Login

Join for Free!
118886 members


Molar extinction coefficient troubles

Discussion of all aspects of biological molecules, biochemical processes and laboratory procedures in the field.

Moderator: BioTeam

Molar extinction coefficient troubles

Postby Teleprion » Wed Aug 04, 2010 3:26 pm

Hi, i've recieved a piece of work to do with beer's law, i've tried to make some headway with it, but the results i get are always either far too large to be probable or far too small.

The question is stated as follows:

You are attempting to determine the amount of haemoglobin present in human blood. To do this, you take 0.005ml of blood and and water to it till the whole volume is 1ml. This causes the cells to lyse releasing theie haemoglobin. You now measure absorbance (540nm) of this solution in a standard 1ml cuvette and obtain a reading of 0.46

The Molar Extinction Coefficient of haemoglobin is 41,700 lotres per mole per cm and the molecular weight of haemoglobin is 64,500.

Using your absorbance measurements calculate how many g of haemoglobin are present in a litre of blood.

The numbers i have had go from 0.0342 g to over a few thousand kilos, please help :S

Thanks in advance
Teleprion
Garter
Garter
 
Posts: 4
Joined: Wed Aug 04, 2010 3:17 pm

Postby JackBean » Wed Aug 04, 2010 5:56 pm

how can it be different? Just put it into the formule, not? How do you calculate it? :roll:
http://www.biolib.cz/en/main/

Cis or trans? That's what matters.
User avatar
JackBean
Inland Taipan
Inland Taipan
 
Posts: 5690
Joined: Mon Sep 14, 2009 7:12 pm

Postby Teleprion » Thu Aug 05, 2010 9:31 am

Well i use the formula A = ε l c and re-arrange it to: C = A/ε.l

And so:

0.49/41,700=1.175x10^-5

but the molar co-efficient is for a litre, and we only used 0.005 which goes into a litre 200,000,000 times

so i multiply that number to make

1.175x10^-5 x 200,000,000 = 2350.12

But here i get stuck and realise i did something wrong as it couldnt be 2350.12 moles per litre of haemoglobin.

:S could you tell me where i am going wrong please?
Teleprion
Garter
Garter
 
Posts: 4
Joined: Wed Aug 04, 2010 3:17 pm


Re:

Postby JackBean » Thu Aug 05, 2010 11:06 am

Teleprion wrote:but the molar co-efficient is for a litre


This is the basic mistake. You read the units wrong, it is l/mol.cm = l . mol^-1 . cm^-1 (^ means index) = (l . mol^-1) . cm^-1 = (mol . l^-1)^-1 . cm^-1 = M^-1 . cm^-1
(I hope this is not too simple:)
so, you can see, it tells you the absorbance per cm of cuvette per molar concentration of solute. So, if you have the 1 ml cuvette, the optical path is probably 1 cm, so the concentration you have calculated should be the actuall concentration.
http://www.biolib.cz/en/main/

Cis or trans? That's what matters.
User avatar
JackBean
Inland Taipan
Inland Taipan
 
Posts: 5690
Joined: Mon Sep 14, 2009 7:12 pm

Re: Molar extinction coefficient troubles

Postby Teleprion » Thu Aug 05, 2010 3:42 pm

:oops: Thanks! I'll keep this in mind for future questions
Teleprion
Garter
Garter
 
Posts: 4
Joined: Wed Aug 04, 2010 3:17 pm

Postby Teleprion » Sat Aug 14, 2010 8:06 am

I know its a bit late on, but i just realised i might have made another foolish mistake, i was wondering, once i have the 1.75x10^-5 do i at least multiply it up by 200 to make it into one ml or just leave it as it is?
Teleprion
Garter
Garter
 
Posts: 4
Joined: Wed Aug 04, 2010 3:17 pm

Postby JackBean » Tue Aug 17, 2010 10:12 am

well, I think, if I recall it correctly, it should be concentration in mol/dm3 and you want g in 1 liter, so you basically just multiply by the mollar mas.
http://www.biolib.cz/en/main/

Cis or trans? That's what matters.
User avatar
JackBean
Inland Taipan
Inland Taipan
 
Posts: 5690
Joined: Mon Sep 14, 2009 7:12 pm


Return to Molecular Biology

Who is online

Users browsing this forum: No registered users and 0 guests