Genetic studies of several marine species with high fecundity have produced "tiny" estimates (≤10^-3) of the ratio of effective population size (N_e) to adult census size (N), suggesting that even very large populations might be at genetic risk. A recent study using close-kin mark-recapture methods estimated adult abundance at N ≈ 2 × 10⁶ for southern bluefin tuna (SBT), a highly fecund top predator that supports a lucrative (~$1 billion/year) fishery. We used the same genetic and life history data (almost 13,000 fish collected over 5 years) to generate genetic and demographic estimates of N_e per generation and N_b (effective number of breeders) per year and the N_e/N ratio. Demographic estimates, which accounted for age-specific vital rates, skip breeding, variation in fecundity at age, and persistent individual differences in reproductive success, suggest that N_e/N is >0.1 and perhaps about 0.5. The genetic estimates supported this conclusion. Simulations using true N_e = 5 × 10⁵ (N_e/N = 0.25) produced results statistically consistent with the empirical genetic estimates, whereas simulations using N_e = 2 × 10⁴ (N_e/N = 0.01) did not. Our results show that robust estimates of N_e and N_e/N can be obtained for large populations, provided sufficiently large numbers of individuals and genetic markers are used and temporal replication (here, 5 years of adult and juvenile samples) is sufficient to provide a distribution of estimates. The high estimated N_e/N ratio in SBT is encouraging and suggests that the species will not be compromised by a lack of genetic diversity in responding to environmental change and harvest.

中文翻译：

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对顶级海洋捕食者南部蓝鳍金枪鱼的高Ne / N比的可靠估计。

具有高繁殖力几个海洋物种的遗传学研究生产的“微小”估计（≤10 ^-3有效群体大小（的比例）ñ _Ë）成人普查规模（ñ），这表明即使是非常大的群体可能是在遗传风险。使用近亲属标记-再捕获方法估计成人丰度最近的一项研究ň ≈2×10 ⁶南部蓝鳍金枪鱼（SBT），一种极为多产的顶级捕食者，它支持一个利润丰厚（〜$ 1十亿/年）渔业。我们使用相同的遗传和生活史数据（在5年中收集了近13,000条鱼）来生成每代N _e和N _b的遗传和人口统计学估计（育种者的有效数量）和每年的N _e / N比。人口估算值说明了特定年龄的生命率，跳过繁殖，年龄时的生殖力变化以及生殖成功的持续个体差异，表明N _e / N > 0.1，也许约为0.5。遗传估计支持这一结论。使用真实N _e = 5×10 ⁵（N _e / N = 0.25）进行的模拟产生的结果与经验遗传估计在统计学上一致，而使用N _e = 2×10 ⁴（N_e / N = 0.01）没有。我们的结果表明，如果使用足够多的个体和遗传标记，并且时间复制（此处为成年和少年样本的5年）足以提供足够的个体和遗传标记，则可以针对大量人群获得N _e和N _e / N的可靠估计。估算值的分布。SBT中高估计的N _e / N比令人鼓舞，这表明该物种不会因缺乏对环境变化和收获的遗传多样性而受到损害。