Do Now: 5/14 (Week 36)

1. Do Now: 5/14 (Week 36) Objectives:1. Define gene pool, phenotype frequency, and genotype frequency. 2. State the Hardy-Weinberg Principle. 3. Describe the conditions required for a population to be in H-W Equilibrium, and define genetic drift and bottlenecking. 4. Identify and explain the equations for f(A), f(a), f(AA), f(Aa), and f(aa). TASK: Pass forward labs & week 35 Do Nows. (Don’t copy) In Cuban tree snails, brown shells (B) are dominant to yellow shells (b). Draw a Punnett square representing a cross between a snail that is homozygous recessive and one that is heterozygous. What % of their offspring will be yellow?

2. Variation within a Population

3. Some Terminology: • Gene Pool: all of the genetic information (“genes”) in a population • Phenotype frequency: How often a particular phenotype is observed in a population (0.00 – 1.00) • Allele frequency: What percentage of the total # of alleles in a gene pool are a certain type. (0.00 – 1.00)

4. Hardy-Weinberg Equilibrium • the frequency of alleles and genotypes in a population will remain constant from generation to generation if the population is stable and in genetic equilibrium.

5. HW Equilibrium: 5 Requirements • Large population. Small populations may experience genetic drift (random changes) or bottlenecking. The “bottleneck” effect

6. HW Equilibrium: 5 Requirements • Random mating: Every phenotype is equally likely to mate with every other phenotype • No net mutation. • No immigration or emigration • No natural selection: there is no survival advantage to certain phenotypes.

7. What to Know • Definitions (gene pool, allele frequency) • 5 conditions for HW equilibrium • Equilibrium is theoretical, not practical. • Disrupting equilibrium = evolution • The rate of allele frequency change measures the rate of evolution. • Microevolution: evolution within a species

8. Hardy-Weinberg Genetic Equilibrium • P = frequency of dominant allele (A) • Q = frequency of recessive allele (a) The Hardy-Weinberg Equation describes the relationship between allele frequencies in a gene pool and phenotype frequencies in a population

9. P + Q = 1 • Consider the following Punnet square, showing a cross between 2 heterozygous individuals for allele A: In the gene pool, f(A) = P f(a) = Q

10. Defining P2 and Q2 • In the population as a whole, the chance of a new individual receiving an “A” allele is equal to that allele’s frequency in the population, represented as p. • Thus, the chance of receiving 2 “A” alleles is p x p, or p2, and the chance of receiving two “a” alleles is q2. In other words, f(AA) = P2 f(aa) = Q2

11. Heterozygotes • The probability for producing a heterozygote = p x q, or pq. • Since there are 2 possible ways to produce a heterozygote, the total probability is 2pq In other words, f(Aa) = 2PQ

12. The Hardy-Weinberg Equation • In any population in equilibrium, • p + q = 1 • Therefore, (p + q)2 = 1 • Expanded… p2 + 2pq + q2 = 1

13. Relationships between allele frequency and phenotype frequency • In a population in equilibrium, • f(AA) = p2 • f(Aa) = 2pq • f(aa) = q2 • Remember: p = f(A) and q = f(a)

14. How it works…

15. Simple Application • Like any formula with 2 variables, if one is known, the other can be determined. • For this type of problem, use the simple form P + Q = 1 • Example: The allele for tongue rolling has a frequency of .95. What is the frequency of the allele for non-tongue rolling?