Osteopathic Medicine

Muscle Contraction Types

Explore active muscle length, sarcomere sliding, ATP cycling, joint moments, and muscle energy technique through visual comparisons and applied reasoning.

A muscle can work while a weight goes down

A cup is descending, but the biceps is still pulling. Is the muscle resting, or is it acting as a brake?

Contraction means active force, not necessarily shortening. With the shoulder held fixed, the biceps bends the elbow. When it brakes a descending cup as the elbow opens, it lengthens under tension: an eccentric action. The triceps straightens the elbow. It can shorten during elbow extension: a concentric action. Classify the named muscle, not the joint-direction word. [3] [13]

Isometric means the specified length is held approximately constant. For a limb task, say which scale is fixed: the joint or muscle-tendon unit. Do not infer that every internal fiber is motionless. [14]

Know the working muscle first. Gastrocnemius produces ankle plantarflexion and crosses both the knee and ankle. Hamstrings flex the knee; hold hip position constant when the named muscles cross both hip and knee. Quadriceps extends the knee; control hip position when its two-joint component is relevant. These controls prevent motion at another joint from confusing the comparison. [12]

Transfer: the triceps resists while the elbow bends. Which length pattern fits?

The triceps normally extends the elbow. Bending the elbow lengthens that extensor. Because it is still producing force, this is eccentric triceps activity, not concentric biceps activity.

Try a question on this topic

A 28-year-old man develops right anterior hip and low-back pain after increasing resistance training. He stands with the right hip slightly flexed, has a positive Thomas test, limited painful hip extension, and focal iliopsoas tenderness. Strength, sensation, reflexes, urinalysis, and abdominal examination are normal. The physician diagnoses iliopsoas hypertonicity and chooses a cooperative direct technique.

Which treatment strategy best fits the diagnosis and examination?

Choose an answer before revealing the reasoning.

Get one of these every morning · Practice more in the QBank

Shorter unit, same-length filaments

A sarcomere is one repeating contractile unit, from one Z-disc to the next. Thin actin filaments attach to the Z-discs. Thick myosin filaments lie near the center. They slide past one another rather than shrinking. [1] [2]

A band: the full length of the thick filament. I band: thin filament without thick overlap. H zone: the central thick filament without thin overlap. During shortening, I and H become smaller; A stays constant. [2]

Two sarcomeres show equal thick and thin filament lengths but different Z-disc separation and overlap. — The same filaments can occupy a shorter unit. Open full-size diagram. Original Bone Wizardry schematic; factual references [1] and [2].

A joint stays fixed, but ultrasound shows fiber shortening. What stretched?

The task can still be externally isometric. Compliant structures in series, including tendon, can lengthen while fibers shorten. Human calf ultrasound has documented changing internal architecture during fixed-joint contraction. That finding does not imply the same measured change in every muscle. [14]

ATP lets myosin release its grip

Myosin is a molecular motor. Before it can grip actin, calcium binds troponin. This changes the position of tropomyosin, the protein covering the myosin-binding sites on actin. The exposed sites permit attachment. [2]

Follow the grip and the nucleotide

Calcium is present throughout this comparison. Select a state; watch the head and the thin filament together.

Inspect a cycle state

Attached: the cocked head carries ADP and phosphate and binds an exposed site on actin.

Power stroke: phosphate release accompanies stronger binding and the force-producing transition. The head pivots; actin shifts.

ADP leaves: the head is still attached after the stroke. Losing ADP does not release the grip.

ATP binds: new ATP lowers myosin's affinity for actin. The head detaches before that ATP is hydrolyzed.

ATP is hydrolyzed: the head returns to the cocked position. ADP and phosphate remain bound; attachment can occur again.

Individual heads cycle asynchronously. No timing, force magnitude, or fixed pivot angle is implied.

One ATP per complete myosin-head cycle. The ATP that binds to detach the head is then hydrolyzed to reset it. These are two events involving the same ATP molecule, not two ATP molecules. Calcium pumps and other transport systems have additional energy requirements. [2] [4]

Myosin remains bound after ADP release; ATP binding detaches it, then hydrolysis restores the cocked ADP-plus-phosphate state. — Separate **detachment** from **resetting**. Open full-size diagram. Original Bone Wizardry schematic; references [2] and [4].

Predict: ATP can bind, but this preparation cannot hydrolyze it. Which event can still happen first?

Detachment. ATP binding releases the grip. Hydrolysis is needed for the subsequent reset, so repeated normal cycling cannot continue.

Predict: stimulation ends, but calcium remains high because its return pump is blocked. What can persist while ATP remains available?

Troponin can remain calcium-bound, keeping actin sites accessible. Continued cross-bridge cycling can delay relaxation. Stopping the electrical stimulus does not itself remove calcium. [2]

Speed, length, and force answer different questions

Isokinetic means constant speed of the measured motion. A dynamometer can regulate joint speed while a muscle either shortens or lengthens. The speed label does not cancel the length label. Real devices have acceleration and deceleration portions; classify the specified constant-speed interval. [10]

A machine bends the knee at fixed speed while the quadriceps resists. Can both eccentric and isokinetic apply?

First identify the working muscle: the quadriceps extends the knee. Knee flexion lengthens it. Active lengthening is eccentric. The independently regulated speed is isokinetic. The complete description includes both.

Torque, or moment, is a turning effect. Multiply force by its perpendicular moment arm. A fixed joint needs balanced moments, not necessarily equal numerical muscle and external forces. [15]

Doubling the external moment arm doubles the required opposing moment. With an unchanged muscle moment arm, required muscle force doubles too. A muscle can therefore exert more force than the external weight even while the joint is stationary.

Under comparable activation and length, a larger opposing load generally slows concentric shortening. Do not turn that relationship into a universal statement that any descending weight must exert more force than its active muscle. Direction, moment arms, and acceleration matter. [9] [15]

Isotonic literally describes constant tension. It is often used broadly for length-changing exercise, but a moving limb is not proof of exactly constant muscle tension. Isolytic is historical osteopathic terminology for clinician-imposed lengthening during active contraction. It overlaps mechanically with eccentric activity; the word alone proves neither scar disruption nor a treatment indication. [3] [7]

Separate where the clinician positions from where the patient pushes

Muscle energy technique, or MET, uses directed patient effort against clinician-defined resistance. In a direct setup, the clinician approaches the assessed restrictive barrier. The named position of ease describes the freer direction, not the direction of correction. [6] [7] [23]

For the described post-isometric relaxation setup, patient effort is directed toward the named ease while the clinician maintains the position with matching resistance. The patient then relaxes. Reassess before selecting a new barrier. Neither a guaranteed gain nor a fixed number of successful repetitions follows from the technique name. Effort and timing depend on the technique, region, and patient; maximal force is not a universal default. [7] [20]

For a neutral thoracic group that rotates right and sidebends left more freely, write N Sl Rr: N is neutral, Sl is sidebent left, and Rr is rotated right. A direct barrier comparison reverses rotation and sidebending while preserving the stated neutral condition. The findings, not a memorized abbreviation alone, determine the setup. [18]

Use anatomy to choose the effort. Psoas contributes hip flexion; a shortened hip-flexor pattern can restrict hip extension. Gluteus medius abducts the hip, so a target-muscle effort is abduction, not adduction. A left sternocleidomastoid contributes right head rotation and left sidebending. [11] [12]

Transfer: a hip-flexor restriction limits extension. For a target-muscle isometric effort, does the patient try to flex or extend?

The target is a hip flexor. The controlled effort is therefore hip flexion, while the clinician maintains the assessed position. An extension effort recruits the opposing action and is not the same target-muscle instruction.

Which muscle contracts, and what does the evidence actually show?

Post-isometric relaxation uses a contraction of the target muscle before relaxation and reassessment. A reciprocal-inhibition approach instead uses the antagonist, the muscle producing the opposing action. Neither label guarantees that the target becomes silent. [7] [17]

Muscle spindles report length-related information using group Ia and II sensory fibers. Golgi tendon organs report tension-related information through group Ib fibers. In classical spinal circuits, spindle input can support a stretch response and inhibition of an antagonist; tendon-organ pathways can influence the motor output to the same muscle. [8] [16] [22]

A spindle lies parallel to contractile fibers and sends Ia and II signals; a tendon organ lies in series and sends Ib signals. — Separate **length sensing** from **tension sensing**. Open full-size diagram. Original Bone Wizardry schematic; receptor anatomy from [8].

Real motor control depends on the task. Human experiments found less reciprocal inhibition during joint-stabilizing co-contraction. That allows opposing muscles to remain active together. A diagram of an inhibitory connection is therefore not an automatic on/off rule for every voluntary effort. [17]

In a randomized study of 40 asymptomatic volunteers, immediate tolerated knee-extension range increased after MET, but range at an equal applied torque did not significantly improve. The authors proposed altered stretch tolerance. This does not establish permanent tissue lengthening, a single GTO mechanism, or benefit in an injured patient. [5]

Does an immediate increase in tolerated range prove that a tendon became permanently longer?

No. The measured endpoint depended on tolerance and applied torque. A structural or durable-change claim would require additional measurements and follow-up.

Respiratory assistance coordinates breathing with an assessed setup; the respiratory motion can itself supply the active contribution. It is not defined by adding a separate isometric hold to every technique. [21] Eye-head assistance concerns the relationship between gaze and neck muscle activity. A physiological association is not proof that gaze alone corrects a cervical dysfunction. [19]

Apply the mechanism to a new problem

Unless a case directly measures fibers, contraction labels describe the joint and muscle-tendon task, not individual fascicles. Use the stated values to reason through the model; published findings are identified separately.

Work through the evidence before opening an explanation. Identify the active muscle or molecular event, determine what the observation actually establishes, then test the proposed next consequence. Every option has its own reasoning; return to any case without losing access to the lesson.

Case 1

Work through your selected option, then compare alternatives

A. Eccentric triceps activity with variable angular speed
What supports the variable-speed description?
The elbow covers unequal angles in equal times.
Which elbow action does triceps produce?
Extension.
What happens to triceps length during the described movement?
It decreases.
Read the complete explanation
The unequal angular displacements support variable speed. Eccentric activity would fit an active triceps resisting elbow flexion, but this elbow extends with the shoulder fixed.
This option does not fit the supplied findings.
B. Concentric triceps activity with variable angular speed
Which elbow action does triceps produce?
Extension.
What happens to the active triceps during this extension?
It shortens.
How do the angular displacements compare across equal time intervals?
The second displacement is larger.
Which combined classification follows?
Concentric activity with variable angular speed.
Read the complete explanation
At a fixed shoulder position, elbow extension shortens the active triceps. The second half-second covers twice the angle of the first, so angular speed cannot have remained constant throughout both intervals.
Best answer.
C. Concentric triceps activity with constant angular speed
Which part correctly classifies triceps length change?
Concentric activity.
What displacement pattern would constant angular speed produce?
Equal angular displacements in equal times.
What supplied measurement excludes that pattern?
The successive displacements are 10 and 20 degrees.
Read the complete explanation
The shortening classification fits the active extensor. Constant cable resistance does not establish constant speed, and the recorded angular displacements contradict that speed description.
This option does not fit the supplied findings.
D. Isometric triceps activity with variable angular speed
Which joint is held fixed?
The shoulder.
Which joint changes angle?
The elbow.
What length behavior does the active elbow extensor exhibit here?
Shortening.
Read the complete explanation
The changing speed is correctly recognized. Muscle activation can occur during an isometric hold, but the fixed shoulder does not make this a fixed-elbow task: the active triceps shortens as the elbow extends.
This option does not fit the supplied findings.

Takeaway: Determine the named muscle's length change separately from the joint's speed.

Case sources: [3] [10] [13]

Case 2

In a calf model, an active gastrocnemius brakes heel lowering while the ankle dorsiflexes and the knee remains fixed in extension. Neglect series compliance when classifying this moving phase. The model then makes two stationary holds at the same ankle and knee angles. A 30 N external force produces a dorsiflexion moment through a perpendicular arm of 0.12 m in hold A and 0.18 m in hold B. Gastrocnemius is the sole opposing actuator, with an unchanged 0.04 m moment arm; neglect gravity, other muscles, and passive joint moments. Its activity adjusts to maintain each hold. A receptor among the tendon collagen is sampled within its nonsaturated tension-sensitive range. Compare its steady signal in B with A. Which pairing fits the initial muscle action and the later sensory change? These are author-created model values, not patient force estimates.

A. Concentric during lowering; tendon-organ Ib signaling decreases in B.
B. Eccentric during lowering; tendon-organ Ib signaling decreases in B.
C. Concentric during lowering; tendon-organ Ib signaling increases in B.
D. Eccentric during lowering; tendon-organ Ib signaling increases in B.

Work through your selected option, then compare alternatives

A. Concentric during lowering; tendon-organ Ib signaling decreases in B.
Which ankle action does gastrocnemius normally produce?
Plantarflexion.
What does dorsiflexion do to its length with the knee fixed?
It increases its length.
What happens to the required opposing moment from A to B?
It increases.
What happens to required tendon tension with the muscle arm unchanged?
It increases.
Read the complete explanation
A descending heel can be mistaken for a shortening calf, and a stationary ankle can be mistaken for an unloaded calf. Here dorsiflexion lengthens the active gastrocnemius, while the longer external arm increases the force required during the second hold.
This option does not fit the supplied findings.
B. Eccentric during lowering; tendon-organ Ib signaling decreases in B.
What external moment acts in A?
3.6 N m.
What external moment acts in B?
5.4 N m.
How must muscle tension change with its arm fixed?
It must increase.
Which sensory prediction does that exclude in the stated range?
Decreased Ib signaling.
Read the complete explanation
The lowering classification is correct. Lower signaling might fit a reduced tendon load, but moving the unchanged external force farther from the ankle increases its moment and therefore the required tendon tension.
This option does not fit the supplied findings.
C. Concentric during lowering; tendon-organ Ib signaling increases in B.
Which ankle action shortens gastrocnemius with the knee fixed?
Plantarflexion.
Which ankle movement occurs during lowering?
Dorsiflexion.
How should its active lengthening be classified?
Eccentric.
Read the complete explanation
The sensory prediction fits the greater load moment. The length classification confuses downward heel motion with muscle shortening: dorsiflexion moves opposite gastrocnemius plantarflexion while its knee attachment is controlled.
This option does not fit the supplied findings.
D. Eccentric during lowering; tendon-organ Ib signaling increases in B.
How does dorsiflexion affect active gastrocnemius length with the knee fixed?
It lengthens the muscle-tendon unit.
What muscle tension balances hold A?
90 N.
What muscle tension balances hold B?
135 N.
What happens to the sampled tendon-organ signal at the higher tension?
Ib signaling increases.
Which paired conclusion follows?
Eccentric lowering followed by greater Ib signaling in B.
Read the complete explanation
Dorsiflexion lengthens the active gastrocnemius with the knee fixed, so it brakes eccentrically. The stationary holds require 30 x 0.12 / 0.04 = 90 N and 30 x 0.18 / 0.04 = 135 N. Greater tension predicts greater Ib signaling within the stated receptor range. Both holds remain externally isometric; unchanged joint angle does not mean unchanged force. Tendon-organ placement and Ib tension signaling are supported by [source 8](https://www.ncbi.nlm.nih.gov/books/NBK10812/); the numerical model is hypothetical.
Best answer.

Takeaway: Active calf braking can precede an externally isometric hold in which altered leverage changes tension and tendon-organ signaling.

Case sources: [3] [8] [12] [14] [15]

Case 3

A hypothetical fixed-knee, fixed-ankle calf model has one active contractile segment in series with one elastic tendon segment. Their total length stays fixed; neglect changes in fascicle angle. A 24 N external force has a perpendicular arm of 0.10 m in stationary state A and 0.15 m in stationary state B. The opposing muscle arm stays 0.03 m. Neglect all other moments, and let activation adjust until each state is stationary. Within this model's linear elastic range, every additional 20 N of tendon tension adds 0.5 mm of tendon length, reversibly. The segments carry the same tension. What are the final muscle tension in B and the segment-length changes from A to B? Use the stated elastic rule to calculate the modeled response; patient measurements require examination.

A. 120 N; tendon lengthens 1 mm and contractile segment shortens 1 mm.
B. 120 N; tendon shortens 1 mm and contractile segment lengthens 1 mm.
C. 40 N; tendon lengthens 1 mm and contractile segment shortens 1 mm.
D. 40 N; tendon shortens 1 mm and contractile segment lengthens 1 mm.

Work through your selected option, then compare alternatives

A. 120 N; tendon lengthens 1 mm and contractile segment shortens 1 mm.
What muscle tension balances state A?
80 N.
What muscle tension balances state B?
120 N.
What tendon-length change follows from the 40 N tension increase?
Lengthening by 1 mm.
What contractile change preserves total series length?
Shortening by 1 mm.
Which complete prediction follows?
120 N, with tendon lengthening and contractile shortening of 1 mm each.
Read the complete explanation
Moment balance gives 80 N in A and 120 N in B. The 40 N increase stretches the modeled tendon by 1 mm. Because total series length is fixed, the contractile segment must shorten by 1 mm. The external task remains isometric even though its internal segments change length; the numerical elastic response is a declared model assumption.
Best answer.
B. 120 N; tendon shortens 1 mm and contractile segment lengthens 1 mm.
How does tension change from A to B?
It increases by 40 N.
What does the elastic rule predict for added tension?
Tendon lengthening.
What must the contractile segment do within the fixed total length?
Shorten.
Read the complete explanation
The force calculation is correct, and the opposite segment changes preserve total length. Their directions fit unloading, however. The model specifies tendon extension with added tension, and state B requires more tension.
This option does not fit the supplied findings.
C. 40 N; tendon lengthens 1 mm and contractile segment shortens 1 mm.
What external moment acts in B?
3.6 N m.
What muscle tension balances that moment through 0.03 m?
120 N.
What does the calculated 40 N represent?
The increase above the initial 80 N.
Read the complete explanation
The length changes correctly use the 40 N increase. The option mistakes that increase for the final tension. Forty newtons through the muscle arm cannot balance the external moment in B.
This option does not fit the supplied findings.
D. 40 N; tendon shortens 1 mm and contractile segment lengthens 1 mm.
What tendon response would a fall from 80 N to 40 N produce?
Shortening by 1 mm.
How does the external moment actually change from A to B?
It increases from 2.4 to 3.6 N m.
What final muscle tension does that require?
120 N.
Which premise of this option therefore fails?
The assumed fall in tension.
Read the complete explanation
If tension actually fell from 80 N to 40 N, these length changes would fit the elastic rule. That unloading premise fails the joint mechanics: the external moment increases when its arm becomes longer.
This option does not fit the supplied findings.

Takeaway: Calculate tension from joint moments before applying tendon elasticity and the fixed total length of series elements.

Case sources: [1] [3] [14] [15]

Case 4

Work through your selected option, then compare alternatives

A. Concentric hamstring activity during an isokinetic interval
What motion does decreasing knee flexion angle represent?
Knee extension.
Which knee action do the hamstrings produce?
Flexion.
What length change occurs when they resist extension at a fixed hip?
Lengthening.
Read the complete explanation
The continuous speed record supports isokinetic testing. The decreasing flexion angle indicates extension, which lengthens the resisting hamstrings when the hip is fixed.
This option does not fit the supplied findings.
B. Isometric hamstring activity during an isokinetic interval
Which variable does the machine hold constant?
Angular speed.
Which variable changes throughout the interval?
Knee angle.
What happens to the active hamstrings as that angle changes?
They lengthen.
Read the complete explanation
The machine holds speed constant, not knee position. The changing knee angle lengthens the active hamstrings despite the controlled velocity.
This option does not fit the supplied findings.
C. Eccentric hamstring activity during an isokinetic interval
What motion is encoded by the falling flexion angle?
Knee extension.
What action do the resisting hamstrings normally produce?
Knee flexion.
What active length behavior follows at the fixed hip?
Eccentric lengthening.
What does the continuous speed record establish?
An isokinetic interval.
Read the complete explanation
The angle sequence represents knee extension. With the hip fixed, the active knee flexors lengthen while resisting that motion. Independently, continuous velocity measurement establishes the isokinetic condition.
Best answer.
D. Eccentric hamstring activity during a variable-speed interval
Which part correctly describes the hamstrings?
Eccentric activity.
What uncertainty would isolated angle samples leave?
Speed could fluctuate between samples.
Which supplied evidence removes that uncertainty?
The continuous constant-speed record.
Read the complete explanation
The hamstring classification fits. Equal-angle samples alone would not exclude speed fluctuations between samples, but the supplied continuous record explicitly verifies constant speed throughout this interval.
This option does not fit the supplied findings.

Takeaway: Verified constant speed can coexist with active lengthening; control the other joint of a multiarticular muscle.

Case sources: [10] [12]

Case 5

Two separate knee-extension intervals use active quadriceps against constant cable tension, with the hip fixed. In interval A, the knee covers progressively larger angles in equal times. In interval B, a continuous record verifies constant nonzero angular speed. During B, the cable's perpendicular moment arm increases from 0.10 to 0.20 m while its tension and the quadriceps moment arm remain unchanged. In a simplified fixed-axis model, neglect all other moments. Which comparison is correct?

A. Both intervals are isokinetic; required quadriceps force rises during B
B. Only interval B is isokinetic; required quadriceps force rises during B
C. Only interval B is isokinetic; required quadriceps force stays constant during B
D. Both intervals are isokinetic; required quadriceps force stays constant during B

Work through your selected option, then compare alternatives

A. Both intervals are isokinetic; required quadriceps force rises during B
What does interval A's displacement pattern imply?
Its angular speed changes.
Which loading quantity remains constant in A?
Cable tension.
Does constant tension establish constant speed?
No.
Read the complete explanation
The increasing cable moment arm raises the opposing moment during B. Constant cable tension, however, does not make A isokinetic: its timed angular displacements show changing speed.
This option does not fit the supplied findings.
B. Only interval B is isokinetic; required quadriceps force rises during B
What does the displacement pattern establish about A?
Its speed is variable.
What does continuous monitoring establish about B?
Its speed is constant.
What happens to cable moment as its arm doubles at constant tension?
It doubles.
What happens to the balancing muscle force with its arm unchanged?
It doubles.
Read the complete explanation
A has changing speed, whereas B has continuously verified constant speed. During B, doubling the cable moment arm doubles its opposing moment. With the quadriceps moment arm unchanged, the required muscle force increases.
Best answer.
C. Only interval B is isokinetic; required quadriceps force stays constant during B
Which cable quantity stays constant?
Tension.
Which cable quantity increases?
Its perpendicular moment arm.
What must increase to balance the resulting moment?
Quadriceps force.
Read the complete explanation
The speed comparison is correct. The force prediction confuses constant cable tension with constant joint moment; the changing cable leverage requires a changing opposing quadriceps moment.
This option does not fit the supplied findings.
D. Both intervals are isokinetic; required quadriceps force stays constant during B
What observation contradicts constant speed in A?
Increasing angular displacements over equal times.
What observation changes external moment during B?
The cable moment arm doubles.
What unchanged muscle property prevents that moment change from being absorbed by leverage?
The quadriceps moment arm.
Read the complete explanation
This applies constant tension to two quantities it does not determine. A's measured speed changes, and B's changing cable leverage changes the required muscle force.
This option does not fit the supplied findings.

Takeaway: Constant cable tension establishes neither constant joint speed nor constant muscle force when leverage changes.

Case sources: [3] [10] [15]

Case 6

A teaching demonstration compares three trials over the same knee arc with the hip fixed. In A, the clinician extends the knee while the hamstrings actively resist; the historical record calls this isolytic work. In B, the clinician repeats extension while the hamstrings are confirmed relaxed. In C, the patient actively flexes the knee with the hamstrings against resistance. Continuous monitoring verifies the same angular-speed magnitude in all three trials. Which pair of comparisons best separates the relevant mechanisms?

A. A–B isolates activation during lengthening; A–C compares eccentric with concentric activity
B. A–C isolates activation during lengthening; A–B compares eccentric with concentric activity
C. A–B isolates activation during shortening; A–C compares eccentric with concentric activity
D. A–B isolates activation during lengthening; B–C compares eccentric with concentric activity

Work through your selected option, then compare alternatives

A. A–B isolates activation during lengthening; A–C compares eccentric with concentric activity
What length behavior occurs in the active hamstrings during A?
Lengthening.
What distinguishes B from A at the same imposed motion?
The hamstrings are relaxed.
What length behavior occurs in the active hamstrings during C?
Shortening.
Which comparison therefore tests activation during lengthening?
A–B.
Which comparison contrasts active lengthening with active shortening?
A–C.
Read the complete explanation
A and B share the imposed lengthening motion but differ in hamstring activity. A and C both involve active hamstrings, but extension lengthens them whereas flexion shortens them. Matching speed does not erase these distinctions.
Best answer.
B. A–C isolates activation during lengthening; A–B compares eccentric with concentric activity
Is hamstring activity absent in C?
No.
What changes between A and C?
The direction of knee motion.
What prevents B from being concentric hamstring activity?
The hamstrings are relaxed.
Read the complete explanation
A–C is a useful active-muscle comparison, but both trials contain activation and their motion directions differ. B cannot supply the concentric member of A–B because its hamstrings are relaxed during extension.
This option does not fit the supplied findings.
C. A–B isolates activation during shortening; A–C compares eccentric with concentric activity
Which knee motion is shared by A and B?
Extension.
Which knee action do the hamstrings produce?
Flexion.
What length direction therefore occurs in A and B?
Lengthening.
Read the complete explanation
The second comparison is correct. The first misreads the common movement: imposed extension lengthens the hamstrings at a fixed hip in both A and B.
This option does not fit the supplied findings.
D. A–B isolates activation during lengthening; B–C compares eccentric with concentric activity
What makes A–B a controlled activation comparison?
Their imposed motion is matched.
Which required feature of eccentric activity is absent in B?
Active muscle force.
What is B's length change therefore classified as?
Passive lengthening.
Read the complete explanation
A–B correctly isolates the presence of activity. B–C contrasts passive lengthening with active shortening; B's changing length does not make it an eccentric contraction without active force.
This option does not fit the supplied findings.

Takeaway: Historical isolytic work includes active lengthening; matching movement alone does not reproduce that active task.

Case sources: [3] [7] [10] [12]

Case 7

In a skeletal-muscle preparation, two images show Z-disc separation decreasing from 2.6 to 2.4 micrometers while both thick- and thin-filament lengths remain unchanged. An H zone remains visible in both images. The preparation is then clamped at the second length. Calcium is held high by the experimental solution independently of cellular pumps. ATP is removed, and selected myosin heads are followed after they complete their power stroke and release ADP. Normal hydrolysable ATP is then restored without changing calcium or the length clamp. Which pairing correctly compares the second image's H zone with the first and predicts the tracked heads' first response to restored ATP? The measurements and sequence are hypothetical.

A. The H zone is wider; the tracked heads detach.
B. The H zone is wider; the tracked heads remain attached.
C. The H zone is narrower; the tracked heads remain attached.
D. The H zone is narrower; the tracked heads detach.

Work through your selected option, then compare alternatives

A. The H zone is wider; the tracked heads detach.
What does a wider H zone represent?
A larger region containing thick filament without thin overlap.
What happens to overlap when these Z-discs approach?
It increases.
What H-zone change follows here?
Narrowing.
Read the complete explanation
ATP restoration can detach the tracked heads. A wider H zone, however, would require less overlap. Bringing the Z-discs closer without shortening either filament increases overlap and narrows the remaining thick-only region.
This option does not fit the supplied findings.
B. The H zone is wider; the tracked heads remain attached.
What happens to overlap at the smaller Z-disc separation?
It increases.
What happens to the remaining H zone?
It narrows.
What permits a post-stroke head to detach?
New ATP binding.
Does maintained calcium prevent that ATP-dependent release?
No.
Read the complete explanation
This combines two tempting errors: treating the central gap as opening during shortening and treating persistent calcium as a lock on attached heads. The fixed-length filaments overlap more, and ATP binding permits release even while calcium remains high.
This option does not fit the supplied findings.
C. The H zone is narrower; the tracked heads remain attached.
Why can the tracked heads remain attached after ADP leaves during depletion?
No new ATP is available to bind.
What changes when ATP is restored?
ATP can bind the heads.
What immediate event does that binding permit?
Detachment.
Read the complete explanation
The geometric prediction is correct. Persistent attachment fits the ATP-depleted interval, but not the first response to restored ATP. High calcium maintains access to actin; it does not prevent ATP from releasing a bound head.
This option does not fit the supplied findings.
D. The H zone is narrower; the tracked heads detach.
What happens to overlap between the two images?
It increases.
What happens to the remaining H zone?
It narrows.
What prevents release of the tracked post-stroke heads during depletion?
The absence of new ATP.
What is their first response when ATP binds after restoration?
Detachment.
Which paired prediction follows?
A narrower H zone and initial detachment of the tracked heads.
Read the complete explanation
The closer Z-discs increase overlap between unchanged filaments, narrowing the residual H zone. Thick-filament length is unchanged, so the A band also remains unchanged. After ADP release, ATP depletion leaves the tracked heads attached; restoring ATP permits detachment before hydrolysis resets them. High calcium permits renewed attachment later, but does not prevent this initial release. The length clamp prevents detachment from being mistaken for an imposed change in sarcomere length.
Best answer.

Takeaway: Filament geometry determines overlap; nucleotide availability independently determines whether an attached myosin head can release.

Case sources: [1] [2] [4]

Case 8

In a supervised MET demonstration, assessment identifies the right psoas as the target and confirms that direct target-muscle work is appropriate and painless. With the pelvis and lumbar spine stabilized, the clinician approaches an assessed hip-extension barrier and requests a small target-muscle effort while intending to hold the starting position. Other hip planes and the opposite limb remain fixed. The hip moves toward flexion without reversals throughout both measured intervals. Psoas activation continues, and the recorded right-hip positions are 8 degrees of extension at 0 seconds, 4 degrees of extension at 2 seconds, and 1 degree of extension at 3 seconds. Extension is measured from neutral, with larger values meaning farther behind neutral. For this length classification, neglect changes in series compliance. Which interpretation fits the actual target-muscle behavior and the mean angular speeds over the two intervals? These are measurements, not a treatment protocol.

A. Eccentric psoas activity; mean angular speed increases.
B. Concentric psoas activity; mean angular speed decreases.
C. Concentric psoas activity; mean angular speed increases.
D. Eccentric psoas activity; mean angular speed decreases.

Work through your selected option, then compare alternatives

A. Eccentric psoas activity; mean angular speed increases.
What motion takes the hip from 8 to 1 degrees of extension?
Hip flexion.
Which hip action does psoas contribute?
Flexion.
What happens to active psoas length during this controlled motion?
It decreases.
Read the complete explanation
Dividing displacement by elapsed time correctly gives an increase in speed. The length label is reversed: the hip moves toward flexion as its extension angle decreases, so the active psoas shortens with the pelvis and lumbar spine fixed.
This option does not fit the supplied findings.
B. Concentric psoas activity; mean angular speed decreases.
What is the first interval's mean angular speed?
2 degrees per second.
What is the second interval's mean angular speed?
3 degrees per second.
Why does the smaller second displacement not establish slower motion?
Its elapsed time is shorter.
Read the complete explanation
The target-muscle classification is correct. The later displacement is smaller, which can tempt a slower-speed answer, but it occurs in half the time. Four degrees over two seconds is slower than three degrees over one second.
This option does not fit the supplied findings.
C. Concentric psoas activity; mean angular speed increases.
Which hip motion is shown by the decreasing extension angles?
Flexion.
How does that motion affect the active psoas with its proximal attachments controlled?
It shortens the psoas.
What is the first interval's mean angular speed?
2 degrees per second.
What is the second interval's mean angular speed?
3 degrees per second.
Which combined interpretation follows?
Concentric psoas activity with increasing mean angular speed.
Read the complete explanation
Decreasing extension angles indicate motion toward hip flexion. With the pelvis and lumbar spine controlled, this shortens the active psoas. Mean angular speed rises from 4/2 = 2 to 3/1 = 3 degrees per second. Thus the intended fixed-position effort was not maintained, and the motion was not constant-speed. Moving toward flexion does not demonstrate a gain in the assessed extension barrier; that would require relaxation and a matched reassessment.
Best answer.
D. Eccentric psoas activity; mean angular speed decreases.
Which motion is represented by decreasing hip extension?
Flexion.
What happens to active psoas length during that motion here?
It shortens.
What is the first interval's mean angular speed?
2 degrees per second.
What is the second interval's mean angular speed?
3 degrees per second.
Read the complete explanation
A progressively smaller extension angle can be mistaken for lengthening of the target, and the smaller second displacement can be mistaken for slower motion. Psoas anatomy determines the length direction, while displacement divided by time determines speed.
This option does not fit the supplied findings.

Takeaway: Use the target's anatomy to interpret actual motion, and divide displacement by time before deciding whether a maintained-position effort or constant-speed movement occurred.

Case sources: [3] [6] [10] [12] [23]

Case 9

Two permeabilized muscle preparations begin with myosin attached after ADP release and ATP depletion. In X, an ATP analogue is experimentally shown to produce ATP-like detachment when it binds myosin, but it cannot be hydrolyzed and remains bound for the test. In Y, ordinary ATP binds and is hydrolyzed normally. Calcium is kept low during these nucleotide manipulations, so actin sites remain covered. Calcium is then restored; the analogue has no effect on thin-filament regulation, and ordinary ATP remains available in Y. Which preparation can begin another normal attachment-and-stroke sequence, and why?

A. Only Y; hydrolysis has prepared ADP-plus-phosphate heads for attachment when sites reopen
B. Only X; retaining an unhydrolyzed nucleotide preserves the cocked state needed for attachment
C. Both X and Y; reopening actin sites supplies the reset needed for another attachment
D. Neither X nor Y; the low-calcium interval leaves both heads uncocked after sites reopen

Work through your selected option, then compare alternatives

A. Only Y; hydrolysis has prepared ADP-plus-phosphate heads for attachment when sites reopen
Which event has the analogue been shown to support in X?
Detachment.
Which subsequent motor event is unavailable in X?
Hydrolysis-dependent resetting.
What nucleotide state can hydrolysis produce in Y?
A cocked head carrying ADP and phosphate.
What does restoring calcium change?
Actin-site accessibility.
Which preparation can therefore begin the next normal sequence?
Y.
Read the complete explanation
The experiment establishes detachment in X but supplies no hydrolysis-dependent reset. Y can reset while actin sites are covered. Restoring calcium then permits the prepared Y heads to attach, whereas calcium cannot replace the missing reset in X.
Best answer.
B. Only X; retaining an unhydrolyzed nucleotide preserves the cocked state needed for attachment
Which nucleotide products remain on a normally cocked head?
ADP and phosphate.
Which reaction produces that state?
ATP hydrolysis.
Which preparation can perform that reaction?
Y.
Read the complete explanation
This treats the cocked head as ATP-loaded. In the taught cycle, hydrolysis produces the prepared ADP-plus-phosphate state; analogue binding alone establishes release, not that reset.
This option does not fit the supplied findings.
C. Both X and Y; reopening actin sites supplies the reset needed for another attachment
What does calcium regulate here?
Access to actin-binding sites.
What event resets the myosin head?
ATP hydrolysis.
Which required event remains blocked in X after calcium returns?
Hydrolysis-dependent resetting.
Read the complete explanation
Calcium restores regulatory permission in both preparations. That permission does not perform the motor's nucleotide-dependent reset, which remains unavailable in X despite its demonstrated detachment.
This option does not fit the supplied findings.
D. Neither X nor Y; the low-calcium interval leaves both heads uncocked after sites reopen
Which process is blocked by covered actin sites?
Attachment to actin.
Which nucleotide reaction occurs normally in Y during the waiting period?
ATP hydrolysis.
What barrier to Y attachment is removed by restoring calcium?
The regulatory covering of actin sites.
Read the complete explanation
Low calcium prevents normal thin-filament access, which can make both preparations appear inactive during the waiting period. It does not prevent the stipulated ATP hydrolysis and motor resetting in Y.
This option does not fit the supplied findings.

Takeaway: A demonstrated binding-induced release does not establish resetting; calcium controls access to the next attachment.

Case sources: [2] [4]

Case 10

A cleared patient's thoracic group moves more freely into left sidebending and right rotation when neutral. With the same right-sidebending comparison held, left rotation reaches its assessed barrier at 20 degrees. After a tolerated effort, a trainee records 30 degrees of left rotation, but the group was flexed during that measurement. A repeat measurement in neutral again reaches 20 degrees. The clinician plans to continue a direct comparison that preserves neutral. Which record and next comparison are justified?

A. Record N Sr Rl; retain the original neutral right-sidebending/left-rotation barrier
B. Record N Sl Rr; retain the original neutral right-sidebending/left-rotation barrier
C. Record N Sl Rr; advance the neutral left-rotation barrier using the 30-degree result
D. Record N Sr Rl; advance the neutral left-rotation barrier using the 30-degree result

Work through your selected option, then compare alternatives

A. Record N Sr Rl; retain the original neutral right-sidebending/left-rotation barrier
Which sidebending direction was freer in neutral?
Left.
Which rotation direction was freer in neutral?
Right.
Which notation represents that ease?
N Sl Rr.
Read the complete explanation
The repeat neutral measurement supports retaining the original barrier. The notation reverses the examined ease: right sidebending and left rotation describe the restricted comparison, not the named freer position.
This option does not fit the supplied findings.
B. Record N Sl Rr; retain the original neutral right-sidebending/left-rotation barrier
Which notation follows from the documented neutral ease?
N Sl Rr.
Which condition changed during the 30-degree measurement?
The group was flexed.
What did the matched neutral retest show?
The original 20-degree limit.
Which neutral barrier remains supported for the next direct comparison?
The original right-sidebending/left-rotation barrier.
Read the complete explanation
The notation follows the examined neutral ease. The apparent additional rotation was measured under a different flexion-extension condition, and the matched neutral retest shows no gain on which to base a farther neutral barrier.
Best answer.
C. Record N Sl Rr; advance the neutral left-rotation barrier using the 30-degree result
What makes the 30-degree result initially tempting?
It exceeds the original rotation measurement.
Which examination condition was different?
Flexion replaced neutral.
What measurement should govern the specified neutral comparison?
The repeated 20-degree neutral limit.
Read the complete explanation
The notation is correct. The larger measurement cannot establish a new neutral barrier because it was obtained in flexion and was not reproduced when neutral was restored.
This option does not fit the supplied findings.
D. Record N Sr Rl; advance the neutral left-rotation barrier using the 30-degree result
What does N Sr Rl describe relative to the original findings?
The restricted directional comparison.
In which condition was 30 degrees measured?
Flexion.
What happened when the required neutral condition was restored?
The limit returned to 20 degrees.
Read the complete explanation
This names the restricted directions as ease and transfers a flexed measurement to neutral. Both the original directional examination and the matched neutral retest oppose that record.
This option does not fit the supplied findings.

Takeaway: Name the examined ease, then require comparable positioning before accepting a new neutral barrier.

Case sources: [6] [7] [18] [23]

Case 11

Work through your selected option, then compare alternatives

A. P resumes repeated cycling; Q resumes repeated cycling
Which obstruction does the intervention remove in P?
Covered actin sites.
Were Q's actin sites obstructed before the intervention?
No.
Which required process remains unavailable in Q?
ATP-dependent resetting through hydrolysis.
Read the complete explanation
Opening actin sites removes P's demonstrated obstruction. Q's sites were already open, so the intervention does not repair the hydrolysis defect needed to reset its detached heads.
This option does not fit the supplied findings.
B. P remains unable to attach; Q resumes repeated cycling
Which motor process is intact in P?
ATP hydrolysis.
What missing condition does P gain?
Accessible actin sites.
What does the intervention leave unchanged in Q?
Its hydrolysis defect.
Read the complete explanation
This assigns recovery to the preparation whose defect the intervention does not address. P retains motor hydrolysis and gains site access; Q gains no missing nucleotide reaction.
This option does not fit the supplied findings.
C. P remains unable to attach; Q remains unable to reset
What originally prevented P from attaching?
Tropomyosin covered the sites.
What happens to that obstruction after the intervention?
It is removed.
What intact process prepares P's heads for another attachment?
ATP hydrolysis.
Read the complete explanation
The prediction for Q fits its persistent motor defect. P's continuing failure to attach does not follow once its measured regulatory obstruction is removed and its motor reactions remain intact.
This option does not fit the supplied findings.
D. P resumes repeated cycling; Q remains unable to reset
Which requirement is missing in P before the intervention?
Actin-site access.
Which requirement is missing in Q?
ATP hydrolysis.
Which missing requirement does moving tropomyosin restore?
Actin-site access.
Which preparation consequently has both access and an intact reset?
P.
Read the complete explanation
P has an intact motor cycle but lacks actin access, so moving tropomyosin addresses its limiting defect. Q already has access; its inability to hydrolyze ATP persists and prevents the normal reset after detachment.
Best answer.

Takeaway: A rescue must address the failed step: restoring site access cannot substitute for restoring motor resetting.

Case sources: [2] [4]

Case 12

Work through your selected option, then compare alternatives

A. Earlier actin-site closure with faster mechanical relaxation
What normally permits actin sites to become covered after stimulation?
Removal of regulatory calcium.
Which calcium process is inhibited?
Return to the sarcoplasmic reticulum.
What direction of relaxation change follows?
Relaxation is delayed.
Read the complete explanation
Faster removal of regulatory calcium could produce this pattern. The intervention blocks its return to the sarcoplasmic reticulum, so it prolongs rather than abbreviates calcium-dependent site availability.
This option does not fit the supplied findings.
B. Prolonged actin-site availability with delayed mechanical relaxation
What signal persists when calcium return is blocked?
Elevated cytosolic calcium.
What regulatory state can that signal maintain?
Calcium-bound troponin.
What happens to actin-site accessibility?
It remains prolonged.
What can the intact ATP-supported motor cycle consequently do?
Continue after the stimulus ends.
Read the complete explanation
Inhibited calcium return can maintain troponin activation after the electrical stimulus ends. With ATP and motor reactions intact, accessible sites can support continued cycling and delay relaxation.
Best answer.
C. Prolonged actin-site availability with ATP-depletion attachment
Which part fits the blocked calcium return?
Prolonged site availability.
What nucleotide condition would cause ATP-depletion attachment?
Unavailable ATP.
What nucleotide condition is explicitly supplied here?
ATP remains available.
Read the complete explanation
Persistent calcium makes prolonged site availability plausible. ATP-depletion attachment describes a different cause of failed relaxation; the stem preserves ATP availability and normal nucleotide reactions.
This option does not fit the supplied findings.
D. Earlier actin-site closure with persistent cycling at covered sites
What does the inhibited pump delay?
Calcium removal.
What calcium-dependent condition permits normal attachment?
Accessible actin sites.
Would earlier site covering support the proposed persistent normal attachment?
No.
Read the complete explanation
This separates the motor from the regulatory permission it needs. The inhibited pump does not accelerate site closure, and covered sites would not support the proposed continued normal attachment sequence.
This option does not fit the supplied findings.

Takeaway: Ending the electrical stimulus does not end contraction until regulatory calcium is removed.

Case sources: [2] [4]

Case 13

In an author-created calibration, at matched initial muscle length and activation, a preparation shortens against 10 N at 8 length units per second and against 20 N at 3 length units per second. A simplified lever now couples that preparation to a 10-N external load. The muscle moment arm is 0.04 m. Setup S uses an external moment arm of 0.04 m; setup L uses 0.08 m. Compare steady-motion portions with balanced moments, negligible other moments, and the same muscle-length and activation conditions as the calibration. Which prediction follows?

A. Both setups produce concentric activity; L shortens more slowly
B. Both setups produce concentric activity; S shortens more slowly
C. S produces concentric activity; L produces eccentric activity
D. S produces concentric activity; L produces isometric activity

Work through your selected option, then compare alternatives

A. Both setups produce concentric activity; L shortens more slowly
What muscle force balances setup S?
10 N.
What muscle force balances setup L?
20 N.
What length direction occurs at both calibrated forces?
Shortening.
Which setup matches the slower calibrated condition?
L.
Read the complete explanation
The lever requires 10 N of muscle force in S and 20 N in L. Both calibrated loads permit shortening, but the 20-N condition shortens more slowly. The same external load therefore produces different muscle behavior through different leverage.
Best answer.
B. Both setups produce concentric activity; S shortens more slowly
Which setup has the larger external moment?
L.
Which setup therefore requires more muscle force?
L.
Which calibrated force corresponds to slower shortening?
20 N.
Read the complete explanation
Both setups map to measured shortening conditions. The speed ranking is reversed: the longer external arm increases the opposing muscle-force requirement, placing L at the slower calibration point.
This option does not fit the supplied findings.
C. S produces concentric activity; L produces eccentric activity
What muscle force does L require?
20 N.
What measured length direction occurs at 20 N?
Shortening.
What observation would instead establish eccentric activity?
Active lengthening.
Read the complete explanation
The longer arm makes L more demanding, but its required force maps to a supplied condition that still shortens. Increased effective load alone does not establish reversal into active lengthening.
This option does not fit the supplied findings.
D. S produces concentric activity; L produces isometric activity
Which calibration point applies to L?
The 20-N point.
What is its measured shortening speed?
3 length units per second.
What speed would a fixed-length muscle condition require?
Zero.
Read the complete explanation
L's higher effective load slows shortening in the calibration. It does not stop shortening: the reported 20-N velocity remains nonzero.
This option does not fit the supplied findings.

Takeaway: Translate external leverage into the muscle's opposing load before applying a measured load–velocity relationship.

Case sources: [3] [9] [15]

Case 14

Work through your selected option, then compare alternatives

A. 45 N
Which quantity must balance at the stationary joint?
Moment.
Are the muscle and external moment arms equal?
No.
What external moment must the muscle balance?
9 N·m.
Read the complete explanation
This matches the new external force but ignores leverage. The muscle acts through a shorter moment arm and must provide the same opposing moment, not the same numerical force.
This option does not fit the supplied findings.
B. 150 N
What was the original required muscle force?
225 N.
Which change alone would reduce it to 150 N?
Reducing the external arm from 0.30 to 0.20 m.
Which simultaneous change must also be included?
Increasing external force from 30 to 45 N.
Read the complete explanation
This follows from applying only the external moment-arm reduction to the original muscle force. The simultaneous increase in external force offsets that reduction.
This option does not fit the supplied findings.
C. 225 N
What was the original external moment?
9 N·m.
What is the new external moment?
9 N·m.
What muscle force produces that moment through 0.04 m?
225 N.
What explains the unchanged required muscle force?
The external force–arm product is unchanged.
Read the complete explanation
The new external moment is 45 × 0.20 = 9 N·m. The required muscle force is 9 ÷ 0.04 = 225 N. The larger external force and shorter external arm leave their product unchanged.
Best answer.
D. 337.5 N
By what factor does external force increase?
1.5.
Which omitted quantity decreases at the same time?
The external moment arm.
What is the resulting external moment after both changes?
9 N·m.
Read the complete explanation
This applies the external-force increase to the original muscle force while retaining the old leverage. The external moment arm also decreases, offsetting the force increase.
This option does not fit the supplied findings.

Takeaway: Combine simultaneous changes in force and perpendicular leverage before calculating the balancing muscle force.

Case sources: [15]

Case 15

A cleared, cooperative patient's examination finds C4 most comfortable in flexion with right rotation and right sidebending; testing confirms restriction toward the opposite position in each plane. A trainee establishes extension with left rotation but leaves right sidebending, then requests a left-rotation effort. The intended technique is a direct setup addressing all three planes, followed by an isometric rotational effort toward the examined ease. Which corrected position and effort satisfy that plan?

A. Extension, left rotation, right sidebending; request right rotation against the maintained position
B. Extension, left rotation, left sidebending; request right rotation against the maintained position
C. Extension, left rotation, left sidebending; request left rotation against the maintained position
D. Extension, left rotation, right sidebending; request left rotation against the maintained position

Work through your selected option, then compare alternatives

A. Extension, left rotation, right sidebending; request right rotation against the maintained position
Which sidebending direction was examined as ease?
Right sidebending.
Which sidebending direction should the direct setup approach?
Left sidebending.
Which part of this option already matches the intended effort?
Right rotation.
Read the complete explanation
The requested effort correctly returns toward rotational ease. The setup still leaves sidebending in its freer direction, so it does not address the full C4 restriction.
This option does not fit the supplied findings.
B. Extension, left rotation, left sidebending; request right rotation against the maintained position
Which flexion-extension component belongs in the direct setup?
Extension.
Which rotational component belongs in the direct setup?
Left rotation.
Which sidebending component corrects the trainee's omission?
Left sidebending.
Which rotational effort returns toward the examined ease?
Right rotation.
What must happen to the position during the isometric effort?
It must remain maintained.
Read the complete explanation
The full direct setup approaches the opposite of the examined ease in all three planes. The subsequent rotational effort has a different purpose: the patient tries to rotate toward rightward ease while the clinician prevents movement.
Best answer.
C. Extension, left rotation, left sidebending; request left rotation against the maintained position
Which direction is used for rotational positioning?
Left rotation.
Which direction was examined as rotational ease?
Right rotation.
Which direction does the intended effort therefore require?
Right rotation.
Read the complete explanation
All three positioning components address the restriction. The requested left-rotation effort repeats the barrier direction rather than following the specified effort toward right-rotation ease.
This option does not fit the supplied findings.
D. Extension, left rotation, right sidebending; request left rotation against the maintained position
Which positioning plane remains uncorrected?
Sidebending.
Which sidebending direction should replace it?
Left sidebending.
Which rotational effort should replace left rotation?
Right rotation.
Read the complete explanation
This retains both trainee errors. Right sidebending leaves one plane in ease, while left-rotation effort follows the barrier rather than the requested rotational ease.
This option does not fit the supplied findings.

Takeaway: A complete C4 setup accounts for every plane, then distinguishes maintained positioning from the subsequent effort.

Case sources: [6] [7] [23]

Case 16

A hypothetical anatomy-based neck model isolates the left sternocleidomastoid (SCM). The torso is fixed, and neck flexion-extension and sidebending are held constant while rotation is studied. An active rotational effort first holds the head stationary. An external controller then moves the head slightly left while left-SCM activation continues. In separate matched trials, the same active muscle turns the head right through the same small angular range against two opposing loads. The second opposing load is larger, but both allow shortening. Muscle activation, starting length, moment arms, and fatigue state are matched; other muscle moments and changing tendon compliance are excluded, and speed is not controlled. Which action sequence fits the first demonstration, and how should rightward shortening speed compare under the larger load? This is a teaching model, not a neck-treatment prescription.

A. Isometric then concentric; rightward shortening is slower under the larger load.
B. Isometric then eccentric; rightward shortening is faster under the larger load.
C. Isometric then eccentric; rightward shortening is slower under the larger load.
D. Isometric then concentric; rightward shortening is faster under the larger load.

Work through your selected option, then compare alternatives

A. Isometric then concentric; rightward shortening is slower under the larger load.
Which rotation does unilateral left SCM contribute?
Right rotation.
How does controlled left rotation affect its length?
It increases its length.
What classification fits that active length change?
Eccentric.
Read the complete explanation
The slower-speed prediction follows the shortening force-velocity relationship. The leftward phase is misclassified: left SCM contributes right rotation, so controlled left rotation lengthens it while it remains active.
This option does not fit the supplied findings.
B. Isometric then eccentric; rightward shortening is faster under the larger load.
Which variable is matched rather than increased between the load trials?
Muscle activation.
Which mechanical variable is increased?
The opposing load.
What speed change is expected during shortening under these controls?
Slower shortening.
Read the complete explanation
The stationary and leftward phases are classified correctly. Greater resistance does not imply greater speed when activation and starting length are matched. In the specified shortening range, a larger opposing load generally reduces shortening velocity.
This option does not fit the supplied findings.
C. Isometric then eccentric; rightward shortening is slower under the larger load.
Which rotation does left SCM contribute?
Right rotation.
What does the controller's leftward motion do to that active muscle?
It lengthens it.
What length change accompanies the separate rightward trials?
Shortening.
How does the larger opposing load affect shortening speed under the matched conditions?
It decreases the speed.
Which full prediction follows?
Isometric then eccentric activity, with slower rightward shortening under the larger load.
Read the complete explanation
Left SCM contributes right rotation and left sidebending. With sidebending and the other planes controlled, imposed left rotation lengthens the active muscle after its external isometric hold. The separate rightward trials involve shortening. At matched activation and length, the larger opposing load generally slows that shortening; no speed controller is present to enforce equal velocities. The stationary label alone does not determine either later result.
Best answer.
D. Isometric then concentric; rightward shortening is faster under the larger load.
Which rotation is produced by left SCM?
Right rotation.
What length change follows imposed left rotation?
Lengthening.
Does the second shortening trial include increased activation?
No.
What does the increased opposing load predict at matched activation?
Slower shortening.
Read the complete explanation
This treats the controller's movement as if it followed the active muscle's action and treats a heavier load as if activation had also increased. Left SCM is lengthened by the leftward motion, and the separate load comparison holds activation constant.
This option does not fit the supplied findings.

Takeaway: For left SCM, controlled rotation determines active length change; a separate matched load comparison determines the expected shortening-speed change.

Case sources: [3] [9] [11]

Case 17

A clinically assessed hamstring restriction is tested after full relaxation at a standardized passive torque. With the hip fixed at 90 degrees of flexion, knee extension stops at 40 degrees of remaining knee flexion. After a tolerated target-muscle effort, a trainee finds 25 degrees of remaining knee flexion, but the hip has moved to 70 degrees. Restoring the hip to 90 degrees reproduces the original 40-degree limit at the same passive torque. Following reassessment, the clinician elects a gentle antagonist-effort trial while maintaining the standardized barrier. Which plan fits?

A. Use the 25-degree knee-flexion position at the standardized hip; request knee extension
B. Use the 40-degree knee-flexion position at the standardized hip; request knee flexion
C. Use the 25-degree knee-flexion position at the standardized hip; request knee flexion
D. Use the 40-degree knee-flexion position at the standardized hip; request knee extension

Work through your selected option, then compare alternatives

A. Use the 25-degree knee-flexion position at the standardized hip; request knee extension
At what hip angle was the 25-degree result obtained?
70 degrees of flexion.
What limit returned at the standardized hip angle?
40 degrees of remaining knee flexion.
Which measurement supports the planned standardized barrier?
The 40-degree result.
Read the complete explanation
Knee extension recruits the antagonist action to the hamstring target. The proposed position imports a measurement obtained at a different hip angle and exceeds the barrier reproduced under the required standardized conditions.
This option does not fit the supplied findings.
B. Use the 40-degree knee-flexion position at the standardized hip; request knee flexion
Which knee action do the hamstrings produce?
Flexion.
Which role did the clinician select for the next effort?
The antagonist.
Which knee effort serves that role?
Extension.
Read the complete explanation
The positioning uses the comparable retest. Knee flexion, however, requests the hamstring target's own action rather than the newly selected antagonist effort.
This option does not fit the supplied findings.
C. Use the 25-degree knee-flexion position at the standardized hip; request knee flexion
What happened to the apparent gain when hip position was restored?
It disappeared.
Which muscle role does knee flexion recruit here?
The hamstring target.
Which role is requested for the new trial?
The antagonist.
Read the complete explanation
The larger apparent extension range came from a changed hip condition, and flexion effort repeats the target-muscle approach. Neither component follows the matched retest and newly selected antagonist plan.
This option does not fit the supplied findings.
D. Use the 40-degree knee-flexion position at the standardized hip; request knee extension
Which retest preserves the original examination conditions?
The hip-at-90-degrees retest.
What knee limit does that retest support?
40 degrees of remaining flexion.
Which knee action belongs to the hamstring target?
Flexion.
Which effort therefore recruits the antagonist action?
Extension.
Read the complete explanation
The matched retest supports the original barrier rather than a gain measured at another hip angle. Because the target hamstrings flex the knee, a maintained knee-extension effort supplies the selected antagonist action.
Best answer.

Takeaway: Use matched reassessment conditions to locate the barrier, then select effort according to the intended muscle role.

Case sources: [7] [12]

Case 18

A hypothetical knee-reflex model samples a receptor around small specialized fibers arranged parallel to ordinary quadriceps fibers. Perturbation P stretches the specialized fibers while tendon force is fixed. Perturbation Q raises tendon force while their length is fixed. Their motor supply is unchanged. In a separate circuit trial, the hip and knee are fixed and the receptor's primary afferent burst is replayed. Before the burst, quadriceps and hamstrings each exert a 2 N m knee moment. With the intact circuit, the burst changes the quadriceps moment to 3 N m and the hamstring moment to 1 N m. The identical burst is then replayed after selectively disabling the inhibitory relay to hamstrings; the direct pathway to quadriceps remains intact. Assume all other inputs and each unaffected response stay the same, and neglect other moments. Which perturbation should increase the sampled receptor's firing, and what net muscle moment should the final replay produce? All numerical responses are hypothetical.

A. Q raises firing; the final replay produces 1 N m of extension moment.
B. P raises firing; the final replay produces 1 N m of extension moment.
C. P raises firing; the final replay produces 2 N m of extension moment.
D. Q raises firing; the final replay produces 2 N m of extension moment.

Work through your selected option, then compare alternatives

A. Q raises firing; the final replay produces 1 N m of extension moment.
Which variable changes in Q?
Tendon force.
Which input to the sampled receptor is unchanged in Q?
Specialized-fiber length.
Which perturbation instead increases its stretch?
P.
Read the complete explanation
The moment prediction correctly removes the hamstring inhibitory effect. Q, however, changes tendon force without stretching the sampled specialized fibers. The parallel-fiber anatomy identifies a spindle rather than a tendon organ.
This option does not fit the supplied findings.
B. P raises firing; the final replay produces 1 N m of extension moment.
Which perturbation stretches the structurally identified spindle?
P.
What quadriceps moment remains with its pathway intact?
3 N m of extension moment.
What hamstring moment remains without the inhibitory effect?
Its baseline 2 N m of flexion moment.
What net muscle moment follows from those opposing moments?
1 N m of extension moment.
Which paired prediction follows?
P raises firing, and the final replay produces 1 N m of extension moment.
Read the complete explanation
The parallel fibers identify a spindle with Ia/II afferents, so P tests its stretch input. Disabling the inhibitory branch leaves the hamstring flexion moment at its 2 N m baseline. The unaffected quadriceps response still supplies 3 N m of extension moment, leaving 1 N m of net extension. This applies [source 8's receptor anatomy](https://www.ncbi.nlm.nih.gov/books/NBK10812/) and [source 16's circuit connections](https://www.ncbi.nlm.nih.gov/books/NBK10809/) to hypothetical moment measurements.
Best answer.
C. P raises firing; the final replay produces 2 N m of extension moment.
Which hamstring moment occurred with the intact circuit?
1 N m of flexion moment.
What hamstring moment applies after its inhibitory relay is disabled?
2 N m of flexion moment.
What net moment remains against 3 N m from quadriceps?
1 N m of extension moment.
Read the complete explanation
P is the correct stimulus. The 2 N m answer carries the intact-circuit hamstring response into the altered trial. Once its inhibitory relay is disabled, the hamstring moment stays at baseline rather than falling to 1 N m.
This option does not fit the supplied findings.
D. Q raises firing; the final replay produces 2 N m of extension moment.
Which perturbation increases specialized-fiber length?
P.
What hamstring moment remains after removing its inhibitory input?
2 N m of flexion moment.
What quadriceps moment remains through the unaffected pathway?
3 N m of extension moment.
What net moment follows?
1 N m of extension moment.
Read the complete explanation
This selects the tendon-loading stimulus and retains the intact-circuit subtraction. The sampled receptor is stretched by P, and the altered circuit no longer reduces the opposing hamstring moment.
This option does not fit the supplied findings.

Takeaway: Identify the receptor's mechanical input, then account for each surviving muscle response before predicting the joint moment.

Case sources: [8] [12] [15] [16]

Case 19

A hypothetical MET teaching model represents a clinician-selected right-biceps target and an appropriate, painless triceps effort. During an initial stationary hold, recordings show ongoing activation of both muscles; the biceps signal has decreased from baseline but has not disappeared. The model assigns active biceps tension of 12 N through a 0.04 m flexion moment arm and active triceps tension of 15 N through a 0.02 m extension moment arm. A clinician support supplies whatever additional moment keeps the elbow stationary. The support is then removed. For the first small movement, assume these active tensions and moment arms persist. The shoulder, forearm rotation, and wrist remain fixed; ignore gravity, passive moments, other muscles, and changing series compliance. Which motion begins, and how does the selected biceps target behave? The forces are model inputs, not values inferred from the activation recordings or patient force estimates.

A. Elbow extension begins; the biceps lengthens actively.
B. Elbow flexion begins; the biceps shortens passively.
C. Elbow flexion begins; the biceps shortens actively.
D. Elbow extension begins; the biceps lengthens passively.

Work through your selected option, then compare alternatives

A. Elbow extension begins; the biceps lengthens actively.
What extension moment does triceps produce?
0.30 N m.
What flexion moment does biceps produce?
0.48 N m.
Which moment dominates after support removal?
The flexion moment.
What motion therefore begins?
Elbow flexion.
Read the complete explanation
Continued biceps activation is recognized correctly. Extension is tempting because the triceps force is numerically larger, but its shorter arm produces the smaller moment. Joint motion follows the net moment rather than the larger force alone.
This option does not fit the supplied findings.
B. Elbow flexion begins; the biceps shortens passively.
Does reduced biceps activation mean its signal disappeared here?
No.
What kind of biceps tension does the model retain after support removal?
Active tension.
How should shortening with that tension be classified?
Concentric activity.
Read the complete explanation
The predicted motion follows the larger biceps moment. Calling its shortening passive mistakes reduced activation during an antagonist effort for complete target silence. The recordings and maintained active-tension assumption show that biceps still generates force.
This option does not fit the supplied findings.
C. Elbow flexion begins; the biceps shortens actively.
What flexion moment does biceps produce?
0.48 N m.
What extension moment does triceps produce?
0.30 N m.
Which direction does the unsupported net moment accelerate the elbow?
Flexion.
What does flexion do to the still-active biceps with the shoulder fixed?
It shortens it.
Which combined prediction follows?
Elbow flexion with concentric biceps activity.
Read the complete explanation
Biceps supplies 12 x 0.04 = 0.48 N m of flexion moment, exceeding triceps' 15 x 0.02 = 0.30 N m of extension moment. Removing the balancing support leaves a 0.18 N m flexion moment. Biceps therefore shortens while still active, so its action is concentric. The active triceps, although recruited as the antagonist, is lengthened and acts eccentrically. The initial hold was externally isometric; an antagonist effort did not guarantee target silence or determine the subsequent movement direction.
Best answer.
D. Elbow extension begins; the biceps lengthens passively.
Which muscle produces the larger joint moment despite its smaller force?
Biceps.
Which motion follows support removal?
Elbow flexion.
What observation excludes a passive-only biceps response?
Its activation continues.
How does the active biceps behave during that motion?
It shortens concentrically.
Read the complete explanation
This assumes both that the larger triceps force controls motion and that antagonist recruitment silences the target. The different moment arms reverse the force-only ranking, and the biceps remains active in the supplied recordings and model assumptions.
This option does not fit the supplied findings.

Takeaway: An antagonist effort can coexist with target activation; after support changes, net joint moments determine which active muscle shortens and which lengthens.

Case sources: [3] [6] [13] [15] [17]

Case 20

Clinical assessment supports a shortened right gluteus medius pattern, excludes contraindications, and confirms restricted adduction with the pelvis controlled. The clinician has gently positioned the hip at the adduction barrier for a target-muscle isometric effort. A trainee asks the patient to push the thigh inward. The clinician also moves the resisting contact farther down the limb, increasing its perpendicular hip moment arm from 0.20 to 0.40 m. In a simplified comparison, the desired resisting hip moment is unchanged and other moments are unchanged. Which corrected effort and resistance adjustment fit?

A. Request an outward abduction effort; halve the clinician's resisting force
B. Request an outward abduction effort; double the clinician's resisting force
C. Request an inward adduction effort; halve the clinician's resisting force
D. Request an inward adduction effort; double the clinician's resisting force

Work through your selected option, then compare alternatives

A. Request an outward abduction effort; halve the clinician's resisting force
Which hip action does gluteus medius produce?
Abduction.
Which effort corrects the trainee's inward instruction?
An outward effort.
What happens to the clinician's perpendicular moment arm?
It doubles.
What force change preserves the same resisting moment?
The force is halved.
Read the complete explanation
Gluteus medius is the selected abductor, so outward effort recruits the target while the position is maintained. Doubling the clinician's moment arm requires half the force to preserve the same resisting hip moment.
Best answer.
B. Request an outward abduction effort; double the clinician's resisting force
Which part correctly recruits gluteus medius?
The abduction effort.
What happens to moment if force stays constant while its arm doubles?
Moment doubles.
What force adjustment instead preserves moment?
Halving force.
Read the complete explanation
The effort recruits the correct target. Moving the contact farther from the hip increases leverage, so doubling force would raise rather than preserve the intended resisting moment.
This option does not fit the supplied findings.
C. Request an inward adduction effort; halve the clinician's resisting force
Which direction defines the current barrier?
Adduction.
Which action belongs to the selected target muscle?
Abduction.
What does an inward effort request?
Adduction.
Read the complete explanation
The force adjustment correctly accounts for the longer moment arm. The inward effort repeats the barrier direction but recruits an opposing action rather than the selected gluteus medius target.
This option does not fit the supplied findings.
D. Request an inward adduction effort; double the clinician's resisting force
Which effort would recruit the gluteus medius target?
Abduction.
What happens to the resistance moment arm?
It doubles.
What force change is needed for an unchanged resisting moment?
A reduction to one half.
Read the complete explanation
The inward effort confuses positioning with target-muscle action. The increased force also compounds the increased leverage instead of preserving the intended hip moment.
This option does not fit the supplied findings.

Takeaway: Select gluteus medius effort from its abduction action, and adjust clinician force when contact leverage changes.

Case sources: [6] [7] [12] [15] [23]

Case 21

Work through your selected option, then compare alternatives

A. Group Ia/II signaling should increase because the required tendon tension rises
Which receptor matches the supplied structure?
A Golgi tendon organ.
Which afferent class serves that receptor?
Group Ib.
Which part of the proposed mechanical prediction is correct?
Required tendon tension rises.
Read the complete explanation
The moment change does require more muscle and tendon force. Ia/II identifies spindle signaling, however, whereas the supplied collagen-in-series structure identifies a tendon organ.
This option does not fit the supplied findings.
B. Group Ib signaling should remain unchanged because the joint angle is fixed
What happens to external moment when its arm doubles?
It doubles.
What happens to required muscle force with its arm unchanged?
It doubles.
Which changing variable is relevant to the tendon organ?
Tension.
Read the complete explanation
The afferent identification is correct. A fixed joint does not fix muscle force when external leverage changes; the tendon organ's tension-related input can rise without joint movement.
This option does not fit the supplied findings.
C. Group Ia/II signaling should remain unchanged because the joint angle is fixed
Which supplied measurement might suggest unchanged length-related input?
Fixed fascicle length.
Which structural feature identifies the actual sampled receptor?
Its position among tendon collagen in series.
What happens to required tension despite the fixed length?
It increases.
Read the complete explanation
The unchanged fascicle length may make a length-sensor interpretation tempting. The sampled structure is a tendon organ, and its force-transmitting pathway experiences increased required tension.
This option does not fit the supplied findings.
D. Group Ib signaling should increase because the required tendon tension rises
Which receptor is identified by the tendon-collagen series arrangement?
A Golgi tendon organ.
Which afferent class carries its signal?
Group Ib.
What happens to the required opposing moment?
It doubles.
What happens to muscle force with its moment arm unchanged?
It doubles.
What qualitative tendon-organ signaling change follows?
An increase related to greater tension.
Read the complete explanation
The structure identifies a Golgi tendon organ and its Ib afferent. The doubled external moment requires doubled opposing muscle force with unchanged muscle leverage, so the tension-related signal is expected to increase despite fixed position. No proportional doubling of discharge is implied.
Best answer.

Takeaway: A stationary joint can generate changing tendon-organ input when leverage changes the required force.

Case sources: [8] [15]

Case 22

A hypothetical follow-up to the lesson's healthy-volunteer flexibility comparison uses degrees of knee extension from a common starting position. Before the intervention, a relaxed participant reaches 40 degrees at 6 N·m, which is also the tolerated endpoint. Immediately afterward, the participant still reaches 40 degrees at 6 N·m but tolerates 50 degrees at 8 N·m. One week later, the tolerated endpoint remains 50 degrees. Which additional one-week observation would most directly challenge a tolerance-only explanation for that maintained endpoint?

A. The relaxed knee reaches 50 degrees at 8 N·m with the original hip position maintained
B. The relaxed knee reaches 50 degrees at 6 N·m with the original hip position maintained
C. The relaxed knee reaches 50 degrees at 6 N·m with the hip held less flexed than originally
D. The knee reaches 50 degrees at 6 N·m with the original hip position and ongoing quadriceps activity

Work through your selected option, then compare alternatives

A. The relaxed knee reaches 50 degrees at 8 N·m with the original hip position maintained
What torque was used at the original 40-degree endpoint?
6 N·m.
What differs at this proposed 50-degree measurement?
Applied torque is higher.
Which explanation remains compatible with that difference?
Increased stretch tolerance.
Read the complete explanation
This reproduces the larger tolerated endpoint under the larger torque. It shows persistence of that endpoint but does not demonstrate a changed passive angle response to the original loading condition.
This option does not fit the supplied findings.
B. The relaxed knee reaches 50 degrees at 6 N·m with the original hip position maintained
Did the immediate range increase at the original torque?
No.
Which immediate change allowed the larger endpoint?
The participant tolerated a higher torque.
What does the proposed one-week test hold constant?
The original passive loading condition.
What new result would weaken the tolerance-only explanation?
More relaxed range at the same torque.
What does the one-week timing add?
Evidence that the changed response persists to that measurement.
Read the complete explanation
The immediate findings showed a larger tolerated endpoint without a range gain at the original torque. A later gain at that original torque, with relaxation and geometry matched, would supply evidence of a changed passive mechanical response. It would not by itself identify a particular structural tissue change.
Best answer.
C. The relaxed knee reaches 50 degrees at 6 N·m with the hip held less flexed than originally
Which loading variable is appropriately matched?
Applied torque.
Which relevant geometric variable changes?
Hip flexion.
What comparison is needed to interpret the extra range mechanically?
A retest at the original hip position.
Read the complete explanation
Matching torque is useful, but changing hip position alters the hamstring examination geometry. The extra knee extension could reflect that changed condition rather than an altered passive response in the original setup.
This option does not fit the supplied findings.
D. The knee reaches 50 degrees at 6 N·m with the original hip position and ongoing quadriceps activity
Which muscle activity is present during the proposed test?
Quadriceps activity.
Which knee action can that activity contribute?
Extension.
Which condition is therefore missing from the passive comparison?
Full muscle relaxation.
Read the complete explanation
Torque and hip position are matched, but active knee extension can contribute to the measured motion. The result would not isolate the relaxed passive response needed to distinguish mechanics from tolerance.
This option does not fit the supplied findings.

Takeaway: A maintained tolerated endpoint and a maintained change in matched-load passive motion are different findings.

Case sources: [3] [5] [12]

Case 23

In an assessed, painless rib-region demonstration, trial A uses a fixed clinician-established fulcrum while the patient voluntarily produces the respiratory excursion. In trial B, the patient does not generate the excursion and the clinician reproduces the rib movement passively. After A, the patient takes a deeper breath and shows a larger active excursion. However, gentle passive reassessment at the same quiet end-expiratory point and the same applied force finds the original limit unchanged. Which interpretation accounts for both the active contribution and the reassessment?

A. A provides respiratory muscle activity; the matched passive barrier has advanced
B. B provides respiratory muscle activity; the matched passive barrier is unchanged
C. A provides respiratory muscle activity; the matched passive barrier is unchanged
D. B provides respiratory muscle activity; the matched passive barrier has advanced

Work through your selected option, then compare alternatives

A. A provides respiratory muscle activity; the matched passive barrier has advanced
What changed during the larger active excursion?
Breath depth.
Which assessment preserved the original measurement conditions?
The passive end-expiratory reassessment.
What happened to the limit in that matched assessment?
It remained unchanged.
Read the complete explanation
A correctly identifies the patient-generated contribution. The larger excursion occurred during a deeper active breath, while the matched passive examination did not show an advanced barrier.
This option does not fit the supplied findings.
B. B provides respiratory muscle activity; the matched passive barrier is unchanged
Who generates the excursion in B?
The clinician.
Who generates the excursion in A?
The patient.
Which trial therefore contains the specified respiratory contribution?
A.
Read the complete explanation
The passive-barrier conclusion fits the repeat examination. In B, however, the clinician supplies the excursion; observing rib movement does not identify a patient-generated respiratory contribution.
This option does not fit the supplied findings.
C. A provides respiratory muscle activity; the matched passive barrier is unchanged
Who supplies the motion in A?
The patient.
Which active process supplies it?
Respiratory muscle activity.
What condition differs during the later larger excursion?
Breath depth.
What does the matched passive reassessment establish?
The original passive limit remains.
Read the complete explanation
A coordinates patient-generated breathing with the assessed setup, providing the active contribution. The larger subsequent breath changes the active task, but the matched passive reassessment supplies no measured barrier gain.
Best answer.
D. B provides respiratory muscle activity; the matched passive barrier has advanced
What is the source of B's rib excursion?
Passive clinician movement.
Under what changed task was the larger later excursion observed?
A deeper active breath.
Which finding directly measures the matched passive response?
The unchanged end-expiratory limit.
Read the complete explanation
This treats clinician-produced movement as the active respiratory contribution and treats a deeper active breath as passive improvement. The supplied motion source and controlled reassessment distinguish both interpretations.
This option does not fit the supplied findings.

Takeaway: Identify who produces the motion, then compare outcomes under the same respiratory and loading conditions.

Case sources: [6] [7] [21]

Case 24

A hypothetical extension of the lesson's gaze experiment records neck-muscle EMG in four matched conditions. During right head rotation, left gaze gives 12 units and right gaze gives 20. During left head rotation, left gaze gives 18 units and right gaze gives 10. Within each head-rotation direction, posture, angular speed, and external resistance are matched. Investigators want to replicate the gaze-related pattern while separating it from the effect of changing head-motion direction. Which prediction and comparison fit?

A. The gaze effect reverses with head direction; compare gaze directions within each matched head-motion task
B. Right gaze raises EMG in both head directions; compare gaze directions within each matched head-motion task
C. The gaze effect reverses with head direction; compare right-rotation/right-gaze with left-rotation/left-gaze
D. Right gaze raises EMG in both head directions; compare right-rotation/right-gaze with left-rotation/left-gaze

Work through your selected option, then compare alternatives

A. The gaze effect reverses with head direction; compare gaze directions within each matched head-motion task
What happens with right rather than left gaze during right head rotation?
EMG increases.
What happens with right rather than left gaze during left head rotation?
EMG decreases.
What pattern relates gaze effect to head-motion direction?
The direction of the gaze effect reverses.
Which variable must remain matched within each gaze comparison?
Head-motion direction.
Read the complete explanation
Right gaze increases EMG during right rotation but decreases it during left rotation. Replication therefore needs gaze contrasts within each head-motion task; a diagonal comparison would change gaze and head direction together.
Best answer.
B. Right gaze raises EMG in both head directions; compare gaze directions within each matched head-motion task
Which row supports an increase with right gaze?
Right head rotation.
What happens from left to right gaze during left head rotation?
EMG falls from 18 to 10 units.
What does that second row exclude?
An increase with right gaze in both tasks.
Read the complete explanation
The proposed comparison is appropriate. The prediction generalizes from the right-rotation row while overlooking the opposite gaze effect in the left-rotation row.
This option does not fit the supplied findings.
C. The gaze effect reverses with head direction; compare right-rotation/right-gaze with left-rotation/left-gaze
Which gaze condition changes in the proposed comparison?
Right gaze changes to left gaze.
Which movement condition also changes?
Head-rotation direction.
What comparison isolates gaze more directly?
Different gaze directions during the same head-motion task.
Read the complete explanation
The pattern is correctly recognized. The proposed diagonal comparison changes both gaze and head-motion direction, preventing the gaze contribution from being isolated.
This option does not fit the supplied findings.
D. Right gaze raises EMG in both head directions; compare right-rotation/right-gaze with left-rotation/left-gaze
Which supplied condition has 10 EMG units?
Left head rotation with right gaze.
How does that compare with left gaze during the same movement?
It is lower.
What confounds the proposed diagonal comparison?
Head-motion direction changes with gaze.
Read the complete explanation
This misses the decrease with right gaze during left rotation and proposes a comparison that changes two task variables. Neither the complete pattern nor a controlled gaze contrast supports it.
This option does not fit the supplied findings.

Takeaway: Interpret the complete task-dependent EMG pattern and hold head motion constant when testing gaze effects.

Case sources: [17] [19]

Case 25

Using the same 0-to-5 manual strength scale, where 5/5 represents normal strength, before a recent fall, documented examination showed stable left-hand strength of 3/5 from an already assessed old hand injury and right-hand strength of 5/5. Two days after the fall, neck pain has improved from 6/10 to 2/10. Current testing shows 3/5 strength in both hands; the right-sided result is reproduced with the arm supported and without hand pain. The patient asks to begin gentle cervical MET because the hands are now equally strong and the neck feels better. Which finding should govern the next action?

A. Proceed with gentle MET because current left- and right-hand strength scores are equal.
B. Proceed with gentle MET because the neck-pain score has improved since the fall.
C. Defer MET for urgent emergency assessment because the longstanding left-hand weakness remains present.
D. Defer MET for urgent emergency assessment because right-hand strength has newly declined.

Work through your selected option, then compare alternatives

A. Proceed with gentle MET because current left- and right-hand strength scores are equal.
What was right-hand strength before the fall?
5/5.
What is right-hand strength now?
3/5.
What explains the new symmetry?
A decline in right-hand strength.
Read the complete explanation
Current symmetry can seem reassuring if the baseline is ignored. Here symmetry arose because right-hand strength declined to the longstanding left-sided level after trauma.
This option does not fit the supplied findings.
B. Proceed with gentle MET because the neck-pain score has improved since the fall.
Which symptom has improved?
Neck pain.
Which objective finding has worsened from baseline?
Right-hand strength.
Which finding prevents pain improvement from settling treatment eligibility?
The new post-traumatic weakness.
Read the complete explanation
The pain trajectory is favorable, but it does not account for the new reproducible strength deficit. The post-traumatic objective change requires evaluation before selecting a manual technique.
This option does not fit the supplied findings.
C. Defer MET for urgent emergency assessment because the longstanding left-hand weakness remains present.
Which hand had weakness documented before the fall?
The left hand.
Has that hand's recorded strength changed?
No.
Which hand demonstrates the new deficit?
The right hand.
Read the complete explanation
Deferring treatment is appropriate, but this identifies the wrong change. The left deficit was assessed previously and remains stable; the new finding is the right-sided decline.
This option does not fit the supplied findings.
D. Defer MET for urgent emergency assessment because right-hand strength has newly declined.
Which strength finding was already present before trauma?
Left-hand strength of 3/5.
Which side has declined from its documented baseline?
The right side.
What does painless supported repetition establish about the right-sided finding?
It is reproducible.
What takes priority before cervical MET?
Urgent emergency assessment takes priority over manual treatment.
Read the complete explanation
Comparing each side with its own baseline identifies new right-hand weakness despite current symmetry. Its reproducibility after trauma outweighs reassurance from improving pain when deciding whether manual treatment should begin. Urgent emergency assessment is required before manual treatment.
Best answer.

Takeaway: Compare objective findings with baseline; improving pain and current symmetry do not erase a new post-traumatic deficit.

Case sources: [7] [24]

Sources

OpenStax Anatomy and Physiology 2e: Skeletal Muscle
Sarcomere organization and muscle-tendon force transmission; factual reference only, not image reuse.
OpenStax Anatomy and Physiology 2e: Muscle Fiber Contraction and Relaxation
Sliding, calcium regulation, nucleotide sequencing, and additional pump energy requirements.
OpenStax Anatomy and Physiology 2e: Nervous System Control of Muscle Tension
Active shortening, active lengthening, and externally isometric tasks.
OpenStax Biology 2e: Muscle Contraction and Locomotion
Supplementary textbook explanation of filament sliding and ATP binding versus hydrolysis, not independent experimental validation.
Ballantyne, Fryer and McLaughlin: The Effect of Muscle Energy Technique on Hamstring Extensibility (2003)
Primary healthy-volunteer experiment: immediate tolerated range, equal-torque comparison, and limits of the stretch-tolerance interpretation.
AACOM Authorized Osteopathic Thesaurus: Muscle Energy, item 253
Historical definition of directed patient effort against practitioner-defined resistance; not efficacy evidence.
Muscle Energy Procedure: Cervical Vertebrae
Cervical positioning, reassessment, historical isolytic terminology, and contraindication concerns; not sole reflex-physiology evidence.
Purves and colleagues: Mechanoreceptors Specialized for Proprioception (2001)
Spindle anatomy and Ia/II afferents; tendon-organ placement and Ib afferents.
On the Shape of the Force-Velocity Relationship in Skeletal Muscles (2019)
Qualified load-versus-shortening-velocity interpretation, not a universal force inequality at a joint.
Kannus: Isokinetic Evaluation of Muscular Performance (1994-01)
Concentric and eccentric isokinetic testing; measurement and device limitations.
Anatomy: Sternocleidomastoid Muscle
Unilateral contribution to contralateral rotation and ipsilateral inclination.
OpenStax: Muscles of the Pelvic Girdle and Lower Limbs
Hip-flexor, gluteus medius, hamstring and gastrocnemius actions in controlled educational examples.
OpenStax: Muscles of the Pectoral Girdle and Upper Limbs
Biceps, triceps and brachialis actions with attention to multiarticular context.
Narici and colleagues: In Vivo Human Gastrocnemius Architecture During Graded Isometric Contraction (1996-10-01)
Primary human ultrasound evidence of internal architectural change during a fixed-joint calf task; no quantitative extrapolation to other muscles.
OpenStax University Physics: Newton's Second Law for Rotation
Fixed-axis moment balance in a declared simplified model with original numbers.
Purves and colleagues: Spinal Cord Circuitry Underlying Muscle Stretch Reflexes (2001)
Classical spinal circuitry, not proof of a unique clinical MET mechanism or complete antagonist silence.
Nielsen and Kagamihara: Regulation of Disynaptic Reciprocal Ia Inhibition During Co-contraction (1992)
Primary human evidence of task-dependent reciprocal inhibition; not a trial of painful-biceps treatment.
AACOM Thesaurus: Neutral Sidebent Rotated, item 265
Historical notation definitions; educational comparisons explicitly supply directional examination findings.
Bexander, Mellor and Hodges: Effect of Gaze Direction on Neck Muscle Activity During Cervical Rotation (2005)
Primary eleven-subject EMG experiment; no inference of treatment efficacy, safety or a uniquely proven pathway.
Phadke and colleagues: Muscle Energy Technique and Static Stretching in Mechanical Neck Pain
Existence of a region-specific 20-percent-effort protocol alongside conventional care, not a universal optimum or isolated treatment effect.
Osteopathic Manipulative Treatment: Muscle Energy Procedure
Respiratory-assistance terminology and active patient contribution only; broad efficacy, force and reflex assertions elsewhere are not imported.
Purves: Other Afferent Feedback that Affects Motor Performance (2001)
Classical homonymous Ib inhibitory circuitry and its other regulatory inputs; not clinical MET efficacy.
AACOM Authorized Osteopathic Thesaurus: Direct Method, item 90
Definition of direct barrier engagement, not a complete regional treatment protocol or efficacy evidence.
NICE NG41: Spinal Injury, Assessment and Initial Management
Urgent assessment of neurological abnormalities following trauma, rather than provocative manual treatment.

Return to lesson title

A muscle can work while a weight goes down

Shorter unit, same-length filaments

Predict what stays the same

ATP lets myosin release its grip

Follow the grip and the nucleotide

Speed, length, and force answer different questions

Same weight, different leverage

Separate where the clinician positions from where the patient pushes

Read all three components

Which muscle contracts, and what does the evidence actually show?

Keep the target fixed; change the recruited muscle

Apply the mechanism to a new problem

Case 1

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Case 5

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Case 7

Case 8

Case 9

Case 10

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Case 18

Case 19

Case 20

Case 21

Case 22

Case 23

Case 24

Case 25

Sources