1. Chapter 23 – Part 1
Part 2 After Break

2. Context Evolution appears to be the typical situation
Sometimes evolution is not occurring
Hardy-Weinberg Equilibrium
Conditions that specify if evolution is NOT happening
Equilibrium = no evolution
More of a theoretical case than a real test
But affords comparison to evolution

3. Definitions
Population Genetics – study of how populations change genetically over time
Population – Group of same species individuals that can and do interbreed successfully
Gene pool – all of the alleles at all loci in all the members of a population
Fixed – only one allele exists for a locus in the population
More fixed alleles = lower species diversity

4. Hardy-Weinberg If gene pools are NOT Evolving, can use Hardy- Weinberg
Describes a population that is NOT evolving
Allelic frequencies will remain constant
Gene frequencies remain constant throughout the generations

5. Conditions for Hardy-Weinberg
1. No mutations
2. Random mating
3. No natural selections
4. Large population size
5. No gene flow
Immigration or emigration
Genes coming into or leaving the population

6. Math in Biology?
p = Frequency of the dominant allele = f(A) q = Frequency of the recessive allele = f(a) There are only 2 alleles (dominant + recessive), so p + q = 1 square both sides: (p + q)2 = 12 expand the binomial: p2 + 2pq + q2 = 1

7. Hardy-Weinberg Math (Page 2)
p2 = pp = f(aa) = Frequency of homozygous dominant
pq = f(Aa) = frequency of heterozygote
However, remember from Punnett Square, there were 2 heterozygotes for the F1 generation; So
2pq = f(Aa)
q2 = qq = f(aa) = frequency of homozygous recessive

8. In Summary Summary: p2 + 2pq + q2 = 1 f(AA) + f(Aa) + f(aa) = 1
f(Homo. Dom.) + f(Hetero.) + f(Homo. Recess.) = 1
Are there any other genotypes?
Makes sense?

9. Problem (Part 1)
A trait has two characters, dominant (A) and recessive (a). In a population of 500 individuals in Hardy- Weinberg equilibrium, 25% display the recessive phenotype (aa).
a) What is the frequency of the dominant allele in the population?
b) What is the frequency of the recessive allele in this population?

10. Answer f(aa) = 0.25 OR q2 = 0.25 q = √(0.25) = 0.5 OR f(a) = 0.5
 So the frequency of the recessive allele = 0.5
From before: p + q = 1 AND q = 0.5, so
p = 0.5 = f(A)
 So the frequency of the dominant allele = 0.5

11. Part 2 of Problem What are the frequencies of
a) homozygous dominant genotype?
b) Heterozygous genotype?
c) homozygous recessive genotype?

12. Part 2 of Problem (Answer)
What are the frequencies of
a) homozygous dominant genotype?
Since p = 0.5, p2 = (0.5)2 = 0.25
b) Heterozygous genotype?
Since p = 0.5 & q = 0.5, 2pq = 2(0.5)(0.5) = 0.5
c) homozygous recessive genotype?
Yep, it was given, but if we must: q = 0.5, q2 = (0.5)2 = 0.25

13. Part 3 of Problem How many individuals have the
a) homozygous dominant genotype?
b) Heterozygous genotype?
c) homozygous recessive genotype?

14. Part 3 of Problem (Answer)
How many individuals have the
a) homozygous dominant genotype?
500 members & p2 = 0.25, (0.25)(500) = 125
b) Heterozygous genotype?
500 members & p = 0.5, q = 0.5 2pq = 0.5
So 500 (0.5) = 250 heterozygotes
c) homozygous recessive genotype?
500 members & q2 = 0.25, (0.25)(500) = 125

15. Part 4 of Problem How many individuals have the a) Dominant phenotype?
b) Recessive phenotype?

16. Part 4 of Problem (Answer)
How many individuals have the
a) Dominant phenotype?
= (Homo. Dom.) + (Hetero.) = = 375
b) Recessive phenotype?
= Homo. Recess. = 125