MCAT Biology : Organic Chemistry, Biochemistry, and Metabolism

Study concepts, example questions & explanations for MCAT Biology

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Example Questions

Example Question #1751 : Mcat Biological Sciences

What is the least reactive acid derivative?

Possible Answers:

Acid chloride

Amide

Ester

Carboxylic acid

Correct answer:

Amide

Explanation:

The reactivity of acid derivatives is largely dependent on the leaving group. If the leaving group is a strong base it is considered a poor leaving group, which makes it less reactive. Amides have a leaving group that is a very strong base, compared to other options. As a result, amides are very stable compounds and are the least reactive out of the options.

Example Question #371 : Organic Chemistry, Biochemistry, And Metabolism

Understanding the function of the sodium-potassium pump, which of the following residues might possibly line the channel?

Possible Answers:

Isoleucine

Lysine

Aspartic acid

Oleic acid

Correct answer:

Aspartic acid

Explanation:

Aspartic acid is the only viable answer. Its R-group contains a second carboxylic acid that is deprotonated in most physiological conditions. The negatively charged aspartate will interact favorably with the positively charged cations moving through the pump. Lysine is a basic amino acid that contains an amine group at the terminal end of its R-group. Isoleucine is an aliphatic and hydrophobic amino acid. Oleic acid is an example of a fatty acid, which can carry a negative charge in some conditions, but it is not an amino acid and thus cannot be incorporated into a protein.

Example Question #371 : Organic Chemistry, Biochemistry, And Metabolism

Carbonic anhydrase is a very important enzyme that is utilized by the body. The enzyme catalyzes the following reaction:

A class of drugs that inhibits this enzyme is carbonic anhydrase inhibitors (eg. acetazolamide, brinzolamide, dorzolamide). These drugs are commonly prescribed in patients with glaucoma, hypertension, heart failure, high altitude sickness and for the treatment of basic drugs overdose.   

In patients with hypertension, carbonic anhydrase inhibitors will prevent the reabsorption of sodium chloride  in the proximal tubule of the kidney. When sodium is reabsorbed back into the blood, the molecule creates an electrical force. This electrical force then pulls water along with it into the blood. As more water enters the blood, the blood volume increase. By preventing the reabsorption of sodium, water reabsorption is reduced and the blood pressure decreases. 

When mountain climbing, the atmospheric pressure is lowered as the altitude increases. As a result of less oxygen into the lungs, ventilation increases. From the equation above, hyperventilation will result in more  being expired. Based on Le Chatelier’s principle, the reaction will shift to the left. Since there is more bicarbonate than protons in the body, the blood will become more basic (respiratory alkalosis). To prevent such life threatening result, one would take a carbonic anhydrase inhibitor to prevent the reaction from shifting to the left.  

Carbonic anhydrase inhibitors are useful in patients with a drug overdose that is acidic. The lumen of the collecting tubule is nonpolar. Due to the lumen's characteristic, molecules that are also nonpolar and uncharged are able to cross the membrane and re-enter the circulatory system. Since carbonic anhydrase inhibitors alkalize the urine, acidic molecules stay in a charged state.

James is designing a protein that is able to reabsorb back into the bloodstream in the presence of a carbonic anhydrase inhibitor. For this protein to be reabsorbed, which of the following animo acid(s) should the protein be abundant in?

I. Glu

II. Asn

III. Gln 

Possible Answers:

II and III

I and III

I only

II only

III only

Correct answer:

II and III

Explanation:

A carbonic anhydrase inhibitor will alkalinize the urine. Amino acids with an acidic side chain will lose their hydrogen and become charged in a basic condition. Amino acids that are basic will stay in an uncharged state and will be able to cross the lumen membrane of the collecting tubule. Of the answer choices, only asparagine (Asn) and glutamine (Gln) will stay in a neutral state due to the base property of the side chain. Glutamic acid (Glu) will lose its proton in the basic environment and will become charged.  

Example Question #44 : Reactions

The transformation of compound B to compound C below is known as what type of reaction?

Mcat_1

Possible Answers:

Dehydration

Hydroboration

Hydration

Hydrogenation

Decarboxylation

Correct answer:

Dehydration

Explanation:

The conversion of compound B to compound C results in the elimination of water, which, by definition, is a dehydration reaction. The hydroxyl group on compound B is protonated by the sulfuric acid, generating an leaving group and allowing the formation of the alkene product in compound C.

Hydroboration is the oxidation of an alkene with a borohydride (usually sodium borohydride) reagent, to produce an alkane. Hydration involves the use of a water reactant, usually producing an alcohol product. Hydrogenation is the oxidation of alkene double bonds with the use of a palladium interface to produce an alkane product. Decarboxylation results in product carbon dioxide from a carboxylic acid or carbonic anhydrase reactant.

Example Question #373 : Organic Chemistry, Biochemistry, And Metabolism

Which alcohol will react most rapidly via an SN1 mechanism?

Possible Answers:

Correct answer:

Explanation:

Tertiary alcohols react most rapidly via SN1 mechanisms because they form stable tertiary carbocations. Primary and secondary alcohols typically react most rapidly via SN2 mechanisms.

Of the available options,  is the only one that contains a tertiary alcohol.

Example Question #45 : Reactions

What type of enzymatic inhibitor binds to an allosteric location on the enzyme with equal affinity for the bound and unbound substrate states?

Possible Answers:

Competitive inhibitor

Suicide inhibitor

Noncompetitive inhibitor

Uncompetitive inhibitor

Correct answer:

Noncompetitive inhibitor

Explanation:

Noncompetitive inhibitors are a specific type of mixed inhibitor that binds to both the free enzyme and the enzyme-substrate complex with equal affinities, resulting in the same binding affinity (Km value) but a decrease in the maximum rate of reaction (Vmax value).

Example Question #51 : Reactions

The enzyme glucose-6-phosphatase plays the most important role in which tissue/organ?

Possible Answers:

Liver

Adipose tissue

Muscle tissue

Brain

Correct answer:

Liver

Explanation:

Glucose-6-phosphatase is responsible for removing the phosphate group from glucose-6-phosphate. The result is free glucose, which can be released into the blood. This process takes place at the end of either glycogenolysis or gluconeogenesis, both processes that are most prominent in the liver due to its large stores of glycogen.

Example Question #52 : Reactions

Which of the following statements is false regarding enzyme function?

Possible Answers:

Enzymes do not affect the thermodynamics of a reaction

Enzymes are a type of catalyst

Enzymes increase the activation energy

All of these are true

Enzymes are not used up during the reaction

Correct answer:

Enzymes increase the activation energy

Explanation:

Enzymes function as a biological catalysts by lowering reactions' activation energies. They are not used up in the reaction mechanism, nor do they affect the thermodynamics of the reaction. They only affect the reaction kinetics.

Example Question #53 : Reactions

The cellular membrane is a very important structure. The lipid bilayer is both hydrophilic and hydrophobic. The hydrophilic layer faces the extracellular fluid and the cytosol of the cell. The hydrophobic portion of the lipid bilayer stays in between the hydrophobic regions like a sandwich. This bilayer separation allows for communication, protection, and homeostasis. 

One of the most utilized signaling transduction pathways is the G protein-coupled receptor pathway. The hydrophobic and hydrophilic properties of the cellular membrane allows for the peptide and other hydrophilic hormones to bind to the receptor on the cellular surface but to not enter the cell. This regulation allows for activation despite the hormone’s short half-life. On the other hand, hydrophobic hormones must have longer half-lives to allow for these ligands to cross the lipid bilayer, travel through the cell’s cytosol and eventually reach the nucleus. 

Cholesterol allows the lipid bilayer to maintain its fluidity despite the fluctuation in the body’s temperature due to events such as increasing metabolism. Cholesterol binds to the hydrophobic tails of the lipid bilayer. When the temperature is low, the cholesterol molecules prevent the hydrophobic tails from compacting and solidifying. When the temperature is high, the hydrophobic tails will be excited and will move excessively. This excess movement will bring instability to the bilayer. Cholesterol will prevent excessive movement.

Epinephrine binds to its receptor on the surface of the cell. Molecule A does bind to the same receptor but is found to bind a different part of the receptor molecule than does epinephrine, causing the receptor to undergo a confirmation change and no longer fits with its associated ligand. What type of regulation is this? 

Possible Answers:

Direct inhibition 

Partial inhibition

Competitive inhibition

None of these

Noncompetitive inhibition

Correct answer:

Noncompetitive inhibition

Explanation:

According to the question, molecule A acts on a different site to inhibit epinephrine's receptor. This is an example of a noncompetitive inhibitor.  

Example Question #1 : Properties Of Hydrocarbons

Which of the following hydrocarbons would have the lowest boiling point?

Possible Answers:

2,2,3-trimethylbutane

Heptane

2,4-dimethylpentane

3-methylhexane

2-methylhexane

Correct answer:

2,2,3-trimethylbutane

Explanation:

Increased branching decreases the intermolecular dispersion forces between hydrocarbon molecules. As a result, it takes less energy (and heat) to overcome these forces, resulting in a lower boiling point. Since 2,2,3-trimethylbutane has the most branching of the five choices, its boiling point should be the lowest.

Hydrocarbon boiling point is also dependent on the length of the carbon chain, and increases as the length of the chain increases. We can conclude that heptane likely has the highest boiling point of the answer choices, as it does not branch and has the longest carbon chain.

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