Question

Given the Henderson-Hasselbalch equation as pH = pK + log {[A-] /[HA]}, what is the probability of finding the form of histidine shown in Figure 2-I at pH 5 assuming the various pK's of histidine are: pK amino = 10.0; pK carboxyl = 2.0; pK imidazole = 6.0

a) 0.001
b) 0.01
c) 0.1
d) 1.0

Answer 1

The probability of finding the form of histidine shown in Figure 2-I at pH 5 is approximately 0.01 (b). Thus the correct option is (b).

Step-by-step explanation:

The Henderson-Hasselbalch equation, pH = pKa +
`log([A^-]/[HA])`, is utilized to determine the ratio of deprotonated
`(A^-)` to protonated (HA) forms of histidine at pH 5. Considering the pKa values provided for histidine:

pKa amino = 10.0

pKa carboxyl = 2.0

pKa imidazole = 6.0

At pH 5, the amino group (pKa = 10.0) is predominantly protonated (HA), while the carboxyl group (pKa = 2.0) is mostly deprotonated (A^-). The imidazole group (pKa = 6.0) falls closer to the pH value, leading to an equilibrium mixture of protonated (HA) and deprotonated (A^-) forms.

Calculating the ratio using the Henderson-Hasselbalch equation:

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

5 = 6 +
`log([A^-]/[HA])`

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

`[A^-]/[HA] = 10^-1` = 0.1

This ratio suggests that at pH 5, there's a 1:10 proportion of the deprotonated (A^-) to protonated (HA) forms. Hence, the probability of finding the form of histidine shown in Figure 2-I at pH 5 would be approximately 0.01, or 1%, as 0.1 represents a 1:10 ratio. This outcome indicates a minor presence of the depicted form of histidine at this pH, aligning closely with option (b) - 0.01.

Answer 2

The probability of finding the form of histidine shown in Figure 2-I at pH 5 is approximately 0.01 (b). Thus the correct option is (b).

Step-by-step explanation:

The Henderson-Hasselbalch equation, pH = pKa + log([A^-]/[HA]), is utilized to determine the ratio of deprotonated (A^-) to protonated (HA) forms of histidine at pH 5. Considering the pKa values provided for histidine:

pKa amino = 10.0

pKa carboxyl = 2.0

pKa imidazole = 6.0

At pH 5, the amino group (pKa = 10.0) is predominantly protonated (HA), while the carboxyl group (pKa = 2.0) is mostly deprotonated (A^-). The imidazole group (pKa = 6.0) falls closer to the pH value, leading to an equilibrium mixture of protonated (HA) and deprotonated (A^-) forms.

Calculating the ratio using the Henderson-Hasselbalch equation:

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

5 = 6 + log([A^-]/[HA])

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

[A^-]/[HA] = 10^-1 = 0.1

This ratio suggests that at pH 5, there's a 1:10 proportion of the deprotonated (A^-) to protonated (HA) forms. Hence, the probability of finding the form of histidine shown in Figure 2-I at pH 5 would be approximately 0.01, or 1%, as 0.1 represents a 1:10 ratio. This outcome indicates a minor presence of the depicted form of histidine at this pH, aligning closely with option (b) - 0.01.