Transcript Hardy-Weinberg Equilibrium

Genetic Change
Hardy
Weinberg
Within Populations:
The HardyWeinberg Rule
Genetic Change Within Populations: The
Hardy-Weinberg Rule
 Population genetics is the study of the properties of genes in a
population
 Genetic variation in populations puzzled scientists
 Dominant alleles were believed to drive recessive alleles out of
populations
 In 1908, G. Hardy and W. Weinberg pointed out that in large
populations with random mating, allele frequencies remain
constant
 Dominant alleles do not, in fact, replace recessive ones
• A population that is in Hardy-Weinberg equilibrium
is not evolving
– Hardy and Weinberg came to their conclusion by analyzing
allele frequencies in successive generations
– Frequency =
Number of individuals
falling within a category
Total number of individuals
being considered
• If a population of 100 cats has 84 black and 16 white
– Then the frequencies of black and white
phenotypes are 0.84 and 0.16, respectively
• B allele  Black color
• b allele  White color
• By convention
– The more common allele (B) is designated p
– The less common allele (b) is designated q
–p+q=1
• The Hardy-Weinberg equilibrium can be written as
an equation
Individuals homozygous
– (p + q)2 = p2 + 2pq + q2
for allele b
Individuals homozygous
for allele B
Individuals heterozygous
for alleles B and b
• The equation allows calculation of allele frequencies
– Frequency of white (bb) cats = 16/100 = 0.16
– => q2 = 0.16
– => q = √ 0.16 = 0.4
– p + q =1
=> p = 1 – q = 1 – 0.4 = 0.6
• What about genotype frequencies?
– Frequency of the homozygous dominant genotype is
36 out of 100 cats are black (BB)
• p2 = (0.6)2 = 0.36
– Frequency of the heterozygous genotype is
48 out of 100 cats are black (Bb)
• 2pq = 2(0.6)(0.4) = 0.48
Hardy-Weinberg Assumptions
 The Hardy-Weinberg equation is true only if the following five
assumptions are met
 1. Large population size
 2. Random mating
 3. No mutation
 4. No migration
 5. No natural selection
Why Allele Frequencies Change
• Five evolutionary forces can significantly alter the
allele frequencies of a population
– 1. Mutation
– 2. Migration
– 3. Genetic drift
– 4. Nonrandom
mating
– 5. Selection
Mutation
• Errors in DNA
replication
• The ultimate source of
new variation
• Mutation rates are too
low to significantly
alter allele frequencies
on their own
Migration
• Movement of
individuals from one
population to another
– Immigration: movement
into a population
– Emigration: movement
out of a population
• A very potent agent of
change
Genetic Drift
• Random loss of alleles
– More likely to occur in
smaller population
• Founder effect
– Small group of individuals
establishes a population
in a new location
• Bottleneck effect
– A sudden decrease in
population size to natural
forces
Nonrandom Mating
• Mating that occurs more
or less frequently than
expected by chance
• Inbreeding
– Mating with relatives
– Increases homozygosity
• Outbreeding
– Mating with non-relatives
– Increases heterozygosity
Selection
• Some individuals leave
behind more offspring
than others
• Artificial selection
– Breeder selects for
desired characteristics
• Natural selection
– Environment selects for
adapted characteristics
Conditions for Hardy-Weinberg
Equilibrium
 Describes a hypothetical population in which these five conditions are
met
 Extremely large population size
 No gene flow
 No mutations
 Random mating
 No natural selection
Hardy-Weinberg Equation
 Hardy-Weinberg Equation
 Can be used to estimate the percentage of a population carrying an
allele for a specific trait
 If p = A
 Then:
And
q=a
p2 + 2pq + q2 = 1
Hardy-Weinberg Equilibrium
 If p = W, wings
And q = w, no wings
 In a population of 100 buffalo
 70 % have wings (WW and Ww)
 30% have no wings (ww)
 Use the equation to figure out the
allele frequencies for W and w
 p2 + 2pq + q2 = 1
 p+q = 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