Class Notes (836,562)
BIOL 1000 (60)
Lecture 46

# BIOL 1000 Lecture 46: Lecture 46 Premium

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School
Department
Biological Sciences
Course
BIOL 1000
Professor
Kevin G.Scott
Semester
Fall

Description
o78P • E.g. imaginary booby population with blue feet: o 500 birds in total o 320 WW (64% non-webbed feet), 160 Ww (32% also non-webbed feet), 20 ww (4% webbed feet) o Since this is a diploid population of 500, there are 1000 alleles for foot type   o 80% W alleles (320 WW 320 x 2 + 160 = 640 + 160 = 8800/1000 x 100 = 80%)  p = .80, q = .20 (p + q = 1) o Probability of producing a WW individual is p x p = .80 x .60 = .64 Ww individual probability is 2pq = 2 x .08 x .02 = .32 Multiply by 2 because this genotype can result from the sperm carrying W and eggs carrying w OR from sperm carrying w and egg carrying W Gene pool remains at state of equilibrium p + 2pq + q = 1 Frequency o homozygous dominants + frequency of heterozygotes + frequency of homozygous  recessives = 1 For a population to be in Hardy-Weinberg equilibrium it must satisfy 5 conditions: 1. Very large population Smaller te population, more likely that allele frequencies will fluctuate by chance between generations 2. No gene flow between populations When people move in/out of populations they add/remove alleles, altering  the gene pool
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