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suter [353]
3 years ago
8

What is the role if DNA in determining an organism's traits

Chemistry
1 answer:
faltersainse [42]3 years ago
3 0
Everything. DNA contains genes and which genes a person has determines which (if any) proteins are formed. Based on how much of a protein is formed determines how strong a trait is expressed. DNA is the genetic material that contains genes. genes contain information about an organisms trait. hope this helps :)
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Carbons from acetyl CoA are transferred to the citric acid cycle. Which is the first round of the citric acid cycle that could p
Zielflug [23.3K]

Answer:

In the third step of the citric acid cycle, the oxidation of isocitrate takes place and one molecule of carbon dioxide is released.

Explanation:

In the first step of citric acid cycle, acetylCoA combines with a four-carbon molecule, oxaloacetate, forming a six-carbon molecule, citrate.

In the second step, the citrate in the presence of enzyme anicotase is converted into isocitrate.

<u>In the third step, the oxidation of isocitrate takes place and one molecule of carbon dioxide is released leaving behind one five-carbon molecule called as α-ketoglutarate. During this step, NAD⁺ is reduced to form NADH. </u>

<u>This is first round of the citric acid cycle that could possibly release a carbon atom originating from this acetyl CoA.</u>

On series of reaction, another carbon dioxide molecule also being relased and oxaloacetate is regenerated again.

7 0
2 years ago
A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a
Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = \frac{x}{100}

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = \frac{10}{100}

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] %  = (90 - x) %

Fractional abundance of middle-weight isotope = \frac{(90-x)}{100}

Now put all the given values in above formula, we get:

48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]

x=78\%

Therefore, the percent abundance of the heaviest isotope is, 78 %

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