appearance of more bands on the gel than would be predicted from a homogenous population of linear DNA molecules. This also causes the partial loss of other fragments. For example, the left most Hin dIII site is 23,130 bp from the left end of the linear λ genome, and the right most site is 4361 bp from the right end. The 4361 bp band is faint in comparison to other bands on the gel of similar size. This indicates that a percentage of the DNA molecules are circular- combining the 4361 bp terminal fragment with the 23,130 bp terminal fragment to produce a 27,491 bp fragment. However, the combined 27,491 bp fragment usually runs as a doublet along with the 23,130 bp fragment from the linear molecule. a. Use a protractor to draw three circles about 3 inches in diameter. These represent λ DNA molecules with base -paired COS sites. b. Label a point at 12:00 on each circle 48/0. This marks the point where the COS sites are joined. c. Using data from the restriction maps of the linear λ genome on Figure 13.1 make a rough map of restriction sites for Hin dIII on one of the circles. Note the situation described above. d. Next make rough restriction maps of Bam HI and Eco RI sites on the remaining two circles. e. What Bam HI and Eco RI fragments are created in the circular molecules? Why (or why not) can you locate each of these fragments on your gel or the ideal gel above? ________________________________________________________ ________________________________________________________ ________________________________________________________
BSC1010L Rev. Win ’0 9 13 - 11 Circular maps
BSC1010L Rev. Win ’0 9 13 - 12 Appendix 13.A Examples of Restriction Enzymes Enzyme Organism from which derived Target sequence (cut at *) 5' -->3' Ava I Anabaena variabilis C* C/T C G A/G G Bam HI Bacillus amyloliquefaciens G* G A T C C Bgl II Bacillus globigii A* G A T C T Eco RI Escherichia coli RY 13 G* A A T T C Eco RII Escherichia coli R245 * C C A/T G G Hae III Haemophilus aegyptius G G * C C Hha I Haemophilus haemolyticus G C G * C Hind III Haemophilus inflenzae Rd A* A G C T T Hpa I Haemophilus parainflenzae G T T * A A C Kpn I Klebsiella pneumoniae G G T A C * C Mbo I Moraxella bovis *G A T C Mbo I Moraxella bovis *G A T C Pst I Providencia stuartii C T G C A * G Sma I Serratia marcescens C C C * G G G SstI Streptomyces stanford G A G C T * C Sal I Streptomyces albus G G * T C G A C Taq I Thermophilus aquaticus T * C G A Xma I Xanthamonas malvacearum C * C C G G G Note: Only one strand of the target DNA is shown, in the interests of clarity. It will be noted that the omitted strand has the same sequence as the one shown, but the nucleotides occur in the reverse order. Thus, for example, the complementary sequence for Sal I (with its point of cleavage indicated as above) is 5' -C A G C T * G-3'. These are hence said to be palindromic sequences, and double-stranded cuts produce short single-stranded cohesive ends. Information from: Biotechnology Made Simple. 4th ed., 1991. PJB Publications. Surrey, U.K.
BSC1010L Rev. Win ’0 9 13 - 13 Review Questions DNA RESTRICTION ANALYSIS 1. What is a restriction enzyme?
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