Chromosome and DNA
Chromosomes are the cell’s packages of DNA, and DNA is the hereditary material that is copied, read and sometimes changed. This unit follows the story: how a chromosome is built, the experiments that proved DNA carries heredity, how DNA replicates, how a gene is expressed through transcription and translation, how gene expression is switched on and off, and how mutations alter genes and chromosomes and cause diseases.
Learning outcomes (22)
- Annotate the detailed structure of a chromosome.
- Narrate the experimental work of Griffith and Hershey-Chase, which proved that DNA is the hereditary material.
- Describe the three models proposed about the mechanism of DNA replication.
- Narrate the work of Meselson and Stahl to justify the semi-conservative replication as the correct method of replication.
- Describe the events of the process of DNA replication.
- Explain DNA stability and variability as two characters of the replicating DNA molecule.
- Describe the characteristics of genetic code (universal, triplet, non-overlapping, degenerate, punctuated).
- Differentiate between the terms genetic code and codon.
- Explain the mechanism of transcription.
- Explain why the length of transcribed m-RNA molecule (in Eukaryotes) shortens as it enters the cytoplasm for translation.
- Describe the mechanism of protein synthesis.
- State the difference between protein synthesis in prokaryotes and eukaryotes.
- Suggest possible ways in which the synthesized protein can be used within or outside a cell that synthesized it.
- State the importance of the regulation of gene expression.
- Describe the negative control of gene expression by repressor proteins.
- Describe the positive control of gene expression by activator proteins.
- Define mutation and identify various sources of mutation.
- Differentiate between natural and induced mutations and mutagens.
- Justify that most mutations are harmful.
- Rationalize that mutations might be a contributing factor towards evolution.
- Describe the symptoms, causes and possible available treatments of some of the chromosomal mutations. (Down's, Klinefelter's and Turner's syndrome)
- Describe the symptoms, causes and possible available treatments of some of the gene mutations.
13.1Morphology of chromosomesOften examined
Mendel's hereditary factors were an abstract idea. At that time no cellular structure was known that could house them. We now know that these factors are genes, that genes lie on chromosomes, and that genes are made of DNA. This chapter studies chromosomes and DNA in detail.
Chromosomes are thick, thread-like structures that appear in the nucleus during cell division. In an interphase cell the chromosomes are uncondensed and look like a very fine network called the chromatin network.
Structure of chromosome
A typical chromosome consists of two strands called chromatids. Each chromatid is made of one long DNA molecule that is highly coiled together with histone proteins, and the chromatids contain the genes for different traits. Both chromatids are attached to each other at a point called the centromere, so each chromosome shows two arms.
- Chromatid
- One of the two strands of a chromosome; it is made of a single, highly coiled DNA molecule with histone proteins and carries genes.
- Centromere (primary constriction)
- The point at which the two chromatids are attached to each other. It divides the chromosome into two arms.
- Arm
- The region of the chromosome from the centromere to an end.
- Secondary constriction (nucleolar organizer)
- Another point of union along the length of the chromatids in some chromosomes. It gives rise to nucleoli during interphase. At least one pair of homologous chromosomes has a nucleolar organizer region.
- Satellite
- The knob-like end that lies beside the secondary constriction. This region has a useless sequence of DNA called junk DNA.
- Telomeres
- The terminal ends of a chromosome. They prevent two chromosomes from attaching to each other by their ends.
Types of chromosomes
On the basis of the position of the centromere along the chromosome, a chromosome may be metacentric, submetacentric, acrocentric or telocentric.
| Type | Position of the centromere | Arms |
|---|---|---|
| Metacentric | In the centre | Two arms of about equal length |
| Submetacentric | Located slightly away from the centre | Two arms of unequal length (one slightly shorter) |
| Acrocentric | Located near the end | One very short arm and one long arm |
| Telocentric | Located at an end | Only one arm, because the centromere is at the end |
13.1Composition and organization of chromosomeOften examined
Generally a chromosome is made of 40% DNA and 60% protein. In a cell that is ready to divide, the chromosome has two identical DNA molecules, so each chromatid has one DNA molecule. An average-sized human chromosome has DNA about 5 cm long, which consists of about 140 million nucleotides. How such a huge molecule fits into a tiny chromosome is explained by the way it is packed.
DNA is a negatively charged molecule because of its phosphate groups. It therefore has a strong affinity for histone proteins, which are positively charged (unlike many other proteins) because they contain many basic amino acids such as arginine and lysine.
During the S phase of the cell cycle, DNA and histones are completely disorganized from each other. After DNA is replicated, DNA and histones organize together again, and the condensation continues until the cell divides and the chromosomes appear. The organization of chromosomes occurs in four levels.
Level 1: nucleosome string
Just after DNA replication, each DNA molecule begins to coil around histone cores. Approximately every 200 nucleotides of the duplex DNA wrap twice around a core of eight histones. This complex is called a nucleosome. One histone molecule (histone H1) is associated with a small segment of DNA, the linker DNA, between every two nucleosomes. In this way the whole DNA (2 nm thick) becomes a chain of beads called the nucleosome string (10 nm thick), the "beads on a string" form of chromatin.
- Use the book's figure: one nucleosome for every 200 nucleotides of DNA.
- Substitute the numbers.
- State the result: the DNA of one average human chromosome is wrapped into about 700 000 nucleosomes, that is, 700 000 beads on a string.
- Link to thickness: winding the DNA round histones turns a 2 nm thread into a 10 nm "beads on a string" nucleosome string, which then coils further into the 30 nm chromatin fibre.
Level 2: chromatin fibre
Immediately afterwards, the nucleosome string coils again about its axis to form a thicker fibre of 30 nm, called the chromatin fibre. During the G₁ and G₂ phases, chromosomes are found at this level of organization. The chromatin fibre shows two regions: heterochromatin and euchromatin.
| Heterochromatin | Euchromatin |
|---|---|
| Highly condensed region of the chromatin fibre | Non-condensed (loosely packed) region of the chromatin fibre |
| Its genes are not expressed (unexpressed region) | Its genes are expressed |
| Stays condensed in the non-dividing cell | Is also condensed when the cell divides, so that a uniform chromatin fibre is established |
Level 3: supercoil
When cell division begins, the higher-order coiling of the chromatin fibre gives rise to the supercoil, which has a diameter of 300 nm.
Level 4: chromatids
The supercoil immediately establishes itself into the chromatid of 700 nm as a result of further coiling around itself. The two chromatids together form the chromosome (about 1400 nm wide).
| Level of organization | Thickness | What it is |
|---|---|---|
| DNA double helix | 2 nm | Bare DNA |
| Nucleosome string | 10 nm | DNA wrapped twice around cores of eight histones every 200 nucleotides ("beads on a string") |
| Chromatin fibre | 30 nm | Nucleosome string coiled again; found in G₁ and G₂ |
| Supercoil | 300 nm | Higher-order coiling of the chromatin fibre at the start of cell division |
| Chromatid | 700 nm | Further coiling of the supercoil; two chromatids make a chromosome (1400 nm) |
13.2DNA as the hereditary materialOften examined
When the chromosomal theory of inheritance was confirmed in 1910, geneticists began to ask in what form the heredity units are found in the chromosome. Chromosomes contain both DNA and protein, so on which of these is the hereditary information written? Over a period of about 30 years, starting in the late 1920s, a series of investigators addressed this question.
Griffith's experiment
In 1928 the British microbiologist Frederick Griffith made a series of unexpected observations while experimenting with Streptococcus pneumoniae. This bacterium is found in two types.
- S-type
- Forms a polysaccharide capsule, so its colony looks smooth and shiny. It is virulent (disease-causing).
- R-type
- Has no polysaccharide capsule, so its colony is rough. It is non-virulent (benign).
- He injected healthy mice with living S-type bacteria: all the mice died.
- He injected similar mice with living R-type bacteria: the mice showed no ill effect. From these two results he made the hypothesis that the virulence of S-type might be associated with the polysaccharide capsule, since R-type lacks a capsule and is non-virulent.
- As a control, he injected heat-killed S-type bacteria into another group of mice, to see if the polysaccharide capsule itself had a virulent effect. The mice remained perfectly healthy.
- As a final control, he blended living R-type with heat-killed S-type cells (both already confirmed as non-virulent) and injected the mixture into healthy mice. Unexpectedly, the mice developed disease symptoms and many of them died.
- The blood of the dead mice contained a large number of living S-type virulent bacteria, which had the surface protein characteristics of the living (previously R) strain.
Griffith concluded that somehow the information specifying the polysaccharide capsule and virulence had passed from the heat-killed S-type bacteria to the living R-type bacteria in the mixture, transforming them into living S-type virulent bacteria that killed the mice. This transfer of genetic material from one organism to another, by which the genetic make-up of the recipient is altered, is called transformation.
Extra reading material: experiment of Avery, Macleod and McCarty
The agent responsible for transforming R-type to S-type was undiscovered until 1944. In a classic series of experiments, Oswald Avery with Colin Macleod and Maclyn McCarty characterized what they called the "transforming principle". They first took the extract (filtrate) of heat-killed S-type bacteria, which contained carbohydrates, lipids, protein, RNA and DNA. They divided this extract into four samples.
| Sample of the extract | Mixed with live R-type cells and injected into mice | Result | Conclusion |
|---|---|---|---|
| Untreated (first sample) | Yes | Transformation occurred | The filtrate of heat-killed S-type bacteria contains the active factor responsible for transformation |
| Treated with proteases (removes all proteins) | Yes | Transformation still occurred | Protein is not the transforming agent |
| Treated with RNase (removes all RNA) | Yes | Transformation still occurred | RNA is not the transforming agent |
| Treated with DNase (removes all DNA) | Yes | No transformation | The active factor must be DNA |
Therefore it was confirmed that the active factor, which caused transformation of live R-type into live S-type in Griffith's experiment, must be DNA.
13.2Hershey and Chase experimentOften examined
Soon after Avery's results, another very convincing experiment on bacteriophages was performed by Alfred Hershey and Martha Chase in 1952.
Bacteriophages are viruses that attack bacteria. Their body consists of DNA and protein. During infection they multiply in the host and many copies of them emerge within 20 to 25 minutes. Until 1952 it was not known whether DNA or protein carries the hereditary information of bacteriophages. Scientists were not even sure whether, during infection, the whole viral particle enters the host or only its DNA or protein gets in. Hershey and Chase set out an experiment for this purpose.
- They labelled the DNA of bacteriophages with a radioactive isotope of phosphorus, (batch 2), and labelled their protein coats with a radioactive isotope of sulphur, (batch 1). DNA contains phosphorus but no sulphur, and protein contains sulphur (in some amino acids) but no phosphorus, so each label marks only one of the two parts.
- The labelled viruses were allowed to infect bacteria.
- Soon after infection, the bacterial cells were separated from the medium (the liquid) by centrifugation. A blender was used to shake the empty protein coats off the surface of the bacteria.
- The medium and the bacterial cells (pellet) were analysed for radioactivity. was found in the bacterial cells, while was found in the medium.
- Conclusion: during infection the -labelled DNA of the bacteriophage was injected into the bacterial cell, while its -labelled protein coat remained outside. Subsequently many viral particles were released outside the host. Hershey and Chase therefore claimed that the virus DNA, not the virus protein, was responsible for directing the production of new viruses.
| Batch | Label and where it is | Found after blending and centrifuging | Meaning |
|---|---|---|---|
| 1 | in the protein coat | In the medium (liquid) | Protein did not enter the bacteria |
| 2 | in the DNA | In the bacterial cells (pellet) | DNA entered the bacteria and directed the making of new viruses |
13.3Models of DNA replicationOften examined
The process of self-synthesis of a DNA molecule is called DNA replication. This process occurs only once, in the S phase of the life cycle of a cell. The DNA molecule that is replicated is called the parent DNA, and the molecules produced in this process are called daughter DNA. One parent DNA molecule gives rise to two daughter DNA molecules.
Models of DNA replication
How can duplex DNA replicate? Scientists tried to answer this after the discovery of the structure of DNA. Three different models were presented to explain the replication process: the semi-conservative, conservative and dispersive models.
- Semi-conservative model
- The two parental DNA strands separate, and each strand serves as a template for the synthesis of a new DNA strand. The result is two DNA double helices, each of which consists of one parental and one new strand.
- Conservative model
- After replication one daughter molecule contains both parental DNA strands, and the other daughter molecule contains DNA strands of all newly synthesized material.
- Dispersive model
- During replication the parental DNA molecule is dispersed (broken) into its nucleotides; these nucleotides combine with new nucleotides to form two new daughter DNA molecules, each a mixture of old and new DNA.
| Point | Semi-conservative | Conservative | Dispersive |
|---|---|---|---|
| Parental strands after replication | Separated, one in each daughter duplex | Stay together in one daughter duplex | Broken up and scattered through both daughters |
| Daughter duplex 1 | One parental + one new strand | Both parental strands | Mixture of parental and new DNA |
| Daughter duplex 2 | One parental + one new strand | Both new strands | Mixture of parental and new DNA |
13.3Meselson and Stahl experimentOften examined
The three models were evaluated by Matthew Meselson and Franklin Stahl of the California Institute of Technology in 1958. They were later awarded a Nobel Prize. From their experiment it was finally concluded that the replication of DNA occurs according to the semi-conservative model.
- They grew bacteria in a medium containing the heavy isotope of nitrogen, , which became incorporated into the bases of the bacterial DNA.
- After several generations the bacteria were shifted to three separate plates, which had been poured with a medium containing the light isotope, .
- Three DNA samples were taken from bacteria after they were shifted: the sample at 0 minutes (just after the transfer), the sample at 20 minutes, and the sample at 40 minutes. In addition, a control sample was taken from bacteria that were grown separately in medium.
- The DNA samples were dissolved in a cesium chloride (CsCl) solution and spun at very high speed in an ultracentrifuge for many hours. The cesium and chloride ions are pushed by the centrifugal force towards the bottom of the tube, so a gradient of and ions is established. DNA molecules settle as bands at the level of their own density.
| Sample | Band in the tube | Reason |
|---|---|---|
| Control ( only) | Lightest: at the top | Both strands contain |
| 0 minutes | Heaviest: at the bottom | Both strands contain |
| 20 minutes (after the first round) | Intermediate level (between control and 0 min) | Each daughter duplex is a hybrid with one strand and one strand |
| 40 minutes (after the second round) | Two bands: one at the top and one at the intermediate level | The hybrid duplex gives a strand to make another hybrid duplex and a strand to make a light duplex containing both strands |
On the basis of these results Meselson and Stahl claimed that DNA replication is semi-conservative.
- Start with one heavy duplex (both strands ) with two parental strands. Semi-conservative replication never destroys a parental strand, so the two strands always stay as two separate strands.
- Every new strand made in the medium is . After generations there are duplexes, and the two strands are in two different duplexes, so exactly 2 duplexes are hybrid.
- The rest are light. Fill in the table below.
| Generation in | Time | Total duplexes | Hybrid duplexes | Light duplexes | Fraction hybrid | Bands in the tube |
|---|---|---|---|---|---|---|
| 0 | 0 min | 1 (all heavy) | 0 | 0 | none (1 heavy band) | heavy |
| 1 | 20 min | 2 | 2 | 0 | all | one hybrid band |
| 2 | 40 min | 4 | 2 | 2 | 1/2 | hybrid and light |
| 3 | 60 min | 8 | 2 | 6 | 1/4 | hybrid (thin) and light (thick) |
| 4 | 80 min | 16 | 2 | 14 | 1/8 | hybrid (thin) and light (very thick) |
After 3 generations: 2 of 8 duplexes (one quarter) are hybrid and 6 of 8 (three quarters) are light. The tube shows two bands, a light band at the top and a hybrid band in the middle, and the light band is three times as thick (dense) as the hybrid band. A heavy band never reappears, because no new is made.
13.3Mechanism of DNA replicationOften examined
Although DNA replication is a continuous process, it is discussed in different phases for convenience: the initiation phase, the extension (polymerization) phase and the termination phase.
Initiation phase
The initiation phase is characterized by the formation of the replication bubble and the replication fork at a particular site called the origin of replication. This is a specific sequence of nucleotides along the length of the DNA from where replication begins. Eukaryotic DNA may have more than one origin of replication, but prokaryotic DNA has only one.
The replication bubble is formed when the enzymes DNA helicase and DNA gyrase work at the origin of replication.
- DNA helicase opens the turns of the DNA duplex by breaking down the base pairs so that the two DNA strands can be separated.
- DNA gyrase works slightly ahead of the helicase and facilitates the unwinding of the DNA duplex by reducing the tension created during unwinding.
- Because of these two enzymes the two strands gradually separate and give a bubble-like appearance at the origin of replication: the replication bubble.
- After the base pairs are broken, the single strands are prevented from pairing up again by specific proteins called single-stranded binding (SSB) proteins.
- Both single strands act as template strands in the next phase and direct the synthesis of the daughter strands along themselves. Each side of the replication bubble is now called a replication fork.
Extension (polymerization) phase
Extension is the formation of the daughter strands (leading or lagging strands) along the template strands. The daughter strands are actually synthesized by the enzyme DNA polymerase. This enzyme cannot work unless some nucleotides are already arranged on the template. For this purpose the enzyme primase arranges some RNA nucleotides on the template strands. Such short fragments of a few RNA nucleotides are called primers; they act as the start site for the activity of DNA polymerase. The primase and DNA helicase enzymes are found together in a complex called the primosome.
After the primers are established, synthesis of the daughter strands begins by DNA polymerase. There are three different forms of this enzyme.
| Enzyme | Function |
|---|---|
| DNA polymerase I | Catalyses the replacement of RNA primers by DNA nucleotides in the termination phase of replication, so it supports DNA polymerase III in the main replication process. |
| DNA polymerase II | Is involved in repairing DNA damage during the life time of a cell. |
| DNA polymerase III | The main enzyme that synthesizes both daughter strands along the template during replication. |
Mechanism of DNA polymerase III activity
DNA polymerase III cannot initiate the replication process. It can add a nucleotide only onto a pre-existing 3′-OH group. Therefore it needs a primer to perform its polymerase activity. It always adds nucleotides at the 3′ end of the primer, so the direction of replication is always 5′ to 3′.
One unit of DNA polymerase III works on one template and continuously synthesizes a daughter strand towards the replication fork. This continuously growing daughter strand is called the leading strand. The second unit works on the other template. It synthesizes the other daughter strand away from the replication fork. As the two units are interlinked, the second unit is allowed to polymerize the daughter strand only up to a specific length (100 to 200 nucleotides in eukaryotes and 1000 to 2000 nucleotides in prokaryotes), and then it has to jump back to a new primer to perform polymerization again. Therefore this daughter strand grows discontinuously, away from the fork, as short fragments interrupted by primers. These fragments are called Okazaki fragments (named after the Japanese scientist who discovered them), and this discontinuously growing strand is the lagging strand.
One subunit of DNA polymerase III also has the ability to remove a wrong nucleotide if it is added by mistake. This ability is called proofreading.
| Point | Leading strand | Lagging strand |
|---|---|---|
| Direction of synthesis | Towards the replication fork | Away from the replication fork |
| Continuity | Synthesized continuously as one long strand | Synthesized discontinuously as short Okazaki fragments |
| Primers needed | One primer at the start | A new primer for every Okazaki fragment |
| Joining of pieces | Not needed (one piece) | Fragments joined by DNA ligase after primers are replaced |
- Each Okazaki fragment is started by one primer, so the number of primers on the lagging strand equals the number of fragments.
- Prokaryote.
- Eukaryote.
- Leading strand: DNA polymerase III makes it continuously towards the fork, so it needs only one primer at the origin side, however long it grows.
- Link to the termination phase: every one of these RNA primers must later be replaced with DNA by DNA polymerase I, and DNA ligase must join the 19 (or 199) gaps between the fragments with phosphodiester bonds.
| Prokaryote | Eukaryote | |
|---|---|---|
| Fragment size (book) | 1000 to 2000 nucleotides | 100 to 200 nucleotides |
| Fragments for 30 000 nucleotides | about 20 | about 200 |
| Primers on the lagging strand | about 20 | about 200 |
| Primers on the leading strand | 1 | 1 |
| Gaps sealed by DNA ligase | about 19 | about 199 |
Termination phase
The termination phase is characterized by the replacement of the RNA primers with DNA nucleotides by DNA polymerase I. The resulting segments are joined by DNA ligase, which forms phosphodiester bonds.
- Initiation: DNA helicase opens the DNA at the origin of replication, breaking the hydrogen bonds between the base pairs, and DNA gyrase reduces the tension ahead of it. A replication bubble forms with a replication fork at each end.
- Single-strand binding (SSB) proteins attach to the separated strands so that they cannot pair up again. Both strands now act as templates.
- Primase makes a short RNA primer on each template strand (the primase and helicase together form the primosome).
- Extension: DNA polymerase III adds nucleotides to the 3′-OH end of the primer, always in the 5′ to 3′ direction. On the leading strand this is continuous towards the fork.
- On the lagging strand polymerase III works away from the fork in short pieces: each Okazaki fragment needs its own primer.
- Proofreading: a subunit of DNA polymerase III removes any wrongly added nucleotide.
- Termination: DNA polymerase I replaces each RNA primer with DNA nucleotides, and DNA ligase joins the fragments by phosphodiester bonds.
- Result: two daughter DNA molecules, each with one parental and one new strand (semi-conservative replication).
| Enzyme or protein | Role in DNA replication |
|---|---|
| DNA helicase | Opens the duplex by breaking the base pairs; forms the replication bubble |
| DNA gyrase | Works ahead of helicase and reduces the tension created during unwinding |
| SSB proteins | Keep the separated single strands from pairing again |
| Primase | Makes short RNA primers on the template strands |
| DNA polymerase III | Main enzyme: adds nucleotides to the 3′ end of a primer, 5′ to 3′; proofreads |
| DNA polymerase I | Replaces RNA primers by DNA nucleotides |
| DNA polymerase II | Repairs DNA damage during the life of the cell |
| DNA ligase | Joins the fragments by phosphodiester bonds |
| Point | DNA helicase | DNA gyrase |
|---|---|---|
| Action | Breaks the base pairs and separates the two strands | Works slightly ahead of helicase and reduces the tension created by unwinding |
| Result | Replication bubble with two forks | Makes unwinding of the duplex easier |
| Phase | Initiation | Initiation |
| Point | Nucleosome | Primosome |
|---|---|---|
| What it is | DNA wrapped twice around a core of eight histones | Complex of primase and DNA helicase enzymes |
| Made of | DNA and histone proteins | Two enzymes (primase + helicase) |
| Role | First level of packing of the chromosome | Starts replication by making the RNA primers |
13.3DNA stability and variability
Replication has to do two opposite jobs. It must copy the base sequence faithfully (stability), and yet occasional errors in copying are the source of new genetic variations (variability). The book describes both as characters of the replicating DNA molecule.
| Point | DNA stability | DNA variability |
|---|---|---|
| Meaning | The DNA molecule remains the same, so DNA sequences are conserved after numerous cell divisions | New genetic variations are produced in the base sequence |
| Cause | DNA polymerase II (repair of damage) and DNA polymerase III (proofreading) correct mistakes | Errors made during DNA replication that are not corrected give mutations in the base sequences |
| Effect | Genetic information is passed on unchanged | Genetic variations; some errors cause breaks in DNA that can lead to cancer |
- DNA stability
- The conservation of DNA sequences through numerous cell divisions, ensured by DNA polymerase II and III.
- DNA variability
- The production of genetic variations by mutations in the base sequence that result from errors during DNA replication.
13.4Gene expressionOften examined
DNA is the genetic material, which can replicate. The genetic material is in the form of specific sequences of nucleotides along the DNA strands. The DNA inherited by an organism forms specific traits by directing the synthesis of a protein that acts as a catalyst and catalyses a specific chemical reaction of the cell. Thus a gene expresses itself in a protein or enzyme that controls the development of a specific character or function. We can say that proteins are the link between the genotype and the phenotype.
Gene expression or protein synthesis includes two processes: transcription and translation.
- Central dogma of molecular genetics
- The idea that DNA makes protein via an intermediate RNA. It involves two processes: transcription (DNA to RNA) and translation (RNA to protein).
- Transcription
- The synthesis of RNA from DNA. It is the first step of gene expression.
- Translation
- The synthesis of a polypeptide (protein) by the ribosome, using the information in the mRNA made by transcription. It is the second step of gene expression.
| Point | Transcription | Translation |
|---|---|---|
| What happens | Synthesis of RNA from DNA | Synthesis of a polypeptide from mRNA |
| Step of gene expression | First step | Second step |
| Where (eukaryote) | Nucleus | Cytoplasm (ribosomes) |
| Enzyme / machinery | One enzyme: RNA polymerase | Ribosomes, tRNA, mRNA and enzymes such as aminoacyl-tRNA synthetase |
| Product | RNA (mRNA, tRNA, rRNA) | Polypeptide chain that folds into a protein |
13.4TranscriptionOften examined
Transcription is the synthesis of RNA from DNA. It is the first step of gene expression. It occurs in the G₀, G₁ and G₂ phases of the cell cycle. It requires only one enzyme, RNA polymerase. It is a continuous process, but for convenience it is divided into three phases: initiation, elongation and termination.
Initiation phase
Transcription begins with the binding of RNA polymerase at the promoter region, a regulatory region of the gene that comprises binding sites for the attachment of RNA polymerase. The names of the sequences (−10, −35, −25, −80, −90) refer to their position in the promoter: about how many nucleotides before the transcription start point of the gene they lie.
| Prokaryotic promoter | Eukaryotic promoter | |
|---|---|---|
| Binding sites | TATAAT (the −10 sequence) and TTGACA (the −35 sequence) | TATA box (TATA, the −25 sequence), CAAT box (CAAT, the −80 sequence) and GC-rich box (the −90 sequence) |
RNA polymerase of prokaryotes consists of four subunits: two alpha (α) and two beta (β) subunits, called the core enzyme, while the sigma factor is required for RNA polymerase to bind on the promoter. The core enzyme and sigma factor together constitute a holoenzyme. Once RNA polymerase is attached to the promoter, the sigma factor is removed and the core enzyme catalyses the remaining process.
- Like DNA polymerase, RNA polymerase adds nucleotides to the 3′ end of the growing chain, but unlike DNA polymerase it does not require a primer to start polymerization.
- In prokaryotes only one type of RNA polymerase is found.
- In eukaryotes there are three types: RNA polymerase I synthesizes rRNA, RNA polymerase II synthesizes mRNA, and RNA polymerase III synthesizes tRNA.
As RNA polymerase binds to the promoter, the DNA duplex unwinds, base pairs are broken and a bubble-like structure, the transcription bubble, appears.
Elongation phase
As RNA polymerase binds to the promoter, it begins to arrange and polymerize ribonucleoside triphosphates (rNTP, ribonucleotides) complementary to the template strand of the DNA. It does not need a primer to start. One of the two strands of the gene acts as the template.
- Template strand (antisense strand)
- The strand of the gene copied by RNA polymerase. It is called antisense because the mRNA is complementary to it.
- Coding strand (sense strand)
- The other strand of the gene. Its sequence is the same as the mRNA (with U in place of T).
In the elongation phase RNA polymerase keeps moving from the 5′ to the 3′ end of the RNA, towards the terminator region of the gene. The transcription bubble also moves along the DNA, leaving the growing RNA strand protruding from the bubble. This continues until the RNA polymerase reaches the terminator region of the gene.
Termination phase
The sequence of the terminator region of the gene stops the synthesis of mRNA. The terminator region consists of a series of GC base pairs followed by a series of AT base pairs. The part of the mRNA transcribed in this region projects to form a loop-like structure called the GC hairpin, followed by a small tail of poly-U nucleotides. The GC hairpin causes the RNA polymerase to stop the synthesis of RNA.
- RNA polymerase (core enzyme + sigma factor) binds the promoter (−35 and −10 sequences in prokaryotes).
- The DNA unwinds and a transcription bubble forms; the sigma factor leaves.
- The polymerase moves 5′ to 3′ along the template strand, adding complementary ribonucleotides, with no primer.
- At the terminator the RNA forms a GC hairpin followed by poly-U, and the polymerase stops.
- The RNA is released; the sigma factor rebinds the core enzyme.
| Point | DNA polymerase | RNA polymerase |
|---|---|---|
| Product | DNA (daughter strands) | RNA |
| Primer | Needed (RNA primer with a 3′-OH) | Not needed |
| Direction | Adds nucleotides to the 3′ end (5′ to 3′) | Adds nucleotides to the 3′ end (5′ to 3′) |
| Number of enzymes | Three forms (I, II, III) | One in prokaryotes; three in eukaryotes (I rRNA, II mRNA, III tRNA) |
13.4Post-transcriptional modification of mRNAOften examined
In prokaryotes there is no delay between transcription and translation: the mRNA emerges from the DNA and is translated at once, because there is no definite nucleus. In eukaryotic cells the pre-mRNA (newly synthesized mRNA) has to be modified into mature or functional mRNA, because transcription occurs in the nucleus while translation occurs in the cytoplasm. The mRNA has to move from the nucleus to the cytoplasm, and during this journey enzymes such as phosphatases and nucleases may degrade it. Eukaryotic mRNA also contains some non-coding regions called introns, which are removed (spliced) during this process.
Post-transcriptional modification therefore involves two events: addition of a cap and tail to protect the mRNA from degradation, and RNA splicing to remove non-coding sequences.
- Cap
- 7-methyl GTP linked from its 5′ to the 5′ end of the mRNA.
- Poly-A tail
- A small chain of 30 to 500 adenine nucleotides attached to the 3′ end of the mRNA. The cap and tail prevent the mRNA from being degraded by phosphatases and nucleases.
- RNA splicing
- The removal of introns and maturation of the primary mRNA to secondary (functional) mRNA. It is catalysed by spliceosomes, a large RNA-protein complex. The exon fragments are then joined with the help of RNA ligase enzyme.
- Exon
- A coding segment that stays in the mature mRNA.
- Intron
- A non-coding segment that is removed during splicing.
| Point | Intron | Exon |
|---|---|---|
| Nature | Non-coding region of the gene and primary mRNA | Coding region of the gene and mRNA |
| Fate in processing | Removed (excised) by spliceosomes | Kept and joined together by RNA ligase |
| In mature mRNA | Absent | Present |
- Transcription makes the primary mRNA (pre-mRNA) with exons and introns.
- A cap is added to the 5′ end and a poly-A tail to the 3′ end for protection.
- Spliceosomes cut out the introns.
- RNA ligase joins the exons to make mature mRNA.
- Mature mRNA leaves the nucleus through a nuclear pore for translation.
13.4Genetic codeOften examined
A gene is a specific segment of DNA that encodes the synthesis of a particular polypeptide (protein). Therefore the order of amino acids in a polypeptide is according to the sequence of nucleotides in a part of DNA. This relationship between the amino acid sequence and the nucleotide sequence is called the genetic code. In other words, the information in DNA for the synthesis of protein, which is transcribed (copied) into mRNA, is called the genetic code.
The mRNA transcript consists of a random sequence of only four kinds of nitrogenous bases (A, G, C and U). How can a code with a four-letter alphabet specify a protein that contains one of twenty different amino acids at any point? A single base cannot specify a single amino acid: only four amino acids could be encoded, and proteins with only four different amino acids would be formed. Two bases are also not feasible, since only 16 amino acids could be encoded (). Three bases are sufficient: a total of combinations are possible, and each is specific for a particular amino acid. Therefore a triplet of bases along the length of mRNA that specifies a particular amino acid is called a codon.
| First base | Third base | Second base U | Second base C | Second base A | Second base G |
|---|---|---|---|---|---|
| U | U | UUU Phenylalanine | UCU Serine | UAU Tyrosine | UGU Cysteine |
| U | C | UUC Phenylalanine | UCC Serine | UAC Tyrosine | UGC Cysteine |
| U | A | UUA Leucine | UCA Serine | UAA Stop | UGA Stop |
| U | G | UUG Leucine | UCG Serine | UAG Stop | UGG Tryptophan |
| C | U | CUU Leucine | CCU Proline | CAU Histidine | CGU Arginine |
| C | C | CUC Leucine | CCC Proline | CAC Histidine | CGC Arginine |
| C | A | CUA Leucine | CCA Proline | CAA Glutamine | CGA Arginine |
| C | G | CUG Leucine | CCG Proline | CAG Glutamine | CGG Arginine |
| A | U | AUU Isoleucine | ACU Threonine | AAU Asparagine | AGU Serine |
| A | C | AUC Isoleucine | ACC Threonine | AAC Asparagine | AGC Serine |
| A | A | AUA Isoleucine | ACA Threonine | AAA Lysine | AGA Arginine |
| A | G | AUG Methionine (start) | ACG Threonine | AAG Lysine | AGG Arginine |
| G | U | GUU Valine | GCU Alanine | GAU Aspartate | GGU Glycine |
| G | C | GUC Valine | GCC Alanine | GAC Aspartate | GGC Glycine |
| G | A | GUA Valine | GCA Alanine | GAA Glutamate | GGA Glycine |
| G | G | GUG Valine | GCG Alanine | GAG Glutamate | GGG Glycine |
- There are 64 codons in total.
- Three of them, UGA, UAG and UAA, are stop codons. One of these must be present at the end of the mRNA to show that the message is over. They do not encode any amino acid, so they are called nonsense codons.
- All the other codons encode specific amino acids and are called sense codons.
- One sense codon, AUG (methionine), also acts as the start codon, which must be present at the beginning of every sequence that codes for an amino acid chain.
Characteristics of the genetic code
- Degenerate. Some amino acids are encoded by only a single codon, while others are encoded by up to four codons, and the amino acids leucine and serine are encoded by six codons. This property is known as code degeneracy.
- Universal. The code is the same in almost all organisms. For example, AGA specifies arginine in bacteria, in humans and in all other organisms whose genetic code has been studied. Because of this universality, genes can be transferred from one organism to another and be successfully transcribed and translated in their new host. The study of the genetic code of mitochondrial DNA, however, showed that the code is not universal: UGA, normally a stop codon, reads as tryptophan in mitochondria; AUA is read as methionine instead of isoleucine; and AGA and AGG are for termination of protein synthesis instead of arginine.
- No punctuation. There is no punctuation between codons: there is no gap between two codons.
- Non-overlapping. The mRNA sequence AUGAGCGCA is not read as AUG / UGA / GAG and so on; it is read as AUG / AGC / GCA.
- Triplet. Each codon is made of three bases.
- The code is a triplet and non-overlapping, so the number of codons is the length divided by 3.
- The last codon is a stop codon (UGA, UAG or UAA), which codes for no amino acid, so it does not add an amino acid.
- Each nucleotide of the mRNA is copied from one base pair of the DNA, so 900 base pairs of the gene are transcribed for this coding region.
- Number of codons: there are four bases, so different codons. One base could code for only amino acids and two bases for only , fewer than the 20 amino acids in proteins, so a triplet is needed (64 is enough, with the extra codons giving degeneracy).
| Length of code word | Combinations | Enough for 20 amino acids? |
|---|---|---|
| 1 base | 4 | No |
| 2 bases | 16 | No |
| 3 bases (triplet) | 64 | Yes |
| Point | Genetic code | Codon |
|---|---|---|
| Meaning | The relationship between the sequence of nucleotides in DNA (copied to mRNA) and the sequence of amino acids in a protein | A triplet of bases in mRNA that specifies one amino acid (or a stop signal) |
| Size | The whole set of rules (64 triplets for 20 amino acids and stops) | Three nucleotides |
| Example | Degenerate, universal, non-overlapping, triplet | AUG codes for methionine and starts translation |
| Point | Start codon | Stop codon | Sense codon | Nonsense codon |
|---|---|---|---|---|
| Codon(s) | AUG | UAA, UAG, UGA | All 61 codons that code for amino acids (AUG included) | The three stop codons |
| Codes for | Methionine; begins the chain | No amino acid | A specific amino acid | No amino acid |
| Position | At the beginning of the message | At the end of the message | Anywhere inside the message | At the end |
13.4TranslationOften examined
Translation is the second step of gene expression. In translation, the messenger RNA (mRNA) produced by transcription is decoded by the ribosome to produce a specific amino acid chain, or polypeptide, that will later fold into an active protein.
Translation is a continuous process, but for convenience it is discussed in four phases: activation of amino acids, formation of the initiation complex, elongation and termination.
Activation of amino acids
Activation of amino acids is the binding of free amino acids dispersed in the cytoplasm to the 3′ end of particular tRNA molecules. In this way a complex called the aminoacyl-tRNA complex is formed. The binding is catalysed by the enzyme aminoacyl-tRNA synthetase (the activation enzyme). The various amino acids that are to take part in polypeptide formation are continuously activated throughout translation.
Formation of the initiation complex
The process of translation actually begins with the formation of the initiation complex, which is formed by the combination of the ribosomal subunits, the mRNA and the first aminoacyl-tRNA complex.
- A tRNA carrying a chemically modified methionine (N-formyl methionine) binds to the smaller ribosomal subunit. This binding is controlled by an enzyme called an initiation factor.
- At the same time the 5′ end of the mRNA molecule binds to the smaller subunit of the ribosome with the help of another initiation factor.
- The initiation complex is completed when the larger ribosomal subunit is placed upon the smaller subunit.
- P site (peptidyl site)
- The region of the ribosome where the first aminoacyl-tRNA complex is attached; peptide bonds are formed here between successive amino acids during elongation.
- A site (aminocyl site)
- The site where successive tRNAs bearing amino acids are attached.
- E site (exit site)
- The site where empty tRNAs leave the ribosome during elongation.
Polypeptide elongation
In this phase the ribosomal units move along the mRNA, amino acids are brought by tRNAs, and they are joined together to form a polypeptide chain. This is accomplished by three steps, which are repeated again and again throughout this phase.
- Whichever codon of mRNA is exposed at the A site, the aminoacyl-tRNA complex bearing the complementary anticodon binds to it with the help of an enzyme, the elongation factor.
- The enzyme peptidyl transferase (from the P site) removes the amino acid (which may be a chain) from the tRNA at the P site and binds it to the newly arrived amino acid with a peptide bond.
- Translocation: the ribosomal subunits move slightly along the mRNA from 5′ to 3′ so that a new codon is exposed. As a result the empty tRNA reaches the E site to leave the ribosome, while the other tRNA bearing the chain of amino acids is shifted from the A site to the P site, and another codon is exposed at the A site.
Termination
Elongation continues until a chain-terminating stop codon (UAG, UAA or UGA) is exposed at the A site. Stop codons do not bind to any tRNA, but they are recognized by release factors that terminate translation. The polypeptide is released from the tRNA, the tRNA is released from the ribosome, and the ribosomal subunits separate from the mRNA.
- Transcription: RNA polymerase builds mRNA 5′ to 3′ complementary to the template (A pairs with U, T with A, C with G, G with C).\text{template }3'\text{-TAC GCA CTT GGT ATT-}5'\ \to\ \text{mRNA }5'\text{-AUG CGU GAA CCA UAA-}3'
- Translation: each tRNA anticodon is complementary to its codon.\text{anticodons: UAC, GCA, CUU, GGU (no tRNA for UAA)}
- Read the codons in the genetic code table: AUG = methionine (start codon), CGU = arginine, GAA = glutamate, CCA = proline, UAA = stop codon.
| Codon 1 | Codon 2 | Codon 3 | Codon 4 | Codon 5 | |
|---|---|---|---|---|---|
| Template (3′ to 5′) | TAC | GCA | CTT | GGT | ATT |
| mRNA codon (5′ to 3′) | AUG | CGU | GAA | CCA | UAA |
| tRNA anticodon | UAC | GCA | CUU | GGU | none (release factor) |
| Amino acid | Methionine | Arginine | Glutamate | Proline | Stop |
- Activation: each amino acid is attached to its own tRNA by aminoacyl-tRNA synthetase.
- Initiation: initiator tRNA (methionine) and the 5′ end of mRNA bind the small subunit; the large subunit joins.
- Elongation: codon recognition at the A site, peptide bond by peptidyl transferase, translocation.
- Termination: a stop codon at the A site, release factors free the polypeptide, subunits separate.
Difference between translation in prokaryotes and eukaryotes
| Point | Prokaryotes | Eukaryotes |
|---|---|---|
| Ribosomes | Smaller: a 50S large subunit and a 30S small subunit bind to form the 70S ribosome | Larger: a 60S large subunit and a 40S small subunit bind to form the 80S ribosome |
| Where ribosomes occur | In the cytosol | In the cytosol and on the rough endoplasmic reticulum |
| Initiating amino acid | Modified: N-formyl methionine | Methionine, not modified |
| Initiation factors | Only one type is used | More than one type is used |
| Release factors | Two release factors are used | A single release factor is used |
13.4Fate of synthesized protein
After termination of translation, the polypeptide chain folds and forms a three-dimensional protein shape. It then enters the Golgi complex for further modifications. The protein can be used within or outside the cell that synthesized it.
| Use of the protein | Examples given in the book |
|---|---|
| Within the cell | Cellular enzymes, membrane proteins, histone proteins |
| Outside the cell (secreted) | Digestive enzymes, protein hormones, albumins, fibrinogens, antibodies |
- The polypeptide folds into a three-dimensional shape.
- It enters the Golgi complex for further modification.
- It is used inside the cell (enzyme, membrane protein, histone) or is secreted from the cell (digestive enzymes, hormones, albumins, fibrinogens, antibodies).
13.5Regulating gene expression
The overall process by which genetic information flows from genes to proteins, that is, from genotype to phenotype, is called gene expression. Regulation of gene expression is the control of the amount and timing of the appearance of a gene's product. All the cells of a living body must continually turn genes ON and OFF in response to signals from their external and internal environment.
The mature human body is composed of about 200 different cell types. The differences between cell types are due not to different genes being present, but to differential gene expression: the expression of different genes by cells with the same genome.
Importance of regulation of gene expression
- It is essential for both prokaryotes and eukaryotes, as it increases the versatility and adaptability of an organism by allowing the cell to express a protein only when it is needed.
- All the cells of a living body must regulate which genes are expressed at any given time, turning genes ON and OFF in response to signals from the external and internal environment.
- It gives the cell control over all structure and function, and is the basis for cellular differentiation and morphogenesis of any organism.
- When the regulation of gene expression is disturbed, serious imbalances and diseases, including cancer, can arise.
13.5Methods of regulation of gene expressionOften examined
There are two possible ways of regulating gene expression: positive regulation and negative regulation. Some genes are controlled by positive regulation and others by negative regulation.
- Positive gene regulation
- The expression of genes is quantitatively increased by the presence of a specific regulatory protein, the activator. When the activator is attached with RNA polymerase, it starts the process of transcription.
- Negative gene regulation
- The expression of genes is diminished by the presence of a specific regulatory protein, the repressor. The repressor protein attaches to the operator, the DNA site next to the promoter, and so stops RNA polymerase from transcribing the gene.
| Point | Positive control | Negative control |
|---|---|---|
| Regulatory protein | Activator | Repressor |
| Effect on gene expression | Increases it: triggers transcription | Diminishes it: shuts transcription down |
| Without the regulatory protein | Little or no transcription (RNA polymerase does not start) | Transcription goes on (nothing blocks RNA polymerase) |
| With the regulatory protein | Transcription takes place | No transcription (RNA polymerase is stopped) |
13.5Lac operon
A classic example of gene regulation in bacteria is the lac operon, which demonstrates both positive and negative regulation. The lac operon is a group of genes responsible for the breakdown of lactose, a sugar, into energy. In bacteria, genes are often organized into operons, which are clusters that encode proteins necessary for coordinated functions. The RNA transcribed from these operons is polycistronic, meaning it can encode multiple proteins within a single transcript.
The lac operon consists of a regulatory gene (the i gene) and three structural genes (z, y and a).
| Gene | Product | Role |
|---|---|---|
| i (regulatory gene) | Repressor protein | When lactose is absent, the repressor binds to the operon and prevents production of the enzymes needed for lactose breakdown |
| z | β-galactosidase | Hydrolyses lactose into glucose and galactose |
| y | Permease | Increases the cell's permeability to β-galactosides, facilitating lactose entry into the cell |
| a | Transacetylase | Transfers acetyl groups to β-galactosides; its exact role in the lac operon is still not known |
This operon shows how cells manage resources efficiently by producing proteins only when necessary, such as in the presence of lactose.
- No lactose: the i gene makes the repressor mRNA and the repressor protein. The repressor binds the operator (o) and prevents RNA polymerase from transcribing the operon, so no enzymes are made.
- Lactose present: lactose (the inducer) binds the repressor and makes it inactive. The operator is free, RNA polymerase transcribes the genes into one lac mRNA.
- The lac mRNA is translated into the three enzymes: β-galactosidase, permease and transacetylase.
- Recall the rule: the repressor (from the i gene) binds the operator and blocks RNA polymerase; lactose (the inducer) binds the repressor and makes it inactive, so transcription of z, y and a can start.
- Normal cell: no lactose, the repressor sits on the operator, no enzymes. With lactose the repressor is inactivated, the genes are transcribed and the enzymes are made.
- Damaged i gene: there is no working repressor to block the operator, so the genes are transcribed whether or not lactose is present. The enzymes are made even when lactose is absent.
- Changed operator: the repressor cannot bind, so again RNA polymerase is never blocked and the genes are transcribed all the time.
| Cell | Lactose absent | Lactose present |
|---|---|---|
| Normal | Genes off: no enzymes | Genes on: enzymes made |
| i gene damaged (no working repressor) | Genes on | Genes on |
| Operator cannot bind repressor | Genes on | Genes on |
13.6MutationsOften examined
A gene mutation is a permanent change in the DNA sequence that makes up a new allele in the population. Mutations range in size from a change in a single DNA nucleotide, to a large segment of a chromosome or a whole chromosome, or sometimes changes in the number of chromosomes.
- Mutation
- A permanent change in the DNA sequence (it may range from one nucleotide to a whole chromosome or a change in chromosome number).
- Mutagen
- An agent that causes mutations.
- Mutant
- The organism or cell in which a mutation occurs.
- Mutagenesis
- The creation or origin of a mutation.
Sources of mutagens
The sources of mutation may be physical, chemical and biological agents.
| Kind of mutagen | Examples given in the book |
|---|---|
| Physical mutagens | Radiation of any kind: UV, gamma rays, X-rays, radioactive rays; high temperature alterations |
| Chemical mutagens | Colchicine, mustard gas, nitrous gas, acridine orange, reactive radioactive isotopes, free oxygen particles |
| Biological mutagens | Certain viruses, transposons, and errors that occur during meiosis or DNA replication |
Types of mutation
Creation or origin of a mutation is called mutagenesis. Based upon mutagenesis there are two types of mutations. On the basis of where the mutations occur and to what extent, there are two further types.
| Point | Natural mutations | Induced mutations |
|---|---|---|
| How they arise | Occur naturally and automatically, due to internal or external factors | Produced artificially, for the establishment of new varieties of organisms |
| Causes | Errors that occur during meiosis or DNA replication | Physical and chemical mutagens are mainly responsible |
| Control | Not planned by people | Deliberately produced |
| Point | Point mutation | Chromosomal mutation (aberration) |
|---|---|---|
| What changes | A single or a few nucleotides in the DNA | The structure or the number of chromosomes |
| Scale | Small (gene level) | Large (whole chromosome level) |
| Examples in the book | Sickle cell anaemia, phenylketonuria | Down syndrome, Klinefelter syndrome, Turner syndrome |
| Point | Mutagen | Mutant |
|---|---|---|
| Meaning | The agent that causes a mutation | The organism or cell in which the mutation has occurred |
| Examples | UV, X-rays, mustard gas, certain viruses | A bacterium or a person carrying a changed allele |
13.6Importance of mutation
A mutation may be harmful or useful.
Harmful aspects of mutation
A mutation that decreases the fitness of the organism in its environment is called a harmful mutation. Examples given in the book:
- A person born without a part of the brain.
- Cancer, which is a form of harmful mutation.
- Developmental abnormalities such as microcephaly, cleft palate, and cases of abnormal number of chromosomes (Down syndrome, Klinefelter syndrome, Turner syndrome and so on).
- Hereditary disorders such as sickle cell anaemia and phenylketonuria.
Useful aspects of mutation
Mutations are also considered as contributing factors towards evolution. Changes in the genes controlling development can have major effects on the morphology of the adult organism. Because these effects are so significant, scientists suspect that changes in developmental genes have helped bring about large-scale evolutionary transformations. Developmental changes may help to explain, for example, how some hoofed mammals evolved into ocean-dwellers, how water plants invaded the land, and how small armoured invertebrates evolved wings.
Mutations are considered the driving force of evolution, where less favourable (or deleterious) mutations are removed from the gene pool by natural selection, while more favourable (beneficial or advantageous) ones tend to accumulate.
| Point | Harmful mutation | Useful (beneficial) mutation |
|---|---|---|
| Effect on fitness | Decreases the fitness of the organism in its environment | Favourable: tends to accumulate in the gene pool |
| Fate in the population | Removed from the gene pool by natural selection | Retained and accumulated by natural selection |
| Examples | Cancer, Down syndrome, sickle cell anaemia, phenylketonuria | Changes in developmental genes that helped bring about large evolutionary transformations (for example, ocean-dwelling hoofed mammals, plants on land, winged invertebrates) |
Critical thinking: how harmful mutations are eradicated by nature
A harmful mutation is a mutation that decreases the fitness of the organism. The eradication of harmful mutations is done by natural selection. The process of DNA repair is an important way in which the body protects itself from disease. Every harmful mutation is eliminated from the population by premature death or reduced reproductive success, a "genetic death".
13.6Chromosomal mutationOften examined
Changes in the structure or number of chromosomes are called chromosomal mutations or chromosomal aberrations. Here the symptoms, causes and possible treatments of three of them are discussed: Down syndrome, Klinefelter syndrome and Turner syndrome.
Down syndrome
- Chromosomes: 2n+1; an extra copy of chromosome 21 (trisomy 21), so 47 chromosomes.
- Symptoms: abnormally shaped, smaller head; rounded inner corner of the eyes instead of pointed; excess skin at the back of the neck; flattened nose; single crease in the palm of the hand; small ears; small mouth; wide, short hands with short fingers; white spots on the coloured part of the eye; excessive space between the large toe and the second toe. The affected people may be male or female.
- Cause: autosomal non-disjunction, in which the 21st chromosomal pair fails to segregate properly during maternal meiosis and results in the formation of an egg having 24 chromosomes. Fertilization of such an egg by a normal sperm containing 23 chromosomes produces a child with 47 chromosomes (2n+1) with three copies of chromosome 21, so it is also called trisomy 21.
- Treatment: there is no treatment for Down syndrome, as it is a chromosomal abnormality. It can be managed to some extent by regular check-ups and screening, medications, surgery, and counselling and support.
Klinefelter syndrome
- Chromosomes: 2n+1 (44+XXY); an extra X chromosome. The patients show trisomy in sex chromosomes.
- Symptoms: several feminine characters although they are males: sparse body hair, enlarged breasts, wide hips, and testicles remain small. The voice may not be as deep. Abnormal body proportions (long legs, short trunk, shoulders equal to hip size).
- Cause: non-disjunction of the sex chromosome pair during maternal meiosis.
- Treatment: testosterone can be given for sexual development, like males, around the age of puberty; it can help a boy have more normal body development. The treatment usually continues throughout a man's life but does not help infertility.
Turner syndrome
- Chromosomes: 2n-1 (44+XO = 45); one of the X chromosomes is absent. The affected individuals are females.
- Symptoms: shorter than average height, infertility, a webbed neck, a low hairline at the back of the neck, abnormal bone development (especially of the bones of the hands and elbows), a larger than usual number of moles on the skin, and oedema (extra fluid) in the hands and feet.
- Cause: non-disjunction of the sex chromosome pair during maternal meiosis.
- Treatment: there is no specific treatment, as it is a chromosomal abnormality. Growth hormone treatment can improve growth and influence a girl's final adult height, especially if started early in childhood. Estrogen replacement treatment helps girls develop the physical changes of puberty, including breast development and menstrual periods; it is often started when a girl reaches about age 12 or 13.
| Point | Down syndrome | Klinefelter syndrome | Turner syndrome |
|---|---|---|---|
| Chromosome constitution | 2n+1, trisomy 21 (47) | 2n+1, 44+XXY (47) | 2n-1, 44+XO (45) |
| Sex of the affected person | Male or female | Male | Female |
| Chromosome involved | Autosome (21) | Sex chromosomes (extra X) | Sex chromosome (X missing) |
| Key symptoms | Small head, rounded inner corner of eyes, single palm crease, flat nose | Feminine characters, enlarged breasts, small testicles, long legs | Short height, infertility, webbed neck, low hairline |
| Cause | Non-disjunction of the 21st pair in maternal meiosis | Non-disjunction of sex chromosomes in maternal meiosis | Non-disjunction of sex chromosomes in maternal meiosis |
| Treatment | None; managed by check-ups, medication, surgery, counselling | Testosterone treatment around puberty (does not help infertility) | Growth hormone; estrogen replacement |
13.6Gene or point mutationsOften examined
A gene or point mutation arises as a result of a base change in an individual gene. It may include base substitution (replacement of one base by another), insertion (addition of one or more bases) and deletion (removal of one or more bases). Frame shift mutations occur when one or more nucleotides are either inserted or deleted from DNA. This results in a completely new sequence of codons and a non-functional protein.
| Type | What happens | Example from Fig. 13.30 |
|---|---|---|
| Base substitution | One base is replaced by another | ATTAGCCTTAGAACT becomes ATTAGCCATAGAACT |
| Insertion | One or more bases are added | ATTAGCCTTAGAACT becomes ATTAGCCATTAGAACT |
| Deletion | One or more bases are removed | ATTAGCCTTAGAACT becomes ATTAGCCTAGAACT |
- Read the original in codons.\text{AUG GAA CCC UUU AAA}\ \to\ \text{Met - Glu - Pro - Phe - Lys}
- (a) GAA becomes GAG, which still codes for glutamate (the code is degenerate). The protein is unchanged.
- (b) GAA becomes GUA, which codes for valine. One amino acid changes (Glu to Val), as in sickle cell anaemia.
- (c) Deleting G shifts the reading frame: AUG AAC CCU UUA AA... gives Met - Asn - Pro - Leu and every codon after the deletion changes (frame shift).
- (d) Inserting U gives AUG UGA ACC...: the new second codon UGA is a stop codon, so the chain ends after one amino acid (a frame shift).
| Change | New mRNA | Polypeptide | Effect |
|---|---|---|---|
| None | AUG GAA CCC UUU AAA | Met - Glu - Pro - Phe - Lys | Normal |
| (a) substitution GAA to GAG | AUG GAG CCC UUU AAA | Met - Glu - Pro - Phe - Lys | No change in the protein |
| (b) substitution GAA to GUA | AUG GUA CCC UUU AAA | Met - Val - Pro - Phe - Lys | One amino acid changed |
| (c) deletion of G | AUG AAC CCU UUA AA | Met - Asn - Pro - Leu ... | Frame shift: new sequence |
| (d) insertion of U | AUG UGA ACC CUU UAA A | Met - stop | Frame shift: chain ends early |
Sickle cell anaemia
Sickle cell anaemia is also known as haemoglobin SS disease (Hb SS) or sickle cell disease.
- Symptoms: painful episodes which can last from hours to days. When the anaemia becomes more severe, symptoms may include fatigue, paleness, fever and rapid heart rate, shortness of breath, jaundice and damage to various organs.
- Cause: a defect in haemoglobin. The sixth amino acid, glutamic acid, in the beta chain of normal haemoglobin is replaced by valine. In the beta chain gene the glutamic acid code (CTT) is substituted by CAT, which codes for valine. The abnormal sickle cell haemoglobin is called Hb^{\mathrm{S}}; it has lost the oxygen-binding capacity, so the red blood cells become sickle shaped and deliver less oxygen to the body's tissues. Sickle cell anaemia is due to a recessive gene: one gets the disease if it is inherited from both parents.
- Treatment: blood transfusions, medicines to relieve pain and plenty of fluids. Bone marrow or stem cell transplants can cure sickle cell anaemia. Folic acid supplements should be taken.
Phenylketonuria (PKU)
A rare condition in which a baby is born without the ability to properly break down an amino acid called phenylalanine is called phenylketonuria.
- Symptoms: lighter skin, hair and eyes; delayed mental and social skills; head size significantly below normal; jerking movements of the arms or legs; mental retardation; seizures and skin rashes.
- Cause: in a healthy baby, phenylalanine is converted into tyrosine by the enzyme phenylalanine hydroxylase. Due to a point mutation, this enzyme becomes defective and converts phenylalanine into toxic phenylketones that accumulate and damage the central nervous system. It is also due to an autosomal recessive gene, and one can get the disease if it is inherited from both parents.
- Treatment: a diet that is extremely low in phenylalanine, particularly when the child is growing. A special infant formula called Lofenalac is made for infants with PKU; it can be used throughout life as a protein source that is extremely low in phenylalanine.
| Point | Sickle cell anaemia | Phenylketonuria |
|---|---|---|
| Type of mutation | Gene (point) mutation: glutamic acid replaced by valine in the beta chain of haemoglobin | Gene (point) mutation: defective phenylalanine hydroxylase enzyme |
| Inheritance | Autosomal recessive | Autosomal recessive |
| Main effect | Sickle-shaped red cells with less oxygen-carrying capacity | Phenylalanine is not converted to tyrosine; toxic phenylketones damage the nervous system |
| Treatment | Blood transfusions, pain medicines, fluids, folic acid; bone marrow or stem cell transplant can cure | Diet extremely low in phenylalanine (Lofenalac formula) |
Key terms
- Chromatid
- One of the two strands of a chromosome; one long, highly coiled DNA molecule with histone proteins.
- Centromere (primary constriction)
- The point where the two chromatids are attached; it divides the chromosome into two arms.
- Secondary constriction (nucleolar organizer)
- A second point of union along the chromatids in some chromosomes; it gives rise to nucleoli during interphase.
- Telomeres
- The terminal ends of chromosomes, which prevent two chromosomes from attaching end to end.
- Metacentric, submetacentric, acrocentric, telocentric
- Chromosome types with the centromere in the centre, slightly away from the centre, near the end and at the end.
- Histone proteins
- Positively charged proteins (rich in arginine and lysine) that bind the negatively charged DNA.
- Nucleosome
- DNA of about 200 nucleotides wrapped twice around a core of eight histones.
- Nucleosome string
- The 10 nm chain of nucleosomes, 'beads on a string', joined by linker DNA.
- Chromatin fibre
- The 30 nm fibre formed by further coiling of the nucleosome string; shows heterochromatin and euchromatin.
- Heterochromatin
- The highly condensed, unexpressed region of the chromatin fibre.
- Euchromatin
- The non-condensed region of the chromatin fibre whose genes are expressed.
- Supercoil
- The 300 nm coil formed from the chromatin fibre at the start of cell division.
- Transformation
- The transfer of genetic material from one organism to another, which alters the genetic make-up of the recipient.
- DNA replication
- The self-synthesis of a DNA molecule, which occurs once in the S phase.
- Semi-conservative model
- Each daughter DNA has one parental and one new strand.
- Conservative model
- One daughter DNA has both parental strands and the other has both new strands.
- Dispersive model
- The parental DNA is dispersed into nucleotides that combine with new ones in both daughters.
- Origin of replication
- The specific nucleotide sequence from where replication begins.
- Replication bubble and replication fork
- The bubble-like opening of DNA at the origin and each side of it where unwinding proceeds.
- DNA helicase
- The enzyme that opens the DNA duplex by breaking the base pairs.
- DNA gyrase
- The enzyme that works ahead of helicase and reduces the tension created by unwinding.
- Single-stranded binding (SSB) proteins
- Proteins that stop the separated DNA strands from pairing again.
- Primase and primer
- The enzyme that arranges short RNA nucleotides (primers) on the template as the start site for DNA polymerase.
- Primosome
- The complex of primase and DNA helicase.
- DNA polymerase I, II and III
- I replaces RNA primers by DNA; II repairs DNA damage; III is the main enzyme that makes both daughter strands.
- Leading strand
- The daughter strand made continuously towards the replication fork.
- Lagging strand and Okazaki fragments
- The daughter strand made discontinuously, away from the fork, as short Okazaki fragments.
- Proofreading
- The ability of a DNA polymerase III subunit to remove a wrongly added nucleotide.
- DNA ligase
- The enzyme that joins DNA segments by phosphodiester bonds.
- Central dogma of molecular genetics
- DNA makes protein through an intermediate RNA by transcription and translation.
- Gene expression
- The flow of genetic information from genes to proteins, genotype to phenotype.
- Transcription
- The synthesis of RNA from DNA; the first step of gene expression.
- Promoter region
- The regulatory part of a gene with the binding sites for RNA polymerase.
- RNA polymerase (core enzyme, holoenzyme)
- The enzyme of transcription; the core enzyme plus the sigma factor is the holoenzyme.
- Transcription bubble
- The bubble formed where the DNA unwinds as RNA polymerase binds.
- Template (antisense) strand and coding (sense) strand
- The strand copied into mRNA and the other strand of the gene.
- Terminator and GC hairpin
- The gene region that stops transcription and the GC-rich loop in the RNA that makes the polymerase stop.
- Introns and exons
- Non-coding parts removed from the pre-mRNA and the coding parts that are joined.
- Cap and poly-A tail
- The 7-methyl GTP cap at the 5′ end and the 30 to 500 adenine tail at the 3′ end that protect mRNA.
- RNA splicing and spliceosome
- The removal of introns and joining of exons, catalysed by the RNA-protein spliceosome.
- Genetic code
- The relationship between the nucleotide sequence of DNA (mRNA) and the amino acid sequence of a protein.
- Codon
- A triplet of mRNA bases that specifies one amino acid.
- Start codon, stop codon
- AUG (methionine) begins a chain; UGA, UAG and UAA end it.
- Sense and nonsense codons
- Codons that code for an amino acid and the three stop codons that do not.
- Degeneracy of the code
- One amino acid can be coded by more than one codon.
- Translation
- The decoding of mRNA by the ribosome into a polypeptide; the second step of gene expression.
- Aminoacyl-tRNA complex
- An amino acid bound to the 3′ end of its tRNA by aminoacyl-tRNA synthetase.
- Initiation complex
- Small subunit, mRNA and first aminoacyl-tRNA, completed by the large subunit.
- P, A and E sites
- The ribosome sites for the peptide bond, for the arriving tRNA and for the exit of empty tRNA.
- Translocation
- The movement of the ribosome along the mRNA by one codon from 5′ to 3′.
- Release factor
- The protein that recognizes a stop codon and ends translation.
- Differential gene expression
- The expression of different genes by cells with the same genome.
- Positive and negative gene regulation
- Control of expression by an activator or by a repressor.
- Operon and lac operon
- A cluster of genes for a coordinated function; the lac operon (i, z, y, a) breaks down lactose.
- Mutation, mutagen, mutant
- A permanent change in DNA; the agent that causes it; the organism that carries it.
- Natural and induced mutations
- Mutations that occur automatically and those produced artificially.
- Point mutation and chromosomal mutation
- A change in a few nucleotides and a change in chromosome structure or number.
- Frame shift mutation
- Insertion or deletion of nucleotides that changes the whole sequence of codons after it.
- Down, Klinefelter and Turner syndromes
- Chromosomal disorders with 47 (trisomy 21), 47 (44+XXY) and 45 (44+XO) chromosomes.
- Sickle cell anaemia and phenylketonuria
- Recessive gene mutations affecting haemoglobin and phenylalanine hydroxylase.
Multiple choice questionsp. 372–373
- ADNA + Pectin
- BRNA + DNA
- CDNA + HistoneAnswer
- DDNA only
A chromosome is about 40% DNA and 60% protein (p.347). The DNA is wound round histone proteins, which bind strongly to it because DNA is negatively charged (phosphate groups) and histones are rich in basic amino acids such as arginine and lysine.
A is wrong because pectin is a plant cell wall polysaccharide. D is tempting, since the genes are DNA, but it leaves out the histones that make up most of the chromosome's mass.
- Aheat-killed pathogenic cell caused pneumonia
- Bsome substance from pathogenic cells was transferred to non-pathogenic cells, making them pathogenic.Answer
- Cthe polysaccharide coat of bacteria caused pneumonia.
- Dbacteriophage injected DNA into bacteria.
When living S (virulent) cells were mixed with harmless R cells, the mice died and living S cells were recovered from their blood (p.350). Griffith concluded that some substance from the dead S cells had transformed the R cells into virulent S cells.
A is wrong: heat-killed S cells alone did not harm mice (control 3, p.349). C is wrong: the capsule is not itself lethal, since heat-killed S cells that still had capsules were benign. D describes the bacteriophage work of Hershey and Chase, not Griffith's experiment.
- ADNA polymerase
- BDNA ligaseAnswer
- COkazaki fragments
- DPrimase
Replication still takes place, so DNA polymerase and primase must be present. The short segments are Okazaki fragments, which are normal products of lagging-strand synthesis (p.355). They stay unjoined because DNA ligase, which forms the phosphodiester bond that seals the gaps between fragments, is missing (p.356).
A is wrong because DNA polymerase is needed for any synthesis to occur. C is wrong because Okazaki fragments are the segments that were formed, so they are present. D is wrong because primase makes the primers that start synthesis, and synthesis is happening.
- Athe origins of replication occur only at the 5' end.
- BDNA ligase works only in the 3' to 5' direction.
- Cpolymerase can only work on one strand at a time.
- DDNA polymerase can join new nucleotides only to the 3' end of a growing strand.Answer
DNA polymerase III can add a nucleotide only to a free 3'-OH end, so every new strand grows in the 5' to 3' direction. The leading strand is copied continuously towards the fork, while the lagging strand is copied away from the fork in short Okazaki fragments, each started by its own primer (p.355).
A is wrong: the origin of replication is a site on the DNA, not a strand end. B is wrong: DNA ligase only joins fragments; it does not work in one direction only. C is wrong: both strands are copied at the same time, by separate units of DNA polymerase III.
- Aan RNA primerAnswer
- BDNA ligase
- Ca DNA primer
- Dan Okazaki fragment
Primase lays down a short primer made of RNA nucleotides, and DNA polymerase can only extend an existing 3'-OH end, so every new strand starts with an RNA primer (p.355).
C is wrong because the primer is RNA, not DNA. B is wrong because DNA ligase joins fragments at the end of replication. D is wrong because an Okazaki fragment is a product that forms after synthesis has begun.
- Ain the centre of chromosome
- Bnear an end of chromosome
- Cslightly away from the centre of a chromosomeAnswer
- Dat the end of a chromosome
In a submetacentric chromosome the centromere lies slightly away from the centre, so the two arms are unequal in length (p.347). A describes a metacentric chromosome, B an acrocentric one and D a telocentric one.
- Aprogresses away from the replication fork
- Boccurs in the 3' to 5' direction
- Cproduces Okazaki fragments
- Ddepends on the action of DNA polymeraseAnswer
The leading strand is synthesised continuously by DNA polymerase III, so it depends on that enzyme (p.355). A is wrong because the leading strand grows towards the replication fork; it is the lagging strand's fragments that are laid down away from the fork. B is wrong because synthesis is always 5' to 3'. C is wrong because Okazaki fragments form on the lagging strand.
- AWatson and Crick
- BMeselson and Stahl
- CHershey and ChaseAnswer
- DBeadle and Tatum
Hershey and Chase labelled phage DNA with ^{32}P and phage protein with ^{35}S. Only the ^{32}P DNA entered the bacterial cell and directed the production of new viruses (p.351).
Watson and Crick, and Meselson and Stahl (who studied replication), did not carry out the bacteriophage experiment described in this chapter, and Beadle and Tatum are not part of this unit. So A, B and D are wrong.
- A^{18}F
- B^{35}SAnswer
- C^{14}C
- D^{32}P
Amino acids such as cysteine and methionine contain sulphur, but no phosphorus, while nucleic acids contain phosphate groups. So ^{35}S labels the protein, not the DNA, and ^{32}P labels the DNA (p.351).
D (^{32}P) labels the DNA, the opposite of what is wanted. C (^{14}C) labels organic molecules in general, so it would label both DNA and protein and could not separate them. A (^{18}F) is not used to label biological molecules in this chapter.
- Athe two DNA strands have completely separated and exposed the promoter
- Bseveral transcription factors have bound to the promoterAnswer
- Cthe DNA introns are removed from the template
- DDNA nucleases have isolated the transcription unit
The DNA unwinds only after RNA polymerase has bound the promoter (p.358), so A is not a precondition. Eukaryotic promoters carry the TATA, CAAT and GC-rich boxes (p.357), and RNA polymerase can only start when these promoter sites are bound by the proteins that position it, which is the sense of B.
C is wrong because introns are removed from the transcript after it is made, by splicing (p.359). D is wrong because nucleases do not isolate the transcription unit.
Note. The book does not name transcription factors. This answer relies on the standard account of eukaryotic initiation, which goes slightly beyond the chapter text.
- Acomplementary to the corresponding triplet in mRNAAnswer
- Bcomplementary to the corresponding triplet in rRNA
- Cthe part of the tRNA bonds to specific amino acid
- Dcatalytic, making the tRNA
The anticodon is the three-base loop of tRNA that pairs with a codon of mRNA by complementary base pairing (p.362). B is wrong because tRNA pairs with mRNA, not rRNA. C describes the amino acid attachment at the 3' end, not the anticodon. D is wrong because tRNA does not catalyse reactions; peptide bonds are formed by the ribosome's enzyme peptidyl transferase (p.364).
Short answer questionsp. 373–374
A chromosome is classed by where its centromere lies along its length, which fixes the length of its two arms (p.347).
| Type | Position of centromere | Arms |
|---|---|---|
| Metacentric | In the centre | Two arms of nearly equal length |
| Submetacentric | Slightly away from the centre | Two arms, one a little longer |
| Acrocentric | Near one end | One very short arm and one long arm |
| Telocentric | At one end | Only one arm |
The chromatin fibre is the 30 nm fibre formed when the 10 nm nucleosome string coils about its own axis (p.348). Chromosomes are found at this level of organisation during G1 and G2.
| Region | Degree of condensation | Genes |
|---|---|---|
| Heterochromatin | Highly condensed | Not expressed (unexpressed region) |
| Euchromatin | Non-condensed (loosely packed) | Expressed |
When the cell divides, the euchromatin is also condensed, so that a uniform chromatin fibre is established (p.348).
| Enzyme | Function |
|---|---|
| DNA polymerase I | Replaces the RNA primers with DNA nucleotides in the termination phase, so supporting DNA polymerase III in the main replication process. |
| DNA polymerase II | Takes part in repairing DNA damage during the life of the cell. |
| DNA polymerase III | Is the main enzyme that synthesises both daughter strands along the template. Its proofreading subunit removes a wrongly added nucleotide. |
The central dogma of molecular genetics is that DNA makes protein through an intermediate RNA. Gene expression, which is the flow of information from gene to protein, has two steps (p.357):
- Transcription: the DNA sequence of a gene is copied into mRNA by RNA polymerase, in the nucleus of eukaryotes.
- Translation: the mRNA is decoded by the ribosome, which links amino acids in the order given by the codons to form a polypeptide.
- The result is that the DNA sequence determines the protein, and so the character or function it controls: DNA \to RNA \to protein.
The genetic code is a triplet code: each codon is three bases, giving 64 combinations (p.361). Its four characteristics are:
- Degeneracy: most amino acids are coded by more than one codon. Leucine and serine have six codons, and methionine and tryptophan have only one.
- Universal: a codon has the same meaning in almost all organisms. AGA codes arginine in bacteria and in humans, so genes can be transferred between organisms. Mitochondrial code differs slightly, for example UGA codes tryptophan there.
- Non-punctuated: there is no gap or punctuation between codons, so the message is read as one continuous string.
- Non-overlapping: each base belongs to only one codon. The sequence AUGAGCGCA is read as AUG/AGC/GCA, not as AUG/UGA/GAG.
In eukaryotes, the newly transcribed pre-mRNA contains introns, which are non-coding regions between the coding exons. The shortening happens in the nucleus, before the mRNA leaves (p.360):
- RNA splicing: spliceosomes, large RNA-protein complexes, cut out the introns.
- RNA ligase then joins the spliced exon fragments into one continuous mature mRNA.
- Only this shorter mature mRNA passes through the nuclear pore into the cytoplasm, where it is translated. The cap and poly-A tail are added at the ends but do not account for the shortening.
The triplet code gives 64 codons (4 \times 4 \times 4, p.361). Of these:
- Three codons (UAA, UAG, UGA) are stop codons. They code no amino acid and are recognised by release factors, not by tRNA (p.364).
- The remaining 61 sense codons each specify one amino acid. Each needs a tRNA whose anticodon pairs with it.
- So a cell needs at least 61 different types of tRNA, one for each sense codon.
Note. This answer assumes one tRNA type for each sense codon, as the triplet code implies. The book does not discuss tRNAs that pair with more than one codon.
The book does not define 'minor' mutations. The examples below are the human mutations the chapter describes (p.367 to 371):
- Sickle cell anaemia: a point mutation in the beta globin gene, giving haemoglobin S.
- Phenylketonuria: a point mutation in the gene for phenylalanine hydroxylase.
- Developmental abnormalities such as microcephaly (small head) and cleft palate.
Note. The question's word 'minor' is not used in the book. The list follows the book's examples of point mutations and developmental abnormalities.
After termination, the polypeptide folds into its three-dimensional shape and passes through the Golgi complex for modification (p.364). It can then be used in two ways:
| Used inside the cell | Used outside the cell |
|---|---|
| Cellular enzymes that speed up reactions | Digestive enzymes |
| Membrane proteins | Protein hormones |
| Histone proteins that help pack the chromosome | Albumins, fibrinogens and antibodies |
Each pair is answered in a table with the points the book uses for comparison.
| Feature | Metacentric | Sub metacentric |
|---|---|---|
| Centromere | In the centre | Slightly away from the centre |
| Arms | Nearly equal | Unequal, one arm longer |
| Feature | Acrocentric | Telocentric |
|---|---|---|
| Centromere | Near one end | At the very end |
| Arms | One very short arm and one long arm | Only one arm |
| Position | Centromere close to the terminal end | Centromere at the terminal end |
| Feature | Nucleosome | Primosome |
|---|---|---|
| Composition | DNA wound about twice round a core of eight histone molecules | A complex of the enzymes primase and DNA helicase |
| Role | Packs the DNA into a nucleosome string | Lays down the RNA primers that start replication |
| Stage | Formed just after DNA replication | Active during replication initiation |
| Feature | Heterochromatin | Euchromatin |
|---|---|---|
| Condensation | Highly condensed | Non-condensed |
| Genes | Not expressed | Expressed |
| Chromatin | Dense, compact region | Loose region |
| Feature | Conservative model | Dispersive model |
|---|---|---|
| Parental DNA | Kept intact in one daughter molecule | Broken into nucleotides or segments |
| Daughter molecules | One has both parental strands; the other has only new strands | Each strand is a mixture of old and new segments |
| Result after two cycles | Old and entirely new molecules | Every molecule contains both old and new material |
| Feature | DNA helicase | DNA gyrase |
|---|---|---|
| Function | Breaks the hydrogen bonds between base pairs so the two strands separate | Relieves the tension built up by unwinding |
| Position | At the fork, unwinding the duplex | Works slightly ahead of the helicase |
| Action | Opens the duplex | Reduces the strain in the DNA |
| Feature | DNA polymerase I | DNA polymerase II |
|---|---|---|
| Main job | Replaces RNA primers with DNA nucleotides in termination | Repairs DNA damage |
| Stage | Termination of replication | Throughout the life of the cell |
| Helps | Supports DNA polymerase III | Keeps the DNA stable |
| Feature | Leading strand | Lagging strand |
|---|---|---|
| Synthesis | Continuous, towards the fork | Discontinuous, away from the fork |
| Primers | One primer at the start | A primer for each Okazaki fragment |
| Products | One long daughter strand | Short Okazaki fragments, joined by DNA ligase |
| Feature | Transcription | Translation |
|---|---|---|
| Makes | mRNA from DNA | A polypeptide from mRNA |
| Place | In the nucleus (eukaryotes) | On ribosomes in the cytoplasm |
| Machinery | RNA polymerase | Ribosome with tRNA and enzymes |
| Units read | Bases of the template strand | Codons of the mRNA |
| Feature | Intron | Exon |
|---|---|---|
| Coding | Non-coding | Coding |
| Fate | Removed by splicing | Kept; joined to form mature mRNA |
| Found | Between exons in the transcript | Between introns in the transcript |
| Feature | Start codon | Stop codon |
|---|---|---|
| Codon | AUG | UAA, UAG or UGA |
| Amino acid | Codes methionine | Codes no amino acid |
| Role | Marks the beginning of the message | Ends translation; recognised by release factors |
| Feature | Non-sense codon | Sense codon |
|---|---|---|
| Number | Three (UAA, UAG, UGA) | Sixty-one |
| Amino acid | None | Each specifies one amino acid |
| Role | Stops translation | Adds an amino acid to the chain |
| Feature | Point mutation | Chromosomal mutation |
|---|---|---|
| Change | One or a few nucleotides | Structure or number of whole chromosomes |
| Example | Sickle cell anaemia | Down syndrome |
| Cause | Base substitution, insertion or deletion | Non-disjunction in meiosis |
| Feature | Harmful | Useful |
|---|---|---|
| Effect | Reduces the fitness of the organism | Gives variation; raw material for evolution |
| Examples | Cancer, sickle cell anaemia, Down syndrome | Favourable changes that accumulate by natural selection |
| Fate | Removed by natural selection | Tend to accumulate in the population |
| Feature | Down syndrome | Klinefelter syndrome |
|---|---|---|
| Chromosomes | 2n + 1 (47): trisomy 21 | 2n + 1 (47): 44 + XXY |
| Sex | Male or female | Male |
| Chromosomal change | Extra chromosome 21 (autosome) | Extra X chromosome (sex chromosome) |
| Feature | Klinefelter syndrome | Turner's syndrome |
|---|---|---|
| Karyotype | 2n + 1: 44 + XXY | 2n - 1: 44 + XO (45) |
| Sex | Male | Female |
| Main features | Small testes, wide hips, sparse body hair | Short stature, webbed neck, infertility |
Extensive answer questionsp. 374–375
A chromosome is a thick, thread-like structure that appears in the nucleus during cell division. It is made of two chromatids held together at the centromere (p.347).
Chromatids
Each chromatid is one long DNA molecule, highly coiled together with histone proteins. The chromatids carry the genes for different traits. In a cell that is ready to divide, the two chromatids are identical, each containing one DNA molecule.
Centromere and arms
The centromere, also called the primary constriction, is the point where the two chromatids are attached. It divides the chromosome into two arms, the regions from the centromere to each end.
Other regions
- Secondary constriction (nucleolar organiser): a second point of union in some chromosomes. At least one pair of homologous chromosomes has it, and it gives rise to the nucleoli in interphase.
- Satellite: the knob-like structure beyond the secondary constriction. It contains a useless sequence of DNA called junk DNA.
- Telomeres: the terminal ends of the chromosome. They stop the chromosomes from attaching to each other.
Diagram
Types by centromere position
| Type | Position of the centromere |
|---|---|
| Metacentric | In the centre |
| Submetacentric | Slightly away from the centre |
| Acrocentric | Near one end |
| Telocentric | At one end |
Composition
A chromosome is about 40% DNA and 60% protein. DNA is negatively charged because of its phosphate groups, so it binds strongly to histones, which are positively charged because they are rich in arginine and lysine (p.347). An average human chromosome contains a DNA molecule about 5 cm long, made of about 140 million nucleotides. It is packed into a chromosome by several levels of coiling.
Four levels of organisation
- Nucleosome: about every 200 nucleotides, the DNA wraps twice round a core of eight histone molecules. One histone (H1) is attached to the linker DNA between nucleosomes. The result is a 10 nm nucleosome string, like beads on a string (p.348).
- Chromatin fibre: the nucleosome string coils about its axis to form a 30 nm fibre. The two regions of this fibre are heterochromatin (highly condensed, genes not expressed) and euchromatin (non-condensed, genes expressed). Euchromatin condenses when the cell divides (p.348).
- Supercoil: the higher-order coiling of the chromatin fibre gives a supercoil with a diameter of 300 nm (p.349).
- Chromatid: further coiling of the supercoil gives a chromatid 700 nm thick. Two chromatids make up the chromosome, about 1400 nm thick.
Organisation through the cell cycle
During the S phase, DNA and histones are separated when the DNA is replicated. After replication, DNA and histones organise again, and condensation continues until the cell divides and the chromosome appears (p.347).
Background
Streptococcus pneumoniae occurs in two forms. S (smooth) cells have a polysaccharide capsule, form smooth colonies and are virulent. R (rough) cells lack the capsule, form rough colonies and are benign (p.349).
Experiment
In 1928 Griffith injected healthy mice with four different treatments and recorded the result of each:
Observations
- Living S cells killed the mice (control 1).
- Living R cells did not harm the mice (control 2).
- Heat-killed S cells did not harm the mice, so the capsule itself is not lethal (control 3).
- A mixture of living R cells and heat-killed S cells killed the mice, and living S cells were recovered from the blood of the dead mice (test).
Conclusion
Some information, which specified the capsule and virulence, passed from the dead S cells into the living R cells. This transformed the harmless R cells into virulent S cells. The transfer of genetic material from one organism to another, which alters the genetic make-up of the recipient, is called transformation (p.350).
Griffith's experiment showed that a transforming factor exists but did not identify it. Avery, MacLeod and McCarty later showed that the transforming factor is DNA (extra reading, p.350).
Background
A bacteriophage is a virus that attacks bacteria. It consists of DNA and protein. After infection, the phage multiplies inside the host and many copies are released within 20 to 25 minutes. Before 1952 it was not known whether the DNA or the protein of the phage entered the host (p.351). Hershey and Chase set out to answer this question.
Labelling the phage
- The DNA was labelled with the radioactive isotope of phosphorus, ^{32}P, which is present in the phosphate groups of DNA but not in protein.
- The protein coat was labelled with the radioactive isotope of sulphur, ^{35}S, which is present in some amino acids but not in DNA.
Procedure
- Grow two batches of phages: batch 1 with ^{35}S (protein labelled) and batch 2 with ^{32}P (DNA labelled).
- Allow the labelled phages to infect separate cultures of bacteria.
- Separate the phages from the bacteria by blending (shaking) the mixture, which strips the empty protein coats off the bacteria.
- Centrifuge the mixture. The bacteria form the pellet at the bottom, and the lighter phage coats stay in the liquid.
- Measure the radioactivity in the liquid and in the pellet.
Results and conclusion
| Batch | Label | Radioactivity found in | Meaning |
|---|---|---|---|
| 1 | ^{35}S (protein) | Liquid (phage coats) | Protein stayed outside the bacterium |
| 2 | ^{32}P (DNA) | Pellet (bacterial cells) | DNA entered the bacterium |
Only the labelled DNA entered the bacterial cell, and it directed the production of new viruses. Therefore the viral DNA, not the protein, carries the hereditary information (p.351).
After the structure of DNA was discovered, scientists proposed three models to explain how a duplex DNA molecule is copied (p.352).
Semi-conservative model
The two parental strands separate, and each acts as a template for a new strand. The result is two double helices, each made of one parental strand and one new strand.
Conservative model
One daughter molecule contains both parental strands, and the other daughter molecule contains two newly synthesised strands.
Dispersive model
The parental DNA is dispersed into its nucleotides or segments. These combine with new nucleotides to form two daughter molecules in which each strand is a mixture of old and new material.
Comparison
| Model | Two daughter molecules after one cycle | Fate of parental DNA |
|---|---|---|
| Semi-conservative | Each has one parental and one new strand | Each parental strand is kept whole |
| Conservative | One has both parental strands, the other both new strands | Parental strands stay together |
| Dispersive | Each strand is a mixture of old and new segments | Parental DNA is broken up and dispersed |
The Meselson and Stahl experiment (question 6) showed that the semi-conservative model is correct.
Aim
Meselson and Stahl (1958) tested the three models of replication to find which one was correct (p.353).
Procedure
- Bacteria were grown for several generations in a medium containing the heavy isotope of nitrogen, ^{15}N, which became part of the bases of their DNA.
- The bacteria were shifted to a medium containing the light isotope, ^{14}N. Samples were taken at 0 minutes (first plate), 20 minutes (one generation) and 40 minutes (two generations). A control was grown in ^{14}N only.
- The DNA was extracted and dissolved in caesium chloride solution, then spun at high speed in an ultracentrifuge for many hours. A density gradient formed, and each DNA molecule settled at the level of its density.
Results
| Sample | Band position | Interpretation |
|---|---|---|
| Control (^{14}N only) | Top, lightest | Both strands are light |
| 0 minutes | Bottom, heaviest | Both strands are heavy (^{15}N) |
| 20 minutes | One intermediate band | One heavy and one light strand (hybrid) |
| 40 minutes | Intermediate band and top band | Hybrids plus molecules with two light strands |
Interpretation
- The conservative model predicts a heavy band and a light band at 20 minutes, with no hybrid. This is not seen.
- The dispersive model predicts a single intermediate band at 40 minutes. Two bands are seen, so this model is rejected.
- The semi-conservative model predicts one hybrid band after one cycle and a hybrid plus a light band after two cycles. This is exactly what was observed.
Conclusion
DNA replication is semi-conservative: each daughter molecule contains one parental strand and one newly synthesised strand.
Overview
DNA replication, the self-synthesis of DNA, occurs once in each S phase. The parent DNA molecule gives two daughter molecules, each with one parental and one new strand (p.351, p.353). The process is described in three phases.
1. Initiation
- The origin of replication is a specific sequence where the process begins. Prokaryotes have one origin; eukaryotic DNA may have several.
- DNA helicase breaks the hydrogen bonds between base pairs and unwinds the duplex, while DNA gyrase works slightly ahead to relieve the tension.
- A replication bubble forms, and each side of it is a replication fork.
- Single-stranded binding (SSB) proteins stop the separated strands from pairing again. The strands now act as templates.
2. Extension (polymerisation)
- Primase lays down short RNA primers. Primase and helicase together form the primosome.
- DNA polymerase III adds nucleotides to the 3' end of each primer, so the new strand grows 5' to 3'.
- On the leading strand, synthesis is continuous towards the fork.
- On the lagging strand, synthesis is discontinuous. Short Okazaki fragments are made away from the fork, each started by its own primer. The fragments are 100 to 200 nucleotides long in eukaryotes and 1000 to 2000 in prokaryotes (p.355).
3. Termination
DNA polymerase I replaces the RNA primers with DNA nucleotides, and DNA ligase joins the resulting segments by forming phosphodiester bonds (p.356).
DNA stability
DNA polymerase II and III keep the DNA molecule stable. The sequences are conserved after many cell divisions, so each new cell has the same genetic information.
DNA variability
Errors during replication produce changes in the base sequence, which are mutations. Some replication errors cause breaks in the DNA that can lead to cancer. These errors give the variability of DNA.
Requirements
DNA polymerase III cannot start a new strand. It can only add a nucleotide onto a pre-existing 3'-OH group, so it needs a primer (p.355).
Direction of synthesis
It always adds the nucleotide at the 3' end of the primer, so the new strand is made in the 5' to 3' direction.
Steps of polymerisation
- A primer sits on the template strand at the replication fork, held by the sliding clamp.
- The polymerase reads the template strand in the 3' to 5' direction.
- It adds the complementary nucleotide to the free 3'-OH end of the growing strand. The next nucleotide is linked to this one by a phosphodiester bond.
- The enzyme moves along the template and repeats step 3, so the chain grows continuously on the leading strand.
- One of its subunits proofreads: if a wrong nucleotide is added, the enzyme removes it before continuing.
Leading and lagging strands
Two units of DNA polymerase III work at the fork. One copies the leading strand continuously. The other copies the lagging strand, forming Okazaki fragments of 100 to 200 nucleotides in eukaryotes, and jumps back to a new primer for each fragment (p.355).
Transcription is the synthesis of RNA from DNA, the first step of gene expression. It needs only one enzyme, RNA polymerase, and it has three phases (p.357 to 359).
1. Initiation
- RNA polymerase binds to the promoter, a regulatory region of the gene.
- In prokaryotes the promoter has the TTGACA (-35) and TATAAT (-10) sequences. In eukaryotes it has the TATA box (-25), the CAAT box (-80) and the GC-rich box (-90).
- The DNA duplex unwinds, the base pairs break and a transcription bubble forms. The sigma factor is released once the enzyme is attached.
2. Elongation
- One strand of the gene is the template (antisense) strand. The other is the coding (sense) strand.
- RNA polymerase adds ribonucleotide triphosphates complementary to the template strand. RNA is made in the 5' to 3' direction, and no primer is needed.
- The enzyme moves along the template, and the transcription bubble moves with it, leaving the growing RNA chain protruding from the bubble (Fig 13.15).
3. Termination
- The terminator region has a series of GC base pairs followed by AT base pairs.
- The transcribed RNA folds back on itself to form a GC hairpin loop, followed by a tail of poly-U nucleotides.
- The hairpin causes RNA polymerase to stop, and the completed RNA chain is released from the DNA.
Translation is the second step of gene expression. The mRNA produced by transcription is decoded by the ribosome to make a polypeptide, which then folds into an active protein (p.362).
1. Activation of amino acids
Each free amino acid is bound to the 3' end of its own tRNA by aminoacyl-tRNA synthetase, using ATP. The product is an aminoacyl-tRNA complex (p.362).
2. Formation of the initiation complex
The small ribosomal subunit binds the mRNA at its 5' end, and the initiator tRNA carrying N-formyl methionine binds at the P site on the AUG start codon. The large subunit is then placed on the small one to complete the initiation complex (p.363).
3. Elongation
- The codon exposed at the A site binds its anticodon-bearing aminoacyl-tRNA with the help of an elongation factor.
- Peptidyl transferase removes the chain from the tRNA at the P site and joins it to the new amino acid by a peptide bond.
- The ribosome moves one codon along the mRNA (translocation). The empty tRNA moves to the E site and leaves, the tRNA carrying the chain moves to the P site, and the next codon is exposed at the A site.
These three steps repeat until a stop codon is reached at the A site (p.364).
4. Termination
Stop codons (UAG, UAA or UGA) do not bind any tRNA. They are recognised by release factors, which release the polypeptide from the tRNA. The tRNA leaves the ribosome and the ribosomal subunits separate from the mRNA (p.364).
After synthesis
The polypeptide folds into a three-dimensional shape and passes to the Golgi complex for further modification. It can then be used inside or outside the cell (p.364).
Translation follows the same basic steps in both groups, but the machinery and some details differ (p.364 to 365).
| Feature | Prokaryotes | Eukaryotes |
|---|---|---|
| Ribosome | Smaller: 70S, made of a 50S large and a 30S small subunit | Larger: 80S, made of a 60S large and a 40S small subunit |
| Location | Cytosol | Cytosol and rough endoplasmic reticulum |
| Initiating amino acid | N-formyl methionine, which is modified | Methionine, which is not modified |
| Initiation factors | Only one type | More than one type |
| Release factors | Two | One |
| Coupling with transcription | No delay: translation starts as the mRNA emerges from the DNA | mRNA is made in the nucleus and processed before it moves to the cytoplasm |
The main reason for the difference in coupling is that a eukaryotic cell separates transcription in the nucleus from translation in the cytoplasm, so the mRNA must be processed and exported before it is translated (p.359).
Regulation of gene expression is the control of the amount and timing of expression of a gene. Cells must continually switch genes on and off in response to signals from their environment (p.365).
- Versatility and adaptability: a cell expresses a protein only when it is needed. In bacteria, the lac operon makes enzymes only when lactose is present (p.366).
- Differentiation: the mature human body has about 200 cell types, all with the same genome. Their differences come from differential gene expression, which is the basis of cell differentiation and morphogenesis.
- Control of structure and function: regulation gives the cell control over all its structure and function, in prokaryotes and eukaryotes alike.
- Prevention of disease: if regulation is disturbed, serious imbalances and diseases, including cancer, can arise (p.365).
Gene expression is controlled by regulatory proteins that either decrease or increase the amount of transcription. The two methods are negative control, by a repressor, and positive control, by an activator (p.365 to 366).
Negative control by repressor proteins
A repressor is a specific regulatory protein. It attaches to the operator, the DNA site next to the promoter, and stops RNA polymerase from transcribing it, so the expression of the gene is diminished (Fig 13.25).
Positive control by activator proteins
An activator is a specific regulatory protein. When it is attached to RNA polymerase or the promoter, it starts the process of transcription, so the expression of the gene is increased quantitatively (Fig 13.24).
Comparison
| Feature | Negative control | Positive control |
|---|---|---|
| Regulatory protein | Repressor | Activator |
| Effect on expression | Diminished: transcription is shut down | Increased: transcription is started |
| Action | Attaches to the promoter or operator and stops RNA polymerase | Attaches to RNA polymerase or the promoter and starts transcription |
Example: the lac operon
The lac operon has the regulatory gene i and the structural genes z, y and a. In the absence of lactose the repressor made by gene i binds the operator and blocks transcription. When lactose (the inducer) is present, it binds the repressor, which becomes inactive, and the operon is transcribed. The operon therefore shows both positive and negative control (p.366, Fig 13.26).
In eukaryotes the transcript is made in the nucleus, but translation takes place in the cytoplasm. The pre-mRNA must therefore be modified into functional mRNA and protected on its way out of the nucleus. It is modified in two events: capping and tailing protect it, and splicing removes the non-coding sequences (p.359 to 360).
Capping
A cap of 7-methyl GTP is linked from its 5' end to the 5' end of the mRNA. It protects the mRNA from being degraded by phosphatases and nucleases.
Poly-A tail
A chain of 30 to 500 adenine nucleotides, the poly-A tail, is added to the 3' end of the transcript. It also protects the mRNA from degradation.
RNA splicing
- Introns, the non-coding regions of the primary mRNA, are recognised.
- The spliceosome, a large RNA-protein complex, cuts out the introns.
- The exon fragments are joined together with RNA ligase, giving a mature mRNA with the exons in order.
- The mature mRNA, with its cap and poly-A tail, passes through a nuclear pore into the cytoplasm for translation.
Note. The book numbers the last items unnumbered after question 14. They appear here as the group following question 14, as printed.
(a) Definition and sources
A gene mutation is a permanent change in the DNA sequence that makes up a new allele in the population. A mutation can range from a change in a single nucleotide to a change in a whole chromosome. The agents that cause mutations are mutagens, and the organism or cell in which a mutation occurs is the mutant (p.367).
- Physical mutagens: radiation of any kind, such as UV, gamma rays, X-rays and radioactive rays, and high temperature.
- Chemical mutagens: colchicine, mustard gas, nitrous gas, acridine orange, reactive radioactive isotopes and free oxygen particles.
- Biological mutagens: certain viruses, transposons, and errors that occur during meiosis or DNA replication.
(b) Natural and induced mutations
| Feature | Natural mutation | Induced mutation |
|---|---|---|
| Origin | Occur naturally and automatically from internal or external factors | Produced artificially |
| Causes | Errors during meiosis or DNA replication | Mainly physical and chemical mutagens |
| Purpose | None; they arise in nature | To establish new varieties of an organism |
(a) Harmful aspects
A harmful mutation decreases the fitness of an organism in its environment. Examples given in the book include (p.367 to 368):
- Being born without part of the brain, and developmental abnormalities such as microcephaly and cleft palate.
- Cancer, which is a form of harmful mutation.
- Chromosomal abnormalities such as Down, Klinefelter and Turner syndromes.
- Heredity disorders such as sickle cell anaemia and phenylketonuria.
Harmful mutations are eliminated by natural selection, through premature death or reduced reproductive success (a 'genetic death') (p.369).
Useful aspects
Some mutations are useful. Changes in the genes that control development can have large effects on the adult form, and the book cites the evolution of hoofed mammals into ocean-dwellers, of water plants that invaded the land, and of armoured invertebrates that evolved wings (p.368).
(b) Mutation and evolution
- Mutations supply new alleles, which is the raw material of variation in a population.
- Less favourable (deleterious) mutations are removed from the gene pool by natural selection.
- More favourable (beneficial) mutations tend to accumulate over generations.
- Mutations are therefore considered the driving force of evolution, and they contribute to the evolution of new forms.
(a) Down syndrome
Down syndrome is characterised by 2n + 1 (47 chromosomes): an extra copy of chromosome 21 (trisomy 21) (p.368).
- Symptoms: smaller, abnormally shaped head; rounded inner corner of the eyes; flattened nose; a single crease in the palm; small ears and mouth; short hands and fingers; white spots on the coloured part of the eye; a gap between the first and second toes.
- Cause: autosomal non-disjunction. The 21st chromosome pair fails to separate in maternal meiosis, so the egg has 24 chromosomes. Fertilisation by a normal sperm (23 chromosomes) gives 47.
- Treatment: there is no cure. It can be managed with regular checkups and screening, medication, surgery where needed, and counselling and support.
(b) Klinefelter's syndrome
Klinefelter's syndrome is characterised by 2n + 1 (44 + XXY). The person is male (p.368).
- Symptoms: several feminine features, such as sparse body hair, enlarged breasts and wide hips; small testes; a voice that may not be deep; abnormal body proportions, with long legs and a short trunk.
- Cause: non-disjunction of the sex chromosome pair during maternal meiosis, giving an extra X chromosome.
- Treatment: testosterone can be given around puberty, which helps the boy's body develop more normally. The treatment usually continues through life, but it does not help infertility.
(c) Turner's syndrome
Turner's syndrome is characterised by 2n - 1 (44 + XO = 45). The affected individuals are female (p.369).
- Symptoms: shorter than average height; infertility; a webbed neck; a low hairline at the back of the neck; abnormal bone development, especially of the hands and elbows; more moles than usual; and oedema (extra fluid) in the hands and feet.
- Cause: non-disjunction of the sex chromosome pair during maternal meiosis, so one X chromosome is absent.
- Treatment: there is no specific treatment. Growth hormone can improve growth and final adult height if started early. Oestrogen replacement helps the girl develop the physical changes of puberty, usually started at about age 12 or 13.
Summary
| Syndrome | Karyotype | Sex | Specific treatment |
|---|---|---|---|
| Down | 2n + 1 (47), trisomy 21 | Male or female | None; management of symptoms |
| Klinefelter | 2n + 1 (44 + XXY) | Male | Testosterone around puberty |
| Turner | 2n - 1 (44 + XO) | Female | Growth hormone and oestrogen replacement |
(a) Sickle cell anaemia
Sickle cell anaemia, also called haemoglobin SS disease (Hb SS), is a point mutation in the beta globin gene. It is inherited as an autosomal recessive trait (p.370, Fig 13.32).
- Symptoms: painful episodes lasting hours to days. When the anaemia is severe, there is fatigue, paleness, fever, a rapid heart rate, shortness of breath, jaundice and damage to various organs.
- Cause: the sixth amino acid of the beta chain, glutamic acid, is replaced by valine. In the gene, the glutamic acid codon is changed to a valine codon by a base substitution: the template triplet CTT becomes CAT, so the mRNA codon GAA becomes GUA. The abnormal haemoglobin (Hb S) forms strands, and the red blood cells become sickle-shaped and block the blood flow. A person is affected if both parents pass on the gene.
- Treatment: blood transfusions, medicines to relieve pain, plenty of fluids and folic acid supplements. A bone marrow or stem cell transplant can cure it.
(b) Phenylketonuria (PKU)
Phenylketonuria is a rare condition in which the baby cannot break down the amino acid phenylalanine. It is an autosomal recessive trait (p.371).
- Symptoms: lighter skin, hair and eyes; delayed mental and social skills; a head much smaller than normal; jerking movements of the arms or legs; mental retardation; seizures and skin rashes.
- Cause: in a healthy baby the enzyme phenylalanine hydroxylase converts phenylalanine into tyrosine. A point mutation makes this enzyme defective, so phenylalanine is converted into toxic phenylketones instead. These accumulate and damage the central nervous system.
- Treatment: a diet extremely low in phenylalanine, especially while the child is growing. A special infant formula, Lofenalac, is used for infants with PKU and can be used throughout life as a low-phenylalanine protein source.
FBISE 2025 and 2026 questions
Questions from the FBISE HSSC-I 2026 (1st Annual) and FBISE HSSC-I 2025 (1st Annual) papers that belong to this unit, answered the way an examiner expects.
- A\alpha (Alpha)
- B\beta (Beta)
- C\gamma (Gamma)Answer
- D\theta (Theta)
The unit that leaves the RNA polymerase once it is attached to the promoter is the **sigma (\sigma) factor**. It is needed only to bind the promoter; then it is removed and the core enzyme (two \alpha and the \beta subunits) carries on. The printed options have no sigma, so the intended letter is probably the one that looks closest in print, option C. The core subunits \alpha and \beta (A and B) stay on the DNA, and there is no \theta subunit.
Note. None of the printed options is sigma. Option C is the likely misprint for \sigma, so C is the answer.
- AAcrocentric
- BSub metacentricAnswer
- CTelocentric
- DMetacentric
A centromere located slightly away from the centre gives two arms of unequal length (one slightly shorter): a submetacentric chromosome. In a metacentric chromosome (D) it is exactly in the centre; in an acrocentric (A) it is near the end; in a telocentric (C) it is at the end.
- AProteins are hereditary material
- BDNA is hereditary materialAnswer
- CRNA is genetic material
- DLipids carry genes
Griffith showed that a heritable factor passes from dead S-type to living R-type bacteria (transformation), and Hershey and Chase showed that only the DNA of a bacteriophage enters the bacterium to direct new viruses, while its protein coat stays outside. Together they point to DNA as the hereditary material. Protein (option A) was the older guess, which these experiments disproved; neither experiment involved RNA or lipids.
Griffith injected mice with different bacterial samples. The control experiments (live R-type alone, heat-killed S-type alone, and the mixture) lead to three conclusions.
| Control experiment | Result | Conclusion |
|---|---|---|
| Live R-type (rough, no capsule) injected alone | Mice stay healthy | R-type is non-virulent: it cannot cause disease by itself |
| Heat-killed S-type injected alone | Mice stay healthy | Killing by heat destroys virulence; the dead capsule material alone does not make the mice ill |
| Live R-type mixed with heat-killed S-type | Mice develop disease and die; living S-type bacteria are found in their blood | Some hereditary information (the transforming principle) passed from the dead S-type to the living R-type and changed it into virulent S-type: transformation |
Point mutation (1 mark)
A point (gene) mutation is a change in a single gene produced by a change of one or a few bases of the DNA. It may be a base substitution (one base replaced by another), an insertion or a deletion of bases.
How it causes sickle cell anaemia (2 marks)
- A single base substitution occurs in the gene for the beta chain of haemoglobin: the DNA triplet CTT for glutamic acid becomes CAT, so the mRNA codon GAA becomes GUA.
- The sixth amino acid of the beta chain is therefore valine in place of glutamic acid. This abnormal haemoglobin is Hb^{\mathrm{S}}.
- Hb^{\mathrm{S}} has lost its normal oxygen-binding capacity and forms strands, so the red blood cells become sickle shaped, block the flow of blood and deliver less oxygen to the tissues. The disease is autosomal recessive: it appears when the gene is inherited from both parents.
Note. The question carries a unit note for units 2 and 12; the base change and the disease are taught in unit 13 (gene or point mutations).
Polypeptide elongation is repeated in three steps while the ribosome moves along the mRNA from the 5′ end towards the 3′ end.
- Codon recognition: the codon exposed at the A site is read by the aminoacyl-tRNA whose anticodon is complementary to it. The tRNA binds there with the help of an elongation factor.
- Peptide bond formation: the enzyme peptidyl transferase takes the amino acid (or the growing chain) from the tRNA at the P site and joins it by a peptide bond to the new amino acid on the tRNA at the A site.
- Translocation: the ribosome moves one codon along the mRNA (5′ to 3′). The empty tRNA reaches the E site and leaves, the tRNA carrying the chain moves from the A site to the P site, and a new codon is exposed at the A site.
Bacteria were grown for many generations in ^{15}\mathrm{N} (heavy) medium, then moved to ^{14}\mathrm{N} (light) medium. DNA was extracted at intervals and spun in a CsCl density gradient; heavier DNA settles lower in the tube.
| Tube | Band position | Interpretation |
|---|---|---|
| 1 (control) | Top (lightest) | DNA with ^{14}\mathrm{N} in both strands, the reference for light DNA |
| 2 (0 min) | Lowest | All DNA is heavy: both strands contain ^{15}\mathrm{N} (no replication has happened in ^{14}\mathrm{N} yet) |
| 3 (20 min, one generation) | One band at the middle | All DNA is a hybrid: each daughter duplex has one old ^{15}\mathrm{N} strand and one new ^{14}\mathrm{N} strand |
| 4 (40 min, two generations) | Two bands: top and middle | Half the molecules are hybrid (one ^{15}\mathrm{N} and one ^{14}\mathrm{N} strand) and half are light (both strands ^{14}\mathrm{N}) |
One middle band after one generation rules out the conservative model (it predicts a heavy and a light band), and the light band at 40 minutes rules out the dispersive model (it predicts a single band that only moves up). The results fit semi-conservative replication.
Three enzymes of DNA replication and their functions (any three):
| Enzyme | Function |
|---|---|
| DNA helicase | Breaks the hydrogen bonds between the base pairs and opens the duplex at the origin to form the replication bubble and forks |
| Primase | Makes short RNA primers on the template strands, which DNA polymerase needs to start |
| DNA polymerase III | Main enzyme: adds nucleotides to the 3′ end of the primer (5′ to 3′) to make the leading and lagging strands, and proofreads |
| DNA polymerase I | Replaces the RNA primers with DNA nucleotides |
| DNA ligase | Joins the Okazaki fragments by forming phosphodiester bonds |
| DNA gyrase | Works ahead of helicase and reduces the tension created by unwinding |
The drawing shows a replication fork moving to the right: the unopened parental DNA is at the right, and the two daughter duplexes trail away at the left. Each part is named below, with its work (1 mark for naming and 1 mark for the explanation in each of the two halves, 3+3).
Identification and explanation (3 + 3 marks)
| Part | Name | What it is and what it does |
|---|---|---|
| A | DNA polymerase III | The main replicating enzyme. It adds nucleotides to the 3′-OH end of a primer, always in the 5′ to 3′ direction, and proofreads. Here it is working on the upper (lagging) arm. |
| B | Lagging strand | The new strand made discontinuously, as short Okazaki fragments, away from the fork. Each fragment needs its own primer; the pieces are later joined by DNA ligase. |
| C | Template strand | The old (parental) strand that is copied. Its sequence decides the order of bases in the new strand, by base pairing. |
| D | Leading strand | The new strand made continuously in one piece towards the fork by DNA polymerase III, after one primer. |
| E | DNA helicase | Works at the fork, breaks the hydrogen bonds between base pairs and opens the duplex so that the two strands separate (gyrase works ahead of it to relieve tension). |
| F | Parental DNA | The original double helix that has not yet been unwound. Each strand of it becomes the template for a daughter duplex. |
In 1953 James Watson and Francis Crick built a model of DNA from the X-ray diffraction pictures of Wilkins and Franklin (a duplex 2 nm wide, with one turn every 3.4 nm) and Chargaff's rule (A = T and G = C). Their model says:
Structure (4 marks)
- DNA is made of two polynucleotide chains wound round each other as a right-handed double helix and held together by base pairs.
- The two chains are antiparallel: one runs 5′ to 3′ downward and the other 5′ to 3′ upward. The sugar-phosphate backbones are on the outside and the bases are inside.
- The helix has a constant width of 2 nm; each turn is 3.4 nm long and contains 10 base pairs.
- Pairing is specific: adenine pairs with thymine (two hydrogen bonds) and guanine pairs with cytosine (three hydrogen bonds), so a purine always pairs with a pyrimidine and the two chains are complementary.
- There is no restriction on the order of nucleotides along one strand; the sequence is specific for each species and individual and carries the genetic information.
Diagram (2 marks)
Note. The double helix is taught in unit 2 (section "Watson and Crick model of DNA") but the paper places this question with unit 13 (chromosomes and DNA).
Translation (the second step of gene expression) is the decoding of the mRNA by the ribosome into a polypeptide. The amino acids have already been activated, that is joined to their own tRNAs by aminoacyl-tRNA synthetase (aminoacyl-tRNA complexes). The three phases that follow are given below.
1. Initiation complex (2 marks)
- The tRNA carrying (N-formyl) methionine binds the smaller ribosomal subunit with an initiation factor.
- The 5′ end of the mRNA binds the small subunit with another initiation factor, the AUG start codon pairing with the initiator anticodon.
- The larger subunit sits on the smaller one. The initiator tRNA is at the P site; the A site is free for the next tRNA and the E site is the exit.
2. Elongation (2 marks)
- Codon recognition: the aminoacyl-tRNA with the complementary anticodon enters the A site, helped by an elongation factor.
- Peptide bond: peptidyl transferase moves the amino acid or chain from the P-site tRNA onto the amino acid at the A site.
- Translocation: the ribosome moves one codon along the mRNA (5′ to 3′); the empty tRNA leaves from the E site, the chain-bearing tRNA shifts from A to P, and a new codon is exposed at the A site. The three steps repeat for every amino acid.
3. Termination (1 mark)
When a stop codon (UAA, UAG or UGA) is exposed at the A site, no tRNA binds. Release factors recognise it, the polypeptide is released from the tRNA, the tRNA leaves, and the ribosomal subunits separate from the mRNA.
Diagrams (2 marks)
- ARecognize enzyme activity
- BBind codon on mRNAAnswer
- CRecognize binding site on ribosome
- DAttach amino acid to CCA
The middle loop of a tRNA is the anticodon loop. It is made of 7 bases, the middle three of which are the anticodon, which pairs with the complementary codon on the mRNA. The amino acid is attached to the CCA at the 3′ end of the molecule (option D), not by the middle loop.
- AHelicaseAnswer
- BPrimase
- CPolymerase
- DGyrase
DNA helicase opens the double helix by breaking the hydrogen bonds between base pairs. Gyrase (D) only works ahead of it to reduce the tension made by unwinding; primase (B) makes RNA primers and polymerase (C) builds new strands.
- ADNAAnswer
- BProtein
- CRNA
- DLipid
Griffith found that something from heat-killed S-type bacteria transformed living R-type bacteria, and Hershey and Chase showed that the DNA of the bacteriophage (labelled with ^{32}\mathrm{P}) entered the bacteria while its protein coat (^{35}\mathrm{S}) stayed outside. Protein (option B) was the tempting older idea, but it was shown not to enter the host cell and not to carry the information.
Why (1 mark)
In eukaryotes transcription occurs in the nucleus but translation in the cytoplasm. The newly made pre-mRNA must be changed into mature, functional mRNA and protected from enzymes while it travels to the cytoplasm. (In prokaryotes there is no nucleus and no such delay.)
Changes during mRNA processing (2 marks)
- Capping: a cap of 7-methyl GTP is added to the 5′ end.
- Poly-A tail: a chain of 30 to 500 adenine nucleotides is added to the 3′ end. Cap and tail protect the mRNA from degradation.
- Splicing: spliceosomes cut out the non-coding introns, and the coding exons are joined by RNA ligase to give mature mRNA, which leaves through a nuclear pore.
Hershey and Chase (1952) labelled the DNA of bacteriophages with ^{32}\mathrm{P} and their protein coats with ^{35}\mathrm{S}, let them infect bacteria, then blended and centrifuged the mixture. Their findings were:
| Finding | Meaning |
|---|---|
| ^{32}\mathrm{P} (DNA label) was found inside the bacterial cells (pellet) | The DNA of the virus enters the bacterium |
| ^{35}\mathrm{S} (protein label) was found in the medium, outside the cells | The protein coat stays outside and does not enter |
| The infected bacteria went on to make new bacteriophages | The injected DNA carries the hereditary information for making new viruses, so DNA is the genetic material |
The initiation complex is formed by the ribosomal subunits, the mRNA and the first aminoacyl-tRNA. It forms in three stages.
- A tRNA carrying (N-formyl) methionine, the initiator, binds to the smaller ribosomal subunit with the help of an initiation factor.
- The 5′ end of the mRNA binds to the same small subunit with the help of another initiation factor, with the start codon AUG paired to the initiator tRNA.
- The larger ribosomal subunit is placed on the smaller one. This completes the complex and sets up the E, P and A sites, with the initiator tRNA at the P site.
Chromosomes are classified by the position of the centromere:
| Type | Position of the centromere | Arms |
|---|---|---|
| Metacentric | In the centre | Two arms of about equal length |
| Submetacentric | Slightly away from the centre | Two unequal arms |
| Acrocentric | Near the end | One very short and one long arm |
| Telocentric | At the end | Only one arm |
A mutagen is an agent that causes mutations. Its sources are physical, chemical and biological agents.
| Source of mutagens | Examples |
|---|---|
| Physical | Radiation of any kind: ultraviolet rays, gamma rays, X-rays, radioactive rays; high temperature alterations |
| Chemical | Colchicine, mustard gas, nitrous gas, acridine orange, reactive radioactive isotopes, free oxygen particles |
| Biological | Certain viruses, transposons, and errors that occur during meiosis or DNA replication |
Note. Mutagen sources are taught under mutations; the unit note also lists unit 12.
The experiment (3 marks)
Matthew Meselson and Franklin Stahl (1958) tested the three models (semi-conservative, conservative, dispersive) using nitrogen isotopes.
- Bacteria were grown for several generations in a medium containing heavy nitrogen, ^{15}\mathrm{N}, which became part of the bases of their DNA.
- The bacteria were shifted to a medium containing light nitrogen, ^{14}\mathrm{N}.
- DNA was sampled at 0, 20 and 40 minutes (and a control grown only in ^{14}\mathrm{N}).
- Each sample was dissolved in cesium chloride (CsCl) and spun in an ultracentrifuge; DNA settles at the level of its own density, so heavy, hybrid and light DNA form separate bands.
Results and explanation (2 marks)
| Sample | Band | Meaning |
|---|---|---|
| Control (^{14}\mathrm{N}) | Top | Light DNA |
| 0 min | Bottom | Heavy DNA, both strands ^{15}\mathrm{N} |
| 20 min | Middle | Hybrid: each molecule has one ^{15}\mathrm{N} and one ^{14}\mathrm{N} strand |
| 40 min | Top and middle | Half hybrid, half light |
One hybrid band after one generation rejects the conservative model (which gives heavy and light bands), and the appearance of a light band after two generations rejects the dispersive model (which gives only intermediate bands). Hence DNA replication is semi-conservative: every daughter molecule keeps one old strand and gets one new strand.
Diagram (2 marks)
The figure shows a replication fork (parental DNA at the left). One new strand is made in one piece and the other in short pieces. Each of the six parts earns 1 mark (name plus its role).
Identification and explanation (6 x 1 mark)
| Part | Name | Explanation |
|---|---|---|
| A | Leading strand | The new strand made continuously, in one piece, towards the replication fork, by DNA polymerase III from a single primer. |
| B | RNA primer | A short stretch of RNA nucleotides made by primase. DNA polymerase III cannot start a chain, so it needs this primer with a free 3′-OH to add the first DNA nucleotide. Later DNA polymerase I replaces it with DNA. |
| C | Okazaki fragments | The short pieces of the lagging strand, made away from the fork, each from its own primer. |
| D | DNA polymerase III | The main enzyme that adds nucleotides at the 3′ end of the primer in the 5′ to 3′ direction and proofreads. |
| E | DNA helicase | Breaks the hydrogen bonds between the base pairs at the fork and opens the double helix (with gyrase reducing the tension, and SSB proteins keeping the strands apart). |
| F | DNA ligase | Joins the Okazaki fragments, after the primers are replaced by DNA, by forming phosphodiester bonds so that the lagging strand becomes continuous. |
In 1953 James Watson and Francis Crick built a model of DNA from the X-ray diffraction pictures of Wilkins and Franklin (a duplex 2 nm wide, with one turn every 3.4 nm) and Chargaff's rule (A = T and G = C). Their model says:
Structure (4 marks)
- DNA is made of two polynucleotide chains wound round each other as a right-handed double helix and held together by base pairs.
- The two chains are antiparallel: one runs 5′ to 3′ downward and the other 5′ to 3′ upward. The sugar-phosphate backbones are on the outside and the bases are inside.
- The helix has a constant width of 2 nm; each turn is 3.4 nm long and contains 10 base pairs.
- Pairing is specific: adenine pairs with thymine (two hydrogen bonds) and guanine pairs with cytosine (three hydrogen bonds), so a purine always pairs with a pyrimidine and the two chains are complementary.
- There is no restriction on the order of nucleotides along one strand; the sequence is specific for each species and individual and carries the genetic information.
Diagram (2 marks)
Note. The double helix is taught in unit 2 (section "Watson and Crick model of DNA") but the paper places this question with unit 13 (chromosomes and DNA).
Exam focus
- Identify parts A to F of a DNA replication fork and explain each (helicase, primer, polymerase III, leading and lagging strands, Okazaki fragments, ligase). (long)Asked in both 2026 papers as the Section C OR (Set B Q.3 OR, 6 marks; Set D Q.3 OR, 6 marks), and the enzymes were asked in 2025 (Set B Q.2 (xii) OR, Set D MCQ 11). Practise labelling a fork from memory.Go to answer
- Meselson and Stahl experiment: procedure, band results and why DNA replication is semi-conservative (with diagram). (long)A 7-mark long question in 2025 (Set D Q.3) and an interpretation of the tubes in 2025 (Set B Q.2 (xi) OR); book long question 6.Go to answer
- Translation: initiation complex, the three elongation steps and termination, with diagrams. (long)2025 Set B Q.4 (7 marks), 2026 Set B Q.2 (x) OR (elongation steps) and 2026 Set D Q.2 (x) OR (initiation complex); tRNA anticodon in 2025 Set D MCQ 1.Go to answer
- Watson and Crick double helix structure of DNA, with a labelled diagram. (long)Section C Q.4 (6 marks) in both 2026 sets. It is taught in unit 2, so revise it with this chapter.Go to answer
- Griffith and Hershey-Chase experiments: controls, conclusions and the finding that DNA is the genetic material. (short)2026 MCQ 17 in both sets, Set B Q.2 (iii) OR (Griffith controls) and Set D Q.2 (iii) OR (Hershey-Chase findings); book long questions 3 and 4.Go to answer
- Point mutation and sickle cell anaemia (CTT to CAT, glutamic acid to valine); sources of mutagens. (short)2025 Set B Q.2 (vi) OR and 2025 Set D Q.2 (xiv); book long questions 15 and 18.Go to answer
- Types of chromosomes by centromere position. (short)2025 Set B MCQ 9 and Set D Q.2 (xii) OR; book exercise MCQ 6 and short question 1.Go to answer
- Transcription and post-transcriptional modification of mRNA (sigma factor, cap, poly-A tail, splicing). (short)2025 Set D Q.2 (ii) OR and 2025 Set B MCQ 4 (sigma factor leaves the promoter); book long questions 9 and 14.Go to answer
- Characteristics of the genetic code and the lac operon. (short)Not asked in the 2025 and 2026 papers; included from the book's short question 5 and long question 13, and from the unit's learning outcomes on gene regulation.Go to answer