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ch4aika [34]
3 years ago
10

What is restriction enzymes and what is it’s purpose?

Chemistry
1 answer:
Eddi Din [679]3 years ago
7 0

Answer:  <u>Endonuclease enzymes used in molecular biology that cut DNA at specified points.</u>

Explanation:

Enzymes are specific protein types which bind to a substrate within a reaction, to increase the rate of reaction within the solution- they speed up the rate of reaction.

Restriction enzymes are bacteria-derived enzymes; these make cuts on deoxyribonucleic acid molecules or DNA. These are also called  restriction endonucleases. They are utilized in molecular biology for DNA cloning and sequencing and cut DNA into smaller pieces called fragments.

Restriction enzymes make directed cuts on DNA molecules. They precisely target sites on DNA to produce mostly identical or homogenous, discrete fragments of equal sizes, producing blunt or sticky ends. In order to do this, they recognize sequences of nucleotides that correspond with a complementary sequence on the endonuclease called restriction sites.

There are several kinds that may require cofactors (chemical or metallic compounds that aid in enzyme activity)  :

  • Type I: cleave far away from the recognition site; require ATP and  SAMe S-Adenosyl-L-Methionine

  • Type II: cleave near to the site; require Magnesium

  • Type III: cleave near to the site; require ATP which is not hydrolysed but  SAMe S-Adenosyl-L-Methionine is optional

  • Type IV: cleavage targeted to DNA that have undergone post transcriptional modification through certain types of methylation (addition of a methyl group)

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Select all of the elements that are classified as metals.
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Answer:

Gold, sodium, aluminum

Explanation:

The metals are located on the left side of the periodic table

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What is different between elastic energy, kinetic energy, and gravitational energy?
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Answer:elastic is stretchy Kinect it is movement and gravitational is vertical

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If a sample contains 70.0 % of the R enantiomer and 30.0 % of the S enantiomer, what is the enantiomeric excess of the mixture?
NeTakaya

Answer:

the enantiomeric excess of the mixture is 40%

Explanation:

The computation of the enantiomeric excess of the mixture is shown below:

As we know that

= |\frac{R - S}{R + S} |\times 100\\\\= |\frac{70 - 30}{70 + 30}| \times 100\\\\= 40\%

Hence, the enantiomeric excess of the mixture is 40%

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3 years ago
The extraction of aluminum metal from the aluminum hydroxide found in bauxite by the Hall-Héroult process is one of the most rem
son4ous [18]

Answer:

10 kg Al(OH)₃

Explanation:

There is some info missing. I think this is the original question.

<em>The extraction of aluminum metal from the aluminum hydroxide found in bauxite by the Hall-Héroult process is one of the most remarkable success stories of 19th-century chemistry, turning aluminum from a rare and precious metal into the cheap commodity it is today. </em>

<em>In the first step, aluminum hydroxide reacts to form alumina (Al₂O₃) and water: 2 Al(OH)₃(s) → Al₂O₃(s) + 3H₂O(g). In the second step, alumina (Al₂O₃ and carbon react to form aluminum and carbon dioxide: 2Al₂O₃(s)+3C(s)→4Al(s)+3CO₂(g). Suppose the yield of the first step is 63% and the yield of the second step is 89%. </em>

<em>Calculate the mass of aluminum hydroxide required to make 2.0 kg of aluminum. Be sure your answer has a unit symbol, if needed, and is rounded to the correct number of significant digits.</em>

<em />

Let's consider the 2 steps in the synthesis of Al.

Step 1: 2 Al(OH)₃(s) → Al₂O₃(s) + 3 H₂O(g)

Step 2: 2 Al₂O₃(s) + 3 C(s) → 4 Al(s) + 3 CO₂(g)

In Step 2, the percent yield of Al is 89% and the real yield is 2.0 kg. The theoretical yield is:

2.0 kg (R) × (100 kg (T) / 89 kg (R)) = 2.2 kg = 2.2 × 10³ g

In Step 2, the mass of Al is 4 × 26.98 g = 107.9 g and the mass of Al₂O₃ is 2 × 101.96 g = 203.92g. The mass of Al₂O₃ that produced 2.2 × 10³ g of Al is:

2.2 × 10³ g Al × (203.92g Al₂O₃ / 107.9 g Al) = 4.2 × 10³ g Al₂O₃

In Step 1, the percent yield of Al₂O₃ is 63% and the real yield is 4.2 × 10³ g. The theoretical yield is:

4.2 × 10³ g (R) × (100 g (T)/ 63 g (R)) = 6.7 × 10³ g

In Step 1, the mass of Al₂O₃ is 101.96 g and the mass of Al(OH)₃ is 2 × 78.00 g = 156.0 g. The mass of Al(OH)₃ that produced 6.7 × 10³ g of Al₂O₃ is:

6.7 × 10³ g Al₂O₃ × (156.0 g Al(OH)₃ / 101.96 g Al₂O₃) = 1.0 × 10⁴ g Al(OH)₃ = 10 kg Al(OH)₃

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8. How much heat will be released when
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Answer:

15 is it

Explanation:

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