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Semmy [17]
3 years ago
6

The aerobic phase of cellular respiration in the mitochondrion produces a net of about 28 to 30 ATP molecules. How does this com

pare to the energy released in glycosis?
Biology
2 answers:
djyliett [7]3 years ago
6 0

Answer:

More ATP is produced in mitochondria

Explanation:

Glycolysis is the first stage of cellular respiration. It is a common process of both anaerobic and aerobic respiration. It occurs in the cytoplasm of a living cell. During glycolysis 4 ATP and 2 NADH is produced and 2 ATP is used up. Thus, there is net gain of 2 ATP. The completion of cellular respiration takes place in mitochondrial matrix and on inner membrane, where all the co-enzymes NADH and FADH₂ oxidised to release ATPs. On oxidation each NADH produce 3ATP and each FADH₂ produces 2 ATP. Total 10 NADH is produced during complete oxudation which yield 30 ATP and 2 FADH₂ yields 4 ATP. Thus, in the mitochondria more ATP is produced as compared to glycolysis.

Mazyrski [523]3 years ago
4 0

In eukaryotic cells, the theoretical maximum yield of ATP generated per glucose is 36 to 38, depending on how the 2 NADH generated in the cytoplasm during glycolysis enter the mitochondria and whether the resulting yield is 2 or 3 ATP per NADH.Jun 26, 2001

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

<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.

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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.

<u>Available data</u>:

  • 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

<u>λ = 0.0001209</u>

Now, let us calculate the first term in the equation

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

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 ≅ <u>6890 years</u>

The bones are approximately<u> 6890 years.</u>

You can learn more about dating organic matter with carbon14 at

brainly.com/question/4149380

#SPJ1

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