Amino acids are the building blocks of proteins. The presence of –NH2 and –COOH groups makes amino a — Analytical Chemistry Chemistry Question
Amino acids and enzymes
Amino acids are the building blocks of proteins. The presence of –NH2 and –COOH groups makes amino acids amphoteric in nature. Certain amino acid side chains in proteins are critically important for their reactivity and catalytic role. Glutamic acid is one such amino acid, whose structure is shown below.
[VISUAL]
Why is the pKa of the α-COOH group lower than that of the γ-COOH ?
Model Answer
The protonated amino group has an electron withdrawing effect. This enhances the release of proton from the neighbouring –COOH, by stabilizing the conjugate base –COO–. This effect is greater when the –COO– is physically closer to NH3+. As NH3+ group is present on the α-carbon, the effect is greater on α-COOH than that on the γ-COOH. So the pKa value of α-COOH is lower than that of γ-COOH.
Calculate the percent of γ-COOH group that remains unionized at pH = 6.3.
Model Answer
The ratio of ionized to unionized γ-COOH group is obtained by using Henderson-Hasselbalch equation,
pH = pKa + log([COO-] / [COOH])
The pH = 6.3 and pKa of γ-COOH group is 4.3. Substituting these values in the above equation we get,
6.3 = 4.3 + log([COO-] / [COOH])
2.0 = log([COO-] / [COOH])
[COO-] / [COOH] = 100
Thus, the percent of unionized [COOH] is 1 / 101 * 100% = 0.99 % at pH 6.3.
Glutamic acid is subjected to paper electrophoresis at pH = 3.25. Will it move towards the anode (+) or cathode (-)? Why ?
Model Answer
Glutamic acid has two pKa values lower than 7.0 and one pKa value higher than 7.0. Thus, the isoelectric point (pI) for glutamic acid will lie between the two acidic pKa values.
pI = (2.2 + 4.3) / 2 = 3.25
At pH = 3.25, net charge on glutamic acid will be zero since this pH coincides with pI of glutamic acid. Hence, glutamic acid will be stationary at pH 3.25.
Hydrolysis of polysaccharides like chitin, cellulose and peptidoglycan is a common biochemical process. This involves the hydrolysis of a glycosidic bond like the β-1,4 linkage shown below.
[VISUAL]
One such hydrolysis reaction is catalysed by lysozyme.
Suppose the lysozyme catalyzed reaction is performed in 18O enriched water, do you expect the 18O to be incorporated into the product? If yes, where?
Model Answer
In the hydrolysis of the glycosidic bond, the glycosidic bridge oxygen goes with C4 of the sugar B. On cleavage, 18O from water will be found on C1 of sugar A.
NOTE: The reaction proceeds with a carbonium ion stabilized on the C1 of sugar A.
The pH-activity profile of lysozyme is shown in the figure:
[VISUAL]
Explain this pH behavior in terms of two carboxylates (Asp-52 and Glu-35) present at the lysozyme active site (note : ionizable groups on the substrate are not involved). Write the ideal state of ionization at the lysozyme active site at pH 5.0.
Model Answer
Most glycosidases contain two carboxylates at the active site that are catalytically important. Lysozyme is active only when one carboxylate is protonated and the other is deprotonated. A descending limb on the alkaline side of the pH profile is due to ionization of –COOH. An ascending limb on the acidic side is due to protonation of –COO–. Thus the enzyme activity drops sharply on either side of the optimum pH. The ideal state of ionization at pH = 5 will be:
- Asp-52: COO-
- Glu-35: COOH
The pKa of Glu-35 in lysozyme active site is 6.0 and not 4.3 as found in the free amino acid. Which of the following local effects is likely to be involved?
- Enhanced negative charge
- Enhanced positive charge
- Enhanced polarity
- Diminished polarity
Model Answer
Answers 2 and 4 are correct. Ionization of COOH leads to generation of a negatively charged species, COO–. This charged species is poorly stabilized by diminished polarity and enhanced negative charge. Hence ionization of COOH group is suppressed and the pKa is elevated.
Organic model reactions have helped to understand many features of enzyme catalytic mechanisms. When a reaction is made intramolecular (like the enzyme catalysts do!), rate acceleration takes place as if the apparent reactant concentration felt at the site is enormously raised. The carboxylate group assisted hydrolysis of three phenylacetates and their rate constants (k) are shown below.
[VISUAL]
Calculate the effective local concentration of the COO- group felt in (2) and (3) above.
Model Answer
The ratios of pseudo-first order rate constant (at c(CH3COO–) = 1 mol dm-3) in (1) to the first order rate constants in (2) and (3) provide the effective local concentrations.
For (2): (0.4) / (0.002) = 200
i.e. the effective concentration = 200 mol dm-3
For (3): (20) / (0.002) = 10,000
i.e. the effective concentration = 10,000 mol dm-3
Why do you see a higher rate in (3) than in (2) ?
Model Answer
In addition to the enhanced local concentration effect, the COO- group in (3) is better oriented to act in catalysis. The double bond restricts the motion of COO- and thus reduces the number of unsuitable orientation of COO-, thereby enhancing the reaction rate.