DNA — Q-Bank
← Back 📄 Notes 🏠 All Units

Unit 23 Question Bank

The double helix · Tm · nucleosomes · telomeres · the replication fork · repair
25 MCQ · five options4 Definitions2 Written answersHarper's verified
Format note: the TMU Biochemistry paper gives five suggested answers (A–E), not four — these MCQs match that. Items tagged TMU 2019 or TMU 2020/21 come from the real papers. Answers are verified against Harper's Illustrated Biochemistry; the "marking schemes" in the source folder are other students' answer sheets, not official, so they are never used as the authority.
0 / 25 answered
1One complete turn of B-DNA contains ( ) base pairs and spans ( ).
A. 6 bp; 2.0 nm (20 Å)
B. 12 bp; 4.6 nm (46 Å)
C. 10 bp; 2.0 nm (20 Å)
D. 10 bp; 3.4 nm (34 Å)
E. 20 bp; 3.4 nm (34 Å)
Answer: D
So the rise is 3.4 Å per base pair, and the width (helical diameter) is 2 nm (20 Å) — option C confuses the length of a turn with the diameter. The B form is usually found under physiologic conditions: low salt, high degree of hydration.Harper's ch.34, p.361
2The two strands of DNA are held together by ( ).
A. hydrogen bonds between the bases alone
B. covalent phosphodiester bonds between strands
C. ionic bonds between the phosphate backbones
D. disulfide bridges linking the two strands
E. hydrogen bonds between the bases AND base stacking
Answer: E
Base stacking is the half candidates forget. Harper's names both explicitly, and stacking matters clinically too — it is the loss of stacking on denaturation that produces hyperchromicity. Option B describes the bonds within a strand; the backbones are both negatively charged and actually repel.Harper's ch.34, p.360
3Chargaff observed that in DNA A = T and G = C. The structural reason is that ( ).
A. a purine must pair with a pyrimidine to keep the diameter constant
B. the phosphodiester backbone requires equal base numbers
C. adenine and thymine have identical molecular weights
D. purines are more abundant than pyrimidines
E. DNA polymerase incorporates bases in pairs
Answer: A
Purine-purine would be too wide and pyrimidine-pyrimidine too narrow — the helix would bulge and pinch instead of holding a uniform 20 Å. Harper's adds the further constraints: restrictions on rotation about the phosphodiester bond, the favored ANTI-configuration of the glycosidic bond, and the predominant tautomers of the four bases.Harper's ch.34, p.360
4G-C-rich DNA has a higher T_m than A-T-rich DNA because ( ).
A. G-C pairs are found only in the major groove
B. the G-C pair has three hydrogen bonds, A-T only two
C. A-T pairs are not hydrogen bonded at all
D. G-C pairs stack rather more tightly
E. G and C are the larger two bases
Answer: B
A useful corollary: the DNA unwinding element (DUE) at a replication origin is an ~80-bp A+T-rich sequence that is easy to unwind — the cell places its origins where the helix is weakest. One number, two facts.Harper's ch.34, p.361 · Harper's ch.35, p.382
5Hyperchromicity of denaturation refers to ( ).
A. the fall in absorbance on denaturation
B. the rise in viscosity on denaturation
C. the rise in absorbance when the strands separate
D. the colour change on ethidium bromide binding
E. the rise in T_m with salt concentration
Answer: C
In the intact helix the stacked bases' electron clouds interact, suppressing absorption at 260 nm; unstack them and the same DNA absorbs more. Note the paired fact in option B is backwards: double-stranded DNA exhibits properties of a rigid rod and in solution is a viscous material that LOSES its viscosity upon denaturation.Harper's ch.34, p.361
6A 10-fold increase in monovalent cation concentration changes T_m by ( ).
A. −16.6 °C
B. +1.66 °C
C. no change
D. +16.6 °C
E. −33 °C
Answer: D
It works by neutralizing the intrinsic interchain repulsion between the highly negatively charged phosphates of the phosphodiester backbone — remove the repulsion and the strands hold together better. Contrast formamide, which destabilizes hydrogen bonding between bases and thereby LOWERS the Tm, allowing strand separation at lower temperature and minimising phosphodiester bond breakage.Harper's ch.34, p.361
7The strand of DNA that is copied during RNA synthesis is ( ).
A. the coding strand, also called the sense strand
B. always the 5′→3′ strand of the duplex
C. both strands simultaneously
D. the strand containing uracil
E. the template strand, also called the noncoding strand
Answer: E
The coding strand is the opposite one, so called because it matches the sequence of the RNA transcript, but with T in place of U. And note the point candidates miss: the template strand for each gene will not necessarily be the same strand of the DNA double helix — “template” is a per-gene designation, not a per-molecule one.Harper's ch.34, p.360
8Which statement about DNA polarity is correct?
A. the strands are antiparallel; sequence written 5′→3′
B. the strands are parallel, both running 5′→3′
C. only the template strand possesses polarity
D. sequences are conventionally written 3′→5′
E. the phosphodiester backbone is uncharged
Answer: A
One strand runs in the 5′ to 3′ direction and the other in the 3′ to 5′ direction. The antiparallel arrangement is not a detail — it is what forces the lagging strand to be built discontinuously in Okazaki fragments. Note also that the phosphodiester backbone is negatively charged, which is why histones must be basic.Harper's ch.34, pp.359–360
9Topoisomerases differ from DNA ligase in that they ( ).
A. they require ATP whereas ligase does not
B. they reseal nicks without any energy input
C. they act only upon single-stranded DNA
D. they remove the RNA primers afterwards
E. they synthesise new DNA at the fork
Answer: B
Their function at the fork is to relieve torsional strain that results from helicase-induced unwinding — unwind a helix whose ends are fixed and the twist piles up ahead of the fork. Their pharmacology follows: they are targets of fluoroquinolone antibiotics and several anticancer drugs.Harper's ch.35, pp.382, 386
10A nucleosome consists of ( ).
A. 200 bp of DNA around a tetramer of H1 and H3
B. 10 bp of DNA around a dimer of H2A and H2B
C. 145–150 bp in 1.75 turns around an H2A/H2B/H3/H4 octamer
D. 145–150 bp around an octamer that includes H1
E. a supercoiled loop anchored to the nuclear matrix
Answer: C
A proven Section I term. Details worth adding: the particles are approximately 10 nm in diameter, and the DNA is supercoiled in a LEFT-HANDED helix over the surface of the disk-shaped histone octamer. H1 is NOT part of the core — option D is the trap.Harper's ch.35, pp.371–373
11Which statement about histone H1 is correct?
A. it forms two of the eight core octamer subunits
B. it is the most tightly bound of the histones
C. it is an acidic rather than a basic protein
D. it is the least tightly bound and not needed for the core
E. it confers nucleosome-like properties on DNA
Answer: D
It is easily removed with a salt solution, and neither H1 nor the nonhistone proteins are necessary for the reconstitution of the nucleosome core. Option E describes the (H3-H4)₂ tetramer, which can itself confer nucleosome-like properties on DNA and thus has a central role in nucleosome formation. Histones are basic, not acidic.Harper's ch.35, pp.371–373
12Histone acetylation promotes transcription because ( ).
A. it increases the positive charge of the tails
B. it links histones covalently to the backbone
C. it methylates the DNA at CpG dinucleotides
D. it recruits DNA polymerase to the promoter
E. it neutralises the tails' charge, loosening the grip
Answer: E
Acetylation occurs on lysine residues in the amino-terminal tails. The logic is purely electrostatic: histones are basic and DNA is acidic, so neutralising the histone charge loosens the grip and gives access of transcription factors to cognate regulatory DNA elements. Histone deacetylation has the opposite effect. Packing and permission are the same thing.TMU Lecture 24 · Harper's ch.35 and ch.38
13The human haploid genome contains about ( ) base pairs and about ( ) nucleosomes.
A. 3 × 10⁹; 1.7 × 10⁷
B. 3 × 10⁶; 1.7 × 10⁴
C. 3 × 10¹²; 1.7 × 10⁹
D. 3 × 10⁹; 1.7 × 10⁹
E. 6 × 10⁹; 3 × 10⁷
Answer: A
A useful check: 3 × 10⁹ ÷ 1.7 × 10⁷ ≈ 176 bp per nucleosome — the 145–150 bp core plus the linker. The numbers are consistent, which is how you can recall one from the other. The DNA must be compressed about 8000-fold to generate a condensed metaphase chromosome.Harper's ch.35, p.375
14The human telomere repeat sequence is ( ).
A. 5′-TTAAGG-3′
B. 5′-TTAGGG-3′
C. 5′-CCCTAA-3′
D. 5′-GGGATT-3′
E. 5′-TATAAA-3′
Answer: B
A proven Section I term. Telomeres consist of short TG-rich repeats, present in a variable number of repeats which can extend for several kilobases. Option E is the TATA box — a promoter element, entirely different (Unit 24).Harper's ch.35, p.374
15Telomerase is best described as ( ).
A. a conventional DNA-dependent DNA polymerase
B. a helicase that unwinds the chromosome end
C. an RNA template-containing reverse transcriptase
D. a nuclease that trims the chromosome ends
E. a topoisomerase of the type II class
Answer: C
It carries its own RNA template, which is precisely why it is a reverse transcriptase rather than an ordinary polymerase. It exists because of the end-replication problem: polymerase cannot initiate a chain, so the last primer on the lagging strand leaves an unfillable gap and every round shortens the chromosome. Telomere shortening is associated with both malignant transformation and aging, making the enzyme an attractive target for cancer chemotherapy.Harper's ch.35, p.375
16The centromere is correctly described as ( ).
A. the region where sister chromatids are transcribed
B. a G-C-rich region at the end of each chromosome
C. a TTAGGG repeat anchoring the nuclear matrix
D. an A-T-rich repeat bound by CENP-A to form the kinetochore
E. the site at which DNA replication originates
Answer: D
Its repeats range in size from 10² (brewers' yeast) to 10⁶ (mammals) base pairs. The kinetochore provides the anchor for the mitotic spindle and is thus an essential structure for chromosomal segregation during mitosis. Note it uses a histone H3 VARIANT, CENP-A — a specialised nucleosome marking a specialised place.Harper's ch.35, p.374
17Semiconservative replication means that each daughter duplex ( ).
A. contains strands that are random mixtures of new and old DNA
B. has a sequence distinct from either parental strand
C. contains solely two newly synthesized strands
D. contains both strands of the parent molecule
E. one parental strand and one newly synthesized strand
Answer: E
This is a real slide question from the TMU deck, reproduced verbatim. The mechanism is a direct consequence of base pairing: because A pairs only with T and G only with C, each strand already contains the full information needed to rebuild the other. The structure IS the copying mechanism.Harper's ch.34, p.363 · TMU Lecture 22 opening question
18DNA polymerases at the replication fork ( ).
A. synthesise only 5′→3′ and cannot start a chain
B. use ribonucleoside triphosphates as substrates
C. can start a chain without any primer at all
D. synthesise in both directions equally well
E. synthesise only in the 3′→5′ direction
Answer: A
These two constraints generate every complication at the fork. 5′→3′ only, on antiparallel strands, gives the continuous leading strand and the discontinuous lagging strand. Cannot initiate gives primase and the RNA primers — and, at a chromosome end, the end-replication problem that telomeres solve.Harper's ch.35, p.383
19Okazaki fragments are ( ).
A. the products of the nucleotide excision repair
B. short 1–5 kb segments on the lagging strand, RNA-primed
C. complexes of single-strand binding protein
D. 20 kb fragments made on the leading strand
E. the RNA primers themselves, before removal
Answer: B
Several Okazaki fragments (up to a thousand) must be sequentially synthesized for each replication fork. The finishing sequence matters: the replication complex removes the RNA primers, fills in the gaps with the proper base-paired deoxynucleotide, and then seals the fragments by DNA ligases.Harper's ch.35, pp.383, 385
20Why does DNA synthesis begin with an RNA primer rather than a DNA one?
A. RNA polymerase is faster than DNA polymerase
B. the RNA primers are in fact never removed
C. RNA is recognisably foreign and can be replaced
D. DNA cannot base-pair with a DNA template
E. RNA is chemically more stable than DNA
Answer: C
A deliberate error-avoidance design. The very first nucleotides of a chain are laid down without the benefit of proofreading, so making them of RNA marks them for later replacement. Note that DNA primase initiates synthesis of RNA primers as a distinct enzyme in the replication machinery.Harper's ch.35, pp.382–385
21Single-strand binding proteins (SSBs) function to ( ).
A. seal the nicks in the backbone
B. relieve the torsional strain
C. unwind the DNA double helix
D. prevent premature reannealing
E. synthesise the RNA primers
Answer: D
Learn the six replication proteins as a set, because examiners like to swap them: polymerases polymerise, helicases unwind (ATP-driven, processive), topoisomerases relieve torsion, primase makes RNA primers, SSBs keep the strands apart, ligase seals nicks.Harper's ch.35, Table 35-5, p.382
22Proofreading during replication is performed by ( ).
A. the 5′→3′ exonuclease activity of DNA ligase
B. mismatch repair enzymes acting during S phase
C. topoisomerase II
D. the origin recognition complex
E. the 3′→5′ exonuclease activity of DNA polymerase
Answer: E
Fidelity is multiplicative, which is why three layers beat one. Base pairing alone gives about one error in 10⁴–10⁵; proofreading adds roughly two orders of magnitude; mismatch repair sweeps up afterwards — together reaching about one error per 10⁹–10¹⁰ bp.Harper's ch.35
23Failure of nucleotide excision repair causes ( ).
A. xeroderma pigmentosum
B. hereditary nonpolyposis colon cancer
C. Lesch-Nyhan syndrome
D. xanthinuria
E. Bloom syndrome only
Answer: A
Nucleotide excision repair removes bulky lesions that distort the helix, notably UV-induced pyrimidine dimers — hence extreme sensitivity to sunlight and early skin cancers. Option B is mismatch repair failure. The general rule: a repair defect is a mutation-rate defect, and a mutation-rate defect is a cancer predisposition.Harper's ch.35
24At the origin of replication in E. coli, the initiator protein is ( ), and in eukaryotes the corresponding complex is ( ).
A. dnaB; the nuclear spliceosome complex
B. dnaA; the origin recognition complex (ORC)
C. primase; the kinetochore complex
D. rho factor; the mediator complex
E. sigma factor; the TFIID complex
Answer: B
ORC homologs have been found in all eukaryotes examined. Adjacent to the origin lies the DNA unwinding element (DUE), an approximately 80-bp A+T-rich sequence that is easy to unwind. Replication then proceeds bidirectionally, generating replication bubbles with a fork at each end.Harper's ch.35, pp.381–382
25Which statement about the metaphase chromosome is correct?
A. each sister chromatid contains two dsDNA molecules
B. metaphase chromatin is the most active form
C. each chromatid holds one dsDNA molecule, and is inactive
D. interphase DNA is more densely packed than metaphase
E. humans have 23 chromosomes in somatic cells
Answer: C
Option D is exactly backwards — during interphase the packing is LESS dense than in the condensed chromosome during metaphase, which is precisely why transcription happens in interphase. And humans have 23 PAIRS: 22 autosomes and two sex chromosomes.Harper's ch.35, pp.370, 375
1 Nucleosomes — 3′ — PROVEN Section I term+
The fundamental packaging unit of chromatin: dense spherical particles, approximately 10 nm in diameter, connected by DNA filaments, composed of DNA wound around an octameric complex of histone molecules.

Structure: the octamer is two each of H2A, H2B, H3 and H4. The DNA is supercoiled in a LEFT-HANDED helix over the surface of the disk-shaped histone octamer; 1.75 superhelical turns are wrapped around it, protecting 145 to 150 bp and forming the nucleosome core particle.

Assembly: the (H3-H4)₂ tetramer can itself confer nucleosome-like properties on DNA; addition of two H2A-H2B dimers stabilizes the particle. H1 is NOT part of the core — it is the histone least tightly bound to chromatin, binds the linker DNA, and is not necessary for reconstitution of the nucleosome core.

Function: to condense DNA — the human haploid genome contains about 3 × 10⁹ bp and about 1.7 × 10⁷ nucleosomes, and must be compressed ~8000-fold into a metaphase chromosome. But histones also integrally participate in gene regulation: acetylation of lysines in the histone tails reduces their positive charge, decreases affinity for DNA and disrupts nucleosomal structure, giving transcription factors access.Harper's ch.35, pp.371–375
2 Telomeres — 3′ — PROVEN Section I term+
The structures at the ends of each chromosome, which consist of short TG-rich repeats. Human telomeres have a variable number of repeats of the sequence 5′-TTAGGG-3′, which can extend for several kilobases.

Why they are needed — the end-replication problem: DNA polymerases synthesise only 5′→3′ and cannot initiate a chain, so on the lagging strand the final RNA primer at the chromosome end is removed and the gap cannot be filled. Every round of replication would therefore shorten the chromosome. Telomeres are expendable non-coding repeats that absorb this loss.

Telomerasea multisubunit RNA template-containing complex related to viral RNA-dependent DNA polymerases (reverse transcriptases) — is the enzyme responsible for telomere synthesis and thus for maintaining telomere length.

Clinical significance: telomere shortening has been associated with both malignant transformation and aging, so telomerase has become an attractive target for cancer chemotherapy and drug development. Most somatic cells lack it; most cancers reactivate it.Harper's ch.35, pp.374–375
3 Melting temperature (Tm) of DNA — 2′+
The midpoint of the temperature range over which the two strands of a DNA molecule separate.

Denaturation is accompanied by hyperchromicity of denaturation — an increase in the optical absorbance of the purine and pyrimidine bases, because the bases unstack — and by a loss of viscosity, since double-stranded DNA exhibits properties of a rigid rod.

Three influences:
Base compositionDNA rich in G-C pairs, which have three hydrogen bonds, melts at a higher temperature than that rich in A-T pairs, which have two.
Salta 10-fold increase of monovalent cation concentration increases the Tm by 16.6 °C, by neutralizing the intrinsic interchain repulsion between the negatively charged phosphates.
Solventformamide destabilizes hydrogen bonding between bases, thereby lowering the Tm.

Separated strands renature when appropriate temperature and salt conditions are achieved; this reannealing is often referred to as HYBRIDIZATION, and it is the basis of Southern (DNA/DNA) and Northern (RNA/DNA) blotting.Harper's ch.34, p.361
4 Semiconservative replication — 2′+
The mode of DNA replication in which each of the two strands of DNA is used as a template for synthesis of a new, complementary strand, so that each daughter duplex is composed of one strand derived from the original parental DNA duplex and one strand that was newly synthesized.

It is a direct consequence of the base-pairing rule: since A pairs only with T and G only with C, each strand already contains the information needed to rebuild the other.

Mechanically: replication starts at an origin (ori) — bound by dnaA in E. coli, by the origin recognition complex (ORC) in eukaryotes — beside an ~80-bp A+T-rich DNA unwinding element (DUE), and proceeds bidirectionally as a replication bubble with a fork at each end.

Because DNA polymerases synthesise only 5′→3′ and the strands are antiparallel, the polymerase functions asymmetrically: the leading strand is synthesized continuously and the lagging strand in short 1–5 kb Okazaki fragments, each primed by RNA and later sealed by DNA ligase.Harper's ch.34, p.363 · Harper's ch.35, pp.381–385
1 Describe the structure of DNA and the forces that stabilise it. 6′

The polymer

The monomeric units — deoxyadenylate, deoxyguanylate, deoxycytidylate and thymidylate — are held in polymeric form by 3′,5′-phosphodiester bonds, with each base attached to its 2-deoxyribose by an N-glycosidic bond. The informational content of DNA resides in the sequence in which these monomers are ordered.

Each strand possesses a polarity — a 5′ and a 3′ end — and by convention a sequence is written 5′→3′. The phosphodiester backbone is negatively charged.

The double helix

Two strands wind around a central axis as a right-handed helix — right-handed because as one looks down the double helix the base residues form a spiral in a clockwise direction. The B form is usually found under physiologic conditions (low salt, high degree of hydration); at least six forms (A–E and Z) exist in the test tube.

Dimensions of B-DNA: 10 bp per turn; 3.4 nm (34 Å) per turn, hence 3.4 Å rise per base pair; helical diameter 2 nm (20 Å). It has a major and a minor groove.

The two strands are ANTIPARALLEL — one runs 5′→3′ and the other 3′→5′.

Base pairing

A pairs with T through two hydrogen bonds, G with C through three. This explains Chargaff's observation that A = T and G = C, and it is required by the geometry: only a purine paired with a pyrimidine keeps the diameter constant. Harper's adds that restrictions on rotation about the phosphodiester bond, the favored anti-configuration of the glycosidic bond, and the predominant tautomers of the four bases allow A to pair only with T, and G only with C.

The stabilising forces — both of them

The two strands are held in register by both hydrogen bonds between the bases AND by van der Waals and hydrophobic interactions between the stacked adjacent base pairs. Base stacking is as important as hydrogen bonding, and it is the loss of stacking that produces hyperchromicity on denaturation.

Consequences of the structure

  • G-C-rich DNA has a higher Tm, because of the third hydrogen bond — which is why replication origins sit beside A+T-rich unwinding elements.
  • The grooves allow sequence recognition without unwinding — the edges of the base pairs are exposed at the floor of the major groove, which is where the helix-turn-helix, zinc finger and leucine zipper motifs bind.
  • The base-pairing rule is the copying mechanism: each strand contains the information to rebuild the other, which is what makes semiconservative replication possible.
Marking guide: phosphodiester backbone and polarity 1 · the three B-DNA dimensions 1.5 · antiparallel strands 0.5 · base pairing with the correct hydrogen bond counts 1.5 · BOTH stabilising forces named 1 · one consequence 0.5.
2 How is DNA packaged into the nucleus, and how does the packaging affect gene expression? 8′

The problem

Human genomic DNA, if extended end-to-end, would be metres in length, yet must fit within a nucleus only microns in diameter — Harper's puts the total at thousands of times the diameter of the cell nucleus. The solution is a hierarchy of folding, each level built on the last.

Level 1 — the nucleosome

Electron microscopic studies have demonstrated dense spherical particles called nucleosomes, approximately 10 nm in diameter and connected by DNA filaments. Each consists of DNA wound around an octameric complex of histone moleculestwo each of H2A, H2B, H3 and H4.

In the nucleosome, the DNA is supercoiled in a LEFT-HANDED helix over the surface of the disk-shaped histone octamer. 1.75 superhelical turns of DNA are wrapped around the octamer, protecting 145 to 150 bp and forming the nucleosome core particle.

Assembly is ordered: the (H3-H4)₂ tetramer itself can confer nucleosome-like properties on DNA and thus has a central role in the formation of the nucleosome; the addition of two H2A-H2B dimers stabilizes the primary particle.

H1 is not part of the core. It is the histone least tightly bound to chromatin, easily removed with a salt solution, and neither H1 nor the nonhistone proteins are necessary for reconstitution of the nucleosome core; it binds the linker DNA between core particles.

Level 2 — chromatin

Strings of nucleosomes form along the linear sequence of genomic DNA to form chromatin, which itself can be more tightly packaged and condensed, ultimately leading to the formation of the chromosomes. The 30-nm fibre is folded into looped domains anchored to the nuclear matrix.

Level 3 — the chromosome

At metaphase, mammalian chromosomes possess a twofold symmetry, with the identical duplicated sister chromatids connected at a centromere. Each sister chromatid contains one dsDNA molecule. The overall compression is about 8000-fold; the human haploid genome of 3 × 10⁹ bp is held in about 1.7 × 10⁷ nucleosomes.

Two specialised regions complete the structure. The centromere is an A-T-rich region of repeated DNA bound by nucleosomes containing the histone H3 variant CENP-A; this complex, the kinetochore, provides the anchor for the mitotic spindle. The telomeres are short TG-rich repeats, in humans 5′-TTAGGG-3′, maintained by telomerase.

Why packaging is regulation

Harper's is explicit that condensation is not the whole story: the histones also integrally participate in gene regulation; indeed histones contribute importantly to all DNA-directed molecular transactions.

The mechanism is electrostatic. Histones are basic — positively charged — and DNA is negatively charged. Acetylation on lysine residues in the amino-terminal tails reduces the positive charge of these tails, decreases the binding affinity of histone for DNA, and disrupts nucleosomal structure, allowing access of transcription factors to cognate regulatory DNA elements and enhancing binding of the basal transcription machinery to the promoter. Histone deacetylation has the opposite effect.

A second layer is methylation of deoxycytidine residues in the sequence 5′-mCpG-3′, which silences genes — in mouse liver, only the unmethylated ribosomal genes can be expressed, and many animal viruses are not transcribed when their DNA is methylated.

A gene wrapped tightly in nucleosomes is off; one whose nucleosomes have been loosened or evicted is available. Packing and permission are the same thing. Metaphase chromosomes are nearly completely transcriptionally inactive — the extreme case of the same principle.

Marking guide: statement of the packaging problem 0.5 · nucleosome structure with the octamer named 2 · 145–150 bp and 1.75 turns 1 · H1 correctly placed outside the core 1 · chromatin and chromosome levels 1 · histone acetylation mechanism 1.5 · DNA methylation 0.5 · a closing statement linking packing to expression 0.5.