Lecture Notes and Example Problems in Biology & Genetics
Problem 2. A bacterium swims in a straight line toward a nutrient source.
Initially, it moves with a constant velocity of 15 µm/s.
After 3.0 seconds, it senses a stronger chemical gradient and begins to accelerate at 2.0 µm/s2 along the same line.
a) How far does the bacterium travel during the first 3.0 seconds?
b) What is its velocity after accelerating for an additional 4.0 seconds?
c) What is the total displacement of the bacterium after the entire motion (7.0 seconds)?
Problem 3. Inside a eukaryotic cell, a motor protein (kinesin) carries a vesicle along a microtubule in a nearly straight line.
The vesicle starts from rest and accelerates uniformly at 0.50 µm/s2
for 6.0 seconds, after which it continues at a constant velocity.
a) What is the velocity of the vesicle at the end of the acceleration phase?
b) How far has the vesicle traveled during the acceleration phase?
c) If the vesicle continues at this constant velocity for another 10 seconds, what is its total displacement from the starting point?
CH3: Motion in 2D Vectors Problem 1.We have the following data for one of the object in the video below (starting from 1:18). Video: Bacterial motility Draw x-y, x-t, y-t, v-t and a-t plots. Solution | Result Time X Y time is in sec ---- --- --- X and Y are in pixel 0.0 449 53 0.5 460 100 1.0 476 145 1.5 525 200 2.0 594 246 2.5 675 260 3.0 774 266 3.5 865 240 4.0 947 224 4.5 1014 218
CH4-5: Newton's Laws of Motion How fast (in rpm) must a centrifuge rotate if a particle 8 cm from the axis of rotation is to experience an acceleration of a = 125000 g's? [Ans: 4 x 104 rpm]
CH7: Work and Energy
A cheetah, one of the fastest land animals, accelerates to catch its prey.
Suppose a cheetah with a mass of 50 kg sprints to reach a top speed of 30 m/s from rest to capture a gazelle.
(a) Calculate the work done by the cheetah's muscles to reach this speed.
(b) Assuming the cheetah's muscles operate with an efficiency of 25% in converting stored chemical energy (from ATP)
into kinetic energy, calculate the total chemical energy that the cheetah must expend during this sprint.
Speed vs Stopping Distance of a Car
CH8: Conservation of Energy
While plants can produce their own energy using the process of photosynthesis, humans, animals
(and other organisms that can’t do photosynthesis) must eat to get energy from food molecules.
Just like energy can be stored in the chemical bond between the second and third phosphate of an ATP molecule,
energy can also be stored in the chemical bonds that make up food molecules.
Most of the energy that we use comes from molecules of glucose, a simple sugar.
In a biological system, adenosine triphosphate (ATP) is often referred to as the "energy currency" of the cell.
During cellular respiration, a glucose molecule is broken down, and the energy released is used to synthesize ATP.
When ATP is hydrolyzed into adenosine diphosphate (ADP) and an inorganic phosphate (Pi),
about 30.5 kJ/mol of energy is released under standard conditions.
(Molar mass of ATP = 507 g/mol, g = 9.8 m/s/s)
(a) If a human body hydrolyzes 1.0 kg of ATP per day, how much energy in kilojoules
is released in a day from ATP hydrolysis? [Ans: 60 kJ/day]
(b) If this energy were to be completely converted into mechanical energy to lift a 70 kg person,
how high could the person be lifted? Assume no energy losses. [Ans: 88 m]
CH4-9: Linear Momentum (Collisions) Two goats having the same mass of 60 kg run towards each other at the same speed of 5 m/s. They collide head-on and the collision lasts for 10 milliseconds, as in this video. Final speed of each goat is zero after collision. Assuming the collision is nearly perfectly inelastic, Estimate the average force exerted by each goat on the other during the collision? [Ans: F = 30 kN] In this application, the audio data of the video is extreacted. Then, a MATLAB script file can read and plot sound intensity vs time data. Using this data, we can roughly estimate the collision time as in this figure.
CH4-10: Rotational Motion
Physics and Biological Motion
When running, a person's legs swing back and forth around the hip joint.
The leg rotates about the hip joint, which acts as a pivot.
Assume a model where the leg can be approximated as a uniform rod of length L = 0.9 m and mass m = 10 kg, and
the leg starts at rest and reaches an angular velocity of w = 6 rad/s in 0.5 s.
Questions:
(a) Calculate the moment of inertia (I) of the leg about the hip joint, treating the leg as a uniform rod pivoted at one end.
[Ans: I = mL2/3 = 2.7kg.m2]
(b) If the angular acceleration of the leg is alpha = 4.0 rad/s2, what is the torque required to produce this acceleration?
[Ans: T = I a = 10.8 N.m]
(c) How much work is done by the muscles to bring the leg to this angular velocity?
[W = dK = 48.6 J]
Discussions:
How does the rotational motion of the leg contribute to energy efficiency while running,
and why is the moment of inertia an important factor?
- Energy efficiency: During running, the legs alternately accelerate and decelerate.
Minimizing the moment of inertia helps reduce the torque needed to swing the leg,
lowering energy expenditure and improving efficiency.
- Moment of inertia: The distribution of mass in the leg significantly affects the rotational dynamics.
A smaller moment of inertia makes it easier for muscles to swing the leg back and forth, critical for maintaining
a high running speed with minimal effort.
From a biomechanical perspective, how do muscles optimize energy use during running?
- Elastic energy storage: Tendons and muscles store elastic potential energy during deceleration and
release it during acceleration, reducing the metabolic energy required for motion.
- Minimizing rotational inertia: Humans naturally flex their knees during the leg swing, effectively shortening the leg
and reducing its moment of inertia. This reduces the torque needed to achieve the same angular acceleration.
- Efficient gait cycles: By optimizing stride frequency and minimizing unnecessary movement,
the body ensures smooth transitions, reducing energy losses due to excessive forces.