Transcript Hardy-Weinberg Equilibrium

Population Genetics: Part I
Hardy-Weinberg Equilibrium
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
The derivation of these
equations is simple
• 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 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 (because0.16 = 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/f
ind.html
• http://www.woodrow.org/teachers/bi/1994/
hwintro.html