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A competitive inhibitor typically competes for the active site with the substrate. In this textbook case, binding of a competitive inhibitor is reversible, because it binds to the active site of the enzyme, but is also released, making way for the substrate to bind. The affinity of the substrate, as well as its concentration determine the amount of ...


7

The answer to the first part of your question is that we don't take the initial 10% of a progress curve (velocity vs time) as a measure of activity, but we measure the initial rate of the reaction. We do this by drawing a tangent at the origin. It is merely a 'rule of thumb' that progress curves are practically linear provided that not more than 10% of ...


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The answer is here, but depending on your level of comfort with the math I'm not sure how enlightening it will be. I think that the reason people tend to stick to one ligand/one receptor models is that they capture all the intuition without the tedious algebra. It's interesting that you are asking for the fraction of ligand bound to receptors ($[RL]/[L]_{...


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Competitive inhibitor competes for the active site. Therefore it will interfere with the binding of the substrate thereby increasing the apparent KM. A strictly non-competitive inhibitor does not compete for the active site. It however inhibits the catalysis by reducing the available molecules of active enzyme, E0 (if it is a perfect inhibitor), thereby ...


5

In biochemistry the Ki is the dissociation constant of a complex and molecules bound to it. It is a measure for the functional strength of the inhibitor. This can for example be an enzyme and its substrate, the Ki defines the stability of the complex. The IC50 on the other hand is the halfmaximal inhibitory concentration of a substance on a biochemical ...


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From the derivation of Michaelis-Menten kinetics you can see that: $$K_m=\frac{k_f + k_{cat}}{k_r}$$ Where $k_f$ and $k_r$ are binding and unbinding rate constants (for Enzyme-Substrate binding), respectively, and $k_{cat}$ is the turnover number. This is for the Quasi-Steady-State approximation (QSSA). For the equilibrium approximation: $$K_m=\frac{k_f}{...


5

I think it is possible to identify the type of inhibition from (initial) velocity vs substrate-concentration curves, but it is difficult. The usual way this is done is by using a linear transformation of the Michaelis-Menten equation, such as the Lineweaver-Burk plot. But you are right: for a reversible inhibitor, the way to identify the inhibition ...


5

In a reaction which follows a saturation kinetics, KM is basically the concentration of substrate/ligand at which the rate of the reaction is half of the maximum rate (or the binding sites are half saturated). The biophysical meaning of KM would depend on the underlying model. For example, in the equilibrium approximation of Michaelis-Menten model, KM is ...


5

Gram-negative bacteria also have peptidoglycan which can be degraded by lysozyme. However they have an outer membrane lying outside the peptidoglycan layer and this will block lysozyme action. Because of this if you intend to treat Gram-negative bacteria with lysozyme it is necessary to add something that will disrupt the outer membrane—the usual choice is ...


4

The reason is that the inhibition models are inherently wrong. Inhibitors bind to enzymes with the same mass action principles as substrates do so the term that should describe their binding should mirror the form of the MM equation i.e. ([S]/([S]+km)). However all the traditional equations use (1+([I]/Ki)) rather than ([I]/([I]+ki)). To compound this ...


4

Short anwer 'Non-competitive active site–binding inhibitors' are called mixed-type inhibitors. These inhibitors exhibit features of both competitive and non-competitive inhibitors, as they increase Km (like a competitive inhibitor) and decrease Vmax (like a non-competitive inhibitor). Background What an interesting question! In theory, a reversible ...


4

This is a tough question. I was reading this paper Patrono, C., et al. "Clinical pharmacology of platelet cyclooxygenase inhibition." Circulation 72.6 (1985): 1177-1184. and they seemed t mention this paragraph in the introduction. Platelet Cycloxygenase or prostaglandin (PG) H synthase (i.e., the enzyme that converts arachidonate released from membrane ...


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I don't think you understand MM kinetics. In your case the $V_{max}$ value depends on the amount of the protein. If we assume 1mg protein, then $V_{max} = 50\frac{\mu{M}}{min}$. According to MM kinetics: $V = V_{max}.\frac{S}{K_M+S}$ so $V_0 = V_{max}.\frac{S_0}{K_M+S_0} = V_{max}.\frac{5\mu{M}}{10\mu{M}+5\mu{M}} = V_{max}.\frac{1}{3}$ or in other ...


4

Since the Michaelis-Menton constant Km is the concentration of substrate at 0.5Vmax, it is an inverse measure of its substrate affinity, because a lower Km indicates that less substrate is needed to reach a certain reaction speed. Hence, a low Km means a high substrate affinity. Your statement "Maybe the maximum velocity (Vmax) of higher-Km enzymes is ...


4

First off, the difference between the types of inhibition: competitive inhibition: The inhibitor only binds to the substrate-free form of the enzyme. (Not necessarily at the active site!) uncompetitive inhibition: The inhibitor only binds to the substrate-bound form of the enzyme. noncompetitive inhibition: The inhibitor binds equally well to both the ...


3

According to this instruction it can be done for 1 hour but this is for large plasmids. I agree with Chris that it should not be of a problem. Alternatively you can setup your 37°C DpnI incubation step in a PCR machine and after however long that is recommended, you can set the PCR machine to cool to 4°C automatically, that way you can leave it there for as ...


3

Alan Boyd's answer covers the mechanistic aspects quite well, but there is another aspect he didn't quite touch on - what's going on at the molecular level. To understand this, you need to think about the actual conditions of the reaction: unless it's taking place inside a cell, which straight biochemical reactions of the kind you're talking about almost ...


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In terms of Michaelis-Menten kinetics, the rate never reaches the maximum rate: $v = V_{max} \times \frac{S}{K_m + S}$ where $S$ is substrate concentration. Notice that however large $S$ is, the term on the bottom line ($K_m + S$) will be larger than $S$, so $v$ will be less than $V_{max}$. You can think about this in terms of the binding equilibrium ...


3

I always used the book "Enzyme Structure and Mechanism" by Alan Fersht. (Structure and Mechanism in Protein Science: A Guide to Enzyme Catalysis and Protein Folding is the latest version, published in 1998.) Both in some courses I did and when working on (complex) enzymatic mechanisms. It's a classic (so a bit old maybe), but it covers all the basics and ...


3

First of all, as you mentioned, there are a lot of membrane / transport proteins with very nice, well defined tunnels. But you're looking for a soluble / cytosolic enzyme. Some oxygen-dependent enzymes have quite long tunnels that oxygen is using. For example this one: http://www.jbc.org/content/283/36/24738.short or this one: http://europepmc.org/articles/...


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Set $ \dfrac{1}{V} = 0$ and solve for $\dfrac{1}{[S]}$: $ 0 = \dfrac{K_m}{V_{max}}\dfrac{1}{[S]}+ \dfrac{1}{V_{max}} $ $ -\dfrac{1}{V_{max}} = \dfrac{K_m}{V_{max}}\dfrac{1}{[S]}$ $ -1 = {K_m}\dfrac{1}{[S]}$ $ -\dfrac{1}{K_m} = \dfrac{1}{[S]} = $ x-intercept


3

Just as $k_{cat}$ represents the rate of reaction at saturating substrate concentration, $k_{cat} / K_m$ represents the rate of the reaction at negligible substrate concentration. If we take a look at the standard one substrate/one product Michaelis–Menten kinetics rate equation: $$v = \frac{k_{cat}[E][S]}{K_m + [S]}$$ We can imagine what happens when $[S]...


3

Probably the big reason you were unsuccessful in your previous searches is that nobody quite knows exactly how enzymes' 3D structures lead to particular kinetic parameters. It's wrapped up in the concept of "how the heck do enzymes work, anyway?" -- surprisingly, it's a question we don't really have a good answer to. The leading concept currently - ...


3

You are correct in stating the relative rates of the two reactions. What may have you confused is the percent sign in the statement. The percent sign is not a unit in the sense that you are taking it. What you have in a simple reaction like you are examining are two opposing rates of reaction. The rate of a catalytic reaction, like others, usually ...


3

Let's do some unit analysis: $\pu t$ = time $\pu U$ = units = $\mathrm{mol_{product}/t}$ $\pu s=$ specific activity $\mathrm{= U/g_{protein}}$ Therefore: $\mathrm{s = mol_{product} / (g_{protien} \times{t})}$ Your sheet says: $\mathrm{k_{cat}= s \times MW_{protien}}$ And we know that: $\mathrm{MW_{protein} = g_{protein} / mol_{protein}}$ ...


2

Phosphofructokinase 1 (PFK 1) is not rate-limiting, its flux control coefficient when measured is invariably small. The fact that textbooks keep suggesting the PFK is rate-limting is just a textbook myth. Here is a selection of papers that support the observation that the flux control coefficient for PFK is invariably small. There are also very strong ...


2

In practice, like Energy calculations, Entropy is a relative numbers and difficult to get. Computational chem and bio have been working on this problem with mixed success. The summary of what I'm going to say here is that when you try to calculate the differences in entropy (or energy for that matter), the Gibbs Free Energy differences in most biological ...


2

Amperometric detection of H2O2 can be performed without HRP, and gives high-resolution real-time readings. You will find many different type of electrodes that are used in the literature to detect H2O2, including for instance MnO2-coated carbon paste microelectrodes or gold electrodes derivatized with cytochrome c. Commercial electrodes also exists. For ...


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