BIOCHEM-ACHIEVE(FIRST DAY DISCOUNTED)
BIOCHEM-ACHIEVE(FIRST DAY DISCOUNTED)
9th Edition
ISBN: 2818000069358
Author: BERG
Publisher: MAC HIGHER
Question
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Chapter 16, Problem 8P
Interpretation Introduction

(a)

Interpretation:

The amount of ATP generated by glucose-6-phosphate when it is processed by glycolysis to lactate should be calculated.

Concept introduction:

Glycolysis is the conversion of one molecule of glucose into two molecules of pyruvate. During this process, two molecules of ATP and two molecules of NADH are produced. Glycolysis is divided into two phases, the preparatory phase and payoff phase. In the preparatory phase, ATP is consumed, while in the payoff phase ATP is produced.

Interpretation Introduction

(b)

Interpretation:

The amount of ATP generated by glycolysis to lactate when this will be processed from dihydroxyacetone phosphate should be calculated.

Concept introduction:

Glycolysis is the conversion of one molecule of glucose into two molecules of pyruvate. During this process, two molecules of ATP and two molecules of NADH are produced. Glycolysis is divided into two phases, the preparatory phase, and payoff phase. In the preparatory phase, ATP is consumed, while in the payoff phase ATP is produced.

Interpretation Introduction

(c)

Interpretation:

The amount of ATP generated by glycolysis to lactate when this will be processed from glyceraldehydes-3-phosphate should be calculated.

Concept introduction:

Glycolysis is the conversion of one molecule of glucose into two molecules of pyruvate. During this process, two molecules of ATP and two molecules of NADH are produced. Glycolysis is divided into two phases, the preparatory phase, and payoff phase. In the preparatory phase, ATP is consumed, while in the payoff phase ATP is produced.

Interpretation Introduction

(d)

Interpretation:

The amount of ATP generated by glycolysis to lactate when this will be processed from fructose should be calculated.

Concept introduction:

Glycolysis is the conversion of one molecule of glucose into two molecules of pyruvate. During this process, two molecules of ATP and two molecules of NADH are produced. Glycolysis is divided into two phases, the preparatory phase and payoff phase. In the preparatory phase, ATP is consumed, while in the payoff phase ATP is produced.

Interpretation Introduction

(e)

Interpretation:

The amount of ATP generated by glycolysis to lactate when this will be processed from sucrose should be calculated.

Concept introduction:

Glycolysis is the conversion of one molecule of glucose into two molecules of pyruvate. During this process, two molecules of ATP and two molecules of NADH are produced. Glycolysis is divided into two phases, the preparatory phase, and payoff phase. In the preparatory phase, ATP is consumed, while in the payoff phase ATP is produced.

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Students have asked these similar questions
The beta-lactamase hydrolyzes the lactam-ring in penicillin. Describe the mechanism  of hydrolysis, insuring to include the involvement of S, D, & K in the reaction sequence. Please help
To map the active site of beta-lactamase, the enzyme was hydrolyzed with trypsin to yield a hexapeptide (P1) with the following amino acids. Glu, Lys, Leu, Phe, Met, and Ser. Treatment of P1 with phenyl isothiocyanate yielded a PTH derivative of phenylalanine and a peptide (P2). Treatment of P1 with cyanogenbromide gave an acidic tetrapeptide (P3) and a dipeptide (P4).Treatment of P2 with 1-fluoro-2,4-dinitrobenzene, followed by complete hydrolysis, yields N-2,4-dinitrophenyl-Glu. P1, P2, and P3 contain the active site serine. Why doesn't D in this hexapeptide not participate in the hydrolysis of the beta-lactam ring even though S, K, and D are involved in the catalyst?
To map the active site of -lactamase, the enzyme was hydrolyzed with trypsin to yield a hexapeptide (P1) with the following amino acids. Glu, Lys, Leu, Phe, Met, and Ser. Treatment of P1 with phenyl isothiocyanate yielded a PTH derivative of phenylalanine and a peptide (P2). Treatment of P1 with cyanogenbromide gave an acidic tetrapeptide (P3) and a dipeptide (P4).Treatment of P2 with 1-fluoro-2,4-dinitrobenzene, followed by complete hydrolysis, yields N-2,4-dinitrophenyl-Glu. P1, P2, and P3 contain the active site serine.  Using the experimental results described above derive the primary sequence of the active site hexapeptide. Please help!
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