Jim Baen’s Universe
Page 71
The gene pool
One of the most useful conventions in population genetics is the model of a gene pool. We assume that each parent contributes equally to a large pool of gametes. Each offspring is regarded as a random sample of one egg and one sperm from this pool. Each gene is chosen randomly, as if we were drawing different coloured beans from a bag. This oversimplified model building is called “beanbag genetics” and has a surprisingly strong predictive power. In 1964, the great British geneticist, J.B.S. Haldane, wrote an amusing and spirited article “A Defence of Beanbag Genetics” (Persp. Biol. Med., 7: 343-359).
Clearly the beanbag model is a crude representation of nature. A real population has individuals of all ages, some dying, some choosing mates, some giving birth, etc, etc. Keeping track of all this information is not only impractical, but often is neither interesting nor important. Most questions of genetic or evolutionary interest do not require minute details. The gene pool model will give us the same kind of insight that simple models in the physical sciences do, and often with predictions that are sufficiently accurate for most uses. Further it is important to be aware of that departure from the simple models may usually be treated with appropriate modifications.
Suppose now that a population consist of individuals with an extremely long life span. In the tradition of “beanbag genetics” the population is homozygous at the locus determining life span; i.e. all genotypes are AA and code for long-life. However, in any population there will be some random death rate basically determined from outer sources like biological (virus, sickness, predation or accidents due, for example, to nature catastrophes). Thus, rather living forever, the long-lived strategy has to tolerate the operation of a random death rate; that is, the survival rate, S, is close to 1. This strategy is called evolutionary stable if the population rejects all kinds of invading alleles that reduce the life span. In order to investigate this stability, we assume that a small fraction e of the gene pool consists of an allele ‘ a’ such that the genotypes Aa induces a shorter life span via changes in the birth rates and survival rates. We must find the requirement that the new allele will increase in the gene pool.
A Hardy- Weinberg type model for analysing the evolutionary stability of infinite long-life
Since we are only interested in the general principles that govern the stability of the long-life strategy we simplify the population dynamics of the mutant allele as follows: On average they live t years and reproduces bAa offspring of which lAa survives the next t years. The probability of surviving t years for the long-lived individuals is L = St = (S multiplied by itself t times). Finally, we ignore the homozygous mutant allele (of genotype aa), since their abundance is of the order e(e in the gene pool.
Thus the initial genotype frequencies are (1-e) of AA and e of Aa where the abundant first group of individuals (AA) have a very long life span and the second potential invader group (Aa) only live t years. In the gene pool the frequency of A is (1-e)*1 + e*(1/2) = 1-e/2 while the frequency of a is q = e-2
We shall now try to answer the question: Under what circumstances is the fraction of the allele a expected to increase in the gene pool?
In order to solve this problem we need another assumption: The population size is so large that the mating may be considered as a random mixture of the genes. We must consider all kind of matings between the genotypes:
1. Consider first a long-lived female (AA) that mates with a long-lived male (AA). During t years she produces bAA offspring (of genoype AA) that have a chance of lAA to survive the next t years. In the population the probability of such matings is (1-e)*(1-e).
2. Consider next a long-lived female (AA) that mates with a mutant short-lived male (Aa). During t years she produces bAA offspring of which 50% are long-lived (AA) with a chance of lAA to survive the next t years, and 50% are short-lived (Aa) with a chance of lAa to survive the next t years. In the population the probability of such matings is (1-e)*e.
3. Then consider a mutant short-lived female (Aa) that mates with a long-lived male (AA). During t years she produces bAa offspring of which 50% are long-lived (AA) with a chance of lAA to survive the next t years, and 50% are short-lived (Aa) with a chance of lAa to survive the next t years. In the population, the probability of such matings is e*(1-e).
4. Finally consider a mutant short-lived female (Aa) that mates with a mutant short-lived male (Aa). During t years she produces bAa offspring of which 25% are long-lived (AA) with a chance of lAA to survive the next t years, and 50% are short-lived (Aa) with a chance of lAa to survive the next t years. In the population the probability of such matings is e*e.
After t years the number of long-lived and short-lived recruits are respectively:
WAA = (1-e)2 bAA lAA + (1-e)*e(1/2)bAA lAA + e*(1-e)(1/2)bAa lAA + e*e(1/4) bAa lAA
WAa = (1-e)*e(1/2) bAA lAa + e*(1-e)e/2) bAa lAa + e*e(1/2) bAa lAa
Now, the number of adult long-lived individuals surviving a period of t years will be (1-e)N*S where N is the total population size. In order to keep the population at a stable size of N individuals, we have to introduce a common death rate m for the recruits of all genotypes:
(1- m)*N* WAA + (1-m)*N* WAa + (1-e)N*L = N
After t years the new fraction of the short-lived allele a in the gene pool is
qt =
Thus we see that the fraction of the allele a will increase in the gene pool if
,i.e.
which is equivalent to
( bAa lAa+(1-e) bAA lAa)» (1- m) WAA+2(1-e) L
Reordering the terms, finally gives the requirement that the mutant allele coding for short-lived may invade the gene pool:
Thus, if the mutant allele a codes for a birth rate bAa and survival rate lAa such that this inequality is satisfied it will invade the gene pool. But what does this inequality essentially tell us? In order to see the essential requirement for invasion we may use standard techniques in numerical analysis. First, since e is a small quantity in the order of 1100, we may use the standard approximations in the final expressions: 1 - e 1 and 1 + e + e2 1 so the inequality is practically the same as
Then we take a closer look at the extra non-specific mortality rate m that was introduced in order to keep the population at its equilibrium value N. In order to understand the magnitude of this quantity we consider the population without the mutant allele a (that is all genotypes are AA), so the equilibrium equation reduces to N* WAA + N* L = N which is seen to imply:
Eq
uilibrium condition: WAA + L = 1
The equilibrium equation simply states that in order to stabilise the population, the recruitment to the stock must be balanced with the death rates. Now according to classical Darwinian thinking, successful new mutants only accomplish small changes. In our case, this means that the new allele a will only slightly modify the recruitment, so the final adjustment of the new short-lived recruits will be small. Hence WAA + WAA + L = 1 so we finally obtain the following essential criteria for successful invasion:
Criteria for successful invasion of the gene pool: lAa» 1
In words this criteria simply says that the new allele a will successfully invade the gene pool if the number of surviving mutant offspring to adulthood exceeds 1. Note that offspring of genotype Aa is created in two different ways: Either the mother is of genotype AA and mates with a male of genotype Aa in which case 50% of the offspring will be Aa, or the mother is of genotype Aa in which case also 50% of the offspring will be Aa. This explains why it is the average litter size of the two genotypes that enters the invasion criteria. Now the expected number of mutant offspring that reaches adulthood is found by multiplying with their survival rate lAa. Since the recruitment of the long-lived genotypes exactly matches the adult mortality 1 - L, the total recruitment to the adult group is exactly 1 (i.e. the equilibrium condition). Note that this is precisely where the long-life enters the equations: the total recruitment is the recruitment of offspring that manage to live up to adulthood PLUS the number of adults surviving the period (as mentioned above, only non-genetic accidents cause a small death rate among the long-lived individuals). So the criteria for successful invasion of the gene pool may be restated in words as follows:
Criteria for successful invasion of the gene pool:
The recruitment of the short-lived individuals must exceed 1
Note that this is good classical Darwinism: The genotype with the highest recruitment will at the end fill up the whole gene pool.
It is now easy to see why Jim Baens hypothesis is supported by this genetic model. The whole issue about why we must die may be formulated the question of whether a long-lived strategy is evolutionary stable. A simple, first approximation, genetic model says that the long-lived strategy will only be evolutionary stable if it is difficult for mutant alternatives to establish a recruitment larger than 1. But as Jim Baen has pointed on classical biological knowledge: there are all sorts of difficulties with inbreeding. Therefore almost any alternative strategy that reduces energy devoted to long-life and put it into clutch size or vitality of offspring will have a better total recruitment than the long-lived. Therefore, the strategy of eternal life is evolutionary unstable. For human beings this has the consequence that 125 years is the maximum age.
Cohort analysis of the evolutionary stability of infinite long-life
In this section we derive the same result in an alternative model that takes all the year-classes (cohorts) into account.
Suppose there are M adult long-lived individuals consisting of 1000 year classes, and let the clutch size per individual be bA, the survival up to maturity is lm and the survival rate of adults is sA. At equilibrium MbA offspring are produced each year by 1000 year classes:
bA[ MbAlA + MbAlAsA1 + MbAlAsA2 + MbAlAsA3 + + MbAlAsA999] = MbA which is seen to reduce to
Equilibrium condition bAlA = 1 which practically states that bAlA = 1 - sA
i.e. the population will be in equilibrium if the recruitment to the adult population (i.e. birth rate multiplied by immature survival) balances the adult death rate.
Consider now a potential invading allele a that codes for living only 100 years as adults, i.e. 10% of the prevailing strategy of 1000 years. Initially there are x0 = e M individuals of these new genotypes having a birth rate of ba, a survivorship of the immature period equal la and an adult survivorship of sa. With these parameters the iteration equation for the adult mutants are
xl+1 = saxx + ( xl-tm) la = saxl + laxl+1- tm
Substituting a general solution of the form xx = rl x0 gives the characteristic equation for the growth of the mutant rtm = sartm-1 + la
In order to see where the root lies, define the two functions f ( r) = rtm and g( r) = sartm-1 + la and notice that f (0) = 0 «la = g(0), but since tm» tm-1, f( r) will eventually become greater than g( r). Thus somewhere they intercept, and the crucial question is whether they intercept to the left of r = 1, in which case the number of invaders will decline, or to the right of r = 1, in which case the number of short lived individuals will increase. Since f (1) = 1 and f (1) = sa + la it follows that the condition for interception to the right of r = 1, i.e. f(1) «g(1), is 1 «sa + la which finally gives
Criteria for successful invasion of the gene pool: la» 1 - sa
In words: The recruitment to the mutant short-lived individuals (=average birth rate of long- and short lived genotypes multiplied with the survival probability up to maturity) must be larger than the adult death rate of the short-lived individuals. Note that this condition is equivalent to that found in our previous Hardy-Weinberg type model, so the same discussion applies.
Let us finally see that if the following approximation of the population dynamics of the mutant xx = ( sa + la )x e M is substituted into the formula of the Hardy-Weinberg model, we arrive at the same criteria for e M mutant to invade a homozygous population of (1-e) M long-lived adults:
where the second term in the denominator is the number of long-lived adult that survived the total life span of the short lived (i.e. 100 years) and the third term is the number of adult long-lived that recruits to the adult population of long lived (since these are prevailing we ignore the contribution of long-lived from the short-lived). Using the equilibrium criterion, bAlA = 1 - sA, we obtain (1 - e)( sa + la )100 «(l - e) which is equivalent to sa + la» l i.e. la» l - sa
Thus, we arrive at precisely the same inequality, demonstrating an inner consistency in our models.
A competition model with trade-off between reproduction and survival.
While the two previous models started with the assumption that long-life was an evolutionary stable strategy, we now develop a model where the different life history strategies are in direct competition with each other. However, if a new mutant code for better survival we shall now explicitly regard this as a strategy to allocate more energy to survival at the cost of energy devoted to reproduction. The simplest equation for such a trade-off mo
del is bl = b( s) = (1 - s)l where l is a positive real number.
A first approximation to the population growth will then be x1/2+| = [ s + (l - s)l] x1/2
which has the solution x1/2 = [ s + (l - s)l]1/2 x0
From this we see that natural selection will favour a strategy which maximises the function f ( S) = S + (l - S)l
Putting the derivative equal to zero, f ( S) = l - l (l - S)l-1 = 0, gives the optimum survival rate as s* = 1 - () implying an evolutionary stable expected life span of T* =
So also in this model will the long-lived be selected against because they do not provide the optimal combination of birth rate and survival rate. Another consequence of tremendous importance is that given a strategy near the optimal, there will be no selection against deleterious genes at ages over T*. This allows death genes, racing-car analogy construction of organs, etc. Now human beings have evolved from species where these mechanisms already have been operating, so this is an evolutionary blind route with respect to long-life unless it is possible to enter our genes directly and reconstruct them. We are an animal that has evolved over many millions of years to be short lived.
James Patrick “Jim” Baen (b. 1943) is a noted science fiction publisher and editor. He started his publishing career, appropriately enough, in the complaints department of Ace Books after stints in the Army, at CCNY, and in Greenwich Village in the 1960s working as the manager of a folk music coffee shop (a “basket house”). After a rapid promotion to the editorial department at ACE, he moved to take Judy-Lynn del Rey’s place at Galaxy Science Fiction, and succeeded Ejler Jakobsson as editor of Galaxy and If in 1974 after a brief trip back to Ace to be assistant Gothics editor. While at Galaxy he published such authors as Jerry Pournelle, Charles Sheffield, Joanna Russ and John Varley, and was nominated for several Hugo Awards. He returned to Ace to head their science fiction line in 1978, working with publisher Tom Doherty. When Doherty left to start Tor in 1980, Baen shortly followed and started the SF line there. In 1984, he had the opportunity to start his own independent company, Baen Books, now a noted SF line. Baen Books has established a large readership, publishing books by authors such as David Weber, John Ringo, Eric Flint, David Drake and Lois McMaster Bujold.