Genetics and DNA

Genetics and DNA

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Board Coverage AQA Paper 2 | Edexcel A Paper 2 | OCR (A) Paper 2 | CIE Paper 4

1. DNA Structure and Replication

1.1 The Structure of DNA

Deoxyribonucleic acid (DNA) is a double-stranded polymer whose monomers are nucleotides. Each nucleotide consists of:

1. Deoxyribose: a pentose sugar with hydrogen at the $2'$ position (distinguishing it from ribose, which has a hydroxyl group at $2'$).
2. Phosphate group: linked to the $5'$ carbon of the sugar.
3. Nitrogenous base: adenine (A), thymine (T), cytosine (C), or guanine (G).

Nucleotides are joined by phosphodiester bonds between the $3'$ carbon of one sugar and the $5'$ carbon of the next, forming a sugar-phosphate backbone with directionality: $5' \to 3'$.

The two strands are antiparallel: one runs $5' \to 3'$ while the other runs $3' \to 5'$. They are held together by hydrogen bonds between complementary base pairs: A pairs with T (2 hydrogen bonds), C pairs with G (3 hydrogen bonds). The double helix twists with approximately 10 base pairs per turn.

1.2 DNA Replication

DNA replication is semi-conservative: each new DNA molecule consists of one original (parent) strand and one newly synthesised strand. This was demonstrated by the Meselson-Stahl experiment (1958) using $^{15}\mathrm{N}$ and $^{14}\mathrm{N}$ isotopes.

Stages:

1. Helicase unwinds and separates the double helix by breaking hydrogen bonds between base pairs. The Y-shaped region where separation occurs is the replication fork.

2. DNA polymerase synthesises the new strand in the $5' \to 3'$ direction by adding complementary nucleotides to the $3'$ end of the growing strand. It catalyses the formation of phosphodiester bonds between nucleotides.

3. Leading strand: synthesised continuously in the $5' \to 3'$ direction towards the replication fork.

4. Lagging strand: synthesised discontinuously in short fragments called Okazaki fragments (100--200 nucleotides in eukaryotes), each initiated by an RNA primer laid down by primase. The fragments are later joined by DNA ligase.

5. DNA polymerase I removes the RNA primers and replaces them with DNA nucleotides.

Proofreading: DNA polymerase III has $3' \to 5'$ exonuclease activity that can remove incorrectly paired nucleotides immediately after they are added, reducing the error rate to approximately $10^{-9}$ per base per replication.

warning

Common Pitfall DNA polymerase can only add nucleotides to the $3'$ end of a growing strand. It cannot initiate synthesis de novo -- it requires a pre-existing $3'-\mathrm{OH}$ group, which is provided by the RNA primer. Students often forget to mention primase and primers when describing replication.

2. Protein Synthesis

2.1 Transcription

Transcription is the synthesis of mRNA from a DNA template. It occurs in the nucleus (in eukaryotes).

1. Initiation: RNA polymerase binds to the promoter region upstream of the gene. In eukaryotes, transcription factors and the TATA box are required for binding.
2. Elongation: RNA polymerase unwinds the DNA and synthesises a complementary mRNA strand in the $5' \to 3'$ direction, using the template (antisense) strand. RNA polymerase adds complementary RNA nucleotides (U pairs with A).
3. Termination: RNA polymerase reaches a terminator sequence and detaches. The pre-mRNA transcript is released.

Post-transcriptional modification (in eukaryotes only):

• 5' capping: addition of a modified guanine nucleotide (7-methylguanosine) to protect the mRNA from degradation and facilitate ribosome binding.
• 3' polyadenylation: addition of a poly-A tail (100--200 adenine nucleotides) to protect the mRNA and aid export from the nucleus.
• Splicing: removal of introns (non-coding regions) and joining of exons (coding regions) by the spliceosome. Alternative splicing allows a single gene to produce multiple mRNA variants and thus multiple protein isoforms.

2.2 Translation

Translation is the synthesis of a polypeptide from an mRNA template. It occurs on ribosomes in the cytoplasm (or on the RER for secreted proteins).

Ribosomes have two subunits: the small subunit (40S in eukaryotes) binds to mRNA, and the large subunit (60S) has three tRNA binding sites: A (aminoacyl), P (peptidyl), and E (exit).

Stages:

1. Initiation: the small ribosomal subunit binds to the $5'$ cap of the mRNA and scans to the start codon (AUG). The initiator tRNA (carrying methionine) binds to the start codon in the P site. The large subunit then joins.

2. Elongation:

   • A charged tRNA carrying the next amino acid enters the A site, matching its anticodon to the mRNA codon.
   • A peptide bond forms between the amino acid in the P site and the amino acid in the A site, catalysed by peptidyl transferase (an rRNA enzyme -- a ribozyme).
   • The ribosome translocates one codon along the mRNA: the empty tRNA moves to the E site and exits, the tRNA carrying the growing polypeptide moves from A to P, and a new tRNA enters A.

3. Termination: the ribosome reaches a stop codon (UAA, UAG, or UGA). There is no tRNA for stop codons. A release factor binds, causing the polypeptide to be released and the ribosome to dissociate.

2.3 The Genetic Code

The genetic code is a triplet code: each codon (three consecutive nucleotides) specifies one amino acid. With 4 bases and codons of length 3, there are $4^3 = 64$ possible codons.

Properties of the genetic code:

• Degenerate: most amino acids are specified by more than one codon (redundancy provides protection against point mutations).
• Universal: the same codons specify the same amino acids in almost all organisms (strong evidence for common ancestry).
• Non-overlapping: each nucleotide is part of only one codon.
• Start codon: AUG (methionine) is both the start codon and codes for methionine.
• Stop codons: UAA, UAG, UGA -- do not code for any amino acid.

3. Mutations

3.1 Types of Mutation

Point mutations (substitutions): a single base is replaced by another.

• Silent mutation: the new codon codes for the same amino acid (due to the degeneracy of the genetic code). No effect on the protein.
• Missense mutation: the new codon codes for a different amino acid. The effect depends on the chemical properties of the substituted amino acid.
• Nonsense mutation: the new codon is a stop codon. This causes premature termination of translation, producing a truncated, usually non-functional protein.

Frameshift mutations: insertion or deletion of a base (or bases not in multiples of three), which shifts the reading frame downstream. All codons after the mutation are changed, typically producing a completely non-functional protein.

3.2 Causes of Mutation

• Spontaneous: errors during DNA replication (despite proofreading); spontaneous deamination of cytosine to uracil; tautomeric shifts in base pairing.
• Induced: exposure to mutagens: ionising radiation (X-rays, gamma rays, UV light) causes pyrimidine dimers (thymine dimers); chemical mutagens (base analogues, alkylating agents, intercalating agents like ethidium bromide) distort DNA structure.

warning

Common Pitfall Students often state that "mutations are always harmful." Most mutations are neutral (silent mutations, mutations in non-coding DNA). Some are harmful (cause genetic disorders), and a few are beneficial (provide the variation upon which natural selection acts). The harmful mutations are eliminated by selection; beneficial ones may increase in frequency.

4. Meiosis

4.1 Purpose and Overview

Meiosis is a form of cell division that produces four genetically distinct haploid ($n$) cells from one diploid ($2n$) parent cell. It is essential for sexual reproduction: it halves the chromosome number so that fertilisation restores the diploid number.

Meiosis consists of two divisions:

• Meiosis I: homologous chromosomes separate (reductional division).
• Meiosis II: sister chromatids separate (equational division, similar to mitosis).

4.2 Meiosis I

Prophase I (the longest and most complex phase):

• Chromatin condenses into visible chromosomes, each consisting of two sister chromatids.
• Homologous chromosomes pair up (synapsis) to form bivalents (tetrads).
• Crossing over occurs: non-sister chromatids exchange corresponding segments at chiasmata. This is the physical basis of genetic recombination.
• The nuclear envelope breaks down; spindle fibres form.

Metaphase I: bivalents align at the metaphase plate. The orientation of each bivalent is random (independent assortment).

Anaphase I: homologous chromosomes (each still consisting of two sister chromatids) are pulled to opposite poles by the spindle.

Telophase I and cytokinesis: two haploid cells are produced (each chromosome still has two chromatids).

4.3 Meiosis II

Prophase II, Metaphase II, Anaphase II, and Telophase II proceed similarly to mitosis but with haploid cells. Sister chromatids separate at Anaphase II. Four haploid daughter cells are produced.

4.4 Sources of Genetic Variation in Meiosis

1. Crossing over (prophase I): creates new combinations of alleles on the same chromosome. The frequency of crossing over between two loci is proportional to the distance between them (used to construct genetic maps).

2. Independent assortment (metaphase I): for $n$ pairs of homologous chromosomes, $2^n$ possible combinations of maternal and paternal chromosomes in gametes.

3. Random fertilisation: any of the $2^n$ gamete types from one parent can fuse with any of the $2^n$ gamete types from the other.

For humans ($n = 23$): $2^{23} \approx 8.4 \times 10^6$ possible gamete types per parent, and $(2^{23})^2 \approx 7 \times 10^{13}$ possible zygote combinations.

5. Genetic Crosses

5.1 Monohybrid Inheritance

A monohybrid cross involves one gene with two alleles. Using the standard notation: uppercase letter for the dominant allele, lowercase for the recessive allele.

Example. In pea plants, tall ($T$) is dominant over dwarf ($t$). Cross two heterozygous plants ($Tt \times Tt$):

	$T$	$t$
$T$	$TT$	$Tt$
$t$	$Tt$	$tt$

Genotypic ratio: $1\ TT : 2\ Tt : 1\ tt$. Phenotypic ratio: $3\ \mathrm{tall} : 1\ \mathrm{dwarf}$.

Test cross: crossing an individual of unknown genotype (showing the dominant phenotype) with a homozygous recessive individual. If any offspring show the recessive phenotype, the unknown parent must be heterozygous.

5.2 Dihybrid Inheritance

A dihybrid cross involves two genes, each with two alleles, on different chromosomes.

Example. In peas, round seeds ($R$) are dominant over wrinkled ($r$), and yellow ($Y$) is dominant over green ($y$). Cross $RrYy \times RrYy$:

Expected phenotypic ratio: $9\ R\_Y\_ : 3\ R\_yy : 3\ rrY\_ : 1\ rryy$ (9:3:3:1).

This ratio arises from the product rule: each gene segregates independently ($3:1 \times 3:1 = 9:3:3:1$).

5.3 Sex-Linked Inheritance

Genes carried on the X chromosome show sex-linked inheritance patterns. Males (XY) have only one copy of X-linked genes and are hemizygous -- a single recessive allele on the X chromosome will be expressed.

Example. Red-green colour blindness is X-linked recessive ($X^c$). A carrier female ($X^CX^c$) crossed with a normal male ($X^CY$):

	$X^C$	$Y$
$X^C$	$X^CX^C$ (normal female)	$X^CY$ (normal male)
$X^c$	$X^CX^c$ (carrier female)	$X^cY$ (colour-blind male)

Offspring: 50% normal females, 50% carrier females, 50% normal males, 50% affected males.

Key patterns of X-linked recessive inheritance:

• More males affected than females.
• Affected males cannot pass the allele to sons (they pass Y to sons).
• Affected males pass the allele to all daughters (who become carriers).
• Carrier females have a 50% chance of passing the allele to sons (who will be affected).

5.4 Co-dominance and Incomplete Dominance

Co-dominance: both alleles are fully expressed in the heterozygote. Example: blood groups -- genotype $I^AI^B$ produces blood group AB, expressing both A and B antigens.

Incomplete dominance: the heterozygote has an intermediate phenotype. Example: snapdragon colour -- $RR$ (red) $\times$ $WW$ (white) gives $RW$ (pink). The pink phenotype is not a blend of pigments but reduced production of red pigment.

warning

Common Pitfall Students often confuse co-dominance and incomplete dominance. In co-dominance, both alleles produce their full product (both A and B antigens are present). In incomplete dominance, the heterozygote produces less of the dominant product (an intermediate phenotype). The genetic ratios for both are 1:2:1, but the phenotypic expression differs.

6. Gene Expression and Epigenetics

6.1 Gene Regulation

Not all genes are expressed in all cells at all times. Gene expression is regulated at multiple levels:

Transcriptional control:

• Transcription factors: proteins that bind to specific DNA sequences (promoters, enhancers) and activate or repress transcription. Activators recruit RNA polymerase; repressors block polymerase binding.
• Lac operon (prokaryotic gene regulation): in E. coli, the lac operon contains genes for lactose metabolism. In the absence of lactose, a repressor protein binds to the operator, blocking transcription. When lactose is present, its isomer allolactose binds to the repressor, causing a conformational change that releases it from the operator, allowing transcription.

Post-transcriptional control:

• mRNA stability (shorter-lived mRNAs produce less protein).
• Alternative splicing (different exons are joined, producing different protein isoforms).

Translational and post-translational control:

• Inhibition of translation by regulatory proteins or microRNAs.
• Protein modification (phosphorylation, acetylation) that activates or deactivates proteins.
• Protein degradation by proteasomes.

6.2 Epigenetics

Definition. Epigenetics is the study of heritable changes in gene expression that do not involve changes to the DNA nucleotide sequence.

Epigenetic modifications alter the accessibility of DNA to transcription machinery:

• DNA methylation: methyl groups ($-\mathrm{CH_3}$) are added to cytosine bases (specifically at CpG islands near gene promoters) by DNA methyltransferases. Methylation typically represses transcription by preventing transcription factor binding or recruiting repressive proteins.
• Histone modification: histone proteins around which DNA is wound can be modified by acetylation, methylation, or phosphorylation. Histone acetylation (addition of acetyl groups to lysine residues) neutralises the positive charge on histones, reducing their affinity for negatively charged DNA. This loosens the chromatin structure, increasing gene expression. Histone deacetylation has the opposite effect.

Epigenetic modifications are heritable during cell division and can be influenced by environmental factors (diet, stress, toxins). This is a mechanism by which environmental experiences can have long-term effects on gene expression without altering the DNA sequence.

Example. The agouti mouse model: maternal diet rich in methyl donors (folic acid, vitamin $\mathrm{B_{12}}$) increases DNA methylation at the agouti gene, silencing it and producing offspring with brown coats and normal weight. A methyl-poor diet produces yellow, obese offspring.

warning

Common Pitfall Students often confuse epigenetic changes with genetic mutations. Epigenetic changes are reversible modifications to gene expression that do not alter the DNA sequence itself. Mutations are permanent changes to the nucleotide sequence. Epigenetic marks can be passed to daughter cells during mitosis (and in some cases meiosis), but they can also be removed by environmental changes.

7. Genetic Technology

7.1 The Polymerase Chain Reaction (PCR)

PCR is a technique for amplifying a specific DNA sequence in vitro. It requires:

1. Template DNA: the DNA containing the target sequence.
2. Primers: short, single-stranded oligonucleotides (typically 18--25 bases) complementary to the sequences flanking the target region. One forward primer and one reverse primer.
3. Taq DNA polymerase: a heat-stable DNA polymerase from the bacterium Thermus aquaticus (optimum temperature $\approx 72\ ^\circ\mathrm{C}$), which does not denature at the high temperatures used in the cycle.
4. Free nucleotides (dNTPs): dATP, dTTP, dCTP, dGTP.
5. Buffer with $\mathrm{Mg^{2+}}$ ions (co-factor for Taq polymerase).

The PCR cycle (repeated 25--35 times):

1. Denaturation ($95\ ^\circ\mathrm{C}$, 30 s): the double-stranded DNA template is heated to separate the strands.
2. Annealing ($55$--$65\ ^\circ\mathrm{C}$, 30 s): the temperature is lowered to allow primers to bind (anneal) to their complementary sequences on the template strands by hydrogen bonding.
3. Extension ($72\ ^\circ\mathrm{C}$, 30--60 s): Taq polymerase extends the primers by adding nucleotides to the $3'$ end, synthesising new DNA strands.

Exponential amplification: after $n$ cycles, the number of copies of the target sequence is approximately $2^n$ (after the first few cycles where the number of target molecules doubles). After 30 cycles, a single molecule yields $2^{30} \approx 1.07 \times 10^9$ copies.

Worked Example. A forensic sample contains 5 copies of a target DNA sequence. After 28 cycles of PCR, how many copies are produced?

Number of copies $= 5 \times 2^{28} = 5 \times 268435456 = 1.34 \times 10^9$ copies.

7.2 Gel Electrophoresis

Gel electrophoresis separates DNA fragments by size. DNA fragments are loaded into wells in an agarose gel and an electric field is applied. DNA is negatively charged (due to phosphate groups) and migrates towards the positive electrode.

• Smaller fragments move faster and travel further through the gel matrix.
• Larger fragments move more slowly and travel less far.

The distance migrated is approximately inversely proportional to the logarithm of fragment size. A DNA ladder (fragments of known size) is run alongside the samples to allow size determination.

For protein electrophoresis, SDS-PAGE (sodium dodecyl sulfate polyacrylamide gel electrophoresis) is used. SDS denatures proteins and gives them a uniform negative charge, so separation is purely by molecular mass.

7.3 DNA Sequencing

Sanger sequencing (chain-termination method): the DNA template is copied using modified nucleotides (dideoxynucleotides, ddNTPs) that lack the $3'-\mathrm{OH}$ group necessary for chain elongation. When a ddNTP is incorporated, synthesis terminates. By running four separate reactions, each with a different ddNTP (A, T, C, G) labelled with a fluorescent dye, and separating the resulting fragments by length, the sequence can be read directly from the electropherogram.

Modern next-generation sequencing (NGS) methods allow massively parallel sequencing of millions of DNA fragments simultaneously, enabling rapid whole-genome sequencing.

7.4 Restriction Enzymes and Recombinant DNA Technology

Restriction endonucleases (restriction enzymes) are enzymes that cut DNA at specific recognition sequences (typically 4--8 base pairs, palindromic). For example, EcoRI cuts at GAATTC.

In recombinant DNA technology:

1. A target gene is cut from donor DNA using a restriction enzyme.
2. A plasmid vector is cut with the same enzyme, producing complementary sticky ends.
3. The target gene and the cut plasmid are mixed with DNA ligase, which joins them by forming phosphodiester bonds.
4. The recombinant plasmid is introduced into host bacteria by transformation.
5. Bacteria carrying the recombinant plasmid are selected using antibiotic resistance markers on the plasmid.

warning

Common Pitfall Students often write that "bacteria are genetically modified" in PCR. PCR does not involve bacteria or modification of living organisms -- it is an in vitro technique. PCR, gel electrophoresis, and DNA sequencing are analytical techniques, while recombinant DNA technology involves the creation of genetically modified organisms (GMOs).

8. Advanced Genetic Crosses

8.1 Epistasis

Epistasis occurs when the expression of one gene is affected by one or more independently inherited genes. The epistatic gene masks or modifies the expression of the hypostatic gene.

Worked Example: Recessive epistasis in coat colour of Labrador retrievers.

Gene $B$ determines pigment colour: $B$ (black) is dominant over $b$ (brown). Gene $E$ determines whether pigment is deposited in the fur: $E$ allows deposition; $ee$ prevents deposition, resulting in a golden coat regardless of the $B/b$ genotype.

Cross: $BbEe \times BbEe$

Expected ratio from independent assortment: 9:3:3:1.

Genotype	Phenotype	Count	Phenotypic class
$B\_E\_$	Black	9	Black
$bbE\_$	Brown	3	Brown
$B\_ee$	Golden	3	Golden
$bbee$	Golden	1	Golden

Combined phenotypic ratio: 9 black : 3 brown : 4 golden (9:3:4).

This is distinct from the standard 9:3:3:1 ratio because the homozygous recessive genotype at gene $E$ masks the effect of gene $B$.

8.2 Autosomal Linkage

When two genes are located on the same chromosome, they do not assort independently (Mendel's second law is violated). The closer the genes are on the chromosome, the less likely they are to be separated by crossing over.

Worked Example. Two genes, $A$ and $B$, are linked on the same chromosome. A test cross is performed: $AB/ab \times ab/ab$ (where $/$ indicates that $AB$ are on one homologous chromosome and $ab$ on the other).

The offspring are:

• $AB/ab$: 42% (parental)
• $ab/ab$: 42% (parental)
• $Ab/ab$: 8% (recombinant)
• $aB/ab$: 8% (recombinant)

The recombination frequency $= \frac{8 + 8}{42 + 42 + 8 + 8} = \frac{16}{100} = 16\%$.

The recombination frequency is approximately equal to the map distance between the two genes in centiMorgans (cM). The genes are approximately 16 cM apart.

If the recombination frequency were 50%, the genes would assort independently (unlinked). The maximum observable recombination frequency is 50% (since greater distances lead to multiple cross-overs that cancel out).

warning

Common Pitfall Students often assume that a 9:3:3:1 ratio always results from a dihybrid cross. If the genes are linked, the observed ratio will deviate from 9:3:3:1, with an excess of parental phenotypes and a deficit of recombinant phenotypes. Always check whether the genes are on the same chromosome before applying Mendel's second law.

8.3 Sex Determination

In mammals, sex is determined by the sex chromosomes: XX = female, XY = male. The SRY gene on the Y chromosome triggers testis development. In the absence of SRY (XX individuals), ovaries develop.

In birds, the system is reversed: ZZ = male, ZW = female. In some reptiles and fish, sex is determined by environmental temperature during embryonic development (temperature-dependent sex determination).

8.4 Chi-Squared Test for Genetic Crosses

The chi-squared test is used to determine whether observed results from a genetic cross fit the expected Mendelian ratio.

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

Worked Example. A dihybrid cross is expected to give a 9:3:3:1 ratio. The observed results are:

Phenotype	Observed	Expected (total = 600)	$(O - E)^2 / E$
Round yellow	352	$600 \times 9/16 = 337.5$	0.62
Round green	106	$600 \times 3/16 = 112.5$	0.38
Wrinkled yellow	98	$600 \times 3/16 = 112.5$	1.87
Wrinkled green	44	$600 \times 1/16 = 37.5$	1.13

$\chi^2 = 0.62 + 0.38 + 1.87 + 1.13 = 4.00$

Degrees of freedom $= 4 - 1 = 3$. Critical value at $p = 0.05$ for 3 df $= 7.82$.

Since $\chi^2 = 4.00 < 7.82$, we accept the null hypothesis: the observed results fit the expected 9:3:3:1 ratio. The genes assort independently.

9. Detailed Mutation Analysis

9.1 Types of DNA Damage

Type of Damage	Cause	Effect
Base substitution	Point mutation; mutagenic chemicals	May change amino acid (or be silent)
Deamination	Spontaneous (C to U)	C:G pair becomes U:A pair after replication
Depurination	Spontaneous loss of purine base	Gap in DNA strand
Pyrimidine dimer	UV radiation	Bends DNA helix; blocks replication
Double-strand break	Ionising radiation	Chromosomal rearrangement; cell death
Insertion/deletion	Replication error; mutagen	Frameshift if not in multiples of 3

9.2 Worked Example: Tracing a Frameshift Mutation

The original mRNA sequence (reading frame starting at the first codon) is:

$\text{AUG-CCU-GAC-UUU-GGC-AAG-UAA}$ $\text{Met - Pro - Asp - Phe - Gly - Lys - Stop}$

A deletion of the third nucleotide (G) produces:

$\text{AUG-CCU-ACU-UUG-GCA-AGU-AA...}$ $\text{Met - Pro - Thr - Leu - Ala - Ser - ...}$

Every amino acid after the deletion site has changed, and the original stop codon is no longer in frame. Translation continues past the original stop codon until a new stop codon is encountered, producing a longer, non-functional protein. This illustrates why frameshift mutations are typically more severe than point mutations.

9.3 Mutagenic Agents and Their Mechanisms

UV radiation (wavelength $< 320\ \mathrm{nm}$): causes thymine dimers by forming covalent bonds between adjacent thymine bases on the same DNA strand. This distorts the helix and blocks replication. Repair mechanism: nucleotide excision repair (NER) -- the damaged section is cut out by endonucleases, DNA polymerase fills the gap, and DNA ligase seals it.

Ionising radiation (X-rays, gamma rays): carries sufficient energy to eject electrons from atoms, causing direct DNA strand breaks (single-strand and double-strand breaks). Double-strand breaks are particularly dangerous because they can lead to chromosomal translocations and deletions.

Chemical mutagens:

• Base analogues (e.g., 5-bromouracil): structurally similar to normal bases; incorporated during replication but pair incorrectly.
• Alkylating agents (e.g., mustard gas, EMS): add alkyl groups to bases, altering their pairing properties.
• Intercalating agents (e.g., ethidium bromide): insert between base pairs, causing insertions or deletions during replication.

10. Gene Technology in Medicine and Agriculture

10.1 Recombinant Human Insulin

Before recombinant DNA technology, insulin for treating diabetes was extracted from pig and cow pancreas. This had problems: slight structural differences from human insulin could cause immune reactions, and supply was limited.

The production of recombinant human insulin involves:

1. Isolation of the human insulin gene.
2. Insertion into a plasmid vector using restriction enzymes and DNA ligase.
3. Transformation of E. coli or yeast cells with the recombinant plasmid.
4. Fermentation: the genetically modified organisms are grown in large bioreactors.
5. Extraction and purification of the insulin protein.

The recombinant insulin is structurally identical to human insulin, eliminating immune reactions, and can be produced in unlimited quantities.

10.2 Genetically Modified Crops

Genetic modification of crops can introduce traits such as:

• Herbicide resistance: insertion of a gene for an enzyme that detoxifies the herbicide (e.g., the BAR gene conferring resistance to glufosinate).
• Pest resistance: insertion of a gene from Bacillus thuringiensis (Bt) that codes for a toxin lethal to specific insect pests but harmless to humans.
• Nutritional enhancement: "Golden Rice" contains genes for beta-carotene (provitamin A) synthesis, addressing vitamin A deficiency in developing countries.
• Disease resistance: insertion of genes conferring resistance to viral or fungal pathogens.

Advantages: increased yield; reduced use of chemical pesticides; improved nutritional content; potential to grow crops in marginal conditions (drought resistance, salinity tolerance).

Concerns: potential gene flow to wild relatives creating "superweeds"; impact on non-target organisms; reduced genetic diversity if only a few GM varieties are planted; ethical concerns about patenting life forms; long-term ecological effects that are difficult to predict.

Practice Problems

Details

Problem 1

A couple, both with normal vision, have a colour-blind son. The woman's father was colour-blind. Determine the genotypes of the parents and the probability that their next child will be a colour-blind daughter.

Answer. Colour blindness is X-linked recessive ($X^c$). The son is $X^cY$, having inherited $X^c$ from his mother. The mother must therefore be a carrier: $X^CX^c$. The woman's father was $X^cY$, so she inherited $X^c$ from him. The father has normal vision and is $X^CY$ (since he is not colour-blind). The cross is $X^CX^c \times X^CY$.

Possible offspring: $X^CX^C$ (25%, normal female), $X^CX^c$ (25%, carrier female), $X^CY$ (25%, normal male), $X^cY$ (25%, colour-blind male).

A colour-blind daughter requires genotype $X^cX^c$, which requires the father to contribute $X^c$ -- but the father is $X^CY$ and cannot pass $X^c$ to any child. Therefore, the probability of a colour-blind daughter is zero.

If you get this wrong, revise: Sex-Linked Inheritance

Details

Problem 2

Explain how the Meselson-Stahl experiment provided evidence for the semi-conservative model of DNA replication. Describe the expected results for conservative, semi-conservative, and dispersive models after one and two generations.

Answer. Meselson and Stahl grew E. coli in a medium containing heavy nitrogen ($^{15}\mathrm{N}$), so all DNA was "heavy." They then transferred the bacteria to a medium with light nitrogen ($^{14}\mathrm{N}$) and allowed one round of replication. After extraction and density gradient centrifugation, all DNA formed a single band at intermediate density. After a second round of replication, two bands appeared: one at intermediate density and one at light density.

Semi-conservative model predicts: Generation 1 -- all intermediate (one heavy + one light strand). Generation 2 -- half intermediate, half light. This matches the observations.

Conservative model predicts: Generation 1 -- one heavy band (original double helix) + one light band (new double helix). This was not observed.

Dispersive model predicts: Generation 1 -- all intermediate. Generation 2 -- all intermediate (slightly lighter). This does not match the observation of two distinct bands.

Only the semi-conservative model is consistent with the experimental results.

If you get this wrong, revise: DNA Replication

Details

Problem 3

In cats, the gene for coat colour is X-linked. Black ($X^B$) is dominant over orange ($X^O$). A calico cat has patches of black and orange fur. Explain the genetic basis of calico coat colour and why calico cats are almost always female.

Answer. A calico cat has genotype $X^BX^O$ -- it is heterozygous for the coat colour gene, carrying one black allele and one orange allele. In female mammals, one X chromosome in each cell is randomly inactivated during early embryonic development (X-inactivation, or the Lyon hypothesis). The inactivated X condenses into a Barr body and its genes are not expressed. Because X-inactivation is random, some cells express $X^B$ (producing black fur) and others express $X^O$ (producing orange fur). The random pattern of inactivation creates patches of black and orange. Males are almost always calico-free because they have only one X chromosome ($X^BY$ or $X^OY$) and therefore cannot be heterozygous for this gene. The rare male calico cats have an abnormal karyotype (XXY, Klinefelter syndrome).

If you get this wrong, revise: Sex-Linked Inheritance and Epigenetics

Details

Problem 4

A dihybrid cross is performed between two pea plants heterozygous for seed shape (round $R$ dominant, wrinkled $r$ recessive) and seed colour (yellow $Y$ dominant, green $y$ recessive). The observed offspring are: 280 round yellow, 95 round green, 85 wrinkled yellow, 40 wrinkled green. Use a chi-squared test to determine whether the observed results fit the expected 9:3:3:1 ratio.

Answer. Total offspring $= 280 + 95 + 85 + 40 = 500$.

Expected (9:3:3:1 ratio): 281.25 round yellow, 93.75 round green, 93.75 wrinkled yellow, 31.25 wrinkled green.

$\chi^2 = \sum \frac{(O - E)^2}{E} = \frac{(280 - 281.25)^2}{281.25} + \frac{(95 - 93.75)^2}{93.75} + \frac{(85 - 93.75)^2}{93.75} + \frac{(40 - 31.25)^2}{31.25}$

$= \frac{1.5625}{281.25} + \frac{1.5625}{93.75} + \frac{76.5625}{93.75} + \frac{76.5625}{31.25}$

$= 0.006 + 0.017 + 0.817 + 2.450 = 3.29$.

Degrees of freedom $= 4 - 1 = 3$. The critical value at $p = 0.05$ for 3 df is $7.82$.

Since $\chi^2 = 3.29 \lt 7.82$, we accept the null hypothesis: the observed results do not differ significantly from the expected 9:3:3:1 ratio. The genes segregate independently.

If you get this wrong, revise: Dihybrid Inheritance

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Problem 5

Explain how epigenetic modifications can influence gene expression without changing the DNA sequence. Use histone acetylation and DNA methylation as examples.

Answer. Epigenetic modifications alter the accessibility of DNA to transcription factors and RNA polymerase. Histone acetylation involves the addition of acetyl groups to lysine residues on histone tails by histone acetyltransferases (HATs). Acetylation neutralises the positive charge on histones, reducing their electrostatic attraction to the negatively charged DNA backbone. This loosens the chromatin structure (euchromatin), allowing transcription factors and RNA polymerase to access the DNA, thereby increasing gene expression. Deacetylation by histone deacetylases (HDACs) has the opposite effect: chromatin becomes more condensed (heterochromatin), and transcription is repressed. DNA methylation involves the addition of methyl groups to cytosine bases at CpG islands near gene promoters by DNA methyltransferases. Methylation physically blocks the binding of transcription factors and recruits repressive proteins that further condense chromatin, thereby reducing gene expression. Both modifications are reversible and can be inherited during cell division, providing a mechanism for environmental influences on gene expression.

If you get this wrong, revise: Epigenetics

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Problem 6

Explain the roles of helicase, DNA polymerase, primase, and DNA ligase in DNA replication. Why is DNA polymerase unable to synthesise the lagging strand continuously?

Answer. Helicase unwinds the double helix by breaking hydrogen bonds between base pairs at the replication fork. DNA polymerase synthesises new DNA strands by adding complementary nucleotides to the $3'$ end of a growing strand, catalysing phosphodiester bond formation. It can only synthesise in the $5' \to 3'$ direction. Primase synthesises short RNA primers complementary to the DNA template, providing the $3'-\mathrm{OH}$ group that DNA polymerase requires to initiate synthesis. DNA ligase joins Okazaki fragments on the lagging strand by forming phosphodiester bonds between adjacent fragments. DNA polymerase cannot synthesise the lagging strand continuously because it can only add nucleotides in the $5' \to 3'$ direction, but the lagging strand template is oriented $3' \to 5'$ relative to the replication fork. As the fork opens, new template is exposed in the $5' \to 3'$ direction (away from the fork), so synthesis must proceed back towards the fork in discontinuous Okazaki fragments, each requiring its own RNA primer.

If you get this wrong, revise: DNA Replication

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Problem 7

A geneticist studies two genes in fruit flies: gene $A$ (normal wings, $A$ dominant over vestigial wings, $a$) and gene $B$ (grey body, $B$ dominant over ebony body, $b$). A test cross is performed: $AaBb \times aabb$. The offspring are:

Phenotype	Observed
Normal, grey	480
Vestigial, grey	20
Normal, ebony	22
Vestigial, ebony	478

(a) Are the genes linked? Justify your answer with a chi-squared test. (b) If linked, calculate the recombination frequency and the map distance between the genes.

Answer. (a) If the genes assorted independently, the expected ratio would be 1:1:1:1, with 250 in each category (total $= 1000$).

$\chi^2 = \frac{(480-250)^2}{250} + \frac{(20-250)^2}{250} + \frac{(22-250)^2}{250} + \frac{(478-250)^2}{250}$

$= \frac{52900}{250} + \frac{52900}{250} + \frac{51984}{250} + \frac{51984}{250}$

$= 211.6 + 211.6 + 207.9 + 207.9 = 839.0$

Degrees of freedom $= 3$. Critical value at $p = 0.05$ is $7.82$.

Since $\chi^2 = 839 \gg 7.82$, the deviation from independent assortment is highly significant. The genes are linked.

(b) Parental (non-recombinant) phenotypes: Normal grey (480) + Vestigial ebony (478) $= 958$. Recombinant phenotypes: Vestigial grey (20) + Normal ebony (22) $= 42$.

Recombination frequency $= \frac{42}{1000} \times 100\% = 4.2\%$.

Map distance $= 4.2\ \mathrm{cM}$.

The genes are 4.2 centiMorgans apart on the same chromosome, which is very close -- they are tightly linked and crossing over between them is rare.

If you get this wrong, revise: Advanced Genetic Crosses

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Problem 8

A PCR reaction starts with 10 copies of a target DNA sequence. After 30 cycles, how many copies are produced? If each cycle takes 90 seconds, how long does the entire PCR process take? State two reasons why the actual number of copies may be lower than the theoretical maximum.

Answer. Theoretical copies after $n$ cycles $= N_0 \times 2^n = 10 \times 2^{30} = 10 \times 1073741824 = 1.07 \times 10^{10}$ copies.

Total time $= 30 \times 90\ \mathrm{s} = 2700\ \mathrm{s} = 45\ \mathrm{minutes}$.

In practice, the actual number may be lower because: (1) the reaction efficiency is rarely 100% -- primers may not anneal perfectly, Taq polymerase may occasionally dissociate, or the template may re-anneal to itself; (2) reagent depletion (nucleotides, primers, or $\mathrm{Mg^{2+}}$ may become limiting in later cycles, reducing the amplification rate); (3) product inhibition at very high concentrations of amplified DNA.

If you get this wrong, revise: The Polymerase Chain Reaction (PCR)

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Problem 9

In a population of 10000 people, the frequency of the recessive allele for cystic fibrosis ($f$) is $0.02$. (a) Calculate the expected number of carriers (heterozygotes) in the population. (b) Calculate the expected number of individuals with cystic fibrosis. (c) A genetic screening programme identifies all carriers and counsels them. If two carriers decide to have a child, what is the probability that the child will have cystic fibrosis?

Answer. (a) $q = 0.02$, $p = 1 - 0.02 = 0.98$.

Frequency of heterozygotes $= 2pq = 2 \times 0.98 \times 0.02 = 0.0392$.

Number of carriers $= 0.0392 \times 10000 = 392$.

(b) Frequency of homozygous recessive $= q^2 = 0.0004$.

Number affected $= 0.0004 \times 10000 = 4$.

(c) If both parents are carriers ($Ff \times Ff$):

	$F$	$f$
$F$	$FF$	$Ff$
$f$	$Ff$	$ff$

Genotypic ratio: $1\ FF : 2\ Ff : 1\ ff$.

Probability of an affected child ($ff$) $= \frac{1}{4} = 25\%$.

If you get this wrong, revise: Monohybrid Inheritance and The Hardy-Weinberg Principle

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11. DNA Replication in Detail

11.1 The Replication Fork

DNA replication is semi-conservative (each new DNA molecule contains one original strand and one new strand), bidirectional (replication proceeds in both directions from each origin of replication), and semi-discontinuous (the leading strand is synthesised continuously, but the lagging strand is synthesised discontinuously as Okazaki fragments).

At each replication fork:

Leading strand:

• Synthesised continuously in the 5' to 3' direction (same direction as fork movement).
• DNA polymerase III extends the leading strand continuously.
• Only one RNA primer is needed.

Lagging strand:

• Synthesised discontinuously in the 5' to 3' direction (opposite to fork movement).
• Multiple RNA primers are laid down by primase at intervals of approximately 1000--2000 nucleotides (in prokaryotes) or 100--200 nucleotides (in eukaryotes).
• DNA polymerase III extends each primer, creating Okazaki fragments.
• DNA polymerase I removes the RNA primers and replaces them with DNA.
• DNA ligase joins the Okazaki fragments by catalysing phosphodiester bond formation between adjacent fragments.

11.2 Enzymes of DNA Replication

Enzyme	Function
Helicase	Unwinds the double helix by breaking hydrogen bonds between bases
DNA gyrase (topoisomerase II)	Relieves supercoiling ahead of the replication fork by cutting, rotating, and resealing DNA
Single-strand binding proteins (SSBs)	Stabilise the separated single strands, preventing them from re-annealing or being degraded by nucleases
Primase	Synthesises short RNA primers (approximately 10 nucleotides) complementary to the template strand
DNA polymerase III	Main replication enzyme; adds nucleotides to the 3' end of the growing strand (5' to 3' synthesis); has 3' to 5' exonuclease (proofreading) activity
DNA polymerase I	Removes RNA primers (5' to 3' exonuclease) and replaces them with DNA
DNA ligase	Joins adjacent DNA fragments by forming phosphodiester bonds

11.3 Calculating the Number of Strands After Multiple Replications Cycles

Meselson-Stahl experiment principle: if DNA is replicated in medium containing only $\mathrm{^{14}N}$ (light nitrogen) after being grown in $\mathrm{^{15}N}$ (heavy nitrogen):

Generation	Strand Composition	Density
0 (parent)	All $\mathrm{^{15}N\text{-}^{15}N}$	Heavy
1	All $\mathrm{^{15}N\text{-}^{14}N}$ (hybrid)	Intermediate
2	50% hybrid, 50% $\mathrm{^{14}N\text{-}^{14}N}$	Two bands
3	25% hybrid, 75% $\mathrm{^{14}N\text{-}^{14}N}$	Two bands
n	$\frac{1}{2^{n-1}}$ hybrid, $1 - \frac{1}{2^{n-1}}$ light	Two bands

After $n$ generations, the proportion of hybrid (intermediate density) DNA is $\frac{1}{2^{n-1}}$.

12. Protein Synthesis in Detail

12.1 Transcription: Initiation, Elongation, Termination

Initiation (in eukaryotes):

1. Transcription factors bind to the promoter region (e.g., TATA box, CAAT box) upstream of the gene.
2. RNA polymerase II binds to the transcription factor complex, forming the transcription initiation complex.
3. The DNA double helix is unwound (approximately 17 base pairs), forming the transcription bubble.

Elongation:

1. RNA polymerase moves along the template strand in the 3' to 5' direction, synthesising mRNA in the 5' to 3' direction.
2. Free RNA nucleotides are added by complementary base pairing (A--U, T--A, G--C, C--G).
3. Behind the RNA polymerase, the DNA re-forms the double helix and the mRNA is released.

Termination:

1. In eukaryotes, RNA polymerase continues past a polyadenylation signal (AAUAAA) in the pre-mRNA.
2. The pre-mRNA is cleaved approximately 10--35 nucleotides downstream of this signal.
3. Poly-A polymerase adds approximately 200 adenine nucleotides to the 3' end (the poly-A tail), which protects the mRNA from degradation and aids in export from the nucleus.

12.2 Post-Transcriptional Modification

In eukaryotes, the primary transcript (pre-mRNA) undergoes three modifications:

1. 5' capping: a modified guanine nucleotide (7-methylguanosine) is added to the 5' end. The cap protects the mRNA from degradation and is recognised by ribosomes during translation initiation.
2. Splicing: introns (non-coding regions) are removed and exons (coding regions) are joined together by the spliceosome (a complex of snRNPs -- small nuclear ribonucleoproteins). Alternative splicing allows a single gene to produce multiple different proteins (e.g., the DSCAM gene in Drosophila can produce 38,000 different protein variants from a single gene through alternative splicing).
3. 3' polyadenylation: described above.

12.3 Translation: Detailed Mechanism

Initiation:

1. The small ribosomal subunit (40S in eukaryotes, 30S in prokaryotes) binds to the 5' cap of the mRNA and scans along the mRNA until it reaches the start codon (AUG).
2. The initiator tRNA (carrying methionine, Met) binds to the start codon via its anticodon (UAC).
3. The large ribosomal subunit (60S in eukaryotes, 50S in prokaryotes) joins, forming the complete ribosome.
4. The initiator tRNA is in the P site (peptidyl site). The A site (aminoacyl site) is empty.

Elongation:

1. An aminoacyl-tRNA carrying the next amino acid enters the A site, where its anticodon base-pairs with the mRNA codon.
2. A peptide bond forms between the amino acid in the P site and the amino acid in the A site. This reaction is catalysed by peptidyl transferase (an rRNA ribozyme in the large subunit).
3. The ribosome translocates (moves 3 nucleotides along the mRNA): the now-empty tRNA in the P site moves to the E site (exit site) and is released; the tRNA carrying the growing polypeptide moves from the A site to the P site; a new codon enters the A site.

Termination:

1. When a stop codon (UAA, UAG, or UGA) enters the A site, no tRNA can bind.
2. Release factor proteins bind to the stop codon.
3. The polypeptide is released from the tRNA in the P site.
4. The ribosomal subunits dissociate, and the mRNA is released.

12.4 Calculating: Number of Nucleotides, Codons, and Amino Acids

Relationship	Calculation
Amino acids in protein =	$\frac◆LB◆\text{mRNA nucleotides}◆RB◆◆LB◆3◆RB◆$ (subtract 3 for stop codon)
mRNA nucleotides =	Number of amino acids $\times 3$ (plus 3 for stop codon)
DNA template strand nucleotides =	Same as mRNA nucleotides (plus introns in eukaryotes)
tRNA molecules required =	Number of amino acids (one per amino acid)
ATP required for translation =	$2 \times$ number of amino acids (1 for tRNA charging, 1 for translocation per elongation step)

13. Epigenetics in Depth

13.1 DNA Methylation

DNA methylation involves the addition of a methyl group ($\mathrm{-CH_3}$) to the 5-carbon of cytosine bases, forming 5-methylcytosine ($5\mathrm{mC}$). This is catalysed by DNA methyltransferases and occurs primarily at CpG dinucleotides (cytosine followed by guanine).

• Hypermethylation of gene promoter regions generally silences gene expression by preventing transcription factors from binding.
• Hypomethylation generally activates gene expression.

DNA methylation patterns are heritable during cell division: after DNA replication, the maintenance methyltransferase (DNMT1) methylates the new strand to match the old (template) strand.

13.2 Histone Modification

Histones are proteins around which DNA is wrapped to form nucleosomes. Histone tails (the N-terminal ends) can be chemically modified:

Modification	Effect on Transcription	Enzyme
Acetylation (of lysine)	Activates (loosens DNA-histone interaction)	Histone acetyltransferase (HAT)
Deacetylation	Represses (tightens DNA-histone interaction)	Histone deacetylase (HDAC)
Methylation	Can activate or repress, depending on which residue	Histone methyltransferase
Phosphorylation	Usually activates	Kinases

Acetylation adds an acetyl group ($\mathrm{-COCH_3}$) to lysine residues, neutralising the positive charge. This reduces the electrostatic attraction between the positively charged histone tails and the negatively charged DNA backbone, loosening the chromatin structure and allowing transcription factors and RNA polymerase to access the DNA.

13.3 Epigenetics and Disease

Condition	Epigenetic Change
Cancer	Global hypomethylation (causing genomic instability) combined with hypermethylation of tumour suppressor gene promoters (silencing them)
Angelman syndrome	Maternal allele of UBE3A is deleted or silenced; paternal allele is normally silenced by imprinting
Prader-Willi syndrome	Paternal allele of SNURF-SNRPN is deleted or silenced; maternal allele is normally silenced by imprinting
Agouti mouse model	Maternal diet (methyl donors: folate, B12) affects offspring coat colour and disease susceptibility via epigenetic changes at the A^vy locus

14. Mutations: Advanced Analysis

14.1 Types of Point Mutations

Mutation Type	Effect on Protein	Example
Silent (same-sense)	No change in amino acid (degeneracy of the genetic code)	GAA $\to$ GAG (both code for Glu)
Missense	One amino acid is changed	GAA $\to$ GUA (Glu $\to$ Val) -- sickle cell mutation in the $\beta$-globin gene
Nonsense	Codon becomes a stop codon; protein is truncated	UAU $\to$ UAA (Tyr $\to$ stop)
Frameshift	Insertion or deletion of nucleotides (not a multiple of 3) shifts the reading frame, changing all downstream amino acids	Deletion of one nucleotide in the CFTR gene causes cystic fibrosis

14.2 Mutagenic Agents

Agent	Type	Mechanism
UV radiation	Physical	Causes thymine dimers (covalent bonds between adjacent thymine bases), distorting the DNA helix and blocking replication
X-rays, gamma rays	Physical	Ionising radiation causes DNA strand breaks and base damage
Benzopyrene (in tobacco smoke)	Chemical	Forms bulky DNA adducts that distort the helix
Nitrous acid	Chemical	Deaminates cytosine to uracil (causing C $\to$ T transition after replication)
Ethidium bromide	Chemical	Intercalates between base pairs, causing insertions or deletions during replication

14.3 Mutation Rates

The spontaneous mutation rate in humans is approximately $1.2 \times 10^{-8}$ mutations per base pair per generation. Given a haploid genome of approximately $3.2 \times 10^9$ base pairs, each human acquires approximately 40--70 new mutations per generation.

Most mutations are neutral (have no effect on fitness) because:

• They occur in non-coding DNA (introns, intergenic regions).
• They are silent mutations (due to the degeneracy of the genetic code).
• The amino acid change does not significantly affect protein function.

A small fraction of mutations are deleterious (reduce fitness) and are removed by natural selection. An even smaller fraction are advantageous (increase fitness) and may be favoured by natural selection.

15. Gene Therapy and Genetic Screening

15.1 Genetic Screening Programmes

Genetic screening tests populations or individuals for specific genetic conditions:

Programme	Test	Purpose
Newborn blood spot test (heel prick)	Screening for sickle cell disease, cystic fibrosis, PKU, hypothyroidism	Early detection allows immediate treatment, preventing disability or death
Antenatal screening	NIPT (non-invasive prenatal testing): cell-free fetal DNA in maternal blood; nuchal translucency ultrasound	Detect chromosomal abnormalities (Down syndrome, Edwards syndrome, Patau syndrome)
Carrier screening	Testing for cystic fibrosis, sickle cell trait, Tay-Sachs carrier status	Inform reproductive decisions; identify couples at risk of having affected children
BRCA testing	Testing for mutations in BRCA1 and BRCA2 genes	Identify increased risk of breast and ovarian cancer; guide surveillance and prophylactic measures

15.2 Ethical Issues in Genetic Screening

Issue	Arguments For	Arguments Against
Newborn screening	Early treatment saves lives; cost-effective	Parental anxiety; false positives; privacy concerns
Prenatal screening	Allows informed decision-making; reduces burden of disease	May lead to termination for non-lethal conditions; disability rights concerns
Carrier screening	Allows informed reproductive choices	Potential discrimination; stigmatisation; psychological impact
Predictive testing (adult-onset)	Allows surveillance and early intervention	No cure for many conditions; psychological burden; insurance discrimination

15.3 PCR in Genetic Diagnosis

PCR is used to diagnose genetic diseases by amplifying the DNA region containing the mutation and analysing the product:

• Gel electrophoresis: detecting insertions/deletions that change PCR product size.
• Allele-specific PCR: primers that only amplify the mutant or wild-type allele.
• Restriction digest: if the mutation creates or destroys a restriction site, different fragment sizes result.
• DNA sequencing: directly reading the sequence to identify the mutation.

15.4 Genetic Fingerprinting: Paternity Testing

The child inherits one allele at each locus from each parent.

Worked Example. At an STR locus: Mother = 12, 15; Child = 12, 18; Alleged Father 1 = 14, 18; Alleged Father 2 = 10, 15.

The child inherited allele 12 from the mother. The child inherited allele 18 from the biological father. Father 1 has allele 18 and could be the father. Father 2 does not have allele 18 and cannot be the father.

With multiple loci tested, the probability of paternity can exceed 99.9%.

16. The Genetic Code: Properties

16.1 Key Features

Property	Description
Triplet	Each codon consists of 3 nucleotides, coding for one amino acid
Degenerate (redundant)	64 codons but only 20 amino acids; most amino acids are coded for by more than one codon
Non-overlapping	Each nucleotide is part of only one codon
Universal	The same genetic code is used by almost all organisms (with minor exceptions in mitochondria and some protozoa)
Unambiguous	Each codon specifies only one amino acid (no codon codes for two different amino acids)
Start and stop signals	AUG = start codon (also codes for methionine); UAA, UAG, UGA = stop codons

16.2 Consequences of Degeneracy

The degeneracy of the genetic code means that some mutations (silent/same-sense mutations) do not change the amino acid sequence of the protein:

• Leucine is coded by 6 codons: UUA, UUG, CUU, CUC, CUA, CUG.
• A mutation from CUU to CUC still produces leucine (silent mutation).
• A mutation from CUU to CCU changes leucine to proline (missense mutation).

Degeneracy is not random: the first two bases of a codon are usually more important than the third. Mutations at the third position (wobble position) are more likely to be silent.

16.3 Calculating the Number of Possible Codons

With 4 bases (A, U, C, G) in groups of 3: $4^3 = 64$ possible codons.

61 codons code for amino acids, 3 are stop codons (UAA, UAG, UGA), and 1 is the start codon (AUG, which also codes for methionine).

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17. Genetic Technology: CRISPR Applications

17.1 CRISPR-Cas9 in Medicine

CRISPR-Cas9 has revolutionised genetic research and holds great promise for treating genetic diseases:

Application	Status	Details
Sickle cell disease	Clinical trials	Editing the BCL11A gene in haematopoietic stem cells to reactivate fetal haemoglobin (HbF) production
Beta-thalassaemia	Clinical trials	Similar approach to sickle cell disease
Duchenne muscular dystrophy	Preclinical	Editing the dystrophin gene in muscle stem cells
Cancer immunotherapy	Clinical trials	Knocking out PD-1 receptor in T cells (CAR-T cells)
HIV	Preclinical	Disrupting the CCR5 co-receptor gene in T cells (making them resistant to HIV entry)
Hereditary blindness	Clinical trials	Editing retinal cells to restore vision

17.2 Ethical Considerations

Issue	For	Against
Germline editing	Could eliminate genetic diseases from future generations	Changes are heritable; unknown long-term effects; "designer babies" concern
Somatic editing	Treats existing patients; not heritable	Expensive; off-target effects; access inequality
Enhancement	Could improve human capabilities	Ethical concerns about fairness, consent, social pressure
Gene drives	Could eliminate disease vectors (mosquitoes)	Unpredictable ecological consequences; irreversible

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18. Advanced DNA Technology: Next-Generation Sequencing

18.1 Sanger vs Next-Generation Sequencing (NGS)

Feature	Sanger Sequencing	NGS (e.g., Illumina)
Throughput	One fragment at a time (up to $\approx 1000\ \mathrm{bp}$)	Millions of fragments simultaneously
Cost per genome	\approx \3000$ (human)	\approx \100$--$600$ (human)
Speed	Days to weeks for a genome	Hours to days
Read length	Up to $\approx 1000\ \mathrm{bp}$	Short ($100$--$300\ \mathrm{bp}$ for Illumina)
Accuracy	Very high ($99.99\%$)	High but with more errors per read (compensated by depth of coverage)
Applications	Single genes, small regions	Whole genomes, transcriptomes, epigenomes

18.2 Applications of NGS

• Whole genome sequencing: identifying mutations in patients with rare genetic diseases.
• Whole exome sequencing: sequencing only the protein-coding regions ($\approx 1.5\%$ of the genome), which is cheaper and more focused than whole genome sequencing.
• RNA-seq: sequencing the entire transcriptome to quantify gene expression levels and identify alternative splicing.
• Metagenomics: sequencing all DNA in an environmental sample (e.g., gut microbiome, soil, ocean water) to identify all the organisms present without culturing them.

19. Epigenetics: Mechanisms and Inheritance

19.1 DNA Methylation

DNA methylation involves the addition of a methyl group ($\mathrm{-CH_3}$) to cytosine bases, specifically at CpG sites (cytosine followed by guanine). Methylation is catalysed by DNA methyltransferases (DNMTs).

• Hypermethylation of promoter regions generally silences gene expression by preventing transcription factor binding or recruiting proteins that compact chromatin.
• Hypomethylation of promoter regions generally activates gene expression.

Pattern maintenance: after DNA replication, the maintenance methyltransferase DNMT1 copies the methylation pattern from the parental strand to the daughter strand. This is how methylation patterns are inherited through cell division (mitosis).

19.2 Histone Modification

Histones are proteins around which DNA is wrapped to form nucleosomes. Histone tails can be chemically modified:

Modification	Enzyme	Effect on Gene Expression
Acetylation (addition of acetyl group to lysine)	Histone acetyltransferases (HATs)	Activates expression: neutralises positive charge on histone, reducing affinity for negatively charged DNA, loosening chromatin
Deacetylation (removal of acetyl group)	Histone deacetylases (HDACs)	Silences expression: increases affinity between histones and DNA, compacting chromatin
Methylation (addition of methyl group to lysine or arginine)	Histone methyltransferases (HMTs)	Can activate or silence depending on which residue is modified (e.g., H3K4me3 activates; H3K9me3 and H3K27me3 silence)

19.3 Epigenetics and Disease

Disease	Epigenetic Mechanism
Cancer	Global hypomethylation (genomic instability) combined with hypermethylation of tumour suppressor gene promoters (silencing tumour suppressors)
Prader-Willi syndrome	Deletion or silencing of paternal genes on chromosome 15 (imprinting disorder)
Angelman syndrome	Deletion or silencing of maternal genes on chromosome 15 (same region as Prader-Willi, but opposite parent)
Beckwith-Wiedemann syndrome	Loss of imprinting at 11p15 (overgrowth disorder)

19.4 Epigenetic Inheritance

Most epigenetic marks are erased during gametogenesis and after fertilisation, but some escape this reprogramming:

• Genomic imprinting: certain genes are expressed in a parent-of-origin-specific manner (e.g., IGF2 is expressed only from the paternal allele; H19 is expressed only from the maternal allele). Imprinting is established during gametogenesis and maintained after fertilisation.
• Transgenerational epigenetic inheritance: in some cases, environmental exposures (e.g., diet, stress, toxins) can cause epigenetic changes that are transmitted across multiple generations. Evidence from animal studies (e.g., Agouti mouse model, where maternal diet affects offspring coat colour and disease risk via methylation of the Agouti gene).

warning

Common Pitfall Students often confuse epigenetic changes with genetic mutations. Epigenetic changes alter gene expression without changing the DNA sequence. They are potentially reversible (unlike mutations). However, epigenetic changes can be inherited through cell division (mitosis) and, in rare cases, across generations (transgenerational epigenetic inheritance). Genetic mutations are permanent changes to the DNA sequence itself.

20. Genetic Disorders: Detailed Analysis

20.1 Cystic Fibrosis (CF)

• Gene: CFTR (cystic fibrosis transmembrane conductance regulator) on chromosome 7.
• Mutation: most common is $\Delta$F508 (deletion of phenylalanine at position 508), a three-nucleotide deletion.
• Inheritance: autosomal recessive. Carriers are heterozygous and unaffected (approximately 1 in 25 people of Northern European descent).
• Protein function: CFTR is a chloride channel in epithelial cell membranes. The mutation causes misfolded CFTR protein, which is degraded before reaching the membrane.
• Symptoms: thick, sticky mucus in the lungs (recurrent infections); pancreatic insufficiency (blocked ducts prevent enzyme delivery to the intestine); male infertility (absence of vas deferens); salty sweat (elevated $\mathrm{Cl^-}$ in sweat -- diagnostic test).
• Treatment: physiotherapy (chest percussion to clear mucus); antibiotics for lung infections; pancreatic enzyme supplements; CFTR modulators (e.g., ivacaftor, lumacaftor) that improve CFTR protein function; gene therapy trials.

20.2 Huntington's Disease

• Gene: HTT (huntingtin) on chromosome 4.
• Mutation: expanded CAG trinucleotide repeat in the coding region (polyglutamine expansion). Normal: 10--35 repeats; disease: > 36 repeats.
• Inheritance: autosomal dominant. Every child of an affected individual has a 50% chance of inheriting the disease.
• Anticipation: the number of CAG repeats tends to increase in successive generations (especially when inherited from the father), causing earlier onset and more severe disease.
• Symptoms: progressive neurodegeneration (especially in the striatum); involuntary movements (chorea); cognitive decline; psychiatric symptoms (depression, psychosis). Onset typically 35--45 years; death 15--20 years after onset.
• Genetic counselling: predictive genetic testing is available. This raises ethical issues: would you want to know you will develop an untreatable disease?

20.3 Sickle Cell Anaemia

• Gene: HBB ($\beta$-globin) on chromosome 11.
• Mutation: point mutation (missense) -- codon GAG (glutamic acid) $\to$ GTG (valine) at position 6 of the $\beta$-globin chain.
• Inheritance: autosomal recessive.
• Effect on protein: valine is hydrophobic (glutamic acid is hydrophilic). The mutation causes haemoglobin S (HbS) to polymerise under low $\mathrm{O_2}$ conditions, deforming red blood cells into a sickle shape.
• Symptoms: chronic haemolytic anaemia; painful vaso-occlusive crises (sickled cells block small blood vessels); increased risk of infection (splenic damage); stroke; organ damage.
• Heterozygote advantage: carriers ($Hb^A Hb^S$) have some resistance to malaria (Plasmodium falciparum has difficulty infecting sickled cells; infected sickled cells are removed more rapidly by the spleen).

21. Protein Synthesis: Detailed Mechanism

21.1 Transcription (in Eukaryotes)

1. Initiation: RNA polymerase II binds to the promoter region (TATA box, approximately 25 bp upstream of the transcription start site) with the help of general transcription factors (TFIID, TFIIA, TFIIB, etc.).
2. Elongation: RNA polymerase II unwinds the DNA double helix and synthesises mRNA in the 5' $\to$ 3' direction, using the antisense (template) strand as a template. The DNA strand that has the same sequence as the mRNA (but with T instead of U) is called the sense strand (coding strand).
3. Termination: RNA polymerase II continues past the polyadenylation signal (AAUAAA). The pre-mRNA is cleaved approximately 20--30 nucleotides downstream, and a poly-A tail is added by polyadenylate polymerase.

21.2 Post-Transcriptional Processing

Before mRNA leaves the nucleus, it undergoes three modifications:

Modification	Description	Function
5' capping	Addition of 7-methylguanosine cap	Protects mRNA from degradation; aids ribosome binding during translation
Splicing	Removal of introns by the spliceosome (snRNPs); joining of exons	Produces mature mRNA with a continuous coding sequence; alternative splicing allows one gene to produce multiple protein isoforms
3' polyadenylation	Addition of approximately 200 adenine nucleotides (poly-A tail)	Protects mRNA from exonuclease degradation; aids export from nucleus; facilitates translation

21.3 Translation

1. Initiation: the small ribosomal subunit (40S in eukaryotes) binds to the 5' cap of the mRNA and scans along the mRNA until it finds the start codon (AUG). The initiator tRNA (carrying methionine) binds to the start codon in the P site. The large ribosomal subunit (60S) joins, forming the complete 80S ribosome.
2. Elongation: aminoacyl-tRNAs enter the A site, where the anticodon pairs with the mRNA codon. A peptide bond forms between the amino acid in the P site and the amino acid in the A site (catalysed by peptidyl transferase, which is rRNA -- a ribozyme). The ribosome translocates by one codon: the empty tRNA moves to the E site and exits; the tRNA with the growing polypeptide moves from A to P.
3. Termination: when a stop codon (UAA, UAG, UGA) enters the A site, a release factor binds (no tRNA has an anticodon for stop codons). The polypeptide is released, and the ribosome subunits dissociate.

21.4 Calculations: From DNA to Protein

Example: A gene has 1,500 base pairs of coding sequence (excluding introns).

Number of codons $= \frac{1500}{3} = 500$ codons.

Number of amino acids in the protein $= 500 - 1 = 499$ (the stop codon does not code for an amino acid).

Molecular weight of the protein $= 499 \times 110 \approx 54,890\ \mathrm{Da}$ (average amino acid molecular weight $\approx 110\ \mathrm{Da}$).

Example: If the mutation rate is $10^{-9}$ mutations per base pair per cell division, and a gene has 1,500 base pairs:

Probability of a mutation in this gene per cell division $= 1500 \times 10^{-9} = 1.5 \times 10^{-6}$ (approximately 1 in 667,000 cell divisions).

22. Genetic Engineering: Detailed Protocols

22.1 Restriction Enzymes (Restriction Endonucleases)

Restriction enzymes are bacterial enzymes that cut DNA at specific recognition sequences (usually palindromic, 4--8 base pairs):

Enzyme	Recognition Sequence	Cut Type	Sticky/Blunt
EcoRI	5'-GAATTC-3'	Cuts between G and A	Sticky ends (5' overhang: AATT)
BamHI	5'-GGATCC-3'	Cuts between G and G	Sticky ends (5' overhang: GATC)
HindIII	5'-AAGCTT-3'	Cuts between A and A	Sticky ends (5' overhang: AGCT)
SmaI	5'-CCCGGG-3'	Cuts between C and C	Blunt ends

Sticky ends are more useful for genetic engineering because complementary sticky ends can base-pair (via hydrogen bonds), allowing DNA fragments from different sources to be joined. DNA ligase then seals the sugar-phosphate backbone.

22.2 Gel Electrophoresis

Gel electrophoresis separates DNA fragments by size:

1. DNA samples are loaded into wells in an agarose gel.
2. An electric current is applied. DNA is negatively charged (phosphate backbone), so it moves towards the positive electrode (anode).
3. Smaller fragments move faster and travel further; larger fragments move slower and stay closer to the wells.
4. The gel is stained with a DNA-binding dye (e.g., ethidium bromide or GelRed) and visualised under UV light.

Applications: separating DNA fragments after restriction digest; analysing PCR products; DNA fingerprinting (forensics, paternity testing); detecting mutations (e.g., sickle cell mutation alters a restriction site).

22.3 Polymerase Chain Reaction (PCR)

PCR amplifies a specific DNA sequence exponentially:

Step	Temperature	Duration	What Happens
Denaturation	95 degrees C	30 s	DNA double helix separates into single strands
Annealing	55--65 degrees C	30 s	Primers (forward and reverse) bind to complementary sequences flanking the target
Extension	72 degrees C	30 s -- 2 min	Taq DNA polymerase synthesises new DNA strands (extends from primers)

After $n$ cycles, the number of copies of the target sequence $= 2^n$. After 30 cycles: $2^{30} \approx 1.07 \times 10^9$ copies.

Requirements for PCR:

• Template DNA (the DNA to be amplified).
• Two primers (short, single-stranded DNA sequences, typically 18--25 nucleotides, complementary to the sequences flanking the target).
• Taq polymerase (thermostable DNA polymerase from Thermus aquaticus, a thermophilic bacterium; active at 72 degrees C).
• Free nucleotides (dATP, dTTP, dCTP, dGTP).
• Buffer (maintains optimal pH for the enzyme).
• $\mathrm{Mg^{2+}}$ ions (cofactor for Taq polymerase).

Applications: forensic DNA analysis; prenatal diagnosis of genetic diseases; detecting pathogens (e.g., COVID-19 RT-PCR); cloning genes; sequencing.

22.4 DNA Fingerprinting (DNA Profiling)

DNA fingerprinting analyses highly variable regions of the genome:

1. Mini-satellites (VNTRs): variable number tandem repeats -- short sequences (10--100 bp) repeated a variable number of times at specific loci. Each person has a unique combination of repeat numbers.
2. Micro-satellites (STRs): short tandem repeats -- sequences of 2--6 bp repeated a variable number of times (e.g., the tetranucleotide repeat GATA). STR analysis is the standard method in forensic DNA profiling (the UK National DNA Database uses 10 STR loci).

Process:

1. Extract DNA from the sample.
2. PCR to amplify the STR loci (using fluorescently labelled primers).
3. Capillary electrophoresis to separate fragments by size.
4. Compare the fragment sizes to a reference database.

Calculating the match probability:

If each STR locus has a probability of a random match of approximately 0.1 (10%), and 10 loci are analysed:

$P(\text{random match at all 10 loci}) = 0.1^{10} = 1 \times 10^{-10} \text{ (1 in 10 billion)}$

This is well below the world population ($\approx 8$ billion), so the match is essentially unique.

23. Mutations: Types and Effects

23.1 Types of Gene Mutation

Type	Description	Effect on Protein
Substitution (point mutation)	One base pair is replaced by another	May change one amino acid (missense), create a stop codon (nonsense), or have no effect (silent)
Deletion	One or more base pairs are removed	Frameshift if not a multiple of 3; all downstream codons are changed
Insertion	One or more base pairs are added	Frameshift if not a multiple of 3
Duplication	A segment of DNA is copied and inserted	May alter gene dosage or protein function
Inversion	A segment of DNA is reversed	May disrupt gene if break occurs within the gene
Translocation	A segment moves to a new location (possibly on a different chromosome)	May create a fusion gene (e.g., Philadelphia chromosome)

23.2 Effects of Mutations

Effect	Description	Example
Neutral (silent)	No change in amino acid sequence (due to degeneracy of the genetic code)	GAA $\to$ GAG (both code for glutamic acid)
Missense	One amino acid is changed	Sickle cell: GAG $\to$ GTG (glutamic acid $\to$ valine at position 6 of $\beta$-globin)
Nonsense	A sense codon is changed to a stop codon	Duchenne muscular dystrophy: premature stop codon in dystrophin gene
Frameshift	Insertion or deletion (not multiple of 3) shifts the reading frame	Cystic fibrosis: $\Delta$F508 (3-bp deletion -- actually NOT a frameshift, but a deletion of one amino acid)
Splice site	Mutation at an exon-intron boundary affects mRNA splicing	$\beta$-thalassaemia: mutations at splice sites cause abnormal mRNA processing

23.3 Mutagenic Agents

Agent	Type	Mechanism	Example
UV radiation	Physical	Causes thymine dimers (covalent bonds between adjacent thymines), which distort the DNA helix and block replication	Xeroderma pigmentosum: inability to repair thymine dimers causes extreme UV sensitivity and skin cancer
Ionising radiation (X-rays, gamma rays)	Physical	Causes double-strand breaks; can delete or rearrange large segments of DNA	Increased cancer risk after radiation exposure
Tobacco smoke	Chemical	Contains over 60 carcinogens, including benzopyrene (causes G $\to$ T transversions)	Lung cancer: mutations in p53 gene
Aflatoxin (from Aspergillus flavus)	Chemical	Causes G $\to$ T transversions, especially in the p53 gene	Liver cancer (hepatocellular carcinoma)
Mustard gas	Chemical	Alkylates guanine (adds alkyl group), causing mispairing	DNA cross-linking; carcinogenic

23.4 DNA Repair Mechanisms

Mechanism	Description	What It Repairs
Mismatch repair	Corrects errors made during DNA replication (wrong base incorporated)	Single base mismatches; small insertion/deletion loops
Base excision repair (BER)	Damaged base is removed by a glycosylase; AP endonuclease cuts the backbone; DNA polymerase fills the gap	Deaminated bases (e.g., uracil in DNA); oxidised bases
Nucleotide excision repair (NER)	A short stretch of DNA containing the damage is excised; DNA polymerase fills the gap	Thymine dimers (UV damage); bulky chemical adducts
Double-strand break repair (HR)	Homologous recombination: uses sister chromatid as template	Double-strand breaks (ionising radiation)
Double-strand break repair (NHEJ)	Non-homologous end joining: broken ends are directly rejoined (error-prone)	Double-strand breaks; can cause small insertions/deletions

---

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24. Genetic Screening and Counselling

24.1 Genetic Screening Programmes

Condition	Screening Method	When	Target Population
Phenylketonuria (PKU)	Guthrie test (blood spot analysed for phenylalanine level)	Days 5--8 after birth	All newborns
Sickle cell disease	Blood test (haemoglobin electrophoresis)	Newborn screening or antenatal carrier screening	Newborns in high-prevalence areas; carriers in at-risk populations
Cystic fibrosis	Immunoreactive trypsin assay (IRT) on blood spot; confirmed by genetic testing	Newborn screening	All newborns
Down syndrome	Non-invasive prenatal testing (NIPT): cell-free fetal DNA in maternal blood; confirmed by amniocentesis/chorionic villus sampling	10--12 weeks (NIPT); 15--20 weeks (amniocentesis)	All pregnant women (NIPT); women with risk factors (amniocentesis)
Huntington's disease	Genetic test (CAG repeat expansion in HTT gene)	Adults with family history (with genetic counselling)	At-risk individuals (presymptomatic testing available)
BRCA1/BRCA2 mutations	Genetic testing (sequencing)	Adults with family history of breast/ovarian cancer	Women with strong family history

24.2 Prenatal Diagnostic Techniques

Technique	Timing	What It Detects	Risk
Amniocentesis	15--20 weeks	Chromosomal abnormalities; genetic disorders (DNA analysis of fetal cells in amniotic fluid)	0.5--1% risk of miscarriage
Chorionic villus sampling (CVS)	11--14 weeks	Chromosomal abnormalities; genetic disorders (DNA analysis of placental cells)	1--2% risk of miscarriage
Non-invasive prenatal testing (NIPT)	10--12 weeks	Chromosomal abnormalities (trisomies 21, 18, 13); sex chromosome aneuploidies	No risk to fetus (blood test)
Ultrasound	12 weeks (nuchal translucency) + 18--20 weeks (anomaly scan)	Structural abnormalities; nuchal translucency (Down syndrome risk)	No risk to fetus

24.3 Genetic Counselling

Genetic counselling provides information and support to individuals and families affected by or at risk of genetic disorders:

1. Taking a family history: constructing a pedigree chart to identify patterns of inheritance.
2. Risk assessment: calculating the probability of an individual being a carrier or being affected.
3. Explaining options: discussing genetic testing, prenatal diagnosis, reproductive options (pre-implantation genetic diagnosis, PGD).
4. Providing emotional support: helping individuals cope with the psychological impact of genetic information.
5. Informed consent: ensuring individuals understand the implications of genetic testing (including the possibility of unexpected findings).

24.4 Pre-implantation Genetic Diagnosis (PGD)

PGD is used during IVF to test embryos for specific genetic conditions before implantation:

1. Eggs are collected from the mother and fertilised with the father's sperm in vitro.
2. Embryos develop to the 8-cell stage (approximately 3 days).
3. One or two cells (blastomeres) are removed from each embryo (biopsy).
4. The cells are tested for the genetic condition (by PCR for single-gene disorders or FISH/NGS for chromosomal abnormalities).
5. Only embryos free of the condition are implanted into the mother's uterus.

Ethical considerations:

• "Designer babies": PGD could potentially be used for non-medical trait selection (sex selection, physical characteristics).
• Discarded embryos: embryos found to carry the condition are destroyed.
• Access and cost: PGD is expensive and not universally available.

25. Genomics and Personalised Medicine

25.1 The Human Genome: Key Statistics

Feature	Data
Total genome size	3.2 billion base pairs
Number of protein-coding genes	Approximately 20,000--25,000
Percentage of genome that codes for proteins	Approximately 1.5%
Percentage that is repetitive (transposons, LINEs, SINEs)	Approximately 50%
Number of common SNPs	Approximately 10 million
Percentage of genome that is conserved (shared with other mammals)	Approximately 5%
Total length of all genes	Approximately 1.5% of genome

25.2 Non-Coding DNA: Functions

Most of the genome does not code for proteins, but this non-coding DNA is not "junk":

Type of Non-Coding DNA	Function
Regulatory sequences	Promoters, enhancers, silencers, insulators -- control when, where, and how much genes are expressed
Introns	Removed during mRNA splicing; may contain regulatory sequences
Telomeres	Protect chromosome ends from degradation; shorten with each cell division; linked to ageing
Centromeres	Site of kinetochore assembly; essential for chromosome segregation during mitosis
Repetitive elements (LINEs, SINEs, transposons)	"Jumping genes" that can move within the genome; some carry regulatory sequences
Non-coding RNAs (miRNA, lncRNA, snRNA, snoRNA)	Regulatory roles in gene expression, chromatin structure, X-inactivation
Pseudogenes	Non-functional copies of genes; evidence for evolution

26. DNA Sequencing Technologies

26.1 Sanger Sequencing (First Generation)

Feature	Details
Principle	Chain-termination method; modified nucleotides (ddNTPs) randomly terminate DNA synthesis
Process	4 separate reactions (one per ddNTP: ddATP, ddCTP, ddGTP, ddTTP); fragments separated by gel electrophoresis; sequence read from band pattern
Read length	500--1000 base pairs per reaction
Throughput	Low (one fragment at a time)
Accuracy	Very high (99.99%)
Use today	Small-scale sequencing; confirming Sanger sequences; clinical diagnostics

26.2 Next-Generation Sequencing (NGS)

Feature	Sanger	NGS (e.g., Illumina)
Cost per genome	~$5,000--10,000	~$100--1,000
Time	Weeks to months	Hours to days
Throughput	One fragment per run	Millions of fragments simultaneously (massively parallel)
Read length	500--1000 bp	50--300 bp (short read)
Applications	Individual gene sequencing; clinical diagnostics	Whole-genome sequencing; metagenomics; transcriptomics; epigenomics

26.3 Bioinformatics

DNA sequencing generates enormous amounts of data. Bioinformatics is the computational analysis of biological data:

Task	Description
Sequence assembly	Piecing together millions of short reads into a complete genome (like solving a jigsaw puzzle)
Gene annotation	Identifying coding sequences, regulatory elements, and functional elements within the genome
Comparative genomics	Comparing genomes between species to identify conserved regions and evolutionary relationships
Variant calling	Identifying SNPs, insertions, deletions, and structural variants compared to a reference genome

27. Genetic Engineering: Restriction Enzymes and Recombinant DNA

27.1 Restriction Endonucleases (Restriction Enzymes)

Restriction enzymes cut DNA at specific recognition sequences (palindromic sequences):

Enzyme	Recognition Sequence	Cut Type	Blunt/Sticky
EcoRI	5'-GAATTC-3'	Cuts between G and A on both strands	Sticky (5' overhang: AATT)
BamHI	5'-GGATCC-3'	Cuts between G and G on both strands	Sticky (5' overhang: GATC)
HindIII	5'-AAGCTT-3'	Cuts between A and A on both strands	Sticky (5' overhang: AGCT)
SmaI	5'-CCCGGG-3'	Cuts between C and G on both strands	Blunt (no overhang)

27.2 DNA Ligase

DNA ligase joins DNA fragments by forming phosphodiester bonds between the 3'-OH of one nucleotide and the 5'-phosphate of the next:

Feature	DNA Ligase	Restriction Enzyme
Function	Joins DNA fragments (seals nicks in the sugar-phosphate backbone)	Cuts DNA at specific recognition sequences
Bond formed	Phosphodiester bond	Breaks phosphodiester bond
Energy source	ATP	None required
Use in genetic engineering	Joining insert DNA to vector DNA (plasmid)	Cutting open the plasmid vector and excising the gene of interest

27.3 Steps in Genetic Engineering

Step	Description
1. Isolation	Gene of interest is identified and cut out of donor DNA using restriction enzymes
2. Vector preparation	Plasmid is cut open with the same restriction enzyme (produces complementary sticky ends)
3. Insertion	Gene of interest is mixed with cut plasmid; DNA ligase seals the recombinant plasmid
4. Transformation	Recombinant plasmid is introduced into host bacteria (heat shock or electroporation)
5. Selection	Bacteria are grown on selective media (e.g., antibiotic resistance gene on plasmid); only transformed bacteria survive
6. Screening	Colonies are screened to confirm the gene has been inserted correctly (e.g., using antibiotic resistance + blue-white screening with lacZ)

28. DNA Replication in Detail

28.1 The Replication Fork

Enzyme/Protein	Function
Helicase	Breaks hydrogen bonds between complementary base pairs; unwinds the double helix; moves in the 5' $\to$ 3' direction along the template strand
DNA polymerase III	Adds nucleotides to the 3' end of the growing strand (5' $\to$ 3' synthesis); proofreads (3' $\to$ 5' exonuclease activity)
DNA polymerase I	Replaces RNA primers with DNA nucleotides
Primase	Synthesises short RNA primers (~10 nucleotides) complementary to the template strand; provides a free 3'-OH for DNA polymerase to start
DNA ligase	Forms phosphodiester bonds between adjacent Okazaki fragments on the lagging strand
Single-strand binding proteins (SSBs)	Stabilise the unwound single-stranded DNA; prevent it from re-annealing or being degraded
Topoisomerase (DNA gyrase)	Relieves tension ahead of the replication fork by cutting, untwisting, and rejoining the DNA double helix

28.2 Leading Strand vs Lagging Strand

Feature	Leading Strand	Lagging Strand
Direction of synthesis	Continuous, in the 5' $\to$ 3' direction (same direction as fork movement)	Discontinuous; synthesised in short Okazaki fragments (100--1000 nucleotides in prokaryotes; 100--200 in eukaryotes)
Primers required	One initial RNA primer	Multiple RNA primers (one per Okazaki fragment)
Speed	Faster (continuous)	Slower (requires repeated primer synthesis and ligation)

28.3 Calculating DNA Replication

If a human cell has 6.4 Gbp (3.2 billion base pairs per haploid genome; 6.4 billion in diploid) and DNA polymerase adds ~50 nucleotides per second:

$\text{Time for leading strand} = \frac◆LB◆3.2 \times 10^9◆RB◆◆LB◆50◆RB◆ = 64 \times 10^6\ \text{seconds} \approx 740\ \text{days}$

In practice, replication takes hours because:

• Multiple replication origins (eukaryotes have ~10,000 origins; bacteria have one).
• Bidirectional replication (two forks per origin).

29. Protein Synthesis: Transcription in Detail

29.1 Transcription Initiation

Step	What Happens
1	RNA polymerase binds to the promoter region upstream of the gene (in eukaryotes, transcription factors must first bind)
2	RNA polymerase unwinds the DNA double helix (about 17 base pairs)
3	RNA polymerase begins synthesising mRNA complementary to the template (antisense) strand in the 5' $\to$ 3' direction

29.2 Transcription Elongation and Termination

Step	What Happens
Elongation	RNA polymerase moves along the template strand; adds complementary RNA nucleotides (A pairs with U on RNA; T pairs with A; G pairs with C; C pairs with G)
Termination	RNA polymerase reaches a terminator sequence; detaches from the DNA; the pre-mRNA is released

29.3 Post-Transcriptional Modification (Eukaryotes Only)

Modification	Description	Why It Matters
5' capping	A modified guanine nucleotide (7-methylguanosine) is added to the 5' end	Protects mRNA from degradation; helps ribosome recognise the mRNA
3' poly-A tail	~200 adenine nucleotides are added to the 3' end	Protects mRNA from degradation; aids export from nucleus
Splicing	Introns (non-coding regions) are removed; exons (coding regions) are joined together by the spliceosome	Produces mature mRNA containing only coding sequences; alternative splicing can produce different proteins from the same gene

30. Mutations and DNA Repair

30.1 Types of Gene Mutations

Mutation Type	Description	Effect on Protein
Substitution (point mutation)	One base pair is replaced by another	May change one amino acid (missense); may not change any amino acid (silent); may introduce a premature stop codon (nonsense)
Insertion	One or more nucleotides are inserted into the sequence	Frameshift; changes all codons downstream of the insertion; usually severe
Deletion	One or more nucleotides are removed from the sequence	Frameshift; changes all codons downstream; usually severe
Duplication	A section of DNA is duplicated	May cause a frameshift or add extra amino acids
Inversion	A section of DNA is reversed	May or may not affect the protein depending on location and whether it disrupts the reading frame

30.2 Causes of Mutation

Cause	Description	Examples
Spontaneous mutations	Occur naturally during DNA replication (errors by DNA polymerase); tautomeric shifts in bases	DNA polymerase occasionally inserts the wrong nucleotide; spontaneous deamination of cytosine to uracil
Mutagens (chemical)	Chemicals that damage or alter DNA	Nitrous acid (deaminates bases); benzopyrene (in tobacco smoke; adds bulky groups to DNA); mustard gas (cross-links DNA strands)
Mutagens (physical)	High-energy radiation that damages DNA	UV light (causes thymine dimers; adjacent thymines become covalently bonded); ionising radiation (X-rays, gamma rays; cause single- and double-strand breaks)
Biological agents	Viruses that integrate into DNA; transposons (jumping genes)	HPV integrates into host DNA; LINEs and SINEs cause mutations when they insert into genes

30.3 DNA Repair Mechanisms

Mechanism	What It Repairs	How It Works
Proofreading (3' $\to$ 5' exonuclease)	Mismatched bases during DNA replication	DNA polymerase III (prokaryotes) or DNA polymerase $\delta$/$\varepsilon$ (eukaryotes) detect and remove mismatched nucleotides immediately
Mismatch repair	Mismatches missed by proofreading	After replication, MutS (bacteria) or MSH proteins (eukaryotes) detect mismatches; the incorrect strand is identified, excised, and replaced
Excision repair	Damaged bases (e.g., thymine dimers, deaminated bases)	Endonuclease cuts out the damaged section; DNA polymerase fills in the correct sequence; DNA ligase seals the gap

31. Genetic Disorders

31.1 Single Gene Disorders

Disorder	Gene	Mutation	Inheritance	Symptoms
Cystic fibrosis	CFTR (chromosome 7)	Deletion of 3 nucleotides ($\Delta$F508); loss of phenylalanine at position 508	Autosomal recessive	Thick mucus in lungs and pancreas; chronic lung infections; malabsorption; male infertility
Sickle cell anaemia	HBB (chromosome 11)	Missense mutation (substitution); glutamic acid $\to$ valine at position 6 of $\beta$-globin	Autosomal recessive	Sickled RBCs; block capillaries; pain crises; organ damage; anaemia
Huntington's disease	HTT (chromosome 4)	CAG trinucleotide repeat expansion (> 35 repeats)	Autosomal dominant	Progressive neurodegeneration; chorea (involuntary movements); dementia; death 15--20 years after onset
Haemophilia A	F8 (X chromosome)	Various mutations (inversions, point mutations, deletions)	X-linked recessive	Deficiency of clotting factor VIII; prolonged bleeding; joint damage (haemarthrosis)
Duchenne muscular dystrophy	DMD (X chromosome)	Deletions/duplications causing frameshift in dystrophin gene	X-linked recessive	Progressive muscle weakness; loss of ambulation by age 12; death by age 20--30 (cardiac/respiratory failure)

31.2 Chromosomal Disorders

Disorder	Chromosomal Abnormality	Karyotype	Symptoms
Down syndrome (trisomy 21)	Extra copy of chromosome 21	47,XX,+21 or 47,XY,+21	Intellectual disability; flat facial profile; single palmar crease; congenital heart defects; increased risk of leukaemia and Alzheimer's
Turner syndrome	Monosomy X (only one X chromosome)	45,X	Female; short stature; webbed neck; underdeveloped ovaries (infertility); no puberty without hormone treatment
Klinefelter syndrome	Extra X chromosome in males	47,XXY	Male; tall; reduced fertility (small testes); gynaecomastia (breast development); learning difficulties (mild)

32. Epigenetics

32.1 What Is Epigenetics?

Epigenetics is the study of heritable changes in gene expression that do NOT involve changes to the DNA sequence.

Mechanism	Description	Effect on Gene Expression
DNA methylation	Methyl group ($-\mathrm{CH_3}$) added to cytosine bases at CpG sites (usually in promoter regions)	Methylated DNA = gene is silenced (switched off); unmethylated DNA = gene is active (switched on)
Histone modification	Acetyl groups, methyl groups, or phosphate groups added to histone tails	Acetylation of histones = chromatin is less condensed (euchromatin) = gene is active; deacetylation = chromatin is more condensed (heterochromatin) = gene is silenced
Non-coding RNA	miRNA binds to mRNA and prevents translation or targets it for degradation	Decreases protein production from the target gene

32.2 Epigenetics and Disease

Disease	Epigenetic Mechanism
Cancer	Global DNA hypomethylation (genomic instability) + hypermethylation of tumour suppressor gene promoters (e.g., p16, BRCA1); histone modification abnormalities
Angelman syndrome	Maternal UBE3A gene is deleted or silenced (methylated) on the maternal chromosome; paternal copy is normally silenced in certain brain regions by genomic imprinting
Prader-Willi syndrome	Paternal region of chromosome 15 is deleted or silenced; maternal copy is normally imprinted (silenced) in this region

33. PCR (Polymerase Chain Reaction)

33.1 What Is PCR?

PCR is a laboratory technique used to amplify a specific region of DNA, producing millions of copies from a small sample.

33.2 The Three Steps (Each Cycle)

Step	Temperature	What Happens
Denaturation	95$\degree$C	DNA double helix is unwound; hydrogen bonds between complementary base pairs are broken; two single strands are produced
Annealing	55--65$\degree$C (depends on primers)	Primers (short, single-stranded DNA sequences, typically 18--25 nucleotides) bind to their complementary sequences at the 3' ends of the target region on each strand
Extension (elongation)	72$\degree$C (optimum for Taq polymerase)	Taq polymerase (thermostable DNA polymerase from Thermus aquaticus) adds nucleotides to the 3' end of each primer, synthesising new DNA strands

33.3 Exponential Amplification

Cycle	Number of DNA Molecules
0 (start)	1
1	2
2	4
3	8
n	$2^n$
30	$2^{30} \approx 1.07 \times 10^9$ (over 1 billion copies)
32	$2^{32} \approx 4.29 \times 10^9$ (over 4 billion copies)

33.4 PCR vs DNA Replication

Feature	PCR	In Vivo DNA Replication
Enzyme	Taq polymerase (thermostable)	DNA polymerase III (prokaryotes) / DNA polymerase $\delta$ (eukaryotes)
Primers	Synthetic, specific to target region	RNA primase synthesises RNA primers
Helicase	Not needed (heat denatures DNA)	Required to unwind the double helix
Temperature	Cycled between 95, 55--65, and 72$\degree$C	Constant 37$\degree$C
Speed	~30 cycles in ~2 hours	~8 hours for the whole human genome

34. Gel Electrophoresis

34.1 Principles

Principle	Description
Separation by size	DNA fragments are separated based on their size (length in base pairs)
Movement towards positive electrode	DNA is negatively charged (phosphate backbone); moves towards the positive electrode (anode)
Agarose gel	Porous gel matrix; acts as a molecular sieve; smaller fragments move faster and further
Staining	DNA is visualised using a fluorescent dye that intercalates between DNA bases (e.g., ethidium bromide, SYBR Safe)

34.2 Procedure

Step	Description
1	Agarose powder is dissolved in buffer and heated to form a gel
2	The gel is poured into a casting tray and a comb is inserted to form wells
3	Once set, the gel is placed in a tank filled with electrophoresis buffer
4	DNA samples are mixed with a loading dye (for visibility) and loaded into the wells
5	An electric current is applied; DNA fragments migrate towards the positive electrode
6	Smaller fragments travel further; larger fragments are retarded by the gel matrix
7	A DNA ladder (standard) of known fragment sizes is run alongside the samples for comparison
8	The gel is stained and viewed under UV light; band positions are compared to the ladder to estimate fragment sizes

35. Genetic Fingerprinting in Practice

35.1 DNA Fingerprinting Process

Step	Description
1. DNA extraction	DNA is isolated from a biological sample (blood, saliva, hair root, semen)
2. PCR amplification	Specific STR loci are amplified using primers complementary to flanking regions
3. Electrophoresis	PCR products are separated by size on an agarose or polyacrylamide gel
4. Analysis	Band pattern is compared between samples; statistical analysis calculates the probability of a match

35.2 Key STR Loci Used in Forensics (UK)

Locus	Chromosome	Repeat Unit	Number of Alleles
D21S11	21	TCTA	~30
D18S51	18	AGAA	~20
D16S539	16	GATA	~10
TH01	11	TCAT	~8
D8S1179	8	TATC	~12
vWA	12	TCTA	~12
FGA	4	TTTC	~25

35.3 Applications

Application	Description
Criminal cases	Matching DNA from a crime scene to a suspect; exonerating the innocent
Paternity testing	Comparing a child's DNA profile to the alleged father; the child must inherit one allele from each parent at each locus
Identification of remains	Comparing DNA from unidentified remains to relatives of missing persons
Immigration disputes	Proving family relationships when documentation is insufficient

36. Non-Mendelian Inheritance

36.1 Codominance

Codominance occurs when both alleles are expressed equally in the heterozygote (neither is dominant):

Example	Genotypes and Phenotypes
Blood groups (ABO system)	$\mathrm{I^AI^A}$ = blood group A; $\mathrm{I^BI^B}$ = blood group B; $\mathrm{I^AI^B}$ = blood group AB (both A and B antigens expressed); $\mathrm{ii}$ = blood group O
Sickle cell trait	$\mathrm{Hb^AHb^A}$ = normal (HbA only); $\mathrm{Hb^SHb^S}$ = sickle cell disease (HbS only); $\mathrm{Hb^AHb^S}$ = sickle cell trait (both HbA and HbS produced; carrier)

36.2 Sex-Linked Inheritance: Worked Example

A woman who is a carrier for colour blindness (X$^N$X$^c$) has a child with a man with normal vision (X$^N$Y).

| Possible gametes from mother | X$^N$ or X$^c$ (50% each) | | Possible gametes from father | X$^N$ or Y (50% each) |

Offspring	Probability	Phenotype
X$^N$X$^N$ (daughter)	25%	Normal vision
X$^N$X$^c$ (daughter)	25%	Carrier (normal vision, but carrier)
X$^N$Y (son)	25%	Normal vision
X$^c$Y (son)	25%	Colour blind

Note: All daughters have normal vision (even carriers), but 50% are carriers. Half of sons are colour blind. This is why sex-linked recessive conditions are much more common in males.

43. Non-Disjunction and Chromosomal Abnormalities

43.1 What Is Non-Disjunction?

Non-disjunction is the failure of homologous chromosomes or sister chromatids to separate properly during meiosis I or meiosis II. This produces gametes with an abnormal number of chromosomes (aneuploidy).

Stage	What Fails to Separate	Result
Meiosis I	Homologous chromosomes	All four gametes are abnormal (two with an extra chromosome, two missing one)
Meiosis II	Sister chromatids	Two gametes are normal, two are abnormal (one with extra, one missing)

43.2 Common Chromosomal Disorders

Disorder	Chromosome	Karyotype	Key Features
Down syndrome (trisomy 21)	Chromosome 21	47,XX,+21 or 47,XY,+21	Learning difficulties; flat facial profile; heart defects; increased risk of leukaemia; incidence increases with maternal age
Turner syndrome	Sex chromosome	45,X	Female; short stature; webbed neck; infertility (streak ovaries); no puberty without hormone treatment
Klinefelter syndrome	Sex chromosome	47,XXY	Male; tall; small testes; infertility; gynaecomastia; mild learning difficulties
Patau syndrome (trisomy 13)	Chromosome 13	47,XX,+13 or 47,XY,+13	Severe intellectual disability; cleft lip/palate; heart defects; most die within first year
Edwards syndrome (trisomy 18)	Chromosome 18	47,XX,+18 or 47,XY,+18	Severe intellectual disability; low birth weight; rocker-bottom feet; most die within first year

44. DNA Technology: Sanger Sequencing

44.1 Principle

Sanger sequencing (chain-termination method) determines the order of nucleotide bases in a DNA fragment by using modified nucleotides (dideoxynucleotides, ddNTPs) that terminate DNA synthesis at specific positions.

44.2 Steps

Step	Description
1. Denaturation	The DNA template is heated to ~95$^\circ$C to separate the double helix into single strands
2. Annealing	A primer binds to the template strand at the 3' end of the region to be sequenced
3. Extension	DNA polymerase extends the primer; the reaction mix contains normal dNTPs plus four types of fluorescently labelled ddNTPs (ddATP, ddCTP, ddGTP, ddTTP)
4. Termination	When a ddNTP is incorporated (instead of a normal dNTP), synthesis stops because the ddNTP lacks the 3'-OH group needed for the next phosphodiester bond
5. Separation	The resulting fragments of different lengths are separated by capillary electrophoresis
6. Detection	A laser detects the fluorescent label on the terminal ddNTP of each fragment; the sequence is read from shortest to longest fragment

44.3 Interpreting Results

Output	Description
Electropherogram (chromatogram)	Shows a series of coloured peaks, one per base; the sequence is read from left to right (5' to 3')
Quality score	Each base is assigned a quality score (Phred score) indicating the confidence of the base call; scores above 20 are considered reliable

45. Genetic Engineering: Steps in Detail

45.1 Overview of Recombinant DNA Technology

Step	Description
1. Identify the desired gene	Locate the gene of interest using DNA probes or genome databases
2. Isolate the gene	Cut the gene out of the donor DNA using restriction endonucleases; alternatively, use reverse transcriptase to make cDNA from mRNA
3. Insert into a vector	The gene is ligated into a plasmid vector (using DNA ligase) that contains a promoter, selectable marker (e.g., antibiotic resistance gene), and origin of replication
4. Transform host cells	The recombinant plasmid is introduced into host cells (e.g., E. coli) by transformation (heat shock or electroporation)
5. Select transformed cells	Cells are grown on agar containing the antibiotic; only cells that have taken up the plasmid (with the resistance gene) survive
6. Identify positive clones	Use gene probes or PCR to confirm that the host cells contain the desired gene (not just the plasmid)
7. Grow and harvest	Positive clones are cultured in large fermenters; the protein product is extracted and purified

45.2 Restriction Endonucleases

Feature	Description
What they do	Cut DNA at specific recognition sequences (usually 4--8 base pairs long); some produce sticky ends, others produce blunt ends
Sticky ends	Single-stranded overhangs; complementary sticky ends on different DNA fragments can base-pair (anneal); DNA ligase seals the nicks to form recombinant DNA
Blunt ends	Cut straight across both strands; no overhang; can be ligated to any other blunt-ended fragment but less efficient than sticky-end ligation
Example	EcoRI recognises GAATTC and cuts between G and A, producing sticky ends with 5'-AATT overhangs

45.3 DNA Ligase

Feature	Description
Function	Catalyses the formation of phosphodiester bonds between adjacent nucleotides; seals nicks in the sugar-phosphate backbone
Use in genetic engineering	Joins the sugar-phosphate backbone of the inserted gene to the plasmid vector after the sticky ends have base-paired
ATP requirement	Uses ATP (or NAD$^+$ in some bacteria) as an energy source to form each phosphodiester bond

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