Quiz Question 5
Radiation Damage in DNA

Background

We begin with a simplified view of the structure of a DNA molecule. For our purposes, a DNA molecule may be viewed as a ladder. The two strands of the DNA molecule are the sides of the ladder. Every point at which a rung joins a side is one of the now-famous DNA nucleotides: Adenine (A), Guanine (G), Thymine (T), or Cytosine (C). (The sequencing of A, G, T, and C determine the genetic information encoded in the DNA molecule.) The side segments between nucleotides are relatively strong phosphodiester bonds between nucleotides (P-E bonds); the rungs represent relatively weak hydrogen bonds.

Radiation can produce a break in a strand by destroying a P-E bond. Think of radiation as blasting away a bond in one side of the ladder. We will call such damage a single-strand break (SSB). While relatively weak, the hydrogen bonds between nucleotides cannot be permanently broken by such a radiation hit. An isolated SSB also does no permanent damage to the molecule; it is soon repaired.

Significant damage to a DNA molecule occurs when SSBs in opposite strands occur sufficiently near to each other. If two SSBs occur directly opposite, the molecule is immediately and cleanly broken. If SSBs are not directly opposite, but still quite close, the hydrogen bonds (rungs) will not be strong enough to hold the molecule together and it will tear apart. Either sort of destruction is called a double-strand break (DSB).

Thus, if SSBs in opposite strands are sufficiently far apart, the several intervening hydrogen bonds will be strong enough to hold the molecule together. How far is "sufficiently" far? The central question here is to find the maximum number h of hydrogen bonds that is not enough to hold a molecule together.

How Radiation Destroys DNA -- A Diagram

The diagram below illustrates the situation. Vertical lines are P-E bonds and horizontal lines are hydrogen bonds. Nucleotides (not shown) are located at vertices where vertical and horizontal lines meet. Asterisks * are radiation hits that destroy P-E bonds. Two curved segments at the right show repaired P-E bonds.

           A MECHANISM OF RADIATION DAMAGE IN DNA

               (How Single-strand Breaks Can 
               Produce Double-strand Breaks)

Damage                                               Result

 ...                                                   ...
 |__|                                                 |__|
 |__|                                                 |__|
 |__|                                                 |__|
 |__|                                                 |__|
 |__|                                                 |__|
 |__|                                                 |__|
 |__|       Double-strand break occurs at once        |__|
 *__*  <--- when single-strand breaks * occur   --->   __   
 |__|       directly opposite one another.            |__|
 |__|                                                 |__|
 |__|                                                 |__|
 |__|                                                 |__|
 |__|                                                 |__|
 |__|                                                 |__|
 *__|  <--- If h = 3, then a double-strand break         |
 |__|       would occur if single-strand breaks      \   |
 |__|       in opposite strands occur within 3  --->  |   \
 |__*  <--- or fewer steps of each other.             |   
 |__|                                                 |__|
 |__|                                                 |__|
 |__|                                                 |__|
 |__|                                                 |__|
 |__*  <---                                     --->  |__( 
 |__|       If h = 3, then no double strand break     |__|
 |__|       will result when single-strand breaks     |__|
 |__|       in opposite strands occur more than 3     |__|
 |__|       steps apart. In this case single-strand   |__|
 *__|  <--- breaks are soon repaired.           --->  )__|
 |__|                                                 |__|
 |__|                                                 |__|
 |__|                                                 |__|
 |__|                                                 |__|
 |__|                                                 |__|
 ...                                                   ...

Experimental Results

The consequence of DSBs is that the free-floating pieces of broken DNA molecules can link together to produce mutant DNA. The number h is important because it determines the stability of DNA molecules. Small values of h would mean that it is relatively difficult for radiation to damage DNA: SSBs in opposite strands need to be very near to produce a DSB. Large values of h would mean that it is easy for radiation to destroy a DNA molecule.

In a particular experimental situation, a sample of DNA was exposed to x-rays, and the following laboratory measurements were made.

• The length of each undamaged DNA molecule (number of P-E bonds per strand) is L = 30,000.
• The surviving fraction F of DNA molecules (i.e., the proportion not sustaining any DSBs) is 0.88. This number was determined by centrifuging the irradiated sample of molecules. (Intact molecules are heaver and are forced to the bottom of the contrifuge tube.)
• The average number b of breaks per strand is 25. This number was determined by chemically breaking all hydrogen bonds. The resulting suspension of single strands (and fragments thereof resulting from SSBs) was centrifuged to find b. (Centrifuging reveals the distribution of fragment sizes, from which we can find the average size, hence the average number of breaks.)

Our purpose here is to find an equation that relates L, F, b, and h; and then to solve for h.

Assumptions

Almost all real-life probability modeling involves assumptions. Often the most difficult phase of research is deciding which assumptions can safely be made. To help you solve the problems below, here is some guidance about assumptions.

A. Simplifying assumptions that we believe to be true. Previous work with x-radiation and DNA supports the following assumptions:

• Radiation creates SSBs completely at random in DNA molecules, not depending on type of nucleotides in the vicinity (A, T, G, or C), not depending on the presence of SSBs nearby, and independent of strand and position within strand.
• The value of h depends on temperature and other factors. Theory and some previous experimentation lead us to believe that in our experimental conditions h is probably less than 4 and certainly less than 6.

B. Simplifying assumptions that we know to be false, but we hope won't matter. We will make the following additional assumptions for purposes of the questions below.

• "End effects" do not exist. The mechanism by which SSBs produce DSBs is probably different very near the end of a molecule, but molecules are so long that this does not matter.
• "Cluster effects" do not exit. Clusters of many nearby SSBs may complicate the mechanism by which SSBs produce DSBs, but the average number of SSBs is so small compared to the length of the molecule that such clusters are too rare to worry about. We can assume the SSBs in any one strand are "widely" separated (several multiples of h apart).

1. What probability distribution governs the total number of SSBs in a randomly chosen molecule. Given the experimental results above, what fraction of molecules have within 10% of the average number of SSBs?

2. Given that there are exactly N SSBs in a molecule, what probability distribution governs the number of SSBs in the "left-hand" strand. If N = 50, what fraction of the molecules have between 20 and 30 breaks (inclusive) in the "left-hand" strand?

3. An "average" molecule is one of length L that sustains exactly 2b SSBs, of which exactly b are in each strand. What is the approximate probability that such an average molecule will survive (have no DSBs). [Everything else being equal, F must decrease as h increases, because large values of h mean a more fragile molecule.]

4. Assuming that all molecules are "average," use the result of Problem 3 and the experimental results to give an approximate numerical value for h.

5. (Advanced; conceptually, but not mathematically) Do not assume that all molecules are average. Find an approximate equation involving L, F, b, and h. Using known values of L, F, and b, how could it be solved for h?

Answers

1. Under assumption A-1 it is reasonable to suppose that the number of SSBs in a randomly chosen molecule has a Poisson distribution. The data show the average number of SSBs per molecule to be 2(25) = 50. Printed cumulative tables of the Poisson distribution do not usually include such a large mean. The normal approximation (with continuity correction gives the probability of getting between 45 and 55 SSBs (inclusive) as 0.563. This agrees with the exact answer obtained from Minitab to three decimal places.

2. Similarly, it is reasonable to suppose that the distribution of SSBs between the two strands is binomial with p = 1/2. For N = 50, the normal approximation gives the requested probability as 0.88. The exact value is 0.881.

3. Imagine that the b SSBs in the left strand have been placed at random. If the molecule is to survive the b SSBs must be placed into the right strand in such a way as to avoid "forbidden regions" -- each of size 2h+1 opposite each break in the left strand. According to assumption B-2 we will ignore possible overlapping of forbidden regions, so that a total of R = b(2h + 1) positions must be avoided.

The probability that the first SSB in the right strand avoids this region (L - R)/L; the probability that the second goes into another legal position is (L-R-1)/(L-1); and so on. The probability of survival is the product of these b ratios. Actually, all b ratios are approximately equal because the adjustments in the numerator and denominator as positions are "used up" are quite small compared to L.

Thus the probability of survival of an average molecule is very nearly F = {[L - b(2h + 1)]/L}^b.

4. Solving the equation just above for h we obtain h = (L/2b)(1 - F^1/b) - 0.5. For our experimental data, this gives h = 2.56.

In the actual experiment, 9 runs gave an average value of h = 2.64. These results are for room temperature and high ionic strength (i.e., a "salty" solution). Values of h are somewhat higher at low ionic strength or at higher temperatures.

5. Taking the Poisson and binomial distributions into account we could proceed as follows:

F is the sum [for N = 0 to 60,000 -- essentially infinity] of P(survival and N SSBs) = P(N SSBs)P(survival | N SSBs), where P(N SSBs) = e^-2b(2b)^N/N!.

In turn, P(survival | N SSBs) is the sum [for n = 0 to N] of P(survival and n of N SSBs in left strand) = P(n in left)P(survival | n of N in left), where P(n in left) is (.5)^NN!/[n!(N-n)!].

Finally, P(survival | n of N in left) is the product [for i = 1 to N-n] of (L-R-i+1)/(L-i+1), where R = n(2h+1). As in the answer to Problem 3, it does little harm to approximate the product as [(L - R)/L]^N-n.

All of these parts can be "telescoped" together into one complex expression with two summations and a product. The result can be solved for h by numerical methods given experimentally determined values of L, F, and b.

• The assumption that forbidden regions do not overlap can be removed by using difference equation methods. This process given an expression of the form (2h + 1)(n - S), where S is an infinite series of terms with small numerators and increasing powers of L in the denominator. Thus this assumption underestimates h by only a negligible amount.
• The assumption in Problems 3 and 4 that all molecules are average slightly underestimates h. The main source of error is to claim that SSBs are evenly divided between strands, because this split is most likely to result in DSBs. (At the extreme, if all SSBs were in the same strand, the molecule would be safe.) However, a solution of the much more complicated expression in Problem 5 by numerical methods shows that the error is much less than the measurement errors in the experiment.
• The assumption that breaks are equally likely regardless of the configuration of A, T, G, and C turns out to be quite harmless -- even in the unlikely event that it is wrong. For example, if P-E bonds between A and T were impervious to radiation damage, the only effect on our equations would be to decrease L and R in equal proportions, with no change in the numerical results.
• Radiation can also damage DNA when a single massive and strongly ionized particle produces a DSB by blasting the molecule asunder. Chemical and analytical tricks made it possible to disregard this mechanism, which produces something like 10% of the DSBs. The numbers we used for F (the surviving fraction) are adjusted to reflect only the damage that occurs by pairing of nearby SSBs in opposite strands.

• This research was originally reported in "Matching of single-strand breaks to form double-strand breaks in DNA" by David Freifelder and Bruce Trumbo, Biopolymers,7 (1969), pp 681-693. For consumption by biophysicists, it assumed that all molecules are average and gave essentially the results in Problems 3 and 4. Many subsequent citations of this article (mostly just quoting the findings) can be found in the Science Citation Index.
• Trumbo wrote the chapter "An approximate probability model for studying radiation damage in DNA" in Statistics by Example, edited by Fred Mosteller, published by Prentice-Hall (1973). For an elementary statistics audience, this chapter dealt with a bit more of the probability theory, but in a somewhat different way than is presented here.