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VIRUS ASSAYS

VIRUS ASSAYS. Animal Cells in Culture. Types of cells Primary cells isolated from embryonic or newborn animal tissue limited life span normal chromosome number and shape Subject to contact inhibition of growth. That is, they stop growing when they touch each other Continuous cell lines

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VIRUS ASSAYS

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  1. VIRUS ASSAYS

  2. Animal Cells in Culture • Types of cells • Primary cells • isolated from embryonic or newborn animal tissue • limited life span • normal chromosome number and shape • Subject to contact inhibition of growth. That is, they stop growing when they touch each other • Continuous cell lines • Usually derived from tumor tissue or treatment of primary cell culture with a mutagen or tumor virus • Essentially immortal when periodically diluted and fed with appropriate nutrients • Have fragmented and reduplicated chromosomes (aneuploid) • Not subject to contact inhibition

  3. Different types of cell cultures used in virology Primary human fibroblasts Continuous mouse fibroblast cell line (3T3) Continuous human cell line (HeLa)

  4. Many viruses cause cytopathic effects (CPE) when they infect cultured cells Example: Poliovirus infection of cells Uninfected 5.5 hr PI 24 hr PI 8 hr PI

  5. Syncytium formation due to infection with HSV-1

  6. Virus infectivity assays • Plaque Assay • Serial ten-fold dilutions of virus stock made and added to monolayers of cells in duplicate • Cells overlaid with agar to inhibit virus spread • Cells fixed and stained and plaques counted. Each plaque arises from single infectious virus Vf = V0 / D Vf=final conc. V0 =orig. conc. D=dilution factor Titer of original stock is ~4 x 107PFU/ml TMTC 22, 46 3,8

  7. Quantal assays or end-point dilution assays • Serial ten-fold dilutions of virus made • Aliquots of diluted virus injected into animals • The animals are scored for illness or death Dilution # dead 10-2 4/4 10-3 4/4 10-4 2/4 10-5 0/4 LD50= 104

  8. Other ways to “count” virus particles • Hemagglutination assay • Measures ability of virus particles to agglutinate red blood cells (it takes a certain number of particles (~104) to agglutinate red blood cells • Serial dilutions of virus are mixed with a constant amount of RBCs RBC Virus 1/8 dilution 1/4 dilution 1/2 dilution Endpoint=1/4. #HA units=reciprocal of endpoint=4 HA units

  9. HEMAGGLUTINATION ASSAY FOR INFLUENZA VIRUS Dilution of virus Hemagglutinin is inhibited at excess virus concentrations. The virions carry a “receptor destroying enzyme” called neuraminidase

  10. COUNTING VIRIONS IN THE ELECTRON MICROSCOPE • In principle, this involves counting the number of particles in a given volume of the virus preparation • In practice, one adds a known number of latex beads and the number of beads and virions in an EM field is counted. From the ratio of virions to beads and the known concentration of beads, the number of virions in the original suspension can be determined. Example: direct counting of poliovirions

  11. In a preparation of virus not all particles are infectious • The ratio of infectious particles to physical particles is called the particle to infectivity ratio or the specific infectivity • This ratio varies from 1/10 to 1/1000 or greater depending on the virus and the quality of the virus preparation.

  12. Multiplicity of infection (moi)= number of infectious virus particles per cell How much virus must be added to a given number of cells in order (i.e., what is the moi) to make sure that nearly all the cells are infected? When cells are mixed with virus some cells are uninfected and other cells receive one, two, three, etc. particles. The distribution of virus particles is best modeled by the Poisson distribution. P(k) = e-mmk/k! where P(k) is the fraction of cells infected by k virus particles. m is the multiplicity of infection For example at an moi=1; P(0) = e-1 (1)/0! = 0.37; P(1) = e-1 (1)/1! = 0.37; > P(2) = 0.26

  13. At an moi=10;P(0) = e-10 (100)/0! = 0.000045; P(1) = e-10 (10)/1! = 0.00045; > P(2) = 0.9995. This means that 99.95% of cells are infected. This is a common moi used in a one step growth experiments. What is the relationship between moi and LD50? An LD50 means that only half the animals are infected. So according to Poisson P(0)= 0.5. Solve for m. P(0) = e-m or 0.5 = e-m. m= -ln 0.5 = 0.7. 1 LD50 = 0.7 pfu Homework problem due 9/12/00. You have infected cells with a herpesvirus. You are using an Ab to detect an an early antigen by immunofluorescence. At 6 hr post-infection (before any virus is released) you determine that 90% of the cells are positive for the protein. What is the moi? What happens if you waited till virus is released?

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