In E. coli DNA, AT//GC ratio is 0.93. If...

In `E. coli DNA,` `AT//GC` ratio is `0.93`. If the number of moles of adenine in its `DNA` sample is `465,000,` calculate the number of moles of guanine present.

Text Solution

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Since the number of moles of adenine must be equal to that of thymine, therefore ,
`(A+T)=465,000+465,000=930,000`.
Since `(A+T)//(C+G)=0.93`,
therefore, number of moles of `C+G=(930,000)/(0.93)=10,00,000`
Further since the number of moles of C=number of molecule of G.
`therefore` No. of moles of guanine (G)`=1000,000//2=500,000`.

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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`

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:

35000

500000

225000

700000

A sample of DNA has the ratio of (AT)/(GC) is 0.8 . If the no. of moles of adenine in a sample is 25000. What is the no. of moles of cytosine in it

50000

40000

31250

62500