You need to carry out a 100 l restriction digest on 5 g of a DNA plasmid using 5

Question

You need to carry out a 100 l restriction digest on 5 g of a DNA plasmid using 5.0 U of the restriction enzyme BamHI. The components of your final reaction mixture include 1x buffer, 1x BSA, and 2 mM MgCl2. Your reagent stock solutions include 10x buffer, 100x BSA, 100 mM MgCl2, 10 U/ l BamHI, and 5 mg/ml DNA.

1) How much of your 10x buffer stock solution would you add?

2) How much of your 100mM MgCl2 stock solution would you add?

3) How much of your DNA stock solution will you add?

4) How much of your BmaHl enzyme stock would you add?

5) How much of your ___ enzyme stock solution will you add?

6) How much water will you add?

Step by Step explanation, please

Biochemistry ll

Explanation / Answer

1) Given 10x buffer , in final total 100microlitre we want it to be 1x.

using: M1*V1 = M2*V2

10x*V1 = 1x*100

V1 = 10 microlitre.

2) Similarly for MgCl2 : initially 100mM, in final 100 microlitre we want it to be 2mM

Using M1*V1 = M2*V2

100*V1 = 2*100

V1 = 2microlitre

3) Our stock contains 5mg DNA in 1ml , finally we want 5 microgram in 100microlitre.

M1*V1 = M2*V2

5mg/ml*V1 = 5microgram/100microlitre*100

V1 = 1microlitre

4) M1*V1 = M2*V2

10U/microl*V1 = 5U/100microl*100microlitre

V1=0.5 microlitre

5) 0.5 microlitre

6) Volume of water = 100-0.5-1-2-10-Vol of BSA to be added= 85.5microlitre

for BSA M1*V1 = M2*V2

100*V1 = 1*100

V1 = 1 microlitre

water = 85.5

BSA = 1

buffer = 10

MgCl2 = 2

enzyme = 0.5

DNA = 1

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