10

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 ...


6

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]_{...


6

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 ...


5

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

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

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 ...


4

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 ...


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 ...


4

You would not expect the same enzyme from different tissues to have different Km values if the enzymes are truly identical. In addition, you would not expect the Km value to change during purification: the Michaelis constant for the purified enzyme should be the same as that determined with a crude sample. If this is not true, it might mean that the ...


4

Temporal kinetics does not refer to the kinetics of kinase phosphorylation reactions and, since it is not standard terminology, you likely wouldn’t find its meaning in any textbook. The authors of the paper quoted in the question use the phrase temporal kinetics to describe the phosphorylation state of a proteome-wide selection of kinase substrates (ie ...


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

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/...


3

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

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: $\mathrm{MW_{protein} = g_{protein} / mol_{protein}}$ Substituting ...


3

It sounds like you are asking about what are commonly referred to as "multifunctional enzymes". For a reasonably recent article covering this subject — see: Cheng, X. Y., Huang, W. J., Hu, S. C., Zhang, H. L., Wang, H., Zhang, J. X., ... & Ji, Z. L. (2012). A global characterization and identification of multifunctional enzymes. PloS one, 7(6), ...


3

Acetylcholinesterase. Chosen because the esterase seems like a low energy reaction so it wouldn't need energetic co-factors. https://en.wikipedia.org/wiki/Acetylcholinesterase Not sure if I should do a new answer or not.


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 ...


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