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The human body ischaracterized by an extremely complex system of interrelated chemical reactions. A large number of enzymes are necessary for many of these reactions to occur at suitable rates. Enzymes are very selective in the reactions they catalyze, and some are absolutely specific. Use the lock-and- key model to account for the specificity of an enzyme.

Short Answer

Expert verified
The lock-and-key model explains enzyme specificity by matching enzyme (lock) and substrate (key) shapes, ensuring enzymes only catalyze specific reactions.

Step by step solution

01

Understand the Lock-and-Key Model

The lock-and-key model explains how enzymes are specific to certain substrates. An enzyme can be thought of as a 'lock' and the substrate it acts on as a 'key'. This model suggests that enzymes have a specific shape that matches the substrate, allowing them to bind together.
02

Explain Enzyme Specificity

Since the enzyme's active site (the 'lock') is uniquely shaped, it only fits specific substrates (the 'keys') that have complementary shapes. This specific binding ensures that enzymes only catalyze reactions for their particular substrates.
03

Demonstrate the Role of Enzymes

Once the substrate and enzyme are bound together, the enzyme can catalyze the chemical reaction, converting the substrate into the product. The enzyme itself remains unchanged and can repeat the process with another substrate molecule.
04

Conclude with Specificity Importance

The lock-and-key model highlights the importance of precise molecular shapes for enzyme function. It ensures that enzymes catalyze only specific reactions, which is vital for maintaining the correct metabolic pathways and homeostasis in the human body.

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Key Concepts

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

The Lock-and-Key Model
Imagine enzymes as a highly sophisticated security system in a high-tech building. The 'lock-and-key model' is one way scientists describe how enzymes function with remarkable specificity to only a particular substrate. Enzymes are the 'locks', and their substrates, the molecules they act upon, are the 'keys'. But what makes this analogy work? It's all about shapes.
Each enzyme has a unique active site, which is like the intricate grooves of a lock. In the same way that a key is crafted to fit only one lock, a substrate is shaped to fit precisely into the enzyme's active site. This specificity is due to the three-dimensional structure, which means that only a particular substrate with a complementary shape can fit into the enzyme.
  • Think of it as a puzzle piece that fits perfectly in only one spot.
  • This specific fit ensures that enzymes bind only to their designated substrate.
Right after binding, the enzyme can start its job of catalysis, guiding the transformation of the substrate into products without being altered itself. This "lock and key" interaction is fundamental to ensuring the precise catalytic action of enzymes.
Enzyme Catalysis
Once the enzyme and substrate form a secure complex—thanks to their perfect fit—the magic of enzyme catalysis begins. Enzymes are biological catalysts, meaning they greatly speed up the rate of chemical reactions. But how do they do it? Like a skillful chef, enzymes orchestrate chemical transformations efficiently without being consumed in the reaction.
When the substrate binds to the enzyme, it forms an enzyme-substrate complex. This arrangement facilitates the conversion of the substrate into its products. The enzyme lowers the activation energy—the energy barrier needed to kickstart the reaction. This means that the reaction can occur swiftly and efficiently at the body's normal temperature.
  • Enzymes enhance reaction speed up to a million times over.
  • They can repeat this process: bind, catalyze, and release, multiple times.
With enzymes acting rapidly and repeatedly, they play an essential role in all biochemical processes, including digestion, DNA replication, and energy production.
Metabolic Pathways
Our bodies are like bustling factories, full of intricate assembly lines known as metabolic pathways. These pathways are sequences of chemical reactions occurring within a cell, orchestrated so that the product of one reaction serves as the starting material for the next. Enzymes are key operators in these pathways, ensuring each step proceeds correctly and efficiently.
The lock-and-key model, coupled with enzyme catalysis, is crucial for maintaining metabolic pathways. Each enzyme in a pathway is specific to a particular substrate, propelling the entire sequence of reactions smoothly and in the correct order. This specificity allows cells to regulate complex biological processes finely.
  • Metabolic pathways include processes like glycolysis, the citric acid cycle, and the electron transport chain.
  • The precise control by enzymes ensures that energy and resources are used efficiently, preventing the chaos of unchecked reactions.
Overall, the specificity and efficiency provided by enzymes ensure that the body's metabolic pathways operate seamlessly, keeping systems balanced and functioning optimally.

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Most popular questions from this chapter

The oxidation of \(\mathrm{SO}_{2}\) to \(\mathrm{SO}_{3}\) is accelerated by \(\mathrm{NO}_{2}\). The reaction proceeds according to: $$ \begin{array}{l} \mathrm{NO}_{2}(g)+\mathrm{SO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{SO}_{3}(g) \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) \end{array} $$ (a) Show that, with appropriate coefficients, the two reactions can be summed to give the overall oxidation of \(\mathrm{SO}_{2}\) by \(\mathrm{O}_{2}\) to give \(\mathrm{SO}_{3}\). (b) Do we consider \(\mathrm{NO}_{2}\) a catalyst or an intermediate in this reaction? (c) Would you classify NO as a catalyst or as an intermediate? (d) Is this an example of homogeneous catalysis or heterogeneous catalysis?

For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product: (a) \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{CO}_{2}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g})\) (b) \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)\) (c) \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)\) (d) \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\)

Platinum nanoparticles of diameter \(-2 \mathrm{nm}\) are important catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face- centered cubic arrangement with an edge length of \(392.4 \mathrm{pm} .\) (a) Estimate how many platinum atoms would fit into a \(2.0-\mathrm{nm}\) sphere; the volume of a sphere is \((4 / 3) \pi r^{3}\). Recall that \(1 \mathrm{pm}=1 \times 10^{-12} \mathrm{~m}\) and \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{~m} .\) (b) Esti- mate how many platinum atoms are on the surface of a \(2.0-\mathrm{nm}\) Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one \(\mathrm{Pt}\) atom can be estimated from its atomic diameter of \(280 \mathrm{pm}\) (c) Using your results from (a) and \((b),\) calculate the percentage of \(\mathrm{Pt}\) atoms that are on the surface of a \(2.0-\mathrm{nm}\) nanoparticle. (d) Repeat these calculations for a \(5.0-\mathrm{nm}\) platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

(a) The reaction \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)+\) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)\) is first order with in \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(a q)\) and zero-order in \(\mathrm{H}_{2} \mathrm{O} .\) At \(300 \mathrm{~K}\) the rate constant equals \(3.30 \times 10^{-2} \mathrm{~min}^{-1}\). Calculate the half- life at this temperature. (b) If the activation energy for this reaction is \(80.0 \mathrm{~kJ} / \mathrm{mol}\), at what temperature would the reaction rate be doubled?

The enzyme urease catalyzes the reaction of urea, \(\left(\mathrm{NH}_{2} \mathrm{CONH}_{2}\right),\) with water to produce carbon dioxide and ammonia. In water, without the enzyme, the reaction proceeds with a first-order rate constant of \(4.15 \times 10^{-5} \mathrm{~s}^{-1}\) at \(100^{\circ} \mathrm{C}\). In the presence of the enzyme in water, the reaction proceeds with a rate constant of \(3.4 \times 10^{4} \mathrm{~s}^{-1}\) at \(21^{\circ} \mathrm{C}\) (a) Write out the balanced equation for the reaction catalyzed by urease. (b) If the rate of the catalyzed reaction were the same at \(100^{\circ} \mathrm{C}\) as it is at \(21^{\circ} \mathrm{C},\) what would be the difference in the activation energy between the catalyzed and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at \(100^{\circ} \mathrm{C}\) as compared to that at \(21^{\circ} \mathrm{C} ?\) (d) On the basis of parts (c) and (d), what can you conclude about the difference in activation energies for the catalyzed and uncatalyzed reactions?

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