MCAT Biology : Molecular Properties

Study concepts, example questions & explanations for MCAT Biology

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Example Questions

Example Question #23 : Functional Groups And Properties

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

Possible Answers:

Isoleucine

Oleic acid

Aspartic acid

Lysine

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 #24 : Functional Groups And Properties

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 only

I and III

III only

II 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 #1 : Properties Of Hydrocarbons

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

Possible Answers:

2-methylhexane

Heptane

3-methylhexane

2,4-dimethylpentane

2,2,3-trimethylbutane

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.

Example Question #2 : Properties Of Hydrocarbons

Which of the following statements is false concerning alkanes?

Possible Answers:

Lengthening the carbon chain raises the boiling point

Branching in an alkane raises the boiling point

Lengthening the carbon chain raises the melting point

Branching in an alkane raises the melting point

Correct answer:

Branching in an alkane raises the boiling point

Explanation:

Adding to the length of an alkane carbon chain will increase the molecular weight. This raises both the boiling and melting point. Branching within the alkane will create steric disturbances, making it difficult for molecules for densely pack together. This separates the molecules more, raising the melting point by making it more difficult to form a solid, but lowering the boiling point by making it easier to exist as a gas.

Example Question #91 : Molecular Properties

Which of the following carbocation intermediates requires the least activation energy?

Possible Answers:

Carbo3

Carbo4

Carbo2

Carbo1

Cannot be determined

Correct answer:

Carbo1

Explanation:

The more stable the carbocation, the lower the activation energy for reaching that intermediate will be. The more substituted a carbocation is, the more stable it is. The carbocation bonded to three alkanes (tertiary carbocation) is the most stable, and thus the correct answer.

Secondary carbocations will require more energy than tertiary, and primary carbocations will require the most energy.

Example Question #2 : Properties Of Hydrocarbons

What is the correct IUPAC name for the molecule shown below?

Name_molecule

Possible Answers:

(E)-4-hydroxy-5-bromo-5-methyl-2-hexene

(E)-2-bromo-2-methyl-4-hexen-3-ol

(E)-2-bromo-2-methyl-3-hydroxy-4-hexene

(Z)-2-bromo-2-methyl-4-hexen-3-ol

(Z)-2-bromo-2-methyl-3-hydroxy-4-hexene

Correct answer:

(E)-2-bromo-2-methyl-4-hexen-3-ol

Explanation:

According to IUPAC rules, the alcohol has the highest priority; therefore the carbons should be counted to give this group the lowest carbon number. This translates to counting the carbon chain from the left side, giving six total carbons (a hexane). At carbon #2 there exist both a -bromo and -methyl group. At carbon #3 we have our alcohol. Finally, at carbon #4 begins an alkene. Thus we have:

Root carbon chain = hexane

Functional groups = 2-bromo, 2-methyl, 3-ol (highest priority), 4-ene (It is a trans alkene, where each carbon branches away from each other, hence the E prefix).

Putting this together, we have (E)-2-bromo-2-methyl-4-hexen-3-ol.

*Note: Because, the hydroxyl group is the highest priority, it is used as X-ol, where X is its position. Had there been a higher priority group, then X-hydroxy would have been used. 

 

Example Question #383 : Organic Chemistry, Biochemistry, And Metabolism

Which of the following cycloalkanes is the least stable?

Possible Answers:

Cycloheptane

Cyclohexane

Cyclobutane

Cyclododecane

Correct answer:

Cyclobutane

Explanation:

Stability in a ring is dependent on the amount of strain placed on carbon atoms when in a cyclic formation. Increased bending in a ring causes the carbons to deviate from their desired bond angle, which increases crowding among the atoms. We typically say that cyclohexane has zero ring strain, and the strain increases as you increase or decrease the number of carbons in the ring. Although strain will increase as the number of carbons in the ring increases to nine, strain will then decrease as the number of carbons exceeds nine in the ring.

Cyclobutane places a great amount of strain on the carbons in the ring, making it the least stable cycloalkane out of all possible options.

Example Question #4 : Reaction Mechanisms

 

  Bromobenzene_labAnisole_labNitrobenzene_lab

Of the aromatic compounds shown above, which would be meta-directing groups for subsequent electrophilic aromatic substitution reactions?

Possible Answers:

Nitrobenzene and anisole

Anisole

Bromobenzene

Nitrobenzene

Correct answer:

Nitrobenzene

Explanation:

Only nitrobenzene would be a meta-directing group for additional electrophilic aromatic substitution (EAS) reactions. Although the bromine of bromobenzene is an electron-withdrawing group, halogens are not meta-directors; therefore, additional EAS reactions with bromobenzene would result in ortho or para attached substituents.

Remember that ortho additions are adjacent to the first substituent, meta additions are two carbons displaced from the first substituent, and para additions are opposite the first substituent.

(Note that these reactions would take place much more slowly than if there was an electron-donating group attached).

Example Question #32 : Organic Chemistry

Which of the aromatic compounds (shown above) would undergo electrophilic aromatic substitution most quickly?

Possible Answers:

Anisole

Bromobenzene

Nitrobenzene

Acetophenone

Correct answer:

Anisole

Explanation:

Electrophilic aromatic substitution occurs most rapidly when the aromatic compound has electron-donating groups attached. Due to their electron affinity, halogens are electron-withdrawing groups. Acetophenone and nitrobenzene both bear partial positive charges on the substituent directly attached to the benzene ring, which pulls electron density out of the ring as well, causing the reaction not to occur.

Anisole is the only compound with an electron-donating group, and is the correct answer. The lone pairs on the oxygen atom can be used to initiate new bonds.

Example Question #3 : Properties Of Hydrocarbons

Which of the following is the most stable form of cyclohexane?

Possible Answers:

Half-chair formation

Boat formation

Chair formation

Twist-boat formation

Correct answer:

Chair formation

Explanation:

The chair formation is the most stable form of cyclohexane because there is little torsional strain, allowing the chair formation to have the lowest overall energy of the cyclohexane forms. In the chair conformation, hydrogen substituents alternate between axial and equatorial orientations, reducing steric hindrance and promoting stability.

The boat conformations have higher energy than the chair formations. There is steric strain caused by the interaction between hydrogens, and torsional strain due to eclipsed carbons. In this conformation, two adjacent hydrogen substituents will be axial in the same plane. The twist-boat conformation is also a high-energy form of cyclohexane for some of the same reasons as the boat conformation. The twist-boat form occurs when the adjacent hydrogen substituents are both axial, but are twisted away from one other to prevent hindrance within a single plane.

The half-chair formation has considerable torsional strain and is higher in energy than the chair formation. The half-chair is essentially formed as a transition state when entering the chair conformation and consists of five carbon atoms essentially in the same plane.

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