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Updated 6 August 2026

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Basic Principles of Inheritance — NCERT Solutions

CBSE · Class 11 · Biotechnology

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EXERCISES — Basic Principles of Inheritance

1Differentiate between the following:
(a) Genotype and Phenotype
(b) Dominant and Recessive characters
(c) Hybrid and Pure individuals
(d) Heterozygous and Homozygous progeny
(e) Monohybrid and Dihybrid cross
(f) Gene and allele
(g) Incomplete dominance and codominanceShow solution

(a) Genotype and Phenotype

Feature	Genotype	Phenotype
Definition	The genetic constitution (allelic combination) of an organism	The observable/expressed characteristics of an organism
Visibility	Cannot be seen directly; determined by molecular/breeding analysis	Can be observed directly (morphology, physiology, behaviour)
Example	$TT$ , $Tt$ , $tt$	Tall, Tall, Dwarf
Stability	Remains constant throughout life	Can be influenced by environment

(b) Dominant and Recessive Characters

Feature	Dominant Character	Recessive Character
Definition	The character that expresses itself in the $F_1$ hybrid (heterozygous condition)	The character that remains suppressed in the presence of the dominant allele
Expression	Expressed in both homozygous ( $AA$ ) and heterozygous ( $Aa$ ) state	Expressed only in homozygous state ( $aa$ )
Example	Tallness ( $T$ ) in pea	Dwarfness ( $t$ ) in pea

(c) Hybrid and Pure Individuals

Feature	Hybrid Individual	Pure Individual
Definition	An individual produced by crossing two genetically different parents; carries two different alleles for a trait	An individual that breeds true for a trait; carries two identical alleles
Genotype	Heterozygous, e.g., $Tt$	Homozygous, e.g., $TT$ or $tt$
Offspring	Produces varied offspring on selfing	Produces identical offspring on selfing

(d) Heterozygous and Homozygous Progeny

Feature	Heterozygous Progeny	Homozygous Progeny
Definition	Progeny carrying two different alleles for a gene locus	Progeny carrying two identical alleles for a gene locus
Genotype	e.g., $Tt$ , $Aa$	e.g., $TT$ , $tt$ , $AA$ , $aa$
Gametes produced	Two types of gametes	Only one type of gamete
Breeding behaviour	Does not breed true	Breeds true

(e) Monohybrid and Dihybrid Cross

Feature	Monohybrid Cross	Dihybrid Cross
Definition	Cross between parents differing in only one pair of contrasting characters	Cross between parents differing in two pairs of contrasting characters
Example	$TT \times tt$ (tall $\times$ dwarf)	$TTRR \times ttrr$ (tall round $\times$ dwarf wrinkled)
$F_2$ Phenotypic ratio	$3:1$	$9:3:3:1$
$F_2$ Genotypic ratio	$1:2:1$	$1:2:1:2:4:2:1:2:1$
Law demonstrated	Law of Dominance and Law of Segregation	Law of Independent Assortment

(f) Gene and Allele

Feature	Gene	Allele
Definition	A specific segment of DNA that codes for a particular protein/trait	Alternative forms of the same gene occupying the same locus on homologous chromosomes
Location	Occupies a specific locus on a chromosome	Present at the same locus but on homologous chromosomes
Example	Gene for seed colour in pea	$R$ (round) and $r$ (wrinkled) are alleles of the seed-shape gene

(g) Incomplete Dominance and Codominance

Feature	Incomplete Dominance	Codominance
Definition	Neither allele is completely dominant; the heterozygote shows an intermediate phenotype	Both alleles are expressed simultaneously and independently in the heterozygote
$F_1$ Phenotype	Intermediate between two parents	Both parental phenotypes expressed together
$F_2$ Phenotypic ratio	$1:2:1$ (same as genotypic ratio)	$1:2:1$ (same as genotypic ratio)
Example	Flower colour in Antirrhinum (snapdragon): Red ( $RR$ ) $\times$ White ( $rr$ ) $\rightarrow$ Pink ( $Rr$ )	ABO blood groups: $I^A I^B$ genotype shows both A and B antigens (AB blood group)

2Mention the genotypic and phenotypic ratio of progeny when there is a cross between:
(a)

F_1

progeny with pure dominant parent
(b)

F_1

progeny with pure recessive parent
(c)

F_1

progeny with

F_1

progenyShow solution

Given: Consider a monohybrid cross for seed shape in pea — Round ( $R$ , dominant) $\times$ Wrinkled ( $r$ , recessive).

Pure dominant parent: $RR$
Pure recessive parent: $rr$
$F_1$ progeny: $Rr$ (heterozygous)

(a) $F_1$ progeny ( $Rr$ ) $\times$ Pure dominant parent ( $RR$ )

$Rr \times RR$

Using a Punnett square:

	$R$	$R$
$R$	$RR$	$RR$
$r$	$Rr$	$Rr$

Genotypic ratio: $RR : Rr = 1:1$ (i.e., $\frac{1}{2}$ homozygous dominant : $\frac{1}{2}$ heterozygous)

Phenotypic ratio: All offspring are Round = $1:0$ (100% Round)

$\boxed{\text{Genotypic ratio} = 1:1 \quad \text{Phenotypic ratio} = \text{All Round (1:0)}}$

(b) $F_1$ progeny ( $Rr$ ) $\times$ Pure recessive parent ( $rr$ ) — This is a Test Cross

$Rr \times rr$

Using a Punnett square:

	$R$	$r$
$r$	$Rr$	$rr$

Genotypic ratio: $Rr : rr = 1:1$

Phenotypic ratio: Round : Wrinkled $= 1:1$

$\boxed{\text{Genotypic ratio} = 1:1 \quad \text{Phenotypic ratio} = 1 \text{ Round} : 1 \text{ Wrinkled}}$

(c) $F_1$ progeny ( $Rr$ ) $\times$ $F_1$ progeny ( $Rr$ ) — This gives the $F_2$ generation

$Rr \times Rr$

Using a Punnett square:

	$R$	$r$
$R$	$RR$	$Rr$
$r$	$Rr$	$rr$

Genotypic ratio: $RR : Rr : rr = 1:2:1$

Phenotypic ratio: Round ( $RR + Rr$ ) : Wrinkled ( $rr$ ) $= 3:1$

$\boxed{\text{Genotypic ratio} = 1:2:1 \quad \text{Phenotypic ratio} = 3 \text{ Round} : 1 \text{ Wrinkled}}$

3Explain test cross through diagrammatic representation.Show solution

Definition: A test cross is a cross between an individual showing a dominant phenotype (but of unknown genotype — either $TT$ or $Tt$ ) and a homozygous recessive individual ( $tt$ ). It is used to determine whether the dominant phenotype individual is homozygous or heterozygous.

Concept: If the individual is homozygous dominant ( $TT$ ), all offspring will show the dominant phenotype. If the individual is heterozygous ( $Tt$ ), offspring will appear in a $1:1$ ratio of dominant to recessive phenotype.

Case 1: Dominant parent is Homozygous ( $TT$ )

$TT \times tt$

$\text{Gametes: } T, T \quad \times \quad t, t$

	$T$	$T$
$t$	$Tt$	$Tt$

Genotypic ratio: All $Tt$ (100% heterozygous)
Phenotypic ratio: All Tall — No recessive offspring appear
Conclusion: Parent is homozygous dominant ( $TT$ )

Case 2: Dominant parent is Heterozygous ( $Tt$ )

$Tt \times tt$

$\text{Gametes: } T, t \quad \times \quad t, t$

	$T$	$t$
$t$	$Tt$	$tt$

Genotypic ratio: $Tt : tt = 1:1$
Phenotypic ratio: Tall : Dwarf $= 1:1$ — Recessive offspring appear
Conclusion: Parent is heterozygous ( $Tt$ )

Summary Table:

Test Cross Result	Genotype of Dominant Parent
All offspring dominant	$TT$ (Homozygous dominant)
$1:1$ dominant : recessive	$Tt$ (Heterozygous)

Significance: The test cross is a powerful tool used by geneticists to determine the genotype of an organism showing a dominant phenotype.

4Explain the following using monohybrid and dihybrid cross:
(a) Law of dominance
(b) Law of segregation
(c) Law of independent assortmentShow solution

(a) Law of Dominance — Explained using Monohybrid Cross

Statement: When two homozygous parents differing in one pair of contrasting characters are crossed, only one character (dominant) expresses itself in the $F_1$ hybrid, while the other character (recessive) remains suppressed.

Monohybrid Cross (Seed shape in pea):

$P: \quad RR \text{ (Round)} \times rr \text{ (Wrinkled)}$

$\text{Gametes: } R \quad \times \quad r$

$F_1: \quad Rr \text{ (Round)}$

In $F_1$ , all plants are Round even though they carry the $r$ allele.
$R$ (Round) is dominant over $r$ (Wrinkled).
The recessive character (wrinkled) is suppressed in $F_1$ .

Conclusion: The law of dominance states that in a heterozygote, one allele (dominant) masks the expression of the other allele (recessive).

(b) Law of Segregation — Explained using Monohybrid Cross

Statement: The two alleles of a gene pair segregate (separate) from each other during gamete formation so that each gamete receives only one allele. The two alleles reunite at fertilisation.

Monohybrid Cross continued to $F_2$ :

$F_1 \times F_1: \quad Rr \times Rr$

	$R$	$r$
$R$	$RR$	$Rr$
$r$	$Rr$	$rr$

$F_2$ Genotypic ratio: $RR : Rr : rr = 1:2:1$
$F_2$ Phenotypic ratio: Round : Wrinkled $= 3:1$

Explanation: In $F_1$ ( $Rr$ ), the two alleles $R$ and $r$ segregate during meiosis. Each gamete receives either $R$ or $r$ (not both). On selfing, the recessive character (wrinkled) reappears in $F_2$ in $\frac{1}{4}$ of the offspring, proving that alleles had separated and remained intact.

Conclusion: Alleles segregate during gamete formation — this is also called the Law of Purity of Gametes.

(c) Law of Independent Assortment — Explained using Dihybrid Cross

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

Dihybrid Cross (Seed shape and seed colour in pea):

Round ( $R$ ) dominant over Wrinkled ( $r$ )
Yellow ( $Y$ ) dominant over Green ( $y$ )

$P: \quad RRYY \text{ (Round Yellow)} \times rryy \text{ (Wrinkled Green)}$

$F_1: \quad RrYy \text{ (Round Yellow — all dominant)}$

$F_1 \times F_1: \quad RrYy \times RrYy$

$F_1$ produces 4 types of gametes: $RY$ , $Ry$ , $rY$ , $ry$ (each with equal frequency $\frac{1}{4}$ )

Punnett Square ( $4 \times 4$ ):

	$RY$	$Ry$	$rY$	$ry$
$RY$	$RRYY$	$RRYy$	$RrYY$	$RrYy$
$Ry$	$RRYy$	$RRyy$	$RrYy$	$Rryy$
$rY$	$RrYY$	$RrYy$	$rrYY$	$rrYy$
$ry$	$RrYy$	$Rryy$	$rrYy$	$rryy$

$F_2$ Phenotypic ratio:

$\text{Round Yellow} : \text{Round Green} : \text{Wrinkled Yellow} : \text{Wrinkled Green} = 9:3:3:1$

9 Round Yellow ( $R\_Y\_$ )
3 Round Green ( $R\_yy$ )
3 Wrinkled Yellow ( $rrY\_$ )
1 Wrinkled Green ( $rryy$ )

Explanation: Two new combinations — Round Green and Wrinkled Yellow — appear in $F_2$ that were not present in the parents. This is possible only if the two gene pairs assort independently of each other during gamete formation.

Conclusion: The $9:3:3:1$ ratio in $F_2$ of a dihybrid cross confirms the Law of Independent Assortment.

5What will be the genotypic and phenotypic ratio when a red and tall homozygous tomato plant is crossed to a red and tall heterozygous plant?

Given:

Let Red colour be controlled by allele $R$ (dominant) and non-red by $r$ .
Let Tall be controlled by allele $T$ (dominant) and dwarf by $t$ .
Parent 1: Red and Tall homozygous → Genotype: $RRTT$
Parent 2: Red and Tall heterozygous → Genotype: $RrTt$

Cross:
$RRTT \times RrTt$

Gametes produced:

$RRTT$ produces only one type of gamete: $RT$
$RrTt$ produces four types of gametes: $RT$ , $Rt$ , $rT$ , $rt$ (each with frequency $\frac{1}{4}$ )

Punnett Square:

Gametes of $RRTT$	$RT$ (from $RrTt$ )	$Rt$ (from $RrTt$ )	$rT$ (from $RrTt$ )	$rt$ (from $RrTt$ )
$RT$	$RRTT$	$RRTt$	$RrTT$	$RrTt$

Offspring genotypes and their proportions:

Genotype	Proportion	Phenotype
$RRTT$	$\frac{1}{4}$	Red, Tall
$RRTt$	$\frac{1}{4}$	Red, Tall
$RrTT$	$\frac{1}{4}$	Red, Tall
$RrTt$	$\frac{1}{4}$	Red, Tall

Genotypic ratio:
$RRTT : RRTt : RrTT : RrTt = 1:1:1:1$

Phenotypic ratio:
All offspring are Red and Tall (since all carry at least one $R$ and one $T$ allele).

$\boxed{\text{Genotypic ratio} = 1:1:1:1}$
$\boxed{\text{Phenotypic ratio} = \text{All Red and Tall} = 1 \text{ (only one phenotypic class)}}$

Conclusion: Although four different genotypes are produced, all offspring express the same phenotype — Red and Tall — because $R$ is dominant over $r$ and $T$ is dominant over $t$ .

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6When one male and one female Drosophila, heterozygous for the two pairs of alleles AaBb, were mated, the offspring's phenotypic ratio 2:1:1:2 was obtained.
(a) Explain how these ratios help in detecting linkages?
(b) How degree of linkage can be determined?

Background:
In a normal dihybrid cross ( $AaBb \times AaBb$ ), the expected $F_2$ phenotypic ratio is $9:3:3:1$ (when genes are on different chromosomes — independent assortment). However, when the observed ratio is $2:1:1:2$ , it indicates linkage between the two genes.

(a) How does the ratio 2:1:1:2 help in detecting linkage?

Expected ratio (no linkage): $9:3:3:1$ (total 16 combinations)

Observed ratio: $2:1:1:2$ (total 6 combinations)

Analysis:

The ratio $2:1:1:2$ deviates significantly from the expected $9:3:3:1$ ratio.
In the ratio $2:1:1:2$ :
The parental combinations ( $AB$ and $ab$ phenotypes) appear in higher frequency (2 each).
The recombinant combinations ( $Ab$ and $aB$ phenotypes) appear in lower frequency (1 each).
This deviation from the expected Mendelian ratio indicates that the two genes $A$ and $B$ are located on the same chromosome (linked) and do not assort independently.
If genes were unlinked, parental and recombinant types would appear in equal frequency. The excess of parental types over recombinant types is the hallmark of linkage.
The greater the deviation from the $9:3:3:1$ ratio (i.e., the greater the excess of parental combinations), the stronger the linkage between the two genes.

Conclusion: The appearance of parental phenotypic classes in higher proportion than recombinant classes, and the deviation from the $9:3:3:1$ ratio, is evidence of genetic linkage between the two loci.

(b) How can the degree of linkage be determined?

The degree of linkage (or the distance between two linked genes) is measured by calculating the recombination frequency (also called crossover frequency or map distance).

Formula:
$\text{Recombination Frequency (RF)} = \frac{\text{Number of recombinant offspring}}{\text{Total number of offspring}} \times 100\%$

Steps:

Perform a test cross ( $AaBb \times aabb$ ) to clearly identify parental and recombinant classes.
Count the number of offspring in each phenotypic class.
Identify parental types (same combination as parents) and recombinant types (new combinations).
Apply the formula above.

Interpretation:

RF close to 0% → genes are very tightly linked (located very close together on the chromosome).
RF close to 50% → genes are far apart on the chromosome (behave as if unlinked).
1% recombination frequency = 1 centimorgan (cM) = 1 map unit of distance on the genetic map.

In the given example:

Ratio $2:1:1:2$ → parental types = $2+2 = 4$ parts; recombinant types = $1+1 = 2$ parts out of 6 total.

$RF = \frac{2}{6} \times 100 = 33.3\%$

This indicates moderate linkage with a map distance of approximately $33.3$ cM between the two genes.

Conclusion: The degree of linkage is inversely proportional to the recombination frequency — lower recombination frequency means stronger (tighter) linkage.

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7Make a close observation with the nature. Do you think that the phenomenon of linkage is absolute?

Answer: No, the phenomenon of linkage is NOT absolute.

Explanation:

Linkage refers to the tendency of genes located on the same chromosome to be inherited together. However, linkage is rarely complete (absolute) in nature. This is because of the phenomenon of crossing over (recombination) that occurs during meiosis I (specifically during the pachytene stage of prophase I).

Reasons why linkage is not absolute:

Crossing Over: During meiosis, homologous chromosomes pair up and exchange segments at points called chiasmata. This physical exchange of chromosomal segments results in recombinant chromosomes carrying new combinations of alleles. Thus, even linked genes can be separated and appear in new combinations in the offspring.

Recombinant offspring: In any cross involving linked genes, a certain percentage of offspring always show recombinant phenotypes (new combinations not seen in parents). This proves that linkage is not absolute.

Distance-dependent: The frequency of crossing over (and hence recombination) depends on the physical distance between genes on the chromosome. Genes that are far apart on the same chromosome show higher recombination frequency (up to 50%) and may appear to assort almost independently.

Observations in nature: In Drosophila, Morgan and his colleagues observed that while some genes showed strong linkage, recombinant offspring were always produced in some proportion, confirming that crossing over breaks linkage.

Types of Linkage:

Complete linkage: Genes are so close together that no crossing over occurs between them (very rare; e.g., genes in male Drosophila). Parental combinations only are produced.
Incomplete linkage: Crossing over occurs between linked genes, producing both parental and recombinant combinations. This is the most common situation in nature.

Conclusion:
Linkage is generally incomplete in nature. The occurrence of crossing over during meiosis ensures that even genes on the same chromosome can be separated and recombined, generating genetic variation. Therefore, linkage is not an absolute phenomenon — it is a tendency, not a certainty, and is broken by recombination. This is why the recombination frequency between two linked genes is always between $0\%$ (complete linkage) and $50\%$ (independent assortment).

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What are the important topics in Basic Principles of Inheritance for CBSE Class 11 Biotechnology?

Key topics in Basic Principles of Inheritance include Introduction to Heredity and Mendel's Work, Monohybrid Cross, Ratios, and Mendel's Laws, Incomplete Dominance, Codominance, and Dihybrid Cross, Linkage, Crossing Over, and Recombination. Study these first, then practise questions on each for Class 11 exams.

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