Transcript What is PCR? - Cobb Learning

PV92 PCR Informatics
Chromosome 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 Mullis

He 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 blood
 Single hair follicle
 One 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 bacteria
 Can tolerate high heat of PCR

• Discovery of this bacterium made PCR
possible
PCR Steps

3 Main Steps to PCR
Denaturing at ~94 °C
 Annealing 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 Informatics
Chromosome 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 size

Therefore 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
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)

5’
Alu
Amplified Region
3’
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…
No insertion: 641 bp

Approximately
500,000 Alu copies
per haploid genome,
representing about
5% of the human
genome.
With Alu: 941 bp
300 bp Alu insert
641 bp
Possible PCR Products
+
-
+/-
S1
S2
S3
S4
941 bp
641 bp
We will use primers that are specific
to the Alu region
 No Alu = 641 bp fragment
 Alu = 941 bp fragment

In the Lab…
1.
2.
3.
4.
5.
We will harvest some of your cells…
Incubate them with Chelex resin (extract
DNA)…
Use PCR to amplify the Alu gene…
Separate Alu fragments on 2% agarose gel…
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.
2.
3.
4.
5.
No selection (= rate of survival for all)
No mutation
No immigration/emigration
Large population
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
?
Hardy-Weinberg Equation
p2 + 2pq + q2 = 1
p = frequency
of + alleles
q = frequency
of - alleles
p
q
p
pp
pq
q
pq
qq
+/+ = p2
+/- =
2pq
-/- = q2
Hardy-Weinberg Equation
p2 + 2pq + q2 = 1
p = frequency of + allele
 q = frequency of - allele
 p2 = frequency of ++
 q2 = frequency of –
 2pq = frequency of +
Hardy-Weinberg Equilibrium
Example
+/+ Genotypic
frequency
=
Number with genotype
Population total (N)
=
25
38
.66
=
Genotype
+/+
+/–
-/-
Total (N)
# of People
25
5
8
38
0.66
0.13
0.21
1.00
Observed
Frequency
Calculating Allelic Frequencies
p = Frequency
of + alleles
=
Number of + alleles
Total number alleles
=
55
76
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 = 0.72; therefore q = 0.28 since p + q = 1.00
=
0.72
p2 + 2pq + q2 = 1
p
q
p
pp
pq
q
pq
qq
+/+ = p2
+/- =
2pq
-/- = q2
This should make sense now!
(p2, 2pq, q2 values)
p2
+
2pq
+
q2
= 1.00
(0.72)2 + 2(0.72)(0.28) + (0.28)2 = 1.00
0.52
+
0.40
p2 = 0.52
+ 0.08
2pq = 0.40
= 1.00
q2 = 0.08
Chi-Square Test

Ok…so…
Genotype
Genotype
frequency
x
Population
total (N)
=
Expected
number
+/+ (p2)
0.52
x
38
=
20
+/– (2pq)
0.40
x
38
=
15
–/–
0.08
x
38
=
3
(q2)
2
X
=∑
Genotype
+/+
+/–
–/–
(Observed – Expected)
2
Expected
Observed
Expected
(O–E)2
E
25
20
1.25
5
15
6.67
8
3
8.33
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 Data
2: 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?