Question

1. The pKa for the nitrogen at position #3 of thymidine is 9.8 A. Using the Henderson-Hasselbalch equation, calculate the percentage of the nitrogen in the protonated form at pH 7.2. B. If the pH in the cell were 10.8, what effect would this have on the protonation state of this nitrogen? Explain. C. What effect would a pH of 10.8 have on the Tm (melting temperature) of the DNA in the cell? Explain your reasoning.

Answer 1

1. Hendersen-Hasselbalck equation,

pH = pKa + log(A-/HA)

A- is deprotonated form and HA is protonated form

7.2 = 9.8 + log(A-/HA)

[A-] = 0.0025[HA]

So the percentage of N in protonated form = (1/1+0.0025) x 100 = 99.75%

B. If pH = 10.8

When the pH goes to 10.8, we would have low value for the protonated N. At this pH concentration of H+ would be low and thus loer amount of N in thymidine would be in protonated form.

C. At pH 10.8 more of free amine is present which can form secondary hydrogen bonds in the DNA structure, thereby higher amount of energy is now required to break these bonds. Thus melting point would also be higher.