Hardy+Weinberg+Equilibrium

# Hardy+Weinberg+Equilibrium - Trouble-shooting on...

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1 Trouble-shooting on Hardy-Weinberg Equilibrium What are the essential parts of Hardy-Weinberg equilibrium? From generation to generation within a population, the genotypes of offspring result from putting alleles together in pairs (because individuals are diploid). If there are only 2 alleles in a population (let's call them A & a ), these alleles can come together in 3 different genotypes: AA , Aa or aa . The frequency of a particular genotype depends on the frequency of the alleles within it. How likely are we to get one particular genotype? The likelihood of getting a particular genotype depends on how likely the alleles are to meet each other. In other words, it all depends on the frequency (commonness or rarity) of each allele. We indicate the frequency of a particular item as f(item). So, let's suppose that 30% of the alleles in the parental generation are A (this is the same as saying f( A ) = 0.30) and 70% of the alleles in the parental generation are a (this is the same as saying f ( a ) = 0.70). Then, we can see that the likelihood of getting two A 's together (the genotype AA in the offspring) would be 30% once for the first allele and 30% again for the second allele . Mathematically, we say the f( AA ) = 0.3 x 0.3 or f( AA ) = (0.3) 2 . The f( AA ) = 0.09. In words, we can say that we expect 9% (0.09/100) of the offspring to have the genotype AA , based on the frequency of this allele in the parental generation. For the genotype aa in the offspring, we make a similar calculation: f( aa ) = 0.7 x 0.7 = (0.7) 2 = 0.49. In words, we can say that we expect 49% of the offspring to have the genotype aa , based on the frequency of this allele in the parental generation. To decide what fraction of the offspring population should have the genotype Aa , we have to consider two situations: we might get A for the first allele, and a for the second allele to make Aa , or we might get a first and then A . Here we add the two possibilities together: f( Aa ) = f( A ) x f( a ) plus f( a ) x f( A ). This is the same as f( Aa ) = 2 (f( A ) x f( a )). In our example f( Aa ) = 2 ((0.7) x (0.3)) = 2 x 0.21 = 0.42. In words, we expect the frequency of the genotype Aa to be 42%, based on the frequency of both alleles in the parental generation. Now, let's look at the frequencies of the genotypes: f( AA ) = 0.09 f( aa ) = 0.49 f( Aa ) = 0.42

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2 If we have done our math correctly, we should be able to account for 100% of the population; the frequencies should sum to one and they do here. The general way to think about these frequencies is to construct a Prout square (similar to a Punnett square, but it works at the population level, not at the level of individual matings). Let's use general terms for the frequencies of the alleles: we can call the frequency of
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## This note was uploaded on 10/08/2009 for the course BIS 2 taught by Professor Schwartzandkeen during the Spring '09 term at UC Davis.

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Hardy+Weinberg+Equilibrium - Trouble-shooting on...

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