dttom

Majority of mature T cells have either CD4 or CD8. An antigen could be bound by more than one antibody as it contains more than one epitope. And I'm sorry I don't quite understand your last two statements, could you explain more?

Na2O becomes hydroxide upon contact with water.

Your reaction is actually an equilibrium shift, so you have concentrated sulfuric acid, chloride binds with proton and form HCl, as the H3O+ concentration is high equilirium position favors HCl side so HCl gas comes out. If you pump the gas to a dilute solution, position would be less intended to the HCl side so more will be dissolved.

When talking about cancer cell, it is presumably that they are different from other normally functioning equivalent, so a cancer cell derived from stomach does not perform stomach functions, so it becomes less important to argue if it is a stomach cell or pulmonary cell if it spreads to the lungs. Cancer cells could spread by individual cell shift, if there is only one cell shifting, lymph vessels are just too wide for it.

That's very clear, thanks.

The geometric sum to infinity of a series with a ratio of r, and started with a, is calculated to be: G(infinity) = a/(1-r) But I am thinking of: Each term in a series could be represented by: a(0) = a; a(n) = ar^n, so why can't we integrate a(n) with respective to dn, giving us (ar^n)/(ln®), but obviously this is the the expected answer. I'm interested in what makes such a difference. Thanks.

There usually are coordinated response from B- and T- populations, for the sake of maximum efficiency evolved. In a rough sense, yes the receptors bind to the same anitgen, but as CharonY has pointed out usually they target for different epitopes. But actually B-cells use mAb to bind antigen while TCR usually recognise antigen presented in peptide-MHC complex, so whether TCR binds antigen depends on your understanding of 'antigen' (whether a digest be still called antigen).

a part of holoenzyme

Enzymes are proteins, some require cofactor while some other not. For those which require one, usually the attacking group originates from the cofactor. To be easy, you may find an example of enzyme itself as a holoenzyme, and as it's a protein, functional group responsible for the attack should be due to functional group found in amino acids. So you may think in the direction which group found in amino acid would grant you basic properties.

Thanks a lot, I got the idea now.

The question is to find to integral of 1/(x(sqrt(x+4))) I tried substitution with x equals 4(tan(x))^2 but not succeed, could anyone help?

I think species containing hydrogenosome is small enough that they don't have breathing. The organelle is found in small species as some fungi and protists, though the former could be potentially very large concerning with its mycelium. Also found in multicellular organism which is also microscopic. So the issue of breathing or not just doesn't make sense as they don't even have a breathing organ (not respiration but breathing).

acid group is much more polar than the anhydride group, possibly by partition?

You can start with 4-chlorobutanal, treated with activated Mg, add 3-oxybutanal, then use a mild oxidising agent like cupric oxide to oxidise hydroxyl group at position 5 to a carbonyl group.

Oh I see, I miss that, thanks. Merged post follows: Consecutive posts mergedBy the way, I got a reddish brown complex when proline is treated with ninhydrin with heating, this indicates an iminium salt formation (usually brown in color). I also got such a brown appearance when I treated histidine. His got 2 nitrogen atom in the side group, one N doubly bonded to a carbon on one hand and singly bonded to another carbon on the other hand, this N atom is basic while another one is not by resonance stabilisation. I want to ask if it is because this basic N atom which behaves as a secondary amine that give rise to the brown complex (probably iminium salt)?

I was told that ninhydrin could react with secondary amine to form a colored complex of iminium salt. However, in an experiment I treat N-acetylglucosamine with ninhydrin, it gave negative result. N-acetylglucosamine contains a secondary amino group, but why it gave negative result? But ninhydrin gives colored complex with proline, which is also a secondary amine... Is it just different iminium salts have different absorption spectrum?

plus there are recovery mecahnsim for DNA repairment. But I do think things get worse with aging, a mutation is not necessarily bad, but it can't be a good stuff for reasons like incompatibility of new change to existing body or the change occurs in tiny portion in a body, bad things got a chance with it like uncontrolled growth caused by inability to express apoptotic genes, hence the probability of getting bad thing with you should be longer if one lives longer.

so you know the pairing rule, there's only two combinations, knowing one of them gives you another.

E=qV if assume constant voltage 1.4V, Cd + 2OH- --> Cd(OH)2 + 2e-; 1 mol Cd consumed, E = (2)(96500)(1.4) = 270kJ

Al ions exist in hexaqua-form, that is, [Al(H2O)6]3+, so you should start with this. Your direction has been correct.

You're correct in recognising monosodium phosphate is an acid salt. There are two ionisable hydrogen atom in H2PO4-, so there should be stepwise hydrogen dissociation.

If, procedures concerned just were just conducted at 4deg, should inhibitors be involved? I personally think they should be.

So could you give an example of when inhibitor is not required?

My question is just stated on title, in the course of hemoglobin isolation from blood sample, why protease inhibitors don't need to be added? Wouldn't proteases degrade the Hb after cell burst? (I was told that proteases are activated after cell burst, though I don't know the actual mechansim)

the clot is the fibrin network that initiated by tissue factors, calcium ion, platelet granule release .etc.