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Hardy-Weinberg Principle
/25
1. Complete the following table.
Term
Definition
The alleles of a particular gene in a
population at a given time
A number between 0 and 1 which describes
how common an allele is
The allele frequency which you can always
work out by looking at the phenotypes
A group of organisms of the same species
living in the same habitat
[4]
2. The ……………………………. of the alleles of a gene will ................................................
from one generation to the next if the following five conditions are met:
a. .......................................................................................................................
b. .......................................................................................................................
c.
.......................................................................................................................
d. .......................................................................................................................
e. .......................................................................................................................
[7]
3. Finish these equations…
a.
p+
b.
p2 +
c. Underline the term in these equations which represents the frequency of the
homozygous dominant genotype.
[3]
4. For which phenotype can you always tell the genotype?
...............................................................................................................................
[1]
5. The banded snail (Cepaea nemoralis) can have pink shells or yellow shells. The shell
colour is controlled by two alleles of a single gene. The pink allele (CP) is dominant
over the yellow allele (Cy). In a population of banded snails, 64 were pink and 36
were yellow.
a. What is the frequency of yellow snails (q2) ?
.......................................................................................................................
b. What is the frequency of the yellow allele (q) ?
.......................................................................................................................
c. What is the frequency of the pink allele (p = 1 – q) ?
.......................................................................................................................
d. What is the frequency of the homozygous pink snails ?
.......................................................................................................................
e. What is the frequency of heterozygous pink snails ?
.......................................................................................................................
f.
If you pick up a pink snail, what is the probability that it will be heterozygous ?
.......................................................................................................................
.......................................................................................................................
[6]
6. Albinism is a metabolic disorder caused by a recessive allele. Sufferers lack
pigmentation. 1 in 10 000 people suffer from the disease. Use the Hardy-Weinberg
equations to calculate…
a. the frequency (q) of the recessive allele. .........................................................
.......................................................................................................................
b. the frequency (p) of the dominant allele. .........................................................
.......................................................................................................................
c. the expected percentage of carriers in the population. ......................................
.......................................................................................................................
d. In a college of 2000 students, how many would you expect to have at least one
allele for albinism ?
.......................................................................................................................
[4]
The Hardy-Weinberg Principle: Kuru .............................. /20
1.
The Fore Tribe of New Guinea used to suffer from a prion disease called Kuru1. It
used to kill 1% of the tribe annually. Before the practice was banned in the 1950s,
women of this tribe in New Guinea used to eat their ancestors in mortuary feasts.
This infected them with the prion protein. It looks as though survival rates were
affected by the genotypes of the islanders. The gene responsible codes for the prion
protein. It has two alleles:
 PM – codes for methionine at codon 129
 PV – codes for valine at codon 129
Two groups of Fore Tribe members were chosen and their genotypes were analysed.
The first group were women who had survived many feasts, the second group were
people who had not taken part in any.
n
P MP M
P MP V
PVPV
Survived feasts
30
0.133
0.767
0.100
Unexposed to
feasts
140
0.221
0.514
0.264
a. Use the PVPV genotype frequency to estimate the frequency (q) of the PV allele
for each group
q (Survivors) ..................................................................................................
q (Unexposed) ...............................................................................................
[2]
b. Use the PMPM genotype frequency to estimate the frequency (p) of the PM allele
for each group
p (Survivors) .................................................................................................
p (Unexposed) ...............................................................................................
[2]
c. Use these values of p and q to calculate the expected frequency of the
heterozygotes
2pq (Survivors)
............................................................................................
2pq (Unexposed) ...........................................................................................
1
Data from Hedrick P. (2005) Genetics of Populations. Jones and Bartlett.
[2]
d. In which group of people is the actual frequency of the heterozygotes much
greater than you expected ?
.......................................................................................................................
[1]
e. One of the assumptions of the Hardy-Weinberg principle is that there is no
selection (all genotypes must have the same reproductive success). One
explanation for these results is that Kuru is killing more of some genotypes than
others. Which genotypes seem to be selected against?
.......................................................................................................................
2.
[1]
When the observed frequencies are different from expected frequencies, something
interesting is going on. What statistical test could you use to find out if the
difference is significant ?
.............................................................................................................................
and what level of probability would be regarded as significant ? ..............................
3.
[2]
Hardy Weinberg relies on four other assumptions. What do you need to check
before you can accept the selection hypothesis? Explain how you might go about it.
a.
.....................................................................................................................
.....................................................................................................................
b.
.....................................................................................................................
.....................................................................................................................
c.
.....................................................................................................................
.....................................................................................................................
d.
.....................................................................................................................
.....................................................................................................................
4.
[8]
Which prion disease has been important in this country during the last 10 years ?
.............................................................................................................................
[1]
What might you look for in people who suffer from this disease compared with
similarly exposed people who don’t ?
.............................................................................................................................
[1]
The Hardy-Weinberg Principle: Rhesus........................... /20
1.
Our red blood cells sometimes have an antigen on their surface which was first seen
in rhesus monkeys. People who have the antigen are called rhesus positive and
those who don’t are called rhesus negative.
The rhesus antigen is coded for by a dominant allele R. Give the genotypes of
people who are…
a. Rhesus positive ..............................................................................................
b. Rhesus negative .............................................................................................
2.
[3]
Before the rhesus system was discovered in 1940, blood transfusions often used to
fail, even when the ABO blood groups were carefully matched.
Explain why the immune system of a rhesus negative person would reject rhesus
positive blood.
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
.............................................................................................................................
3.
[3]
When a rhesus negative mother has a rhesus positive fetus, there are sometimes
problems towards the end of pregnancy. This is because fragments of fetal blood
cells can cross the placenta into the mother’s blood. The mother’s immune system
responds to these fragments and forms memory cells. The first fetus is usually
unaffected because pregnancy is nearly over. During later pregnancies, any rhesus
positive fetus is attacked by the mother’s antibodies much more strongly. These
antibodies destroy fetal red blood cells and, without a number of fetal blood
transfusions, the fetus will die.
What is the probability (0 to 1) of a rhesus positive child in each of the following
cases…
a. rr mother x rr father ......................................................................................
b. rr mother x Rr father .....................................................................................
c. rr mother x RR father ....................................................................................
[3]
4.
If hospitals know that a Rh- mother has a Rh+ fetus, they can stop the mother
responding to its antigens. They do this by injecting rhesus antibodies into the
mother after the birth. This destroys the antigens before the mother’s own immune
system can respond to them.
Obviously, it is important that the hospital knows how often they need to do this.
They need to know how often Rh+ babies are born to Rh- mothers.
The Hardy Weinberg equations make this very easy to work out.
a.
Write out the two equations you will use.
.....................................................................................................................
[2]
b.
16% of mothers are Rh-. What is q2 ? .........................................................
[1]
c.
What is q ? ..................................................................................................
[1]
d.
What is p ? ..................................................................................................
[1]
e.
If fathers are just as likely to be Rh+ as mothers, what is the probability that a
rhesus positive man is
i. homozygous (p2) ? ...................................................................................
ii. heterozygous (2pq) ? ..............................................................................
f.
[2]
What is the probability that a baby has
i. a mother who is rr AND a father who is RR
.....................................................................................................................
ii. a mother who is rr AND a father who is Rr AND the sperm was R
.....................................................................................................................
g.
What is the probability that a baby will have EITHER of these two possibilities ?
.....................................................................................................................
h.
[2]
[1]
So… What percentage of babies will be rhesus positive and have rhesus
negative mothers?
.....................................................................................................................
[1]
Answers
Hardy-Weinberg Principle
/25
1. Complete the following table.
Term
Definition
Gene pool
The alleles of a particular gene in a
population at a given time
Allele frequency
A number between 0 and 1 which describes
how common an allele is
q
The allele frequency which you can always
work out by looking at the phenotypes
Population
A group of organisms of the same species
living in the same habitat
[4]
2. The frequency of the alleles of a gene will remain constant
from one generation to the next if the following five conditions are met:
a. Large Population
b. No Immigration / Emigration
c.
Random mating
d. All genotypes have same reproductive success (no selection)
e. There is no gene mutation
[7]
3. Finish these equations…
a.
p+ q=1
b.
p2 + 2pq + q2 = 1
c.
Underline the term in these equations which represents the frequency of the
homozygous dominant genotype.
[3]
4. For which phenotype can you always tell the genotype?
phenotype expressed by the homozygous recessive
[1]
5. The banded snail (Cepaea nemoralis) can have pink shells or yellow shells. The shell
colour is controlled by two alleles of a single gene. The pink allele (CP) is dominant
over the yellow allele (Cy). In a population of banded snails, 64 were pink and 36
were yellow.
a. What is the frequency of yellow snails (q2) ?
0.36
b. What is the frequency of the yellow allele (q) ?
0.6
c.
What is the frequency of the pink allele (p = 1 – q) ?
0.4
d. What is the frequency of the homozygous pink snails ?
0.16
e. What is the frequency of heterozygous pink snails ?
2 x 0.4 x 0.6 = 0.48
f.
If you pick up a pink snail, what is the probability that it will be heterozygous ?
0.48 : 0.16 = 0.75 or 3/4
.......................................................................................................................
[6]
6. Albinism is a metabolic disorder caused by a recessive allele. Sufferers lack
pigmentation. 1 in 10 000 people suffer from the disease. Use the Hardy-Weinberg
equations to calculate…
a.
the frequency (q) of the recessive allele. 0.01
b.
the frequency (p) of the dominant allele. 0.99
c.
the expected percentage of carriers in the population.
2pq x 100 = 0.0198 x 100 = 1.98%
d.
In a college of 2000 students, how many would you expect to have at least one
allele for albinism ?
1.98% + 0.01% = 39.8 students … should round up to 40
[4]
Answers
The Hardy-Weinberg Principle: Kuru
1.
/20
The Fore Tribe of New Guinea used to suffer from a prion disease called Kuru2. It
used to kill 1% of the tribe annually. Before the practice was banned in the 1950s,
women of this tribe in New Guinea used to eat their ancestors in mortuary feasts.
This infected them with the prion protein. It looks as though survival rates were
affected by the genotypes of the islanders. The gene responsible codes for the prion
protein. It has two alleles:
 PM – codes for methionine at codon 129
 PV – codes for valine at codon 129
Two groups of Fore Tribe members were chosen and their genotypes were analysed.
The first group were women who had survived many feasts, the second group were
people who had not taken part in any.
n
P MP M
P MP V
PVPV
Survived feasts
30
0.133
0.767
0.100
Unexposed to
feasts
140
0.221
0.514
0.264
a. Use the PVPV genotype frequency to estimate the frequency (q) of the PV allele
for each group
q (Survivors) 0.316
q (Unexposed) 0.514
[2]
b. Use the PMPM genotype frequency to estimate the frequency (p) of the PM allele
for each group
p (Survivors) 0.365
p (Unexposed) 0.470
c.
[2]
Use these values of p and q to calculate the expected frequency of the
heterozygotes
2pq (Survivors) 0.231
2pq (Unexposed) 0.483
2
Data from Hedrick P. (2005) Genetics of Populations. Jones and Bartlett.
[2]
d. In which group of people is the actual frequency of the heterozygotes much
greater than you expected ?
Survivors
[1]
e. One of the assumptions of the Hardy-Weinberg principle is that there is no
selection (all genotypes must have the same reproductive success). One
explanation for these results is that Kuru is killing more of some genotypes than
others. Which genotypes seem to be selected against?
Homozygotes – so under-represented in sample
2.
[1]
When the observed frequencies are different from expected frequencies, something
interesting is going on. What statistical test could you use to find out if the
difference is significant ?
Chi-squared
and what level of probability would be regarded as significant ? p <0.05
3.
[2]
Hardy Weinberg relies on four other assumptions. What do you need to check
before you can accept the selection hypothesis? Explain how you might go about it.
a.
Large population
Count population of Fore Tribe
b.
Random mating
Survey tribe and find out
c.
No immigration / Emigration
Survey tribe / ask elders / carry out genetic tests / look at any census data
d.
No gene mutation
This would probably need a long term genetic study
4.
[8]
Which prion disease has been important in this country during the last 10 years ?
BSE / Mad Cow Disease
[1]
What might you look for in people who suffer from this disease compared with
similarly exposed people who don’t ?
Genetic differences which correlate with susceptibility
[1]
Answers
The Hardy-Weinberg Principle: Rhesus
1.
/20
Our red blood cells sometimes have an antigen on their surface which was first seen
in rhesus monkeys. People who have the antigen are called rhesus positive and
those who don’t are called rhesus negative.
The rhesus antigen is coded for by a dominant allele R. Give the genotypes of
people who are…
a. Rhesus positive RR or Rr
b. Rhesus negative rr
2.
[3]
Before the rhesus system was discovered in 1940, blood transfusions often used to
fail, even when the ABO blood groups were carefully matched.
Explain why the immune system of a rhesus negative person would reject rhesus
positive blood.
Immune response; Phagocyte; Macrophage; Antigen Presenting Cell;
B Lymphocyte; Antigen matches receptors; Clonal Selection;
Differentiation to form plasma cells; which secrete antibodies;
Depth required will depend on your syllabus!
3.
[3]
When a rhesus negative mother has a rhesus positive fetus, there are sometimes
problems towards the end of pregnancy. This is because fragments of fetal blood
cells can cross the placenta into the mother’s blood. The mother’s immune system
responds to these fragments and forms memory cells. The first fetus is usually
unaffected because pregnancy is nearly over. During later pregnancies, any rhesus
positive fetus is attacked by the mother’s antibodies much more strongly. These
antibodies destroy fetal red blood cells and, without a number of fetal blood
transfusions, the fetus will die.
What is the probability (0 to 1) of a rhesus positive child in each of the following
cases…
a. rr mother x rr father 0
b. rr mother x Rr father 0.5
c. rr mother x RR father 1
[3]
4.
If hospitals know that a Rh- mother has a Rh+ fetus, they can stop the mother
responding to its antigens. They do this by injecting rhesus antibodies into the
mother after the birth. This destroys the antigens before the mother’s own immune
system can respond to them.
Obviously, it is important that the hospital knows how often they need to do this.
They need to know how often Rh+ babies are born to Rh- mothers.
The Hardy Weinberg equations make this very easy to work out.
a.
Write out the two equations you will use.
p+q=1
p2 + 2pq + q2 = 1
[2]
b.
16% of mothers are Rh-. What is q2 ? 0.16
[1]
c.
What is q ? 0.4
[1]
d.
What is p ? 0.6
[1]
e.
If fathers are just as likely to be Rh+ as mothers, what is the probability that a
rhesus positive man is
i. homozygous (p2) ? 0.36
ii. heterozygous (2pq) ? 0.48
f.
[2]
What is the probability that a baby has
i. a mother who is rr AND a father who is RR
0.16 x 0.36 = 0.0576
ii. a mother who is rr AND a father who is Rr AND the sperm was R
0.16 x 0.48 x 0.5 = 0.0384
g.
What is the probability that a baby will have EITHER of these two possibilities ?
0.0576 + 0.0384 = 0.096
h.
[2]
[1]
So… What percentage of babies will be rhesus positive and have rhesus
negative mothers?
9.6% of all babies
[1]