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Table 21.1 Free Energies of Hydrolysis of Some Phosphates R-O-P-O- + H?O \longrightarrow ROH + HO-P-O- P Compound Name Function ?G (kcal/mol) ?G (kJ/mol) Phosphoenol pyruvate Final intermediate in conversion of glucose to pyruvate (glycolysis)- Stage 2, Figure 21.5 -14.8 -61.9 1, 3-Bisphosphoglycerate Another intermediate in glycolysis -11.8 -49.4 Creatine phosphate Energy storage in muscle cells -10.3 -43.1 ATP ($\longrightarrow$ ADP) Principal energy carrier -7.3 -30.5 Glucose 1-phosphate First intermediate in breakdown of carbohydrates stored as starch or glycogen -5.0 -20.9 Glucose 6-phosphate First intermediate in glycolysis -3.3 -13.8 Fructose 6-phosphate Second intermediate in glycolysis -3.3 -13.8 4. In carbohydrate metabolism, you start by using energy as glucose is phosphorylated to form glucose-6-phosphate, where the alcohol group on the number 6 carbon is modified by the addition of a phosphate group. Draw the reaction of glucose being converted to glucose-6-phosphate, with the glucose (substrate/reactant) and glucose-6-phosphate (product) as Haworth projections.

          Table 21.1 Free Energies of Hydrolysis of Some Phosphates
R-O-P-O- + H?O \longrightarrow ROH + HO-P-O-
P
Compound Name	Function	?G (kcal/mol)	?G (kJ/mol)
Phosphoenol pyruvate	Final intermediate in conversion of
glucose to pyruvate (glycolysis)-
Stage 2, Figure 21.5	-14.8	-61.9
1, 3-Bisphosphoglycerate	Another intermediate in glycolysis	-11.8	-49.4
Creatine phosphate	Energy storage in muscle cells	-10.3	-43.1
ATP ($\longrightarrow$ ADP)	Principal energy carrier	-7.3	-30.5
Glucose 1-phosphate	First intermediate in breakdown of
carbohydrates stored as starch or
glycogen	-5.0	-20.9
Glucose 6-phosphate	First intermediate in glycolysis	-3.3	-13.8
Fructose 6-phosphate	Second intermediate in glycolysis	-3.3	-13.8
4. In carbohydrate metabolism, you start by using energy as glucose is phosphorylated to
form glucose-6-phosphate, where the alcohol group on the number 6 carbon is modified
by the addition of a phosphate group. Draw the reaction of glucose being converted to
glucose-6-phosphate, with the glucose (substrate/reactant) and glucose-6-phosphate
(product) as Haworth projections.

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Table 21.1 Free Energies of Hydrolysis of Some Phosphates
R-O-P-O- + H?O ⟶ROH + HO-P-O-
P
Compound Name Function ?G (kcal/mol) ?G (kJ/mol)
Phosphoenol pyruvate Final intermediate in conversion of
glucose to pyruvate (glycolysis)-
Stage 2, Figure 21.5 -14.8 -61.9
1, 3-Bisphosphoglycerate Another intermediate in glycolysis -11.8 -49.4
Creatine phosphate Energy storage in muscle cells -10.3 -43.1
ATP (⟶ ADP) Principal energy carrier -7.3 -30.5
Glucose 1-phosphate First intermediate in breakdown of
carbohydrates stored as starch or
glycogen -5.0 -20.9
Glucose 6-phosphate First intermediate in glycolysis -3.3 -13.8
Fructose 6-phosphate Second intermediate in glycolysis -3.3 -13.8
4. In carbohydrate metabolism, you start by using energy as glucose is phosphorylated to
form glucose-6-phosphate, where the alcohol group on the number 6 carbon is modified
by the addition of a phosphate group. Draw the reaction of glucose being converted to
glucose-6-phosphate, with the glucose (substrate/reactant) and glucose-6-phosphate
(product) as Haworth projections.

Added by Ashlee M.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Shaiju T

03/29/2022

Step 1

Step 1: Start with the Haworth projection of glucose, which is a 6-membered ring with an oxygen atom at the top and the carbon atoms numbered from 1 to 6 in a clockwise direction. Show more…

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Table 21.1 Free Energies of Hydrolysis of Some Phosphates Compound Name: Phosphoenol pyruvate Function: Final intermediate in conversion of glucose to pyruvate [glycolysis) - Stage 2, Figure 21.5 G(kcal/mol): -14.8 G(kJ/mol): -61.9 Compound Name: 1,3-Bisphosphoglycerate Function: Another intermediate in glycolysis G(kcal/mol): -11.8 G(kJ/mol): -49.4 Compound Name: Creatine phosphate Function: Energy storage in muscle cells G(kcal/mol): -10.3 G(kJ/mol): -43.1 Compound Name: ATP Function: Principal energy carrier G(kcal/mol): -7.3 G(kJ/mol): -30.5 Compound Name: ADP Function: Principal energy carrier G(kcal/mol): -5.0 G(kJ/mol): -20.9 Compound Name: Glucose 1-phosphate Function: First intermediate in breakdown of carbohydrates stored as starch or glycogen G(kcal/mol): -3.3 G(kJ/mol): -13.8 Compound Name: Glucose 6-phosphate Function: First intermediate in glycolysis G(kcal/mol): -3.3 G(kJ/mol): -13.8 Compound Name: Fructose 6-phosphate Function: Second intermediate in glycolysis G(kcal/mol): -3.3 G(kJ/mol): -13.8 4. In carbohydrate metabolism, you start by using energy as glucose is phosphorylated to form glucose-6-phosphate, where the alcohol group on the number 6 carbon is modified by the addition of a phosphate group. Draw the reaction of glucose being converted to glucose-6-phosphate, with the glucose (substrate/reactant) and glucose-6-phosphate (product) as Haworth projections.

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00:01 Hello everyone let's solve this question this question we have to find out delta g not value for the given reaction.

00:08 So let us choose the first reaction that is phosphoenol pyruid, phosphoenol pyruid.

00:19 It dissociate to form pyruid and a phosphate bond and delta g not value for this is minus 61 .9 kilojou per moon.

00:34 And let us consider the second reaction that is dissociation of atp into adp plus phosphate bond.

00:44 So here the delta g value for this reaction is minus 30 .5 kilojoule per moon.

00:52 Now we will reverse equation 1 and adding it to equation 2 that will give us the required equation.

01:13 So let us write it.

01:15 It will be pyruid plus phosphate bond gives phospho in all pyruid and we are adding it to the second that is atp gives atp plus pi.

01:37 So the pi is getting cancelled here and we will get the required reaction that is pyruvate plus atp gives phospho enol fireweight.

01:59 So, let us do the same thing with the delta g values.

02:03 So delta g not for the reaction, the above reaction will be, we have just reverse and added.

02:12 So it is 61 .9 kilojoule per mole plus minus 30 .5 kilojoule per mole and that will give the value 31 .4.

02:25 Kilojoules per moon.

02:27 So that is delta g not for the reaction.

02:31 This is the answer to the first part of the question.

02:36 Now let us move to the second one.

02:39 That is we have to find out the delta g not is minus r t log k e.

02:52 So we can write log k e equilibrium constant can be written as minus r t log k e.

03:00 So we can be written as minus r t log k e.

03:00 So we can be written as minus r t log k e.

03:00 Minus delta g not divided by r t so here we can substitute the value minus 31 .4 in multiplied by 10 to power 3 that is jules per mole we have converted kilojoules in jules divided by 8 .314 joules per kelvin mole that is value of r multiplied by 298 kelvin because temperature is 25 degrees so here it is minus 12 .673 so this is the value of log of k equilibrium so we can write k equilibrium is equals to e raised to power minus 12 .673 and we in when we solve it it is 3 .13 multiplied by 10 to power minus 6 so that will give us the value of equilibrium constant this completes the first part now, let us come to the second part of the question that is we have to find out the ratio at equilibrium...

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