Lab 11

Two equations:

-p + q = 1 A + a = 1, where A and a equal allele percentages (as decimal) All dominant alleles plus all recessive alleles add up to 100% of alleles for a particular gene in a population Allele frequencies -p2 + 2pq + q2 = 1 AA + 2Aa + aa = 1 For a particular gene, all homozygous dominant individuals plus all heterozygous individuals plus all homozygous recessive individuals add up to 100% of the individuals in the population Genotype frequencies

Another Example:

A population (Observed): SS=1100/1778 Ss=500/1778 ss=178/1778 Allele Frequencies: q=(500+(2x178))/3556=0.241 p=1-q=0.759 Genotype Frequencies: SS=0.619 Ss=0.281 ss=0.100

Given: In a population of 100 individuals (200 alleles), sixteen exhibit a recessive trait. Problem: Find the allele frequencies for A and a. Find the genotypic frequencies of AA, Aa, and aa.

Allele frequency p + q = 1 or A + a = 1 ?/200 + 32/200 = 200/200 (You need total alleles) ?% + 0.16 = 1 or 0.16 = aa and 0.84 = AA +Aa aa = qq or q2 = 0.16 or q = 0.4 1 - q = p 1 - 0.4 = 0.6 or A = 0.6 and a = 0.4

Calculate the allele frequencies from this data:

Directly from the numbers given the allele frequencies of p (S) and q (s): 2914/3556 S alleles = 0.82 S frequency 2914= 2x1194 + 526 then also for q.............. (1) 642 / 3556 s alleles = 0.18 s frequency OR (2) q = 1-p = 1-0.82 = 0.18

SS =1194/1778=0.67 Ss= 526/1778=0.30 ss= 58/1778 = 0.03 Is this population in HW equilibrium?

First figure actual genotype frequencies from the actual numbers given in the problem. Then count the actual A alleles in the population and the actual a alleles in the population to get the allele frequencies. Then calculate the genotype frequencies predicted by the Hardy Weinberg equilibrium equations and compare to the actual numbers.

Fraggles are mythical, mouse-like creatures that live beneath flower gardens. Of the 100 fraggles in a population, 91 have green hair (Ff or FF) and 9 have grey hair (ff). Assuming genetic equilibrium: What are the gene frequencies of F and f? What are the genotypic frequencies?

Gene frequencies: F = 0.7 and f = 0.3 q2=9/100=0.09 q=0.3 and 1-0.3=p=0.7 Genotypic frequencies FF + Ff = 0.91 FF = 49% or 0.49 (q2) Ff = 42% or 0.42 (2pq) ff = 9% or 0.09 (p2)

ALLELE FREQUENCY VARIATIONS

Hardy-Weinberg applies only if there is genetic equilibrium or NO allele frequency changes

A population that is not changing genetically is said to be at

Hardy-Weinberg equilibrium

Therefore, in the population:

Homozygous dominant = 36/100 or 36% Heterozygous dominant = 48/100 or 48% Recessive = 16/100 or 16%

Phenotypic frequencies

If: p = 0.6 and q = 0.4, then p2 = (0.6)(0.6) = 0.36 q2 = (0.4)(0.4) = 0.16 2pq = 2(0.6)(0.4) = 0.48

Causes of allele frequency variations

Mutation Migration Non-random mating Genetic drift Natural selection (Remember these...we will look at the effects of some of these when we do simulations for lab 11) How often in nature do NONE of these occur? Rarely, if ever.

Causes of allele frequency variations

Mutation Migration Non-random mating Genetic drift Natural selection How often in nature do NONE of these occur? Rarely, if ever.

Directly from the numbers given The genotype frequencies are:

SS =1194/1778=0.67 Ss= 526/1778=0.30 ss= 58/1778 = 0.03

The gene pool:

The sum total of all alleles for all genes which exist in a population. This means that genotype numbers can be used to calculate allele frequencies. Example: In a population of 100 people (with 200 alleles):36 are AA; 48 are Aa; 16 are aa.

SS genotype frequency would be predicted by p2 = (0.82)2= 0.67 and Ss frequency would be 2pq or 2 (0.82)(0.18) = 0.30; and ss frequency would be q2 = (0.18)2= 0.03.

These numbers almost exactly match the measured genotype frequencies - so this population appear to be in Hardy-Weinberg equilibrium.

Using Chi Square to determine H-W

Use numerical values NOT frequencies or percentages df = For Mendelian Complete Dominance= # genotypes-# alleles df for incomplete dominance, codominance etc. df= # phenotypes - #alleles p=0.05 X2 = ∑ (O-E)2 E Value smaller than critical value: observed and expected do not differ; population in HW equilibrium Value greater than critical value: observed and expected values differ; population is not in HW equilibrium

Two Alleles in this Population

W (dominant) w (recessive) WW, Ww, ww are possible genotypes

In a population at Hardy-Weinberg equilibrium, allele frequencies remain the same from generation to generation

and genotype frequencies remain in the proportions p2 + 2pq + q2 = 1.

Use actual numbers?

do not use frequencies with chi-square Examples: p2: 0.67x1178 ~789 2pq: 0.3x1178 ~354 q2: 0.03x1178 ~35

Hardy-Weinberg applies only

if there is genetic equilibrium or NO allele frequency changes

Expected (if in HWE):

p2=(0.759)2=0.576 (1024 individuals) 2pq=2*0.759*0.241=0.366 (651 individuals) q2=(0.241)2=0.058 (103 individuals)

The assumptions that underlie the Hardy-Weinberg equilibrium are

population is large mating is random There is no migration (no immigration or emigration) There is no mutation of the alleles natural selection is not acting on the population. (all genotypes have an equal chance of surviving and reproducing)

Conversely, if there have been changes in the frequencies

then evolution has occurred.

If there is no change in frequencies

there is no evolution

Biologists can determine

whether an agent of evolution is acting on a population by comparing the population's genotype frequencies with Hardy-Weinberg equilibrium frequencies.

SS (1100-1024)2/1024=5.64 Ss (500-651)2/651=35.02 Ss(178-103)2/103=54.61

∑(sum)=95.27 Compare to critical value (df=1)=3.841 Larger than critical value so the population is not in HWE