Download Unit 8: Problem 14.9

Unit 8: Mendelian Genetics (from Unit 2)
Mendel (1865) discovered that:
1. Traits (at least some traits) are governed by genes (he called them
“factors”) that are passed down from parent to offspring.
2. Individuals have a pair of each gene with one version (we calle
different versions, “alleles”) coming from mom and the other allele
coming from dad.
Knowing this, we can predict the results of any particular mating:
AA (mom) x aa (dad) = 100% Aa offspring
Possible Eggs
A
a
Aa (mom) x Aa (da) = 25% AA
50% Aa
25% aa
Possible
Sperm
A
AA
Aa
a
Aa
aa
Unit 8: Hardy (1908) and Weinberg (1908)
Based on Mendel’s discoveries, Hardy (a British mathematician) and
Weinberg (a German physician) independently (in 1908) came up with
a hypothesis about how alleles act in populations overall (not just in one
female-male pair). Their simultaneous discovery is known as the HardyWeinberg Principle. They showed that:
1. The dominant allele will not eventually become the only allele in a
population (contrary to common belief at the time).
2. Allele and genotype frequencies become stable and unchanging (that
is, are at equilibrium) after one generation. This assumes:
a. Random mating
b. No gene flow (immigration or emigration)
c. No natural selection
d. No mutation
e. No meiotic drive
f. Non-overlapping generations (this just keeps the math simple)
Unit 8: Hardy-Weinberg Equilibrium 1
Suppose a gene has two alleles, A and a. Let
p = Frequency of A allele in a population
q = Frequency of a allele in the population
The frequencies of the three genotypes will be:
Frequency of AA = p2
Frequency of Aa = 2pq
Frequency of aa = q2
Example: Suppose p = 0.6 and q = 0.4. What are the H-W genotype
frequencies?
Unit 8: Hardy-Weinberg Equilibrium 2
Suppose a gene has two alleles, A and a. Let
p = Frequency of A allele in a population
q = Frequency of a allele in the population
The frequencies of the three genotypes will be:
Frequency of AA = p2
Frequency of Aa = 2pq
Frequency of aa = q2
Example: Suppose p = 0.6 and q = 0.4. What are the H-W genotype
frequencies?
Freq of AA = 0.6 x 0.6 = 0.36
Freq of Aa = 2 x 0.6 x 0.4 = 0.48
Freq of aa = 0.4 x 0.4 = 0.16
Unit 8: Hardy-Weinberg Equilibrium 3
Example: Suppose a population has the following genotype
frequencies:
Freq of AA = 0.6 x 0.6 = 0.36
Freq of Aa = 2 x 0.6 x 0.4 = 0.48
Freq of aa = 0.4 x 0.4 = 0.16
What are the allele frequencies, p and q?
Unit 8: Hardy-Weinberg Equilibrium 4
Example: Suppose a population has the following genotype
frequencies:
Freq of AA = 0.6 x 0.6 = 0.36
Freq of Aa = 2 x 0.6 x 0.4 = 0.48
Freq of aa = 0.4 x 0.4 = 0.16
What are the allele frequencies, p and q?
p = Freq of A allele = (0.36 + 0.36 + 0.48) / 2 = 0.6
q = Freq of a allele = (0.48 + 0.16 + 0.16) / 2 = 0.4
Unit 8: Problem 14.9
Which of the following genotype frequencies of AA, Aa, and aa,
respectively, satisfy the Hardy-Weinberg principle?
(a) 0.25, 0.50, 0.25 [Antoine, Ben, Brandy]
(b) 0.36, 0.55, 0.09 [Carin, Courtney, Giselle]
(c) 0.49, 0.42. 0.09 [Janina, Kimberly, Laura W.]
(d) 0.64, 0.27, 0.09 [Reba, Lawanda, Maria]
(e) 0.29, 0.42, 0.29 [Melissa, Laura Y.]
Unit 8: Solution 14.9a
Which of the following genotype frequencies of AA, Aa, and aa,
respectively, satisfy the Hardy-Weinberg principle (HWP)?
(a) 0.25, 0.50, 0.25 [Antoine, Ben, Brandy]
Allele Frequencies
p = Freq of A = (0.25 + 0.25 + 0.5) / 2 = 0.5
q = Freq of a = (0.25 + 0.25 + 0.5) / 2 = 0.5
Genotype Frequencies in Offspring
Freq of AA = p2 = 0.5 * 0.5 = 0.25
Freq of Aa = 2pq = 2 * 0.5 * 0.5 = 0.5
Freq of aa = q2 = 0.5 * 0.5 = 0.25
Offspring genotype frequencies = Parental genotype frequencies.
Thus, Parental genotype frequencies do satisfy HWP.
Unit 8: Solution 14.9b
Which of the following genotype frequencies of AA, Aa, and aa,
respectively, satisfy the Hardy-Weinberg principle?
(b) 0.36, 0.55, 0.09 [Carin, Courtney, Giselle]
Allele Frequencies
p = Freq of A = (0.36 + 0.36 + 0.55) / 2 = 0.635
q = Freq of a = (0.55 + 0.09 + 0.09) / 2 = 0.365
Genotype Frequencies in Offspring
Freq of AA = p2 = 0.635 * 0.635 = 0.403
Freq of Aa = 2pq = 2 * 0.635 * 0.365 = 0.464
Freq of aa = q2 = 0.365 * 0.365 = 0.133
Offspring genotype frequencies ≠ Parental genotype frequencies.
Thus, Parental genotype frequencies do NOT satisfy HWP.
Unit 8: Solution 14.9c
Which of the following genotype frequencies of AA, Aa, and aa,
respectively, satisfy the Hardy-Weinberg principle?
(c) 0.49, 0.42. 0.09 [Janina, Kimberly, Laura W.]
Allele Frequencies
p = Freq of A = (0.49 + 0.49 + 0.42) / 2 = 0.7
q = Freq of a = (0.42 + 0.09 + 0.09) / 2 = 0.3
Genotype Frequencies in Offspring
Freq of AA = p2 = 0.7 * 0.7 = 0.49
Freq of Aa = 2pq = 2 * 0.7 * 0.3 = 0.42
Freq of aa = q2 = 0.3 * 0.3 = 0.09
Offspring genotype frequencies = Parental genotype frequencies.
Thus, Parental genotype frequencies do satisfy HWP.
Unit 8: Solution 14.9d
Which of the following genotype frequencies of AA, Aa, and aa,
respectively, satisfy the Hardy-Weinberg principle?
(d) 0.64, 0.27, 0.09 [Reba, Lawanda, Maria]
Allele Frequencies
p = Freq of A = (0.64 + 0.64 + 0.27) / 2 = 0.775
q = Freq of a = (0.27 + 0.09 + 0.09) / 2 = 0.225
Genotype Frequencies in Offspring
Freq of AA = p2 = 0.775 * 0.775 = 0.601
Freq of Aa = 2pq = 2 * 0.775 * 0.225 = 0.349
Freq of aa = q2 = 0.225 * 0.225 = 0.506
Offspring genotype frequencies ≠ Parental genotype frequencies.
Thus, Parental genotype frequencies do NOT satisfy HWP.
Unit 8: Solution 14.9e
Which of the following genotype frequencies of AA, Aa, and aa,
respectively, satisfy the Hardy-Weinberg principle?
(e) 0.29, 0.42, 0.29 [Melissa, Laura Y.]
Allele Frequencies
p = Freq of A = (0.29 + 0.29 + 042) / 2 = 0.5
q = Freq of a = (0.42 + 0.29 + 0.29) / 2 = 0.5
Genotype Frequencies in Offspring
Freq of AA = p2 = 0.5 * 0.5 = 0.25
Freq of Aa = 2pq = 2 * 0.5 * 0.5 = 0.5
Freq of aa = q2 = 0.5 * 0.5 = 0.25
Offspring genotype frequencies ≠ Parental genotype frequencies.
Thus, Parental genotype frequencies do NOT satisfy HWP.
Unit 8: HWP Application 1
Cystic fibrosis is a genetic disorder of the lungs and the pancreas that is
relatively common among the caucasian population of the United
States.
According to the CDC, what is the prevalence of cystic fibrosis? What is
the carrier rate?
Unit 8: HWP Application 2
Cystic fibrosis is a genetic disorder of the lungs and the pancreas that is
relatively common among the caucasian population of the United
States.
According to the CDC, what is the prevalence of cystic fibrosis? What is
the carrier rate?
Prevalence of Cystic Fibrosis = 1/3,300
Carrier Rate = 1/30
How do they know the Carrier Rate when carriers of cystic fibrosis
show no symptoms?
Unit 8: HWP Application 3
Prevalence of Cystic Fibrosis = 1/3,300
Carrier Rate = 1/30
How do they know the Carrier Rate when carriers of cystic fibrosis
show no symptoms?
If A = normal allele and a = cystic fibrosis allele
Then AA = normal Aa = carrier aa = cystic fibrosis
Prevalence of Cystic Fibrosis (aa genotype) = q2 = 1/3300
q = square root (q2) = 0.017
p = 1 - q = 0.992
Carrier (Aa genotype) Rate = 2pq = 2 * 0.992 * 0.017 = 0.034 = 1/30
Unit 8: HWP with More Than Two Alleles 1
What if there are three alleles for a gene within a population: A1, A2,
and A3? Is Hardy-Weinberg Equilibrium possible? What are the HardyWeinberg allele and genotype frequencies?
Unit 8: HWP with More Than Two Alleles 2
What if there are three alleles for a gene within a population: A1, A2,
and A3? Is Hardy-Weinberg Equilibrium possible? What are the HardyWeinberg allele and genotype frequencies?
Let p = Freq of A1; q = Freq of A2; r = Freq of A3
The genotype frequencies will be:
Freq of A1 A1 = p2
Freq of A1 A2 = 2pq
Freq of A2 A2 = q2
Freq of A2 A3 = 2qr
Freq of A3 A3 = r2
Freq of A1 A3 = 2pr
Unit 8: Problem 14.7
In a pygmy group in Central Africa, the frequencies of alleles
determining the ABO blood groups were estimated as 0.74 for Io, 0.16
for IA, and 0.10 for IB. Assuming random mating, what are the expected
frequencies of ABO genotypes and phenotypes?
Unit 8: Solution 14.7.1
In a pygmy group in Central Africa, the frequencies of alleles
determining the ABO blood groups were estimated as 0.74 for Io, 0.16
for IA, and 0.10 for IB. Assuming random mating, what are the expected
frequencies of ABO genotypes and phenotypes?
Io codes for no protein
IA codes for Type A protein
IB codes for Type B protein
Genotype
Phenotype
Io I o
Type O
IA Io
Type A
IA I A
Type A
IB Io
Type B
IB I B
Type B
IA I B
Type AB
Unit 8: Solution 14.7.2
In a pygmy group in Central Africa, the frequencies of alleles
determining the ABO blood groups were estimated as 0.74 for Io, 0.16
for IA, and 0.10 for IB. Assuming random mating, what are the expected
frequencies of ABO genotypes and phenotypes?
Genotype
Phenotype
Frequency
Io Io
Type O
p2
IA Io
Type A
2pq
IA IA
Type A
q2
IB Io
Type B
2pr
IB IB
Type B
r2
IA IB
Type AB
2qr
Unit 8: Solution 14.7.3
In a pygmy group in Central Africa, the frequencies of alleles
determining the ABO blood groups were estimated as 0.74 for Io, 0.16
for IA, and 0.10 for IB. Assuming random mating, what are the expected
frequencies of ABO genotypes and phenotypes?
Genotype
Phenotype
Frequency
Io Io
Type O
p2 = 0.74 * 0.74 = 0.548
IA Io
Type A
2pq = 2 * 0.74 * 0.16 = 0.237
IA IA
Type A
q2 = 0.16 * 0.16 = 0.026
IB Io
Type B
2pr = 2 * 0.74 * 0.10 = 0.148
IB IB
Type B
r2 = 0.10 * 0.10 = 0.01
IA IB
Type AB
2qr = 2 * 0.16 * 0.10 = 0.032
Unit 8: HWP Assumptions
1. The dominant allele will not eventually become the only allele in a
population (contrary to common belief at the time).
2. Allele and genotype frequencies become stable and unchanging (that
is, are at equilibrium) after one generation. This assumes:
a. Random mating
b. No gene flow (immigration or emigration)
c. No natural selection
d. No mutation
e. No meiotic drive
f. Non-overlapping generations (this just keeps the math simple)
Unit 8: Problem 14.11
How does the frequency of heterozygotes in an inbred population
compare with that in a randomly mating population with the same allele
frequencies?
Unit 8: Solution 14.11 (Nonrandom mating 1)
How does the frequency of heterozygotes in an inbred population
compare with that in a randomly mating population with the same allele
frequencies?
Inbreeding is mating between genetically-related individuals. This leads
to an accumulation of homozygotes and a deficit of heterozygotes
(relative to Hardy-Weinberg expectations). This is why:
Suppose there are three genotypes: AA, Aa, and aa.
In the first generation, AA individuals will mate with AA, Aa, or aa. The
offspring of these matings will be either AA or Aa.
Unit 8: Solution 14.11 (Nonrandom mating 2)
Inbreeding is mating between genetically-related individuals. This leads
to an accumulation of homozygotes and a deficit of heterozygotes
(relative to Hardy-Weinberg expectations). This is why:
Suppose there are three genotypes: AA, Aa, and aa.
In the first generation, AA individuals will mate with AA, Aa, or aa. The
offspring of these matings will be either AA or Aa.
In the second generation, these AA and Aa offspring will mate with
each other (that’s inbreeding). This will make more AA and fewer
Aa.
Unit 8: Solution 14.11 (Nonrandom mating 3)
Inbreeding is mating between genetically-related individuals. This leads
to an accumulation of homozygotes and a deficit of heterozygotes
(relative to Hardy-Weinberg expectations). This is why:
Suppose there are three genotypes: AA, Aa, and aa.
In the first generation, AA individuals will mate with AA, Aa, or aa. The
offspring of these matings will be either AA or Aa.
In the second generation, these AA and Aa offspring will mate with
each other (that’s inbreeding). This will make more AA and fewer
Aa.
In the third generation, these mostly AA and fewer Aa offspring
will mate with each other (that’s inbreeding). This will make even
more AA and even fewer Aa. This continues, generation after
generation.
Unit 8: Solution 14.11 (Nonrandom mating 4)
Suppose there are three genotypes: AA, Aa, and aa. The same thing is
going to happen to the aa lineages:
In the first generation, aa individuals will mate with AA, Aa, or aa. The
offspring of these matings will be either aa or Aa.
In the second generation, these aa and Aa offspring will mate with
each other (that’s inbreeding). This will make more aa and fewer
Aa.
In the third generation, these mostly aa and fewer Aa offspring
will mate with each other (that’s inbreeding). This will make even
more aa and even fewer Aa. This continues, generation after
generation.
Unit 8: Solution 14.11 (Nonrandom mating 5)
So overall, the AA lineages will become mostly AA.
the aa lineages will become mostly aa.
the Aa lineages will still produce HW genotype frequencies.
Added altogether, inbreeding produces more homozygotes (AA and aa)
and fewer heterozygotes (Aa) than predicted by HWP.
Unit 8: Discussion Question (Natural Selection 1)
If the genotype AA is an embryonic lethal and the genotype aa is fully viable but
sterile, what genotype frequencies would be found in adults in an equilibrium
population containing the A and a alleles? Is it necessary to assume random
mating?
Unit 8: Discussion Question (Natural Selection 2)
If the genotype AA is an embryonic lethal and the genotype aa is fully viable but
sterile, what genotype frequencies would be found in adults in an equilibrium
population containing the A and a alleles? Is it necessary to assume random
mating?
Under Hardy-Weinberg assumptions:
Freq of AA = p2
Freq of Aa = 2pq
Freq of aa = q2
But natural selection is operating:
WAA (fitness of AA) = 0
Waa (fitness of aa) = 0
Thus, the only successful matings are Aa x Aa. This yields the following
offspring genotype frequencies:
Genotype
Freq of Fertilized Eggs
Freq of Live Individuals
AA
0.25
0
Aa
0.5
0.5/(0.5+0.25) = 0.67
aa
0.25
0.25/(0.5+0.25) = 0.33