1. Measuring Evolution of Populations SLIDE SHOW MODIFIED FROM KIM FOGLIA@explorebiology.com

2. 5 Agents of evolutionary change Mutation Gene Flow Genetic Drift Selection Non-random mating

9. Using Hardy-Weinberg equation What are the genotype frequencies? q 2 (bb): 16/100 = .16 q (b): √.16 = 0.4 p (B): 1 - 0.4 = 0.6 population: 100 cats 84 black, 16 white How many of each genotype? bb Bb BB p 2 =.36 2pq =.48 q 2 =.16 Must assume population is in H-W equilibrium!

10. Using Hardy-Weinberg equation bb Bb BB p 2 =.36 2pq =.48 q 2 =.16 Assuming H-W equilibrium Sampled data p 2 =.74 2pq =.10 q 2 =.16 How do you explain the data? p 2 =.20 2pq =.64 q 2 =.16 How do you explain the data? Null hypothesis bb Bb BB

13. Malaria Single-celled eukaryote parasite ( Plasmodium ) spends part of its life cycle in red blood cells 1 2 3

15. HARDY-WEINBERG PRACTICE PROBLEMS p + q = 1 p 2 + 2 pq + q 2 = 1

19. Within a population of butterflies, the color brown (B) is dominant over the color white (b). And, 40% of all butterflies are white. Image from: BIOLOGY by Miller and Levine; Prentice Hall Publishing ©2006 q 2 = 0.4 q = 0.4 = 0.6324 p = 1 - 0.6324 = 0.3676 aa = 0.4 = 40% Aa = 2 (0.632) (0.368) = 0.465 =46.5% AA = (0.3676) (0.3676) = .135 = 13.5%

20. PRACTICE HARDY WEINBERG 1 in 1700 US Caucasian newborns have cystic fibrosis. C for normal is dominant over c for cystic fibrosis. Calculate the allele frequencies for C and c in the population Image from: BIOLOGY by Miller and Levine; Prentice Hall Publishing ©2006

21. 1/1700 have cystic fibrosis q 2 = 1/1700 q = 0.00059 q = 0.024 p = 1 – 0.024 = 0.976 Frequency of C = 97.6% Frequency of c = 2.4% NOW FIND THE GENOTYPIC FREQUENCIES

22. CC or p 2 = (0.976) 2 = .953 Cc or 2pq = 2 (0.976) (0.024) = 0.0468 cc = 1/1700 = 0.00059 CC = 95.3% of population Cc = 4.68% of population cc = .06% of population

23. Now you can answer questions about the population: How many people in this population are heterozygous? 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? 0.0468 (1700) = 79.5 ~ 80 people are Cc Cc more likely to survive than CC. c will increase in population

24. The gene for albinism is known to be a recessive allele. In Michigan, 9 people in a sample of 10,000 were found to have albino phenotypes. The other 9,991 had skin pigmentation normal for their ethnic group. Assuming hardy-Weinberg equilibrium, what is the allele frequency for the dominant pigmentation allele in this population? q 2 = 9/10000 q = 0.0009 q = 0.03 p = 1 – 0.03= 0.97 Frequency of C = 97% Frequency of c = 3%

25. CC or q 2 = (0.976) 2 = .953 Cc or 2pq = 2 (0.976) (0.024) = 0.0468 cc = 1/1700 = 0.00059 CC = 95.3% of population Cc = 4.68% of population cc = .06% of population