Download Evolutionary Mechanisims and Hardy

Laboratory 2 EVOLUTIONARY MECHANISMS
& Hardy Weinberg Equilibrium
Revised September 30th-changes in red and underlined
Before lab
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•
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Review the mechanisms of evolution and the Hardy-Weinberg principle in your textbook
(Freeman Chapter 24, especially section 24.2).
Do the web/CD tutorial 24.1
Read this chapter from the BIO152 electronic lab manual
Objectives After completing this exercise you should be able to:
1. Determine the allele frequencies for a gene in a model population.
2. Calculate observed and expected ratios of genotypes based on Hardy-Weinberg
proportions.
3. Test hypotheses about the effect of evolutionary agents (natural selection, gene flow,
genetic drift, or mutation) on allele frequencies in a population.
4. Explain why the Hardy-Weinberg principle serves as a null hypothesis.
Timeline
2:10- 2:25 Quiz (15 minutes)
2:30 – 3:00 Intro and Exercise 1
3:00 – 3:40 Exercise 2 (40 minutes)
3:45 – 4: 20 group presentations (Each bench gets 10 minutes MAX)
4:30 – 4:50 each group measures BACTERIA plates from lab 1
Evaluation
10 points quiz (done individually)
10 points presentation of experiment (group work, group mark)
Introduction
The Hardy-Weinberg Principle says that heredity itself cannot cause changes in the
frequencies of alternate forms of the same gene (alleles). If certain conditions are met, then
the proportions of genotypes that make up a population of organisms should remain constant
generation after generation according to Hardy- Weinberg equilibrium:
Genotype frequencies: p2 + 2pq + q2 = 1.0 (for two alleles)
If p is the frequency of one allele (A), and q is the frequency of the other allele (a), then
Allele frequencies: p + q=1.0
In nature, however, the frequencies of genes in populations are not static (that is, not
unchanging). Natural populations never meet all of the assumptions for Hardy-Weinberg
equilibrium. The assumptions for Hardy-Weinberg equilibrium are:
1. The organism in question is diploid.
2. Reproduction is sexual.
3. Mating is random.
BIO152 2005 University of Toronto at Mississauga
2-2 Evolutionary Mechanisms
4.
5.
6.
7.
Population size is very large.
Migration is negligible. (i.e., no immigration or emigration occurs)
No net changes in the gene pool due to mutation.
Natural selection does not affect the locus under consideration (i.e., all genotypes are
equally likely to reproduce).
Your text lists five conditions which must be met (which cover the same assumptions as listed
above):
1. No natural selection at the gene in question.
2. No genetic drift or random allele frequency changes affecting the gene in question.
3. No gene flow.
4. No mutation.
5. Random mating.
►Be able to explain the Hardy-Weinberg equilibrium and the reason for each condition.
Evolution is a process resulting in changes in the genetic makeup of populations through time;
therefore, factors that disrupt Hardy-Weinberg equilibrium are referred to as evolutionary
agents. In random mating populations, natural selection, gene flow, genetic drift, and mutation
can all result in a shift in gene frequencies predicted by the Hardy-Weinberg formula.
Nonrandom mating can also result in such changes. The exercises in this lab will demonstrate
the effect of these agents on the genetic structure of a simplified model population.
If a population is in Hardy-Weinberg equilibrium, then evolution is NOT happening;
therefore, the Hardy-Weinberg principle may be called the null hypothesis for evolution.
Hypothesis
In a large, randomly mating population with no mutation, migration, or selection, the allelic and
genotypic frequencies should remain at equilibrium.
Prediction
If a population is at Hardy-Weinberg equilibrium, then the frequencies of the alleles
(represented by beads in this lab) should not change.
Summary of Materials
Per student group of 4:
•
•
•
•
•
•
•
•
•
•
plastic dishpan (12" x 7" x 2")
100 large (10-mm diameter) white beads
100 large red/brown beads
100 large pink beads
4,000 small (8-mm diameter) white beads
pair of long forceps
coarse sieve (9.5-mm)
small bowl
2 small paper/opaque plastic bags
1 plastic bottle
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2-3 Evolutionary Mechanisms
Exercise 1 Testing Hardy-Weinberg (HW) equilibrium (work in pairs)
Materials
Plastic/paper bag with a 50 beads (mixture of white and red/brown—different groups are given
a different proportion of red/brown and white beads)
Introduction
Simulate a population using coloured beads and test whether this population is in HW
equilibrium. The bag of beads represents the gene pool for the population; each bead is a
single allele (in a single gamete) and the two colours represent the two alleles for that gene in
the population. For this simulation we will call red/brown (R) dominant to white (r)
The code number of the outside of your bag is ___________
______ # red/brown beads
______ # white beads
‘allele’ frequency of R =___________
‘allele’ frequency of r = ___________
______ total # beads
How many diploid individuals are represented in this population? ______
Hypothesis: (re-state the Hardy-Weinberg theory)
Prediction: Predict the genotype frequencies of the population in future generations (If/then)
Procedure
1. Without looking, randomly remove two beads from the bag. These two beads represent
one diploid individual in the next generation. Record the genotype in Table 1.
2. Return the beads to the bag, shake before selecting two more beads. Why? (see
explanation below-‘sampling with replacement’)
3. Continue steps 1 & 2 until you have recorded the genotypes for 20 individuals.
Results
1. Determine the expected frequencies of genotypes and alleles for the population from
the original allele frequency: allele frequencies: p + q=1.0
p (freq of R) = ______ q (freq of r) = ________
2
2
Calculate expected genotype frequencies ( p + 2pq + q = 1.0 (for two alleles)) and the
expected number of individuals of each genotype.
Expected freq of RR______Rr_______ rr______
# individuals
RR______Rr_______ rr______
2. Calculate the observed frequencies from the observed number of individuals of each
genotype obtained from your experiment.
# individuals
RR______Rr_______ rr______
Observed freq
RR ______ Rr _______ rr ______
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2-4 Evolutionary Mechanisms
3. To determine whether the observed frequencies were consistent with what was
expected,use the Chi-Square statistical test. (See sample calculation in Table 1b)
Table 1a The # of individuals with the three genotypes after one generation—comparing the
observed versus expected # of individuals
# RR
#Rr
#rr
Observed (o)
Expected (e)
Deviation (o-e)= d
d2
d2//e
Chi-square = sum d2//e
_____
Discussion
4. Do the results of your statistical analysis suggest that the frequencies of genotypes in
your population are significantly different from expected?
5. Were your results consistent with your hypothesis and prediction?
6. What would you expect to happen to the frequencies if you continued the simulation for
25 generations?
7. Is this population evolving? Explain your answer.
8. How does this simulation meet the conditions for the population to be in HW
equilibrium? Answer yes or no for each condition: Random mating____; Large
population ____; no gene flow ____; no selection____
Sample data, calculations, and discussion: (you are NOT responsible for
doing this statistical test on your results in this lab)
Table 1b Example for 50 individuals randomly selected with an initial allele frequency or p=q
# RR
#Rr
#rr
Observed (o)
22
22
6
Expected (e)
18
24
8
Deviation (o-e)= d
4
2
2
d2
16
4
4
d2//e
0.89
0.16
0.5
Chi-square = sum d2//e
1.55 (degrees of freedom = 2 because 3
genotypes -1 = 2),
level of significance p=0.05) NOT Significant
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Table 1c Critical values for the χ2 - distribution
Degrees
Level of probability
of
0.10
0.05
0.01
freedom
6.64
3.84
2.71
1
9.21
5.99
4.61
2
11.34
7.82
6.25
3
13.28
9.49
7.78
4
Discussion: The observed results are consistent with the expected results. The data support
the hypothesis; the observed distribution is not significant from expected under the HardyWeinberg theory. Any minor differences can be attributed to change of sampling error. (This
population is not evolving.)
Sampling with replacement
By replacing the beads each time after sampling, the size of the gene pool stays the same and
the probability of selecting any allele should remain equal to its frequency.
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Exercise 2 Simulating Evolutionary Change
Under the conditions specified by the HW-model, the genetic frequencies should not change,
and evolution should not occur. In this exercise, you will modify each of the conditions to
determine the effect on gene frequencies in subsequent generations.
Your TA will assign your bench to one of the Projects below. Work with your partner to do the
specified experiment and then COMPARE data with the other 3 groups at your bench. Prepare
a COMMON presentation.
Your task is to determine the changes in genetic frequencies over several generations using
the bead model. Each bench will present their findings on the experiments to the class and
compare the effect on allele and genotype frequency over time for the specific mechanism.
Though your group is presenting your findings for your assigned experiments, you are
responsible for understanding the results for the others as well.
Projects (one per bench)
1. genetic drift:
Bottleneck and gene flow (2 pairs)
Founder and gene flow (2 pairs)
2. gene flow (migration)
3. natural selection
Results & Discussion Presentations (GROUP)
One member of your team will write out the following to give your TA BEFORE your talk.
BENCH ____ Full names of all people at the bench:
______________ ________________ _____________
_______________
______________
_______________
________________
_____________
Your experiment [genetic drift-bottleneck, genetic drift-founder effect, gene flow, natural
selection] ________________________
1. State your hypothesis:
2. State your prediction:
3. Which of the conditions necessary for Hardy-Weinberg equilibrium were met?
4. Which condition (s) changed?
5. Describe the changes in allele frequencies of p and q over time and compare the end
frequencies to starting frequencies.
6. Describe the changes in genotype frequencies and compare the end frequencies to
starting frequencies
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The General Model
1. The populations you will be working with are composed of coloured beads representing
diploid individuals as follows:
• Red/brown beads are homozygous for the red allele (RR).
• White beads are homozygous for the white allele (rr)
• Pink beads are heterozygotes (Rr).
The alleles in our population are codominant. Thus, each white bead contains two white
alleles; each pink bead, one white and one redbrown; and each redbrown bead, two red
alleles. The total number of colour alleles in a population of twenty individuals is forty. If such
a population contains five white beads and ten pink beads, the frequency of the white allele is:
p = (2 x 5) + 10
40
= 0.5
Because p + q = 1.0, the frequency of the red allele (q) must also be 0.5 if there are only two
colour alleles in this population.
2. SAMPLING: Sampling with replacement: After removing a bead from the bag and
recording the colour (representing the genotype of one individual), replace the bead
before removing the next bead. Sampling without replacement: do not replace the bead
before taking the next.
3. Reestablishing a population with new allelic frequencies
In some cases, the number of individuals will decrease as a result of the simulation. In those
cases, return the population to 100 total alleles with the new allelic frequencies. For example, if
you eliminate by selection all homozygous recessive (ww) individuals in your simulation, then
the resulting frequencies would be
# individuals: 14 RR 24 Rr 0 rr
# alleles:
28 R + 24R, 24r
Total # alleles: 76
Frequency of R = (28 + 24)/76 = 52/76 = 0.68
Frequency of r = 24/76 = 0.32
(notice that p+q=1)
To reestablish a population of 100, then the number of beads should reflect these new
frequencies. ADJUST THE NUMBER OF BEADS so that W is now 68/100 and w is 32/100
before starting the next round of the simulation.
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2-8 Evolutionary Mechanisms
7.
Experiment A Simulation of genetic drift—founder and
bottleneck effects
Materials
Paper bag containing 60 beads: 20 redbrown, 20 white, 20 pink
Bottle with a narrow neck (bottleneck effect)
Introduction
Genetic drift is the change in allele frequencies in a small population as a result of chance
alone not selection based on a heritable trait. In a small population, combinations of gametes
may not be random due to sampling error. (If you pull 50 beads out of the bag without looking,
you are more likely to get the same ratio of the two colours than if you only pull out 10).
Genetic fixation is the loss of all but one possible allele at a gene locus in a population.
Genetic fixation is a common result of genetic drift in a small population.
A-1 Founder effect
When a small group of individuals becomes separated from the larger parent population, the
allele frequencies in this small gene pool may be different from those in the original population
as a result of chance. This separation may occur when a group establishes a new area such
as the colonization of an island; thus a small group establishing a new area is called the
founder effect.
Hypothesis to explain the founder effect:
Prediction of what type of change you expect to occur in a small population:
Null Hypothesis (re-state the Hardy-Weinberg principle)
Predication as a result of Hardy-Weinberg
Procedure
1. Establish a starting population with 60 individuals with a starting allele frequency of 0.5
for each allele.
Alleles
#R
60
frequency (p) 60/120 = 0.5
#r
60
frequency (q) 60/120 = 0.5
Total#: 120
Genotypes:
# RR 15
#Rr
30
#rr
15
Total#: 60
frequency 15/60 = 0.25
frequency 30/60 = 0.50
frequency 15/60 = 0.25
Record the number of each allele and genotype for Generation 0 in Table 2a; record the
frequency of each genotype and allele for Generation 0 in Table 2b.
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2. Without replacement randomly select 5 individuals (5 beads), to establish a new
founder population. Record the genotypes below:
Individual genotype
1
2
3
4
5
Count the number of each genotype and the numbers of each allele (record below & in
Table 2b Generation 1); then calculate the new genotype frequencies; these are your
observed frequencies. Record this information below and in Table 2b as Generation
1.
Alleles
# R ____
# r ____
Total#: 10
frequency (p) ____
frequency (q) ____
Genotypes:
# RR ___
#Rr ___
#rr ___
Total#: 5
frequency ____
frequency ____
3. Using the new observed allele frequencies, calculate the expected genotype
frequencies for Generation 1. Record these results in Table 2b.
4. Reestablish the population to 50 diploid individuals using the new allele frequencies.
5. Follow the founder population through 5 generations in the new population. From this
point, each new generation will be produced by sampling 50 individuals with
replacement. After each generation reestablish the new population based on the new
allele frequencies.
Table 2a Changes in the number of alleles and genotypes due to the founder effect.
# of each Genotype
# of each Allele
Generatio
RR
Rr
rr
R
r
n
0
1
2
3
4
5
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15
30
15
0.5
0.5
2-10 Evolutionary Mechanisms
Table 2b Changes in the frequencies of alleles and genotypes over time for simulations of the
founder effect.
Genotype freq observed
Allele freq
Genotype freq. expected
observed
Generatio
RR
Rr
rr
R(p)
r(q)
p2
2pq
q2
n
0
1
2
3
4
5
---
---
---
0.5
0.5
0.25
0.50
0.25
A-2 Bottleneck effect
A bottleneck occurs when a population undergoes a drastic reduction in size as a result of
chance events (not differential selection) such as a tsunami, hurricane, volcanic eruption, or
even the effects of an ice age. (Bad luck, not bad genes). Often this effect is illustrated by
showing beads passing through the narrow mouth of a bottle, which results in an unpredictable
combination of ‘beads’ passing through (living). These beads would represent the beginning of
the next generation.
Hypothesis to address the bottleneck effect
Prediction to predict the type of change that one expects to occur in a small population.
Procedure
1. Establish a starting population with 60 individuals with a frequency of 0.5 for each
allele.—Record the numbers for Generation 0 in Table 3a and the frequencies in Table
3b.
2. Without replacement randomly select 5 individuals (10% of the population). Record the
genotypes and # of alleles for this new population:
Individual genotype
1
2
3
4
5
3. Count the number of each genotype and the number of each allele (record in Table 3a)
and then calculate the new genotype frequencies for the bottleneck population; these
are your observed frequencies. Record this information as Generation 1 in Table 3b.
4. Using the new observed allele frequencies, calculate the expected genotype
frequencies. Record these results in Table 4B, Generation 1
5. Reestablish the population to 50 diploid individuals using the new allele frequencies.
6. Repeat steps 2, 3, 4 and 5. Record your results in the appropriate generation in Tables
3a & 3b. After the first round, sample 50 individuals with replacement (unlike what you
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2-11 Evolutionary Mechanisms
did with the bottleneck effect)
7. Graph the change in p and q over time (you should have two lines, one for each allele.)
8. Did one allele go to fixation in that time period? If so which one.
Table 3a Changes in the number of alleles and genotypes over time due to the bottleneck
effect.
# of each Genotype
# of each Allele
Generatio
n
RR
Rr
rr
R
r
0
1
2
3
4
5
15
30
15
60
60
Population
s size
60
50
50
50
50
50
Table 3b Changes in frequencies of alleles and genotypes over time for simulations of the
bottleneck effect.
Genotype observed
Generatio
n
RR
Rr
rr
Allele freq
observed
R (p)
r (q)
0
1
2
3
4
5
---
---
---
0.5
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0.5
Genotype freq. expected
p2
2pq
q2
0.25
0.5
0.25
Population
size
60
50
50
50
50
50
2-12 Evolutionary Mechanisms
Experiment B. Simulation of Migration: gene flow
Materials
2 paper bags each containing 100 beads of two colours
Introduction
The migration of individuals between populations results in gene flow. In a natural population,
gene flow can be the result of the immigration and emigration of individuals or gametes (e.g.,
pollen movement). The rate and direction of migration and the starting allele frequencies for
the two populations can affect the rate of genetic change.
In the first simulation, which all students will do, the migration rate is equal in the two
populations; the starting allele frequencies for the two populations are different.
Hypothesis (addressing migration):
Predication (the type of change you expect to observe as a result of migration):
Procedure
1. Establish two populations of 50 beads each. Choose different starting allele
frequencies: population 1 90R and 1r; population 2 50R and 50r). Record the allele
frequencies for each starting population in Table 4.
2. Randomly select 10 individuals from each population to ‘migrate’ to the other
population.
3. Calculate the new allelic and genotypic frequencies in the two populations following
migration. Record your results in Table 4b.
4. Repeat steps 2-4 for 4 more generations (thus migration is occurring in each
generation).
5. Graph the change in allele frequencies over time. What is happening in the two
populations?
What would happen if the migration from population 1 to population 2 is not the same as from
population 2 to population 1? Some teams might want to design an experiment to test the
outcome.
Table 4a Changes in the number of alleles and genotypes over time due to gene flow.
Population 1
# of each Genotype
Generatio
n
0
1
2
3
4
5
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RR
Rr
# of each Allele
rr
R
r
Population
s size
50
50
50
50
50
50
2-13 Evolutionary Mechanisms
Population 2
# of each Genotype
Generatio
n
RR
Rr
# of each Allele
rr
R
r
0
1
2
3
4
5
Population
s size
50
50
50
50
50
50
Table 4b Changes in frequencies of alleles and genotypes over time for simulations of gene
flow.
Population 1
Genotype observed
Generatio
n
RR
Rr
rr
0
1
2
3
4
5
---
---
---
Allele freq
observed
R (p)
r (q)
Genotype freq. expected
p2
2pq
Population
size
q2
50
50
50
50
50
50
Population 2
Genotype observed
Generatio
n
RR
Rr
rr
Allele freq
observed
R (p)
r (q)
0
1
2
3
4
5
---
---
---
0.5
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0.5
Genotype freq. expected
p2
2pq
q2
0.25
0.5
0.25
Population
size
50
50
50
50
50
50
2-14 Evolutionary Mechanisms
Experiment C. Natural Selection
Natural selection disturbs Hardy-Weinberg equilibrium by discriminating between individuals
with respect to their ability to produce young. Those individuals that survive and reproduce will
perpetuate more of their genes in the population. These individuals are said to exhibit greater
fitness than those who leave no offspring or fewer offspring. We will model the effect of natural
selection by simulating predation on our population.
Hypothesis to address natural selection
Prediction: [Predict the type of change your would expect to observe as a result of natural
selection]
This time your beads exist in "ponds" that are represented by plastic dishpans filled with
smaller beads. The smaller beads can be strained to retrieve all the "individuals" that make up
the model population. When the individuals are recovered, the frequencies of the colour alleles
can be determined using the Hardy-Weinberg formula.
Procedure
1. The initial population:
• ten large white beads,
• ten large red/brown beads
• twenty large pink beads
Put into a dishpan filled with small white beads (to a depth of at least 5 cm).
2. One student is the predator. After the beads are mixed, the predator searches the pond
and removes as many prey items (large beads) as possible in 30 seconds. In order to
more closely model the handling time required by real predators, you must search for and
remove beads with a pair of long forceps.
3. Because some of the large beads are cryptically coloured (they blend into the
environment), the proportions of beads taken may not reflect the original proportions. Sift
the pond with the sieve, count the number of large white, pink, and red beads, and record
the totals in Table 5A . Use these counts to calculate the frequencies of the white (p) and
red (q) alleles remaining in the population after selection and record them in Table 5B. For
example, if five white, eight pink, and eight red beads remain, the frequency of the white
allele is:
p = (2 x 5) + 8 = 0.43
40
4. Using the new values for allele frequencies, calculate genotype frequencies for
homozygous white (p2), heterozygous pink (2pq), and homozygous red (q2) individuals,
and record them in Table 5B. For example, if p now equals 0.43, the frequency of
homozygous white individuals is:
p2 = (0.43)2 = 0.18
5. Reestablish the population to 50 diploid individuals using the new allele
frequencies. Using these numbers, construct a new pond.
6. Repeat steps 2 – 5 for three more rounds. Stop when Tables 5A and B are filled in
completely. A different student should be the predator in each round. Graph your data.
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2-15 Evolutionary Mechanisms
Thought question 1
How would the gene frequencies change, if you started with small red beads as a background?
Table 5A Counts of large beads before and after four rounds of simulated predation .
Beads
Initial Population
Before
White
Pink
Red
Total
10
20
10
40
After
Second Population
Before
50
After
Third Population
Before
50
After
Fourth Population
Before
50
After
Table 5B Allele and genotype frequencies due to selection by simulated predation (Part B).
Population
Initial population
First
generation after selection
Second
generation after selection
Third
generation after selection
Fourth
generation after selection
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p
q
p2
2pq
q2
0.5
0.5
0.25
0.5
0.25
Frequency of red allele (q)
2-16 Evolutionary Mechanisms
1.0
0.5
1
3
2
Time (in generations)
4
Figure 1 Effects of predation on allele frequencies.
Selection that favors one extreme phenotype over the other and causes allele frequencies to
change in a predictable direction is known as directional selection. When selection favors an
intermediate phenotype rather than one at the extremes, it is known as stabilizing selection.
Selection that operates against the intermediate phenotype and favors the extreme ones is
called disruptive selection.
It is important to realize that selection operates on the entire phenotype so that the overall
fitness of an organism is based on the result of interactions of thousands of genes.
The model presented here is very simple. Occasionally simple genetic differences like the one
you have modeled are critical to the survival of different phenotypes
Thought question 2
If two identical populations were in different environments (such as in our red and white ponds),
how would the frequency of the colour genes in each pond compare after many generations?
As two populations become genetically different through time (divergence), individuals from
these populations may lose the ability to interbreed (=reproductive isolation). If this happens,
two species form from one ancestral species. This process is called speciation.
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2-17 Evolutionary Mechanisms
Review of Chi-square (χ2; pronounced kye-square) test
The chi-square (χ2; pronounced kye-square) test is one way to test an hypothesis in an
experiment in which the data collected are frequency data, rather than continuous data ( t -test).
Genetic and behavioural experiments often measure the frequency of occurrence of events; for
instance, an experimenter may want to compare the number of animals responding to some
stimulus to the number of animals that did not respond. Note that the χ2; analysis uses raw data
only, not percentages or proportions.
The χ2 formula is
χ2 = ∑
(observed - expected) 2
expected
Table 6. Critical values for the χ2 - distribution
Level of probability
Degrees
of
0.10
0.05
0.01
freedom
6.64
3.84
2.71
1
9.21
5.99
4.61
2
11.34
7.82
6.25
3
13.28
9.49
7.78
4
Study Questions
1. Based on results from this lab and your text book complete the following table:
Table Consequence of evolutionary mechanisms on genetic variation and biological fitness
Process
Definition
Effect on
Genetic variation
Mutation
Gene flow
Genetic drift
Selection
2. Define evolution in terms of gene frequencies:
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Effect on
Fitness
Example
2-18 Evolutionary Mechanisms
3. Why is non-random mating NOT an evolutionary mechanism?
4. True or false? Heterozygous advantage refers to the tendency for heterozygous
individuals to have better fitness than homozygous individuals. This higher fitness
results in less genetic variation in the population. HINT: What is genetic variation?
5. Which of the following are basic components of the Hardy-Weinberg equation?
HINT: What were Hardy and Weinberg trying to determine about the consequences of
matings among all the individuals in a population?
a. Allele frequencies in a subset of the population
b. Frequencies of two alleles in a gene pool before and after many random matings
c. Allele frequencies, phenotype frequencies
d. Allele frequencies, number of individuals in the population
6. Which of the following statements is not a part of the Hardy-Weinberg principle?
HINT:
a. When alleles are transmitted according to the rules of Mendelian inheritance, their
frequencies do not change over time.
b. If allele frequencies in a population are given by p and q, then genotype frequencies will
be given by p2, 2pq, and q2 for generation after generation.
c. Even if allele A1 is dominant to allele A2, it does not increase in frequency.
d. The genotype frequencies in the offspring generation must add up to 2.
7. True or false? The Hardy-Weinberg model makes the following assumptions: no
selection at the gene in question; no genetic drift; no gene flow; no mutation; random
mating.
Sample Problems
Problem 1.step-by-step see http://science.nhmccd.edu/biol/hwe/qld.html
In the United States 1 in 1700 Caucasian newborns have cystic fibrous (C for normal is
dominant over c for cystic fibrous).
1-1. When counting the phenotypes in a population why is homozygous recessive (cc in this
example) the most significant?
1-2. What percent of the above population have cystic fibrous (cc or q2)? ________
1-3.From the above numbers you should be able to calculate the expectant frequencies of all
the following (assuming a Hardy-Weinberg equilibrium):
ALLELE FREQUENCY CALCULATIONS:
a) Why calculate "q" first?
b) (ƒ)c = q = _______________
c) Why is it now easy to find "p"?
d) (ƒ)C = p = ____________
1-4. Now that you know that p and q, the following genotypes can be found.
BIO152 2005
2-19 Evolutionary Mechanisms
GENOTYPE FREQUENCY CALCULATIONS:
a) (ƒ)CC- Normal homozygous dominate = p2 = _____________(show calculations)
b) (ƒ)Cc -carriers of cystic fibrous = 2pq = ____________
1-5. How many of the 1700 of the population are homozygous Normal? ________
1-6. How many of the 1700 in the population are heterozygous (carrier)? ________
1-7. It has been found that a carrier is better able to survive diseases with severe diarrhea.
What would likely happen to the frequency of the "c" if there was a epidemic of cholera or other
type of diarrhea producing disease? EXPLAIN your prediction
1-8. Explain the difference between the ALLELE frequency and the GENOTYPE frequency.
[Problems 2-4 are from the text web tutorial 24.1]
Problem 2 What is the frequency of the A1A2 genotype in a population composed of 20 A1A1
individuals, 80 A1A2 individuals, and 100 A2A2 individuals?
a.
b.
c.
d.
The frequency of the A1A2 genotype is 0.1.
The frequency of the A1A2 genotype is 0.5.
The frequency of the A1A2 genotype is 0.4.
The frequency of the A1A2 genotype is 80.
80 A1A2 individuals ÷ (20 + 80 + 100) total individuals = ____ the frequency of the A1A2 genotype.
Problem 3
What is the frequency of the A1 allele in a population composed of 20 A1A1
individuals, 80 A1A2 individuals, and 100 A2A2 individuals?
a.
b.
c.
d.
The frequency of the A1 allele is 0.3.
The frequency of the A1 allele is 0.5.
The frequency of the A1 allele is 0.7.
The frequency of the A1 allele is 0.1.
p = the frequency of the A1 allele = (number of A1 alleles) ÷ (total of all alleles) = [(2 × 20) + 80] ÷ [(2 × 20) + (2 ×
80) + (2 × 100)] = _____.
Problem 4
What genotype frequencies are expected under Hardy-Weinberg equilibrium for
a population with allele frequencies of p = 0.8 and q = 0.2? HINT: When you calculate the
expected genotype frequencies for a real-life population, you can then compare them to the
actual genotype frequencies in that population to determine if the population is evolving or at
equilibrium.
a.The expected genotype frequencies are 0.33, 0.33, and 0.33 for A1A1, A1A2, and A2A2,
respectively.
b.The expected genotype frequencies are 0.25, 0.5, and 0.25 for A1A1, A1A2, and A2A2,
respectively.
c.The expected genotype frequencies are 0.64, 0.32, and 0.04 for A1A1, A1A2, and A2A2,
respectively.
d. The expected genotype frequencies are 0.32, 0.64, and 0.04 for A1A1, A1A2, and A2A2, respectively.
The expected frequency of the A1A1 genotype is p2 = (0.8)(0.8) = ___;
The expected frequency of the A1A2 genotype is 2pq = 2(0.8)(0.2) = ____;
The expected frequency of the A2A2 genotype is q2 = (0.2)(0.2) = ____.
To verify your calculations, make sure that the three frequencies add up to 1!
BIO152 2005
2-20 Evolutionary Mechanisms
Problem 5 If the frequency of allele A1 in a population is 0.4 and the frequency of allele A2 is
0.6 in the same population, what is the frequency of the heterozygotes A1A2 in the next
generation? HINT: Review the frequency of genotypes in the Hardy-Weinberg principle.
a. 0.16
b. 1
c. 0.36
d. 0.48
Problem 6 Researchers studying a small milkweed population note that some plants produce
a toxin and other plants do not. They identify the gene responsible for toxin production. The
dominant allele (T) codes for an enzyme that makes the toxin, and the recessive allele (t)
codes for a nonfunctional enzyme that cannot produce the toxin. Heterozygotes produce an
intermediate amount of toxin. The genotypes of all individuals in the population are determined
(see chart) and used to determine the allele frequencies in the population.
Genotype Frequencies
Allele Frequencies
Observed
1.
2.
3.
4.
TT
0.56
Tt
0.28
tt
0.16
T
T
Is this population in Hardy-Weinberg equilibrium?
Yes
No; there are more heterozygotes than expected.
No; there are more homozygotes than expected.
More information is needed in order to answer this question.
Hint: Calculate allele frequencies and use the Hardy-Weinberg principle to calculate
expected genotype frequencies.
BIO152 2005