Show Bid Request Graphing Enzymes in VB6 Bid Request Id: 58087 Bookmark in my 'To Do' list Posted by: Prez808 (1 ratings) (Software buyer rating 10) Non-action Ratio: Above Average - 33.33% Buyer Security Verifications: Unverified Approved on: Apr 16, 2003 9:40:21 AM EDT Bidding Closes: Apr 22, 2003 12:00:00 PM EDT Viewed (by coders): 107 times Deadline: 4/25/2003 TIME EXPIRED Phase: 100% of work completed and accepted. Coder has been paid. Max Accepted Bid: Bidding is closed Project Type: Personal Project / Homework Help Bidding Type: Open Auction Categories: Windows, Visual Basic, Graphics Enter chat room for this bid request (0 active users at Jul 16, 2003 3:40:31 AM EDT) Description: Hello, I have an assignment that requires me to graph the action of enzymes in Visual Basic 6. An example was given which graphs normal action. The problem as stated is as follows: Simulate a metabolic pathway with substrates A though E over 1,000 time intervals in which an inhibitor "K" of the forward reaction from B to C is applied at time t1 and is removed at time t2. Graph the concentractions of metabolites A though E. The given code shows what normally happens without any inhibitor. A is the top graph, and E is the last. The forward reactions from A to E are F1-F5 and the reverse reactions are R1-R4 (working from E backward) So the area of concern is from B-C which is "F2". I'm not sure how to add an inhibitor at that point. All that needs to be done is to add on to the existing sample program another button which would graph an inhibitor's activity. I imagine this should be pretty basic for a vb6 expert to do quickly. I will include my latest files as well as the code Does anyone have any ideas? Thanks! Deliverables: 1) Complete and fully-functional working program(s) in executable form as well as complete source code of all work done in .frm, .vbp and .vbw files. 2) A screen shot of the gui. Here is the original code: 'Here are the calculated metabolite concentrations A-E: Dim A(1000), B(1000), C(1000), D(1000), E(1000) As Single 'Here are the forward and reverse reaction rates Dim F1, F2, F3, F4, F5 As Single Dim R1, R2, R3, R4, R5 As Single 'Here is the independent variable, time Dim t As Integer 'Feed is the rate of addition to A 'bsln is the baseline for the (inverted) vertical axis origin Dim Hscale, Vscale, feed, bsln As Single Private Sub Command1_Click() 'Set some reasonable values Hscale = 6.5: Vscale = 100: feed = 2 F1 = 0.1: F2 = 0.1: F3 = 0.1: F4 = 0.1: F5 = 0.1 R1 = 0.07: R2 = 0.07: R3 = 0.07: R4 = 0.07 bsln = Picture1.ScaleHeight For t = 1 To 1000 'calculate updates of metabolite concentrations A(t) = A(t - 1) - (F1 * A(t - 1)) + (R1 * B(t - 1)) + feed B(t) = B(t - 1) + (F1 * A(t - 1)) - (R1 * B(t - 1)) - (F2 * B(t - 1)) + (R2 * C(t - 1)) C(t) = C(t - 1) + (F2 * B(t - 1)) - (R2 * C(t - 1)) - (F3 * C(t - 1)) + (R3 * D(t - 1)) D(t) = D(t - 1) + (F3 * C(t - 1)) - (R3 * D(t - 1)) - (F4 * D(t - 1)) + (R4 * E(t - 1)) E(t) = E(t - 1) + (F4 * D(t - 1)) - (R4 * E(t - 1)) - (F5 * E(t - 1)) 'display each metabolite concentration Picture1.Circle (t * Hscale, bsln - (A(t) * Vscale)), 15 Picture1.Circle (t * Hscale, bsln - (B(t) * Vscale)), 15 Picture1.Circle (t * Hscale, bsln - (C(t) * Vscale)), 15 Picture1.Circle (t * Hscale, bsln - (D(t) * Vscale)), 15 Picture1.Circle (t * Hscale, bsln - (E(t) * Vscale)), 15 'Waste some time so that we can watch the plot grow For w = 1 To 20000: q = q + 1: Next w Next t End Sub Private Sub Command2_Click() End End Sub Platform: Windows 98 Must be 100% finished and received by buyer on: Apr 25, 2003 EDT Deadline legal notes: All times are expressed in the time zone of the site EDT (UT - 5). 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See all rejected bids (and all comments) Name   Bid Amount    Date   Coder Rating   This bid was accepted by the buyer! SmartCoders (34 ratings) in Bay Harbor Islands, Florida United States Bid id: 626,739   $15 (USD) Apr 16, 2003 8:55:07 PM EDT  8.73 (Superb)     Hello, here is the example of the modified application as I understand your description. What I do is to supress the - (F2 * B(t - 1)) and the + (F2 * B(t - 1)) from the B(t) and C(t) equations in the t1-t2 interval. Regards, Attached File           See 1 private reply(ies) to/from SmartCoders.