Mechanics of Motor Proteins and the Cytoskeleton Jonathon Howard - - PowerPoint PPT Presentation

Mechanics of Motor Proteins and the Cytoskeleton

Jonathon Howard

Chapter 3: Mass, Stiffness, and Damping of Proteins

Reviewed by: Dr. David Tees Ameneh Mohammadalipour Sneha Pandya

Outline

Purpose:

• What proteins are like as mechanical devices.
• How rigid they are.
• How quickly they move and change shape.
• What is the quality of their motion.
• …

We need to know:

• Material properties of protein; density, elasticity ….
• Frictional forces that damp their motion.

Summary:

• Proteins have similar densities and rigidities to hard plastics and Plexiglas.
• Because of their small sizes, the viscous forces from the surrounding fluid are

large, compared to their inertial forces.

• The motion of proteins are overdamped: they relax monotonically into new

conformations.

Mass

m = ρV

Molecular Mass (Da): The mass in grams of a mole

• f the molecules.

1 Da= 1.66×10-27kg

Average density= 1.38×103kg/m3

Alberts et al, Molecular Biology of the Cell, 2002.

Elasticity

F = κΔL

A solid is homogenous if its mechanical properties are identical throughout, and it is isotropic if these properties do not depend on direction. Young’s modulus Stress (pressure) Strain (relative length change)

F A = E ΔL L

stiffness

Examples of Tensile Stress

F = κΔL ⇒ κ = F ΔL F A = E ΔL L ⇒ F ΔL = EA L ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ κ = EA L

Young’s modulus is independent of size and shape Stiffness depends on size, shape and Young’s modulus

d2y dx 2 = M(x) EI M(x) = F(L − x) ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ y(x) = F EI (Lx 2 2 − x 3 6 )

To solve the small- angle equation:

Bending moment second moment

y(L) = FL3 3EI ⇒ κ = F y(L) = 3EI L3 I = π 4 r4 ⎫ ⎬ ⎪ ⎪ ⎭ ⎪ ⎪ ⇒ κ = 3π 4 Er4 L3

Shear Modulus

F A = Gθ

shear modulus: G = E /2(1+σ)

Δw w = σ ΔL L G = E /3

Incompressible:

A homogeneous and isotopic material has two elastic parameters:

• Young’s modulus
• Poisson's ratio

Poisson’s ratio (-1≤σ≤0.5)

The Molecular Basis of Elasticity

U(r) = −U0 2 r r ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

6

− r r ⎛ ⎝ ⎜ ⎞ ⎠ ⎟

12

⎡ ⎣ ⎢ ⎤ ⎦ ⎥

U(r) ≅U0 + 1 2 κ(r − r

0)2

At the bottom of the well:

F(r) = dU /dr ≅ κ(r − r

0)

Required force to stretch the bond a small distance (r-r0): : Hooke’s law Lennard- Jones Potential:

attractive, dispersion energy –C/r6 repulsive, exchange energy +C/r12

• Strong covalent chemical bonds: Electrostatic

bonds (ion pairs and hydrogen bonds)

• Weak noncovalent physical bonds: Van der

Waals bonds Bond energy Stiffness of the bond

Models of Materials

F = κΔr ⇒ F r

2 = κ

r Δr r F A = E ΔL L ⎫ ⎬ ⎪ ⎪ ⎭ ⎪ ⎪ ⇒ E = κ /r

Rigidity of proteins is primarily limited by the rigidity of the van der Waals bonds.

Viscous Damping

F A = η dv dx

Two damping forces:

Solvent friction: From the viscosity of the surrounding fluid Protein friction: From transitory interactions between amino acids that slide with respect to one another as the protein changes shape.

Viscosity (Pa.s) Shear rate (velocity gradient) applied stress (pressure)

Newton’s Law of Viscosity:

Newtonian and Non-Newtonian fluids

Non-Newtonian fluid: viscosity dependents on the velocity gradient

Shear Thinning: – High molecular weight (polymers) – Higher shear rate aligns molecules and viscosity decreases Shear Thickening: – Large particles suspended in smaller particles – Higher shear rate pushes out smaller molecules and viscosity increases

Newtonian fluid: viscosity is independent of the velocity gradient

Reynolds Number

Re = ρLv η Fd = −γv = −6πηrv γ = 6πηr

Te drag force on object depends on the pattern of the fluid flow around the object: Reynolds number=inertial forces/viscous forces

Density

• f

the liquid Characteristic length

• f the object

Speed of the

• bject

viscosity

At low Reynolds numbers (Re<1), the drag force is proportional to the speed: :

Drag force for a sphere of radius r, moving at constant velocity v

Stokes’ law

• Laminar flow
• Turbulent flow

Coefficient of viscosity: