PV92 PCR Informatics Chromosome 16 Day #1: What is PCR? Day #2: Alu Insertion & PCR.
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PV92 PCR Informatics Chromosome 16 Day #1: What is PCR? Day #2: Alu Insertion & PCR
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Transcript of PV92 PCR Informatics Chromosome 16 Day #1: What is PCR? Day #2: Alu Insertion & PCR.
PV92 PCR InformaticsChromosome 16
Day #1: What is PCR?
Day #2: Alu Insertion & PCR
PCR
Polymerase chain reaction (PCR) enables researchers to produce millions of copies of a specific DNA sequence in a relatively short period of time The specific sequence is called the
“target sequence”
PCR
1st described in 1985 by Kary MullisHe won the Nobel in 1993 for this
Made it possible for researchers in a variety of biological fields to
incorporate molecular biology (genetics) into their research.
•Pathology
•Botany
•Zoology
•Pharmacology
What Is It & Why Did it Revolutionize Research
PCR produces exponentially large amounts of DNA from trace amounts.Drop of bloodSingle hair follicleOne cheek cell
What’s Needed For PCR? Free nucleotides – building blocks DNA primers
A strand of nucleic acid that serves as a starting point for DNA replication
Complementary to target sequence Taq polymerase
Comes from hot springs bacteriaCan tolerate high heat of PCR
• Discovery of this bacterium made PCR possible
PCR Steps
3 Main Steps to PCRDenaturing at ~94 °CAnnealing at ~54 °C
• Binding of a primer to DNA strand.
Extension at ~72 °C• The bases (complementary to the
template) are coupled to the primer on the 3' side.
Lots of Copies!
PCR Animation
http://www.sumanasinc.com/webcontent/animations/content/pcr.html
Any Questions?
PV92 PCR InformaticsChromosome 16
Day #1: What is PCR?
Day #2: Alu Insertion & PCR
What is PV92?
PV92 – A human-specific Alu insertion on C16
Alu’s are a type of transposon Recall that transposons are pieces of
DNA that can move around within the genome of a cell
Barbara McClintock discovered transposons
and won the Nobel in 1983.
More About Alu Transposons
Also known as ‘jumping genes’ Copies itself and inserts into
new locations on the chromosome
No evidence that ‘parent’ Alu segments are ever excised…what does this mean in terms of evolution & human population genetics?
Alu is a SINE as well as a Retrotransposon
Alu is ~300 bp in sizeTherefore known as a SINE (short
interspersed repetitive element) Highly Conserved Alu endonuclease is called so because it
was isolated from Anthrobacter luteus Retrotransposon because it uses
reverse transcriptase to copy itself
But Alu a Defective Transposon
Can’t make it’s own reverse transcriptase (RT)…so…it hijacks another gene
Hijacked gene is known as L1 (a LINE) and is basically a non-functional retrovirus that can make RT
Say WHAT??????
It’s Like This…
1. Alu is transcribed into mRNA via RNA polymerase
2. mRNA hijacks L1 which converts Alu to ds DNA by way of RT
3. DNA copy of Alu is integrated into new chromosome site
PV92 Alu Insertion
5’5’ 3’3’Alu
Amplified RegionAmplified Region
A member of Alu repeat family Human-specific Alu insertion Found in a non-coding region of your DNA Not diagnostic for any disease or disorder
(usually)
Introns, Exons and Alu Inserts
Only ~5% of our genome consists of coding DNA (exons).
• This is where our functional proteins come from.
Out of the remaining 95%, ~40% is intron based, or non-coding.
• We pass introns and exons on from one generation to the next.
Introns, Exons and Alu Inserts Introns are full of SINES, Alu being
one of them. • Origins of Alu unknown.
We are targeting a specific locus on C16 that is known to carry the Alu sequence, or not.
• Some of you will have the Alu insert and others will not.
The Alu Insert is Dimorphic
The PV92 Alu is dimorphic so there are two possible PCR products: 641 bp and 941 bp
If you have the Alu insert on both of your homologous chromosomes = +,+
If you have Alu on one chromosome = +, - If you don’t have the insert = -,-
To Alu Or Not…
Approximately 500,000 Alu copies per haploid genome, representing about 5% of the human genome.
No insertion: 641 bpNo insertion: 641 bp
300 bp Alu insert
641 bp 641 bp
With Alu: 941 bp With Alu: 941 bp
Possible PCR Products
941 bp941 bp641 bp641 bp
-- +/-+/-++
We will use primers that are specific to the Alu region
No Alu = 641 bp fragment Alu = 941 bp fragment
S1S1 S2S2 S3S3 S4S4
In the Lab…1. We will harvest some of your cells…
2. Incubate them with Chelex resin (extract DNA)…
3. Use PCR to amplify the Alu gene…
4. Separate Alu fragments on 2% agarose gel…
5. Use Chi-Square or Hardy Weinberg to calculate population frequency of (+,+), (+,-) and (-,-).
Any Questions?
Calculating Observed Genotypic Frequencies
Hardy-Weinberg Equilibrium
The HW equilibrium describes what happens to alleles in an ‘ideal’ population.
1. No selection (= rate of survival for all)
2. No mutation
3. No immigration/emigration
4. Large population
5. Random mating
Hardy-Weinberg Equilibrium
We know that when we cross two individuals who are heterzygous for Alu we will see:
+
+
-
-
+ + + -
+ - - -
So there is a 25% chance of being ++, 25% being - - and
50% + -.
Hardy Weinberg
?
+/+ = p+/+ = p22
+/- = +/- = 2pq2pq -/- = q-/- = q22
pppp pqpq
qqqqpqpq
p p
pp
Hardy-Weinberg Equation
pp22 + 2pq + q + 2pq + q22 = 1 = 1
p = frequency of + alleles
q = frequency of - alleles
p = frequency of + allele q = frequency of - allele p2 = frequency of ++ q2 = frequency of – 2pq = frequency of +-
Hardy-Weinberg Equation
p2 + 2pq + q2 = 1p2 + 2pq + q2 = 1
Hardy-Weinberg Equilibrium Example
Genotype +/+ +/– -/- Total (N)
# of People 25 5 8 38
Observed 0.66 0.13 0.211.00
Frequency
+/+ Genotypic frequency
Number with genotype Population total (N)
=
25
38
.66
=
=
Calculating Allelic Frequencies
Number of + alleles
25 individuals with two + alleles = 50 + alleles
5 individuals with one + allele = 5 + alleles
Total = 55 + alleles
Total number of alleles
2N = 2(38) = 76
p Frequency
of + alleles
Number of + alleles
Total number alleles
= 55
76
= 0.72==
p = 0.72; therefore q = 0.28 since p + q = 1.00
p2 + 2pq + q2 = 1p2 + 2pq + q2 = 1
+/+ = p+/+ = p22
+/- = +/- = 2pq2pq -/- = q-/- = q22
pppp pqpq
qqqqpqpq
p p
pp
This should make sense now!
(p2, 2pq, q2 values)
p2 + 1.00=2pq + q2
(0.72)2 + 1.00=2(0.72)(0.28) + (0.28)2
0.52 + 1.00=0.40 + 0.08
p2 = 0.52 2pq = 0.40 q2 = 0.08
Chi-Square Test
Ok…so…
+/+ (p2)
+/– (2pq)
–/– (q2)
0.52
Genotype frequency
x =Population total (N)
Expected number
x =
x =
x =
38 20
0.40 38 15
0.08 38 3
Gen
otyp
e
X2=∑(Observed – Expected)2
Expected
+/+
+/–
–/–
25 20 1.25
5 15 6.67
8 3 8.33
Genoty
pe
Observed Expected (O–E)2
E
X2 = 16.25
Allele Server(1 of 17)
Cold Springs Harbor Laboratory DNA Learning Center
Web site:http://www.dnalc.org/
Allele Server(2 of 17) Scroll through DNALC internet sites
until BioServers Link appears
Allele Server(3 of 17)
Click on Bioservers
Allele Server(4 of 17)
Enter the Allele Server
Allele Server(5 of 17)
Click on Manage Groups
Allele Server(6 of 17)
Select Group
Allele Server(7 of 17)
Scroll Down to Select “Your Group”
Allele Server(8 of 17)
Fill Out Form
Allele Server(9 of 17)
Click on Edit Group
Allele Server(10 of 17)
Edit Your Group Information
Allele Server(11 of 17)
Click on Individuals Tab
Allele Server(12 of 17)
Add Each Student’s Information
Add as much information as possible:
• Genotype (+/+, +/–. –/–)• Gender• Personal Information
Allele Server(13 of 17)
Click on Done
Allele Server(14 of 17)
Select and then Click OK
Allele Server(15 of 17)
Analyze Data2: Then Click Here
1: Click Here First
Allele Server(16 of 17)
Click on the Terse and Verbose Tabs to Review Data Results
Any Questions?
