Each ionizable group of an amino acid can exist in one of two states, charged or neutral. The electric charge on the functional group is determined by the relationship between its pK₂ and the pH of the solution. This relationship is described by the Henderson-Hasselbalch equation. Histidine has three ionizable functional groups. Complete the equilibrium equations by assigning the proper pK₁ for each ionization and the net charge on the histidine molecule in each ionization state. H H₂N-C-C-OH 1 CH₂ + HN O || -1 -NH PK₁ = 1.8 +2 H || H₂N-C-C CH₂ + HN -NH pK₂ = 9.17 -O- Answer Bank pK₂ = 7.59 H || H₂N-C-C -O- I CH₂ -NH pK₂ = 6.0 -2 O H || H₂N-C-C-0- CH₂ +1 -NH 0

Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
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Chapter1: Biochemistry: An Evolving Science
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Each ionizable group of an amino acid can exist in one of two states, charged or neutral. The electric charge on the functional
group is determined by the relationship between its pKa and the pH of the solution. This relationship is described by the
Henderson-Hasselbalch equation.
a
Histidine has three ionizable functional groups. Complete the equilibrium equations by assigning the proper pK₁ for each
ionization and the net charge on the histidine molecule in each ionization state.
H ||
H₂N-C-C-OH
|
CH₂
2
+
HN
-1
-NH
pKa
= 1.8
+2
||
H₂N-C-C-0-
1
CH₂
+
HN
-NH
pKa = 9.17
Answer Bank
pK₂ = 7.59
H||
H₂N-C-C-0-
|
CH₂
N
pKa
-NH
= 6.0
-2
н II
H₂N-C-C-0-
|
CH₂
+1
N
C
-NH
0
Transcribed Image Text:Each ionizable group of an amino acid can exist in one of two states, charged or neutral. The electric charge on the functional group is determined by the relationship between its pKa and the pH of the solution. This relationship is described by the Henderson-Hasselbalch equation. a Histidine has three ionizable functional groups. Complete the equilibrium equations by assigning the proper pK₁ for each ionization and the net charge on the histidine molecule in each ionization state. H || H₂N-C-C-OH | CH₂ 2 + HN -1 -NH pKa = 1.8 +2 || H₂N-C-C-0- 1 CH₂ + HN -NH pKa = 9.17 Answer Bank pK₂ = 7.59 H|| H₂N-C-C-0- | CH₂ N pKa -NH = 6.0 -2 н II H₂N-C-C-0- | CH₂ +1 N C -NH 0
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