Advanced Protein Models - Clemson CECAScecas.clemson.edu/.../L37-AdvancedProteinModelsOne.pdf · Starting Calculus for Biologists Advanced Protein Models Now that we have a bit more

Starting Calculus for Biologists

Advanced Protein Models

James K. Peterson

Department of Biological Sciences and Department of Mathematical SciencesClemson University

March 27, 2014


Outline

1 Advanced Protein Models

2 The Bound Fraction

3 Transcription Regulation



Abstract

This lecture is going to talk about advanced protein models.



Now that we have a bit more mathematics under our belts,let’s go back and look at proteins again. A great review ofthis, with lots of great pictures and diagrams, is in the paperby Kim Sneppen, Sandeep Krishna and Szabolcs Semsey,“Simplified Models of Biological Networks” in the AnnualReview of Biophysics, 2010. It is well worth your time tocheck it out.

We will do some of what is discussed in this paper and useappropriate MatLab to get some insight, but view this chapteras a pointer to more fascinating stuff!

As you know from your first biology courses, the inside of thecell is quite crowded. There are protein filaments that guidethe movement of other proteins, proteins that help the aminoacid strands coming out of the ribosome to coil properly sothat they can take their correct shape and so on.













Water is a very complicated substance with interesting andcomplex abilities to assemble into cages because its electriccharge distribution is not symmetric. Oxygen and twohydrogens give a molecule with a definite minus to plus look.You should read another survey paper on this to get a betterunderstanding. Look at Pascale Mentre’s paper on water inthe cell, “Water in the orchestration of the cell machinery.Some misunderstandings: a short review”, The Journal ofBiological Physics, 2012, to get a nice overview.

Regulation of protein transcription is very dynamic astranscription factors TF’s rapidly associate and disassociate tothe sites on the portion of DNA that has been exposed fromthe genome.

We looked at some of this action earlier and we called thisspecial site the promoter. Now a single transcription factormoves slowly across a cell because it slowly drifts by bouncingoff other molecules.













Imagine a tiny robotic submarine being driven in a pond justchock full of debris, living plants and so on. If it was smallenough it would even bounce off of tadpoles and small fish! Ifit was smaller yet, it would bounce off of small multi cellularcreatures too. You can see the progress of the tiny submarinewould be slow and halting and even though the captain of thesubmarine is trying to move in a straight line towards hertarget, she can’t really do that. Instead the submarinebounces around and finds the target in an indirect way. Well,the transcription factor is like the tiny submarine and wemeasure the speed at which the TF moves in the cell withwhat is called the diffusion coefficient, D.

This coefficient is measured in units ofmolecules of TF × (µm)2/ second where the symbol µm is

called a micrometer and has the value µm = 10−6m.Typically D ≈ 5− 50molecules(µm)2/s.



Imagine a tiny robotic submarine being driven in a pond justchock full of debris, living plants and so on. If it was smallenough it would even bounce off of tadpoles and small fish! Ifit was smaller yet, it would bounce off of small multi cellularcreatures too. You can see the progress of the tiny submarinewould be slow and halting and even though the captain of thesubmarine is trying to move in a straight line towards hertarget, she can’t really do that. Instead the submarinebounces around and finds the target in an indirect way. Well,the transcription factor is like the tiny submarine and wemeasure the speed at which the TF moves in the cell withwhat is called the diffusion coefficient, D.

This coefficient is measured in units ofmolecules of TF × (µm)2/ second where the symbol µm is

called a micrometer and has the value µm = 10−6m.Typically D ≈ 5− 50molecules(µm)2/s.



The transcription factor is trying to hit a specific promoterwhose diameter, a, is on average is about 5 nm. Here, theunit n m is called a nano meter and it is 10−3 µm is size. Thepromoter is hidden in the cell which has volume V. So theproduct D a has units of molecules(µm)3/sec and sorepresents a kind of search speed.

If we divide this by the volume of the cell, we divide by theunits (µm)3 and so the fraction V

D a gives an estimate of howlong it would take to search through the entire cell volume tofind the target promoter. A typical cell has volume 1 (µm)3).So, the time to find the target is roughly τon ∝ V

D a .

and careful reasoning which you can find in biophysics textstells us the proportionality constant is 1/4π – not toosurprising as the TF is moving through a sphere and thevolume of a sphere is (4/3)πa3 here. So we have τon = V

4π D a .






D a .


4π D a .






D a .


4π D a .



Plugging in our values, since V = 4/3π(13) (µm))3 orV = 1.33(µm)3, we have

τon = 523.64π (5 to 50) (5× 10−3) sec

=1.33

(0.100 to 1.0)sec

= 1.33 to 13.3 sec.

Next, if you think about it, it is easy to see that it costs energy forthe transcription factor TF to break off the promoter. We won’t gothrough how we come with a quantitative estimate here. Instead,we will just say the the time it takes for TF to break off isproportional to τon. We can show τoff ∝ τon

K . where K is a numberwhich is related to the binding energy. The proportionality constantthen turns out to be 1/(6× 108). For a reasonably strong bindingenergy of about −13 kcal / mole , where kcal is a unit of kilocalories which is a measure of energy used and mole is a term youprobably saw in your chemistry class which represents 6.02× 1023

molecules, we find K = 10−9.



Plugging in our values, since V = 4/3π(13) (µm))3 orV = 1.33(µm)3, we have

τon = 523.64π (5 to 50) (5× 10−3) sec

=1.33

(0.100 to 1.0)sec

= 1.33 to 13.3 sec.

Next, if you think about it, it is easy to see that it costs energy forthe transcription factor TF to break off the promoter. We won’t gothrough how we come with a quantitative estimate here. Instead,we will just say the the time it takes for TF to break off isproportional to τon. We can show τoff ∝ τon

K . where K is a numberwhich is related to the binding energy. The proportionality constantthen turns out to be 1/(6× 108). For a reasonably strong bindingenergy of about −13 kcal / mole , where kcal is a unit of kilocalories which is a measure of energy used and mole is a term youprobably saw in your chemistry class which represents 6.02× 1023

molecules, we find K = 10−9.



So we have in general

τoff =τon

6× 108 K

And for a binding energy of 13 kcal / mole, K = 10−9, we get

τoff =τon

0.6= 1.67 τon.

So both τon and τoff have nominal ranges of 1− 23 seconds. Ourpurpose here is just to make you see the rough time order for the TFsearch and bind to the promoter and its release from the promoter.The back and forth bind and release takes place a small fraction ofa minute. We also know from experiments and theoretical reasoningthat protein transcription takes time on the order of minutes.

So the transcription factor binding and releasing is occurring on atime scale that is significantly less than the time scale we have forfull protein transcription. Signal changes to the promoter, i.e. TFmovements due to external signals that release the moleculesneeded to release the TF, occur much faster than the time scale wesee when protein production is at equilibrium.




τoff =τon

6× 108 K


τoff =τon

0.6= 1.67 τon.






τoff =τon

6× 108 K


τoff =τon

0.6= 1.67 τon.




The Bound Fraction

Let the promoter be denoted by P and the TF plus promotercomplex by TFP. We have the following reaction

k[P] + [TF] −→ [TFP]

where k is the rate at which P and TF combine to form thecomplex TFP. The complex also breaks apart at the rate k ′ whichwe denote in equation form as

k ′

[TFP] −→ [P] + [TF] .

Combining, we have the model

k[P] + [TF] ←→ [TFP]

k ′


The Bound Fraction


k[P] + [TF] −→ [TFP]


k ′

[TFP] −→ [P] + [TF] .


k[P] + [TF] ←→ [TFP]

k ′


The Bound Fraction


k[P] + [TF] −→ [TFP]


k ′

[TFP] −→ [P] + [TF] .


k[P] + [TF] ←→ [TFP]

k ′


The Bound Fraction

At equilibrium, the rate at which TFP forms must equal the rate atwhich TFP breaks apart. The concentration of TFP is written as[TFP]. The concentrations of P and TF are then [P] and [TF].The amount of TFP depends on how much of the needed recipeingredients are available. Hence the amount made is k [P] [TF].

The amount of P and TF made because TFP breaks apart dependson how much TFP is available which is k ′ [TFP]. So atequilibrium, because the formation rate and disassociation rate arethe same that we must have a balance

k ′ [TFP] = k [P] [TF]

Solving we find the relationship

[TFP] =k

k ′[P] [TF]

We call the fraction k′

k the disassociation constant K and this isthe same variable we use earlier which had to do with the bindingenergy!


The Bound Fraction



k ′ [TFP] = k [P] [TF]


[TFP] =k

k ′[P] [TF]




The Bound Fraction



k ′ [TFP] = k [P] [TF]


[TFP] =k

k ′[P] [TF]




The Bound Fraction

We see the fraction of TF bound to the promoter can bewritten as

bound fraction =[TFP]

[P] + [TFP]

=1K [P] [TF]

[P] + [TF].

Now simplify a bit to get


[P] + [TFP]=

K [TFP]

K [P] + K [TFP]

=[P] [TF]

K [P] + [P] [TF]=

[TF]

K + [TF]

=[TF] /K

1 + [TF] /K.


The Bound Fraction

We see the fraction of TF bound to the promoter can bewritten as


[P] + [TFP]

=1K [P] [TF]

[P] + [TF].

Now simplify a bit to get


[P] + [TFP]=

K [TFP]

K [P] + K [TFP]

=[P] [TF]

K [P] + [P] [TF]=

[TF]

K + [TF]

=[TF] /K

1 + [TF] /K.


The Bound Fraction

So we express the disassociation constant K in two ways to getinsight. The first uses energy and movement ideas to estimate howmuch time it takes for the transcription factor and the promoterbinding to reach equilibrium – tens of seconds at most. Thesecond uses familiar reaction ideas to estimate the fraction oftranscription factor that is bound to the promoter. If we let Rdenote the concentration of our transcription factor and K denoteits disassociation constant, we have a generic relationship.

bound fraction =RK

1 + RK

.


The Bound Fraction

We can do a similar bit of analysis to figure out the unboundfraction.

unbound fraction =[P]

[P] + [TFP]=

K [P]

K [P] + K [TFP]

=K [P]

K [P] + [P] [TF]=

K

K + [TF]

=1

1 + [TF]K

.

Hence, the unbound fraction, using R to denote the concentrationof the transcription factor is

unbound fraction =1

1 + RK


The Bound Fraction

It turns out we can do more. Sometimes transcription factors bindcooperatively or other factors can bind to the transcription factor prior tobinding to the promoter. We can model this sort of thing too, but thedetails are not for us here. In such cases, the bound and unboundfractions can be represented by adding a power b to the term R

K . Hence,the more general fractions are

bound fraction =

(RK

)b

1 +

(RK

)b.

unbound fraction =1

1 +

(RK

)b

where b ≥ 1 is called a Hill coefficient.


The Bound Fraction

We think of the ratio RK in this way. If the amount of [TF] exactly

matched the disassociation rate K, then the ratio would be 1 andthe bound and unbound fraction are both 1/2. Just like theprobability of being bound and unbound is exactly the same.However, the mixture can easily be skewed by simply letting [TF]be rK for different values of r . If r � 1, the bound fraction movespast 1/2 and is closer to saturation: the bound fraction is 1.Similarly, if r � 1, the bound fraction is quite small with theunbound fraction close to 1.

Example

Calculate the bound and unbound fractions assuming the fractionR/K is 0.2 for a Hill’s coefficient b = 1.


The Bound Fraction

Solution

We know

bound fraction =

(RK

)b

1 +

(RK

)b.

=0.2

1 + 0.2= 0.1667

unbound fraction =1

1 +

(RK

)b

=1

1 + 0.2= 0.8333

That’s all there is to it! So most of the transcription factor isunbound here.


The Bound Fraction

Homework 57

57.1 Calculate the bound and unbound fractions assuming thefractionR/K is 0.6 for a Hill’s coefficient b = 1.





Transcription Regulation

The transcription factor can activate or repress a promoter. If thereis activation, then

activation ∝ bound fraction

∝

(RK

)b

1 +

(RK

)b.

but if there is repression, then

repression ∝ unbound fraction

∝ 1

1 +

(RK

)b.

For convenience, let x = RK . Let’s graph the some of these

activation and repression functions. We’ll use MatLab. Here is asimple session.





∝

(RK

)b

1 +

(RK

)b.



∝ 1

1 +

(RK

)b.







∝

(RK

)b

1 +

(RK

)b.



∝ 1

1 +

(RK

)b.





f = @( b , x ) ( ( x ) . ˆ b . / ( 1+ ( x ) . ˆ b ) ) ;X = l i n s p a c e ( 0 , 1 0 , 1 0 1 ) ;b = 1 ;p l o t (X , f ( 1 ,X) ) ;x l a b e l ( ’ x = [ TF ] /K’ ) ;y l a b e l ( ’ a c t i v a t i o n ’ ) ;t i t l e ( ’ a c t i v a t i o n w i t h b = 1 ’ ) ;

You can see the plot in the next figure. This curve will beasymptotic to 1 as [TF] /K get large. Now if you think of [TF] asfixed, varying K means we are changing the effectiveness of thebinding. If K is large, the fraction is small and the activation issmall. On the other hand, as K decreases, the fraction increasesand the activation increases towards 1.





Consider the activation graphs for b = 2, b = 3 and b = 4 on the samegraph as shown in in the graph below. Note that all pass through thesame pair (1, 1/2) but approach the saturation value of 1 in differentways. Effectively, increasing b ramps protein production up to itsmaximum value quicker. Another way of looking at it is that a largerHill’s coefficient b makes the activation more sensitive to variations inthe [TF] level. As [TF] moves past K, production ramps quickly.



Indeed, if b is very large, the activation is essentially a step function whichmoves from 0 to 1 very fast indeed. This is shown in the next figure.



Given all we have discussed, we are now ready to show youour new protein transcription model. We start with the genebeing repressed.

C ′ = Λ +γ

1 +

(RK

)b− 1

τC , C (0) = C0; (1)

The new terms in our equation need to be defined:

Λ is the leakage production; i.e. it is possible that the genecannot be fully repressed and so there is some productionalways.γ is the maximum production rate over the base rate set bythe leakage value Λ.τ is the lifetime of the protein which is essentially the termthat includes all the losses that occur due to degradation, cellvolume increase and so forth.



Given all we have discussed, we are now ready to show youour new protein transcription model. We start with the genebeing repressed.

C ′ = Λ +γ

1 +

(RK

)b− 1

τC , C (0) = C0; (1)

The new terms in our equation need to be defined:

Λ is the leakage production; i.e. it is possible that the genecannot be fully repressed and so there is some productionalways.γ is the maximum production rate over the base rate set bythe leakage value Λ.τ is the lifetime of the protein which is essentially the termthat includes all the losses that occur due to degradation, cellvolume increase and so forth.



This is still a standard x ′ = −αx + β model though. We know thesteady state C repression

SS is β/α. Here

α =1

τ, β = Λ +

γ

1 +

(RK

)b

and so

C repressionSS = τ

Λ +γ

1 +

(RK

)b

. (2)

and the response time is therefore tR = τ ln(2).




SS is β/α. Here

α =1

τ, β = Λ +

γ

1 +

(RK

)b

and so

C repressionSS = τ

Λ +γ

1 +

(RK

)b

. (2)





SS is β/α. Here

α =1

τ, β = Λ +

γ

1 +

(RK

)b

and so

C repressionSS = τ

Λ +γ

1 +

(RK

)b

. (2)




The production is fully repressed when R � K , say R = 20Kat which point the production model isC ′ = Λ + 0− C/τ = Λ− C/τ . This fully repressed model has

steady state C full repressionSS = τ Λ.

On the other hand, if there is no repression, i.e. R � K , sayR = .01K , then the minimally repressed model isC ′ = Λ + γ − C/τ with steady state Cno repression

SS = τ(Λ + γ).

How quickly production switches between these extremesdepends on the Hill coefficient b as can be seen in the figurebelow.






SS = τ(Λ + γ).







SS = τ(Λ + γ).




The code to generate this figure is as follows:

g = @( b , tau , Lambda , gamma , x ) tau . ∗ ( Lambda+gamma . / ( 1 + x . ˆ b ) ) ;

X = l i n s p a c e ( 0 , 2 0 , 1 0 1 ) ;b = 2 ;Lambda = 1 ;gamma = 1 0 ;tau = 0 . 5 ;p l o t (X , g ( b , tau , Lambda , gamma , X) ) ;

We see in that the production is quickly shut off.





We can also activate protein production by allowing thetranscription factor to activate production. The model is similar.

C ′ = Λ + γ

(RK

)b

1 +

(RK

)b− 1

τC , C (0) = C0. (3)

We know the steady state C activationSS is β/α. Here

α =1

τ, β = Λ + γ

(RK

)b

1 +

(RK

)b



We can also activate protein production by allowing thetranscription factor to activate production. The model is similar.

C ′ = Λ + γ

(RK

)b

1 +

(RK

)b− 1

τC , C (0) = C0. (3)

We know the steady state C activationSS is β/α. Here

α =1

τ, β = Λ + γ

(RK

)b

1 +

(RK

)b



So

C activationSS = τ

Λ + γ

(RK

)b

1 +

(RK

)b

. (4)



Example

Calculate C repressionSS and C activation

SS for Λ = 4, γ = 2, b = 4, τ = 1 andthe ratio R/K = 0.1, 1, 2, 5, 20. Comment on your results.

Solution

R/K = 0.1.

C repressionSS = τ

Λ +γ

1 +

(RK

)b

= 4 +2

1 + (.1)4= 5.9998

C activationSS = τ

Λ + γ

(RK

)b

1 +

(RK

)b

= 4 + 2(.1)4

1 + (.1)4= 4.0002.



Solution

R/K = 1.

C repressionSS = τ

Λ +γ

1 +

(RK

)b

= 4 +2

1 + (1)4= 5

C activationSS = τ

Λ + γ

(RK

)b

1 +

(RK

)b

= 4 + 2(1)4

1 + (1)4= 5.



Solution

R/K = 2.

C repressionSS = τ

Λ +γ

1 +

(RK

)b

= 4 +2

1 + (2)4= 4.11765

C activationSS = τ

Λ + γ

(RK

)b

1 +

(RK

)b

= 4 + 2(2)4

1 + (2)4= 5.88235.



Solution

R/K = 5.

C repressionSS = τ

Λ +γ

1 +

(RK

)b

= 4 +2

1 + (5)4= 4.00001

C activationSS = τ

Λ + γ

(RK

)b

1 +

(RK

)b

= 4 + 2(5)4

1 + (5)4= 5.9984.



Solution

R/K = 20.0.

C repressionSS = τ

Λ +γ

1 +

(RK

)b

= 4 +2

1 + (20)4= 4.0000

C activationSS = τ

Λ + γ

(RK

)b

1 +

(RK

)b

= 4 + 2(20)4

1 + (20)4= 6.000.

Note both the unbound and bound are in the range [4, 6]



Example

Plot the steady state for the repression model for Λ = 0, γ = 5,b = 1 and τ = 1 as a function of R/K . Comment on your results.We don’t show the plot here.

Solution

g = @( b , tau , Lambda , gamma , x ) tau . ∗ ( Lambda+gamma . / ( 1 + x . ˆ b ) ) ;

X = l i n s p a c e ( 0 , 2 0 , 1 0 1 ) ;b = 1 . 0 ;Lambda = 0 ;gamma = 5 ;tau = 1 . 0 ;p l o t (X , g ( b , tau , Lambda , gamma , X) ) ;



Homework 58

58.1 Calculate C repressionSS and C activation

SS for Λ = 2, γ = 3, b = 2,τ = 2 and the ratio R/K = 0.1, 1, 2, 5, 20. Comment on yourresults.

58.2 Calculate C repressionSS and C activation

SS for Λ = 5, γ = 2, b = 3,τ = 1 and the ratio R/K = 0.1, 1, 2, 5, 20. Comment on yourresults.

58.3 Plot the steady state for the repression model for Λ = 2,γ = 3, b = 2 and τ = 2 as a function of R/K . Comment onyour results.

58.4 Plot the steady state for the repression model for Λ = 0,γ = 1, b = 1 and τ = 1 as a function of R/K . Comment onyour results.

Documents

Advanced Protein Models - Clemson CECAScecas.clemson.edu/.../L37-AdvancedProteinModelsOne.pdf · Starting Calculus for Biologists Advanced Protein Models Now that we have a bit more