Question
The formula that is circle as seen above, why does it exists?
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: 12,1x106 bp. f= 5kbp/12100 kbp = 4.13x10-4 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
Explanation / Answer
The number of clones depends upon the size of the genome f and average size of the cloned DNA.
If f is the fraction of the genome size compared to the average individual cloned fragment size and thus would represent the lowest possible no. of clones
f=genome size/fragment size
The no. of independent recombinants must be greater than f.In 1976 Clarke and Carbon derived a formula to calculate probablity(P) of including any DNA sequence in a random library of N independent recombinants.
The actual no. of clones to be calculated
N=In (1-P) / In(1-1/f)
N=no.of clones
P=probablity
Bigger the library better will be the chance of finding the gene of interest.