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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.

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Unit 8: Mendelian Genetics (from Unit 2)

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  1. 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 Unit 8: Mendelian Genetics (from Unit 2) Possible Eggs A a A AA Aa Possible Sperm a Aa aa

  2. 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: Unit 8: Hardy (1908) and Weinberg (1908) 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)

  3. 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 Unit 8: Hardy-Weinberg Equilibrium 1 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?

  4. 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 Unit 8: Hardy-Weinberg Equilibrium 2 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

  5. 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 3

  6. 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: Hardy-Weinberg Equilibrium 4

  7. Which of the following genotype frequencies of AA, Aa, and aa, respectively, satisfy the Hardy-Weinberg principle? Unit 8: Problem 14.9 0.25, 0.50, 0.25 [Antoine, Ben, Brandy] 0.36, 0.55, 0.09 [Carin, Courtney, Giselle] 0.49, 0.42. 0.09 [Janina, Kimberly, Laura W.] 0.64, 0.27, 0.09 [Reba, Lawanda, Maria] 0.29, 0.42, 0.29 [Melissa, Laura Y.]

  8. Which of the following genotype frequencies of AA, Aa, and aa, respectively, satisfy the Hardy-Weinberg principle (HWP)? Unit 8: Solution 14.9a • 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.

  9. Which of the following genotype frequencies of AA, Aa, and aa, respectively, satisfy the Hardy-Weinberg principle? Unit 8: Solution 14.9b • 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.

  10. Which of the following genotype frequencies of AA, Aa, and aa, respectively, satisfy the Hardy-Weinberg principle? Unit 8: Solution 14.9c • 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.

  11. Which of the following genotype frequencies of AA, Aa, and aa, respectively, satisfy the Hardy-Weinberg principle? Unit 8: Solution 14.9d • 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.

  12. Which of the following genotype frequencies of AA, Aa, and aa, respectively, satisfy the Hardy-Weinberg principle? Unit 8: Solution 14.9e (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.

  13. 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 1

  14. 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 2

  15. 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 Application 3

  16. What 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 1

  17. What 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: HWP with More Than Two Alleles 2

  18. 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: Problem 14.7

  19. 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 Io codes for no protein IA codes for Type A protein IB codes for Type B protein

  20. 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 14.7.2

  21. 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 14.7.3

  22. 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: Unit 8: HWP Assumptions 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)

  23. 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: Problem 14.11

  24. 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 1)

  25. 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 2)

  26. 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 3)

  27. 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 4)

  28. 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: Solution 14.11 (Nonrandom mating 5)

  29. 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 1)

  30. 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: Unit 8: Discussion Question (Natural Selection 2)

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