Question

4. For a ligation reaction, it is generally recommended to add Insert to Vector at a molecular ratio of 3:1 (three molecules of insert to one molecule of vector). If your vector and insert preparations have the same DNA concentration (in ng/ul), but your vector is 4500 base pairs long and your insert only 1500 base pairs, which volume ratio of Insert: Vector would you use in your ligation reaction? Explain. (5 points)

I have no idea what clarification you need the question is as follows I don't understand how to get the answer

Answer 1

• Genetic engineering is a technique which is used to change the genetic makeup of a specific organism by adding new traits to it and finally producing an entirely new variety of organisms.
• The desired trait or gene is ripped from the donor gene and glued to the carrying vector. Thus formed recombinant DNA is transported into the host organism which has to be altered genetically.
• In this process two enzymes Restriction enzymes and ligases are used.
• Restriction enzymes are known as molecular scissors are specifically cut the desired DNA and its carrying vector at precise sites
• Ligase enzyme is known as the Molecular glue affixes these DNA fragments into the carrying vector.
• The process of ligase enzyme for joining the two ends of DNA strands is called ligation.
• There occurs a fusion of phosphodiester bond linking 3’hydroxyl of one nucleotide and 5’phosphate of another.  
• so for the Ligation process, there is a need for adding an insert (vector) to the parent vector
• Ligation Process protocol recommends the addition of 1 molecule of the parent vector to 3 molecules of the insert.

Considering the fact that vector and insert preparation has the same DNA concentration as well as the vector is 4500 base pairs long and insert is 1500 has only base pair. The volume ratio of insert used is as below:

Molar ::

insert mass in ng = molar concentration (insert/vector)x  (insert length in base pairs/vector length in base pairs)

molar concentration = 3:1

length of insert = 1500 bp

length of vector = 4500 bp

insert mass (ng) = 3 x (1500/4500)

=1 ng

volumatic::

Vv = T/ [(Vc x IL x Ir) / (Ic x VL ) + 1]

Vv = volume of vector

Vi = volume of insert

T = Total volume (T= Vv +Vi) of DNA solution

Vc = concentration of vector(μg/l) = Ic = concentration of insert (μg/l) (STATED in question)

IL = insert length in kb = 1.5

VL= Vector length in kb = 4.5

Ir = ratio insert:vector 3:1 = 3/1 = 3

As T is not mentioned, taking the below into consideration:

ligation reaction is usually carried out in a total volume of 10–20 μl at 4–37°C

Assuming T to be 10 μl

Vv = 10 / [(1 x 1.5 x 3) / (1 x 4.5) + 1]

Vv = 5μl = Vector volume

Vi = vector volume = 10μl - 5μl = 5μl

volume ratio for insert : vector = 1:1