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Transcription and Translation

BiologyMolecular Basis of InheritanceFor NEET aspirants

Transcription and translation are the two steps by which the information in a gene becomes a protein. This page covers the transcription unit, template and coding strands, cistrons, exons and introns, the three types of RNA, and transcription in bacteria with the sigma and rho factors. It then explains the three eukaryotic RNA polymerases, splicing, capping and tailing, and translation on the ribosome, from charging of tRNA to the release factor. NEET often tests strand polarity, the roles of RNA polymerases and RNA processing in transcription and translation.

On this page1Transcription2Transcription unit3Gene and cistron4Bacteria5Eukaryotes6Translation7Exam essentials8Quick revision9Solved examples10Practice
Key Points at a Glance
  1. ★ Must learn Transcription: copying genetic information from one strand of DNA into RNA; adenine pairs with uracil.
  2. Transcription unit = promoter + structural gene + terminator.
  3. ★ Must learn Template strand: polarity , copied. Coding strand: polarity , same sequence as RNA (T in place of U), displaced.
  4. Promoter: upstream (towards the 5′ end of the coding strand), binds RNA polymerase. Terminator: downstream (3′ end).
  5. Cistron: DNA segment coding for a polypeptide. Monocistronic genes mostly in eukaryotes, polycistronic mostly in bacteria.
  6. ★ Must learn Bacteria: one RNA polymerase, which needs the sigma () factor to initiate and the rho () factor to terminate.
  7. ★ Must learn Eukaryotes: RNA polymerase I (28S, 18S, 5.8S rRNA), II (hnRNA), III (tRNA, 5S rRNA, snRNAs).
  8. hnRNA processing: splicing (introns out), capping (methyl guanosine triphosphate at the 5′ end), tailing (200-300 adenylates at the 3′ end).
  9. Translation: amino acids are activated with ATP and linked to tRNA (charging); the ribosome joins them by peptide bonds.
  10. ★ Must learn Translation starts at AUG with the initiator tRNA and ends when a release factor binds a stop codon; 23S rRNA is a ribozyme.

1. Transcription

  • ★ Exam imp Transcription: the process of copying genetic information from one strand of DNA into RNA.
  • The principle of complementarity governs transcription too, except that adenine now pairs with uracil instead of thymine.
  • In replication, once it begins, the total DNA of an organism is duplicated.
  • In transcription, only a segment of DNA, and only one of the strands, is copied into RNA.
  • So the boundaries of the region, and the strand to be transcribed, must be defined.

1.1 Why both strands are not copied

  1. If both strands acted as templates, they would code for RNAs with different sequences, since complementary does not mean identical.
  2. These RNAs would code for proteins with different amino acid sequences, so one DNA segment would code for two different proteins. This would complicate genetic information transfer.
  3. The two RNAs, made at the same time, would be complementary and would pair into a double-stranded RNA.
  4. Double-stranded RNA cannot be translated into protein, so transcription would become futile.
Key idea
Transcription copies only one strand of only one segment, so that each gene gives one usable RNA.

2. The Transcription Unit

  • ★ Exam imp A transcription unit in DNA is defined mainly by three regions: a promoter, the structural gene and a terminator.
  • The two DNA strands of the structural gene have opposite polarity.
  • DNA-dependent RNA polymerase also polymerises in only one direction, .
  • So the strand with polarity acts as the template, and is called the template strand.
  • The other strand, with polarity , has the same sequence as the RNA (except thymine in place of uracil). It is displaced during transcription.
  • Strangely, this strand, which does not code for anything, is called the coding strand.
  • All reference points of a transcription unit are defined with respect to the coding strand.
StrandHypothetical sequence
Template strand3′-ATGCATGCATGCATGCATGCATGC-5′
Coding strand5′-TACGTACGTACGTACGTACGTACG-3′
RNA transcribed5′-UACGUACGUACGUACGUACGUACG-3′
  • The promoter and terminator flank the structural gene.
  • ★ Exam imp The promoter lies towards the 5′ end (upstream) of the structural gene, judged by the polarity of the coding strand.
  • It is a DNA sequence that provides the binding site for RNA polymerase.
  • The presence of a promoter also defines which strand is the template and which is the coding strand.
  • If the promoter switched places with the terminator, the definitions of the coding and template strands would be reversed.
  • The terminator lies towards the 3′ end (downstream) of the coding strand and usually defines the end of transcription.
  • Additional regulatory sequences may lie further upstream or downstream of the promoter; they matter in the regulation of gene expression.
Schematic structure of a transcription unit Two DNA strands run in opposite directions. The upper strand, running 3 prime to 5 prime from left to right, is the template strand; the lower strand, running 5 prime to 3 prime, is the coding strand. A promoter box on the left and a terminator box on the right flank the structural gene. A bent arrow at the right edge of the promoter marks the transcription start site. Labels: promoter, transcription start site, structural gene, template strand, coding strand, terminator, 3 prime and 5 prime ends. Transcription start site Promoter Terminator Structural gene Template strand Coding strand 3′ 5′ 5′ 3′
Figure 1: Schematic structure of a transcription unit. The promoter and terminator flank the structural gene; the template strand runs 3′ to 5′ and the coding strand 5′ to 3′.
★ Very important The RNA has the same sequence as the coding strand, with U in place of T, and is complementary to the template strand. The template strand runs , so the RNA grows .
Memory Trick Coding strand = Copy of the RNA (both start with C); the Template strand is Turned round (3′ to 5′). Promoter comes Prior (upstream); Terminator comes Trailing (downstream).
Tips and Tricks To write mRNA from the coding strand, just change every T to U. From the template strand, write the complement and use U for A. Always write the RNA , the same direction as the coding strand.
Key idea
Promoter upstream, terminator downstream; the 3′-to-5′ template is copied and the 5′-to-3′ coding strand matches the RNA.

3. Transcription Unit and the Gene

  • ★ Exam imp A gene is defined as the functional unit of inheritance.
  • Genes are certainly located on DNA, but a gene is hard to define as a literal DNA sequence.
  • DNA sequences coding for a tRNA or rRNA molecule also define genes.
  • Cistron: a segment of DNA coding for a polypeptide.
  • ★ Exam imp So the structural gene of a transcription unit is monocistronic (mostly in eukaryotes) or polycistronic (mostly in bacteria or prokaryotes).
  • In eukaryotes, the monocistronic structural genes have interrupted coding sequences: the genes are split.
  • Exons: the coding or expressed sequences, which appear in mature or processed RNA.
  • Introns: intervening sequences that interrupt the exons and do not appear in mature RNA.
  • The split-gene arrangement makes defining a gene as a DNA segment even harder.
  • The inheritance of a character is also affected by the promoter and regulatory sequences of a structural gene.
  • So regulatory sequences are sometimes loosely called regulatory genes, even though they do not code for any RNA or protein.
  • In simple terms, a segment of DNA that codes for an RNA may be called a gene.
ExonsCoding (expressed) sequences.
Appear in mature RNA.
Joined in a defined order by splicing.
IntronsIntervening sequences.
Do not appear in mature RNA.
Removed by splicing.
Memory Trick EXons EXit the nucleus in mature mRNA; INtrons stay IN and are cut out.
Key idea
A cistron codes for one polypeptide; eukaryotic genes are monocistronic and split into exons and introns.

4. Types of RNA and Transcription in Bacteria

  • In bacteria there are three major types of RNA, and all three are needed to synthesise a protein.
RNARole in protein synthesis
mRNA (messenger RNA)Provides the template
tRNA (transfer RNA)Brings amino acids and reads the genetic code
rRNA (ribosomal RNA)Plays structural and catalytic roles during translation
  • ★ Exam imp A single DNA-dependent RNA polymerase transcribes all types of RNA in bacteria.
  1. Initiation: RNA polymerase binds the promoter and starts transcription.
  2. Elongation: it uses nucleoside triphosphates as substrate and polymerises them on the template, following complementarity. It also helps open the helix as it moves on.
  3. Only a short stretch of RNA stays bound to the enzyme.
  4. Termination: when the polymerase reaches the terminator region, the nascent RNA falls off, and so does the RNA polymerase.
  • How does one enzyme manage all three steps? The RNA polymerase itself can only catalyse elongation.
  • ★ Exam imp It associates transiently with the initiation factor (sigma, ) to initiate and the termination factor (rho, ) to terminate.
  • These factors alter the specificity of the RNA polymerase, so that it either initiates or terminates.
Process of transcription in bacteria Three stages on a DNA double helix that carries a promoter (amber) and a terminator (pink). Initiation: RNA polymerase joined to the sigma factor binds the promoter. Elongation: the polymerase opens the helix and the growing RNA leaves it; the sigma factor has come off. Termination: at the terminator the RNA is released and the polymerase leaves with the rho factor. Labels: promoter, RNA polymerase, sigma factor, DNA helix, terminator, RNA, rho factor, 3 prime and 5 prime ends. σ σ ρ Initiation Elongation Termination 3′ 5′ 5′ 3′ 3′ 5′ 5′ 3′ 3′ 5′ 5′ 3′ Promoter RNA polymerase Sigma factor DNA helix RNA Terminator RNA RNA polymerase Rho factor
Figure 2: Process of transcription in bacteria. RNA polymerase uses the sigma factor to start at the promoter and the rho factor to stop at the terminator; by itself it only elongates.
  • In bacteria, the mRNA needs no processing to become active.
  • Transcription and translation also take place in the same compartment, since there is no separation of cytosol and nucleus.
  • So translation often begins well before the mRNA is fully transcribed: transcription and translation can be coupled in bacteria.
Memory Trick Sigma Starts, Rho Retires: helps initiation and helps termination; the core enzyme does the elongation in between.
Key idea
One bacterial RNA polymerase elongates; sigma and rho let it start and stop, and its mRNA is translated even while being made.

5. Transcription in Eukaryotes

  • Eukaryotes add two more complexities to transcription.
  • ★ Exam imp First, there are at least three RNA polymerases in the nucleus, besides the RNA polymerase found in the organelles. There is a clear division of labour.
EnzymeTranscribes
RNA polymerase IrRNAs: 28S, 18S and 5.8S
RNA polymerase IIPrecursor of mRNA, the heterogeneous nuclear RNA (hnRNA)
RNA polymerase IIItRNA, 5S rRNA and snRNAs (small nuclear RNAs)
Memory Trick I, II, III = r, m, t: polymerase I makes rRNA, II makes mRNA (as hnRNA), III makes tRNA. The odd one out, 5S rRNA, goes with III, along with snRNAs.
  • Second, the primary transcripts contain both exons and introns and are non-functional.
  1. Splicing: the introns are removed and the exons are joined in a defined order.
  2. Capping: an unusual nucleotide, methyl guanosine triphosphate, is added to the 5′ end of hnRNA.
  3. Tailing: 200-300 adenylate residues are added at the 3′ end, in a template-independent manner.
  4. The fully processed hnRNA, now called mRNA, is transported out of the nucleus for translation.
Process of transcription in eukaryotes Top: RNA polymerase transcribes a split gene into a primary transcript. Capping: a methyl guanosine triphosphate cap is added at the 5 prime end of the transcript, which has exons (orange) separated by introns (grey). RNA splicing: the introns loop out and are removed, and polyadenylation adds a poly A tail (brown) at the 3 prime end. Bottom: the messenger RNA has only joined exons, the cap and the tail. Labels: capping, cap, exon, intron, RNA splicing, polyadenylation, poly A tail, messenger RNA. mGppp mGppp mGppp Capping RNA splicing Messenger RNA 3′ 5′ 5′ 3′ 5′ 3′ Cap Exon Intron 5′ 3′ Polyadenylation Poly A tail 5′ 3′
Figure 3: Process of transcription in eukaryotes. The primary transcript is capped at the 5′ end, its introns are spliced out and a poly A tail is added at the 3′ end to give mRNA.
  • The meaning of these complexities is now beginning to be understood.
  • The split-gene arrangement is probably an ancient feature of the genome; the presence of introns is reminiscent of antiquity.
  • Splicing represents the dominance of the RNA world.
  • In recent times, RNA and RNA-dependent processes in living systems have gained more importance.
NEET Focus Polymerase questions hide the trap in one product: 5S rRNA is made by RNA polymerase III, while the 28S, 18S and 5.8S rRNAs come from polymerase I. Also check the ends: the cap goes on the 5′ end and the tail on the 3′ end, and tailing does not need a template.
Quick Recall: tap to check
In Figure 3, what is added at the 5′ end of the transcript?
The cap, an unusual nucleotide, methyl guanosine triphosphate.
Which RNA polymerase makes hnRNA?
RNA polymerase II.
Why can translation start before transcription ends in bacteria?
Their mRNA needs no processing, and both processes occur in the same compartment.
Key idea
Eukaryotes use three nuclear RNA polymerases, then splice, cap and tail hnRNA into mRNA before it leaves the nucleus.

6. Translation

  • Recall: mRNA is read three bases at a time, and each codon stands for one amino acid. A tRNA reads a codon through its complementary anticodon and carries a specific amino acid.
  • ★ Exam imp Translation: the process of polymerisation of amino acids to form a polypeptide.
  • The order and sequence of amino acids are defined by the sequence of bases in the mRNA.
  • The amino acids are joined by a peptide bond, and forming it needs energy.

6.1 Charging of tRNA

  • In the first phase, amino acids are activated in the presence of ATP and linked to their cognate tRNA (the tRNA meant for that amino acid).
  • ★ Exam imp This is called charging of tRNA, or more precisely aminoacylation of tRNA.
  • When two charged tRNAs are brought close enough, peptide bond formation between them is favoured energetically.
  • A catalyst would increase the rate of peptide bond formation.

6.2 The ribosome

  • The ribosome is the cellular factory that synthesises proteins.
  • It consists of structural RNAs and about 80 different proteins.
  • In its inactive state, it exists as two subunits: a large subunit and a small subunit.
  • Translation begins when the small subunit meets an mRNA.
  • The large subunit has two sites where successive amino acids bind, close enough to form a peptide bond.
  • ★ Exam imp The ribosome also acts as the catalyst for peptide bond formation. In bacteria, the 23S rRNA is this enzyme, a ribozyme.
★ Very important A ribozyme is an RNA enzyme. The 23S rRNA of the bacterial ribosome catalyses peptide bond formation, which shows that translation evolved around RNA and that life began around RNA.

6.3 The translational unit

  • A translational unit in mRNA is the RNA sequence flanked by the start codon (AUG) and the stop codon. It codes for a polypeptide.
  • An mRNA also has extra sequences that are not translated: the untranslated regions (UTRs).
  • UTRs lie at the 5′ end (before the start codon) and at the 3′ end (after the stop codon).
  • They are needed for an efficient translation process.

6.4 Steps of translation

  1. Initiation: the ribosome binds the mRNA at the start codon (AUG), which is recognised only by the initiator tRNA.
  2. Elongation: complexes of an amino acid linked to tRNA bind one after another to the right codon in mRNA, by complementary base pairing with the tRNA anticodon.
  3. The ribosome moves from codon to codon along the mRNA, and amino acids are added one by one in the sequence dictated by DNA and represented by mRNA.
  4. Termination: at the end, a release factor binds the stop codon, ends translation and releases the complete polypeptide from the ribosome.
Translation A ribosome sits on an mRNA read from 5 prime to 3 prime: AUG UCU CUG GGC AAA GCU UGG UUC. Two charged tRNAs sit in the two sites of the large subunit; their anticodons UUU and CGA pair with the codons AAA and GCU. The growing polypeptide chain Met-Ser-Leu-Gly-Lys hangs from the tRNA in the first site, and alanine sits on the tRNA in the second. An empty tRNA leaves on the left and a tRNA carrying tryptophan arrives on the right. Labels: growing polypeptide chain, tRNA, mRNA, ribosome, 5 prime and 3 prime ends. G C A U G U C U C U G G G C A A A G C U U G G U U C A A U U U Lys C G A Ala Gly Leu Ser Met C C G A C C Trp Growing polypeptide chain tRNA tRNA tRNA tRNA mRNA Ribosome 5′ 3′
Figure 4: Translation. Each charged tRNA pairs its anticodon with a codon on the mRNA, and the ribosome joins the amino acids one by one into a polypeptide.
FeatureTranscriptionTranslation
ProductRNAPolypeptide
TemplateOne DNA strand (template strand)mRNA
Main machineryDNA-dependent RNA polymeraseRibosome with charged tRNAs
Start signalPromoterStart codon AUG with the initiator tRNA
Stop signalTerminatorStop codon with a release factor
Memory Trick Charge, Start, Stretch, Stop: charge tRNA with ATP, start at AUG, stretch the chain codon by codon, stop when the release factor meets a stop codon.
Tips and Tricks The two essential roles of the ribosome are easy to tell apart: it is the platform that binds mRNA and holds two charged tRNAs side by side, and it is the catalyst (23S rRNA) that forms the peptide bond.
Quick Recall: tap to check
In Figure 4, which anticodon pairs with the codon GCU?
CGA, written under the codon so that G faces C, C faces G and U faces A.
What are the UTRs, and where are they?
Untranslated regions of mRNA, before the start codon at the 5′ end and after the stop codon at the 3′ end.
Key idea
Charged tRNAs read codons on mRNA while the ribosome, through its 23S rRNA ribozyme, links the amino acids from AUG to a stop codon.

7. Exam Essentials

Pairs to Match

List IList II
PromoterBinding site for RNA polymerase, upstream
TerminatorDefines the end of transcription, downstream
Coding strandSame sequence as RNA, with T in place of U
CistronDNA segment coding for a polypeptide
ExonsSequences that appear in mature RNA
Sigma factorInitiation of transcription in bacteria
Rho factorTermination of transcription in bacteria
RNA polymerase I28S, 18S and 5.8S rRNA
RNA polymerase IIhnRNA, the precursor of mRNA
RNA polymerase IIItRNA, 5S rRNA and snRNAs
CappingMethyl guanosine triphosphate at the 5′ end
Tailing200-300 adenylate residues at the 3′ end
Charging of tRNAAmino acid linked to its tRNA using ATP
23S rRNARibozyme for peptide bond formation
Release factorBinds the stop codon and ends translation
Exceptions
  • In transcription, adenine pairs with uracil, not thymine.
  • The coding strand, despite its name, does not code for anything; it is displaced.
  • Bacteria have a single RNA polymerase; eukaryotes have at least three in the nucleus, besides those in organelles.
  • Only bacterial mRNA needs no processing, so transcription and translation can be coupled.
  • 5S rRNA is made by RNA polymerase III, not by polymerase I.
  • Tailing is the only processing step described as template independent.
  • Regulatory sequences, loosely called regulatory genes, do not code for any RNA or protein.
  • The start codon is recognised only by the initiator tRNA.

Numbers to Remember

  • 3 regions in a transcription unit; 3 major RNAs; at least 3 nuclear RNA polymerases.
  • rRNAs 28S, 18S and 5.8S (polymerase I); 5S rRNA (polymerase III); 23S rRNA, the bacterial ribozyme.
  • 200-300 adenylate residues in the tail.
  • About 80 different proteins in a ribosome; 2 subunits; 2 sites in the large subunit.

8. Quick Revision

  • Transcription copies one strand of one DNA segment into RNA; A pairs with U.
  • Copying both strands would give two proteins per segment, or double-stranded RNA that cannot be translated.
  • Transcription unit: promoter (upstream), structural gene, terminator (downstream).
  • Template strand ; coding strand , same as RNA with T for U.
  • Gene: functional unit of inheritance; cistron codes for a polypeptide.
  • Monocistronic genes mostly in eukaryotes; polycistronic mostly in prokaryotes.
  • Eukaryotic genes are split: exons stay in mature RNA, introns do not.
  • Bacteria: mRNA, tRNA and rRNA; one RNA polymerase; for initiation and for termination.
  • Bacterial transcription and translation can be coupled.
  • Eukaryotes: polymerase I (28S, 18S, 5.8S rRNA), II (hnRNA), III (tRNA, 5S rRNA, snRNA).
  • hnRNA is spliced, capped at 5′ (methyl guanosine triphosphate) and tailed at 3′ (200-300 adenylates) to give mRNA.
  • Translation: charging (aminoacylation) of tRNA with ATP; peptide bonds need energy.
  • Ribosome: about 80 proteins and structural RNAs; two subunits; two sites; 23S rRNA is a ribozyme.
  • Translational unit lies between AUG and a stop codon; UTRs flank it.
  • Initiation at AUG with initiator tRNA, elongation codon by codon, termination by a release factor.

9. Solved Examples

Solved Example 1
The coding strand of a transcription unit reads 5′-ATGCTTAGC-3′. What is the sequence of the mRNA?
(A) 5′-AUGCUUAGC-3′
(B) 5′-UACGAAUCG-3′
(C) 3′-AUGCUUAGC-5′
(D) 5′-GCUAAGCAU-3′
Solution:

Answer: (A). mRNA has the same sequence and polarity as the coding strand, with U in place of T.

Solved Example 2
Match List I with List II.
List I: A. RNA polymerase I, B. RNA polymerase II, C. RNA polymerase III, D. Sigma factor
List II: I. hnRNA, II. Initiation in bacteria, III. 28S, 18S and 5.8S rRNA, IV. tRNA, 5S rRNA and snRNA
Choose the correct answer:
(A) A-III, B-I, C-IV, D-II
(B) A-I, B-III, C-IV, D-II
(C) A-III, B-IV, C-I, D-II
(D) A-IV, B-I, C-III, D-II
Solution:

Answer: (A). Polymerase I makes the large rRNAs (III), II makes hnRNA (I), III makes tRNA, 5S rRNA and snRNAs (IV), and sigma helps initiation (II).

Solved Example 3
Read the statements about a transcription unit.
A. The promoter lies upstream of the structural gene.
B. The template strand has the polarity .
C. The coding strand is copied into RNA.
D. The terminator lies towards the 3′ end of the coding strand.
E. The promoter is the binding site for RNA polymerase.
Choose the correct answer:
(A) A, B, D and E only
(B) A, B and C only
(C) B, C and D only
(D) A, C and E only
Solution:

Answer: (A). C is wrong: the template strand is copied; the coding strand has the same sequence as the RNA and is displaced.

Solved Example 4
Arrange the events of translation in the correct order.
A. A release factor binds the stop codon
B. Amino acids are linked to their tRNAs using ATP
C. The initiator tRNA recognises AUG on the mRNA
D. The ribosome moves from codon to codon adding amino acids
Choose the correct answer:
(A) B, C, D, A
(B) C, B, D, A
(C) B, D, C, A
(D) C, D, B, A
Solution:

Answer: (A). Charging comes first (B), then initiation at AUG (C), elongation (D) and termination (A).

Solved Example 5
Which of the following is NOT true of transcription in bacteria?
(A) A single RNA polymerase makes all types of RNA
(B) The rho factor helps termination
(C) The mRNA must be spliced before translation
(D) Translation can begin before transcription is complete
Solution:

Answer: (C). Bacterial mRNA needs no processing. Splicing is a feature of eukaryotic hnRNA.

Solved Example 6
Statement I: In eukaryotes, the primary transcript is non-functional.
Statement II: The primary transcript contains introns as well as exons.
(A) Both Statement I and Statement II are correct
(B) Statement I is correct, Statement II is incorrect
(C) Statement I is incorrect, Statement II is correct
(D) Both are incorrect
Solution:

Answer: (A). Because the primary transcript still contains introns, it is non-functional until splicing, capping and tailing turn it into mRNA.

10. Practice Questions

Practice Questions
  1. The template strand of a gene reads 3′-TACGGA-5′. Write the mRNA.Answer: 5′-AUGCCU-3′. The mRNA is complementary to the template, with U facing A.
  2. Match List I with List II.
    List I: A. Capping, B. Tailing, C. Splicing, D. UTR
    List II: I. Removal of introns, II. Untranslated sequence at an mRNA end, III. Methyl guanosine triphosphate at the 5′ end, IV. Adenylate residues at the 3′ end
    (A) A-III, B-IV, C-I, D-II (B) A-IV, B-III, C-I, D-II (C) A-III, B-I, C-IV, D-II (D) A-II, B-IV, C-I, D-IIIAnswer: (A). Capping adds the cap at 5′, tailing adds adenylates at 3′, splicing removes introns, and UTRs are untranslated end regions.
  3. Statement I: The promoter defines which strand is the template.
    Statement II: If the promoter and terminator switched places, the coding and template strands would be reversed.
    (A) Both Statement I and Statement II are correct (B) Statement I is correct, Statement II is incorrect (C) Statement I is incorrect, Statement II is correct (D) Both are incorrectAnswer: (A). The promoter's position sets the direction of transcription, and so decides the template strand.
  4. Which statement about the ribosome is NOT correct? (A) It has about 80 different proteins (B) Its subunits stay joined in the inactive state (C) The large subunit has two sites for amino acids (D) The 23S rRNA in bacteria acts as a ribozymeAnswer: (B). In the inactive state, the ribosome exists as two separate subunits.
  5. Arrange the events of bacterial transcription in order.
    A. The nascent RNA and the polymerase fall off
    B. RNA polymerase with the sigma factor binds the promoter
    C. The polymerase reaches the terminator
    D. The RNA chain elongates using nucleoside triphosphates
    (A) B, D, C, A (B) B, C, D, A (C) D, B, C, A (D) B, D, A, CAnswer: (A). Binding at the promoter, elongation, arrival at the terminator, then release.
  6. If the sequence of the coding strand in a transcription unit is 5′-ATGCATGCATGCATGCATGCATGCATGC-3′, write down the sequence of mRNA.Answer: 5′-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′: the same as the coding strand, with U in place of T.
  7. Depending on the chemical nature of the template (DNA or RNA) and of the nucleic acid made from it, list the types of nucleic acid polymerases.Answer: DNA-dependent DNA polymerase (replication); DNA-dependent RNA polymerase (transcription); RNA-dependent DNA polymerase, or reverse transcriptase (some viruses); and RNA-dependent RNA polymerase, which copies the RNA genome of some RNA viruses.
  8. Differentiate between (a) mRNA and tRNA, (b) template strand and coding strand.Answer: (a) mRNA carries the code and is the template for translation; tRNA brings a specific amino acid and reads the code with its anticodon. (b) The template strand () is copied into RNA; the coding strand () has the same sequence as the RNA (T for U) and is displaced.
  9. List two essential roles of the ribosome during translation.Answer: It binds the mRNA and provides two sites where charged tRNAs bring amino acids close together; and it catalyses peptide bond formation through its rRNA (23S rRNA in bacteria, a ribozyme).
  10. Explain in one or two lines the function of (a) promoter, (b) tRNA, (c) exons.Answer: (a) The promoter is the DNA sequence where RNA polymerase binds; it marks the start and decides the template strand. (b) tRNA reads a codon with its anticodon and brings the matching amino acid. (c) Exons are the coding sequences that remain in mature RNA.
  11. Briefly describe (a) transcription, (b) translation.Answer: (a) Transcription copies one strand of a DNA segment into RNA using DNA-dependent RNA polymerase, from promoter to terminator. (b) Translation joins amino acids into a polypeptide on the ribosome, in the order set by mRNA codons, from AUG to a stop codon.

Common Mistakes to Avoid

Watch out
  • Saying the coding strand is copied. Correct: the template strand () is copied; the coding strand has the same sequence as the RNA.
  • Placing the promoter downstream. Correct: the promoter is upstream (5′ side of the coding strand); the terminator is downstream.
  • Giving 5S rRNA to RNA polymerase I. Correct: 5S rRNA is made by RNA polymerase III.
  • Putting the cap at the 3′ end. Correct: the cap goes on the 5′ end and the poly A tail on the 3′ end.
  • Saying exons are removed. Correct: introns are removed, and exons are joined in a defined order.
  • Thinking RNA polymerase starts and stops by itself. Correct: it only elongates; it needs the sigma factor to initiate and the rho factor to terminate.
  • Saying bacterial mRNA is spliced. Correct: it needs no processing, so transcription and translation can be coupled.
  • Crediting a ribosomal protein with peptide bond formation. Correct: the 23S rRNA, a ribozyme, catalyses it in bacteria.

Frequently Asked Questions

What is a transcription unit?

A transcription unit is the stretch of DNA that is transcribed, defined by three regions: a promoter, the structural gene and a terminator. The promoter lies upstream and binds RNA polymerase; the terminator lies downstream and marks the end of transcription.

What is the difference between the template strand and the coding strand?

The template strand has the polarity 3 prime to 5 prime and is copied into RNA. The coding strand has the polarity 5 prime to 3 prime and the same sequence as the RNA, with thymine in place of uracil. It is displaced during transcription and does not code for anything.

Why is only one strand of DNA transcribed?

If both strands were copied, one DNA segment would code for two different RNAs and proteins, which would confuse the transfer of information. The two RNAs would also be complementary and pair into double-stranded RNA, which cannot be translated.

What are the roles of the sigma and rho factors?

The bacterial RNA polymerase can only catalyse elongation. It associates for a short time with the sigma factor to initiate transcription at the promoter, and with the rho factor to terminate it at the terminator. These factors change the specificity of the enzyme.

Which RNAs do the three eukaryotic RNA polymerases make?

RNA polymerase I transcribes the 28S, 18S and 5.8S rRNAs. RNA polymerase II transcribes hnRNA, the precursor of mRNA. RNA polymerase III transcribes tRNA, 5S rRNA and small nuclear RNAs. Organelles have their own RNA polymerase as well.

How is hnRNA processed into mRNA?

The primary transcript contains exons and introns and is non-functional. In splicing, introns are removed and exons are joined in order. Capping adds methyl guanosine triphosphate at the 5 prime end, and tailing adds 200 to 300 adenylate residues at the 3 prime end without a template.

Why is the ribosome called a ribozyme?

The ribosome catalyses peptide bond formation, and in bacteria the catalyst is its 23S rRNA, not a protein. An RNA that acts as an enzyme is called a ribozyme. This is one sign that translation, and life itself, evolved around RNA.

What are the steps of translation?

First, amino acids are activated with ATP and linked to their tRNAs, which is charging of tRNA. The ribosome then binds the mRNA at the start codon AUG with the initiator tRNA, adds amino acids codon by codon during elongation, and releases the polypeptide when a release factor binds a stop codon.

Previous year questions on Transcription and Translation

10 questions from past papers, each with a step-by-step solution.

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