Correct Orientation: 768bp + 15bp = 783bp fragment
Incorrect Orientation: 495bp + 15bp= 510bp fragment
Send comments, questions, and suggestions to: sahenry@davidson.edu
Digestion of DNA with Restriction Enzymes
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It is important to verify that all of the IDH Genes have been cloned into the plasmid in the correct orientation before proceeding to study the expression of the proteins made by these genes. When the plasmid is cut and the gene inserted, the gene can be inserted in a forwards or backwards direction. To verify that the genes were inserted in the forwards orientation, the plasmids will be digested with restriction enzymes. The following restriction enzymes were available: BamH I, Bgl II, Cla I, EcoR I, EcoR V, Hind III, Kpn I, Pst I, and Sal I. Using the restriction maps located on the web page for each gene and the map of the pQE-30 UA vector, a restriciton enzyme was found for each gene that cut a fragment of that gene and also cut a portion of the polylinker of the vector. Gel electrophoresis can then be performed to determine the size fragment of DNA that was cut. Below are the expected fragment sizes for the corrent and incorrent orientation of each gene.
IDH1: cut with EcoRV: Cuts the polylinker 16bp upstream of the PCR product and cuts the gene at the 340bp
The gene is 1083bp long, therefore, we expect:
Correct Orientation: 16bp + 340bp= 356bp fragment
Incorect Orientation: 16bp+ 743bp= 759bp fragment
The 743bp length is the amount of DNA that would be located within the restriciton site if the gene were inserted backwards. 1083bp-340bp= 743bp.
IDH2: cut with BglII: Cuts the polylinker 15bp downstream (after) the PCR product and cuts the gene at the 335bp
The gene is 1110bp long, therefore, we expect:
Correct Orientation: 775bp + 15bp = 790bp fragment
Incorrect Orientation: 335bp + 15bp = 350bp fragment
The 775bp is the length of the gene from the restriction site to the end of the gene (1110bp-335bp=775bp). Since the restriction site in the polylinker is located after the PCR insert, the calculation must be done this way.
IDP1: cut with BglII: Cuts the polylinker 15bp downstream of the PCR product and cuts the gene at the 1208bp
The gene is 1287bp long, therefore, we expect:
Correct Orientation: 79bp + 15bp = 94bp fragment
Incorrect Orientation: 1208bp + 15bp = 1223bp fragment
IDP2: cut with BglII: Cuts the polylinker 15bp downstream of the PCR product and cuts the gene at the 496bp
The gene is 1239bp long, therefore, we expect:
Correct Orientation: 743bp + 15bp= 758bp fragment
Incorrect Orientation: 496bp + 15bp= 511bp fragment
IDP3: Cut with ClaI and BglII: ClaI cuts the gene at the 495bp and BglII cuts the polylinker 15bp downstream of the PCR product
The gene is 1263bp long, therefore, we expect:
Correct Orientation: 768bp + 15bp = 783bp fragment
Incorrect Orientation: 495bp + 15bp= 510bp fragment
Send comments, questions, and suggestions to: sahenry@davidson.edu