1 · Mendel and the Garden Pea
P1Why Pisum sativum was the perfect choice
- Many true-breeding varieties available
- Clear-cut contrasting characters — no intermediates
- Bisexual flowers, naturally self-pollinating (cleistogamous) → pure lines maintained on their own
- Easy to cross-pollinate artificially (emasculation + bagging)
- Short life cycle and a large number of seeds per plant → good statistics
- Mendel worked 1856–1863, published 1865; rediscovered in 1900 by de Vries, Correns and von Tschermak.
- He was the first to apply mathematics and statistics to biology, and to keep large samples across generations.
The seven characters — dominant trait first
| Character | Dominant | Recessive |
|---|---|---|
| Stem height | Tall | Dwarf |
| Flower colour | Violet | White |
| Flower position | Axial | Terminal |
| Pod shape | Inflated | Constricted |
| Pod colour | Green | Yellow |
| Seed shape | Round | Wrinkled |
| Seed colour | Yellow | Green |
Trap — pod colour vs seed colour
Green pod is dominant, but yellow seed is dominant. The colours flip between the two
characters. NEET has repeatedly used this pair to build a wrong option.
Vocabulary you must not blur
Gene = unit of inheritance. Allele = alternative form of a gene.
Genotype = genetic constitution. Phenotype = observable trait.
Homozygous = identical alleles (TT, tt). Heterozygous = different alleles (Tt).
Hybrid = heterozygote. Phenotypic ratio counts appearances, genotypic ratio counts letter combinations.
2 · Monohybrid Cross & Mendel's Laws
P1
P: TT (tall) × tt (dwarf) → F₁: all Tt, tall
F₁ selfed → F₂ phenotypic 3 : 1 · genotypic 1 TT : 2 Tt : 1 tt
| Law | Statement | Cytological basis |
|---|---|---|
| Law of Dominance | Characters are controlled by factors that occur in pairs; in a dissimilar pair one factor is dominant and expresses itself, the other is recessive and stays hidden. Not a universal law — incomplete dominance and codominance break it. | — |
| Law of Segregation (law of purity of gametes) | The two alleles of a pair separate during gamete formation, so each gamete carries only one. Universally applicable — no exception. | Separation of homologous chromosomes in anaphase I of meiosis |
| Law of Independent Assortment | Alleles of one gene assort independently of the alleles of another gene. Fails for linked genes on the same chromosome. | Random alignment of different bivalents at the metaphase I plate |
Trap — which law has no exception?
Only the Law of Segregation. Dominance is broken by incomplete dominance/codominance;
independent assortment is broken by linkage. This exact question appears almost every year.
Counting rules that save time
Number of gamete types from a heterozygote with n gene pairs = 2ⁿ
Number of F₂ genotypes = 3ⁿ · number of F₂ phenotypes = 2ⁿ (complete dominance)
Size of the Punnett square = 4ⁿ boxes
Fraction of F₂ that is fully homozygous recessive = (1/4)ⁿ
Fraction of F₂ that resembles the F₁ genotype = (1/2)ⁿ
n = 2 → 4 gametes, 9 genotypes, 4 phenotypes, 16 boxes. n = 3 → 8 gametes, 27 genotypes, 8 phenotypes, 64 boxes.
3 · Punnett Square Lab
Build the ratio yourselfPick a cross — the square and the ratios build themselves
Work through every option once. If you can predict the ratio before the grid draws, you own this chapter.
4 · Test Cross vs Back Cross vs Carrier Cross
Fix this tonight| Test cross | Back cross | Carrier × carrier | |
|---|---|---|---|
| Cross | unknown × homozygous recessive | F₁ × either parent | Aa × Aa |
| Purpose | reveal the genotype of the dominant-looking individual | recover a parental genotype / fix a trait in breeding | predict risk of a recessive disorder in children |
| Result | Tt × tt → 1 : 1; TT × tt → all dominant | depends which parent; the cross with the recessive parent is a test cross | 3 : 1 phenotypic; 25% affected, 50% carriers, 25% normal |
| Relationship | Every test cross is a back cross only when the recessive parent is one of the actual parents. Not every back cross is a test cross. | neither — both parents are heterozygous | |
Trap — the one you have lost marks on before
Read the question for the words "homozygous recessive" or "carrier".
- Tt × tt is a test cross → 1 : 1 → 50% affected
- Tt × Tt is a carrier cross → 3 : 1 → 25% affected, 50% carriers
Why the recessive parent is used
A homozygous recessive parent contributes only recessive alleles, so every offspring's phenotype
is a direct readout of the gamete it received from the unknown parent. The recessive parent is a
"silent partner" that hides nothing.
5 · Dihybrid Cross
P1
P: RRYY (round yellow) × rryy (wrinkled green) → F₁ all RrYy, round yellow
F₂ phenotypic ratio: 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green
F₂ genotypic ratio: 1:2:2:4:1:2:1:2:1 (nine genotypes)
Dihybrid test cross RrYy × rryy → 1 : 1 : 1 : 1 — proof of independent assortment
Reading the 9:3:3:1 without drawing 16 boxes
Treat each gene separately and multiply the probabilities:
P(round) = 3/4 · P(yellow) = 3/4 → P(round yellow) = 9/16
P(round green) = 3/4 × 1/4 = 3/16 · P(wrinkled yellow) = 1/4 × 3/4 = 3/16
P(wrinkled green) = 1/4 × 1/4 = 1/16
This multiplication method works for three or more genes too, where a Punnett square would take 64 boxes.
Trap — "how many are homozygous?" vs "how many show the recessive phenotype?"
In a 16-box dihybrid F₂: 1 box is fully homozygous recessive (rryy),
4 boxes are homozygous at both loci (RRYY, RRyy, rrYY, rryy),
4 boxes have the F₁ genotype RrYy, and 9 boxes show both dominant traits.
Read the question word by word before choosing.
6 · Deviations from Mendelism
P1| Phenomenon | Definition | Example | F₂ ratio |
|---|---|---|---|
| Incomplete dominance | Heterozygote shows an intermediate phenotype; neither allele is fully dominant | Mirabilis jalapa (four o'clock plant) and snapdragon — red × white → pink | 1 : 2 : 1 (pheno = geno) |
| Codominance | Both alleles express themselves fully and independently in the heterozygote | ABO blood group — AB; roan coat in cattle; MN blood group | 1 : 2 : 1 |
| Multiple alleles | More than two alleles of the same gene exist in the population, though any individual carries only two | IA, IB, i for ABO | — |
| Pleiotropy | One gene controls several apparently unrelated phenotypes | Phenylketonuria, sickle cell anaemia, starch synthesis in pea (seed shape + starch grain size) | — |
| Polygenic inheritance | A trait controlled by three or more genes, each contributing additively; strongly influenced by environment | Human skin colour (A, B, C), kernel colour in wheat, human height | continuous variation (bell curve) |
Incomplete dominance — the molecular reason
The dominant allele produces a functional enzyme; the recessive allele produces none. In a heterozygote a single copy makes only half the product, which is not enough for the full phenotype — so the trait appears intermediate. In true dominance, one functional copy is already enough.
Trap — incomplete dominance vs codominance
Both give a 1 : 2 : 1 phenotypic ratio, and both make phenotypic ratio = genotypic ratio.
The difference is the heterozygote: a blend (pink) means incomplete dominance;
both traits shown side by side (AB blood, roan coat with distinct red and white hairs) means codominance.
7 · ABO Blood Groups — the standard question
P1| Blood group (phenotype) | Possible genotypes | Antigen on RBC | Antibody in plasma |
|---|---|---|---|
| A | IAIA, IAi | A | anti-B |
| B | IBIB, IBi | B | anti-A |
| AB | IAIB | A and B | none — universal recipient |
| O | ii | none — universal donor | anti-A and anti-B |
ABO illustrates three concepts at once, and NEET has asked for all three:
- Multiple alleles — three alleles (IA, IB, i) exist in the population, but any person has only two.
- Codominance — IA and IB are codominant, giving group AB.
- Dominance — both IA and IB are completely dominant over i.
- The gene I codes for a sugar-transferring enzyme; allele i produces no functional enzyme.
The classic puzzle
An AB parent (IAIB) and an O parent (ii) can have children of group
A or B only — never AB and never O. That single line answers a whole family of questions.
8 · Chromosomal Theory of Inheritance
P2- 1902 — Sutton and Boveri independently noticed that the behaviour of chromosomes during meiosis exactly parallels the behaviour of Mendel's factors, and united the two ideas.
- Sutton is credited with combining chromosome movement with Mendelian segregation.
- T. H. Morgan then proved the theory experimentally using Drosophila melanogaster (fruit fly).
Why Drosophila?
- Grown on simple synthetic medium; life cycle of about two weeks
- A single mating produces a large number of progeny
- Sexes are easily distinguished; males and virgin females are readily separated
- Many hereditary variations visible under a low-power microscope
- Only four pairs of chromosomes (2n = 8)
| Mendelian factor | Chromosome |
|---|---|
| Occur in pairs | Occur in homologous pairs |
| Segregate during gamete formation, only one per gamete | Homologues separate at anaphase I |
| Assort independently of another pair | Different bivalents align randomly at metaphase I |
9 · Linkage and Recombination
P1- Linkage — the physical association of genes located on the same chromosome, causing them to be inherited together and reducing recombinants. Coined by Morgan.
- Recombination — the generation of non-parental combinations by crossing over during pachytene of prophase I.
- Recombination frequency is always less than 50% for linked genes. At exactly 50% the genes behave as if unlinked (independent assortment).
- Tightly linked genes are close together → low recombination. Loosely linked genes are far apart → high recombination.
- Recombination frequency is directly proportional to the distance between genes. 1 map unit (centimorgan) = 1% recombination frequency.
- Alfred Sturtevant, Morgan's student, used recombination frequency to construct the first genetic map of a chromosome.
Recombination frequency (%) = (number of recombinant progeny / total progeny) × 100
Morgan's dihybrid test crosses in Drosophila — the numbers
- Yellow body (y) and white eye (w): tightly linked, recombination only 1.3%
- White eye (w) and miniature wing (m): loosely linked, recombination 37.2%
- Both pairs lie on the X chromosome, so neither shows the expected 1:1:1:1 test-cross ratio.
10 · Sex Determination
P1| System | Female | Male | Heterogametic sex | Examples |
|---|---|---|---|---|
| XX–XY | XX | XY | Male | Humans, Drosophila, most mammals |
| XX–XO | XX | XO (one X, no Y) | Male | Grasshopper, roundworm, many insects |
| ZZ–ZW | ZW | ZZ | Female | Birds, butterflies, moths, some reptiles |
| Haplodiploidy | diploid (fertilised egg) | haploid (unfertilised egg, parthenogenesis) | — | Honeybee |
Honeybee — the details that get asked
- Female (queen or worker) = diploid, 32 chromosomes, from a fertilised egg
- Male drone = haploid, 16 chromosomes, from an unfertilised egg by parthenogenesis
- Drones produce sperm by mitosis, not meiosis — they have no father, but they do have a grandfather
- Queen vs worker is decided by diet (royal jelly), not by genotype
Trap — who decides the sex of the child?
In humans the father determines the sex, because he is the heterogametic parent: the mother
can only give X. All eggs carry X; half the sperms carry X and half carry Y. The probability of a
son or a daughter is 50% each, in every pregnancy, independent of previous children.
11 · Mutation
P2- Mutation = a sudden, heritable change in the DNA sequence, producing variation.
- Point mutation — change in a single base pair. The textbook example is sickle cell anaemia (GAG → GUG, glutamic acid replaced by valine at position 6 of the β-globin chain).
- Frameshift mutation — insertion or deletion of bases, shifting the whole reading frame downstream. Far more damaging than a substitution unless the number is a multiple of three.
- Chromosomal aberrations — deletion, duplication, inversion, translocation; common in cancer cells.
- Aneuploidy — gain or loss of a chromosome, caused by failure of separation (non-disjunction) during meiosis. Polyploidy — gain of a whole chromosome set, common in plants.
- Mutagens — UV radiation, X-rays, gamma rays, and chemical mutagens.
12 · Pedigree Analysis
P1 · guaranteed diagram question| Clue in the pedigree | Conclusion |
|---|---|
| Unaffected parents have an affected child | Recessive, and both parents are carriers |
| Trait skips generations | Recessive |
| Every affected child has at least one affected parent; appears in every generation | Dominant |
| An affected female exists with unaffected father | Rules out X-linked recessive → autosomal recessive |
| Mostly males affected; passed from carrier mother to son | X-linked recessive (haemophilia, colour blindness) |
| Affected father → all daughters affected, no sons | X-linked dominant |
| Affected father → all sons affected, no daughters | Y-linked (holandric) |
13 · Genetic Disorders
P1 · high yieldA · Mendelian disorders — single gene
| Disorder | Inheritance | Defect | Key facts |
|---|---|---|---|
| Haemophilia | X-linked recessive | missing clotting factor | Even a minor cut bleeds continuously. Transmitted from an unaffected carrier female to her son. Female homozygotes are rarely seen — such a mother is usually not viable (dies before birth). Seen in the royal families of Europe. |
| Colour blindness (red–green) | X-linked recessive | defective cone pigment gene | About 8% of males and only 0.4% of females affected. A daughter is colour blind only if her father is affected and her mother is at least a carrier. |
| Sickle cell anaemia | Autosomal recessive | β-globin gene, GAG → GUG, glutamic acid → valine at position 6 | Only HbˢHbˢ is diseased; HbᴬHbˢ is a carrier and is resistant to malaria. RBCs become sickle-shaped under low oxygen tension. A point mutation and an example of pleiotropy. |
| Phenylketonuria (PKU) | Autosomal recessive | lacks the enzyme converting phenylalanine → tyrosine | Phenylalanine accumulates and is converted to phenylpyruvic acid; causes mental retardation; excreted in urine. Example of pleiotropy. |
| Thalassemia | Autosomal recessive | reduced synthesis of globin chains | α-thalassemia: HBA1/HBA2 on chromosome 16. β-thalassemia: HBB on chromosome 11. Anaemia due to fewer globin molecules. |
| Cystic fibrosis, Huntington's | autosomal recessive / dominant | — | Huntington's is a standard example of an autosomal dominant disorder. |
Trap — thalassemia vs sickle cell anaemia
Both are autosomal recessive blood disorders, but the defect is different:
sickle cell = a qualitative defect (a structurally wrong globin, from a point mutation);
thalassemia = a quantitative defect (structurally normal globin made in too little quantity).
NEET has repeatedly used "qualitative vs quantitative" as the deciding word.
B · Chromosomal disorders
| Syndrome | Karyotype | Cause | Features |
|---|---|---|---|
| Down's syndrome | 47, trisomy of chromosome 21 (2n + 1 = 45 + XX or XY) | non-disjunction; risk rises with maternal age | Short stature, small round head, furrowed protruding tongue, partially open mouth, palm crease, broad flat hands, congenital heart disease, mental retardation. First described by Langdon Down (1866). |
| Klinefelter's syndrome | 47, XXY (2n + 1) | non-disjunction of sex chromosomes | Male phenotype with feminine development — gynaecomastia (breast development), sparse body hair, sterile. Barr body present. |
| Turner's syndrome | 45, XO (2n − 1) | loss of one X | Female, sterile, rudimentary ovaries, short stature, webbed neck, lack of secondary sexual characters. |
Memory hook — one line each
Down = extra 21, mental retardation.
Klinefelter = extra X in a male, gynaecomastia.
Turner = missing X in a female, webbed neck.
Klinefelter and Turner are both sterile; only Down's is an autosomal trisomy.
Aneuploidy vs polyploidy
Aneuploidy = failure of segregation of a chromatid during cell division, so one extra or one fewer
chromosome (Down's, Turner's, Klinefelter's). Polyploidy = an extra whole set of chromosomes,
caused by failure of cytokinesis after telophase — common in plants, rare and lethal in animals.
60-Second Recall — read this in the car
Monohybrid F₂3:1 pheno · 1:2:1 geno
Dihybrid F₂9:3:3:1
Test crossTt × tt → 1:1 → 50% affected
Carrier crossAa × Aa → 3:1 → 25% affected
Dihybrid test crossRrYy × rryy → 1:1:1:1
No-exception lawSegregation only
Gametes from n hets2ⁿ · genotypes 3ⁿ · boxes 4ⁿ
Incomplete dominanceMirabilis, pink, 1:2:1
CodominanceAB blood, roan cattle
PleiotropyPKU, sickle cell, pea starch
Polygenicskin colour, wheat kernel, height
AB × Ochildren A or B only
Chromosomal theorySutton & Boveri 1902
Linkage proved byMorgan · maps by Sturtevant
1 map unit1% recombination = 1 cM
Morgan's numbersy–w 1.3% · w–m 37.2%
Heterogameticmale in XY/XO · female in ZW
Honeybee♀ 32 diploid · ♂ 16 haploid
Sickle cellGAG→GUG, Glu→Val, position 6
Thalassemiaquantitative · α chr16 · β chr11
Down'strisomy 21, 47
KlinefelterXXY 47, gynaecomastia, sterile
TurnerXO 45, webbed neck, sterile
Colour blindness8% males · 0.4% females