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Chapter 12: Problem 6

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?

Short Answer

Expert verified
The source of free energy for monosaccharide transport is ATP hydrolysis.

Step by step solution

01

Understanding the Transport Process

Monosaccharides are absorbed by the cells lining the intestine through a process called active transport. This process involves the movement of monosaccharides against their concentration gradient.
02

Identifying the Energy Source

To move monosaccharides against their concentration gradient, energy is required. This energy comes from the transport system, which utilizes ATP (adenosine triphosphate) as the primary energy source.
03

ATP Role in Transport

ATP provides energy by undergoing hydrolysis, which releases phosphate groups and energy that is used to change the conformation of the transport proteins, facilitating the movement of monosaccharides into the cells.
04

Mechanism of ATP in Transport Systems

In the transport system, ATP hydrolysis generates free energy sufficient to maintain an electrochemical gradient (such as the sodium gradient), which is then used to drive the transport of monosaccharides through co-transport or symport systems.

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Key Concepts

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

Active Transport
The process by which monosaccharides enter the cells lining the intestine is known as active transport. This is a vital mechanism, especially considering that these molecules need to traverse against their concentration gradient to be absorbed efficiently. In simpler terms, active transport functions like a biological pump. It requires energy to "push" or "pull" the monosaccharides from an area of lower concentration to an area of higher concentration, contrary to what typically occurs during simple diffusion, which is passive.

Unlike passive processes, such as diffusion, where substances move from higher to lower concentrations naturally, active transport is crucial for moving essential nutrients into cells even when they are in lower concentrations outside the cells compared to inside. This ensures that cells receive critical nutrients and maintain homeostasis, a stable internal environment required for proper functioning.
ATP Hydrolysis
Energy for active transport comes chiefly from the molecule called ATP, or adenosine triphosphate. But how exactly does ATP supply the energy required for moving monosaccharides? Well, this is possible through a process known as ATP hydrolysis.

Hydrolysis refers to a chemical breakdown involving water. In ATP hydrolysis, a water molecule helps break one of the phosphate bonds in ATP, resulting in adenosine diphosphate (ADP) and an inorganic phosphate (P_i).
  • This bond-breaking process occurs in a controlled manner, releasing a significant amount of energy.
  • The energy released is utilized to change the shape or conformation of transport proteins.
  • It facilitates the movement of monosaccharides against their concentration gradient into the cell.
This energy transformation is essential as it enables transport proteins to perform work, ensuring that key nutrients like monosaccharides are adequately taken up by the cells.
Electrochemical Gradient
The culmination of ATP hydrolysis is not just about providing energy; it also helps in establishing what is known as an electrochemical gradient. This gradient is pivotal for active transport processes like the absorption of monosaccharides.

An electrochemical gradient encapsulates both the difference in ion concentration and electric charge across a membrane. Being in a biological system, ions such as sodium (Na^+) play a critical role in this process.
  • Due to the energy released by ATP hydrolysis, these ions are actively transported to create a high concentration on one side of the membrane.
  • This setup generates both an electrical and a concentration (chemical) difference between the two sides of the membrane.
  • It acts as stored energy or a potential energy source.
Transport proteins then use this electrochemical gradient to drive the uptake of monosaccharides into the cell via co-transport or symport systems. By coupling the movement of sodium ions with monosaccharides, it becomes possible to transport them efficiently against their gradient without directly expending more ATP. This indirect use of energy derived from ATP hydrolysis is a hallmark of efficient biological systems.

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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?

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.
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