Biochemistry
Biochemistry
8th Edition
ISBN: 9781464126109
Author: Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr., Lubert Stryer
Publisher: W. H. Freeman
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Chapter 20, Problem 39P
Interpretation Introduction

(a)

Interpretation:

The data were most likely to be generated by the C4 plant and C3 plant should be determined.

Concept introduction:

When O2content is high in the atmosphere, the Rubisco enzyme(carbon-fixing enzyme of Calvin cycle) binds with O2 instead of CO2. The O2binding initiates oxygenase reaction rather than carboxylase reaction. This process is called photorespiration.

Interpretation Introduction

(b)

Interpretation:

Possible explanations for the decrease in photosynthetic activity at higher temperature should be determined.

Concept introduction:

Multiple factors affect enzyme activity such as temperature, pH, enzyme concentration, substrate concentration, and inhibitors or activators.

Interpretation Introduction

(c)

Interpretation:

Why C4 plants thrive at CO2 concentrations that do not support the growth of C3 plants should be explained.

Concept introduction:

C3 plants have no specific adaptations to minimize photorespiration. But C4 plants have a mechanism to reduce photorespiration. In C4, cells light-dependent reactions and Calvin cycles occur in two different types of cell

Interpretation Introduction

(d)

Interpretation:

A possible explanation of why C3 plants continue to increase photosynthetic activity at higher CO2 concentrations, whereas C4 plants reach a plateau should be suggested.

Concept introduction:

In C4 plants, atmospheric CO2 is first fixed in one type of cell as oxaloacetate and then converted to malate. Malate is then broken down to produce CO2 in another type of cells. This CO2 is then fixed to produce glucose via the Calvin cycle like C3 plants.

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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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Photosynthesis & Respiration | Reactions | Chemistry | FuseSchool; Author: FuseSchool - Global Education;https://www.youtube.com/watch?v=3XIyweZg6Sw;License: Standard YouTube License, CC-BY