The roles of genes in determining the phenotype

· Phenotype = observable features of an organism, produced by genotype + environment.
· Genotype = the alleles an organism has for a gene or genes.
· Gene = a sequence of DNA nucleotides that codes for a polypeptide or functional product.
· Locus = fixed position of a gene on a chromosome.
· Allele = different version of the same gene, found at the same locus.
· Homozygous = two identical alleles, e.g. AA or aa.
· Heterozygous = two different alleles, e.g. Aa.
· Dominant allele = expressed in the phenotype when present in one or two copies.
· Recessive allele = expressed only when no dominant allele is present.
· Codominant alleles = both alleles are expressed in the heterozygous phenotype.
· F1 generation = first offspring generation from a cross.
· F2 generation = offspring produced by crossing F1 individuals.

Monohybrid crosses

· Monohybrid cross = inheritance of one gene.
· Standard dominant/recessive heterozygous cross: Aa × Aa.
· Expected genotype ratio = 1 AA : 2 Aa : 1 aa.
· Expected phenotype ratio = 3 dominant : 1 recessive if complete dominance.
· Always state: parental phenotypes, parental genotypes, gametes, Punnett square, offspring genotypes, offspring phenotypes/ratios.
· Use clear allele symbols: uppercase for dominant, lowercase for recessive.
· Avoid using the same letter where uppercase/lowercase are hard to distinguish, e.g. avoid S/s if handwritten unclearly.

This diagram shows how a dominant phenotype appears in the F1 generation and how the recessive phenotype reappears in the F2 generation. It is useful for visualising the typical 3:1 phenotypic ratio in a monohybrid cross. Source

Dihybrid crosses

· Dihybrid cross = inheritance of two genes.
· If genes are unlinked, alleles assort independently during meiosis.
· A heterozygous dihybrid such as AaBb can produce four gamete types: AB, Ab, aB, ab.
· Standard unlinked dihybrid cross: AaBb × AaBb.
· Expected phenotype ratio with complete dominance and independent assortment = 9 : 3 : 3 : 1.
· In exam answers, list all possible gametes before constructing the Punnett square.
· If the question mentions linkage or genes on the same chromosome, do not assume a 9:3:3:1 ratio.

This diagram shows how two genes can be followed together in a dihybrid cross. It helps students understand why independently assorting genes can produce the classic 9:3:3:1 F2 phenotypic ratio. Source

Codominance, multiple alleles and sex linkage

· Codominance = both alleles contribute to the phenotype in a heterozygote.
· Use superscripts for codominant alleles, e.g. Iᴬ, Iᴮ, Iᴼ for blood groups.
· Multiple alleles = more than two alleles exist for a gene in the population, but each individual still has only two alleles.
· ABO blood group example: Iᴬ and Iᴮ are codominant; Iᴼ is recessive.
· Sex linkage = a gene is located on a sex chromosome, usually the X chromosome.
· Males are more likely to show X-linked recessive conditions because they have only one X chromosome: XY.
· Write sex-linked genotypes with sex chromosomes, e.g. XᴴXʰ, XʰY, not just Hh or hY.
· For X-linked recessive inheritance, a male with the recessive allele on his X chromosome expresses the condition because the Y chromosome lacks the corresponding allele.

This pedigree shows how a sex-linked condition can pass through generations. It is useful for recognising why X-linked recessive phenotypes often appear more frequently in males. Source

Linkage and autosomal linkage

· Linkage = two or more genes are found on the same chromosome and tend to be inherited together.
· Autosomal linkage = linked genes are found on an autosome, not a sex chromosome.
· Linked genes do not show independent assortment unless separated by crossing over.
· Parental combinations = allele combinations already present in the parents.
· Recombinant combinations = new allele combinations produced by crossing over.
· The closer two loci are on the same chromosome, the less likely crossing over occurs between them.
· Linked genes often produce more parental phenotypes than recombinant phenotypes.
· In genetic diagrams for linkage, write linked alleles together on the same chromosome, e.g. AB / ab or Ab / aB.

This page illustrates how genes on the same chromosome can be inherited together unless crossing over occurs. It is useful for understanding why linked genes produce offspring ratios that differ from independent assortment. Source

Epistasis

· Epistasis = one gene affects or masks the expression of another gene.
· The gene doing the masking is the epistatic gene.
· The gene being masked is the hypostatic gene.
· Epistasis changes expected dihybrid ratios because the phenotype depends on interaction between genes.
· CIE requires you to interpret and construct genetic diagrams for epistasis, but specific epistasis ratios do not need to be memorised.
· In exam questions, identify which gene controls whether the second gene can be expressed.
· Explain phenotypes by linking genotype to the biochemical pathway, e.g. one gene may control pigment production while another controls pigment type.

Test crosses

· Test cross = cross an organism showing a dominant phenotype with a homozygous recessive organism.
· Purpose: determine whether the dominant-phenotype organism is homozygous dominant or heterozygous.
· If AA × aa, all offspring show the dominant phenotype.
· If Aa × aa, expected phenotype ratio = 1 dominant : 1 recessive.
· Test crosses can also be used in dihybrid inheritance and linkage to identify parental and recombinant offspring.
· A test cross is powerful because the recessive parent contributes only recessive alleles, so offspring phenotypes reveal the unknown genotype.

Chi-squared test in genetics

· Chi-squared test checks whether differences between observed and expected results are likely due to chance.
· Used when comparing experimental offspring numbers with predicted genetic ratios.
· Formula is provided in the exam, but you must know how to use it.
· Observed values = actual data from the cross.
· Expected values = predicted numbers from the genetic ratio.
· Calculate expected numbers using: total offspring × expected proportion.
· Use degrees of freedom = number of categories − 1.
· Compare calculated χ² with the critical value from the table.
· If χ² is less than critical value, difference is not significant; accept that results fit the expected ratio.
· If χ² is greater than critical value, difference is significant; reject the expected ratio.
· Use probability language carefully: the test does not prove a hypothesis; it indicates whether deviation is likely due to chance.

Genes, proteins and phenotype

· Genes determine phenotype by coding for polypeptides/proteins that affect cell structure, enzymes, transport proteins or regulatory pathways.
· A mutation can alter the amino acid sequence, changing protein structure and function.
· A changed protein can lead to a changed phenotype.

· TYR gene → tyrosinase → albinism
· TYR codes for tyrosinase, an enzyme involved in melanin production.
· Non-functional tyrosinase reduces or prevents melanin synthesis.
· Result: albinism, with reduced pigmentation in skin, hair and eyes.

· HBB gene → haemoglobin → sickle cell anaemia
· HBB codes for the β-globin polypeptide in haemoglobin.
· A mutation can cause production of abnormal haemoglobin S.
· Haemoglobin S can cause red blood cells to become sickle-shaped, reducing oxygen transport and blocking capillaries.

· F8 gene → factor VIII → haemophilia
· F8 codes for factor VIII, a blood-clotting protein.
· A faulty F8 allele can produce absent or non-functional factor VIII.
· Result: haemophilia, where blood clotting is impaired and bleeding is prolonged.

· HTT gene → huntingtin → Huntington’s disease
· HTT codes for the huntingtin protein.
· A dominant mutant allele causes abnormal huntingtin protein.
· Result: Huntington’s disease, a progressive nervous system disorder.
· This is an example where a dominant allele can cause a harmful phenotype.

This diagram links a DNA base substitution to a change in the haemoglobin protein. It is directly useful for explaining the relationship between gene mutation, protein structure and phenotype in sickle cell anaemia. Source

Gibberellin, the Le gene and stem elongation

· Gibberellin is a plant hormone involved in stem elongation.
· In peas, the dominant allele Le codes for a functional enzyme in the gibberellin synthesis pathway.
· Functional enzyme → gibberellin is produced → stem elongation occurs → tall plant phenotype.
· The recessive allele le codes for a non-functional enzyme.
· Non-functional enzyme → reduced gibberellin synthesis → reduced stem elongation → dwarf plant phenotype.
· LeLe and Lele plants are tall because at least one functional allele produces enough enzyme.
· lele plants are dwarf because they lack a functional enzyme for gibberellin synthesis.
· This is a clear example of: gene → enzyme → hormone synthesis → phenotype.

Exam technique for genetic diagrams

· Define allele symbols clearly before starting, e.g. A = normal pigment, a = albinism.
· Write parental genotypes and phenotypes.
· Show gametes produced by each parent.
· Use a Punnett square or clear genetic diagram.
· State offspring genotype ratio and phenotype ratio.
· For sex linkage, include X and Y chromosomes in the genotype.
· For codominance, use correct superscript notation.
· For linked genes, keep alleles on the same chromosome together.
· For chi-squared questions, show expected values and degrees of freedom.