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Analysis

Analysis. www.le.ac.uk/genie. Full length. one site / one cut. Plasmid. A. A. two sites / two cuts. A + B = Full length. B. Plasmid. B. X. Z + Y. X. +. Y. Z + X. Plasmid. +. Y. Z. X + Y. Z . +. Restriction enzyme mapping. large. small. M. Restriction enzyme mapping.

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Analysis

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  1. Analysis www.le.ac.uk/genie

  2. Full length one site / one cut Plasmid A A two sites / two cuts A + B = Full length B Plasmid B X Z + Y X + Y Z + X Plasmid + Y Z X + Y Z + Restriction enzyme mapping

  3. large small M Restriction enzyme mapping B Y+Z X+Z X+Y Z Y X A

  4. Consider plasmid 1 digests first ……. 1B 1P 1E 1EP 1BP 1EB

  5. Plasmid 1 1B 1P 1E 1EP 1BP 1EB • 1B 3.56kbp • 1P 3.56 • 1E 3.56 • 1EP 2.81 0.75 • 1BP 2.43 1.13 • 1EB 3.18 0.38 (NB sum of lengths for each digest is approx. 3.5 kbp = size of plasmid)

  6. Plasmid 1 E Choose one restriction site to start at e.g. EcoRI

  7. E P Plasmid 1 0.75 Now locate the PstI site….. This will be 0.75 kbpfrom the EcoRI site. This is the size of the smaller EP fragment.

  8. E B? P Plasmid 1 0.38 Now locate the BamHI site….. This will be 0.38 kbp from the EcoRI site. This is the size of the smaller EB fragment. BUT there is another possibility…….

  9. E B? B? P Plasmid 1 0.38 …….. We now need to eliminate one of these possibilities ……..

  10. E B? B? P Plasmid 1 If orange is correct then the BP fragment will be smaller than the EP fragment Alternatively if green is correct then the BP fragment will be bigger than the EP fragment We need to go back to the gel for the answer …..

  11. Plasmid 1 1B 1P 1E 1EP 1BP 1EB • 1B 3.56kbp • 1P 3.56 • 1E 3.56 • 1EP 2.81 0.75 • 1BP 2.43 1.13 • 1EB 3.18 0.38 The small BP is bigger than the small EP so the map is ……

  12. E B 0.75 P 1B 1P 1E 1EP 1BP 1EB Plasmid 1 0.38 1.13

  13. Plasmid 2 2B 2P 2E 2EP 2BP 2EB • 2B 4.99kbp • 2P 3.56 1.43 • 2E 4.99 • 2EP 2.81 1.43 0.75 • 2BP 2.43 1.43 1.13 • 2EB 4.61 0.38 NB the sum of the fragments for each of these digests is the same (about 5.0kbp)

  14. Plasmid 2 2B 2P 2E 2EP 2BP 2EB This time the PstI digest (2P) gives two fragments The larger is the same size as the 1P (and 1E and 1B) fragment and is the plasmid vector. The smaller is 1.43kbpand is the insert DNA

  15. Plasmid 2 2B 2P 2E 2EP 2BP 2EB we also see the same 1.43kbpfragment in the 2EP and 2BP digests as well

  16. Now compare the 1EP and 2EP digests … 1B 1P 1E 1EP 1BP 1EB 2B 2P 2E 2EP 2BP 2EB Both share the 2.81 and 0.75 fragments but the 2EP has the extra 1.43 fragment

  17. Similarly compare the 1BP and 2BP digests … 1B 1P 1E 1EP 1BP 1EB 2B 2P 2E 2EP 2BP 2EB Both share the 2.43 and 1.13 fragments but the 2BP has the extra 1.43 fragment

  18. So we can conclude that… • Plasmid 2 is constructed from plasmid 1 • The inserted DNA fragment is 1.43kbp • This fragment was inserted at the PstI site • We can now draw a map of plasmid 2

  19. P Plasmid 1 insert E E B B P P Plasmid 2

  20. E B Plasmid 13.6kbp P E B insert P P Plasmid 2 5.0kbp

  21. Plasmid 2 2B 2P 2E 2EP 2BP 2EB We can confirm our map: 1EB gave 3.18 and 0.38 fragments 2EB gave 4.61 (3.18 + 1.43 insert) and 0.38 (not visible on this photo) as predicted so our map is correct.

  22. Strain A (Plasmid 1) Ampicillin resistant Tetracycline resistant Kanamycin sensitive Strain B (Plasmid 2) Ampicillin sensitive Tetracycline resistant Kanamycin resistant Phenotypes

  23. Conclusions Inserted fragment carries the Kanamycin resistance gene which is inserted into the PstI site of plasmid 1 This fragment has been inserted into the Ampicillin resistance gene which is disrupted in plasmid 2 We cannot draw many conclusions about the position of the Tetracycline resistance gene except that it is NOT at the PstI site (in fact it spans the BamHI site)

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