against a recessive lethal in lab experiment Dawson (1970) Genetica flour beetle (Tribolium castaneum) Why doesn’t the lethal allele go to 0 after 12 generations?  Demonstration of directional selection against lethal alleles in lab experiment Dawson (1970) Genetica Lethal alleles at two loci “l” and “Sa” Initial populations all heterozygous (+/ l or +/ Sa) Freq of l gene  Demonstration of directional selection against lethal alleles in lab experiment Dawson (1970) Genetica Lethal alleles at two loci “l” and “Sa” Initial populations all heterozygous (+/ l or +/ Sa) Freq of l gene Freq of Sa gene “l” allele is “Sa” allele is recessive partially additive (expressed in het)  Effects of BENEFICIAL selection on different types of alleles E&Z Fig. 6.18 Note: This is selection for a beneficial allele, selection against a harmful allele would be the inverse of this  Key points 1. Selection occurs when genotypes differ in fitness 2. Use relative fitness to compare different genotypes 3. In the general selection model genotypes change in frequency according to their relative fitness 4. Can calculate genotype frequencies after selection, and predict allele frequencies in the next generation 5. How completely & how quickly a beneficial allele is fixed depends on its expression compared to the alternate allele – understand this figure (6.18)!