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Lecture 2 &acirc;€“ Evolution and Genetics

# Lecture 2 &acirc;€“ Evolution and Genetics -...

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Lecture 2 – Evolution and Genetics A. Genotype - The genetic makeup of an individual - For every characteristic (such as eye color), then you will either be of the genotype BB, Bb, or bb B. Gene pool - The genetic makeup of a population - The sum total of all alleles (genes) for all individuals in a population (the whole world for humans) C. Hardy-Weinberg equations - p + q = 1 Gene pool, defines the gene frequencies p = the frequency of the dominant allele q = the frequency of the recessive allele the gene pool should remain unchanged as long as you meet certain conditions (there are 5) o microevolution is a change in the gene pool - p 2 + 2pq + q 2 = 1 genotype frequencies p 2 = frequency of homozygous dominant genotype 2pq = frequency of heterozygous genotype q 2 = frequency of homozygous recessive genotype In a population of 50, the total number of alleles would be 100 The dominant allele frequency is 70% p = .7, therefore q = .3 because p + q = 1 If there were 50 people to start with, how many people are heterozygous?

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o (.7) 2 + (2)(.7)(.3) + (.3) 2 = 1 o .49 + .42 + .09 = 1 o Then multiply 50 by .42, because 2pq is the heterozygous number Bb = 21 BB = 25 bb = 4 Free earlobe (EE, Ee), attached earlobe (ee) In the class, 12 people have an attached earlobe and 41 people have a free earlobe, there are 53 people in the class o What are the allele frequencies? q 2 = 12/53 = .23 q = .48 1 - .48 = .5 2 = p2 2pq = .50 How many people are homozygous dominant? .27 x 53 = 27 - p = 2N AA + N Aa 2N - q = 2N aa + N Aa 2N Out of 53 people, all of the homozygous recessives were knocked out. To solve the equations, we need to use the bottom equations, you would use the p one Hardy-Weinberg is a starting point D. Conditions -
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