Ribose-bind protein in vitro reconstitution
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- Ribose-bind protein in vitro reconstitution
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Ribose-bind protein in vitro reconstitution
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[ID] => 560318
[post_author] => 12815
[post_date] => 2025-01-14 07:05:02
[post_date_gmt] => 2025-01-14 12:05:02
[post_content] => Practice Passage (Question 1-7)
*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.
Ribose-binding protein (RBP) from Thermotoga maritima (a thermophile) is a well characterized protein containing two distinct domains. RBP can be split into fragments representing its constituent domains and can be restored in vitro. It serves as an excellent model system for studying the evolution of multi-domain proteins that could have arisen from a gene duplication or a domain swapping event.
Researchers separated full-length RBP into two fragments, RBP-N (residues 30–153 of RBP) and RBP-Trunc (residues 142–310), then reconstituted in vitro using the individually expressed domains. Figure 1 shows the far-UV circular dichroism (CD) spectrum and intrinsic fluorescence spectrum of native RBP, RBP-N, RBP-trunc, and reconstituted protein. The far-UV CD spectrum provides information about the secondary structural elements, while the intrinsic fluorescence spectrum indicates the environment of aromatic residues, providing insights into the folding state of the proteins.
Figure 1. Biophysical characterization of native RBP, individual domains, and in vitro reconstituted hetero-dimer
Researchers then evaluated the efficacy (ribose binding capacity) of each protein using differential scanning calorimetry (DSC), which measures the heat absorbed (CP) during protein denaturation. Protein binding to ribose alters DSC results by increasing the thermal stability of the protein to which ribose is complexed. DSC results in the presence and absence of ribose are shown in Figure 2.
Figure 2. Thermodynamic characterization of the RBP constructs and their interaction with ribose by differential scanning calorimetry (DSC)
Adapted from Michel F, Romero-Romero S, Höcker B. Retracing the evolution of a modern periplasmic binding protein, 2023.
[post_title] => Ribose-bind protein in vitro reconstitution
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[post_modified] => 2025-01-14 07:05:02
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[questions] => Array
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[0] => Array
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[quiz_unique_key] => 578908434
[question] => Which of the following statements does NOT represent the data from Figure 2?
[value] => Array
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[answer] => 4
[description] => Reason for the Correct Answer:
Figure 2 shows heat absorbed (CP) by the proteins at various temperatures. The peaks therefore represent the melting points of each protein. (The statement in Choice A is correct.)
Figure 2a shows that the melting temperature of the native RBP (labeled RBP) was greater than the reconstituted protein. However, Figure 2b show that they had similar melting temperatures when ribose was present. (The statement in Choice B is correct.)
Comparison of Figure 2a and 2b shows that ribose binding increased the stability (and therefore the melting point) of the dimers but not the monomers. This suggests that ribose does not bind to the monomers. (The statement in Choice C is correct.)
The native protein had the highest melting point in Figure 2a and therefore highest stability, so Choice D is an incorrect statement.
)
[answers] => Array
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[0] => Array
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[each_answer] => A.Sharp rises in CP indicate the melting temperature of the proteins tested.
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[1] => Array
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[each_answer] => B.Native and reconstituted RBP proteins have similar thermal stability upon ribose binding.
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[2] => Array
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[each_answer] => C.The individual protein monomers do not appear to bind ribose.
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[3] => Array
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[each_answer] => D.RBP halves folded in vitro have greater stability than native protein.
)
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)
[1] => Array
(
[quiz_unique_key] => 3873426850
[question] => How does ΔG relate to the equilibrium constant K, and what happens to the ΔG of protein folding at the melting temperature of the protein?
[value] => Array
(
[answer] => 2
[description] => Reason for the Correct Answer:
The Gibbs free energy change (ΔG) is related to the equilibrium constant (K) by the equation: ΔG = −RTlnK. Here, R is the universal gas constant (8.314 J/mol·K), and T is the temperature in Kelvin.
K is the equilibrium constant, or ratio of products to reactants.
ΔG tells you whether the reaction is spontaneous (negative ΔG) or nonspontaneous (positive ΔG).
At the melting point, protein folding transitions from an unfavorable, nonspontaneous process to a favorable, spontaneous process. Here, Keq is 1 and ΔG is 0.
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[answers] => Array
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[0] => Array
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[each_answer] => A.ΔG = −RTlnK; at the melting temperature, ΔG equals 1.
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[1] => Array
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[each_answer] => B.ΔG = −RTlnK; at the melting temperature, ΔG equals 0.
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[2] => Array
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[each_answer] => C.ΔG = +RTlnK; at the melting temperature, ΔG is negative.
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[3] => Array
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[each_answer] => D.ΔG = +RTlnK; at the melting temperature, ΔG equals 1.
)
)
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[2] => Array
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[quiz_unique_key] => 2377279144
[question] => Which of the following statements best describes the scientific basis of using intrinsic fluorescence spectroscopy to assess protein tertiary structure?
[value] => Array
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[answer] => 2
[description] => Reason for the Correct Answer:
The passage states that “the intrinsic fluorescence spectrum indicates the environment of aromatic residues, providing insights into the folding state of the proteins.”
It thus measures the fluorescence of aromatic residues (e.g., tryptophan, tyrosine), providing insight into tertiary structure (protein folding).
Aromatic amino acid residues possess conjugated π-electron systems in their aromatic rings. When these aromatic residues are excited by UV light, their π-electrons undergo electronic transitions to higher energy states. The subsequent relaxation of these electrons to their ground state results in the emission of fluorescence.
The emitted fluorescence is sensitive to the local microenvironment surrounding the aromatic residues, including factors like solvent exposure, local polarity, hydrogen bonding, and interactions with neighboring amino acids. Changes in these environmental factors alter the intensity and wavelength of the emitted fluorescence, providing valuable insights into the tertiary structure, folding dynamics, and conformational changes of proteins. Therefore, intrinsic fluorescence spectroscopy is a powerful tool for studying protein tertiary structure based on the behavior of their conjugated π-electrons.
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[answers] => Array
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[0] => Array
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[each_answer] => A.Intrinsic fluorescence spectroscopy measures the absorbance of UV light by peptide bonds, providing information about protein secondary structure.
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[1] => Array
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[each_answer] => B.Intrinsic fluorescence provides insights into the tertiary structure and folding state of the protein through differences in light emitted from their conjugated π-electrons.
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[2] => Array
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[each_answer] => C.Intrinsic fluorescence spectroscopy measures the circular dichroism of UV light by chiral amino acids, revealing details about the protein’s tertiary interactions and stability.
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[3] => Array
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[each_answer] => D.Intrinsic fluorescence spectroscopy assesses the absorbance of visible light by protein-bound cofactors, offering information about the protein’s enzymatic activity and kinetics.
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[3] => Array
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[quiz_unique_key] => 1723550837
[question] => What changes describe the process of protein denaturation?
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[answer] => 1
[description] => Reason for the Correct Answer:
Protein denaturation will occur spontaneously at high temperatures.
Remember that ΔG = ΔH – TΔS.
Spontaneous processes occur when ΔG is negative.
A process that becomes spontaneous at high temperatures is characterized by a positive enthalpy change and a positive entropy change, such that the process becomes spontaneous when the magnitude of TΔS is greater than the magnitude of ΔH and ΔG becomes negative.
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[answers] => Array
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[0] => Array
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[each_answer] => A.Positive enthalpy change, positive entropy change
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[1] => Array
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[each_answer] => B.Positive enthalpy change, negative entropy change
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[2] => Array
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[each_answer] => C.Negative enthalpy change, negative entropy change
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[3] => Array
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[each_answer] => D.Negative enthalpy change, positive entropy change
)
)
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[4] => Array
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[quiz_unique_key] => 2261298308
[question] => Thermodynamic data (melting temperatures and enthalpy changes) measured during DSC analysis are represented in the table below.
Does this data confirm the interaction between the reconstituted hetero-dimer and ribose?
[value] => Array
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[answer] => 2
[description] => Reason for the Correct Answer:
The reconstituted hetero-dimer has a Tm of 99.7, and ΔH of +355.7 upon denaturation.
When bound to ribose, it has a Tm of 113.5, and ΔH of +484.8 upon denaturation.
Tm is the melting point and the positive ΔH gives you the amount of energy absorbed during denaturation.
The increase in Tm and ΔH for the denaturation of the heterodimer-ribose complex suggests that more energy is required to break the dimer-ribose interaction, as more energy is required to denature the protein when it’s bound to ribose.
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[answers] => Array
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[0] => Array
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[each_answer] => A.Yes, because less energy is required to break this interaction.
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[1] => Array
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[each_answer] => B.Yes, because more energy is required to break this interaction.
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[2] => Array
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[each_answer] => C.No, because less energy is required to break this interaction.
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[3] => Array
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[each_answer] => D.No, because more energy is required to break this interaction.
)
)
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[5] => Array
(
[quiz_unique_key] => 83407773
[question] => What is the enthalpy change for the binding of RBP-N and RBP-Trunc in vitro?
[value] => Array
(
[answer] => 3
[description] => Reason for the Correct Answer:
In this case, we’re looking for ΔH(binding), or the enthalpy change of binding of subunits RBC-N and RBC-Trunc. We’re given the enthalpy change for denaturation (unfolding) of each subunit and the bound protein, RBP-N/RBP-Trunc.
Enthalpy is a state function, which means you can calculate overall change from adding or subtracting individual steps.
We could express the enthalpy change of the folding of RBP-N/RBP-Trunc as the sum of the folding of each subunit plus the binding of the subunits:
ΔH(folding, RBPN/RBPTrunc) = ΔH(binding) + ΔH(folding, RBPN) + ΔH(folding, RBPTrunc)
However, we are only given unfolding values, which equal the negative of the folding values, so let’s substitute those in:
-ΔH(unfolding, RBPN/RBPTrunc) = ΔH(binding) – ΔH(unfolding, RBPN) – ΔH(unfolding, RBPTrunc)
Rearranging:
ΔH(binding) = ΔH(unfolding), RBPN/RBP – Trunc) – ΔH(unfolding, RBPN) – ΔH(unfolding, RBPTrunc)
Substituting values from the table data:
ΔH(binding) = 355.7 – 35.2 – 79.9 = 240.6
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[answers] => Array
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[0] => Array
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[each_answer] => A.-115.8
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[1] => Array
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[each_answer] => B.44.7
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[2] => Array
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[each_answer] => C.240.6
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[each_answer] => D.355.7
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[6] => Array
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[quiz_unique_key] => 3637642455
[question] => Below is a picture of SDS PAGE representing an in vitro binding assay of RBP fragments. The results support which conclusion?
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[answer] => 1
[description] => Reason for the Correct Answer:
The native RBP band appears around 30 kDa.
This is also around where the reconstituted (RBP-N/RBP-Trunc) should appear.
There is no 30 kDa band visualized for RBP-N/RBP-Trunc in the absence of ribose.
However, a 30kDa band corresponding to the full length RBP is observed in the last two lanes in the presence of 1mM and 10mM ribose
This indicates that the hetero dimer complex is not particularly stable under reducing conditions, at least for the reconstituted protein, but it is stabilized and therefore observable in the presence of higher concentrations of ribose.
)
[answers] => Array
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[0] => Array
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[each_answer] => A.The hetero dimer complex is stabilized in the presence ribose.
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[each_answer] => B.Hetero dimer formation is not observed irrespective of presence or absence of ribose.
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[2] => Array
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[each_answer] => C.Under denaturing gel conditions the hetero dimer complex formed is stable.
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[3] => Array
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[each_answer] => D.The two individual domains of RBP interact with ribose independently of each other.
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[560318|1] => D
[560318|2] => B
[560318|3] => B
[560318|4] => A
[560318|5] => B
[560318|6] => C
[560318|7] => A
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