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Nucleic Acids (AHL) - Transcription and gene expression

Grade 12IBBiology

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

🔑Concepts

Transcription occurs in a 535' \rightarrow 3' direction. RNA polymerase adds the 55' phosphate group of a free RNA nucleotide to the 33' hydroxyl group of the growing mRNA strand.

Nucleosomes help regulate transcription in eukaryotes. The tails of histone proteins can be modified; for example, acetylation of lysine residues neutralizes positive charges, loosening the binding between DNA and histones and increasing transcription rates.

Eukaryotic DNA contains non-coding sequences called introns and coding sequences called exons. Post-transcriptional modification involves the removal of introns and the splicing together of exons to form mature mRNAmRNA.

The promoter is a non-coding DNA sequence located near the start of a gene. It provides a binding site for RNA polymerase and determines which DNA strand is the antisense (template) strand.

Gene expression is regulated by proteins that bind to specific DNA sequences. Transcription factors, such as activators and repressors, bind to enhancers and silencers respectively to modulate the rate of transcription.

The environment can impact gene expression through epigenetic modifications. Chemical tags such as methyl groups (CH3-CH_3) added to DNA (cytosine bases) typically inhibit gene expression by preventing transcription factor binding.

Alternative splicing allows a single gene to code for multiple proteins. By including or excluding different exons during mRNAmRNA processing, different mature mRNAmRNA transcripts are produced from the same primary transcript.

In prokaryotes, transcription and translation can occur simultaneously because there is no nuclear membrane to separate the processes, and no post-transcriptional modification (splicing) is required.

📐Formulae

Antisense Strand (Template): 35\text{Antisense Strand (Template): } 3' \rightarrow 5'

Sense Strand (Coding): 53\text{Sense Strand (Coding): } 5' \rightarrow 3'

mRNA Transcript: 53\text{mRNA Transcript: } 5' \rightarrow 3'

Base Pairing Rules: AU,TA,CG\text{Base Pairing Rules: } A \leftrightarrow U, \, T \rightarrow A, \, C \leftrightarrow G

💡Examples

Problem 1:

Given an antisense (template) DNA strand with the sequence 3TACGGCATTCGA53' - TAC \, GGC \, ATT \, CGA - 5', determine the sequence of the resulting mRNAmRNA strand and specify its directionality.

Solution:

5AUGCCGUAAGCU35' - AUG \, CCG \, UAA \, GCU - 3'

Explanation:

Transcription follows complementary base pairing rules where AA pairs with UU, TT pairs with AA, CC pairs with GG, and GG pairs with CC. The mRNAmRNA is synthesized antiparallel to the template strand, resulting in a 535' \rightarrow 3' orientation.

Problem 2:

Explain the effect of histone acetylation on the interaction between DNA and the histone complex. Use the charge properties of LysineLysine in your answer.

Solution:

Acetylation of the LysineLysine residues on histone tails removes the positive charge (++). This reduces the electrostatic attraction between the histone and the negatively charged phosphate groups (-) of the DNA backbone.

Explanation:

DNA is negatively charged. Normally, the positive charge of histone tails causes tight wrapping (heterochromatin). Acetylation leads to a neutral charge, resulting in a looser structure (euchromatin) that is accessible to RNA polymerase.

Problem 3:

A primary mRNAmRNA transcript contains 4 exons and 3 introns. If Exon 1 is 100 bp, Exon 2 is 150 bp, Exon 3 is 200 bp, and Exon 4 is 120 bp, calculate the length of the mature mRNAmRNA if Exon 3 is skipped during alternative splicing (exclude the cap and tail).

Solution:

100+150+120=370 nucleotides100 + 150 + 120 = 370 \text{ nucleotides}

Explanation:

Mature mRNAmRNA consists only of exons. If Exon 3 is excluded through alternative splicing, the length is the sum of the remaining exons: 100(E1)+150(E2)+120(E4)=370100 \, (E1) + 150 \, (E2) + 120 \, (E4) = 370. Introns are always removed and do not contribute to the final length.

Transcription and gene expression - Revision Notes & Key Diagrams | IB Grade 12 Biology