MCAT 2025 EXAM Questions AND Correct Answers
[H+] = - ✔✔10^-pH
0.0005 kilomorgans to centimorgans - ✔✔50 centimorgans
1 joule equals - ✔✔1 Nm (newton * meter)
1 Newton equals - ✔✔1 kg m/s^2
1 pascal is equal to - ✔✔1 N/m^2
2 ways to form ATP - ✔✔1. substrate level phosphorylation
2. oxidative phosphorylation
3 categories of passive transport - ✔✔simple diffusion, facilitated diffusion,
osmosis
3 major types of organizations - ✔✔utilitarian, coercive, normative
3 power formulas - ✔✔P = IV
P = I^2R
P = V^2/R
,4 colligative properties of solutions - ✔✔vapor pressure lowering, boiling point
elevation, freezing point depression, osmotic pressure
4 phases of bacteria growth - ✔✔1. lag phase (bacteria adapt to a new
environment, growth lags)
2. log phase (exponential growth of population)
3. stationary phase (maximum population size is reached, growth limited by
nutrients and space)
4. death phase (bacteria begin to die off)
a 1 M solution of glucose contains - ✔✔6.02 × 1023 molecules of glucose in each
liter of solution.
A 5 kg box is resting on a ramp with an incline of 45 degrees. What can we say
about the value of the coefficient of static friction between the ramp and the
box? - ✔✔For this problem it is crucial to realize that the force parallel along the
incline, a component of gravity, must be equal to the force of static friction
currently being exerted. If this were not the case, the box would not be at rest.
From this we can setup an inequality, knowing that static friction increases in
response to an applied force until the applied force is greater than the
maximum possible static friction. Knowing this we can setup the equations
below:
mgsin(theta) <= usN
N = mgcos(theta)
mgsin(theta) <= us mg cos (theta)
,tan(theta) =< us
1<= us
And thereby determine that our coefficient of static friction must be greater
than or equal to 1, or the box would be in motion. Notice that although you
were provided with the weight of the box, it turned out to be irrelevant.
A car is continuously driving in a circle of radius 30 meters in a parking lot. How
is the centripetal force for this circular motion being generated? - ✔✔Friction
between the tires and the road
When a car turns, friction between the tires and the road is responsible for the
change in the car's velocity vector. Despite being less intuitive, this is an
identical scenario to all other circular motion scenarios we have discussed:
Centripetal force is pointing inwards, directly at the center of the circle, while
the car's velocity vector is perpendicular to the centripetal force. The result is
circular motion of the vehicle.
A cell in a hypertonic solution will - ✔✔shrink and lose water
a cell in a hypotonic solution - ✔✔-Gains water, water travels from environment
into cell
-Ruptures or lyse (hemolysis of red blood cells)
A child is sliding a toy block (with mass = m) down a ramp. The *coefficient of
static friction between the block and the ramp is 0.25*. When the block is
halfway down the ramp, the child pushes down on the block perpendicular to
the plane, halting it. What is the minimum force the child must apply to keep
the block from starting to slide down the ramp? - ✔✔{[mgsin(θ)] / 0.25} -
mgcos(θ)
, Gravitational force pulling block down the ramp is mgsin(θ)
To stop the block sliding down the ramp, we must have an equal and opposite
frictional force Ff = μFN (FN = normal force)
Since these forces are equal and opposite, they must be set equal to eachother
The block itself has a mass, m, and generates a normal force of FN = mgcos(θ)
Thus Ff = 0.25 * mgcos(θ)
The force exerted by the child (Fa) will add to the force created by the mass of
the block. Our total FN = mgcos(θ) + Fa.
Setting this force equal to Ff = mgsin(θ), we get:
Ff = 0.25 * (mgcos(θ) +Fa) = mgsin(θ)
mgcos(θ) +Fa = mgsin(θ) / 0.25
Solve for Fa:
Fa = (mgsin(θ) / 0.25) - mgcos(θ)
A circuit is constructed with a 12-V battery and four identical resistors, each
with a resistance of 16 Ω, and connected in parallel. What is the total power
dissipated by the circuit? - ✔✔36 W
[H+] = - ✔✔10^-pH
0.0005 kilomorgans to centimorgans - ✔✔50 centimorgans
1 joule equals - ✔✔1 Nm (newton * meter)
1 Newton equals - ✔✔1 kg m/s^2
1 pascal is equal to - ✔✔1 N/m^2
2 ways to form ATP - ✔✔1. substrate level phosphorylation
2. oxidative phosphorylation
3 categories of passive transport - ✔✔simple diffusion, facilitated diffusion,
osmosis
3 major types of organizations - ✔✔utilitarian, coercive, normative
3 power formulas - ✔✔P = IV
P = I^2R
P = V^2/R
,4 colligative properties of solutions - ✔✔vapor pressure lowering, boiling point
elevation, freezing point depression, osmotic pressure
4 phases of bacteria growth - ✔✔1. lag phase (bacteria adapt to a new
environment, growth lags)
2. log phase (exponential growth of population)
3. stationary phase (maximum population size is reached, growth limited by
nutrients and space)
4. death phase (bacteria begin to die off)
a 1 M solution of glucose contains - ✔✔6.02 × 1023 molecules of glucose in each
liter of solution.
A 5 kg box is resting on a ramp with an incline of 45 degrees. What can we say
about the value of the coefficient of static friction between the ramp and the
box? - ✔✔For this problem it is crucial to realize that the force parallel along the
incline, a component of gravity, must be equal to the force of static friction
currently being exerted. If this were not the case, the box would not be at rest.
From this we can setup an inequality, knowing that static friction increases in
response to an applied force until the applied force is greater than the
maximum possible static friction. Knowing this we can setup the equations
below:
mgsin(theta) <= usN
N = mgcos(theta)
mgsin(theta) <= us mg cos (theta)
,tan(theta) =< us
1<= us
And thereby determine that our coefficient of static friction must be greater
than or equal to 1, or the box would be in motion. Notice that although you
were provided with the weight of the box, it turned out to be irrelevant.
A car is continuously driving in a circle of radius 30 meters in a parking lot. How
is the centripetal force for this circular motion being generated? - ✔✔Friction
between the tires and the road
When a car turns, friction between the tires and the road is responsible for the
change in the car's velocity vector. Despite being less intuitive, this is an
identical scenario to all other circular motion scenarios we have discussed:
Centripetal force is pointing inwards, directly at the center of the circle, while
the car's velocity vector is perpendicular to the centripetal force. The result is
circular motion of the vehicle.
A cell in a hypertonic solution will - ✔✔shrink and lose water
a cell in a hypotonic solution - ✔✔-Gains water, water travels from environment
into cell
-Ruptures or lyse (hemolysis of red blood cells)
A child is sliding a toy block (with mass = m) down a ramp. The *coefficient of
static friction between the block and the ramp is 0.25*. When the block is
halfway down the ramp, the child pushes down on the block perpendicular to
the plane, halting it. What is the minimum force the child must apply to keep
the block from starting to slide down the ramp? - ✔✔{[mgsin(θ)] / 0.25} -
mgcos(θ)
, Gravitational force pulling block down the ramp is mgsin(θ)
To stop the block sliding down the ramp, we must have an equal and opposite
frictional force Ff = μFN (FN = normal force)
Since these forces are equal and opposite, they must be set equal to eachother
The block itself has a mass, m, and generates a normal force of FN = mgcos(θ)
Thus Ff = 0.25 * mgcos(θ)
The force exerted by the child (Fa) will add to the force created by the mass of
the block. Our total FN = mgcos(θ) + Fa.
Setting this force equal to Ff = mgsin(θ), we get:
Ff = 0.25 * (mgcos(θ) +Fa) = mgsin(θ)
mgcos(θ) +Fa = mgsin(θ) / 0.25
Solve for Fa:
Fa = (mgsin(θ) / 0.25) - mgcos(θ)
A circuit is constructed with a 12-V battery and four identical resistors, each
with a resistance of 16 Ω, and connected in parallel. What is the total power
dissipated by the circuit? - ✔✔36 W
