Kinesin Molecular Motors and Force Calculation

Kinesin Molecular Motors

  • Kinesins are tiny molecular motors responsible for transporting organelles within cells along microtubules.

Key Concepts

  • Organelles: Small structures within cells with specific functions, carrying proteins synthesized within the cell.
  • Microtubules: Part of the cytoskeleton in eukaryotic cells, serving as tracks for kinesin transport.
  • Molecular Motors: Proteins that convert chemical energy into mechanical work, enabling movement at the molecular scale.

Problem Statement

  • A question posed involves calculating the force that a kinesin molecule needs to exert in order to accelerate an organelle.
  • Specifics of the Problem:
    • Mass of the organelle (m): 1.00imes10−17extkg1.00 imes 10^{-17} ext{ kg}
    • Initial velocity (v_i): 0extμm/s0 ext{ μm/s}
    • Final velocity (v_f): 5.60extμm/s5.60 ext{ μm/s}
    • Time (t): 10.0extμs10.0 ext{ μs}

Required Calculation

  • To find the force exerted by the kinesin molecule, we first need to calculate the acceleration of the organelle.
Step 1: Calculate Acceleration
  • Acceleration (a) can be calculated using the formula:

exta=extchangeinvelocityexttime=v<em>f−v</em>itext{a} = \frac{ ext{change in velocity}}{ ext{time}} = \frac{v<em>f - v</em>i}{t}

Substituting the values:

  • Initial velocity ($v_i$): 0 μm/s
  • Final velocity ($v_f$): 5.60 μm/s
  • Time ($t$): 10.0 μs = 10.0imes10−6exts10.0 imes 10^{-6} ext{ s}

exta=5.60extμm/s−0extμm/s10.0imes10−6extsext{a} = \frac{5.60 ext{ μm/s} - 0 ext{ μm/s}}{10.0 imes 10^{-6} ext{ s}}

  • Convert units for $ ext{a}$ for proper calculation: 1extμm/s=10−6extm/s1 ext{ μm/s} = 10^{-6} ext{ m/s}

Thus:
exta=5.60imes10−6extm/s10.0imes10−6exts=560extm/s2ext{a} = \frac{5.60 imes 10^{-6} ext{ m/s}}{10.0 imes 10^{-6} ext{ s}} = 560 ext{ m/s}^2

Step 2: Calculate Force
  • Now that we have the acceleration, we can calculate the force using Newton's Second Law:

F=mimesaF = m imes a

Substituting the values we’ve calculated:

  • Mass ($m$): 1.00imes10−17extkg1.00 imes 10^{-17} ext{ kg}
  • Acceleration ($a$): 560extm/s2560 ext{ m/s}^2

Thus:
F=(1.00imes10−17extkg)imes(560extm/s2)F = (1.00 imes 10^{-17} ext{ kg}) imes (560 ext{ m/s}^2)

F=5.60imes10−15extNF = 5.60 imes 10^{-15} ext{ N}

Conclusion

  • The force required by the kinesin to accelerate the organelle with the specified parameters is 5.60imes10−15extN5.60 imes 10^{-15} ext{ N}. This calculation underscores the tiny forces at play within cellular processes, showcasing the intricate mechanics of molecular motors.