Advanced Protein Models

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1 Advanced Protein Models James. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University March 27, 2014

2 Outline 1 Advanced Protein Models 2 The Bound Fraction 3 Transcription egulation

3 Advanced Protein Models Abstract This lecture is going to talk about advanced protein models.

4 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 of this, with lots of great pictures and diagrams, is in the paper by im Sneppen, Sandeep rishna and Szabolcs Semsey, Simplified Models of Biological Networks in the Annual eview of Biophysics, It is well worth your time to check it out.

5 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 of this, with lots of great pictures and diagrams, is in the paper by im Sneppen, Sandeep rishna and Szabolcs Semsey, Simplified Models of Biological Networks in the Annual eview of Biophysics, It is well worth your time to check it out. We will do some of what is discussed in this paper and use appropriate MatLab to get some insight, but view this chapter as a pointer to more fascinating stuff!

6 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 of this, with lots of great pictures and diagrams, is in the paper by im Sneppen, Sandeep rishna and Szabolcs Semsey, Simplified Models of Biological Networks in the Annual eview of Biophysics, It is well worth your time to check it out. We will do some of what is discussed in this paper and use appropriate MatLab to get some insight, but view this chapter as a pointer to more fascinating stuff! As you know from your first biology courses, the inside of the cell is quite crowded. There are protein filaments that guide the movement of other proteins, proteins that help the amino acid strands coming out of the ribosome to coil properly so that they can take their correct shape and so on.

7 Advanced Protein Models Water is a very complicated substance with interesting and complex abilities to assemble into cages because its electric charge distribution is not symmetric. Oxygen and two hydrogens give a molecule with a definite minus to plus look. You should read another survey paper on this to get a better understanding. Look at Pascale Mentré s paper on water in the cell, Water in the orchestration of the cell machinery. Some misunderstandings: a short review, The Journal of Biological Physics, 2012, to get a nice overview.

8 Advanced Protein Models Water is a very complicated substance with interesting and complex abilities to assemble into cages because its electric charge distribution is not symmetric. Oxygen and two hydrogens give a molecule with a definite minus to plus look. You should read another survey paper on this to get a better understanding. Look at Pascale Mentré s paper on water in the cell, Water in the orchestration of the cell machinery. Some misunderstandings: a short review, The Journal of Biological Physics, 2012, to get a nice overview. egulation of protein transcription is very dynamic as transcription factors TF s rapidly associate and disassociate to the sites on the portion of DNA that has been exposed from the genome.

9 Advanced Protein Models Water is a very complicated substance with interesting and complex abilities to assemble into cages because its electric charge distribution is not symmetric. Oxygen and two hydrogens give a molecule with a definite minus to plus look. You should read another survey paper on this to get a better understanding. Look at Pascale Mentré s paper on water in the cell, Water in the orchestration of the cell machinery. Some misunderstandings: a short review, The Journal of Biological Physics, 2012, to get a nice overview. egulation of protein transcription is very dynamic as transcription factors TF s rapidly associate and disassociate to the sites on the portion of DNA that has been exposed from the genome. We looked at some of this action earlier and we called this special site the promoter. Now a single transcription factor moves slowly across a cell because it slowly drifts by bouncing off other molecules.

10 Advanced Protein Models Imagine a tiny robotic submarine being driven in a pond just chock full of debris, living plants and so on. If it was small enough it would even bounce off of tadpoles and small fish! If it was smaller yet, it would bounce off of small multi cellular creatures too. You can see the progress of the tiny submarine would be slow and halting and even though the captain of the submarine is trying to move in a straight line towards her target, she can t really do that. Instead the submarine bounces around and finds the target in an indirect way. Well, the transcription factor is like the tiny submarine and we measure the speed at which the TF moves in the cell with what is called the diffusion coefficient, D.

11 Advanced Protein Models Imagine a tiny robotic submarine being driven in a pond just chock full of debris, living plants and so on. If it was small enough it would even bounce off of tadpoles and small fish! If it was smaller yet, it would bounce off of small multi cellular creatures too. You can see the progress of the tiny submarine would be slow and halting and even though the captain of the submarine is trying to move in a straight line towards her target, she can t really do that. Instead the submarine bounces around and finds the target in an indirect way. Well, the transcription factor is like the tiny submarine and we measure the speed at which the TF moves in the cell with what is called the diffusion coefficient, D. This coefficient is measured in units of molecules of TF µm) 2 / second where the symbol µm is called a micrometer and has the value µm = 10 6 m. Typically D 5 50 molecules µm) 2 /s.

12 Advanced Protein Models The transcription factor is trying to hit a specific promoter whose diameter, a, is on average is about 5 nm. Here, the unit n m is called a nano meter and it is 10 3 µm is size. The promoter is hidden in the cell which has volume V. So the product D a has units of moleculesµm) 3 /sec and so represents a kind of search speed.

13 Advanced Protein Models The transcription factor is trying to hit a specific promoter whose diameter, a, is on average is about 5 nm. Here, the unit n m is called a nano meter and it is 10 3 µm is size. The promoter is hidden in the cell which has volume V. So the product D a has units of moleculesµm) 3 /sec and so represents a kind of search speed. If we divide this by the volume of the cell, we divide by the units µm) 3 and so the fraction V D a gives an estimate of how long it would take to search through the entire cell volume to find the target promoter. A typical cell has volume 1 µm) 3 ). So, the time to find the target is roughly τ on V D a.

14 Advanced Protein Models The transcription factor is trying to hit a specific promoter whose diameter, a, is on average is about 5 nm. Here, the unit n m is called a nano meter and it is 10 3 µm is size. The promoter is hidden in the cell which has volume V. So the product D a has units of moleculesµm) 3 /sec and so represents a kind of search speed. If we divide this by the volume of the cell, we divide by the units µm) 3 and so the fraction V D a gives an estimate of how long it would take to search through the entire cell volume to find the target promoter. A typical cell has volume 1 µm) 3 ). So, the time to find the target is roughly τ on V D a. and careful reasoning which you can find in biophysics texts tells us the proportionality constant is 1/4π not too surprising as the TF is moving through a sphere and the volume of a sphere is 4/3)πa 3 here. So we have τ on = V 4π D a.

15 Advanced Protein Models Plugging in our values, since V = 4/3π1 3 ) µm)) 3 or V = 1.33µm) 3, we have τ on = π 5 to 50) ) sec = to 1.0) sec = 1.33 to 13.3 sec.

16 Advanced Protein Models Plugging in our values, since V = 4/3π1 3 ) µm)) 3 or V = 1.33µm) 3, we have τ on = π 5 to 50) ) sec = 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 for the transcription factor TF to break off the promoter. We won t go through how we come with a quantitative estimate here. Instead, we will just say the the time it takes for TF to break off is proportional to τ on. We can show τ off τon. where is a number which is related to the binding energy. The proportionality constant then turns out to be 1/ ). For a reasonably strong binding energy of about 13 kcal / mole, where kcal is a unit of kilo calories which is a measure of energy used and mole is a term you probably saw in your chemistry class which represents molecules, we find = 10 9.

17 Advanced Protein Models So we have in general τ off = τ on And for a binding energy of 13 kcal / mole, = 10 9, we get τ off = τ on 0.6 = 1.67 τ on.

18 Advanced Protein Models So we have in general τ off = τ on And for a binding energy of 13 kcal / mole, = 10 9, we get τ off = τ on 0.6 = 1.67 τ on. So both τ on and τ off have nominal ranges of 1 23 seconds. Our purpose here is just to make you see the rough time order for the TF search and bind to the promoter and its release from the promoter. The back and forth bind and release takes place a small fraction of a minute. We also know from experiments and theoretical reasoning that protein transcription takes time on the order of minutes.

19 Advanced Protein Models So we have in general τ off = τ on And for a binding energy of 13 kcal / mole, = 10 9, we get τ off = τ on 0.6 = 1.67 τ on. So both τ on and τ off have nominal ranges of 1 23 seconds. Our purpose here is just to make you see the rough time order for the TF search and bind to the promoter and its release from the promoter. The back and forth bind and release takes place a small fraction of a minute. We also know from experiments and theoretical reasoning that protein transcription takes time on the order of minutes. So the transcription factor binding and releasing is occurring on a time scale that is significantly less than the time scale we have for full protein transcription. Signal changes to the promoter, i.e. TF movements due to external signals that release the molecules needed to release the TF, occur much faster than the time scale we see when protein production is at equilibrium.

20 The Bound Fraction Let the promoter be denoted by P and the TF plus promoter complex by TFP. We have the following reaction k [P] + [TF] [TFP]

21 The Bound Fraction Let the promoter be denoted by P and the TF plus promoter complex by TFP. We have the following reaction k [P] + [TF] [TFP] where k is the rate at which P and TF combine to form the complex TFP. The complex also breaks apart at the rate k which we denote in equation form as k [TFP] [P] + [TF].

22 The Bound Fraction Let the promoter be denoted by P and the TF plus promoter complex by TFP. We have the following reaction k [P] + [TF] [TFP] where k is the rate at which P and TF combine to form the complex TFP. The complex also breaks apart at the rate k which we denote in equation form as Combining, we have the model k [TFP] [P] + [TF]. k [P] + [TF] [TFP] k

23 The Bound Fraction At equilibrium, the rate at which TFP forms must equal the rate at which 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 recipe ingredients are available. Hence the amount made is k [P] [TF].

24 The Bound Fraction At equilibrium, the rate at which TFP forms must equal the rate at which 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 recipe ingredients are available. Hence the amount made is k [P] [TF]. The amount of P and TF made because TFP breaks apart depends on how much TFP is available which is k [TFP]. So at equilibrium, because the formation rate and disassociation rate are the same that we must have a balance k [TFP] = k [P] [TF]

25 The Bound Fraction At equilibrium, the rate at which TFP forms must equal the rate at which 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 recipe ingredients are available. Hence the amount made is k [P] [TF]. The amount of P and TF made because TFP breaks apart depends on how much TFP is available which is k [TFP]. So at equilibrium, because the formation rate and disassociation rate are the same that we must have a balance Solving we find the relationship k [TFP] = k [P] [TF] [TFP] = k k [P] [TF] We call the fraction k k the disassociation constant and this is the same variable we use earlier which had to do with the binding energy!

26 The Bound Fraction We see the fraction of TF bound to the promoter can be written as bound fraction = = [TFP] [P] + [TFP] 1 [P] [TF] [P] + [TF].

27 The Bound Fraction We see the fraction of TF bound to the promoter can be written as bound fraction = = [TFP] [P] + [TFP] 1 [P] [TF] [P] + [TF]. Now simplify a bit to get bound fraction = = = [TFP] [P] + [TFP] = [P] [TF] [P] + [P] [TF] = [TF] / 1 + [TF] /. [TFP] [P] + [TFP] [TF] + [TF]

28 The Bound Fraction So we express the disassociation constant in two ways to get insight. The first uses energy and movement ideas to estimate how much time it takes for the transcription factor and the promoter binding to reach equilibrium tens of seconds at most. The second uses familiar reaction ideas to estimate the fraction of transcription factor that is bound to the promoter. If we let denote the concentration of our transcription factor and denote its disassociation constant, we have a generic relationship. bound fraction = 1 +.

29 The Bound Fraction We can do a similar bit of analysis to figure out the unbound fraction. unbound fraction = = = [P] [P] + [TFP] = [P] [P] + [TFP] [P] [P] + [P] [TF] = + [TF] [TF]. Hence, the unbound fraction, using to denote the concentration of the transcription factor is unbound fraction = 1 1 +

30 The Bound Fraction It turns out we can do more. Sometimes transcription factors bind cooperatively or other factors can bind to the transcription factor prior to binding to the promoter. We can model this sort of thing too, but the details are not for us here. In such cases, the bound and unbound fractions can be represented by adding a power b to the term. Hence, the more general fractions are bound fraction = unbound fraction = where b 1 is called a Hill coefficient

31 The Bound Fraction We think of the ratio in this way. If the amount of [TF] exactly matched the disassociation rate, then the ratio would be 1 and the bound and unbound fraction are both 1/2. Just like the probability of being bound and unbound is exactly the same. However, the mixture can easily be skewed by simply letting [TF] be r for different values of r. If r 1, the bound fraction moves past 1/2 and is closer to saturation: the bound fraction is 1. Similarly, if r 1, the bound fraction is quite small with the unbound fraction close to 1. Example Calculate the bound and unbound fractions assuming the fraction / is 0.2 for a Hill s coefficient b = 1.

32 The Bound Fraction Solution We know bound fraction = = unbound fraction = = = = That s all there is to it! So most of the transcription factor is unbound here.

33 The Bound Fraction Homework Calculate the bound and unbound fractions assuming the fraction/ is 0.6 for a Hill s coefficient b = Calculate the bound and unbound fractions assuming the fraction/ is 1.2 for a Hill s coefficient b = Calculate the bound and unbound fractions assuming the fraction/ is 10.5 for a Hill s coefficient b = Calculate the bound and unbound fractions assuming the fraction/ is 1.6 for a Hill s coefficient b = 4.

34 Transcription egulation The transcription factor can activate or repress a promoter. If there is activation, then activation bound fraction. 1 +

35 Transcription egulation The transcription factor can activate or repress a promoter. If there is activation, then but if there is repression, then activation bound fraction 1 +. repression unbound fraction

36 Transcription egulation The transcription factor can activate or repress a promoter. If there is activation, then but if there is repression, then activation bound fraction 1 +. repression unbound fraction For convenience, let x =. Let s graph the some of these activation and repression functions. We ll use MatLab. Here is a simple session.

37 Transcription egulation f b, x ) x ). ˆ b. / 1+ x ). ˆ b ) ) ; X = l i n s p a c e 0, 1 0, ) ; b = 1 ; p l o t X, f 1,X) ) ; x l a b e l x = [ TF ] / ) ; 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 be asymptotic to 1 as [TF] / get large. Now if you think of [TF] as fixed, varying means we are changing the effectiveness of the binding. If is large, the fraction is small and the activation is small. On the other hand, as decreases, the fraction increases and the activation increases towards 1.

38 Transcription egulation

39 Transcription egulation Consider the activation graphs for b = 2, b = 3 and b = 4 on the same graph as shown in in the graph below. Note that all pass through the same pair 1, 1/2ut approach the saturation value of 1 in different ways. Effectively, increasing b ramps protein production up to its maximum value quicker. Another way of looking at it is that a larger Hill s coefficient b makes the activation more sensitive to variations in the [TF] level. As [TF] moves past, production ramps quickly.

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

41 Transcription egulation Given all we have discussed, we are now ready to show you our new protein transcription model. We start with the gene being repressed. C γ = Λ τ C, C0) = C 0; 1)

42 Transcription egulation Given all we have discussed, we are now ready to show you our new protein transcription model. We start with the gene being repressed. C γ = Λ τ C, C0) = C 0; 1) The new terms in our equation need to be defined: Λ is the leakage production; i.e. it is possible that the gene cannot be fully repressed and so there is some production always. γ is the maximum production rate over the base rate set by the leakage value Λ. τ is the lifetime of the protein which is essentially the term that includes all the losses that occur due to degradation, cell volume increase and so forth.

43 Transcription egulation This is still a standard x = αx + β model though. We know the steady state C repression SS is β/α. Here α = 1 τ, β = Λ + γ 1 +

44 Transcription egulation This is still a standard x = αx + β model though. We know the steady state C repression SS is β/α. Here α = 1 τ, β = Λ + γ 1 + and so C repression SS = τ Λ + γ )

45 Transcription egulation This is still a standard x = αx + β model though. We know the steady state C repression SS is β/α. Here α = 1 τ, β = Λ + γ 1 + and so C repression SS = τ Λ + γ ) and the response time is therefore t = τ ln2).

46 Transcription egulation The production is fully repressed when, say = 20 at which point the production model is C = Λ + 0 C/τ = Λ C/τ. This fully repressed model has full repression steady state CSS = τ Λ.

47 Transcription egulation The production is fully repressed when, say = 20 at which point the production model is C = Λ + 0 C/τ = Λ C/τ. This fully repressed model has full repression steady state CSS = τ Λ. On the other hand, if there is no repression, i.e., say =.01, then the minimally repressed model is C no repression = Λ + γ C/τ with steady state C = τλ + γ). SS

48 Transcription egulation The production is fully repressed when, say = 20 at which point the production model is C = Λ + 0 C/τ = Λ C/τ. This fully repressed model has full repression steady state CSS = τ Λ. On the other hand, if there is no repression, i.e., say =.01, then the minimally repressed model is C no repression = Λ + γ C/τ with steady state C = τλ + γ). How quickly production switches between these extremes depends on the Hill coefficient b as can be seen in the figure below. SS

49 Transcription egulation 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, ) ; 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.

50 Transcription egulation

51 Transcription egulation We can also activate protein production by allowing the transcription factor to activate production. The model is similar. C = Λ + γ 1 τ C, C0) = C 0. 3) 1 +

52 Transcription egulation We can also activate protein production by allowing the transcription factor to activate production. The model is similar. C = Λ + γ τ C, C0) = C 0. 3) We know the steady state C activation SS is β/α. Here α = 1 τ, β = Λ + γ 1 +

53 Transcription egulation So C activation SS = τ Λ + γ )

54 Transcription egulation Example Calculate C repression SS and CSS activation for Λ = 4, γ = 2, b = 4, τ = 1 and the ratio / = 0.1, 1, 2, 5, 20. Comment on your results. Solution / = 0.1. C repression SS = τ C activation SS = τ Λ + Λ + γ γ = ) 4 = = ) ) 4 =

55 Transcription egulation Solution / = 1. C repression SS = τ C activation SS = τ Λ + Λ + γ γ = ) 4 = 5 = ) ) 4 = 5.

56 Transcription egulation Solution / = 2. C repression SS = τ C activation SS = τ Λ + Λ + γ γ = ) 4 = = ) ) 4 =

57 Transcription egulation Solution / = 5. C repression SS = τ C activation SS = τ Λ + Λ + γ γ = ) 4 = = ) ) 4 =

58 Transcription egulation Solution / = C repression SS = τ C activation SS = τ Λ + Λ + γ γ = ) 4 = = ) ) 4 = Note both the unbound and bound are in the range [4, 6]

59 Transcription egulation Example Plot the steady state for the repression model for Λ = 0, γ = 5, b = 1 and τ = 1 as a function of /. 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, ) ; b = 1. 0 ; Lambda = 0 ; gamma = 5 ; tau = 1. 0 ; p l o t X, g b, tau, Lambda, gamma, X) ) ;

60 Transcription egulation Homework Calculate C repression SS and CSS activation for Λ = 2, γ = 3, b = 2, τ = 2 and the ratio / = 0.1, 1, 2, 5, 20. Comment on your results Calculate C repression SS and CSS activation for Λ = 5, γ = 2, b = 3, τ = 1 and the ratio / = 0.1, 1, 2, 5, 20. Comment on your results Plot the steady state for the repression model for Λ = 2, γ = 3, b = 2 and τ = 2 as a function of /. Comment on your results Plot the steady state for the repression model for Λ = 0, γ = 1, b = 1 and τ = 1 as a function of /. Comment on your results.