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pKa 's in a protein. Suppose the protein in problem 1 has an aspartic acid residue and a lysine residue at its surface. In bulk water, the pKa of the aspartic acid side group is 3.9 and the pKa of the lysine NH2 side group is 10.8. What are the pKas of these groups in a protein if it has a net charge of +20 e?

Short Answer

Expert verified
The pKa of aspartic acid will decrease (below 3.9), and the pKa of lysine will increase (above 10.8).

Step by step solution

01

- Understand the pKa Values

The pKa values are a measure of the acidity of a functional group. Here, we have pKa=3.9 for aspartic acid and pKa=10.8 for lysine. In water, these values indicate their tendency to gain or lose protons.
02

- Analyze the Protein Environment

In a protein with a net charge of +20e, the environment around the residues is different compared to bulk water. This could influence the pKa values of the aspartic acid and lysine residues.
03

- Predict the Effect of the Net Charge

A highly positive environment (net charge of +20e) will tend to decrease the pKa of acidic residues (like aspartic acid) and increase the pKa of basic residues (like lysine). This is because such an environment stabilizes the deprotonated form of acidic groups and the protonated form of basic groups.
04

- Estimate the New pKa Values

Given the highly positive environment:For the aspartic acid residue, the pKa will likely decrease from 3.9 to a lower value.For the lysine residue, the pKa will likely increase from 10.8 to a higher value.Without exact values for these changes, we can only predict the direction of the shift.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Proteins Environment
The environment within a protein significantly impacts the behavior of its residues. This is because the protein’s internal conditions differ drastically from bulk water.
Proteins can create unique microenvironments around their residues, influenced by factors like charge, polarity, and hydrogen bonding. When a protein has a high net charge, such as +20e, it usually indicates a highly charged and perhaps polar environment.
Such an environment can affect the protonation state of acidic and basic residues, which in turn influences their pKa values.
Acidic and Basic Residues
Aspartic acid and lysine are two examples of acidic and basic residues, respectively.
Each of these amino acids has a side chain with a specific pKa, which describes the pH at which half of the side chain molecules are protonated.
  • Aspartic acid has a carboxylate side chain with a pKa of 3.9 in water. It is acidic because its side chain tends to donate protons.
  • Lysine has an amino side chain with a pKa of 10.8 in water. It is basic because its side chain readily accepts protons.
In different environments, the pKa of these side chains can change, altering their tendency to be protonated or deprotonated.
Protonation and Deprotonation
Protonation and deprotonation refer to the gain and loss of protons, respectively.
These processes are critical because they affect the charge and function of protein residues.
  • When the pH of the environment is below the pKa, the residue is more likely to be protonated (gain a proton).
  • When the pH is above the pKa, the residue is more likely to be deprotonated (lose a proton).
In the context of a protein with a +20e charge, acidic residues like aspartic acid are more likely to lose protons (deprotonated) because the environment favors stabilizing the negative charge. Conversely, basic residues like lysine are more likely to retain protons (protonated) due to the positively charged environment. This shifts the pKa values for these residues, usually decreasing for acidic residues and increasing for basic residues.

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Most popular questions from this chapter

Free energy of charging a sphere. A sphere with radius a in a medium with dielectric constant D is uniformly filled with a charge of volume density ρ. Derive the electrical free energy of this sphere in two different ways: (a) Derive the potential field and then charge the sphere from charge density 0 to ρ. (b) Add fully charged shells of thickness dr from r=0 to r=a.

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A charged protein. Model a protein as a sphere with a radius of 20 A and a charge of 20e in water at 25C and Dw=78.54. Assume the sphere is uniformly charged on its surface. (a) In units of kT, what is the potential at a distance 30\AA from the protein surface? (b) What is the electrostatic free energy of the charge distribution on the protein in kcal mol1 ?

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Oil/water interfacial potential. Consider an uncharged oil/water interface. On the aqueous side, the proximity of the oil phase biases the orientation of the water molecules in an unknown way. Calculate the resulting potential across the boundary layer for the maximum bias, the complete line-up of the first layer of water dipoles perpendicular to the interface. Treat this layer as a parallel-plate capacitor, with one water molecule occupying 10 A2 of the interfacial area. The dipole moment is μ=1.85 debye (1 debye =3.336×1030Cm) per water molecule. Take two values of the dielectric constant D between the capacitor plates: (a) D=2, as for oil, and (b) D=80, as for bulk water. (If water is perfectly oriented-a situation called dielectric saturation-then D=2 is more likely.)

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