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Problem Set 5 (Due Dec. 9 th , Tuesday, 8pm EST) Please make sure to show your work and calculations and state any assumptions you make in answering the following questions. Include the names of the people you worked with at the top of your problem set. Here’s a summary of files you need to submit: If your name is John Harvard and you’re in Lan Zhang’s section: JohnHarvard_ps5_LZ.doc JohnHarvard_ps5_LZ.pdb JohnHarvard_ps5_LZ.xls 1. Protein structure (35 points total) You may find the following resources useful in answering this part: The Protein Data Bank (PDB) - http://www.rcsb.org/pdb/ MacroMolecular DataBase (MMDB) - http://www.ncbi.nlm.nih.gov/Structure/ Rasmol - http://www.umass.edu/microbio/rasmol/getras.htm (for v.2.6b2a) or http://www.bernstein-plus-sons.com/software/rasmol/ (for v. 2.7.2.1) Biology WorkBench - http://workbench.sdsc.edu/ Swiss-PdbViewer - http://us.expasy.org/spdbv/ In 1999, the atomic structure of GCSF-receptor complex was resolved. The following paper describes discovery of this new cytokine-receptor recognition method and will be useful in answering questions in this part. Aritomi M, Kunishima N, Okamoto T, Kuroki R, Ota Y, Morikawa K. Atomic structure of the GCSF-receptor complex showing a new cytokine-receptor recognition scheme. N ature. 1999 Oct 14;401(6754):713-7. 1.1. The Protein Data Bank (PDB) and MacroMolecular DataBase (MMDB) (14 pts) Page 1 of 25

Transcript of Problem Set 5 - HUIT Sites Hostingsites.fas.harvard.edu/~bphys101/problemsets/PS5_answer... · Web...

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Problem Set 5

(Due Dec. 9th, Tuesday, 8pm EST)

Please make sure to show your work and calculations and state any assumptions you make in answering the following questions. Include the names of the people you worked with at the top of your problem set.

Here’s a summary of files you need to submit:If your name is John Harvard and you’re in Lan Zhang’s section:

JohnHarvard_ps5_LZ.docJohnHarvard_ps5_LZ.pdbJohnHarvard_ps5_LZ.xls

1. Protein structure (35 points total)

You may find the following resources useful in answering this part: The Protein Data Bank (PDB) - http://www.rcsb.org/pdb/ MacroMolecular DataBase (MMDB) - http://www.ncbi.nlm.nih.gov/Structure/ Rasmol - http://www.umass.edu/microbio/rasmol/getras.htm (for v.2.6b2a) or

http://www.bernstein-plus-sons.com/software/rasmol/ (for v. 2.7.2.1) Biology WorkBench - http://workbench.sdsc.edu/ Swiss-PdbViewer - http://us.expasy.org/spdbv/

In 1999, the atomic structure of GCSF-receptor complex was resolved. The following paper describes discovery of this new cytokine-receptor recognition method and will be useful in answering questions in this part. Aritomi M, Kunishima N, Okamoto T, Kuroki R, Ota Y, Morikawa K. Atomic structure of the GCSF-receptor complex showing a new cytokine-receptor recognition scheme. N ature. 1999 Oct 14;401(6754):713-7.

1.1. The Protein Data Bank (PDB) and MacroMolecular DataBase (MMDB) (14 pts)

1.1.1 Complex of GCSF with its Receptor. According to the Aritomi et al. paper, unlike the 1:2 ligand-receptor stoichiometry found in GH and EPO complexes with their respective receptors, the stoichiometry of GCSF ligand with its receptor is 2:2. Go to the Protein Data Bank (PDB). Using the accession code found at the very end of the Aritomi et al. paper, download the .pdb crystal structure file that has no more than one 2:2 complex in asymmetric unit to your local computer. (Hint: Examine Table 1 in the article to determine which form satisfies this criterion and then match the accession code with that form.)

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1.1.1.1 What is the accession code of .pdb crystal structure file that you retrieved from PDB (1pt)?

1cd9

Title: 2:2 Complex Of G-CSF With Its ReceptorAuthors: M. Aritomi, N. Kunishima, T. Okamoto, R. Kuroki, Y. Ota, K. MorikawaSpace group: I4122 (as Table 1 in the article indicated, this has one 2:2 complex in asymmetric unit)

1.1.1.2 Open the .pdb file you downloaded in a text editor and examine it. Since you downloaded the crystal structure which had no more than one 2:2 complex in asymmetric unit, and applying what you already learned about GCSF and its Receptor, how many distinct protein chains do you expect to find in this .pdb file and how many do you find (1pt)? For each chain, indicate the chain label and whether it is ligand or receptor (2 pts).

As expected, one finds 4 distinct chains in one 2:2 complex in asymmetric unit, 2 GCSF ligand chains labeled A and C, and 2 GCSF receptor chains labeled B and D, thus forming the 2:2 dimer. Note that in examining the format of the .pdb file, one column in the ATOM coordinates is dedicated to indicate chain.

1.1.1.3 Examine the .pdb file further in the text editor. A helpful Guide to the PDB File Format can be found at: http://www.rcsb.org/pdb/docs/format/pdbguide2.2/guide2.2_frame.html

Scroll down to the “ATOM” coordinate section. There are twelve columns of data, including the leftmost section label ATOM. What does each of these columns indicate? List the 12 definitions used in our .pdb file. Be careful, since CA indicates something different in the protein from C, the Guide indicates the type of data that should be found in a given column, and some columns in the Guide may not be in our .pdb. (6pts total, 0.5 pt per column)

Taken from Chapter 9. Coordinate Section, ATOM entry, of the Guide provided above:1. Section label ATOM2. Atom serial number3. Atom name AND alternate location indicator (CA = Cα carbon, C is the name and α the alternate location)4. Amino acid (AA) residue name to which atom belongs5. Chain identifier to which atom belongs6. AA residue sequence number to which atom belongs7. External/Orthogonal X coordinate in Angstroms.8. External/Orthogonal Y coordinate in Angstroms.9. External/Orthogonal Z coordinate in Angstroms.10. Occupancy. 11. Temperature factor.

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12. Element symbol.

Note that in our 1cd9.pdb, columns for code for insertion of residues, segment identifier, and charge on the atom are missing.

1.1.2 Related domains of GCSF and GCSF Receptor. Go to the MacroMolecular DataBase (MMDB) and enter the accession code in the search field to obtain an MMDB structure summary. What are the conserved and 3D domains for each of the chains in our .pdb file? List the chain name and any conserved and 3D domains that belong to it by number and description by filling in the chart below (4 pts total, 1 pt per row). If for one chain, there is no conserved or 3D domain indicated on the MMDB page, just leave it blank. [Hint: Notice the symmetry!]

Chain Name

Conserved Domain (CD) Number

and Description

3D Domain Number and Description

A IL6, Interleukin-6/G-CSF/MGF family

B 1, 2:2 Complex Of G-Csf With Its Receptor2, 2:2 Complex Of G-Csf With Its Receptor

C IL6, Interleukin-6/G-CSF/MGF family

D 1, 2:2 Complex Of G-Csf With Its Receptor2, 2:2 Complex Of G-Csf With Its Receptor

1.2 Rasmol (12 pts)

Binding of GCSF to its Receptor. Go to the Rasmol site, download and install the Rasmol program appropriate for your computer architecture to your local computer. Start the Rasmol program (which opens two windows, Rasmol display window and Rasmol command line window) and open our .pdb file. Use the Rasmol Reference Manual for help on how to use the command line or using mouse controls to view and manipulate the crystal structures, as well as other operations.

1.2.1 Let’s examine the overall structure of the 2:2 dimer. Change the Display to “Cartoons” and Colors to “Structure.” Under Options, enable “Labels” so you know what structural features are found on which chains. What is the predominant secondary structural feature of GCSF and of its GCSF Receptor? For each chain, indicate the Chain Name, Secondary Structural Feature, and whether it is GCSF Ligand or GCSF Receptor by filling in the chart below (4 pts total, 1 pt each row). [Hint: Notice, again, the symmetry!]

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Chain Name Secondary Structural Feature GCSF Ligand or GCSF Receptor?A predominantly alpha helix GCSF ligandB predominantly beta sheet GCSF receptorC predominantly alpha helix GCSF ligandD predominantly beta sheet GCSF receptor

1.2.2 Now let’s examine the interfaces between these chains, which are sites of interest for binding and dimerization. Reset your view by typing “reset” at the command line and then “”. Change the Display to “Wireframe” and Colors to “Chain.” Under Options, disable “Hetero atoms.” We are particularly interested in amino acids with charged side groups no more than 2.0 Angstroms from the centers of the interfaces between these chains. Assume the centers of these interfaces are at the following residues:

A ASP113 B TYR143 C ASP113 D TYR143

If you’re clever and use various commands found in the Rasmol Reference Manual, under the sections called “Predefined Sets” and “Within Expressions” you can:

1. select everything but the set of charged amino acids and hide it (same as coloring everything but the charged amino acids black).

2. select and highlight with another color residues no more than 2.0 Angstroms from the residues above.

3. label only the residues that pass requirements 1 (charged amino acid) AND 2 ( 2.0 Angstroms from ASP113:A / TYR143:B / ASP113:C / TYR143:D) above and fill in the chart below with those labels.

If you do not manage to select the appropriate residues using the command line commands, under Options, enable “Slab Mode” and “Labels.” While in “Slab Mode,” methodically slice through the .pdb file slowly from front to back using Ctrl-Left Mouse button, see how to measure distance from atom to atom and fill in the table below (8 pts total, 2 pts per row). Indicate residue name, number and chain, e.g. GLY123:A, and sort them by chain, e.g. the GLY123:A would go in the row with Chain Name A. [Hint: Once again, symmetry!]

Chain Name Charged Residues No More Than 2.0Å from Center of InterfacesA ASP113:AB GLU142:BC ASP113:CD GLU142:D

The RasMol command line commands used (but student does not have to include) were:

RasMol> select not charged4873 atoms selected!RasMol> color blackRasMol> select within ( 2.0, ASP113 )20 atoms selected!

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RasMol> color yellowRasMol> label onRasMol> select within ( 2.0, TYR143 )28 atoms selected!RasMol> color redRasMol> label on

Thus, chains A and C with residues of interest will be yellow while chains B and D with residues of interest will be red.

Note that the series of commands above also highlight ALA112:A, PHE114:A, TYR143:B, MET144:B, ALA112:C, PHE114:C, TYR143:D, MET144:D, but they are NOT charged but rather neutral amino acids. See the Venn Diagram below, taken from the RasMol Reference Manual:

1.3 Swiss-PdbViewer (9 pts)

Examination and modification of the structure of GCSF bound to its Receptor. Go to the Swiss-PdbViewer site. Download and install the Swiss-PdbViewer program appropriate for your computer architecture to your local computer. Start the Swiss-

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PdbViewer program (starts one window, DeepView / Swiss-PdbViewer 3.7) and open our .pdb file.

Swiss-PdbViewer works similarly to RasMol, in that atoms or residues have to be selected before they can be manipulated. The Control Panel lists the .pdb file being manipulated, whether that .pdb file is visible in the viewport, whether it can be moved in the viewport and characteristic features (like chain, whether alpha helix (h) or beta sheet (s), whether backbone is visible, side chain is visible, label is visible, VDW radius electron density, ribbon is visible, color of selected pulldown (default B for backbone color)). Clicking on a chain letter highlights all amino acids in that chain, similarly for h and s). Peruse the Help menu and the User Guide for information on the menus, manipulation, display and rendering.

The Biophysics 101 Teaching Staff was able to crystallize a 13 amino acid residue protein fragment of interest, which they named B101, from the GCSF-GCSF Receptor system. However, due to miscalculation and experimental error during the x-ray diffraction process, the resulting B101.pdb coordinate file was slightly damaged. With a December 9, 2003 8PM EST deadline looming, they are asking your help in identifying, space transforming and repairing the B101.pdb.

Download and open B101.pdb in Swiss-PdbViewer. It may look as though nothing has happened. In fact, B101 has been centered in the same spatial grid as our original .pdb file. In the Control Panel, uncheck “visible” and “can move” when our original .pdb is selected. Now you should be able to see B101.

Let us now highlight our entire protein fragment so we can monitor any manipulations more easily Change the protein to B101 in the Control Panel by click on the Control Panel heading and pressing TAB. Pressing TAB again changes the Control Panel back to our original .pdb file. Select all of B101’s amino acid residues (Ctrl+A or Select→All from the menu), click on the word “col” in the Control Panel, and select a color for B101. Now if you rotate and zoom, you should be able to see B101 more clearly as a solid color wireframe among our original .pdb file, which incidentally is colored CPK.

Note that B101 does not have the same coordinate system and origin as our original .pdb file, since it was crystallized independently. Our first task is to modify the coordinate system of B101 to the same coordinate system of our original .pdb file. We can randomly move and rotate B101 (by checking “visible” and “can move” for B101) while keeping our original .pdb file fixed (by checking “visible” and unchecking “can move” for our original .pdb file) and try to match them up by eye. But we have no idea where B101 belongs in our original .pdb file and have a deadline to meet! Fortunately, Swiss-PdbViewer has a built-in iterative magical fitting algorithm, which attempts to spatially overlap one protein structure on top of another.

Also note that B101 could be a fragment from any part of the GCSF-GCSF Receptor system, since it was obtained independently. Our second task is to figure out where B101 is in our original .pdb file. Taking a cue from ideas learned earlier in the semester, you

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can perform a protein sequence alignment between these two proteins to determine where the B101 fragment most likely came from. We can certainly use other alignment programs, used during the semester. But to do so, we would have to meticulously extract the amino acid sequences from both B101 and our original .pdb file, perhaps by a Perl script or by hand. Fortunately, Swiss-PdbViewer has a built-in structural alignment feature to expedite this process. Click on Window and select Alignment to open the Alignment window.

By default B101 rests in a different coordinate space and is aligned to the very beginning of our original .pdb file. Once we perform an Iterative Magic Fit and Generate a Structural Alignment, we can space transform and identify B101 with respect to our original .pdb file.

1.3.1 First let’s perform an Iterative Magic Fit:

Make sure both proteins are visible and can be moved. Select Iterative Magic Fit (Shift+Ctrl+M or Fit→Iterative Magic Fit from

the menu). Click OK to use just the CA (carbon alpha) atoms in performing the fit, as

they are least prone to be inaccurate in B101. The structure of B101 should have moved to fit some of the atoms in our

original .pdb file. Rotate and zoom in to the area and determine its location with respect to our original .pdb file by selectively viewing only the amino acids from one chain while hiding the rest and iterating this until you find B101 in the visible chain.

Where has B101 been placed? Specify in the table below whether it is near GCSF ligand or receptor (1pt) and near which chain from our original .pdb file (1 pt).

GCSF ligand or receptor Near which chain from our original .pdb fileGCSF ligand Chain C

1.3.2 Next let’s generate a Structural Alignment: Make sure both proteins are visible and can be moved and our

original .pdb file is under consideration in the Control Panel. Select Generate Structural Alignment (Ctrl+G or Fit→Generate Structural

Alignment from the menu). The sequence of B101 should have aligned somewhere in the Alignment

window. Mouse over the amino acids in the Alignment window to determine its location with respect to our original .pdb file.

Where did B101 align with respect to the amino acids in our original .pdb file? Specify in the table below the start amino acid name, number and chain (1pt) and the end amino acid name, number and chain of the original .pdb file (1pt), using the same naming convention, e.g. GLY123:A.

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Start: LEU107:CEnd: TRP119:C

1.3.3 Note that as you zoom into the structural alignment of B101, one residue side chain fails to overlap our original .pdb file. You can identify the side chain by clicking on the

identity toolbar icon and then clicking on the amino acid of interest. Alternately, as you mouse over B101’s residue letters in the Alignment window the structures will blink in the viewport.

Which amino acid residue in B101 has this non-overlapping side chain (1pt)? What is the corresponding amino acid residue in our original .pdb file (1pt)? State in the table below the residue names, numbers and chains using the same naming convention above.

B101 residue name, number and chain with nonoverlapping sidechain:

PHE8:Z

The corresponding amino acid residue from original .pdb file’sresidue name, number and chain:

PHE114:C

1.3.4 Use the torsion feature to make this sidechain overlap our original .pdb file as follows. After you are done, submit a new .pdb file, called FirstnameLastname_ps5_TFinitials.pdb (3 pts).

It may help if you hide everything else from our original .pdb file but the amino acid that you wish to overlap. To do so, make sure the original .pdb file is under consideration in the Control panel. Select the corresponding amino acid residue that you found in above in IV.1.c by clicking on it in the Control Panel. Now click “show” and “side” in the Control Panel header. Also uncheck “can move” since you want this side chain to remain static.

Now to align the nonoverlapping sidechain. Change the Control Panel so now B101 is under consideration.

Click on the torsion toolbar icon and then select the B101 residue that you want to twist in the viewport. Remember the B101 residues are one solid color, which you specified earlier in the PS. Avoid selecting the original .pdb file residue by clicking on the part of the B101 sidechain, since it does not overlap the original .pdb file like the backbones.

Four left-right arrows should appear to the right of and below the torsion toolbar icon. Use the topmost left-right arrows to rotate the sidechain so that it overlaps the original .pdb files side chain.

Once you have a close overlap, make sure once again that B101 is under consideration in the Control Panel and save your structure (Ctrl+S or File→Save→Layer…) using the naming convention above.

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Please see PS5_answer_2003.pdb. Below are screenshots of the area of overlap between 1cd9.pdb (in CPK, white labels) and B101.pdb (in green, no labels) before and after PHE8:Z (red arrow) torsion bond modification.

BEFORE TORSION MODIFICATION

AFTER TORSION MODIFICATION

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2. Mass spectroscopy (26 pts)

2.1 Mass spec basics (13 pts)

2.1.1 What are the three essential modules for every mass spectrometer? Explain what the MALDI technique is to which of the three aforementioned subunits does it belong? (6 pts)

Ion-source, mass analyzer, and detector. 1 point each.

MALDI stands for Matrix Assisted Laser Desorption Ionization. It is an ionization method (i.e. part of ion source) where your molecule is dissolved in a matrix solution containing a UV absorbing molecule. The matrix mixture is added to a target source and dried to produce crystals. The target is placed in voltage potential field and is hit with a laser; the matrix molecules absorb energy from the laser and energize and charge the test molecule. The embedded molecules are now ionized and the energy causes them to transfer to the gas phase. Since the molecules are in a voltage gradient and they are charged they are accelerated through the field.

1 point for spelling it out, 2 points for description based on accuracy.

2.1.2 Magnets are a classic type of mass analyzer used by mass spectrometers, name two newer mass analyzer types and briefly explain (three sentences each) how they resolve ions of different masses. (4 pts)

One type of mass analyzer is the time of flight (TOF) mass analyzer. It works by measuring the time an ion takes to fly through a tube. At one end of the tube is the ionization source and at the other is a detector. A voltage gradient is needed to accelerate the ions. The kinetic energy is applied uniformly to all ions so those with a higher mass to charge ratio accelerate more slowly and take longer to travel through the tube.

Another mass analyzer is the ion-trap mass analyzer. The trap is made from three hyperbolic electrodes that create an electric field capable of focusing and trapping ions within stable orbits. Ions are injected into the trap and sequentially released by adjusting the voltages and frequencies to destabilize the orbits of particular ions at specific m/z values. The ions are ejected out of the trap as their orbits destabilize and detected by a electron multiplier detector.

2 points for 2 examples, 2 points for their descriptions.

2.1.3 How can you use mass spectrometry to locate a disulfide bridge between two peptides? (3 pts)

You can look for the mass of the dipeptide to be equal to the combined mass of two separate peptides minus 2 hydrogens masses lost during disulfide arrangement.

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2.2 MS/MS (13 pts)

2.2.1 Describe the MS/MS technique in three sentences or less. (3 pts)

MS/MS refers to the practice of selecting a molecule of interest within the ion-source of mass analyzer and then fragmenting the molecule to produce fragments that can be detected and used to help determine the sequence structure of the intact molecule.

2.2.2 What’s the difference between b and y ions? (2 pts)

A b-ion retains its charge on the N-terminal of the fragmented peptide and a y-ion retains the charge on the C-Terminal of the peptide.

1 points each.

2.2.3 You isolate a peptide through co-immunoprecipitation with the human growth hormone. After trypsin cleavage, you obtain a mass spectrum for one of the resulting fragments. The spectrum you obtained has b-ion series values as follows (5 pts)

263.09 378.12 475.17 588.25 701.34 802.38 903.43 990.46 1089.53 1186.59 1285.65 1448.72 1535.75 1648.83

Use the provide values to solve the partial peptide sequence (if there is a position that is ambiguous between amino acid X and Y, mark it X/Y):

http://i-mass.com/guide/aamass.html

DP(I/L)(I/L)TTSVPVYS(I/L)

2.2.4 After performing other enzymatic cleavages, obtaining many more spectra, obtaining side chain fragmentation data, and piecing everything together, you find a more complete and polished peptide sequence:

netkwkmmdpilttsvpvyslkvdkeyevrvrskqrnsgnygefsevlyvtlpqmsqftceedfyfpwlliifgifgltvmlfvflfskqqrikmlilppvpvpkikgidpdllkegkleevntilaihdsykpefhsddswvefieldidepdekteesdtdrllssdhekshsnlg

Use the tools you’ve learned in this course to find what the protein is. (3 pts)

The protein is somatotropin receptor precursor or growth hormone receptor precursor.

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3. Metabolic network, chemical kinetics, Flux Balance Analysis (39 pts)

For this part, the following readings might be helpful in addition to the lecture and section notes:

Schilling, CH, Edwards, JS, and Palsson, BO . Towards metabolic phenomics: Analysis of genomic data using flux balances. Biotechnol Prog 15: 288-295 (1999).

Schilling, CH et al. Metabolic pathway analysis: Basic concepts and scientific applications in the post-genomic era. Biotechnol Prog 15: 296-303 (1999).

Segre, D, Vitkup, D, and Church, GM. Analysis of optimality in natural and perturbed metabolic networks. Proc. Nat. Acad. Sci USA 99: 15112-7 (2002).

To make it clearer, things you’re asked to do are highlighted in bold.

3.1 Using the formalism of chemical kinetics, formulate a system of differential equations to describe the change of concentration with respect to time of each of the species A through F in the figure below. The ith rate constant is denoted symbolically as ki. (6 pts)

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B

A

F

2D

E

Ck3

k2

k9

k5

k1

k4

k7k6 k8D

2Bk- 2

k- 7

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3.2 Instead of focusing on the enzyme kinetic constants and concentrations of metabolites, we can use a more tractable method of analyzing the fluxes (reaction rates). Given the labeling of fluxes in the figure below, write a system of differential equations to describe the change of concentration with respect to time of each of the species A through F. Note a) the reaction rate incorporates all relevant rate constants; b) v2 and v7 are the net reaction rates of reversible reactions. (6 pts)

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B

A

F

2D

E

Cv3

v2

v9

v5

v1

v4

v7v6 v8D

2B

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3.3 We are going to examine this system more carefully at steady state. Which two fundamental assumptions allow us to simplify this model for analysis at steady state (each in a single sentence) (2 pts)? Rewrite the equations from question 3.2 under the steady state condition (2 pts).

Assumptions: 1) There is no net accumulation of metabolites in the system over time.2) The time constants for metabolic reactions are very rapid as compared to cellular processes such as growth and can be ignored.

3.4 Rewrite the equations from question 3.3 in matrix notation, i.e.

v 9 x 1 matrix of reaction rates S 6 x 9 matrix of stoichiometric coefficients (5 pts)

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3.5 Assume that v1 is limited between 0 and 15 mmol/hr. Formulate a linear programming model to maximize the production of metabolite D (i.e. v8) and use Excel to solve it. Note that in a real system, D may be biomass, ATP production, etc. Provide below (a) the explicit mathematical model in a format similar to the following example; and (b) the optimal solution found using Excel. Please submit your Excel file, named FirstnameLastname_ps5_TFinitials.xls, in which only the answer report should be generated. Name the corresponding worksheet and answer report (NOT the Excel file) as “3.5-FBA-Wildtype” and “3.5-FBA-Wildtype-Answer Report”, respectively. [Hint: The constraints include the mass balance of metabolites, non-negativity constraints, and the input limit constraint. Refer to the answer key to the Excel bonus question in Problem Set 1 for how to solve linear programming problems using Excel.] (7 pts)

Example of a mathematical programming model :

Example of an optimal solution : objective = 5[x1, x2, ……] = [2, 5, … …]

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Optimal solution:Objective v8= 30 mmol/hr[v1, v2, … … , v9] = [15, 15, 0, 15, 15, 0, 0, 30, 0].

See Excel file PS5_answer_2003.xls

3.6. Suppose that the system undergoes a mutation that removes the reaction of converting B and E to D, as shown in the figure below. How can we model such a mutation in the scheme of flux balance analysis? Assuming the same limit on the availability of A as in question 3.5, what is now the possible maximum production rate of metabolite D? Provide (a) the explicit mathematical model; and (b) the optimal solution found using Excel. Submit your Excel file (the same file for question 3.5, named FirstnameLastname_ps5_TFinitials.xls), in which only the answer report should be generated. Name the corresponding worksheet and answer report (NOT the Excel file) as “3.6-FBA-Mutant” and “3.6-FBA-Mutant-Answer Report”, respectively. (4 pts)

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Page 17: Problem Set 5 - HUIT Sites Hostingsites.fas.harvard.edu/~bphys101/problemsets/PS5_answer... · Web viewIn 1999, the atomic structure of GCSF-receptor complex was resolved. The following

Additional constraint: v4 = 0

Optimal solution:Objective v8= 15 mmol/hr[v1, v2, … … , v9] = [15, 0, 15, 0, 0, 0, 15, 15, 0].

See Excel file PS5_answer_2003.xls

3.7 For the perturbed metabolic network in the previous question, an alternative assumption one can make, instead of maximizing the production of D, is the minimization of metabolic adjustment (MOMA). That is, the deviation of the metabolic fluxes from their wild type values is minimized in a mutant. Find the new flux distribution under this assumption. Provide (a) the explicit mathematical model; and (b) the optimal solution. Submit your Excel file (the same file for question 3.5, named FirstnameLastname_ps5_TFinitials.xls), in which only the answer report should be

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B

A

F

2D

E

Cv3

v2

v9

v5

v1

v4

v7v6 v8D

2B STOP

Page 18: Problem Set 5 - HUIT Sites Hostingsites.fas.harvard.edu/~bphys101/problemsets/PS5_answer... · Web viewIn 1999, the atomic structure of GCSF-receptor complex was resolved. The following

generated. Name the corresponding worksheet and answer report (NOT the Excel file) as “3.7-MOMA-Mutant” and “3.7-MOMA-Mutant-Answer Report”, respectively. Hint: You will need the answer from question 3.5 as input for this part; use the Euclidean distance between two vectors to define the deviation and solve a quadratic programming (QP) problem. (7 pts)

Optimal solution:Objective = 1050[v1, v2, … … , v9] = [15, 10, 5, 0, 0, 5, 10, 10, 5].

See Excel file PS5_answer_2003.xls

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