Recent Question/Assignment

Question 1: HIV Protease Enzyme System
Human immunodeficiency virus (HIV) was discovered in 1983 and instigates acquired immunodeficiency syndrome (AIDS). Found within the virus is HIV Protease that cleaves two 55 kDa and 160 kDa precursor polypeptides that are essential to the virus and involved in the viral life cycle. As a result of this, HIV Protease has been the target of therapeutics for many years.
Figure 1: Ribbon diagram of HIV Protease (1hvr.pdb)
Figure 1 is a ribbon structure representation of HIV Protease with the bonds of the catalytic residues Asp25 and Asp25’ (called Asp25-prime) shown as sticks in the centre.
An example polypeptide substrate for HIV-Protease is shown below in Figure 2 together with the active site side chain residues drawn below it. R and R’ corresponds to extensions of the polypeptide chain beyond the residues and bonds of interest.
Hydrolysis of the substrate peptide bond is initiated by a reaction with a single water molecule. The water molecule attacks the peptide bond because it is influenced by base catalysis provided by the side chain of Asp25’.

Figure 2: HIV-1 Protease Active Site and Substrate
(1a) Using Figure 2 as your starting point, draw the enzyme mechanism that describes the catalysed hydrolysis of the substrate to form two products. Label clearly throughout your mechanism both catalytic aspartic acid residues.
Use the following points to guide you:
i. Use Figure 2 as your starting point for the mechanism
ii. Insert a water molecule to be initiated in the base catalysis step with residue 25’
iii. Identify the products formed by hydrolysing the substrate in the figure
iv. Remember that residue 25 is there for a reason: your mechanism should use it
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(1b) How would you chemically describe the roles of Asp25 and Asp25’ in the mechanism?
[2]

Question 2: Hen Egg White Lysozyme Enzyme System
Hen egg white (HEW) lysozyme is a small enzyme of 129 amino acids with a molecular weight of 14.5 kDa. It was the first enzyme structure to be solved (using x-ray crystallography in 1967).
Lysozyme acts as a mild anti-bacterial agent that works by breaking the glycosidic linkage between N-acetylmuramic acid (NAM) and N-acetylglucosamine (NAG): components found in bacterial cell walls. The structure of lysozyme is shown below.

Figure 3: HEW lysozyme structure, key residues A and B are shown in the active site cleft.
A pH dependence study of kcat/KM highlighted that the reaction rate was linked to the ionised state of two carboxyl side chain groups: one with a pKa ~ 6 and one with a pKa ~ 4. These amino acids are known to be at positions (A) and (B) in the figure above.
(2a) If lysozyme operates at pH 5, what is the predominant carboxyl side chain species for the pKa ~ 6 side chain and the pKa ~ 4 side chain?
[2]

The pKa values of all carboxyl side chain groups in HEW lysozyme were determined using 2D NMR spectroscopy and are listed below together with the secondary structure element they are found in:
Amino Acid pKa Secondary structure element where amino acid resides
Asp 18 4.1 turn
Asp 48 6.2 turn
Asp 52 3.7 ?
Asp 66 4.0 turn
Asp 87 4.1 turn
Asp 101 3.9 turn
Asp 119 4.0 ?

Glu 7 4.2 ?
Glu 35 6.2 ?
(2b) As described earlier, key residues (A) and (B) in figure 3 are known to be either/or aspartic acid or glutamic acid. Using all other information you have been given, deduce the amino acids that are key residues (A) and (B).
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(2c) Give reasons for your deduction above in (2b).
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(2d) Parts of the mechanism for lysozyme are shown below. The catalytic lysozyme residue side chains can be identified as being above and below the polysaccharide chain shown at point A. Complete the missing structures and curly arrows in the mechanism. Structures are missing at points B and E and curly arrows are missing from points B, C, D and E. Point E is the Enzyme + Products point and the end of the reaction.
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(2e) From the knowledge you have gained, label the catalytic amino acids at point A in your mechanism with their correct name and residue number.
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Question 3: Analytical NMR
Look carefully at compound X below:

Figure 4: Compound X
(3a) Using letters (a, b, c, d etc.) label each unique carbon chemical environment on the structure above with a different letter.
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(3b) Sketch the 1D 1H NMR spectrum you would expect for the compound X (including spin-spin couplings). On your spectrum, label each 1H peak with the correct letter from part (3a) that defines the carbon to which each 1H is attached.
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(3c) Sketch the 1D 13C NMR spectrum you would expect for compound X above (assume it is 1H decoupled). On your spectrum, label each 13C peak with the correct letter from part (3a) that defines the carbon in the spectrum.
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(3d) Draw the DEPT-90 and DEPT-135 13C spectra for compound X. Label each carbon peak using your notation from (3a).
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(3e) Sketch what you would think the 13C, 1H HSQC spectrum of compound X would look like. Label each carbon peak using your notation from (3a).
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Question 4: Binding and protein NMR
Look at the 15N, 1H HSQC spectra shown below. Both spectra were obtained two samples of the same 14.5 kDa protein at 25°C. SDS-PAGE of samples A and B show only pure protein that run at 14.5 kDa on the gel. Samples A and B samples were also analysed by protein sequencing and electrospray mass spectrometry and both techniques confirm the same protein present by weight and sequence.

(4a) Ring (or highlight) all the NH2 side-chain groups you can see on spectrum (A) and spectrum (B).
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(4b) What might have caused the differences in the NMR spectra in A and B?
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Compound Y is a potential drug candidate that is designed to inhibit the activity of protein in sample B above. In an attempt to determine the affinity of the binding interaction of compound Y to the protein in B, a series of 15N, 1H HSQC spectra were collected at a protein concentration of 0.1 mM, and compound Y concentrations up to 5 mM. Some of the peaks are seen to shift as function of compound Y concentration as show below:

The overall changes in the chemical shift of one of the peaks is shown in the table below:
Concentration of compound Y / mM NMR chemical shift change / 103 • ppm
0.0 0.0
0.2 6.1
0.4 14.7
0.6 18.6
0.8 23.8
1.0 23.5
1.2 28.0
1.4 30.2
1.6 28.5
1.8 32.0
2.0 36.2
2.2 34.3
2.4 34.4
2.6 36.7
2.8 36.0
3.0 38.1
3.2 37.6
3.4 38.4
3.6 39.9
3.8 42.1
4.0 40.7
4.2 43.5
4.4 36.5
4.6 40.0
4.8 41.7
5.0 39.9
(4c) Draw the experimental binding isotherm data in a labelled figure showing the NMR chemical shift change as function of the concentration of the compound Y added.
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(4d) Based on the above experimental results, what is the likely affinity (Kd) of compound Y binding to the protein in sample B? Show in the same graph how the binding curve will look like with your estimated Kd value in a graph, and calculate how much of the protein is saturated with compound Y at 5 mM addition of Y.
You can use the following points to guide you:
i. Using all experimental information available, guess a possible value of Kd as well as a possible scaling factor linking the chemical shift change and the concentration of the complex or % saturation.
ii. Plot how such a binding curve would look like in the same graph as the data and compare. The use of graphing software such as Excel will simplify this task.
iii. Refine your guess until the binding curve represent the data as closely as possible.
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(4e) What is the standard free energy of association (?G°) for the above binding reaction? Show how you have calculated the ?G° value.
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(4f) Is compound Y a good drug candidate? Explain why or why not.
[2]
End of assessment