What is the number of fragments that must be cloned to ensure a high probability that desired sequence is present at least once in genomic library? -Allows to pr characterize single amino the protein ar roteins de no P = 1-(1-f)” P probability f- fraction of desired sequence in genome N- number of fragments Green-gene to Mismatched pri mutation) is ext N- Log(1-P)/Log(1-f) polymerase How many yeast DNA fragments of average length 5 kbp has to be clone to in order to have 99% probability that a genomic library contains certain sequence. Yeast genome: 12,1×106 bp. f= 5kbp/12100 kbp = 4.13×10-4 of yeast genome Altered gene is vector and expr bacteria, where serves as a temp new strands that mutated nucleoti N = log(1.99)/log(1-4.13104)-11,148 The Example of chi GST-fusion proteir protein of intere GST sequence is in alongside the ge Cassette mutation – Plasmid DNA is cut by restriction enzymes to remove a short sequence of bases and the new is added and ligated