WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases

Detailed explanations in West Bengal Board Class 10 Life Science Book Solutions Chapter 3B Some Common Genetic Diseases offer valuable context and analysis.

WBBSE Class 10 Life Science Chapter 3B Question Answer – Some Common Genetic Diseases

Short Answer Type Questions : 2 Marks

Question 1.
What is genetical disease?
Answer:
The disease which arises due to mutation of genes or chromosomes and is inherited, then the disease is known as genetical disease.

Question 2.
What do you mean by autosomal and sex linked inheritance.
Answer:
Autosomal inheritance : When the gene is transmitted through autosomes from one generation to the next, then the situation is known as autosomal inheritance.
Sex linked inheritance: When the gene is transmitted through sex chromosome from one generation to the next therr the situation is known as sex linked inheritance.

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases

Question 3.
Give examples of autosomal and sex-chromosomal disease.
Answer:

  • Autosomal : Thalassemia
  • Sex-chromosomal : Haemophilia

Question 4.
What is the cause of Thalassemia?
Answer:
Mutational change in 16th and 11th autosome causes defective structure of haemoglobin.

Question 5.
Write two symptoms of thalassemia.
Answer:

  1. Digestive disorder and frequent fever.
  2. Enlargement of spleen.

Question 6.
What is causes of haemophilia?
Answer:
Recessive gene of ‘X’ chromosome does not produce anti-coagulant in the blood, causing not to clot in the body any time.

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases

Question 7.
Write two symptoms of ‘haemophilia?
Answer:

  1. Blood does not coagulate.
  2. After birth body may be died due to excessive bleeding.

Question 8.
What do you mean by haemophilia A and B ?
Answer:
Haemophilia A : Due to the absence of anti-haemophilic gene globulin (AHG) or factor (VII) ; the blood does not clot, there is injury in the body. 80% of the patient is of this type.
Haemophilia B : Due to the absence of christmas factor, or factor IX, the blood can’t coagulate when there forms an injury in or outside the body.

Question 9.
What is colour blindness?
Answer:
Colour blind is a genetical disease caused by a sex linked recessive gene in human being. The normal gene is C+ and its mutant recessive form, which hampers the activity of C+. The male carries C in his ‘ X ‘ chromosome and becomes colour blind. He is unable to detect red and green light.

Question 10.
What is genetic counselling?
Answer:
Consultation about genetic inheritance and probability of occurring genetical disease like thalassemia is known as genetic counselling.

Long Answer Type Questions : 5 Marks

Question 1.
What do you mean by autosomal and X linked inheritance in Human.
Answer:
Autosomal: The gene responsible for the phenotype is located on one of the 22 pairs of autosomes (non-sex determining chromosomes). Dominant: conditions that are manifest in heterozygotes (individuals with just one copy of the mutant allele). Recessive: conditions are only manifest in individuals who have two copies of the mutant allele (are homozygous).
Autosomal Dominant Conditions:

  • Huntington Disease
  • Acondroplasia (short-limbed dwarfism)
  • Polycystic kidney disease

Affected individuals are indicated by solid black symbols and unaffected carriers are indicated by the half black symbols. Autosomal recessive diseases:

  • Cystic fibrosis
  • Tay-Sachs
  • Haemochromatosis
  • Phenylketonuria (PKU)

X linked: The gene that encodes for the trait is located on the X chromosome. X linked Recessive Disorders:

  • Duchenne muscular dystrophy
  • haemophilia A
  • X linked severe combined immune disorder (SCID)
  • some forms of congenital deafness

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases

Question 2.
What is Thalassemia trait?
Answer:
Thalassemia Trait : Thalassemia is a genetic disease. This means that a person can only get thalassemia disease or trait by inheriting the genes for thalassemia from their parents. Genes determine what we look like, such as hair colour, and are also responsible for many diseases. Inheritance of thalassemia happens purely by chancethere is nothing that parents do, or do not do, that will cause their child to inherit thalassemia. Thalassemia is never ‘caught’ by another person in the way that a cold or flu is transmitted. People with thalassemia disease and trait are born with it.

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases 1

Thalassemia trait is not and never will become thalassemia disease. There are two main types of thalassemia trait: Alpha Thalassemia trait and Beta Thalassemia trait. Most types of thalassemia trait cause the red blood cells to be smaller in size than usual, but there is no scientific evidence that thalassemia trait causes health problems. Individuals with thalassemia trait have some level of protection from malaria. Therefore, although thalassemia trait is found in all populations, it is most common in people from region where malaria occurs.

Question 3.
What do mean by the Inheritance of Thalassemia?
Answer:
The Inheritance of Thalassemia

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases 2

Haemoglobin is made of heme, alpha globins, and beta globins. At least 9 different genes direct the production of heme. Changes in these genes may lead to disorders of heme production, a group of conditions separate from the thalassemias. Alpha thalassemia occurs when a mutation in the gene that codes for alpha globin results in reduced or absent production of alpha globins. Beta thalassemia occurs with a corresponding change in the beta globin gene. Therefore, the thalassemias are a result of quantitative mutations in the globin genes.

Below is a karyotype, a picture of all of an individual’s genetic information as seen through a microscope. There are 22 pairs of numbered chromosomes (autosomes) and one pair of sex determining chromosomes. A male carries both an X- and a Ychromosome, whereas a female has two X-chromosomes. Because a man can pass down either an X – or a Y-chromosome, he is the one who randomly determines the sex of the baby. The instructions for alpha globin production are present in duplicate, two genes on each chromosome 16 for a total of four. The instructions for beta globin production are on chromosome 11 , one gene on each chromosome for a total of two.

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases 3

Half of a woman’s genetic information goes into each egg, including one chromosome 11 and one chromosome 16. The same is true in the formation of sperm. At conception, the total of 46 chromosomes is restored in which chromosome 11 or 16 is passed down is determined randomly. There is nothing that a mother or father does (or does not do) to direct which chromosome, and therefore which allele, is transmitted to his or her children.

Recessive Inheritance : In recessive conditions, a benign trait or carrier state can exist in which an individual has both a normal and a mutated copy of the gene. The term “normal” is a convention to simply describe the copy of the gene, called an allele, that is seen most often in the general population. The normal allele is able to compensate for the missing or altered function of the mutant allele. Therefore, the individual with trait does not have the symptoms seen in the disease. It is generally thought that each one of us carries 7-8 of these recessive traits, which would cause disease if present in a “double dose.”

It is not uncommon for a person with a recessive disease to be the first individual in his or her family to have the condition. Typically, this occurs when both parents are carriers of the trait. This is different from a dominantly inherited condition in which members of a family in several consecutive generations are affected. Some Recessive Conditions:

  • Attached earlobes
  • Sickle cell anemia
  • Cystic Fibrosis
  • Thalassemia

Some Dominant Conditions :

  • Tongue rolling
  • Huntington’s disease

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases

Question 4.
Write about the inheritance of Haemophilia.
Answer:
The genes – or instructions – for blood clotting are found on the X chromosome, which means that haemophilia is described as an X-linked disorder. Since girls have two X chromosomes, they usually receive a healthy gene on the second X chromosome. The healthy gene will be dominant, meaning that girls who inherit the gene for haemophilia remain carriers, but do not usually show symptoms as their healthy gene can generally produce enough clotting factor for their blood to clot properly. However, this is not the case for boys who inherit the gene for haemophilia, because they only inherit one X chromosome. This means that they have no “back up” set of instructions for the clotting factor so are unable to produce very much, or in some cases none at all.

It can be quite complicated to visualise how haemophilia is inherited and what the chances are of someone having haemophilia or being a carrier, so it is worth looking at the diagram and reading the explanation. And remember, with the same parents, the likelihood of having a child with haemophilia is the same for each pregnancy.
Diagram of haemophilia inheritance (where is the gene for haemophilia A)

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases 4

When a man with an affected gene on his X chromosome and a woman with two healthy genes have a child (see part (a) of the inheritance diagram), a daughter will receive one affected gene (from their father) and one healthy gene (from their mother), making them a ‘carrier’ of the condition, although they will not suffer from it. No son receives the affected gene as they must inherit the Y chromosome (not the X

chromosome) from their father. If the mother is a carrier and the father has healthy genes (see part (b) of the inheritance diagram), each son has a 50 per cent chance of receiving the affected gene from the mother. Any daughter has 50 per cent chance of receiving one affected gene and one healthy gene (therefore becoming a carrier) or of receiving two healthy genes. It is worth to note that in 30 per cent of newly diagnosed children their haemophilia is caused by a new mutation in the X chromosome, either in the mother or the child, where there is no previous family history of haemophilia. If the altered gene occurs in the mother it can also be passed to other children that she has, as detailed above. In addition, it may have arisen in her mother so her sisters, aunts and cousins may also be carriers.

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases

Question 5.
Write about the inheritance pattern of Haemophilia.
Answer:
Inheritance Pattern of Haemophilia
The following diagrams show how the haemophilia gene can be inherited. It is important to note that in one-third of people with haemophilia, there is no family history of the disorder.
Mother is a carrier

  • 50% chance that each son will have haemophilia.
  • 0% chance that each daughter will be a carrier of the haemophilia gene.

Father has haemophilia

  • All daughters will carry the haemophilia gene.
  • No sons will have haemophilia.

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases 5

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases 6

Mother is a carrier and
Father has haemophilia

  • 50% chance that each son will have haemophilia.
  • 50% chance that each daughter will be a carrier of the haemophilia gene.
  • 50% chance that each daughter will have haemophilia.

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases

Question 6.
What do you mean by Genetic Counselling? State its necessity, component and limitation in relation to Thalassemia.
Genetic Counselling
Genetic counselling is the communication process of providing information and support to individuals and families with a diagnosis and/or risk of occurrence of an inherited disorder. Culturally sensitive genetic counselling, with an emphasis on reproductive issues, is an integral and necessary component of comprehensive care for patients and parents affected by all forms of thalassemia disease and trait. Services should be provided by a licensed genetic counselor in states with licensed legislation and by an ABGC board-certified or board-eligible genetic counselor in all other states.
Genetic counselling is needed :

  • at diagnosis and during adolescence
  • prior to and after any genetic testing
  • prior to pregnancy and/or as early in pregnancy as possible
  • Annual follow-ups are needed to reinforce teaching.

Critical components of genetic counselling include :

obtaining a three-generation genetic family history (pedigree)
assessing risk for thalassemia in family members
identifying risk factors impacting medical management (e.g., family history of other haemoglobin traits or diseases, hereditary haemochromatosis, G6 PD deficiency, inherited thrombophilia, cardiovascular disease or its risk factors, cardiac conduction defects, diabetes, renal disease, ophthalmologic disorders, hearing loss, allergies, ethnicity, consanguinity)
incorporating psychosocial information impacting the family system and relationships (e.g., location of residence, disclosure/nondisclosure of diagnosis, reliable source of emotional/social support)
assisting patients in conveying information about genetic risk to other family members
providing informed consent, pre-and post-counselling for all genetic testing
alpha-globin genotyping: haemoglobin H-Constant Spring and other structural alpha-globin variants, possible modifying effects of alpha-globin deletions/triplications on beta-thalassemia
beta-globin genotyping: beta0/beta+, S, D, E, O, and other structural variants

The limitations of drawing genotype/phenotype correlations include:

  • developmentally appropriate consent/education for minors
  • reproductive genotype post-stem cell transplant or bone marrow transplant
  • the possibility of revealing undisclosed adoption or alternative paternity
  • discussing/facilitating appropriate screening and diagnostic tests for relatives

Genetic Testing

If HLA typing is performed when stem cell transplant or bone marrow transplant is an option, genetic counselling and education is vital due to ethical implications. A genetic counselor should provide initial and ongoing teaching regarding natural history and clinical manifestations; signs and symptoms of disease that warrant immediate medical attention; and available emotional and social support services. Genetic counselors should also provide available resources in collaboration with outreach coordinators and social workers (e.g. research studies, support groups, advocacy organizations, and patient-to-patient or parent-to-parent connections).

Multiple Choice Questions : 1 Mark

Question 1.
Both husband and wife have normal vision though their fathers were colour blind. The probability of their daughter becoming colour blind is
(a) 0%
(b) 25%
(c) 50%
(d) 75%
Answer:
(a) 0%

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases

Question 2.
Which one is a hereditary disease?
(a) cataract
(b) leprosy
(c) blindness
(d) phenylketonuria.
Answer:
(d) phenylketonuria.

Question 3.
Haemophilia is more common in males because it is a
(a) recessive character carried by Y-chromosome
(b) dominant character carried by Y-chromosome
(c) dominant trait carried by X-chromosome
(d) recessive trait carried by X-chromosome.
Answer:
(d) recessive trait carried by X-chromosome.

Question 4.
A colour blind girl is rare because she will be born only when
(a) her mother and maternal grandfather were colour blind
(b) her father and maternal grandfather were colour blind
(c) her mother is colour blind and father has normal vision
(d) parents have normal vision but grandparents were colour blind.
Answer:
(b) her father and maternal grandfather were colour blind

Question 5.
A colour blind mother and normal father would have
(a) colour blind sons and normal/carrier daughters
(b) colour blind sons and daughters
(c) all colour blind
(d) all normal.
Answer:
(a) colour blind sons and normal/carrier daughters

Question 6.
The colour blindness is more likely to occur in males than in females because
(a) the Y-chromosome of males have the genes for distinguishing colours
(b) genes for characters are located on the sex-chromosomes
(c) the trait is dominant in males and recessive in females
(d) none of the above.
Answer:
(b) genes for characters are located on the sex-chromosomes

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases

Question 7.
The genes, which remain confined to differential region of Y-chromosome, are
(a) autosomal genes
(b) holandric genes
(c) completely sex-linked genes
(d) mutant genes.
Answer:
(b) holandric genes

Question 8.
A genetically diseased father (male) marries with a normal female and gives birth to 3 carrier girls and 5 normal sons. It may be which type of genetic disease?
(a) sex-influenced disease
(b) blood group inheritance disease
(c) sex-linked disease
(d) sex-recessive disease.
Answer:
(c) sex-linked disease

Question 9.
A person whose father is colour blind marries a lady whose mother is a daughter of a colour blind man. Their children will be
(a) all sons colour blind
(b) some sons normal and some colour blind
(c) all colour blind
(d) all daughters normal.
Answer:
(d) all daughters normal.

Question 10.
A woman with two genes for haemophilia and one gene for colour blindness on one of the ‘X’ chromosomes marries a normal man. How will the progeny be?
(a) 50% haemophilic colour-blind sons and 50% normal sons
(b) 50% haemophilic daughters (carrier) and 50% colour blind daughters (carrier)
(c) all sons and daughters haemophilic and colour blind
(d) haemophilic and colour blind daughters.
Answer:
(b) 50% haemophilic daughters (carrier) and 50% colour blind daughters (carrier)

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases

Question 11.
Haemophilic man marries a normal woman. Their offsprings will be
(a) all haemophilic
(b) all boys haemophilic
(c) all girls haemophilic
(d) all normal.
Answer:
(d) all normal.

Question 12.
A marriage between normal visioned man and colour blind woman will produce offspring
(a) colour blind sons and 50% carrier daughter
(b) 50% colour blind sons and 50% carrier daughter
(c) normal males and carrier daughters
(d) colour blind sons and carrier daughters.
Answer:
(d) colour blind sons and carrier daughters.

Question 13.
A diseased man marries a normal woman. They get three daughters and five sons. All the daughters were diseased and sons were normal. The gene of this disease is
(a) sex linked dominant
(b) sex linked recessive
(c) sex limited character
(d) autosomal dominant.
Answer:
(a) sex linked dominant

Question 14.
The recessive genes located on X-chromosome humans are always
(a) lethal
(a) sub-lethal
(c) expressed in males
(d) expressed in females.
Answer:
(c) expressed in males

Question 15.
A normal woman, whose father was colour blind is married to a normal man. The sons would be
(a) 75% colour blind
(b) 50% colour blind
(c) all normal
(d) all colour blind.
Answer:
(b) 50% colour blind

Question 16.
A woman with normal vision, but whose father was colour blind, marries a colour blind man. Suppose that the fourth child of this couple was a boy. This boy
(a) may be colour blind or may be of normal vision
(b) must be colour blind
(c) must have normal colour vision
(d) will be partially colour blind since he is heterozygous for the colour blind mutant allele.
Answer:
(a) may be colour blind or may be of normal vision

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases

Question 17.
Which of the following is not a hereditary disease?
(a) cystic fibrosis
(b) thalassemia
(c) haemophilia
(d) cretinism.
Answer:
(d) cretinism.

Question 18.
Haemophilia is more commonly seen in human males than in human females because
(a) a greater proportion of girls die in infancy
(b) this disease is due to a Ylinked recessive mutation
(c) this disease is due to an X-linked recessive mutation
(d) this disease is due to an X-linked dominant mutation.
Answer:
(c) this disease is due to an X-linked recessive mutation

Question 19.
If a colour blind woman marries a normal visioned man, their sons will be
(a) all colour blind
(b) all normal visioned
(c) one-half colour blind and onehalf normal
(d) three-fourths colour blind and one-fourth normal
Answer:
(a) all colour blind

Very Short Answer Type Questions : 1 Mark

Question 1.
When the disease transmitted from one generation to next, the disease known as-
Answer:
Genetical disease.

Question 2.
Give example of one autosomal disease.
Answer:
Thalassemia.

Question 3.
Give two examples of sex linked disease.
Answer:
Haemophilia, Colour blindness.

Question 4.
Which chromosomes are involved in case of thalassemia?
Answer:
16th and 11th autosome.

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases

Question 5.
Write the vision defect for colour blindness.
Answer:
Unable to detect red and green colour.

Question 6.
Which disease is also known as sea of blood?
Answer:
Thalassemia.

Question 7.
Name one disease for which genetic counselling is necessary.
Answer:
Thalassemia.

Question 8.
Excess iron causes diseases.
Answer:
Thalassemia.

Question 9.
Which factor is absent in haemophilia-A?
Answer:
Anti-haemophilia gene globulin (AHG) or factor VIII.

Question 10.
Which factor is absent in haemophilia-B?
Answer:
Christmas factor or factor IX.

WBBSE Class 10 Life Science Solutions Chapter 3B Some Common Genetic Diseases

Question 11.
Which disease is also known as haemoglobinopathy.
Answer:
Thalassemia.

Question 12.
What is the nature of gene for haemophilia?
Answer:
Recessive X chromosome linked.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Detailed explanations in West Bengal Board Class 10 Life Science Book Solutions Chapter 3A Heredity offer valuable context and analysis.

WBBSE Class 10 Life Science Chapter 3A Question Answer – Heredity

Short answer type questions : 2 Marks

Question 1.
Define heredity. Give some common instances of inherited traits in Humans.
Answer:
Heredity may be defined as the transmission of characteristics through generations. Common instances of inherited traits may include, colour of skin, colour of eyes, diseases like, diabetes and conditions like baldness.

Question 2.
Name the two laws of inheritance postulated by Mendel?
Answer:
1. The Law of Segregation
2. The law of Independent Assortment

Question 3.
How do Mendel’s experiments show that traits mav be dominant or recessive?
Answer:
Mendel took one tall pea (TT) plant and one short pea (tt) plant and produced progeny from them. The plants grown from F1 seeds represent first filial (F1) generation. All F1 plants were tall. Then Mendel self-pollinated F1 plants and found that all plants obtained in F2 progeny were not tall. Instead 75% were tall (dominant) while remaining 25% were dwarf (recessive traits). From the experiment, he concluded the F1 progeny is not a true breed but carries both traits. He concluded one trait is dominant over the other, that’s why plants appear tall. It is also called the law of Segregation.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 4.
How do Mendel’s experiments show that traits are inherited independentily?
Answer:
When a pea plant of round green (RRyy) seeds is crossed with a pea plant having wrinkled yellow seeds (rrYY), plants in F1 progeny have round and yellow seeds. However in F2 generation, all traits appear independently. In F2, he got the following result (called dihybrid ratio):
Round Yellow-9
Wrinkled Yellow – 3
Round Green – 3
Wrinkled Green – 1
He concluded alleles controlling different traits are not linked and are inherited independently. It is also known as Law of Independent Assortment.

Question 5.
A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enoagh to tell you which of the traits-blood group A or O – is dominant? Why or why not?
Answer:
No, the information is not sufficient. The blood group is determined by a pair of genes. It might happen the gene responsible for O group in the daughter might be inherited from mother or from father. For example, one possibility is father having AO pair and mother has O O pair of genes.

Question 6.
Define allelomorph.
Answer:
The contrasting pair of genes or alleles constitute an allelomorph. Examples : Tall and dwarf plants, wrinkled and smooth seed coat, white and violet coloured flower.

Question 7.
How is the sex of the child determined in human beings?
Answer:
Normal human somatic cells are diploid. They have 46 chromosomes made up of two sets of 23-one set from each parent. In human diploid cells, there are 22 homologous pairs of autosomes, each with a maternal and a paternal homolog. The 23rd pair, the sex chromosomes, determines whether the person is female (XX) or male (XY). Mother provides only X chromosomes. The sex of the child is determined by the type of chromosome (X or Y) received from male gamete.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 8.
Define Homozygous.
Answer:
An organism that has a pair of identical alleles for a character is said to be homozygous for the gene controlling that character. Homozygous plants “breed true” because all of their gametes contain the same allele-either TT or t t for example.

Question 9.
Define Heterozygous.
Answer:
An organism that has two different alleles for a gene is said to be heterozygous for that gene. Unlike homozygotes, heterozygotes are not true-breeding because they produce gametes with different alleles; for example T t.

Question 10.
A normal pea plant bearing coloured flowers suddenly start producing white flowers. What could be the possible cause?
Answer:
Because of change in genetic sequence (called mutation).

Question 11.
What is mutation?
Answer:
Organisms have evolved in number of ways to protect their DNA from changes. In spite of these mechanisms, however, changes in the DNA occasionally do occur. Any change in the DNA sequence is called a mutation. Mutations can be caused by errors in replication, transcription, cell division, or by external agents. e.g. nuclear radiation can lead to mutation.

Question 12.
Why are human beings who look so different from each other in terms of size, colour and looks, said to belong to the same species?
Answer:
Although human beings look so different from each other in terms of size, colour and looks, but all of them belong to same species (Homo sapiens) because of the following reasons:

  1. All human beings belong to same species and are able to interbreed.
  2. Fossil evidences prove that Archaic Homo sapiens arose in South Africa and moved across continents and developed into distinct races during the ice age.
  3. Fossil evidence shows that humans have not changed much anatomically over the last 200000 years.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 13.
If the sperm bearing Y-chromosome fertilizes the egg, the child born will not be entirely like his father. Why is it so?
Answer:
It is because the other sex chromosome i.e. X-chromosome also has its effect on the fertilized egg.

Question 14.
A normal girl baby receives her X chromosomes from which?
Answer:
Both from her mother and father (One X from father and another X from mother).

Question 15.
List the seven pairs of contrasting characters or traits selected by Mendel while doing experiment on garden sweet pea (Pisum sativum).
Answer:
seven pairs of contrasting characters or traits of garden sweet pea (Pisum sativum):

Traits α Characters Dominant Recessive
1. Form of Seed Round(R) Wrinkle(r)
2. Colour of Seed Coats Coloured(C) White(w)
3. Colour of Cotyledons Yellow(Y) Green(y)
4. Colour of pod Green(G) Yellow(g)
5. Form of Pod Inflated(I) Constricted(c)
6. Position of flower Axial(A) Terminal(a)
7. Height of Plant Tall(T) Dwarf(t)

Question 16.
Define gene. What are its important characteristics?
Answer:
Gene is the unit of inheritance.
1. Genes are found in chromosomes at fixed positions.
2. Each gene is responsible for a providing particular trait to an organism.
3. Chemically gene is a segment of a large polynucleotide molecule called DNA (deoxy-ribonucleic acid) and RNA (ribonucleic acid)

Question 17.
Who was Gregor Mendel and what was his contribution to science?
Answer:
Gregor Mendel was an Austrian monk who first started the study of inheritance in an organized manner. This was the beginning of genetics and therefore, he is called the Father of Genetics.

Question 18.
Which plant did Mendel select for his studies and why?
Answer:
Mendel selected the purebred garden pea plant for his studies. These plants showed many contrasting traits that were easy to track through the generations. These plants are naturally self-pollinated, allowed him to easily get pure lines for his studies.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 19.
What are hybrids? How does one distinguish a monohybrid from a dihybrid cross?
Answer:
The plants produced as a result of cross between two different plants are called hybrids. Monohybrid cross is the cross between two organisms in which the inheritance of only one trait is followed. The crosses that study two traits together are called the dihybrid crosses.

Question 20.
What are phenotype and genotype?
Answer:
Phenotype refers to the externally exhibited trait, whereas genotype refers to the traits developed by genes.

Question 21.
What is the physical basis of heredity? What are alleles?
Answer:
Genes are the physical basis of heredity. The different expressions of the same genes are called alleles.

Question 22.
What are homozygous and heterozygous plants?
Answer:
Homozygous plants are those with similar alleles for a trait. Heterozygous plants have different alleles for a trait.

Question 23.
What are the phenotypic and genotypic ratio in the F2 generation in monohybrid cross?
Answer:
The phenotypic ratio in the F2 generation in monohybrid cross is 3 : 1 and the genotypic ratio is 1 : 2 : 1.

Question 24.
Which are the allelic forms of the gene for height?
Answer:
The two allelic forms of the gene for height are T, the dominant gene for tallness and t, the recessive gene for dwarfness.

Question 25.
What is the law of unit characters?
Answer:
According to the law of unit characters, all the traits are separate entities or units by themselves whose inheritance is controlled by factors, now called genes.

Question 26.
Where are genes present and what are they made up of?
Answer:
Genes are present arranged in a linear manner on the chromosomes. They are made up of series of nucleotides in a particular sequence.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 27.
Which are the two types of chromosomes and what is the difference between them?
Answer:
The two types of chromosomes are : autosomes and sex chromosomes. While autosomes have homologous chromosomes as pairs, sex chromosomes are of two different types – X and Y.

Question 28.
How do Mendel’s experiments show that traits may be dominant or recessive?
Answer:
The trait which appears in all the members of F1 generation and also in 75% numbers of F2 generation obtained by self fertilization of F1 generation is dominant character.
The trait which does not appear in F1 generation but after self-fertilization of F1 generation, reappears in 25% of F2 generation is known as recessive.

Question 29.
How do Mendel’s experiments show that traits are inherited independently?
Answer:
Mendel crossed pure breeding tall plants having round seeds with pure breeding short plants having wrinkled seeds. The plants of F1 generation were all tall with round seeds indicating that the traits of tallness and round seeds were dominant. Self breeding of F1 yielded plants with characters of 9 tall round seeded, 3 tall wrinkled seeded , 3 short round seeded and 1 short wrinkled seeded. Tall wrinkled seeded and short round seeded plants are new combinations which can develop only when the traits are inherited independently.

Question 30.
A man with blood group A marries a woman with blood group O and their daughter has blood group O. Is this information enough to tell you which of the traits- blood group A or O – is dominant? Why or why not?
Answer:
No. This information is not sufficient to determine which of the traits – blood group A or O – is dominant. This is because we do not know about the blood group of all the progeny.
Blood group A can be genotypically AA or AO. Hence, the information is incomplete to draw any such conclusion.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 31.
A study found that children with light-coloured eyes are likely to have parents with light-coloured eyes. On this basis, can we say anything about whether the light eye colour trait is dominant or recessive? Why or why not?
Answer:
This information is not sufficient. For considering a trait as dominant or recessive, we need data of at least three generations. This data is about only two generations.

Question 32.
Outline a project which aims to find the dominant coat colour in dogs.
Answer:
Dogs have a variety of genes that govern coat colour. There are at least eleven identified gene series (A, B, C, D, E, F, G, M, P, S, T) that influence coat colour in dog.

A dog inherits one gene from each of its parents. The dominant gene gets expressed in the phenotype. For example, in the B series, a dog can be genetically black or brown.
Let us assume that one parent is homozygous black (BB), while the other parent is homozygous brown (bb).

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 1

In this case, all the offsprings will be heterozygous (Bb).
Since black (B) is dominant, all the offsprings will be black. However, they will have both B and b alleles.

If such heterozygous pups are crossed, they will produce 25% homozygous black (BB), 50% heterozygous black (Bb) and 25% homozygous brown (bb) offsprings.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 2

Question 33.
How is the equal genetic contribution of male and female parents ensured in the progeny?
Answer:
In human beings, equal genetic contribution of male and female parents is ensured in the progeny through inheritance of equal number of chromosomes from both parents. There are 23 pairs of chromosomes. All human chromosomes are not paired. Out of these 23 pairs, the first 22 pairs are known as autosomes and the remaining one pair is known as sex chromosomes represented as X and Y. Females have a perfect pair of two X sex chromosomes and males have a mismatched pair of one X and one Y sex chromosome.

During the course of reproduction, as fertilization process takes place, the male gamete (haploid) fuses with the female gamete (haploid) resulting in formation of the diploid zygote. The zygote in the progeny receives an equal contribution of genetic material from the parents. Out of 23 pairs of chromosomes in progeny, male parent contributes 22 autosomes and one X or Y chromosome and female parent contributes 22 autosomes and one X chromosome.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 34.
Guineapig having black colour when crossed with guineapig having same colour produced 80 offsprings, out of which 60 were black and 20 were white. Now find out : (a) What is the possible genotype of the guineapigs? (b) Which trait is dominant and which trait is recessive? (c) What is this cross called as and what is its phenotypic ratio?
Answer:
(a) Bb × Bb
(b) Black is dominant and white is recessive.
(c) Monohybrid cross, phenotypic ratio = 3 : 1

Question 35.
Why did Mendel choose pea plant for his experiments?
Answer:
Mendel chose pea plant for his experiments because it is
(a) Easy to grow.
(b) Short lifespan.
(c) Easily distinguishable characters.
(d) Larger size of flower.
(e) Self-pollinated.

Question 36.
What do you mean by genotype?
Answer:
The complete set of genes in an organism’s genome is called genotype.

Question 37.
What do you mean by phenotype?
Answer:
The observable characters in an organism make the phenotype. Phenotype is a result of genotype’s interaction with the environment. Due to this reason, many phenotypes are not inheritable.

Question 38.
What are the possible reasons of pea plants used by Mendel?
Answer:
Pea can be termed as biennial plant, i.e. two generations of a pea plant can grow in a given year. This means that Mendel could get enough time to observe a larger number of generations.

  • Many easily identifiable and contrasting characters are present in pea plants.
  • Cross pollination can be easily induced in pea plants.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 39.
State the Law of Independent Assortment.
Answer:
Mendel’s Second Law : Law of Independent Assortment : Alleles of different characters separate independent from each other during gamete formation.

Question 40.
What do you mean by sex determination in human?
Answer:
Somatic cells in human beings contain 23 pairs of chromosomes. Out of them the 23 rd pair is composed of different types of chromosomes, which are named as X and Y chromosomes. The 23 rd pair contains one X and one Y chromosome in a male. On the other hand, the 23rd pair in a female contains X chromosomes. This means that all the eggs would have X chromosome as the 23 rd chromosome, while a sperm may have either X or Y chromosome as the 23 rd chromosome. When a sperm with X chromosome fertilizes the egg, the resulting zygote would develop into a female child. When a sperm with Y chromosome fertilizes the egg, the resulting zygote would develop into a male child.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 3

Question 41.
What are the components of chromosome?
Answer:
Each chromosome is double or replicated and consists of two thread like structures called chromatids attached to each other at centromere. Each chromatid has a number of dense areas arranged linearly, which are known as chromomeres. They represent the areas of active genes. Internally each chromosome has two spirally coiled threads called chromonemata embedded in matrix. These are DNA molecules.

Question 42.
How is sex determined in human beings?
Answer:
Sex is determined at the time of fertilization and the two sexes are produced in approximately equal numbers. A human male has XY sex-chromosomes and produces two types of sperms (heterogametic). Either with X-chromosome or with Ychromosome.
A human female has XX sex-chromosomes and produces ova of one type (homogametic) all with X-chromosomes.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 4

Long Answer Type Questions : 5 Marks

Question 1.
What do mean by heredity and variation?
Answer:
The offsprings of all the organisms (plants and animals) resemble their parents in several aspects. This is only due to the phenomenon of heredity. Hereditary literally means “tendency of like begets like” i.e. all living organisms tend to produce offsprings like themselves.

Hereditary may be defined as the transmission of characters from one generation to successive generations or from parents to their offsprings. Thus, heredity is the cause of similarities between the offsprings, so that the individuals of the same parents resemble each other in many aspects. Heredity involves the transfer of genetic characters from parents to the offsprings via the egg and sperm. These transferable characters are called “hereditary characters”.

Variation : Though offsprings receive all the characters from their parents, they are not exactly alike. Differences are found even between the offsprings of the same parents. It is difficult to find out the identical individuals. The progeny differs not only in itself but also with the parents. These differences are called variations. Thus, variations may be defined as the visible differences between the parents and the offsprings or between the offsprings of the same parents.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 2.
Describe different types of Variation.
Answer:
The variations may be classified into two types:
i. Hereditary variation : The variations which arise as a result of any change in the structure and function of the gene and are inherited from one generation to another are called hereditary variation.

ii. Environmental variations : Two individuals with the same genotype may become different in phenotype when they come in contact with different conditions of food, temperature, light, humidity and other external factors. Such differences among organisms of similar heredity are known as environmental variation. These are not heritable.

Based on the type of cells, variation is classified into two types.
i. Somatic variation : The variation which occurs in somatic cells is called somatic variation. It is generally insignificant, because it is not inherited from parents. It is acquired by the organisms during their own lifetime and is lost with death. Hence, it is also called acquired variation.

ii. Germinal variation : The variation which affects the germinal or reproductive cells is called germinal variation. It is heritable and genetically significant. It provides raw materials for evolution.

Based on the degree of differences, variation is classified into two types:
i. Continuous variation : Small and indistinct variations are called continuous variation.
(a) These are fluctuating with environmental conditions.
(b) These are non-heritable.
(c) They have no role in evolution.
(d) They are most common and occur in all organisms.

ii. Discontinuous variation : Large, distinct and sudden variations are called discontinuous variation.
(a) These are relatively unaffected by environmental conditions.
(b) These are heritable.
(c) They provide raw materials for evolution on which selection is based.
(d) They are not common and appear suddenly.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 3.
How is the sex of the child determined in human beings?
Answer:
WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 5
In human beings, the females have two X chromosomes and the males have one X and one Y chromosome. Therefore, the females are XX and the males are XY.

The gametes, as we know, receive half of the chromosomes. The male gametes have 22 autosomes and either X or Y sex chromosome.
Type of male gametes : 22+X Or 22+Y.
However, since the females have XX sex chromosomes, their gametes can only have X sex chromosome.
Type of female gamete : 22+X
Thus, the mother provides only X chromosomes. The sex of the baby is determined by the type of male gamete (X or Y) that fuses with the X chromosome of the female.

Question 4.
Explain Mendel’s experiment of monohybrid cross on pea plant?
Answer:
Mendel selected for his experiment two types of bisexual sweet pea plants-one tall pure plant and one dwarf pure plant. Each tall and dwarf plants are made true breading by self pollination. These true breeding tall and dwarf plants are called parent plants and the generation is known as parental generation. First and second filial generations are arranged through cross pollination and fertilization among the hybrid plants.

T = gene for tallness of pea plant
t = gene for dwarfness of pea plant
T, t = genes are alleles for each other
T = gene is dominant over t gene

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 6

Cross pollination performed among all hybrid plants produced during first filial generation.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 7

F2 generation is shown through checker board or Punnet square

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 8

Phenotype : Tall pea plant : Dwarf pea plant = 3 : 1
Genotype : TT : Tt : tt = 1 : 2 : 1
Explantion : In monohybrid cross, hybrid pea plant is produced at the first filial generation. There is no dwarf plant in this generation. But at the second filial generation, there is 25% dwarf plant. Dwarfness character is again developed at the second generation.

When in a homologous chromosome pair a pair of contrasting characters are present in a cross, the contrasting characters never get blended but they separate out during gamete formation in the next generation. This is known as Law of Segregation or First Law of Mendel.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 5.
Explain Mendel’s experiment on dihybrid cross with the help of two pairs of characters in pea plant.
Answer:
Mendel selected bisexual pea plant for dihybrid cross. He has separated one pea plant with pure, smooth, yellow seeded pea plant and one pea plant with pure, winkle, green needed pea plant.
R = gene for smooth seeded character.
r = gene for wrinkle seeded character.
Y = gene for yellow coloured seed.
y = gene for green coloured seed.
R, r are both alleles and R gene is dominant over r gene.
Y, y are both alleles and Y gene is dominant over y gene.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 9

Phenotype : 100% pea plant with smooth, yellow coloured seed.
Genotype : Hybrid plant (Heterozygous)
Cross pollination is performed among all hybrid plants produced during first filial generation.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 10

F2 generation is shown through checker board or Punnet square

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 11
WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 12.1

Explanation : In the first filial generation, hybrid plants with smooth yellow seeds are developed. Four types of gamete are produced in this generation. Any gamete is fused with any other type of gametes to form sixteen types of plants.

When two or more pairs of contrasting characteristics present in a pair of homologous chromosomes are involved in a cross, segregation of one pair of characters is independent that of the others and one can fuse any of the others independently. This is known as Law of Independent Assortment or, Second Law of Mendel.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 6.
Explanation : Mendel’s monohybrid cross in guineapig.
Answer:
Pure black coloured hair and pure white coloured hair are selected, when monohybrid cross is experimented in case of guineapig. Gene for black colour hair (B) is dominant over gene for white colour hairs (b). Both B, b genes are both allele.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 13

Phenotype : 100 % guineapig with black coloured hair.
Genotype : Hybrid guineapig (Heterozygous)
Sexual reproduction is performed among all hybrids produced during first filial generation.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 14

F2 generation is shown through the checker board or punnet square

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 15

Phenotype : Black coloured hair : White coloured hair
= 3 : 1
Genotype : BB : Bb bb
= 1 : 2 : 1

Explanation : Recessive characters-white coloured hair of guineapig remain suppressed in the first filial generation. Dominant character is expressed. But in the second generation, the recessive character is segregated separately. This is the Law of Segregation.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 7.
Explain dihybrid cross with the help of two pairs of characters present in guinea pig.
Answer:
Two type of guinea pigs are selected for dihybrid cross guinea pig with rough (RR) and black coloured (BB) hair is crossed with smoth (rr), white coloured (bb) guinea pig.
B = gene for black coloured hair.
b = gene for white coloured hair.
B, b are both alleles; B gene is dominant over b gene.
R = gene for rough hair
r = gene for smooth hair.
R, r are both alleles; R gene is dominant over r gene.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 16

Phenotype : 100% rough, black coloured guinea pig.
Genotype : Hybrid guinea pig (Heterozygous)
Sexual reproduction is performed among all hybrid guinea pigs produced in the first filial generation.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 17

F2 generation is shown through checker board or Punnet square

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 18

Explanation : In the first filial generation, hybrid guineapigs with rough, black colour are producted. Gametes developed from the male and female hybrid guineapigs. Any gamete are fused with other type of other gametes to form sixteen types of guineapigs. This is the Law of Independent Assortment or Second Law of Mendel.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 8.
What is incomplete dominance? Explain it with an example.
Answer:
Incomplete dominance : It is the deviation of Mendel’s law of heredity. When pure homozygous dominant alleles are crossed with pure homozygous recessive allele, the heterozygous hybrids are produced during first generation, with a few phenotypic character. The phenotypic character is not the dominant but near to the dominant character. It is known as incomplete dominance, e.g. when homozygous dominants red flowered 4 o’ clock plant (Mirabilis jalapa) is crossed with homozygous recessive flower plant, the heterozygous plant with pink colour flowers are developed in the first filial generation. The homozygous recessive characters become reappeared in the next generation without blending each other.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 19

F2 generation is shown through checker board or Punnet square. Cross pollination among the plants of F1 generation develops two type of gametes

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 20

Phenotype : Red flower : Pink flower : White Flower
= 1 : 2 : 1
Genotype : RR : Rr : rr = 1 : 2 : 1
Explanation : Dominant character i.e. red colour is not expressed in the F1 generation but new character, the pink coloured flower is developed. The dominant and recessive character with the new character are again expressed in the F1 generation. It is the deviation of Mendel’s law. The phenotype and genotypic ratios are not same as Mendel’s experiment.

Multiple Choice Questions : 1 Mark

Question 1.
Alternative form of gene are called ________.
(a) Loci
(b) Multiples
(c) Chromosomes
(d) Alleles
Answer:
(d) Alleles

Question 2.
Heredity or inheritance of specific traits became clearer due to
(a) Lamarck’s theory
(b) Mendel worked on garden peas
(c) Darwinism
(d) Neo-Darwinism
Answer:
(b) Mendel worked on garden peas

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 3.
Which of these is homozygous recessive?
(a) Ss
(b) ss
(c) SS
(d) s
Answer:
(b) ss

Question 4.
What will be the genotypic ratio of the cross between Rr and rr ?
(a) 1 : 2 : 1
(b) 3 : 1
(c) 1 : 1
(d) 1 : 1 : 1
Answer:
(c) 1 : 1

Question 5.
What will be the genotypic ratio of the cross between Rr and Rr ?
(a) 1 : 1
(b) 3 : 1
(c) 1 : 2 : 1
(d) 1 : 1 : 1
Answer:
(c) 1 : 2 : 1

Question 6.
The offspring resulting from a cross between two pure homozygous recessives would be ________.
(a) 50% homozygous recessive and 50% homozygous dominant
(b) 75% homozygous recessive and 25% heterozygous dominant
(c) 75% homozygous recessive and 25% homozygous dominant
(d) 100% homozygous recessive
Answer:
(d) 100% homozygous recessive

Question 7.
On what cellular structures are genes in eukaryotes carried?
(a) Endoplasmic reticulum
(b) Nuclear membrane
(c) Chromosomes
(d) Mitochondria
Answer:
(c) Chromosomes

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 8.
In man the chromosome number is 46. How many chromosomes are present in man’s muscle cells?
(a) 23
(b) 46
(c) 96
(d) variable
Answer:
(b) 46

Question 9.
The component of a chromosome that controls heredity is ________.
(a) proteins
(b) histones
(c) DNA
(d) RNA
Answer:
(c) DNA

Question 10.
Number of chromosomes in a human male is ________.
(a) 23
(b) 23 pairs
(c) 22 pairs +XY
(d) 22 pairs
Answer:
(c) 22 pairs +XY

Question 11.
In dogs the chromosome number is 78 . How many chromosomes are present in the dog sperm?
(a) 78
(b) 39
(c) cannot determine from this information
(d) variable
Answer:
(a) 78

Question 12.
Homologous chromosomes which are similar in both the sexes are called ________.
(a) sex chromosomes
(b) autosomes
(c) allosomes
(d) androsomes
Answer:
(b) autosomes

Question 13.
How many pairs of contrasting characters of pea were selected by Mendel for cross fertilization?
(a) six
(b) seven
(c) five
(d) twelve
Answer:
(b) seven

Question 14.
RrYy yield gamete of which type?
(a) four
(b) five
(c) two
(d) six
Answer:
(a) four

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 15.
Genetic information is carried out by long chain of molecules made up of :
(a) phosphates
(b) MRNA
(c) nucleotides
(d) none of these
Answer:
(c) nucleotides

Question 16.
The character which predominates and clearly seen in F1 Generation is said to be :
(a) Dominant
(b) Recessive
(c) Inherited
(d) None of these
Answer:
(a) Dominant

Question 17.
The Genotypic Ratio of a mono hybrid cross in F2 Generation is :
(a) 3 : 1
(b) 2 : 1
(c) 1 : 2 : 1
(d) 1 : 3
Answer:
(c) 1 : 2 : 1

Question 18.
When two individuals are similar in external appearance but different in their genetic makeup they are called as?
(a) Allele
(b) Dominant
(c) Homozygous
(d) Heterozygous
Answer:
(d) Heterozygous

Question 19.
The phenotypic ratio of a di-hybrid cross in F2 Generation is :
(a) 1 : 4 : 6 : 1
(a) 1 : 2 : 1
(c) 9 : 3 : 3 : 1
(d) 2 : 1 : 4
Answer:
(c) 9 : 3 : 3 : 1

Question 20.
A couple with blood types A and B may have children with blood types
(a) A & B only
(b) AB
(c) A, B, AB & O
(d) A and O
Answer:
(a) A & B only

Question 21.
A gamete contains :
(a) only one allele of a gene
(b) two alleles
(c) three alleles
(c) none of these
Answer:
(a) only one allele of a gene

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 22.
A character which is expressed in a hybrid is called
(a) Dominant
(b) Recessive
(c) Co-dominant
(d) Epistatic
Answer:
(a) Dominant

Question 23.
How many different kinds of gametes will be produced by a plant having the genotype AABbCC?
(a) Three
(b) Four
(c) Nine
(d) Two
Answer:
(d) Two

Question 24.
If a colour blind woman marries a normal visioned man, their sons will be
(a) Three-fourths colour blind and one-fourth normal
(b) One-half colour blind and one-half normal
(c) All normal visioned
(d) All colour blind
Answer:
(d) All colour blind

Question 25.
The term Genetics first coined by
(a) Mendel
(b) Darwin
(c) Purret
(d) Bateson
Answer:
(d) Bateson

Question 26.
Genes for cytoplasmic male sterility in plants are generally located in:
(a) Nuclear-genome
(b) Mitochondrial genome
(c) Chloroplast genome
(d) Cytosol
Answer:
(b) Mitochondrial genome

Question 27.
A polygenic trait is controlled by 3 genes A, B and C. In a cross A aBbCc × AaBbCc, the phenotypic ratio of the “offsprings was observed as 1 : 6 : x : 20 : x : 6 : 1. What is the possible value of x ?
(a) 3
(b) 9
(c) 15
(d) 25
Answer:
(c) 15

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 28.
A common test to find the genotype of a hybrid is by:
(a) Crossing of one F1 progeny with male parent
(b) Crossing of one F2 progeny with male parent
(c) Crossing of one F2 progeny with female parent
(d) Studying the sexual behaviour of F1 progenies
Answer:
(a) Crossing of one F1 progeny with male parent

Question 29.
A man and a women, who do not show any apparent signs of a certain inherited disease, have seven children (2 daughter and 5 sons). Three of the sons suffer from the given disease but none of the daughters are affected. Which of the following mode of inheritance do you suggest for this disease?
(a) Sex-linked recessive
(b) Autosomal dominant
(c) Sex-limited recessive
(d) Sex-linked dominant
Answer:
(a) Sex-linked recessive

Question 30.
A person with unknown blood group under ABO system, has suffered much blood loss in an accident and needs immediate blood transfusion. His one friend who has a valid certificate of his own blood type, offers for blood donation without delay. What would have been the type of blood group of the donor friend?
(a) Type A
(b) Type B
(c) Type AB
(d) Type O
Answer:
(d) Type O

Question 31.
A self-fertilizing trihybrid plant forms :
(a) 8 different gametes and 32 different zygotes
(b) 8 different gametes and 16 different zygotes
(c) 8 different gametes and 64 different zygotes
(d) 4 different gametes and 16 different zygotes
Answer:
(c) 8 different gametes and 64 different zygotes

Question 32.
Assertion : An organism with lethal mutation may not even develop beyond the zygote stage.
Reason : All types of gene mutations are lethal.
(a) Both assertion and reason are true and the reason is the correct explanation of the assertion.
(b) Both assertion and reason are true but the reason is not the correct explanation of the assertion.
(c) Assertion is true statement but reason is false.
(d) Both assertion and reason are false.
Answer:
(c) Assertion is true statement but reason is false.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 33.
Excessive growth of hair on the pinna is a feature found only in males because
(a) The gene responsible for the character is recessive in females and dominant only in males
(b) The character is induced in males as males produce testosterone
(c) The female sex hormone estrogen suppresses the character in females
(d) The gene responsible for the character is present on the Y chromosome only
Answer:
(d) The gene responsible for the character is present on the Y chromosome only

Question 34.
Given below is a pedigree chart of a family with five children. It shows the inheritance of attached ear lobes as opposed to the free ones. The squares represent the male individuals and circles the female individuals. Which one of the following conclusions drawn is correct?
WBBSE Class 10 Life Science Solutions Chapter 3A Heredity 21
(a) The parents are homozygous recessive
(b) The trait is Y-linked
(c) The parents are homozygous dominant
(d) The parents are heterozygous
Answer:
(d) The parents are heterozygous

Question 35.
Grain colour in wheat is determined by three pairs of polygene. Following the cross AABBCC (dark colour) × aabbcc (light colour), in F2-generation what proportion of the progeny is likely to resemble either parent
(a) One fourth
(b) Less than 5 percent
(c) One third
(d) None of these
Answer:
(b) Less than 5 percent

Question 36.
Haemophilia is more commonly seen in human males than in human females because
(a) This disease is due to a Y-linked recessive mutation
(b) This disease is due to an X-linked recessive mutation
(c) This disease is due to an X-linked dominant
(d) A greater proportion of girls die in infancy
Answer:
(b) This disease is due to an X-linked recessive mutation

Question 37.
Gregor Mendel proposed which of the following concepts?
I. The law of independent assortment
III. The theory of evolution
II. The law of segregation
IV. Natural selection
(a) I, II and IV
(b) I and II
(c) I, II, III and IV
(d) II, III and IV
Answer:
(b) I and II

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 38.
Hemophilia is a trait carried by the mother and passed to her sons. The allele for hemophilia, therefore,
(a) is carried on one of the ‘mother’s autosomal chromosomes.
(b) can be carried on the X or Y chromosome.
(c) is on the X chromosome and can only be inherited by the son if the mother is a carrier (heterozygous).
(d) is carried in the mitochondrial genome because sons inherit this allele from their mothers.
Answer:
(c) is on the X chromosome and can only be inherited by the son if the mother is a carrier (heterozygous).

Question 39.
Which of the following statements is/are true?
I. A human with two X chromosomes and a Y chromosome is male.
II. A human with one X chromosome is female.
III. All children of a colour blind woman and a man who is not colour blind will carry a colour blind allele.
(a) I and II
(b) II
(c) II and III
(d) I, II and III
Answer:
(a) I and II

Question 40.
True-breeding plants show what phenotype?
(a) They have no mutations
(b) They result from a monohybrid cross
(c) They produce the same offspring when crossed for many generations.
(a) They result from a dihybrid cross.
Answer:
(c) They produce the same offspring when crossed for many generations.

Very Short Answer Type Questions : 1 Mark

Question 1.
Name the plant on which Mendel performed his experiments?
Answer:
Garden Pea (Pisum sativum.)

Question 2.
Define a gene?
Answer:
Gene is a small segment of DNA on a chromosome occupying specific position in which is a hereditary determinant or unit of a biological function.

Question 3.
Write the expanded form of DNA?
Answer:
DNA – Deoxyribonucleic Acid.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 4.
Define : Heredity.
Answer:
Heredity is defined as ‘the transmission of characters from parents to the offsprings’.

Question 5.
Give the meaning of variation from biological point of view.
Answer:
From biological point of view, variation means some individual difference(diversity) in the organism of one generation from that of the parental generation.

Question 6.
What is Genetics?
Answer:
Genetics is the branch of biology which studies important aspects of heredity and variation.

Question 7.
What did Mendel propose ?
Answer:
Mendel proposed that for regulation of every character(in an organism) there is a pair of factors.

Question 8.
What happens when a cross is made between two pea plants ?
Answer:
When a cross is made between two pea plants showing different characters, only one character is seen in the offsprings but the successive generations show both characters.

Question 9.
Who showed that gene is a part of chromosome ?
Answer:
T. Boveri and W.S. Sutton showed that gene is a part of chromosome.

Question 10.
Where are genes located ?
Answer:
Genes are located on chromosomes.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 11.
Of what is a gene the segment ?
Answer:
A gene is a segment of DNA(deoxyribonucleic acid).

Question 12.
What is meant by homologous chromosomes?
Answer:
Homologous chromosomes are chromosomes occurring in pair and the members of a pair of chromosomes are exactly identical.

Question 13.
What is the difference between prokaryotic and eukaryotic chromosomes ?
Answer:
In prokaryotes only one chromosome occurs whereas in eukaryotes many chromosomes occur in pairs.

Question 14.
What is the difference in number of chromosomes of somatic cells and reproductive cells?
Answer:
The number of chromosomes in reproductive cells is half(haploid) of that in somatic cells(diploid).

Question 15.
When do chromosomes get separated from each other ?
Answer:
Chromosomes get separated from each other during metaphase of cell division.

WBBSE Class 10 Life Science Solutions Chapter 3A Heredity

Question 16.
What do you study in Genetics ?
Answer:
We study important aspects of heredity and variation in Genetics.

WBBSE Class 10 Life Science Solutions Chapter 2D Growth and Development

Detailed explanations in West Bengal Board Class 10 Life Science Book Solutions Chapter 2D Growth and Development offer valuable context and analysis.

WBBSE Class 10 Life Science Chapter 2D Question Answer – Growth and Development

Short Snswer Type Questions : 2 Marks

Question 1.
Define growth.
Answer:
Growth may be defined as an irreversible permanent increase in size, volume or mass of a cell or organ or whole organism accompanied by an increase in dry weight.

Question 2.
Differentiate between differentiation, dedifferentiation and redifferentiation of cells.
Answer:
Differentiation is a process during which cells undergoes structural changes in the cell wall and protoplasm. An undividable differentiated cell sometimes regains the power of division. This process is called dedifferentiation. The new cells produced by dedifferentiated cells again loose the power of division and become a part of permanent tissue. This process is called redifferentiation.

WBBSE Class 10 Life Science Solutions Chapter 2D Growth and Development

Question 3.
What is a sigmoid curve?
Answer:
It is an ‘S’ shaped curve obtained when growth is plotted against time. It is also called growth curve.

Question 4.
Name the phases of growth curve.
Answer:
The four phases of growth curve are lag phase, log phase, diminishing phase and steady phase.

Question 5.
What are phytohormones?
Answer:
Plant hormones are chemically diverse substances produced in minute quantity and they involve in most of the plant cell activities. They regulate the growth process both by promoting and inhibiting growth. They are produced in certain parts of the plant and transported to other tissues where their action is shown.

Question 6.
Name growth promoters and growth inhibitors. List their functions. Answer:
Answer:
WBBSE Class 10 Life Science Solutions Chapter 2D Growth and Development 1

Question 7.
What is photoperiodism?
Answer:
This flowering response in plants to the relative length of light and dark periods within a 24 hour cycle is called ‘Photoperiodism’.

WBBSE Class 10 Life Science Solutions Chapter 2D Growth and Development

Question 8.
Classify the plants based on their photoperiodic response.
Answer:
There are three types of plant based on their photoperiodic response. They are long day plants, short day plants and day neutral plants.

Question 9.
Define vernalization.
Answer:
Some biennials like carrot, cabbage and sugar beet need a cold/low temperature treatment before they flower. This is called ‘vernalization’.

Long Answer Type Questions : 5 Marks

Question 1.
Define growth. Explain types of growth in plants.
Answer:
Grow th may be defined as an irreversible permanent increase in size, volume or mass of a cell or organ or whole organism accompanied by an increase in dry weight.
Types of growth :
Primary and secondary growth: The mitotic division of meristematic cells present at the root and shoot apex increases the length of the plant body. This is called the primary growth. The secondary meristem increases the diameter of the plant body and it is called the secondary growth.
Life Sc. Comp. (x) : 8

Unlimited Growth: The root and the shoot system of plants grow continuously from germination stage to the death or throughout the life span of the plant. It is called ‘Unlimited’ or ‘indeterminate’ type of growth.
Limited growth: The leaves, fruits and flowers stop growing after attaining certain size. This is called ‘limited’ or ‘determinate’ type of growth.

WBBSE Class 10 Life Science Solutions Chapter 2D Growth and Development 2

Vegetative growth : The earlier growth of plant producing leaves, stem and branches without flowers is called Vegetative growth/ phase.
Reproductive growth : After the vegetative growth, plants produce flowers which is the reproductive part of the plant. This is called reproductive growth/phase.

WBBSE Class 10 Life Science Solutions Chapter 2D Growth and Development

Question 2.
What do you mean by growth curve? Explain with a graph.
Answer:
WBBSE Class 10 Life Science Solutions Chapter 2D Growth and Development 3
Growth curve : It is an ‘S’ shaped curve obtained when we plot growth against time. It is also called ‘sigmoid ‘curve. This curve mainly shows four phases of growth- (i) initial slow growth (Lag phase), (ii) the rapid period of growth (log phase/grand period of growth/ exponential phase) where maximum growth is seen in a short period and (iii) The diminishing phase where growth will be slow and (iv) Stationary / steady phase where finally growth stops.

Question 3.
What do you mean by differentiation, dedifferentiation and redifferentiation of cells ?
Answer:
The three phases of cell growth are cell division, cell enlargement and cell differentiation. The first two stages increase the size of the plant cell while the 3rd stage brings maturity to the cells. Differentiation, is a process during which cells undergoes structural changes in the cell wall and protoplasm (Fig). A differentiated cell cannot divide.

WBBSE Class 10 Life Science Solutions Chapter 2D Growth and Development 4

An undividable differentiated cell sometimes regains the power of division. This process is called dedifferentiation. Dedifferentiation is a common process in plants during secondary growth and in wound healing mechanisms. A dedifferentiated cell can divide and produce new cells. Thus produced new cells again loose the power of division and become a part of permanent tissue. This process is called “redifferentiation”. Tumour cells form good example for redifferentiated cells.

WBBSE Class 10 Life Science Solutions Chapter 2D Growth and Development

Question 4.
What is development? What are plant hormones regulating development.
Answer:
Development : Development is defined as sum total of growth and differentiation. Development is governed by both environmental and internal factors. One of the internal factors that regulate growth and development is ‘plant hormones’.
Plant hormones/ Phytohormones/ Growth regulators : Plant hormones are chemically diverse substances produced in minute quantity and they involve in most of the plant cell activities. They regulate the growth process both by promoting and inhibiting growth. They are produced in certain parts of the plant and transported to other tissues where their action is shown. Based on their promotory and inhibitory activity they have been classified into growth promoters and growth inhibitors.
e.g. Growth Promoters : Auxins
Growth inhibitors : Ethylene.

Question 5.
What do you mean by human development?
Answer:
Human development is a lifelong process of physical, behavioral, cognitive, and emotional growth and change. In the early stages of life-from babyhood to childhood, childhood to adolescence, and adolescence to adulthood-enormous changes take place. Throughout the process, each person develops attitudes and values that guide choices, relationships, and understanding.

Sexuality is also a lifelong process. Infants, children, teens, and adults are sexual beings. Just as it is important to enhance a child’s physical, emotional, and cognitive growth, so it is important to lay foundations for a child’s sexual growth. Adults have a responsibility to help young people understand and accept their envolving sexuality. Read on to learn more about human growth and development.

  • Infants & Toddlers-Ages 0 to 3
  • Preschoolers-Ages 4 to 5
  • Grade Schoolers -Ages 6 to 8
  • Preteens-Ages 9 to 12
  • Teens-Ages 13 to 17 young
  • Adults-Ages 18 and Over

WBBSE Class 10 Life Science Solutions Chapter 2D Growth and Development

Question 6.
Describe different stages of growth in human.
Answer:
Different stages of growth.

  • There are stages of growth that a person goes through.
  • Infancy
  • Early childhood
  • Childhood
  • Late childhood
  • Adolescence
  • Adulthood

Infancy :

    • Infancy : from birth to 18th months
  • Physical changes :
    • Bones are still soft and flexible
    • Learns to sit, to crawl, to stand and to sit.
  • Mental changes :
    • Gets what it needs by crying
    • Can recognize parents and siblings

Early childhood :

    • 18th month to 3 years
  • Physical changes :
    • Learns to walk and to talk
    • Arms and legs get longer
  • Mental changes :
    • Learns to use the toilet
  • Social changes :
    • Not ready to share or to play interactively with others

Childhood :

    • 3 to 6 years
  • Physical changes :
    • Begins to lose milk teeth
  • Mental changes :
    • Learns to behave while in a group
  • Social changes
    • Learns to play interactively and to make friends
    • Emotional changes
    • Physical contact becomes less frequent

Late childhood :

  • Physical changes :
    • Appetite increases
    • Process of sexual maturity begins
  • Mental changes :
    • Acquires high-level thinking skills
  • Emotional changes :
    • Self-centeredness lessens
    • Choose friends of the same sex

Adolescence :

    • The period of gradual change from childhood to adulthood the time of life between childhood and adulthood.
  • Physical changes :
    • Rapid physical growth
    • Body starts to look more like an adult
  • Mental changes :
    • Begins to question oneself
    • Starts to enjoy participating in adult conversation

Adulthood :

    • Period from the twenties onward
  • Physical changes :
    • A slow weakening of the five senses
    • A slow loss of calcium in the bones
    • Bones become brittle
    • Stiffening of the joints and weakening of the muscles
  • Emotional changes :
    • Feeling very quickly from being happy to being lonely
    • Sees people as having needs like his
    • Social changes
    • Friends’ views become more important

Multiple Choice Questions : 1 Mark

Question 1.
The phase of growth which is the first phase and represents lag phase of growth curve is
(a) Formative phase
(b) Cell enlargement phase
(c) Maturation phase
(d) Stationary phase
Answer:
(a) Formative phase

WBBSE Class 10 Life Science Solutions Chapter 2D Growth and Development

Question 2.
The lateral meristem in plant is responsible for
(a) Primary growth
(b) Secondary growth
(c) Exponential growth
(d) Growth in elongation
Answer:
(b) Secondary growth

Question 3.
Plant growth can be measured by
(a) Horizontal microscope
(b) Crescograph
(c) Auxanometer
(d) All of these
Answer:
(d) All of these

Question 4.
The rate of growth is called as
(a) Growing index
(b) Increasing index
(c) Effective index
(d) Efficiency index
Answer:
(d) Efficiency index

Question 5.
Which of the following shows secondary growth?
(a) Monocots
(b) Dicots
(c) Ferns
(d) Mosses
Answer:
(b) Dicots

Question 6.
Growth can be measured in terms of
(a) Fresh or fry weight increase
(b) Increase in girth of stem
(c) Increase is surface area of leaf
(d) All of these
Answer:
(d) All of these

WBBSE Class 10 Life Science Solutions Chapter 2D Growth and Development

Question 7.
Auxins were first isolated from the plants by
(a) Darwin
(b) F.W.Wend
(c) Boysen-Jensen
(d) Sachs
Answer:
(b) F.W.Wend

Question 8.
Apical dominance is due to
(a) Abcisic acid
(b) Gibberelic acid
(c) Auxin
(d) Cytokinin
Answer:
(c) Auxin

Question 9.
Hormone related with cell divisions is
(a) NAA
(b) IAA
(c) Cytokinin
(d) GAS
Answer:
(c) Cytokinin

Question 10.
Which of the following hormone is mainly concerned with root Initiation?
(a) Kinetin
(b) GAB
(c) IAA
(d) ABA
Answer:
(c) IAA

Question 11.
Primary precursor of IAA is
(a) Methionine
(b) Adenine
(c) Tryptophan
(d) Alanine
Answer:
(c) Tryptophan

WBBSE Class 10 Life Science Solutions Chapter 2D Growth and Development

Question 12.
Avena-Curvature biossay was conducted by
(a) Went
(b) Pal
(c) Darwin
(d) Boysen-Jenson
Answer:
(a) Went

Question 13.
Which one of the following groups comprises synthetic auxin?
(a) 2, 4-D; 2, 4, 5,-T and Zeatin
(b) Gibberellic acid, ABA, IAA
(c) 2, 4-D, 2, 4, 5-T and NAA
(d) Cytokinin, IAA, IBA
Answer:
(c) 2, 4-D, 2, 4, 5-T and NAA

Very short answer type questions : 1 Mark

Question 1.
Define growth, differentiation, development, dedifferentiation, redifferentiation, determinate growth, meristem and growth rate.
Answer:
Growth : A permanent and irreversible increase in size of an organ or its part or even of an individual cell is called growth.
Differentiation : The process which leads to maturation of cells is called differentiation. During differentiation, a few or major changes happen in protoplasm and cell walls of the cells.
Development : All the changes which an organism goes through during its life cycle become parts of development. In case of a flowering plant, right from seed germination to seed bearing, each stage is a part of the development process.
Dedifferentiation : A differentiated cell can regain its capacity for cell division under certain conditions. This phenomenon is called dedifferentiation. Formation of interfascicular cambium and cork cambium from fully differentiated parenchyma cells is an example of dedifferentiation.
Redifferentiation : A dedifferentiated plant cell once again loses its capacity to divide and becomes mature. This phenomenon is called redifferentiation.
Determinate Growth : When growth stops after a certain phase, this type of growth is called determinate growth.
Meristem : The plant tissue which has the ability to divide is called meristematic tissue. The region with such tissues is called meristem.
Growth Rate : The increased growth per unit time is called growth rate.

WBBSE Class 10 Life Science Solutions Chapter 2D Growth and Development

Question 2.
Why is not any one parameter good enough to demonstrate growth throughout the life of a flowering plant?
Answer:
Different parts of a plant grow in different ways. There could be one way of measuring growth for a stem, while there could be a different way of measuring growth in a leaf. Different plants can be of different sizes and measurement of growth should be done in different ways in them. Hence, there cannot be one parameter to measure growth in different plants or even at different stages of the life cycle of a flowering plant.

WBBSE Class 10 Life Science Solutions Chapter 2C Sexual Reproduction in Flowering Plants

Detailed explanations in West Bengal Board Class 10 Life Science Book Solutions Chapter 2C Sexual Reproduction in Flowering Plants offer valuable context and analysis.

WBBSE Class 10 Life Science Chapter 2C Question Answer – Sexual Reproduction in Flowering Plants

Short Answer Type Questions : 2 Marks

Question 1.
Define asexual and sexual reproduction.
Answer:
Asexual Reproduction : It is a type of multiplication in which a young one is formed from a specialized or unspecialized part of a parent without the formation and fusion of sex cells, gametes.
Sexual Reproduction : It is a type of reproduction which takes place by the formation and fusion of gametes. It involves two major processes;
(a) Meiosis (reductional division) by which diploid sporophytic cells give rise to haploid gametes and
(b) Fertilization, which reconstitutes the sporophytic diploid generation through gametic fusion.

Question 2.
What are the two kinds of pollination?
Answer:
Two Kinds of pollination-
i. Self pollination : It is the transfer of pollen from the anther to the stigma of the same flower or to the stigma of another flower of the same plant.
ii. Cross-pollination : It is the transfer of pollen from the anther of flowers of the one plant to the stigma of a flower of another plant.

WBBSE Class 10 Life Science Solutions Chapter 2C Sexual Reproduction in Flowering Plants

Question 3.
Define the term unisexual and bisexual giving one example of each.
Answer:
Unisexual Organism : Male and female sex organs are present in different individuals. Such organisms are called unisexual. Example- Human.
Bisexual organism : Single individual having both male and female sex organs. Examples: Most of the plants, Tapeworm, Earthworm.

Question 4.
Mention the reproductive parts of a flower.
Answer:
Reproductive parts of a flower :
Stamens : It consists of stalk and a flattened top called anther. Anthers produce pollen grains. The pollen grains produce two male gametes.
Carpels : It has a swollen ovary at the base and an elongated middle style and terminal stigma. The ovary contains, ovules. Each ovule has an egg or female gamete.

Question 5.
Differentiate ovule and ovary.
Answer:
Ovule : A structure in the ovary of a plant that develops into a seed after fertilization.
Ovary : The enlarged basal portion of a pistil that bears the ovules in angiosperms.

Question 6.
What are the ways in which pollination happens?
Answer:
Pollination may happen in any of the following ways :

  1. The pollen of the same flower may fall on its stigma by itself.
  2. The pollen of another flower of the same plant may fall on the stigma.
  3. The pollen of a flower of another plant of the same species may land on the stigma. This transference can occur through wind, insects or other agents.

WBBSE Class 10 Life Science Solutions Chapter 2C Sexual Reproduction in Flowering Plants

Question 7.
Where do the molds on bread come from?
Answer:
The spores of the molds are present in the air. The spores of the molds under suitable conditions settle on the bread and grow.

Long Answer Type Questions : 5 Marks

Question 1.
Draw the chart for the fertilization of flowering plant.
Answer:
WBBSE Class 10 Life Science Solutions Chapter 2C Sexual Reproduction in Flowering Plants 1

Question 2.
What do mean by Pollination?
Answer:
Pollination is the process of transfer of pollen grains from the anther to the stigma of a flower, it is of two types :
(i) Self-pollination : If the pollen grains from the anther of a flower are transferred to the stigma of the same flower, it is termed as self-pollination or autogamy (auto : self; gamy: marriage), e.g. pea and china rose.

(ii) Cross-pollination : If the pollen grains from anther of one plant reach the stigma of a flower on another plant of the same species, then this is called as cross-pollination or allogamy (alias: other; gamy: marriage). Cross-pollination has the advantage of increasing the chances of variations.

WBBSE Class 10 Life Science Solutions Chapter 2C Sexual Reproduction in Flowering Plants

Question 3.
Write a short note on asexual and sexual reproduction in plant.
Answer:
ASEXUAL : DNA is identical to the parent (clones). Plants use this method naturally, they have developed special structures for asexual “propagation.” Bulbs, rhizomes, tubers, corns, runners are all structures that can form new plants identical to the parent.

SEXUAL : DNA from male (sperm) and female (egg) is recombined to produce offspring genetically different than the parents. Flowers develops as the sexual structures in flowering plants. Egg and sperm come together (egg is fertilized) to produce seeds.

Question 4.
Mention different parts of flower with their role.
Answer:
Parts of a flower :
Stamen : Male part, Anther : produces pollen (sperm), Filament : positions anther,
Pistil : Female part, Stigma : traps pollen (sperm), Style : positions stigma, Ovary : produces egg cells, develops into fruit, Egg cells : get fertilized by sperm develops into seeds.

Question 5.
How pollination occurs in flowering plants?
Answer:
Pollination: “Pollen landing on the stigma of a pistil”- The first step in the development of seed.
Reasons for Pollination methods :

  1. Sperm must travel to the egg (Pistil) of a flower
  2. Plants’ movement is limited
  3. Must use other methods to move pollen

Methods : Animal Wind

WBBSE Class 10 Life Science Solutions Chapter 2C Sexual Reproduction in Flowering Plants

Question 6.
What are the adaptations required for animal and wind pollination?
Answer:
Adaptation for Animal Pollination :
i. Size and Colour of petals :
Red colour attracts some birds, like Humming birds. White colour flowers open at night and moths are attracted.
Pattern of petals sometimes reflects UV light and helps the insects where to land.
ii. Fragrance of flowers :
Sweet : some animals are attracted like bees.
Rotten : Some animals like flies are attracted.
iii. Nectar : contains sugar water which helps in pollination.
Adaptations for Wind Pollination :

  1. Lack of petals : Energy is not spent on producing petals
  2. Lack of Fragrance : Energy is not spent on producing these chemicals
  3. Lack of Nectar : Energy is not spent on producing sugars
  4. Lots of pollen : Only by chance that pollen lands on the stigma, produces lots of pollen to increase lots of success
  5. Modification of stigma : Large, Feathery, Sticky- all to increase chance of catching pollen.

Question 7.
Write the various steps involved in the formation of a plant seed, starting from pollination.
Answer:
The transfer of pollen grains from anther to stigma is called pollination. The stamens in flower have anthers which produce pollen grains. Gynaecium or pistil bears three parts ovary, style and stigma. After pollination pollen grains form pollen tubes. Inside the pollen tube the nucleus divides into vegetative nucleus and generative nucleus. This generative nucleus gives rise to two male gametes inside the pollen tube. Pollen tube after piercing through the style reaches the ovary. One of which fuses with egg to form zygote. The second male gamete fuses with two polar nuclei to form triploid endosperm nucleus which finally gives rise to endosperm. So, the higher plants (angiosperms) show the process of double fertilization.

Soon after fertilization the petals, stamens, stigma and style fall off. The sepals wither and hold on to the ovule. The zygote, divides in a fixed fashion to form embryo. It may bear one or two cotyledons. An embryo axis made up of plumule and radicle. Radicle gives rise to root and plumule to shoot after germination of seed. Cotyledons contain food reserve.

WBBSE Class 10 Life Science Solutions Chapter 2C Sexual Reproduction in Flowering Plants

Question 8.
Explain the different parts of the flower with the help of a labelled diagram.
Answer:
A complete flower contains sepals, petals, androecium or stamen and gynoecium or carpel or pistils. Sepals are green in colour and protect the flower in bad condition. Petals are large and variously coloured to attract insects for pollination. Stamen is the male reproductive part. It consists of two parts-a long narrow stalk like filament and anther. The anther lobes consist of pollen sacs that contain millions of pollen grains. The male gametes are produced inside the pollen grains.

A pistil is the female reproductive part and consists of a swollen stigma at the top, a slender tube like style and a swollen ovary at the bottom. Inside the ovary, ovules are present which contain embryo sac. The unripe seed or eggs are present inside this sac.

WBBSE Class 10 Life Science Solutions Chapter 2C Sexual Reproduction in Flowering Plants 2

Question 9.
Describe the various post fertilization changes in flowers.
Answer:
Various post fertilization changes in flowers are :
(a) Bright colour of flower is lost.
(b) Except ovary, all the parts of flower fall off.
(c) Ovules in the ovary supplies food to the zygote which becomes embryo.
(d) Meanwhile, the walls of ovules become hard.
(e) Thus, the ovule transforms itself into seed.
(f) As the seeds form, the ovary increases in size and becomes fruit.
(g) The ovary wall becomes fruit wall.

Question 10.
Discuss the three different methods of vegetative propagation. Give suitable examples to support your answer.
Answer:
When new plants are produced from parts of the parent plant such as the root, stem or leaves, without the help of any reproductive organs, the process is known as vegetative propagation. Three types of vegetative propagation are:
(a) Vegetative propagation by stem : Aerial weak stems like runners and stolons, when they touch the ground, give off adventitious roots. When the connection with the parent plant is broken, the portion with the newly struck roots develops into an independent plant. Example-Some species of grass and strawberry. Underground stems can also produce several new plants from their buds. Example : Tuber of potato, bulb of onion.

(b) Vegetative propagation by root : Roots also help in vegetative propagation. For example, dahlia and sweet potato have roots that help in vegetative propagation. Root tuber of sweet potato grow into a new plant when sown in well prepared soil. Weeds like dandelion are capable of regenerating into a new plant, if some parts of its root are left behind in the soil.

(c) Vegetative propagation by leaves : It can be seen in very few plants like bryophyllum and begonia which are produced from leaves. They have buds on the notches in their margins. These buds, after falling on the ground or coming in contact with the soil, grow into new plants.

Multiple Choice Questions : 1 Mark

Question 1.
Flowers with both Androecium and the Gynoecium are called
(a) Anthers
(b) Stamens
(c) Bisexual flowers
(d) Unisexual flowers
Answer:
(c) Bisexual flowers

WBBSE Class 10 Life Science Solutions Chapter 2C Sexual Reproduction in Flowering Plants

Question 2.
Flowers with either the Androecium or the Gynoecium are called
(a) unisexual flowers
(b) Stigma
(c) Anther
(d) Bisexual flowers
Answer:
(a) unisexual flowers

Question 3.
Which of the following have unisexual flowers?
(a) Papaya
(b) Coconut
(c) Palm
(d) All of these
Answer:
(d) All of these

Question 4.
The Anthers of the stamens when ripe produce fine dust like particles called
(i) Sperms
(b) Egg Cells
(c) Stigma
(d) Pollen grains
Answer:
(d) Pollen grains

Question 5.
The transfer of pollen grains from the anther to the stigma of a flower is known as
(a) Fertilization
(b) Pollination
(c) Diffusion
(d) Adoption
Answer:
(b) Pollination

Question 6.
The pollen grains of a flower are transferred to the stigma of the same flower in
(a) Diffusion
(b) Cross Pollination
(c) Self pollination
(d) Fertilization
Answer:
(c) Self pollination

WBBSE Class 10 Life Science Solutions Chapter 2C Sexual Reproduction in Flowering Plants

Question 7.
Self pollination occurs in plants like
(a) Wheat
(b) Tobacco
(c) Peas
(d) All of these
Answer:
(d) All of these

Question 8.
Cross pollination takes place in
(a) Tomato
(b) Hibiscus and Lady Finger
(c) Brinjal
(d)All of these
Answer:
(d)All of these

Question 9.
Which of the following helps in cross pollination in unisexual flowers?
(it) Wind
(b) Bees
(c) Butterflies
(d) All of these
Answer:
(d) All of these

Question 10.
Pollination by wind takes place in Maize, Paddy and grass
(a) Maize
(b) Paddy
(c) Grass
(d) All of these
Answer:
(d) All of these

Question 11.
In Vallisneria, Hydrilla pollination takes place by
(a). Wind
(b) Water
(c) Insects
(d) All of these
Answer:
(d) All of these

WBBSE Class 10 Life Science Solutions Chapter 2C Sexual Reproduction in Flowering Plants

Question 12.
Flowers like night queen, Jasmine emit a sweet scent to attract insects like
(a) Honey bees
(b) Butterflies
(c) Beetles and moths
(d) All the above
Answer:
(d) All the above

Question 13.
The fusion of the male reproductive nucleus and the female reproductive nucleus is called
(a) Fertilization
(b) Adoption
(c) Regeneration
(d) Excretion
Answer:
(d) Excretion

Question 14.
At the end of pollen tube there are two nuclei called
(a) Sperm and Ovum
(b) Tube nucleus and generatvie nucleus
(c) Tube nucleus and Sperm
(d) Generative nucleus and Stigma
Answer:
(b) Tube nucleus and generatvie nucleus

Question 15.
Which of the following are the two nuclei inside the embryo sac
(a) Egg cell and Secondary nucleus
(b) Micropyle and Egg cell
(c) Stamen and Stigma
(d) Anther and style
Answer:
(a) Egg cell and Secondary nucleus

Question 16.
The tip of the pollen tube breaks when the pollen tube passes into the ovule through
(a) Micropyle
(b) Embryo
(c) Calyx
(d) Corolla
Answer:
(a) Micropyle

WBBSE Class 10 Life Science Solutions Chapter 2C Sexual Reproduction in Flowering Plants

Question 17.
The zygote develops into
(a) Pedicel
(b) Embryo
(c) Stigma
(d) Style
Answer:
(b) Embryo

Question 18.
One of the male nuclei fuses with the egg cell and forms
(a) Stigma
(b) Pedicel
(c) Corolla
(d) Zygote
Answer:
(d) Zygote

Question 19.
One of the nucleus of the pollen tube fuses with the secondary nucleus of the ovum and grows into
(a) Stamen
(b) Stigma
(c) Anther
(d) Endosperm
Answer:
(d) Endosperm

Question 20.
Generative nucleus divides forming
(a) Three female nuclei
(b) Two male nuclei
(c) Three male nuclei
(d) Two female nuclei
Answer:
(b) Two male nuclei

Question 21.
The number of outer layers present in each ovule are
(a) 2
(b) 3
(c) 4
(d) 5
Answer:
(a) 2

Question 22.
At the upper end of the outer layer there is an opening known as
(a) Stigma
(b) Style
(c) Micropyle
(d) Anther
Answer:
(c) Micropyle

WBBSE Class 10 Life Science Solutions Chapter 2C Sexual Reproduction in Flowering Plants

Question 23.
Embryo sac is present inside the
(a) Ovule
(b) Stigma
(c) Micropyle
(d) Style
Answer:
(a) Ovule

Question 24.
The number of nuclei present inside the embryo sac are
(a) 4
(b) 2
(c) 5
(d) 3
Answer:
(b) 2

Question 25.
Pollen from one flower is carried to the stigma of another flower of the same plant or a different plant of same species in
(a) Self pollination
(b) Cross pollination
(c) Fertilization
(d) Diffusion
Answer:
(b) Cross pollination

Very Short Answer Type Questions : 1 Mark

Question 1.
Name the reporductive whorls of flowering plants.
Answer:
Male sex organ : Stamens
Female sex organ : Carpels.

Question 2.
Name the elements of egg apparatus.
Answer:
Egg nucleus, definitive nucleus, synergids, antipodal cells.

Question 3.
Give example for self and cross pollination
Answer:
Self pollination : Vinca
Cross pollination : Cucurbita

WBBSE Class 10 Life Science Solutions Chapter 2C Sexual Reproduction in Flowering Plants

Question 4.
Give examples of the following : Anemophilous, Hydrophilous. Entomophilous, Ornithophilous Plant
Answer:

  • Anemophilous : Paddy flower
  • Hydrophilous : Ceratophyllum
  • Entomophilous : Mangifera indica
  • Ornithophilous : Erythrina

Question 5.
What are synergids ?
Answer:
There are three nuclei adjacent to micropyle. Out of these three nuclei, the central one is egg and other two nuclei are known as synergids.

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Detailed explanations in West Bengal Board Class 10 Life Science Book Solutions Chapter 2B Reproduction offer valuable context and analysis.

WBBSE Class 10 Life Science Chapter 2B Question Answer – Reproduction

Long Answer Type Questions : 5 Marks

Question 1.
Describe different types of asexual reproduction in plants.
Answer:
(a) Fission : Fission is of two types: Binary fission and multiple fission.
Binary fission : In binary fission, two individuals are formed from a single parent. This type of reproduction is found in organisms like bacteria, yeast and Amoeba (Fig. Binary fission in Amoeba).

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction 1

Multiple fission : In multiple fission, many individuals are formed from a single parent. This type of reproduction by multiple fission occurs during unfavourable conditions.

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction 2

In this type of reproduction, the unicellular organism develops a protective covering called cyst over the cell. The nucleus of the cell divides repeatedly producing many nuclei. Many daughter cells are produced within the cyst. The cyst breaks and small offsprings are liberated. This type of reproduction is seen in many algae and in some protozoans, such as the malarial parasite (Plasmodium) (Fig. Binary fission in Plasmodium).

(b) Budding : In this type of reproduction, a bulb-like projection or outgrowth arises from the parent body known as bud, which detaches and forms a new organism.
For example, Hydra reproduces by budding (Fig. multiple fission in Plasmodium). A small protuberance arises from one side of its body, which grows, develops tentacles and gets detached to lead an independent life.

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction 3

(c) Regeneration or Fragmentation : In this type of reproduction, the body of an individual breaks up into two or more parts and each part develops into a complete individual. Example : Spirogyra, and Planaria.

(d) Spore formation : In lower forms of life like the alga, Chlamydomonas, the protoplast of the cell divides to form 4-8 spores. These being motile are termed as zoospores. When spores are released in the surrounding medium they develop into new plants.

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

(e) Vegetative propagation or vegetative reproduction in plants : Vegetative reproduction (or vegetative propagation) is a form of asexual reproduction in plants in which a bud grows and develops into a new plant. In this type of reproduction, any vegetative part of the plant body like leaf, stem or root develops into a complete new plant. Vegetative reproduction can take place by two methods-natural and artificial.

I) Vegetative reproduction by natural methods : This type of vegetative reproduction can involve roots, stem or leaves. Some common modes of vegetative reproduction are given below:
i. By roots : The roots of sweet potato and mint bear adventitious buds. When these roots are planted in the soil, new plants are produced.
ii. By stem : In many plants the stem develops buds on it. The part of the stem that bears buds serves as an organ for vegetative multiplication, e.g. the modified parts of stem, such as runners of grass, suckers of mint and Chrysanthemum, bulbs of onion and tulip, rhizomes of ginger, corns of gladiolus and Colocasia, and tubers of potato etc.

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction 4

iii. By leaves : In some plants, e.g. in Bryophyllum and Bigonia, adventitious buds are developed in the margins of their leaves. When the leaf falls on moist soil, these buds develop into small plantlets, which can be separated and grown into independent plants.

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

(b) Vegetative propagation by artificial methods : Some plants can be propagated artificially. The methods of artificial propagation include grafting, layering, cutting and tissue culture.

i. Grafting : It is the method of obtaining a superior quality plant from two different plants, taking the root system of one plant and the shoot system of another plant. The plant whose root system is taken is called stock. The plant whose shoot system is taken is called scion.

The ends to be grafted, of the stock and the scion, are cut obliquely and placed face to face and are bound firmly with tape. The stock supplies all the desired nutrients to the scion. This technique has been used in raising superior quality plants of mango, apples, roses, rubber and citrus.

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction 5

ii. Cutting : In some plants like rose, sugarcane, Bougainvillaea etc. this method is used quite frequently. Stem cuttings with nodes and internodes are placed in moist soil which give rise to adventitious roots, and grow into new plants.

iii. Layering : Layering is the development of roots on a stem while it is still attached to the parent plant. The stem or the branch that develops adventitious roots while still attached to the parent plant is called a layer. It is a means of reproduction in black raspberries, jasmine (Jasminum), Magnolia etc.

iv. Micropropa galion Tissue culture : This is a modern technique of vegetative propagation. In this technique, a small piece of tissue is cut from a plant and is transferred to a container with nutrient medium under aseptic conditions. The tissue utilizes nutrients from the medium, divides and re-divides, and forms a callus. Small portions of this callus are transferred to another medium which induces differentiation and plantlets are produced. These plantlets are transplanted in soil to form an adult plant. Orchids, Chrysanthemum, Asparagus and many other plants are now being grown by using plant tissue culture technique.

Multiple Choice Questions : 1 Mark

Question 1.
______ is (are) the basis for asexual reproduction.
(a) Mitosis
(b) Meiosis
(c) Sex hormones
(d) All A, B and C
Answer:
(a) Mitosis

Question 2.
What is the advantage of asexual reproduction?
(a) Organisms increases rapidly
(b) Organisms are morphologically alike
(c) Organisms are genetically alike
(d) All A, B and C
Answer:
(d) All A, B and C

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 3.
What is the advantage of sexual reproduction?
(a) Genetically varied individuals are born
(b) Chances of survival increases
(c) Chances of evolution increases
(d) All A, B and C
Answer:
(d) All A, B and C

Question 4.
Natural method(s) of asexual reproduction in plants is (are)
(a) Spores
(b) Vegetative propagation
(c) Apomixis
(d) All A, B and C
Answer:
(d) All A, B and C

Question 5.
Artificial method(s) of asexual reproduction in plants is (are)
(a) Cuttings
(b) Tissue culture
(c) Both A and B
(d) Alternation of generation
Answer:
(c) Both A and B

Question 6.
In apomixis, an embryo is created from a diploid cell in the
(a) Pollen tube
(b) Leaf
(c) Ovule
(d) All of the choices are correct
Answer:
(c) Ovule

Question 7.
Asexual reproduction differs from sexual reproduction, in that it does not require
(a) 1 parent
(b) 2 parents
(c) Spores
(d) Vegetative parts
Answer:
(b) 2 parents

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 8.
Asexual reproduction does not introduce
(a) Variation
(b) Similarity between parents & offsprings
(c) Same chromosomal number in offsprings
(d) All of the choices are incorrect
Answer:
(a) Variation

Question 9.
Which of the following statements is true of clones?
(a) Clones show variation
(b) Clones have DNA identical to parent
(c) Clones are formed by meiotic division
(d) All the choices are incorrect
Answer:
(b) Clones have DNA identical to parent

Question 10.
Vegetative propagation does not involve
(a) Root parts
(b) Stem parts
(c) Leaf parts
(d) Flower parts
Answer:
(d) Flower parts

Question 11.
At the cut end of shoot a mass of dividing undifferentiated cell is called
(a) Callus
(b) Periblem
(c) Dermatogen
(d) Pericycle
Answer:
(a) Callus

Question 12.
One of the following is NOT a method of asexual reproduction.
(a) Sporulation
(b) Gametogenesis
(c) Apomixis
(d) Parthenogenesis
Answer:
(b) Gametogenesis

Question 13.
Tissue culture is a technique used to produce a large number of ______ plants quickly.
(a) Variable
(b) Unicellular
(c) Identical
(d) All A, B and C
Answer:
(c) Identical

Question 14.
A flagellated motile sperm fertilizing a non-motile egg, this syngamy is called
(a) Isogamy
(b) Anisogamy
(c) Oogamy
(d) All of the choices are correct
Answer:
(c) Oogamy

Question 15.
A type of syngamy in which both fusing gametes are flagellated but different in size are known as
(a) Isogamy
(b) Anisogamy
(c) Oogamy
(d) All of the choices are correct
Answer:
(b) Anisogamy

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 16.
A type of syngamy in which both fusing gametes are flagellated and same in size are known as
(a) Isogamy
(b) Anisogamy
(c) Oogamy
(d) All of the choices are correct
Answer:
(a) Isogamy

Question 17.
If someone gives you a plant and tells you that it is an angiosperm, you know that during its life cycle it will produce
(a) Swimming sperm
(b) A prothallus
(c) Flowers
(d) Cones
Answer:
(c) Flowers

Question 18.
The transfer of pollen grains to the female part of the plant is called
(a) Germination
(b) Reproduction
(c) Pollination
(d) Fertiliztion
Answer:
(c) Pollination

Question 19.
The production of new plants from underground stems is an example of ______ reproduction.
(a) Two parents
(b) Asexual
(c) Zygote
(d) Sexual
Answer:
(b) Asexual

Question 20.
Which one of the following is the male reproductive part of a flower?
(a) Stamen
(b) Sepal
(c) Petal
(d) Pistils
Answer:
(a) Stamen

Question 21.
In seed plants, sperm travels down a(an) _______ to reach the egg.
(a) Stigma tube
(b) Ovule tube
(c) Pollen tube
(d) Stamen tube
Answer:
(c) Pollen tube

Question 22.
A carpel is a leaf which has been modified to produce
(a) Microsporania
(b) 2 male gametes
(c) Pollen grains
(d) Ovules
Answer:
(d) Ovules

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 23.
The stamens are leaves modified for the production of
(a) Microspores
(b) Megaspores
(c) Ovules
(d) Seed
Answer:
(a) Microspores

Question 24.
The sepals and petals are
(a) Reproductive parts of flower
(b) Non-reproductive parts of flower
(c) Parts of Gametophyte
(d) Both A and C
Answer:
(b) Non-reproductive parts of flower

Question 25.
The ovule contains
(a) Microsporangium
(b) Male gametophyte
(c) Embryo sac
(d) All A, B and C
Answer:
(c) Embryo sac

Question 26.
The unisexual flowers are called
(a) Staminate
(b) Carpellate
(c) A and B
(d) Monoecious
Answer:
(c) A and B

Question 27.
If staminate and carpellate flowers are present on same plant it is termed as
(a) Monoecious
(b) Dioecious
(c) Unisexual
(d) Neuter
Answer:
(a) Monoecious

Question 28.
Angiosperms, double fertilization produces two distinct portions of the seed. The endosperm portion’s role is to
(a) Develop into the embryo
(b) Nourish the embryo
(c) Develop into the mature sperm
(d) Serve as a reservoir for extra DNA
Answer:
(b) Nourish the embryo

Question 29.
Each of the following is a part of a seed except the
(a) Embryo
(b) Endosperm
(c) Seed-coat
(d) Gametophyte
Answer:
(d) Gametophyte

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 30.
The _______ ultimately matures into a fruit
(a) Integument
(b) Ovary
(c) Archegonium
(d) Ovule
Answer:
(b) Ovary

Question 31
Which of the following is a part of the third whorl?
(a) Calyx
(b) Corolla
(c) Petal
(d) Stamen
Answer:
(d) Stamen

Question 32.
Which of the following is formed in the double fertilization and becomes an endosperm?
(a) Synergid cells
(b) Antipodal cells
(c) Primary endosperm nucleus
(d) Triploid (3n) nucleus
Answer:
(d) Triploid (3n) nucleus

Question 33
A pollen grain is a
(a) Immature male gametophyte
(b) Spore
(c) Fruiting body
(d) Mature male gametophyte
Answer:
(a) Immature male gametophyte

Question 34.
In plants, spores are formed by $I$, whereas gametes are formed by 2
(a) I-meiosis, 2-mitosis
(b) I -fission, 2-fusion
(c) I-meiosis, 2-meiosis
(d) I-mitosis, 2-mitosis
Answer:
(a) I-meiosis, 2-mitosis

Question 35.
From life cycle point of view the most important part of a plant is
(a) Flower
(b) Leaf
(c) Stem
(d) Root
Answer:
(a) Flower

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 36.
The main embryo develops from the structure formed as result of fusion of
(a) 2 polar nuclei of embryo sac
(b) Definitive nucleus and male gamete
(c) Egg cell and male gamete
(d) Male gamete and synergids
Answer:
(c) Egg cell and male gamete

Question 37.
The fertilization occurs in
(a) Ovary
(b) Ovule
(c) Embryo sac
(d) Nucleus
Answer:
(c) Embryo sac

Question 38.
The process in which fruit develops without fertilization is called
(a) Parthenogenesis
(b) Parthenocarpy
(c) Viviparous germination
(d) Apomixis
Answer:
(a) Parthenogenesis

Question 39.
The first organ to emerge from the germinating seed is
(a) Radicle
(b) Plumule
(c) Cotyledon
(d) Epicotyl
Answer:
(a) Radicle

Question 40.
A form of asexual reproduction in which new individual grows out as small outgrowth and eventually separates from parent body is called
(a) Forming a spore
(b) Budding
(c) Regeneration
(d) Fission
Answer:
(b) Budding

Question 41.
Reproduction of egg without fertilization by sperm is termed as
(a) Parthenogenesis
(b) Parthenocarpy
(c) Regeneration
(d) Budding
Answer:
(a) Parthenogenesis

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 42.
The technique of producing a genetically identical copy of an organism by replacing the nucleus of an unfertilized ovum with the nucleus of a body cell from the organism is
(a) Budding
(b) Cloning
(c) Parthenocarpy
(d) Fission
Answer:
(b) Cloning

Question 43.
Reproduction is important to avoid
(a) Variation
(b) Chances of survival
(c) Genetic monotony
(d) All A, B and C
Answer:
(c) Genetic monotony

Question 44.
Which of the following cell type is haploid?
(a) Primary spermatocyte
(b) Spermatogonium
(c) Sertoli cell
(d) Secondary spermatocyte
Answer:
(d) Secondary spermatocyte

Question 45.
The animals which lay eggs are called
(a) Oviparous
(b) Viviparous
(c) Dioecious
(d) Neuter
Answer:
(a) Oviparous

Question 46.
In terrestrial conditions which type of fertilization is more common
(a) External
(b) Internal
(c) Self
(d) None of these
Answer:
(b) Internal

Question 47.
Which of the following is not a unicellular organism ?
(a) Amoeba
(b) Paramecium
(c) Hydra
(d) Yeast
Answer:
(c) Hydra

Question 48.
Which of the following shows budding ?
(a) Yeast
(b) Amoeba
(c) Paramecium
(d) Plasmodium
Answer:
(a) Yeast

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 49.
Which of the following shows multiple fission ?
(a) Hydra
(b) Yeast
(c) Spirogyra
(d) Plasmodium
Answer:
(d) Plasmodium

Question 50.
Which is the most common method of reproduction in majority of fungi and bacteria?
(a) Binary fission
(b) Multiple fission
(c) Budding
(d) Spore formation
Answer:
(d) Spore formation

Question 51.
Which of the following shows spore formation?
(a) Amoeba
(b) Mucor
(c) Plasmodium
(d) Paramecium
Answer:
(b) Mucor

Question 52.
Regeneration is observed in ______.
(a) Amoeba
(b) Planaria
(c) Spirogyra
(d) Yeast
Answer:
(b) Planaria

Question 53.
Which of the following does not show regeneration?
(a) Mucor
(b) Planaria
(c) Sponge
(d) Starfish
Answer:
(a) Mucor

Question 54.
Which of the following is not an artificial method of vegetative propagation?
(a) Cutting
(b) Layering
(c) Grafting
(d) Hybridization
Answer:
(d) Hybridization

Question 55.
Testes produce _______ hormone.
(a) Estrogen
(b) Testosterone
(c) Progesterone
(d) Both estrogen and testosterone
Answer:
(b) Testosterone

Question 56.
Which of the following is not produced in the ovary ?
(a) Ovum
(b) Estrogen
(c) Progesterone
(d) Testosterone
Answer:
(d) Testosterone

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 57.
The main reproductive organ of human male is ______.
(a) a pair of testes
(b) vas deferens
(c) urethra
(d) penis
Answer:
(a) a pair of testes

Question 58.
The main female reproductive organ is _______.
(a) Fallopian tubes
(b) uterus
(c) vagina
(d) a pair of ovaries
Answer:
(d) a pair of ovaries

Question 59.
Which of the following shows budding ?
(a) Hydra
(b) Amoeba
(c) Paramecium
(d) Spirogyra
Answer:
(a) Hydra

Question 60.
Several new individuals are produced
(a) Binary fission
(b) Multiple fission
(c) Both binary fission and multiple fission
(d) Fertilization
Answer:
(b) Multiple fission

Question 61.
In grafting the plant rooted in the soil is called _______.
(a) stock
(b) sucker
(c) scion
(d) rhizome
Answer:
(a) stock

Question 62.
Which of the following plants shows vegetative propagation ?
(a) Wheat
(b) Radish
(c) Sunflower
(d) Bryophyllum
Answer:
(d) Bryophyllum

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 63.
In the process of sexual reproduction ______
(a) only males are needed
(b) both male and female are needed
(c) only females are needed
(d) only mitosis is involved
Answer:
(b) both male and female are needed

Question 64.
The unicellular fungus which shows bud formation is
(a) mucor
(b) yeast
(c) amoeba
(d) none of these
Answer:
(a) mucor

Question 65.
The animal which shows regeneration is ______.
(a) amoeba
(b) paramecium
(c) hydra
(d) rhizopus
Answer:
(b) paramecium

Question 66.
The method by which desired characters of two plants can be combined is ______.
(a) cutting
(b) layering
(c) budding
(d) grafting
Answer:
(c) budding

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 67.
Yeast reproduces by _____.
(a) budding
(b) fission
(c) regeneration
(d) spore formation
Answer:
(d) spore formation

Question 68.
A chain of buds is formed in _____.
(a) hydra
(b) yeast
(c) mucor
(d) plasmodium
Answer:
(b) yeast

Question 69.
Vegetative propagation is observed in _____.
(a) yeast
(b) hydra
(c) mucor
(d) bryophyllum
Answer:
(d) bryophyllum

Question 70.
_____ is useful where seeds have long dormancy period and poor germination capacity _____.
(a) Cutting
(b) Layering
(c) Grafting
(d) Regeneration
Answer:
(c) Grafting

Question 71.
_____ is a very useful method for developing garden.
(a) Cutting
(b) Layering
(c) Grafting
(d) Budding
Answer:
(a) Cutting

Very Short Answer Type Questions : 1 Mark

Question 1.
Name any two types of asexual reproduction.
Answer:
Budding, Fragmentation.

Question 2.
Name the process by which ovam gives rise to plant without fertilization.
Answer:
Parthenogenesis.

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 3.
Adventitious root is seen in which plant?
Answer:
Ipomea sp.(sweet potato).

Question 4.
In which plant leaf buds are seen?
Answer:
Bryophyllum.

Question 5.
Name two process for artificial vegetative reproduction.
Answer:
Cutting, Grafting.

Question 6.
What is callus?
Answer:
In case cell culture, from callus organ (eg. root, stem), tiny plant is developed.

Question 7.
Name the phases of alternation of generation.
Answer:
It consists of two phases-gametophyte and sporophyte.

Question 8.
Name the different types of floral leaves.
Answer:
Floral leaves are of four types: Calyx, Corolla, androecium and gynoecium.

Question 9.
Name the rooted plant and cutting branch in case of grafting.
Answer:
Rooted plant is known as stalk and cutting branch is known as scion.

Question 10.
Give example of plants which show reproduction through spore formation and fragmentation.
Answer:
Spore formation : moss, fern,
Fragmentation : Spirogyra.

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 11.
What is reproduction?
Answer:
Reproduction: It is the ability of an organism to produce the young ones of its own kind.

Question 12.
What is the role of reproduction?
Answer:
i. Propagation of species
ii. Evolution of species.

Question 13.
Name the two major kinds of reproduction.
Answer:
i. Asexual reproduction and
ii. Sexual reproduction.

Question 14.
Name the various methods of vegetative propagation in plants.
Answer:

  1. Cutting
  2. Layering
  3. Grafting.

Question 15.
Name common method for vegetative propagation of Rose and Sugarcane.
Answer:
Cutting.

Question 16.
How do Spirogyra and Mucor reproduce asexually?
Answer:
Spirogyra-Fragmentation and Regeneration; Mucor – Spore formation.

Question 17.
Name the mode of vegetative propagation in (i) Begonia (ii) Mint.
Answer:
(i) Begonia – Leaf buds (ii) Mint – Runners.

Question 18.
How do yeast, sponges and hydra reproduce asexually?
Answer:
All the three reproduce by budding.

Question 19.
Which part of Bryophyllum can be used for vegetative propagation?
Answer:
Leaf of Bryophyllum.

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 20.
Give one example of each : Vegetative propagation by (i) root (ii) stem.
Answer:
i. Vegetative propagation by roots, e.g.-Sweet potato.
ii. Vegetative propagation by stem, e.g.-Potato.

Question 21.
List two advantages of vegetative propagation.
Answer:
Advantages of vegetative propagation-
i. It is usually a means of propagating such plants which do not produce viable seeds.
ii. It is rapid method of producing young ones.

Question 22.
What is pollination?
Answer:
Pollination: It is the transference of pollen grains from the anther of a flower to the stigma of a flower.

Question 23.
What term is used if the pollen is transferred to the stigma of same flower?
Answer:
Self-pollination.

Question 24.
Where are pollens and ovules present in flower?
Answer:
i. Pollens – Anther lobes
ii. Ovules – Ovary.

Question 25.
Which floral part is very attractive and coloured?
Answer:
Petal.

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 26.
What is fruit?
Answer:
A ripened ovary.

Question 27.
What is reproduction ?
Answer:
The process by which the organisms produce new organisms similar to them is known as reproduction.

Question 28.
Why is reproduction essential ?
Answer:
Reproduction is essential for the perpetuation of species and thereby life. 29. Mention the types of reproduction.
Answer: Asexual reproduction and sexual reproduction are the two types of reproduction.

Question 30.
What is asexual reproduction ?
Answer:
Asexual reproduction is the method of reproduction involving only a single individual (organism),i.e. only one organism is needed for reproduction.

Question 31.
Mention different types of asexual reproduction.
Answer:

  1. Fission
  2. Budding
  3. Spore Formation
  4. Regeneration
  5. Vegetative propagation etc. are the types of asexual reproduction.

Question 32.
Mention two types of reproduction by fission.
Answer:
Binary fission and multiple fission are the two types of fission.

Question 33.
What is binary fission?
Answer:
When two independent adult organisms are formed after the division of the nucleus and the cytoplasm of one (mother) cell into two (daughter) cells, the process of reproduction is called binary fission.

Question 34.
Give examples of organisms showing binary fission.
Answer:
Amoeba and Paramecium show binary fission.

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 35.
What is multiple fission ?
Answer:
When the nucleus of one (mother) cell divides several times into many (daughter) nuclei, each forming an individual organism, the process is called multiple fission.

Question 36.
Give examples of organism showing multiple fission.
Answer:
Plasmodium and Amoeba show multiple fission.

Question 37.
What is budding ?
Answer:
Budding is a process of reproduction in which a small protuberance (bud) appearing on an adult cell gradually grows bigger and then behaves as an independent organism after being detached from the (mother) cell.

Question 38.
Give examples of organism showing budding.
Answer:
Yeast and Hydra show budding.

Question 39.
What is the similarity between Hydra and Yeast ?
Answer:
Both Hydra and Yeast exhibit budding as the method of reproduction.

Question 40.
Mention some organisms showing spore formation.
Answer:
Mucor, Rhizopus and Penicillium show spore formation.

Question 41.
What is sporangium ?
Answer:
Sporangium is a structure developing from fungal hypha and it contains a nucleus (spore) which divides several times forming a large number of spores which develop into new hyphae after falling on the ground.

Question 42.
What is regeneration?
Answer:
The ability of an organism to replace the lost parts of its body is called regeneration.

Question 43.
What is fragmentation ?
Answer:
When an organism breaks into two or more pieces after being mature and then each piece grows as an individual organism, the process is known and fragmentation.

Question 44.
Give examples of organism showing fragmentation.
Answer:
Oscillatoria and spirogyra show fragmentation.

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 45.
What is vegetative propagation ?
Answer:
Vegetative propagation is a method of reproduction in which a new plant developes from a part of a root, stem or leaf.

Question 46.
Give examples of plant showing vegetative propagation.
Answer:
Potato, sweet potato, Bryophyllum etc. show vegetative propagation.

Question 47.
Mention the name of artificial methods of vegetative propagation.
Answer:
Cutting, layering, grafting etc. are artificial methods of vegetative propagation.

Question 48.
Give examples of some plants in which cutting is used as a method of vegetative propagation.
Answer:
Cutting is used as a method of vegetative propagation for sugarcane, grapes, rose, phalsa etc.

Question 49.
Give examples of some plants in which layering is used as a method of vegetative propagation.
Answer:
layering is used as a method of vegetative propagation for lemon, guava, Hibiscus, jasmine, bougainvillaea, eie.

Question 50.
What is grafting ?
Answer:
Grafting is a method of vegetative propagation in which two parts of two different plants are joined together in a specific manner so that they unite to grow as one plant.

Question 51.
What is ‘scion’?
Answer:
The portion of the plant which is grafted on other plant is called scion.

Question 52.
What is ‘stock’?
Answer:
The plant on which grafting is performed is called stock.

WBBSE Class 10 Life Science Solutions Chapter 2B Reproduction

Question 53.
In what circumstances is grafting a better method ?
Answer:
Grafting is a better method for plants where seeds are having long dormancy period and poor germination capacity.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Detailed explanations in West Bengal Board Class 10 Life Science Book Solutions Chapter 2A Cell Division and Cell Cycle offer valuable context and analysis.

WBBSE Class 10 Life Science Chapter 2A Question Answer – Cell Division and Cell Cycle

Short Answer Type Questions : 2 Marks

Question 1.
What is prokaryotic cell? Mention two differences between Prokaryotic and Eukaryotic cell.
Answer:
Prokaryotic cell : The most primitive type of cell are small, lacking membrane around the nucleus and the nuclear contents as protein and nuclic acid is termed as prokaryotic cell (Pro-primitive, karyon-nucleus)

Prokaryotic cell Eukaryotic cell
i. Cell wall composed of amino sugar and muramic acid. i. Plant cell wall composed of cellulose.
ii. Cytoplasmic organellesendoplasmic reticulum, mitochondria, golgi bodies etc. are absent. ii. Cytoplasmic organelles endoplasmic reitculum, mitochondria, golgi bodies etc. are present.

Question 2.
What is amitosis? Give example What is cell cycle?
Answer:
Amitosis : The process of cell division in which the nucleus first constricts in the middle to divide into two nuclei and then the cell body undergoes changes without spindle formation and ultimately divides into two daughter cells are known as amitosis.
e.g. Chara, Amoeba etc. show amitosis.
Cell Cycle : The cycle through which a cell passes involving different phases of changes, when it undergoes division is called cell cycle.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 3.
Compare ‘cleavage’ and ‘cell plate formation’ during cell division. Mention the importance features of G1 of cell cycle.
Dissimilarities :
Answer:

Cleavage Cell plate formation
i. Occurs in animal cell i. Occurs in plant cell.
ii. Formed by furrowing. ii. Formed by aggression of golgi vesicles.

Similarity : Both helps in cytokinesis either in animal or in plant cell.
Importance of G1 phase : Cell grows in size, RNA and protein synthesis occurs, DNA content remains unchanged.

Question 4.
Write two differences between plant and animal mitosis.
Answer:

Plant mitosis Animal mitosis
i. Spindle is formed by microtubules of nucleoplasm. i. Spindle is formed from astral rays of the centrosome.
ii. Absence of centromere. ii. During division centromere is separated out and centrioles move towards opposite poles.

Question 5.
Write two major differences between plant cell cytokinesis and animal cell cytokinesis.
Answer:

Plant cytokinesis Animal Cytokinesis
i. Cytokinesis takes place by cell plate formation. i. Cytokinesis takes place by furrowing cleavage.
ii. In plant cell cytokinesis begins from telophase. ii. Cytokinesis begins from anaphase.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 6.
What are Phragmoplasts?
Answer:
Phragmoplast is the aggregation of vesicles from golgi body which helps in plant cytokinesis.

Question 7.
Why mitosis is called equational division and meiosis is called reductional division? When meiosis cell division occurs?
Answer:
Equational division is the division in which chromosome number of parent cell and daughter cell is equal. In Mitosis, two daughter cells are produced (diploid or 2n) remain indentical to the diploid (2n) parent cell.
Reductional division is the division in chromosome. Number of daughter cell becomes half of that of mother cell. In meiosis, the daughter cells which are produced are haploid cells (n) while that of parent is a diploid cell (2n).

Question 8.
What is mitosis? What is the importance of mitosis?
Answer:
Mitosis is the process in which one eukaryotic cell divides into two cells identical to the parent cell (generally identical, since alterations in genetic material can occur, more or less organelles may be distributed between the daughter cells, etc.)
Mitosis is fundamental for asexual reproduction of eukaryotes, for the embryonic development and for tissue renewal.

Question 9.
Why in some cases is mitosis a synonym of reproduction?
Answer:
In some living beings asexual reproduction occurs by many means: binary division, schizogony, budding, grafting, etc. In asexual reproduction of eukaryotes mitosis is the mechanism by which the constituent cells of the new beings are made.
The term mitosis does not apply to prokaryotes since it involves nuclear division and eukaryotic structures.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 10.
What is the importance of mitosis for the embryonic development?
Answer:
Every embryo grows from a single cell that suffers mitosis and generates other cells that also divide themselves by mitosis forming tissues and complete organs. The perfect regulation and control of each of those cell divisions are fundamental for the creation of a normal individual. Without mitosis the embryonic development would be impossible.

Question 11.
What are some examples of organ and tissue where mitosis is more frequent, less frequent or practically absent?
Answer:
Generally in vertebrates mitosis is more frequent in tissues that require intense renewing due to their functions, like epithelial tissues and the bone marrow. In plants the meristem tissue has numerous cells undergoing mitosis. Mitosis takes place with low frequency in tissues of slow renovation, like the bones in adults and the connective tissues. In some adult tissues mitosis is almost absent, like the nervous tissue and the striated muscle tissue (skeletal and cardiac). The nervous tissue develops from stimulus by development of new electrical networks between cells and the striated muscle tissue grows by cellular hypertrophy.

Question 12.
How does mitosis participate in the growth of pluricellular organisms?
Answer:
All pluricellular beings grow with the increase in quantity of their cells. This increase is produced by mitosis (although some types of growth occur by cellular hypertrophy or by deposition of substances in interstitial spaces).

Question 13.
What is the uncontrolled mitotic process that occurs as disease in pluricellular beings called?
Answer:
Uncontrolled mitotic cell division is called neoplasia. Neoplasia (the formation of new strange tissues) occurs when a cell suffers mutation in its genetic material, loses the ability to control its own division and the failure is transmitted to its descendants. Cancers are malignant neoplasias. The term malignant means that neoplastic cells can disseminate to distant sites invading other organs and tissues. Neoplasias whose cells cannot disseminate to distant sites are called benign neoplasias.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 14.
What is cellular regeneration? How is mitosis related to this process?
Answer:
Some tissues are able to regenerate when injured. The liver, for example, regenerates when small pieces of hepatic tissue are removed, bones make new tissues in fracture regions, etc. Some animals, like planarias, are capable of regenerating their bodies when sectioned. In tissue regeneration cellular proliferation happens by mitosis.

Question 15.
What is cell cycle?
Answer:
Cell cycle, or mitotic cycle, is the time period that begins when the cell is created and finishes when it is divided by mitosis creating two daughter cells. The cell cycle is divided into interphase and the mitotic phase.

Question 16.
Is cell division happening during the entire cell cycle? What is interphase?
Answer:
Cell division properly occurs during the mitotic phase of the cell cycle. During interphase processes that are a preparation to cell division take place, like the duplication of DNA and centrioles. Interphase is the preceding phase and the mitotic is the following phase.

Question 17.
What are the three periods into which interphase is divided?
Answer:
Interphase is the preceding phase to the mitotic division. It is divided into three periods, G1, S and G2 (the letter G comes from “gap”, meaning interval or breach, and the letter S comes from “synthesis”, indicating the period in which DNA replicates).
In fact, “gap” is not totally appropriate for the periods immediately before and after the DNA synthesis. The idea of “growth” would be more adequate since in those periods (G1 and G2) the cell is growing to divide later in mitosis.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 18.
What are the events that mark the beginning and the end of the second interphase period? What happens in the cell in this period?
Answer:
The second interphase period is the S. It starts with the beginning of DNA replication and finishes with the end of that process. The main event in this period is the synthesis of new polynucleotide chains, each bound to each DNA chain that served as a template, i.e., the duplication of the original set of DNA molecules.

Question 19.
What are the events that mark the beginning and the end of the third interphase period? What happens in the cell in this period?
Answer:
The third interphase period is the G2. It begins with the end of DNA replication and ends with the beginning of the first period of the mitotic phase. On G2 the cell is growing too and the duplication of centrioles occurs (only in cells that have these structures).

Question 20.
What are centrioles? In which type of cell are they present?
Answer:
Centrioles are tiny cylindrical structures made of nine microtubule triplets. They appear in pairs in the cell. Centrioles participate in the making of cytoskeleton and of cilia and flagella. In cell division they play a role in the formation of the aster fibers. Centrioles are structures present in animal cells, in most protists and in some primitive fungi. There are no centrioles in cells of superior plants and in general it is considered that plant cells do not have centrioles (although this is not entirely correct since some plants have centriole-containing cells).
The region where the centrioles are located is called the centrosome of the cell.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 21.
What are the main events of the first mitotic period?
Answer:
The first mitotic period is prophase. During prophase the following events occur : migration of each centriole pair (centrioles were duplicated in interphase) to opposite cell poles; aster formation around the centriole pairs; formation of the spindle fibres between the two centriole pairs; end of chromosome condensation; disintegration of the nucleolus; breaking of the karyotheca; dispersion of condensed chromosomes in the cytoplasm; binding of chromosomes to the spindle fibres.

Question 22.
What is the mitotic apparatus?
Answer:
Mitotic apparatus is the set of aster fibres, radial structures around each centriole pair, plus the spindle fibres, fibres that extend across the cell between the two centriole pairs located in opposite cell poles. The mitotic apparatus appears in prophase and has important role in the orientation and gripping of chromosomes and other cellular elements causing them to separate and migrate to opposite cell poles.

Substances that disallow the formation of the mitotic apparatus, like colchicine, a molecule that binds to tubulin molecules and prevents the synthesis of microtubules, interrupt cell division. Colchicine is used to study chromosomes since it paralyzes mitosis when chromosomes are condensed and so are more easily viewed under the microscope.

Question 23.
What are the main events of the second mitotic period?
Answer:
The second mitotic period is metaphase. In metaphase the following events occur : condensed chromosomes bind (in their centromere region) to the spindle fibres and get concentrated in the middle of the cell; the formation of the mitotic apparatus is completed. Metaphase ends with the breaking of the binding of identical chromatids and then anaphase begins.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 24.
What are the main events of the third mitotic period?
Answer:
The third mitotic period is anaphase. In anaphase the following events occur: duplication and breaking of centromeres with separation of identical chromatids; traction (by the spindle fibres) of identical chromatids each to opposite cell poles; beginning of chromosome decondensation.

Question 25.
During mitotic anaphase is there separation of homologous chromosomes or separation of identical chromatids?
Answer:
In the anaphase of mitosis the identical chromatids separate and complete pairs of homologous chromosomes continue to exist in each daughter cell. The separation of the homologous chromosomes occurs in the anaphase of the cell division by meiosis.

Question 26.
What are the main events of the final mitotic period?
Answer:
The final mitotic phase is telophase. In telophase the following events occur: decondensation of chromosomes, each set located in opposite cell poles; karyotehca formation around each set of chromosomes forming two nuclei; destruction of the mitotic apparatus; reappearing of the nucleoli; beginning of cytokinesis (the division of cytoplasm to ultimately separate the new cells).

Question 27.
How does the quantity of genetic material vary within the cell during the sequential phases of the cell cycle?
Answer:
The first period of the first phase (interphase) of the cell cycle is the G1, followed by S and G2 and then by the mitotic phase.
In G1 the ploidy (the quantity of DNA molecules in the cell) can be represented by the formula 2n ( n is the number of DNA molecules in a gamete cell of a given species). In S DNA duplicates and the quantity of genetic material increases from 2n to 4n. In G2 that quantity is constant: 4n. After the mitotic phase the quantity of genetic material decreases to 2n in each daughter cell.

Question 28.
Concerning their final products (daughter cells and their ploidies) what are the differences between mitosis and meiosis?
Answer:
In mitosis one cell, for example, with 2n chromosomes, duplicates its chromosomal set and divides generating two other cells, each with 2n chromosomes too. In meiosis, one diploid cell (2n) duplicates its chromosomes too, but four cells with n chromosomes are generated.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 29.
Why is meiosis important for the maintenance of the normal quantity of chromosomes of a species with sexual reproduction?
Answer:
A reduction to a half of the maximum normal quantity of chromosomes is mandatory in some phase of the life cycle of a species that reproduces sexually. If that could not happen in each generation, whenever a zygote is formed by fusion of gametes there would be duplication in the quantity of chromosomes in a geometric progression.

Question 30.
What are the two divisions of meiosis? What are the main events that occur in those divisions?
Answer:
Meiosis is divided into first meiotic division, or meiosis I, and second meiotic division, or meiosis II. During meiosis I the separation of homologous chromosomes occurs, with formation of two haploid cells. In meiosis II there is separation of identical chromatids of each of the two haploid cells created in meiosis I, giving birth to four haploid cells.
Meiosis II is a process identical to mitosis.

Question 31.
In which meiotic division does the separation of the homologous occur? What are the ploidies of the generated cells after the end of that process?
Answer:
The separation of the homologous chromosomes occurs in the first division of meiosis, or meiosis I. After the end of this cell division two haploid cells are made, each having different chromosomes (with no set of homologous). Note that in the cells generated after meiosis I each chromosome is still duplicated since the homologous chromosomes and not the identical chromatids were separated.

Question 32.
In which meiotic division does the separation of identical chromatids occur? After the end of this process what are the ploidies of the new cells?
Answer:
The separation of identical chromatids occurs in the second meiotic division, or meiosis II. After this cell division (similar to mitosis and that does not alter ploidy) the cells are still haploid (they have become haploid after meiosis I).

Question 33.
In which period of meiosis does the pairing of homologous chromosomes occur?
Answer:
The pairing of homologous chromosomes is a vital step for meiosis because the rightness of the homologous separation depends on the process. This event occurs in prophase I of the cell division.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 34.
What is crossing over? In which period of meiosis does this event occur?
Answer:
Crossing over is the eventual.exchange of chromosomal fragments between homologous chromosomes. The phenomenon occurs in prophase I when homologous chromosomes are paired. Crossing over is of great importance for evolution and biodiversity since it provides recombination of alleles (of different genes) linked in the same chromosome during cell divison by meiosis.

Question 35.
What are the ‘chiasms’ of homologous chromosomes seen in prophase I?
Answer:
Chiasms are intersections of two tracts in the form of ‘x’.
The chiasms seen in prophase I are chromosome arms crossing over same arms of their homologous. In fact when chiasms are seen under the microscope chromatids exchange chromosomal segments with other chromatids of its homologous.

Question 36.
What are the respective functions of the separation of homologous chromosomes and of the separation of identical chromatids in meiosis?
Answer:
The separation of homologous chromosomes in meiosis I has two main functions: to reduce to half the total number of chromosomes, generating haploid daughter cells at the end of the process, and to make possible genetic recombination since the separation is aleatory, i.e., each pair of daughter cells can be different from the other pair relating chromosomal combination from paternal and maternal origins.
The separation of identical chromatids in meiosis II has the same function it has in mitosis: to separate the chromosomes already duplicated to the daughter cells.

Long Answer Type Questions : 5 Marks

Question 1.
State two significance of each of mitotic and amitotic cell division. What is gene?
Answer:
Significance of mitotic cell division :
i. Helps in growth and development of organs and body of organism.
ii. Helps in maintaining proper size.

Significance of amitetic cell division :
i. Helps in multiplication of cells in foetal membranes.
ii. Easy process.

Gene : Gene is the heriditary unit consisting of a particular sequence of basis and amino acids in DNA and specifying the production of distinct enzymatic protein in the cell.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 2.
Describe anaphase stage of mitosis. Write differences between Mitosis and Meiosis.
Answer:
Anaphase stage of mitosis :

  1. The centromeres of the chromosomes divide and the two chromatids of each pair separate. They are called-daughter chromosomes.
  2. The daughter chromosomes migrate towards poles due to shortening of spindle fibres attached to the centromeres.
  3. Pulling causes the chromosomes to assume their characteristic V-shape of L-shape etc.
Mitosis Meiosis
i. Occurs in somatic cell i. Occurs in germ mother cell or spare mother cell.
ii. No crossing over takes place. ii. Crossing over takes place.
iii. Two diploid cells are formed from a diploid cell. iii. Four haploid cells are formed from a diploid cell.
iv. Cell divides only once. iv. Cell divides twice.

Question 3.
Discuss characteristics of prophase of mitotic cell division in an animal cell.
Answer:

  1. Chromosomes are condensed, thicken and become stainable.
  2. Chromatids coil around each other and are held together at centromere.
  3. The chromosomes are evenly distributed in the neucleoplasm.
  4. Nucleoli gradually deminishes in size and disappear.
  5. Centrioles duplicate and move towards opposite poles.
  6. The movement of centrioles are due to they are being pushed by the growth of spindle fibres.
  7.  Asters are formed from cytoplasm.

Question 4.
When does DNA synthesis occur during interphase? Discuss process of cytokinesis in plant cell and in animal cell.
Answer:
S-phase : DNA synthesis occurs.
Cytokinesis for plant cell : Cytokinesis in plant cell involves the formation of a cell wall in between the daughter nuclei. This begins as a cell plate or phragmoplast formed by aggregation of vesicles from Golgi bodies. These vesicles fuse with each other to form cell membranes and cell walls, there by the cell divides into two parts. The spindle which is also involved in the cell plate formation, then disintegrates and cell division is completed.

Cytokinesis in animal cell : A cleavage furrow appears at the outer edges of the cell and midway between the poles at the beginning. This furrow of constriction becomes progressively deeper as the spindle breaks down. Finally the ingrowing constriction joins and cleves the cell into two parts.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 5.
What is the name of the cytoplum divioion in the end of mitosis? What are the differences in this process ‘bétween animal and plant cells?
Answer:
Cytoplasm division occurs after telophase and it is called cytokinesis. In animal cells an invagination of the plasma membrane toward the cell centre appears in the equator of the parent cell and then the cell is strangulated in that region and divided into two daughter cells. This type of division is called centripetal cytokinesis (from outside).

In plant cells the cytokinesis is not centripetal since the division happens from the inside. Membranous sacs full of pectin concentrate in the internal central region of the cell and propagate to the periphery toward the plasma membrane. The pectin-containing sacs fuse themselves and form a central structure called phragmoplast. On the phragmoplast cellulose deposition occurs and a true cell wall is created to separate the daughter cells. Plant cells thus present centrifugal cytokinesis.
The phragmoplast has ‘failures’ or pores to permit cytoplasmic communications between the daughter cells. These openings are called plasmodesma.

Question 6.
Why is it important for chromosomes to be condensed during mitosis and decondensed during interphase?
Answer:
During mitosis the main problem to be solved is the correct separation of chromosome sets between daughter cells. If chromosomes were decondensed long, tiny fibres of DNA would be dispersed in cytoplasm after the karyotheca breaking and chromosomes could not be easily organized and pulled by the spindle fibres.

During interphase the function of chromosomes, i.e, of DNA molecules, is the synthesis of RNA and thus of proteins. For this task it is necessary for functional molecular regions to be decondensed (these regions form the euchromatin). During interphase in addition DNA replication occurs as a preparatory step for cell division. In this process it is fundamental for the exposition of DNA molecules to serve as templates to new DNA chains under production.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 7.
Concerning their biological function what is the difference between mitosis and meiosis?
Answer:
The main biological function of mitosis is cellular multiplication, a fundamental process for the growth and development of multicellular organisms, tissue renewing, asexual reproduction etc. The biological function of meiosis is gamete formation (in gametic meiosis) or spore formation (in sporic meiosis), i.e., the production of cells qualified for sexual reproduction with half the quantity of chromosomes compared to the original cell.

There is a special type of meiosis that happens in zygotes of some algae, protozoans and fungi. This meiosis, called zygotic meiosis, has the function of reducing half the number of chromosomes of adult individuals that will be formed from the zygote. In species with zygotic meiosis the adult individuals are haploid and they form gametes by mitosis. These gametes fuse in pairs with others and generate a diploid zygote. Then, undergoes meiosis to restitute the normal ploidy of adult individuals.

Question 8.
For the biological diversity is mitosis or meiosis the more important process?
Answer:
Meiosis is the cell division process that allows the formation of gametes to sexual reproduction, with aleatory separation of each chromosome of the individual homologous pairs. These gametes can fecundate gametes from other individuals promoting combination of homologous chromosomes from different individuals. In that manner the chromosomal recombination provided by meiosis and sexual reproduction creates individuals with dissimilar genetic patrimony from their fathers and thus promotes biological diversity.

Some fungi species and plants, for example-present sporic meiosis, i.e., a structure where half of the chromosomes of the species is generated from meiosis. This structure, by mitosis, forms gametes. Even in this case diversity comes from meiosis. Meiosis then is the cell division process that in conjunction with genetic mutations is responsible for the biological diversity. Even in species having zygotic meiosis the aleatory separation of homologous chromosomes in meiosis creates biological diversity.

Question 9.
During which meiosis division does ploidy reduction occur? Does ploidy reduction occur in mitosis?
Answer:
In the cell division by meiosis ploidy reduction occurs in meiosis II. Initially, taking as example a 2n somatic cell, ploidy increases to 4n (duplication of DNA) during interphase. During meiosis I, since homologous chromosomes are separated, ploidy falls to 2n (the original number) and then during meiosis II ploidy finally falls to n in the resulting daughter cells.

Ploidy reduction does not occur in mitosis. This fact shows that, although in meiosis ploidy is decreased from its original number, in meiosis II, a process similar to mitosis, the cause of that reduction is what happens in meiosis I, i.e., the separation of the homologous chromosomes.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 10.
In order for organisms to grow they must produce new cells and each cell rnus maintain the same genetic identity as the original cell. This will insure that the new cells will function properly and survive.
(a) Starting with interphase, explain how the events of the cycle insure that each new cell will be genetically identical to the original?
(b) What features of the prokaryotic genetic material eliminates the need for mitosis to insure identical daughter celis?
Answer:
(a) Replication during S phase results in two identical chromatids. They remain attached which insures proper sorting later. Chromosomes coil and condense which insures easier maneuvering of the chromosomes during mitosis. Nuclear membrane dissolves to allow greater area for movement of chromosomes. Kinetochore proteins on each chromatid allows for attachment to spindle fibers which will pull chromatids apart. Centromeres do not divide until they are in alignment to insure one chromatid moves to each new cell.

(b) The prokaryotic genome consists of only one circular chromosome. Upon replication there are only two chromosomes to sort. Each chromosome attaches to a specific site on the membrane and growth of the cell separates them. There is no need for the mitotic process to sort the chromosomes.

Question 11.
Write a short note on Cell cycle.
Answer:
Howard and Pelc (1953) first time described it. A series of changes or sequential events which occur regularly in a dividing cell is called cell cycle. The cell cycle includes the period from the beginning of one cell division to the beginning of the next cell division.
The time interval between two successive divisions is called generation time.
The cell cycle consists of two main phases i.e.
(i) Interphase or I – phase includes G1, S, G2 stages.
(ii) Mitosis or M – phase represents only a small part of life cycle of a cell i.e. about an hour in most cells.
i. Interphase or I-phase :
Resting phase.
Preparatory and synthetic phase of cell division. During this phase, the cell is metabolically active. An active phase for replication of DNA and synthesis of histone

proteins. This phase also involves synthesis of energy-rich compounds to supply energy. The interphase is further divided into three sub-phases : (a) G1 phase (b) S phase (c) G2 phase

G1 phase : First phase. Also called the post-mitotic gap phase. The cell grows in size due to synthesis of RNA, nucleotides, amino acids and proteins. In some cases, cell may not divide and it enters a quiescent stage called GG0 phase. It is also called as restriction point.

S phase : It is synthetic phase. It occurs between G1 phase and G2 phase. Replication of DNA takes place during this stage. DNA content doubles and a duplicate set of chromosomes is formed.

G2 phase: Second phase. Also called the pre-mitotic gap phase. Synthesis of spindle proteins and RNA. The cell prepares to enter mitotic phase. Duplication of cell organelles also takes place.

ii. Mitosis or M-phase : Shortest phase in the cell cycle. A cell divides to form two daughter cells, identical to each other and also to the mother cell. The division of nucleus and cytoplasm occurs only once.

During mitosis, the number of chromosomes remains the same. Mitosis can take place in both haploid and diploid cells.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle 1

Also Called :
Somatic cell division : Cell division results in the formation of somatic body cells.

Equational division: Mitosis maintains chromosome number in daughter cells equal to that of the mother cell.

Question 12.
Describe about mitosis with proper diagram.
Answer:
The term ‘Mitosis’ was coined by Walther Flemming, a German biologist who for the first time observed dividing cell in 1878. Mitosis takes place in two stages: Karyokinesis and Cytokinesis.
(i) Karyokinesis (Nuclear division) : Nucleus undergoes a series of changes to form two daughter nuclei. Various stages of Karyokinesis are prophase, metaphase, anaphase and telophase.

a. Prophase (pro = first; phase = stage) :
Longest phase of mitosis. During early prophase, chromosomes become short and thick due to condensation and coiling. It appears as single-stranded (monad chromosome), but by the end of prophase, it appears double-stranded (dyad chromosome). It consists of two identical chromatids joined by a centromere. In late prophase, the nuclear membrane and nucleolus begin to disappear. The centrioles (in animal cells) start moving in opposite direction, till they occupy polar position.

b. Metaphase (meta = middle; phase = stage) :
Nuclear membrane and nucleolus completely disappear. The short and thick chromosomes get organized along the equatorial plane of the cell. Formation of spindle fibres takes place. Spindle fibres are fine thread-like structures formed by the proteins called tubulin. Tubulin is organized into protein units called microtubules. Spindle fibres are made up of microtubules. The spindle fibres are of two types.
Continuous fibres or non-chromosomal fibres : Fibres extend from pole to pole. Discontinuous or chromosomal fibres: Fibres extend from pole to centromere. Metaphase is the best stage to count the number of chromosomes during mitosis.
c. Anaphase (ana = up; phase = stage) :
It is the shortest phase of karyokinesis in which the centromere divides into two, resulting in the separation of chromatids.

Each separated chromatid is now called the daughter chromosome. The sister chromatids repel each other.

The chromosomal spindle fibres undergo contraction and pull the daughter chromosomes to the respective poles to form two groups of chromosomes.
d. Telophase ( telo = end; phase = stage ) :
The separated daughter chromosomes undergo uncoiling to form long thin chromosomes which become indistinct to form chromatin network.

The nuclear membrane begins to reappear around each group of chromosomes to form a daughter nucleus.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle 2

The spindle fibres disappear while nucleolus reappears in each newly formed nucleus.
Cytokinesis :
Cytokinesis: Division of cytoplasm of the mother cell. Cytokinesis starts at the end of telophase.

In animal cells : Cytokinesis occurs by furrowing of plasma membrane (cleavage). In animal cell, cytokinesis takes place by cleavage constriction of cell cytoplasm. It begins peripherally and centripetally.

In plant cells : Cytokinesis takes place by formation of cell plate. Cell plate is first laid down in centre and then proceeds towards periphery. Vesicles provided by Golgi apparatus join to form phragmoplasts, which further join to form cell plate.
Cytokinesis results in formation of two daughter cells.
Significance of Mitosis :
Growth and development of multicellular organisms take place by forming new somatic cells.

  • It maintains surfaces volume ratio of the cell.
  • It maintains chromosome number of parent cell in the daughter cells.
  • It plays a significant role in the reproduction of unicellular organisms.
  • It plays an important role in healing and repairs by producing new cells.
  • Mitosis distributes and carries hereditary materials or genes to the next generation.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 13.
Write a short note on Amitosis?
Answer:
Discovered by Remark. No synthesis of spindle fibres. No condensation of chromatin fibres. No disappearing of nuclear membrane.
Occurrence : In prokaryotes, in yeast cells, in eukaryotes Amoeba, meganucleus of Paramecium, endosperm cells, foetal membrane cells etc.
Mechanism : Two steps; Karyokinesis (karyon = nucleus; kinesis = movement): Nucleus divides into two.
Cytokinesis (kytos = cell; kinesis = movement): Cytoplasm divides and two cells are formed. Unequal distribution of chromatin in the daughter cells, leads to structural and functional irregularities.

Multiple Choice Questions : 1 Mark

Question 1.
The phase of mitosis cell division in which two sister chromatids are separated is _______.
(a) Prophase
(b) Metaphase
(c) Anaphase
(d) Telophase
Answer:
(c) Anaphase

Question 2.
In a DNA molecule complementary base pair of guanine is _______.
(a) Adenine
(b) Cytosine
(c) Thymine
(d) Uracil
Answer:
(b) Cytosine

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 3.
Number of chromosome present in germ cell of man is _______.
(a) 46
(b) 47
(c) 23
(d) 22
Answer:
(c) 23

Question 4.
In RNA molecule the complementary base of adenine is _______.
(a) Guanine
(b) Thymine
(c) Uracil
(d) Cytosine
Answer:
(c) Uracil

Question 5.
The phase of mitosis cell division in which two sister chromatids are separated is
(a) Prophase
(b) Metaphase
(c) Anaphase
(d) Telophase
Answer:
(c) Anaphase

Question 6.
Who discovered Meiosis?
(a) Flemming
(b) Weldeyar
(c) Farmar
(d) Pavlov
Answer:
(c) Farmar

Question 7.
The number of chromosomes of pollen mother cell is 10 . The number of chromosomes of the flower of some plant will be
(a) 10
(b) 20
(c) 5
(d) 30
Answer:
(b) 20

Question 8.
Who named the chromatins as chromosomes?
(a) Flemming
(b) Waldeyar
(c) Farmar
(d) Pavlov
Answer:
(b) Waldeyar

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 9.
In which of the stages of mitotic cell division does the nuclear membrane reappear?
(a) Prophase
(b) Metaphase
(c) Telophase
(d) Anaphase
Answer:
(d) Anaphase

Question 10.
In the specific part of DNA chain, if there be adenine and guanine in one chain then which N2 bases will be present in another chain as a complementary base respectively?
(a) Thymine and cytosine
(b) Thymine and adenine
(c) Guanine and cytosine
(d) Adenine and thymine
Answer:
(a) Thymine and cytosine

Question 11.
Chromosomes move towards the polar region of spindle during
(a) Prophase
(b) Metaphase
(c) Telophase
(d) Anaphase
Answer:
(b) Metaphase

Question 12.
Chromosome number of human gamet is _______.
(a) 32 pairs
(b) 23 pairs
(c) 32
(d) 23
Answer:
(d) 23

Question 13.
Chromosomes can be best at the stage of
(a) Late anaphase
(b) Telophase
(c) Metaphase
(d) Late prophase
Answer:
(c) Metaphase

Question 14.
In mitosis chromosome duplication occurs during
(a) Interphase
(b) Prophase
(c) Late prophase
(d) Late telophase
Answer:
(a) Interphase

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 15.
In mitosis the number of chromosome sets in daughter cells
(a) half the number in the parent cell
(b) twice the number
(c) one fourth the number
(d) same as in parent cell
Answer:
(d) same as in parent cell

Question 16.
A complete set of chromosome inherited as a unit from one parent is known as
(a) genotype
(b) karyotype
(c) genepool
(d) genome
Answer:
(d) genome

Question 17
Separation of sister chromatids takes place in
(a) anaphase of mitosis
(b) anaphase I of meiosis
(c) anaphase II of meiosis
(d) anaphase of mitosis and anaphase II of meiosis
Answer:
(d) anaphase of mitosis and anaphase II of meiosis

Question 18.
Homologous chromosomes are separated during
(a) anaphase of mitosis
(b) metaphase of meiosis
(c) anaphase I of meiosis
(d) anaphase I of meiosis and anaphase of mitosis
Answer:
(d) anaphase I of meiosis and anaphase of mitosis

Question 19.
Longest phase in cell cycle
(a) G1
(b) S
(c) G2
(d) M
Answer:
(a) G1

Question 20.
Which is the longest phase in Meiosis I?
(a) Prophase I
(b) Metaphase I
(c) Anaphase I
(d) Telophase I
Answer:
(a) Prophase I

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 21
Synapsis is characteristic of?
(a) Leptotene
(b) Zygotene
(c) Pachytene
(d) Diplotene
Answer:
(b) Zygotene

Question 22.
Which is the longest stage of Prophase I of Meiosis 1 ?
(a) Leptotene
(b) Zygotene
(c) Pachytene
(d) Diplotene
Answer:
(c) Pachytene

Question 23.
Synaptonemal complex completed düring
(a) Leptotene
(b) Zygotene
(c) Pachytene
(d) Diplotene
Answer:
(c) Pachytene

Question 24.
Terminalisation of chiasma occurs at
(a) Diakinensis
(b) Pachytene
(c) Zygotene
(d) Diplotene
Answer:
(a) Diakinensis

Question 25.
Chiasmata in cell division is seen in
(a) Leptotene
(b) Zygotene
(c) Pachytene
(d) Diplotene
Answer:
(d) Diplotene

Question 26.
Synapsis is the pairing of
(a) Homologous chromosomes
(b) Non homologous chromosomes
(c) Analogous chromosomes
(d) Paralogous chromosomes
Answer:
(a) Homologous chromosomes

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 27
The chromosomes appear as long thin threads in
(a) Leptotene
(b) Zygotene
(c) Pachytene
(d) Diplotene
Answer:
(a) Leptotene

Question 28.
A cell in G1 of interphase has 12 chromosomes how many chromatids will be found during metaphase two of meiosis?
(a) 6
(b) 12
(c) 18
(d) 24
Answer:
(b) 12

Question 29.
How many gametes would be formed from a genotype AaBBccDdEe?
(a) 8
(b) 16
(c) 32
(d) 64
Answer:
(a) 8

Question 30
Amitosis is _______.
(a) cleavage of nucleus without recognizable chromosomes
(b) a division in which chromosomes are unequally distributed
(c) a division in which chromosome bridge are formed
(d) a division in which spindles are formed
Answer:
(a) cleavage of nucleus without recognizable chromosomes

Question 31.
Homologous chromosome includes
(a) one smaller and one bigger chromosome
(b) one chromosome from each parent
(c) one complete and one incomplete chromosome
(d) none of these
Answer:
(b) one chromosome from each parent

Question 32.
How many mitotic division must occur in a cell of the root tip to form \mathbf{1 2 8 cells?
(a) 7
(b) 8
(c) 64
(d) 128
Answer:
(a) 7

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 33.
The minimum number of meiotic division to obtain 100 pollen grains of wheat
(a) 25
(b) 50
(c) 75
(d) 100
Answer:
(a) 25

Question 34.
The minimum number of chiasmata in a bivalent is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(a) 1

Question 35.
Chromosome exhibits minimum coiling during
(a) Prophase
(b) Metaphase
(c) Anaphase
(d) Telophase
Answer:
(a) Prophase

Question 36.
Interkinesis is a period between
(a) Prophase I and metaphase I
(b) Metaphase I and anaphase I
(c) Anaphase I and telophase I
(d) Meiosis I and meiosis II
Answer:
(d) Meiosis I and meiosis II

Question 37.
Down syndrome is caused due to nondisjunction of chromosomes at
(a) Anaphase of mitosis
(b) Metaphase I of meiosis
(c) Anaphase I of meiosis
(d) Telophase I of meiosis
Answer:
(c) Anaphase I of meiosis

Question 38.
Cell cycle progression from one phase to another is primarily controlled by
(a) phosphorylation of CDKs
(b) proteolysis of cyclin
(d) dephosphorylation of cyclin
(e) proteolysis of CDKs
Answer:
(a) phosphorylation of CDKs

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 39.
CDK associated with S phase
(a) CDK 2
(b) CDK 1
(c) CDK 3
(d) CDK 4
Answer:
(a) CDK 2

Question 40.
DNA sequence is responsible for chromatid separation
(a) centromere
(b) telomere
(c) kinetochore
(d) satellite
Answer:
(a) centromere

Question 41.
During cell cycle sister chromatids are pulled apart during
(a) Metaphase
(b) Anaphase
(c) Prophase
(d) Interphase
Answer:
(b) Anaphase

Question 42.
Among the following the most variable stage of cell cycle is
(a) G1
(b) S
(c) G2
(d) M
Answer:
(a) G1

Question 43.
The Mendelian law of independent assortment is due to the arrangement of chromosome during
(a) anaphase I
(b) anaphase II
(c) S phase
(d) cytokinesis
Answer:
(a) anaphase I

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 44.
If cell has 46 chromosomes at the beginning of mitosis, then at anaphase there would be a total of
(a) 23 chromatids
(b) 46 chromatids
(c) 46 chromosomes
(d) 92 chromosomes
Answer:
(d) 92 chromosomes

Question 45.
If a cell has 46 chromosomes at the beginning of meiosis, then at anaphase I there would be a total of
(a) 23 chromatids
(b) 46 chromatids
(c) 46 chromosomes
(d) 92 chromosomes
Answer:
(b) 46 chromatids

Question 46.
Approximately how many cells are present in the body of an adult person ?
(a) 1014
(b) 1015
(c) 1018
(d) 1021
Answer:
(d) 1021

Question 47.
The period between two successive cell divisions is called
(a) Duplication
(b) Growth phase
(c) Cell cycle
(d) Interphase
Answer:
(d) Interphase

Question 48.
Which is fundamental property of all living organisms ?
(a) Respiration
(b) Germination
(c) Growth
(d) Photosynthesis
Answer:
(a) Respiration

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 49.
Which factors are required for growth ?
(a) An increase in group of cells, a duplication of genetic material
(b) An increase in group of cells, production of daughter cells by mitosis
(c) A dupliction of genetic material and a division assuring that daughter cells receive an equal complement of genetic material.
(d) An increase in cell mass, a duplication of genetic material, a division assuring that each daughter cell receives an equal complement of the genetic material
Answer:
(d) An increase in cell mass, a duplication of genetic material, a division assuring that each daughter cell receives an equal complement of the genetic material

Question 50.
With how many cell, reproduction starts ?
(a) Two cells
(b) Single cell
(c) Many cells
(d) Somatic cell
Answer:
(b) Single cell

Question 51.
Which of the following is present in maximum number in an adult person ?
(a) Somatic cell
(b) Gamete
(c) Reproductive cell
(d) Zygote
Answer:
(a) Somatic cell

Question 52.
At the end of which stage does cell enter mitosis ?
(a) G1 – phase
(b) S – phase
(c) M – phase
(d) G2– phase
Answer:
(d) G2– phase

Question 53.
What is synthesized during G2 – phase ?
(a) Protein
(b) Micro tubules
(c) RNA
(d) (a)and(b)
Answer:
(b) Micro tubules

Question 54.
The sequence in the cell cycle is _______.
(a) S, G1 G2 M
(b) G1, S G, M
(c) S, MG1, G2, M
(d) G2, S, M, G1
Answer:
(b) G1, S G, M

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 55.
Synthesis of RNA and protein takes place in which phase of the cell cycle?
(a) S-phase
(b) M-phase
(c) G1-phase ,
(d) Metaphase
Answer:
(a) S-phase

Question 56.
During mitosis, ER and nucleolus begin to disappear at
(a) Late prophase
(b) Early prophase
(c) Late metaphase
(d) Early metaphase
Answer:
(b) Early prophase

Question 57.
Mitosis is similar to
(a) Meiosis I
(b) Meiosis II
(c) Both a and b
(d) None of these
Answer:
(b) Meiosis II

Question 58.
Which of the following represents the best stage to view the shape, size and number of chromosomes?
(a) Interphase
(b) Prophase
(c) Metaphase
(d) Telophase
Answer:
(c) Metaphase

Question 59.
During metaphase mitosis chromosomes
(a) undergo coiling
(b) line up at the equator
(c) break and disintegrate
(d) none of these
Answer:
(b) line up at the equator

Question 60.
Number of chromatids at metaphase is
(a) Two each in mitosis and meiosis
(b) Two in mitosis and one in meiosis
(c) Two in mitosis and four in meiosis
(d) One in mitosis and two in meiosis
Answer:
(a) Two each in mitosis and meiosis

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 61.
During cell division in apical meristem, the nuclear membrane appears in
(a) Metaphase
(b) Telophase
(c) Anaphase
(d) Cytokinesis
Answer:
(b) Telophase

Question 62.
Mitotic anaphase differs from metaphase in possessing
(a) Same number of chromosomes and same number of chromatids
(b) Half number of chromosomes and half number of chromatids
(c) Half number of chromosomes and same number of chromatids
(d) Same number of chromosomes and half number of chromatids
Answer:
(d) Same number of chromosomes and half number of chromatids

Question 63.
How many mitotic divisions are needed for a single cell to make 128 cells?
(a) 7
(b) 14
(c) 28
(d) 32
Answer:
(a) 7

Question 64.
Which aspect of mitosis is affected by colchicine in inducing polyploidy?
(a) DNA replication.
(b) Spindle formation
(c) Formation of cell plate
(d) Chromosome doubling
Answer:
(b) Spindle formation

Question 65.
If you are provided with root tips of onion in your class and are asked to count the chromosomes, which of the following stages can you most conveniently look into?
(a) Prophase
(b) Metaphase
(c) Anaphase
(d) Telophase
Answer:
(b) Metaphase

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 66.
Meiosis takes place
(a) Only in haploid individuals
(b) Only in diploid individuals
(c) Both in haploid and diploid individuals
(d) None of these
Answer:
(b) Only in diploid individuals

Question 67.
In one of the following stages of cell division, the DNA content is doubled. It is
(a) G1 phase
(b) G2 phase
(c) Metaphase
(d) S phase
Answer:
(d) S phase

Question 68.
Amitosis is the usual process of cell division in
(a) Meristematic cells
(b) Prokaryotic cells
(c) Eukaryotic cells
d) Spore mother cells
Answer:
(b) Prokaryotic cells

Question 69.
During synapsis the number of thread (Chromonemata) in each chromosome is
(a) 2
(b) 4
(c) 8
(d) Many
Answer:
(a) 2

Question 70.
Homologous chromosomes are
(a) Morphologically and genetically similar
(b) Morphologically similar
(c) Those which pair during synapsis
(d) None of these
Answer:
(c) Those which pair during synapsis

Question 71.
Cell division of mitosis is a normal process in a living cell but sudden and abnormal mitosis in an organ will sometime result in
(a) Cancer
(b) New organ
(c) Zygote
(d) Gastrula
Answer:
(a) Cancer

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 72.
Synapsis is characteristic of
(a) Leptotene
(b) Diplotene
(c) Zygotene
(d) Pachytene
Answer:
(c) Zygotene

Question 73.
The mechanism ensuring genetic continuity in mitosis is
(a) Having of chromosome number between the two new cells
(b) Formation of cells with 8 chromosomes
(c) Formation of two daughter cells
(d) Formation of two cells with identical kind of DNA
Answer:
(d) Formation of two cells with identical kind of DNA

Question 74.
During meiosis, crossing over occurs at
(a) Diplotene
(b) leptotene
(c) Pachytene
(d) Diakinesis
Answer:
(a) Diplotene

Question 75.
How many generations of mitotic division must occur in a cell of root tip to form 256 cells?
(a) 8
(b) 32
(c) 64
(d) 128
Answer:
(a) 8

Question 76.
Which of the following is not true concerning mitosis?
(a) Animal cells have centrioles while plant cells do not.
(b) Both plant and animal cells undergo cytokinesis.
(c) Mitosis allows growth and increases in size in both plants and animals.
(d) Animal cells form a cell plate during cytokinesis while plant cells do not
Answer:
(d) Animal cells form a cell plate during cytokinesis while plant cells do not

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 77.
The term that refers to division of the cytoplasm is
(a) cytokinesis
(b) anaphase
(c) apoptosis
(d) karyokinesis
Answer:
(a) cytokinesis

Question 78.
Cancer cells require lot of nutrients, which are supplied by blood vessels. The growth of new blood vessels to cancerous tissue is called
(a) angiogenesis
(b) metastasis
(c) carcinogenesis
(d) apoptosis
Answer:
(a) angiogenesis

Question 79.
Which is not true about bacterial chromosomes?
(a) There is generally only one chromosome in each bacterial cell.
(b) A bacterial chromosome is present in a single copy per cell.
(c) A bacterial chromosome is attached to the plasma membrane.
(d) A bacterial chromosome contains both DNA and associated histones.
Answer:
(d) A bacterial chromosome contains both DNA and associated histones.

Question 80.
The structure that contains the genetic information in a bacterial cell is called the
(a) nucleus
(b) nuçleoid
(c) nucleolus
(d) nucleosome
Answer:
(b) nuçleoid

Question 81.
The term “nucleoid” is built from the Latin and Greek root word meaning
(a) “center” and “outside.”
(b) “kernel-like.”
(c) “true-center.”
(d) “master control.”
Answer:
(b) “kernel-like.”

Question 82.
Virtually all specialized cells of multicellular organisms
(a) develop through mutation from less specialized cells of the organism.
(b) contain more genetic material than less specialized cells of the same organism.
(c) contain less genetic material than less specialized cells of the same organism.
(d) have the same amount of DNA and the same number of chromosomes as all other cells of the organism.
Answer:
(d) have the same amount of DNA and the same number of chromosomes as all other cells of the organism.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 83.
The term used for DNA in a nucleus before it becomes condensed in preparation for mitosis is
(a) eukaryote
(b) chromosome
(c) chromatin.
(d) cytokinesis
Answer:
(c) chromatin.

Question 84.
Generally, animals build an organism using the diploid number of chromosomes. However, insects in the order of ants, wasps and bees can use a haploid-diploid system where adults of one sex are formed with a haploid number of chromosomes. This would mean that
(a) a single set of chromosomes is sufficient to code for a functional individual.
(b) a female could determine the sex of the offspring by fertilizing or not fertilizing an egg.
(c) the males and females are not equally “related” to their mothers considering the proportion of genes held in common.
(d) not all animals must be diploid in order to be viable organisms.
Answer:
(d) not all animals must be diploid in order to be viable organisms.

Question 85.
Which is not true about the chromosomes of a multicellular organism?
(a) They are made up of DNA and protein
(b) Each chromosome is replicated into two chromatids during the S phase of interphase
(c) Each chromosome separates into two daughter chromosomes by binary fission
(d) All cells contain chromosomes that carry the same genetic information.
Answer:
(c) Each chromosome separates into two daughter chromosomes by binary fission

Question 86.
In prokaryotes the chromosome is held in a region called the
(a) nucleoid
(b) centriole
(c) centrosome
(d) kinetochore
Answer:
(a) nucleoid

Question 87.
Generally, complex organisms do require more genes to control their synthesis and organization than do primitive organisms. However, the numbers of chromosomes vary from ants with 2 , molds with 8-14, humans with 46, potatoes with 100 and the crayfish with 200 . Given this, then
(a) there must be no relationship between amount of genetic information and complexity of the organism
(b) the number of genes per chromosome may vary among organisms, preventing a simple relationship between chromosome number and complexity.
(c) mitosis must differ from organism to organism.
(d) simpler organisms have more DNA than more complex organisms.
Answer:
(b) the number of genes per chromosome may vary among organisms, preventing a simple relationship between chromosome number and complexity.

Question 88.
Which statement is not true about eukaryotic chromosomes?
(a) There is only one chromosome of each type in each body cell.
(b) Chromosomes contain both DNA and associated histones.
(c) Chromosomes condense from chromatin at the start of mitosis.
(d) Chromosomes disperse back into chromatin at the end of mitosis.
Answer:
(a) There is only one chromosome of each type in each body cell.

Question 89.
The diploid (2n) number of chromosomes for humans is
(a) 23
(b) 24
(c) 44
(d) 46
Answer:
(d) 46

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 90.
The haploid (n) number of chromosomes for humans is
(a) 23
(b) 24
(c) 44
(d) 46
Answer:
(a) 23

Question 91.
Which statement is NOT true about mitosis?
(a) Mitosis is a process that duplicates and divides the nuclear contents only.
(b) Mitosis produces two daughter cells that contain the same number of chromosomes as the parent cell.
(c) Mitosis produces two daughter cells that contain the same kinds of chromosomes as the parent cell.
(d) Mitosis uses a 2n parent cell to form daughter cells containing n chromosomes
Answer:
(d) Mitosis uses a 2n parent cell to form daughter cells containing n chromosomes

Question 92.
Which represents the correct sequence of stages in the cell cycle?
(a) G1, G2, S, M
(b) G1, G2, M, S
(c) G1, M, G2, S
(d) G1, S, G2, M
Answer:
(d) G1, S, G2, M

Question 93.
Below the skin are “stem cells” that divide with some cells continuing the stem cell line and others being pushed toward the surface to flatten and sloughed off. In the bone marrow are other stem cells producing cells such as the erythrocytes that lose their nucleus and function for a few months in the bloodstream before they too die. Such “dead end” cells that reproduce no further
(a) leave the cell cycle in a GG0 phase, which immediately follows telophase.
(b) leave the cell cycle in G3 phase, which immediately follows G2
(c) halt in the midst of the S phase
(d) continually cycle but simply fail to go through cytokinesis
Answer:
(a) leave the cell cycle in a G0 phase, which immediately follows telophase.

Question 94.
At the completion of mitosis which of the following is not true?
(a) The cell has not yet undergone the G2 phase
(b) Each replicated chromosome has been separated into single chromatids.
(c) The daughter cell nuclei have acquired the same number and kinds of chromosomes as the parent cell nucleus.
(d) The cell may not have yet undergone cytokinesis
Answer:
(a) The cell has not yet undergone the G2 phase

Question 95.
The critical checkpoints that control the cell cycle are at the
(a) G1 to S stage and G2 to M stage
(b) S to G2 stage and G2 to M stage
(c) M to G1 stage and G2 to M stage.
(d) M to G1 stage and S to G2 stage.
Answer:
(a) G1 to S stage and G2 to M stage

Question 96.
During which stage of the cell cycle is cell growth and replication of organelles most significant?
(a) M phase
(b) G1 phase
(c) G2 phase
(d) S phase
Answer:
(d) S phase

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 97.
Which stage is most associated with a cell that is unable to divide again, such as a muscle or nerve cell?
(a) M phase
(b) G1 phase
(c) G2 phase
(d) G0 phase
Answer:
(d) G0 phase

Question 98.
Which sequence of stages in mitosis is correct?
(a) prophase, anaphase, prometaphase, metaphase, telophase
(b) prophase, telophase, anaphase, prometaphase, metaphase
(c) prophase, prometaphase, metaphase, anaphase, telophase
(d) telophase, anaphase, prophase, prometaphase, metaphase
Answer:
(c) prophase, prometaphase, metaphase, anaphase, telophase

Question 99.
Which occurs in metaphase?
(a) Centrioles move to opposite poles.
(b) Chromosomes line up along the equator of the dividing cell
(c) Chromosomes move to opposite poles.
(d) The nuclear envelope disappears.
Answer:
(b) Chromosomes line up along the equator of the dividing cell

Question 100.
Which occurs in anaphase?
(a) Centrioles move to opposite poles.
(b) Chromosomes line up along the equator of the dividing cell
(c) Chromosomes move to opposite poles
(d) The nuclear envelope disappears.
Answer:
(c) Chromosomes move to opposite poles

Question 101.
Plant cells differ from animal cells in mitosis in all except which of these ways?
(a) Plants lack centrioles, but animals have them
(b) Plants lack microtubules and spindles, but animals have them
(c) Plants lack cell furrows, but animals have them
(d) Plants form a cell plate, but animals do not
Answer:
(b) Plants lack microtubules and spindles, but animals have them

Question 102.
Which does not occur in telophase?
(a) Cytokinesis usually gets under way
(b) The nuclear envelope is being constructed
(c) The centromeres split apart
(d) Chromosomes decondense into chromatin
Answer:
(c) The centromeres split apart

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 103.
Cytokinesis in plant cells differs from animal cells in the following way
(a) The plant endoplasmic reticulum forms a cell plate.
(b) Microtubules are constructed into a cell plate pattern
(c) The Golgi apparatus produces vesicles that migrate along microtubules and fuse to become a cell plate
(d) The inner plasma membrane divides by cytokinesis as in animal cells and then secretes a cellulose cell wall.
Answer:
(c) The Golgi apparatus produces vesicles that migrate along microtubules and fuse to become a cell plate

Question 104.
Binary fission by bactería differs from mitosis because
(a) the chromosome copies attach to the plasma membrane and are pulled apart by cell growth
(b) the chromosome is a simple DNA ‘strand without complex proteins and no spindle forms
(c) there is no nuclear membrane to break down and rebuild
(d) All of the choices are correct.
Answer:
(d) All of the choices are correct.

Question 105.
In multicellular organisms, mitosis is
(a) the means of tissue growth and repair
(b) a way of generating new kinds of mutant or recombinant organisms
(c) the means of sexual reproduction
(d) not useful in stem cell lines that constantly replace skin etc.
Answer:
(a) the means of tissue growth and repair

Question 106.
Which of the following is not true about cancer cells?
(a) They never fully differentiate.
(b) They exhibit contact inhibition.
(c) They exhibit uncontrolled growth
(d) They exhibit disorganized growth.
Answer:
(b) They exhibit contact inhibition.

Question 107.
Which is not correctly associated with cancer?
(a) Angiogenesis forms new blood vessels and brings nutrients and oxygen to the tumor
(b) The disorganized mass of cells is encapsulated and does not invade adjacent tissue
(c) Metastasis establishes new tumors distant from the site of the primary tumor
(d) Cells have receptors to adhere to basement membranes, then secrete proteinase enzymes to invade underlying tissues
Answer:
(b) The disorganized mass of cells is encapsulated and does not invade adjacent tissue

Question 108.
Development of a tumor from a benign form to a malignant form is dependent upon
(a) growth of blood vessels into the tumor
(b) growth of the tumor
(c) Movement of tumor cells from the original location to another part of the body
(d) Lack of function of the cells in the tumor
Answer:
(c) Movement of tumor cells from the original location to another part of the body

Question 109.
Although cancer may originate in many regions of the body, many patients die from cancerous growth in the lungs, lymph glands, or liver. This is most readily explained as
(a) spreading of cancer by angiogenesis
(b) metastasis occurring more commonly in organs that have a filter effect
(c) proteinase enzymes making cancer particularly damaging to these tissues
(d) growth anywhere but in these organs is called benign
Answer:
(b) metastasis occurring more commonly in organs that have a filter effect

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 110.
Apoptosis refers to cell death and
(a) is always biologically detrimental to an organism
(b) is merely the accumulation of genetic errors
(c) can be programmed and is essential to normal development
(d) is a failure in the translation or transcription mechanism
Answer:
(c) can be programmed and is essential to normal development

Question 111.
Stages of the cell cycle include
(a) G1 stage
(b) mitotic stage
(c) cytokinesis
(d) All of the choices are correct.
Answer:
(d) All of the choices are correct.

Question 112.
Most of the cell cycle is spent in
(a) prophase
(b) cytokinesis
(c) interphase
(d) anaphase
Answer:
(c) interphase

Question 113.
Apoptosis
(a) is programmed cell death
(b) is a process that acts to decrease the number of somatic cells
(c) works to oppose the effects of mitosis
(d) All of the choices are correct.
Answer:
(d) All of the choices are correct.

Question 114.
Eukaryotic chromosomes
(a) consist of both DNA and protein.
(b) may occur as chromatin
(c) contain histones responsible for packaging DNA to fit into a small space
(d) All of the choices are correct.
Answer:
(d) All of the choices are correct.

Question 115.
The overall process of mitosis functions to ensure
(a) growth and tissue repair
(b) that each new cell receives a complete set of genetic information
(c) asexual reproduction in some species
(d) All of the choices are correct
Answer:
(d) All of the choices are correct

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 116.
The event that signals the start of anaphase is
(a) division of the centromeres to separate sister chromatids
(b) migration of the centrioles to opposite poles of the nuclear space
(c) a cleavage furrow starts to form
(d) asters disappear
WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle 3
Answer:
(a) division of the centromeres to separate sister chromatids

Question 117.
Chrcimosomes typically appear like the above diagram in which stage of the : cell cycle?
(a) G1
(b) G2
(c) S
(d) Late prophase and metaphase
Answer:
(d) Late prophase and metaphase

Question 118.
What is the term for the structure labelled D ?
(a) centromere
(b) chromosome
(c) kinetochore
(d) chromatid
Answer:
(d) chromatid

Question 119.
Which structure is responsible for attaching to a spindle fibre?
(a) structure “A”
(b) structure ” B ”
(c) structure “C”
(d) structure “D”
Answer:
(c) structure “C”

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 120.
Which of the statements is not true of the above chromosome?
(a) The chromosome is only highly coiled like this during mitosis
(b) Each half of the chromosome is genetically different from the other
(c) Each half will split from the other during anaphase.
(d) Each half of the above chromosome will again be called a chromosome after they split from each other
Answer:
(b) Each half of the chromosome is genetically different from the other

Very Short Answer Type Questions : 1 Mark

Question 1.
What is chromosome?
Answer:
Chromosome : Intranuclear gene bearing fixed numbered, rod shaped structures which become clearly visible during metaphase and anaphase to cell division.

Question 2.
Where does meiosis occur?
Answer:
Germ mother cell.

Question 3.
Mention different stages of cell cycle.
Answer:
(i) G1 phase (ii) S Phase (iii) G2 phase
(iv) Mitotic phase [Prophase, Metaphase, Anaphase, Telophase, Cytokinesis, GG0]

Question 4.
Name the cytoplasmic organelle that helps in the formation of cell plate.
Answer:
Golgi vesicles.

Question 5.
Where is centromere situated?
Answer:
Within primary constriction.

Question 6.
Name the N2 base which is only present in RNA but not in tpNA.
Answer:
Uracil

Question 7.
In which phase of mitotic cell division nucleus reappears?
Answer:
Telophase.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 8.
In which type of cell division the nuclear membrane does not disappè ar? State the importance of S-phase and G2– phase.
Answer:
Amitosis.
Importance of S-phase : New DNA is synthesized
Importance of G2 phase : Synthesis of RNA and protein continues and DNA content is doubled but paired dipoid condition exists.

Question 9.
Mention the number of autosomes present in human ovum.
Answer:
22

Question 10.
In which phase of cell division does a centromere divide.
Answer:
Anaphase.

Question 11.
What is autosome?
Answer:
The chromosomes in the cells which are responsible for the determination of the body (somatic) structure and functions are called autosomes.

Question 12.
Write the nitrogen bases of an RNA molecule.
Answer:
Adenine, guanine, cytosine, uracil.

Question 13.
In which type of cell division do the nucleus and cytoplasm divide directly?
Answer:
Amitosis.

Question 14.
What type of sex-chromosomes are found in the cells of human body?
Answer:
x and y.

Question 15.
In which type of cell division spindle fibre is not formed?
Answer:
Amitosis.

Question 16.
How many daughter cells are produced during one mitosis cell division?
Answer:
Two.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 17.
By which method cytokinesis occurs in animal cell?
Answer:
Cleavage or by furrowing.

Question 18.
What will happen to a cell if karyokinesis is not followed by cytokinesis? From where do the spindle fibres develop during the cell divison of a plant cell and an animal cell?
Answer:
Spindle fibres are formed from microtubules of nucleoplasm in plant cell. Spindle fibres are formed from astral rays in animal cell.

Question 19.
By which part an Eukaryotic chromosome remains attached with spindle fibre?
Answer:
Centromere.

Question 20.
Mention two differences between DNA and RNA :
Answer:

DNA RNA
i. Sugar is deoxyribose sugar. i. Sugar is ribose sugar.
ii. N2 base is thymine. ii. N2 base is uracil.

Question 21.
What is the function of centromeres telomere?
Answer:
Centromere : Helps in attachment of spindle fibre during cell division. Telomere : Prevents the ends of chromosomes from sticking together.

Question 22.
In the chromosome number in pea plant is 2n = 18, what will be the chromosome number in leaf, pollen grain endosperm nucleus and egg of the plant?
Answer:
Chromosome number in leaf = 18(2n)
Chromorsome number in pollen grain = 9(n)
Chromosome number in endosperm nucleus = 27(3n)

Question 23.
What is cytokinesis?
Answer:
The process of cleavage and separation of the cytoplasm and the final stage of mitosis is known as cytokinesis.

Question 24.
What is mitotic apparatus?
Answer:
The aster, the centrioles and the spindle together make up the structure called mitotic apparatus.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 25.
What is the chromosome number in human female gamet?
Answer:
Chromosome number present in human female gamet is 23.

Question 26.
Write two significance of meiosis cell division.
Answer:
i. Meiosis maintains a definite and constant number of chromosome in the cell of a particular species.
ii. In animals meiosis leads to the formation of sexual gamets, the egg ovum and the sperm with haploid number of chromosome.

Question 27.
Which allosomes are present in human male?
Answer:
y

Question 28.
Where is haploid chromosome seen?
Answer:
Germ mother cell.

WBBSE Class 10 Life Science Solutions Chapter 2A Cell Division and Cell Cycle

Question 29.
When is chromosome distinctly visible?
Answer:
Metaphase.

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Detailed explanations in West Bengal Board Class 10 Life Science Book Solutions Chapter 1E Locomotion as a Type of Response in Animals offer valuable context and analysis.

WBBSE Class 10 Life Science Chapter 1E Question Answer – Locomotion as a Type of Response in Animals

Short Answer Type Questions : 2 Marks

Question 1.
What are the roles of fins in locomotion of fishes?
Answer:
The main function of fins are to maintain stability and to change the direction during swimming of fishes. The dorsal and anal fins prevent the fishes from rolling and caudal fins helps to change the direction by action or rudder during swimming.

Question 2.
What are the motivations behind animal locomotion?
Answer:
The motivations behind animal locomotion are to find food, a mate, a suitable microhabitat and to escape form predators.

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 3.
Name the different components of diarthroses?
Answer:
The different components of diarthroses are :
i. Hinge joint.
ii. Ball and Socket joint.

Question 4.
What are hinge joints?
Answer:
Hinge joints allow movement in one plane only. It helps to the parts in all sides.

Question 5.
What are abduction and adduction?
Answer:
Abduction : The process which move parts of the body away from the central axis of the body is called abduction.
Adduction : It refers to a motion that pulls a structure or part towards the midline of the body.

Question 6.
What are the different types of rotational movement?
Answer:
The different types of rotational movement : (i) Medial rotation (ii) Lateral rotation.

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 7.
What is meant by flexion and extension?
Answer:
Flexion : The process by which joint flexure occurs is called flexion.
Extension : The process by which joints are extended is known as extension.

Long Answer Type Questions : 5 Marks

Question 1.
What are the different type of movements exhibited by the cells of human body?
Answer:
Cells of the human body exhibit three main types of movements, namely, amoeboid, ciliary and muscular.
Amoeboid movement : Some specialised cells in our body like macrophages and leucocytes in blood exhibit amoeboid movement. It is effected by pseudopodia formed by the streaming of protoplasm (as in Amoeba). Cytoskeletal elements like microfilaments are also involved in amoeboid movement.

Ciliary Movement : Ciliary movement occurs in most of our internal tubular organs which are lined by ciliated epithelium. The coordinated movements of cilia in the trachea help us in removing dust particles and some of the foreign substances inhaled along with the atmospheric air. Passage of ova through the female reproductive tract is also facilitated by the ciliary movement.

Muscular Movement : Movement of our limbs, jaws, tongue etc require muscular movement. The contractile property of muscles are effectively used for locomotion and other movements by human beings and majority of multicellular organisms. Locomotion requires a perfect coordinated activity of muscular, skeletal and neural systems.

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 2.
How do you distinguish between a skeletal muscle and a cardiac muscle?
Answer:
Both cardiac and skeletal muscles are striated. Cardiac and skeletal muscles are ‘striated’ in that they contain sarcomeres and are packed into highly-regular arrangements of bundles. While skeletal muscles are arranged in regular, parallel bundles, cardiac muscle connects at branching irregular angles (called intercalated discs). Striated muscle contracts and relaxes in short, intense bursts, whereas smooth muscle sustains longer or even near-permanent contractions.

Multiple Choice Questions : 1 Mark

Question 1.
Skeletal muscle fibre is
(a) multinucleated
(b) uninucleated
(c) binucleated
(d) anucleated
Answer:
(a) multinucleated

Question 2.
Muscle of heart is known as
(a) striated
(b) cardiac
(c) voluntary
(d) non striated
Answer:
(b) cardiac

Question 3.
The number of bones in the appendicular skeleton is
(a) 206
(b) 80
(c) 120
(d) 126
Answer:
(d) 126

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 4.
Total number of bones in man
(a) 206
(b) 205
(c) 204
(d) 201
Answer:
(a) 206

Question 5.
Number of verterae in man is
(a) 32
(b) 33
(c) 26
(d) 28
Answer:
(c) 26

Question 6.
Alternative source of energy for muscle contraction is
(a) ATP
(b) creatine phosphate
(c) glycolysis
(d) all of these
Answer:
(c) glycolysis

Question 7.
Muscles are connected to the bones by
(a) ligament
(b) tendons
(c) cartilage
(d) bone
Answer:
(b) tendons

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 8.
Where do you find osteocytes
(a) cartilage
(b) blood
(c) bone
(d) liver
Answer:
(c) bone

Question 9.
Which of the following is a freely movable joint?
(a) ball and socket
(b) fibrous joint
(c) cartilaginous joint
(d) both a and b
Answer:
(a) ball and socket

Question 10.
Elbow joint is a
(a) saddle joint
(b) ball and socket joint
(c) pivot joint
(d) hinge joint
Answer:
(d) hinge joint

Very Short Answer Type Questions : 1 Mark

Question 1.
What are the main functions of the skeletal system?
Answer:
Functions of the skeletal system are as follows :

  1. It gives shape and provides supports to the body.
  2. It forms the framework of the body.
  3. It protects some internal organs.

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 2.
Name the various organs of the skeletal system.
Answer:
Various organs of the skeletal system are : skull, vertebral column, rib cage, bones in arms and legs, shoulder bone and hip bone.

Question 3.
Give an example of pivot joint in the human body.
Answer:
The joint between skull and the first vertebrae.

Question 4.
What is slithering movement?
Answer:
Movement of a snake is called slithering movement.

Question 5.
How many vertebrae are found in the human backbone?
Answer:
There are 33 vertebrae in the human backbone.

Question 6.
How many ribs are found in the rib cage of humans?
Answer:
There are 12 pairs of ribs in the rib cage of humans.

Question 7.
How does a snake move?
Answer:
A snake moves by making loops on its sides.

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 8.
How does a fish swim in water?
Answer:
When a fish swims, its front part curves to one side and the tail part stays in the opposite direction. In the next move, the front part curves to the opposite side and the tail part also changes its position to another side. The tail fin helps in changing direction.

Question 9.
How does the snail move?
Answer:
A snail moves using its muscular foot.

Question 10.
How many bones are found in the human skeletal system?
Answer:
There are 206 bones in the human skeleton.

Question 11.
Name the type of joint between the following :
(a) atlas/axis
(b) carpal/metacarpal of thumb
(c) between phalanges.
(d) femur/acetabulum
(e) between cranial bones
(f) between pubic bones in the pelvic girdle
Answer:
(a) Pivot joint
(b) Saddle joint
(c) Gliding joint
(d) Ball and socket joint
(e) Fibrous joint
(f) Cartilaginous joint.

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 12.
The surface area of wings is increased bird by the help of
Answer:
Remiges.

Question 13.
Which gland is situated in kidney?
Answer:
Adrenal gland.

Question 14.
The name of the locomotary organ of Amoeba is ________.
Answer:
Pseudopodia.

Question 15.
Bipedal locomotion is found in
Answer:
Man.

Question 16.
Triceps help in ______.
Answer:
Extension.

Question 17.
What is a ball and socket joint?
Answer:
In this joint the rounded end of one bone fits into the cavity (hollow space) of another bone. It permits movements in all directions.

Question 18.
Which of the skull bones are movable?
Answer:
Lower jaw bone or mandible is the only movable bone in the skull.

Question 19.
Why can our elbow not move backwards?
Answer:
Elbow joint is composed of hinge joint. These joints allow movement in one plane and up to 180° only. Hence, we cannot move our elbows backwards.

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 20.
What is the nature of hip joint?
Answer:
In this joint rounded head of a bone lies within the cuplike groove of another bone.

Question 21.
What are the movement pattern in human legs?
Answer: Rotation.

Question 22.
What are the different locomotory organs of Protozoa?
Answer:
Flagellum, cilium, pseudopodium.

Question 23.
How locomotion differs from movement?
Answer:

Locomotion Movement
i. It takes place at organism level i. It takes place at any biological level from cellular to organisms.
ii. It is found mainly in animals ii. It is found mainly in Plants.

Question 24.
What is ciliary movement?
Answer:
The animal body is covered by cilia. The effective and recovery movements of cilia help the animal to take ciliary movement. e.g. Paramecium sp.

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 25.
What is the nature of myotomes?
Answer:
It helps in contraction and relaxation of antagonistic muscles of fish.

Question 26.
What is bipedalism?
Answer:
It involves in movement of two feet.

Question 27.
What is arthrology?
Answer:
It is the study of science that deals with the study of joints.

Question 28.
What are diarthrosis joint?
Answer:
The joints that have a space between the adjoining bones are called diarthrosis joints.

Question 29.
What is locomotion?
Answer: The process by means of which the living organism can change their position wholly from one place to another in response to stimuli is known as locomotion.

Question 30.
Name the locomotory organ of Euglena?
Answer:
Flagella

Question 31.
Give one example of ‘Ball and Socket’ joint?
Answer:
Hip joint.

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 32.
What is abduction?
Answer:
The process which moves parts of the body away from the central axis of the body is called abduction.

Question 33.
What are the locomotory organ of Amoeba?
Answer:
Pseudopodia.

Question 34.
Name the locomotory organ of Paramecium?
Answer: Cilia.

Question 35.
State the name of locomotory muscles of fish?
Answer:
Myotomes.

Question 36.
What is the other name for tail fin?
Answer:
Caudal fin.

Question 37.
Name the wing feathers of bird.
Answer: Remiges.

Question 38.
What is the other name for studing joint?
Answer:
Arthrology.

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 39.
What is the nature of knee joint?
Answer:
In this joint, ends of two bones remain arranged like hinge.

Question 40.
Pseudopodia is the locomotory structure of
Answer:
Ameoba.

Question 41.
Synovial joints are ______.
Answer:
Freely movable.

Question 42.
Moving away from the midline is ______.
Answer:
Abduction.

Question 43.
Cilia is the locomotary organ of ______.
Answer:
Paramecium.

Question 44.
Myotomes are the locomotory organ of ______.
Answer:
Fish.

Question 45.
Rectrices in bird are located in
Answer:
Tail feather.

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 46.
The semicircular canals are located in
Answer:
Ear.

Question 44.
The synovial joints are also called
Answer:
Diarthroser.

Question 48.
The elbow joint is
Answer:
Elbow.

Question 49.
Adduction and Abduction is a common phenomenon found in ______.
Answer:
Leg.

Question 50.
What is articular cartilage?
Answer:
A layer of haylne cartilage is called articular cartilage.

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 51.
What are synovial joints?
Answer:
Synovial joints are the only joints that have a space between the adjoining bones. These are freely movable joints mostly present between the limb bones.

Fill in the blanks :

Question 1.
Joints of the bone help in the ______ of the body.
Answer:
Movement

Question 2.
A combination of bones and cartilages forms the ______ of the body.
Answer:
Skeleton

Question 3.
The bones at the elbow are joined by a ______ joint.
Answer:
Hinge joint.

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 4.
The contraction of the pulls the bones during movements.
Answer:
Muscles.

True or False :

Question 5.
The movement and locomotion of all animals is exactly the same.
Answer:
False

Question 6.
The cartilages are harder than bones.
Answer:
False

Question 7.
The finger bones do not have joints.
Answer:
False

WBBSE Class 10 Life Science Solutions Chapter 1E Locomotion as a Type of Response in Animals

Question 8.
The forearm has two bones.
Answer:
True

Question 9.
Cockroaches have an outer skeleton.
Answer:
True

Question 10.
Match the items in coloumn I with items of column II.

Column I Column II
Upper jaw Have fins on the body
Fishes Has outer skeleton
Ribs Is an immovable joint
Snail Protect the heart
Cockroach Shows very slow movement.

Answer:

Column I Column II
Upper jaw Is an immovable joint
Fishes Have fins on the body
Ribs Protect the heart
Snail Shows very slow movement.
Cockroach Has outer skeleton

WBBSE Class 9 Physical Science Notes Chapter 6 Heat

Comprehensive WBBSE Class 9 Physical Science Notes Chapter 6 Heat can help students make connections between concepts.

Heat Class 9 WBBSE Notes

Heat : It is an external agency whose absorption turns a body hot and extraction turns a body cold.

Sensible heat : It is the type of heat that is perceptible by senses. This causes rise of temperature of a body that can be measured with a thermometer.

Latent heat : It is the quantity of heat absorbed or released when a substance of unit mass undergoes a physical change of state at a constant temperature and pressure.

Radiant heat : It transmits from a source of heat in all directions with the speed of light and in the form of waves.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat

Temperature : The temperature of a body is the thermal condition which determines whether it will absorb heat from other body or release heat to that body when they are kept in contact.

Lower fixed point of a thermometer : At normal atmospheric pressure, the temperature at which ice melts into water or pure water freezes into ice is called the lower fixed point of a thermometer.

Upper fixed point of a thermometer : It is the temperature at which pure water boils and transforms into steam under normal atmospheric pressure.

Fundamental interval : The range of temperature between the upper and lower fixed points is called fundamental interval.

Celsius scale of temperature : In this scale, the lower fixed point is 0°C and the upper fixed point is 100°C. The fundamental difference is divided into hundred equal parts and each part represents 1°C.

Fahrenheit scale of temperature : In this scale, the lower fixed point is 32°F and the upper fixed point is 212°F. The fundamental difference is divided into 180 equal parts and each part represents 1°F.

Kelvin scale of temperature : In this scale the lower fixed point is 273K and the upper fixed point is 373K, the fundamental interval is divided into 100 equal divisions like that in the Celsius scale. Each division represents 1K.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat

Calorie : It is the amount of heat required to raise the temperature of one gram of pure water through 1°C.

Specific heat : It is defined as the quantity of heat required to raise the temperature of unit mass of it by 1 degree.

Specific heat capacity of a substance : Specific heat capacity of a substance is the quantity of heat required to raise the temperature of 1kg of the substance through 1K.

Units of specific heat :

  • CGS : calg-10 C-1
  • SI : Jkg-1K-1

Thermal capacity or heat capacity of a body : It is defined as the quantity of heat needed to raise the temperature of the body for unit degree rise of temperature.

Units of heat capacity :

  • CGS : calg-10 C-1
  • SI : Jk-1

Water equivalent : The water equivalent of a body is the mass of water which will be heated through one degree by the amount of heat required to raise the temperature of the body through one degree.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat

Units of water equivalent:

  • CGS: gram (g)
  • SI: kilogram (kg)

Fundamental principle of calorimetry :

Conditions :

  • During the process of heat transfer, there is no heat exchange with the surrounding.
  • No chemical reaction takes place between the bodies.

Principle :
Heat lost by the hotter body = Heat gained by the colder body.

First law of thermodynamics : Whenever mechanical work is converted into heat energy, the heat so obtained is directly proportional to the work done.
If, ‘W’ be the mechanical work done and ‘H’ be the corresponding heat energy produced, then from first law of thermodynamics,
W α H or W = JH
where J is the constant of proportionality and is known as mechanical equivalent of heat.

Mechanical equivalent of heat : The mechanical equivalent of heat is numerically equal to the work done to produce unit quantity of heat.

Unit of mechanical equivalent of heat :

  • CGS: 4.8 x 107 erg cal-1
  • SI : In SI, both heat and work are measured in Joule, so mechanical equivalent of heat has no unit and its value is 1.

Change of state : There are three states of matter – solid, liquid and gas. Substances can be changed from one state to other by application or removal of heat from them. This process is known as change of state.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat

Mathematical expression of Latent heat : Q = mL
where Q is the amount of heat supplied (or extracted) to change the state of ‘m’ mass of substance and L is latent heat

Unit of Latent heat :

  • CGS : cal g-1
  • SI : J kg-1

Types of Latent heat:

  • Latent heat of fusion
  • Latent heat of solidification
  • Latent heat of vaporisation
  • Latent heat of condensation

Latent heat of fusion of ice and latent heat of solidification of water are both equal:

  • CGS : 80 cal g-1
  • SI: 336 × 105 Kg-1

Latent heat of vaporisation of water and latent heat of condensation of steam are both equal:

  • CGS: 537cal g-1
  • SI : 226 × 106 Jkg-1

Melting point : The melting point of a solid is the definite temperature at which it melts on heating under normal atmospheric pressure and the temperature remains constant until the melting is
completed.

Boiling point : The boiling point of a liquid is the temperature at which a liquid boils under normal atmospheric pressure. The boiling point of a liquid depends upon:

  • The nature of the liquid
  • Presence of dissolved substance
  • The superincumbent pressure

pressure: Whenever a liquid evaporates at any temperature, the vapour exerts a definite pressure on everything in contact. This pressure is called vapour pressure of the liquid at that temperature.

Saturated vapour : If a liquid is allowed to evaporate in a closed space, after some time it is found that evaporation stops i.e. at a given temperature. there is a maximum limit to the amount of
vapour the space can hold. The space is said to be then saturated with vapour and the vapour is then called saturated vapour.

Unsaturated vapour : If at a given temperature, a space contains a smaller amount of vapour than the maximum possible amount that the space can hold, the space is then said to be unsaturated and the vapour is called unsaturated vapour.

Saturated vapour does not obey Charles’ law or Boyle’s law, while unsaturated vapour obeys charles’ and Boyle’s law.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat

Critical temperature : For every substance in gaseous state there is a certain temperature such that if the substance is below this temperature, it can be liquefied by the application of suitable pressure,
and above this temperature, it cannot be liquefied, however large the pressure may be, then the temperature is called the critical temperature for that substance.

Gas : A substance at a temperature above critical temperature is called gas.

Vapour : When the substance is at a temperature below its critical temperature it is called vapour.

Anomalous expansion of water: Usually liquids expand on heating. But in the case of water we find deviation from this general behaviour of the liquids within a certain range of temperature. It is found that if certain mass of water at an temperature about 10°C be taken and allowed to cool, its volume is observed to fall gradually till it reaches the temperature of 4°C, when with further cooling the volume of water will increase instead of decreasing.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat 1
The behaviour of water is some how different from other liquids. The volume of water is minimum at 4°C and its density is maximum at 4°C. This phenomenon is called the anomalous expansion of water. This phenomenon is called the Ienipentturi anomalous expansion of water.

Effect on marine life : Anomalous expansion of water has great practical importance on the marine life in the cold countries. In these countries when air above the surface of water of pond falls below 0°C, the water on the surface on cooling, become denser than that below and gradually sinks downwards.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat 2

WBBSE Class 9 Physical Science Notes Chapter 6 Heat

This continues until the temperature of water falls to 4°C. As the surface temperature falls further it becomes less dense than the water below, which is at 4°C and is the densest. It thus remains at the top, cooling C more and more and ultimately a layer of ice is formed there and remains there as ice is lighter than water. The layer of ice forms a poor conducting layer on the top, which prevents the heat from water below to pass into colder atmosphere above.

WBBSE Class 9 Physical Science Notes Chapter 6 Heat 3

The temperature of the deeper layers of the water in the pond remains nearly at 4°C and falls gradually to 0°C upwards till the layer of ice is reached. The marine life in water is thus saved.