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# Population Genetics - The Hardy-Weinberg Principle

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Introduction

Experiment 6 Title: Population Genetics - The Hardy-Weinberg Principle, Evolution and Computer Simulations Objectives: > Compare phenotypic frequencies of inherited human traits in a class. > Determine if the inherited traits conform to the Hardy-Weinberg equilibrium. Results: Activity 1: Human Populations - Single Gene Traits Traits/ Phenotypes Class data (total = 26) Allele frequency Dominant Recessive p q Tongue rolling roller = 18 non-roller (rr) = 8 0.692 0.308 Widow's peak present = 11 absent (ww) = 15 0.423 0.577 Hitchhiker's thumb present = 11 absent (hh) = 15 0.423 0.577 Little finger bent = 13 straight (pp) = 13 0.5 0.5 Ear lobes free = 13 attached (ee) = 13 0.5 0.5 Dimples present = 10 absent (dd) = 16 0.385 0.615 PTC testing taster = 24 Non-taster (tt) = 2 0.923 0.077 Table 1: Data for the inherited human traits In the table above, the frequency of the dominant and recessive alleles are counted based on the formula below, For the dominant alleles, it is assigned with the symbol p, p = No. of dominant alleles / No. of total alleles For the dominant alleles, it is assigned with the symbol q, q = No. of recessive alleles / No. of total alleles Activity II: Human Population Stimulation Simulation 1 - Hardy-Weinberg Equilibrium Generation Class Data AA Aa aa p (AA + 1/2 Aa) ...read more.

Middle

q is 0.55 with a total sample of 20 individuals, the expected genotype frequency of p2 = AA = 0.1225 Expected number of AA individual= expected genotype frequency of AA x Sample = 0.5625 x 20 = 2.45 Expected genotype frequency of Aa = 2pq = 0.455 Expected no. of Aa individuals = Expected genotype frequency of AA � Sample = 0.455 � 20 = 9.10 Expected genotype frequency of aa = q2 = 0.4225 Expected no. of aa individuals = Expected genotype frequency of aa � Sample = 0.4225 � 20 = 8.45 Genotypes O E O - E ( O - E ) 2 ( O - E ) 2/ E AA 4 2.45 -1.55 2.40 0.98 Aa 6 9.10 -3.10 9.61 1.06 aa 10 8.45 1.55 2.40 0.28 Total: 20 20 X2 = 2.32 Table 4: Chi-square test for the fifth (F5) generation Hence the calculated value of X2 = 2.32 is smaller than the critical value of 5.99. Therefore we accept the null hypothesis and state that the dominant and recessive alleles of the parental generation of this sample and subsequent generation conform to the Hardy-Weinberg equilibrium of genotypes. Simulation 2 - Selection against Homozygous Recessives Generation Class Data AA Aa aa p (AA + 1/2 Aa) q (aa + 1/2 Aa) ...read more.

Conclusion

with a total sample of 20 individuals, the expected genotype frequency of p2 = AA = 0.5625 Expected number of AA individual= expected genotype frequency of AA x Sample = 0.5625 x 20 = 11.25 Expected genotype frequency of Aa = 2pq = 0.375 Expected no. of Aa individuals = Expected genotype frequency of AA � Sample = 0.375 � 20 = 7.50 Expected genotype frequency of aa = q2 = 0.0625 Expected no. of aa individuals = Expected genotype frequency of aa � Sample = 0.0625� 20 = 1.25 Genotypes O E O - E ( O - E ) 2 ( O - E ) 2/ E AA 10 11.25 -1.25 1.56 0.14 Aa 10 7.50 2.50 6.25 0.83 aa 0 1.25 -1.25 1.56 1.25 Total: 20 20 X2 = 2.22 Table 7: Chi-square test for the fifth (F5) generation Hence the calculated value of X2 = 2.22 is smaller than the critical value of 5.99. Therefore we accept the null hypothesis and state that the dominant and recessive alleles of this sample, the fifth generation and subsequent generation conform to the Hardy-Weinberg equilibrium of genotypes. Conclusion Objective of the experiment was achieved successfully. A population represented by student in the class has different phenotypic expression and inherited traits. The fifth generation and subsequent generation sample conform to the Hardy-Weinberg equilibrium of genotypes. The allele frequency is found to move more towards dominant allele. ...read more.

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