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2. Trauma MCQs

Practice Set 330 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 6581

Topic: 2. Trauma

An active 76-year-old woman sustains a highly comminuted 4-part proximal humerus fracture with a head-split component and severe osteopenia. Which surgical treatment provides the most predictable restoration of forward elevation?

. Open reduction internal fixation with a locked plate
. Intramedullary nailing
. Hemiarthroplasty
. Reverse total shoulder arthroplasty
. Latissimus dorsi transfer

Correct Answer & Explanation

. Reverse total shoulder arthroplasty

Explanation

Reverse total shoulder arthroplasty provides the most reliable functional outcome and forward elevation for elderly patients with complex, unreconstructible proximal humerus fractures. Hemiarthroplasty relies on predictable tuberosity healing, which is poor in osteopenic bone.

Question 6582

Topic: 2. Trauma

A 32-year-old woman sustains an isolated capitellum fracture that extends medially to include the lateral trochlear ridge (Dubberley Type 1B). What is the preferred surgical approach for optimal open reduction and internal fixation?

. Medial approach (Hotchkiss)
. Posterior triceps-split approach
. Anterior approach (Henry)
. Extensile lateral approach (Kocher or Kaplan)
. Posterior olecranon osteotomy

Correct Answer & Explanation

. Extensile lateral approach (Kocher or Kaplan)

Explanation

An extensile lateral approach (such as Kocher or Kaplan) is preferred for capitellar shear fractures. It provides excellent direct visualization of the anterior articular surface of the radiocapitellar joint for precise reduction.

Question 6583

Topic: 2. Trauma
A 22-year-old rugby player sustained a closed avulsion of the flexor digitorum profundus (FDP) tendon of the ring finger. Radiographs show no fracture. The tendon is palpable in the palm. What is the Leddy-Packer classification and maximum recommended time frame for primary repair?
. Leddy-Packer Type I; repair within 7-10 days
. Leddy-Packer Type II; repair within 3 weeks
. Leddy-Packer Type III; repair within 6 weeks
. Leddy-Packer Type I; repair within 3 weeks
. Leddy-Packer Type III; repair within 7-10 days

Correct Answer & Explanation

. Leddy-Packer Type I; repair within 7-10 days

Explanation

A Leddy-Packer Type I FDP avulsion involves retraction of the tendon into the palm, disrupting both the vincula longus and brevis. Due to compromised blood supply, surgical repair must be performed within 7-10 days to prevent tendon necrosis.

Question 6584

Topic: 2. Trauma

An orthopedic surgeon is applying a unilateral external fixator for a comminuted tibia fracture. Which of the following modifications will most significantly increase the bending stiffness of the construct?

. Decreasing the distance between the fracture site and the innermost pins
. Increasing the distance between the bone and the external longitudinal bar
. Increasing the pin diameter from 4.0 mm to 5.0 mm
. Adding an additional longitudinal bar of the same diameter
. Using stainless steel instead of titanium pins of the same diameter

Correct Answer & Explanation

. Increasing the pin diameter from 4.0 mm to 5.0 mm

Explanation

Increasing the pin diameter has the greatest effect on bending stiffness, as the stiffness is proportional to the radius of the pin to the fourth power (r^4). While decreasing the bone-to-bar distance and using stiffer materials also help, pin diameter is the most mathematically powerful variable.

Question 6585

Topic: 2. Trauma

When using a locking compression plate (LCP) in a bridging technique for a comminuted diaphyseal fracture, omitting screws in the holes immediately adjacent to the fracture site achieves which of the following?

. Increases the bending stiffness of the construct
. Decreases the working length of the plate
. Increases the working length, allowing symmetrical stress distribution and secondary healing
. Induces absolute stability and primary bone healing
. Decreases the load shared by the intact opposite cortex

Correct Answer & Explanation

. Increases the working length, allowing symmetrical stress distribution and secondary healing

Explanation

Omitting screws near the fracture site increases the working length of the plate, which decreases the overall construct stiffness. This allows for controlled interfragmentary motion (relative stability), promoting secondary bone healing via callus formation.

Question 6586

Topic: 2. Trauma

An orthopaedic surgeon uses an unreamed, solid intramedullary nail to treat a diaphyseal femur fracture. The torsional rigidity of this solid nail is proportional to its radius raised to which power?

. Radius to the 1st power (r)
. Radius to the 2nd power (r^2)
. Radius to the 3rd power (r^3)
. Radius to the 4th power (r^4)
. Radius to the 5th power (r^5)

Correct Answer & Explanation

. Radius to the 4th power (r^4)

Explanation

The torsional rigidity of a solid cylinder, such as an intramedullary nail, is proportional to its polar moment of inertia, which scales with the radius to the fourth power (r^4). Therefore, small increases in the nail's diameter result in exponential increases in its resistance to torsional forces.

Question 6587

Topic: Lower Extremity Trauma

A surgeon decides to upsize a solid tibial intramedullary nail from 10 mm to 12 mm. What is the approximate proportional increase in the bending stiffness of the implant?

. 1.2 times
. 1.44 times
. 1.7 times
. 2.1 times
. 4.0 times

Correct Answer & Explanation

. 2.1 times

Explanation

The bending stiffness of a solid cylinder is proportional to the radius to the fourth power based on the area moment of inertia. Upsizing the nail from 10 mm to 12 mm increases the stiffness by (12/10)^4, which equals approximately 2.07.

Question 6588

Topic: 2. Trauma

When constructing an external fixator for a highly comminuted tibial shaft fracture, which of the following modifications will most effectively increase the axial stiffness of the construct?

. Decreasing the distance between the fracture and the nearest pins
. Increasing the distance between the bone and the connecting rod
. Decreasing the pin diameter
. Placing pins in a uniplanar, unilateral configuration
. Increasing the distance between the two innermost pins

Correct Answer & Explanation

. Decreasing the distance between the fracture and the nearest pins

Explanation

Moving the pins closer to the fracture site decreases the working length of the bone-pin-rod construct, significantly increasing axial stiffness. Increasing pin diameter or moving the rod closer to the bone also increases stiffness.

Question 6589

Topic: 2. Trauma

In a locked plating construct used for a comminuted metaphyseal fracture, how is the construct's working length defined?

. The total length of the plate
. The distance from the outermost proximal screw to the outermost distal screw
. The distance between the innermost proximal and innermost distal screws spanning the fracture
. The diameter of the core screw threads
. The length of plate strictly in contact with the bone

Correct Answer & Explanation

. The distance between the innermost proximal and innermost distal screws spanning the fracture

Explanation

The working length of a plate is the distance between the two innermost screws on either side of the fracture. Increasing the working length in a comminuted fracture distributes stress over a longer segment, reducing the risk of fatigue failure.

Question 6590

Topic: 2. Trauma

During the initial inflammatory phase of fracture healing, which growth factor, released immediately by degranulating platelets, is most responsible for the initial chemotaxis of mesenchymal stem cells?

. Insulin-like growth factor-1 (IGF-1)
. Platelet-derived growth factor (PDGF)
. Matrix metalloproteinase-13 (MMP-13)
. Fibroblast growth factor-2 (FGF-2)
. Osteocalcin

Correct Answer & Explanation

. Platelet-derived growth factor (PDGF)

Explanation

PDGF is released immediately from alpha granules of platelets within the fracture hematoma. It serves as a potent chemoattractant and mitogen for mesenchymal stem cells, macrophages, and fibroblasts to initiate the healing cascade.

Question 6591

Topic: 2. Trauma

When using a dynamic compression plate (DCP) for a transverse radius fracture, eccentric screw placement within the oval plate holes creates which mechanical effect at the fracture site?

. Gap distraction
. Axial compression
. Torsional instability
. Shear displacement
. Increased working length

Correct Answer & Explanation

. Axial compression

Explanation

As the eccentrically placed spherical screw head engages the sloped contour of the DCP hole, it glides down the incline. This translates the bone relative to the plate, generating axial compression across the transverse fracture site.

Question 6592

Topic: 2. Trauma

When applying a locking compression plate (LCP) for a comminuted diaphyseal fracture, an orthopaedic surgeon intentionally leaves three screw holes empty directly over the fracture site. What is the primary biomechanical effect of increasing this working length?

. Decreases the bending stiffness of the construct
. Increases the torsional rigidity of the construct
. Promotes primary bone healing directly at the fracture
. Increases the risk of screw pullout
. Decreases interfragmentary strain at the near cortex

Correct Answer & Explanation

. Decreases the bending stiffness of the construct

Explanation

Increasing the working length of a bridging plate decreases the overall bending stiffness of the construct. This allows for controlled micromotion, which stimulates secondary bone healing through callus formation.

Question 6593

Topic: 2. Trauma

An external fixator is applied to stabilize an open tibia fracture. Which of the following modifications will most significantly increase the bending stiffness of the fixator construct?

. Increasing the distance between the bar and the bone
. Decreasing the diameter of the half-pins
. Increasing the diameter of the half-pins
. Decreasing the distance between the innermost pins and the fracture site
. Placing the pins strictly in a single plane

Correct Answer & Explanation

. Increasing the diameter of the half-pins

Explanation

The bending stiffness of a pin is proportional to the radius to the fourth power. Therefore, increasing the pin diameter provides an exponential and most significant increase in the overall stiffness of the external fixator construct.

Question 6594

Topic: 2. Trauma

An intramedullary nail is inserted for a transverse midshaft femur fracture. Using a solid nail instead of a slotted (split) nail of the exact same diameter most significantly improves which biomechanical property?

. Bending stiffness
. Torsional rigidity
. Fatigue strength
. Compressive strength
. Axial pullout strength

Correct Answer & Explanation

. Torsional rigidity

Explanation

The torsional rigidity of an intramedullary nail is dramatically reduced if the nail has a longitudinal slot. A solid, closed-section nail has significantly higher torsional rigidity compared to a slotted nail of identical material and size.

Question 6595

Topic: 2. Trauma

Following a tightly reamed intramedullary nailing of a diaphyseal tibia fracture, the local diaphyseal blood supply undergoes a significant physiologic alteration. Which of the following best describes the resulting flow?

. Centrifugal flow from the medullary cavity to the periosteum
. Centripetal flow from the periosteum to the medullary cavity
. Complete dependence on epiphyseal vessels
. Reversal of flow entirely to the nutrient artery
. Unaltered centrifugal flow

Correct Answer & Explanation

. Centripetal flow from the periosteum to the medullary cavity

Explanation

Normal diaphyseal blood flow in adult long bones is centrifugal, moving from the medullary cavity to the periosteum. After reaming destroys the endosteal supply, the flow temporarily reverses to a centripetal direction to maintain cortical bone viability.

Question 6596

Topic: 2. Trauma

When evaluating the biomechanics of an intramedullary nail for a diaphyseal femur fracture, increasing the diameter of a solid titanium nail from 10 mm to 11 mm will increase its torsional rigidity by approximately what percentage?

. 10 percent
. 21 percent
. 46 percent
. 100 percent
. 146 percent

Correct Answer & Explanation

. 46 percent

Explanation

The torsional rigidity of a solid cylinder is proportional to the radius to the fourth power (r^4). Increasing the diameter from 10 mm to 11 mm increases the rigidity by a factor of (11/10)^4, which is approximately 1.46, representing a 46% increase.

Question 6597

Topic: 2. Trauma

Which of the following biomechanical principles best distinguishes a locking plate construct from a conventional non-locking compression plate?

. It relies heavily on friction at the plate-bone interface for construct stability.
. It functions as a fixed-angle single beam construct independent of underlying bone quality.
. It requires precise contouring to the periosteal surface to avoid fracture malalignment.
. It routinely exhibits sequential screw failure when subjected to mechanical overload.
. It is specifically designed to create absolute stability via dynamic axial compression.

Correct Answer & Explanation

. It functions as a fixed-angle single beam construct independent of underlying bone quality.

Explanation

Locking plates function as fixed-angle constructs where the screw heads lock into the plate, acting as a single beam. They do not rely on plate-to-bone friction, thereby preserving periosteal blood supply and providing superior hold in osteoporotic bone.

Question 6598

Topic: 2. Trauma

When treating a highly comminuted diaphyseal femur fracture with bridge plating, increasing the working length of the plate has which of the following biomechanical effects?

. Increases construct stiffness and promotes primary bone healing
. Decreases construct stiffness and promotes secondary bone healing
. Increases the risk of screw pullout by concentrating stress at the innermost screws
. Requires the use of exclusively locking screws to maintain construct stability
. Decreases interfragmentary motion leading to hypertrophic nonunion

Correct Answer & Explanation

. Decreases construct stiffness and promotes secondary bone healing

Explanation

The working length is the distance between the two innermost screws spanning the fracture gap. Increasing this length decreases the overall bending stiffness of the construct, allowing micromotion that promotes robust secondary bone healing (callus formation).

Question 6599

Topic: 2. Trauma

An orthopaedic surgeon elects to use a fracture plate that is exactly twice as thick as the originally planned plate, while maintaining the same width. According to the area moment of inertia for a rectangular cross-section, how much will the bending stiffness of the new construct increase?

. 2 times
. 4 times
. 8 times
. 16 times
. It will remain unchanged

Correct Answer & Explanation

. 8 times

Explanation

The bending stiffness of a rectangular plate is proportional to the area moment of inertia, which is calculated as (width x thickness^3) / 12. Doubling the thickness increases the bending stiffness by a factor of 2^3, resulting in an 8-fold increase.

Question 6600

Topic: 2. Trauma

When applying a uniplanar external fixator to a tibial shaft fracture, which biomechanical adjustment will most dramatically increase the overall stiffness of the construct?

. Increasing the distance between the bar and the bone
. Decreasing the pin diameter
. Decreasing the distance between the pins within a single fragment
. Increasing the pin diameter
. Placing the pins further away from the fracture site

Correct Answer & Explanation

. Increasing the pin diameter

Explanation

External fixator stiffness is most sensitive to pin diameter because bending stiffness is proportional to the radius to the fourth power (r^4). Decreasing the bar-to-bone distance or spreading pins out also increases stability, but altering pin diameter has an exponential effect.

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