Structure of DNA and its packaging
A human cell holds DNA about two metres long inside a nucleus smaller than a speck of dust. It fits by folding thread around spools, like winding wool.
DNA is a long polymer of deoxyribonucleotides. Each nucleotide has a nitrogenous base (A, G, C, T), a deoxyribose sugar and a phosphate. In 1953 Watson and Crick proposed the double helix model, using X-ray data of Wilkins and Franklin and Chargaff's rule that A = T and G = C. In a cell, DNA is wrapped around histones as nucleosomes.
In the figure : Two antiparallel strands joined by hydrogen bonds between A–T and G–C.
In the figure : Histones are positively charged proteins and DNA is negatively charged, so they stick together.
Rule 1: Features of the double helix — Two polynucleotide chains run in opposite directions (antiparallel); the backbone is sugar–phosphate and the bases project inside. A pairs with T by two hydrogen bonds, G with C by three. The helix is right-handed; the pitch is 3.4 nm, with about 10 base pairs per turn, so the distance between two base pairs is 0.34 nm.
- Easy : A = T and G = C pairing.
- Medium : Because of 0.34 nm per base pair, the length of a DNA molecule can be estimated from its number of base pairs.
- Exam-type : Q: What is the distance between two successive base pairs? A: 0.34 nm.
Rule 2: Packaging — Histones (H2A, H2B, H3, H4 — two molecules of each, forming the histone octamer) have positively charged lysine and arginine. About 200 bp of DNA wrap around one octamer to form a nucleosome. Nucleosomes form chromatin, which is packed into chromatin fibres and then chromosomes. Euchromatin is loosely packed and active; heterochromatin is dense and inactive.
- Easy : Histone octamer has eight histone molecules.
- Medium : Euchromatin stains light and is transcriptionally active.
- Exam-type : Q: Why does DNA bind tightly to histones? A: DNA is negatively charged and histones are rich in positively charged amino acids.
Human haploid DNA: 3.3 × 10⁹ bp; E. coli: 4.6 × 10⁶ bp. DNA is also more stable than RNA because of the missing 2′-OH group and the use of thymine.
Thinking RNA has thymine. RNA has uracil, and DNA has thymine.
Memory clue: A-T, G-C: "Apple Tree, Car Garage" — Adenine–Thymine; Guanine–Cytosine.
Spot the difference
| Point | DNA | RNA |
|---|---|---|
| Sugar | Deoxyribose | Ribose |
| Bases | A, G, C, T | A, G, C, U |
| Strands | Double (usually) | Single (usually) |
Practice questions
- 1Who proposed the double helix model of DNA?
- 2How many hydrogen bonds join G and C?
- 3Name the proteins around which DNA is wound.
- 4What is the pitch of the DNA helix?
- 5What is a nucleosome?
- 6Which kind of chromatin is transcriptionally active?
- 7Why are the strands of DNA called antiparallel?
- 8If a DNA piece has 20 per cent adenine, what are the percentages of T, G and C?
Show answers (try them yourself first)
- 1James Watson and Francis Crick (1953).
- 2Three.
- 3Histones.
- 43.4 nm.
- 5About 200 bp of DNA wrapped around a histone octamer.
- 6Euchromatin.
- 7One strand runs 5′→3′ and the other 3′→5′.
- 8T = 20, G = 30 and C = 30 per cent (A = T, G = C).
DNA is the genetic material
Which part of a cell carries the instruction book: protein or DNA? Three classic experiments settled the question.
Griffith (1928) saw that heat-killed S-type pneumococci changed live R-type bacteria into virulent S type (transformation). Avery, MacLeod and McCarty showed that DNA was the transforming principle. Hershey and Chase (1952) labelled phage DNA with radioactive ³²P and protein with ³⁵S and showed that only DNA enters the bacterium.
In the figure : Some transforming principle passed from dead S bacteria to live R bacteria.
In the figure : Only the DNA of the virus enters the bacterial cell: DNA is the genetic material.
Rule 1: Griffith and Avery — Avery and co-workers purified biochemicals from heat-killed S cells. Protein-digesting enzymes (proteases) and RNase did not stop transformation, but DNase did. So DNA was the transforming principle.
- Easy : DNase stops transformation.
- Medium : Protease and RNase do not stop transformation.
- Exam-type : Q: What did the DNase experiment show? A: That DNA, not protein or RNA, causes transformation.
Rule 2: Hershey–Chase — Bacteriophages were grown in ³²P (labels DNA) or ³⁵S (labels protein). After infection, the coats were removed in a blender and the cells were separated in a centrifuge. Radioactive ³²P was found inside the bacteria.
- Easy : Phosphorus is in DNA, sulphur in protein.
- Medium : The protein coats remained outside the cell.
- Exam-type : Q: Why was ³²P used for DNA and ³⁵S for protein? A: DNA has phosphorus but no sulphur; proteins have sulphur but no phosphorus.
DNA is the genetic material in most organisms, but in some viruses (TMV, Qβ phage) RNA is the genetic material. RNA likely evolved first; DNA is more stable.
Saying Hershey and Chase used the same label for DNA and protein. They used ³²P for DNA and ³⁵S for protein.
Memory clue: P for DNA, S for protein — ³²P = DNA; ³⁵S = protein.
Spot the difference
| Point | Griffith | Hershey–Chase |
|---|---|---|
| Organism | Streptococcus pneumoniae | Bacteriophage and E. coli |
| Year | 1928 | 1952 |
| Shows | Transformation | DNA enters the cell |
Practice questions
- 1Name the bacterium used by Griffith.
- 2What is transformation?
- 3What did Hershey and Chase use to label DNA?
- 4Which enzyme stopped transformation in the experiment of Avery and co-workers?
- 5What is the capsule of S-type bacteria made of?
- 6Name one virus that has RNA as genetic material.
- 7Why were mice not killed by heat-killed S bacteria alone?
- 8Why was a blender used in the Hershey–Chase experiment?
Show answers (try them yourself first)
- 1Streptococcus pneumoniae.
- 2Change in the genotype of a cell by the entry of foreign DNA.
- 3Radioactive phosphorus (³²P).
- 4DNase.
- 5Polysaccharide (mucous coat).
- 6Tobacco mosaic virus (or Qβ bacteriophage).
- 7The bacteria were dead and could not multiply.
- 8To remove viral coats stuck outside the bacterial cells.
DNA replication
Before a cell divides, its entire instruction book must be copied once and only once. The copying follows a simple rule: each old strand serves as a template.
Replication is the copying of DNA. It is semi-conservative: each daughter DNA has one old and one new strand, as shown by Meselson and Stahl (1958) with heavy nitrogen (¹⁵N) in E. coli. The main enzyme is DNA-dependent DNA polymerase, which makes the new strand in the 5′→3′ direction. It starts at the origin of replication and forms a replication fork.
In the figure : In Meselson–Stahl results, after one generation all DNA was of intermediate density.
In the figure : DNA polymerase can add nucleotides only in the 5′→3′ direction.
Rule 1: Steps and enzymes — Helicase and topoisomerase unwind and open the DNA at the origin. DNA polymerase adds nucleotides complementary to the template. Because the two strands are antiparallel, one is copied continuously (leading) and the other in pieces (lagging) joined by DNA ligase.
- Easy : The leading strand is made continuously.
- Medium : Okazaki fragments are on the lagging strand.
- Exam-type : Q: Which enzyme joins Okazaki fragments? A: DNA ligase.
Rule 2: Meselson–Stahl experiment — Cells grown in ¹⁵N (heavy) medium were shifted to ¹⁴N (light) medium; DNA was spun in a CsCl density gradient. After one generation all DNA had intermediate density and after two generations equal amounts of intermediate and light DNA appeared.
- Easy : After 1 generation: all hybrid.
- Medium : After 2 generations: 50 per cent hybrid, 50 per cent light.
- Exam-type : Q: What did the result prove? A: DNA replicates semi-conservatively.
Replication in E. coli takes about 18 minutes for 4.6 million bp; in humans there are many origins on each chromosome, so a long DNA is copied faster.
Thinking both strands are made continuously. Only the leading strand is continuous.
Memory clue: Lead → Long, Lag → Lumps — Leading strand is continuous; lagging strand is made in lumps (Okazaki fragments).
Spot the difference
| Point | Leading strand | Lagging strand |
|---|---|---|
| Synthesis | Continuous | Discontinuous |
| Fragments | None | Okazaki fragments |
| Direction | Towards the fork | Away from the fork |
Practice questions
- 1What is meant by semi-conservative replication?
- 2Name the main enzyme of replication.
- 3What are Okazaki fragments?
- 4Who proved semi-conservative replication?
- 5Which isotope of nitrogen was heavy?
- 6What does DNA ligase do?
- 7Why does DNA polymerase need a template?
- 8Why is replication said to be highly accurate?
Show answers (try them yourself first)
- 1Each daughter DNA has one parental and one new strand.
- 2DNA polymerase.
- 3Short DNA pieces made on the lagging strand.
- 4Meselson and Stahl.
- 5¹⁵N.
- 6It joins the DNA pieces together.
- 7It adds the base complementary to the template strand; it cannot decide the base on its own.
- 8The proof-reading action of DNA polymerase and complementary base pairing keep errors very low.
Transcription
A librarian does not carry the master copy of a book out of the library; she makes a copy. A gene is copied into RNA in the same way.
Transcription is the formation of RNA from a DNA template. A transcription unit has a promoter, the structural gene and a terminator. RNA polymerase binds to the promoter and builds RNA complementary to the template strand (3′→5′); the other strand is the coding strand. Uracil replaces thymine.
In the figure : The promoter is upstream (5′ side) and the terminator is downstream (3′ side) of the coding strand.
In the figure : In eukaryotes the primary transcript is processed before leaving the nucleus.
Rule 1: Steps in bacteria — Initiation: RNA polymerase with the σ factor binds the promoter. Elongation: nucleotides are added to the growing RNA. Termination: at the terminator, with the help of the ρ (rho) factor, the RNA is released. In bacteria mRNA is functional at once, and translation can begin even before transcription ends.
- Easy : σ factor starts transcription.
- Medium : ρ factor ends it.
- Exam-type : Q: What is the function of the promoter? A: It gives the binding site for RNA polymerase.
Rule 2: Eukaryotic transcription — Eukaryotes have three RNA polymerases: pol I (rRNAs), pol II (hnRNA/mRNA), pol III (tRNA, 5S rRNA). The gene is split: coding exons are interrupted by non-coding introns, which are removed by splicing. A cap (methyl guanosine triphosphate) is added at the 5′ end and a poly-A tail (200–300 adenylates) at the 3′ end.
- Easy : Introns are removed; exons are joined.
- Medium : Capping and tailing give mature mRNA.
- Exam-type : Q: What is hnRNA? A: The unprocessed, heterogeneous nuclear RNA with exons and introns.
Template strand 3′→5′; RNA made 5′→3′. The sequence of mRNA matches the coding strand, except T is replaced by U.
Writing that mRNA is complementary to the coding strand. It is complementary to the template strand and identical to the coding strand (with U for T).
Memory clue: PST — Promoter → Structural gene → Terminator (5′ to 3′ of the coding strand).
Spot the difference
| Point | Exon | Intron |
|---|---|---|
| Meaning | Coding sequence | Non-coding sequence |
| In mature mRNA | Present | Removed |
| Found in | Split genes of eukaryotes | Eukaryotes |
Practice questions
- 1Name the enzyme that makes RNA.
- 2What is a promoter?
- 3What is splicing?
- 4Which base is present in RNA instead of thymine?
- 5How is hnRNA changed into mRNA?
- 6What are the two strands of a gene called?
- 7The coding strand reads 5′-ATGCAT-3′. Write the mRNA sequence.
- 8Why does a bacterial mRNA not need processing?
Show answers (try them yourself first)
- 1RNA polymerase.
- 2The DNA sequence where RNA polymerase binds to start transcription.
- 3Removal of introns and joining of exons.
- 4Uracil.
- 5By splicing, capping and tailing.
- 6Template strand and coding strand.
- 75′-AUGCAU-3′.
- 8Bacterial genes are not split into exons and introns, and there is no nucleus.
Genetic code and tRNA
Four letters make 64 three-letter words, and those words write all proteins. The code is the dictionary and tRNA is the translator.
The genetic code is the relationship between mRNA base triplets (codons) and amino acids. Har Gobind Khorana prepared artificial RNA molecules and Marshall Nirenberg used cell-free systems to decode it. Of the 64 codons, 61 code for amino acids and 3 (UAA, UAG, UGA) are stop codons; AUG codes for methionine and is the start codon.
In the figure : These properties help the cell read every mRNA in the same way.
Rule 1: Properties — The code is read in triplets without gaps or overlaps. More than one codon can code for one amino acid (degeneracy), but each codon codes for only one amino acid (specificity). Almost all organisms use the same code (universal), which allows bacteria to make human insulin.
- Easy : AUG is the start codon.
- Medium : Methionine and tryptophan have a single codon each (AUG, UGG).
- Exam-type : Q: How many codons code for amino acids? A: 61.
Rule 2: tRNA, the adapter — tRNA has a clover-leaf shape in two dimensions and an L shape in three. Its anticodon loop reads the codon by base pairing; its amino acid acceptor end (CCA, 3′) carries the amino acid. There is at least one tRNA for each amino acid; the initiator tRNA carries methionine. The enzyme aminoacyl tRNA synthetase attaches the amino acid to the tRNA (charging).
- Easy : tRNA is the adapter molecule.
- Medium : Charging needs ATP.
- Exam-type : Q: Where is the anticodon found? A: In the anticodon loop of tRNA.
Mutations: a point mutation that changes the reading frame by insertion or deletion of a base is a frame-shift mutation; sickle-cell anaemia is a point mutation (GAG → GUG, glutamic acid → valine).
Saying 64 codons code for 64 amino acids. There are only 20 amino acids, and 3 of the codons are stop signals.
Memory clue: 61 + 3 → 64 — 61 sense codons plus 3 stop codons.
Spot the difference
| Point | Codon | Anticodon |
|---|---|---|
| Found in | mRNA | tRNA |
| Pairs with | Anticodon | Codon |
| Number | 64 | At least one tRNA per amino acid |
Practice questions
- 1What is a codon?
- 2Name the start codon.
- 3Name the three stop codons.
- 4Who decoded the genetic code with the chemical method?
- 5What is the meaning of degeneracy of the code?
- 6What is attached to the 3′ end of a tRNA?
- 7Why is the genetic code called universal?
- 8An mRNA codon is GCU. What is the anticodon on tRNA?
Show answers (try them yourself first)
- 1A triplet of bases in mRNA that codes for an amino acid.
- 2AUG.
- 3UAA, UAG and UGA.
- 4Har Gobind Khorana.
- 5More than one codon can code for the same amino acid.
- 6An amino acid.
- 7Almost all organisms use the same codons for the same amino acids.
- 8CGA.
Translation: protein synthesis
A factory reads a work order, brings the parts one by one and fixes them in the right order. The ribosome does this with amino acids.
Translation is the synthesis of a polypeptide using the sequence of the mRNA. It takes place on ribosomes (with a small and a large subunit). The steps are charging of tRNA, initiation, elongation and termination. A stop codon is recognised by a release factor. One of the rRNAs acts as an enzyme (ribozyme) for peptide bond formation.
In the figure : The ribosome moves along the mRNA from the 5′ end to the 3′ end.
Rule 1: Steps — The ribosome has two sites, P and A. The first tRNA (carrying methionine) sits at the P site. The next tRNA enters the A site and a peptide bond forms between the two amino acids; the ribosome moves by one codon (translocation). The cycle repeats until a stop codon is reached. Then the polypeptide is released.
- Easy : Peptide bond joins amino acids.
- Medium : The initiator tRNA brings methionine.
- Exam-type : Q: What ends translation? A: A stop codon and the release factor.
Rule 2: Untranslated regions and polysomes — mRNA has extra sequences at both ends, the untranslated regions (UTRs), that help in efficient translation. Many ribosomes can move along one mRNA together, forming a polysome, so more copies of the protein are made quickly.
- Easy : UTRs are not translated.
- Medium : A polysome is a string of ribosomes on one mRNA.
- Exam-type : Q: Why are polysomes useful? A: Many copies of a protein are made from a single mRNA at the same time.
Transcription and translation are costly in energy, so cells regulate them. Translation proceeds in the 5′→3′ direction of mRNA.
Saying that the ribosome reads from 3′ to 5′ of mRNA. mRNA is read 5′→3′.
Memory clue: I-E-T — Initiation, Elongation, Termination.
Spot the difference
| Point | Transcription | Translation |
|---|---|---|
| Product | RNA | Polypeptide |
| Place (eukaryote) | Nucleus | Cytoplasm (ribosome) |
| Template | DNA | mRNA |
Practice questions
- 1Where does translation take place?
- 2Which amino acid starts a polypeptide?
- 3What is a polysome?
- 4What is a ribozyme?
- 5What joins amino acids together?
- 6Name the three stages of translation.
- 7Why is the start codon AUG always present at the beginning of the coding region?
- 8Why do cells need so much energy for protein synthesis?
Show answers (try them yourself first)
- 1On ribosomes.
- 2Methionine.
- 3Several ribosomes attached to a single mRNA.
- 4An RNA molecule that acts as an enzyme (the rRNA in the ribosome).
- 5Peptide bonds.
- 6Initiation, elongation and termination.
- 7It is the codon for the initiator methionine tRNA, which starts translation.
- 8Each charging of tRNA and each peptide bond use ATP/GTP.
Regulation of gene expression: the lac operon
A shop keeps its lights off when nobody is inside and switches them on when a customer walks in. Bacteria do the same with the genes for digesting lactose.
An operon is a group of genes controlled together. The lac operon of E. coli (Jacob and Monod) has a regulator gene i (makes the repressor), a promoter, an operator and three structural genes: z (β-galactosidase), y (permease) and a (transacetylase). Lactose is the inducer.
In the figure : The repressor sits on the operator and blocks RNA polymerase when lactose is absent.
In the figure : Regulation is negative: the repressor switches the operon off.
Rule 1: How it works — Without lactose the repressor binds the operator and blocks the promoter. When lactose (allolactose) is present, it binds the repressor, which can no longer attach to the operator; RNA polymerase transcribes the three structural genes together as one polycistronic mRNA.
- Easy : Lactose is the inducer.
- Medium : The lac operon makes three enzymes with one mRNA.
- Exam-type : Q: What is the role of the repressor? A: To bind to the operator and stop transcription of the lac genes.
Rule 2: Why regulation matters — Gene expression can be controlled at transcription, processing, mRNA transport and translation. The primary level is transcriptional control. Regulation of the lac operon is a case of the enzyme synthesis being controlled by its substrate.
- Easy : Glucose is also a preferred sugar.
- Medium : Bacteria save energy by not making enzymes they do not need.
- Exam-type : Q: What does the z gene code for? A: β-galactosidase, which splits lactose into glucose and galactose.
lac operon genes z, y, a make: β-galactosidase, permease and transacetylase. The i gene is not part of the operon proper but regulates it.
Thinking lactose is the repressor. Lactose is the inducer; the repressor is the protein made by the i gene.
Memory clue: Z-Y-A → Gal, Perm, Trans — Z → β-Galactosidase, Y → Permease, A → Transacetylase.
Spot the difference
| Point | Repressor | Inducer |
|---|---|---|
| What is it | Protein from the i gene | Lactose (allolactose) |
| Binds to | Operator | Repressor |
| Effect | Switches the operon off | Switches the operon on |
Practice questions
- 1What is an operon?
- 2Who proposed the lac operon model?
- 3Name the inducer of the lac operon.
- 4What does the permease do?
- 5What does the i gene code for?
- 6What is a polycistronic mRNA?
- 7Why is the lac operon described as negative regulation?
- 8What would happen if the i gene is deleted?
Show answers (try them yourself first)
- 1A group of genes regulated together as one unit.
- 2Jacob and Monod.
- 3Lactose (allolactose).
- 4It increases the permeability of the cell to lactose.
- 5The repressor protein.
- 6An mRNA that codes for more than one polypeptide.
- 7The regulatory protein (repressor) switches the genes off by binding to the operator.
- 8No repressor is formed, so the lac genes would always be ON.
Human Genome Project and DNA fingerprinting
Police match a tiny blood stain at a crime scene to a suspect with DNA. This works because every person's DNA has a unique repeat pattern.
The Human Genome Project (HGP), started in 1990 and completed in 2003, aimed at finding all the genes and writing out the sequence of the roughly 3 billion base pairs of human DNA. DNA fingerprinting is a technique (developed by Alec Jeffreys) that uses variations in repetitive DNA sequences, called VNTRs, to identify individuals.
In the figure : Repeated sequences make up a very large part of the human genome.
In the figure : Bands of an individual are compared with those of a sample from the crime scene or a relative.
Rule 1: Human genome — The genome has 3 billion base pairs spread over 24 different chromosomes (22 autosomes, X and Y). About 1.4 million places show single base differences called SNPs. Bioinformatics is needed to store and analyse the huge data.
- Easy : Human genome has about 3 × 10⁹ bp.
- Medium : The project involved many countries, not only the USA.
- Exam-type : Q: What are SNPs? A: Single nucleotide polymorphisms: single base differences among individuals.
Rule 2: DNA fingerprinting — The principle is polymorphism in repetitive DNA (VNTRs or microsatellites). The number of repeats differs from person to person, but an offspring inherits half the pattern from each parent. It is used in forensics, paternity tests, and studying genetic diversity of populations.
- Easy : The DNA pattern of a child matches both parents.
- Medium : Identical twins have identical DNA fingerprints.
- Exam-type : Q: Name two uses of DNA fingerprinting. A: Forensic investigation and paternity testing.
HGP: 1990–2003, 3 billion bp. Less than 2 per cent of the genome codes for proteins. 99.9 per cent of the bases are identical in all humans.
Thinking that DNA fingerprinting uses all of the DNA. It uses only variable repeated segments.
Memory clue: VNTR → Very Nice To Recognise — Variable Number of Tandem Repeats identify people.
Spot the difference
| Point | HGP | DNA fingerprinting |
|---|---|---|
| Aim | Sequence the whole genome | Identify individuals |
| Based on | Sequencing methods | Polymorphism in repeats |
| Used in | Medicine, research | Forensics, paternity |
Practice questions
- 1Expand HGP.
- 2In which year was the HGP completed?
- 3What is the approximate size of the human genome?
- 4What are VNTRs?
- 5Who developed DNA fingerprinting?
- 6What per cent of genome codes for proteins?
- 7Why is bioinformatics important for HGP?
- 8Why can DNA fingerprinting be used in a paternity test?
Show answers (try them yourself first)
- 1Human Genome Project.
- 22003.
- 3About 3 billion base pairs.
- 4Variable number of tandem repeats, the repeated DNA sequences used in fingerprinting.
- 5Alec Jeffreys.
- 6Less than 2 per cent.
- 7The huge amount of sequence data must be stored, sorted and analysed with computers.
- 8Half of a child's DNA bands come from the father, so the pattern can be matched.