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Harlamova29_29 [7]
4 years ago
7

Which of the following treatments would enhance the level of the Pfr form of phytochrome?A) exposure to far-red lightB) exposure

to red lightC) long dark periodD) inhibition of protein synthesisE) synthesis of phosphorylating enzymes
Physics
1 answer:
puteri [66]4 years ago
4 0

Answer:

B) exposure to red light

Explanation:

Plants use a phytochrome system to sense the level, intensity, duration, and color of environmental light do as to adjust their physiology.

The phytochromes are a family of chromoproteins with a linear tetrapyrrole chromophore, similar to the chlorophyll. Phytochromes have two photo-interconvertible forms: Pr and Pfr. Pr absorbs red light (~667 nm) and is immediately converted to Pfr. Pfr absorbs far-red light (~730 nm) and is quickly converted back to Pr. Absorption of red or far-red light causes a massive change to the shape of the chromophore, altering the conformation and activity of the phytochrome protein to which it is bound. Together, the two forms represent the phytochrome system.

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We can find the answer step-by-step:

1) The electric charges on a conductor must lie entirely on its surface. This is because the charges have same sign, so the force acting between each other is repulsive therefore the charges must be as far apart as possible, i.e. on the surface of the conductor.

2) We consider a cylinder perpendicular to the surface of the conductor, that crosses the surface with its section. We then apply Gauss law, which states that the flux of the electric field through this cylinder is equal to the total charge inside it divided the electrical permittivity:
\Phi =  \frac{Q}{\epsilon_0}

3) The electric field outside the surface is perpendicular to the surface itself (otherwise there would be a component of the electric force parallel to the surface, which would move the charge, violating the condition of equilibrium). The electric field inside the conductor is instead zero, because otherwise charges would move violating again equilibrium condition. Therefore, the only flux is the one crossing the section A of the cylinder outside the surface: 
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4) The total charge contained in the cylinder is the product between the section, A, and the charge density \sigma on the surface of the conductor:
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5) Substituting the flux and the charge density inside Gauss law, we can find the electric field just outside the surface of the conductor:
EA= \frac{\sigma A}{\epsilon_0}
therefore
E= \frac{\sigma}{\epsilon_0}
4 0
3 years ago
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This question is incorrect.The correct question is here

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