Q11: How is sex determined in human beings?

Sex determination in human beings is of the XX-XY type. This mechanism is based on the presence of specific sex chromosomes in an individual. 1. Chromosomal Basis: Human beings have 23 pairs of chromosomes. 22 pairs are autosomes, and the 23rd pair are the sex chromosomes. Females have a pair of…

Principles of Inheritance and Variation Class 12 Biology NCERT Solutions

Q1EXERCISES

Mention the advantages of selecting pea plant for experiment by Mendel.

Solution

Mendel selected the garden pea plant (Pisum sativum) for his experiments due to several advantages it offered:

Well-defined Contrasting Traits: The pea plant has many easily observable and contrasting characteristics. Mendel studied seven such pairs, for example, tall/dwarf stem height and round/wrinkled seeds.
Bisexual Flowers: Pea flowers are bisexual (containing both male and female reproductive parts) and are naturally self-pollinating. This made it easy to obtain pure-line parent plants through continuous self-pollination.
Easy Cross-Pollination: While they self-pollinate naturally, the flowers can be easily cross-pollinated artificially. This allowed Mendel to conduct controlled crosses between plants with different traits.
Short Life Cycle: The pea plant has a short life span and completes its life cycle in one season. This enabled Mendel to study several generations of the plant within a short period.
Large Number of Offspring: A single cross-pollination event produces a large number of seeds (offspring). This large sample size provided reliable data and allowed for statistical analysis of the results.
Easy to Cultivate: Pea plants are easy to grow and maintain, and do not require special care.

Q2EXERCISES

Differentiate between the following -

(a)

Dominance and Recessive

(b)

Homozygous and Heterozygous

(c)

Monohybrid and Dihybrid.

Solution

(a) Dominance and Recessive

Feature	Dominance	Recessive
Definition	It is an allele that expresses its phenotypic effect in both homozygous (e.g., TT) and heterozygous (e.g., Tt) conditions.	It is an allele that expresses its phenotypic effect only in the homozygous condition (e.g., tt).
Expression	It masks the effect of the recessive allele in a heterozygous individual.	Its effect is masked by the dominant allele in a heterozygous individual.
Example	In a pea plant, the allele for tallness (T) is dominant.	In a pea plant, the allele for dwarfness (t) is recessive.

(b) Homozygous and Heterozygous

Feature	Homozygous	Heterozygous
Definition	An individual possessing two identical alleles for a particular trait.	An individual possessing two different alleles for a particular trait.
Allelic Pair	The alleles are identical, e.g., TT (homozygous dominant) or tt (homozygous recessive).	The alleles are different, e.g., Tt.
Gamete Production	Produces only one type of gamete for that trait.	Produces two different types of gametes for that trait.
Purity	It is a pure line for that trait.	It is a hybrid for that trait.

(c) Monohybrid and Dihybrid

Feature	Monohybrid Cross	Dihybrid Cross
Definition	A cross between two parents differing in only one pair of contrasting characters.	A cross between two parents differing in two pairs of contrasting characters.
Characters Studied	Inheritance of a single character is studied at a time.	Inheritance of two characters is studied at a time.
F2 Phenotypic Ratio	The typical phenotypic ratio in the F2 generation is 3:1.	The typical phenotypic ratio in the F2 generation is 9:3:3:1.
Purpose	It is used to study the inheritance of a single gene pair and to determine the dominance relationship.	It is used to study the principle of independent assortment of two different gene pairs.

Q3EXERCISES

A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?

Solution

To determine the number of different types of gametes that can be produced by a diploid organism heterozygous for a certain number of loci, we use a specific formula.

Given: The organism is diploid. Number of heterozygous loci,

n = 4

.

To Find: The total number of types of gametes that can be produced.

Formula: The number of types of gametes is given by the formula

2^n

, where 'n' is the number of heterozygous loci.

Calculation: Here,

n = 4

. Substituting the value of 'n' into the formula: Number of gamete types =

2^4

Number of gamete types =

2 \times 2 \times 2 \times 2 = 16

Final Answer: The diploid organism can produce 16 different types of gametes.

Q4EXERCISES

Explain the Law of Dominance using a monohybrid cross.

Solution

The Law of Dominance is one of the principles of inheritance proposed by Gregor Mendel. This law states that:

Characters are controlled by discrete units called factors (now known as genes).
Factors occur in pairs.
In a dissimilar pair of factors (a heterozygous condition), one member of the pair dominates (dominant) the other (recessive).

Explanation using a Monohybrid Cross:

A monohybrid cross involves the study of the inheritance of a single pair of contrasting traits. Let us consider the example of stem height in pea plants.

Parental Generation (P): A cross is made between a pure tall pea plant (genotype TT) and a pure dwarf pea plant (genotype tt).

Parental Genotypes: TT (Tall)

\times

tt (Dwarf)

Gamete Formation: The homozygous tall parent (TT) produces only one type of gamete, containing the allele 'T'. The homozygous dwarf parent (tt) produces only one type of gamete, containing the allele 't'.

Gametes: T and t

First Filial Generation (F1): During fertilization, the gametes fuse to form the zygote. All offspring in the F1 generation will have the genotype Tt.

F1 Genotype: Tt

Phenotype of F1 Generation: According to the Law of Dominance, when two different alleles are present in an individual (Tt), only the dominant allele expresses itself. In this case, the allele for tallness (T) is dominant over the allele for dwarfness (t). Therefore, all plants in the F1 generation are tall.

F1 Phenotype: All Tall

Conclusion: The appearance of only the tall phenotype in the F1 generation, despite the presence of the allele for dwarfness, demonstrates the Law of Dominance. The recessive trait (dwarfness) does not appear in the F1 generation because its expression is masked by the dominant trait (tallness). This experiment shows that 'T' is the dominant factor and 't' is the recessive factor.

Q5EXERCISES

Define and design a test-cross.

Solution

Definition of a Test Cross:

A test cross is a genetic cross performed to determine the genotype of an individual that exhibits a dominant phenotype. This is achieved by crossing the individual of unknown genotype with another individual that is homozygous recessive for the trait in question.

The phenotypic ratio of the offspring produced from this cross reveals whether the dominant parent was homozygous dominant or heterozygous.

Design of a Test Cross:

Let us design a test cross to determine the genotype of a tall pea plant. A tall pea plant can have one of two possible genotypes: homozygous dominant (TT) or heterozygous (Tt). The recessive parent will always be a dwarf plant, which is homozygous recessive (tt).

Case 1: The unknown tall plant is homozygous dominant (TT).

Cross: The tall plant (TT) is crossed with a dwarf plant (tt). Parental Genotypes: TT $\times$ tt
Gametes: The tall parent produces only 'T' gametes. The dwarf parent produces only 't' gametes.
Offspring: All offspring will have the genotype Tt.
```
  t      t
```
T Tt Tt T Tt Tt
Result: All offspring will be heterozygous (Tt) and will exhibit the tall phenotype.
Conclusion: If all the offspring from a test cross display the dominant phenotype, the unknown parent plant is homozygous dominant (TT).

Case 2: The unknown tall plant is heterozygous (Tt).

Cross: The tall plant (Tt) is crossed with a dwarf plant (tt). Parental Genotypes: Tt $\times$ tt
Gametes: The heterozygous tall parent produces two types of gametes, 'T' and 't', in equal proportions. The dwarf parent produces only 't' gametes.
Offspring: Two types of genotypes, Tt and tt, will be produced in equal proportion.
```
  t      t
```
T Tt Tt t tt tt
Result: 50% of the offspring will be heterozygous (Tt) and exhibit the tall phenotype, while the other 50% will be homozygous recessive (tt) and exhibit the dwarf phenotype. The phenotypic ratio of tall to dwarf plants will be 1:1.
Conclusion: If the offspring from a test cross display both dominant and recessive phenotypes in a 1:1 ratio, the unknown parent plant is heterozygous (Tt).

Q6EXERCISES

Using a Punnett Square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.

Solution

A Punnett square is a graphical representation to calculate the probability of all possible genotypes of offspring in a genetic cross.

Let us consider a single locus for the height of a plant, where Tall (T) is dominant over dwarf (t).

Parents:

Homozygous female: We will assume the female is homozygous recessive (dwarf) with the genotype tt.
Heterozygous male: The male will be phenotypically tall with the genotype Tt.

Cross: tt (female)

\times

Tt (male)

Gametes:

The homozygous recessive female (tt) will produce only one type of gamete: t.
The heterozygous male (Tt) will produce two types of gametes in equal proportion: T and t.

Punnett Square: We can now set up the Punnett square to determine the genotypes and phenotypes of the first filial (

F_1

) generation.

| Gametes from Male

\rightarrow

Gametes from Female $\downarrow$	T	t
t	Tt	tt

Distribution of Features in $F_1$ Generation:

Genotypic Distribution:
- Tt (Heterozygous tall): 1 out of 2, or 50%
- tt (Homozygous dwarf): 1 out of 2, or 50% The genotypic ratio is 1 Tt : 1 tt.
Phenotypic Distribution:
- Tall (from genotype Tt): 1 out of 2, or 50%
- Dwarf (from genotype tt): 1 out of 2, or 50% The phenotypic ratio is 1 Tall : 1 Dwarf.

Therefore, in the first filial generation, 50% of the offspring will be tall and 50% will be dwarf.

(Note: If the homozygous female was dominant (TT), all offspring would have the genotype Tt and would be phenotypically tall.)

Q7EXERCISES

When a cross in made between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to be

(a)

tall and green.

(b)

dwarf and green.

Solution

This is a dihybrid cross involving two traits: plant height and seed color.

Alleles:

T (Tall) is dominant over t (dwarf).
Y (Yellow seeds) is dominant over y (green seeds).

Parental Genotypes:

Parent 1: Tall plant with yellow seeds = TtYy
Parent 2: Tall plant with green seeds = Ttyy

Gametes Produced:

Parent 1 (TtYy) produces four types of gametes: TY, Ty, tY, ty.
Parent 2 (Ttyy) produces two types of gametes: Ty, ty.

Punnett Square: We can create a Punnett square to find the genotypes of the offspring.

| Gametes from Parent 1

\rightarrow

Gametes from Parent 2 $\downarrow$	TY	Ty	tY	ty
Ty	TTYy	TTyy	TtYy	Ttyy
ty	TtYy	Ttyy	ttYy	ttyy

Phenotypes of the Offspring: From the 8 possible combinations in the Punnett square, we can determine the phenotypes:

TTyy, Ttyy, Ttyy: Tall, green (3)
ttyy: dwarf, green (1)
TTYy, TtYy, TtYy: Tall, yellow (3)
ttYy: dwarf, yellow (1)

The overall phenotypic ratio is 3 (Tall, green) : 1 (dwarf, green) : 3 (Tall, yellow) : 1 (dwarf, yellow).

Now, we can find the proportions for the specific phenotypes asked in the question:

(a) tall and green:

The genotypes corresponding to this phenotype are TTyy and Ttyy.
From the Punnett square, there are 3 such offspring (TTyy, Ttyy, Ttyy) out of a total of 8.
The proportion is $\frac{3}{8}$ .

(b) dwarf and green:

The genotype corresponding to this phenotype is ttyy.
From the Punnett square, there is 1 such offspring out of a total of 8.
The proportion is $\frac{1}{8}$ .

Final Answer: (a) The proportion of offspring expected to be tall and green is 3/8. (b) The proportion of offspring expected to be dwarf and green is 1/8.

Q8EXERCISES

Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in $F_{1}$ generation for a dibybrid cross?

Solution

In a dihybrid cross where the two loci are linked, the genes do not assort independently. Instead, they tend to be inherited together because they are located on the same chromosome. The distribution of phenotypes in the

F_1

generation will, therefore, differ significantly from the typical Mendelian 9:3:3:1 ratio.

The outcome depends on two factors: (1) the arrangement of alleles on the chromosomes of the heterozygous parents (cis or trans arrangement) and (2) whether crossing over occurs (incomplete linkage) or not (complete linkage).

Let's assume the simplest case of complete linkage (no crossing over) and a cis arrangement of alleles in the heterozygous parents.

Parental Genotypes: AaBb

\times

AaBb

Allele Arrangement (Cis): In this arrangement, the dominant alleles (A and B) are on one chromosome, and the recessive alleles (a and b) are on the homologous chromosome. The parental genotypes can be represented as AB/ab.

Gamete Formation: Due to complete linkage, the alleles on each chromosome are inherited as a single unit. Therefore, each parent will produce only two types of gametes, corresponding to the parental combinations:

AB
ab

No recombinant gametes (Ab and aB) will be formed.

Punnett Square for the $F_1$ Generation:

Gametes	AB	ab
AB	AABB	AaBb
ab	AaBb	aabb

Distribution of Phenotypic Features:

Genotypic Ratio: The genotypes of the offspring will be AABB, AaBb, and aabb in a 1:2:1 ratio.
Phenotypic Ratio:
- AABB and AaBb: These genotypes will express the dominant phenotype for both traits (Phenotype A, Phenotype B).
- aabb: This genotype will express the recessive phenotype for both traits (Phenotype a, Phenotype b).

Combining these, the phenotypic ratio in the

F_1

generation will be 3 (Dominant, Dominant) : 1 (Recessive, Recessive).

Conclusion: When two heterozygous parents are crossed and the two loci are completely linked in a cis arrangement, the dihybrid cross behaves like a monohybrid cross, resulting in a phenotypic ratio of 3:1 in the

F_1

generation, instead of the 9:3:3:1 ratio expected with independent assortment.

Q9EXERCISES

Briefly mention the contribution of T.H. Morgan in genetics.

Solution

Thomas Hunt Morgan (1866-1945) was an American evolutionary biologist and geneticist whose work provided crucial evidence for the chromosomal theory of inheritance. His contributions revolutionized the field of genetics. His major contributions are:

Use of Drosophila melanogaster: Morgan chose the fruit fly, Drosophila melanogaster, as a model organism for his genetic studies. Its short generation time (about two weeks), high reproductive rate, simple culture requirements, and easily observable traits made it ideal for genetic research.
Discovery of Sex-Linked Inheritance: Morgan's work with a white-eyed male mutant fly led to the discovery of sex-linkage. He demonstrated that the gene for eye color in Drosophila is located on the X chromosome and is inherited along with the sex of the individual. This was a key piece of evidence that genes reside on specific chromosomes.
Discovery of Linkage and Crossing Over: By performing dihybrid crosses, Morgan observed that certain genes did not assort independently as predicted by Mendel's laws. He concluded that these genes were 'linked' because they were physically located on the same chromosome. He also found that this linkage was incomplete; the parental gene combinations could be broken and new combinations (recombinants) could be formed. He correctly proposed that this was due to 'crossing over', a physical exchange of segments between homologous chromosomes during meiosis.
Gene Mapping: Based on the concept of crossing over, Morgan's student, Alfred Sturtevant, proposed that the frequency of recombination between linked genes is a measure of the distance between them on the chromosome. This principle was used to construct the first-ever genetic maps, showing the linear arrangement of genes on a chromosome.

In summary, T.H. Morgan's experimental work with Drosophila firmly established the chromosomal theory of inheritance, discovered the phenomena of linkage, crossing over, and sex-linkage, and laid the foundation for modern genetics.

Q10EXERCISES

What is pedigree analysis? Suggest how such an analysis, can be useful.

Solution

Pedigree Analysis

Pedigree analysis is a method of studying the inheritance of a particular genetic trait or disease in human beings over several generations. It involves constructing a family tree, or pedigree chart, that uses standardized symbols to represent family members, their relationships, and their status with respect to the trait in question.

In a pedigree chart:

Males are represented by squares and females by circles.
Affected individuals (those expressing the trait) are shaded.
Mating between individuals is shown by a horizontal line connecting a square and a circle.
Offspring are connected to their parents by a vertical line.
Generations are designated by Roman numerals (I, II, III, etc.).

Usefulness of Pedigree Analysis

Since controlled crosses cannot be performed in humans, pedigree analysis is an essential tool for human genetics. Its usefulness includes:

Determining the Mode of Inheritance: By analyzing the pattern of inheritance across generations, it is possible to determine whether a trait is dominant or recessive, and whether it is autosomal (linked to a non-sex chromosome) or sex-linked (linked to X or Y chromosome).
- Autosomal Dominant: The trait appears in every generation.
- Autosomal Recessive: The trait often skips generations.
- Sex-Linked: The trait shows a different pattern of inheritance in males and females.
Genetic Counseling and Risk Prediction: Pedigree analysis helps genetic counselors advise families about the probability of an inherited disease appearing in future offspring. For example, if a couple has a child with a recessive genetic disorder, a pedigree can confirm that both parents are heterozygous carriers and can predict the risk (25%) for subsequent children.
Understanding Gene Action: It provides information about whether a gene is expressed and how it is passed on, which is fundamental to understanding gene function and regulation in humans.
Mapping Disease Genes: In research, pedigrees of large families with a high incidence of a particular genetic disease are used to help locate the specific chromosome and position of the gene responsible for the disease.

Q11EXERCISES

How is sex determined in human beings?

Solution

Sex determination in human beings is of the XX-XY type. This mechanism is based on the presence of specific sex chromosomes in an individual.

Chromosomal Basis:
- Human beings have 23 pairs of chromosomes. 22 pairs are autosomes, and the 23rd pair are the sex chromosomes.
- Females have a pair of X-chromosomes (XX).
- Males have one X-chromosome and one Y-chromosome (XY).
Gamete Formation:
- During meiosis, females produce only one type of egg (ovum), and each egg carries a single X-chromosome. Therefore, females are homogametic.
- Males produce two types of sperm in approximately equal numbers. 50% of the sperm carry an X-chromosome, and the other 50% carry a Y-chromosome. Therefore, males are heterogametic.
Fertilization and Determination of Sex:
- The sex of the child is determined at the time of fertilization. It depends on which type of sperm fertilizes the ovum.
- If a sperm carrying the X-chromosome fertilizes the ovum (which always has an X-chromosome), the resulting zygote will have the genotype XX and will develop into a female child.
- If a sperm carrying the Y-chromosome fertilizes the ovum, the resulting zygote will have the genotype XY and will develop into a male child.

Since the father produces both X- and Y-carrying sperm in equal proportions, there is a 50% probability of having a male child and a 50% probability of having a female child in each pregnancy. The sex of the child is determined by the father, not the mother.

Punnett Square for Sex Determination:

	Sperm from Father (X)	Sperm from Father (Y)
Ovum from Mother (X)	XX (Female)	XY (Male)
Ovum from Mother (X)	XX (Female)	XY (Male)

Q12EXERCISES

A child has blood group O . If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.

Solution

The inheritance of blood groups in humans is an example of multiple alleles and co-dominance. The gene for blood group is denoted by 'I' and has three alleles:

I^A

,

I^B

, and

i

.

Alleles $I^A$ and $I^B$ are dominant over the allele $i$ .
Alleles $I^A$ and $I^B$ are co-dominant when present together.
The genotype for blood group O is $ii$ .
The genotype for blood group A can be $I^A I^A$ or $I^A i$ .
The genotype for blood group B can be $I^B I^B$ or $I^B i$ .

Step 1: Determine the genotypes of the parents.

The child has blood group O, which means the child's genotype is $ii$ .
A child inherits one allele from each parent. Therefore, the child must have received one $i$ allele from the father and one $i$ allele from the mother.
The father has blood group A. Since he passed the $i$ allele to his child, his genotype cannot be homozygous ( $I^A I^A$ ). It must be heterozygous, which is $I^A i$ .
The mother has blood group B. Similarly, since she passed the $i$ allele to her child, her genotype cannot be homozygous ( $I^B I^B$ ). It must be heterozygous, which is $I^B i$ .

Parental Genotypes:

Father (Blood Group A): $I^A i$
Mother (Blood Group B): $I^B i$

Step 2: Work out the possible genotypes of other offspring. We can use a Punnett square to determine the possible genotypes and phenotypes of the offspring from a cross between the parents (

I^A i \times I^B i

).

	Gamete from Father ( $I^A$ )	Gamete from Father ( $i$ )
Gamete from Mother ( $I^B$ )	$I^A I^B$	$I^B i$
Gamete from Mother ( $i$ )	$I^A i$	$ii$

Possible Genotypes and Phenotypes of the Offspring: Based on the Punnett square, the possible offspring are:

Genotype $I^A I^B$ : Phenotype - Blood Group AB
Genotype $I^A i$ : Phenotype - Blood Group A
Genotype $I^B i$ : Phenotype - Blood Group B
Genotype $ii$ : Phenotype - Blood Group O

Therefore, the other offspring can have genotypes

I^A I^B

,

I^A i

,

I^B i

, or

ii

, which correspond to blood groups AB, A, B, and O, respectively.

Q13EXERCISES

Explain the following terms with example

(a)

Co-dominance

(b)

Incomplete dominance

Solution

(a) Co-dominance

Definition: Co-dominance is a pattern of inheritance where both alleles in a heterozygous individual are fully and independently expressed in the phenotype. Neither allele is dominant or recessive to the other. As a result, the phenotype of the heterozygote is not an intermediate blend but a combination of both parental traits.

Explanation: In co-dominance, the F1 generation resembles both parents. Both alleles produce their respective functional proteins or products.

Example: ABO blood group in humans. The ABO blood group system is controlled by the gene 'I', which has three alleles:

I^A

,

I^B

, and

i

. Alleles

I^A

and

I^B

are co-dominant with each other and are both dominant over the allele

i

.

An individual with genotype $I^A I^B$ has blood group AB.
In this case, both allele $I^A$ (which produces A-type sugar antigen) and allele $I^B$ (which produces B-type sugar antigen) express themselves completely.
As a result, the red blood cells of an AB individual have both A and B antigens on their surface. This is a clear example of co-dominance.

(b) Incomplete Dominance

Definition: Incomplete dominance is a pattern of inheritance where the phenotype of a heterozygous individual is an intermediate blend between the phenotypes of the two homozygous parents. Neither allele completely dominates the other.

Explanation: This happens when the dominant allele does not produce enough protein to completely mask the effect of the recessive allele. The F1 generation shows a phenotype that is different from either parent.

Example: Flower color in Snapdragon (Antirrhinum majus) or Four o'clock plant (Mirabilis jalapa).

When a homozygous red-flowered plant (genotype RR) is crossed with a homozygous white-flowered plant (genotype rr), the F1 generation consists of all pink-flowered plants (genotype Rr).
The pink color is an intermediate phenotype, a blend of red and white. This demonstrates that the R allele is incompletely dominant over the r allele.
When the F1 pink flowers (Rr) are self-pollinated, the F2 generation shows a phenotypic ratio of 1 Red (RR) : 2 Pink (Rr) : 1 White (rr). The genotypic and phenotypic ratios are the same (1:2:1), which is a characteristic feature of incomplete dominance.

Q14EXERCISES

What is point mutation? Give one example.

Solution

Point Mutation

A point mutation is a type of gene mutation that arises due to a change in a single base pair of DNA. This change can be a substitution (one base is replaced by another), an insertion (one base is added), or a deletion (one base is removed) at a single point in the DNA sequence.

These mutations can have a range of effects on protein synthesis. If the change in the DNA base results in a change in the mRNA codon, it may lead to the incorporation of a different amino acid into the polypeptide chain. This can alter the structure and function of the resulting protein, sometimes leading to genetic disorders.

Example: Sickle-cell Anaemia

Sickle-cell anaemia is a classic example of a genetic disorder caused by a point mutation. It is an autosomal recessive disease.

The Gene: The mutation occurs in the gene responsible for coding the beta-globin chain of the haemoglobin molecule.
The Mutation: It is caused by a single base substitution at the sixth codon of the beta-globin gene. The DNA base Adenine (A) is replaced by Thymine (T).
- Normal DNA sequence: GAG
- Mutated DNA sequence: GTG
Effect on mRNA: This substitution in the DNA template strand leads to a corresponding change in the mRNA codon during transcription.
- Normal mRNA codon: GAG (codes for Glutamic acid)
- Mutated mRNA codon: GUG (codes for Valine)
Effect on Protein: During translation, this altered mRNA causes the amino acid Valine to be incorporated into the beta-globin chain at the sixth position, instead of the normal Glutamic acid.
Phenotypic Consequence: This single amino acid substitution alters the properties of the haemoglobin molecule. Under low oxygen conditions, the mutant haemoglobin (HbS) undergoes polymerization, causing the red blood cells to change from their normal biconcave disc shape to a rigid, elongated sickle-like shape. These sickled cells can block blood vessels and are more fragile, leading to the symptoms associated with sickle-cell anaemia.

Q15EXERCISES

Who had proposed the chromosomal theory of the inheritance?

Solution

The chromosomal theory of inheritance was proposed independently by Walter Sutton and Theodor Boveri in 1902.

They observed the parallel behavior between chromosomes during meiosis and Gregor Mendel's laws of inheritance. Based on these observations, they postulated that genes, the units of heredity, are located on chromosomes.

The key points of their theory were:

Genes are carried on chromosomes.
The segregation of a pair of homologous chromosomes during meiosis and the subsequent separation of sister chromatids lead to the segregation of the pair of alleles they carry. This provides a physical basis for Mendel's law of segregation.
The independent assortment of different chromosome pairs during meiosis provides the physical basis for Mendel's law of independent assortment.

This theory united the fields of cytology (the study of cells) and genetics. It was later experimentally verified by the work of Thomas Hunt Morgan and his colleagues on the fruit fly, Drosophila melanogaster.

Q16EXERCISES

Mention any two autosomal genetic disorders with their symptoms.

Solution

Two autosomal genetic disorders with their symptoms are:

1. Sickle-cell Anaemia: This is an autosomal recessive disorder. It is caused by a point mutation in the gene controlling the beta-globin chain of haemoglobin. The mutation causes the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the protein chain. The genotype of the affected individual is homozygous recessive (

\text{Hb}^S \text{Hb}^S

).

Symptoms:

The mutant haemoglobin molecule undergoes polymerisation under low oxygen tension, causing the biconcave red blood cells (RBCs) to change into an elongated sickle-like shape.
This change in shape reduces the oxygen-carrying capacity of the blood, leading to anaemia.
The sickle-shaped RBCs can clog capillaries and small blood vessels, causing severe pain, fatigue, and damage to organs like the spleen, brain, and kidneys.

2. Down's Syndrome: This is an autosomal chromosomal disorder caused by the presence of an additional copy of chromosome number 21. This condition is also known as Trisomy of 21.

Symptoms:

The affected individual is short-statured with a small, round head.
They have a furrowed tongue and a partially open mouth.
The palm is broad with a characteristic single palmar crease (Simian crease).
Physical, psychomotor, and mental development is retarded.
They may also have congenital heart disease.

Principles of Inheritance and VariationClass 12 Biology NCERT Solutions

Mention the advantages of selecting pea plant for experiment by Mendel.

Solution

Differentiate between the following -

Solution

A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?

Solution

Explain the Law of Dominance using a monohybrid cross.

Solution

Define and design a test-cross.

Solution

Using a Punnett Square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.

Solution

When a cross in made between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to be

Solution

Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F1F_{1}F1​ generation for a dibybrid cross?

Solution

Briefly mention the contribution of T.H. Morgan in genetics.

Solution

What is pedigree analysis? Suggest how such an analysis, can be useful.

Solution

How is sex determined in human beings?

Solution

A child has blood group O . If the father has blood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.

Solution

Explain the following terms with example

Solution

What is point mutation? Give one example.

Solution

Who had proposed the chromosomal theory of the inheritance?

Solution

Mention any two autosomal genetic disorders with their symptoms.

Solution

Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in $F_{1}$ generation for a dibybrid cross?