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GET OUT YOUR CALCULATORS

GET OUT YOUR CALCULATORS. WE’RE GOING TO DO A QUICK TEST OF YOUR MATH SKILLS . GENOTYPE & ALLELE FREQUENCIES. GENOTYPE & ALLELE FREQUENCIES. GENOTYPE & ALLELE FREQUENCIES. GENOTYPE & ALLELE FREQUENCIES. GENOTYPE & ALLELE FREQUENCIES. HARDY - WEINBERG.

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GET OUT YOUR CALCULATORS

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  1. GET OUT YOUR CALCULATORS WE’RE GOING TO DO A QUICK TEST OF YOUR MATH SKILLS 

  2. GENOTYPE & ALLELE FREQUENCIES

  3. GENOTYPE & ALLELE FREQUENCIES

  4. GENOTYPE & ALLELE FREQUENCIES

  5. GENOTYPE & ALLELE FREQUENCIES

  6. GENOTYPE & ALLELE FREQUENCIES

  7. HARDY - WEINBERG • In a population at Hardy–Weinberg equilibrium (no evolution occurring), allele frequencies remain the same from generation to generation, and genotype frequencies remain in the proportions p2 + 2pq + q2 = 1. • Two equations • Allele frequencies • p + q = 1 • A + a = 1, where A and a equal gene percentages • All dominant alleles plus all recessive alleles add up to all of the alleles for a particular gene in a population • Genotype frequencies • p2 + 2pq + q2 = 1 • AA + 2Aa + aa = 1 • For a particular gene, all homozygous dominant individuals plus all heterozygous individuals plus all homozygous recess individuals add up to all of the individuals in the population

  8. H-W ProblemsExample 1 The gene for albinism is known to be a recessive allele. In Lakeville, 9 people in a sample of 10,000 were found to have albino phenotypes. The other 9,991 had skin pigmentation normal for their ethnic group. Assuming Hardy-Weinberg equilibrium, what is the allele frequency for the DOMINANTpigmentation allele in this population? HINT: PLUG IN GIVEN INFORMATION USE MRS. MACWILLIAMS CHART TO MAKE IT EASIER!!!

  9. Example 1 9/10,000 = 0.09% 9991/10000 = 99.91%

  10. H-W ProblemsExample 1 The gene for albinism is known to be a recessive allele. In Lakeville, 9 people in a sample of 10,000 were found to have albino phenotypes. The other 9,991 had skin pigmentation normal for their ethnic group. Assuming Hardy-Weinberg equilibrium, what is the allele frequency for the DOMINANTpigmentation allele in this population? ALWAYS TRY TO FIND “q” FIRST Then find “p” where p = 1-q

  11. Example 1 9/10,000 = 0.09% p = 1 – q p = 1 – 0.03 p = 0.97 q2 = 0.0009 q = √0.0009 q = 0.03 9991/10000 = 99.91%

  12. H-W ProblemsExample 1 The gene for albinism is known to be a recessive allele. In Lakeville, 9 people in a sample of 10,000 were found to have albino phenotypes. The other 9,991 had skin pigmentation normal for their ethnic group. Assuming Hardy-Weinberg equilibrium, How many out of the 10,000 people in the sample above were expected to be HETEROZYGOUS for pigmentation?

  13. Example 1 2pq = 2 (.97)(.03) = .06 = 6% 9/10,000 = 0.09% p = 1 – q p = 1 – 0.03 p = 0.97 q2 = 0.0009 q = √0.0009 q = 0.03 6% of the population are heterozygous (.06 x 10,000 = 600 people) 9991/10000 = 99.91%

  14. H-W ProblemsExample 2 A similar survey was carried out in Wisconsin, but only 2,500 people were surveyed. If allele frequencies are the same in Wisconsin as they are in Lakeville, how many people would you expect to have the albino phenotypes in this sample? WHAT DO WE NEED TO FIND? Need to find q2 and then calculate percentage of 2,500 people.

  15. Example 2 q2 = (.03)(.03) = 0.0009 = 0.09% p = 0.97 q = 0.03 If .09% of Wisconsin population are albino, then .0009 x 2500 = 2.25 or 2 people

  16. HW PROBLEMExample 3 • Given: In a population of 100 individuals (200 alleles), 16 exhibit a recessive trait. • Problem: • Find the allele frequencies for A and a. • Find the genotypic frequencies of AA, Aa, and aa.

  17. Example 3 p2 = (.6)(.6) = .36 = 36% 2pq = 2(.6)(.4) = 0.48 = 48% 16/100 = 16% (given) p = 1-0.4= 0.6 q = √.16 q = 0.4 100 – 16 = 84%

  18. HW ProblemsExample 4 • Fraggles are mythical, mouselike creatures that live beneath flower gardens. • Of the 100 Fraggles in a population, 91 have green hair (F) and 9 have grey hair (f). • Assuming genetic equilibrium: • What are the allele frequencies of F and f? • What are the genotypic frequencies? • Work this out at your desk (HINT: the 91 with green hair could be either FF or Ff therefore find the frequency of f first)

  19. Example 4 p2 = (.7)(.7) = .49 = 49% 2pq = 2(.7)(.3) = 0.42 = 42% q = f = √.09 = 0.3 p = F = 1-.3 = 0.7 9/100 = 9% (given) 91/100 = 91% (given)

  20. NOW LET’S GET TO WORK ON SOME PROBLEMS!!!!

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