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8_murik_8 [283]
3 years ago
10

reddit The amount of NE released by sympathetic nerve terminals will be most strongly influenced by a change in which of the fol

lowing? Alpha receptor sensitivity Alpha receptor density Extracellular [Ca2+] COMT activity
Chemistry
1 answer:
erastovalidia [21]3 years ago
8 0

Answer: Extracellular [Ca2+]

Explanation:

The sensitivity and density of the alpha receptors serve to <em>enhance the response to the release of</em> <em>norepinephrine (NE)</em> . However, they do not exert a strong influence as the concentration of calcium ions on the amount of <em>norepinephrine (NE)</em> released by sympathic nerve terminals.

The release of neurotransmitters depends more on either an external or internal stimulus.This results in an action potential which on reaching a nerve terminal, results in the opening of Ca²⁺ channels in the neuronal membrane. Because the extracellular concentration of Ca²⁺ is greater than the intracellular Ca²⁺ concentration, Ca²⁺ flows into the nerve terminal. This triggers a series of events that cause the vesicles containing <em>norepinephrine (NE)</em> to fuse with the plasma membrane and release <em>norepinephrine (NE)</em> into the synapse. The higher the action potential, the higher the Ca²⁺ flow into the terminals resulting in higher amount of <em>norepinephrine (NE)</em> into the synapse, and vice versa.

Catechol-O-methyltransferase (COMT) is one of several enzymes that degrade catecholamines such as dopamine, epinephrine, and norepinephrine. It serves a regulatory purpose to lower the concentration of norepinephrine upon its release from nerve terminals.

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What did aristotle believe about the nature of matter
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Aristotle believes that all material substances are matter and form.

Explanation:

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3 0
2 years ago
Calculate the pH for the following weak acid. A solution of HCOOH has 0.12M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4.
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Answer:

the pH of HCOOH solution is 2.33

Explanation:

The ionization equation for the given acid is written as:

HCOOH\leftrightarrow H^++HCOO^-

Let's say the initial concentration of the acid is c and the change in concentration x.

Then, equilibrium concentration of acid = (c-x)

and the equilibrium concentration for each of the product would be x

Equilibrium expression for the above equation would be:

\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}

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From given info, equilibrium concentration of the acid is 0.12

So, (c-x) = 0.12

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1.8*10^-^4=\frac{x^2}{0.12}

Let's solve this for x. Multiply both sides by 0.12

2.16*10^-^5=x^2

taking square root to both sides:

x=0.00465

Now, we have got the concentration of [H^+] .

[H^+] = 0.00465 M

We know that, pH=-log[H^+]

pH = -log(0.00465)

pH = 2.33

Hence, the pH of HCOOH solution is 2.33.

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