Lecture 3 & 4: DNA, Genes, Chromatin

Learning Objectives• Understand the structure of DNA, including the structure of the bases, nucleosides,

nucleotides, and the structure of the DNA double helix • Understand the significance of the specificity of base pairing and the complementarity of the

DNA strands • Understand the nature of the forces contributing to the stability of the DNA double helix • Understand the process of DNA denaturation and the relationship between melting

temperature and the base composition of DNA • Understand the principles of the DNA reassociation reaction • Understand the relationship between the rate of DNA reassociation and DNA complexity • Understand what repetitive sequences are and how they are arranged in the human genome • Understand the mechanism by which Alu sequences have affected the LDL receptor gene • Understand basic gene structure and the basic characteristics of human nuclear and

mitochondrial DNA • Understand basic chromosome structure and how DNA is packaged into chromosomes • Understand the significance of telomeres to aging and cancer and the use of telomerase in

cancer diagnosis

GENETIC DOGMA

• Genetic diseases occur because of mutations in DNA.• Many of these mutations affect the repair of other

mutations that occur during DNA replication or at other times, which in turn affect the flow of genetic information from DNA to RNA (transcription and processing) and from RNA to protein synthesis (translation).

• Many of these mutations also affect the structures of the resulting proteins, affecting their functions.

THE FLOW OF GENETIC INFORMATION

DNA RNA PROTEIN

DNA

1

2 3

1. REPLICATION (DNA SYNTHESIS)2. TRANSCRIPTION (RNA SYNTHESIS)3. TRANSLATION (PROTEIN SYNTHESIS)

DNA Structure and Chemistry

a). Evidence that DNA is the genetic informationi). DNA transformation – know this termii). Transgenic experiments – know this processiii). Mutation alters phenotype – be able to define

genotype and phenotypeb). Structure of DNA

i). Structure of the bases, nucleosides, and nucleotidesii). Structure of the DNA double helixiii). Complementarity of the DNA strands

c). Chemistry of DNAi). Forces contributing to the stability of the double helixii). Denaturation of DNA

a). Evidence that DNA is the genetic information

i). DNA transformation – know this termii). Transgenic experiments – know this processiii). Mutation alters phenotype – be able to define

genotype and phenotype

DNA transformation DNA transformation experiments show that DNA is the carrier of the

genetic information. • Injecting mice with both a heat-killed virulent strain of Streptococcus and a non-

heated, non-virulent strain of Streptococcus. • Something (DNA) from the heat-killed virulent strain of Streptococcus was able to

alter the (still viable) non-virulent strain, converting some of the cells to virulent bacteria and killing the host.

• Purified DNA from Type S (smooth colony) Strep cells is able to be taken up by Type R (rough colonies) Strep cells.

• The process of getting functionally active DNA into cells is called DNA transformation.

• Transformation by Type S DNA alters the "genotype" of host cells, since new genes are introduced into these cells thus altering their genetic constitution. The expression of this Type S DNA changes the "phenotype“ of the transformed cells, making their colonies look "smooth" instead of "rough.“

• Genotype is an organism’s genetic constitution. • Phenotype is the observed characteristics of an organism as determined by the

genetic makeup and the environment.

ii). Transgenic experiments – know this process

• Transfer of a specific gene into the nucleus of a fertilized egg.

• Specific phenotypic traits can be conferred by specific genes, and thus that DNA is the carrier of genetic information.

• Poducing mutation of specific genes in the mouse to determine the functions of those genes and to create mouse models of human genetic disease.

• The mutation of a gene that eliminates the gene's function, is called a knockout mutation and the mouse carrying that mutation is called a knockout mouse.

iii). Mutation alters phenotype – be able to definegenotype and phenotype

• Phenotypic differences between individuals are due to differences between genes.

• One-third of our genes are polymorphic-differences in the nucleotide sequences

• Differences occurred by mutation of DNA • All of this evidence indicates that DNA is the

carrier of the genetic information.• Genetic differences between individuals can have

many clinical implications.( aging , drug metabolism).

b). Structure of DNA

i). Structure of the bases, nucleosides, and nucleotides

ii). Structure of the DNA double helix

iii). Complementarity of the DNA strands

Thymine (T)

Guanine (G) Cytosine (C)

Adenine (A)

Structures of the bases

Purines Pyrimidines

5-Methylcytosine (5mC)

5-methylcytosine (5mC).

• A common base modification in DNA results from the methylation of cytosine, giving rise to 5-methylcytosine (5mC).

• 5mC is highly mutagenic. It is believed that this methylation functions to regulate gene expression because 5-methylcytosine (5mC) residues are often clustered near the promoters of genes in so-called "CpG islands.“ (Along one strand of DNA the nucleotides are sometimes indicated by the base followed by a phosphate or “p” such as ApTpCpCpGpApCpTpGpGp - this sequence contains one CpG site.)

• The problem that arises from these methylations is that subsequent deamination of a 5mC results in the production of thymine, which is not foreign to DNA. As such, 5'-mCG-3' sites (or mCpG sites) are "hot-spots" for mutation, and when mutated are a common cause of cancer.

[structure of deoxyadenosine]

Nucleoside

Nucleotide

Nomenclature

Purinesadenine adenosineguanine guanosinehypoxanthine inosine

Pyrimidinesthymine thymidinecytosine cytidine

+ribose uracil uridine

Nucleoside NucleotideBase +deoxyribose +phosphate

• polynucleotide chain• 3’,5’-phosphodiester bond

ii). Structure of the DNA double helix

Structure of the DNApolynucleotide chain

5’

3’

ii). Structure of the DNA double helix

• The DNA double helix requires that the two polynucleotide chains be base-paired to each other.

A-T base pair

G-C base pair

Chargaff’s rule: The content of A equals the content of T, and the content of G equals the content of C in double-stranded DNA from any species

Hydrogen bonding of the bases

Double-stranded DNA

• Double-stranded DNA, is composed of two base-paired, complementary polynucleotide chains.

• Base-pairing between the complementary strands is required for two important functions of DNA: 1) DNA replication involves an unwinding of the double helix (right) followed by synthesis of a complementary strand from each of the unpaired template strands, and 2) DNA serves as a template for RNA synthesis by utilizing the information in one strand to code for a complementary RNA strand.

• DNA in the "B" form has a major groove and a minor groove, and has 10 base pairs per one turn of the double helix. DNA that is overwound or underwound, with fewer than or more than 10 base pairs per turn, is said to be "supercoiled".

• antiparallel with respect to each other.• Each polynucleotide chain has a 5' end and a 3' end. • Deoxyribonucleases (or DNases) are enzymes that cleave phosphodiester bonds.

Some are used for constructive purposes, such as proofreading during DNA replication, whereas others are used to degrade DNA. There are two basic classes of DNases: exonucleases and endonucleases. Exonucleases remove only the terminal nucleotide, whereas endonucleases cleave anywhere within the DNA double helix.

Double-stranded DNA

Major groove

Minor groove

5’ 3’

5’ 3’3’ 5’

“B” DNA

Chemistry of DNA

Forces affecting the stability of the DNA double helix

• hydrophobic interactions - stabilize - hydrophobic inside and hydrophilic outside

• stacking interactions - stabilize - relatively weak but additive van der Waals forces

• hydrogen bonding - stabilize - relatively weak but additive and facilitates stacking

• electrostatic interactions - destabilize - contributed primarily by the (negative) phosphates - affect intrastrand and interstrand interactions - repulsion can be neutralized with positive charges

(e.g., positively charged Na+ ions or proteins)

Stacking interactions

Charge repulsion

Ch

arg

e re

pu

lsio

n

• Next slide shows a side view of three base pairs in the DNA double helix. Note the base-pair stacking interactions, the hydrophobic interior, and the phosphates on the exterior.

Model of double-stranded DNA showing three base pairs

Denaturation of DNA

Double-stranded DNA

A-T rich regions denature first

Cooperative unwinding of the DNA strands

Extremes in pH or high temperature

Strand separationand formation ofsingle-strandedrandom coils

Electron micrograph of partially melted DNA

• A-T rich regions melt first, followed by G-C rich regions

Double-stranded, G-C rich DNA has not yet melted

A-T rich region of DNAhas melted into asingle-stranded bubble

Hyperchromicity

• When a solution of double-stranded DNA is placed in a spectrophotometer cuvette and the absorbance of the DNA is determined across the electromagnetic spectrum, it characteristically shows an absorbance maximum at 260 nm (in the UV region of the spectrum).

• If the same DNA solution is melted, the absorbance at 260 nm increases approximately 40%. This property is termed "hyperchromicity." The hyperchromic shift is due to the fact that unstacked bases absorb more light than stacked bases.

Hyperchromicity

The absorbance at 260 nm of a DNA solution increases when the double helix is melted into single strands.

260

Ab

sorb

ance

Absorbance maximumfor single-stranded DNA

Absorbancemaximum fordouble-stranded DNA

220 300

DNA melting curve

• Hyperchromicity can be used to follow the denaturation of DNA as a function of increasing temperature. As the temperature of a DNA solution gradually rises above 50 degrees C, the A-T regions will melt first giving rise to an increase in the UV absorbance. As the temperature increases further, more of the DNA will become single-stranded, further increasing the UV absorbance, until the DNA is fully denatured above 90 degrees C. The temperature at the mid-point of the melting curve is termed "melting temperature" and is abbreviated Tm.

• The Tm for a DNA depends on its average G+C content: the higher the G+C content, the higher the Tm.

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7050 90

Temperature oC

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DNA melting curve

• Tm is the temperature at the midpoint of the transition

Average base composition (G-C content) can bedetermined from the melting temperature of DNA

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Temperature oC

Tm is dependent on the G-C content of the DNA

Pe

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E. coli DNA is 50% G-C

Genomic DNA, Genes, Chromatin

a). Complexity of chromosomal DNAi). DNA reassociationii). Repetitive DNA and Alu sequencesiii). Genome size and complexity of genomic DNA

b). Gene structurei). Introns and exonsii). Properties of the human genome iii). Mutations caused by Alu sequences

c). Chromosome structure - packaging of genomic DNAi). Nucleosomes

ii). Histonesiii). Nucleofilament structureiv). Telomeres, aging, and cancer

Complexity of chromosomal DNA: DNA reassociation (renaturation)

• Denaturation of a DNA double helix produces single-stranded DNA. The reverse reaction (the formation of double-stranded DNA) can be carried out if the complementary DNA strands are incubated under conditions that will promote renaturation (reassociation).

• This process involves two steps. The first step is a slower, rate-limiting reaction in which the complementary strands attempt to find each other. The single strands randomly interact until complementary base pairing can occur. This may occur over only a short region of each strand, but may be enough to "nucleate" the reaction.

• Because the two reacting strands are at the same concentration in solution, the reaction is second-order, with the rate of the reaction being defined by the second-order rate constant, k2.

• Once nucleation has occurred, the complementary strands rapidly zipper up. The complexity of a DNA is a function of how many base pairs it has. In other words, a low complexity DNA may consist of a few hundred or a few thousand base pairs, in contrast to a high complexity DNA that may contain millions of base pairs.

• Low complexity sequences are able to find each other much faster during a renaturation reaction, than are high complexity sequences, since the low complexity sequences have to make fewer collisions to find a base-pairing partner.

• Thus, the rate of a renaturation reaction is a function of the complexity of the DNA. Or in other words, the complexity of a DNA can be determined by measuring its second-order reassociation rate constant.

DNA reassociation (renaturation)

Double-stranded DNA

Denatured,single-strandedDNA

Slower, rate-limiting,second-order process offinding complementarysequences to nucleatebase-pairing

k2

Faster,zipperingreaction toform longmoleculesof double-strandedDNA

DNA reassociation kinetics for human genomic DNA

• This figure shows a DNA renaturation (reassociation) reaction or "Cot curve" for human genomic DNA. The reaction is graphed as a semi-log plot with "log Cot" on the x-axis and "% DNA reassociated" on the y-axis (0% at the top and 100% at the bottom). The reaction follows ideal second-order kinetics. Cot is the product of Co (initial DNA concentration) and t (time). Thus, if DNA concentration is constant, the x-axis is simply the time course of the reaction. The reaction would proceed as follows: at time-zero the DNA is denatured into single strands and the conditions are adjusted to promote DNA base pairing to allow the renaturation of the DNA; during the reaction the various kinetic fractions find partners (complements) and reassociate until the reaction is complete (100% reassociation). This DNA shows complex reassociation kinetics because it is actually a mixture of three different species or populations of DNA: fast-reassociating, intermediate-reassociating, and slow-reassociating fractions. The fast and intermediate fractions are composed of repeated sequences in the genome, and the slow fraction is composed of single-copy (unique) sequences. Single-copy sequences are those that are present in one copy per haploid genome. The next slide illustrates the concept of how sequence complexity affects the rate of DNA reassociation.

• The Cot1/2 is equal to the inverse of the second-order reassociation rate constant, k2. In other words, the rate constant for a sample of DNA can be determined experimentally by measuring the Cot1/2 of the reaction. Each reassociating component has its characteristic rate constant and Cot1/2. The rate constant provides the complexity of the DNA by comparison to the rate constants of other DNA samples of known complexity.

Cot1/2

DNA reassociation kinetics for human genomic DNA

Cot1/2 = 1 / k2 k2 = second-order rate constant Co = DNA concentration (initial) t1/2 = time for half reaction of each

component or fraction

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% D

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log Cot

fast (repeated)

intermediate (repeated)

slow (single-copy)

Kinetic fractions: fast intermediate slow

Cot1/2

Cot1/2

sequence complexity and the rate of DNA reassociation

• This illustrates the concept of how sequence complexity affects the rate of DNA reassociation. Imagine two different DNA sequences in a genome, one present one time per haploid genome (right) and the other present 1,000,000 times per haploid genome (left). They would be present at a 1:1,000,000 ratio with respect to each other. If these sequences were mixed together (which is what would happen if total genomic DNA was isolated for analysis), then fragmented, denatured and allowed to reassociate, the repeated sequences would reassociate much more rapidly because it would be much easier for them to find complementary strands to base pair with. The repeated sequences would reassociate with a very low Cot1/2 and therefore with a very high k2, consistent with a rapid rate of reassociation.

high k2

106 copies per genome ofa “low complexity” sequence

of e.g. 300 base pairs

1 copy per genome ofa “high complexity” sequence

of e.g. 300 x 106 base pairs

low k2

Types of human DNA

• The human genome consists of three populations of DNA: the fast and intermediate fractions make up about 10% and 15% of the genome, respectively, and the slow fraction makes up about 75% of the genome. Most of the genes in the human genome are in the single-copy fraction. As shown in the next slide, repeated sequences can be of two types: those that are interspersed throughout the genome or those that are tandemly repeated satellite DNAs. Among the interspersed repetitive sequences are so-called "Alu" sequences, which are about 300 base pairs in length and are repeated about 300,000 times in the genome. They can be found adjacent to or within genes, and as illustrated later, their presence can sometimes lead to the occasional disruption of genes. The interspersed repetitive sequences also include VNTRs (variable numbers of tandem repeats), which are comprised of short repeated sequences of only a few base-pairs, but of variable lengths. They, too, are interspersed throughout the genome, and are quite useful as landmarks for mapping genes because they are highly polymorphic (they differ in length or number of repeats from individual to individual).

Type of DNA % of Genome Features

Single-copy (unique) ~75% Includes most genes 1

Repetitive Interspersed ~15% Interspersed throughout genome between

and within genes; includes Alu sequences 2

and VNTRs or mini (micro) satellites Satellite (tandem) ~10% Highly repeated, low complexity sequences

usually located in centromeres and telomeres

2 Alu sequences are about 300 bp in length and are repeated about 300,000 times in the genome. They can be found adjacent to or within genes in introns or nontranslated regions.

1 Some genes are repeated a few times to thousands-fold and thus would be in the repetitive DNA fraction

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fast ~10%

intermediate ~15%

slow (single-copy) ~75%

Classes of repetitive DNA

• Interspersed repeats are sequences that are repeated many times and scattered throughout the genome. In contrast, tandem repeats are sequences that are repeated many times adjacent to each other. The latter are usually found in the centromeres and telomeres of chromosomes (the sequence above TTAGGG comprises human teleomeric DNA – see last slide of this series).

Classes of repetitive DNA

Interspersed (dispersed) repeats (e.g., Alu sequences)

TTAGGGTTAGGGTTAGGGTTAGGG

Tandem repeats (e.g., microsatellites)

GCTGAGG GCTGAGGGCTGAGG

Genome sizes in nucleotide pairs (base-pairs)

• On June 26, 2000, the Human Genome Project and Celera Genomics Corp. jointly announced that the sequencing of the human genome was all but completed. A so-called rough draft of approximately 90% of the genome was completed and ready for release to scientists and medical researchers at that time. A rough draft was released to make the sequence available as soon as possible while completion of the remaining sequence took place. What still needs to be done is to fill in some difficult-to-sequence gaps and to find "typographical errors" in the sequence. Knowing the complete sequence of the human genome will allow medical researchers to more easily find disease-causing genes. In addition, it should become possible to understand how differences in our DNA sequences from individual to individual may affect our predisposition to diseases and our ability to metabolize drugs. Because the human genome has ~3 billion bp of DNA and there are 23 pairs of chromosomes in diploid human cells, the average metaphase chromosome has ~130 million bp DNA.

viruses

plasmids

bacteria

fungi

plants

algae

insects

mollusks

reptiles

birds

mammals

Genome sizes in nucleotide pairs (base-pairs)

104 108105 106 107 10111010109

The size of the humangenome is ~ 3 X 109 bp;almost all of its complexityis in single-copy DNA.

The human genome is thoughtto contain ~30,000 to 40,000 genes.

bony fish

amphibians

Gene structure

• This slide shows the structure of a typical human gene and its corresponding messenger RNA (mRNA).

• Most genes in the human genome are called "split genes" because they are composed of "exons" separated by "introns."

• The exons are the regions of genes that encode information that ends up in mRNA. The transcribed region of a gene (double-ended arrow) starts at the +1 nucleotide at the 5' end of the first exon and includes all of the exons and introns (initiation of transcription is regulated by the promoter region of a gene, which is upstream of the +1 site). RNA processing (the subject of a another lecture) then removes the intron sequences, "splicing" together the exon sequences to produce the mature mRNA.

• The translated region of the mRNA (the region that encodes the protein) is indicated in blue. Note that there are untranslated regions at the 5' and 3‘ ends of mRNAs that are encoded by exon sequence but are not directly translated.

5’ 3’

promoter region

exons (filled and unfilled boxed regions)

introns (between exons)

transcribed region

translated region

mRNA structure

+1

Gene structure

The (exon-intron-exon)n structure of various genes

• This figure shows examples of the wide variety of gene structures seen in the human genome. Some (very few) genes do not have introns. One example is the histone genes, which encode the small DNA-binding proteins, histones H1, H2A, H2B, H3, and H4.

• Shown here is a histone gene that is only 400 base pairs (bp) in length and is composed of only one exon.

• The beta-globin gene has three exons and two introns. • The hypoxanthine-guanine phosphoribosyl transferase (HGPRT or HPRT)

gene has nine exons and is over 100-times larger than the histone gene, yet has an mRNA that is only about 3-times larger than the histone mRNA (total exon length is 1,263 bp). This is due to the fact that introns can be very long, while exons are usually relatively short.

• An extreme example of this is the factor VIII gene which has numerous exons (the blue boxes and blue vertical lines).

The (exon-intron-exon)n structure of various genes

-globin

HGPRT(HPRT)

total = 1,660 bp; exons = 990 bp

histone

factor VIII

total = 400 bp; exon = 400 bp

total = 42,830 bp; exons = 1263 bp

total = ~186,000 bp; exons = ~9,000 bp

Properties of the human genome

Nuclear genome

• the haploid human genome has ~3 X 109 bp of DNA• single-copy DNA comprises ~75% of the human genome• the human genome contains ~30,000 to 40,000 genes• most genes are single-copy in the haploid genome• genes are composed of from 1 to >75 exons• genes vary in length from <100 to >2,300,000 bp• Alu sequences are present throughout the genome

Mitochondrial genome

• circular genome of ~17,000 bp• contains <40 genes

Alu sequences can be “mutagenic”

• The rather common (~1 in 500) autosomal dominant disease, familial hypercholesterolemia (FH), is caused by mutations in the LDL (low density lipoprotein) receptor gene . Plasma LDL, which carries circulating cholesterol, is cleared from the serum by binding to the LDL receptor on liver cells and is internalized. Normal plasma cholesterol levels average below 200 mg/dl. Individuals who have one defective LDL receptor gene (heterozygous) have approximately double this amount, and those with two defective genes (homozygous) have approximately four times this amount. Heterozygous individuals are predisposed to cardiovascular disease, with males having a 50% risk of myocardial infarction by age 50. There are many ways that the LDL receptor gene has been mutated rendering it inactive or abnormal. As shown in the next figure, one mechanism has involved Alu sequences.

Familial hypercholesterolemia• autosomal dominant• LDL receptor deficiency

Alu sequences can be “mutagenic”

From Nussbaum, R.L. et al. "Thompson & Thompson Genetics in Medicine," 6th edition (Revised Reprint), Saunders, 2004.

LDL receptor gene

• Here you see the structure of the LDL receptor gene (which has 18 exons). Six Alu sequences are present within three of the introns and two of the exons. Because of the close proximity of the two Alu repeats located within introns 4 and 5, unequal crossing over can occur during meiosis. Crossing over (the topic of a future lecture) requires homologous sequences, which base pair with each other during the process of meiosis. The homologous sequences can be provided by the Alu repeats, which can cause an out-of-register misalignment and subsequent crossing over deleting exon 5 from one of the two products of crossing over. This exon 5 in-frame deletion can be inherited and is currently a cause of FH. This deletion affects the LDL binding region of the receptor. Thus, while Alu sequences have no known function in our genomes, there are a lot of them scattered throughout our genomes, within and around genes, and they can be quite disruptive.

LDL receptor gene

Alu repeats present within introns

Alu repeats in exons

4

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5

5

5 6

6

6

Alu Alu

AluAlu

X

4 6Alu

unequalcrossing over

one product has a deleted exon 5(the other product is not shown)

Chromatin structure

• This high power electron micrograph shows the detailed structure of chromosome threads following a gentle preparation technique that involves removal of loosely bound chromosomal proteins while preserving the more tightly bound DNA-binding proteins. The appearance of a "beads on a string" structure is due to regularly spaced nucleosomes (see next slide). "Chromatin" is the biochemical term for DNA-protein complexes that are isolated from eukaryotic chromosomes.

Chromatin structure

EM of chromatin shows presence ofnucleosomes as “beads on a string”

Nucleosome structure

• Each nucleosome is composed of a core (left) consisting of two each of the histones, H2A, H2B, H3, and H4, around which the DNA winds 1 3/4 times. The DNA undergoes negative supercoiling as a consequence of being wound around the core histones. Histones are positively charged proteins and thus interact with the negatively charged phosphates along the backbone of the DNA double helix. While the core has 146 bp of DNA, the nucleosome proper (right) has approximately 200 bp of DNA and also includes one histone H1 monomer lying on the outside of the structure. Nucleosomes are regularly spaced along eukaryotic chromosomal DNA every ~200 bp, giving rise to the "beads on a string" structure.

Nucleosome structure

Nucleosome core (left)• 146 bp DNA; 1 3/4 turns of DNA• DNA is negatively supercoiled• two each: H2A, H2B, H3, H4 (histone octomer)

Nucleosome (right)• ~200 bp DNA; 2 turns of DNA plus spacer• also includes H1 histone

Histones (H1, H2A, H2B, H3, H4)

• Histones are small, positively charged proteins that can be extensively modified posttranslationally, in general to make them less positively charged. Histone deacetylases (HDACs) are associated with transcriptional repression because they make histones better able to bind DNA, thus making DNA less accessible to the transcription machinery. Histone deacetylases are recruited to the chromosome by transcriptional repressors such as the retinoblastoma (Rb) protein (the subject of another lecture). Histone acetylases are recruited to chromosomes by transcription factors (TFs). Histone acetylases reduce the positive charges on histones, causing them to loosen their grip on the DNA to allow transcription factors to bind.

Histones (H1, H2A, H2B, H3, H4)• small proteins• arginine or lysine rich: positively charged• interact with negatively charged DNA• can be extensively modified - modifications in

general make them less positively chargedPhosphorylationPoly(ADP) ribosylationMethylationAcetylation

Hypoacetylation by histone deacetylase (facilitated by Rb)

“tight” nucleosomes assoc with transcriptional repression

Hyperacetylation by histone acetylase (facilitated by TFs)“loose” nucleosomes assoc with transcriptional activation

Nucleofilament structure

• The orderly packaging of DNA in the cell is essential for the process of DNA replication, as well as for the process of transcription. Packaging of DNA into nucleosomes is only the first step, foreshortening chromosomal DNA somewhat by virtue of its being wrapped around the core histones 1 3/4 times. However, if the average human genomic DNA molecule is ~130 million bp in length, its length would be an astounding 44 mm. All this DNA X 23 chromosomes has to packaged in the nucleus of a cell that is too small to be seen with the unaided eye. Thus, the DNA needs to be packaged in higher-order structures such as shown above, first into closely packed arrays of nucleosomes called nucleofilaments, which are then coiled into thicker and thicker filaments.

Nucleofilament structure

Condensation and decondensation of a chromosome in the cell cycle

• The interphase nucleus contains loosely packed, filamentous chromosomes, whose DNA is available for gene transcription. During each round of cell division, the chromosomal DNA is replicated and then condensed into metaphase chromosomes for segregation into the daughter cells, followed by decondensation as the interphase nucleus is formed.

Condensation and decondensation of a chromosome in the cell cycle

Telomeres and aging

• The chromosome contains a single, long molecule of double stranded DNA, and thus has two ends. These ends create two problems: they are difficult to replicate and they have a tendency to fuse with other chromosome ends causing karyotypic rearrangements. To prevent these problems, chromosomes have protective ends called "telomeres" that are composed of tandemly repeated, 5-8 bp sequences up to 12 kb in length. In germline cells and in the cells of young individuals, telomeres are of maximal length, but with every round of somatic cell division telomeres get a little shorter. After many rounds of replication and cell division, telomeres become too short to protect the chromosome ends from fusing with other chromosomes. At this stage, cells are said to be "senescent." Telomere length is maintained in germline cells by an enzyme called "telomerase," which can restore any shortening that has occurred. Tumor cells also have telomerase and thus are immortal and can grow indefinitely.

Telomeres and aging

Metaphase chromosome

centromere

telomere telomere

telomere structure

young

senescent

Telomeres are protective“caps” on chromosomeends consisting of short5-8 bp tandemly repeatedGC-rich DNA sequences,that prevent chromosomesfrom fusing and causingkaryotypic rearrangements.

(TTAGGG)many

(TTAGGG)few

• telomerase (an enzyme) is required to maintain telomere length in germline cells

• most differentiated somatic cells have decreased levels of telomerase and therefore their chromosomes shorten with each cell division

<1 to >12 kb