Midterm II Flashcards
What is an elementary step?
A step with a single transition state
For an elementary step the molecularity (number of ‘different’ molecules involved) and the rate order are the same
Eg. V = [A]^a[B]^b
What is Km?
Michaelis Constant
Km = [S] at Vmax/2
Only if Sm = Ks (Kp at steady state) is it meaningful as binding affinity. Otherwise it is a collection of rate constants.
What is Kcat?
Kp
Vmax = Kcat[E]T
It is the turnover number. In other words, the time it takes one enzyme molecule to convert one substrate to one product per active site.
What is the equation for enzyme efficiency?
kcat/km
The efficiency of an enzyme can be expressed in terms of kcat/Km. This is also called the specificity constant and incorporates the rate constants for all steps in the reaction up to and including the first irreversible step. Because the specificity constant reflects both affinity and catalytic ability, it is useful for comparing different enzymes against each other, or the same enzyme with different substrates. The theoretical maximum for the specificity constant is called the diffusion limit and is about 108 to 109 (M−1 s−1). At this point every collision of the enzyme with its substrate will result in catalysis, and the rate of product formation is not limited by the reaction rate but by the diffusion rate. Enzymes with this property are called catalytically perfect or kinetically perfect.
Why is dialysis sometimes useful for determining if an enzyme inhibitor is reversible?
Recovering activity by dialysis over` time can show that a very strong reversible inhibitor is indeed reversible
What are four types of enzyme inhibition?
- Competitive
- Uncompetitive
- Noncompetitive
- Linear mixed type
List the four interactions that stabilize a protein
- Ion pair interactions
- Van der Waals interactions
- H bonds
- Hydrophobic Interactions
The orientation of H bonds is important. Is this the case for charge-charge electrostatic interactions?
No
U = Electrostatic force
What shows us that electrostatic ion-pair interactions aren’t ‘that’ important for stabilizing proteins?
- Majority of ion pairs (salt bridges) are not found in the protein core. They are found in solvent accessible regions of the protein.
- In the hydrophobic core, the dielectric constant is low (around 3-4), and so electrostatic force (U) is high.
- In the solvent (~water) the dielectric constant is around 80, and therefore the electrostatic force is pretty weak.
- The net effect is that this type of energetic stabilization is pretty low
Give the three classes of van der waals interactions, rated in order from strongest to weakest.
- Dipole-dipole (eg. C=O, -9.3 kJ/mol)
- Dipole-induced dipole
- London dispersion (0.5-2 kJ/mol)
What is the net effect of london dispersion forces?
Highly stabilizing due to many small interactions
- Eg. interatomic contact in protein core and protein ligand binding providing steric interactions (contributing to binding energy)
How do H bonds contribute to protein stability?
Contribute for secondary structure, but not very stabilizing otherwise. About -2-8 kJ/mol per internal H bonding
- This is because water is just as good at forming H bonds. If there was no water, it would be about -20 kJ/mol of dipole-dipole interaction per H bond, which would be a lot!
How does hydrophobicity contribute to protein stability?
- Increases entropy, but no enthalpic contribution
- Net effect is negative ΔG
The negligible enthalpy change is due to the fact that there are similar interactions in nonpolar solvent and water.
BUT, entropy change is from the fact that water forms ordered structures to satisfy H-bonding and this increases temp requirement for energy (higher heat capacity) and a resulted lower entropy. So putting solute in non-polar solvent (hydrophobic) resultants in high entropic gain
How is ΔG practically independent of temperature when hydrocarbons are transferred from liquid (non-polar) to aqueous solution?
Because both enthalpy and entropy increase with temp and dependence to temperature arises from difference in Cp between phases (hydrophobic effect)
Hydrophobic effect decreases as temperature decreases, due to increased water ordering (which increases the solubility of nonpolar molecules).
How does ΔG change as a function of buried surface area?
More positive as buried area increases
H doesn’t change
Entropy becomes more negative as buried area increases (not favourable)
