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Offline Lizzy

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Lab question
« on: September 04, 2013, 01:21:00 AM »
How would you make 100 ml of 3.2 mg/ml isocitrate dehydrogenase (IDH) dissolved in a solution of 20 mM Tris buffer (use a 1.0 M stock of Tris)

This is a practice question for a test so if someone could walk me through it that would be great!

Offline Borek

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Re: Lab question
« Reply #1 on: September 04, 2013, 02:59:18 AM »
You have to show your attempts at solving the question to receive help. This is a forum policy.
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Offline Lizzy

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Re: Lab question
« Reply #2 on: September 04, 2013, 04:45:40 PM »
Okay so here is how I attempted to solve it.
 3.2 mg/ml of IDH 3.2 * 100= 320 mg of IDH
 20 mM Tris = 0.2 M Tris
 0.2M = xmoles/ 0.1 L x=0.02 moles
 0.02 moles= x/121.1grams x= 2.422 grams
 So I would add 2.422 grams of Tris and 320 mg of IDH to 100ml of H2O?

Is this right? If not can you show me where  I am going wrong?
 

Offline Babcock_Hall

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Re: Lab question
« Reply #3 on: September 04, 2013, 07:01:18 PM »
You are given a stock solution of Tris buffer in the problem.  Your method, which employs only Tris base, will produce a relatively alkaline solution that may harm the enzyme.

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