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If E.coli containing N^(15)-N^(15) DNA w...

If E.coli containing `N^(15)-N^(15)` DNA was allowed to grow for 80 minutes in medium containing `N^(14)` then, what would be the percentage of light density and hybrid density DNA molecule ?

A

87.5% light density , 12.5% hybrid density

B

75% light density , 25% hybrid density

C

25% light density , 75% hybrid density

D

12.5% light density , 87.5% hybrid density

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem regarding the percentage of light density and hybrid density DNA molecules in E. coli after growing in a medium containing \(N^{14}\) for 80 minutes, we can follow these steps: ### Step 1: Understand the Initial Conditions Initially, the E. coli contains only \(N^{15}\) DNA, which is heavy due to the presence of the heavy nitrogen isotope. **Hint:** Remember that \(N^{15}\) is heavier than \(N^{14}\). ### Step 2: Growth in Medium with \(N^{14}\) After the initial growth, the E. coli is transferred to a medium containing \(N^{14}\). During replication, the DNA will incorporate \(N^{14}\) into the newly synthesized strands. **Hint:** DNA replication involves pairing of bases and incorporation of nucleotides from the medium. ### Step 3: Determine the DNA Density Types 1. **Light Density DNA (L)**: This will consist entirely of \(N^{14}\) DNA. 2. **Hybrid Density DNA (H)**: This will consist of one strand of \(N^{15}\) and one strand of \(N^{14}\) (i.e., one old strand and one new strand). ### Step 4: Calculate the Proportions After One Round of Replication After one round of replication in the \(N^{14}\) medium: - Each original \(N^{15}\) DNA molecule will produce two daughter molecules: - One will be \(N^{14}/N^{15}\) (hybrid). - The other will be \(N^{14}/N^{14}\) (light). Thus, after one generation: - 50% of the DNA will be hybrid density (H). - 50% of the DNA will be light density (L). **Hint:** Consider how many generations have occurred and how DNA replication works. ### Step 5: Determine the Percentage After 80 Minutes Assuming that the E. coli divides approximately every 20 minutes, in 80 minutes, there will be about 4 generations. 1. **After 1st generation**: - 50% Hybrid (H) - 50% Light (L) 2. **After 2nd generation**: - Hybrid (H) from the first generation will produce: - 50% Hybrid (H) - 50% Light (L) - Light (L) remains 50%. - Total: - 25% H (from H) + 50% L = 25% H + 75% L 3. **After 3rd generation**: - Repeat the process: - 50% of 25% H = 12.5% H - 50% of 75% L = 37.5% L - Total: - 12.5% H + 37.5% L = 12.5% H + 87.5% L 4. **After 4th generation**: - Repeat again: - 50% of 12.5% H = 6.25% H - 50% of 87.5% L = 43.75% L - Total: - 6.25% H + 43.75% L = 6.25% H + 93.75% L ### Final Calculation After 4 generations: - **Hybrid Density DNA (H)**: 6.25% - **Light Density DNA (L)**: 93.75% ### Conclusion Thus, after 80 minutes, the percentage of light density DNA molecules is 93.75%, and the percentage of hybrid density DNA molecules is 6.25%.
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