Heredity – Case-based Questions with Answers
Syllabus → Learning objectives → Topic-wise cases → Analytical Q&A
These 20 case-based questions with clear, exam-focused answers are aligned strictly to the NCERT Class 10 Biology Chapter 8: Heredity. Each case is followed by focused questions and concise model answers for quick learning and board-style revision.
Case 1 — Mendel’s Pea Plants
Mendel crossed pure-breeding tall pea plants (TT) with pure-breeding dwarf plants (tt). All F1 plants were tall. On selfing F1, the F2 generation showed both tall and dwarf plants in a 3:1 ratio.
- Explain why all F1 were tall though dwarf trait appeared in F2.
- State the genotypic and phenotypic ratios in F2.
- Tall allele (T) is dominant over dwarf (t). F1 (Tt) shows tall phenotype because T masks t in heterozygote.
- Genotypic ratio in F2 = 1 TT : 2 Tt : 1 tt. Phenotypic ratio = 3 tall : 1 dwarf.
Case 2 — Unknown Genotype
A pea plant with round seeds (dominant R) is crossed with a tester plant with wrinkled seeds (rr). Half the offspring are round and half wrinkled.
- What is the genotype of the unknown round plant?
- Explain the test-cross principle illustrated.
- The unknown plant is heterozygous (Rr) because crossing with rr produced 1:1 ratio of phenotypes.
- Test cross: crossing with homozygous recessive reveals whether unknown dominant phenotype is homozygous or heterozygous; appearance of recessive phenotype among progeny indicates heterozygosity.
Case 3 — Co-dominant Blood Group
A couple has blood groups A and B and their child has blood group AB.
- Which alleles could the parents carry?
- Explain co-dominance in this context.
- Possible genotypes: Parent with A could be IAi or IAIA; parent with B could be IBi or IBIB. To produce AB child, one parent must contribute IA and the other IB (e.g., IAi × IBi → possible IAIB).
- Co-dominance: IA and IB alleles both express equally in IAIB heterozygote, producing AB phenotype (both A and B antigens present).
Case 4 — Sex Determination
A family is tracking sex of newborns. Father provides sperm with either X or Y; mother provides eggs with X only.
- Explain how the sex of a child is determined genetically.
- If a father’s sperm all carried only X, what would be the result?
- Sex determined by combination of sex chromosomes: XX = female, XY = male. Father’s sperm (X or Y) determines sex because mother supplies X.
- If father’s sperm carried only X, all zygotes would be XX and all children would be genetically female.
Case 5 — X-linked Colour Blindness
A woman who is a carrier of red-green colour blindness (XcX) marries an unaffected man (XY). They have four children.
- What is the probability a son will be colour blind?
- Why are sons more likely to express this trait than daughters?
- Each son has a 50% chance of being colour blind (mother passes Xc in half of her eggs; sons receiving Xc and Y are affected).
- Males have only one X chromosome; a single recessive allele on X causes expression. Females need two copies (XcXc) to be affected, so the trait is rarer in females.
Case 6 — Incomplete Dominance
In snapdragons, red (RR) crossed with white (rr) produces pink (Rr) F1 flowers.
- What phenotype would you expect in F2 after selfing the pink F1?
- Explain how this is different from simple dominance.
- F2 genotypes: 1 RR (red) : 2 Rr (pink) : 1 rr (white). Phenotypic ratio: 1 red : 2 pink : 1 white.
- Incomplete dominance: heterozygote shows intermediate phenotype (pink) rather than showing only the dominant trait as in simple dominance.
Case 7 — Pedigree Interpretation
You are shown a pedigree where a trait appears only in males across generations and is transmitted through unaffected females.
- Which mode of inheritance is most likely?
- How would you confirm your hypothesis with family data?
- Likely X-linked recessive inheritance (affected males, females usually carriers).
- Confirm by checking if affected males have carrier mothers and if affected fathers do not pass trait to sons but may pass carrier X to daughters; check ratios consistent with X-linked patterns.
Case 8 — Multiple Alleles & Blood Donation
A child with blood group O (ii) is born to parents with groups A (IAi) and B (IBi).
- Explain how this genotype combination is possible.
- What are possible blood groups of other children from this couple?
- Parents IAi × IBi can produce ii (i from each parent), giving blood group O in child—this follows Mendelian segregation with multiple alleles.
- Possible genotypes: IAIB (AB), IAi (A), IBi (B), ii (O). So children can be A, B, AB, or O.
Case 9 — Linked Genes (Conceptual)
Two traits in certain plants are frequently inherited together and do not show a 9:3:3:1 ratio in dihybrid crosses.
- What genetic phenomenon can explain this observation?
- How might recombination affect this pattern?
- Gene linkage—genes located close together on the same chromosome tend to be inherited together and deviate from independent assortment.
- Recombination (crossing over) during meiosis can separate linked genes at a certain frequency, producing some recombinant offspring; the closer the genes, the lower the recombination frequency.
Case 10 — Environmental Influence
Two genetically identical plants (clones) are grown—one in nutrient-rich soil and one in poor soil. They show different heights.
- Is this variation genetic or environmental? Explain.
- How does this illustrate the genotype–environment interaction?
- This is environmental variation—genetically identical individuals show different phenotype due to environmental conditions (nutrition).
- It shows phenotype results from genotype interacting with environment: same genotype produces different phenotypes under different environmental conditions (phenotypic plasticity).
Case 11 — Carrier Detection
A couple has a child affected by an autosomal recessive disorder. They want to know the risk for their next child.
- What are the likely genotypes of the parents?
- What is the probability the next child will be affected?
- To have an affected child (aa), both parents must at least be carriers (Aa) or affected. If parents are not affected, most likely genotypes are Aa × Aa (both carriers).
- For Aa × Aa, probability of affected child (aa) = 1/4 (25%). There is 50% chance child will be a carrier (Aa) and 25% chance unaffected non-carrier (AA).
Case 12 — Pedigree & Autosomal Dominant Trait
A certain trait appears in every generation of a family and both sexes are affected equally.
- Which inheritance pattern fits this description?
- What would you expect about affected parents' offspring risk?
- Autosomal dominant inheritance is likely—trait appears in each generation and affects both sexes equally.
- If one parent is heterozygous affected (Aa) and other is unaffected (aa), each child has a 50% chance of being affected. If parent is homozygous dominant (AA), all children affected.
Case 13 — Hybrid Vigour in Crops
Two inbred crop lines are crossed producing hybrid offspring with higher yield and disease resistance.
- What is this phenomenon called and why does it occur?
- How is it useful in agriculture?
- Hybrid vigour (heterosis): hybrids show superior traits due to combination of diverse alleles mitigating deleterious recessives and producing favourable gene interactions.
- Useful for producing high-yield, disease-resistant varieties; farmers use hybrids to increase productivity though hybrid seeds may need to be purchased each season (F1 advantage).
Case 14 — Colour of Pea Flowers
A gardener crosses purple-flowered pea plants with white-flowered ones. All F1 purple. On selfing, F2 yields 75% purple and 25% white.
- Which allele is dominant? Explain with genotypes.
- How would you present this cross in the exam (steps)?
- Purple is dominant (P) over white (p). Parental: PP × pp → F1 all Pp (purple). F2 from Pp × Pp gives PP : Pp : pp = 1:2:1 → 3 purple : 1 white.
- In exam: state allele symbols, write parental genotypes, list gametes, draw Punnett square, fill offspring genotypes, give genotypic & phenotypic ratios and conclude.
Case 15 — Mutation and New Trait
A spontaneous mutation in a gene in a plant population confers resistance to a fungus. Over generations, resistant individuals increase.
- Explain the role of mutation in evolution here.
- How does natural selection act on this variation?
- Mutation introduced a new allele providing fungal resistance—new genetic variation that was not present before.
- Natural selection favours resistant individuals (higher survival/reproduction), increasing allele frequency over generations—this is adaptive evolution.
Case 16 — Polygenic Trait Observation
A class measures heights of 200 students and finds a continuous range rather than discrete classes.
- Which type of inheritance explains this pattern?
- Why is environment important for such traits?
- Polygenic inheritance—many genes each contribute small effects producing continuous variation in height.
- Environment (nutrition, health) influences the expression of polygenic traits strongly, so genotype and environment together determine phenotype.
Case 17 — Trait Skipping Generations
In a family, a recessive disorder appears in grandchildren though parents are unaffected.
- Explain how this can occur genetically.
- How could genetic counselling help this family?
- Parents could be carriers (Aa) without showing trait; when both carriers have children, there's a 25% chance of aa (affected) offspring—thus trait can 'skip' a generation.
- Counselling can determine carrier status (tests), estimate recurrence risk, discuss reproductive options and early diagnosis to manage or reduce disease impact.
Case 18 — Test of Independent Assortment
A dihybrid cross doesn't produce expected 9:3:3:1 ratio; instead parental combinations are much more frequent.
- What genetic explanation fits this observation?
- What additional experiment could you do to test linkage?
- Likely gene linkage—genes located close on same chromosome do not assort independently, producing more parental-type offspring.
- Perform test crosses and calculate recombination frequency (cross heterozygote with double recessive); low recombination frequency indicates linkage and map distance between genes can be estimated.
Case 19 — Conservation of Traits in Pure Lines
A seed company maintains pure-breeding lines for a desirable trait across generations.
- What does pure-breeding mean genetically?
- Why are pure lines useful in breeding programs?
- Pure-breeding individuals are homozygous for the trait and produce offspring with the same phenotype generation after generation.
- They provide predictable parental traits for breeding; crossing pure lines helps produce hybrids with desired uniformity and known performance (and F1 heterosis if needed).
Case 20 — Exam-style Application
An examiner asks: "Describe Mendel’s laws and give one example each where they are not strictly followed."
- Provide a concise model answer suitable for CBSE board exam.
Model Answer: Mendel’s first law (Segregation) states that allele pairs segregate during gamete formation so each gamete gets one allele. Example: monohybrid cross of tall (T) and dwarf (t) gives F2 genotypic ratio 1:2:1 and phenotypic 3:1. Mendel’s second law (Independent Assortment) states alleles of different genes assort independently during gamete formation (dihybrid cross gives 9:3:3:1). Exceptions: linked genes (do not assort independently) and co-dominance/incomplete dominance (do not show classic dominant–recessive pattern; e.g., AB blood group and pink snapdragon). Conclude with significance: Mendel’s laws provide a framework but molecular mechanisms and chromosomal behaviour explain exceptions.