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Molecular Basis of Inheritance - Transcription

Grade 12CBSEBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Transcription is the process of copying genetic information from one strand of the DNADNA into RNARNA. It is governed by the principle of complementarity, except that adenosine forms a base pair with UracilUracil (UU) instead of ThymineThymine (TT).

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A Transcription Unit in DNADNA is defined primarily by three regions: a Promoter (located towards 5′5'-end of the coding strand), the Structural gene, and a Terminator (located towards 3′3'-end of the coding strand).

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The enzyme DNADNA-dependent RNARNA polymerase catalyzes the polymerization in only one direction, i.e., 5′→3′5' \rightarrow 3'. Consequently, the strand with 3′→5′3' \rightarrow 5' polarity acts as the Template Strand.

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The strand with 5′→3′5' \rightarrow 3' polarity is called the Coding Strand. Its sequence is identical to the synthesized RNARNA, except for the substitution of TT by UU.

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In Prokaryotes, transcription requires the initiation factor (σ\sigma factor) and termination factor (ρ\rho factor) to start and stop the process respectively. Translation can often begin before the mRNAmRNA is fully transcribed.

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In Eukaryotes, there are three types of RNARNA polymerases: RNA polymerase IRNA \ polymerase \ I (transcribes rRNAsrRNAs like 28S28S, 18S18S, and 5.8S5.8S), RNA polymerase IIRNA \ polymerase \ II (transcribes hnRNAhnRNA), and RNA polymerase IIIRNA \ polymerase \ III (transcribes tRNAtRNA, 5S rRNA5S \ rRNA, and snRNAssnRNAs).

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Post-transcriptional modifications in eukaryotes involve Splicing (removal of non-coding intronsintrons), Capping (addition of methyl guanosine triphosphatemethyl \ guanosine \ triphosphate at 5′5' end), and Tailing (addition of 200−300200-300 adenylateadenylate residues at 3′3' end).

📐Formulae

3′→Template Strand5′3' \xrightarrow{\text{Template Strand}} 5'

5′→Coding Strand3′5' \xrightarrow{\text{Coding Strand}} 3'

5′→mRNA/Transcript3′5' \xrightarrow{\text{mRNA/Transcript}} 3'

Base Pairing Rules: A→U, T→A, G⇌C\text{Base Pairing Rules: } A \rightarrow U, \ T \rightarrow A, \ G \rightleftharpoons C

💡Examples

Problem 1:

If the sequence of the coding strand in a transcription unit is written as follows: 5′−ATGCATGCATGCATGCATGCATGCATGC−3′5'-ATGCATGCATGCATGCATGCATGCATGC-3'. Write down the sequence of mRNAmRNA.

Solution:

5′−AUGCAUGCAUGCAUGCAUGCAUGCAUGC−3′5'-AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3'

Explanation:

The sequence of mRNAmRNA is exactly the same as the coding strand, with the only difference being that ThymineThymine (TT) is replaced by UracilUracil (UU). The polarity remains 5′→3′5' \rightarrow 3'.

Problem 2:

Given a template strand sequence 3′−TACGTACGTACGTACG−5′3'-TACGTACGTACGTACG-5', deduce the sequence of the transcribed RNARNA.

Solution:

5′−AUGCUGCUGCUGCUGC−3′5'-AUGCUGCUGCUGCUGC-3' (Corrected: 5′−AUGCUGCUGCUGCUGC−3′5'-AUGCUGCUGCUGCUGC-3' based on complementarity: AA pairs with UU, TT pairs with AA, GG with CC, CC with GG)

Explanation:

Since the template strand is 3′→5′3' \rightarrow 5', the RNARNA polymerase will synthesize a complementary strand in the 5′→3′5' \rightarrow 3' direction. Using the base pairing rule (T→AT \rightarrow A, A→UA \rightarrow U, C→GC \rightarrow G, G→CG \rightarrow C), we obtain the sequence.

Transcription - Revision Notes & Key Diagrams | CBSE Class 12 Biology