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2. Trauma MCQs

Practice Set 63 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 1241

Topic: 2. Trauma

A 68-year-old male with severe osteoporosis sustains a proximal humerus fracture. A locking compression plate is selected over a conventional non-locking plate. Biomechanically, what is the primary advantage of a locked plate construct in osteoporotic bone?

. It relies on friction between the plate and bone for stability.
. It permits dynamic compression across the fracture site.
. It functions as a single fixed-angle construct that resists sequential screw toggle.
. It relies on bicortical fixation for torsional resistance.
. It requires precise contouring to the native bone anatomy to avoid displacement.

Correct Answer & Explanation

. It functions as a single fixed-angle construct that resists sequential screw toggle.

Explanation

Locking plates function as single fixed-angle constructs where the screws lock into the plate. This prevents sequential screw toggle and failure in poor-quality osteoporotic bone, without relying on plate-to-bone friction.

Question 1242

Topic: 2. Trauma

According to Perren's strain theory, what happens when a fracture fixed with a conventional rigid plate has a microscopic gap (less than 1 mm) but lacks absolute stability?

. The strain exceeds the tolerance of granulation tissue, leading to nonunion.
. Secondary bone healing will bridge the small gap rapidly.
. Primary osteonal healing will occur due to the proximity of the fragments.
. The construct will dynamize over time, promoting callus formation.
. The small gap size prevents stress concentration on the plate.

Correct Answer & Explanation

. The strain exceeds the tolerance of granulation tissue, leading to nonunion.

Explanation

In a very small fracture gap with microscopic motion, deformation leads to a massively high strain percentage (Strain = change in length / original length). Granulation tissue cannot form if strain exceeds 100%, leading to nonunion.

Question 1243

Topic: 2. Trauma

A 25-year-old male receives a statically locked intramedullary nail for a transverse tibial shaft fracture. Which of the following technical modifications will maximize the torsional stiffness of this intramedullary construct?

. Increasing the distance between the proximal and distal interlocking screws.
. Using a nail with a longitudinal slot.
. Decreasing the working length by placing interlocking screws closer to the fracture site.
. Over-reaming the canal by 2 mm larger than the nail diameter.
. Using unicortical rather than bicortical interlocking screws.

Correct Answer & Explanation

. Decreasing the working length by placing interlocking screws closer to the fracture site.

Explanation

The working length of an IM nail is the distance between the nearest points of fixation across the fracture. Decreasing this working length (placing screws closer to the fracture) significantly increases both bending and torsional stiffness.

Question 1244

Topic: 2. Trauma

During the fixation of a transverse radius fracture using a Dynamic Compression Plate (DCP), the surgeon places screws eccentrically in the oval holes. What biomechanical principle allows this technique to generate interfragmentary compression?

. Cantilever bending
. Spherical gliding principle
. Tension band principle
. Fixed-angle stability
. Stress shielding

Correct Answer & Explanation

. Spherical gliding principle

Explanation

DCPs utilize the spherical gliding principle. As an eccentrically placed screw with a spherical head is driven into the oval plate hole, it glides down the slope of the hole, translating the bone fragment axially to compress the fracture.

Question 1245

Topic: 2. Trauma

A surgeon plans to use Far Cortical Locking (FCL) screws for a distal femur fracture fixed with a lateral locking plate. What is the intended biomechanical effect of FCL screws compared to standard locking screws?

. They increase construct stiffness to promote primary bone healing.
. They provide absolute stability by compressing the plate to the bone.
. They permit parallel interfragmentary motion to stimulate symmetric callus formation.
. They act as a tension band to neutralize bending forces.
. They increase the pullout strength in the near cortex.

Correct Answer & Explanation

. They permit parallel interfragmentary motion to stimulate symmetric callus formation.

Explanation

Far cortical locking screws have a reduced shaft diameter at the near cortex, allowing slight motion (biphasic stiffness). This permits parallel interfragmentary micromotion, which stimulates robust and symmetric callus formation.

Question 1246

Topic: 2. Trauma

When applying a tension band plate to a transverse fracture of the olecranon, what is the fundamental biomechanical requirement for the construct to function properly?

. The plate must be placed on the compressive surface of the bone.
. The fracture must have a large gap to allow for dynamic closure.
. The opposite cortex must be able to withstand compressive loads.
. The plate must utilize locking screws exclusively.
. The plate working length must be maximized.

Correct Answer & Explanation

. The opposite cortex must be able to withstand compressive loads.

Explanation

A tension band converts tensile forces into compressive forces at the fracture site. For this to occur, the plate must be on the tension side, and the opposite (compressive) cortex must be intact to act as a buttress against compression.

Question 1247

Topic: 2. Trauma

A short 60-year-old female sustains a subtrochanteric femur fracture and undergoes cephalomedullary nailing. Postoperatively, she has anterior cortical penetration of the distal femur by the nail tip. What biomechanical mismatch most likely caused this complication?

. The nail's radius of curvature is greater than the femur's radius of curvature.
. The nail's radius of curvature is smaller than the femur's radius of curvature.
. The nail selected had too small of an outer diameter.
. The interlocking screws were placed in a dynamic mode.
. The entry point was chosen too posteriorly in the greater trochanter.

Correct Answer & Explanation

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

Explanation

A higher radius of curvature means a straighter nail. If a straight nail (e.g., ROC 1.5-2.0m) is placed in a highly bowed femur (e.g., ROC 1.2m), the distal tip will impinge upon and potentially penetrate the anterior cortex.

Question 1248

Topic: 2. Trauma

A 34-year-old male with an open tibia fracture is managed with an unreamed solid intramedullary nail. Biomechanically, what is the most significant disadvantage of a solid unreamed nail compared to a hollow reamed nail of a larger diameter?

. Higher risk of endosteal blood supply disruption.
. Decreased torsional and bending stiffness.
. Increased rate of intramedullary infection.
. Increased risk of fat embolism syndrome.
. Inability to place static interlocking screws.

Correct Answer & Explanation

. Decreased torsional and bending stiffness.

Explanation

Unreamed nails typically necessitate a smaller diameter. Since stiffness is exponentially related to the radius, a smaller unreamed solid nail has significantly decreased bending and torsional stiffness compared to a larger reamed hollow nail.

Question 1249

Topic: 2. Trauma

A surgeon removes the proximal static interlocking screw in a tibial intramedullary nail for a nonunion at 5 months post-op. The distal screws remain intact, and the nail has a slotted proximal hole. What biomechanical change occurs due to this dynamization?

. It allows axial translation and compression while restricting torsional forces.
. It decreases the bending stiffness of the intramedullary nail.
. It increases the working length of the nail for both bending and torsion.
. It converts the construct from relative to absolute stability.
. It permits early rotational micromotion to stimulate callus.

Correct Answer & Explanation

. It allows axial translation and compression while restricting torsional forces.

Explanation

Dynamization by removing a static screw allows the nail to slide along the dynamic slot during weight-bearing. This permits controlled axial compression at the fracture site while still providing rotational (torsional) and bending stability.

Question 1250

Topic: 2. Trauma

A 40-year-old male sustains an oblique distal fibula fracture. The surgeon utilizes a lag screw followed by a lateral one-third tubular plate. What is the primary biomechanical function of the plate in this construct?

. To provide dynamic axial compression.
. To act as a tension band.
. To neutralize torsional and bending forces protecting the lag screw.
. To increase the working length of the fixation.
. To force secondary bone healing through micromotion.

Correct Answer & Explanation

. To neutralize torsional and bending forces protecting the lag screw.

Explanation

In this construct, the plate functions as a neutralization (protection) plate. Its primary role is to resist torsional, bending, and shear forces that would otherwise cause the interfragmentary lag screw to fail.

Question 1251

Topic: 2. Trauma

When managing a proximal third extra-articular tibia fracture with an intramedullary nail, the surgeon places a Poller (blocking) screw. To effectively prevent the typical apex anterior and valgus deformity, where should the blocking screws be placed relative to the nail?

. On the convex side of the anticipated deformity.
. On the concave side of the anticipated deformity.
. Directly orthogonal to the fracture plane.
. Parallel to the interlocking screws.
. In the distal fragment only.

Correct Answer & Explanation

. On the concave side of the anticipated deformity.

Explanation

Blocking (Poller) screws artificially narrow the medullary canal to direct the nail. To correct or prevent deformity, they must be placed on the concave side of the expected deformity (e.g., posterior and lateral in the proximal tibia).

Question 1252

Topic: 2. Trauma

A patient with a highly comminuted midshaft clavicle fracture is treated with an anterior-inferior reconstruction plate. Postoperatively, the plate fails due to excessive fatigue bending. Biomechanically, why is a reconstruction plate prone to this specific failure?

. The deep notches between holes significantly decrease the area moment of inertia.
. Reconstruction plates are exclusively made of titanium, which lacks fatigue resistance.
. The plate relies solely on locking screws, concentrating stress.
. It prevents primary bone healing by blocking periosteal blood flow.
. It has an excessively large working length by design.

Correct Answer & Explanation

. The deep notches between holes significantly decrease the area moment of inertia.

Explanation

Reconstruction plates have deep scallops (notches) between holes to allow bending in multiple planes. These notches significantly reduce the cross-sectional area and the area moment of inertia, making them much weaker in bending and prone to fatigue failure.

Question 1253

Topic: 2. Trauma

During closed reduction and intramedullary nailing of a femur fracture, the surgeon decides to over-ream the canal by 1.5 mm larger than the selected nail. What is the primary mechanical benefit of reaming the medullary canal?

. It creates a thermal necrosis zone that paradoxically stimulates osteogenesis.
. It acts as an internal bone graft to speed up cortical healing.
. It allows for insertion of a larger diameter nail, exponentially increasing construct stability.
. It forces the fracture into an absolute stability paradigm.
. It decreases the risk of postoperative fat embolism.

Correct Answer & Explanation

. It allows for insertion of a larger diameter nail, exponentially increasing construct stability.

Explanation

Reaming enlarges the medullary isthmus, permitting the insertion of a larger-diameter nail. Since a nail's bending and torsional stiffness are proportional to its radius to the 4th and 3rd powers respectively, this massively increases construct stability.

Question 1254

Topic: 2. Trauma

A patient has two rigid locking plates on the same bone (e.g., a proximal plate and a distal plate) separated by a 2 cm gap of unplated diaphyseal bone. What is the major biomechanical risk of this configuration?

. Stress shielding leading to diffuse osteopenia.
. Galvanic corrosion between the two distinct plates.
. Creation of a severe stress riser in the gap, predisposing to peri-implant fracture.
. Excessive dynamization leading to hypertrophic nonunion.
. Interference with the intramedullary blood supply.

Correct Answer & Explanation

. Creation of a severe stress riser in the gap, predisposing to peri-implant fracture.

Explanation

Leaving a short gap of normal bone between two highly rigid implants creates an abrupt transition in structural stiffness. This acts as a severe stress riser, heavily concentrating bending forces in the gap and increasing the risk of peri-implant fractures.

Question 1255

Topic: 2. Trauma

A surgeon is performing intramedullary nailing for a proximal one-third extra-articular tibia fracture. To counteract the deforming forces that commonly lead to apex anterior (procurvatum) and valgus malalignment, where should the blocking (Poller) screws be placed in the proximal segment relative to the planned nail trajectory?

. Anterior and medial
. Posterior and lateral
. Anterior and lateral
. Posterior and medial
. Directly central

Correct Answer & Explanation

. Posterior and lateral

Explanation

Blocking (Poller) screws are biomechanically placed on the concave side of the expected deformity to narrow the canal and direct the nail. To prevent apex anterior and apex medial (valgus) deformities, screws should be placed posteriorly and laterally in the proximal segment.

Question 1256

Topic: 2. Trauma

During open reduction and internal fixation of a diaphyseal fracture, a surgeon decides to switch from a 3.5 mm thick conventional plate to a 4.5 mm thick plate of the same width and material. According to the area moment of inertia, the bending rigidity of the plate increases by a factor proportional to which of the following?

. The square of the plate thickness
. The cube of the plate thickness
. The fourth power of the plate thickness
. The square root of the plate thickness
. The cube of the plate width

Correct Answer & Explanation

. The cube of the plate thickness

Explanation

The bending stiffness of a rectangular plate is proportional to the base times the height (thickness) cubed (bh^3/12). Therefore, increasing plate thickness has a cubic effect on its bending rigidity.

Question 1257

Topic: Lower Extremity Trauma

While up-sizing a solid titanium intramedullary nail from 10 mm to 12 mm in diameter, the torsional rigidity of the nail increases significantly. Biomechanically, the torsional stiffness of a solid cylinder is proportional to which of the following parameters?

. Radius squared
. Radius cubed
. Radius to the fourth power
. Square root of the radius
. Diameter cubed

Correct Answer & Explanation

. Radius to the fourth power

Explanation

The torsional and bending stiffness of a solid cylinder, such as a solid intramedullary nail, is determined by its polar moment of inertia, which is proportional to the radius to the fourth power (r^4).

Question 1258

Topic: 2. Trauma

A surgeon utilizes minimally invasive plate osteosynthesis (MIPO) with a locking plate for a comminuted distal femur fracture. To encourage secondary bone healing via callus formation, the surgeon deliberately leaves the plate holes immediately adjacent to the fracture empty. What biomechanical effect does this technical modification achieve?

. It decreases the plate stand-off distance
. It increases the construct's interfragmentary strain beyond 10%
. It decreases the plate's bending moment
. It increases the working length, thereby decreasing longitudinal stiffness
. It converts the construct from relative to absolute stability

Correct Answer & Explanation

. It increases the working length, thereby decreasing longitudinal stiffness

Explanation

Leaving screw holes empty adjacent to the fracture increases the working length of the plate. This decreases the longitudinal stiffness of the construct, allowing for controlled interfragmentary micromotion that stimulates callus formation.

Question 1259

Topic: 2. Trauma

A transverse subtrochanteric femur fracture is treated with plate osteosynthesis. To effectively utilize the tension band principle and convert bending forces into compressive forces across the fracture, the plate must be placed on which surface, and what type of stability does it achieve?

. Medial surface; relative stability
. Medial surface; absolute stability
. Lateral surface; relative stability
. Lateral surface; absolute stability
. Anterior surface; relative stability

Correct Answer & Explanation

. Lateral surface; absolute stability

Explanation

The lateral surface of the femur is under tension due to the eccentric mechanical axis and muscular forces. Applying a tension band plate on the lateral (tension) side converts bending forces into compression, providing absolute stability for primary healing.

Question 1260

Topic: 2. Trauma

When employing the lag screw technique across a fracture plane to achieve dynamic interfragmentary compression, the surgeon drills a gliding hole in the near cortex and a thread hole in the far cortex. Biomechanically, the diameter of the gliding hole must perfectly match which parameter of the screw?

. The core (inner) diameter
. The outer (thread) diameter
. The pitch distance
. The head diameter
. Slightly less than the core diameter

Correct Answer & Explanation

. The outer (thread) diameter

Explanation

The gliding hole must equal the outer (thread) diameter of the screw to prevent the threads from purchasing in the near cortex. This allows the screw head to pull the near fragment toward the far fragment, generating compression.

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