Question

4. After primers have been designed and ordered, they are received in very small quantities as a dry solid and need to be resuspended in highly purified water. A volume of highly purified water is added to the dried primer to create the 10uM working solution. How many ul of highly purified water needs to be added to check your discussion forum for your value) nMoles of dried primer in order to create a 10uM working solution? Show all calculations. Hint - you might need the value from question 3 during your calculation. (3 marks) 37.1 nmol → 0.0371 umol 0.0001 umu (ul) = 0.0371 umol 10 uM = 10 umol/L = 0.00001 umul 1x ul-0.0371 umol distilled water - ul. 0.0000lumuluR c=1 0 .000olumlue = 0.0371 umol xlll - 3,710 x ul : volume of distillad water is 3710 je 16 Grade TA Initials
the value that is (check your discussion forum for your value) is 37.1 nmol.

~0.0371umol.

Answer 1

We have 37.1 nmoles of dried primer. We need to make a 10 $\mu M$ solution.

A 10 $\mu M$ solution has 10 micromoles in 1 litre of solution. However, we only have 37.1 nanomoles, therefore, we need to calculate how much water is to be added to make a 10 micromolar solution with 37.1 nanomoles.

Now,

Molarity = No. of Moles/Volume (in Litre)
10 micromolar = 37.1 nanomoles/x L
=> x Litre = 37.1 nanomoles/10 micromolar
=> x = 37.1 * 10^-3 micromoles/10 micromolar = 3.71 * 10^-3 litre

=> x = 3.71 mL

To make a 10 micromolar solution with 37.1 nmoles of primer, we need to add 3.71 mL of pure water.