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german
3 years ago
9

In the classic Meselson and Stahl experiment, E. coli are first grown in 15N-enriched media (0th generation) and, subsequently,

switched to 14N-enriched media. A DNA sample isolated during the first generation of E. coli grown in 14N media is subjected to a centrifugation step in a dense CsCl gradient.
Assuming that replication occurs in a semiconservative fashion, the DNA sample will:____________
a) run as two bands (of equal intensity), one at the same position as a uniformly 15N-labeled DNA, and one at the same position as a uniformly 14N-labeled DNA.
b) run as one band, at the same position as a uniformly 15N-labeled DNA.
c) run as one band, at the same position as a uniformly 14N-labeled DNA.
d) run as one band, at a position expected for a DNA containing a 14N:15N ratio equal to 7.
e) run as one band, at a position expected for a DNA containing a 14N:15N ratio equal to 1.
Biology
1 answer:
mars1129 [50]3 years ago
6 0

Answer:

The correct answer is - option e.)

Explanation:

The experiment that was performed by Meselson and Stahl was created to prove that DNA replication is a semiconservative process.

When Bacteria grow on 15N media they incorporate the isotype of nitrogen and after few generations, the culture moved to 14N media where it incorporates as they replicate in semi-conservative mode and one DNA strand comes from the Parent DNA.

The right option for question one is Option (e) Run as one band, at a position expected for DNA containing a 14N:15N ratio equal to 1.

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Answer:

d) 2v

Explanation:

Since, root mean square speed of a molecule,

V_{rms}=\sqrt{\frac{3kT}{m}}

Where,

k = Boltzmann constant,

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Also, the temperature of a gas,

T=\frac{pV}{nR}

Where,

p = pressure of the gas,

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If V = 2V and p = 2P

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V_{rms}=\sqrt{\frac{12kpV}{mnR}

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Hence, if rms speed of a gas molecule under initial conditions is v then rms speed of a molecule will be 2v

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3 years ago
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