Each of three botanists made a hypothesis regarding specimens of a particular plant species: Botanist 1: Any individual specimen possessing...

TPA Solution: Botanist Hypotheses

Visual Representation of Hypotheses

Let me translate the botanists' hypotheses into logical statements:

• Botanist 1 (B1): \(\mathrm{CS} \rightarrow (\mathrm{LR} \lor \mathrm{PF})\)
  If curly stems, then long roots OR purple flowers (or both)
• Botanist 2 (B2): \(\mathrm{LR} \rightarrow (\mathrm{FL} \lor \mathrm{RS})\)
  If long roots, then flat leaves OR round seeds (or both)
• Botanist 3 (B3): \((\mathrm{CS} \lor \mathrm{FL}) \rightarrow \neg\mathrm{PF}\)
  If curly stems OR flat leaves, then NOT purple flowers

Where:

• CS = curly stems
• FL = flat leaves
• LR = long roots
• PF = purple flowers
• RS = round seeds

Understanding the Question

We need to find characteristics 1 and 2 such that:

• A specimen with gene for characteristic 1
• But NOT the gene for characteristic 2
• Would prove at least one hypothesis is incorrect

Seeking the Critical Insight

The key is finding a combination where having one gene but not another creates a logical contradiction between the hypotheses.

Processing the Solution

Testing Key Combinations

Let me test the most promising combination:

Test: 1 = curly stems, 2 = long roots

Suppose a specimen has CS but NOT LR:

1. From B1: \(\mathrm{CS} \rightarrow (\mathrm{LR} \lor \mathrm{PF})\)
   - Since the specimen has CS, it must have either LR or PF
   - Since it does NOT have LR, it must have PF
2. From B3: \((\mathrm{CS} \lor \mathrm{FL}) \rightarrow \neg\mathrm{PF}\)
   - Since the specimen has CS, it satisfies the condition \((\mathrm{CS} \lor \mathrm{FL})\)
   - Therefore, it must NOT have PF

The Contradiction

• B1 requires: If CS and not LR, then must have PF
• B3 requires: If CS, then cannot have PF

These requirements directly contradict each other! A specimen cannot both have and not have purple flowers.

Verification

This contradiction proves that at least one of the hypotheses (B1 or B3) must be incorrect. No specimen could exist that satisfies all three hypotheses while having CS but not LR.

Final Solution Synthesis

Step-by-step recap:

1. Identified that we need a logical contradiction between hypotheses
2. Found that CS without LR creates opposing requirements for PF
3. B1 demands PF must be present, B3 demands PF must be absent
4. This impossibility proves at least one hypothesis is wrong

Answer specification:

• Column 1: curly stems
• Column 2: long roots

Key insight: The contradiction arises from the interaction between B1's inclusive condition (CS implies LR or PF) and B3's exclusive condition (CS implies not PF). When LR is absent, these create mutually exclusive requirements for PF.

Exam strategy: In TPA questions involving logical conditions, look for combinations that create contradictions between different rules. These contradictions often directly answer questions about impossibility or disproof.