Boolean Modeling of the Lac Operon in E. coli: Understanding Gene Regulation, Lecture notes of Logic

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Boolean models of the lac

operon in E. coli

Matthew Macauley Clemson University

Gene expression

— Gene expression is a process that takes gene info and creates a functional gene product (e.g., a protein). — Some genes code for proteins. Others (e.g., rRNA, tRNA) code for functional RNA. — Gene Expression is a 2-step process:

1. transcription of genes (messenger RNA synthesis)
2. translation of genes (protein synthesis) — DNA consists of bases A, C, G, T. — RNA consists of bases A, C, G, U. — Proteins are long chains of amino acids. — Gene expression is used by all known life forms.

Translation

• During translation, the mRNA is read by ribosomes.
• Each triple of RNA bases codes for an amino acid.
• The result is a protein: a long chain of amino acids.
• Proteins fold into a 3-D shape which determine their function

Gene expression

— The expression level is the rate at which a gene is being expressed. — Housekeeping genes are continuously expressed, as they are essential for basic life processes. — Regulated genes are expressed only under certain outside factors (environmental, physiological, etc.). Expression is controlled by the cell. — It is easiest to control gene regulation by affecting transcription. — One way to block repression is for repressor proteins bind to the DNA or RNA. — Goal : Understand the complex cell behaviors of gene regulation, which is the process of turning on/off certain genes depending on the requirements of the organism.

Lactose and β−galactosidase

— When a host consumes milk, E. coli is exposed to lactose (milk sugar). — Lactose consists of one glucose sugar linked to one galactose sugar. — If both glucose and lactose are available, then glucose is the preferred energy source. — Before lactose can used as energy, the β−galactosidase enzyme is needed to break it down. — β−galactosidase is encoded by the LacZ gene on the lac operon. — β−galactosidase also catalyzes lactose into allolactose.

Transporter protein

— To bring lactose into the cell, a transport protein, called lac permease, is required. — This protein is encoded by the LacY gene on the lac operon. — If lactose is not present, then neither of the following are produced:

1. β−galactosidase (LacZ gene) 2) lac permease (LacY gene) — In this case, the lac operon is OFF.

with lactose and no gluclose

— Lactose is brought into the cell by the lac permease transporter protein — β−galactosidase breaks up lactose into glucose and galactose.. — β−galactosidase also converts lactose into allolactose. — Allolactose binds to the lac repressor protein, preventing it from binding to the operator region of the genome. — Transcription begins: mRNA encoding the lac genes is produced. — Lac proteins are produced, and more lactose is brought into the cell. (The operon is ON.) — Eventually, all lactose is used up, so there will be no more allolactose. — The lac repressor can now bind to the operator, so mRNA transcription stops. (The operon has turned itself OFF.)

An ODE lac operon model

— M: mRNA — B: β−galactosidase — A: allolactose — P: transporter protein — L: lactose

A Boolean approach

— Let’s assume everything is “Boolean” (0 or 1): o Gene products are either present or absent o Enzyme concentrations are either high or low. o The operon is either ON or OFF. — mRNA is transcribed (M=1) if there is no external glucose (G=0), and either internal lactose (L=1) or external lactose (Le=1) are present. — The LacY and LacZ gene products (E=1) will be produced if mRNA is available (M=1). — Lactose will be present in the cell if there is no external glucose (Ge=0), and either of the following holds: ü External lactose is present (Le=1) and lac permease (E=1) is available. ü Internal lactose is present (L=1), but β−galactosidase is absent (E=0). x M ( t + 1 ) = f M ( t + 1 ) = G e ∧( L ( t )∨ L e

x E ( t + 1 ) = f E ( t + 1 ) = M ( t ) x L ( t + 1 ) = f L ( t + 1 ) = G e

∧ ( L

e ∧ E ( t ))∨( L ( t )∧ E ( t ))

Comments on the Boolean model

— We have two “types” of Boolean quantities: o mRNA (M), lac gene products (E), and internal lactose (L) are variables. o External glucose (Ge) and lactose (Le) are parameters (constants). — Variables and parameters are drawn as nodes. — Interactions can be drawn as signed edges. — A signed graph called the wiring diagram describes the dependencies of the variables. — Time is discrete: t = 0, 1, 2, …. — Assume that the variables are updated synchronously. x M ( t + 1 ) = f M ( t + 1 ) = G e ∧( L ( t )∨ L e

x E ( t + 1 ) = f E ( t + 1 ) = M ( t ) x L ( t + 1 ) = f L ( t + 1 ) = G e

∧ ( L

e ∧ E ( t ))∨( L ( t )∧ E ( t ))

How to analyze a Boolean model

— We can plot the state space using the software: Analysis of Dynamical Algebraic Models (ADAM), at adam.plantsimlab.org. — First, we need to convert our logical functions into polynomials. — Here is the relationship between Boolean logic and polynomial algebra: Boolean operations logical form polynomial form o AND o OR o NOT

• Also, everything is done modulo 2, so 1+1=0, and x^2 =x, and thus x(x+1)=0. x M ( t + 1 ) = f M ( t + 1 ) = G e ∧( L ( t )∨ L e

x E ( t + 1 ) = f E ( t + 1 ) = M ( t ) x L ( t + 1 ) = f L ( t + 1 ) = G e

∧ ( L

e ∧ E ( t ))∨( L ( t )∧ E ( t ))

z = x ∧ y z = x ∨ y z = x z = xy z = x + y + xy z = 1 + x

x M ( t + 1 ) = f M ( t + 1 ) = G e ∧( L ( t )∨ L e

x E ( t + 1 ) = f E ( t + 1 ) = M ( t ) x L ( t + 1 ) = f L ( t + 1 ) = G e

∧ ( L

e ∧ E ( t ))∨( L ( t )∧ E ( t ))

x M ( t + 1 ) = f M ( t + 1 ) = G e ∧( L ( t )∨ L e

x E ( t + 1 ) = f E ( t + 1 ) = M ( t ) x L ( t + 1 ) = f L ( t + 1 ) = G e

∧ ( L

e ∧ E ( t ))∨( L ( t )∧ E ( t ))

State space when (Ge, Le) = (0, 0). The operon is OFF.

x M ( t + 1 ) = f M ( t + 1 ) = G e ∧( L ( t )∨ L e

x E ( t + 1 ) = f E ( t + 1 ) = M ( t ) x L ( t + 1 ) = f L ( t + 1 ) = G e

∧ ( L

e ∧ E ( t ))∨( L ( t )∧ E ( t ))

State space when (Ge, Le) = (1, 0). The operon is OFF.