CHAPTER 23.1 Population Genetics

Quick Review: Natural Selection Variation  Natural Selection  Speciation Organisms better suited to the environment SURVIVE & REPRODUCE at 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 level AKA: 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 = = 800  W alleles = = 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 =.2 p + 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 = 1 p =.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  p 2 +2pq + q 2 = Snapdragon Example  p =.8 & q =.2  (.8) 2 +2(.8*.2) + (.2) 2 = 1  = 1  = 500

Applying H.W.E. p 2 = homozygous dominant condition 2pq = heterozygous condition q 2 = homozygous recessive condition p 2 +2pq + q 2 = 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? p 2 +2pq + q 2 = 1

H.W.E. Conditions Our equations are great for:  Finding allele frequencies: p + q = 1  Finding genotype frequencies: p 2 +2pq + q 2 = 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