Question

A dipeptide (50 μM concentration) is transferred into a Deuterium oxide (D2O) solution, maintained at pH 6. At this pH, the pseudo first order rate constant for the amide hydrogen exchange (kobs) is 0.06 s-1

1. What is the velocity of amide hydrogen exchange (μM/s)?

2. If the pH was shifted to 8, by what factor would the velocity of exchange be altered ?

3. 25 seconds after transfer into D2O what % of the amide groups still carry protium?

Answer 1

Part 1.

At pH = 6: The velocity of amide hydrogen exchange = [H⁺] * [dipeptide] * kobs

= 10^-6* 50 μM * 0.06 s^-1

= 3*10^-6 μM/s

Part 2.

At pH = 8: The velocity of amide hydrogen exchange = 3*10^-8 μM/s, which is altered by a factor of 10^-2 = 0.01.

Part 3.

kobs = 1/t ln([dipeptide]/[dipeptide]_t)

i.e. 0.06 s^-1 = 1/25 s ln(50 μM/[dipeptide]_t)

i.e. [dipeptide]_t = 11.16 M

Therefore, the percent amide groups that still carry protium = (11.16/50)*100 = 22.3%