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.
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 → F1RrYy → 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).
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.
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.
Gene I; alleles IA IB i; 6 genotypes, 4 phenotypes; i makes no sugar
Honey bee
Female 32 (diploid), male 16 (haploid); drone from unfertilised egg by parthenogenesis; sperm by mitosis; no father, no sons, has grandfather and grandsons
Incomplete: 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 dominance
Only the phenotypic ratio (3:1 → 1:2:1). The genotypic ratio stays 1:2:1.
Which law came from which cross
Dominance and Segregation from the monohybrid cross; Independent Assortment from the dihybrid cross.
Test cross vs self-pollination
Test cross = dominant phenotype of unknown genotype × homozygous recessive, not a self-cross.
Single base pair change = point (sickle-cell). Insertion/deletion of base pairs = frame-shift.
Aneuploidy vs polyploidy
Aneuploidy: 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 sex
Humans and Drosophila: the sperm. Birds: the egg (female is ZW).
Multiple alleles
More than two alleles for one gene in a population; any individual still carries only two.
Dominance is not absolute
Same gene, different phenotype examined: Bb is dominant for seed shape (round) but incompletely dominant for starch grain size.
Pleiotropy vs polygenic
Pleiotropy: one gene → many traits (PKU). Polygenic: many genes → one trait (skin colour, height).
SET 1Day 1 check — Mendel and one gene
Mendel carried out his hybridisation experiments on garden pea for
(a) 5 years
(b) 7 years
(c) 9 years
(d) 12 years
The number of true-breeding pea varieties Mendel selected was
(a) 7
(b) 12
(c) 14
(d) 22
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
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
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
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
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
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
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
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
A diploid organism heterozygous at 4 loci can produce how many types of gametes?
(a) 4
(b) 8
(c) 16
(d) 32
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
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
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
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
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
Recombination frequency was first used as a measure of distance between genes by
(a) Morgan
(b) Sturtevant
(c) Punnett
(d) Boveri
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
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
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
The "X body" was described in 1891 by
(a) Morgan
(b) Henking
(c) Sutton
(d) Correns
Grasshopper shows which type of sex determination?
(a) XY
(b) XO
(c) ZW
(d) Haplodiploid
In birds, the sex of the chick is determined by the
(a) sperm
(b) egg
(c) number of autosomes
(d) temperature
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
A female honey bee (queen or worker) develops from
(a) an unfertilised egg
(b) a fertilised egg
(c) a diploid sperm
(d) parthenogenesis
Sickle-cell anaemia is a classical example of
(a) frame-shift mutation
(b) point mutation
(c) aneuploidy
(d) polyploidy
Insertion or deletion of base pairs in DNA causes
(a) point mutation
(b) frame-shift mutation
(c) trisomy
(d) monosomy
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
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
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
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
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
The codon change responsible for sickle-cell anaemia is
(a) GUG → GAG
(b) GAG → GUG
(c) GAA → GUA
(d) CTC → CAC
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
Thalassemia differs from sickle-cell anaemia in being a
(a) qualitative defect
(b) quantitative defect
(c) sex-linked defect
(d) chromosomal disorder
Down's syndrome is caused by
(a) monosomy of 21
(b) trisomy of 21
(c) trisomy of X
(d) deletion in chromosome 21
The karyotype 47, XXY corresponds to
(a) Turner's syndrome
(b) Klinefelter's syndrome
(c) Down's syndrome
(d) a normal male
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
Failure of cytokinesis after telophase leads to
(a) aneuploidy
(b) polyploidy
(c) monosomy
(d) point mutation
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)
4–8 AUGUSTConsolidation, where the last four marks live
Day
Work
4 Aug
Re-solve every question she got wrong on 3 August, from scratch, without looking at the key. Then 20 pedigree and blood-group problems.
5 Aug
Fact vault recited cold, twice. Full 45-question mock #2 in 50 minutes. Error log updated.
6 Aug
NEET 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 Aug
Mock #3. Then re-read NCERT p. 53–78 end to end, out loud, marking any sentence she cannot immediately convert into a question.
8 Aug
Light 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.