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Molecular Biology - Gene expression (HL)

Grade 11IBBiology

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

🔑Concepts

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Transcription occurs in a 5′→3′5^{\prime} \rightarrow 3^{\prime} direction. RNA polymerase binds to the promoter region and adds nucleoside triphosphates (NTPs) to the growing 3′3^{\prime} end of the mRNA strand.

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Gene expression is regulated by proteins that bind to specific DNA sequences. Transcription factors, activators (binding to enhancers), and repressors (binding to silencers) control the rate of transcription.

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Nucleosomes help regulate transcription in eukaryotes. The addition of acetyl groups (acetylation) to histone tails neutralizes positive charges, loosening the DNA-histone interaction and increasing transcription. Conversely, DNA methylation (addition of a methyl group to cytosine) typically inhibits gene expression.

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Eukaryotic cells modify mRNA after transcription (post-transcriptional modification). This includes the addition of a 5′5^{\prime} cap, a poly-A tail, and the removal of non-coding introns through splicing.

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Alternative splicing allows a single gene to code for multiple proteins. By including different combinations of exons in the mature mRNA, a variety of functional proteins can be produced from the same gene sequence.

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Translation consists of initiation, elongation, and termination. The process begins when the small ribosomal subunit binds to the mRNA at the AUGAUG start codon. tRNA molecules carry specific amino acids to the ribosome based on the codon-anticodon interaction.

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tRNA-activating enzymes (aminoacyl-tRNA synthetases) attach specific amino acids to tRNA molecules using energy from ATPATP. There are 2020 different enzymes, one for each amino acid, ensuring high specificity.

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Ribosomes are composed of two subunits (large and small) and contain three binding sites for tRNA: the A (aminoacyl) site, the P (peptidyl) site, and the E (exit) site.

📐Formulae

Number of Nucleotides in mRNA=(Number of Amino Acids×3)+3 (for stop codon)\text{Number of Nucleotides in mRNA} = (\text{Number of Amino Acids} \times 3) + 3 \text{ (for stop codon)}

Direction of synthesis: 5′→3′\text{Direction of synthesis: } 5^{\prime} \rightarrow 3^{\prime}

Base Pairing (RNA Transcription): A→U,T→A,C→G,G→C\text{Base Pairing (RNA Transcription): } A \rightarrow U, T \rightarrow A, C \rightarrow G, G \rightarrow C

💡Examples

Problem 1:

A segment of the DNA template strand has the sequence 3′−TAC GGG ATT−5′3^{\prime}-\text{TAC GGG ATT}-5^{\prime}. Determine the sequence of the mRNA produced and the corresponding anticodons on the tRNA.

Solution:

  1. The mRNA sequence is complementary to the template strand and synthesized 5′→3′5^{\prime} \rightarrow 3^{\prime}: 5′−AUG CCC UAA−3′5^{\prime}-\text{AUG CCC UAA}-3^{\prime}.
  2. The tRNA anticodons are complementary to the mRNA codons: 3′−UAC−5′3^{\prime}-\text{UAC}-5^{\prime}, 3′−GGG−5′3^{\prime}-\text{GGG}-5^{\prime}, and 3′−AUU−5′3^{\prime}-\text{AUU}-5^{\prime}.

Explanation:

RNA polymerase reads the template strand in the 3′→5′3^{\prime} \rightarrow 5^{\prime} direction to build mRNA in the 5′→3′5^{\prime} \rightarrow 3^{\prime} direction. tRNA anticodons then pair antiparallel to the mRNA codons.

Problem 2:

If a mature mRNA molecule consists of 450450 nucleotides (excluding the stop codon and non-coding regions), how many amino acids will be present in the resulting polypeptide chain?

Solution:

Number of Amino Acids=Number of Nucleotides3\text{Number of Amino Acids} = \frac{\text{Number of Nucleotides}}{3} Number of Amino Acids=4503=150\text{Number of Amino Acids} = \frac{450}{3} = 150

Explanation:

Since each codon consists of a triplet of nucleotides (33 bases) and codes for one amino acid, we divide the total number of coding nucleotides by 33.

Problem 3:

Explain the effect of histone acetylation on the charge of the histone tail and its impact on gene expression.

Solution:

Histone tails are naturally positively charged due to lysine residues, attracting negatively charged DNA (PO43−PO_{4}^{3-} groups). Acetylation (CH3CO−CH_{3}CO^{-}) removes these positive charges.

Explanation:

When the positive charge is neutralized, the attraction between the histone and DNA weakens, leading to a less condensed DNA structure (euchromatin). This makes the DNA more accessible to RNA polymerase, thereby increasing gene expression.