@Arishtat you are correct to say that the volumes are 300 uL because the dilution is always calculated with the total volume (30 + 270) then that is intern used for the stock
@Arishtat Yes, so that is why you have to think ahead. For example if I was using this dilution series for a standard curve and I need triplicates of 100 uL, it wouldn't work.
Thank you Greg. Clear explanation! I have one question (may be it sounds stupid). How does well 1-7 have 300 ul final volume. We started with 270 ul and added 30 ul to well 1. But, after that we pipetted out 30 ul to well 2 from well 1. Then, well 1 should have 270 ul as before, right ? Please enlighten me.
Thanks for the fantastic help with the video but I hope you don't mind it's I just have a few questions to ask: Does the dilution factor form a geometric sequence in the form 10 x 10^n-1 for each beaker you diluted? Secondly, shouldn't the last dilution be 0.000001ug/uL (1x10^-6) because you started from 100ug/uL and diluted it by 10x for 8 turns? If you could provide answers to this then that would be fantastic! Thank you :)
Hello nice video. I understand most of it apart from where u talked about the other way of doing it instead of pippeting it out using the serial dilution. Then u mention 1ul of stock/100 L H20. I was wondering if u could explain how u derived those numbers. Thanks
We're supposed to find out that if I take 5mL from a 10mL sample substance, mix it into 10mL of water (the outcome would be 1:3), then take 5mL of that 1:3 solution and mix it with the original sample substance, then which one would have a higher concentration based on it's original substance?
So you multiplied the dilution factor of 1/10 by the original stock to get 10 microliter/ mL in the first box? And then proceeded to multiply 10 by 1/10 to get the second box? Also, did you change tips after each dilution? thnx.
what if the amount of sample you will serial transfer from one well to the next well is not consistent? e.g from first well is 30 micro-liter then the next is 20 micro-liter how will you compute for the final diluting factor?
Hi there. I want to dilute sodium hypochlorite 5%/1litre to 1.5%/1litre but I m struggling with the maths. Could you please write down your workings I think the answer is 4 parts water and 1part sodium hypochlorite Kind regards Steve
This really helped me a lot to understand the topic! I appreciate the application. However, I just noticed a little mistake of what you did in the illustration which confused the viewers. You said that the final volume (which is the volume of the last test tube) was 300 uL yet you drew a line which was misunderstood by your viewers because they thought all the test tubes have 300 uL when the rest has 270 uL except the last test tube. Anyway, great tutorial!!! Please post more.
Can you make a video showing how to calculate it going backwards. For example, what if your stock solution was an "Unknown concentration" and the concentration of the final test tube was 10. How would you do the math to compute the concentration of the substance inside the stock solution?
In this example, the dilution factor is 10. All you have to do is multiply your final concentration by the dilution factor (10) to calculate the concentration of the preceding solution. You would repeat this step, calculating backwards until you finally get to your stock solution.
Hey I have a question for you. This may be silly. I know you initially placed 30ul into the first mircrotube. but then you took out 30 ul. why is the total volume still 300 ul? wouldn't it be 270? and the last tube contain 300ul?
-I think your issue is in calculating concentration. Which volume do you use? 270 or 300. If i'm right keep reading :) -Yes. the new volume for that container is 270ul. However, when you made the mixture the total volume was 300. Removing volume after the fact doesn't change the concentration. If you make a new mixture - freeze. calculate concentration with that. -analogy: When you pour a glass of orange juice, the orange juice in the jug has the same concentration there is just less volume. you didn't dilute the orange juice. -hopefully that helps :)
This is a satisfactory explanation. When any tube was first diluted after addition of 30 uL from the previous tube and was for a moment at 300 ul. After the then diluted product was removed from the 300 uL the final volume was 270 ul. At both 300 ul level after the initial 30 was added AND at 270 AFTER the 30 was removed for the next tube the concentration remain the same. The exact volumes after dilutions are complete might be useful for downstream analyses, but does not affect concentrations. You could take 30 from the very last dilution and discard it, and then all tubes would have appropriate concentrations and all would have 270 ul of sample remaining.
I do have unclear idea on this. when we serially dilute each tube with the same amount, but the final tube becomes 10ml and the other will be 9ml. so how can we make the last tube 9ml?
That does not really matter because the concentration does not change. If you reduce the last volume to 270microL it is still the same concentration. So why waste 30microL?
OK so basically that means that in serial dilution it doesn't matter for the concentrations of all the tubes to be of the same concentration?!! that because the volume of water remains the same ....ryte?
Can somebody please explain why the concentration of tube 1 is 10ug/uL and tube tube concentration is 1ug/uL. Shouldn’t the concentration of tube 1 be 0.1ug/uL??
how total volume of solution in each tube became 300micro liter you added 30 micro liter to 270 micro liter then subsequently you removed 30 from each so it should become 270 again but you have written 300
i beleive that what you have done is wrong coz 1- final volum is 270 it is not 300 since you have subsequently removed 30 from the first tube and diluted it in each tube and you should have throw it after last tube. 2- to do 1/10 dilution of total volum 300 you should add 33 of stock and 333 of diluent (solut/slolvent+solute=33/300+33 =0.099 =approx 1/10=0.1) 3- then you can add 300 of diluent and then 33 of stock and subsquent dilution so the total volum will be 300 in each tube 4- what you have done is good for one step dilution not serial! 1/10 = 300/10 = 30 - 300 =270+30 which is coreect for one step dilution.
best video I ever see. Really I saw a lot of viideos but could not understand dilution which you teach me in one minute. Awesome. thanks a lot
You have an extra zero on your 7th tube as a result you also have an extra zero in your 8th tube . You went from 3 zeros to five
Best teacher for the lab calculation lesson . Thanks a lot Sir .
my right ear feels lonely. besides that great video!
best explanation I've seen so far. Thanks
Greate series. I'm a little confused about the total volumes, though. In the end, doesn't well 8 contain a total of 300µl and all other wells 270µl?
@Arishtat you are correct to say that the volumes are 300 uL because the dilution is always calculated with the total volume (30 + 270) then that is intern used for the stock
Beautifully explained---clear, and easy to follow. Excellent! Thank you! :)
I have a practical test tomorrow, and this really helped! Thanks.
This was very helpful! Thank you so much!!
0.0001->0.000001?Isn't it 0.00001
The Hh yes its a mistake
@Arishtat Yes, so that is why you have to think ahead. For example if I was using this dilution series for a standard curve and I need triplicates of 100 uL, it wouldn't work.
how do you find the concentration of each well?
is it possible to reverse the equation to produced concentrated solutions?
How would you determine the concentration now?
Thank you very much. It helped me with my Research Project. I needed to construct a calibration curve for my project.
I dont get the 1microliter/100L at the end part. How did you get that?
You probably don't need the explanation now, but.... 1micoL/100L is equal to (1x10^-6)/100L= 0.00000001microG/microL or 100,000,000x dilution.
Sir, in stock the conc is 100 microgm/ 1micro liter. From 1microliter how u take 30 microliter?
How did you know how many tubes to use? especially if you aren't given wells to pipette in.
What is the concentration for each of the dilution
Can you explain what happens if you were to use different dilutions from well 2 on?
How to calculate the concentration of dye present in each dilution?
Thank you Greg. Clear explanation! I have one question (may be it sounds stupid). How does well 1-7 have 300 ul final volume. We started with 270 ul and added 30 ul to well 1. But, after that we pipetted out 30 ul to well 2 from well 1. Then, well 1 should have 270 ul as before, right ? Please enlighten me.
you're right
+Samiksha Ghimire Once we make require concentration , then we can take any volume from the vial, it does not matter...
but during measurment of turbidity it will
absolutely right
Thanks for the fantastic help with the video but I hope you don't mind it's I just have a few questions to ask:
Does the dilution factor form a geometric sequence in the form 10 x 10^n-1 for each beaker you diluted?
Secondly, shouldn't the last dilution be 0.000001ug/uL (1x10^-6) because you started from 100ug/uL and diluted it by 10x for 8 turns?
If you could provide answers to this then that would be fantastic!
Thank you :)
Hello nice video. I understand most of it apart from where u talked about the other way of doing it instead of pippeting it out using the serial dilution. Then u mention 1ul of stock/100 L H20. I was wondering if u could explain how u derived those numbers. Thanks
We're supposed to find out that if I take 5mL from a 10mL sample substance, mix it into 10mL of water (the outcome would be 1:3), then take 5mL of that 1:3 solution and mix it with the original sample substance, then which one would have a higher concentration based on it's original substance?
the best explanation!!! you are GOD!!! I am Spanish and you explain better than my stupid teacher jajajaja xx .
If solutions contain 100ug/ul and then you take 30 ul from it so how much amount of substance in 30 ul?
Formula for Dilution factor is Final volume by Aliquot volume.
This is correct or not please let me know.
So you multiplied the dilution factor of 1/10 by the original stock to get 10 microliter/ mL in the first box? And then proceeded to multiply 10 by 1/10 to get the second box? Also, did you change tips after each dilution? thnx.
There's no need to change tips. You are literally transferring the same solution from one beaker to the next.
what if the amount of sample you will serial transfer from one well to the next well is not consistent? e.g from first well is 30 micro-liter then the next is 20 micro-liter how will you compute for the final diluting factor?
Then it's not a serial dilution.
Hi there.
I want to dilute sodium hypochlorite 5%/1litre to 1.5%/1litre but I m struggling with the maths.
Could you please write down your workings
I think the answer is 4 parts water and 1part sodium hypochlorite
Kind regards
Steve
why do you jump from microlitres to ml, say at 4:38?
If you watch for 10 more seconds he fixes it to Micro liter
Final volume is 270 microliter. 30:270=1:9. Isn't it 9 times? not net times?
The total volume of solution is 300 um once the 30 um of stock solution is added to the 270 um of water. 30:300 = 1:10
thank you for this video !
I have a question for you. Is stock undiluted solution...?plz reply T.T
This really helped me a lot to understand the topic! I appreciate the application. However, I just noticed a little mistake of what you did in the illustration which confused the viewers. You said that the final volume (which is the volume of the last test tube) was 300 uL yet you drew a line which was misunderstood by your viewers because they thought all the test tubes have 300 uL when the rest has 270 uL except the last test tube. Anyway, great tutorial!!! Please post more.
Why cant u just discard the extra 30ul from the last tube?
Can you make a video showing how to calculate it going backwards. For example, what if your stock solution was an "Unknown concentration" and the concentration of the final test tube was 10. How would you do the math to compute the concentration of the substance inside the stock solution?
In this example, the dilution factor is 10. All you have to do is multiply your final concentration by the dilution factor (10) to calculate the concentration of the preceding solution. You would repeat this step, calculating backwards until you finally get to your stock solution.
If I have the following solution: 3x capture antigen (dilute with 1x PBS). Does this mean that I have to dilute it or is it already diluted?
Hey I have a question for you. This may be silly. I know you initially placed 30ul into the first mircrotube. but then you took out 30 ul. why is the total volume still 300 ul? wouldn't it be 270? and the last tube contain 300ul?
I had this same question as I was watching.
-I think your issue is in calculating concentration. Which volume do you use? 270 or 300. If i'm right keep reading :)
-Yes. the new volume for that container is 270ul. However, when you made the mixture the total volume was 300. Removing volume after the fact doesn't change the concentration. If you make a new mixture - freeze. calculate concentration with that.
-analogy: When you pour a glass of orange juice, the orange juice in the jug has the same concentration there is just less volume. you didn't dilute the orange juice.
-hopefully that helps :)
thanks it does,
This is a satisfactory explanation. When any tube was first diluted after addition of 30 uL from the previous tube and was for a moment at 300 ul. After the then diluted product was removed from the 300 uL the final volume was 270 ul. At both 300 ul level after the initial 30 was added AND at 270 AFTER the 30 was removed for the next tube the concentration remain the same. The exact volumes after dilutions are complete might be useful for downstream analyses, but does not affect concentrations. You could take 30 from the very last dilution and discard it, and then all tubes would have appropriate concentrations and all would have 270 ul of sample remaining.
This was helpful! Thank you so much for this video
I do have unclear idea on this. when we serially dilute each tube with the same amount, but the final tube becomes 10ml and the other will be 9ml. so how can we make the last tube 9ml?
just take out the 1ml of the final tube
Hello..excellent videos more examples PLEASE..
I believe you made a mistake. you would need to get rid of the 30microL from the last tube to keep the volume 270ml in all tubes.
That does not really matter because the concentration does not change. If you reduce the last volume to 270microL it is still the same concentration. So why waste 30microL?
but in case of elisa .....concentration matters a lot
OK so basically that means that in serial dilution it doesn't matter for the concentrations of all the tubes to be of the same concentration?!! that because the volume of water remains the same ....ryte?
Can somebody please explain why the concentration of tube 1 is 10ug/uL and tube tube concentration is 1ug/uL. Shouldn’t the concentration of tube 1 be 0.1ug/uL??
why would it be 1 microlitres of stock to 100 L of H20 why isn't is 100 microlitres
Thanks. I get the solution to dilute
So the DF is the 30=300/Vi? And that’s how he got the 1/10? Sense it’s volume it would be DF=Vf/Vi And that’s how he got the DF as 1/10?
THANK YOU FOR THIS :) I totally understand now!!!
how total volume of solution in each tube became 300micro liter
you added 30 micro liter to 270 micro liter then subsequently you removed 30 from each so it should become 270 again but you have written 300
300 um was the final volume of the new solution produced. You calculate the concentration of that solution, then dilute it to make the next solution.
Thanks very helpful. More. Samples ..
Serial dilution for bacteria was given by..................... Pls give me answer as soon as
Great stuff, thanks!
Your right, thanks. I will annotate that later. Good catch.
i mean like I kinda get it, but uggh the whole like 10 to the negative exponent thingy still confuses me, you didn't touch base on that :(
Thank you
No need to change tips in between wells here (I think).
yh, you can keep the same tip in this case
The final concentration is 1uL in 10L, not 100L
wow! I get it now! thank you!
Excellent,explanation,thank,u,sir,
The seventh dilution is wrong.
You changed units from micro liters to milli liters when you were writing out the dilution, good explanation though.
You're awesome! thanks!
thank you for the upload
THANKYOU
Thank you very much.
Science IV - Well done!
i beleive that what you have done is wrong coz
1- final volum is 270 it is not 300 since you have subsequently removed 30 from the first tube and diluted it in each tube and you should have throw it after last tube.
2- to do 1/10 dilution of total volum 300 you should add 33 of stock and 333 of diluent (solut/slolvent+solute=33/300+33 =0.099 =approx 1/10=0.1)
3- then you can add 300 of diluent and then 33 of stock and subsquent dilution so the total volum will be 300 in each tube
4- what you have done is good for one step dilution not serial! 1/10 = 300/10 = 30 - 300 =270+30 which is coreect for one step dilution.
Very helpful,thanks :)
GREAT VID
but each time you take 30 you will not have at the end 300 in all the tube!
thanks
Excelente!!! Gracias
U THE BEST...THANK U
Thank you!
You added an extra zero in your 7th tube. The final stock dilution should be 0.000001 ug/uL
nice video
Thank u so much !!!
merci de france
you added an extra zero going from well 6 to 7
THANKS!!
My left ear enjoyed that
👍💛
اهلين ياحلوين
whoops, spoke to soon i see you corrected it
There are so many mistakes in this video... all you're doing is confusing people. You should just take this video down.
I can see why you find it confusing (maybe the fact that he's adding the 270microL first?) but the dilutions are correct.
Also, maybe instead of just saying that there are mistakes, how about just pointing them out?
Thank you!
Thank you kind sir.