Question

Hi fellow tutor,

Please note that from part A-J have been answered. I would like some help please with parts K-N.

Thank you

Question 5: Applying Mathematical Analysis to Evolutionary Change (17 points)

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
AA	460	403
Aa	436	325
aa	104	22

a) Calculate the frequency of the recessive allele in the population in 2002. (1 point)

= 0.322

b) Calculate the frequency of the dominant allele in the population in 2002. (1 point)

= 0.678

c) Using the allelic frequencies from 2002, calculate the expected number of individuals in the population carrying the AA genotype if the population was at Hardy Weinberg equilibrium. (1 point)

= 460

d) Using the allelic frequencies from 2002, calculate the expected number of individuals in the population carrying the Aa genotype if the population was at Hardy Weinberg equilibrium. (1 point)

= 437

e) Using the allelic frequencies from 2002, calculate the expected number of individuals in the population carrying the aa genotype if the population was at Hardy Weinberg equilibrium. (1 point)

= 104

f) Calculate the frequency of the recessive allele in the population in 2012. (1 point)

= 0.2455

g) Calculate the frequency of the dominant allele in the population in 2012. (1 point)

= 0.7545

h) 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. (1 point)

= 0.427

i) 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. (1 point)

= 278

j) 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. (1 point)

= 45

k) Use chi-square analysis to calculate the chi square value corresponding to the 2002 data from parts C through E above. (1 point)

l) Analyze whether the snake population in 2002 was in genetic equilibrium based on the set of data available. Explain your reasoning. (2 points)

m) Use chi-square analysis to calculate the chi square value corresponding to the 2012 data from parts H through J above. Enter the chi-square value you obtain from this analysis. (1 point)

n) Using both sets of results from 2002 and 2012, analyze whether the snake population is evolving. Explain your analysis, including suggesting reasons for any change. (3 points)

Answer 1

Genotype	2002	2012
AA	460	403
Aa	436	325
aa	104	22
	Total = 1000	Total = 750

a.Allele frequency = Number of particular allele/Total number of allele

Number of a allele = 2*104+436 = 644

Total number of alleles = 2*1000 = 2000

Allele frequency of a = 644/2000 = 0.322

b. Number of A allele= 2*460+436 = 1356

Allele frequency of A = 1356/2000 = 0.678

c.According to Hardy Weinberg, p²+2pq+q² = 1

p² = Frequency of dominant homozygous genotype

2pq = Frequency of heterozygous individual

q² = Frequency of recessive homozygous genotype

p = Frequency of dominant allele

q = Frequency of recessive allele

here p = 0.678 so p²= 0.46,

So expected number of individual carrying AA genotype in a population is 460 (1000*0.46 = 460)

d. 2pq = 2*0.678*0.32 = 0.4366

So expected number of individual carrying Aa genotype in a population is 437 (1000*0.4366= 437)

e. q² = 0.104

So expected number of individual carrying aa genotype in a population is 104 (1000*0.104 = 104)

k.

Genotype	O	expected frequency	E	O-E	(O-E)2	(O-E)2/E
AA	460	0.460	460	0	0	0
Aa	436	0.437	437	1	1	0.0023
aa	104	0.104	104	0	0	0
						Total = 0.0023

We have two alleles so degree of freedom is 2-1 = 1

The table value for 1 d.f. is 3.841 and caculated value is 0.0023

l. Here calculated value is less than table value so we can accept null hypothesis and we can say that snake population in year 2002 is in genetic equilibrium.

m.

Genotype	O	expected frequency	E	O-E	(O-E)2	(O-E)2/E
AA	403	0.568	426	-23	529	1.24
Aa	325	0.37	278	47	2209	7.94
aa	22	0.006	5	17	289	57.8
						Total = 66.98

We have two alleles so degree of freedom is 2-1 = 1

The table value for 1 d.f. is 3.841 and caculated value is 66.98

l. Here calculated value is greater than table value so we can reject null hypothesis and we can say that snake population in year 2012 is not in genetic equilibrium.