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

Aa (mom) x Aa (da) = 25% AA

50% Aa

25% aa

Possible EggsA a

Possible Sperm

A

a

AA

Aa

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 Hardy-Weinberg 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 1Suppose 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 2Suppose 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 3Example: 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 4Example: 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 14.9Which 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 14.9aWhich 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 14.9bWhich 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 14.9cWhich 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 14.9dWhich 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 14.9eWhich 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 1Cystic 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 2Cystic 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 3Prevalence 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 1What if there are three alleles for a gene within a population: A1, A2, and A3? Is Hardy-Weinberg Equilibrium possible? What are the Hardy-Weinberg allele and genotype frequencies?

Unit 8: HWP with More Than Two Alleles 2What if there are three alleles for a gene within a population: A1, A2, and A3? Is Hardy-Weinberg Equilibrium possible? What are the Hardy-Weinberg 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 A1 A3 = 2pr

Freq of A2 A2 = q2 Freq of A2 A3 = 2qr

Freq of A3 A3 = r2

Unit 8: Problem 14.7In 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 14.7.1In 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

Io Io Type O

IA Io Type A

IA IA Type A

IB Io Type B

IB IB Type B

IA IB Type AB

Io codes for no proteinIA codes for Type A proteinIB codes for Type B protein

Unit 8: Solution 14.7.2In 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 14.7.3In 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 Assumptions1. 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 14.11How 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