Current Affairs updated daily
KalviGavaniKalviGavani
Class 12

Tamil Nadu Class 12 Biology — Genetics and Heredity Notes & Important Questions

Complete Samacheer Kalvi Class 12 Biology Genetics guide — Mendel's laws, monohybrid and dihybrid crosses, codominance, linkage, mutations, and board exam Q&A.

Updated 22 May 202612 min read

Why Genetics Matters for NEET

Genetics contributes 6–8 questions in NEET Biology (360 marks total). Monohybrid and dihybrid cross ratios, blood group genetics, sex-linked disorders, and mutations are frequently tested. Master the ratios and cross diagrams first.


1. Mendel's Experiments — The Foundation

Why Mendel Chose Pea Plants (Pisum sativum)

  • Short life cycle — multiple generations in one season
  • Clear contrasting characters — easy to score
  • Naturally self-pollinating — pure lines easy to maintain
  • Large number of offspring — statistically reliable

Mendel's Seven Contrasting Characters

CharacterDominantRecessive
Seed shapeRound (R)Wrinkled (r)
Seed colourYellow (Y)Green (y)
Pod shapeInflated (I)Constricted (i)
Pod colourGreen (G)Yellow (g)
Flower colourPurple (P)White (p)
Flower positionAxial (A)Terminal (a)
Plant heightTall (T)Dwarf (t)

Law 1 — Law of Segregation (Monohybrid Cross)

Statement: During gamete formation, the two alleles of a gene separate (segregate) so that each gamete carries only one allele for each gene.

Cross: Tall × Dwarf pea plant

P generation: TT × tt (Tall) (Dwarf) F1 generation: Tt (Tall) — all plants tall F1 × F1: Tt × Tt Gametes: T, t T, t Punnett Square: T t T | TT | Tt | t | Tt | tt | F2 result: TT : Tt : tt = 1 : 2 : 1 (genotypic ratio) Tall : Dwarf = 3 : 1 (phenotypic ratio)

Key insight: The recessive trait (dwarf) disappears in F1 but reappears in F2 in 25% of plants — this proves alleles segregate and recessive alleles are not lost.

Law 2 — Law of Independent Assortment (Dihybrid Cross)

Statement: Genes for different characters assort independently during gamete formation (applies only to genes on different chromosomes).

Cross: Round Yellow × Wrinkled Green

P: RRYY × rryy F1: RrYy (all Round Yellow) F1 × F1: RrYy × RrYy F1 gametes: RY, Ry, rY, ry (4 types, equal frequency) F2 phenotypic ratio: 9 Round Yellow : 3 Round Green : 3 Wrinkled Yellow : 1 Wrinkled Green = 9:3:3:1

Memory tip for 9:3:3:1: "Nine three three one — first class gets the most, last class gets the least."


2. Extensions of Mendelian Genetics

Incomplete Dominance

Neither allele is completely dominant. Heterozygote shows intermediate phenotype.

Classic example: Mirabilis jalapa (4 o'clock plant) P: RR × rr (Red) (White) F1: Rr (Pink) — intermediate F1 × F1: Rr × Rr F2: RR : Rr : rr = 1 : 2 : 1 Red : Pink : White = 1 : 2 : 1

Note: In incomplete dominance, phenotypic ratio = genotypic ratio (both 1:2:1) — unlike Mendel's 3:1.

Codominance

Both alleles are fully expressed in the heterozygote. Neither masks the other.

Example: ABO Blood Groups

Blood GroupGenotypeAntigen on RBC
AI^A I^A or I^A iA antigen
BI^B I^B or I^B iB antigen
ABI^A I^BBoth A and B antigens — codominant
OiiNo antigen
  • I^A and I^B are codominant to each other
  • Both I^A and I^B are dominant over i (recessive)

Cross: A × B → possible offspring?

I^A i × I^B i Offspring: I^A I^B (AB), I^A i (A), I^B i (B), ii (O) All four blood groups possible from one couple!

Multiple Alleles

A gene with more than two alleles in the population (any individual has only two).

ABO blood group is the classic example — three alleles: I^A, I^B, i.

Pleiotropy

One gene, multiple effects. A single gene influences more than one phenotypic trait.

Example: Sickle Cell Anaemia (HbS gene)

  • Abnormal haemoglobin → sickle-shaped RBCs
  • This single change causes: anaemia, organ damage, jaundice, poor growth, low immunity

Polygenic Inheritance

Multiple genes, one trait. Several genes act together to produce a single character, giving a continuous range of phenotypes.

Examples: Skin colour (3–6 gene loci), human height, intelligence.

  • F2 shows a bell-curve (normal distribution) of phenotypes — no sharp classes.

3. Chromosomal Theory of Inheritance

Sutton and Boveri (1902)

Chromosomes behave exactly like Mendel's factors during meiosis:

  • Chromosomes occur in pairs → alleles occur in pairs
  • Chromosomes separate at meiosis → alleles segregate
  • Chromosome pairs assort independently → genes assort independently

Morgan's Drosophila Experiments

T.H. Morgan worked with Drosophila melanogaster (fruit fly) — chosen because:

  • 4 pairs of chromosomes only
  • Short generation time (2 weeks)
  • Large number of offspring
  • Easy to distinguish sex

Sex-linked Inheritance (White eye in Drosophila):

P: X^W X^w × X^w Y (Red-eyed female, carrier) (White-eyed male) Offspring: X^W X^w — Red-eyed female X^w X^w — White-eyed female X^W Y — Red-eyed male X^w Y — White-eyed male

Sex Determination

SystemSpeciesFemalesMales
XX-XYHumans, DrosophilaXXXY
ZW-ZZBirds, butterfliesZWZZ
XOGrasshopperXXXO

In humans: Sex determined by father's sperm (X-bearing or Y-bearing). Y chromosome carries SRY gene → testes development.

Human X-linked Recessive Conditions

ConditionGene LocationMales AffectedFemales Affected
Colour blindnessX chromosomeX^b YX^b X^b only
Haemophilia AX chromosomeX^h YX^h X^h only

Why males are more commonly affected: Males have only one X chromosome — one recessive allele is enough. Females need two copies of the recessive allele.

Carrier female: X^B X^b — normal phenotype but 50% chance of passing recessive allele to sons.


4. Linkage and Crossing Over

Linkage

Genes located on the same chromosome tend to be inherited together — they do not independently assort.

  • Linked genes give ratios different from 9:3:3:1
  • Morgan first demonstrated linkage in Drosophila

Complete linkage: Linked genes always transmitted together — only parental combinations in offspring.

Incomplete linkage: Crossing over occasionally separates linked genes — some recombinant types appear.

Crossing Over

Exchange of chromosome segments between homologous chromosomes during prophase I of meiosis I.

Before crossing over: Chromosome 1: A ——— B Chromosome 2: a ——— b After crossing over at one point: Recombinant 1: A ——— b Recombinant 2: a ——— B

Recombination frequency: Percentage of recombinant offspring. Used to calculate map distance.

  • 1% recombination frequency = 1 centimorgan (cM) = 1 map unit
  • Genes far apart on chromosome → more crossing over → higher recombination frequency

5. Mutations

Gene Mutations (Point Mutations)

Changes in DNA base sequence:

TypeDescriptionExample
TransitionPurine → Purine or Pyrimidine → PyrimidineA→G or C→T
TransversionPurine → Pyrimidine or vice versaA→C or G→T
InsertionAddition of a base pair—
DeletionRemoval of a base pair—

Frameshift mutations: Insertions and deletions shift the reading frame → all downstream codons change → usually non-functional protein.

Chromosomal Mutations

Change in number (Aneuploidies):

ConditionChromosome CountDescription
Down syndrome47 (trisomy 21)Three copies of chromosome 21
Turner syndrome45 (45,X)Monosomy — only one X
Klinefelter syndrome47 (XXY)Extra X in male
Patau syndrome47 (trisomy 13)Severe intellectual disability

Change in structure:

  • Deletion: segment lost
  • Duplication: segment repeated
  • Inversion: segment reversed
  • Translocation: segment moves to non-homologous chromosome

Polyploidy: Entire genome duplicated. Common in plants (wheat: hexaploid 6n=42). Induced by colchicine (inhibits spindle formation).

Mutagens

TypeExampleMechanism
PhysicalUV radiationThymine dimers
ChemicalColchicineDisrupts spindle → polyploidy
ChemicalAlkylating agentsAdd alkyl groups to bases
BiologicalTransposonsJump into genes

Board Examination Questions

1-Mark Questions

Q1: What is the genotypic ratio of F2 in a monohybrid cross? 1:2:1 (AA : Aa : aa)

Q2: Which enzyme does colchicine inhibit? Colchicine does not inhibit an enzyme — it inhibits spindle fibre formation by binding tubulin, preventing chromosome separation during cell division.

Q3: Name the scientist who demonstrated semiconservative replication. Meselson and Stahl (1958).

Q4: What is a carrier in genetics? A carrier is a heterozygous individual (e.g., X^B X^b) who has one copy of a recessive allele but shows no symptoms of the associated disease.

Q5: Define pleiotropy. Pleiotropy is the phenomenon where a single gene controls or influences multiple phenotypic traits. Example: sickle cell anaemia gene affects RBC shape, oxygen transport, and many organ systems.


2-Mark Questions

Q1: Distinguish between dominance and codominance.
Answer
DominanceCodominance
One allele masks the otherBoth alleles fully expressed
Heterozygote resembles dominant homozygoteHeterozygote shows both phenotypes
Example: Tall pea plantExample: AB blood group
Q2: Why are males more commonly affected by X-linked recessive disorders than females?
Answer

Males are hemizygous for the X chromosome (XY) — they have only one X chromosome. If the X carries a recessive allele, there is no second X allele to mask it, so the disease phenotype appears. Females (XX) need two copies of the recessive allele to be affected, which is statistically less likely.

Q3: What are Okazaki fragments? How are they relevant to chromosome inheritance?
Answer

Okazaki fragments are short DNA segments synthesised discontinuously on the lagging strand during DNA replication. They are joined by DNA ligase to form a continuous strand. Accurate replication of DNA is essential for faithful transmission of genetic information from parent to daughter cells during cell division — the basis of Mendelian inheritance.

Q4: What is the significance of Mendel's test cross?
Answer

A test cross involves crossing an individual showing the dominant phenotype (unknown genotype) with the homozygous recessive (aa). If offspring ratio is 1 dominant : 1 recessive, the parent was heterozygous (Aa). If all offspring are dominant, the parent was homozygous dominant (AA). Test crosses reveal unknown genotypes.


5-Mark Questions

Q1: Describe a dihybrid cross with the expected F2 ratio. Why does the 9:3:3:1 ratio depend on independent assortment?
Answer

Dihybrid Cross: Round Yellow (RRYY) × Wrinkled Green (rryy)

P generation: RRYY × rryy F1 generation: All RrYy — Round Yellow (dominant phenotypes)

F1 × F1: RrYy × RrYy

F1 gametes: Four types produced in equal frequency — RY, Ry, rY, ry

F2 Punnett Square (16 combinations):

  • 9 R_Y_ = Round Yellow
  • 3 R_yy = Round Green
  • 3 rrY_ = Wrinkled Yellow
  • 1 rryy = Wrinkled Green

Phenotypic ratio: 9:3:3:1

Why this ratio requires independent assortment: If R and Y genes were linked (on same chromosome), gametes would be mainly RY and ry — only two types, not four. The F2 ratio would be approximately 3 Round Yellow : 1 Wrinkled Green instead of 9:3:3:1. The 9:3:3:1 ratio is only possible when genes on different chromosomes assort independently, producing all four gamete types in equal frequency.


Q2: Describe the types of chromosomal mutations. Give examples of each.
Answer

Chromosomal mutations fall into two categories: changes in chromosome number and changes in chromosome structure.

Changes in Number:

Aneuploidy — loss or gain of individual chromosomes:

  • Trisomy (2n+1): Down syndrome has three copies of chromosome 21 (47 chromosomes total). Caused by non-disjunction during meiosis I or II. Symptoms: intellectual disability, flat facial features, short stature, heart defects.
  • Monosomy (2n−1): Turner syndrome has only one X chromosome (45,X). Symptoms: short stature, absence of secondary sexual characters, infertility.

Polyploidy — multiplication of entire chromosome sets:

  • Triploid (3n), Tetraploid (4n) etc. Rare in animals, common in plants.
  • Wheat (Triticum aestivum) is hexaploid (6n = 42) — produced by natural hybridisation.
  • Induced by colchicine, which blocks spindle formation so chromosomes do not separate.

Changes in Structure:

  • Deletion: Loss of a chromosome segment. Cri-du-chat syndrome is caused by deletion of chromosome 5p.
  • Duplication: Repetition of a chromosome segment. Can increase gene dosage.
  • Inversion: Segment breaks and reinserts in reversed orientation.
  • Translocation: Segment moves from one chromosome to a non-homologous chromosome. Chronic myeloid leukaemia (CML) involves translocation between chromosomes 9 and 22 (Philadelphia chromosome).

NEET Focus Points

2. Blood groups: Know all 6 genotypes, universal donor (O/ii), universal recipient (AB/I^A I^B) 3. Sex-linked: Colour blindness and haemophilia — X-linked recessive, why males affected more 4. Chromosomal disorders: Down (trisomy 21), Turner (45,X), Klinefelter (47,XXY) — chromosome numbers 5. Mutation types: Point mutation (sickle cell = one amino acid change), frameshift, chromosomal

1. Ratios to memorise: Monohybrid F2 = 3:1 phenotypic, 1:2:1 genotypicDihybrid F2 = 9:3:3:1Incomplete dominance F2 = 1:2:1

Frequently Asked Questions

In a monohybrid cross between two heterozygotes (Aa × Aa), the phenotypic ratio is 3:1 (three dominant : one recessive) and the genotypic ratio is 1:2:1 (AA : Aa : aa). This demonstrates Mendel's law of segregation.

Related Notes