Biochemistry (Looseleaf)
9th Edition
ISBN: 9781319114800
Author: BERG
Publisher: MAC HIGHER
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Question
Chapter 23, Problem 28P
Interpretation Introduction
Interpretation:
The way by which deficiency of arginine succinate synthetase can be removed should be determined.
Name the molecules that eliminate the nitrogen out of the body should be determined.
Concept introduction:
The urea cycle is a series of biochemical reactions which involve the formation of urea (NH2)2CO from the ammonia (NH3). It is also known as the ornithine cycle. It helps in the excretion of toxic ammonia by converting into urea. This cycle takes place in ureotelic organisms.
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Amino AcidMetabolism.
a. What are the SIXprecursors used in amino acid biosynthesis?
b. What are the SEVEN metabolic intermediates that result from amino acid degradation?
c. Circle the metabolites found in your answers to both a) & b) above.
d) What is the difference between glucogenic and ketogenic amino acids? List the glucogenic amino acids, the ketogenic amino acids & the amino acids that are both glucogenic & ketogenic.
I. Active site analysis.
Below is a diagram of a putative active site for Monoamine oxidase. As we learned, the purpose of
tertiary structure is to form a scaffold so you can orient just a few amino acids in the right orientation
to promote binding and/or catalysis. The position where this occurs is the active site. The amino acid
architecture of an active site is designed to bind substrates. Amino acid side chains are capable of
hydrogen bonding, ionic and hydrophobic interactions. Fill in each amino acid that you think is
suitable for interacting with the part of the substrate it is closest to. Assume the pH will be at 7.0
a.a.#1
a.a.#2
a.a.#6
HO
Lond
NH₂
НО
a.a.#5
OH
a.a.#3
a.a.#4
elearn.squ.edu.om/mod/qui
ystem (Academic)
The enzyme asparaginase is used to reduce the level of asparagine in blood in the treatment of leukemia.
Which of the following forms of Asparaginase would be most useful if the blood asparagine level is 0.2 mM?
Select one:
O a. Km = 2.0 mM; Vmax = 0.1 mM/hour
O b. Km = 0.1 mM; Vmax = 0.5 mM/hour
%3D
О с. Кm
0.2 mM; Vmax = 0.1 mM/hour
O d. Km = 0.1 mM; Vmax = 0.1 mM/hour
O e. Km = 0.2 mM; Vmax = 0.5 mM/hour
Clear my choice
In a steady state (of ES formation and ES breakdown)
Select one:
a The rate of formation of ES is equal to the rate of its degradation during the reaction
Chapter 23 Solutions
Biochemistry (Looseleaf)
Ch. 23 - Prob. 1PCh. 23 - Prob. 2PCh. 23 - Prob. 3PCh. 23 - Prob. 4PCh. 23 - Prob. 5PCh. 23 - Prob. 6PCh. 23 - Prob. 7PCh. 23 - Prob. 8PCh. 23 - Prob. 9PCh. 23 - Prob. 10P
Ch. 23 - Prob. 11PCh. 23 - Prob. 12PCh. 23 - Prob. 13PCh. 23 - Prob. 14PCh. 23 - Prob. 15PCh. 23 - Prob. 16PCh. 23 - Prob. 17PCh. 23 - Prob. 18PCh. 23 - Prob. 19PCh. 23 - Prob. 20PCh. 23 - Prob. 21PCh. 23 - Prob. 22PCh. 23 - Prob. 23PCh. 23 - Prob. 24PCh. 23 - Prob. 25PCh. 23 - Prob. 26PCh. 23 - Prob. 27PCh. 23 - Prob. 28PCh. 23 - Prob. 29PCh. 23 - Prob. 30PCh. 23 - Prob. 31PCh. 23 - Prob. 32PCh. 23 - Prob. 33PCh. 23 - Prob. 34PCh. 23 - Prob. 35PCh. 23 - Prob. 36PCh. 23 - Prob. 37PCh. 23 - Prob. 38PCh. 23 - Prob. 39PCh. 23 - Prob. 40PCh. 23 - Prob. 41PCh. 23 - Prob. 42PCh. 23 - Prob. 43PCh. 23 - Prob. 44PCh. 23 - Prob. 45PCh. 23 - Prob. 46PCh. 23 - Prob. 47PCh. 23 - Prob. 48PCh. 23 - Prob. 49P
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