Chromosome and DNA

Class 11 Biology, unit 13, book pages 346–375

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.
Structure of a typical chromosome. The centromere (accent) joins the two chromatids and divides each into a short arm and a long arm.
Types of chromosomes

On the basis of the position of the centromere along the chromosome, a chromosome may be metacentric, submetacentric, acrocentric or telocentric.

Figure 13.1: Types of chromosomes. The position of the centromere (accent) decides the type.
Types of chromosomes by the position of the centromere
TypePosition of the centromereArms
MetacentricIn the centreTwo arms of about equal length
SubmetacentricLocated slightly away from the centreTwo arms of unequal length (one slightly shorter)
AcrocentricLocated near the endOne very short arm and one long arm
TelocentricLocated at an endOnly 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.

Figure 13.2: Levels of organization of a chromosome. Each level winds up the one before it; the width of the structure (accent) increases from 2 nm to 1400 nm.
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.

A nucleosome: about 200 nucleotides of DNA wrapped twice around a core of eight histones, with linker DNA between neighbouring nucleosomes (histone H1 is associated with the linker DNA).
Worked example: how many nucleosomes in a human chromosome?
An average human chromosome has about 140 million nucleotides in its DNA, and about every 200 nucleotides are wrapped around one histone core. About how many nucleosomes does the chromosome have? How long is one nucleosome string compared with the 2 nm DNA it is made from (write one sentence)?
  1. Use the book's figure: one nucleosome for every 200 nucleotides of DNA.
  2. Substitute the numbers.
  3. 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.
  4. 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 and euchromatin
HeterochromatinEuchromatin
Highly condensed region of the chromatin fibreNon-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 cellIs 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).

Levels of organization of a chromosome
Level of organizationThicknessWhat it is
DNA double helix2 nmBare DNA
Nucleosome string10 nmDNA wrapped twice around cores of eight histones every 200 nucleotides ("beads on a string")
Chromatin fibre30 nmNucleosome string coiled again; found in G₁ and G₂
Supercoil300 nmHigher-order coiling of the chromatin fibre at the start of cell division
Chromatid700 nmFurther 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).
Griffith's experiment, step by step
  1. He injected healthy mice with living S-type bacteria: all the mice died.
  2. 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.
  3. 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.
  4. 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.
  5. 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.
Figure 13.3: Griffith's experiment with the S-type (smooth, capsule) and R-type (rough, no capsule) strains of Streptococcus pneumoniae. Mice that die are shown in the accent colour.

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.

Avery, Macleod and McCarty's four samples
Sample of the extractMixed with live R-type cells and injected into miceResultConclusion
Untreated (first sample)YesTransformation occurredThe filtrate of heat-killed S-type bacteria contains the active factor responsible for transformation
Treated with proteases (removes all proteins)YesTransformation still occurredProtein is not the transforming agent
Treated with RNase (removes all RNA)YesTransformation still occurredRNA is not the transforming agent
Treated with DNase (removes all DNA)YesNo transformationThe active factor must be DNA
Figure 13.4: Avery, Macleod and McCarty's experiment. Only the DNase-treated extract failed to transform living R cells (accent).

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.

Hershey and Chase experiment, step by step
  1. 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.
  2. The labelled viruses were allowed to infect bacteria.
  3. 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.
  4. The medium and the bacterial cells (pellet) were analysed for radioactivity. was found in the bacterial cells, while was found in the medium.
  5. 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.
Figure 13.5: The Hershey and Chase experiment. Radioactive label is shown in the accent colour: ³⁵S in the protein coat of batch 1 and ³²P in the DNA of batch 2.
Results of the Hershey and Chase experiment
BatchLabel and where it isFound after blending and centrifugingMeaning
1 in the protein coatIn the medium (liquid)Protein did not enter the bacteria
2 in the DNAIn 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.
Figure 13.6: Models of DNA replication. Plain lines are parental strands, coloured lines are strands made in the first round and grey lines are strands made in the second round.
Comparison of the three models
PointSemi-conservativeConservativeDispersive
Parental strands after replicationSeparated, one in each daughter duplexStay together in one daughter duplexBroken up and scattered through both daughters
Daughter duplex 1One parental + one new strandBoth parental strandsMixture of parental and new DNA
Daughter duplex 2One parental + one new strandBoth new strandsMixture 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.

Meselson and Stahl experiment, step by step
  1. They grew bacteria in a medium containing the heavy isotope of nitrogen, , which became incorporated into the bases of the bacterial DNA.
  2. After several generations the bacteria were shifted to three separate plates, which had been poured with a medium containing the light isotope, .
  3. 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.
  4. 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.
Figure 13.7 and 13.8: The Meselson-Stahl experiment. After one generation all the DNA is a hybrid band (accent); after two generations there are two bands, hybrid and light.
Results and interpretation
SampleBand in the tubeReason
Control ( only)Lightest: at the topBoth strands contain
0 minutesHeaviest: at the bottomBoth 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 levelThe 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.

Worked example: predicting the DNA bands for later generations
Bacteria were grown for many generations in medium and then moved to medium, where they divide once every 20 minutes. If DNA replication is semi-conservative, what fraction of the DNA duplexes is hybrid () and what fraction is light () after 3 generations (60 minutes) in medium? Which bands appear in the centrifuge tube?
  1. 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.
  2. 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.
  3. The rest are light. Fill in the table below.
Expected duplexes and bands
Generation in TimeTotal duplexesHybrid duplexesLight duplexesFraction hybridBands in the tube
00 min1 (all heavy)00none (1 heavy band)heavy
120 min220allone hybrid band
240 min4221/2hybrid and light
360 min8261/4hybrid (thin) and light (thick)
480 min162141/8hybrid (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.

Figure 13.9: Origin of replication, replication bubbles and replication forks. Three origins on one eukaryotic DNA molecule open into bubbles, which grow and join to give two daughter DNA molecules.

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.

The three DNA polymerases
EnzymeFunction
DNA polymerase ICatalyses 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 IIIs involved in repairing DNA damage during the life time of a cell.
DNA polymerase IIIThe 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′.

Figure 13.10: Process of DNA replication at a replication fork. The fork moves to the left; new DNA (accent) is made as the leading strand (continuous) and the lagging strand (Okazaki fragments).

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.

Leading strand and lagging strand
PointLeading strandLagging strand
Direction of synthesisTowards the replication forkAway from the replication fork
ContinuitySynthesized continuously as one long strandSynthesized discontinuously as short Okazaki fragments
Primers neededOne primer at the startA new primer for every Okazaki fragment
Joining of piecesNot needed (one piece)Fragments joined by DNA ligase after primers are replaced
Worked example: how many Okazaki fragments and primers?
The lagging-strand template of a fork has to be copied over a stretch of 30 000 nucleotides. In a prokaryote each Okazaki fragment is about 1500 nucleotides long (the book gives 1000 to 2000); in a eukaryote it is about 150 nucleotides (the book gives 100 to 200). How many Okazaki fragments, and so how many RNA primers, are needed for the lagging strand in each case? How many primers does the leading strand of this fork need?
  1. Each Okazaki fragment is started by one primer, so the number of primers on the lagging strand equals the number of fragments.
  2. Prokaryote.
  3. Eukaryote.
  4. 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.
  5. 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.
The same stretch copied in a prokaryote and a eukaryote
ProkaryoteEukaryote
Fragment size (book)1000 to 2000 nucleotides100 to 200 nucleotides
Fragments for 30 000 nucleotidesabout 20about 200
Primers on the lagging strandabout 20about 200
Primers on the leading strand11
Gaps sealed by DNA ligaseabout 19about 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.

Figure 13.11: Overview of DNA replication. Replication runs in both directions from one origin. At each fork one new strand is leading (continuous) and the other is lagging (Okazaki fragments).
The events of DNA replication in order
  1. 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.
  2. Single-strand binding (SSB) proteins attach to the separated strands so that they cannot pair up again. Both strands now act as templates.
  3. Primase makes a short RNA primer on each template strand (the primase and helicase together form the primosome).
  4. 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.
  5. On the lagging strand polymerase III works away from the fork in short pieces: each Okazaki fragment needs its own primer.
  6. Proofreading: a subunit of DNA polymerase III removes any wrongly added nucleotide.
  7. Termination: DNA polymerase I replaces each RNA primer with DNA nucleotides, and DNA ligase joins the fragments by phosphodiester bonds.
  8. Result: two daughter DNA molecules, each with one parental and one new strand (semi-conservative replication).
Enzymes of replication
Enzyme or proteinRole in DNA replication
DNA helicaseOpens the duplex by breaking the base pairs; forms the replication bubble
DNA gyraseWorks ahead of helicase and reduces the tension created during unwinding
SSB proteinsKeep the separated single strands from pairing again
PrimaseMakes short RNA primers on the template strands
DNA polymerase IIIMain enzyme: adds nucleotides to the 3′ end of a primer, 5′ to 3′; proofreads
DNA polymerase IReplaces RNA primers by DNA nucleotides
DNA polymerase IIRepairs DNA damage during the life of the cell
DNA ligaseJoins the fragments by phosphodiester bonds
DNA helicase and DNA gyrase
PointDNA helicaseDNA gyrase
ActionBreaks the base pairs and separates the two strandsWorks slightly ahead of helicase and reduces the tension created by unwinding
ResultReplication bubble with two forksMakes unwinding of the duplex easier
PhaseInitiationInitiation
Nucleosome and primosome
PointNucleosomePrimosome
What it isDNA wrapped twice around a core of eight histonesComplex of primase and DNA helicase enzymes
Made ofDNA and histone proteinsTwo enzymes (primase + helicase)
RoleFirst level of packing of the chromosomeStarts 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.

DNA stability and variability as two outcomes of replication: correct copying keeps the sequence stable (accent); uncorrected errors produce variation.
Stability and variability of replicating DNA
PointDNA stabilityDNA variability
MeaningThe DNA molecule remains the same, so DNA sequences are conserved after numerous cell divisionsNew genetic variations are produced in the base sequence
CauseDNA polymerase II (repair of damage) and DNA polymerase III (proofreading) correct mistakesErrors made during DNA replication that are not corrected give mutations in the base sequences
EffectGenetic information is passed on unchangedGenetic 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.
The central dogma of molecular genetics: DNA makes protein through an intermediate RNA, by transcription and then translation. Proteins link the genotype to the phenotype.
Transcription and translation
PointTranscriptionTranslation
What happensSynthesis of RNA from DNASynthesis of a polypeptide from mRNA
Step of gene expressionFirst stepSecond step
Where (eukaryote)NucleusCytoplasm (ribosomes)
Enzyme / machineryOne enzyme: RNA polymeraseRibosomes, tRNA, mRNA and enzymes such as aminoacyl-tRNA synthetase
ProductRNA (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.

Promoter sequences
Prokaryotic promoterEukaryotic promoter
Binding sitesTATAAT (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)
Figure 13.12: Structure of the prokaryotic promoter region. The upper strand is the coding strand and the lower strand is the template strand; the −35 and −10 sequences (accent) are the binding sites for RNA polymerase.
Figure 13.13: Structure of the eukaryotic promoter region with the GC-rich box (−90), the CAAT box (−80) and the TATA box (−25), counted back from the start point +1.

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.

Figure 13.14: Structure of prokaryotic RNA polymerase. The core enzyme has two α, a β and a β′ subunit; with the σ (sigma) factor it is the holoenzyme.
  • 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.

Figure 13.15: Process of transcription (prokaryote) in four stages. The RNA transcript is drawn in the accent colour and the thicker DNA segments are the promoter and the terminator.
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.

Figure 13.16 and 13.17: Termination of transcription. The GC-rich end of the new RNA folds into a hairpin (a stem of G–C base pairs with a loop), followed by a tail of U nucleotides; the hairpin makes RNA polymerase stop.
Transcription in order
  1. RNA polymerase (core enzyme + sigma factor) binds the promoter (−35 and −10 sequences in prokaryotes).
  2. The DNA unwinds and a transcription bubble forms; the sigma factor leaves.
  3. The polymerase moves 5′ to 3′ along the template strand, adding complementary ribonucleotides, with no primer.
  4. At the terminator the RNA forms a GC hairpin followed by poly-U, and the polymerase stops.
  5. The RNA is released; the sigma factor rebinds the core enzyme.
DNA polymerase and RNA polymerase
PointDNA polymeraseRNA polymerase
ProductDNA (daughter strands)RNA
PrimerNeeded (RNA primer with a 3′-OH)Not needed
DirectionAdds nucleotides to the 3′ end (5′ to 3′)Adds nucleotides to the 3′ end (5′ to 3′)
Number of enzymesThree 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.
Figure 13.18: mRNA processing in eukaryotes. The primary transcript receives a 5′ cap and a poly-A tail, then the spliceosomes remove the introns and join the exons; the mature mRNA leaves the nucleus through a nuclear pore.
Intron and exon
PointIntronExon
NatureNon-coding region of the gene and primary mRNACoding region of the gene and mRNA
Fate in processingRemoved (excised) by spliceosomesKept and joined together by RNA ligase
In mature mRNAAbsentPresent
Processing of eukaryotic mRNA
  1. Transcription makes the primary mRNA (pre-mRNA) with exons and introns.
  2. A cap is added to the 5′ end and a poly-A tail to the 3′ end for protection.
  3. Spliceosomes cut out the introns.
  4. RNA ligase joins the exons to make mature mRNA.
  5. 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.

Figure 13.19: Types of codons (the genetic code table). Each cell gives the codon and the amino acid it codes for.
First baseThird baseSecond base USecond base CSecond base ASecond base G
UUUUU PhenylalanineUCU SerineUAU TyrosineUGU Cysteine
UCUUC PhenylalanineUCC SerineUAC TyrosineUGC Cysteine
UAUUA LeucineUCA SerineUAA StopUGA Stop
UGUUG LeucineUCG SerineUAG StopUGG Tryptophan
CUCUU LeucineCCU ProlineCAU HistidineCGU Arginine
CCCUC LeucineCCC ProlineCAC HistidineCGC Arginine
CACUA LeucineCCA ProlineCAA GlutamineCGA Arginine
CGCUG LeucineCCG ProlineCAG GlutamineCGG Arginine
AUAUU IsoleucineACU ThreonineAAU AsparagineAGU Serine
ACAUC IsoleucineACC ThreonineAAC AsparagineAGC Serine
AAAUA IsoleucineACA ThreonineAAA LysineAGA Arginine
AGAUG Methionine (start)ACG ThreonineAAG LysineAGG Arginine
GUGUU ValineGCU AlanineGAU AspartateGGU Glycine
GCGUC ValineGCC AlanineGAC AspartateGGC Glycine
GAGUA ValineGCA AlanineGAA GlutamateGGA Glycine
GGGUG ValineGCG AlanineGAG GlutamateGGG 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
  1. 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.
  2. 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.
  3. No punctuation. There is no punctuation between codons: there is no gap between two codons.
  4. Non-overlapping. The mRNA sequence AUGAGCGCA is not read as AUG / UGA / GAG and so on; it is read as AUG / AGC / GCA.
  5. Triplet. Each codon is made of three bases.
The genetic code is read as non-overlapping triplets with no punctuation (no gap) between codons.
Worked example: from the length of an mRNA to the length of a protein
The coding region of an mRNA, from the start codon to the end of the stop codon, is 900 nucleotides long. (a) How many codons is this? (b) How many amino acids are in the polypeptide? (c) How many base pairs of DNA code for it? (d) How many different codons are there altogether, and why is a triplet and not a single base or a pair enough?
  1. The code is a triplet and non-overlapping, so the number of codons is the length divided by 3.
  2. 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.
  3. 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.
  4. 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).
Why the code is a triplet
Length of code wordCombinationsEnough for 20 amino acids?
1 base4No
2 bases16No
3 bases (triplet)64Yes
Genetic code and codon
PointGenetic codeCodon
MeaningThe relationship between the sequence of nucleotides in DNA (copied to mRNA) and the sequence of amino acids in a proteinA triplet of bases in mRNA that specifies one amino acid (or a stop signal)
SizeThe whole set of rules (64 triplets for 20 amino acids and stops)Three nucleotides
ExampleDegenerate, universal, non-overlapping, tripletAUG codes for methionine and starts translation
Start, stop, sense and nonsense codons
PointStart codonStop codonSense codonNonsense codon
Codon(s)AUGUAA, UAG, UGAAll 61 codons that code for amino acids (AUG included)The three stop codons
Codes forMethionine; begins the chainNo amino acidA specific amino acidNo amino acid
PositionAt the beginning of the messageAt the end of the messageAnywhere inside the messageAt 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.

Figure 13.20: Activation of amino acids. An amino acid and ATP are joined by aminoacyl-tRNA synthetase (releasing PPᵢ); the amino acid is then attached to the 3′ end of its tRNA (releasing AMP) to make the aminoacyl-tRNA complex.
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.

Initiation complex
  1. 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.
  2. 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.
  3. 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.
Figure 13.21: Formation of the initiation complex. The initiator tRNA (carrying methionine) and the 5′ end of the mRNA bind the small subunit; then the large subunit is placed on it, giving the E, P and A sites.
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.

Three steps of elongation
  1. 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.
  2. 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.
  3. 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.
Figure 13.22: Chain elongation in translation, in three repeated steps. The growing polypeptide chain (accent) is passed from the P-site tRNA to the amino acid on the A-site tRNA, and the ribosome then moves one codon along the mRNA (5′ to 3′).
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.

Figure 13.23: Termination of translation. A release factor recognises the stop codon at the A site, the polypeptide (accent) is released and the ribosome falls apart into its two subunits.
Worked example: from a template strand to a polypeptide
A segment of the template (antisense) strand of a gene reads 3′-TAC GCA CTT GGT ATT-5′. Write (a) the mRNA made from it, (b) the anticodons of the tRNAs that read it, and (c) the amino acids of the polypeptide, using the genetic code table.
  1. 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'
  2. Translation: each tRNA anticodon is complementary to its codon.
    \text{anticodons: UAC, GCA, CUU, GGU (no tRNA for UAA)}
  3. Read the codons in the genetic code table: AUG = methionine (start codon), CGU = arginine, GAA = glutamate, CCA = proline, UAA = stop codon.
Gene to protein
Codon 1Codon 2Codon 3Codon 4Codon 5
Template (3′ to 5′)TACGCACTTGGTATT
mRNA codon (5′ to 3′)AUGCGUGAACCAUAA
tRNA anticodonUACGCACUUGGUnone (release factor)
Amino acidMethionineArginineGlutamateProlineStop
\text{Met - Arg - Glu - Pro}\ \ (\text{stop at UAA})
Protein synthesis in brief
  1. Activation: each amino acid is attached to its own tRNA by aminoacyl-tRNA synthetase.
  2. Initiation: initiator tRNA (methionine) and the 5′ end of mRNA bind the small subunit; the large subunit joins.
  3. Elongation: codon recognition at the A site, peptide bond by peptidyl transferase, translocation.
  4. Termination: a stop codon at the A site, release factors free the polypeptide, subunits separate.
Difference between translation in prokaryotes and eukaryotes
Translation in prokaryotes and eukaryotes
PointProkaryotesEukaryotes
RibosomesSmaller: a 50S large subunit and a 30S small subunit bind to form the 70S ribosomeLarger: a 60S large subunit and a 40S small subunit bind to form the 80S ribosome
Where ribosomes occurIn the cytosolIn the cytosol and on the rough endoplasmic reticulum
Initiating amino acidModified: N-formyl methionineMethionine, not modified
Initiation factorsOnly one type is usedMore than one type is used
Release factorsTwo release factors are usedA 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.

Fate of the synthesized protein
Use of the proteinExamples given in the book
Within the cellCellular enzymes, membrane proteins, histone proteins
Outside the cell (secreted)Digestive enzymes, protein hormones, albumins, fibrinogens, antibodies
From polypeptide to working protein
  1. The polypeptide folds into a three-dimensional shape.
  2. It enters the Golgi complex for further modification.
  3. 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.
Figures 13.24 and 13.25: Positive control (left): the regulatory protein (activator) helps RNA polymerase and triggers transcription. Negative control (right): the regulatory protein (repressor) shuts transcription down.
Positive and negative control of gene expression
PointPositive controlNegative control
Regulatory proteinActivatorRepressor
Effect on gene expressionIncreases it: triggers transcriptionDiminishes it: shuts transcription down
Without the regulatory proteinLittle or no transcription (RNA polymerase does not start)Transcription goes on (nothing blocks RNA polymerase)
With the regulatory proteinTranscription takes placeNo 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).

Genes of the lac operon
GeneProductRole
i (regulatory gene)Repressor proteinWhen lactose is absent, the repressor binds to the operon and prevents production of the enzymes needed for lactose breakdown
zβ-galactosidaseHydrolyses lactose into glucose and galactose
yPermeaseIncreases the cell's permeability to β-galactosides, facilitating lactose entry into the cell
aTransacetylaseTransfers 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.

Figure 13.26: The lac operon. Without the inducer the repressor (made by the i gene) sits on the operator (accent) and the genes z, y and a are off; with the inducer (lactose) the repressor is inactive and the three genes are transcribed.
Working of the lac operon
  1. 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.
  2. 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.
  3. The lac mRNA is translated into the three enzymes: β-galactosidase, permease and transacetylase.
Worked example: predicting what happens when the lac operon is changed
In a bacterium the genes of the lac operon are normally switched off when lactose is absent. Predict whether the enzymes of lactose breakdown are made, with and without lactose, in (a) a normal cell, (b) a cell with a damaged i gene that cannot make a working repressor, and (c) a cell whose operator has changed so that the repressor cannot bind to it.
  1. 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.
  2. 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.
  3. 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.
  4. Changed operator: the repressor cannot bind, so again RNA polymerase is never blocked and the genes are transcribed all the time.
Predictions for the lac operon
CellLactose absentLactose present
NormalGenes off: no enzymesGenes on: enzymes made
i gene damaged (no working repressor)Genes onGenes on
Operator cannot bind repressorGenes onGenes 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.

Sources of mutation
Kind of mutagenExamples given in the book
Physical mutagensRadiation of any kind: UV, gamma rays, X-rays, radioactive rays; high temperature alterations
Chemical mutagensColchicine, mustard gas, nitrous gas, acridine orange, reactive radioactive isotopes, free oxygen particles
Biological mutagensCertain 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.

Classification of mutations used in this chapter: by how they arise (natural or induced) and by where they occur and how far (point or chromosomal).
Natural and induced mutations
PointNatural mutationsInduced mutations
How they ariseOccur naturally and automatically, due to internal or external factorsProduced artificially, for the establishment of new varieties of organisms
CausesErrors that occur during meiosis or DNA replicationPhysical and chemical mutagens are mainly responsible
ControlNot planned by peopleDeliberately produced
Point mutations and chromosomal mutations
PointPoint mutationChromosomal mutation (aberration)
What changesA single or a few nucleotides in the DNAThe structure or the number of chromosomes
ScaleSmall (gene level)Large (whole chromosome level)
Examples in the bookSickle cell anaemia, phenylketonuriaDown syndrome, Klinefelter syndrome, Turner syndrome
Mutagen and mutant
PointMutagenMutant
MeaningThe agent that causes a mutationThe organism or cell in which the mutation has occurred
ExamplesUV, X-rays, mustard gas, certain virusesA 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.

Harmful and useful aspects of mutation
PointHarmful mutationUseful (beneficial) mutation
Effect on fitnessDecreases the fitness of the organism in its environmentFavourable: tends to accumulate in the gene pool
Fate in the populationRemoved from the gene pool by natural selectionRetained and accumulated by natural selection
ExamplesCancer, Down syndrome, sickle cell anaemia, phenylketonuriaChanges 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.

Non-disjunction: a pair of chromosomes fails to separate in meiosis, so two gametes have one chromosome too many (n + 1) and two have one too few (n − 1).
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.
The chromosome constitution of the three syndromes (schematic, in the spirit of Figures 13.27 to 13.29): an extra chromosome 21 (accent), an extra X in a male, and a missing sex chromosome in a female.
Chromosomal mutations compared
PointDown syndromeKlinefelter syndromeTurner syndrome
Chromosome constitution2n+1, trisomy 21 (47)2n+1, 44+XXY (47)2n-1, 44+XO (45)
Sex of the affected personMale or femaleMaleFemale
Chromosome involvedAutosome (21)Sex chromosomes (extra X)Sex chromosome (X missing)
Key symptomsSmall head, rounded inner corner of eyes, single palm crease, flat noseFeminine characters, enlarged breasts, small testicles, long legsShort height, infertility, webbed neck, low hairline
CauseNon-disjunction of the 21st pair in maternal meiosisNon-disjunction of sex chromosomes in maternal meiosisNon-disjunction of sex chromosomes in maternal meiosis
TreatmentNone; managed by check-ups, medication, surgery, counsellingTestosterone 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.

Figure 13.30: Types of point mutation in a stretch of DNA: deletion, insertion and base substitution. The changed base is shown in the accent colour.
Types of point mutation
TypeWhat happensExample from Fig. 13.30
Base substitutionOne base is replaced by anotherATTAGCCTTAGAACT becomes ATTAGCCATAGAACT
InsertionOne or more bases are addedATTAGCCTTAGAACT becomes ATTAGCCATTAGAACT
DeletionOne or more bases are removedATTAGCCTTAGAACT becomes ATTAGCCTAGAACT
Worked example: what a point mutation does to a protein
A short mRNA reads 5′-AUG GAA CCC UUU AAA-3′. Use the genetic code to find the polypeptide it makes, and then the effect of each change: (a) the third base of GAA changes to G; (b) the middle base of GAA changes to U; (c) the first G after AUG is deleted; (d) one U is inserted straight after AUG.
  1. Read the original in codons.
    \text{AUG GAA CCC UUU AAA}\ \to\ \text{Met - Glu - Pro - Phe - Lys}
  2. (a) GAA becomes GAG, which still codes for glutamate (the code is degenerate). The protein is unchanged.
  3. (b) GAA becomes GUA, which codes for valine. One amino acid changes (Glu to Val), as in sickle cell anaemia.
  4. (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).
  5. (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).
Effect of point mutations
ChangeNew mRNAPolypeptideEffect
NoneAUG GAA CCC UUU AAAMet - Glu - Pro - Phe - LysNormal
(a) substitution GAA to GAGAUG GAG CCC UUU AAAMet - Glu - Pro - Phe - LysNo change in the protein
(b) substitution GAA to GUAAUG GUA CCC UUU AAAMet - Val - Pro - Phe - LysOne amino acid changed
(c) deletion of GAUG AAC CCU UUA AAMet - Asn - Pro - Leu ...Frame shift: new sequence
(d) insertion of UAUG UGA ACC CUU UAA AMet - stopFrame 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.
Figure 13.31: Normal red blood cells and sickle cells. Abnormal haemoglobin forms strands (accent) that bend the cell into a sickle shape, and the sticky sickle cells block the flow of blood.
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.
Figure 13.32: Inheritance pattern of sickle cell anaemia and phenylketonuria as an autosomal recessive trait. Half-shaded symbols are carriers (Aa); fully shaded symbols are affected (aa).
Sickle cell anaemia and phenylketonuria
PointSickle cell anaemiaPhenylketonuria
Type of mutationGene (point) mutation: glutamic acid replaced by valine in the beta chain of haemoglobinGene (point) mutation: defective phenylalanine hydroxylase enzyme
InheritanceAutosomal recessiveAutosomal recessive
Main effectSickle-shaped red cells with less oxygen-carrying capacityPhenylalanine is not converted to tyrosine; toxic phenylketones damage the nervous system
TreatmentBlood transfusions, pain medicines, fluids, folic acid; bone marrow or stem cell transplant can cureDiet 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

1
Chromosomes are made of:
  1. ADNA + Pectin
  2. BRNA + DNA
  3. CDNA + HistoneAnswer
  4. DDNA only
Why

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.

2
In his work with pneumonia-causing bacteria and mice, Griffith found that:
  1. Aheat-killed pathogenic cell caused pneumonia
  2. Bsome substance from pathogenic cells was transferred to non-pathogenic cells, making them pathogenic.Answer
  3. Cthe polysaccharide coat of bacteria caused pneumonia.
  4. Dbacteriophage injected DNA into bacteria.
Why

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.

3
A biochemist isolates and purifies molecules needed for DNA replication. When DNA is added, replication occurs, but each DNA molecule consists of a normal strand paired with numerous segments of DNA a few hundred nucleotides long. What has probably been left out of the mixture?
  1. ADNA polymerase
  2. BDNA ligaseAnswer
  3. COkazaki fragments
  4. DPrimase
Why

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.

4
What is the basis for the difference in how the leading and lagging strands of DNA molecules are synthesized?
  1. Athe origins of replication occur only at the 5' end.
  2. BDNA ligase works only in the 3' to 5' direction.
  3. Cpolymerase can only work on one strand at a time.
  4. DDNA polymerase can join new nucleotides only to the 3' end of a growing strand.Answer
Why

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.

5
Synthesis of a new DNA strand usually begins with:
  1. Aan RNA primerAnswer
  2. BDNA ligase
  3. Ca DNA primer
  4. Dan Okazaki fragment
Why

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.

62025 Annual I, Set B, MCQ 9
A sub metacentric chromosome has centromere:
  1. Ain the centre of chromosome
  2. Bnear an end of chromosome
  3. Cslightly away from the centre of a chromosomeAnswer
  4. Dat the end of a chromosome
Why

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.

7
The elongation of the leading strand during DNA synthesis:
  1. Aprogresses away from the replication fork
  2. Boccurs in the 3' to 5' direction
  3. Cproduces Okazaki fragments
  4. Ddepends on the action of DNA polymeraseAnswer
Why

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.

8
Whose pioneering work on the gene used the bacteriophage to show that DNA was the part of this virus which went into the bacterial cell for replication?
  1. AWatson and Crick
  2. BMeselson and Stahl
  3. CHershey and ChaseAnswer
  4. DBeadle and Tatum
Why

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.

9
If you want to label amino acids but not DNA, which of the following radioactive isotopes would you use?
  1. A^{18}F
  2. B^{35}SAnswer
  3. C^{14}C
  4. D^{32}P
Why

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.

10
In eukaryotic cells, transcription cannot begin until:
  1. Athe two DNA strands have completely separated and exposed the promoter
  2. Bseveral transcription factors have bound to the promoterAnswer
  3. Cthe DNA introns are removed from the template
  4. DDNA nucleases have isolated the transcription unit
Why

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.

11
The anticodon of a particular tRNA molecule is:
  1. Acomplementary to the corresponding triplet in mRNAAnswer
  2. Bcomplementary to the corresponding triplet in rRNA
  3. Cthe part of the tRNA bonds to specific amino acid
  4. Dcatalytic, making the tRNA
Why

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

12025 Annual I, Set D, Q.2 (xii) OR3 marks
On the basis of position of centromere describe the four types of chromosomes.
Solution

A chromosome is classed by where its centromere lies along its length, which fixes the length of its two arms (p.347).

TypePosition of centromereArms
MetacentricIn the centreTwo arms of nearly equal length
SubmetacentricSlightly away from the centreTwo arms, one a little longer
AcrocentricNear one endOne very short arm and one long arm
TelocentricAt one endOnly one arm
23 marks
What is chromatin fibre? What is the difference between the two regions of chromatin fibre?
Solution

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.

RegionDegree of condensationGenes
HeterochromatinHighly condensedNot expressed (unexpressed region)
EuchromatinNon-condensed (loosely packed)Expressed

When the cell divides, the euchromatin is also condensed, so that a uniform chromatin fibre is established (p.348).

33 marks
Write the functions of DNA polymerase I, II and III.
Solution
Functions of the three DNA polymerases (p.355 to 356)
EnzymeFunction
DNA polymerase IReplaces the RNA primers with DNA nucleotides in the termination phase, so supporting DNA polymerase III in the main replication process.
DNA polymerase IITakes part in repairing DNA damage during the life of the cell.
DNA polymerase IIIIs the main enzyme that synthesises both daughter strands along the template. Its proofreading subunit removes a wrongly added nucleotide.
43 marks
Describe central dogma of gene expression.
Solution

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):

  1. Transcription: the DNA sequence of a gene is copied into mRNA by RNA polymerase, in the nucleus of eukaryotes.
  2. Translation: the mRNA is decoded by the ribosome, which links amino acids in the order given by the codons to form a polypeptide.
  3. The result is that the DNA sequence determines the protein, and so the character or function it controls: DNA \to RNA \to protein.
53 marks
Describe the four characteristics of genetic code.
Solution

The genetic code is a triplet code: each codon is three bases, giving 64 combinations (p.361). Its four characteristics are:

  1. Degeneracy: most amino acids are coded by more than one codon. Leucine and serine have six codons, and methionine and tryptophan have only one.
  2. 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.
  3. Non-punctuated: there is no gap or punctuation between codons, so the message is read as one continuous string.
  4. Non-overlapping: each base belongs to only one codon. The sequence AUGAGCGCA is read as AUG/AGC/GCA, not as AUG/UGA/GAG.
63 marks
Explain why the length of transcribed m-RNA (in Eukaryotes) shortens as it enters the cytoplasm for translation.
Solution

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):

  1. RNA splicing: spliceosomes, large RNA-protein complexes, cut out the introns.
  2. RNA ligase then joins the spliced exon fragments into one continuous mature mRNA.
  3. 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.
73 marks
Interpret how many types of t-RNA molecules are necessary for a living cell, if the genetic code is triplet code.
Solution

The triplet code gives 64 codons (4 \times 4 \times 4, p.361). Of these:

  1. Three codons (UAA, UAG, UGA) are stop codons. They code no amino acid and are recognised by release factors, not by tRNA (p.364).
  2. The remaining 61 sense codons each specify one amino acid. Each needs a tRNA whose anticodon pairs with it.
  3. 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.

83 marks
Make a list of some commonly occurring minor mutations in human.
Solution

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.

93 marks
Suggest possible ways in which the synthesized protein can be used within or outside a cell that synthesized it.
Solution

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 cellUsed outside the cell
Cellular enzymes that speed up reactionsDigestive enzymes
Membrane proteinsProtein hormones
Histone proteins that help pack the chromosomeAlbumins, fibrinogens and antibodies
10
Write the differences between:
Solution

Each pair is answered in a table with the points the book uses for comparison.

(a) metacentric and sub metacentric chromosome
FeatureMetacentricSub metacentric
CentromereIn the centreSlightly away from the centre
ArmsNearly equalUnequal, one arm longer
(b) acrocentric and telocentric chromosome
FeatureAcrocentricTelocentric
CentromereNear one endAt the very end
ArmsOne very short arm and one long armOnly one arm
PositionCentromere close to the terminal endCentromere at the terminal end
(c) nucleosome and primosome
FeatureNucleosomePrimosome
CompositionDNA wound about twice round a core of eight histone moleculesA complex of the enzymes primase and DNA helicase
RolePacks the DNA into a nucleosome stringLays down the RNA primers that start replication
StageFormed just after DNA replicationActive during replication initiation
(d) heterochromatin and euchromatin
FeatureHeterochromatinEuchromatin
CondensationHighly condensedNon-condensed
GenesNot expressedExpressed
ChromatinDense, compact regionLoose region
(e) conservative and dispersive model of DNA replication
FeatureConservative modelDispersive model
Parental DNAKept intact in one daughter moleculeBroken into nucleotides or segments
Daughter moleculesOne has both parental strands; the other has only new strandsEach strand is a mixture of old and new segments
Result after two cyclesOld and entirely new moleculesEvery molecule contains both old and new material
(f) DNA helicase and DNA gyrase
FeatureDNA helicaseDNA gyrase
FunctionBreaks the hydrogen bonds between base pairs so the two strands separateRelieves the tension built up by unwinding
PositionAt the fork, unwinding the duplexWorks slightly ahead of the helicase
ActionOpens the duplexReduces the strain in the DNA
(g) DNA polymerase I and II
FeatureDNA polymerase IDNA polymerase II
Main jobReplaces RNA primers with DNA nucleotides in terminationRepairs DNA damage
StageTermination of replicationThroughout the life of the cell
HelpsSupports DNA polymerase IIIKeeps the DNA stable
(h) leading strand and lagging strand of replication fork
FeatureLeading strandLagging strand
SynthesisContinuous, towards the forkDiscontinuous, away from the fork
PrimersOne primer at the startA primer for each Okazaki fragment
ProductsOne long daughter strandShort Okazaki fragments, joined by DNA ligase
(i) translation and transcription in gene expression
FeatureTranscriptionTranslation
MakesmRNA from DNAA polypeptide from mRNA
PlaceIn the nucleus (eukaryotes)On ribosomes in the cytoplasm
MachineryRNA polymeraseRibosome with tRNA and enzymes
Units readBases of the template strandCodons of the mRNA
(j) intron and exon
FeatureIntronExon
CodingNon-codingCoding
FateRemoved by splicingKept; joined to form mature mRNA
FoundBetween exons in the transcriptBetween introns in the transcript
(k) start codon and stop codon
FeatureStart codonStop codon
CodonAUGUAA, UAG or UGA
Amino acidCodes methionineCodes no amino acid
RoleMarks the beginning of the messageEnds translation; recognised by release factors
(l) non sense codon and sense codon
FeatureNon-sense codonSense codon
NumberThree (UAA, UAG, UGA)Sixty-one
Amino acidNoneEach specifies one amino acid
RoleStops translationAdds an amino acid to the chain
(m) point mutation and chromosomal mutation
FeaturePoint mutationChromosomal mutation
ChangeOne or a few nucleotidesStructure or number of whole chromosomes
ExampleSickle cell anaemiaDown syndrome
CauseBase substitution, insertion or deletionNon-disjunction in meiosis
(n) harmful and useful aspects of mutation
FeatureHarmfulUseful
EffectReduces the fitness of the organismGives variation; raw material for evolution
ExamplesCancer, sickle cell anaemia, Down syndromeFavourable changes that accumulate by natural selection
FateRemoved by natural selectionTend to accumulate in the population
(o) Down syndrome and Klinefelter syndrome
FeatureDown syndromeKlinefelter syndrome
Chromosomes2n + 1 (47): trisomy 212n + 1 (47): 44 + XXY
SexMale or femaleMale
Chromosomal changeExtra chromosome 21 (autosome)Extra X chromosome (sex chromosome)
(p) Klinefelter syndrome and Turner's syndrome
FeatureKlinefelter syndromeTurner's syndrome
Karyotype2n + 1: 44 + XXY2n - 1: 44 + XO (45)
SexMaleFemale
Main featuresSmall testes, wide hips, sparse body hairShort stature, webbed neck, infertility

Extensive answer questionsp. 374–375

18 marks
Describe the structure of chromosome.
Solution

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

Figure: a chromosome with its labelled regions.

Types by centromere position

Fig 13.1: the four types of chromosome.
TypePosition of the centromere
MetacentricIn the centre
SubmetacentricSlightly away from the centre
AcrocentricNear one end
TelocentricAt one end
28 marks
Describe in detail the composition and organization of chromosome.
Solution

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

  1. 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).
  2. 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).
  3. Supercoil: the higher-order coiling of the chromatin fibre gives a supercoil with a diameter of 300 nm (p.349).
  4. Chromatid: further coiling of the supercoil gives a chromatid 700 nm thick. Two chromatids make up the chromosome, about 1400 nm thick.
Figure: the levels of organisation of a chromosome, from the DNA double helix to the chromosome.

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).

38 marks
Describe the experiment of Griffith to prove that DNA is the hereditary material.
Solution

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:

Figure: Griffith's experiment, the four treatments, their results and conclusions.

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).

48 marks
Describe the experiment of Hershey - Chase to prove that DNA is the hereditary material.
Solution

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

  1. Grow two batches of phages: batch 1 with ^{35}S (protein labelled) and batch 2 with ^{32}P (DNA labelled).
  2. Allow the labelled phages to infect separate cultures of bacteria.
  3. Separate the phages from the bacteria by blending (shaking) the mixture, which strips the empty protein coats off the bacteria.
  4. Centrifuge the mixture. The bacteria form the pellet at the bottom, and the lighter phage coats stay in the liquid.
  5. Measure the radioactivity in the liquid and in the pellet.
Figure: the Hershey and Chase experiment with the two batches.

Results and conclusion

BatchLabelRadioactivity found inMeaning
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).

58 marks
Describe the three models proposed about the mechanism of DNA replication.
Solution

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.

Figure: the three models of DNA replication over two cycles.

Comparison

ModelTwo daughter molecules after one cycleFate of parental DNA
Semi-conservativeEach has one parental and one new strandEach parental strand is kept whole
ConservativeOne has both parental strands, the other both new strandsParental strands stay together
DispersiveEach strand is a mixture of old and new segmentsParental DNA is broken up and dispersed

The Meselson and Stahl experiment (question 6) showed that the semi-conservative model is correct.

62025 Annual I, Set D, Q.38 marks
Describe the work of Meselson and Stahl to justify the semi-conservative replication as the correct method of replication.
Solution

Aim

Meselson and Stahl (1958) tested the three models of replication to find which one was correct (p.353).

Procedure

  1. 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.
  2. 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.
  3. 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.
Figure: the Meselson and Stahl density gradient results.

Results

SampleBand positionInterpretation
Control (^{14}N only)Top, lightestBoth strands are light
0 minutesBottom, heaviestBoth strands are heavy (^{15}N)
20 minutesOne intermediate bandOne heavy and one light strand (hybrid)
40 minutesIntermediate band and top bandHybrids 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.

78 marks
Describe the events of the process of DNA replication and explain DNA stability and variability as two characters of the replicating DNA molecule.
Solution

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

  1. The origin of replication is a specific sequence where the process begins. Prokaryotes have one origin; eukaryotic DNA may have several.
  2. DNA helicase breaks the hydrogen bonds between base pairs and unwinds the duplex, while DNA gyrase works slightly ahead to relieve the tension.
  3. A replication bubble forms, and each side of it is a replication fork.
  4. Single-stranded binding (SSB) proteins stop the separated strands from pairing again. The strands now act as templates.

2. Extension (polymerisation)

  1. Primase lays down short RNA primers. Primase and helicase together form the primosome.
  2. DNA polymerase III adds nucleotides to the 3' end of each primer, so the new strand grows 5' to 3'.
  3. On the leading strand, synthesis is continuous towards the fork.
  4. 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).
Figure: the replication fork, leading and lagging strands and Okazaki fragments.

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.

88 marks
Describe the mechanism of polymerase III activity.
Solution

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

  1. A primer sits on the template strand at the replication fork, held by the sliding clamp.
  2. The polymerase reads the template strand in the 3' to 5' direction.
  3. 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.
  4. The enzyme moves along the template and repeats step 3, so the chain grows continuously on the leading strand.
  5. One of its subunits proofreads: if a wrong nucleotide is added, the enzyme removes it before continuing.
Figure: DNA polymerase III adding nucleotides to a primer in the 5' to 3' direction.

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).

98 marks
Explain the mechanism of transcription.
Solution

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

  1. RNA polymerase binds to the promoter, a regulatory region of the gene.
  2. 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).
  3. 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

  1. One strand of the gene is the template (antisense) strand. The other is the coding (sense) strand.
  2. RNA polymerase adds ribonucleotide triphosphates complementary to the template strand. RNA is made in the 5' to 3' direction, and no primer is needed.
  3. 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

  1. The terminator region has a series of GC base pairs followed by AT base pairs.
  2. The transcribed RNA folds back on itself to form a GC hairpin loop, followed by a tail of poly-U nucleotides.
  3. The hairpin causes RNA polymerase to stop, and the completed RNA chain is released from the DNA.
Figure: initiation, elongation and termination of transcription.
102025 Annual I, Set B, Q.48 marks
Describe the mechanism of protein synthesis.
Solution

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).

Figure: the stages of translation, with the main events in each.

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

  1. The codon exposed at the A site binds its anticodon-bearing aminoacyl-tRNA with the help of an elongation factor.
  2. Peptidyl transferase removes the chain from the tRNA at the P site and joins it to the new amino acid by a peptide bond.
  3. 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).

118 marks
Write the difference between protein synthesis in prokaryotes and eukaryotes.
Solution

Translation follows the same basic steps in both groups, but the machinery and some details differ (p.364 to 365).

Difference between translation in prokaryotes and eukaryotes (p.364 to 365).
FeatureProkaryotesEukaryotes
RibosomeSmaller: 70S, made of a 50S large and a 30S small subunitLarger: 80S, made of a 60S large and a 40S small subunit
LocationCytosolCytosol and rough endoplasmic reticulum
Initiating amino acidN-formyl methionine, which is modifiedMethionine, which is not modified
Initiation factorsOnly one typeMore than one type
Release factorsTwoOne
Coupling with transcriptionNo delay: translation starts as the mRNA emerges from the DNAmRNA 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).

128 marks
What is the importance of the regulation of gene expression?
Solution

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).
138 marks
Describe the negative control of gene expression by repressor proteins and the positive control of gene expression by activator proteins.
Solution

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).

Figure: negative control by a repressor and positive control by an activator.

Comparison

FeatureNegative controlPositive control
Regulatory proteinRepressorActivator
Effect on expressionDiminished: transcription is shut downIncreased: transcription is started
ActionAttaches to the promoter or operator and stops RNA polymeraseAttaches 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).

142025 Annual I, Set D, Q.2 (ii) OR8 marks
Explain post-transcription modification of mRNA.
Solution

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

  1. Introns, the non-coding regions of the primary mRNA, are recognised.
  2. The spliceosome, a large RNA-protein complex, cuts out the introns.
  3. The exon fragments are joined together with RNA ligase, giving a mature mRNA with the exons in order.
  4. The mature mRNA, with its cap and poly-A tail, passes through a nuclear pore into the cytoplasm for translation.
Figure: processing of the pre-mRNA in eukaryotes.

Note. The book numbers the last items unnumbered after question 14. They appear here as the group following question 14, as printed.

(a) and (b) after 148 marks
(a) Define mutation and identify various sources of mutations. (b) Differentiate between natural and induced mutations.
Solution

(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

FeatureNatural mutationInduced mutation
OriginOccur naturally and automatically from internal or external factorsProduced artificially
CausesErrors during meiosis or DNA replicationMainly physical and chemical mutagens
PurposeNone; they arise in natureTo establish new varieties of an organism
(a) and (b) after 14, second pair8 marks
(a) Explain the harmful and useful aspects of mutations. (b) Rationalize that mutations might be a contributing factor towards evolution.
Solution

(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.
158 marks
Describe the symptoms, causes and possible treatment of the following chromosomal mutations: (a) Down syndrome (b) Klinefelter's syndrome (c) Turner's syndrome
Solution

(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.
Figure: how non-disjunction of chromosome 21 produces Down syndrome.

(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

SyndromeKaryotypeSexSpecific treatment
Down2n + 1 (47), trisomy 21Male or femaleNone; management of symptoms
Klinefelter2n + 1 (44 + XXY)MaleTestosterone around puberty
Turner2n - 1 (44 + XO)FemaleGrowth hormone and oestrogen replacement
168 marks
Describe the symptoms, causes and possible treatment of the following genetic mutations: (a) Sickle cell anaemia (b) Phenylketonuria
Solution

(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.
Figure: normal and sickled red blood cells, with the codon change.

(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.
Figure: the defective step in phenylketonuria.

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.

2025 Annual I, Set B, MCQ 41 mark
When RNA polymerase binds to promoter for transcription, the unit which detaches form it is:
  1. A\alpha (Alpha)
  2. B\beta (Beta)
  3. C\gamma (Gamma)Answer
  4. D\theta (Theta)
Why

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.

2025 Annual I, Set B, MCQ 91 mark
The condition in which centromere of a chromosome is located slightly away from the center is termed as:
  1. AAcrocentric
  2. BSub metacentricAnswer
  3. CTelocentric
  4. DMetacentric
Why

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.

2026 Annual I, Set B, MCQ 171 mark
Griffith and Hershey-Chase experiments proved that:
  1. AProteins are hereditary material
  2. BDNA is hereditary materialAnswer
  3. CRNA is genetic material
  4. DLipids carry genes
Why

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.

2026 Annual I, Set B, Q.2 (iii) OR1x3 = 3 marks
State three conclusions of Griffith control experiments.
Solution

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.

Griffith control experiments
Control experimentResultConclusion
Live R-type (rough, no capsule) injected aloneMice stay healthyR-type is non-virulent: it cannot cause disease by itself
Heat-killed S-type injected aloneMice stay healthyKilling by heat destroys virulence; the dead capsule material alone does not make the mice ill
Live R-type mixed with heat-killed S-typeMice develop disease and die; living S-type bacteria are found in their bloodSome hereditary information (the transforming principle) passed from the dead S-type to the living R-type and changed it into virulent S-type: transformation
2025 Annual I, Set B, Q.2 (vi) OR1+2 = 3 marks
What is point mutation? How it causes sickle cell anaemia?
Solution
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)
  1. 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.
  2. The sixth amino acid of the beta chain is therefore valine in place of glutamic acid. This abnormal haemoglobin is Hb^{\mathrm{S}}.
  3. 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).

2026 Annual I, Set B, Q.2 (x) OR1x3 = 3 marks
Describe three steps of Polypeptide elongation during protein synthesis.
Solution

Polypeptide elongation is repeated in three steps while the ribosome moves along the mRNA from the 5′ end towards the 3′ end.

  1. 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.
  2. 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.
  3. 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.
2025 Annual I, Set B, Q.2 (xi) OR3 marks
Interpret the results of Meselson and Stahl experiment when they grew bacteria in N_{15} and then in N_{14} medium for tube 2, 3 and 4. [Figure]
The figure as set: bands of DNA in four tubes (1 control, 2 at 0 min, 3 at 20 min, 4 at 40 min).
Solution

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.

TubeBand positionInterpretation
1 (control)Top (lightest)DNA with ^{14}\mathrm{N} in both strands, the reference for light DNA
2 (0 min)LowestAll 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 middleAll 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 middleHalf 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.

2025 Annual I, Set B, Q.2 (xii) OR3 marks
Name any three enzymes with their functions involved in DNA replication.
Solution

Three enzymes of DNA replication and their functions (any three):

EnzymeFunction
DNA helicaseBreaks the hydrogen bonds between the base pairs and opens the duplex at the origin to form the replication bubble and forks
PrimaseMakes short RNA primers on the template strands, which DNA polymerase needs to start
DNA polymerase IIIMain enzyme: adds nucleotides to the 3′ end of the primer (5′ to 3′) to make the leading and lagging strands, and proofreads
DNA polymerase IReplaces the RNA primers with DNA nucleotides
DNA ligaseJoins the Okazaki fragments by forming phosphodiester bonds
DNA gyraseWorks ahead of helicase and reduces the tension created by unwinding
2026 Annual I, Set B, Q.3 OR3+3 = 6 marks
Identify parts A - F in DNA replication process and briefly explain each part.
The figure as set: a replication fork with parts A to F.
Solution

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).

The answer figure: parts A to F of the replication fork named.
Identification and explanation (3 + 3 marks)
PartNameWhat it is and what it does
ADNA polymerase IIIThe 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.
BLagging strandThe 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.
CTemplate strandThe old (parental) strand that is copied. Its sequence decides the order of bases in the new strand, by base pairing.
DLeading strandThe new strand made continuously in one piece towards the fork by DNA polymerase III, after one primer.
EDNA helicaseWorks 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).
FParental DNAThe original double helix that has not yet been unwound. Each strand of it becomes the template for a daughter duplex.
2026 Annual I, Set B, Q.44+2 = 6 marks
Explain the double helical structure of DNA as proposed by Watson and Crick. Also draw diagram.
Solution

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)
Watson and Crick model of DNA: (a) key features of the double helix; (b) partial chemical structure with antiparallel backbones and hydrogen-bonded base pairs.

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).

2025 Annual I, Set B, Q.42+2+1+2 = 7 marks
Explain translation process starting from initiation complex till termination. Also draw diagram of initiation complex and elongation steps.
Solution

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)
  1. The tRNA carrying (N-formyl) methionine binds the smaller ribosomal subunit with an initiation factor.
  2. The 5′ end of the mRNA binds the small subunit with another initiation factor, the AUG start codon pairing with the initiator anticodon.
  3. 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)
  1. Codon recognition: the aminoacyl-tRNA with the complementary anticodon enters the A site, helped by an elongation factor.
  2. Peptide bond: peptidyl transferase moves the amino acid or chain from the P-site tRNA onto the amino acid at the A site.
  3. 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)
Initiation complex: the initiator tRNA (methionine) and the 5′ end of the mRNA bind the small subunit, then the large subunit joins and gives the E, P and A sites.
Chain elongation in three repeated steps: codon recognition at the A site, peptide bond formation, and translocation.
2025 Annual I, Set D, MCQ 11 mark
In tRNA molecule, the main function of middle loop is to:
  1. ARecognize enzyme activity
  2. BBind codon on mRNAAnswer
  3. CRecognize binding site on ribosome
  4. DAttach amino acid to CCA
Why

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.

2025 Annual I, Set D, MCQ 111 mark
During unwinding of DNA double helix, the hydrogen bonds between base pairs are broken by enzyme:
  1. AHelicaseAnswer
  2. BPrimase
  3. CPolymerase
  4. DGyrase
Why

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.

2026 Annual I, Set D, MCQ 171 mark
According to Griffith and Hershy - Chase genetic material proved to be:
  1. ADNAAnswer
  2. BProtein
  3. CRNA
  4. DLipid
Why

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.

2025 Annual I, Set D, Q.2 (ii) OR1+2 = 3 marks
Why post-transcriptional modifications occur in eukaryotic mRNA? Summarize the changes which occur during mRNA processing.
Solution
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)
  1. Capping: a cap of 7-methyl GTP is added to the 5′ end.
  2. 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.
  3. 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.
2026 Annual I, Set D, Q.2 (iii) OR1x3 = 3 marks
What were the findings of Hershey-Chase experiment? Write briefly.
Solution

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:

Findings of the Hershey-Chase experiment
FindingMeaning
^{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 cellsThe protein coat stays outside and does not enter
The infected bacteria went on to make new bacteriophagesThe injected DNA carries the hereditary information for making new viruses, so DNA is the genetic material
2026 Annual I, Set D, Q.2 (x) OR1x3 = 3 marks
Describe three stages of formation of initiation complex during translation process.
Solution

The initiation complex is formed by the ribosomal subunits, the mRNA and the first aminoacyl-tRNA. It forms in three stages.

  1. A tRNA carrying (N-formyl) methionine, the initiator, binds to the smaller ribosomal subunit with the help of an initiation factor.
  2. 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.
  3. 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.
2025 Annual I, Set D, Q.2 (xii) OR3 marks
Name the types of chromosomes on the basis of position of centromere mentioning its location in each type. (Any three)
Solution

Chromosomes are classified by the position of the centromere:

TypePosition of the centromereArms
MetacentricIn the centreTwo arms of about equal length
SubmetacentricSlightly away from the centreTwo unequal arms
AcrocentricNear the endOne very short and one long arm
TelocentricAt the endOnly one arm
2025 Annual I, Set D, Q.2 (xiv)3 marks
Write down the sources of mutagens with their examples. (Any three)
Solution

A mutagen is an agent that causes mutations. Its sources are physical, chemical and biological agents.

Source of mutagensExamples
PhysicalRadiation of any kind: ultraviolet rays, gamma rays, X-rays, radioactive rays; high temperature alterations
ChemicalColchicine, mustard gas, nitrous gas, acridine orange, reactive radioactive isotopes, free oxygen particles
BiologicalCertain 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.

2025 Annual I, Set D, Q.33+2+2 = 7 marks
Explain Meselson-Stahl experiment and results to prove that DNA replication is semi conservative. Also draw the diagram.
Solution
The experiment (3 marks)

Matthew Meselson and Franklin Stahl (1958) tested the three models (semi-conservative, conservative, dispersive) using nitrogen isotopes.

  1. Bacteria were grown for several generations in a medium containing heavy nitrogen, ^{15}\mathrm{N}, which became part of the bases of their DNA.
  2. The bacteria were shifted to a medium containing light nitrogen, ^{14}\mathrm{N}.
  3. DNA was sampled at 0, 20 and 40 minutes (and a control grown only in ^{14}\mathrm{N}).
  4. 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)
SampleBandMeaning
Control (^{14}\mathrm{N})TopLight DNA
0 minBottomHeavy DNA, both strands ^{15}\mathrm{N}
20 minMiddleHybrid: each molecule has one ^{15}\mathrm{N} and one ^{14}\mathrm{N} strand
40 minTop and middleHalf 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)
Meselson-Stahl experiment: bacteria grown in ^{15}\mathrm{N} are moved to ^{14}\mathrm{N}; the tubes show a heavy band at 0 minutes, one hybrid band at 20 minutes and hybrid plus light bands at 40 minutes.
2026 Annual I, Set D, Q.3 OR1x6 = 6 marks
Identify parts A - F in DNA replication process and briefly explain each part.
The figure as set: a replication fork with parts A to F.
Solution

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).

The answer figure: parts A to F of the replication fork named.
Identification and explanation (6 x 1 mark)
PartNameExplanation
ALeading strandThe new strand made continuously, in one piece, towards the replication fork, by DNA polymerase III from a single primer.
BRNA primerA 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.
COkazaki fragmentsThe short pieces of the lagging strand, made away from the fork, each from its own primer.
DDNA polymerase IIIThe main enzyme that adds nucleotides at the 3′ end of the primer in the 5′ to 3′ direction and proofreads.
EDNA helicaseBreaks 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).
FDNA ligaseJoins the Okazaki fragments, after the primers are replaced by DNA, by forming phosphodiester bonds so that the lagging strand becomes continuous.
2026 Annual I, Set D, Q.44+2 = 6 marks
Explain the double helical structure of DNA as proposed by Watson and Crick. Draw diagram as well.
Solution

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)
Watson and Crick model of DNA: (a) key features of the double helix; (b) partial chemical structure with antiparallel backbones and hydrogen-bonded base pairs.

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

  1. 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
  2. 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
  3. 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
  4. 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
  5. 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
  6. 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
  7. 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
  8. 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
  9. 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

Mark this unit as done to fill the Class 11 Biology ring.

Next, unit 14
Evolution

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