Question

5. Dr. Smith has another data set that he would like you to analyze and interpret....

5. Dr. Smith has another data set that he would like you to analyze and interpret. Once he had a graduate student researching western whiptails in the eastern population. His student was quite adept at finding the nests of females in the population. His student collected blood samples from female individuals after they had laid their eggs. He wanted to know if the genotype at the 6-phosphogluconate dehydrogenase (6PGD) locus had any relationship with adaptability to the local environmental. He also counted the average number of eggs laid by females of each genotype and the average number of offspring that survived after week. He assumed that after this time the rate of survival was similar for all genotypes. The student noticed that females with each genotype laid a different number of eggs on average. He also noticed a difference in the number of offspring that survive after one week for each genotype. Based upon this information calculate the percentage survivability for each genotype. After that, calculate the new allele and genotypic frequencies in the next generation given the difference in survival between the three genotypes. Predict the allele and genotypic frequencies after a second generation. Please show Dr. Smith all of your calculations. Please minimize rounding. Please tell him your conclusions from this data. (14 points)

Genotype        Number           Egg number    Offspring number after 1 week

AA                   28                    9                      5

AB                   44                    7                      4

BB                   24                    8                      5

Homework Answers

Answer #1

percentage survivability;

AA=5/14*100=38% similarlly find the rest of the two percentage survivability

Allelic frequency ,AA=33 ,AB=48,BB=29

now add 48/2=24 to 33 and 29 to get 57 and 53

allele frequency of A=57/33+48+29=0.52

allele frequency of B=53/110=0.48

Genotypic frequency of AA=33/110=0.3

AB=48/110=0.43

BB=29/110=0.26

This population might be in hardy weinberg equilibrium,so that the allele and genotypic frequencies for the corresponding generations will remaine same.

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