Michaelis-Menten Plot

Transcription

1 NAM: HAITI CHMISTRY, SPRING, 17 Circle Section Number: 1 11 amination, April 15, 17 Answer each question in the space provie; use back o page i etra space is neee. Answer questions so the graer can RADILY unerstan your work; only work on the eam sheet will be consiere. Write answers, where appropriate, with reasonable numbers o signiicant igures. You may use only the "Stuent Hanbook," a calculator, an a straight ege points The table below gives the initial velocity o a reaction catalyze by a ecarboylation enzyme measure by the rate o CO ormation. Determine the Michaelis-Menten constant an the maimum velocity base on these ata by plotting the appropriate ata in the graph provie. Show your work clearly on the graph, the answer without eplanation will be given no points. Substrate concentration [M] Initial velocity [µmol CO/ min] One has to plot 1/v vs. 1/[S]. It is convenient to write the velocity on a per-minute basis, rather than on a per-two-minute basis. Michaelis-Menten Plot DO NOT WRIT IN THIS SPAC p. 1 /15 p. /15 p. /1 p. /1 p. 5 /15 p. 6 /1 1/v min/µmol CO y R² /[S] L/mol p. 7 /15 p. 8 /1 p. 9 /5 tra creit TOTAL PTS /1 From the graph intercept, 1/Vma.99 [min/µmol o CO], an Vma. [µmol CO/min] From the slope KM/Vma 1.5 [min/µmol CO/L/mol] KM 1.5 [min/µmol CO/L/mol]. µmol CO/min. mol/l. M

2 NAM: CHM, am, Spring, 17, page. 15 points Fill in the blanks. 1 The physical meaning o the Michaelis constant is that this is the substrate concentration require to reach hal maimum velocity. A critical parameter in evaluating surace asorption is the ractional coverage, eine as the ratio o the number o asorption sites occupie to the total number o asorption sites. In a boune quantum mechanical system, only certain values o energy are allowe, an such a system has a iscrete energy spectrum. I any well-behave wave unction o a system can be represente as a linear combination o a set o unctions φn, this set o unctions is complete. 5 In quantum mechanics, in any single measurement o the observable that correspons to the operator A, the only values that will ever be measure are the eigenvalues o that operator. Score or Page

3 Score or Page NAM: CHM, am, Spring, 17, page. 1 points Fin the simplest orm or the commutator o the operators. Put answer in the space provie; show work below the equation. a an, Thus,, is the simplest orm o this commutator an the operators o not commute. b an, Thus,, is the simplest orm o this commutator an the operators o not commute.

4 NAM: CHM, am, Spring, 17, page. 1 points From the statements in column B, select the best match or each phrase in column A an put its letter in the ajacent blank. There is only one best match or each phrase. Column A 1. I an operator acting upon a wave unction returns the wave unction multiplie by a constant, this unction is calle an m o the operator.. All quantum mechanical operators are Hermitian, meaning that they have q eigenvalues.. ψ ψ a ; ψ or > a an or < is a list o i or a particle in a one-imensional bo o length a.. h is an acceptable wave unction over the interval 1 < < 1 5. I several istinctly ierent wave unctions o the total energy operator correspon to the same energy, the corresponing energy level is k Column B a eigenequation b borerline conitions c A ep-, where A is a normalization constant egenerative e borer conitions unrealistic g comple h A sin, where A is a normalization constant i bounary conitions j split k egenerate l A/, where A is a normalization constant m eigenunction n A 1/, where A is a normalization constant o realistic p eigenvalue q real r Snoopy Score or Page

5 NAM: CHM, am, Spring, 17, page points Determine the normalization constant or the groun state o the harmonic oscillator, given by Ψ AA eeeeee αα where α is the characteristic istance an A is the normalization constant. To normalize the wave unction, one requires its integral over all space be 1. Then Ψ Ψ 1 AA AA ee αα eeeeee αα AA AA eeeeee αα 1 Because o the symmetry o the unction about, one may write the equation as: AA AA eeeeee αα This integral is oun in the integral table in your hanbook. Substituting or it gives the ollowing relation: AA AA αα ππ 1 Assuming that A is real, one may reuce this equation. AA αα ππ 1 Hence, one ins a value or the normalization constant. 1 The ull, normalize wave unction is AA 1 ππ αα Ψ 1 ππ αα eeeeee αα Score or Page

6 NAM: CHM, am, Spring, 17, page points A particle in a one-imensional bo can be in a state speciie by the normalize wave unction: Ψ nn aa ssssss nnnnnn aa where aa is the with o the bo that stretches rom to aa. nn is the quantum number o the state. Fin the probability that the particle will be oun in the let quarter o the bo. Show your work. The probability o ining a particle in a range is By substitution, one has, or this wave unction PP, aa aa/ Ψ Ψ aa/ PP, aa aa ssssss nnnnnn aa Let us convert this equation to a unction o yy nnnnnn/aa. PP, aa nnnn ssssss nnnn/ yy This integral is given in Table. o the Hanbook: PP, aa nnnn yy 1 nnnn/ ssssssyy 1 nnnn nnnn 1 ssssss nnnn 1 sin This can be simpliie to PP, aa 1 1 nnnn ssssss nnnn This answer is acceptable, however one may go urther by realizing that, or nn even, the secon term is always zero. Thus, one may write PP nn eeeeeeee, aa 1 For nn o, there are two possibilities. I the quantum number is 1, 5, 9,.., the secon term is 1 PP 1,5,9,1,, aa 1 1 nnnn nnnn, an the probability is I the quantum number is, 7, 11, 15,.., the secon term is 1 nnnn, an the probability is PP,7,11,, aa 1 1 nnnn Score or Page

7 NAM: CHM, am, Spring, 17, page points The π electrons o metal porphyrins can be reasonably easily escribe using a simple moel o a two-imensional bo. a Obtain the energy levels o a ree electron in a two-imensional square bo o length a in the units o h /8ma, where is Planck s constant an mm is the electron mass. n n n n h n 1 1 8ma b For a porphyrin-like hemin that contains 6π electrons, sketch the energy iagram o the groun state accounting or all o them within the particle in a two-imensional bo moel. Remember that every orbital can it up to two electrons even i several orbitals have the same energy o 1 5; egeneracy 8o 1 1; egeneracy 1; egeneracy 1 17; egeneracy 18o ; egeneracy 5; egeneracy 51 6; egeneracy 5 9; egeneracy o c The porphyrin structure measures about 1 nm on a sie. Calculate the lowest-energy absorption ban position or this molecule the eperimental observation is 6 nm. This transition woul correspon to the minimum energy, or transition J s kg 1 1 m 5h ma or 8 c hc J s m / s 7 hν h an λ m nm 19 λ.1 1 J 19 J Score or Page

8 NAM: CHM, am, Spring, 17, page points For the phrases in column A, in the appropriate einition in column B an insert its letter in the blank net to the phrase. 1. h Ban gap Column A. i Corresponence principle Column B a. Properties o couple particles are relate an can be inerre, no matter how separate the particles are. b. Conition in which a quantum particle is in a region a classical particle cannot be. e Degeneracy c. Quantum moel or rotational motion. a ntanglement. Wave unctions mae to represent probability 5. j Heisenberg s principle e. Multiple states with the same value o a parameter 6. Normalization. Non-zero minimum energy 7. c Rigi rotor g. Probability ensity 8. b Tunneling h. Region with no allowe eigenunctions o the total energy 9. g Ψ Ψ i. Speciication o operators 1. Zero-point energy j. Statement o the impossibility o measuring two noncommuting properties o a quantum system with ininite precision Score or Page

9 NAM: CHM, am, Spring, 17, page points, etra creit Consier the ollowing potential energy proile in one imension: Suppose that a particle is travelling rom the let sie an encounters this barrier, with a momentum, kk. a What is the wave unction or this particle i V is zero? Ψ llllllll AA eeeeeeiiiiii This is the equation o a ree particle travelling rom the let to the right with momentum kk. b I the barrier is not zero, an i the energy o the particle is, write an equation or the particle in the right region when the potential is eective. Give a solution o this equation or the conition that VV. This is just a writing own o Schroeinger s equation in this region: mm This equation may be written in many ways. Rewrite it this way: Ψ rrrrrrhtt VV Ψ rrrrrrhtt Ψ rrrrrrhtt The solution to this equation is Ψ rrrrrrhtt mmvv Ψ rrrrrrhtt Ψ rrrrrrhtt AA eeeeeessss, where A is a normalization constant an ss mmvv. Note that there coul mathematically be a secon particular solution Ψ rrrrrrhtt BB eeeeeessss, however it must be rejecte because it iverges as goes to ininity, which violates the postulates o quantum mechanics. Thus, B woul be a requirement, yieling the wave unction above as the only acceptable solution in this region. Score or Page