Step 1 : Understanding the Question:
The question asks to determine the genotypes of the homozygous, true-breeding parental lines (R1, R2, R3, R4) that produced the $F_1$ dihybrids based on the ratios of gametes produced by those $F_1$ plants.
Step 2 : Key Formulas and Approach:
The approach relies on the principles of genetic linkage and recombination.
In a linked dihybrid system, parental gametes are always produced in higher frequencies ($> 25\%$ each) due to linkage, while recombinant gametes are produced in lower frequencies ($< 25\%$ each) due to crossing over.
Step 3 : Detailed Explanation:
• Analyzing Cross 1 (R1 $\times$ R2):
The $F_1$ dihybrid produces gametes in the proportions: $35\%\ pQ$, $35\%\ Pq$, $15\%\ PQ$, and $15\%\ pq$.
The gametes with the highest frequencies ($35\%$ each) represent the non-recombinant (parental) configurations: $pQ$ and $Pq$.
This indicates the $F_1$ alleles are in the repulsion (trans) phase, represented as $pQ/Pq$.
Since the parents R1 and R2 are true-breeding (homozygous), one must have contributed the $pQ$ chromosome and the other the $Pq$ chromosome.
Thus, the parental genotypes must be $ppQQ$ and $PPqq$.
• Analyzing Cross 2 (R3 $\times$ R4):
The $F_1$ dihybrid produces gametes in the proportions: $35\%\ PQ$, $35\%\ pq$, $15\%\ pQ$, and $15\%\ Pq$.
Here, the high-frequency parental gametes are $PQ$ and $pq$ ($35\%$ each).
This indicates the $F_1$ alleles are in the coupling (cis) phase, represented as $PQ/pq$.
The homozygous parents R3 and R4 must have contributed the $PQ$ and $pq$ chromosomes.
Thus, their genotypes are $PPQQ$ and $ppqq$.
Step 4 : Final Answer:
Matching these findings, we get: R1 = ppQQ, R2 = PPqq, R3 = PPQQ, R4 = ppqq.
This is represented by Option (A).