FSU PCB 4674 - Chapter 6 – Lectures 6 and 7

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What you need to be able to do for Exam 2Chapter 6 – Lectures 6 and 7Be able to use the equation for selection in haploids to predict genetic change in the next generation.Selection in haploid organisms- A locus has two alleles: A1 and A2.- Each allele has a frequency p and q = 1-p- Each allele has an associated fitness W1 and W2- How will the allele frequencies change?What is meant by “fitness”?Fitness allows you to see what is happening in frequencies rather than absolute numbers. Removes potential confusion of population increase or decrease. Can scale by populationmean fitness OR set one of the alleles as having fitness of 1 and scale the other appropriately. - W = absolute fitness, number of progeny in a generation (assuming non-overlapping generations).- So (for example) N1’=N1W1- p’ = N1’/(N1’+N2’) = N1W1/(N1W1 +N2W2) =- pw1/ωbar- Using same logic q’ = qw2/wbar- Selection coefficient = s = w1-w2Change in allele frequency- Δp = p’ – p = p(w1 – wbar)/ωbar- Change in allele frequency is equal to the final allele frequency minus the initial- ωbar = mean fitness in the populationBe able to estimate allele frequencies from genotype frequenciesExample: a population with genotype frequencies of 0.36(AA), 0.48(Aa), and 0.16(aa).Given genotype frequencies, use the following equations to estimate the frequency of the A and a allelesFreq(A) = F(AA)+0.5F(Aa) = 0.6Freq(a) = F(aa)+0.5F(Aa) = 0.4Be able to calculate allele frequencies from phenotype frequencies for a population in Hardy Weinberg equilibrium.Here is an example that I found online for the question. It shows the genotype frequencies given and how to convert them to allelic frequencies- It is important to remember that p and q represent the dominant and recessive alleles, respectively, for a trait. Also that in HW equilibrium, homozygousdominant genotype is represented by p2, heterozygous genotype is 2pq, and homozygous recessive is q2.1)@A study on blood types in a population found the following genotypic distribution among the people sampled: 1101 were MM, 1496 were MN and 503 were NN. Calculate the allele frequencies of M and N, the expected numbers of the three genotypic classes (assuming random mating). Using/X2, determine whether or not this population is in Hardy-Weinberg equilibrium.OBSERVED GENOTYPE FREQUENCIES:MM (p2) = 1101/3100 = 0.356MN (2pq) = 1496/3100 = 0.482NN (q2) = 503/3100 = 0.162ALLELE FREQUENCIES:Freq of M = p = p2@+ 1/2 (2pq) = 0.356 + 1/2 (0.482) = 0.356 + 0.241 = 0.597Freq of N = q = 1-p = 1 - 0.597 = 0.403.Be able to test a population, given genotype frequencies to see if it is in Hardy Weinberg equilibrium, using a chi-square test.The next step in the equation above is to convert the OBSERVED genotype frequencies into the EXPECTED genotype frequencies, or expected frequencies of the individuals, assuming HW equilibrium. This is done by taking the allele frequencies calculated above, and plugging them in as we talked about before. We must calculate p2, 2pq, and q2, which all represent the genotypes that we are measuring. Once we are done with this, we will be able to use the chi-square equation and actually see if a population is in HW equilibrium or not.EXPECTED GENOTYPE FREQUENCIES (assuming Hardy-Weinberg):MM (p2) = (0.597)2@= 0.357MN (2pq) = 2 (0.597)(0.403) = 0.481NN (q2) = (0.403)2@= 0.162EXPECTED NUMBER OF INDIVIDUALS of EACH GENOTYPE:# MM = 0.357 X 3100 = 1107# MN = 0.481 X 3100 = 1491# NN = 0.162 X 3100 =502CHI - SQUARE (X2):X2@=@Σ (Observed genotype – Expected genotype)2@/@Expected genotypeX2@= (1101-1107)2@/1107 + (1496-1491)2@/1491 + (502-503)2@/503= (-6)2@/1107 + (5)2@/1491 + (-1)2@/503= 0.0325 + 0.0168 + 0.002= 0.0513X2@(calculated)@<@X2@(table)@[3.841, 1 df, 0.05 ls].Therefore, conclude that there is no statistically significant difference between what you observed and what you expected under Hardy-Weinberg. That is, you fail to reject the null hypothesis and conclude that the population is in HWE.Be able to use the equation for selection in diploids to predict genetic change in the next generation.Modeling selection in diploid organisms- Assign fitnesses to genotypes rather than allelesOne-locus selection diploids- Two alleles A1, A2- p = frequency of A1, q = frequency of A2- p2 = frequency of A1A1 genotype, etc.FitnessesGenotype Relative FitnessA1A1w11A1A2w12A2A2w22- p’ = (p2ω11 + pqω12)/ωbar = pω1/ωbar- Where ω1 = pω11 + qω12 WHY??- Dp = p’ – p = p(ω1 – ωbar)/ωbarBe able to calculate changes in allele frequency due to mutationModel of mutation:- p’ = p – μp- μ = rate of mutation from p allele to q allele- q’ = q +μp- After n generations pn = p0e-μnBe able to calculate the equilibrium frequency due to mutation selection balanceMutation selection balance: mutation is adding variation at the same rate that selection is taking it out. Why is there so much genetic variation within natural populations? One way that this can happen is through mutation selection balance.- Say that we have a recessive deleterious alleleo ω22 = 1-s Fitnesses of other genotypes = 1- System with selection followed by mutation- From last lecture (p* = p following selection but before mutation)- p* = p/ (1 – s(1-p)2)Now we add mutation- p’= (1-μ)p* = (1-μ)p/(1 – s(1-p)2)- At equilibrium- p’ = p- (1-p)2 = μ/s q(hat) = squareroot (μ/s)Be able to explain the general pattern of evolution under selection against recessive alleles and contrast it to the pattern under selection against dominance alleles.Pattern of evolution under selection against dominant alleles:- Evolution of the dominant allele will be rapid at first, but will slow as the experiment proceeds, as shown by the graph- Frequency of the dominant allele rises- wAA = wAa = 1-s, waa = 1- p’ = (p2w11 + pqw12)/wbar- Substituting as before- p’ = p(1-s)/(1-2sp + sp2)- If s = 1, then dominant allele is gone in one generation. If s is large then p declines rapidly.Pattern of evolution under selection against recessive alleles:- Evolution of the recessive allele will be rapid at first, but will slow as the experiment proceeds, as shown by the graph- Frequency of the recessive allele falls- wAA = wAa = 1, waa = 1-s- s is the selection coefficient, strength of selection against the aa genotype.- Fitnesses of wAA and wAa are rescaled to one by dividing by wAA. Fitness of recessive homozygote rescaled in the same manner.Experiment shows that dominance and allele


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FSU PCB 4674 - Chapter 6 – Lectures 6 and 7

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