Algorithms for peptide and PTM identification using tandem mass spectrometry

... concentration on algorithms for peptide and PTM identification using tandem mass spectrometry The main concern for algorithms in peptide identification is achieving fast and accurate peptide identification ... SOM Peptide (and PTM) identification algorithm based on tags and SOM PTM PSP PSPi = m1t1m2t2 mntnmn+1 mi and ti refer to mass difference...

Ngày tải lên: 12/09/2015, 08:18

... If the state labels at and are the same, but the words at and are different, then Else if the state labels at and are the same, and the words at and are the same, then There are a couple of ... Ư D $ Â  D " " " e$ D Vb Vb c D $ Vb x Vb c x  D f " Rb f x " $ x Rb x c x xx " $ " x " Ê$ fi D f " " c" D fi NP N the b) D % saw D...

Ngày tải lên: 23/03/2014, 20:20

... screw-cap vial for HPLC analysis and 100 µL of the working solution of the internal standard were added to each sample Then the samples were vortex mixed and a 900 µL aliquot was injected into the LC/MS/MS ... some of the structural isomers of prostaglandins, which resulted in the same parent-daughter ion transitions The whole analytical run time, including the...

Ngày tải lên: 20/06/2014, 00:20

... DISCUSSION We have addressed the problem of tracking a maneuvering target that moves over a large 2-dimensional space using a limited-size network of dynamic sensors (with motion capabilities) that produce ... use a relatively small network to follow a target over a large area by making the sensors move according to the estimated track 3.1 Signal model We can d...

Ngày tải lên: 22/06/2014, 23:20

... operation of combining is obtained by choosing two elements within the survivors and by genetically combining them The genetic combination is a numerical operation that can be performed in many ... Engineering (EEE) of the University of Adelaide under a postdoctoral contract, and the Department of Information Technology and Electrical Engineering (ITEE) of the University...

Ngày tải lên: 22/06/2014, 23:20

... (x1(n+1) ^2) ); x2(n+1)=b2*y2(n)+a*x2(n) +2* (1-a)*(x2(n)) ^2/ (1+ (x2(n) ^2) ); y2(n+1)=-x2(n)+a*x2(n+1) +2* (1-a)*(x2(n+1) ^2) /(1+ (x2(n+1) ^2) ); end subplot (2, 1,1); plot(x1,y1,'.') title('a =-0 .99 b=1') ... y1(n+1)=b*x1(n)+a*(y1(n )-( x1(n)) ^2) ; x2(n+1)=a*x2(n)-b*(y2(n )-( x2(n)) ^2) ; y2(n+1)=b*x2(n)+a*(y2(n )-( x2(n)) ^2) ; end plot(x1,y1,'ro',x2,y2,'bx') 2. 8 .2. 1 Demon...

Ngày tải lên: 13/08/2014, 02:21

... theory © 2001 by CRC Press LLC FIGURE 3. 4 Profile of the signal of Pb 3. 31 FIGURE 3. 5 Profile of the first input to Pb 3. 32 Pb 3. 33 In DSB-AM (double-sideband amplitude modulation), the amplitude ... following sub-units: mF = 10 3 F (milli-Farad); nF = 10 −9 F (nano-Farad); fF = 10 −15 F (femto-Farad); µF = 10 –6 F (micro-Farad); pF = 10 −12 F (pico-Farad); aF = 10 −18 F (atto-Farad)...

Ngày tải lên: 13/08/2014, 02:21

... +a*D2(k-1)+u(k)); D(k)=(2/dt)*(y(k)-y(k-1))-D(k-1); D2(k)= (4/ dt^2)*(y(k)-y(k-1) )-( 4/ dt)*D(k-1)-D2 (k-1); end plot(t,y,t,u,' ') The dashed curve is the temporal profile of the source term In-Class ... D2(1)=(1/a(1))*(-b(1)*D(1)-c(1)*y(1)+u(1)); for k=2:N y(k)=( (4* a(k)/dt^2+2*b(k)/dt+c(k))^ (-1 ))* (y(k-1)* (4* a(k)/dt^2+2*b(k)/dt)+D(k-1) * (4* a(k)/dt+b(k))+a(k)*D2(k-1)+u(k)); D(k...

Ngày tải lên: 13/08/2014, 02:21

... a(1)) (5. 25) a(2) = c(1) = a(1) + r(b(1) − a(1)) (5. 26) b(2) = b(1) (5. 27) Now, replacing the values of a(2) and b(2) from Eqs (5. 26) and (5. 27) into Eq (5. 25) , we are led to a second-degree ... The command is poly poly(r) In-Class Exercise Pb 5. 14 Find the roots of the polynomial p = [1 pute their sum and product 3], and com- Pb 5. 15 Consider the two polynomials: p1 = [1...

Ngày tải lên: 13/08/2014, 02:21

... LLC (6. 67) and where ˜ ˜ ˜ Vtot = Atot e jφtot = V1 + V2 (6. 68) Preparatory Exercise Pb 6. 35 Write the analytical expression for Atot and φtot in Eq (6. 68) as functions of the amplitudes and phases ... signals? Pb 6. 42 Two- and three-phase power can be extended to N-phase power In such a scheme, the N-110-V /60 -Hz signals are given by: © 2001 by CRC Press LLC πn   Vn = 110...

Ngày tải lên: 13/08/2014, 02:21

... Replacing l by l – in Eq (7. 94) and eliminating Pl′ ( x) from Eq − (7. 95), we find that: (1 − x ) dPl ( x) = lPl−1 ( x) − lxPl ( x) dx (7. 97) Differentiating Eq (7. 97) and using Eq (7. 95), we obtain: ... unique For example, in 4-D space, the canonical four-unit orthonormal basis vectors are given, respectively, by: ê1 = [1 0 0] (7. 15) ê2 = [0 0] (7. 16) ê3 = [0 0] (7. 17) ê4 = [0...

Ngày tải lên: 13/08/2014, 02:21

... n v m = λ n v m (8. 83) On the other hand, if we dotted Eq (8. 78) on the left with the bra-vector , we obtain: H v m = λ m v m = λ m v m (8. 84) Now compare Eqs (8. 83) and (8. 84) They are equal, ... ilamp=circuit872(RL) M=[1 0 0 0;1 -1 -5 0 0;0 -1 -1 00 0; 0 -3 00;0 -RL 0;0 0 -1 -1 ]; Vs=[5;0;0;0;0;0]; VI=M\Vs; ilamp=VI(5); Then, from the command window, we proceed b...

Ngày tải lên: 13/08/2014, 02:21

... transformations, in 2-D, are respectively: s Sx =  x 0 0 1  (9. 8) 1 Sy =  0 0 sy   (9. 9) In-Class Exercises Pb 9. 12 Find the transformation matrix for simultaneously compressing the x-coordinate ... 2, while expanding the y-coordinate by a factor of Apply this transformation to the trapezoid of Example 9. 1 and plot the result Pb 9. 13 Find the inverse matrices for Sx...

Ngày tải lên: 13/08/2014, 02:21

... Solution: The number of ways of selecting 10 items from a batch of 100 items is: N= 100 ! 100 ! 100 = = C10 10! (100 − 10) ! 10! 90! n where Ck is the binomial coefficient and represents the number of combinations ... (10. 43) Again, using Eqs (10. 37) and (10. 43), we have: P( Ai B) = P(B Ai )P( Ai ) (10. 44) P(B) Now, substituting Eq (10. 41) in the denominator of Eq (10. 44),...

Ngày tải lên: 13/08/2014, 02:21