Hardy-Weinburg equilibrium and Microevolution (L12)

Exam 2 Lecture 12

Population Genetics

The study of the distribution and frequencies of alleles in populations + how and why frequencies change

Genetic Locus

location of a specific gene or DNA sequence on a chromosome

Homozygous

An Individual possessing 2 copies of the same allele at a locus

• AA or aa

Heterozygous

An individual possessing 2 different alleles at a locus

• Aa

Hardy-Weinberg conditions

• no mutation

• random mating

• no selection

• very large population size

• no gene flow

Considered the null hypothesis for evolution

Hardy-Weinberg equations

p + q = 1

p² + 2pq + q² =1

• p - frequency of dominant allele

• q - frequency of recessive allele

• p² - frequency of homozygous dominant genotype

• 2pq - frequency of heterozygous genotype

• q² - frequency of homozygous recessive genotype

In an isolated human population (assume H-W), the frequency of the Tay-Sachs allele t is 0.02.

What is the chance that any one individual is a carrier (Tt)?

1. q = .02

2. p = (1 - .02) = .98

3. 2pq = (2 x .02 x .98) = 0.04 or 4%

An island population of butterflies is in H-W equilibrium. 64% have black stripes, a trait due to an autosomal dominant allele B.

What are the allele frequencies?

What % of the population is homozygous dominant?

if 64% have a dominant phenotype, then 36% must be homozygous recessive

1. q² = .36

2. q = (sqrt .36) = 0.6 or 60%

3. p = (1 - 0.6) = 0.4 or 40%

1. p² = (0.4²) = .16 or 16%

A population of 1000 ducks has:

500 AA, 200 Aa, 300 aa

Is this population evolving?

Observed ratios:

• 0.5 AA (500/1000) (p²)

• 0.2 Aa (200/1000) (2pq)

• 0.3 aa (300/1000) (q²)

Expected ratios:

1. p = (1200/2000) = 0.6

2. q = (800/2000) = 0.4

• p² = (0.6²) = 0.36 AA

• 2pq = (2 x .6 x .4) = 0.48 Aa

• q² = (0.4²) = 0.16 aa

Observed ≠ Expected, so the population is evolving

Solve for the genetic structure (% y , % Y) of a population with 12 homozygous recessive individuals (yy), 8 homozygous dominant individuals (YY), and 4 heterozygous individuals (Yy). Don’t assume H-W equilibrium!

Don’t assume HW = don’t use p² + 2pq + q²

1. y = ((12 × 2) + 4) = (28/48) = 0.58 or 58%

2. Y = (1 - 0.58) = 0.42 or 42%

Microevolution

change in allele frequencies from one generation to the next