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Chapter 12: Problem 23

The phosphorolysis reaction that removes glucose residues from glycogen yields as its product glucose-1-phosphate. Glucose-1-phosphate is isomerized to glucose-6-phosphate; then the phosphate group is removed in a hydrolysis reaction. Why is it necessary to remove the phosphate group before the glucose exits the cell to enter the circulation?

Short Answer

Expert verified
Removing the phosphate group allows glucose to exit the cell and enter the bloodstream.

Step by step solution

01

Understanding Glucose-1-phosphate Formation

The breakdown of glycogen through phosphorolysis leads to the production of glucose-1-phosphate, which is an essential step in converting glycogen into a usable form of glucose for the body.
02

Isomerization to Glucose-6-phosphate

Glucose-1-phosphate undergoes isomerization to form glucose-6-phosphate. This transformation is crucial because glucose-6-phosphate is a more versatile intermediate that can enter various pathways, such as glycolysis.
03

The Role of the Phosphate Group

The presence of the phosphate group on glucose-6-phosphate ensures that glucose remains within the cell, as phosphorylated glucose cannot easily cross cell membranes. This prevents glucose from leaving the cell prematurely.
04

Hydrolysis Reaction

The hydrolysis reaction removes the phosphate group from glucose-6-phosphate, converting it into free glucose. This step is necessary to prepare glucose for export out of the cell.
05

Importance of Phosphate Removal

After the phosphate group is removed, glucose can freely exit the cell and enter the circulation. It is necessary to remove the phosphate group, or else the glucose would remain trapped within the cell and unable to be used by the body.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Glucose-1-Phosphate: The Starting Point
Glucose-1-phosphate is an important molecule in the process of glycogen metabolism. As glycogen is broken down, glucose residues are released, forming glucose-1-phosphate through a reaction called phosphorolysis. This step is critical, as it helps in converting glycogen鈥攐ur body's stored form of glucose鈥攊nto a usable format.
Understanding the role of glucose-1-phosphate is crucial. It starts as an attachment point for additional enzymes that facilitate further transformations, ultimately leading to energy production or replenishment of glucose levels in the blood.
Isomerization: Shifting Structures
Isomerization is a fascinating process that involves rearranging the molecular structure of glucose-1-phosphate to form glucose-6-phosphate. Although the two molecules have the same atomic makeup, their structural difference is key.
  • Isomerization allows glucose-6-phosphate to participate in several cellular pathways, including glycolysis, where it is prepared for energy extraction.
  • This transformation enhances the flexibility of glucose utilization, making it a versatile intermediate in cellular metabolism.
The shift from glucose-1-phosphate to glucose-6-phosphate occurs thanks to enzymes like phosphoglucomutase, which catalyzes the rearrangement efficiently within the cell, thereby maintaining metabolic balance.
Phosphorolysis: Breaking Down Glycogen
Phosphorolysis is akin to a biochemical hammer that breaks the bonds in glycogen. This reaction involves the cleavage of glycogen's glucose units by adding a phosphate group, freeing them as glucose-1-phosphate. This process is vital for energy mobilization, especially during fasting or strenuous activity when the body's demand for glucose increases.
The creation of glucose-1-phosphate rather than free glucose during phosphorolysis is efficient because it conserves energy. Directly producing glucose-1-phosphate rather than free glucose saves the body from requiring additional energy for phosphorylation, keeping metabolic processes smooth and efficient.
Glucose-6-Phosphate: A Gateway Molecule
Glucose-6-phosphate holds a central position in cellular metabolism. After its formation through isomerization, this molecule can enter numerous pathways, playing a key role in maintaining the cell's energy balance.
  • Its phosphorylated state ensures it remains within the cell, minimizing the risk of glucose loss through membranes, which are generally impermeable to charged molecules like phosphorylated sugars.
  • Participating in glycolysis, it serves as a precursor for pyruvate, contributing to ATP production, the energy currency of the cell.
In situations where glucose must exit the cell to serve the body's systemic needs, the phosphate group is removed through hydrolysis, producing free glucose. This transformation is needed for glucose to cross the cell membrane and circulate in the bloodstream, aiding in glucose delivery to tissues that require it for energy.

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Most popular questions from this chapter

Vitamin \(\mathrm{B}_{12}\) is synthesized by certain gastrointestinal bacteria and is also found in foods of animal origin such as meat, milk, eggs, and fish. When vitamin \(\mathbf{B}_{12}\)-containing foods are consumed, the vitamin is released from the food and binds to a salivary vitamin \(B_{12}-\) binding protein called haptocorrin. The haptocorrin-vitamin \(B_{12}\) complex passes from the stomach to the small intestine, where the vitamin is released from the haptocorrin and then binds to intrinsic factor (IF). The IF-vitamin \(\mathrm{B}_{12}\) complex then enters the cells lining the intestine by receptor-mediated endocytosis. Using this information, make a list of individuals most at risk for vitamin \(\mathrm{B}_{12}\) deficiency.

a. The complete oxidation of glucose releases a considerable amount of energy. The \(\Delta G^{\circ r}\) for the reaction shown is \(-2850 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\). $$ \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}+6 \mathrm{O}_{2} \longrightarrow 6 \mathrm{CO}_{2}+6 \mathrm{H}_{2} \mathrm{O} $$ How many moles of ATP could be produced under standard conditions from the oxidation of one mole of glucose, assuming about \(33 \%\) efficiency? b. The oxidation of palmitate, a 16 -carbon saturated fatty acid, releases \(9781 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\). $$ \mathrm{C}_{16} \mathrm{H}_{32} \mathrm{O}_{2}+23 \mathrm{O}_{2} \rightarrow 16 \mathrm{CO}_{2}+16 \mathrm{H}_{2} \mathrm{O} $$ How many moles of ATP could be produced under standard conditions from the oxidation of one mole of palmitate, assuming \(33 \%\) efficiency? c. Calculate the number of ATP molecules produced per carbon for glucose and palmitate. Explain the reason for the difference.

Citrate is isomerized to isocitrate in the citric acid cycle (Chapter 14). The reaction is catalyzed by the enzyme aconitase. The \(\Delta G^{\circ}\) of the reaction is \(5 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\). The properties of the reaction are studied in vitro, where \(1 \mathrm{M}\) citrate and \(1 \mathrm{M}\) isocitrate are added to an aqueous solution of the enzyme at \(25^{\circ} \mathrm{C}\). a. What is the \(K_{\text {eq }}\) for the reaction? b. What are the equilibrium concentrations of the reactant and product? c. What is the preferred direction of the reaction under standard conditions? d. The aconitase reaction is the second step of an eight-step pathway and occurs in the direction shown in the figure. How can you reconcile these facts with your answer to part c?

Monosaccharides, the products of polysaccharide and disaccharide digestion, enter the cells lining the intestine via a specialized transport system. What is the source of free energy for this transport process?

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