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I am a graduate student in biochemistry working on cancer. I am currently looking at epidemiology models of the disease. Epidemiologists have developed formulas that predict the frequency of various cancers as a function of age. The disease incidence and mortality rates occurs as a function of $a^4$ to $a^7$. I am reading some papers, but I don't understand what the authors are saying:

1) If the probability of an outcome is indicated by $a^n$, this means that $n+1$ independent events, each occurring randomly and with comparable probability per unit time, must take place before the ultimate outcome (e.g. lethal cancer). I understand that one multiplies the probabilities together to follow $a^n$ (e.g. flipping a coin three times to get a sequence). However, why is it $n +1$ and not simply $n$?

2) The author introduces the log-log plot where the slope of the regression represents $n$. Basically, you take the log of the age (x-axis) and the log of the incidence rate and put a regression line. For example, I have used Canadian statistics data on cancer mortality in men and generated a log-log plot.

enter image description here

So, the slope represents $n$, so there should be 5.5 events that occurred to create a lethal cancer, no? I would greatly appreciate the community's feedback. Thank you!

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  • $\begingroup$ In your plot the x axis is age in years rather than "events" (presumably mutations). You would need to get to either directly measure, or at least get some estimate of "events" per year of age to test the theory. $\endgroup$ – Livid Apr 7 '15 at 22:29
  • $\begingroup$ Sorry for the previous comment, I did not read the description correctly. I do not understand from your description how the exponent can be interpreted as number of events required to create a lethal cancer. Can you share the paper and data source? $\endgroup$ – Livid Apr 8 '15 at 2:12
  • $\begingroup$ As Livid mentioned, could you please provide a reference for what you're pulling this from. I'm a trained epidemiologist, and what you're describing is...odd, and makes me think there's potentially some misunderstanding going on. $\endgroup$ – Fomite Apr 8 '15 at 5:04
  • $\begingroup$ @Fomite I believe it is this: ncbi.nlm.nih.gov/pmc/articles/PMC2007940 There it looks like they calculate the expected number of times a specific sequence of n events had occurred by time t. However, from Armitage and Doll's description, I would think they wanted the probability that sequence occurred at least once, before any others (eg out of sequence mutations kill the burgeoning cancer cell). I'd think that if q=1-p and p1=p2=..pn, then P(CancerCell)=((1-q^t)^n)/n! $\endgroup$ – Livid Apr 8 '15 at 11:52
  • $\begingroup$ @Johnathan Can you provide the data used for that chart? I want to use it to ask a follow up question based on that A & D paper. I still do not think that they calculated the same thing they discuss in the text. Also, "there should be 5.5 events that occurred to create a lethal cancer" is incorrect based on that legend. They interpret the exponent as number of required events. You need to convert from log: log(Incidence)=n*x-log(Constant). So 4.51x-5.51 means Incidence=exp(-5.51)*x^4.51. $\endgroup$ – Livid Apr 10 '15 at 17:54
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Armitage and Doll assumed that some total number of mutations are required for a cancer to develop. It's simplest to explain if the probability of any mutation occurring over a short time interval is low, independent of age, and equal for all cancer-causing mutations. I find it simpler to understand if you drop the Armitage-Doll requirement of a particular required order of mutations, but the relation between the slope of your plot and the number of mutations will be the same in either case.

If only 1 mutation were needed for a cancer, then the incidence rate would be constant over age, equal to the mutation probability. That would be a slope of 0 on your plot.

If 2 mutations are needed, the age-dependent issue becomes the probability that 1 prior mutation occurred before the (age-independent) second mutation. The probability of the cancer developing at any age is then the probability of already having the first mutation (first power of age) times the (age-independent) probability of getting the second mutation. Slope of 1 on your plot. And so on for larger numbers of required mutations: the slope of incidence rate versus age represents the probability of already having all except the last mutation, with the probability of the last mutation independent of age. So a slope of n means that n+1 mutations are ultimately needed.

The value of 5.5 on your plot represents the difference between cancers at young ages and older ages. Childhood and youth cancers typically occur when an individual is already born with a cancer-related mutation, so that fewer additional mutations are required for a cancer to develop. In adults, cancer incidence represents the accumulation of acquired mutations. Armitage and Doll reviewed literature suggesting that analysis should be restricted to individuals over 25 years of age, in which case your plot would have a slope closer to 6.

You should know that the simple interpretation in terms of "mutations" has changed somewhat over the 60 years since that paper appeared. It's probably best to think about cancer-related events rather than mutations in DNA as we understand them today. (The Watson-Crick DNA-structure paper was less than 1 year old when Armitage and Doll wrote.) The nature of these events is nicely explained by Hanahan and Weinberg in a 2011 review.

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  • $\begingroup$ >"not necessarily in any particular order" Then what is that (r-1)! doing in equation 1 in the appendix? After reading that paper I am also confused as to the reason for this "n+1". Actually, I think there are number of points of confusion here due to the way they 1) refer to mutation rates as probabilities, 2) lump everything but time together into a constant, and 3) interpret expected number of events (p*t) as a probability. Can you explain precisely what the number represents that A & D have derived in that paper? $\endgroup$ – Livid Apr 8 '15 at 16:19
  • $\begingroup$ The (r-1)! specifically corrects for the number of different orders in which the (r-1) mutations might have occurred, allowing all orders of occurrence to be treated the same. The text below Equation 1 explains that. What might be confusing you about probabilities in this paper is that they are to be understood as probabilities per unit time. So the probability of an event over a specified period of time t is the product of the probability p per unit time and the elapsed time t. Technically, mutations are being treated as Poisson processes. $\endgroup$ – EdM Apr 8 '15 at 16:39
  • $\begingroup$ Why do they write (page 2): "It will be assumed further that the changes must proceed in a unique order"? Also, if "the probability of an event over a specified period of time t is the product of the probability p per unit time and the elapsed time t", then the "probability" (p*t) will be >1 when t gets large. This is why I say that p*t is not a "probability", it is the expected number of events. It does not make sense to me to calculate this for a single cell (each mutation can only happen once). I am sure I am just confused, thank you for your patience. $\endgroup$ – Livid Apr 8 '15 at 16:56
  • $\begingroup$ Sorry, it had been a while since I read this paper so I had forgotten their assumption of a particular order. You are correct, and I will edit my answer accordingly. (The slope of the relation between log(incidence rate) and log(time) will still be one less than the number of required events, as (r-1)! contributes to the constant term.) An assumption in modeling with a Poisson process is that the probability per unit time is so low that more than one occurrence of the same event over the time course of interest (human life span) is negligible. $\endgroup$ – EdM Apr 8 '15 at 17:20
  • $\begingroup$ "more than one occurrence of the same event over the time course of interest (human life span) is negligible." How would you interpret the intercept? From the A & D paper, I gather it is (p1*p2*...pn)/(n-1)!. From the plot the value is ~exp(-5.51)~.004. If n=7 then (p1*p2*...pn)=.004*(n-1)!=2.88. Doesn't seem right. $\endgroup$ – Livid Apr 8 '15 at 18:06

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