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20 July, 17:16

You would like to examine the product of a PCR reaction using gel electrophoresis. You have about 50 µL of PCR product, but the well can only hold 10 µL of sample. You have a 5X stock solution of loading dye which must be mixed with the PCR product to a final concentration of 1X. What is the volume of loading dye and PCR product that you will need in order to make 10 µL of the final mix? Round each to the nearest microliter.

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  1. 20 July, 20:51
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    You need to mix 2 µL of the 5x loading dye and 8 µL of the PCR product.

    Explanation:

    To get a final concentration of 1X of the loading dye, you make the next calculations:

    (V1) (C1) = (V2) (C2) (Volume1) (Concentration1) = (Volume2) (Concentration2)

    where:

    (V1) (5x) = (10 µL) (1x)

    V1=[ (10 µL) (1x) ]/5x

    V1 = 2 µL
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