Higher Tier: 証拠に基づく遺伝子交配図を作成する
導入済み| English | 意味 |
|---|---|
| Punnett square/ˈpʌnɪt skweə/ | ポンネット方眼 |
| probability/ˌprɒbəˈbɪlɪti/ | 確率 |
What would explain this observation?
- Higher Tier: Two dominant-looking parents have a recessive offspring in a stated simple model. Their appearance alone was not enough to identify their genotypes.
- Start with a prediction. State the quantities or features you would compare, then decide what evidence could distinguish two explanations.
Build the model
- Start with the allele key. In a completely dominant single-gene model, a recessive offspring has genotype aa and must receive an a allele from each parent. If both parents show the dominant phenotype, each also has A, so both are Aa. List each parent’s possible gametes, A and a, before filling the cross.
- Punnett square 庞尼特方格: A grid combining possible parental gametes; probability 概率: The chance of an outcome under the stated model.
提示されたモデルにおいて、両親が優性である場合、劣性の子孫は何を示しているか？
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
劣性仔（aa）は両親からそれぞれa対立遺伝子を受け取る必要があるため、両親とも劣性を隠して持つ優性表現者でなければならない（Aa）。AA×AAの組み合わせでaaを生むことはできない。
各専門用語を正確な意味と対応させよ。
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
関連する物理量やプロセスを区別するために定義を使用せよ。
Choose evidence that can test it
- Each box combines one allele from each parent: AA, Aa, Aa and aa. State genotype and phenotype probabilities separately. The inference depends on the stated model: one gene, complete dominance and no new mutation. Do not apply it unchanged to characteristics controlled by many genes or to a scenario with incomplete dominance.
- Write the key, infer parental genotypes using the evidence, label gametes along the grid edges and combine them. Show each step so a mistaken parental assumption is visible. Compare the predicted distribution with supplied observed counts; small samples often deviate from expected proportions without disproving the model.
この場合、調査やモデルをより妥当にする2つの習慣は何か。
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
条件や仮定を明記することで他者が正確に再現でき、根拠となる証拠を記録することで結論を検証できます。相反する結果を隠すと、誤った結論を疑う証拠を失いますし、単一の測定でより広い主張を成立させることはできません。
Work from known quantities
- State the known values and their units. Choose the relation because its assumptions fit this case, then rearrange before substitution.
- Known: an Aa parent crossed with aa produces Aa, aa, Aa, aa. Recessive probability is 1/2. In 60 independent model offspring the expected count is 30. For two independent offspring, probability both are recessive is 1/2×1/2 = 1/4. Multiplication applies because the prompt states independent outcomes.
An Aa×aa model gives recessive probability 1/2. Find the expected number in 90 offspring. Use the same sequence: known quantities → model → relation → substitution → unit and interpretation.
Aa×aa のモデルにおける劣性表現の確率は 1/2 です。90 個の子孫における期待値を求めよ。
難易度 2/5 · 140 コイン + 今日初回完了時に小ボーナス付与
結果は45個の仔となる。既知：Aa × aaの交配で劣性の確率は1/2；90仔。期待される劣性個体数 = 90 × 1/2 = 45個。期待値は、実際の産仔について保証するものではない。
Check the conclusion and its limits
- Writing only a 3:1 ratio earns no explanation of where it came from. Aa×aa does not give 3:1. A recessive outcome is possible again even when the previous offspring was recessive; the model has no memory of earlier births.
- Return to the original observation. Explain what the result supports, which conditions it assumes, and one way to test a competing explanation.
Aa×aa gives the same phenotype ratio as Aa×Aa. This claim is false: Aa x Aa gives a 3:1 phenotype ratio, while Aa x aa gives 1:1. The crosses differ in the second parent's genotype, so the ratios cannot be the same.
Higher Tier: construct a genetic cross from evidence: Each box combines one allele from each parent: AA, Aa, Aa and aa. State genotype and phenotype probabilities separately. The inference depends on the stated model: one gene, complete dominance and no new mutation. Do not apply it unchanged to characteristics controlled by many genes or to a scenario with incomplete dominance.
Aa×aa は Aa×Aa と同じ表現型比を与える。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
Aa × Aaでは表現型比が3:1となるのに対し、Aa × aaでは1:1となる。第二親の遺伝子型が異なるため、比率は一致しない。
可能な親の.Authenticationを組み合わせるグリッド: 専門用語を記せ。
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
ポンネット方阵とは、可能な親の.Authenticationを組み合わせるグリッドを意味する。