Hardy Weinberg Lab

On Friday September 21st we went over the Population Genetics Lab. In this lab we had four cases. Each case represented one of the agents of evolution. It was supposed to represent Hardy Weinber's theory of equilibrium but our population wasn't big enough to correctly test whether or not his theory of evolution is accurate or not. In case #4 it represented the evolution agent of genetic drift, because we were only allowed to mate with our population and not migrate anywhere to reproduce. This agent went against Hardy's theory because he saw that evolution happened through migrating with different individuals and not just individuals in your population.

-- This website explains Hardy Weinberg's theory and what agents of evolution make it false, which is what we discussed while going through this lab: Hardy Weinberg

For the first 3 cases we had 23 individuals which meant 46 alleles, and then for the last case we had 8 individuals and 16 alleles. After collecting your individual data we combined the whole classes data together and calculated the frequency of the genotypes.

For the ones that were absent here are the frequencies:

Formula used to calculate this data:

Case #1:
Generation# :
p2 (AA)- parental=.25 F5=.04

2pq(Aa)- parental=.5 F5=.43

q2(aa)- parental=.25 F5=.52

Frequency of alleles:
p(A)-.5 q(a)-.5

p(A)-.26 q(a)-.74

Case #2:
Generation# :
p2 (AA)- parental=.25 F5=.78
2pq(Aa)- parental=.5 F5=.22
q2(aa)- parental=.25 F5=.0

Frequency of alleles:
p(A)-.5 q(a)-.5
p(A)-..89 q(a)-.11

Case #3:
Generation# :
p2 (AA)- parental=.25 F5=.30
2pq(Aa)- parental=.5 F5=.70
q2(aa)- parental=.25 F5=.0

Frequency of alleles:
p(A)-parental=.5 q(a)-parental= .5
p(A)-F5= .65 q(a)-F5= .30 (didn't do F10)

Case #4:
Generation# :
p2 (AA)-
parental=.25 F5

(Group 1)=.375
(Group 2)=.125
(Group 3)=.0

2pq(Aa)-
parental=.5
F5
(Group 1)=.125
(Group 2)=.1875

(Group 3)= .1875

q2(aa)-
parental=.25
F5
(Group 1).25
(Group 2)=.0625

(Group 3)=.1875

Frequency of alleles:
p(A)-.5
q(a)-.5
p(A)(Group 1)= .875

(Group 2).4375
(Group 3)=.5625

q(a)
(Group 1)=.125

(Group 2)=.5625
(Group 3)=.25

In this lab the change in allele frequency wasn't representing that the white homozygous recessive cats died but representing evolution making white homozygous recessive cats less common than black cats. There were white cats made but they didn't get to live to the age of mating. Alleles determine the type of genes a creature has, whether the parents pass on dominanttraits so that the offspring survives in their environment or a parent passes on a recessive trait which gives that offspring a less percentage for survival.

- A cute cartoon on the evolution of cats.