You are preparing a large scale bacterial virus (phage) lysate. In your refriger

Question

You are preparing a large scale bacterial virus (phage) lysate. In your refrigerator you have 100 ml of a primary suspension of the bacterial virus to be amplified. The titer of the 100 ml lysate is 3x1010 pfu/ml. It will be necessary to add this virus at an multiplicity of infection (moi) of 0.1 to 24 liters of culture cells grown to an OD575 = 0.1 (i.e., about 1x108 cfu/ml). The 24 liters of culture cells are evenly distributed into 12 two liter portions, each shaking in 12 large culture flasks. What volume of the phage suspension is to be added to each of the 12 shaking flasks?

Infection step calculations:
Determining the volume of phage added at moi = 0.1 to each of 12 flasks containing 2 liters of growing cells:

cell titer per ml at OD575 = about 1X 108 cfu/ml
i) determine # cells in 2 liter = 2000ml of cells per flask

ii) determine # of phage at an moi of 0.1 to be added /flask containing 2000 ml of cells, where moi = # phage added/# cells

iii) determine volume of original phage lysate required to be added to infect 2 liter flask of cells
= # phage needed to be added divided by the original phage lysate titer (note, you are given 100 ml of this)

Explanation / Answer

For example, if 2x106 cells is infected by 50 ml of virus with a titer of 108 pfu/ml. The moi will be 0.05*108/2*106 = 2.5.

no.of cells to which the lysate is added= 108 * 24,000

titre of the virus= 3*1010 pfu/ml ( i hope this is 3* 10 to the power 10 given in youe question. if not simply substitute the correct value)

m.o.i= 0.1

therefore the volume of phage added= 108 * 24,000 * 0.1/ 3*1010 pfu/ml = 8.64 microlitres or 8.64* 10 -3 ml

1.) no. of cells = 108* 2000= 216000

2) As shown above 8.64* 0.001 vol. of phage is added, that is there are (3*1010 ) * (8.64*0.001) phages..the answer is 25.92* 107

3)( 8.64* 10 -3 ml * 2000) / 24000

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