This question asks about the expected phenotypic segregation ratios for a transgenic plant containing two unlinked copies of T-DNA, one functional and one silenced. We need to determine the ratio of transgenic to non-transgenic plants in progeny from backcrossing and self-pollination.

Let's define the genetic makeup of the plant and how the phenotype is determined.

Genotype of the Transgenic Plant:

The plant has two unlinked copies of T-DNA. This means the inserts are located at two different positions (loci) in the genome. Let's represent the locus with the functional T-DNA as gene A, and the locus with the silenced T-DNA as gene B. Since it's a transgenic plant created through transformation, it is likely heterozygous for the presence of the inserted DNA at each locus. So, the genotype of the transgenic plant can be represented as AaBb, where 'A' is the allele with functional T-DNA, 'a' is the allele without T-DNA at the first locus, 'B' is the allele with silenced T-DNA, and 'b' is the allele without T-DNA at the second locus.

Phenotype Determination:

The "transgenic phenotype" is defined by the presence of the functional T-DNA copy. The silenced T-DNA copy does not contribute to this specific phenotype. Therefore, a plant is considered transgenic if it has at least one copy of the 'A' allele (which carries the functional T-DNA). A plant is non-transgenic if it lacks the 'A' allele (i.e., has the genotype aa). The alleles at locus B (silenced T-DNA) do not influence whether the plant is considered transgenic or non-transgenic based on this definition.

• Plants with genotype AA or Aa are Transgenic.
• Plants with genotype aa are Non-transgenic.

Now let's analyze the segregation ratios in the specified crosses.

Backcrossing Analysis (i)

Backcrossing typically involves crossing the transgenic plant with a non-transgenic parent. Assuming the non-transgenic parent is the original wild-type line used for transformation, it would lack both T-DNA inserts. Its genotype would be aabb.

Cross: Transgenic plant (AaBb) × Non-transgenic plant (aabb)

Gametes produced by AaBb: AB, Ab, aB, ab

Gametes produced by aabb: ab

We can determine the progeny genotypes and phenotypes:

The progeny genotypes AaBb, Aabb, aaBb, and aabb are produced in equal proportions (1/4 each). Based on our phenotype definition:

• Transgenic progeny genotypes: AaBb, Aabb (Total proportion = \(1/4 + 1/4 = 2/4\))
• Non-transgenic progeny genotypes: aaBb, aabb (Total proportion = \(1/4 + 1/4 = 2/4\))

The phenotypic ratio of transgenic to non-transgenic progeny in the backcross is \(2:2\), which simplifies to \(1:1\).

Self-pollination Analysis (ii)

Self-pollination involves crossing the transgenic plant (genotype AaBb) with itself.

Cross: Transgenic plant (AaBb) × Transgenic plant (AaBb)

This is a dihybrid cross between two individuals heterozygous for two unlinked genes. The expected genotypic ratio in the progeny is \(9:3:3:1\) for A_B_ : A_bb : aaB_ : aabb.

Let's determine the phenotype for each genotype class based on the presence of the 'A' allele:

• A_B_: This includes genotypes AABB, AABb, AaBB, and AaBb. All have at least one 'A'. They are Transgenic. Proportion = \(9/16\).
• A_bb: This includes genotypes AAbb and Aabb. All have at least one 'A'. They are Transgenic. Proportion = \(3/16\).
• aaB_: This includes genotypes aaBB and aaBb. They have no 'A'. They are Non-transgenic. Proportion = \(3/16\).
• aabb: This genotype has no 'A'. It is Non-transgenic. Proportion = \(1/16\).

Now, let's sum the proportions for each phenotype:

• Total Transgenic progeny = Proportion of A_B_ + Proportion of A_bb = \(9/16 + 3/16 = 12/16\)
• Total Non-transgenic progeny = Proportion of aaB_ + Proportion of aabb = \(3/16 + 1/16 = 4/16\)

The phenotypic ratio of transgenic to non-transgenic progeny in the F1 generation from self-pollination is \(12:4\), which simplifies to \(3:1\).

Therefore, the segregation of the transgenic to non-transgenic phenotype would occur in a \(1:1\) ratio in progeny obtained by backcrossing and in a \(3:1\) ratio in F1 progeny obtained by self-pollination.

The correct combination is (i) - 1 ∶ 1 and (ii) - 3 ∶ 1.