/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 The enzyme hexokinase adds a pho... [FREE SOLUTION] | 91Ó°ÊÓ

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The enzyme hexokinase adds a phosphate to D-glucose but ignores its mirror image, L- glucose. Suppose that you were able to synthesize hexokinase entirely from D-amino acids, which are the mirror image of the normal L-amino acids. A. Assuming that the "D" enzyme would fold to a stable conformation, what relationship would you expect it to bear to the normal "L" enzyme? B. Do you suppose the "D" enzyme would add a phosphate to L-glucose, and ignore D-glucose?

Short Answer

Expert verified
A mirror image 'D' enzyme should interact with L-glucose, ignoring D-glucose, due to structural mirror symmetry.

Step by step solution

01

Understanding the Original Enzyme

Hexokinase is an enzyme made from L-amino acids (the natural form) that catalyzes the phosphorylation of D-glucose but not its enantiomer, L-glucose.
02

Considering the Mirror Image Enzyme

When supposed to be made entirely from D-amino acids, a 'D' enzyme would be the mirror image of a normal 'L' enzyme.
03

Predicting the Folding of the Mirror Image

Given a stable conformation, the 'D' enzyme should fold into a structure that is the mirror image of the 'L' enzyme.
04

Substrate Specificity of the Enzymes

Since enzymes are highly specific to their substrates, the mirror image 'D' enzyme should be specific to the mirror images of the substrates of the original enzyme.
05

Predicting the Activity on L-Glucose

The 'D' enzyme, being the mirror image, would likely add a phosphate to L-glucose, since it is the mirror image of the naturally phosphorylated D-glucose.
06

Conclusion on Substrate Ignorance

Conversely, just as the 'L' enzyme ignores L-glucose, the 'D' enzyme would ignore D-glucose, maintaining substrate specificity due to its mirror-image nature.

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Key Concepts

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

Stereochemistry in Enzymes
Enzymes are nature's efficient catalysts, and their activity is deeply rooted in their 3-dimensional structures. This 3D structure relies on the specific orientation of atoms within the molecules, known as stereochemistry. Stereochemistry is crucial in enzymes because it dictates how enzymes interact with substrates. Enzymes have active sites that provide a precise fit, like a lock and key, for their specific substrates. This specificity often depends on the chiral nature of the enzyme and its substrate. The chiral nature means that the enzyme can distinguish between different isomers of a substrate. Chiral molecules have non-superimposable mirror images, much like our left and right hands. Due to this property, enzymes are often selective and will only interact with one enantiomer of a substrate, reinforcing the snug fit required for catalysis to occur. In sum, the stereochemistry in enzymes plays an integral role in enzyme specificity, ensuring that enzymes interact, react, and transform the correct substrates.
Hexokinase Action
Hexokinase is an enzyme that initiates glycolysis by catalyzing the phosphorylation of glucose. Specifically, it adds a phosphate group to D-glucose, converting it into glucose-6-phosphate. This reaction is one of the first steps in breaking down glucose for energy production. Hexokinase is highly specific for D-glucose and does not catalyze the same reaction for L-glucose due to the engagement of the enzyme's active site with the particular stereochemical configuration of D-glucose. This high specificity is similar to having a key that only fits a specific lock. Because of this specificity, when hexokinase is made from alternative amino acid arrangements, such as D-amino acids, its activity can change drastically. An enzyme synthesized from D-amino acids would have a mirror image structure compared to the one composed of L-amino acids. Thus, instead of acting on D-glucose, it might indeed find L-glucose as a suitable substrate if it maintains a stable conformation. This transformation showcases the delicate balance and precision required in enzymatic functions.
D-amino acids vs L-amino acids
Amino acids serve as the building blocks of proteins, including enzymes. They exist in two optical isomeric forms: D-amino acids and L-amino acids. Most naturally occurring proteins, including enzymes, are composed of L-amino acids. Both D- and L-amino acids have the same molecular formula but differ in the spatial arrangement of atoms. This spatial orientation leads to significantly different properties in peptides and proteins. While L-amino acids are integral in cellular functions, D-amino acids are relatively rare in nature and usually found in specific bacterial cell walls or as part of certain antibiotics. In enzymatic contexts, replacing L-amino acids with D-amino acids can alter the enzyme's folding and activity. Such an enzyme might function as the mirror image of its L-amino acid counterpart, possibly interacting with the mirror images of substrates it would typically engage with. For instance, a hexokinase synthesized from D-amino acids might act on L-glucose instead of D-glucose. This concept illustrates the importance of amino acid configuration in determining enzyme specificity and activity, highlighting the elegant complexity of biochemical systems.

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Most popular questions from this chapter

Consider the following statement. "To produce one molecule of each possible kind of polypeptide chain, 300 amino acids in length, would require more atoms than exist in the universe." Given the size of the universe, do you suppose this statement could possibly be correct? since counting atoms is a tricky business, consider the problem from the standpoint of mass. The mass of the observable universe is estimated to be about \(10^{80}\) grams, give or take an order of magnitude or so. Assuming that the average mass of an amino acid is 110 daltons, what would be the mass of one molecule of each possible kind of polypeptide chain 300 amino acids in length? Is this greater than the mass of the universe?

A common strategy for identifying distantly related homologous proteins is to search the database using a short signature sequence indicative of the particular protein function. Why is it better to search with a short sequence than with a long sequence? Do you not have more chances for a "hit" in the database with a long sequence?

An antibody binds to another protein with an equilibrium constant, \(K\), of \(5 \times 10^{9} \mathrm{M}^{-1}\). When it binds to a second, related protein, it forms three fewer hydrogen bonds, reducing its binding affinity by \(11.9 \mathrm{kJ} / \mathrm{mole}\). What is the \(K\) for its binding to the second protein? (Free-energy change is related to the equilibrium constant by the equation \(\Delta G^{\circ}=-2.3 R T \log K,\) where \(R\) is \(8.3 \times 10^{-3} \mathrm{kJ} /(\text { mole } \mathrm{K})\) and \(T \text { is } 310 \text { K. })\)

The protein \(\mathrm{SmpB}\) binds to a special species of tRNA, tmRNA, to eliminate the incomplete proteins made from truncated mRNAs in bacteria. If the binding of SmpB to tmRNA is plotted as fraction tmRNA bound versus SmpB concentration, one obtains a symmetrical S-shaped curve as shown in Figure \(Q 3-3 .\) This curve is a visual display of a very useful relationship between \(K_{\mathrm{d}}\) and concentration, which has broad applicability. The general expression for fraction of ligand bound is derived from the equation for \(K_{\mathrm{d}}\left(K_{\mathrm{d}}=[\operatorname{Pr}][\mathrm{L}] /[\mathrm{Pr}-\mathrm{L}]\right)\) by substituting \(\left([\mathrm{L}]_{\mathrm{TOT}}-[\mathrm{L}]\right)\) for \([\mathrm{Pr}-\mathrm{L}]\) and rearranging. Because the total concentration of ligand ( \(\left.[\mathrm{L}]_{\mathrm{TOT}}\right)\) is equal to the free ligand ([L]) plus bound \(\operatorname{ligand}([\mathrm{Pr}-\mathrm{L}])\) $$\text { fraction bound }=[\mathrm{Pr}-\mathrm{L}] /[\mathrm{L}]_{\mathrm{TOT}}=[\mathrm{Pr}] /\left([\mathrm{Pr}]+K_{\mathrm{d}}\right)$$ For \(\mathrm{SmpB}\) and tmRNA, the fraction bound \(=[\mathrm{SmpB}\) \(\mathrm{tmRNA}] /[\mathrm{tmRNA}]_{\mathrm{TOT}}=[\mathrm{SmpB}] /\left([\mathrm{SmpB}]+K_{\mathrm{d}}\right) .\) Using this relationship, calculate the fraction of tmRNA bound for SmpB concentrations equal to \(10^{4} K_{\mathrm{d}}, 10^{3} K_{\mathrm{d}}, 10^{2} K_{\mathrm{d}}\) \(10^{1} K_{\mathrm{d}}, K_{\mathrm{d}}, 10^{-1} K_{\mathrm{d}}, 10^{-2} K_{\mathrm{d}}, 10^{-3} K_{\mathrm{d}},\) and \(10^{-4} K_{\mathrm{d}}\)

Many enzymes obey simple Michaelis-Menten kinetics, which are summarized by the equation $$\text { rate }=V_{\max }[\mathrm{S}] /\left([\mathrm{S}]+K_{\mathrm{m}}\right)$$ where \(V_{\max }=\operatorname{maximum}\) velocity, \([\mathrm{S}]=\) concentration of substrate, and \(K_{\mathrm{m}}=\) the Michaelis constant. It is instructive to plug a few values of \([\mathrm{S}]\) into the equation to see how rate is affected. What are the rates for [S] equal to zero, equal to \(K_{\mathrm{m}}\), and equal to infinite concentration?

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