Download Evolution Problem Drill – 02: Genetic Variation and the Hardy

Evolution Problem Drill – 02: Genetic Variation and the Hardy-Weinberg Principle
Question No. 1 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 1. Which of the following statements about genetic variation is not correct?
Question #01
(A)
(B)
(C)
(D)
Genetic variation is necessary for evolution to occur.
Genetic variation that occurs in the somatic cells of an individual results in evolution.
Genetic variation results from mutations.
Genetic variation refers to the genetic differences between individuals of a population.
A. Incorrect!
Without genetic variation evolution would not occur.
B. Correct!
Genetic variation that occurs in the germ cells of individuals, not the somatic cells, may result in
evolution.
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Choice
C. Incorrect!
Mutations result in genetic variation.
D. Incorrect!
Genetic variation describes the genetic differences between individuals of a population.
Genetic variation is the basis of evolution. If there were not variation in the genetic material of
individuals that comprise a population, selection would not have anything to act on. Mutations are the
source of genetic variation, but only mutations that affect the germ cells, because only those cells will be
passed on to the next generation. Somatic cells are not passed to offspring.
Solution
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Question No. 2 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 2. How do mutations result in evolution? Choose the correct order of events.
Question #02
(A)
(B)
(C)
(D)
Mutation – Phenotypic difference – Change in allele frequency – Natural selection.
Mutation - Change in allele frequency – Phenotypic difference - Natural selection.
Mutation – Phenotypic difference – Natural selection – Change in allele frequency.
Mutation – Natural selection – Phenotypic difference – Change in allele frequency.
A. Incorrect!
Natural selection must operate before a change in allele frequency is observed.
B. Incorrect!
Mutations do not directly result in a change in allele frequency. They must first result in a phenotypic
difference.
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Choice
C. Correct!
Random mutations occur first, which result in a phenotypic difference that undergoes natural selection,
ultimately resulting in changes in allele frequency.
D. Incorrect!
Natural selection can not act unless the mutation results in a phenotypic difference.
The correct order of events that must occur for a mutation to result in evolution is as follows: mutation,
phenotypic difference, natural selection, and change in allele frequency. It is important to understand
that a mutation that occurs in somatic cells do not result in evolution because these mutations are not
passed onto the individual’s offspring. Only mutations that occur in germ cells can result in the
phenotypic differences that may undergo natural selection, because these cells are passed on to the
next generation. Allele frequencies change over generations, not within generations.
Solution
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Question No. 3 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 3. Which of the following statements about mutations is not correct?
Question #03
(A)
(B)
(C)
(D)
Mutations are random.
Mutation rates are similar across species.
Most mutations are deleterious.
Mutations that occur in non-coding regions are neutral.
A. Incorrect!
Mutation randomly occur, they are not directed by the individual or anything else.
B. Correct!
Mutation rates are variable across species and across loci.
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Choice
C. Incorrect!
Most mutations are deleterious; very few are beneficial.
D. Incorrect!
Mutations that occur in non-coding regions are neutral because they do not result in a phenotypic
difference.
Note that this question asks you to identify the statement about mutations that is NOT correct. It is true
that mutations are random, that most mutations are deleterious, and that mutations that occur in noncoding regions are neutral and do not result in evolution. However, it is not true that mutation rates are
similar across species, nor are they similar across loci. Some species and certain loci acquire mutations
at a much faster rate than others. Thus, the rate of evolution can vary between species and between loci
within the same species. Remember, mutations contribute to genetic variation, which is required for
evolution.
Solution
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Question No. 4 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3) Pick
the answer (4) Go back to review the core concept tutorial as needed.
Question 4. PKU is a metabolic disease due to a single recessive gene. The incidence of PKU is
approximately 1/13,000 births in a population that is in Hardy-Weinberg equilibrium. What is the
estimated frequency of the normal or dominant PKU allele?
Question #04
(A)
(B)
(C)
(D)
0.999.
0.991.
9.9%.
0.991%.
A. Incorrect!
You would get this number if you subtracted the genotype frequency, q2, from 1, rather than q, the allele
frequency to obtain p.
B. Correct!
The estimated frequency of the dominant allele, p, is determined by subtracting q from 1.
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Choice
C. Incorrect!
Allele frequencies are not given as percentages.
D. Incorrect!
Allele frequencies are not given as percentages.
The frequency of q2 is equal to the incidence of PKU and the estimated frequency of q can be determined
by taking the square root of q2, p or the dominant allele frequency, is determined by subtracting q from 1
because p + q = 1.
Incidence of PKU = 1/13000 = 0.0000769 = q2
q = square root of q = 0.008769
p+q=1
p=1–q
p = 1 – 0.008769 = 0.991
Solution
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Question No. 5 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 5. Given the following data for height in pea plants (a two allele locus), calculate the genotype
frequencies? Number of TT individuals = 75; Number of Tt individuals = 35; Number of tt individuals =
8.
Question #05
(A)
(B)
(C)
(D)
p2 = 0.07; 2pq = 0.30; q2 = 0.63.
p2 = 63%; 2pq = 30%; q2 = 7%.
p2 = 0.63; 2pq = 0.30; q2 = 0.07.
p2 = 0.75; 2pq = 0.35; q2 = 0.08.
A. Incorrect!
Recheck the homozygous genotypes.
B. Incorrect!
Genotype frequencies are not expressed as percentages.
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Choice
C. Correct!
The genotype frequencies are 0.63 for the homozygous dominant genotype, 0.30 for the heterozygous
genotype, and 0.07 for the homozygous recessive genotype.
D. Incorrect!
Genotype frequency is not determined by dividing the number of individuals for each genotype by 100.
Remember that the homozygous dominant genotype frequency is expressed as p2, the heterozygote
frequency as 2pq, and the homozygous recessive genotype frequency as q2. The frequency for each
genotype is determined by taking the number of individuals that have that genotype divided by the total
number of individuals for all genotypes. The frequencies of all genotypes should sum to 1.
Frequency of TT = p2 = 75/118 = 0.63
Frequency of Tt = 2pq = 35/118 = 0.30
Frequency of tt = q2 = 8/118 = 0.07
Solution
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Question No. 6 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 6. Given the following data for height in pea plants (a two allele locus), calculate the allele
frequencies? Number of TT individuals = 75; Number of Tt individuals = 35; Number of tt individuals =
8.
Question #06
(A)
(B)
(C)
(D)
p = 0.78; q = 0.22
p = 0.22; q = 0.78
p = 0.63; q = 0.07
p = 0.47; q = 0.53
A. Correct!
The allele frequencies are 0.78 for the dominant T allele and 0.22 for the recessive t allele.
B. Incorrect!
Almost, but you are mixing up your p’s and q’s.
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Choice
C. Incorrect!
These are the genotype frequencies for the homozygous dominant and recessive genotypes,
respectively.
D. Incorrect!
If you got these numbers, you are forgetting that the homozygous dominant allele contributes two
alleles, rather than one.
The allele frequencies can be calculated using the already determined genotype frequencies, or by
counting alleles.
Using the genotype frequencies:
p = p2 + ½(2pq)
p = 0.63 + ½(0.30)
p = 0.78
p
q
q
q
Solution
+
=
=
=
q=1
1–p
1 – 0.78
0.22
Counting alleles:
p = [2(number of individuals TT) + (number of individuals Tt)] / total number of alleles
p = [2(75) + 35] / 236
p = 0.78
q is determined as above.
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Question No. 7 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 7. Tay-Sachs is a recessive genetic disease found mostly in populations of Ashkenazi Jews. It’s
frequency in these populations is 1/3900. What is the estimated frequency of the recessive allele?
Question #07
(A)
(B)
(C)
(D)
1.6%.
0.984.
0.00026.
0.016.
A. Incorrect!
Allele frequencies are not expressed as percentages.
B. Incorrect!
This is the estimated allele frequency of the dominant allele, p.
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Choice
C. Incorrect!
This is the homozygous recessive genotype frequency, q2, or the incidence of individuals who have TaySachs.
D. Correct!
The frequency of the recessive allele that causes Tay-Sachs is 0.016.
The estimated frequency of the recessive allele, q, for Tay-Sachs can be determined from the incidence
of the disease as follows.
Incidence of Tay-Sachs = 1/3900 = q2
q
q
q
q
=
=
=
=
square root of q2
square root of 1/3900
square root of 0.000256
0.016
Solution
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Question No. 8 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 8. The following conditions must be met for a population to be in Hardy-Weinberg equilibrium
except for which of the following?
Question #08
(A)
(B)
(C)
(D)
No migration or gene flow.
Non-random mating.
No selection.
No mutation.
A. Incorrect!
No migration or gene flow is one of the conditions that must be met for a population to be in HardyWeinberg equilibrium.
B. Correct!
Non-random mating would cause a population to be out of Hardy-Weinberg equilibrium.
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Choice
C. Incorrect!
Selection cannot be operating if a population is in Hardy-Weinberg equilibrium.
D. Incorrect!
Mutations cannot be occurring if a population is in Hardy-Weinberg equilibrium.
Note that this question is asking for the condition that would cause a population to NOT be in HardyWeinberg equilibrium. A population that is in Hardy-Weinberg equilibrium is not evolving. Therefore,
none of the forces that cause evolution can be operating. Non-random mating is a selective force that
would cause a population to be out of Hardy-Weinberg equilibrium and evolution to occur. The following
conditions must be met: no migration or gene flow, random mating, no selection, no mutation, and large
population size or no genetic drift.
Solution
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Question No. 9 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 9. Albinism is a recessive trait. In a sample of 100,000 humans from a population that is in
Hardy-Weinberg equilibrium, 6 individuals are albino. How many individuals in this sample would you
expect to carry the recessive albinism allele, but not express the albinism phenotype?
Question #09
(A)
(B)
(C)
(D)
1537 individuals
153 individuals
775 individuals
99,225 individuals
A. Correct!
This is the estimated number of individuals that would be carrying the albinism allele without expressing
albinism.
B. Incorrect!
The number of individuals is 10 times higher.
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Each Answer
Choice
C. Incorrect!
Remember to use the Hardy-Weinberg equation to determine the heterozygote frequency at the albinism
locus.
D. Incorrect!
Those individuals who are homozygous dominant at this locus do not carry the allele that causes
albinism.
This is a tough question and requires you to combine your knowledge of basic genetics with what you
have learned about the Hardy-Weinberg principle. If 6 out of 100,000 individuals express the albinism
phenotype, this means that q2 is equal to 6/100,000, or 0.00006. In order to determine the number of
individuals who carry the albinism allele, but do not express the phenotype, you must first calculate the
heterozygous genotype frequency, 2pq. q can be determined by taking the square root of q2 to get
0.007746. p is equal to 1-q, or 0.99231. From this we can calculate 2pq by plugging in our p and q
values to get 0.01537. To determine the number of individuals in the sample that should have this
genotype you simply multiply 100,000 by the frequency 0.01537 to get 1537 individuals. Note that this
is just an estimate because you are determining this based on the original q2.
Solution
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Question No. 10 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 10. There is a population of insects that have either black, gray, or white eyes. The number of
individuals with black eyes, BB, is 70, gray eyes, Bb, is 60, and white eyes, bb, is 120. Determine the
Chi-squared value and choose the correct answer below (Chi-squared significance at 5%).
Question #10
(A)
(B)
(C)
(D)
The Chi-squared value is 62.5 and the population is in Hardy-Weinberg equilibrium.
The Chi-squared value is 62.5 and the population is not in Hardy-Weinberg equilibrium.
The Chi-squared value is 0.625 and the population is not in Hardy-Weinberg equilibrium.
The Chi-squared value is 0.625 and the population is in Hardy-Weinberg equilibrium.
A. Incorrect!
With a Chi-squared value this high, the population is not in Hardy-Weinberg equilibrium.
B. Correct!
The Chi-squared value is much higher than the critical value of 3.84, so the population is not in HardyWeinberg equilibrium.
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Choice
C. Incorrect!
With a Chi-squared value this low, the population would be in Hardy-Weinberg equilibrium.
D. Incorrect!
If this were the correct Chi-squared value, the population would be in Hardy-Weinberg equilibrium as
stated. However, this value is incorrect.
The first thing you must do is determine the allele frequencies from the information you are given about
the insect population using the Hardy-Weinberg principle.
p = (2)(70) + (1)(60) = 0.40
(2)(250)
q = (2)(120) + (1)(60) = 0.60
(2)(250)
Using this information, you can then determine the expected number of individuals for each genotype.
Exp. number BB = (0.4)(0.4)(250) = 40
Exp. number Bb = (2)(0.4)(0.6)(250) = 120
Solution
Exp. number bb = (0.6)(0.6)(250) = 90
Then use the Chi-squared test to compare the expected numbers with the observed numbers.
Chi square: = (70 – 40)2 + (60 – 120)2 + (120– 90)2
40
120
90
= 90 + 360 + 90
40 120 90
= 22.5 + 30 + 10
= 62.5
Since this value is larger than 3.841, this population is not in HW equilibrium.
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