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    A researcher is required to perform a biochemical assay that will contain a purified adenylyl cyclase signalling enzyme. The researcher must add 1 ug of the adenylyl cyclase to the assay. The researcher has been provided with a 12.5 uM solution of the adenylyl cyclase. The adenylyl cyclase has a molecular weight of 20,000. How many ul of the adenylyl cyclase will be added to the assay?
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    What's the density (g/L) of the adenylyl cyclase solution? (Molarity = moles/L) From there, you know you need 1g of adenylyl cyclase, so what is the final volume required?
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    (Original post by zombiejon)
    What's the density (g/L) of the adenylyl cyclase solution? (Molarity = moles/L) From there, you know you need 1g of adenylyl cyclase, so what is the final volume required?
    Hey ... its wasn't stated in the question this is all the information that I have
    I'm so lost right now!
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    (Original post by amielle)
    Hey ... its wasn't stated in the question this is all the information that I have
    I'm so lost right now!
    Ok. I was wording those as prompts for help you think a bit.

    You have the following information:
    Final is 1 ug of the adenylyl cyclase
    Stock solution is at 12.5 uM
    Adenylyl cyclyase mw is 20000 (or g/mol)
    Volume is ??uL

    So there are two ways to go about it - convert everything into g/L, or convert everything into mol. For ease of use, we'll convert everything into g/L

    We know that stock soln (solution) is at 12.5uM (or umol/L).
    Adenylyl cyclase mw is 20000 g/mol.
    Molarity (mol/L) / mw (g/mol) = g/L
    So 12.5 umol/L * 20000 g/mol = 250000 ug/L (note how the moles cancel out)


    Now our updated information is as follows:
    Final is 1 ug of the adenylyl cyclase
    Stock solution is at 12.5 uM (or 250000 ug/L)
    Adenylyl cyclyase mw is 20000 (or g/mol) <<< no longer necessary
    Volume is ??uL

    Since our final is still 1 ug of adenylyl cyclase, we need x volume of stock solution, or the following:
    250000 ug/L (stock solution) x needed volume = 1 ug
    From here we can understand that the following equation is needed: 1 ug / 250000 ug/L = needed volume (units cancel out again)
    needed volume = 0.000004 L
    Since there is 1000 mL/L, the needed volume can be converted in 0.004mL
    Since there is 1000uL/mL, the volume can also be converted into 4uL
 
 
 
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