73. Microsatellite are used as marker for identifying individuals via DNA fingerprinting os the alleles may differ in the number of repeals. From the Southern blot shown below identify the progeny (A, B, C and D) for the given parents (M = mother, F = father).

(1) A, B, C and D     (2) A, B and D
(3) A and D only      (4) B, C and D

Introduction:
Microsatellite DNA fingerprinting questions in CSIR NET Life Science often use Southern blot band patterns to test your understanding of parentage and progeny identification. In the June 2016 paper, Question 39 presents a Southern blot for a microsatellite locus in parents (M = mother, F = father) and four individuals (A, B, C and D) and asks which of them can be progeny of this parental pair.

To solve this, focus on three key ideas:

• Each individual has two alleles (bands) at a microsatellite locus (diploid organism).

• Every progeny must inherit exactly one band from the mother and one band from the father at that locus.

• Any individual with a band not present in either parent, or missing both parental bands, cannot be their biological offspring.

Step 1: Decode parental band sizes from the blot
From the schematic at the top of the slide and the gel at the bottom (lanes labelled M, F, A, B, C, D with size positions 3, 4, 5, 6), the alleles are interpreted as follows (one locus, multiple microsatellite alleles differing in size):

• Mother (M):

  • One band at size 4

  • One band at size 5
    ⇒ Maternal genotype: 4 / 5

• Father (F):

  • One band at size 3

  • One band at size 6
    ⇒ Paternal genotype: 3 / 6

These band sizes represent different microsatellite alleles defined by repeat number and separated by Southern blot after restriction digestion and hybridization with a locus‑specific probe.

Step 2: List all possible progeny genotypes
With mother 4/5 and father 3/6, the possible zygotic combinations (child genotypes) are:

• 4 (from M) with 3 (from F) → 3 / 4

• 4 (from M) with 6 (from F) → 4 / 6

• 5 (from M) with 3 (from F) → 3 / 5

• 5 (from M) with 6 (from F) → 5 / 6

Therefore, any true progeny must show exactly one maternal band (4 or 5) plus one paternal band (3 or 6), and must not contain any additional bands.

Step 3: Read band patterns of A, B, C and D
From the lower Southern blot panel in the question image:

• Lane A:

  • Bands at 3 and 5
    ⇒ Genotype: 3 / 5

• Lane B:

  • Bands at 4 and 6
    ⇒ Genotype: 4 / 6

• Lane C:

  • Bands at 3 and 4
    ⇒ Genotype: 3 / 4

• Lane D:

  • Bands at 5 and 6
    ⇒ Genotype: 5 / 6

All four individuals show exactly two bands, each corresponding to valid alleles (3, 4, 5, 6) present in the parents, and each genotype matches one of the four possible progeny genotypes listed above.

Therefore:

• A has one maternal band (5) and one paternal band (3) → possible child

• B has one maternal band (4) and one paternal band (6) → possible child

• C has one maternal band (4) and one paternal band (3) → possible child

• D has one maternal band (5) and one paternal band (6) → possible child

Step 4: Evaluate each option in the MCQ

Question options:
(1) A, B, C and D
(2) A, B and D
(3) A and D only
(4) B, C and D

Explanation of each option:

• Option (1) A, B, C and D:

  • As shown, genotypes of all four (3/5, 4/6, 3/4, 5/6) can arise from the cross 4/5 × 3/6.

  • None of them carry any “new” band absent in both parents and all satisfy the rule of one allele from each parent.

  • Hence this option is correct.

• Option (2) A, B and D:

  • Excludes C.

  • However, C has genotype 3/4, which is a perfectly valid combination (4 from mother, 3 from father).

  • Excluding C is unjustified, so this option is incorrect.

• Option (3) A and D only:

  • Excludes B and C.

  • Both B (4/6) and C (3/4) are valid progeny genotypes, so this option is too restrictive.

  • Therefore, it is incorrect.

• Option (4) B, C and D:

  • Excludes A.

  • A has genotype 3/5, also a valid progeny genotype (5 from mother, 3 from father).

  • Excluding A is not justified, so this option is incorrect.

Thus, the only option consistent with microsatellite inheritance rules and the Southern blot banding pattern is:

Option (1) A, B, C and D.

Key takeaway concepts for exam:

• Microsatellites are short tandem repeats whose allele sizes vary between individuals and can be resolved as distinct bands by Southern blotting or PCR-based assays.

• In diploid organisms, each child must display exactly one allele (band) from each parent at a microsatellite locus, enabling parentage testing through DNA fingerprinting.

• When solving such CSIR NET questions, first deduce parental genotypes from their bands, then enumerate all possible progeny combinations and match each candidate’s band pattern to these possibilities.