Tension in the muscles
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Tension in the muscles
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[post_content] => Practice Passage (Question 1-5)
*This passage is the property of Khan Academy and has been reformatted into an AAMC-style interface in their entirety by MedLife Mastery. MedLife Mastery does not endorse and is not an affiliate of Khan Academy.
Human skeletal muscle tissue is uniquely structured by specialized layers. Muscles are composed of a collection of muscle fibers. In turn, these fibers are made of closely packed myofibrils, or a subunit made of tubular muscle cells. These myofibrils are surrounded by sarcoplasm, a substance similar to cytoplasm in other cells. Finally, myofibrils are composed of repeated sections of sarcomeres, which contain actin and myosin filaments.
Physiologists interested in testing muscle cell tension capacity and musculoskeletal dynamics removed a sarcomere from a primate and placed it in physiological laboratory conditions. They analyzed the sarcomere’s movements under a number of different scenarios, including movement. A simplified model of a section of this sarcomere is given in Figure 1.
Figure 1. A section of the sarcomere used by the scientists.
The tension in the actin between each myosin section are summed up and referred to as a single tension. The tension following myosin section 1 (M1) is referred to as TA, and the tension following myosin section 2 (M2) is TB. This section of actin is directly connected to the sarcomere’s end.
Scientists noted that M1 contained an abnormal mass, making its mass greater than the mass of M2. The mass of M1 = 60kDa, and the mass of M2 = 30kDa.
1.6x10⁻²⁴ kg = 1kDa. Assume that frictional forces are negligible.
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[question] => If the mass of M1 > the mass of M2, and the sarcomere is moving with constant velocity, what can be said about the tension in both actin sections?
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[description] => Reason for the Correct Answer:
The sarcomere is moving with constant velocity
Since the sarcomere is moving with constant velocity there is no acceleration
Since there is no acceleration the net force on the both myosin sections must be zero
Since the net force must be zero, and there’s no friction, the tension in actin A must be zero, which means the tension in actin B must also be zero
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[each_answer] => A. Both tensions are the same and are zero
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[each_answer] => B. TA is larger than TB since the mass of M1 is larger than the mass of M2
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[each_answer] => C. Both tensions TA and TB are equal and are not equal to zero
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[each_answer] => D. TA is smaller than TB since actin B has to pull the mass of both boxes
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[question] => If the mass of M1 > the mass of M2 and the sarcomere is accelerating to the right, how will the tensions in the actin sections compare?
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[answer] => 4
[description] => Reason for the Correct Answer:
Use ΣF=ma for the horizontal forces on the masses
The only horizontal force on the 60kDa mass is the tension from the actin A
Actin B has to pull the mass of both myosin sections so TB is greater than TA
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[each_answer] => A. TA is larger than TB since the mass of M1 is larger than the mass of M2
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[1] => Array
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[each_answer] => B. Both tensions TA and TB are equal and are not equal to zero
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[each_answer] => C. Both tensions are the same and are zero
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[each_answer] => D. TA is smaller than TB since actin B has to pull the mass of both boxes
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[question] => During one trial, the acceleration is 2m/s2 to the right. What calculation will give the tensions in the actin filaments during this trial?
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[answer] => 4
[description] => Reason for the Correct Answer:
Think about Newton’s second law ΣF=ma for each box and draw a force diagram for each box
Newton’s second law requires the use of kg, not kDa. The conversion between the two is 1.6×10⁻²⁴ kg = 1kDa
The mass for TA is the mass of M1. The mass for TB is the mass of M1 plus the mass of M2.
For M1, ΣF=ma gives TA=60kDa x 1.6×10⁻²⁴ kg x 2m/s²
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[each_answer] => A. TA = 60 x 2, TB = (30+60) x 2
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[each_answer] => B. TA= 1.6×10⁻²⁴ x 60 x 2, TB= 1.6×10⁻²⁴ x 30 x 2
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[each_answer] => C. TA = 60 x 2, TB = 30 x 2
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[each_answer] => D. TA= 1.6×10⁻²⁴ x 60 x 2, TB= 1.6×10⁻²⁴ x (60+30) x 2
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[question] => If the maximum force an actin section can maintain before tearing is 9pN, what is the maximum acceleration the scientists could make the sarcomere go while maintaining its structural integrity?
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[answer] => 2
[description] => Reason for the Correct Answer:
Use ΣF=ma and isolate acceleration algebraically.
F = ma. We know the force is 9 picoNewtons, or 9 x 10⁻¹²N
You must use the mass of M1, as this will be able to handle less acceleration than M2. This conclusion can be drawn by the fact that mass and acceleration are inversely proportional in F = ma.
Convert the mass of M1 from kDa to kg: 60kDa x 1.6 x 10⁻²⁴kg
Plug in the variables: 9×10⁻¹²N = (60kDa x 1.6 x 10⁻²⁴kg) x a
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[each_answer] => A. 9×10⁻⁹N = (30kDa x 1.6 x 10⁻²⁴kg) x a
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[each_answer] => B. 9×10⁻¹²N = (60kDa x 1.6 x 10⁻²⁴kg) x a
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[each_answer] => C. 9×10⁻⁹N = (60kDa x 1.6 x 10⁻²⁴kg) x a
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[each_answer] => D. 9×10⁻¹²N = (30kDa x 1.6 x 10⁻²⁴kg) x a
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[question] => Assume tensions are distributed equally amongst actin filaments in TA. What is the force F1 on one filament of the first actin section if acceleration is a?
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[answer] => 4
[description] => Reason for the Correct Answer:
Use F = ma to solve for the maximum force possible
Note that there are 4 actin filaments in each actin section. So, the tension will be divided equally amongst these filaments.
The calculation to convert kDa to kg is 60kDa * 1.6 * 10⁻²⁴
Plugging in our numbers and dividing by 4, we get: F₁ = (60 * 1.6 * 10⁻²⁴) a / 4
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[each_answer] => A. F₁ = 60 a / 4
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[each_answer] => B. F₁ = (60 * 1.6 * 10⁻²⁴) a
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[each_answer] => C. F₁ = (60 * 1.6 * 10⁻²⁴) a * 4
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[each_answer] => D. F₁ = (60 * 1.6 * 10⁻²⁴) a / 4
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The tension in the actin between each myosin section are summed up and referred to as a single tension. The tension following myosin section 1 (M1) is referred to as TA, and the tension following myosin section 2 (M2) is TB. This section of actin is directly connected to the sarcomere’s end.
Scientists noted that M1 contained an abnormal mass, making its mass greater than the mass of M2. The mass of M1 = 60kDa, and the mass of M2 = 30kDa.
1.6x10⁻²⁴ kg = 1kDa. Assume that frictional forces are negligible.
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