The HardyWeinberg Theorem Hardy WHO States that frequencies

The Hardy-Weinberg Theorem Hardy WHO? ? ?

�States that… frequencies of alleles & genotypes in a population’s gene pool remain constant between generations �Theorem shows how Mendelian inheritance can be preserved in populations that are NOT evolving BUT… �It also shows how evolutionary change can happen over time What the heck is the Hardy. Weinberg Theorem?

�You need a VERY large population �There can be NO gene flow �There can be NO mutations �There MUST BE random mating �There can be NO natural selection In order for a stable H-W equilibrium to work…

How many populations remain in a stable H-W equilibrium for a long time? ? Are the five conditions ever met longterm? What types of situations will throw a population OUT of H-W equilibrium? Some questions to ponder…

�If populations show the SAME allele frequencies generation to generation, they are said to be in a state of Hardy. Weinberg Equilibrium Let’s take a look at the H-W Equation: p 2 + 2 pq + q 2 = 1

p = dominant (A) allele frequency q = recessive (a) allele frequency So… p 2 = homozygous dominant frequency (AA) q 2 = homozygous recessive frequency (aa) 2 pq = heterozygous frequency (Aa) p 2 + 2 pq + q 2 = 1

�If we know the frequency of one of the alleles, we can calculate the frequency of the other allele: p+q=1 p = 1 – q q = 1 - p

The hardest part of the H-W Theorem is applying it to mathematical problem sets. Guess what we’re doing next… (da da da duuum!) a. b. c. d. Learn how to weave baskets Make daisy chains Draw stick men Do some H-W problems!

1 in 1700 US Caucasian newborns have cystic fibrosis. C for normal is dominant over c for cystic fibrosis. 1. When counting the phenotypes in a population why is cc the most significant? When dominance is the case. CC and Cc are both normal. The recessive genotype cc is the only one that will result in cystic fibrosis. 2. What percent of the above population have cystic fibrosis (cc or q 2)? 1/1700 are cc. This is 0. 00059 or 0. 059% of the population. Problem #1

ALLELE FREQUENCY CALCULATIONS Now we know q 2 we can calculate q. q = square root of 0. 00059 = 0. 024 or 2. 4% We know that p = 1 – q so… p = 1 – 0. 024 = 0. 976 or 97. 6%

GENOTYPE FREQUENCY CALCULATIONS CC homozygous dominant (normal) = p 2 = (0. 976)2 = 0. 953 or 95. 3% Cc heterozygous dominant (carrier) = 2 pq = 2(0. 976)(0. 024) = 0. 0468 or 4. 68% How many of the 1700 of the population are homozygous normal? (0. 953)(1700) = ~1620 individuals How many of the 1700 in the population are heterozygous carrier? (0. 0468)(1700) = ~79 individuals