collet@SCRIPPS.EDU (Thomas Collet) (02/23/91)
I am trying to clone immunoglobulins from hybridoma cells and am wondering whether I am using the best possible restriction enzymes. Does anyone have a compilation of the cutting frequencies of restriction enzymes in immunoglobulin genes (or maybe just mammalian genes in general ?) Or maybe a literature reference ? Thanks Thomas
BIOCOL@UNIPAD.UNIPD.IT (02/23/91)
In BioNews Thomas Collet says: >I am trying to clone immunoglobulins from hybridoma cells and >am wondering whether I am using the best possible restriction >enzymes. Does anyone have a compilation of the cutting >frequencies of restriction enzymes in immunoglobulin genes (or >maybe just mammalian genes in general ?) Or maybe a literature >reference ? >Thanks I think you can try by yourself using stochastic computer methods (as I did), for example RND in BASIC. The frequence of cutting for each enzyme must be 1/4 exp n for each class with n bases of recogniction. I have no litterature but if you want I can send you, with hard mail, some table for 6 and 4 base-cut ters. Best Re: and Re:gards. Argenton Francesco Dip. di Biologia Universita' di Padova Via Trieste 75, 35100 Padova, ITALIA
eesnyder@boulder.Colorado.EDU (Eric E. Snyder) (02/24/91)
BIOCOL@UNIPAD.UNIPD.IT writes: >In BioNews Thomas Collet says: >>Does anyone have a compilation of the cutting >>frequencies of restriction enzymes in immunoglobulin genes (or >>maybe just mammalian genes in general ?) Or maybe a literature >>reference ? >>Thanks >I think you can try by yourself using stochastic computer methods (as I did), >for example RND in BASIC. The frequence of cutting for each enzyme must be >1/4 exp n for each class with n bases of recogniction. A better way to estimate the cutting frequency is to adjust for genomic nucleotide composition. For instance, in AT rich genomes, restriction enzymes that have AT rich recognition sequences will cut more frequently that estimated by (.25)^n, where n is the length of the recognition sequence. For example, A = .30 T = .30 C = .20 G = .20 GAATTC = (.2)(.3)(.3)(.3)(.3)(.2) = 3.24x10^-4 vs. A = C = G = T = .25 GAATTC = .25^6 = 2.44x10^-4 --------------------------------------------------------------------------- TTGATTGCTAAACACTGGGCGGCGAATCAGGGTTGGGATCTGAACAAAGACGGTCAGATTCAGTTCGTACTGCTG Eric E. Snyder Department of MCD Biology We are not suspicious enough University of Colorado, Boulder of words, and calamity strikes. Boulder, Colorado 80309-0347 LeuIleAlaLysHisTrpAlaAlaAsnGlnGlyTrpAspLeuAsnLysAspGlyGlnIleGlnPheValLeuLeu ---------------------------------------------------------------------------
levy@CELLBIO.STANFORD.EDU ("SHOSHANA LEVY") (02/26/91)
Thomas, See Chaudhary et al. PNAS 87, 1066-1070,1990. However, you might consider cloning without the need to cut. Shoshana Levy ------