the peptide hond vs the degree of rotation abstit the (j) Spatial arrangement of
ID: 197650 • Letter: T
Question
the peptide hond vs the degree of rotation abstit the (j) Spatial arrangement of arnino acid residues that are nearby in the sequence, such as a-helical and -sheet structures 3- 2. (14 pts) The pkx valu the three residues of ue of His-57 at the active site of chymotrypsin is 6.8. This histidine residue is one of the catalytice triad essential for catalysis. Answer the following questions. (a) Calculate the ratio of positively charged to neutral side chain of this histidine residue at physiological pH of 7.4 b) What is the percentage (%) of the neutral side chain? e) Does this histidine residue more likely to behave as an acid or a base under this condition? Esplain Calculate this from the answer in (a) whyExplanation / Answer
The ratio of deprotonated to protonated histidine can be abruptly estimated by using Henderson-Hasselbalch equation, pH=pKa+log([A]/[HA+]). It is given that pH=7.4 and pKa=6.8, [A]/[HA+]= 4, that means there is a 4 times more unprotonated compared to protonated.
Moreever, pKa value will change in a protein, because the pKa is controled by the local environment of the side chain of histidine. Due to this reason, many biochemical mechanisms shows that both protonated as well as deprotonated states are reachable at physiological pH. Histidines are frequently participated in acid/base chemistry, that is one of the mainr strategies used by an enzymes for catalysis.
By using the Henderson-Hasselbach equation, we can find the percentage of neutral histidines which are protonated when the pKa is 6.8.
At the pKa = 7.4, the pKa = pH; so the fraction of protonated histidines is 0.5.
When the pKa is 6.8, fraction of protonated H is:
7.4 = 6.8 + log[A]/[HA+], log[A]/[HA+] = 0.6, [A]/[HA+] = 3.981 =4
fraction of [HA+] = 1/1+4 = 1/5 = 0.2
The number of protons if the pKa goes from 6.8 to 7.4 is given by the difference 0.5 - 0.2 = 0.3
therefore, 70% chance of neutral side chain.
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