This MCQ module is based on: Dihybrid Laws of Inheritance
TOPIC 16 OF 64
Dihybrid Laws of Inheritance
🎓 Class 12
Biology
CBSE
Theory
Ch 4 – Principles of Inheritance and Variation
⏱ ~14 min
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Dihybrid Laws of Inheritance
5.5 The Dihybrid Cross — Two Pairs of Traits Together
Having seen monohybrid crosses, Mendel asked: what happens if I track two traits at the same time? A cross involving two pairs of contrasting characters is a dihybrid cross. Mendel crossed pea plants differing in seed shape (Round R / wrinkled r) and seed colour (Yellow Y / green y).
P: Round Yellow (RRYY) × Wrinkled green (rryy)
F1: All Round Yellow (RrYy) — both dominant traits expressed.
F2 from F1 × F1 (RrYy × RrYy): Phenotypic ratio = 9 : 3 : 3 : 1
| Phenotype | Genotype combinations | Ratio |
|---|---|---|
| Round Yellow | R_Y_ (1 RRYY + 2 RRYy + 2 RrYY + 4 RrYy) | 9 |
| Round green | R_yy (1 RRyy + 2 Rryy) | 3 |
| Wrinkled Yellow | rrY_ (1 rrYY + 2 rrYy) | 3 |
| Wrinkled green | rryy | 1 |
Law of Independent Assortment: When two pairs of traits are inherited together, the alleles of one pair segregate independently of the alleles of the other pair during gamete formation. This law is true only for genes on different chromosomes (or far apart on the same chromosome).
5.6 Beyond Mendel — Inheritance Patterns That Bend the Rules
5.6.1 Incomplete Dominance
In incomplete dominance, the heterozygote shows a phenotype intermediate between the two homozygous parents. Classic example: Snapdragon (Antirrhinum) flower colour.
P: RR (Red) × rr (white) → F1: Rr (Pink) → F2: 1 Red : 2 Pink : 1 white
Note: Phenotype ratio = Genotype ratio = 1 : 2 : 1. Each genotype shows its own visible phenotype.
5.6.2 Codominance
In codominance, both alleles are fully and simultaneously expressed in the heterozygote — both phenotypes appear together. Classic example: ABO blood groups in humans.
| Genotype | Phenotype (Blood group) | Antigen on RBC |
|---|---|---|
| IAIA or IAi | A | Antigen A |
| IBIB or IBi | B | Antigen B |
| IAIB | AB | Both A and B (codominance) |
| ii | O | None |
5.6.3 Multiple Alleles
While Mendel's pea height had only two alleles (T, t), many genes have multiple alleles in the population. The ABO blood group gene has 3 alleles: IA, IB, and i. IA and IB are both dominant over i but codominant to each other.
5.6.4 Pleiotropy — One Gene, Many Effects
Pleiotropy is the phenomenon where a single gene affects several phenotypic traits. Examples include phenylketonuria (PKU) and sickle cell anaemia. In sickle cell, one mutation in haemoglobin causes anaemia, joint pain, organ damage, and even partial malaria resistance.
🧬 Interactive: ABO Blood Group Cross Predictor
Choose parental genotypes to see all possible blood group offspring:
📐 Activity 5.2 — Snapdragon Cross
Setup: A pure-breeding red snapdragon (RR) is crossed with a pure-breeding white snapdragon (rr). The F1 are pink (Rr).
Predict: If 200 F2 plants are grown from F1 × F1, how many should be red, pink, and white?
F2 ratio: 1 Red (RR) : 2 Pink (Rr) : 1 White (rr).
Out of 200: Red = ¼ × 200 = 50; Pink = ½ × 200 = 100; White = ¼ × 200 = 50.
Note: Unlike Mendelian dominance, here phenotype ratio (1:2:1) equals genotype ratio — each genotype shows a unique colour.
Worked Examples
Worked Example 1: Dihybrid Test Cross
An F1 dihybrid plant (RrYy) is test crossed with a homozygous recessive (rryy). What phenotypic ratio is expected?
Step 1: RrYy gametes: RY, Ry, rY, ry (each 1/4).
Step 2: rryy gametes: only ry.
Step 3: Offspring: 1 RrYy (round yellow) : 1 Rryy (round green) : 1 rrYy (wrinkled yellow) : 1 rryy (wrinkled green).
Phenotype ratio = 1 : 1 : 1 : 1. This is the classic dihybrid test cross — each gamete type appears equally.
Step 2: rryy gametes: only ry.
Step 3: Offspring: 1 RrYy (round yellow) : 1 Rryy (round green) : 1 rrYy (wrinkled yellow) : 1 rryy (wrinkled green).
Phenotype ratio = 1 : 1 : 1 : 1. This is the classic dihybrid test cross — each gamete type appears equally.
Worked Example 2: ABO Inheritance
A father has blood group AB. A mother has blood group O. What blood groups are possible in their children, and which are NOT possible?
Father: IAIB — gametes IA or IB.
Mother: ii — gametes only i.
Offspring: IAi (A) or IBi (B), each 50%.
Possible: Blood groups A and B only.
NOT possible: AB (would need both IA and IB from mother — she has none) and O (would need i from father — he has none).
This is a useful tool in disputed paternity cases (though now replaced by DNA testing).
Mother: ii — gametes only i.
Offspring: IAi (A) or IBi (B), each 50%.
Possible: Blood groups A and B only.
NOT possible: AB (would need both IA and IB from mother — she has none) and O (would need i from father — he has none).
This is a useful tool in disputed paternity cases (though now replaced by DNA testing).
Worked Example 3: Predict Phenotype
In snapdragons, R = red, r = white, with incomplete dominance. Cross Rr × rr. What is the phenotype ratio?
Gametes: Rr → R, r (1:1); rr → only r.
Offspring: Rr (Pink, 50%) : rr (White, 50%).
Phenotype ratio = 1 Pink : 1 White. No red appears because no F1 contributed two R alleles.
Offspring: Rr (Pink, 50%) : rr (White, 50%).
Phenotype ratio = 1 Pink : 1 White. No red appears because no F1 contributed two R alleles.
🎯 Competency-Based Questions
Q1. The dihybrid F2 ratio of 9:3:3:1 indicates that:L1 Remember
Answer: (b). The 9:3:3:1 ratio results when alleles of two genes segregate independently into gametes during meiosis. Linked genes would NOT give 9:3:3:1.
Q2. Fill in the blank: A blood group AB person owes their phenotype to the phenomenon of _____. L2 Understand
Answer: Codominance. Both IA and IB alleles produce their respective antigens (A and B) on the red blood cell surface — both fully expressed simultaneously, neither one masks the other.
Q3. A man with blood group A marries a woman with blood group B. They have a child with blood group O. Determine the genotype of each parent. L3 Apply
Answer: Both parents are heterozygous: Father = IAi and Mother = IBi. The child is ii (O), which means each parent must have contributed a recessive i allele. Therefore both parents must carry one i allele.
Q4. Compare: Distinguish between incomplete dominance and codominance with one example each. L4 Analyse
Incomplete dominance: Heterozygote shows an intermediate phenotype — alleles "blend" at the phenotype level. Example: Pink snapdragon (Rr) is intermediate between Red (RR) and White (rr).
Codominance: Heterozygote shows BOTH phenotypes simultaneously and distinctly — no blending. Example: AB blood group has BOTH antigen A and antigen B on red blood cells.
Key difference: Incomplete dominance creates a new intermediate phenotype; codominance shows both parental phenotypes side-by-side.
Codominance: Heterozygote shows BOTH phenotypes simultaneously and distinctly — no blending. Example: AB blood group has BOTH antigen A and antigen B on red blood cells.
Key difference: Incomplete dominance creates a new intermediate phenotype; codominance shows both parental phenotypes side-by-side.
Q5. HOT (Create): A geneticist crosses two RrYy plants and gets only 100 F2 instead of an expected 1600. What problems might arise from such a small sample? Design a strategy to minimize error. L6 Create
Problems with small sample:
- Random chance fluctuations — observed ratio may deviate from 9:3:3:1 (e.g., observed 60:18:16:6 due to chance).
- Rare phenotype (1/16) might be entirely missed.
- Statistical tests (chi-square) may falsely accept or reject Mendelian inheritance.
- Increase sample size — at minimum, 10 plants per smallest expected class (so >160 plants).
- Pool data from multiple independent crosses (technical replicates).
- Apply chi-square test to assess goodness-of-fit.
- Test cross F1 with double recessive (1:1:1:1) to confirm independent assortment.
🧠 Assertion–Reason Questions
Choose: (A) Both true, R explains A. (B) Both true, R doesn't explain A. (C) A true, R false. (D) A false, R true.
A: The F1 hybrid in a dihybrid cross between RRYY and rryy produces 4 types of gametes in equal frequency.
R: The two genes assort independently during meiosis if they are on different chromosomes.
Answer: (A). Both true; R explains A. Independent assortment of two unlinked genes during meiosis produces gametes in 1:1:1:1 ratio (RY, Ry, rY, ry).
A: A child of two heterozygous parents (IAi × IBi) has a 1/4 chance of being blood group O.
R: Probability of homozygous recessive offspring from two heterozygous parents is 1/4.
Answer: (A). Both true; R explains A. P(ii from IAi × IBi) = ½ × ½ = ¼ — same logic as any heterozygous × heterozygous cross for the recessive phenotype.
A: Pleiotropy is a violation of Mendel's law of independent assortment.
R: In pleiotropy, one gene controls multiple traits.
Answer: (D). A is FALSE — pleiotropy is NOT a violation of independent assortment (which concerns multiple genes, not multiple traits per gene). Independent assortment governs how multiple genes segregate; pleiotropy is a single-gene phenomenon. R is TRUE — that's the definition of pleiotropy.
Frequently Asked Questions - Dihybrid Laws of Inheritance
What is the main concept covered in Dihybrid Laws of Inheritance?
In NCERT Class 12 Biology Chapter on Principles of Inheritance and Variation, "Dihybrid Laws of Inheritance" covers the core biological structures, processes, and pathways students need for board exam success. The MyAiSchool lesson explains the topic with definitions, labelled diagrams, comparison tables, and interactive simulations. Scientific terminology and physiological/genetic significance are highlighted throughout to build conceptual depth aligned with CBSE 2025-26 syllabus.
How is Dihybrid Laws of Inheritance useful in real-life or applied biology?
Real-life applications of "Dihybrid Laws of Inheritance" from NCERT Class 12 Biology Principles of Inheritance and Variation include medical diagnostics, agriculture, biotechnology, public health, evolutionary insights, and ecological monitoring. The MyAiSchool lesson links every biological concept to a tangible application so students see biology as a problem-solving framework for living systems and real-world challenges.
What are the key terms students should memorize for Dihybrid Laws of Inheritance?
Key terms in "Dihybrid Laws of Inheritance" (NCERT Class 12 Biology Principles of Inheritance and Variation) are tabulated in the MyAiSchool key-terms grid. Students should memorize each term with its precise definition, function, and example. Terminology is high-yield in CBSE board exams — 1-mark MCQs and 2-mark short answers test definitions directly. The Summary section provides a printable quick-reference card.
How does this part connect to other parts of the chapter?
NCERT Class 12 Biology Principles of Inheritance and Variation is structured so each part builds biological understanding sequentially. "Dihybrid Laws of Inheritance" connects to neighbouring parts via shared mechanisms, structural hierarchies, and physiological processes. The MyAiSchool lesson cross-references related concepts with internal links so students can navigate the whole chapter as one connected biological story rather than disconnected fragments.
What types of CBSE board questions come from Dihybrid Laws of Inheritance?
CBSE board questions from "Dihybrid Laws of Inheritance" typically include: (1) 1-mark MCQs on definitions and processes, (2) 2-mark short-answer differences/comparisons, (3) 3-mark labelled-diagram questions, (4) 5-mark long-answer essays combining mechanism + diagram + significance. The MyAiSchool lesson tags each Competency-Based Question (CBQ) with Bloom level (L1-L6) so students know how to study for each weight.
How can students use the interactive simulation effectively?
The interactive simulation in the "Dihybrid Laws of Inheritance" lesson allows students to explore biological processes, classifications, or pathways using selectors and sliders, with live visual feedback. To use it effectively: (1) explore each option/state, (2) compare with textbook diagrams, (3) note the function/outcome changes, (4) try the integrated practice quiz. The simulation reinforces visual-spatial understanding that pure text-based study cannot.
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