/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q. 7.2 In a real hemoglobin molecule, t... [FREE SOLUTION] | 91影视

91影视

In a real hemoglobin molecule, the tendency of oxygen to bind to a heme site increases as the other three heme sites become occupied. 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 -0.55eV, and the energy of the doubly occupied state to be -1.3eV(so the change in energy upon binding the second oxygen is -0.75eV). 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?

Short Answer

Expert verified

Occupancy rises gradually for smaller values of P. This shows that with cooperative binding, hemoglobin reacts with oxygen much fast in cells when P is small.

Step by step solution

01

Step 1. Formula Gibb's Factor

Gibb's Factor is given by formula:

Gibb'sFactor=e-(-n)kT

where, kis Boltzmann's constant, Tis temperature, is chemical potential, and is energy of state .

As, in hemoglobin molecule, oxygen can have four different states given by:

localid="1647144603783" o=0eV1=-0.55eV(twosinglyoccupied)2=-1.3eV

The average number of oxygen molecules is given by,

N=SN(S)=2e-(1-)kTZ+2e-(2-2)kTZ

02

Step 2. Formula occupancy

Occupancy of system is given by:

n=NZ=1Ze-(1-)kT+-(2-2)kT

Chemical Potential formula is :

=-kTlnVZinNQ

Substitute VN=kTPin above formula,

=-kTlnkTZinPQ

ekT=PQkTZin

Gibb's Factor can be written as:

e-(1-)kT=e-1kTPQkTZin=PQkTZine1kT=PPo

03

Step 3. Calculation

Oxygen molecule quantum volume is:

Q=h2蟺尘办罢3

Substitute the values of the variables in above formula,

Q=6.6310-34Js2(321.6610-27kg)(1.3810-23J/K)(310K)3=5.310-33m3

In equation, n=1Ze-(1-)kT+-(2-2)kT

Substitute eKT=1790 and Po=2.03barat room temperature,

n=P2.03+1790P2.0321+P2.03+1790P2.032

Occupancy in this model is nearly 100%near lungs given P=0.20bar

04

Step 4. Table and Graph

The table is created between Pressure Pand fraction of occupied sites:

P
n
0.01
0.0461
0.02
0.1554
0.03
0.2899
0.04
0.4204
0.05
0.5332
0.06
0.6257
0.07
0.7001
0.08
0.7598
0.09
0.8080

The graph between fraction of occupied sites and oxygen partial pressure is created as,

We see that the occupancy rises gradually for smaller values of P. This shows that with cooperative binding, hemoglobin reacts with oxygen much fast in cells when Pis small.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The sun is the only star whose size we can easily measure directly; astronomers therefore estimate the sizes of other stars using Stefan's law.

(a) The spectrum of Sirius A, plotted as a function of energy, peaks at a photon energy of2.4eV, while Sirius A is approximately 24times as luminous as the sun. How does the radius of Sirius A compare to the sun's radius?

(b) Sirius B, the companion of Sirius A (see Figure 7.12), is only role="math" localid="1647765883396" 3%as luminous as the sun. Its spectrum, plotted as a function of energy, peaks at about7eV. How does its radius compare to that of the sun?

(c) The spectrum of the star Betelgeuse, plotted as a function of energy, peaks at a photon energy of 0.8eV, while Betelgeuse is approximately10,000times as luminous as the sun. How does the radius of Betelgeuse compare to the sun's radius? Why is Betelgeuse called a "red supergiant"?

The argument given above for why CvTdoes not depend on the details of the energy levels available to the fermions, so it should also apply to the model considered in Problem 7.16: a gas of fermions trapped in such a way that the energy levels are evenly spaced and non-degenerate.

(a) Show that, in this model, the number of possible system states for a given value of q is equal to the number of distinct ways of writing q as a sum of positive integers. (For example, there are three system states for q = 3, corresponding to the sums 3, 2 + 1, and 1 + 1 + 1. Note that 2 + 1 and 1 + 2 are not counted separately.) This combinatorial function is called the number of unrestricted partitions of q, denoted p(q). For example, p(3) = 3.

(b) By enumerating the partitions explicitly, compute p(7) and p(8).

(c) Make a table of p(q) for values of q up to 100, by either looking up the values in a mathematical reference book, or using a software package that can compute them, or writing your own program to compute them. From this table, compute the entropy, temperature, and heat capacity of this system, using the same methods as in Section 3.3. Plot the heat capacity as a function of temperature, and note that it is approximately linear.

(d) Ramanujan and Hardy (two famous mathematicians) have shown that when q is large, the number of unrestricted partitions of q is given approximately by

p(q)e2q343q

Check the accuracy of this formula for q = 10 and for q = 100. Working in this approximation, calculate the entropy, temperature, and heat capacity of this system. Express the heat. capacity as a series in decreasing powers of kT/, assuming that this ratio is large and keeping the two largest terms. Compare to the numerical results you obtained in part (c). Why is the heat capacity of this system independent of N, unlike that of the three dimensional box of fermions discussed in the text?

Consider a degenerate electron gas in which essentially all of the electrons are highly relativistic mc2so that their energies are =pc(where p is the magnitude of the momentum vector).

(a) Modify the derivation given above to show that for a relativistic electron gas at zero temperature, the chemical potential (or Fermi energy) is given by =

=hc(3N/8V)1/3

(b) Find a formula for the total energy of this system in terms of N and .

A ferromagnet is a material (like iron) that magnetizes spontaneously, even in the absence of an externally applied magnetic field. This happens because each elementary dipole has a strong tendency to align parallel to its neighbors. At t=0the magnetization of a ferromagnet has the maximum possible value, with all dipoles perfectly lined up; if there are Natoms, the total magnetization is typically~2eN, where 碌a is the Bohr magneton. At somewhat higher temperatures, the excitations take the form of spin waves, which can be visualized classically as shown in Figure 7.30. Like sound waves, spin waves are quantized: Each wave mode can have only integer multiples of a basic energy unit. In analogy with phonons, we think of the energy units as particles, called magnons. Each magnon reduces the total spin of the system by one unit of h21rand therefore reduces the magnetization by ~2e. However, whereas the frequency of a sound wave is inversely proportional to its wavelength, the frequency of a spin-wave is proportional to the square of 1.. (in the limit of long wavelengths). Therefore, since=hfand p=h.. for any "particle," the energy of a magnon is proportional

In the ground state of a ferromagnet, all the elementary dipoles point in the same direction. The lowest-energy excitations above the ground state are spin waves, in which the dipoles precess in a conical motion. A long-wavelength spin wave carries very little energy because the difference in direction between neighboring dipoles is very small.

to the square of its momentum. In analogy with the energy-momentum relation for an ordinary nonrelativistic particle, we can write =p22pm*, wherem* is a constant related to the spin-spin interaction energy and the atomic spacing. For iron, m* turns out to equal 1.241029kg, about14times the mass of an electron. Another difference between magnons and phonons is that each magnon ( or spin-wave mode) has only one possible polarization.

(a) Show that at low temperatures, the number of magnons per unit volume in a three-dimensional ferromagnet is given by

NmV=22mkTh2320xex-1dx.

Evaluate the integral numerically.

(b) Use the result of part (a) to find an expression for the fractional reduction in magnetization, (M(O)-M(T))/M(O).Write your answer in the form (T/To)32, and estimate the constantT0for iron.

(c) Calculate the heat capacity due to magnetic excitations in a ferromagnet at low temperature. You should find Cv/Nk=(T/Ti)32, where Tidiffers from To only by a numerical constant. EstimateTifor iron, and compare the magnon and phonon contributions to the heat capacity. (The Debye temperature of iron is 470k.)

(d) Consider a two-dimensional array of magnetic dipoles at low temperature. Assume that each elementary dipole can still point in any (threedimensional) direction, so spin waves are still possible. Show that the integral for the total number of magnons diverge in this case. (This result is an indication that there can be no spontaneous magnetization in such a two-dimensional system. However, in Section 8.2we will consider a different two-dimensional model in which magnetization does occur.)

Consider an isolated system of Nidentical fermions, inside a container where the allowed energy levels are nondegenerate and evenly spaced.* For instance, the fermions could be trapped in a one-dimensional harmonic oscillator potential. For simplicity, neglect the fact that fermions can have multiple spin orientations (or assume that they are all forced to have the same spin orientation). Then each energy level is either occupied or unoccupied, and any allowed system state can be represented by a column of dots, with a filled dot representing an occupied level and a hollow dot representing an unoccupied level. The lowest-energy system state has all levels below a certain point occupied, and all levels above that point unoccupied. Let be the spacing between energy levels, and let be the number of energy units (each of size 11) in excess of the ground-state energy. Assume thatq<N. Figure 7 .8 shows all system states up to q=3.

(a) Draw dot diagrams, as in the figure, for all allowed system states with q=4,q=5,andq=6. (b) According to the fundamental assumption, all allowed system states with a given value of q are equally probable. Compute the probability of each energy level being occupied, for q=6. Draw a graph of this probability as a function of the energy of the level. ( c) In the thermodynamic limit where qis large, the probability of a level being occupied should be given by the Fermi-Dirac distribution. Even though 6 is not a large number, estimate the values of and T that you would have to plug into the Fermi-Dirac distribution to best fit the graph you drew in part (b).

A representation of the system states of a fermionic sytern with evenly spaced, nondegen erate energy levels. A filled dot rep- resents an occupied single-particle state, while a hollow dot represents an unoccupied single-particle state . {d) Calculate the entropy of this system for each value of q from 0to6, and draw a graph of entropy vs. energy. Make a rough estimate of the slope of this graph near q=6, to obtain another estimate of the temperature of this system at that point. Check that it is in rough agreement with your answer to part ( c).

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.