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

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

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

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

5. Plasmid 1 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) 1B 1P 1E 1EP 1BP 1EB

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

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

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

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

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

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

13. Plasmid 2 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) 2B 2P 2E 2EP 2BP 2EB

14. Plasmid 2 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 1.43kbp and is the insert DNA 2B 2P 2E 2EP 2BP 2EB

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

16. 1B 1P 1E 1EP 1BP 1EB Now compare the 1EP and 2EP digests … 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. 1B 1P 1E 1EP 1BP 1EB Similarly compare the 1BP and 2BP digests … 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. Plasmid 1 E P B Plasmid 2 E P B P insert

20. Plasmid 1 3.6kbp E P B Plasmid 2 5.0kbp E P B P insert

21. Plasmid 2 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. 2B 2P 2E 2EP 2BP 2EB

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

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)