Chapter 19 : Inheritance
Inheritance Notes
Tutorial 21: Inheritance (JPJC/JC2 H2 Biology, 2025). Covers Mendelian genetics, epistasis, linkage, sex-linked traits, pedigree analysis, polygenic/continuous variation, chi-squared testing in genetics, and applied breeding problems (plants and animals).
Explanations, worked examples, and typical exam question formats are summarized with key formulas and ratios in LaTeX.
All numerical results, ratios, and probability expressions are included in LaTeX format where shown or useful for study.
1) Basic Gamete Formation and Unlinked Genes
When an organism is heterozygous at n unlinked loci (e.g., AaBbCcDd), a gamete receives one allele from each gene locus.
For each locus, there is a ½ chance of the gamete containing a specific allele (e.g., a from Aa, b from Bb, c from Cc, d from Dd).
For four unlinked loci, the probability of a gamete with genotype abcd is:
Key takeaway: independence of assortment for unlinked genes leads to multiplicative probabilities across loci.
2) Gene Interaction and Epistasis (examples)
Coat pigmentation in dogs (epistasis among pigment genes):
Alleles: aW (white) > aS (sable) > ac (copper)
White is dominant to sable and copper; sable is dominant to copper.
Crosses aWac x aSac yield 4 genotypes and 3 phenotypes (white, sable, copper) in the offspring.
Genotype-phenotype mapping (example shown):
aW aC aW aS → white
aW aS → white
aW ac → white
aS aC → sable or white depending on dominance relations; the typical outcome in the example yields 4 genotypes and 3 phenotypes overall.
Epistasis summary: the effect of one gene (epistatic) masks or modifies the effect of another gene (hypostatic).
3) Sex Determination Systems and Sex-Linked Traits
Birds have ZW sex-determination: Males ZZ, Females ZW.
If a lethal recessive allele is on the Z chromosome (Za), a cross between a male heterozygote (ZAZa) and a normal female (ZAW) yields certain sex ratios depending on Z-linked lethality.
Example logic (simplified): Za on ZZ vs ZAW on ZW yields some offspring where Z with lethal allele loses viability; the remaining viable offspring produce a skewed sex ratio.
Red-green color blindness is X-linked recessive in humans.
A color-blind man (XbY) has sons who are color blind (XbY) and sons who are not (XBY) depending on the X-bearing sperm; the wife must be a carrier (XBXb) to pass both alleles.
If the baby is a girl, the chance she will be color blind is 50% if the mother is a carrier and the father is color-blind (XbY) because daughters inherit one X from each parent.
Practical rule: X-linked traits show distinct patterns in pedigrees and often affect males more severely.
4) Pedigrees and Modes of Inheritance
Autosomal vs sex-linked patterns (qualitative cues):
If affected females have affected sons and affected males do not pass to sons, that suggests X-linked inheritance (sex-linked).
If males and females are equally affected, or if affected individuals appear in every generation, autosomal dominant or recessive depending on the pattern.
Examples (summarized):
Pedigree with a trait where affected males do not pass it to sons indicates a sex-linked trait.
If affected parents have affected offspring regardless of the child's sex, an autosomal pattern is implicated.
Important caveat: Pedigree patterns give clues, but experimental or lecture-based context is often required to state the mode accurately.
5) Autosomal Dominant/Autosomal Recessive Traits in Pedigrees
Problem context: A dominant mutant allele can cause early death if homozygous; if both parents are affected and produce unaffected children, the trait may be recessive or the affected condition may affect viability.
In autosomal traits, both sexes are affected with similar frequency, and affected individuals can have unaffected children if they’re carriers in certain contexts.
When a pedigree indicates autosomal traits with lethal outcomes, the following interpretation arises:
The trait is autosomal because both sexes are affected and transmission is not restricted by sex.
Lethality of homozygotes explains absence of certain genotypes in offspring.
6) Epistasis: Types and Phenotypic Ratios
Epistasis is gene interaction where one gene masks/modulates the effect of another:
Dominant epistasis: a dominant allele at one locus masks the effect of another locus. Example ratio: for phenotypes in the F2 when two genes interact (dominant epistasis).
Recessive epistasis: recessive alleles at one locus mask another gene’s effect. Classic example yields a 9:3:4 ratio.
Duplicate recessive epistasis: both gene pairs need at least one dominant allele to express phenotype; homozygous recessive at either locus leads to a recessive phenotype. Often yields a 9:7 ratio.
Duplicate dominant epistasis: dominant alleles at either gene give a phenotype; only double recessive yields alternate phenotype, commonly 15:1.
Example cross discussions in material show several epistasis patterns, with explicit genotype-phenotype maps and resulting F2 phenotypes.
7) Dihybrid and Polygenic Inheritance in Plants and Animals
Dihybrid crosses (AaBb x AaBb) for two unlinked genes typically yield a 9:3:3:1 phenotypic ratio in the F2 (for complete dominance and independent assortment).
In some cases, trait interactions produce more complex ratios (epistasis), e.g., 12:3:1 (dominant epistasis) or 9:7 (duplicate/recessive/dominant interactions).
Polygenic/continuous variation concept:
Continuous variation arises from the additive effects of many genes (polygenes) and is strongly influenced by the environment.
Discontinuous variation arises from one or a few genes with clear-cut phenotypes and is less influenced by the environment (exceptions exist).
Practical example (coffee leaf length graph): continuous variation is environment-influenced; the graph shows a gradient rather than discrete categories.
8) Chi-Squared Tests in Genetics
Purpose: determine whether observed offspring phenotypes/genotypes differ significantly from expected Mendelian ratios.
Formula: where Oi are observed counts and E_i are expected counts under a hypothesis (e.g., 9:3:3:1).
Decision rule: compare computed chi-squared value to critical value for given degrees of freedom (df). If chi-squared < critical value, fail to reject the null hypothesis (no significant difference).
Example outcomes in the notes show several chi-squared calculations:
F2 dihybrid 9:3:3:1 example yielded
chi-squared value = with p-value > 0.05 at appropriate df, leading to no significant deviation.For a 12:3:1 test (dominant epistasis), chi-squared like result: vs a critical value of (p = 0.05); conclusion: not significantly different; supports dominant epistasis as explanation.
For a 9:7 epistasis scenario, chi-squared value 1.60 with critical 3.841 at p = 0.05; conclusion: data consistent with epistasis as explanation.
Practical tips:
Always state df (degrees of freedom) when interpreting chi-squared values.
Use chi-squared to test for fit to expected Mendelian ratios and to compare observed to expected under hypotheses like epistasis or linkage.
9) Linkage, Recombination, and Mapping Genes
Linkage: genes on the same chromosome tend to be inherited together (non-independent assortment) unless crossing over occurs.
Parental types vs recombinant types in a test cross: higher frequency of parental gametes indicates linkage; recombinant gametes arise from crossing over.
Measures:
Recombination frequency (RF) = (number of recombinant offspring / total offspring) × 100%. RF provides map units (centiMorgans, cM) distance between genes.
Classic problem types: two linked genes with a dihybrid parent produce a majority of parental phenotypes and a minority of recombinants in the offspring; chi-squared can test whether observed proportions fit expected linkage patterns.
Example (Drosophila/ linked gene cross): test crosses yield 8:8:1:1 phenotypic ratios under certain linkage configurations; analysis requires identifying parental and recombinant combinations and using chi-squared to support linkage vs independent assortment.
10) Polygenic Traits and Environmental Influence (Continued)
Fast Plants (Brassica): demonstrates rapid generation time and utility in teaching polygenic traits and selection.
Observed F2 phenotypes after two crosses included: purple stems with green leaves, purple stems with yellow leaves, white stems with green leaves, white stems with yellow leaves.
Expected dihybrid ratio under independent assortment would be 9:3:3:1 for two unlinked genes.
Observed counts led to chi-squared testing to see if deviation from expectation is due to chance.
The discussion highlights the value of model organisms with rapid generation times, enabling quick cycles of crossing and selection for desired traits.
11) Plant and Animal Genetics: Worked Examples and Key Takeaways
Maize (corn) dihybrid crosses (texture and color) illustrate classic dihybrid inheritance with a 9:3:3:1 expectation under normal dihybrid cross.
Observed F2 results can be analyzed for linkage or epistasis by mapping phenotypes to genotypes and applying chi-squared tests.
Labrador coat color (two unlinked genes with epistasis) illustrates how dominant and recessive alleles in separate loci interact to produce multiple phenotypes (yellow, black, brown) with specific genotype requirements.
Dihybrid and test-cross problems in Drosophila demonstrate X-linked inheritance and the usefulness of test crosses to reveal linkage and gene order.
In many examples, the correct answer requires:
Identifying whether genes are linked or unlinked
Recognizing epistasis patterns (dominant, recessive, duplicate)
Translating between genotype and phenotype in multi-gene interactions
Applying chi-squared tests to support or refute hypotheses about linkage and interaction
12) Quick Reference: Common Ratios and Concepts
Independent assortment (unlinked genes):
Dihybrid cross (AaBb) yields F2 phenotype ratio under complete dominance.
Epistasis patterns and typical ratios:
Dominant epistasis:
Recessive epistasis:
Duplicate recessive epistasis:
Duplicate dominant epistasis:
Sex-linked inheritance: males & females can show different patterns; X-linked recessive traits frequently affect males more due to having a single X chromosome.
Polygenic traits: continuous variation caused by multiple genes plus environmental factors; example: leaf length in hybrid coffee plants and plant height in peas.
Linkage and recombination: RF between genes determines map distance; lower RF → closer linkage; higher RF → farther apart.
Chi-squared basics: ; compare to critical values to decide significance.
13) Selected Problem Variants (Summary of Key Points from Transcript)
Page 1: Unlinked four loci: abcd probability in gametes = .
Page 2: Hair-color gene alleles in dogs: 4 genotypes, 3 phenotypes; white dominant to sable and copper; sable dominant to copper; cross yields specific genotype-phenotype mapping.
Page 3: Sex determination in birds (ZW); lethal Z-linked allele calculations; color blindness X-linked inheritance and 50% female risk if baby is a girl when the mother is a carrier.
Page 4-6: Pedigree interpretation for autosomal vs sex-linked; autosomal dominant vs recessive evidence; hypophosphatemia pedigree indicates autosomal recessive pattern; I-1 and I-2 genotypes discussed.
Page 7-8: Epistasis in mice coat color; epistatic interactions for pigment (dd morphs lead to white); codominance/epistasis in plant coloration; cotyledon color examples.
Page 9-11: Drosophila and coffee leaf data; chi-squared calculations; interpretation of continuous vs discontinuous variation; concept of epistasis in maize and labrador retrievers.
Page 12-13: Labrador coat color cross: two genes (E/e and B/b) with epistasis; dihybrid cross; progeny phenotypes with ratios; epistasis explanation.
Page 14-15: Maize dihybrid results and chi-squared test; F2 phenotypic ratio; concluding normal dihybrid inheritance in a chi-squared context.
Page 16-18: Morgan’s crosses for X-linked eye color in Drosophila; creation of cross tables; male/female progeny expectations; instructions for completing cross tables.
Page 19-20: Honey bees behavior controlled by genes A/a and B/b; offspring phenotypes under different parental genotypes; selection to fix traits in workers; diet/environment in bees for queen vs worker development.
Page 21-22: Fast Plants (Brassica) data; chi-squared interpretation; discussion of continuous vs discontinuous variation and selection implications; epistasis in F2 (
9:3:4 example).Page 23-25: Mice and epistasis problems; inheritance of fur color and cross outcomes, including test crosses and ratios.
Page 26-27: Tobacco corolla tube length (continuous variation); discussion of polygenic inheritance and experimental design; chi-squared interpretation.
Page 28-29: Drosophila test crosses with linkage and recombination; explanation of parental vs recombinant outcomes; mapping and cross diagrams.
Page 30-31: Dihybrid and independent assortment explanation via pea genetics (R/r and A/a); scenario where two loci independently assort to produce a 9:3:3:1 ratio; height in peas and the effect of the T/t allele on gibberellin synthesis; heterozygous Tt shows same phenotype as TT due to functional enzyme production.
Page 32-33: ABO blood typing system with IA, IB, i alleles; Rh factor genetics (Rh+ and Rh-); independence of ABO and Rh loci; di- and tri-allelic inheritance; cross examples with Type A Rh+ x Type B Rh+ yielding type O Rh- offspring under certain scenarios; genetic-diagram practice.
Page 34-35: Epistasis in pea flowers (9:7 ratio for duplicate recessive epistasis); F2 phenotypes and linking genotype to phenotype; chi-squared interpretation for epistasis; discussion of pea plant strengths for genetic studies.
Page 36-38: Tobacco continuous variation; locus definition; test cross data for linkage; definition of linkage; use of test crosses to deduce linkage vs independent assortment.
Page 39-41: Drosophila gene mapping with test crosses; multiple tables showing parental vs recombinant gametes and phenotypes; steps to determine gene order and map distances via recombinant frequency.
Page 42-45: Fruiting flies cross data; chi-squared test utility; vestigial wing and ebony body as a mutational event; recessive mutations and why they are less likely to be found in natural populations.
Page 46-50: Haemophilia in Queen Victoria’s family; genetic pedigree exercises; epistasis and X-linked inheritance in human pedigrees; chicken comb genetics and Bateson/Punnett demonstration of gene interaction (epistasis in comb types); walnut vs rose vs pea vs single combs; mapping and cross diagrams; two- and multi-gene interaction demonstrations.
14) Practical Takeaways for Exam Preparation
Always identify whether genes are unlinked or linked when interpreting ratios and crosses.
Map gene interactions by writing down genotype-to-phenotype relationships, then derive expected phenotypic ratios for F2 under the given interaction (dominant/recessive/duplicate epistasis).
Use chi-squared tests to assess fit to expected Mendelian ratios and to justify conclusions about linkage and epistasis.
For sex-linked traits, remember that males are XY and females XX in many organisms (e.g., Drosophila), and cross outcomes often show different patterns for sons vs daughters.
Distinguish continuous vs discontinuous variation. Continuous variation indicates polygenic inheritance with environmental influence; discontinuous variation suggests monogenic or oligogenic controls.
For plant/j breeding problems (maize, peas, tobacco, fast plants), map genotypes to phenotypes carefully, and consider linkage, epistasis, and environmental effects when interpreting data.
Practice constructing and interpreting Punnett squares, dihybrid/tri-hybrid crosses, test crosses, and pedigree analyses as standard exam patterns.
Appendix: Notation Quick-Reference
Alleles: uppercase for dominant (e.g., A, B, E, R) and lowercase for recessive (a, b, e, r).
Epistasis notation conventions in explanations often label dominant epistatic allele as E (or A) and recessive epistatic allele as e (or a), with hypostatic alleles for the visible trait as e.g., B/b or R/r depending on the example.
Common genotype-phenotype mappings in epistasis problems:
Recessive epistasis: aa masks whether B/b is present (e.g., 9:3:4).
Dominant epistasis: presence of A- masks B/b effects (e.g., 12:3:1).
Duplicate recessive epistasis: aa or bb yields the same phenotype; often 9:7.
Duplicate dominant epistasis: A- or B- each can produce the same phenotype; only aabb yields another phenotype (often 15:1).
Quick problem-solving reminders:
If a problem involves linked genes, first identify parental types and recombinants, then decide whether to apply a linkage-based ratio or a 9:3:3:1 expectation.
In sex-linked problems, draw X and Y chromosomes explicitly, and track paternal and maternal alleles across generations.
In chi-squared problems, list observed and expected counts for all phenotype/genotype classes, compute chi-squared, and compare with the appropriate critical value for the given df.
If you’d like, I can tailor these notes to a specific sub-topic (e.g., epistasis, linkage, or chi-squared calculations) and expand the corresponding sections with more worked examples and step-by-step solutions.