It seems a good time to speak about qPCR

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

Why do you want to use qPCR?

- 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)

Questions are more important than answers

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

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:

The curve depends on the initial DNA 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

\[X(0) = X(C)⋅2^{-C}\]

DNA concentration crosses 50% at 13.73 cycles

Start with a large concentration of template, and dilute it several times. Measure the CT of each dilution

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

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

\[\frac{X_B(0)}{X_A(0)} = \frac{X_B(CT_B)⋅2^{-CT_B}}{X_A(CT_A)⋅2^{-CT_A}}\]

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

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)

A single pipet, at the same time

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

\[\frac{X_B(0)}{R_B(0)}÷\frac{X_A(0)}{R_A(0)} = \frac{K_B}{K_A} ⋅ 2^{-(\Delta CT_B-Δ CT_A)}\]

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

\[\frac{X_B(0)}{R_B(0)}÷\frac{X_A(0)}{R_A(0)} = 2^{-(\Delta CT_B-Δ CT_A)}\]

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*

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)

In that case the *standard curve slope* changes

If the primer efficiency is \(E_X,\) the correct formula is

\[\frac{X_B(0)}{X_A(0)} = \frac{X_B(CT_{BX})⋅(1+E_X)^{-CT_{BX}}}{X_A(CT_{AX})⋅(1+E_X)^{-CT_{AX}}} =(1+E_X)^{-(CT_{BX}-CT_{AX})}\]

The efficiencies may be different for the housekeeping gene

\[\frac{R_B(0)}{R_A(0)} = \frac{R_B(CT_{BR})⋅(1+E_R)^{-CT_{BR}}}{R_A(CT_{AR})⋅(1+E_R)^{-CT_{AR}}} =(1+E_R)^{-(CT_{BR}-CT_{AR})}\]

\[\frac{X_B(0)}{X_A(0)}÷\frac{R_B(0)}{R_A(0)}= \frac{(1+E_X)^{-(CT_{BX}-CT_{AX})}}{(1+E_R)^{-(CT_{BR}-CT_{AR})}}\]

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*

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 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 \(E\)

For the first set of primers, we have

```
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%

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

**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

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

Since the curve is nearly flat in the lower signals, a small error in the signal has a large impact in the concentration

Thr | CT | Concentration | Error.ratio |
---|---|---|---|

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%

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.