Irreversible: Competitive or Non-Competitive?
This question had sparked off a furore of discussion so let me summarize and make some adjustment to the statement made in class.
Under the concept of enzymes, we recognise that competitive inhibition refers to competition at the active site of an enzyme betw an inhibitor and the substrate.
Non-competitive inhibitor refers to an inhibitor that binds to a site other than the active site (allosteric site) to inhibit the catalysis process.
I think that much we got it. Weak bonds (H- and ionic bonds) are involved above and thus all interactions are transient and thus reversible.
What is the issue with Irreversible Binding?
Irreversible binding by an inhibitor, because of the formation of strong covalent bonds, regardless of whether it is at the active site or allosteric site, will effectively render the enzyme non-functional (a decrease in enzyme no.) And in this situation, we may also refer the binding as non-competitive.
I suppose this kind of binding usually do occur at the active site because most enzymes do not have an allosteric site anyway . But if we extrapolate, a irreversible binding at the allosteric site will have the same effect.
PS: This whole idea is not uncommon in proteins really (not only enzymes). In most other proteins, there is usually just a BINDING site (NOT 'active site' which is specifically used for enzymes)
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Allosteric Site & Allosteric Enzyme - Are they the same?
Nope. Shocked?
An allosteric site is just a site other than the active site and which an inhibitor can bind to. And we call the inhibitor a non-competitive one.
Such an inhibitor can bind to both the enzyme and enzyme-substrate complex. The main idea here is that the inhibitor does not prevent the binding of substrate but prevent catalysis from taking place (diagram in notes is very accurate/clear) and thus can also bind to the ES complex to prevent catalysis. This will explain why you can never reach Vmax.
An allosteric enzyme is simply an enzyme whose activity is affected by an activator or inhibitor binding to its allosteric site. When I wrote "affected", i refer to whether the reaction is enhanced or impeded. Do you think the Vmax will be reached then?
(do not have to worry about this. we will not be touching on this)
Thus an enzyme with an allosteric site may not be an allosteric enzyme.
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Acid-Base Catalysis
As I mentioned to some of you and I am convinced myself on:
"...facilitate the transfer of reactants to and from the reactants"
I think it is best rephrased to
"....facilitate the transfer of charges to stabilize the formation of a transition state (nucleophile or electrophile) in reactants"
Such a transition state is necessary so that catalysis in the presence of other molecule e.g. water can take place (you will learn it all in chemistry later :) so just take note first!!)
See Diagram (Zn ion is involved as the charged particle but the idea is the same):
Hydrophobicity
According to thermodynamics, matter seeks to be in a low-energy state, and bonding reduces chemical energy. Water is electrically polarized, and is able to form hydrogen bonds internally, which gives it many of its unique physical properties. But, since hydrophobes are not electrically polarized, and because they are unable to form hydrogen bonds, water repels hydrophobes, in favour of bonding with itself. It is this effect that causes the hydrophobic interaction — which in itself is incorrectly named as the energetic force comes from the hydrophilic molecules.[2] Thus the two immiscible phases (hydrophilic vs. hydrophobic) will change so that their corresponding interfacial area will be minimal. This effect can be visualized in the phenomenon called phase separation.
http://en.wikipedia.org/wiki/Hydrophobic
Equation for Fehling’s Test
The chemical test is best looked upon at the half-eqn level for simplicity.
Aliphatic aldehyde reduces Cu(II) to Cu(I). Ketones and aromatic aldehyde give no reaction.
Half Eqn: 2Cu2+ + OH- + 2e → Cu2O + H+
RCHO + H2O → RCOOH + 2H+ + 2e
Overall : RCHO + 2Cu2+ + 5OH- + H2O → RCOOH + 2H+ + 2e [+4OH-]
Fructose can undergo keto-enol tautomeric shift to exist as an aldehyde under basic conditions.
Isn't milk alkaline? Aren't we recommend to drink milk when experiencing gastric?
Gastric juice, which consists of hydrochloric acid and an enzyme, pepsin, which breaks down protein, can digest any living tissue, including your stomach and duodenum. Normally, both your stomach and duodenum are bathed constantly in gastric acid. But protective mechanisms, including the work of prostaglandins, which govern secretion of mucus from your stomach lining, and your food and saliva's ability to dilute acid, prevent your stomach from digesting itself. The pH of the human stomach is about 1.8. The pH of fresh milk is between 6.5 and 6.75. After drinking a glass of milk, the stomach's acidity changes. A powerful acid environment is buffered up to a 6.0, so that everything ferments and putrefies for the next four hours. Initially, milk does dilute stomach acid -- but then, acting on the rebound, it prompts the production of even more.
For years, ulcer patients had to survive on a bland diet of boiled fish, rice, milk, and cream. Now we know that while milk coats your stomach and may relieve the ulcer pain temporarily, it may retard the ulcer's healing. The calcium in milk can make you feel worse in the long run by stimulating the production of gastric acid. So can fried foods, citrus fruits, alcohol, caffeine in beverages or in chocolate, decaffeinated coffee, and smoking. Tea seems to particularly stimulate production of gastric juice.
http://www.otherhealth.com/homeopathy-discussion/2364-x-c-2.html
Fat and Carbohydrate as Energy Source
If we take a look at the diagram, we will realise that carbohydrates and triglycerides undergoes different pathways to yield ATP although both will converge at Krebs Cycle. Thus in comparison, carbo has to undergo more reactions (in glycolysis) where energy can be lost before getting to Krebs Cycle. Having C-H rather than C=O and C-OH bonds avoid that pathway.
Mistakes? Pls check
General formula of polysaccharides: Cn(H2O)n-1
Why glutamine is polar and not basic?
Despite the presence of NH2, we have to look at the neighboring molecules as well. If you notice, there is a C=O beside the NH2. Thus we are actually looking at an amide group and not an amine side group and former is not basic because C=O draws electrons away from NH2, thus the lone pair on N is less likely to be donated to a H+.
That's all folks!
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