Lecture 7: Histone Tail Modification Learning Objectives

Explain the significance of the histone tails – where they lie on the nucleosome, what fraction of its mass they account for, and why they carry most of the modifications

• Location: The unstructured amino-terminal tails extrude from the nucleosome core and stick out between the DNA gyres.

• Mass: They account for about a quarter (one-fourth) of the nucleosome mass.

• Modifications: They carry most of the post-translational modifications because they host the vast majority of known covalent modification sites.

Name the writers, erasers and readers of histone modifications and give two examples of each

• Writers: Add epigenetic marks. Examples: Histone acetyltransferases (HATs), Histone methyltransferases (HMTs), Protein arginine methyltransferases (PRMTs), or Kinases.

• Erasers: Remove epigenetic marks. Examples: Histone deacetylases (HDACs), Lysine demethylases (KDMs), or Phosphatases.

• Readers: Recognize and bind to specific epigenetic marks. Examples: Proteins containing bromodomains, chromodomains, or Tudor domains.

Explain how acetylation of lysine affects chromatin, and name the reader domain that binds acetylated lysine

• Chromatin Effect: Acetylation of lysine decreases the net positive charge of the histone tail, which prevents the tails from binding to those of histones in nearby nucleosomes and forces a transcriptionally active euchromatic state.

• Reader Domain: Acetylated lysine is bound by proteins carrying bromodomains.

Say how many methyl groups a lysine can carry, name each state, and say what happens to the charge of the residue

• Number of States: A lysine residue can carry up to three methyl groups.

• State Names: Monomethyl-lysine (one group), dimethyl-lysine (two groups), and trimethyl-lysine (three groups).

• Residue Charge: Unlike acetylation, methylation does not neutralize the lysine residue; mono-, di-, and trimethyl-lysine all remain positively charged.

Give one active and one repressive lysine methylation mark, and explain why the same chemical mark can mean opposite things

• Active Mark: H3K4me3 (or H3K36, H3K79).

• Repressive Mark: H3K9me3 (or H3K27me3, H4K20).

• Why the Same Chemical Mark Can Differ: The effect depends on the location of the methylation on the histone tail (e.g., H3K4 vs. H3K9) and the specific number of methyl groups added (e.g., H3K4me2 vs. H3K4me3). Different spatial contexts recruit different proteins specifying alternate transcriptional responses.

Explain how H3K9me3 recruits HP1 to silence a locus, and why other chromodomain proteins reading methylated lysine can instead activate transcription

• H3K9me3 & HP1: H3K9me3 binds to proteins containing a chromodomain, such as Heterochromatin Protein 1 (HP1), which recruits further factors to drive a suppressed, transcriptionally silent chromatin state.

• Activating Chromodomains: Other chromodomain-containing proteins can activate transcription instead because the overall functional outcome depends on the other components (domains) of the specific protein carrying the chromodomain.

Name the histone modifications covered here other than acetylation and methylation, and give the main effect of each

• Phosphorylation (Serine): Controls the expression of specific genes (e.g., EGF-responsive genes) and triggers chromosome condensation before mitosis (via H3S10).

• Ubiquitination (Lysine): H2BK120 ubiquitination licenses/prerequisites gene activation (by enabling H3K4 and H3K79 methylation), whereas H2AK119 ubiquitination suppresses gene expression.

• SUMOylation (Lysine): Generally leads to the inhibition/interference of transcription and interferes with acetylation.

• Biotinylation (Lysine): Enriched in heterochromatin (especially H4K8 and H4K12)

Define a bivalent domain, name the two marks that form it, and explain what it does for a pluripotent cell

• Definition: Promoters that simultaneously carry both an active mark (H3K4me3) and a repressive mark (H3K27me3).

• Function: They keep developmental genes silent but poised rather than permanently shut off. Upon cell differentiation, the bivalent domain resolves (losing one mark), allowing the pluripotent cell to keep its options open until it commits to an active or repressed developmental fate.

Explain why marks do not act independently

• Steady State: Writer and eraser enzymes act continuously and in opposition. The level of any mark is a dynamic steady-state balance (not a simple on/off switch), allowing marks to shift in either direction without waiting for DNA replication.

• Licensing: Marks do not act independently; one mark can license another. For example, H2BK120 ubiquitination is a prerequisite that licenses subsequent di- and trimethylation of H3K4 and H3K79.

Define the histone code hypothesis and give two reasons it remains a hypothesis rather than a principle, including the difficulty of separating cause from correlation

• Definition: The hypothesis states that a combination of multiple histone modifications can be interpreted by the cell to produce a specific biological outcome (like a distinct transcriptional state).

• Why it is a Hypothesis (not a Principle):

  1. Inconsistent Outcomes: Certain combinations of histone modifications do not always produce the same outcome in experimental systems.

  2. Cause vs. Correlation: The code assumes marks instruct (cause) transcription, but for many marks, it is difficult to prove causality rather than mere correlation (e.g., removing H3K4me3 often does not silence the active gene, suggesting some marks are consequences of transcription rather than its cause).

Summary — the marks covered in this lecture