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Problem B.1.
A transition is the exchange between either two purines (C → T or T → C) or between two pyrimidines (A →
G or G → A). A transversion, on the other hand, is the exchange of one purine by one pyrimidine or vice
versa.
Taking this into account, and since the probability of transitions is twice that of a transversion, the resulting
matrix grid for the transitions and transversions between the different nucleic bases is given by:
A
C
T
G
A
100%
33%
33%
67%
C
33%
100%
67%
33%
T
33%
67%
100%
33%
G
67%
33%
33%
100%
Problem B.2. (a)
i)
An inversion consists on the reordering of the subsequence from the 3→5 order into a 5→3 order. A
deletion consists on the removal of at least one nucleotide from the subsequence. A transpose consists on
the removal of the subsequence and its insertion in other region of the DNA sequence. A duplication
consists on the replacement of one nucleotide by two identical nucleotides. Finally, a point mutation
includes both transitions and transversions where one nucleotide is replaced by another.
Taking this into account, the resulting mutated sequences would be:
a) Parent sequence = …TTAAGGGGGGCCTTTTGAAA…
• Inverted sequence: …TTAAGGGGGGCCTTTTGAAA… (note that the sequence is identical
to the parent one since the subsequence being inverted is symmetric)
• Sample deleted sequence: …TTAAGGGGGCCTTTTGAAA…
• Sample transposed sequence: …TTAAGGCCGGGGTTTTGAAA…
• Sample duplicated sequence: …TTAAGGGGGGGCCTTTTGAAA…
• Sample point mutation sequence: …TTAAGGGGAGCCTTTTGAAA…
b) Parent sequence = …AAAAGGGGGGCCTTGGGACC…
• Inverted sequence: …GAAAAGGGGGCCTTGGGACC…
• Sample deleted sequence: …AAAGGGGGGCCTTGGGACC…
• Sample transposed sequence: …GAAAAGGGGGCCTTGGGACC…
• Sample duplicated sequence: …AAAAGGGGGGGCCTTGGGACC…
• Sample point mutation sequence: …AATAGGGGGGCCTTGGGACC…
c) Parent sequence = …CCAAGGGTGTCCTTTTGAGG…
• Inverted sequence: …CCAAGGTGTGCCTTTTGAGG…
• Sample deleted sequence: …CCAAGGGTTCCTTTTGAGG…
• Sample transposed sequence: …CCAAGGCCTGTGTTTTGAGG…
• Sample duplicated sequence: …CCAAGGGTGGTCCTTTTGAGG…
• Sample point mutation sequence: …CCAAGGGTATCCTTTTGAGG…
ii) The pre-RNA sequence is built by considering the pairing rules of the nucleic bases stated in the following
table:
Nucleotide of the DNA strand
Adenine (A)
Thymine (T)
Cytosine (C)
Guanine (G)
Equivalent nucleotide in the pre-RNA strand
Uracyl (U)
Adenine (A)
Guanine (G)
Cytosine (C)
Taking this into account, the corresponding pre-RNA strands will be:
1. DNA sequence: GGCG
Pre-RNA sequence: CCGC
2. DNA sequence: AAATAA
Pre-RNA sequence: UUUAUU
3. DNA sequence: TCCT
Pre-RNA sequence: AGGA
4. DNA sequence: CGTTA
Pre-RNA sequence: GCAAU
Problem B.2. (b)
1)
Parent sequence
Mutation
Single substitution
No change
Multiple substitution
ACCCTAC
Change involved
C→G
None
A→T→C
Back substitution
C→T→C
Parallel substitution
G→ A or T
Coincidental
substitution
Convergent substitution
C→G
A or T → G
Intermediate sequence
AC(C→G)CTAC
ACGCT(A→T)C
ACGCT(T→C)C
ACG(C→T)TCC
ACG(T→C)TCC
AC(G→A)CTCC
AC(G→T)CTCC
A(C→G)A(C→G)TCC
A(C→G)T(C→G)TCC
AG(A→G)GTCC
AG(T→G)GTCC
Taking this sequence into account, the final sequence (1) would be AGGGTCC.
Final sequence
ACGCTAC
ACGCTAC
ACGCTCC
ACGCTCC
ACACTCC
ACTCTCC
AGAGTCC
AGTGTCC
AGGGTCC
AGGGTCC
2)
Parent sequence
Mutation
Single substitution
No change
Multiple substitution
ACCCTAC
Change involved
A→T
None
C→A→T
Back substitution
C→G→C
Parallel substitution
C→ T or A
Coincidental
substitution
Convergent substitution
T→A
Intermediate sequence
ACCCT(A→T)C
AC(C→A)CTTC
AC(A→T)CTTC
A(C→G)TCTTC
A(G→C)TCTTC
ACT(C→T)TTC
ACT(C→A)TTC
AC(T→A)TTTC
AC(T→A)ATTC
ACA(T→G)TTC
ACA(A→G)TTC
A or T → G
Final sequence
ACCCTTC
ACCCTTC
ACTCTTC
ACTCTTC
ACTTTTC
ACTATTC
ACATTTC
ACAATTC
ACAGTTC
ACAGTTC
Taking this sequence into account, the final sequence (2) would be ACAGTTC.
Problem B.3.
Parent sequence (X):
5’ – TAC GGA TCG AAT GCT CCC GTA ATC – 3’
Complementary sequence (Y):
3’ – ATG CCT AAC TTA CGG CAT CAT CAT TAG – 5’
As instructed, single point mutations are highlighted in yellow, triplet deletions in red and triplet
duplications in green
Problem B.4.
Aligned test pair:
F
|
F
[Q
D]
[D
Q]
[I
:
[L
V
L]
.
V]
V
S
[F]
.
[V]
V
R
|
R
[R]
.
[E]
V
[D]
.
[N]
V
D
|
D
[I
.
[D
I]
.
D]
V
[I]
.
[N]
V
[F
Q]
[Q
F]
L
:
I
V
S
Taking this into account, a total of 8 transversion mutations and 2 transition mutations are taking place in
the aligned test pair.
Problem B.5.
The alignment providing the highest degree of interaction between the two DNA sequences is given below:
A
:
T
–
–
–
C
C
C
A
:
T
–
–
A
A
C
:
G
C
A
:
T
–
G
T
–
–
A
:
T
A
T
–
–
Problem B.6.
U:
V:
A
|
A
G
|
G
C
C
|
C
–
A
|
A
T
A
|
A
–
–
T
|
T
T
A
|
A
This alignment has three gaps with three total spaces. It also has 7 matches and three mismatches
corresponding to the spaces. Taking this into account, the score for this alignment would be (7*100) –
(3*50) = 550
Problem B.7.
i)
X*
Y*
A
:
T
G
:
C
T
|
T
G
.
A
G
|
G
G
.
A
C
A
–
–
T
|
T
T
|
T
C
|
C
C
.
T
G
T
|
T
T
|
T
Levenshtein distance = 8
Score of this alignment = (7*100) + (2*75) + (3*50) – (3*50) = 850
It is the alignment providing the highest score and the lowest LD taking into account that only 8 changes
are necessary. Any other alignment will increase the Levenshtein distance and hence decrease the score as
more gaps are introduced and need to be filled.
ii)
Pair 1
X*
Y*
T
0
–
C
1
C
G
0
C
V
C
0
T
T
0
C
G
0
A
S
S
S
G
1
G
C
0
G
V
G
1
G
C
0
T
S
A
0
T
V
A
0
G
S
A
0
C
V
C
0
A
V
C
1
C
G
0
A
S
Pair 2
X*
A
0
–
Y*
A
1
A
G
0
C
V
C
0
T
A
0
C
V
S
G
0
A
T
0
C
S
S
C
0
G
V
T
1
T
C
0
T
A
0
T
V
S
A
0
G
A
0
G
S
S
C
0
G
V
S
iii)
X
T
.
C
Y
A
C
|
C
C
C
:
G
A
:
U
G
.
A
T
:
A
Levenshtein distance = 6
Problem B.8.
a)
C
U
T
I
M
|
M
B
|
B
E
|
E
R
|
R
L
|
L
A
|
A
N
|
N
D
|
D
Hamming distance = 2 (C/T, U/I)
b)
B
I
O
I
N
|
|
|
|
|
B
I
O
I
N
Levenshtein distance = 2
F
|
F
O
|
O
R
|
R
M
|
M
A
|
A
T
|
T
I
|
I
C
↓
O
S
↓
N
T
E
L
E
↓ ↓ ↓ ↓ |
H
A
I
L
Levenshtein distance = 8
I
↓
N
↓
F
|
F
O
|
O
R
|
R
M
|
M
A
|
A
T
|
T
I
|
I
C
↓
O
T
E
L
E
I
↓ ↓ ↓ ↓ ↓
C
O
Levenshtein distance = 8
N
↓
N
F
|
F
O
|
O
R
|
R
M
|
M
A
|
A
T
|
T
I
|
I
C
↓
O
S
↓
N
G
0
A
S
↓
N
G
0
C
V
0
C
Problem B.9.
a)
s
A
G
|
↓
t
A
Levenshtein distance = 2
C
|
C
A
|
A
C
|
C
A
|
A
C
|
C
↓
U
A
|
A
G
|
G
U
|
U
G
|
G
U
|
U
G
|
G
↓
A
U
|
U
A
G
C
|
↓
↓
t
A
C
A
Original Levenshtein distance = 5
A
↓
C
A
|
A
C
|
C
C
↓
A
A
↓
T
A
|
A
A
↓
C
C
↓
A
C
|
C
A
|
A
b)
s
U
C
|
↓
t
U
Levenshtein distance = 2
Problem B.10.
s
Assuming 1 insertion and 1 deletion:
s
A
G
|
↓
t
A
Levenshtein distance = 4
C
|
C
↓
T
Assuming 2 deletions:
s
A
G
|
↓
t
A
Levenshtein distance = 6
C
|
C
A
|
A
A
↓
–
C
|
C
C
↓
A
A
↓
C
↓
A
↓
T
Problem B.11.
a)
s
T
G
|
↓
t
T
C
Original Hamming distance = 6
C
↓
A
A
↓
C
C
↓
A
A
↓
C
C
↓
T
C
|
C
s
C
|
C
A
|
A
C
|
C
A
|
A
C
|
C
↓
T
T
G
|
↓
t
T
Final Hamming distance = 2
C
|
C
b)
s
T
G
|
|
t
T
G
Original Hamming distance = 0
s
T
G
|
↓
t
T
Final Hamming distance = 3
C
|
C
C
↓
G
A
|
A
↓
C
A
|
A
A
|
A
C
|
C
A
|
A
T
|
T
C
|
C
C
|
C
T
|
T
C
|
C
Problem B.12.
Window number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
X
0110
1110
0101
1010
0110
1100
0111
1111
0100
1010
1111
1011
1100
1101
0101
1100
0110
1110
0010
1001
1010
0011
1110
0111
1
Y
1010
0010
1001
1011
1001
0110
0011
1111
1010
0101
0111
1101
1110
0110
1010
0001
0110
1000
1111
1100
0101
1100
1111
0101
111
HD
2
2
2
1
4
2
1
0
3
4
1
2
1
3
4
3
0
2
3
2
4
4
1
2
2
The resulting graph is shown in figure 1. Additionally, figure 2 represents a pie chart with the relative
abundance of the Hammings distance of the different windows. As can be observed, the most abundant HD
is 2, which accounts for a 36% of the windows.
Figure 1. HD obtained for the different windows
Figure 2. Pie chart
Problem B.13.
Window
number
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
X
Y
P(0),X
Q(1),X
KL1
P(0),Y
Q(1),Y
KL2
KL
0110
1110
0101
1010
0110
1100
0111
1111
0100
1010
1111
1011
1100
1101
0101
1100
0110
1110
0010
1001
1010
0011
1110
0111
1
1010
0010
1001
1011
1001
0110
0011
1111
1010
0101
0111
1101
1110
0110
1010
0001
0110
1000
1111
1100
0101
1100
1111
0101
111
0.50
0.25
0.50
0.50
0.50
0.50
0.25
0.00
0.75
0.50
0.00
0.25
0.50
0.25
0.50
0.50
0.50
0.25
0.75
0.50
0.50
0.50
0.25
0.25
0.00
0.50
0.75
0.50
0.50
0.50
0.50
0.75
1.00
0.25
0.50
1.00
0.75
0.50
0.75
0.50
0.50
0.50
0.75
0.25
0.50
0.50
0.50
0.75
0.75
1.00
0.00
-0.27
0.00
0.00
0.00
0.00
-0.27
0.50
0.75
0.50
0.25
0.50
0.50
0.50
0.00
0.50
0.50
0.25
0.25
0.25
0.50
0.50
0.75
0.50
0.75
0.00
0.50
0.50
0.50
0.00
0.50
0.00
0.50
0.25
0.50
0.75
0.50
0.50
0.50
1.00
0.50
0.50
0.75
0.75
0.75
0.50
0.50
0.25
0.50
0.25
1.00
0.50
0.50
0.50
1.00
0.50
1.00
0.00
0.82
0.00
-0.27
0.00
0.00
0.00
0.00
0.55
0.00
-0.27
0.00
0.00
-0.27
0.00
0.82
0.00
-0.27
-0.55
-0.27
-0.27
0.00
0.82
0.00
0.55
0.82
0.00
0.00
0.00
-0.27
-0.27
0.00
0.82
0.00
-0.27
0.00
-0.27
0.00
0.00
0.00
-0.27
0.82
0.00
0.00
0.00
-0.27
-0.27
Taking these data into account, the corresponding plot is presented in figure 3:
Figure 3. KL plot
0.00
0.00
-0.27
-0.27
-0.27
0.00
0.00
0.82
0.00
0.82
0.00
0.00
0.00
0.00
As can be observed in this plot, the most abundant KL value is 0, which corresponds to a binary code
containing 2 zeroes and 2 ones such that the probability of obtaining a zero or a one in the window is
equivalent: 2/4 = 0.50
Problem B.14.
J7
J6
J5
J4
J3
J2
J1
I8
G
1
0
1
0
0
0
0
G
A
G
C
A
I7
A
0
1
0
0
1
0
0
I6
T
0
0
0
0
0
0
0
I5
T
0
0
0
0
0
0
0
I4
C
0
0
0
1
0
0
0
I3
A
0
1
0
0
1
0
0
I2
I1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
I6
T
2
2
2
1
0
0
0
I5
T
2
2
2
1
0
0
0
I4
C
1
1
1
2
0
0
0
I3
A
0
1
0
0
1
0
0
I2
I1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Calculation shown in the attached xls file
I8
G
J7
G
4
J6
A
3
J5
G
3
J4
C
1
J3
A
0
J2
0
J1
0
The resulting alignment is then:
I7
A
2
3
2
1
1
0
0
X: G A G – C A
| | .
| |
Y: G A T T C A
Problem B.15.
J12
J11
J10
J9
J8
J7
J6
J5
J4
J3
J2
J1
S
S
T
Q
F
S
E
D
A
I
0
0
I12
W
0
0
0
0
0
0
0
0
0
0
0
0
I11
F
0
0
0
0
1
0
0
0
0
0
0
0
I10
G
0
0
0
0
0
0
0
0
0
0
0
0
I9
Q
0
0
0
1
0
0
0
0
0
0
0
0
I8
F
0
0
0
0
1
0
0
0
0
0
0
0
I7
T
0
0
1
0
0
0
0
0
0
0
0
0
I6
S
1
1
0
0
0
1
0
0
0
0
0
0
I5
A
0
0
0
0
0
0
0
0
1
0
0
0
I4
I
0
0
0
0
0
0
0
0
0
1
0
0
I3
W
0
0
0
0
0
0
0
0
0
0
0
0
I2
I1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Calculation shown in the attached xls file
J12
J11
J10
J9
J8
J7
J6
J5
J4
J3
J2
J1
S
S
T
Q
F
S
E
D
A
I
0
0
I12
W
5
5
5
4
3
2
2
2
1
0
0
0
I11
F
5
5
5
4
4
2
2
2
1
0
0
0
I10
G
5
5
5
4
3
2
2
2
1
0
0
0
I9
Q
4
4
4
5
3
2
2
2
1
0
0
0
I8
F
4
4
3
3
4
2
2
2
1
0
0
0
I7
T
3
3
4
3
3
2
2
2
1
0
0
0
I6
S
3
3
2
2
2
3
2
2
1
0
0
0
I5
A
1
1
1
1
1
1
1
1
2
0
0
0
I4
I
0
0
0
0
0
0
0
0
0
1
0
0
I3
W
0
0
0
0
0
0
0
0
0
0
0
0
I6
C
0
0
0
0
0
0
0
1
0
0
0
0
I5
G
0
0
1
0
1
0
0
0
0
0
0
0
I4
C
0
0
0
0
0
0
0
1
0
0
0
0
I3
A
1
0
0
1
0
1
1
0
0
0
0
0
I2
I1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
The resulting alignment is then:
X: W F G Q F T S – – A I W
| |
|
| |
Y: S S T Q F – S E D A I –
Problem B.16.
J12
J11
J10
J9
J8
J7
J6
J5
J4
J3
J2
J1
A
U
G
A
G
A
A
C
U
U
I10
C
0
0
0
0
0
0
0
1
0
0
0
0
I9
U
0
1
0
0
0
0
0
0
1
1
0
0
I8
U
0
1
0
0
0
0
0
0
1
1
0
0
I7
A
1
0
0
1
0
1
1
0
0
0
0
0
I2
I1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Calculation shown in the attached xls file
J12
J11
J10
J9
J8
J7
J6
J5
J4
J3
J2
J1
I10
C
4
3
3
2
2
2
2
3
1
0
0
0
A
U
G
A
G
A
A
C
U
U
I9
U
4
4
3
2
2
2
1
1
2
1
0
0
I8
U
3
4
3
2
2
2
1
0
1
1
0
0
I7
A
3
2
2
3
1
2
2
0
0
0
0
0
I6
C
2
2
2
2
1
1
1
1
0
0
0
0
I5
G
1
1
2
1
2
1
1
0
0
0
0
0
I4
C
1
1
1
1
1
1
0
1
0
0
0
0
I3
A
1
0
0
1
0
1
1
0
0
0
0
0
I2
I1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
–
–
–
A
A
A
C
|
C
U
U
The resulting alignment is then:
s:
C
t:
A
U
|
U
U
–
–
G
A
|
A
C
–
G
|
G
Problem B.17.
J7
J6
J5
J4
J3
J2
J1
C
T
C
G
T
I8
C
1
0
1
0
0
0
0
I7
T
0
1
0
0
1
0
0
I6
A
0
0
0
0
0
0
0
I5
A
0
0
0
0
0
0
0
I4
G
0
0
0
1
0
0
0
I3
T
0
1
0
0
1
0
0
I2
I1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
I6
A
2
2
2
1
0
0
0
I5
A
2
2
2
1
0
0
0
I4
G
1
1
1
2
0
0
0
I3
T
0
1
0
0
1
0
0
I2
I1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Calculation shown in the attached xls file
J7
J6
J5
J4
J3
J2
J1
C
T
C
G
T
I8
C
4
3
3
1
0
0
0
I7
T
2
3
2
1
1
0
0
The resulting alignment is then:
U: C T C – G T
| |
| |
V: C T A A G T
Problem B.18.
J12
J11
J10
J9
J8
J7
J6
J5
J4
J3
J2
J1
M
S
T
A
L
P
G
L
G
S
0
0
I12
M
1
0
0
0
0
0
0
0
0
0
0
0
I11
A
0
0
0
1
0
0
0
0
0
0
0
0
I10
V
0
0
0
0
0
0
0
0
0
0
0
0
I9
R
0
0
0
0
0
0
0
0
0
0
0
0
I8
K
0
0
0
0
0
0
0
0
0
0
0
0
I7
L
0
0
0
0
1
0
0
1
0
0
0
0
I6
S
0
1
0
0
0
0
0
0
0
1
0
0
I5
L
0
0
0
0
1
0
0
1
0
0
0
0
I4
E
0
0
0
0
0
0
0
0
0
0
0
0
I3
G
0
0
0
0
0
0
1
0
1
0
0
0
I2
I1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
I9
R
3
3
3
3
2
2
2
1
1
0
0
0
I8
K
3
3
3
3
2
2
2
1
1
0
0
0
I7
L
3
2
2
2
3
2
2
2
1
0
0
0
I6
S
2
3
2
2
2
2
2
1
0
1
0
0
I5
L
1
1
1
1
2
1
1
2
0
0
0
0
I4
E
1
1
1
1
1
1
1
1
0
0
0
0
I3
G
0
0
0
0
0
0
1
0
1
0
0
0
I2
I1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Calculation shown in the attached xls file
J12
J11
J10
J9
J8
J7
J6
J5
J4
J3
J2
J1
M
S
T
A
L
P
G
L
G
S
0
0
I12
M
5
4
4
3
2
2
2
1
1
0
0
0
I11
A
3
3
3
4
2
2
2
1
1
0
0
0
I10
V
3
3
3
3
2
2
2
1
1
0
0
0
The resulting alignment is then:
X: M – – A V R K L S L E G |
|
|
|
Y: M S T A L P G L – – – G S
Problem B.19.
J12
J11
J10
J9
J8
J7
J6
J5
J4
J3
J2
J1
S
F
T
Q
F
S
E
D
A
I
0
0
I12
W
0
0
0
0
0
0
0
0
0
0
0
0
I11
E
0
0
0
0
0
0
1
0
0
0
0
0
I10
G
0
0
0
0
0
0
0
0
0
0
0
0
I9
Q
0
0
0
1
0
0
0
0
0
0
0
0
I8
E
0
0
0
0
0
0
1
0
0
0
0
0
I7
T
0
0
1
0
0
0
0
0
0
0
0
0
I6
S
1
0
0
0
0
1
0
0
0
0
0
0
I5
A
0
0
0
0
0
0
0
0
1
0
0
0
I4
I
0
0
0
0
0
0
0
0
0
1
0
0
I3
S
1
0
0
0
0
1
0
0
0
0
0
0
I2
I1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
I9
R
4
4
3
4
3
3
2
2
1
0
0
0
I8
K
4
4
3
3
3
2
3
2
1
0
0
0
I7
L
3
3
4
3
3
2
2
2
1
0
0
0
I6
S
3
2
2
2
2
3
2
2
1
0
0
0
I5
L
1
1
1
1
1
1
1
1
2
0
0
0
I4
E
1
1
1
1
1
0
0
0
0
1
0
0
I3
G
1
0
0
0
0
1
0
0
0
0
0
0
I2
I1
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
Calculation shown in the attached xls file
J12
J11
J10
J9
J8
J7
J6
J5
J4
J3
J2
J1
M
S
T
A
L
P
G
L
G
S
0
0
I12
M
4
4
4
3
3
3
2
2
1
0
0
0
I11
A
4
4
4
3
3
3
3
2
1
0
0
0
I10
V
4
4
4
3
3
3
2
2
1
0
0
0
The resulting alignment is then:
X: W E G Q E T S – – A I S
|
|
| |
Y: S F T Q F – S E D A I –
Problem B.20.
X
A
A
U
G
C
C
A
U
U
G
A
C
G
G
X
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
A
0
1
1
A
A
U
G
C
C
A
U
U
G
A
C
G
G
A
1.0
1.0
0.0
0.0
0.0
0.0
1.0
0.0
0.0
0.0
1.0
0.0
0.0
0.0
C
0.0
0.7
0.7
0.0
1.0
1.0
0.0
0.7
0.0
0.0
0.0
2.0
0.0
0.0
C
0
A
0
1
1
G
0
C
0
C
0
U
0
C
0
G
0
C
0
1
1
1
1
U
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
A
1.0
1.0
0.4
0.4
0.0
0.7
2.0
0.0
0.4
0.0
1.0
0.0
1.7
0.0
G
0.0
0.7
0.7
1.4
0.1
0.0
0.4
1.7
0.0
1.4
0.0
0.7
1.0
2.7
C
0.0
0.0
0.4
0.4
2.1
1.1
0.0
0.1
1.4
0.0
1.1
1.0
0.4
0.7
1
1
1
C
0.0
0.0
0.0
0.1
1.4
3.1
0.8
0.0
0.0
1.1
0.0
2.1
0.8
0.1
U
0.0
0.0
1.0
0.0
0.0
1.1
2.8
1.8
1.0
0.0
0.8
0.0
1.8
0.5
C
0.0
0.0
0.0
0.7
1.0
1.0
0.8
2.5
1.5
0.7
0.0
1.8
0.0
1.5
G
0.0
0.0
0.0
1.0
0.4
0.7
0.7
0.5
2.2
2.5
0.4
0.0
2.8
1.0
1
1
C
0.0
0.0
0.0
0.0
2.0
1.4
0.4
0.4
0.2
1.9
2.2
1.4
0.0
2.5
U
0.0
0.0
1.0
0.0
0.0
1.7
1.1
1.4
1.4
0.0
1.6
1.8
1.1
0.0
U
0.0
0.0
1.0
0.7
0.0
0.0
1.4
2.1
2.4
1.1
0.0
1.3
1.5
0.8
A
1.0
1.0
0.0
0.7
0.4
0.0
1.0
1.1
1.8
2.1
2.1
0.0
1.0
1.2
The resulting alignment is then:
X:
Y:
…G
|
…G
C
|
C
G
0
1
1
1
1
1
A
0
1
1
1
1
1
1
U
0
C
|
C
A
U
|
U
C
U
G…
|
G…
G
0.0
0.7
0.7
1.0
0.4
0.1
0.0
0.7
0.8
2.5
1.8
1.8
1.0
2.0
Problem B.21.
X
W
G
Q
E
G
S
I
E
A
X
0
0
0
0
0
0
0
0
0
0
W
0
1
W
G
Q
E
G
S
I
E
A
W
1.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
R
0.0
0.7
0.0
0.0
0.0
0.0
0.0
0.0
0.0
R
0
N
0
D
0
C
0
Q
0
E
0
G
0
S
0
A
0
1
1
1
1
1
1
N
0.0
0.0
0.4
0.0
0.0
0.0
0.0
0.0
0.0
D
0.0
0.0
0.0
0.1
0.0
0.0
0.0
0.0
0.0
C
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
Q
0.0
0.0
1.0
0.0
0.0
0.0
0.0
0.0
0.0
E
0.0
0.0
0.0
2.0
0.0
0.0
0.0
0.0
0.0
G
0.0
1.0
0.0
0.0
3.0
0.0
0.0
0.0
0.0
S
0.0
0.0
0.7
0.0
0.0
4.0
0.0
0.0
0.0
A
0.0
0.0
0.0
0.4
0.0
0.0
3.7
0.0
1.0
The resulting alignment is then:
X:
Y:
…Q
|
…Q
E
|
E
G
|
G
S
|
S
–
–
I
E
A…
|
A…
Problem B.22.
X
A
A
S
R
N
P
S
C
W
T
T
W
H
T
X
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
A
0
1
1
A
A
S
R
N
P
S
C
W
T
T
W
H
T
A
1.0
1.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
A
1
2.0
0.7
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
A
0
1
1
S
0
T
0
H
0
E
0
C
0
W
0
C
0
T
0
W
0
H
0
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
S
0.0
0.7
3.0
0.4
0.0
0.0
1.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
T
0.0
0.0
0.4
2.7
0.1
0.0
0.0
0.7
0.0
1.0
1.0
0.0
0.0
1.0
1
H
0.0
0.0
0.0
0.1
2.4
0.0
0.0
0.0
0.4
0.0
0.7
0.7
1.0
0.0
E
0.0
0.0
0.0
0.0
0.0
2.1
0.0
0.0
0.0
0.1
0.0
0.4
0.4
0.7
C
0.0
0.0
0.0
0.0
0.0
0.0
1.8
1.0
0.0
0.0
0.0
0.0
0.1
0.1
W
0.0
0.0
0.0
0.0
0.0
0.0
0.0
1.5
2.0
0.0
0.0
1.0
0.0
0.0
C
|
C
C
0.0
0.0
0.0
0.0
0.0
0.0
0.0
1.0
1.2
1.7
0.0
0.0
0.7
0.0
T
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.7
2.2
2.7
0.0
0.0
0.4
W
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
1.0
0.4
1.9
3.7
0.0
1.0
C
T
|
T
H
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.0
0.7
0.1
1.5
4.7
0.0
The resulting alignment is then:
X:
Y:
…A A
| |
…A A
S
|
S
T
H
E
–
R
N
P
S
W
|
W
T
W
|
W
H…
|
H…
…
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