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7 Solved Problems on Biophysics - Examination | BIOP 401, Exams of Biology

Material Type: Exam; Class: Introduction to Biophysics; Subject: Biophysics; University: University of Illinois - Urbana-Champaign; Term: Fall 2008;

Typology: Exams

Pre 2010

Uploaded on 03/16/2009

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November 3rd 2008. BIOP-401 2nd 1-hr exam Name:
1
1. (10 points) A typical kinesin couples the energy of ATP hydrolysis with directional movement
along microtubules. In the presence of microtubules, it hydrolyzes ATP with the kcat and KM of 115 s-1
and 30µM, respectively. Estimate the translocation rate for this kinesin at
1.1 1 mM ATP
Solution:
Translocation rate = 115 s-1*8 nm 920 nm*s-1
1.2 100 nM ATP
Solution:
Translocation rate = 0.38 s-1*8 nm 3 nm*s-1
(assume that ADP and Pi are at very low concentrations and do not inhibit ATP hydrolysis; Mg2+ is
present in sufficiently high concentration to saturate all ATP)
Bonus 10 points: Why (and under what circumstances) the answer you obtained may represent a
drastic underestimation or overestimation of the actual kinesin velocity?
Solution:
Underestimation: Because the ATPase activity was measured at the steady-state situation, it averages
ATP hydrolysis of actively moving as that of the motor that has dissociated from the track and has to
rebound to it. If the latter state takes up a significant portion of the overall cycle of the motor, and the
ATPase activity in the detached motor is low, the kcat of ATP hydrolysis will be an underestimation of
the ATP hydrolysis rate by actively walking motor.
Overestimation: We assumed that 1 ATP molecule is hydrolyzed per one step and that all steps are
forward. Imperfect coupling and/or sliding or backward stepping will result in overestimation of the
velocity.
2. (total 30 points) For a chemical reaction whose
reaction coordinate is depicted on the right (T=300K), find:
2.1 (5 points) The ratio of Product to Substrate at
equilibrium
2.2 (5 points) k forward (forward rate constant)
2.3 (5 points) k reverse (reverse rate constant)
pf3
pf4

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  1. (10 points) A typical kinesin couples the energy of ATP hydrolysis with directional movement along microtubules. In the presence of microtubules, it hydrolyzes ATP with the kcat and KM of 115 s-^1 and 30μM, respectively. Estimate the translocation rate for this kinesin at 1.1 1 mM ATP Solution: Translocation rate = 115 s
    • 1 8 nm ≈ 920 nms - 1 1.2 100 nM ATP Solution: Translocation rate = 0.38 s
    • 1 8 nm ≈ 3 nms - 1 (assume that ADP and Pi are at very low concentrations and do not inhibit ATP hydrolysis; Mg 2+ is present in sufficiently high concentration to saturate all ATP) Bonus 10 points: Why (and under what circumstances) the answer you obtained may represent a drastic underestimation or overestimation of the actual kinesin velocity? Solution: Underestimation: Because the ATPase activity was measured at the steady-state situation, it averages ATP hydrolysis of actively moving as that of the motor that has dissociated from the track and has to rebound to it. If the latter state takes up a significant portion of the overall cycle of the motor, and the ATPase activity in the detached motor is low, the kcat of ATP hydrolysis will be an underestimation of the ATP hydrolysis rate by actively walking motor. Overestimation: We assumed that 1 ATP molecule is hydrolyzed per one step and that all steps are forward. Imperfect coupling and/or sliding or backward stepping will result in overestimation of the velocity.
  2. (total 30 points) For a chemical reaction whose reaction coordinate is depicted on the right (T=300K), find: 2.1 (5 points) The ratio of Product to Substrate at equilibrium 2.2 (5 points) k (^) forward (forward rate constant) 2.3 (5 points) k (^) reverse (reverse rate constant)

2.4 (15 points) The pre-exponential factor in Arrhenius relation describing this reaction is 10 12 s

  • 1 . Find the entropy and the enthalpy of activation.
  1. (total 15 points) 3.1 (5 points) Sketch a typical force-extension curve for dsDNA and ssDNA 3.2 (5 points) Would you expect an upward or downward shift in the force corresponding to the overstretch transition with increase in temperature? Will go down 3.3 (5 points) Would you expect an upward or downward shift in the force corresponding to the overstretch transition with increasing salt concentration? (note that 88% of phosphates are neutralized in dsDNA and only 71% in ssDNA leading to release of counterions upon DNA melting) Will go up
  2. (10 points) Actin filament polymerizes against an applied force of 2 pN ( in vitro ). The load is increased to 4 pN. If you want to maintain the polymerization rate the same as it was at 2 pN load, which experimental parameter will you change and by how much? Solution: The only parameter amenable to experimental alteration is the concentration of monomers, M. To achieve the same polymerization rate as at 2 pN, M should be increased by 3.7 fold:

EQUATIONS:

CONSTANTS:

R = 8.31 J/K/mol ħ = 6.626*

  • 34 Js kB = 1.3810-^23 J/K MOTOR STEP SIZES/DIMENSIONS: Kinesin – 8 nm center of the mass steps; Myosin (walking) – 34 nm center of the mass steps