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.

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) AA Aa

a (q) Aa aa

In the gene pool,

f(A) = Pf(a) = Q

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.

A (p) a (q)

A (p) AA Aa

a (q) Aa aa

In other words,

f(AA) = P2

f(aa) = Q2

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) AA Aa

a (q) Aa aa

In other words,

f(Aa) = 2PQ

The Hardy-Weinberg Equation

• In any population in equilibrium,

• p + q = 1

• Therefore, (p + q)2 = 1

• Expanded… p2 + 2pq + q2 = 1

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)

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?