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# 100 micro molar to 250 micro molar

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### #1 uzalive

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Posted 25 September 2012 - 11:31 AM

1. You are provided with an antibody solution Ab that has a concentration 600 ug/ul. For lab it is necessary to make thefollowing dilutions.

a.  10 ul of 600 ug/ul Ab + 190ul of buffer to make a 1:20 dilution at _________ ug/ul.

b.  20 ul of 1:20 +40 ul of buffer to make a 1:60 dilution at __________ ug/ul.

c.  5ul of 1:60 + 5ul of  buffer to make a ________ dilution at _________ug/ul.

It seems I am missing dome concepts.  Could someone help?

### #2 bob1

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Posted 25 September 2012 - 12:45 PM

This is homework - tell us what you think and we will help correct/expand your answers.

### #3 ascacioc

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Posted 25 September 2012 - 01:45 PM

This is the second one today... as I said: on our times we did our HWs alone or googled for the answers. Not asked directly others. How will you learn?

### #4 uzalive

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Posted 25 September 2012 - 02:08 PM

I did not answer the question at first this is how I did it

a.  10ul times 600ug/ul =6000 ug

add the buffer, which is 190 ul, to 10ul to get 200 total ul.

6000ug/200= 30 ug/ul

I don't understand how do we know that it is 1:20 dilution.

b.  600ug/ul/ 20=30 ug/ul

30 ug/ul times 20ul=600ug/ 60 ul= 10 ug/ul

I don't understand why the it is a 1:60 dilution.  I don't understand why 1:60 dilution was given.  I did not use it in solving the question, but my answer is correct.

Am I missing something.

c.  600ug/ul/60=10ug/ul times 5ul=50 ug

50 ug/10ul= 5ug/ul   I could not figure out the dilution.

### #5 bob1

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Posted 25 September 2012 - 03:07 PM

uzalive, on 25 September 2012 - 02:08 PM, said:

I did not answer the question at first this is how I did it

a.   30 ug/ul

I don't understand how do we know that it is 1:20 dilution.
Correct.  Dilution ratios work like fractions 1:20 is a 1 part in 20 therefore 600/20 = 30ug/ul...

uzalive, on 25 September 2012 - 02:08 PM, said:

b.  10 ug/ul

I don't understand why the it is a 1:60 dilution.  I don't understand why 1:60 dilution was given.  I did not use it in solving the question, but my answer is correct.

Am I missing something.
1:60 of the original solution. 600/60 = 10 ug/ul.

uzalive, on 25 September 2012 - 02:08 PM, said:

c.  600ug/ul/60=10ug/ul times 5ul=50 ug

50 ug/10ul= 5ug/ul   I could not figure out the dilution.
To calculate the ratio do the inverse of what you would do for working out the concentration- so 600ug/ul / 5 ug/ul =...

### #6 El Crazy Xabi

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Posted 25 September 2012 - 08:44 PM

One comment, 1/20 is not 1:20

1/20 -> one part of x IN 20 parts

1:20 -> one part of x AND 20 parts of y

### #7 bob1

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Posted 26 September 2012 - 01:30 AM

Most people would use 1 part with 19 parts diluent to give 1:20, as in the example questions above.

### #8 Astilius

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Posted 26 September 2012 - 03:14 AM

El Crazy Xabi, on 25 September 2012 - 08:44 PM, said:

One comment, 1/20 is not 1:20

1/20 -> one part of x IN 20 parts

1:20 -> one part of x AND 20 parts of y

You would think so, but no.   1:20 when talking about dilutions is 1 part in 20 (or 1 part X + 19 parts Y)
It looks like a mathematical ratio but it's never used like one.
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### #9 Trof

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Posted 26 September 2012 - 06:22 AM

1:20 is a confusing notation and should be abandoned. Because you're never really sure if it's a dilution or a ratio of solute and solvent.
Why can't just people say '20 times diluted' or '1 in 20 dilution'?
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### #10 El Crazy Xabi

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Posted 27 September 2012 - 07:53 PM

Astilius, on 26 September 2012 - 03:14 AM, said:

El Crazy Xabi, on 25 September 2012 - 08:44 PM, said:

One comment, 1/20 is not 1:20

1/20 -> one part of x IN 20 parts

1:20 -> one part of x AND 20 parts of y

You would think so, but no.   1:20 when talking about dilutions is 1 part in 20 (or 1 part X + 19 parts Y)
It looks like a mathematical ratio but it's never used like one.

Maybe you don't, but I use both notations. It's not the first time finding mixes with 70:25:5 of x,y and z and similar. That's why I keep the ration and fraction notation separated. Also because I learned it in that way. It can also be a cultural stuff like the number of continents or the number of lives of cats

Textbook: http://www.hofstra.e...d_dilutions.pdf
see page 29 (It's not what I think, it's also in a book)

There is an old thread about and each people use one or another
http://www.protocol-...osts/31255.html

### #11 Astilius

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Posted 28 September 2012 - 04:57 AM

Interesting.   Though, component ratios (e.g. 1:2:1:5) are not the same as dilutions.
The obvious difficulty lies if it's a two component ratio, the resulting notation could easily be ambiguious.

Looking at everyone's favourite, Wikipedia, I see this:  http://en.wikipedia....#Dilution_ratio

But yes, there is often much confusion with this, but it should be clear, either by declaration or context if a ratio or dilution were involved.

I also think Trof is onto something -- I have seen many, many people confused over this and the notation isn't helping.   Someone want to write up a letter and submit a suggestion of new notation to Nature?

Edited by Astilius, 28 September 2012 - 04:59 AM.

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### #12 bob1

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Posted 28 September 2012 - 03:00 PM

There's also the point that after a certain level of dilution, it really doesn't matter whether you are using the fraction or ratio.  For example, 1:5 (one part and 5 parts) is 0.166, whereas 1/5 is 0.2, sure biggish difference, but when you get out to say 1:100 (0.0099) vs 1/100 (0.01) really, what's the difference?

### #13 Trof

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Posted 28 September 2012 - 10:31 PM

Difference is when people don't understand dilution factor notation at all, they may actually use it wrong when it matters.
Besides, for some applications it does matter even then, for example if you're doing standard curve by serial dilution, you better have the exactly correct ratio, otherwise your results would be screwed.
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### #14 uzalive

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Posted 29 September 2012 - 10:02 AM

You have an antibody solution that has a concentration of 600 ug/ul.  You want to make th following dilutions.

10 ul of 600 ug/ul Ab +190 ul of buffer to make a 1:20 dilution at ________ ug/ul.

This is how I did it.

1.  600 ug/ul times 10 ul= 6000 ug

2.  6000 ug/200ul= 30ug/ul

I know that this answer is correct because I checked.  I don't understand how this gives you a 1:20 dilution.  Why is the 1:20 given?  I did not use it in my calculations.  It seems that I am missing some concept.  Can someone clarify this.  I would have thought that you would divide 30 ug/ul by 20.

Uzalive

### #15 pito

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Posted 29 September 2012 - 10:07 AM

if you go from 6000µg to 30µg, then you diluted it 20 times.
6000/20 = 30

and where does the 1:20 comes from (without making the math of the antibody) 10 µl and 190 µl , is 10 µl in 200 µl in total, its 1:20 , 1 part of 20 parts in total or 10 parts of 200 parts in total.
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