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Test Prep MCAT Test Exam
Medical College Admission Test: Verbal Reasoning, Biological Sciences, Physical Sciences, Writing Sample (Page 24 )

Updated On: 30-Jan-2026
Page 24 of 164
PDF with 811 Questions

Which of the following neurotransmitters could be imbalanced if someone woke up and realized they could not move any muscle?

  1. Catecholamine
  2. Acetylcholine
  3. Frataxin
  4. Cortisol

Answer(s): B

Explanation:

B is correct. Acetylcholine is a neurotransmitter. It is released by the neurons. It then synapses at the muscular junctions and provokes the contraction of the skeletal motor cells. It is critical for motor movement. Its absence or blockage (for example, via antagonist drugs) would cause paralysis.
E). This is incorrect. Catecholamines are neurotransmitters (such as adrenaline, dopamine, and noradrenaline) that have been found to have a role in the flight-or-fight response to stressful stimulI). Specifically, they are involved in the flight response and responsible for the activation of the sympathetic nervous system. They are not critical for skeletal muscle control.
F). This is incorrect. Frataxin is one protein which may cause paralysis. It is not a neurotransmitter.
G). This is incorrect. Cortisol is a hormone that has been found to have a role in the flight-or-fight response to stressful stimulI). Specifically, it is involved in the flight response and responsible for the activation of the sympathetic nervous system. It is not critical for skeletal muscle control.



Which one of the following chemicals would UNLIKELY be involved in the onset of a panic attack?

  1. Melatonin
  2. Epinephrine
  3. Norepinephrine
  4. Serotonin

Answer(s): A

Explanation:

A is correct. Melatonin helps to sustain people's circadian rhythms. It is mainly involved in the sleep-wake cycle regulation. It is secreted into the bloodstream at night, inducing drowsiness and making the person feel sleepy.
It has been found to be effective in the treatment of anxiety due to its calming effects. These effects suggest that melatonin is not involved in the outset of acute anxiety episodes. Moreover, no findings have yet associated low or high levels of melatonin with anxiety disorders.
B and C. These are incorrect. The hormone epinephrine (or adrenaline) and the neurotransmitter norepinephrine (or noradrenaline) are released in greater quantities during stressing, threatening situations.
Anxiety disorders involve experiencing normal situations as menacing. This perception prompts fight-or-flight, stress-related responses. For example, the person's blood flow, arousal, alertness, attention and memory are boosted; and the chemicals epinephrine and norepinephrine are released in great quantities. Thus, these substances could be present during the onset of a panic attack.
E. This is incorrect. Serotonin is a neurotransmitter. It is mainly involved in the regulation of homeostatic processes, such as sleep and appetite. It also plays an important role in the regulation of mood. Low levels of serotonin can result in the development of both depressive and anxiety disorders. Thus, serotonin levels might decrease during the onset of panic attack.



The equation of state of an ideal gas is given by the ideal gas law:
= nRT
PV
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas. The gas particles in a container are constantly moving at various speeds. These speeds are characterized by the Maxwell shown in the figure below.


If two particles collide, their velocities change. However, if the gas is in thermal equilibrium, the velocity distribution of the gas as a whole will remain unchanged by the collision.
The average kinetic energy (E) of a gas particle is given by:

Equation 1
where m is the mass of one particle and u is the root mean square speed (rms speed) of the gas particles:

where N is the number of gas particles; this is different from the average speed. For an ideal gas, the kinetic energy of all the particles is:

Equation 2
where n is the number of moles of gas. Combining these equations gives:

Equation 3
where M is the molar mass of the gas particles.
The average distance a particle travels between collisions is known as the mean free path l. Intuitively, the mean free path (mfp) could be expected to be larger for gases at low pressure, since there is a lot of space between particles. Similarly, the mfp should be larger when the gas particles are small. The following expression for the mfp shows this to be correct.

Equation 4
In this equation, s is the atomic diameter (typically on the order of 10-8), k is the Boltzmann constant, and P is the pressure.
In addition to colliding with one another, gas particles also collide with the walls of their container. If the

container wall has a pinhole that is small compared to the mfp of the gas, and a pressure differential exists across the wall, the particles will effuse (or escape) through this pinhole without disturbing the Maxwellian distribution of the particles. The rate of effusion can be described by:

Equation 5
Where neff is the number of moles of effusing particles, A is the area of the pinhole, P and P1 are the pressures on the inside and outside of the container wall respectively, and P > P1.
Which of the following gives values for both standard temperature and pressure?

  1. 273 K and 760 Torr
  2. 273 K and 1 atm
  3. 0° C and 760 mm Hg
  4. All of the above

Answer(s): D

Explanation:

Standard temperature is usually expressed in either Celsius or Kelvin; for pressure, there are 3 different commonly used units. These are Torr, atmospheres, and millimeters of mercury. Conversion to other commonly used units is possible given standard temperature and pressure in one set of units. Conversion factors were not given in this problem because it is general knowledge that 1 atmosphere of pressure is equal to 760 Torr and 760 millimeters of mercury. To convert from Celsius to Kelvin, simply add 273. If you recall that standard temperature is 0 degrees Celsius, then it is a simple matter to convert to Kelvin to get choice D.



The equation of state of an ideal gas is given by the ideal gas law:
= nRT
PV
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas. The gas particles in a container are constantly moving at various speeds. These speeds are characterized by the Maxwell shown in the figure below.

If two particles collide, their velocities change. However, if the gas is in thermal equilibrium, the velocity distribution of the gas as a whole will remain unchanged by the collision.
The average kinetic energy (E) of a gas particle is given by:

Equation 1
where m is the mass of one particle and u is the root mean square speed (rms speed) of the gas particles:

where N is the number of gas particles; this is different from the average speed. For an ideal gas, the kinetic energy of all the particles is:

Equation 2
where n is the number of moles of gas. Combining these equations gives:

Equation 3
where M is the molar mass of the gas particles.
The average distance a particle travels between collisions is known as the mean free path l. Intuitively, the mean free path (mfp) could be expected to be larger for gases at low pressure, since there is a lot of space between particles. Similarly, the mfp should be larger when the gas particles are small. The following expression for the mfp shows this to be correct.

Equation 4
In this equation, s is the atomic diameter (typically on the order of 10-8), k is the Boltzmann constant, and P is the pressure.
In addition to colliding with one another, gas particles also collide with the walls of their container. If the container wall has a pinhole that is small compared to the mfp of the gas, and a pressure differential exists across the wall, the particles will effuse (or escape) through this pinhole without disturbing the Maxwellian distribution of the particles. The rate of effusion can be described by:

Equation 5
Where neff is the number of moles of effusing particles, A is the area of the pinhole, P and P1 are the pressures on the inside and outside of the container wall respectively, and P > P1.
If a pinhole were made in a container containing a mixture of equal amounts of H2, O2, N2 and CO2, which gas would have the fastest effusion rate?

  1. H2
  2. O2
  3. N2
  4. Co2

Answer(s): A

Explanation:

According to Equation 5, the rate of effusion depends on the area of the pinhole, the pressures inside and outside the container, the molecular weight of the particles, the gas constant R, and the temperature. Since all 4 gases are in one container, subject to the same conditions, the only one of these factors that is different for the 4 gases is molecular weight. Since the rate of effusion is inversely proportional to the square root of the molecular weight, the rate of effusion increases as the molecular weight decreases. Thus, the lightest particle will effuse through the pinhole fastest. Molecular hydrogen is the lightest of the four species, with a molecular weight of 2 grams per mole, so it will effuse the fastest and choice A is the correct answer.



The equation of state of an ideal gas is given by the ideal gas law:
= nRT
PV
where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature of the gas. The gas particles in a container are constantly moving at various speeds. These speeds are characterized by the Maxwell shown in the figure below.

If two particles collide, their velocities change. However, if the gas is in thermal equilibrium, the velocity distribution of the gas as a whole will remain unchanged by the collision.
The average kinetic energy (E) of a gas particle is given by:

Equation 1
where m is the mass of one particle and u is the root mean square speed (rms speed) of the gas particles:

where N is the number of gas particles; this is different from the average speed. For an ideal gas, the kinetic

energy of all the particles is:

Equation 2
where n is the number of moles of gas. Combining these equations gives:

Equation 3
where M is the molar mass of the gas particles.
The average distance a particle travels between collisions is known as the mean free path l. Intuitively, the mean free path (mfp) could be expected to be larger for gases at low pressure, since there is a lot of space between particles. Similarly, the mfp should be larger when the gas particles are small. The following expression for the mfp shows this to be correct.

Equation 4
In this equation, s is the atomic diameter (typically on the order of 10-8), k is the Boltzmann constant, and P is the pressure.
In addition to colliding with one another, gas particles also collide with the walls of their container. If the container wall has a pinhole that is small compared to the mfp of the gas, and a pressure differential exists across the wall, the particles will effuse (or escape) through this pinhole without disturbing the Maxwellian distribution of the particles. The rate of effusion can be described by:

Equation 5
Where neff is the number of moles of effusing particles, A is the area of the pinhole, P and P1 are the pressures on the inside and outside of the container wall respectively, and P > P1.
The mean free path of a gas will be longer if the :

  1. pressure of the gas is increased.
  2. number of gas particles per unit volume is increased.
  3. distance between collisions is decreased.
  4. pressure of the gas is decreased.

Answer(s): D

Explanation:

The mean free path of a particle is the average distance the particle can travel before it collides with another gas particle or the wall of the container. The longer the distance that a gas particle travels between collisions, the further apart the individual gas molecules must be. That means that the volume of the gas must increase to increase the distance between collisions. One way to increase the volume is to decrease the pressure of the
gas, so D is correct. Choice A is incorrect because an increase in pressure leads to a decrease in the volume and thus a decrease in mean free path as well. Thus the gas particles are closer together and are more likely to collide, decreasing the average distance they travel between collisions. Choice B is incorrect because if the number of particles is increased while the volume remains constant, the likelihood of a collision will also increase. The pressure also increases as the number of particles per unit volume increases. Choice C is incorrect because the mean free path is analogous to the distance between collisions and if this is decreased then the mean free path decreases s well. Basically, choices A, B and C will all decrease the mean free path of a gas, making them all incorrect.



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