المستوى الأعلى: بناء تقاطع جيني من الأدلة
مُقدَّم| English | المعنى |
|---|---|
| Punnett square/ˈpʌnɪt skweə/ | مربع بونيت |
| probability/ˌprɒbəˈbɪlɪti/ | احتمال |
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.
تحت النموذج المذكور، ماذا يظهر نسل انسحابي من أبوين سائدين؟
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النسل المتنحي (aa) يحتاج إلى أليل a من كل أب، لذا يجب أن يحمل كلا الأبوين اللذين يظهر عليهما السمة أليلاً متنحياً مخفياً - وهما Aa. لم يستطع أبوان AA إنتاج aa.
طابق كل مصطلح تقني مع معناه الدقيق.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
استخدم التعريفات للتمييز بين الكميات والعمليات ذات الصلة.
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.
أي عادة من العادتين تجعل التحقيق أو النموذج في هذه الحالة أكثر قابلية للدفاع؟
Difficulty 2/5 · 140 coins + a small bonus on first completion today
توضيح الشروط والافتراضات يسمح للآخرين بتكرار العمل بدقة، وتسجيل الأدلة الداعمة يتيح لهم التحقق من الاستنتاج. إخفاء النتائج المتعارضة يزيل الأدلة التي قد تتحدى استنتاجًا خاطئًا، ولا يمكن لقياس واحد أن يثبت ادعاءً أوسع.
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.
يعطي نموذج Aa×aa احتمالية تنحية 1/2. أوجد العدد المتوقع في 90 نسل.
Difficulty 2/5 · 140 coins + a small bonus on first completion today
الناتج هو 45 نسل. معلوم: Aa × aa يعطي احتمالية متنحية 1/2؛ 90 نسل. المتوقعة المتنحية = 90 × 1/2 = 45 نسل. التوقع ليس ضماناً لأي مجاميع فعلية.
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.
يعطي Aa×aa نفس نسبة الظواهر مثل Aa×Aa.
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Aa × Aa يعطي نسبة نمط ظاهري 3:1، بينما Aa × aa يعطي 1:1. الاختلافات بين التزاوجين تكمن في جينotype الأب الثاني، لذا لا يمكن أن تكون النسب متساوية.
شبكة تجمع أمشاج الآباء المحتملة: اكتب المصطلح التقني.
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مربع بانيت يعني شبكة تجمع أمشاج الآباء المحتملة.