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There have been 7 items by .Bioforum. (Search limited from 06-August 19)


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#109847 ELISA- Limit of Detection(LOD) - just need an idea PLEASE

Posted by .Bioforum. on 14 May 2011 - 05:53 PM in General Biology Discussion

Ok, I know limit of detection (LOD) is the lowest concentration at which any substance can be detected....blah blah BUT when it comes to determining the LOD from a calibration curve, I always get it wrong...i think I'm missing a way to determine the LOD here..

Eg. in the picture below, I'd say LOD is 0.001% for goat and for buffalo and sheep i'd say it's 0.001% but this is wrong....as the right answer is 0.01% in goat and -0.1% in buffalo and sheep :(

Posted Image

(http://img641.images...3180/elisao.jpg)

Could you kindly suggest what I need to look at...anything in x-values?....currently i look at the lowest y-value and read the corresponding x-value where the line touches!

PLEASE HELP!



#103902 Drawing this Graph with wide values?!

Posted by .Bioforum. on 16 March 2011 - 05:53 PM in General Biology Discussion

My values are:-

x-axis(time): 0, 1, 3, 7
y-axis(colonies): 2890000, 8910000, 13000, 450, 50, 1700
(there are more of these but this includes the highest and lowest values!)

I can't figure out how to contruct the divisions on y-axis!! Please help!

(I can't do any log graphs for this. So, I'm currently doing this in Microsoft Excel and its producing a useless huge graph where i can't see the values! The numbers are spread so hugely that the graph ignores small values. But I'm sure there's a way to construct it nicely. Hence, require your help please.)



#103585 CALCULATING no of bacteria in original sample...after dilutions? [Very SHORT Que

Posted by .Bioforum. on 14 March 2011 - 11:07 AM in -Microbiology and Virology-

I am confused again.

For Plate 1, which was 10^3 dilution: I found 176 counts. So, I was told that colony forming units = (counts multiplied by dilution factor) divided by amount transferred...so here:
total counts = 176
dilution factor = 3 (from 10^3)
amount transferred in each dilution = 0.1ml

Hence, the no of organism/ml in original sample was= (176 x 3)/0.1= 5.28 x 10^3

I can see why you have multipled by 1000, which is because we made 3 serial dilution (1/10, 1/100, 1/1000)...BUT i don't get it which I should follow...

The whole thing with this exercise is that we need to compare the no of organisms in the sample over three weeks. We are incubating a sample for 3 weeks where, every week we take the sample and plate it and count....then compare the no of organisms in that sample for that week with the no of organism in the next week..so we look at the change in number of bacteria over three weeks. But I haven't been able to work out the number of organisms from these plates I have cultured!

Is colony forming units the same as number of bacteria in that sample?

I am very grateful to you as you are helping me with something that I should and need to understand and apply all my life. Really thanks a lot for this.


A 'Thank you' is such a small word for the favour you have just done...you explained me the whole thing so clearly...my tutors should quit their jobs and stay at home...i'm a final year student doing biomedical sciences and i was embarrassed that i cudn't even do this calculation :(

Yeah, I meant 10^3 on 5.28 x 10^3 and not 103 :)

So, about my plates:

Only plate 2 (45) and 1 (176) are ok.
So,
in Plate 1: there were 176 counts meaning, the no of organism/ml in original sample was= (176 x 3)/0.1= 5.28 x 10^3
in Plate 2: there were 45 counts meaning, the no of organism/ml in original sample was= (45 x 4)/0.1= 1.80 x 10^3

So, if I average 5.28 x 10^3 and 1.80 x 10^3 , I would get 3.54 x 10^3.
So am I right saying there were 3.54 x 10^3 (or 3540) bacteria in my original sample?


Once again, thanks for the detailed information..i was i had you as my tutor!

I allready did the calculation for your plates.. check my post again.


I dont understand why you multiply with 3 or 4 ...

If you have 176 bacteria in your dilution (10^-3 or 1/1000) then you have 176 x 1000 bacteria in the non diluted sample.

I really dont understand how you calculate it with the 3 or the 4 ...?




#103478 Finding GLUCOSE concentration ...Values for dilutions to make standard graph?

Posted by .Bioforum. on 13 March 2011 - 07:44 PM in General Biology Discussion

Lab exercise on Finding Glucose Concentration:

We are given a 20 mM(micro-molar) solution of Glucose. Our tutor asked to chose several dilutions to make a standard graph of concentrations(x-axis) versus absorbance (y-axis) so we could compare this with another calibration graph when we try to find concentration of glucose in a sample of blood.

For standard curve, what we did was we got dilutions like this and measured the absorbance:
http://img683.imageshack.us/i/glucosetable.jpg/

I ABSOLUTELY DON'T UNDERSTAND THE LOGIC HERE...WHY IS THERE 0.1, 0.2 and why is this value going up by 0.1 and why until 1.0?? And why is a value of 0.4ml glucose corresponding to 8mM Glucose Solution Concentration?

How is this working...? (i know the whole thing to calculate the concentrate from a sample using absorbance from graph and calculating the concentration...but I just don't get the whole idea of this dilution and how they link with glucose concentration...i just don't get these figures for glucose and water concentration and why they need to make up 1ml and why not 2ml or 3ml...?!?!?!

Any help in explaining this would be a massive favour...thanks a lot in advance.

[If the table doesn't load, its here:

Tube no. Glucose Solution Conc. (mM) Glucose (ml) Water (ml) Absorbance
1 0 0.0 1.0 0.020
2 2 0.1 0.9 0.035
3 4 0.2 0.8 0.053
4 6 0.3 0.7 0.067
5 8 0.4 0.7 0.088
6 10 0.5 0.6 0.096
7 12 0.6 0.5 0.112
8 14 0.7 0.4 0.143
9 16 0.8 0.3 0.164
10 18 0.9 0.2 0.167
11 20 1.0 0.1 0.182



#103473 CALCULATING no of bacteria in original sample...after dilutions? [Very SHORT Que

Posted by .Bioforum. on 13 March 2011 - 05:24 PM in -Microbiology and Virology-

A 'Thank you' is such a small word for the favour you have just done...you explained me the whole thing so clearly...my tutors should quit their jobs and stay at home...i'm a final year student doing biomedical sciences and i was embarrassed that i cudn't even do this calculation :(

Yeah, I meant 10^3 on 5.28 x 10^3 and not 103 :)

So, about my plates:

Only plate 2 (45) and 1 (176) are ok.

So,
in Plate 1: there were 176 counts meaning, the no of organism/ml in original sample was= (176 x 3)/0.1= 5.28 x 10^3
in Plate 2: there were 45 counts meaning, the no of organism/ml in original sample was= (45 x 4)/0.1= 1.80 x 10^3

So, if I average 5.28 x 10^3 and 1.80 x 10^3 , I would get 3.54 x 10^3.

So am I right saying there were 3.54 x 10^3 (or 3540) bacteria in my original sample?


Once again, thanks for the detailed information..i was i had you as my tutor!



#103417 CALCULATING no of bacteria in original sample...after dilutions? [Very SHORT Que

Posted by .Bioforum. on 12 March 2011 - 06:16 PM in -Microbiology and Virology-

We transferred 0.1ml from original sample and diluted with peptone and got various dilutions. then did colony counts and got these results:

No. of bacteria / ml:

in dilution 10-^3 = 176 (so no of organism/ml is (176 x 3)/0.1= 5.28 x 103)
in dilution 10^-4 = 45 (so no of organism/ml is (45 x 4)/0.1= 1.80 x 103)
in dilution 10^-5 = 3 (so no of organism/ml is (3 x 5)/0.1= 1.50 x 102)

BUT what was the number of bacteria in original sample??

Do I average the no of organism/ml from the 3 dilutions and would that be the total no of bacteria in the original sample?

Please help!



#95784 Measuring/detecting C3 Nephritic Factor?!

Posted by .Bioforum. on 22 December 2010 - 06:30 PM in Immunology

Where can I get schedules to measuring/detect C3 Nephritic Factor?

Can't find description of any methods to serve the purpose, so any help/suggestions would be brilliant. thanks.




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