Question

A plasmid is digested with a restriction enzyme to produce three fragments of sizes 7 kb, 3 kb and 2 kb. To obtain 500 ng of a 3 kb fragment, the amount of plasmid required to be digested in µg is ..........

Hint:

To calculate the amount of plasmid needed for a specific fragment size, use the fraction of the fragment's size relative to the total plasmid size.

Answer 1

Correct Answer: 1.9

Step 1: Calculate the amount of plasmid required.
The size of the 3 kb fragment is 3 kb, and we need to obtain 500 ng of this fragment. The plasmid is digested into fragments of sizes 7 kb, 3 kb, and 2 kb, so the total size of the plasmid is 12 kb (7 + 3 + 2 kb).

Step 2: Find the fraction of the 3 kb fragment.
The fraction of the 3 kb fragment is: \[ \frac{3 \, \text{kb}}{12 \, \text{kb}} = \frac{1}{4}. \]

Step 3: Use the fraction to calculate the total amount of plasmid.
To obtain 500 ng of the 3 kb fragment, the total amount of plasmid required is: \[ \text{Amount of plasmid} = \frac{500 \, \text{ng}}{1/4} = 2000 \, \text{ng} = 2 \, \mu\text{g}. \]

Step 4: Adjust for the specific 3 kb fragment amount.
Since we only need 500 ng of the 3 kb fragment, we divide the total plasmid amount by the size of the fragment, resulting in 0.5 µg.