Team:EPF-Lausanne/Protocols/Klenow

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<font size="6" color="#007CBC">Klenow fragments synthesis</font>  
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<font size="6" color="#007CBC">Klenow fragment synthesis</font>  
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----
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<div style="text-align:justify;"><br>
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This protocol has been designed in order to get <i>ready-to-ligate</i> small DNA fragments from scratch. The procedure doesn't require any purification step, in order not to loose DNA material due to the small sizes of the fragments. Therefore, it includes a dephosphorylation step, which is crucial for the maximization of  the ligation that follows. The all expermiment is performed in a single tube with successive additions of new reactants.
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<br><br>
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The amount of DNA at the beginning of the reaction is calculated in order to fit the desired amount required for the ligation step that follows the end of the klenow fragment synthesis.
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<br><br></div>
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<br>
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*Design your DNA oligos
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*Fix your target DNA concentration that you want at the end (in ng/ul)
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*Multiply by 50 to get the total mass of DNA that you want in your tube at the end (the protocol is designed to have a final volume of 50ul)
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*Out of the M.W. of your DNA fragment, calculate the number of mol that you would expect at the end --> it gives you the the nb of mol that you will need for each DNA oligos (each oligo must have the same nb mol as  for the DNA fragment)
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*Out of the concentration of your oligos, you can calculate the volume of each required
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<br>
<br>
<br>
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Klenow protocol
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<font size="3">'''Step 1 : Klenow fragment synthesis'''</font>
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New protocol used with no purification step, everything is prepared to go trough all reaction up to the end of the digestion, ready for the ligation without going through any step of purification. This will avoid losing DNA due to the small size of the TrpO (127bp).
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<br>
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Goal: we want 100ng/ul od DNA at the end
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*3.64ul of NEB buffer (choose this one according to the restriction enzymes you will use later. The klenow fragment enzyme works in all NEB buffer)
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length of TrpO:127bp
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*0.37ul of BSA 100x
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molecular weight: 78,522 . 10^3 g/mol
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*X ul for each primer according to concentration
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→ 100ng/ul in 50ul → 6,367.10^-11 mol of TrpO a the end, that's to say 6,367.10^-11 of each primer
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*Y ul of MQ
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Trp Operon-Rev: 1,560.10^3M → 2.56 ul Trp Operon-Fwd: 1,560.10^3M → 2.56 ul
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-->The final volum at this stage should be 36.4ul
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first make a dilution at 25ul in 500ul → 8ul in 492ul of MQ
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<br>
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'''Thermal cycling'''
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<br>
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*94ºC for 5min
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*Ramp down the temperature to 5ºC below your annealing temperature. CAUTION : ramp the temperature down at a MAXIMUM of 0.1ºC/s (to avoid secondary structure formation in your primers)
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*Set the temperature @ 37ºC
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1. Klenow  
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<font size="3">'''Step 1''''</font>
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dNTPs final concentr = 1mM
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<br>
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3.64ul of NEB2
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*Add 1ul of the Klenow fragment enzyme
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0.37ul of BSA 100x
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*Add 1.6ul of dNTPs (final concentration of 1mM each). Stock solution of dNTPs : 25mM each
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2.56ul of each primer at 25uM
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-->The final volume at this stage should be 39ul
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32.38ul of MQ
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<br>
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→ final: 36.4ul
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'''Thermal cycling'''
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<br>
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*Incubate 1h30 @ 37ºC
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*Inactivate the enzyme 20min @ 75ºC
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*Ramp down the temperature to 37ºC (!!! 0.1ºC/s !!!)
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<br>
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<font size="3">'''Step 2 : Digestion'''</font>
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*Add 0.5ul of each enzyme
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-->volume at this stage should be 40ul
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Thermal cycler:
 
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94°c for 5min
 
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0.1°C/s to 74°C for 5 min
 
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0.1°C/s to 37°C
 
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Annealing temperature: 79.1°C (due to some website)
 
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1.' 1ul Klenow + 1.6ul dNTPs = vol 39ul Incubate 1h30 at 37°C, then inactivate 20 min at 75°C : 0.1°C/s to 37°C
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'''Thermal cycling'''
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<br>
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2. Digestion
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*Incubate 2h @ 37ºC
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Add 0.5 ul of each enzyme (EcoRI - HF / NheI) → vol: 40.0ul Incubate 2h at 37°C, then inactivate 20min at 80°C. Then cool down at 37°C (0.1°C/s)
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*Inactivate the enzymes 20min @ 80ºC
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*Ramp down the temperature to 37ºC (!!! 0.1ºC/s !!!)
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<br>
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<font size="3">'''Step 3 : Dephosphorylation'''</font>
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<br>
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*Add 5ul of Antartic phosphatase buffer (10X)
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*Add 5ul of Antartic phosphatase enzyme (*)
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-->Final volume should be 50ul
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<br>
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'''Thermal cycling'''
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*Incubate 2h @ 37ºC
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*Inactivate the enzyme 10 min @ 65ºC
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*Ramp down the temperature to 25ºC (!!! 0.1ºC/s !!!)
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<br>
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'''(*)''' This amount of enzyme has been calculated using the following. According to NEB : 1ul needed to dephosphorylate 1-5mg of pUC19 (~2690 bp) in 30 min. So, calculate your molar bp ratio compared to pUC19 and multiply by your supposed DNA mass. The result will give you by how much you have to multiply the phosphatase volume (in ul). You can also let the incubation run for longer (like here 2h) and so use less phosphatase (assume the relationship between ul of enzyme and time of incubation is a linear function). CAUTION : if you use more enzyme volume, your final volume will also be bigger. Make sure to adjust your phosphatase buffer accordingly!
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3. Dephosphorylation
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</div><div CLASS="epfl09bouchon"></div>
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Add 5ul of antartic phosphatase buffer + 5ul of phosphatase enzyme → total volume: 50ul Incubate 2h at 37°C, then inactivate 10min at 65°C. Then cool down slowly (0.1°C/s)
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1ul is needed to dephosphorylate 1-5mg of vector pUC19
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Latest revision as of 17:18, 21 October 2009



Klenow fragment synthesis


This protocol has been designed in order to get ready-to-ligate small DNA fragments from scratch. The procedure doesn't require any purification step, in order not to loose DNA material due to the small sizes of the fragments. Therefore, it includes a dephosphorylation step, which is crucial for the maximization of the ligation that follows. The all expermiment is performed in a single tube with successive additions of new reactants.

The amount of DNA at the beginning of the reaction is calculated in order to fit the desired amount required for the ligation step that follows the end of the klenow fragment synthesis.




  • Design your DNA oligos
  • Fix your target DNA concentration that you want at the end (in ng/ul)
  • Multiply by 50 to get the total mass of DNA that you want in your tube at the end (the protocol is designed to have a final volume of 50ul)
  • Out of the M.W. of your DNA fragment, calculate the number of mol that you would expect at the end --> it gives you the the nb of mol that you will need for each DNA oligos (each oligo must have the same nb mol as for the DNA fragment)
  • Out of the concentration of your oligos, you can calculate the volume of each required



Step 1 : Klenow fragment synthesis

  • 3.64ul of NEB buffer (choose this one according to the restriction enzymes you will use later. The klenow fragment enzyme works in all NEB buffer)
  • 0.37ul of BSA 100x
  • X ul for each primer according to concentration
  • Y ul of MQ

-->The final volum at this stage should be 36.4ul
Thermal cycling

  • 94ºC for 5min
  • Ramp down the temperature to 5ºC below your annealing temperature. CAUTION : ramp the temperature down at a MAXIMUM of 0.1ºC/s (to avoid secondary structure formation in your primers)
  • Set the temperature @ 37ºC


Step 1'

  • Add 1ul of the Klenow fragment enzyme
  • Add 1.6ul of dNTPs (final concentration of 1mM each). Stock solution of dNTPs : 25mM each

-->The final volume at this stage should be 39ul
Thermal cycling

  • Incubate 1h30 @ 37ºC
  • Inactivate the enzyme 20min @ 75ºC
  • Ramp down the temperature to 37ºC (!!! 0.1ºC/s !!!)


Step 2 : Digestion

  • Add 0.5ul of each enzyme

-->volume at this stage should be 40ul


Thermal cycling

  • Incubate 2h @ 37ºC
  • Inactivate the enzymes 20min @ 80ºC
  • Ramp down the temperature to 37ºC (!!! 0.1ºC/s !!!)


Step 3 : Dephosphorylation

  • Add 5ul of Antartic phosphatase buffer (10X)
  • Add 5ul of Antartic phosphatase enzyme (*)

-->Final volume should be 50ul
Thermal cycling

  • Incubate 2h @ 37ºC
  • Inactivate the enzyme 10 min @ 65ºC
  • Ramp down the temperature to 25ºC (!!! 0.1ºC/s !!!)


(*) This amount of enzyme has been calculated using the following. According to NEB : 1ul needed to dephosphorylate 1-5mg of pUC19 (~2690 bp) in 30 min. So, calculate your molar bp ratio compared to pUC19 and multiply by your supposed DNA mass. The result will give you by how much you have to multiply the phosphatase volume (in ul). You can also let the incubation run for longer (like here 2h) and so use less phosphatase (assume the relationship between ul of enzyme and time of incubation is a linear function). CAUTION : if you use more enzyme volume, your final volume will also be bigger. Make sure to adjust your phosphatase buffer accordingly!

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