(B. Sc. Biochemistry) Fourth Semester 2005 Subject: Biochem 225 (Bioinformatics)
Full Marks: 100 Pass Marks: 45 Time: 3 hours
Candidates are required to give their answers in their own words as far as practicable. The figure in the margin indicates full marks. (3× ×14 = 42)
Group (A) Long Questions (Any three)
1. Comment on principle of DNA sequencing. What are different types of DNA sequence? Briefly comment on each. 2. How is quality control maintained in GeneBank, given that thousands of individual investigators submit data? 3. Various methods are employed for phylogenetic tree analysis. Among them distance method, maximum parsimony method and maximum likelyhood method are the major one for constructing rooted tree. What criteria are required for choosing these methods. Explain it by displaying a flow-chart designed by oneself. 4. The Protein Data Bank (PDB) is the central repositary of protein structure data. What do databases such as SCOP and CATH offer that PDB Lacks? Explain. Group (B) Short Questions: (Any six) (6× ×7 = 42) 1. What is the scope of bioinformatics? Why is it a multi-disciplinary field? 2. Explain the role of computers in bioinformatics. 3. What is the difference between structural and functional genomics. 4. Differentiate between SWISS-PROT & TrEMBL Protein databases. 5. What is the difference between local and global sequence alignment methods? 6. Why do you need multiple sequence alignment? Can they be done globally? 7. Is it possible to predict the structure and function of a protein from sequence information? Explain. 8. How can we obtain a three dimensional image of the protein? Group (C) Very Short Question: (Any eight) 1. What is database? 2. What does WWW mean? 3. Mention few search engines. 4. What is contig? 5. What is DNA micro arrey? 6. What is sequence format? 7. Mention few sequence submission tools. 8. What is substitution score matrice? 9. What is hidden Markov Model? 10. What is RMSD? 11. Mention few molecular modeling software. 12. Mention few molecular graphic package to view protein structure. ***
(2× ×8 = 16)