Question

3. Dr. Jack Smith of the Fisheries and Wildlife Department at NMSU has hired you to analyze some data that was collected several years and he has not had time to do it himself. He knows that you are enrolled in GENE 320 and have a basic understanding of population genetics. He had done some research on two separate populations of western whiptail lizards. Population 1 was found on the western side of the Organ mountains while population 2 was located on the eastern side of the Organ mountains. At the time, he collected blood samples from individuals in both populations. He analyzed the samples and determined the alleles of the gene for the enzyme α-glycerol-phosphate dehydrogenase (α-GPD). Based upon the number of individuals below, he would like you to calculate the allele and genotypic frequencies of both populations and determine if both populations are in Hardy-Weinberg equilibrium. Please show him all of your work. Please minimize rounding. (8 points)

Genotype        Population 1                Population 2

CC                   36                                54

CD                  92                                108

DD                  52                                38

Answer 1

Population 1

Total population =36+92+52=180

Allele freq of C allele=q=CC genotype freq+1/2(CD genotype freq)

C allele freq=(36/180)+1/2(92/180)=0.45

D allele frequency =p=DD genotype freq+1/2(CD genotype freq)

D allele freq=p=(52/180)+1/2(92/180)=0.54

P+q=0.45+0.54=0.99 ~1

Ideal population is in hardy Weinberg equilibrium follow

Sum of allele freq =p+q=1

Here sum is 0.99 that is approximately equal to 1

So population1 is in hardy weinberg equilibrium .

Population 2

Total popln=54+108+38=200

C allele freq in popln2=q'=(54/200)+1/2(108/200)=0.54

D allele freq in popln2= p'=(38/200)+1/2(108/200)=0.46

p'+q'=0.46+0.54=1.0

As sum of allele is equal to 1 so population 2 is also in hardy weinberg equilibrium.