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muminat
3 years ago
14

Meiosis I is being stimulated with a pair of homologous that are red and yellow. why will one resulting daughter cell cintains o

nly red-long and the other daughter cell contain only yellow-long? what occured if each dauggter cell contains a red and yellow-long?
Biology
1 answer:
il63 [147K]3 years ago
6 0
In meiosis I, the pair of homologous chromosomes separate from each other so that the entire red-long homolog will move one direction and the entire yellow-long homolog will move in the other direction. 

<span>If a cell ended up with both homologs, that would be the result of nondisjunction. The gametes formed from that cell will up with 2 copies of the long chromosome. The gametes formed from the other cell from meiosis I will end up with no copies of that chromosome.</span>
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<h3>What is Carbon 14?</h3>

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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

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