I am doing a yeast two hybrid library screen using a protein of no
homology to known proteins as by bait. Screening about 500,000
cotransformants gave about 300 clones that remain His+ LacZ+ after
restreaking. Once I have grouped the clones through restriction digest,
does anyone have any idea of what proportion of these clones (using one
repesentative clone per group of clones with same restriction pattern) I
can expect to eliminate in the first test for false positives
(auto-activation). I am using a bone marrow library.
Because initial digests are revealing very few clones with similar
pattern, I was wondering if it would be more energy efficient to
eliminate self activators first (through cycloheximide selection against
my bait). I suspect this would only be the case if large numbers of
clones are always self activating, which is the main reason for my
Thanks for your help.