[molecular bio]calculation

scholesmu · Post by **scholesmu** » Sat Aug 15, 2009 9:48 am

[molecular bio]calculation

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Post by scholesmu » Sat Aug 15, 2009 9:48 am

1. A reaction has been design to be carried out in 4ml total volume. It requires a final phenylalanine concentration of 5mM. If the protocol allows 50 ul for the phenylalanine addition, what concentration would you make up the stock solution?

2.A particular cell line will grow well in a standard medium (DMEM) provided it is supplemented with extra glutamine. The standard medium contains 1 mM glutamine but this cell line grows better with 5 mM glutamine. How much more glutamine would you need to weigh out and add to 750 mL of the DMEM medium? The molecular weight of glutamine is 120.

i hv no idea how to do it...

canalon · Post by **canalon** » Sat Aug 15, 2009 1:16 pm

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Post by canalon » Sat Aug 15, 2009 1:16 pm

1 - use the well known formula: C1*V1=C2*V2
you know C2 (5mM), V2 (4ml) and V1 (50ul) solve for C2

2 - Calculate how many (milli)moles of Glutamine are needed to adjust the concentration in 1L (1mM=1mol.L) correct for the volume (express in liter...) and then you can easily figure out how much powder you need to weight.

Patrick

Science has proof without any certainty. Creationists have certainty without
any proof. (Ashley Montague)

scholesmu · Post by **scholesmu** » Sun Aug 16, 2009 11:49 am

Re: [molecular bio]calculation

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Post by scholesmu » Sun Aug 16, 2009 11:49 am

thanks, please i have one more... :cry:

stock solutions and powders available: NaCl powder (molecular weight 58.4), 1 M potassium phosphate, and a 5 mg/ml stock of cholesterol (mol. wt. 386.7)

Acetic acid (CH3COOH) is a liquid at room temperature. When making acetate buffers (as in the question above) you must add specific molar quantities of acetic acid and sodium acetate to achieve the required pH. How would you make up 1 L of the acetic acid stock solution (0.4 M CH3COOH)? The density of the liquid is 1.05 g/mL and the molecular weight is 60.

MrMistery · Post by **MrMistery** » Mon Aug 17, 2009 11:42 am

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Post by MrMistery » Mon Aug 17, 2009 11:42 am

huh? well if you have 1M solution and wanna make 1L of a 0.4M solution you just add 400 mL of your 1M solution and 600 mL of water. You don't even need to know molecular weight, density or anything else. You can just derive it out of C1*V1=C2*V2 (1M*V1=0.4M*1L and solve for V1)

"As a biologist, I firmly believe that when you're dead, you're dead. Except for what you live behind in history. That's the only afterlife" - J. Craig Venter