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Chapter 7: Problem 38

The enzyme hexokinase acts on both glucose and fructose. Using the \(K_{\mathrm{M}}\) and \(V_{\max }\) values in the table, compare and contrast the interaction of hexokinase with each substrate. $$ \begin{array}{lcc} \text { Substrate } & K_{\mathbf{M}}(\mathbf{M}) & \boldsymbol{V}_{\max } \text { (relative) } \\ \hline \text { Glucose } & 1.0 \times 10^{-4} & 1.0 \\ \text { Fructose } & 7.0 \times 10^{-4} & 1.8 \end{array} $$

Short Answer

Expert verified
Hexokinase has a higher affinity for glucose but a higher \(V_{\max}\) for fructose.

Step by step solution

01

Understanding the Variables

We have two key metrics: the Michaelis constant \(K_{M}\) and the maximum velocity \(V_{\max}\) for each substrate. \(K_{M}\) indicates the substrate concentration at which the reaction velocity is half of \(V_{\max}\). A lower \(K_{M}\) implies higher affinity since less substrate is needed to reach half \(V_{\max}\). \(V_{\max}\) is the rate of reaction when the enzyme is fully saturated with the substrate.
02

Analyze Glucose Interaction

For glucose, \(K_{M} = 1.0 \times 10^{-4} M\) and \(V_{\max} = 1.0\). The low \(K_{M}\) value suggests that hexokinase has a high affinity for glucose as the enzyme reaches half of its maximum speed at a lower glucose concentration. The \(V_{\max}\) indicates that when glucose saturates hexokinase, the relative maximum velocity is achieved.
03

Analyze Fructose Interaction

Fructose has \(K_{M} = 7.0 \times 10^{-4} M\) and \(V_{\max} = 1.8\). The higher \(K_{M}\) value compared to glucose means hexokinase has a lower affinity for fructose, as it requires a higher concentration of fructose to reach half \(V_{\max}\). However, the \(V_{\max}\) is higher for fructose (1.8) than glucose, indicating that when saturated, hexokinase processes fructose at a higher relative maximum velocity.
04

Comparison and Conclusion

Hexokinase has a higher affinity for glucose than fructose as shown by the lower \(K_{M}\). However, the enzyme reaches a higher relative \(V_{\max}\) with fructose, meaning that when enough substrate is present, the enzyme can catalyze the reaction faster with fructose than with glucose.

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Key Concepts

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

Hexokinase
Hexokinase is an enzyme that plays a crucial role in metabolism by catalyzing the phosphorylation of six-carbon sugars such as glucose and fructose. This process is essential for cellular energy production, as it is the first step in glycolysis.
Hexokinase facilitates the transfer of a phosphate group from adenosine triphosphate (ATP) to the sugar molecule, effectively converting it into a form that can be further processed by the cell to generate energy.
  • Hexokinase exhibits a broad specificity, enabling it to act on various sugars, but it shows different affinities for each substrate.
  • Understanding how hexokinase interacts with different sugars, like glucose and fructose, can provide insights into metabolism and potentially influence the development of therapeutic strategies for metabolic diseases.
Due to its vital function, hexokinase activities are tightly regulated according to the cell's energy needs and the availability of substrates. This makes it a key enzyme in balancing energy production and consumption in cells.
Michaelis Constant (K_M)
The Michaelis Constant, denoted as \( K_{\mathrm{M}} \), is a fundamental parameter in enzyme kinetics that tells us about the affinity between an enzyme and its substrate.
Essentially, \( K_{\mathrm{M}} \) is the substrate concentration at which the reaction rate is half of the enzyme's maximum velocity (\( V_{\max} \)).
  • A smaller \( K_{\mathrm{M}} \) value indicates a higher affinity of the enzyme for the substrate, meaning that only a lower concentration of substrate is required to reach an effective rate of reaction.
  • Conversely, a larger \( K_{\mathrm{M}} \) value suggests a lower affinity, as more substrate is needed to achieve the same rate.
In the exercise, hexokinase exhibited a lower \( K_{\mathrm{M}} \) for glucose compared to fructose, signifying a higher affinity for glucose. This behavior impacts the cell's metabolism because enzymes are more efficient under conditions where they readily bind with their substrates.
Maximum Velocity (V_max)
The maximum velocity, often symbolized as \( V_{\max} \), is the rate of an enzyme-catalyzed reaction when the enzyme is saturated with substrate.
Simply put, it is the fastest rate at which the enzyme can convert substrate into product.
  • \( V_{\max} \) reflects not only the enzymatic efficiency but also hints at how many substrate molecules can be processed per unit time.
  • Different substrates can lead to different \( V_{\max} \) values even for the same enzyme, due to variations in substrate-enzyme interactions.
In the given exercise, although hexokinase had a higher affinity for glucose (as indicated by \( K_{\mathrm{M}} \)), it achieved a higher \( V_{\max} \) with fructose. This suggests that when excess substrate is available, the enzyme processes fructose faster than glucose.Understanding \( V_{\max} \) is essential for comprehending how enzyme concentrations affect metabolic pathways and how alterations in these values could inform treatment strategies for enzyme-related disorders.

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Most popular questions from this chapter

An alternative way to calculate \(K_{1}\) for the vanadate inhibitor in the PTP1B reaction (see Problem 63) is to measure the velocity of the enzyme- catalyzed reaction in the presence of increasing amounts of inhibitor and a constant amount of substrate. These data are shown in the table for a substrate concentration of \(6.67 \mu \mathrm{M}\). To calculate \(K_{\mathrm{l}}\), rearrange Equation \(7.28\) and solve for \(\alpha\). Then construct a graph plotting \(\alpha\) versus \([\mathrm{I}]\). Since \(\alpha=1+[\mathrm{I}] / K_{\mathrm{I}}\), the slope of the line is equal to \(1 / K_{\mathrm{I}}\). Determine \(K_{\mathrm{I}}\) for vanadate using this method. $$ \begin{array}{cc} \text { [Vanadate] }(\boldsymbol{\mu} \mathbf{M}) & \boldsymbol{v}_{0}\left(\mathbf{n M} \cdot \mathbf{s}^{-1}\right) \\ \hline 0.0 & 5.70 \\ 0.2 & 3.83 \\ 0.4 & 3.07 \\ 0.7 & 2.35 \\ 1.0 & 2.04 \\ 2.0 & 1.18 \\ 4.0 & 0.71 \end{array} $$

A bacterial enzyme catalyzes hydrolysis of the disaccharide maltose to produce two glucose monosaccharides. During an interval of one minute, the concentration of maltose decreases by \(65 \mathrm{mM}\). What is the rate of disappearance of maltose in the enzyme-catalyzed reaction?

The rate of hydrolysis of trehalose to its constituent glucose monomers in the absence of a catalyst is even slower than hydrolysis of sucrose (see Problem 3) and may indicate that the two sugars are hydrolyzed by different mechanisms. If the initial concentration of trehalose is \(0.050 \mathrm{M}\), it takes \(6.6 \times 10^{6}\) years for the concentration to decrease by half. What is the rate of disappearance of trehalose in the absence of a catalyst?

Brain glutaminase has a \(V_{\max }\) of \(1.1 \mu \mathrm{mol} \cdot \mathrm{min}^{-1} \cdot \mathrm{mL}^{-1}\) and a \(K_{\mathrm{M}}\) of \(0.6 \mathrm{mM}\). What is the substrate concentration when the velocity is \(0.3 \mu \mathrm{mol} \cdot \min ^{-1} \cdot \mathrm{mL}^{-1}\) ?

Explain why some drugs must be administered intravenously rather than swallowed in the form of a pill.
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