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11111nata11111 [884]
3 years ago
12

In a real hemoglobin molecule, the tendency of oxygen to bind to a heme site increases as the other three heme sites become occu

pied. To model this effect in a simple way, imagine that a hemoglobin molecule has just two sites, either or both of which can be occupied. This system has four possible states (with only oxygen present). Take the energy of the unoccupied state to be zero, the energies of the two singly occupied states to be -0.55 eV, and the energy of the doubly occupied state to be -1.3 eV (so the change in energy upon binding the second oxygen is -0.75 eV). As in the previous problem, calculate and plot the fraction of occupied sites as a function of the effective partial pressure of oxygen. Compare to the graph from the previous problem (for independent sites). Can you think of why this behavior is preferable for the function of hemoglobin?

Physics
1 answer:
sasho [114]3 years ago
7 0
Since there are four states, then the grand partition function of the system is

Z = 1+2 e^{(0.55eV+ \alpha )/kT} +e^{(1.3eV+ 2\alpha )/kT}

where α is the chemical potential

Then, the occupancy of the system is

bar \ n=(e^{(0.55eV+ \alpha )/kT} +e^{(1.3eV+ 2\alpha )/kT})/Z

Then using this equation, \alpha =-kT(V Z_{int} /N v_{Q}) and approximating Z_int to be kT/0.00018 eV, the model would look as that attached in the figure. That is the occupancy vs. pressure graph. 

There are more occupancies when the oxygen is high (high pressure) especially in the lungs. Heme sites tend to be occupied by oxygen. 

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3 years ago
A system consists of three particles with these masses and velocities: mass 3.00 kg, moving north at 3.00 m/s; mass 4.00 kg, mov
lions [1.4K]

Answer:

the total momentum is 8 .2 kg m/s in north direction.

Explanation:

given,

mass(m₁) 3.00 kg, moving north at v₁ = 3.00 m/s

mass(m₂) 4.00 kg, moving south at v₂ =  3.70 m/s

mass(m₃) 7.00 kg, moving north at v₃ = 2.00 m/s

north as the positive axis

south as the negative axis

now

total momentum = m₁v₁ + m₂ v₂ + m₃ v₃

total momentum = 3 x 3 - 4 x 3.7 + 7 x 2

                           = 9 - 14.8 + 14

                           = 8 .2 kg m/s

hence, the total momentum is 8 .2 kg m/s in north direction.

7 0
3 years ago
As the shuttle bus comes to a sudden stop to avoid hitting a dog, it accelerates uniformly at -4.1 m/s2 as it slows from 9.0 m/s
Dennis_Churaev [7]

Answer:

The time interval of acceleration for the bus is 2.20 seconds

Explanation:

Acceleration is the rate of change of velocity

→ a=\frac{v-u}{t}

where a is the acceleration, v is the final velocity, u is the initial velocity

and t is the time

The given is:

The uniform acceleration = -4.1 m/s²

The bus slows from 9 m/s to 0 m/s

We need to find the time interval of acceleration for the bus

Lets use the rule above

→ a = -4.1 m/s² , v = 0 m/s , u = 9 m/s

→ -4.1=\frac{0-9}{t}

Multiply both sides by t

→ -4.1 t = -9

Divide both sides by -4.1

∴ t = 2.20 seconds

<em>The time interval of acceleration for the bus is 2.20 seconds</em>

3 0
4 years ago
a body is thrown vertically upward from the earth's surface and it took 8 seconds to return to its original position . find out
Mnenie [13.5K]

Answer:

The initial velocity with which the body was thrown up is 39.2 m/s

Explanation:

The given parameters for the body are;

The time it takes the body to return back to its initial position = 8 seconds

To answer the question, we make use of the kinematic equation of motion, v = u - g·t

Where

v = The final velocity of the body = 0 m/s at the maximum height

u = The initial velocity

g = The acceleration due to gravity = 9.8 m/s²

t = The time in which the body spends in the air

Therefore, at maximum height, we have;

v = 0 = u - g·t

u = g·t

t = u/g

From h = 1/2gt², which gives t = √(2·h/g), the time the body takes to maximum height = The time the body takes to return to its original position from maximum height.

Therefore, the total time in which the body is in the air = 2 × t = 2× u/g

∴

The total time in which the body is in the air = The time it takes the body to return back to its initial position after being thrown = 2 × t =  8 seconds

∴ 2 × t = 8 s = 2 × u/g

8 s = 2 × u/g

u = (8 s × g)/2

∴ u = (8 s × 9.8 m/s²)/2 = 39.2 m/s

The initial velocity with which the body was thrown up = u = 39.2 m/s.

4 0
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