16 O&C of Eukaryotic Genome Notes 2024

Created by joo sheng

p.20

Why is proteolytic cleavage important? Give an example.

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p.20

Proteolytic cleavage can activate proteins by removing signal peptides or pro‑regions and enabling proper folding; example: insulin maturation from pre‑proinsulin -> proinsulin -> active insulin via removal of signal peptide and C‑chain plus disulfide bond formation.

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p.20

Why is proteolytic cleavage important? Give an example.

Proteolytic cleavage can activate proteins by removing signal peptides or pro‑regions and enabling proper folding; example: insulin maturation from pre‑proinsulin -> proinsulin -> active insulin via removal of signal peptide and C‑chain plus disulfide bond formation.

p.2

Describe the general structure of a eukaryotic genome.

Eukaryotic genomes are generally much larger than prokaryotic or viral genomes, consist of two or more linear chromosomes located in the nucleus, each chromosome is a single double‑stranded linear DNA molecule associated with histones, with multiple origins of replication; genes have introns interrupting coding sequences and each gene has its own promoter; eukaryotes also have smaller circular mitochondrial genomes (and chloroplast genomes in plants).

p.2

What is the first level of DNA packing in eukaryotes and what structure does it form?

The first level is winding DNA around groups of eight core histone proteins (histone octamer) to form nucleosomes, producing a bead‑like structure; linker DNA joins nucleosomes and histone H1 stabilises DNA entry/exit.

p.2

What is the 30-nm chromatin fibre?

The 30‑nm chromatin fibre is the second level of packing where nucleosome 'beads' coil into a helical structure, which then forms looped domains with scaffold proteins and further coils into condensed chromosomes during mitosis.

p.3

Differentiate euchromatin and heterochromatin.

Euchromatin is less condensed, appears lightly stained, contains actively transcribed genes and may be depleted of nucleosomes in promoters; heterochromatin is highly condensed, appears darkly stained, contains repetitive sequences and is transcriptionally inactive; ~10% of mammalian genome is in heterochromatin and it's concentrated at centromeres and telomeres.

p.3

How does modification of histone proteins affect gene expression?

Chemical modification (e.g., methylation or acetylation) changes histone charge and alters DNA–histone association, affecting chromatin condensation: modifications can make DNA more or less accessible to transcription factors, thereby regulating transcription.

p.5

Describe the mitotic chromosome structure.

A mitotic chromosome is a condensed structure of two identical sister chromatids joined at the centromere; the tips are telomeres, which protect chromosome ends.

p.5

What are introns and what role do they play?

Introns are non‑coding sections within eukaryotic genes; they enable alternative splicing, allowing one gene to code for multiple polypeptides. Introns commonly begin with 'GT' and end with 'AG'.

p.5

Define promoter, enhancer and silencer.

Promoter: nucleotide sequence just upstream of a gene where general transcription factors and RNA polymerase bind to initiate transcription. Enhancer: nucleotide sequence that may lie thousands of base pairs away where activator proteins bind to increase transcription. Silencer: distal sequence where repressor proteins bind to suppress transcription.

p.6

What is the function of the centromere?

Centromere is a constricted chromosome region (~220 nucleotides, highly repeated) that mediates sister chromatid cohesion and kinetochore formation, enabling spindle fibre attachment for chromatid separation; centromere dysfunction can cause aneuploidy.

p.7

List three functions of telomeres.

Telomeres protect genes from erosion during replication, maintain chromosomal end integrity by preventing end‑to‑end fusion, and limit cell lifespan by causing apoptosis after critical shortening.

p.8

Explain how telomerase extends telomeres.

Telomerase carries a short RNA template complementary to the telomeric repeat (e.g., 5'‑TTAGGG‑3' in humans); telomerase reverse transcriptase (TERT) uses this RNA to extend the parental strand by RNA‑templated DNA synthesis, then DNA polymerase fills in the complementary strand, leaving a single‑stranded overhang. Telomerase is active in germ cells but inactive in most somatic cells.

p.9

Why is gene regulation necessary in multicellular eukaryotes?

Gene regulation directs development and maintains homeostasis; organisms have thousands of genes but do not express them all at once or in every cell—differential expression enables different cell types and developmental stages to express needed proteins.

p.9

Name the five main methods of controlling gene expression in eukaryotes.

Chromatin remodelling, transcriptional control, post‑transcriptional control, translational control, and post‑translational control.

p.10

How does chromatin remodelling regulate transcription?

By changing DNA packaging (via histone modifications and DNA methylation), chromatin remodelling alters accessibility of promoters to general transcription factors and RNA polymerase, enabling or preventing transcription initiation.

p.11

What effect does histone acetylation have on chromatin and transcription?

Histone acetylation (by histone acetyltransferases) decreases histone positive charge, lowers histone–DNA affinity, unravels chromatin into euchromatin, increases accessibility for transcription machinery, and generally activates transcription.

p.11

What is the effect of histone deacetylation?

Histone deacetylation (by histone deacetylases) restores positive charge, increases affinity between histones and DNA, condenses chromatin into heterochromatin, reduces accessibility of transcription machinery, and represses transcription.

p.12

Describe two ways DNA methylation silences genes.

DNA methylation (by DNA methyltransferase) can block transcription factor binding at CpG‑rich sites, preventing transcription initiation, and can recruit proteins (e.g., histone deacetylases) that induce chromatin condensation into transcriptionally inactive heterochromatin.

p.13

Compare prokaryotic and eukaryotic chromatin‑level control.

Prokaryotes lack chromatin packaging with histones and thus lack histone modifications; while they have DNA methylation for other purposes, they do not use chromatin remodelling for gene regulation. Eukaryotes use histone modification and DNA methylation to control chromatin structure and gene expression.

p.14

What roles do general and specific transcription factors play?

General transcription factors bind promoters (e.g., TATA box) and recruit RNA polymerase to form a basal transcription initiation complex; specific transcription factors bind control elements (enhancers/silencers) to modulate transcription rate by interacting with the initiation complex.

p.14

How do enhancers and silencers affect transcription over long distances?

Activators binding to enhancers or repressors binding to silencers induce DNA looping (aided by DNA‑bending proteins) to bring bound factors into contact with the promoter and transcription initiation complex, thereby increasing or decreasing RNA polymerase activity.

p.15

Describe the two domains of a transcription factor.

A transcription factor has a DNA‑binding domain for recognition and binding to specific DNA sequences and an activation domain for interacting with other proteins of the transcriptional machinery.

p.15

List mechanisms by which eukaryotic repressors act (give examples).

Eukaryotic repressors can: block activator binding to enhancers; mask activator activation domains; interact with general transcription factors to block assembly or release of RNA polymerase; package regions into heterochromatin; recruit histone deacetylases to condense chromatin.

p.16

What is combinational control of transcription in eukaryotes?

Combinational control is when a gene is regulated by multiple enhancers/silencers and the unique combination of control elements and available activator/repressor proteins in a cell determines precise spatial and temporal gene expression.

p.16

Explain the cell‑type specific transcription example of albumin and crystallin.

Both liver and lens cells have genes for albumin and crystallin, but only liver cells express albumin and lens cells express crystallin because the required activator proteins are present only in the appropriate cell type; different repressors also prevent expression in other cell types.

p.17

Contrast transcriptional control between prokaryotes and eukaryotes.

Prokaryotes often use a single operator proximal to genes (operon) producing polycistronic mRNA and can use positive/negative regulation; eukaryotes use unique combinations of distal control elements (enhancers/silencers) for each gene with monocistronic mRNA and each gene has its own promoter.

p.18

What post‑transcriptional modifications occur in eukaryotic pre‑mRNA?

5' capping, splicing (including removal of introns), and 3' polyadenylation; these produce mature mRNA that exits the nucleus for translation.

p.18

Define alternative splicing and its significance.

Alternative splicing is a regulated process where different combinations of exons are included/excluded from mature mRNAs derived from the same pre‑mRNA, allowing one gene to produce multiple proteins and increasing proteome diversity beyond the ~20,000 protein‑coding genes.

p.18

How does alternative splicing contribute to cell‑type protein diversity?

Different cell types can include or exclude particular exons during splicing, producing protein isoforms with different amino acid sequences and functions from the same gene, enabling tissue‑specific protein expression.

p.18

Compare post‑transcriptional control in prokaryotes and eukaryotes.

Prokaryotes lack a nuclear envelope so transcription and translation are simultaneous and post‑transcriptional modifications do not occur; eukaryotes separate transcription and translation with nuclear processing steps (capping, splicing, polyadenylation) and can perform alternative splicing.

p.19

How does mRNA half‑life affect protein production?

Longer mRNA half‑life allows it to remain in the cytoplasm longer and be translated more times, producing more protein; mRNA stability is influenced by the 5' cap, 3' poly‑A tail length, and binding proteins or hormones that affect degradation.

p.19

What roles do the 5' cap and 3' poly‑A tail play in mRNA stability?

The 5' cap protects mRNA from 5' exonucleases and aids translation initiation; the poly‑A tail buffers against 3' exonucleases so a longer tail increases mRNA stability and translation potential.

p.19

How can translation initiation be regulated?

Translational repressors (sequence‑specific RNA‑binding proteins) can bind the 5' UTR to prevent ribosome binding; complementary RNA molecules can bind the 5' UTR to block ribosome access, inhibiting formation of the translation initiation complex.

p.19

Compare translational control in prokaryotes and eukaryotes.

Prokaryotic mRNAs lack 5' cap and poly‑A tail and are generally less stable, though stability can be enhanced by 3' stem‑loops; both use translational repressors or complementary RNAs to block ribosome binding, while eukaryotes use 5' cap and poly‑A tail to regulate half‑life and translation.

p.20

What is post‑translational modification and give examples.

Post‑translational modification adds chemical groups to proteins after synthesis to regulate function; examples include glycosylation (carbohydrate addition), phosphorylation (phosphate addition), acetylation, methylation, and hydroxylation.

p.21

Describe the ubiquitin‑proteasome pathway for protein degradation.

Proteins destined for degradation are tagged with ubiquitin covalently attached to their N‑terminus; ubiquitinated proteins are recognised and degraded by the proteasome, allowing removal of unneeded or damaged proteins and recycling of amino acids.

p.20

How do eukaryotes regulate the amount of functional proteins?

By post‑translational modifications that control protein activity (e.g., phosphorylation) and by targeted degradation via ubiquitinylation and proteasome action to control protein quantity and remove faulty proteins.

p.21

Compare post‑translational control between prokaryotes and eukaryotes.

Prokaryotes generally do not have post‑translational modifications like eukaryotes; eukaryotes use biochemical modifications and proteolytic cleavage to regulate activity and ubiquitin‑mediated proteasomal degradation to control protein levels.

p.20

Give examples of biochemical additions in post‑translational modification and one functional role each.

Glycosylation: adds carbohydrates to membrane proteins to form glycoproteins; Phosphorylation: adds phosphates, activating many transcription factors and signalling proteins; Acetylation: e.g., histone acetylation and activation of p53; Methylation: activates some enzymes (e.g., phosphatase); Hydroxylation: modifies collagen residues for stability.

p.2

Why might eukaryotic genes not be regulated together like prokaryotic operons?

Each eukaryotic gene typically has its own promoter and distinct control elements, allowing independent regulation; genes are not necessarily co‑regulated because eukaryotic genomes have complex, cell‑type and temporally specific regulation.

p.2

What is the role of histone H1?

Histone H1 binds to the site where DNA enters and exits the nucleosome and stabilises the nucleosome structure.

p.3

Where is repetitive satellite DNA commonly found and what is its transcriptional status?

Repetitive satellite DNA is concentrated in heterochromatin regions such as centromeres and telomeres and is transcriptionally inactive.

p.10

How does chromatin remodelling contribute to cell differentiation?

Chromatin remodelling switches expression of certain genes off and others on during differentiation by altering chromatin condensation via histone modifications and DNA methylation, enabling tissue‑specific gene expression.

p.12

Explain how DNA methylation can indirectly lead to histone deacetylation.

Proteins that recognise methylated DNA recruit histone deacetylases, which remove acetyl groups from histones, leading to chromatin condensation and gene silencing.

p.5

What sequence features characterise introns?

Introns tend to begin with 'GT-' and end with '-AG'.

p.17

What is an operon and how does it relate to mRNA type?

An operon is a cluster of multiple structural genes under the control of a single promoter and operator in prokaryotes, producing polycistronic mRNA that encodes multiple proteins.

p.19

How can hormones affect mRNA half‑life?

Certain hormones can stimulate or retard mRNA degradation rates, altering mRNA half‑life and thus protein production.

p.7

Why are telomeres important in preventing chromosomal fusion?

Telomeres cap chromosome ends, preventing DNA repair machinery from recognising chromosome ends as breaks and fusing them; without telomeres broken ends may be joined, disrupting gene regulation.

p.8

What happens to cells when telomeres reach a critical length?

Cells undergo apoptosis after a limited number (~50) of divisions when telomeres shorten to a critical length, limiting mutation accumulation and cancer development.

p.2

What is the significance of multiple origins of replication in eukaryotic chromosomes?

Multiple origins allow efficient replication of large linear chromosomes during S phase, ensuring timely duplication of the genome.

p.12

How do CpG‑rich sites relate to transcription factor binding?

Many transcription factors bind CpG‑rich recognition sites; methylation of these CpG sites blocks their binding and suppresses transcription initiation.

p.14

What mechanisms ensure that enhancers can communicate with promoters despite distance?

DNA looping mediated by DNA‑bending proteins brings enhancer‑bound activators (or silencer‑bound repressors) into proximity with the promoter and transcription initiation complex, modulating RNA polymerase activity.

p.17

Explain why most eukaryotic genes are monocistronic.

Each eukaryotic gene has its own promoter and is transcribed into a separate mRNA, producing monocistronic mRNAs that code for a single polypeptide, enabling independent regulation of each gene.

p.3

How does packaging DNA into heterochromatin affect genes located there?

Genes packaged into heterochromatin are usually transcriptionally inactive and resistant to expression due to tight compaction limiting access to transcription machinery.

p.2

What role do scaffold proteins have in chromosome organisation?

Chromatin fibres associate with nuclear matrix scaffold proteins to form looped domains and undergo higher‑order coiling, aiding condensation into mitotic chromosomes.

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