8

What you have given is: $$\begin{array}{c|c} \hline \dfrac{v}{[S]}, \ \pu{s^{-1}} & v, \ \pu{mol dm^-3 s^-1} \\ \hline 0.257 & 5.15 \times 10^{-2}\\ 0.895 & 4.48 \times 10^{-2}\\ 2.00 & 3.35 \times 10^{-2}\\ 3.59 & 1.8 \times 10^{-2}\\ 4.82 & 0.48 \times 10^{-2}\\ \hline \end{array}$$ Michaelis-Menten equation for enzyme ...


6

The turnover number or catalytic constant $ k_{\mathrm{cat}}$ in the Michaelis-Menten model is the rate constant for the productive dissociation of intermediate $\ce{ES}$: $$\nu = k_{\mathrm{cat}}[\ce{ES}]$$ The constant $k_{\mathrm{cat}}$ says how much product forms from intermediate but does not say how much intermediate forms in the first place. It is ...


6

The most basic kinetic scheme for enzymes is represented as $$\ce{E + S <=>[K_m] ES ->[k_{cat}] E + P}$$ As should be clear, the $k_{cat}$ is the rate constant for the reaction that occurs after substrate is bound to the enzyme. The resulting rate (kcat[E]tot) is only achieved when every molecule of enzyme essentially always is in the act of ...


5

Cysteine is a dispensable amino acid required for synthesis of protein and non-protein compounds. These non-protein compounds include pyruate (hence acetyl coenzyme A), taurine, sulfate, and glutatione (GSH). The fate of the cycteine is hence sketched in following diagram (Ref.1): As evidence from the diagram, the sulfur atom in cysteine has been converted ...


4

Michaelis-Menten kinetics is given by the equation: $$V = V_\mathrm{max}\frac{S}{K_M+S} \tag1$$ Where $V_\mathrm{max} = k_\mathrm{cat}\cdot [\ce{E_T}]$ and $K_M = \dfrac{k_\mathrm{on} + k_\mathrm{cat}}{k_\mathrm{off}}$ (using the conventional description of Briggs and Haldane's derivation of the Michaelis-Menten equation; Ref.1): $$\ce{E + S <=>[$k_\...


4

Michaelis-Menten kinetics is given by $$V = V_{\max} \frac{S}{K_M + S}$$ Taking the inverse of both sides to linearise the equation, $$\frac{1}{V} = \frac{1}{V_\max} +\frac{K_M}{V_{\max} S}$$ Taking $y = 1/V$ and $x = 1/S$ in $y = A + Bx$ leads by linear regression to $A = 0.4767$, $B = 4.4754$ (not including units). Therefore, $$V_\max = \frac{1}{A} = \pu{2....


4

By saying Catalysts are not consumed by reactions. is meant there is no stoichimetric ratio to reactants, consuming catalysts and forming from them catalytically inactive compound. By other words, some of reaction steps regenerates the original form of a catalyst, consumed by a prior step, so the net consumption is negligible. The transition between active ...


3

I believe you have misread the first referenced paper. What it says is that the $V_{max}$ and $K_m$ values for the forward and reverse reactions can be varied to optimize a property such as the maximum velocity (represented by $V_{max}$) in one direction, but are always constrained by the fixed value of $K_{eq}$. The reason that the $K_m$ values (which are ...


3

The conceptually easiest case is that of a positively charged active site with a negatively charged substrate. The substrate (i.e. reactant) enters the active site with kinetics that are faster than diffusion because there is a long-range electrostatic interaction. Catalase works near the diffusion limit. The document the OP cites is a hypothesis, without ...


3

According to https://www.uniprot.org/uniprot/P25325, 3-mercaptopyruvate sulfurtransferase is also involved in the catabolism of cysteine: Transfer of a sulfur ion to cyanide or to other thiol compounds. Also has weak rhodanese activity. Detoxifies cyanide and is required for thiosulfate biosynthesis. Acts as an antioxidant. In combination with cysteine ...


3

In the steady-state reaction, the intermediate concentration [ES] is assumed to remain at a small constant value. So in this case only if k2 >> k1 and similar for the second reaction. ES is now a reactive intermediate and there is no stable equilibrium between S, E and P. \begin{align} \frac{d[\ce{S1E}]}{dt} &= \ce{k1}[\ce{S1}][\ce{E}] - k_{-1} [\ce{S1E}...


3

Based on your description, it sounds like you are measuring the production of hydrogen peroxide by an oxidase enzyme. In that case, it is not necessary for the Amplex Red to be at the same concentration as the substrate. You only need enough reagent to determine the initial rate of reaction at each substrate concentration, ie when only a small portion of the ...


3

Well mostly it does but there can be exceptions. Eg Consider the reaction $$\ce{2NO + O_2 -> 2NO_2}$$ It has a negative temperature dependence. The rate decrease in increasing the temperature. There is a simple reason to why this happens. We have a RDS with an intermediate species. The rate determining step must have a positive temperature dependence. ...


3

To see why the $K_m$ is equivalent to the substrate concentration at which the rate is equal to $V_{max}/2$, we need to look at the derivation of the Michaelis-Menten, which is usually included in detail in any introductory biochemistry book. Here is a brief version. We will consider an enzyme-catalyzed reaction described by the following scheme: $$\ce{E + S ...


3

As this seems like a homework question (and may be closed), as an example consider the reaction $\ce{H2 + I2 = 2HI}$. The rate, r is $\displaystyle r=\frac{1}{2}\frac{d\mathrm{[HI]}}{dt}=k_1\mathrm{[H_2][I_2]}-k_2\mathrm{[HI]}^2$ where the rate constants $k_1,\;k_2$ are for the forward and reverse reactions respectively. (This assumes that the reaction ...


3

In this paper, a similar reduction in rate is observed at concentrations of H2O2 above 0.03 M final H2O2 concentration (which is about 0.1 wt %). They attribute it to $\ce{HO.}$ reacting with H2O2 (to yield $\ce{HOO.}$) more quickly than it reacts with MB. The result is that $\ce{HOO.}$ is much more abundant than $\ce{HO.}$ and its lower reactivity means a ...


3

Some enzyme-catalyzed reactions always go in the same direction in a live cell (e.g. reactions of citric acid cycle, RNA polymerase) and others go in either direction depending on circumstances (for example, some enzymes of glycolysis are also involved in the reverse pathway that makes glucose). As for reasons why a reaction might go in one direction only, ...


3

The legal limit in some states of 0.08% BAC corresponds to about 17 mM ethanol in the blood. The $K_M$ value for human liver alcohol dehydrogenase is 0.5 mM according to this paper. So at this particular legal limit, the enzyme works at 97% of its maximal capacity. ($v = v_{max} \frac{[S]}{[S] + K_M}$). Above or very much above the legal limit, the rate ...


2

There is no obvious direct connection between catalytic properties and stability of an enzyme. You are trying to make a connection, but let's consider your statement: Then the Arrhenius equation can be used to describe how the rate constant of an enzymatic reaction changes with temperature. According to the Arrhenius equation, all reactions (catalyzed ...


2

Mechanism of cooperative enzyme activity The easiest way to explain cooperativity is the example of hemoglobin, where four nearly identical subunits switch cooperatively between two states T and R of different binding affinity. When no ligand (L) is bound, hemoglobin is in the T state; once at least one ligand is bound, it is predominately in the R state. $...


2

You can also solve for $\mathrm{[S]} $ using Lineweaver-Burk plot (or double reciprocal plot). One way of writing the Michaelis Menten kinetic equation given in Wikipedia is: $$v_{\circ} = V_\mathrm{Max} \frac {[S]}{K_\mathrm{M} + [S] } = = k_\text{cat} \mathrm{[E]_{\circ}} \frac {[S]}{K_\mathrm{M} + [S]} $$ If you take the reciprocal of both sides of the ...


2

You can solve for [S]. There are two common ways to write the equation, like you have it: $$v_0 = k_\text{cat} [E_0] \frac{[S]}{K_\text{m} + [S]}$$ or after dividing both numerator and denominator by [S]: $$v_0 = k_\text{cat} [E_0] \frac{1}{\frac{K_\text{m}}{[S]} + 1}$$ In this form, it is easier to see that [S] occurs once and you can solve for it. Get ...


2

A nonfluorescent derivative of dihydroresorufin, 10-acetyl-3,7-dihydroxyphenoxazine has been marketed under the name Amplex® Red (AR). AR is regarded as the best fluorogenic substrate for peroxidase because it is highly specific and stable. It has been used to detect Reactive Oxygen Species (ROS) such as $\ce{H2O2}$ in various experimental enzymatic systems (...


2

[OP] The catalytic efficiency of an enzyme is given by $k_\mathrm{cat}/K_\mathrm{M}$ where $k_\mathrm{cat}$ is the turnover number, or the number of molecules that can be produced per second per active site of an enzyme. The last part is not quite accurate. $k_\mathrm{cat}$ is the rate of the reaction under saturating conditions divided by the enzyme ...


2

If the substrate still binds to the protein, other parts of the mechanism are still in place. Specifically, the main chain amide groups that help to stabilize the tetrahedral intermediate are still present. In the absence of serine, water will attack the carbonyl. In the presence of serine and the absence of the histidine, serine will still be the ...


1

Some proteins can be denatured mechanically, e.g. when whisking egg whites. This denaturing may or may not be reversible. I don't know how to estimate whether these exact enzymes under your conditions will denature or not and if so whether it will be reversible. You can add them after the blending step to be on the safe side.


1

It is correct to say: "If nothing else changes, an increase in temperature will increase the rate of an elementary step of a chemical reaction". All kinds of things could go wrong. If the temperature is so high that your reaction vessel melts, spilling a liquid reactant and removing it from a solid reactant, the rate will not increase. If the temperature is ...


1

Without attempting to elaborate (this is not my area of expertise) I rely on the abstract to an article [1] reviewing the subject for an answer: The Arrhenius equation has been widely used as a model of the temperature effect on the rate of chemical reactions and biological processes in foods. Since the model requires that the rate increase ...


1

OP's concern: Monooxygenases always require the presence of a regenerating system ($\mathrm{NADPH}$). It is true that most Monooxygenases are nicotinamide adenine dinucleotide ($\mathrm{NADP^+}$ and its reduced form, $\mathrm{NADPH}$) dependent (Wikipedia). Monooxygenases catalyzes the incorporation of a hydroxyl group into a variety of substrates by ...


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