• Double-Strand Break Repair (DSBR) Model of Recombination (Szostak et al., 1983):
• Two Holliday Junctions Formed: Repair of a double-strand DNA break involves $5'\rightarrow 3'$ resection, strand invasion by $3'$ single-stranded overhangs, DNA synthesis, and ligation, generating an intermediate with two Holliday junctions.
• Four Topologically Equivalent Ways of Resolution: Each of the two Holliday junctions can be independently cleaved by resolvase enzymes in either the horizontal or vertical plane ($2 \times 2 = 4$ possible cleavage combinations). Cleavage of both junctions in the same plane produces non-crossover (gene conversion) products, whereas cleavage in opposite planes produces crossover (recombinant) products.
• Incorrect Statement: Option 2 asserts that Holliday junctions can be resolved by only "two topologically equivalent ways", which is biologically incorrect because four distinct topological resolution combinations exist.
• Conclusion: Option 2 is the statement that is NOT correct.

Correct option: (2) Holliday Junctions can be resolved by two topologically equivalent ways

• Statement (I) is Correct: In prokaryotes, when the codon GUG functions as the translation initiation codon (start codon), it is recognized by initiator tRNA ($tRNA_f^{Met}$) carrying $N$-formylmethionine (fMet), directing the incorporation of fMet as the first amino acid of the nascent polypeptide chain.
• Statement (II) is Correct: When the GUG codon occurs at internal (elongation) positions within the mRNA open reading frame, it is decoded by standard valyl-tRNA ($tRNA^{Val}$) and specifies valine.
• Conclusion: Both Statement (I) and Statement (II) are correct (Option 2).

• Allelic Fixation in Doubled Haploids:
• Doubled haploid (DH) lines are produced via chromosome doubling of haploid cells, achieving immediate and total homozygosity ($F = 1.0$) across all loci in a single generation.
• At any single biallelic locus ($A/a$), a DH line is completely fixed: gene frequency is either $p = 1.0$ (homozygous $AA$) or $p = 0.0$ (homozygous $aa$).
• When comparing any two DH inbred lines:
• If both carry the same allele: $y = |1 - 1| = 0$ (or $|0 - 0| = 0$).
• If they carry alternate alleles: $y = |1 - 0| = 1$.
• Conclusion: The difference in gene frequency $y$ can only be 0 or 1. (Correct answer: Option 3)