2008 PGSAS G-nomes
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Transcript of 2008 PGSAS G-nomes
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The Human Genome Project
➢ June 26, 2000: Successful completion of the first ‘draft’ of the entire human genome!!!
➢ The race between Celera and NIH is finished. The private company appears to have won.
•But,What the heck is a ‘genome’?What did they/we win?
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The Chicken Genome Project
➢ An initiative begun by NIH in 2002➢ Completed in 2004➢
➢ Other species considered:➢ Cats, cows, sheep, horses, dogs
➢ Cow begun in 2004➢ Pig begun in 2005➢ Who's next?
➢ Look here: Ensembl
➔But,What the heck is a ‘genome’?What did they/we win?
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The Genome (?)
➢ G-nomes;Grumpy and Sleepy?
➢ With apologies to Dr. Dean Snow
✔Not really.✔A genome is a complete sequence of all the known genes of an organism; including their structure and function
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Maps and markers
➢ What’s a genetic map?
➢ With apologies to Dr. David Bottstein.
✔WELL,Let’s start with a simpler question:✔How do you get to Penn State??
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One kind of map of Penn State
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Here’s a better view
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Now I know this will be helpful
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Perhaps we need a different kind of map?
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How about this?
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Or, this?
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Or, even this?
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The Genome (among friends)
➢ Chromosomes➢ Each chromosome is
one molecule of DNA.➢ 107 to 108 base pairs➢ A structural gene,
coding for a polypeptide/protein, is between 103 to 104 bp.
➢ Approximately 10% of the genome is coding.
➢ DO THE MATH!!➢ A chromosome contains
1,000 to 10,000 genes.➢ Vertebrate genomes contain
approximately 50,000 to 100,000 genes.
➢ These are generalizations and are highly species specific.
➢ Indeed, calculations from the human genome project suggest that there are approx. 35,000 genes
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Genes and Markers and Maps
➢ Gene Mapping➢ The location of genes to specific positions (e.g.,
loci) on specific chromosomes.
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Structural Genes
➢ Consider Hemoglobin!➢ Normal adult hemoglobin consists of 2 molecules
each of 2 different polypeptides.➢ α (141 aa) and β (146 aa)➢ On chromosomes 16 and 11
➢ Given 3 bp per aa➢ the β chain has 4438 possible single bp variants➢ This number exceeds the total number of fundamental
particles in the universe.
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Hemoglobin-β mutations
Non-sense
Nil-STOPUAGATCMutant
Mis-sense
ValineGUGCACMutant
Same-sense
GlutamateGAACTTMutant
Wild-type
GlutamateGAGCTCNormal
TypeAmino Acid
mRNA codon
DNA codon
Allele
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Mapping➢ Prior to the 1980’s all mapping was accomplished
using major genes of obvious phenotypic effect.➢ The advent of RFLP’s, AFLP’s, microsatellites and
other molecular markers, we can identify large numbers of segregating loci, simultaneously in the same cross.
➢ Remember that these markers are not true genes and are really ‘framework maps’, since they provide the ‘road map’ to locate genes of interest.➢ Useful for locating and studying QTL / MAS.➢ Invaluable to investigating genomic organization across
related species/genera.
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Power Supply
Electrode
BufferSolution
Gel
Electrophoresis ApparatusElectrophoresis Apparatus
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Restriction EndonucleasesRestriction Endonucleases
Eco
RI
BamHI
Xho
I
Hae
III
Hha
I
Alu
I
5' - C C G C - 3'3' - C G C G - 5'
5' - G G C C - 3'3' - C C G G - 5'
5' - G A A T T C - 3'3' - C T T A A G - 5'
5' - G G A T C C - 3'3' - C C T A G G - 5'
5' - C T C G A C - 3'3' - G A G C T C - 5'
5' - A G C T - 3'3' - T C G A - 5'
These enzymes cleave These enzymes cleave DNA at sites of DNA at sites of specific, short specific, short nucleotide sequencesnucleotide sequences
··
More than 500 are More than 500 are commercially commercially availableavailable
··
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Southern Transfer ProcedureSouthern Transfer ProcedureGel Filter
DNA restrictionfragments
(a) (b) (c)
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Restriction Length PolymorphismsRestriction Length Polymorphisms
AA Aa aa
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Design for Interpreting a Marker Locus and Design for Interpreting a Marker Locus and PhenotypePhenotype
Hypothetical FHypothetical F11 Genotype Genotype
qq mm
rr MM F1 Gametes Frequencies
QM
qm
Qm
qM
(1 - r)/2
(1 - r)/2
r/2
r/2
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FF22 Genotypic Array Genotypic ArrayGenotypes Frequency Value
QQMM
QqMM
qqMM
QQMm
QqMm
qqMm
QQmm
Qqmm
qqmm
(1 - r)2/4
(r - r2)/2
r2/4
(r - r2)/4
[r2 + (1 - r)2]/2
(r - r2)/2
r2/4
(r - r2)/2
(1 - r)2/4
+a
d
-a
+a
d
-a
+a
d
-a
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Estimation of additive and dominance effectsEstimation of additive and dominance effects
Marker class Mean expression
MM
Mm
mm
(1-2r)a + 2r(1 - r)d[(1 - r)2 + r2]d
(1 - 2r)(-a) + 2r(1 - r)d
Marker-locus class means (frequency adjusted)
Additivity: (Additivity: (MMMM - - mmmm)/2 =)/2 = a a(1 - 2(1 - 2rr))Dominance: Dominance: Mm Mm - (- (MM MM + + mmmm)/2 = )/2 = d d(1 - 2(1 - 2rr))22
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Gene Order and Arrangements
➢ Now that we’ve talked about structure and function …
➢ How do we figure out their placement on the map?
➢ We take advantage of a violation of the law.
✔ Specifically,Mendel’s law of independent assortment.
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Consequences of crossing over (1)
Chiasma
A B
A Ba b
a b A B
a b
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Meiosis
Linkage between a mutant gene and a marker
Mutant gene
DNA marker
Wild-type gene
Variant DNA marker
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Consequences of crossing over (2a)
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Consequences of crossing over (2b)
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Chiasma frequency and distance between loci
0102030405060708090
0 0.5 1 1.5 2 2.5 3 3.5
Chiasma Frequency
cM o
r R
F%
MAP DISTANCE
RECOMBINATIONFREQUENCY
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Using the test-cross
135Total
3aBaaBbaB
4AbAabbAbRecombinants
60abAabbab
68ABAaBbABParentals
NumberProgeny Phenotype
Progeny Genotype
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Calculating Recombination Frequency
➢ Number of ‘A’ individuals:
➢ 68 + 4 = 72➢ Number of ‘a’
individuals:➢ 60 + 3 = 63
χ2=0.6; ns
➢ Number of ‘B’ individuals:
➢ 68 + 3 = 71➢ Number of ‘b’
individuals:➢ 60 + 4 = 64
χ2=0.37; ns.
✔Test for single factor ratios:
✔RF = (4+3)/135 = 0.0518 or 5.18%
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What if you had 3 genes of interest?
➢ Start with an F1 produced by 2 pureline parents (AABBCC x aabbcc).
➢ Backcross the F1 to the triple-recessive parent.➢ Check that all alleles are segregating in a 1:1 ratio
in the backcross➢ Altered segregation will give a poor estimate of RF%.
➢ differential survival➢ misclassification
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Here’s how to determine gene order
400TOTAL
32aBc
24AbC4
7aBC
14Abc3
49abC
51ABc2
120abc
103ABC1F1 gametes
Number of progeny
Progeny phenotypes
Class
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Calculate RF% as before
➢ ALL χ2 are non-significant.
➢ A – B = (14+7+24+32)/400 = 0.1925 or 19.25%
➢ A – C = (51+49+14+7)/400 =0.3025 or 30.25%
➢ B – C = (51+49+24+32)/400 =0.3900 or 39.00%
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And the answer is:
B A C
Since B-C is the largest RF, genes B and C must be the furthest apart; while A is in between.
39.00%
19.25% 30.25%
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Genes and Markers and Maps
➢ Gene Mapping➢ The location of genes to specific positions (e.g.,
loci) on specific chromosomes.➢ Linkage
➢ Genes that are located on the same chromosome are ‘linked’.
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The Human Map
11263 Mb263 Mb
171792 Mb92 Mb
212150 Mb50 Mb
XX164 Mb164 Mb