BIOL 202 Chapter Notes - Chapter 18: Haplotype, Neutral Mutation, Linkage Disequilibrium

156 views17 pages

Document Summary

Answer: individual 3 has three heterozygous loci (microsatellites 1, 3 and 4), while individual 1 has none. Answer: gd (gene diversity) is calculated as: 1- (p12 + p22 + pn2) where p is allele frequency. 2/7 chromosomes are different at the indel site, so gd = 1 0. 59 = 0. 41. 4/7 at the microsatellite locus, so gd = 0. 49. At position 29, g is present in 6 of 7 haplotypes for a frequency of 0. 857. A is present at position 33 in ut. 3 ut o of 7 haplotypes for a frequency of 0. 429. A random combination of g and a would be expected to occur with a frequency of 0. 857 0. 429 = 0. 368, so p the ga haplotype, is 3 out of 7, which equals 0. 429. Pab papb equals 0. 429 0. 368 = 0. 061= d. d is different than 0, so population is in linkage disequilibrium for the given haplotypes at the positions 29 and 33.

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 Documents