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2. Trauma MCQs

Practice Set 331 of 640

This practice set contains high-yield board review questions covering key concepts in 2. Trauma. Each clinical scenario is designed to test your diagnostic and management skills relevant to this subspecialty.

Question 6601

Topic: 2. Trauma

Unlike conventional compression plates, a locked plating construct provides fracture stabilization primarily through which biomechanical principle?

. Friction between the undersurface of the plate and the cortical bone
. Rigid fixed-angle screw-to-plate interfaces acting as a single construct
. Dynamic compression across the fracture gap via eccentric drilling
. The lag screw principle utilizing the plate as a washer
. Load-sharing interface matching the modulus of the intact bone

Correct Answer & Explanation

. Rigid fixed-angle screw-to-plate interfaces acting as a single construct

Explanation

Locked plates function as an 'internal external fixator' relying on the threaded fixed-angle interaction between the screw head and the plate hole. They do not require friction between the plate and bone, thus preserving periosteal blood supply.

Question 6602

Topic: 2. Trauma

A surgeon is performing bridge plating for a comminuted femoral shaft fracture. By intentionally leaving empty screw holes over the fracture zone to increase the 'working length' of the plate, the surgeon biomechanically achieves:

. Increased rigidity of the construct
. Decreased construct strain at the fracture site
. Increased bending stiffness
. Elimination of axial micromotion
. Promotion of primary osteonal bone healing

Correct Answer & Explanation

. Decreased construct strain at the fracture site

Explanation

Increasing the working length (distance between the innermost screws) decreases the construct's stiffness, distributing deformation over a longer segment. This decreases the overall strain percentage at the fracture gap, facilitating secondary bone healing.

Question 6603

Topic: 2. Trauma

Achieving absolute stability through anatomic reduction and interfragmentary compression with a lag screw and neutralization plate specifically bypasses callus formation to promote which biological process?

. Endochondral ossification
. Chondroid intramembranous healing
. Primary (osteonal) bone healing
. Secondary bone healing
. Woven bone hypertrophic nonunion

Correct Answer & Explanation

. Primary (osteonal) bone healing

Explanation

Absolute stability reduces interfragmentary strain to less than 2%, preventing the formation of soft callus. This mechanical environment promotes primary (osteonal) bone healing, characterized by osteoclastic cutting cones crossing the fracture directly.

Question 6604

Topic: 2. Trauma

An intramedullary nail is being selected for the fixation of a diaphyseal femur fracture. Biomechanically, the torsional stiffness of a solid cylindrical intramedullary nail is directly proportional to the radius raised to what power?

. First power
. Second power
. Third power
. Fourth power
. Fifth power

Correct Answer & Explanation

. Fourth power

Explanation

The torsional stiffness of a solid cylinder is determined by its polar moment of inertia, which is proportional to the radius to the fourth power (r^4). Therefore, a small increase in the nail's radius drastically increases its resistance to torsion.

Question 6605

Topic: 2. Trauma

When inserting a cortical screw for fracture fixation, the ultimate pull-out strength of the screw is directly proportional to which of the following design parameters?

. Inner (core) diameter of the screw
. Outer (thread) diameter of the screw
. Pitch of the screw
. Length of the unthreaded shaft
. Insertion torque applied

Correct Answer & Explanation

. Outer (thread) diameter of the screw

Explanation

The pull-out strength of a screw is directly proportional to its outer (thread) diameter and the total length of thread engagement in the bone. It is inversely proportional to the pitch (the distance between adjacent threads).

Question 6606

Topic: 2. Trauma

An orthopaedic surgeon decides to use a thicker plate to stabilize a comminuted fracture. According to the biomechanical principles of rectangular plates, if the thickness of the plate is doubled, how does its bending stiffness change?

. It increases by a factor of 2
. It increases by a factor of 4
. It increases by a factor of 8
. It increases by a factor of 16
. It increases by a factor of 32

Correct Answer & Explanation

. It increases by a factor of 8

Explanation

The bending stiffness (area moment of inertia) of a rectangular plate is proportional to its width and to the cube of its thickness (h^3). Therefore, doubling the thickness increases the bending stiffness by a factor of 8 (2^3).

Question 6607

Topic: 2. Trauma

A 35-year-old man undergoes closed intramedullary nailing for a transverse midshaft femur fracture. Callus formation is noted at 6 weeks. Which of the following mechanical environments optimally promotes this type of secondary bone healing?

. Absolute stability with no interfragmentary strain (< 2%)
. Moderate interfragmentary strain (2% to 10%)
. High interfragmentary strain (11% to 30%)
. Extremely high interfragmentary strain (> 30%)
. Rigid plate fixation with anatomic reduction

Correct Answer & Explanation

. Moderate interfragmentary strain (2% to 10%)

Explanation

Secondary bone healing (callus formation) is stimulated by moderate interfragmentary strain (2% to 10%). Strain below 2% leads to primary bone healing, while strain above 10% promotes fibrous tissue or nonunion.

Question 6608

Topic: 2. Trauma

A dynamic hip screw used to treat a basicervical femoral neck fracture fails 6 months post-operatively. The fracture is a nonunion, and the lag screw has broken despite the patient being non-weight bearing. Which of the following describes failure of an implant due to repetitive sub-maximal cyclic loading?

. Yield failure
. Ultimate tensile failure
. Fatigue failure
. Creep
. Stress shielding

Correct Answer & Explanation

. Fatigue failure

Explanation

Fatigue failure occurs when an implant is subjected to repetitive cyclic loads below its ultimate tensile strength, eventually leading to the propagation of microcracks and sudden fracture.

Question 6609

Topic: Lower Extremity Trauma

The torsional rigidity of a solid cylindrical intramedullary nail is proportional to its radius raised to which of the following powers?

. First power
. Second power
. Third power
. Fourth power
. Fifth power

Correct Answer & Explanation

. Fourth power

Explanation

The torsional rigidity of a solid cylinder is proportional to its polar area moment of inertia, which scales with the radius to the fourth power (r^4). Therefore, small increases in nail radius dramatically increase torsional stiffness.

Question 6610

Topic: Lower Extremity Trauma

In the design of a solid intramedullary nail, doubling the radius of the nail will increase its bending stiffness by what factor?

. 2
. 4
. 8
. 16
. 32

Correct Answer & Explanation

. 16

Explanation

The bending stiffness of a solid cylinder is proportional to its area moment of inertia, which is calculated as (pi * r^4) / 4. Therefore, doubling the radius increases the bending stiffness by a factor of 16 (2^4 = 16).

Question 6611

Topic: 2. Trauma

When using a locked plating construct to stabilize a comminuted diaphyseal fracture, which of the following biomechanical principles makes it superior to a conventional non-locking plate in osteoporotic bone?

. It relies on friction between the plate and the bone surface
. It acts as a single-beam construct with fixed-angle stability
. It applies dynamic compression across the fracture site
. It depends entirely on the pullout strength of individual screws
. It undergoes minimal elastic deformation during loading

Correct Answer & Explanation

. It acts as a single-beam construct with fixed-angle stability

Explanation

Locked plates act as fixed-angle constructs, meaning stability is dependent on the screw-plate interface rather than plate-to-bone friction. This greatly improves pullout resistance and stability in poor-quality, osteoporotic bone.

Question 6612

Topic: 2. Trauma

A surgeon is planning plate fixation for a comminuted diaphyseal femur fracture using a bridge plating technique. To decrease the stiffness of the construct and promote secondary bone healing via callus formation, which modification should the surgeon make?

. Decrease the distance between the innermost screws
. Increase the distance between the innermost screws
. Use a thicker, stronger plate
. Fill every screw hole in the plate
. Use a longer plate but place all screws close to the fracture site

Correct Answer & Explanation

. Increase the distance between the innermost screws

Explanation

The working length of a plate is the distance between the two closest screws on either side of the fracture. Increasing the working length decreases the stiffness of the construct, allowing for more micromotion to promote secondary bone healing.

Question 6613

Topic: 2. Trauma

When using a bridge plating technique for a comminuted diaphyseal fracture, increasing the "working length" of the plate has which of the following biomechanical effects?

. Increases the torsional stiffness
. Decreases the construct flexibility
. Decreases the interfragmentary strain at the fracture site
. Increases the axial stiffness
. Decreases the pullout strength of the individual screws

Correct Answer & Explanation

. Decreases the interfragmentary strain at the fracture site

Explanation

Increasing the working length (the distance between the innermost screws on either side of the fracture) decreases construct stiffness. This distributes the deformation over a longer segment, lowering the interfragmentary strain to promote secondary bone healing.

Question 6614

Topic: 2. Trauma

In the biomechanics of bridge plating for a comminuted diaphyseal fracture, the "working length" of the construct is defined as the distance between:

. The first and last screws in the entire plate
. The two innermost screws adjacent to the fracture gap
. The proximal end of the plate and the fracture site
. The fracture gap and the farthest distal screw
. The bone surface and the plate undersurface

Correct Answer & Explanation

. The two innermost screws adjacent to the fracture gap

Explanation

The working length of a plate is the distance between the two innermost screws on opposite sides of the fracture. Increasing the working length makes the construct more flexible, which promotes secondary bone healing through callus formation.

Question 6615

Topic: Lower Extremity Trauma

Regarding the gross anatomy and biomechanics of the knee menisci, the medial meniscus differs from the lateral meniscus in that the medial meniscus is:

. More mobile during terminal knee flexion
. C-shaped and covers less of the tibial plateau articular surface
. O-shaped and covers more of the tibial plateau articular surface
. Devoid of any direct attachment to the joint capsule
. Primarily composed of Type II collagen rather than Type I

Correct Answer & Explanation

. O-shaped and covers more of the tibial plateau articular surface

Explanation

The medial meniscus is C-shaped, larger in AP dimension, and firmly attached to the deep medial collateral ligament, making it less mobile than the O-shaped lateral meniscus. Both menisci are primarily composed of Type I collagen.

Question 6616

Topic: 2. Trauma
An orthopedic surgeon is selecting a solid intramedullary titanium nail for a diaphyseal femur fracture. If the radius of the solid nail is increased by a factor of 2, how does this affect its bending stiffness?
. Increases by a factor of 2
. Increases by a factor of 4
. Increases by a factor of 8
. Increases by a factor of 16
. Increases by a factor of 32

Correct Answer & Explanation

. Increases by a factor of 16

Explanation

The bending stiffness of a solid cylinder is proportional to its area moment of inertia, which scales with the radius to the fourth power (r^4). Therefore, doubling the radius of a solid nail increases its bending stiffness by 2^4, or a factor of 16.

Question 6617

Topic: 2. Trauma

When applying a tension band construct for a transverse olecranon fracture, optimal biomechanical stability is achieved by placing the implant on the:

. Compression side of the bone to convert compressive forces into tension
. Tension side of the bone to convert tensile forces into compression at the opposite cortex
. Neutral axis to minimize bending moments
. Cortical surface with the thickest periosteum to enhance biology
. Medullary canal to maximize the area moment of inertia

Correct Answer & Explanation

. Tension side of the bone to convert tensile forces into compression at the opposite cortex

Explanation

The tension band principle involves applying a fixation device eccentrically on the tension (convex) side of a loaded bone. This converts physiological tensile forces into compressive forces at the opposite (concave) cortex, dynamically compressing the fracture.

Question 6618

Topic: 2. Trauma

In a rectangular fracture plate, bending stiffness is dictated by the area moment of inertia (I = bh^3 / 12). Increasing the thickness (h) of the plate by a factor of 2 increases its bending stiffness by a factor of:

. 2
. 4
. 6
. 8
. 16

Correct Answer & Explanation

. 8

Explanation

The area moment of inertia for a rectangular structure subjected to bending is proportional to the base multiplied by the height (thickness) cubed. Doubling the thickness increases the bending stiffness by a factor of 8 (2^3).

Question 6619

Topic: 2. Trauma

According to Perren's strain theory, what level of interfragmentary strain is ideal for promoting secondary bone healing through fracture callus formation?

. Less than 2%
. 2% to 10%
. 10% to 30%
. 30% to 50%
. Greater than 50%

Correct Answer & Explanation

. 2% to 10%

Explanation

Secondary bone healing (endochondral ossification) with callus formation occurs when interfragmentary strain is maintained between 2% and 10%. Strain less than 2% promotes primary (direct) bone healing, while strain exceeding 10% typically results in a fibrous nonunion.

Question 6620

Topic: Lower Extremity Trauma

If the radius of a solid cylindrical intramedullary nail is doubled, by what factor does its theoretical bending stiffness increase?

. 2
. 4
. 8
. 16
. 32

Correct Answer & Explanation

. 16

Explanation

The area moment of inertia for a solid cylinder, which dictates its bending stiffness, is proportional to the radius raised to the fourth power (r^4). Therefore, doubling the radius of a solid intramedullary nail increases its bending stiffness by a factor of 16 (2^4).

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