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Lecture 16

BIOC15Fall2013 Lecture 16 and Lecture 17.docx

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Department
Biological Sciences
Course
BIOC15H3
Professor
Karen Williams
Semester
Fall

Description
BIOC15Fall2013 Lecture 16: Population Genetics and Lecture 17: Hardy-Weinberg Equilibrium Gene mutation o Gene mutation  a gene can mutate from one allelic form to another o Forward mutation: A+ to a, D+ to D o Reverse mutation  a to A+, D to D+ o Mutations can be used to study the process of mutation or to permit the dissection of a biological function o Chromosome mutations can be used as physical markers o We can use gene mutations to study population variation Tobiano phenotype with a SNP of KIT and an inversion o Inversion in chromosome 3, which has only been found in Tobiano horses, is believed to disrupt the normal functioning of the KIT gene and cause the Tobiano pattern o o KIT intron 13 SNP  KM1 associated with Tobiano o Equine chromosome 3 ECA3 inversion  To (inversion) o DNA segments from homologous chromosomes can be run on gel after being mixed with specific restrictions enzymes  more than one band = organism is heterozygous o Imaginary Example w/ Physical Markers o O/O or O/o gives black eyes aliens o oo = orange eyes o O/O x o/o  black eyes (all O/o) o O/o x o/o  O/o or o/o (black eyes or orange eyes) o If there was an association between the alleles and a short tandem repeat DNA marker then we could write the cross by the STR (short tandem repeat) genotype: o 6/6 x 10/10  10/6 o 10/6 x 6/6  10/6 and 6/6 progeny 1 o o o Having 10 STR = black eyes o Having o and 6 STR = orange eyes  the region of 6 STR is a marker for the orange eye colour C1B Chromosome o In 1928 Muller devised a method of searching for any lethal mutation on the X-chromosome (chromosome 1 in Drosophila) o Constructed chromosome called C1B which carries and inversion cross over suppressor (C – no recombinations), a lethal (l) and a dominant mutation Bar-eye (B) marker 2 o o The F1 male will have white eyes, and the females will be carrying the induced mutation in the parent male o o all the F2 females without the Bar eye mutation will have the mutated chromosome from the original male parent Balancers o since Muller’s CIB chromosome many others have been constructed o chromosomes bearing multiple inversions are used to suppress crossing over and are often marked with recessive lethal mutations o only offspring that are heterozygous for the inversion AND the lethal mutation will survive o examples of balancers: o FM7  first chromosome balancer o SM6  second chromosome balancer marked with Cy (curly wings) o TM3  third chromosome balancer marked with Stubble bristles Inversions o A chromosomal rearrangement in which the chromosome is broken 2x, and flipped 180 degrees then rejoined o The loops in polytene chromosome reveal the breakpoints of the inversion o paracentric inversion  does not include the centromere o pericentric inversion  includes the centromere o 3 o where the loop is = no crossing over o Inversions identify mosquito populations o Anopheles gambiae M and Anopheles gambiae S o The different variants exist in different areas  with an overlap on the West coast of Africa o o with increasing aridity, the frequency of the inversion increases Drosophila problem o For two genes, dumpy and cinnabar, 44 m.u. apart how many recombinants would you expect from 511 progeny? o RF = recomb/total progeny o 0.44 = recomb/total progeny  recomb = 224.8 = 225 approx o But: you cross wt x dumpy cinnabar mutant  F1 wt heterozygotes o You test cross F1 and get the following progeny: o o so you only get 12 recombinants, not nearly 225 as expected o the gametes derived from crossing over in an inversion heterozygote are not viable so the only viable progeny result from the gametes where the chromosome were not involved in crossing over Recombination reduced within the inversion (back to inversion in mosquitos) o o the first one resembles the mapping function  the mapping function assumes no interference 4 Allele frequencies o p = frequency (A) = (2x count of AA homozygotes) + (count of Aa heterozygotes) / 2 x total number of individuals o q = frequency (a) = (2x count of aa homozygotes) + (count of Aa heterozygotes) / 2 x total number of individuals o Assume an aridity locus with two alleles (A and a within the 2La inversion in the mosquito example. The following are numbers for each genotype in a population of 1000 mosquitos: o AA = 353, Aa = 494, aa = 153 o Therefore the allele frequencies are: o p = frequency(A) = 2(353) + 494 / 2000 = 0.6 o q = frequency (a) = 2(153) +492 / 2000 = 0.4 o p + q = 1 The Hardy-Weinberg law o is a binomial equation o o in a large population of randomly breeding individuals, with no new mutations, no migration, and no differences in fitness based on genotype: o p + q + 2pq = 1 Genes in a pool and allele frequencies o gene pool  the sum of the total of all alleles carried in all members of a population o allele frequency  proportion of gene copies in a population tat are of a given allele type o Example: you have a population with 12 AA individuals, 4 AB and 4BB individuals o Genotype frequencies: o AA = 12/20 = 0.6, AB = 4/20 = 0.2, BB = 4/20 = 0.5 o Allele frequencies: o F(A) = 12x2 + 4 / 20x2 = 0.7 o F(B) = 4x2 + 4 / 40 = 0.3 What are the genotype frequencies of Tobiano from the ECA3 inversion data? o Genotype frequency = individuals with tobiano genotype / total o  Cline o allele frequencies changing in response to a change in geographic factors 5 o o black part of the pie chart occurs more frequently the farther south you go o Example: (Thr-GLy)20 allele of period o o Latitudinal cline  gradual change in a character over the geographic range of the species that is associated with a change in environmental variables o Frequency of (Thr-Gly)20 increases with latitude in both Europe and Australia Allele frequencies of an X-linked locus o o p = f(X ) = (2x X X females + X X females + X Y male) / (2x number of females + number males) o q = f(X ) = (2x X X females + X X females + X Y male) / (2x number of females + number males) o Example 2: per(Thr-Gly)20 (X linked) o o p=f(XA)= (2 x XAXA females)+ (XA Xa females)+ (XA Y males) / (2x number of females) + number of males o q=f(Xa)= (2 x Xa Xa females) + (XA Xa females)+(Xa Y males) / (2 x number of females) + number of males o the frequency of 20,20 genotype, p
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