Question
Trace the complete pathway from a gene in the DNA to a functional protein. Include transcription, RNA processing (in eukaryotes), and translation. Name the key enzymes and molecules involved at each step.
(NEET + CBSE 12 pattern)
Solution — Step by Step
RNA polymerase binds to the promoter region of the gene and synthesises mRNA using the template (antisense) strand of DNA.
Key points:
- RNA polymerase reads template strand 3’ 5’
- mRNA is synthesised 5’ 3’
- The mRNA sequence is complementary to the template strand and identical to the coding (sense) strand (except U replaces T)
- Transcription ends at the terminator sequence
In eukaryotes: RNA polymerase II transcribes mRNA. In prokaryotes: a single RNA polymerase does all transcription.
The primary transcript (pre-mRNA) undergoes three modifications:
- 5’ capping — a methylated guanine cap is added to the 5’ end (protects from degradation, aids ribosome binding)
- 3’ polyadenylation — a poly-A tail (200+ adenines) is added to the 3’ end (stability, export from nucleus)
- Splicing — introns (non-coding sequences) are removed by the spliceosome; exons (coding sequences) are joined together
The mature mRNA then exits the nucleus through nuclear pores.
Occurs on ribosomes in the cytoplasm. Three phases:
Initiation: Small ribosomal subunit binds to mRNA at the 5’ cap, scans for the start codon AUG. The initiator tRNA (carrying methionine) binds. Large subunit joins.
Elongation: Codons are read 5’ 3’. Each codon is matched by an anticodon on a tRNA carrying the appropriate amino acid. Peptide bonds form between adjacent amino acids. The ribosome moves one codon at a time.
Termination: A stop codon (UAA, UAG, or UGA) is reached. Release factors bind, the polypeptide is released, and the ribosome dissociates.
flowchart TD
A["Gene on DNA"] --> B["Transcription by RNA polymerase"]
B --> C["Pre-mRNA (eukaryotes)"]
C --> D["5' capping + 3' poly-A tail"]
D --> E["Splicing: introns removed"]
E --> F["Mature mRNA exits nucleus"]
F --> G["Ribosome binds at AUG"]
G --> H["tRNAs bring amino acids"]
H --> I["Peptide bond formation"]
I --> J["Polypeptide chain grows"]
J --> K["Stop codon reached"]
K --> L["Functional protein after folding"]Why This Works
The central dogma of molecular biology — DNA RNA Protein — describes the flow of genetic information. DNA stores the instructions, mRNA carries a temporary copy to the ribosome, and tRNA translates the nucleotide language into the amino acid language of proteins.
The genetic code is a triplet code — every 3 nucleotides (codon) code for one amino acid. With 4 bases, there are possible codons coding for only 20 amino acids, making the code degenerate (multiple codons per amino acid). The code is nearly universal — the same in almost all organisms.
Alternative Method — Quick Summary Table
| Stage | Location | Template | Product | Key Enzyme |
|---|---|---|---|---|
| Transcription | Nucleus | DNA template strand | Pre-mRNA | RNA polymerase |
| Processing | Nucleus | Pre-mRNA | Mature mRNA | Spliceosome, capping enzymes |
| Translation | Ribosome (cytoplasm) | mRNA | Polypeptide | Ribosome, aminoacyl-tRNA synthetase |
For NEET, remember the key differences between prokaryotic and eukaryotic gene expression: prokaryotes have no introns (no splicing needed), transcription and translation are coupled (occur simultaneously), and there is no 5’ cap or poly-A tail. In eukaryotes, transcription occurs in the nucleus and translation in the cytoplasm — they are separated in space and time.
Common Mistake
Students confuse the template strand and coding strand. RNA polymerase reads the template (antisense) strand 3’ to 5’, producing mRNA that is complementary to the template strand. The mRNA sequence matches the coding (sense) strand (with U instead of T). When a question gives you the “DNA sequence” and asks for the mRNA, first check whether they gave you the coding strand or the template strand. If it is the coding strand, just replace T with U. If it is the template strand, write the complement and replace T with U.