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Summary

Summary Protein Science (AM_470145)

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This document consists of a summary of the course Protein Science, corresponding to the Master Biomolecular Science.

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  • October 26, 2023
  • 46
  • 2023/2024
  • Summary
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Available practice questions

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Some examples from this set of practice questions

1.

Attractive interactions can occur even between two non-polar groups within a protein molecule. These attractive interactions are called van der Waals interactions or London dispersion forces. Explain how these interactions originate and can lead to attraction between two non-polar groups.

Answer: Spontaneous dipole occurs due to temporary, stochastic, uneven electron distribution. The temporary dipole induces a dipole in neighboring molecule/group. The induced dipole has the same orientation as the temporary dipole, so the two dipoles attract each other.

2.

What does the term “electronegativity” mean? With regard to electronegativity, explain the differences in electron distribution between a C-C, a C-H, and a C-O single bond

Answer: Electronegativity describes the ability to attract electrons within a chemical bond. C-H slightly polar, partial negative charge at C Strongly polar, partial negative charge at O C-C unpolar, no difference in electronegativity

3.

Explain what the term “dielectric constant” means. What are typical values for the dielectric constant in water and in a bio-membrane?

Answer: The dielectric constant quantifies the influence of the medium on the electrostatic interaction energy. Typical values are 80 for water and 4 for a biomembrane

4.

The equilibrium constant describing a chemical reaction is related to the free energy change of this reaction. How exactly does the equilibrium constant depend on the free energy change (not at all, linearly, exponentially, logarithmically,…..) and what is an important consequence of this dependency?

Answer: The equilibrium constant depends exponentially on the free energy of the reaction. As a consequence, small energy changes can drastically change the equilibrium.

5.

The energy difference between properly folded (native) and completely unfolded state is quite small (20-30 kJ/mol, corresponding to energy of ~5 H-bonds, compare to covalent bond >200 kJ/mol). Explain this phenomenon and state two important consequences of the low free energy difference.

Answer: Residues can form non-covalent interactions with other residues, stabilizing the protein. However, residues can alternatively form bonds with water, which gives nearly the same energy. So, the interactions with other residues in the native structure and interactions with water in the unfolded state nearly cancel out. Moreover, the entropy typically decreases when the protein is folded (less flexible), but the entropy of the water molecules that were arranged in the water cage in the unfolded state (and are released into bulk wate upon folding) increases. Consequences: 1. Proteins quite unstable, to be handled with care 2. A single interaction may lead to small conformation change

6.

Several algorithms are available for predicting the secondary structure of a protein based on its known primary amino acid sequence, e.g. the Chou-Fasman approach. Explain the principle of how this approach works.

Answer: Each residue has its characteristic likelihood to be located in a helix/sheet/loop etc (calculated based on information from known structures). For a chosen stretch of amino acid residues in the new sequence (10-20 residues) these probabilities are summed up and normalized. This procedure is repeated for the next stretch of residues and so on.

7.

Does a hydrophobic (as compared with water) environment increase or decrease the pKa value of a basic amino acid side-chain? Give reasons for your answer.

Answer: In a hydrophobic environment charges are disfavored, therefore the protonated/deprotonated equilibrium of a basic reside will be shifted towards the deprotonated state. Consequently, the pKa will decrease.

8.

Serine proteases represent an important class of proteases. Next to the catalytically active serine residue these proteases display an oxyanion hole. Which role do you expect for the oxyanion hole related to the transition state and the proteolysis reaction?

Answer: The oxyanion hole binds to the oxygen atom of the peptide bond that is about to be split. This oxygen carries a stronger (partial) negative charge in the transition state (“oxyanion”) as compared to the substrate. Consequently the oxyanion hole stabilizes the transition state.

9.

Interactions between an inhibitor and a protein can be characterized by the KI value. What is a KI value? Give an equation and explain in words. The inhibitor I binds to the enzyme E in a 1:1 ratio. You know the KI value, 10^-5 M. What percentage of enzyme molecules has the inhibitor bound if [I] = 10^-4 M or 10^-7 M, respectively?

Answer: KI = [I[E]/[IE] It represents the equilibrium dissociation constant of the inhibitor. With [I] = 10^-4 90 % of the enzyme molecules have the Inhibitor bound With [I] = 10^-7 1 % of the enzyme molecules have the inhibitor bound

10.

Explain the molecular mechanism of the reversible switch employed by G-proteins.

Answer: These proteins switch back and forth between a GTP and a GDP bound state. If GTP is bound the terminal phosphate forms hydrogen bonds that move a protein domain, if GTP is hydrolyzed and replaced by GDP then this phosphate is not present, the hydrogen bond not formed and the domain moves back. This structure change can be used for signal transduction.

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