Bioinformatics Assignment: Sequence Analysis and Alignment Solutions

This document presents solutions to a bioinformatics assignment focusing on sequence analysis and alignment techniques. The assignment covers a range of topics, including the use of Hidden Markov Models (HMMs) for sequence alignment, the application of BLAST for sequence searching, and the scoring matrices used in these processes. It also addresses the differences between global and local alignment methods, the relationship between genes and proteins, and the analysis of CpG islands in DNA sequences. The solutions include detailed explanations of the concepts and calculations involved, such as the calculation of dinucleotide frequencies. Overall, the assignment provides a comprehensive overview of key bioinformatics principles and their practical application in sequence analysis.

Running head: BUSINESS INTELLIGENCE 1
Business Intelligence
Name of Student
Name of University

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Business Intelligence
Mid Term Exam
1. After building the profile HMM, each sequence will have a unique alignment with that
HMM
Any sequence can be represented by a path through any profile HMM model
2. A T T C
A 1 -1 -1 -1
G 0 -2 -2 -2
A 1 -3 -3 -3
C 0 -4 -4 -2
3. D D A T C
D 1 1 -1 -1 -1
A 0 0 0 -2 -2
- -2 -2 -2 -4 -4
T -3 -3 -3 -3 -5
C -4 -4 -4 -4 -4
4. DNA contains the genetic information of an organism
The process called slicing cuts exon sequences and joins introns together
5. Compared to PAM 100, PAM 250 is more proper to score diverse sequence alignments
Two sequences with 100 residues each can be 2 PAM units apart, but with only one
position different from each other
6. By using a pseudo count you will incorporate more information into the construction of
an alignment that exists in the sequences themselves
Pseudo count will bias the sequences used in building a HMM
7. Blast is a tool for sequence alignment
One can search for a protein sequences using BLAST against a nucleic acid database

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BLAST search results depend on the scoring matrix used
8. The number of genes is always more than the number of proteins in a cell
mRNA is a single stranded RNA
Genes comprise a piece of the DNA sequence and encode proteins
9. Global alignment measures sequence similarity of the whole genome
Local alignment can be used to detect functionally important regions
10. 0.25+0.25+0.25+0.25=1.00
11. A = 0.8
B =0.5
C=0.2
D=0.5
E=0.8
F= 0.2
G= 0.5
H=1
I= 0
J=0.8
12. A G T C A G C T
A 1 -1 -1 -1 1 -1 -1 -1
T 0 -2 0 -2 0 -2 -2 0
A 1 -3 -1 -3 1 -3 -3 -1
C 0 -4 -2 -2 0 -4 -2 -2
A 1 -5 -3 -3 1 -5 -3 -3
C 0 -6 -4 -2 0 -6 -2 -4
T -1 -7 -3 -3 -1 -7 -3 -3
Global alignment TCA-T-A---
Local alignment TCTGTCAGCT
13.

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Where n±βγ is number of dinucleotides
= 8 _
0.25(0.18+0.368+0.274+0.188)
= 8 _
0.25(1.01)8
=29.551
The sequence does not come from a CpG island

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