8.5 Allelic Frequencies and
population genetics
Learning objectives:
1. Define the terms gene pool and allele frequency
2. Explain the Hardy-Weinberg equation and use it
to calculate allele frequencies and genotype
frequencies from data concerning the phenotypes
within a population.
STARTER
1. Define allele. Give an example of a gene with
two alleles.
2. Describe dominant and recessive alleles.
3. Define gene flow. Give a specific example of
gene flow and describe how it causes changes in
allele frequencies over time.
4. List and briefly describe examples of two real
barriers to gene flow.
5. Describe how natural selection affects the allele
frequencies of a population.
Allele Frequency
• Population genetics is the study of genetic changes in populations.
• Gene Pool is the sum total of the genes in a population. This term
usually refers to one phenotypic character at a time.
• Allelic frequency is the number of times an allele occurs within the
gene pool.
E.g. Huntingdon’s disease is an autosomal dominant degenerative
neurological disease. Usually, people who have Huntington's disease
don't start showing any symptoms until they are in their 30s or 40s.
Mood changes, depression, irritability, or paranoia are often the first
signs of the disease and are followed by a decline in coordination and
an unsteady gait. WRITE DOWN ALL THE POSSIBLE GENOTPYES FOR
THIS DISEASE.
When looking at the gene pool each person has two alleles so if there
are 1000 individuals then there will be twice as many alleles of this
gene. How many will there be?
Allele Frequency
When one allele is dominant to another the symbols p and q are used
to represent their frequency.
H = dominant allele
p = its frequency
H = recessive allele
q = its frequency
Frequency of
dominant allele
p
+
+
Frequency of
recessive allele
q
=
gene pool
=
1
i)
If the frequency of the dominant allele for Huntington’s is 0.4,
what is the frequency of the recessive allele?
ii) If everyone in a population was heterozygous (Hh), what would
the frequency of the dominant allele be?
iii) …. And the recessive?
BUT populations are not made up of just one genotype!!!
1.
2.
3.
4.
5.
6.
7.
8.
Work in groups of three.
Each group is to take 100 beads – 80 red (dominant allele), 20 white (recessive
lethal allele). These represent the gene pool of a population.
Mix them and spill them out on the table. Work out the allele frequencies (p and
q).
With your eyes, each take a pair of alleles.
The recessive allele is lethal so if you picked two white beads then you must
discard these. This represents selection.
With the remaining gene pool work out the allele frequency.
Assuming your population reproduces top up your alleles to the percentages you
have just calculated. This will keep the population at 100.
Repeat the process ten times (or less if all the white alleles are gone).
Generation
no
Dominant
Alleles
Recessive
Alleles
No of lethal
alleles
Total Alleles
in gene pool
1. Starting no
80
20
e.g. 2
e.g. 98
Ending %
80/98
20/98
2. Starting no
Discussion
1. Which genotypes were selected to ‘survive’
and reproduce?
2. What are the effects of natural selection on a
population?
3. What else could affect the gene pool?
4. How could you modify this modelling activity
to look at the effect of gene flow on
populations?
Timed Essay
You have thirty minutes to write an essay. Spend
five minutes planning out the essay.
“Discuss monohybrid inheritance in humans”
(25 marks)
Hardy-Weinberg Equation
A Mendelian population is a group of sexually reproducing organisms
of the same specie residing within defined geographic boundaries
wherein interbreeding occurs.
Gene Pool is the sum total of the genes in a population. This term
usually refers to one phenotypic character at a time.
The gene pool changes over time as new alleles enter of exit through
events such as natural selection, migration or mutation.
Allele frequency is not something that can be measured directly in
populations.
What can be seen or measured is the phenotype of each individual of
the population.
Possible chance combinations of gametes
If we know the relative frequencies of dominant and
recessive alleles in a population (p and q) then we can
calculate the expected frequencies of progeny
genotypes and phenotypes (assuming random fusion of
gametes).
Male/Female
p (A)
q (a)
p (A)
p² AA
pq Aa
q (a)
pq Aa
q² aa
NB. p + q = 1
i.e. the percentage of A and a gametes must add up to 100% in
order to account for all the gametes in the gene pool.
Formula for the expected genotypes in
the next generation
The rule is named
after G. H. Hardy
and W. Weinberg
who independently
formulated it in
1908.
Several assumptions are needed:
1.
2.
3.
4.
The population is infinitely large and mates at random.
No selection pressures
The population is closed (no immigration or emigration)
There is no mutation of alleles (unless mutation rates
are equal)
5. Meiosis is normal so that chance is the only factor in
producing gametes.
No population is infinitely large, mutations cannot be
prevented, selection and migration pressures usually exist
in most natural populations etc. so it may be surprising to
note that despite this many populations do conform to
equilibrium conditions between two generations.
Worked Example
For example, if the frequency of A = 0.6 and the
frequency of a = 0.4 then the expected frequencies would
be
p² = (0.6)² = 0.36
2pq = 2 x 0.6 x 0.4 = 0.48
q² = (0.4)² = 0.16
You can use this equation to calculate allele frequencies if
you know the frequency of a least one of the genotypes.
You can identify homozygous recessive individuals to
calculate allele frequencies.
Using the HW equation
150 plants produce long pollen grains and 50 plants produce
round pollen grains in a population of peas. The allele frequencies
(p and q) can be found by calculating the frequency of
homozygous recessive plants (round pollen grains).
Number of round pollen grains = 50
Total number of plants = 200
q² = 50/200 = 0.25
Frequency of recessive allele (q) = √ 0.25 = 0.50
Since p + q = 1 then p must be 1-0.50 = 0.50
The frequency of the pollen grain genotypes may also be
calculated;
p² = (0.5)² = 0.25 (AA) or 50 individuals in a population of 200
2pq = 2 x 0.5 x 0.5 = 0.50 (Aa) or 100 individuals
q² = (0.5)² = 0.25 (aa) or 50 individuals
Questions
1. A population of Drosophilia contains 64 long-winged flies
and 36 vestigial winged flies.
a) What is the frequency of vestigial winged flies (q²) in this
population?
b) What is the frequency of vestigial-wings allele (q)?
c) What is the frequency of long-wings allele?
d) What is the frequency of homozygous long-winged flies?
e) What proportion of long-winged flies are homozygous and
what proportion are heterozygous?
2. Humans suffering from insulin-dependent mellitus cannot
secrete the hormone insulin. This disorder is inherited as a
recessive allele at a single locus. If the frequency of allele (q)
in a human population is 0.07, calculate the frequency of
a) The normal allele
b) People who suffer from diabetes
c) Heterozygous carriers of the diabetes allele.
Homework
• Complete the exam questions on HW.
Due next Thursday.
STARTER – 6 marks
Huntington’s disease is a human inherited condition
resulting in gradual degeneration of nerve cells in the
brain. It is caused by a dominant allele, but usually no
symptoms are evident until the person is at least 30 years
old. It is very rare in most populations. However, in one
isolated area in Venezuela, 48% of the population possess
a genotype which gives rise to Huntington’s disease.
Many of the inhabitants can trace their origins to a
common ancestor 200 years ago.
a) Use the HW equation to estimate the % of this
Venezuelan population which is heterozygous for
Huntington’s disease. Show your working.
b) Suggest why i) there is such a high incidence of
Huntington’s disease in this population.
ii) Huntington’s disease has not been eliminated from this
population by natural selection
Answers – Self Assess
a) 48% = 0.48
1 – 0.48 = 0.52 = frequency of homozygous recessives.
Square root of 0.52 = 0.72 = frequency of recessive allele.
Frequency of dominant allele = 1-0.72 = 0.28
Frequency of heterozygotes = 2pq = 2 x 0.28 x 0.72 = 0.403 =
40%
b) i) Common ancestor/in-breeding/no migration/genetic
isolation/small gene pool
ii) Reproduction occurs before the symptoms of disease are
apparent/possessors of the dominant allele are not at a
selective disadvantage/Hh x hh … 50% offspring affected/
Hh x Hh …. 75% of offspring affected.