Menu

2. Trauma MCQs

Practice Set 64 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 1261

Topic: 2. Trauma

During internal fixation of an osteoporotic ankle fracture, screw pull-out is a significant concern. Based on screw mechanics, which of the following design alterations most significantly increases a screw's pull-out strength?

. Increasing the core diameter while maintaining the thread diameter
. Increasing the outer (thread) diameter while maintaining the core diameter
. Decreasing the thread pitch (fewer threads per inch)
. Decreasing the length of engagement in the bone
. Increasing the size of the screw head

Correct Answer & Explanation

. Increasing the outer (thread) diameter while maintaining the core diameter

Explanation

Pull-out strength is directly proportional to the outer diameter of the thread, the length of engagement, and the shear strength of the bone. Increasing the outer diameter provides a larger volume of bone between the threads, maximizing pull-out resistance.

Question 1262

Topic: Lower Extremity Trauma

A 65-year-old female undergoes antegrade intramedullary nailing for a distal-third femoral shaft fracture. Intraoperatively, the nail tip abuts the anterior cortex of the distal femur. What inherent geometric mismatch between the intramedullary nail and the native femur causes this complication?

. The radius of curvature of the nail is smaller than that of the femur
. The radius of curvature of the nail is larger than that of the femur
. The nail diameter is disproportionately large
. The proximal anteversion angle of the nail is incorrect
. The working length of the nail is excessively short

Correct Answer & Explanation

. The radius of curvature of the nail is larger than that of the femur

Explanation

A larger radius of curvature denotes a straighter object. Many modern IM nails have a larger radius of curvature (e.g., 1.5m to 2.0m) compared to the native femur (~1.2m), making the nail too straight and leading to anterior cortical impingement distally.

Question 1263

Topic: 2. Trauma

A patient presents with a delayed union 4 months after statically locked intramedullary nailing of an isthmic, transverse femoral shaft fracture. The surgeon performs dynamization by removing the static interlocking screws furthest from the fracture. Biomechanically, how does dynamization promote union in this setting?

. By increasing the torsional rigidity of the construct
. By allowing controlled axial compression at the fracture site
. By increasing the bending stiffness of the nail
. By converting the construct from relative to absolute stability
. By decreasing the overall working length of the nail

Correct Answer & Explanation

. By allowing controlled axial compression at the fracture site

Explanation

Dynamization of an intramedullary nail involves removing static locking screws to permit the bone fragments to compress axially during weight-bearing. This increased axial micro-motion stimulates robust callus formation in axially stable fracture patterns.

Question 1264

Topic: 2. Trauma

When placing interlocking screws for an intramedullary nail in an unstable, comminuted midshaft tibia fracture, a surgeon wishes to maximize the torsional stiffness of the nail-bone construct. How should the locking screws be positioned to achieve this goal?

. As close to the fracture site as possible, minimizing working length
. As far from the fracture site as possible, maximizing working length
. Exclusively in the metaphyseal bone of the proximal segment
. With a maximum stand-off distance from the endosteum
. Screw position relative to the fracture has no effect on torsional stiffness

Correct Answer & Explanation

. As close to the fracture site as possible, minimizing working length

Explanation

The torsional stiffness of an intramedullary nail construct is inversely proportional to its working length. Therefore, placing locking screws as close to the fracture site as possible minimizes the working length and maximizes torsional rigidity.

Question 1265

Topic: 2. Trauma

A 45-year-old male sustains a simple transverse olecranon fracture. To effectively counteract the strong tensile forces of the triceps mechanism and convert them into dynamic articular compression, a contoured plate should ideally be applied to which surface of the proximal ulna?

. Anterior
. Posterior
. Medial
. Lateral
. Intramedullary

Correct Answer & Explanation

. Posterior

Explanation

The olecranon experiences significant tensile forces on its posterior surface due to the pull of the triceps. Applying a plate to the posterior (tension) surface acts as a tension band, converting these distraction forces into beneficial compression across the articular surface.

Question 1266

Topic: 2. Trauma

A 32-year-old male sustains a highly comminuted midshaft femoral fracture. The surgeon elects to use a bridge plating technique rather than attempting anatomic reduction with absolute stability. According to Perren's strain theory, what is the primary biomechanical advantage of this technique for this fracture pattern?

. It maximizes load-sharing between the plate and the comminuted bone fragments.
. It distributes motion across multiple fracture gaps, keeping interfragmentary strain low enough to allow secondary bone healing.
. It decreases the overall working length of the construct, thereby increasing absolute stability.
. It increases the pullout strength of the locking screws by engaging both cortices.
. It creates a high-strain environment that directly induces primary intramembranous bone formation.

Correct Answer & Explanation

. It distributes motion across multiple fracture gaps, keeping interfragmentary strain low enough to allow secondary bone healing.

Explanation

Bridge plating provides relative stability, leaving a highly comminuted fracture zone undisturbed. According to Perren's theory, distributing the overall motion across multiple gaps ensures the strain at any single gap remains below the 2-10% threshold required for callus formation.

Question 1267

Topic: 2. Trauma

A 65-year-old female of short stature undergoes intramedullary nailing for a transverse midshaft femur fracture. Intraoperatively, the tip of the standard nail impinges on and nearly perforates the anterior cortex of the distal femur. What is the primary biomechanical cause of this complication?

. The nail's outer diameter is too large for the medullary canal.
. The working length of the nail is too short for the fracture pattern.
. The radius of curvature of the intramedullary nail is greater than the radius of curvature of the patient's femur.
. Over-reaming of the distal medullary canal displaced the trajectory.
. The insertion starting point in the piriformis fossa was too posterior.

Correct Answer & Explanation

. The radius of curvature of the intramedullary nail is greater than the radius of curvature of the patient's femur.

Explanation

A higher radius of curvature means the nail is straighter. Native femurs, especially in shorter individuals, have a smaller radius of curvature (more pronounced anterior bow), causing a straighter nail to impact the anterior cortex distally.

Question 1268

Topic: 2. Trauma

An orthopedic surgeon is planning the fixation of a nonunion and decides to substitute a 2 mm thick plate with a 4 mm thick plate of the same material and width. By what factor does the bending stiffness of the construct increase?

. 2
. 4
. 8
. 16
. 32

Correct Answer & Explanation

. 8

Explanation

The bending stiffness of a plate is proportional to its area moment of inertia, calculated as (base x height^3) / 12. Doubling the thickness (height) increases the bending stiffness by a factor of 8 (2 cubed).

Question 1269

Topic: Lower Extremity Trauma

When comparing a solid intramedullary nail to a slotted (open-section) intramedullary nail of the same outer diameter and material, how do their torsional rigidities differ biomechanically?

. Both have torsional rigidities proportional to the radius to the fourth power.
. The solid nail's rigidity is proportional to the radius to the fourth power, while the slotted nail's is proportional to the radius cubed.
. The solid nail's rigidity is proportional to the radius cubed, while the slotted nail's is proportional to the radius squared.
. The slotted nail has higher torsional rigidity due to its ability to deform elastically.
. Both have torsional rigidities proportional to the radius cubed, but the solid nail has a higher coefficient.

Correct Answer & Explanation

. The solid nail's rigidity is proportional to the radius to the fourth power, while the slotted nail's is proportional to the radius cubed.

Explanation

A closed-section (solid) nail's torsional rigidity is proportional to the radius to the fourth power (r^4). An open-section (slotted) nail's torsional rigidity drops significantly, becoming proportional to the radius cubed (r^3).

Question 1270

Topic: 2. Trauma

A surgeon is inserting cortical screws to secure a neutralization plate. Which of the following alterations to the screw geometry will most significantly increase its pullout strength?

. Decreasing the outer (major) diameter of the screw.
. Increasing the inner (minor) diameter of the screw.
. Increasing the outer (major) diameter of the screw.
. Increasing the thread pitch (fewer threads per inch).
. Decreasing the length of thread engagement.

Correct Answer & Explanation

. Increasing the outer (major) diameter of the screw.

Explanation

Screw pullout strength is directly proportional to the major (outer) diameter, the length of engagement, and the shear strength of the bone. Increasing the inner diameter primarily increases the bending strength of the screw, not the pullout strength.

Question 1271

Topic: 2. Trauma

During bridge plating of a transverse femoral diaphysis fracture, the surgeon fills every screw hole adjacent to the fracture site on both sides. What is the most likely biomechanical consequence of this screw configuration?

. Decreased construct stiffness and decreased interfragmentary strain.
. Increased construct stiffness and increased interfragmentary strain.
. Decreased construct stiffness and increased interfragmentary strain.
. Increased construct stiffness and decreased interfragmentary strain.
. No effect on construct stiffness or strain.

Correct Answer & Explanation

. Decreased construct stiffness and increased interfragmentary strain.

Explanation

Placing screws immediately adjacent to the fracture drastically decreases the plate's working length, resulting in a highly stiff construct. In a narrow gap (transverse fracture), this forces all motion into a small area, yielding critically high interfragmentary strain that can cause nonunion or hardware failure.

Question 1272

Topic: 2. Trauma

In the application of an intramedullary nail, what is the biomechanical effect of utilizing distal interlocking screws that are positioned closer to the fracture site?

. It eliminates the need for reaming.
. It increases the working length and decreases construct stiffness.
. It decreases the working length and increases overall construct stiffness.
. It increases the bending moment on the proximal interlocking screws.
. It decreases the torsional rigidity of the nail itself.

Correct Answer & Explanation

. It decreases the working length and increases overall construct stiffness.

Explanation

The working length of an intramedullary nail is the unsupported span between the closest proximal and distal fixation points. Moving the interlocking screws closer to the fracture decreases this working length, thereby increasing the overall stiffness of the construct.

Question 1273

Topic: 2. Trauma

A patient with a midshaft femur fracture is treated with a plate applied to the lateral cortex. What biomechanical principle is primarily utilized by placing the plate on this specific side of the bone?

. Neutralization principle
. Buttress principle
. Tension band principle
. Bridge plating principle
. Dynamic compression principle

Correct Answer & Explanation

. Tension band principle

Explanation

The anatomical bow of the femur places the lateral cortex under tension and the medial cortex under compression during weight-bearing. Applying a plate to the convex (tension) side acts as a tension band, converting tensile forces into compressive forces at the fracture.

Question 1274

Topic: 2. Trauma

Far cortical locking (FCL) screws are occasionally utilized in locking plate constructs for distal femur fractures. What is the primary biomechanical rationale for their use?

. To maximize absolute stability and eliminate all micromotion.
. To increase the axial pullout strength in severely osteoporotic bone.
. To decrease the effective working length of the plate.
. To provide symmetric biphasic flexibility, thereby promoting callus formation.
. To securely compress the plate flush against the near cortex.

Correct Answer & Explanation

. To provide symmetric biphasic flexibility, thereby promoting callus formation.

Explanation

FCL screws bypass the near cortex and lock only into the far cortex and the plate. This reduces the inherently high stiffness of standard locked plating, allowing symmetric parallel micromotion that promotes secondary bone healing.

Question 1275

Topic: 2. Trauma

A surgeon is treating a proximal third tibia fracture with an intramedullary nail. To prevent a procurvatum (apex anterior) deformity, a Poller (blocking) screw is placed. Where is the optimal placement of this screw relative to the nail?

. Posterior to the nail in the distal fragment.
. Anterior to the nail in the proximal fragment.
. Posterior to the nail in the proximal fragment.
. Medial to the nail in the proximal fragment.
. Lateral to the nail in the distal fragment.

Correct Answer & Explanation

. Posterior to the nail in the proximal fragment.

Explanation

Proximal tibia fractures frequently deform into procurvatum (apex anterior). Placing a Poller screw posterior to the nail in the proximal fragment narrows the metaphyseal canal, blocking the nail from displacing posteriorly and thereby correcting the deformity.

Question 1276

Topic: 2. Trauma

A patient is 6 months post-operative from a statically locked intramedullary nail for a femoral shaft fracture, showing signs of delayed union. The surgeon decides to dynamize the nail by removing the distal interlocking screws. What is the primary biomechanical effect of this procedure?

. It converts the fixation from relative to absolute stability.
. It completely eliminates the torsional stability of the construct.
. It allows axial translation and compression at the fracture site while maintaining bending alignment.
. It decreases the bending stiffness of the nail by exactly 50%.
. It shortens the working length of the remaining construct.

Correct Answer & Explanation

. It allows axial translation and compression at the fracture site while maintaining bending alignment.

Explanation

Dynamization removes static locking from one side, allowing the bone fragments to slide axially along the nail and compress under physiologic load. The nail continues to act as a load-sharing device that maintains bending and, depending on cortical interdigitation, rotational alignment.

Question 1277

Topic: Lower Extremity Trauma

An engineer evaluates a cannulated intramedullary nail and a solid intramedullary nail of the exact same outer diameter. What is the effect of the inner cannulation on the nail's bending stiffness?

. It reduces bending stiffness proportionally to the inner radius squared.
. It reduces bending stiffness proportionally to the inner radius to the fourth power.
. It increases bending stiffness by allowing internal elastic deformation.
. It has zero effect on the area moment of inertia.
. It reduces bending stiffness linearly with the volume of removed material.

Correct Answer & Explanation

. It reduces bending stiffness proportionally to the inner radius to the fourth power.

Explanation

The bending stiffness of a cylinder is proportional to its area moment of inertia. For a hollow cylinder (cannulated nail), this is proportional to (outer radius^4 - inner radius^4). Therefore, the loss in stiffness corresponds to the inner radius to the fourth power.

Question 1278

Topic: 2. Trauma

A 78-year-old female with severe osteoporosis sustains an extra-articular distal femur fracture. The surgeon elects to use a locking compression plate (LCP). Biomechanically, what is the primary advantage of a locked construct over a conventional non-locked plate in this osteoporotic patient?

. It relies heavily on friction between the plate and bone for stability.
. It increases the dynamic compressive forces directly across the fracture site.
. It resists pull-out via fixed-angle screw-to-plate interfaces rather than friction.
. It decreases the working length of the construct regardless of screw placement.
. It mandates anatomical reduction and absolute stability prior to plate application.

Correct Answer & Explanation

. It resists pull-out via fixed-angle screw-to-plate interfaces rather than friction.

Explanation

Locked plates do not rely on friction between the plate and the underlying bone. Instead, the screw heads thread into the plate to create a fixed-angle construct, which significantly improves pull-out strength and stability in poor-quality, osteoporotic bone.

Question 1279

Topic: Lower Extremity Trauma

A 32-year-old male undergoes antegrade intramedullary nailing for a distal third femoral shaft fracture. Intraoperatively, the distal tip of the nail impinges on and perforates the anterior cortex of the distal femur. Which of the following biomechanical mismatches is the most likely cause of this complication?

. The nail's radius of curvature is smaller than the native femur's.
. The nail's radius of curvature is larger than the native femur's.
. The nail possesses a lower bending stiffness than the cortical bone of the femur.
. The entry point was placed excessively posterior in the piriformis fossa.
. The medullary canal was over-reamed by more than 2 millimeters.

Correct Answer & Explanation

. The nail's radius of curvature is larger than the native femur's.

Explanation

Anterior cortical perforation in the distal femur is typically caused by using a nail with a larger radius of curvature (meaning it is straighter) than the native femur. As the straight nail advances, its distal tip is driven anteriorly into the cortex.

Question 1280

Topic: 2. Trauma

A 45-year-old patient requires open reduction and internal fixation for a transverse, non-comminuted fracture of the proximal olecranon. The surgeon plans to apply a plate utilizing the tension band principle. To function optimally, where must the plate be applied and what must the opposite cortex do?

. Volar surface; opposite cortex must distract under functional load.
. Dorsal surface; opposite cortex must be able to resist compressive forces.
. Medial surface; opposite cortex must shear under dynamic motion.
. Lateral surface; opposite cortex must remain completely unloaded.
. Dorsal surface; opposite cortex must be highly comminuted to allow settling.

Correct Answer & Explanation

. Dorsal surface; opposite cortex must be able to resist compressive forces.

Explanation

The tension band principle requires a plate or wire to be applied to the tension side of the bone (the dorsal surface of the ulna). For this to convert tensile forces into compressive forces at the fracture, the opposite (volar) cortex must be intact to resist compression.

Test Yourself

Switch to an interactive, timed exam simulation to truly master this topic.

On this Page