Hardy – Weinberg Equilibrium

1. Hardy – Weinberg Equilibrium Unit 8: Evolutionary Biology

2. Hardy – Weinberg Equilibrium • In the absence of factors that alter them, the frequencies of gametes, genotypes, and phenotypes remain constant generation after generation.

3. Hardy – Weinberg Equilibrium • The original proportions of the genotypes in a population will remain constant from generation to generation, as long as the following assumptions are met: • No mutations takes place. • No genes are transferred to or from other sources (no immigration or emigration takes place) • Random mating is occurring • The population size is very large • No selection occurs.

4. Hardy – Weinberg Equilibrium

5. Hardy – Weinberg Principle of Equilibrium • In a large population mating at random and in the absence of other forces that would change allele proportions, the process of sexual reproduction alone will not change these proportions.

6. Hardy – Weinberg Equilibrium Equation • p2 + 2 pq + q2 = 1 • p = the allele frequency of the dominant allele • q = the allele frequency of the recessive allele. • p2 = genotype frequency homozygous dominant • q2 = genotype frequency homozygous recessive • 2pq = genotype frequency of heterozygous

7. DEMO Problem #1 p2 + 2pq + q2 = 1 p + q = 1Problem: 1 in 1700 US Caucasian newborns have cystic fibrosis. C for normal is dominant over c for cystic fibrosis. When counting the phenotypes in a population why is cc the most significant?

8. What percent of the above population have cystic fibrosis (cc or q2)? q2 = 1 /1700 = 0.00059 or .059%

9. From the above numbers you should be able to calculate the expectant frequencies of all the following (assuming a Hardy-Weinberg equilibirum): ALLELE FREQUENCY CALCULATIONS Why calculate "q" first? q2 = cc So: The square root of q2 = 0.024 q = 0.024 p + q = 1 SO 1 – 0.024 = 0.976 Now that we have the allele frequencies we can calculate the genotype frequencies.

10. GENOTYPE FREQUENCY CALCULATIONS: CC- Normal homozygous dominant = p2 p2 = .953 Cc –carriers of cystic fibrosis = 2pq = 2pq = 0.0468

11. How many of the 1700 of the population are homozygous Normal? 953 x 1700 = 1620.1 individuals are CC How many of the 1700 in the population are heterozygous (carrier)? 0.0468 x 1700 = 79.56 are carriers (Cc)

12. It has been found that a carrier is better able to survive diseases with severe diarrhea. What would happen to the frequency of the "c" if there was a epidemic of cholera or other type of diarrhea producing disease?Would "c" Increase? or would it Decrease?

13. ALBINISM: A SAMPLE HARDY-WEINBERG PROBLEM  Albinism is a rare genetically inherited trait that is only expressed in the phenotype of homozygous recessive individuals (aa).  The most characteristic symptom is a marked deficiency in the skin and hair pigment melanin.  This condition can occur among any human group as well as among other animal species.  The average human frequency of albinism in North America is only about 1 in 20,000.

14. Genotype Frequency q² = 1/20,000 = .00005 So q = 0.007 (allele frequency) 1 – 0.007 = 0.993 = p (allele frequency) So we can now calculate the genotype frequencies: p2 = 0.986 and 2pq = 0.014

15. Hardy – Weinberg