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Researchers investigating the regulation of neurotransmitter release from presynaptic neurons proposed a model in which CDK5, a

GrogVix [38]

Answer:

"GFP contained in synaptic vesicles moved into the synaptic cleft by exocytosis."

Explanation:

Synaptic vesicles are small membrane sacs that carry neurotransmitters from the cell body where they are produced, to the presynaptic membrane of the terminal button where they are released. The release zone of the presynaptic membrane contains voltage-dependent calcium channels. When an action potential depolarizes the presynaptic plasma membrane, $Ca^{2+}$ -channels open, and $Ca^{2+}$ flows into the nerve terminal to trigger the exocytosis of synaptic vesicles, thereby releasing their neurotransmitters into the synaptic cleft

3 0

3 years ago

The bones from an animal found at an archaeological dig have a C614 activity of 0.10 Bq per gram of carbon. The half-life of C61

erastova [34]

C14 is an isotope used in radiocarbon dating techniques to date organic matter remains. The age of these bones is approximately 6890 years.

<h3>What is Carbon 14?</h3>

Carbon 14, also known as radiocarbon, is a radioactive carbon isotope.

Isotopes are the atoms of the same element -carbon- that vary in neutrons and, hence, in their massic number. They are alternative forms of the same element.

The radioactive C14 nucleus contains 6 protons and 8 neutrons and has a half-life of 5730 years.

The term half-life is a reference. It means that an organism that has been dead for 5730 years has half the C14 amount or concentration than the same organism had when it was alive.

Knowing the half-life of an element is useful to determine the age of the dead matter.

C14 is used in radiocarbon dating techniques or methods to estimate the age of fossils. This is a reliable technique used for dating organic samples that are less than 50,000 years old.

Available data:

The half-life of C14 is 5730 years
Bones activity of 0.10 Bq per gram of carbon

To answer this question we can make use of the following equation

Ln (C14T₁/C14 T₀) = - λ T₁

Where,

C14 T₀ ⇒ Amount of carbon in a living body. We know, by bibliography, that living organism activity is 0.23 Bq per gram of carbon. So, C14 T₀ = 0.23 Bq/g

C14T₁ ⇒ Amount of carbon in the dead body. C14T₁ = 0.1 Bq/g

λ ⇒ radioactive decay constant = (Ln2)/T₀,₅

T₀,₅ ⇒ The half-life of carbon 14 = 5730 years

T₀ = Time when the organism was alive

T₁ = Age of bones

Let us first calculate the radioactive decay constant.

λ = (Ln2)/T₀,₅

λ = 0.693/5730

λ = 0.0001209

Now, let us calculate the first term in the equation

Ln (C14T₁/C14 T₀) = Ln (0.1/0.23) = Ln 0.4347 = - 0.833

Finally, let us replace the terms, clear the equation, and calculate the value of T₁.

Ln (C14T₁/C14 T₀) = - λ T₁

- 0.833 = - 0.0001209 x T₁

T₁ = - 0.833 / - 0.0001209

T₁ = 6889.99 ≅ 6890 years

The bones are approximately 6890 years.

You can learn more about dating organic matter with carbon14 at

brainly.com/question/4149380

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6 0

2 years ago

After a natural disaster such as a hurricane or a drought a population

GREYUIT [131]

After A Natural Disaster , Population Usually Decreases Due To Death Or People Fleeing From Their Home .

3 0

3 years ago

How society would provide services of the ecosystem couldnt

dangina [55]

In the course of recent years, 60 percent of all biological community administrations have declined as an immediate consequence of the change of area to the creation of nourishments, powers and strands. This ought to shock no one, say seven of the world's driving ecological researchers, who met to all things considered study the pitfalls of using markets to prompt individuals to assess the natural expenses of their conduct and arrangements. We are getting what we pay for.

5 0

3 years ago

The type of soil that has the least amount of space between the particles is _______.

Alex_Xolod [135]

The type of soil that has the least amount of space between the particles is clay.

6 0

3 years ago

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Good morning friends