1. Consider glutamate in its fully protonated form (e.g. in a pH = 1 solution) 1) Draw all the forms of glutamate at various pH 2) Calculate the pl of this amino acid 3) Sketch a titration curve showing pH as a function of added [OH-] and locate the predominant forms of histidine in the curve pka carboxyl = 2.19 pkaamino = 9.67 pkasidechain = 4.25 STEPS: 1. Find the H atoms that can be removed on the molecule 2. Associated a pka value to each removable H. Draw the Aa structure at: 3. i) pH < pka₁ ii) pka₁ < pH
1. Consider glutamate in its fully protonated form (e.g. in a pH = 1 solution) 1) Draw all the forms of glutamate at various pH 2) Calculate the pl of this amino acid 3) Sketch a titration curve showing pH as a function of added [OH-] and locate the predominant forms of histidine in the curve pka carboxyl = 2.19 pkaamino = 9.67 pkasidechain = 4.25 STEPS: 1. Find the H atoms that can be removed on the molecule 2. Associated a pka value to each removable H. Draw the Aa structure at: 3. i) pH < pka₁ ii) pka₁ < pH
Biochemistry
9th Edition
ISBN:9781319114671
Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.
Chapter1: Biochemistry: An Evolving Science
Section: Chapter Questions
Problem 1P
Related questions
Question
![Consider the positively charged amino acid lysine
Lys2+
21
COOH
I
H&N-C-H
I
pH
14
12
10
8
6
4
2
0
CH₂
I
CH₂
I
CH₂
I
CH₂
T
NH₂+
0
Nelson p85
2.18
=
2.18
PK₁
Lys+
COO™
I
H₂N-C-H H₂N-C-H
ī
I
-----)
8.95
Lysº
8.95
pK₂
pka carboxyl = 2.19
pkaamino = 9.67
pka sidechain = 4.25
COO™
I
CH₂
I
CH₂
I
CH₂
I
CH₂
I
NH₂¹
1.0
2.0
Equivalents of OH added-
COO™
I
H₂N-C-H
I
10.79
1
10.79
pk
Isoelectric
point
Lys
CH₂2
I
CH₂
I
CH₂
I
CH₂
T
NH₂
3.0
+H3N
NH3+
T
CH₂
T
CH₂
CH₂
CH₂
-COO™
H
Lysine
(Lys, K)
Physiological pH = 7.4 < pl
→ Amino acid is positively
charged at physiological pH
1. Consider glutamate in its fully protonated form (e.g. in a pH = 1 solution)
1) Draw all the forms of glutamate at various pH
2) Calculate the pl of this amino acid
3) Sketch a titration curve showing pH as a function
of added [OH-] and locate the predominant forms
of histidine in the curve
STEPS:
1.
Find the H atoms that can be removed on the molecule
2.
Associated a pka value to each removable H.
3. Draw the Aa structure at:
i) pH < pka₁
ii) pka₁ < pH <pkaR
iii) pKar < pH <pka₂
iv) pka₂ < pH
4.
Indicate the net charge of each structure
5. Calculate the pl with the pka's that bracket the neutral form of glutamate.
6. Sketch a titration curve and locate the dominant forms of glutamate on the curve](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f7442e5-ed1a-4154-ba5d-4813abad6cff%2Ff5172d3e-d623-47f2-a3ca-2207637efcb4%2Fr05aj5t_processed.png&w=3840&q=75)
Transcribed Image Text:Consider the positively charged amino acid lysine
Lys2+
21
COOH
I
H&N-C-H
I
pH
14
12
10
8
6
4
2
0
CH₂
I
CH₂
I
CH₂
I
CH₂
T
NH₂+
0
Nelson p85
2.18
=
2.18
PK₁
Lys+
COO™
I
H₂N-C-H H₂N-C-H
ī
I
-----)
8.95
Lysº
8.95
pK₂
pka carboxyl = 2.19
pkaamino = 9.67
pka sidechain = 4.25
COO™
I
CH₂
I
CH₂
I
CH₂
I
CH₂
I
NH₂¹
1.0
2.0
Equivalents of OH added-
COO™
I
H₂N-C-H
I
10.79
1
10.79
pk
Isoelectric
point
Lys
CH₂2
I
CH₂
I
CH₂
I
CH₂
T
NH₂
3.0
+H3N
NH3+
T
CH₂
T
CH₂
CH₂
CH₂
-COO™
H
Lysine
(Lys, K)
Physiological pH = 7.4 < pl
→ Amino acid is positively
charged at physiological pH
1. Consider glutamate in its fully protonated form (e.g. in a pH = 1 solution)
1) Draw all the forms of glutamate at various pH
2) Calculate the pl of this amino acid
3) Sketch a titration curve showing pH as a function
of added [OH-] and locate the predominant forms
of histidine in the curve
STEPS:
1.
Find the H atoms that can be removed on the molecule
2.
Associated a pka value to each removable H.
3. Draw the Aa structure at:
i) pH < pka₁
ii) pka₁ < pH <pkaR
iii) pKar < pH <pka₂
iv) pka₂ < pH
4.
Indicate the net charge of each structure
5. Calculate the pl with the pka's that bracket the neutral form of glutamate.
6. Sketch a titration curve and locate the dominant forms of glutamate on the curve
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