Question

1) The pKa for the imidazole group in histidine in Papain's catalytic centers is 8.3.

What is the ratio of protonated and deprotonated His-R groups at pH 8.2?

		a) 5 NH+ / 4 N:
		b) 1 NH2+ / 1 NH
		c) 4 N: / 5 NH+
		d) 1 NH / 1 NH2+

2) What is the ratio of protonated vs deprotonated Cysteine sulfhydryl groups in the catalytic center of papain, pKa 3.4, at optimal pH of 6.2?

		a) 1 S- / 631 SH
		b) 1 SH- / 631 SH2
		c) 631 S- / 1 SH
		d) 631 SH- / 1 SH2

3) What is the ratio of protonated vs deprotonated Histidine R-groups in papain within the catalytic center, pKa = 8.3, at optimal pH of 6.2?

		a) 2 N: / 250 NH+
		b) 2 NH: / 250 NH2+
		c) 250 N: / 2 NH+
		d) 250 NH: / 2 NH2+

Answer 1

To calculate the ratio of protonated to deprotonated groups at given pH, we can use the HendersonHasselbalch equation to solve for the ratio of [HA] to [A- ], as we are given both the pH and pKa of the compound.

As given pH and pKa are fixed values, we're free to simply compare the ratio in form of [A−] or deprotonated form and [HA] protonated form .

HendersonHasselbalch equation

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

or, log ([HA] / [A- ]) = pKa – pH

or, [HA] / [A- ] = 10 ^ ( pKa – pH ) ------- we can use this form to solve the problems.

If pKa=pH+1 then we have [HA] / [ A- ] = 10. means the solution would be mildly basic as HA or protonated form is 10 times more than the deprotonated form.

on the other hand, If pKa=pH - 1 then we have [HA] / [ A- ] = 10 ^ (-1) . means the solution would be mildly acidic as [A-] or deprotonated form is 10 times more than the protonated form.

Now We solve Question no. 2

In the catalytic center of papain, Cysteine sulfhydryl group's deprotonated form would be (S-) like (A-) and Protonated form (SH) like (HA)

We have the equation [HA] / [A- ] = 10 ^ ( pKa – pH )

or, (SH) / (S-) = 10 ^ ( pKa – pH )

here pKa 3.4 and pH of 6.2. so, ( pKa – pH ) = ( 3.4-6.2) = ( - 2.8 )

Now, (SH) / (S-) = 10 ^  ( - 2.8 ) = 0.00158

or,  (S-) / (SH) = 1 / 0.00158 = 632 ( approx)

So, choosing nearest answer would be a) 1 S- / 631 SH

Now, question no. 3.

ratio of protonated vs deprotonated Histidine R-groups in papain within the catalytic center.

here pKa 8.3 and pH of 6.2. so, ( pKa – pH ) = ( 8.3 - 6.2) = 2.1

Histidine R- group's deprotonated form would be (N:) and Protonated form (NH+)

From the equation [HA] / [A- ] = 10 ^ ( pKa – pH )

or, (NH+) / (N:) = 10 ^ ( pKa – pH ) = 10 ^ 2.1 = 125.89

or, 2 x (NH+) = 250 (N:) ( approx)

So, choosing the nearest Answer, it would be  c) 250 N: / 2 NH+

Now Question no. 1

The pKa for the imidazole group in histidine in Papain's catalytic centers is 8.3. and given pH 8.2.

so, ( pKa – pH ) = ( 8.3 - 8.2) = 0.1

Histidine R- group's deprotonated form would be (N:) and Protonated form (NH+)

From the equation [HA] / [A- ] = 10 ^ ( pKa – pH )

or, (NH+) / (N:) = 10 ^ ( pKa – pH ) = 10 ^ 0.1 = 1.2589

or, 4 X (NH+) = 4 x 1.2589 X (N:)

or, 4 X (NH+) = 5 X  (N:) [ approx]

We known that if pKa > pH then protonated form would be more than the deprotonated form. But there is no matched answer. Possibly that would be a Typing error i presume.

So the answer would be 4(NH+) / 5(N:).

That is why, i have answered this one at last.

Thanks for asking.

+) / 5(N:)