Part 1: Rate of Acceleration by Trypsin
The rate of a reaction is related to the Gibbs free energy of activation (ΔG‡ΔG‡) by the Arrhenius equation and the Eyring equation. The Gibbs free energy of activation is given by:
ΔG‡=ΔH‡−TΔS‡ΔG‡=ΔH‡−TΔS‡
Here, we have:
ΔH‡ΔH‡ (enthalpy change of activation)
ΔS‡ΔS‡ (entropy change of activation)
TT is the temperature in Kelvin
Given that trypsin lowers the standard enthalpy of the transition state by 4.8 kcal/mol and increases the standard entropic favorability by 20 cal/mol·K, we can write:
ΔHcat‡=ΔHuncat‡−4.8 kcal/molΔHcat‡=ΔHuncat‡−4.8 kcal/mol
ΔScat‡=ΔSuncat‡+20 cal/mol\cdotpKΔScat‡=ΔSuncat‡+20 cal/mol\cdotpK
The change in Gibbs free energy of activation (ΔΔG‡ΔΔG‡) due to the catalyst can be expressed as:
ΔΔG‡=ΔGuncat‡−ΔGcat‡ΔΔG‡=ΔGuncat‡−ΔGcat‡
Since:
ΔGuncat‡=ΔHuncat‡−TΔSuncat‡ΔGuncat‡=ΔHuncat‡−TΔSuncat‡
ΔGcat‡=ΔHuncat‡−4.8−T(ΔSuncat‡+20)ΔGcat‡=ΔHuncat‡−4.8−T(ΔSuncat‡+20)
Therefore:
ΔΔG‡=[ΔHuncat‡−TΔSuncat‡]−[ΔHuncat‡−4.8−T(ΔSuncat‡+20)]ΔΔG‡=[ΔHuncat‡−TΔSuncat‡]−[ΔHuncat‡−4.8−T(ΔSuncat‡+20)]
ΔΔG‡=4.8+T⋅20ΔΔG‡=4.8+T⋅20
At standard temperature (25°C or 298 K):
ΔΔG‡=4.8 kcal/mol+(298 K)⋅(20 cal/mol\cdotpK)ΔΔG‡=4.8 kcal/mol+(298 K)⋅(20 cal/mol\cdotpK)
ΔΔG‡=4.8 kcal/mol+5960 cal/molΔΔG‡=4.8 kcal/mol+5960 cal/mol
ΔΔG‡=4.8 kcal/mol+5.96 kcal/molΔΔG‡=4.8 kcal/mol+5.96 kcal/mol
ΔΔG‡=10.76 kcal/molΔΔG‡=10.76 kcal/mol
The rate of acceleration due to the catalyst is given by the ratio of the rate constants, which is related to the exponential of the Gibbs free energy difference:
Rate acceleration=eΔΔG‡RTRate acceleration=eRTΔΔG‡
where R=1.987 cal/(mol\cdotpK)R=1.987 cal/(mol\cdotpK):
Rate acceleration=e10760 cal/mol(1.987 cal/(mol\cdotpK))⋅(298 K)Rate acceleration=e(1.987 cal/(mol\cdotpK))⋅(298 K)10760 cal/mol
Rate acceleration=e10760592.126Rate acceleration=e592.12610760
Rate acceleration=e18.17Rate acceleration=e18.17
Therefore, the rate of acceleration achieved by trypsin compared to the uncatalyzed reaction is approximately:
e18.17≈7.44×107e18.17≈7.44×107
Part 2: Concentration of Trypsin Bound to Its Substrate
Given:
Dissociation constant, Kd=1.4 mMKd=1.4 mM
Substrate concentration, [S]=2.8 mM[S]=2.8 mM
Trypsin concentration, [E]=3μM[E]=3μM
The concentration of the enzyme-substrate complex ([ES][ES]) can be determined using the following relationship:
[ES]=[E][S]Kd+[S][ES]=Kd+[S][E][S]
Plugging in the given values:
[ES]=(3×10−3 mM)(2.8 mM)1.4 mM+2.8 mM[ES]=1.4 mM+2.8 mM(3×10−3 mM)(2.8 mM)
[ES]=(3×2.8)×10−3 mM4.2 mM[ES]=4.2 mM(3×2.8)×10−3 mM
[ES]=8.4×10−34.2[ES]=4.28.4×10−3
[ES]=2×10−3 mM[ES]=2×10−3 mM
[ES]=2μM[ES]=2μM
Therefore, the concentration of trypsin bound to its substrate is 2μM2μM.
