| Nick Bennett 2004-08-25, 11:10 am |
| j.umber@ac-nancy-metz.fr (Jean) wrote in message news:<5eb4d562.0408082142.494b850@posting.google.com>...
> The mutation T215Y or T215F needs two changes of adjacent bases (in
> the ARNm):
>
> ACU -> UAU / ACC -> UAC for T215Y and ACU -> UUU / ACC -> UUC fr T215F
>
> Threfore we must found at the same place the mutations :
>
> T215S (f. i. ACU -> UCU) and T215N. Where are these mutations ?
It's true that a double-base substitution is required, but it's also
clear that any and all mutations will exist at all sites in a mixture
of HIV in vivo. ON average every genome will have at least mutation
in it. The fact of the matter is though, that not all mutations have
a selective advantage. Some will be disadvantageous, some will be
neutral. You cannot expect a necessarily orderly progression,
In the same way, while we descended from apes there are no
middle-steps between us and the bonobo chimp. That doesn't detract
that historically some intermediates are likely to have existed. The
issue is clouded somewhat by the fact that us and chimps and evolved
seperately from a common ancesestor.
>
> p 1673, I read :
> The pocket can probably accommodate two or three water molecules in
> addition to the 3'-OH, and it clearly has room for the azido group of
> AZTTP.
>
> and p 1674 :
> Second, the sites of mutations conferring resistance to the dideoxy
> class of inhibitors (including 3TC) nearly all impinge on the dNTP
> from the "front" (Fig. 6A), whereas sites for resistance to AZT (with
> a 3' substituent larger than a hydroxyl group) impinge on the 3'
> pocket from the "rear" (Fig. 6B).
>
> It is greatly contradictory, because in the first case, the authors
> talk us over the hydrogen bounded water, which surround the 3'-OH and
> confer to it a larger size as he 3'-azide group of AZT,
No, they're saying that in addition to the OH group, there is room for
3 H2O molecules. As they say, ample room for an azido, presumeably
through the loss of one or more water molecules.
and in the
> second case, water has disappeared, to have an 3'-azide group greater
> as the 3'-OH. Is it chemically not possible, because the hydrogen
> bonds between the alcohol OH and the water OH remain the same.
Why can't the azide group replace the water OH bonding? H-bonds are
stronger than some, but they are non-covelant. They are changing all
the time: if they weren't water wouldn't be a liquid at room
temperature! I see no problem with H-bonds being replaced, altered,
swapped out, unused....they're dynamic bonds by their very nature.
It makes sense that with AZT versus 3TC, the mutations are in an area
which interacts with the modified bit of the nucleotide found in AZT
but not 3TC. All this supports the proposed mechanism of viral
resistance.
>
> I think that this research is flawed here, an can not demonstrate that
> the T215Y (or T215F) mutation prevent the bond of AZT to RT
>
> And it is clear that the London positive interaction (VdW) are much
> better between an aromatic cycle and the azide group, and the hydrogen
> bond is better between two hydroxyl compounds.
Not quite so clearly - at least to me. I can see no difference
between a hydrogen bond formed between hydroxyl groups and azide
groups: it's all just a matter of dipoles after all and electron
clouds. In certain situations (taking into account all the other
bonds being formed and steric conformation alterations due to bonding)
I can see no reason why a classically "less favoured" reaction cannot
occur. After all, glucose can be converted to starch can it not,
under the right conditions, even though that is energticaly
unvavourable. That's part of the difference between biology and
chemistry, in my mind.
Cheers
Bennett
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