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Read Chapter 6 of text

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We saw in chapter 5 that a cross between two individuals heterozygous for a dominant allele produces a 3:1 ratio of individuals expressing the dominant phenotype: to those expressing the recessive phenotype. For example brachydachtyly (shortening of the digits) displays this pattern of inheritance.

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In the early 1900’s when Mendel’s work was rediscovered there was confusion about how these simple patterns of inheritance affected populations. Why, for example, was not 3 of every 4 people a person with brachdactyly? Why did not dominant alleles replace recessive alleles?

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The confusion stemmed from confusing what was happening at the level of the individual with what occurs at the population level. Individual-level thinking enables us to figure out the result of particular crosses.

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Population level thinking however is needed to figure out how the genetic characteristics of populations change over time. It enables us to figure out quantitatively what is happening in a population as a result of evolution. Remember, evolution occurs when genotype frequencies change over time.

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Study of the distribution of alleles in populations and causes of allele frequency changes

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Diploid individuals carry two alleles at every locus › Homozygous: alleles are the same › Heterozygous: alleles are different Remember Evolution: change in allele frequencies from one generation to the next

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Hardy-Weinberg serves as the fundamental null model in population genetics

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Null models provide us with a baseline. They tell us what we expect to be the case if certain forces are not operating. The Hardy-Weinberg equilibrium tells us what we expect to happen to genotype frequencies when forces such as natural selection are not operating on a population.

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The Hardy-Weinberg model enables us to determine what allele and genotype frequencies we would expect to find in a population if all that is happening is alleles are being randomly assigned to gametes when gametes are made (during meiosis) and those gametes meet up at random.

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The Hardy-Weinberg model examines a situation in which there is one gene with two alleles A 1 and A 2. Recall that alleles are different versions of a gene. There are three possible genotypes A 1 A 1, A 2 A 2,and A 1 A 2

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Hardy and Weinberg used their model to predict what would happen to allele frequencies and genotype frequencies in a population in the absence of any evolutionary forces. Their model produced three important conclusions

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The three conclusions of the H-W model. In the absence of evolutionary processes acting on them: 1. The frequencies of the alleles A 1 and A 2 do not change over time. 2. If we know the allele frequencies in a population we can predict the equilibrium genotype frequencies (frequencies of A 1 A 1, A 2 A 2,and A 1 A 2 ).

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3. A gene not initially at H-W equilibrium will reach H-W equilibrium in one generation.

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1. No selection. › If individuals with certain genotypes survived better than others, allele frequencies would change from one generation to the next.

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2. No mutation › If new alleles were produced by mutation or alleles mutated at different rates, allele frequencies would change from one generation to the next.

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3. No migration › Movement of individuals in or out of a population would alter allele and genotype frequencies.

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4. Large population size. › Population is large enough that chance plays no role. Eggs and sperm collide at same frequencies as the actual frequencies of p and q. › If assumption was violated and by chance some individuals contributed more alleles than others to next generation allele frequencies might change. This mechanism of allele frequency change is called Genetic Drift.

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5. Individuals select mates at random. › Individuals do not prefer to mate with individuals of a certain genotype. If this assumption is violated allele frequencies will not change, but genotype frequencies might.

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Assume two alleles A 1 and A 2 with known frequencies (e.g. A 1 = 0.6, A 2 = 0.4.) Only two alleles in population so their allele frequencies add up to 1.

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Can predict frequencies of genotypes in next generation using allele frequencies. Possible genotypes: A 1 A 1, A 1 A 2 and A 2 A 2

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Assume alleles A 1 and A 2 enter eggs and sperm in proportion to their frequency in population (i.e. 0.6 and 0.4) Assume sperm and eggs meet at random (one big gene pool).

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Then we can calculate genotype frequencies. A 1 A 1 : To produce an A 1 A 1 individual, egg and sperm must each contain an A 1 allele. This probability is 0.6 x 0.6 or 0.36 (probability sperm contains A 1 times probability egg contains A 1 ).

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Similarly, we can calculate frequency of A 2 A 2. 0.4 x 04 = 0.16.

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Probability of A 1 A 2 is given by probability sperm contains A 1 (0.6) times probability egg contains A 2 (0.4). 0.6 x 04 = 0.24.

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But, there’s a second way to produce an A 1 A 2 individual (egg contains A 1 and sperm contains A 2 ). Same probability as before: 0.6 x 0.4= 0.24. Overall probability of A 1 A 2 = 0.24 + 0.24 = 0.48.

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Genotypes in next generation: A 1 A 1 = 0.36 A 1 A 2 = 0.48 A 2 A 2 = 0.16 Adds up to one.

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General formula for Hardy-Weinberg. Let p= frequency of allele A 1 and q = frequency of allele A 2. p 2 + 2pq + q 2 = 1.

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If three alleles with frequencies P 1, P 2 and P 3 such that P 1 + P 2 + P 3 = 1 Then genotype frequencies given by: P 1 2 + P 2 2 + P 3 2 + 2P 1 P 2 + 2P 1 P 3 + 2P 2 P 3

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Allele frequencies in a population will not change from one generation to the next just as a result of assortment of alleles and zygote formation. If the allele frequencies in a gene pool with two alleles are given by p and q, the genotype frequencies will be given by p 2, 2pq, and q 2.

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The frequencies of the different genotypes are a function of the frequencies of the underlying alleles. The closer the allele frequencies are to 0.5 the greater the frequency of heterozygotes.

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You need to be able to work with the Hardy-Weinberg equation. For example, if 9 of 100 individuals in a population suffer from a homozygous recessive disorder can you calculate the frequency of the disease causing allele? Can you calculate how many heterozygotes are in the population?

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p 2 + 2pq + q 2 = 1. The terms in the equation represent the frequencies of individual genotypes. P and q are allele frequencies. It is vital that you understand this difference.

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9 of 100 (frequency = 0.09) of individuals are homozygotes. What term in the H-W equation is that equal to?

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It’s q 2. If q 2 = 0.09, what’s q? Get square root of q 2, which is 0.3. If q=0.3 then p=0.7. Now plug p and q into equation to calculate frequencies of other genotypes.

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p 2 = (0.7)(0.7) = 0.49 2pq = 2 (0.3)(0.7) = 0.42 Number of heterozygotes = 0.42 times population size = (0.42)(100) = 42.

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There are three alleles in a population A 1, A 2 and A 3 whose frequencies respectively are 0.2, 0.2 and 0.6 and there are 100 individuals in the population. How many A 1 A 2 heterozygotes will there be in the population?

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Just use the formulae P 1 + P 2 + P 3 = 1 and P 1 2 + P 2 2 + P 3 2 + 2P 1 P 2 + 2P 1 P 3 + 2P 2 P 3 = 1 Then substitute in the appropriate values for the appropriate term 2P 1 P 2 = 2(0.2)(0.2) = 0.08 or 8 people out of 100.

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Hardy Weinberg equilibrium principle identifies the forces that can cause evolution. If a population is not in H-W equilibrium then one or more of the five assumptions is being violated.

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