An SDSU junior has grown a bacteriophage stock during a BIOL 350, General Microb

Question

An SDSU junior has grown a bacteriophage stock during a BIOL 350, General Microbiology, lab. He determines it has a concentration of 3.2 x 1013 phage/mL. He is given a Salmonella typhimurium host culture with a concentration of 6.5 x 1011 cells/mL and adds a 500 uL aliquot of his bacteriophage stock to 3 mL of Salmonella cells. please show work

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.

Explanation / Answer

number of phage = 3.2 x 1013 phage/mL * 500 * 10-3 ml = 1.6 x 1013 phage

number of cells=6.5 x 1011 cells/mL * 3 ml=19.5x 1011 cells

The MOI is determined by simply dividing the number of phage added (ml added x PFU/ml) by the number of bacteria added (ml added x cells/ml)

MOI = number of phage /number of cells=8.2

What is the total volume of his phage + host cell mixture

==0.5ml + 3ml=3.5ml

MOI 1 / MOI 2 = PHAGE V 1 /PHAGE V2

8.2 /0.7 = 0.5 / (0.5 + x )

457 microlitres

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