Folding nobody designed

A domain too short to be unique

A staple holds the scaffold by pairing with a stretch of it, and two arguments decide how long that stretch has to be. One is combinatorics — a stretch of seven bases has about four hundred other places in a 7,249-base strand it would also match. The other is thermodynamics, and it is the one that binds: a duplex that is unique at eleven base pairs still comes apart at the temperature the design is held at, and staying paired takes seventeen. Rounded up to the crossover period, that is three periods on both lattices, and the lattice the helix prefers is the one whose three-period domain leaves some of its staples unattached.

Assumes The helix chooses the lattice and A sheet that routes itself.

Six essays here have treated a DNA origami as a graph. A helix is a cell, two helices that sit side by side can be joined, and the design problem is to find a route that visits every cell once — which is where the counting argument came from, where the two ceilings were measured, and where every cheap test turned out to miss a shape. None of it knows that the strand is a sequence. A cell has no letters in it.

The staples are where the letters arrive. A staple is a short synthetic strand that holds two or three helices together by pairing with a stretch of the scaffold in each of them, and the stretch it pairs with is a domain. How long a domain has to be is not a question about shape at all, and it has two answers that were arrived at from different subjects.

The first is combinatorial. A domain of kk bases is one word out of 4k4^k, and the scaffold offers about seven thousand places for that word to occur. Short enough and the domain is not an instruction; it is an ambiguity.

The second is thermodynamic. A duplex of kk base pairs comes apart above a temperature that rises with kk, and a staple that is not paired at the temperature the design is held at is not holding anything.

Neither argument mentions the other’s terms, and the lengths they demand are not the same length. The one that turns out to bind is not the one these essays’ habits would predict.

How long a staple's domain has to be, and whyTwo demands on the length of a staple's binding domain drawn on one axis of bases: the length at which a domain is expected to have no second place in the scaffold it also matches, and the length at which its duplex is still paired at the temperature the design is held at. Below them, the base positions at which a crossover may sit on each lattice, and the first admissible domain length that falls inside both bands.the two things a domain has to be long enough forthe bar starts where the demand is first met and runs to the rightunique in the whole design11 bases and upstill paired at 45 degrees17 bases and upwhere a honeycomb crossover may sitfirst one inside both bands: 21 baseswhere a square-lattice crossover may sitfirst one inside both bands: 24 bases4812162024283236a domain runs between two crossovers, so its length is a whole multiple of the period
Fig. 1 The two demands on a domain’s length, drawn on one axis of bases. Being unique across a whole design is met from eleven bases up; staying paired at forty-five degrees is met from seventeen. Below, the positions at which a crossover may sit on each lattice — every seventh base on the honeycomb, every eighth on the square — and the first admissible domain length that falls inside both bands.

How short is too short to mean one thing

The counting is the same argument that Maekawa’s condition is, run over an alphabet instead of over creases: count the things that could happen and compare with the things that do.

A scaffold of 7,249 bases has 7,249 windows a domain could be read from, minus the few at the end. One of them is the site the designer intends. Every other one is a place the domain would bind if the letters happened to agree, and in a sequence with no preferred composition the chance that they agree over kk bases is 4k4^{-k}. So the expected number of second homes for one domain is about 7,249/4k7{,}249/4^{k}, and for a design with several hundred domains it is that again multiplied by the count.

That multiplication is the part worth being careful about, because it moves the answer by two and a half orders of magnitude. A designer does not care whether this domain has somewhere else to go; they care whether any of the design’s domains does, and a full-scaffold design has something like four hundred and fifty of them.

How short a domain has to be before it has somewhere else to goThe expected number of binding domains in a design that match a second place in the scaffold, against the length of a domain in bases, on a logarithmic scale. At one crossover period the count is in the hundreds; it falls below one at eleven bases.a scaffold of 7,249 bases, a design of 454 domains-6-4-2024812162024bases in the domaindomains of a 454-domain design expected to have a second place to bindone domain of the design — first cleared at 11 basesa honeycomb perioda square periodthe count is the whole design's, not one domain's, which is where the two and a half orders of magnitude between them go
Fig. 2 The expected number of a design’s domains that match a second place in the scaffold, against the length of a domain, on a logarithmic scale. At one crossover period the expectation is in the hundreds. It falls below one domain in the whole design at eleven bases, and the two orders of magnitude between the two marked periods is what one extra base buys.

At seven bases — one crossover period on the honeycomb lattice — a design expects two hundred of its domains to have somewhere else to go. At eight, fifty. At eleven, fewer than one. The curve is a straight line on this scale because each added base divides the expectation by four, which is the whole of the argument and is worth stating in that form: a domain buys its specificity at two bits a base, and it needs about twenty-two bits to be alone in a scaffold this size.

What one actual sequence has in it

That is an expectation over sequences, and a design has one sequence. The distinction matters here in the same way it mattered when a population of crease patterns was drawn rather than assumed: a statistic computed over an ensemble says nothing certain about the member in hand.

So the count is checked against a sequence. Every window of a 7,249-base string is hashed and the repeats are counted directly, which is a cheap thing to do and is not the same calculation as the one above.

Words that repeat in one scaffoldFor domain lengths of seven to sixteen bases, how many windows of a seeded scaffold sequence are words that occur somewhere else in it, beside the number the counting argument expects. At seven bases a third of the sequence repeats; at fourteen none of it does.every window of one 7,249-base sequence, countedthe bar is how many windows hold a word that occurs more than once7 bases2,6195,825 distinct words of 7,243 · 3,202 expected to repeat8 bases7916,837 distinct words of 7,242 · 800 expected to repeat11 bases147,232 distinct words of 7,239 · 12 expected to repeat14 bases07,236 distinct words of 7,236 · 0.20 expected to repeat16 bases07,234 distinct words of 7,234 · 0.01 expected to repeatthe sequence is drawn from a seeded stream with equal base frequencies, so it is a typical sequence of that length and not a genome
Fig. 3 Every window of one seeded scaffold sequence, counted: how many of them hold a word that occurs somewhere else in the same sequence, beside the number the counting argument expects. At seven bases more than a third of the sequence is repeated somewhere; at fourteen, none of it is.

The measured counts sit below the expected ones, and the reason is arithmetic rather than a disagreement. The expectation counts pairs of matching windows, and a word occurring three times is three pairs and only three windows, so a sequence with clustered repeats reports fewer repeated windows than the pair count predicts. The two numbers agree in the only sense that matters: the length at which repetition stops. Fourteen bases is already past it, and every longer domain in this essay is past it by a wide margin.

The temperature nobody counted

The second requirement has nothing to do with how many words there are. A duplex is held together by base pairing and by stacking between neighbouring pairs, and the energy of that scales with the number of pairs while the entropy lost on binding two strands does not. The result is a melting temperature that rises steeply at short lengths and flattens out, and short means exactly the lengths a crossover period supplies.

Where a domain stops coming apartThe melting temperature of a duplex against its length in base pairs, with the temperature a design is held at drawn across it, and the domain lengths each lattice admits marked. Two crossover periods melts below the hold on both lattices and three melts above it.a duplex at half its bases guanine or cytosine-25025508141621242832base pairs in the domainmelting temperature, degreesheld at 45 degrees — below this line a staple is not holding anythingthe round marks are the domain lengths a lattice actually admits: multiples of seven above, of eight below
Fig. 4 The melting temperature of a duplex against its length in base pairs, with the temperature a design is held at drawn across it and the domain lengths each lattice admits marked. A domain of one crossover period melts below room temperature on either lattice. Two periods is still below the hold. Three is above it on both.

An eight-base duplex melts at about one degree. A seven-base duplex, on the same approximation, melts below zero — which is not a temperature a design is ever at, and the sentence to take from it is that a domain one crossover period long is not a bond at all. It is a transient contact that the anneal will undo as fast as it makes it.

The hold temperature is where the requirement comes from. A DNA origami is annealed: heated until nothing is paired, then cooled slowly so that staples find their sites rather than each other, and held at some temperature where the correct pairings are stable and the incorrect ones are not. Forty-five degrees is a representative figure and the argument does not depend on the exact value, because the curve is steep over the lengths in question. The length at which a half-and-half duplex melts at forty-five is seventeen base pairs.

Seventeen against eleven. Uniqueness is met six bases before stability is, and at the length stability demands, uniqueness is slack by four orders of magnitude. The requirement everybody states first is the one that decides nothing.

That shape — two independent bounds on one quantity, of which only one is ever the operative one — is the same shape the substrate’s two ceilings has, and the comparison is worth making because there the answer depends on the regime. A model’s layer count is bounded by how thin the paper is and by how fine a division a pair of hands can place, the two improve with different quantities, and which binds crosses over at a computable sheet size. Here there is no crossing. Stability rises with length and so does uniqueness, they never trade, and the gap between them is six bases at every hold temperature worth using. A pair of bounds can be a competition or a formality, and which it is has to be computed rather than assumed from the fact that there are two of them.

The period rounds both numbers up

Neither threshold is a length a design can use, because a domain is not free to be any length. It runs from one crossover to the next, and the crossover positions are fixed by the helix: a strand can cross to a neighbouring helix only where its backbone faces that neighbour, which on a lattice with three neighbours a third of a turn apart happens every seventh base, and on one with four neighbours a quarter of a turn apart happens every eighth — with the residual that essay measured.

So the admissible lengths are the multiples of seven and the multiples of eight, and both thresholds have to be rounded up to one.

The domain lengths each lattice has on offerFor each lattice, the domain lengths that are whole multiples of its crossover period: how many bases, the temperature at which the duplex melts, how many second binding sites a design of that domain length expects, and which of the two demands the length meets. Three periods is the first that meets both, on either lattice.every admissible domain length, to four periodsagainst a hold of 45 degrees and a scaffold of 7,249 basesperiodsbasesmelts atsecond sites expectedmeetshoneycomb × 17-10.7201neitherhoneycomb × 21437.41.2e-2unique, and comes aparthoneycomb × 32153.47.5e-7bothhoneycomb × 42861.44.5e-11bothsquare × 181.450neithersquare × 21643.47.6e-4unique, and comes apartsquare × 32457.41.2e-8bothsquare × 43264.41.8e-13botha domain has to run from one crossover to another, so only these lengths are on offer at all
Fig. 5 Every admissible domain length to four crossover periods on each lattice: how many bases, the temperature at which the duplex melts, how many second binding sites a design of that domain length expects, and which of the two demands the length meets. Two periods meets uniqueness on both lattices and stability on neither. Three meets both.

Two periods is fourteen bases on the honeycomb and sixteen on the square lattice. Both clear the uniqueness threshold comfortably. Neither clears the stability one: fourteen melts at thirty-seven degrees and sixteen at forty-three, and a design held at forty-five has both of them coming apart. Three periods is twenty-one and twenty-four, and both are above.

That is the result stated as a designer would use it. A staple’s domain is three crossover periods long because of a temperature, and the fact that it is also unique is a side effect it got for nothing. A design that shortened its domains to the length the counting argument demands would have every staple falling off.

A sequence has a weakest domain

The stability figure above is drawn for a duplex at half guanine and cytosine, which is a composition and not a sequence. A real scaffold is one molecule, its domains have whatever composition they have, and a domain rich in adenine and thymine binds more weakly than one rich in the other two. So a design has a weakest domain as well as a mean one, and the anneal has to be held below the weakest, not below the mean.

What 21 bases actually buys, domain by domainThe melting temperatures of every 21-base domain of one seeded scaffold sequence, as a histogram, with the hold temperature marked. The mean sits well above the hold and the weakest domains sit below it, because a domain's temperature depends on its own composition and a sequence has whatever composition it has.345 domains of 21 bases, mean 53.4 degreesweakest 38.7, strongest 66.1 — a spread of 27.3 degrees050100405060melting temperature of one domain, degreesdomainsheld at 4513 of 345 domains melt below the hold, and every one of them is a staple that is not attached
Fig. 6 The melting temperatures of every twenty-one-base domain of one seeded scaffold, as a histogram, with the hold temperature marked. The mean sits eight degrees above the hold and the weakest domains sit below it, because the temperature of a domain depends on its own composition and a sequence has whatever composition it has.

Three hundred and forty-five domains of twenty-one bases have a mean melting temperature of 53.4 degrees, a spread of twenty-seven degrees from weakest to strongest, and thirteen of them below the hold. The mean is eight degrees clear and thirteen staples are not attached.

Thirteen out of three hundred and forty-five is under four per cent, and four per cent of the staples missing is not a small defect in an object whose whole rigidity comes from being stapled everywhere. It is also exactly the kind of failure that does not announce itself: the structure forms, it is the right shape, and it is softer than it was designed to be in places nobody can point at.

The lattice the helix wants is the one that fails

This is where the two halves of the essays here meet, and they disagree.

The helix chooses the lattice found that a three-neighbour lattice is the one a double helix’s pitch actually fits: three neighbours a third of a turn apart are faced exactly every seven bases, and four neighbours a quarter of a turn apart are never faced exactly by any whole number, missing by 4.3 degrees a crossover and accumulating. On that argument the honeycomb wins outright, and the square lattice is a convenience that costs a twisting residual.

Round the stability threshold to each lattice’s period and the verdict inverts.

The lattice the helix wants is the one whose domains come apartFor each lattice, the shortest domain length it admits whose mean melting temperature clears the hold, and how many domains of four seeded scaffold sequences fall below the hold at that length. The honeycomb's twenty-one leaves some domains unpaired on every sequence; the square lattice's twenty-four leaves none.the shortest domain each lattice admits that melts above the hold on averagethe bar is how many of its domains are not paired at the holdhoneycomb · 21 bases · sequence 188 of 345 below the hold · weakest 38.7 degreeshoneycomb · 21 bases · sequence 288 of 345 below the hold · weakest 40.7 degreeshoneycomb · 21 bases · sequence 31313 of 345 below the hold · weakest 38.7 degreeshoneycomb · 21 bases · sequence 41515 of 345 below the hold · weakest 40.7 degreessquare · 24 bases · sequence 100 of 302 below the hold · weakest 47.1 degreessquare · 24 bases · sequence 200 of 302 below the hold · weakest 45.4 degreessquare · 24 bases · sequence 300 of 302 below the hold · weakest 45.4 degreessquare · 24 bases · sequence 400 of 302 below the hold · weakest 47.1 degreesfour seeded sequences, because one sequence's weakest domain is one draw and the question is about the length
Fig. 7 The shortest domain each lattice admits whose mean melting temperature clears the hold, and how many domains of four seeded scaffold sequences fall below the hold at that length. The honeycomb’s twenty-one leaves between eight and fifteen of its domains unpaired on every sequence tried. The square lattice’s twenty-four leaves none.

Twenty-one bases is the honeycomb’s three periods and it clears the hold on average by eight degrees, which is not enough to carry the tail: between eight and fifteen domains fall below it, on every sequence tried. Twenty-four is the square lattice’s three periods and it clears the hold on average by twelve, which is enough: on four seeded sequences, not one domain of twenty-four bases melts below forty-five.

The honeycomb could of course use four periods, at twenty-eight bases — but that is a third more scaffold between crossovers, which is a third fewer crossovers holding the structure, and the crossover density is what makes a bundle of helices stiff rather than a bundle of helices. The lattice the molecule prefers has its admissible lengths in the wrong places, and the design has to pay for that either in unpaired staples or in stiffness.

The choice between those two payments is the one a deployable makes between compaction and the chance of opening, arriving from the other end. There, finer folding buys smaller packing and spends the probability that every hinge works, because a structure needing all of its hinges has a reliability that is a product. Here, denser crossovers buy stiffness and spend the probability that every staple is attached, because a structure needing all of its staples has the same arithmetic. The two subjects share no vocabulary and the optimum in both is below the point where the thing being bought stops improving — which is what a product of probabilities does to any quantity that is proportional to the count of its terms.

That is a connection these essays had no reason to expect. The lattice question looked like geometry — where does a backbone face — and the answer to it turns out to be decided by a temperature and by the granularity a periodicity imposes on a length. Two arguments about entirely different things meet because the second one is only allowed to answer in multiples of the first one’s period.

What none of this shows

The melting temperature is an approximation and is named as one. The figure uses the salt-adjusted formula for a short oligonucleotide, which knows a domain’s composition and nothing about the order of its bases. Stacking energies depend on which base follows which, so a nearest-neighbour calculation would move any individual domain by a few degrees. The argument turns on a gap of twenty degrees between two periods and on a spread of twenty-seven within one length, so it survives that error comfortably — but no single number here should be read as the temperature of a particular domain.

The scaffold is drawn from a seeded stream and is not a genome. M13mp18 is a real molecule with a real composition, and it is not a quarter of each base. It has repeats no random sequence has and it lacks coincidences a random sequence would have. What a drawn sequence can say is what a typical sequence of that length does, which is the right comparison when the question is whether a length is sufficient in principle. It is not a statement about that virus, and the essay makes none.

A domain is treated as either paired or not. Binding is a population, not a switch: at the hold temperature a domain a little below its melting point is bound most of the time, and one a little above is bound some of the time. The count of thirteen unpaired domains is therefore a count of domains whose occupancy is under a half, not of thirteen staples lying in the tube. The effect the essay names — a structure softer in places nobody can point at — is if anything better described that way.

And the whole argument assumes the staple finds its site. Nothing here models the kinetics of an anneal, which is where the difference between a sequence with repeats and one without would actually show up: a domain with a second home spends time in the wrong place and has to leave it, and that is a rate rather than an equilibrium. The uniqueness calculation says how many wrong homes exist and says nothing about how long a staple sits in one.

Nothing here is a fact about a protein. The word folding covers two different things and this essay is about the second of them throughout: a designed object whose parts are specified one at a time. The reason a domain can be given a length at all is that somebody chose its sequence, and the reason the calculation is short is that the only interaction modelled is a domain pairing with its complement. A chain that folds because every part of it interacts with every other part has no quantity in it corresponding to a domain length, which is the difference that essay found and this one relies on.

How the numbers were got

The uniqueness count is closed-form and the sequence count is not. The expectation is (Lk)/4k(L-k)/4^{k} per domain, multiplied by the number of domains, and it is checked against a direct count of repeated windows in a seeded string of the same length — two calculations sharing no code, which is the same arrangement the routing search and the colour count have had since these essays started.

The melting temperatures come from one formula and the composition from the sequence. Each domain’s guanine and cytosine content is counted from the drawn string and put through the same approximation, so the spread in the histogram is a property of the sequence rather than of the formula.

Four seeds, not one. A single sequence’s weakest domain is one draw from a distribution, and the question is about a length rather than about that draw. The lattice comparison is run on four independent seeded sequences and the honeycomb’s twenty-one-base domain fails on all four.

And nothing here is an exhaustive search, which is worth saying because the rest of these essays is nothing else. Listing every shape and searching the survivors is how the routing results were got, and it is what makes them statements about every shape rather than about the ones somebody drew. The calculations here are closed forms over a length, checked against one drawn sequence. They are weaker in kind, and the compensation is that a closed form says what happens at every length rather than at the lengths a search could afford.

Still open: what a scaffold could be chosen for

The scaffold in every design is a viral genome, and the reason is that a long single strand of known sequence is easy to get from a phage and hard to get any other way. Nobody chose its sequence for this purpose, which is the same observation the strand’s length invited and is a sharper one here, because a sequence has more structure to it than a length does.

A scaffold could be designed rather than borrowed, and the criterion is computable. The quantity to minimise is the spread of melting temperatures across the domains a given routing cuts it into — not the mean, which any sequence can be made to hit, but the weakest. A sequence whose composition is uniform over every window of twenty-one bases would have almost no spread, and the thirteen unpaired domains would be none. Whether a sequence can be both uniform at that window length and free of the repeats a synthesiser would introduce is a constraint-satisfaction question of exactly the kind these essays keep meeting, and it has a free parameter nobody has spent: the scaffold does not have to be one molecule’s worth of anything.

The second direction is the one the lattice result opens. The crossover period is fixed by the helix’s pitch and the number of neighbours, and both thresholds have to be rounded up to it — so a lattice’s usefulness depends on where its period’s multiples fall relative to a temperature, which is a criterion nobody applies when choosing between the square and the honeycomb. A lattice with a period of nine or ten would put its second multiple at eighteen or twenty, just above the stability threshold, and would need half as much scaffold between crossovers as the honeycomb’s three periods do. Whether such a lattice can be realised — whether there is an arrangement of neighbours a backbone faces every nine or ten bases — is a question about the helix and not about the design.

A third direction is the one these essays have been circling since the odd honeycomb blocks. Every result about routing has been about the graph, and the sequence has now turned out to constrain the graph: a lattice is not a free choice between two arrangements of neighbours, it is a choice between two granularities of length. Whether the routing results change when the route has to respect where a crossover may sit — a route that steps between helices only at multiples of seven, or of eight — is a question with the same shape as the one local conditions keep failing to answer, and it has not been asked here at all. The graph the six routing essays searched permits a crossing anywhere along a helix.

The habit worth carrying is about requirements that are stated in the wrong order. When two independent conditions bound the same quantity, compute both before deciding which one the design is about. Uniqueness is the requirement everybody names first, it is real, and it is met so far in advance of the other one that it has never once been the reason a staple is the length it is.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

What links here

Every essay whose body links to this one.

The objects this essay names

Each one links to every other essay that touches it.

DNA origamiIdealisationPeriodicityScaffoldSelf assemblyThreshold