6 Pages

18_Translation & Mutation2

Course: BOTN 2460, Winter 2008
School: Manitoba
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& Translation Mutation Intro (2006, Ch. 14:355-363; 2002, Ch. 6:138-146) Proteins Genetic code Mutations at nucleotide level (2006, Ch. 15:377-382; 2002, Ch. 19:541-548) General Background Direct Consequences of Mutation Translation Process (2006, Ch. 14:363-370; 2002, Ch. 6:147-154) Fig. 19.3a-d Types of base-pair substitution mutations purines A pyrimidines C transistion tranversion G T Peter...

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& Translation Mutation Intro (2006, Ch. 14:355-363; 2002, Ch. 6:138-146) Proteins Genetic code Mutations at nucleotide level (2006, Ch. 15:377-382; 2002, Ch. 19:541-548) General Background Direct Consequences of Mutation Translation Process (2006, Ch. 14:363-370; 2002, Ch. 6:147-154) Fig. 19.3a-d Types of base-pair substitution mutations purines A pyrimidines C transistion tranversion G T Peter J. Russell, iGenetics: Copyright Pearson Education, Inc., publishing as Benjamin Cummings. 1 Fig. 19.3c-d Types of base-pair substitution mutations Sequence from normal gene Sequence from mutated gene Peter J. Russell, iGenetics: Copyright Pearson Education, Inc., publishing as Benjamin Cummings. Fig. 19.3e-g Types of base-pair substitution mutations Sequence from normal gene sequence from mutated gene Peter J. Russell, iGenetics: Copyright Pearson Education, Inc., publishing as Benjamin Cummings. 2 Translation & Mutation Intro (2006, Ch. 14:355-363; 2002, Ch. 6:138-146) Proteins Genetic code Mutations at nucleotide level (2006, Ch. 15:377-382; 2002, Ch. 19:541-548) General Background Direct Consequences of Mutation Translation Process (2006, Ch. 14:363-370; 2002, Ch. 6:147-154) tRNA's & Ribosomes process: initiation, elongation, termination Fig. 6.9 Example of base-pairing wobble Peter Russell, J. iGenetics: Copyright Pearson Education, Inc., publishing as Benjamin Cummings. 3 Overview of Gene Expression (see also Fig. 5.8/13.9) after Fig. 5.2 / 13.3 5' DNA 3' promoter transcription start RNA-coding sequence 5' terminator transcription stop +1 3' Fig. 5.7 / 13.8 polypeptide N-terminus Peter J. Russell, iGenetics: Copyright Pearson Education, Inc., publishing as Benjamin Cummings. C-terminus Fig. 6.12 Initiation of protein synthesis in prokaryotes Peter J. Russell, iGenetics: Copyright Pearson Education, Inc., publishing as Benjamin Cummings. 4 Fig. 6.16 The formation of a peptide bond between the first two amino acids of a polypeptide chain is catalyzed on the ribosome by peptidyl transferase Peter J. Russell, iGenetics: Copyright Pearson Education, Inc., publishing as Benjamin Cummings. Fig. Elongation stage of translation in prokaryotes Peter J. Russell, iGenetics: Copyright Pearson Education, Inc., publishing as Benjamin Cummings. 5 Fig. 6.18 Termination of translation Peter J. Russell, iGenetics: Copyright Pearson Education, Inc., publishing as Benjamin Cummings. Fig. 6.17 Diagram of a polysome, a number of ribosomes each translating the same mRNA sequentially Peter J. Russell, iGenetics: Copyright Pearson Education, Inc., publishing as Benjamin Cummings. 6
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Manitoba - BOTN - 2460
DNA TechnologyPolymerase Chain Reaction (PCR)(2006, Ch. 16:444-447; 2002, Ch. 7:190-192)DNA sequencing(2006, Ch. 16:441-444; 2002, Ch. 7:187-190)Fig. 7.23 The polymerase chain reaction (PCR) for selective amplification of DNA sequencesPeter
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Chemical ID number CHECK 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39Kow 1555000 5 364 111 100 1000 423 1900 51 809 174 300 46 1380 6925 40 40 1066 36 190 180000 145 76344 460000 10542
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Assignment #7 Solutions page ECSE-2410 Signals & Systems Fri 09/30/05function hw7 clf % Define the time vector% Define the original triangular function (using sums of steps) y = (-1-t).*u(-1-t)+(2*t+2).*(u(t+1)-u(t)+(2-t).*(u(t)-u(t-2)+ .% Pl
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Assignment #11- Solutions ECSE-2410 Signals & Systems - Fall 2005 Tue 10/18/05Assignment #11- Solutions ECSE-2410 Signals & Systems - Fall 2005 Tue 10/18/05Assignment #11- Solutions ECSE-2410 Signals & Systems - Fall 2005 Tue 10/18/05Assignment
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Assignment #12 p 1 ECSE-2410 Signals & Systems - Fall 2005 Fri 10/28/05Assignment #12 p 2 ECSE-2410 Signals & Systems - Fall 2005 Fri 10/28/05Assignment #12 p 3 ECSE-2410 Signals & Systems - Fall 2005 Fri 10/28/05
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Assignment #15 p 1 ECSE-2410 Signals & Systems - Fall 2005 Fri 11/11/05-4, -5 + 5j, -5 -5jAssignment #15 p 2 ECSE-2410 Signals & Systems - Fall 2005 Fri 11/11/05Assignment #15 p 3 ECSE-2410 Signals & Systems - Fall 2005 Fri 11/11/05Assignm
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Assignment #16 -#16 Solutions Assignment Solutions - page 1ECSE-2410 Fall 2005ECSE~2410 Signals & Systems - Fa112003Tuesday 11/16/05 Fri 11/7/031.Ca.)~IX.t) =-0~JKt:s,+2.)~.\<LS+z)LJ2f,)~~L~--SLSi-I)--.:y/<:.[5>+
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.Assignment #18 - Solutions - pageAssignment #18 Solutions - page 1~I~ ~ATC"~IFri 12/02/05 Fri 11/14/03ECSE-2410 Signals & Systems - Fall 2005 ECSE-2410 Signals & Systems (90,.,"~Cf-)I)fl (')lfT0\.)(.l-L-fUa t, 17 2.-I-h
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Assignment #19 p 1 ECSE-2410 Signals & Systems - Fall 2005 Fri 12/09/05Assignment #19 p 2 ECSE-2410 Signals & Systems - Fall 2005 Fri 12/09/05Assignment #19 p 3 ECSE-2410 Signals & Systems - Fall 2005 Fri 12/09/05Assignment #19 p 4 ECSE-241
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Assignment #19 ECSE-2410 Signals & Systems - Fall 2005 1(10). Find the z-transform of x[n] (See Text, Example 10.4.)1 n 3Fri 12/09/05cos 4 n u[n] , and sketch the pole-zero diagram of X (z) .2(20). Using tables 10.1 and 10.2, find the inverse
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Assignment #1 ECSE-2410 Signals & Systems - Fall 2005 1(24). Given signal x(t ) shown, (a) write the equation for x(t ) . (b) find the equation fortDue Fri 09/02/05x(t )2 1the running integral y (t )x( )d versus t,0 -1 1 2 3 4 5 6tfor
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Assignment #3 ECSE-2410 Signals & Systems - Fall 2005 Due Fri 09/09/051(30). For each system below, let y denote the output of the system when the input is x. Note that some parts deal with a continuous-time system, while others deal with a discret
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Assignment #4 ECSE-2410 Signals & Systems - Fall 2005 Tue 09/13/051(15). Suppose the step response of a discrete-time, LTI system is y step [n] (a) Sketch y step [n] .0, n 0 n, 0 n 2 . 3, 2 n(b) Suppose the input to this system is x1[n] 1,1,1,1
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Correction and clarification for Assignment #04.Problem 1.0, n 0 n, 0 n 2 . 3, 2 nThe correct y step [n] is y step [n]In part (b) of problem1, last sentence: y[n] should be y step [n] . Note: the step response y step [n] in the homework is simp
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