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Class 10 Science
Heredity and Evolution - Mendelian inheritance principles, monohybrid and dihybrid crosses, and mechanism of sex determination in human beings
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ScienceClass 10Heredity and Evolution

Heredity and Evolution - Mendelian inheritance principles, monohybrid and dihybrid crosses, and mechanism of sex determination in human beings

2026-09-2819 min readRHS Academic Faculty
Overview & Key Summary:Class 10 Science: Heredity and Evolution Mendelian Inheritance and Sex Determination Heredity is the process by which physical, physiological, or behavioral traits are passed do...

Class 10 Science: Heredity and Evolution - Mendelian Inheritance and Sex Determination

Heredity is the process by which physical, physiological, or behavioral traits are passed down from parents to their offspring. While offspring resemble their parents, they are not exact carbon copies; they display subtle differences known as variations. Understanding how traits are inherited and expressed is one of the most significant triumphs of modern biological science.

The systematic study of inheritance began in the mid-19th century with Gregor Johann Mendel, widely recognized as the "Father of Genetics." By conducting controlled hybridization experiments on the garden pea plant (Pisum sativum), Mendel formulated the fundamental principles of inheritance long before the discovery of DNA, genes, or chromosomes.

In this study guide, we will explore Mendel's groundbreaking experiments, the laws of inheritance derived from monohybrid and dihybrid crosses, the cellular mechanisms controlling gene expression, and the biological process of sex determination in human beings.


1. Fundamental Terminology in Genetics

To analyze Mendelian crosses effectively, one must master the basic terminology used in classical genetics:

  • Gene: The functional unit of heredity located on a chromosome. It comprises a specific sequence of DNA that encodes instructions for synthesizing a specific protein, which subsequently determines a biological trait.
  • Allele: Alternative forms of a single gene that control contrasting expressions of a trait. For instance, the gene for plant height has two primary alleles: tallness (TTT) and dwarfness (ttt).
  • Dominant Allele: An allele that expresses its phenotypic effect even in the presence of a contrasting allele (i.e., in both homozygous and heterozygous conditions). Represented by a capital letter (e.g., TTT).
  • Recessive Allele: An allele whose phenotypic effect is masked in the presence of a dominant allele and is expressed only when present in a duplicate, identical state (i.e., homozygous condition). Represented by a lowercase letter (e.g., ttt).
  • Homozygous: An organism possessing two identical alleles for a particular gene (e.g., TTTTTT for pure tall or tttttt for pure dwarf).
  • Heterozygous: An organism possessing two different alleles for a particular gene (e.g., TtTtTt).
  • Genotype: The genetic constitution or allelic makeup of an organism regarding a specific trait (e.g., TTTTTT, TtTtTt, or tttttt).
  • Phenotype: The observable physical appearance or functional expression of a trait in an organism (e.g., Tall or Dwarf).
  • F1\text{F}_1F1​ Generation (First Filial Generation): The progeny produced by crossing two pure-breeding parents showing contrasting traits.
  • F2\text{F}_2F2​ Generation (Second Filial Generation): The progeny produced by self-pollinating or intercrossing the individuals of the F1\text{F}_1F1​ generation.

2. Mendelian Inheritance Principles

Why Mendel Chose the Garden Pea (Pisum sativum)

Mendel selected the garden pea plant for his experiments due to several distinct biological advantages:

  1. Clear, Contrasting Traits: Pea plants possess easily identifiable contrasting characters (e.g., Tall vs. Dwarf height, Round vs. Wrinkled seed shape, Yellow vs. Green seed color).
  2. Short Life Cycle: The plant completes its lifecycle within a few months, enabling the collection of data across multiple generations in a short timeframe.
  3. Self-Pollination & Bisexual Flowers: Naturally, pea flowers self-pollinate, making it straightforward to maintain pure lines (homozygous parents).
  4. Ease of Artificial Cross-Pollination: Emasculation (removal of anthers) and dusting of desired pollen allow controlled cross-breeding.
  5. Large Sample Size: A single plant produces numerous seeds, yielding statistically reliable data.

A. Monohybrid Cross: Inheritance of a Single Trait

A cross involving two plants that differ in only one pair of contrasting characters is known as a monohybrid cross.

Experimental Procedure:

Mendel crossed a pure-breeding Tall pea plant (TTTTTT) with a pure-breeding Dwarf pea plant (tttttt).

  1. Parental Generation (P\text{P}P):

    • Phenotypes: Tall ×\times× Dwarf
    • Genotypes: TT×ttTT \times ttTT×tt
    • Gametes produced: TTT from tall parent; ttt from dwarf parent.
  2. First Filial Generation (F1\text{F}_1F1​):

    • All F1\text{F}_1F1​ offspring had the genotype TtTtTt.
    • Phenotypically, 100%100\%100% of the plants were Tall.
    • The dwarf trait appeared to have completely vanished in the F1\text{F}_1F1​ generation.
  3. Second Filial Generation (F2\text{F}_2F2​):

    • Mendel allowed the F1\text{F}_1F1​ plants to self-pollinate (Tt×TtTt \times TtTt×Tt).
    • Gametes produced by F1\text{F}_1F1​: 50%50\%50% carrying allele TTT, 50%50\%50% carrying allele ttt.

Punnett Square Representation (F2\text{F}_2F2​ Generation):

GametesTTT (0.50.50.5)ttt (0.50.50.5)
TTT (0.50.50.5)TTTTTT (Tall)TtTtTt (Tall)
ttt (0.50.50.5)TtTtTt (Tall)tttttt (Dwarf)

Ratios of the Monohybrid F2\text{F}_2F2​ Generation:

  • Phenotypic Ratio: Tall:Dwarf=3:1\text{Tall} : \text{Dwarf} = 3 : 1Tall:Dwarf=3:1
  • Genotypic Ratio: TT:Tt:tt=1:2:1TT : Tt : tt = 1 : 2 : 1TT:Tt:tt=1:2:1
DIAGRAM
Swipe sideways ↔Scrollable
Parental Generation (P):      TT (Tall)   ×   tt (Dwarf)
                                  │               │
Gametes:                          T               t
                                  └───┬───────┬───┘
                                      │       │
F1 Generation:                       Tt (All Tall)
                                (Self-pollination)
                                      │
F2 Generation:               TT  :   Tt  :   tt
Phenotypic Ratio:            [  3 Tall  ] : [1 Dwarf]
Genotypic Ratio:                 1  :  2  :  1

Laws Derived from the Monohybrid Cross

1. Law of Dominance

  • Characters are controlled by discrete units called factors (now known as genes), which occur in pairs in diploid organisms.
  • In a dissimilar pair of factors (heterozygous condition, e.g., TtTtTt), one factor dominates or masks the expression of the other.
  • The factor that expresses itself is the Dominant factor (TTT), and the one that remains hidden is the Recessive factor (ttt).

2. Law of Segregation (Principle of Purity of Gametes)

  • Alleles of a gene pair do not blend or mix when present together in a heterozygous individual (TtTtTt).
  • During gamete formation (meiosis), the two alleles segregate (separate) from each other such that each gamete receives only one allele with equal probability (50%50\%50%).
  • Consequently, gametes are always "pure" for a given trait (they carry either TTT or ttt, never both).

B. Dihybrid Cross: Inheritance of Two Traits Simultaneously

A cross involving two plants that differ in two pairs of contrasting characters is known as a dihybrid cross.

Experimental Procedure:

Mendel crossed a plant having Round and Yellow seeds with a plant having Wrinkled and Green seeds.

  • Seed Shape Alleles: Round (RRR) is dominant over Wrinkled (rrr).
  • Seed Color Alleles: Yellow (YYY) is dominant over Green (yyy).
  1. Parental Generation (P\text{P}P):

    • Phenotypes: Round Yellow ×\times× Wrinkled Green
    • Genotypes: RRYY×rryyRRYY \times rryyRRYY×rryy
    • Gametes: RYRYRY (from RRYYRRYYRRYY) and ryryry (from rryyrryyrryy)
  2. First Filial Generation (F1\text{F}_1F1​):

    • Genotype: RrYyRrYyRrYy
    • Phenotype: 100%100\%100% Round Yellow seeds
  3. Second Filial Generation (F2\text{F}_2F2​):

    • Self-pollination of F1\text{F}_1F1​: RrYy×RrYyRrYy \times RrYyRrYy×RrYy
    • Types of gametes formed by each RrYyRrYyRrYy plant (due to independent segregation): RYRYRY, RyRyRy, rYrYrY, ryryry in equal proportions (1:1:1:11:1:1:11:1:1:1).

Punnett Square Representation (F2\text{F}_2F2​ Generation):

GametesRYRYRYRyRyRyrYrYrYryryry
RYRYRYRRYYRRYYRRYY <br> (Round Yellow)RRYyRRYyRRYy <br> (Round Yellow)RrYYRrYYRrYY <br> (Round Yellow)RrYyRrYyRrYy <br> (Round Yellow)
RyRyRyRRYyRRYyRRYy <br> (Round Yellow)RRyyRRyyRRyy <br> (Round Green)RrYyRrYyRrYy <br> (Round Yellow)RryyRryyRryy <br> (Round Green)
rYrYrYRrYYRrYYRrYY <br> (Round Yellow)RrYyRrYyRrYy <br> (Round Yellow)rrYYrrYYrrYY <br> (Wrinkled Yellow)rrYyrrYyrrYy <br> (Wrinkled Yellow)
ryryryRrYyRrYyRrYy <br> (Round Yellow)RryyRryyRryy <br> (Round Green)rrYyrrYyrrYy <br> (Wrinkled Yellow)rryyrryyrryy <br> (Wrinkled Green)

Ratios of the Dihybrid F2\text{F}_2F2​ Generation:

  • Phenotypic Ratio: Round Yellow:Round Green:Wrinkled Yellow:Wrinkled Green=9:3:3:1\text{Round Yellow} : \text{Round Green} : \text{Wrinkled Yellow} : \text{Wrinkled Green} = 9 : 3 : 3 : 1Round Yellow:Round Green:Wrinkled Yellow:Wrinkled Green=9:3:3:1

  • Detailed Breakdown of Phenotypes:

    • Round, Yellow: 999
    • Round, Green: 333
    • Wrinkled, Yellow: 333
    • Wrinkled, Green: 111
  • Genotypic Ratio (1:2:1:2:4:2:1:2:11:2:1:2:4:2:1:2:11:2:1:2:4:2:1:2:1):

    • RRYY=1RRYY = 1RRYY=1
    • RRYy=2RRYy = 2RRYy=2
    • RRyy=1RRyy = 1RRyy=1
    • RrYY=2RrYY = 2RrYY=2
    • RrYy=4RrYy = 4RrYy=4
    • Rryy=2Rryy = 2Rryy=2
    • rrYY=1rrYY = 1rrYY=1
    • rrYy=2rrYy = 2rrYy=2
    • rryy=1rryy = 1rryy=1

Law Derived from the Dihybrid Cross

Law of Independent Assortment

When two pairs of traits are combined in a hybrid, the segregation or inheritance of one pair of characters is completely independent of the segregation of the other pair of characters during gamete formation.

This principle explains why brand-new recombinant combinations of traits (Round Green and Wrinkled Yellow) appear in the F2\text{F}_2F2​ generation alongside parental combinations (Round Yellow and Wrinkled Green).


3. How Genes Express Traits (Cellular Mechanism)

Genes do not directly build physical structures; rather, they serve as information manuals. The pathway from a gene to a physical trait works via molecular mechanisms:

DNA (Gene)→TranscriptionCellular RNA→TranslationFunctional Protein / Enzyme→Catalyzes ReactionSpecific Trait\text{DNA (Gene)} \xrightarrow{\text{Transcription}} \text{Cellular RNA} \xrightarrow{\text{Translation}} \text{Functional Protein / Enzyme} \xrightarrow{\text{Catalyzes Reaction}} \text{Specific Trait}DNA (Gene)Transcription​Cellular RNATranslation​Functional Protein / EnzymeCatalyzes Reaction​Specific Trait

Example: Control of Plant Height

  1. A specific region of nuclear DNA contains the gene for plant height.
  2. This gene encodes instructions to produce a specific enzyme.
  3. The enzyme catalyzes the biochemical synthesis of a plant growth hormone (e.g., gibberellin).
  4. If the allele for tallness (TTT) is present:
    • It produces a fully functional, efficient enzyme.
    • Large quantities of plant hormone are produced.
    • The plant grows tall.
  5. If the allele for dwarfness (ttt) is mutated/recessive:
    • It produces a non-functional or less efficient enzyme.
    • Insufficient plant hormone is produced.
    • The plant remains dwarf.

Thus, genes control traits by controlling the synthesis and efficiency of specific proteins and enzymes.


4. Mechanism of Sex Determination in Human Beings

Sex determination is the biological system that determines the development of sexual characteristics in an organism. In human beings, sex is determined genetically at the precise moment of fertilization.

Chromosomal Composition in Humans

  • Human cells contain 232323 pairs of chromosomes (464646 individual chromosomes) inside the nucleus.
  • Autosomes (222222 pairs / 444444 chromosomes): These control general somatic/body traits and are identical in both males and females.
  • Sex Chromosomes / Allosomes (111 pair / 222 chromosomes): These determine the biological sex of the individual.
    • Human Females possess two identical, normal-sized sex chromosomes designated as XXXXXX.
    • Human Males possess two distinct sex chromosomes: one normal-sized XXX chromosome and one smaller YYY chromosome, designated as XYXYXY.
DIAGRAM
Swipe sideways ↔Scrollable
Human Cell (46 Chromosomes / 23 Pairs)
 ├── Autosomes (22 Pairs / 44 Chromosomes) ──> Determines general body traits
 └── Sex Chromosomes (1 Pair / 2 Chromosomes)
      ├── Female: XX (Homogametic)
      └── Male:   XY (Heterogametic)

Gametogenesis and Determination Process

  1. Female Gametes (Eggs/Ova):

    • During meiosis, the female pair of sex chromosomes (XXXXXX) segregates.
    • Every egg produced carries 222222 autosomes + one XXX chromosome (22+X22 + X22+X).
    • Because all eggs are uniform in their chromosomal composition, females are termed homogametic.
  2. Male Gametes (Sperm):

    • During meiosis, the male sex chromosome pair (XYXYXY) segregates.
    • 50%50\%50% of the sperm carry 222222 autosomes + one XXX chromosome (22+X22 + X22+X).
    • 50%50\%50% of the sperm carry 222222 autosomes + one YYY chromosome (22+Y22 + Y22+Y).
    • Because males produce two distinct types of gametes, males are termed heterogametic.
  3. Fertilization Events:

    • If an XXX-bearing egg is fertilized by an XXX-bearing sperm: Genotype=44+XX  ⟹  Offspring is Female (Girl)\text{Genotype} = 44 + XX \implies \text{Offspring is Female (Girl)}Genotype=44+XX⟹Offspring is Female (Girl)
    • If an XXX-bearing egg is fertilized by a YYY-bearing sperm: Genotype=44+XY  ⟹  Offspring is Male (Boy)\text{Genotype} = 44 + XY \implies \text{Offspring is Male (Boy)}Genotype=44+XY⟹Offspring is Male (Boy)

Genetic Cross Diagram for Sex Determination

DIAGRAM
Swipe sideways ↔Scrollable
Parents:                Mother (Female)        ×        Father (Male)
Genotype:                    44 + XX                       44 + XY
                                │                             │
Gametes (Eggs/Sperm):        (22 + X)             (22 + X)        (22 + Y)
                                │                    │               │
                                ├─── Fertilization ──┘               │
                                │    (50% Probability)               │
                                │   Genotype: 44 + XX                │
                                │   Phenotype: Female Child          │
                                │                                    │
                                └─────── Fertilization ──────────────┘
                                     (50% Probability)
                                    Genotype: 44 + XY
                                    Phenotype: Male Child

Key Biological Conclusions:

  1. The statistical probability of having a male or female child is precisely 1:11:11:1 (50%50\%50% chance for each pregnancy).
  2. The father (male parent) determines the sex of the child, because only the sperm can contribute either an XXX or a YYY chromosome. The mother contributes an XXX chromosome in all instances.

5. Real-World Applications & Conceptual Connections

1. Plant and Animal Breeding (Agriculture)

Mendel’s principles allow agricultural scientists to conduct selective breeding. By crossing plants with desirable contrasting traits (e.g., high yield and pest resistance), breeders can produce hybrid crops that combine dominant beneficial traits.

2. Pedigree Analysis and Medical Genetics

In medical science, Mendelian genetics forms the foundation for mapping human genetic disorders (e.g., Hemophilia, Thalassemia, Sickle Cell Anemia, Cystic Fibrosis). By building pedigree charts, genetic counselors can calculate the exact probability of an offspring inheriting a recessive genetic defect from heterozygous carrier parents.

3. Debunking Social Stigmas Surrounding Gender

In many traditional societies, women are erroneously blamed for giving birth to female children. Understanding the genetic mechanism of sex determination proves that female eggs carry only XXX chromosomes, while male sperm carries either XXX or YYY. Hence, sex is determined solely by the type of male gamete that fertilizes the egg, dispelling harmful gender-bias myths.


6. Step-by-Step Solved Examples

Example 1: Monohybrid Inheritance of Flower Color

Problem: In garden pea plants, the allele for violet flower color (VVV) is dominant over the allele for white flower color (vvv). A pure-breeding violet-flowered plant is crossed with a pure-breeding white-flowered plant.

  1. What will be the phenotype and genotype of the F1\text{F}_1F1​ generation?
  2. If the F1\text{F}_1F1​ generation plants are self-pollinated, determine the phenotypic and genotypic ratios of the F2\text{F}_2F2​ generation using a Punnett square.

Solution:

Step 1: Identify Parental Genotypes

  • Pure violet parent = VVVVVV
  • Pure white parent = vvvvvv

Step 2: Determine F1\text{F}_1F1​ Generation Cross: VV×vv\text{Cross: } VV \times vvCross: VV×vv Gametes: V and v\text{Gametes: } V \text{ and } vGametes: V and v F1 Genotype: Vv ( Heterozygous )\text{F}_1 \text{ Genotype: } Vv \text{ ( Heterozygous )}F1​ Genotype: Vv ( Heterozygous ) F1 Phenotype: 100% Violet flowers\text{F}_1 \text{ Phenotype: } 100\% \text{ Violet flowers}F1​ Phenotype: 100% Violet flowers

Step 3: Determine F2\text{F}_2F2​ Generation through Self-Pollination (Vv×VvVv \times VvVv×Vv) Gametes from each VvVvVv parent: VVV (50%50\%50%) and vvv (50%50\%50%).

Punnett Square (F2\text{F}_2F2​):

GametesVVVvvv
VVVVVVVVV (Violet)VvVvVv (Violet)
vvvVvVvVv (Violet)vvvvvv (White)

Step 4: Compute Ratios

  • Phenotypic Ratio: Violet : White = 3:13 : 13:1 (75%75\%75% Violet, 25%25\%25% White)
  • Genotypic Ratio: VV:Vv:vv=1:2:1VV : Vv : vv = 1 : 2 : 1VV:Vv:vv=1:2:1 (25%:50%:25%25\% : 50\% : 25\%25%:50%:25%)

Example 2: Calculating Phenotypic Outcomes in Dihybrid Offspring

Problem: A researcher crosses two heterozygous round-seeded, yellow-seeded pea plants (RrYy×RrYyRrYy \times RrYyRrYy×RrYy). If a total of 1,6001,6001,600 seeds are harvested in the F2\text{F}_2F2​ generation, calculate the expected numerical count of:

  1. Round and Yellow seeds
  2. Wrinkled and Green seeds
  3. Round and Green seeds

Solution:

Step 1: Recall the standard phenotypic ratio of a Mendelian Dihybrid Cross Round Yellow:Round Green:Wrinkled Yellow:Wrinkled Green=9:3:3:1\text{Round Yellow} : \text{Round Green} : \text{Wrinkled Yellow} : \text{Wrinkled Green} = 9 : 3 : 3 : 1Round Yellow:Round Green:Wrinkled Yellow:Wrinkled Green=9:3:3:1 Total ratio parts=9+3+3+1=16\text{Total ratio parts} = 9 + 3 + 3 + 1 = 16Total ratio parts=9+3+3+1=16

Step 2: Calculate expected numbers out of 1,6001,6001,600 total seeds

  1. Round and Yellow seeds: Fraction=916\text{Fraction} = \frac{9}{16}Fraction=169​ Expected Count=916×1600=900 seeds\text{Expected Count} = \frac{9}{16} \times 1600 = 900 \text{ seeds}Expected Count=169​×1600=900 seeds

  2. Wrinkled and Green seeds: Fraction=116\text{Fraction} = \frac{1}{16}Fraction=161​ Expected Count=116×1600=100 seeds\text{Expected Count} = \frac{1}{16} \times 1600 = 100 \text{ seeds}Expected Count=161​×1600=100 seeds

  3. Round and Green seeds: Fraction=316\text{Fraction} = \frac{3}{16}Fraction=163​ Expected Count=316×1600=300 seeds\text{Expected Count} = \frac{3}{16} \times 1600 = 300 \text{ seeds}Expected Count=163​×1600=300 seeds

Final Answer Highlights:

  • Round and Yellow = 900900900 seeds
  • Wrinkled and Green = 100100100 seeds
  • Round and Green = 300300300 seeds

Example 3: Test Cross Determination

Problem: A pea plant displaying the dominant phenotype (Tall) could have either a homozygous (TTTTTT) or a heterozygous (TtTtTt) genotype. How can a breeder determine its exact genotype? Show the crosses involved.

Solution:

To determine the genotype of a dominant phenotype, a Test Cross is performed. The individual with the unknown genotype is crossed with a pure recessive individual (tttttt).

Case A: If the unknown tall plant is Homozygous Dominant (TTTTTT)

  • Cross: TT×ttTT \times ttTT×tt
  • Gametes: TTT and ttt
  • Offspring: All TtTtTt (100%100\%100% Tall)
  • Conclusion: If all progeny are tall, the unknown plant is TTTTTT.

Case B: If the unknown tall plant is Heterozygous (TtTtTt)

  • Cross: Tt×ttTt \times ttTt×tt
  • Gametes from tall parent: TTT and ttt; Gametes from dwarf parent: ttt
  • Offspring Genotypes: 50%50\%50% TtTtTt and 50%50\%50% tttttt
  • Offspring Phenotypes: 50%50\%50% Tall and 50%50\%50% Dwarf (1:11:11:1 ratio)
GametesTTTttt
tttTtTtTt (Tall)tttttt (Dwarf)
  • Conclusion: If dwarf plants appear in a 1:11:11:1 ratio, the unknown plant is TtTtTt.

7. Common Student Mistakes to Avoid

1. Confusing Phenotypic and Genotypic Ratios

  • Mistake: Writing the monohybrid F2\text{F}_2F2​ genotypic ratio as 3:13:13:1.
  • Correction: Always remember that 3:13:13:1 is the Phenotypic ratio (physical appearance: Tall vs Dwarf). The Genotypic ratio reflects the precise allelic combination (TT:Tt:tt=1:2:1TT : Tt : tt = 1 : 2 : 1TT:Tt:tt=1:2:1).

2. Errors in Writing Gametes for Dihybrid Crosses

  • Mistake: Writing gametes as pairs of identical alleles like RRRRRR or YyYyYy.
  • Correction: According to the Law of Segregation, a gamete must contain one allele from each gene pair. For a parent with genotype RrYyRrYyRrYy, each gamete must contain one letter for shape (RRR or rrr) AND one letter for color (YYY or yyy). Correct gametes are: RYRYRY, RyRyRy, rYrYrY, ryryry.

3. Misinterpreting Ratios as Exact Absolute Numbers

  • Mistake: Assuming that if a pea plant produces 444 seeds, exactly 333 must be tall and 111 must be dwarf.
  • Correction: Mendelian ratios (3:13:13:1 or 9:3:3:19:3:3:19:3:3:1) represent statistical probabilities, not fixed quantities. Large sample sizes are necessary to observe these exact mathematical ratios.

4. Misunderstanding the Role of Parents in Sex Determination

  • Mistake: Stating that the mother contributes XXX or YYY chromosomes to the child.
  • Correction: Mothers carry XXXXXX chromosomes and can only contribute an XXX chromosome through the egg. The male father carries XYXYXY chromosomes and contributes either an XXX or a YYY chromosome via the sperm. Thus, the male parent alone determines sex.

8. Practice Questions for Self-Assessment

Question 1

A tall pea plant with red flowers (TTRRTTRRTTRR) is crossed with a dwarf pea plant with white flowers (ttrrttrrttrr).

  1. What is the phenotype of the F1\text{F}_1F1​ generation?
  2. What are the types of gametes produced by the F1\text{F}_1F1​ generation?
  3. Calculate the fraction of F2\text{F}_2F2​ plants that will be dwarf with red flowers.
<details> <summary>Click to view Detailed Solution</summary>

Solution:

  1. F1\text{F}_1F1​ Genotype will be TtRrTtRrTtRr. Since Tall (TTT) and Red (RRR) are dominant, 100%100\%100% of the F1\text{F}_1F1​ generation will be Tall with Red flowers.

  2. Gametes produced by F1\text{F}_1F1​ (TtRrTtRrTtRr): TRTRTR, TrTrTr, tRtRtR, trtrtr.

  3. In a dihybrid cross, the phenotypic ratio is:

    • Tall Red = 9/169/169/16
    • Tall White = 3/163/163/16
    • Dwarf Red = 3/163/163/16
    • Dwarf White = 1/161/161/16

    Answer: 316\frac{3}{16}163​ of the F2\text{F}_2F2​ plants will be dwarf with red flowers.

</details>

Question 2

In humans, sex is determined genetically. A couple has four daughters. What is the probability that their fifth child will be a son? Explain your reasoning biologically.

<details> <summary>Click to view Detailed Solution</summary>

Solution:

Probability: 12\frac{1}{2}21​ or 50%50\%50%.

Biological Explanation:

  1. Each pregnancy is an independent biological event.
  2. The sex of a child is determined by whether an XXX-bearing sperm or a YYY-bearing sperm fertilizes the egg.
  3. Males produce equal proportions (50%50\%50% each) of XXX-bearing and YYY-bearing sperm during meiosis.
  4. Previous births have no influence on the outcome of subsequent fertilizations. Therefore, the chance of having a male child remains strictly 50%50\%50% for every individual pregnancy.
</details>

Question 3

When a pure tall pea plant (TTTTTT) is crossed with a hybrid tall pea plant (TtTtTt), what will be the phenotypic and genotypic percentages of the resulting offspring? Show the Punnett square.

<details> <summary>Click to view Detailed Solution</summary>

Solution:

  • Parents: TT×TtTT \times TtTT×Tt
  • Gametes from parent 1 (TTTTTT): TTT
  • Gametes from parent 2 (TtTtTt): TTT and ttt

Punnett Square:

GametesTTT
TTTTTTTTT (Tall)
tttTtTtTt (Tall)
  • Genotypes produced: 50%50\%50% TTTTTT, 50%50\%50% TtTtTt
  • Phenotypes produced: 100%100\%100% Tall

Answer:

  • Phenotypic Percentage: 100%100\%100% Tall (0%0\%0% Dwarf)
  • Genotypic Percentage: 50%50\%50% Homozygous Tall (TTTTTT), 50%50\%50% Heterozygous Tall (TtTtTt)
</details>

9. Board Exam Revision & Frequently Asked Questions (FAQs)

FAQ 1: Why is the Law of Segregation considered universal and without exception, unlike the Law of Dominance?

Answer: The Law of Segregation is based on the universal biological process of meiosis. During gamete formation, homologous chromosomes (and therefore alleles) must physically separate into different gametes so that the species maintains a constant chromosome number across generations. This physical separation occurs in all diploid sexually reproducing organisms without exception. In contrast, the Law of Dominance has exceptions, such as incomplete dominance or codominance, where neither allele is completely dominant over the other.


FAQ 2: Differentiate between Homogametic and Heterogametic organisms with examples.

Answer:

FeatureHomogametic SexHeterogametic Sex
DefinitionProduces only one type of gamete regarding sex chromosomes.Produces two different types of gametes regarding sex chromosomes.
Sex ChromosomesSimilar pair (e.g., XXXXXX).Dissimilar pair (e.g., XYXYXY).
Human ExampleHuman Females (All eggs carry an XXX chromosome).Human Males (50%50\%50% sperm carry XXX, 50%50\%50% carry YYY).

FAQ 3: How do new combinations of traits arise in the F2\text{F}_2F2​ generation of a dihybrid cross?

Answer: New combinations (recombinants like Round Green and Wrinkled Yellow) arise due to the Law of Independent Assortment.

During gamete formation in the F1\text{F}_1F1​ hybrid (RrYyRrYyRrYy), the segregation of the gene pair controlling seed shape (R/rR/rR/r) occurs completely independently of the segregation of the gene pair controlling seed color (Y/yY/yY/y). This independent alignment of chromosome pairs during meiosis allows alleles from different parents to combine freely, producing four distinct types of gametes (RYRYRY, RyRyRy, rYrYrY, ryryry) in equal proportions (25%25\%25% each), which leads to novel trait combinations in the F2\text{F}_2F2​ offspring.


FAQ 4: Outline the steps explaining how a gene directs the trait "dwarfness" in a pea plant.

Answer:

  1. A gene is a segment of DNA that holds instructions for synthesizing a specific enzyme.
  2. The specific enzyme catalyzes the metabolic pathway responsible for producing plant growth hormone (gibberellin).
  3. The allele for dwarfness (ttt) is an altered/mutated form of the gene that codes for a non-functional or inefficient enzyme.
  4. Due to the lack of functional enzyme, inadequate growth hormone is synthesized inside the plant cells.
  5. Consequently, cell elongation is limited, and the plant fails to grow tall, resulting in a dwarf phenotype.
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Class 10 NCERT Guides21 chapters
  • Statistics
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  • Control and Coordination
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Class 9 NCERT Guides12 chapters
  • Tissues
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Class 8 NCERT Guides13 chapters
  • Practical Geometry
  • Crop Production and Management
  • Comparing Quantities
  • Algebraic Expressions and Identities
  • Friction
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  • Sound
  • Combustion and Flame
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  • Microorganisms: Friend and Foe
  • Linear Equations in One Variable
  • Understanding Quadrilaterals
  • Rational Numbers
Class 7 NCERT Guides8 chapters
  • Acids, Bases and Salts
  • Heat
  • Nutrition in Animals
  • Nutrition in Plants
  • Perimeter and Area
  • Integers
  • Rational Numbers
  • Simple Equations
Class 6 NCERT Guides9 chapters
  • Playing With Numbers
  • Algebra
  • Decimals
  • Fractions
  • Knowing Our Numbers
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  • Components of Food
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  • Separation of Substances
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