- A researcher compares the chromatin of a developmentally regulated gene that is silenced in muscle cells with the chromatin of a centromeric satellite repeat in the same cells. Both loci are transcriptionally inert. Which of the following correctly distinguishes the epigenetic state of the silenced gene from the centromeric repeat?
Answer: The gene is likely in facultative heterochromatin marked by H3K27me3, while the centromere is in constitutive heterochromatin marked by H3K9me3.
When analyzing transcriptionally silent chromatin, you need to distinguish between two fundamentally different types of silencing: constitutive heterochromatin (permanently silenced) and facultative heterochromatin (conditionally silenced).
A developmentally regulated gene that's silenced in muscle cells represents facultative heterochromatin. This gene could potentially be active in other cell types or developmental stages, so it's conditionally silenced through the Polycomb repressive system, which deposits H3K27me3 marks. This allows for potential reactivation when developmental signals change.
Centromeric satellite repeats, however, are constitutively heterochromatic—they're permanently silenced across all cell types to maintain chromosomal stability. This constitutive silencing is maintained by H3K9me3 marks, which create stable, long-term repression that's not meant to be reversed.
Answer A reverses these relationships incorrectly, placing the developmentally regulated gene in permanent silencing (H3K9me3) and the centromere in conditional silencing (H3K27me3). Answer C focuses on DNA methylation patterns, but both regions can actually be methylated—the key distinction lies in histone modifications, not methylation status. Answer D discusses histone variants: while centromeres do contain CENP-A, this variant is specifically at the kinetochore for chromosome segregation, not in the heterochromatic satellite repeats being compared. H2A.Z is associated with active or poised chromatin, not silenced genes.
Remember: constitutive = H3K9me3 (permanent), facultative = H3K27me3 (conditional). This distinction is crucial for understanding how cells maintain stable silencing versus reversible developmental control.
- Vorinostat is a drug used in cancer therapy that inhibits histone deacetylases (HDACs). Its therapeutic goal is to re-activate silenced tumor suppressor genes. Which of the following describes the most direct biochemical mechanism that leads to chromatin decondensation upon HDAC inhibition?
Answer: Neutralization of the positive charge on lysine residues reduces their affinity for DNA.
When you encounter questions about epigenetic regulation and chromatin structure, focus on the direct molecular interactions between histones and DNA. The key principle here is understanding how chemical modifications to histone tails affect the electrostatic forces that hold chromatin together.
Histone deacetylases (HDACs) remove acetyl groups from lysine residues on histone tails. When vorinostat inhibits HDACs, acetyl groups remain on these lysines. This is crucial because lysine residues are normally positively charged, creating strong electrostatic attraction to the negatively charged DNA backbone. Acetylation neutralizes this positive charge by adding the acetyl group, which significantly weakens the histone-DNA interaction. This charge neutralization is the most direct mechanism causing chromatin to decondense and allowing transcriptional machinery to access previously silenced genes.
Looking at the wrong answers: (A) incorrectly focuses on bromodomain proteins, which actually bind to acetylated histones rather than being prevented from binding. (B) confuses the mechanism - HDAC inhibition doesn't directly block DNA methyltransferases, and DNA methylation isn't the primary factor in the immediate chromatin decondensation. (D) misrepresents the relationship between acetylation and methylation - acetylation doesn't sterically hinder H3K9 methylation, and this wouldn't explain the direct decondensation effect.
Study tip: For epigenetics questions, always trace the direct molecular cause-and-effect. Acetylation = charge neutralization = weaker DNA binding = open chromatin. Don't get distracted by secondary effects or other epigenetic modifications that might occur downstream.
- A locus contains an enhancer located 50 kb upstream of Gene A's promoter and 20 kb downstream of Gene B's promoter. A functional CTCF binding site, acting as an insulator, is located between the enhancer and Gene B. If a small deletion removes only this CTCF binding site, what is the most likely outcome?
Answer: Inappropriate activation of Gene B, with normal expression of Gene A.
CTCF binding sites often function as insulators, creating boundaries that block enhancer-promoter communication. In this setup, the insulator prevents the enhancer from acting on Gene B, while allowing it to act on Gene A. Removing the insulator (the CTCF site) eliminates this boundary. Consequently, the enhancer is now free to interact with the closer promoter of Gene B, leading to its inappropriate activation. The interaction with Gene A, which was not blocked by the insulator, should remain unaffected.
- A cell line has a temperature-sensitive mutation in a core subunit of the SWI/SNF chromatin remodeling complex, rendering it inactive at a restrictive temperature. After shifting to the restrictive temperature, a researcher induces a gene that requires SWI/SNF for activation. What is the most direct and immediate consequence on the gene's promoter?
Answer: Target nucleosomes will not be repositioned, preventing transcription factor access.
SWI/SNF is an ATP-dependent chromatin remodeling complex. Its primary function is to alter chromatin structure by sliding, evicting, or restructuring nucleosomes. This action is often required to expose promoter and enhancer DNA sequences so that transcription factors and the general transcriptional machinery can bind. If SWI/SNF is inactive, the nucleosomes covering these critical regions will not be moved, physically blocking access and preventing transcription initiation. The other options describe functions of different enzymes (histone demethylases, HATs) or a different step in transcription initiation (Mediator-Pol II interaction).
- To activate a target gene, a specific transcription factor must first acetylate H3K27 via its intrinsic histone acetyltransferase (HAT) domain. This modification then serves as a binding site for a large co-activator complex that contains a bromodomain subunit. What is the specific role of the bromodomain in this activation sequence?
Answer: To recognize and bind to the acetylated H3K27 created by the transcription factor.
When you encounter questions about histone modifications and transcriptional activation, focus on the specific molecular interactions that create regulatory cascades. This question tests your understanding of how chromatin-modifying enzymes and reader proteins work together in gene regulation.
The correct answer is C because bromodomains are specialized protein modules that specifically recognize and bind to acetylated lysine residues on histones. In this scenario, after the transcription factor acetylates H3K27 through its HAT domain, the bromodomain subunit within the co-activator complex acts as a "reader" protein. It detects the newly created acetyl mark and binds to it, allowing the entire co-activator complex to be recruited to the chromatin. This binding is essential for the activation cascade to proceed.
Let's examine why the other options are incorrect. Option A describes adding another acetyl group, but bromodomains are reader domains, not writer domains with enzymatic activity. Option B suggests acetyl group removal, which would be the function of a histone deacetylase (HDAC), not a bromodomain. Option D describes H3K4 methylation, which would require a histone methyltransferase activity that bromodomains don't possess.
Remember this key principle: histone modifications work through "writer-reader-eraser" mechanisms. Writers add marks, readers recognize them, and erasers remove them. Bromodomains are always readers for acetyl marks. When you see bromodomain in a question, think "acetyl recognition" - this will help you identify the correct function quickly on genetics exams.
- A researcher performs ChIP-seq in a human cell line for two histone modifications, H3K4me3 and H3K27me3, across a specific gene locus. The results are shown in the figure. Based on this data, what is the most likely status of Gene B?
Answer: Repressed through the formation of facultative heterochromatin.
The ChIP-seq data for Gene B shows a broad domain of H3K27me3 enrichment across the gene locus and a lack of the H3K4me3 mark at its promoter. H3K4me3 is a hallmark of active promoters, so its absence indicates the gene is not active. H3K27me3 is the key modification deposited by the Polycomb Repressive Complex 2 (PRC2), and it defines facultative heterochromatin—a form of silencing that is developmentally regulated and potentially reversible. Constitutive heterochromatin is marked by H3K9me3. A bivalent domain, characteristic of poised genes in stem cells, would show the co-occurrence of both H3K4me3 and H3K27me3 at the promoter.
- In many cases of gene silencing, there is synergistic crosstalk between DNA methylation and histone modification. Proteins containing a methyl-CpG-binding domain (MBD), such as MeCP2, are key mediators of this link. What is a primary function of MBD proteins after they bind to methylated DNA?
Answer: To recruit co-repressor complexes that include histone deacetylases (HDACs).
When you encounter questions about epigenetic gene silencing, focus on the interconnected network of DNA methylation and histone modifications that work together to repress transcription.
MBD proteins like MeCP2 function as crucial molecular bridges in this silencing network. After binding to methylated CpG sites through their methyl-CpG-binding domains, these proteins serve as recruitment platforms that bring in co-repressor complexes containing histone deacetylases (HDACs). The HDACs then remove acetyl groups from nearby histones, creating a more condensed, transcriptionally silent chromatin state. This establishes the synergistic crosstalk the question mentions—DNA methylation guides MBD protein binding, which then triggers histone modifications that reinforce the silenced state.
Option A is incorrect because MBD proteins don't guide de novo methylation; they respond to existing methylation patterns. Option B misrepresents the mechanism—while MBD proteins contribute to transcriptional repression, they don't directly block basal transcription factors at the TATA box. Instead, they work through chromatin remodeling. Option D confuses the roles: MBD proteins themselves lack enzymatic activity to remove acetyl groups; they recruit the HDACs that perform this function.
The correct answer is C because MBD proteins act as adaptor molecules that translate the DNA methylation signal into histone modifications by recruiting co-repressor complexes.
Remember this pattern: in epigenetic silencing, look for proteins that serve as "molecular matchmakers"—they rarely perform the silencing directly but instead recruit the enzymatic complexes that do the actual chromatin remodeling work.
- The enzyme LSD1 (KDM1A) is a histone demethylase that is overexpressed in some cancers, where it contributes to silencing tumor suppressor genes. Based on its known substrates, what molecular change would be expected at a target promoter if cancer cells were treated with a highly specific LSD1 inhibitor?
Answer: An increase in H3K4me2 levels, promoting gene expression.
When you encounter questions about histone-modifying enzymes in cancer, focus on understanding what happens when you block the enzyme's normal function. LSD1 (KDM1A) is a lysine-specific demethylase that removes methyl groups from specific histone residues, particularly H3K4me1/me2 and H3K9me1/me2.
LSD1's role in cancer involves removing activating marks like H3K4me2 from tumor suppressor gene promoters, contributing to their silencing. When you inhibit LSD1 with a specific inhibitor, the enzyme can no longer remove these methyl groups. This means H3K4me2 levels would accumulate at target promoters. Since H3K4me2 is associated with active transcription and open chromatin, this accumulation would promote gene expression of previously silenced tumor suppressor genes.
Choice A is incorrect because LSD1 doesn't target H3K9me3 - this repressive mark is modified by different enzymes. Choice B is wrong because LSD1 doesn't act on H3K27me3, which is removed by demethylases in the KDM6 family. Choice D is incorrect because H3K36me3 is not a known LSD1 substrate and is typically associated with transcribed gene bodies rather than promoters where LSD1 primarily acts.
The correct answer is C - LSD1 inhibition would increase H3K4me2 levels, promoting expression of tumor suppressor genes.
Remember: when analyzing enzyme inhibitors, always consider what substrate accumulates when the enzyme is blocked, then determine whether that substrate promotes or represses the biological process in question.
- A tumor suppressor gene (TSG) is found to be silenced in a cancer cell line, despite having no mutations in its coding or promoter sequence. Analysis shows heavy CpG methylation in its promoter. Treatment with 5-azacytidine, a demethylating agent, restores expression. Which enzyme is most likely responsible for maintaining the silenced state of this TSG as the cancer cells divide?
Answer: DNMT1
The question asks about maintaining a pre-existing methylation pattern through cell division. DNMT1 (DNA methyltransferase 1) is the maintenance methyltransferase. After DNA replication, it recognizes hemi-methylated DNA (where the parent strand is methylated but the new strand is not) and methylates the new strand, thus faithfully propagating the methylation pattern. DNMT3A and DNMT3B are de novo methyltransferases that establish new methylation patterns. TET1 is involved in demethylation. HDAC1 is a histone deacetylase; while involved in silencing, it does not maintain DNA methylation patterns.
- A researcher performs bisulfite sequencing on a DNA sample. For one specific gene promoter, the results from multiple cloned DNA molecules show that at a particular CpG site, approximately 50% of the sequences read 'CG' and 50% read 'TG'. What is the most accurate conclusion?
Answer: This CpG site is methylated in about half of the cells in the original sample.
When you encounter bisulfite sequencing questions, remember that this technique specifically detects DNA methylation patterns. Bisulfite treatment converts unmethylated cytosines to uracil (which reads as thymine), while methylated cytosines remain unchanged.
The key insight here is understanding what the 50/50 pattern represents. Since you're looking at multiple cloned DNA molecules from the same sample, and seeing roughly half showing 'CG' and half showing 'TG' at this CpG site, this indicates that approximately half the cells in the original sample had methylated cytosines at this position (reading as 'CG'), while half had unmethylated cytosines (converted to 'TG'). This makes answer D correct.
Let's examine why the other options are wrong. Answer A suggests imprinting, but imprinting would affect all cells equally since it's determined by parental origin - you wouldn't see this mixed population unless you had a heterogeneous cell sample. Answer B misinterprets the chemistry: if bisulfite conversion were only 50% efficient, you'd see incomplete conversion across all unmethylated sites, not this clean 50/50 split. Answer C confuses a SNP with methylation detection - a true C-to-T SNP would show consistent patterns that aren't dependent on bisulfite treatment.
For genetics exams, remember that bisulfite sequencing reveals cell-to-cell variation in methylation status. When you see mixed readouts from cloned molecules, think about the heterogeneity of methylation patterns across different cells in the original sample, not technical failures or genetic variants.
- In Drosophila, an inversion places the euchromatic white gene adjacent to pericentromeric heterochromatin, causing a variegated eye phenotype. A screen identifies a loss-of-function mutation in a gene, E(var)3-9, that causes a nearly all-white eye phenotype. What is the most likely function of the wild-type E(var)3-9 gene product?
Answer: A histone acetyltransferase that maintains an open chromatin state.
Position effect variegation in Drosophila occurs when a euchromatic gene is relocated near heterochromatin, causing the repressive chromatin structure to spread and silence the gene in some cells but not others. This creates a mosaic phenotype where some eye cells express white (pigmented) while others don't (white patches).
The key insight is understanding what E(var)3-9 mutations do: they suppress variegation by making eyes nearly all-white. This means the wild-type protein normally promotes the silencing that creates variegation. When you lose this protein's function, you lose most of the pigmentation, suggesting it normally helps maintain active gene expression in the face of nearby heterochromatin.
Answer D is correct because histone acetyltransferases add acetyl groups to histones, creating open chromatin that promotes transcription. The wild-type E(var)3-9 protein likely counteracts heterochromatin spreading by maintaining acetylated, transcriptionally active chromatin around the white gene.
Answer A is wrong because nucleosome ejection would increase rather than decrease gene expression. Answer B incorrectly identifies the protein as a barrier - while barriers do exist, the phenotype suggests E(var)3-9 actively promotes transcription rather than just blocking heterochromatin spread. Answer C describes the opposite function - H3K9me3 methyltransferases create repressive marks associated with heterochromatin, which would enhance rather than suppress variegation.
Remember: when analyzing variegation suppressors or enhancers, always consider whether the mutation increases or decreases the silencing effect, then work backward to deduce the wild-type protein's normal function.
- Histone variants can replace canonical histones, conferring specialized functions. Which of the following correctly pairs a histone variant with its primary role and typical genomic location?
Answer: CENP-A; kinetochore assembly; located at centromeres.
CENP-A is a histone H3 variant that is the epigenetic hallmark of centromeres. It replaces canonical H3 in centromeric nucleosomes and is essential for the assembly of the kinetochore, the protein structure to which spindle microtubules attach during cell division. H3.3 is associated with active transcription, not DNA repair primarily. H2A.Z is associated with transcriptional regulation at promoters and enhancers and can act as a barrier to heterochromatin spreading, but it is not located primarily at telomeres. macroH2A is associated with transcriptional repression, most notably on the inactive X chromosome (Barr body).
- A mutation in a gene encoding a chromodomain-containing protein prevents it from binding to its target histone modification. This protein normally recruits a complex that compacts chromatin. Which of the following is the most likely consequence of this mutation on the expression of the protein's target genes?
Answer: Inappropriate activation of genes normally marked by H3K9me3.
Chromodomains are protein motifs that function as 'readers' for methylated lysine residues on histones, particularly H3K9me3 and H3K27me3, which are repressive marks. The stem states the protein recruits a compacting complex, consistent with a role in silencing. If the chromodomain protein cannot bind to its target (e.g., H3K9me3), it cannot recruit the silencing machinery. This would lead to a failure to establish or maintain the heterochromatic state, resulting in de-repression or inappropriate activation of the target genes.
- While DNA methylation in promoter CpG islands is a well-established silencing mark, dense methylation is also frequently observed within the gene bodies of actively transcribed genes. What is a widely accepted hypothesis for the function of this gene-body methylation?
Answer: To prevent transcription initiation from cryptic promoters within the gene body.
When you encounter questions about DNA methylation, remember that context matters enormously. While promoter methylation typically silences genes, methylation within actively transcribed gene bodies serves a completely different purpose and follows different rules.
Gene-body methylation primarily functions as a quality control mechanism to prevent aberrant transcription initiation from cryptic promoters embedded within genes. These internal sequences can accidentally recruit RNA Polymerase II, leading to truncated transcripts or antisense RNA production that interferes with normal gene expression. The dense methylation in gene bodies blocks these unwanted initiation events while allowing the legitimate promoter to function normally.
Option A is incorrect because methylation doesn't enhance RNA Polymerase II elongation speed. If anything, heavily methylated DNA can create obstacles for the transcription machinery. Option C misrepresents the mechanism—while gene-body methylation correlates with active transcription (and thus splicing), it doesn't directly recruit splicing factors to exon-intron boundaries. The splicing machinery recognizes specific RNA sequences and structures, not DNA methylation marks. Option D contradicts the fundamental observation in the question: gene-body methylation is found in actively transcribed genes, not genes being prepared for silencing.
For genetics exams, always distinguish between different types of epigenetic marks based on their genomic location. Promoter methylation usually means silencing, but gene-body methylation in actively transcribed genes serves a protective function, preventing transcriptional chaos from internal cryptic promoters.
- The three-dimensional folding of the genome into topologically associating domains (TADs) is critical for regulating gene expression by constraining enhancer-promoter interactions. The boundaries of TADs are highly enriched for binding sites of the insulator protein CTCF. Which protein complex functions as the motor that extrudes chromatin loops to form TADs, a process that is halted by CTCF?
Answer: Cohesin
When you encounter questions about three-dimensional genome organization and TAD formation, focus on the mechanical process of chromatin loop extrusion and the key proteins involved.
TADs form through a process called loop extrusion, where a motor protein complex actively moves along chromatin fibers, extruding loops of DNA until it encounters a boundary element. Cohesin (B) is the ring-shaped protein complex that serves as this molecular motor. It slides along chromatin and extrudes loops by reeling in DNA, similar to how you might pull a rope through your hands. This process continues until cohesin encounters CTCF proteins bound at specific sites, which act as roadblocks that halt the extrusion process and establish TAD boundaries.
The wrong answers represent other important nuclear complexes with different functions. The Mediator complex (A) facilitates transcription by bridging enhancers and promoters but doesn't extrude chromatin loops. The SWI/SNF complex (C) is an ATP-dependent chromatin remodeling complex that moves nucleosomes to make DNA accessible, but it doesn't create large-scale chromosomal loops. The anaphase-promoting complex (D) is a cell cycle regulator that targets proteins for degradation during mitosis and has no role in interphase chromatin organization.
Study tip: Remember that cohesin has dual roles - it holds sister chromatids together during cell division AND organizes interphase chromatin through loop extrusion. When you see questions about TAD formation or chromatin looping, think "cohesin extrudes, CTCF stops."
- The Igf2 gene is maternally imprinted (silenced) and paternally expressed. This is governed by an imprinting control region (ICR) that, on the maternal chromosome, is unmethylated and binds the insulator protein CTCF. On the paternal chromosome, this ICR is heavily methylated. What is the direct consequence of this paternal-specific ICR methylation?
Answer: It prevents CTCF from binding, allowing a downstream enhancer to activate Igf2.
This question describes the well-characterized Igf2/H19 locus. On the paternal chromosome, the ICR is methylated. This methylation directly blocks the binding of the CTCF insulator protein. Without CTCF bound, the ICR can no longer function as an enhancer-blocking insulator. This allows a downstream enhancer to loop over and interact with the Igf2 promoter, leading to its expression. On the maternal allele, CTCF binds to the unmethylated ICR and blocks this interaction, silencing Igf2 and allowing H19 expression.
- The Polycomb group (PcG) and Trithorax group (TrxG) proteins maintain cellular memory by regulating key developmental genes. A cell has a loss-of-function mutation in a core component of the Polycomb Repressive Complex 2 (PRC2), such as EZH2. Which is a direct molecular consequence?
Answer: Reduced levels of H3K27 trimethylation at target gene loci.
When you encounter questions about Polycomb and Trithorax proteins, focus on their specific molecular functions in chromatin modification. These protein complexes maintain gene expression states through histone modifications—PcG proteins repress genes while TrxG proteins activate them.
PRC2 (Polycomb Repressive Complex 2) has a very specific enzymatic function: its catalytic subunit EZH2 is a histone methyltransferase that adds methyl groups to histone H3 at lysine 27, creating the repressive H3K27me3 mark. When EZH2 is lost due to mutation, PRC2 cannot perform this methylation reaction, directly resulting in reduced H3K27 trimethylation at target loci. This makes answer D correct—it describes the immediate, direct molecular consequence of losing the enzyme responsible for this specific modification.
Let's examine why the other options are wrong: A) DNA methylation is performed by DNA methyltransferases (DNMTs), not PRC2, so EZH2 loss wouldn't directly affect DNA methylation patterns. B) H2A ubiquitination is carried out by PRC1, not PRC2—these are distinct Polycomb complexes with different functions. C) H3K4 demethylation is performed by specific demethylases like LSD1, not by PRC2, and EZH2 loss wouldn't prevent this process.
For genetics exams, remember that each chromatin-modifying complex has a specific enzymatic activity. Always connect the protein mentioned to its direct molecular function—if a methyltransferase is lost, methylation of its specific target will decrease. Don't confuse the various Polycomb complexes or assume indirect effects.
- The positioning of nucleosomes is influenced by the underlying DNA sequence. Certain sequences are intrinsically disfavored for nucleosome formation and are often found in nucleosome-depleted regions (NDRs) at active promoters. Which DNA sequence feature is most likely to create an NDR?
Answer: A long adenine-thymine (A-T) rich sequence, such as a poly(A) tract.
When you encounter questions about nucleosome positioning, remember that DNA packaging isn't random—certain sequence features make DNA either more or less favorable for wrapping around histone octamers. Nucleosome-depleted regions (NDRs) form when DNA sequences are intrinsically difficult to package into nucleosomes.
Long A-T rich sequences, particularly poly(A) tracts, are exceptionally rigid and resist the bending required for nucleosome formation. The multiple hydrogen bonds between A-T base pairs in these regions create structural inflexibility that makes it energetically unfavorable for DNA to wrap around histones. This is why poly(A) tracts are commonly found in promoter regions where chromatin must remain accessible for transcription factor binding.
Let's examine why the other options don't create NDRs: Option A describes sequences with alternating GC/AT content every 5 base pairs, which actually facilitates nucleosome formation since this periodicity matches the helical repeat and allows favorable histone-DNA contacts. Option C, high CpG density regions, can affect nucleosome positioning through methylation but don't inherently prevent nucleosome formation—CpG islands are often associated with active promoters for regulatory reasons, not structural ones. Option D, palindromic hairpin-forming sequences, might create local structural distortions but don't systematically exclude nucleosomes the way rigid A-T tracts do.
For genetics exams, remember that nucleosome positioning questions often test your understanding of DNA structural properties. A-T rich regions = rigid = nucleosome exclusion, while sequences that can bend appropriately or have favorable periodicity promote nucleosome formation.
- To provide the strongest evidence for transgenerational epigenetic inheritance of a phenotype induced by an environmental exposure in an F0 pregnant female mouse, the phenotype must be observed in which generation?
Answer: F3
When analyzing transgenerational epigenetic inheritance, you need to distinguish between direct environmental exposure and true inherited effects. The key is identifying which generations were directly exposed to the environmental factor versus those that inherited epigenetic marks.
In this scenario, an F0 pregnant female experiences environmental exposure. This exposure directly affects three generations simultaneously: the F0 female herself, her developing F1 offspring in utero, and the germ cells within those F1 offspring that will become the F2 generation. Since epigenetic marks can persist from direct exposure, observing the phenotype in F1 or F2 generations doesn't prove inheritance—it could simply reflect direct exposure effects.
The F3 generation provides the strongest evidence because it's the first generation with no direct exposure to the original environmental factor. Any phenotype observed in F3 must result from inherited epigenetic modifications passed down through the germline, not from direct environmental effects.
Looking at the wrong answers: F1 (B) and F2 (C) were directly exposed as described above, so phenotypes could result from exposure rather than inheritance. F4 (A) would also demonstrate transgenerational inheritance, but F3 is the earliest generation that provides definitive evidence, making it the strongest proof since it rules out direct exposure effects most clearly.
Remember this pattern: for maternal exposure during pregnancy, count three generations of direct exposure (F0, F1, F2), making F3 the critical generation for proving true epigenetic inheritance.
- In female mammals, one X chromosome is inactivated by being coated with Xist RNA, leading to heterochromatin formation. However, about 15% of genes on the inactive X manage to escape silencing. Which set of epigenetic features would most likely be found at the promoter of such an 'escaper' gene?
Answer: Enrichment for H3K4me3 and absence of dense CpG island methylation.
Genes that escape X-inactivation must maintain features of active chromatin despite being on an otherwise silenced chromosome. Active chromatin is characterized by activating histone marks like H3K4me3 at the promoter and a lack of repressive DNA methylation in the promoter's CpG island. The other options describe features of silenced chromatin: dense DNA methylation, the repressive histone variant macroH2A, the repressive PRC2 complex and its mark H3K27me3, and the repressive mark H3K9me3 coupled with a lack of activating acetylation are all hallmarks of the inactive X chromosome.
