69. In normal individuals, there are three Mstll restrictionsites. two flanking the ß-globin gene and one within thegene. In individuals affected by a disease, a singlenucleotide polymorphism in the ß-globin gene abolishesthe internal Mstll recognition site. The RFLP pattern forthis locus, obtained by hybridization using a probeinternal to the flanking Mstll sites, from three siblings ofa family is shown below: Based on the above profile, what is the nature of the genetic disorder? (1) X-linked Recessive (2) Autosomal Dominant (3) Autosomal Recessive (4) X-linked Dominant

69. In normal individuals, there are three Mstll restrictionsites. two flanking the ß-globin gene and one within thegene. In individuals affected by a disease, a singlenucleotide polymorphism in the ß-globin gene abolishesthe internal Mstll recognition site. The RFLP pattern forthis locus, obtained by hybridization using a probeinternal to the flanking Mstll sites, from three siblings ofa family is shown below:

Based on the above profile, what is the nature of the genetic disorder?
(1) X-linked Recessive
(2) Autosomal Dominant
(3) Autosomal Recessive
(4) X-linked Dominant


Introduction

Restriction Fragment Length Polymorphism (RFLP) of the β‑globin gene using the enzyme MstII is a classic exam problem for understanding how point mutations change restriction patterns and inheritance of hemoglobin disorders such as sickle‑cell anemia or β‑thalassemia.
In this CSIR NET‑style question, the disappearance of an internal MstII site produces altered fragment sizes (1.35 kb vs 1.15 + 0.2 kb), allowing deduction that the underlying disease is autosomal recessive based on the pattern observed in three siblings.


Concept: MstII sites in the β‑globin gene

In the normal β‑globin allele, there are three MstII restriction sites: two flanking the gene and one inside the gene; digestion yields fragments of 1.15 kb and 0.2 kb within the probed region.
A single‑nucleotide mutation (for example the sickle or some β‑thalassemia mutations) abolishes the internal MstII site, so the two smaller fragments (1.15 + 0.2 kb) fuse into a single 1.35 kb fragment when cut with MstII.

Thus:

  • Normal allele (N) → two bands: 1.15 kb and 0.2 kb

  • Mutant allele (D) (disease) → one band: 1.35 kb

A heterozygote (N/D) shows three bands (1.35, 1.15, 0.2 kb), whereas a homozygous normal has two bands (1.15, 0.2) and a homozygous affected has one band (1.35).


Interpreting the siblings’ RFLP profile

From the table in the question (bands scored for each child):

  • Normal son: bands at 1.15 kb and 0.2 kb only → genotype N/N (homozygous normal).

  • Normal daughter: bands at 1.35, 1.15, and 0.2 kb → genotype N/D (carrier, phenotypically normal).

  • Affected son: band at 1.35 kb only → genotype D/D (homozygous mutant).

This pattern exactly matches the expected segregation for a single autosomal locus with a recessive disease allele.
Both parents must therefore be heterozygous carriers (N/D), because they produced an unaffected homozygote (N/N), a carrier (N/D), and an affected homozygote (D/D), which conforms to the classical 1:2:1 Mendelian genotypic ratio for an autosomal recessive trait.

To visualize how each option fits (or fails), the following table summarizes typical expectations:

Inheritance type Who usually shows disease in sibship with carrier parents Compatibility with: N son (N/N), carrier daughter (N/D), affected son (D/D)?
Autosomal recessive Both sexes; only homozygous mutants affected Fully compatible: parents N/D, children N/N, N/D, D/D.
Autosomal dominant Both sexes; heterozygotes already affected Incompatible: heterozygous daughter would be affected, not normal.
X‑linked recessive Mostly males affected; females usually carriers Incompatible: affected son hemizygous mutant; daughter carrier, but autosomal RFLP here (β‑globin on chr 11).
X‑linked dominant Both sexes, but heterozygous females usually affected Incompatible: normal daughter with one mutant allele would still be affected.

Detailed evaluation of each option

Option (1) X‑linked Recessive

X‑linked recessive traits typically involve a pathogenic allele on the X chromosome, causing hemizygous males (XᵈY) to be affected and heterozygous females (XᴺXᵈ) to be carriers but usually asymptomatic.
However, the β‑globin gene (HBB) is located on chromosome 11, which is autosomal, not X‑linked, and the RFLP pattern reflects two alleles (N and D) present in both sexes, contradicting X‑linkage.
Moreover, an X‑linked recessive model would not explain the homozygous normal son (N/N) and homozygous affected son (D/D) using the same two‑allele system seen in both son and daughter with identical fragment sizes; therefore option (1) is incorrect.

Option (2) Autosomal Dominant

In autosomal dominant disorders, heterozygous individuals (A/a) already express the disease phenotype, so each affected person usually has an affected parent and carriers are themselves diseased.
Here, the daughter clearly shows three bands (1.35, 1.15, 0.2 kb), proving that she carries both normal and mutant alleles (N/D), yet she is described as normal, which rules out a dominant effect of the mutant allele.
Therefore, the banding pattern and phenotype combination do not fit an autosomal dominant model, making option (2) incorrect.

Option (3) Autosomal Recessive – Correct

Autosomal recessive inheritance requires two copies of the mutant allele (d/d) for disease; heterozygotes (D/d or N/D) are phenotypically normal carriers.
The genotypes inferred from the RFLP bands—normal son N/N (1.15 + 0.2), normal daughter N/D (1.35 + 1.15 + 0.2), and affected son D/D (1.35 only)—fit perfectly with this model, indicating that the internal MstII‑site‑abolishing mutation behaves as a recessive autosomal allele.
Thus, both parents must be carriers (N/D), and the disorder is best characterized as autosomal recessive, so option (3) is correct.

Option (4) X‑linked Dominant

X‑linked dominant conditions show clinical manifestations in heterozygous females and hemizygous males, often with females more frequently affected due to having two X chromosomes.
If the mutant β‑globin allele were X‑linked dominant, the carrier daughter with one mutant allele should be affected, but she is normal; furthermore, the gene analyzed (β‑globin) is autosomal, and the same two‑allele RFLP pattern appears in both sexes, which is inconsistent with X‑linkage and dominance.
Therefore, option (4) is also incorrect.


By carefully mapping MstII‑generated RFLP fragments (1.35, 1.15, 0.2 kb) to underlying genotypes and considering the chromosomal location of the β‑globin gene, the inheritance pattern in this CSIR NET question is conclusively identified as autosomal recessive.

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