1 / 30

The RNA Maximum Base Pair Matching Problem

The RNA Maximum Base Pair Matching Problem. 報告 者 : 99366076 顧家源. Introduce. G. A. C. C. U. RNA 由 4 種鹼基構成 : {A, G, C, U} Base Pair: 可降低 RNA 的自由能量 , 使其結構 更 穩定 {A – U, G – C, G - U} RNA Sequence: R = A – G – G – C – C – U – U – C – C – U Secondary Structure:. In 3D Presentation.

conway
Télécharger la présentation

The RNA Maximum Base Pair Matching Problem

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. The RNA Maximum Base Pair Matching Problem 報告者:99366076顧家源

  2. Introduce • G • A • C • C • U • RNA由4種鹼基構成: {A, G, C, U} • Base Pair: 可降低RNA的自由能量, 使其結構更穩定{A – U, G – C, G - U} • RNASequence:R = A – G – G – C – C – U – U – C – C – U • Secondary Structure:

  3. In 3D Presentation

  4. Constraint • 每組Base Pair的兩個鹼基必須間隔4以上 • Example:

  5. Structure & Free Energy

  6. Problem Statement • 已知一RNA序列R • 找出含有最多Base Pair的SecondaryStructureof R • 設定參數: • A– U, G – C, G – U的權重均為+1

  7. Max. #Base Pairs

  8. Step. 1 – Compute ρ(ri, rj)

  9. Step. 2– Set M(i, i~i+3) = 0

  10. Case 1 No match! ri rj-1 rj Si,j-1

  11. Case 2 ri match rj ri rj-1 rj ri+1 Si+1,j-1

  12. Case 3 rk match rj ri rj-1 rj rk Si,k-1 Sk+1,j-1 k = i + 1 to j - 4

  13. Step. 3– Compute all M(i, j)

  14. Step. 3– Compute all M(i, j)

  15. Step. 3 - M1,10

  16. Result

  17. Route 1

  18. Route 2

  19. Route 3

  20. Algorithm • Input:R = r1r2…rn • Output: a secondary structure of R with the maximum #base pairs. • Step1: /* Computation of ρ(ri, rj), 1 ≤ i < j ≤ n */WW = {(A, U), (U, A), (G, C), (C, G), (G, U), (U, G)};fori = 1 tondoforj = itondoif (ri, rj) WW thenρ(ri, rj) = 1; elseρ(ri, rj) = 0;end forend for Time Complexity = O(n2)

  21. Algorithm • Step2: /* Initialization of Mi,j for j – i≤ 3 */fori = 1 tondoforj = itoi + 3 doif j ≤ nthenMi,j = 0;end forend for Time Complexity = O(n)

  22. Algorithm • Step3: /* Calculation of Mi,j for j – i > 3 */forh = 4 ton - 1 dofori = 1 ton - hdoj = i + h;case1 = Mi,j-1;case2 = (1 + Mi+1,j-1) ρ(ri, rj);case3 = ;Mi,j = max{ case1, case2, case3 };end forend for Time Complexity = O(n3)

  23. Case 1 No match! ri rj-1 rj Si,j-1

  24. Case 2 ri match rj ri rj-1 rj ri+1 Si+1,j-1

  25. Case 3 rk match rj ri rj-1 rj rk Si,k-1 Sk+1,j-1 k = i + 1 to j - 4

  26. Time Complexity • Step. 1 • Step. 2 • Step. 3

  27. Errata • P. 273 • Since we want to find the k between i + 1 and j + 4 such that Mi,j is the maximum, … • “j + 4” correct to “j - 4”. • P. 282 • Next, we analyze the time complexity… • Step 1 is O(n2) • Step 2 is O(n) • Step 3 is O(n3)

  28. Comparison • 前面幾節所介紹的Dynamic Programming，其運作模式如右： • Ai,j表示a提供a1~aib提供b1~bj的local optimal

  29. Comparison • 本節的方式則如右： • Mi,j表示ri~rj的local optimal • 主要的差異在於RNA的「折疊」 • 時間複雜的的差異來自較多的候補

  30. Resources • 線上RNA結構預測網頁(CGI):http://rna.tbi.univie.ac.at/cgi-bin/RNAfold.cgi • 視覺化RNA結構呈現軟體－RnaDv:http://rna-dv.sourceforge.net/

More Related