Question

A student observes 5,000 RFUs in a 200μl aliquot from a G3-500 ml culture. How many total RFUs will be present in a G3-15ml sample? During rGFP purification in lab#4, assume they recover 80% of the fluorescing protein. What would be the maximum total number of RFUs that they collected in the purification procedure? Show your calculations for full credit.

Answer 1

Answer:

The maximum no. of RFUs collected in the purification procedure $= 3.05 \times 10^5$ RFUs

Explanation:

No. of RFUs in a 200 ul (0.2ml) aliquot = 5000 RFUs

So, No. of RFUs in G3-500 ml culture $= 5000 \times \frac{500 ml}{0.2 ml}$ RFUs

                                                               $= 5000 \times 2500$ RFUs  

                                                                $= 1.25 \times 10^7$ RFUs

Similarly, No. of RFUs in a G3-15ml sample $= 5000 \times \frac{15 ml}{0.2 ml}$ RFUs

                                                                          $= 5000 \times 75$ RFUs

                                                                          $= 3.75 \times 10^5$ RFUs

The maximum no. of RFUs collected in the purification procedure

$= 3.75 \times 10^5 \times 80$% RFUs

$= 3.75 \times 10^5 \times 0.80$ RFUs

$= 3.05 \times 10^5$ RFUs