12.1 Cutting and cloning #
12.1.1 Restriction enzymes and the logic of a join #
Type II restriction enzymes cut both strands at one short defined site. Many sites are palindromes. EcoRI recognises GAATTC and leaves a 5′ AATT overhang. EcoRV recognises GATATC and cuts flush, leaving blunt ends. In random DNA a six-base site is expected about once every 4^6 bp, about 4 kb, and a four-base site about once every 256 bp. Real genomes are not random, so these are expectations, not maps.
Compatible sticky ends base-pair and hold insert and vector together. DNA ligase, using ATP, restores the phosphodiester bonds. Blunt ends can be ligated, with nothing to hold them in register, so the join is less efficient and orientation is not enforced. The product replicates only if the origin survived. It does not show that the insert is the intended one, or that it is expressed, until the DNA is cut again or sequenced.
12.1.2 Plasmids and selectable markers #
A plasmid vector needs an origin, a selectable marker, and a cloning site that spares both. The marker is commonly antibiotic resistance. On selective medium, only cells that took up a plasmid form colonies. It does not report an insert, because an empty plasmid is resistant too.
Amino Acid Codon Table
A screen is a second test on colonies that already survived. In the blue-white screen the cloning site lies in a fragment of lacZ. An empty vector complements a host mutation, so the colony is blue on a chromogenic substrate. An insert interrupts the fragment, and the colony is white. Both colours grow. The usual host is a laboratory strain that will not destroy the incoming DNA with its own restriction system. Growing that strain is microbiology.
Example Plasmid
12.1.3 cDNA and genomic libraries #
| Genomic library | cDNA library | |
|---|---|---|
| Source | fragmented genomic DNA | mRNA copied by reverse transcriptase |
| Introns and promoters | present | introns already removed; promoters generally absent |
| Which genes | essentially all | those expressed in the sampled cells |
| Use in E. coli | a split eukaryotic gene is not expressed correctly | the coding sequence is continuous if a bacterial promoter and Shine–Dalgarno site are supplied |
| Bias | depends on how the DNA was broken | abundant mRNAs dominate; oligo(dT) priming misses 5′ ends when reverse transcriptase stops, and misses histone mRNAs, which lack poly(A) |
Reverse transcriptase is the laboratory form of RNA-dependent DNA synthesis. Covering a human genome needs large inserts. Bacterial artificial chromosomes, of about 100–200 kb, were a main vehicle for the reference assembly, built at several-fold coverage because one genome equivalent still misses fragments by chance.
12.2 Reading sequence #
12.2.1 PCR #
PCR repeats three steps. Near 95 °C the strands separate. At a lower temperature, often about 50–65 °C, two primers anneal, one on each strand, 3′ ends facing. Near 72 °C a polymerase extends. Taq polymerase, from Thermus aquaticus, survives denaturation, so it need not be added every cycle. It has no proofreading exonuclease, errs on the order of once in 10^4 to 10^5 bases, and often leaves an unpaired A at the product’s 3′ end. A proofreading polymerase is used when the copy must be exact.
A primer base-pairs at one intended site and supplies the 3′-OH that DNA polymerase cannot create for itself. The two primers define the product. It is not a sequence, and PCR of DNA does not show that the region is transcribed. The cycle at which product crosses a detection threshold ranks starting amounts, and only with standards.
12.2.2 Sanger sequencing and short reads #
Sanger sequencing uses chain termination. Polymerase copies one template with ordinary nucleotides plus a few dideoxynucleotides, which have no 3′-OH. The nested products, separated by length and labelled by which terminator ended them, spell the sequence. A clean read is about 500–900 nucleotides of one plasmid or one PCR product.
Short-read sequencing of the Illumina type amplifies fragments as clusters on a surface and reads them by synthesis with reversible terminators, millions to billions at once. Reads are commonly about 100–300 nucleotides, often from both ends of a fragment. RNA-seq counts cDNA reads as a measure of RNA abundance.
12.2.3 Long reads #
Single-molecule long-read methods commonly return tens of kilobases, and sometimes much more. In one method a polymerase is watched as it copies a single template. In the other, a strand is drawn through a pore and the bases perturb an ionic current. Length is what they buy: a read can cross a repeat, an entire splice isoform, or a rearrangement that short reads shatter. A single pass has a higher raw error than a finished Sanger read. Consensus of many passes reduces that error. Long reads do not replace short reads for every counting experiment.
12.3 Detecting a molecule #
12.3.1 Southern, northern, and western blots #
| Blot | Molecule on the gel | Probe | Question |
|---|---|---|---|
| Southern | DNA, usually cut | a labelled nucleic acid | Is this sequence present, and on a fragment of what size? |
| Northern | RNA | a labelled nucleic acid | Is this RNA present, roughly how much, and how long is it? |
| Western | protein | an antibody | Is this polypeptide present, roughly how much, and what is its apparent mass? |
A band is size plus the real specificity of the probe. A Southern band is not a finished sequence. A northern band is an RNA, not the protein it might encode. A western band is an antibody bound to a polypeptide, not an enzyme activity, and a cross-reactive antibody will mislead.
12.3.2 Reporter genes #
A reporter replaces a hard-to-see product with an easy one. lacZ yields colour or fluorescence from a substrate. Luciferase yields light from a substrate. GFP and its derivatives fluoresce with no substrate at all.
The reporter reports the construct. It misses regulation that lived in a region left out of the clone, and a GFP fusion can mis-localise or harm the cell. Agreement with a northern blot, a western blot, or a careful RNA-seq count is what makes the light evidence about the endogenous gene.
12.4 A nuclease directed by RNA #
12.4.1 CRISPR–Cas9, NHEJ, HDR, and what a knockout does not prove #
CRISPR–Cas systems are adaptive immunity. A fragment of an invader is stored as a spacer and later guides a Cas nuclease to matching nucleic acid. The laboratory enzyme in wide use is Cas9, programmed with a single guide RNA that does the work of the two natural RNAs. Pairing is not enough. The protein also demands a short adjacent PAM. For this Cas9 the PAM is NGG, and both strands are cut a few base pairs to the guide side of it. A match without a PAM is not a substrate. The PAM is part of the mechanism.
Repair uses Chapter 5. Non-homologous end joining often leaves a small insertion or deletion. A frameshift or a broken splice site may destroy the protein, and that is a knockout only if both alleles are actually inactivated and the protein is actually gone. The junction is whatever end joining produced. It was not designed.
Homology-directed repair can copy a supplied template that has homology on both sides of the break, which is how a chosen base change or a tag is written. In many mammalian cells this is the minority path. It does not guarantee the designed allele.
A confirmed frameshift and a lost phenotype suggest that the gene was required for what disappeared. They do not say which function of the protein mattered, and an off-target cut elsewhere can be the real cause. A frameshift need not be a null: translation may restart, a fragment may still act, and nonsense-mediated decay can provoke compensation by a paralogue. RNA interference, which destroys the mRNA and leaves the gene intact, is partial and has different off-targets. It is not a mild knockout. Sequencing the intended site is necessary and does not survey the genome. Base editors and prime editors can change a sequence with less reliance on a double-strand break. A knockout begins an argument about function. By itself it does not finish one.
12.5 Key terms, summary, and self-check #
Key terms. Sticky end; selectable marker; cDNA; Taq; Sanger; short read; long read; Southern; northern; western; reporter; Cas9; PAM.
Summary. Restriction enzymes cut defined sites and ligase joins compatible ends to an origin. Antibiotic resistance selects the plasmid, not the insert; a screen such as blue-white only suggests one. Genomic libraries keep introns and silent genes. cDNA libraries keep spliced transcripts of one sample. PCR cycles denaturation, annealing, and extension, and a primer’s job is a specific 3′-OH. Sanger sequencing reads one template for about 500–900 bases by chain termination. Short reads sequence vast numbers of short clusters. Long reads cross repeats and isoforms. Southern, northern, and western blots detect DNA, RNA, and protein. A reporter reports a construct. Cas9 cuts where a guide pairs next to a PAM. End joining may knock a gene out with an unplanned indel. Homology-directed repair may copy a template. Neither result, unchecked, proves what the protein does.
Self-check.
1. Why do sticky ends ligate more readily than blunt ends, and what does an antibiotic-resistance marker select?
2. A split human gene is to be expressed in E. coli. Which library supplies the coding sequence, and what has already been removed?
3. Why did a heat-stable polymerase make PCR practical, and what must a primer provide?
4. Which blot uses an antibody, and what can a northern blot say about an mRNA that a western blot cannot?
5. A Cas9 cut, a frameshift, and a lost phenotype are in hand. Give two reasons this does not, by itself, prove the protein’s normal job.
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# Closing: how the chapters fit together
The course is one argument. Nucleic acid, and in cells DNA, is the genetic material (Chapter 1). B-form dimensions, supercoiling, and nucleosomes decide how that DNA is read and how it fits in a nucleus (Chapter 2). Real genomes are not strings of genes: bacteria often use operons, eukaryotic genes are split, and most of a human C-value is repetitive DNA (Chapter 3).
Replication is semiconservative and 5′→3′ on both strands, so one strand is made in pieces, primers are removed, and linear ends need telomerase (Chapter 4). Proofreading and the repair pathways keep the copy honest, with a choice between end joining and recombination once both strands are broken (Chapter 5).
Transcription in bacteria is one core plus a sigma, ended by a hairpin or by Rho. Eukaryotes split the work among three polymerases, couple processing to the pol II CTD, and let enhancers act through a loop and Mediator (Chapter 6). A pol II transcript becomes an mRNA only after capping, splicing, and polyadenylation, with editing as a rare extra change, and only the finished particle is exported (Chapter 7). Only cells without a nuclear envelope translate while the RNA is still being made (Chapter 8). The new chain folds with chaperones or is destroyed when a K48 ubiquitin chain sends it to the proteasome. Where it is sent is the other course (Chapter 9).
Regulation adds decisions to the same machines. Lac and trp show negative control, positive control, and an RNA switch read by a ribosome (Chapter 10). Eukaryotes add nucleosome remodelling, modular DNA-binding proteins, heritable CpG methylation, small RNAs, and, in Xist, one long RNA that silences a chromosome (Chapter 11). Each claim is only as strong as the method behind it: a join, an amplification, a read, a blot, a reporter, or a cut whose repair was actually checked (Chapter 12).
Revision checklist #
1. In one sentence each, state what Griffith showed, what Avery, MacLeod, and McCarty added, and what Hershey and Chase added, including one limit.
2. Give the B-form rise and base pairs per turn, the DNA in a nucleosome core, and why a static 30 nm fibre is the wrong picture of a living nucleus.
3. Compare E. coli and the human haploid genome in size, chromosome number and shape, protein-coding gene count, and the share that is exons.
4. Place helicase, primase, leading polymerase, lagging polymerase, clamp, and ligase on a fork, and say why only the lagging strand is discontinuous.
5. Name the pathway for uracil, for a thymine dimer, for a mismatch, and for a double-strand break in G1, and say what Rad51 does that Ku does not.
6. Contrast an intrinsic terminator with Rho, and say which CTD marks accompany capping and 3′-end processing.
7. Trace a eukaryotic pre-mRNA to a polypeptide with the words cap, lariat, poly(A), scanning, Kozak, and ribozyme.
8. Give the four rows of the lac glucose-and-lactose table, and say which trp hairpin forms when the ribosome stalls.
9. Say what DNMT1 copies, what complementarity decides inside RISC, and what Xist coats.
10. For a PCR band, a northern blot, a Sanger read, and a Cas9 frameshift, state one thing each result does not prove.
Selected sources #
These notes are an original teaching summary, not a transcript of any textbook.
- Watson JD, Baker TA, Bell SP, Gann A, Levine M, Losick R. Molecular Biology of the Gene. Cold Spring Harbor Laboratory Press.
- Krebs JE, Goldstein ES, Kilpatrick ST. Lewin’s Genes. Jones and Bartlett.
- Alberts B, Johnson A, Lewis J, Morgan D, Raff M, Roberts K, Walter P. Molecular Biology of the Cell. Garland Science / W. W. Norton. Information chapters.
- Meselson M, Stahl FW. The replication of DNA in Escherichia coli. Proceedings of the National Academy of Sciences 44:671–682 (1958).
- Jacob F, Monod J. Genetic regulatory mechanisms in the synthesis of proteins. Journal of Molecular Biology 3:318–356 (1961).
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# Part C. Visual Figures & Scientific Diagrams Gallery
Comprehensive high-resolution visual reference repository for Molecular Biology from the B10logy Visual Atlas.
201704 Chromatin Structure
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# Part C. Visual Figures & Scientific Diagrams Gallery
Essential high-resolution reference diagrams for Molecular Biology from the B10logy Visual Atlas.
