A biochemist isolates and purifies various molecules needed for DNA replication. When she adds these molecules to DNA, replication occurs, but half of the DNA produced consists of a normal DNA strand paired with numerous segments of DNA a few hundred nucleotides long. What has she probably left out of the mixture

Answers 2

Answer:

DNA ligase

Explanation:

The biochemist must have left out DNA ligase enzyme.

The DNA ligase enzyme is able to catalyze the formation of phosphodiester bonds and as such, capable of joining strands of DNA together to form a single strand.

The numerous DNA segments of a few nucleotides long observed by the biochemist must have been the replicated product of the lagging DNA strand. The lagging strand is replicated discontinuously in short strands because the DNA polymerase enzyme can only elongate primers in 5' to 3' direction. The short segments are known as Okazaki segments and are usually joined together to form a whole strand by the DNA ligase enzyme.

Hence, the missing component is the DNA ligase.

AI generated Answer

Answer: She has probably left out the enzyme helicase. Explanation: During the process of DNA replication, several molecules are needed in order for it to occur. These include enzymes known as helicase, topoisomerase, primase, DNA polymerase, and DNA ligase. The biochemist has likely left out the enzyme called helicase, which is responsible for unwinding the two strands of DNA. Without it, only a partial replication is possible, with mismatched segments of DNA being produced. This is likely why half of the DNA produced consists of a normal DNA strand paired with numerous segments of DNA a few hundred nucleotides long.

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in order to end the recession in the early 1990s, president Clinton​

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He found that global changes will bring to the world economy and so his administration began to push for the kind of solutions our country would require in order to prosper in the 21st century.

Answer:

decreased government spending

Explanation:

took the test

In Niagara-on-the-Lake, you can ride a horse-drawn carriage for a fixed price plus a variable amount that depends on the length of the trip. The variable cost is $10/km and a 2.5-km trip costs $40. a) Determine the equation relating cost, C, in dollars, and distance, d, in kilometres. b) Use your equation to find the cost of a 6.5-km ride. c) Graph this relation. d) Use the graph to find the cost of a 6.5-km ride.

Answer:

Step-by-step explanation:

a) The cost of a product is given as:

Cost = a + bx

Where a is the fixed cost, b is the variable cost, x is the independent variable.

In this case, the independent variable is the distance(d), therefore:

Cost (C) = a + bd

C = a + bd

But the variable cost is $10/km, therefore:

C = a + 10d

A 2.5 km trip cost $40, therefore:

40 = a + 10(2.5)

40 = a + 25

a = 40 - 25 = 15

The fixed cost is $15. Therefore the equation for cost is:

C = 15 + 10d  

b) C = 15 + 10d  

For 6.5 km ride:

C = 15 + 10(6.5) = 15 + 65 = 80

It would cost $80 for a 6.5 km drive.

 

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Draw structural formulas for the major organic product(s) of the reaction shown below.• You do not have to consider stereochemistry.If no reaction occurs, draw the organic starting material.Remember to include all of the formal charges on the atoms of any nitro groups.​
question img

Answer:

3-bromobenzoic acid

Explanation:

In this case, we have to remember that the Br_2/FeBr_3  is a reaction in which we add Br into the molecule an electrophilic aromatic substitution. Additionally, we have a carboxylic acid in the benzene. This carboxylic acid is an ortho director because is a deactivating group (it removes electrons from the benzene ring). With this in mind, a "Br" atom would be added in an ortho position respect to the COOH group and we will obtain 3-bromobenzoic acid.

See figure 1.

I hope it helps!

answer img
When ethanol, C2H5OH (a component in some gasoline mixtures) is burned in air, one molecule of ethanol combines with three oxygen molecules to form two CO2 molecules and three H2O molecules. A) Write the balanced chemical equation for the reaction described. B) How many molecules of CO2 and H2O would be produced when 2 molecules ethanol are consumed? Equation? C) How many H2O molecules are formed, then 9 O2 molecules are consumed? What conversion factor did you use? Explain! D) If 15 ethanol molecules react, how many molecules O2 must also react? What conversion factor did you use? Explain!

Answer:

1) C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)

2) four molecules of CO2 will be produced and six molecules of water

3)9 molecules of water are formed when 9 molecules of oxygen are consumed.

4) 45 molecules of oxygen

Explanation:

The balanced chemical reaction equation is shown here and must guide our work. When ethanol is burned in air, it reacts as shown;

C2H5OH(l)+3O2(g)⟶2CO2(g)+3H2O(l)

Hence, if we use 2 molecules of ethanol, the balanced reaction equation will look like this;

2C2H5OH(l)+6O2(g)⟶4CO2(g)+6H2O(l)

Hence four molecules of CO2 are formed and six molecules of water are formed

From the balanced stoichiometric equation;

3 molecules of oxygen yields 3 molecules of water

Therefore, 9 molecules of oxygen will yield 9 × 3/3 = 9 molecules of water

Therefore, 9 molecules of water are formed when 9 molecules of oxygen are consumed.

From the reaction equation;

1 molecule of ethanol reacts with 3 molecules of oxygen

Therefore 15 molecules of ethanol will react with 15 × 3/1 = 45 molecules of oxygen

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