Biochemical Thermodynamics Flashcards
protein-ligand binding equilibrium
(for 1:1 binding)
P + L ⇌ PL
protein-ligand binding equilibrium
(for 1:n binding)
P + nL ⇌ PLn
note: subscritp n
note: for 1:1 binding, n=1 so equilbrium is P + L ⇌ PL
association constant
Ka = [PLn]/[P][L]ⁿ
note: for 1:1 binding = [PL]/[P][L]
dissocation constant
Kd = [P][L]ⁿ/[PLn]
note: for 1:1 binding [P][L]/[PL]
units for Ka
1/M
units for Kd
M
relationship between Ka and Kd
Kd = 1/Ka
stronger binding affinity means…
larger Ka
smaller Kd
B
fractional occupancy
= [bound proteins]/[total proteins]
= [PLn]/[PLn]+[P]
(assuming that ligands bind togetger - cooperative binding)
note: for 1:1 binding = [PL]/[PL]+[P]
langmuir isotherm
plot of B against [L]
B=[L]/Kd+[L]
linearised langmuir isotherm
ln(B/1-B) = ln[L] - lnKd
y axis = ln(B/1-B)
x axis = ln[L]
gradient = 1 (1:1 binding)
intercept = -lnKd
when binding is 1:1, what is the concentraion of L when B=0.5
1/2
quick determination of Kd
(when binding is 1:1)
Kd = [L]1/2
shift in isotherm when binding affinity is stronger
stronger affinity = smaller Kd
isotherm shifts to the left
hill isotherm
B=[L]ⁿ/Kd+[L]ⁿ
linearised hill plot
ln(B/1-B) = nln[L] - lnKd
y axis = ln(B/1-B)
x axis = ln[L]
gradient = n (1:n binding)
intercept = -lnKd
what is the preferred measure of affinity?
Kd
Thermal energy
RT (J mol-1)
(molar gas constant (J K-1 mol-1) x temperature (K))
value of R
8.314 JK-1mol-1
why can’t protein unfolding happen at lower temperatures?
at lower temperatures, RT is less than the enthalpy gap between the folded and unfolded proteins states
Wpl and Wp+l
number of conformations of the protein-ligand complex
and
number of conformations for the dissociated protein and ligand
there are generally more possible coformations for P+L than PL
relate Kd to the enthalpy gap and number of conformations
Kd = (Wp+l/Wpl) e^-(Hp+l-Hpl/RT)
boltzmanns law of entropy
Si = R ln Wi
the entropy of a state, Si, can be expressed in terms of its number of conformational states, Wi.
e.g. entropy of PL : Spl = R ln Wpl
entropy of dissociation
ΔSd = R ln Wp+l/Wpl
Δ because dissociation involves a change in entropy
enthalpy of dissociation
ΔHd = Hp+l - Hpl
Δ because dissociation involves a change in enthalpy
gibbs free energy expresses…
… stability as a balance between sufficiently low enthalpy and sufficiently large entropy
G = H-TS
change of gibbs free energy
ΔG = ΔH - TΔS
gibbs free energy of dissocation
ΔGd = -RTlnKd
Δ because dissociation involves a change in gibbs free energy
a favourable change …
K is large
ΔG is negative
ΔH is negative
ΔS is positive
van’t Hoff plot
ln K = -ΔH/RT + ΔS/R
y axis = lnK
x axis = 1/T
intercept = ΔS/R
gradient = -ΔH/R
what is the purpose of calorimetry
to determine the enthalpy chane of protein denaturation as a measure of protein stability
proteins with greater stability in the folded state have higher ΔHm (enthalpy of denaturation / melting)
Cp
heat capacity - the amount of thermal energy (RT) required to raise the temperature by 1 K
Cp = Δq/ΔT
units: Jg-1K-1
output of calorimetry
graph of Cp against T
gives the value of Cp at each temperature whilst heating through a range
how can ΔH be found from a plot of Cp against T?
the area under the line = ΔH
Differential Scanning Calorimetry
Cp(protein) = Cp(protein + water) - Cp(water)
DCS performed on aqueous solutions of proteins. the Cp of water must be removed to account for the heat used to break H-bonds between water molecules, rather than increase temperature or melt protein
what does the peak in DCS output correspond to?
the temperature at which protein unfolding/denaturing/melting takes place
= Tm
large peak at Tm = high input of heat needed to raise temperature by 1K at this point. because energy input is used to overcome non-covalent interactions in the protein rather than raise temperature.
why are base lines included on a DCS output?
base lines represent the heat input per 1K temp increase for the native (before peak) and denatured (after peak) proteins, when denaturation is not occuring
find ΔHm on a DCS output
the area between the curve and the baselines
ΔHm = the amount of heat/energy required to overcome non-covalent interactions during denaturation
entropy of unfolding
ΔSm = ΔHm / Tm
