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8090 [49]
4 years ago
11

In 1958, Meselson and Stahl conducted an experiment to determine which of the three proposed methods of DNA replication was corr

ect. Identify the three proposed models for DNA replication. Conservative Semiconservative Dispersive Answer Bank The Meselson and Stahl experiment starts with E. coli containing 15 N / 15 N labeled DNA grown in 14 N media. Which result did Meselson and Stahl observe by sedimentation equilibrium centrifugation to provide strong evidence for the semiconservative model of DNA replication? Both the first and second generation have both 15 N / 15 N DNA and 14 N / 14 N DNA. No hybrid 15 N / 14 N DNA was observed. The first generation has hybrid 15 N / 14 N DNA and the second generation has both hybrid 15 N / 14 N DNA and 14 N / 14 N DNA. No 15 N / 15 N DNA was observed. The first generation has hybrid 15 N / 14 N DNA and the second generation has hybrid 15 N / 14 N DNA. No 15 N / 15 N DNA nor 14 N / 14 N DNA was observed.
Physics
1 answer:
Korolek [52]4 years ago
8 0

Answer:

Explanation:

The original has hybrid 15N/14N DNA, and the second generation has both hybrid 15N/14N DNA and 14N/14N DNA. No 15N/15N DNA was observed. In this experiment:  

Nitrogen is a significant component of DNA. 14N is the most bounteous isotope of nitrogen, however, DNA with the heavier yet non-radioactive and 15N isotope is likewise practical.  

E. coli was developed for several generations in a medium containing NH4Cl with 15N. When DNA is extracted from these cells and centrifuged on a salt density gradient, the DNA separates at which its density equals to the salt arrangement. The DNA of the cells developed in 15N medium had a higher density than cells developed in typical 14N medium. After that, E. coli cells with just 15N in their DNA were transferred to a 14N medium.

DNA was removed and compared to pure 14N DNA and 15N DNA. Immediately after only one replication, the DNA was found to have an intermediate density. Since conservative replication would result in equal measures of DNA of the higher and lower densities yet no DNA of an intermediate density, conservative replication was eliminated. Moreso, this result was consistent with both semi-conservative and dispersive replication. Semi conservative replication would result in double-stranded DNA with one strand of 15N DNA, and one of 14N DNA, while dispersive replication would result in double-stranded DNA with the two strands having mixtures of 15N and 14N DNA, either of which would have appeared as DNA of an intermediate density.  

The DNA from cells after two replications had been completed and found to comprise of equal measures of DNA with two different densities, one corresponding to the intermediate density of DNA of cells developed for just a single division in 14N medium, the other corresponding to DNA from cells developed completely in 14N medium. This was inconsistent with dispersive replication, which would have resulted in a single density, lower than the intermediate density of the one-generation cells, yet at the same time higher than cells become distinctly in 14N DNA medium, as the first 15N DNA would have been part evenly among all DNA strands. The result was steady with the semi-conservative replication hypothesis. The semi conservative hypothesis calculates that each molecule after replication will contain one old and one new strand. The dispersive model suggests that each strand of each new molecule will possess a mixture of old and new DNA.

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A 16.7 kg block is dragged over a rough, horizontal surface by a constant force of 132 N acting
Alisiya [41]

The magnitude of the work done by the force of friction is 1045.68 joules

Explanation:

Work done = force × distance

  • Work done by friction force (W) = friction force (F) × distance (d)
  • Friction force (F) = coefficient of  kinetic friction (μ) × normal reaction force (R)

A 16.7 kg block is dragged over a rough, horizontal surface by a constant force of 132 N acting  at an angle of 25.8. above the horizontal. The block is displaced 57,8 m, and the coefficient of  kinetic friction is 0.229

To find the normal reaction force of the block distribute The constant force into two component, horizontal component of 132 cos(25.8°) and vertical component 132 sin(25.8°), Then equate ∑vertical forces by 0 to find R

The weight of the block = mg, where m is its mass and g is 9.8 m/s²

→ ∑Vertical forces = R + 132 sin(25.8°) - mg

→ ∑Vertical forces = R + 132 sin(25.8°) - (16.7)(9.8)

→ ∑Vertical forces = R - 106.2

Equate it by 0

→ R - 106.2 = 0 ⇒ add 106.2 to both sides

→ R = 106.2 N

Now let us find the friction force

→ F = μ R

→ μ = 0.229

→ F = 0.229 (106.2) = 24.3198 N

Let us calculate the work don by the force of friction

→ W = F × d

→ d = 57.8 meters

→ W = 24.3198 × 57.8 = 1045.68 joules

The magnitude of the work done by the force of friction is 1045.68 joules

Learn more:

You can learn more about force of friction in brainly.com/question/6217246

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6 0
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Answer:

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If Susan exerted a force of 5 newtons and the output force of the machine is 15
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Answer:

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Explanation:

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