CHAPTER 23 .1

Population Genetics

Quick Review: Natural Selection

Variation Natural Selection Speciation

Organisms better suited to the environment

SURVIVE & REPRODUCEat a greater rate than those

less suited to the environment and this is how…

SPECIES EVOLVE

Quick Review: DNA & Mutations

A change in genetic material may change a protein

Mutation Variation Natural Selection Speciation

We can study variation at the molecular levelAKA: mutations

Hardy & Weinberg

Mutation Variation Natural Selection Speciation

I love math!

Gene Pools

Collection of genes within a population of a species

Hardy & Weinberg

We can calculate allele frequencies based upon the

genotypes

Thus, a math equation will

show if evolution is occurring

We can study changes in phenotypes of one trait in a population

over time

We can convert

phenotypes into

genotypes

Ex: Allele Frequencies in Snapdragons

Collect data of phenotypes of a population 320 red flowers, 160 pink flowers, & 20 white flowers

Convert phenotypes to genotypes 320 RR 160 RW 20 WW

Calculate allele frequencies R alleles = 320 +320 + 160 = 800 W alleles = 20 + 20 + 160 = 200

Ex: Allele Frequencies in Snapdragons

Allele Frequency 800 R alleles / 1000 total alleles (80% or 0.8) 200 W alleles / 1000 total alleles (20% or 0.2)

Ex: Allele Frequencies in Snapdragons

Hardy-Weinberg: If evolution is not occurring in this population Then allele frequency will remain constant over time Therefore at any moment the population will have:

80% R alleles & 20% W alleles

If 10 years later: 50% R alleles 50 % W alleles Then microevolution is occurring

Applying H.W.E.

This happens to nearly all populations for all traits p represents the dominant allele (R) q represents the recessive allele (W)

p = .8 & q = .2p + q = 1

Solve this story problem

In certain Native American groups, albinism is due to a homozygous recessive condition. The frequency of the allele for this condition is currently .06 of the Native American population.

What is the frequency of the dominant allele?p + q = 1

P + .06 = 1p = .94

Extrapolating H.W.E.

H.W.E. Equation 1: p + q = 1 (shows allele frequencies)

H.W.E. Equation 2: (1) * (1) = 1 (p + q) * (p + q) = 1 p2 +2pq + q2 = 1

500 Snapdragon Example p = .8 & q = .2 (.8)2 +2(.8*.2) + (.2)2 = 1 .64 + .32 + .04 = 1 320 + 160 + 20 = 500

Applying H.W.E.

p2 = homozygous dominant condition2pq = heterozygous conditionq2 = homozygous recessive condition

p2 +2pq + q2 = 1 RR + 2RW + WW = 1

Solve this story problem

In a certain flock of sheep, 4 percent of the population has black wool (recessive condition) and 96 percent has white wool.

What % of sheep are heterozygous for wool color?

p2 +2pq + q2 = 1

H.W.E. Conditions

Our equations are great for: Finding allele frequencies: p + q = 1 Finding genotype frequencies: p2 +2pq + q2 = 1 Showing microevolution if values change over time

When would allele frequencies not change over time?

H.W.E. Conditions

No Mutations No new genotypes/phenotypes

Very large population size No minor population disruptions (genetic drift)

Isolation from other populations No immigration/emigration (gene flow)

Random Mating No picky females choosing one allele over another

No natural selection No environmental pressures selecting one allele over

another