All MCAT Biology Resources
Example Questions
Example Question #1 : Other Properties Of Biological Compounds
How do the bond angles in ammonia compare to the bond angles in methane?
Ammonia has smaller bond angles compared to methane.
Methane has smaller bond angles compared to ammonia.
More information is required to answer this question.
Both compounds have the same bond angles.
Ammonia has smaller bond angles compared to methane.
Both ammonia and methane display sp3 hybridization, however, methane is surrounded by four hydrogens, while nitrogen is surrounded by three hydrogens and a lone electron pair. Lone electron pairs in a molecule require more room compared to bonding atom pairs, as they generate more electron repulsion. As a result, the lone pair in ammonia will make the bond angles in ammonia smaller than the bond angles in the methane molecule.
*Extra information: the bond angles in methane are 109.5o, while ammonia has bond angles of 107o.
Example Question #57 : Molecular Properties
Which of the following molecules would have a trigonal pyramidal molecular structure?
Recall that electronic structure refers to the geometry of the atoms and lone pairs around the central atom, while the molecular structure refers strictly to the geometry of the atoms of the molecule. , , and all adopt a tetrahedral electronic geometry; however, when considering just the atoms, only NH3 has a trigonal pyramidal geometry. is trigonal planar, while is linear.
Example Question #2 : Other Properties Of Biological Compounds
What is the molecular geometry around the carbonyl carbon in acetic acid?
Trigonal planar
Tetrahedral
Trigonal pyramidal
Bent
Trigonal planar
Acetic acid has the formula . This compound is common enough that the IUPAC name will rarely be given, and you will need to recognize it by common name alone.
The carbonyl carbon exhibits hybridization, and has no lone electron pairs. There are three atoms bonded around the carbon. In order for the atoms to be farthest away from one another, they will all orient themselves at angles within the same plane. This results in a trigonal planar geometry.
An atom with four bonds and no lone pairs will show tetrahedral geometry. An atom with two bonds and one lone pair will be bent. Trigonal pyramidal geometry results from three bonded atoms and one lone pair.
Example Question #1546 : Mcat Biological Sciences
Which compound does not have any atoms with bond angles of ?
By drawing the Lewis structures of these molecules, we can determine the bond angles between the atoms. The following molecular geometries contain bond angles of : octahedral, trigonal bipyramidal, and square planar.
has a tetrahedral geometry, which means that there are only bond angles of .
will be square planar. will be octahedral. will be trigonal bipyramidal.
Example Question #61 : Molecular Properties
Which of the following compounds is the most stable?
Benzene
Cyclopentane
2,4-cyclobutadiene
Cyclohexane
1,3-dichlorocyclooctane
Benzene
Aromatic compounds are highly stable due to resonance stability. The two aromatic choices are benzene and 2,4-cyclobutadiene. Because a six-carbon ring is more stable than a four-carbon ring, benzene is the best answer. Ring strain will distort a four-carbon ring, increasing its bond energy and decreasing its stability.
Example Question #62 : Molecular Properties
Which of the following statements is not true concerning resonance structures?
Resonance atoms are in the same plane
There is a constant number of unpaired electrons
The number of double bonds is constant between structures
Atoms do not move between structures
The number of double bonds is constant between structures
There is no requirement that the number of double bonds in a molecule be kept constant. Electrons can be moved in order to convert between triple bonds, double bonds, and single bonds when transitioning between resonance structures.
Example Question #1 : Other Properties Of Biological Compounds
Nuclear transport is a very important concept of study in modern cellular biology. Transport of proteins into the nucleus of an organism requires energy in the form of GTP, which is attached to a protein called Ras-related Nuclear protein (RAN).
RAN is a monomeric G protein found in both the cytosol as well as the nucleus and its phosphorylation state plays an important role in the movement of proteins into and out of the nucleus. Specifically, RAN-GTP and RAN-GDP binds to nuclear import and export receptors and carries them into or out of the nucleus. They also play a role in dropping off cargo that import and export receptors hold onto. RAN's functions are controlled by two other proteins: RAN guanine exchange factor (RAN-GEF) and RAN GTPase activating protein (GAP). RAN-GEF binds a GTP onto RAN, while RAN-GAP hydrolyzes GTP into GDP. As a result, there is a RAN-GTP and RAN-GDP concentration gradient that forms between the cytosol and nucleus.
Suppose an experiment was done and the concentrations of RAN-GEF and RAN-GAP were determined, as shown below:
Nucleus:
Cytosol:
Given these results, what can be said about the charge of RAN in the cytosol and nucleus?
RAN's charge state cannot be determined without further experimentation
RAN has equal charge in both the nucleus and cytosol
RAN is more positively charged in the cytosol than in the nucleus
RAN's charge may fluctuate depending on the metabolic state of the cell
RAN is more positively charged in the nucleus than in the cytosol
RAN is more positively charged in the cytosol than in the nucleus
First, we have to understand which of the proteins exist in higher concentrations and the impact of both proteins. From the information, we can see that RAN-GEF has a higher concentration in the nucleus than in the cytosol, and that RAN-GAP has a higher concentration in the cytosol than in the nucleus. RAN-GEF will add on a GTP group onto RAN, whereas RAN-GAP will hydrolyze the RAN-GTP to RAN-GDP.
We can therefore conclude that RAN-GTP exists at a higher concentration in the nucleus, and that RAN-GDP exists at a higher concentration in the cytosol. Since GTP has a more negative charge than GDP, RAN will have a more negative charge in the nucleus, or conversely that RAN will have a more positive charge in the cytosol.
Example Question #1554 : Mcat Biological Sciences
The central nervous system consists of the brain and the spinal cord. In general, tracts allow for the brain to communicate up and down with the spinal cord. The commissures allow for the two hemispheres of the brain to communicate with each other. One of the most important commissures is the corpus callosum. The association fibers allow for the anterior regions of the brain to communicate with the posterior regions. One of the evolved routes from the spinal cord to the brain is via the dorsal column pathway. This route allows for fine touch, vibration, proprioception and 2 points discrimination. This pathway is much faster than the pain route. From the lower limbs, the signal ascends to the brain via a region called the gracile fasciculus. From the upper limbs, the signal ascends via the cuneate fasciculus region in the spinal cord.
One of the most common neurotransmitters is acetylcholine. Which of the following statements is/are true?
I. The neurotransmitter is water soluble
II. The neurotransmitter is lipid soluble
III. The neurotransmitter has a charge
III only
None of these
I only
II only
I and III
I and III
Acetylcholine is water soluble, which is in opposition to claim II. Furthermore, most water soluble molecules have some degree of charge polarity. Acetylcholine does indeed have a charged amino group, which contributes to its polarity and solubility in water.
Example Question #2 : Other Properties Of Biological Compounds
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.
Which of the following hormones utilizes cholesterol as its precursor?
I. Cortisol
II. Testosterone
III. Dihydroxytestosterone
None of these
I only
I, II and III
III only
II only
I, II and III
Cortisol and testosterone both utilizes cholesterol to the precursor. Dihydrotestosterone is a potent form of testosterone.
Example Question #3 : Other Properties Of Biological Compounds
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.
Based on the passage, cholesterol has which of the following properties?
I. Hydrophilic
II. Hydrophobic
III. Can cross the lipid bilayer
II only
II and III
I only
I, II, and III
III only
II and III
According to the passage, cholesterol interacts with the hydrophobic region of the lipid bilayer. Therefore, cholesterol must by hydrophobic. As mentioned in the passage that hydrophobic molecules are able to cross the bilayer.
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