A Simple Protein Synthesis Model

Transcription

1 A Simple Protein Synthesis Model James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University September 3, 213 Outline A Simple Protein Synthesis Model Sample Protein Models Example One Example Two

2 Abstract This lecture is going to talk a lot a simple protein synthesis model. We can model protein synthesis very simply as follows: A signal S binds to the promoter of the gene that controls the construction of a protein. After binding, the nucleotides in the gene are transcribed into a messenger RNA fragment which is the complement of the original nucleotide chain. The messenger RNA fragment is sent to the ribosome where it is used to assemble the protein which is a string of amino acids. The signal S can thus be used to start and stop the transcription process.

3 A mathematical model: Let P(t) denote the concentration of the protein in the cell s cytoplasm. Growth Decay After a short period of time, usually milliseconds or less, protein transcription reaches full production at a constant rate of β (millimoles/ cc). So we assume P growth = β. There are enzymes that break apart proteins for reuse and as living cells grow their volume increases causing the protein s concentration to drop. Combine these two effects and assume an exponential decay model: P decay = αp where α is a positive number. The net rate of change is P = P growth + P decay. This gives the IVP P (t) = α P(t) + β P() = P >. How do we solve this model? How do we handle sequences of signal on and signal off patterns?

4 So we have a model of protein transcription, but we have only mentioned in passing the molecular machinery that is involved. It s time to look at the underlying biology more closely so you can have a better idea what is going on. Note how much of this detail we actually do not use when we build the model, yet the model still gives a lot of insight. Let s examine how a single gene Y is activated to produce its product which is a protein. We will discuss this using simple cartoons to represent a lot of complicated biology. We intend to show you how the mathematics we are learning has a lot of explanatory power. So remember, Protein Modeling here, is an approximation of the underlying complex biological processes which we are making in order to gain insight. The gene Y is a string of nucleotides (A, C, T and G) with a special starting string in front of it called the promoter. We will draw this as shown here Figure: Promoter and Gene

5 The nucleotides in the gene Y are read three at a time to create the amino acids which form the protein Y corresponding to the gene. The process is this: a special RNA polymerase, RNAp, which is a complex of several proteins, binds to the promoter region as shown here Figure: Promoter Binding Once RNAp binds to the promoter, messenger RNA, mrna, is synthesized that corresponds to the specific nucleotide triplets in the gene Y. The process of forming this mrna is called transcription. Once the mrna is formed, the protein Y is then made. The protein creation process is typically regulated. A single regulator works like this. An activator called X is a protein which increases the rate of mrna creation when ti binds to the promoter. The activator X switches between and active and inactive version due to a signal S X. We let the active form be denoted by X.

6 Here we show X moving to state X and back. Figure: Activator and Inactivator Switching If X binds in front of the promoter, mrna creation increases implying an increase in the creation of the protein Y also. Figure: Signal binding and protein transcription

7 We indicate all of this by the simple interaction arrow X Y. Once the signal S X appears, X rapidly transitions to its state X, binds with the front of the promoter and protein Y begins to accumulate. We let β denote the rate of protein accumulation which is constant once the signal S X begins. However, proteins also degrade and this process is modeled by exponential decay, αy. The net rate of change of the protein concentration is then our familiar model dy dt = β }{{} α}{{ Y } constant growth loss term We usually do not make a distinction between the gene Y and its transcribed protein Y. We usually treat the letters Y and Y as the same even though it is not completely correct. Hence, we just write as our model Y = β α Y Y () = Y and then solve it using the integrating factor method even though, strictly speaking, Y is the gene!

8 Example Solve the following problem: x (t) =.2 x(t) + 5; x() = 75 Solution Step 1: Rewrite as x (t) +.2x(t) = 5. Step 2a: Multiply both sides by the integrating factor e.2t. ) e (x.2t (t) +.2x(t) = 5 e.2t. Step 2b: Recognize this is a product rule and rewrite as ( e.2t x(t)) = 5 e.2t. Solution Step 3 Integrate both sides from to t: t ( t e x(s)).2s ds = 5 e.2s ds. Step 4 This gives Step 5 This gives ( ) t e.2s x(s) e.2t x(t) x() = 5.2 = 5 t.2 e.2s ( ) e.2t 1.

9 Solution Step 6 Simplify: e.2t x(t) = x() + 5 ( ) e.2t 1.2 ( = x() 5 ) e.2t. Step 7 Solve for x(t): multiply both sides by e.2t. ( ( x(t) = e.2t x() 5 ).2 x(t) = e.2t ( x() 5 ) e.2t ) Solution Step 8 Use the IC: x(t) = ( ) e.2t + 25 Interpretation: = 175 e.2t As t, e.2t and so x(t) 5.2 = 25. This is a horizontal asymptote and we call the fraction β α = 5.2 = 25 the steady state value or SS. The IC is 75 which is below the SS so x(t) approachs the SS from below. We can also do this for the IC 375. The pictures for the two cases are on the next slide.

10 Here is what it looks like. How long does it take x(t) to rise from its IC of 75 to half way to the SS of 25? This time is like a half life but in this context it is called the response time. Denote it by t r. Half way to 25 from 75 is the average ( )/2. This is 325/2. We find = 175 e.2tr = 175 e.2tr 1 2 = e.2tr ln(2) =.2t r t r = ln(2).2 This is the same formula we had for the half life in an exponential decay.

11 We can draw using the response time. Example Solve the following problem: x (t) =.3 x(t) + 2; x() = 5 Solution Step 1: Rewrite as x (t) +.3x(t) = 2. Step 2a: Multiply both sides by the integrating factor e.3t. ) e (x.3t (t) +.3x(t) = 2 e.3t. Step 2b: Recognize this is a product rule and rewrite as ( e.3t x(t)) = 2 e.3t.

12 Solution Step 3 Integrate both sides from to t: t ( t e x(s)).3s ds = 2 e.3s ds. Step 4 This gives ( ) t e.3s x(s) Step 5 This gives e.3t x(t) x() = = 2 t.3 e.3s ( ) 2 e.3t 1..3 Solution Step 6 Simplify: e.3t x(t) = x() + 2 ( ) e.3t 1.3 ( = x() 2 ) e.3t. Step 7 Solve for x(t): multiply both sides by e.3t. ( ( x(t) = e.3t x() 2.3 x(t) = ) e.3t ( x() 2 ) e.3t )

13 Solution Step 8 Use the IC: x(t) = ( ) e.3t = e.3t Interpretation: As t, e.3t and so x(t) 2.3 = The SS is β α = 2.3 The IC is below the SS so x(t) approachs the SS from below. We can also do this for the IC 8. We show both cases in the next slide. Here is what it looks like.

14 How long does it take x(t) to rise from its IC of 5 to half way to the SS of 2/3? This is the response time, t r. Half way to 2/3 from 5 is the average.5(5 + 2/3). This is 175/3. We find = 5 3 e.3tr = 5 3 e.3tr 2 = e.3tr ln(2) =.3t r t r = ln(2).3 This is the same formula we had for the half life in an exponential decay. We can draw using the response time.

15 In general, for the protein synthesis model P (t) = αp(t) + β P() = P > α is the rate of protein degradation, i.e. decay modeled as exponential decay, P decay = αp, β is the constant rate of protein production. The steady state value, SS, is β α. The response time is t r = ln(2) α. Homework 43 Solve the following in great detail using the integrating factor method. Tell me the value of the steady state and the response time. Do a careful graph using the response time for 3 response times and then indicate how the curve is drawn past the third response time P =.2P + 15; P() = Q =.3Q + 8; Q() = W =.3W + 3; W () = 3.