13.6 to 13.8

PopulationSpecies  A group of interacting individuals belonging to one species and living in the same geographic area  A group whose members possess similar anatomical characteristics and have the ability to interbreed

Population GeneticsModern Synthesis  The study of genetic changes in populations; the science of microevolutionary changes in a population  A comprehensive theory of evolution that incorporates genetics and includes most of Darwin’s original ideas, focusing on populations as the fundamental units of evolution (individuals don’t evolve – populations do)

Gene PoolExample  All of the alleles for all of the loci in all individuals in a population  Each allele has a frequency in the population  Example: you have a wild boar population in which 50 percent of the alleles for a particular gene are dominant (B) and 50 percent of the alleles for the gene are recessive (b).

Microevolutionexample  A change in a populations gene pool over a succession of generations; evolutionary changes in species over relatively brief periods of geologic time  Change in the allele frequency over time

Hardy Weinberg Equilibrium 5 Conditions of Hardy Weinberg  Named for 2 men who figured out that the shuffling of genes that occurs during sexual reproduction, by itself, cannot change the overall genetic make-up of a population  p + q = 1 (p = dominant allele frequency / q = recessive allele frequency)  p 2 + 2pq + q 2 = 1 ( p 2 = homo dominant; 2pq = hetero; q 2 = homo recessive) 1. Large population 2. No migration in or out 3. Mutations do not alter gene pool 4. Random mating 5. Natural selections does not occur (all have equal chance to survive)

PKU – autosomal recessive trait  About 1 in 10,000 babies born in the US have PKU  How many people are carriers?  First step: calculate q 2 (individuals with PKU / homo recessive) ◦ q 2 = 1/10,000 =  Solve for q (the frequency of the recessive allele in the population) ◦ q =  q 2 = 0.01  Use the equation “p + q = 1” to solve for p. ◦ p = 0.99  Use the equation “p 2 + 2pq + q 2 = 1” and solve for carrier genotype. ◦ 2pq = 2(.99)(.01) =.0198 ◦ ~2% of the population are carriers

Try it yourselfAnswer  You have sampled a population in which you know that the percentage of the homozygous recessive genotype (aa) is 36%. Using that 36%, calculate the following: ◦ The frequency of the "aa" genotype. ◦ The frequency of the "a" allele. ◦ The frequency of the "A" allele. ◦ The frequencies of the genotypes "AA" and "Aa." ◦ The frequencies of the two possible phenotypes if "A" is completely dominant over "a."  q 2 =.36  q =.6; frequency of “a” allele is 60%  p =.4; frequency of “A” allele is 40%  p 2 =.16; frequency of AA is 16%  2pq =.48; frequency of Aa is 48%  Frequency of A phenotype is 64%  Frequency of a phenotype is 36%