Population Genetics
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Chapter 2
Lecture structure
Population Genetics
Population genetics is the study of genetic variation in populations.
Basic concepts of population genetics allow us to understand how and why the prevalence of various genetic diseases differs among populations.
Genotype and allele frequencies
An essential step in understanding genetic variation is to measure it in populations.
This is done by estimating genotype and allele frequencies.
For a given locus, the genotype frequency measures the proportion of each
genotype in a population. Suppose that
a population of 100 individuals has been assayed
for an autosomal restriction fragment length
polymorphism. If the RFLP has 2 possible alleles,
labeled 1 and 2, there are 3 possible genotypes:
1-1, 1-2, and 2-2.
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1-1 2-2 1-2
Genotype Frequencies
Visualization of a Southern blot allows us to determine the genotype of each individual in our population, and we find that the genotypes are distributed as follows:
The genotype frequency is then obtained by dividing the count for each genotype by the total number of individuals. Thus, the frequency of genotype 1-1 is 49/100 = 0.49, and the frequencies of genotypes 1-2 and 2-2 are 0.42 and 0.09, respectively.
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Allele Frequencies�
The allele frequency measures the proportion of chromosomes that contain a specific allele. To continue the earlier RFLP example, we wish to estimate the frequencies of alleles 1 and 2 in our population. Each individual with the 1-1 genotype has 2 copies of allele 1, and each heterozygote (1-2 genotype) has one copy of allele 1. Because each diploid somatic cell contains 2 copies of each autosome, our denominator is 200. Thus, the frequency of allele 1 in the population is:
(The same approach can be used to estimate the frequency of allele 2, which is 0.3. A convenient shortcut is to remember that the allele frequencies for all of the alleles of a given locus must add up to 1. Therefore, we can obtain the frequency of allele 2 simply by subtracting the frequency of allele 1 (0.7) from 1.
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HARDY-WEINBERG EQUILIBRIUM
If a population is large and if individuals mate at random with respect to their genotypes at a locus, the population should be in Hardy-Weinberg equilibrium.
This means that there is a constant and predictable relationship between genotype frequencies and allele frequencies. This relationship, expressed in the Hardy-Weinberg equation, allows one to estimate genotype frequencies if one knows allele frequencies, and vice versa.
The Hardy-Weinberg Equation:
In this equation:
p = frequency of allele 1 (conventionally the most common, normal allele)
q = frequency of allele 2 (conventionally a minor, disease-producing allele)
p2 = frequency of genotype 1-1 (conventionally homozygous normal)
2pq = frequency of genotype 1-2 (conventionally heterozygous)
q2 = frequency of genotype 2-2 (conventionally homozygous affected)
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HARDY-WEINBERG EQUILIBRIUM
In most cases where this equation is used, a simplification is possible. Generally p, the normal allele frequency in the population, is very close to 1 (e.g., most of the alleles of this gene are normal). In this case, we may assume that p ~ 1, and the equation simplifies to:
The frequency of the disease-producing allele in question, q, is a very small
fraction. This simplification would not necessarily be used in actual medical genetics practice, but for answering test questions, it works quite well. However, if the disease prevalence >1/100, e.g., q >1/10, the complete Hardy-Weinberg equation should be used to obtain an accurate answer.
In this case, p = 1 - q.
dominant and recessive alleles, genotypes, and diseases, the equation is most
frequently used with autosomal recessive conditions.
In these instances, a large percentage of the disease-producing allele is “hidden” in heterozygous carriers who cannot be distinguished phenotypically (clinically) from homozygous normal individuals.
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Practical Application of Hardy-Weinberg
A 20-year-old college student is taking a course in human genetics. She is
aware that she has an autosomal recessive genetic disease that has required
her lifelong adherence to a diet low in natural protein with supplements of
tyrosine and restricted amounts of phenylalanine. She also must avoid foods
artificially sweetened with aspartame (Nutrasweet™). She asks her genetics
professor about the chances that she would marry a man with the disease producing allele.
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The geneticist tells her that the known prevalence of PKU in the population is 1/10,000 live births, but the frequency of carriers is much higher, approximately 1/50. Her greatest risk comes from marrying a carrier for two reasons. First, the frequency of carriers for this condition is much higher than the frequency of affected homozygotes, and second, an affected person would be identifiable clinically. The geneticist used the Hardy-Weinberg equation to estimate the carrier frequency from the known prevalence of the disease in the following way:
Disease prevalence = q2 = 1/10,000 live births
Carrier frequency = 2q (to be calculated)
q = square root of 1/10,000, which is 1/100
2q = 2/100, or 1/50, the carrier frequency
Practical Application of Hardy-Weinberg
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The woman now asks a second question: “Knowing that I have a 1/50 chance
of marrying a carrier of this allele, what is the probability that I will have a
child with PKU?”
The geneticist answers, “The chance of you having a child with PKU is 1/100.”
This answer is based on the joint occurrence of two nonindependent events:
• The probability that she will marry a heterozygous carrier (1/50), and
• If he is a carrier, the probability that he will pass his PKU allele versus
the normal allele to the child (1/2).
These probabilities would be multiplied to give:
• 1/50 × 1/2 = 1/100, the probability that she will have a child with PKU.
If events are nonindependent, multiply the probability of one event by the probability of the second event, assuming that the first has occurred.
Practical Application of Hardy-Weinberg
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In summary, there are 3 major terms one usually works with in the Hardy-Weinberg equation applied to autosomal recessive conditions:
• q2, the disease prevalence
• 2q, the carrier frequency
• q, the frequency of the disease-causing allele
of heterozygous carriers in populations when we know only the prevalence of the recessive disease.
heterozygous carriers is much higher than the prevalence of affected
homozygotes. In effect, the vast majority of recessive genes are hidden
in the heterozygotes.
Hardy-Weinberg Equilibrium for Dominant Diseases
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The calculations for dominant diseases must acknowledge that most of the
affected individuals will be heterozygous.
For example,
• 1/500 people in the United States have a form of LDL-receptor deficiency
and are at increased risk for cardiovascular disease and myocardial infarction.
• Taking 2q = 1/500, one can calculate that q2 = 1/106, or one in a million live births are homozygous for the condition. These individuals have greatly elevated LDL-cholesterol levels, a much-higher risk for cardiovascular disease than heterozygotes, and are more likely to present with characteristic xanthomas, xanthelasmas, and corneal arcus.
In contrast, in Huntington disease (autosomal dominant), the number of triplet
repeats correlates much more strongly with disease severity than does heterozygous or homozygous status.
Sex Chromosomes and Allele Frequencies
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most cases occur in hemizygous males (xY).
allele frequency for the disease-producing allele: 1/10,000.
• q2 = prevalence of disease in females (1/108, or 1/100,000,000)
• 2q = prevalence of female carriers (1/5,000)
This exercise demonstrates that:
• As with autosomal recessive traits, the majority of X-linked recessive
genes are hidden in female heterozygous carriers (although a considerable number of these genes are seen in affected males).
• X-linked recessive traits are seen much more commonly in males than
in females.
Factors responsible for genetic variation�in/among populations
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for most loci, deviations from equilibrium can be produced by:
are: Mutation, Natural selection, Genetic drift, and Gene flow
Mutation
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Branched Chain Ketoacid Dehydrogenase Deficiency
Branched chain ketoacid dehydrogenase deficiency (maple syrup urine
disease) occurs in 1/176 live births in the Mennonite community of
Lancastershire, Pennsylvania.
In the U.S. population at large, the disease occurs in only 1/180,000 live births.
The predominance of a single mutation (allele) in the branched chain dehydrogenase gene in this group suggests a common origin of the mutation. This may be due to a founder effect.
Natural Selection
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Sickle Cell Disease and Malaria
Sickle cell disease affects 1/600 African Americans and up to 1/50 individuals in some parts of Africa. How could this highly deleterious disease-causing mutation become so frequent, especially in Africa? The answer lies in the fact that the falciparum malaria parasite, which has been common in much of Africa, does not survive well in the erythrocytes of sickle cell heterozygotes. These individuals, who have no clinical signs of sickle cell disease, are thus protected against the lethal effects of malaria. Consequently, there is a heterozygote advantage for the sickle cell mutation, and it maintains a relatively high frequency in some African populations.
Natural Selection
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There is now evidence for heterozygote advantages for several other recessive diseases that are relatively common in some populations.
Examples include:
• Cystic fibrosis (heterozygote resistance to typhoid fever)
• Hemochromatosis (heterozygote advantage in iron-poor environments)
• Glucose-6-phosphate dehydrogenase deficiency, hemolytic anemia
(heterozygote resistance to malaria)
Genetic Drift
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Genetic Drift
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Genetic drift begins. In both examples the frequency of affected persons in generation III is 2/3, higher than the 1/2 predicted by statistics.
Genetic Drift
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Gene Flow
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Consanguinity and Its Health Consequences
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Consanguinity and Its Health Consequences
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• Siblings (II-2 and II-3 or II-4) share 1/2 of their genes.
• First cousins (III-3 and III-4) share 1/8 of their genes (1/2 × 1/2 × 1/2).
• Second cousins (IV-1 and IV-2) share 1/32 of their genes (1/8 × 1/2 × 1/2).
A Pedigree Illustrating Consanguinity
Review Questions
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1. A population has been assayed for a 4-allele polymorphism, and the
following genotype counts have been obtained:
On the basis of these genotype counts, what are the gene frequencies of
alleles 1 and 2?
A. 0.38, 0.28
B. 0.19, 0.14
C. 0.095, 0.07
D. 0.25, 0.25
E. 0.38, 0.20
Review Questions
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2. Which of the following best characterizes Hardy-Weinberg equilibrium?
A. Consanguinity has no effect on Hardy-Weinberg equilibrium.
B. Genotype frequencies can be estimated from allele frequencies, but
the reverse is not true.
C. Natural selection has no effect on Hardy-Weinberg equilibrium.
D. Once a population deviates from Hardy-Weinberg equilibrium, it takes many generations to return to equilibrium.
E. The frequency of heterozygous carriers of an autosomal recessive
mutation can be estimated if one knows the incidence of affected homozygotes in the population.
Review Questions
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3. In a genetic counseling session, a healthy couple has revealed that they
are first cousins and that they are concerned about health risks for their
offspring. Which of the following best characterizes these risks?
A. Because the couple shares approximately half of their genes, most of the
offspring are likely to be affected with some type of genetic disorder.
B. The couple has an increased risk of producing a child with an autosomal
dominant disease.
C. The couple has an increased risk of producing a child with an autosomal
recessive disease.
D. The couple has an increased risk of producing a child with Down
syndrome.
E. There is no known increase in risk for the offspring.
Review Questions
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4. An African American couple has produced two children with sickle cell
disease. They have asked why this disease seems to be more common in
the African American population than in other U.S. populations. Which
of the following factors provides the best explanation?
A. Consanguinity
B. Genetic drift
C. Increased gene flow in this population
D. Increased mutation rate in this population
E. Natural selection
Review Questions
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5. If the incidence of cystic fibrosis is 1/2,500 among a population of
Europeans, what is the predicted incidence of heterozygous carriers of a
cystic fibrosis mutation in this population?
A. 1/25
B. 1/50
C. 2/2,500
D. 1/2,500
E. (1/2,500)2
Review Questions
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6. A man is a known heterozygous carrier of a mutation causing hyperprolinemia, an autosomal recessive condition. Phenotypic expression is variable and ranges from high urinary excretion of proline to neurologic manifestations including seizures. Suppose that 0.0025% (1/40,000) of the population is homozygous for the mutation causing this condition.
If the man mates with somebody from the general population, what is the probability that he and his mate will produce a child who is homozygous for the mutation involved?
A. 1% (1/100)
B. 0.5% (1/200)
C. 0.25% (1/400)
D. 0.1% (1/1,000)
E. 0.05% (1/2,000)
Review Questions
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7. The incidence of Duchenne muscular dystrophy in North America is
about 1/3,000 males. On the basis for this figure, what is the gene
frequency of this X-linked recessive mutation?
A. 1/3,000
B. 2/3,000
C. (1/3,000)2
D. 1/6,000
E. 1/9,000