An enzyme requires both aspartate (pKa of side chain = 4.5) and histid...
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.