Genetics | DSE - Wyatt's Notes

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.

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.

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:

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%.

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):

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.

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.

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

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.

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.

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:

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:

Six possible genotypes and four phenotypes:

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

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).

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.

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).

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

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 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)

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 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).

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.

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

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

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.

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)

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

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 Cand 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 Aand B produce triangular seeds; aabb produces oval seeds

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

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

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 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

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 lethal in humans.

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 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, 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

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)

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.

• 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)

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.

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, 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 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 ( 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.

PCR amplifies a specific segment of DNA exponentially.

Components:

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 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 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.

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.

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

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”

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)

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

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”

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

• 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.

Nature’s instruction manual: Genetics is like a recipe book — DNA contains the instructions, genes are individual recipes, and proteins are the dishes that get made. Mutations are typos in the recipe.

Why it matters: From genetic diseases to crop improvement, genetics explains inheritance and variation. Understanding genetics helps diagnose and treat genetic disorders.

The key insight: Genes don’t work alone — most traits are polygenic (controlled by many genes) and influenced by the environment.

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.

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

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

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

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

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

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

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

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)

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)

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

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

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

Components:

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

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

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

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).

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

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

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