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MCAT: Chemistry and Physics practice exam

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This free MCAT practice exam simulates the actual test experience to help you prepare for exam day. Questions are designed to match the real MCAT exam format, difficulty, and time constraints. Once you select an answer, it's locked in. Your progress is saved automatically, so you can return anytime. Good luck!

Solve each of the MCAT practice test questions below to get a feel for what to expect on the actual MCAT exam. Achievable's free MCAT practice questions are scored instantly, providing the correct answer along with an easy to understand explanation. Get started on the path to passing the MCAT exam by solving 59 MCAT exam sample questions.
Question 1

A pharmaceutical company is developing a new drug derived from plant extracts. They need to separate and collect a specific bioactive compound from a complex mixture of other plant metabolites for further testing. Which of the following techniques would be most suitable for isolating this compound based on its polarity?

A.
Distillation
B.
Polymerase chain reaction (PCR)
C.
Thin-layer chromatography (TLC)
D.
Centrifugation
Question 2

During ATP hydrolysis, which specific bond is broken to release energy?

A.
The phosphoanhydride bond between the γ\gamma and β\beta phosphate groups
B.
The bond connecting the α\alpha phosphate directly to the adenine base
C.
The phosphoanhydride bond that joins the β\beta and α\alpha phosphate groups
D.
One of the bonds between a phosphate and its hydroxyl group in the terminal γ\gamma phosphate
Question 3

Chemical reactions can be described by how their free energy changes. If a reaction has a ΔG\Delta G of –598kcal/mol–598 kcal/mol, what kind of energy profile does it follow and how would introducing a catalyst affect the ΔG\Delta G of this process?

A.
Energy-absorbing; catalyst increases ΔG\Delta G
B.
Energy-releasing; catalyst decreases ΔG\Delta G
C.
Energy-absorbing; catalyst decreases ΔG\Delta G
D.
Energy-releasing; catalyst has no effect on ΔG\Delta G
Question 4

A chemical system is at equilibrium. If the volume of the container is halved, what happens to the equilibrium if the reaction is N2(g)+3H2(g)⇌2NH3(g)N_2(g) + 3H_2(g) ⇌ 2NH_3(g)?

A.
Shifts left
B.
Shifts right
C.
No shift
D.
Reaction stops
Question 5

A medical student is preparing phosphate-buffered saline (PBS) for a lab experiment. She recalls that a buffer solution resists changes in pH upon the addition of small amounts of acid or base. While selecting chemical reagents, she considers combining the following pairs of compounds to create a buffer that functions around physiological pH (~7.4). Which of the following combinations is most likely to produce an effective buffer?

A.
HClHCl and NaClNaCl
B.
NaOHNaOH and NaClNaCl
C.
NH3NH_3 and NH4ClNH_4Cl
D.
CH3COONaCH_3COONa and NaOHNaOH
Question 6

A chemist is analyzing how different substitution reactions affect molecular structure. She starts with a pure sample of (S)-2-bromo-2-methylbutane and runs several reactions. After one, she finds the product is a racemic mixture of ( R)- and (S)-2-methylbutanol. Which reaction mechanism most likely produced this result?

A.
SN2 reaction on a primary halide
B.
SN1 reaction on a tertiary halide
C.
E2 reaction on a trans alkyl halide
D.
Addition of Br2Br_2 to a cis-alkene
Question 7

A ball is thrown vertically upward with an initial velocity of 20 m/s20 \, m/s. Neglecting air resistance, what is the maximum height the ball reaches? (Assume g=10 m/s2g = 10 \, m/s^2)

A.
10 m
B.
20 m
C.
40 m
D.
80 m
Question 8

A 1200 W hair dryer is plugged into a 120 V outlet. What is the current flowing through the hair dryer?

A.
5 A
B.
15 A
C.
20 A
D.
10 A
Question 9

A sound wave has a frequency of 500500 Hz and a wavelength of 0.70.7 meters. What is the speed of the sound wave?

A.
280 m/sm/s
B.
420 m/sm/s
C.
350 m/sm/s
D.
571 m/sm/s
Question 10

A lens has a focal length of +20+20 cm. If an object is placed 3030 cm in front of the lens, where is the image formed?

A.
6060 cm behind the lens.
B.
1010 cm behind the lens
C.
1010 cm in front of the lens
D.
6060 cm in front of the lens
Question 11

A radioactive isotope has a half-life of 8 days. If a sample initially contains 100 grams of the isotope, how much of the isotope remains after 24 days?

A.
25 grams
B.
50 grams
C.
75 grams
D.
12.5 grams
Question 12

A uniform, 4-meter-long horizontal beam is balanced on a fulcrum positioned at its midpoint. The beam itself is massless and extends 2 meters on either side of the fulcrum. A 40 N mass is suspended exactly 1 meter from the left end of the beam. To keep the beam perfectly horizontal and prevent it from rotating, a second mass of 20 N must be suspended from the right side. Where should the 20 N mass be placed on the right side of the fulcrum in order to maintain equilibrium?

A.
1 meter from the right end
B.
0.5 meters from the right end
C.
1.5 meters from the center
D.
2 meters from the center
Question 13

A vertical cylindrical tank is filled completely with water to a height of 3.0 meters. A pressure gauge is installed at the very bottom of the tank and is used to measure the pressure exerted by the water column alone (gauge pressure). Assume standard conditions, neglect any atmospheric pressure, and use the following constants: ρwater=1000kg/m3ρ_{water}=1000kg/m^3, g=10m/s2g = 10m/s^2. What is the pressure recorded by the gauge at the bottom of the tank?

A.
30,000 Pa
B.
1000 Pa
C.
3000 Pa
D.
300,000 Pa
Question 14

A proton is launched into a region of space where a uniform magnetic field points into the page. The proton enters the magnetic field perpendicularly, with an initial speed of v₀ and immediately begins to travel in a circular path due to the magnetic force acting on it. Later, the speed of the proton is increased to 2v02v_0, while the strength of the magnetic field remains unchanged. What will happen to the radius of the circular path the proton travels?

A.
It remains unchanged
B.
It increases
C.
It decreases
D.
It becomes infinite
Question 15

A radioactive isotope undergoes beta decay. Which of the following occurs during this process?

A.
A proton is converted into a neutron.
B.
An electron is captured by the nucleus.
C.
A positron is emitted from the nucleus.
D.
A neutron is converted into a proton
Question 16

Vehicle accidents on icy roads result in significant property damage and injuries annually. National transportation statistics indicate that winter-related collisions cost millions in repairs and lost productivity. A local municipality is investigating methods to improve vehicle traction on a particularly hazardous, inclined bridge during freezing conditions. A diagram of the bridge is shown below.

Diagram of icy bridge friction

One approach to reducing these accidents is to implement advanced road surface treatments that enhance tire grip.

Test vehicles, equipped with specialized winter tires, were used to assess the frictional properties of the bridge surface. The coefficients of friction between the winter tires and the icy bridge surface are as follows:

Coefficient of static friction = 0.2 Coefficient of kinetic friction = 0.1


A 1500 kg vehicle is stationary on the icy bridge. The coefficient of static friction between the tires and the ice is 0.4. If the component of the gravitational force acting parallel to the bridge surface is 5000 N, what is the minimum normal force required to prevent the vehicle from sliding?

A.
10,000 N
B.
12,500 N
C.
15,000 N
D.
20,000 N
Question 17

Vehicle accidents on icy roads result in significant property damage and injuries annually. National transportation statistics indicate that winter-related collisions cost millions in repairs and lost productivity. A local municipality is investigating methods to improve vehicle traction on a particularly hazardous, inclined bridge during freezing conditions. A diagram of the bridge is shown below.

Diagram of icy bridge friction

One approach to reducing these accidents is to implement advanced road surface treatments that enhance tire grip.

Test vehicles, equipped with specialized winter tires, were used to assess the frictional properties of the bridge surface. The coefficients of friction between the winter tires and the icy bridge surface are as follows:

Coefficient of static friction = 0.2 Coefficient of kinetic friction = 0.1


A vehicle is traveling at a constant velocity up the icy bridge. Conceptually, how does the driving force required from the vehicle's engine relate to the forces of gravity and kinetic friction?

A.
The driving force must be less than the sum of gravity and friction.
B.
The driving force must be greater than the sum of gravity and friction.
C.
The driving force must be equal to the sum of gravity and friction.
D.
The driving force must be equal to gravity minus friction.
Question 18

Vehicle accidents on icy roads result in significant property damage and injuries annually. National transportation statistics indicate that winter-related collisions cost millions in repairs and lost productivity. A local municipality is investigating methods to improve vehicle traction on a particularly hazardous, inclined bridge during freezing conditions. A diagram of the bridge is shown below.

Diagram of icy bridge friction

One approach to reducing these accidents is to implement advanced road surface treatments that enhance tire grip.

Test vehicles, equipped with specialized winter tires, were used to assess the frictional properties of the bridge surface. The coefficients of friction between the winter tires and the icy bridge surface are as follows:

Coefficient of static friction = 0.2 Coefficient of kinetic friction = 0.1


A vehicle starts from rest at the top of the icy bridge and accelerates down. If the bridge has a length of 20 meters and the vehicle reaches a speed of 10 m/s at the bottom, what is the average acceleration of the vehicle?

A.
2 m/s2m/s^2
B.
2.5 m/s2m/s^2
C.
5 m/s2m/s^2
D.
10 m/s2m/s^2
Question 19

Vehicle accidents on icy roads result in significant property damage and injuries annually. National transportation statistics indicate that winter-related collisions cost millions in repairs and lost productivity. A local municipality is investigating methods to improve vehicle traction on a particularly hazardous, inclined bridge during freezing conditions. A diagram of the bridge is shown below.

Diagram of icy bridge friction

One approach to reducing these accidents is to implement advanced road surface treatments that enhance tire grip.

Test vehicles, equipped with specialized winter tires, were used to assess the frictional properties of the bridge surface. The coefficients of friction between the winter tires and the icy bridge surface are as follows:

Coefficient of static friction = 0.2 Coefficient of kinetic friction = 0.1


Due to a temperature increase, the coefficient of static friction on the icy bridge decreases. Conceptually, how does this affect the maximum incline at which a stationary vehicle can remain at rest?

A.
The maximum incline increases.
B.
The maximum incline decreases.
C.
The maximum incline remains the same.
D.
The change depends on the vehicle's mass.
Question 20

Vehicle accidents on icy roads result in significant property damage and injuries annually. National transportation statistics indicate that winter-related collisions cost millions in repairs and lost productivity. A local municipality is investigating methods to improve vehicle traction on a particularly hazardous, inclined bridge during freezing conditions. A diagram of the bridge is shown below.

Diagram of icy bridge friction

One approach to reducing these accidents is to implement advanced road surface treatments that enhance tire grip.

Test vehicles, equipped with specialized winter tires, were used to assess the frictional properties of the bridge surface. The coefficients of friction between the winter tires and the icy bridge surface are as follows:

Coefficient of static friction = 0.2 Coefficient of kinetic friction = 0.1


A vehicle with a mass of 1500kg is on the bridge. A sensor detects that the normal force on the vehicle is 12000 N. If the coefficient of static friction is 0.3, what is the maximum static friction force that can act on the tires?

A.
4000 N
B.
3600 N
C.
12000 N
D.
15000 N
Question 21

Energy Losses on a Swinging-Ship Ride

A swinging-ship ride is a giant pendulum: a boat-shaped gondola hangs from a pivot and swings back and forth along a circular arc. Each time the gondola passes through the bottom of the arc, a motor-driven tire presses against the hull and pushes it, adding energy until the gondola reaches its operating height. Between pushes, the gondola's mechanical energy shifts back and forth between gravitational potential energy at the ends of the arc and kinetic energy at the bottom, while air resistance and friction at the pivot convert some of it into thermal energy.

Engineers tested a ride whose loaded gondola has a mass of 5,000 kg and whose center of mass hangs 16 m below the pivot. They drove the gondola until its center of mass rose 7.20 m above its lowest point at the end of each swing, then pulled the tire away and let the gondola coast. Table 1 gives the greatest height the center of mass reached at the end of each full swing that followed, where a full swing is one complete back-and-forth cycle, lasting one period. For calculations, treat the ride as a simple pendulum whose period, about 8.0 s, does not depend on the size of the swing, and use g = 10 m/s².

Table 1. Maximum height of the gondola's center of mass while coasting

Full swings completedMaximum height above lowest point (m)
07.20
16.56
26.00
35.52
45.10

Ignoring energy losses, what is the gondola's speed when its center of mass is 2.20 m above its lowest point, after starting from rest at 7.20 m?

A.
6.6 m/s
B.
7.1 m/s
C.
12 m/s
D.
10 m/s
Question 22

Energy Losses on a Swinging-Ship Ride

A swinging-ship ride is a giant pendulum: a boat-shaped gondola hangs from a pivot and swings back and forth along a circular arc. Each time the gondola passes through the bottom of the arc, a motor-driven tire presses against the hull and pushes it, adding energy until the gondola reaches its operating height. Between pushes, the gondola's mechanical energy shifts back and forth between gravitational potential energy at the ends of the arc and kinetic energy at the bottom, while air resistance and friction at the pivot convert some of it into thermal energy.

Engineers tested a ride whose loaded gondola has a mass of 5,000 kg and whose center of mass hangs 16 m below the pivot. They drove the gondola until its center of mass rose 7.20 m above its lowest point at the end of each swing, then pulled the tire away and let the gondola coast. Table 1 gives the greatest height the center of mass reached at the end of each full swing that followed, where a full swing is one complete back-and-forth cycle, lasting one period. For calculations, treat the ride as a simple pendulum whose period, about 8.0 s, does not depend on the size of the swing, and use g = 10 m/s².

Table 1. Maximum height of the gondola's center of mass while coasting

Full swings completedMaximum height above lowest point (m)
07.20
16.56
26.00
35.52
45.10

Based on Table 1, how much mechanical energy did the gondola lose during the first two full swings after the tire was pulled away?

A.
6.0 kJ
B.
300 kJ
C.
60 kJ
D.
360 kJ
Question 23

Energy Losses on a Swinging-Ship Ride

A swinging-ship ride is a giant pendulum: a boat-shaped gondola hangs from a pivot and swings back and forth along a circular arc. Each time the gondola passes through the bottom of the arc, a motor-driven tire presses against the hull and pushes it, adding energy until the gondola reaches its operating height. Between pushes, the gondola's mechanical energy shifts back and forth between gravitational potential energy at the ends of the arc and kinetic energy at the bottom, while air resistance and friction at the pivot convert some of it into thermal energy.

Engineers tested a ride whose loaded gondola has a mass of 5,000 kg and whose center of mass hangs 16 m below the pivot. They drove the gondola until its center of mass rose 7.20 m above its lowest point at the end of each swing, then pulled the tire away and let the gondola coast. Table 1 gives the greatest height the center of mass reached at the end of each full swing that followed, where a full swing is one complete back-and-forth cycle, lasting one period. For calculations, treat the ride as a simple pendulum whose period, about 8.0 s, does not depend on the size of the swing, and use g = 10 m/s².

Table 1. Maximum height of the gondola's center of mass while coasting

Full swings completedMaximum height above lowest point (m)
07.20
16.56
26.00
35.52
45.10

In normal operation, the tire keeps the gondola swinging to 7.20 m. Based on Table 1, about how much average power must the tire deliver to the gondola to replace the energy lost?

A.
0.25 kW
B.
32 kW
C.
4.0 kW
D.
45 kW
Question 24

Energy Losses on a Swinging-Ship Ride

A swinging-ship ride is a giant pendulum: a boat-shaped gondola hangs from a pivot and swings back and forth along a circular arc. Each time the gondola passes through the bottom of the arc, a motor-driven tire presses against the hull and pushes it, adding energy until the gondola reaches its operating height. Between pushes, the gondola's mechanical energy shifts back and forth between gravitational potential energy at the ends of the arc and kinetic energy at the bottom, while air resistance and friction at the pivot convert some of it into thermal energy.

Engineers tested a ride whose loaded gondola has a mass of 5,000 kg and whose center of mass hangs 16 m below the pivot. They drove the gondola until its center of mass rose 7.20 m above its lowest point at the end of each swing, then pulled the tire away and let the gondola coast. Table 1 gives the greatest height the center of mass reached at the end of each full swing that followed, where a full swing is one complete back-and-forth cycle, lasting one period. For calculations, treat the ride as a simple pendulum whose period, about 8.0 s, does not depend on the size of the swing, and use g = 10 m/s².

Table 1. Maximum height of the gondola's center of mass while coasting

Full swings completedMaximum height above lowest point (m)
07.20
16.56
26.00
35.52
45.10

Engineers propose a children's version of the ride with a pivot arm one-quarter as long and a loaded gondola one-half the mass. Treated as a simple pendulum, its period would be closest to:

A.
4.0 s
B.
2.0 s
C.
2.8 s
D.
5.7 s
Question 25

Energy Losses on a Swinging-Ship Ride

A swinging-ship ride is a giant pendulum: a boat-shaped gondola hangs from a pivot and swings back and forth along a circular arc. Each time the gondola passes through the bottom of the arc, a motor-driven tire presses against the hull and pushes it, adding energy until the gondola reaches its operating height. Between pushes, the gondola's mechanical energy shifts back and forth between gravitational potential energy at the ends of the arc and kinetic energy at the bottom, while air resistance and friction at the pivot convert some of it into thermal energy.

Engineers tested a ride whose loaded gondola has a mass of 5,000 kg and whose center of mass hangs 16 m below the pivot. They drove the gondola until its center of mass rose 7.20 m above its lowest point at the end of each swing, then pulled the tire away and let the gondola coast. Table 1 gives the greatest height the center of mass reached at the end of each full swing that followed, where a full swing is one complete back-and-forth cycle, lasting one period. For calculations, treat the ride as a simple pendulum whose period, about 8.0 s, does not depend on the size of the swing, and use g = 10 m/s².

Table 1. Maximum height of the gondola's center of mass while coasting

Full swings completedMaximum height above lowest point (m)
07.20
16.56
26.00
35.52
45.10

Which conclusion about the coasting gondola is best supported by Table 1?

A.
The mechanical energy lost per swing decreased as the swings became smaller.
B.
The mechanical energy lost per swing stayed the same on every swing.
C.
The mechanical energy lost per swing increased as the swings became smaller.
D.
No mechanical energy was lost, since the period stayed the same on every swing.
Question 26

Water Pressure, Lung Gas and Buoyancy in Breath-Hold Diving

Because the chest wall is flexible, the gas in a submerged person's lungs is squeezed to nearly the pressure of the surrounding water. In seawater (density 1020 kg/m³), that pressure rises by about 10 kPa for every meter of depth. Take atmospheric pressure at the surface as 100 kPa and g as 9.8 m/s².

Water pressure limits snorkeling. A snorkeler draws in air at atmospheric pressure while the water presses on her chest. In one test in seawater, a volunteer could expand her chest to inhale only when the water pressure on her chest exceeded the pressure of the inhaled air by no more than 8.0 kPa.

Water pressure also compresses a breath-hold diver's lungs. Researchers modeled a diver who inhales 6.0 L of air at the surface and then descends. The model assumes the lung gas stays at body temperature and exchanges no gas with the blood (Table 1). Together, the diver and her equipment have a mass of 78.5 kg and, not counting the lung gas, a volume of 74.0 L.

Table 1. Modeled lung gas volume during a descent in seawater

Depth (m)Absolute pressure (kPa)Lung gas volume (L)
01006.0
102003.0
203002.0
304001.5

In a real dive, the blood also takes up oxygen from the lung gas. When the diver reached the bottom of a 30-m dive, oxygen made up 5.0% of her lung gas, and this fraction changed little during her rapid ascent. Divers can lose consciousness from lack of oxygen in the last few meters of an ascent, even when they felt well at the bottom.


Under the model's assumptions, if the diver had inhaled only 4.5 L of air at the surface, what would her lung gas volume be at a depth of 20 m?

A.
1.5 L
B.
0.50 L
C.
1.1 L
D.
2.3 L
Question 27

Water Pressure, Lung Gas and Buoyancy in Breath-Hold Diving

Because the chest wall is flexible, the gas in a submerged person's lungs is squeezed to nearly the pressure of the surrounding water. In seawater (density 1020 kg/m³), that pressure rises by about 10 kPa for every meter of depth. Take atmospheric pressure at the surface as 100 kPa and g as 9.8 m/s².

Water pressure limits snorkeling. A snorkeler draws in air at atmospheric pressure while the water presses on her chest. In one test in seawater, a volunteer could expand her chest to inhale only when the water pressure on her chest exceeded the pressure of the inhaled air by no more than 8.0 kPa.

Water pressure also compresses a breath-hold diver's lungs. Researchers modeled a diver who inhales 6.0 L of air at the surface and then descends. The model assumes the lung gas stays at body temperature and exchanges no gas with the blood (Table 1). Together, the diver and her equipment have a mass of 78.5 kg and, not counting the lung gas, a volume of 74.0 L.

Table 1. Modeled lung gas volume during a descent in seawater

Depth (m)Absolute pressure (kPa)Lung gas volume (L)
01006.0
102003.0
203002.0
304001.5

In a real dive, the blood also takes up oxygen from the lung gas. When the diver reached the bottom of a 30-m dive, oxygen made up 5.0% of her lung gas, and this fraction changed little during her rapid ascent. Divers can lose consciousness from lack of oxygen in the last few meters of an ascent, even when they felt well at the bottom.


Based on the passage, what was the partial pressure of oxygen in the diver's lung gas at the bottom of the 30-m dive and on reaching the surface?

A.
5.0 kPa at the bottom and 20 kPa at the surface
B.
20 kPa at the bottom and 20 kPa at the surface
C.
20 kPa at the bottom and 5.0 kPa at the surface
D.
5.0 kPa at the bottom and 5.0 kPa at the surface
Question 28

Water Pressure, Lung Gas and Buoyancy in Breath-Hold Diving

Because the chest wall is flexible, the gas in a submerged person's lungs is squeezed to nearly the pressure of the surrounding water. In seawater (density 1020 kg/m³), that pressure rises by about 10 kPa for every meter of depth. Take atmospheric pressure at the surface as 100 kPa and g as 9.8 m/s².

Water pressure limits snorkeling. A snorkeler draws in air at atmospheric pressure while the water presses on her chest. In one test in seawater, a volunteer could expand her chest to inhale only when the water pressure on her chest exceeded the pressure of the inhaled air by no more than 8.0 kPa.

Water pressure also compresses a breath-hold diver's lungs. Researchers modeled a diver who inhales 6.0 L of air at the surface and then descends. The model assumes the lung gas stays at body temperature and exchanges no gas with the blood (Table 1). Together, the diver and her equipment have a mass of 78.5 kg and, not counting the lung gas, a volume of 74.0 L.

Table 1. Modeled lung gas volume during a descent in seawater

Depth (m)Absolute pressure (kPa)Lung gas volume (L)
01006.0
102003.0
203002.0
304001.5

In a real dive, the blood also takes up oxygen from the lung gas. When the diver reached the bottom of a 30-m dive, oxygen made up 5.0% of her lung gas, and this fraction changed little during her rapid ascent. Divers can lose consciousness from lack of oxygen in the last few meters of an ascent, even when they felt well at the bottom.


Based on the snorkel test, what is the greatest depth in seawater at which the volunteer's chest could be held while she inhales air from the surface through a long snorkel?

A.
0.080 m
B.
7.8 m
C.
80 m
D.
0.80 m
Question 29

Water Pressure, Lung Gas and Buoyancy in Breath-Hold Diving

Because the chest wall is flexible, the gas in a submerged person's lungs is squeezed to nearly the pressure of the surrounding water. In seawater (density 1020 kg/m³), that pressure rises by about 10 kPa for every meter of depth. Take atmospheric pressure at the surface as 100 kPa and g as 9.8 m/s².

Water pressure limits snorkeling. A snorkeler draws in air at atmospheric pressure while the water presses on her chest. In one test in seawater, a volunteer could expand her chest to inhale only when the water pressure on her chest exceeded the pressure of the inhaled air by no more than 8.0 kPa.

Water pressure also compresses a breath-hold diver's lungs. Researchers modeled a diver who inhales 6.0 L of air at the surface and then descends. The model assumes the lung gas stays at body temperature and exchanges no gas with the blood (Table 1). Together, the diver and her equipment have a mass of 78.5 kg and, not counting the lung gas, a volume of 74.0 L.

Table 1. Modeled lung gas volume during a descent in seawater

Depth (m)Absolute pressure (kPa)Lung gas volume (L)
01006.0
102003.0
203002.0
304001.5

In a real dive, the blood also takes up oxygen from the lung gas. When the diver reached the bottom of a 30-m dive, oxygen made up 5.0% of her lung gas, and this fraction changed little during her rapid ascent. Divers can lose consciousness from lack of oxygen in the last few meters of an ascent, even when they felt well at the bottom.


As the diver descends from the surface to 10 m, her lung gas is compressed at constant temperature (Table 1). Which statement about work during this compression is correct?

A.
No work is done on the gas, because its temperature and internal energy stay constant.
B.
The gas does positive work on its surroundings, because its pressure doubles.
C.
The surroundings do positive work on the gas, equal to the area under its P–V curve.
D.
The work done on the gas equals its final pressure times its final volume, or 600 J.
Question 30

Water Pressure, Lung Gas and Buoyancy in Breath-Hold Diving

Because the chest wall is flexible, the gas in a submerged person's lungs is squeezed to nearly the pressure of the surrounding water. In seawater (density 1020 kg/m³), that pressure rises by about 10 kPa for every meter of depth. Take atmospheric pressure at the surface as 100 kPa and g as 9.8 m/s².

Water pressure limits snorkeling. A snorkeler draws in air at atmospheric pressure while the water presses on her chest. In one test in seawater, a volunteer could expand her chest to inhale only when the water pressure on her chest exceeded the pressure of the inhaled air by no more than 8.0 kPa.

Water pressure also compresses a breath-hold diver's lungs. Researchers modeled a diver who inhales 6.0 L of air at the surface and then descends. The model assumes the lung gas stays at body temperature and exchanges no gas with the blood (Table 1). Together, the diver and her equipment have a mass of 78.5 kg and, not counting the lung gas, a volume of 74.0 L.

Table 1. Modeled lung gas volume during a descent in seawater

Depth (m)Absolute pressure (kPa)Lung gas volume (L)
01006.0
102003.0
203002.0
304001.5

In a real dive, the blood also takes up oxygen from the lung gas. When the diver reached the bottom of a 30-m dive, oxygen made up 5.0% of her lung gas, and this fraction changed little during her rapid ascent. Divers can lose consciousness from lack of oxygen in the last few meters of an ascent, even when they felt well at the bottom.


Based on the passage and Table 1, at approximately what depth is the diver neutrally buoyant?

A.
0 m
B.
20 m
C.
10 m
D.
30 m
Question 31

Arteriole Radius and Flow in an Isolated Vessel

Arterioles account for most of the resistance to blood flow in the systemic circulation, and their smooth muscle adjusts that resistance by narrowing or widening the lumen. For steady, laminar flow of a fluid of constant viscosity through a rigid tube, Poiseuille's law gives the volume flow rate:

Q=π ΔP r48 η LQ = \dfrac{\pi\,\Delta P\,r^4}{8\,\eta\,L}

where ΔP\Delta P is the pressure difference between the ends of the tube, rr is its inner radius, η\eta is the fluid's viscosity and LL is its length.

In one study, researchers removed a 2.0-mm segment of arteriole from a rat's skeletal muscle and mounted it between two glass cannulas. They held the pressure difference along the segment at 1.0 mmHg and perfused it with a salt solution. They measured the flow rate at baseline, after adding either a vasodilator (adenosine) or a vasoconstrictor (phenylephrine) to the bath, and, in a final trial with no drug, after adding dextran, a large sugar polymer, to raise the perfusate's viscosity. In each trial they measured the vessel's inner radius through a microscope (Table 1).

Table 1. Flow through the arteriole at a constant pressure difference of 1.0 mmHg

ConditionInner radius (μm)Perfusate viscosity (mPa·s)Flow rate (μL/min)
Baseline401.04.0
Adenosine501.09.8
Phenylephrine301.01.3
Dextran, no drug402.02.0

In every trial, the measured flow rate matched the value Poiseuille's law predicts from the measured radius and viscosity.


If phenylephrine were applied while the arteriole was perfused with the dextran solution, the flow rate would be closest to:

A.
1.1 μL/min
B.
0.63 μL/min
C.
1.5 μL/min
D.
2.5 μL/min
Question 32

Arteriole Radius and Flow in an Isolated Vessel

Arterioles account for most of the resistance to blood flow in the systemic circulation, and their smooth muscle adjusts that resistance by narrowing or widening the lumen. For steady, laminar flow of a fluid of constant viscosity through a rigid tube, Poiseuille's law gives the volume flow rate:

Q=π ΔP r48 η LQ = \dfrac{\pi\,\Delta P\,r^4}{8\,\eta\,L}

where ΔP\Delta P is the pressure difference between the ends of the tube, rr is its inner radius, η\eta is the fluid's viscosity and LL is its length.

In one study, researchers removed a 2.0-mm segment of arteriole from a rat's skeletal muscle and mounted it between two glass cannulas. They held the pressure difference along the segment at 1.0 mmHg and perfused it with a salt solution. They measured the flow rate at baseline, after adding either a vasodilator (adenosine) or a vasoconstrictor (phenylephrine) to the bath, and, in a final trial with no drug, after adding dextran, a large sugar polymer, to raise the perfusate's viscosity. In each trial they measured the vessel's inner radius through a microscope (Table 1).

Table 1. Flow through the arteriole at a constant pressure difference of 1.0 mmHg

ConditionInner radius (μm)Perfusate viscosity (mPa·s)Flow rate (μL/min)
Baseline401.04.0
Adenosine501.09.8
Phenylephrine301.01.3
Dextran, no drug402.02.0

In every trial, the measured flow rate matched the value Poiseuille's law predicts from the measured radius and viscosity.


The dextran trial models the effect of more viscous blood. Which condition would, like dextran, reduce blood flow through a vascular bed held at a constant pressure difference?

A.
Anemia, an abnormally low red blood cell count
B.
Hemodilution after a rapid infusion of saline
C.
Dilation of the arterioles that supply the bed
D.
Polycythemia, an abnormally high red blood cell count
Question 33

Arteriole Radius and Flow in an Isolated Vessel

Arterioles account for most of the resistance to blood flow in the systemic circulation, and their smooth muscle adjusts that resistance by narrowing or widening the lumen. For steady, laminar flow of a fluid of constant viscosity through a rigid tube, Poiseuille's law gives the volume flow rate:

Q=π ΔP r48 η LQ = \dfrac{\pi\,\Delta P\,r^4}{8\,\eta\,L}

where ΔP\Delta P is the pressure difference between the ends of the tube, rr is its inner radius, η\eta is the fluid's viscosity and LL is its length.

In one study, researchers removed a 2.0-mm segment of arteriole from a rat's skeletal muscle and mounted it between two glass cannulas. They held the pressure difference along the segment at 1.0 mmHg and perfused it with a salt solution. They measured the flow rate at baseline, after adding either a vasodilator (adenosine) or a vasoconstrictor (phenylephrine) to the bath, and, in a final trial with no drug, after adding dextran, a large sugar polymer, to raise the perfusate's viscosity. In each trial they measured the vessel's inner radius through a microscope (Table 1).

Table 1. Flow through the arteriole at a constant pressure difference of 1.0 mmHg

ConditionInner radius (μm)Perfusate viscosity (mPa·s)Flow rate (μL/min)
Baseline401.04.0
Adenosine501.09.8
Phenylephrine301.01.3
Dextran, no drug402.02.0

In every trial, the measured flow rate matched the value Poiseuille's law predicts from the measured radius and viscosity.


A muscle is supplied by 10 arterioles arranged in parallel, each identical to the baseline vessel and each with the same 1.0-mmHg pressure difference. If adenosine dilates 5 of them as in Table 1 and the other 5 are unchanged, the total flow is closest to:

A.
45 μL/min
B.
51 μL/min
C.
69 μL/min
D.
98 μL/min
Question 34

Arteriole Radius and Flow in an Isolated Vessel

Arterioles account for most of the resistance to blood flow in the systemic circulation, and their smooth muscle adjusts that resistance by narrowing or widening the lumen. For steady, laminar flow of a fluid of constant viscosity through a rigid tube, Poiseuille's law gives the volume flow rate:

Q=π ΔP r48 η LQ = \dfrac{\pi\,\Delta P\,r^4}{8\,\eta\,L}

where ΔP\Delta P is the pressure difference between the ends of the tube, rr is its inner radius, η\eta is the fluid's viscosity and LL is its length.

In one study, researchers removed a 2.0-mm segment of arteriole from a rat's skeletal muscle and mounted it between two glass cannulas. They held the pressure difference along the segment at 1.0 mmHg and perfused it with a salt solution. They measured the flow rate at baseline, after adding either a vasodilator (adenosine) or a vasoconstrictor (phenylephrine) to the bath, and, in a final trial with no drug, after adding dextran, a large sugar polymer, to raise the perfusate's viscosity. In each trial they measured the vessel's inner radius through a microscope (Table 1).

Table 1. Flow through the arteriole at a constant pressure difference of 1.0 mmHg

ConditionInner radius (μm)Perfusate viscosity (mPa·s)Flow rate (μL/min)
Baseline401.04.0
Adenosine501.09.8
Phenylephrine301.01.3
Dextran, no drug402.02.0

In every trial, the measured flow rate matched the value Poiseuille's law predicts from the measured radius and viscosity.


Compared with the baseline trial, the average speed of the perfusate through the vessel in the adenosine trial is:

A.
about 1.6 times as fast
B.
about 0.64 times as fast
C.
about the same
D.
about 2.4 times as fast
Question 35

Arteriole Radius and Flow in an Isolated Vessel

Arterioles account for most of the resistance to blood flow in the systemic circulation, and their smooth muscle adjusts that resistance by narrowing or widening the lumen. For steady, laminar flow of a fluid of constant viscosity through a rigid tube, Poiseuille's law gives the volume flow rate:

Q=π ΔP r48 η LQ = \dfrac{\pi\,\Delta P\,r^4}{8\,\eta\,L}

where ΔP\Delta P is the pressure difference between the ends of the tube, rr is its inner radius, η\eta is the fluid's viscosity and LL is its length.

In one study, researchers removed a 2.0-mm segment of arteriole from a rat's skeletal muscle and mounted it between two glass cannulas. They held the pressure difference along the segment at 1.0 mmHg and perfused it with a salt solution. They measured the flow rate at baseline, after adding either a vasodilator (adenosine) or a vasoconstrictor (phenylephrine) to the bath, and, in a final trial with no drug, after adding dextran, a large sugar polymer, to raise the perfusate's viscosity. In each trial they measured the vessel's inner radius through a microscope (Table 1).

Table 1. Flow through the arteriole at a constant pressure difference of 1.0 mmHg

ConditionInner radius (μm)Perfusate viscosity (mPa·s)Flow rate (μL/min)
Baseline401.04.0
Adenosine501.09.8
Phenylephrine301.01.3
Dextran, no drug402.02.0

In every trial, the measured flow rate matched the value Poiseuille's law predicts from the measured radius and viscosity.


Which additional result would most weaken the assumption that the baseline arteriole behaves as a rigid tube whose radius changes only when a drug is applied?

A.
Raising the pressure difference to 2.0 mmHg increased the flow rate to 8.0 μL/min.
B.
Raising the pressure difference to 2.0 mmHg increased the flow rate to 9.5 μL/min.
C.
Lowering the pressure difference to 0.50 mmHg reduced the flow rate to 2.0 μL/min.
D.
Perfusing a 4.0-mm segment of the same vessel at 1.0 mmHg gave a flow rate of 2.0 μL/min.
Question 36

Maintaining a stable blood pH is crucial for survival and involves various regulatory systems working together. Under normal conditions, the pH of blood plasma is tightly regulated around 7.4. Deviations below 6.9 or above 7.7 can be fatal.

One major contributor to this balance is the enzyme carbonic anhydrase, which speeds up the reaction converting carbon dioxide (CO2CO_2) and water into carbonic acid. This carbonic acid can dissociate into hydrogen ions and bicarbonate, forming the carbonic acid–bicarbonate buffer system shown below:

CO2(g)+H2O(l)↔H2CO3(aq)↔H+(aq)+HCO3−(aq)CO_2(g) + H_2O(l) \leftrightarrow H_2CO_3(aq) \leftrightarrow H^+(aq) + HCO_3^-(aq)

In the absence of a catalyst, CO2CO_2 and H+H^+ in the blood can bind to hemoglobin after it releases oxygen in peripheral tissues. Once hemoglobin reaches the lungs, it rebinds oxygen and releases CO2CO_2 and H+H^+, reversing the prior binding. This continuous exchange of gases is part of the process of respiration.


Which of the following statements regarding respiratory control of blood pH is INCORRECT?

A.
Conditions like COPD can raise blood [H+H^+] due to reduced ventilation.
B.
Increased respiratory rate during physical activity helps raise blood pH.
C.
Reduced ventilation leads to CO2CO_2 buildup in the bloodstream.
D.
Hyperventilation increases bicarbonate ion levels in the blood.
Question 37

Maintaining a stable blood pH is crucial for survival and involves various regulatory systems working together. Under normal conditions, the pH of blood plasma is tightly regulated around 7.4. Deviations below 6.9 or above 7.7 can be fatal.

One major contributor to this balance is the enzyme carbonic anhydrase, which speeds up the reaction converting carbon dioxide (CO2CO_2) and water into carbonic acid. This carbonic acid can dissociate into hydrogen ions and bicarbonate, forming the carbonic acid–bicarbonate buffer system shown below:

CO2(g)+H2O(l)↔H2CO3(aq)↔H+(aq)+HCO3−(aq)CO_2(g) + H_2O(l) \leftrightarrow H_2CO_3(aq) \leftrightarrow H^+(aq) + HCO_3^-(aq)

In the absence of a catalyst, CO2CO_2 and H+H^+ in the blood can bind to hemoglobin after it releases oxygen in peripheral tissues. Once hemoglobin reaches the lungs, it rebinds oxygen and releases CO2CO_2 and H+H^+, reversing the prior binding. This continuous exchange of gases is part of the process of respiration.


Which of the following best characterizes carbonic acid?

A.
It acts as both an acid and a base.
B.
It is a strong acid.
C.
It serves as the conjugate acid of CO2CO_2.
D.
It can donate more than one proton.
Question 38

Maintaining a stable blood pH is crucial for survival and involves various regulatory systems working together. Under normal conditions, the pH of blood plasma is tightly regulated around 7.4. Deviations below 6.9 or above 7.7 can be fatal.

One major contributor to this balance is the enzyme carbonic anhydrase, which speeds up the reaction converting carbon dioxide (CO2CO_2) and water into carbonic acid. This carbonic acid can dissociate into hydrogen ions and bicarbonate, forming the carbonic acid–bicarbonate buffer system shown below:

CO2(g)+H2O(l)↔H2CO3(aq)↔H+(aq)+HCO3−(aq)CO_2(g) + H_2O(l) \leftrightarrow H_2CO_3(aq) \leftrightarrow H^+(aq) + HCO_3^-(aq)

In the absence of a catalyst, CO2CO_2 and H+H^+ in the blood can bind to hemoglobin after it releases oxygen in peripheral tissues. Once hemoglobin reaches the lungs, it rebinds oxygen and releases CO2CO_2 and H+H^+, reversing the prior binding. This continuous exchange of gases is part of the process of respiration.


If CO2CO_2 is steadily introduced into a container of water, forming carbonic acid, which statements are accurate?

I. Adding carbonic anhydrase will raise the KaK_a of the system.

II. Carbonic acid will accumulate until equilibrium is reached.

III. Adding bicarbonate will cause the pH to rise.

A.
I only
B.
II and III
C.
II only
D.
III only
Question 39

Maintaining a stable blood pH is crucial for survival and involves various regulatory systems working together. Under normal conditions, the pH of blood plasma is tightly regulated around 7.4. Deviations below 6.9 or above 7.7 can be fatal.

One major contributor to this balance is the enzyme carbonic anhydrase, which speeds up the reaction converting carbon dioxide (CO2CO_2) and water into carbonic acid. This carbonic acid can dissociate into hydrogen ions and bicarbonate, forming the carbonic acid–bicarbonate buffer system shown below:

CO2(g)+H2O(l)↔H2CO3(aq)↔H+(aq)+HCO3−(aq)CO_2(g) + H_2O(l) \leftrightarrow H_2CO_3(aq) \leftrightarrow H^+(aq) + HCO_3^-(aq)

In the absence of a catalyst, CO2CO_2 and H+H^+ in the blood can bind to hemoglobin after it releases oxygen in peripheral tissues. Once hemoglobin reaches the lungs, it rebinds oxygen and releases CO2CO_2 and H+H^+, reversing the prior binding. This continuous exchange of gases is part of the process of respiration.


Estimate the pH of a buffer solution prepared by mixing 40 mL of 0.040 M acetic acid (Ka=2.2×10−5K_a = 2.2 × 10^{-5}) with 20 mL of 0.10 M sodium acetate.

A.
pH = 2.0
B.
pH = 3.6
C.
pH = 5.0
D.
pH = 4.2
Question 40

Maintaining a stable blood pH is crucial for survival and involves various regulatory systems working together. Under normal conditions, the pH of blood plasma is tightly regulated around 7.4. Deviations below 6.9 or above 7.7 can be fatal.

One major contributor to this balance is the enzyme carbonic anhydrase, which speeds up the reaction converting carbon dioxide (CO2CO_2) and water into carbonic acid. This carbonic acid can dissociate into hydrogen ions and bicarbonate, forming the carbonic acid–bicarbonate buffer system shown below:

CO2(g)+H2O(l)↔H2CO3(aq)↔H+(aq)+HCO3−(aq)CO_2(g) + H_2O(l) \leftrightarrow H_2CO_3(aq) \leftrightarrow H^+(aq) + HCO_3^-(aq)

In the absence of a catalyst, CO2CO_2 and H+H^+ in the blood can bind to hemoglobin after it releases oxygen in peripheral tissues. Once hemoglobin reaches the lungs, it rebinds oxygen and releases CO2CO_2 and H+H^+, reversing the prior binding. This continuous exchange of gases is part of the process of respiration.


In addition to this system, another buffering mechanism involves phosphoric acid (H3PO4H_3PO_4), which plays a lesser but still important role. As a triprotic acid, H3PO4H_3PO_4 can release three protons in a stepwise manner, shown below with their respective acid dissociation constants:

1. H3PO4(aq)H_3PO_4(aq) ⇌ H+(aq)H^+(aq) + H2PO4−(aq)H_2PO_4^-(aq), Ka1=6.8×10−3K_{a1} = 6.8 \times 10^{-3}

2. H2PO4−(aq)H_2PO_4^-(aq) ⇌ H+(aq)H^+(aq) + HPO42−(aq)HPO_4^{2-}(aq), Ka2=5.4×10−8K_{a2} = 5.4 \times 10^{-8}

3. HPO42−(aq)HPO_4^{2-}(aq) ⇌ H+(aq)H^+(aq) + PO43−(aq)PO_4^{3-}(aq), Ka3=2.1×10−12K_{a3} = 2.1 \times 10^{-12}

The decreasing pattern in acid dissociation constants for phosphoric acid (Ka1>Ka2>Ka3K_{a1} > K_{a2} > K_{a3}) is mostly due to:

A.
Increasing difficulty removing protons from increasingly negative species.
B.
Equilibrium favoring reactants in later steps due to initial proton loss.
C.
Smaller H+H^+ ion in the first step compared to later ones.
D.
A slower dissociation rate in the later steps.
Question 41

Carnosine: A Built-In Buffer for Sprinting Muscle

During a sprint, fast-twitch muscle makes lactic acid faster than it can export it, and the hydrogen ions released can push intracellular pH from about 7.1 toward 6.5, low enough to inhibit enzymes such as phosphofructokinase-1. Besides bicarbonate, phosphate and proteins, these fibers buffer acid with carnosine (β-alanyl-L-histidine), a dipeptide stored at high concentrations in the muscles of sprinting animals. Between pH 5.5 and 7.5, the only group of carnosine that gains or loses a proton is the imidazole ring of its histidine, written below as Im:

ImH+⇌Im+H+pKa=6.83 at 37 ∘C\text{ImH}^+ \rightleftharpoons \text{Im} + \text{H}^+ \qquad \text{p}K_a = 6.83 \text{ at } 37\ ^\circ\text{C}

To measure this group's buffering, researchers prepared a 30 mM carnosine solution in 150 mM KCl at 37 °C, adjusted it to pH 7.13, and added HCl in steps, recording the pH after each addition. A KCl solution with no carnosine, also adjusted to pH 7.13, served as the control (Table 1).

Table 1. pH after adding HCl (mmol per liter of solution)

HCl added (mM)30 mM carnosineKCl only
07.137.13
56.832.30
106.532.00
156.131.82
185.681.74

Before any HCl is added, what are the concentrations of the two forms of the imidazole group in the carnosine solution?

A.
20 mM ImH+\text{ImH}^+ and 10 mM Im
B.
10 mM ImH+\text{ImH}^+ and 20 mM Im
C.
15 mM ImH+\text{ImH}^+ and 15 mM Im
D.
3 mM ImH+\text{ImH}^+ and 27 mM Im
Question 42

Carnosine: A Built-In Buffer for Sprinting Muscle

During a sprint, fast-twitch muscle makes lactic acid faster than it can export it, and the hydrogen ions released can push intracellular pH from about 7.1 toward 6.5, low enough to inhibit enzymes such as phosphofructokinase-1. Besides bicarbonate, phosphate and proteins, these fibers buffer acid with carnosine (β-alanyl-L-histidine), a dipeptide stored at high concentrations in the muscles of sprinting animals. Between pH 5.5 and 7.5, the only group of carnosine that gains or loses a proton is the imidazole ring of its histidine, written below as Im:

ImH+⇌Im+H+pKa=6.83 at 37 ∘C\text{ImH}^+ \rightleftharpoons \text{Im} + \text{H}^+ \qquad \text{p}K_a = 6.83 \text{ at } 37\ ^\circ\text{C}

To measure this group's buffering, researchers prepared a 30 mM carnosine solution in 150 mM KCl at 37 °C, adjusted it to pH 7.13, and added HCl in steps, recording the pH after each addition. A KCl solution with no carnosine, also adjusted to pH 7.13, served as the control (Table 1).

Table 1. pH after adding HCl (mmol per liter of solution)

HCl added (mM)30 mM carnosineKCl only
07.137.13
56.832.30
106.532.00
156.131.82
185.681.74

In Table 1, the last 3 mM of HCl (from 15 to 18 mM) lowers the carnosine solution's pH by more than the first 5 mM did. Which statement best explains this?

A.
Carnosine's pKa falls as the solution acidifies, so its imidazole accepts protons less readily
B.
Little Im remains, so each added proton changes the ratio of Im to ImH+\text{ImH}^+ by a larger factor
C.
The KCl in the medium starts releasing hydrogen ions once the pH falls below about 6.5
D.
The pH is now close to the pKa, where a buffer resists a change in pH least effectively
Question 43

Carnosine: A Built-In Buffer for Sprinting Muscle

During a sprint, fast-twitch muscle makes lactic acid faster than it can export it, and the hydrogen ions released can push intracellular pH from about 7.1 toward 6.5, low enough to inhibit enzymes such as phosphofructokinase-1. Besides bicarbonate, phosphate and proteins, these fibers buffer acid with carnosine (β-alanyl-L-histidine), a dipeptide stored at high concentrations in the muscles of sprinting animals. Between pH 5.5 and 7.5, the only group of carnosine that gains or loses a proton is the imidazole ring of its histidine, written below as Im:

ImH+⇌Im+H+pKa=6.83 at 37 ∘C\text{ImH}^+ \rightleftharpoons \text{Im} + \text{H}^+ \qquad \text{p}K_a = 6.83 \text{ at } 37\ ^\circ\text{C}

To measure this group's buffering, researchers prepared a 30 mM carnosine solution in 150 mM KCl at 37 °C, adjusted it to pH 7.13, and added HCl in steps, recording the pH after each addition. A KCl solution with no carnosine, also adjusted to pH 7.13, served as the control (Table 1).

Table 1. pH after adding HCl (mmol per liter of solution)

HCl added (mM)30 mM carnosineKCl only
07.137.13
56.832.30
106.532.00
156.131.82
185.681.74

A second solution contains 60 mM carnosine in the same medium, also at pH 7.13. How much HCl, in mmol per liter, would lower its pH to 6.53?

A.
10 mmol/L
B.
20 mmol/L
C.
40 mmol/L
D.
5 mmol/L
Question 44

Carnosine: A Built-In Buffer for Sprinting Muscle

During a sprint, fast-twitch muscle makes lactic acid faster than it can export it, and the hydrogen ions released can push intracellular pH from about 7.1 toward 6.5, low enough to inhibit enzymes such as phosphofructokinase-1. Besides bicarbonate, phosphate and proteins, these fibers buffer acid with carnosine (β-alanyl-L-histidine), a dipeptide stored at high concentrations in the muscles of sprinting animals. Between pH 5.5 and 7.5, the only group of carnosine that gains or loses a proton is the imidazole ring of its histidine, written below as Im:

ImH+⇌Im+H+pKa=6.83 at 37 ∘C\text{ImH}^+ \rightleftharpoons \text{Im} + \text{H}^+ \qquad \text{p}K_a = 6.83 \text{ at } 37\ ^\circ\text{C}

To measure this group's buffering, researchers prepared a 30 mM carnosine solution in 150 mM KCl at 37 °C, adjusted it to pH 7.13, and added HCl in steps, recording the pH after each addition. A KCl solution with no carnosine, also adjusted to pH 7.13, served as the control (Table 1).

Table 1. pH after adding HCl (mmol per liter of solution)

HCl added (mM)30 mM carnosineKCl only
07.137.13
56.832.30
106.532.00
156.131.82
185.681.74

Which finding would most weaken the conclusion that carnosine's buffering in Table 1 comes from its imidazole group?

A.
β-Alanyl-alanine, which lacks an imidazole ring, titrates exactly like carnosine
B.
Free histidine, whose imidazole pKa is about 6.0, buffers most strongly near pH 6.0
C.
The carnosine solution's pH drops sharply once the added HCl exceeds 20 mM
D.
A 60 mM carnosine solution needs twice the HCl of the 30 mM solution per pH step
Question 45

Carnosine: A Built-In Buffer for Sprinting Muscle

During a sprint, fast-twitch muscle makes lactic acid faster than it can export it, and the hydrogen ions released can push intracellular pH from about 7.1 toward 6.5, low enough to inhibit enzymes such as phosphofructokinase-1. Besides bicarbonate, phosphate and proteins, these fibers buffer acid with carnosine (β-alanyl-L-histidine), a dipeptide stored at high concentrations in the muscles of sprinting animals. Between pH 5.5 and 7.5, the only group of carnosine that gains or loses a proton is the imidazole ring of its histidine, written below as Im:

ImH+⇌Im+H+pKa=6.83 at 37 ∘C\text{ImH}^+ \rightleftharpoons \text{Im} + \text{H}^+ \qquad \text{p}K_a = 6.83 \text{ at } 37\ ^\circ\text{C}

To measure this group's buffering, researchers prepared a 30 mM carnosine solution in 150 mM KCl at 37 °C, adjusted it to pH 7.13, and added HCl in steps, recording the pH after each addition. A KCl solution with no carnosine, also adjusted to pH 7.13, served as the control (Table 1).

Table 1. pH after adding HCl (mmol per liter of solution)

HCl added (mM)30 mM carnosineKCl only
07.137.13
56.832.30
106.532.00
156.131.82
185.681.74

The 30 mM carnosine solution at pH 7.13 is diluted tenfold with the same KCl medium. Compared with the undiluted solution, the diluted solution has:

A.
A pH about one unit lower and the same buffer capacity
B.
A pH about one unit higher and one-tenth the buffer capacity
C.
About the same pH and the same buffer capacity
D.
About the same pH and about one-tenth the buffer capacity
Question 46

Three Alcohols, One Enzyme: Why Methanol Is Dangerous

Methanol, ethanol and 2-propanol are all oxidized in the liver by alcohol dehydrogenase (ADH), which removes a hydride ion from the carbon bearing the –OH group and passes it to NAD+\text{NAD}^+. Aldehyde products are then oxidized further, by aldehyde dehydrogenases, to carboxylic acids. For methanol, this path ends at formic acid (pKa=3.75\text{p}K_a = 3.75). Humans convert formate to CO2\text{CO}_2 only slowly, so after a large methanol dose, formic acid accumulates and blood pH falls.

In one study, researchers purified a human liver ADH and measured its kinetics with each alcohol at saturating NAD+\text{NAD}^+ (Table 1).

Table 1. Kinetic constants of purified liver ADH at 37 °C

SubstrateKm (mM)Vmax (relative to ethanol)Product
Ethanol0.8100Acetaldehyde
Methanol2412Formaldehyde
2-Propanol935Acetone

The researchers then measured how much of each carbonyl product exists as its hydrate, a geminal diol, at equilibrium in water at 25 °C (Table 2):

R2C=O+H2O⇌R2C(OH)2\text{R}_2\text{C=O} + \text{H}_2\text{O} \rightleftharpoons \text{R}_2\text{C(OH)}_2

Table 2. Hydration equilibria of the carbonyl products

CompoundKhydK_\text{hyd}Hydrate at equilibrium (%)
Formaldehyde2×1032 \times 10^{3}99.95
Acetaldehyde1.152
Acetone1.4×10−31.4 \times 10^{-3}0.14

A patient who drank windshield-washer fluid arrives with a blood methanol concentration of 15 mM. Until dialysis can remove the methanol, clinicians infuse ethanol to hold blood ethanol near 22 mM.


Based on Table 1, why does holding blood ethanol near 22 mM sharply slow the patient's production of formaldehyde?

A.
Ethanol's much lower Km lets it outcompete methanol for most ADH active sites
B.
Ethanol binds a separate regulatory site on ADH, lowering the enzyme's Vmax for methanol
C.
Ethanol's higher Km makes ADH release methanol before oxidizing it to formaldehyde
D.
Ethanol at 22 mM denatures ADH, halting oxidation of every alcohol until it clears
Question 47

Three Alcohols, One Enzyme: Why Methanol Is Dangerous

Methanol, ethanol and 2-propanol are all oxidized in the liver by alcohol dehydrogenase (ADH), which removes a hydride ion from the carbon bearing the –OH group and passes it to NAD+\text{NAD}^+. Aldehyde products are then oxidized further, by aldehyde dehydrogenases, to carboxylic acids. For methanol, this path ends at formic acid (pKa=3.75\text{p}K_a = 3.75). Humans convert formate to CO2\text{CO}_2 only slowly, so after a large methanol dose, formic acid accumulates and blood pH falls.

In one study, researchers purified a human liver ADH and measured its kinetics with each alcohol at saturating NAD+\text{NAD}^+ (Table 1).

Table 1. Kinetic constants of purified liver ADH at 37 °C

SubstrateKm (mM)Vmax (relative to ethanol)Product
Ethanol0.8100Acetaldehyde
Methanol2412Formaldehyde
2-Propanol935Acetone

The researchers then measured how much of each carbonyl product exists as its hydrate, a geminal diol, at equilibrium in water at 25 °C (Table 2):

R2C=O+H2O⇌R2C(OH)2\text{R}_2\text{C=O} + \text{H}_2\text{O} \rightleftharpoons \text{R}_2\text{C(OH)}_2

Table 2. Hydration equilibria of the carbonyl products

CompoundKhydK_\text{hyd}Hydrate at equilibrium (%)
Formaldehyde2×1032 \times 10^{3}99.95
Acetaldehyde1.152
Acetone1.4×10−31.4 \times 10^{-3}0.14

A patient who drank windshield-washer fluid arrives with a blood methanol concentration of 15 mM. Until dialysis can remove the methanol, clinicians infuse ethanol to hold blood ethanol near 22 mM.


Which ranking of methanol and its two oxidation products, from highest to lowest boiling point, is correct?

A.
Methanol > formic acid > formaldehyde
B.
Formic acid > formaldehyde > methanol
C.
Formic acid > methanol > formaldehyde
D.
Formaldehyde > methanol > formic acid
Question 48

Three Alcohols, One Enzyme: Why Methanol Is Dangerous

Methanol, ethanol and 2-propanol are all oxidized in the liver by alcohol dehydrogenase (ADH), which removes a hydride ion from the carbon bearing the –OH group and passes it to NAD+\text{NAD}^+. Aldehyde products are then oxidized further, by aldehyde dehydrogenases, to carboxylic acids. For methanol, this path ends at formic acid (pKa=3.75\text{p}K_a = 3.75). Humans convert formate to CO2\text{CO}_2 only slowly, so after a large methanol dose, formic acid accumulates and blood pH falls.

In one study, researchers purified a human liver ADH and measured its kinetics with each alcohol at saturating NAD+\text{NAD}^+ (Table 1).

Table 1. Kinetic constants of purified liver ADH at 37 °C

SubstrateKm (mM)Vmax (relative to ethanol)Product
Ethanol0.8100Acetaldehyde
Methanol2412Formaldehyde
2-Propanol935Acetone

The researchers then measured how much of each carbonyl product exists as its hydrate, a geminal diol, at equilibrium in water at 25 °C (Table 2):

R2C=O+H2O⇌R2C(OH)2\text{R}_2\text{C=O} + \text{H}_2\text{O} \rightleftharpoons \text{R}_2\text{C(OH)}_2

Table 2. Hydration equilibria of the carbonyl products

CompoundKhydK_\text{hyd}Hydrate at equilibrium (%)
Formaldehyde2×1032 \times 10^{3}99.95
Acetaldehyde1.152
Acetone1.4×10−31.4 \times 10^{-3}0.14

A patient who drank windshield-washer fluid arrives with a blood methanol concentration of 15 mM. Until dialysis can remove the methanol, clinicians infuse ethanol to hold blood ethanol near 22 mM.


A second patient drinks rubbing alcohol, which is mostly 2-propanol. As the liver metabolizes it, which finding is most likely?

A.
Blood propanoic acid rises and pH falls, because the aldehyde is oxidized to an acid
B.
Blood formic acid rises and pH falls, because ADH splits off a one-carbon aldehyde
C.
Blood acetone rises with little fall in pH, since a ketone is not oxidized to an acid
D.
Blood propanal rises with little fall in pH, because ADH stops at the aldehyde stage
Question 49

Three Alcohols, One Enzyme: Why Methanol Is Dangerous

Methanol, ethanol and 2-propanol are all oxidized in the liver by alcohol dehydrogenase (ADH), which removes a hydride ion from the carbon bearing the –OH group and passes it to NAD+\text{NAD}^+. Aldehyde products are then oxidized further, by aldehyde dehydrogenases, to carboxylic acids. For methanol, this path ends at formic acid (pKa=3.75\text{p}K_a = 3.75). Humans convert formate to CO2\text{CO}_2 only slowly, so after a large methanol dose, formic acid accumulates and blood pH falls.

In one study, researchers purified a human liver ADH and measured its kinetics with each alcohol at saturating NAD+\text{NAD}^+ (Table 1).

Table 1. Kinetic constants of purified liver ADH at 37 °C

SubstrateKm (mM)Vmax (relative to ethanol)Product
Ethanol0.8100Acetaldehyde
Methanol2412Formaldehyde
2-Propanol935Acetone

The researchers then measured how much of each carbonyl product exists as its hydrate, a geminal diol, at equilibrium in water at 25 °C (Table 2):

R2C=O+H2O⇌R2C(OH)2\text{R}_2\text{C=O} + \text{H}_2\text{O} \rightleftharpoons \text{R}_2\text{C(OH)}_2

Table 2. Hydration equilibria of the carbonyl products

CompoundKhydK_\text{hyd}Hydrate at equilibrium (%)
Formaldehyde2×1032 \times 10^{3}99.95
Acetaldehyde1.152
Acetone1.4×10−31.4 \times 10^{-3}0.14

A patient who drank windshield-washer fluid arrives with a blood methanol concentration of 15 mM. Until dialysis can remove the methanol, clinicians infuse ethanol to hold blood ethanol near 22 mM.


Which statement best explains the trend in Table 2?

A.
Alkyl groups withdraw electron density from the carbonyl carbon and destabilize the hydrate
B.
Acetone hydrogen-bonds to water more strongly, which holds it in the carbonyl form
C.
Acetone is only slightly soluble in water, so little of it meets water to react
D.
Alkyl groups donate electron density to the carbonyl carbon and crowd the hydrate
Question 50

Three Alcohols, One Enzyme: Why Methanol Is Dangerous

Methanol, ethanol and 2-propanol are all oxidized in the liver by alcohol dehydrogenase (ADH), which removes a hydride ion from the carbon bearing the –OH group and passes it to NAD+\text{NAD}^+. Aldehyde products are then oxidized further, by aldehyde dehydrogenases, to carboxylic acids. For methanol, this path ends at formic acid (pKa=3.75\text{p}K_a = 3.75). Humans convert formate to CO2\text{CO}_2 only slowly, so after a large methanol dose, formic acid accumulates and blood pH falls.

In one study, researchers purified a human liver ADH and measured its kinetics with each alcohol at saturating NAD+\text{NAD}^+ (Table 1).

Table 1. Kinetic constants of purified liver ADH at 37 °C

SubstrateKm (mM)Vmax (relative to ethanol)Product
Ethanol0.8100Acetaldehyde
Methanol2412Formaldehyde
2-Propanol935Acetone

The researchers then measured how much of each carbonyl product exists as its hydrate, a geminal diol, at equilibrium in water at 25 °C (Table 2):

R2C=O+H2O⇌R2C(OH)2\text{R}_2\text{C=O} + \text{H}_2\text{O} \rightleftharpoons \text{R}_2\text{C(OH)}_2

Table 2. Hydration equilibria of the carbonyl products

CompoundKhydK_\text{hyd}Hydrate at equilibrium (%)
Formaldehyde2×1032 \times 10^{3}99.95
Acetaldehyde1.152
Acetone1.4×10−31.4 \times 10^{-3}0.14

A patient who drank windshield-washer fluid arrives with a blood methanol concentration of 15 mM. Until dialysis can remove the methanol, clinicians infuse ethanol to hold blood ethanol near 22 mM.


At a blood pH of 7.40, approximately what is the ratio of formate to formic acid?

A.
About 3.7 to 1
B.
About 4.5×1034.5 \times 10^3 to 1
C.
About 1 to 4.5×1034.5 \times 10^3
D.
About 45 to 1
Question 51

Penicillin is widely regarded as one of the most iconic and influential antibiotics in medical history. Its chemical structure is defined by a key component called a beta-lactam—a four-membered ring containing an amide group. Figure 1 demonstrates the beta-lactam structure in several different antibiotics. This beta-lactam structure allows the drug to form a covalent bond with Penicillin Binding Proteins (PBPs), which play a crucial role in bacterial cell wall integrity by facilitating the cross-linking of peptidoglycan chains. When penicillin or related compounds attach to PBPs, this cross-linking process is disrupted, weakening the cell wall and leading to bacterial death through lysis.

Beta-lactam structure in antibiotics

The beta-lactam ring is especially reactive due to its geometric strain; the four-membered ring creates 90° bond angles, which deviate significantly from ideal angles for sp2^2 or sp3^3 hybridized atoms. This strain contributes to the ring's high reactivity and ability to acylate PBPs effectively. In contrast, larger lactam rings, which are less strained, show markedly lower reactivity. To counteract the effects of beta-lactam antibiotics, some bacteria produce an enzyme called beta-lactamase. This enzyme cleaves the beta-lactam ring, neutralizing the antibiotic's activity by preventing it from interacting with PBPs.


When a beta-lactam forms a permanent covalent bond with a Penicillin Binding Protein (PBP), what type of enzyme inhibition is this an example of?

A.
Feedback inhibition
B.
Competitive inhibition
C.
Noncompetitive inhibition
D.
Uncompetitive inhibition
Question 52

Penicillin is widely regarded as one of the most iconic and influential antibiotics in medical history. Its chemical structure is defined by a key component called a beta-lactam—a four-membered ring containing an amide group. Figure 1 demonstrates the beta-lactam structure in several different antibiotics. This beta-lactam structure allows the drug to form a covalent bond with Penicillin Binding Proteins (PBPs), which play a crucial role in bacterial cell wall integrity by facilitating the cross-linking of peptidoglycan chains. When penicillin or related compounds attach to PBPs, this cross-linking process is disrupted, weakening the cell wall and leading to bacterial death through lysis.

Beta-lactam structure in antibiotics

The beta-lactam ring is especially reactive due to its geometric strain; the four-membered ring creates 90° bond angles, which deviate significantly from ideal angles for sp2^2 or sp3^3 hybridized atoms. This strain contributes to the ring's high reactivity and ability to acylate PBPs effectively. In contrast, larger lactam rings, which are less strained, show markedly lower reactivity. To counteract the effects of beta-lactam antibiotics, some bacteria produce an enzyme called beta-lactamase. This enzyme cleaves the beta-lactam ring, neutralizing the antibiotic's activity by preventing it from interacting with PBPs.


The beta-lactam ring undergoes a reaction with the serine hydroxyl (−CH2OH-CH_2OH) group in the active site of PBPs. Considering the strain on the four-membered ring, which type of chemical reaction best characterizes this interaction?

A.
Formation of an anhydride
B.
Nucleophilic substitution
C.
Transesterification
D.
Loss of a carboxyl group
Question 53

Penicillin is widely regarded as one of the most iconic and influential antibiotics in medical history. Its chemical structure is defined by a key component called a beta-lactam—a four-membered ring containing an amide group. Figure 1 demonstrates the beta-lactam structure in several different antibiotics. This beta-lactam structure allows the drug to form a covalent bond with Penicillin Binding Proteins (PBPs), which play a crucial role in bacterial cell wall integrity by facilitating the cross-linking of peptidoglycan chains. When penicillin or related compounds attach to PBPs, this cross-linking process is disrupted, weakening the cell wall and leading to bacterial death through lysis.

Beta-lactam structure in antibiotics

The beta-lactam ring is especially reactive due to its geometric strain; the four-membered ring creates 90° bond angles, which deviate significantly from ideal angles for sp2^2 or sp3^3 hybridized atoms. This strain contributes to the ring's high reactivity and ability to acylate PBPs effectively. In contrast, larger lactam rings, which are less strained, show markedly lower reactivity. To counteract the effects of beta-lactam antibiotics, some bacteria produce an enzyme called beta-lactamase. This enzyme cleaves the beta-lactam ring, neutralizing the antibiotic's activity by preventing it from interacting with PBPs.


Which of the following would not be a viable bacterial defense mechanism against beta-lactam antibiotics?

A.
Possessing an outer membrane that limits drug permeability
B.
Producing proteins to displace beta-lactams already attached to PBPs
C.
Altering the structure of PBPs to reduce antibiotic binding
D.
Increasing expression of beta-lactamase enzymes
Question 54

Penicillin is widely regarded as one of the most iconic and influential antibiotics in medical history. Its chemical structure is defined by a key component called a beta-lactam—a four-membered ring containing an amide group. Figure 1 demonstrates the beta-lactam structure in several different antibiotics. This beta-lactam structure allows the drug to form a covalent bond with Penicillin Binding Proteins (PBPs), which play a crucial role in bacterial cell wall integrity by facilitating the cross-linking of peptidoglycan chains. When penicillin or related compounds attach to PBPs, this cross-linking process is disrupted, weakening the cell wall and leading to bacterial death through lysis.

Beta-lactam structure in antibiotics

The beta-lactam ring is especially reactive due to its geometric strain; the four-membered ring creates 90° bond angles, which deviate significantly from ideal angles for sp2^2 or sp3^3 hybridized atoms. This strain contributes to the ring's high reactivity and ability to acylate PBPs effectively. In contrast, larger lactam rings, which are less strained, show markedly lower reactivity. To counteract the effects of beta-lactam antibiotics, some bacteria produce an enzyme called beta-lactamase. This enzyme cleaves the beta-lactam ring, neutralizing the antibiotic's activity by preventing it from interacting with PBPs.


In the structure of a beta-lactam ring, which atom is most likely to exhibit sp2^2 hybridization?

A.
The nitrogen within the ring
B.
The carbon adjacent to the nitrogen (β-carbon)
C.
The carbon of the carbonyl group
D.
The carbon next to the carbonyl (α-carbon)
Question 55

A Cyclist Who Cannot Sprint: The Isozymes of Lactate Dehydrogenase

Lactate dehydrogenase (LDH) catalyzes the last step of anaerobic glycolysis:

pyruvate+NADH+H+⇌lactate+NAD+ΔG∘′=−25 kJ/mol\text{pyruvate} + \text{NADH} + \text{H}^+ \rightleftharpoons \text{lactate} + \text{NAD}^+ \qquad \Delta G^{\circ\prime} = -25\ \text{kJ/mol}

Active LDH is a tetramer built from two kinds of subunit: M, encoded by the LDHA gene and dominant in skeletal muscle, and H, encoded by LDHB and dominant in heart. Heart muscle takes up lactate from the blood, runs the reaction in reverse, and oxidizes the resulting pyruvate in its mitochondria. In one study, the two pure tetramers were compared at low pyruvate concentrations (Table 1).

Table 1. Pyruvate affinity of the two pure LDH tetramers

IsozymeMain tissueKm for pyruvate (mM)
M4Skeletal muscle0.30
H4Heart0.08

A 19-year-old cyclist reports cramps and dark urine after short sprints, yet can ride for hours at an easy pace. Clinicians performed an ischemic forearm test: with a cuff stopping blood flow to the forearm, the patient squeezed a hand grip hard for one minute; the cuff was then released. Blood was drawn from a forearm vein before exercise and two minutes after it. Ammonia, which muscle releases by deaminating AMP during intense effort, rises whether or not glycolysis works, so it confirms that the effort was adequate. A healthy volunteer served as the control (Table 2). A later muscle biopsy found no M subunits.

Table 2. Venous blood in the ischemic forearm test

AnalytePatient, beforePatient, afterControl, beforeControl, after
Lactate (mM)1.01.31.05.5
Pyruvate (mM)0.080.600.080.25
Ammonia (µM)3015030140

Suppose that in the control subject's exercised muscle, cytosolic [NAD+]/[NADH][\text{NAD}^+]/[\text{NADH}] is 45 and [lactate]/[pyruvate][\text{lactate}]/[\text{pyruvate}] equals the post-exercise venous ratio in Table 2. At pH 7.0 and 37 °C, where 2.303RT≈5.9 kJ/mol2.303RT \approx 5.9\ \text{kJ/mol}, what is ΔG\Delta G for the LDH reaction as written?

A.
About −7 kJ/mol
B.
About −43 kJ/mol
C.
About +18 kJ/mol
D.
About −22 kJ/mol
Question 56

A Cyclist Who Cannot Sprint: The Isozymes of Lactate Dehydrogenase

Lactate dehydrogenase (LDH) catalyzes the last step of anaerobic glycolysis:

pyruvate+NADH+H+⇌lactate+NAD+ΔG∘′=−25 kJ/mol\text{pyruvate} + \text{NADH} + \text{H}^+ \rightleftharpoons \text{lactate} + \text{NAD}^+ \qquad \Delta G^{\circ\prime} = -25\ \text{kJ/mol}

Active LDH is a tetramer built from two kinds of subunit: M, encoded by the LDHA gene and dominant in skeletal muscle, and H, encoded by LDHB and dominant in heart. Heart muscle takes up lactate from the blood, runs the reaction in reverse, and oxidizes the resulting pyruvate in its mitochondria. In one study, the two pure tetramers were compared at low pyruvate concentrations (Table 1).

Table 1. Pyruvate affinity of the two pure LDH tetramers

IsozymeMain tissueKm for pyruvate (mM)
M4Skeletal muscle0.30
H4Heart0.08

A 19-year-old cyclist reports cramps and dark urine after short sprints, yet can ride for hours at an easy pace. Clinicians performed an ischemic forearm test: with a cuff stopping blood flow to the forearm, the patient squeezed a hand grip hard for one minute; the cuff was then released. Blood was drawn from a forearm vein before exercise and two minutes after it. Ammonia, which muscle releases by deaminating AMP during intense effort, rises whether or not glycolysis works, so it confirms that the effort was adequate. A healthy volunteer served as the control (Table 2). A later muscle biopsy found no M subunits.

Table 2. Venous blood in the ischemic forearm test

AnalytePatient, beforePatient, afterControl, beforeControl, after
Lactate (mM)1.01.31.05.5
Pyruvate (mM)0.080.600.080.25
Ammonia (µM)3015030140

At the resting venous pyruvate concentration in Table 2, and assuming Michaelis–Menten kinetics, at about what fraction of its Vmax would each isozyme in Table 1 work?

A.
H4 at about 50% and M4 at about 27%
B.
M4 at about 50% and H4 at about 21%
C.
H4 at about 100% and M4 at about 27%
D.
H4 at about 50% and M4 at about 21%
Question 57

A Cyclist Who Cannot Sprint: The Isozymes of Lactate Dehydrogenase

Lactate dehydrogenase (LDH) catalyzes the last step of anaerobic glycolysis:

pyruvate+NADH+H+⇌lactate+NAD+ΔG∘′=−25 kJ/mol\text{pyruvate} + \text{NADH} + \text{H}^+ \rightleftharpoons \text{lactate} + \text{NAD}^+ \qquad \Delta G^{\circ\prime} = -25\ \text{kJ/mol}

Active LDH is a tetramer built from two kinds of subunit: M, encoded by the LDHA gene and dominant in skeletal muscle, and H, encoded by LDHB and dominant in heart. Heart muscle takes up lactate from the blood, runs the reaction in reverse, and oxidizes the resulting pyruvate in its mitochondria. In one study, the two pure tetramers were compared at low pyruvate concentrations (Table 1).

Table 1. Pyruvate affinity of the two pure LDH tetramers

IsozymeMain tissueKm for pyruvate (mM)
M4Skeletal muscle0.30
H4Heart0.08

A 19-year-old cyclist reports cramps and dark urine after short sprints, yet can ride for hours at an easy pace. Clinicians performed an ischemic forearm test: with a cuff stopping blood flow to the forearm, the patient squeezed a hand grip hard for one minute; the cuff was then released. Blood was drawn from a forearm vein before exercise and two minutes after it. Ammonia, which muscle releases by deaminating AMP during intense effort, rises whether or not glycolysis works, so it confirms that the effort was adequate. A healthy volunteer served as the control (Table 2). A later muscle biopsy found no M subunits.

Table 2. Venous blood in the ischemic forearm test

AnalytePatient, beforePatient, afterControl, beforeControl, after
Lactate (mM)1.01.31.05.5
Pyruvate (mM)0.080.600.080.25
Ammonia (µM)3015030140

Heart muscle converts lactate to pyruvate even though ΔG∘′\Delta G^{\circ\prime} for the LDH reaction favors lactate. Which statement best explains this?

A.
H4's lower Km for pyruvate shifts the equilibrium constant of the reaction toward pyruvate
B.
High lactate and NAD+\text{NAD}^+, plus removal of pyruvate, make ΔG\Delta G negative for the reverse reaction
C.
H4 couples lactate oxidation to ATP hydrolysis, which makes the reverse reaction exergonic
D.
H4 lowers the activation energy of the reverse reaction only, leaving the forward reaction slow
Question 58

A Cyclist Who Cannot Sprint: The Isozymes of Lactate Dehydrogenase

Lactate dehydrogenase (LDH) catalyzes the last step of anaerobic glycolysis:

pyruvate+NADH+H+⇌lactate+NAD+ΔG∘′=−25 kJ/mol\text{pyruvate} + \text{NADH} + \text{H}^+ \rightleftharpoons \text{lactate} + \text{NAD}^+ \qquad \Delta G^{\circ\prime} = -25\ \text{kJ/mol}

Active LDH is a tetramer built from two kinds of subunit: M, encoded by the LDHA gene and dominant in skeletal muscle, and H, encoded by LDHB and dominant in heart. Heart muscle takes up lactate from the blood, runs the reaction in reverse, and oxidizes the resulting pyruvate in its mitochondria. In one study, the two pure tetramers were compared at low pyruvate concentrations (Table 1).

Table 1. Pyruvate affinity of the two pure LDH tetramers

IsozymeMain tissueKm for pyruvate (mM)
M4Skeletal muscle0.30
H4Heart0.08

A 19-year-old cyclist reports cramps and dark urine after short sprints, yet can ride for hours at an easy pace. Clinicians performed an ischemic forearm test: with a cuff stopping blood flow to the forearm, the patient squeezed a hand grip hard for one minute; the cuff was then released. Blood was drawn from a forearm vein before exercise and two minutes after it. Ammonia, which muscle releases by deaminating AMP during intense effort, rises whether or not glycolysis works, so it confirms that the effort was adequate. A healthy volunteer served as the control (Table 2). A later muscle biopsy found no M subunits.

Table 2. Venous blood in the ischemic forearm test

AnalytePatient, beforePatient, afterControl, beforeControl, after
Lactate (mM)1.01.31.05.5
Pyruvate (mM)0.080.600.080.25
Ammonia (µM)3015030140

Before the biopsy, which result in Table 2 best indicated that the patient's block lies at the LDH step rather than at an earlier step of glycolysis?

A.
Pyruvate rose more than in the control, while lactate barely rose
B.
Ammonia rose as much as in the control, showing a full effort
C.
Lactate barely rose, showing that glycolysis made little end product
D.
Resting lactate and pyruvate matched the control's resting values
Question 59

A Cyclist Who Cannot Sprint: The Isozymes of Lactate Dehydrogenase

Lactate dehydrogenase (LDH) catalyzes the last step of anaerobic glycolysis:

pyruvate+NADH+H+⇌lactate+NAD+ΔG∘′=−25 kJ/mol\text{pyruvate} + \text{NADH} + \text{H}^+ \rightleftharpoons \text{lactate} + \text{NAD}^+ \qquad \Delta G^{\circ\prime} = -25\ \text{kJ/mol}

Active LDH is a tetramer built from two kinds of subunit: M, encoded by the LDHA gene and dominant in skeletal muscle, and H, encoded by LDHB and dominant in heart. Heart muscle takes up lactate from the blood, runs the reaction in reverse, and oxidizes the resulting pyruvate in its mitochondria. In one study, the two pure tetramers were compared at low pyruvate concentrations (Table 1).

Table 1. Pyruvate affinity of the two pure LDH tetramers

IsozymeMain tissueKm for pyruvate (mM)
M4Skeletal muscle0.30
H4Heart0.08

A 19-year-old cyclist reports cramps and dark urine after short sprints, yet can ride for hours at an easy pace. Clinicians performed an ischemic forearm test: with a cuff stopping blood flow to the forearm, the patient squeezed a hand grip hard for one minute; the cuff was then released. Blood was drawn from a forearm vein before exercise and two minutes after it. Ammonia, which muscle releases by deaminating AMP during intense effort, rises whether or not glycolysis works, so it confirms that the effort was adequate. A healthy volunteer served as the control (Table 2). A later muscle biopsy found no M subunits.

Table 2. Venous blood in the ischemic forearm test

AnalytePatient, beforePatient, afterControl, beforeControl, after
Lactate (mM)1.01.31.05.5
Pyruvate (mM)0.080.600.080.25
Ammonia (µM)3015030140

During the ischemic test, which glycolytic enzyme in the patient's forearm muscle is most directly slowed by the failure to regenerate NAD+\text{NAD}^+ through LDH?

A.
Hexokinase, which uses NAD+\text{NAD}^+ to phosphorylate incoming glucose
B.
Phosphofructokinase-1, which uses NADH to phosphorylate fructose 6-phosphate
C.
Glyceraldehyde-3-phosphate dehydrogenase, which uses NAD+\text{NAD}^+ to oxidize its substrate
D.
Pyruvate kinase, which uses NADH to transfer a phosphate group to ADP
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