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NCERT Class 12 Biology · Chapter 4 · Botany

Principles of Inheritance
and Variation — 4-day mastery plan

Built for one target: 45 questions, 180 marks, ILTS on 9 August. Every hour below is spent on something the chapter can actually ask.

HOW TO USE THISThe scoring maths first

If the paper is 45 questions on this one chapter at +4 / −1, then a wrong guess costs you 5 marks relative to a correct one. Chapter 4 is not a memory chapter — roughly half the questions are numerical/logic (crosses, ratios, gamete counts, blood groups, pedigrees) and half are hard-recall (names, years, percentages, chromosome numbers). Full marks means both halves are automatic, not just familiar.

So each day has the same shape: Learn (read NCERT lines aloud, no notes-making) → Lock (memorise the numbers and names) → Drill (solve, timed) → Check (10 questions, closed book, mark honestly).

DAY 1 · 31 JulMendel → one gene
DAY 2 · 1 AugTwo genes, linkage
DAY 3 · 2 AugSex determination, mutation
DAY 4 · 3 AugDisorders + full mock
4–8 AUGFive consolidation cycles
9 AUGILTS · 180 marks
You have ten days, not four. Days 1–4 build the chapter; 4–8 August is where full marks actually get made, through repetition of the fact vault and re-solving every question she got wrong. Don't skip that half.

THE FOUR DAYSDay-by-day plan

1
31 July · Sections 4.1 – 4.2.2.2 · NCERT p. 53–62

Mendel and the inheritance of one gene

Learn (2.5 h)

  • Why pea was the ideal choice: true-breeding lines, contrasting traits, self-pollinating but easy to cross, short life cycle, large progeny.
  • The 7 pairs of contrasting traits — memorise the table, both columns.
  • Steps in making a cross: emasculation → pollination. Know which parent is female.
  • Monohybrid cross: F1 all tall, F2 3:1 phenotypic and 1:2:1 genotypic, no blending.
  • Terminology, precisely: factor, gene, allele, homozygous, heterozygous, genotype, phenotype, monohybrid.
  • Punnett square mechanics + the binomial form (½T + ½t)².
  • Test cross: dominant phenotype of unknown genotype × homozygous recessive. All dominant → homozygous; 1:1 → heterozygous.
  • Law of Dominance (3 clauses) and Law of Segregation — know which observation each explains.
  • Incomplete dominance: Antirrhinum, RR red × rr white → Rr pink, F2 1:2:1. Genotypic ratio unchanged; only the phenotypic ratio changes.
  • Explanation of dominance via enzyme logic: modified allele → normal/less efficient enzyme, non-functional enzyme, or no enzyme.
  • Co-dominance and multiple alleles: gene I, alleles IA, IB, i; 6 genotypes, 4 phenotypes.
  • Dominance is not autonomous — the Bb starch grain example.

Lock

  • 1856–1863, 7 years, garden pea
  • 14 true-breeding varieties, 7 trait pairs
  • Punnett square → Reginald C. Punnett (British)
  • ABO: 6 genotypes, 4 phenotypes
  • i produces no sugar — that is why it is recessive

Drill (1.5 h)

  • Draw, by hand, 10 monohybrid Punnett squares including 2 incomplete-dominance ones. No shortcuts — the hand-drawing is what makes it automatic under time pressure.
  • 8 blood-group problems: given parents' groups, list all possible and all impossible children's groups.
  • NCERT exercises 1, 2, 4, 5, 6, 12, 13.

Check (25 min)

Set 1 below, closed book. Target: 10/10. Anything wrong goes into the error log with the NCERT line that answers it.

2
1 August · Sections 4.3 – 4.5 · NCERT p. 62–69

Two genes, chromosomes, linkage, polygenes

Learn (2.5 h)

  • Dihybrid cross RRYY × rryy → F1 RrYy → F2 9:3:3:1. Fill the 16-cell square yourself at least twice.
  • 9:3:3:1 as (3:1)(3:1). Each trait pair still segregates 3:1 on its own.
  • Law of Independent Assortment — statement in NCERT's exact words.
  • Four gamete types RY, Ry, rY, ry at 25% each; general rule 2n gamete types for n heterozygous loci.
  • Cytological basis: independent alignment of the two chromosome pairs at the metaphase plate (Possibility I / II figure).
  • Why Mendel's work was ignored till 1900 — the four reasons — and who rediscovered it.
  • Chromosomal theory of inheritance: Sutton and Boveri paralleled chromosome behaviour with gene behaviour; Sutton united it with Mendelian principles and named it.
  • Morgan and Drosophila melanogaster: why it suited the work (synthetic medium, ~2-week life cycle, many progeny, sexes easily told apart, visible variations).
  • Linkage vs recombination; tight linkage → low recombination. Cross A: y–w. Cross B: w–m.
  • Sturtevant used recombination frequency to map genes; maps later used in genome sequencing.
  • Polygenic inheritance: additive alleles, environment matters, human skin colour with A/B/C, AABBCC darkest, aabbcc lightest.
  • Pleiotropy: one gene, many phenotypes, usually via metabolic pathways — phenylketonuria.

Lock

  • Rediscovery 1900: de Vries, Correns, von Tschermak
  • Chromosome movement in meiosis worked out by 1902
  • y–w recombination 1.3%; w–m 37.2%
  • Mendel published 1865
  • Chromosomes = "coloured bodies" (stain-visible)
  • Dihybrid F2: 9 genotypes, 4 phenotypes, 16 squares

Drill (1.5 h)

  • Two full 4×4 dihybrid squares from scratch; then tabulate all 9 genotypes with their phenotypes and counts (1:2:1:2:4:2:1:2:1).
  • Practise the shortcut method: solve dihybrid ratios as a product of two monohybrid ratios. Do 6 problems both ways to prove they match.
  • Gamete-count questions: AaBbCcDd, AABbCc, AaBBCcDdEe — how many types, and write them out for the first.
  • NCERT exercises 3, 7, 8, 9.

Check (25 min)

Set 2 below, closed book.

3
2 August · Sections 4.6 – 4.8.1 · NCERT p. 69–73

Sex determination, mutation, pedigree analysis

Learn (2.5 h)

  • Henking's X body (1891) → recognised later as the X chromosome; origin of the terms sex chromosome and autosome.
  • XO type (grasshopper): male has one X, female has XX; some sperms carry X, some don't. Male heterogamety.
  • XY type (humans, Drosophila): male XY, female XX. Male heterogamety. Equal chromosome number in both sexes.
  • ZW type (many birds): female ZW, male ZZ. Female heterogamety — the egg decides the chick's sex.
  • Sex determination in humans: 22 pairs autosomes + XY / XX; 50% X-sperm, 50% Y-sperm; the sperm determines the child's sex, and every pregnancy is a fresh 50:50.
  • Honey bee haplodiploidy: fertilised egg → female (queen or worker); unfertilised egg → drone by parthenogenesis. Female 32, male 16. Drones make sperm by mitosis. A drone has no father and can have no sons, but has a grandfather and can have grandsons.
  • Mutation: alteration of DNA sequence → change in genotype and phenotype; along with recombination it generates variation.
  • Deletion/insertion/duplication of segments → chromosomal aberrations (common in cancer cells). Single base pair change → point mutation (sickle-cell anaemia). Insertions/deletions of base pairs → frame-shift.
  • Mutagens: chemical and physical; UV radiation is a mutagen.
  • Pedigree analysis: purpose, and every symbol in Figure 4.13 — including the diamond for unspecified sex, the double line for consanguineous mating, and the numbered diamond for multiple unaffected offspring.

Lock

  • Henking, 1891 — X body
  • Honey bee: 32 / 16; drone sperm by mitosis
  • Human: 23 pairs, 22 autosomal pairs
  • Grasshopper = XO; birds = ZW/ZZ
  • Sickle-cell = point mutation, not frame-shift

Drill (1.5 h)

  • Redraw the honey bee sex-determination flow chart (Figure 4.13, p. 71) from memory, with all chromosome numbers.
  • Build a comparison table: XO / XY / ZW — who is heterogametic, chromosome numbers, example organisms.
  • 10 pedigree problems. Use this decision routine: affected in every generation? → dominant. Skips generations, both sexes equally, consanguinity present? → autosomal recessive. Almost only males affected, through unaffected mothers? → X-linked recessive.
  • NCERT exercises 10, 11, 14.

Check (25 min)

Set 3 below, closed book.

4
3 August · Sections 4.8.2 – 4.8.3 + full revision · NCERT p. 73–78

Genetic disorders, then the first full mock

Learn (2 h)

  • The two categories: Mendelian disorders (single gene) vs chromosomal disorders (number/arrangement of chromosomes).
  • Colour blindness — X-linked recessive, red/green cone defect, 8% of males and 0.4% of females; a daughter is affected only if mother is a carrier and father is colour blind.
  • Haemophilia — X-linked recessive; a clotting-cascade protein is affected; carrier mother → affected sons; affected females extremely rare; Queen Victoria's pedigree.
  • Sickle-cell anaemia — autosome-linked recessive; alleles HbA/HbS; only HbSHbS is diseased; HbAHbS is a carrier showing sickle-cell trait; Glu → Val at the 6th position of the β-globin chain; codon GAG → GUG; polymerisation under low oxygen tension changes the biconcave disc to a sickle shape.
  • Phenylketonuria — autosomal recessive; missing enzyme converts phenylalanine → tyrosine; phenylalanine accumulates as phenylpyruvic acid and derivatives; also excreted in urine.
  • Thalassemia — autosome-linked recessive; reduced synthesis of a globin chain. α: genes HBA1 and HBA2 on chromosome 16. β: gene HBB on chromosome 11. Thalassemia is a quantitative problem; sickle-cell is a qualitative one.
  • Aneuploidy (failure of chromatid segregation) vs polyploidy (failure of cytokinesis; common in plants). Trisomy vs monosomy.
  • Down's syndrome — trisomy of 21, total 47; described by Langdon Down, 1866; short stature, small round head, furrowed tongue, partially open mouth, broad palm with characteristic crease, retarded development.
  • Klinefelter's — 47, XXY; masculine build plus gynaecomastia; sterile. Turner's — 45, X0; rudimentary ovaries; sterile.

Revise (1 h)

  • Read the NCERT Summary (p. 77–78) aloud, twice.
  • Recite the whole fact vault below without looking. Anything she hesitates on gets a star and goes on a card.

Mock (1.5 h)

  • Set 4 below, then a full 45-question chapter test in one sitting, phone away, strict 50 minutes, with negative marking applied.
  • Then the part most students skip: for every wrong or guessed question, write the exact NCERT sentence that settles it. That list is her 8 August revision sheet.
  • Colour blindness 8% / 0.4%
  • Down's 1866, trisomy 21, 47
  • Klinefelter 47, XXY · Turner 45, X0
  • Sickle-cell: 6th position, β-chain, GAG→GUG, Glu→Val
  • α-thal chr 16 · β-thal chr 11

HIGH-YIELDThe fact vault

These are the lines that decide a 176 from a 180. She should be able to recite the whole table in under four minutes by 8 August.

FactValue to recite
Mendel's experimentsGarden pea, 1856–1863, 7 years, 14 true-breeding varieties, 7 contrasting trait pairs; first use of statistics in biology
Published / rediscovered1865 / 1900 by de Vries, Correns, von Tschermak
Punnett squareReginald C. Punnett
Chromosomal theorySutton and Boveri; Sutton united it with Mendelian principles and named it (chromosome movement mapped by 1902)
MorganDrosophila melanogaster; discovered linkage and recombination; F2 deviated from 9:3:3:1
Recombination valuesy–w 1.3% (tight); w–m 37.2% (loose)
Genetic mappingAlfred Sturtevant, using recombination frequency as distance
X bodyHenking, 1891
RatiosMonohybrid 3:1 (geno 1:2:1) · Incomplete dominance 1:2:1 (geno 1:2:1) · Dihybrid 9:3:3:1 · Test cross 1:1
Gamete types2n for n heterozygous loci (4 loci → 16)
ABOGene I; alleles IA IB i; 6 genotypes, 4 phenotypes; i makes no sugar
Honey beeFemale 32 (diploid), male 16 (haploid); drone from unfertilised egg by parthenogenesis; sperm by mitosis; no father, no sons, has grandfather and grandsons
Human chromosomes46 total, 23 pairs, 22 autosomal pairs; sperm decides sex; 50:50 every pregnancy
Colour blindnessX-linked recessive; 8% males, 0.4% females
Sickle-cell anaemiaAutosome-linked recessive; Glu→Val, 6th position of β-globin; GAG→GUG; polymerises at low O2
Thalassemiaα: HBA1 + HBA2, chromosome 16; β: HBB, chromosome 11; quantitative defect
PhenylketonuriaAutosomal recessive; phenylalanine hydroxylase missing; phenylpyruvic acid; also the NCERT example of pleiotropy
Down's syndromeTrisomy 21, total 47; Langdon Down, 1866
Klinefelter / Turner47, XXY (gynaecomastia, sterile) / 45, X0 (rudimentary ovaries, sterile)
Polygenic skin colourGenes A, B, C; AABBCC darkest, aabbcc lightest; effects additive; environment contributes

EXAM TRAPSThe pairs that cost marks

ConfusionWhat NCERT actually says
Incomplete dominance vs co-dominanceIncomplete: F1 is intermediate (pink) — neither allele fully expressed. Co-dominant: F1 resembles both parents, both products present (A and B sugars in AB blood).
Which ratio changes in incomplete dominanceOnly the phenotypic ratio (3:1 → 1:2:1). The genotypic ratio stays 1:2:1.
Which law came from which crossDominance and Segregation from the monohybrid cross; Independent Assortment from the dihybrid cross.
Test cross vs self-pollinationTest cross = dominant phenotype of unknown genotype × homozygous recessive, not a self-cross.
Autosomal vs X-linked recessiveSickle-cell, thalassemia, PKU → autosomal. Haemophilia, colour blindness → X-linked.
Sickle-cell vs thalassemiaSickle-cell = qualitative (wrong globin). Thalassemia = quantitative (too little globin).
Point vs frame-shift mutationSingle base pair change = point (sickle-cell). Insertion/deletion of base pairs = frame-shift.
Aneuploidy vs polyploidyAneuploidy: chromatids fail to segregate → one chromosome gained or lost (Down's, Turner's). Polyploidy: cytokinesis fails after telophase → whole extra set; common in plants.
Who determines sexHumans and Drosophila: the sperm. Birds: the egg (female is ZW).
Multiple allelesMore than two alleles for one gene in a population; any individual still carries only two.
Dominance is not absoluteSame gene, different phenotype examined: Bb is dominant for seed shape (round) but incompletely dominant for starch grain size.
Pleiotropy vs polygenicPleiotropy: one gene → many traits (PKU). Polygenic: many genes → one trait (skin colour, height).

SET 1Day 1 check — Mendel and one gene

  1. Mendel carried out his hybridisation experiments on garden pea for
    • (a) 5 years
    • (b) 7 years
    • (c) 9 years
    • (d) 12 years
  2. The number of true-breeding pea varieties Mendel selected was
    • (a) 7
    • (b) 12
    • (c) 14
    • (d) 22
  3. Reappearance of the recessive trait in the F2 generation is best explained by the Law of
    • (a) Dominance
    • (b) Segregation
    • (c) Independent Assortment
    • (d) Linkage
  4. A violet-flowered plant of unknown genotype is test-crossed and gives 50% white-flowered offspring. Its genotype is
    • (a) VV
    • (b) Vv
    • (c) vv
    • (d) Cannot be decided
  5. In Antirrhinum, a cross between pink and pink flowers gives an F1 in the ratio
    • (a) 3 red : 1 white
    • (b) 1 red : 2 pink : 1 white
    • (c) All pink
    • (d) 1 red : 1 white
  6. In pea seeds, the Bb genotype produces round seeds but starch grains of intermediate size. This shows that
    • (a) B is always incompletely dominant
    • (b) dominance depends on the phenotype examined
    • (c) two genes control starch synthesis
    • (d) b is a lethal allele
  7. The total number of genotypes and phenotypes possible for the human ABO blood group system is
    • (a) 4 and 6
    • (b) 6 and 4
    • (c) 6 and 6
    • (d) 3 and 4
  8. A man of blood group AB marries a woman of blood group O. The possible blood groups of their children are
    • (a) A and B only
    • (b) AB and O only
    • (c) A, B, AB, O
    • (d) O only
  9. Allele i of the ABO system is recessive because it
    • (a) is present in low frequency
    • (b) produces no sugar
    • (c) produces a different sugar
    • (d) is located on the Y chromosome
  10. In a monohybrid cross, the F2 genotypic ratio can be obtained by expanding
    • (a) (½T + ½t)
    • (b) (½T + ½t)²
    • (c) (¼T + ¾t)²
    • (d) (⅓T + ⅔t)²

SET 2Day 2 check — two genes and linkage

  1. A diploid organism heterozygous at 4 loci can produce how many types of gametes?
    • (a) 4
    • (b) 8
    • (c) 16
    • (d) 32
  2. In the F2 of a dihybrid cross, the number of genotypes and phenotypes respectively is
    • (a) 9 and 4
    • (b) 4 and 9
    • (c) 16 and 4
    • (d) 9 and 9
  3. Mendel's results were rediscovered in 1900 by
    • (a) Sutton, Boveri and Morgan
    • (b) de Vries, Correns and von Tschermak
    • (c) Punnett, Bateson and Morgan
    • (d) Sturtevant, Morgan and Henking
  4. Who united the knowledge of chromosomal segregation with Mendelian principles and named it the chromosomal theory of inheritance?
    • (a) Boveri
    • (b) Sutton
    • (c) Morgan
    • (d) Sturtevant
  5. Morgan's dihybrid crosses in Drosophila deviated significantly from 9:3:3:1 because the genes studied were
    • (a) on different chromosomes
    • (b) on the same chromosome
    • (c) pleiotropic
    • (d) polygenic
  6. Genes w and m showed 37.2% recombination while y and w showed 1.3%. This means
    • (a) w and m are more tightly linked
    • (b) y and w are more tightly linked
    • (c) all three are unlinked
    • (d) y and m lie on different chromosomes
  7. Recombination frequency was first used as a measure of distance between genes by
    • (a) Morgan
    • (b) Sturtevant
    • (c) Punnett
    • (d) Boveri
  8. In the cross TtYy × Ttyy, the proportion of offspring that are tall and green is
    • (a) 1/8
    • (b) 3/8
    • (c) 1/4
    • (d) 9/16
  9. Assuming three genes A, B and C control human skin colour additively, an individual with three dominant and three recessive alleles will have
    • (a) the darkest skin
    • (b) the lightest skin
    • (c) intermediate skin colour
    • (d) the same colour as AABBCC
  10. Phenylketonuria is cited in NCERT as an example of
    • (a) polygenic inheritance
    • (b) pleiotropy
    • (c) co-dominance
    • (d) linkage

SET 3Day 3 check — sex determination, mutation, pedigree

  1. The "X body" was described in 1891 by
    • (a) Morgan
    • (b) Henking
    • (c) Sutton
    • (d) Correns
  2. Grasshopper shows which type of sex determination?
    • (a) XY
    • (b) XO
    • (c) ZW
    • (d) Haplodiploid
  3. In birds, the sex of the chick is determined by the
    • (a) sperm
    • (b) egg
    • (c) number of autosomes
    • (d) temperature
  4. Which statement about a honey bee drone is incorrect?
    • (a) It has 16 chromosomes
    • (b) It produces sperm by mitosis
    • (c) It has no father
    • (d) It can have sons
  5. A female honey bee (queen or worker) develops from
    • (a) an unfertilised egg
    • (b) a fertilised egg
    • (c) a diploid sperm
    • (d) parthenogenesis
  6. Sickle-cell anaemia is a classical example of
    • (a) frame-shift mutation
    • (b) point mutation
    • (c) aneuploidy
    • (d) polyploidy
  7. Insertion or deletion of base pairs in DNA causes
    • (a) point mutation
    • (b) frame-shift mutation
    • (c) trisomy
    • (d) monosomy
  8. In a human pedigree chart, a double horizontal line between two individuals indicates
    • (a) affected individuals
    • (b) consanguineous mating
    • (c) unspecified sex
    • (d) five unaffected offspring
  9. A trait that skips generations, affects both sexes equally, and appears more often in families with cousin marriages is most likely
    • (a) autosomal dominant
    • (b) autosomal recessive
    • (c) X-linked dominant
    • (d) Y-linked
  10. Which of these is a mutagen mentioned in the chapter?
    • (a) Infrared radiation
    • (b) UV radiation
    • (c) Radio waves
    • (d) Visible light

SET 4Day 4 check — genetic disorders

  1. Red-green colour blindness occurs in about
    • (a) 8% males, 0.4% females
    • (b) 0.4% males, 8% females
    • (c) 4% males, 0.8% females
    • (d) 8% of both sexes
  2. In sickle-cell anaemia, glutamic acid is replaced by valine at position
    • (a) 5 of the α-chain
    • (b) 6 of the β-chain
    • (c) 6 of the α-chain
    • (d) 11 of the β-chain
  3. The codon change responsible for sickle-cell anaemia is
    • (a) GUG → GAG
    • (b) GAG → GUG
    • (c) GAA → GUA
    • (d) CTC → CAC
  4. β-thalassemia is controlled by which gene and chromosome?
    • (a) HBA1, chromosome 16
    • (b) HBA2, chromosome 11
    • (c) HBB, chromosome 11
    • (d) HBB, chromosome 16
  5. Thalassemia differs from sickle-cell anaemia in being a
    • (a) qualitative defect
    • (b) quantitative defect
    • (c) sex-linked defect
    • (d) chromosomal disorder
  6. Down's syndrome is caused by
    • (a) monosomy of 21
    • (b) trisomy of 21
    • (c) trisomy of X
    • (d) deletion in chromosome 21
  7. The karyotype 47, XXY corresponds to
    • (a) Turner's syndrome
    • (b) Klinefelter's syndrome
    • (c) Down's syndrome
    • (d) a normal male
  8. Turner's syndrome results from
    • (a) an extra X chromosome
    • (b) loss of one X chromosome
    • (c) an extra chromosome 21
    • (d) loss of a Y chromosome
  9. Failure of cytokinesis after telophase leads to
    • (a) aneuploidy
    • (b) polyploidy
    • (c) monosomy
    • (d) point mutation
  10. The enzyme deficient in phenylketonuria converts
    • (a) tyrosine to phenylalanine
    • (b) phenylalanine to tyrosine
    • (c) phenylalanine to phenylpyruvic acid
    • (d) tyrosine to melanin

Answer keys

Set 1 — 1 (b) · 2 (c) · 3 (b) · 4 (b) test cross giving 1:1 means heterozygous · 5 (b) pink × pink is Rr × Rr · 6 (b) dominance is not autonomous · 7 (b) · 8 (a) parents are IAIB × ii · 9 (b) · 10 (b)

Set 2 — 1 (c) 2⁴ · 2 (a) · 3 (b) · 4 (b) · 5 (b) both genes were on the X · 6 (b) lower recombination = tighter linkage · 7 (b) · 8 (b) tall = 3/4, green (yy) = 1/2 → 3/8 · 9 (c) · 10 (b)

Set 3 — 1 (b) · 2 (b) · 3 (b) the female is ZW, so the egg carries Z or W · 4 (d) a drone has no sons but can have grandsons · 5 (b) · 6 (b) · 7 (b) · 8 (b) · 9 (b) · 10 (b)

Set 4 — 1 (a) · 2 (b) · 3 (b) · 4 (c) · 5 (b) · 6 (b) · 7 (b) · 8 (b) 45, X0 · 9 (b) · 10 (b)

4–8 AUGUSTConsolidation, where the last four marks live

DayWork
4 AugRe-solve every question she got wrong on 3 August, from scratch, without looking at the key. Then 20 pedigree and blood-group problems.
5 AugFact vault recited cold, twice. Full 45-question mock #2 in 50 minutes. Error log updated.
6 AugNEET previous-year questions from this chapter, last 10 years, in one sitting. These repeat heavily — Mendelian ratios, sex determination in honey bee, sickle-cell codon, karyotypes.
7 AugMock #3. Then re-read NCERT p. 53–78 end to end, out loud, marking any sentence she cannot immediately convert into a question.
8 AugLight only: fact vault, trap table, error log, all figures (4.1, 4.4, 4.7, 4.11, 4.13, 4.15, 4.17). No new material, no new mock. Early night.
One rule worth holding to: in this chapter, the questions she gets wrong are worth more than the ones she gets right. The error log is the whole method — the four sets above exist to generate it, not to score well on.