Question

The base ratio of DNA is 0.03 and the amount of A-T,N2 base contents is 30000; then what is the amount of G-C, N2 base contents in this DNA?​

Answer 1

A+T/G+C

Given ratio=0.03

Amount of A+T is 3000

Then, A+T/G+C=0.03

3000/G+C=3/100

G+C=100000

Hope this may help you.....

Answer 2

Answer:

The base ratio of the DNA is 0.03%, amount of A-T N2 base content is 30,000 .

Explanation:

In a DNA base sequence, there are 4 nucleotides A, T, G & C. These are all three, follows the base-pairing rule which states that the amount of adenine (A) in the DNA is the same as the amount of Thymine (T) and the amount of Guanine (G) is the same as that of cytosine (C).

The amount of adenine = $\frac{0,03}{100} \times 30000=9$

According to the base pairing rule, (A - T + G - C = 100%)

$A-T=9 \%+9 \%=\frac{9}{100} \times \frac{9}{100}=1-0.18=0.8=82 \%$

$A-T= 18%$

From the above equation,

$\mathrm{G}-\mathrm{C}=100 \%-18 \% \frac{100}{100} \times \frac{18}{100}=1-0.18=0.8=82 \%$

$\mathrm{G}-\mathrm{C}=82 \% \text { where }(\mathrm{G}=41 \% \& \mathrm{C}=41 \%)$