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Molecular Genetics

Molecular Genetics . Worksheet 2. Question 1. 1a ) Study the map of pUC18 .

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Molecular Genetics

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  1. Molecular Genetics Worksheet 2

  2. Question 1 • 1a) Study the map of pUC18. • Which restriction enzymes could you use to clone the TetR gene (Fig 1) into the pUC18 plasmid so that the start of the gene is in the correct orientation for the M13 Forward Sequencing Primer?

  3. So you have the choice of: • Pstl • Kpnl • Xhol • EcoRI  would cut gene • HindIII would cut gene • BamHI

  4. Before you even look at the plasmid sequencing primer, you can narrow the options down to • Pstl • Kpnl • Xhol • BamHl

  5. Now look at the sequencing primer on the plasmid • You can see where each enzyme cuts • For part a) you need to use the Forward Sequencing Primer, 5’ to 3’

  6. Reading the enzymes 5’ to 3’, the first one we have is KpnI • This must be the first enzyme to use (the most 5’ option) • Carrying on from this one, the next enzyme we have is BamHI • This must be the second enzyme

  7. Start Stop

  8. For part b), you need the Reverse Sequencing Primer, i.e. 3’ to 5’ • This time, look for the first enzyme we have from 3’ (PstI) • Then look for the next enzyme we have when moving in a 5’ direction (BamHI)

  9. Stop Start

  10. Question 2 • Study the restriction map of the erythromycin resistance gene in Fig 2. • Explain how you could clone this gene into pBluescript so that the start of the gene is in the correct orientation for the T7 promoter

  11. The answer for this one can be found in the lecture notes • There is more than one answer! • Xha I & Bgl II can be used to cut the resistance gene, however Bgl II can’t cut the T7 promoter. • What do we do? • We can simply use another enzyme that will create complementary sticky ends.

  12. We know that Bgl II will cut TAG and thus leave ATC. • From the T7 map, we can find an enzyme that will complement this. In this case it is BamHI. • BamHI cut’s ATC leaving TAG. • T7 and the gene can then be ligated together. Start Stop

  13. Question 3 • This question is based on your understanding of the lac-operon and how it works. • You have to understand how the operon works with and without an inducer. • How the repressor protein works. • How the structural genes are transcribed and what they create. • How the CAP-cAMP positive feedback mechanism works.

  14. A mutation which inactivated the lacI promoter would have what effect on the lac operon? • Constitutive expression, as the repressor is broken. • Genes will be produced even when there is no lactose present. • As the repressor protein will no longer be able to bind the promoter.

  15. Question 4 • 100 µl of E.coliDH5α cells were transformed with 100ng of pUC19 DNA, then 0.9ml of LB broth was added and the cells incubated for 1 hour at 37 ºC. 100 µl aliquots of this incubated culture were plated out onto LB agar plates containing ampicillin and it was found that an average of 252 colonies grew on the 10-2 plate.

  16. Calculate the number of cells transformed per µg of pUC19DNA • Average number of colonies =252 = 2.25x102 • 10-2plate so need to multiply by 102to get back to original concentration • 0.9 ml of broth were added to 100µl of cells. Then 100µl of this solution was used to plate out. • Thus we have diluted by a factor of 10. • We also want an answer in µg-1 and 100ng have been used. Therefore, we need to convert it.

  17. Now that we know this, we simply have to apply the information into one calculation. • (2.25x102 )x 100 x 10 x 10 =2.52x106 Transformants µg-1 Reverting back to original conc (10-2) Converting from Ng to µg-1 Colonies grown Accounting for the 1:10 dilution in the BROTH

  18. 4b) After the 1 hour incubation a sample of the mixture from 3a was diluted several times and 100 µl aliquots were onto LB agar containing no antibiotics. • It was found that 197 colonies were present on the 10-5 dilution plate.

  19. i) The total number of viable cells in the transformation reaction • Average number of colonies=197=1.97x102 • Multiply by 10 to get an answer in ml • Multiply by 105 as 10-5 plate • (1.97x102) x 10 x 105= 1.97x108 cells/ ml-1

  20. ii) The percentage of cells transformed with plasmid. • Total number of cells = 1.97x108 cells/ ml-1 Number of transformed cells = 2.52x106 Transformants µg-1 • x100 = 0.13% 2.52x106 1.97x108

  21. Question 5 15 kb fragment in total This fragment doesn’t add up to 15 as the two 3 kb fragments will be at the same place on the gel, so will only be counted as one 3 kb fragment

  22. It doesn’t matter where you start, so let’s just start at the top of the list • HindIII- 7 kb, 8 kb • Think of kb as cm, it’s a lot easier to draw then! • This fragment adds up to 15 kb, so draw a line that’s 15 cm long and split it into a 7 cm section and an 8 cm section

  23. HindIII 7 kb 8 kb • That’s the first enzyme placed! • Moving on to the second on the list • BamHI- 3 kb, 4 kb, 8 kb • These fragment needs to be placed in a way which fit in with the first enzyme • There’s an 8 kb fragment, so that must be the same as the HindIII fragment • 3 + 4 = 7kb so those fragments fit too

  24. HindIII • For now, we’ll put the 3 and 4 kb fragments in this order, but this might have to change if this doesn’t work • As we’ve now placed 2 enzymes, we can use the fragment sizes produced when they are both used together to check what we have so far works • HindIII+ BamHI- 3 kb, 3 kb, 4 kb, 5kb 4 kb 8 kb 3 kb

  25. HindIII HindIII+ BamHI- 3 kb, 3 kb, 4 kb, 5kb • This doesn’t work as there needs to be 2 BamHI sites as 3 fragments are produced. • If the second BamHI site was in the same place as HindIII, the 3 kb and 5 kb fragments would fit into the 8 kb fragment. • But, this configuration would not match the fragments created by BamHI alone • Ie, we need to fit in these fragments so that they can so produce fragments of 3 kb, 4 kb and 8 kb 4 kb 8 kb 3 kb

  26. HindIII HindIII • To do this, there are three possible solutions: 3 kb 4 kb 5 kb 3 kb These options create all of the fragments necessary for the combined enzymes, but there are no 8 kb fragments for BamHI, so they can’t be right 3 kb 4 kb 5 kb 3 kb

  27. HindIII BamHI BamHI This option creates all of the fragments necessary for the combined enzymes, and for BamHI, so we can assume this is right • Now that we have the HindIII and BamHI sites, we can move onto XbaI 3 kb 4 kb 3 kb 5 kb

  28. HindIII • XbaI- 0.5 kb, 14.5 kb • We need to create 0.5 kb and 14.5 kb fragments from the map. • These fragments also need to agree with the combined enzyme fragments • HindIII+ XbaI- 0.5 kb, 7 kb, 7.5 kb • BamHI+ XbaI- 0.5 kb, 2.5 kb, 4kb, 8kb, BamHI BamHI 3 kb 4 kb 3 kb 5 kb

  29. HindIII • Looking at the fragments needed should be enough for this final enzyme (no need for trial and error) • HindIII + XbaI- 0.5 kb, 7 kb, 7.5 kb • Using HindIII, we have a 7 kb fragment and a 8 kb fragment, so it must be the 8 kb fragments that is cut by the Xbal into 0.5 kb and 7.5 kb • Doing this still needs to produce fragments that agree with BamHI + XbaI • BamHI + XbaI- 0.5 kb, 2.5 kb, 4 kb, 8 kb • BamHI cutting produces the 8 kb fragment • The only way to produce the other fragments is to cut the 3 kb fragment. BamHI BamHI 3 kb 4 kb 3 kb 5 kb

  30. HindIII BamHI BamHI XbaI 2.5 kb 0.5 kb 4 kb 5 kb 3 kb

  31. Question 6 • The overall goal here is to produce an expected electrophoresis gel after running all of the lanes. • The plasmid is 5500 bases in total

  32. HindIII HindIII creates 2 fragments For fragment 1 (red), calculate the difference between the two enzyme cuts: 1730-375= 1355 Fragment 2 (black) is the rest of the plasmid: 5500-1355= 4135 4000 Linear DNA Markers 3000 5000 1500 HindIII 4135 500 2000 250 1355

  33. KpnI 4000 Linear DNA Markers 3000 5000 1500 Again there are 2 fragments to work out Fragment 1 (red): 5108-3185= 1293 Fragment 2 (black): 5500- 1293= 4207 HindIII 4207 4135 500 KpnI 3077 2000 250 2423 2423 2477 2085 1355 1293 700 1668 1367 1293 1355 1293 755 600 417

  34. PstI 4000 Linear DNA Markers 3000 5000 1500 HindIII 4207 4135 500 PstI KpnI 3077 Fragment 1 (red): 3398-975=2423 Fragment 2 (black): 5500-2423=3077 2000 250 2423 2423 2477 2085 1355 1293 700 1668 1367 1293 1355 1293 755 600 417

  35. HindIII and PstI 4 bands are produced Band 1 (red) is formed between HindIII and PstI: 975-375=600 Band 2 (blue) is formed between PstI and HindIII: 1730-975= 755 Band 3 (green) is formed between HindIII and PstI: 3398-1730= 1668 Band 4 (black) is what’s left of the plasmid 5500-(600+755+1668)=2477 When 2 enzymes are used, you’ll get a cut at all the sites used by the enzymes. Just work your way around the circle

  36. Band 1 (red) :975-375=600 Band 2 (blue):1730-975= 755 Band 3 (green):3398-1730= 1668 Band 4 (black):5500-(600+755+1668)=2477 4000 KpnI and PstI KpnI and HindIII Linear DNA Markers HindIII and PstI 3000 5000 1500 HindIII 4207 4135 500 PstI KpnI 3077 2000 250 2423 2423 2477 2085 1355 1293 700 1668 1367 1293 1355 1293 755 600 417

  37. KpnI and HindIII Band 1(red): 1730-375=1355 Band 2 (black): 3815-1730= 2085 Band 3 (blue): 5108-3815=1293 Band 4 (green): 5500-(1355+2085+1293)=767

  38. Band 1(red): 1730-375=1355 Band 2 (black): 3815-1730= 2085 Band 3 (blue): 5108-3815=1293 Band 4 (green):5500-(1355+2085+1293)=767 KpnI and HindIII 4000 KpnI and PstI Linear DNA Markers HindIII and PstI 3000 5000 1500 HindIII 4207 4135 500 PstI KpnI 3077 2000 250 2423 2477 2085 1355 1293 700 1668 1355 1293 755 600 600

  39. KpnI and PstI Band 1(green): 3398-975= 2423 Band 2 (blue): 3815-3398= 417 Band 3 (black): 5108-3815= 1293 Band 4 (red): 5500-(2423+417+1293)=1367

  40. Band 1(green):3398-975= 2423 Band 2 (blue): 3815-3398= 417 Band 3 (black): 5108-3815= 1293 Band 4 (red):5500-(2423+417+1293)=1367 4000 KpnI and PstI KpnI and HindIII Linear DNA Markers HindIII and PstI 3000 5000 1500 HindIII 4207 4135 500 PstI KpnI 3077 2000 250 2423 2423 2477 2085 1355 1293 700 1668 1367 1293 1355 1293 755 600 417

  41. And it’s over……….  Any Questions?

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