An E. coli whose DNA is fully labeled by ${ }^{15} \mathrm{~N}$ is grown in a medium containing ${ }^{14} \mathrm{NH}_4 \mathrm{Cl}$. What will be the percentage of hybrid $\left({ }^{15} \mathrm{~N} /{ }^{14} \mathrm{~N}\right)$ and light $\left({ }^{14} \mathrm{~N} /{ }^{14} \mathrm{~N}\right)$ DNA molecules at the end of 80 minutes (assume that doubling time of $E$. coli is 20 minutes)?

• A

  $25 \%$ hybrid and $75 \%$ light
• B

  $50 \%$ hybrid and $50 \%$ light
• C

  $0 \%$ hybrid and $100 \%$ light
• Correct answer
  D

  $12.5 \%$ hybrid and $87.5 \%$ light

Hence, the correct answer is option: 'D'.