Heredity – Long Answer Type Questions
Syllabus mapping → Learning outcomes → Topic-wise long-answer practice → Diagram & problem solving
These 40 long answer type questions with detailed, exam-focused answers are aligned strictly to the NCERT Class 10 Biology Chapter 8: Heredity. Use them for in-depth revision, board exam practice, and classroom assessments.
1. Explain the terms heredity and variation. Give examples and describe their importance in living organisms.
Answer: Heredity is the process by which traits are transmitted from parents to offspring through units of inheritance called genes. For example, eye colour or seed shape may be inherited. Variation refers to differences among individuals of the same species—these may be genetic (different alleles) or environmental (nutrition affecting height). For instance, siblings may inherit different combinations of alleles and therefore show different traits. Heredity ensures continuity of species characteristics, while variation provides the raw material for natural selection and evolution, and is essential for adaptation, selective breeding and survival in changing environments.
2. Define gene and allele. Explain how alleles cause variation in a trait.
Answer: A gene is a specific segment of DNA that codes for a particular trait or protein. Alleles are alternative forms of the same gene that occupy the same position (locus) on homologous chromosomes. Different alleles may code for slightly different versions of the trait (for example, allele R for round seeds and r for wrinkled seeds). When an organism inherits different alleles from its parents (heterozygous condition), the interaction of these alleles (dominance, recessivity, co-dominance, etc.) produces variation in observable traits (phenotype). Thus allelic differences are a primary source of inherited variation.
3. What is genotype and phenotype? Illustrate with an example and explain how environment can influence phenotype.
Answer: Genotype is the genetic constitution of an organism—the specific alleles present (e.g., TT, Tt, tt). Phenotype is the observable characteristic or trait resulting from the genotype and environmental influences (e.g., tall or dwarf). For example, plants with genotype Tt (where T is tall and t is dwarf) have the genotype Tt but the phenotype may be tall if T is dominant. Environment—such as soil quality, temperature or nutrition—can modify how a genotype is expressed; for instance, poor nutrition can limit growth so genetically tall plants may be shorter in poor soil. Therefore phenotype = genotype + environment (interaction).
4. Distinguish between homozygous and heterozygous conditions. Why is the distinction important in inheritance studies?
Answer: Homozygous refers to having two identical alleles for a gene (e.g., AA or aa), while heterozygous means having two different alleles (e.g., Aa). This distinction matters because homozygous individuals breed true—producing offspring with the same trait—whereas heterozygotes may produce offspring with either phenotype depending on allele combinations. Knowing whether an organism is homozygous or heterozygous is crucial for predicting outcomes of crosses (e.g., using test crosses) and understanding how recessive traits can be carried silently through generations until two carriers mate.
5. Explain the role of chromosomes in heredity and how genes are arranged on chromosomes.
Answer: Chromosomes are long DNA-protein complexes that carry many genes arranged linearly along their length. Each chromosome pair (homologous chromosomes) carries the same genes at the same loci but may carry different alleles. During sexual reproduction, chromosomes are halved in gametes (meiosis) and restored at fertilisation; this ensures each offspring receives one chromosome of each homologous pair from each parent, carrying genes that determine traits. Thus chromosomes are the physical carriers of hereditary information and their behaviour during cell division underlies Mendelian segregation and assortment.
6. What is mutation? Discuss its role in producing new variation and give one example.
Answer: A mutation is a change in the DNA sequence of a gene or chromosome. Mutations can create new alleles and thereby introduce genetic variation into a population. Some mutations may be harmful, some neutral, and some beneficial—if beneficial, they may spread through the population by natural selection. For example, a mutation that confers resistance to a disease in a plant or animal can be advantageous. Mutations are thus one of the basic sources of genetic variation necessary for evolution and breeding programmes.
7. Differentiate between continuous and discrete variation with examples.
Answer: Continuous variation shows a range of phenotypes with no clear categories—examples include human height or skin colour, usually controlled by multiple genes (polygenic). Discrete variation shows distinct categories with no intermediates—examples include pea flower colour (purple or white) or seed shape (round or wrinkled) where single genes with alternative alleles produce clear classes. Continuous variation is typically influenced by many genes plus environment, while discrete variation is often due to single genes with different alleles.
8. Why is variation important in agriculture and selective breeding?
Answer: Variation provides the traits that breeders select to improve crops and livestock—such as higher yield, disease resistance or drought tolerance. Without genetic variation, selective breeding cannot produce improved varieties. By crossing individuals with desirable traits and selecting offspring with the best combinations, breeders use heredity and variation to develop improved strains. Maintaining genetic diversity also helps prevent vulnerability to pests and environmental change.
9. Describe in detail Mendel’s experimental approach and why his method was significant.
Answer: Mendel used garden pea plants and selected pure-breeding lines differing in discrete traits (e.g., tall vs dwarf, purple vs white flowers). He performed controlled cross-pollinations, carefully recorded large numbers of progeny across generations (P, F1, F2) and quantified results to derive ratios. His method—controlled crosses, use of pure lines, focus on single traits first, counting offspring and mathematical analysis—was significant because it transformed qualitative observations into predictable, quantitative principles of inheritance. This rigorous, statistical approach allowed Mendel to infer the existence of discrete hereditary units (alleles) and formulate laws that predict offspring ratios.
10. State and explain Mendel’s law of segregation with a monohybrid example and Punnett square.
Answer: Mendel’s law of segregation states that the two alleles for a trait separate (segregate) from each other during gamete formation, so each gamete carries only one allele. Example: Cross RR (round) × rr (wrinkled). F1 are all Rr (round). When F1 (Rr) self-pollinate, gametes are R and r; combining these gives genotypes RR, Rr, Rr, rr with genotypic ratio 1:2:1 and phenotypic ratio 3:1 (round:wrinkled). A Punnett square (R and r across axes) demonstrates these outcomes clearly, illustrating segregation of alleles.
11. Explain Mendel’s law of independent assortment with an example of a dihybrid cross and the expected phenotypic ratio.
Answer: The law of independent assortment states that alleles of different genes assort independently during gamete formation when the genes are on different chromosomes. Example: Consider two traits—seed shape (R/r) and seed colour (Y/y). Crossing RRYY × rryy yields F1 RrYy; selfing F1 produces F2 with phenotypic ratio 9:3:3:1 (9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green) if the genes assort independently. This ratio arises because gametes form with all combinations of alleles (RY, Ry, rY, ry) equally.
12. What were the key reasons Mendel selected particular traits and pea plants for his study?
Answer: Mendel chose traits that displayed clear, alternative forms (discrete variation) to avoid ambiguous intermediate phenotypes and facilitate counting. Pea plants were ideal because they have a short generation time, produce many offspring, can be easily cross- and self-pollinated, and possess several distinct traits on different chromosomes. These features allowed Mendel to perform controlled experiments and obtain statistically meaningful numbers, enabling him to deduce reliable ratios and formulate his laws.
13. Mendel’s experiments were quantitative. Explain what this means and how it helped him generalise inheritance patterns.
Answer: Being quantitative means Mendel counted and recorded the numbers of offspring showing each trait, then analysed the numerical ratios. Instead of anecdotal evidence, his statistical data (e.g., 3:1 ratios) provided reproducible patterns that could be generalised. This numerical approach revealed consistent mathematical relationships underlying inheritance, enabling Mendel to propose laws (segregation and independent assortment) rather than just descriptive observations, and to predict outcomes of future crosses.
14. Discuss limitations and exceptions to Mendel’s laws with brief examples relevant to Class 10 syllabus.
Answer: Mendel’s laws assume genes are on different chromosomes and act independently; exceptions include linked genes (inherited together), incomplete dominance (heterozygote shows intermediate phenotype), co-dominance (both alleles express, e.g., AB blood group), multiple alleles (more than two alleles, e.g., ABO system) and polygenic inheritance (many genes contribute). For instance, genes close together on the same chromosome may not assort independently, and AB blood group demonstrates co-dominance where both A and B antigens are expressed.
15. How would you design a simple experimental outline to demonstrate Mendel’s law of segregation in the classroom?
Answer: Choose a model organism with distinct traits (e.g., fast-growing pea plant or even a visible trait in Drosophila in advanced labs). Use pure-breeding parental strains showing contrasting traits (e.g., round vs wrinkled seeds). Cross-pollinate to obtain F1 and record phenotype. Allow F1 to self-pollinate to obtain F2 and count phenotypes. Demonstrate that F1 show the dominant trait and F2 segregate in ~3:1 ratio. Use Punnett squares to compare observed counts with predicted ratios and discuss any deviations or environmental influences.
16. Why is Mendel often called the 'Father of Genetics'?
Answer: Mendel is called the 'Father of Genetics' because his experiments established the fundamental laws of inheritance—segregation and independent assortment—introducing the concept of discrete hereditary units (later called genes). His systematic, quantitative approach laid the foundation for modern genetics, enabling later scientists to discover chromosomes, DNA structure, and gene function. Mendel’s work converted heredity into a scientific discipline with predictive power.
17. Explain how to perform and present a monohybrid cross (with symbols), show the Punnett square, and compute genotypic and phenotypic ratios.
Answer: Choose allele symbols (e.g., T = tall, t = dwarf). For cross Tt × Tt, list parental genotypes and their gametes (T and t). Draw a 2×2 Punnett square with gametes along top and side. Fill in combinations: TT, Tt, Tt, tt. Genotypic ratio = 1 TT : 2 Tt : 1 tt (1:2:1). Phenotypic ratio (if T is dominant) = 3 tall : 1 dwarf (3:1). Present the work neatly with labels (parents, gametes, offspring genotypes, ratios) for clarity in exams.
18. A tall pea plant (TT) is crossed with a dwarf plant (tt). What are the genotypes and phenotypes of F1 and F2 generations? Explain with ratios.
Answer: Parental cross: TT × tt. Gametes: T and t. F1 genotypes: all Tt (heterozygous) and phenotypically tall. When F1 self-pollinate (Tt × Tt), F2 genotypes: TT, Tt, Tt, tt → genotypic ratio 1:2:1. Phenotypically, tall to dwarf = 3:1 because TT and Tt are tall. Thus F1 all tall (100%), F2 shows both tall and dwarf in 3:1 proportion.
19. How do you perform a test cross to determine the genotype of a dominant phenotype individual? Provide an example and explain the interpretation.
Answer: To determine if an individual with dominant phenotype is homozygous dominant (AA) or heterozygous (Aa), cross it with a homozygous recessive (aa). Example: unknown tall (T?) × tt. If unknown is TT, all offspring will be tall (Tt). If unknown is Tt, half the offspring will be tall (Tt) and half dwarf (tt) — phenotypic ratio 1:1. Thus presence of any recessive phenotype among progeny indicates the tested parent was heterozygous.
20. Explain how a dihybrid cross is set up and why a 9:3:3:1 ratio appears in F2 under independent assortment.
Answer: In a dihybrid cross consider two separately inherited traits, e.g., seed shape (R/r) and seed colour (Y/y). Cross RRYY × rryy → F1 all RrYy. When F1 self-pollinate, gametes produced include RY, Ry, rY, ry in equal proportions. Combining these gametes in a 4×4 Punnett square gives phenotypes in F2 with expected ratio 9 (both dominant) : 3 (dominant for first only) : 3 (dominant for second only) : 1 (both recessive). This 9:3:3:1 arises because alleles for different genes assort independently and combine randomly.
21. A homozygous round yellow pea (RRYY) is crossed with homozygous wrinkled green (rryy). Show F1 and F2 phenotypes and ratios.
Answer: Parental cross RRYY × rryy → F1 all RrYy (round yellow). Selfing F1 (RrYy × RrYy) produces F2 with phenotypes: 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green (9:3:3:1). This demonstrates independent assortment of the two gene pairs, each segregating into gametes independently to form combinations in F2.
22. How do you compute expected numerical counts from phenotypic ratios? Give a worked example using 320 F2 offspring and a 3:1 ratio.
Answer: For a 3:1 ratio (total 4 parts) and 320 offspring, calculate each part as 320 ÷ 4 = 80. Thus expected dominant phenotype = 3 × 80 = 240; recessive = 1 × 80 = 80. So out of 320, expect 240 dominant and 80 recessive. This method scales ratios to actual counts and helps compare observed vs expected values for statistical checks.
23. Describe how you would document and present genetic cross answers in a board exam to gain maximum marks.
Answer: Start by stating the alleles with clear symbols (uppercase for dominant), write parental genotypes, list gametes for each parent, draw a labelled Punnett square, fill in offspring genotypes, derive genotypic and phenotypic ratios, and state final conclusion in words (e.g., "3:1 phenotypic ratio"). Neat layout, correct symbols and brief explanation of dominance or segregation principles demonstrate understanding and maximise marks.
24. Explain the significance of the 1:2:1 genotypic ratio and how it relates to phenotype in monohybrid crosses.
Answer: The 1:2:1 genotypic ratio (TT : Tt : tt) in F2 of a monohybrid cross indicates genetic diversity among offspring—one homozygous dominant, two heterozygotes, and one homozygous recessive. Phenotypically, if the trait shows simple dominance, TT and Tt display the dominant phenotype while tt shows the recessive, producing a 3:1 phenotypic ratio. The 1:2:1 ratio also highlights that heterozygotes carry recessive alleles that may appear in later generations.
25. Explain co-dominance and incomplete dominance with one clear example each.
Answer: Co-dominance occurs where both alleles in a heterozygote express fully and simultaneously—for example human blood group AB where IA and IB alleles both express, producing AB phenotype. Incomplete dominance occurs when heterozygotes exhibit an intermediate phenotype between the two homozygotes—for example red (RR) × white (rr) flowers producing pink (Rr). Both illustrate departures from simple dominant-recessive patterns and show that allele interactions determine phenotype in different ways.
26. Describe the ABO blood group system as an example of multiple alleles and co-dominance.
Answer: The ABO system is governed by three alleles—IA, IB and i—at a single gene locus. IA and IB are co-dominant: individuals with genotype IAIB express both A and B antigens (blood group AB). The i allele is recessive: ii results in blood group O. Genotypes IAIA or IAi give blood group A, IBIB or IBi give B. Thus ABO demonstrates multiple alleles within a population and co-dominant expression of certain allele pairs.
27. What is polygenic inheritance? Give two human traits that show polygenic inheritance and explain why they produce continuous variation.
Answer: Polygenic inheritance involves multiple genes contributing additively to a single trait. Examples in humans include height and skin colour. Because many genes each contribute small effects, the combined influence produces a continuous range of phenotypes rather than discrete categories—the result is a bell-shaped distribution of trait values in the population. Environmental factors further modify expression, enhancing the continuous nature of variation.
28. How are single-gene disorders and chromosomal abnormalities different? Give one example of each.
Answer: Single-gene disorders are caused by mutations in a single gene and are inherited according to Mendelian patterns (e.g., X-linked colour blindness). Chromosomal abnormalities involve changes in chromosome number or large structural changes (e.g., Down syndrome caused by trisomy 21, an extra copy of chromosome 21). Single-gene disorders affect protein function or regulation, while chromosomal abnormalities change gene dosage across many genes and often have broader developmental impacts.
29. Discuss the genetic basis of colour blindness and explain why it is more common in males.
Answer: Red-green colour blindness is typically an X-linked recessive condition caused by mutations affecting photopigments on the X chromosome. Males (XY) have only one X chromosome, so a single mutant allele on their X will cause the disorder. Females (XX) require two copies of the mutant allele to express the trait; heterozygous females are usually carriers but phenotypically normal. Hence the trait is more common in males due to their hemizygous state for X-linked genes.
30. Explain how genetic counselling can help families with risk of inherited disorders. Mention two services provided during counselling.
Answer: Genetic counselling assesses the risk of inherited disorders in a family, interprets genetic test results, and explains inheritance patterns and options to prospective parents. Services include risk estimation (using pedigree analysis and genetic tests) and advice on reproductive options (prenatal diagnosis, carrier screening, assisted reproductive technologies) as well as psychological support and information about management and prevention. It helps families make informed decisions about health and reproduction.
31. Describe chromosomal mechanism of sex determination in humans and explain why sex ratio at fertilisation is approximately 1:1.
Answer: In humans, females are XX and males XY. Females produce eggs carrying only X chromosomes; males produce sperm carrying either X or Y. At fertilisation, an X-bearing sperm produces an XX zygote (female) while a Y-bearing sperm produces an XY zygote (male). Because males produce roughly equal numbers of X- and Y-bearing sperm, the probability of XX or XY zygote is about equal, leading to an approximate 1:1 sex ratio at conception. Environmental and demographic factors can slightly alter the ratio at birth.
32. What does 'heterogametic' and 'homogametic' mean? Which is the heterogametic sex in humans and why?
Answer: 'Heterogametic' refers to the sex that produces two types of gametes with respect to sex chromosomes (e.g., X and Y), while 'homogametic' produces one type (e.g., X only). In humans, males are heterogametic (XY) because their sperm can carry either X or Y; females are homogametic (XX) because eggs carry only X. The heterogametic sex typically determines the sex of the offspring.
33. Explain a cross between a carrier mother of an X-linked recessive trait and an unaffected father. Predict the possible genotypes and phenotypes of children.
Answer: Let Xc be mutant allele (recessive), XC normal. Mother is carrier (XCXc), father is unaffected (XCY). Possible children: daughters: X C X C (unaffected), XC Xc (carrier)—50% daughters carriers, none affected; sons: X C Y (unaffected) and Xc Y (affected)—50% sons affected. Thus daughters are typically carriers without phenotype, while sons have a 50% chance of being affected because they receive the single X from mother.
34. Discuss inheritance pattern and societal implications of sex-linked disorders with an example.
Answer: Sex-linked (often X-linked) disorders show different inheritance patterns in males and females; males often express recessive X-linked traits because they have only one X. Example: haemophilia—affected males bleed excessively due to clotting factor deficiency. Societal implications include the need for carrier screening, genetic counselling, awareness of gendered disease prevalence, and ethical considerations around testing and reproductive choices. Management includes medical care and informed reproductive decision-making to reduce disease burden.
35. How can pedigrees be used to trace inheritance of traits in families? Provide a brief description of symbols and interpretation.
Answer: Pedigrees are family trees showing relationships and presence/absence of a trait across generations. Standard symbols: circles = females, squares = males, shaded symbols = affected, half-shaded = carriers, horizontal line = mating, vertical line = offspring. By analysing patterns (e.g., trait appears only in males → likely X-linked recessive; trait appears every generation → likely dominant), geneticists and counsellors can infer mode of inheritance and estimate carrier status or risk for future children.
36. Explain why some traits skip generations while others do not, using genetic terminology.
Answer: Traits skip generations often because they are recessive—carriers (heterozygotes) do not show the phenotype but can pass the recessive allele; when two carriers mate, the recessive phenotype appears in offspring. Dominant traits typically appear in every generation because a single copy of the dominant allele produces the phenotype. Also sex-linked patterns can cause apparent skipping or sex-specific expression (e.g., X-linked recessive traits may appear in males and seem to skip female carriers).
37. Describe how to draw and label a neat Punnett square diagram for a monohybrid cross in an exam and what to include in the answer text.
Answer: Draw a clear 2×2 square, label parental genotypes above and to the left, list gametes (single letters) on respective margins, fill in four offspring genotype cells, and calculate genotypic and phenotypic ratios. In the answer text, briefly state parental genotypes, explain gamete formation, show ratios and conclude (e.g., "25% homozygous dominant, 50% heterozygous, 25% homozygous recessive; phenotype 3:1 dominant:recessive"). Neat labels and short explanation gain marks.
38. Provide a worked long-answer example: Cross a homozygous dominant (AA) plant with a heterozygous (Aa) plant and predict offspring genotypes and phenotypes. Explain steps.
Answer: Parental genotypes: AA × Aa. Gametes: parent1 → A, parent2 → A or a. Punnett square: cells = AA and Aa (two A from parent1 with A and a from parent2). Offspring genotypes: 50% AA, 50% Aa (genotypic ratio 1:1). Phenotypically, if A is dominant, all offspring show dominant trait (100% dominant phenotype) though half are heterozygous carriers. Steps: state parents, list gametes, draw square, derive ratios and conclude.
39. Summarise the chapter 'Heredity' in a way that helps a student answer a 10-mark board question efficiently.
Answer: For a 10-mark answer, begin with definitions (heredity, variation), mention genes and alleles, describe Mendel’s approach and two laws with clear examples (monohybrid and dihybrid crosses), include a labelled Punnett square example and state ratios, briefly discuss exceptions (co-dominance, incomplete dominance, multiple alleles, polygenic traits), explain sex determination and one example of human heredity (blood groups or colour blindness), and conclude on the importance of heredity and variation. Use headings, diagrams and concise numbered steps to enhance clarity and marks.
40. Recommend an effective revision plan and exam strategy for this chapter to score well in CBSE Class 10 board exams.
Answer: Revision plan: review definitions and key terms daily, practice 10–15 Punnett squares (monohybrid and dihybrid), memorise Mendel’s laws with examples, solve NCERT and previous-year long-answer questions, and attempt timed mock answers. Exam strategy: read question carefully (identify if diagram/calculation required), allocate time (10–12 minutes for 5-mark answers, 20–25 minutes for 10-mark), label diagrams clearly, show all working steps, write concise conclusions and underline final ratios or results. Regular practice and neat presentation improve scoring.