Genetics

Mendelian Genetics

Terminology

Before proceeding, the following terms must be understood precisely. Ambiguity in terminology is the single most common source of lost marks in DSE genetics questions.

Term	Definition
Gene	A specific sequence of DNA that codes for a specific polypeptide or functional RNA
Allele	One of two or more alternative forms of a gene at the same locus on homologous chromosomes
Locus	The fixed position of a gene on a chromosome
Genotype	The genetic constitution of an organism with respect to a particular trait (e.g., Bb)
Phenotype	The observable physical or biochemical characteristics of an organism (e.g., brown eyes)
Homozygous	Having two identical alleles at a locus (e.g., BB or bb)
Heterozygous	Having two different alleles at a locus (e.g., Bb)
Dominant allele	An allele that is expressed in the phenotype even in the presence of a different allele
Recessive allele	An allele that is only expressed in the phenotype when two copies are present (homozygous)
Carrier	A heterozygous individual who carries a recessive allele but does not express the phenotype
Test cross	Crossing an individual of unknown genotype with a homozygous recessive individual

Mendel's Laws

Mendel's work with pea plants (Pisum sativum) established three fundamental laws:

1. Law of Segregation (First Law):

Each individual possesses two alleles for any given trait. During gamete formation, the two alleles segregate so that each gamete carries only one allele. This is a direct consequence of the separation of homologous chromosomes during meiosis I (anaphase I).

2. Law of Independent Assortment (Second Law):

Alleles of different genes on different chromosomes assort independently of one another during gamete formation. This occurs because the orientation of one homologous pair on the metaphase plate during meiosis I is independent of the orientation of other pairs. This law only holds for genes on different chromosomes or genes that are far apart on the same chromosome (linked genes that undergo frequent crossing over approximate independent assortment).

3. Law of Dominance:

In a heterozygote, one allele may mask the expression of another. The allele that is expressed is dominant; the allele that is masked is recessive.

info

Mendel's laws are idealisations that assume: (a) genes are on different chromosomes, (b) complete dominance, (c) no gene interaction, and (d) no linkage. Real organisms frequently violate one or more of these assumptions. The DSE syllabus expects you to recognise when Mendelian ratios do not hold and to explain why.

Monohybrid Crosses

A monohybrid cross involves a single pair of contrasting traits.

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

Punnett square:

	T	t
T	TT	Tt
t	Tt	tt

Genotypic ratio: 1 TT : 2 Tt : 1 tt

Phenotypic ratio: 3 tall : 1 dwarf

Worked calculation:

If two carriers of cystic fibrosis (a recessive disorder, Cc) have a child, what is the probability that the child will have cystic fibrosis?

The cross is Cc x Cc. The probability of cc = 1/4 = 25%.

Dihybrid Crosses

A dihybrid cross involves two pairs of contrasting traits simultaneously.

Example: In pea plants, round seeds (R) are dominant over wrinkled (r), and yellow seeds (Y) are dominant over green (y). Cross two double heterozygotes (RrYy x RrYy).

The gametes from each parent are: RY, Ry, rY, ry (four types, produced in equal frequency assuming independent assortment).

Punnett square (4 x 4):

	RY	Ry	rY	ry
RY	RRYY	RRYy	RrYY	RrYy
Ry	RRYy	RRyy	RrYy	Rryy
rY	RrYY	RrYy	rrYY	rrYy
ry	RrYy	Rryy	rrYy	rryy

Genotypic ratio: 1 RRYY : 2 RRYy : 1 RRyy : 2 RrYY : 4 RrYy : 2 Rryy : 1 rrYY : 2 rrYy : 1 rryy

Phenotypic ratio: 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green (9:3:3:1)

Forked-line (branching) method for dihybrid ratios:

Rather than constructing a 4 x 4 Punnett square, the two monohybrid ratios can be multiplied:

$\mathrm{P(round)} = \frac{3}{4}, \quad \mathrm{P(wrinkled)} = \frac{1}{4}$

$\mathrm{P(yellow)} = \frac{3}{4}, \quad \mathrm{P(green)} = \frac{1}{4}$

$\mathrm{P(round, yellow)} = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}$

$\mathrm{P(round, green)} = \frac{3}{4} \times \frac{1}{4} = \frac{3}{16}$

$\mathrm{P(wrinkled, yellow)} = \frac{1}{4} \times \frac{3}{4} = \frac{3}{16}$

$\mathrm{P(wrinkled, green)} = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}$

This yields the familiar 9:3:3:1 ratio and is significantly faster on an exam.

The Test Cross

A test cross determines the genotype of an individual showing the dominant phenotype by crossing it with a homozygous recessive individual.

If the unknown genotype is homozygous dominant (BB x bb):

All offspring will be Bb (100% dominant phenotype).

If the unknown genotype is heterozygous (Bb x bb):

Offspring will be 50% Bb (dominant phenotype) and 50% bb (recessive phenotype).

The appearance of any recessive offspring proves that the parent was heterozygous. The absence of recessive offspring suggests (but does not prove with certainty, especially in small sample sizes) that the parent is homozygous.

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Non-Mendelian Inheritance

Mendel's laws assume complete dominance. In reality, dominance relationships are more nuanced.

Incomplete Dominance

In incomplete dominance, the heterozygote has an intermediate phenotype that is distinct from either homozygote. Neither allele is fully dominant.

Classic example: Snapdragon flower colour.

• RR = red
• rr = white
• Rr = pink (intermediate)

Cross: Rr x Rr yields 1 RR (red) : 2 Rr (pink) : 1 rr (white).

The phenotypic ratio (1:2:1) matches the genotypic ratio directly. This is a hallmark of incomplete dominance.

Worked Example: Incomplete Dominance vs Codominance

A red-flowered snapdragon (RR) is crossed with a white-flowered snapdragon (rr). The F1 offspring are self-pollinated.

(a) State the genotype and phenotype of the F1 generation.

(b) What phenotypic ratio is expected in the F2 generation?

(c) If a roan shorthorn cow (RW) is crossed with a white cow (WW), what phenotypic ratio is expected? Explain why this is different from the snapdragon cross.

Solution

(a) The F1 genotype is Rr. Since this is incomplete dominance, the phenotype is pink (intermediate between red and white).

(b) F2 cross: Rr x Rr. The phenotypic ratio is 1 red (RR) : 2 pink (Rr) : 1 white (rr). This 1:2:1 ratio matches the genotypic ratio because the heterozygote has a distinct intermediate phenotype.

(c) RW x WW produces: 1 roan (RW) : 1 white (WW). This is a 1:1 ratio, NOT 1:2:1. The difference is that coat colour in shorthorn cattle is codominance, not incomplete dominance. In codominance, both alleles are fully expressed simultaneously -- roan cattle have both red and white hairs. In incomplete dominance (snapdragons), the heterozygote is a blend (pink, a genuinely intermediate colour).

Codominance

In codominance, both alleles are fully expressed simultaneously in the heterozygote. Neither allele masks the other.

Classic example: Human ABO blood groups and coat colour in shorthorn cattle.

• AB blood group: Both A and B antigens are present on the red blood cell surface. The genotype is I^A I^B, and the phenotype is blood group AB. This is codominance.
• Shorthorn cattle: Red (RR) x White (WW) = Roan (RW), where both red and white hairs are present in the coat.

Key distinction between incomplete dominance and codominance:

Feature	Incomplete Dominance	Codominance
Heterozygote phenotype	Intermediate (blend) of the two homozygotes	Both phenotypes are expressed simultaneously
Molecular mechanism	One functional allele produces less product	Both alleles produce functional, distinct products
Example	Pink snapdragons (Rr from red x white)	AB blood group (I^A I^B), roan cattle (RW)
Biochemical basis	Reduced gene product dosage	Two different gene products are both detectable

Multiple Alleles

A gene may have more than two alleles in the population, although any individual can only carry two alleles (one on each homologous chromosome).

Human ABO blood groups:

Three alleles exist at the I locus:

Allele	Type	Antigen on RBC	Antibody in Plasma
I^A	Dominant to i	A	Anti-B
I^B	Dominant to i	B	Anti-A
i	Recessive	None	Anti-A and Anti-B

Six possible genotypes and four phenotypes:

Genotype	Phenotype	Antigens	Antibodies
I^A I^A	A	A	Anti-B
I^A i	A	A	Anti-B
I^B I^B	B	B	Anti-A
I^B i	B	B	Anti-A
I^A I^B	AB	A and B	None
ii	O	None	Anti-A, Anti-B

warning

A common error is writing "three alleles" when the question asks about an individual's genotype. An individual always carries exactly two alleles. The phrase "multiple alleles" refers to the gene pool of the entire population, not to a single organism.

Worked calculation:

A woman with blood group A (whose father was blood group O) has a child with a man who is blood group B (whose mother was blood group O). What are the possible blood groups of their child?

The woman is I^A i (her father was ii, so she must carry i). The man is I^B i (his mother was ii, so he must carry i).

Cross: I^A i x I^B i

Gametes from woman: I^A, i

Gametes from man: I^B, i

	I^B	i
I^A	I^A I^B	I^A i
i	I^B i	ii

Possible genotypes: I^A I^B (AB), I^A i (A), I^B i (B), ii (O)

Possible blood groups of child: A, B, AB, or O (all four are possible).

Worked Example: Epistasis Ratio Identification

In a plant, gene A controls flower colour (A = pigment produced, a = no pigment) and gene B controls petal shape (B = normal petals, b = narrow petals). A plant with genotype AaBb is self-pollinated, and the following phenotypic ratio is observed: 9 coloured, normal : 3 coloured, narrow : 4 white. Identify the type of epistasis and explain.

Solution

This is recessive epistasis (9:3:4 ratio).

The gene A locus is epistatic over the B locus. When the plant is aa (homozygous recessive for A), no pigment is produced regardless of the B/b genotype, so the flowers are white. This masks the expression of petal shape.

The 9:3:4 ratio breaks down as:

• 9 A_B_ = coloured, normal petals
• 3 A_bb = coloured, narrow petals
• 4 aa__ = white (includes 1 aaBB, 2 aaBb, 1 aabb)

The 4 white category combines all genotypes that are aa, regardless of the B/b genotype. The aa genotype is epistatic because it masks the effect of the B locus.

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Sex-Linked Inheritance

Sex Determination in Humans

Humans have 23 pairs of chromosomes: 22 pairs of autosomes and 1 pair of sex chromosomes.

• Females: XX (homogametic)
• Males: XY (heterogametic)

The sex of the offspring is determined by the sperm: if an X-bearing sperm fertilises the egg, the result is XX (female); if a Y-bearing sperm fertilises the egg, the result is XY (male).

$P(\mathrm{female}) = \frac{1}{2}, \quad P(\mathrm{male}) = \frac{1}{2}$

X-Linked Inheritance

Genes located on the X chromosome exhibit X-linked inheritance. Males have only one X chromosome, so a single recessive allele on the X chromosome will be expressed in the male phenotype (hemizygous condition). Females need two copies of the recessive allele to express the trait.

Notation: X-linked alleles are written with the chromosome letter as a superscript.

• X^H = normal allele (dominant)
• X^h = allele for haemophilia (recessive)
• X^N = normal allele (dominant)
• X^n = allele for colour blindness (recessive)

Haemophilia

Haemophilia is an X-linked recessive bleeding disorder caused by a deficiency in clotting factor VIII (haemophilia A) or factor IX (haemophilia B). Affected individuals have prolonged bleeding after injury because their blood cannot clot properly.

Cross: Carrier female (X^H X^h) x Normal male (X^H Y)

	X^H	Y
X^H	X^H X^H	X^H Y
X^h	X^H X^h	X^h Y

Offspring: 1 normal female (X^H X^H) : 1 carrier female (X^H X^h) : 1 normal male (X^H Y) : 1 affected male (X^h Y)

Key observations:

• Affected males inherit the allele from their carrier mother (never from their father, since fathers pass Y to sons)
• Affected males pass the allele to ALL their daughters (who become carriers) but to NONE of their sons
• Females are rarely affected (would require an affected father and a carrier mother)

Red-Green Colour Blindness

Red-green colour blindness is another X-linked recessive condition. Affected individuals cannot distinguish between red and green wavelengths of light. The gene codes for opsins (photopigments) in the cone cells of the retina.

The inheritance pattern is identical to haemophilia. The cross diagrams use X^N (normal) and X^n (colour blind).

Why X-Linked Recessive Disorders Are More Common in Males

Factor	Explanation
Hemizygosity	Males have only one X chromosome; a single recessive allele is sufficient to produce the phenotype
Females are protected by the second X allele	A female with one recessive allele is a carrier but is phenotypically normal (heterozygous advantage)
Fathers do not pass X-linked alleles to sons	X-linked alleles cannot pass directly from father to son (father gives Y to sons)

Probability calculations for X-linked traits:

A carrier woman (X^N X^n) marries a normal man (X^N Y).

Probability that their son is colour blind: 1/2 (the son receives X^n from the mother with probability 1/2).

Probability that their daughter is colour blind: 0 (the daughter would need to receive X^n from both parents, but the father only has X^N).

Probability that their daughter is a carrier: 1/2.

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Genetic Diagrams and Punnett Squares

Conventions

When constructing genetic diagrams for DSE examinations, follow these conventions:

1. Write the parental phenotypes at the top
2. Write the parental genotypes below the phenotypes
3. Show the gametes produced by each parent (in a circle or brackets)
4. Use a Punnett square to show all possible offspring genotypes
5. State the genotypic ratio and the phenotypic ratio
6. Answer the specific question asked

Steps for Solving Any Genetics Problem

1. Identify whether one or two traits are involved (monohybrid or dihybrid)
2. Assign symbols to alleles (use the first letter of the dominant phenotype, capital for dominant, lowercase for recessive)
3. Determine the genotypes of the parents from the information given
4. Work out the gametes each parent can produce
5. Construct a Punnett square or use the forked-line method
6. Read off the genotypic and phenotypic ratios
7. Answer the question using the ratios

info

info standard Mendelian ratios. Use a Punnett square when the genes are linked or when you need to show all individual genotypes. For X-linked crosses, always use a Punnett square since the gamete combinations differ between males and females.

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Gene Interaction (Epistasis)

Definition

Epistasis occurs when the expression of one gene is modified or masked by the expression of one or more other genes. The gene that masks is the epistatic gene; the gene that is masked is the hypostatic gene.

Epistasis differs from dominance. Dominance is an interaction between alleles at the same locus. Epistasis is an interaction between alleles at different loci.

Types of Epistasis

1. Recessive epistasis (9:3:4 ratio):

The recessive allele of the epistatic gene (when homozygous) masks the expression of the hypostatic gene.

Example: Coat colour in Labrador retrievers.

• Gene B: B (black) is dominant over b (brown)
• Gene E: E allows pigment deposition; ee prevents any pigment deposition (epistatic)

Genotype	Phenotype
BE	Black
bbE_	Brown
_ _ ee	Yellow (epistatic)

Cross: BbEe x BbEe yields a 9:3:4 phenotypic ratio (9 black : 3 brown : 4 yellow).

The 4 includes 1 BBee, 2 Bbee, 1 bbee (all are yellow because ee is epistatic and masks the B/b locus).

2. Dominant epistasis (12:3:1 ratio):

A dominant allele of the epistatic gene masks the hypostatic gene.

Example: Fruit colour in summer squash.

• Gene W: W (white) is epistatic; when present, fruit is white regardless of Y/y
• Gene Y: Y (yellow) is dominant over y (green) -- only expressed when ww

Genotype	Phenotype
W_ _ _	White
wwY_	Yellow
wwyy	Green

Cross: WwYy x WwYy yields a 12:3:1 ratio (12 white : 3 yellow : 1 green).

3. Complementary gene action (9:7 ratio):

Two genes work together to produce a trait. A dominant allele of each gene is required. If either gene is homozygous recessive, the trait is not expressed.

Example: Flower colour in sweet peas.

• Gene C and gene P: both C* and P* are required for purple flowers
• cc or pp results in white flowers

Cross: CcPp x CcPp yields 9 purple (CP) : 7 white (3 Cpp + 3 ccP + 1 ccpp).

4. Duplicate gene action (15:1 ratio):

Two dominant alleles at either gene produce the same phenotype. Only the double recessive shows the alternate phenotype.

Example: Seed shape in shepherd's purse.

• Both A* and B* produce triangular seeds; aabb produces oval seeds

Cross: AaBb x AaBb yields 15 triangular : 1 oval.

Summary of Epistatic Ratios

Type	Ratio	Description
Standard dihybrid	9:3:3:1	No epistasis; independent assortment
Recessive epistasis	9:3:4	Homozygous recessive at one locus masks the other
Dominant epistasis	12:3:1	Dominant allele at one locus masks the other
Complementary	9:7	Both dominant alleles required for phenotype expression
Duplicate	15:1	Either dominant allele sufficient; only double recessive is different

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Chromosome Mutations

Chromosome mutations involve changes to the structure or number of whole chromosomes. These are distinct from gene mutations, which affect individual genes.

Structural Changes

1. Deletion:

Loss of a segment of a chromosome. The deleted segment may contain one or more genes.

• Consequences: loss of genetic material; often lethal if the deletion is large; heterozygous deletions can cause disorders (e.g., Cri du chat syndrome -- deletion on chromosome 5)
• Mechanism: breakage of a chromosome during meiosis; the fragment without a centromere is lost

2. Duplication:

A segment of a chromosome is repeated. The extra copy may be adjacent (tandem duplication) or elsewhere.

• Consequences: gene dosage effects; can alter phenotype if the duplicated gene is dosage-sensitive
• Mechanism: unequal crossing over between misaligned homologous chromosomes during meiosis

3. Inversion:

A segment of a chromosome is reversed end-to-end. The genes are present but in the wrong order.

• Paracentric inversion: The inversion does not include the centromere.
• Pericentric inversion: The inversion includes the centromere.
• Consequences: may cause problems during meiosis if crossing over occurs within the inverted region (produces abnormal gametes)
• The inversion carrier is phenotypically normal (no genetic material is lost)

4. Translocation:

A segment of one chromosome breaks off and attaches to a non-homologous chromosome.

• Reciprocal translocation: Two non-homologous chromosomes exchange segments. No genetic material is lost overall, but the arrangement is altered.
• Robertsonian translocation: The long arms of two acrocentric chromosomes fuse at the centromere. The short arms are lost. This is the most common form of chromosomal rearrangement in humans and is involved in some cases of Down syndrome (translocation trisomy 21).
• Consequences: carriers are phenotypically normal but may produce unbalanced gametes, leading to recurrent miscarriages or offspring with chromosomal disorders

Numerical Changes

Aneuploidy: Gain or loss of individual chromosomes (e.g., trisomy 21 -- Down syndrome, having three copies of chromosome 21).

Polyploidy: Having extra complete sets of chromosomes (e.g., 3n, 4n). Common in plants, rare and usually lethal in humans.

Disorder	Type	Chromosomal Change	Key Features
Down syndrome	Trisomy 21	Extra copy of chromosome 21	Intellectual disability, flat facial profile, heart defects
Turner syndrome	Monosomy X	XO (only one X chromosome)	Female, short stature, sterile, no puberty without treatment
Klinefelter syndrome	XXY	Extra X chromosome in male	Male, tall, sterile, reduced testosterone, gynaecomastia
Cri du chat syndrome	Deletion	Deletion on short arm of chromosome 5	High-pitched cry, intellectual disability, microcephaly

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Gene Mutations

Types of Gene Mutations

1. Point mutations (substitution):

A single nucleotide 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 change to the protein.
• Missense mutation: The new codon codes for a different amino acid. The effect depends on the chemical properties of the new amino acid (conservative vs non-conservative substitution).
• Nonsense mutation: The new codon is a stop codon (UAA, UAG, UGA). Translation terminates prematurely, producing a truncated (and usually non-functional) protein.

2. Frameshift mutations:

Insertion or deletion of one or more nucleotides that is NOT a multiple of three. This shifts the reading frame of all codons downstream of the mutation, typically producing a completely non-functional protein.

• Insertion: One or more nucleotides are added.
• Deletion: One or more nucleotides are removed.

3. Large-scale mutations:

• Gene duplication: entire gene is duplicated
• Expansion of repeating triplets: e.g., CGG repeat expansion in fragile X syndrome

Effects of Mutations

Effect	Description
Neutral	No effect on the organism's fitness (silent mutations, mutations in non-coding regions)
Harmful (deleterious)	Reduce the organism's fitness (most mutations fall into this category)
Beneficial	Increase the organism's fitness and may be selected for (rare, but provides raw material for evolution)

Mutation Rate

Spontaneous mutation rates in humans are approximately $1 \times 10^{-8}$ to $1 \times 10^{-6}$ per gene per generation. Mutations can be caused by:

• Spontaneous: Errors during DNA replication; spontaneous base changes (e.g., deamination of cytosine to uracil)
• Induced: Exposure to mutagens -- UV radiation, ionising radiation (X-rays, gamma rays), chemical mutagens (e.g., mustard gas, nitrous acid, benzopyrene in tobacco smoke)

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The Human Genome Project

Overview

The Human Genome Project (HGP) was an international scientific research project (1990-2003) that mapped and sequenced the entire human genome. The project determined the sequence of approximately 3 billion base pairs of DNA and identified approximately 20,000-25,000 protein-coding genes.

Key Findings

• The human genome contains approximately 3.2 billion base pairs
• Only approximately 1.5% of the genome codes for proteins
• Approximately 50% of the genome consists of repetitive sequences (transposable elements, satellite DNA)
• Humans share approximately 99.9% of their DNA sequence with each other
• The genetic variation between any two humans is approximately 0.1% (about 3 million base pairs)

Applications

Application	Description
Medicine	Identification of disease-causing genes; development of targeted therapies; pharmacogenomics
Forensic science	DNA fingerprinting; identification of victims and suspects
Genetic screening	Prenatal diagnosis; carrier testing; pre-symptomatic testing for inherited disorders
Gene therapy	Correcting defective genes to treat genetic diseases
Evolutionary biology	Comparing human genome with other species to understand evolutionary relationships
Agriculture	Genomic selection in crop and livestock breeding

Pharmacogenomics

The study of how an individual's genetic makeup affects their response to drugs. This allows for personalised medicine -- tailoring drug choice and dosage to an individual's genotype to maximise efficacy and minimise adverse effects.

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Genetic Engineering Techniques

Recombinant DNA Technology

Recombinant DNA is DNA formed by combining DNA sequences from two different organisms. The basic process:

1. Isolation of the gene of interest:

The target gene is identified and extracted from the donor organism's DNA. For eukaryotic genes, mRNA is often used as the template, and reverse transcriptase produces complementary DNA (cDNA). This avoids the intron problem (bacteria cannot splice out introns).

2. Cutting with restriction enzymes:

Restriction endonucleases (restriction enzymes) are enzymes that cut DNA at specific recognition sequences (palindromic sequences, typically 4-8 base pairs long).

• EcoRI recognises GAATTC and cuts between G and A
• BamHI recognises GGATCC and cuts between G and G
• Produces "sticky ends" (single-stranded overhangs) or "blunt ends"

3. Ligation with DNA ligase:

DNA ligase catalyses the formation of phosphodiester bonds, sealing the gene into the vector (carrier) DNA. The vector is typically a plasmid that has been cut with the same restriction enzyme, producing complementary sticky ends.

4. Transformation:

The recombinant DNA is introduced into host cells (typically E. coli bacteria). Methods include:

• Heat shock (calcium chloride treatment)
• Electroporation (electric pulse)
• Microinjection

5. Selection and expression:

Transformed cells are selected using marker genes (e.g., antibiotic resistance). Successfully transformed cells are cultured and the desired protein product is harvested.

Polymerase Chain Reaction (PCR)

PCR amplifies a specific segment of DNA exponentially.

Components:

Component	Function
DNA template	The DNA to be copied
Taq polymerase	Heat-stable DNA polymerase from Thermus aquaticus; synthesises new DNA strands
Primers	Short single-stranded DNA sequences complementary to regions flanking the target
Free nucleotides	dATP, dTTP, dCTP, dGTP -- building blocks
Thermal cycler	Machine that cycles through the required temperatures

PCR cycle (repeated 25-35 times):

1. Denaturation (94-96 degrees C): Hydrogen bonds break; double-stranded DNA separates into two single strands.
2. Annealing (50-65 degrees C): Primers bind to complementary sequences on the single-stranded DNA.
3. Extension (72 degrees C): Taq polymerase adds nucleotides to the 3-prime end of each primer, synthesising new DNA strands.

$\mathrm{After } n \mathrm{ cycles: } N = N_0 \times 2^n$

Gel Electrophoresis

Gel electrophoresis separates DNA fragments by size.

Process:

1. DNA samples are loaded into wells in an agarose gel
2. An electric current is applied across the gel
3. DNA fragments are negatively charged (phosphate backbone) and move towards the positive electrode
4. Smaller fragments move faster and travel further; larger fragments move more slowly
5. The DNA is visualised using a fluorescent dye (e.g., ethidium bromide) that binds to DNA and fluoresces under UV light

Applications: DNA profiling, analysis of PCR products, checking the results of restriction digests.

DNA Fingerprinting (DNA Profiling)

DNA fingerprinting identifies individuals based on their unique DNA pattern.

Principle: Non-coding regions of DNA contain Short Tandem Repeats (STRs) -- short sequences (2-6 base pairs) repeated a variable number of times. The number of repeats at each locus varies between individuals.

Process:

1. Extract DNA from a sample (blood, saliva, hair root)
2. Amplify specific STR loci using PCR with primers designed for each locus
3. Separate the amplified fragments by gel electrophoresis (or capillary electrophoresis)
4. Compare the pattern of bands (the "fingerprint") between samples

Applications:

• Forensic science (matching crime scene DNA to suspects)
• Paternity and maternity testing
• Identifying victims of disasters
• Immigration cases (proving family relationships)
• Establishing evolutionary relationships between species

Reliability: The probability of two unrelated individuals having the same DNA profile at 13 or more STR loci is less than 1 in a billion. This makes DNA fingerprinting extremely reliable for identification.

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Genetically Modified Organisms (GMOs)

Definition

A genetically modified organism (GMO) is an organism whose genome has been altered using genetic engineering techniques. This differs from conventional selective breeding because specific genes can be inserted, deleted, or modified precisely.

Examples of GMOs

GMO	Modification	Purpose
Bt cotton	Contains Bacillus thuringiensis toxin gene	Produces insecticidal protein; reduces pesticide use
Golden rice	Contains beta-carotene synthesis genes	Produces vitamin A precursor; combats vitamin A deficiency
Herbicide-resistant soybeans	Resistant to glyphosate herbicide	Farmers can spray herbicide without harming the crop
GM salmon (AquAdvantage)	Contains growth hormone gene from Chinook salmon	Grows faster than conventional salmon
Insulin-producing bacteria	Contains human insulin gene	Produces human insulin for medical use

Benefits of GMOs

1. Increased crop yield: Pest-resistant and disease-resistant varieties reduce losses
2. Reduced pesticide use: Bt crops produce their own insecticide
3. Enhanced nutritional content: Golden rice addresses vitamin A deficiency in developing countries
4. Tolerance to abiotic stress: Drought-resistant, salt-tolerant varieties
5. Pharmaceutical production: Bacteria and plants engineered to produce vaccines, antibodies, and other drugs
6. Environmental benefits: Reduced fuel use from fewer pesticide applications; conservation tillage with herbicide-resistant crops reduces soil erosion

Risks and Concerns

1. Gene flow: Transferred genes may escape to wild relatives through cross-pollination, potentially creating "superweeds" (herbicide-resistant weeds)
2. Impact on non-target organisms: Bt toxin may affect beneficial insects (though evidence suggests it is specific to certain pest groups)
3. Development of resistance: Pests may evolve resistance to Bt toxin, similar to antibiotic resistance
4. Biodiversity reduction: Large-scale monocultures of GM varieties reduce genetic diversity
5. Human health concerns: Potential allergenicity of novel proteins; long-term effects not fully known (though extensive testing is required before approval)
6. Corporate control: Patents on GM seeds create dependency on biotechnology companies; farmers cannot legally save seeds
7. Ethical concerns: Some consider genetic modification of organisms to be "unnatural"

Regulation

Different countries have different approaches to GMO regulation:

• European Union: Highly restrictive; mandatory labelling; precautionary principle
• United States: Generally permissive; voluntary labelling (changing); substantial equivalence standard
• Hong Kong: No specific GMO labelling requirements (but pre-packaged food must be accurately described)

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Gene Therapy

Definition

Gene therapy is the treatment of genetic disorders by introducing, removing, or altering genetic material within a patient's cells.

Types of Gene Therapy

1. Somatic gene therapy:

• Targets body cells (somatic cells)
• Changes are NOT passed to offspring
• The treated individual benefits, but the genetic change is not inherited
• Currently the main approach in clinical trials
• Example: Treating severe combined immunodeficiency (SCID) by introducing a functional copy of the defective gene into bone marrow cells

2. Germline gene therapy:

• Targets gametes (sperm or egg) or early embryos
• Changes ARE passed to all future generations
• Currently not permitted in most countries due to ethical concerns
• Raises issues of consent (future generations cannot consent), equity, and the possibility of "designer babies"

Methods of Delivery

Method	Description
Viral vectors	Modified viruses (retroviruses, adenoviruses) carry the therapeutic gene into target cells
Liposomes	Fatty spheres that fuse with the cell membrane, delivering DNA into the cell
Naked DNA injection	Direct injection of plasmid DNA into the target tissue
Electroporation	Electric pulse creates temporary pores in the cell membrane for DNA entry

Challenges and Limitations

1. Delivery: Getting the therapeutic gene to the correct cells in sufficient quantity
2. Immune response: The body may attack the viral vector or the newly expressed protein
3. Duration: The therapeutic effect may be temporary if the modified cells do not persist or divide
4. Off-target effects: The gene may insert into an unintended location, disrupting normal gene function (insertional mutagenesis)
5. Cost: Gene therapy treatments are extremely expensive (e.g., Zolgensma for spinal muscular atrophy costs approximately USD 2.1 million per treatment)
6. Ethical concerns: Germline editing, enhancement vs treatment, access and equity

Successful Examples

• Luxturna (2017): Gene therapy for a form of inherited retinal dystrophy that causes blindness. A working copy of the RPE65 gene is delivered to retinal cells using an adeno-associated virus vector.
• Zolgensma (2019): Gene therapy for spinal muscular atrophy (SMA). A functional copy of the SMN1 gene is delivered via an adeno-associated virus.

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Common Pitfalls

1. Confusing incomplete dominance with codominance: Incomplete dominance produces an intermediate phenotype (pink from red and white). Codominance produces a phenotype showing both parental traits simultaneously (AB blood group, roan coat). These are fundamentally different mechanisms.

2. Forgetting that males are hemizygous for X-linked genes: A male has only one X chromosome, so he cannot be "heterozygous" or "homozygous" for an X-linked gene. He is either affected or normal. This is why X-linked recessive disorders are far more common in males.

3. Writing that a father passes an X-linked allele to his son: Fathers pass their X chromosome to daughters and their Y chromosome to sons. An X-linked allele cannot pass directly from father to son.

4. Confusing epistatic ratios with Mendelian ratios: If a dihybrid cross does not produce a 9:3:3:1 ratio, consider epistasis. The modified ratios (9:3:4, 12:3:1, 9:7, 15:1) each have a specific biological explanation. Learn them.

5. Stating that gene mutations are always harmful: Most are neutral or harmful, but a small fraction are beneficial and provide the raw material for natural selection. "Mutation is always bad" is a false statement.

6. Confusing chromosome mutations with gene mutations: Chromosome mutations affect whole chromosomes or large segments (deletion, duplication, inversion, translocation). Gene mutations affect individual nucleotides or small regions (point mutations, frameshifts).

7. Writing that DNA ligase cuts DNA: DNA ligase JOINS DNA fragments (seals the sugar-phosphate backbone). Restriction enzymes CUT DNA. Students frequently confuse these two enzymes.

8. Stating that PCR amplifies the entire genome: PCR amplifies a specific segment of DNA defined by the primers. It does not copy the entire genome.

9. Confusing DNA fingerprinting with whole-genome sequencing: DNA fingerprinting analyses specific STR loci in non-coding regions. It does not sequence the entire genome.

10. Forgetting that genetic engineering uses cDNA from eukaryotes: Bacterial cells cannot splice out introns from eukaryotic genes. Therefore, the mRNA is extracted and reverse transcriptase produces cDNA (which has no introns). Direct insertion of genomic DNA containing introns would not produce a functional protein in bacteria.

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Problem Set

Problem 1: Cystic fibrosis is an autosomal recessive disorder. A couple has a child with cystic fibrosis. Neither parent has the disorder.

(a) What are the genotypes of the parents?

(b) What is the probability that their next child will have cystic fibrosis?

(c) What is the probability that their next child will be a carrier?

If you get this wrong, revise: Mendelian Genetics -- Monohybrid Crosses; Terminology

Solution

(a) Both parents are carriers: Cc. They must each carry one recessive allele (c) because their affected child is cc, and each parent contributed one allele.

(b) Cross: Cc x Cc. Punnett square gives: 1 CC : 2 Cc : 1 cc. P(cc) = 1/4 = 25%.

(c) P(Cc) = 2/4 = 1/2 = 50%.

Problem 2: A woman with normal vision whose father was colour blind marries a man with normal vision. Colour blindness is X-linked recessive.

(a) What is the woman's genotype?

(b) What proportion of their sons will be colour blind?

(c) What proportion of their daughters will be carriers?

If you get this wrong, revise: Sex-Linked Inheritance -- X-Linked Inheritance; Red-Green Colour Blindness

Solution

(a) The woman's father was colour blind (X$^n$Y), so he gave her X$^n$. Her mother could be X$^N$X$^N$ or X$^N$X$^n$. The woman is X$^N$X$^n$ (carrier).

(b) The man is X$^N$Y. Cross: X$^N$X$^n$ x X$^N$Y. Sons receive X from their mother: P(X$^n$Y) = 1/2. So 50% of sons will be colour blind.

(c) Daughters receive X$^N$ from their father. P(X$^N$X$^n$) = 1/2. So 50% of daughters will be carriers.

Problem 3: In pea plants, round seeds (R) are dominant over wrinkled (r), and yellow seeds (Y) are dominant over green (y). Two double heterozygotes (RrYy) are crossed. Use the forked-line method to determine the probability of producing a wrinkled, green offspring.

If you get this wrong, revise: Mendelian Genetics -- Dihybrid Crosses; The Forked-Line Method

Solution

Using the forked-line method:

P(wrinkled) = 1/4 (rr from Rr x Rr)

P(green) = 1/4 (yy from Yy x Yy)

P(wrinkled, green) = 1/4 x 1/4 = 1/16

This is consistent with the 9:3:3:1 dihybrid ratio, where wrinkled green corresponds to the 1/16 category.

Problem 4: In Labrador retrievers, gene B controls coat colour (B = black, b = brown) and gene E controls pigment deposition (E = pigment deposited, e = no pigment). A BbEe x BbEe cross produces a 9:3:4 phenotypic ratio. Explain this ratio and identify the epistatic gene.

If you get this wrong, revise: Gene Interaction (Epistasis) -- Recessive Epistasis

Solution

The 9:3:4 ratio indicates recessive epistasis. Gene E is epistatic over gene B.

• 9 B_E_ = black (pigment deposited, black colour)
• 3 bbE_ = brown (pigment deposited, brown colour)
• 4 __ee = yellow (no pigment deposited regardless of B/b genotype)

The 4 yellow includes: 1 BBee, 2 Bbee, 1 bbee. When the genotype is ee (homozygous recessive for E), no pigment is deposited, so the coat is yellow regardless of whether the B/b locus would normally produce black or brown. The E locus is epistatic because ee masks the expression of the B locus.

Problem 5: Describe the steps involved in producing human growth hormone (HGH) in bacteria. Explain why cDNA is used rather than genomic DNA.

If you get this wrong, revise: Genetic Engineering Techniques -- Recombinant DNA Technology

Solution

1. mRNA for HGH is extracted from human pituitary gland cells.
2. Reverse transcriptase produces cDNA from the mRNA (removes introns).
3. The cDNA and a plasmid are both cut with the same restriction enzyme, producing complementary sticky ends.
4. DNA ligase seals the cDNA into the plasmid, creating a recombinant plasmid.
5. The recombinant plasmid is introduced into E. coli by transformation.
6. Transformed bacteria are selected using a marker gene (antibiotic resistance).
7. Bacteria are cultured; they transcribe and translate the HGH gene.

cDNA is used because genomic DNA contains introns. Bacteria lack spliceosomes and cannot remove introns from pre-mRNA. If genomic DNA were inserted, the ribosome would translate intron sequences, producing a non-functional protein.

Problem 6: A forensic scientist has a tiny blood stain. After 30 cycles of PCR starting from 5 DNA molecules, how many DNA molecules will be present? Explain why PCR is necessary before DNA fingerprinting.

If you get this wrong, revise: Genetic Engineering Techniques -- Polymerase Chain Reaction (PCR); DNA Fingerprinting

Solution

After 30 cycles: $N = 5 \times 2^{30} = 5 \times 1\,073\,741\,824 = 5\,368\,709\,120$ DNA molecules.

PCR is necessary because the amount of DNA in a tiny blood stain is far too small for direct analysis. PCR amplifies specific STR loci to produce millions of copies, providing sufficient DNA for gel electrophoresis and comparison. Without PCR, the DNA profile could not be obtained from trace evidence.

Problem 7: Explain the difference between somatic and germline gene therapy. Why is germline gene therapy currently not permitted in most countries?

If you get this wrong, revise: Gene Therapy -- Types of Gene Therapy

Solution

Somatic gene therapy targets body cells (somatic cells). Changes are NOT passed to offspring. The treated individual benefits but the genetic change is not inherited. Example: treating SCID by introducing a functional gene into bone marrow cells.

Germline gene therapy targets gametes or early embryos. Changes ARE passed to all future generations. It raises ethical concerns about consent (future generations cannot consent), equity (access to enhancement), and the possibility of "designer babies." Currently not permitted in most countries due to these unresolved ethical issues and the risk of off-target effects being inherited.

Problem 8: A man with blood group AB has a child with a woman of blood group O. What are the possible blood groups of their child? Show your working with a Punnett square.

If you get this wrong, revise: Non-Mendelian Inheritance -- Multiple Alleles (ABO Blood Groups)

Solution

The man is I$^A$I$^B$ (genotype AB). The woman is ii (genotype O).

Gametes from man: I$^A$ or I$^B$. Gametes from woman: i only.

	I$^A$	I$^B$
i	I$^A$i	I$^B$i

Possible genotypes: I$^A$i (blood group A) or I$^B$i (blood group B).

Possible blood groups of the child: A or B. Each has a 50% probability. Blood group AB and O are not possible.

Problem 9: A person has 46 chromosomes in their somatic cells. How many chromosomes and chromatids are present in each cell during (a) prophase of mitosis, (b) metaphase I of meiosis, (c) after meiosis II is complete?

If you get this wrong, revise: Chromosome Mutations -- Numerical Changes; Cell Division (Meiosis)

Solution

(a) Prophase of mitosis: 46 chromosomes, 92 chromatids. Each chromosome has been replicated into two sister chromatids joined at the centromere (replication occurs during S phase, before mitosis begins).

(b) Metaphase I of meiosis: 46 chromosomes (23 homologous pairs), 92 chromatids. Homologous pairs are aligned at the metaphase plate. Each chromosome still consists of two chromatids.

(c) After meiosis II: Each of the four daughter cells has 23 chromosomes, 23 chromatids (single-chromatid chromosomes). Sister chromatids separated during anaphase II, halving the chromosome number.

Problem 10: Explain the difference between a point mutation and a frameshift mutation. Give an example of how each type of mutation can affect a protein.

If you get this wrong, revise: Gene Mutations -- Types of Gene Mutations

Solution

A point mutation (substitution) replaces a single nucleotide with another. A missense point mutation changes one codon, potentially replacing one amino acid with another (e.g., sickle cell mutation: GAG to GTG, glutamic acid to valine in haemoglobin).

A frameshift mutation is an insertion or deletion of nucleotides that is NOT a multiple of three. This shifts the reading frame of all codons downstream, typically changing every subsequent amino acid and often creating a premature stop codon. The resulting protein is usually completely non-functional (e.g., a 1-bp insertion in the CFTR gene causes cystic fibrosis).

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Gene Regulation

Why Gene Regulation Matters

Every cell in an organism contains the same DNA (same genes). However, not all genes are expressed in every cell at all times. Gene regulation allows cells to:

• Differentiate into specialised cell types (e.g., muscle cells express muscle proteins; nerve cells express neurotransmitter-related genes)
• Respond to changes in the environment (e.g., producing enzymes only when a substrate is available)
• Maintain homeostasis (e.g., regulating insulin production based on blood glucose levels)
• Progress through the cell cycle in a controlled manner

Gene Regulation in Prokaryotes: The Lac Operon

The lac operon in E. coli is the classic model of gene regulation. It controls the production of enzymes needed to metabolise lactose.

Components:

Component	Description
Structural genes	lacZ (beta-galactosidase, breaks down lactose into glucose and galactose), lacY (permease, transports lactose into the cell), lacA (transacetylase, function less well understood)
Promoter	DNA sequence where RNA polymerase binds to transcribe the structural genes
Operator	DNA sequence between the promoter and structural genes; acts as an on/off switch controlled by a repressor protein
Regulator gene	lacI gene, located upstream; produces the lac repressor protein

Mechanism of the lac operon:

When lactose is ABSENT (repression):

1. The regulator gene (lacI) is continuously transcribed and translated, producing the lac repressor protein
2. The repressor binds to the operator, physically blocking RNA polymerase from transcribing the structural genes
3. The structural genes (lacZ, lacY, lacA) are NOT transcribed
4. No beta-galactosidase or permease is produced (wasting energy on unnecessary enzymes is avoided)

When lactose is PRESENT (induction):

1. Lactose enters the cell (in small amounts via constitutively produced permease)
2. Lactose is converted to allolactose (an isomer of lactose) by beta-galactosidase
3. Allolactose acts as an inducer: it binds to the repressor protein, changing its shape
4. The repressor can no longer bind to the operator
5. RNA polymerase binds to the promoter and transcribes the structural genes
6. Beta-galactosidase and permease are produced, enabling efficient lactose metabolism

Positive control by glucose (catabolite repression):

When both glucose AND lactose are present, E. coli preferentially uses glucose (more efficient energy source). The lac operon is not fully activated because:

1. Low glucose leads to high cAMP (cyclic AMP) levels
2. cAMP binds to the CAP protein (catabolite activator protein)
3. The cAMP-CAP complex binds to a site upstream of the promoter, enhancing RNA polymerase binding
4. When glucose is HIGH, cAMP is LOW; the cAMP-CAP complex does not form; the lac operon is transcribed at a low rate even if lactose is present

Gene Regulation in Eukaryotes

Eukaryotic gene regulation is more complex than prokaryotic regulation, involving multiple levels of control:

Level of Regulation	Description
Transcriptional	Control of whether a gene is transcribed into mRNA (transcription factors, enhancers, silencers, chromatin structure)
Post-transcriptional	mRNA processing (splicing, capping, poly-A tail); alternative splicing produces different proteins from the same gene
Translational	Control of whether mRNA is translated into protein (microRNAs, RNA interference, initiation factors)
Post-translational	Modification of proteins after synthesis (phosphorylation, glycosylation, proteolytic cleavage, ubiquitin-mediated degradation)

Transcription factors:

• Proteins that bind to specific DNA sequences (promoters, enhancers) and regulate transcription
• Activators: Enhance transcription by helping RNA polymerase bind to the promoter
• Repressors: Inhibit transcription by blocking RNA polymerase binding or recruiting chromatin-modifying enzymes
• Hormones can act as transcription factors: steroid hormones (e.g., oestrogen, testosterone) enter the cell, bind to intracellular receptors, and the hormone-receptor complex acts as a transcription factor, binding to specific DNA sequences and regulating gene expression

Epigenetics

Epigenetics refers to heritable changes in gene expression that do NOT involve changes to the DNA sequence itself.

Epigenetic Mechanism	Description
DNA methylation	Addition of methyl groups (-CH$_3$) to cytosine bases, typically at CpG islands near gene promoters; usually associated with gene silencing
Histone modification	Acetylation, methylation, or phosphorylation of histone proteins; affects how tightly DNA is wound around histones (chromatin structure)
Non-coding RNA	MicroRNAs (miRNAs) and small interfering RNAs (siRNAs) that bind to mRNA and inhibit translation or promote degradation

info

Epigenetic changes are REVERSIBLE (unlike mutations) and can be influenced by environmental factors (diet, stress, toxins). This has important implications for understanding how environmental exposures affect gene expression and disease risk across generations. However, epigenetic changes should NOT be confused with Lamarckian inheritance of acquired characteristics -- the DNA sequence itself is unchanged.

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Chromosome Mutations

Changes in Chromosome Number

Type	Description	Example
Polyploidy	Having extra complete sets of chromosomes (3n, 4n, etc.)	Common in plants (wheat is hexaploid, 6n)
Aneuploidy	Having one extra or one missing chromosome	Down syndrome (trisomy 21, 47 chromosomes)
Trisomy	Having three copies of a chromosome instead of two	Trisomy 21 (Down syndrome), Trisomy 18 (Edwards syndrome)
Monosomy	Having only one copy of a chromosome instead of two	Monosomy X (Turner syndrome, 45 chromosomes)

Causes of aneuploidy -- non-disjunction:

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

If a gamete with an extra chromosome (n+1) is fertilised by a normal gamete (n), the resulting zygote has 2n+1 chromosomes (trisomy). If a gamete missing a chromosome (n-1) is fertilised, the zygote has 2n-1 chromosomes (monosomy).

Changes in Chromosome Structure

Type	Description	Consequence
Deletion	A segment of a chromosome is lost	Genes on the deleted segment are lost; often lethal
Duplication	A segment is repeated	Extra copies of genes; may affect gene dosage
Inversion	A segment is reversed (flipped end-to-end)	Genes are present but in wrong order; may affect meiosis
Translocation	A segment breaks off and attaches to a non-homologous chromosome	Can create fusion genes; e.g., Philadelphia chromosome in CML

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Additional Problem Set

Problem 11: Explain how the lac operon in E. coli is regulated when (a) only glucose is present in the medium, (b) only lactose is present, and (c) both glucose and lactose are present.

If you get this wrong, revise: Gene Regulation -- Gene Regulation in Prokaryotes: The Lac Operon

Solution

(a) Only glucose present: The lac repressor (produced by lacI) binds to the operator, blocking transcription of lacZ, lacY, and lacA. Additionally, high glucose means low cAMP, so the cAMP-CAP complex does not form, providing no positive stimulation. The lac operon is OFF. E. coli uses glucose directly.

(b) Only lactose present: Lactose enters the cell (via low basal permease) and is converted to allolactose. Allolactose binds to the repressor, causing it to detach from the operator. Additionally, low glucose means high cAMP; the cAMP-CAP complex binds upstream of the promoter, enhancing RNA polymerase binding. The lac operon is fully ON. E. coli produces beta-galactosidase and permease to metabolise lactose.

(c) Both glucose and lactose present: Lactose/allolactose inactivates the repressor (operator is unblocked), but high glucose means low cAMP, so the cAMP-CAP complex does not form. The lac operon is transcribed at a LOW rate (not fully activated). E. coli preferentially uses glucose and only slowly metabolises lactose. Once glucose is depleted, cAMP levels rise, cAMP-CAP forms, and the lac operon becomes fully active.

Problem 12: A couple has a child with Down syndrome (trisomy 21). Explain the chromosomal event that most likely caused this condition. Why does the risk of Down syndrome increase with maternal age?

If you get this wrong, revise: Chromosome Mutations -- Changes in Chromosome Number

Solution

Down syndrome is most commonly caused by non-disjunction during meiosis. Non-disjunction is the failure of homologous chromosome 21 to separate during anaphase I of meiosis (or the failure of sister chromatids of chromosome 21 to separate during anaphase II). This produces an egg (or sperm) with two copies of chromosome 21 instead of one. When this gamete fuses with a normal gamete (carrying one copy of chromosome 21), the resulting zygote has three copies of chromosome 21 (trisomy 21), giving a total of 47 chromosomes.

The risk increases with maternal age because:

• Oocytes begin meiosis during foetal development and arrest at prophase I for decades until ovulation
• The longer an oocyte remains arrested, the greater the chance that cohesin proteins (which hold homologous chromosomes together) deteriorate
• Deteriorated cohesin may fail to hold homologous chromosomes together properly, increasing the risk of non-disjunction during the first meiotic division
• Additionally, accumulated damage to the meiotic machinery over time may contribute

The risk is approximately 1 in 1,500 at age 20, 1 in 350 at age 35, and approximately 1 in 100 at age 40.

Problem 13: Compare the structure and function of mRNA, tRNA, and rRNA. Explain why all three types of RNA are necessary for protein synthesis.

If you get this wrong, revise: Protein Synthesis; Gene Regulation

Solution

Type	Structure	Function
mRNA	Single-stranded; carries a copy of the genetic code from DNA; has a 5' cap and 3' poly-A tail	Carries the genetic information from the nucleus to the ribosome; determines the amino acid sequence of the protein
tRNA	Cloverleaf shape in 2D; has an anticodon loop and an amino acid attachment site (3' end)	Carries specific amino acids to the ribosome; anticodon base-pairs with the codon on the mRNA, ensuring the correct amino acid is inserted
rRNA	Single-stranded with folded regions; combined with proteins to form ribosomes	Forms the structural and catalytic core of the ribosome; catalyses peptide bond formation between amino acids

All three are necessary because: mRNA provides the template (the "message"), tRNA brings the correct building blocks (amino acids) based on the codon-anticodon matching, and rRNA forms the ribosome (the "factory") where the mRNA is read and amino acids are linked together. Without any one of these, protein synthesis cannot occur.

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tip

Diagnostic Test Ready to test your understanding of Genetics? The diagnostic test contains the hardest questions within the DSE specification for this topic, each with a full worked solution.

Unit tests probe edge cases and common misconceptions. Integration tests combine Genetics with other biology topics to test synthesis under exam conditions.

See Diagnostic Guide for instructions on self-marking and building a personal test matrix.

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Genetic Diseases

Autosomal Dominant Disorders

Disorder	Gene / Chromosome	Description	Inheritance Pattern
Huntington's disease	HTT gene on chromosome 4	Progressive neurodegenerative disorder; caused by expanded CAG trinucleotide repeat (codes for glutamine); symptoms typically appear after age 35 (chorea, dementia, personality changes); lethal	Autosomal dominant; 50% chance of inheriting if one parent affected; homozygous dominant is rare (often more severe, earlier onset)
Marfan syndrome	FBN1 gene on chromosome 15	Defect in fibrillin-1 protein (connective tissue); tall stature, long limbs, dislocated eye lenses, aortic aneurysm	Autosomal dominant
Achondroplasia	FGFR3 gene on chromosome 4	Most common form of dwarfism; mutation in fibroblast growth factor receptor 3 causes impaired cartilage formation; short limbs, normal trunk	Autosomal dominant; 75% of cases are new mutations (not inherited)
Familial hypercholesterolaemia	LDLR gene on chromosome 19	Defective or absent LDL receptors; high blood cholesterol from birth; increased risk of early heart disease	Autosomal dominant; codominant (heterozygotes have ~2x normal cholesterol; homozygotes have ~4x)

Autosomal Recessive Disorders

Disorder	Gene / Chromosome	Description	Carrier Frequency
Cystic fibrosis	CFTR gene on chromosome 7	Defective chloride ion channel; thick, sticky mucus in lungs, pancreas, and digestive system; recurrent lung infections; pancreatic insufficiency; shortened life expectancy (~40-50 years with treatment)	~1 in 25 in Caucasian populations
Sickle cell anaemia	HBB gene on chromosome 11	Mutation in beta-globin gene (GAG $\rightarrow$ GTG, Glu $\rightarrow$ Val at position 6); produces abnormal haemoglobin (HbS); RBCs become sickle-shaped under low oxygen; block capillaries; pain crises, organ damage, anaemia	~1 in 10 in African populations
Phenylketonuria (PKU)	PAH gene on chromosome 12	Deficiency of phenylalanine hydroxylase; cannot convert phenylalanine to tyrosine; phenylalanine accumulates, causing brain damage and intellectual disability if untreated; treated with low-phenylalanine diet from birth	~1 in 50 in Caucasian populations
Tay-Sachs disease	HEXA gene on chromosome 15	Deficiency of hexosaminidase A; lipid (GM2 ganglioside) accumulates in nerve cells; progressive neurological deterioration; death usually by age 4-5	~1 in 27 in Ashkenazi Jewish populations
Albinism	Various genes (TYR, OCA2)	Deficiency of melanin production; lack of pigment in skin, hair, and eyes; vision problems (photophobia, nystagmus); increased risk of skin cancer	Varies by population

Sex-Linked Disorders

Disorder	Gene / Chromosome	Description	Affected Sex
Haemophilia A	F8 gene on X chromosome	Deficiency of clotting factor VIII; blood does not clot properly; prolonged bleeding after injury; spontaneous internal bleeding into joints (haemarthrosis)	Primarily males (X$^h$Y); females rarely affected (need X$^h$X$^h$)
Red-green colour blindness	Opsin genes on X chromosome	Defective photopigments in cone cells; cannot distinguish between red and green; most common in males (~8% of males vs ~0.5% of females in Caucasian populations)	Primarily males
Duchenne muscular dystrophy	DMD gene on X chromosome	Absence of dystrophin protein; progressive muscle weakness and degeneration; affected boys typically lose the ability to walk by age 12; life expectancy ~20-30 years	Males only (X-linked recessive)

Pedigree Analysis

Feature	Autosomal Dominant	Autosomal Recessive	X-Linked Recessive
Affected children from unaffected parents	No (at least one parent must be affected)	Yes (both parents can be unaffected carriers)	No (mother must be a carrier or father affected)
Both sexes affected	Yes (approximately equal)	Yes (approximately equal)	Mostly males; females rarely affected
Skips generations	No (appears in every generation)	Yes (can skip generations through carriers)	Yes (can skip through carrier females
Father-to-son transmission	Possible	Possible	No (father passes Y to son, not X)
Affected female with unaffected male: all children affected?	Yes (she passes the dominant allele to all children)	No (children are carriers at most)	No (sons receive Y from father; daughters receive one normal X)

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Karyotyping and Chromosomal Analysis

Human Karyotype

• A karyotype is a photograph of the chromosomes of a cell, arranged in pairs according to size and banding pattern
• Human somatic cells are diploid (2n = 46): 22 pairs of autosomes + 1 pair of sex chromosomes (XX or XY)
• Karyotyping is performed by:
  1. Collecting cells (e.g., from blood, amniotic fluid, or chorionic villus sampling)
  2. Stimulating the cells to divide
  3. Arresting cell division in metaphase (using colchicine) -- when chromosomes are most condensed and visible
  4. Staining the chromosomes (Giemsa stain produces characteristic banding patterns)
  5. Photographing and arranging the chromosomes in homologous pairs

Chromosomal Abnormalities

Numerical abnormalities (aneuploidy):

Condition	Karyotype	Description	Symptoms
Down syndrome (trisomy 21)	47,XX,+21 or 47,XY,+21	Extra copy of chromosome 21; most common chromosomal abnormality; incidence increases with maternal age	Intellectual disability; characteristic facial features; heart defects; increased risk of leukaemia
Edwards syndrome (trisomy 18)	47,XX,+18 or 47,XY,+18	Extra copy of chromosome 18; severe developmental abnormalities; most affected infants die within the first year	Severe intellectual disability; heart defects; clenched fists; rocker-bottom feet
Patau syndrome (trisomy 13)	47,XX,+13 or 47,XY,+13	Extra copy of chromosome 13; severe abnormalities; most die within the first year	Severe intellectual disability; cleft lip/palate; polydactyly; heart defects
Klinefelter syndrome	47,XXY	Extra X chromosome in a male	Tall stature; small testes; infertility; gynaecomastia; mild learning difficulties
Turner syndrome	45,X0	Missing one X chromosome in a female	Short stature; webbed neck; lack of ovarian development; infertility; no puberty without hormone treatment
Triple X syndrome	47,XXX	Extra X chromosome in a female	Usually normal; tall stature; may have learning difficulties
Jacob's syndrome	47,XYY	Extra Y chromosome in a male	Usually normal; tall stature; may have mild learning difficulties

Structural abnormalities:

Type	Description	Effect
Deletion	A segment of a chromosome is lost	Cri-du-chat syndrome: deletion on chromosome 5 (short arm); severe intellectual disability; cat-like cry in infants
Duplication	A segment of a chromosome is duplicated (present twice)	May cause developmental abnormalities depending on the genes involved
Inversion	A segment of a chromosome is reversed (flipped 180 degrees)	May disrupt gene function if the breakpoint occurs within a gene; may cause problems during meiosis
Translocation	A segment of one chromosome breaks off and attaches to another chromosome	Reciprocal translocation: two chromosomes exchange segments; Robertsonian translocation: two acrocentric chromosomes (13, 14, 15, 21, 22) fuse at the centromere
Robertsonian translocation	A type of translocation involving acrocentric chromosomes where the long arms fuse and the short arms are lost	Can cause Down syndrome if a Robertsonian translocation involves chromosome 21; a carrier (45 chromosomes) is phenotypically normal but at risk of having children with Down syndrome

Pre-Natal Diagnosis

Method	Timing	Description	Risks
Amniocentesis	15-20 weeks	A needle is inserted through the abdomen into the amniotic sac; a sample of amniotic fluid (containing foetal cells) is withdrawn; cells are cultured and karyotyped	~0.5-1% risk of miscarriage
Chorionic villus sampling (CVS)	10-13 weeks	A sample of chorionic villi (placental tissue) is obtained via a needle through the abdomen or a catheter through the cervix; cells are karyotyped directly	~1-2% risk of miscarriage
Non-invasive prenatal testing (NIPT)	10+ weeks	Cell-free foetal DNA is extracted from the mother's blood and analysed for chromosomal abnormalities using sequencing technology	No risk to the foetus (blood test only); high sensitivity for trisomy 21, 18, 13

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Common Pitfalls

• Dominant does NOT mean more common. A dominant allele is expressed when only one copy is present, but it can still be rare in the population (e.g., Huntington's disease is dominant but rare; polydactyly is dominant but uncommon)
• A carrier is a heterozygote who shows no symptoms. This term is used for recessive disorders and X-linked disorders, NOT for dominant disorders (a person with a dominant disorder cannot be an unaffected carrier)
• X-linked recessive disorders affect males more frequently than females. Males have only one X chromosome, so a single recessive allele on the X chromosome will be expressed. Females need two copies (homozygous recessive) to be affected
• Down syndrome can be caused by trisomy 21 OR Robertsonian translocation. Most cases (95%) are caused by non-disjunction during meiosis, but some are caused by a translocation involving chromosome 21 -- in these cases, one parent may be a balanced carrier (45 chromosomes, phenotypically normal)

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DNA Technology in Detail

Restriction Enzymes (Restriction Endonucleases)

• Proteins produced by bacteria as a defence mechanism against bacteriophages (viruses that infect bacteria)
• They recognise specific short DNA sequences (recognition sites, typically 4-8 base pairs long) and cut the DNA at or near these sites
• The recognition sites are palindromic -- the sequence reads the same on both strands in the 5' to 3' direction (e.g., GAATTC on one strand reads CTTAAG on the complementary strand)

Type	Cut Pattern	Result
Blunt end cutters	Cut straight through both strands at the same position in the recognition site	Flat ends; can be joined to any other blunt end, but less efficient and less specific
Sticky end cutters	Cut at staggered positions on the two strands, creating short single-stranded overhangs (sticky ends)	The single-stranded overhangs can base-pair with complementary overhangs from another fragment; more efficient and more specific joining

Enzyme	Recognition Site (5' $\rightarrow$ 3')	Cut Type
EcoRI	GAATTC	Sticky ends (5' overhang: AATT)
BamHI	GGATCC	Sticky ends (5' overhang: GATC)
HindIII	AAGCTT	Sticky ends (5' overhang: AGCT)
SmaI	CCCGGG	Blunt ends

DNA Ligase

• An enzyme that joins DNA fragments by forming phosphodiester bonds between adjacent nucleotides
• Used to join:
  • Insert DNA and plasmid vector in genetic engineering
  • Okazaki fragments during DNA replication (natural role)
• DNA ligase is more efficient at joining sticky ends than blunt ends because the complementary overhangs hold the fragments in the correct position

PCR (Polymerase Chain Reaction) in Detail

PCR is a technique used to amplify a specific region of DNA in vitro (outside a living cell).

Steps in each cycle:

Step	Temperature	What Happens	Duration (approximate)
1. Denaturation	94-96 degrees C	Double-stranded DNA template is heated to separate the two strands (breaks hydrogen bonds between base pairs)	30 seconds
2. Annealing	50-65 degrees C	Temperature is lowered; short DNA primers (forward and reverse) bind to complementary sequences on the target DNA; the annealing temperature depends on the primer sequence	30 seconds
3. Extension	72 degrees C	Taq polymerase (heat-stable DNA polymerase from Thermus aquaticus) adds nucleotides to the 3' end of each primer, synthesising new DNA strands complementary to the template	1-2 minutes (depending on length of target DNA)

Key features of PCR:

Feature	Description
Primers	Short, single-stranded DNA sequences (typically 18-25 bases) that are complementary to the regions flanking the target DNA; they define the start and end points of the amplified region
Taq polymerase	Heat-stable DNA polymerase from Thermus aquaticus (a thermophilic bacterium); does not denature at 95 degrees C (unlike human DNA polymerase); essential because the high temperatures in each cycle would destroy most enzymes
Nucleotides	Free dNTPs (deoxynucleoside triphosphates: dATP, dTTP, dCTP, dGTP) provide the building blocks for new DNA strands
Exponential amplification	Each cycle doubles the amount of DNA; after 30 cycles, there are approximately $2^{30}$ ($\approx$ 1 billion) copies of the target region
Thermocycler	A programmable machine that automatically cycles through the temperature changes

Gel Electrophoresis

Feature	Description
Principle	DNA fragments are negatively charged (due to phosphate groups); when placed in an electric field, they migrate towards the positive electrode (anode)
Gel medium	Agarose gel (a polysaccharide from seaweed); acts as a molecular sieve -- smaller fragments move through the pores more easily and travel further
Separation	DNA fragments are separated by SIZE (not by charge -- all fragments have the same charge-to-mass ratio); smaller fragments travel further, larger fragments travel less far
Staining	DNA is not visible; the gel is stained with a DNA-binding dye (e.g., ethidium bromide, GelRed) that fluoresces under UV light
DNA ladder (marker)	A set of DNA fragments of known sizes run in a separate lane; used to estimate the size of unknown fragments by comparing their distance travelled with the ladder
Applications	Comparing DNA fragments (e.g., in genetic fingerprinting, forensic analysis, paternity testing); checking the results of PCR and restriction digestion; diagnosing genetic diseases (e.g., detecting the sickle cell mutation)

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Genetic Fingerprinting (DNA Profiling)

Process

1. Extract DNA: DNA is extracted from a sample (blood, saliva, hair root, semen)
2. PCR amplification: Specific regions of DNA are amplified using primers that target short tandem repeats (STRs) or variable number tandem repeats (VNTRs)
3. Gel electrophoresis: The amplified fragments are separated by size using gel electrophoresis (or capillary electrophoresis)
4. Visualisation: The DNA bands are visualised and compared between samples

STRs (Short Tandem Repeats)

• STRs are short sequences of DNA (typically 2-6 base pairs) that are repeated multiple times in tandem
• The number of repeats varies between individuals (e.g., one person might have ATCG repeated 7 times at a particular locus; another might have it repeated 12 times)
• The FBI uses a panel of 13 standard STR loci (CODIS) for forensic DNA profiling in the United States
• The probability of two unrelated individuals having the same DNA profile at all 13 loci is less than 1 in a trillion

Applications

Application	Description
Forensic science	Matching DNA from crime scenes to suspects or victims; DNA evidence can confirm presence at a scene, establish innocence, or link serial crimes
Paternity testing	Comparing the STR profile of a child with that of the alleged father; the child must inherit one allele at each locus from each biological parent
Identifying disaster victims	DNA profiles from remains are compared with profiles from family members or personal belongings (e.g., toothbrush)
Immigration cases	Confirming family relationships when documentation is disputed
Conservation biology	Monitoring genetic diversity in endangered populations; identifying illegal trade in wildlife products (e.g., elephant ivory, rhino horn)

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Common Pitfalls

• Restriction enzymes cut at specific recognition sites; they do NOT cut at random. The recognition sequence must be present in the DNA for the enzyme to cut it. Different enzymes recognise different sequences
• Sticky ends are more useful than blunt ends in genetic engineering because the complementary overhangs ensure that the DNA fragments are joined in the correct orientation. Blunt ends can be joined in either orientation, and the process is less efficient
• PCR amplifies a SPECIFIC region of DNA, not the entire genome. The region amplified is determined by the primers used
• Gel electrophoresis separates DNA by SIZE (length), not by charge. All DNA fragments carry the same charge per unit length (negative charge from phosphate groups), so charge does not affect separation; smaller fragments travel further because they move through the gel pores more easily
• DNA profiling compares the NUMBER OF REPEATS at specific loci, NOT the sequence of bases. The bases in the repeat are the same between individuals; what varies is how many times the repeat occurs**