Question

Use information in the following paragraph to answer questions 1-3:

An SDSU junior has grown a bacteriophage stock during a BIOL 350, General Microbiology, lab. He determines it has a concentration of 7.5 x 1014 phage/mL. He is given a Salmonella typhimurium host culture with a concentration of 8.1 x 1012 cells/mL and adds a 300 uL aliquot of his bacteriophage stock to 5 mL of Salmonella cells

1. What is the m.o.i. in phage/cell? Answer to one decimal place.

2. What is the total volume of his phage + host cell mixture? Answer to one decimal place.

3. What volume, in microliters, of bacteriophage stock should he add to 3 mL of host cells to have a 0.7 moi? Answer to the nearest whole number.

I have the answers but would like to see how they are solved

Answers. 1. 5.6 phage/cell

2. 5.3 mL

3. 1.701 x 1013 phage should be added to give an moi of 0.7. The volume of stock with this amount of phage is 23 uL.

Answer 1

1) To find out the multiplicity of infection , you need to divide the number of bacteriophages that is used by the number of bacterial cells that has been inoculated with the phage. In this case :

MOI = [(0.3 ml * 7.5 * 1014) / ( 5 * 8.1 * 1012) ] * 100 phage/cell

= 5.6 phage/cell

2) Total volume of phage and host cell mixture is = 300uL + 5000uL = 5300uL = 5.6ml

3) Let the stock of bacteriophage be x uL

0.7 = [ ( x* 7.5 * 1014 ) / (3000uL * 8.1 * 1012 )]  

From here, calculate the value of x.