About This Section
This is a collection of 30+ solved examples covering every topic of Chapter 8 (Heredity) — variation, Mendel's monohybrid and dihybrid crosses, dominant/recessive, genotype/phenotype, Punnett squares, genes-proteins-chromosomes, sex determination and blood groups. All NCERT in-text and exercise questions are worked here.
Tips for Board Exam Preparation:
- Try each cross yourself with a Punnett square before reading the solution.
- Memorise the key ratios: monohybrid 3:1, dihybrid 9:3:3:1, test cross 1:1.
- Learn genotype vs phenotype cold.
- Practise every NCERT exercise.
Let's begin!
Example 1: NCERT — Which trait arose earlier?
Trait A is in 10% and trait B in 60% of an asexually reproducing population. Which arose earlier?
Solution: Trait B. In asexual reproduction a trait spreads over generations, so a more widespread trait (B, 60%) has usually existed longer than a rarer one (A, 10%).
Example 2: NCERT — Variation and survival
How does the creation of variations promote survival?
Solution: Variations give individuals different advantages. When the environment changes, some variants are suited to the new conditions and survive, ensuring the survival of the species even if many individuals die.
Example 3: Asexual vs sexual variation
Why does sexual reproduction create more variation?
Solution: Asexual offspring differ only by minor DNA-copying errors. Sexual offspring combine DNA from two parents, each with its own accumulated variations, producing new combinations and far more diversity.
Example 4: Define heredity and trait
Define heredity and trait; give an inherited human trait.
Solution: Heredity = transfer of traits from parents to offspring via DNA. A trait = a characteristic feature. Example: free/attached earlobes.
Example 5: Two versions of a trait
Why does each child carry two versions of every trait?
Solution: Because both parents contribute equal DNA, the child gets one copy of each gene from each parent — two versions per trait, which may be identical or different.
Example 6: Why the pea plant?
Why did Mendel choose the garden pea?
Solution: It has clear contrasting characters (round/wrinkled, tall/short, violet/white) and can be self- or cross-pollinated, allowing controlled experiments.
Example 7: The monohybrid F1
In a tall × short cross, what is the F1 and what does it show?
Solution: All F1 are tall (Tt) — no medium plants. Only the dominant trait (tallness) is expressed; traits do not blend.
Example 8: The monohybrid F2
What is the F2 ratio on self-pollinating the F1, and its meaning?
Solution: F2 = 3 tall : 1 short. Shortness reappears, proving it was hidden (recessive) in the F1 and that each trait has two gene copies.
Example 9: NCERT — Dominant or recessive?
How do Mendel's experiments show that traits may be dominant or recessive?
Solution: In F1 (Tt) only tallness appears (dominant); shortness is hidden. In F2 shortness reappears in 1/4 of the plants — so it was recessive, expressed only when both copies are t (tt). Thus traits can be dominant (T) or recessive (t).
Example 10: NCERT — Independent inheritance
How do Mendel's experiments show traits are inherited independently?
Solution: In the dihybrid cross, the F2 shows new combinations (round-green, wrinkled-yellow) not present in the parents. This proves seed shape and colour are inherited independently (9:3:3:1).
Example 11: Genotype vs phenotype
Give the phenotype of TT, Tt, tt (T tall dominant).
Solution: TT → tall, Tt → tall, tt → short. TT and Tt share a phenotype (tall) but differ in genotype.
Example 12: The Tt × Tt cross
Give the genotype and phenotype ratios of Tt × Tt.
Solution: Punnett square → TT, Tt, Tt, tt. Genotype 1:2:1; phenotype 3 tall : 1 short.
Example 13: The test cross Tt × tt
What offspring result from Tt × tt?
Solution: Gametes T,t × t,t → Tt, Tt, tt, tt. Genotype 1 Tt : 1 tt; phenotype 1 tall : 1 short (1:1).
Example 14: Reasoning from a 3:1 ratio
A cross gives a 3:1 phenotype ratio. What were the parents' genotypes?
Solution: A 3:1 ratio comes from Tt × Tt (two heterozygotes), with 1/4 of the offspring recessive.
Example 15: Dihybrid F2 ratio
State the F2 phenotype ratio of a dihybrid cross and its four classes.
Solution: 9 : 3 : 3 : 1 — 9 round-yellow, 3 round-green, 3 wrinkled-yellow, 1 wrinkled-green.
Example 16: Gametes of a dihybrid
What gametes does RrYy make?
Solution: Four types in equal numbers: RY, Ry, rY, ry (each gene assorts independently).
Example 17: Fraction double recessive
In a dihybrid F2, what fraction is wrinkled and green?
Solution: The double-recessive class is 1/16 (genotype rryy).
Example 18: Gene to trait
How does a gene control plant height?
Solution: Gene → makes an enzyme (protein) → enzyme controls the amount of growth hormone → more hormone = tall; an altered gene → less-efficient enzyme → less hormone → short. Genes control traits by controlling proteins.
Example 19: Two copies of each gene
Why does each organism carry two copies of every gene?
Solution: Because it inherits one set from each parent. Body cells therefore have two copies of each gene (on paired chromosomes).
Example 20: Why one set in germ cells?
Why must a germ cell carry only one set of genes?
Solution: If gametes carried a full double set, the number of sets would double every generation. A single set per gamete keeps the number constant: fertilisation restores two sets.
Example 21: Chromosomes and independence
How do chromosomes explain independent inheritance?
Solution: Genes lie on separate chromosomes, not one thread. Each chromosome pair is sorted independently into gametes, so genes on different chromosomes form new combinations — independent inheritance.
Example 22: NCERT — Equal genetic contribution
How is the equal genetic contribution of both parents ensured?
Solution: Genes are on paired chromosomes; each germ cell gets one of each pair. At fertilisation the offspring receives one set from the father and one from the mother — an equal contribution.
Example 23: NCERT — How is sex determined?
How is the sex of a human child determined?
Solution: Females are XX, males XY. The mother gives an X to every child; the father gives X (→ girl) or Y (→ boy). So the father's chromosome determines the sex.
Example 24: Autosomes vs sex chromosomes
Differentiate autosomes and sex chromosomes in humans.
Solution: Autosomes = 22 pairs, alike in both sexes. Sex chromosomes = 1 pair (XX female, XY male) that determines sex.
Example 25: The 1:1 sex ratio
Why are boys and girls born in roughly equal numbers?
Solution: The father makes equal numbers of X and Y sperm; X gives a girl and Y a boy, so the chances are equal — about 1:1.
Example 26: Who determines sex?
Is it scientifically correct to blame the mother for the child's sex?
Solution: No. The mother can give only X; the father's X or Y decides the sex. So the father's sperm determines the child's sex.
Example 27: NCERT exercise — genotype of tall parent
Tall violet × short white gives all violet, ~half short. Genotype of the tall violet parent? (a)TTWW (b)TTww (c)TtWW (d)TtWw
Solution: All violet → WW; ~half short → Tt (1:1 test cross). Answer: (c) TtWW.
Example 28: NCERT exercise — light eye colour
Can "light-eyed children have light-eyed parents" tell us if light is dominant/recessive?
Solution: No. Such similarity does not reveal dominance; we would need to see whether the trait can be hidden and reappear, or use crosses with known genotypes.
Example 29: NCERT exercise — dog coat colour project
Outline a project to find the dominant coat colour in dogs.
Solution: Cross pure-breeding dogs of two colours; the colour seen in all F1 is dominant. Confirm by breeding F1 and checking the recessive colour reappears in ~1/4 of F2 (3:1).
Example 30: NCERT exercise — blood group dominance
A-group father × O-group mother → O-group daughter. Enough to decide which of A/O is dominant?
Solution: No. The A father may be I^A i; I^A i × ii gives A or O children regardless of dominance. One family cannot establish which allele is dominant.
Example 31: Blood group cross
Parents are I^A i (A) and I^B i (B). List the possible blood groups of the children.
Solution: Gametes: I^A, i × I^B, i → I^A I^B (AB), I^A i (A), I^B i (B), i i (O). Children can be AB, A, B or O — all four groups possible.
Example 32: Fraction double dominant
In RrYy × RrYy, what fraction is round-yellow?
Solution: The double-dominant class is 9/16 of the offspring.
Example 33: Monohybrid counting
A Tt × Tt cross produces 200 offspring. About how many are expected to be short?
Solution: Short (tt) = 1/4. So about 1/4 × 200 = 50 are expected to be short (and ~150 tall).