352_Outline16_S11 - Biology 352 Lecture 16 LECTURE 16:...

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Unformatted text preview: Biology 352 Lecture 16 LECTURE 16: NATURAL SELECTION (PART 2) Spring 2011 I. Last Time: The General Selection Model. A. Predicting allele frequencies and amount of change from initial frequencies and fitness values. 2 2 B. Mean population fitness = W = p wAA + 2pqwAa + q waa C. Mean allele fitnesses. 1. W a = pwAa + qwaa 2. W A = pwAA + qw Aa D. q after one generation of selection: q' = q( W a/ W ) E. Change in q after one generation of selection: Δq = [pq( W a - W A)]/ W II. The General Selection Model: An Example. A. Gene with alleles A and a. 1. p = 0.3 and q = 0.7. 2. allele a is recessive. 3. Recessive homozygote has a relative fitness waa = 0.6. B. What is the value of q after one generation of selection? 1. Since this is a dominant-recessive system, wAA = wAa = 1. 2. W a = pwAa + qwaa = (0.3)(1) + (0.7)(0.6) = 0.3 + 0.42 = 0.72 3. W = p2wAA + 2pq wAa + q2 waa. 1 Biology 352 2 Lecture 16 2 Fall 2011 W = (0.3) (1) + 2(0.3)(0.7)(1) + (0.7) (0.6) W = 0.09 + 0.42 + 0.29 = 0.80 4. q' = q( W a/ W ) = (0.7)(0.72/0.80) = 0.63 III. Applying the General Selection Model. A. Different sets of fitnesses can be used to study different selection systems. B. Some selection systems. 1. Selection against a recessive. 2. Selection against a dominant. 3. Selection for the heterozygotes (heterozygote advantage, overdominance). Type of Selection Against a Recessive Against a Dominant* *sAA = sAa Selection for the Heterozygote Partial Dominance AA 1 1 - sAA Genotype Fitness Aa 1 1 – sAa 1 1-0.5 saa aa 1 - saa 1 1 – sAA 1 1 - saa 1-saa Under Dominance 1 1- s 1 C. Equations of general model can be modified by substituting fitness terms from the table. 2 Biology 352 Lecture 16 Spring 2011 D. Example: Selection against a recessive. 2 2 1. W = p wAA + 2pqwAa + q waa W = p + 2pq + q (1 - s ) W = p + 2pq + q - s q W =1-s q aa 2 2 2 2 2 aa 2 aa 2. W a = pwAa+ qwaa W a = p + q(1 - s ) aa W a = (1 - q) + q - s q aa Wa = 1 - s q aa 3. W A = pwAA + qwAa W A= p + q = 1 4. q' = q( W a / W ) q' = q(1-saaq)/ (1 - saaq ) 5. Δq = pq( W a - W A)]/ W Δq = pq(1 - saaq - 1)/(1 - saaq2) Δq = - saaq2 (1 - q)/( 1 - saaq2) 2 E. Two things to notice. 1. Δq is always negative. 2. System with trivial equilibria: when s = 0; when q = 0 or q = 1. IV. Other Systems of Selection. A. Behavior of models studied by computer simulation. B. Selection against a recessive. 3 Biology 352 Lecture 16 Fall 2011 1. Keep q constant (e.g. q = 0.9). 2. Vary the selection coefficient for the homozygous recessive (saa). 3. Generate a family of curves. 4. Things to note. a. Steepness of curves determined by intensity of selection (s). b. As q gets small, the rate of change in q per generation gets small. C. Selection against a dominant. 1. Keep q constant (q = 0.9) and vary values for s. 2. Family of curves. 4 Biology 352 Lecture 16 Spring 2011 3. Things to note. a. Steepness of curve determined by the intensity of selection. b. Curves do not approach 0 asymptotically. c. A dominant allele will be eliminated in a single generation. D. Selection for the heterozygote. 1. Sometimes called overdominance or heterozygote advantage. 2. Could also be called selection against homozygotes. 3. Fitnesses: wAA = 1 - sAA; wAa = 1 and waa = 1 - saa 4. This system has a stable equilibrium. 5. Δq = pq(sAAp - saaq)/ W 6. Equilibrium will result if p or q = 0 (trivial) or if (sAAp - saaq) = 0 7. qeq = sAA/(sAA + saa) 5 Biology 352 Lecture 16 Fall 2011 8. Things to note. a. Equilibrium determined by selection coefficients, not p and q. b. Equilibrium is stable -- graphical analysis. c. Model maintains both alleles in the population. 1). A mechanism that produces stable polymorphisms. 2). Genetic load. D. Partial Dominance: w = 1 " spq " sq2 Only stable equilibrium is fixation for p (A allele) E. Underdominance: w = 1 " s2 pq If heterozygotes have the lowest fitness….Then stable equilibria exist at fixation of p or q (A or a)…if both alleles are present in a population the equilibrium is! nstable u V. The Selection-Mutation Model. A. Models can be constructed to study joint effects of forces. B. Here's a model that studies the joint effects of selection and mutation. ! 6 Biology 352 Lecture 16 Spring 2011 1. Allele A (frequency p) is dominant to allele a (frequency q). 2. Selection is against the aa homozygote (waa = 1 – saa). 3. There is forward mutation of A alleles to a alleles at a rate µ. C. This model has two forces operating in opposite directions. D. The selection-mutation equilibrium: qeq = (µ/s)1/2 E. Model can be used to estimate mutation rates. F. Example: Tay-Sachs disease. 1. Tay-Sachs disease produced by a recessive allele. 2. Disease is lethal, so saa = 1. 3. Frequency of Tay-Sachs allele is q = 0.001. 4. Assuming selection-mutation equilibrium, then 2 q = µ/s µ = q2s 2 µ = (0.001) (1.0) µ = 1 X 10-6 Next time: Quantitative Genetics (Part I) 7 ...
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This note was uploaded on 05/07/2011 for the course BIOLOGY 352 taught by Professor Townsend during the Spring '08 term at San Diego State.

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