The following table is a partial protocol for a double digest that must be done
ID: 140335 • Letter: T
Question
The following table is a partial protocol for a double digest that must be done on a DNAplasmid. Please fill out the missing information in the table to successfully complete your experiment. The final volume of your reaction needs to be 50uL. Show your calculations below (12 points). For the restriction enzymes, the unit “U/mL” means Units of enzyme per mL. A Unit of an enzyme is usually the amount of an enzyme required to convert 1 ug of substrate to product per minute.
Stock []
Final [] or amount
Volume needed
Rxn Buffer
10x
Plasmid
MluI
10,000 U/mL
200 U/mL
XhoI
5,000 U/mL
BSA
10 mg/mL
100 ug/mL
MQH2O
--
Stock []
Final [] or amount
Volume needed
Rxn Buffer
10x
5 uLPlasmid
37 ng/uL 1 ugMluI
10,000 U/mL
200 U/mL
XhoI
5,000 U/mL
1 uLBSA
10 mg/mL
100 ug/mL
MQH2O
----
Page of 3 -ZOOM+ 2. The following table is a partial protocol for a double digest that must be done on a DNA plasmid. Please fill out the missing information in the table to successfully complete your experiment. The final volume of your reaction needs to be 50 HL. Show your calculations below (12 points). For the restriction enzyme Nol the unit U/uL means Units of enzyme per ul. A Unit of an enzyme is usually the amount of an enzyme required to convert 1 umol of substrate to product per minuteExplanation / Answer
Final volume = 50 uL
Buffer stock concentration = 10X
Working concentration = 1X
Buffer required volume = 5 uL
Plasmid stock concentration = 37 ng/uL = 0.037 ug/uL
Final amount = 1 ug
1 uL ----> 0.037 ug
? uL ----> 1 ug
Final volume = 1/0.037 = 27.02 uL = 27 uL
Mlu I stock concentration = 10000 u/mL
Required concentration = 200 U/mL
C1V1 = C2V2
V1 = C2V2/C1
= (200 X 50)/10000
Required volume = 1 uL
Xho I stock concentration = 5000 u/mL
Required volume = 1 uL
C1V1 = C2V2
C2 = C1V1/V2
= (5000 X 1)/50
Required concentration = 100 U/mL
BSA stock concentration = 10 mg/mL = 10000 ug/mL
Final concentration = 100 ug/mL
C1V1 = C2V2
V1 = C2V2/C1
= (100 X 50)/10000
Required volume = 0.5 uL
Total volume = 50 uL
Volume of water = 50 - (5+27+1+1+0.5) = 15.5 uL
Please rate the solution. Your efforts are highly appreciated. Thank you
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