Protein Binding

Question 1

Rate of association and dissociation equations? Relationship between them at equilibrium

Answer 1

Rate of association = k1[P][L]; k1is association rate constant
Rate of dissociation = k-1[PL]; k-1 is dissociation rate constant

At equilibrium: k1[P][L] = k-1[PL]

Question 2

Association and dissociation equilibrium constants (Ka and Kd)

Answer 2

Association equilibrium constant Ka = [PL] / ([P][L]) = k1 / k-1
Dissociation equilibrium constant Kd = [P][L] / [PL] = k-1 / k1

Question 3

Ka x Kd = ?

Answer 3

1

Question 4

Can you get from [PL] to Kd

Answer 4

[PL] = Ka[P][L] = [P][L] / Kd

Question 5

Ligand bound fraction

Answer 5

θ = [PL] / ([P]+[PL]) = Ka[P][L]/([P]+Ka[P][L]) = Ka[L]/(1+Ka[L]) = [L]/(Kd+[L])[Ptotal] = [P]+[PL]

θ= bound fraction
θ= [PL]/[Ptot]

Question 6

what kind of curve is the ligand binding curve

Answer 6

hyperbolic curve

Question 7

what does the Kd value correspond to

Answer 7

corresponds to the [L] at which bound fraction is 50%

Question 8

How can we measure [L], [P] or [PL]

Answer 8

1. Free L and bound PL have different absorbance or fluorescence at a certain wavelength
2. Free P and bound PL have different absorbance or fluorescence at a certain wavelength

Question 9

Scatchard linear equation and what can it be used on

Answer 9

Deals with homogenous binding sites

Starts with the
θ = [L]/ (Kd+[L])
then replace θ with θ/n

Scatchard Linear equation:
θ/[L] = n/Kd - θ/Kd
or
θ/[L] = nKa - θKa

Question 10

Average occupancy per ligand-binding site

Answer 10

θ/n

Question 11

Scatchard linear plot slope, y-intercept, x-intercept

Answer 11

Slope = -Ka or -1/Kd
y-intercept= nKa or n/Kd
X-intercept= n

Question 12

Double reciprocal equation? When does it work?

Answer 12

Works when there is only n= 1 binding site; θ is occupancy 0-1

1/[PL] = Kd/([Ptotal][L]) +1/[Ptotal]

Question 13

Double reciprocal linear plot slope, y-intercept, x-intercept

Answer 13

Slope= Kd/[Ptotal]

y-intercept is 1/[Ptotal]

x-intercept is -1/Kd or -1Ka

Question 14

Cooperative binding rate of association and dissociation? Equilibrium relationship?

Answer 14

rate of association = k1 [P][L]^n
Rate of dissociation= k-1 [PLn]

These two rates are equal at equilibrium

Question 15

Association and dissociation equilibrium constant (Ka & Kd)

Answer 15

Association equilibrium constant Ka = [PLn]/([P[L]n) = k1/ k-1 Dissociation equilibrium constant Kd = [P][L]n/[PLn] = k-1 / k1

Question 16

Fraction of Ligand-bound protein for cooperative binding

Answer 16

Question 17

Hill plot

Answer 17

Question 18

Equilibrium dialysis

Answer 18

Establish equilibrium. Protein is retained in the membrane bag. Measure free [L] outside; and bound and free [PL] + [L] inside bag. Free [L] is the same on both side of dialysis membrane. The difference must be [PL]

Kd can therefore be determined with a series of initial [L] inside.

Question 19

Isothermal Titration Calorimetry (ITC)

Answer 19

ITC measures heat exerted or absorbed by adjusting the electric current to the sample cell to maintain a constant temperature with the reference cell.
Protein is in the sample cell.
Concentrated ligand is injected to the sample cell. Binding gives or takes heat

ITC directly measures DH, Ka, stoichiometry, and indirectly derives DG and DS.

Question 20

Surface Plasmon Resonance (SPR)

Answer 20

SPR measures both k1 and k-1 so also gets kinetic information

Question 21

MicroScale Thermophoresis (MST)

Answer 21

Upon binding ligand Fnorm increases because diffusion becomes slower (larger mass of the PL complex).
Therefore, plot Fnorm as a function of ligand concentration

Kd can be derived from the midpoint of the curve!!!