a) An aliquot of the 24 liter lysate was diluted 10-2, 10-2, 10-2, 10-2 and 0.1
ID: 179620 • Letter: A
Question
a)An aliquot of the 24 liter lysate was diluted 10-2, 10-2, 10-2, 10-2 and 0.1 ml was plated in triplicate on sensitive cells using the agar overlay technique. Upon incubation, 14, 15, and 16 plaques appeared on the plates. What is the titer of phage virus particles in the 24 liter lysate? How many viral particles [total pfu] are contained in the 24 liter lysate?
Part a) calculations: tittering -- gives you pfu/ml; then calculate # pfu present in 24 liters
= total viable phage particles in the 24 liters of phage lysate
b)The virus in the 24 liter lysate are purified by a differential centrifugation technique. First the cell debris and unlysed cells are removed by a low speed spin. Second, PEG (polyethylene glycol) is added and the virus particles are pelleted by spinning the first supernatant at a higher speed. The PEG-virus pellet is resuspended in buffer to 30 ml and titered. 1.0x1013 pfu/ml are observed. What is the percent phage recovery in PEG pelleted cells?
Explanation / Answer
a) here you are asked to calculate the viral titter. viral titter is a quantitative measurement of the biological activity of a virus and it is expressed as plaque forming units "pfu" per ml. After you find the no. of pfu s in 1 ml, you have to multiply it with 24×1000 ml to find the total...
Use the following equation:
Average no. of plaques ÷ (dilution factor) (volume of diluted virus added) = pfu/ml
First find the average no. of plaques= (14+15+16) ÷ 3 = 15
Dilution factor is 10-2 and vol. of diluted virus added is 0.1 ml
The solution wil be = 15÷ (10-2) (0.1) = 15,000 pfu/ml
Therefore in 24 litres it is = 15000 × 24000 = 360,000,000
b) what you have to understand here is that during the purification process some of the cells will be lost. In the question it clearly states that unlysed cells are removed. secondly PEG is added to form a virus pellet. In this case too all the viruses might not form the pellet. Therefore all the viruses will not be recovered.
The no. of pfu/ml is 1013 (i hope the number is correct. the printed number is not clear. if it is not simply substitute the correct value intead of 1013)
Then the pfu's in 30 ml is = 1013 × 30 = 30390
Therefore percentage recovery is = (30390) ÷ (360,000,000) × 100% = 8.44 × 10-3 = 0.00844
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