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Principles of Inheritance and Variation

Principles of Inheritance and VariationNEET Botany · Class 12 · NCERT Chapter 2

Introduction to Genetics

Genetics is the branch of biology that studies the inheritance of traits from parents to offspring. It explains why offspring resemble their parents and why they are also slightly different. The principles of inheritance and variation underlie modern medicine, agriculture, evolutionary biology, and biotechnology.

Key vocabulary

  • Gene: a unit of inheritance; a stretch of DNA that codes for a specific trait.
  • Allele: different versions of the same gene (e.g., T for tall, t for dwarf).
  • Genotype: the genetic makeup (e.g., TT, Tt, tt).
  • Phenotype: the visible trait (e.g., tall or dwarf).
  • Homozygous: two identical alleles (TT or tt).
  • Heterozygous: two different alleles (Tt).
  • Dominant: the allele that masks the recessive in a heterozygote (T).
  • Recessive: the allele masked by the dominant (t).

Mendel and His Garden Pea Experiments

Gregor Johann Mendel (1822-1884) was an Austrian monk and the founder of modern genetics. He worked with garden pea (Pisum sativum) in his monastery garden between 1856 and 1863, performing thousands of crosses and recording data carefully. His work, published in 1866, was largely ignored until 1900 when three botanists (de Vries, Correns, Tschermak) independently rediscovered his laws.

Why Garden Pea?

  • True-breeding lines: homozygous lines that always produced the same trait, generation after generation.
  • Sharp contrasting traits: seven characters with two clear-cut alternatives each.
  • Controllable pollination: normally self-pollinating (cleistogamous), but cross-pollination could be done by emasculation + manual pollen transfer.
  • Short life cycle: a few months per generation, allowing many generations in a few years.
  • Large progeny: hundreds of seeds per cross, allowing reliable statistical analysis.

The 7 Contrasting Characters Mendel Studied

  • Seed shape: round vs wrinkled
  • Seed colour: yellow vs green
  • Flower colour: violet/purple vs white
  • Pod shape: full/inflated vs constricted
  • Pod colour: green vs yellow
  • Flower position: axial vs terminal
  • Plant height: tall vs dwarf

Monohybrid Cross (3:1 ratio)

A monohybrid cross involves a SINGLE character (one gene). Mendel crossed true-breeding tall pea plants (TT) with true-breeding dwarf plants (tt).

Cross results

  • Parental cross: TT × tt → all F1 are Tt (heterozygous, ALL TALL).
  • F1 self: Tt × Tt → F2 shows TWO different phenotypes!
  • F2 phenotypic ratio: 3 TALL : 1 DWARF = 3:1.
  • F2 genotypic ratio: 1 TT : 2 Tt : 1 tt = 1:2:1.

What did this prove?

The recessive trait DISAPPEARS in F1 but REAPPEARS in F2. This proved that the alleles are NOT lost or blended. They are stored separately and can re-emerge in the next generation. This was the key insight that broke the prevailing "blending" theory of inheritance.

Genetics

Punnett square: monohybrid cross simulator

Pick the genotypes of two parents (T = tall, dominant; t = dwarf, recessive). The Punnett square shows all possible offspring genotypes with their ratios.

Parent 1 (mother): Tt

TT
Tt
tt

Heterozygous (carrier)

Parent 2 (father): Tt

TT
Tt
tt

Heterozygous (carrier)

Tt × Tt

T
t
T

TT

Tall

Tt

Tall

t

Tt

Tall

tt

Dwarf

Genotypic ratio

1 TT : 2 Tt : 1 tt

(out of 4)

Phenotypic ratio

3 Tall : 1 Dwarf

(out of 4)

Try these classic crosses

P x P (homozygous dom x recessive)
F1 x F1 (heterozygous self) → 3:1
TEST CROSS → 1:1
Pure tall x pure tall

NEET key facts

!

F1 of TT x tt = all Tt (heterozygous, all show DOMINANT phenotype).

!

F2 of Tt x Tt = 1 TT : 2 Tt : 1 tt (genotypic) and 3 tall : 1 dwarf (phenotypic) = 3:1.

!

TEST CROSS (Tt x tt): if heterozygous → 1:1 ratio. If homozygous (TT x tt) → all tall.

!

Mendel's laws: Dominance + Segregation (from monohybrid). Independent Assortment (from dihybrid).

Try this

  • Try Tt × Tt: see the classic 3:1 phenotypic ratio (Mendel's monohybrid F2). Genotypic is 1:2:1.
  • Try Tt × tt (TEST CROSS): notice the 1:1 ratio - this is how Mendel verified the heterozygous genotype.
  • Try TT × tt: F1 - all Tt. All offspring look tall but are HETEROZYGOUS. This is why heterozygotes are sometimes called "carriers".

Test Cross

Looking at a tall pea plant, you cannot tell whether it is homozygous (TT) or heterozygous (Tt) - both look the same. To find out, Mendel devised the TEST CROSS: cross the unknown with a homozygous recessive (tt).

  • If unknown is TT: TT × tt → all offspring are Tt → ALL TALL.
  • If unknown is Tt: Tt × tt → 1 Tt : 1 tt → 1 TALL : 1 DWARF (1:1 ratio).

NEET trap

Test cross specifically uses HOMOZYGOUS RECESSIVE as the test parent. Crossing with a homozygous DOMINANT would not reveal anything (all offspring would look dominant regardless of the unknown\'s genotype).

Dihybrid Cross (9:3:3:1 ratio)

A dihybrid cross involves TWO characters (two genes) at the same time. Mendel\'s classic dihybrid cross used round-yellow seeds (RRYY) crossed with wrinkled-green seeds (rryy).

  • Parental cross: RRYY × rryy → F1 = all RrYy (round yellow).
  • F1 self: RrYy × RrYy → F2 shows FOUR different phenotypes.
  • F2 phenotypic ratio: 9 round-yellow : 3 round-green : 3 wrinkled-yellow : 1 wrinkled-green = 9:3:3:1.
  • The TWO new phenotypes (round-green and wrinkled-yellow) are RECOMBINANTS not present in the parents.

Why 9:3:3:1?

The 9:3:3:1 ratio is the PRODUCT of TWO independent monohybrid 3:1 ratios:

(3:1) × (3:1) = 9:3:3:1

This product rule WORKS only if the two genes assort INDEPENDENTLY. Mendel\'s observation of the 9:3:3:1 ratio is the basis of his 3rd law (independent assortment). Note: this only holds for genes on DIFFERENT chromosomes. Linked genes give different ratios.

Genetics

Dihybrid cross: Mendel's classic 9:3:3:1

Mendel's classic dihybrid cross with pea (round/wrinkled, yellow/green seeds). The F2 generation from RrYy x RrYy gives the famous 9:3:3:1 phenotypic ratio.

F1 self-cross: RrYy × RrYy

R = round (dominant), r = wrinkled; Y = yellow (dominant), y = green

Show 4x4 grid
Hide grid
RY
Ry
rY
ry
RY

RRYY

Round
Yellow

RRYy

Round
Yellow

RrYY

Round
Yellow

RrYy

Round
Yellow

Ry

RRyY

Round
Yellow

RRyy

Round
Green

RryY

Round
Yellow

Rryy

Round
Green

rY

rRYY

Round
Yellow

rRYy

Round
Yellow

rrYY

Wrinkled
Yellow

rrYy

Wrinkled
Yellow

ry

rRyY

Round
Yellow

rRyy

Round
Green

rryY

Wrinkled
Yellow

rryy

Wrinkled
Green

Phenotypic ratio (the famous 9:3:3:1)

Round Yellow

9/16

Both dominant; parental type

Round Green

3/16

Round dom + green rec; recombinant

Wrinkled Yellow

3/16

Wrinkled rec + yellow dom; recombinant

Wrinkled Green

1/16

Both recessive; parental type

9 : 3 : 3 : 1

(Round-Yellow : Round-Green : Wrinkled-Yellow : Wrinkled-Green)

Why 9:3:3:1?

The 9:3:3:1 ratio is the PRODUCT of two independent 3:1 ratios (one for each gene). Mathematically:

(3:1) × (3:1) = 9:3:3:1

This works ONLY when the two genes are on DIFFERENT chromosomes (independent assortment). For LINKED genes, the ratio is different.

NEET key facts

!

Dihybrid F2 phenotypic ratio = 9:3:3:1.

!

Genotypic ratio = 1:2:1:2:4:2:1:2:1 (9 genotype classes - more complex).

!

Mendel's 3rd law: INDEPENDENT ASSORTMENT - alleles of different genes assort independently into gametes.

!

Test cross of RrYy x rryy gives 1:1:1:1 phenotypic ratio (4 gamete types revealed).

!

For LINKED genes, the 9:3:3:1 ratio fails. Linked genes show parental types more than recombinants.

Try this

  • Count the squares: 9 round-yellow, 3 round-green, 3 wrinkled-yellow, 1 wrinkled-green. This is the 9:3:3:1 ratio.
  • Notice ROUND-GREEN and WRINKLED-YELLOW are RECOMBINANT phenotypes - new combinations not seen in the parents.
  • NEW combinations only appear because Mendel's 3rd law (independent assortment) splits the alleles into all 4 gamete types in equal proportions.

Mendel\'s Laws of Inheritance

Law 1: Law of Dominance

When two contrasting alleles for a gene are present in a heterozygote, ONLY the dominant allele is expressed in the phenotype. The recessive allele is masked.

Law 2: Law of Segregation (Law of Purity of Gametes)

During gamete formation, the two alleles of a gene SEPARATE so that each gamete gets only ONE allele. When fertilisation occurs, the alleles combine again. This is why F2 shows the recessive trait reappearing in 1/4 of offspring.

Law 3: Law of Independent Assortment

During gamete formation, the alleles of one gene assort INDEPENDENTLY of the alleles of another gene. This produces all possible gamete combinations in equal proportions, giving the 9:3:3:1 dihybrid F2 ratio.

Limitations of Mendel\'s laws

Mendel\'s laws assume: (a) complete dominance, (b) independent assortment, (c) two alleles per gene. Real biology often violates these:

  • INCOMPLETE DOMINANCE (Mirabilis: Rr is pink, not red).
  • CODOMINANCE (ABO blood: IA + IB = AB).
  • MULTIPLE ALLELES (ABO has 3 alleles).
  • LINKAGE (genes on the same chromosome).
  • PLEIOTROPY (one gene affects multiple traits).
  • POLYGENIC INHERITANCE (one trait controlled by multiple genes).

Incomplete Dominance

When neither allele is fully dominant, the heterozygote shows an INTERMEDIATE / blended phenotype.

  • Example: Mirabilis jalapa (4 o\'clock plant). RR (red) × rr (white) → F1 = all Rr (PINK, intermediate). F2 ratio = 1 red : 2 pink : 1 white = 1:2:1.
  • Phenotypic ratio EQUALS genotypic ratio (each genotype has a unique phenotype).
  • Other examples: snapdragon (Antirrhinum), 4 o\'clock plant.

Codominance and Multiple Alleles (ABO Blood Group)

In CODOMINANCE, BOTH alleles are equally expressed in the heterozygote. The two parental phenotypes are visible side by side, not blended (different from incomplete dominance).

The ABO Blood Group System

  • Three alleles: IA, IB, i. (Multiple alleles - more than 2 alleles in the population.)
  • IA and IB are CODOMINANT to each other.
  • Both IA and IB are DOMINANT over i.
  • Four blood groups: A (IA IA or IA i), B (IB IB or IB i), AB (IA IB - codominance), O (ii).
  • The ABO gene is on chromosome 9 and encodes a glycosyltransferase that adds sugar groups to the H antigen.

Other Examples

  • MN blood group (M, N alleles - both codominant).
  • Roan coat in cattle (red x white = roan, mix of red and white hairs visible).
Genetics

Inheritance pattern classifier

Four major inheritance patterns: complete dominance, incomplete dominance, codominance, and multiple alleles. Click each to compare heterozygotes, F2 ratios, and examples.

Complete Dominance
Incomplete Dominance
Codominance
Multiple Alleles

Complete Dominance

Heterozygote looks like the dominant parent

When two contrasting alleles are present, only the DOMINANT allele is expressed. The recessive allele is masked. Heterozygote is phenotypically identical to the homozygous dominant. Mendel's discoveries are based on this.

Heterozygote phenotype:

Same as dominant homozygote

F2 ratio:

F2 = 3:1 phenotypic; 1:2:1 genotypic

Classic examples:

Mendel's pea: tall x dwarf → tall heterozygote

Mendel's 7 traits in pea

Most simple Mendelian crosses

Quick comparison

Pattern
Heterozygote
F2 ratio
Complete dominance
= dominant homozygote
3:1 phenotypic
Incomplete dominance
Intermediate (new colour)
1:2:1 (= genotypic)
Codominance
Both phenotypes visible
1:2:1 (= genotypic)
Multiple alleles
3+ alleles in population
Up to 4 phenotypes

NEET key facts

!

Complete dominance: heterozygote = dominant homozygote. Mendel's pea (3:1).

!

Incomplete dominance: heterozygote = INTERMEDIATE (e.g., pink Mirabilis). 1:2:1 ratio.

!

Codominance: heterozygote = BOTH phenotypes (e.g., AB blood, roan coat). 1:2:1 ratio.

!

Multiple alleles: more than 2 alleles in population. ABO has 3 (IA, IB, i).

!

IA and IB are CODOMINANT to each other. Both are DOMINANT over i. So O blood = ii (recessive).

Try this

  • Compare INCOMPLETE DOMINANCE (Mirabilis pink) vs CODOMINANCE (AB blood). Both have 1:2:1 ratio but different heterozygote appearance.
  • Pink in Mirabilis = NEW colour (blended). AB blood = BOTH antigens (no blending; two distinct molecules side by side).
  • Multiple alleles: ABO has 3 alleles. Any individual has only 2 (one from each parent). The population has 3.
ABO Blood Group

ABO blood group: multiple alleles + codominance

The ABO system has 3 alleles (IA, IB, i) giving 4 blood groups (A, B, AB, O). IA and IB are codominant; i is recessive to both. Pick parents to see possible offspring blood groups.

The 3 alleles

IA

A antigen

IB

B antigen

i

No antigen

IA and IB are CODOMINANT to each other. Both DOMINANT over i.

Parent 1 (mother): IAi → Group

A

IAIA (A)
IAIB (AB)
IAi (A)
IBIB (B)
IBi (B)
ii (O)

Parent 2 (father): IBi → Group

B

IAIA (A)
IAIB (AB)
IAi (A)
IBIB (B)
IBi (B)
ii (O)

IAi × IBi

IB
i
IA

IAIB

Group AB

IAi

Group A

i

iIB

Group B

ii

Group O

Offspring blood group probabilities

A

1/4

= 25%

B

1/4

= 25%

AB

1/4

= 25%

O

1/4

= 25%

ABO blood group reference

Group
Genotypes
Antigens on RBC
Antibodies in plasma
Frequency in India
A
IA IA, IA i
A
Anti-B
~22%
B
IB IB, IB i
B
Anti-A
~33%
AB
IA IB
A + B (both)
No antibodies
~7%
O
ii
None (just H)
Both anti-A and anti-B
~38%

NEET key facts

!

ABO has 3 alleles: IA, IB, i. Multiple alleles example.

!

IA and IB CODOMINANT (both expressed in IA IB → AB blood). Both DOMINANT over i.

!

Genotype to phenotype: IA IA or IA i = A. IB IB or IB i = B. IA IB = AB. ii = O.

!

O is universal donor (no A or B antigens to react with recipient antibodies).

!

AB is universal recipient (no antibodies; can accept any blood type).

!

AB man + O woman → only A or B children possible. NEET classic problem.

Try this

  • Try IA IB (AB) × ii (O): the children are 2 IAi (A) + 2 IBi (B). NO AB or O children possible. Classic NEET problem.
  • Try IA i (A) × IB i (B): the children can be A, B, AB, OR O. Both parents heterozygous → all 4 groups possible.
  • Try IA IA (A pure) × IB IB (B pure): all children are IA IB (AB) - codominance gives the heterozygote phenotype.

Pleiotropy

PLEIOTROPY is when a SINGLE GENE affects MULTIPLE traits. The opposite of polygenic inheritance.

  • Phenylketonuria (PKU): mutations in PAH gene (autosomal recessive). Prevents conversion of phenylalanine to tyrosine. Causes mental retardation, lighter pigmentation, musty body odour, eczema. ONE gene, MANY symptoms.
  • Sickle-cell anaemia: mutation in beta-globin gene. Affects RBC shape (sickled), causes anaemia, joint pain, organ damage, increased malaria resistance.
  • Marfan syndrome: mutation in fibrillin gene. Affects connective tissue: tall stature, long limbs, eye problems, heart defects.

Polygenic Inheritance

POLYGENIC INHERITANCE is when a SINGLE TRAIT is controlled by MULTIPLE GENES (the opposite of pleiotropy). The traits show CONTINUOUS variation rather than distinct categories.

  • Human skin colour: controlled by ~4-6 genes. Each contributes additively. Result: a wide spectrum from very dark to very light.
  • Human height: influenced by hundreds of genes plus environment.
  • Eye colour intensity: graded; multiple genes contribute.
  • Wheat kernel colour: classic example - 3 genes, additive contribution.

Key features of polygenic traits

  • Continuous variation (no clear categories).
  • Bell-shaped distribution in populations.
  • Each gene contributes a small additive effect.
  • Often affected by environment (height = genes + nutrition).

Sex Determination

Different organisms use different chromosomal systems to determine sex. The HETEROGAMETIC sex (the one producing 2 types of gametes) determines offspring sex.

  • XY (humans, Drosophila): male XY (heterogametic), female XX. SRY gene on Y is the male-determining gene.
  • XO (grasshoppers): male XO (only ONE X, no Y, heterogametic), female XX.
  • ZW (birds, butterflies): female ZW (heterogametic), male ZZ. OPPOSITE of XY.
  • ZO (some moths): female ZO (heterogametic), male ZZ.

In humans

Total chromosomes: 22 pairs of autosomes + 1 pair of sex chromosomes = 46. Females: 44 + XX. Males: 44 + XY. The FATHER determines the child\'s sex (X-bearing sperm gives a girl, Y-bearing gives a boy). Probability of son = 50%.

Sex Determination

Sex determination systems across organisms

Four major chromosomal sex determination systems. The HETEROGAMETIC sex (which produces 2 types of gametes) determines the offspring sex. Click each to see who is heterogametic.

XY system (humans)
XO system (grasshopper)
ZW system (birds)
ZO system (some moths)

XY system (humans)

The classic mammal system. The Y chromosome carries male-determining genes (SRY in humans). Male is heterogametic. Father determines sex of children (X-bearing sperm = girl, Y-bearing = boy). 22 pairs of autosomes + XX or XY = 46 chromosomes in humans.

FEMALE

XX

MALE

XY

Heterogametic sex (determines offspring sex):

MALE (produces X and Y sperm)

Homogametic sex:

FEMALE (all eggs are X)

Examples:

Humans

Most mammals

Drosophila (fruit fly)

FEMALEXXMALEXY×Eggs:Sperm:XXXY↓ Possible offspring ↓XXFemaleXYMaleXXFemaleXYMale

NEET key facts

!

XY system (humans, Drosophila): male XY (heterogametic), female XX. SRY gene on Y determines maleness.

!

XO system (grasshopper): male XO (only ONE X, no Y), female XX. Males produce X-bearing and no-chromosome sperm.

!

ZW system (birds, butterflies): FEMALE ZW (heterogametic), male ZZ. Reverse of XY.

!

ZO system (some moths): female ZO, male ZZ.

!

In humans, the FATHER determines the child's sex (he provides X or Y). Probability of son = 50%.

!

NEET memory: XY = male heterogametic, ZW = female heterogametic.

Try this

  • Click "XY system": humans / mammals. The father determines sex (X sperm = girl, Y sperm = boy).
  • Click "ZW system": birds. Notice that the FEMALE is heterogametic - opposite of mammals.
  • Click "XO system": grasshoppers. Notice the male has only ONE X (and no Y at all). Males are heterogametic.

Sex-Linked Inheritance

Genes located on the SEX CHROMOSOMES (X or Y) show sex-linked inheritance. Most sex-linked traits in humans are X-LINKED (the X chromosome is large and has many genes; the Y is small).

Why X-Linked Recessive Traits Affect Males More

  • Males have only ONE X (XY). They are HEMIZYGOUS for X-linked genes - a single recessive allele expresses immediately.
  • Females have TWO X (XX). They need TWO recessive alleles to show the disease (homozygous recessive). Heterozygous females are CARRIERS.
  • This is why haemophilia and colour blindness are far more common in males.

Examples of X-Linked Recessive Traits

  • Haemophilia: blood clotting disorder. Famous Queen Victoria pedigree.
  • Colour blindness: red-green confusion. Affects ~8% of males, ~0.5% of females.
  • Duchenne muscular dystrophy (DMD): progressive muscle wasting.
Sex-Linked Inheritance

Sex-linked (X-linked) inheritance simulator

X-linked recessive traits like haemophilia and colour blindness are more common in males. Pick parents and see offspring outcomes for each sex.

Haemophilia
Colour blindness
DMD

X = normal allele. Xc = recessive disease allele (e.g., haemophilia). Y has no allele for this trait.

Mother: XcX (Carrier (heterozygous))

XX
XcX
XcXc

Father: XY (Normal)

XY
XcY
X
Y
Xc

XXc

CARRIER daughter

XcY

AFFECTED son

X

XX

Normal daughter

XY

Normal son

Why are males more affected by X-linked recessive traits?

Males have only ONE X chromosome (XY). If they inherit the recessive allele Xc, there is NO second X to mask it. So a single recessive allele is enough to cause the disease. Males are HEMIZYGOUS for X-linked genes.

Females have TWO X chromosomes (XX). They need TWO recessive alleles (XcXc) to express the disease. With one recessive allele (XcX), they are CARRIERS - they have the allele but do not show the disease.

NEET key facts

!

X-linked recessive: haemophilia, colour blindness, Duchenne muscular dystrophy (DMD).

!

Affected MALES have only ONE X (XcY) - hemizygous - single allele expresses.

!

CARRIER females are XcX (heterozygous) - phenotypically normal but pass the allele.

!

A father's X-linked allele is passed to ALL daughters (never to sons; sons get Y from father).

!

A carrier mother (XcX) x normal father (XY): 50% sons affected, 50% daughters carriers.

Try this

  • Try carrier mother (XcX) × normal father (XY): notice 1 in 4 daughters is carrier, 1 in 4 sons is affected. Classic Queen Victoria pedigree pattern.
  • Try affected mother (XcXc) × normal father (XY): ALL daughters become carriers (XcX), ALL sons become affected (XcY). Severe situation.
  • Try carrier mother × affected father (XcY): some daughters become affected (XcXc), some carriers (XcX); 50% sons affected.

Linkage and Recombination

T.H. MORGAN, working with the fruit fly (Drosophila melanogaster), discovered that genes on the SAME chromosome are inherited TOGETHER (linked) - violating Mendel\'s 3rd law of independent assortment.

  • Linkage: physical proximity of genes on the same chromosome causes them to be inherited together.
  • Crossing over: during meiosis I, homologous chromosomes exchange segments. This separates linked genes and creates RECOMBINANT gametes.
  • Recombination frequency (RF): the percentage of recombinant offspring. Maximum = 50% (effectively unlinked).
  • Map distance: 1 centimorgan (cM) = 1% recombination frequency. Used to map gene positions on chromosomes.
  • Sturtevant (Morgan\'s student) created the first genetic map.
  • Morgan won the Nobel Prize in 1933 for this work.
Linkage

Linkage and recombination calculator

When two genes are on the same chromosome, they tend to be inherited together. Crossing over creates recombinants. The recombination frequency tells you how close the genes are.

Out of 100 test cross offspring, how many are PARENTAL (non-recombinant) types?

85

So 15 are recombinant types (new combinations).

Recombination frequency

15.0%

= 15.0 cM (centimorgans)

MEDIUM linkage

The formula

Recombination frequency = (recombinants / total) × 100

1 cM (centimorgan) = 1% recombination frequency.

MAXIMUM possible RF = 50% (genes effectively unlinked, on different chromosomes or very far apart).

Linkage strength scale

0 (perfect linkage)
10
25
50% (unlinked)

T.H. Morgan and Drosophila

Thomas Hunt Morgan and his students (Sturtevant, Bridges, Muller) discovered linkage in Drosophila melanogaster (fruit fly). They found that some genes did NOT assort independently as Mendel had predicted - they tended to be inherited together. They concluded these genes are on the SAME chromosome (linked). Crossing over during meiosis I separates linked genes, creating recombinants. Morgan won the Nobel Prize in 1933 for this work.

NEET key facts

!

Linked genes: located on the SAME chromosome. Tend to be inherited together.

!

Crossing over during meiosis I creates RECOMBINANT gametes.

!

Recombination frequency (RF) = (recombinants / total) × 100. Maximum = 50%.

!

1 cM (centimorgan) = 1% RF. Used to map gene positions on chromosomes.

!

Mendel's law of independent assortment WORKS only for genes on different chromosomes.

!

Morgan studied this in Drosophila. Sturtevant created the first genetic map.

Try this

  • Set parental = 100 (recombinant = 0): RF = 0%. The genes are PERFECTLY linked (no crossing over between them).
  • Set parental = 50 (recombinant = 50): RF = 50%. The genes behave INDEPENDENTLY (effectively unlinked - 9:3:3:1 ratio).
  • Set parental = 90 (recombinant = 10): RF = 10%. Tight linkage, genes are 10 cM apart.

Chromosomal Disorders

Errors in chromosome number cause significant disorders. Most are caused by NON-DISJUNCTION (failure of chromosomes to separate properly during meiosis).

Aneuploidy vs Polyploidy

  • Aneuploidy: abnormal number of one specific chromosome (extra = trisomy, missing = monosomy).
  • Polyploidy: extra full chromosome SETS (e.g., 4n, 6n). Common in plants; lethal in most animals.

Major Aneuploidies

  • Down\'s syndrome: trisomy 21 (47, +21). Autosomal. Short stature, intellectual disability, heart defects. Risk increases with maternal age.
  • Klinefelter\'s syndrome: 47, XXY. Sex chromosomal. Tall, sterile males with some female features.
  • Turner\'s syndrome: 45, XO. Sex chromosomal. Short, sterile females with no secondary sexual characteristics.
  • Edward\'s syndrome: trisomy 18.
  • Patau syndrome: trisomy 13.
Chromosomal Disorders

Chromosomal disorders explorer

Five major chromosomal disorders. All caused by ANEUPLOIDY (abnormal chromosome number from non-disjunction). Click each to see karyotype, symptoms, and discovery.

Down's syndrome
Klinefelter's syndrome
Turner's syndrome
Edward's syndrome
Patau syndrome

Down's syndrome

Karyotype

Trisomy 21 (47, +21)

Total chromosomes

47

Type

Autosomal

Cause:

Extra copy of chromosome 21 (3 copies instead of 2). Caused by NON-DISJUNCTION of chromosome 21 during meiosis I or II.

Key symptoms / features:

Short stature

Characteristic facial features (small round face, flat nasal bridge, slanted eyes)

Intellectual disability (mild to moderate)

Congenital heart defects (~50% of cases)

Single deep transverse palm crease (simian crease)

Increased risk with maternal age over 35

Discovered: Langdon Down (1866) - clinical description; cause identified by Lejeune in 1959

What is non-disjunction?

NON-DISJUNCTION is the failure of homologous chromosomes (in meiosis I) or sister chromatids (in meiosis II) to separate properly. Result: one daughter cell gets an EXTRA chromosome (n+1 gamete) and another gets ONE FEWER (n-1 gamete). When these abnormal gametes fertilise normal ones, the zygote has extra (trisomy) or missing (monosomy) chromosomes.

ANEUPLOIDY = abnormal chromosome number. POLYPLOIDY = extra full chromosome SETS (common in plants, e.g., 4n).

NEET key facts

!

Down's syndrome = TRISOMY 21 (47 chromosomes). Autosomal aneuploidy. Risk increases with maternal age.

!

Klinefelter's syndrome = 47, XXY. Extra X in MALE. Sterile, tall, gynecomastia.

!

Turner's syndrome = 45, XO. MISSING X in female. Short, sterile, no secondary sexual characteristics.

!

Edward's syndrome = trisomy 18. Patau syndrome = trisomy 13. Both severe, usually fatal.

!

All caused by NON-DISJUNCTION (failure of chromosomes to separate during meiosis).

!

ANEUPLOIDY = abnormal number. POLYPLOIDY = extra full sets (4n, 6n in plants).

Try this

  • Click "Down's syndrome": notice it is AUTOSOMAL (chromosome 21, not sex). Karyotype 47.
  • Click "Klinefelter's" and "Turner's": both are SEX CHROMOSOMAL. Klinefelter has EXTRA X in male; Turner has MISSING X in female.
  • Compare karyotypes: Down 47, Klinefelter 47, Turner 45. The "47" disorders (Down, Klinefelter) have one EXTRA chromosome; Turner has one FEWER.

Pedigree Analysis

A pedigree is a family tree showing the inheritance of a trait. Standard symbols are used: square = male, circle = female, filled = affected, half-filled = carrier (for recessive traits). Pedigrees help determine the inheritance pattern of a trait.

Common Inheritance Patterns

  • Autosomal dominant: appears in every generation; both sexes affected equally; vertical pattern. Examples: Marfan, Huntington\'s.
  • Autosomal recessive: may skip generations; carriers do not show; both sexes affected equally; horizontal pattern (siblings affected, parents not). Examples: PKU, sickle cell, albinism.
  • X-linked recessive: mostly males affected; skips through carrier mothers; father-to-son transmission impossible. Examples: haemophilia, colour blindness.
  • X-linked dominant: father-to-son impossible; affected father gives to ALL daughters.
Pedigree

Pedigree analysis: 4 inheritance patterns

A pedigree is a family tree showing inheritance of a trait. Identifying the pattern (autosomal dominant, recessive, X-linked dominant, or recessive) helps determine genotypes and predict offspring.

Pedigree symbols

Unaffected male

Affected male

Unaffected female

Affected female

Carrier male (rare)

Carrier female

Marriage line

|

Parent-offspring line

Autosomal Dominant
Autosomal Recessive
X-Linked Recessive
X-Linked Dominant

Autosomal Recessive

Pedigree features:

May SKIP generations (carriers do not show the trait)

Often appears in offspring of unaffected parents (both carriers)

Both sexes equally affected

Increased frequency with consanguineous marriage

Inheritance rules:

Two unaffected carriers (Aa x Aa) have ~25% affected children

Two affected parents (aa x aa) have ALL affected children

Horizontal pattern in pedigree (siblings affected, parents not)

Examples:

Sickle cell anaemia

Cystic fibrosis

Phenylketonuria (PKU)

Albinism

Tay-Sachs disease

How to identify the pattern from a pedigree

  1. Does the trait SKIP generations? Yes → recessive. No → dominant.
  2. Are MALES disproportionately affected? Yes → X-linked recessive likely.
  3. Does the trait show FATHER-to-SON transmission? Yes → cannot be X-linked (must be autosomal or Y-linked).
  4. Are TWO UNAFFECTED parents producing affected children? Yes → recessive (they are carriers).
  5. Does an AFFECTED FATHER pass to ALL daughters but NO sons? → X-linked dominant.

NEET key facts

!

Pedigree symbols: square = male, circle = female, filled = affected, half-filled = carrier.

!

Autosomal recessive: skips generations, two carriers can have affected child. Sickle cell, PKU, albinism.

!

Autosomal dominant: appears in every generation. Marfan, Huntington's, polydactyly.

!

X-linked recessive: more in MALES, skips through carrier mothers. Haemophilia, colour blindness, DMD.

!

NEET trick: father-to-son transmission RULES OUT X-linked (sons get Y from father, not X).

Try this

  • If you see two unaffected parents with an affected child: it is RECESSIVE (parents are carriers).
  • If a trait appears in EVERY generation: it is likely DOMINANT.
  • If MALES are mostly affected and the trait skips generations: think X-LINKED RECESSIVE (haemophilia pattern).

Worked NEET Problems

1

NEET-style problem · Monohybrid

Question

In a monohybrid cross between TT (tall) and tt (dwarf) pea plants, all F1 are tall. F1 is selfed. What are the genotypic and phenotypic ratios in F2?

Solution

Step 1: Parental cross. TT (tall, homozygous) × tt (dwarf, homozygous). Gametes: T from one parent, t from the other. F1: All Tt (heterozygous, all tall - law of dominance). Step 2: F1 selfed. Tt × Tt. Gametes from each parent: T (1/2) or t (1/2). Punnett square: T t T TT Tt t Tt tt Step 3: Count the offspring. - TT (tall): 1 - Tt (tall): 2 - tt (dwarf): 1 GENOTYPIC ratio: 1 TT : 2 Tt : 1 tt = 1:2:1. PHENOTYPIC ratio: 3 tall (TT and Tt look the same) : 1 dwarf = 3:1. This is Mendel's classic monohybrid F2 ratio. The 3:1 phenotypic ratio is among the most important numbers in genetics.
2

NEET-style problem · Dihybrid

Question

In Mendel\'s dihybrid cross with peas, RRYY (round-yellow) is crossed with rryy (wrinkled-green). F1 is selfed to produce F2. Show how the 9:3:3:1 ratio arises and identify the recombinant phenotypes.

Solution

Step 1: Parental cross. RRYY × rryy → F1 = all RrYy (round-yellow). Step 2: F1 selfed. RrYy × RrYy. Each parent produces 4 gamete types in equal proportions (1:1:1:1): RY, Ry, rY, ry. Step 3: Punnett square (4x4 = 16 boxes). Step 4: Phenotype counts (out of 16): - 9 round-yellow (R_Y_): both dominant. Genotypes: 1 RRYY + 2 RRYy + 2 RrYY + 4 RrYy. - 3 round-green (R_yy): R dominant, y recessive. Genotypes: 1 RRyy + 2 Rryy. - 3 wrinkled-yellow (rrY_): r recessive, Y dominant. Genotypes: 1 rrYY + 2 rrYy. - 1 wrinkled-green (rryy): both recessive. F2 PHENOTYPIC RATIO = 9 : 3 : 3 : 1. RECOMBINANT phenotypes: round-green (3) and wrinkled-yellow (3). These NEW combinations were NOT present in the original parents (parents were round-yellow and wrinkled-green). Why 9:3:3:1: it is the product of two independent 3:1 ratios. (3:1) × (3:1) = 9:3:3:1. This works only if the two genes are on different chromosomes (independent assortment, Mendel's 3rd law).
3

NEET-style problem · ABO blood group

Question

A man with blood group AB married a woman with blood group O. They have a child. What blood groups are possible for this child? Why is it impossible for the child to have blood group AB or O?

Solution

Father: blood group AB → genotype IA IB. Mother: blood group O → genotype ii. Gametes from father: IA (1/2) or IB (1/2). Gametes from mother: i (always; she has only i alleles). Possible offspring: - IA + i → IA i (blood group A) - IB + i → IB i (blood group B) Probabilities: - Group A: 50% - Group B: 50% - Group AB: 0% - Group O: 0% WHY NOT AB: AB requires BOTH IA and IB. The mother has neither (only i), so children cannot inherit IA AND IB at the same time. WHY NOT O: O requires homozygous ii. The father has IA IB - no i alleles to pass. Children always inherit either IA or IB from him. This is a classic NEET problem demonstrating that the ABO system has multiple alleles (IA, IB, i) with codominance between IA and IB.
4

NEET-style problem · Sex-linked

Question

A normal man marries a woman who is a CARRIER for haemophilia (XhX). What are the probabilities for their offspring being affected, carriers, or normal?

Solution

Father: normal man, XY (no haemophilia allele). Mother: carrier XhX. Father's gametes: X (1/2) or Y (1/2). Mother's gametes: Xh (1/2) or X (1/2). Punnett square: Xh X X XhX XX Y XhY XY Offspring (4 combinations, each 1/4): - XhX: carrier daughter (phenotypically normal) - XX: normal daughter - XhY: AFFECTED son (haemophilia - has Xh on his only X) - XY: normal son Probabilities: - Daughters: 50% carrier, 50% normal. NONE affected. - Sons: 50% affected, 50% normal. Overall: - Affected: only sons - 25% of all children (50% of sons) - Carriers: only daughters - 25% of all children (50% of daughters) - Normal: 50% (25% normal sons + 25% normal daughters) Key NEET insight: from a CARRIER mother, half her sons will be AFFECTED, half her daughters will be CARRIERS. This is the classic Queen Victoria haemophilia pedigree pattern.
5

NEET-style problem · Pedigree analysis

Question

A pedigree shows: an affected father, an affected son, but unaffected daughters. The father has unaffected parents but his sister is affected. What is the most likely inheritance pattern?

Solution

Step 1: Look at the family pattern. - Affected father → affected son (vertical transmission). - Affected father has UNAFFECTED parents. - Affected SISTER too. Step 2: Test each pattern. AUTOSOMAL DOMINANT? At least one parent should be affected. But the affected father's parents are UNAFFECTED. Does NOT fit. AUTOSOMAL RECESSIVE? Both unaffected parents must be carriers (Aa x Aa) → 25% affected children. Fits the parents being unaffected and having affected son and daughter (sister of affected father). Both sexes affected equally. Plausible. X-LINKED RECESSIVE? Affected father (XhY) x carrier mother (XhX). Daughters: XhX (carrier - unaffected) or XhXh (very rare, only if both parents carry). Sons: XhY (affected) or XY (normal). The affected SISTER would need XhXh (mother must be carrier AND grandfather affected on mother's side). Possible but requires more specific carrier conditions. X-LINKED DOMINANT? Affected father → all daughters affected (since they inherit his X). But here daughters are UNAFFECTED. Does NOT fit. Most LIKELY: AUTOSOMAL RECESSIVE. Both unaffected parents are carriers; produce 1/4 affected children (regardless of sex). Affected father and affected sister fit this pattern. Key clues used: (1) unaffected parents producing affected children (recessive), (2) both sexes affected (autosomal not X-linked).

Cheat Sheet

Mendel

7 traits in pea (Pisum). 3 laws: dominance, segregation, independent assortment. Worked 1856-1863, rediscovered 1900.

Monohybrid F2

3:1 phenotypic, 1:2:1 genotypic. Tt × Tt classic.

Test cross

Cross unknown × homozygous recessive (tt). 1:1 = heterozygous; all dominant = homozygous.

Dihybrid F2

9:3:3:1 phenotypic. RrYy × RrYy. Mendel's 3rd law (independent assortment).

Incomplete dominance

Mirabilis: Rr = pink (intermediate). F2 = 1:2:1 phenotypic = genotypic.

Codominance

AB blood group: IA + IB = both antigens. Roan cattle. Heterozygote shows BOTH phenotypes.

Multiple alleles

ABO: 3 alleles (IA, IB, i). 4 phenotypes: A, B, AB, O. Gene on chromosome 9.

Pleiotropy

One gene affects multiple traits. PKU, sickle cell, Marfan.

Polygenic

One trait controlled by many genes. Skin colour, height. Continuous variation.

Sex determination

XY (humans) - male heterogametic. XO (grasshopper). ZW (birds) - female heterogametic. ZO (some moths).

Sex-linked recessive

X-linked: more in MALES. Haemophilia, colour blindness, DMD. Father-to-son impossible.

Linkage

Genes on same chromosome inherited together. Morgan, Drosophila, Nobel 1933. RF = recombinants/total × 100. Max 50%. 1cM = 1% RF.

Down's syndrome

Trisomy 21 (47). Autosomal. Risk with maternal age. Non-disjunction.

Klinefelter

47, XXY. Sex chromosomal. Tall, sterile male.

Turner

45, XO. Sex chromosomal. Short, sterile female.

Pedigree

Square=male, circle=female, filled=affected. Auto recessive skips, X-linked rec male-biased.

Frequently asked questions

How often does Principles of Inheritance and Variation appear in NEET?

Principles of Inheritance and Variation is a Very High Weightage chapter with 6 to 8 questions in most NEET exams. Questions focus on Mendel's laws, monohybrid (3:1) and dihybrid (9:3:3:1) crosses, test cross, incomplete dominance vs codominance, ABO blood groups (multiple alleles), pleiotropy, polygenic inheritance, sex determination types (XY, XO, ZW, ZO), sex-linked inheritance (haemophilia, colour blindness), linkage and recombination (Morgan's work), chromosomal disorders (Down, Klinefelter, Turner), and pedigree analysis. This chapter is one of the highest-yield chapters in NEET Biology.

Why did Mendel choose garden pea (Pisum sativum) for his experiments?

Mendel chose garden pea because: (1) Many TRUE-BREEDING varieties existed (homozygous lines that always produced the same trait). (2) The plant has SHARP CONTRASTING TRAITS (e.g., tall vs short, round vs wrinkled, yellow vs green). (3) Pea flowers are normally SELF-POLLINATED (autogamous) due to cleistogamous structure, so Mendel could prevent unwanted cross-pollination. He could also do controlled CROSS-POLLINATION by removing anthers (emasculation) and dusting pollen by hand. (4) Pea has a SHORT LIFE CYCLE (a few months), allowing many generations in a few years. (5) It produces LARGE NUMBERS of offspring per cross, allowing reliable statistical analysis. Mendel studied 7 contrasting characters: seed shape, seed colour, flower colour, pod shape, pod colour, flower position, plant height.

What is the difference between incomplete dominance and codominance?

Both are exceptions to complete dominance, but they differ in how the heterozygote looks: (1) INCOMPLETE DOMINANCE: the heterozygote shows an INTERMEDIATE / BLENDED phenotype - neither dominant allele fully expressed. Example: Mirabilis jalapa flower colour: RR red x rr white → Rr PINK (intermediate). F2 ratio: 1 red : 2 pink : 1 white (= 1:2:1 phenotypic = genotypic). (2) CODOMINANCE: the heterozygote shows BOTH parental phenotypes EQUALLY and INDEPENDENTLY (no blending). Example: ABO blood group: IA + IB = AB (both A and B antigens present). Roan coat in cattle: red x white = roan (mixture of red and white hairs, both visible). NEET trap: pink (incomplete) is a NEW colour; AB (codominance) is BOTH alleles expressed.

How is the ABO blood group system inherited?

The ABO blood group is controlled by THREE alleles at one gene (ABO gene on chromosome 9): IA, IB, and i. (1) IA and IB are CODOMINANT to each other (both express their antigens when together). (2) Both IA and IB are DOMINANT over i (recessive). The four blood groups: A (IA IA or IA i), B (IB IB or IB i), AB (IA IB - codominance, both antigens), O (ii - no antigens). NEET facts: this is an example of MULTIPLE ALLELES (more than 2 alleles in the population). The IA and IB alleles produce antigens A and B respectively; the i allele produces no antigen. The ABO gene encodes a glycosyltransferase enzyme that adds sugar groups to the H antigen. People with O blood are universal donors (no A or B antigens to react). People with AB are universal recipients.

What is the difference between sex determination in humans, grasshoppers, and birds?

Three main systems for NEET: (1) XY (humans, Drosophila): males are XY (heterogametic - produce 2 types of gametes: X and Y), females are XX (homogametic). The Y chromosome carries male-determining genes (SRY in humans). (2) XO (grasshopper): males are XO (have only ONE X, no Y), females are XX. Males are heterogametic (X or no chromosome). (3) ZW (birds, butterflies): FEMALES are ZW (heterogametic), males are ZZ (homogametic). The reverse of human XY system. (4) ZO (some moths, fishes): females are ZO, males are ZZ. NEET trap: in birds, the FEMALE is the heterogametic sex (ZW), opposite to humans where male (XY) is heterogametic. Sex chromosomes in humans: 22 pairs of autosomes + 1 pair of sex chromosomes (XX in female, XY in male).

What is linkage and how is it different from independent assortment?

LINKAGE: when two or more genes are located on the SAME CHROMOSOME, they tend to be inherited TOGETHER (they do not segregate independently as Mendel's 3rd law would predict). INDEPENDENT ASSORTMENT happens only when genes are on DIFFERENT chromosomes. Discovered by T.H. MORGAN in Drosophila experiments. Linked genes: alleles on the same chromosome stay together unless crossing over separates them. Recombination frequency: the percentage of offspring that show new (non-parental) gene combinations = (number of recombinants / total offspring) x 100. The MAP DISTANCE between two genes on a chromosome is measured in CENTIMORGANS (cM); 1 cM = 1% recombination frequency. Genes very close together: low recombination, tightly linked. Genes far apart: high recombination, almost independent. Morgan's work showed that Mendel's 3rd law (independent assortment) only works for genes on different chromosomes. He won the Nobel Prize in 1933 for this work.

What is the difference between Down's syndrome, Klinefelter's syndrome, and Turner's syndrome?

All three are CHROMOSOMAL DISORDERS caused by ANEUPLOIDY (abnormal chromosome number from non-disjunction during meiosis). (1) DOWN'S SYNDROME: trisomy 21 (extra copy of chromosome 21). Total chromosomes: 47 (22 pairs of autosomes + chromosome 21 has 3 copies + XX or XY). Symptoms: short stature, characteristic facial features, intellectual disability, heart defects. AUTOSOMAL aneuploidy. (2) KLINEFELTER'S SYNDROME: XXY (extra X chromosome in a male). Total chromosomes: 47 (22 pairs of autosomes + XXY). Symptoms: tall stature, sterile, develops some female features (gynecomastia). SEX CHROMOSOMAL aneuploidy. (3) TURNER'S SYNDROME: XO (missing X chromosome in a female). Total chromosomes: 45 (22 pairs of autosomes + only one X). Symptoms: short stature, sterile, lack of secondary sexual characteristics, webbed neck. SEX CHROMOSOMAL aneuploidy. All three are caused by NON-DISJUNCTION (failure of chromosomes to separate) during meiosis I or II.

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