The burst size experiment involved infecting 2 x 108 cells with phage at an MOI of 0.1. Unadsorbed phage were removed and infected cells were incubated for 90 minutes. 200 infective centers (ICs) were counted initially. After incubation, 150 plaques were counted, corresponding to 3 x 109 PFU/ml. The burst size was calculated as the PFUs released (3 x 109 PFU/ml) divided by the original ICs (2 x 107 ICs/ml), which is 150 PFUs/IC.

1. MEASURING BURST SIZE
DR. S. SIVASANKARA NARAYANI., M.SC., M.PHIL., PH.D., MRSB (UK)

2. INTRO
• The number of mature virus particles released from an infected cell is called the burst
size.
• The plaques arising from this titration represent the number of infected cells (termed
infective centers (ICs)).
• Each infected cell gives rise to one plaque. After a period of time sufficient for cell lysis to
occur, phage are titered again for PFUs.
• The burst size is calculated as PFUs released per IC (PFUs/IC).

3. In a burst size experiment, phage are added to 2 x 108 cells at an moi of 0.1 and allowed to adsorb to
susceptible cells for 20 minutes. Dilution and centrifugation is performed to remove unadsorbed phage.
The pelleted infected cells are resuspended in 10 mL of tryptone broth. At this point, ICs are assayed by
diluting an aliquot of the infected cells as follows
0.1mL 0.1mL
X X 0.1mL
10mL 10mL
The resuspended infected cells are then shaken at 37°C for 90 min. The number of PFUs is then assayed
by diluting an aliquot as follows:
0.1mL 0.1mL 1ml 0.5ml
X X X X 0.1mL
10mL 10mL 10ml 10ml
From the assay of ICs, 200 plaques are counted on the bacterial lawn. From the assay of PFUs
following the 90 min incubation, 150 plaques are counted from the diluted sample
What is the burst size, expressed as PFUs/IC?

4. The number of ICs is determined by dividing the number of plaques obtained following dilution by the
dilution factor. The dilution scheme to obtain ICs is
0.1mL 0.1mL
X X 0.1mL
10mL 10mL
This is equivalent to the expression
(1x 10-2) x (1x 10-2) x (1x 10-1mL) = 1x 10-5mL
The number of ICs is then equal to
200PFUs
1x10-5ml
= 2x107PFUs/ml = 2x107ICs/ml

5. The number of PFUs following the 90 min incubation is obtained by diluting the sample in the following
manner:
0.1mL 0.1mL 1ml 0.5ml
X X X X 0.1mL
10mL 10mL 10ml 10ml
(1x 10-2) x (1x 10-2) x (1x 10-1) x (5x10-2) x (x10-1 ml) = 5x 10-8mL
Since 150 plaques were obtained from this dilution, the concentration of phage in the culture after 90 min
is
150PFUs
5x10-8ml
= 3x109PFU/ml

6. The burst size is then calculated as the concentration of phage following 90 min incubation divided by
the concentration of ICs:
3x109PFUs/ml
2x107 Ics/ml
= 150PFUs/Ic

7. REFERENCE
• Calculations for molecular biology and Biotechnology – A Guide
to mathematics in the laboratory, Second Edition – Frank. H.
Stephenson
29-09-2021
sivasan91@gmail.com
DR.SS

8. 29-09-2021
DR.SS