CIE AS/A Level Biology -16.2 The roles of genes in determining the phenotype- Study Notes- New Syllabus
CIE AS/A Level Biology -16.2 The roles of genes in determining the phenotype- Study Notes- New Syllabus
Ace A level Biology Exam with CIE AS/A Level Biology -16.2 The roles of genes in determining the phenotype- Study Notes- New Syllabus
Key Concepts:
- explain the terms gene, locus, allele, dominant, recessive, codominant, linkage, test cross, F1, F2, phenotype, genotype, homozygous and heterozygous
- interpret and construct genetic diagrams, including Punnett squares, to explain and predict the results of monohybrid crosses and dihybrid crosses that involve dominance, codominance, multiple alleles and sex linkage
- interpret and construct genetic diagrams, including Punnett squares, to explain and predict the results of dihybrid crosses that involve autosomal linkage and epistasis (knowledge of the expected ratios for different types of epistasis is not expected)
- interpret and construct genetic diagrams, including Punnett squares, to explain and predict the results of test crosses
- use the chi-squared test to test the significance of differences between observed and expected results (the formula for the chi-squared test will be provided, as shown in the Mathematical requirements)
- explain the relationship between genes, proteins and phenotype with respect to the:
- TYR gene, tyrosinase and albinism
- HBB gene, haemoglobin and sickle cell anaemia
- F8 gene, factor VIII and haemophilia
- HTT gene, huntingtin and Huntington’s disease
- explain the role of gibberellin in stem elongation including the role of the dominant allele, Le, that codes for a functional enzyme in the gibberellin synthesis pathway, and the recessive allele, le, that codes for a non-functional enzyme
Key Genetics Terms
🌱 Basic Definitions
- Gene → a length of DNA that codes for a polypeptide or functional RNA.
- Locus → the specific position of a gene on a chromosome.
- Allele → different versions of the same gene, found at the same locus.
⚖️ Relationships Between Alleles
- Dominant allele → expressed in the phenotype even if only one copy is present (heterozygous).
- Recessive allele → only expressed in the phenotype if two copies are present (homozygous).
- Codominant alleles → both alleles are expressed in the phenotype when present together (e.g., blood group AB).
🔗 Linkage
- Linkage → when genes are located on the same chromosome, they tend to be inherited together (unless separated by crossing over).
🧪 Crosses and Generations
- Test cross → crossing an individual with the dominant phenotype (unknown genotype) with a homozygous recessive individual to determine the unknown genotype.
- F1 generation → the first generation of offspring from a cross between two parents (P generation).
- F2 generation → offspring produced when F1 individuals are crossed together.
👀 Expression and Genetic Makeup
- Phenotype → observable characteristics of an organism, influenced by genotype and environment.
- Genotype → genetic makeup of an organism (the alleles it carries).
🧬 Homozygosity vs Heterozygosity
- Homozygous → both alleles at a locus are the same (e.g., AA or aa).
- Heterozygous → alleles at a locus are different (e.g., Aa).
✅ Summary Box – Quick Recall
- Gene = DNA unit of inheritance.
- Locus = position of a gene.
- Allele = gene variant.
- Dominant / Recessive / Codominant = how alleles interact.
- Linkage = genes inherited together.
- Test cross = to find unknown genotype.
- F1 / F2 = generations in breeding.
- Phenotype = appearance; Genotype = alleles.
- Homozygous vs Heterozygous = same vs different alleles.
Genetic Diagrams & Crosses
🌱 Key Tool – Punnett Squares
- A Punnett square shows how alleles from parents combine in offspring.
- Each gamete contributes one allele per gene.
- Helps predict genotype and phenotype ratios.
1. Monohybrid Cross (Single Gene, Simple Dominance)
Example: Tall (T) is dominant over dwarf (t).
Parental genotypes: TT × tt
| T | T | |
|---|---|---|
| t | Tt | Tt |
| t | Tt | Tt |
Genotype ratio = 100% Tt
Phenotype ratio = 100% Tall
2. Monohybrid with Codominance
Example: Flower colour – Red (CR), White (CW).
Cross: CRCW × CRCW
| CR | CW | |
|---|---|---|
| CR | CRCR (Red) | CRCW (Pink) |
| CW | CRCW (Pink) | CWCW (White) |
Genotype ratio = 1 CRCR : 2 CRCW : 1 CWCW
Phenotype ratio = 1 Red : 2 Pink : 1 White
3. Monohybrid with Multiple Alleles
Example: Human ABO blood group system (IA, IB, i).
IA and IB are codominant, both dominant over i.
Cross: IAi × IBi
| IA | i | |
|---|---|---|
| IB | IAIB (AB) | IBi (B) |
| i | IAi (A) | ii (O) |
Phenotypes = A : B : AB : O (1:1:1:1)
4. Dihybrid Cross (Two Genes, Independent Assortment)
Example: Seed shape (R = round, r = wrinkled) and seed colour (Y = yellow, y = green).
Cross: RrYy × RrYy
Gametes = RY, Ry, rY, ry
Phenotypic ratio = 9 Round Yellow : 3 Round Green : 3 Wrinkled Yellow : 1 Wrinkled Green
Classic 9:3:3:1 ratio
5. Sex-Linkage (X-linked Genes)
Genes on the X chromosome behave differently in males (XY) and females (XX).
Example: Haemophilia (recessive) – allele h.
Cross: Carrier female (XHXh) × Normal male (XHY)
| XH | Y | |
|---|---|---|
| XH | XHXH (Normal female) | XHY (Normal male) |
| Xh | XHXh (Carrier female) | XhY (Haemophiliac male) |
Phenotypes:
- 25% Normal female
- 25% Carrier female
- 25% Normal mal
- 25% Haemophiliac male
✅ Summary Box – Key Ratios to Remember
- Monohybrid (dominance) = 3:1 (F2)
- Codominance = 1:2:1
- Dihybrid = 9:3:3:1
- Sex-linkage = males more often affected
Genetic Diagrams: Autosomal Linkage & Epistasis
🌱 1. Autosomal Linkage
- Definition: When two genes are located on the same autosome (non-sex chromosome).
- They are said to be linked and tend to be inherited together because they are not separated by independent assortment (unless crossing over occurs).
- Key Point: Linked genes do not produce the 9:3:3:1 ratio expected in a normal dihybrid cross.
Example:
- Genes A and B are linked on the same chromosome.
- Parent genotype: AB/ab × AB/ab (no crossing over).
- Gametes: AB, ab
| AB | ab | |
|---|---|---|
| AB | AABB | AaBb |
| ab | AaBb | aabb |
- Genotypes: 1 AABB : 2 AaBb : 1 aabb
- Phenotypes: Only 2 combinations instead of 4.
- If crossing over occurs → recombinants appear (but at a lower frequency).
Takeaway: Linked genes reduce genetic variation compared to independent assortment.
🌱 2. Epistasis
- Definition: Interaction where one gene masks or affects the expression of another gene at a different locus.
- Epistatic gene = the one that masks.
- Hypostatic gene = the one being masked.
Types:
- Recessive epistasis – two copies of a recessive allele at one locus mask expression of a gene at another locus.
Example: Coat colour in mice → allele aa prevents pigment production, so the other gene (B/b for colour intensity) has no effect if aa is present. - Dominant epistasis – a single dominant allele at one locus masks expression at another.
Example: Plant colour → allele A prevents pigment deposition, so regardless of B/b, plants stay white.
🧪 Worked Example: Recessive Epistasis
- Gene 1 (C/c) → controls pigment production (C = pigment, cc = no pigment).
- Gene 2 (A/a) → controls pigment colour (A = black, aa = brown).
- Cross: CcAa × CcAa
- Without epistasis, expect 9:3:3:1 ratio.
- But cc masks A/a → all cc individuals are white (no pigment).
Phenotypes:
- Black (C_A_)
- Brown (C_aa)
- White (cc__)
Ratio changes from 9:3:3:1 to 9:3:4.
✅ Key Recap Box
- Autosomal linkage: Genes on the same chromosome are inherited together → fewer combinations than 9:3:3:1.
- Epistasis: One gene masks another. Ratios deviate from 9:3:3:1 (e.g., 9:3:4 or 12:3:1).
- Punnett squares: Still used, but need to group genotypes into phenotypes carefully.
Test Crosses
1. What is a Test Cross?
- A test cross is used to determine the genotype of an organism showing a dominant phenotype.
- The individual is crossed with a homozygous recessive (which always shows the recessive trait).
- If offspring include recessives → the unknown parent was heterozygous.
2. Example: Monohybrid Test Cross
Trait: Tall (T) dominant over dwarf (t).
Case 1: Unknown Parent = TT (homozygous dominant)
Cross: TT × tt
Gametes: (T) × (t)
| t | t | |
|---|---|---|
| T | Tt | Tt |
| T | Tt | Tt |
Offspring: 100% Tall (Tt).
No recessives appear.
Case 2: Unknown Parent = Tt (heterozygous)
Cross: Tt × tt
Gametes: (T, t) × (t)
| t | t | |
|---|---|---|
| T | Tt | Tt |
| t | tt | tt |
Offspring: 50% Tall, 50% Dwarf.
Recessives appear → confirms heterozygous.
Conclusion: If any recessive offspring appear, the tested parent must be heterozygous.
3. Dihybrid Test Cross
Trait: Seed shape (R = round, r = wrinkled) & seed colour (Y = yellow, y = green).
Unknown parent shows Round, Yellow.
Cross with rryy (homozygous recessive for both traits).
Case 1: Unknown Parent = RRY Y (homozygous dominant for both)
Cross: RRY Y × rryy
Gametes: RY × ry
Offspring: 100% RrYy (Round, Yellow).
Case 2: Unknown Parent = RrYy (heterozygous for both)
Cross: RrYy × rryy
Gametes: (RY, Ry, rY, ry) × (ry)
| ry | |
|---|---|
| RY | RrYy (Round, Yellow) |
| Ry | Rryy (Round, Green) |
| rY | rrYy (Wrinkled, Yellow) |
| ry | rryy (Wrinkled, Green) |
Phenotypic ratio = 1:1:1:1.
Equal proportions of all 4 combinations.
✅ Key Recap Box
- Test cross: Used to identify if dominant phenotype = homozygous or heterozygous.
- Always crossed with homozygous recessive.
- Monohybrid test cross results: 100% dominant (if homozygous) OR 1:1 (if heterozygous).
- Dihybrid test cross (heterozygous unknown) → produces 1:1:1:1 ratio of offspring.
Chi-Squared Test (χ²) in Genetics
1. Purpose of the Test
- Used to compare observed results (from an experiment) with expected results (from genetic ratios).
- Tells us if the differences are:
- Significant → due to another factor (not just chance).
- Not significant → differences are small and likely due to chance.
2. Formula
\[ \chi^2 = \sum \frac{(O – E)^2}{E} \]
- O = observed value
- E = expected value
- Σ = sum across all categories
3. Steps to Perform χ² Test
- Write the hypothesis
- Null hypothesis (H₀): There is no significant difference between observed and expected results.
- Any differences are due to chance.
- Work out expected ratios from the genetic cross (e.g. 3:1, 9:3:3:1).
- Calculate χ² using the formula.
- Find degrees of freedom (df):
\[ df = \text{number of categories} – 1 \]
- Compare χ² value with the critical value (usually at p = 0.05).
4. Decision Rule
- If χ² < critical value → Accept H₀ → no significant difference (fits ratio).
- If χ² ≥ critical value → Reject H₀ → significant difference (doesn’t fit ratio).
5. Example: Monohybrid Cross
- Cross: Tt × Tt
- Expected ratio: 3 Tall : 1 Dwarf
- Observed results: 76 Tall, 24 Dwarf (total 100).
Step 1: Expected numbers
- Tall = 100 × (3/4) = 75
- Dwarf = 100 × (1/4) = 25
Step 2: Apply formula
\[ \chi^2 = \frac{(76-75)^2}{75} + \frac{(24-25)^2}{25} \] \[ \chi^2 = \frac{1}{75} + \frac{1}{25} = 0.013 + 0.04 = 0.053 \]
Step 3: Degrees of freedom
\[ df = 2 – 1 = 1 \]
Step 4: Compare with critical value
- At df = 1 and p = 0.05 → critical value = 3.84
- Since 0.053 < 3.84, we accept H₀.
The difference is not significant; results fit the 3:1 ratio.
Chi-squared test checks if observed ≈ expected.
Formula: \((O – E)^2 / E\).
df = categories – 1.
Compare with critical value at p = 0.05.
Accept H₀ if χ² < critical value.
Genes, Proteins and Phenotype
1. General Principle
- A gene is a section of DNA that codes for a protein.
- The protein’s function (or malfunction) determines the phenotype (observable characteristic).
- Mutations in genes can change the structure or function of proteins → leading to disease phenotypes.
2. Examples
(a) TYR Gene → Tyrosinase → Albinism
- Gene (TYR): Codes for the enzyme tyrosinase.
- Protein: Tyrosinase is required for the production of melanin (pigment) from tyrosine.
- Mutation effect: Faulty TYR gene → non-functional tyrosinase → no melanin.
- Phenotype: Albinism → very pale skin, hair, eyes; sensitive to sunlight.
(b) HBB Gene → Haemoglobin → Sickle Cell Anaemia
- Gene (HBB): Codes for the β-globin chain of haemoglobin.
- Protein: Normal haemoglobin (HbA) carries oxygen efficiently.
- Mutation effect: A single base substitution (GAG → GTG) → valine instead of glutamic acid in β-chain → abnormal haemoglobin (HbS).
- Phenotype: Red blood cells become sickle-shaped → block blood vessels, reduced oxygen transport, anaemia, pain crises.
(c) F8 Gene → Factor VIII → Haemophilia
- Gene (F8): Codes for clotting factor VIII.
- Protein: Factor VIII is essential in the blood clotting cascade.
- Mutation effect: Faulty or missing factor VIII → inability to form stable clots.
- Phenotype: Haemophilia A → excessive bleeding, bruising, risk of internal haemorrhage.
(d) HTT Gene → Huntingtin → Huntington’s Disease
- Gene (HTT): Codes for huntingtin protein.
- Protein: Normal huntingtin has a role in nerve cell function.
- Mutation effect: Expansion of CAG triplet repeats → abnormally long huntingtin protein → toxic build-up in neurons.
- Phenotype: Huntington’s disease → progressive neurodegeneration, movement disorders, cognitive decline.
✅ Summary Box
- Gene mutation → altered protein → altered phenotype.
- TYR mutation → no melanin → Albinism.
- HBB mutation → abnormal haemoglobin → Sickle cell anaemia.
- F8 mutation → missing clotting factor → Haemophilia.
- HTT mutation → toxic huntingtin protein → Huntington’s disease.
Role of Gibberellin in Stem Elongation
🔬 Gibberellin and Plant Growth
- Gibberellins (GA): Plant hormones that stimulate stem elongation, seed germination, and flowering.
- In stems, GA promotes cell elongation and cell division in the internodes (spaces between leaves).
🧬 Genetic Control (Le and le alleles)
- Le gene: Codes for an enzyme in the gibberellin synthesis pathway.
- Dominant allele (Le): Produces a functional enzyme → normal gibberellin synthesis → normal stem elongation (tall plants).
- Recessive allele (le): Produces a non-functional enzyme → little or no gibberellin made → reduced stem elongation (dwarf plants).
🌱 Phenotypes
- LeLe or Lele (at least one dominant allele):
- Functional enzyme produced.
- Sufficient gibberellin → tall phenotype.
- lele (homozygous recessive):
- No functional enzyme.
- Very little gibberellin → dwarf phenotype.
📌 Key Point
- Gibberellin’s role in stem elongation is dependent on the Le gene.
- Without gibberellin (in lele plants), stems remain short, but adding external gibberellin can restore elongation.
✅ Summary:
- Gibberellin stimulates stem elongation by promoting cell growth.
- Le allele = functional enzyme = gibberellin produced = tall plants.
- le allele = non-functional enzyme = no gibberellin = dwarf plants.