An enzyme requires both aspartate (pKa of side chain = 4.5) and histidine (pKa of side chain = 6.5) residues in the catalytic site to be protonated for activity. The expected enzyme activity (in %) at a pH of 5.5 would be closed toa)90b)78c)50d)10Correct answer is option 'D'. Can you explain this answer?

Question

Answer

Explanation:

To understand why the expected enzyme activity is close to 10% at pH 5.5, we need to consider the pKa values of the aspartate and histidine residues in the catalytic site.

At pH 5.5, both the aspartate and histidine residues will be partially protonated. The pKa of aspartate is 4.5, so at pH 5.5, the aspartate residue will be mostly protonated but some of it will be deprotonated. The pKa of histidine is 6.5, so at pH 5.5, the histidine residue will be mostly deprotonated but some of it will be protonated.

The enzyme requires both the aspartate and histidine residues to be fully protonated for activity. Since both residues are only partially protonated at pH 5.5, the enzyme activity will be reduced.

The degree of protonation of the aspartate and histidine residues can be calculated using the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

where pH is the pH of the solution, pKa is the pKa of the acid, [A-] is the concentration of the deprotonated form of the acid, and [HA] is the concentration of the protonated form of the acid.

Using this equation, we can calculate the degree of protonation of the aspartate and histidine residues at pH 5.5:

For aspartate:

pH = 4.5 + log([A-]/[HA])

5.5 = 4.5 + log([A-]/[HA])

log([A-]/[HA]) = 1

[A-]/[HA] = 10

This means that at pH 5.5, there will be 10 times more deprotonated aspartate than protonated aspartate.

For histidine:

pH = 6.5 + log([A-]/[HA])

5.5 = 6.5 + log([A-]/[HA])

log([A-]/[HA]) = -1

[A-]/[HA] = 0.1

This means that at pH 5.5, there will be 10 times more protonated histidine than deprotonated histidine.

Since both the aspartate and histidine residues need to be fully protonated for enzyme activity, the enzyme activity will be reduced when both residues are only partially protonated. Therefore, the expected enzyme activity at pH 5.5 will be close to 10%.

Answer

Any chemical group that presents in a environment below the Pka points is highly pronated and above the pka is highly/mostly depronated state. With the above point as reference, both the chemical groups/molecules are needed for proceeding the reaction to forward by the enzyme.

In pH 5.5, available aspartate % for the reaction is,

The ratio of acid of a molecule with pKa 4.5 and solution pH of 5.5 can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-] / [HA])

Where:

pH = solution pH

pKa = acid dissociation constant

[A-] = concentration of the conjugate base

[HA] = concentration of the acid

To solve for the ratio of [A-] to [HA], we can rearrange the equation:

[A-] / [HA] = 10^(pH - pKa)

Substituting the values given, we get:

[A-] / [HA] = 10^(5.5 - 4.5) = 10

Therefore, the available aspartate is 10 %

Then, Available histidine for the reaction is,

The ratio of acid of a molecule with pKa 4.5 and solution pH of 5.5 can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-] / [HA])

Where:

pH = solution pH

pKa = acid dissociation constant

[A-] = concentration of the conjugate base

[HA] = concentration of the acid

To solve for the ratio of [A-] to [HA], we can rearrange the equation:

[A-] / [HA] = 10^(pH - pKa)

Substituting the values given, we get:

[A-] / [HA] = 10^(6.5 - 4.5) = 100

Therefore, the available Histidine is 100 %

Finally, the requires both the molecules for the reaction. Even though, most of the histidine molecules are available(100%), aspartate is only 10 %. So, the reaction will be entirely depending on the aspartate % for the enzyme reaction.

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