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FCH 532 Lecture 7
Chapter 5: DNA sequencing
Chapter 7
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Genome sequencing
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In order to sequence entire genomes, segments need to be assembled into
contigs (contiguous blocks) to establish the correct order of the sequence.
Chromosome walking may be one way to do so, but is prohibitively expensive.
Two methods have been used recently:
1. Conventional genome sequencing-low resolution maps made by
identifying “landmarks” in ~250 kb inserts in YACs.
Landmarks are 200-300 bp segments, aka sequence tagged sites(STSs)-2
clones with the same STS overlap.
STS-containing inserts are sheared randomly into ~40kB segments and
cloned into cosmid vectors-used to create high resolution maps.
The cosmid inserts are fragmented to smaller sizes and sequenced.
Cosmid inserts are assembled by using the STS sequence overlaps and
cosmid walking.
Cannot be used effectively with sequences containing high amounts of
repetitive sequence. (Use expressed sequence tags (ESTs)).
Genome sequencing
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2. Shotgun strategy– genome library is randomly fragmented
– large amount of cloned fragments are sequenced.
– Genome is assembled by identifying overlaps between pairs of fragments.
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The probability that a base is not sequenced is e-c,
c is the redundancy of coverage, c = LN/G,
– where L is the average length of the cloned inserts in base pairs,
– N is the number of inserts sequenced,
– and G is the length of the genome in base pairs.
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The aggregate length of the gaps between contigs is G e-c and the
average gap size is G/N.
Bacterial genomes-shotgun strategy is straightforward. Gaps are filled in
by synthesizing PCR primers and finishing a genome.
Eukaryotic genomes-larger size so it must be carried out in stages using
BACs and then identifying ~500 bp sequences from each to yield
sequence tagged connnectors (STCs or BAC ends)
This allows assembly via the overlapping of STCs.
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Figure 7-17
Genome sequencing strategies.
Human genome
• 2.2 billion nucleotide sequence ~90% complete because of highly
repetitive sequence.
• About half of the human genome consists of various repeating
sequences.
• Only ~28% of the genome is transcribed to RNA
• Only 1.1% to 1.4% of the genome (~5% of the transcribed RNA)
encodes protein.
• Only ~30,000 protein encoding genes (open reading frames or
ORFs) identified. Predicted 50,000 - 140,000 ORFs.
• Only a small fraction of human protein families are unique to
vertebrates; most occur in other life forms.
• Two randomly selected human genomes differ, on average, by only
1 nucleotide per 1250; that is, any 2 people are likely to be >99.9%
identical.
Human genome
• 2.2 billion nucleotide sequence ~90% complete because of highly
repetitive sequence.
• About half of the human genome consists of various repeating
sequences.
• Only ~28% of the genome is transcribed to RNA
• Only 1.1% to 1.4% of the genome (~5% of the transcribed RNA)
encodes protein.
• Only ~30,000 protein encoding genes (open reading frames or
ORFs) identified. Predicted 50,000 - 140,000 ORFs.
• Only a small fraction of human protein families are unique to
vertebrates; most occur in other life forms.
• Two randomly selected human genomes differ, on average, by only
1 nucleotide per 1250; that is, any 2 people are likely to be >99.9%
identical.
Chemical evolution
• Evolutionary aspects of amino acid sequences.
• Change stem from random mutational events that alter
a protein’s primary structure.
• Mutational change must offer a selective advantage or
at least, not decrease fitness.
• Most mutations are deleterious and often lethal so they
are not reproduced.
• Sometimes mutations occur that increase fitness of the
host in its natural environment.
• Example: Sickle-cell anemia.
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Figure 7-18a Scanning electron microscope of human
erythrocytes. (a) Normal human erythrocytes revealing their
biconcave disklike shape.
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Figure 7-18b Scanning electron microscope of human
erythrocytes. (b) Sickled erythrocytes from an individual with
sickle-cell anemia.
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Figure 7-20 A map indicating the regions of the world
where malaria caused by P. falciparum was prevalent before
1930.
Chemical evolution
• Pauling and co-workers showed that normal human
hemoglobin (HbA) is more electronegative than sicklecell hemoglobin (HbS).
• Sickle-cell anemia is inherited according to the laws of
Mendelian genetics.
• Homozygous for HbS is almost all HbS,
phenotype=sickle cell anemia.
• Heterozygous for HbS is ~40% HBs, phenotype=sickle
cell trait.
• Homozygous for HbA, normal human hemoglobin.
Mutations in a- or b-globin genes
can cause disease state
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Sickle cell anemia – E6 to V6
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Causes V6 to bind to hydrophobic pocket in
deoxy-Hb
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Polymerizes to form long filaments
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Cause sickling of cells
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Sickle cell trait offers advantage against malaria
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Cells sickle under low oxygen conditions and if
infected with Plasmodium falciparum.
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Causes the preferential removal of infected
erythrocytes from circulation.
Variations in homologous proteins
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Similar proteins from related species likely derived from the same ancestor.
A protein that is well adapted to its function will continue to evolve.
Neutral drift-mutational changes in a protein that don’t affect its function over time.
Homologous proteins-evolutionarily related proteins.
Comparison of the primary structures of homologous structures can be used
to identify which residues are essential to its function, lesser significance,
and little function.
Invariant residue-the same side chain at a particular position in the amino acid
sequence of related proteins.
If an invariant residue is observed between related proteins, it is likely necessary to
some essential function of the protein.
Other amino acids may have less stringent side chain requirements-where amino
acids may be conservatively substituted-(be substituted with an amino acid with
similar properties).
If many amino acids tolerated at a specific position - hypervariable.
Cytochrome c
• Cytochrome c is nearly universal eukaryotic protein
necessary for electron transport.
• Vertebrates 103-104 residues; up to 8 more aas in
other phyla.
• Similarities are observed in an alignment.
• 38 of 105 residues are invariant and the others are
conservatively substituted.
• 8 positions are hypervariable.
• His 18 and Met 80 form bonds with the redox Fe of the
heme group.
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Table 7-4a
Amino Acid Sequences of Cytochromes c
from 38 species.
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Cytochrome c
• Evolutionary differences between two homologous
proteins are determined by counting the amino acid
differences between them.
• Order of differences parallels taxonomy and can be put
into a table.
• This data can be used to construct a phylogenetic
tree-a tree that indicates ancestral relationships among
organisms and their proteins.
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Figure 7-21
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Each branch point
indicates a possible
common ancestor to
everything above it.
Relative evolutionary
distances between
neighboring branch
points are expressed
as the number of
amino acid differences
per 100 residues of the
protein (percentage of
accepted point
mutations or PAM
units).
Phylogenic tree of cytochrome c.
Evolutionary rates
• Evolutionary distances between various species can be plotted
against the time when species diverged.
• Each protein has a characteristic rate of change-unit evolutionary
period-the time required for the amino acid sequence of the
protein to change by 1% after two species have diverged.
• Acceptance rate of mutations depends on the extent to which
amino acid affects function.
• Amino acid substitutions in a protein mostly result from single base
changes in the gene specifying the protein (point mutations).
• Point mutations in DNA accumulate at a constant rate with timeresulting from random chemical change rather than errors from the
replication process.
Figure 7-22
Rates of evolution of four unrelated proteins.
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Tolerant
Intolerant of changes
Evolutionary rates
• Amino acid substitutions in a protein mostly
result from single base changes in the gene
specifying the protein (point mutations).
• Point mutations in DNA accumulate at a
constant rate with time-resulting from random
chemical change rather than errors from the
replication process.
• Know this based on generation times of
different organisms.
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Figure 7-23 A phylogenetic tree for cytochrome c.
Evolutionary rates
• Protein evolution is not the basis for organismal evolution.
• Rapid divergence is likely due to mutational changes in DNA that
control gene expression.
• Some proteins have extensive sequence similarity in the same
organism resulting from a gene duplication.
• Gene duplication is an efficient mode of evolution because the new
gene can evolve a new functionality while the original directs
synthesis of the ancestral protein.
• Globins-an example-see Ch. 7.
• Paralog-homologous proteins in the same organism
• Orthologs-homologous proteins/genes in different organisms that
arose through species divergence.
Bioinformatics
• Biotechnology meets computer science.
• Sequence databases
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Sequence alignment
• Sequence similarity of two polypeptides or two DNAs
can be quantified by determining the number of aligned
residues that are identical.
• Human and dog cytochromes c differ in 11 or 104
residues [(104-11)/104] X 100=89% identical.
• Human and yeast are [(104-45)/104] X 100 = 57%
identical.
• When determining percent identity, the length of the
shorter peptide/DNA is by convention, used in the
denominator.
• Must also decide which amino acid residues are
considered similar (e.g. Asp and Glu).
Homology of distantly related proteins
• Hypothetical example: Assume that we have a 100 residue protein
in which all point mutations have equal probability of being
accepted and happen at a constant rate.
• At an evolutionary distance of 1 PAM unit, the original and evolved
proteins are 99% identical.
• At an evolutionary distance of 2 PAM units, they are (0.99)2 X 100
= 98% identical
• At 50 PAM units, they are (0.99)50 X 100 = 60.5% identical.
• This is due to the stochastic (random) process of mutation. Every
residue has an equal chance of mutating.
• These can be plotted as percent identity vs. evolutionary distance.
Figure 7-25a Rate of sequence change in evolving proteins.
(a) Protein evolving at random and that initially consists of
5% each of the 20 “standard” amino acid residues.
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Approaches but never = 0!
Homology of distantly related proteins
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Real proteins are more complex.
Certain amino acids are more likely to be accepted than others.
Distribution of amino acids in proteins is not uniform (9.5% are Leu on
average and only 1.2% are Trp).
Can also be affected by shifts in the sequence resulting from insertion or
deletion of one or more residues within a chain.
Example is different lengths of cytochrome c peptides.
If the amino acid sequence is allowed to shift, the best alignment will
increase.
Homology of distantly related proteins
SQMCILFKAQMNYGH
MFYACRLPMGAHYWL
Unlimited
gapping
SQMCILFKAQMNYGH
--M---F-----YACRLPMGAHYWL
• Unlimited gapping because of insertions and deletions (indels)
cannot be allowed because we won’t get the proper alignment.
• At the same time we need to allow for gapping (cyt c).
• There must be a penalty for gaps.
• Unrelated proteins will exhibit sequence identities (15-25%) which
will be the same as distantly related proteins.
• Requires more sophisticated algorithms to describe
Figure 7-25b Rate of sequence change in evolving proteins.
(b) A protein of average amino acid composition evolving as
is observed in nature.
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Area in which unrelated
and distantly related
proteins have sequence
identity (15%-25%)
Sequence alignments
• Pairwise sequence alignments can be done by using a
dot matrix.
• One sequence is plotted horizontally and the other
vertically. Whenever there is an identical residue you
place a dot on the chart.
• Dot plot of a peptide against itself results in a square
matrix with a row of dots along the diagonal and
scattered dots for chance identities.
• If the peptides are conserved, there are only a few
absences along the diagonal.
• Distantly related peptides will have a number of gaps
along the diagonal.
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Figure 7-26
Sequence alignment with dot matrices.
Sequence alignments
• An alignment score (AS) is used to determine if there
is any relationship.
• 10 for every identity except Cys which scores 20
• Subtract 25 for every gap.
• The normalized alignment score (NAS) by dividing
the AS by the number of residues of the shortest of the
two polypeptides in the alignment and multiplying by
100.
• Example Human hemoglobin and myoglobin.
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Figure 7-27 The optical alignments of human myoglobin
(Mb, 153 residues) and the human hemoglobin a chain (Hba,
141 residues).
Hemoglobina is 141 and myoglobin is 153
AS = number of identities X 10 + 20 for Cys -number of gaps
= 37 identities X 10 + 20 (Cys) - (1gap X 25)= 365
NAS = AS/number of residues for shortest polypeptide
=365/141 = 259
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Figure 7-28 A guide to the significance of normalized
alignment scores (NAS) in the comparison of peptide
sequences.
Alignments are weighted according to
the likelihood of substitution
• Realistic way of assigning the probability of occurrence
(weight) for a substitution is to look at the physical similarity
of amino acids.
• Dayhoff measured a number of residue exchanges for
closely related proteins and determined their relative
frequency of the 20 X 19/2 = 190 different possible residue
changes.
• This number is divided by 2 to account for the fact that A 
B and B  A are equally likely.
• These data can be used to create a square matrix (20 X 20)
• The elements (20 properties per side) Mij, indicate the
probability that, in a related sequence, amino acid i will
replace amino acid j after an evolutionary interval (usually
one PAM unit).
• PAM-1 matrix.
PAM matrix
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Mutation probability can be determined for other evolutionary distances.
PAM-N matrix is made bt multiplying the matrix by itself N times ([M]N).
Relatedness odds matrix - Rij = Mij/fi
fi = probability that the amino acid i will occur in the second sequence by
chance.
Rij = probability that amino acid i will replace amino acid j or vice versa
every time i or j is encountered in the sequence.
When two polypeptides are compared with each other, the Rij values for
each position are multiplied to give the relatedness odds.
For example A-B-C-D-E-F and P-Q-R-S-T-U, relatedness odds = RAP X
RBQ X RCR X RDS X RET X RFU
Log odds substitution matrix - is made by taking the log of the
relatedness odds.
Log odds need to be maximized to get the best alignment.
Table 7-7
The PAM-250 Log Odds Substitution Matrix.
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All elements multiplied by 10.
Each diagonal element indicates
the mutability of the
corresponding amino acid.
Neutral score = 0.
Sequence alignment
• Make a matrix with the log odds values associated with
the amino acids at the appropriate positions.
• Example use a PAM-250 log odds matrix with a 10
peptide horizontal and 11 peptide vertical.
• The alignment of these two peptides must have at least
one gap assuming a significant alignment can be found.
• This is called a comparison matrix
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Figure 7-29a Use of the Needleman-Wunsch alignment
algorithm [alignment of 10-residue peptide (horizontal) with
11-residue peptide (vertical)]. (a) Comparison matrix.
Needleman-Wunsch algorithm
• Needleman and Wunsch constructed an algorithm to
find the best alignment between 2 polypeptides.
• Start at the lower right corner of the matrix (C-termini)
at position M and N (these correspond to the 10th and
11th amino acid residues) and add the value to the
position M-1, N-1 in the matrix.
• Add to each element of the matrix the largest number
from the row or column to the lower right of each
element proceeding right to left, bottom to top.
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Figure 7-29b Use of the Needleman-Wunsch alignment
algorithm [alignment of 10-residue peptide (horizontal) with
11-residue peptide (vertical)]. (b) Transforming the matrix.