Heritable variants change frequency when they contribute unequally to later generations.
By the end: Calculate a simple selection step and distinguish fitness from survival, strength, and deliberate effort.
Natural selection requires differences among organisms, an association between those differences and reproductive contribution, and inheritance that connects generations. Selection on a nonheritable difference need not produce an inherited response. Selection can act through survival, mating success, fertility, or offspring survival.
Fitness in this context means expected reproductive contribution under specified conditions. Living a long time is not enough if no descendants inherit the variant. Relative fitness compares contributions; it is not a moral judgment or a permanent ranking of organisms across all environments.
Selection connects an inherited difference to an unequal share of descendants. Begin with two types in the same specified environment and ask what reaches a later census. Survival to breeding can matter, but mating success, offspring number, and offspring survival can matter too. The census stage fixes what the calculation counts. Comparing adults with newborns differs from comparing adults with adults. A fitness value without a life stage and environment leaves out information needed to interpret the proposed reproductive difference.
Expected contribution is an average over possible reproductive outcomes, not a promise about each individual. Suppose type A has probability 0.8 of surviving to breeding and produces two surviving offspring on average conditional on survival. Its expected contribution per starting individual is 0.8 times two, or 1.6. If type a survives with probability 0.5 but produces four surviving offspring per survivor, its corresponding expectation is two. Type A survives more often, but type a contributes more across the whole specified cycle.
The multiplication follows from the two possible routes. An individual that dies before breeding contributes zero under these assumptions. A survivor contributes the stated conditional average. Weighting zero by the chance of dying adds nothing; weighting the survivor average by the chance of surviving gives the total expectation. This is why multiplying an unconditional offspring average by survival again would be wrong. If the offspring average already includes deaths before breeding, it already accounts for that stage. Check whether each reported rate is conditional before combining life-cycle measurements.
Inheritance determines whether this reproductive sorting changes inherited composition. Imagine larger leaves surviving drought better, but suppose all leaf-size differences in this example came from recent watering and parents do not transmit the relevant difference. The adult survivors can have larger leaves without offspring inheriting a tendency toward larger leaves. In contrast, if inherited variants influence leaf size under those conditions, differential contribution can change their representation. A correlation between a trait and survival alone establishes neither the transmission rule nor total reproductive contribution.
Relative fitness removes a common scale factor. In the survival example, divide contributions 1.6 and two by two. Relative fitness is then 0.8 for A and one for a. Dividing both by 1.6 instead gives one and 1.25. These descriptions predict the same compositional change because their ratio is unchanged. Neither choice says that a particular individual must leave a fractional offspring. The fractions summarize expected contributions and provide weights for population arithmetic. Absolute counts and relative representation remain different questions even when both calculations start with the same observations.
A favorable relative value also need not mean population growth. Suppose two haploid types contribute 0.6 and 0.4 descendants per starting individual on average. Both decline in absolute expectation, but the first declines less and gains relative frequency if other processes are absent. At equal starting numbers, their next shares are 60% and 40%. Selection can therefore accompany demographic decline. Reversing the environment could reverse the contribution values, so the claim that one type is favored must stay attached to the conditions under which those values were measured or assumed.
| Haploid type | Starting count | Expected descendants per individual | Expected descendants |
|---|---|---|---|
| A | 50 | 0.6 | 30 |
| a | 50 | 0.4 | 20 |
| Total | 100 | 0.5 overall | 50 |
| A share | 50% | Not a descendant count | 60% |
Consider a haploid population with one copy per organism. A starts at frequency 0.2 and contributes 1.5 times as much as a, whose relative fitness is one. The weighted contributions are 0.2 × 1.5 = 0.3 and 0.8 × 1 = 0.8. A therefore makes up 0.3/1.1, about 27.27%, of the next generation.
The denominator matters: it is total weighted contribution, not the original population size. A relative advantage does not mean every A individual succeeds or that every a individual fails. The lab treats a large population deterministically and holds relative fitness constant, with no drift, mutation, migration, or overlapping generations.
The haploid model avoids a separate pairing calculation because every organism carries one copy at the modeled locus. Start with 100 organisms, 20 A and 80 a. Assign expected contributions of three descendants per A and two per a. A then contributes 60 descendants and a contributes 160, giving 220 in total expectation. The expected-pool A share is 60/220, or three elevenths, about 27.27%. Its starting share was 20%. The increase follows from unequal contributions to a newly calculated total, not from adding an advantage directly to 20%.
Divide both contributions by two to recover the lesson's relative fitness values, 1.5 for A and one for a. Using starting frequencies instead of counts gives weighted shares 0.3 and 0.8. Their sum, 1.1, is the mean relative contribution across the starting population. Dividing 0.3 by 1.1 produces the same three elevenths. The apparent change in units is harmless because a common multiplier cancels in the ratio. This deterministic calculation forms proportions from expected contributions; it does not enumerate uncertain realized family sizes in a finite population.
One way to detect a wrong denominator is to carry both types through the arithmetic. If A's next frequency were reported as 0.3 while a remained 0.8, the two frequencies would sum to 1.1. Frequencies at this two-allele haploid locus must sum to one. Normalization divides each weighted contribution by the same total, producing 3/11 and 8/11. The difference between A's starting and ending shares is about 7.27 percentage points. Its proportional frequency increase is about 36.36%, a different quantity from the 50% relative reproductive advantage.
| Type | Starting count × contribution | Expected descendants | Frequency × relative fitness |
|---|---|---|---|
| A | 20 × 3 | 60 | 0.2 × 1.5 = 0.3 |
| a | 80 × 2 | 160 | 0.8 × 1 = 0.8 |
| Total | 100 individuals initially | 220 | 1.1 |
| A next share | 60/220 | 3/11 ≈ 27.27% | 0.3/1.1 ≈ 27.27% |
The general rule uses p for the current A frequency and one minus p for a. Multiply p by A fitness and one minus p by a fitness, add those contributions, then divide A's contribution by their sum. If both fitness values equal one, the denominator equals one and the result is p. If A is absent, multiplying zero by an advantage still gives zero. Selection alone cannot supply a missing allele. A source such as mutation or genetic contribution from elsewhere would require an additional model step.
Starting composition changes the size of the frequency response even when relative fitness stays fixed. At p equal to one half, A fitness 1.5 gives weights 0.75 and 0.5, so the next share is 0.75/1.25, or 60%. At p equal to 0.9, the weights are 1.35 and 0.1, giving about 93.10%. The same advantage produces different percentage-point changes because the remaining alternative and the total weighted contribution differ. A constant advantage is therefore not a constant increment added at every generation.
The model's algebra explains the direction too. With a fitness fixed at one and A fitness written as one plus s, the change in A frequency is s times p times one minus p, divided by one plus s times p. For interior frequencies and positive fitness, the denominator is positive. Positive s gives an increase, zero s gives no deterministic change, and negative s gives a decrease. At either frequency boundary, the change is zero. This direction result assumes no mutation, immigration, transfer, or random sampling in the step being calculated.
| Starting A share | A weight | a weight | Next A share | Change in percentage points |
|---|---|---|---|---|
| 20% | 0.30 | 0.80 | 27.27% | +7.27 |
| 50% | 0.75 | 0.50 | 60% | +10 |
| 90% | 1.35 | 0.10 | 93.10% | +3.10 |
Baseline A fitness is 1.5 relative to a fitness of one. Frequencies have no units.
p is the current A frequency; 1 + s is A fitness relative to a = 1. The lab keeps s fixed and excludes drift and genetic input.
Generation 0: A = 20%. Neutral reference remains 20%.
| Type | Starting frequency | Relative fitness | Contribution |
|---|---|---|---|
| A | 0.2 | 1.5 | 0.3 |
| a | 0.8 | 1 | 0.8 |
| A next generation | 0.3/1.1 ≈ 27.27% |
The selection lab begins every scenario at 20% A and follows eight generations. Choose the neutral preset first and predict the endpoint before playback. Because both types contribute equally in this deterministic model, the blue values should remain on the gold reference at 20%. This is an expectation model with random sampling omitted, not a claim that every neutral finite population stays unchanged. A drifting population could move above or below the reference. The lab does not draw those alternative trajectories or assign probabilities to them.
Next choose the positive preset, where A contributes 1.5 relative to a at one. The first update is three elevenths. For the second update, multiply that current share by 1.5 and the remaining eight elevenths by one. A's weighted contribution is 4.5 elevenths and the total is 12.5 elevenths, so A reaches 4.5/12.5, or 36%. The denominator from generation one is not reused. Each update weights the composition that actually begins that step. The plotted connecting segments do not represent continuous changes within a generation.
A compact independent check uses the ratio of A to a rather than their separate shares. Initially that ratio is 0.2/0.8, or one quarter. A relative fitness of 1.5 multiplies the ratio by 1.5 each generation. After two generations the ratio is 0.5625, and converting back gives 0.5625/1.5625, or 36%. After eight generations the ratio is about 6.4072, giving about 86.50% A. These calculations check the repeated normalization without implying that the frequency itself is multiplied by 1.5 on every step.
| Generation | A-to-a ratio | A share |
|---|---|---|
| 0 | 0.25 | 20% |
| 1 | 0.375 | 27.27% |
| 2 | 0.5625 | 36% |
| 8 | 6.4072265625 | 86.50% |
Compare the disadvantage preset, with A fitness 0.8. Its first weighted contributions are 0.16 and 0.8, so A becomes one sixth, about 16.67%. After eight generations the ratio method gives one quarter times 0.8 to the eighth power, about 0.04194. Convert the ratio to a share by dividing by one plus that ratio, giving about 4.03%. The curves do not reach zero or one in these eight steps. Approaching a boundary in a deterministic model is different from losing the last physical copy in a finite population.
Use the controls to compare a stated assumption rather than to search for a preferred curve. Changing the advantage or selecting a preset resets playback to generation zero while retaining the fixed initial 20%. Reset keeps the chosen advantage. The complete generation values are also available without playback. Pause after the second generation to compare the 36% calculation with the displayed result, then examine the endpoint. Playback speed changes viewing time only. It does not change the reproductive contribution, number of biological generations, or model outcome.
Finally, ask what a neutral reference can establish. A rising selected curve demonstrates the arithmetic consequence of an assumed positive contribution difference. It does not demonstrate that a real population has that difference, nor that mutation caused its observed increase. A measured rise could also involve genetic contribution from another population or finite sampling. To use this recurrence as an empirical explanation, one would need starting composition, appropriate reproductive measurements, and evidence that omitted mechanisms do not account for the pattern. Agreement with one endpoint alone leaves many possible histories.
| Observation or lab behavior | Supported interpretation | Missing information |
|---|---|---|
| Neutral lab stays at 20% | Equal deterministic contributions | Finite realized sampling |
| Positive lab rises | Consequence of assumed advantage | Measured real-world fitness |
| Real population rises | Inherited composition changed if properly measured | Selection versus genetic input or drift |
| Same endpoint as the lab | Compatibility with this model | Whether other histories also fit |
An allele beneficial under one food supply, predator regime, or drug exposure can be costly elsewhere. Selection can also depend on how common a type is. A model with one fixed advantage is useful for isolating the mechanism, not a complete forecast of a real population.
Diploid selection requires fitness values for AA, Aa, and aa, so dominance can affect how quickly a rare allele changes. A favorable recessive allele may spend much of its rare phase in heterozygotes where its effect is not expressed. The haploid lab does not model that hiding effect or infer human trait outcomes.
Fitness conditional on environment means the same allele can have different relative contributions under different resources, competitors, or exposures. Suppose A has relative fitness 1.5 in environment E and 0.8 in environment F, with a fixed at one in both. At p equal to 0.2, one generation in E gives about 27.27%, whereas one in F gives about 16.67%. These are constructed alternatives, not measured universal properties of A. The allele has no permanent advantage or intention to improve.
Alternating environments illustrate why adding advantages is misleading. Apply one generation in E and then one in F. The A-to-a ratio starts at one quarter, becomes 0.375, and then becomes 0.3. The final A share is 0.3/1.3, about 23.08%. The product 1.5 times 0.8 equals 1.2, so the ratio increased by 20% over the two steps. The fixed-environment lab does not simulate this switch. This separate calculation assumes each environmental value stays applicable regardless of frequency and excludes other life-cycle changes or genetic input.
| Constructed history | A-to-a ratio | A share |
|---|---|---|
| Start | 0.25 | 20% |
| E only, A fitness 1.5 | 0.25 × 1.5 = 0.375 | 27.27% |
| F only from start, A fitness 0.8 | 0.25 × 0.8 = 0.2 | 16.67% |
| E followed by F | 0.25 × 1.5 × 0.8 = 0.3 | 23.08% |
Diploidy adds a different limitation. The effect of a rare allele can depend on whether it is present in one copy or two. Consider 100 individuals containing two Aa and 98 aa, with no AA. There are two A copies among 200 copies, so p is 1%. If Aa and aa have equal reproductive contribution because the favorable effect requires AA, there is no current contribution advantage for these carriers in this example. An immediate haploid advantage would misrepresent these genotype effects.
Make the contrasting assumption that both AA and Aa have relative contribution 1.5 while aa has one. In the same starting population, the two Aa individuals contribute weighted total three and the 98 aa contribute 98. With ordinary equal segregation, A occupies half the gametes from Aa, yielding weighted A contribution 1.5 among 101 total gamete contributions. The resulting A frequency is about 1.485%. This calculation concerns the gamete pool, not the fraction of offspring expressing a phenotype. Predicting offspring genotypes also requires specifying how those gametes unite.
Finite reproduction imposes another limitation even when the contribution advantage is real. Begin with one A among four haploid copies and let it contribute twice as much as each a. The weighted pool contains A fraction 2/5. If four offspring copies are drawn independently from that pool, each draw misses A with probability 3/5. All four miss it with probability 0.6 to the fourth power, or 12.96%. Selection changes the expectation, while drift can still remove the allele. An advantage does not guarantee persistence.
| A copies k in four draws | Probability |
|---|---|
| 0, complete loss | 12.96% |
| 1 | 34.56% |
| 2 | 34.56% |
| 3 | 15.36% |
| 4 | 2.56% |
A responsible prediction therefore names its scope. Specify the environment and census stage, distinguish genotype effects from allele labels, and state whether the result is an expected pool or a realized finite sample. If a variant is useful now, the calculation supports a possible present reproductive effect under the assumptions. It does not reconstruct why the sequence originated or show that earlier selection favored the same function. Inherited constraints and changing conditions can preserve traits with different earlier roles. Historical adaptive explanations require evidence beyond a plausible advantage inserted into today's recurrence.
A begins at frequency 0.4 with relative fitness two; a has fitness one. Find the next A frequency.
Divide A’s weighted contribution by the total contribution.
| Stage | Value |
|---|---|
| A contribution | 0.8 |
| a contribution | 0.6 |
| A fraction of total | 0.8/1.4 ≈ 57.14% |
A contributes 0.4 × 2 = 0.8; a contributes 0.6. The next frequency is 0.8/1.4 = 4/7, about 57.14%. The answer is not 80%, because total contribution changed too.
Type A survives to breeding with probability 0.8 and each survivor produces two surviving offspring. Type a survives with probability 0.5 and each survivor produces four. Which contributes more per initial individual?
Multiply survival by offspring contribution conditional on survival.
| Type | Survival | Offspring per survivor | Expected contribution |
|---|---|---|---|
| A | 0.8 | 2 | 1.6 |
| a | 0.5 | 4 | 2.0 |
A contributes 0.8 × 2 = 1.6 expected offspring per initial individual. Type a contributes 0.5 × 4 = 2.0. Despite lower survival, a has higher total contribution under these assumptions. No further mating or inheritance differences are included.
The explanations and exercises are complete without these optional links. The references cover inheritance, population mechanisms, and evidence for ancestry. Source annotations distinguish textbook explanations, secondary history, and a primary paper whose bibliographic identity alone was checked. The examples are constructed teaching models, not empirical estimates or professional decision advice.
Descent with modification, selection, evidence, and common misconceptions. Numerical examples here are invented, not field measurements.
Forms of selection, sexual selection, frequency dependence, and constraints on adaptation.