Southern Blotting for Polymorphism Detection in Fetal Cells

A protein D is encoded by a gene, which is 5 Kb long and has three Hind Ill restriction enzyme sites. The first one is 0.5 Kb from the transcription start site, the second one is 2.5 Kb from the first site and the third one is 0.5 Kb internal to the stop codon. The second site is polymorphic. In order to find out whether fetal cells contain the normal or the mutated gene, total genomic DNA from fetal cells was isolated, completely digested with Hind Ill, separated in an agarose gel, transferred to membrane and detected by a probe against the region between the second and third restriction site. Which one of the following band patterns will be obtained if the fetal cell is heterozygous?

Understanding the Southern Blot Experiment

We are analyzing a 5 kb-long gene encoding Protein D, which contains three HindIII restriction sites:

First site – 0.5 kb from the transcription start site
Second site – Polymorphic site located 2.5 kb from the first site
Third site – 0.5 kb inside the stop codon

A Southern blot experiment is performed to determine whether fetal cells contain the normal or mutated version of this gene. The probe hybridizes to the region between the second and third restriction sites.

How Polymorphism Affects Band Patterns

Homozygous Normal (No Polymorphism)
- HindIII cuts at all three sites.
- Fragments:
  - 2.5 kb (between the first and second site)
  - 2.5 kb (between the second and third site)
- Bands: 2.5 kb, 2.5 kb
Homozygous Mutant (Polymorphic Site Not Present)
- The second restriction site is missing, leading to only two cuts.
- Fragments:
  - 4 kb (from the first site to the third site)
  - 1.5 kb (from the third site onward)
- Bands: 4 kb, 1.5 kb
Heterozygous (One Normal and One Mutant Allele)
- One chromosome follows the normal restriction pattern (2.5 kb, 2.5 kb).
- The other chromosome follows the mutant pattern (4 kb, 1.5 kb).
- Bands Expected: 4 kb, 2.5 kb, 1.5 kb (Answer A).

Why Is (B) the Correct Answer?

The probe used in the experiment hybridizes only to the region between the second and third restriction site.

In the normal allele, this region is 2.5 kb.
In the mutated allele, the second site is missing, resulting in a 4 kb fragment.
The probe will not detect the 1.5 kb fragment because it is outside the probe’s hybridization region.

Thus, the visible bands in the Southern blot will be 4 kb and 1.5 kb, making (B) the correct answer.

Conclusion

Southern blotting is a powerful tool for detecting gene polymorphisms. In this case, the fetal cells are heterozygous, but due to the probe’s hybridization specificity, only 4 kb and 1.5 kb bands appear, making (B) the correct answer.