[Histonet] equation problem PLEASE help

[Anthony Reilly]

Hello Eva
 
Unfortunately some of the script did not transfer in the original mail and 10-9 should read 10 to the minus 9 as the superscript was lost.  The same for 10-8.  Also there should have been a division line between 1.0 and 0.01064 in the calculation.  I hope this helps.
 
regards
 
Tony Reilly

Chief Scientist
Anatomical Pathology
Pathology Queensland
Level 1, Building 15
Princess Alexandra Hospital
Ipswich Rd, 
Woolloongabba Q 4102
Australia
Ph: 07 32402412
Fax:07 32402930
tony_reilly <@t> health.qld.gov.au

>>> "Anthony Reilly" <Tony_Reilly <@t> health.qld.gov.au> 8/01/2009 12:39 pm >>>
Hello Eva

Put simply


1nmol is 10-9 moles 
Therefore 10nmol is 10-8 moles
1 mole = 1,064g
Therefore 10-8 moles = 0.00001064g    or     0.01064mg
Therefore your working solution of 10nmol/ml = 0.01064mg/ml

Your stock solution is 1mg/ml
Therefore your dilution factor is             1.0                   =          94 (93.984962 exactly)
                                                            0.01064

Therefore you can add 1ml stock to 93ml of diluent to get your desired concentration..

I hope this is useful

regards



Tony Reilly

Chief Scientist
Anatomical Pathology
Pathology Queensland
Level 1, Building 15
Princess Alexandra Hospital
Ipswich Rd, 
Woolloongabba Q 4102
Australia
Ph: 07 32402412
Fax:07 32402930
tony_reilly <@t> health.qld.gov.au 


>>> <eca9 <@t> georgetown.edu> 8/01/2009 7:46 am >>>
Good afternoon,
I am hoping someone out there will take pity on a "mathematically" challenged individual such as myself. I have been trying for hours to wrap my head around this equation and am now getting to the point where I am more confused than ever. Please help me.
The protocol calls for 10nmol of a substance per 1ml needed. It comes in a 1mg/ml solution and has a molecular weight of 1064g/mol. How do I do this? If for example I needed 2ml of the solution...
The clearer the explanation the better. I really want to understand the calculation not just have an answer. PLEASE HELP ME.
Thank you,
Eva

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