biol 3301 - genetics ch6b - map construction students
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BIOL 3301 - Genetics Ch6B - Map Construction StudentsTRANSCRIPT
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Map Construction
• Need more than two genes to determine the orientation of genes on the chromosome
• If you have three genes, can orient them on the chromosome based on pair linkage
• Not only single crossovers will occur
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Map Construction
• Non-crossovers (NCO), single crossovers (SCO) and double crossovers (DCO)
• Need to investigate three gene pairs• AaBbCc x aabbcc• A and B (SCO) – 20% recombinants (rf=0.20)• B and C (SCO) – 30% recombinants (rf=0.30)• DCO – A/B and B/C rf expected = 0.20 x 0.30 = 0.06 or
6%• The expected frequency of DCO is always much lower
than SCO
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Single and Double Crossovers
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Three-Point Mapping In Drosophila
• All three genes should be heterozygous• All phenotypes should be observed (usually
test cross is performed)• A sufficient number of progeny should be
produced• The double crossover genotype of the least
frequent classes is in the middle of the other two flanking genes.
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Calculation of the Distances Between Genes
• Drosophila cross– Scute bristles– Echinus eyes– Crossveinless wings
•
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Class Phenotype Genotype # observed
1 Scute, echinus, crossveinless
s e c 1158
2 Wild type + + + 14553 Scute s + + 1634 Echinus, crossveinless + e c 1305 Scute, echinus s e + 1926 Crossveinless + + c 1487 Scute, crossveinless s + c 18 Echinus + e + 1
Total 3248
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Steps to Calculate the Distance Between Genes
• Step 1. Determine the number of the parental (noncrossover, NCO) types
• Step 2. Determine the phenotype and number of the single crossover products
• Step 3. Determine the phenotypes and number of the double crossover products
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Steps to Calculate the Distance Between Genes
• To calculate a distance between two genes in a three-point mapping, add a number of single crossovers between the two genes plus the number of both double crossovers; divide by total number and multiply by 100
• Repeat for the second pair of genes• To verify, you may calculate the distance
between the third pair of genes
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Map Construction
• sc- ec = 163 + 130 + 2 = 295/3248x 100=9%
• ec – cv =192 + 148 + 2 = 342/3248 x 100 = 11%
• sc – cv = 163 +192 +130 + 148 = 633/3248= 0.194x100=19.4%
• Map: sc----9cM---ec---11cM---cv
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Another Cross
• Drosophila– Singed bristles sn– Crossveinless wings cv– Vermilion eyes v– On X-chromosome
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Class phenotype genotype # observed
1 singed crossveinless vermilion S c v 32 crossveinless vermilion + c v 3923 vermilion + + v 344 crossveinless + c + 615 singed crossveinless S c + 326 singed vermilion S + v 657 singed S + + 4108 wild type + + + 3
total 1000
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Map Construction
• Identify parental classes (NCO): - s + c v and s c+ v+
• Then mother’s genotype is s+ c v/s c+ v+• To determine the gene order, identify
double crossovers – less numerous – s c v and s+ c+ v+
• Suggest order: c+ v/ +s+
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Map Construction
_____________________________________ • Between c and s:
34+32+3+3=72/1000x100=7.2%• Between s and v:
61+65+3+3=132/1000x100=13.2%c---7.2cM---s---13.2cM---v
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Determining The Gene Sequence
• Method 1. There are only three possible orders- Determine the arrangement of alleles on each homolog of the heterozygous parent- Determine whether a double-crossover will produce the observed phenotype- If suggested order does not produce the observed phenotype, try another order
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Determining The Gene Sequence
• Method 2.- Determine the arrangement of alleles on the homologs of the heterozygote parent- Determine the actual double-crossover phenotypes- Select the single allele that has been switched this one will be in the middle
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Figure 5-10b Copyright © 2006 Pearson Prentice Hall, Inc.
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Interference
• Interference reduces the expected number of multiple crossovers when a crossover event in one region of the chromosome inhibits a second event nearby.
• Interference is positive if fewer double-crossover events than expected occur and negative if more double-crossover events than expected occur.
• The coefficient of coincidence (C) is the observed number of DCOs divided by the expected number of DCOs
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Interference• Inhibition of a crossover event in one region of
chromosome by another crossover nearby• Usually observed number of DCOs is less than
calculated from the distance between genes• Coefficient of coincidence, C:
C = Observed DCO/Expected DCO• Interference I = 1 - C
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Coefficient of Coincidence
• For the problem we solved,
observed frequency
c= of double haploids = 0.006 = 0.63expected frequency 0.072 x 0.132
of double crossovers
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Genetic Problem
• A woman has two dominant traits, cataract, which she inherited from her father, and polydactyly, which she inherited from her mother. Her husband has neither trait.
• If genes for these two traits are 15 cM apart on the same chromosome, what is the chance that the first child of this couple will have both traits?
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Example 2. Predict the progeny phenotypes and numbers for this cross
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Genetic Maps
• There are some problems with preparing genetic maps of chromosomes.
• The probability of a crossover is not uniform along
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Genetic Maps
• Some regions are "hot spots" for recombination (for reasons that are not clear). Approximately 80% of genetic recombination in humans
• In humans, the frequency of recombination of loci on most chromosomes
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Genetic Maps
• Chromosome maps prepared by counting phenotypes are called genetic maps.
• They have been prepared for many eukaryotes, including corn. Drosophila, the mouse , and the tomato.
• Genes that are present on the same chromosome are called syntenic.
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Modern Methods of Map Construction
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Mapping Using DNA Markers
• Polymorphic DNA markers instead of phenotypic traits
• Double haploids
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Genetic versus Physical Maps• Chromosome mapping by counting the number of
recombinants produces a genetic map of the chromosome.
• But all the genes on the chromosome are incorporated in a single molecule of DNA.
• Genes are simply portions of the molecule (open reading frames of ORFs) encoding products that create the observed trait (phenotype)
• The rapid progress in DNA sequencing has produced complete genomes for many prokaryotes and several eukaryotes.
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Genetic versus Physical Maps
• Having the complete sequence makes it possible to determine directly the order and spacing of the genes. Maps drawn in this way are called physical maps
• What is the relationship between the genetic map and the physical map of a chromosome?
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Mapping in Humans
• Using pedigrees• The basic difficulty with mapping genes in
humans is that it is hard to get big enough pedigrees, with enough informative families.
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Mapping Using Hybridoma Cells
Figure 5-23 Copyright © 2006 Pearson Prentice Hall, Inc.
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Mapping in Humans
• Other strategies include:
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Human Genome Project
• Chromosome viewer:– http://www.ornl.gov/sci/techresources/
Human_Genome/posters/chromosome/chooser.shtml
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