Biochemical Thermodynamics

Question 1

protein-ligand binding equilibrium
(for 1:1 binding)

Answer 1

P + L ⇌ PL

Question 2

protein-ligand binding equilibrium
(for 1:n binding)

Answer 2

P + nL ⇌ PLn

note: subscritp n
note: for 1:1 binding, n=1 so equilbrium is P + L ⇌ PL

Question 3

association constant

Answer 3

Ka = [PLn]/[P][L]ⁿ

note: for 1:1 binding = [PL]/[P][L]

Question 4

dissocation constant

Answer 4

Kd = [P][L]ⁿ/[PLn]

note: for 1:1 binding [P][L]/[PL]

Question 5

units for Ka

Answer 5

1/M

Question 6

units for Kd

Answer 6

M

Question 7

relationship between Ka and Kd

Answer 7

Kd = 1/Ka

Question 8

stronger binding affinity means…

Answer 8

larger Ka
smaller Kd

Question 9

B

Answer 9

fractional occupancy
= [bound proteins]/[total proteins]
= [PLn]/[PLn]+[P]

(assuming that ligands bind togetger - cooperative binding)

note: for 1:1 binding = [PL]/[PL]+[P]

Question 10

langmuir isotherm

Answer 10

plot of B against [L]
B=[L]/Kd+[L]

Question 11

linearised langmuir isotherm

Answer 11

ln(B/1-B) = ln[L] - lnKd

y axis = ln(B/1-B)
x axis = ln[L]
gradient = 1 (1:1 binding)
intercept = -lnKd

Question 12

when binding is 1:1, what is the concentraion of L when B=0.5

Answer 12

1/2

Question 13

quick determination of Kd

Answer 13

(when binding is 1:1)
Kd = [L]1/2

Question 14

shift in isotherm when binding affinity is stronger

Answer 14

stronger affinity = smaller Kd
isotherm shifts to the left

Question 15

hill isotherm

Answer 15

B=[L]ⁿ/Kd+[L]ⁿ

Question 16

linearised hill plot

Answer 16

ln(B/1-B) = nln[L] - lnKd

y axis = ln(B/1-B)
x axis = ln[L]
gradient = n (1:n binding)
intercept = -lnKd

Question 17

what is the preferred measure of affinity?

Answer 17

Kd

Question 18

Thermal energy

Answer 18

RT (J mol-1)
(molar gas constant (J K-1 mol-1) x temperature (K))

Question 19

value of R

Answer 19

8.314 JK-1mol-1

Question 20

why can’t protein unfolding happen at lower temperatures?

Answer 20

at lower temperatures, RT is less than the enthalpy gap between the folded and unfolded proteins states

Question 21

Wpl and Wp+l

Answer 21

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

Question 22

relate Kd to the enthalpy gap and number of conformations

Answer 22

Kd = (Wp+l/Wpl) e^-(Hp+l-Hpl/RT)

Question 23

boltzmanns law of entropy

Answer 23

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

Question 24

entropy of dissociation

Answer 24

ΔSd = R ln Wp+l/Wpl

Δ because dissociation involves a change in entropy

Question 25

enthalpy of dissociation

Answer 25

ΔHd = Hp+l - Hpl ## Footnote Δ because dissociation involves a change in enthalpy

Question 26

gibbs free energy expresses...

Answer 26

... stability as a balance between sufficiently low enthalpy and sufficiently large entropy G = H-TS

Question 27

change of gibbs free energy

Answer 27

ΔG = ΔH - TΔS

Question 28

gibbs free energy of dissocation

Answer 28

ΔGd = -RTlnKd ## Footnote Δ because dissociation involves a change in gibbs free energy

Question 29

a favourable change ...

Answer 29

K is large ΔG is negative ΔH is negative ΔS is positive

Question 30

van't Hoff plot

Answer 30

ln K = -ΔH/RT + ΔS/R ## Footnote y axis = lnK x axis = 1/T intercept = ΔS/R gradient = -ΔH/R

Question 31

what is the purpose of calorimetry

Answer 31

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)

Question 32

Cp

Answer 32

heat capacity - the amount of thermal energy (RT) required to raise the temperature by 1 K Cp = Δq/ΔT units: Jg-1K-1

Question 33

output of calorimetry

Answer 33

graph of Cp against T gives the value of Cp at each temperature whilst heating through a range

Question 34

how can ΔH be found from a plot of Cp against T?

Answer 34

the area under the line = ΔH

Question 35

Differential Scanning Calorimetry

Answer 35

Cp(protein) = Cp(protein + water) - Cp(water) ## Footnote 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

Question 36

what does the peak in DCS output correspond to?

Answer 36

the temperature at which protein unfolding/denaturing/melting takes place = Tm ## Footnote 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.

Question 37

why are base lines included on a DCS output?

Answer 37

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

Question 38

find ΔHm on a DCS output

Answer 38

the area between the curve and the baselines ΔHm = the amount of heat/energy required to overcome non-covalent interactions during denaturation

Question 39

entropy of unfolding

Answer 39

ΔSm = ΔHm / Tm