Question

In an experiment, bacteria were grown for 500 generations in a medium containing $^{14}\text{N}$ (light isotope), then transferred to a medium with $^{15}\text{N}$ (heavy isotope) for one generation, and finally transferred back to the $^{14}\text{N}$ medium for one more generation. Assuming all cells divide synchronously and replication is semi-conservative, the ratio of $^{14}\text{N}^{15}\text{N}$ to $^{14}\text{N}^{14}\text{N}$ double-stranded DNA at the end of the experiment is:

• A. 1:1
• B. 1:2
• C. 1:3
• D. 1:4

Hint:

To quickly solve Meselson-Stahl type problems, keep track of the total number of individual strands. Any $^{15}\text{N}$ strand introduced in the second step will always remain in the population, but can only form hybrid DNA ($^{14}\text{N}^{15}\text{N}$) once the cells are returned to $^{14}\text{N}$ medium.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the ratio of $^{14}\text{N}^{15}\text{N}$ (hybrid) to $^{14}\text{N}^{14}\text{N}$ (light) double-stranded DNA after a specific sequence of bacterial growth generations in different nitrogen isotopes.

Step 2: Key Formula or Approach:
Semiconservative replication means that during division, each of the two parental DNA strands serves as a template for the synthesis of a new complementary strand.

Step 3: Detailed Explanation:

• Initially, bacteria are grown for 500 generations in a $^{14}\text{N}$ medium. At this point, virtually $100\%$ of the DNA strands are light ($^{14}\text{N}$), forming $^{14}\text{N}^{14}\text{N}$ double-stranded molecules. Let us denote a $^{14}\text{N}$ strand as $L$ and a $^{15}\text{N}$ strand as $H$.

• These bacteria are then transferred to a $^{15}\text{N}$ medium for exactly one generation.

• During this replication cycle, the parent $L-L$ double helices separate. Each $L$ strand template is paired with a newly synthesized heavy $H$ strand. This results in $100\%$ hybrid $L-H$ ($^{14}\text{N}^{15}\text{N}$) double-stranded DNA molecules.

• Finally, these bacteria are transferred back to a $^{14}\text{N}$ medium for one more generation.

• During this replication cycle, the hybrid $L-H$ parent molecules separate into individual $L$ and $H$ strands.

• Since the medium contains $^{14}\text{N}$, all newly synthesized complementary strands will be light ($L$).

• The parental $L$ strand pairs with a new $L$ strand, forming an $L-L$ ($^{14}\text{N}^{14}\text{N}$) double-stranded molecule.

• The parental $H$ strand pairs with a new $L$ strand, forming an $L-H$ ($^{14}\text{N}^{15}\text{N}$) double-stranded molecule.

• Thus, for every starting hybrid molecule, we obtain one light molecule ($^{14}\text{N}^{14}\text{N}$) and one hybrid molecule ($^{14}\text{N}^{15}\text{N}$).

• The ratio of $^{14}\text{N}^{15}\text{N}$ to $^{14}\text{N}^{14}\text{N}$ double-stranded DNA molecules at the end of the experiment is therefore $1:1$.


Step 4: Final Answer:
Therefore, the ratio of $^{14}\text{N}^{15}\text{N}$ to $^{14}\text{N}^{14}\text{N}$ double-stranded DNA is 1:1.