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2. Trauma MCQs

Practice Set 413 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 8241

Topic: Lower Extremity Trauma

During a primary TKA, a surgeon utilizes spacer blocks to assess gap kinematics after performing standard bone cuts. The extension gap is symmetric but tight, requiring significant force to insert the block. The flexion gap is symmetric and rectangular, and accepts the spacer block with 2 mm of balanced laxity. What is the most appropriate next step to achieve balanced gaps?

. Resect more posterior femur
. Resect more proximal tibia
. Resect more distal femur
. Downsize the femoral component
. Perform a posterior capsular release

Correct Answer & Explanation

. Resect more distal femur

Explanation

The patient has a tight extension gap and an acceptable/balanced flexion gap. Resecting more distal femur will increase the extension gap without affecting the flexion gap. Resecting the proximal tibia would increase both gaps. Downsizing the femoral component or resecting more posterior femur would increase only the flexion gap.

Question 8242

Topic: Lower Extremity Trauma

During a posterior-stabilized TKA, the knee is found to be symmetrically tight in flexion and well-balanced in extension. Which of the following modifications is the most appropriate step to balance the knee?

. Recut the distal femur
. Downsize the femoral component
. Release the posterior capsule
. Upsize the femoral component
. Decrease the posterior tibial slope

Correct Answer & Explanation

. Downsize the femoral component

Explanation

A knee that is symmetrically tight in flexion and balanced in extension has an isolated tight flexion gap. Downsizing the femoral component reduces the posterior condylar offset, thereby increasing the flexion gap without affecting the extension gap. Recutting the distal femur would increase the extension gap. Decreasing the posterior tibial slope would further tighten the flexion gap.

Question 8243

Topic: 2. Trauma
A 72-year-old man is brought to the emergency department after sustaining a fall. He underwent a primary total knee arthroplasty 5 years ago. Radiographs reveal a displaced, short oblique supracondylar femur fracture located entirely proximal to the femoral component. The femoral component shows no radiographic signs of loosening and remains well-fixed. Which of the following is the most appropriate surgical treatment for this periprosthetic fracture?
. Nonoperative management in a long-leg cast
. Open reduction and internal fixation with a locked lateral plate
. Revision to a distal femoral replacement endoprosthesis
. Revision of the femoral component using a long-stemmed component and structural allograft
. Placement of an external fixator spanning the knee

Correct Answer & Explanation

. Open reduction and internal fixation with a locked lateral plate

Explanation

This is a Lewis-Rorabeck Type II periprosthetic distal femur fracture, characterized by a displaced fracture with a well-fixed femoral component. The standard of care for a Type II fracture is surgical stabilization, typically achieved with open reduction and internal fixation (ORIF) utilizing a locked lateral plate or a retrograde intramedullary nail (if the femoral component design allows for nail passage). Revision arthroplasty (e.g., distal femoral replacement) is generally reserved for Type III fractures, where the femoral component is loose.

Question 8244

Topic: Pelvic & Acetabular Trauma
A 24-year-old female presents with symptomatic developmental dysplasia of the hip (DDH) characterized by a lateral center-edge angle of 15 degrees and a Tรถnnis angle of 18 degrees. She is scheduled to undergo a Bernese periacetabular osteotomy (PAO). Which of the following represents a primary biomechanical or structural advantage of the Bernese PAO compared to the Salter innominate osteotomy?
. It requires a complete division of the posterior column to maximize correction.
. It maintains an intact posterior column, preserving pelvic ring stability.
. It alters the true pelvic dimensions, expanding the birth canal.
. It achieves correction through a purely extra-articular sub-gluteal approach.
. It primarily relies on a single hinge point at the pubic symphysis.

Correct Answer & Explanation

. It maintains an intact posterior column, preserving pelvic ring stability.

Explanation

The Bernese periacetabular osteotomy (PAO) involves a series of osteotomies (ischial, pubic, and iliac) that completely free the acetabulum while leaving the posterior column of the hemipelvis intact. This is a major advantage because it preserves intrinsic pelvic stability, allowing for earlier mobilization without the need for prolonged casting. Furthermore, because the posterior column is intact, it does not alter the shape or dimensions of the true pelvis, making it advantageous for women of childbearing age regarding future vaginal deliveries.

Question 8245

Topic: 2. Trauma
An 82-year-old female sustains a mechanical fall 8 years following a primary total knee arthroplasty. Radiographs demonstrate a displaced periprosthetic distal femur fracture. Careful radiographic and clinical evaluation reveals that the femoral component is definitively loose and has migrated proximally. According to the Lewis-Rorabeck classification, which of the following is the most appropriate definitive management?
. Open reduction and internal fixation utilizing a lateral distal femoral locking plate
. Retrograde intramedullary nailing of the femur
. Revision total knee arthroplasty with a stemmed or distal femoral replacement component
. Nonoperative management with a hinged knee brace and restricted weight-bearing
. Excision arthroplasty (resection of the knee joint)

Correct Answer & Explanation

. Revision total knee arthroplasty with a stemmed or distal femoral replacement component

Explanation

This patient has a Lewis-Rorabeck Type III periprosthetic distal femur fracture, defined as a fracture with a loose prosthesis. The definitive management for a loose femoral component in the setting of a periprosthetic fracture is revision arthroplasty (utilizing stems, augments, or a distal femoral replacement depending on bone stock). Open reduction and internal fixation (ORIF) and retrograde nailing are appropriate for Lewis-Rorabeck Type II fractures, where the fracture is displaced but the prosthesis remains solidly fixed.

Question 8246

Topic: 2. Trauma

Following a high-energy trauma, a patient exhibits a peripheral nerve injury characterized by loss of axonal continuity and endoneurial tube disruption, but with preservation of the perineurium and epineurium. According to the Sunderland classification, what grade is this nerve injury?

. First degree
. Second degree
. Third degree
. Fourth degree
. Fifth degree

Correct Answer & Explanation

. Fourth degree

Explanation

According to the Sunderland nerve injury classification: First degree (Neuropraxia) involves local conduction block with intact axons. Second degree (Axonotmesis) involves axon disruption but an intact endoneurium. Third degree involves disruption of the axon and endoneurium, with an intact perineurium (intrafascicular scarring often blocks functional regeneration). Fourth degree involves disruption of the axon, endoneurium, and perineurium, leaving only the epineurium intact (neuroma-in-continuity). Fifth degree (Neurotmesis) is complete transection of the nerve.

Question 8247

Topic: 2. Trauma



When applying a bridge plate to a comminuted diaphyseal fracture to promote secondary bone healing, increasing the 'working length' of the plate will have which of the following biomechanical effects?

. Increases the axial stiffness of the construct
. Decreases the torsional rigidity of the construct
. Increases the strain at the fracture site by promoting rigid fixation
. Decreases the bending forces experienced by the hardware
. Shifts the mode of failure from plate pullout to plate breakage

Correct Answer & Explanation

. Decreases the torsional rigidity of the construct

Explanation

The 'working length' of a plate is defined as the distance between the two innermost screws closest to the fracture on either side. Increasing the working length decreases the stiffness of the construct in axial loading, bending, and torsion (decreased torsional rigidity). This calculated reduction in stiffness permits increased interfragmentary motion (micromotion), which is necessary to stimulate callus formation for secondary bone healing. It does not increase axial stiffness or rigid fixation.

Question 8248

Topic: 2. Trauma
A patient sustains a closed fracture resulting in a peripheral nerve injury characterized by loss of axonal continuity and subsequent Wallerian degeneration. However, the endoneurium, perineurium, and epineurium remain completely intact. According to the Sunderland classification, what grade is this injury, and what is the expected clinical outcome?
. Grade I; complete recovery without Wallerian degeneration
. Grade II; excellent prognosis for spontaneous recovery at approximately 1 mm/day
. Grade III; unpredictable recovery heavily dependent on the degree of intraneural fibrosis
. Grade IV; no spontaneous recovery without surgical intervention
. Grade V; complete nerve transection requiring primary repair

Correct Answer & Explanation

. Grade II; excellent prognosis for spontaneous recovery at approximately 1 mm/day

Explanation

This injury describes a Sunderland Grade II nerve injury, which directly corresponds to Seddon's 'axonotmesis'. In a Grade II injury, the axon is disrupted leading to Wallerian degeneration distally, but the intact endoneurial tubes (along with the perineurium and epineurium) guide the regenerating axons. This architectural preservation leads to an excellent prognosis for complete spontaneous recovery, which typically proceeds at a rate of 1 mm per day (or 1 inch per month). Grade III involves disruption of the endoneurium, Grade IV involves disruption of the perineurium, and Grade V is a complete transection.

Question 8249

Topic: 2. Trauma

In a locking plate construct, decreasing the 'working length' (the distance between the two innermost screws on either side of the fracture) has which of the following biomechanical effects?

. Increases the axial stiffness of the construct
. Decreases the torsional stiffness of the construct
. Increases interfragmentary strain at the fracture site
. Promotes enhanced secondary bone healing
. Decreases the pull-out strength of the peripheral screws

Correct Answer & Explanation

. Increases the axial stiffness of the construct

Explanation

The working length of a plate is defined as the distance between the two closest screws on opposite sides of the fracture. Decreasing the working length significantly increases the construct's overall stiffness (axial, bending, and torsional), thereby reducing interfragmentary strain. While highly stable, excessive stiffness may suppress callus formation and delay secondary bone healing in comminuted fractures.

Question 8250

Topic: 2. Trauma

During internal fixation of a diaphyseal fracture, a surgeon aims to maximize the pull-out strength of a cortical screw. Which of the following alterations in screw design or insertion technique most significantly increases the screw's pull-out strength?

. Decreasing the outer (thread) diameter of the screw
. Increasing the inner (core) diameter of the screw while maintaining the outer diameter
. Decreasing the thread pitch of the screw
. Increasing the pilot hole size to match the outer diameter
. Decreasing the total length of screw engagement in the cortical bone

Correct Answer & Explanation

. Decreasing the thread pitch of the screw

Explanation

The pull-out strength of a screw is a measure of its resistance to axial pull-out forces. It is directly proportional to the outer (thread) diameter, the length of engagement in the bone, and the volume of bone caught between the threads. Decreasing the thread pitch (the distance between adjacent threads) increases the number of threads engaged per unit of length, which subsequently increases the pull-out strength. Increasing the inner (core) diameter while maintaining the outer diameter decreases the thread depth, which would reduce pull-out strength (though it increases the screw's torsional strength).

Question 8251

Topic: Lower Extremity Trauma

A surgeon decides to upsize a solid titanium intramedullary nail for a tibial shaft fracture to achieve greater stability.

According to the principles of biomechanics, the bending rigidity (area moment of inertia) of a solid cylinder is proportional to the 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)
. Diameter squared (d^2)

Correct Answer & Explanation

. Radius to the fourth power (r^4)

Explanation

For a solid cylinder, both the area moment of inertia (resistance to bending) and the polar moment of inertia (resistance to torsion) are proportional to the radius to the fourth power (r^4). Therefore, even small increases in the diameter (and thus radius) of a solid intramedullary nail result in dramatic increases in its bending and torsional rigidity.

Question 8252

Topic: 2. Trauma

According to Perren's strain theory of fracture healing, primary bone healing (direct osteonal remodeling via cutting cones without callus formation) occurs exclusively when interfragmentary strain is below what critical threshold?

. 2%
. 10%
. 25%
. 50%
. 100%

Correct Answer & Explanation

. 2%

Explanation

Perren's strain theory dictates that primary (direct) bone healing requires absolute stability, which corresponds to an interfragmentary strain of less than 2%. At strains between 2% and 10%, secondary bone healing (endochondral ossification with callus formation) occurs. If the strain exceeds 10%, the granulation tissue cannot differentiate into cartilage or bone, ultimately resulting in a nonunion.

Question 8253

Topic: 2. Trauma
When evaluating the biomechanical properties of a solid titanium intramedullary nail designed for femoral shaft fractures, how is the torsional rigidity of the implant mathematically affected if the radius of the solid nail is doubled?
. 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 16.

Explanation

For a solid cylinder, torsional rigidity is directly proportional to the polar area moment of inertia (J). The formula for the polar area moment of inertia of a solid circular cross-section is J = (ฯ€ * r^4) / 2. Because torsional rigidity relies on the radius to the fourth power (r^4), doubling the radius (2^4) increases the torsional rigidity by a factor of 16.

Question 8254

Topic: 2. Trauma

A comminuted midshaft femur fracture is treated with a bridge plate technique using a locking compression plate. To promote secondary bone healing by maximizing interfragmentary motion (strain) while mitigating the risk of implant failure, which of the following mechanical configurations is most appropriate?

. Increasing the plate working length
. Decreasing the distance between the innermost screws
. Placing lag screws directly across the fracture site
. Using absolute rigid fixation with dynamic compression
. Placing the plate solely on the tension side with eccentric screws

Correct Answer & Explanation

. Increasing the plate working length

Explanation

In bridge plating for comminuted fractures, the goal is relative stability to induce secondary bone healing (callus formation). Increasing the plate working length (the span of the plate left empty of screws over the fracture zone) decreases the overall construct stiffness. This accommodates controlled, resilient interfragmentary micro-motion (strain) necessary for callus induction, while distributing bending stresses over a longer segment of the plate to prevent fatigue failure.

Question 8255

Topic: 2. Trauma

During internal fixation of a severely osteoporotic proximal humerus fracture, a surgeon wishes to optimize the biomechanics of the construct to prevent screw pullout. According to the mathematical principles of screw mechanics, which of the following modifications is most proportional to an increase in screw pullout strength?

. Increasing the minor diameter of the screw
. Increasing the major diameter of the screw
. Decreasing the thread pitch
. Increasing the length of the unthreaded shaft
. Decreasing the inner root diameter

Correct Answer & Explanation

. Increasing the major diameter of the screw

Explanation

Screw pullout strength is directly proportional to the major (outer) diameter of the screw, the length of thread engagement, and the shear strength of the host bone. While decreasing thread pitch increases the number of threads engaged, its effect is less proportionally significant than increasing the major diameter. The minor (inner root) diameter determines the tensile and bending strength of the screw itself, not its pullout resistance.

Question 8256

Topic: 2. Trauma

A multicenter randomized controlled trial is designed to evaluate a novel intramedullary nail versus standard plating for distal tibia fractures. The investigators determine that a sample size of 150 patients is required to achieve 80% power to detect a clinically significant difference in union rates at a significance level of 0.05. Which of the following modifications to the study design would strictly increase the statistical power of this trial?

. Decreasing the sample size
. Decreasing the significance level (alpha) to 0.01
. Increasing the sample size
. Decreasing the expected effect size
. Increasing the measurement variance

Correct Answer & Explanation

. Increasing the sample size

Explanation

Statistical power (1 - Beta) represents the probability of correctly rejecting the null hypothesis when it is indeed false (i.e., successfully detecting a true difference between groups). Power increases when you increase the sample size, increase the true effect size (the clinical difference between groups), decrease the variance of the data, or increase the significance level (alpha, e.g., moving from 0.05 to 0.10). Decreasing alpha (e.g., to 0.01) makes it harder to reject the null hypothesis, thereby decreasing power. Increasing the sample size gives more precision to the estimates, directly increasing statistical power.

Question 8257

Topic: 2. Trauma

A resident is inserting a cortical screw for plate fixation of a transverse radius fracture. To mathematically maximize the pull-out strength of the screw in the diaphyseal bone, which of the following alterations to the screw design would be most effective?

. Decreasing the outer diameter (thread diameter) of the screw
. Increasing the inner diameter (core diameter) of the screw while maintaining the outer diameter
. Decreasing the thread pitch
. Decreasing the length of thread engagement in the bone
. Over-drilling the pilot hole to match the outer diameter

Correct Answer & Explanation

. Decreasing the thread pitch

Explanation

The pull-out strength of a bone screw is a critical factor in fracture fixation. It is directly proportional to the outer (thread) diameter of the screw, the length of thread engagement (cortical thickness), and the shear strength of the bone. It is inversely proportional to the thread pitch (the distance between adjacent threads). Decreasing the thread pitch means the threads are closer together, resulting in more threads engaging a given length of bone, which increases the total volume of bone caught between the threads and maximizes pull-out strength. Increasing the core diameter while keeping the outer diameter constant actually decreases thread depth, thereby reducing pull-out strength.

Question 8258

Topic: 2. Trauma
An orthopedic surgeon is evaluating different titanium fracture plates for the internal fixation of a comminuted femoral shaft fracture. Bending stiffness is a critical factor in preventing hardware failure. If the surgeon chooses a plate that has double the thickness (height) of a standard plate, how does this specifically alter the area moment of inertia for bending?
. 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 remains unchanged

Correct Answer & Explanation

. It increases by a factor of 8

Explanation

The bending stiffness of a rectangular plate is directly proportional to its area moment of inertia (I). The formula for the area moment of inertia of a rectangular cross-section is I = (b ร— h^3) / 12, where 'b' is the base (width) and 'h' is the height (thickness). Doubling the thickness (height) increases the area moment of inertia by 2 cubed (2^3), which equals 8. Therefore, the bending stiffness increases by a factor of 8.

Question 8259

Topic: 2. Trauma

An orthopedic surgeon is considering intramedullary nailing for a tibial shaft fracture. According to the principles of biomechanics, if the diameter of a solid cylindrical nail is increased by a factor of 2, how does its bending stiffness change?

. 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 cylindrical structure is directly proportional to its area moment of inertia. For a solid cylinder, the area moment of inertia is proportional to the radius (or diameter) raised to the fourth power (r^4). Therefore, increasing the diameter by a factor of 2 increases the bending stiffness by 2^4, which equals 16.

Question 8260

Topic: 2. Trauma

A 35-year-old male undergoes bridge plating for a comminuted midshaft femur fracture. The surgeon opts to leave four empty screw holes in the plate directly overlying the comminuted fracture segment. How does this increased 'working length' biomechanically alter the construct?

. It increases the torsional stiffness of the construct.
. It decreases the bending stiffness of the construct.
. It increases the axial stiffness of the construct.
. It eliminates interfragmentary motion, promoting primary bone healing.
. It increases the pull-out strength of the screws closest to the fracture.

Correct Answer & Explanation

. It decreases the bending stiffness of the construct.

Explanation

Working length in bridge plating is defined as the distance between the two innermost screws on either side of the fracture. Increasing the working length makes the construct less rigid (more flexible), which correspondingly decreases both the bending and torsional stiffness of the plate. This deliberate flexibility allows for controlled interfragmentary micromotion (strain), which stimulates callus formation and secondary bone healing. It does not increase pull-out strength, and high flexibility prevents primary bone healing.

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