KVPY SX SB Exam 2012 Question Paper

Question : 102 of 120

Marks: +1, -0

In the DNA of E. Coli the mole ratio of adenine to cytosine is 0.7. If the number of moles of adenine in the DNA is 350000, the number of moles of guanine is equal to

350000
500000
225000
700000

Validate

Solution:

\A/\C=0.7

{35×10^4}/\C=0.7

\C={35×10^4}/0.7=5×10^5

Moles of cytosine = moles of guanine

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