STAT151 Lecture Notes - Lecture 13: Allele Frequency, Genetic Drift, Population Genetics

45 views5 pages
4erikapadilla and 37146 others unlocked
STAT151 Full Course Notes
4
STAT151 Full Course Notes
Verified Note
4 documents

Document Summary

Total individuals p= 2(aa) + 1(aa) / 2(total individuals) q= 2(aa) + 1(aa) / 2(total individuals) p= 2(1125) + 1(750) / 2(2000) p = 0. 75 q=2(125) + 1 (750) / 2(2000) q=0. 25: note !!! Knowing allele frequencies alone does not tell you anything about the population structure, or fitness advantage/disadvantage of various alleles e. g. each of these populations each has p=0. 8, q=0. 2. Hw formula can be derived from simple probability (punnet square) p2= frequency of aa genotype. 2pq = frequency of aa genotype q2= frequency of aa genotype. Use hw formula with observed allelic frequencies as inputs to calculate expected genotypic frequencies as outputs. = 1125 aa individuals in a population of 2000. = 125 aa individuals in a population of 2000. = 750 aa individuals in a population of 2000 e. g. if p=0. 75, q=0. 25. Note that in this case, the expected genotypic ratios matched the observed genotypic ratios; this population is in hardy-weinberg equilibrium .

Get access

Grade+20% off
$8 USD/m$10 USD/m
Billed $96 USD annually
Grade+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
40 Verified Answers
Class+
$8 USD/m
Billed $96 USD annually
Class+
Homework Help
Study Guides
Textbook Solutions
Class Notes
Textbook Notes
Booster Class
30 Verified Answers

Related textbook solutions

Related Documents