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2. Trauma MCQs

Practice Set 92 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 1821

Topic: 2. Trauma

A comminuted midshaft femur fracture is treated with a bridge plate. To maximize the fatigue life of the plate and prevent early breakage, which surgical strategy regarding screw placement is most effective?

. Decreasing the working length of the plate
. Leaving multiple screw holes empty directly over the comminuted fracture zone
. Placing screws as close to the fracture gap as possible on both sides
. Using a shorter overall plate length
. Using only conventional cortical screws instead of locking screws

Correct Answer & Explanation

. Leaving multiple screw holes empty directly over the comminuted fracture zone

Explanation

Leaving screw holes empty over the fracture zone increases the working length of the plate. A longer working length distributes bending stresses over a larger segment of the plate, significantly reducing the strain per unit area and thereby increasing the implant's fatigue life.

Question 1822

Topic: 2. Trauma

A surgeon uses a single lag screw to stabilize an oblique tibial shaft fracture. To achieve the maximal amount of interfragmentary compression, what must be the orientation of the lag screw relative to the fracture anatomy?

. Parallel to the long axis of the bone
. Perpendicular to the long axis of the bone
. Perpendicular to the fracture plane
. Parallel to the fracture plane
. Bisecting the angle between the long axis and the fracture plane

Correct Answer & Explanation

. Perpendicular to the fracture plane

Explanation

To maximize interfragmentary compression and prevent shear forces that could cause fracture displacement (gliding), a lag screw must be placed perfectly perpendicular to the fracture plane.

Question 1823

Topic: Pelvic & Acetabular Trauma

A supra-acetabular external fixator is placed for a pelvic ring injury. To optimally decrease the stresses at the pin-bone interface and minimize the risk of pin loosening, which biomechanical adjustment is best?

. Decreasing the pin diameter
. Increasing the distance between the connecting rod and the skin
. Using a single connecting rod instead of two stacked rods
. Increasing the diameter of the Schanz pins
. Placing the pins perfectly parallel to the joint surface

Correct Answer & Explanation

. Increasing the diameter of the Schanz pins

Explanation

Pin loosening is primarily caused by excessive stress at the pin-bone interface. Increasing the diameter of the Schanz pins exponentially increases their bending stiffness (proportional to r^4), drastically reducing micromotion and stress at the bone interface.

Question 1824

Topic: 2. Trauma

Increasing the thickness of a fracture plate has what mathematical effect on its bending stiffness?

. Directly proportional to the thickness
. Proportional to the square of the thickness
. Proportional to the cube of the thickness
. Proportional to the fourth power of the thickness
. Inversely proportional to the thickness

Correct Answer & Explanation

. Proportional to the cube of the thickness

Explanation

The bending stiffness of a rectangular plate is proportional to its area moment of inertia, calculated as (base x height^3) / 12. Therefore, increasing the thickness (height) increases the stiffness by the cube of that change.

Question 1825

Topic: 2. Trauma

A surgeon is applying a unilateral external fixator to a tibial shaft fracture. Which of the following modifications will most effectively increase the bending stiffness of the construct?

. Decreasing the pin diameter
. Increasing the distance between the bone and the connecting bar
. Decreasing the distance between the near and far pins in each fragment
. Increasing the diameter of the half-pins
. Decreasing the number of half-pins per segment

Correct Answer & Explanation

. Increasing the diameter of the half-pins

Explanation

External fixator stiffness is highly dependent on pin diameter, as stiffness is proportional to the radius to the fourth power (r^4). Increasing the pin diameter is the single most effective way to increase the overall stiffness of the construct.

Question 1826

Topic: 2. Trauma

In a locked plating construct, how is the stability of the fracture primarily achieved compared to conventional plating?

. Friction between the plate and the bone surface
. Compression of the plate to the bone
. Fixed-angle construct acting as a single load-bearing beam
. Dynamic compression of the fracture gap
. Primary bone healing through high gap strain

Correct Answer & Explanation

. Fixed-angle construct acting as a single load-bearing beam

Explanation

Locked plates act as fixed-angle constructs where the screws thread directly into the plate, functioning as a single load-bearing beam. Unlike conventional plates, they do not rely on friction between the plate and bone for stability.

Question 1827

Topic: 2. Trauma

For a tension band construct of the patella to function properly, where must the wire be placed relative to the joint line and the bending axis?

. On the articular surface side to compress the joint line
. Exactly on the neutral axis to prevent bending
. On the tension surface (anteriorly) to convert tensile forces into compressive forces at the articular surface
. On the compression surface to augment native compressive forces
. Equidistant from the anterior and posterior cortices

Correct Answer & Explanation

. On the tension surface (anteriorly) to convert tensile forces into compressive forces at the articular surface

Explanation

The tension band principle relies on placing the implant on the tension side of a bone (anterior patella). This converts the tensile forces generated during knee flexion into dynamic compressive forces at the articular surface.

Question 1828

Topic: 2. Trauma

When placing a cortical lag screw across an oblique fracture, the gliding hole in the near cortex must be overdrilled. What biomechanical consequence occurs if the near cortex is not overdrilled?

. The screw will strip the far cortex prematurely
. The fracture fragments will undergo massive dynamic compression
. The threads will engage both cortices, locking the fracture gap and preventing compression
. The screw will have exponentially increased pullout strength
. The bending stiffness of the screw will increase

Correct Answer & Explanation

. The threads will engage both cortices, locking the fracture gap and preventing compression

Explanation

If the near cortex is not overdrilled to create a gliding hole, the screw threads engage both cortices simultaneously. This locks the fragments at their current distance and prevents interfragmentary compression.

Question 1829

Topic: 2. Trauma

A manufacturer redesigns an intramedullary nail by replacing its open, slotted cross-section with a closed, solid tubular cross-section of the same outer diameter. What is the most significant biomechanical change?

. Decreased bending stiffness
. Increased flexibility in compression
. Dramatically increased torsional rigidity
. Decreased risk of thermal necrosis during insertion
. Decreased resistance to bending fatigue

Correct Answer & Explanation

. Dramatically increased torsional rigidity

Explanation

A closed-section intramedullary nail has dramatically higher torsional rigidity compared to an open (slotted) nail. Slotted nails are more flexible in torsion, which can lead to malrotation if the fracture is not adequately locked.

Question 1830

Topic: 2. Trauma

In the context of intramedullary nailing, how is the "working length" of the nail construct defined?

. The distance between the entry portal and the distal tip of the nail
. The distance between the most proximal and most distal locking screws
. The distance between the fracture site and the distal locking screw
. The distance between the two closest points of firm fixation proximal and distal to the fracture
. The length of the unreamed portion of the medullary canal

Correct Answer & Explanation

. The distance between the two closest points of firm fixation proximal and distal to the fracture

Explanation

The working length of an IM nail is the distance between the two closest points of firm fixation (or locking screws) immediately proximal and distal to the fracture site. Decreasing this working length increases the stiffness of the nail construct.

Question 1831

Topic: Lower Extremity Trauma

A surgeon exchanges a 10 mm intramedullary nail for a 12 mm intramedullary nail of the same material. Assuming a solid cylindrical design, by approximately what factor does the bending stiffness increase?

. 1.2 times
. 1.4 times
. 2.1 times
. 3.6 times
. 4.8 times

Correct Answer & Explanation

. 2.1 times

Explanation

The bending stiffness of a solid cylinder is proportional to the radius to the fourth power (r^4). Increasing the radius from 5 mm to 6 mm increases the stiffness by a factor of (6/5)^4, which is approximately 2.07 times.

Question 1832

Topic: 2. Trauma

In external fixation, increasing the distance between the near and far pins within a single bone fragment (pin spread) has what biomechanical effect on the construct?

. Decreases the bending stiffness
. Increases the overall stability and stiffness of the construct
. Increases the stress at the bone-pin interface
. Decreases the torsional rigidity exclusively
. Promotes higher interfragmentary gap strain

Correct Answer & Explanation

. Increases the overall stability and stiffness of the construct

Explanation

Increasing the distance between the pins within a single fragment maximizes the mechanical advantage of the frame. This distributes the load more effectively, thereby increasing the overall stability and stiffness of the external fixator construct.

Question 1833

Topic: 2. Trauma

When applying a titanium locking plate in a bridge plating technique for a comminuted fracture, what is the biomechanical consequence of placing screws in the holes immediately adjacent to the fracture gap?

. It increases the working length and increases flexibility
. It decreases the working length, making the construct excessively rigid and increasing strain at the fracture gap
. It promotes primary bone healing by achieving absolute stability
. It converts the construct into a tension band
. It prevents galvanic corrosion

Correct Answer & Explanation

. It decreases the working length, making the construct excessively rigid and increasing strain at the fracture gap

Explanation

Placing screws immediately adjacent to the fracture site significantly decreases the plate's working length. In a bridging construct, this makes the segment too rigid, concentrating stress over a small area and increasing the risk of implant fatigue failure or nonunion.

Question 1834

Topic: 2. Trauma

A 45-year-old male undergoes open reduction and internal fixation of a transverse patella fracture using a tension band construct. For this construct to function optimally and convert tensile forces into compressive forces at the articular surface, where must the tension band be primarily placed?

. On the articular (compression) surface of the patella
. On the anterior (tension) surface of the patella
. Through the central longitudinal axis of the patella
. Circumferentially around the equator of the patella
. Subchondrally within the distal pole of the patella

Correct Answer & Explanation

. On the anterior (tension) surface of the patella

Explanation

According to tension band principles, the implant must be applied to the tension side of the bone. When the knee flexes, the anterior surface of the patella experiences tensile forces, which the tension band wire converts to compression at the articular side.

Question 1835

Topic: 2. Trauma

According to Perren's strain theory, what range of interfragmentary strain is optimally tolerated by a fracture gap to promote secondary bone healing (callus formation) without progressing to primary healing or nonunion?

. Less than 2%
. Between 2% and 10%
. Between 10% and 30%
. Between 30% and 50%
. Greater than 50%

Correct Answer & Explanation

. Less than 2%

Explanation

Perren's strain theory dictates that absolute stability (strain <2%) results in primary bone healing without callus. Strain between 2% and 10% promotes secondary bone healing via callus formation, while strain >10% typically leads to fibrous tissue and nonunion.

Question 1836

Topic: 2. Trauma

A 65-year-old female sustains a comminuted distal femur fracture treated with a lateral locking plate. The surgeon intentionally leaves three empty screw holes directly over the fracture site. What is the primary biomechanical effect of this technique?

. It increases the axial stiffness of the construct
. It decreases the bending stiffness of the construct
. It severely limits interfragmentary motion at the fracture site
. It promotes direct (primary) intramembranous bone healing
. It increases the risk of screw pullout in the diaphyseal segment

Correct Answer & Explanation

. It decreases the bending stiffness of the construct

Explanation

Increasing the working length (the distance between the two closest screws across a fracture) decreases the construct's bending stiffness, allowing for more flexible fixation. This controlled micro-motion stimulates secondary bone healing via callus formation.

Question 1837

Topic: 2. Trauma

An orthopedic resident is applying a bridging unilateral external fixator for a highly comminuted tibial plateau fracture. Which of the following technical modifications will most effectively increase the overall bending stiffness of this construct?

. Increasing the distance between the bone and the longitudinal bar
. Decreasing the diameter of the half-pins
. Increasing the distance between the innermost pins across the fracture gap
. Increasing the distance between the innermost and outermost pins within each bone fragment
. Decreasing the number of longitudinal bars connecting the pins

Correct Answer & Explanation

. Increasing the distance between the innermost and outermost pins within each bone fragment

Explanation

Increasing the pin spread (the distance between the innermost and outermost pins within the same bone fragment) significantly increases the stiffness of an external fixator. Moving the bar closer to the bone and increasing pin diameter are other effective ways to increase stiffness.

Question 1838

Topic: 2. Trauma

A surgeon is applying a dynamic compression plate (DCP) for a diaphyseal forearm fracture. When comparing traditional non-locking plate constructs to fixed-angle locking constructs, what is the primary mode of failure for a non-locking construct subjected to excessive bending loads?

. Simultaneous screw fracture
. Screw pullout (toggle) from the bone
. Cold welding of the screw head to the plate hole
. Plate fracture strictly at the screw hole interface
. Sequential shear failure of the screw heads

Correct Answer & Explanation

. Screw pullout (toggle) from the bone

Explanation

Traditional non-locking constructs rely on friction between the plate and bone for stability and primarily fail sequentially via screw pullout (toggle) when overloaded. In contrast, locking constructs function as single fixed-angle devices and typically fail by simultaneous screw fracture or bone failure.

Question 1839

Topic: 2. Trauma

During the fixation of a spiral fibula fracture, a lag screw is inserted perpendicular to the fracture plane. If the surgeon mistakenly drills the near cortex with a drill bit matching the core diameter of the screw instead of the outer thread diameter, what is the expected biomechanical consequence?

. Optimal interfragmentary compression will be achieved
. The threads will engage the near cortex, preventing compression
. The screw will fail in shear prior to engaging the far cortex
. The holding power in the far cortex will increase significantly
. The working length of the screw will be effectively shortened

Correct Answer & Explanation

. The threads will engage the near cortex, preventing compression

Explanation

For a lag screw to effectively compress two fragments, it must glide freely through the near cortex (gliding hole) and only engage the far cortex. If the near cortex is drilled to the core diameter, the threads will engage both cortices simultaneously, preventing interfragmentary compression.

Question 1840

Topic: 2. Trauma

An intramedullary nail is selected for a subtrochanteric femur fracture. Biomechanically, the torsional rigidity of a solid cylindrical intramedullary nail is proportional to the nail's radius raised to which power?

. Radius squared (r^2)
. Radius cubed (r^3)
. Radius to the fourth power (r^4)
. Radius to the fifth power (r^5)
. Inversely proportional to the radius

Correct Answer & Explanation

. Radius to the fourth power (r^4)

Explanation

The torsional rigidity and bending stiffness of a solid cylinder are proportional to its polar area moment of inertia, which is dictated by the radius to the fourth power (r^4). Therefore, even a small increase in nail diameter profoundly increases its mechanical rigidity.

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