2. POINT > Define Hardy-Weinberg Equilibrium POINT > Use Hardy-Weinberg to determine allele frequencies POINT > Define “heterozygous advantage” POINT > Describe the five causes of evolution

3. POINT > Define Hardy-Weinberg Equilibrium Hardy-Weinberg Principle: Allele frequencies in a population remain constant unless one or more factors cause those frequencies to change Hardy-Weinberg analyses are used for making predictions (like Punnett squares) Used for traits with only two alleles

4. POINT > Define Hardy-Weinberg Equilibrium + p2p2 2pq+q2q2 = 1 p+q=1 p = Frequency of dominant allele (A) q = Frequency of recessive allele (a) (Frequency of AA) + (Frequency of Aa) + (Frequency of aa) = 1 (Frequency of A) + (Frequency of a) = 1 (100%) This is the Hardy-Weinberg Equation***

5. POINT > Define Hardy-Weinberg Equilibrium + p2p2 2pq + q2q2 = 1 If you know that an allele is recessive, then the recessive phenotype must be aa If you know the percentage of a population that has the recessive phenotype (q 2 or aa) you can determine the allele frequencies in the population For example....

6. POINT > Use Hardy-Weinberg to determine allele frequencies Example 1 In one generation, the frequency of allele A is 40% (p = 0.40) and the frequency of allele a is 60% (q = 0.60). Use the Hardy-Weinberg Principle to show the genotypes in the population.

7. POINT > Use Hardy-Weinberg to determine allele frequencies Population genotypes: 16% AA 48% Aa 36% aa p = 0.4 q = 0.6 p 2 + 2pq + q 2 = 1 (or) A 2 + 2Aa + a 2 = 1 p 2 = 0.16 2pq = 0.48 q 2 = 0.36

8. POINT > Use Hardy-Weinberg to determine allele frequencies Example 2: 1 in 1700 US Caucasian newborns have cystic fibrosis. C is the dominant, normal allele (p) c is the recessive allele for cystic fibrosis (q) Determine: 1) allele frequencies for C and c 2) the genotype percentages in the population

9. POINT > Use Hardy-Weinberg to determine allele frequencies What percent of the population have cystic fibrosis? cc = 1/1700 = 0.00059 0.00059 = q 2 As a percentage… 0.059% are cc (homozygous recessive)

10. POINT > Use Hardy-Weinberg to determine allele frequencies 0.00059 = q 2 taking the square root: q =.024 if q =.024 then p =.976 (p + q = 1) So, allele frequency of C is.976 and c is.024 p 2 =.953 (CC) 2pq =.047 (Cc) So: 95.3% are CC 4.7% are Cc.059% are cc

11. POINT > Use Hardy-Weinberg to determine allele frequencies How many out of 1700 in the population are homozygous normal? 1700 x.953 = How many of the 1700 are carriers (heterozygous)? 1700 x.047 = ~ 1620 ~ 79

12. POINT > Define heterozygous advantage A cystic fibrosis carrier (Cc) is better able to survive diseases involving severe dehydration. What would happen to the frequency of the c allele if there was an epidemic of cholera or another dehydration- producing disease? Expect c to increase in the population as Cc would survive better than CC This is known as heterozygous advantage: the heterozygote is more “fit”

13. WB CHECK: What is the Hardy-Weinberg Equation? Which term in the H-W equation represents heterozygotes? Which term in the H-W equation represents homozygous dominant individuals? What is heterozygous advantage?

14. POINT > Describe the five causes of evolution Five conditions can disturb genetic equilibrium, causing evolution to occur: 1. Nonrandom Mating (sexual selection) 2. Genetic Drift (Small Population Size) 3. Gene Flow (Immigration or Emigration)

15. POINT > Describe the five causes of evolution Five conditions can disturb genetic equilibrium, causing evolution to occur: 4. Mutations 5. Natural Selection All real populations have one or more of these five…so, evolution happens in all populations

16. POINT > Describe the five causes of evolution Genetic drift changes allele frequencies due to random chance. This occurs only in small populations (i.e. cheetah)

17. Gene flow moves alleles from one population to another through migration POINT > Describe the five causes of evolution

18. Mutations produce most of the genetic variation needed for evolution. POINT > Describe the five causes of evolution

19. Sexual selection selects for traits that improve mating success. POINT > Describe the five causes of evolution

20. Natural selection selects for traits advantageous for survival. POINT > Describe the five causes of evolution

21. WB CHECK: Which of the 5 factors of evolution occurs only in small populations? Which of the factors produces the most variation? Which of the factors is associated with migration of individuals? Which of the factors is often described as “survival of the fittest”?

22. WB CHECK: Albinism (aa) is a recessive condition causing a lack of melanin production in cells. This condition is relatively rare in North America, as it only affects 1 in 20,000 individuals.

23. WB CHECK: Determine the following: 1. Frequency of the A and a alleles 2. All three genotypic frequencies (AA, Aa and aa) 3. Probability of being a carrier (Aa) 4. Expectant number of albino individuals in a population of 3,280,000 (Twin Cities metro)

24. WB CHECK: 1. q 2 = 0.00005 so q = 0.0071 = allele frequency of a p = 0.9929 = allele frequency of A 2. p 2 = AA = 0.986 98.6% are homozygous normal 3. 2(p)(q) = Aa = 0.014 1.4% are carriers 4. q 2 = 0.00005 = aa frequency of albinos 0.00005 x 3,280,000 = 164 expected in the Twin cities

25. Homework: Read 17.2 pages 487-492 Assess page 492 #2-3 Workbook pages 331, 342-344