NEET · NCERT Class 12 Biology

Principles of
Inheritance and
Variation Revision Mini-Book · Trap-Solved Question Bank · Master Ratio Sheet

Part 1 — Concept capsules: every NCERT line that becomes a question
Part 2 — 30 trap questions solved in full (NEET / NEET-pattern)
Part 3 — Master revision sheet: all ratios, crosses, disorders, exceptions
Part 4 — NCERT diagram pack, redrawn
Part 5 — 10-minute last look + 25 blunders to never repeat

Aligned to NTA NEET Syllabus Prepared for Aamirah Fathima · Chapter 5 (old NCERT) = Chapter 4 (rationalised NCERT)

Orientation

Syllabus map & how to use this book

The NTA NEET syllabus lists this chapter under Heredity and Variation. Everything the paper can ask sits inside the eight blocks below. Tick a block only when you can teach it out loud without the book.

#	Block (as the syllabus words it)	What NEET actually asks from it	Weight
1	Mendelian inheritance	Seven pea characters (which trait is dominant), monohybrid and dihybrid ratios, test cross, the three laws, gamete counting	High
2	Deviations from Mendelism	Incomplete dominance, codominance, multiple alleles and ABO blood groups, pleiotropy	Very high
3	Polygenic inheritance	Skin colour, height; "number of dark alleles" numericals	Medium
4	Chromosomal theory; chromosomes and genes	Sutton and Boveri; the parallel behaviour of genes and chromosomes	Medium
5	Sex determination — human, bird, honey bee	XX-XY, XX-XO, ZZ-ZW, haplodiploidy; who decides the sex of the child	High
6	Linkage and crossing over	Morgan's Drosophila crosses, 1.3% vs 37.2%, Sturtevant's genetic maps	High
7	Sex-linked inheritance — haemophilia, colour blindness	Pedigree logic, carrier mothers, why females are rarely affected	Very high
8	Mendelian and chromosomal disorders	Thalassemia, sickle-cell anaemia, PKU; Down, Turner, Klinefelter and their chromosome counts	Very high

How to use the five parts Day 1: read Part 1 with NCERT open beside you — it is written line-parallel to the text. Day 2: attempt Part 2 with the answers covered; every question there exists because students lose a mark on it. Day 3 onwards: revise only Part 3 and Part 5, and re-draw the Part 4 diagrams from memory on blank paper twice.

Old NCERT vs rationalised NCERT Nothing examinable was deleted from this chapter in the rationalised edition — the chapter number changed from 5 to 4, that is all. Both editions carry Mendel, the deviations, linkage, sex determination, mutation and the disorders. Use whichever copy you own; page numbers differ, content does not.

Part 1

Concept capsules — the examinable core

1. Mendel: the man, the plant, the method

Point	Exam-ready detail
Period of work	1856–1863 (seven years), in the monastery garden at Brno
Material	Garden pea, Pisum sativum
Characters studied	Seven contrasting characters, each with two contrasting traits
Lines used	14 true-breeding pea lines (7 pairs)
Why pea?	Many contrasting traits; bisexual flowers that are naturally self-pollinating; easy to cross-pollinate artificially (emasculation, then dusting pollen); short life cycle; large progeny; the hybrids are fully fertile
Why he succeeded where others failed	He counted — used mathematics and statistics on large samples, studied one character at a time, and maintained records over several generations
Fate of the work	Presented 1865, published 1866, then ignored; rediscovered in 1900 independently by de Vries, Correns and von Tschermak
Terminology	Mendel called them "factors"; the word gene was coined much later. "Dominant" and "recessive" are Mendel's own terms.

The seven pairs — memorise the dominant side

Character	Dominant trait	Recessive trait
Stem height	Tall	Dwarf
Flower colour	Violet	White
Flower position	Axial	Terminal
Pod shape	Inflated (full)	Constricted
Pod colour	Green	Yellow
Seed shape	Round	Wrinkled
Seed colour	Yellow	Green

Trap — the colour swap Pod colour: GREEN is dominant. Seed colour: YELLOW is dominant. They are deliberately opposite, and options are built on this swap again and again. Memory hook: "a green pod holds a yellow seed" — the dominant pair is exactly what a ripe pea looks like from outside to inside. Second hook: Axial flower position is dominant, not terminal ("A" comes first).

Vocabulary that gets swapped inside options

Allele — a slightly different form of the same gene. Homozygous = identical alleles; heterozygous = different alleles.
Genotype = genetic constitution; phenotype = observable expression.
Hybrid = offspring of genetically dissimilar parents. Monohybrid cross = one character; dihybrid = two characters.
The Punnett square was devised by Reginald C. Punnett, not by Mendel.
Emasculation = removing the anthers from a bisexual flower before they dehisce; it is done on the female parent.

2. Monohybrid cross and the first two laws

Tall (TT) × Dwarf (tt) gives an F₁ that is entirely Tall (Tt) — the dwarf trait disappears but is not lost. Selfing the F₁ brings it back in the F₂.

F₁ × F₁	T	t
T	TT Tall	Tt Tall
t	Tt Tall	tt Dwarf

Fig. 1 — F₂ of a monohybrid cross. Phenotypic ratio 3 : 1 · Genotypic ratio 1 : 2 : 1

Law	Statement	Exception?
Law of Dominance	Characters are controlled by discrete units called factors, which occur in pairs; in a dissimilar pair, one member dominates and is expressed.	Has exceptions — incomplete dominance, codominance
Law of Segregation	The two alleles of a pair separate during gamete formation, so a gamete receives only one allele — gametes are always pure.	No exception — universally true
Law of Independent Assortment	When two pairs of traits are combined, the segregation of one pair is independent of the other pair.	Has an exception — linkage (genes on the same chromosome)

Trap — "which law has no exception?" The answer is always the Law of Segregation, also called the Law of Purity of Gametes. Segregation depends only on homologues separating in anaphase I, which always happens. Dominance fails in Mirabilis; independent assortment fails under linkage. Note also that independent assortment needs two or more genes on different chromosomes — the law is meaningless for a monohybrid cross.

Test cross vs back cross — the most abused pair of terms

	Back cross	Test cross
Definition	Cross of F₁ with either parent	Cross of an individual of unknown genotype with the homozygous recessive parent
Relationship	Every test cross is a back cross, but every back cross is not a test cross.
Purpose	To recover the parental type / improve a variety	To find whether a dominant-looking individual is homozygous or heterozygous
How to read it	—	Even one recessive offspring ⇒ the parent was heterozygous. All offspring dominant ⇒ homozygous.
Ratios	—	Monohybrid 1 : 1 · Dihybrid 1 : 1 : 1 : 1

Tt × tt	t	t
T	Tt · Tall	Tt · Tall
t	tt · Dwarf	tt · Dwarf

Fig. 2 — Test cross of a heterozygote. Phenotypic ratio 1 Tall : 1 Dwarf; here the genotypic ratio is also 1 : 1

3. Dihybrid cross and the 9 : 3 : 3 : 1

Round-Yellow (RRYY) × wrinkled-green (rryy) gives an F₁ of RrYy, all round and yellow. Selfing that F₁ fills sixteen boxes.

♀ / ♂	RY	Ry	rY	ry
RY	RRYY	RRYy	RrYY	RrYy
Ry	RRYy	RRyy	RrYy	Rryy
rY	RrYY	RrYy	rrYY	rrYy
ry	RrYy	Rryy	rrYy	rryy

Fig. 3 — F₂ dihybrid checkerboard. Green = round-yellow (9), amber = round-green (3), pink = wrinkled-yellow (3), grey = wrinkled-green (1)

Quantity	Value
F₂ phenotypic ratio	9 round-yellow : 3 round-green : 3 wrinkled-yellow : 1 wrinkled-green
F₂ genotypic ratio	1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1 — that is 9 genotypes in 16 combinations
Commonest genotype	RrYy — 4/16, the double heterozygote
Parental vs new types in F₂	Parental (9 + 1) = 10/16; new / recombinant (3 + 3) = 6/16
Fully homozygous individuals	4/16 = 1/4 (RRYY, RRyy, rrYY, rryy)
Test cross of the F₁ (RrYy × rryy)	1 : 1 : 1 : 1

The counting formulas — learn these three lines For an individual heterozygous at n gene pairs, with genes unlinked and complete dominance:

Gamete types	F₂ genotypes	F₂ phenotypes	F₂ phenotypic ratio	Punnett boxes
2ⁿ	3ⁿ	2ⁿ	(3 : 1)ⁿ expanded	4ⁿ

Count only the heterozygous pairs. AaBBCcDd is heterozygous at A, C and D, so n = 3 and it makes 8 gamete types, not 16.

Faster than a Punnett square — the multiplication (forked-line) method The probability of any genotype or phenotype is the product of the separate monohybrid probabilities.
In AaBbCc × AaBbCc: P(A_B_cc) = ¾ × ¾ × ¼ = 9/64, and P(aabbcc) = ¼ × ¼ × ¼ = 1/64. Never draw a 64-box square in the exam — multiply.

4. Deviations from Mendelism — the highest-yield section in the chapter

(a) Incomplete dominance

In Mirabilis jalapa (four o'clock plant) and Antirrhinum majus (snapdragon or dog flower): red (RR) × white (rr) gives a pink F₁ (Rr). The F₁ is intermediate because one dose of the functional enzyme is not enough to make the full red pigment.

Rr × Rr	R	r
R	RR · Red	Rr · Pink
r	Rr · Pink	rr · White

Fig. 4 — Incomplete dominance in Mirabilis. F₂ = 1 Red : 2 Pink : 1 White

Trap — the 1:2:1 that examiners love In incomplete dominance the phenotypic ratio equals the genotypic ratio (1 : 2 : 1), because every genotype now has its own appearance. The genotypic ratio never changes — it is 1 : 2 : 1 in a normal monohybrid cross too. What changed is only the phenotypic ratio (3 : 1 → 1 : 2 : 1). If an option says "incomplete dominance alters the genotypic ratio", it is wrong.

Second trap: the law of segregation is NOT violated here. The alleles still separate cleanly; only dominance fails. NEET has asked exactly this.

(b) Codominance and the ABO blood group system

In codominance both alleles express themselves fully and independently in the heterozygote — you can see both, not a blend. The textbook case is human ABO blood grouping.

Feature	Detail
Gene	Gene I — it codes for a glycosyl transferase enzyme that adds a sugar to the surface of the RBC
Alleles	Three: I^A, I^B, i — this makes it a case of multiple allelism
What each does	I^A adds sugar A (N-acetylgalactosamine); I^B adds sugar B (galactose); i adds no sugar at all
Dominance pattern	I^A and I^B are codominant to each other; both are completely dominant over i
Counts	6 genotypes, 4 phenotypes; any one person carries only 2 of the 3 alleles

Blood group (phenotype)	A	B	AB	O
Genotypes	I^AI^A , I^Ai	I^BI^B , I^Bi	I^AI^B	ii
Antigen on RBC	A	B	A and B	none

Trap — three things wrongly said about ABO 1. "Multiple alleles means an individual has three alleles." False. Multiple alleles exist in the population; a diploid individual still carries only two, one on each homologue.
2. "ABO is only codominance." Incomplete. It is multiple allelism + codominance together, and I^A/I^B over i is plain complete dominance. One system, three phenomena.
3. Parent–child impossibilities. An AB × O cross can give only A or B children — never AB and never O. An O child needs an i from each parent, so an AB parent can never have an O child.

(c) Pleiotropy — one gene, many effects

A single gene producing multiple phenotypic effects. It usually happens when the gene controls an early step of a metabolic pathway, so a defect ripples outward.

Phenylketonuria (PKU) — one defective gene causes mental retardation and reduced hair and skin pigmentation.
Sickle-cell anaemia — one substitution causes anaemia, sickling, vaso-occlusive crises and organ damage.
Starch synthesis in pea seeds (the NCERT example that is most often missed): allele B makes large starch grains and round seeds; bb makes small grains and wrinkled seeds; Bb makes round seeds but intermediate-sized starch grains.

The starch-grain lesson NEET keeps testing For seed shape, B is completely dominant (Bb looks round). For starch grain size, the same Bb is intermediate — incomplete dominance. Conclusion, in NCERT's own words: dominance is not an autonomous property of a gene or its product; it depends on which level of the phenotype you examine. The same allele pair can be "dominant" at one level and "incompletely dominant" at another.

(d) Polygenic (quantitative) inheritance

A trait controlled by three or more genes, each contributing a small additive effect, and influenced by the environment. The result is a continuous range with a bell-shaped distribution, not discrete classes.

Feature	Detail
NCERT examples	Human skin colour and human height
Model used	Three genes A, B, C. Each capital (dark) allele adds one unit of melanin.
Extremes	AABBCC = darkest (6 dark alleles) · aabbcc = lightest (0) · AaBbCc = intermediate (3)
Phenotype depends on	The number of dark alleles, not on which particular genes they sit in. AABbcc, AaBbCc and aaBBCc with the same count look alike.
F₂ of AaBbCc × AaBbCc	7 phenotypic classes in the ratio 1 : 6 : 15 : 20 : 15 : 6 : 1 (out of 64). Only 1/64 is as dark as AABBCC and 1/64 as light as aabbcc.

Trap — polygeny vs pleiotropy They are opposites and options love to swap them. Polygeny = many genes → one trait (skin colour). Pleiotropy = one gene → many traits (PKU). Hook: "poly-genic" has the genes in the plural on the input side. Also do not confuse polygenic inheritance with multiple allelism: multiple alleles are many versions of one gene at one locus; polygenes are different genes at different loci.

5. Chromosomal theory of inheritance

Point	Detail
Proposed by	Walter Sutton and Theodore Boveri, 1902
Sutton's contribution	He united the knowledge of chromosome segregation with Mendel's principles and called it the chromosomal theory of inheritance
Core idea	Chromosomes and genes occur in pairs; both members of a pair segregate at gamete formation so that a gamete gets only one of each; and the pairs assort independently of one another
Experimental proof	Given later by Thomas Hunt Morgan working on Drosophila melanogaster

Trap — who did what Sutton and Boveri = proposed the chromosomal theory. Morgan = experimentally proved it (and discovered linkage). Sturtevant = made the first genetic map. Henking = discovered the X body. Options routinely credit Morgan with the proposal.

6. Linkage and recombination — Morgan's Drosophila work

Why Drosophila? It grows on simple synthetic medium; completes its life cycle in about two weeks; a single mating gives a large number of progeny; males and females are easily distinguished; it has many visible hereditary variations that can be seen with a low-power microscope.

The finding. Morgan carried out dihybrid crosses in Drosophila and found the F₂ ratio deviated sharply from 9 : 3 : 3 : 1. Genes that sat on the same chromosome tended to stay together in the parental combinations. He called this physical association linkage, and the generation of non-parental combinations recombination.

Morgan's cross	Genes involved	Recombination frequency	Interpretation
Cross A	white eye (w) and yellow body (y)	1.3 %	Tightly linked — very close on the chromosome
Cross B	white eye (w) and miniature wing (m)	37.2 %	Loosely linked — far apart on the chromosome

The rule that turns this into a map Recombination frequency is proportional to the physical distance between two genes. Closer genes → fewer crossovers between them → lower recombination frequency → stronger linkage. Alfred Sturtevant, Morgan's student, used recombination frequency to determine gene order and construct the first genetic map, with 1 % recombination = 1 map unit (centimorgan).

Trap — five ways this gets asked wrong

Linked genes are not "always inherited together." Crossing over can separate them; only completely linked genes (recombination ≈ 0) always travel together.
Recombination frequency has a ceiling of 50 %. At 50 % the genes behave as if unlinked — either on different chromosomes, or so far apart on the same one that they assort independently.
Higher recombination = weaker linkage = greater distance. Students invert this constantly. 37.2 % is the looser pair, not the tighter one.
Crossing over occurs in the pachytene stage of prophase I, between non-sister chromatids of homologous chromosomes — not between sister chromatids.
Linkage is an exception to independent assortment only. Segregation still holds.

7. Sex determination

Henking (1891) traced a specific nuclear structure through spermatogenesis in some insects. It went into 50 % of the sperm and not the other 50 %. He called it the X body; later work identified it as a chromosome, hence X chromosome.

System	Found in	Male	Female	Heterogametic sex
XX – XO	Grasshopper and many insects	Only one X, no Y. Produces two sperm types: with X and without any sex chromosome	XX; all eggs carry one X	Male
XX – XY	Humans, Drosophila, most mammals	XY; two sperm types: X-bearing and Y-bearing	XX; one type of egg (X)	Male
ZZ – ZW	Birds; also moths and butterflies	ZZ	ZW — two egg types	Female
Haplodiploidy	Honey bee, and other Hymenoptera	Haploid drone, 16 chromosomes, from an unfertilised egg (parthenogenesis)	Diploid, 32 chromosomes, from a fertilised egg — queen or worker	—

Trap — the honey bee, an almost guaranteed question Because a drone develops from an unfertilised egg:

A drone has no father but does have a grandfather (his mother's father).
A drone cannot have sons, but can have grandsons (through his daughters).
Drones produce sperm by mitosis, not meiosis — they are already haploid, so meiosis would halve them again.
Whether a female becomes a queen or a worker is decided by diet (royal jelly), not by genotype. Both are diploid females.

Trap — "who determines the sex of the child?" The father. The mother produces only X-bearing eggs, so she cannot influence the outcome; the sperm supplies either X or Y. The chance of a son or a daughter is 50 : 50 in every pregnancy, independent of previous children. Watch for the option that says the mother is responsible, and for the one that says the ratio changes after several children of one sex.

Also: in the XX–XO system the male has an odd chromosome number and the female an even one. Grasshopper males are XO, not XY.

8. Mutation

A mutation is a change in the DNA sequence that leads to a change in genotype and often phenotype. It is one of the sources of variation and is central to evolution.

Type	What happens	Example
Point mutation	Change in a single base pair of DNA	Sickle-cell anaemia
Frameshift mutation	Insertion or deletion of one or two bases, shifting the reading frame from that point onwards	Many severe loss-of-function disorders
Deletion / insertion of a whole segment	Loss or gain of a chromosome segment — a chromosomal aberration	Often seen in cancer cells
Chromosomal aberrations	Deletion, duplication, inversion, translocation	—

Mutagens — agents that induce mutations: UV radiation, X-rays and other ionising radiation, and chemical mutagens.

Trap — insertion or deletion of THREE bases A frameshift needs one or two bases added or removed. If three (or a multiple of three) bases are inserted or deleted, the reading frame is preserved and only one amino acid is gained or lost — it is not a frameshift mutation. This distinction is a favourite.

9. Pedigree analysis

A pedigree is a family-tree record of a trait across generations. It is the human substitute for a controlled cross, since we cannot arrange matings in people.

Pattern	Tell-tale signature in a pedigree	Examples
Autosomal dominant	Appears in every generation; an affected child always has at least one affected parent; both sexes equally affected	Myotonic dystrophy
Autosomal recessive	Skips generations; can appear from two unaffected (carrier) parents; both sexes equally affected	Sickle-cell anaemia, PKU, cystic fibrosis, thalassemia
X-linked recessive	Far more males than females affected; never passes father → son; an affected male gets it from his mother	Haemophilia, red-green colour blindness

10. Mendelian disorders in humans

Caused by an alteration or mutation in a single gene; they follow Mendelian transmission and their pattern can be traced by pedigree analysis.

Disorder	Inheritance	Defect and key facts
Haemophilia	X-linked recessive, sex-linked	A single protein in the blood-clotting cascade is not made, so a minor cut bleeds continuously. Transmitted from an unaffected carrier female to some of her sons. Famous in the royal families of Europe — the "royal disease" traced to Queen Victoria.
Colour blindness (red-green)	X-linked recessive	Defect in the red or green cone of the eye. About 8 % of males and only 0.4 % of females are affected.
Sickle-cell anaemia	Autosomal recessive	Substitution of glutamic acid by valine at the 6th position of the beta-globin chain, caused by a single base substitution in the gene (the codon GAG becomes GUG). Only HbS HbS individuals are diseased; HbA HbS heterozygotes are carriers with sickle-cell trait, who show sickling only under low oxygen tension. The mutant haemoglobin polymerises at low O₂, distorting the biconcave RBC into a sickle shape.
Thalassemia	Autosomal recessive	Reduced or absent synthesis of one of the globin chains due to mutation or deletion. α-thalassemia — two closely linked genes HBA1 and HBA2 on chromosome 16 (four gene copies in a diploid); β-thalassemia — the HBB gene on chromosome 11. The result is an abnormal ratio of α to β chains and anaemia.
Phenylketonuria (PKU)	Autosomal recessive	Lack of the enzyme phenylalanine hydroxylase, so phenylalanine is not converted to tyrosine. It accumulates and is converted to phenylpyruvic acid and other derivatives, which build up in the brain causing mental retardation; they are also excreted in the urine because the kidney reabsorbs them poorly. Also pleiotropic: reduced hair and skin pigmentation.
Cystic fibrosis, myotonic dystrophy	Recessive / dominant respectively	Listed by NCERT among Mendelian disorders — know the names.

Trap — sickle cell vs thalassemia, the classic confusion pair Sickle-cell anaemia is a QUALITATIVE problem — the right amount of globin is made, but it is structurally wrong (one wrong amino acid). Thalassemia is a QUANTITATIVE problem — the globin that is made is structurally normal, but too little of it is made. Hook: "Thala = Too Little." Both are autosomal recessive blood disorders, so options mix them freely.

Trap — the sickle-cell details that get swapped 6th position (not 5th or 7th) · of the beta chain (not alpha) · glutamic acid → valine (not the reverse) · caused by substitution, a point mutation (not a frameshift or deletion) · the heterozygote is a carrier, not a patient, and is protected against malaria — which is why the allele persists at high frequency in malarial regions.

11. Chromosomal disorders in humans

Caused by the absence, excess or abnormal arrangement of one or more chromosomes, not by a single-gene mutation.

Term	Meaning
Aneuploidy	Gain or loss of one or a few chromosomes, caused by the failure of chromatids to segregate during cell division (non-disjunction). Trisomy = 2n + 1; monosomy = 2n – 1.
Polyploidy	An increase in the whole set of chromosomes, from failure of cytokinesis after the telophase stage. Common in plants.

Disorder	Karyotype	Description
Down syndrome described by Langdon Down, 1866	47, trisomy of chromosome 21 (45 autosomes + XX or XY)	Short statured with a small round head; furrowed tongue and partially open mouth; palm is broad with a characteristic palm crease; physical, psychomotor and mental development is retarded. Risk rises with increasing maternal age.
Klinefelter syndrome	47, XXY (44 + XXY)	Overall masculine build, but with feminine development including development of the breast (gynaecomastia). Such individuals are sterile.
Turner syndrome	45, X0 (44 + X0)	Sterile female with rudimentary ovaries and a lack of other secondary sexual characters.

Trap — counting chromosomes and Barr bodies Down = 47 and it is an AUTOSOMAL trisomy. Klinefelter (47) and Turner (45) are sex-chromosomal. Turner is the only one of the three with fewer than 46.

Barr bodies = (number of X chromosomes) – 1. Normal female XX → 1 · normal male XY → 0 · Klinefelter XXY → 1 · Turner X0 → 0. So a Klinefelter male has a Barr body and a Turner female has none — a favourite reversal in options.

One more: Turner is 44 + X0, written as 45, and students frequently write 45 + X0 by mistake.

One-line separators for the whole disorder block Mendelian disorder = single-gene mutation, traced by pedigree analysis (haemophilia, colour blindness, sickle cell, thalassemia, PKU, cystic fibrosis, myotonic dystrophy).
Chromosomal disorder = whole chromosome excess or absence, seen in a karyotype (Down, Klinefelter, Turner).
Sickle cell, thalassemia, PKU, cystic fibrosis are autosomal recessive. Haemophilia and colour blindness are X-linked recessive. Myotonic dystrophy is autosomal dominant.

Part 2

Trap-solved question bank — 30 questions

About this set These are NEET past-year questions and NEET-pattern questions built directly on NCERT lines — chosen because each one has a specific trap that makes good students lose the mark. Cover the answer box, attempt the question, then read the full working. The TRAP line tells you what the question was actually testing.

Q1 · Conditional probability

In a monohybrid cross between a tall and a dwarf pea plant, the F₂ generation shows a 3 : 1 ratio. If one tall plant is picked at random from the F₂, the probability that it is heterozygous is:

(a) 1/4
(b) 1/2
(c) 2/3
(d) 3/4

Answer: (c) 2/3

Working. F₂ genotypes are 1 TT : 2 Tt : 1 tt. You are told the plant is tall, so tt is eliminated and the sample space shrinks to 3 plants: 1 TT and 2 Tt. P(heterozygous | tall) = 2/3.
TRAP. The tempting answer is 1/2 (Tt is 2 out of 4). That would be right only if the question said "picked at random from the F₂". The words "one tall plant" convert this into a conditional probability and change the denominator from 4 to 3. Any question that first gives you a phenotype and then asks for a genotype probability is doing this.

Q2 · NCERT table recall

Which of the following combinations of dominant traits in Pisum sativum is entirely correct?

(a) Green seed, yellow pod, terminal flower
(b) Yellow seed, green pod, axial flower
(c) Yellow seed, yellow pod, axial flower
(d) Green seed, green pod, terminal flower

Answer: (b)

Working. Dominant traits: seed colour yellow, pod colour green, flower position axial, seed shape round, pod shape inflated, flower colour violet, stem height tall.
TRAP. Option (c) traps anyone who assumes "yellow is dominant everywhere", and (a)/(d) trap the reverse assumption. The seed and the pod behave oppositely, and this single fact is the whole question. Hook: a ripe pea is a green pod holding a yellow seed — both dominant.

Q3 · Gamete counting

How many different types of gametes can be produced by an individual with the genotype AaBBCcDdEE?

(a) 4
(b) 8
(c) 16
(d) 32

Answer: (b) 8

Working. Only heterozygous pairs generate variety. Here A, C and D are heterozygous (n = 3); BB and EE are homozygous and contribute only one kind of allele each. Gamete types = 2ⁿ = 2³ = 8.
TRAP. Students count all five gene pairs and answer 2⁵ = 32. Circle the heterozygous pairs with your pencil before you compute anything. Same rule for the follow-ups: number of F₂ phenotypes on selfing = 2³ = 8, genotypes = 3³ = 27.

Q4 · Linkage changes the ratio

A dihybrid F₁ (AaBb) is test-crossed. If the two genes lie on the same chromosome and are completely linked, the expected phenotypic ratio in the progeny is:

(a) 9 : 3 : 3 : 1
(b) 1 : 1 : 1 : 1
(c) 1 : 1
(d) 3 : 1

Answer: (c) 1 : 1

Working. With complete linkage there is no crossing over between the two loci, so the F₁ makes only the two parental gamete types (AB and ab), in equal numbers. Crossed to the double recessive (ab), the progeny are AaBb and aabb in a 1 : 1 ratio — the two recombinant classes are absent.
TRAP. 1 : 1 : 1 : 1 is the answer for an unlinked dihybrid test cross, and it is the reflex choice. Read for the words "same chromosome", "linked" or "linkage" — they are the signal that independent assortment has been switched off. If linkage were incomplete, you would get all four classes, but with the two parental types in large excess and the two recombinants rare.

Q5 · Blood group logic

A man of blood group AB marries a woman of blood group A whose father was of blood group O. What is the probability that their first child will be of blood group O?

(a) 0
(b) 1/4
(c) 1/2
(d) 3/4

Answer: (a) 0 — it is impossible

Working. The woman is blood group A but her father was O (ii), so he could give her only an i. She must be I^Ai. The man is I^AI^B. His gametes are I^A or I^B — he has no i allele at all. Blood group O requires ii, one i from each parent. Therefore P(O) = 0. The possible children are I^AI^A (A), I^Ai (A), I^AI^B (AB), I^Bi (B) — that is 1/2 A, 1/4 AB, 1/4 B.
TRAP. The detail "whose father was O" is placed there to make you spend time deducing the mother's genotype, and it does matter for the A : AB : B split — but it is irrelevant to the O question, because the AB father alone already rules O out. Learn the shortcut: an AB parent can never have an O child, and an O parent can never have an AB child.

Q6 · Sex-linked inheritance

A colour-blind man marries a woman who is homozygous normal for colour vision. Regarding their children:

(a) All sons will be colour blind
(b) All daughters will be colour blind
(c) All daughters will be carriers and all sons will be normal
(d) Half the sons and half the daughters will be colour blind

Answer: (c)

Working. Father is X^cY, mother is X^CX^C. Daughters get X^c from the father and X^C from the mother → X^CX^c, carriers with normal vision. Sons get their single X from the mother (X^C) and the Y from the father → X^CY, normal.
TRAP. The governing rule is that a father gives his X only to his daughters and his Y only to his sons. So an X-linked recessive condition never passes from father to son. Any option offering affected sons from an affected father and a normal-homozygous mother is automatically wrong. This is also called criss-cross inheritance: the father's X-linked trait reaches his grandsons through his daughters.

Q7 · Pedigree deduction

A phenotypically normal couple has a haemophilic son. The genotypes of the father and mother respectively are:

(a) X^hY and X^HX^H
(b) X^HY and X^HX^h
(c) X^hY and X^HX^h
(d) X^HY and X^HX^H

Answer: (b)

Working. The son is X^hY. His Y came from his father, so his X^h must have come from his mother. The mother is phenotypically normal, so she is heterozygous — a carrier, X^HX^h. The father is normal, therefore X^HY. Note his genotype had no influence on this son at all.
TRAP. Students instinctively hunt for the defective allele in the father. For an X-linked recessive condition in a son, the father is genetically irrelevant — always trace the son's X back to the mother. A daughter is different: she would need X^h from both parents, which is why affected females are rare and require an affected father plus at least a carrier mother.

Q8 · Deviation from Mendelism

In Mirabilis jalapa, a cross between red-flowered and white-flowered plants gives pink F₁. On selfing the F₁, the F₂ shows:

(a) Phenotypic ratio 3 : 1 and genotypic ratio 1 : 2 : 1
(b) Phenotypic ratio 1 : 2 : 1 and genotypic ratio 1 : 2 : 1
(c) Phenotypic ratio 1 : 2 : 1 and genotypic ratio 3 : 1
(d) Phenotypic ratio 2 : 1 : 1 and genotypic ratio 1 : 2 : 1

Answer: (b)

Working. Incomplete dominance gives the heterozygote its own appearance, so each of the three genotypes is a separate phenotype: 1 RR red : 2 Rr pink : 1 rr white.
TRAP. The exam sentence to hold on to is: "the genotypic ratio of a monohybrid F₂ is ALWAYS 1 : 2 : 1 — dominance affects only the phenotypic ratio." Options like (c) exist purely to catch students who think incomplete dominance somehow changes the genotypes. Nothing about segregation has changed; only the visible expression has.

Q9 · Which law survives?

Which of Mendel's laws is not violated in cases of incomplete dominance and codominance?

(a) Law of dominance
(b) Law of segregation
(c) Law of independent assortment
(d) Both (a) and (c)

Answer: (b)

Working. Incomplete dominance and codominance are exceptions to the law of dominance only. The alleles still separate cleanly into different gametes at anaphase I, so the law of segregation holds perfectly — indeed the clean 1 : 2 : 1 F₂ is direct proof of it. Independent assortment is not being tested at all in a monohybrid situation.
TRAP. Remember the exception map once and for all: dominance → broken by incomplete dominance and codominance; independent assortment → broken by linkage; segregation → never broken. Option (d) is designed for the student who half-remembers "two of the three laws have exceptions" without knowing which two.

Q10 · Pleiotropy at two levels

In pea, the gene for starch synthesis shows that a heterozygote (Bb) produces round seeds but starch grains of intermediate size. This demonstrates that:

(a) The gene shows codominance for seed shape
(b) Dominance is not an autonomous feature of a gene; it depends on the level of the phenotype examined
(c) Two different genes control seed shape and starch grain size
(d) The trait is polygenic

Answer: (b)

Working. The same allele pair behaves as completely dominant when you score seed shape (Bb looks round like BB) and as incompletely dominant when you measure starch grain size (Bb is intermediate). One gene, several effects — this is pleiotropy — and the apparent kind of dominance depends on which effect you choose to observe.
TRAP. Option (c) is very tempting because two different characters are being described, but NCERT is explicit that it is a single gene. This is one of the few places where the textbook makes a conceptual argument rather than stating a fact, and it therefore gets converted into a question often.

Q11 · Polygenic numerical

Human skin colour is controlled by three genes (A, B, C), each dominant allele adding one unit of pigment. If AaBbCc × AaBbCc, the fraction of the offspring expected to be as dark as the darkest possible individual is:

(a) 1/8
(b) 1/16
(c) 1/64
(d) 20/64

Answer: (c) 1/64

Working. The darkest genotype is AABBCC (six dark alleles). P(AA) = 1/4, P(BB) = 1/4, P(CC) = 1/4, so P = 1/4 × 1/4 × 1/4 = 1/64. Likewise the lightest, aabbcc, is 1/64. The full F₂ distribution over 0–6 dark alleles is 1 : 6 : 15 : 20 : 15 : 6 : 1 out of 64.
TRAP. Option (d), 20/64, is the count of the intermediate class (three dark alleles) — the most frequent class, and the correct answer to a differently worded question. Also remember the concept the numerical is protecting: the phenotype depends on the total number of dark alleles, not on which genes they belong to, so AABbcc, AaBbCc and aaBBCc are all equally pigmented.

Q12 · Morgan's numbers

Morgan found a recombination frequency of 1.3 % between the genes for white eye and yellow body, and 37.2 % between white eye and miniature wing. This means:

(a) White eye and miniature wing are more tightly linked
(b) White eye and yellow body are more tightly linked and lie closer together
(c) White eye and yellow body lie on different chromosomes
(d) Miniature wing and yellow body are completely linked

Answer: (b)

Working. Recombination frequency is proportional to the distance between two genes. A low value (1.3 %) means very few crossovers occur between them, so they are close together and tightly linked. A high value (37.2 %) means they are far apart and loosely linked.
TRAP. The relationship gets inverted under exam pressure — a "big number" feels like "strong linkage". Fix it with: big recombination = big distance = weak linkage. Also note the ceiling: recombination frequency cannot exceed 50 %. At 50 % two genes behave exactly as if they were on different chromosomes, so a value near 50 % does not prove they are on different chromosomes — it only means you cannot tell them apart from unlinked genes by this test. Sturtevant used these frequencies to build the first genetic maps.

Q13 · Honey bee

Which of the following statements about a honey bee drone is incorrect?

(a) It develops from an unfertilised egg by parthenogenesis
(b) It is haploid with 16 chromosomes
(c) It produces sperm by meiosis
(d) It has a grandfather but no father

Answer: (c) — this is the incorrect statement

Working. A drone is already haploid (n = 16). Meiosis would halve the number again and produce non-viable gametes, so drones make sperm by mitosis. Statements (a), (b) and (d) are all true: he has no father because no sperm was involved in his origin, but his diploid mother did have a father, so he has a grandfather. Extending the same logic — a drone cannot have sons (his sons would need to come from unfertilised eggs of a female) but he can have grandsons through his daughters.
TRAP. The "no father but has a grandfather" line is famous enough that students expect it to be the answer. Read the word "incorrect" in the stem. Also keep separate: queen vs worker is decided by diet (royal jelly), male vs female is decided by fertilisation.

Q14 · Sex determination systems

In which of the following is the female the heterogametic sex?

(a) Human beings
(b) Grasshopper
(c) Birds
(d) Drosophila

Answer: (c) Birds

Working. Birds (and moths and butterflies) follow the ZZ–ZW system: the male is ZZ (homogametic, one type of sperm) and the female is ZW (heterogametic, two types of egg). Humans and Drosophila are XX–XY and the grasshopper is XX–XO — in all three the male is heterogametic.
TRAP. Two things to keep straight. First, "heterogametic" means producing two kinds of gamete, which is not the same as "having two different chromosomes" — a grasshopper male (XO) has only one sex chromosome, yet he is heterogametic because half his sperm carry X and half carry none. Second, the letters Z and W are used precisely so that you do not read them as X and Y; ZW is female, and writing the male as ZW is the single most common slip here.

Q15 · Chromosome counting

A person with Turner syndrome has a total chromosome number of:

(a) 47
(b) 45
(c) 46
(d) 44

Answer: (b) 45

Working. Turner syndrome is 44 autosomes + a single X (X0), giving 45 in total — a monosomy. Such an individual is a sterile female with rudimentary ovaries and a lack of other secondary sexual characters.
TRAP. The written form "44 + X0" makes students add and get 45 + something. The 44 is autosomes only; the X is the 45th. Line up the three disorders once: Down = 47 (trisomy 21, autosomal) · Klinefelter = 47 (XXY, sex chromosomal) · Turner = 45 (X0, sex chromosomal). Turner is the only one below 46, and Down is the only autosomal one of the three.

Q16 · Barr bodies

The number of Barr bodies in the somatic cells of a person with Klinefelter syndrome and a person with Turner syndrome respectively is:

(a) 0 and 1
(b) 1 and 0
(c) 1 and 1
(d) 2 and 0

Answer: (b) 1 and 0

Working. Use Barr bodies = (number of X chromosomes) – 1, because all X chromosomes except one are inactivated. Klinefelter is XXY → 2 X's → 1 Barr body. Turner is X0 → 1 X → 0 Barr bodies. For reference, a normal female (XX) has 1 and a normal male (XY) has 0.
TRAP. The reversal in option (a) exploits the intuition that "a male shouldn't have a Barr body and a female should". Barr bodies count X chromosomes, not the person's sex. So a Klinefelter male is Barr-body positive and a Turner female is Barr-body negative — which is exactly why the test was historically used.

Q17 · Sickle-cell anaemia

Sickle-cell anaemia is caused by the substitution of:

(a) Valine by glutamic acid at the 6th position of the alpha chain
(b) Glutamic acid by valine at the 6th position of the beta chain
(c) Glutamic acid by valine at the 6th position of the alpha chain
(d) Valine by glutamic acid at the 6th position of the beta chain

Answer: (b)

Working. A single base substitution in the gene changes the codon GAG → GUG, so glutamic acid is replaced by valine at the sixth position of the beta-globin chain. The mutant haemoglobin (HbS) polymerises under low oxygen tension and distorts the biconcave red cell into a sickle shape.
TRAP. All four options contain the same words in different orders — the question is pure precision. Fix three things: beta chain, 6th position, and the direction glu → val (normal to mutant, never the reverse). Memory hook: "Glu leaves, Val arrives — on Beta 6." Also note this is a point mutation by substitution, not a deletion or frameshift.

Q18 · Two blood disorders compared

Which statement correctly distinguishes thalassemia from sickle-cell anaemia?

(a) Thalassemia is X-linked while sickle-cell anaemia is autosomal
(b) Thalassemia reduces the amount of globin synthesised, while sickle-cell anaemia produces a structurally abnormal globin
(c) Thalassemia is caused by a point mutation and sickle-cell anaemia by gene deletion
(d) Both are qualitative defects of haemoglobin

Answer: (b)

Working. Thalassemia is a quantitative problem — mutation or deletion reduces the synthesis of an α or β globin chain, so too little of a structurally normal protein is made and the α : β ratio becomes abnormal. Sickle-cell anaemia is a qualitative problem — normal amounts of globin are made, but one amino acid is wrong. Both are autosomal recessive, which kills option (a).
TRAP. Because both are inherited anaemias affecting haemoglobin, options mix their features freely. Hook: "Thala = Too Little" (quantity), "Sickle = Shape" (quality). Genes to remember: β-thalassemia involves HBB on chromosome 11; α-thalassemia involves HBA1 and HBA2 on chromosome 16.

Q19 · Phenylketonuria

In phenylketonuria, mental retardation results because:

(a) Tyrosine accumulates in the brain
(b) The kidney fails to filter phenylalanine
(c) Phenylalanine is not converted to tyrosine and its derivatives accumulate in the brain
(d) Melanin is overproduced in nerve tissue

Answer: (c)

Working. The affected individual lacks the enzyme phenylalanine hydroxylase, so phenylalanine cannot be converted to tyrosine. Phenylalanine accumulates and is diverted into phenylpyruvic acid and other derivatives, which build up in the brain and cause mental retardation. They are also excreted in the urine because the kidney reabsorbs them poorly.
TRAP. Option (a) swaps the two amino acids — it is the substrate (phenylalanine) that accumulates, not the product. Option (b) misstates the kidney's role: the kidney is not failing to filter, it is failing to reabsorb, which is why the compounds appear in urine. Remember that PKU is also a standard example of pleiotropy (mental retardation plus reduced hair and skin pigmentation) and is autosomal recessive.

Q20 · Mutation type

Insertion or deletion of which of the following will not cause a frameshift mutation?

(a) One base pair
(b) Two base pairs
(c) Three base pairs
(d) Four base pairs

Answer: (c) Three base pairs

Working. The genetic code is read in triplets. Adding or removing a multiple of three keeps the downstream reading frame intact — only one amino acid is gained or lost. Adding or removing one, two or four bases shifts every codon after that point, so the whole downstream protein is changed. NCERT states it as: insertion or deletion of one or two bases changes the reading frame from that point onwards — these are frameshift mutations.
TRAP. Students memorise "insertions and deletions cause frameshift" as an unqualified rule and never test it against multiples of three. Note also the vocabulary pairing: a point mutation is a change in a single base pair (sickle cell); a frameshift is an indel. Do not call sickle-cell anaemia a frameshift mutation.

Q21 · Aneuploidy vs polyploidy

Failure of chromatids to segregate during cell division leads to, and failure of cytokinesis after telophase leads to, respectively:

(a) Polyploidy and aneuploidy
(b) Aneuploidy and polyploidy
(c) Aneuploidy and monosomy
(d) Trisomy and aneuploidy

Answer: (b)

Working. Non-disjunction — the failure of chromatids to separate — gives a gamete one chromosome too many or too few, producing aneuploidy (2n + 1 trisomy, as in Down syndrome, or 2n – 1 monosomy, as in Turner syndrome). Failure of cytokinesis after telophase leaves the whole doubled chromosome set in one cell, producing polyploidy, which is common in plants.
TRAP. The two causes are stated in adjacent sentences in NCERT and are simply swapped in the options. Anchor them by scale: aneuploidy = a few chromosomes off (one chromatid pair misbehaved); polyploidy = whole sets extra (the entire cell failed to divide).

Q22 · Down syndrome

Which of the following is not a feature of Down syndrome?

(a) Presence of an additional copy of chromosome 21
(b) Broad palm with a characteristic palm crease
(c) The affected individual is sterile with rudimentary gonads
(d) Furrowed tongue and partially open mouth

Answer: (c) — this belongs to Turner syndrome

Working. NCERT's description of Down syndrome: short statured with a small round head, furrowed tongue and partially open mouth, broad palm with a characteristic palm crease, and retarded physical, psychomotor and mental development. Rudimentary ovaries and sterility describe Turner syndrome; sterility with gynaecomastia describes Klinefelter.
TRAP. All three syndromes are taught in one block, so features get pooled in the mind. Keep one signature each: Down → palm crease and furrowed tongue; Klinefelter → gynaecomastia; Turner → rudimentary ovaries. Also remember the cause is non-disjunction, the risk rises with maternal age, and the total count is 47. Langdon Down described it in 1866.

Q23 · Who did what

The chromosomal theory of inheritance was proposed by, and experimentally proved by, respectively:

(a) Morgan and Sturtevant
(b) Mendel and Punnett
(c) Sutton and Boveri; Morgan
(d) Henking and Sutton

Answer: (c)

Working. Sutton and Boveri (1902) proposed the chromosomal theory of inheritance, Sutton having united the behaviour of chromosomes with Mendel's principles. Thomas Hunt Morgan supplied the experimental proof using Drosophila melanogaster, and in doing so discovered linkage. Sturtevant, his student, built the first genetic maps. Henking discovered the X body.
TRAP. Morgan's name dominates the chapter, so he is wrongly credited with the proposal. Fix the sequence as a chain: Henking (X body, 1891) → Sutton & Boveri (theory, 1902) → Morgan (proof + linkage) → Sturtevant (maps). And keep Punnett out of it — his contribution is the checkerboard, nothing more.

Q24 · Reading the dihybrid F₂

In the F₂ generation of a dihybrid cross RrYy × RrYy, the proportion of individuals that are homozygous for both genes and the proportion showing new (recombinant) phenotypic combinations are respectively:

(a) 1/16 and 9/16
(b) 4/16 and 6/16
(c) 4/16 and 10/16
(d) 9/16 and 6/16

Answer: (b)

Working. Fully homozygous genotypes in the 16 boxes are RRYY, RRyy, rrYY and rryy — one box each, so 4/16 = 1/4. The four phenotypic classes are 9 round-yellow : 3 round-green : 3 wrinkled-yellow : 1 wrinkled-green. The parental combinations are round-yellow and wrinkled-green (9 + 1 = 10/16); the new combinations are the two middle classes, 3 + 3 = 6/16.
TRAP. Option (c) is the parental fraction, offered to whoever reads "new" as "old". Option (a) counts only rryy as homozygous, forgetting that RRYY, RRyy and rrYY are homozygous too. Read the phrase carefully: "homozygous for both genes" ≠ "recessive for both genes."

Assertion–Reason: use these four options for Q25–Q27 (a) Both A and R are true and R is the correct explanation of A · (b) Both A and R are true but R is not the correct explanation of A · (c) A is true but R is false · (d) A is false but R is true

Q25 · Assertion–Reason

Assertion: Genes located on the same chromosome do not always assort independently.
Reason: Linked genes can be separated from one another by crossing over during meiosis.

Answer: (b) — both true, but R is not the correct explanation of A

Working. The Assertion is true: genes on the same chromosome tend to be inherited together, which is linkage, the exception to independent assortment. The Reason is also true: crossing over does separate linked genes and generate recombinants. But crossing over is the mechanism that partially restores independent assortment; it is not the reason independent assortment fails. The correct explanation of the Assertion is physical linkage on a single chromosome.
TRAP. Both statements are lifted from the chapter and both are correct, which pushes students straight to (a). In assertion–reason questions always ask the third question: does R explain A, or does R describe something that works against A? Here R works in the opposite direction.

Q26 · Assertion–Reason

Assertion: Haemophilia is far more common in males than in females.
Reason: A female becomes haemophilic only when she inherits the recessive allele from both parents, which requires a haemophilic father and at least a carrier mother.

Answer: (a) — both true and R correctly explains A

Working. Haemophilia is X-linked recessive. A male has a single X, so one recessive allele is enough to make him haemophilic. A female has two X's and needs the recessive allele on both, which requires an affected father (who gives his only X to every daughter) plus a mother who is at least a carrier. Such a mating is rare, and NCERT notes that this situation is extremely rare. So the Reason is exactly why the Assertion holds.
TRAP. The wrong instinct is to write the reason as "because the gene is on the Y chromosome" or "because males are hemizygous for Y". The relevant fact is that males are hemizygous for X — they have only one copy of every X-linked gene, so recessive X-linked alleles are never masked. The same logic explains why 8 % of males but only 0.4 % of females are red-green colour blind.

Q27 · Assertion–Reason

Assertion: The sickle-cell allele has remained at a high frequency in some human populations despite being harmful in the homozygous state.
Reason: Heterozygous individuals (HbA HbS) show sickling of red blood cells only under low oxygen tension and have a survival advantage in malaria-endemic regions.

Answer: (a) — both true and R correctly explains A

Working. Only HbS HbS individuals are affected by the disease. The heterozygote HbA HbS is a carrier who is essentially healthy, showing sickling only when oxygen tension falls. Because the carrier state gives partial protection against malaria, the allele is retained in populations where malaria is common, even though homozygotes are severely affected. This is the standard example of heterozygote advantage.
TRAP. Two things to keep straight. First, the heterozygote is a carrier, not a patient — options describing HbA HbS as "mildly diseased at all times" are wrong; the qualifier "under low oxygen tension" is essential. Second, the disorder is inherited as an autosomal recessive condition, so both parents must be at least carriers; two carriers give 1/4 affected, 1/2 carriers, 1/4 normal.

Q28 · Match the following — scientists

Match Column I with Column II and choose the correct option.

Column I	Column II
(i) Henking	(A) First genetic map using recombination frequency
(ii) Sutton and Boveri	(B) Discovery of the X body
(iii) Sturtevant	(C) Chromosomal theory of inheritance
(iv) Morgan	(D) Discovery of linkage in Drosophila

(a) i-A, ii-C, iii-B, iv-D
(b) i-B, ii-C, iii-A, iv-D
(c) i-B, ii-D, iii-A, iv-C
(d) i-C, ii-B, iii-D, iv-A

Answer: (b) i-B, ii-C, iii-A, iv-D

Working. Henking (1891) traced a nuclear structure through insect spermatogenesis and named it the X body. Sutton and Boveri (1902) proposed the chromosomal theory of inheritance. Sturtevant used recombination frequency to construct the first genetic map. Morgan proved the chromosomal theory experimentally and discovered linkage in Drosophila.
TRAP. Sturtevant and Morgan are the pair that gets swapped, because Sturtevant worked in Morgan's lab. Attach the map to the student, the linkage discovery to the teacher. In matching questions, lock the one pair you are certain of first and eliminate — here fixing i-B alone removes options (a) and (d) immediately.

Q29 · Match the following — disorders

Match Column I with Column II and choose the correct option.

Column I	Column II
(i) Klinefelter syndrome	(A) 45, X0
(ii) Down syndrome	(B) Autosomal recessive, single amino acid substitution
(iii) Turner syndrome	(C) 47, XXY
(iv) Sickle-cell anaemia	(D) 47, trisomy of chromosome 21

(a) i-A, ii-D, iii-C, iv-B
(b) i-C, ii-A, iii-D, iv-B
(c) i-C, ii-D, iii-A, iv-B
(d) i-D, ii-C, iii-A, iv-B

Answer: (c) i-C, ii-D, iii-A, iv-B

Working. Klinefelter = 47, XXY (masculine build with gynaecomastia, sterile). Down = 47, trisomy 21. Turner = 45, X0 (sterile female, rudimentary ovaries). Sickle-cell anaemia = autosomal recessive, glutamic acid replaced by valine at the 6th position of the beta chain.
TRAP. Both Klinefelter and Down are 47, so the number alone does not separate them — you must read whether the extra chromosome is a sex chromosome (XXY) or an autosome (21). Options (a) and (d) are built precisely on that ambiguity.

Q30 · Match the following — inheritance patterns

Match Column I with Column II and choose the correct option.

Column I	Column II
(i) Mirabilis jalapa flower colour	(A) Multiple allelism and codominance
(ii) ABO blood groups	(B) Pleiotropy
(iii) Human skin colour	(C) Incomplete dominance
(iv) Phenylketonuria	(D) Polygenic inheritance

(a) i-C, ii-A, iii-B, iv-D
(b) i-C, ii-A, iii-D, iv-B
(c) i-A, ii-C, iii-D, iv-B
(d) i-C, ii-B, iii-D, iv-A

Answer: (b) i-C, ii-A, iii-D, iv-B

Working. Mirabilis red × white → pink is the standard case of incomplete dominance. ABO is the standard case of multiple allelism together with codominance. Human skin colour is polygenic (three genes, additive effect). PKU is pleiotropic — one gene causing mental retardation as well as reduced hair and skin pigmentation, and it is also autosomal recessive.
TRAP. The B and D swap in option (a) is the whole point: polygeny = many genes → one trait; pleiotropy = one gene → many traits. They sound similar and sit only a page apart in the textbook. Say the definition in the direction of the arrow every time you meet either word.

Part 3

Master revision sheet

This is the sheet to revise on the last day. Everything numerical or list-like in the chapter is here in one place.

A. Every ratio in the chapter

Cross	Phenotypic ratio	Genotypic ratio	Note
Monohybrid F₂ (Tt × Tt)	3 : 1	1 : 2 : 1	Complete dominance
Monohybrid test cross (Tt × tt)	1 : 1	1 : 1	Identifies a heterozygote
Back cross to dominant parent (Tt × TT)	All dominant	1 TT : 1 Tt	Not a test cross
Incomplete dominance F₂ (Rr × Rr)	1 : 2 : 1	1 : 2 : 1	Mirabilis, Antirrhinum
Codominance F₂	1 : 2 : 1	1 : 2 : 1	Heterozygote shows both
I^Ai × I^Bi	1 A : 1 B : 1 AB : 1 O	1 : 1 : 1 : 1	All four blood groups from one couple
Dihybrid F₂ (RrYy × RrYy)	9 : 3 : 3 : 1	1:2:1:2:4:2:1:2:1 9 genotypes	Unlinked genes
Dihybrid test cross (RrYy × rryy)	1 : 1 : 1 : 1	1 : 1 : 1 : 1	Gamete ratio read directly
Dihybrid test cross, complete linkage	1 : 1	1 : 1	Only parental types
Dihybrid test cross, incomplete linkage	Parental ≫ recombinant	—	4 classes, very unequal
Trihybrid F₂ (AaBbCc × AaBbCc)	27:9:9:9:3:3:3:1	27 genotypes	8 gametes, 64 boxes
Trihybrid test cross	1:1:1:1:1:1:1:1	—	8 equal classes
Polygenic, 3 genes (AaBbCc selfed)	1:6:15:20:15:6:1	—	7 classes out of 64

Formulas for n heterozygous gene pairs Gamete types 2ⁿ · Punnett boxes 4ⁿ · F₂ genotypes 3ⁿ · F₂ phenotypes 2ⁿ · F₂ phenotypic ratio = expansion of (3 : 1)ⁿ · fully homozygous individuals in F₂ = (1/4)ⁿ · number of phenotypic classes in polygenic inheritance with n genes = 2n + 1.

B. Blood group cross-reference

Parents	Possible children	Impossible children
AB × O	A, B	AB and O
AB × AB	A, B, AB (1 : 1 : 2)	O
O × O	O only	A, B, AB
I^Ai × I^Bi	A, B, AB, O (1:1:1:1)	—
I^AI^A × anything	never O	O

Quick rule: an O child needs an i from each parent; an AB child needs I^A from one parent and I^B from the other. Test every option against these two lines.

C. Sex-linked cross outcomes — learn these four rows cold

Cross (X-linked recessive)	Daughters	Sons
Affected father × normal homozygous mother	All carriers, none affected	All normal
Normal father × carrier mother	1/2 carriers, 1/2 normal	1/2 affected, 1/2 normal
Affected father × carrier mother	1/2 affected, 1/2 carriers	1/2 affected, 1/2 normal
Affected father × affected mother	All affected	All affected

Three lines that solve almost every sex-linkage question 1. A father passes his X to every daughter and his Y to every son — so no father-to-son transmission of X-linked traits.
2. An affected son always got the allele from his mother.
3. An affected daughter needs it from both parents, so her father must be affected — which is why affected females are rare.

D. Disorders at a glance

Disorder	Type	Chromosome / gene	Signature fact
Haemophilia	Mendelian, X-linked recessive	X chromosome	A single clotting-cascade protein missing; the "royal disease"
Colour blindness	Mendelian, X-linked recessive	X chromosome	Red or green cone defective; 8 % of males, 0.4 % of females
Sickle-cell anaemia	Mendelian, autosomal recessive	β-globin gene	Glu → Val at position 6 of the beta chain; GAG → GUG; heterozygote is a carrier
Thalassemia	Mendelian, autosomal recessive	α: HBA1, HBA2 on chr 16 β: HBB on chr 11	Quantitative — too little globin made
Phenylketonuria	Mendelian, autosomal recessive	Phenylalanine hydroxylase gene	Phenylpyruvic acid accumulates in brain, excreted in urine; also pleiotropic
Cystic fibrosis	Mendelian, autosomal recessive	—	Listed by NCERT among Mendelian disorders
Myotonic dystrophy	Mendelian, autosomal dominant	—	The dominant one in the list
Down syndrome	Chromosomal, autosomal trisomy	47, trisomy 21	Palm crease, furrowed tongue; risk rises with maternal age
Klinefelter syndrome	Chromosomal, sex chromosomal	47, XXY	Gynaecomastia; sterile; 1 Barr body
Turner syndrome	Chromosomal, sex chromosomal	45, X0	Rudimentary ovaries; sterile; 0 Barr bodies

E. Every exception you must be able to name

The general rule	The exception	Example
One allele dominates the other	Incomplete dominance	Mirabilis jalapa, Antirrhinum majus — pink F₁
Only one allele is expressed in a heterozygote	Codominance	I^AI^B — both A and B antigens present
A gene has only two alleles	Multiple allelism (in the population)	I^A, I^B, i — but any individual still has only two
One gene controls one trait	Pleiotropy	PKU; starch synthesis in pea; sickle-cell anaemia
One trait is controlled by one gene	Polygenic inheritance	Human skin colour and height
Genes assort independently	Linkage	Morgan's Drosophila crosses; ratio deviates from 9 : 3 : 3 : 1
Alleles segregate cleanly	No exception	The law of segregation is universally true
Traits fall into discrete classes	Continuous variation	Quantitative traits, influenced by the environment
The male is the heterogametic sex	ZZ–ZW system	Birds, moths, butterflies — the female is ZW
Every animal is diploid	Haplodiploidy	Honey bee drone — haploid, 16 chromosomes, no father
Gametes are made by meiosis	Drone sperm by mitosis	He is already haploid
Humans have 46 chromosomes	Aneuploidy	Down 47 · Klinefelter 47 · Turner 45
Recombination frequency can be anything	Ceiling of 50 %	At 50 % the genes appear unlinked

F. Numbers, names and dates worth one mark each

Fact	Value
Mendel's period of experimentation	1856–1863 (seven years); published 1866
Rediscovery of Mendel's work	1900 — de Vries, Correns, von Tschermak
Pea characters and true-breeding lines	7 characters, 14 lines
Henking's X body	1891
Chromosomal theory of inheritance	Sutton and Boveri, 1902
Morgan's recombination frequencies	white eye – yellow body 1.3 % · white eye – miniature wing 37.2 %
Drosophila life cycle	About two weeks; grows on simple synthetic medium
Colour blindness frequency	8 % of males, 0.4 % of females
Sickle-cell mutation	6th position, beta chain, Glu → Val; codon GAG → GUG
Honey bee chromosome numbers	Female (queen/worker) 32 · drone 16
Down syndrome	Described by Langdon Down, 1866; 47, trisomy 21
Barr body rule	Number of X chromosomes – 1
Human skin colour model	3 genes (A, B, C); 7 phenotypic classes; 1:6:15:20:15:6:1

Beyond NCERT — read once, do not memorise deeply Coaching test papers sometimes include gene-interaction ratios (9:7 complementary genes, 9:3:4 recessive epistasis, 12:3:1 dominant epistasis, 13:3, 15:1 duplicate genes) and the Bombay phenotype (an hh individual cannot make the H antigen and types as O despite carrying I^A or I^B). None of these appears in the NCERT chapter or in the NTA syllabus list. Recognise the names so an option does not startle you, but spend no revision time here until everything above is secure.

Part 4

Diagram pack — redraw each one from memory

These are the figures NCERT uses in this chapter, redrawn. Cover the labels and re-draw each on blank paper; anything you cannot reproduce is the topic to re-read.

Diagram 1 — Pedigree symbols

Fig. 5 — Standard pedigree symbols. Squares are males, circles are females; shading indicates the affected individual.

Diagram 2 — The three pedigree patterns

Fig. 6 — How to recognise the three inheritance patterns from a family tree.

Diagram 3 — Sex determination: the four systems

Fig. 7 — The four sex-determination systems. Note which sex is heterogametic in each.

Diagram 4 — Linkage, crossing over and the genetic map

Fig. 8 — Crossing over produces recombinants; recombination frequency is proportional to distance, which is how Sturtevant built the first gene map.

Diagram 5 — Sickle-cell anaemia, from base to red cell

Fig. 9 — Sickle-cell anaemia: one base changes one amino acid, which changes the shape of the whole cell. Autosomal recessive.

Diagram 6 — Non-disjunction and Down syndrome

Fig. 10 — Aneuploidy arises from the failure of chromatids to segregate. Polyploidy, by contrast, arises from the failure of cytokinesis after telophase.

Part 5

Ten-minute last look & the blunder list

The 10-minute read-through

Say each of these out loud once

Mendel: pea, 7 characters, 14 lines, 1856–63, ignored until 1900. Dominant side: tall, violet, axial, inflated, green pod, round, yellow seed.
Monohybrid F₂: 3 : 1 phenotype, 1 : 2 : 1 genotype. Dihybrid F₂: 9 : 3 : 3 : 1, 9 genotypes.
Test cross = cross with the homozygous recessive. Mono 1 : 1, di 1 : 1 : 1 : 1. Every test cross is a back cross; not every back cross is a test cross.
Gametes = 2ⁿ, counting only heterozygous pairs.
Segregation has no exception. Dominance is broken by incomplete dominance and codominance. Independent assortment is broken by linkage.
Incomplete dominance: Mirabilis, pink F₁, F₂ 1 : 2 : 1 for both ratios.
ABO: gene I, glycosyl transferase, three alleles, six genotypes, four phenotypes; I^A/I^B codominant, both dominant over i.
Pleiotropy = one gene, many traits (PKU, starch grain in pea). Polygeny = many genes, one trait (skin colour, height).
Sutton and Boveri proposed the chromosomal theory; Morgan proved it; Sturtevant mapped; Henking found the X body.
Linkage: 1.3 % tight, 37.2 % loose; recombination frequency ∝ distance; ceiling 50 %; crossing over at pachytene, between non-sister chromatids.
Heterogametic sex: male in XX-XY and XX-XO, female in ZZ-ZW (birds). Honey bee: drone haploid 16, female 32, sperm by mitosis, no father but a grandfather.
Father decides the child's sex; 50 : 50 every time.
X-linked recessive: never father to son; affected son gets it from his mother; colour blindness 8 % males, 0.4 % females.
Sickle cell: Glu → Val, position 6, beta chain, GAG → GUG, point mutation, autosomal recessive, heterozygote is a carrier.
Thala = Too Little (quantitative); sickle = shape (qualitative).
PKU: phenylalanine hydroxylase missing; phenylpyruvic acid in brain and urine.
Down 47, trisomy 21, autosomal, maternal age · Klinefelter 47 XXY, gynaecomastia, 1 Barr body · Turner 45 X0, rudimentary ovaries, 0 Barr bodies.
Barr bodies = X count minus one. Aneuploidy from non-disjunction; polyploidy from failure of cytokinesis, common in plants.
Frameshift needs one or two bases; three bases keeps the reading frame.
Mendelian disorder → pedigree analysis. Chromosomal disorder → karyotype.

25 blunders to never repeat

#	The mistake	The correction
1	Saying yellow pods and green seeds are dominant	Green pod, yellow seed are dominant
2	Answering 1/2 for "a tall F₂ plant is heterozygous"	2/3 — the tall condition removes tt from the sample space
3	Counting homozygous pairs when finding gamete types	2ⁿ where n = heterozygous pairs only
4	Thinking incomplete dominance changes the genotypic ratio	Genotypic ratio stays 1 : 2 : 1; only the phenotypic ratio changes
5	Saying segregation is violated in incomplete dominance	Only the law of dominance is violated
6	Saying an individual carries three alleles in ABO	Three alleles exist in the population; an individual has two
7	Allowing an O child from an AB parent	Impossible — an AB parent has no i allele
8	Confusing pleiotropy with polygeny	Pleiotropy = one gene → many traits; polygeny = many genes → one trait
9	Crediting Morgan with proposing the chromosomal theory	Sutton and Boveri proposed it; Morgan proved it
10	Reading 37.2 % as strong linkage	High recombination = large distance = weak linkage
11	Saying linked genes are always inherited together	Only if completely linked; crossing over separates them
12	Quoting a recombination frequency above 50 %	50 % is the maximum
13	Saying crossing over occurs between sister chromatids	Between non-sister chromatids of homologues, at pachytene
14	Making the grasshopper male XY	Grasshopper male is XO
15	Making the male bird ZW	Male bird is ZZ; the female is ZW and heterogametic
16	Saying drones make sperm by meiosis	By mitosis — they are already haploid
17	Saying the mother determines the sex of the child	The father does; the mother makes only X-bearing eggs
18	Passing an X-linked trait from father to son	A son gets Y from his father; the X comes from his mother
19	Writing Glu → Val at the 6th position of the alpha chain	Beta chain
20	Calling HbA HbS a patient	A carrier; sickles only under low oxygen tension
21	Calling thalassemia a structural defect	Quantitative — reduced synthesis of a normal chain
22	Giving Turner 45 + X0	44 + X0 = 45 total
23	Giving Klinefelter 0 Barr bodies and Turner 1	Klinefelter 1, Turner 0
24	Calling a 3-base deletion a frameshift	Multiples of three preserve the reading frame
25	Calling Down syndrome a sex-chromosomal disorder	It is an autosomal trisomy (chromosome 21)

Self-check before you close the book

Answer these from memory — no looking

Which two of Mendel's laws have exceptions, and what are those exceptions?
An individual is AaBbCC. How many gamete types, and how many phenotypes in the F₂ on selfing?
Why can an AB × O couple never have an AB child?
A woman's father was colour blind. She marries a normal man. What fraction of their sons will be colour blind?
Name the disorder for each: 47 XXY, 45 X0, 47 trisomy 21.
Why does a drone have a grandfather but no father?
State two ways in which thalassemia differs from sickle-cell anaemia.
What is the maximum recombination frequency and what does that value mean?
Why is the heterozygote Bb round-seeded but intermediate for starch grain size?
How many Barr bodies would a hypothetical XXXY individual have?

Answers: 1 — dominance (incomplete dominance, codominance) and independent assortment (linkage). 2 — 4 gamete types; 4 phenotypes. 3 — an AB child needs I^A from one parent and I^B from the other, and an O parent supplies neither. 4 — she is a carrier, so half her sons. 5 — Klinefelter, Turner, Down. 6 — he develops from an unfertilised egg, but his mother was diploid and had a father. 7 — thalassemia is quantitative and involves reduced globin synthesis, sickle cell is qualitative and involves one wrong amino acid; thalassemia arises by mutation or deletion affecting HBA1/HBA2 or HBB, sickle cell by a single substitution. 8 — 50 %, meaning the genes behave as if unlinked. 9 — pleiotropy: the same gene affects two different levels of phenotype and dominance is not autonomous. 10 — XXXY carries three X chromosomes, so 2 Barr bodies (X count minus one).

A note on sources Everything in Parts 1, 3, 4 and 5 is taken from the NCERT Class 12 Biology chapter Principles of Inheritance and Variation (Chapter 5 in the older edition, Chapter 4 in the rationalised edition), scoped to the NTA NEET syllabus. Part 2 mixes genuine NEET past-year questions with NEET-pattern questions constructed on NCERT lines; the questions have been kept in the exam's own option style, and years have deliberately not been printed against them so that you judge each question on its concept rather than on how recently it appeared. The only material flagged as outside NCERT is the gene-interaction box at the end of Part 3.

Principles ofInheritance andVariation Revision Mini-Book · Trap-Solved Question Bank · Master Ratio Sheet

Syllabus map & how to use this book

Concept capsules — the examinable core

1. Mendel: the man, the plant, the method

The seven pairs — memorise the dominant side

2. Monohybrid cross and the first two laws

Test cross vs back cross — the most abused pair of terms

3. Dihybrid cross and the 9 : 3 : 3 : 1

4. Deviations from Mendelism — the highest-yield section in the chapter

(a) Incomplete dominance

(b) Codominance and the ABO blood group system

(c) Pleiotropy — one gene, many effects

(d) Polygenic (quantitative) inheritance

5. Chromosomal theory of inheritance

6. Linkage and recombination — Morgan's Drosophila work

7. Sex determination

8. Mutation

9. Pedigree analysis

10. Mendelian disorders in humans

11. Chromosomal disorders in humans

Trap-solved question bank — 30 questions

Master revision sheet

A. Every ratio in the chapter

B. Blood group cross-reference

C. Sex-linked cross outcomes — learn these four rows cold

D. Disorders at a glance

E. Every exception you must be able to name

F. Numbers, names and dates worth one mark each

Diagram pack — redraw each one from memory

Diagram 1 — Pedigree symbols

Diagram 2 — The three pedigree patterns

Diagram 3 — Sex determination: the four systems

Diagram 4 — Linkage, crossing over and the genetic map

Diagram 5 — Sickle-cell anaemia, from base to red cell

Diagram 6 — Non-disjunction and Down syndrome

Ten-minute last look & the blunder list

The 10-minute read-through

25 blunders to never repeat

Self-check before you close the book

Principles of
Inheritance and
Variation Revision Mini-Book · Trap-Solved Question Bank · Master Ratio Sheet