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Lecture 6

Lecture 6 - February 8, 2013.docx
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Department
Biology
Course
BIOLOGY 2C03
Professor
Joe Kim
Semester
Winter

Description
Course Code Year Lecture 6 – Feb 8 Gene Mapping – Linkage Maps  Observed exceptions to Mendel’s principle of independent assortment  “linkage” between genes; genes that are on the same chromosome  linkage maps – Morgans student, Alfred Cturtevant  Lobo, I. & Shaw, K (2008) Discovery and types of genetic linkage o “soon after the recovery of Mendel’s work, several scientists noted traits in their crosses seemed “coupled”. But this deviated from Mendel’s principles, so how did they explain this?” o Hypothesis: genes are “coupled” o Upon the establishment of the chromosomal theory of inheritance – genes are physical entities that are on chromosomes – perhaps two genes can reside on a single chromosome and be attached to one another  Are linked alleles inherited together?  No o Two genes – both on the X chromosome = linked o w = white eye gene o m = miniature wings o F2  Half white eyes, miniature wings = parental phenotype  Half red eyes, long wings = a parental phenotype o The following results were obtains: o 36.9% of F2 were recombinant = new combinations of phenotypes  Linked genes do not sort independently o Another pair of linked genes on an autosome o pr = eye color gene o vg = small wings ‘ o F2 Option 1 – different chromosomes, independent sorting 1 Course Code Year  recombination of unlinked genes due to independent assortment; recombination frequency = 50% o F2 Option 2 – on same chromosome, linked  No recombination; recombination frequency = 0% o Neither of these are observes  instead  Note: parentals appear most frequently (genotypes of true-breeding grandparents)  Recombinants occur less frequently (genotypic combination not seen in true breeding grandparents)  Conclusions o Genes often appear to be linked in a genetic analysis; i.e. they do not sort independently  Can be tested with a χ test  These genes are on the same chromosome o This linkage can be broken  This suggests that homologous chromosomes can exchange DNA = physically recombine  Crossover – physical even that can occur randomly across the chromosome o Figure – tetrad of the grasshopper Chothippus parallelus that shows 5 chiasmata  Can the frequency with which this event occurs be used to create a map that represents the distance between genes o Paint ball analogy – use the frequency of hits to define a distance between two points  Linkage maps – using recombination frequencies (RF) to create the first genetic maps (A. Sturtevant) o Distances between genes are defined by the frequency of recombination between them which represents chiasmata frequency o 1 map unit (m.u) = a recombinant frequency of 1% = 1 centimorgan (cM) o if you see recombinants between genes A and B at a frequency of 5%, the distance is defined as 5 m.u o Genes are arranged in linear order o Eg/ Drosophila melanogaster o Location of a gene is referred to as the gene locus and is known by the shorthand for the first mutant allele + o Eg/ pr = purple; represents the site of pr , pr etc vg = vestigial; represents the site of vg , vg etc 2 Course Code Year  Three-point testcross – linkages between three genetic loci o v = vermillion o cv = crossveinless o ct = cut o Largest 2 classes are parental; all res of recombinant o Chromosomes from P generation: v cv ct and v cv ct + o Place genes on a single map  Each RF is less than 50% - therefore each pair of genes are linked  Recombination of unlinked genes by random assortment (Mendelian ratios) RF = 50%  Recombination of linked genes by crossing-over; RF < 50%  RF between v and cv is the largest – therefore they are the farthest apart 3 Course Code Year  Re-write genotypes in correct order:  You have determined a gene order and a linkage map  Location on the chromosome is not known (are they all on one arm or spread across the chromosomes?)  There is no inherent polarity (v ct cv is the same as cv ct v)  Why is there a discrepancy?  we have not accounted for double recombinants  Accounting for double recombinants: o Two chiasmata between loci a and b o Eg/ parental genotypes v ct cv and v ct cv+ double recombinants v ct cv and v ct cv + + + o Chromosomes from P generation
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