652
E DNA Splitting by Radiation
while, if M >> m, Eq. (E.17) gives
r 2 t =
1
2
ρvD
ε 2
2
.
(E.21)
Since, under the second scenario, two or more energy deposits are needed for
cleavage, the average ε 2 should be about half of ε 1 . The expressions for rt, given
in Eqs. (E.16) and (E.17), determine the dependence of the DNA fragmentation
numbers, N l (Eqs. (E.9)–(E.10)), on radiation dose. In terms of the frequency of
cleavage α (even if ρvD/ε is not small),
α 1 = 1 − e
−ρvD/ε 1
and
α 2 = 1 − e
−ρvD/ε 2 (1 + ρvD/ε 2 ).
(E.22)
If ρvD/ε << 1 (low doses), then α 1 ∼ ρvD/ε 1 and α 2 ∼ (ρvD/ε 2 ) 2 /2. The ratio
α/D will be constant with dose for a given sample only for small doses and only in
the case of single-event cleavages.
To appreciate the size of these numbers, let us estimate the magnitude of the
expression in Eq. (E.20). Suppose DNA molecules are in a water solution and are
given a dose of 100 Gy. The density ρ is approximately that of water, 10 3 kg/m
3 .
Note that the volume v is not the size of the primary and possible secondary
ionization volumes covering an ionization track, but rather the volume surrounding
a DNA site which, if sufficient energy is deposited within, causes cleavage. If we
take the radius of the interaction volume to be 5 nm, then v will be approximately
10 −25 m 3 . Now take the effective energy ε needed in the volume v to cause a doublestrand break to be 25 eV. (The threshold energy has been measured to be about
8 eV for photons, with 20–30 eV needed for electrons.) Then r 1 t ∼ 0.01. With the
cleavage number rt much less than one, rt will be close to the average fraction of
cuts α. Even so, the number of possible cleavage sites n along a DNA molecule
can be much larger than 10 2 , so that the exponents in Eqs. (E.5) must be used for
the large fragments (l ∼ n), rather than their small rt approximation. Measured
values for gamma ray irradiation of mammalian DNA give rather small values for α.
Measurement gives α/D in the range (6 ± 2) × 10 −9 double-strand breaks per Gray
per base-pair. With an effective energy deposit of 8 eV, the radius of the interaction
volume if breakage had been dominated by a single-hit would be under 0.2 nm while
a double-hit breakage at 100 Gy would give an interaction radius near 2 nm.
E DNA Splitting by Radiation
while, if M >> m, Eq. (E.17) gives
r 2 t =
1
2
ρvD
ε 2
2
.
(E.21)
Since, under the second scenario, two or more energy deposits are needed for
cleavage, the average ε 2 should be about half of ε 1 . The expressions for rt, given
in Eqs. (E.16) and (E.17), determine the dependence of the DNA fragmentation
numbers, N l (Eqs. (E.9)–(E.10)), on radiation dose. In terms of the frequency of
cleavage α (even if ρvD/ε is not small),
α 1 = 1 − e
−ρvD/ε 1
and
α 2 = 1 − e
−ρvD/ε 2 (1 + ρvD/ε 2 ).
(E.22)
If ρvD/ε << 1 (low doses), then α 1 ∼ ρvD/ε 1 and α 2 ∼ (ρvD/ε 2 ) 2 /2. The ratio
α/D will be constant with dose for a given sample only for small doses and only in
the case of single-event cleavages.
To appreciate the size of these numbers, let us estimate the magnitude of the
expression in Eq. (E.20). Suppose DNA molecules are in a water solution and are
given a dose of 100 Gy. The density ρ is approximately that of water, 10 3 kg/m
3 .
Note that the volume v is not the size of the primary and possible secondary
ionization volumes covering an ionization track, but rather the volume surrounding
a DNA site which, if sufficient energy is deposited within, causes cleavage. If we
take the radius of the interaction volume to be 5 nm, then v will be approximately
10 −25 m 3 . Now take the effective energy ε needed in the volume v to cause a doublestrand break to be 25 eV. (The threshold energy has been measured to be about
8 eV for photons, with 20–30 eV needed for electrons.) Then r 1 t ∼ 0.01. With the
cleavage number rt much less than one, rt will be close to the average fraction of
cuts α. Even so, the number of possible cleavage sites n along a DNA molecule
can be much larger than 10 2 , so that the exponents in Eqs. (E.5) must be used for
the large fragments (l ∼ n), rather than their small rt approximation. Measured
values for gamma ray irradiation of mammalian DNA give rather small values for α.
Measurement gives α/D in the range (6 ± 2) × 10 −9 double-strand breaks per Gray
per base-pair. With an effective energy deposit of 8 eV, the radius of the interaction
volume if breakage had been dominated by a single-hit would be under 0.2 nm while
a double-hit breakage at 100 Gy would give an interaction radius near 2 nm.