A Simple Protein Synthesis Model - Clemson CECAScecas.clemson.edu/~petersj/Courses/StartingCalcForBio/Presentations/L24-Protein...A Simple Protein Synthesis Model We can model protein

Starting Calculus for Biologists

A Simple Protein Synthesis Model

James K. Peterson

Department of Biological Sciences and Department of Mathematical SciencesClemson University

September 30, 2013


Outline

1 A Simple Protein Synthesis Model

2 Sample Protein ModelsExample OneExample Two

3 The General Picture


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 controlsthe construction of a protein.

After binding, the nucleotides in the gene are transcribed intoa messenger RNA fragment which is the complement of theoriginal nucleotide chain.

The messenger RNA fragment is sent to the ribosome whereit is used to assemble the protein which is a string of aminoacids.

The signal S can thus be used to start and stop thetranscription process.



A mathematical model: Let P(t) denote the concentration of theprotein in the cell’s cytoplasm.

Growth

After a short period of time, usually millisecondsor less, protein transcription reaches fullproduction at a constant rate of β (millimoles/cc).So we assume P ′growth = β.

Decay

There are enzymes that break apart proteins forreuse and as living cells grow their volumeincreases causing the protein’s concentration todrop.Combine these two effects and assume anexponential decay model: P ′decay = −αP whereα 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(0) = P0 > 0.

How do we solve this model?

How do we handle sequences of signal on and signal offpatterns?



So we have a model of protein transcription, but we have onlymentioned in passing the molecular machinery that isinvolved. It’s time to look at the underlying biology moreclosely so you can have a better idea what is going on. Notehow much of this detail we actually do not use when we buildthe model, yet the model still gives a lot of insight.

Let’s examine how a single gene Y is activated to produce itsproduct which is a protein. We will discuss this using simplecartoons to represent a lot of complicated biology. We intendto show you how the mathematics we are learning has a lot ofexplanatory power. So remember, Protein Modeling here, isan approximation of the underlying complex biologicalprocesses which we are making in order to gain insight.



So we have a model of protein transcription, but we have onlymentioned in passing the molecular machinery that isinvolved. It’s time to look at the underlying biology moreclosely so you can have a better idea what is going on. Notehow much of this detail we actually do not use when we buildthe model, yet the model still gives a lot of insight.

Let’s examine how a single gene Y is activated to produce itsproduct which is a protein. We will discuss this using simplecartoons to represent a lot of complicated biology. We intendto show you how the mathematics we are learning has a lot ofexplanatory power. So remember, Protein Modeling here, isan approximation of the underlying complex biologicalprocesses which we are making in order to gain insight.



The gene Y is a string of nucleotides (A, C, T and G) with aspecial starting string in front of it called the promoter. We willdraw this as shown here

Figure: Promoter and Gene



The nucleotides in the gene Y are read three at a time to createthe amino acids which form the protein Y∗ corresponding to thegene. The process is this: a special RNA polymerase, RNAp,which is a complex of several proteins, binds to the promoterregion as shown here

Figure: Promoter Binding



Once RNAp binds to the promoter, messenger RNA, mRNA,is synthesized that corresponds to the specific nucleotidetriplets in the gene Y. The process of forming this mRNA iscalled transcription. Once the mRNA is formed, the proteinY∗ is then made.

The protein creation process is typically regulated. A singleregulator works like this. An activator called X is a proteinwhich increases the rate of mRNA creation when ti binds tothe promoter. The activator X switches between and activeand inactive version due to a signal SX. We let the activeform be denoted by X∗.



Once RNAp binds to the promoter, messenger RNA, mRNA,is synthesized that corresponds to the specific nucleotidetriplets in the gene Y. The process of forming this mRNA iscalled transcription. Once the mRNA is formed, the proteinY∗ is then made.

The protein creation process is typically regulated. A singleregulator works like this. An activator called X is a proteinwhich increases the rate of mRNA creation when ti binds tothe promoter. The activator X switches between and activeand inactive version due to a signal SX. We let the activeform be denoted by X∗.



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 increasesimplying an increase in the creation of the protein Y∗ also.

Figure: Signal binding and protein transcription



We indicate all of this by the simple interaction arrow

X −→ Y.

Once the signal SX appears, X rapidly transitions to its stateX∗, binds with the front of the promoter and protein Y∗

begins to accumulate.

We let β denote the rate of protein accumulation which isconstant once the signal SX begins.




X −→ Y.







X −→ Y.






However, proteins also degrade and this process is modeled byexponential decay, −αY∗.

The net rate of change of the protein concentration is thenour familiar model

dY∗

dt= β︸︷︷︸

constant growth

− α Y∗︸︷︷︸loss term

We usually do not make a distinction between the gene Y andits transcribed protein Y∗. We usually treat the letters Y andY∗ as the same even though it is not completely correct.Hence, we just write as our model

Y ′ = β − α Y

Y (0) = Y0

and then solve it using the integrating factor method eventhough, strictly speaking, Y is the gene!





dY∗

dt= β︸︷︷︸

constant growth



Y ′ = β − α Y

Y (0) = Y0






dY∗

dt= β︸︷︷︸

constant growth



Y ′ = β − α Y

Y (0) = Y0



Sample Protein Models

Example One

Example

Solve the following problem:

x ′(t) = −.02 x(t) + 5; x(0) = 75

Solution

Step 1: Rewrite as x ′(t) + .02x(t) = 5.

Step 2a: Multiply both sides by the integrating factor e .02t .

e .02t(x ′(t) + .02x(t)

)= 5 e .02t .

Step 2b: Recognize this is a product rule and rewrite as(e .02tx(t)

)′= 5 e .02t .



Example One

Solution

Step 3 Integrate both sides from 0 to t:∫ t

0

(e .02sx(s)

)′ds =

∫ t

05 e .02s ds.

Step 4 This gives (e .02sx(s)

)∣∣∣∣t0

=5

.02e .02s

∣∣∣∣t0

Step 5 This gives

e .02tx(t)− x(0) =5

.02

(e .02t − 1

).



Example One

Solution

Step 6 Simplify:

e .02tx(t) = x(0) +5

.02

(e .02t − 1

)=

(x(0)− 5

.02

)+

5

.02e .02t .

Step 7 Solve for x(t): multiply both sides by e−.02t .

x(t) = e−.02t

((x(0)− 5

.02

)+

5

.02e .02t

)

x(t) =

(x(0)− 5

.02

)e−.02t +

5

.02.



Example One

Solution

Step 8 Use the IC:

x(t) =

(75− 250

)e−.02t + 250

= −175 e−.02t + 250.

Interpretation:

As t −→∞, e−.02t −→ 0 and so x(t) −→ 5.02 = 250. This is

a horizontal asymptote and we call the fraction βα = 5

.02 = 250the steady state value or SS.

The IC is 75 which is below the SS so x(t) approachs the SSfrom below.

We can also do this for the IC 375. The pictures for the twocases are on the next slide.



Example One

Here is what it looks like.



Example One

How long does it take x(t) to rise from its IC of 75 to half way tothe SS of 250? This time is like a half life but in this context it iscalled the response time. Denote it by tr . Half way to 250 from75 is the average (75 + 250)/2. This is 325/2. We find

325

2= −175 e−.02tr + 250

−175

2= −175 e−.02tr

1

2= e−.02tr

− ln(2) = −.02tr

tr =ln(2)

.02

This is the same formula we had for the half life in an exponentialdecay.



Example One

We can draw using the response time.



Example Two

Example

Solve the following problem:

x ′(t) = −.003 x(t) + 2; x(0) = 500

Solution

Step 1: Rewrite as x ′(t) + .003x(t) = 2.

Step 2a: Multiply both sides by the integrating factor e .003t .

e .003t(x ′(t) + .003x(t)

)= 2 e .003t .

Step 2b: Recognize this is a product rule and rewrite as(e .003tx(t)

)′= 2 e .003t .



Example Two

Solution

Step 3 Integrate both sides from 0 to t:∫ t

0

(e .003sx(s)

)′ds =

∫ t

02 e .003s ds.

Step 4 This gives(e .003sx(s)

)∣∣∣∣t0

=2

.003e .003s

∣∣∣∣t0

Step 5 This gives

e .003tx(t)− x(0) =2

.003

(e .003t − 1

).



Example Two

Solution

Step 6 Simplify:

e .003tx(t) = x(0) +2

.003

(e .003t − 1

)=

(x(0)− 2

.003

)+

2

.003e .003t .

Step 7 Solve for x(t): multiply both sides by e−.003t .

x(t) = e−.003t

((x(0)− 2

.003

)+

2

.003e .003t

)

x(t) =

(x(0)− 2

.003

)e−.003t +

2

.003.



Example Two

Solution

Step 8 Use the IC:

x(t) =

(500− 666.67

)e−.003t + 666.67

= −166.67 e−.003t + 666.67.

Interpretation:

As t −→∞, e−.003t −→ 0 and so x(t) −→ 2.003 = 666.67.

The SS is βα = 2

.003

The IC is below the SS so x(t) approachs the SS from below.

We can also do this for the IC 800. We show both cases inthe next slide.



Example Two

Here is what it looks like.



Example Two

How long does it take x(t) to rise from its IC of 500 to half way tothe SS of 2000/3? This is the response time, tr . Half way to2000/3 from 500 is the average .5(500 + 2000/3). This is 1750/3.We find

1750

3= −500

3e−.003tr +

2000

3

−250

3= −500

3e−.003tr

1

2= e−.003tr

− ln(2) = −.003tr

tr =ln(2)

.003

This is the same formula we had for the half life in an exponentialdecay.



Example Two

We can draw using the response time.


The General Picture

In general, for the protein synthesis model

P ′(t) = −αP(t) + β

P(0) = P0 > 0

α is the rate of protein degradation, i.e. decay modeled asexponential decay, P ′decay = −αP,

β is the constant rate of protein production.

The steady state value, SS, is βα .

The response time is tr = ln(2)α .


The General Picture

Homework 43

Solve the following in great detail using the integrating factormethod. Tell me the value of the steady state and the responsetime. Do a careful graph using the response time for 3 responsetimes and then indicate how the curve is drawn past the thirdresponse time.

43.1 P ′ = −.2P + 15;P(0) = 20.

43.2 Q ′ = −.03Q + 8;Q(0) = 100.

43.3 W ′ = −.3W + 30;W (0) = 30.

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A Simple Protein Synthesis Model - Clemson CECAScecas.clemson.edu/~petersj/Courses/StartingCalcForBio/Presentations/L24-Protein...A Simple Protein Synthesis Model We can model protein