Hardy-Weinberg Equilibrium

Population Genetics: Part I

Outline

• Introduction

• The problem of genetic variation and natural selection

• Why do allele frequencies stay constant for long periods ?– Hardy-Weinberg Principle

Population Genetics

• The study of the properties of genes in populations

• Genetic variation within natural populations was a puzzle to Darwin and his contemporaries

• The way in which meiosis produces genetic segregation among the progeny of a hybrid had not yet been discovered

• They thought that Natural Selection should always favour the optimal form and eliminate variation

• Also, blending inheritance was widely accepted – if correct, then the effect of any new genetic variant would be quickly diluted and disappear in subsequent generations

They thought selection should always favour the optimal form

and eliminate variation

Yet there is lots of variation in natural populations !

Hardy and Weinberg independently solved the puzzle of why genetic

variation exists

Background

• Hardy & Weinberg showed that the frequency of genotypes in a population will stay the same from one generation to the next.

• Dominant alleles do not, in fact, replace recessive ones.

• We call this a Hardy-Weinberg equilibrium• This means that if 23% of the population has

the genotype AaTTRR in a generation, 23% of the following generation will also have that genotype.

There are, however, a number of conditions that must be met for a population to exhibit the

Hardy-Weinberg equilibrium.

These are: 1) A large population, to ensure no statistical

flukes 2) Random mating (i.e. organisms with one

genotype do not prefer to mate with organisms with a certain genotype)

3) No mutations, or mutational equilibrium 4) No migration between populations (i.e. the

population remains static) 5) No natural selection (i.e. no genotype is

more likely to survive than another)

• In a population exhibiting the Hardy-Weinberg equilibrium, it is possible to determine the frequency of a genotype in the following generation without knowing the frequency in the current generation.

• Hardy and Weinberg determined that the following equations can determine the frequency when p is the frequency of allele A and q the frequency of allele a:

• The Hardy-Weinberg equation can be expressed in terms of what is known as a binomial expansion:

• p + q = 1 • p2 + 2pq + q2 = 1

• For the first equation, if allele A has a frequency of say 46%, then allele a must have a frequency of 54% to maintain 100% in the population.

• For the latter equation, a monohybrid Punnett square will prove its validity.

• Set up the Punnett square so that two organisms with genotype pq (or Aa) are mated.

The derivation of these equations is simple

• The Punnett square results in pp, pq, pq, and qq.

• Because these are probabilities for genotypes, each square has a 25% chance.

• This means that all four should equal 100%, or one.

• To make things easier, convert pp and qq to p2 and q2 (elementary algebra, p*p = p2).

• If the results are added, the equation p2 + pq + pq + q2 = 1 emerges.

• By simplifying, it is p2 + 2pq + q2 = 1.

Applications

Sample problem 01

• Consider a population of 100 jaguars, with 84 spotted jaguars and 16 black jaguars. The frequencies are 0.84 and 0.16.

• Based on these phenotypic frequencies, can we deduce the underlying frequencies of genotypes ?

• If the black jaguars are homozygous recessive for b (i.e. are bb) and spotted jaguars are either homozygous dominant BB or heterozygous Bb, we can calculate allele frequencies of the 2 alleles.

• Let p = frequency of B allele and q = frequency of b allele.

• (p+q)2 = p2 + 2pq + q2

• where p2 = individuals homozygous for B

• pq = heterozygotes with Bb• q2 = bb homozygotes

• If q2 = 0.16 (frequency of black jaguars), then q = 0.4 (because0.16 = 0.4)

• Therefore, p, the frequency of allele B, would be 0.6 (because 1.0 – 0.4 = 0.6).

• The genotype frequencies can be calculated:

• There are p2 = (0.6)2 X 100 (number of jaguars in population) = 36 homozygous dominant (BB) individuals

• The heterozygous individuals (Bb) = 2pq = (2 * 0.6 * 0.4) * 100 = 48 heterozygous Bb individuals

Applications

Sample problem 02

• In 1986, Henry Horn counted 133 Gray Squirrels and 25 Black Squirrels (16% Black) at Princeton University.

• In 1994, he counted 43 Gray Squirrels and 9 Black Squirrels (17% Black).

• This indicates that the ratio of Gray:Black squirrels may be in Hardy-Weinberg equilibrium

• Assuming that the gene for the black morph is autosomal dominant (not sex-linked) what are the gene frequencies in the population ?

• Black squirrels, which could be Black homozygotes (BB) or black heterozygotes (Bb) = 0.17 of the population in 1994

• Thus, the proportion of gray recessives (bb) in the population is q2 = 1 – 0.17 = 0.83 (or 83%)

• q = 0.83 = 0.91 (or 91%) of the genes in this gene pool

• Therefore, the frequency of the dominant allele (for Black) = 1 – 0.91 = 0.09

• Thus, black individuals that are homozygous dominants (BB) = p2 = 0.09 * 0.09 = 0.0081 (=0.08% of the population).

• The black individuals that are heterozygotes (Bb) are 2pq = 2 * 0.09 * 0.91 = 0.162 (= 16.2% of the population)

The population therefore consists of:

– 83% homozygous (recessive) gray

– 16.2% heterozygous black

– 0.08% homozygous (dominant) black

This example shows that dominants can be less common than recessives and that there is no evidence that the dominant character will eliminate the recessive

Sample problem 03• The allele for cystic fibrosis is present in

Caucasians at a frequency q of 22 per 1000 individuals (= 0.022)

• What proportion of Caucasians is expected to express this trait ?

• The frequency of double recessives (q2) is 0.022 * 0.022 = 0.000484, or approx. 1 in every 2000 individuals.

• What proportion is expected to be heterozygous carriers ?

• If the frequency of the recessive allele q is 0.022, then the frequency of the dominant allele p is 1 – 0.022 = 0.978

• The frequency of heterozygous individuals (2pq) is 2 * 0.978 * 0.022 = 0.043, or 43 people in every 1000

Another way of stating the Hardy-Weinberg principle

• In a large population mating at random and in the absence of other forces that would change the proportions of the different alleles at a given locus, the process of sexual reproduction (meiosis and fertilization) alone will not change these proportions.

Why do allele frequencies change ?

• According to the Hardy-Weinberg principle, allele and genotype frequencies will remain the same from generation to generation in a large, random mating population IF no mutation, no gene flow and no selection occur.

• In fact, allele frequencies often change in natural populations, with some alleles increasing in frequency and others decreasing.

• The Hardy-Weinberg principle establishes a convenient baseline against which to measure such changes

• By examining how various factors alter the proportions of homozygotes and heterozygotes, we can identify the forces affecting the particular situation we study.

Reference

• Winterer, J. 2001. A lab exercise explaining Hardy-Weinberg equilibrium and evolution effectively. American Biology Teacher 63:678-687.

• http://www.woodrow.org/teachers/bi/1994/find.html

• http://www.woodrow.org/teachers/bi/1994/hwintro.html