Dear All, we discovered a small error really at the end of this video. To calculate PFU/ml please multiply 2.9×10^6 because PFU/ml=0.69*TCID50/ml. Thank you.
At the beginning, from A to B, we just dilute 2 times (1:25 --> 1:50), but in the column of "Dilution" in the figure, the dilution from A to B is 10 times. I was wondering if this dilution factor will affect the final result. Could you please explain it to me?
thank you for the video, but the inoculum volume was 100µl and not 8µl in my opinion, which means that 2371373/0.1ml=23713730 or 2.4x10^7 ---> is the TCID50/ml
Hello there, Thanks for your comment. Unfortunately No. We took 8 ul of the virus and added 192 ul of the media. So it is 8 ul of the virus. Please go through the example we provided in the video again for the clarification.
You prepare 8ul virus + 192ul of media for 200ul virus solution but you only add 100ul of them. Does it mean you take 4ul of virus instead of 8ul? So you should divided by 0.004 when calculating TCID50/ml?
Thanks so much for the explanation! I just wonder how do you determine which wells are +/- using colorimetric assay such as MTS or MTT as they would give use absorbance reading of each well. Do I compare the absorbance to the negative control's absorbance and if it's like 50% lower than the negative control then it's a -, else a +? Really hope to hear your suggestion!
Hi Vie Tran, Thank you very much for asking this questions and apologies for responding little late. In fact, you have answered in your question. We would recommend the following article: www.ncbi.nlm.nih.gov/pmc/articles/PMC3550799/
Assuming that the same cell system is used, that the virus forms plaques on those cells, and that no procedures are added which would inhibit plaque formation, 1 ml of virus stock would be expected to have about half of the number of plaque forming units (PFUs) as TCID[50]. This is only an estimate but is based on the rationale that the limiting dilution which would infect 50% of the cell layers challenged would often be expected to initially produce a single plaque in the cell layers which become infected. In some instances, two or more plaques might by chance form, and thus the actual number of PFUs should be determined experimentally. Mathematically, the expected PFUs would be somewhat greater than one-half the TCID[50], since the negative tubes in the TCID[50] represent zero plaque forming units and the positive tubes each represent one or more plaque forming units. A more precise estimate is obtained by applying the Poisson distribution. Where P(o) is the proportion of negative tubes and m is the mean number of infectious units per volume (PFU/ml), P(o) = e(-m). For any titer expressed as a TCID[50], P(o) = 0.5. Thus e(-m) = 0.5 and m = -ln 0.5 which is ~ 0.7. Therefore, one could multiply the TCID[50] titer (per ml) by 0.7 to predict the mean number of PFU/ml. When actually applying such calculations, remember the calculated mean will only be valid if the changes in protocol required to visualize plaques do not alter the expression of infectious virus as compared with expression under conditions employed for TCID[50]. Thus as a working estimate, one can assume material with a TCID[50] of 1x 10(5) TCID[50]/ml will produce 0.7 x 10(5) PFUs/ml. www.lgcstandards-atcc.org/Global/FAQs/4/8/Converting%20TCID50%20to%20plaque%20forming%20units%20PFU-124.aspx?geo_country=be
Dear All, we discovered a small error really at the end of this video. To calculate PFU/ml please multiply 2.9×10^6 because PFU/ml=0.69*TCID50/ml. Thank you.
At the beginning, from A to B, we just dilute 2 times (1:25 --> 1:50), but in the column of "Dilution" in the figure, the dilution from A to B is 10 times. I was wondering if this dilution factor will affect the final result. Could you please explain it to me?
Great explanation. Thank you!
You're welcome!
thank you for the video, but the inoculum volume was 100µl and not 8µl in my opinion, which means that 2371373/0.1ml=23713730 or 2.4x10^7 ---> is the TCID50/ml
Hello there, Thanks for your comment. Unfortunately No. We took 8 ul of the virus and added 192 ul of the media. So it is 8 ul of the virus. Please go through the example we provided in the video again for the clarification.
@@BiologyLectures Yes but we added only half of it to the cells i.e. 100ul. So aren't we inoculating with an initial titer of 4ul?
You prepare 8ul virus + 192ul of media for 200ul virus solution but you only add 100ul of them. Does it mean you take 4ul of virus instead of 8ul? So you should divided by 0.004 when calculating TCID50/ml?
No. since our stock virus solution was perepared using 8 ul of virus plus 192 ul of media, we must divide by 0.008 while calculating TCID50/ml.
Excellent explanation sir. You are awesome ❤️
You
e most welcome 😃
Thanks so much for the explanation! I just wonder how do you determine which wells are +/- using colorimetric assay such as MTS or MTT as they would give use absorbance reading of each well. Do I compare the absorbance to the negative control's absorbance and if it's like 50% lower than the negative control then it's a -, else a +?
Really hope to hear your suggestion!
Hi Vie Tran,
Thank you very much for asking this questions and apologies for responding little late. In fact, you have answered in your question. We would recommend the following article: www.ncbi.nlm.nih.gov/pmc/articles/PMC3550799/
Kindly give lecture video of influenza virus rapid screaming and cultivation method tests and their principle, materials and procedure
Excellent explanation 🤗
Glad it was helpful!
what is the different between neutralization antibody test and TCID50 and IC50
How is 6.375 is equal to 2.37 x 10^6? Could you please explain in step 3?
Why did you multiply 0.69 by 2.37× 10^6 ???. Shouldn't you ve multiplying by 2.9×10^6 ???
You are absolutely correct. Thank you very much for pointing this error out.
Could you please explain the calculation of number 3 again ? Thank you very much.
Why do you multiply by 0.69 to calculate PFU/ml?
Assuming that the same cell system is used, that the virus forms plaques on those cells, and that no procedures are added which would inhibit plaque formation, 1 ml of virus stock would be expected to have about half of the number of plaque forming units (PFUs) as TCID[50]. This is only an estimate but is based on the rationale that the limiting dilution which would infect 50% of the cell layers challenged would often be expected to initially produce a single plaque in the cell layers which become infected. In some instances, two or more plaques might by chance form, and thus the actual number of PFUs should be determined experimentally. Mathematically, the expected PFUs would be somewhat greater than one-half the TCID[50], since the negative tubes in the TCID[50] represent zero plaque forming units and the positive tubes each represent one or more plaque forming units.
A more precise estimate is obtained by applying the Poisson distribution. Where P(o) is the proportion of negative tubes and m is the mean number of infectious units per volume (PFU/ml), P(o) = e(-m). For any titer expressed as a TCID[50], P(o) = 0.5. Thus e(-m) = 0.5 and m = -ln 0.5 which is ~ 0.7. Therefore, one could multiply the TCID[50] titer (per ml) by 0.7 to predict the mean number of PFU/ml. When actually applying such calculations, remember the calculated mean will only be valid if the changes in protocol required to visualize plaques do not alter the expression of infectious virus as compared with expression under conditions employed for TCID[50]. Thus as a working estimate, one can assume material with a TCID[50] of 1x 10(5) TCID[50]/ml will produce 0.7 x 10(5) PFUs/ml.
www.lgcstandards-atcc.org/Global/FAQs/4/8/Converting%20TCID50%20to%20plaque%20forming%20units%20PFU-124.aspx?geo_country=be
-6 plus 0.375 euqals to -5.625
Please put log in front of all those values and you will get value exactly what has been shown in the video. Thanks
No.we don't get it.
could i get whatsapp to learn more from you,and second think couldyou make graph of differenttime interval of virus titer