57. A 1 Kb insert (I) is ligated to a 5 Kb vector (V) in a molar ratio of I:V of 5:1, in a reaction volume of 1 mL at a
final DNA concentration at 10 μ g/mL. How much of the insert and vector was used in the ligation mixture?
1. 5 μg I and 5 μg V
2. 2 μg I and 8 μg V
3. 8 μg I and 2 μg V
4. 3 μg I and 7 μg V
How to Calculate Insert and Vector DNA Amounts in a Ligation Reaction
In molecular cloning, ligation reactions involve combining vector and insert DNA in specific molar ratios, not just weight ratios. Correct calculations ensure high cloning efficiency.
Let’s break down the question and solve it step-by-step.
Question Summary:
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Insert size: 1 kb
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Vector size: 5 kb
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Molar ratio (Insert:Vector): 5:1
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Final DNA concentration: 10 µg/mL
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Total volume: 1 mL → Total DNA = 10 µg
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We need to find: How many micrograms of insert and vector were used
Step-by-Step Calculation:
To match the molar ratio, we must use the formula:
Number of moles=masslength (in kb)\text{Number of moles} = \frac{\text{mass}}{\text{length (in kb)}}
Let:
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xx = mass of vector (in µg)
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10−x10 – x = mass of insert (since total DNA = 10 µg)
Now set up the molar ratio:
(10−x)/1x/5=5(Insert : Vector molar ratio)\frac{(10 – x)/1}{x/5} = 5 \quad \text{(Insert : Vector molar ratio)} ⇒10−xx/5=5\Rightarrow \frac{10 – x}{x/5} = 5 ⇒(10−x)⋅5=5x\Rightarrow (10 – x) \cdot 5 = 5x ⇒50−5x=5x\Rightarrow 50 – 5x = 5x ⇒50=10x\Rightarrow 50 = 10x ⇒x=5 μg(vector)\Rightarrow x = 5 \, \mu g \quad \text{(vector)} ⇒10−x=5 μg(insert)\Rightarrow 10 – x = 5 \, \mu g \quad \text{(insert)}
Final Answer:
5 μg of Insert and 5 μg of Vector → Option 1
Why It Matters:
Understanding this type of calculation is crucial when setting up ligation reactions in molecular cloning. Using the correct molar ratio rather than just equal masses ensures efficient ligation, minimizing empty vectors or multi-insert clones.
Correct Answer: 1. 5 μg I and 5 μg V
