The nucleotide sequence of the trna anticodon will be; GUA against the CAU codon sequence of nucleotide on mRNA
<h3>
What is the nucleotide sequence?</h3>
According to the base-pairing rule A form base pair with T with two hydrogen bond and G pairs with C with 3 hydrogen bond in DNA. But in RNA T is replaced by U that means A pairs with U in mRNA.
Thus, the nucleotide sequence of the trna anticodon will be GUA against the CAU codon sequence of nucleotide on mRNA and this tRNA must be attached with histidine amino acid.
Read more about Nucleotide Sequence at; brainly.com/question/2570717
Answer:
5' side on top, broken arrow leading out, 3' side on the bottom, continuous arrow leading in
Explanation:
C = The new strand is built 5′ to 3′, and the Okazaki fragments form, moving away from the replication fork.
The answer is William Parker.
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;
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}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D%20-%5Cdfrac%7B1%7D%7B2%7D%5Cdfrac%7Bd%5BNO%5D%7D%7Bdt%7D%20%3D%20-%5Cdfrac%7B1%7D%7B1%7D%5Cdfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D%20%2B%5Cdfrac%7B1%7D%7B2%7D%5Cdfrac%7Bd%5BNO_2%5D%7D%7Bdt%7D%7D)
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}}](https://tex.z-dn.net/?f=%5Cmathbf%7B%3D%20-%5Cdfrac%7B1%7D%7B2%7D%5Cdfrac%7Bd%5BNO%5D%7D%7Bdt%7D%20%3D%20-%5Cdfrac%7B1%7D%7B1%7D%5Cdfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%7D)
Thus;
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.
Learn more about the rate law here:
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