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Within a very smelly, mixed bacterial community, there exists 1,000 Michiganeous wolverenious, and 1,000,000 Ohious statious. If it takes 25 rounds of PCR to detect O.statious, approximately how many rounds of PCR would it take to detect M. wolverenious from the same smelly microbial community?
A. 15
B. 35
C. 10
D. 25
E. 100

Respuesta :

Answer:

The answer is b) 35

Explanation:

Since the DNA is double-stranded, a base equal to 2 will be taken:

[tex]1000*2^{n} =1000000*2^{25}[/tex]

Applying the logarithm to both sides of the equation:

[tex]log1000+nlog2=6log10+25log2\\3log10+nlog2=6log10+25log2\\nlog2=3log10+25log2\\nlog2=3+25log2\\n=\frac{3+25log2}{log2} \\n=\frac{3+7.55}{0.301} =35.04=35[/tex]

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