64. Doubled haploids (DH) are plants derived from single immature pollen and doubled artificially to form diploids. A DH population was created from progeny derived from a cross between two parents (P1 and P2), one resistant (R) and the other sensitive (S) to white rust. The parents, Fi and DH population were profited with a set of co-dominant markers, which is represented below.

The following table summarizes the proposed percentage of the 4 different types (1 to 4) of DH progeny, assuming that the DNA marker is (i) unlinked to the gene governing the trait and (n) linked at a distance of 10 cM.

Which one of the following options correctly represents the expected ratio for unlinked and linked, respectively?
(1) A, ii         (2) A, i
(3) B, i          (4) A, iii

Doubled haploids (DH) are produced from F1 plants and then doubled to form completely homozygous diploids in a single generation, so each DH line represents one of the possible meiotic products from the F1. In this CSIR NET–style question, the marker–trait segregation in DH progeny is used to decide whether a DNA marker is linked to a rust‑resistance gene and to identify the correct expected percentage profiles under unlinked and linked conditions.​

Understanding the genetic setup

• Parents: P1 is resistant (R) to white rust and P2 is sensitive (S); both are completely homozygous at the DNA marker, but for opposite alleles (e.g., P1 = M, P2 = m), so the F1 is heterozygous (Mm) and phenotypically resistant or sensitive depending on dominant/recessive nature of the trait locus.​

• Doubled haploids: Gametes from this F1 are fixed by chromosome doubling, so each DH line is homozygous at both the marker locus and the trait locus, carrying one of the four possible combinations of parental or recombinant chromatids.​

Because all DHs are homozygous, they can be classified into four phenotype–marker profiles (1 to 4), typically:

• Profile 1: R with P1 marker allele

• Profile 2: R with P2 marker allele

• Profile 3: S with P1 marker allele

• Profile 4: S with P2 marker allele.​

Expected DH ratios when marker is unlinked (option profiles A–C)

When the marker and the rust‑resistance gene are on different chromosomes (unlinked), independent assortment gives each of the four gamete types with equal probability (25%). Therefore, in a DH population, each of the four profiles (1–4) is expected at 25% frequency, producing the ratio 25 : 25 : 25 : 25 for the four classes.​

• Profile A (unlinked column): 25, 25, 25, 25 – this exactly matches the expectation under independent assortment and therefore correctly represents the unlinked case.​

• Profile B (unlinked column): 37.5, 37.5, 12.5, 12.5 – this suggests some profiles are more frequent, which would only happen if there is linkage or segregation distortion, so it does not represent an unlinked marker.​

• Profile C (unlinked column): 37.5, 12.5, 12.5, 37.5 – again, unequal class frequencies contradict the 1:1:1:1 ratio expected for unlinked loci, so it is not correct for the unlinked situation.​

Thus, only “A” is correct for the unlinked part of the question.

Expected DH ratios when marker is linked at 10 cM

If the marker is linked to the rust‑resistance gene at 10 cM, the recombination frequency $r$ is 0.10 and the parental (non‑recombinant) gametes occur with total frequency 0.90, while recombinant gametes occur with total frequency 0.10. Each parental type appears with frequency $0.45$ and each recombinant type with $0.05$, giving expected DH percentages of 45%, 45%, 5%, 5% distributed appropriately across the four profiles.​

From the table in the question:

• In the linked section of Profile A (i): the classes 1 and 4 (or equivalently, two opposite profiles) have 45% and the remaining two have 5%, which matches the expected pattern of two parental and two recombinant classes (45, 45, 5, 5) for a 10 cM distance.​

• In the linked section of Profile B (ii): the 5% and 45% entries are arranged so that the same phenotype–marker combinations are not grouped as parental pairs; this pattern does not correspond to a simple 10 cM linkage with two clear parental and two recombinant classes.​

• In the linked section of Profile C (iii): the 37.5 and 12.5 percentages correspond to a 25 cM distance (0.25 recombination) rather than the stated 10 cM, so this does not fit the question condition.​

Thus, “i” correctly represents the linked case at 10 cM.

Evaluating all answer options

The final MCQ asks which combination correctly represents the expected ratio for unlinked and linked marker–trait relationships, respectively.​

• Option (1) A, ii – A is correct for unlinked, but ii is wrong for 10 cM linkage, so the whole option is incorrect.

• Option (2) A, i – A gives the equal 25% profile for unlinked loci, and i gives the 45%/5% pattern consistent with 10 cM linkage, so this option is entirely correct.

• Option (3) B, i – B is incorrect for the unlinked case, so this option is rejected even though i is correct for linkage.

• Option (4) A, iii – A is correct for unlinked, but iii corresponds to a larger recombination distance (around 25 cM), not 10 cM, so this option is incorrect.​

Therefore, the correct answer is: Option (2) A, i.