2. Local Alignment (BLAST) and Statistics

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visit MIT OpenCourseWare at ocw.mit.edu. PROFESSOR: All right. So today, we're going
to briefly review classical sequencing and
next-gen or second-gen sequencing, which sort of
provides a lot of the data that the analytical methods
we'll be talking about work on. And we'll then introduce
local alignment a la BLAST and some of
the statistics associated with that. So just a few brief
items on topic one. All right. So today, we're going to
talk about sequencing first. Conventional– or Sanger
sequencing– then next-gen or second-gen
sequencing briefly. And then talk about
local alignments. So background for the sequencing
part, the Metzger review covers everything you'll need. And for the alignment– we'll
talk about local alignment today, global
alignment on Tuesday– then chapters four and five of
the text cover it pretty well.

So here's the text. If you haven't decided
whether to get it or not, I'll have it up here. You can come flip
through it after class. Sequencing is mostly
done at the level of DNA. Whether the original
material was RNA or not, usually convert to DNA and
sequence at the DNA level. So we'll often think about
DNA as sort of a string. But it's important to
remember that it actually has a three dimensional
structure as shown here. And often, it's helpful to
think of it in sort of a two dimensional
representation where you think about the bases and
their hydrogen bonding and so forth as
shown in the middle. My mouse is not working
today for some reason, but hopefully, we won't need it. So the chemistry of sequencing
is very closely related to the chemistry of
the individual bases.

And there are really
three main types that are going to
be relevant here. Ribonucleotides,
deoxyribonucleotides, then for Sanger sequencing,
dideoxyribonucleotides. So who can tell me which
of these structures corresponds to which
of those names? And also, please let
me know your name and I'll attempt
to remember some of your names toward the end
of the semester probably. So, which are which? Yes, what's your name? AUDIENCE: I'm Simona. So the ribonucleotide
is the top right.

The deoxy is the one below it. And the dideoxy is
the one to the left. PROFESSOR: OK, so
that is correct. So one way to keep
these things in mind is the numbering of the bases. So the carbons in the ribo
sugar are numbered one, so carbon 1 is the one
where the base is attached. Two is here, which has an OH
in RNA and just an H in DNA. And then three is
very important. Four, and then five. So five connects
to the phosphates, which then will connect the
base to the sugar phosphate backbone. And three is where you extend. That's where you're going to
add the next base in a growing chain. And so what will happen if you
give DNA polymerase a template and some dideoxy nucleotides? It won't be able to extend
because there's no 3-prime OH. And all the chemistry
requires the OH. And so that's the basis
of classical or Sanger sequencing, which Fred
Sanger got the Nobel Prize for in the 1980s– I think
it was developed in the '70s– and it's really
the basis of most of the sequencing, or
pretty much all the DNA sequencing up until
the early 2000s before some newer
technologies came about.

And it takes advantage
of this special property of dideoxy nucleotides that they
terminate the growing chain. So imagine we have
a template DNA. So this is the
molecule whose sequence we want to determine
shown there in black. We then have a primer. And notice the primer's
written in 5-prime to 3-prime direction. The ends would be
primer sequences and then primer complimentary
sequences in the template. So you typically will have
your template cloned– this is in conventional
sequencing– cloned into some vector
like a phage vector for sequencing so you know
the flanking sequences.

And then you do four
sequencing reactions in conventional
Sanger sequencing. And I know some of you have
probably had this before. So let's take the first
chemical reaction. The one here with a DDGTP. So what would you
put in that reaction? What are all the
components of that reaction if you wanted to do
conventional sequencing on, say, an acrylonitrile? Anyone? What do you need and
what does it accomplish? Yeah, what's your name? AUDIENCE: I'm Tim. PROFESSOR: Tim? Oh yeah, I know you, Tim. OK, go ahead. AUDIENCE: So you need
the four nucleotides– the deoxynucleotides. You will need the
dideoxy P nucleotides. In addition, you need all
the other [INAUDIBLE]. You need polymerase. Generally, you need a buffer
of some sort, [INAUDIBLE], to [INAUDIBLE]. PROFESSOR: Yeah,
primary template. Yeah. Great. That's good. It sounds like Tim could
actually do this experiment. And what ratio would you put in? So you said you're
going to put in all four conventional
deoxynucleotides and then one dideoxynucleotide. So let's say dideoxy G
just for simplicity here.

So in what ratio would you
put the dideoxynucleotide compared to the
conventional nucleotides? AUDIENCE: To lower
the concentration. PROFESSOR: Lower? Like how much lower? AUDIENCE: Like, a lot lower. PROFESSOR: Like maybe 1%? AUDIENCE: Yeah. PROFESSOR: Something like that. You want to put it a lot lower. And why is that so important? AUDIENCE: Because you want the
thing to be able to progress. Because you need enough of the
ribonucleotide concentration so that [INAUDIBLE]
every [INAUDIBLE] equivalent or excess and you're
going to terminate [INAUDIBLE]. PROFESSOR: Right. So if you put equamolar
deoxy G and dideoxy G, then it's going to be a
50% chance of terminating every time you hit
a C in the template. So you're going to
have half as much of the material at the second
G, and a quarter as much as the third, and you're going to
have vanishingly small amounts.

So you're only going
to be able to sequence the first few C's
in the template. Exactly. So that's a very good point. So now let's imagine you do
these four separate reactions. You typically would
have radiolabeled primer so you can see your DNA. And then you would run
it on some sort of gel. This is obviously not a real
gel, but an idealized version. And then in the lane
where you put dideoxy G, you would see the
smallest products. So you read these guys
from the bottom up. And in this lane there
is a very small product that's just one base longer
than the primer here.

And that's because there was a
C there and it terminated there. And then the next C appears
several bases later. So you have sort of a gap here. And so you can see that the
first base in the template would be a complement
of T, or C. And the second base
would be, you can see, the next smallest product
in this dideoxy T lane, therefore it would
be A. And you just sort of snake your
way up through the gel and read out the sequence. And this works well. So what does it actually
look like in practice? Here are some actual
sequencing gels.

So you run four lanes. And on big polyacrylamide
gels like this. Torbin, you ever
run one of these? AUDIENCE: Yes. PROFESSOR: Yes? They're a big pain to cast. Run for several hours, I think. And you get these
banding patterns. And what limits the
sequence read length? So we normally call
the sequence generated from one run of a
sequencer as a read. So that one attempt to sequence
the template is called a read. And you can see
it's relatively easy to read the sequence
toward the bottom, and then it gets
harder as you go up. And so that's really what
fundamentally limits the read length, is that the bands get
closer and closer together. So they'll run inversely
proportional to size with the small ones
running faster. But then the difference between
a 20 base product and a 21 might be significant. But the difference between a
500 base product and a 501 base product is going
to be very small. And so you basically can't
order the lanes anymore.

And therefore, that's sort of
what fundamentally limits it. All right. So here we had to run
four lanes of a gel. Can anyone think of
a more efficient way of doing Sanger sequencing? Is there any way to
do it in one lane? Yeah, what's your name? AUDIENCE: Adrian. You can use four different
types of the entities. Maybe like four
different colors. AUDIENCE: Four different colors. OK, so instead of using radio
labeling on the primary, you use fluorophore on your
dideoxy entities, for example. And then you can run them. Depending where that
strand terminated, it'll be a different color. And you can run them
all in one lane.

OK, so that looks like that. And so this was an
important development called terminator
sequencing in the '90s. That was the basis of the ABI
3700 machine, which was really the workhorse of
genome sequencing in the late '90s
and early 2000s. Really what enabled the
human genome to be sequenced. And so one of the other
innovations in this technology was that instead of having a
big gel, they shrunk the gel. And then they just had
a reader at the bottom. So the gel was shrunk to as thin
as these little capillaries. I don't know if you
can see these guys. But basically it's like
a little thread here.

And so each one of these
is effectively– oops! Oh no. No worries, this
is not valuable. Ancient technology that I
got for free from somebody. So the DNA would be
loaded at the top. There would be a
little gel in each of these– it's called
capillary sequencing. And then it would
run out the bottom and there would be
a detector which would detect the
four different flours and read out the sequence. So this basically condensed the
volume needed for sequencing. Any questions about
conventional sequencing? Yes? AUDIENCE: Where
are the [INAUDIBLE] where you'd put the
fluorescent flags? Like the topic from
the [INAUDIBLE]? PROFESSOR: Yeah,
that's a good question. I don't actually remember. I think there are different
options available. And sometimes with some
of these reactions, you need to use modified
polymerases that can tolerate these
modified nucleotides. Yeah, so I don't remember that. It's a good question. I can look that up.

So how long can a
conventional sequencer go? What's the read length? Anyone know? It's about, say, 600 or so. And so that's reasonably long. How long is a typical
mammalian mRNA? Maybe two, three kb? So you have in a typical
exon, maybe 150 bases or so. So you have a chunk. You don't generally
get full length cDNA. But you get a chunk
of a cDNA that's say, three, four exons in length. And that is actually
generally sufficient to uniquely identify the gene
locus that that read came from. And so that was the
basis of EST sequencing– so-called Expressed
Sequence Tag sequencing. And millions of these 600 base
chunks of cDNA were generated and they have been quite
useful over the years. All right. So what is next-gen sequencing? So in next-gen sequencing, you
only read one base at a time. So it's often a
little bit slower. But it's really
massively parallel. And that's the big advantage. And it's orders of
magnitude cheaper per base than
conventional sequencing. Like when it first
came out it, it was maybe two orders
of magnitude cheaper. And now it's probably another
four orders of magnitude.

So it really blows away
conventional sequencing if the output that
you care about is mostly proportional to
number of bases sequence. If the output is proportional
to the quality of the assembly or something, then
there are applications where conventional
sequencing still is very useful Because the
next-gen sequencing tends to be shorter. But in terms of just volume, it
generates much, much more bases in one reaction. And so the basic ideas are
that you have your template DNA molecules. Now typically, tens of thousands
for technologies like PacBio or hundreds of millions for
technologies like Illumina that are immobilized on
some sort of surface– typically a flow
cell– and there are either single
molecule methods where you have a single
molecule of your template or there are methods that
locally amplify your template and produce, say, hundreds
of identical copies in little clusters.

And then you use
modified nucleotides, often with
fluorophores attached, to interrogate the next base at
each of your template molecules for hundreds and hundreds
of millions of them. And so there are several
different technologies. We won't talk about all of them. We'll just talk
about two or three that are interesting
and widely used. And they differ depending
on the DNA template, what types of modified
nucleotides are used, and to some extent, in the
imaging and the image analysis, which differs for single
molecule methods, for example, compared to the ones
that sequence a cluster. So there's a table in
the Metzger review. And so I've just told you
that next-gen sequencing is so cheap. But then you see how
much these machines cost and you could buy lots of
other interesting things with that kind of money. And I also want to emphasize
that that's not even the full cost.

So if you were to buy an
Illumina GA2– this would be like a couple years
ago when the GA2 was the state of the art– for
half a million dollars, the reagents to run
that thing, if you're going to run it continuously
throughout the year, the reagents to run it
would be over a million. So this actually
underestimates the cost. However, the cost per
base is super, super low. Because they generate
so much data at once. All right, So we'll talk
about a couple of these. The first next-gen
sequencing technology to be published and
still used today was from 454–
now Roche– and it was based on what's
called emulsion PCR. So they have these little
beads, the little beads have adapter DNA molecules
covalently attached.

You incubate the beads
with DNA, and you actually make an emulsion. So it's an oil water emulsion. So each bead, which
is hydrophilic, is in the little bubble
of water inside oil. And the reason for that
is so that you do it at a template
concentration that's low enough that only a
single molecule of template is associated with each bead. So the oil then
provides a barrier so that the DNA can't get
transferred from one bead to another. So each bead will have a
unique template molecule. You do sort of a local
PCR-like reaction to amplify that DNA
molecule on the bead, and then you do sequencing
one base at a time using a luciferase based
method that I'll show you on the next slide. So Illumina technology differs
in that instead of an emulsion, you're doing it on the
surface of a flow cell. Again, you start with a
single molecule of template. Your flow cell has
these two types of adapters covalently attached.

The template anneals to
one of these adapters. You extend the adapter molecule
with dNTPs and polymerase. Now you have the complement of
your template, your denature. Now you have the
inverse complement of your template
molecule covalently attached to the cell surface. And then at the other end
there's the other adapter. And so what you
could do is what's called bridge amplification
where that now complement of the template molecule
will bridge over hybridized to the other
adapter, and then you can extend that adapter. And now you've regenerated
your original template. And so now you have the
complementary strand, and the original
strand, your denature. And then each of those molecules
can undergo subsequent rounds of bridge amplification to
make clusters of typically several hundred
thousand molecules. Is that clear? Question.

Yeah, what's your name? AUDIENCE: Stephanie. How do they get the adapters
onto the template molecules? PROFESSOR: How do
you get the adapters onto the template molecules? So that's typically
by DNA ligation. So we may cover
that in later steps. It depends. There's a few
different protocol. So for example, if you're
sequencing microRNAs, you typically would
isolate the small RNAs and use RNA litigation
to get the adapters on. And then you would do
an RT step to get DNA. With most other applications
like RNA-seq or genome sequencing– so with RNA-seq,
you're starting from mRNA, you typically will isolate
total RNA, do poly(A) selection, you fragment your RNA
to reduce the effects of secondary structure,
you random prime with, like, random hexamers RT enzyme.

So that'll make little bits
of cDNA 200 bases long. You use second strand synthesis. Now you have double
stranded cDNA fragments. And then you do, like, blunt end
ligation to add the adapters. And then you denature so
you have single strand. AUDIENCE: I guess my
question is how do you make sure that the two
ends sandwiching the DNA are different as opposed to– PROFESSOR: That the
two ends are different. Yeah, that's a good question. I'll post some stuff about–
It's a good question. I don't want to sweep
it under the rug.

But I kind of want to move on. And I'll post a
little bit about that. All right so we
did 454 Illumina. Helicos is sort of like
Illumina sequencing except single molecule. So you have your
template covalently attached to your substrate. You just anneal primer and
just start sequencing it And there's major pros and cons
of single molecule sequencing, which we can talk about.

And then the PacBio
technology is fundamentally different in that the
template is not actually covalently attached
to the surface. The DNA polymerase is covalently
attached to the surface and the template is sort of
threaded into the polymerase. And this is a phage polymerase
that's highly processive and strand displacing. And the template is often
a circular molecule. And so you can
actually read around the template multiple
times, which turns out to be really useful in PacBio
because the error rate is quite high for the sequencing. So in the top, in
the 454, you're measuring luciferase
activity– light.

In Illumina, you're
measuring fluorescence. Four different
fluorescent tags, sort of like the four different tags
we saw in Sanger sequencing. Helicose, it's single
tag one base at a time. And in PacBio, you actually
have a fluorescently labeled dNTP that has
the label on– it's actually
hexaphosphate– it's got the label on the
sixth phosphate. So the dNTP is labeled. It enters the active site
of the DNA polymerase. And the residence
time is much longer if the base is actually
going to get incorporated into that growing chain. And so you measure how much time
you have a fluorescent signal. And if it's long, that
means that that base must have incorporated
into the DNA.

But then, the extension
reaction itself will cleave off the
last five phosphates and the fluorophore tag. And so you'll
regenerate native DNA. So that's another difference. Whereas in Illumina
sequencing, as we'll see, there's this reversible
terminator chemistry. So the DNA is not native
that you're synthesizing. So this is just a
little bit more on 454. Just some pretty pictures. I think I described that before. The key chemistry here is that
you add one dNTP at a time. So only a subset of the wells–
perhaps a quarter of them– that have that next base,
the complementary base free– as the next
one after the primer– will undergo synthesis. And when they undergo synthesis,
you release pyrophosphate. And they have these
enzymes attached to these little micro beads–
the orange beads– sulfurylase and luciferase, that use
pyrophosphate to basically generate light. And so then you have one of
these beads in each well. You look at which wells
lit up when we added dCTP.

And they must have had G as
the next base and so forth. And there's no termination here. The only termination
is because you're only adding one base at a time. So if you have a single
gene in the template, you'll add one base. But if you have two Gs in the
template, you'll add two Cs. And in principle, you'll
get twice as much light. But then you have to sort of
do some analysis after the fact to say, OK how much
light do we have? And was that one G,
two G, and so forth. And the amount of
light is supposed to be linear up to
about five or six Gs.

But that's still a
more error-prone step. And the most common
type of error in 454 is actually insertions
and deletions. Whereas in Illumina
sequencing, it's substitutions. David actually encouraged
me to talk more about sequencing errors
and quality scores. And I need to do a little
bit more background. But I may add that a little
bit later in the semester. OK, so in Illumina
sequencing, you add all four dNTPs
at the same time. But they're non-native.

They have two major
modifications. So one is that they're
three prime blocked. That means that
the OH is not free, I'll show the chemical
structure in a moment. So you can't extend
more than one base. You incorporate that one
base, and the polymerase can't do anything more. And they're also tagged
with four different fluors. So you add all
four dNTPs at once. You let the polymerase
incorporate them. And then you image
the whole flow cell using two lasers
and two filters. So basically, to
image the four fluors. So you have to sort of take
four different pictures of each portion of the flow
cell and then the camera moves and you scan the whole cell.

And so then, those clusters that
incorporated a C, let's say, they will show up in the
green channel as spots. And those incorporated
in A, and so forth. So you basically have these
clusters, each of them represents a distinct
template, and you read one base at a time. So, first you read the
first base after the primer. So it's sequencing
downwards into the template. And you read the
first base so you know what the first base
of all your clusters is. And then you reverse
the termination. You cleave off
that chemical group that was blocking the 3-prime
OH so now it can extend again. And then you add the
four dNTPs again, do another round of extension,
and then image again, and so forth. And so it takes a little while. Each round of imaging
takes about an hour. So if you want to do 100 base
single and Illumina sequencing, it'll be running on the machine
for about four days or so. Plus the time you have to
build the clusters, which might be several hours
on the day before.

So what is this? So actually the whole idea
of blocking termination– basically Sanger's idea–
is carried over here in Illumina sequencing
with a little twist. And that's that you can
reverse the termination. So if you look down
here at the bottom, these are two different
3-prime terminators. Remember your base counting. Base one, two, three. So this was the
3-prime OH, now it's got this methyl [INAUDIBLE],
or whatever that is. I'm not much of a chemist,
so you can look that one up.

And then here's another version. And this is sort
of chemistry that can cleave this off
when you're done. And then this whole thing
here, hanging off the base, is the fluor. And you cleave that off as well. So you add this big complicated
thing, you image it, and then you cleave
off the fluor and cleave off
the 3-prime block. These are some actual
sequencing images you would image in
the four channels. They're actually
black and white. These are pseudocode. And then you can merge
those and you can see then all the clusters
on the flow cell. So this is from a GA2 with
the recommended cluster density back in the day,
like a few years ago. And nowadays, the image
now since the software has gotten a lot better,
so you can actually load the clusters more
densely and therefore get more sequence
out of the same area.

But imagine just
millions and millions of these little
clusters like this. Notice the clusters are
not all the same size. Basically, you're
doing PCR in situ, and so some molecules are easier
to amplify by PCR than others. And that probably accounts
for these variations in size. So what is the
current throughput? These data are accurate as
of about, maybe, last year. So the HiSeq 2000 instrument
is the most high performance, widely used instrument. Now there's a 2500, but I
think it's roughly similar. You have one flow cell. So a flow cell looks sort
of like a glass slide, except that it has
these tunnels carved in it like eight little
tubes inside the glass slide. And on the surfaces
of those tubes is where the adapters
are covalently attached. And so you have
eight lanes and so you can sequence eight different
things in those eight lanes. You could do yeast genome
in one and fly RNA-seq in another, and so forth. And these days, a single
lane will produce something like 200 million reads. And this is typically routine
to get 200 million reads from a lane.

Sometimes you can get more. You can do up to 100 bases. You can do 150 these
days on a MiSeq, which is a miniature version. You can do maybe 300 or more. And so that's a whole
lot of sequence. So that's 160 billion bases of
sequence from a single lane. And that will cost you– that
single lane– maybe $2,000 to $3,000, depending
where you're doing it. And the cost doesn't
include the capital cost, that's just the reagent
cost for running that. So 160 billion– the
human genome is 3 billion, so you've now sequenced the
human genome over many times there. You can do more. So you can do
paired-end sequencing, where you sequence both
ends of your template. And that'll basically double
the amount of sequence you get. And you can also,
on this machine, do two flow cells at once.

So you can actually
double it beyond that. And so for many applications,
160 billion bases is overkill. It's more than you need. Imagine you're doing
bacterial genome sequencing. Bacterial genome might
be five megabases or so. This is complete overkill. So you can do bar coding
where you add little six base tags to
different libraries, and then mix them together,
introduce them to the machine, sequence the tags
first or second, and then sequence the templates. And then you effectively
sort them out later. And then do many
samples in one lane. And that's what people
most commonly do. So, questions about
next-gen sequencing? There's a lot more to learn. I'm happy to talk about it more. It's very relevant
to this class. But I'm sure it'll come up
later in David's sections, so I don't want to take
too much time on it. So, now once you generate
reads from an Illumina instrument or some
other instrument, you'll want to align
them to the genome to determine, for
example, if you're doing RNA-seq mapping
reads that come from mRNA, you'll want to know what
genes they came from.

So you need to map those
reads back to the genome. What are some other reasons you
might want to align sequences? Just in general, why
is aligning sequences– meaning, matching them up
and finding individual bases or amino acid residues that
match– why is that useful? Diego? AUDIENCE: You can
assemble them if you want. PROFESSOR: You
can assemble them? Yes. So if you're doing
genome sequencing, if you align them to
each other and you find a whole stack that
sort of align this way, you can then assemble
and infer the existence of a longer sequence. That's a good point. Yes, your name? AUDIENCE: Julianne. Looking at homologs.

PROFESSOR: Looking at homologs. Right. So if you, for example, are
doing disease gene mapping, you've identified a human
gene of unknown function that's associated
with a disease. Then you might want to search
it against, say, the mouse database and find
a homolog in mouse and then that might be what you
would want to study further. You might want to then
knock it out in mouse or mutate it or something. So those are some good reasons. There's others. So we're going to first talk
about local alignment, which is a type of alignment
where you want to find shorter stretches
of high similarity. You don't require alignment
of the entire sequence. So there are certain
situations where you might want to do that.

So here's an example. You are studying a recently
discovered human non-coding RNA. As you can see, it's 45 bases. You want to see if
there's a mouse homolog. You run it through NCBI
BLAST, which as we said is sort of the Google search
engine of mathematics– and you're going get a chance
to do it on pump set one, and you get a hit
that looks like this. So notice, this is
sort of BLAST notation. It says Q at the top. Q is for "query," that's
the sequence you put in. S is "subject,"
that's the database you were searching against. You have coordinates,
so 1 to 45. And then, in the subject, it
happened to be base 403 to 447 in some mouse
chromosome or something. And you can see that
it's got some matching.

But it also has some mismatches. So in all, there are 40
matches and five mismatches in the alignment. So is that significant? Remember, the mouse genome
is 2.7 billion bases long. It's big. So would you get a match
this good by chance? So the question is really,
should you trust this? Is this something you
can confidently say, yes mouse is a
homolog, and that's it? Or should you just
be like, well, that's not better
than I get by chance so I have no
evidence of anything? Or is it sort of
somewhere in between? And how would you tell? Yeah, what's your name? AUDIENCE: Chris. You would want to
figure out a scoring function for the alignment. And then, with that
scoring function, you would find whether or not
you have a significant match.

PROFESSOR: OK. So Chris says you want to
define a scoring system and then use the
scoring system to define statistical significance. Do want to suggest
a scoring system? What's the simplest
one you can think of? AUDIENCE: Just if there's a
match, you add a certain score. If it's a mismatch, you
subtract a certain score. PROFESSOR: So let's do
that scoring system. So the notation that's
often used is Sii. So that would be a match
between nucleotide i and then another
copy of nucleotide i.

We'll call that 1,
plus 1 for a match. And sij, where i
and j are different, we'll give that
a negative score. Minus 1. So this is i not equal to j. So that's a scoring matrix. It's a four by four matrix
with 1 on the diagonal and minus 1 everywhere else. And this is commonly
used for DNA. And then there's a
few other variations on this that are also used. So good, a scoring system. So then, how are we going
to do the statistics? Any ideas? How do we know
what's significant? AUDIENCE: The higher
score would probably be a little more significant
than a lower score. But the scale, I'm not sure– PROFESSOR: The scale
is not so obvious. Yes, question? AUDIENCE: My name is Andrea.

So if you shuffled the RNA,
like permute the sequence, then we'll get the
[INAUDIBLE] genome you get with that
shuffled sequence. And the score is about
the same as you'd get with the non-shuffled
sequence [INAUDIBLE] about very significant scores. PROFESSOR: Yeah, so
that's a good idea. BLAST, as it turns
out– is pretty fast. So you could shuffle
your RNA molecule, randomly permute the
nucleotides many times, maybe even like
1,000 times, search each one against
the mouse genome, and get a distribution
of what's the best score– the top score– that
you get against a genome, look at that
distribution and say whether the score
of the actual one is significantly higher
than that distribution or just falls in the
middle of that somewhere. And that's reasonable. You can certainly do that, and
it's not a bad thing to do. But it turns out there is
an analytical theory here that you can use. And so that you can determine
significance more quickly without doing so
much computation. And that's what
we'll talk about. But another issue, before
we get to the statistics, is how do you actually
find that alignment? How do you find the top scoring
match in a mouse genome? So let's suppose
this guy is your RNA.

OK, of course, we're
using T's, but that's just because you usually
sequences it at the DNA level. But imagine this is your RNA. It's very short. This is like 10 or so, I think. And this is your database. But it goes on a
few billion more. Several more blackboards. And I want to come up
with an algorithm that will find the highest scoring
segment of this query sequence against this database. Any ideas? So this would be like
our first algorithm. And it's not terribly
hard, so that's why it's a good
one to start with. Not totally obvious either. Who can think of an algorithm
or something, some operation that we can do on
this sequence compared to this sequence–
in some way– that will help us find the
highest scoring match? I'm sorry. Yeah? AUDIENCE: You have to consider
insertion and deletion. PROFESSOR: Yeah, OK. So we're going to
keep it simple.

That's true, in general. But we're going
to keep it simple and just say no
insertions and deletions. So we're going to look for
an ungapped local alignment. So that's the
algorithm that I want. First, no gaps. And then we'll do
gaps on Tuesday. Tim? AUDIENCE: You could just
compare your [INAUDIBLE] to [INAUDIBLE] all
across the database and turn off all the
[INAUDIBLE] on that [INAUDIBLE], and then figure out [INAUDIBLE]. PROFESSOR: Yeah, OK. Pretty much. I mean, that's
pretty much right. Although it's not quite as
much of a description as you would need if you want
to actually code that.

Like, how would you
actually do that? So, I want a
description that is sort of more at the
level of pseudocode. Like, here's how you would
actually organize your code. So, let's say you
entertain the hypothesis that the alignment can be
in different registers. The alignment can correspond to
base one of the query and base one of the subject. Or it could be shifted. It could be an alignment
where base 1 of the query matches these two, and so forth.

So there's sort of
different registers. So let's just consider
one register first. The one where base 1 matches. So let's just look
at the matches between corresponding bases. I'm just going to make these
little angle bracket guys here. Hopefully I won't
make any mistakes. I'm going to take this. This is sort of implementing
Tim's idea here. And then I'm going to
look for each of these– so consider it going down here. Now we're sort of looking
at an alignment here. Is this a match or a mismatch? That's a mismatch.

That's a match. That's a mismatch. That's a mismatch. That's a match. Match Match. Mismatch. Mismatch. Mismatch. So where is the
top scoring match between the query
and the subject? Tim? Anyone? AUDIENCE: 6, 7, 8. PROFESSOR: 6, 7, 8. Good. Oh– AUDIENCE: 5, 6, 7. PROFESSOR: 5, 6, 7. Right. Right here. You can see there's
three in a row. Well, what about this? Why can't we add
this to the match? What's the reason why
it's not 2, 3, 4, 5, 6, 7? AUDIENCE: Because the
score for that is lower. PROFESSOR: Because the
score for that is lower.

Right. We defined top scoring segment. You sum up the scores
across the map. So you can have
mismatches in there, but this will have a score of 3. And if you wanted to
add these three bases, you would be adding
negative 2 and plus 1, so it would reduce your score. So that would be worse. Any ideas on how to do this in
an automatic, algorithmic way? Yeah? What's your name? AUDIENCE: Simon. So if you keep shifting the
entire database, [INAUDIBLE]. PROFESSOR: OK so you
keep shifting it over, and you generate
one of these lines. But imagine my query was
like 1,000 or something. And my database
is like a billion. How do I look along here? And here it was obvious what
the top scoring match is. But if I had two
matches here, then we would've actually had
a longer match here.

So in general, how do I
find that that top match? For each of those
registers, if you will, you'll have a thousand
long diagonal here with 1's and minus 1's on it. How do I process those scores
to find the top scoring segment? What's an algorithm to do that? It's kind of
intuitively obvious, but I want to do something
with, you define a variable and you update it, and you
add to it, and subtract.

Something like that. But, like a computer
could actually handle. Yeah? What was your name? Julianne? AUDIENCE: Could you keep track
of what the highest total score is, and then you keep
going down the diagonal, And then you update it? PROFESSOR: OK. You keep track of what the
highest total score was? AUDIENCE: Yeah. The highest test score. PROFESSOR: The
highest segment score? OK. I'm going to put this up here. And we'll define max s. That's the highest segment
score we've achieved to date. And we'll initialize
that to zero, let's say. Because if you had
all mismatches, zero would be the
correct answer. If your query was A's
and your subject was T's.

And then what do you do? AUDIENCE: As you go down the
diagonal, you keep track of– PROFESSOR: Keep track of what? AUDIENCE: So you
look at 1 in 1 first. And then you go 1 in 2, and
you find a score of zero. But that's higher
than negative 1. PROFESSOR: But the score of the
maximum segment at that point, after base 2, is not zero. It's actually 1. Because you could have a
segment of one base alignment. The cumulative score is zero. I think you're onto something
here that may be also something useful
to keep track of. Let's do the cumulative score
and then you tell me more. We'll define cumulative
score variable. We'll initialize that to zero. And then we'll have some for
loops that, as some of you have said, you want to
loop through the subject. All the possible
registers of the subject.

So that would be
maybe j equals 1 to subject length
minus query length. Something like that. Don't worry too much about this. Again, this is not
real code, obviously. It's pseudocode. So then this will be,
say, 1 to query language. And so this will be
going along our diagonal. And we're going to plot
the cumulative score.

So here you would you have an
update where cumulative score plus equals the score
of query position i matched against
subject position j. And update that. So that's just cumulative score. So what will it look like? So in this case, I'll
just use this down here. So you have zero, 1,
2, minus 1, minus 2. So you'll start at position
zero in the sequence. At position 1 you're
down here at minus 1 because it was a mismatch. Then at position 2, as you
said, we're back up to zero. And then what happens? Go down to minus
1, down to minus 2. Then we go up three times in a
row until we're up here to 1. And then we go down after that. So where is your highest scoring
match in this cumulative score plot? People said it was from 5 to 7.

Yeah, question? AUDIENCE: So would it be
from like a local minimum to a local maximum? PROFESSOR: Yeah. Exactly. So, what do you want
to keep track of? AUDIENCE: You want to keep track
of the minimum and the maximum. And look for the range which
you maximize to different– PROFESSOR: Yeah, so
this is now sort of more what I was looking
for in terms of– so this was the local minimum,
and that's the local maximum. This is the score. That's your mass s there. And you also want to keep
track of where that happened in both the query
and the subject. Does that make sense? So you would keep track of
this running cumulative score variable.

You keep track of
the last minimum. The minimum that
you've achieved so far. And so that would then
be down here to minus 2. And then when your cumulative
score got up to plus 1, you always take that
cumulative score, minus the last minimum
cumulative score. That gives you a
potential candidate for a high scoring segment. And if that is bigger than
your current max high scoring segment, then you update it
and you would update this. And then you would
also have variables that would store where you are.

And also, where did
that last minimum occur. So I'm not spelling it all out. I'm not going to give
you all the variables. But this is an algorithm that
would find the maximum score. Yeah, question? AUDIENCE: So you're
keeping track of the global maximum,
local minimum, so that you can accept the most
recent local minimum following the global maximum? PROFESSOR: I'm not
sure I got all that. But you're keeping track
of the cumulative score.

The minimum that that
cumulative score ever got to. And the maximum
difference, the maximum that you ever in the
past have gone up. Where you've had a
net increment upwards. Like here. So this variable
here, this max s, it would be initialized to zero. When you got to here, your last
minimum score would be minus 1. Your cumulative
score would be zero. You would take the difference
of those, and you'd be like, oh I've got a high scoring
segment of score one. So I'm going to update that. So now, that variable
is now 1 at this point. Then you're going down, so
you're not getting anything. You're just lowering this
minimum cumulative score down to minus 2 here. And then when you
get to here, now you check the cumulative score
minus the last minimum. It's 1. That's a tie. We won't keep track of ties. Now at here, that
difference is 2. So now we've got a new record.

So now we update this maximum
score to 2 in the locations. And then we get here, now
it's 3, and we update that. Does that make sense? AUDIENCE: Imagine the first
dip– instead of going down to negative 1, it went
down to negative 3. PROFESSOR: Negative 3? AUDIENCE: That first dip. PROFESSOR: Right here? So we started back a little bit. So back here, like this? AUDIENCE: No. PROFESSOR: Down to negative 3? AUDIENCE: No. PROFESSOR: But how do
we get to negative 3? Because our scoring is this way. You want this dip to minus 3? AUDIENCE: No PROFESSOR: This one minus 3? Imagine we are at minus 3 here? AUDIENCE: Yeah.

Imagine it dipped to minus 3. And then the next one dipped to
higher than that, to minus 2. And then it went up to 1. And so, would the difference
you look at be negative 2 to 1, or negative 3 to 1? PROFESSOR: Like that, right? So, minus 3, let's
say, minus 2, 1. Something like that. What do people think? Anyone want to–? AUDIENCE: Minus 3 to 1. PROFESSOR: Minus 3 to 1. It's the minimum
that you ever got to. This might be a stronger match,
but this is a higher scoring match. And we said we want
higher scoring. So you would count that. AUDIENCE: So you keep track
of both the global minimum and the global
maximum, and you take the difference between them.

PROFESSOR: You keep track
of the global minimum and the current
cumulative score, and you take the difference. AUDIENCE: The global maximum– PROFESSOR: It's not
necessarily global maximum because we could be
well below zero here. We could do like this. From here to here. So this is not the
global maximum. This just happens
to be, we went up a lot since our last minimum. So that's your high
scoring segment. Does that make sense? I haven't completely
spelled it out.

But I think you guys have given
enough ideas here that there;s sort of the core
of an algorithm. And I encourage you to think
this through afterwards and let me know if
there are questions. And we could add an
optional homework where I ask you to do
this, that we've sometimes had in the past. It is a useful thing to look at. This is not exactly how
the BLAST algorithm works. It uses some tricks
for faster speed. But this is sort of
morally equivalent to BLAST in the sense that it has the
same order of magnitude running time. So this algorithm– what is the
running time in Big-O notation? So just for those who
are non-CS people, when you use this Big-O notation,
then you're asking, how does the running
time increase in the size of the input? And so what is the input? So we have two inputs. We have a query of length. And let's say
subject of length n. So clearly, if those are bigger,
it'll take longer to run. But when you compare
different algorithms, you want to know how the run
time depends on those lengths.

Yes. What's your name? AUDIENCE: Sally. m times n. PROFESSOR: So with
this, this is what you would call an
order mn algorithm. And why is that? How can you see that? AUDIENCE: You have two for
loops And for each length, essentially, you're
going through everything in the query. And then, for everything that
you go through in the query, you would [INAUDIBLE].

PROFESSOR: Right. In this second for loop here,
you're going through the query. And you're doing that
nested inside of a for loop that's basically the
length of the subject. And eventually
you're going to have to compare every
base in the query to every base in the subject. There's no way around that. And that takes
some unit of time. And so the actual time will
be proportional to that. So the bigger n gets
and m gets, it's just proportional
to the product. Does that make sense? Or another way to
think about it is, you're clearly going to have to
do something on this diagonal. And then you're going
to have to do something on this diagonal, and
this one, and this one. And actually, you have to
also check these ones here. And in the end, the total
number of computations there is going to
be this times that. You're basically doing
a rectangle's worth of computations. Does that makes sense? So that's not bad, right? It could be worse. It could be, like, mn squared
or something like that.

So that's basically
why BLAST is fast. So what do these things
look like, in general? And what is the condition on
our score for this algorithm to work? What if I gave a score
of plus 1 for a match, and zero for a mismatch? Could we do this? Joe, you're shaking your head. AUDIENCE: It would
just be going up. PROFESSOR: Yeah. The problem is, it might
be flat for a while, but eventually it would go up. And it would just
go up and up and up. And so your highest
scoring segment would, most of the
time, be something that started very
near the beginning and ended very near the end. So that doesn't work. So you have to have
a net negative drift. And the way that's formalized
is the expected score has to be negative.

So why is the expected score
negative in this scoring system that has plus 1 for a match,
and minus 1 for a mismatch? Why does that work? AUDIENCE: It should be wrong
three quarters of the time. PROFESSOR: Yeah. You'll have a mismatch
three quarters of the time. So on average, you
tend to drift down. And then you have these
little excursions upwards, and those are your
high scoring segments. Any questions about that? AUDIENCE: Question. Is there something
better than m times n? PROFESSOR: We've got some
computer scientists here. David? Better than m times n? I don't think so,
because you have to do all those comparisons. And so there's no way around
that, so I don't think so.

All right. But the constant–
you can do better on the constant than
this algorithm thing. AUDIENCE: With
multiple queries– PROFESSOR: With
multiple queries, yeah. Then you can maybe
do some hashing or find some– to speed it up. OK, so what about the
statistics of this? So it turns out that Karlin and
Altschul developed some theory for just exactly this problem. For searching a query sequence. It can be nucleotide or protein
as long as you have integer scores and the average– or the
expected– score is negative, then this theory
tells you how often the highest score of all–
across the entire query database comparison–
exceeds a cut off x using a local alignment
algorithm such as BLAST.

And it turns out
that these scores follow what's called an extreme
value or Gumbel distribution. And it has this kind of
double exponential form here. So x is some cut off. So usually x would be the
score that you actually observed when you searched your
query against the database. That's the one you care about. And then you want
to know, what's the probability we would've
seen something higher than that? Or you might do x is one less
than the score you observed.

So what's the chance we
observed something the same, as good as this, or better? Does that make sense? And so this is going to
be your P value then. So the probability
of S. The score of the highest segment
under a model where you have a random query
against a random database of the same length is 1
minus e to the minus KMN e to the minus lambda x. Where M and N are the lengths
of the query and the database. x is the score. And then K and lambda are
two positive parameters that depend actually on the
details of your score matrix and the composition
of your sequences.

And it turns out that lambda
is really the one that matters. And you can see
that because lambda is up there in that
exponent multiplying x. So if you double
lambda, that'll have a big effect on the answer. And K, it turns
out, you can mostly ignore it for most purposes. So as a formula, what
does this thing look like? It looks like that. Kind of a funny shape. It sort of looks like
an umlauf a little bit, but then has a different shape
on the right than the left.

And how do you
calculate this lambda? So I said that lambda is
sort of the key to all this because of its uniquely
important place in that formula,
multiplying the score. So it turns out that lambda is
the unique positive solution to this equation here. So now it actually depends
on the scoring matrix. So you see there's sij there. It depends on the
composition of your query. That's the pi's. The composition of your
subject, that's the rj's. You sum over the i
and j equal to each of the four nucleotides. And that sum has to be 1. So there's a unique positive
solution to this equation.

So how would we solve
an equation like this? First of all, what
kind of equation is this, given that we're
going to set the sij, and we're going to just
measure the pi and the rj? So those are all
known constants, and lambda is what we're
trying to solve for here. So what kind of an
equation is this in lambda? Linear? Quadratic? Hyperbolic? Anybody know what this is? So this is called a
transcendental equation because you have
different powers. That sounds kind of unpleasant. You don't take a class in
transcendental equations probably. So in general, they're not
possible to solve analytically when they get complicated. But in simple cases, you
can solve them analytically. And in fact, let's just do one. So let's take the
simplest case, which would be that all the
pi's are a quarter.

All the ri's are a quarter. And we'll use the scoring system
that we came up with before, where sii is 1,
and sij is minus 1. If i does not equal j. And so when we plug those in to
that sum there, what do we get? We'll get four terms that are
one quarter, times one quarter, times e to the lambda. There's four possible
types of matches, right? They have probability one
quarter times a quarter. That's pi and rj. And the e to the lambda
sii is just e to the lambda because sii is 1. And then there's 12 terms that
are one quarter, one quarter, e to the minus lambda.

Because there's
the minus 1 score. And that has to equal 1. So cancel this, we'll
multiply through by 4, maybe. So now we get e to
the lambda plus 3. e to the minus lambda equals 1. It's still a
transcendental equation, but it's looking
a little simpler. Any ideas how to
solve this for lambda? Sally? AUDIENCE: Wouldn't the 1 be 4? PROFESSOR: I'm sorry. 4. Thank you. Yeah, what's your name? AUDIENCE: [INAUDIBLE] I think
[INAUDIBLE] quadratic equation. If you multiply both sides by
[INAUDIBLE] then [INAUDIBLE]. PROFESSOR: OK, so
the claim is this is basically a
quadratic equation. So you multiply both
sides by e to the lambda. So then you get e to
the 2 lambda plus 3. And then it's going to
move this over and do minus 4 e to the
lambda equals zero. Is that good? So how is it quadratic? What do you actually
do to solve this? AUDIENCE: Well, [INAUDIBLE]. PROFESSOR: Change the variable,
x equals e to the lambda. Then it's quadratic in x.

Solve for x. We all know how to solve
quadratic equations. And then substitute
that for lambda. OK, everyone got that? If you use 16 different
scores to represent all the different types
of matches and mismatches, this will be very unpleasant. It's not unsolvable,
it's just that you have to use computational
numerical methods to solve it. But in simple cases
where you just have a couple different
types of scores, it will often be a
quadratic equation. All right. So let's suppose that we have
a particular scoring system– particular pi's, rj's– and
we have a value of lambda that satisfies those. So we've solved this
quadratic equation for lambda. I think we get lambda
equals natural log 3, something like that. Remember, it's a unique
positive solution. Quadratic equations
are two solutions, but there's going to be
just one positive one. And then we have that value. It satisfies this equation.

So then, what if we
double the scores? Instead of plus 1 minus
1, we use plus 2 minus 2? What would then happen? You can see that the
original version of lambda wouldn't necessarily still
satisfy this equation. But if you think
about it a little bit, you can figure out what
new value of lambda would satisfy this equation. We've solved for the lambda
that solves with these scores. Now we're going to
have new scores. sii prime equals 2. sij prime equals minus 2. What is lambda prime? The lambda that goes
with these scores? Yeah, go ahead. AUDIENCE: Half of the original? PROFESSOR: Half of the original? Right. So you're saying that lambda
prime equals lambda over 2. And why is that? Can you explain? AUDIENCE: Because
of the [INAUDIBLE]. PROFESSOR: Yeah, if you think
about these terms in the sum, the s part is all doubling. So if you cut the lambda apart,
and the product will equal what it did before. And we haven't changed the pi's
and rj's, so all those terms will be the same. So therefore, it will still
satisfy that equation.

So that's another way
of thinking about it. Yes, you're correct. So if you double
the scores, lambda will be reduced
by a factor of 2. So what does that
tell us about lambda? What is it? What is its meaning? Yeah, go ahead, Jeff. AUDIENCE: Scale of
the distribution to the expectant score? Or the range score? PROFESSOR: Yeah. It basically scales the scores. So we can have the
same equation here used with arbitrary scoring.

It just scales it. You can see the way it appears
as a multiplicative factor in front of the score. So if you double all
the scores, will that change what the highest
scoring segment is? No, it won't change
it because you'll have this cumulative thing. It just changes how
you label the y-axis. It'll make it bigger, but it
won't change what that is. And if you look
at this equation, it won't change the
statistical significance. The x will double in value,
because all the matches are now worth twice as much as
what they were before.

But lambda will be half
as big, and so the product will be the same and therefore,
the final probability will be the same. So it's just a scaling factor
for using different scoring systems. Everyone got that? All right. So what scoring matrix
should we use for DNA? How about this one? So this is now a
slight generalization. So we're going to keep
1 for the matches. You don't lose any generality
by choosing 1 here for matches, because if you use 2,
then lambda is just going to be reduced
to compensate. So 1 for matches. And then we're going to
use m for mismatches.

And m must be negative in
order to satisfy this condition for this theory to work,
that the average score has to be negative. Clearly, you have to have
some negative scores. And the question then
is, should we use minus 1 like we used before? Or should we use like minus 2
or minus 5, or something else? Any thoughts on this? Or does it matter? Maybe it doesn't matter. Yeah, what's your name? AUDIENCE: [INAUDIBLE].

Would it make sense to
not use [INAUDIBLE], because [INAUDIBLE]. PROFESSOR: Yeah, OK. So you want to use a more
complicated scoring system. What particular
mismatches would you want to penalize more and less? AUDIENCE: [INAUDIBLE]
I think [INAUDIBLE] needs to be [INAUDIBLE]. PROFESSOR: Yeah, you are
correct in your intuition. Maybe one of the
biologists wants to offer a suggestion here. Yeah, go ahead. AUDIENCE: So it's a mismatch
between purine and pyrimidine [INAUDIBLE]. PROFESSOR: OK so now we've
got purines and pyrimidines.

So everyone
remember, the purines are A and G. The
pyrimidines are C and T. And the idea is that
this should be penalized, or this should be penalized
less than changing a purine to a pyrimidine. And why does that makes sense? AUDIENCE: Well,
structurally they're– PROFESSOR: Structurally, purines
are more similar to each other than they are to pyrimidines. And? More importantly, I think. In evolution? AUDIENCE: [INAUDIBLE]. PROFESSOR: I'm sorry,
can you speak up? AUDIENCE: C to C mutations
happen spontaneously in [INAUDIBLE] chemistry. PROFESSOR: Yes. So C to C mutations
happen spontaneously. So basically, it's
easier because they look more similar structurally. The DNA polymerase is more
likely to make a mistake and substitute another purine. The rate of purine,
purine or pyrimidine, pyrimidine to
transversions which switch the type is
about three to one, or two to one in
different systems. So yeah, that's a good idea. But for simplicity, just
to keep the math simple, we're just going to go
with one mismatch penalty. But that is a good point. In practice, you
might want to do that. So now, I'm saying
I'm going to limit you to one mismatch penalty.

But I'm going to let you
choose any value you want. So what value should you choose? Or does it matter? Or maybe different applications? Tim, yeah? AUDIENCE: I've just
got a question. Does it depend on pi and ri? For example, we could
use all these numbers. But if the overall
wants to be negative, then you couldn't
use negative .1. PROFESSOR: Right,
that's a good point. You can't make it too weak. It may depend on what your
expected fraction of matches is, which actually
depends on pi and ri.

So if you have very biased
sequences, like very AT rich, your expected fraction of
matches is actually higher. When you're researching
an AT rich sequence against another
AT rich sequence, it's actually higher
than a quarter. So even minus one might
not be sufficient there. You might need to go
down more negative. So you may need to use a
higher negative value just to make sure that the
expected value is negative. That's true. And yeah, you may want to adjust
it based on the composition. So let's just do a bit more. So it turns out that the
Karlin and Altschul theory, in addition to telling you what
the p value is of your match– the statistical significance–
it also tells you what the matches will
look like in terms of what fraction of
identity they will have.

And this is the so-called
target frequency equation. The theory says that if
I search a query with one particular composition, p,
subject meta-composition r– here, I've just assumed
they're the same, both p just for simplicity– with a
scoring matrix sij, which has a corresponding of lambda. Then, when I take those
very high scoring matches– the ones that are
statistically significant– and I look at those
alignments of those matches, I will get values qij,
given by this formula. So look at the formula. So it's qij. So pipj e to the lambda sij.

So it's basically
the expected chance that you would have base
i matching based j just by chance. That's pipj. But then weighted by
e to the lambda sij. So we notice for a match,
s will be positive, so e to the lambda
will be positive. So that will be bigger than 1. And you'll have more
matches and you'll have correspondingly
less mismatches because the mismatch
has a negative. So get the target value score. And that also tells you that
the so-called natural scores are actually determined
by the fraction of matches that you want in your
high scoring segments. If we want 90% matches,
we just set qii to be 0.9, and use
this equation here. Solve for sij. For example, if you want to
find regions with R% identities. Little r is just the
r as a proportion. qii is going to be r over 4. This assumes unbiased
base composition. A quarter of the
matches are acgt. Qij, then, is 1 minus r over 12. 1 minus r is a fraction
of non-matching positions. They're 12 different types. Set sii equal to 1, that's
what we said we normally do.

And then you do a little
bit of algebra here. m is sij. And you sort of plug in
this equation twice here. And you get this equation. So it says that m equals
log of 4 1 minus r over 3 over log 4 r. And for this to be
true, this assumes that both the query
and the database have uniform composition
of a quarter, and that r is between
a quarter and 1. The proportion of matches in
your high scoring segment– you want it to be
bigger than a quarter. A quarter is what you
would see by chance. There's something wrong with
your scoring system if you're considering those
to be significant. So it's something above 25%. And so it's just
simple algebra– you can check my work at
home– to solve for m here. And then this equation
then tells you that if I want to find
75% identical matches in a nucleotide
search, I should use a mismatch penalty of minus 1. And if I want 99%
identical matches, I should use a
penalty of minus 3. Not minus 5, but minus 3. And I want you to think
about, does that make sense? Does that not make sense? Because I'm going to ask you
at the beginning of class on Tuesday to
explain and comment on this particular phenomenon
of how when you want higher percent identities, you want a
more negative mismatch score.

Any last questions? Comments? .

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