2016.09.28TOPIC REVIEW• Exam • PS2 Sequence Alignment .docx

2016.09.28
TOPIC REVIEW
• Exam
• PS2 Sequence Alignment
• Command Line Blast
• PS1 Molecular Biology
• Personal Microbiome Project
CURRENTLY
LET’S NEGOTIATE
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• Microbiome project - 20%
• Exam (1) - 20%
• Research project - 45%
• Participation - 5%
OR
• Problem sets (4) - 10%
• Microbiome project - 20%
• Exam 1 - 15%
• Exam 2 - 15%
• Research project - 35%
• Participation - 5%
PS2 SEQUENCE ALIGNMENT

PS2 SEQUENCE ALIGNMENT
RefSeqs, protein (experimentally supported)
On chromosome 17
Reverse strand
PRCD Progressive rod-cone degeneration
PS2: GLOBAL ALIGNMENT
BLOSUM62
• substitutions less penalized and are
preferred to gaps. There is also a
decrease in the level of identity.
BLOSUM80
• Substitutions more penalized and
gaps are favored.

PAM60
• Substitutions more penalized and gaps
are favored.
PAM250
• substitutions less penalized and are
preferred to gaps. There is also a
decrease in the level of identity.
PS2: LOCAL ALIGNMENT
SEQ1 A L S C V W M I P
SEQ2 A I S C M I P T
9 residues
8 residues
Create Matrix: length of seq1 + 1
x
length of seq2 + 1
Matrix 10 x 9

A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2
-4
-6
-8
-10
-12
-14
-16
A
I
S
C
M
I
P
T
Exercise: fill the scores of the alignment matrix
using the BLOSUM62 substitution matrix.
Gap opening penalty: -5
Gap extension penalty: -1
S V E T D
T
S
I
N

Q
E
T
Ala A 4
Arg R -1 5
Asn N -2 0 6
Asp D -2 -2 1 6
Cys C 0 -3 -3 -3 9
Gln Q -1 1 0 0 -3 5
Glu E -1 0 0 2 -4 2 5
Gly G 0 -2 0 -1 -3 -2 -2 6
His H -2 0 1 -1 -3 0 0 -2 8
Ile I -1 -3 -3 -3 -1 -3 -3 -4 -3 4
Leu L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4
Lys K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5
Met M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5
Phe F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6
Pro P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7
Ser S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4
Thr T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5
Trp W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11
Tyr Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7
Val V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4
A
la
A
rg
A
sn

A
sp
C
y
s
G
ln
G
lu
G
ly
H
is
Il
e
L
e
u
L
y
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M
e
t
P
h

e
P
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T
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A R N D C Q E G H I L K M F P S T W Y V
Dynamical programming - global alignment
83
BLOSUM62
GAP COST: -2

At each cell, 3 scores are calculated:
• match score = diagonal cell score +
score from the substitution matrix.
• Vertical gap score = upper neighbor
+ gap cost
• Horizontal gap score = left neighbor
+ gap cost
• The highest score is retained and
the arrow is labelled
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2
-4
-6
-8
-10
-12
-14
-16
A
I
S
C
M
I
P
T

S V E T D
T
S
I
N
Q
E
T
Ala A 4
Arg R -1 5
Asn N -2 0 6
Asp D -2 -2 1 6
Cys C 0 -3 -3 -3 9
Gln Q -1 1 0 0 -3 5
Glu E -1 0 0 2 -4 2 5
Gly G 0 -2 0 -1 -3 -2 -2 6
His H -2 0 1 -1 -3 0 0 -2 8
Ile I -1 -3 -3 -3 -1 -3 -3 -4 -3 4
Leu L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4
Lys K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5
Met M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5
Phe F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6

Pro P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7
Ser S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4
Thr T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5
Trp W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11
Tyr Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7
Val V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4
A
la
A
rg
A
sn
A
sp
C
y
s
G
ln
G
lu
G
ly
H
is
Il

e
L
e
u
L
y
s
M
e
t
P
h
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P
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S
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h
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T
rp
T
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V
a
l
83
BLOSUM62
GAP COST: -2
+ gap cost
+ gap cost
4 comes from the
substitution matrix.
Match score = 0 + (4) = 4

A
sp
C
y
s
G
ln
G
lu
G
ly
H
is
Il
e
L
e
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L
y
s
M
e
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P

h
e
P
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S
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T
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T
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a
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83
BLOSUM62
GAP COST: -2

Vertical gap score = -2 + (-2) = -4
+ gap cost
+ gap cost
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2
-4
-6
-8
-10
-12
-14
-16
A
I
S
C
M

I
P
T
S V E T D
T
S
I
N
Q
E
T
Ala A 4
Arg R -1 5
Asn N -2 0 6
Asp D -2 -2 1 6
Cys C 0 -3 -3 -3 9
Gln Q -1 1 0 0 -3 5
Glu E -1 0 0 2 -4 2 5
Gly G 0 -2 0 -1 -3 -2 -2 6
His H -2 0 1 -1 -3 0 0 -2 8
Ile I -1 -3 -3 -3 -1 -3 -3 -4 -3 4

Leu L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4
Lys K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5
Met M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5
Phe F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6
Pro P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7
Ser S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4
Thr T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5
Trp W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11
Tyr Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7
Val V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4
A
la
A
rg
A
sn
A
sp
C
y
s
G
ln
G
lu
G
ly

H
is
Il
e
L
e
u
L
y
s
M
e
t
P
h
e
P
ro
S
e
r
T
h
r
T
rp

T
y
r
V
a
l
83
BLOSUM62
GAP COST: -2
Horizontal gap score = -2 + (-2) = -4
+ gap cost
+ gap cost

A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4
-4
-6
-8
-10
-12
-14
-16
A
I
S
C
M
I
P
T
S V E T D
T

S
I
N
Q
E
T
Ala A 4
Arg R -1 5
Asn N -2 0 6
Asp D -2 -2 1 6
Cys C 0 -3 -3 -3 9
Gln Q -1 1 0 0 -3 5
Glu E -1 0 0 2 -4 2 5
Gly G 0 -2 0 -1 -3 -2 -2 6
His H -2 0 1 -1 -3 0 0 -2 8
Ile I -1 -3 -3 -3 -1 -3 -3 -4 -3 4
Leu L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4
Lys K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5
Met M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5
Phe F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6
Pro P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7
Ser S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4
Thr T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5
Trp W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11
Tyr Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7
Val V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4
A
la

A
rg
A
sn
A
sp
C
y
s
G
ln
G
lu
G
ly
H
is
Il
e
L
e
u
L
y
s

M
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P
h
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P
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83
BLOSUM62
GAP COST: -2
+ gap cost
+ gap cost
S V E T D

T
S
I
N
Q
E
T
Ala A 4
Arg R -1 5
Asn N -2 0 6
Asp D -2 -2 1 6
Cys C 0 -3 -3 -3 9
Gln Q -1 1 0 0 -3 5
Glu E -1 0 0 2 -4 2 5
Gly G 0 -2 0 -1 -3 -2 -2 6
His H -2 0 1 -1 -3 0 0 -2 8
Ile I -1 -3 -3 -3 -1 -3 -3 -4 -3 4
Leu L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4
Lys K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5
Met M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5
Phe F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6
Pro P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7
Ser S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4
Thr T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5
Trp W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11
Tyr Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7
Val V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4
A

la
A
rg
A
sn
A
sp
C
y
s
G
ln
G
lu
G
ly
H
is
Il
e
L
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L
y

s
M
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P
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a
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83
BLOSUM62
GAP COST: -2
+ gap cost
+ gap cost
-1 comes from the
Match score = -2 + (-1) = -3
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4
-4
-6
-8
-10
-12
-14

-16
A
I
S
C
M
I
P
T
+ gap cost
+ gap cost
S V E T D
T

83
BLOSUM62
GAP COST: -2
Match score = -2 + (-1) = -3
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4
-4
-6
-8
-10
-12
-14
-16
A
I
S
C
M
I
P
T

+ gap cost
+ gap cost
S V E T D
T
S
I
N
Q
E
T
Ala A 4
Arg R -1 5
Asn N -2 0 6
Asp D -2 -2 1 6

Cys C 0 -3 -3 -3 9
Gln Q -1 1 0 0 -3 5
Glu E -1 0 0 2 -4 2 5
Gly G 0 -2 0 -1 -3 -2 -2 6
His H -2 0 1 -1 -3 0 0 -2 8
Ile I -1 -3 -3 -3 -1 -3 -3 -4 -3 4
Leu L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4
Lys K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5
Met M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5
Phe F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6
Pro P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7
Ser S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4
Thr T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5
Trp W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11
Tyr Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7
Val V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4
A
la
A
rg
A
sn
A
sp
C
y
s
G
ln

G
lu
G
ly
H
is
Il
e
L
e
u
L
y
s
M
e
t
P
h
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P
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T

h
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T
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a
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83
BLOSUM62
GAP COST: -2
Match score = -2 + (-1) = -3
Horizontal gap score = 4 + (-2) = 2
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4
-4
-6

-8
-10
-12
-14
-16
A
I
S
C
M
I
P
T
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4 2
-4
-6
-8
-10
-12
-14
-16
A
I
S
C
M
I
P
T

S V E T D
T
S
I
N
Q
E
T
Ala A 4
Arg R -1 5
Asn N -2 0 6
Asp D -2 -2 1 6
Cys C 0 -3 -3 -3 9
Gln Q -1 1 0 0 -3 5
Glu E -1 0 0 2 -4 2 5
Gly G 0 -2 0 -1 -3 -2 -2 6
His H -2 0 1 -1 -3 0 0 -2 8
Ile I -1 -3 -3 -3 -1 -3 -3 -4 -3 4
Leu L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4
Lys K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5
Met M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5

Phe F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6
Pro P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7
Ser S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4
Thr T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5
Trp W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11
Tyr Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7
Val V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4
A
la
A
rg
A
sn
A
sp
C
y
s
G
ln
G
lu
G
ly
H
is

Il
e
L
e
u
L
y
s
M
e
t
P
h
e
P
ro
S
e
r
T
h
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T
rp
T
y
r

V
a
l
83
BLOSUM62
GAP COST: -2
+ gap cost
+ gap cost
Match score = -2 + (-1) = -3

A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4 2
-4
-6
-8
-10
-12
-14
-16
S V E T D
T
S
I
N
Q
E
T
Ala A 4

Arg R -1 5
Asn N -2 0 6
Asp D -2 -2 1 6
Cys C 0 -3 -3 -3 9
Gln Q -1 1 0 0 -3 5
Glu E -1 0 0 2 -4 2 5
Gly G 0 -2 0 -1 -3 -2 -2 6
His H -2 0 1 -1 -3 0 0 -2 8
Ile I -1 -3 -3 -3 -1 -3 -3 -4 -3 4
Leu L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4
Lys K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5
Met M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5
Phe F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6
Pro P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7
Ser S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4
Thr T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5
Trp W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11
Tyr Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7
Val V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4
A
la
A
rg
A
sn
A
sp
C
y
s

G
ln
G
lu
G
ly
H
is
Il
e
L
e
u
L
y
s
M
e
t
P
h
e
P
ro
S
e

r
T
h
r
T
rp
T
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V
a
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83
BLOSUM62
GAP COST: -2
+ gap cost

+ gap cost
+1 comes from the
Match score = -4 + (1) = -3
A
I
S
C
M
I
P
T
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4 2
-4
-6
-8
-10
-12
-14
-16

+ gap cost
+ gap cost
S V E T D
T
S
I
N
Q
E
T
Ala A 4
Arg R -1 5

Asn N -2 0 6
Asp D -2 -2 1 6
Cys C 0 -3 -3 -3 9
Gln Q -1 1 0 0 -3 5
Glu E -1 0 0 2 -4 2 5
Gly G 0 -2 0 -1 -3 -2 -2 6
His H -2 0 1 -1 -3 0 0 -2 8
Ile I -1 -3 -3 -3 -1 -3 -3 -4 -3 4
Leu L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4
Lys K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5
Met M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5
Phe F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6
Pro P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7
Ser S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4
Thr T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5
Trp W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11
Tyr Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7
Val V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4
A
la
A
rg
A
sn
A
sp
C
y
s
G

ln
G
lu
G
ly
H
is
Il
e
L
e
u
L
y
s
M
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P
h
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P
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S
e
r

T
h
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T
rp
T
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a
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83
BLOSUM62
GAP COST: -2
Match score = -4 + (1) = -3
A
I
S
C
M

I
P
T
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4 2
-4
-6
-8
-10
-12
-14
-16
+ gap cost
+ gap cost

S V E T D
T
S
I
N
Q
E
T
Ala A 4
Arg R -1 5
Asn N -2 0 6
Asp D -2 -2 1 6
Cys C 0 -3 -3 -3 9
Gln Q -1 1 0 0 -3 5
Glu E -1 0 0 2 -4 2 5
Gly G 0 -2 0 -1 -3 -2 -2 6
His H -2 0 1 -1 -3 0 0 -2 8
Ile I -1 -3 -3 -3 -1 -3 -3 -4 -3 4
Leu L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4
Lys K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5
Met M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5
Phe F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6
Pro P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7
Ser S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4
Thr T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5
Trp W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11
Tyr Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7

Val V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4
A
la
A
rg
A
sn
A
sp
C
y
s
G
ln
G
lu
G
ly
H
is
Il
e
L
e
u

L
y
s
M
e
t
P
h
e
P
ro
S
e
r
T
h
r
T
rp
T
y
r
V
a
l

83
BLOSUM62
GAP COST: -2
Match score = -4 + (1) = -3
A
I
S
C
M
I
P
T
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4 2 0
-4
-6
-8
-10
-12
-14

-16
A
I
S
C
M
I
P
T
S V E T D
T
S
I
N
Q
E
T
Ala A 4
Arg R -1 5
Asn N -2 0 6

Asp D -2 -2 1 6
Cys C 0 -3 -3 -3 9
Gln Q -1 1 0 0 -3 5
Glu E -1 0 0 2 -4 2 5
Gly G 0 -2 0 -1 -3 -2 -2 6
His H -2 0 1 -1 -3 0 0 -2 8
Ile I -1 -3 -3 -3 -1 -3 -3 -4 -3 4
Leu L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4
Lys K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5
Met M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5
Phe F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6
Pro P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7
Ser S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4
Thr T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5
Trp W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11
Tyr Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7
Val V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4
A
la
A
rg
A
sn
A
sp
C
y
s
G
ln

G
lu
G
ly
H
is
Il
e
L
e
u
L
y
s
M
e
t
P
h
e
P
ro
S
e
r

T
h
r
T
rp
T
y
r
V
a
l
83
BLOSUM62
GAP COST: -2

+ gap cost
+ gap cost
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4 2 0 -2 -4 -6 -8 -10 -12
-4 2 6 4 2 1 -1 -3 -4 -6
-6 0 0 10 8 6 4 2 0 -2
-8 -2 -1 -1 19 17 15 13 11 9
-10 -4 0 -2 17 20 18 20 18 16
-12 -6 -2 -2 15 20 18 19 24 22
-14 -8 -4 -3 13 18 16 17 22 31
-16 -10 -6 -3 11 16 16 15 20 29
A
I
S
C
M
I
P
T

S V E T D
T
S
I
N
Q
E
T
Ala A 4
Arg R -1 5
Asn N -2 0 6
Asp D -2 -2 1 6
Cys C 0 -3 -3 -3 9
Gln Q -1 1 0 0 -3 5
Glu E -1 0 0 2 -4 2 5
Gly G 0 -2 0 -1 -3 -2 -2 6
His H -2 0 1 -1 -3 0 0 -2 8
Ile I -1 -3 -3 -3 -1 -3 -3 -4 -3 4
Leu L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4
Lys K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5
Met M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5
Phe F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6
Pro P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7
Ser S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4
Thr T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5

Trp W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11
Tyr Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7
Val V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4
A
la
A
rg
A
sn
A
sp
C
y
s
G
ln
G
lu
G
ly
H
is
Il
e
L

e
u
L
y
s
M
e
t
P
h
e
P
ro
S
e
r
T
h
r
T
rp
T
y
r
V
a
l

83
BLOSUM62
GAP COST: -2
+ gap cost
+ gap cost
-
T
Seq1
Seq2

A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4 2 0 -2 -4 -6 -8 -10 -12
-4 2 6 4 2 1 -1 -3 -4 -6
-6 0 0 10 8 6 4 2 0 -2
-8 -2 -1 -1 19 17 15 13 11 9
-10 -4 0 -2 17 20 18 20 18 16
-12 -6 -2 -2 15 20 18 19 24 22
-14 -8 -4 -3 13 18 16 17 22 31
-16 -10 -6 -3 11 16 16 15 20 29
A
I
S
C
M
I
P
T
S V E T D
T
S
I
N

+ gap cost
+ gap cost
P -
P T
Seq1
Seq2
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4 2 0 -2 -4 -6 -8 -10 -12
-4 2 6 4 2 1 -1 -3 -4 -6
-6 0 0 10 8 6 4 2 0 -2
-8 -2 -1 -1 19 17 15 13 11 9
-10 -4 0 -2 17 20 18 20 18 16
-12 -6 -2 -2 15 20 18 19 24 22
-14 -8 -4 -3 13 18 16 17 22 31
-16 -10 -6 -3 11 16 16 15 20 29

A
I
S
C
M
I
P
T
S V E T D
T
S
I
N
Q
E
T
Ala A 4
Arg R -1 5
Asn N -2 0 6
Asp D -2 -2 1 6
Cys C 0 -3 -3 -3 9

Gln Q -1 1 0 0 -3 5
Glu E -1 0 0 2 -4 2 5
Gly G 0 -2 0 -1 -3 -2 -2 6
His H -2 0 1 -1 -3 0 0 -2 8
Ile I -1 -3 -3 -3 -1 -3 -3 -4 -3 4
Leu L -1 -2 -3 -4 -1 -2 -3 -4 -3 2 4
Lys K -1 2 0 -1 -3 1 1 -2 -1 -3 -2 5
Met M -1 -1 -2 -3 -1 0 -2 -3 -2 1 2 -1 5
Phe F -2 -3 -3 -3 -2 -3 -3 -3 -1 0 0 -3 0 6
Pro P -1 -2 -2 -1 -3 -1 -1 -2 -2 -3 -3 -1 -2 -4 7
Ser S 1 -1 1 0 -1 0 0 0 -1 -2 -2 0 -1 -2 -1 4
Thr T 0 -1 0 -1 -1 -1 -1 -2 -2 -1 -1 -1 -1 -2 -1 1 5
Trp W -3 -3 -4 -4 -2 -2 -3 -2 -2 -3 -2 -3 -1 1 -4 -3 -2 11
Tyr Y -2 -2 -2 -3 -2 -1 -2 -3 2 -1 -1 -2 -1 3 -3 -2 -2 2 7
Val V 0 -3 -3 -3 -1 -2 -2 -3 -3 3 1 -2 1 -1 -2 -2 0 -3 -1 4
A
la
A
rg
A
sn
A
sp
C
y
s
G
ln
G

lu
G
ly
H
is
Il
e
L
e
u
L
y
s
M
e
t
P
h
e
P
ro
S
e
r
T
h

r
T
rp
T
y
r
V
a
l
83
BLOSUM62
GAP COST: -2
+ gap cost
+ gap cost

I P -
I P T
Seq1
Seq2
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4 2 0 -2 -4 -6 -8 -10 -12
-4 2 6 4 2 1 -1 -3 -4 -6
-6 0 0 10 8 6 4 2 0 -2
-8 -2 -1 -1 19 17 15 13 11 9
-10 -4 0 -2 17 20 18 20 18 16
-12 -6 -2 -2 15 20 18 19 24 22
-14 -8 -4 -3 13 18 16 17 22 31
-16 -10 -6 -3 11 16 16 15 20 29
A
I
S
C
M
I
P
T

83
BLOSUM62
GAP COST: -2
+ gap cost
+ gap cost
M I P -
M I P T
Seq1
Seq2

+ gap cost
+ gap cost
W M I P -
- M I P T
Seq1
Seq2
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4 2 0 -2 -4 -6 -8 -10 -12
-4 2 6 4 2 1 -1 -3 -4 -6
-6 0 0 10 8 6 4 2 0 -2
-8 -2 -1 -1 19 17 15 13 11 9
-10 -4 0 -2 17 20 18 20 18 16
-12 -6 -2 -2 15 20 18 19 24 22
-14 -8 -4 -3 13 18 16 17 22 31
-16 -10 -6 -3 11 16 16 15 20 29

V W M I P -
- - M I P T
Seq1
Seq2
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4 2 0 -2 -4 -6 -8 -10 -12
-4 2 6 4 2 1 -1 -3 -4 -6
-6 0 0 10 8 6 4 2 0 -2
-8 -2 -1 -1 19 17 15 13 11 9
-10 -4 0 -2 17 20 18 20 18 16
-12 -6 -2 -2 15 20 18 19 24 22
-14 -8 -4 -3 13 18 16 17 22 31
-16 -10 -6 -3 11 16 16 15 20 29
A
I
S
C
M
I
P
T

83
BLOSUM62
GAP COST: -2
+ gap cost
+ gap cost
C V W M I P -
C - - M I P T
Seq1
Seq2

+ gap cost
+ gap cost
S C V W M I P -
S C - - M I P T
Seq1
Seq2
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4 2 0 -2 -4 -6 -8 -10 -12
-4 2 6 4 2 1 -1 -3 -4 -6
-6 0 0 10 8 6 4 2 0 -2
-8 -2 -1 -1 19 17 15 13 11 9
-10 -4 0 -2 17 20 18 20 18 16
-12 -6 -2 -2 15 20 18 19 24 22
-14 -8 -4 -3 13 18 16 17 22 31
-16 -10 -6 -3 11 16 16 15 20 29

L S C V W M I P -
I S C - - M I P T
Seq1
Seq2
A L S C V W M I P
0 -2 -4 -6 -8 -10 -12 -14 -16 -18
-2 4 2 0 -2 -4 -6 -8 -10 -12
-4 2 6 4 2 1 -1 -3 -4 -6
-6 0 0 10 8 6 4 2 0 -2
-8 -2 -1 -1 19 17 15 13 11 9
-10 -4 0 -2 17 20 18 20 18 16
-12 -6 -2 -2 15 20 18 19 24 22
-14 -8 -4 -3 13 18 16 17 22 31
-16 -10 -6 -3 11 16 16 15 20 29
A
I
S
C
M
I
P
T

83
BLOSUM62
GAP COST: -2
+ gap cost
+ gap cost
A L S C V W M I P -
A I S C - - M I P T
Seq1
Seq2

+ gap cost
+ gap cost
A L S C V W M I P -
A I S C - - M I P T
Seq1
Seq2
4 2 4 9-2-2 5 4 7
alignment score: 31
COMMAND LINE BLAST
command
•Resource: https://www.ncbi.nlm.nih.gov/books/
NBK279675/
•Log onto compile

source /usr/local/ncbi-blast-2.4.0+/blast_env.sh
export BNFO=/home/bnfo301/assignments/2016-09-28
COMMAND LINE BLAST
• Find the documentation for makeblastdb.
• Create blast database
• Run blastp with query1.faa file
• Run blastp with query2.faa file
makeblastdb -help
makeblastdb -in $BNFO301/protein-db.faa -dbtype prot
-parse_seqids -out /home/bnfo301/huangb2/protein-
db.faa
blastp -query $BNFO301/query1.faa -db /home/bnfo301/
huangb2/protein-db.faa -out query1-out.txt
huangb2/protein-db.faa -out query1-out.txt
COMMAND LINE BLAST
• Change the output format
• Change the evalue

• Change the substitution matrix
huangb2/protein-db.faa -out query2-out.txt -outfmt 7
-evalue 1e-05
-evalue 1e-05 -matrix BLOSUM80
PS1 MOLECULAR BIOLOGY
BNFO301: PS Molecular Biology
DUE:
1. Complete the following table:
2. Below is the double-stranded DNA sequence of a
hypothetical genome, which happens to
have a very small gene.
a. Which strand of DNA shown, the top or the bottom, is the
template strand?

b. What is the sequence of the mRNA produced from this gene?
c. What is the sequence of the protein produced from the
mRNA?
d. If a mutation were found where a T/A (top/bottom) base pair
were added immediately
after the T/A base pair shown in bold, what would be the
sequence of the mRNA? What
would be the sequence of the protein? What type of mutation is
it?
Translation Problem Set - 1
BNFO 301: Introduction to Bioinformatics
Introduction to Molecular Biology: Translation - Problem Set
1. Complete the following table:
DNA
double helix
A G A
T G T
mRNA
transcribed 5' A U

Appropriate
tRNA anticodon U G 5'
Amino acids
incorporated into protein
met
2. List the changes that can be produced by a single basepair
mutation in the AGA codon
encoding arginine and label each silent (no effect on protein
structure), conservative (mild
effect on protein structure), hydrophobic-to-hydrophilic,
hydrophilic-to-hydrophobic, or
other.
3. Hemophilia A is an X-linked disease associated with the
absence of an essential blood clotting
factor, factor VIII (if you don't have any idea what an X-linked
trait is, don't worry about it).
Factor VIII is encoded by the gene called FACTOR8. This gene
was cloned from several
individuals -- some affected, some not -- and sequenced. A
portion of each sequence that
you're sure contains the beginning of the gene (i.e., the start
codon) was compared with the
same portion of the wild-type sequence, as shown below. Each
sequence contains only one
mutation, shown emphasized.
Wild-type 5'-
GGAGTTGAGTCATGGACTCTAAGCAGCGATCCACAAAG...

Individual a 5'-
GGAGTTTAGTCATGGACTCTAAGCAGCGATCCACAAAG...
Individual b 5'-
GGAGTTGAGTCATTGACTCTAAGCAGCGATCCACAAAG...
Individual c 5'-
GGAGTTGAGTCATGGACTCTTAGCAGCGATCCACAAAG...
Individual d 5'-
GGAGTTGAGTCATGGACTCTAAGCAGCTATCCACAAAG...
Individual e 5'-
GGAGTTGAGTCATGGACTCTAAGCAGCGATCCACTAAG...
For each individual, choose from the list below to describe what
you predict would be the
severity of the phenotype, and give the reason for your choice.
A. Severe hemophilia
B. Mild hemophilia
C. No hemophilia
A U G
U A C
T A C
A T G A
U
A
T T
A A
U
C C T A
C G A T
GC C

GC
AA
A U U3’
3’
3’
3’
5’
5’
Lys Ala Stop
PS1 MOLECULAR BIOLOGY
• Which strand of DNA shown, the top or the bottom, is the
template strand?
• What is the sequence of the mRNA produced from this gene?
• What is the sequence of the protein produced from the mRNA?
• If a mutation were found where a T/A (top/bottom) base pair
were added
immediately after the T/A base pair shown in bold, what would
be the
sequence of the mRNA? What would be the sequence of the
protein? What
type of mutation is it?
Bottom

5’- CTATAAAGAGCCATG CAT TAT CTA GAT AGT AGG
CTC TGA GAATTTATCTCACT - 3’
||||||||||||||| ||| ||| ||| ||| ||| ||| ||| ||| |||||
3’- GATATTTCTCGGTAC GTA ATA GAT CTA TCA TCC
GAG ACT CTTAAATAGAGTGA - 5’
PROMOTER TERMINATOR
mRNA 5’- GAGCCAUG CAU UAU CUA GAU AGU AGG CUC
UGA GAAUUUAUCUC -3’
protein 5’- met his tyr leu asp ser arg leu stp -3’
Standard Score Problems
Assuming a population mean of 500 and a population standard
deviation of 100 for the verbal subtest of the SAT exam:
1. What percentage of the student population has SAT-V scores
greater than 600?
greater than 700?
lower than 420?
between 300 and 520?

7. A student gets a 620 on this test. Convert this to a percentile.
8. 7. A student gets a 340 on this test. Convert this to a
percentile.
BNFO301: Exam 1
1. List all the changes that can be produced by a single base
pair mutation in the AGA
codon encoding arginine and label the resulting amino acid. In
addition label each
mutation as silent, missense or nonsense. (4pts)
2. What would be the value of using a dot plot to compare a
sequence to its own reverse
complement? (2 pts)
Sketch the dot plot o

3. f a 1 kb sequence in which a motif of approximately 50
consecutive bases appears six
times in the N terminal region of the sequence. (4 pts)
4. Use the PAM250 matrix to answer question 4.
a. Give the score for aligning two alanines (A) (1 pt)
b. Give the score for aligning two tryptophans (W) (1 pt)
c. Both of these alignments constitute “matches”, so why are the
scores so different? (2
pts)

Use the BLOSUM62 matrix for questions 5 and 6.
5. Calculate the dynamic programming matrix and an optimal
GLOBAL alignment for the
protein sequences FKHMEDPLE and FMDTPLNE , scoring -2
for a gap (i.e. 2 is the gap
penalty). Use the BLOSUM62 substitution matrix (given
above).
a. Fill out the matrix. (6 pts)
b. Highlight the traceback alignment. (1 pt)
c. Write out the final alignment. (2 pts)
d. Score the final alignment. (1 pt)

6. Calculate the dynamic programming matrix and an optimal
LOCAL alignment for the
protein sequences FKHMEDPLE and FMDTPLNE . Use the
BLOSUM62 matrix (provided
above).
a. Fill out the matrix. (6 pts)
b. Highlight the traceback local alignments. (1 pt)
c. Write out the final alignment. (2 pts)
d. Score the final alignment. (1 pt)

7. What is 16S rRNA and what is its function inside a cell? (2
pts)
8. 16s rRNA is widely used in microbiome studies. List two
strengths and two limitations of
16S rRNA sequencing. (4 pts)

9. Can 16S rRNA be used to classify viruses? Why or why not?
(2 pts)
10. Which of the following amino acids is least mutable
according to the PAM scoring
matrix? (2 pts)
a. Alanine
b. Glutamine
c. Methionine
d. Cysteine
1. Which of the following sentences BEST describes the

difference between a global
alignment and a local alignment between two sequences? (2 pts)
a. Global alignment is usually used for DNA sequences, while
local alignment is usually
used for protein sequences.
b. Global alignment has gaps, while local alignment does not
have gaps.
c. Global alignment finds the global maximum, while local
alignment finds the local
maximum.
d. Global alignment aligns the whole sequence, while local
alignment finds the best
subsequence that aligns.
2. How does the BLOSUM scoring matrix differ most notably
from the PAM scoring matrix?
(2 pts)
a. It is best used for aligning very closely related proteins.
b. It is based on global multiple alignment from closely related
proteins.
c. It is based on local multiple alignments from distantly related
proteins.
d. It combines local and global alignment information.

3. A global alignment algorithm (such as Needleman-Wunsch
algorithm) is guaranteed to
find an optimal alignment. Such an algorithm: (2 pts)
a. puts the two proteins being compared into a matrix and finds
the optimal score by
exhaustively searching every possible combination of
alignments.
b. puts the two proteins being compared into a matrix and finds
the optimal score by
iterative recursions.
c. puts the two proteins being compared into a matrix and finds
the optimal alignment
by finding optimal subpaths that define the best alignment(s)
d. can be used for proteins but not for DNA sequences.
4. What are the basic concepts of library preparation? (4 pts)
5. List 3 applications of next-generation sequencing. (2 pts)
6. How many reads do you need to get 30x coverage of your

genome if your read length is
300bp and your genome size is 10Mb? (2 pts)
Command line
Log in to compile. Navigate to the bnfo301 (home/bnfo301 )
directory. There is a folder
called exam1 where you will find all the files you need to
answer the next set of questions.
Instructions for this section:
• Write your output files to your user specific folder in
/home/bnfo301 (ex. my user specific
folder is /home/bnfo301/huangb2 ). You will be graded on the
files found in your specific
folder. If the files are not in that folder you will not get credit
for your answers. No
exceptions.
• Make sure you name your output file as instructed in each
question. I will take off 1 point
for each output file that is not correctly named.
• Code is typically written using a fixed width font. Use a fixed
width font to type your
commands in this section (ex. courier, inconsolata, menlo,
monaco).
• For each question, provide the command when specified, or
the command and answer. All

output files from this section should be written to you user
specific folder on compile. I will
access your user specific folder to grade this section.
1. List the files in the exam1 folder. command only (2 pts)
2. Count how many sequences are in the protein-db.faa file?
command and answer (2
pts)
3. You have an unknown1.faa sequence that you want to blast
against sequences in the
protein-db.faa file.
a. Copy the protein-db.faa to your user specific folder.
command only
b. Create a blast database for protein-db.faa . command only (2
pts)
c. Blast unknown1.faa against the database you just created.
Name your blast output
file 3b-unknown-output.txt . command only, leave output file on
Compile (2 pts)
d. Filter your blast results for hits with an evalue greater than
1e-05. Name your blast
output file 3c-unknown-output.txt . command only, leave output
file on Compile
(2 pts)
e. What is the percent identity and alignment length of the best

hit in your blast results
when you filter based on an evalue greater than 1e-05? Hint:
you may need to change
your output format. (8 pts)
f. What is the percent identity and alignment length of the worst
hit in your blast results
when you filter based on an evalue greater than 1e-05? Hint:
you may need to change
your output format. (4 pts)
7. BLAST is a tool that can be used to query multiple databases.
It is not always necessary
to create your own database. One of the most common blast
databases is the
non-redundant database (nr).
a. Blast the unknown1.faa sequence against the nr database
(/home/norrissw/bin/I-TASSER4.2/lib/nr/nr ) to find out what it
is. Name
your blast output file 4a-unknown-nr-output.txt . NOTE: you do
not need to run
the makeblastdb command. Also, it can take a few minutes for
your blast to run
because the nr database is very big. command only, leave output
file on Compile (2
pts)
b. Filter your blast results for hits with an e-value greater than
1e-10. Name your blast
output file 4b-unknown-nr-output.txt . command only, leave
output file on
Compile (2 pts)

c. Based on the best hit from nr, take the accession number and
identify what that
protein is. (4 pts)
8. The next set of questions involve the pipeline.py script
a. Copy the pipeline.py script to your /home/bnfo301/vcuid (2
pts)
b. Rename the pipeline.py script to 5b-pipeline.py . (2 pts)
c. Describe in detail what the script is doing, including what the
output from each step
is. (4 pts)
d. Modify the script so it filters the blast results using an e-
value cut off of 1e-05. Save
the modified script as 5d-pipeline.py . You do not need to run
the script, just add
in your modification. leave output file on Compile (2 pts)

2016.09.28TOPIC REVIEW• Exam • PS2 Sequence Alignment .docx

2016.09.28TOPIC REVIEW• Exam • PS2 Sequence Alignment .docx

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