Cloning-Problem

C Jamison jamisoncsnospam at atallaccessam.com
Fri Mar 12 22:03:51 EST 1999


Arnd,

A little more information would help me to solve your problem.  I'll offer
the basics.  The idea in cloning is to achieve a high insert to vector ratio
(in terms of number of molecules).  A 6 kb fragment into a 12kb vector can
be more difficult than a smaller fragment cloned into the same size vector.
Also, your vector is a bit on the large size, so you'll probably have to do
a number of attempts to clone the fragment.  One solution to getting a
higher number of molecules for the ligation step is to cut your insert into
2-4 pieces and clone them into the vector sequentially.

The other suggestions are obvious, but always bear repeating:  1)  you need
to make sure that your ends are complementary, 2) try to do a directed
ligation (different, non-complementary ends on the vector so that the vector
cannot reanneal and the tranformation will only form colonies if the insert
ligates), 3) try not to do blunt end cloning unless you have no other
choice, 4) you need to make sure that the ends really are cut (run controls
for the restriction enzyme reactions and make sure you are using a newly
purchased enzyme), 5) you need to make sure that your selection system in
transformation will allow you to find positive colonies, 6) you need to make
sure that your colony identification system will find the positives (a quick
Southern blot on a colony lift works nicely in order to find colonies with
inserts).

Good luck,

Scott Jamison
jamisoncsnospam at atallaccessam.com

Arnd Richardt wrote in message <36E507FB.10EB4F26 at ruhr-uni-bochum.de>...
>I´ve tried to ligate a 6kb-fragment into a Gal4-Transformation-Vektor
>(p221-4; 12kb long), but failed several times. Does anybody have a
>protocol or reference for such a cloning problem ?? I hope, it´s not
>impossible
>
>Thanks
>
>--
>Arnd Richardt
>
>Arndt.Richardt at ruhr-uni-bochum.de
>Ruhr-Universität Bochum
>Fakultät Chemie
>Molekulare Zellbiochemie
>Gebäude NC 5 / 171
>44780 Bochum
>
>





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