Protein binding
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Protein binding
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[ID] => 560313
[post_author] => 12815
[post_date] => 2025-01-14 07:00:01
[post_date_gmt] => 2025-01-14 12:00:01
[post_content] => Practice Passage (Question 1-6)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
Like protein folding, protein binding obeys the laws of thermodynamics and kinetics. The strength of the interaction between a protein and its target is described by the size of the dissociation constant (Kd), which is directly related to the standard free energy of binding (ΔGo). Protein interaction strength spans a wide spectrum from weak to strong and this range of binding affinities mediate a range of functions.
Transient interactions are reversible, usually short-lived, and may be weak or strong. For example, some weak binding transient interactions include reversible cell-to-cell contacts or transient signalling complexes. Other transient interactions are extremely strong (e.g., hormone–receptor interactions or protease–inhibitor interactions) with much longer half-lives of up to days.
While reversible transient interactions are abundant, arguably, the most important protein interactions are obligatory. Proteins involved in these obligatory interactions do not exist on their own. Instead they self-associate to permanently exist as multiprotein assemblies or oligomers that define a protein’s quaternary structure.
Figure 1 on the left shows the formation of an obligatory homodimer between two folded monomers, which creates an active site between the subunits. On the right is an example of a transient multifunctional complex between the homodimer and an unstructured region containing two phosphorylated residues within a second multi-domain protein. The formation of the multifunctional complex causes the two monomers within the dimer to rotate to reveal secondary binding sites and subsequently close the original active site.
Figure 1. Obligatory and transient interactions Given that non-specific targets may also bind to a given protein, we must also define the free energy of ‘specificity’ for the interaction. The ‘specificity’ of the interaction is a relative thermodynamic term that tells us how well a given protein binds to its target compared with other potential “non-targets” in the cell. More favourable energy changes upon binding are indicative of more favourable binding and therefore higher target specificity. Figure 2 shows the energy changes upon the binding of protein P to a non-target (A) and two different targets (B and C).
Figure 2. Thermodynamics of protein-target specificity
Passage and images adapted from Morris R, Black KA, Stollar EJ. Uncovering protein function: from classification to complexes. Essays Biochem. 2022 Aug 10;66(3):255-285. doi: 10.1042/EBC20200108. PMID: 35946411; PMCID: PMC9400073.
[post_title] => Protein binding
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[question] => When an additional hydrogen bond is added to an interaction, it is most likely to result in:
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[answer] => 1
[description] => Reason for the Correct Answer:
The addition of a hydrogen bond would strengthen a protein-protein interaction. The question is asking you how this would affect ΔG° or Kd and to what extent.
Changing the strength of the interaction will influence both ΔG° or Kd. The extent to which is will change these parameters can be approximated by the mathematical relationship between ΔG° or Kd, a formula you’re expected to know for the MCAT.
The relationship between ΔG° and Ka, the association constant, is:
ΔG° = –RT ln Ka
Since Kd = 1/Ka, the relationship between ΔG° and Kd is:
ΔG° = –RT ln (1/Kd)
This can also be described as:
ΔG° = +RT ln Kd
Or:
Kd = e^[(ΔG°/ RT)]
Because ΔG° is proportional to the natural logarithm of the association and dissociation changes, relatively small changes in ΔG° correspond to much larger changes in Kd.
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[answers] => Array
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[each_answer] => A.a small change in ΔG° but a larger change in Kd
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[each_answer] => B.a small change in Kd but a larger change in ΔG°
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[each_answer] => C.a significant change in ΔG° only.
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[each_answer] => D.a small change in Kd only.
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[1] => Array
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[quiz_unique_key] => 3873426850
[question] => Figure 1 is an example of allosteric regulation. Why?
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[answer] => 1
[description] => Reason for the Correct Answer:
Allosteric regulation occurs when a molecule binds to a site other than the active site on an enzyme, causing a conformational change that affects the enzyme’s activity.
Neither C nor D would help to explain why something is an allosteric regulator.
B mentions steric hindrance, but this does not necessarily involve binding at an allosteric site.
Choice A clearly describes the mechanism of allosteric regulation by indicating that the binding of phospho groups at a secondary site causes the closing of the active site, fitting the definition of allosteric regulation.
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[each_answer] => A.Phospho groups binding at a secondary site cause closing of the active site of the homodimer.
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[each_answer] => B.The active site of the homodimer becomes inaccessible due to steric hindrance following phospho group binding.
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[each_answer] => C.The allosteric site is similar to the active site of the homodimer.
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[each_answer] => D.Phospho groups on heterodimer are not responsible for closing the active site.
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[quiz_unique_key] => 2377279144
[question] => Interactions between which amino acids are LEAST likely to contribute to obligatory interactions?
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[answer] => 3
[description] => Reason for the Correct Answer:
Obligatory interactions are strong and often essential for protein function, usually involving specific, complementary properties of amino acids that stabilize protein structures.
Cysteine and cysteine can form disulfide bonds, which are strong covalent bonds that stabilize protein structures, making them likely to participate in obligatory interactions.
Arginine is positively charged, and aspartic acid is negatively charged. Their interaction can form strong ionic bonds, making them likely to contribute to obligatory interactions.
Aspartic acid is negatively charged, and lysine is positively charged. Their interaction can form strong ionic bonds, which are likely to be involved in obligatory interactions.
Valine is a hydrophobic amino acid, and glutamic acid is negatively charged. Their interaction is unlikely to result in strong, stable interactions needed for obligatory binding because they do not complement each other’s properties.
See the full chart of amino acids below.
https://commons.wikimedia.org/wiki/File:Homemade_Chart_of_the_20_Amino_Acids.jpg
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[each_answer] => A.Cysteine and cysteine
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[each_answer] => B.Arginine and aspartic acid
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[each_answer] => C.Valine and glutamic acid
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[each_answer] => D.Aspartic acid and lysine
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[quiz_unique_key] => 1723550837
[question] => In Figure 2 from the passage, which target is more specific for protein P and why?
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[answer] => 3
[description] => Reason for the Correct Answer:
While binding at multiple sites can contribute to specificity, the key factor in this context is the overall stability (ΔG) of the complex.
The energy diagram shows the ΔG values for the complexes PA, PB, and PC.
ΔG (Gibbs Free Energy change) indicates the stability and specificity of a protein-ligand complex.
The complex with the most negative ΔG (lowest energy) is the most stable and has the strongest, most specific binding.
This is complex PC, so target C binds the most strongly.
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[each_answer] => A.Target A because the Gibbs free energy of the protein complex PA is the greatest.
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[each_answer] => B.Target C because it is binding to protein P at two sites.
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[each_answer] => C.Target C because the ΔG of the protein complex PC is most negative.
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[each_answer] => D.Target A because ΔGspecificity is 0.
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[quiz_unique_key] => 2261298308
[question] => Which of the following represents a very strong transient interaction?
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[answer] => 4
[description] => Reason for the Correct Answer:
The dissociation constant (Kd) is a measure of the affinity between two molecules, with a lower Kd indicating a higher affinity.
The half-life of a complex indicates how long the complex remains intact before dissociating, with a longer half-life suggesting a more stable interaction.
In magnitude, mM (millimolar) > µM (micromolar) > pM (picomolar), and ms (milliseconds) > µs (microseconds) > ns (nanoseconds).
A half life of 12 minutes would indicate a stronger interaction than one of only milliseconds or nanoseconds.
The option that combines a very low dissociation constant (pM Kd), indicating a high affinity between the hormone and the receptor, with a long half-life (12 minutes), represents the interaction that is stable and persists for a considerable duration. This makes it a very strong transient interaction.
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[each_answer] => A.A signalling complex with a half-life of milliseconds and a micromolar Kd
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[each_answer] => B.A protein-protein interaction with a half-life of nanoseconds and a picomolar Kd
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[each_answer] => C.A protein nucleic acid complex with a half-life of milliseconds and a micromolar Kd
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[each_answer] => D.A hormone-receptor interaction with a half-life of 12 minutes and picomolar Kd
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[quiz_unique_key] => 83407773
[question] => When a protein with net charge approaches close to its target that has an opposite charge, the peptide first docks in multiple ways onto the surface of the protein via long-range/loose interactions resulting in an encounter complex. In a second step, the peptide orients itself correctly and goes on to bind as the final complex. Which is NOT true regarding the encounter complex?
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[answer] => 3
[description] => Reason for the Correct Answer:
The encounter complex is an intermediate formed when a protein and a peptide first interact through long-range, loose interactions. This complex is an essential step in the pathway from the free peptide/domain to the final, stable domain-peptide complex.
The encounter complex is a relatively stable intermediate, as it is lower energy than the free domain and peptide and transition states on the energy diagram.
The initial protein-peptide contact does last long enough for rearrangement, because the encounter complex allows the peptide to correctly orient itself on the protein surface, leading to the final stable binding.
The encounter complex facilitates faster binding of the protein to the peptide, as the encounter complex accelerates the formation of the final complex by allowing the peptide to find its correct orientation more efficiently.
Long-range electrostatic interactions ARE useful in encounter complex formation, as they are crucial for the initial docking of the peptide onto the protein surface, facilitating the formation of the encounter complex.
Long-range electrostatic interactions refer to the forces between charged groups that can act over relatively large distances, typically up to several nanometers, in biological systems. As a protein approaches its target, long-range electrostatic interactions play a crucial role in bringing the two molecules close enough to form an encounter complex.
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[each_answer] => A.It is a relatively stable intermediate formed during a pathway from free peptide/domain to the final domain-peptide complex.
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[each_answer] => B.The initial protein peptide contact lasts long enough for the peptide to rearrange on the surface of the protein to form the correct final complex.
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[each_answer] => C.Long-range electrostatic interactions are not useful in the encounter complex formation.
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[each_answer] => D.It facilitates the protein to bind to the peptide at a functional faster rate.
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