Question

All the questions refer to an experimental drug (RDRPkill), designed to treat COVID-19. The drug is a specific inhibitor of the coronavirus specific RNA-directed RNA-polymerase. Although it does not dissolve in pure water, RDRPkill is soluble in 10% (w/v) ethanol or pure dimethyl sulfoxide (DMSO). It is supplied in vials that generally contain 5 mg of powder. Liquid is transferred in and out of these vials by puncturing the rubber seal at the top with a syringe. 5. You are setting up multiple lung cell incubations. These lung cells adhere to and grow on the bottom of a plastic well as a monolayer. This picture shows a 24-well plate. The culture medium, which comes pre-made from a commercial supplier) is added into the wells, onto the cells. Media supplements such as glucose, virus and drug are later added into the culture medium In your experiment you will set up a 24-well plate, with 1 mL of culture medium in each of the wells. In the culture medium you will need to add glucose to a final concentration of 5 mM (to make the lung cells happy). 1,000 COVID-19 viruses will be added to 20 of the wells on each plate (this won’t make the cells so happy). The incubations will take place in the absence or presence of RDRPkill (concentrations ranging from 0 to 50 μM). The glucose is supplied as a 500 mM stock solution. The virus is supplied as a suspension which contains 1 million viruses per μL. You have your RDRPkill stock solution which is 10 mM.

a. How would you set up ONE well containing 1 mL culture medium, 5 mM glucose, 1,000 COVID-19 viruses, and 50 μM RDRPkill?

b. When setting up the entire experiment, your supervisor suggests that you make up a Master Mix containing culture medium, virus and glucose – thus meaning that you only need add about 1 mL of this to each well How would you make up that master mix?

c. Explain, in general terms and perhaps using a picture of the plate or a table, how you would organise the plate.

Answer 1

a) Add 1 ml culture media to the well

Stock Solutions

Glucose = 500mM, Virus = 1 million viruses/ul, RDRPKill = 10 mM

Using M1V1=M2V2, 5mM Glucose is taken as

500 X V1 = 5 X 1000 (1ml of culture = 1000ul)

V1 = 10 ul

Therefore 10 ul of glucose is to be taken from 500mM stock to make a final conc. of 5mM

Now for RDRPKill; M1 = 10mM, M2= 0.05mM (50um=0.05mM), V2 = 1000ul

So 10 x V1 = 0.05 x 1000

V2 = 5ul

Therefore 5ul of RDRPKill is to be taken from 10mM stock

For Virus:

1ul contains 1 million (10^6) viruses

or 1 virus is in (1/10^6) ul

1000 viruses are in (1/10^6) x 1000 ul = 1/1000 = 0.001ul

Therefore from the suspension of 1million/ul 0.001 ul is to be taken. it is difficult to take, better is to dilute the virus suspension in 1;100 and then take accordingly, for example adding 100ul to 1ul of 1million/ul suspension will give 10^4 viruses/ul and then 1/10^4 x 1000 = 0.1ul is to be taken (which can be taken via a micropipette)

b) Make master mix for 20 plates (i.e for 20ml)

Glucose: 500 x V1 = 5 x 20

V1 = 0.2ml = 200ul

RDRD: 10 x V1 = 0.05 x 20

V1 = 0.1ml = 100ul

Virus: 0.001 x 20 = 0.02ul (from the diluted one it is 0.1 x 20 = 2ul)

Culture media to be taken = 20ml x 1000 = 20000 minus (200+100+2) = 19698ul (19.698ml)

Then add 1 ml of the master mix to each of the well.