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Register a new userSexual reproduction from the male (men) point of view
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To counterbalance the views presented here by Suzana Moss de Oliveira, we explain here the truth: How men are oppressed by Mother Nature, who may have made an error inventing us, and by living women, who could get rid of most of us. Why do women live longer than us? Why is the Y chromosome for men so small? What are the dangers of marital fidelity? In an appendix we mention the demographic challenges of the future with many old and few young people.
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The question as to why most higher organisms reproduce sexually has remained open despite extensive research, and has been called the queen of problems in evolutionary biology. Theories dating back to Weismann have suggested that the key must lie in the creation of increased variability in offspring, causing enhanced response to selection. Rigorously quantifying the effects of assorted mechanisms which might lead to such increased variability, and establishing that these beneficial effects outweigh the immediate costs of sexual reproduction has, however, proved problematic. Here we introduce an approach which does not focus on particular mechanisms influencing factors such as the fixation of beneficial mutants or the ability of populations to deal with deleterious mutations, but rather tracks the entire distribution of a population of genotypes as it moves across vast fitness landscapes. In this setting simulations now show sex robustly outperforming asex across a broad spectrum of finite or infinite population models. Concentrating on the additive infinite populations model, we are able to give a rigorous mathematical proof establishing that sexual reproduction acts as a more efficient optimiser of mean fitness, thereby solving the problem for this model. Some of the key features of this analysis carry through to the finite populations case.
This paper develops a simplified set of models describing asexual and sexual replication in unicel- lular diploid organisms. The models assume organisms whose genomes consist of two chromosomes, where each chromosome is assumed to be functional if it is equal to some master sequence $ sigma_0 $, and non-functional otherwise. The first-order growth rate constant, or fitness, of an organism, is determined by whether it has zero, one, or two functional chromosomes in its genome. For a population replicating asexually, a given cell replicates both of its chromosomes, and splits its genetic material evenly between the two cells. For a population replicating sexually, a given cell first divides into two haploids, which enter a haploid pool, fuse into diploids, and then divide via the normal mitotic process. Haploid fusion is modeled as a second-order rate process. When the cost for sex is small, as measured by the ratio of the characteristic haploid fusion time to the characteristic growth time, we find that sexual replication with random haploid fusion leads to a greater mean fitness for the population than a purely asexual strategy. However, independently of the cost for sex, we find that sexual replication with a selective mating strategy leads to a higher mean fitness than the random mating strategy. This result is based on the assumption that a selective mating strategy does not have any additional time or energy costs over the random mating strategy, an assumption that is discussed in the paper. The results of this paper are consistent with previous studies suggesting that sex is favored at intermediate mutation rates, for slowly replicating organisms, and at high population densities.
This paper develops mathematical models describing the evolutionary dynamics of both asexually and sexually reproducing populations of diploid unicellular organisms. We consider two forms of genome organization. In one case, we assume that the genome consists of two multi-gene chromosomes, while in the second case we assume that each gene defines a separate chromosome. If the organism has $ l $ homologous pairs that lack a functional copy of the given gene, then the fitness of the organism is $ kappa_l $. The $ kappa_l $ are assumed to be monotonically decreasing, so that $ kappa_0 = 1 > kappa_1 > kappa_2 > ... > kappa_{infty} = 0 $. For nearly all of the reproduction strategies we consider, we find, in the limit of large $ N $, that the mean fitness at mutation-selection balance is $ max{2 e^{-mu} - 1, 0} $, where $ N $ is the number of genes in the haploid set of the genome, $ epsilon $ is the probability that a given DNA template strand of a given gene produces a mutated daughter during replication, and $ mu = N epsilon $. The only exception is the sexual reproduction pathway for the multi-chromosomed genome. Assuming a multiplicative fitness landscape where $ kappa_l = alpha^{l} $ for $ alpha in (0, 1) $, this strategy is found to have a mean fitness that exceeds the mean fitness of all of the other strategies. Furthermore, while the other reproduction strategies experience a total loss of viability due to the steady accumulation of deleterious mutations once $ mu $ exceeds $ ln 2 $, no such transition occurs in the sexual pathway. The results of this paper demonstrate a selective advantage for sexual reproduction with fewer and much less restrictive assumptions than previous work.
We study the genetic behaviour of a population formed by haploid individuals which reproduce asexually. The genetic information for each individual is stored along a bit-string (or chromosome) with L bits, where 0-bits represent the wild-type allele and 1-bits correspond to harmful mutations. Each newborn inherits this chromosome from its parent with some few random mutations: on average a fixed number m of bits are flipped. Selection is implemented according to the number N of 1-bits counted along the individuals chromosome: the smaller N the higher the probability an individual has to survive a new time step. Such a population evolves, with births and deaths, and its genetic distribution becomes stabilised after many enough generations have passed. The question we pose concerns the procedure of increasing L. The aim is to get the same distribution of relative genetic loads N/L among the equilibrated population, in spite of a larger L. Should we keep the same mutation rate m/L for different values of L? The answer is yes, which intuitively seems to be plausible. However, this conclusion is not trivial, according to our simulational results: the question involves also the population size.
This paper studies the mutation-selection balance in three simplified replication models. The first model considers a population of organisms replicating via the production of asexual spores. The second model considers a sexually replicating population that produces identical gametes. The third model considers a sexually replicating population that produces distinct sperm and egg gametes. All models assume diploid organisms whose genomes consist of two chromosomes, each of which is taken to be functional if equal to some master sequence, and defective otherwise. In the asexual population, the asexual diploid spores develop directly into adult organisms. In the sexual populations, the haploid gametes enter a haploid pool, where they may fuse with other haploids. The resulting immature diploid organisms then proceed to develop into mature organisms. Based on an analysis of all three models, we find that, as organism size increases, a sexually replicating population can only outcompete an asexually replicating population if the adult organisms produce distinct sperm and egg gametes. A sexual replication strategy that is based on the production of large numbers of sperm cells to fertilize a small number of eggs is found to be necessary in order to maintain a sufficiently low cost for sex for the strategy to be selected for over a purely asexual strategy. We discuss the usefulness of this model in understanding the evolution and maintenance of sexual replication as the preferred replication strategy in complex, multicellular organisms.
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