chapter 23 – part 1 part 2 after break. slide 2 of 15 context evolution appears to be the typical...

Chapter 23 – Part 1

Part 2 After Break

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

of 15

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

of 15

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

of 15

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

of 15

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

of 15

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

of 15

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?

of 15

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?

of 15

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

of 15

Part 2 of Problem

What are the frequencies of

a) homozygous dominant genotype?

b) Heterozygous genotype?

c) homozygous recessive genotype?

of 15

Part 2 of Problem (Answer)

What are the frequencies of


Since p = 0.5, p2 = (0.5)2 = 0.25


Since p = 0.5 & q = 0.5, 2pq = 2(0.5)(0.5) = 0.5


Yep, it was given, but if we must: q = 0.5, q2 = (0.5)2 = 0.25

of 15

Part 3 of Problem

How many individuals have the




of 15




500 members & p2 = 0.25, (0.25)(500) = 125


500 members & p = 0.5, q = 0.5 2pq = 0.5

So 500 (0.5) = 250 heterozygotes


500 members & q2 = 0.25, (0.25)(500) = 125

of 15

Part 4 of Problem


a) Dominant phenotype?

b) Recessive phenotype?

of 15



a) Dominant phenotype?

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

b) Recessive phenotype?

= Homo. Recess. = 125

chapter 23 – part 1 part 2 after break. slide 2 of 15 context evolution appears to be the typical...

Documents