Polymerase Chain Reaction
Polymerase Chain Reaction
The Polymerase Chain Reaction (PCR) is a method used to synthesize millions of copies of a given DNA sequence.
A typical PCR reaction consists of series of cycles:
- template DNA denaturation,
- primer annealing, and
- extension of the annealed primers by DNA polymerase.
This loop is repeated between 25 and 30 times
What is the question?
Why do you want to use qPCR?
qPCR can answer several questions
- Presence of a specific mRNA molecule in the sample
- Concentration of that mRNA in the sample
- Change in concentration of that mRNA between two samples
(you can replace “mRNA” for other keywords)
Always declare what is the question you want to answer
Questions are more important than answers
How to bake your cake
Use the following ingredients
- Template: the thing we want to study
- Primers for that template, in high concentration
- dNTP, in high concentration
- Taq polymerase
- Some marker, such as a fluorophore
- Salt & pepper (Na++, K+, Mg++)
- Cooking machine: thermocycler with sensors
The thermocycler is a computer.
You program it to cook your cake
Thermocyclers are kitchen robots
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PCR model
We simplify and we forget about the polymerase and the dNTP
They both will be represented by primers
The system is represented by this diagram:
How does it work
qPCR depends on initial concentration
The curve depends on the initial DNA concentration
Finding the initial concentration
Finding the initial concentration
We care only about the exponential phase
The signal increases 2 times on every cycle
\[X(C) = X(0)⋅2^C\]
So we can find the initial concentration
\[X(0) = X(C)⋅2^{-C}\]
CT: cycle where Signal crosses threshold
DNA concentration crosses 50% at 13.73 cycles
Standard curve: CT changes with concentration
Start with a large concentration of template, and dilute it several times. Measure the CT of each dilution
Standard curve gives concentration
If everything is right, we get
\[X(0) = X(CT)⋅2^{-CT}\]
But sometimes we do not know \(X(CT)\)
because the Signal is not 100% DNA concentration
It depends on the fluorophore assimilation
Delta CT
We still can measure change of concentration
Let’s say we extract mRNA before and after a shock
\[\begin{aligned}
X_B(0) & = X_B(CT_B)⋅2^{-CT_B}\\
X_A(0) &= X_A(CT_A)⋅2^{-CT_A}\end{aligned}\]
therefore the fold change of expression is
\[\frac{X_B(0)}{X_A(0)} = \frac{X_B(CT_B)⋅2^{-CT_B}}{X_A(CT_A)⋅2^{-CT_A}}\]
Fold change of expression
If we assume that the DNA concentrations are the same when the signal crosses the threshold, i.e.
\[X_B(CT_B)=X_A(CT_A)\] then \[\frac{X_B(0)}{X_A(0)} = 2^{-(CT_B-CT_A)} = 2^{-Δ CT}\]
Here \(Δ CT\) means the change in CT for one gene in two conditions
Why does the CT change?
This change of concentration has two components
- The real biological change
- The variability of the RNA extraction protocol
To avoid the second component, we use an endogenous reference
(typically, a housekeeping gene)
The reference must be taken together with the target
A single pipet, at the same time
Delta Delta CT
We normalize each sample
\[\begin{aligned}
\frac{X_B(0)}{R_B(0)} &=
\frac{X_B(CT_{XB})⋅2^{-CT_{XB}}}{R_B(CT_{RB})⋅2^{-CT_{RB}}}=
K_B⋅ 2^{-(CT_{XB}-CT_{RB})}\\
\frac{X_A(0)}{R_A(0)} &=
\frac{X_A(CT_{XA})⋅2^{-CT_{XA}}}{R_A(CT_{RA})⋅2^{-CT_{RA}}}=
K_A⋅ 2^{-(CT_{XA}-CT_{RA})}
\end{aligned}\]
\(K_A\) and \(K_B\) are constants that depend on the target and reference genes, and how each Signal changes with concentration
Ratio of relative expressions
\[\frac{X_B(0)}{R_B(0)}÷\frac{X_A(0)}{R_A(0)} =
\frac{K_B}{K_A} ⋅ 2^{-(\Delta CT_B-Δ CT_A)}\]
We can assume that \(K_B=K_A,\) because we are comparing the same pair of genes every time
In that case the change in relative expression is
\[\frac{X_B(0)}{R_B(0)}÷\frac{X_A(0)}{R_A(0)} = 2^{-(\Delta CT_B-Δ CT_A)}\]
Log fold change
It is usually a good idea to take logarithms
Using \(\log_2\) we get log fold change, which can be written as
\[\begin{aligned}
Δ CT_A - Δ CT_B &= \overbrace{(CT_{BX}-CT_{BR})}^{\text{before}}
- \overbrace{(CT_{AX}-CT_{AR})}^{\text{after}}\\
&=\underbrace{(CT_{BX}-CT_{AX})}_{\text{target}}
- \underbrace{(CT_{BR}-CT_{AR})}_{\text{reference}}
\end{aligned}\]
This works very well with a linear model
We can change the order of deltas
The reordering of deltas is equivalent to \[\frac{X_B(0)}{R_B(0)}÷\frac{X_A(0)}{R_A(0)} =
\frac{X_B(0)}{R_B(0)}⋅ \frac{R_A(0)}{X_A(0)} =
\frac{X_B(0)}{X_A(0)}÷\frac{R_B(0)}{R_A(0)}\]
In other words, the ratio of normalized values is also the ratio between ratios of change
(I think the formula is more clear than the text)
Primers may not be 100% efficient
In that case the standard curve slope changes
If any efficiency is not 100%
\[\frac{X_B(0)}{X_A(0)}÷\frac{R_B(0)}{R_A(0)}=
\frac{(1+E_X)^{-(CT_{BX}-CT_{AX})}}{(1+E_R)^{-(CT_{BR}-CT_{AR})}}\]
The log ratio is \[-F_X(CT_{BX}-CT_{AX})+ F_R(CT_{BR}-CT_{AR})\] where \(F_X=-\log_2 (1+E_X)\) and \(F_R=-\log_2 (1+E_R)\) are the slopes of the corresponding standard curves
Calculating the efficiencies
We know the dilutions, the CT values, and their relationship \[X(0) = X(C)⋅(1+E)^{-C}\] That is \[\text{initial}_i =
\text{threshold}⋅(1+E)^{-CT_i}\] Taking logarithms, we have \[\log(\text{initial}_i) =
\log(\text{threshold}) + \log(1+E)⋅ -CT_i\] Notice that threshold concentration should be the same for all \(i\)
We care about the slope coefficient
We can find \(E\) using a linear model like \[\log_2(\text{initial}_i) = β_0 + β_1 ⋅ CT_i + e_i\]
Fitting the linear models gives us \[β_1 = -log(1+E)\] and from there we can find the primers efficiency \(E\)
Results
For the first set of primers, we have
model <- lm(log(DNA_ini) ~ CT, CT_X)
intervals <- confint(model)
2.5 % 97.5 %
(Intercept) 0.288 0.332
CT -0.694 -0.692
We get intervals for \(\beta_0\) and \(\beta_1.\) The efficiency is
2.5 % 97.5 %
1.000 0.997
In this case the real value is 100%
Second set of primers
For the second set of primers, we have
2.5 % 97.5 %
(Intercept) 0.108 0.566
CT -0.606 -0.582
The efficiency is in the interval
2.5 % 97.5 %
0.833 0.790
The real value is 80%
It is recommended to always do the standard curve by triplicate
Threshold in the machine is not perfect
All our measuring devices have a margin of error.
There may be a small error measuring the signal.
That will affect the resulting CT value
Lower thresholds are more sensitive
When the initial concentration is small, the signal may not reach 50% in 30 or 40 cycles
Then we will use a lower threshold, and pay the price
Margin of Error
Since the curve is nearly flat in the lower signals, a small error in the signal has a large impact in the concentration
| 0.05 |
10.00 |
1025.1 |
|
| 0.06 |
10.21 |
1186.4 |
0.1573 |
| 0.50 |
13.73 |
13624.5 |
|
| 0.51 |
13.78 |
14054.3 |
0.0315 |
Thus, an error between signal 0.05 and 0.06 results in an apparent increase of 16% in concentration. But an error between signal 0.5 and 0.51 results in an error of 3%
References
Livak KJ, Schmittgen TD. “Analysis of relative gene expression data using real-time quantitative PCR and the 2(-Delta Delta C(T)) Method”. Methods. 2001 Dec; 25(4):402-8. doi: 10.1006/meth.2001.1262
Pfaffl, Michael W. “Relative Quantification.” In Real-Time PCR, Published by International University Line (Editor: T. Dorak), 64–82, 2007.
