Normalize viability selection before counting adult alleles

1. Orient: Normalize viability selection before counting adult alleles

Natural selection requires differences in contribution to subsequent generations associated with heritable variation. Survival is one component; mating success and fertility can also matter. Relative fitness compares contributions under a defined environment and life cycle, not an organism’s strength or moral worth.

2. Explain the population mechanism

The original viability model starts from random-union genotype frequencies p²,2pq,q². AA survival weight is1, Aa is1−0.4h and aa is0.6. Multiplying initial frequencies by weights gives unnormalized survivor contributions. Their sum is mean relative viability; divide each contribution by that sum before counting adult alleles.

Adult A frequency is fAA+fAa/2 after normalization. The coefficient h determines how much of the aa disadvantage appears in Aa for this particular viability phenotype. It is not a universal property of the allele across traits or environments. A dominant phenotype can be rare, and a recessive variant need not be harmful.

The calculation describes one stage, not every generation of a complete ecosystem. Equal subsequent reproduction is stipulated. Age structure, sex-specific success, density dependence, linkage, migration and random sampling could change the prediction. A real inference would measure initial cohorts and survival independently rather than assume the weights from an endpoint.

3. Work a complete evidence calculation

At p=0.5 and h=0.5, initial frequencies are0.25,0.5,0.25 and weights1,0.8,0.6. Survivor contributions0.25,0.4,0.15 sum0.8; adult frequencies0.3125,0.5,0.1875 give adult p=0.5625.

4. Predict and investigate

Predict first: Can you count adult allele frequency directly from weighted contributions without first dividing by total survival?

Hold p=0.5 and compare h0,0.5 and1. Retain the initial frequencies, weights, normalization denominator and adult allele count. Explain which additional reproduction measurement is needed before equating adult frequency with next-generation frequency.

Retain the original population definition, first prediction, controls and full numerical ledger, including conflicting or null results. These are original bounded teaching calculations, not measured populations.

5. Transfer and explain

Apply the mechanism to a changed case

With p=0.5, h=1 and the same AA1/aa0.6 weights, calculate adult A frequency.

Retain your calculation, units and assumptions in the lesson notebook before opening this public practice response.

Compare a worked response

Contributions are0.25,0.30,0.15, total0.70. Adult A frequency=(0.25+0.30/2)/0.70=4/7≈0.57143. The numerator alone0.40 is not a normalized frequency.

6. Reflect and retain evidence

State a competing mechanism, an observation that would distinguish it, and a limitation of the supplied sampling or life-cycle model. Preserve first reasoning and a criterion-linked human review in the General Biology II workspace. Public practice feedback and simulated evidence do not certify an independently demonstrated comparative argument, field investigation or practical skill.

Supporting scientific reference, checked 30 October 2026. All prose, numerical packets and diagrams are original. No organism collection, culture, breeding, release, personal genetic inference or clinical procedure is requested.