Question

1- Calculate the number of grams needed to make 1 mL of a 1M solution of the oligonucleotide (32nt)

2- You have a 300 μM solution of a primer (45nt) in a volume of 400μL, how many μg of primer do you have total?

3- You received a primer (25nt) which came to you as a powder with a mass of 0.31 mg

a- what concentration of primer do you have when you add 100μL water?

b- How would you make a 5 pmol/μL solution with a total of 100 μL?

PLEASE HELP!

Answer 1

1. Moles ssDNA = mass of ssDNA (g)/molecular weight of ssDNA (g/mol)
molecular weight of Oligonucleotide = ( 32 x average molecular weight of a deoxynucleotide monophosphate) + 18.02 g/mol = 9905.38g/mol
(average molecular weight of a deoxynucleotide monophosphate = 308.97 g/mol, excluding the water molecule removed during polymerization. The 18.02 g/mol accounts for the -OH and -H added back to the ends.)

thats why 1 Molar solution will contain 9905.38g, means 9905.38g in 1000 ml, so 1ml solution will contain 9.90538 g.

2. 300uM = moles/400uL, Moles = 300*400 = 1200uML. since mole = weight(g)/molecular weight.

therefore weight = 1200pML*45*308.97 = 16.684380ug.

3. a. molarity = w/mwt*volume = 0.00031/7712*100uL = 0.000401M. Therefore solution will be 4*10^-4 M

3. b, 38.560 ng of powder will make 5pmol. therefore we have to dissolve 100*38.56ng powder in 100uL.