The question reads, Using T7 RNA polymerase to transcribe in vitro a 100-nt RNA off a DNA template. This RNA contains 19 Adenosines. In your 100 uL transcription reaction you added 1.00 µL of α32P•[ATP] that had a specific activity on its reference date (May 6th) of 10.0 µCi/µL. The concentration of "cold" (non-radioactive) ATP in your transcription reaction was .500 mM. At the end of a 2.00-hour transcription reaction, you run a calibration curve using autoradiography on standards of radioactive ATP plus 2 µL of your reaction to obtain the following data:


1:1000 dilution of α32P•[ATP] Stock (10.0µL)-------------3.211x10^7

1:10,000 dilution of α32P•[ATP] Stock (10.0µL)-----------2,483,565

1:100,000 dilution of α32P•[ATP] Stock (10.0µL)---------299,215

1:1,000,000 dilution of α32P•[ATP] Stock (10.0µL)-------31,444

Your transcription reaction (2.00 µL)-------------------------102,784


Estimate the concentration of RNA (in pmol/uL) that exists in your remaining 98 uL of the reaction.

I am using the equation: S.A = (Chn*(.5)^d/14.3 *Vhn * N * cf)/ Vt*Ccn

where Chn = concentration of the S.A. of the α32P•[ATP] in µCi/µL =10 µCi/µL

Vhn = volume of α32P•[ATP] used in the labeling reaction =1.00 µL

N = the number of A nucleotides in the NA= 19

cf = calibration factor from the standard curve (“pixels”/µCi)= 10278.4 pixels/uCi

Vt = volume of the reaction= 100µL

d = number of days past your isotope’s “reference date” = 7 (May 6th to May 13th given in problem)

Ccn = concentration of cold ATP in the labeling reaction (pmol/µL) =.500 mM =500pmol

After I calculate the S.A. how do use that to estimate the concentration of RNA that exists in my remaining 98 uL of the reaction. I am beyond confused, any help would be greatly appreciated.


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