Lecture 20: Protein Chains

Lecture 20: Protein Chains


PROFESSOR: Today
we’re going to talk about protein folding and its
relation to linkage folding. We’re going to look at a
mechanical model of proteins. This is an example of a
protein from lecture one. There’s a ton out
there in this place called the protein data
bank, all freely available. It’s really hard to
get pictures like this. But you get some
idea that there’s a linkage embedded in here. You see various little
spheres and edges. That’s, of course, not reality. Those spheres are
actually atoms and they’re kind of amorphous blobs. The edges are chemical bonds. And those are connections. We don’t know whether they’re–
it’s not really matter, but it’s force. This is a rather messy picture. This is what a protein
folds into, some 3D shape. Most proteins fold
consistently into one shape. We don’t really know
how that happens. We can’t watch it happen. So the big challenge is
to know how proteins fold. Given a protein, what
does it fold into? That’s the protein
folding problem. Major unsolved problem
in biology, biochemistry. The protein design problem is
I want to make a particular 3D shape so that it docks into
something, binds to a virus, whatever, what protein should
I synthesize in order for it to fold into that shape? That is potentially an easier
question algorithmically and it’s the really useful one
from a drug design standpoint. Some new virus comes
along, you design a drug to attack it and only
it, you build it. Usually you would manufacture
some synthetic DNA, you feed it into the
cell, DNA goes to the RNA, goes to the mRNA,
goes to the protein. You all remember
biology 101, hopefully. We don’t need to
know much about it. If you look at what’s called
the backbone of the protein, a protein’s basically a chain
and attached to the chain are various amino acids. Today I’m going to ignore
the amino acids, which is a little crazy, and just
think about the backbone chain. Backbone chain looks
something like this. One of the challenges of
video recording a class is I can only use copyright free
or Creative Commons images. This one, I couldn’t get
one, so I had to draw it. There’s various
measurements here in certain numbers of angstroms. Those are the chemical bonds. Various atoms here– nitrogen,
carbon, and so on, hydrogen. But basically, it’s a chain
that zigzags back and forth. You can also see
the angles here. They’re not quite all the
same, but they’re very similar. All the lengths in
the angles are close. It zigzags– this is
really in three dimensions. I tried to draw the spheres
so you could see the three dimensionality, but
it’s a little tricky. And then attached on the
sides are the amino acids. I’m going to focus
just on the backbone. The way this thing is allowed
to fold– these lengths, as far as we know,
are pretty static. They probably would
jiggle a little bit. But you can think
of them as edges. So you can think of
this as a linkage. The catch is, also,
the angles are fixed because the way this atom
wants to bind to other things has very fixed angle patterns. If you ever played with a
chemistry construction set, that’s how they work. They have holes at
just particular angles. So if you think of
like a robotic arm, normally– like here, I have
it two edge robotic arm, let’s say. Normally, you have two degrees
of freedom in three dimensions. You can change the angle and
you can spin around this edge. Now, it’s saying the angle
is fixed– for example, here it’s, say, at 90 degrees. All I can do is spin. I’m not allowed to flex
my muscle in this way. So that is the model. All of– in this case, we have
a tree– all of the angles here are fixed. But you can still, for example,
take this entire sub chain and spin it around this edge. That’ll preserve all the
angles and all the lengths. That’s all you’re allowed to do. You take an edge,
you spin it– spin one half of the edge
relative to the other half. These are called
fixed angle linkages. And they have been studied
quite a lot because of their connection
to protein folding. So embedded in the
term linkage, we assume that the edge
lengths are fixed, and then we add the constraint
that the angles are fixed. And the motivation
is the backbone is something like– the
backbone of a protein is something like
a fixed angle tree. Of course, it’s
not much of a tree. Most of it is a chain. There’s just small
objects hanging off, and if you add the
amino acid there are bigger things hanging
off, but still constant size. They’ll have some cycles. They’re not trees. But it’s slightly
more approximately– I should draw wavier lines. It’s a chain. Usually an open chain although
occasionally a closed chain. So we think a lot about fixed
angle chain and sometimes about fixed angle trees. Now, fixed angle linkages
are harder to think about than universal
joints– that’s the usual kind of linkage. So we know 3D linkages
are kind of tough. Nonetheless, we found
lots of really interesting mathematical problems
to solve here, and that is the topic of today. At some level, we are
thinking about the mechanics of protein folding. We’re throwing away energy. We’re throwing
away the actuators in real life that
make proteins fold. We’re just imagining,
given this mechanical model of how a protein might
fold, what’s possible. So in some sense, it’s
broader than reality. And the hope is you find
an interesting algorithm for how to fold
these protein chains. And maybe that’s the algorithm
that nature is implementing. That’s the kind of
general picture. We’re not constrained
by reality, and by how nature
actually folds things. So I’m going to talk today
about four main problems here. The first one’s called span. Second one’s called flattening. Third one is flat
state connectivity. And the fourth one is locked,
our good friend locked chains. And of course there
are locked chains because we’re constraining
linkages even more than before. So you could take knitting
needles, it’ll still be locked. So you add extra constraints. Makes it harder to fold. But there are actually some
interesting positive results we can give of chains that
are not locked in some sense. And flat state connectivity is
about the same kind of thing, where instead of worrying
about getting from anywhere to anywhere, we just
worry about getting from one flat state
to another flat state. Flat means lying in a plane. Flattening is about is
there such a configuration. And span is about
given robotic arm– like a more complicated one,
like with multiple edges– how far apart can the endpoints
get, and how close can the endpoints get. The universal chain
is not very exciting. Farthest it can get is when
it’s straight, and the least far it can get is when it’s closed. You can always do that, I think. Well, no, I guess you
can’t always close it up. That’s a little nontrivial. But for fixed
angle linkages, you can’t straighten out because
you have to preserve the angles. So it’s kind of what is the
straightest like configuration, given that the angles are fixed. So let’s start with span. So the span of a
configuration is the distance between the endpoints. And in general, you’ll find
the max span and the min span. This search was
begun by a guy named Mike Soss, who was a
PhD student at McGill. And he proved that
if you want to find, for example, a flat
state that lives in two dimensions with the
minimum or the maximum span, this is NP-hard. This is in his PhD thesis. Question? AUDIENCE: If you have a
linkage or [INAUDIBLE] chain that actually loops
around, is there a span because there is no endpoint? PROFESSOR: Oh, here I’m
assuming open chain– I should say that–
which most proteins are. I’ve been talking
about trees and stuff. Here I mean chain, otherwise
there aren’t two end points to think about. Good. So here are his
NP-hardness proofs. [INAUDIBLE] the problems
are NP-complete. They’re pretty simple. The problem here we’re
reducing from is partition. I give you a bunch of integers. I want to divide them into
two halves of equal sum. And the top example is
minimum flat span problem. So you make an orthogonal chain
where the horizontal edges are long and they’re proportional
to the integers you’re given, the vertical edges
are really tiny. And so what you’d like to do–
all you can do is sort of flip because you have to
stay in the plane, you can flip one of
the vertical edges, say, and make any of these
edges go left or right. You get that freedom. So each integer,
you get to choose. Do I go right by that amount and
or do I go left by that amount? And if the amount you go left
is equal to the amount you go right– in other
words, is it partitioned into two equal sums– then
those endpoints will be aligned, and then their distance
will be very tiny. Otherwise, it will
be quite large because the horizontal
distances are all big. So it’s kind of a very
easy NP-hardness proof. To maximize your
flat span, instead of mapping your
integers on to lengths, you map them on to
angles– return angles. I won’t specify
that too precisely. But again, if you make
your total counterclockwise turn equal to your
total clockwise turn, then the two end edges,
which are super, super long, will be parallel. And to maximize the distance
between the endpoints, you want them to be parallel. If you make them go some
other angle, they’re closer. Now, both of these proofs
rely on the requirement that you want a
flat configuration with minimum or maximum span. Now, there’s a claim
that flat configurations matter for proteins, so
it’s a natural constraint. But what about the
general problem? What about, I have something
in three dimensions, I want to maximize– I have
a fixed angle chain in 3D, maximize or minimize the span? Both of those problems are open. Can you solve them
in polynomial time? For 3D max span, so the non flat
version just for maximization, there’s been a lot of work. And there are two
papers on the subject. One of them is by
Nadia and Joe O’Rourke. Another one is by
Borcea and Streinu. And I just want to
quickly summarize that because there’s
a lot of stuff there. But essentially, they find what
the structure of those spans look like. I have an early figure
that’s in our book before all this work was done. The simple chain, this black guy
at 1, 2 3, 4 bars open chain, and in that black three
dimensional state, it maximizes the span,
the green span there. And if you look
from above, which is this picture– of course,
the end points look much closer in projection– and the red
configuration is the max span if you restrict to
flat configurations. So here, of course,
3D buys you something. In general, it always will. An interesting thing
is that this max span– the green line–
passes through another vertex. It seems kind of weird. And in fact, there’s a
general theorem there sort of characterizing the
structure of these chains. It’s still not
known whether we can solve this problem
in polynomial time. But for orthogonal chains, where
all the angles are 90 degrees, we can solve that in
linear time, I guess. And here’s what it looks like. Suppose you have some
orthogonal chain. Orthogonal chains are nice
because you can draw them in the plane as a staircase. So there’s a nice
canonical configuration. One way to think about
how to find the max span configuration– I’m just
going to give a high level overview here, this won’t
be a complete algorithm– is you triangulated
that staircase in this sort of obvious
way of connecting every endpoint to
the one, two ahead. And think about this as like a
body that’s hinging around here because I can spin– if I spin
the left part of this chain around this edge, it’s
like hinging that triangle around that hinge. Same thing. You could think
of these triangles as just being hinged together,
like in rigid origami. It’s the same class of motions. And now you can–
what I’m going to do is compute a shortest path in
this surface from here to here. Confusingly, this is called
a geodesic shortest path although it’s not really
related to geodesics from polyhedral surfaces. But if I compute
a shortest path, it’s going to go
like to this vertex and then probably
to that vertex. But I’m constrained to
stay inside the union of those triangles. I want to go from one
endpoint to another. Then I claim that–
OK, these two edges will stay planar, of course
they form a triangle– I claim these four edges
will stay planar, and in the orthogonal
case they’ll stay zigzag. And then also these two guys
will stay in their own plane. And then I claim that actually
this wiggly line, which is not straight
because it bends here and it bends here, the total
length of that wiggly line is the max span. And you achieve that by folding
this planar part with respect to this planar part with
respect to this planar part so that the wiggly lines
become aligned and straight. And that’s very hard to draw. But it can be done,
and that’s what you do in the orthogonal
case and that gives you the answer in linear
time with enough work. For non orthogonal
though, it’s open whether you can do this
in polynomial time. Maybe it’s NP-hard, actually. I don’t know. That’s all I want
to say about span. Next, we go to flattening. The first question
about flattening, and the main one
we’ll talk about here until we get to flat
state connectivity, is does a fixed angle chain
have a flat state at all? Can you even draw it in
the plane without crossing? So we’re restricted here to
have no self intersections. We want flat state,
no self intersection. Then there would be a question
of given some configuration, can I actually continuously
get to a flat state? But the simplest question
is ignore getting there. Just, is there a flat state? And this problem is NP-hard. Again, Mike Soss and his
advisor Godfried Toussaint. It’s a little more
complicated, but it’s basically the same idea as that very
simple proof, which was just to map integers to a little
zigzag staircase here. So the goal is to
force x to end up being– the two endpoints
of the green curve– to be aligned with each other. That will exist if
and only if there is a partition of
given integers. And there’s all this
infrastructure that’s sort of– there’s little
lock here and a key, and some structure on the left. Basically forces the
picture to look like that. So the first claim is
that the black stuff is basically unique. I think there’s one
global reflection you can do that doesn’t
affect anything. But you try any of
the other flips. Again, we’re restricted
to flat states here. So there’s only sort of a
bounded number of things you can do, a finite number
of things you can do. You try all of them,
they self intersect. So the black thing
is basically forced, and it forces the
endpoint– this endpoint x– from the black
side to be aligned with this very narrow spike. And because the
angles are preserved, that red guy’s going
to be vertical. It can’t go down
so it must go up. And so only if this thing is
aligned in the center, aligned with x– in other
words, this problem has a partition– will
this have a flat state. So it’s not the most
exciting example. This is only a weak
NP-hardness proof. Lots of interesting
questions still open here, like if all the
links are the same, if they’re all equal, then we don’t know. Or if all the links are
even polynomially bounded, this needs really, really long
lengths verses really, really tiny links exponentially–
exponential in ratio. All these problems are open. And that’s flattening. So we’re going very quickly
because there isn’t– well, partly because I’m more
excited about this– but there’s more work
in these two parts. So I’m going to focus on that. Next topic is flat
state connectivity. So the idea is to think
about the configuration space of these fixed
angle chains, let’s say. And we kind of know
that it’s going to be disconnected because
there are knitting needles, there are nasty things. So there’s maybe various
connective components. But let’s say that we really
care about flat states. And the question is, are
they connected to each other? So in other words, do
all the flat states– mark them with x’s–
do they all appear? There’s only finitely many. So configurations,
there’s this continuum that there are these messy
blobs, semi algebraic sets. The flat states, those
are discrete things. Because we have
fixed angles, you can flip or not flip every edge. So [INAUDIBLE] most
exponentially many of them, so finite. Are they all in one component? So I can get– if I pick two
of my favorite flat states, there’s a path between them? Or are some of them in
multiple components? So in this case, we call
it flat state disconnected. And if they’re all like this,
we call it flat state connected. And we’d just like to
know which chains, which fixed angle trees, whatever,
are flat state connected versus flat state disconnected. I would say, the
big open problem here is every fixed angle
chain, open chain flat state connected? That is still open. We have lots of results
in that direction. So the top four results
are about open chains, but they have an
extra constraint. For example, open chains that
have a monotone configuration, like the staircase. Those are flat state connected. In fact, whenever the
angles between the edges are either orthogonal
or obtuse, then they’re flat state connected. When the angles are acute,
we’re not really sure. If all the angles are equal
and acute, then we can do it. But if they’re different
and acute, we don’t know. Unless the edges
are all unit length and the angles are in this
funny range, then we can do it. So there’s all these
special cases we can solve. The most relevant to proteins
is actually obtuse chains, so we’ve solved sort
of the main problem with this second result. But there’s a natural
theoretical question here is, are all open
chains flat state connected or do we get disconnectivity? I will show you that– I’ll
show the orthogonal case in a little bit. We can do some stuff if you
have multiple chains that are attached to some
blob like a cell. Closed chains is a little
bit– for disconnected, we don’t have very interesting
examples, I would say. This is funny because
locked examples are easy to come by but flat
state disconnected examples are little trickier because
flat is so constrained. So let me just show
you these examples. This is what we call a partially
rigid fixed angle tree. So not only are
the angles fixed, but also the black
edges are not– in fact, only the blue edges here
are allowed to spin. Everything else is held rigid. So these arms are somehow forced
to be in exactly that geometry. I can spin it
around this edge, so spin it up into 3D, for example. These are two different flat
states of the same linkage. The only difference between
these two– I haven’t rotated or anything– is that I’ve
taken each of these arms and flipped it
around a blue axis. If I do all four of them,
I would get this picture. But the claim is, you cannot do
that without self intersection. The intuition is, when
there aren’t very– oh, one other thing that makes
it slightly more interesting. It’s weird to say, well why
did you force some of the edges to be rigid and not others? One way to force that is
to use a general graph. If you add some extra
edges to sort of brace this and all these angles are
fixed, then this linkage will behave exactly
like that one. So that at least is
somewhat more natural, although what we really care
about are chains, maybe trees. But we don’t know
whether there’s a– we also don’t know
whether all fixed angle trees are flat state connected. These are the worst
examples we know. Let me give you an idea
of why it doesn’t work. This is a little animation
of just a couple of moves attempted, and it’s just
going cycle through that. And these are some static images
of the same kind of thing. So the intuition is the
following– you have four arms. You have two sides to the
plane, there’s up and down. The four arms and two
sides, at least two of them are going to have to
go to the same side. The best you can do is two
and two, or three and one. But in either case,
you have two sides go to the– two arms
that go on the same side. Now, it could be,
like in this image, that they’re opposite arms. So there’s this arm here
and there’s this arm here. So they’re connected
by a 180 degree angle. And those guys,
when they fold up, actually these edges will
just hit each other dead on. So that’s kind of obvious
from a geometric standpoint. Maybe you call it cheating
for them to hit dead on. You can twiddle the edge lengths
so that they will properly intersect without
dead on collision, without being
degenerate basically. The alternative
is that– and this is a little harder
to see geometrically, and that’s why we drew that
animation– is that you have one arm and you have an adjacent
arm connected by a 90 degree angle. Now here, there’s clearly
some collision going on. And if you happen to fold
it up 90 degrees like that and then fold the other guy,
obviously you get stuck. But maybe you fold it a
little bit and the other guy goes a little bit
more and there could be some dance between those two
degrees of freedom, those two arms, that somehow gets
them both to pass over to the other side. It’s obviously not possible. How do you prove it? Well, you could prove it with
topology– knot theory or link theory. So it’s a very cute proof. You start with– so
here’s the full example, but I’ve highlighted
the two arms in red that are going to move. And I imagine
connecting the endpoints of each arm with these
little blue ropes underneath the plane. They’re both going
on the same side. Let’s say they somehow
pass through each other on the top side. Then I’m free to connect
stuff on the bottom, and I shouldn’t
collide with that. So if somehow,
both of these guys flip over– so arm on
the left, A3 flips over. A3 stays where it
is but now the arm is on the top, the north side
instead of the south side. And the other guy, from B
to B3, used to go like this and now it goes like this. If that happens somehow,
then these ropes could remain intact
during that whole motion. On the top, you have two closed
loops that are not interlocked. On the bottom, you
have two closed loops that are interlocked. So there’s no way to
get from there to there without colliding somewhere. The blue stuff didn’t
move, so the red stuff must have collided. So even just topologically,
you are screwed. That is their only
negative example. Lots of interesting
open questions here. On the positive
side, let me show you for orthogonal chains–
and the same algorithm works for obtuse
chains, all the angles are obtuse– how they
are flat state connected. So in order to show it’s
flat state connected, I want to think
about two flat states and show that I
can fold from one to the other via some
intermediate 3D stuff. Let’s start with one
of the flat states. So it’s orthogonal. So in two dimensions,
all the edges will be horizontal or vertical. In 3D, they can kind of be in
many, many different angles, many different
dihedral triangles. In 2D, it’s pretty simple. And all I need to do is
sort of pick up that chain, and I’m going to try to
pick it up into a staircase because there’s
only one staircase. If I can make it a
staircase, I make flat configuration A a
staircase, flat configuration B a staircase, and just
FedEx in the middle. Once they’re both
staircases, I play one motion and the other one backwards,
get from anywhere to anywhere. So here’s all you do you. You take the first
edge and you just rotate it up to the red line A.
And then you take the next edge and you take both
of those edges, and you just rotate
them like this, so you get that little
2-step staircase. Now I’d really like
to pick up this edge, but I want to
first get these two edges in a plane with that edge. So I rotate this flag over to
the left, I get those two guys. And now they’re in a plane with
this, and I just lift that up. Then I’m going to
flip, then rotate up. Flip, rotate, flip, rotate. Here’s some more examples. So if at this point, I have
this staircase– sorry, I guess originally I have
from V3 to D up there. it’s not in plane with this guy,
so I just rotate it like that. I’m spinning around this edge. So now I have from
B3 to #, and then I rotate it up along
that green arc. And I get a bigger
staircase above the chain and because everything’s
staying above, it will never
penetrate the plane and will never hit anybody else. And I’m building a
staircase by design. I always rotate this–
there’s actually two ways I could be in
plane– but I always rotate it so that when
I pick an edge up, it’ll be in a staircase. So this is actually really easy. And slight generalization
is to obtuse chains, then instead of making a
staircase, we make a monotone. Let me get this right. Yeah, sum z monotone state. So it goes monotone and z,
out of the plane, that’s enough to avoid
collision, and you get a canonical configuration. Also, if you have acute angles
but all the angles are equal, then there’s a
natural conical state, which is just like a
compressed staircase. And that will work here, too. That takes more effort. That was in a separate paper. But big open question is,
chains with arbitrary angles. We have no idea. It is very hard to do
an operation like this. Wow, we are burning
through this. This is fun. So the next topic is
about locked chains. Now as I said, you can
take a knitting needles example, which has five edges. And that will still
be locked if you force the angles to be
fixed because it was locked without the angles being fixed. Now, it required a length
ratio of 3:1, I think. This edge had to be longer
than the sum of those three. So let me put down
some open problems. So you may recall in the
case of universal chains– universal joints, I should
say– the big open question was, can you lock a
universal joint 3D chain with unit edge lengths? So, equilateral– every
edge is the same length. Is there a locked chain
like the knitting needles when all the edge
lengths are the same. And one of the motivations
for that is in proteins, the edge lengths are all
within like 50% of each other. So it’s pretty
natural, of course. We don’t have universal
joints with proteins. We have fixed angle joints. So the big open problem
for fixed angle joints– I guess we’ll do this in parts–
is there a locked 3D fixed angle chain that’s equilateral? I’m going to add
some conditions here. So that’s the first
natural question. Knitting needles
doesn’t suffice. We need a 3:1 length
ratio, as far as we know. Turns out that question’s
not very interesting. I need to do slightly
nonlinear editing here. So you take your
knitting needles example, and you just subdivide the edges
into lots of little tiny bars. It doesn’t have to
be this extreme. You could not subdivide
these edges at all, and make these
guys subdivide them into like three or four parts. Because the angles
are fixed, these guys act as a single [INAUDIBLE]. There’s really no difference. Maybe you make a
slight curve there and then they can bend a little
bit, but really not much. So if you just say, oh, I
want it to be unit length. I don’t constrain what the
angles are but I fix them, then it’s trivial to come
up with locked examples. So that’s not very interesting. What if I make it not
only equilateral– the lengths are the same–
if I make it equiangular. Because, again, in proteins,
all the angles are similar. They’re around 110, 108,
something like that. They’re all pretty close,
I think within 10 to 20% of each other. Well, there’s also
a locked example. And just to show you how
research was done back at the turn of the century,
this is pre-web 2.0, pre-Ajax and all that fancy stuff. We used Ascii Art. Email was the tool of choice. I know it’s hard to imagine
a time– 2002, so long ago. And I tracked this down. This is the original
claim it looks– we call this the
crossed legs example because it’s like two legs
crossed around each other. And this is the first time
we thought, oh, maybe it could be done, unit length. This is Stefan Langerman. And here, for the
first time ever– this is not the first model,
but this is the first photograph of any model I’m aware of– this
is the crossed legs example. This is made with a
construction toy that used to be sold around here
but is no longer in production. So they’re pretty hard to get. It’s straws– nicely colored
straws– and the cool part are these connectors. So the connectors force
particular angles. In this case, every
angle is 45 degrees. So this is equiangular
and equilateral because all the straws, I’m
told, are the same length. That’s how they’re sold. And you can do edge spins. Whoops, that’s called cheating. It’s not totally obvious
that this is locked. The problem with the model
is that the edges can bend. But if you treat it properly
and only spin around the edges, then you’re stuck. Now, there is one
thing you can do. See if I– yeah, like this. So here, I’m almost in a plane. I’ve got the purple edge
right against the pink one. Easier to see from that angle? I don’t know. So here, this guy can
come out and this guy can barely go on the edge. So actually, this doesn’t
quite work for equilateral. It works for one plus epsilon. That’s why I added these
little nubs at the end. So if they’re all
exactly equal length and you allow just
abrasion of the endpoint, then this could go
around like that and then you’d be unlocked. But if you just add
slightly– either you change the angles
to be not quite equal, so make this a little smaller,
or you make the lengths a little bit longer
at the ends– then the claim is it’s locked. We don’t actually have
a formal proof of this. We’re just remembering, hey, we
should probably write this up. I was talking to
Stefan last night. So someday we’ll prove
that this is locked. But it certainly looks like it. So this isn’t open yet. I mean, modulo the
details of that proof. Equilateral and
equiangular seems easy to lock with
fixed angle chains. In fact, even easier, this
example only has four edges. So even less than
the knitting needles. Fixed angles make for
complicated motions, I guess. Make it hard to unlock things. So I need to add
one more constraint, and the constraint is obtuse. So again, all of
these properties are enjoyed by proteins. Protein backbones have
all these properties. Even if you looked
at fixed angle trees, is there something like
this that’s locked? And now, we don’t know. And this seems quite tricky. I guess the intuition is that
obtuse– and usually we think about orthogonal,
just cause it’s easier to draw the
pictures, but reality is more like 108 degrees–
the conjecture is obtuse fixed angle chains behave
kind of like universal joints. And with universal
joints, we don’t know whether
equilateral is enough. So it’s tricky. Yeah, question. AUDIENCE: In the
previous example, you showed the ribbon thing– PROFESSOR: The subdivided. AUDIENCE: Subdivided into a
bunch of little interconnecting pieces. What if you, instead,
made your ribbon lengths basically a bunch of little
unit obtuse angle connectors, and then when you hit
the big terms it’s just obtuse, obtuse, obtuse, obtuse. PROFESSOR: Yeah, you
can make this example be entirely obtuse. You can make every angle obtuse. Here, you could
arc a little bit. Here, you could arc some
more, but not too sharp. And because here,
we actually know that this part can
be made a string. We don’t really care
what it looks like. So you can make
it fairly obtuse. It’s just that these guys
should not bend much. They have to be long no
matter how you fold them. So if you want equilateral
and obtuse, that’s also easy. But to make all the
angles actually be equal, as far as we know you cannot
take that knitting needles subdivided. Make all the lengths equal,
and all the angles equal, and make them obtuse. That’s open. But any two out of
the three, it’s easy. Of course, in reality they’re
not quite equilateral. They’re not quite equiangular. But it’s still open for those. If you have like a small
range for the lengths and a small range for
the angles, this is open. We pose it this way because
it’s the cleanest geometrically. But the real question
you care about is when these are
fuzzy constraints. Obtuse is real, but
these guys are fuzzier. So if you think about proteins,
which fold very well in nature, there are a couple of
reasons they might fold well. We know, as far as
fixed angle chains go, it’s actually quite easy
to find locked examples. And, this is somewhat
intuitive but bear with me, because they are locked examples
in this configuration space, we believe these configuration
spaces are really ugly nasty. So it would be very hard–
even if you know, oh I only need to fold something
in my component– if these guys are
highly disconnected and flat states are
all over the place, it’s probably even within
this connected component, it looks really ugly. And so it’s very hard to find a
path from one state to another. Probably pieced based
complete, although we don’t know that for sure. But that’s the intuition. Locked equals messy. When there are no
locked configurations, like carpenter’s rules, we
get really nice algorithms. It’s super easy to get from
state A to state B. Now, if you’re nature or
you’re designing nature, let’s say, or you’re building
your own virtual world, Second Life, and you want
to design proteins, you would like to design
them in such a way that they fold easily because
it happens all the time. Every thing that is being acted
on by our body, every living thing that we know has
tons of little proteins that are doing all the work. They are folded into their
shape and they do something. That’s proteins plus
RNA, but mostly proteins. So to understand life, we
should understand proteins. Now, how to proteins
fold so well when we know there are all these
locked configurations? One possible answer is that
proteins have extra structure, namely these three
things, which somehow make it very easy to
algorithmically go from A to B. Notice I’m not
assuming anything about how proteins fold in terms of what
is the mechanism that drives them because we don’t really
understand those mechanisms. There’s hydrophobia, which we
don’t really know how it works. So all these little forces
that we don’t fully understand. We understand lots of
parts of the story, but not the whole story. And what’s convenient about
these kinds of problems is you don’t need
to assume anything about how it actually happens. All we’re assuming is
the mechanical behavior of the proteins, and how
they could possibly fold. And the idea is if there’s
locked configurations, that’s probably the wrong model
because then everything’s messy. Now there’s also
evolution coming into play and maybe some proteins
are easy to fold, some proteins are hard to fold. That’s an interesting
question which should be experimented with. But let’s hope that
there’s a model. Things are mutating randomly. You really like
everything to fold nicely. Maybe it’s because
you have all three of these properties,
approximately, in real proteins. So general idea is that nature
has some extra constraints that make protein folding easy. Just have to figure
out what they are and why it makes them easy. Unfortunately, this is
still an open problem. If this had an algorithm, that
would be a natural candidate for what nature’s doing
using its mechanical– or using it’s energies
and forces, and so on. This would be a rather
unsatisfactory ending if this was– if the
climax was an open problem. We have a theorem too. And this is what I’ll
cover in most detail. And it’s a paper called
producible protein chains. Protein chains just means fixed
angle chains, open chains. And the idea is well, yeah,
there are these constraints or there are these
extra features. We don’t know how
to exploit them, so let’s not even
worry about them. Suppose they don’t even exist. Maybe I’m going to assume
obtuse, but none of the others. There’s another
constraint in how proteins fold, or really how
proteins are created. They’re created by a machine,
a molecular machine made up of a whole bunch of proteins
and RNA, called the ribosome. You may have heard of. It translates messenger
RNA into proteins. So there’s some mRNA
around here, maybe. Don’t know exactly how
this machine works. But there are actually very
accurate three dimensional reconstructions of the ribosome. With no copyright
free images, you’re going to have to– there’s
a link on the slide that goes to the cool and 3D
models of the ribosome, with a slice away. So you can see there’s
a tunnel down here, and the protein get
sort of created here. The background
gets created here. And it starts going
through this tunnel. There’s a bend in
the tunnel around here, where it’s conjectured
and an amino acid gets attached. And then it goes out the
tunnel and the protein starts spewing out here
and presumably folding at the same time. We don’t really know. So this is how
proteins are created. The birds and bees,
I guess, of proteins. So what’s interesting
about this is it’s not like a protein
exists and then folds, which is how a lot
of people might think about it at first glance. That’s the natural way to
model protein folding– start with a protein, say,
and just zigzag configuration. If it’s obtuse, there’s a nice
zigzag monotone configuration. Then you see what is
the best configuration I could fold into, for
some notion of best. And that’s sort of what
this configuration space picture is about. It’s if I already
have a protein, what configurations can
I reach by motions? And that is interesting. That’s important
because you’re still going to have to
reach by a motion. But, it’s actually
more flexible than that because the protein could
just be partially built. The rest of the protein
hasn’t been built. And it could start
folding already. It might be easier
to fold when you don’t have the obstacles
of your existing protein. So that’s both a worry, but
it’s also a convenient structure because this ribosome
is a giant obstacle. Bigger than most proteins. If your protein’s really long,
maybe it could go over here. But most the time, it’s going to
stay on one side of this plane because locally, this thing
is basically flat, if you look at the real 3D pictures,
not the schematic. Now, this is good news for a
geometer because there’s this giant obstacle– think
of it as a half space– which the protein cannot
penetrate while it’s being produced over here. That’s it. That half space constraint
is enough to get really good algorithms
for folding your chain. It’s weird because we’ve
made a problem both harder because the protein is only
partially produced at any time and it can fold,
which is part of it, but we’ve also made
our life easier because there’s
this big obstacle. AUDIENCE: [INAUDIBLE]
makes sense why there’s only obtuse
angles there, right? PROFESSOR: Yeah, right. Out of this, we’re going to get
that the angles and the protein are constrained. And in particular, for
this angle– it depends. I mean, in this picture because
it’s perpendicular here, yeah, the sharpest angle
you could make is 90 degrees, more or less. That’s a good point. So it’s a convenient match
between the chemistry, which also forces the angles
to be obtuse, I guess. I don’t know a ton of chemistry. But also, the ribosome
just geometrically forces. We’re going to use a
property like that. Our model is going to be a
little bit more– both more general and simpler. We’re going to imagine that
the ribosome is a cone. It’s part of the
upper come here. This is like a mirror image. And in reality, that
cone is actually a plane and everything
above the plane. But to be more
general, we’re going to allow some angle alpha here. It’s also just easier to think
about when alpha is smaller than 90, but everything I say
will work when alpha equals 90 and that is sort of
the reality case. So the model is– so
the ribosome is a cone. We call this the half
angle of the cone. From the vertical axis to the
edge of the cone is alpha. So if you were going from
one axis to the other, it would be 2 alpha. The model is, you start with
one link of your chain, which is inside the cone. It’s spews out through the apex. That’s the exit of the tunnel. Here, we’re allowing the tunnel
to be actually quite free. Doesn’t have to be
perpendicular to the apex or the plane of the apex. So the edge comes
out, and as soon as the endpoint of the
chain reaches here, then a new link is created. This is like a very simple model
for how a chain can come out of a cone without
worrying about what’s happening inside the cone. Imagining everything’s
totally free. This is like you can
allow self intersection in the cone, who knows what. But once you come
outside the cone, you’re not allowed
to self intersect and you’re not allowed
to intersect the cone. Once you come out,
you can’t go back in. So that is a model of
producing protein chains. And if you have a
cone of angle alpha, we call this an alpha
producible chain. For whatever reason, we often
call it beta producible chain. Just change the variable. So if you think of the ribosome
as a cone with half angle beta, you can produce it like this. That is beta producible. Now this is a pretty
powerful model because you only have to
worry about it link by link. You don’t have to worry
about the rest of the chain until it spews
outside of the cone. But it’s restrictive in that
you cannot penetrate the cone. All right. One thing we can
talk about is angles. So I’m going to
write call a chain a less than or
equal to alpha chain if all the turn angles are
less than or equal to alpha. I don’t know if I’ve used
turn angles in this class. Probably. If I have two edges,
the angle would be this. The turn angle would be
this, the supplement. Yeah. I guess we used
turn angles way back in origami land, Kawasaki’s
theorem and so on. It’s just, if you’re
going straight, how much do you have to turn
to get to the next edge. So we’d like fairly
obtuse things. So alpha is going to be small. There isn’t a ton of turn. But in general, less than or
equal to alpha chain for some alpha. Now there’s a relation–
as Jason was mentioning, there’s a relation between
alpha and beta in the ribosome because you always
exited orthogonally to the plane that was your cone. The sharpest angle you could
get was 90 degree turn angle. Here, we’re a little freer
because this edge can wiggle around as long as
it touches the apex. So if you’re up
against the cone, you have to slide out into
the complimentary cone– that was the previous picture–
and as soon as you get there, you could create a new
edge which is like this. So the sharpest angle you can
get is actually twice beta. In general, we’re going to
have alpha over 2 is less than or equal to beta. That is– you can get up
to beta equals 2 alpha. Get that right. And also in the obtuse case,
this is not too exciting. But it’s true. There’s actually
some problems here. When you have that
full flexibility and you set alpha to two beta,
not the other way around. I’m going to assume here
that alpha equals beta. This will be convenient. And it’s the interesting case
because, in reality, the cone has a half angle of 90 degrees. So beta is 90. And the sharpest angle
we’re going to make was always obtuse. So saying that you have a
less than or equal to 90 chain is just fine. But on the mathematical
side, I think we saw the case when alpha is
less than or equal to beta, but not when alpha over 2 is
less than or equal to beta. That’s a weaker constraint. So there is a range
where it’s not so easy. Now, what do I claim
about these chains other than their angles
are not so sharp? I claim they’re good
algorithms for folding them. What could I possibly mean? There are still
locked configurations. Is that true? Well, I mean presumably–
this is acute– but you take the obtuse
versions of this guy. Because I didn’t constrain
the edge lengths or anything, I just said that the
angles are obtuse. So I could just sort of round
these corners, make it obtuse. You know, add lots of
dots just at the corners that would be obtuse. And a chain like this
will be producible. A chain with these angles
and these edge lengths can be produced from a cone. But this configuration of
this chain cannot be produced, I claim. I claim anything
that can be produced is in one connected component. So while I can make a
linkage that is locked and that there are bad
configurations you can’t get out of, the things
you can actually make, you can always get out of. So there’s going to be the space
of producible configurations. Maybe there’s some stuff
that’s unproducible but still connected to it. I don’t know. It doesn’t matter too much. I won’t worry about this stuff. There’s other bad locked
configurations that cannot reach here. But everything that’s producible
is in one connected component of the configuration space. That’s property one. That’s kind of nice. Also, all the flat states
are going to be in here. This is actually pretty easy. I just need to prove that
flat states are producible, which we’ll worry about later. So in particular, these guys
are flat state connected. All the producible
protein chains are flat state connected. That’s interesting
because we don’t even know that all chains are
flat state connected. But here– I guess we know
that obtuse chains are flat state connected, so
maybe it’s not so surprising. But what’s important is not only
are the flat states connected to each other and
the producible states are connected to each
other, but producible is connected to flat states. Everything is together here. I might have more properties. But that’s already
some good news. And there’s algorithms
to do all of this. How do we prove it? Well, as usual we
use the FedEx method. And in some sense,
one of the challenges is what is the natural canonical
state for protein chains? In fact, we’re just going to
assume that our chain is– in reality, we’re
going to assume that it’s an orthogonal chain–
an obtuse chain, I should say. But in general, for any
less than or equal to alpha chain, for whatever
alpha you like– and it will be the half angle of the
cone, so alpha equals beta– we will define a
canonical configuration. I think we called
it the alpha CCC. So it’s going to be
kind of like a helix. I think I have an example,
an actual computed example. That’s not the best
picture because you can’t see everything
that’s going on. But this is an actual
canonical configuration of a particular chain. Let me tell you how
it works in general. So in general, we have
some chain, v1, v2– sorry, starting at v0, v1, v2. I want to define a
canonical– and there’s defined lengths between the two. And there’s defined angles
between every triple in sequence. So I’m going to start
with v0 somewhere. Doesn’t really matter
by translation. Say, the origin of space. And what I’m going to do is draw
a cone whose apex is that v0. And the half angle here is
going to be alpha over two, not alpha. This is a smaller cone
than– by a factor of two– than the ribosome. That’s important. So there’s this vertical line. And to pick v1, I’m just going
to use the right edge, which is– let’s say this
is the x direction. This is the z direction. Maximum x-coordinate. It’s going to lie on the
cone, maximum x-coordinate. That’s v1. Now, v1– let me redraw
this picture a little lower. So there was v0, v1. Now I want to draw v2, and
I want to draw it above. So what I’ll do is
draw a vertical column whose half angle here
is alpha over two. I want to draw v2
on the cone here. Of course, the
height of the cone is the length of the
edge, not the height. You could think of
the cone as infinite. And then I just clip this to
when it has the right length. So again, this might be
a different height cone. I clip it to whatever
the length v1, v2 is. I want it to be
somewhere on this cone, but now I’m constrained to
have the correct angle at v1. I can’t just put it over here,
because then the angle here would be 180. Presumably, I don’t want
to make a 180 degree angle. So in reality, what
happens is that there’s a cone which is– whose
axis is the edge v0, v1. So I extend v0 v1 out
here, which, in this case, happens to lie here. And I make a cone like that. Let me draw it slightly
more accurately. In this case, the
center axis of the cone would go right here, whatever
the extension of v0, v1 was. And to have the right angle at
v1, v2 must be on that cone. Conveniently, there are
two intersection points between those two cones. I could choose either
one of them to be v2. And I will choose the
counterclockwise most one, which is this one. This is going to be v2. So I draw that edge. Now I repeat. So for v2, I’m going to
draw a vertical cone whose half angle– here, the half
angle is always alpha over two. Half angle is alpha over two. This cone had half angle,
whatever the angle v0, v1, v2 was. Was that the half
angle or angle? The half angle. I think this is right. OK. And then I take the
intersection of those two cones, and that will give
me where v3 is. So I do the same thing
for v2, for v3, and so on. I have a unique choice
at every moment. Yeah, basically. And the only exception
was at the beginning here, when I had
a vertical line. These two cones could
actually be equal, and then the intersection
is the entire cone. And that case, I
guess I’d choose the maximum
x-coordinate one again. And then I have a
canonical choice of everything along the way. It will always go up. And it sort of
spinals around because of the counter
clockwise most choice. And the result is a
picture like this. Anything else I
need to say here? I claim this canonical
configuration lies in an alpha over
2 half angle cone. That’s true by construction. The challenge is, does the
construction really work? So I start, obviously,
with one cone, and I can think of this as
actually an infinite cone that goes out to infinity here. And I claim the
entire construction will lie inside that cone. And it’s kind of
obvious because v1– I chose v2 to lie in the
same cone, just translated up to start at v1 instead
of starting at v0. Of course, this cone is
contained in this bigger one. And by induction, in fact,
the entire rest of the chain will lie in this smaller cone. Therefore, it lies
in the big one, also. OK. Fine. So by construction it will
lie in alpha over two cone. The worry is that these
two cones don’t intersect. Here we have an angle. The half angle of
the cone is whatever angle the angle is at v1. Sorry, this should
not be the angle. This should be the turn angle. That angle is how much
you turn from v0, v1. Now, we know that– we’re
assuming that– the turn angles are all, at most, alpha. The turn angle
cone could actually be twice as big as the vertical
cone that we were always using. We always use a vertical cone,
half angle alpha over two. But it’s OK because we
always keep these edges, like v0, v1, was on
the edge of the cone. And so when we
extend it, it lies on the edge of
this vertical cone. So its angle– in the most
extreme case– its half angle is alpha, which
would look like this. We’ll go all the way over from
the right side of the cone to the left side of the cone. In general, it’s not going
to be right and left, but it’s going to be some
side and the antipodal point. And because the double
angle of the cone is alpha, it’s still OK. You will intersect
somewhere on the cone. This is a subtle detail,
but it’s really crucial because we start
with a chain that has relatively
large angles, alpha. And we get it
into– we squeeze it into– a cone that
still has double– twice its angle is alpha,
but we kind of compress it into something
of half angle alpha over two. You might think, oh, I’m
just changing the definition and calling it half angle. Therefore, it gets
to an alpha over two. But it’s a little
tricky to actually get it to fit in a vertical
alpha over two cone. Once we have this,
it’s really easy to canonicalize a chain,
a producible chain. So let me tell you
how to do that. So we’re going use
the FedEx method of taking some configuration
and canonicalizing it, and then uncanonicalizing
to something else. We’re also going to use a
new method, which I just came up with the term. Is called the
momento method which is you play the
movie in reverse. So, I guess, also
the Merlin method. That’s more complicated. So we have a movie
here in mind, which is how was the chain produced? So what I want to
show is that if– I want to start with a producible
configuration and chain. Somehow, it got produced. So you had your
cone and the thing starts spewing out and
folding, and doing whatever. That’s an animation, in some
sense, of one edge coming out and stuff is folding
at the same time. Then an edge is created,
then another edge comes out, and so on. What I want to do is play
that movie backwards. It’s a pretty intuitive idea. I just want to start feeding
the edges back into the cone, and just keep stuffing them in. Now, what happens
out here is easy because we know it doesn’t
penetrate the cone. That’s the assumption. And we know whatever was
created here could be uncreated, as long as you can afford
to erase edges one by one. That’s the tricky part. How do I erase an edge? And usually, I can’t. But I don’t have
to erase any edge. Like, if I had to
erase this one, that would be hard
because some motion here might penetrate where that
edge ought to have been. And erasing the
edge will make it– adding the edge makes
it harder to fold. So I can’t erase it. But the edges I have
to erase are the ones that have been fully
inserted into the cone. So if I can somehow do something
inside the cone, I would be OK. All this work, defining a
canonical configuration, was about forcing a chain to
stay inside a cone– and not only an alpha cone,
which is what we’re going to have as the ribosome,
but an alpha over two cone. This is smaller than my
ribosome cone by factor of two. And I need that. Why do I need that? Because this thing, I have no
control over the outside chain. So the way that it
approaches the cone could be as sharp
as like this, where I have the first edge that
is– the current edge that is– inside the cone. As far as the
movie is concerned, there’s only one edge. There’s nothing up here. I want to put something up
here, in a cone, naturally. But I don’t get to control
the first angle because that is controlled by this motion. Maybe it really needed
to go sharp like that so that it could make a
sharper angle or whatever. So you might say, well, of
course I can put it inside and alpha cone, which is
the same as this alpha. Sorry– bad picture. This is alpha. This would be a problem because
I have this weird angle coming in, and now suddenly I have
to bend back like that. Maybe the turn angle here
is not so sharp as alpha. Maybe it’s one degree. So I really can’t force
the rest of my chain to lie in this cone. But I claim I can force
it to lie in this cone, with half angle alpha over two. This is a little more subtle. Oh, right. I wanted to mention that
helices appear in nature. They appear in
proteins, but they also appear in this
crazy climber plant. Marty, have you seen
this in Costa Rica? AUDIENCE: Yeah. PROFESSOR: Yeah. There’s some really incredible
wildlife in Costa Rica. I’ve never been there, but
I’ve seen lots of pictures and this is spirals in practice. Also, proteins tend
to form these things. They call them out alpha
helices because they spin like an alpha, I guess. So it’s kind of neat that the
canonical configuration, which is totally
geometrically motivated, also appears in biology. Not that kind of biology though. So here is the picture. I have the big cone. That is my ribosome,
and here I’m going to write beta for
the– beta for big, I guess. And then we know that we
can canonicalize a chain, or there at least exists
a canonical configuration of the chain, where
everything lies inside a cone of half
angle alpha over two. Now the problem is that cone, we
want it to be at a funny angle. Let me draw a real picture. Here’s a ribosome. It has a big angle here. We’ll call alpha. Now, I’m going to do
the not extreme case, try to be a little more general. There’s some edge that
is currently entering, and we have no control over that
edge or the rest of the chain. That’s determined
by the movie which we’re trying to play backwards. What we have to control, and
what we’re free to control, is the rest of the
chain because as far as the movie is concerned,
that hasn’t been created yet or it’s already been destroyed,
depending on whether you’re playing forwards or backwards. We need to say
what happens to it. And what I want to happen
is so that by the time this edge is inside,
that edge plus the rest is in the canonical
configuration. If I can achieve
that, then as edges come in they become
canonicalized, and then everything will be
canonical and inside the cone and we’re done. That’s our goal. Canonicalization. So, what’s the deal? Well, in reality,
there’s some cone– yeah, it can penetrate
like that– could penetrate the outside cone. This is getting messy. So if I extend this line,
there’s a cone of half angle here, which is equal to
whatever the turn angle is at that vertex, which
is, again, specified. We’re not free to set
it to whatever we want. We know the next edge
must lie on this cone. What I do– now, on the
other hand, off to the side, I have in mind a
vertical cone whose half angle is alpha over two. So it’s quite small. And I know– how do
we do it– initially, the first edge was along
the maximum x direction, and then it spirals
up from there. OK. Here’s what I’m going to do. There are sort of
two situations. What I’d like to do is put a
cone here that is vertical. Something like that. Just like I have here. The trouble is, the
right side of this cone does not intersect the
boundary of this cone. And need it to in order to
form the right angle here. It’s going to– it might go
through the middle of the cone like it does here. So then what I do is I take
this canonical configuration and I rotate it so that–
this is hard to draw– it’ll be something like this. There’s no hope of seeing this. So it’s on the
surface of this cone. It’s not on the far right edge. It’s going to be some
intermediate point. And that’s exactly where
it intersects this cone. It’s just like the previous
picture, just harder to see. I’m taking the intersection
of these two cones. If I just rotate
the picture, then it will lie in the intersection–
if they intersect. But they might not intersect. So let me go here
and try it again. So the easy case is when
I can draw a vertical cone and it intersects the cone
that I need to intersect. The harder case is–
maybe it’s more extreme. Maybe it’s a very
tight angle here. So there’s a very
small turn angle. I have to intersect this cone. Because the cone
that I’m working with is actually smaller–
it’s half of this angle– it might not
intersect this cone. In that case, I’m going
to rotate the cone to fall over a little bit. So instead of being like that,
it’s going to be like this. OK. So I want the intersection
of these two cones, which I’ve conveniently
made this edge here. So the first edge
will lie along here, and then it’s going to
spiral out from there. So I still have the
canonical configuration. I’ve just tilted it. Tilting is going to be necessary
because I have this angle to match up. The convenient thing about
the canonical– the alpha CCC, canonical
configuration– is that I have this half
angle of alpha over two, so I can afford to tilt it
by up to alpha over two. It will still stay
within the ribosome, which is of half angle alpha. And all I need to show here–
and once you think about it for a while, it’s
obvious– that ideally, I don’t tilt it at all. I’ve got tons of room. Huge amount of room here. But sometimes I’ll have to
tilt it, but by at most alpha over two. So I will stay
inside the ribosome, because this apex was inside
the cone and the smaller cone, even if I tilt it all the
way to meet this edge, it will stay inside the cone. In fact, it’ll stay
inside the big cone. In fact, this cone that
I tilt– the one that contains the rest of the
canonical configuration– will always contain
the up direction. That’s how to see it. If it always contains the
up direction, then at most, it’s that big. And that will– because that’s
an angle of alpha because that was the half angle
of the big cone. So that will be a half
angle of alpha over two. As long as you contain the
up direction at all times, you will not fall
outside the big cone. So the rest is just momento. So you play this
movie backwards. As things come in
here, this cone is going to wiggle
back and forth, depending on how
this angle changes. Once the edge gets
all the way in, you absorb it into the
canonical configuration. Little bit of work there
but you just sort of twist the cone around
until that algorithm that we described for producing
canonical configuration would actually produce
what’s inside the cone. Then the next edge comes
in, cone wiggles around until the edge gets
all the way in, then you canonicalize what’s
inside the cone, and repeat. So if you had a
way to get it out, you could put it
back in and keep track of all the stuff
that happens on the inside. That’s what this theorem says. And then there’s just
slightly more to say. If you have a flat
state, we want to prove these things
are flat state connected. So take some flat
state– I really should have only obtuse angles. I claim this flat
state can be produced using a cone of the
appropriate angle. If the sharpest turn
angle here is alpha, then you need an alpha cone. And the way to
think about that is to think of the cone moving
instead of as the chain moving. It’s a lot easier. So you want the chain
to lie around here. So you start by moving the
cone, I guess like this. So that it just barely
touches the plane, and this edge spews out. Is that the right way
to think about it? I don’t know. Should be like this. So you take the cone. This is like putting
frosting on a cake. So you move your cone like here. You squirt out some– this edge. Then, you do it so
that when you’re here and the new edge is created,
it’s still in the plane. And then you just sort
of move around there. I’m just going to leave
it as a sketch like that. Once you know that
you can produce any flat state, of course,
you can reorient yourself by relativity so that the chain
is moving instead of the cone. So you can produce this
thing with a ribosome. Once you know all flat
states can be produced and you know all producible
configurations can be canonicalized, then you
know it’s flat state connected and you know all canonical
things are flattenable and vice versa, by continuous motions
without self intersection. And all of this is algorithmic. It can tell you how to go
from one place to another. And so this is a
candidate algorithm, I would say, for how
nature folds proteins. Just thinking about
the mechanics, not worrying about
how it’s implemented. Maybe you take
this model and then you try to make it
physical forces, and you get a way
to fold proteins. That, of course,
remains a mystery. But next time we will talk about
some very simple models that are motivated more closely via
biology of how proteins might actually fold, and talk
about the complexities that you get there.