NEET Biology: Principles of Inheritance & Variation – Complete Revision Guide & Solved PYQs

Introduction to Principles of Inheritance and Variation for NEET UG

Genetics is one of the most vital, fascinating, and high-scoring units in the NTA NEET UG Biology syllabus. The chapter Principles of Inheritance and Variation forms the backbone of classical genetics and molecular biology. It explains how characters are passed from parents to offspring (Inheritance) and why offspring differ from their parents and among themselves (Variation).

For a medical aspirant, understanding genetic principles is not just about clearing NEET—it is fundamental to human pathology, medical diagnostics, gene therapy, and clinical genetics. From understanding why hereditary conditions like Haemophilia or Sickle-Cell Anemia run in families to predicting genetic probability using pedigree charts, this chapter seamlessly connects pure biology with modern clinical medicine.

NEET Weightage & Expected Questions

In the NEET UG examination, Principles of Inheritance and Variation holds an \textraordinarily high weightage. On average, NTA asks 3 to 5 questions directly from this chapter every year, contributing 12 to 20 marks to your total score.

• Total Questions Expected: 3–5 MCQs
• Marks Contribution: 12–20 Marks
• Question Types: Conceptual direct statements from NCERT, cross-based numerical problems, pedigree analysis, matching-type questions, and disease mechanism identifications.
• Importance for AIR (All India Rank): Because questions are conceptual and application-based, mastering this chapter gives top rankers a decisive competitive edge over candidates who rely purely on rote learning.

Core Concepts & Key Mechanisms Explained

1. Mendel's Laws of Inheritance

Gregor Johann Mendel, known as the 'Father of Genetics', conducted hybridization experiments on garden peas (Pisum sativum) for seven years (1856–1863). He selected 7 pairs of contrasting traits:

• 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

A. Monohybrid Cross & First Two Laws

A cross involving a single pair of contrasting traits is a monohybrid cross. Crossing pure tall ($TT$) and pure dwarf ($tt$) pea plants yields all tall ($Tt$) plants in the F1 generation. Selfing F1 ($Tt \times Tt$) gives the F2 generation:

• Phenotypic Ratio: $3 : 1$ (3 Tall : 1 Dwarf)
• Genotypic Ratio: $1 : 2 : 1$ ($1\ TT : 2\ Tt : 1\ tt$)

Law of Dominance: Characters are controlled by discrete units called factors (genes) which occur in pairs. In a dissimilar pair of factors ($Tt$), one dominates (dominant factor) and the other is masked (recessive factor).

Law of Segregation (Purity of Gametes): Alleles do not show any blending. During gamete formation, the two alleles of a gene pair segregate from each other such that a gamete receives only one of the two alleles. This law is universal with no exceptions.

B. Dihybrid Cross & The Third Law

A cross studying two traits simultaneously (e.g., Seed shape and Seed colour). Crossing Round-Yellow ($RRYY$) with Wrinkled-Green ($rryy$) yields all Round-Yellow ($RrYy$) in F1.

• F2 Phenotypic Ratio: $9 : 3 : 3 : 1$ (9 Round Yellow, 3 Round Green, 3 Wrinkled Yellow, 1 Wrinkled Green)
• F2 Genotypic Ratio: $1 : 2 : 1 : 2 : 4 : 2 : 1 : 2 : 1$ (9 distinct genotypes)

Law of Independent Assortment: When two pairs of traits are combined in a hybrid, segregation of one pair of characters is independent of the other pair of characters. Note: This law applies only to genes located on different chromosomes or far apart on the same chromosome.

2. Deviations from Mendelian Inheritance

A. Incomplete Dominance

When neither of two alleles is completely dominant, resulting in an intermediate phenotype in heterozygous condition. Seen in snapdragon (Antirrhinum majus) and Mirabilis jalapa (4 o'clock plant).

• Crossing Red ($RR$) and White ($rr$) flowers gives Pink ($Rr$) flowers in F1.
• F2 Ratio: Phenotypic Ratio = Genotypic Ratio = $1 : 2 : 1$ (1 Red : 2 Pink : 1 White).

B. Codominance

Both alleles express themselves fully in the heterozygous state without blending. Example: ABO Blood Grouping in Humans.

The gene $I$ controls ABO blood types and has three alleles: $I^A$, $I^B$, and $i$. Both $I^A$ and $I^B$ are dominant over $i$, but $I^A$ and $I^B$ are codominant to each other. Heterozygous $I^A I^B$ individuals express both A and B sugar polymers on RBC surfaces, resulting in blood group AB.

C. Multiple Allelism

When more than two alleles govern the same character in a population. Example: ABO blood group system has 3 alleles ($I^A, I^B, i$). Total possible genotypes in population = $\frac{n(n+1)}{2} = \frac{3(3+1)}{2} = 6$ genotypes, giving 4 distinct phenotypes (A, B, AB, O).

D. Pleiotropy vs. Polygenic Inheritance

• Pleiotropy: A single gene controls multiple phenotypic traits. Example: Phenylketonuria (PKU) in humans—a mutation in the PAH gene causes mental retardation, reduction in hair, and skin pigmentation. Another classic example is starch synthesis in pea seeds ($BB$ large round seeds, $bb$ small wrinkled seeds, $Bb$ intermediate size seeds).
• Polygenic Inheritance: Multiple genes collectively influence a single phenotypic trait. Phenotype reflects the contribution of each additive allele. Examples: Human skin colour (controlled by three genes A, B, C; quantitative inheritance) and human height.

3. Chromosomal Theory of Inheritance & Linkage

Proposed independently by Walter Sutton and Theodor Boveri in 1902. They noted that the behavior of chromosomes during meiosis was parallel to the behavior of Mendel's factors (genes).

Thomas Hunt Morgan provided experimental verification using Drosophila melanogaster (fruit fly). Drosophila was chosen because:

• They can be grown on simple synthetic medium in laboratory.
• They complete their life cycle in about two weeks.
• A single mating produces a large number of progeny.
• Clear sexual dimorphism (male and female easily distinguishable).
• Have few pairs of chromosomes ($2n = 8$) easily visible under low power microscope.

Linkage & Recombination: Morgan discovered that genes situated on the same chromosome do not assort independently. Physical association of genes on a chromosome is called Linkage. Non-parental gene combinations produced during crossing over in meiosis are called Recombination.

• Tight Linkage $\rightarrow$ Low Recombination Frequency (e.g., Yellow body and White eye in Drosophila: 1.3% recombination).
• Loose Linkage $\rightarrow$ High Recombination Frequency (e.g., White eye and Miniature wing: 37.2% recombination).
• Alfred Sturtevant used recombination frequencies between gene pairs to map their position on chromosomes (1% recombination = 1 map unit or centimorgan, cM).

4. Sex Determination Systems

• XX-XY Type: Females homogametic ($XX$), males heterogametic ($XY$). Examples: Humans, Drosophila.
• XX-XO Type: Females homogametic ($XX$), males heterogametic ($XO$). Examples: Grasshoppers, insects.
• ZZ-ZW Type: Females heterogametic ($ZW$), males homogametic ($ZZ$). Examples: Birds, reptiles, butterflies.
• Haplodiploidy: Females are diploid ($2n = 32$, develop from fertilized egg), males are haploid ($n = 16$, develop via parthenogenesis from unfertilized egg). Example: Honeybees.

5. Genetic Disorders in Humans

A. Mendelian Disorders (Gene Mutations)

• Haemophilia: Sex-linked recessive disorder. Defect in blood clotting factor cascade. Unaffected carrier female ($X^H X^h$) passes disease to 50% of sons. Queen Victoria was a carrier.
• Sickle-Cell Anemia: Autosomal recessive disorder. Point mutation at 6th codon of $\beta$-globin gene changing $GAG$ to $GUG$, substituting Glutamic acid with Valine. Causes RBCs to become sickle-shaped under low oxygen tension, causing ischemia and anemia. Genotypes: $Hb^A Hb^A$ (Normal), $Hb^A Hb^S$ (Carrier/Trait), $Hb^S Hb^S$ (Diseased).
• Phenylketonuria (PKU): Autosomal recessive inborn error of metabolism. Lack of enzyme phenylalanine hydroxylase converts Phenylalanine to Tyrosine. Phenylalanine accumulates and converts into phenylpyruvic acid, causing mental retardation.
• Thalassemia: Autosomal recessive quantitative blood disorder. Reduced synthesis of $\alpha$-chains ($\alpha$-Thalassemia, genes $HBA1$ & $HBA2$ on chromosome 16) or $\beta$-chains ($\beta$-Thalassemia, gene $HBB$ on chromosome 11).

B. Chromosomal Disorders (Aneuploidy)

• Down’s Syndrome: Trisomy of chromosome 21 ($2n + 1 = 47$). Features: Short stature, small round head, furrowed tongue, open mouth, broad palm with palm crease, mental retardation.
• Klinefelter’s Syndrome: Trisomy of sex chromosome ($47, XXY$). Overall masculine development with feminine traits like enlarged breasts (Gynecomastia), sterile individual.
• Turner’s Syndrome: Monosomy of sex chromosome ($45, XO$). Females with rudimentary ovaries, lack of secondary sexual characters, sterile, short stature.

Important Formulas & Key Terms Table

Solved Step-by-Step Previous Years Questions (PYQs)

PYQ 1 (NEET 2021)

Question: If a human female with blood group 'AB' marries a male with blood group 'O', what are the possible blood groups of their children?

• (A) Only A and B
• (B) A, B, and AB
• (C) A, B, AB, and O
• (D) Only O

Solution & Explanation:

• Mother's genotype: $I^A I^B$ (Blood group AB)
• Father's genotype: $ii$ (Blood group O)
• Gametes from Mother: $I^A$ and $I^B$
• Gametes from Father: $i$
• Possible offspring genotypes: $I^A i$ (Blood group A) and $I^B i$ (Blood group B).
• Probability: 50% Blood group A, 50% Blood group B. Neither AB nor O children can be produced from this marriage.
• Correct Answer: (A) Only A and B

PYQ 2 (NEET 2020)

Question: How many true-breeding pea plant varieties did Mendel select as pairs which were similar except for one character with contrasting traits?

• (A) 7
• (B) 14
• (C) 2
• (D) 4

Solution & Explanation:

• Mendel selected 7 pairs of contrasting traits, which means he chose 14 true-breeding pea plant varieties in total (e.g., true-breeding tall, true-breeding dwarf, true-breeding yellow seed, true-breeding green seed, etc.).
• Read NEET options carefully: 7 pairs = 14 varieties.
• Correct Answer: (B) 14

PYQ 3 (NEET 2019)

Question: What trigger causes conversion of inactive protoxin into active Bt toxin in Bacillus thuringiensis in the gut of insects, or in genetic disease context: What substitution occurs at the 6th position of $\beta$-globin chain causing Sickle-cell anemia?

• (A) Glutamic acid to Valine
• (B) Valine to Glutamic acid
• (C) Glutamine to Valine
• (D) Lysine to Valine

Solution & Explanation:

• In Sickle-cell anemia, point mutation at the 6th codon of $\beta$-globin gene changes $GAG$ (coding for Glutamic acid) to $GUG$ (coding for Valine).
• This substitutes the hydrophilic amino acid Glutamic acid with hydrophobic Valine.
• Correct Answer: (A) Glutamic acid to Valine

PYQ 4 (NEET 2018)

Question: A gene showing codominance has:

• (A) One allele dominant on the other
• (B) Alleles tightly linked on the same chromosome
• (C) Alleles that are recessive to each other
• (D) Both alleles independently expressed in the heterozygote

Solution & Explanation:

• In codominance, both alleles express themselves equally and fully in the heterozygous condition (e.g., $I^A I^B$ produces both A and B surface antigens).
• Correct Answer: (D) Both alleles independently expressed in the heterozygote

Common NEET Traps & Mistakes to Avoid

• Trap 1: Confusing Test Cross with Back Cross. A Test Cross is specifically crossing an unknown genotype ($F_1$ dominant) with a homozygous recessive parent ($Tt \times tt$). All test crosses are back crosses, but not all back crosses are test crosses!
• Trap 2: Phenotypic vs. Genotypic Ratio in Incomplete Dominance. Remember that for incomplete dominance (e.g., Snapdragon), both phenotypic and genotypic ratios are identical: $1 : 2 : 1$. Do not mark $3 : 1$ in a hurry!
• Trap 3: Counting Traits vs. Varieties. Mendel studied 7 contrasting characters/traits, but selected 14 true-breeding plant varieties. Pay close attention to wording in the question.
• Trap 4: Sickle-Cell Anemia Nucleotide Substitution. Codon changes from $GAG \rightarrow GUG$. In DNA, substitution is adenine ($A$) by thymine ($T$) on template strand. In mRNA, it becomes uracil ($U$). Glutamic acid (hydrophilic) is replaced by Valine (hydrophobic).
• Trap 5: Pedigree Chart Interpretation. If the disease skips generations and affects males and females equally, suspect Autosomal Recessive. If affected fathers pass to all daughters but no sons, it is X-linked Dominant. If fathers pass only to sons, it is Y-linked.

High-Yield NEET Practice MCQs with Answer Keys

Q1. A plant with genotype $AaBbCc$ is self-pollinated. Assuming independent assortment, what proportion of the progeny will show the genotype $AABBCC$?

• (A) $1/8$
• (B) $1/16$
• (C) $1/64$
• (D) $1/32$

Solution: Break down gene pair by gene pair:

• Probability of getting $AA$ from $Aa \times Aa = 1/4$
• Probability of getting $BB$ from $Bb \times Bb = 1/4$
• Probability of getting $CC$ from $Cc \times Cc = 1/4$
• Total probability of $AABBCC = 1/4 \times 1/4 \times 1/4 = 1/64$.
• Answer: (C) $1/64$

Q2. Distance between genes A and B is 10 cM, B and C is 15 cM, and A and C is 5 cM. What is the correct order of these genes on the chromosome?

• (A) A - B - C
• (B) B - A - C
• (C) A - C - B
• (D) C - B - A

Solution: Match map unit distances:

• Distance A to C = 5 cM.
• Distance A to B = 10 cM.
• Distance B to C = 15 cM (which equals $10 + 5$).
• Therefore, gene A is in the middle: B --- (10) --- A --- (5) --- C. The sequence is B - A - C (or C - A - B).
• Answer: (B) B - A - C

Q3. Which of the following conditions represents Klinefelter's syndrome?

• (A) $45, XO$
• (B) $47, XXY$
• (C) $47, \text{Trisomy } 21$
• (D) $47, XYY$

Solution: Klinefelter's syndrome is caused by non-disjunction of sex chromosomes resulting in an \textra X chromosome in males ($47, XXY$). $45, XO$ is Turner's syndrome; Trisomy 21 is Down's syndrome.

• Answer: (B) $47, XXY$

Q4. Which of the following genetic diseases is an autosomal dominant disorder?

• (A) Haemophilia
• (B) Myotonic Dystrophy
• (C) Thalassemia
• (D) Cystic Fibrosis

Solution: As explicitly mentioned in NCERT Biology Class XII, Myotonic dystrophy is an autosomal dominant Mendelian disorder. Haemophilia (Sex-linked recessive), Thalassemia (Autosomal recessive), and Cystic Fibrosis (Autosomal recessive).

• Answer: (B) Myotonic Dystrophy

Q5. In grasshoppers, sex determination is of which type?

• (A) XX - XY
• (B) ZZ - ZW
• (C) XX - XO
• (D) Haplodiploid

Solution: In grasshoppers and several insects, male sex determination is XX-XO type. Males have only one X chromosome along with autosomes ($A + XO$), while females have two X chromosomes ($A + XX$).

• Answer: (C) XX - XO

Summary & Final NEET Revision Tips

• NCERT Diagrams are Gold: Memorize the Punnett square representations, Morgan's fly cross diagrams (Cross A and Cross B comparing yellow-white vs. white-miniature), and Pedigree Symbols in NCERT.
• Formula Mastery: Keep $2^n$ (for gametes/phenotypes) and $3^n$ (for genotypes) on your fingertips where $n$ represents the number of heterozygous locus pairs.
• Disease Classification Table: Always maintain a quick flashcard table distinguishing Mendelian vs. Chromosomal disorders and Autosomal vs. Sex-linked inheritance.
• Strict Rule for Numerical Questions: Never do genetics probability calculations in your head. Write down gametes explicitly step-by-step using branch diagrams or Punnett squares to eliminate silly arithmetic mistakes.
• Pedigree Shortcut: Dominant traits never skip generations (affected child always has an affected parent). Recessive traits can appear in children of unaffected heterozygous parents.