NADP NADPH 3. The amino acid proline can be made from glutamate-5-semialdehyde in a two-step process. LOH NH2 ntramolecular imine bond formation NH OH HO OH molecule X proline glutamate glutamate-5-semialdehyde a. Glutamic acid (glutamate) has three acidic groups and three corresponding pKa values: 2.1, 4.0, 9.5. Redraw glutamic acid with every acidic group in acid state and assign pKa values. Briefly justify your assignments. b. The first step in proline synthesis is the intramolecular reaction of glutamate-5- semialdehyde to form an imine (molecule X). Draw the mechanism for this reaction, including any relevant intermediate, using the arrow formalism. Assume that the reaction is in H2O at pH 7. The second step is the reduction of molecule X by NADPH. Draw the mechanism for this reaction, including any relevant intermediates, using the arrow formalism. You may abbreviate NADPH and NADPp* as the following: C. H H NADPH NADP* d. Assume that AG°under cellular conditions for the reaction NADPH + X = NADP* + proline %3D is -15 kJ/mol, and that at steady-state in the cell [NADPH] [NADP+] fixed due to the activities of other cellular reactions. What's the concentration ratio of = 20 and that this ratio stays %3D proline to molecule X in the cell when the reaction is at equilibrium?
NADP NADPH 3. The amino acid proline can be made from glutamate-5-semialdehyde in a two-step process. LOH NH2 ntramolecular imine bond formation NH OH HO OH molecule X proline glutamate glutamate-5-semialdehyde a. Glutamic acid (glutamate) has three acidic groups and three corresponding pKa values: 2.1, 4.0, 9.5. Redraw glutamic acid with every acidic group in acid state and assign pKa values. Briefly justify your assignments. b. The first step in proline synthesis is the intramolecular reaction of glutamate-5- semialdehyde to form an imine (molecule X). Draw the mechanism for this reaction, including any relevant intermediate, using the arrow formalism. Assume that the reaction is in H2O at pH 7. The second step is the reduction of molecule X by NADPH. Draw the mechanism for this reaction, including any relevant intermediates, using the arrow formalism. You may abbreviate NADPH and NADPp* as the following: C. H H NADPH NADP* d. Assume that AG°under cellular conditions for the reaction NADPH + X = NADP* + proline %3D is -15 kJ/mol, and that at steady-state in the cell [NADPH] [NADP+] fixed due to the activities of other cellular reactions. What's the concentration ratio of = 20 and that this ratio stays %3D proline to molecule X in the cell when the reaction is at equilibrium?
Chapter24: Amines And Heterocycles
Section24.4: Basicity Of Arylamines
Problem 6P
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![NADP+
NADPH
3.
The amino acid proline can be made from glutamate-5-semialdehyde in a two-step process.
NH,
он
Intramolecular imine
bond formation
NH,
OH
HO
???
molecule X
proline
glutamate
glutamate-5-semialdehyde
Glutamic acid (glutamate) has three acidic groups and three corresponding pKa values:
2.1, 4.0, 9.5. Redraw glutamic acid with every acidic group in acid state and assign pKa
values. Briefly justify your assignments.
b. The first step in proline synthesis is the intramolecular reaction of glutamate-5-
semialdehyde to form an imine (molecule X). Draw the mechanism for this reaction,
including any relevant intermediate, using the arrow formalism. Assume that the
reaction is in H20 at pH 7.
a.
The second step is the reduction of molecule X by NADPH. Draw the mechanism for this
reaction, including any relevant intermediates, using the arrow formalism. You may
abbreviate NADPH and NADP* as the following:
C.
H.
NADPH
NADP*
d. Assume that AG°under cellular conditions for the reaction
NADPH + X = NADP* + proline
is -15 kJ/mol, and that at steady-state in the cell
fixed due to the activities of other cellular reactions. What's the concentration ratio of
proline to molecule X in the cell when the reaction is at equilibrium?
[NADPH]
[NADP+]
= 20 and that this ratio stays](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fce768285-628a-4724-9f2b-b5d3bc5907ae%2F264b61b4-d044-4e2d-832b-7ba4e1132fa3%2F6qmbx4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:NADP+
NADPH
3.
The amino acid proline can be made from glutamate-5-semialdehyde in a two-step process.
NH,
он
Intramolecular imine
bond formation
NH,
OH
HO
???
molecule X
proline
glutamate
glutamate-5-semialdehyde
Glutamic acid (glutamate) has three acidic groups and three corresponding pKa values:
2.1, 4.0, 9.5. Redraw glutamic acid with every acidic group in acid state and assign pKa
values. Briefly justify your assignments.
b. The first step in proline synthesis is the intramolecular reaction of glutamate-5-
semialdehyde to form an imine (molecule X). Draw the mechanism for this reaction,
including any relevant intermediate, using the arrow formalism. Assume that the
reaction is in H20 at pH 7.
a.
The second step is the reduction of molecule X by NADPH. Draw the mechanism for this
reaction, including any relevant intermediates, using the arrow formalism. You may
abbreviate NADPH and NADP* as the following:
C.
H.
NADPH
NADP*
d. Assume that AG°under cellular conditions for the reaction
NADPH + X = NADP* + proline
is -15 kJ/mol, and that at steady-state in the cell
fixed due to the activities of other cellular reactions. What's the concentration ratio of
proline to molecule X in the cell when the reaction is at equilibrium?
[NADPH]
[NADP+]
= 20 and that this ratio stays
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