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clueless clueless
wrote...
Posts: 10
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10 years ago
I thought i was on a role with my biochem buffer problems until i got stuck with this enzyme problem.
I have no clue as to where to begin =/

"An enzyme contains an active site histidine residue involved in the catalytic mechanism. The optimal pH of the enzyme is pH 6.0. What is the percentage of imidizole and imidizoleum at the active site?"

I know that histidine, is a triprotic amino acid and has pK1 = 1.8, pK2 = 6.0, pK3 = 9.3, but yea other than that im pretty lost... =/
Any help would be greatly appreciated!
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wrote...
Donated
Valued Member
10 years ago
I aced biochem and I'm dying to see this answer because I haven't the faintest clue.
Pretty fly for a SciGuy
wrote...
Staff Member
Educator
10 years ago
Some information that I suppose could get your thinking in the right path. In terms of percentage, I have never answered something like that Undecided

The imidazole sidechain of histidine has a pKa of approximately 6.0, and, overall, the amino acid has a pKa of 6.5. This means that, at physiologically relevant pH values, relatively small shifts in pH will change its average charge. Below a pH of 6, the imidazole ring is mostly protonated as described by the Henderson–Hasselbalch equation. When protonated, the imidazole ring bears two NH bonds and has a positive charge. The positive charge is equally distributed between both nitrogens and can be represented with two equally important resonance structures.
Mastering in Nutritional Biology
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clueless Author
wrote...
10 years ago
Heres what i did, but im not sure if its right...
I used the henderson hasslebalch equation, pH=pKa +log [imidizoleum]/[imidizole].

6.0=6.1 +log [imidizoleum]/[imidizole]
-0.1=log [imidizoleum]/[imidizole]
10^(-0.1)= [imidizoleum]/[imidizole]
[imidizoleum]/[imidizole]=0.794
[imidizoleum]/[imidizole]=0.794/1.794=0.44=44% imidizoleum
1/1.794=0.56=56% Imidazole

If anyone figures it out or knows if im right or now, lemme know cuz its due tomorrow =/
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