Hardy Weinberg Principle:
The work of
mathematician G.H. Hardy and German doctor W. Weinberg, it explains some basic
properties of populations and how to describe them mathematically. The Law:
"The
frequencies of alleles that make up a gene will remain the same in a
stable population. When p and q stand for the frequency of each allele
in a gene then:
p + q= 1 (100% for homozygous matings)
p 2 + 2pq +
q 2 = 1
( heterozygous matings)
In a sexually
reproducing population, the frequencies will stay the same in generation after
generation provided these conditions are met:
1. mutations can not occur
2. matings are at
random
3. natural
selection can not occur
4. no genes may
enter or leave the population
5. the population
must be large.
Hardy- Weinberg
Worksheet
The Hardy-
Weinberg model is much easier to teach if the students calculate gene
frequencies along with the instructor. This means that you (me) must pause
frequently to allow plenty of time for students (you) to actively process the
information and practice the calculations.
In the absence of
other factors, the segregation and recombination of alleles during meiosis and
fertilization will not alter the overall genetic makeup of a population.
·
The
frequencies of alleles in the gene pool will remain constant unless acted upon
by other agents; this is known as the Hardy- Weinberg theorem.
The Hardy-Weinberg model describes the genetic
structure of nonevolving populations. This theorem can be tested with
theoretical population models.
To test the
Hardy-Weinberg theorem, imagine an isolated population of wildflowers with the
following characteristics:
•· It
is a diploid species with both pink and white flowers.
The population size is 500 plants: 480_ plants have
pink flowers, 20 plants have white flowers.
·
Pink flower color is coded for by the dominant allele "A," white flower color is coded for by the
recessive allele "a."
Of the 480 pink-flowered plants, 320 are homozygous
(AA) and 160 are heterozygous (Aa). Since white color is recessive, all white
flowered plants are homozygous aa.
•· There are 1000 genes for
flower color in this population, since each of the 500 individuals has two
genes (this is a diploid species).
A total of 320 genes are present in the 160
heterozygotes (Aa): half are dominant (160 A) and half are recessive (160 a).
·800 of
the 1000 total genes are dominant.
The frequency of
the A allele is 80% or 0.8 (800/1000).
•·
200 of the 1000 total genes are recessive.
The frequency of the a
allele is 20% or 0.2 (200/1000).
Assuming that
mating in the population is completely random (all male-female mating
combinations have equal chances), the frequencies of A and a will remain the
same in the next generation.
•· Each gamete will
carry one gene for flower color, either A or
a.
Since mating is random, there is an 80% chance that
any particular gamete will carry the A
allele and a 20% chance that any particular gamete will carry the a allele.
The frequencies of the three possible genotypes of
the next generation can be calculated using the rule of multiplication.
The probability of two A
alleles joining is 0.8 x 0.8 = 0.64; thus, 64% of the next generation will be AA.
The probability of two a
alleles joining is 0.2 x 0.2 = 0.04; thus, 4% of the next generation will be aa.
·
Heterozygotes can be produced in two ways, depending upon whether the sperm or
ovum contains the dominant allele (Aa or aA). The probability of a heterozygote
being produced is thus (0.8 x 0.2) + (0.2 x 0.8) = 0.16 + 0.16 = 0.32.
The frequencies of
possible genotypes in the next generation are 64% AA, 32% Aa and 4% aa.
The frequency of
the A allele in the new generation is 0.64 +
(0.32/2) = 0.8, and the frequency of the a
allele is 0.04 + (0.32/2) = 0.2. Note that the alleles are present in the gene
pool of the new population at the same frequencies they were in the
original gene pool.
•· Continued sexual
reproduction with segregation, recombination and random mating would not
alter the frequencies of these two alleles: the gene pool of this
population would be in a state of equilibrium referred to as Hardy-Weinberg
equilibrium.
If our original population had not been in
equilibrium, only one generation would have been necessary for equilibrium to
become established.
From this
theoretical wildflower population, a general formula, called the HardyWeinberg
equation, can be derived to calculate allele and genotype frequencies.
The Hardy-Weinberg
equation can be used to consider loci with three or more alleles.
·
By way of example, consider the simplest case with only two alleles with one
dominant to the other.
·
In our wildflower population, let p represent allele A
and q represent allele a, thus p = 0.8 and q
= 0.2.
The sum of
frequencies from all alleles must equal 100% of the genes for that locus in the
population: p + q = 1.
Where only two alleles exist, only the
frequency of one must be known since the other can be derived:
1 -p=q or 1 -q=p
When gametes fuse
to form a zygote, the probability of producing the AA
genotype is p2; the probability of producing aa is q2; and the probability of producing an Aa heterozygote is 2pq (remember heterozygotes may
be formed in two ways Aa or aA).
·
The sum of these frequencies must equal 100%, thus:
p2 +
2pq + q2 = 1
Frequency Frequency Frequency
of AA
of Aa of aa
The Hardy-Weinberg
equation permits the calculation of allelic frequencies in a gene pool, if the
genotype frequencies are known. Conversely, the genotype can be calculated from
known allelic frequencies.
For example, the
Hardy-Weinberg equation can be used to calculate the frequency of inherited
diseases in humans (e.g., phenylketonuria):
·
1 of every 10,000 babies in the United States is born with phenylketonuria
(PKU), a metabolic disorder that, if left untreated, can result in mental
retardation.
•· The allele for
PKU is recessive, so babies with this disorder are homozygous recessive = q2.
Thus q2 = 0.0001,
with q = 0.01
(the square root of 0.0001).
· The
frequency of p can be determined since p = 1 - q:
p = I - 0.01 = 0.99
•· The
frequency of carriers (heterozygotes) in the population is 2pq.
2pq = 2(0.99)(0.01) = 0.0198
Thus, about 2% of the U.S. population are carriers
for PKU.