Medical
Hypotheses
8: 297-301,
1982
y-RAY INDUCED LUNG ADENOCARCINOMA IN THE MOUSE P. Rosen, Hasbrouck Laboratory, University of Massachusetts Amherst, Massachusetts, 01003, U.S.A. ABSTRACT A theoretical calculation of y-ray induced adenocarcinoma of mouse lung at low dose rates is performed. It is assumed that double strand breaks are the major damage to DNA and whose repair is about 99%, From the data of Ullrich and Storer, comparisons between theory and experiment yield the immunological efficiency which is relatively constant over the dose range O-200 rads is about 50% of that for humans. Key words:
Tumor cell, y-radiation, lung, immune surveillance. INTRODUCTION
It is desirable in a theory of carcinogenesis to compare theoretical calculations with experimental data. I have developed a model of carcinogenesis (1) (2) (3), in which the operator region of the operon controlling mitosis is damaged by a carcinogenic agent and subsequently sustains a mutation. Using a lac operon model the operator is derepressed such that uncontrolled mitosis takes place. An experiment of Ullrich and Storer (4) on mice using y-rays to produce adenocarcinoma of the lung at low dose rate gives a reasonable straight line of incidence vs dose. The high dose rate curve is a mixture of linear and quadratic components. Storer (5) points out that low dose rate should give a linear dose response. I shall consider only the low dose rate. In Table I I have given the data of Ullrich and Storer, I have corrected for zero dose incidence in calculating the average number of tumors per mouse. Because of Poisson's statistics, the fraction, F, of mice with tumors is related to the average number of tumors per mouse !i ITC by: -'ITC F = 1 - (1 - Fo) e (1) where F is the zero dose fraction. The average number of tumors per mouse $&C = zITC + zITC (0) where xITC is only due to radiation. 297
TABLE I Incidence of Adenocarcinoma of the Lung in Mouse vs Dose Incidence %
D(rads) 0 50 100 200
=ITC 0
12.8 14,5 1605 21.4
.02 .04 .lO
DAMAGE TO DNA I am not sure whether base damage or double strand breaks are the important damage here. I will use a process of elimination to show that double strand breaks are the important damage. For mammalian cells repair of t ' base damage is excellent (6)(7). I can look at the mutation frequency per rad per locus of gametes from the graph of Abrahamson et al (8). For mouse, F, (mutation frequency per rad perlocu$ is 2.26 x 1O'70 I can subtract the effect of double strand breaks if I consider (2) that there are about 6.7 x lo-11 such lesions per rad per base pair. The average gene in mammals is 1000 base pairs (9). Also one could expect the probability for mutation per unrepaired double strand break to be of the order of unity since this damage is very severe. Thus the contribution to mutation frequency is .67 x 10-7 for double strand breaks. I am left with a mutation frequency per locus per rad of about 1.67 x lo-7NUMBER OF CELLS AT RISK The number of lung alveoli in the mouse is 2.66 x lo8 (10). There are two kinds of epithelial cells in the lung. Type I epithelial cells are incapable of cell division (ll), The type II cells seem to be the cells at risk and we have about 2 such cells per alveoli (11). Thus the number of cells at risk per mouse, No = 5,32 x 108. CALCULATION OR NlJM@EROF INITIAL TlJl$GR CELLS A mutation caused by base damage would lead to the following number zITC: dNITC
= Nor (1.6 x 10B7) 2
D(l-r)(l-I)
(2)
where L is the length of the operator region, L is an average gene size, D is thg dose, r is the surviving fraction, R is the fraction of damaged base pairs excised and I is the immunological efficiency (1-I) is the fraction of neoplastic cells that escape the immune surveillance. If instead of 100 % repair I assume 99% for the sake of argument there results:
298
=ITc= No(1.6xlO
-9
)m
20
D(l-I)
(3)
For the survival fraction at 200 rads I can take almost unity (12), For 200 rads I can use equation (3) to find (1-I). Taking the result from Table I we have .l.=
(5.32 x 108) (1.6 x lo-'> (4) (1 -I> (1-I) = 3 .03
(4) (5)
or I = 0.97 which is higher than the result for humans. According to Pitot (13), for humans (1-I) = l/25 or .04. I therefore believe that base damage is not the responsible lesion in this case. DOUBLE STRAND BREAKS I am left with the possibility that double strand breaks are responsible for the neoplastic damage. Previously Corry and Cole (14) for hamster cells, stated that for this damage at least 80% or perhaps all are repaired. In a more recent paper Cole et al (15), it is stated that almost complete repair occurs. It has been suggested by Hutchinson (16) that 99% repair is a good figure for mammalian cells. Lehmann and Stevens (17) suggest in humans this damage is completely repaired. They give a value of 1 double strand break per 1,3 x lOlo daltons per Krad. Thus the number of double strand breaks is 5 x lo-"/rad B.P, If these are 99% repaired and have a probability of order unity per unexcised lesion for mutation, the following result for z ITC results: EITC
= 5.32 x lo8 (5x10-11)(10-2)(20)D(l-I)
(6)
Using Table I,1 have for 200 rads: .l = 1.06 (1-I)
(7)
or I '? .92 and (l- I)-l = 10.6. This compares to (1-1)-l = 25 for humans. This is a reasonable result because for mice we have (l-1)-1= 10.6 which is less than for humans. I can repeat the calculations for 50 and 100 rad. For 50 rad (1-1)-l = 13.5 and for 100 rad (1-1)-l = 13,25. These last two figures show that about twice as many neoplastic cells escape in mice as in humans. The results are summarized in Table II.
299
TABLE II Immunological surveillance factor (1-I) D 50 100 200
-1
as a function of D
(1-1)-l 13.5 13.25 10,6 CONCLUSIONS
I have calculated the number of y-ray induced adenocarcinomas in the mouse, assuming the major damage to be that of double strand breaks with an excision repair of 99%. From this calculation I have found the fraction of nascent neoplastic cells not escaping the immune surveillance. For doses from O-200 rads that fraction is about twice the value for humans which is about l/25. REFERENCES 10
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2.
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8.
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9.
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13.
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15.
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301