FL 5 - Chem/Phys

Question 1

Which of the following refers to the mechanism of peptide bond formation?

I. Nucleophilic substitution

II. Condensation

III. Hydrolysis

Answer 1

I and II only. Peptide bond formation occurs as the lone pair of electrons on the nitrogen atom on the amine attacks, and the amine adds to the carbonyl carbon, making it a nucleophilic substitution (I). In the process of bond formation, the hydroxyl group of the carboxylic acid leaves as a water molecule. Therefore, this is a condensation reaction (II).

Question 2

What is Dynein?

Answer 2

Dynein is a motor protein that walks towards the minus end of the microtubules, which is oriented towards the center of the cell. Movement toward the center of the cell is described as retrograde.

Question 3

Energy is measured in joules (J). What are the SI units of a joule?

Answer 3

Energy is measured in joules (J). The SI units for a joule are kilograms•(meters²/seconds²).

Question 4

If a 1 ng red blood cell has 5 x 10^-13 J of kinetic energy, how fast is it travelling in the circulatory system?

Answer 4

Kinetic energy = (0.5)(1 x 10^-12 kg)(v²). In this case, we are given the kinetic energy of the red blood cell and asked to find the velocity, so we can rearrange the kinetic energy equation as follows:

5 x 10^-13 J / [(0.5)(1 x 10^-12 kg)] = v²

5 x 10^-13 J / (5 x 10^-13 kg) = v²

1 J / kg = v²

Taking the square root of both sides to isolate “v” gives us 1 m/s. In scientific notation this would be written as 1 x 10⁰ m/s because 10⁰ = 1.

Question 5

What is the gravitational potential energy and the kinetic energy of a 5 g stone that is three-quarters of the way down a 7000 cm drop off a cliff in an ideal system?

Answer 5

Originally the stone had all of its energy stored as gravitational potential energy.

PE = mgh = (5 x 10^-3 kg)(10 m/s²)(70 m) = 3.5 J.

Remember to convert 5 g into kilograms and 7000 cm into meters before plugging it into the equation for gravitational potential energy. When the stone is three-quarters of the way down a 70 m drop, it will be 17.5 m off the ground and 25 % of its original 3.5 J of energy will be stored as gravitational potential energy and 75 % will be in the form of kinetic energy. This means that three-quarters of the way down the drop, the stone will have approximately 0.9 J of gravitational potential energy and approximately 2.6 J of kinetic energy. Because the question stem specifies this is an ideal system, there is no energy lost from the system to the surroundings.

Question 6

What is the gravitational potential energy and the kinetic energy of a 5 g stone that is three-quarters of the way down an 80 m drop off a cliff in a non-ideal system?

Answer 6

Originally the stone had all of its energy stored as gravitational potential energy.

PE = mgh = (5 x 10^-3 kg)(10 m/s²)(80 m) = 4 J

Remember to convert 5 g into kilograms before plugging it into the equation for gravitational potential energy. When the stone is three quarters of the way down an 80 m drop, it will be 20 m off the ground and 25% of its original 4 J of energy will be stored as gravitational potential energy and 75% will be in the form of kinetic energy. This means that three quarters of the way down the drop, the stone will have approximately 1 J of gravitational potential energy and approximately 3 J of kinetic potential energy. Because the question stem specifies this is a non-ideal system, some energy will be lost from the system due to the presence of non-conservative forces like air resistance and friction. Therefore, the correct answer should be 1 J of gravitational potential energy and slightly less than 3 J of kinetic energy because the transfer of gravitational potential energy to kinetic energy is not 100% efficient.

Question 7

If a 7 kg object moving at some velocity collides with a 5 kg object, which then hits a third object, compare the amount of work done by each of the first two objects. Assume complete transfer of energy.

Answer 7

According to the work-energy theorem, the amount of work done by an object is equal to the change in kinetic energy.

In other words, an object’s kinetic energy will increase in direct proportion to the amount of work we put in, and it will decrease proportionally to the amount of work the object expends. The heavier object will transfer all of its kinetic energy to the lighter object, which will, in turn, transfer all of its kinetic energy to the third object. The heavier object will be moving at a slightly lower velocity than the lighter object, but the amount of total kinetic energy transferred each time remains the same.

Question 8

A 1000 kg red car traveling at 25 m/s hits a stopped 2000 kg orange car. This causes the orange car to be propelled forward into a stopped 1500 kg yellow car in front of it and causes the yellow car to move forward. How fast will the orange car be moving when it hits the yellow car? Assume the system is ideal.

Answer 8

The question stem says the system is ideal, so we can assume 100 % efficient transfer of kinetic energy between the cars.

KE = (0.5)mv², so KE_{red car} = (0.5)(1000 kg)(25 m/s)² = approximately 312 kJ.

We can use this total amount of kinetic energy to find the velocity of the orange car because we know its mass.

312,000 J = (0.5)(2000 kg)(v²), so velocity = approximately 17 m/s.

We do not need to consider the yellow car in this scenario because it will not have an effect on the velocity of the orange car.

Question 9

What is the kinetic energy of a 2 kg bird that is 100 m off the ground and flying at 3 m/s?

Answer 9

The formula for calculating kinetic energy (KE) is KE = (1/2)mv², where m is the object’s mass and v is the object’s velocity. Plugging in 2 kg for the mass and 3 m/s for the velocity gives us:

KE = (1/2)(2 kg)(3 m/s)²= (1/2)(2 kg)(9 m²/s²) = (1/2)(18 kg m²/s²) = 9 J.

Question 10

What are 4 different types of potential energy?

Answer 10

Chemical

Elastic

Gravitational

Electric

Question 11

What property must an object have for it to be impossible to have elastic potential energy?

Answer 11

It must be incompressible.

We usually think of elastic potential energy as the energy stored in a spring, but more generally it refers to energy stored in objects due to some kind of compression. This could be a spring, or it could be a rubber band that you’ve stretched out to snap someone, or the legs of a super jumpy animal like a rabbit, or even some kind of stretchy protein or biological material.

Question 12

What happens to the kinetic energy of an object when its velocity is tripled?

Answer 12

Kinetic energy (KE) = (0.5)mv², where m is the object’s mass and v is its velocity. Note from this equation that kinetic energy is directly proportional to mass, and that, since velocity is squared, kinetic energy increases exponentially with how fast the object is moving. If the velocity of the object increases by a factor of 3,the kinetic energy will increase by a factor of 9 because 3² = 9.

Question 13

True or false: The law of conservation of energy states that equal amounts of energy that are created must eventually be destroyed so that the total amount of energy in the universe remains constant.

Answer 13

This statement is false.

The law of conservation of energy states that energy is neither created nor destroyed, only transformed to different forms (i.e. kinetic, gravitational potential, elastic potential, electric, heat, etc). This law of conservation of energy is the first law of thermodynamics.

Question 14

What is the gravitational potential energy and the kinetic energy of a 1 g stone that is halfway down an 800 m drop off a cliff in an ideal system? (Assume that the stone has zero gravitational potential energy at the bottom of the cliff.)

Answer 14

Originally the stone had all of its energy stored as gravitational potential energy.

PE = mgh = (1 x 10^-3 kg)(10 m/s²)(800 m)= 8 J

Remember to convert 1 g into kilograms before plugging it into the equation for gravitational potential energy. When the stone is halfway down an 800 m drop, it will be 400 m off the ground and half of its original 8 J of energy will be stored as gravitational potential energy and the other half will be in the form of kinetic energy. This means that halfway down the drop, the stone will have 4 J of gravitational potential energy and 4 J of kinetic potential energy. Because the question stem specifies this is an ideal system, there is no energy lost from the system to the surroundings.

Question 15

How much kinetic energy can a person gain if they are attached to a bungee cord that has been horizontally stretched 3 m and they are suddenly released? Assume the stiffness of the bungee cord is 5 N/m.

Answer 15

Elastic potential energy = (0.5)kx², where k is the spring constant, which is an indicator of stiffness that’s unique to each spring, and x is the distance by which the spring (or bungee cord in this case) is compressed or extended. The maximum amount of kinetic energy this person could gain is equal to the amount of elastic potential energy stored in the bungee.

Elastic potential energy = (0.5)(5 N/m)(3 m)²= (0.5)(5 N/m)(9 m²) = (0.5)(45 J)

One half of 45 J is between 22.5 J.

Question 16

A 6 kg object, in a non-ideal system, is initially at rest and then falls from a cliff with a height of 250 m. What is the kinetic energy of that object shortly before impact?

Answer 16

Because energy is always conserved, the final energy of the system will be equal to the initial energy of the system. The initial energy of the system is stored as gravitational potential energy, which we can calculate using the following formula:

PE = mgh= (6 kg)(10 m/s²)(250 m)= 15000 J

However, we need to account for the fact that some of this energy will be lost as heat due to non-conservative forces like friction and air resistance. Because this object is falling, it will experience air resistance. We can organize these concepts into the following formula:

initial gravitational potential energy = final kinetic energy + energy lost as heat

15000 J = final kinetic energy + energy lost as heat

15000 J – energy lost as heat = final kinetic energy

This means that the final kinetic energy must be some value less than 15000 J when we account for the energy that is lost as heat.

Question 17

A driver in a 2,000 kg SUV moving at 15 m/s crashes into a stopped car, causing it to be propelled forward and the truck to be slowed to 10 m/s. What is the work done on the SUV by the stopped car?

Answer 17

We are asked to solve for the amount of work done, and we are given information about one car’s mass and velocity. We can use this to solve for initial kinetic energy of the SUV and final kinetic energy of the SUV:

KE_initial = (0.5)(2 x 10³ kg)(15 m/s)² = (0.5)(2 x 10³ kg)(225 m²/s²) = 2.25 x 10⁵ J

KE_final = (0.5)(2 x 10³ kg)(10 m/s)² = (0.5)(2 x 10³ kg)(100 m²/s²) = 1 x 10⁵ J

There are two things to be careful about here. First, it’s important to always remember to subtract INITIAL energy from FINAL energy so that the sign is positive when energy is gained, and the sign is negative when energy is lost by the object we’re considering. Second, we need to know which object we’re talking about. The question stem specifies “work done on the SUV by the stopped car”. Intuitively we know that, because the SUV lost energy (KE_initial > KE_final), the work done on it must be negative. We can back this up mathematically with the following:

Work done = KE_final - KE_initial

Plugging in our values for final and initial kinetic energy gives us:

(1 x 10⁵ J) - (2.25 x 10⁵ J) = -1.25 x 10⁵ J

We can assume this is an ideal system with no work lost and the only type of energy at play is kinetic energy unless otherwise indicated.

Question 18

An aide is rigging a pulley system to elevate a patient’s leg. The leg weighs 12 kg and the aide needs to raise it by 0.6 m. How much work does the aide need to put into the system in order to elevate the patient’s leg?

Answer 18

The patient’s leg is starting off motionless in the bed with 0 J of energy and is going to be raised 0.6 m, giving it gravitational potential energy (PE). We can calculate:

PE_final = mgh = (12 kg)(10 m/s²)(0.6 m)= 72 J

It’s important to always remember to subtract INITIAL energy from FINAL energy so that the sign is positive when energy is gained, and the sign is negative when energy is lost by the object we’re considering.

PE_final - PE_inital = 72 J - 0 J = 72 J

The aide must supply 72 J of energy to elevate the patient’s leg 0.6 m. Because energy is being added to the system (comprised of the patient’s leg), we know the final work value should be positive according to the sign conventions for work used in physics.

You may be wondering why we modeled the leg as a uniform block undergoing translational motion, rather than an object with a moment of inertia rotating about a pivot the way a real leg might. The answer is simple: always solve problems within the scope of the MCAT. As well as practicing the work-energy theorem, helping you to develop such an MCAT-question answering skill is the lesson of this practice question.

Question 19

True or false: The work-energy theorem represents an exception to the law of conservation of energy because, in the real world, energy transfer between systems is not 100% efficient.

Answer 19

False.

The work-energy theorem is NOT an exception to the law of conservation of energy. The law of conservation is interested in the energy of an entire isolated system - that system, at least in a theoretical sense, can be as small as just the bowling ball and pins, or as big as the entire universe. The point is that individual objects in that system can gain or lose energy, so energy can be transferred from one object to another, and energy can also be transformed from one type to another. But all of that energy still stays within the system in some capacity or another. This same rule applies to scenarios when we say energy is “lost as heat”. The energy is not technically lost because if we were able to measure the amount of heat energy that was added to the system, it would account for the amount of kinetic energy or other types of energy that was lost. We only use the terminology “lost as heat” because for real-world purposes, the energy that does not do work isn’t all that useful to us.

Question 20

If a red blood cell has 5 x 10^-13J of kinetic energy and is moving at 1m/s, what is the mass of the cell?

Answer 20

Kinetic energy = (0.5)(m)(1 m/s)². In this case, we are given the kinetic energy of the red blood cell and asked to find the mass so we can rearrange the kinetic energy equation as follows:

5 x 10^-13 J / [(0.5)(1 m/s)²] = m

5 x 10^-13 J / (0.5 m²/ s²) = m

1 x 10^-12 kg = m

1 x 10^-12 kg = 1 x 10^-9 g

Question 21

What is the gravitational potential energy and the kinetic energy of a 5 g stone that is one-quarter of the way down a 7000 cm drop off a cliff in an ideal system?

Answer 21

Originally the stone had all of its energy stored as gravitational potential energy.

PE = mgh = (5 x 10^-3 kg)(10 m/s²)(70 m) = 3.5 J

Remember to convert 5 g into kilograms and 7000 cm into meters before plugging it into the equation for gravitational potential energy. When the stone is one-quarter of the way down a 70 m drop, it will be 52.5 m off the ground and 75 % of its original 3.5 J of energy will be stored as gravitational potential energy and 25 % will be in the form of kinetic energy. This means that one-quarter of the way down the drop the stone will have approximately 2.6 J of gravitational potential energy and approximately 0.9 J of kinetic energy. Because the question stem specifies this is an ideal system, there is no energy lost from the system to the surroundings.

Question 22

What is the gravitational potential energy and the kinetic energy of a 5 g stone that is one-quarter of the way down an 80 m drop off a cliff in a non-ideal system?

Answer 22

Originally the stone had all of its energy stored as gravitational potential energy.

PE = mgh = (5 x 10^-3 kg)(10 m/s²)(80 m) = 4 J

Remember to convert 5 g into kilograms before plugging it into the equation for gravitational potential energy. When the stone is one quarter of the way down an 80 m drop, it will be 60 m off the ground and 75% of its original 4 J of energy will be stored as gravitational potential energy and 25% will be in the form of kinetic energy. This means that one quarter of the way down the drop, the stone will have approximately 3 J of gravitational potential energy and approximately 1 J of kinetic energy. Because the question stem specifies this is a non-ideal system, some energy will be lost from the system due to the presence of non-conservative forces like air resistance and friction. Therefore, the correct answer should be 3 J of gravitational potential energy and slightly less than 1 J of kinetic energy because the transfer of gravitational potential energy to kinetic energy is not 100% efficient.

Question 23

If a 7 kg object and a 5 kg object are each pushed until they have the same amount of kinetic energy, which object had more work done on it?

Answer 23

According to the work-energy theorem, the amount of work done by an object is equal to the change in its kinetic energy.

In other words, an object’s kinetic energy will increase in direct proportion to the amount of work put in, and it will decrease proportionally to the amount of work the object expends. The two objects here will move at different velocities since they have different masses, but the amount of total kinetic energy transferred to each of them remains the same.

Question 24

True or false: An object in motion stays in motion if there are no forces acting on it.

Answer 24

This statement is true. An object in motion stays in motion if no forces are acting on it. An outside force must compel the object to stop or change its course of motion.

Question 25

A heterogenous block is made up of two homogenous portions: A wood portion at the top, and a steel portion at the bottom. The horizontal center of mass will be located where?

Answer 25

On the midline

The horizontal center of mass will not shift in this setup. The vertical center of mass and the overall center of mass of the object would indeed be shifted downward, but not the horizontal center of mass.

Question 26

You are about to enter your first street race. Your cousin, Vincente, tells you the following: “Just floor it when you start, spin your wheels! The more gas you give it, the faster your car will go off the starting line!” Assuming you want to go as fast as possible, this advice is: true or false?

Answer 26

This statement is false.

Car engines are able to transmit more force to the wheels than the force of static friction (F_S = μ_S*N), resulting in a condition where only kinetic friction (F_k = μ_k*N) is acting between the tire and the road as the wheels spin. While this may still propel the car forward, the coefficient of kinetic friction is always less than that of static friction, and the car will accelerate more slowly than it could otherwise.

This is what causes wheelspin during the launch of high-performance cars, and when done intentionally is usually called a ‘burnout’.

Question 27

Needle thoracostomy is an emergency procedure used to decompress tension pneumothorax. Part of this procedure involves penetrating the distressed patient’s chest wall with a large needle along the second intercostal space. The force opposing movement of the needle is greatest at what time?

Answer 27

The instant before the needle breaks skin

Shortly before penetrating the skin, before static friction is overcome, the force applied is greatest. As the needle is moving, only kinetic friction is acting - and this force must be less than that of static friction. This is similar to the phenomenon of trying to penetrate a rubber skin or the seal of a juicebox - the force required to break the seal can be so much larger than the force required to move the straw through the opening, that one pushes the straw further than intended.

Question 28

Two masses are suspended across a pulley and each mass rests on an incline, of the exact angles below.

This system is friction-less. If the system is at rest, what can be said of m₁ and m₂?

Answer 28

If the system is at rest, meaning F_net = 0, we can quickly discover that F_{g on m2} = F_{g on m1}. Substituting in our expressions for both, we get m₂sin(60°) = m₁sin(30°). Solving this for m₁, we get m₁ = m₂sin(60°)/sin(30°) ≅ m₂(0.87/0.5) ≅ m₂(1.7) . This tells us that m₁ > m₂.

For the MCAT™, it can be useful to build an intuition for this sort of problem to, for example, realize that the force generated by gravity acting on m₁ must be greater than the force acting by gravity on m₂, but that the component vectors parallel to the plane must be identical in magnitude if the system is at rest.

Question 29

A particle is in a circular orbit around a fixed point at a distance of 50 cm. It completes one revolution per second and weighs 10 grams. What is the magnitude of the centripetal force?

Answer 29

To answer this we need to:

convert rotational velocity (⍵ = 2𝜋 rad/time ≅ 6.3 rad/s) to linear velocity, using v = r*⍵ (where v is the linear velocity and ⍵ the angular velocity).

This yields a linear velocity of (6.3 rad/s)(0.5 m) = 3.15 m/s. From there, we can simply substitute into our known expression for centripetal force: F_centripetal = mv²/r = (0.01 kg)(10 m²/s² ) / 0.5 m = 0.2 N.

Question 30

An automotive spring clutch is a mechanism that pushes two circular, rotating plates (the clutch and the flywheel) together with the purpose of keeping them locked together and rotating at the same angular velocity. Assuming a spring constant of about 100000 N/m and a compression of 28 cm, what is the maximum force static friction can exert between a clutch and flywheel with a coefficient of static friction of 0.30 between them?

Answer 30

To answer this, it is important to realize that the force of the clutch springs will be the source of the normal force (contact force) between the clutch and the flywheel. And we can determine the maximum force exerted by static friction (before the clutch and flywheel begin slipping) by F_static ≦ μ_static*N. Since we know that N = F_spring = -kx, we can substitute our expression to get F_static ≦ μ_static *( -kx). Solving, that gives us F_static ≦ 0.30*(-10⁵ N/m*-0.28 m) ⪅ 2.8*10³*3 ⪅ 8400 N.

Question 31

For the pulley system pictured, m₁ has a mass of 10 kg and m₂ has a mass of 5kg. The incline and the ground form a 30 degree angle. The rope and pulley are ideal, meaning both massless and frictionless. In this system, m₁ is what?

Answer 31

Not accelerating

This system is in static equilibrium.

F_{g on m1} = m₁gsin(θ)

F_{g on m2} = m₂g

sin(30 deg.) = 0.5

So F_{g on m1} = (10kg)(10m/s²)(0.5) = 50N

and F_{g on m2} = (5kg)(10m/s²) = 50N

The system is experiencing a 50N force to the left (or “counterclockwise”) and a 50N force to the right (“clockwise”), so there is zero net force and therefore zero acceleration.

Question 32

Assuming no air resistance, how fast would a horizontally thrown baseball need to be pitched to orbit the Earth? Assume the Earth is 12600 kilometers in diameter and g is 10 m/s².

Answer 32

We know that for orbital motion, v²/r = a. And since we also know that the acceleration for an orbit near surface level must be g, we can say that a = g. Therefore v²/r = g, and we can now solve this equation for v, giving us gr = v². Note that this equation uses radius, not diameter, so we must divide 12600 km by 2, yielding 6300 km as the radius. Converting this to meters results in 6.3 x 10⁶ m. Plugging in the values given, we get (10 m/s²)(6.3 x 10⁶ m) = v², or 63 x 10⁶ m²/s² = v². Using some estimation, we take the square root of both sides and we can say that the root of 63 is approximately 8 and the root of 10⁶ is 10³. This gives us about 8 x 10³m/s as our final estimation of the minimal velocity that a frictionless projectile must be launched at to orbit the earth at ground level.

Question 33

A fish is placed on a massless spring with a spring constant of 140 N/m. The spring was initially at rest and is now compressed by 3 centimeters. What is the mass of the fish?

Answer 33

Using Hooke’s law, we know F_spring= kxfor the forces acting upwards on the fish, while F_grav = mg for the downward forces. When the fish is in its final position, we know that F_grav = F_spring, therefore kx = mg. Solving for m, this yields m = kx/g = 140 N/m * 0.03 m / 10 m/s² ≅ (1.4 * 3/10) kg ≅ 0.42 kg. The fish weighs approximately 420 grams.

Question 34

Your grandfather hands you the keys to his 1967 Fastback Mustang. Before you take her for a spin, he gives you some advice: “When you want to stop quickly, hit the brakes hard and lock the wheels (which means to stop the rotation of the wheels entirely). When the tires squeal, you’re stopping as fast as you can!” Is this true or false? (Hint: the coefficient of static friction is greater than the coefficient of kinetic friction between the tires and the road.)

Answer 34

This statement is false.

Your grandfather in this story is a dangerous man! When the wheels of a car are locked while the car remains in motion, the tire and road begin to glide across each other - meaning kinetic friction is acting between the two and opposing motion. While the wheels are still rotating, static friction acts between the road and the tires. Since the coefficient of static friction for a material is always larger than the coefficient of kinetic friction, it is very likely that a skidding car is braking far more slowly than a car whose tires remain in motion.

The squealing noise occurs during this maneuver because the act of exerting kinetic friction involves the generation of large amounts of heat and sound as the rubber abrades against the road surface.

For this reason, modern cars employ the maximum amount of braking power without exceeding the force of static friction.

Question 35

Needle thoracostomy is an emergency procedure used to decompress tension pneumothorax. Part of this procedure involves penetrating the distressed patient’s chest wall with a large needle along the second intercostal space. After penetration, the needle is held in place and does not slide further down into the patient’s chest due to what force?

Answer 35

Static friction

Two objects in contact without sliding past each other with a normal force acting between them (meaning, any contact force) will be exerting static frictional forces on each other. The needle does not move because these frictional forces are exactly equal and opposite the action of gravity. Note that if more force is applied, static friction can be overcome and the needle can continue to move.

Question 36

A 15 kg box is sliding down a ramp with an incline of 25 degrees. What is the magnitude of the force from gravity on the box parallel to the incline?

Answer 36

The force from gravity on a box pointing parallel to an incline is equal to (m)(g)sin(theta). sin(25°) is a little less than sin(30°) = 1/2; therefore, (mg)sin(25°) would be a little less than (mg)/2. Since (mg)/2 = 75 N

Question 37

An automotive clutch spring is a mechanism that pushes two circular, rotating plates (the clutch and the flywheel) together with the purpose of keeping them locked together and rotating at the same angular velocity. Assume a maximum force of static friction exerted of about 6000 N between the clutch and flywheel and a coefficient of friction of 0.25 and a compression of the spring of 0.4 meters, what is the spring constant of the clutch spring?

Answer 37

To answer this, it is important to realize that the force of the clutch spring will be the source of the normal force (contact force) between the clutch and the flywheel. And we can determine the maximum force exerted by static friction (before the clutch and flywheel begin slipping) by F_static ≦ μ_static*N. Since we know that N = F_spring = -kx and we want to solve for k, we can substitute our expression to get F_{static max} = μ_static *( -kx) and then isolate k, resulting in -k = F_{static max} /(μ_static*x) or k = -F_{static max} /(μ_static*x).

Solving this, we arrive at k = -6000N /(0.25*-0.4 m) = 60000 N/m. Notice that we did not formulate static friction as an inequality in this case because we were dealing with the boundary condition, the maximum possible static friction, and therefore designated our variable F_{static max}.

Question 38

True or false: The overwhelming majority of biologically active carbohydrates are D-isomers.

Answer 38

This statement is true.

Carbohydrates can be classified as to what direction they rotate plane polarized light (a property of chiral carbons). This can be visually determined by seeing which direction the hydroxyl group on the bottommost chiral carbon points (Right = D-sugar, Left = L-sugar). D and L isomers of the same sugar are slightly different structurally, and thus will differentially interact with enzymes involved in metabolism. The enzymes in the body involved in metabolism primarily interact with D-isomers.

Question 39

The straight chain sugar D-galactose has what functional group located on its C1 carbon?

Answer 39

Galactose is an aldose – by definition, this means that it has a terminal C1 aldehyde in its straight chain form.

Question 40

True or false: Glycosidic linkages are strong covalent bonds formed through enzymatic catalysis.

Answer 40

This statement is true.

Glycosidic linkages are covalent carbon/oxygen/carbon linkages formed between two sugar molecules. These linkages are formed through the action of particular enzymes specific for the structure of particular reactants.

Question 41

The reaction that breaks glycosidic linkages between anomeric carbons can be classified as what kind of reaction?

Answer 41

A hydrolysis reaction

During hydrolysis, water is added to an atom in order to break a bond. This describes how glycosidic linkages are broken.

Question 42

Lactase will breakdown what kind of glycosidic linkage?

Answer 42

Lactase breaks down molecules of lactose, specifically by hydrolyzing a beta-1,4 bond between a molecule of galactose and a molecule of glucose.

Question 43

True or false: Glycosidic linkages are formed at the sole epimeric carbon of a sugar.

Answer 43

This statement is false.

Glycosidic linkages are formed at an anomeric carbon (the carbon of a monosaccharide which contains a carbonyl group).

An epimeric carbon is a designation between two similar sugars that indicates that the two sugars have opposite R/S designations at the same position.

Question 44

True or false: Consuming the carbohydrate polymer peptidoglycan may result in an immune response.

Answer 44

This statement is true.

Peptidoglycan is a carbohydrate polymer which is found in the cell walls of prokaryotes. It has unique structural features which differentiate it from other carbohydrates. Like many substances not normally found in the human body (antigenically foreign), immune cells will react against the substance, causing an immune response.

Question 45

True or false: Ribose is the six carbon aldose used in the production of RNA nucleotides.

Answer 45

This statement is false.

Ribose is a five carbon aldose, otherwise known as an aldopentose. D-Ribose is an easy and useful structure to memorize and be able to recognize on the MCAT™, because all of its hydroxyl groups point to the right.

Question 46

Deoxy-ribose sugars have lost a molecule of oxygen at which carbon position, compared to their ribose counterparts?

Answer 46

Deoxyribose differs from ribose in that it is missing a hydroxyl group at the carbon 2 position.

Question 47

True or false: Glucose exists in predominantly the pyranose form when placed in aqueous solution.

Answer 47

This statement is true.

The straight-chain form of many monosaccharides will form a ring structure when placed in aqueous solution. Pyranose is the term for a six-membered ring composed of 5 carbon atoms and 1 oxygen atom within the ring structure, and furanose is the term for a five-membered ring with 4 carbon atoms and 1 oxygen atom within the ring structure. Glucose predominantly exists as a six-membered (a pyranose) ring when placed in an aqueous solution (such as bodily fluids).

Question 48

The following image depicts the structure of 4 unique monosaccharides.

What is the name of monosaccharide A?

Answer 48

Ribose

Question 49

The following image depicts the structure of 4 unique monosaccharides.

What is the name of monosaccharide B?

Answer 49

D-glucose

Question 50

The following image depicts the structure of 4 unique monosaccharides.

What is the name of monosaccharide C?

Answer 50

Galactose

Question 51

The following image depicts the structure of 4 unique monosaccharides.

What is the name of monosaccharide D?

Answer 51

L-glucose

Question 52

The reaction that creates glycosidic linkages between anomeric carbons can be classified as a(n):

Answer 52

Dehydration reaction

Glycosidic linkages between sugars are created when a free hydroxyl group on one sugar is linked to the anomeric carbon of another sugar, which in turn loses a bond to water (dehydration).

Question 53

Describe the function of lactase

Answer 53

Lactase breaks down the disaccharide lactose into its two constituent monomers, glucose and galactose.

Question 54

Describe the function of glycoside hydrolase

Answer 54

Glycoside hydrolase is the general class of enzyme which catalyzes the hydrolysis, or breakage, of glycosidic bonds.

Lactase and maltase are specific kinds of glycoside hydrolases.

Question 55

Describe the function of glycosyltransferases

Answer 55

Glycosyltransferases are a class of enzyme which function opposite glycoside hydrolases – they catalyze the formation of glycosidic linkages.

Question 56

Describe the function of maltase

Answer 56

Maltase breaks down the disaccharide maltose into its two constituent monomers, glucose and glucose.

Question 57

The two monosaccharides below are galactose (left) and glucose (right).

What kind of glycosidic linkage may be formed between them during the formation of lactose?

Answer 57

Beta-1,4

Galactose and glucose monomers can be linked to form the disaccharide lactose. The beta-glycosidic bond between the two monomers is created between the 1 position carbon of galactose, and the 4 position of glucose.

Question 58

True or false: In order to be formed, glycosidic bonds require two anomeric carbons.

Answer 58

This statement is false.

Glycosidic bonds only require at least one anomeric carbon, although sometimes two anomeric carbons are used in the formation of a glycosidic bond.

Question 59

The glycosidic linkage shown here depicts what kind of bond?

Answer 59

Beta-1,4

The structure shown in the question stem is lactose – a disaccharide composed of galactose and glucose. With or without knowing this, one can identify the glycosidic bond as a “beta-1,4” bond. The bond exists between the 1-carbon of galactose (on the left, pointing upward in a beta orientation) and the 4 carbon of glucose (on the right).

Question 60

Sucrose, also known as table sugar, is a disaccharide composed of glucose (a pyranose) and fructose (a furanose). The molecule is shown below.

What is the correct shorthand glycosidic bond notation to describe this bond?

Answer 60

Glu (a-1 → b-2) Fru

Shorthand bond notation takes into account the 3-letter abbreviation for the first sugar, followed by the glycosidic bond notation in parentheses, followed by a 3-letter abbreviation for the second sugar. Sucrose is glucose and fructose connected by an alpha-1, beta-2 bond.

Question 61

Starch is composed of what two polysaccharides?

Answer 61

Amylose and Amylopectin

The polymer starch is composed of two distinct polysaccharides. One, amylose, is long chains of glucose connected by alpha-1,4 bonds. The other, amylopectin, is similar but additionally has branches of alpha-1,6 bonds.

Question 62

True or false: Human skeletal muscle stores glycogen and readily breaks it down for release into the bloodstream during activity.

Answer 62

This statement is false.

Two things are wrong with this statement. Human skeletal muscle does store glycogen and break it down during activity, but it uses the resultant glucose itself. Secondly, the liver is the primary storage site of glycogen polymers which results in increased blood glucose after breakdown.

Question 63

Glycogen is a branched polymer of glucose. The core of a granule of glycogen is the protein ________. Chains of glucose molecules connected by alpha-1,4 bonds extend off of this. Each glucose chain allows for much greater packing because the chains are branched through the use of alpha-1,6 bonds. At the terminal end of each branch of glycogen, hydrolysis of glycosidic bonds allows for the release of usable glucose monomers.

Answer 63

Glycogen is a branched polymer of glucose. The core of a granule of glycogen is the protein glycogenin. Chains of glucose molecules connected by alpha-1,4 bonds extend off of this. Each glucose chain allows for much greater packing because the chains are branched through the use of alpha-1,6 bonds. At the terminal end of each branch of glycogen, hydrolysis of glycosidic bonds allows for the release of usable glucose monomers.

Question 64

Glycogen is a branched polymer of glucose. The core of a granule of glycogen is the protein glycogenin. Chains of glucose molecules connected by _______________ extend off of this. Each glucose chain allows for much greater packing because the chains are branched through the use of alpha-1,6 bonds. At the terminal end of each branch of glycogen, hydrolysis of glycosidic bonds allows for the release of usable glucose monomers.

Answer 64

Glycogen is a branched polymer of glucose. The core of a granule of glycogen is the protein glycogenin. Chains of glucose molecules connected by alpha-1,4 bonds extend off of this. Each glucose chain allows for much greater packing because the chains are branched through the use of alpha-1,6 bonds. At the terminal end of each branch of glycogen, hydrolysis of glycosidic bonds allows for the release of usable glucose monomers.

Question 65

Glycogen is a branched polymer of glucose. The core of a granule of glycogen is the protein glycogenin. Chains of glucose molecules connected by alpha-1,4 bonds extend off of this. Each glucose chain allows for much greater packing because the chains are branched through the use of _____________. At the terminal end of each branch of glycogen, hydrolysis of glycosidic bonds allows for the release of usable glucose monomers.

Answer 65

Glycogen is a branched polymer of glucose. The core of a granule of glycogen is the protein glycogenin. Chains of glucose molecules connected by alpha-1,4 bonds extend off of this. Each glucose chain allows for much greater packing because the chains are branched through the use of alpha-1,6 bonds. At the terminal end of each branch of glycogen, hydrolysis of glycosidic bonds allows for the release of usable glucose monomers.

Question 66

Glycogen is a branched polymer of glucose. The core of a granule of glycogen is the protein glycogenin. Chains of glucose molecules connected by alpha-1,4 bonds extend off of this. Each glucose chain allows for much greater packing because the chains are branched through the use of alpha-1,6 bonds. At the ___________ of each branch of glycogen, hydrolysis of glycosidic bonds allows for the release of usable glucose monomers.

Answer 66

Glycogen is a branched polymer of glucose. The core of a granule of glycogen is the protein glycogenin. Chains of glucose molecules connected by alpha-1,4 bonds extend off of this. Each glucose chain allows for much greater packing because the chains are branched through the use of alpha-1,6 bonds. At the terminal end of each branch of glycogen, hydrolysis of glycosidic bonds allows for the release of usable glucose monomers.

Question 67

Glycogen and amylopectin differ structurally based on:

Answer 67

The frequency with which they contain alpha-1,6 bonds

Glycogen contains many more alpha-1,6 bonds than amylopectin. This gives glycogen a greater degree of branching than amylopectin, allowing it to better serve its purpose as a storage molecule.

Question 68

Is the carbohydrate polymer cellulose metabolically useful for the human body?

Answer 68

No, because the human body does not contain suitable enzymes for its digestion.

For a macromolecule to be metabolically useful to the body, it must be digestible and absorbable so that it can be brought from the digestive tract into the body. The human body does not have enzymes suited to digest cellulose’s beta-1,4 bonds. This does not mean cellulose is not useful outside of a metabolic context – for instance, cellulose (aka fiber) has useful functions in the digestive tract, such as helping us poop.

Question 69

True or false: Peptidoglycan is a unique carbohydrate polymer which composes the cell membrane of bacteria.

Answer 69

This statement is false.

Peptidoglycan composes the cell wall outside of the cell membrane of bacteria, not the cell membrane itself.

Question 70

True or false: the anomeric carbon of a monosaccharide is always located at the C-1 position.

Answer 70

This statement is false.

The anomeric carbon of a monosaccharide is the carbon at which a carbonyl functional group is located. On some sugars, such as aldoses like glucose, this is at the C-1 position. On other sugars, such as ketoses like fructose, this is at the C-2 position.

Question 71

In the structure of a monosaccharide depicted below, carbons 1 through 6 are labeled.

Which carbon will participate in both ring-structure formation AND glycosidic bond formation?

Answer 71

C1

In the molecule of D-glucose depicted, the anomeric carbon, carbon-1 (bearing an aldehyde group in its straight chain form), can be attacked by a hydroxyl group in order to form a ring structure. In the ring form of D-glucose, carbon-1 will participate in glycosidic linkages to other sugars.

Question 72

True or false: The structure shown below is D-galactose

Answer 72

This statement is true.

Galactose’s structure always features the C2/C5 carbons pointing in one direction, and the C3/C4 carbons pointing in the opposite direction. D-Galactose is the form of galactose where the hydroxyl group positioned on C5 (the bottommost chiral carbon) points to the right. On the other hand, in L-galactose, that hydroxyl group points left.

Question 73

The monosaccharide below is named:

Answer 73

The molecule pictured is L-glucose

Question 74

The ___ checkpoint takes place before cell division and checks for DNA damage after DNA replication, and if damage is detected, serves to “pause” cell division until the damage is repaired.

Answer 74

The G₂checkpoint takes place before cell division and checks for DNA damage after DNA replication, and if damage is detected, serves to “pause” cell division until the damage is repaired.

Question 75

The ____ checkpoint, also known as the restriction point, is when a cell commits to division. The presence of DNA damage or other external factors can cause a cell to fail this checkpoint and not divide.

Answer 75

The G₁/S checkpoint, also known as the restriction point, is when a cell commits to division. The presence of DNA damage or other external factors can cause a cell to fail this checkpoint and not divide.

Question 76

the ____ checkpoint is located after the S phase (during which DNA replication occurs) but before entry into mitosis. Specifically, this checkpoint ensures that DNA has been replicated properly before the cell begins to divide.

Answer 76

the G₂/M checkpoint is located after the S phase (during which DNA replication occurs) but before entry into mitosis. Specifically, this checkpoint ensures that DNA has been replicated properly before the cell begins to divide.

Question 77

The protein component of microtubules is ______. In contrast, the protein component of microfilaments is ______

Answer 77

The protein component of microtubules is tubulin. In contrast, the protein component of microfilaments is actin.

Question 78

A newly-developed VDA is found to target only capillaries, effectively destroying them entirely. Which of the following cell or tissue types will directly be affected by treatment with this drug?

A. Endothelial cells

B. Smooth muscle cells

C. Collagen-based connective tissue

D. Elastin-based connective tissue

Answer 78

A. Endothelial cells

The question stem states that this new VDA destroys capillaries (but no other vessels) entirely. Capillaries are the smallest blood vessels in the body, with walls that are composed only of a single layer of endothelial cells. Thus, the new VDA must affect endothelial cells, but we have no reason to believe that any of the other listed cell types will be impacted.

Question 79

Are collagen or elastin found in capillaries?

Answer 79

No, neither collagen nor elastin are present in capillaries

Collagen is found in the walls of larger vessels, especially arteries, but it is not present in capillaries.

Elastin is found in the walls of larger vessels, especially arteries, but it is not present in capillaries.

Question 80

Gluconeogenesis is (upregulated/downregulated) by glucagon and by the presence of surplus pyruvate/acetyl-CoA.

Answer 80

Gluconeogenesis is upregulated by glucagon and by the presence of surplus pyruvate/acetyl-CoA.

Question 81

Is acetyl-CoA an appropriate substrate for gluconeogenesis?

Answer 81

No

Acetyl-CoA cannot, when produced from the oxidation of even-chain fatty acids or from the oxidation of pyruvate to acetyl-CoA, serve as a substrate for gluconeogenesis. This is because the conversion of pyruvate to acetyl-CoA is irreversible

Question 82

In eukaryotic cells, where is the highest concentration of phosphorus likely to be found?

Answer 82

The nucleus

The nucleus contains most of the DNA of a eukaryotic cell (remember, some is also found in the mitochondria). The DNA backbone is made up of phosphate groups and deoxyribose sugars. Of the four major groups of macromolecules, nucleic acids contain the most phosphorus. Therefore, it is likely that the nucleus has the highest concentration of phosphorus in the cell.

Question 83

Why is there a higher concentration of phosphorous in the nucleus than in the cell membrane?

Answer 83

The cell membrane is primarily composed of phospholipids, which have one phosphate group per molecule.

In contrast, DNA has a phosphate group for every base.

Question 84

All of the following are functions of cholecystokinin EXCEPT:

A. suppressing hunger.

B. inhibiting somatostatin.

C. inhibiting gastric emptying.

D. stimulating pancreatic acinar cells.

Answer 84

B. inhibiting somatostatin.

Cholecystokinin (CCK) acts in the small intestine upon the entry of food into the duodenum from the stomach. This peptide hormone functions to aid a series of processes involved in digestion. Among these are stimulating pancreatic acinar cells to release digestive enzymes, stimulating feelings of satiety (fullness) to suppress hunger, inhibiting stomach emptying, and lowering gastric acid secretion.

Somatostatin, aka growth hormone inhibiting hormone, is the hormone that inhibits the release of CCK.

Question 85

A hiker out in the woods consumes a large amount of L. sativus seeds, which contain neurotoxic β-ODAP acid. 12 hours later, the hiker is having difficulty maintaining balance and opening her water bottle. What region of the brain is likely being attacked by β-ODAP?

Answer 85

The cerebellum receives information from sensory systems and then regulates fine motor movements. Thus, it coordinates posture, balance, fine motor coordination (like opening a bottle), and speech. It is located at the bottom back of the skull

Question 86

The pons is involved in ______ and _______.

Answer 86

The pons is involved in sleeping and respiration.

Question 87

What is a transversion?

Answer 87

A transversion is the result of a mutation that substitutes a purine for a pyrimidine or a pyrimidine for a purine.

Question 88

Calcitriol has a similar function to ____ in that it increases serum calcium levels, but it does so primarily through a different mechanism: it promotes the absorption of Ca²⁺ from the gastrointestinal tract.

Answer 88

Calcitriol has a similar function to PTH in that it increases serum calcium levels, but it does so primarily through a different mechanism: it promotes the absorption of Ca²⁺ from the gastrointestinal tract.

Question 89

What is the role of calcitriol?

Answer 89

calcitriol increases serum calcium levels by promoting the absorption of Ca²⁺ from the gastrointestinal tract.

Question 90

What are dendrocytes, or dendritic cells?

Answer 90

These bone marrow-derived cells are abundant in normal human and murine dermis and occupy the perivascular space, where they are closely associated with mast cells. They are a kind of dermal macrophage.

Question 91

At the beginning of an animal study, one of the female mice had ovaries that failed to form and had to be excluded. If the failure of the ovaries to form resulted from a defect early in the differentiation of one of the three germ layers, what other organs could the researchers evaluate for additional defects?

Answer 91

Kidneys, bone marrow, spleen

The ovaries arise from the mesoderm, and these 3 structures also arise from the mesoderm.

Question 92

As the zygote travels to the uterus, it undergoes a series of mitotic cell divisions known as cleavage. Once the zygote has cleaved into a mass of 16 cells by three to four days after fertilization, it is known as the ________.

Answer 92

As the zygote travels to the uterus, it undergoes a series of mitotic cell divisions known as cleavage. Once the zygote has cleaved into a mass of 16 cells by three to four days after fertilization, it is known as the morula.

Question 93

By three to five days after fertilization, the morula develops some degree of internal structure and becomes a blastocyst, with a fluid-filled cavity in the middle known as the blastocoel. The blastocyst implants in the uterine endometrium and further differentiates into the ________.

Answer 93

By three to five days after fertilization, the morula develops some degree of internal structure and becomes a blastocyst, with a fluid-filled cavity in the middle known as the blastocoel. The blastocyst implants in the uterine endometrium and further differentiates into the gastrula.

Question 94

The _______ has three layers: the ectoderm, the mesoderm, and the endoderm. These layers eventually go on to form specific organs and components in the body.

Answer 94

The gastrula has three layers: the ectoderm, the mesoderm, and the endoderm. These layers eventually go on to form specific organs and components in the body.

Question 95

What structures arise from the ectoderm?

Answer 95

The ectoderm primarily gives rise to the nervous system and epidermis (skin), as well as related structures like hair, nails, and sweat glands, and the linings of the mouth, anus, and nostrils. The process through which the nervous system is formed from the ectoderm is known as neurulation.

Question 96

What structures arise from the mesoderm?

Answer 96

The mesoderm generates many of the structures present within the body, including the musculature, connective tissue (including blood, bone, and cartilage), the gonads, the kidneys, and the adrenal cortex.

Question 97

True or false:

The mesoderm gives rise to the entire adrenal gland

Answer 97

False.

The mesoderm gives rise to the adrenal cortex.

The ectoderm gives rise to the adrenal medulla.

Question 98

What structures arise from the endoderm?

Answer 98

The endoderm is basically responsible for the interior linings of the body, including the linings of the gastrointestinal system, the pancreas and part of the liver, the urinary bladder and part of the urethra, and the lungs.

Question 99

Why would it be difficult for a prokaryote to express and entire genomic sequence?

Answer 99

Expression of an entire genome in prokaryotes is hard for a number of reasons, including the fact that there is no splicing mechanism in prokaryotes to remove the non-coding regions.

Question 100

What is cDNA?

Answer 100

cDNA is complementary DNA that is extracted from mRNA, therefore it only contains coding regions.

It does not contain and introns, promoters, or other noncoding regions.

Question 101

The ____________ shunts glucose 6-phosphate away from the glycolysis/gluconeogenesis pathway and uses it to do other structurally important things.

Answer 101

The pentose phosphate pathway shunts glucose 6-phosphate away from the glycolysis/gluconeogenesis pathway and uses it to do other structurally important things.

Question 102

What occurs in the oxidative phase of the pentose phosphate pathway?

Answer 102

During the oxidative phase, each mole of glucose 6-phosphate is converted to one mole of ribulose 5-phosphate, and one mole of carbon is lost in the form of CO₂.

Question 103

What are ribozymes?

Answer 103

Ribozymes are RNA molecules that can catalyze reactions, similarly to how enzymes are proteins that are biological catalysts. Ribozymes in the ribosomes are involved in various reactions that contribute to the translation process, and ribozymes can also catalyze reactions involved in various forms of RNA processing.

Question 104

True or False:

The sense strand is transcribed into RNA

Answer 104

FALSE

The sense strand is never transcribed into RNA; only the antisense strand is.

Question 105

What is the purpose of southern blotting?

Answer 105

The purpose of Southern blotting is to look for specific nucleotide sequences in the DNA being tested and/or to identify the size of the DNA fragments that contain the sequence of interest.

The Southern blot procedure breaks DNA into fragments using a restriction enzyme, runs the fragments on a gel, and then uses a specific labeled sequence as a probe. The DNA sequences are spaced out on the gel by length, where distance traveled is inversely proportional to fragment length.

Question 106

What is the direct use of RT-PCR techniques?

Answer 106

RT-PCR techniques are best used to detect RNA expression

Question 107

Where on a single codon would a point mutation most likely result in the same amino acid as the original codon?

Answer 107

A point mutation that happened in the third position of a codon.

for example, GAG → GAA

Question 108

The cytochrome oxidase complex, encoded in part by COI, requires the transfer of electrons from how many molecules of cytochrome C in order to catalyze the formation of two molecules of water from molecular oxygen?

Answer 108

4

The cytochrome c oxidase complex (complex IV) is the last enzyme of the electron transport chain. It receives one electron from each of four soluble cytochrome C molecules, transferring them to a single oxygen molecule and thus converting one O₂ molecule into two molecules of water.

Question 109

Patients entering the ICU with acute liver failure often also display dramatically lowered plasma glucose concentrations. Which of the conditions below could promote the same effect in individuals without liver damage?

A. Inhibition of the initial step of glycolysis

B. Understimulation of pancreatic alpha cells

C. Lowered sensitivity to insulin

D. Upregulation of gluconeogenesis

Answer 109

Understimulation of pancreatic alpha cells

Pancreatic alpha cells release glucagon, a peptide hormone that increases blood glucose levels. If the activity of these cells is lowered, less glucagon will be released, and plasma glucose levels will remain lower than is typical. This matches the condition described in the question stem.

Question 110

Which phases of the pentose phosphate pathway should be activated to produce the greatest amount of nucleotides and NADPH, respectively?

A. Non-oxidative phase, non-oxidative phase

B. Non-oxidative phase, oxidative phase

C. Oxidative phase, non-oxidative phase

D. Oxidative phase, oxidative phase

Answer 110

B. Non-oxidative phase, oxidative phase

NADPH is produced only in the oxidative phase of the pentose phosphate pathway. Therefore, the answer to the second part of this question must be the oxidative phase. Remember that the oxidative phase is irreversible, while the non-oxidative phase is reversible. Ribulose 5-phosphate, a product of both the oxidative and non-oxidative pathways, can be converted to ribose 5-phosphate, which in turn can be used to make nucleotides. Therefore, both the oxidative and non-oxidative phases could be utilized for nucleotide production. In the oxidative phase, a carbon dioxide molecule is lost when converting the six-carbon glucose 6-phosphate to the five-carbon ribulose 5-phosphate. However, no carbon dioxide is lost in the non-oxidative phase. Thus, the non-oxidative phase yields more moles of ribulose 5-phosphate per mole of glucose 6-phosphate than the oxidative phase.

Question 111

Which cytoskeletal component forms the majority of the outer layer of skin?

A. Microfilaments

B. Microtubules

C. Intermediate filaments

D. Elastins

Answer 111

C. Intermediate filaments

The outer layer of skin is made up of keratin accumulated in dead cells. Keratin is an intermediate filament that has great strength