BioChem Exam 2 - SU 2013

evan406's version from 2016-03-21 23:55


Question Answer
List the posttranslational modifications that take place at the N-terminus.-Removal of Met
- Acetylation
- Fatty acyl groups
List and describe the posttranslational modification "Removal of Met". Describe the function of these changes.-The initiating amino acid in all proteins is Met, which is formylated in prokaryotes.
-In many proteins the formyl group and the Met residue is cleaved off.
-This creates a wider variety of amino acids at this position which affect the half-life of the protein.
List and describe the posttranslational modification "Acetylation". Describe the function of these changes.The N-terminus is acetylated.
50% of eukaryotic proteins are modified in this way, 90% of human cytosolic proteins
List and describe the post translational modification "Fatty acyl groups". Describe the function of these changes.Fatty-acid-associated proteins can be attached to the lipid bilayer of the membrane and are usually membrane proteins


Question Answer
How does urea destroy native protein structure?Urea acts primarily by disrupting hydrophobic interactions.
How does high temperature destroy native protein structure?High temperature provides thermal energy greater than the strength of the weak interactions (hydrogen bonds, electrostatic interactions, hydrophobic interactions, and van der Waals forces) breaking these interactions.
How does detergent destroy native protein structure?Detergents bind to hydrophobic regions of the protein, preventing hydrophobic interactions among several hydrophobic patches on the native protein.
How does low pH destroy native protein structure?Low pH causes protonation of the side chains of Asp, Glu, and His, preventing electrostatic interactions.


Question Answer
Describe in detail the mechanism of folding by the chaperones DnaJ/DnaK1) DnaJ binds to regions of unfolded hydrophobic residues on the target polypeptide preventing inappropriate aggregation.

2) The polypeptide is then transferred to the low affinity ATP-DnaK and DnaJ is released

3) This stimulates ATPase activity of DnaK causing a conformational change into a high affinity stable ADP-DnaK- substrate complex

4) GrpE binding to DnaK results in the dissociation of ADP and binding of ATP.

5) This destabilizes the interaction between DnaK and the substrate protein causing the release of substrate from the chaperone.
Draw the ligand-binding plot (θ vs. [L]) for a protein which displays:
-tight binding of L
-weak binding of L
also label Kd values
Normally the reaction of O2 with Fe2+ results in the formation of Fe3+ and the superoxide radical. Explain how the structure of heme c and the binding sites on myoglobin and hemoglobin are designed to prevent this from happening.1) The heme structure is designed to prevent oxidation of Fe2+.

2) The coordinating nitrogen atoms (which have an electron-donating character) help prevent conversion of the heme iron to the 3+ state.

3) Iron in the 2+ state binds oxygen reversibly; in the 3+ state it does not bind oxygen.

4) A nitrogen of a His residue is the fifth ligand to the heme iron.

5) The protein structure is designed to increase heme c affinity for O2.

6) Hydrogen bond formation between the distal His and O2 highly affects the ligand affinity.


Question Answer
Explain in detail the allosteric effect in hemoglobin.-An allosteric protein is one in which the binding of a ligand to one site affects the binding properties of another site on the same protein.

-Hemoglobin exists in two conformations, the R(elaxed) state and the T(ense) state.

-Oxygen can bind to hemoglobin in either state, but has a higher affinity for hemoglobin in the R state.

-The binding of oxygen to a hemoglobin subunit in the T state triggers a change in conformation to the R state.

-In the T state, the porphyrin is slightly puckered, causing the heme-iron to protrude somewhat on the proximal His side. The binding of oxygen causes the heme to assume a more planar conformation, shifting the position of the proximal His and the attached F helix. These changes lead to the T → R transition.

-This change starts in one subunit, which is sensed by the other three subunits that as a result also change to the R state resulting in the binding of three more oxygen molecules:

-The first molecule of oxygen that interacts with deoxyhemoglobin binds weakly, because it binds to a subunit in the T state. Its binding, however, leads to conformational changes that are communicated to adjacent subunits, making it easier for additional molecules of oxygen to bind.
Characteristics of allosteric enzymes?-may have binding sites for regulatory molecules that are separate from active sites

- generally have more than one subunit

-interconvert between a more active form and less active form


Question Answer
What is the effect on purified hemoglobin of the following treatment on the affinity for oxygen?

Increase in pH from 7.2 to 7.4
-Due to the Bohr Effect the affinity will go up.
-The increase in pH causes a deprotonation of the HC3 His residue that in the protonated form stabilizes the T-state.
-equilibrium between the T- and R-state moves towards the R-state which has a higher affinity for oxygen.
What is the effect on purified hemoglobin of the following treatment on the affinity for oxygen?

Increase in pCO2 from 10 to 40 torr.
-Due to the Bohr Effect the affinity will go down.
-CO2 will react with the amino terminus of all four subunits, forming carbamate intermediates.
-protons are released in side reaction, lowering the pH and causing the protonation of the HC3 His residue and therefore a shift towards the T-state.
What is the effect on purified hemoglobin of the following treatment on the affinity for oxygen?

Dissociation of the α2β2 protein into monomeric subunits.
-The separation of the subunits causes the loss of the allosteric binding effects.
-If the separate subunits are in the T-state the affinity will go up in comparison to the tetrameric form.
-If the separate subunits are in the R-state the affinity will go down in comparison to the tetrameric form.


Question Answer
Give the Michaelis Menten equation and the reactions used to derive this equation.
List the important assumptions used by Michaelis and Menten to derive a rate equation for this reaction.1) [P] = 0, so that the rate of the reaction depends exclusively on the breakdown of ES and is not influenced by the reverse reaction; that is, k-2 can be ignored and V0 = k2 [ES]. This condition is possible only if early reaction times are measured; the velocity, therefore, is an initial velocity.

2) the rate of ES formation equals the rate of ES breakdown; in other words, the reaction is at a steady state.

3) S] >> [Et], so that total [S], which equals free substrate and enzyme-bound substrate, is essentially equal to [S].
The Michaelis-Menten constant, Km, is actually a summary of three terms. What are they?Km = (k2 + k-1)/ k1,

k-1: rate constant for the breakdown of the ES complex to form E + S
k1: rate constant for the association of the ES complex
k2: rate constant for the breakdown of ES to form E + P.
How is Km determined graphically?1) on a plot of V0 vs. [S] by finding the [S] at which V0 = 1⁄2 Vmax.

2) on a lineweaver burk plot the x-axis intercept = –1/ Km


Question Answer
The following set of data is for an enzyme that is known to follow Michaelis-Menten kinetics.

a) What is the Vmax for the enzyme. Explain your answer.
b) What is the Km for the enzyme.
c) When [S] = 4, what will V0 be?
(a) Vmax is about 700.

(b) Km is about 8 μM, the [S] at which the velocity equals 1/2 Vmax. The [S] at ~350 is about 8 μM.

(c) V0=233 μmol/min
An enzyme follows Michaelis-Menten kinetics. Indicate (with an "x") which of the kinetic parameters at the left would be altered by the following factors.
Explain in detail what an uncompetitive inhibitor is. Explain how it in theory interacts with a protein.An uncompetitive inhibitor interacts solely with the ES complex:

It effectively lowers the concentration of ES available for the reaction, changing k2 and therefore Vmax:
Explain how the theoretical kinetic measurements of enzyme activity are affected by the presence of an uncompetitive inhibitorVmax/α’, with α’ = 1 + [I]/KI
The apparent Km value is also affected: Km/ α’.

The new Michaelis-Menten equation becomes:

The Lineweaver-Burk plots can be drawn as such: