Question 5: Applying Mathematical Analysis to Evolutionary Change Coloration pat

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Question 5: Applying Mathematical Analysis to Evolutionary Change Coloration patterns of a species of snake are controlled by a single gene with two alleles (A- dominant, a-recessive). One population of snakes living in North Carolina was sampled and analyzed using RFLP analysis (restriction fragment length polymorphism), a technique that cuts DNA in specific regions that allows detection of genotypes. Sampling and analysis were done at two different times 10 years apart. The distribution of genotypes corresponding to the coloration trait within this population is summarized in the table below Genotype 2002 2012 460 40:3 436 325 104 22 Chi-Square formula Hardy-Weinberg equations: Degrees of Freedom 6 0.05 3.84 5.99 7.82 9.49 .07 12.59 14.07 15.51 0.01 6.64 9.32 11.34 13.28 15.09 16.81 18.48 20.09 a) Calculate the frequency of the recessive allele in the population in 2012. b) Calculate the frequency of the dominant allele in the population in 2012 c) Using the allelic frequencies from 2012, calculate the expected number of individuals in the population carrying the AA genotype if the population was at Hardy Weinberg equilibrium. d) Using the allelic frequencies from 2012, calculate the expected number of individuals in the population carrying the Aa genotype if the population was at Hardy Weinberg equilibrium e) Using the allelic frequencies from 2012, calculate the expected number of individuals in the population carrying the aa genotype if the population was at Hardy Weinberg equilibrium. f) Use chi-square analysis to calculate the chi square value corresponding to the data from parts C through E above g) Analyze whether the snake population in 2012 was in genetic equilibrium based on the set of data available. Explain your reasoning h) Repeat the analysis from parts A through F using data from 2002. Enter the chi-square value you obtain from this analysis i Using both sets of results from 2002 and 2012, analyze whether the snake population is evolving. Explainvouc analysis, including suggesting reasons for any change

Explanation / Answer

A. q2 = 22/750 = 0.029

B. q2 = 0.029. So, q = 0.17
p=1-q = 1-0.17 = 0.83
p2 = 0.688

c. For AA genotype 750 X 0.688 = 517

D. For Aa genotype 2pq = 0.28
750 X 0.28 = 211

E. For aa genotype 750 X 0.029 = 22

F.

G. No it is not in genetic equillibrium. DOF n=2. P value = 5.99. Chi value >P value.

F. A. q2 = 104/1000 = 0.1

B. q2 = 0.1. So, q = 0.31
p=1-q = 1-0.31 = 0.69
p2 = 0.47

c. For AA genotype 1000 X 0.48 = 480

D. For Aa genotype 2pq = 0.42
1000 X 0.42 = 420

E. For aa genotype 1000 X 0.1 = 100

O E O-E (O-E)2 (O-E)2 / E AA 403 517 114 12996 25.13 Aa 325 211 114 12996 25.13 aa 22 22 0 0 0 750 50.26

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