Palindromes in Viral Genomes. Ming-Ying Leung Department of Mathematical Sciences The University of Texas at El Paso (UTEP)

Transcription

1 Palindromes in Viral Genomes Ming-Ying Leung Department of Mathematical Sciences The University of Texas at El Paso (UTEP)

2 Outline: Palindromes Viral Genomes Roles of palindromes in DNA and RNA viruses Cytomegalovirus (CMV) Particle SARS Coronavirus Particle

3 Palindromes in Letter Sequences Odd Palindrome: A nut for a jar of tuna ANUTFORAJAROFTUNA remove spaces and capitalize Even Palindrome: Step on no pets STEPON NOPETS

4 DNA and RNA DNA is deoxyribonucleic acid, made up of 4 nucleotide bases Adenine, Cytosine, Guanine, and Thymine. RNA is ribonucleic acid, made up of 4 nucleotide bases Adenine, Cytosine, Guanine, and Uracil. The bases A and T form a complementary pair, so are C and G. A C G T A T A C G U C G

5 DNA/RNA Palindromes

6 CMV Particle Virus and Eye Diseases Genome size ~ 230 kbp CMV Retinitis inflammation of the retina triggered by CMV particles may lead to blindness

7 Palindromes and Replication Origins in Herpesviruses High concentration of palindromes exists around replication origins of herpesviruses Locating clusters of palindromes (above a minimal length) on herpesvirus genome sequences might reveal likely locations of its replication origins.

8 Human Herpesvirus (HHV-6) Genome Replication origin (orilyt) in non-coding region Two binding sites for origin-binding protein (OBP) Two 200-bp DNA unwinding elements (DUE)

9 Computational Prediction of Replication Origins Poisson approximation for palindrome distribution in an i.i.d. random sequence model Criterion for identifying statistically significant palindrome clusters using scan statistics Improve prediction accuracy

10 Poisson Process Approximation of Palindrome Distribution

11 Use of the Scan Statistic to Identify Clusters of Palindromes

12 Improve Prediction Accuracy Using a Markov random sequence model Taking the lengths of palindromes and base composition of the genomes into consideration when counting palindromes. [Chew et al. 2005, Nucleic Acids Research 33(15):e134] Using machine learning approaches to utilize knowledge about replication origin locations in closely related genomes.

13 SARS Viral Particles

14 SARS Virus Particle Host Cell Spike Glycoprotein Single-stranded RNA genome with ~30 kilobases Only about 7% the genome size of Cytomegalovirus (CMV) with double-stranded DNA

15 SARS Virus Genome Map ORF1a ORF1b S X1 X2 Replicase (1a and 1b), spike glycoprotein (S), X1 and X2 occupy 87% of the genome Two pairs of overlapping ORFs,1a & 1b and X1 & X2 (as designated by Rota et al. 2003), are predicted in this region 1a and 1b are standard in all coronaviruses, X1 and X2 are unique to SARS. Whether X1 and X2 do code for proteins is still unconfirmed

16 A Long Palindrome in X1 and X2 TCTTTAACAAGCTTGTTAAAGA Positions: (22 bases) Found in SARS but not in other 6 coronavirus genomes (Chew et al. 2004) The next longest palindrome in SARS is 14 bases long In the overlapping region of X1 and X2

17 Palindrome: A string of nucleotide bases that reads the same as its reverse complement. A palindrome must be even in length, e.g. palindrome of length 10: 5.. GCAATATTGC..3 j - L +1 j j + 1 j +L b 1 b 2 b L b L+1 b 2L-1 b 2L We say that a palindrome of length 2L occurs at position j when the (j-i+1)st and the (j+i)th bases are complementary to each other for i=1,, L. In an i.i.d. sequence model this occurs with probability 2( p p + p p ) L A T C G.

18 Probability of Observing a Length 22 Palindrome Approximate the palindrome distribution by a Poisson process with rate λ = np = Here n = genome length = 29727, and [ 2( ˆ ˆ ˆ ˆ )] 11 p= p p + p p A T C G The probability of the occurrence of at least one length 22 palindrome in the genome is e 0.01

19 Expression of Overlapping Genes Requires Frameshifting in Reading Frameshifting must have the following elements: Slippery Sequence - a mechanism that allows the reader (called ribosome) to slip Stimulatory Element - a pseudoknot or stem-loop structure that blocks the ribosome Pseudoknot: Short binding sequence Stem-loop or hairpin loop: Slippery Sequences Palindrome-like sequences

20 -1 Frameshifting Pseudoknot start reading CGGGTTT Ribosome Reading with default frame: GGG then TTT

21 -1 Frameshifting Pseudoknot start reading CGGGTTT Ribosome Reading with default frame: GGG then TTT Reading after -1 frameshifting: CGG then GTT

22 Heptanucleotide Slippery Sequences A string in the form of XXXYYYN where X = A,T, or G; Y = A or T; and N = A, T, or C ORF1a and ORF1b (Overlap: 13398, 1 base only) X1 and X2 (Overlap: , 401 bases)

23 Locations of Slippery Sequences ORF1a ORF1b S X1 X2 Slippery Sequences: TTTAAAC TTGAAAA? Right preceding the overlapping base between 1a and 1b, there is a slippery sequence followed by a pseudoknot (Theil et al 2003) Possible slippery sequences are detected in the overlapping region of X1 and X2; any pseudoknot or stem-loop structure in close proximity downstream?

24 Pseudoknot Predicted by PknotsRG G = kcal/mol.

25 RNA Secondary Structure Prediction Prediction RNA secondary structures including pseudoknots is a computational intensive problem. Use of heterogeneous grid computing to provide computing power. Use of palindrome content evaluation to improve prediction consistency.

26 Palindrome counts in random nucleotide sequences Define the indicator random variable 1 if palindrome of length 2 L occurs at base j I j = 0 otherwise Then X L n L = I is the total count of palindromes of length at least 2L in a sequence of length n. j= L k

27 Mean and variance of palindrome counts n L µ L = EX ( L) = E Ik = ( n 2L+ 1) EI ( L) j= L σ n L n L 1 n L 2 L = X L = I j + I j Ik j= L j= L k= j+ 1 var( ) var( ) 2 cov(, ) If we let γ (0) = PI ( = 1) for L j n L ( d) = P( I = 1, I = 1) for 1 d n L γ j j+ d j j then cov( I, I ) EI ( ) = γ (0) j var( I ) = γ(0)(1 γ(0)) j j j+ d = γ( d) γ(0) 2

28 Mean and variance of palindrome counts (cont d) 2 L n L n L 1 n L ( ) µ = E( X ) = ( n 2L+ 1) γ 0 σ L L = var( X ) L = var( I ) + 2 cov( I, I ) j j k j= L j= L k= j+ 1 ( γ ) = ( n 2L+ 1) γ(0) 1 (0) n 2L d = 1 ( ) + 2 n 2L+ 1 d γ( d) γ(0) 2

29 How to find the γ s? Under a Markov sequence model, Chew et al. (INFORMS Journal of Computing, 16: , 2004) have obtained computable formulas for the γ s, expressed in terms of the transition and stationary probabilities of the Markov chain. These can be estimated by the observed base frequencies and dinucleotide frequencies. Let s look at a special case, namely the i.i.d. random sequence model where the nucleotide bases are generated independently with probabilities p A, p C, p G, p T.

30 Finding γ(0) for the i.i.d. sequence model γ (0) = PI ( = 1) = [2( p p + p p )] L j A T C G j - L +1 j j + 1 j +L b 1 b 2 b L b L+1 b 2L-1 b 2L

31 Finding γ(d) for the i.i.d. sequence model: Case 1: d 2L Case 2: L d < 2L Case 3: 1 d < L

32 Palindromes, and More Palindromes i - L i - 2 i - 1 i i + 1 i + 2 i + 3 i +L+1 b 1 b 2 b L - 1 b L b L b L - 1 b 2 b 1 i - L i - 3 i 2 I - 1 i i + 1 i + 2 i + 3 i +L+1 b 1 b 2 b L - 1 b L b L b L - 1 b 2 b 1

33 Acknowledgements Collaborators Louis H. Y. Chen (National University of Singapore) David Chew (National University of Singapore) Kwok Pui Choi (National University of Singapore) Raul Cruz-Cano (The University of Texas at El Paso) Hans Heidner (The University of Texas at San Antonio) Michela Taufer (The University of Texas at El Paso) Aihua Xia (University of Melbourne, Australia) Funding Support NIH Grants S06GM , 2G12RR and 3T34GM S1. THECB Advanced Research Program