Answer:
1. C- Three.
2. A- Methionine
3. D- Translocation.
4. C- OH.
5. A - 5'
6. A - 3' carbon
7. A. adenine and guanine
Explanation:
1. A codon is a group of three nucleotide sequence that encodes or specifies an amino acid. This means that, during translation (second stage of gene expression), when a CODON is read, an amino acid is added to the growing peptide chain.
2. The codon that initiates the translation process is called a start codon. It has a sequence: AUG and it specifies Methionine amino acid. Hence, during translation where a tRNA binds to the mRNA codon to read it and add its corresponding amino acid, a tRNA with a complementary sequence of AUG (start codon) binds to it and carries Methionine amino acid.
3. Translocation is a process during translation whereby the mRNA-tRNA moeity moves forward in the ribosome to allow another codon to move into the vacant site for translation process to continue.
4. The sugar component of a nucelotide that makes up the nucleic acid (DNA or RNA) i.e. ribose or deoxyribose, contains an hydroxyll functional group (-OH).
5. A nucleotide consists of a pentose (five carbon) sugar, phosphate group and a nitrogenous base. The phosphate group (PO43-) is attached to the 5' carbon of the sugar molecule.
6. The free hydroxyll group (-OH) of the five carbon sugar molecule in DNA is attached to its 3' carbon.
7. Nitrogenous bases are the third component of a nucleotide, the other two being pentose sugar and phosphate group. The nitrogenous bases are four viz: Adenine, Guanine, Cytosine, and Thymine. These bases are classified into Purines and Pyrimidines based on the similarity in their structure. Adenine (A) and Guanine (G) are Purines because they possess have two carbon-nitrogen rings, as opposed to one possessed by Pyrimidines (Thymine and Cytosine).
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:
Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
Larger elements are able to form in a supernova explosion because the star releases very large amounts of energy as well as neutrons, which allows elements heavier than iron to be produced.
<h3>What is Supernova?</h3>
This is referred to the explosion of a star and it resulting in larger elements being formed through a process known as nucleosynthesis and is usually accompanied by an increase in the brightness of the star.
The elements produced are usually larger than elements such as iron and examples include uranium, gold etc.
This is therefore the reason why it was chosen as the most appropriate choice.
Read more about Supernova here brainly.com/question/27492871
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Answer:
The method which is used to separate a mixture of oil and vinegar is separation by density. Density is defined as the ratio of mass to the amount of matter i.e. volume.
Explanation:
