Suppose a liver extract capable of carrying out normal metabolic reactions (including gluconeogenesis) is incubated with pyruvate partially labeled with deuterium at the methyl group (see below: D= deuterium, i.e., H-2) a) Explain why the oxaloacetate formed in step 1 of gluconeogenesis is only partially labeled (i.e., some of the oxaloacetate is labeled and some is not). b) Where would the label be found in glucose?
Suppose a liver extract capable of carrying out normal metabolic reactions (including gluconeogenesis) is incubated with pyruvate partially labeled with deuterium at the methyl group (see below: D= deuterium, i.e., H-2) a) Explain why the oxaloacetate formed in step 1 of gluconeogenesis is only partially labeled (i.e., some of the oxaloacetate is labeled and some is not). b) Where would the label be found in glucose?
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
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Question
Suppose a liver extract capable of carrying out normal
a) Explain why the oxaloacetate formed in step 1 of gluconeogenesis is only partially labeled (i.e., some of the oxaloacetate is labeled and some is not).
b) Where would the label be found in glucose?
Transcribed Image Text:H
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Expert Solution
Step 1
The formation of glucose from pyruvate occurs through the process of gluconeogenesis which occurs in the cytoplasm of the cell. The gluconeogenesis the first reaction is the carboxylation of pyruvate into oxaloacetate. The carboxylation step is catalyzed by the enzyme pyruvate carboxylase.
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