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 : 1.Pass forward labs & week 35 Do Nows. 2. (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?

Variation within a Population

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)

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.alleles

HW Equilibrium: 5 Requirements 1.Large population. Small populations may experience genetic drift (random changes) or bottlenecking. The “bottleneck” effect

HW Equilibrium: 5 Requirements 2.Random mating: Every phenotype is equally likely to mate with every other phenotype 3.No net mutation. 4.No immigration or emigration 5.No natural selection: there is no survival advantage to certain phenotypes.

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

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

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

Defining P 2 and Q 2 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 p 2, and the chance of receiving two “a” alleles is q 2. A (p)a (q) A (p)AAAa a (q)Aaaa In other words, f(AA) = P 2 f(aa) = Q 2

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 A (p)a (q) A (p)AAAa a (q)Aaaa In other words, f(Aa) = 2PQ

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

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

How it works…

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?