Heredity and Common Genetic Diseases: WBBSE Class 10 Life Science Chapter 3 Solutions & Question Bank

WBBSE Class 10 Life Science Heredity and Common Genetic Diseases Chapter 3 Solutions & Question Bank - Vision Institute

Syllabus Units Included in This Chapter

As prescribed in the official WBBSE Class 10 English Medium syllabus (Pages 68–89), this single comprehensive chapter contains all questions and detailed solutions from the following two interconnected units:

Unit 3.A Heredity (Mendel's Laws, Variation, Monohybrid & Dihybrid Crosses, Sex Determination)
Unit 3.B Some common genetic Diseases (Thalassemia, Haemophilia, Colour Blindness & Genetic Counselling)
Section A: Multiple Choice Questions [Mark-1]

Textbook Exercise MCQs (Questions 1 to 31)

Q.1Mendel's choice contrasting characters from pea plant are—
A 7 pairs
B 9 pairs
C 6 pairs
D 8 pairs
Correct Answer: (A) 7 pairs

Explanation: Gregor Johann Mendel selected 7 pairs of true-breeding contrasting characters in garden pea (Pisum sativum): stem height, flower colour, flower position, pod shape, pod colour, seed shape, and seed colour.

Q.2The ratio of genotype in $F_2$ generation of monohybrid cross of Mendel was— MP '18
A 1:2:1
B 3:1
C 9:3:3:1
D 2:1:2
Correct Answer: (A) 1:2:1

Explanation: In the $F_2$ generation of a monohybrid cross ($Tt \times Tt$), the genotypic ratio is 1 TT (pure tall) : 2 Tt (hybrid tall) : 1 tt (pure dwarf), i.e., 1 : 2 : 1.

Q.3When a hybrid black guineapig is crossed with a pure white guineapig the offspring of $F_1$ will be—
A 1:2:1
B 1:1
C 3:1
D None of them
Correct Answer: (B) 1:1

Explanation: This is a classic test cross: heterozygous black ($Bb$) $\times$ homozygous recessive white ($bb$) $\rightarrow$ $1\ Bb\text{ (Black)} : 1\ bb\text{ (White)}$ (50% Black and 50% White, ratio 1 : 1).

Q.4Phenotype result of Mendel's dihybrid cross will be—
A 9:3:3:1
B 1:2:2:4:1:2:1:2:1
C 1:2:1
D 1:2:2:4
Correct Answer: (A) 9:3:3:1

Explanation: In $F_2$ generation of a dihybrid cross ($RrYy \times RrYy$), the four phenotypic categories appear in the classic Mendelian ratio of 9 Round-Yellow : 3 Round-Green : 3 Wrinkled-Yellow : 1 Wrinkled-Green.

Q.5Protanopia is a type of—
A Haemophilia
B Thalassemia
C Colourblindness
D None of them
Correct Answer: (C) Colourblindness

Explanation: Protanopia is red colour blindness resulting from defective long-wavelength cone opsin photopigment on the X chromosome.

Q.6Among the following which one is applicable to a normal male in man?—
A 44A + XY
B 44A + XYY
C 44A + XX
D 44A + XXY
Correct Answer: (A) 44A + XY

Explanation: A healthy human male possesses 46 chromosomes consisting of 44 autosomes and one heterogametic pair of sex chromosomes ($44A + XY$).

Q.7Chromosome theory of inheritance was propose by—
A Morgan
B Sutton and Boveri
C Mendel
D Lamarck
Correct Answer: (B) Sutton and Boveri

Explanation: In 1902, Walter Sutton and Theodor Boveri independently formulated the Chromosome Theory of Inheritance, demonstrating that Mendelian factors (genes) are carried on chromosomes.

Q.8The genetic constitution for which a characteristic feature in an organism is expressed is called—
A Phenotype
B Genotype
C Dominance
D Hybrid
Correct Answer: (B) Genotype

Explanation: The actual genetic makeup or allelic composition of an individual for a particular trait is termed its genotype, whereas the observable physical expression is the phenotype.

Q.9The process of fertilization of female gamete with male gamete of the same plant, is called—
A Selfing
B Back cross
C Test cross
D Hybrid
Correct Answer: (A) Selfing

Explanation: Self-fertilization or selfing involves the fusion of gametes produced by the same individual or genetically identical clones.

Q.10When both the alleles of a gene are equally expressive, is called—
A Codominance
B Dominance
C Recessiveness
D Phenotype
Correct Answer: (A) Codominance

Explanation: In codominance, both alleles in a heterozygote express themselves fully and simultaneously without blending (e.g., $I^A$ and $I^B$ alleles producing AB blood group).

Q.11The $F_2$ generation of a cross produced identical phenotypic and genotypic ratio. It is not an expected Mendelian result, and can be attributed to—
A Homologous pairs
B Independent assortment
C Linkage
D Incomplete dominance
Correct Answer: (D) Incomplete dominance

Explanation: Under incomplete dominance, the heterozygous phenotype is an intermediate blending (e.g. Pink in *Mirabilis jalapa*), giving both phenotypic and genotypic ratios of 1 Red : 2 Pink : 1 White (1:2:1).

Q.12The first work on genetics was done by—
A Lamarck
B Mendel
C Vries
D Darwin
Correct Answer: (B) Mendel

Explanation: Gregor Johann Mendel conducted pioneering hybridization experiments on garden peas from 1856 to 1863, establishing the basic principles of heredity.

Q.13The discipline which deals with the study of inheritance of characters is—
A Cytology
B Genetics
C Darwinism
D Evolution
Correct Answer: (B) Genetics

Explanation: Genetics (term coined by William Bateson in 1905) is the biological science studying heredity and variation in living organisms.

Q.14Position of gene on chromosome is called—
A Locus
B Nucleosome
C Factor
D Cistron
Correct Answer: (A) Locus

Explanation: The specific, fixed physical position occupied by a gene on a chromosome is called its locus.

Q.15Which of the following crosses would produce a genotype ratio 1 : 2 : 1 in $F_2$?—
A $Ab \times Ab$
B $ab \times ab$
C $Ab \times ab$
D $AB \times AB$
Correct Answer: (A) $Ab \times Ab$

Explanation: Selfing or crossing of two monohybrid heterozygotes ($Aa \times Aa$ or here designated $Ab \times Ab$) yields the classical $1 : 2 : 1$ genotypic ratio.

Q.16Mendel is popular for postulating—
A Cell theory
B Laws of inheritance
C Linkage theory
D Origin of species
Correct Answer: (B) Laws of inheritance

Explanation: Mendel formulated the Law of Segregation and the Law of Independent Assortment.

Q.17The phenomenon which defines the independent assortment is—
A Segregation
B Dominance
C Crossing over
D Linkage
Correct Answer: (C) Crossing over

Explanation: Crossing over and random alignment of non-homologous chromosomes at metaphase-I ensure the independent assortment of non-allelic genes.

Q.18Which of the following traits in pea plant is recessive?— MP '17
A Wrinkled seed
B Yellow coloured seed
C Purple coloured flower
D Axial flower
Correct Answer: (A) Wrinkled seed

Explanation: In garden pea, Round seed shape ($R$) is dominant over wrinkled seed shape ($r$).

Q.19How many types of gametes are formed from pea plant having genotype YyRr?—
A 1
B 4
C 2
D 3
Correct Answer: (B) 4

Explanation: Number of gametes $= 2^n$ (where $n = \text{number of heterozygous pairs}$). For $YyRr$, $n = 2$, so $2^2 = 4$ types of gametes: $YR$, $Yr$, $yR$, and $yr$.

Q.20The probability of haemophilic girl children born to a haemophilia carrier mother and normal father is— MP '17
A 75%
B 50%
C 100%
D 0%
Correct Answer: (D) 0%

Explanation: Cross: Mother ($X^H X^h$) $\times$ Father ($X^H Y$). Female progeny receive the dominant normal $X^H$ allele from the father ($X^H X^H$ normal or $X^H X^h$ carrier). Thus, the probability of a diseased haemophilic girl ($X^h X^h$) is 0%.

Q.21Which one of the following is not controlled by the autosomal gene of human?— MP '18
A Roller tongue
B Haemophilia
C Thalassemia
D Attached ear lobe
Correct Answer: (B) Haemophilia

Explanation: Haemophilia is an X-linked recessive sex-linked disorder, whereas tongue rolling, thalassemia, and attached ear lobes are governed by autosomal genes.

Q.22Who was father of genetics?—
A Darwin
B Lamarck
C Morgan
D Mendel
Correct Answer: (D) Mendel

Explanation: Gregor Johann Mendel is universally honored as the Father of Genetics.

Q.23How many genes control rolling of tongue?—
A One gene
B A line of genes
C Multiple gene
D More than one gene
Correct Answer: (A) One gene

Explanation: The ability to roll the lateral edges of the tongue upward into a tube is controlled by a single dominant autosomal gene.

Q.24Alternative form of gene is called—
A Dominant
B Recessive
C Allele
D Hybrid
Correct Answer: (C) Allele

Explanation: Alternative, contrasting forms of a gene occupying the same locus on homologous chromosomes are known as alleles (e.g. $T$ and $t$).

Q.25When none of the two opposite characters are expressed, instead an intermediate character is expressed, is called—
A Dominance
B Hybrid
C Incomplete dominance
D None of these
Correct Answer: (C) Incomplete dominance

Explanation: When neither allele is completely dominant over the other, the heterozygous condition produces a blending intermediate phenotype (e.g. Pink flowers in Snapdragon).

Q.26What was the phenotypic ratio in $F_2$ generation of monohybrid cross?—
A 1:2:1
B 3:1
C 1:3
D 2:1:1
Correct Answer: (B) 3:1

Explanation: In Mendel's monohybrid cross ($Tt \times Tt$), 3/4 show the dominant phenotype and 1/4 show the recessive phenotype, giving a ratio of 3 : 1.

Q.27Which one of the following disease is caused by recessive gene?—
A Haemophilia
B Malaria
C Goitre
D Typhoid
Correct Answer: (A) Haemophilia

Explanation: Haemophilia is an inherited congenital bleeding disorder caused by an X-linked recessive gene mutation.

Q.28In which disease a carrier mother's 50% sons become diseased?—
A Nightblindness
B Thalassemia
C Haemophilia
D All of these
Correct Answer: (C) Haemophilia

Explanation: A carrier mother ($X^H X^h$) transmits either $X^H$ or $X^h$ to her sons with equal probability. Since sons receive their Y chromosome from the father, 50% of sons ($X^h Y$) inherit the defective allele and suffer from haemophilia.

Q.29What we can detect by test cross?—
A Phenotype
B Genotype
C Phenotype and genotype
D None of these
Correct Answer: (B) Genotype

Explanation: A test cross (crossing an individual showing dominant phenotype with a homozygous recessive parent) is specifically designed to determine whether the dominant organism is homozygous ($TT$) or heterozygous ($Tt$).

Q.30What percentage of black and white guineapig will be produced if a hybrid black is crossed with pure white guineapig?—
A 50% black, 50% white
B 75% black, 75% white
C 75% white, 25% black
D 100% black, 0% white
Correct Answer: (A) 50% black, 50% white

Explanation: Heterozygous black ($Bb$) crossed with pure white ($bb$): $$\text{Gametes: } B, b \times b \rightarrow 50\% Bb\text{ (Black)} : 50\% bb\text{ (White)}$$

Q.31Alleles for a pair of contrasting character though unite in $F_1$ hybrid but are never blended together rather they separate from each other during gamete formation in pure form. What is the name of this law?—
A Law of natural selection
B Law of segregation
C Law of independent assortment
D Law of dominance
Correct Answer: (B) Law of segregation

Explanation: This is the statement of Mendel's First Law (Law of Segregation / Law of Purity of Gametes), derived from monohybrid crosses.

Section B.1: Fill in the Blanks [Mark-1]
1Phenotypic ratio in monohybrid cross was ______.
Answer
3 : 1
2In ______, the affected person fails to recognise red colour.
Answer
Protanopia (Red colour blindness).
3Incomplete dominance found in ______ plant.
Answer
Mirabilis jalapa (Four O'clock plant) / Antirrhinum majus (Snapdragon).
4Mendel chosen ______ contrasting characteristics in pea plant.
Answer
7 pairs (seven pairs).
5______ is a disease created by sex linked gene. MP '17
Answer
Haemophilia (or Colour blindness).
6The different ______ of pea plant may show same phenotype.
Answer
Genotypes (e.g. $TT$ and $Tt$ both produce Tall phenotype).
7The sexual reproduction that occurs among two genotypically different organisms belonging to the same species, is called ______. MP '19
Answer
Hybridization (or Cross-breeding).
8Fundamental laws of heredity is known as ______.
Answer
Mendel's Laws of Inheritance.
9Differences in characteristic features of organisms are called ______.
Answer
Variations.
10In ______, the affected person fails to recognise green colour.
Answer
Deuteranopia (Green colour blindness).
11Pea plant is a ______ plant.
Answer
Bisexual (self-pollinating).
12Mendel's law of ______ found from monohybrid cross.
Answer
Segregation.
13______ combination in human male is XY and in female is XX.
Answer
Sex chromosome (Heterosome / Allosome).
14Sex determination in a child depends on ______ gene / chromosome.
Answer
Y chromosome (specifically SRY gene of father).
Section B.2: Mention True or False [Mark-1]
1The another name of haemophilia-B disease is christmas disease.
Answer: True
Haemophilia B (Factor IX clotting deficiency) is termed Christmas disease after Stephen Christmas, the first patient diagnosed.
2Pure tall is denoted by 'TT'.
Answer: True
'TT' represents homozygous dominant pure tall.
344A + XXY is normal man's chromosome number.
Answer: False
Correction: Normal man is 44A + XY. $44A + XXY$ is Klinefelter's Syndrome.
4Protanopia is a type of haemophilia.
Answer: False
Correction: Protanopia is a type of red colour blindness, not haemophilia.
5Self pollination or cross pollination can be exercised in flowers of pea plant according to the need.
Answer: True
Pea flowers are naturally self-pollinating but can easily be artificially cross-pollinated via emasculation and dusting.
6In his monohybrid cross experiment, Mendel obtained 75% pure tall pea plants in the first filial generation. MP '18
Answer: False
Correction: In the first filial ($F_1$) generation, Mendel obtained 100% hybrid tall ($Tt$) plants. In $F_2$, only 25% were pure tall ($TT$).
7Mendel used the term gene while describing his experiments related with heredity. MP '20
Answer: False
Correction: Mendel used the term 'Factor' (Elemente). The word 'gene' was coined later by Wilhelm Johannsen in 1909.
8Phenotype result of Mendel's dihybrid cross will be 9:3:3:1.
Answer: True
Classical $F_2$ dihybrid phenotypic ratio is 9:3:3:1.
944A +XXY is a normal man.
Answer: False
Correction: It causes Klinefelter's Syndrome with sterile hypogonadism.
10Seven characters studied by Mendel in pea plant.
Answer: True
Mendel analyzed 7 morphological traits.
11Sex determination of a child is depends on father.
Answer: True
The father produces heterogametic sperm (50% X-bearing and 50% Y-bearing). Fertilization by a Y sperm produces a son, while an X sperm produces a daughter.
12Blood did not coagulate in thalassemia disease.
Answer: False
Correction: Failure of blood coagulation occurs in Haemophilia. Thalassemia is characterized by defective hemoglobin chain synthesis causing microcytic anemia.
Section B.3: Column Matching [Mark-1 each]
Match 1Column Matching
Swipe table horizontally →
Left ColumnRight Column (Given)Correct Matching & Scientific Reason
A. Pure tall(i) TtA → (iii) TT (Homozygous dominant tall genotype)
B. Hybrid tall(ii) Test crossB → (i) Tt (Heterozygous tall genotype)
C. Pure dwarf(iii) TTC → (iv) tt (Homozygous recessive dwarf genotype)
D. $TT \times tt$(iv) tt
(v) Back cross
D → (ii) Test cross / Back cross (Parental monohybrid cross)
Match 2Column Matching
Swipe table horizontally →
Left ColumnRight Column (Given)Correct Matching & Scientific Reason
A. Haemophilia(i) Red colourblindnessA → (iii) Christmas disease (Haemophilia B caused by clotting factor IX deficiency)
B. Thalassemia(ii) Hereditary unitB → (iv) Genetic counselling (Essential preventative measure against carrier marriage)
C. Protanopia(iii) Christmas diseaseC → (i) Red colourblindness (Specific X-linked defect in red cone photopigments)
D. Gene(iv) Genetic counselling
(v) Genetic code
D → (ii) Hereditary unit (Fundamental functional unit of inheritance on DNA)
Match 3Column Matching
Swipe table horizontally →
Left ColumnRight Column (Given)Correct Matching & Scientific Reason
A. Homozygous(i) Dominant featureA → (ii) TT (Organism with identical alleles at a given locus)
B. Green coloured pea seed(ii) TTB → (iv) Recessive character (Yellow seed colour 'Y' is dominant over green 'y')
C. Free earlobe(iii) EmasculationC → (i) Dominant feature (Free earlobe is autosomal dominant over attached earlobe)
D. Snapdragon(iv) Recessive character
(v) Incomplete dominance
D → (v) Incomplete dominance (Antirrhinum majus shows pink flowers in heterozygotes)
Match 4Column Matching
Swipe table horizontally →
Left ColumnRight Column (Given)Correct Matching & Scientific Reason
A. Criss-cross inheritance(i) Dihybrid crossA → (iii) Colourblindness (Transmission of X-linked trait from father to daughter to grandson)
B. Factor(ii) ThalassemiaB → (iv) Gene (Term used by Gregor Mendel for hereditary determiners)
C. Iron deposition(iii) ColourblindnessC → (ii) Thalassemia (Hemosiderosis caused by repeated blood transfusions)
D. Law of independent assortment(iv) Gene
(v) Bagging
D → (i) Dihybrid cross (Mendel's second law deduced from 2-trait hybridization)
Section B.4: Very Short Answer (VSA) Questions [Mark-1]
1Who is the father of genetics?
Answer
Gregor Johann Mendel.
2'Tt' is homozygous or heterozygous?
Answer
Heterozygous (contains two different alleles: dominant $T$ and recessive $t$).
3What is the genotypic ratio?
Answer
The numerical ratio representing the exact genetic and allelic constitution of offspring in a generation (e.g., $1\ TT : 2\ Tt : 1\ tt = \mathbf{1 : 2 : 1}$ in monohybrid $F_2$).
4Give the ratio of phenotypes in monohybrid cross?
Answer
3 : 1 (3 Dominant : 1 Recessive).
5What do you mean by trait?
Answer
A trait is a distinct phenotypic manifestation of an inherited characteristic (e.g., tallness or dwarfness for the character plant height).
6When a cross takes place between two hybrid black guineapigs, what will be the phenotype ratio of black and white guineapig?
Answer
3 Black : 1 White (3 : 1) (Cross: $Bb \times Bb \rightarrow 1\ BB, 2\ Bb, 1\ bb$).
7What do you mean by $F_1$ generation?
Answer
First Filial Generation ($F_1$) is the immediate generation of hybrid offspring produced by cross-pollinating or mating two genetically distinct parental varieties.
8Write a character of the gene responsible for colourblindness. MP '23
Answer
It is an X-linked recessive allele that shows criss-cross inheritance.
9What is self?
Answer
Selfing (self-fertilization) is the fusion of male and female gametes originating from the same individual or genetically identical individuals.
10What do you mean by parental generation?
Answer
The initial generation of pure-breeding homozygous individuals chosen to initiate a cross-breeding experiment, designated as 'P' generation.
11Give an example of a plant which shows incomplete dominance in flower colour.
Answer
Mirabilis jalapa (Four O'clock plant) or Antirrhinum majus (Snapdragon).
12Give the two forms of thalassemia.
Answer
$\alpha$-Thalassemia (defect on chromosome 16) and $\beta$-Thalassemia (defect on chromosome 11).
13Who proposed the chromosomal theory of inheritance?
Answer
Walter Sutton and Theodor Boveri (1902).
14Give an example of a variation inherited in man along the generation.
Answer
Roller tongue (dominant) vs Non-roller tongue (recessive).
15Give another example of a variation inherited in man along the generation.
Answer
Free earlobe (dominant) vs Attached earlobe (recessive).
16Which type of chromosome in human carries gene responsible for the disease thalassemia.
Answer
Autosomes (Chromosome 16 for $\alpha$-globin genes $HBA1, HBA2$; Chromosome 11 for $\beta$-globin gene $HBB$).
17In case of guineapig whether the phenotype of the two genotypes bbRR and bbRr is same? MP '18
Answer
Yes, identical: Both express White coat colour with Rough hair, because 'R' (rough) is completely dominant over 'r' (smooth).
18What is the cause of expression of haemophilia disease only at homozygous condition? MP '19
Answer
Because haemophilia is an X-linked recessive trait. In females with two X chromosomes, a single mutant allele ($X^h$) is masked by the normal dominant allele ($X^H$), expressing disease only when homozygous ($X^h X^h$).
19What is the fate of a thalassemic patient. Write what kind of measures they can take to eradicate this disease from the population. MP '18
Answer
Fate: Severe microcytic anemia, bone deformities, hepatosplenomegaly, and fatal organ failure from iron overload (hemosiderosis).
Eradication measures: Pre-marital genetic screening (Hb electrophoresis) to prevent carrier-to-carrier marriages, and prenatal genetic counselling.
20A daughter is born to a woman carrier for the colourblind disease who married a colourblind man. What would be the probability of expression of colourblindness in that girl child? MP '22
Answer
50% (or 0.5). Cross: Mother ($X^C X^c$) $\times$ Father ($X^c Y$). Daughters produced are $X^C X^c$ (Carrier, 50%) and $X^c X^c$ (Colourblind, 50%).
21State the second law of Mendel related to inheritance. MP '22
Answer
Law of Independent Assortment: When two pairs of contrasting traits are combined in a hybrid, segregation of one pair of characters is completely independent of the other pair during gamete formation.
Section C: Short Answer Type Questions [Marks-2]
1What is heredity?
Answer

Heredity is the biological transmission of distinct genetic morphological, physiological, and behavioural characters from parents to offspring across successive generations through germ cells.

2What is mutation?
Answer

A mutation is a sudden, discontinuous, heritable change in the nucleotide sequence of DNA or chromosomal structure that alters genetic information and serves as the ultimate source of new alleles and variation.

3What is variation?
Answer

Variation refers to the morphological, physiological, or biochemical differences exhibited by individuals of the same species, caused by genetic recombination (crossing over), independent assortment, or environmental influences.

4What is hybridization?
Answer

Hybridization is the process of interbreeding or artificial crossing between two genetically dissimilar individuals of the same or closely related species to combine desirable characteristics in the resulting hybrid offspring.

5What is monohybrid cross?
Answer

A monohybrid cross is a hybridization experiment where two parent organisms differing in only a single pair of contrasting allelic traits are mated (e.g. pure tall $TT \times$ pure dwarf $tt$).

6What do you mean by dihybrid cross?
Answer

A dihybrid cross is a genetic cross between two pure-breeding parents considering two pairs of contrasting alleles simultaneously (e.g. Yellow Round $YYRR \times$ Green Wrinkled $yyrr$).

7What is recessive gene?
Answer

A recessive gene (allele) is an allele whose phenotypic expression is completely suppressed or masked in the presence of its dominant counterpart, and is phenotypically expressed only in homozygous state (e.g. $t$ in $tt$).

8What is incomplete dominance?
Answer

Incomplete dominance is a non-Mendelian inheritance pattern where neither allele is dominant, producing an intermediate blending phenotype in heterozygotes (e.g. Red $RR \times$ White $rr \rightarrow$ Pink $Rr$ in Mirabilis jalapa).

9What is dominant gene?
Answer

A dominant gene is an allele that masks the phenotypic expression of the alternative allele and expresses its trait in both homozygous ($TT$) and heterozygous ($Tt$) states.

10What do you mean by homozygous organism?
Answer

An organism that possesses two identical alleles for a particular trait at the same locus on homologous chromosomes (e.g. pure tall $TT$ or pure dwarf $tt$), breeding true for that character.

11What do you mean by heterozygous organism?
Answer

An organism possessing two different, contrasting alleles for a specific trait at the same locus on homologous chromosomes (e.g. hybrid tall $Tt$), producing two distinct types of gametes upon meiosis.

12What is pure breeding variety?
Answer

A pure breeding (true-breeding) variety is a homozygous strain that has undergone repeated self-pollination or inbreeding, consistently producing offspring with the exact same parental phenotype generation after generation.

13What is test cross?
Answer

A test cross is the cross of an individual exhibiting a dominant phenotype with its homozygous recessive parent ($T? \times tt$) to determine whether the dominant organism is homozygous ($TT$) or heterozygous ($Tt$).

14State about the Mendel's law of segregation.
Answer

Law of Segregation: The two alleles of a contrasting character remain together in a hybrid without blending or altering each other, and segregate (separate) cleanly from each other during gametogenesis so that each gamete receives only one allele in pure form.

15State two symptoms of thalassemia.
Answer
  • Severe Microcytic Hypochromic Anemia: Pale skin, extreme fatigue, and breathlessness due to destruction of defective erythrocytes.
  • Hepatosplenomegaly & Hemosiderosis: Enlarged liver and spleen, accompanied by toxic iron accumulation in heart and endocrine glands from repeated blood transfusions.
16State with example how dominant trait is expressed in the experiment of hybridization.
Answer

When Mendel crossed a pure tall pea plant ($TT$) with a pure dwarf plant ($tt$), all resulting $F_1$ progeny ($Tt$) were completely tall. The allele for tallness ($T$) completely masked the dwarf allele ($t$), demonstrating the Law of Dominance.

17What is Mendelism?
Answer

Mendelism encompasses the fundamental principles and laws of heredity discovered by Gregor Johann Mendel (Law of Segregation and Law of Independent Assortment), which laid the foundation of classical genetics.

18Show with the help of a cross, how colourblindness is inherited?
Answer

Cross: Carrier Female ($X^C X^c$) $\times$ Normal Male ($X^C Y$):

Sperm \ Egg$X^C$ (Normal)$X^c$ (Carrier allele)
$X^C$$X^C X^C$ (Normal Daughter, 25%)$X^C X^c$ (Carrier Daughter, 25%)
$Y$$X^C Y$ (Normal Son, 25%)$X^c Y$ (Colourblind Son, 25%)

This demonstrates criss-cross inheritance, where 50% of sons inherit the disease from their carrier mother.

19Write the help of a cross show how the sex of offsprings of humans is determined?
Answer

Cross: Mother ($44A + XX$) $\times$ Father ($44A + XY$):

  • Mother produces one type of ovum: $22A + X$ (Homogametic).
  • Father produces two types of sperm in equal ratio: $22A + X$ (Gynosperm, 50%) and $22A + Y$ (Androsperm, 50%).
Sperm \ Ovum$22A + X$
$22A + X$$44A + XX$ (Female Child — 50%)
$22A + Y$$44A + XY$ (Male Child — 50%)

Thus, sex determination is entirely governed by whether the fertilizing sperm carries an X or Y chromosome from the father.

20One day students read an article in newspaper on Thalassemia and were very scared to know the fate of a thalassemic patient. Write what kind of measures they can take to eradicate this disease from the population. MP '18
Answer
  1. Pre-marital Genetic Screening: Undergoing Hemoglobin HPLC or electrophoresis before marriage to detect whether prospective partners are thalassemia carriers (Thalassemia Minor). Marriage between two carriers must be avoided.
  2. Genetic Counselling & Prenatal Diagnosis: Providing professional guidance and conducting Chorionic Villus Sampling (CVS) or Amniocentesis in pregnant carrier mothers to identify Thalassemia Major before birth.
Section D: Long Answer Type Questions [Marks-5]
1Why did Mendel choose pea plants as his experimental organism? [5]
Answer
  1. Distinct Contrasting Characters: Pea plant (Pisum sativum) displays easily observable, sharply contrasting alternative traits (e.g. Tall vs Dwarf, Round vs Wrinkled seeds) with no ambiguous intermediate forms.
  2. Bisexual Flowers & Natural Self-Pollination: Flowers are naturally cleistogamous/closed, ensuring strict self-pollination and pure true-breeding lines.
  3. Ease of Artificial Cross-Pollination: Large floral structures make manual emasculation (removal of anthers) and controlled hybridization straightforward.
  4. Short Life Cycle & High Fecundity: Annual plant that completes its generation in 3-4 months, producing abundant seeds per cross for statistically reliable data.
  5. Easy Cultivation: Can easily be grown in small garden plots or pots without complex maintenance.
3What are the reasons behind the Mendel's success? [5]
Answer
  1. Focus on One or Two Traits at a Time: Unlike previous hybridization researchers, Mendel studied the inheritance of single characters (monohybrid) before analyzing complex combinations (dihybrid).
  2. Use of True-Breeding Homozygous Parents: Mendel verified the purity of parental lines through selfing for multiple generations before crossing.
  3. Quantitative & Statistical Rigour: Maintained meticulous numerical records of every offspring across $F_1, F_2, F_3$ generations and applied probability ratios.
  4. Careful Avoidance of Foreign Pollen: Practiced diligent emasculation and protective bagging to prevent unintended insect pollination.
  5. Absence of Gene Linkage: By sheer biological good fortune, the seven characters chosen were situated on different chromosomes (or sufficiently far apart on chromosomes 1, 4, 5, and 7), avoiding complicating linkage.
4What are seven characters of pea plant? State their contrasting features in pea plant. [2+3]
Answer

The seven contrasting pairs of characters studied by Mendel in Pisum sativum are:

No.CharacterDominant TraitRecessive Trait
1Stem HeightTall ($T$)Dwarf ($t$)
2Seed ShapeRound ($R$)Wrinkled ($r$)
3Seed (Cotyledon) ColourYellow ($Y$)Green ($y$)
4Pod ShapeInflated / Full ($I$)Constricted ($i$)
5Pod ColourGreen ($G$)Yellow ($g$)
6Flower PositionAxial ($A$)Terminal ($a$)
7Flower Colour (Seed Coat)Purple / Violet ($P$)White ($p$)
5What is law of segregation? Describe the events of monohybrid cross of pea plant with checker boards. [2+3]
Answer

Law of Segregation (Mendel's First Law): Alleles of a gene do not blend in a hybrid but remain distinct, and segregate from each other during gamete formation so that each gamete receives only one allele in pure condition.

Monohybrid Cross Description: Pure Tall ($TT$) is crossed with Pure Dwarf ($tt$). The $F_1$ hybrids are 100% Tall ($Tt$). When $F_1$ plants are selfed ($Tt \times Tt$), $F_2$ progeny yield 3 Tall : 1 Dwarf (Phenotypic) and 1 TT : 2 Tt : 1 tt (Genotypic).

Fig: Checkerboard of Mendel's Monohybrid Cross (Pea Stem Height)
P: Pure Tall (TT) × Pure Dwarf (tt) F1: All Hybrid Tall (Tt) [Selfed] ♂ \ ♀ T (Gamete) t (Gamete) T TT Pure Tall Tt Hybrid Tall t Tt Hybrid Tall tt Pure Dwarf F2 Phenotypic Ratio = 3 Tall : 1 Dwarf (3:1) • Genotypic Ratio = 1 TT : 2 Tt : 1 tt (1:2:1)
7What are the explanation of monohybrid cross? [5]
Answer
  1. Unit Factors in Pairs: Each trait is governed by discrete particulate factors (genes) present in pairs in diploid cells.
  2. Principle of Dominance: When two unlike unit factors are present in an individual ($Tt$), one factor expresses itself (Dominant) while the other remains unexpressed (Recessive).
  3. Law of Segregation: Factors do not blend; during meiosis, homologous chromosome pairs disjoin, sending only one factor of each pair into a gamete.
  4. Random Fertilization: Female gametes ($T, t$) fuse with male gametes ($T, t$) with equal probability ($25\%$ chance for each of the 4 zygotic combinations), producing the $3:1$ phenotypic and $1:2:1$ genotypic ratios.
8What are the explanation of dihybrid cross? [5]
Answer
  1. Independent Inheritance: Two separate gene pairs controlling seed shape ($R/r$) and seed colour ($Y/y$) assort independently during meiosis.
  2. Gametic Combinations: The dihybrid $F_1$ ($RrYy$) produces four types of pollen and four types of ova in equal numbers ($25\%$ each): $RY$, $Ry$, $rY$, and $ry$.
  3. 16-Square Random Syngamy: Unbiased fertilization produces 16 zygotic combinations, giving 9 genotypes and 4 phenotypes in the ratio of 9 Round-Yellow : 3 Round-Green : 3 Wrinkled-Yellow : 1 Wrinkled-Green.
  4. Dihybrid as Two Combined Monohybrids: $(3\text{ Round} : 1\text{ Wrinkled}) \times (3\text{ Yellow} : 1\text{ Green}) = 9 : 3 : 3 : 1$.
9What is dihybrid cross? Describe the events of dihybrid cross of plant with checkerboard. [2+3]
Answer

Definition (2 Marks): A dihybrid cross is a breeding experiment analyzing the inheritance patterns of two distinct pairs of contrasting traits simultaneously.

Events (3 Marks): Parent cross between Pure Round-Yellow ($RRYY$) and Pure Wrinkled-Green ($rryy$). $F_1$ generation produces all Round-Yellow plants ($RrYy$). Self-pollination of $F_1$ yields 16 combinations shown in the Punnett checkerboard below:

Fig: 16-Cell Checkerboard of Mendel's Dihybrid Cross ($F_2$)
F1 Dihybrid Selfing: RrYy × RrYy → 4 Gamete Types Each ♂ \ ♀ RY Ry rY ry RY RRYYR-Yellow RRYyR-Yellow RrYYR-Yellow RrYyR-Yellow Ry RRYyR-Yellow RRyyR-Green RrYyR-Yellow RryyR-Green rY RrYYR-Yellow RrYyR-Yellow rrYYWr-Yellow rrYyWr-Yellow ry RrYyR-Yellow RryyR-Green rrYyWr-Yellow rryyWr-Green Phenotypic Ratio: 9 Round-Yellow : 3 Round-Green : 3 Wrinkled-Yellow : 1 Wrinkled-Green (9:3:3:1)
10Describe the experiments of monohybrid cross in guineapig with checkerboard. [5]
Answer

Parental Generation (P): Pure homozygous Black guinea pig ($BB$) is mated with pure homozygous White guinea pig ($bb$).

$F_1$ Generation: All progeny are heterozygous Black ($Bb$).

$F_2$ Generation (Inbreeding $Bb \times Bb$): Male gametes ($B, b$) and female gametes ($B, b$) produce 3 phenotypic categories in a 3 Black : 1 White ratio.

♂ \ ♀B (Black allele)b (White allele)
B$BB$ (Pure Black, 25%)$Bb$ (Hybrid Black, 25%)
b$Bb$ (Hybrid Black, 25%)$bb$ (Pure White, 25%)

Phenotypic Ratio: 3 Black : 1 White • Genotypic Ratio: 1 BB : 2 Bb : 1 bb (1:2:1).

11What are the deviation from Mendel's law of heredity? [5]
Answer
  1. Incomplete Dominance: Heterozygote displays an intermediate phenotype (e.g. Mirabilis jalapa flower color: $1\text{ Red} : 2\text{ Pink} : 1\text{ White}$).
  2. Codominance: Both alleles express themselves fully in the phenotype (e.g. human ABO blood group allele $I^A I^B \rightarrow \text{AB blood group}$).
  3. Multiple Allelism: A gene existing in more than two allelic forms in a population (e.g. $I^A, I^B, i$ alleles controlling ABO blood groups).
  4. Linkage: Tendency of genes situated closely on the same chromosome to be inherited together, preventing $9:3:3:1$ independent assortment.
  5. Polygenic (Quantitative) Inheritance: Traits controlled by multiple additive genes (e.g. human skin pigmentation and human height).
12What is law of independent assortment? Briefly describe the event of dihybrid cross in guineapig with checkerboard. [2+3]
Answer

Law of Independent Assortment (2 Marks): Factors controlling two or more pairs of contrasting characters segregate independently of each other during gamete formation and combine randomly during fertilization.

Dihybrid Cross in Guinea Pig (3 Marks): Pure Black Rough ($BBRR$) $\times$ Pure White Smooth ($bbrr$). $F_1$ is all Black Rough ($BbRr$). When $F_1$ is crossed, gametes $BR, Br, bR, br$ produce 16 combinations resulting in: $$\mathbf{9\text{ Black-Rough} : 3\text{ Black-Smooth} : 3\text{ White-Rough} : 1\text{ White-Smooth}}$$

13With the help of a checkerboard show the types of offsprings that might be produced in a cross between a hybrid black guineapig and a pure white guineapig. State the law of segregation as proposed by Mendel. [3+2] MP '17
Answer

Test Cross Checkerboard (3 Marks):

Parents: Hybrid Black ($Bb$) $\times$ Pure White ($bb$)

Gametesb (from white parent)
B (from hybrid parent)$Bb$ — Black Guinea Pig (50%)
b (from hybrid parent)$bb$ — White Guinea Pig (50%)

Phenotypic & Genotypic Ratio: 1 Black : 1 White (1:1).

Law of Segregation (2 Marks): Alleles remain unblended in hybrids and separate cleanly during gamete formation so each gamete contains only one allele.

14What is colourblindness? What are the symptoms of thalassemia? [2+3]
Answer

What is Colour Blindness? (2 Marks):
An X-chromosome-linked congenital visual defect where individuals fail to perceive or distinguish between specific primary colours (most commonly Red and Green) due to defective cone cell opsin photopigments.

Symptoms of Thalassemia (3 Marks):

  • Severe Chronic Anemia: Pale skin, weakness, lethargy, and stunted somatic growth due to continuous hemolysis.
  • Hepatosplenomegaly & Skeletal Deformities: Splenomegaly from overactive RBC removal, along with maxillary bone expansion (chipmunk facies).
  • Iron Overload (Hemosiderosis): Excessive toxic accumulation of iron in heart, liver, and pancreas from frequent blood transfusions.
15Brief notes on thalassemia as an autosomal chromosomal disorder and its genetic counselling. [5]
Answer

Thalassemia as an Autosomal Disorder (3 Marks):
Thalassemia is an autosomal recessive blood disorder resulting from mutations or deletions in genes coding for hemoglobin polypeptide chains:

  • $\alpha$-Thalassemia: Caused by deletion of one or more of the 4 alpha-globin genes on Chromosome 16.
  • $\beta$-Thalassemia: Caused by mutation in the $HBB$ gene on Chromosome 11, reducing beta-globin synthesis. When homozygous ($Hb\beta^T Hb\beta^T$), it causes life-threatening Thalassemia Major (Cooley's Anemia) requiring lifelong transfusions.

Role of Genetic Counselling (2 Marks):

  1. Pre-marital Carrier Detection: Blood test (HPLC / Hb electrophoresis) identifies asymptomatic heterozygous carriers (Thalassemia Minor). If two carriers marry, their child has a $25\%$ risk of Thalassemia Major.
  2. Prenatal Diagnosis: Amniocentesis or chorionic villus sampling (CVS) between 10–16 weeks of gestation detects fetal gene status, guiding medical advice and preventing the birth of diseased children.
16Briefly state about the types of colourblindness. Describe haemophilia in brief. [2+3]
Answer

Types of Colour Blindness (2 Marks):

  1. Protanopia (Red Blindness): Inability to distinguish red light/spectrum; red appears dark grayish.
  2. Deuteranopia (Green Blindness): Inability to perceive green light (most common sex-linked form).
  3. Tritanopia (Blue-Yellow Blindness): Rare autosomal condition where blue cone pigments are non-functional.

Haemophilia in Brief (3 Marks):
An X-linked recessive hereditary bleeding coagulopathy:

  • Haemophilia A (Royal Disease): Deficiency of Anti-Hemophilic Factor (Factor VIII), accounting for ~85% of cases.
  • Haemophilia B (Christmas Disease): Deficiency of Plasma Thromboplastin Component (Factor IX).
  • Symptoms: Prolonged, uncontrolled bleeding from minor cuts, spontaneous internal joint bleeding (hemarthrosis), and hematomas.
18What is variation? Explain about the two variable features in man. [2+3]
Answer

What is Variation? (2 Marks):
Morphological, anatomical, physiological, or behavioural differences observed between individual members of the same biological species, originating through crossing over, independent assortment, and mutations.

Two Inheritable Variable Features in Man (3 Marks):

  1. Tongue Rolling: Controlled by a single autosomal gene. The dominant allele ($R$) enables rolling the lateral margins of the tongue upward into a 'U' tube; homozygous recessive individuals ($rr$) are non-rollers.
  2. Ear Lobe Attachment: The allele for free/detached earlobes ($E$) is dominant over the allele for attached earlobes ($e$) directly connected to the side of the head.
19Define mutation. State about the development of variation. [2+3]
Answer

Definition of Mutation (2 Marks):
A sudden, stable, heritable change in the genetic material (DNA nucleotide sequence or chromosomal count/structure) of an organism that is not caused by genetic segregation or recombination.

Development of Variation (3 Marks):

  • Mutational Innovations: Spontaneous errors during DNA replication create new alleles, introducing fresh genetic diversity into a gene pool.
  • Recombination during Meiosis: Crossing over during pachytene shuffles linked alleles, while random orientation of maternal and paternal chromosomes in metaphase-I generates billions of unique gametic combinations.
  • Random Fertilization: Amphimixis of genetically diverse sperm and ova creates novel combinations in offspring, providing raw material for natural selection and speciation.
20Mention the symptoms of thalassemia disease. In many families mothers are labelled as responsible for the birth of daughter child. Demonstrate with the help of a cross, that this belief is not justified. [2+3] MP '17
Answer

Symptoms of Thalassemia (2 Marks): Chronic severe anemia, pale complexion, jaundice, bone deformities of skull and face, hepatosplenomegaly, and endocrine failure from iron deposition.

Demonstration that Mothers are NOT Responsible for Female Births (3 Marks):
Human females are homogametic ($44A + XX$) and produce only one type of egg carrying an X chromosome ($22A + X$). Human males are heterogametic ($44A + XY$) and produce two types of sperms in equal numbers ($50\% \text{ with X}$ and $50\% \text{ with Y}$).

Fig: Scientific Mechanism of Human Sex Determination Proving Father Determines Gender
MOTHER: 44A + XX (Homogametic • Only X Ovum) FATHER: 44A + XY (Heterogametic • 50% X : 50% Y) 22A + X 22A + X 22A + Y 44A + XX (Female Child, 50%) Egg X + Sperm X → DAUGHTER 44A + XY (Male Child, 50%) Egg X + Sperm Y → SON

Conclusion: The mother always contributes only an X chromosome. It is solely the father's sperm carrying either X or Y that determines the sex of the child. Therefore, blaming mothers for the birth of female children is completely unscientific, false, and socially unjust.

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