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Botany · Class 12 · Chapter 5

Principles of Inheritance & Variation

Mendel's ratios through to pedigree analysis, with a working Punnett builder, a recombination model and three readable pedigrees.

NEET 20273–4 questions/yearMost reasoning-heavy chapter

01Chapter map & why Mendel chose the pea

This chapter is worth 3–4 NEET questions every year and it is the most reasoning-heavy chapter in Botany. Nothing here can be bluffed: ratios, crosses and pedigrees are either worked correctly or not at all.

Priority 1

  • Monohybrid and dihybrid ratios; test cross
  • Incomplete dominance, codominance, ABO blood groups
  • Sex determination (XX-XY, XX-XO, ZZ-ZW, honeybee)
  • Mendelian and chromosomal disorders

Priority 2

  • Chromosomal theory, Morgan's Drosophila work
  • Linkage, recombination frequency, genetic maps
  • Pleiotropy and polygenic inheritance
  • Pedigree analysis

Why Pisum sativum?

CharacterDominantRecessive
Stem heightTallDwarf
Flower colourVioletWhite
Flower positionAxialTerminal
Pod shapeInflatedConstricted
Pod colourGreenYellow
Seed shapeRoundWrinkled
Seed colourYellowGreen

Two facts NEET repeats

Mendel worked from 1856 to 1863 and published in 1865 (Proceedings of the Natural History Society of Brünn). His work was ignored and was rediscovered in 1900 independently by de Vries, Correns and von Tschermak. He used statistics and large samples — that was the real novelty.

02Monohybrid cross & Mendel's laws

Tall (TT) × dwarf (tt) → F1 all Tt, tall. Selfing the F1 gives F2 phenotypic ratio 3 : 1 and genotypic ratio 1 : 2 : 1.

Law of dominance (not a universal law)

Characters are controlled by discrete units called factors, which occur in pairs. In a dissimilar pair one member dominates and expresses itself; the other is recessive and is masked in the heterozygote but reappears unchanged in F2.

Law of segregation (universal — no exception)

The two alleles of a pair separate during gamete formation (at anaphase I of meiosis) so that each gamete receives only one of them. Gametes are always pure for a character.

Live: Punnett square builder set the parents and the dominance type

Test cross vs back cross — the distinction you have lost marks on

Test crossBack cross
Crossed withthe homozygous recessive parent onlyeither parent (dominant or recessive)
Purposeto find whether a dominant-looking individual is TT or Ttgeneral breeding technique
Result if heterozygous1 : 1 ratio in the progenydepends on which parent was used
Result if homozygousall offspring show the dominant trait

Where it went wrong

Every test cross is a back cross, but not every back cross is a test cross. If the question says “crossed with the recessive parent” it is a test cross; if it just says “crossed with a parent”, do not assume. This exact confusion (test-cross vs carrier-cross) has cost marks on more than one paper.

03Dihybrid cross & independent assortment

Round yellow (RRYY) × wrinkled green (rryy) → F1 RrYy, all round yellow. F2 gives 9 : 3 : 3 : 1 — 9 round yellow, 3 round green, 3 wrinkled yellow, 1 wrinkled green.

Law of independent assortment

When two pairs of traits are considered together, the segregation of one pair is independent of the other pair. Valid only for genes on different chromosomes (or very far apart on the same chromosome) — linkage is the exception.

Gametes from a heterozygote for n genes = 2n  |  F2 phenotypes = 2n  |  F2 genotypes = 3n  |  Punnett square boxes = 4n
n genesGamete typesF2 phenotypesF2 genotypesRatio
12233 : 1
24499 : 3 : 3 : 1
3882727 : 9 : 9 : 9 : 3 : 3 : 3 : 1

Do it without the square

Use the product rule. For RrYy × RrYy, probability of wrinkled green = P(rr) × P(yy) = ¼ × ¼ = 1/16. For a test cross RrYy × rryy, all four classes come out 1 : 1 : 1 : 1. Multiplying single-gene probabilities is far faster and far less error-prone than filling sixteen boxes under time pressure.

Two equally likely alignments at metaphase I → four gamete types RY / ry Ry / rY
Independent assortment is a meiotic event: which way each bivalent faces is decided independently.

04Deviations from Mendelism

Incomplete dominance

Antirrhinum majus (snapdragon) and Mirabilis jalapa (four o'clock plant): red (RR) × white (rr) → pink F1 (Rr). F2 is 1 red : 2 pink : 1 white — the phenotypic ratio now equals the genotypic ratio. The dominant allele produces only enough pigment for an intermediate phenotype.

Codominance

Both alleles express themselves fully and independently in the heterozygote. Human AB blood group is the standard case: IA and IB both make their own sugar antigen. Also roan cattle (red and white hairs both present, not blended).

Multiple alleles & the ABO system

Three alleles, IA, IB, i, but any one person carries only two. IA and IB are codominant to each other and both dominant over i.

Blood groupGenotype(s)Antigen on RBCAntibody in plasma
AIAIA, IAiAanti-B
BIBIB, IBiBanti-A
AB (universal recipient)IAIBA and Bnone
O (universal donor)iinoneanti-A and anti-B

Pleiotropy

A single gene affecting several apparently unrelated traits. Phenylketonuria: one defective enzyme → mental retardation, reduced hair and skin pigmentation. Also sickle-cell anaemia and starch synthesis in pea (round vs wrinkled seed involves both starch grain size and water content).

Polygenic (quantitative) inheritance

Live: polygenic skin colour how many dominant alleles?

Do not mix these up

Incomplete dominance = blended, intermediate phenotype (pink). Codominance = both phenotypes visible separately and fully (AB blood, roan coat). Both give a 1 : 2 : 1 F2 ratio, so the ratio alone will not tell you which is which — the description of the heterozygote will.

06Sex determination

SystemFemaleMaleHeterogametic sexExamples
XX – XYXXXYmalehumans, Drosophila, most mammals
XX – XOXXXO (one X, no Y)malegrasshopper, roundworm, many insects
ZZ – ZWZWZZfemalebirds, some reptiles, moths, butterflies
Haplodiploidydiploid (fertilised egg)haploid (unfertilised egg, parthenogenesis)honeybee

Honeybee — asked almost every year

Human specifics

Sex of the child is decided by the father's gamete: 50% of sperm carry X, 50% carry Y; all ova carry X. The Y chromosome carries the SRY gene for maleness. In Drosophila the Y is not male-determining — the ratio of X chromosomes to autosomes decides sex, which is why XO Drosophila is a sterile male but XO human (Turner's) is female.

07Mutation

A sudden, heritable change in the DNA sequence, and hence in the phenotype. Mutation is the ultimate source of new variation.

normal β-globin C A T — G T A — C T C — A A A → Glu (normal Hbḥ) C A T — G T A — C A C — A A A → Val (Hbˢ, sickle) One base changed on the template strand → one amino acid changed → whole-body disease. This is pleiotropy too.
Sickle-cell anaemia: autosomal recessive; HbS/HbS is diseased, HbA/HbS is a carrier with sickle-cell trait and resistance to malaria.

08Genetic disorders & pedigree analysis

Mendelian disorders

DisorderInheritanceKey facts
HaemophiliaX-linked recessiveA protein in the clotting cascade is missing; a simple cut bleeds on and on. Transmitted from an unaffected carrier female to some sons. A female is affected only if both parents carry it — the homozygous condition is usually lethal before birth. Queen Victoria's family is the classic pedigree.
Colour blindnessX-linked recessiveRed–green confusion; about 8% of men and 0.4% of women.
Sickle-cell anaemiaAutosomal recessiveHbS/HbS affected, HbA/HbS carrier. Glu → Val at position 6 of the β chain; the RBC becomes sickle-shaped under low oxygen tension.
PhenylketonuriaAutosomal recessiveEnzyme converting phenylalanine to tyrosine is missing; phenylalanine and its derivatives accumulate, damaging the brain, and are excreted in urine.
ThalassaemiaAutosomal recessiveReduced synthesis of a globin chain (quantitative problem); α-thalassaemia involves HBA1/HBA2 on chromosome 16, β-thalassaemia HBB on chromosome 11.
Cystic fibrosisAutosomal recessiveThick mucus in lungs and pancreas.

The comparison NEET loves

Thalassaemia is quantitative — too few normal globin chains are made. Sickle-cell anaemia is qualitative — the right number of chains is made but one is structurally wrong. Same organ, different defect.

Chromosomal disorders

DisorderKaryotypeFeatures
Down's syndrometrisomy 21, 45 + XX or XY = 47Described by Langdon Down (1866). Short stature, small round head, furrowed protruding tongue, partially open mouth, palm crease, retarded physical and mental development. Risk rises sharply with maternal age.
Klinefelter's syndrome47, XXYOverall masculine build but feminine development (gynaecomastia); sterile.
Turner's syndrome45, X0Sterile female; ovaries rudimentary, lack of other secondary sexual characters.

Aneuploidy = failure of chromatids to separate during cell division (non-disjunction) → gain or loss of a chromosome. Polyploidy = failure of cytokinesis after the replication of DNA → an extra whole set; common in plants.

Live: read the pedigree switch the inheritance pattern

Pedigree rules that settle most questions

  1. Two affected parents with an unaffected child → the trait is dominant.
  2. Two unaffected parents with an affected child → the trait is recessive (and both parents are carriers).
  3. If a recessive trait appears far more often in males, and affected males have carrier mothers, it is X-linked.
  4. An affected father passing the trait to every daughter but no son is X-linked dominant; no father-to-son transmission is ever X-linked.

09Last-hour recall sheet & rapid drill

Ratio / numberWhat it means
3 : 1monohybrid F2, complete dominance
1 : 2 : 1monohybrid genotypic ratio; also the F2 phenotypic ratio for incomplete dominance and codominance
9 : 3 : 3 : 1dihybrid F2
1 : 1test cross of a monohybrid heterozygote
1 : 1 : 1 : 1test cross of a dihybrid heterozygote (independent assortment)
2n / 3n / 4ngamete types / F2 genotypes / Punnett boxes for n heterozygous genes
1.3% and 37.2%Morgan's recombination values: white–yellow (tight) and white–miniature wing (loose)
47, XXY / 45, X0 / trisomy 21Klinefelter / Turner / Down
Rapid drill — 10 one-liners tap to reveal

Final checklist

  1. Did I read whether the cross is F1 × F1, a test cross, or a back cross?
  2. Is the question asking for a phenotypic or genotypic ratio?
  3. For blood-group problems, did I write all possible genotypes for each parent before eliminating?
  4. Did I check whether the disorder is X-linked before assuming a 3 : 1 outcome?
  5. Assertion–reason: judge each statement separately, then ask whether the reason really explains the assertion.