Question

You want to set-up a restriction enzyme digestion of a plasmid DNA sample. You would like to digest 0.75ug of this DNA with 50 units of BamHI and 50 units of HindIII in a total volume of 25uL containing 1x buffer A.

How much (in uL) of the following stock solutions and water are needed?

a) 200ng/uL plasmid DNA
b) 10x buffer A
c) 50,000 units/mL BamHI
d) 25,000 units/mL HindIII
e) water

Answer 1

the final volume needed is 25 uL we have 10 X buffer

so M1V1= M2V2

M1= stock concentration

V1=volume of stock needed ( we have to find V1)

M2= final concentration needed

V2= final volume ( here final volume is 25 uL)

V1= M2V2/M1

1) for buffer

M1= 10X V1=? M2=1X V2=25uL

V1=25*1/10

=2.5 uL

we need 0.75 ug of DNA and we have a stock of 200ng/uL

the volume of the DNA solution to the take= amount of DNA needed/ concentration of the stock solution

= 0.75 ug/200ng/ul

(1ug= 1000 ng, so 0.75ug=750 ng)

=750ng/200ng/uL

=3.75 uL

We need 50 units of BamHI and stock is 50,000 units/mL BamHI so

in 1 mL 50000 units is there ( 1 mL=1000 uL)

so 1 unit of enzyme is present in 1000 /50000 uL= 0.02 uL

we need 50 units of Hind III and stock is 25,000 units/mL HindIII

so in 1 mL 25000 is there in 1 mL so 1 unit of enzyme is present in 1000/25000 ul = 0.04 uL

a) volume of DNA solution to be taken is 3.75 uL

b) 2.5 uL 10X buffer A

c) volume of BamH I is 0.02 uL

d) volume of Hind III is 0.04 uL

e) rest is water that is 25-0.04-0.02-3.75-2.5=18.69 uL