1. 12. 4 Population Genetics

2.  Definition = study of genetics of groups of interbreeding individuals  Gene pool = all of the genes in a population at any one time  Relative frequency = number of times the allele occurs in the gene pool  Evolution = a change in the relative frequency of alleles All the b’s plus all the B’s equals all of the alleles in the gene pool

3.  Short def- idealized model of gene pools  Long def- A biological theorem that states both allele and genotype frequencies within a population remain constant from one generation to the next unless specific disturbing influences are introduced…

4.  The organism is diploid  Generations are nonoverlapping  Reproduction is only sexual  NO natural selection  Mutation does NOT occur  Population size is INFINITELY large  Random mating  NO migration Could any population actually meet ALL of these criteria???

5.  Natural Selection  Mutations  Limited Population Size  Non-random mating  Migration of Individuals Due to these influences, the Hardy-Weinberg Equation DOES NOT reflect the reality of nature! So why is it used???

6.  Can ALL of the H-W criteria always be met?  NO!  Therefore, evolution is the inevitable result  Using H-W calculations, we can compare what a population’s allele frequencies were to what they are now and measure real change (evolution in action)!

7.  p + q = 1  p 2 +2pq + q 2 = 1Where: p = allele frequency of dominant allele p = allele frequency of dominant allele q = allele frequency of recessive allele q = allele frequency of recessive allele

8.  In a flock of sheep, 9% have black wool and 91% have white wool. Black wool is recessive. Use the Hardy-Weinberg equation to determine:  A - the allele frequency for black wool  B - the % of the population that would be heterozygous for white wool

9.  In a flock of sheep, 9% have black wool and 91% have white wool. Black wool is recessive.  q 2 = 0.09 (9% are homozygous recessive)  q = √0.09 = 0.3  allele frequency for black wool

10. Step 1:  p + q =1  q = 0.3  p + 0.3 = 1  p = 0.7 Step 2: p 2 = (0.7) 2 = 0.49 = 49% WW p 2 = (0.7) 2 = 0.49 = 49% WW p 2 +2pq + q 2 = 1 p 2 +2pq + q 2 = 1 2pq = 1 – (.49 +.09) =.42 = 42% OR 2pq = 2(0.7)(0.3) = 0.42 = 42% Ww 2pq = 2(0.7)(0.3) = 0.42 = 42% Ww In a flock of sheep, 9% have black wool and 91% have white wool. Black wool is recessive.