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What is feedback inhibition? Why is it a useful property?

Short Answer

Expert verified
Feedback inhibition is a process where the end product of a pathway inhibits an enzyme for regulation, conserving resources and maintaining balance in the cell.

Step by step solution

01

Understanding Feedback Inhibition

Feedback inhibition is a regulatory mechanism where the end product of a metabolic pathway inhibits an enzyme involved earlier in the pathway. This helps the cell conserve resources by halting the pathway when the end product is abundant.
02

Mechanism of Action

In feedback inhibition, the end product binds to a specific site on an enzyme (often called an allosteric site) that is not the active site. This binding alters the enzyme's shape, reducing its activity and thus slowing down the production of the end product.
03

Benefits of Feedback Inhibition

Feedback inhibition is useful because it allows the cell to maintain homeostasis by preventing the overproduction of substances. It efficiently controls metabolic pathways and conserves cellular energy and resources.

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Key Concepts

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

Metabolic Regulation
Metabolic regulation is crucial for the survival and efficiency of living organisms. It refers to the processes that govern the rate and direction of metabolic pathways, ensuring that cells operate efficiently. Ensuring that these pathways produce the right amount of metabolites is vital for cellular function and energy conservation. By adjusting the activity of enzymes in response to changes in the cellular environment, cells can prioritize certain pathways over others. This is often achieved through mechanisms such as feedback inhibition, which ensures that an excess of products does not accumulate. Metabolic regulation thus plays a pivotal role in maintaining balance within the cell and optimizing resource use.
Allosteric Regulation
Allosteric regulation involves the modification of an enzyme's activity through the binding of molecules at specific sites distinct from the enzyme's active site. This type of regulation is significant for controlling the metabolic pathways in cells. When a molecule binds to an allosteric site, it induces a conformational change in the enzyme. This can either enhance or inhibit the enzyme's activity, depending on whether the binding acts as an activator or inhibitor.
  • Feedback inhibition is a common example where the end product of a metabolic pathway binds to an allosteric site on an enzyme in the pathway.
  • Allosteric interactions allow for fine control over enzyme activity which can be adjusted according to the cell’s needs.
These regulations ensure that enzymes can respond precisely to the concentration of different metabolites and adjust metabolic processes accordingly.
Enzyme Activity
The activity of enzymes is essential for the regulation of biological reactions. Enzymes act as catalysts that speed up the rate of reactions by lowering the activation energy required. The modulation of enzyme activity is a primary mechanism of metabolic regulation. This is often achieved through various means, including changes in enzyme concentration, covalent modifications, and binding of regulatory molecules.
  • Enzymes have active sites where substrates bind and undergo a chemical transformation.
  • Allosteric sites provide additional control by allowing inhibitors or activators to modify enzyme shape and activity.
Through these mechanisms, cells can regulate the speed and outcomes of metabolic pathways to meet their dynamic needs. Enzyme activity is directly linked to efficient metabolic functioning and energy conservation.
Homeostasis
Homeostasis is the state of steady internal conditions maintained by living organisms. It is crucial for ensuring that the physiological processes occur within an optimal range. Metabolic pathways and their regulation, such as through feedback inhibition, are central to maintaining homeostasis. By preventing the overproduction or underproduction of metabolic products, these regulatory mechanisms keep cellular environments stable.
  • Feedback inhibition exemplifies how homeostatic balance is achieved; it halts pathways when products are in excess.
  • This dynamic balance allows organisms to successfully respond to external environmental changes.
Homeostasis relies on the precise regulation of metabolic activities to sustain life processes, making feedback inhibition an invaluable aspect of this control system.

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Most popular questions from this chapter

What is the biochemical advantage of having a \(K_{\mathrm{M}}\) approximately equal to the substrate concentration normally available to an enzyme?

Draw a double-reciprocal plot for a typical Michaelis-Menten enzyme and an allosteric enzyme that have the same $V_{\max },

Penicillin is hydrolyzed and thereby rendered inactive by penicillinase (also known as \(\beta\) -lactamase), an enzyme present in some resistant bacteria. The mass of this enzyme in Staphylococcus aureus is \(29.6 \mathrm{kDa}\). The amount of penicillin hydrolyzed in 1 minute in a \(10-\mathrm{ml}\) solution containing \(10^{-9} \mathrm{g}\) of purified penicillinase was measured as a function of the concentration of penicillin. Assume that the concentration of penicillin does not change appreciably during the assay. $$\begin{array}{cc}\text { [Penicillin] } \mu \mathrm{M} & \text { Amount hydrolyzed (nanomoles) } \\\\\hline 1 & 0.11 \\\3 & 0.25 \\\5 & 0.34 \\\10 & 0.45 \\\30 & 0.58 \\\50 & 0.61 \\ \hline\end{array}$$ (a) Plot \(V_{0}\) versus \([\mathrm{S}]\) and \(1 / V_{0}\) versus \(1 /[\mathrm{S}]\) for these data. Does penicillinase appear to obey Michaelis- Menten kinetics? If so, what is the value of \(K_{\mathrm{M}} ?\) (b) What is the value of \(V_{\max } ?\) (c) What is the turnover number of penicillinase under these experimental conditions? Assume one active site per enzyme molecule.

Differentiate between a first-order rate constant and a second-order rate constant.

The hydrolysis of pyrophosphate to orthophosphate is important in driving forward biosynthetic reactions such as the synthesis of DNA. This hydrolytic reaction is catalyzed in Escherichia coli by a pyrophosphatase that has a mass of \(120 \mathrm{kDa}\) and consists of six identical subunits. For this enzyme, a unit of activity is defined as the amount of enzyme that hydrolyzes \(10 \mu \mathrm{mol}\) of pyrophosphate in 15 minutes at \(37^{\circ} \mathrm{C}\) under standard assay conditions. The purified enzyme has a \(V_{\max }\) of 2800 units per milligram of enzyme. (a) How many moles of substrate are hydrolyzed per second per milligram of enzyme when the substrate concentration is much greater than \(K_{M} ?\) (b) How many moles of active sites are there in \(1 \mathrm{mg}\) of enzyme? Assume that each subunit has one active site. (c) What is the turnover number of the enzyme? Compare this value with others mentioned in this chapter.

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