The equilibrium concentrations of fructose-1,6-bisphosphate, dihydroxyacetone (DHAP), and glyceraldehyde-3-phosphate (G3P) under given conditions should be determined. Concept introduction: By the action of aldolase enzyme, fructose-1,6-bisphopshate is converted into DHAP and G3P. Aldolase uses zinc ion as its cofactor. This is the fourth step in the glycolysis process, and the last step of the preparatory phase.

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Answer 1

Question

Chapter 16, Problem 25P

Interpretation Introduction

Interpretation:

The equilibrium concentrations of fructose-1,6-bisphosphate, dihydroxyacetone (DHAP), and glyceraldehyde-3-phosphate (G3P) under given conditions should be determined.

Concept introduction:

By the action of aldolase enzyme, fructose-1,6-bisphopshate is converted into DHAP and G3P. Aldolase uses zinc ion as its cofactor. This is the fourth step in the glycolysis process, and the last step of the preparatory phase.

Expert Solution & Answer

Answer to Problem 25P

Equilibrium concentration of fructose-1,6-bisphosphate = $7.8 ×10−4 M$

Equilibrium concentration of DHAP = $2.2×10−4M$

Equilibrium concentration of G3P = $2.2×10−4M$

Explanation of Solution

Given information:

1 mM (0.001M) of fructose-1,6-bisphosphate is incubated with aldolase under standard conditions.

The reaction is:

$Fructose-1,6-bisphosphate⇌aldolaseDHAP+G3P ; ΔG∘=23.9kJ/mol = 23.9 × 103J/mol$

Creating the ICE table as:

$Reaction - Fructose-1,6-bisphosphate⇌aldolaseDHAP+G3P ; ΔG∘=23.9kJ/mol = 23.9 × 103J/molInitial concentration- 0.001M 0M 0MDifference in conc - -xM +xM +xMFinal Concentration - (0.001-x)M xM xM$

Calculation of equilibrium constant, $Keq$ :

$ΔG∘=−RTlnKeq23.9 × 103J/mol=−8.314J/mol K × 298K×ln KeqKeq = 6.5×10−5$

Now,

$Keq = concentration of productsconcentration of substratesKeq = [DHAP]×[G3P][fructose-1,6-bisphosphate]6.5×10−5=x × x0.001−x6.5×10−5(0.001−x) = x26.5×10−8−6.5×10x = x2x2+6.5×10−5x−6.5×10−8=0$

Solving for x,

$x = −b± b2 −4ac2ax = −(6.5× 10 −5)± (6.5× 10 −5 )2 −4×1×(−6.5× 10 −8 )2×1x = 2.2×10−4M$

So, equilibrium concentrations of DHAP and G3P = $2.2×10−4M$

Equilibrium concentration of fructose-1,6-bisphosphate = $(0.001-2.2×10−4) M = 7.8 ×10−4 M$

Conclusion

Equilibrium concentration of fructose-1,6-bisphosphate = $7.8 ×10−4 M$

Equilibrium concentration of DHAP = $2.2×10−4M$

Equilibrium concentration of G3P = $2.2×10−4M$

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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!

Answer 2

Question

Chapter 16, Problem 25P

Interpretation Introduction

Interpretation:

The equilibrium concentrations of fructose-1,6-bisphosphate, dihydroxyacetone (DHAP), and glyceraldehyde-3-phosphate (G3P) under given conditions should be determined.

Concept introduction:

By the action of aldolase enzyme, fructose-1,6-bisphopshate is converted into DHAP and G3P. Aldolase uses zinc ion as its cofactor. This is the fourth step in the glycolysis process, and the last step of the preparatory phase.

Expert Solution & Answer

Answer to Problem 25P

Equilibrium concentration of fructose-1,6-bisphosphate = $7.8 ×10−4 M$

Equilibrium concentration of DHAP = $2.2×10−4M$

Equilibrium concentration of G3P = $2.2×10−4M$

Explanation of Solution

Given information:

1 mM (0.001M) of fructose-1,6-bisphosphate is incubated with aldolase under standard conditions.

The reaction is:

$Fructose-1,6-bisphosphate⇌aldolaseDHAP+G3P ; ΔG∘=23.9kJ/mol = 23.9 × 103J/mol$

Creating the ICE table as:

$Reaction - Fructose-1,6-bisphosphate⇌aldolaseDHAP+G3P ; ΔG∘=23.9kJ/mol = 23.9 × 103J/molInitial concentration- 0.001M 0M 0MDifference in conc - -xM +xM +xMFinal Concentration - (0.001-x)M xM xM$

Calculation of equilibrium constant, $Keq$ :

$ΔG∘=−RTlnKeq23.9 × 103J/mol=−8.314J/mol K × 298K×ln KeqKeq = 6.5×10−5$

Now,

$Keq = concentration of productsconcentration of substratesKeq = [DHAP]×[G3P][fructose-1,6-bisphosphate]6.5×10−5=x × x0.001−x6.5×10−5(0.001−x) = x26.5×10−8−6.5×10x = x2x2+6.5×10−5x−6.5×10−8=0$

Solving for x,

$x = −b± b2 −4ac2ax = −(6.5× 10 −5)± (6.5× 10 −5 )2 −4×1×(−6.5× 10 −8 )2×1x = 2.2×10−4M$

So, equilibrium concentrations of DHAP and G3P = $2.2×10−4M$

Equilibrium concentration of fructose-1,6-bisphosphate = $(0.001-2.2×10−4) M = 7.8 ×10−4 M$

Conclusion

Equilibrium concentration of fructose-1,6-bisphosphate = $7.8 ×10−4 M$

Equilibrium concentration of DHAP = $2.2×10−4M$

Equilibrium concentration of G3P = $2.2×10−4M$

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Answer 3

Question

Chapter 16, Problem 25P

Interpretation Introduction

Interpretation:

The equilibrium concentrations of fructose-1,6-bisphosphate, dihydroxyacetone (DHAP), and glyceraldehyde-3-phosphate (G3P) under given conditions should be determined.

Concept introduction:

By the action of aldolase enzyme, fructose-1,6-bisphopshate is converted into DHAP and G3P. Aldolase uses zinc ion as its cofactor. This is the fourth step in the glycolysis process, and the last step of the preparatory phase.

Expert Solution & Answer

Answer to Problem 25P

Equilibrium concentration of fructose-1,6-bisphosphate = $7.8 ×10−4 M$

Equilibrium concentration of DHAP = $2.2×10−4M$

Equilibrium concentration of G3P = $2.2×10−4M$

Explanation of Solution

Given information:

1 mM (0.001M) of fructose-1,6-bisphosphate is incubated with aldolase under standard conditions.

The reaction is:

$Fructose-1,6-bisphosphate⇌aldolaseDHAP+G3P ; ΔG∘=23.9kJ/mol = 23.9 × 103J/mol$

Creating the ICE table as:

$Reaction - Fructose-1,6-bisphosphate⇌aldolaseDHAP+G3P ; ΔG∘=23.9kJ/mol = 23.9 × 103J/molInitial concentration- 0.001M 0M 0MDifference in conc - -xM +xM +xMFinal Concentration - (0.001-x)M xM xM$

Calculation of equilibrium constant, $Keq$ :

$ΔG∘=−RTlnKeq23.9 × 103J/mol=−8.314J/mol K × 298K×ln KeqKeq = 6.5×10−5$

Now,

$Keq = concentration of productsconcentration of substratesKeq = [DHAP]×[G3P][fructose-1,6-bisphosphate]6.5×10−5=x × x0.001−x6.5×10−5(0.001−x) = x26.5×10−8−6.5×10x = x2x2+6.5×10−5x−6.5×10−8=0$

Solving for x,

$x = −b± b2 −4ac2ax = −(6.5× 10 −5)± (6.5× 10 −5 )2 −4×1×(−6.5× 10 −8 )2×1x = 2.2×10−4M$

So, equilibrium concentrations of DHAP and G3P = $2.2×10−4M$

Equilibrium concentration of fructose-1,6-bisphosphate = $(0.001-2.2×10−4) M = 7.8 ×10−4 M$

Conclusion

Equilibrium concentration of fructose-1,6-bisphosphate = $7.8 ×10−4 M$

Equilibrium concentration of DHAP = $2.2×10−4M$

Equilibrium concentration of G3P = $2.2×10−4M$

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Answer 4

Question

Chapter 16, Problem 25P

Interpretation Introduction

Interpretation:

The equilibrium concentrations of fructose-1,6-bisphosphate, dihydroxyacetone (DHAP), and glyceraldehyde-3-phosphate (G3P) under given conditions should be determined.

Concept introduction:

By the action of aldolase enzyme, fructose-1,6-bisphopshate is converted into DHAP and G3P. Aldolase uses zinc ion as its cofactor. This is the fourth step in the glycolysis process, and the last step of the preparatory phase.

Expert Solution & Answer

Answer to Problem 25P

Equilibrium concentration of fructose-1,6-bisphosphate = $7.8 ×10−4 M$

Equilibrium concentration of DHAP = $2.2×10−4M$

Equilibrium concentration of G3P = $2.2×10−4M$

Explanation of Solution

Given information:

1 mM (0.001M) of fructose-1,6-bisphosphate is incubated with aldolase under standard conditions.

The reaction is:

$Fructose-1,6-bisphosphate⇌aldolaseDHAP+G3P ; ΔG∘=23.9kJ/mol = 23.9 × 103J/mol$

Creating the ICE table as:

$Reaction - Fructose-1,6-bisphosphate⇌aldolaseDHAP+G3P ; ΔG∘=23.9kJ/mol = 23.9 × 103J/molInitial concentration- 0.001M 0M 0MDifference in conc - -xM +xM +xMFinal Concentration - (0.001-x)M xM xM$

Calculation of equilibrium constant, $Keq$ :

$ΔG∘=−RTlnKeq23.9 × 103J/mol=−8.314J/mol K × 298K×ln KeqKeq = 6.5×10−5$

Now,

$Keq = concentration of productsconcentration of substratesKeq = [DHAP]×[G3P][fructose-1,6-bisphosphate]6.5×10−5=x × x0.001−x6.5×10−5(0.001−x) = x26.5×10−8−6.5×10x = x2x2+6.5×10−5x−6.5×10−8=0$

Solving for x,

$x = −b± b2 −4ac2ax = −(6.5× 10 −5)± (6.5× 10 −5 )2 −4×1×(−6.5× 10 −8 )2×1x = 2.2×10−4M$

So, equilibrium concentrations of DHAP and G3P = $2.2×10−4M$

Equilibrium concentration of fructose-1,6-bisphosphate = $(0.001-2.2×10−4) M = 7.8 ×10−4 M$

Conclusion

Equilibrium concentration of fructose-1,6-bisphosphate = $7.8 ×10−4 M$

Equilibrium concentration of DHAP = $2.2×10−4M$

Equilibrium concentration of G3P = $2.2×10−4M$

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Answer 5

Chapter 16, Problem 25P

Interpretation Introduction

Interpretation:

The equilibrium concentrations of fructose-1,6-bisphosphate, dihydroxyacetone (DHAP), and glyceraldehyde-3-phosphate (G3P) under given conditions should be determined.

Concept introduction:

By the action of aldolase enzyme, fructose-1,6-bisphopshate is converted into DHAP and G3P. Aldolase uses zinc ion as its cofactor. This is the fourth step in the glycolysis process, and the last step of the preparatory phase.

Expert Solution & Answer

Answer to Problem 25P

Equilibrium concentration of fructose-1,6-bisphosphate = $7.8 ×10−4 M$

Equilibrium concentration of DHAP = $2.2×10−4M$

Equilibrium concentration of G3P = $2.2×10−4M$

Explanation of Solution

Given information:

1 mM (0.001M) of fructose-1,6-bisphosphate is incubated with aldolase under standard conditions.

The reaction is:

$Fructose-1,6-bisphosphate⇌aldolaseDHAP+G3P ; ΔG∘=23.9kJ/mol = 23.9 × 103J/mol$

Creating the ICE table as:

$Reaction - Fructose-1,6-bisphosphate⇌aldolaseDHAP+G3P ; ΔG∘=23.9kJ/mol = 23.9 × 103J/molInitial concentration- 0.001M 0M 0MDifference in conc - -xM +xM +xMFinal Concentration - (0.001-x)M xM xM$

Calculation of equilibrium constant, $Keq$ :

$ΔG∘=−RTlnKeq23.9 × 103J/mol=−8.314J/mol K × 298K×ln KeqKeq = 6.5×10−5$

Now,

$Keq = concentration of productsconcentration of substratesKeq = [DHAP]×[G3P][fructose-1,6-bisphosphate]6.5×10−5=x × x0.001−x6.5×10−5(0.001−x) = x26.5×10−8−6.5×10x = x2x2+6.5×10−5x−6.5×10−8=0$

Solving for x,

$x = −b± b2 −4ac2ax = −(6.5× 10 −5)± (6.5× 10 −5 )2 −4×1×(−6.5× 10 −8 )2×1x = 2.2×10−4M$

So, equilibrium concentrations of DHAP and G3P = $2.2×10−4M$

Equilibrium concentration of fructose-1,6-bisphosphate = $(0.001-2.2×10−4) M = 7.8 ×10−4 M$

Conclusion

Equilibrium concentration of fructose-1,6-bisphosphate = $7.8 ×10−4 M$

Equilibrium concentration of DHAP = $2.2×10−4M$

Equilibrium concentration of G3P = $2.2×10−4M$

Answer 6

Chapter 16, Problem 25P

Interpretation Introduction

Interpretation:

The equilibrium concentrations of fructose-1,6-bisphosphate, dihydroxyacetone (DHAP), and glyceraldehyde-3-phosphate (G3P) under given conditions should be determined.

Concept introduction:

By the action of aldolase enzyme, fructose-1,6-bisphopshate is converted into DHAP and G3P. Aldolase uses zinc ion as its cofactor. This is the fourth step in the glycolysis process, and the last step of the preparatory phase.

Expert Solution & Answer

Answer to Problem 25P

Equilibrium concentration of fructose-1,6-bisphosphate = $7.8 ×10−4 M$

Equilibrium concentration of DHAP = $2.2×10−4M$

Equilibrium concentration of G3P = $2.2×10−4M$

Explanation of Solution

Given information:

1 mM (0.001M) of fructose-1,6-bisphosphate is incubated with aldolase under standard conditions.

The reaction is:

$Fructose-1,6-bisphosphate⇌aldolaseDHAP+G3P ; ΔG∘=23.9kJ/mol = 23.9 × 103J/mol$

Creating the ICE table as:

$Reaction - Fructose-1,6-bisphosphate⇌aldolaseDHAP+G3P ; ΔG∘=23.9kJ/mol = 23.9 × 103J/molInitial concentration- 0.001M 0M 0MDifference in conc - -xM +xM +xMFinal Concentration - (0.001-x)M xM xM$

Calculation of equilibrium constant, $Keq$ :

$ΔG∘=−RTlnKeq23.9 × 103J/mol=−8.314J/mol K × 298K×ln KeqKeq = 6.5×10−5$

Now,

$Keq = concentration of productsconcentration of substratesKeq = [DHAP]×[G3P][fructose-1,6-bisphosphate]6.5×10−5=x × x0.001−x6.5×10−5(0.001−x) = x26.5×10−8−6.5×10x = x2x2+6.5×10−5x−6.5×10−8=0$

Solving for x,

$x = −b± b2 −4ac2ax = −(6.5× 10 −5)± (6.5× 10 −5 )2 −4×1×(−6.5× 10 −8 )2×1x = 2.2×10−4M$

So, equilibrium concentrations of DHAP and G3P = $2.2×10−4M$

Equilibrium concentration of fructose-1,6-bisphosphate = $(0.001-2.2×10−4) M = 7.8 ×10−4 M$

Conclusion

Equilibrium concentration of fructose-1,6-bisphosphate = $7.8 ×10−4 M$

Equilibrium concentration of DHAP = $2.2×10−4M$

Equilibrium concentration of G3P = $2.2×10−4M$

Answer 7

Chapter 16, Problem 25P

Interpretation Introduction

Interpretation:

The equilibrium concentrations of fructose-1,6-bisphosphate, dihydroxyacetone (DHAP), and glyceraldehyde-3-phosphate (G3P) under given conditions should be determined.

Concept introduction:

By the action of aldolase enzyme, fructose-1,6-bisphopshate is converted into DHAP and G3P. Aldolase uses zinc ion as its cofactor. This is the fourth step in the glycolysis process, and the last step of the preparatory phase.

Answer 8

Expert Solution & Answer

Answer to Problem 25P

Equilibrium concentration of fructose-1,6-bisphosphate = $7.8 ×10−4 M$

Equilibrium concentration of DHAP = $2.2×10−4M$

Equilibrium concentration of G3P = $2.2×10−4M$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given information:

1 mM (0.001M) of fructose-1,6-bisphosphate is incubated with aldolase under standard conditions.

The reaction is:

$Fructose-1,6-bisphosphate⇌aldolaseDHAP+G3P ; ΔG∘=23.9kJ/mol = 23.9 × 103J/mol$

Creating the ICE table as:

$Reaction - Fructose-1,6-bisphosphate⇌aldolaseDHAP+G3P ; ΔG∘=23.9kJ/mol = 23.9 × 103J/molInitial concentration- 0.001M 0M 0MDifference in conc - -xM +xM +xMFinal Concentration - (0.001-x)M xM xM$

Calculation of equilibrium constant, $Keq$ :

$ΔG∘=−RTlnKeq23.9 × 103J/mol=−8.314J/mol K × 298K×ln KeqKeq = 6.5×10−5$

Now,

$Keq = concentration of productsconcentration of substratesKeq = [DHAP]×[G3P][fructose-1,6-bisphosphate]6.5×10−5=x × x0.001−x6.5×10−5(0.001−x) = x26.5×10−8−6.5×10x = x2x2+6.5×10−5x−6.5×10−8=0$