101. Four altered meromyosin preparations show the following properties.
| Sample | LMM tail present | Head \(\mathrm{ATP}\)ase active | Actin binding |
|---|
| P | Yes | No | Yes |
| Q | No | Yes | Yes |
| R | No | No | No |
| S | Yes | Yes | No |
Which sample retains its tail region and \(\mathrm{ATP}\)ase activity but cannot attach to actin?
ⓐ. Sample P
ⓑ. Sample Q
ⓒ. Sample R
ⓓ. Sample S
Correct Answer: Sample S
Explanation: The required preparation must satisfy three separate conditions: an LMM tail must be present, the globular head must retain \(\mathrm{ATP}\)ase activity, and actin binding must be absent. Sample S is the only row with that combination. Sample P retains the tail and actin binding but lacks enzymatic activity. Sample Q retains both head functions but lacks the LMM tail, while Sample R lacks all three listed properties. The table demonstrates that tail structure, \(\mathrm{ATP}\) hydrolysis and actin attachment can be evaluated as distinct features of one meromyosin molecule. Preservation of one head function does not automatically establish preservation of its other binding function. With the limiting condition included, each myosin head possesses ATPase activity together with binding sites for ATP and actin. In interpreting the biological outcome, loss of a binding or ATPase site impairs a specific head function even when the rest of the thick filament remains present.
102. Assertion: The cross arms of a thick filament are formed by HMM portions of meromyosin.
Reason: The globular heads and short arms project outward from the filament surface at regular distances and angles.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion
Explanation: HMM consists of the globular head and its short arm. During assembly of the thick filament, these HMM regions project outward from the filament surface instead of remaining buried within the elongated tail region. Their regular spacing and angular arrangement produce the structures described as cross arms. The assertion identifies the molecular component forming the cross arms, and the reason states the projecting arrangement that gives those components their cross-arm appearance and position. LMM forms the tail and does not supply the outward globular projections. The reason provides the structural basis for the assertion rather than merely giving an unrelated property of thick filaments. For a consistent biological interpretation, regular projection of cross arms places functional myosin heads near neighbouring actin sites.
103. A polymerised thick filament is described as having an elongated central portion formed mainly by aligned tail regions. At regular intervals, structures consisting of globular heads and short arms project outward. The two regions are best identified as:
ⓐ. central F-actin and projecting G-actin
ⓑ. central LMM and projecting HMM
ⓒ. central troponin and projecting tropomyosin
ⓓ. central HMM and projecting LMM
Correct Answer: central LMM and projecting HMM
Explanation: The tail of each meromyosin molecule forms LMM and contributes to the elongated structural region of the polymerised thick filament. The globular head and short arm form HMM. These HMM components project outward at regular distances and angles, producing the cross arms visible on the filament surface. Reversing the two would incorrectly place the functional heads within the central tail region and make the tails project outward. F-actin, G-actin, tropomyosin and troponin belong to thin-filament organisation and do not fit the described meromyosin architecture. The spatial arrangement connects the structural tail component with the projecting region that carries the active head functions.
104. A treatment selectively removes HMM from meromyosin molecules but leaves most LMM tails associated with the thick filament. What functional change is most directly expected?
ⓐ. Thin filaments lose G-actin but retain their regulatory proteins
ⓑ. Z lines lose their role as boundaries of each sarcomere
ⓒ. Thick filaments retain LMM tails but lose HMM motor functions
ⓓ. LMM tails preserve normal ATPase and actin binding after HMM removal
Correct Answer: Thick filaments retain LMM tails but lose HMM motor functions
Explanation: HMM contains the globular myosin head and short arm. The head carries the active \(\mathrm{ATP}\)ase as well as binding sites for \(\mathrm{ATP}\) and actin. Selective removal of HMM eliminates these projecting functional regions and prevents normal cross-arm interaction with the thin filament. The remaining LMM tails preserve part of the elongated meromyosin structure, but they cannot replace the missing head activities. The treatment does not directly remove G-actin, alter sarcomere boundaries or convert tropomyosin into an enzyme. The predicted combination of retained tail material and lost energy-processing and actin-binding capacity follows from the division of functions between LMM and HMM.
105. Consider the following statements about the sliding-filament theory.
I. Thin filaments slide over thick filaments during contraction.
II. Actin and myosin filaments both become shorter.
III. Sarcomere shortening can occur without shortening either filament type.
IV. Contraction requires thick filaments to leave the A band.
ⓐ. I and II only
ⓑ. I and III only
ⓒ. II and IV only
ⓓ. I, III and IV only
Correct Answer: I and III only
Explanation: Sliding-filament theory explains muscle contraction through relative movement between the thin actin and thick myosin filaments. Thin filaments move farther over the thick filaments, increasing their overlap and shortening the sarcomere. The protein filaments themselves do not become shorter during this process. The A band corresponds to the full length of the thick filaments, so thick filaments do not need to leave that band for contraction to occur. Statements I and III express the essential distinction between filament sliding and filament shortening. This distinction prevents sarcomere shortening from being incorrectly interpreted as physical shortening of actin or myosin molecules. The underlying principle shows that contraction increases filament overlap, bringing Z lines closer and narrowing the I band and H zone without shortening actin or myosin.
106. A resting sarcomere is compared with the same sarcomere after contraction. The distance between two marked points on one thin filament is unchanged, and the distance between two marked points on one thick filament is also unchanged. However, the overlap between the two filament types has increased. This observation supports:
ⓐ. shortening of both filament types within the sarcomere
ⓑ. conversion of thick filaments into thin filaments
ⓒ. breakdown of myofilaments during force generation
ⓓ. sliding of thin filaments over thick filaments
Correct Answer: sliding of thin filaments over thick filaments
Explanation: The unchanged distance between marks placed on each individual filament shows that neither the thin nor the thick filament has shortened. At the same time, the increased region of overlap demonstrates that their relative positions have changed. This is the central prediction of the sliding-filament mechanism: thin filaments move inward over thick filaments while both retain their original lengths. Sarcomere shortening results from this rearrangement rather than destruction, conversion or compression of the contractile proteins. The use of internal marks provides direct evidence that distinguishes movement of an intact filament from shortening of that filament itself.
107. Assertion: Skeletal-muscle contraction can shorten a sarcomere without shortening actin or myosin filaments.
Reason: Thin filaments slide over thick filaments and increase their overlap during contraction.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion
Explanation: Sarcomere shortening reflects a change in the relative positions of actin and myosin rather than a reduction in the length of either protein filament. As thin filaments move toward the central region of the sarcomere, they overlap the thick filaments more extensively. This inward sliding draws the sarcomere boundaries closer even though each filament remains structurally the same length. The reason states the mechanism that directly accounts for the assertion. The relation is central to sliding-filament theory and separates molecular movement from molecular shortening, two processes that can produce very different structural predictions. The defining boundary becomes clear when contraction increases filament overlap, bringing Z lines closer and narrowing the I band and H zone without shortening actin or myosin.
108. In an experiment, fluorescent marks are placed at both ends of selected thin and thick filaments. During contraction, the length of each marked filament remains constant, but the distance between the sarcomere boundaries decreases. Which inference is best supported?
ⓐ. Sarcomere boundaries move apart while filaments remain fixed
ⓑ. Myosin filaments are digested to permit contraction
ⓒ. Relative filament sliding produces sarcomere shortening
ⓓ. Actin monomers are removed from thin filaments
Correct Answer: Relative filament sliding produces sarcomere shortening
Explanation: Fluorescent marks at the filament ends allow direct comparison of filament length before and during contraction. Since those lengths remain constant, shortening cannot be attributed to loss of actin monomers or digestion of myosin. The reduced distance between sarcomere boundaries shows that the contractile unit has nevertheless become shorter. The strongest interpretation is that thin and thick filaments have changed their degree of overlap through sliding. The experiment distinguishes the structural state of the filaments from the overall length of the sarcomere, supporting a mechanism in which intact filaments move relative to one another to produce contraction. The result is consistent with the fact that the A band spans the full length of thick filaments and remains constant when filament length does not change. The classification depends on recognising that successive Z lines delimit one sarcomere, and each Z line bisects an I band while anchoring thin filaments.
109. Motor neuron P supplies muscle fibres \(1\), \(2\), \(3\) and \(4\), while motor neuron Q supplies fibres \(5\), \(6\) and \(7\). Which group constitutes one motor unit?
ⓐ. Motor neuron P together with fibres \(1\text{–}3\)
ⓑ. Motor neuron Q together with fibres \(4\text{–}7\)
ⓒ. Motor neuron P together with fibres \(1\text{–}4\)
ⓓ. Motor neurons P and Q together with fibres \(1\text{–}7\)
Correct Answer: Motor neuron P together with fibres \(1\text{–}4\)
Explanation: A motor unit consists of one motor neuron and all the muscle fibres supplied by that neuron. Motor neuron P and fibres \(1\text{–}4\) satisfy both parts of this definition. The fibres supplied by neuron Q belong to a different motor unit even though they may lie within the same muscle. A collection of fibres without its supplying neuron is incomplete, while two motor neurons cannot form one motor unit under this definition. The organisation links one neural source with the complete group of muscle fibres under its supply, allowing neural activity to activate a defined set of fibres together.
110. Damage to motor neuron P produces the observations shown below.
| Muscle fibre | Response after damage to neuron P |
|---|
| Fibre \(1\) | Loses neural activation |
| Fibre \(2\) | Loses neural activation |
| Fibre \(3\) | Loses neural activation |
| Fibre \(4\) | Remains normally activated by neuron Q |
Which set most likely belongs to the motor unit of neuron P?
ⓐ. Neuron P with fibre \(4\) only
ⓑ. Neuron P with fibres \(1\), \(2\) and \(3\)
ⓒ. Neurons P and Q with fibres \(1\text{–}4\)
ⓓ. Fibres \(1\), \(2\) and \(3\) without neuron P
Correct Answer: Neuron P with fibres \(1\), \(2\) and \(3\)
Explanation: Fibres \(1\), \(2\) and \(3\) lose activation when neuron P is damaged, indicating that they receive their neural supply from that neuron. Fibre \(4\) remains active through neuron Q and belongs to a different motor unit. One motor unit includes the supplying motor neuron as well as every muscle fibre that it supplies, so the affected set must include neuron P together with fibres \(1\), \(2\) and \(3\). The pattern of simultaneous functional loss provides evidence for a shared neural source. It also shows that neighbouring fibres in the same muscle need not belong to the same motor unit. The prediction follows from recognising that a motor unit consists of one motor neuron and every skeletal muscle fibre supplied by that neuron. The relationship is preserved when the neuromuscular junction is the contact between a motor-neuron ending and the sarcolemma of a muscle fibre.
111. A motor neuron supplying one group of muscle fibres is severed, while adjacent motor neurons and their fibres remain intact. What result is most likely?
ⓐ. Only adjacent motor units lose activation while the damaged unit remains active
ⓑ. Neighbouring motor neurons activate fibres of the severed motor unit
ⓒ. All fibres in the affected fascicle lose activation regardless of their neurons
ⓓ. Muscle fibres belonging to the damaged motor unit lose normal neural activation
Correct Answer: Muscle fibres belonging to the damaged motor unit lose normal neural activation
Explanation: Each motor unit is organised around one motor neuron and the particular muscle fibres supplied by it. Severing that neuron interrupts neural activation to the fibres within its own unit. Adjacent fibres supplied by different intact motor neurons retain their separate neural connections under the stated condition. The injury does not merge motor units or spread automatically to every muscle fibre in the body. The predicted pattern follows the defined neuron-to-fibre relationship: loss of one neural source affects the complete group dependent on that source while sparing fibres controlled by other motor neurons.
112. Match each structure or unit with its defining relation. A Column II entry is used once.
| Column I | Column II |
|---|
| P. Motor unit | 1. Contact between a motor-neuron ending and muscle-fibre sarcolemma |
| Q. Neuromuscular junction | 2. Membrane excitation travelling along a muscle fibre |
| R. Action potential | 3. One motor neuron and all muscle fibres supplied by it |
| S. Sarcoplasmic reticulum | 4. Internal store releasing \(\mathrm{Ca^{2+}}\) during activation |
ⓐ. P-1, Q-3, R-4, S-2
ⓑ. P-3, Q-1, R-2, S-4
ⓒ. P-4, Q-2, R-1, S-3
ⓓ. P-3, Q-4, R-2, S-1
Correct Answer: P-3, Q-1, R-2, S-4
Explanation: A motor unit contains one motor neuron together with every muscle fibre supplied by it. The neuromuscular junction, also called the motor end plate, is the specialised contact where the ending of that neuron meets the sarcolemma of a muscle fibre. Acetylcholine released at this contact initiates an action potential, which spreads as membrane excitation along the fibre. The sarcoplasmic reticulum is an internal membrane system that stores \(\mathrm{Ca^{2+}}\) and releases it when the fibre is activated. The mapping follows the progression from neural organisation to junctional communication, electrical excitation and intracellular calcium mobilisation. The reasoning proceeds from the fact that the neuromuscular junction is the contact between a motor-neuron ending and the sarcolemma of a muscle fibre.
113. Use the arrangement described below. Structure P is the terminal ending of a motor neuron. It lies next to Region Q, a specialised part of the sarcolemma of a skeletal muscle fibre. The contact between P and Q is the:
ⓐ. neuromuscular junction or motor end plate
ⓑ. cartilaginous joint between skeletal elements
ⓒ. M line holding thick filaments together
ⓓ. Z line anchoring thin filaments
Correct Answer: neuromuscular junction or motor end plate
Explanation: The described contact joins a motor-neuron ending with the sarcolemma of a muscle fibre. This specialised communication site is the neuromuscular junction, also known as the motor end plate. It is not a skeletal joint, since it does not connect bones or cartilage. The M line and Z line are internal landmarks of sarcomeres and have roles in organising thick and thin filaments. At the neuromuscular junction, the arriving neural signal is converted into chemical signalling through acetylcholine and then into electrical excitation of the muscle-fibre membrane. The locations of P and Q identify the junction independently of any unseen diagram.
114. A learner calls the neuromuscular junction a movable joint between a neuron and a bone. Which correction is accurate?
ⓐ. a synaptic contact between the endings of two neighbouring motor neurons
ⓑ. a structural contact between thick filaments and the central M line
ⓒ. a synapse between a motor-neuron ending and the muscle-fibre sarcolemma
ⓓ. a connective-tissue attachment between a muscle fibre and its tendon
Correct Answer: a synapse between a motor-neuron ending and the muscle-fibre sarcolemma
Explanation: The term junction in this context refers to a specialised site of communication, not to a skeletal articulation. A neuromuscular junction forms where the terminal ending of a motor neuron contacts the sarcolemma of a muscle fibre. Neural arrival at this site leads to acetylcholine release and excitation of the muscle membrane. Bones, tendons and synovial cavities are not the structures defining this junction. Similarly, the M line lies inside a sarcomere and organises thick filaments rather than connecting a neuron to a muscle cell. The correction identifies both participating structures and the communication function that links neural activity with muscle activation.
115. A motor neuron and its muscle fibre are tested under two conditions. Stimulating the motor neuron fails to activate the fibre, but direct stimulation of the sarcolemma produces a normal action potential and contraction. The most likely defect lies at the:
ⓐ. actin polymerisation within the thin filament
ⓑ. neuromuscular contact between neuron and fibre
ⓒ. M-line anchorage of central thick filaments
ⓓ. collagenous fascia around the muscle fascicle
Correct Answer: neuromuscular contact between neuron and fibre
Explanation: Direct sarcolemmal stimulation produces a normal action potential and contraction, showing that the muscle membrane, calcium-release system and contractile machinery can function. Failure occurs only when activation must pass from the motor neuron to the fibre. This localises the defect to the communication site between them, the neuromuscular junction. Damage to actin or the M line would interfere with contraction even after direct membrane stimulation. Fascia organises muscle bundles and does not transmit the neural signal into an individual fibre. Comparing neural stimulation with direct sarcolemmal stimulation separates a junctional transmission failure from a defect within the muscle fibre itself.
116. Arrange the events initiating excitation of a skeletal muscle fibre.
P. Acetylcholine is released from the motor-neuron ending.
Q. An action potential is generated in the sarcolemma.
R. A neural signal reaches the motor-neuron ending.
S. Acetylcholine acts at the neuromuscular junction.
ⓐ. \(P \rightarrow R \rightarrow S \rightarrow Q\)
ⓑ. \(R \rightarrow S \rightarrow P \rightarrow Q\)
ⓒ. \(S \rightarrow P \rightarrow R \rightarrow Q\)
ⓓ. \(R \rightarrow P \rightarrow S \rightarrow Q\)
Correct Answer: \(R \rightarrow P \rightarrow S \rightarrow Q\)
Explanation: The pathway begins when a neural signal reaches the terminal ending of the motor neuron. This arrival causes acetylcholine to be released at the neuromuscular junction. The transmitter then acts on the muscle-fibre side of the junction and initiates electrical excitation of the sarcolemma. Generation of the muscle action potential must come after transmitter release and action, not before the neural signal reaches the ending. The sequence represents conversion from an electrical signal in the neuron to chemical communication at the junction and back to an electrical signal in the muscle-fibre membrane. Each transition depends on the preceding one. A transmitter cannot act before its release, and a muscle-fibre action potential cannot be generated through this pathway until acetylcholine has crossed the junction and acted on the sarcolemma. This preserves the direction of information flow from neuron to muscle.
117. Two identical nerve-muscle preparations receive the same neural stimulation. In Preparation P, acetylcholine remains available at the neuromuscular junction and a sarcolemmal action potential develops. In Preparation Q, acetylcholine is destroyed immediately after release and no muscle action potential develops. The result indicates that:
ⓐ. acetylcholine stores \(\mathrm{Ca^{2+}}\) inside the muscle fibre
ⓑ. acetylcholine directly shortens actin and myosin filaments
ⓒ. acetylcholine is required to initiate sarcolemmal excitation
ⓓ. acetylcholine forms the structural boundary of a motor unit
Correct Answer: acetylcholine is required to initiate sarcolemmal excitation
Explanation: Both preparations receive an equivalent neural signal, so the difference in muscle-fibre excitation is linked to the availability of acetylcholine at the junction. Preparation P retains the transmitter and develops an action potential, whereas immediate destruction of the transmitter in Preparation Q prevents that electrical response. The comparison supports the role of acetylcholine in initiating sarcolemmal excitation after neural arrival. Acetylcholine does not serve as the intracellular calcium store or shorten the contractile filaments directly. It acts at an earlier communication stage, transferring the signal from the motor-neuron ending to the muscle-fibre membrane. The observed pattern is explained by recognising that acetylcholine released at the motor end plate initiates a sarcolemmal action potential. Loss of the muscle action potential only when acetylcholine is removed shows that transmitter availability is required to carry excitation across the neuromuscular junction.
118. Assertion: Arrival of a neural signal can produce an action potential in the sarcolemma of a muscle fibre.
Reason: The signal causes acetylcholine release at the neuromuscular junction, and the transmitter initiates muscle-fibre excitation.
ⓐ. Both Assertion and Reason are true, and Reason correctly explains Assertion
ⓑ. Both Assertion and Reason are true, but Reason does not correctly explain Assertion
ⓒ. Assertion is true, but Reason is false
ⓓ. Assertion is false, but Reason is true
Correct Answer: Both Assertion and Reason are true, and Reason correctly explains Assertion
Explanation: Neural and muscle-fibre electrical events are linked through chemical transmission at the neuromuscular junction. When the neural signal reaches the motor-neuron ending, acetylcholine is released. Its action on the muscle-fibre membrane generates an action potential in the sarcolemma. The reason supplies the intermediate communication step connecting neural arrival with muscle excitation and directly accounts for the assertion. The transmitter is released from the neural ending rather than from the contractile filaments, and its immediate role is initiation of membrane excitation rather than production of mechanical force. Mechanical contraction follows only after later intracellular events have occurred. The key structural point is that the neuromuscular junction is the contact between a motor-neuron ending and the sarcolemma of a muscle fibre.
119. Which sequence correctly links sarcolemmal excitation with exposure of actin sites:
P. \(\mathrm{Ca^{2+}}\) binds to troponin.
Q. Myosin-binding sites on actin become exposed.
R. The action potential spreads through the muscle fibre.
S. \(\mathrm{Ca^{2+}}\) is released into the sarcoplasm.
ⓐ. \(S \rightarrow R \rightarrow P \rightarrow Q\)
ⓑ. \(R \rightarrow P \rightarrow S \rightarrow Q\)
ⓒ. \(P \rightarrow S \rightarrow R \rightarrow Q\)
ⓓ. \(R \rightarrow S \rightarrow P \rightarrow Q\)
Correct Answer: \(R \rightarrow S \rightarrow P \rightarrow Q\)
Explanation: Once generated, the action potential spreads through the muscle fibre and activates the intracellular calcium-release system. The sarcoplasmic reticulum then releases \(\mathrm{Ca^{2+}}\) into the sarcoplasm. The increased calcium concentration allows calcium to bind troponin on the thin filament. This interaction alters the resting regulatory arrangement so that the myosin-binding sites on actin are no longer masked. Each stage creates the condition needed for the next: membrane excitation precedes calcium release, free calcium precedes binding to troponin, and troponin activation precedes exposure of the actin sites used in cross-bridge formation. The key functional point is that spread of excitation through the fibre triggers release of \(\mathrm{Ca^{2+}}\) from sarcoplasmic stores into the sarcoplasm. A complete interpretation recognises that calcium binds troponin and removes the masking of myosin-binding sites on actin before productive cross bridges can form.
120. An action potential spreads normally through a skeletal muscle fibre, but the sarcoplasmic reticulum fails to release \(\mathrm{Ca^{2+}}\). Which immediate consequence is expected?
ⓐ. Actin sites become exposed because calcium release is unnecessary
ⓑ. Actin sites remain masked and contraction is impaired
ⓒ. Myosin binds normally because the action potential alone is sufficient
ⓓ. Troponin shifts tropomyosin without binding calcium
Correct Answer: Actin sites remain masked and contraction is impaired
Explanation: The action potential supplies the electrical signal that normally triggers calcium release from the sarcoplasmic reticulum. If calcium is not released, its concentration in the sarcoplasm does not rise sufficiently for binding to troponin. The resting masking of myosin-binding sites on actin is maintained, preventing normal cross-bridge formation and impairing contraction. Neural acetylcholine release occurs earlier in the pathway and is not predicted to become excessive from this intracellular defect. Thick filaments do not shorten as a substitute mechanism, and troponin neither stores nor releases \(\mathrm{ATP}\). The blocked step separates successful membrane excitation from unsuccessful activation of the contractile proteins.
