Topology in Molecular Biology

5 From Tangle Fractions to DNA 107

there arep′,q′,r′such that for somen(or a range of values ofnto be specified
below)Kn=N((p′+q′r′n)/q′).
Ifn= 0 then we haveN(p/q)=N(p′/q′).Hence by the classification
theorem we know thatp=p′and thatqandq′are arithmetically related.
Note that the same argument shows that if the equality holds for any two
consecutive values ofn,thenp=p′.We shall assume henceforth thatp=p′.
With this assumption in place, we see that if the equality holds for anyn=0
thenqr=q′r′.Hence we shall assume this as well from now on.
If|p+qrn|is sufficiently large, then the congruences for the arithmetical
relation ofqandq′must beequalities over the integers. Sinceqq′=1over
the integers can hold only ifq=q′ =1or−1 we see that it must be the
case thatq=q′if the equality is to hold for sufficiently largen. From this
and the equationqr=q′r′it follows thatr=r′.It remains to determine a
bound onn. In order to be sure that|p+qrn|is sufficiently large, we need
that|qq′|≤|p+qrn|.Sinceq′r′=qr, we also know that|q′|≤|qr|.Hencen
is sufficiently large if|q^2 r|≤|p+qrn|.
Ifqr >0 then, sincep> 0 ,we are asking that|q^2 r|≤p+qrn.Hence

n≥(|q^2 r|−p)/(qr)=|q|−(p/qr).

Ifqr <0 then fornlarge we will have|p+qrn|=−p−qrn.Thus we want
to solve|q^2 r|≤−p−qrn, whence

n≥(|q^2 r|+p)/(−qr)=|q|−(p/qr).

Since these two cases exhaust the range of possibilities, this completes the
proof of the theorem.

Here is a special case of Theorem 7. (See Fig. 5.33.) Suppose that we were
given a sequence of knots and linksKnsuch that

Kn=N

(

1

[−3]

+ [1] + [1] +...+ [1]

)

=N

(

1

[−3]

+n[1]

)

.

We haveF(1/[−3] +n[1]) = (3n−1)/3 and we shall writeKn=N([(3n−
1)/3]).We are told that each of these rational knots is in fact the numerator
closure of a rational tangle denoted

[p/q]+n[r]

for some rational numberp/qand some integerr.That is, we are told that
they come from a DNA knitting machine that is using rational tangle patterns.
But we only know the knots and the fact that they are indeed the closures
forp/q=− 1 /3andr=1.By this analysis, the uniqueness is implied by the
knots and links{K 1 ,K 2 ,K 3 ,K 4 }.This means that a DNA knitting machine
Kn=N(S+nR) that emits the four specific knotsKn=N([(3n−1)/3]) for
n=1, 2 , 3 ,4 must be of the formS=1/[−3] andR= [1]. It was in this way
(with a finite number of observations) that the structure of recombination in
Tn 3 resolvase was determined [37].