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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