---
title: "Chromatin and Epigenetics"
book: "University Biology — Year 3"
subject: biology
language: en
chapter: 1
exercises: 12
source: https://one-course.com/books/biology/5/en/chapter/1-chromatin-and-epigenetics
---

# Chapter 1 — Chromatin and Epigenetics

Two mice from the same litter, genetically identical down to the last base, sit side by side: one is slim and brown, the other plump and yellow, and it will become diabetic. A calico cat wears patches of orange and black although every cell of her skin carries the same two X chromosomes. A child inherits the same deleted stretch of chromosome 15 as another child, and one has Prader–Willi syndrome while the other has Angelman syndrome — the difference is only which parent the deletion came from. None of this is written in the DNA sequence. It is written *on* it: in the way two metres of DNA are packed into a nucleus ten micrometres across, in the small chemical groups attached to the packing proteins and to the cytosines themselves, and in the machinery that copies those marks from a cell to its daughters. This chapter is about that second layer of information, its molecular carriers, and the honest limits of what it can remember.

## 1.1 Chromatin: DNA on a spool

**Definition 1.1 (Nucleosome and chromatin).**

*Chromatin* is the complex of DNA with proteins in which the genome exists in a eukaryotic nucleus. Its repeating unit is the *nucleosome*: $147\,\mathrm{bp}$ of DNA wound in $1.65$ left-handed turns around a *histone* octamer — two copies each of the small, very basic proteins H2A, H2B, H3 and H4 — followed by a *linker* of $20\text{ to }60\,\mathrm{bp}$ to the next nucleosome. A fifth histone, H1, binds the linker where it enters and leaves the core and helps the beads pack against one another. Each core histone carries an unstructured amino-terminal *tail* of some $20\text{ to }40$ residues that projects out of the particle. Chromatin that stains lightly, is loosely packed and is transcribed is *euchromatin*; densely packed, late-replicating and largely silent chromatin, at centromeres, telomeres and repeated sequences, is *heterochromatin*.

**Evidence.** Hewish and Burgoyne (1973) digested rat liver nuclei with an endogenous nuclease and found that the DNA came out not as a smear but as a ladder of fragments whose lengths were multiples of about $200\,\mathrm{bp}$: the DNA was protected at regular intervals and cut only between them. Kornberg (1974) showed that the protected unit was a particle of eight [histones](#def-b3-chromatin-epigenetics-nucleosome) with about $200\,\mathrm{bp}$, and electron micrographs of gently spread [chromatin](#def-b3-chromatin-epigenetics-nucleosome) showed the “beads on a string”. Luger and colleagues (1997) solved the crystal structure of the core particle at $2.8\,\text{Å}$: $147\,\mathrm{bp}$, an octamer with the tails passing out between the DNA gyres. ∎

![The 10\, nm fibre: nucleosome cores (blue) on a continuous DNA (red), each bead a histone octamer with its tails (orange) projecting out, linker histone H1 (green) at the entry–exit point. About six nucleosomes occupy the length that 1200\, bp of naked DNA would.](https://one-course.com/images/onecourse/chapters/biology-5/b3-chromatin-epigenetics/fig-f9a1934868a0.svg)

*The $10\,\mathrm{nm}$ fibre: [nucleosome](#def-b3-chromatin-epigenetics-nucleosome) cores (blue) on a continuous DNA (red), each bead a [histone](#def-b3-chromatin-epigenetics-nucleosome) octamer with its tails (orange) projecting out, linker [histone](#def-b3-chromatin-epigenetics-nucleosome) H1 (green) at the entry–exit point. About six [nucleosomes](#def-b3-chromatin-epigenetics-nucleosome) occupy the length that $1200\,\mathrm{bp}$ of naked DNA would.*

![Chromatin at the molecular scale: the double helix wound around successive histone cores, the “beads on a string” that the nuclease ladder revealed.](https://one-course.com/images/onecourse/chapters/biology-5/b3-chromatin-epigenetics/img-4a3089c69564.jpg)

*[Chromatin](#def-b3-chromatin-epigenetics-nucleosome) at the molecular scale: the double helix wound around successive [histone](#def-b3-chromatin-epigenetics-nucleosome) cores, the “beads on a string” that the nuclease ladder revealed.*

**Proposition 1.2 (Levels of compaction).**

The *[packing ratio](#prop-b3-chromatin-epigenetics-compaction)* of a [chromatin](#def-b3-chromatin-epigenetics-nucleosome) structure is the length of the DNA it contains divided by the length of the structure. Wrapping into the $10\,\mathrm{nm}$ fibre gives a ratio of about $6$: the $68\,\mathrm{nm}$ of a $200\,\mathrm{bp}$ repeat is compressed into one $11\,\mathrm{nm}$ bead. In the interphase nucleus the fibre is not wound into a regular thicker fibre, as textbooks long drew it, but folded into *loops* of tens to hundreds of kilobases that are held at their bases by ring-shaped cohesin complexes and by the DNA-binding protein CTCF; the loops group into *[topologically associating domains](#prop-b3-chromatin-epigenetics-compaction)* (TADs) of about a megabase within which a gene meets its regulators far more often than it meets anything outside. A mitotic chromosome, the most compact state, has a [packing ratio](#prop-b3-chromatin-epigenetics-compaction) near $10\,000$: the $4\,\mathrm{cm}$ of DNA in an average human chromosome are folded into a rod $4\,\text{µ}\mathrm{m}$ long.

**Example 1.3 (Two metres in ten micrometres).**

A human diploid nucleus holds $6.4\times 10^{9}$ base pairs. At $0.34\,\mathrm{nm}$ per base pair that is $2.2\,\mathrm{m}$ of double helix, in a nucleus about $10\,\text{µ}\mathrm{m}$ across. It is packed into $6.4\times 10^{9}/200 \approx 3.2\times 10^{7}$ [nucleosomes](#def-b3-chromatin-epigenetics-nucleosome), each carrying about $108\,\mathrm{kDa}$ of [histone](#def-b3-chromatin-epigenetics-nucleosome) against $96\,\mathrm{kDa}$ of DNA ($147\,\mathrm{bp}$ at $650\,\mathrm{Da}$ per pair): the nucleus contains roughly as much [histone](#def-b3-chromatin-epigenetics-nucleosome) as DNA by mass. The DNA itself, treated as a cylinder $2\,\mathrm{nm}$ across, has a volume of only about $7\,\text{µ}\mathrm{m}^{3}$, some $1\,\%$ of a spherical nucleus of radius $5\,\text{µ}\mathrm{m}$; the packing problem is one of entanglement and access, not of room.

## 1.2 The histone code

**Definition 1.4 (Histone modifications).**

The residues of the [histone](#def-b3-chromatin-epigenetics-nucleosome) tails are covalently modified: *acetylation* of lysines (which neutralises their positive charge and loosens the grip on DNA), *methylation* of lysines and arginines (one, two or three methyl groups, with no change of charge), phosphorylation of serines, ubiquitination of lysines. A modification is named by [histone](#def-b3-chromatin-epigenetics-nucleosome), residue and group: H3K4me3 is [histone](#def-b3-chromatin-epigenetics-nucleosome) H3, lysine 4, trimethylated; H3K27ac is H3 lysine 27 acetylated. The *[histone](#def-b3-chromatin-epigenetics-nucleosome) code* is the correlation between combinations of marks and the state of the underlying gene: H3K4me3 and H3K27ac at active promoters and enhancers, H3K36me3 along transcribed gene bodies, H3K9me3 on constitutive [heterochromatin](#def-b3-chromatin-epigenetics-nucleosome), H3K27me3 on genes silenced during development. Enzymes that add a mark are *writers* ([histone](#def-b3-chromatin-epigenetics-nucleosome) acetyltransferases, HATs; methyltransferases), enzymes that remove it are *erasers* ([histone](#def-b3-chromatin-epigenetics-nucleosome) deacetylases, HDACs; demethylases), and proteins that bind a mark and act on it are *readers*: a *bromodomain* binds acetyl-lysine, a *chromodomain* methyl-lysine.

**Proposition 1.5 (Marks are read, not merely felt).**

[Acetylation](#def-b3-chromatin-epigenetics-histone-code) acts partly by electrostatics — an acetylated tail holds its DNA less tightly and the fibre opens — but most marks act by recruiting [readers](#def-b3-chromatin-epigenetics-histone-code). H3K9me3 is bound by the chromodomain of the *[heterochromatin](#def-b3-chromatin-epigenetics-nucleosome) protein HP1*, which dimerises, bridges neighbouring [nucleosomes](#def-b3-chromatin-epigenetics-nucleosome) and recruits the very methyltransferase (SUV39H) that writes H3K9me3 on the next [nucleosome](#def-b3-chromatin-epigenetics-nucleosome): the mark spreads along the fibre until it meets a boundary, and compacts what it covers. H3K27me3 is written by *Polycomb repressive complex 2* (PRC2, catalytic subunit EZH2), read by PRC1, which ubiquitinates H2A and compacts the fibre, and the complex is itself stimulated by the mark it makes; the *Trithorax* group of proteins writes the opposing H3K4me3 and keeps the developmental genes of a given cell type on. Each such loop — [reader](#def-b3-chromatin-epigenetics-histone-code) recruits [writer](#def-b3-chromatin-epigenetics-histone-code) — is a positive feedback that lets a state maintain itself once established, which is what a heritable mark needs.

![The writer–reader–eraser logic. A reader bound to a mark on one nucleosome recruits the writer that puts the same mark on the next, so the state spreads and copies itself; erasers oppose it. Acetyl groups (green) open the fibre; H3K9me3 (red) closes it.](https://one-course.com/images/onecourse/chapters/biology-5/b3-chromatin-epigenetics/fig-a2c2527572c0.svg)

*The writer–reader–eraser logic. A [reader](#def-b3-chromatin-epigenetics-histone-code) bound to a mark on one [nucleosome](#def-b3-chromatin-epigenetics-nucleosome) recruits the [writer](#def-b3-chromatin-epigenetics-histone-code) that puts the same mark on the next, so the state spreads and copies itself; [erasers](#def-b3-chromatin-epigenetics-histone-code) oppose it. Acetyl groups (green) open the fibre; H3K9me3 (red) closes it.*

**Method 1.6 (Chromatin immunoprecipitation (ChIP)).**

To map where a mark or a protein sits on the genome: (1) cross-link proteins to DNA in living cells with formaldehyde; (2) shear the [chromatin](#def-b3-chromatin-epigenetics-nucleosome) into fragments of a few hundred base pairs; (3) precipitate the fragments with an antibody against the mark (say H3K27me3) bound to beads; (4) reverse the cross-links and purify the DNA; (5) sequence it (ChIP-seq) and count reads along the genome: peaks are the places where the mark was. A control of input DNA without antibody sets the background. The resolution is that of the fragments, about one [nucleosome](#def-b3-chromatin-epigenetics-nucleosome).

**Remark 1.7 (Code or correlation).**

The word “code” overstates the case. Many marks are consequences of transcription rather than causes of it (H3K36me3 is deposited by a methyltransferase riding with the polymerase), and removing a [writer](#def-b3-chromatin-epigenetics-histone-code) often changes expression far less than the mark’s distribution suggests. What is well established is that a few marks — H3K9me3, H3K27me3, and [DNA methylation](#def-b3-chromatin-epigenetics-methylation) below — carry self-propagating silencing that outlives the signal that set it up.

## 1.3 DNA methylation and the copying of a mark

**Definition 1.8 (DNA methylation).**

In vertebrates a methyl group is added to carbon 5 of cytosine almost only in the dinucleotide *CpG* (a C followed by a G on the same strand), a sequence that is its own complement, so that a methylated CpG is normally methylated on both strands. Some $70\text{ to }80\,\%$ of the CpGs of a human cell are methylated, and methylated promoters are silent: the methyl groups are read by methyl-CpG-binding proteins (MeCP2, MBD1–4) that recruit HDACs and H3K9 methylases, and they block several transcription factors directly. The exceptions are the *CpG islands*, stretches of about a kilobase, rich in G and C and in CpG, that sit at the promoters of most housekeeping genes and stay unmethylated in normal cells. The methyl group is written by *DNA methyltransferases*: DNMT3A and DNMT3B perform *de novo methylation* of unmethylated sites, and *DNMT1*, which prefers a hemimethylated CpG — one strand methylated, the other not — performs *maintenance methylation* behind the replication fork. Removal is passive, by replication without maintenance, or active, by the TET enzymes, which oxidise 5-methylcytosine to 5-hydroxymethylcytosine and beyond until base-excision repair replaces it with a plain cytosine.

**Example 1.9 (Why the genome is short of CpG).**

The human genome is $42\,\%$ G$+$C, so if bases were independent a [CpG](#def-b3-chromatin-epigenetics-methylation) would occur with frequency $0.21\times 0.21 \approx 0.044$, about $2.8\times 10^{8}$ times in the diploid genome. The observed number is about $5.6\times 10^{7}$, one fifth of that. The reason is the mark itself: a methylated cytosine that loses its amino group becomes thymine, a normal base that repair cannot recognise as wrong, whereas an unmethylated cytosine deaminates to uracil, which is excised. Over evolutionary time methylated [CpGs](#def-b3-chromatin-epigenetics-methylation) have mutated into TpG and CpA, and only the unmethylated islands have kept their [CpGs](#def-b3-chromatin-epigenetics-methylation). The mark has left its signature in the sequence.

**Theorem 1.10 (Maintenance of a methylation state through divisions).**

Consider one [CpG](#def-b3-chromatin-epigenetics-methylation) site, followed along a lineage of cells. At each division the new strand’s cytosine is methylated with probability $\mu$ if the template strand’s cytosine is methylated (maintenance efficiency) and with probability $\delta$ if it is not (de novo rate). Let $p_{n}$ be the probability that the site is methylated after $n$ divisions. Then

$$
p_{n+1} = \mu\, p_{n} + \delta\,(1 - p_{n}),
\qquad
p_{n} = p^{*} + (p_{0} - p^{*})\,(\mu - \delta)^{n},
\qquad
p^{*} = \frac{\delta}{1 - \mu + \delta}.
$$

Every site drifts toward the same equilibrium $p^{*}$, whatever its starting state, and the memory of the initial state decays geometrically with ratio $\mu - \delta$. A mark is remembered for about $1/(1 - \mu + \delta)$ divisions; faithful memory of a *difference* between sites therefore requires $\mu$ very close to $1$ and $\delta$ very close to $0$, or a feedback that makes $\mu$ and $\delta$ depend on the state of the neighbourhood.

**Proof.** The recurrence is the law of total probability over the two states of the template. Its fixed point solves $p^{*} = \mu p^{*} + \delta(1 -
p^{*})$, giving $p^{*}(1 - \mu + \delta) = \delta$. Subtracting the fixed-point equation from the recurrence, $p_{n+1} - p^{*} = (\mu -
\delta)(p_{n} - p^{*})$, which iterates to the closed form. Since $0 \le \delta \le \mu \le 1$, the ratio $\mu - \delta$ lies in $[0,1]$ and the deviation shrinks; it is halved every $\ln 2 / (-\ln(\mu -
\delta)) \approx \ln 2/(1 - \mu + \delta)$ divisions when $\mu -
\delta$ is near $1$. ∎

**Example 1.11 (Numbers).**

Measured values for mammalian cells are $\mu \approx 0.99$ and $\delta \approx 0.02$ per division at a typical [CpG](#def-b3-chromatin-epigenetics-methylation). Then $p^{*} =
0.02/0.03 \approx 0.67$, close to the genome-wide methylated fraction, and $\mu - \delta = 0.97$: a fully methylated site left to the machinery alone would be half-way back to the average after $\ln 2 /
0.03 \approx 23$ divisions. A silenced promoter that stays silent for the hundreds of divisions of a lifetime therefore needs more than [DNMT1](#def-b3-chromatin-epigenetics-methylation): the [methyl-CpG](#def-b3-chromatin-epigenetics-methylation) [readers](#def-b3-chromatin-epigenetics-histone-code) recruit H3K9 methylation and the H3K9 [readers](#def-b3-chromatin-epigenetics-histone-code) recruit DNMT3, so that a dense island of marks raises its own $\mu$ toward $1$ and lowers the neighbouring $\delta$ toward $0$. Conversely, a cell that loses [DNMT1](#def-b3-chromatin-epigenetics-methylation) loses the mark passively: with $\mu = \delta = 0$ the fraction of methylated strands halves at each division.

![Left: maintenance methylation. After replication every CpG is hemimethylated; DNMT1 copies the mark onto the new strand with probability per site. Right: the fate of a site over divisions for = 0.99, = 0.02 (blue from methylated, red from unmethylated) — both converge on p* = 0.67 — and for a site whose neighbourhood feedback raises to 0.999 (green).](https://one-course.com/images/onecourse/chapters/biology-5/b3-chromatin-epigenetics/fig-c3379454f641.svg)

*Left: [maintenance methylation](#def-b3-chromatin-epigenetics-methylation). After replication every [CpG](#def-b3-chromatin-epigenetics-methylation) is hemimethylated; [DNMT1](#def-b3-chromatin-epigenetics-methylation) copies the mark onto the new strand with probability $\mu$ per site. Right: the fate of a site over divisions for $\mu = 0.99$, $\delta = 0.02$ (blue from methylated, red from unmethylated) — both converge on $p^{*} = 0.67$ — and for a site whose neighbourhood feedback raises $\mu$ to $0.999$ (green).*

**Method 1.12 (Reading methylation: bisulfite sequencing).**

Sequencing alone cannot see a methyl group. Treat denatured DNA with sodium bisulfite: unmethylated cytosine is deaminated to uracil, which reads as T after PCR, while 5-methylcytosine is protected and still reads as C. Sequence the converted DNA and compare with the reference: every C that remains C was methylated, every C that became T was not. Done genome-wide (whole-genome [bisulfite sequencing](#met-b3-chromatin-epigenetics-bisulfite)) this gives the methylation state of each of the $2.8\times 10^{7}$ [CpGs](#def-b3-chromatin-epigenetics-methylation) of a haploid genome at single-base resolution, as a fraction of the cells in the sample.

## 1.4 X inactivation and genomic imprinting

**Definition 1.13 (X-chromosome inactivation).**

In female mammals one of the two X chromosomes of every somatic cell is transcriptionally silenced early in development, a form of *dosage compensation* that equalises X-linked gene expression between XX females and XY males. The silenced chromosome is compacted into a dense body against the nuclear envelope, the *Barr body*. The choice of which X to silence is random in the embryo proper and, once made, is inherited clonally by all descendants of the cell. Inactivation is initiated by *Xist*, a $17\,\mathrm{kb}$ non-coding RNA transcribed from the X-inactivation centre of the chromosome that will be silenced: it coats that chromosome in cis, recruits Polycomb (H3K27me3), the [histone](#def-b3-chromatin-epigenetics-nucleosome) variant macroH2A and finally [DNA methylation](#def-b3-chromatin-epigenetics-methylation) of the promoters, and the silence is thereafter maintained without Xist being needed for each division. About $15\,\%$ of human X-linked genes escape inactivation, in part or wholly.

**Evidence.** Barr and Bertram (1949) noticed a dense body in the nuclei of female but not male cat neurons. Ohno (1959) showed it was one whole X chromosome, condensed. Lyon (1961) put the pieces together: female mice heterozygous for X-linked coat-colour genes have a mosaic coat of patches expressing one allele or the other, so each patch is a clone in which one X, always the same one, is silent; the same holds for the orange-and-black patches of a tortoiseshell or calico cat, whose orange gene is X-linked, and for the mosaic sweat glands of women carrying a mutant allele of an X-linked gene. A male calico cat is almost always XXY. ∎

![Left: a female nucleus stained for DNA, with the compact inactive X — the Barr body — pressed against the nuclear edge. Right: a calico cat. Each orange or black patch is a clone of skin cells that silenced the same X chromosome; the white is a separate, autosomal, spotting gene.](https://one-course.com/images/onecourse/chapters/biology-5/b3-chromatin-epigenetics/img-3f3811e47dc3.jpg)

![Left: a female nucleus stained for DNA, with the compact inactive X — the Barr body — pressed against the nuclear edge. Right: a calico cat. Each orange or black patch is a clone of skin cells that silenced the same X chromosome; the white is a separate, autosomal, spotting gene.](https://one-course.com/images/onecourse/chapters/biology-5/b3-chromatin-epigenetics/img-f88c4a5d803d.jpg)

*Left: a female nucleus stained for DNA, with the compact inactive X — the [Barr body](#def-b3-chromatin-epigenetics-x-inactivation) — pressed against the nuclear edge. Right: a calico cat. Each orange or black patch is a clone of skin cells that silenced the same X chromosome; the white is a separate, autosomal, spotting gene.*

**Definition 1.14 (Genomic imprinting).**

A gene is *imprinted* when only one of its two alleles is expressed, and which one depends on the parent it came from: some genes are expressed only from the paternal copy, others only from the maternal. Mammals have some $100\text{ to }200$ imprinted genes, clustered, each cluster governed by an *imprinting control region* (ICR) that is methylated on one parental allele and not the other. The methylation is set in the germ line — one pattern in sperm, another in eggs — erased in the germ cells of the next generation and set again according to the sex of that individual. A child who receives both copies of a chromosome from the same parent (*uniparental disomy*) therefore has the wrong dose of the imprinted genes on it although its sequence is entirely normal.

**Evidence.** McGrath and Solter, and independently Surani, Barton and Norris (1984), made mouse zygotes carrying two maternal pronuclei, or two paternal ones, by transferring pronuclei between fertilised eggs. Neither kind developed to term, although each had two complete sets of chromosomes: the gynogenetic embryos formed a reasonable embryo with a poor placenta, the androgenetic ones a large placenta and almost no embryo. The two parental genomes are not equivalent and both are required. Cattanach and Kirk (1985) showed with chromosome-specific uniparental disomies that the non-equivalence is confined to particular regions of particular chromosomes, which are where the imprinted clusters were later found. ∎

**Example 1.15 (The Igf2–H19 cluster).**

*Igf2*, a fetal growth factor, is expressed from the paternal allele; its neighbour *H19*, a non-coding RNA, from the maternal. Between them lies the ICR and, beyond *H19*, a shared set of enhancers. On the maternal chromosome the ICR is unmethylated and binds CTCF, which acts as an insulator: the enhancers can reach *H19* but not *Igf2*. On the paternal chromosome the ICR is methylated (set in the sperm), CTCF cannot bind, and the enhancers reach across to *Igf2*; the methylation also spreads to silence the *H19* promoter. One mark, set in one germ line, decides both genes.

![Imprinting at Igf2–H19. Maternal allele: the unmethylated ICR binds CTCF, an insulator, so the downstream enhancers activate H19 only. Paternal allele: the ICR was methylated in the sperm, CTCF cannot bind, the enhancers reach Igf2, and the methylation silences H19.](https://one-course.com/images/onecourse/chapters/biology-5/b3-chromatin-epigenetics/fig-b2fe663aab35.svg)

*Imprinting at *Igf2*–*H19*. Maternal allele: the unmethylated ICR binds CTCF, an insulator, so the downstream enhancers activate *H19* only. Paternal allele: the ICR was methylated in the sperm, CTCF cannot bind, the enhancers reach *Igf2*, and the methylation silences *H19*.*

**Example 1.16 (Prader–Willi and Angelman).**

The same deletion of about $5\,\mathrm{Mb}$ in chromosome 15 (region 15q11–q13) causes *Prader–Willi syndrome* — poor muscle tone at birth, then insatiable appetite and obesity — when it is on the paternal chromosome, and *Angelman syndrome* — severe intellectual disability, absent speech, a happy demeanour and seizures — when on the maternal. The region carries paternally expressed genes (*SNRPN* and a cluster of small RNAs) whose only active copy is lost in the first case, and the maternally expressed *UBE3A*, lost in the second. Maternal [uniparental disomy](#def-b3-chromatin-epigenetics-imprinting) of 15, with no deletion at all, also gives Prader–Willi: two maternal copies, silent for the paternal genes.

**Remark 1.17 (Why imprint at all).**

Imprinted genes are enriched for control of fetal growth and of placental nutrient transfer, with paternally expressed genes (*Igf2*, *Peg3*) tending to promote growth and maternally expressed ones (*H19*, *Igf2r*, *Cdkn1c*) to restrain it. The *kinship theory* reads this as a conflict: a father’s genes gain from extracting more from a mother who may carry other males’ offspring later, while the mother’s genes gain from rationing across all her litters. Imprinting occurs in placental mammals and in flowering plants, whose endosperm is likewise fed by the mother, and not in birds or reptiles, which is what the theory predicts.

## 1.5 Epigenetic inheritance and its limits

**Definition 1.18 (Epigenetics).**

*Epigenetics* is the study of heritable changes in gene activity that are not changes in DNA sequence: states carried by [DNA methylation](#def-b3-chromatin-epigenetics-methylation), [histone](#def-b3-chromatin-epigenetics-nucleosome) marks, or self-sustaining transcription-factor circuits, and copied through cell division. Two alleles of identical sequence but different stable epigenetic state are *epialleles*. Inheritance through mitosis is the rule and is what makes a differentiated cell stay differentiated. Inheritance through meiosis, from parent to offspring, is the exception in mammals, because the marks are erased in two waves of *reprogramming*: in the primordial germ cells, where imprints and most methylation are wiped and then set afresh according to sex, and again in the zygote and early embryo, where the paternal genome is actively demethylated by TET3 and the maternal passively, before the embryo’s own patterns are laid down by DNMT3.

**Example 1.19 (The agouti viable yellow mouse).**

In the $A^{vy}$ allele a retrotransposon has inserted upstream of the *agouti* coat-colour gene. Where the transposon’s promoter is unmethylated it drives *agouti* constantly in every cell; the mouse is yellow, obese and prone to diabetes. Where it is methylated the gene follows its normal control and the mouse is brown (“pseudoagouti”). Genetically identical littermates range across the whole spectrum, and a yellow mother has more yellow pups than a brown one: the [epiallele](#def-b3-chromatin-epigenetics-epigenetics) is incompletely erased through the maternal germ line. Waterland and Jirtle (2003) fed pregnant $A^{vy}$ mothers a diet rich in methyl donors (folate, vitamin B12, choline, betaine) and shifted the pups’ coats toward brown. The environment of one generation had set a mark that the next generation carried.

![Two Avy littermates of identical genotype: a yellow, obese mouse in which the transposon upstream of agouti is unmethylated and a brown pseudoagouti in which it is methylated.](https://one-course.com/images/onecourse/chapters/biology-5/b3-chromatin-epigenetics/img-f80cae50b46a.jpg)

*Two $A^{vy}$ littermates of identical genotype: a yellow, obese mouse in which the transposon upstream of *agouti* is unmethylated and a brown pseudoagouti in which it is methylated.*

**Proposition 1.20 (Where heritable epigenetic states are found).**

Three well-analysed systems illustrate the mechanisms above. *[Position-effect variegation](#prop-b3-chromatin-epigenetics-examples)* in *Drosophila*: a gene relocated by a chromosome rearrangement next to [heterochromatin](#def-b3-chromatin-epigenetics-nucleosome) is silenced in some cells and not others, giving a mottled eye; each patch is a clone, the silencing spreads from the [heterochromatin](#def-b3-chromatin-epigenetics-nucleosome) by the HP1 loop, and the genes whose mutation suppresses the variegation (*Su(var)* genes) are HP1 and the H3K9 methyltransferase. *Vernalization* in *Arabidopsis*: weeks of winter cold silence the flowering repressor *FLC* by Polycomb-deposited H3K27me3, the silence holds through months of spring growth and thousands of cell divisions, and it is reset in every generation so that each plant must experience its own winter. *Paramutation* in maize: one allele of the *b1* pigment gene, when it meets a particular other allele in a heterozygote, converts it to its own silent state, and the converted allele passes the state to its descendants for generations, through a small-RNA-directed methylation taken up in [Chapter 2](https://one-course.com/books/biology/5/en/chapter/2-non-coding-rnas-and-post-transcriptional-regulation#ch-b3-rna-regulation). In each case the inheritance is mitotic and robust; transmission across generations is either actively reset (plants) or leaky and partial (mice).

**Remark 1.21 (The limits of transgenerational claims).**

Claims that a grandparent’s famine or stress is “inherited epigenetically” by human grandchildren are common and mostly unproven. Two waves of erasure stand between a somatic mark and a grandchild, the imprinted genes and a few transposons are the only documented survivors of erasure in mammals, and in human cohorts the effects of a shared environment, of the uterine environment, and of genetic differences are hard to separate from a mark carried in the germ line. The demonstrated cases — $A^{vy}$, some transposon-linked [epialleles](#def-b3-chromatin-epigenetics-epigenetics), RNA-mediated states in worms and plants — are real, molecularly understood, and narrow. The chapter’s theorem says why: without a feedback that pushes $\mu$ to $1$, a mark forgets itself in a few dozen divisions, and the germ line adds an active reset on top.

## 1.6 Exercises

**Exercise 1.1 ★.**

Give the composition of a [nucleosome](#def-b3-chromatin-epigenetics-nucleosome) core particle, the length of DNA it binds, and the typical linker length. Why did nuclease digestion of [chromatin](#def-b3-chromatin-epigenetics-nucleosome) give a ladder of fragments rather than a smear?

**Solution of Exercise 1.1.**

Two copies each of H2A, H2B, H3 and H4 (an octamer) with $147\,\mathrm{bp}$ of DNA in $1.65$ turns; linkers of $20\text{ to }60\,\mathrm{bp}$, giving a repeat near $200\,\mathrm{bp}$. The nuclease cuts only the exposed linker DNA, never inside a core, so every fragment is a whole number of repeats: a ladder of $200$, $400$, $600\,\mathrm{bp}$, not a smear.

**Exercise 1.2 ★.**

A frog genome has $3.0\times 10^{9}$ base pairs per haploid set. How many [nucleosomes](#def-b3-chromatin-epigenetics-nucleosome) does a diploid nucleus contain, taking a repeat of $190\,\mathrm{bp}$, and how long is its DNA?

**Solution of Exercise 1.2.**

Diploid: $6.0\times 10^{9}$ bp; $6.0\times 10^{9}/190 \approx
3.2\times 10^{7}$ [nucleosomes](#def-b3-chromatin-epigenetics-nucleosome); length $6.0\times 10^{9}\times
0.34\,\mathrm{nm} = 2.0\,\mathrm{m}$.

**Exercise 1.3 ★.**

Decode H3K27me3, H4K16ac and H3S10ph. For each say whether it is associated with active or silent [chromatin](#def-b3-chromatin-epigenetics-nucleosome), and name a [reader](#def-b3-chromatin-epigenetics-histone-code) for the first.

**Solution of Exercise 1.3.**

H3K27me3: [histone](#def-b3-chromatin-epigenetics-nucleosome) H3, lysine 27, trimethylated — silent [chromatin](#def-b3-chromatin-epigenetics-nucleosome) of developmentally repressed genes, written by PRC2 and read by the chromodomain proteins of PRC1. H4K16ac: H4, lysine 16, acetylated — active, open [chromatin](#def-b3-chromatin-epigenetics-nucleosome) (it breaks a nucleosome–nucleosome contact). H3S10ph: H3, serine 10, phosphorylated — a mitotic mark associated with chromosome condensation (and, transiently, with gene activation by some signals); not a stable silencing or activating mark.

**Exercise 1.4 ★.**

Explain why a calico cat is almost always female, and what a male calico must be.

**Solution of Exercise 1.4.**

The orange gene is X-linked with an orange and a black allele; a patched coat needs one of each, on two different X chromosomes, one of which is silenced per cell. A male has one X and is either orange or black all over. A calico male carries two X chromosomes and a Y (XXY), and is sterile.

**Exercise 1.5 ★★.**

Compute the [packing ratio](#prop-b3-chromatin-epigenetics-compaction) of the $10\,\mathrm{nm}$ fibre from the numbers of [Definition 1.1](#def-b3-chromatin-epigenetics-nucleosome) (repeat $200\,\mathrm{bp}$, bead $11\,\mathrm{nm}$), and the ratio needed to fit the $4\,\mathrm{cm}$ of DNA of an average human chromosome into a mitotic chromatid $4\,\text{µ}\mathrm{m}$ long. By what further factor must the fibre be folded?

**Solution of Exercise 1.5.**

$200\,\mathrm{bp}\times 0.34\,\mathrm{nm} = 68\,\mathrm{nm}$ per bead of $11\,\mathrm{nm}$: ratio $6.2$. Required: $4\,\mathrm{cm}/4\,\text{µ}\mathrm{m} =
10^{4}$. Further folding by $10^{4}/6.2 \approx 1600$.

**Exercise 1.6 ★★.**

In the model of [Theorem 1.10](#thm-b3-chromatin-epigenetics-maintenance) with $\mu = 0.98$ and $\delta = 0.01$, find $p^{*}$ and the number of divisions after which a site that started methylated has lost half of its excess over $p^{*}$. Repeat for $\mu = 0.999$, $\delta = 0.001$.

**Solution of Exercise 1.6.**

$\mu = 0.98$, $\delta = 0.01$: $p^{*} = 0.01/0.03 = 0.33$; ratio $\mu -
\delta = 0.97$; half-life $\ln 2/(-\ln 0.97) = 22.8 \approx 23$ divisions. $\mu = 0.999$, $\delta = 0.001$: $p^{*} = 0.5$; ratio $0.998$; half-life $\ln 2 / 0.002 \approx 350$ divisions.

**Exercise 1.7 ★★.**

A ChIP-seq experiment on liver cells gives an H3K4me3 peak and an H3K27ac peak at the promoter of gene $A$, a broad H3K27me3 domain over gene $B$, and H3K9me3 over gene $C$. Predict the transcriptional state of each, and say which of $B$ and $C$ is more likely to be switched on in another cell type and why.

**Solution of Exercise 1.7.**

$A$: active (promoter H3K4me3 with [acetylation](#def-b3-chromatin-epigenetics-histone-code)). $B$: silent, Polycomb-repressed — facultative [heterochromatin](#def-b3-chromatin-epigenetics-nucleosome), the state of a developmental gene switched off in this lineage but used in others, so $B$ is the one likely to be active in another cell type. $C$: silent, constitutive [heterochromatin](#def-b3-chromatin-epigenetics-nucleosome) (HP1 pathway), typically repeats or genes silenced in every cell type.

**Exercise 1.8 ★★.**

A woman carries a deletion of the *UBE3A* region on one chromosome 15 but is healthy. Her father carried the same deletion and was healthy too. What syndrome, if any, is each of her children at risk of, and with what probability? What if the deletion had come from her mother?

**Solution of Exercise 1.8.**

*UBE3A* is expressed only from the maternal allele in neurons. Her deletion came from her father, on the allele that is silent anyway, so she is healthy; when she transmits it, it becomes a maternal deletion: each child has probability $1/2$ of inheriting it and then has Angelman syndrome. Had the deletion come from her mother, she would herself have Angelman syndrome; a healthy carrier can only have inherited it paternally. (A paternal deletion of the whole region would give Prader–Willi; *UBE3A* alone is not enough for that.)

**Exercise 1.9 ★★.**

Predict the consequence, in a female mouse embryo, of deleting the *[Xist](#def-b3-chromatin-epigenetics-x-inactivation)* gene from one X chromosome; from both. Predict what a *[Xist](#def-b3-chromatin-epigenetics-x-inactivation)* transgene inserted into an autosome does to that autosome.

**Solution of Exercise 1.9.**

One X lacking *[Xist](#def-b3-chromatin-epigenetics-x-inactivation)* cannot be inactivated, so the other X is silenced in every cell: inactivation is complete but no longer random (skewed), and the embryo is viable. Both X lacking *[Xist](#def-b3-chromatin-epigenetics-x-inactivation)*: no inactivation, two active X chromosomes, twofold X-linked expression, and the female embryo dies early. A *[Xist](#def-b3-chromatin-epigenetics-x-inactivation)* transgene on an autosome coats that autosome in cis and silences its genes, causing loss of function of one autosomal copy — the experiment that showed *[Xist](#def-b3-chromatin-epigenetics-x-inactivation)* is sufficient for silencing in cis.

**Exercise 1.10 ★★★.**

Explain, with [Example 1.9](#ex-b3-chromatin-epigenetics-cpg-deficit), why [CpG islands](#def-b3-chromatin-epigenetics-methylation) have kept their [CpGs](#def-b3-chromatin-epigenetics-methylation) while the rest of the genome has lost four fifths of them, and what this implies about the methylation state of islands in the germ line over evolutionary time. Why does the argument fail for a gene methylated only in liver?

**Solution of Exercise 1.10.**

Methylated cytosines deaminate to thymine, which repair does not recognise, so a methylated [CpG](#def-b3-chromatin-epigenetics-methylation) mutates to TpG or CpA at a high rate. Only sequences unmethylated in the *germ line*, generation after generation, keep their [CpGs](#def-b3-chromatin-epigenetics-methylation); islands have kept theirs, so they have been unmethylated in the germ line for most of vertebrate history. A gene methylated only in liver is unmethylated in sperm and egg; mutations that arise in liver cells are not inherited, so somatic methylation leaves no footprint in the sequence.

**Exercise 1.11 ★★★.**

Show, using the writer–reader loop of [Proposition 1.5](#prop-b3-chromatin-epigenetics-readers), how a domain of H3K9me3 can be stable through replication although each division dilutes the marked [histones](#def-b3-chromatin-epigenetics-nucleosome) twofold with new unmarked ones. What determines where the domain stops spreading? Propose an experiment to test your account.

**Solution of Exercise 1.11.**

At replication the old, marked [histones](#def-b3-chromatin-epigenetics-nucleosome) are shared between the two daughter duplexes and interleaved with new, unmarked ones, so each daughter domain is about half marked. HP1 bound to a retained marked [nucleosome](#def-b3-chromatin-epigenetics-nucleosome) recruits SUV39H, which methylates the unmarked neighbours; as long as marked [nucleosomes](#def-b3-chromatin-epigenetics-nucleosome) are dense enough to bring a [writer](#def-b3-chromatin-epigenetics-histone-code) to every gap, the domain is refilled before the next division — the same feedback as the theorem’s high $\mu$. Spreading stops at boundaries: insulator proteins (CTCF), highly transcribed genes and tRNA genes, [nucleosomes](#def-b3-chromatin-epigenetics-nucleosome) carrying incompatible active marks (H3K4me3, [acetylation](#def-b3-chromatin-epigenetics-histone-code)) that the [writer](#def-b3-chromatin-epigenetics-histone-code) cannot use, and the point where [erasers](#def-b3-chromatin-epigenetics-histone-code) (demethylases, HDAC turnover) outpace the [writer](#def-b3-chromatin-epigenetics-histone-code). Test: recruit HP1 transiently to a reporter (a fusion with a DNA-binding domain whose binding is drug-controlled), release it, and follow the reporter’s silence and H3K9me3 over divisions; predict persistence that depends on the [writer](#def-b3-chromatin-epigenetics-histone-code), and loss if the [writer](#def-b3-chromatin-epigenetics-histone-code) or the [reader](#def-b3-chromatin-epigenetics-histone-code)’s dimerisation is mutated, or if an insulator is placed within the domain.

**Exercise 1.12 ★★★.**

A study reports that the grandchildren of people who lived through a famine have altered methylation at a growth gene and a higher body mass. List the alternative explanations that must be excluded before concluding that a methylation mark was transmitted through the germ line, and say what experiment in mice would settle the question.

**Solution of Exercise 1.12.**

Exclude: a shared environment (the grandchildren’s own diet and poverty); a uterine effect — the famine-exposed mother’s daughter was a fetus, with her germ cells already present, so the grandchild’s germline was exposed directly and the effect is not transgenerational; genetic differences between families that survived and those that did not; reverse causation (body mass itself alters methylation in blood); differences in blood cell composition; and multiple testing across many [CpGs](#def-b3-chromatin-epigenetics-methylation). Mouse experiment: expose inbred F$_0$ *males* (no uterine route), mate them to unexposed females or use in vitro fertilisation and embryo transfer, and follow F$_2$ and F$_3$ through the paternal line, measuring methylation in the sperm of each generation; a mark that persists in F$_3$ sperm and phenotype, in a strain with no genetic variation, is germline transmission.

## 1.7 Problem: The Memory of a Silenced Gene

**Problem 1.1.**

Weekend problem — a human genome packed into its nucleus, a methylation mark followed through divisions, a calico cat read as a clonal map, and an imprinted region’s diseases counted, ending on the number of divisions a mark survives without feedback and the size of the cell population that chose its X

Data: a human diploid nucleus holds $6.4\times 10^{9}$ base pairs, at $0.34\,\mathrm{nm}$ and $650\,\mathrm{Da}$ per pair; the [nucleosome](#def-b3-chromatin-epigenetics-nucleosome) repeat is $200\,\mathrm{bp}$, a [histone](#def-b3-chromatin-epigenetics-nucleosome) octamer weighs $108\,\mathrm{kDa}$, and a core particle is a disc $11\,\mathrm{nm}$ across and $5.5\,\mathrm{nm}$ thick. The nucleus is a sphere of diameter $10\,\text{µ}\mathrm{m}$. The genome is $42\,\%$ G$+$C. Methylation at an ordinary [CpG](#def-b3-chromatin-epigenetics-methylation): $\mu = 0.99$, $\delta = 0.02$ per division.

**Part I — Packing.**

1. Compute the total length of DNA in the nucleus and the number of [nucleosomes](#def-b3-chromatin-epigenetics-nucleosome) .
2. Compute the total mass of [histone](#def-b3-chromatin-epigenetics-nucleosome) (octamers only) and of DNA, in picograms ( $1\,\mathrm{Da} = 1.66 \times 10^{-24}\,\mathrm{g}$ ).
3. Compute the volume of the nucleus and the total volume of the core particles, treated as cylinders. What fraction of the nucleus do they fill?
4. If the [nucleosomes](#def-b3-chromatin-epigenetics-nucleosome) were laid end to end as a $10\,\mathrm{nm}$ fibre, how long would the fibre be, and what is the [packing ratio](#prop-b3-chromatin-epigenetics-compaction) relative to naked DNA?
5. The fibre is folded into loops of $200\,\mathrm{kb}$ anchored by cohesin. How many loops, and how many [nucleosomes](#def-b3-chromatin-epigenetics-nucleosome) per loop?
6. Under the independence assumption, how many [CpG](#def-b3-chromatin-epigenetics-methylation) dinucleotides would the diploid genome contain? The observed number is $5.6\times 10^{7}$ . What fraction has been lost, and by what mutational mechanism?

**Part II — A mark through divisions.**

7. Write the recurrence for $p_{n}$ and compute $p^{*}$ .
8. A promoter [CpG](#def-b3-chromatin-epigenetics-methylation) is fully methylated at $n = 0$ . Compute $p_{n}$ after $10$ , $50$ and $200$ divisions.
9. After how many divisions is the excess $p_{n} - p^{*}$ halved? A tissue stem cell divides about once a week: how many years is that?
10. A drug inhibits [DNMT1](#def-b3-chromatin-epigenetics-methylation) completely ( $\mu = 0$ ) and DNMT3 as well ( $\delta = 0$ ). Show that the fraction of methylated strands in the population halves at each division, and find the number of divisions to bring $80\,\%$ methylation below $5\,\%$ .
11. In a silenced island the [methyl-CpG](#def-b3-chromatin-epigenetics-methylation) [readers](#def-b3-chromatin-epigenetics-histone-code) recruit H3K9 methylation, which recruits DNMT3, so that within the island $\mu = 0.9995$ and $\delta = 0.0005$ . Recompute the half-life of the excess, in divisions and in years at one division a week.
12. Explain in one sentence why the result of question 11 and not that of question 9 is what a lifelong silenced gene requires, and what kind of molecular arrangement produces it.

**Part III — The calico map.**

13. A cat is heterozygous for the X-linked orange gene. Explain why her coat has orange and black patches and what one patch represents.
14. [X inactivation](#def-b3-chromatin-epigenetics-x-inactivation) occurred when the cells destined to form the skin numbered $N$ . Each chose an X at random. What is the probability that all $N$ chose the same one, giving a single-colour cat?
15. No such single-coloured heterozygote has been documented among perhaps a billion cats; take the probability as $10^{-9}$ and estimate $N$ .
16. The cat’s skin surface is about $0.3\,\mathrm{m}^{2}$ and her patches average $15\,\mathrm{cm}^{2}$ . Estimate the number of patches, and explain why it is larger than $N$ .
17. An XXY tomcat is also calico. How many Barr bodies do his cells show? How many in an XXX female and in an XY male?
18. A gene that escapes inactivation is present in two active copies in females and one in males. Give one reason such escape may be tolerated and one situation where it matters.

**Part IV — Chromosome 15.**

19. Prader–Willi syndrome affects about one birth in $15\,000$ ; $70\,\%$ of cases are a de novo deletion of the paternal 15q11–q13 and $25\,\%$ a maternal [uniparental disomy](#def-b3-chromatin-epigenetics-imprinting) . Compute the birth frequency of each cause.
20. Angelman syndrome has about the same frequency; $70\,\%$ are maternal deletions, $5\,\%$ paternal disomies and $10\,\%$ point mutations of *UBE3A* . Why are point mutations a cause of Angelman but almost never of Prader–Willi?
21. A father carries a balanced translocation that gives $20\,\%$ of his sperm a 15q11–q13 deletion. What fraction of his children have Prader–Willi, and what fraction Angelman?
22. [Uniparental disomy](#def-b3-chromatin-epigenetics-imprinting) arises when a trisomic zygote loses one chromosome. If a trisomy-15 zygote loses one of its three chromosomes at random, and the trisomy came from the mother (two maternal copies), what is the probability that the survivor has maternal disomy? Which syndrome?
23. The Prader–Willi genes are paternally expressed and the Angelman gene maternally. Using the kinship theory, say which of the two syndromes’ symptoms (feeding difficulty at birth versus voracious appetite) fit the theory and which are harder to fit.
24. A method is proposed to treat Angelman by reactivating the silent paternal *UBE3A* . Its silence is due to a paternally expressed antisense RNA. Propose the molecular target and say why the therapy would not disturb the other imprinted genes.
25. Summarise: the half-life of a mark without feedback (question 9) and with it (question 11), in divisions; and the founder population $N$ of the calico coat (question 15).

**Solution of Problem 1.1.**

**1.** $6.4\times 10^{9}\times 0.34\,\mathrm{nm} = 2.2\,\mathrm{m}$; $6.4\times 10^{9}/200 = 3.2\times 10^{7}$ [nucleosomes](#def-b3-chromatin-epigenetics-nucleosome). **2.** [Histone](#def-b3-chromatin-epigenetics-nucleosome): $3.2\times 10^{7}\times 1.08\times 10^{5} =
3.5\times 10^{12}$ Da $= 5.7\,\mathrm{pg}$. DNA: $6.4\times 10^{9}\times
650 = 4.2\times 10^{12}$ Da $= 6.9\,\mathrm{pg}$. Comparable masses. **3.** Nucleus: $\tfrac{4}{3}\pi\,(5\,\text{µ}\mathrm{m})^{3} =
520\,\text{µ}\mathrm{m}^{3}$. Core: $\pi\,(5.5)^{2}\times 5.5 =
520\,\mathrm{nm}^{3} = 5.2\times 10^{-7}\,\text{µ}\mathrm{m}^{3}$; total $3.2\times 10^{7}\times 5.2\times 10^{-7} = 17\,\text{µ}\mathrm{m}^{3}$: about $3\,\%$ of the nucleus. **4.** $3.2\times 10^{7}\times 11\,\mathrm{nm} = 0.35\,\mathrm{m}$; [packing ratio](#prop-b3-chromatin-epigenetics-compaction) $2.2/0.35 \approx 6$. **5.** $6.4\times 10^{9}/2\times 10^{5} = 32\,000$ loops of $1000$ [nucleosomes](#def-b3-chromatin-epigenetics-nucleosome) each. **6.** $0.21^{2}\times 6.4\times 10^{9} = 2.8\times 10^{8}$; observed $5.6\times 10^{7}$, so $80\,\%$ lost, by deamination of 5-methylcytosine to thymine (unrepaired C$\to$T transitions at methylated [CpGs](#def-b3-chromatin-epigenetics-methylation)). **7.** $p_{n+1} = 0.99\,p_{n} + 0.02\,(1 - p_{n}) = 0.97\,p_{n} +
0.02$; $p^{*} = 0.02/0.03 = 0.67$. **8.** $p_{n} = 0.67 + 0.33\times 0.97^{n}$: $n = 10$: $0.91$; $n = 50$: $0.74$; $n = 200$: $0.67$ (the excess is $7\times 10^{-4}$). **9.** $\ln 2/(-\ln 0.97) = 22.8 \approx 23$ divisions; at one a week, about $23$ weeks — five months. **10.** Each replication pairs one old strand, still methylated, with one new strand that no enzyme methylates; the old strands are neither gained nor lost, the number of strands doubles, so the methylated fraction halves. $0.80/2^{n} < 0.05$ requires $2^{n} > 16$: five divisions ($0.025$; four give exactly $5\,\%$). **11.** Ratio $0.9995 - 0.0005 = 0.999$; half-life $\ln
2/(-\ln 0.999) \approx 690$ divisions, about $13$ years at one a week. **12.** A gene silent for a lifetime of some $4000$ weekly divisions needs a half-life of hundreds of divisions, which only the feedback of question 11 gives: [readers](#def-b3-chromatin-epigenetics-histone-code) of the mark (MeCP2, HP1) recruit its [writers](#def-b3-chromatin-epigenetics-histone-code) (H3K9 methylase, DNMT3), so a dense domain of marks copies itself with an efficiency no single enzyme has. **13.** One X carries the orange allele, the other the black one; every skin cell has silenced one X and expresses only the other; the descendants of a cell keep its choice, so a patch is a clone (or a group of adjacent clones) that made the same choice. **14.** All $N$ choose the orange-bearing X, or all choose the other: $2\times (1/2)^{N} = 2^{1-N}$. **15.** $2^{1-N} = 10^{-9}$: $N - 1 = \log_{2} 10^{9} = 29.9$, $N \approx 30$ cells. **16.** $3000\,\mathrm{cm}^{2}/15\,\mathrm{cm}^{2} = 200$ patches, far more than $30$ founders: during growth the descendants of a founder are dispersed by cell mixing and migration, so one clone forms several islands (and neighbours with the same choice merge, which works the other way but less). **17.** One [Barr body](#def-b3-chromatin-epigenetics-x-inactivation) per X beyond the first: XXY one, XXX two, XY none. **18.** Tolerated when the gene has a functional Y homologue, so males also have two copies (the pseudoautosomal genes and pairs such as *KDM5C*/*KDM5D*); it matters in X aneuploidies — in Turner syndrome (XO) the escapee genes are present in a single copy, which is why an XO individual is not simply a normal female (*SHOX* haploinsufficiency and short stature), and in XXY the escapees are overdosed. **19.** $1/15\,000 = 6.7\times 10^{-5}$; deletions $0.70\times
6.7\times 10^{-5} = 4.7\times 10^{-5}$ (one in $21\,000$); maternal disomy $1.7\times 10^{-5}$ (one in $60\,000$). **20.** Angelman is the loss of one gene, *UBE3A*, so a single point mutation in the maternal allele suffices. Prader–Willi is the loss of several paternally expressed genes at once (*SNRPN*, the small-RNA cluster); a point mutation inactivates one of them and gives at most a partial phenotype, never the whole syndrome. **21.** $20\,\%$ of the children receive a paternal deletion and have Prader–Willi; none has Angelman, which needs a maternal deletion. **22.** Three chromosomes, two maternal and one paternal, one lost at random: the paternal is lost with probability $1/3$, leaving maternal disomy and Prader–Willi; with probability $2/3$ a maternal copy is lost and the child is normal. **23.** Paternal genes should promote demand on the mother; their loss (Prader–Willi) should reduce demand — which fits the poor suckling and failure to thrive of the newborn — while the later insatiable appetite is the opposite and is harder to fit (a proposed reading is that the paternal genes normally switch the child toward weaning). Angelman, the loss of a maternal gene, should increase demand; the prolonged sucking and frequent night waking of Angelman children fit, the seizures and speech loss are not about resources at all. **24.** Target the paternal antisense transcript (*UBE3A-ATS*) with an antisense oligonucleotide that degrades it or blocks its elongation: the paternal *UBE3A* is intact and is silenced only in cis by that RNA, so removing the RNA restores its expression. The [imprinting control region](#def-b3-chromatin-epigenetics-imprinting)’s methylation is untouched, so *SNRPN* and the other imprinted genes keep their normal parental pattern. **25.** Half-life of a mark: about $23$ divisions without feedback, about $690$ with it; the calico coat was decided by a founder population of about $30$ cells.