Class Notes (836,562)
Canada (509,854)
BIOL 1000 (60)
Lecture 46

BIOL 1000 Lecture 46: Lecture 46
Premium

3 Pages
28 Views
Unlock Document

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
More Less

Related notes for BIOL 1000

Log In


OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


OR

By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.


Submit