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Some substances do exist that slow down reactions and they're different from catalytic inhibitors/poisons. Such substances are called negative catalysts. Here is one example I can think of right now:
$$\ce{2H2O2 \xrightarrow{\large\mathrm{glycerol}} 2H2O + O2}$$
Glycerol, in this case behaves as a negative catalyst and slows down the reaction.
Another ...
9
The turnover frequency is a rate. (It's actually the kinetic rate of the reaction in saturating substrate concentration, normalized to the amount of enzyme.)
It might be helpful to think about something like a factory that makes a certain number of widgets in a certain number of time. How would you naturally describe how fast the factory could make widgets? ...
8
The inhibition comes about because:
Uracil and $5$-fluorouracil are sufficiently similar that both moieties have a strong binding affinity for the active site of thymidylate synthase.
The presence of the fluorine prevents $5$-methylation of the uracil core by a folate cofactor, which otherwise would turn it into a thymine moiety.
This inability of the ...
8
all 10 steps in glycolysis require an enzyme
We count it as 10 steps because there are 10 enzymes. There might be other steps that are fast (like a product acting as an acid or base) that we don't count because they don't need an enzyme.
I’d have thought that evolution would have selected for “simpler” pathways.
What is simple is that everything is ...
7
[…] there's the same number of catalase molecule within an undiluted and a diluted solution. Also, seeing as water is not an enzyme inhibitor, […]
Q: Why is dating between a fixed number of guys and girls more difficult on a large square packed with nuns than in a small club (without the nuns)? ;)
A: It's more difficult to bump into each ...
answered Apr 6 '15 at 6:02
Klaus-Dieter Warzecha
41.3k7 gold badges85 silver badges148 bronze badges
7
The major reason of the ‘hydrophobic environment’ is to block the enzyme-complex from accessing the solvent (water), which might otherwise enter the active site and hydrolyse ATP.
The reduction of polarity also helps to speed up the subsequent nucleophilic substitution.
Note: Hexokinase is a soluble, cytosolic protein
The reasoning behind this statement ...
6
This is a question that is relatively easy to answer mathematically.
To keep the algebra a little simpler, let me use a simplified Michaelis-Menten mechanism, where we assume that the enzyme-substrate complex (ES) and the free enzyme are in equilibrium. (We could use the less restrictive pseudo-steady state hypothesis on (ES) and obtain the same result ...
6
A few things first:
Enzymes are proteins (usually) that promote or accelerate a biochemical reaction. The enzyme lipoxygenase in particular, promotes the dioxygenation of some lipids called Polyunsaturated Fatty Acids (PUFAs); for this, the enzyme requires molecular oxygen (which is present in the air, and gets dissolved in the soy milk) . You can think of ...
6
Your question is a little bit all over the place, but I believe I can answer it anyway. First, though, allow me to point out that your sum formula for cellulose is wrong. While glucose is indeed a single $\ce{C6}$ compound with the exhaustive sum formula of $\ce{C6H12O6}$, cellulose is, in fact, a poorly defined polymer consisting of multiple $\ce{C6H10O5}$ ...
6
Enzymes act as catalysts to increase the rate of reactions. In the absence of enzymes, most biological reactions would be incredibly slow, in some cases many many orders of magnitude slower than when catalyzed.
Given the large number of biological reactions that require enzymes, it seems that there might be an advantage to reactions that are slow in the ...
5
20% hydrogen peroxide sounds HUGE. But something can happen that is called substrate inhibition. It can be detected if the substrate acts as a non-competitive reversible inhibitor (i.e. binds only to the active ES Michaelis complex). In the case I studied during my PhD, the inhibition constant ($K_\mathrm{i}$) was roughly 1000 times higher that the Michaelis ...
5
There are two enantiomeric forms of glucose, D- and L, with D-glucose being the naturally occurring form. Dextrose is just another common name for D-glucose, so if a microorganism can metabolise D-glucose then it can by definition metabolise dextrose as they are just two words for the same thing.
As for the question of whether the enzyme would recognise ...
5
tl;dr: The active site catalytic residues (BH+ and B’) are thought to be Lys and a His–Glu dyad, respectively. These facilitate the abstraction of proton from C2 and subsequent protonation on C-1.
Some important points to note:
"shows histidine protonating c5 oxygen and lysine deprotonating c1
oxygen to form an open chain aldolase
”.
The opening of ...
5
Your equation
$$\ce{Enzyme + Substrate <=>[k_1] ES complex ->[k_2] Enzyme + Product}$$
contains only two rate constants, $k_1$ and $k_2$, but not $k_{cat}$ you refer to.
The correct schema for the Michaelis-Menten kinetics would be
$$\ce{Enzyme + Substrate <=>[k_1][k_{-1}] ES complex ->[k_2 = k_{cat}] Enzyme + Product}$$
Note that $...
5
The ultimate reason is that cellulose is designed to be hard to digest
The specific chemical reason is well covered in Jan's answer, but there is an explanation that is simpler and more fundamental: cellulose is designed that way. Nature has created some organisms that need a structural component that isn't easy for other organisms to break down (for the ...
5
Chitin and protein are completely unrelated.
The only common thing is that they are polymers.
Chitin is a polymer of amino sugars while protein is a polymer of amino acids.
Both monomers are very different and are not converted one to the other.
In fact, chitin, like cellulose, is not fragmented by animals, so it is not absorbed.
Moreover, chitin is a ...
5
I think I got it. The diagram in PubChem is not bromelain molecule, but a tiny fragment of it. This thing is indeed too small to be an enzyme. Also, it has nothing to do with proteins. Look at the formula, there is not a single amino acid residue here. Indeed, scroll down to 3.4.2 and read: "carbohydrate moiety of bromelain". That's what it is, and not the ...
5
The authors mention three residues that they consider "second shell", 12, 77, and 102. They are shown in the figure below in green:
Residues Trp50 and Glu101 are directly involved in ligand binding, shown as space-filling in yellow. As you can see, they are part of the central beta barrel. The residues in green are also part of the beta barrel, but the side ...
5
We speak of reactions that
build complex molecules
or break complex molecules at a specific position
they must happen at constant 37°C
all of which must be rather low energy
or be very well controlled, like photosynthesis
no fancy other reagents are available
many would never happen at all without stringent control
and/or the result could only be achieved ...
4
Given $\ce{E + S <=>ES -> E + P}$
$\frac{d[\ce{ES}]}{dt}=-\frac{d[\ce{E}]}{dt}= k_1[\ce{E}][\ce{S}]-k_2[\ce{ES}] -k_3[\ce{ES}] $
is the slope of $[ES]$ equal to $k_2$ ?
no
is the slope of $[E]$ equal to $k_1 + k_3$
no
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This is a somewhat complex question to answer because even your ‘simple’ enzyme example isn’t as simple as it looks. Let me discuss enzyme optimum temperatures first.
For any reaction that is exergonic — and that explicitly includes enzyme catalysis — an increase in temperature raises the reaction rate. This is due to at least two microscopic features:
...
4
The main challenge is the macroscopic structure of cellulose. It's a solid, enzymes obviously can only start working on the surface, and hydrolases will work only on a few random loose chains.
This is too slow, even if you have a lot of time and are not trying to make any money. So you need to break apart your cellulose first using a process that's cheap ...
3
The enzyme catalysis reaction following MM kinetics can be depicted as:
$$\Large\ce{E + S<=>[k_f][k_r]ES->[k_{cat}]E + P }$$
Note that $V_{max} = k_{cat}.[E_0]$ i.e. conversion rate constant (turnover number) times the total amount of enzyme.
You should note that the enzyme here is p38 and substrate is MSK1 (which in turn is an enzyme). When MSK1 ...
3
You need to find a linear combination of the time derivatives such that they add up to zero on the right hand side. This will tell you the corresponding linear combination of concentrations that is conserved. For this problem $s+e+2c_1+3c_2+p$ is conserved.
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The reaction scheme is:
$$\ce{S + E ->[k_1] C1} \tag{1} $$ $$\ce{C1 ->[k_2] E + P} \tag{2}$$ $$\ce{S + C1 <=>[k_3][k_4]C2} \tag{3} $$
Let's define a bit of nomenclature.
S = compound S in chemical reaction equation
s = $[S]_x$, the concentration of S at t=x.
$S^*$ = ds/dt
The whole set of differential equations has 5 equations with 5 unknowns....
3
If you diluted the catalase, while keeping the peroxide concentration fixed (by adding both water and peroxide to the solution), then you're right, the total reaction rate (measured in reactions catalyzed per second, within the whole solution) would not change.
However, by adding just water to the solution, you're diluting both the enzyme and the peroxide. ...
3
For the first subquestion:
Using unit analysis, the answer is very quickly found: The unit of the slope $m$, assuming dimensionless values for the absorbance, is
$$[m] = \frac{[\Delta y]}{[\Delta x]} = \frac{[\Delta A]}{[\Delta t]} = \frac{1}{\text{min}}$$
Now you need a volume in the denominator as well, so dividing by a volume seems like a good idea. ...
3
The difference between biodegradable polymers and 'regular' polymers is the bonds that make up the backbone. The backbone is the part of the polymer that extents the brackets. Anything attached to that chain is a side group and is less important.
The two examples you show here are very dissimilar. The polypropylene is only made up of Carbon-Carbon bonds. ...
3
You were actually describing the turnover frequency of a catalyst, and frequency has the unit hertz which is 1/s. Turnover number is a dimensionless value: number of molecules of reagent that a catalytic site can convert before becoming inactivated.
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