๐ Contents
01 โ Introduction
Gregor Mendel โ The Father of Genetics
Gregor Johann Mendel (1822โ1884), an Augustinian monk from Brno (now Czech Republic), conducted experiments on Pisum sativum (garden pea) from 1856โ1863 and published his findings in 1866. His work remained unrecognized until 1900, when it was independently rediscovered by Hugo de Vries, Carl Correns, and Erich von Tschermak.
Why Mendel Chose Pea Plants
Short Life Cycle
Results within one growing season โ allowed study of many generations.
Contrasting Traits
7 easily distinguishable characters with clear dominant/recessive forms.
Self-Fertilization
Naturally self-fertilizing โ easy to maintain pure lines.
Cross-Fertilization
Could be cross-fertilized artificially by emasculation.
Large Numbers
Produced many offspring โ statistically reliable ratios.
7 Characters / 7 Chromosomes
By chance, his 7 characters were on separate chromosomes โ giving independent assortment!
Mendel’s 7 Contrasting Characters in Pea
| Character | Dominant Trait | Recessive Trait |
|---|---|---|
| Seed shape | Round (Smooth) | Wrinkled |
| Seed color (cotyledon) | Yellow | Green |
| Flower color | Violet/Purple | White |
| Flower position | Axial | Terminal |
| Pod shape | Inflated/Full | Constricted |
| Pod color (unripe) | Green | Yellow |
| Stem height | Tall | Dwarf |
๐ก Mnemonic โ 7 Traits of Mendel’s Pea
“Silly Sam Fell Flat โ People Get Dizzy”
Seed shape ยท Seed color ยท Flower color ยท Flower position ยท Pod shape ยท Pod color ยท Dwarf/Tall (stem height)
02 โ Laws
Mendel’s Three Laws of Inheritance
โ๏ธ Law 1 โ Law of Dominance
When two parents differing in one character are crossed, the character that appears in the Fโ generation is dominant; the one that disappears is recessive. In a heterozygous individual, only the dominant allele is expressed.
๐ Law 2 โ Law of Segregation (Law of Purity of Gametes)
The two alleles of a gene separate (segregate) during gamete formation so that each gamete carries only one allele. This is the most fundamental law โ it is universal and has no exceptions. This is based on the separation of homologous chromosomes during meiosis I.
๐ฒ Law 3 โ Law of Independent Assortment
Alleles of different genes assort into gametes independently of one another (applies only to genes on different chromosomes or far apart on the same chromosome). Explains dihybrid 9:3:3:1 ratio. Exceptions exist when genes are linked.
โ ๏ธ EXAM TRAP
The Law of Segregation has NO exceptions โ it applies to all sexually reproducing diploid organisms. The Law of Independent Assortment, however, does NOT apply to linked genes.
03 โ Monohybrid Cross
Monohybrid Cross & Punnett Square
A monohybrid cross involves parents differing in a single character. When Mendel crossed Tall (TT) ร Dwarf (tt) plants, all Fโ plants were Tall (Tt). On selfing Fโ, he obtained a 3:1 phenotypic ratio in Fโ.
Fโ Cross: Tt ร Tt
| T | t | |
|---|---|---|
| T | TT | Tt |
| t | Tt | tt |
Genotypic ratio: 1 TT : 2 Tt : 1 tt
Phenotypic ratio: 3 Tall : 1 Dwarf
Phenotypic ratio: 3 Tall : 1 Dwarf
| Cross Type | Phenotypic Ratio | Genotypic Ratio |
|---|---|---|
| Monohybrid (Fโ) | 3 : 1 | 1 : 2 : 1 |
| Test Cross (Aa ร aa) | 1 : 1 | 1 Aa : 1 aa |
| Back Cross (Fโ ร either parent) | 3:1 or 1:1 | Depends on parent |
๐ฌ Test Cross โ Exam Favourite
Crossing an individual of unknown genotype with the homozygous recessive (aa). If offspring ratio is 1:1 โ individual is heterozygous (Aa). If all offspring show dominant phenotype โ individual is homozygous dominant (AA). Mendel himself used this to confirm his results.
๐ก Mnemonic โ Monohybrid Fโ Ratios
“3 Show, 1 Hides โ Genotype: 1 Pure, 2 Mixed, 1 Pure”
Phenotypic: 3 dominant : 1 recessive
Genotypic: 1 homozygous dominant : 2 heterozygous : 1 homozygous recessive
Genotypic: 1 homozygous dominant : 2 heterozygous : 1 homozygous recessive
04 โ Dihybrid Cross
Dihybrid Cross โ Two Characters at Once
Mendel crossed Round Yellow (RRYY) ร Wrinkled Green (rryy) seeds. Fโ = All Round Yellow (RrYy). On selfing Fโ ร Fโ, he obtained the famous 9:3:3:1 ratio in Fโ.
Dihybrid Fโ โ RrYy ร RrYy
| RY | Ry | rY | ry | |
|---|---|---|---|---|
| RY | RRYY | RRYy | RrYY | RrYy |
| Ry | RRYy | RRyy | RrYy | Rryy |
| rY | RrYY | RrYy | rrYY | rrYy |
| ry | RrYy | Rryy | rrYy | rryy |
9 Round Yellow : 3 Round Green : 3 Wrinkled Yellow : 1 Wrinkled Green
| Phenotype (Fโ) | Fraction | Genotypes |
|---|---|---|
| Round Yellow (R_Y_) | 9/16 | RRYY, RRYy, RrYY, RrYy |
| Round Green (R_yy) | 3/16 | RRyy, Rryy |
| Wrinkled Yellow (rrY_) | 3/16 | rrYY, rrYy |
| Wrinkled Green (rryy) | 1/16 | rryy |
๐ก Mnemonic โ Dihybrid Gametes
“For Dihybrid (AaBb), gametes = 2โฟ types”
AaBb โ 4 gamete types (2ยฒ = 4): AB, Ab, aB, ab
AaBbCc โ 8 gamete types (2ยณ = 8)
General formula: heterozygous pairs (n) โ 2โฟ gamete types โ Fโ box = 4โฟ combinations
AaBbCc โ 8 gamete types (2ยณ = 8)
General formula: heterozygous pairs (n) โ 2โฟ gamete types โ Fโ box = 4โฟ combinations
05 โ Deviations from Mendelism
When Mendel’s Rules Don’t Fully Apply
Incomplete Dominance
Neither allele is completely dominant โ Fโ shows an intermediate phenotype. Classic example: Antirrhinum majus (Snapdragon) โ Red (RR) ร White (rr) โ Pink (Rr) in Fโ. Fโ gives 1 Red : 2 Pink : 1 White โ phenotypic ratio = genotypic ratio (1:2:1).
๐ Incomplete Dominance vs Codominance
Incomplete Dominance: Intermediate phenotype (blending). Phenotypic = genotypic ratio (1:2:1).Codominance: Both alleles expressed simultaneously, no blending. Example: ABO blood groups (Iแดฌ and Iแดฎ are codominant). AB blood type shows both A and B antigens on RBCs.
Epistasis
Epistasis = one gene masks the expression of another non-allelic gene. Several modified ratios arise:
| Type of Epistasis | Modified Fโ Ratio | Example |
|---|---|---|
| Dominant epistasis | 12 : 3 : 1 | Fruit color in squash |
| Recessive epistasis | 9 : 3 : 4 | Coat color in mice |
| Dominant & recessive | 13 : 3 | Feather color in poultry |
| Duplicate recessive | 9 : 7 | Flower color in sweet pea |
| Duplicate dominant | 15 : 1 | Capsule shape in Shepherd’s purse |
| Supplementary genes | 9 : 3 : 4 | Coat color in Labrador dogs |
๐ก Mnemonic โ Modified Ratios add up to 16
“All ratios in digenic interaction total = 16”
12+3+1 = 16 ยท 9+3+4 = 16 ยท 13+3 = 16 ยท 9+7 = 16 ยท 15+1 = 16
Because Fโ always has 16 boxes in a 4ร4 Punnett square for two genes.
Because Fโ always has 16 boxes in a 4ร4 Punnett square for two genes.
Pleiotropy
A single gene affects multiple phenotypic traits. Examples: Phenylketonuria (PKU) โ one gene affects skin, hair, brain development. Sickle cell anaemia โ one mutation causes multiple symptoms (anemia, organ damage, sickling).
06 โ Blood Groups
ABO Blood Groups โ Multiple Alleles & Codominance
ABO blood groups (discovered by Karl Landsteiner, 1901) are controlled by a single gene with three alleles (Iแดฌ, Iแดฎ, i) โ this is multiple allelism. Iแดฌ and Iแดฎ are codominant; both are dominant over i.
| Blood Group | Genotype(s) | Antigen on RBC | Antibody in Serum | Can Donate To | Can Receive From |
|---|---|---|---|---|---|
| A | IแดฌIแดฌ or Iแดฌi | A antigen | Anti-B | A, AB | A, O |
| B | IแดฎIแดฎ or Iแดฎi | B antigen | Anti-A | B, AB | B, O |
| AB | IแดฌIแดฎ | A & B antigens | None (Universal Receptor) | AB only | All groups |
| O | ii | None (H antigen) | Anti-A & Anti-B (Universal Donor) | All groups | O only |
โ ๏ธ NEET Exam Trick โ Blood Group O
Blood Group O is called “Universal Donor” in emergency settings only โ in reality, Rh factor (+ or โ) also must match. Blood Group AB is the “Universal Receptor.” Never call O the universal receptor or AB the universal donor!
Rh Blood Group System
Discovered by Landsteiner and Wiener (1940) using Rhesus monkey. About 80% of humans are Rhโบ. The Rh factor is determined by a single gene โ Rhโบ is dominant. Erythroblastosis foetalis (haemolytic disease of newborn) occurs when an Rhโป mother carries an Rhโบ foetus โ maternal anti-Rh antibodies attack fetal RBCs in the second pregnancy.
๐ก Mnemonic โ ABO Antibody Rule
“Your serum attacks what you DON’T have”
Blood group A โ no B antigen โ makes Anti-B antibody
Blood group B โ no A antigen โ makes Anti-A antibody
Blood group O โ neither โ makes both Anti-A and Anti-B
Blood group AB โ has both โ makes neither antibody
Blood group B โ no A antigen โ makes Anti-A antibody
Blood group O โ neither โ makes both Anti-A and Anti-B
Blood group AB โ has both โ makes neither antibody
07 โ Sex Determination
Sex Determination Mechanisms
| Type | Female | Male | Example |
|---|---|---|---|
| XX-XY (Male heterogamety) | XX | XY | Humans, Drosophila, most mammals |
| XX-XO | XX | XO | Grasshoppers, bugs (Orthoptera) |
| ZW-ZZ (Female heterogamety) | ZW | ZZ | Birds, moths, butterflies, some fish |
| ZO-ZZ | ZO | ZZ | Some moths |
| Haploid-Diploid | Diploid (2n) | Haploid (n) | Honey bee (Apis mellifera) |
๐งฌ Human Sex Determination โ Key Points
The Y chromosome determines maleness in humans via the SRY gene (Sex-determining Region of Y). A person with XXY is Klinefelter syndrome (male, sterile). A person with XO is Turner syndrome (female, sterile). Sex of offspring is determined by the sperm (X-bearing or Y-bearing) โ the father determines sex, not the mother.
Sex-Linked Inheritance โ X-Linked Traits
Genes located on the X chromosome show sex-linked inheritance. Recessive X-linked disorders are more common in males (hemizygous โ only one X copy).
Colour Blindness
X-linked recessive. Genotypes: X^c X^c (affected female), X^c X (carrier female), X^c Y (affected male). ~8% males affected, ~0.5% females.
Haemophilia A
X-linked recessive. Factor VIII deficiency. Royal disease โ affected Queen Victoria’s descendants. Carrier females transmit to sons.
Duchenne MD
X-linked recessive. Dystrophin protein absent. Progressive muscle wasting. Affects males; females are carriers.
Y-Linked (Holandric)
Genes on Y passed father โ all sons. Example: Hypertrichosis of ears (hairy ear). Affects only males, never skips generation.
08 โ Linkage & Crossing Over
Linkage & Crossing Over
Linkage (coined by T.H. Morgan) is the tendency of genes located on the same chromosome to be inherited together. Linked genes violate the Law of Independent Assortment โ they do NOT give 9:3:3:1 in dihybrid crosses.
| Feature | Complete Linkage | Incomplete Linkage |
|---|---|---|
| Recombination? | No crossing over | Some crossing over |
| New combinations | None produced | Recombinant gametes produced |
| Fโ ratio | 1:2:1 (two phenotypes) | Modified ratio (some parental types dominate) |
| Example | Theoretical only | Morgan’s Drosophila experiments |
๐งซ Morgan’s Contribution
T.H. Morgan worked on Drosophila melanogaster (fruit fly) โ ideal for genetics due to: short life cycle (2 weeks), large number of offspring, only 4 pairs of chromosomes, and easily observable phenotypic traits. Morgan coined the terms linkage, crossing over, sex-linkage, and established the concept of the linkage group. He won the Nobel Prize in 1933.
Recombination Frequency & Gene Mapping
Recombination frequency (%) = (Number of recombinant offspring / Total offspring) ร 100. One centimorgan (cM) = 1% recombination frequency. Genes >50 cM apart assort independently.
๐ก Mnemonic โ Crossing Over Location
“Closer genes = Fewer chances to cross over = Lower recombination frequency”
The farther apart two genes are on a chromosome, the higher the chance of a crossover occurring between them โ higher recombination frequency โ they appear less linked. Genes >50 cM apart behave as if unlinked (independent assortment).
09 โ Pedigree Analysis
Pedigree Analysis โ Reading Family Trees
| Inheritance Pattern | Key Clues in Pedigree |
|---|---|
| Autosomal Recessive | Affected individuals can appear from unaffected parents (carriers). Affects both sexes equally. Skips generations. Parents of affected = both carriers (Aa ร Aa). |
| Autosomal Dominant | At least one parent is always affected. Condition appears in every generation (no skipping). Both sexes equally affected. |
| X-Linked Recessive | More males affected than females. Affected males have carrier mothers. Son of affected father is never affected (father gives Y to son). |
| X-Linked Dominant | Affected father passes to ALL daughters but NO sons. Affected mother passes to 50% of children. |
| Y-Linked (Holandric) | Only males affected. Every son of an affected father is affected. No daughters affected. Father-to-son transmission only. |
๐ก Golden Rule for NEET Pedigree
If two unaffected parents have an affected child โ the trait is recessive. If the condition is sex-linked recessive, the mother is a carrier and the affected child is almost always male.
10 โ Mutations
Mutations โ Heritable Changes in DNA
| Type | Description | Example |
|---|---|---|
| Point Mutation | Change in single nucleotide base | Sickle cell anaemia (GAG โ GUG; Glu โ Val) |
| Missense | Codon codes for different amino acid | Sickle cell anaemia |
| Nonsense | Codon changed to stop codon | Truncated protein; often non-functional |
| Silent | Changed codon still codes same amino acid | No phenotypic effect (due to degeneracy) |
| Frameshift | Insertion/deletion changes reading frame | Usually non-functional protein |
| Chromosomal | Change in chromosome structure/number | Deletion, duplication, inversion, translocation |
| Aneuploidy | Abnormal number of individual chromosomes | Trisomy 21 (Down syndrome); Monosomy X (Turner) |
| Polyploidy | Extra sets of entire genome | Common in plants (wheat = hexaploid) |
Sickle Cell Anaemia
Autosomal recessive. Point mutation in ฮฒ-globin gene. GAGโGTG at codon 6 (GluโVal). Sickling under low Oโ. Heterozygotes are resistant to malaria.
Phenylketonuria (PKU)
Autosomal recessive. Phenylalanine hydroxylase deficiency. Accumulation of phenylalanine. Mental retardation if untreated. Treated by low-Phe diet.
Down Syndrome
Trisomy 21 (2n+1 = 47). Risk increases with maternal age. Flat face, intellectual disability, short stature, simian crease. Discovered by John Langdon Down.
Turner Syndrome (XO)
Monosomy โ 45 chromosomes (44+XO). Phenotypically female. Short stature, webbed neck, sterile (streak gonads), no secondary sexual characters.
11 โ Previous Year Questions
Practice MCQs โ Test Yourself
Click an option to check instantly. Based on real NEET, CSIR NET & GATE questions.
Q1. Which of Mendel’s laws is also known as the “Law of Purity of Gametes”?
โ
B โ Law of Segregation. Also called the “Law of Purity of Gametes” because each gamete is pure โ it carries only ONE allele for any gene. During meiosis, the two alleles segregate so that each gamete receives just one copy.
Q2. In a monohybrid cross, Fโ generation showed 787 tall and 277 dwarf plants. The ratio is approximately:
โ
C โ 3 : 1. 787 : 277 โ 2.84 : 1 โ 3 : 1. This is Mendel’s actual experimental data! The small deviation from exact 3:1 is normal statistical variation. This is the classic phenotypic ratio of a monohybrid Fโ cross (Tt ร Tt โ 3 Tall : 1 Dwarf).
Q3. A woman of blood group ‘A’ (heterozygous) married a man of blood group ‘B’ (heterozygous). Which blood groups are possible in their children?
โ
C โ A, B, AB and O. IแดฌI ร IแดฎI cross gives: IแดฌIแดฎ (AB), IแดฌI (A), IแดฎI (B), II (O). All four blood groups in 1:1:1:1 ratio. This is only possible when BOTH parents are heterozygous for their respective alleles.
Q4. A colour blind man marries a woman with normal vision whose father was colour blind. What is the probability of their son being colour blind?
โ
B โ 50%. Woman’s father was colour blind โ she is a carrier (X^c X). Colour blind man = X^c Y. Cross: X^c X ร X^c Y โ sons: X^c Y (colour blind) and XY (normal) = 50% of sons are colour blind. Daughters: X^c X^c (colour blind) and X^c X (carrier) = 50% of daughters colour blind too!
Q5. In snapdragon (Antirrhinum), red flower (RR) ร white flower (rr) gives pink Fโ. The Fโ phenotypic ratio will be:
โ
C โ 1 : 2 : 1 (1 Red : 2 Pink : 1 White). This is incomplete dominance โ Fโ pink (Rr) ร pink (Rr) gives RR (red) : Rr (pink) : rr (white) = 1:2:1. KEY POINT: In incomplete dominance, phenotypic ratio = genotypic ratio (1:2:1), not 3:1 as in complete dominance.
Q6. In a cross AaBb ร AaBb, if A is epistatic to B (dominant epistasis), the expected phenotypic ratio in Fโ would be:
โ
B โ 12 : 3 : 1. In dominant epistasis, A_ (with at least one dominant A allele) masks the expression of B gene โ A_B_ + A_bb = 9+3 = 12 (phenotype A); aaB_ = 3 (phenotype B); aabb = 1 (double recessive). Total = 16. All epistasis ratios sum to 16.
Q7. Two genes A and B are 20 cM apart on the same chromosome. In a test cross of AaBb ร aabb, what percentage of offspring would be recombinant?
โ
B โ 20%. 1 centimorgan (cM) = 1% recombination frequency. If two genes are 20 cM apart, 20% of offspring will be recombinants (10% Ab + 10% aB gametes). The remaining 80% are parental types (40% AB + 40% ab).
Q8. In a pedigree, two unaffected parents have an affected son and daughter. The trait is most likely:
โ
A โ Autosomal recessive. Unaffected parents โ affected children = RECESSIVE. Both son AND daughter affected โ autosomal (not X-linked, because X-linked recessive rarely affects daughters unless mother is also carrier and father is affected). This is the most classic NEET pedigree question pattern.
Q9. Sickle cell anaemia is caused by a substitution of which amino acid in the beta chain of haemoglobin?
โ
B โ Glutamic acid replaced by Valine. In sickle cell anaemia, a point mutation (GAGโGTG) at the 6th codon of the beta-globin gene changes Glutamic acid (charged, hydrophilic) to Valine (nonpolar, hydrophobic). This causes HbS to aggregate under low Oโ, forming long fibers that distort RBCs into a sickle shape.
Q10. A child with Down syndrome (Trisomy 21) has a chromosome number of:
โ
C โ 47. Normal humans have 46 chromosomes. Trisomy 21 (Down syndrome) = 3 copies of chromosome 21 instead of 2 โ total 47 chromosomes. It is the most common chromosomal disorder (~1 in 700 live births). Turner syndrome has 45 (44+XO); Klinefelter has 47 (44+XXY).
Q11. In a dihybrid cross AaBb ร AaBb, how many offspring out of 160 are expected to be homozygous dominant (AABB)?
โ
C โ 10. Probability of AABB = P(AA) ร P(BB) = 1/4 ร 1/4 = 1/16. In 160 offspring: 160 ร 1/16 = 10. Formula: In dihybrid, AABB appears in 1/16 of Fโ. This is a classic GATE-style calculation question.
Q12. In birds, which sex is heterogametic?
โ
B โ Female (ZW) is heterogametic in birds. In birds (and moths, butterflies, some fish), females are ZW (heterogametic) and males are ZZ (homogametic). This is the opposite of mammals. Mnemonic: “Birds and Moths โ Females are the W-ild ones (W = heterogametic).”
12 โ Quick Revision
โก Last-Minute Revision Points
๐ฑ Mendel โ Key Numbers
7 characters ยท 7 years (1856โ1863) ยท 3:1 (monohybrid Fโ) ยท 9:3:3:1 (dihybrid Fโ) ยท 1:1 (test cross) ยท 1:2:1 (incomplete dominance Fโ) ยท Rediscovered in 1900 by 3 scientists.
๐ฉธ Blood Groups โ Must Memorise
O = Universal Donor ยท AB = Universal Receptor ยท Iแดฌ and Iแดฎ = Codominant ยท i = recessive ยท Rh system: 80% Rhโบ ยท Erythroblastosis foetalis = 2nd pregnancy Rhโป mother + Rhโบ foetus.
๐งฌ Mutations โ Top Exam Points
Sickle cell = Glu โ Val (GAG โ GTG) ยท Point mutation ยท Autosomal recessive ยท Down = Trisomy 21 = 47 chromosomes ยท Turner = XO = 45 chromosomes ยท Klinefelter = XXY = 47 chromosomes.
๐งซ Morgan’s Drosophila โ Remember
4 pairs of chromosomes ยท 2 week life cycle ยท Discovered sex linkage ยท Coined “linkage,” “crossing over,” “linkage group” ยท Nobel Prize 1933 ยท 1 cM = 1% recombination frequency ยท Genes >50 cM = independent assortment.
๐ก Final Mnemonic โ All Modified Ratios Sum to 16
“No matter how genes interact โ they always give 16 Fโ combinations”
9:3:3:1 (no epistasis) = 16 | 12:3:1 (dominant epistasis) = 16 | 9:3:4 (recessive epistasis) = 16
9:7 (duplicate recessive) = 16 | 15:1 (duplicate dominant) = 16 | 13:3 = 16
All come from the same 4ร4 Punnett square โ just grouped differently!
9:7 (duplicate recessive) = 16 | 15:1 (duplicate dominant) = 16 | 13:3 = 16
All come from the same 4ร4 Punnett square โ just grouped differently!