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## Genetics Week 15

by: Lauran Notetaker

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5

# Genetics Week 15 BIOL/PBIO 3333

Lauran Notetaker
OU

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Week 15
COURSE
Genetics
PROF.
Dr. Jim Thompson
TYPE
Class Notes
PAGES
5
WORDS
CONCEPTS
Genetics
KARMA
25 ?

## Popular in Biology

This 5 page Class Notes was uploaded by Lauran Notetaker on Saturday May 14, 2016. The Class Notes belongs to BIOL/PBIO 3333 at University of Oklahoma taught by Dr. Jim Thompson in Winter 2016. Since its upload, it has received 10 views. For similar materials see Genetics in Biology at University of Oklahoma.

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Date Created: 05/14/16
April 25, 2016 XIV. Genes in Populations A. Gene Pool B. Hardy-Weinberg Equilibrium C. Exceptions to Assumptions 1. Non-random mating 2. Drift and inbreeding Assumptions: Sexual reproduction Diploid No selection Mating is at random (panmixis) Population size is large (mathematically inﬁnite) No migration No mutation Maternal Genotypes AA(0.64) Aa(0.32) aa(0.04) AA a 0.64x0.64 b 0.64x0.32 c (0.64) Aa b d e (0.32) aa c e f (0.04) genotype frequencies: p^2 + 2pq + q^2 = 1 allele frequencies: p + q = 1 This is stable over time, unless some outside force is acting - an exception to one of the model assumptions. Sample question: For a certain trait, 1/10,000 newborns shows a certain recessive trait. What is the frequency of heterozygous carriers in that population? aa = q^2 = 1/10,000 or 0.0001 q = sqrt of 0.0001 since p + q = 1 p = 1-q = 1 - 0.01 = .99 so freq of 2pq Sample question: For a trait of interest, 64% show the dominant (H). What proportion of these are actually homozygous (HH)? HH Hh hh 1 - 0.64 = 0.36 hh = 0.36 = q^2 q = 0.6 p = 1 - 0.6 = 0.4 freq of HH = p^2 = (0.4)^2 = 0.16 Special cases: Sex-linkage Frequency in females: Since there are two X chromosomes, the genotype and allele frequencies are determined just as we have done for autosomal traits AA Aa aa p^2 2pq q^2 But, males have only one X, so Freq of A = p a = q Sample question: Imagine that an island is settled by a population of 100 people with the following blood types for the locus: What are the allele frequencies in this population? MM MN NN total 34 58 8 100 (200 alleles) M 68 + 58 58 +16 N p = freq of M = 2(34) + 58 = 0.63 200 N = 2(8) + 58 = 0.37 200 After at least one generation of random mating, when the population reaches 2000 people, how many will be homozygous “MM”? since p = freq of M = 0.63 then MM = p^2 = (0.63)^2 = 0.3969 out of 2000 people we expect 0.3969 (2000) = 793.8 Cystic ﬁbrosis ( c c ) affects about 1/1600 newborns q^2 = 1/1600 = q = 1/40 p - 1/40 = 39/40 2pq = 2(39/40)(1/40) ~1/20 1/20 x 1/20 = 1/400 of this mating 1/400 x 1/4 = 1/1600 Cc x Cc of CC child Rh factor = rhesus blood type April 27, 2016 XIV. Genes in Populations A. Gene Pool B. Hardy-Weinberg Equilibrium C. Exceptions to Assumptions 1. Non-random mating 2. Drift and inbreeding 3. Mutation 4. Migration 5. Selection Participation #8 of 8 Founder Effect settling on an uninhabited island establishing a new population by migration to a new habitat of populations AA Aa aa Generation 100% 25% 50% 25% non-random mating Positive assortative mating - like mating with like - Increase homozygosity for genes on which mate choice is based - or genes linked to them Negative assortative mating - mating with unlike - increase heterozygosity for genes on which mate choice is based Inbreeding - Increase homozygosity for entire genome p = 0.8 p = 0.4 q = 0.2 q = 0.6 recipient population (r) migrant population (m) q’r = qr (1-m) + qm (m) = qr - mqr + mqm Deltaq = q’r - qr = qr - mqr +mgm - qr = m(qm-qr) April 29, 2016 Exam

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