This chapter covers Advanced Higher Biology content, extending beyond Higher level.

Molecular Genetics

DNA Structure

DNA is a double-stranded polymer of nucleotides. Each nucleotide consists of:

• A deoxyribose sugar
• A phosphate group
• A nitrogenous base: adenine (A), thymine (T), cytosine (C), or guanine (G)

Base pairing: A pairs with T (2 hydrogen bonds); C pairs with G (3 hydrogen bonds).

Antiparallel strands: The two strands run in opposite directions (5’ to 3’ and 3’ to 5’).

Double helix: The two strands coil around each other, with bases on the inside and Sugar-phosphate backbones on the outside.

DNA Replication

Semi-conservative replication: Each new DNA molecule contains one original strand and one new Strand.

Meselson-Stahl experiment (1958): Used heavy (^{15}\mathrm{N) and light (^{14}\mathrm{N) Nitrogen to demonstrate semi-conservative replication.

Process:

1. Helicase unwinds and separates the double helix by breaking hydrogen bonds
2. DNA polymerase synthesises the new strand in the 5’ to 3’ direction
3. Primase synthesises RNA primers to initiate replication
4. Ligase joins Okazaki fragments on the lagging strand
5. On the leading strand, replication is continuous; on the lagging strand, it is discontinuous (Okazaki fragments)

Worked Example: The Meselson-Stahl experiment.

Bacteria are grown in medium containing ^{15}\mathrm{N (heavy nitrogen) for many generations. All Their DNA is heavy (^{15}\mathrm{N/^{15}\mathrm{N). They are then transferred to medium Containing ^{14}\mathrm{N (light nitrogen) and allowed to replicate once.

If replication is semi-conservative: each new DNA molecule contains one heavy strand and one light Strand (^{15}\mathrm{N/^{14}\mathrm{N). After one generation, all DNA is intermediate density.

After two generations: 50% intermediate (^{15}\mathrm{N/^{14}\mathrm{N) and 50% light (^{14}\mathrm{N/^{14}\mathrm{N). This is exactly what Meselson and Stahl observed, confirming Semi-conservative replication.

Protein Synthesis

Transcription: DNA $\to$ mRNA (in the nucleus).

1. RNA polymerase binds to the promoter region of the gene
2. The DNA double helix unwinds
3. RNA polymerase synthesises mRNA using the template strand (3’ to 5’)
4. MRNA is processed: 5’ cap added, poly-A tail added, introns removed (splicing)
5. Mature mRNA exits the nucleus through nuclear pores

Translation: mRNA $\to$ protein (at ribosomes).

1. MRNA binds to the ribosome
2. The ribosome reads the mRNA in codons (triplets of bases)
3. TRNA molecules carry amino acids to the ribosome
4. The anticodon on the tRNA pairs with the codon on the mRNA
5. Peptide bonds form between adjacent amino acids
6. Translation continues until a stop codon is reached (UAA, UAG, UGA)
7. The polypeptide chain is released and folds into its functional shape

The genetic code:

• Degenerate: most amino acids are coded for by more than one codon
• Universal: the same codons code for the same amino acids in almost all organisms
• Non-overlapping: each base is part of only one codon

Gene Regulation

Operon model (Jacob and Monod, 1961):

The lac operon in E. Coli controls the metabolism of lactose:

• Structural genes: lacZ (beta-galactosidase), lacY (permease), lacA (transacetylase)
• Promoter: Where RNA polymerase binds
• Operator: Where the repressor binds
• Regulatory gene (lacI): Produces the repressor protein

When lactose is absent:

• Repressor binds to the operator, blocking RNA polymerase
• Structural genes are not transcribed

When lactose is present:

• Lactose (allolactose) binds to the repressor, changing its shape
• Repressor cannot bind to the operator
• RNA polymerase transcribes the structural genes
• Enzymes for lactose metabolism are produced

Eukaryotic gene regulation:

• Transcription factors (activators and repressors) bind to enhancer or silencer regions
• Epigenetic modifications (DNA methylation, histone acetylation) affect gene expression
• Post-transcriptional regulation (miRNA, RNA interference)
• Post-translational modifications (phosphorylation, ubiquitination)

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Inheritance

Mendelian Genetics

Monohybrid inheritance: Inheritance of a single characteristic.

Example: In pea plants, tall (T) is dominant over dwarf (t). Cross a heterozygous tall plant With a dwarf plant.

Parents: \mathrm{Tt \times \mathrm{tt

Gametes: \mathrm{T, \mathrm{t \times \mathrm{t

	T	t
t	Tt	tt
t	Tt	tt

Genotype ratio: 1 Tt : 1 tt

Phenotype ratio: 1 tall : 1 dwarf

Test cross: Crossing an organism with a dominant phenotype (unknown genotype) with a homozygous Recessive individual to determine the genotype.

Dihybrid inheritance: Inheritance of two characteristics simultaneously.

Example: In pea plants, round seeds (R) are dominant over wrinkled (r), and yellow seeds (Y) are Dominant over green (y). Cross \mathrm{RrYy \times \mathrm{RrYy.

Gametes: RY, Ry, rY, ry (four types)

Phenotype ratio: 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green.

Sex-Linked Inheritance

Genes carried on the X chromosome show sex-linked inheritance. Males have only one X chromosome, so A single recessive allele will be expressed.

Example: Haemophilia is caused by a recessive allele on the X chromosome.

If a carrier female (\mathrm{X^H\mathrm{X^h) marries a normal male (\mathrm{X^H\mathrm{Y):

	\mathrm{X^H	\mathrm{Y
\mathrm{X^H	\mathrm{X^H\mathrm{X^H (normal female)	\mathrm{X^H\mathrm{Y (normal male)
\mathrm{X^h	\mathrm{X^H\mathrm{X^h (carrier female)	\mathrm{X^h\mathrm{Y (haemophiliac male)

Probability of a haemophiliac son: 1/4 (25%).

Non-Mendelian Genetics

Incomplete dominance: Heterozygote has an intermediate phenotype.

Example: \mathrm{CRCR (red) $\times$ \mathrm{CWCW (white) $\to$ \mathrm{CRCW (pink).

Codominance: Both alleles are expressed in the heterozygote.

Example: Blood groups. \mathrm{I^\mathrm{A\mathrm{I^\mathrm{B = blood group AB.

Multiple alleles: More than two alleles for a gene (e.g., blood groups: \mathrm{I^\mathrm{A \mathrm{I^\mathrm{B, $i$).

Epistasis: One gene affects the expression of another gene.

Polygenic inheritance: Characteristics controlled by many genes (e.g., height, skin colour). Shows continuous variation.

Genetic Crosses with Chi-Squared

The chi-squared test determines whether observed results differ significantly from expected Mendelian ratios.

$\chi^2 = \sum \frac{(O - E)^2}{E}$

Example: A cross is expected to give a 3:1 ratio. Observed: 72 dominant, 28 recessive. Is this Consistent with a 3:1 ratio?

Expected: 75 dominant, 25 recessive (total 100).

$\chi^2 = \frac{(72 - 75)^2}{75} + \frac{(28 - 25)^2}{25} = \frac{9}{75} + \frac{9}{25} = 0.12 + 0.36 = 0.48$

Degrees of freedom: $2 - 1 = 1$.

Critical value at 5% for 1 df: 3.841.

Since $0.48 \lt 3.841$The results are consistent with a 3:1 ratio. The difference is not Statistically significant.

Worked Example: Chi-squared for a dihybrid cross.

A dihybrid cross is expected to give a 9:3:3:1 ratio. Observed: 520, 180, 170, 30 (total 900).

Expected: 506.25, 168.75, 168.75, 56.25.

$\chi^2 = \frac{(520-506.25)^2}{506.25} + \frac{(180-168.75)^2}{168.75} + \frac{(170-168.75)^2}{168.75} + \frac{(30-56.25)^2}{56.25}$

$= \frac{189.06}{506.25} + \frac{126.56}{168.75} + \frac{1.56}{168.75} + \frac{689.06}{56.25}$

$= 0.37 + 0.75 + 0.009 + 12.25 = 13.38$

Degrees of freedom: $4 - 1 = 3$. Critical value at 5% for 3 df: 7.815.

Since $13.38 \gt 7.815$The observed results differ significantly from the expected 9:3:3:1 ratio. The null hypothesis is rejected.

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Advanced Higher Genetics

Mutation

Types of mutations:

Point mutations:

• Substitution: One base replaced by another
• Silent mutation: No change to the amino acid (due to degenerate code)
• Missense mutation: Change to a different amino acid
• Nonsense mutation: Change to a stop codon

Frameshift mutations:

• Insertion: Extra base added
• Deletion: Base removed
• These shift the reading frame and have severe effects

Chromosomal mutations:

• Deletion, duplication, inversion, translocation

Worked Example: Effect of a frameshift mutation.

Original DNA sequence (coding strand): ATG GCA TAC CAG

Reading frame: ATG GCA TAC CAG

Amino acids: Met - Ala - Tyr - Gln

After a deletion of the first base (A):

New sequence: TGG CAT ACC AG…

Reading frame: TGG CAT ACC AG…

Amino acids: Trp - His - Thr - …

The reading frame has shifted, and every amino acid after the deletion is different. The protein Will almost certainly lose its function.

Genetic Engineering

Recombinant DNA technology:

1. Identify the gene of interest
2. Cut the gene from the DNA using restriction enzymes (restriction endonucleases)
3. Cut the vector DNA (plasmid) with the same restriction enzyme
4. Join the gene to the plasmid using DNA ligase
5. Introduce the recombinant plasmid into host cells (transformation)
6. Select transformed cells using marker genes (e.g., antibiotic resistance)
7. Grow the transformed cells to produce the desired product

Applications:

• Production of human insulin by genetically modified bacteria
• Production of human growth hormone
• GM crops (herbicide resistance, pest resistance, improved nutrition)
• Gene therapy (treating genetic disorders)

PCR (Polymerase Chain Reaction): Amplifies a specific segment of DNA.

Steps:

1. Denaturation: Heat to 95$^\circ$C to separate DNA strands
2. Annealing: Cool to 55-65$^\circ$C for primers to bind
3. Extension: Heat to 72$^\circ$C for Taq polymerase to synthesise new DNA

Each cycle doubles the amount of DNA. After $n$ cycles: $2^n$ copies.

Worked Example: Calculating PCR products.

Starting with 1 DNA molecule, how many copies after 30 cycles?

$2^{30} = 1,073,741,824$ copies (over 1 billion).

If each cycle takes 5 minutes, the total time is $30 \times 5 = 150$ minutes = 2.5 hours.

Genomics and Bioinformatics

Genome sequencing: Determining the complete DNA sequence of an organism.

Bioinformatics: The use of computational tools to analyse biological data (DNA sequences, Protein structures).

Applications:

• Identifying disease-causing genes
• Pharmacogenomics (personalised medicine)
• Evolutionary relationships (comparing genomes)
• Forensic analysis (DNA fingerprinting)

Evolution and Population Genetics

Hardy-Weinberg Principle:

For a population in equilibrium (no evolution):

$p^2 + 2pq + q^2 = 1$

Where $p$ = frequency of dominant allele, $q$ = frequency of recessive allele.

Conditions for Hardy-Weinberg equilibrium:

• Large population (no genetic drift)
• No mutations
• No migration (no gene flow)
• Random mating
• No natural selection

Example: In a population, 16% of individuals show the recessive phenotype. Find the allele Frequencies.

$q^2 = 0.16$So $q = 0.4$.

$p = 1 - q = 0.6$.

Genotype frequencies: $p^2 = 0.36$ (homozygous dominant), $2pq = 0.48$ (heterozygous), $q^2 = 0.16$ (homozygous recessive).

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Worked Examples

See the examples integrated throughout the sections above.

Common Pitfalls

1. DNA replication direction: DNA polymerase only synthesises in the 5’ to 3’ direction. The lagging strand is synthesised in fragments.

2. Transcription vs. Translation: Transcription makes mRNA from DNA; translation makes protein from mRNA.

3. Codons and anticodons: Codons are on mRNA (5’ to 3’); anticodons are on tRNA (3’ to 5’).

4. Hardy-Weinberg: $p^2$ is the frequency of the homozygous dominant genotype, not the dominant allele frequency.

5. Chi-squared test: The null hypothesis is that there is no significant difference between observed and expected values.

6. Frameshift mutations: A single insertion or deletion changes ALL amino acids downstream, not just one.

7. Restriction enzymes cut at specific recognition sequences. They do not cut randomly; each enzyme recognises a specific palindromic sequence.

8. PCR requires primers. DNA polymerase cannot start synthesis from scratch; it needs a short primer to bind to the template.

9. Confusing the promoter and the operator. The promoter is where RNA polymerase binds; the operator is where the repressor binds. The operator is between the promoter and the structural genes.

10. Thinking that all mutations are harmful. Most are neutral or harmful, but beneficial mutations provide the raw material for natural selection.

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Practice Questions

1. Describe the process of DNA replication, naming the key enzymes involved.

2. Explain how the lac operon allows E. Coli to regulate lactose metabolism.

3. In a dihybrid cross between \mathrm{RrYy \times \mathrm{rryyDetermine the expected phenotype ratio and the probability of an offspring with the genotype \mathrm{RrYy.

4. A population has 9% of individuals showing a recessive genetic disorder. Calculate the frequency of the carrier genotype.

5. Describe the steps involved in producing human insulin using recombinant DNA technology.

6. Explain the difference between missense, nonsense, and frameshift mutations, giving an example of each.

7. In a cross between red (\mathrm{CRCW) and white (\mathrm{CWCW) flowers where codominance operates, predict the phenotype ratio of the F1 and F2 generations.

8. Explain why the Hardy-Weinberg equilibrium rarely holds in natural populations.

9. Describe the role of epigenetic modifications in gene regulation, giving two examples.

10. Explain how PCR works and calculate the number of copies of DNA produced after 25 cycles.

11. A dihybrid cross produces the following offspring: 410, 130, 145, 15. Use a chi-squared test to determine whether these results fit a 9:3:3:1 ratio.

12. Explain why DNA polymerase can only synthesise in the 5’ to 3’ direction and describe how the lagging strand is replicated.

13. Compare and contrast the processes of transcription and translation, including their locations, the enzymes involved, and the molecules produced.

14. Explain how genetic engineering could be used in gene therapy to treat cystic fibrosis.

15. A population has the following genotype frequencies: AA = 0.64, Aa = 0.32, aa = 0.04. Is this population in Hardy-Weinberg equilibrium? Show your calculations.

16. Describe the Meselson-Stahl experiment and explain why it provided evidence for semi-conservative DNA replication.

17. Explain the role of RNA interference (RNAi) in regulating gene expression.

18. Describe the process of DNA extraction from a biological sample, explaining the purpose of each reagent used.

19. Explain how bioinformatics is used to compare the genomes of different species and what this tells us about their evolutionary relationships.

20. A student carries out a test cross and obtains the following results: 48 dominant phenotype, 52 recessive phenotype. Use a chi-squared test to determine whether these results fit a 1:1 ratio.

21. Explain the difference between the leading strand and the lagging strand during DNA replication, including the role of Okazaki fragments.

22. Describe how gel electrophoresis separates DNA fragments and explain how this technique is used in DNA fingerprinting.

23. Explain how the trp operon in E. Coli differs from the lac operon in terms of its regulatory mechanism.

24. A population of 10,000 individuals has 8100 with the dominant phenotype and 1900 with the recessive phenotype. Calculate the allele frequencies and the expected number of heterozygous individuals.

25. Describe the process of CRISPR-Cas9 gene editing and explain two potential applications and one ethical concern.

Review: DNA Replication in Detail

DNA replication is a complex process involving multiple enzymes that work together with high Fidelity.

Enzymes involved:

Enzyme	Function
Helicase	Unwinds the double helix by breaking hydrogen bonds
DNA polymerase III	Adds nucleotides to the 3’ end; has proofreading ability
DNA polymerase I	Removes RNA primers and replaces them with DNA
Primase	Synthesises short RNA primers to initiate DNA synthesis
DNA ligase	Joins Okazaki fragments on the lagging strand
Topoisomerase	Relieves tension ahead of the replication fork
Single-strand binding proteins	Prevent re-annealing of separated strands

Leading vs lagging strand:

DNA polymerase can only add nucleotides in the 5’ to 3’ direction. Since the two strands of DNA are Antiparallel, only one strand (the leading strand) can be synthesised continuously in the direction Of the replication fork. The other strand (the lagging strand) must be synthesised discontinuously In short fragments called Okazaki fragments, each initiated by an RNA primer. The RNA primers are Later removed and replaced with DNA, and the fragments are joined by DNA ligase.

Proofreading: DNA polymerase III has 3’ to 5’ exonuclease activity, meaning it can detect and Remove a mismatched nucleotide and replace it with the correct one. This proofreading mechanism, Along with the accuracy of base pairing, gives DNA replication an error rate of approximately 1 in $10^9$ base pairs.

Review: Protein Synthesis in Detail

Transcription (in the nucleus):

1. RNA polymerase binds to the promoter region of the gene.
2. The DNA double helix is unwound in the region to be transcribed.
3. RNA polymerase synthesises a complementary mRNA strand using the DNA template strand, adding nucleotides in the 5’ to 3’ direction.
4. RNA processing: the pre-mRNA is modified by adding a 5’ cap (protects the mRNA and aids ribosome binding) and a poly-A tail (protects the mRNA from degradation). Introns (non-coding regions) are removed by splicing, and exons (coding regions) are joined together.
5. The mature mRNA exits the nucleus through a nuclear pore.

Translation (at the ribosome):

1. The mRNA binds to the small ribosomal subunit.
2. The initiator tRNA carrying methionine binds to the start codon (AUG) on the mRNA.
3. The large ribosomal subunit binds, forming the complete ribosome with three sites: A (aminoacyl), P (peptidyl), and E (exit).
4. TRNAs bring amino acids to the ribosome. Each tRNA has an anticodon that is complementary to the codon on the mRNA.
5. Peptide bonds form between adjacent amino acids (catalysed by peptidyl transferase, which is actually a ribosomal RNA — a ribozyme).
6. The ribosome moves along the mRNA, reading one codon at a time, until it reaches a stop codon (UAA, UAG, or UGA).
7. The polypeptide is released and folds into its functional three-dimensional structure.

Worked Example: Determining the amino acid sequence from an mRNA sequence.

Given the mRNA sequence: 5’-AUG CCA GAC UUU GAA UAA-3’

Reading from the start codon (AUG):

• AUG = methionine (Met)
• CCA = proline (Pro)
• GAC = aspartic acid (Asp)
• UUU = phenylalanine (Phe)
• GAA = glutamic acid (Glu)
• UAA = stop codon (translation terminates)

The polypeptide is: Met-Pro-Asp-Phe-Glu

Review: PCR (Polymerase Chain Reaction)

PCR is a technique used to amplify a specific region of DNA. It is widely used in forensic science, Medical diagnosis, and research.

Steps (each cycle doubles the amount of DNA):

1. Denaturation (94—96$\degree$C): The DNA is heated to separate the two strands.
2. Annealing (50—65$\degree$C): Primers (short, single-stranded DNA sequences complementary to the target region) bind to their complementary sequences on each strand.
3. Extension (72$\degree$C): Taq polymerase (a heat-stable DNA polymerase from Thermus aquaticus) extends the primers, synthesising new DNA strands.

After $n$ cycles, the number of copies of the target DNA is $2^n$.

Worked Example: After 25 cycles, the number of copies is $2^{25} = 33,554,432$ copies. Starting From a single DNA molecule, PCR can produce over 33 million copies in just a couple of hours.

Review: Genetic Engineering Techniques

Restriction enzymes (restriction endonucleases): Cut DNA at specific recognition sequences ( palindromic). For example, EcoRI cuts at GAATTC, producing sticky ends (single-stranded Overhangs).

DNA ligase: Joins DNA fragments by forming phosphodiester bonds between adjacent nucleotides. Used to insert a gene of interest into a plasmid vector.

Gel electrophoresis: Separates DNA fragments by size. Smaller fragments move faster through the Agarose gel and travel further. DNA fragments can be visualised using a fluorescent dye that binds To DNA. This technique is used in DNA fingerprinting, analysing PCR products, and checking the Success of restriction enzyme digestion.

Worked Example: Designing a genetic engineering experiment.

A scientist wants to insert the human insulin gene into a bacterial plasmid:

1. Extract the insulin gene from human DNA using restriction enzyme EcoRI (which cuts at specific sequences flanking the gene).
2. Cut the plasmid vector with the same restriction enzyme, producing complementary sticky ends.
3. Mix the insulin gene with the cut plasmid and add DNA ligase to join them.
4. Transform the recombinant plasmid into bacterial cells.
5. Grow the bacteria on a medium containing an antibiotic (the plasmid carries an antibiotic resistance gene, so only bacteria that have taken up the plasmid will survive).
6. Screen the surviving bacteria to identify those carrying the insulin gene.
7. Grow the recombinant bacteria in large fermenters to produce human insulin.

Review: CRISPR-Cas9 Gene Editing

CRISPR-Cas9 is a revolutionary gene-editing tool derived from a bacterial immune defence system.

Mechanism:

1. A guide RNA (gRNA) is designed to be complementary to the target DNA sequence.
2. The gRNA directs the Cas9 protein to the target site.
3. Cas9 cuts both strands of the DNA at the target site.
4. The cell’s own DNA repair mechanisms then either:

• Disable the gene (non-homologous end joining, or NHEJ)
• Replace the gene with a new sequence (homology-directed repair, or HDR)

Applications:

• Treating genetic disorders (e.g., sickle cell disease, cystic fibrosis)
• Creating genetically modified organisms
• Studying gene function by knocking out specific genes
• Cancer research (targeting oncogenes)

Ethical concerns:

• Off-target effects (cutting DNA at unintended sites)
• Germline editing (changes would be inherited by future generations)
• Designer babies (selecting for non-medical traits)
• Ecological concerns (gene drives in wild populations)

Review: Gel Electrophoresis in Detail

Gel electrophoresis separates DNA fragments based on size by applying an electric field across a gel Matrix.

Procedure:

1. An agarose gel is prepared ( 0.8-2% agarose, depending on the size of fragments to be separated).
2. The DNA samples are loaded into wells at one end of the gel.
3. An electric current is applied. DNA is negatively charged (due to phosphate groups), so it migrates towards the positive electrode.
4. Smaller fragments move through the gel more quickly than larger fragments, so they travel further.
5. After electrophoresis, the DNA is visualised using a fluorescent dye (e.g., ethidium bromide or SYBR Safe) that binds to DNA and fluoresces under UV light.

DNA fingerprinting (DNA profiling):

DNA fingerprinting uses gel electrophoresis to compare DNA samples from different individuals.

1. DNA is extracted from the samples.
2. PCR is used to amplify specific regions of DNA that are highly variable between individuals (short tandem repeats, or STRs).
3. The amplified fragments are separated by gel electrophoresis or capillary electrophoresis.
4. The resulting banding pattern is unique to each individual (except identical twins).
5. Banding patterns are compared to determine identity, paternity, or to match forensic samples.

Worked Example: Interpreting a DNA fingerprint.

A crime scene sample and three suspects are analysed using DNA fingerprinting at three STR loci. The Banding pattern at each locus is compared:

• Locus 1: Crime scene has bands at 8 and 12 repeat units. Suspect A matches; Suspect B has bands at 6 and 10; Suspect C has bands at 8 and 12 (matches).
• Locus 2: Crime scene has bands at 5 and 9. Suspect A has bands at 5 and 9 (matches); Suspect C has bands at 7 and 11.
• Locus 3: Crime scene has bands at 3 and 7. Suspect A has bands at 3 and 7 (matches).

Only Suspect A matches at all three loci, providing strong evidence that the DNA at the crime scene Came from Suspect A. The probability of a random match at all three loci is the product of the Individual match probabilities, which is less than 1 in a billion.

Review: The trp Operon — A Repressible Operon

The trp operon in E. Coli controls the synthesis of tryptophan and is an example of a repressible Operon. Unlike the lac operon (inducible), the trp operon is normally ON and is switched OFF when Tryptophan is abundant.

Components:

• Structural genes: trpE, trpD, trpC, trpB, trpA (encode enzymes for tryptophan biosynthesis).
• Promoter: Where RNA polymerase binds.
• Operator: Where the repressor binds.
• Regulatory gene (trpR): Produces the repressor protein.
• Leader sequence (trpL): Contains a short open reading frame with two consecutive tryptophan codons, which acts as a sensor for tryptophan availability.

When tryptophan is absent:

• The repressor is inactive (cannot bind to the operator).
• RNA polymerase transcribes the structural genes.
• Tryptophan is synthesised.

When tryptophan is present:

• Tryptophan acts as a corepressor, binding to the repressor and activating it.
• The activated repressor binds to the operator, blocking RNA polymerase.
• Transcription of the structural genes stops.
• This is an example of negative feedback: the end product of the pathway inhibits its own synthesis.

Comparison of lac and trp operons:

Feature	lac operon (inducible)	trp operon (repressible)
Default state	OFF (repressor bound)	ON (repressor inactive)
Inducer	Allolactose (inactivates repr)	Tryptophan (activates repr)
Function	Catabolise lactose	Synthesise tryptophan
Regulatory logic	Turn ON when substrate present	Turn OFF when product abundant
Type of control	Negative (repressor)	Negative (repressor)

Review: Epigenetics in Gene Regulation

Epigenetics refers to heritable changes in gene expression that do not involve changes to the DNA Sequence itself.

DNA methylation:

• Methyl groups (-\mathrm{CH_3) are added to cytosine bases, particularly at CpG islands near gene promoters.
• Methylation generally silences gene expression by preventing transcription factors from binding to the promoter.
• Methylation patterns are inherited during cell division (maintained by DNA methyltransferase DNMT1, which copies the methylation pattern to the new strand during replication).

Histone modification:

• Histone proteins can be acetylated, methylated, or phosphorylated, affecting how tightly DNA is wound around them.
• Acetylation (by histone acetyltransferases, HATs) loosens chromatin structure, allowing transcription factors to access DNA. This generally promotes gene expression.
• Deacetylation (by histone deacetylases, HDACs) tightens chromatin structure, reducing access to DNA. This generally represses gene expression.

RNA interference (RNAi):

• Small RNA molecules (microRNAs, miRNAs, and small interfering RNAs, siRNAs) can regulate gene expression post-transcriptionally.
• miRNAs bind to complementary sequences on target mRNAs, leading to either degradation of the mRNA or inhibition of translation.
• RNAi is an important mechanism for controlling gene expression during development and for defending against viral infections.

Worked Example: Epigenetics in cancer.

Cancer cells often show abnormal epigenetic patterns:

• Tumour suppressor genes are frequently silenced by DNA hypermethylation (e.g., the BRCA1 gene in some breast cancers).
• Oncogenes may be activated by DNA hypomethylation.
• HDAC inhibitors are being investigated as cancer treatments because they can reactivate tumour suppressor genes by increasing histone acetylation.

Review: Population Genetics — Selection and Fitness

Fitness ($w$): The relative reproductive success of a genotype compared to the most fit genotype In the population. The most fit genotype has $w = 1$; less fit genotypes have $w \lt 1$.

Selection coefficient ($s$): The reduction in fitness of a genotype: $s = 1 - w$.

Worked Example: Selection against a recessive allele.

In a population, the recessive allele $a$ has frequency $q = 0.4$. Individuals with genotype $aa$ Have reduced fitness ($w = 0.5$So $s = 0.5$).

After one generation of selection:

$q' = \frac{q^2 w_{aa} + pq \cdot 1}{\bar{w}}$

Where $\bar{w}$ is the mean fitness of the population.

$\bar{w} = p^2(1) + 2pq(1) + q^2(0.5) = (0.6)^2 + 2(0.6)(0.4) + (0.4)^2(0.5) = 0.36 + 0.48 + 0.08 = 0.92$.

$q' = \frac{(0.4)^2(0.5) + (0.6)(0.4)}{0.92} = \frac{0.08 + 0.24}{0.92} = \frac{0.32}{0.92} = 0.348$

The frequency of the recessive allele has decreased from 0.4 to 0.348 after one generation of Selection. The recessive allele decreases slowly because it is “hidden” in heterozygous individuals, Who are not affected by selection.

Heterozygote advantage: When heterozygotes have higher fitness than either homozygote, both Alleles are maintained in the population. This is also called balanced polymorphism.

Example: Sickle cell anaemia and malaria. In regions where malaria is endemic:

• $H^AH^A$ (normal): susceptible to malaria (fitness reduced).
• $H^AH^S$ (carrier): resistant to malaria (highest fitness).
• $H^SH^S$ (sickle cell disease): severe anaemia (fitness greatly reduced).

The heterozygote advantage maintains both the normal and sickle cell alleles in the population. This Explains why sickle cell anaemia remains relatively common in malaria-endemic regions despite being A severe genetic disorder.

Summary

This topic covers the biological principles of genetics and adaptation, including key concepts, experimental evidence, and real-world applications.

Key concepts include:

• Mendelian inheritance
• gene expression and regulation
• mutations and genetic variation
• genetic engineering (PCR, gel electrophoresis)
• genome projects

Success requires the ability to recall specific factual content, apply knowledge to novel scenarios, and evaluate experimental evidence critically.