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Principles of Inheritance and Variation

NCERT Class 12 · Biology Based on NCERT Class 12 Biology textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter introduces the fundamental principles of genetics as discovered by Gregor Mendel, explaining how traits are inherited from parents to offspring. It covers Mendelian laws, the chromosomal basis of inheritance, sex determination, mutations, and genetic disorders. The one big idea a student must walk away with is that inheritance follows predictable patterns governed by discrete units (genes) that segregate and assort independently during gamete formation, leading to both stability and variation in living organisms.
Key Concepts (30)
Genetics
The branch of biology that deals with the inheritance, as well as the variation of characters from parents to offspring.
e.g. Studying why siblings look similar or different from each other and from their parents.
Inheritance
The process by which characters are passed on from parent to progeny; it is the basis of heredity.
e.g. A child inheriting eye colour from their parents.
Variation
The degree by which progeny differ from their parents.
e.g. Siblings having different heights or hair colours.
True-breeding line
A line that, having undergone continuous self-pollination, shows the stable trait inheritance and expression for several generations.
e.g. Mendel's pure tall pea plants that always produced tall offspring when self-pollinated.
Alleles
Genes which code for a pair of contrasting traits; they are slightly different forms of the same gene.
e.g. T (tall) and t (dwarf) are alleles of the gene for plant height in pea.
Genotype
The allelic composition of an organism for a particular trait.
e.g. TT, Tt, or tt for height in pea plants.
Phenotype
The observable or expressed characteristic of an organism.
e.g. Tall or dwarf appearance of a pea plant.
Dominant factor
In a pair of dissimilar factors, the one that dominates (is expressed) in the F1 generation.
e.g. T (tall) is dominant over t (dwarf) in pea plants.
Recessive factor
The factor that remains unexpressed in the F1 generation but reappears in the F2 generation.
e.g. t (dwarf) is recessive to T (tall) in pea plants.
Monohybrid cross
A cross between two individuals that differ in one pair of contrasting traits.
e.g. Cross between a tall (TT) and a dwarf (tt) pea plant.
Dihybrid cross
A cross between two individuals that differ in two pairs of contrasting traits.
e.g. Cross between a pea plant with yellow round seeds (RRYY) and one with green wrinkled seeds (rryy).
Punnett Square
A graphical representation developed by Reginald C. Punnett to calculate the probability of all possible genotypes of offspring in a genetic cross.
e.g. A 2x2 grid used to predict offspring from a monohybrid cross.
Test cross
A cross in which an organism showing a dominant phenotype (whose genotype is to be determined) is crossed with the recessive parent.
e.g. Crossing a tall F2 pea plant (unknown genotype TT or Tt) with a dwarf (tt) plant.
Incomplete dominance
A phenomenon where the F1 generation has a phenotype that is in between the two parental phenotypes.
e.g. Cross between red-flowered (RR) and white-flowered (rr) snapdragon produces pink-flowered (Rr) F1.
Co-dominance
A condition in which the F1 generation resembles both parents, and both alleles are expressed simultaneously.
e.g. ABO blood group: when IA and IB are present together, both A and B sugars are expressed on red blood cells.
Multiple alleles
When more than two alleles govern the same character in a population.
e.g. The three alleles IA, IB, and i for ABO blood grouping in humans.
Linkage
The physical association of genes on a chromosome, leading to their tendency to be inherited together.
e.g. Genes for white eye and yellow body in Drosophila are linked on the X chromosome.
Recombination
The generation of non-parental gene combinations due to crossing over during meiosis.
e.g. Offspring with new combinations of traits not seen in either parent.
Polygenic inheritance
Inheritance controlled by three or more genes, where each allele has an additive effect on the phenotype.
e.g. Human skin colour, controlled by multiple genes like A, B, C.
Pleiotropy
A phenomenon where a single gene exhibits multiple phenotypic expressions.
e.g. Phenylketonuria (PKU) in humans: a single gene mutation causes mental retardation and reduced skin/hair pigmentation.
Sex determination
The genetic or chromosomal mechanism that determines the sex of an individual.
e.g. In humans, XX females and XY males.
Male heterogamety
A sex determination mechanism where males produce two different types of gametes (e.g., X and Y).
e.g. In humans and Drosophila, males produce X- and Y-bearing sperm.
Female heterogamety
A sex determination mechanism where females produce two different types of gametes (e.g., Z and W).
e.g. In birds, females have ZW chromosomes and males have ZZ.
Haplodiploid sex-determination
A system where sex is determined by the number of sets of chromosomes; females are diploid and males are haploid.
e.g. In honey bees, fertilised eggs develop into females (queen/worker) and unfertilised eggs develop into males (drones).
Mutation
A phenomenon that results in alteration of DNA sequences, leading to changes in genotype and phenotype.
e.g. Sickle cell anaemia caused by a point mutation in the beta-globin gene.
Mutagen
Chemical or physical factors that induce mutations.
e.g. UV radiation is a mutagen.
Pedigree analysis
The study of the inheritance of a particular trait in several generations of a family, represented in a family tree.
e.g. Tracing the inheritance of haemophilia in the royal families of Europe.
Mendelian disorders
Genetic disorders mainly determined by alteration or mutation in a single gene, transmitted according to Mendelian principles.
e.g. Haemophilia, sickle cell anaemia, colour blindness, phenylketonuria, thalassemia.
Chromosomal disorders
Genetic disorders caused by alterations in chromosome number or structure.
e.g. Down syndrome (trisomy 21).
X-linked recessive trait
A recessive trait whose gene is located on the X chromosome, more commonly expressed in males.
e.g. Colour blindness: a son of a carrier mother has a 50% chance of being colour blind.
Formulas & Equations (5)
Monohybrid cross phenotypic ratio: 3:1
Reads as: Three dominant to one recessive
Variables: 3 = proportion of dominant phenotype; 1 = proportion of recessive phenotype
Use when: Predicting offspring phenotypes from a cross between two heterozygotes (Tt x Tt) for a single trait with complete dominance.
⚠ Writing the ratio as 1:2:1 (which is the genotypic ratio) instead of 3:1 for phenotypes.
Monohybrid cross genotypic ratio: 1:2:1
Reads as: One homozygous dominant, two heterozygous, one homozygous recessive
Variables: 1 = TT; 2 = Tt; 1 = tt
Use when: Determining the genotypic composition of offspring from a monohybrid cross (Tt x Tt).
⚠ Confusing this with the phenotypic ratio; forgetting that TT and Tt both show the dominant phenotype.
Dihybrid cross phenotypic ratio: 9:3:3:1
Reads as: Nine double dominant, three dominant-recessive, three recessive-dominant, one double recessive
Variables: 9 = Round Yellow; 3 = Wrinkled Yellow; 3 = Round Green; 1 = Wrinkled Green
Use when: Predicting offspring from a cross between two double heterozygotes (RrYy x RrYy) for two independent traits.
⚠ Forgetting that the 9:3:3:1 ratio only applies when genes are on different chromosomes (independent assortment).
Binomial expansion for monohybrid cross: (1/2T + 1/2t)^2 = 1/4 TT + 1/2 Tt + 1/4 tt
Reads as: One half T plus one half t, squared, equals one quarter TT plus one half Tt plus one quarter tt
Variables: 1/2 = frequency of each allele in gametes; TT, Tt, tt = zygote genotypes
Use when: Mathematically deriving genotypic ratios from gamete frequencies in a monohybrid cross.
⚠ Not squaring the expression correctly or misinterpreting the coefficients as percentages.
Recombination frequency = (Number of recombinant progeny / Total progeny) x 100
Reads as: Recombination frequency equals number of recombinant progeny divided by total progeny, multiplied by 100
Variables: Recombinant progeny = offspring with non-parental gene combinations; Total progeny = total number of offspring
Use when: Calculating the distance between linked genes on a chromosome (1% recombination = 1 map unit).
⚠ Using parental progeny instead of recombinant progeny in the numerator.
How It Works
Monohybrid Cross (Mendel's experiment with tall and dwarf pea plants)
  1. Step 1: Cross true-breeding tall (TT) and true-breeding dwarf (tt) parent plants.
  2. Step 2: Collect seeds and grow F1 generation — all plants are tall (Tt).
  3. Step 3: Self-pollinate the F1 tall plants.
  4. Step 4: Observe F2 generation — 3/4 tall and 1/4 dwarf (phenotypic ratio 3:1).
  5. Step 5: Genotypic ratio in F2 is 1 TT : 2 Tt : 1 tt.
💡 The recessive trait (dwarf) disappears in F1 but reappears unchanged in F2, showing no blending of traits.
Test Cross
  1. Step 1: Take an organism showing a dominant phenotype (e.g., tall plant) whose genotype (TT or Tt) is unknown.
  2. Step 2: Cross it with a homozygous recessive parent (e.g., dwarf tt).
  3. Step 3: Analyse the offspring phenotypes.
  4. Step 4: If all offspring are tall, the unknown parent is homozygous dominant (TT).
  5. Step 5: If half the offspring are tall and half are dwarf, the unknown parent is heterozygous (Tt).
💡 A test cross always uses the recessive parent, not self-crossing.
Dihybrid Cross (Mendel's experiment with seed colour and shape)
  1. Step 1: Cross true-breeding yellow round (RRYY) and green wrinkled (rryy) parents.
  2. Step 2: F1 generation all yellow round (RrYy).
  3. Step 3: Self-pollinate F1 plants (RrYy x RrYy).
  4. Step 4: Gametes produced: RY, Ry, rY, ry in equal proportion (1/4 each).
  5. Step 5: Use Punnett square (16 squares) to find F2 combinations.
  6. Step 6: Phenotypic ratio: 9 Round Yellow : 3 Wrinkled Yellow : 3 Round Green : 1 Wrinkled Green.
💡 The 9:3:3:1 ratio is derived from (3:1) x (3:1) and only holds when genes are on different chromosomes.
Sex Determination in Humans (XY type)
  1. Step 1: Females have 22 pairs of autosomes + XX; males have 22 pairs of autosomes + XY.
  2. Step 2: During spermatogenesis, males produce two types of sperm: 50% with X and 50% with Y.
  3. Step 3: Females produce only one type of ovum: all with X chromosome.
  4. Step 4: Fertilisation of ovum with X-bearing sperm results in female (XX).
  5. Step 5: Fertilisation of ovum with Y-bearing sperm results in male (XY).
  6. Step 6: There is always a 50% probability of a male or female child in each pregnancy.
💡 It is the genetic makeup of the sperm that determines the sex of the child, not the egg.
Sex Determination in Honey Bee (Haplodiploid system)
  1. Step 1: Fertilised eggs (union of sperm and egg) develop into females (queen or worker) — diploid (32 chromosomes).
  2. Step 2: Unfertilised eggs develop into males (drones) by parthenogenesis — haploid (16 chromosomes).
  3. Step 3: Males produce sperms by mitosis (not meiosis).
  4. Step 4: Males have no father and cannot have sons, but can have a grandfather and grandsons.
💡 Males are haploid and produce sperm by mitosis, which is unique to this system.
Pedigree Analysis
  1. Step 1: Collect family history data about the inheritance of a particular trait over several generations.
  2. Step 2: Represent the data using standard symbols (e.g., squares for males, circles for females, shaded for affected).
  3. Step 3: Analyse the pattern to determine if the trait is dominant or recessive.
  4. Step 4: Determine if the trait is autosomal or sex-linked (e.g., X-linked recessive).
  5. Step 5: Predict the probability of inheritance in future offspring.
💡 Pedigree analysis is the only ethical way to study human genetics since controlled crosses are not possible.
Exam Traps — Don't Make These Mistakes
⚠ Confusing phenotypic and genotypic ratios in monohybrid cross
✗ The F2 ratio in a monohybrid cross is 1:2:1.
→ 1:2:1 is the genotypic ratio (TT:Tt:tt), not the phenotypic ratio. The phenotypic ratio is 3:1 (tall:dwarf) when dominance is complete.
✓ In a monohybrid cross with complete dominance, the F2 phenotypic ratio is 3:1 and the genotypic ratio is 1:2:1.
⚠ Using T and d for alleles instead of T and t
✗ Tall = T, dwarf = d.
→ Using different letters (T and d) makes it difficult to remember that they are alleles of the same gene. The correct notation uses the same letter in capital and lowercase (T and t).
✓ Use T for tall (dominant) and t for dwarf (recessive) — both are alleles of the same gene.
⚠ Assuming 9:3:3:1 ratio always applies in dihybrid crosses
✗ All dihybrid crosses give a 9:3:3:1 ratio in F2.
→ The 9:3:3:1 ratio only applies when the two genes are on different chromosomes (independent assortment). If genes are linked on the same chromosome, the ratio deviates significantly from 9:3:3:1.
✓ The 9:3:3:1 ratio is observed only when the two genes assort independently, i.e., they are located on different chromosomes.
⚠ Thinking incomplete dominance and co-dominance are the same
✗ In incomplete dominance, both alleles are expressed equally.
→ In incomplete dominance, the F1 phenotype is intermediate (e.g., pink from red and white). In co-dominance, both alleles are fully expressed simultaneously (e.g., AB blood group).
✓ In incomplete dominance, the phenotype is a blend (intermediate); in co-dominance, both parental phenotypes appear together.
⚠ Blaming the mother for the sex of the child
✗ The mother determines the sex of the child because she provides the egg.
→ The mother always provides an X chromosome. The father's sperm determines the sex: X-bearing sperm gives a female (XX), Y-bearing sperm gives a male (XY).
✓ The sex of the child is determined by the type of sperm (X or Y) that fertilises the egg. The mother contributes only X chromosomes.
⚠ Confusing linkage with independent assortment
✗ Linked genes assort independently and produce a 9:3:3:1 ratio.
→ Linked genes are physically associated on the same chromosome and do not assort independently. They produce a higher proportion of parental combinations and fewer recombinants.
✓ Linked genes are located on the same chromosome and tend to be inherited together, leading to deviation from the 9:3:3:1 ratio.
Compare & Contrast
Expression of alleles in heterozygote
Incomplete dominance
Co-dominance
In incomplete dominance, the heterozygote shows an intermediate phenotype (e.g., pink flowers from red and white parents). In co-dominance, both alleles are fully expressed simultaneously (e.g., AB blood group has both A and B antigens).
Which sex produces two types of gametes
Male heterogamety (XY type)
Female heterogamety (ZW type)
In male heterogamety (humans, Drosophila), males produce X and Y sperm; females are XX (homogametic). In female heterogamety (birds), females produce Z and W eggs; males are ZZ (homogametic).
Cause of the disorder
Mendelian disorders
Chromosomal disorders
Mendelian disorders are caused by mutation in a single gene (e.g., sickle cell anaemia, haemophilia). Chromosomal disorders are caused by alterations in chromosome number or structure (e.g., Down syndrome).
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total approximately 12 marks
Most asked: Mendel's laws of inheritance (Law of Dominance and Law of Segregation) · Monohybrid and dihybrid cross ratios (3:1, 1:2:1, 9:3:3:1) · Test cross and its significance · Incomplete dominance and co-dominance (with examples like snapdragon and ABO blood group) · Sex determination in humans and honey bees · Pedigree analysis and genetic disorders (haemophilia, colour blindness, sickle cell anaemia) · Linkage and recombination
Don't skip: Explain Mendel's Law of Segregation with a monohybrid cross (5 marks); Dihybrid cross and the Law of Independent Assortment (5 marks); Sex determination in humans — why the sperm determines the sex (3 marks); Pedigree analysis of an X-linked recessive trait like haemophilia (3 marks); Difference between incomplete dominance and co-dominance with examples (3 marks)
One-Line Revision (read aloud before the exam)
  1. Genetics is the branch of biology dealing with inheritance and variation of characters from parents to offspring.
  2. Mendel's Law of Dominance states that in a dissimilar pair of factors, one dominates the other.
  3. Mendel's Law of Segregation states that alleles segregate during gamete formation so that each gamete receives only one allele.
  4. In a monohybrid cross, the F2 phenotypic ratio is 3:1 and genotypic ratio is 1:2:1.
  5. A test cross involves crossing a dominant phenotype individual with a homozygous recessive parent to determine its genotype.
  6. Incomplete dominance produces an intermediate phenotype in F1, as seen in snapdragon flower colour (RR x rr = Rr pink).
  7. Co-dominance results in both alleles being expressed, as in ABO blood group where IA and IB produce both A and B antigens.
  8. ABO blood groups are an example of multiple alleles (IA, IB, i) and co-dominance.
  9. The Law of Independent Assortment states that when two pairs of traits are combined, segregation of one pair is independent of the other.
  10. A dihybrid cross between RrYy parents gives a 9:3:3:1 phenotypic ratio.
  11. Linkage is the physical association of genes on the same chromosome, leading to higher parental combinations.
  12. Recombination frequency is used to map genes on chromosomes (1% recombination = 1 map unit).
  13. In humans, sex is determined by the sperm: X-bearing sperm gives a female (XX), Y-bearing sperm gives a male (XY).
  14. In honey bees, females are diploid (from fertilised eggs) and males are haploid (from unfertilised eggs by parthenogenesis).
  15. Mutation is an alteration in DNA sequence; point mutation causes sickle cell anaemia.
  16. Pedigree analysis traces inheritance of traits in families using standard symbols.
  17. Mendelian disorders like haemophilia and colour blindness are X-linked recessive traits.
  18. Colour blindness affects about 8% of males and 0.4% of females due to X-linked recessive inheritance.
  19. Pleiotropy occurs when a single gene affects multiple phenotypes, e.g., phenylketonuria causes mental retardation and reduced pigmentation.
  20. Polygenic inheritance involves multiple genes with additive effects, e.g., human skin colour.

MCQs — 10 Questions with Answers

Q1. A true-breeding tall pea plant is crossed with a true-breeding dwarf plant. What will be the phenotype of all F1 offspring?

  • A. All tall ✓
  • B. All dwarf
  • C. Half tall, half dwarf
  • D. All intermediate height

Answer: A — In the F1 generation, all offspring are heterozygous (Tt) and show only the dominant tall phenotype; the recessive dwarf trait is masked.

Q2. When F1 plants from a monohybrid cross (Tt) are self-pollinated, the F2 generation shows a 3:1 phenotypic ratio. What is the underlying genotypic ratio?

  • A. 3:1
  • B. 1:2:1 ✓
  • C. 1:1
  • D. 2:1

Answer: B — Self-pollination of Tt × Tt produces 1 TT : 2 Tt : 1 tt genotypic ratio, where TT and Tt both show the dominant phenotype, giving 3:1 phenotypic ratio.

Q3. Why did Mendel's observation of contrasting traits in F2 (tall and dwarf appearing together) disprove the blending hypothesis of inheritance?

  • A. Because blending would predict intermediate heights, not discrete tall or dwarf plants ✓
  • B. Because the dwarf trait was lost in F1 and could not reappear in F2
  • C. Because true-breeding lines always produce identical offspring
  • D. Because statistical analysis showed random variation

Answer: A — If traits blended, F2 would show a range of intermediate heights; instead, Mendel observed discrete tall and dwarf plants, proving traits remain as separate units.

Q4. Mendel selected 14 true-breeding pea varieties representing 7 contrasting trait pairs. Which of the following is NOT a contrasting pair he studied?

  • A. Smooth or wrinkled seeds
  • B. Yellow or green seeds
  • C. Tall or dwarf plants
  • D. Red or white flower petals ✓

Answer: D — The seven contrasting pairs Mendel studied were: stem height, flower colour (violet/white), flower position, pod shape, pod colour, seed shape, and seed colour; red/white petals were not among them.

Q5. In Mendel's tall × dwarf cross, the dwarf trait disappears in F1 but reappears in F2 in a 1:3 ratio. Which statement correctly explains this observation?

  • A. The dwarf allele is dominant and present in all F1 plants
  • B. The dwarf allele is recessive, hidden in heterozygous F1, and segregates in F2 from heterozygous parents ✓
  • C. The dwarf trait is lost during F1 and newly appears by mutation in F2
  • D. Environmental factors prevent dwarf expression in F1 only

Answer: B — F1 plants (Tt) carry the recessive dwarf allele but do not express it; when F1 plants self-pollinate, segregation produces 1/4 homozygous recessive (tt) plants with the dwarf phenotype.

Q6. Mendel's use of statistical analysis and large sample sizes in his pea breeding experiments was significant because it:

  • A. Proved that inheritance follows mathematical patterns, not random chance ✓
  • B. Eliminated the need for multiple generations of crosses
  • C. Showed that all plant species follow the same inheritance laws
  • D. Replaced the need for controlled artificial pollination

Answer: A — Large sample sizes and statistical analysis gave credibility to Mendel's data and proved his results reflected general laws of inheritance rather than unsubstantiated observations or random variation.

Q7. Consider a cross between a true-breeding yellow-seeded pea plant (YY) and a true-breeding green-seeded plant (yy). If the F1 plants are self-pollinated, which statement is correct about the F2 generation?

  • A. All F2 plants will produce yellow seeds only
  • B. The F2 will show 3 yellow : 1 green seed ratio ✓
  • C. The F2 will show equal numbers of yellow and green seeds
  • D. No green seeds will appear in F2 because the trait is lost

Answer: B — F1 (Yy) self-pollination produces F2 with genotype 1 YY : 2 Yy : 1 yy, giving a 3:1 phenotypic ratio of yellow to green seeds.

Q8. Which of the following characteristics made garden pea an ideal model organism for Mendel's inheritance studies? (i) Distinct contrasting traits with no intermediate forms, (ii) Ability to self-pollinate naturally, (iii) Short generation time, (iv) Availability of true-breeding varieties.

  • A. Only (i) and (ii)
  • B. Only (ii) and (iii)
  • C. (i), (ii), (iii), and (iv) ✓
  • D. Only (iii) and (iv)

Answer: C — All four features made pea ideal: distinct traits allowed clear observation, natural self-pollination enabled controlled crosses, rapid generations allowed multi-generational studies, and available pure lines provided starting material.

Q9. A monohybrid cross produces an F1 generation where 100% of offspring show the dominant phenotype. When F1 plants are self-pollinated, the F2 generation shows 75% dominant and 25% recessive phenotypes. This 3:1 ratio supports which of Mendel's principles?

  • A. The principle that dominant traits always persist in all generations
  • B. The principle of segregation, where allele pairs separate during reproduction ✓
  • C. The principle that recessive traits are always eliminated in F2
  • D. The principle that traits blend during inheritance

Answer: B — The 3:1 ratio in F2 directly demonstrates segregation: alleles separate during gamete formation in F1, producing a 1:1 gamete ratio that recombines to give 1:2:1 genotypic and 3:1 phenotypic ratios.

Q10. [HOTS] Mendel observed that when tall F1 plants (Tt) from a cross between TT and tt were self-pollinated, they produced both tall and dwarf offspring in an exact 3:1 ratio, not a range of intermediate heights. What does this observation reveal about the nature of genetic factors compared to earlier theories of inheritance? Explain how Mendel's findings contradicted the blending hypothesis.

  • A. Genetic factors are discrete units that do not blend; blending would predict intermediate offspring ✓
  • B. Genetic factors merge permanently and cannot be separated in F2
  • C. Tall and dwarf are the only possible phenotypes because intermediate forms cannot exist
  • D. Environmental factors determine whether traits blend or remain discrete

Answer: A — Mendel's 3:1 ratio with no intermediates proves genetic factors are discrete, independent units; the blending hypothesis would predict a continuous range of heights in F2, which was NOT observed.

Flashcards

What is a true-breeding line in Mendel's experiments?

A plant line that has undergone continuous self-pollination and shows stable trait inheritance for several generations without variation.

Why did all F1 plants from a tall × dwarf cross appear tall?

Because the tall trait is dominant and masks the recessive dwarf trait in heterozygous F1 plants.

What is the phenotypic ratio observed in the F2 generation of a monohybrid cross?

The F2 generation shows a 3:1 ratio (three dominant : one recessive phenotype).

What key observation showed that traits do NOT blend in pea plants?

F2 offspring were either tall or dwarf with no intermediate heights, proving traits remain discrete units.

Why did Mendel choose garden pea plants for his inheritance experiments?

Pea plants had easily distinguishable contrasting traits, were easy to cross-pollinate artificially, and produced many offspring quickly.

How many contrasting trait pairs did Mendel select for his initial studies?

Mendel selected 14 true-breeding pea plant varieties as 7 pairs with contrasting traits.

What does the disappearance of the dwarf trait in F1 and its reappearance in F2 suggest?

The recessive allele for the dwarf trait is present but hidden in F1 heterozygotes and segregates in F2.

Why was statistical analysis and large sample size important in Mendel's work?

Large sample sizes and mathematical logic gave credibility to data and proved results reflected general rules, not chance observations.

What is meant by inheritance of one gene in the context of Mendel's monohybrid cross?

A cross tracking the transmission of a single contrasting trait (one gene with two alleles) from parents through F1 and F2 generations.

What key inference can be drawn from Mendel's observation that F1 always resembled one parent?

One allele is dominant and masks the expression of the recessive allele in heterozygous individuals.

Important Board Questions

Define the terms 'inheritance' and 'variation'. Give one example of each from living organisms. [2 marks]

Inheritance = passage of characters from parents to offspring (heredity basis). Variation = difference in offspring traits from parents. Example: Sahiwal cows inherit milk-producing trait; offspring may vary in coat colour or height.

Mendel conducted a monohybrid cross between tall (TT) and dwarf (tt) pea plants. Describe the F1 and F2 generations, explaining why the dwarf trait disappears in F1 but reappears in F2. Draw a Punnett square for the F2 generation and label the genotypes and phenotypes. [5 marks]

F1: All Tt (tall) because tall is dominant. F2: From Tt × Tt self-pollination using Punnett square → 1 TT : 2 Tt : 1 tt genotypes → 3 tall : 1 dwarf phenotypes. Segregation of alleles during F1 gamete formation causes reappearance.

Explain why Mendel chose garden pea plants for his inheritance experiments and why his use of statistical analysis and large sample sizes was crucial for establishing the laws of inheritance. How do Mendel's findings contradict the blending hypothesis of inheritance? [6 marks]

Pea advantages: distinct contrasting traits, artificial pollination control, true-breeding varieties available, rapid generations, large offspring numbers. Statistics proved patterns were laws, not chance. Blending predicts intermediate F2; Mendel observed discrete 3:1 ratio with no intermediates, proving traits remain as separate units that segregate.

Assertion-Reason Questions for Principles of Inheritance and Variation

Q1. Assertion (A):

In Mendel's monohybrid cross between tall (TT) and dwarf (tt) plants, the F1 generation exhibits only the tall phenotype.

Reason (R):

The tall allele is dominant over the dwarf allele, and the F1 plants are heterozygous (Tt), expressing only the dominant phenotype.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true, and the reason correctly explains why only the tall phenotype appears in F1. Dominance of the tall allele in heterozygous F1 plants masks the recessive dwarf allele's expression.

Q2. Assertion (A):

The appearance of dwarf plants in the F2 generation (absent in F1) proves that the dwarf trait was not lost but was masked in F1 by the dominant tall trait.

Reason (R):

The dwarf allele was mutated into a tall allele during the formation of F1 gametes.

Answer: C — A is true but R is false — The assertion is correct—the reappearance of dwarf plants in F2 demonstrates the segregation of alleles, not mutation. The reason is false; alleles segregate during meiosis in F1; they do not mutate, and the 3:1 ratio in F2 is explained by segregation, not by mutation events.

Q3. Assertion (A):

Mendel selected pea plants with clearly contrasting traits (tall/dwarf, yellow/green seeds) for his hybridisation experiments.

Reason (R):

Contrasting traits are more frequently inherited than intermediate traits and show higher expression rates across generations.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is correct—Mendel did choose contrasting traits, which were scientifically advantageous. However, the reason is incorrect. Contrasting traits were selected not because they're more frequently inherited, but because they are easily distinguishable and provide unambiguous phenotypic classifications, making ratios countable without ambiguity.

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Case-Based Questions (CBQ) for Principles of Inheritance and Variation

Case Study:

A farmer in Punjab wishes to breed an improved variety of pea plants that combines tall height (from Variety A) with yellow seeds (from Variety B). Both Variety A and B are true-breeding lines. The farmer performs a cross: Tall, yellow seeds (Variety A) × Dwarf, green seeds (Variety B). All F1 plants are tall with yellow seeds. When the F1 plants are self-pollinated, the farmer observes the following F2 generation results: 300 tall with yellow seeds, 100 tall with green seeds, 100 dwarf with yellow seeds, and 33 dwarf with green seeds (approximately).

Q1. What are the most likely genotypes of the true-breeding parents?

  • A. Variety A: TtYy; Variety B: ttyy
  • B. Variety A: TTyy; Variety B: ttYY
  • C. Variety A: TTYy; Variety B: ttyy
  • D. Variety A: TTYY; Variety B: ttyy ✓

Answer: D

Q2. The F2 ratio of approximately 9:3:3:1 in the farmer's experiment demonstrates:

  • A. Mendel's Law of Segregation with only one gene
  • B. Mendel's Law of Independent Assortment, showing two genes assorting independently ✓
  • C. That tall and yellow are recessive traits in pea plants
  • D. That the F1 plants must be homozygous for both traits

Answer: B

Q3. If the farmer wants to produce pure-breeding tall plants with yellow seeds for commercial cultivation, what should he do?

  • A. Select dwarf plants with yellow seeds from F2 and self-pollinate them
  • B. Select tall plants with yellow seeds from F2 and self-pollinate them repeatedly until only homozygotes are produced ✓
  • C. Cross all F2 tall, yellow plants together to concentrate the desired alleles
  • D. Use only F1 plants for breeding, as they show both desired traits

Answer: B

Q4. Among the F2 plants, what fraction of the tall plants with yellow seeds are expected to be homozygous for both traits?

  • A. 1/9 ✓
  • B. 1/16
  • C. 3/16
  • D. 1/3

Answer: A

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Previous Year Questions (PYQ) for Principles of Inheritance and Variation

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 57/1/1 · 1 mark

In the following figure, two ways of pairing of two homologous pairs of chromosomes are shown. Which of the following phenomena is expressed?

  • Linkage of genes
  • Independent assortment of genes
  • Multiple alleles
  • Incomplete dominance

CBSE 2026 · Set 57/1/1 · 2 marks

(a) Analyse the following two cases of sex determination in different organisms: Case I: Males have (XO) sex chromosomes and females have two copies of same sex chromosome (XX). Case II: Females have two different sex chromosomes (ZW) and males have two copies of same sex chromosome (ZZ). Identify the type of heterogamety in each case, giving one example of each. (1 + 1 marks)

CBSE 2026 · Set 57/1/1 · 3 marks

(a) What is pedigree analysis? Mention its importance in human genetics. (1 + 1 marks) (b) Analyse the following pedigree and write the (i) Pattern of inheritance. (0.5 marks) (ii) Give one example of disease showing such an inheritance pattern. (0.5 marks)

CBSE 2026 · Set 57/1/1 · 4 marks

A student performed some crosses in plants and represented the result in the form of bar graphs as shown below. Each graph displays the phenotypic proportion of the progeny.

(a) What can you infer about the genotype of parents in crosses I and II? (1 mark)

CBSE 2026 · Set 57/1/2 · 2 marks

(a) Analyse the following two cases of sex determination in different organisms: Case I: Males have (XO) sex chromosomes and females have two copies of same sex chromosome (XX). Case II: Females have two different sex chromosomes (ZW) and males have two copies of same sex chromosome (ZZ). Identify the type of heterogamety in each case, giving one example of each. (1+1 marks) OR (b) (i) Differentiate between a DNA and RNA nucleotide. (1 mark) (ii) Name the two types of bonds present within a single nucleotide. (1 mark)

CBSE 2026 · Set 57/1/2 · 3 marks

(a) What is pedigree analysis? Mention its importance in human genetics. (1+1 marks) (b) Analyse the following pedigree and write the (i) Pattern of inheritance. (½ mark) (ii) Give one example of disease showing such an inheritance pattern. (½ mark)

CBSE 2026 · Set 57/1/2 · 4 marks

A student performed some crosses in plants and represented the result in the form of bar graphs as shown below. Each graph displays the phenotypic proportion of the progeny. Study the graphs and answer the questions.

(a) What can you infer about the genotype of parents in crosses I and II? (1 mark) OR (a) Which genetic cross is represented by these crosses? (1 mark)

CBSE 2026 · Set 57/1/3 · 2 marks

Mendel published his work on inheritance in 1865, but it remained unnoticed for over three decades. State the reasons why his work was not recognised during his life time. (4 points)

CBSE 2026 · Set 57/1/3 · 3 marks

What is pedigree analysis? Mention its importance in human genetics. Analyse the following pedigree and write the: (i) Pattern of inheritance. (ii) Give one example of disease showing such an inheritance pattern.

CBSE 2026 · Set 57/1/3 · 4 marks

A student performed some crosses in plants and represented the result in the form of bar graphs as shown below. Each graph displays the phenotypic proportion of the progeny.

What can you infer about the genotype of parents in crosses I and II?

CBSE 2026 · Set 57/2/1 · 1 mark

In a dihybrid cross, 2400 individuals are produced in F2 generation. Approximately how many will be phenotypically similar to parents ?

  • 2000
  • 1500
  • 580
  • 450

CBSE 2026 · Set 57/2/1 · 3 marks

(a) Give the scientific name of the organism used by Morgan to study sex-linked traits. (b) Write any four reasons why this organism made it an ideal experimental model.

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Class 12 Biology Previous Year Questions, Chapter-wise

All 1074 CBSE board questions from 2022–2026, sorted by chapter and question type. Free PDF download.

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