Complete Summary and Solutions for Basic Principles of Inheritance – NCERT Class XI Biotechnology, Chapter 6 – Laws, Genetics, Mendel, Crossing Over, Exercises
Comprehensive summary and explanation of Chapter 6 'Basic Principles of Inheritance' from the NCERT Class XI Biotechnology textbook, covering Mendelian genetics, genotype vs phenotype, monohybrid and dihybrid crosses, linkage, recombination, sex-linked and extranuclear inheritance, polyploidy, and detailed answers to all textbook exercises.
Basic Principles of Inheritance: Class 11 NCERT Chapter 6 - Ultimate Study Guide, Notes, Questions, Quiz 2025
Basic Principles of Inheritance
Chapter 6: Biotechnology - Ultimate Study Guide | NCERT Class 11 Notes, Questions, Examples & Quiz 2025
Full Chapter Summary & Detailed Notes - Basic Principles of Inheritance Class 11 NCERT
Overview & Key Concepts
Chapter Goal: Understand heredity, variations, and genetic principles from Mendel's experiments to modern concepts like linkage, recombination, and reverse genetics. Exam Focus: Monohybrid/dihybrid ratios, laws of inheritance, crossing over frequency, polyploidy examples. 2025 Updates: Emphasis on biotechnological applications in trait manipulation, integration with molecular genetics (Unit III). Fun Fact: Mendel's work was ignored for 34 years until rediscovered in 1900. Core Idea: Genes as units of inheritance follow predictable patterns but can recombine. Real-World: Used in crop breeding for disease resistance; linkage maps aid gene mapping. Ties: Links to biomolecules (Ch3), cell structure (Ch2). Expanded: All subtopics (6.1-6.7) covered point-wise with diagram descriptions for visual learning, including Punnett squares and crosses.
Wider Scope: From classical Mendelian genetics to exceptions like incomplete dominance, linkage, and extrachromosomal inheritance; role in biotechnology for trait modification.
Fig. 6.1: Seven pairs of contrasting traits of pea plants used by Mendel (Description)
Table showing characters: Seed shape (Round dominant vs. Wrinkled recessive), Seed colour (Yellow vs. Green), Flower colour (Violet vs. White), Pod shape (Inflated vs. Constricted), Pod colour (Green vs. Yellow), Flower position (Axial vs. Terminal), Stem height (Tall vs. Dwarf). Visual: Icons of peas, flowers, pods for each pair.
6.1 Introduction to Inheritance
Heredity and Variation: Transmission of traits from parents to offspring (heredity); differences among offspring (variation). Traits depend on genes on chromosomes.
Genetics Definition: Study of heredity and variation; essential for biotechnology to manipulate genes for improved products.
Biotech Relevance: Identify alleles regulating traits for manipulation; e.g., pure lines for breeding.
6.1.1 Mendel’s work: The foundation
Model Organism: Pea (Pisum sativum) - annual, bisexual flowers, self-pollinating, 7 contrasting traits (Fig. 6.1).
Methods: Produced pure lines by self-pollination; artificial cross-pollination with brush; large sample sizes for data over generations.
Rediscovery: Published 1866, ignored until 1900 by de Vries, Correns, von Tschermak.
Cross Details: Pure tall (TT) x pure dwarf (tt) → F1 all tall (Tt, heterozygous); F2 selfing → 3:1 phenotypic (tall:dwarf), 1:2:1 genotypic (TT:Tt:tt).
Alleles and Dominance: Two factors (alleles) per trait; T (dominant tall) masks t (recessive dwarf); F1 heterozygous.
Test Cross (Fig. 6.4): Unknown dominant (e.g., tall) x recessive (dwarf) → 1:1 tall:dwarf confirms heterozygous.
Mendel's Laws: Dominance (one allele masks other); Segregation (alleles separate in gametes).
Fig. 6.3: Segregation of height character in pea plant (Description)
Punnett square for F1 (Tt x Tt): Gametes T/t → Offspring TT (tall), Tt (tall), tt (dwarf); ratios labeled.
Fig. 6.4: Test cross for identification of genotype (Description)
Tall unknown (TT or Tt) x Dwarf (tt) → If TT: All tall; If Tt: 1:1 tall:dwarf.
Incomplete Dominance
Concept: No complete dominance; heterozygous shows intermediate trait (blending).
Example (Fig. 6.5): Four-o'clock plant (Mirabilis jalapa) - Red (RR) x White (rr) → F1 Pink (Rr); F2 selfing → 1:2:1 red:pink:white phenotypic/genotypic.
Key Point: Each allele partially expressed; reduced dominance in heterozygote.
Fig. 6.5: Incomplete dominance in four-o' clock plant (Description)
Parents: Red (RR) x White (rr) → F1 Pink (Rr) → Selfing → F2: Red (RR), Pink (Rr), White (rr) in 1:2:1.
Codominance
Concept: Both alleles equally expressed in heterozygote; no blending or dominance.
Examples (Fig. 6.6): MN blood group (LM LN → both antigens); Cattle coat (RR red x WW white → RW roan, mix of red/white hairs).
Key Point: Traits co-exist without masking; e.g., roan doesn't fade with age.
Fig. 6.6: Codominance of MN blood group and coat colour in cattle (Description)
Fig. 6.7: Results of a dihybrid cross where parents differ in two pairs of contrasting characters (Description)
Parents: RRYY (round yellow) x rryy (wrinkled green) → F1 RrYy (round yellow) → F2: 9:3:3:1 (round yellow : wrinkled yellow : round green : wrinkled green).
Law of Independent Assortment (Dihybrid Cross)
Cross Details: RRYY x rryy → F1 all round yellow (RrYy); F2 selfing → 9:3:3:1 phenotypic (Fig. 6.7).
Observation: New combinations (round green, wrinkled yellow) indicate independent inheritance.
Genotypic Ratio: 1:2:1:2:4:2:1:2:1 (9 types).
Principle: Genes for different traits assort independently if on separate chromosomes.
6.2 Linkage and Crossing Over
Linkage Concept: Genes on same chromosome inherited together as linkage group; retain parental combinations.
Evidence (Fig. 6.8): Bateson & Punnett sweet pea - Red long x White short → F1 red long; F2 more parental (red long/white short) than recombinants.
Morgan's Drosophila Experiment: Grey vestigial (BBvv) x Black long (bbVV) → F1 grey long (BbVv); test cross → 83% parental, 17% non-parental (recombinants).
Crossing Over (Fig. 6.10): Exchange between non-sister chromatids during meiosis; produces recombinants; frequency indicates distance (1% = 1 map unit/cM).
Linear Arrangement (Fig. 6.9): Genes in linear order on chromosome; closer genes = tighter linkage.
Fig. 6.8: Bateson and Punnett experiment on sweet pea to study linkage (Description)
Parents: Red long x White short → F1 Red long → F2: Parental (red long/white short) > Non-parental (red short/white long).
Fig. 6.9: Linkage and crossing over among genes for flower colour (R and r) and pollen shape (L and l) (Description)
Fig. 6.11: Experimental evidence of crossing over (Description)
(a) Abnormal chromosome (knob + piece) with markers; (b) Test cross results: Non-CO parental, CO recombinant cytology.
6.4 Sex-linked Inheritance
Concept: Genes on sex chromosomes (X/Y); e.g., hemophilia, color blindness in humans (X-linked recessive).
Morgan's Example (Fig. 6.12): Drosophila white-eyed male (X^w Y) x red female (X^W X^W) → F1 all red; F2: Red females, red/white males (3:1 overall, sex-specific).
All terms from chapter; detailed with examples, relevance. Expanded: 30+ terms with depth for easy learning; grouped by subtopic. Added linkage/recombination terms.
Heredity
Transmission of traits from parents to offspring. Relevance: Basis of genetics. Ex: Eye color inheritance. Depth: Gene-based.
Tip: Group by Mendelian/exceptions; examples link to experiments. Depth: Ratios tie to laws. Errors: Confuse dominance/codominance. Historical: Morgan linkage. Interlinks: Ch7 molecular. Advanced: Map functions. Real-Life: Sex-linked in pedigree analysis. Graphs: Ratio bars. Coherent: Laws → Exceptions → Applications. For easy learning: Flashcard per term with ratio/example.
60+ Questions & Answers - NCERT Based (Class 11) - From Exercises & Variations
Based on chapter content + expansions. Part A: 10 (1 mark short, one line each), Part B: 10 (4 marks medium, five lines each), Part C: 10 (6 marks long, eight lines each). Answers point-wise, step-by-step for marks. Easy learning: Structured, concise. Additional 30 Qs follow similar pattern in full resource.
Part A: 1 Mark Questions (10 Qs - Short from Content)
1. Who is known as the father of genetics?
1 Mark Answer: Gregor Johann Mendel.
2. What is the phenotypic ratio in a monohybrid cross F2 generation?
1 Mark Answer: 3:1 (dominant:recessive).
3. Define allele.
1 Mark Answer: Alternative forms of a gene.
4. What is a test cross used for?
1 Mark Answer: To determine if a dominant phenotype is homozygous or heterozygous.
5. In incomplete dominance, what is the F2 phenotypic ratio?
1 Mark Answer: 1:2:1 (parental:intermediate:parental).
6. What is the dihybrid cross F2 phenotypic ratio?
1 Mark Answer: 9:3:3:1.
7. Define linkage.
1 Mark Answer: Inheritance of genes on the same chromosome together.
8. What causes recombination?
1 Mark Answer: Crossing over between homologous chromosomes.
9. What is polyploidy?
1 Mark Answer: More than two sets of chromosomes in a cell.
10. Differentiate forward and reverse genetics briefly.
1 Mark Answer: Forward: Phenotype to gene; Reverse: Gene to phenotype.
Part B: 4 Marks Questions (10 Qs - Medium, Exactly 5 Lines Each)
1. Explain Mendel's monohybrid cross with ratios.
4 Marks Answer:
Pure tall (TT) x dwarf (tt) → F1 all tall (Tt).
F1 selfing → F2: 3 tall:1 dwarf phenotypic.
Genotypic: 1 TT:2 Tt:1 tt.
Law of segregation: Alleles separate.
Dominance: Tall masks dwarf.
2. Describe incomplete dominance with example.
4 Marks Answer:
Heterozygote intermediate phenotype.
Red (RR) x white (rr) → pink (Rr) F1.
F2: 1 red:2 pink:1 white.
Partial expression of both alleles.
Example: Mirabilis jalapa flowers.
3. State and explain law of independent assortment.
4 Marks Answer:
Genes for different traits assort independently.
Dihybrid: RRYY x rryy → F1 RrYy.
F2: 9:3:3:1 ratio with new combinations.
Applies if on separate chromosomes.
Explains recombination.
4. What is linkage? Give evidence.
4 Marks Answer:
Genes on same chromosome inherited together.
Bateson sweet pea: More parental in F2.
Morgan Drosophila: 83% parental combinations.
Linkage group per chromosome.
Deviation from independent assortment.
5. Explain crossing over and its significance.
4 Marks Answer:
Exchange between non-sister chromatids in meiosis.
Produces recombinants (non-parental).
Frequency measures gene distance (1% = 1 cM).
Evidence: Creighton-McClintock maize cytology.
Breaks linkage, increases variation.
6. Describe sex-linked inheritance in Drosophila.
4 Marks Answer:
Eye color gene on X chromosome.
White male x red female → F1 all red.
F2: Red females, red/white males.
Males hemizygous (X^w Y white).
Criss-cross inheritance.
7. What is extrachromosomal inheritance? Example.
4 Marks Answer:
Maternal inheritance via cytoplasm (mito/plastids).
Example: Snapdragon red x white → pink F1; F2 1:2:1 red:pink:white.
Both exceptions to complete dominance.
MN blood group codominance.
Biotech: Marker traits.
10. Integrate Mendel's laws with linkage and recombination in modern genetics.
6 Marks Answer:
Mendel's laws assume independent genes; linkage violates for same chromosome.
Segregation holds, but assortment not for linked.
Recombination via CO restores independence partially.
Morgan: Linked genes map distances.
Applications: QTL mapping in breeding.
Exceptions explain deviations from ratios.
Reverse genetics uses maps for function.
Foundation for chromosome theory.
Tip: Use diagrams/ratios for marks; practice crosses. Easy learning: Short for recall, long for essays. Additional 30 Qs: Variations on polyploidy, sex-linkage pedigrees.
Key Concepts - In-Depth Exploration
Core ideas with examples, pitfalls, interlinks. Expanded: All concepts from 6.1-6.7 with steps/examples for easy learning. Added depth with cross steps, mapping calculations.
Advanced: RF calculation: (Recombinants/total) x100. Pitfalls: Ratio memorization without understanding. Interlinks: Ch8 disorders. Real: GMO crops linkage. Depth: 7 Mendel traits details. Examples: Hemophilia royal. Graphs: F2 pie charts. Errors: Homo/hetero mixup. Tips: Steps for crosses; compare tables for dominance types.
Historical Perspectives - Detailed Guide
Timeline of inheritance discoveries; expanded with points for easy learning; links to key experiments. Added Mendel details, Morgan era.
Mendel's Era (19th C)
1822-84: Mendel monk, pea experiments 1856-63.
1866: Published laws in Brno society.
Ignored until 1900 rediscovery.
Depth: 7 traits, 28k plants; statistical ratios.
Rediscovery (1900)
de Vries, Correns, Tschermak: Independent pea work.
Confirmed Mendel's ratios.
Sutton 1902: Chromosome theory link.
Depth: Correns snapdragon incomplete dominance.
Linkage & Mapping (1910s)
Morgan 1910: Drosophila sex-linkage.
1911: Linkage in body/wing genes.
Sturtevant 1913: First gene map.
Depth: White eye mutant; fly room culture.
Cytological Proof (1930s)
Creighton-McClintock 1931: Maize CO evidence.
Bridges 1916: Nondisjunction aneuploidy.
McClintock 1940s: Transposons (Nobel 1983).
Depth: Knob/translocation markers.
Modern Genetics (1950s+)
Watson-Crick 1953: DNA structure.
1960s: Operon model (Jacob-Monod).
1970s: Recombinant DNA.
Depth: Reverse genetics post-HGP 2003.
Polyploidy & Extranuclear
1907: De Vries coins polyploidy.
1960s: Plastid inheritance studies.
1970s: mtDNA maternal.
Depth: Wheat hybrid history.
Tip: Link to tools (Drosophila for fast generations). Depth: Bateson 1905 coupling. Examples: 1900 triple rediscovery. Graphs: Timeline with milestones. Advanced: QTL 1990s. Easy: Chronological bullets with impacts.
Solved Examples - From Text with Simple Explanations
Expanded with more examples, steps for easy understanding; focus on crosses, ratios. Added dihybrid, test cross, recombination calc.
Example 1: Monohybrid Cross Ratio (Fig 6.2)
Simple Explanation: Predicts F2 reappearance of recessive.
Step 1: TT x tt → Tt (all dominant).
Step 2: Tt gametes T/t (1:1).
Step 3: Punnett: TT, Tt, Tt, tt.
Step 4: Phenotypic 3:1, genotypic 1:2:1.
Simple Way: Hidden dwarf waits for homozygous pair.
Example 2: Incomplete Dominance Flowers
Simple Explanation: Mix colors like paint blending.
Step 1: RR red x rr white → Rr pink.
Step 2: Rr gametes R/r (1:1).
Step 3: F2: RR red, Rr pink (x2), rr white.
Step 4: 1:2:1 ratio all visible.
Simple Way: Half red + half white = pink.
Example 3: Dihybrid Cross (Fig 6.7)
Simple Explanation: Two traits mix freely.
Step 1: RRYY x rryy → RrYy.
Step 2: Gametes RY, Ry, rY, ry (1:1:1:1).
Step 3: 16 combinations → 9 RY dom, 3 ry dom, etc.
Step 4: 9:3:3:1 phenotypes.
Simple Way: Dice roll for each trait independently.
Example 4: Test Cross Identification
Simple Explanation: Reveals hidden recessive.
Step 1: Unknown tall x tt dwarf.
Step 2: If TT: All tall offspring.
Step 3: If Tt: 50% tall, 50% dwarf.
Step 4: Ratio confirms genotype.
Simple Way: Cross with known to unmask.
Example 5: Recombination Frequency Calculation
Simple Explanation: Measures gene distance.
Step 1: Total progeny 1000.
Step 2: Recombinants 170 (non-parental).
Step 3: RF = (170/1000) x 100 = 17%.
Step 4: Distance = 17 cM.
Simple Way: % surprises = map units apart.
Example 6: Sex-linked Cross in Humans (Hypothetical)
Simple Explanation: Males show recessive more.
Step 1: Carrier female X^H X^h x normal male X^H Y.
Step 2: Daughters: 50% carrier, 50% normal.
Step 3: Sons: 50% normal, 50% affected.
Step 4: No father-to-son transmission.
Step 5: Pedigree skips generations in females.
Step 6: Simple Way: X from mom decides son's trait.