Question

You have a sample of small single-stranded DNA (ssDNA) molecule that you will use as a primer for Polymerase Chain Rection (PCR). you dissolved the dried ssDNA pellet in Tris-EDTA (TE) buffer to a total volume of 400 ul (microliters)to make the "Stock DNA sample". you now need to determine the concentration (ug / ul) of this primer DNA in solution. to do this you took 4 ul from the "Stock DNA sample" and added these to 996 ul of TE buffer in a spectrophotometric cuvette, making a total volume 1000 ul in a cuvette. next, you determined the Absorbance of 260 nm light (A₂₆₀) for this 1 ml of diluted primer DNA in the cuvette, and ​A₂₆₀ VALUE IS 0.014.

​(i) what is the concentration (ug / ul) of the primer DNA solution in the cuvette (​must show calculations)?

​​(ii) what is the concentration (ug / ul​) of the primer DNA in the original Stock DNA sample (must show calculations)?

​​(iii) what is the ​​total amount (​ug) of DNA in the stock DNA sample tube ​(show calculations)?

(iv) how many ul (microliters) ​of the original "Stock DNA sample" would you need to use in a PCR experiment that required a total of 0.25 ug ​of DNA from this sample ​(show calculations)?

Answer 1

1) what is the concentration of the DNA solution in the cuvette?

An OD of 1 at A260 corresponds 40 ug/ml for single-stranded DNA

0.014 OD---->40 x 0.014 =0.56ug/mL

2) what is the concentration of the primer DNA in the original Stock DNA sample?

Since the dilution factor is = (4 + 996)/4 =250

Concentration of stock DNA sample = 0.56 x 250 = 140 ug/mL = 140ug/1000uL = 0.14ug/uL

3) what is the total amount (ug) of DNA in the stock DNA sample tube?

Stock DNA sample has 400uL volume with concentration 0.14ug/uL

The total DNA is = 400 x 0.14 ug= 56ug

4) how many ul of the original "stock DNA sample" would you need to use in a PCR experiment that required a total of 0.25 ug of DNA from this sample?

Required DNA= 0.25 ug

Concentration of stock DNA sample= 0.14ug/uL

= 0.25/0.14

=0.8uL