Team:Carnegie Mellon/Modelling/Documentation
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<li><a href="#section1-2">1.2 Experimental Data Analysis</a></li> | <li><a href="#section1-2">1.2 Experimental Data Analysis</a></li> | ||
<li><a href="#section1-3">1.3 Equilibrium Constants</a></li> | <li><a href="#section1-3">1.3 Equilibrium Constants</a></li> | ||
+ | <li><a href="#section1-4">1.4 Degradation</a></li> | ||
+ | <li><a href="#section1-5">1.5 mRNA Expression</a></li> | ||
+ | <li><a href="#section1-6">1.6 Protein Expression</a></li> | ||
+ | <li><a href="#section1-7">1.7 Polymerase Per Second</a></li> | ||
</ul> | </ul> | ||
</li> | </li> | ||
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</p> | </p> | ||
<p> | <p> | ||
- | We can solve for [Rf ] using a quadratic equation. | + | We can solve for [Rf ] using a quadratic equation. |
</p> | </p> | ||
<p> | <p> | ||
Line 91: | Line 95: | ||
<p> | <p> | ||
2·KDR | 2·KDR | ||
+ | </p> | ||
+ | <h1 id = "section1-4" align="center" /><div class="text-glow"><b>Degradation</b></div><br /><br /></h1> | ||
+ | |||
+ | <p> | ||
+ | Degradation occurs for both mRNA and protein. After shutting off production of mRNA/protein, one can measure the degradation coefficient. Some intuition | ||
+ | reveals that the amount that is degraded is proportional to the amount of mRNA/protein that is present. | ||
+ | </p> | ||
+ | <p> | ||
+ | d[R] =−α·[R] dt | ||
+ | </p> | ||
+ | <p> | ||
+ | Protein often has another constant attached to degradation, labeled maturation. Matu- ration (a) takes into account the time it takes for a protein to | ||
+ | mature before fluorescence can actually occur. Maturation is also dependent on the amount of protein available. In this case, the equation would be | ||
+ | </p> | ||
+ | <p> | ||
+ | d[P] =−(a+β)·[P] dt | ||
+ | </p> | ||
+ | <p> | ||
+ | However, since the fluorogen activated protein (FAP) takes a small amount of time to fold and to bind to the dye, one can make a reasonable assumption that | ||
+ | maturation is 0. So the simplified equation is | ||
+ | </p> | ||
+ | <p> | ||
+ | d[P] =−β·[P] dt | ||
+ | </p> | ||
+ | <p> | ||
+ | These equations can be solved by first order linear differential equation techniques.PROTEIN MODEL 5 | ||
+ | </p> | ||
+ | <p> | ||
+ | [R] = [R]max · e−α·t [P] = [P]max · e−β·t | ||
+ | </p> | ||
+ | <p> | ||
+ | From these equations α and β can be determined easily. | ||
+ | </p> | ||
+ | |||
+ | <h1 id = "section1-5" align="center" /><div class="text-glow"><b>mRNA Expression</b></div><br /><br /></h1> | ||
+ | |||
+ | <p> | ||
+ | From the mRNA expression equations, we know that | ||
+ | </p> | ||
+ | <p> | ||
+ | d[R] =Ts·[D]−α·[R] dt | ||
+ | </p> | ||
+ | <p> | ||
+ | Where Ts is the transcriptional efficiency and α is the degradation constant associated with mRNA degradation, and R is the mRNA concentration or amount. | ||
+ | </p> | ||
+ | <p> | ||
+ | We see next that this is a first order linear equation, as Ts, [D] and α are constants. Rearranging, we get | ||
+ | </p> | ||
+ | <p> | ||
+ | d[R] +α·[R]=Ts·[D] dt | ||
+ | </p> | ||
+ | <p> | ||
+ | The small integrating factor is thus eα·t | ||
+ | </p> | ||
+ | <p> | ||
+ | Simplifying | ||
+ | </p> | ||
+ | <p> | ||
+ | d[R] ·eα·t +α·[R]·eα·t =Ts·[D]·eα·t dt | ||
+ | </p> | ||
+ | <p> | ||
+ | d([R]·eα·t) = Ts · [D] · eα·t dt | ||
+ | </p> | ||
+ | <p> | ||
+ | [R]·eα·t =Ts·[D]·eα·t dt6 | ||
+ | </p> | ||
+ | |||
+ | <p> | ||
+ | [R]·eα·t = Ts·[D] ·eα·t +C α | ||
+ | </p> | ||
+ | <p> | ||
+ | C = −Ts·[D] α | ||
+ | </p> | ||
+ | <p> | ||
+ | [R] · eα·t = Ts·[D] · eα·t − Ts·[D] αα | ||
+ | </p> | ||
+ | <p> | ||
+ | [R] = T s·[D] − T s·[D] · e−α·t αα | ||
+ | </p> | ||
+ | <p> | ||
+ | [R]= Ts·[D] ·(1−e−α·t) α | ||
+ | </p> | ||
+ | |||
+ | <h1 id = "section1-6" align="center" /><div class="text-glow"><b>Protein Expression</b></div><br /><br /></h1> | ||
+ | |||
+ | <p> | ||
+ | The protein model is a bit different from the mRNA model due to the fact that the amount of protein depends on the amount of mRNA, which is variable. mRNA | ||
+ | is only dependent on D, which is invariable. | ||
+ | </p> | ||
+ | <p> | ||
+ | The basic equation looks like: | ||
+ | </p> | ||
+ | <p> | ||
+ | d[P] =[R]·Tl−β·[P] dt | ||
+ | </p> | ||
+ | <p> | ||
+ | Where [P] is the protein concentration or amount, [R] is still mRNA, Tl is the transla- tional efficiency, and β is the degradation constant associated | ||
+ | with the protein. | ||
+ | </p> | ||
+ | <p> | ||
+ | Conveniently, we have already solved for our only hurdle to a first order linear equation, the mRNA amount. We will substitute in for mRNA now. | ||
+ | </p> | ||
+ | <p> | ||
+ | d[P] =(1−e−α·t)·Ts·[D] ·Tl−β·[P] dt α | ||
+ | </p> | ||
+ | <p> | ||
+ | Now we can solve the first order linear equation. | ||
+ | </p> | ||
+ | <p> | ||
+ | d[P] +β·[P]=(1−e−α·t)·Ts·[D] ·Tl dt α | ||
+ | </p> | ||
+ | <p> | ||
+ | It can be seen that the integrating factor is eβ·t d[P] ·eβ·t+β·[P]·eβ·t =eβ·t·(1−e−α·t)·Ts·[D] ·Tl | ||
+ | </p> | ||
+ | <p> | ||
+ | dt α | ||
+ | </p> | ||
+ | <p> | ||
+ | d([P]·eβ·t) =eβ·t·(1−e−α·t)·Ts·[D] ·Tl dt α | ||
+ | </p> | ||
+ | <p> | ||
+ | [P]·eβ·t =(1−e−α·t)·Ts·[D] ·eβ·t·Tldt α | ||
+ | </p> | ||
+ | <p> | ||
+ | [P]·eβ·t =Tl·Ts·[D] ·eβ·t dt−Tl·Ts·[D] ·e(−α+β)·t dt αα | ||
+ | </p> | ||
+ | <p> | ||
+ | [P]·eβ·t =Tl·Ts·[D] ·eβ·t−Tl· Ts·[D] ·e(−α+β)·t+C α·β α·(−α+β) | ||
+ | </p> | ||
+ | <p> | ||
+ | Now we solve for C. When t = 0, P = 0 | ||
+ | </p> | ||
+ | <p> | ||
+ | C=−Tl·Ts·[D]·(1− 1 ) α β −α+β | ||
+ | </p> | ||
+ | <p> | ||
+ | [P]·eβ·t =Tl·Ts·[D] ·eβ·t−Tl· Ts·[D] ·e(−α+β)·t−Tl·Ts·[D] ·(1 − 1 ) α·β α·(−α+β) α β −α+β | ||
+ | </p> | ||
+ | <p> | ||
+ | |||
+ | </p> | ||
+ | <p> | ||
+ | FInally, we solve for Tl. | ||
+ | </p> | ||
+ | <p> | ||
+ | [P]=Tl·(Ts·[D]− Ts·[D] ·e−α·t−Ts·[D]·(1− 1 )·e−β·t) α·β α·(−α+β) α β −α+β | ||
+ | </p> | ||
+ | <p> | ||
+ | Tl = [P] T s·[D] − T s·[D] ·e−α·t− T s·[D] ·( 1 − 1 )·e−β·t | ||
+ | </p> | ||
+ | <p> | ||
+ | α·β α·(−α+β) α β −α+β | ||
+ | </p> | ||
+ | <p> | ||
+ | Tl is the translational efficiency, which is the second characteristic we were trying to solve for. | ||
+ | </p> | ||
+ | <p> | ||
+ | Tl = [P] T s·[D] ·(1−e−β·t)− T s·[D] ·(e−α·t−e−β·t) | ||
+ | </p> | ||
+ | <p> | ||
+ | (α·β) α·(−α+β) | ||
+ | </p> | ||
+ | |||
+ | <h1 id = "section1-7" align="center" /><div class="text-glow"><b>Polymerase Per Second</b></div><br /><br /></h1> | ||
+ | |||
+ | |||
+ | <p> | ||
+ | Taking inspiration from ”Measuring the activity of BioBrick promoters using an in vivo reference standard” by Kelly et al., we can derive our own equation | ||
+ | for polymerase per second (PoPS). | ||
+ | </p> | ||
+ | <p> | ||
+ | mRNA is produced by the number of promoters times the rate of initiations of poly- merase onto the promoters, or n · P oP S. mRNA is degraded by the | ||
+ | degradation equation we derived earlier, which is −α · [R]. | ||
+ | </p> | ||
+ | <p> | ||
+ | d[R] =n·PoPS−α·[R] dt | ||
+ | </p> | ||
+ | <p> | ||
+ | Where n is the number of promoters in a cell, PoPS is the rate of initiations of RNA polymerase onto the promoters. | ||
+ | </p> | ||
+ | <p> | ||
+ | Protein is produced by the translational efficiency times the mRNA, which is [R] · T l. Protein is degraded by the degradation equation we derived above, | ||
+ | which is −β · [P ]. | ||
+ | </p> | ||
+ | <p> | ||
+ | d[P] =[R]·Tl−β·[P]dt | ||
+ | </p> | ||
+ | <p> | ||
+ | At steady state, it can be assumed that d[R] = 0 and d[P ] = 0. | ||
+ | </p> | ||
+ | <p> | ||
+ | So we have | ||
+ | </p> | ||
+ | <p> | ||
+ | Substituting leaves: | ||
+ | </p> | ||
+ | <p> | ||
+ | dt dt | ||
+ | </p> | ||
+ | <p> | ||
+ | PoPS = α·[R] n | ||
+ | </p> | ||
+ | <p> | ||
+ | [R] = β·[P] Tl | ||
+ | </p> | ||
+ | <p> | ||
+ | PoPS = α·β·[P] n·T l | ||
+ | </p> | ||
+ | <p> | ||
+ | The output of the model is polymerase per second, which is what we have found here. It is important to realize that the purpose of finding polymerase per | ||
+ | second is that for the current environment of a promoter and the specific type of promoter, it can be characterized using polymerase per second. | ||
+ | Experiments can thus easily be conceived by running two experiments on the same promoter under different conditions to see how a promoter is affected, or | ||
+ | by running two experiments on different promoters under the same conditions to see which is a stronger promoter. | ||
</p> | </p> | ||
</body> | </body> |
Latest revision as of 23:04, 9 September 2012
Documentation Preface
The documentation of the model consists of the derivations of all the equations used to create the model. Each equation contributes a piece of the picture which ultimately results in the calculations of important cell characteristics. These equations live in the matlab model that can be found here.
Let fluorescent mRNA and protein concentration be represented by [Rf] and [Pf] respectively. They are related directly to the fluorescence level, which we will label Fr and Fp. The mRNA and protein concentration levels can be measured by just the fluorescence. To do this, we will abide by the assumptions that
Where Sr and Sp are scaling factors for mRNA and protein respectively and kr and kp are constants that transform fluorescence to mRNA and protein concentrations.
In the experiment, one uses a plate reader with varying concentration of the dyes in rows and varying time measurements in columns. The following image represents this.
We will also have another row for in vitro measurements. From this row we will graph the fluorescence versus the dye concentration, and the fluorescence will level off at some saturation point. Because the saturation point in vitro will be greater than the saturation point in vivo, we must scale all the fluorescence measurements we find in vivo, which is the importance of Sr and Sp.
At this point we will find out the scaling factors Sr and Sp. Step 1 is to put samples into the plate reader and take more samples of the same concentration and measure them in vitro. Then, we will measure all the wells at the same time point, and find the saturation fluorescence of the in vitro and the in vivo wells. Dividing the two gives us the Sr and Sp.
At each time point we will graph the in vivo fluorescence vs dye concentrations and find the first dye concentration where saturation occurs. This dye concentration is thus the mRNA/protein total concentration, as we will assume that there will be a 1-1 correspon- dence of dye and mRNA/protein. We then multiply each by the scaling factor Sr or Sp to get the actual mRNA.
To check, we can find the fluorescent mRNA concentrations from the mRNA values we obtained in vivo. General first order chemical reactions begin (theoretically):
Where [AB] = 0. So then, the equation becomes:
(α[A] − γ[AB]) + (β[B] − γ[AB]) ↔ γ[AB]
We will assume that α, β, and γ are all equal to 1. Our [A] will be mRNA/protein and [B] will be the dye concentrations. mRNA dye, which is DFHBI, will be [DD] and protein dye, which is malachite green, will be [DM ]. Our equations are thus:
([R]i −[Rf])+([DD]−[Rf])↔([Rf]) ([P]i −[Pf])+([DM]−[Pf])↔([Pf])
Then, the equilibrium constant KD is then defined as: KD = [AB]
R ([A]0−[AB])([B]0−[AB]) Now inputting our variables for mRNA expression,
KD = [Rf] R ([R]0−[Rf ])([DR]0−[Rf ])
We can solve for [Rf ] using a quadratic equation.
[Rf]2 ·KDR −[Rf]·[KDR([R]+DR)+1]+KDR ·[R]·DR =0 [Rf ] = [KDR ([R][DR ])+1]±√[KDR ([R][DR ])+1]2 −4·(KDR )·(KDR [R][DR ])
And similarly for protein.
2·KDR
Degradation occurs for both mRNA and protein. After shutting off production of mRNA/protein, one can measure the degradation coefficient. Some intuition reveals that the amount that is degraded is proportional to the amount of mRNA/protein that is present.
d[R] =−α·[R] dt
Protein often has another constant attached to degradation, labeled maturation. Matu- ration (a) takes into account the time it takes for a protein to mature before fluorescence can actually occur. Maturation is also dependent on the amount of protein available. In this case, the equation would be
d[P] =−(a+β)·[P] dt
However, since the fluorogen activated protein (FAP) takes a small amount of time to fold and to bind to the dye, one can make a reasonable assumption that maturation is 0. So the simplified equation is
d[P] =−β·[P] dt
These equations can be solved by first order linear differential equation techniques.PROTEIN MODEL 5
[R] = [R]max · e−α·t [P] = [P]max · e−β·t
From these equations α and β can be determined easily.
From the mRNA expression equations, we know that
d[R] =Ts·[D]−α·[R] dt
Where Ts is the transcriptional efficiency and α is the degradation constant associated with mRNA degradation, and R is the mRNA concentration or amount.
We see next that this is a first order linear equation, as Ts, [D] and α are constants. Rearranging, we get
d[R] +α·[R]=Ts·[D] dt
The small integrating factor is thus eα·t
Simplifying
d[R] ·eα·t +α·[R]·eα·t =Ts·[D]·eα·t dt
d([R]·eα·t) = Ts · [D] · eα·t dt
[R]·eα·t =Ts·[D]·eα·t dt6
[R]·eα·t = Ts·[D] ·eα·t +C α
C = −Ts·[D] α
[R] · eα·t = Ts·[D] · eα·t − Ts·[D] αα
[R] = T s·[D] − T s·[D] · e−α·t αα
[R]= Ts·[D] ·(1−e−α·t) α
The protein model is a bit different from the mRNA model due to the fact that the amount of protein depends on the amount of mRNA, which is variable. mRNA is only dependent on D, which is invariable.
The basic equation looks like:
d[P] =[R]·Tl−β·[P] dt
Where [P] is the protein concentration or amount, [R] is still mRNA, Tl is the transla- tional efficiency, and β is the degradation constant associated with the protein.
Conveniently, we have already solved for our only hurdle to a first order linear equation, the mRNA amount. We will substitute in for mRNA now.
d[P] =(1−e−α·t)·Ts·[D] ·Tl−β·[P] dt α
Now we can solve the first order linear equation.
d[P] +β·[P]=(1−e−α·t)·Ts·[D] ·Tl dt α
It can be seen that the integrating factor is eβ·t d[P] ·eβ·t+β·[P]·eβ·t =eβ·t·(1−e−α·t)·Ts·[D] ·Tl
dt α
d([P]·eβ·t) =eβ·t·(1−e−α·t)·Ts·[D] ·Tl dt α
[P]·eβ·t =(1−e−α·t)·Ts·[D] ·eβ·t·Tldt α
[P]·eβ·t =Tl·Ts·[D] ·eβ·t dt−Tl·Ts·[D] ·e(−α+β)·t dt αα
[P]·eβ·t =Tl·Ts·[D] ·eβ·t−Tl· Ts·[D] ·e(−α+β)·t+C α·β α·(−α+β)
Now we solve for C. When t = 0, P = 0
C=−Tl·Ts·[D]·(1− 1 ) α β −α+β
[P]·eβ·t =Tl·Ts·[D] ·eβ·t−Tl· Ts·[D] ·e(−α+β)·t−Tl·Ts·[D] ·(1 − 1 ) α·β α·(−α+β) α β −α+β
FInally, we solve for Tl.
[P]=Tl·(Ts·[D]− Ts·[D] ·e−α·t−Ts·[D]·(1− 1 )·e−β·t) α·β α·(−α+β) α β −α+β
Tl = [P] T s·[D] − T s·[D] ·e−α·t− T s·[D] ·( 1 − 1 )·e−β·t
α·β α·(−α+β) α β −α+β
Tl is the translational efficiency, which is the second characteristic we were trying to solve for.
Tl = [P] T s·[D] ·(1−e−β·t)− T s·[D] ·(e−α·t−e−β·t)
(α·β) α·(−α+β)
Taking inspiration from ”Measuring the activity of BioBrick promoters using an in vivo reference standard” by Kelly et al., we can derive our own equation for polymerase per second (PoPS).
mRNA is produced by the number of promoters times the rate of initiations of poly- merase onto the promoters, or n · P oP S. mRNA is degraded by the degradation equation we derived earlier, which is −α · [R].
d[R] =n·PoPS−α·[R] dt
Where n is the number of promoters in a cell, PoPS is the rate of initiations of RNA polymerase onto the promoters.
Protein is produced by the translational efficiency times the mRNA, which is [R] · T l. Protein is degraded by the degradation equation we derived above, which is −β · [P ].
d[P] =[R]·Tl−β·[P]dt
At steady state, it can be assumed that d[R] = 0 and d[P ] = 0.
So we have
Substituting leaves:
dt dt
PoPS = α·[R] n
[R] = β·[P] Tl
PoPS = α·β·[P] n·T l
The output of the model is polymerase per second, which is what we have found here. It is important to realize that the purpose of finding polymerase per second is that for the current environment of a promoter and the specific type of promoter, it can be characterized using polymerase per second. Experiments can thus easily be conceived by running two experiments on the same promoter under different conditions to see how a promoter is affected, or by running two experiments on different promoters under the same conditions to see which is a stronger promoter.