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Hardy-Weinberg
Textbook
What do we mean by the following terms?
• Gene pool
• Allele frequencies
DEFINITIONS
• Gene pool
•
All the alleles of all the genes of all the individuals of an
entire population
• Allele frequencies
•
The number of times an allele for a particular gene occurs
within a gene pool
The Hardy-Weinberg Principle
A mathematical model which predicts that
allele frequencies will not change from one
generation to the next.
Relies on a number of conditions:
• A large, isolated population
• No mutations
• No selection of one allele over another
• Mating is random
• No net migration
The Hardy-Weinberg Principle
Where a gene has two alleles:
Frequency of dominant allele in a population =
Frequency of recessive allele in a population =
p
q
In any population, the total frequency of alleles is 1,
therefore:
p+q=1
The Hardy-Weinberg Principle
Each person has pairs of alleles, so the possible combinations
of alleles are:
pp
p2
pq
qp
2pq
qq
q2
In any population, the total frequency of genotypes is 1,
therefore:
p2 + 2pq + q2 = 1
Equations
2
p
+ 2pq +
2
q
=1
p2 = frequency of homozygous dominant
individuals
2pq = frequency of heterozygotes
q2 = frequency of homozygous recessive individuals
Remember:
p+q=1
Use for allele frequencies.
p2 + 2pq + q2 = 1
Use for genotype/phenotype frequencies.
Worked example
 1 in 25000 show a recessive condition (aa)
 How many people in a population of 25000 are carriers of the condition
(Aa or aA)?
1 / 25000 = 0.00004 are aa (q2)
q2 = 0.00004
q = √0.00004
q = 0.00632 (3 sig fig)
p + q = 1.0
p = 1.0 – 0.00632
p = 0.99368
Aa or aA = 2pq
2pq = 2 x 0.99368 x 0.00063
2pq = 0.0126
• 0.0126 of 10,000 people
• 0.0126 x 10,000 = 126
• 126 in 10,000 people carry
the recessive allele
• But what about in 25,000?
• 0.0126 X 25,000 = 315
• 315 people carry the
recessive allele in a
population of 25,000
Question 1
 If 98 out of 200 individuals in a population
express the recessive phenotype, what
percent of the population would you predict
would be heterozygotes?
p2 + 2pq + q2 = 1
Answer 1
 98/200 = (q2)
 0.49 = q2
 0.7 = q
 p+q=1
 p = 1 – 0.7
 p = 0.3
 2pq = 2(0.3)(0.7) = 0.42 = 42% heterozygotes
Question 2
 2. Your original population of 200 was hit
by a tidal wave and 100 organisms were
wiped out, leaving 36 homozygous recessive
out of the 100 survivors. If we assume that
all individuals were equally likely to be
wiped out, how did the tidal wave affect the
predicted frequencies of the alleles in the
population?
Answer 2
 36/100 = q2
 0.6 = q
 p+q=1
 p = 0.4
 Heterozygous = 2 (0.4)(0.6) = 0.48 = 48%
 Homozygous dominant = (0.4)(0.4) = 0.16 = 16%
Question 3
 Lets say that brown fur coloring is dominant
to grey fur coloring in mice. If you have 168
brown mice in a population of 200 mice........
What is the predicted frequency of



Homozygous dominants
Heterozygotes
Homozygous recessives
Answer 3
 200 mice in total
 168 = brown = p2 + 2pq
 32/200 = grey fur = q2
 0.16 = q2
 0.4 = q
 p = 0.6 (p + q = 1)
 p2 = 0.36 = 36%
 2pq = 0.48 = 48%
 q2 = 0.16 = 16%
Question 4
 If 81% of a population is homozygous
recessive for a given trait. Calculate



Frequency of homozygous dominant
Frequency of heterozygotes
Frequency of dominant and recssive alleles
Answer 4
 q2 = 0.81
 q = 0.9
 p = 0.1
 p2 = 0.01
 2pq = 0.18
Question 5
 If 51% of the population carries at least one
copy of the recessive allele

what is the predicted frequency of the population
expressing the dominant phenotype
Answer 5
 51% = 2pq + q2
 49% = 0.49 = p2
 0.7 = p
 0.3 = q
 p2 + 2pq =
 0.49 + 0.42 = 0.91 have dominant phenotype
Question 8
 1 in 1700 US Caucasian new borns have cystic
fibrosis.
 calculate the frequency of the recessive cystic
fibrosis allele and the dominant allele in the
population
 calculate the frequency of non cystic fibrosis
sufferers in the population
Answer 8
 q2 = 1/1700
 q = 0.0243
 p = 0.09757
 p2 + 2pq
 (0.09757)(0.09757) + 2(0.09757)(0.0243)
 0.9567
Question 9
 If 9% of an African population is born with a
severe form of sickle-cell anemia (ss), what
percentage of the population will be more
resistant to malaria because they are
heterozygous(Ss) for the sickle-cell gene?
Answer 9
 q2 = 9%
 q2 = 0.09
 q = 0.3
 p = 0.7
 2pq = 2(0.3)(0.7) = 0.42 = 42%
Question 10
 The allele y occurs with a frequency of 0.8 in a
population of clams. Give the frequency of
 genotypes YY, Yy, and yy. Show your work!
Answer 10
 The allele y (recessive) has a frequency q = 0.8.
 p + q = 1, then p = 1 – 0.8 = 0.2
 genotype:
 YY genotype frequency = p2 = 0.04
 Yy genotype frequency = 2pq = 0.32
 yy genotype frequency = q2 = 0.64.
Question 11
 In the year 2374, humans finally developed the technology
necessary for time travels. You are a scientist interested in the
population genetics of extinct animals. Taking advantage of
this technological advance, you decide to go to the past 8
million years to conduct a field work in Venezuela to study a
population of Phoberomys pattersoni*, the world’s largest
extinct rodent weighing approximately 700 kg (1500 lb) and
looking vaguely like a giant guinea pig.
 The coat color of this rodent varies between tan (dominant)
and brown (recessive). Assume the population is in HardyWeinberg equilibrium. You observed 336 tan Phoberomys and
64 brown Phoberomys during your study.



What is the frequency of the homozygous recessive genotype
What is the allelic frequency of the dominant (tan) allele in the population?
Of the animals you observed, how many were heterozygous?
Answer 11
 There are 336 + 64 = 400 animals in the population.
 64 are homozygous recessive (brown)
 Frequency of homozygous recessive = q2 = 64/400 = 0.16
 Since q2 = 0.16, take the square root to get q = 0.4
 p + q = 1 (formula for allele frequencies)
 Frequency of the dominant allele p = 0.6
 Since q2 = 0.16, take the square root to get q = 0.4
 Remember that p + q = 1 (formula for allele frequencies)
 Frequency of the dominant allele p = 0.6
Question 12
 You make another trip to Venezuela and this time
you observe 650 animals.



How many of the 650 animals would you expect to be tan,
assuming the population is still in Hardy-Weinberg
equilibrium?
How many of these tan animals are homozygous for the
dominant allele?
How many of these 650 animals would you expect to be brown,
assuming the population is still in Hardy-Weinberg
equilibrium?
Answer 12
 If the population is still in H-W equilibrium, then the
allele frequencies would be the same: p = 0.6, q = 0.4
 The tan phenotype = p2 + 2pq
 (0.6)2 + (2)*(0.6)*(0.4) = 0.84
 0.84 * 650 = 546 tan
 p2 = (0.6)2 = 0.36,
 (0.36)*(650) = 234
 Brown animals are homozygous recessive
 Frequency of brown is q2 = (0.4)2 = 0.16
 (0.16)*(650) = 104
Question 13
 As you observe the animals, you count 200 brown
Phoberomys and 450 tan.
 Conduct a chi-square test to determine if your
observations are significantly different from what
you expect.
Chi square
In every χ2-test the calculated χ2 value will either be
(i) less than or equal to the critical χ2 value
(ii) greater that the critical χ2 value.
• If calculated χ2 ≤ critical χ2,
no statistically significant difference
Greater than 5% Probability that (differences in) results are due to
chance;
Accept null hypothesis
• If calculated χ2 > critical χ2,
there is a statistically significant difference
less than 5% probability that (differences in) results are due to
chance;
reject null hypothesis
Answer 13
 The calculated X2 is 105.5
 There are 2 phenotypes (brown and tan), so there is 1
degree of freedom (2 – 1 = 1)
 The theoretical X2 for 1 degree of freedom is 3.841, which is
much smaller than our calculated one. Therefore, we reject
the null hypothesis that the population of 650 is in
 H-W equilibrium. Our observations are significantly
different from our espectation, assuming H-W equilibrium.
Question 14 past paper
1.
In a study of people living in India, the frequency of
the IO allele was found to be 0.55 and that of the
IA allele, 0.18. What was the frequency of the IB
allele in this population?(1)
2. In a village with a population of 500, there were 8
people who were homozygous for the sickle–cell
allele and 96 who were heterozygous. Calculate the
frequency of the HbS allele in the village. Show
your working.
Answer
0.55 + 0.18 + x = 1
x = 0.27
1.
2. 500 people = 100 alleles in total
8 people homozygous recessive = 16 copie sof sickle
cell allele
96 heterozygous = 96 copies of sickle cell allele
total occurrences of the allele = 96 + 16 = 112
112/1000 = frequency = 0.0112
Question 15 past paper
Warfarin is a substance which inhibits blood clotting. Rats which eat warfarin are killed due to
internal bleeding. Some rats are resistant to warfarin as they have the allele WR. Rats have
three possible genotypes:
WRWR: resistant to warfarin
WRWS: resistant to warfarin
WSWS: susceptible (not resistant) to warfarin.
In addition, rats with the genotype WRWR require very large amounts of vitamin K in their
diets. If they do not receive this they will die within a few days due to internal bleeding.
1) A population of 240 rats was reared in a laboratory. They were all fed on a diet containing an
adequate amount of vitamin K. In this population, 8 rats had the genotype WSWS, 176 had the
genotype WRWS and 56 had the genotype WRWR.
Use these figures to calculate the actual frequency of the allele WR in this population. Show
your working.
2) The diet of the rats was then changed to include only a small amount of vitamin K. The rats
were also given warfarin. How many rats out of the population of 240 would be likely to die
within a few days? (1)
3) In a population of wild rats, 51% were resistant to warfarin.
Use the Hardy-Weinberg equation to estimate the percentage of rats in this population which
would be heterozygous for warfarin resistance. Show your working.
Answer
240 rats = 480 alleles
Total WR = 56*2 + 176 = 288
WR frequency = 288/480 = 0.6
56 + 8 = 64
49% = q2
0.7 = q
p = 0.3
2pq = 2(0.3)(0.7) = 0.42 = 42%
Question 16 past paper
 In the flour beetle, the allele for red body colour (R)
is dominant to the allele for black body colour (r). A
mixed culture of red beetles and black beetles was
kept in a container in the laboratory under optimal
breeding conditions. After one year, there were 149
red beetles and 84 black beetles in the container.

Use the Hardy-Weinberg equation to calculate the expected
percentage of heterozygous red beetles in this population
Answer
total number beetles = 233
frequency of black (q2) = 84/233 = 0.36
q = 0.6
p = 0.4
2pq = 0.48
Question past paper
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