πΈ Chapter 5
Molecular Basis Of Inheritance
( 5 Marks)
1. Describe the structure of DNA as
proposed by Watson and Crick.
Answer:
- DNA is a double-stranded
helical molecule made of nucleotides.
- Each nucleotide = phosphate
+ deoxyribose sugar + nitrogen base.
- Bases pair by hydrogen bonds:
A–T (2 bonds), G–C (3 bonds).
- Strands are antiparallel
(5′→3′ and 3′→5′).
- Pitch = 3.4 nm, 10 base pairs per
turn.
- Sugar-phosphate backbone forms the outer part; bases
face inward.
- This structure explains replication
and heredity accurately.
2. Explain Griffith’s experiment and
how it led to the discovery of genetic material.
Answer:
- Organism used: Streptococcus pneumoniae
(S and R strains).
- S strain: Smooth, virulent; R strain:
Rough, non-virulent.
- Experiment:
- Live S → mouse dies.
- Live R → mouse lives.
- Heat-killed S → mouse lives.
- Heat-killed S + live R → mouse
dies.
- Conclusion: Some factor (“transforming
principle”) from dead S transformed R into S.
- Later found: Avery, MacLeod, McCarty proved
DNA is the transforming principle.
3. Describe the Hershey–Chase
experiment proving DNA is the genetic material.
Answer:
- Used bacteriophage T₂
with ³²P-labeled DNA and ³⁵S-labeled protein.
- After infection of E. coli:
- ³²P (DNA) entered bacteria.
- ³⁵S (protein) remained outside.
- Progeny viruses contained ³²P,
not ³⁵S.
- Conclusion: DNA, not protein, carries
genetic information.
4. Explain Meselson and Stahl’s
experiment to prove semiconservative replication of DNA.
Answer:
- E. coli grown in ¹⁵N medium,
then transferred to ¹⁴N medium.
- DNA samples centrifuged in CsCl
gradient:
- First generation: Intermediate (hybrid) DNA.
- Second generation: Half hybrid, half light DNA.
- Conclusion: Each daughter molecule
contains one parental and one new strand — semiconservative replication.
5. Describe the enzymes involved in
DNA replication and their functions.
Answer:
1.
Helicase: Unwinds DNA helix.
2.
Topoisomerase: Removes supercoils.
3.
Primase: Synthesizes RNA primer.
4.
DNA
polymerase:
Adds nucleotides (5′→3′).
5.
Ligase: Joins Okazaki fragments.
6.
SSB
proteins: Prevent
rejoining of strands.
Result: Accurate replication ensuring genetic stability.
6. Explain the process of DNA
replication in detail.
Answer:
- Initiation: Helicase opens double helix →
replication fork forms → RNA primer added.
- Elongation: DNA polymerase adds
nucleotides to 3′ end of primer.
- Leading strand: Continuous synthesis.
- Lagging strand: Discontinuous synthesis
(Okazaki fragments).
- Termination: Ligase joins fragments;
proofreading corrects errors.
- Result: Two identical DNA
molecules, each with one old and one new strand.
7. Explain how RNA differs from DNA
structurally and functionally.
Answer:
|
Feature |
DNA |
RNA |
|
Sugar |
Deoxyribose |
Ribose |
|
Bases |
A, T, G, C |
A, U, G, C |
|
Strands |
Double |
Single |
|
Stability |
More stable |
Less stable |
|
Function |
Genetic information |
Protein synthesis, catalytic activity |
- Types of RNA: mRNA (messenger), tRNA
(transfer), rRNA (ribosomal).
- RNA is also the genetic
material in some viruses (e.g., TMV).
8. Discuss transcription in
prokaryotes with stages and enzymes involved.
Answer:
1.
Initiation:
o RNA polymerase (core + Ο factor)
binds to promoter region.
2.
Elongation:
o RNA polymerase adds complementary
ribonucleotides to DNA template.
3.
Termination:
o Rho-dependent or independent
mechanism stops transcription.
o RNA transcript released.
Result: Formation of mRNA, ready for translation.
9. How is transcription in
eukaryotes different from prokaryotes?
Answer:
- Enzymes: 3 RNA polymerases (I: rRNA,
II: mRNA, III: tRNA).
- Promoter & regulatory
sequences
more complex.
- mRNA processing:
- Capping: Methyl guanosine cap at 5′
end.
- Polyadenylation: Poly-A tail at 3′ end.
- Splicing: Introns removed, exons
joined.
- Compartmentalization: Transcription in nucleus;
translation in cytoplasm.
10. Explain post-transcriptional
modifications in eukaryotic mRNA.
Answer:
1.
Capping:
o 7-methyl guanosine cap added to 5′
end → prevents degradation.
2.
Splicing:
o Introns removed; exons joined →
functional mRNA.
3.
Polyadenylation:
o Poly-A tail added at 3′ end →
stability.
Final result: Mature mRNA ready for translation.
11. Describe the process of
translation in prokaryotes.
Answer:
1.
Initiation:
o mRNA binds to small ribosomal
subunit.
o tRNA (fMet) binds start codon (AUG).
o Large subunit joins (70S ribosome
formed).
2.
Elongation:
o New tRNAs bring amino acids to A
site; peptide bond forms at P site.
3.
Termination:
o Stop codon reached; release factor
ends translation.
4.
Product: Polypeptide chain forms a
functional protein.
12. Explain the structure and
function of tRNA.
Answer:
- Cloverleaf model:
- Anticodon loop: Pairs with mRNA codon.
- Amino acid attachment site: 3′ end (CCA).
- D loop: Recognizes aminoacyl tRNA
synthetase.
- TΨC loop: Binds to ribosome.
- Function: Adapter between codon and amino
acid.
- Ensures correct sequence during
translation.
13. Explain the concept of genetic
code and its characteristics.
Answer:
- Triplet code: 3 bases = 1 codon
= 1 amino acid.
- Universal: Same for all organisms.
- Degenerate: More than one codon per amino
acid.
- Non-overlapping: Read continuously.
- Start codon: AUG; Stop codons: UAA,
UAG, UGA.
Significance: Provides the rules for protein synthesis.
14. Explain the process of gene
expression through the lac operon model.
Answer:
- Genes: lacZ, lacY, lacA (structural);
regulated by promoter, operator, regulator gene.
- Without lactose: Repressor binds operator →
transcription blocked.
- With lactose: Lactose binds repressor →
inactivates it → RNA polymerase transcribes genes.
Result: Enzymes Ξ²-galactosidase, permease, and transacetylase produced → lactose metabolized.
→ Inducible system.
15. Describe the structure of
nucleosome and its role in DNA packaging.
Answer:
- Histone octamer: 2 each of H2A, H2B, H3, H4.
- DNA (~200 bp) wraps around the octamer.
- H1 histone seals DNA entry/exit.
- Nucleosomes coil → chromatin →
chromosome.
- Euchromatin: Active; Heterochromatin:
Inactive.
→ Efficient packaging and regulation of gene expression.
16. What are the main objectives and
outcomes of the Human Genome Project (HGP)?
Answer:
Objectives:
- Identify all human genes
(~30,000).
- Determine complete DNA sequence
(~3 × 10⁹ bp).
- Store data for analysis.
- Study genetic disorders and
evolution.
Findings: - <2% codes for proteins.
- Repetitive DNA common.
- Genes unevenly distributed.
Significance: - Helped in disease diagnosis,
forensic science, and bioinformatics.
17. Describe the technique of DNA
fingerprinting and its applications.
Answer:
Principle: VNTRs (Variable Number Tandem Repeats) differ among
individuals.
Steps:
1.
DNA
isolation.
2.
Restriction
digestion.
3.
Gel
electrophoresis.
4.
Blotting
and hybridization with radioactive probe.
5.
Autoradiography
to visualize bands.
Applications:
- Forensic identification,
paternity testing, genetic diversity studies.
Scientist: Alec Jeffreys (1985).
18. Describe the salient features of
the human genome.
Answer:
- ~3.2 billion base pairs.
- ~30,000 genes.
- Only 2% codes for proteins.
- Repetitive sequences common.
- Chromosome 1 has max genes
(~2968); Y chromosome has fewest (~231).
- Understanding it aids in disease
gene mapping and evolutionary studies.
19. Explain the flow of genetic
information as per the central dogma.
Answer:
Proposed by Francis Crick:
DNA → RNA → Protein
- Transcription: DNA → RNA.
- Translation: RNA → Protein.
In some viruses: RNA → DNA → RNA → Protein (Reverse transcription).
→ Explains how genetic information dictates protein synthesis.
20. Compare the steps of
replication, transcription, and translation.
Answer:
| Feature | Replication | Transcription | Translation |
|----------|-------------|---------------|
| Template | DNA | DNA | mRNA |
| Product | DNA | RNA | Protein |
| Enzyme | DNA polymerase | RNA polymerase | Ribosome |
| Base pairing | A–T, G–C | A–U, G–C | Codon–anticodon |
| Location | Nucleus | Nucleus | Cytoplasm |
All three ensure faithful transfer of genetic information.
(Due to space limits, I’ll continue with Questions 21–50
in the next message — covering topics like mutations, genome organization, RNA
world hypothesis, regulation, and diagram-based answers.)
Would you like me to continue with Q.21–50 now (the
remaining 5-mark answers)?
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Would you like me to continue with Q.21–50 now (the
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ChatGPT said:
Perfect π Let's continue with the remaining 5-mark
questions (Q.21–50) from Chapter 6: Molecular Basis of Inheritance
(NCERT Class 12 Biology) — prepared for both CBSE Boards & NEET.
𧬠5-Mark Questions with
Answers (Q.21 – Q.50)
Q21. Explain the process of
Transcription in Prokaryotes in detail.
Answer:
- Definition: Transcription is the process
by which RNA is synthesized from a DNA template.
- Steps:
1.
Initiation:
RNA polymerase binds to the promoter region on DNA (–35 and –10 sequences).
The Ο (sigma) factor helps in promoter recognition.
2.
Elongation:
The enzyme moves along the template strand, synthesizing RNA in the 5’→3’
direction using ribonucleoside triphosphates (rNTPs).
3.
Termination:
Transcription ends when a terminator sequence is reached.
It may be rho-dependent or rho-independent.
- Result: Formation of a single-stranded
RNA (mRNA, tRNA, or rRNA) molecule.
Q22. Describe the process of
Transcription in Eukaryotes.
Answer:
- Occurs in nucleus and involves three RNA
polymerases:
- RNA Pol I → rRNA
- RNA Pol II → mRNA
- RNA Pol III → tRNA
- Steps:
1. Initiation: RNA Pol II binds to promoter (TATA
box).
Transcription factors are required.
2. Elongation: mRNA strand grows complementary to
template DNA.
3. Termination: A termination signal releases the
RNA molecule.
Post-transcriptional
Modifications:
o 5’ capping (addition of methyl
guanosine triphosphate)
o Splicing (removal of introns)
o 3’ polyadenylation (poly-A tail
addition)
Final
Product: Mature
mRNA ready for translation.
Q23. Explain the Central Dogma of
Molecular Biology.
Answer:
- Proposed by Francis Crick.
- Concept: Flow of genetic information
occurs as:
DNA → RNA → Protein - In Prokaryotes: Direct transcription and
translation occur in cytoplasm.
- In Eukaryotes:
- DNA (in nucleus) transcribed
to mRNA.
- mRNA transported to cytoplasm
for translation.
- Reverse Transcription: RNA → DNA (in retroviruses
like HIV using reverse transcriptase).
- Importance: Explains how genes control
traits by directing protein synthesis.
Q24. What is the difference between
Transcription and Translation?
Answer:
|
Feature |
Transcription |
Translation |
|
Definition |
Synthesis of RNA from DNA |
Synthesis of protein from mRNA |
|
Enzyme |
RNA Polymerase |
Ribosome |
|
Template |
DNA |
mRNA |
|
Product |
mRNA, tRNA, rRNA |
Polypeptide chain |
|
Location |
Nucleus (Eukaryotes) |
Cytoplasm/Ribosome |
|
Significance |
Transfers genetic code to RNA |
Expresses genetic code as protein |
Q25. Explain the process of
Translation in Prokaryotes.
Answer:
- Definition: Process of polypeptide
synthesis from mRNA sequence.
- Steps:
1.
Activation
of Amino Acids:
Amino acids attach to respective tRNAs (aminoacyl-tRNA synthetase).
2.
Initiation:
Ribosome binds to mRNA at Shine-Dalgarno sequence.
Start codon (AUG) recognized by tRNA^fMet.
3.
Elongation:
New aminoacyl-tRNA binds at A site. Peptide bond forms between amino acids.
4.
Translocation:
Ribosome moves one codon ahead; empty tRNA exits from E site.
5.
Termination:
Stop codon (UAA, UAG, UGA) releases the polypeptide.
- Product: Functional polypeptide chain.
Q26. Explain the mechanism of DNA
replication in detail.
Answer:
- Definition: Semi-conservative process
where each new DNA molecule has one old strand and one new strand.
- Enzymes Involved:
- Helicase: Unwinds DNA double helix.
- Primase: Synthesizes RNA primer.
- DNA Polymerase III: Adds nucleotides.
- DNA Ligase: Joins Okazaki fragments.
- DNA Polymerase I: Removes RNA primer.
- Steps:
1. Initiation: At ori site.
2. Elongation: Continuous on leading strand;
discontinuous on lagging strand.
3. Termination: Replication forks meet.
Result: Two identical DNA molecules.
Q27. What is a Genetic Code? Mention
its main features.
Answer:
- Definition: Sequence of nucleotides on
mRNA that determines amino acid sequence.
- Features:
1.
Triplet
code (3 bases → 1 amino acid)
2.
Universal
(same in all organisms)
3.
Degenerate
(more than one codon for an amino acid)
4.
Non-overlapping
and Commaless
5.
Start
codon: AUG (Methionine)
6.
Stop
codons: UAA, UAG, UGA
- Example:
AUG → Methionine
UUU → Phenylalanine
Q28. What are Mutations? Explain
types with examples.
Answer:
- Definition: Sudden heritable change in DNA
sequence.
- Types:
1.
Point
Mutation: Change in
a single base pair.
Example: Sickle cell anemia (GAG → GTG).
2.
Frameshift
Mutation: Insertion
or deletion of bases.
Example: Thalassemia.
- Causes: Errors in replication,
chemicals, radiation.
- Effect: Can alter protein structure
and function.
Q29. Explain the process of RNA
processing in Eukaryotes.
Answer:
1.
Capping:
Addition of methyl guanosine triphosphate at 5’ end (for stability and ribosome
recognition).
2.
Splicing:
Removal of introns and joining of exons (by spliceosome).
3.
Polyadenylation:
Addition of poly-A tail (200 adenine residues) at 3’ end.
4.
Result:
Formation of mature mRNA ready for export to cytoplasm.
Q30. Explain the structure and role
of tRNA.
Answer:
- Structure:
- Cloverleaf model.
- Three arms:
1.
Anticodon
arm: Base pairs with codon on mRNA.
2.
Amino
acid acceptor arm:
Attaches specific amino acid.
3.
TΟC
and DHU arms:
Help in binding to ribosome.
- Role:
- Brings amino acids during
translation.
- Decodes codons on mRNA.
Q31. What are the differences
between Prokaryotic and Eukaryotic Transcription?
Answer:
|
Feature |
Prokaryotes |
Eukaryotes |
|
Enzyme |
One RNA polymerase |
Three RNA polymerases |
|
Location |
Cytoplasm |
Nucleus |
|
RNA Processing |
Absent |
Present (splicing, capping, tailing) |
|
Transcription & Translation |
Coupled |
Separate |
|
Promoter |
Pribnow box |
TATA box |
Q32. Describe Hershey and Chase
experiment.
Answer:
- Aim: To prove DNA is the genetic
material.
- Organism: Bacteriophage (virus infecting
bacteria).
- Method:
- Used radioactive isotopes:
³²P → DNA, ³⁵S → Protein. - Infection of E. coli → Blender
→ Centrifugation.
- Observation:
- ³²P found inside bacteria.
- ³⁵S remained outside.
- Conclusion: DNA is the genetic material.
Q33. Explain Meselson and Stahl’s
experiment.
Answer:
- Aim: To prove semi-conservative
replication of DNA.
- Method:
- Grew E. coli in ¹⁵N
medium → transferred to ¹⁴N medium.
- DNA extracted after each
generation and centrifuged.
- Result:
- 1st generation → hybrid
(¹⁵N–¹⁴N)
- 2nd generation → half hybrid,
half light (¹⁴N–¹⁴N)
- Conclusion: Replication is
semi-conservative.
Q34. Explain the structure of a
nucleosome.
Answer:
- Definition: Structural unit of chromatin.
- Structure:
- DNA (146 bp) wrapped around
histone octamer (H2A, H2B, H3, H4 × 2).
- Linked by linker DNA (H1
histone).
- Function: DNA packaging and gene
regulation.
Q35. Explain the packaging of DNA in
Eukaryotes.
Answer:
- Step 1: DNA + histones → Nucleosome
(beads on a string)
- Step 2: Nucleosomes coil → Solenoid
(30 nm fiber)
- Step 3: Loops attached to scaffold
proteins → Chromatin fiber
- Step 4: Supercoiling forms visible
chromosomes during cell division.
Q36. What is the role of regulatory
genes in lac operon?
Answer:
- Regulatory gene (i): Produces repressor protein.
- When lactose absent:
Repressor binds to operator → transcription blocked. - When lactose present:
Lactose acts as inducer → binds repressor → detaches → transcription occurs. - Result:
Structural genes (Z, Y, A) express to metabolize lactose.
Q37. Differentiate between
Euchromatin and Heterochromatin.
Answer:
|
Property |
Euchromatin |
Heterochromatin |
|
Structure |
Loosely packed |
Densely packed |
|
Function |
Transcriptionally active |
Inactive |
|
Location |
Central part of nucleus |
Near nuclear membrane |
|
Staining |
Light |
Dark |
Q38. Explain the structure of DNA as
proposed by Watson and Crick.
Answer:
- Double helix model:
- Two antiparallel
polynucleotide chains.
- Sugar-phosphate backbone
outside, bases inside.
- Base pairing rule:
- A = T (2 H-bonds)
- G = C (3 H-bonds)
- Helix: Right-handed, 10 bp per turn.
- Diameter: 2 nm.
- Pitch: 3.4 nm.
Q39. Describe Griffith’s
transformation experiment.
Answer:
- Organism: Streptococcus pneumoniae.
- Types: Smooth (S – virulent), Rough
(R – non-virulent).
- Observation:
1.
S
→ killed mice
2.
R
→ alive
3.
Heat-killed
S → alive
4.
Heat-killed
S + R → killed mice
- Conclusion:
Some “transforming principle” from dead S converted R → S.
Later, Avery identified it as DNA.
Q40. Explain the role of DNA
polymerase in replication.
Answer:
- Adds nucleotides complementary
to the template.
- Proofreads new strand (3’→5’
exonuclease activity).
- Synthesizes continuously
(leading) and discontinuously (lagging).
- Ensures fidelity and repair of
DNA.
Q41. Explain the structure and
function of a Lac Operon.
Answer:
- Concept: Proposed by Jacob and Monod
in E. coli — it’s a model for gene regulation.
- Components:
1.
Structural
genes:
§ lacZ → Ξ²-galactosidase (breaks lactose →
glucose + galactose)
§ lacY → Permease (helps lactose entry)
§ lacA → Transacetylase
2.
Regulatory
gene (i): Produces repressor
protein.
3.
Promoter
(P): Binding site for RNA polymerase.
4.
Operator
(O): Binding site for repressor.
- Working:
o Without lactose:
Repressor binds to operator → transcription blocked.
o With lactose:
Lactose acts as inducer, binds repressor → repressor inactive →
structural genes transcribed.
Result: Enzymes for lactose metabolism are
produced only when needed.
Significance: Shows how genes are turned “ON” or
“OFF” in prokaryotes.
Q42. What are the types of RNA and
their functions?
Answer:
|
Type |
Full Form |
Function |
|
mRNA |
Messenger RNA |
Carries genetic code from DNA to ribosome |
|
tRNA |
Transfer RNA |
Brings amino acids during protein synthesis |
|
rRNA |
Ribosomal RNA |
Forms structural & functional part of ribosome |
- Additional notes:
- mRNA is linear and
short-lived.
- tRNA is cloverleaf-shaped; has
anticodon and amino acid site.
- rRNA binds proteins to form
ribosomes.
- Together, they translate genetic
information into proteins.
Q43. Explain how protein synthesis
occurs in eukaryotes.
Answer:
Steps:
1.
Transcription
(in nucleus):
DNA → pre-mRNA → mature mRNA (via splicing, capping, tailing)
2.
Export
to cytoplasm:
mRNA leaves nucleus via nuclear pores.
3.
Translation
(in cytoplasm):
o Initiation: Ribosome binds mRNA at start codon
(AUG).
First tRNA brings Methionine.
o Elongation:
tRNAs bring amino acids sequentially; peptide bonds form.
o Termination:
Stop codon reached → polypeptide released.
4.
Post-translational
Modifications:
Folding, cleavage, or adding chemical groups for final protein structure.
Result:
Functional protein synthesized as per gene code.
Q44. Explain the flow of genetic
information in a retrovirus like HIV.
Answer:
- Normal flow: DNA → RNA → Protein
- Retrovirus flow: RNA → DNA → RNA → Protein
- Mechanism:
1.
Virus
enters host cell and releases RNA.
2.
Uses
reverse transcriptase enzyme to make complementary DNA (cDNA).
3.
cDNA
integrates with host DNA → forms provirus.
4.
Host
machinery produces viral RNA and proteins.
5.
New
virus particles assemble and exit cell.
- Significance: Reverse flow of information
defies the central dogma and explains retroviral infection.
Q45. Explain how gene expression is
regulated in eukaryotes.
Answer:
Levels of regulation:
1.
Transcriptional
control:
o Promoters, enhancers, repressors
control RNA synthesis.
2.
Post-transcriptional
control:
o Splicing variations, mRNA capping,
and tailing affect stability.
3.
Translational
control:
o Regulatory proteins bind mRNA to
control translation speed.
4.
Post-translational
control:
o Proteins activated/degraded as
needed.
5.
Epigenetic
control:
o DNA methylation and histone
modification alter gene activity.
Significance:
Helps cells differentiate and maintain specialized functions.
Q46. Differentiate between
replication, transcription, and translation.
Answer:
|
Feature |
Replication |
Transcription |
Translation |
|
Definition |
DNA → DNA |
DNA → RNA |
RNA → Protein |
|
Enzyme |
DNA Polymerase |
RNA Polymerase |
Ribosome |
|
Template |
DNA |
DNA |
mRNA |
|
Location |
Nucleus (Eukaryotes) |
Nucleus |
Cytoplasm |
|
Direction |
5′→3′ |
5′→3′ |
5′→3′ |
|
Product |
Two DNA molecules |
mRNA, tRNA, rRNA |
Polypeptide |
- All three processes ensure accurate gene
expression and inheritance.
Q47. Explain how mutations can
affect protein synthesis.
Answer:
- Mutation: Change in DNA sequence →
altered codons → altered protein.
- Effects:
1.
Silent
Mutation: No change
in amino acid → no effect.
2.
Missense
Mutation: Changes
one amino acid (e.g., Sickle Cell Anaemia).
3.
Nonsense
Mutation: Converts
amino acid codon → stop codon → truncated protein.
4.
Frameshift
Mutation:
Insertion/deletion changes reading frame → nonfunctional protein.
- Example:
Sickle Cell Anaemia → substitution of Glu by Val due to base change (GAG → GTG).
Q48. Explain how chromatin organization
affects gene expression.
Answer:
- Chromatin = DNA + Histone
proteins.
- Types:
- Euchromatin: Loosely packed → active
genes.
- Heterochromatin: Tightly packed → inactive
genes.
- Histone modification:
- Acetylation: Opens chromatin →
transcription active.
- Methylation: Closes chromatin →
transcription repressed.
- Epigenetic regulation:
Modifications inherited without DNA sequence change. - Example: X-chromosome inactivation in
females (Barr body).
Q49. Discuss the concept of the
Human Genome Project (HGP).
Answer:
- Launched: 1990; completed in 2003.
- Objectives:
1.
Identify
~20,000–25,000 human genes.
2.
Determine
complete DNA sequence.
3.
Store
and analyze data (bioinformatics).
4.
Study
variations and diseases.
- Findings:
o Human genome: ~3 billion base pairs.
o Only 2% codes for proteins.
o Many genes are shared with other
organisms.
Applications:
o Gene therapy, diagnosis,
personalized medicine.
Importance: Foundation for genomics and
biotechnology.
Q50. What is DNA fingerprinting?
Explain its principle and applications.
Answer:
- Definition: Technique to identify
individuals based on unique DNA sequences.
- Principle:
Based on polymorphism in VNTRs (Variable Number Tandem Repeats). - Steps:
1.
DNA
extraction.
2.
Restriction
enzyme digestion.
3.
Gel
electrophoresis separation.
4.
Hybridization
using radioactive probes.
5.
Autoradiography
to visualize band pattern.
- Applications:
o Forensic identification.
o Paternity testing.
o Genetic disorder detection.
o Evolutionary studies.
Developed
by: Alec Jeffreys (1985).
