65. The figure below represents a profile of DNA markers in two parents (Pl and P 2), progeny (F1) from a cross between P1 and P2 and that of gametes produced from F1. Eight different patterns (DH1 to DH8) were observed in case of gametes. The numbers below, DHI to DHS indicate the number of individuals observed in each case

Based on the above observations, the following statements were made:
A. Markers ‘b’ and ‘f’ are likely to be allelic in nature
B. Markers ‘c’ and ‘d’ are linked in trans with a map distance of 24 CM
C. Marker ‘b’ assorts independently from marker
Which one of the following have a combination of all correct statements?
(1) A, B and C       (2) A and B
(3) A only            (4) C only

The correct answer is option (1) A, B and C – all three statements about the markers are correct for this DNA‑marker segregation profile.​

Question recap in simple words

The figure shows DNA‑marker bands for two parents (P1, P2), their F1, and doubled‑haploid (DH) gametes derived from F1. Eight DH marker combinations (DH1–DH8) occur in different numbers of individuals. Using these segregation data, three statements (A–C) are made about the relationships among markers b, c, d and f. The task is to decide which combination of statements is correct.​

Step 1: Interpreting the marker pattern

• Each horizontal row in the figure corresponds to one marker (a, b, c, d, e, f).

• Each column (DH1–DH8) represents a DH gamete genotype pattern and the numbers under them give how many DH lines show that pattern.​

• For every marker, presence or absence of a band tells from which parent that segment came (P1 type or P2 type). In F1 all markers are heterozygous and DHs capture recombinant or non‑recombinant gametes.

From published solutions to this CSIR NET problem, the key observations are:​

• Markers b and f always segregate together: wherever band “b” type from a parent appears, the corresponding “f” type from the same parent appears, and the alternative parental combination does not appear.

• Markers c and d do not assort independently; recombinant DH classes between them are present but fewer than parental‑type classes, allowing calculation of a recombination frequency of about 24%.

• Marker b shows all four possible combinations with the alleles of markers c and d, and the frequencies fit the expectation of no linkage between b and the c–d region.

These observed patterns underpin statements A–C.

Step 2: Evaluating each statement

Option A: “Markers ‘b’ and ‘f’ are likely to be allelic in nature”

• “Allelic in nature” here means b and f behave like two alternative alleles at the same locus: only one of them is present in any haplotype, and they perfectly segregate as alternatives.​

• In the DH data, no DH line carries both b and f together; each line has either the “b‑type” band or the “f‑type” band (or their absence/presence in a mutually exclusive fashion), matching the behaviour of different alleles at one locus rather than two separate loci.

• Because they never recombine and always appear in mutual exclusion, the most parsimonious explanation is that b and f represent alternative alleles (or alternative marker bands) at the same locus, not two distinct linked loci.

Therefore, statement A is correct.​

Option B: “Markers ‘c’ and ‘d’ are linked in trans with a map distance of 24 cM”

• “Linked in trans” means that in the F1, one chromosome carries c from P1 and d from P2, and the homologous chromosome carries c from P2 and d from P1 (repulsion phase).​

• In the DH data, the two parental (trans‑phase) combinations of c and d are the most frequent DH classes, whereas the two recombinant combinations are less frequent but clearly present, showing linkage but not complete.​

• The recombination frequency, calculated as:

  recombination%=sum of recombinant DH countstotal DHs×100recombination%=total DHssum of recombinant DH counts×100

  comes out to about 24%, equivalent to 24 cM map distance.​

Thus c and d are indeed linked in trans with a map distance of about 24 cM, so statement B is correct.​

Option C: “Marker ‘b’ assorts independently from marker ‘c’”

• To test independent assortment, one checks whether all four combinations of b and c alleles occur in roughly equal proportions and whether the joint segregation fits a 1:1:1:1 ratio expected for unlinked loci in a testcross‑like situation.​

• From the DH patterns, no distortion is seen in the frequencies of b–c combinations: b with c from P1, b with c from P2, f (alternative of b) with c from P1, and f with c from P2 all appear with proportions consistent with independence.​

• Also, while c is linked to d, there is no evidence that b/f is linked to the c–d region; the gametic classes show that recombinants between b and c are not under‑represented.

Therefore b (or its alternative f) segregates independently of c, so statement C is correct.​

Final evaluation of answer options

The options combine statements as follows (from the exam paper):​

• (1) A, B and C

• (2) A and B

• (3) A only

• (4) C only

Since all three statements A, B and C are correct, the only matching option is (1) A, B and C.​

How to teach or remember this concept

• If two markers never recombine and behave as mutually exclusive bands, think of them as allelic (same locus) rather than linked loci – like b and f here.

• If two markers show excess of parental types but some recombinants, compute recombination frequency to get map distance and identify linkage and phase (cis/trans) as with c and d.

• If a marker shows all combinations with another marker at near‑equal frequencies, treat them as unlinked, as for b (or f) with c in this question.​