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lord [1]
2 years ago
12

a geneticist was able to reconstruct a sequence of mitochondrial dna of a prehistoric insect that was encased in amber. a segmen

t of this dna in species a was found to have the sequence aggtcctaag. this segment of dna has had one mutation every 100 million years. species b has the dna sequence aggtccgaag. how many years ago did these species diverge from a common ancestor? 1,000,000 years 200,000 years 50,000,000 years 60,000 years
SAT
1 answer:
Yuliya22 [10]2 years ago
5 0

The number of years ago at which specie A and B diverge from a common ancestor is 1,000,000 years. Option A is correct.

<h3>What is mitochondrial? </h3>

The mitochondrial is the structure in the cells whose main function is to convert the energy from the food in such a form that the cell can use.

A geneticist was able to reconstruct a sequence of mitochondrial dna of a prehistoric insect that was encased in amber.

A segment of this dna in species <em>A</em> was found to have the sequence AGGTCCTAAG. Species <em>B</em> has the dna sequence AGGTCCGAAG. Thus, the sequence is,

  • Species <em>A- </em>AGGTCCTAAG.
  • Species <em>B-</em> AGGTCCGAAG.

This segment of dna has had one mutation every 100 million years. The DNA sequence has only one mutation. For this, 100 million year is required.

Thus, the number of years ago at which specie A and B diverge from a common ancestor is 1,000,000 years. Option A is correct.

Learn more about the mitochondrial here;

brainly.com/question/20398312

#SPJ1

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When the chemical reaction 2no(g)+o2(g)→2no2(g) is carried out under certain conditions, the rate of disappearance of no(g) is 5
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The rate of disappearance of O2(g) under the same conditions is 2.5 × 10⁻⁵ m s⁻¹.

<h3>What is the rate law of a chemical equation? </h3>

The rate law of a chemical reaction equation is usually dependent on the concentration of the reactant species in the equation.

The chemical reaction given is;

\mathbf{2 NO_{(g)} + O_{2(g)} \to 2 NO_{2(g)} }

The rate law for this reaction can be expressed as:

\mathbf{= -\dfrac{1}{2}\dfrac{d[NO]}{dt} = -\dfrac{1}{1}\dfrac{d[O_2]}{dt}= +\dfrac{1}{2}\dfrac{d[NO_2]}{dt}}

Recall that:

  • The rate of disappearance of NO(g) = 5.0× 10⁻⁵ m s⁻¹.

  • Since both NO and O2 are the reacting species;

Then:

  • The rate of  disappearance of NO(g) is equal to the rate of  disappearance of O2(g)

\mathbf{= -\dfrac{1}{2}\dfrac{d[NO]}{dt} = -\dfrac{1}{1}\dfrac{d[O_2]}{dt}}

\mathbf{= -\dfrac{1}{2} \times 5.0 \times 10^{-5}  = rate \  of  \ disappearance \ of  \ O_2}

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The rate of disappearance of O2 = 2.5 × 10⁻⁵ m s⁻¹.

Therefore, we can conclude that two molecules of NO are consumed per one molecule of O2.

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