Chapter 23 – Part 1 Part 2 After Break. Slide 2 of 15 Context Evolution appears to be the typical situation Sometimes evolution is not occurring Hardy-Weinberg.

Chapter 23 – Part 1

Part 2 After Break

Slide 2 of 15

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

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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

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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

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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

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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

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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

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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?

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Problem (Part 1)

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?

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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

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Part 2 of Problem

What are the frequencies of

a) homozygous dominant genotype?

b) Heterozygous genotype?

c) homozygous recessive genotype?

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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

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Part 3 of Problem

How many individuals have the

a) homozygous dominant genotype?

b) Heterozygous genotype?

c) homozygous recessive genotype?

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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

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Part 4 of Problem

How many individuals have the

a) Dominant phenotype?

b) Recessive phenotype?

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Part 4 of Problem (Answer)

How many individuals have the

a) Dominant phenotype?

= (Homo. Dom.) + (Hetero.) = 125 + 250 = 375

b) Recessive phenotype?

= Homo. Recess. = 125