EBK LEHNINGER PRINCIPLES OF BIOCHEMISTR
EBK LEHNINGER PRINCIPLES OF BIOCHEMISTR
7th Edition
ISBN: 8220103662253
Author: nelson
Publisher: MAC HIGHER
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Chapter 10, Problem 3P

(a)

Summary Introduction

To determine: The structural aspect of the 18-carbon fatty acids can be correlated to the melting point

Introduction:

Fatty acids are a class of lipids. They are non-polar compound this means they can only dissolve in non-polar solvents such as benzene, diethyl ether, and hexane. Fatty acids can be saturated or unsaturated. Saturated fatty acids are those which do no possess any double bond in its structure, For example: Palmitic acid and stearic acid. While unsaturated fatty acids possess one or more double bonds, For example: Oleic acid and linoleic acid.

(b)

Summary Introduction

To draw: The possible triacyglycerols that can be constructed from glycerol, palmitic acid, and oleic acid, and also rank them in order of increasing melting point.

Introduction:

Fatty acids are a class of lipids. They are non-polar compound this means they can only dissolve in non-polar solvents such as benzene, diethyl ether, and hexane. Fatty acids can be saturated or unsaturated. Saturated fatty acids are those which do no possess any double bond in its structure, For example: Palmitic acid and stearic acid. While unsaturated fatty acids possess one or more double bonds, For example: Oleic acid and linoleic acid.

Summary Introduction

(c)

To determine: The presence or absence of branched-chain fatty acid increase or decrease the fluidity of the membrane.

Introduction:

Fatty acids are a class of lipids. They are non-polar compound this means they can only dissolve in non-polar solvents such as benzene, diethyl ether, and hexane. Fatty acids can be saturated or unsaturated. Saturated fatty acids are those which do no possess any double bond in its structure, For example: Palmitic acid and stearic acid. While unsaturated fatty acids possess one or more double bonds, For example: Oleic acid and linoleic acid.

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The standard free energy, AGO, of hydrolysis of inorganic polyphosphate, polyP, is about −20 kJ/mol for each P; released. In a cell, it takes about 50 kJ/mol of energy to synthesize ATP from ADP and Pi. ○ P O Inorganic polyphosphate (polyP) Is it feasible for a cell to use polyP to synthesize ATP from ADP? Why or why not? No. The reaction is unidirectional and always proceeds in the direction of polyP synthesis from ATP. Yes. If [ADP] and [polyP] are kept high, and [ATP] is kept low, the actual free-energy change would be negative. No. The synthesis of ATP from ADP and P; has a large positive G'o compared to polyP hydrolysis. Yes. The hydrolysis of polyP has a sufficiently negative AG to overcome the positive AGO of ATP synthesis. Correct Answer
In the glycolytic pathway, a six-carbon sugar (fructose 1,6-bisphosphate) is cleaved to form two three-carbon sugars, which undergo further metabolism. In this pathway, an isomerization of glucose 6-phosphate to fructose 6-phosphate (as shown in the diagram) occurs two steps before the cleavage reaction. The intervening step is phosphorylation of fructose 6-phosphate to fructose 1,6-bisphosphate. H H | H-C-OH H-C-OH C=0 HO-C-H HO-C-H phosphohexose isomerase H-C-OH H-C-OH H-C-OH H-C-OH CH₂OPO CH₂OPO Glucose 6-phosphate Fructose 6-phosphate What does the isomerization step accomplish from a chemical perspective? Isomerization alters the molecular formula of the compound, allowing for subsequent phosphorylation. Isomerization moves the carbonyl group, setting up a cleavage between the central carbons. Isomerization causes the gain of electrons, allowing for the eventual release of NADH. Isomerization reactions cause the direct production of energy in the form of ATP.
From data in the table, calculate the AG value for the reactions. Reaction AG' (kJ/mol) Phosphocreatine + H₂O →>> creatine + P -43.0 ADP + Pi → ATP + H₂O +30.5 Fructose +P → fructose 6-phosphate + H₂O +15.9 Phosphocreatine + ADP creatine + ATP AG'O ATP + fructose → ADP + fructose 6-phosphate AG'° kJ/mol kJ/mol
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