BIOCHEMISTRY W/1 TERM ACHEIVE ACCESS
9th Edition
ISBN: 9781319425746
Author: BERG
Publisher: MAC HIGHER
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Question
Chapter 17, Problem 39P
Interpretation Introduction
Interpretation:
The reason for the reaction of a symmetric molecule with an enzyme in an asymmetric way should be explained.
Concept introduction:
The asymmetric catalysis can be achieved via chiral reagents, asymmetric catalyst or via an enzyme that can induce asymmetry.
All the molecules present in biological systems are essentially single enantiomers. For example, all the amino acids are S- amino acids. Therefore, the biological system can serve as a chiral environment.
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Macmillan Learning
Cholesterol synthesis begins with the formation of mevalonate from acetyl CoA. This
process activates mevalonate and converts it to isopentenyl pyrophosphate.
Identify the atoms in mevalonate and isopentenyl pyrophosphate that will be labeled
from acetyl CoA labeled with 14C in the carbonyl carbon. Place 14C atoms and C atoms
to denote which carbon atoms are labeled and which are not labeled.
H₂C
COA
14C-labeled acetyl-CoA
HHH
[c]
H
H
OH
014C
- OH
H
HH
H
Mevalonate
CH3
H H
14C
H
Η
H
H
Incorrect Answer
of o
-P-O-P-0-
Isopentenyl pyrophosphate
с
Answer Bank
14C
Chapter 17 Solutions
BIOCHEMISTRY W/1 TERM ACHEIVE ACCESS
Ch. 17 - Prob. 1PCh. 17 - Prob. 2PCh. 17 - Prob. 3PCh. 17 - Prob. 4PCh. 17 - Prob. 5PCh. 17 - Prob. 6PCh. 17 - Prob. 7PCh. 17 - Prob. 8PCh. 17 - Prob. 9PCh. 17 - Prob. 10P
Ch. 17 - Prob. 11PCh. 17 - Prob. 12PCh. 17 - Prob. 13PCh. 17 - Prob. 14PCh. 17 - Prob. 15PCh. 17 - Prob. 16PCh. 17 - Prob. 17PCh. 17 - Prob. 18PCh. 17 - Prob. 19PCh. 17 - Prob. 20PCh. 17 - Prob. 21PCh. 17 - Prob. 22PCh. 17 - Prob. 23PCh. 17 - Prob. 24PCh. 17 - Prob. 25PCh. 17 - Prob. 26PCh. 17 - Prob. 27PCh. 17 - Prob. 28PCh. 17 - Prob. 29PCh. 17 - Prob. 30PCh. 17 - Prob. 31PCh. 17 - Prob. 32PCh. 17 - Prob. 33PCh. 17 - Prob. 34PCh. 17 - Prob. 35PCh. 17 - Prob. 36PCh. 17 - Prob. 37PCh. 17 - Prob. 38PCh. 17 - Prob. 39PCh. 17 - Prob. 40PCh. 17 - Prob. 41PCh. 17 - Prob. 42PCh. 17 - Prob. 43PCh. 17 - Prob. 44P
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