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

The maximum no. of RFUs collected in the purification procedure
`= 3.05 * 10^5` RFUs

Step-by-step explanation:

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

So, No. of RFUs in G3-500 ml culture
`= 5000 * (500 ml)/(0.2 ml)` RFUs

`= 5000 * 2500` RFUs

`= 1.25 * 10^7` RFUs

Similarly, No. of RFUs in a G3-15ml sample
`= 5000 * (15 ml)/(0.2 ml)` RFUs

`= 5000 * 75` RFUs

`= 3.75 * 10^5` RFUs

The maximum no. of RFUs collected in the purification procedure

`= 3.75 * 10^5 * 80`% RFUs

`= 3.75 * 10^5 * 0.80` RFUs

`= 3.05 * 10^5` RFUs