6. Today we amplified 150 ng of Bos taurus (calf) DNA by PCR. This amount of DNA contains about 46324 molecules of the insulin gene [150 ng DNA= 7.69 x 10-20 mol; (7.69x10-20 )x(6.023×1023) = 46324 molecules]. performed PCR for 35 cycles to amplify the amount of this gene. a. What is the theoretical fold amount of DNA amplified by 35 cycles of PCR (remember the 2N formula)?
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![Today we amplified 150 ng of Bos taurus (calf) DNA by PCR. This amount of DNA contains about 46324
molecules of the insulin gene [150 ng DNA= 7.69 x 10-2⁰ mol; (7.69x10-20 )x(6.023×10²³) = 46324 molecules].
performed PCR for 35 cycles to amplify the amount of this gene.
a. What is the theoretical fold amount of DNA amplified by 35 cycles of PCR (remember the 2N formula)?
6.
b. How many molecules of the insulin gene would therefore be present after PCR?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf8c917b-7d25-4fea-9b3f-90a987505563%2Fce52aa9b-16d7-4a07-b91a-f6b966497013%2F9fn1xit_processed.png&w=3840&q=75)
Given,
Number of molecules of insulin gene in 150ng of Bos taurus DNA= 46324 molecules
Number of PCR cycles (n) = 35
- Theoretical fold amount of DNA = 2n (where "n" is the number of cycles)
= 235
= 3.436 x 1010 fold
Therefore , we get total fold amount of DNA= 3.436 x 1010
b. Number of molecules of insulin gene present after PCR = Number of molecules before PCR x Total fold amount
= 46324 x 3.436 x 1010
= 1.5916 x 1015 molecules
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