Class 12

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 6 Molecular Basis of Inheritance Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 6 Molecular Basis of Inheritance

PSEB 12th Class Biology Guide Molecular Basis of Inheritance Textbook Questions and Answers

Question 1.
Group the following as nitrogenous bases and nucleosides:
Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:
Nitrogenous Bases: Adenine, thymine, uracil, and cytosine.
Nucleosides: Cytidine and guanosine.

Question 2.
If a double stranded DNA has 20 per cent of cytosine, calculate the per cent of adenine in the DNA.
Answer:
According to Chargaffs rule, the DNA molecule should have an equal ratio of pyrimidine (cytosine and thymine) and purine (adenine and guanine). It means that the number of adenine molecules is equal to thymine molecules and the number of guanine molecules is equal to cytosine molecules.
% A = % T and % G = % C
If double stranded DNA has 20% of cytosine, then according to the law, it would have 20% of guanine.
Thus, percentage of G + C content = 40%
The remaining 60% represents both A + T molecule. Since adenine and guanine are always present in equal numbers, the percentage of adenine molecule is 30%.

Question 3.
If the sequence of one strand of DNA is written as follows: 5-ATGCATGCATGCATGCATGCATGCATGC-3′
Write down the sequence of complementary strand in 5′ → 3′ direction.
Answer:
The DNA strands are complementary to each other with respect to base sequence. Hence, if the sequence of one strand of DNA is
5′- ATGCATGCATGCATGCATGCATGCATGC – 3′
Then, the sequence of complementary strand in 5′-3′ direction will be
3′- TACGTACGTACGTACGTACGTACGTACG – 5′
Therefore, the sequence of nucleotides on DNA polypeptide in 5′-3′ direction is
5′- GCATGCATGCATGCATGCATGCATGCAT – 3′

Question 4.
If the sequence of the coding strand in a transcription unit is written as follows:
5′ – ATGCATGCATGCATGCATGCATGCATGC-3 Write down the sequence of mRNA.
Answer:
If the coding strand in a transcription unit is
5′ – ATGCATGCATGCATGCATGCATGCATGC-3′
Then, the template strand in 3′ to 5′ direction would be
3′ – TACGTACGTACGTACGTACGTACGTACG-5′
It is known that the sequence of mRNA is same as the coding strand of DNA.
However, in RNA, thymine is replaced by uracil.
Hence, the sequence of mRNA will be
5′ – AUGCAUGCAUGCAUGCAUGCAUGCAUGC-3′

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer:
Watson and Crick observed that the two strands of DNA are f anti-parallel and complementary to each other with respect to their base sequences. This type of arrangement in DNA molecule led to the hypothesis that DNA replication is semi-conservative. It means that the double stranded DNA molecule separates and then, each of the separated strand acts as a template for the synthesis of a new complementary strand. As a result, each DNA molecule would have one parental strand and a newly synthesised daughter strand.

Since only one parental strand is conserved in each daughter molecule, it is known as semi-conservative mode of replication.

Quetion 6.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
Answer:
There are two different types of nucleic acid polymerases.

1. DNA-dependent DNA polymerases
2. DNA-dependent RNA polymerases

The DNA-dependent DNA polymerases use a DNA template strand for synthesising a new strand of DNA, whereas DNA-dependent RNA polymerases use a DNA template strand for synthesising a new strand of RNA.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer:
Hershey and Chase worked with bacteriophage and E.coli to prove that DNA is the genetic material. They used different radioactive isotopes to label DNA and protein coat of the bacteriophage.

They grew some bacteriophages on a medium containing radioactive phosphorus (³²) to identify DNA and some on a medium containing radioactive sulphur (³⁵S) to identify protein. Then, these radioactive labelled phages were allowed to infect E.coli bacteria. After infecting, the protein coat of the bacteriophage was separated’from the bacterial cell by blending and then subjected to the process of centrifugation. Since the protein coat was lighter, it was found in the supernatant while the infected bacteria got settled at the bottom of the centrifuge tube. Hence, it was proved that DNA is the genetic material as it was transferred from virus to bacteria.

Question 8.
Differentiate between the following :
(a) Repetitive DNA and Satellite DNA
(b) mRNA and tRNA
(c) Template strand and Coding strand Arts,
Answer:
(a) Repetitive DNA and Satellite DNA

Repetitive DNA	Satellite DNA
Repetitive DNA are DNA sequences that contain small segments, which are repeated many times.	Satellite DNA are DNA sequences that contain highly repetitive DNA.

(b) mRNA and tRNA

mRNA	tRNA
1. mRNA or messenger RNA acts as a template for the process of transcription.	tRNA or transfer RNA acts as an adaptor molecule that carries a specific amino acid to mRNA for the synthesis of polypeptide.
2. It is a linear molecule.	It has clover leaf shape.

(c) Template strand and Coding strand

Template strand	Coding strand
1. Template strand of DNA acts as a template for the synthesis of mRNA during transcription.	Coding strand is a sequence of DNA that has the same base sequence as that of mRNA (except thymine that is replaced by uracil in DNA).
2. It runs from 3′ to 5′.	It runs from 5′ to 3′.

Question 9.
List two essential roles of ribosome during translation.
Answer:
The important functions of ribosome during translation are as follows :
(a) Ribosome acts as the site where protein synthesis takes place from individual amino .acids. It is made up of two subunits.
The smaller subunit comes in contact with mRNA and forms a protein synthesising complex whereas the larger subunit acts as an amino acid binding site.

(b) Ribosome acts as a catalyst for forming peptide bond. For example, 23s r-RNA in bacteria acts as a ribozyme.

Question 10.
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after addition of lactose in the medium?
Answer:
Lac operon is a segment of DNA that is made up of three adjacent structural genes, namely, an operator gene, a promoter gene, and a regulator gene. It works in a coordinated manner to metabolise lactose into glucose and galactose.

In lac operon, lactose acts as an inducer. It binds to the repressor and inactivates it. Once the lactose binds to the repressor, RNA polymerase binds to the promoter region. Hence, three structural genes express their product and respective enzymes are produced. These enzymes act on lactose so that lactose is metabolised into glucose and galactose. After sometime, when the level of inducer decreases as it is completely metabolised by enzymes, it causes synthesis of the repressor from regulator gene. The repressor binds to the operator gene and prevents RNA polymerase from transcribing the operon. Hence, the transcription is stopped. This type of regulation is known as negative regulation.

Question 11.
Explain (in one or two lines) the function of the following:
(a) Promoter
(b) tRNA
(c) Exons
Answer:
(a) Promoter: Promoter is a region of DNA that helps in initiating the process of transcription. It serves as the binding site for RNA polymerase.

(b) tRNA: tRNA or transfer RNA is a small RNA that reads the genetic code present on mRNA. It carries specific amino acid to mRNA on ribosome during translation of proteins.

(c) Exons: Exons are coding sequences of DNA in eukaryotes that transcribe for proteins.

Question 12.
Why is the Human Genome project called a mega project?
Answer:
Human genome project was considered to be a mega project because it had a specific goal to sequence every base pair present in the human genome. It took around 13 years for its completion and got accomplished in year 2006. It was a large scale project, which aimed at developing new technology and generating new information in the field of genomic studies. As a result of it, several new areas and avenues have opened up in the field of genetics, biotechnology, and medical sciences. It provided clues regarding the understanding of human biology.

Question 13.
What is DNA fingerprinting? Mention its application.
Answer:
DNA fingerprinting is a technique used to identify and analyse the variations in various individuals at the level of DNA. It is based on variability and polymorphism in DNA sequences.
Applications

1. It is used in forensic science to identify potential crime suspects.
2. It is used to establish paternity and family relationships.
3. It is used to identify and protect the commercial varieties of crops and livestock.
4. It is used to find out the evolutionary history of an organism and trace out the linkages between groups of various organisms.

Question 14.
Briefly describe the following:
(a) Transcription
(b) Polymorphism
(c) Translation
(d) Bioinformatics
Answer:
(a) Transcription: It is the process of synthesis of RNA from DNA template. A segment of DNA gets copied into mRNA during the process. The process of transcription starts at the promoter region of the template DNA and terminates at the terminator region. The segment of DNA between these two regions is known as transcription unit. The transcription requires RNA polymerase enzyme, a DNA template, four types of ribonucleotides, and certain cofactors such as Mg²⁺.
The three important events that occur during the process of transcription are as follows:

1. Initiation
2. Elongation
3. Termination

The DNA-dependent RNA polymerase and certain initiation factors bind at the double stranded DNA at the promoter region of the template strand and initiate the process of transcription. RNA polymerase moves along the DNA and leads to the unwinding of DNA duplex into two separate strands. Then, one of the strands, called sense strand, acts as template for mRNA synthesis. The enzyme, RNA polymerase, utilises nucleoside triphosphates (dNTPs) as raw material and polymerises them to form mRNA according to the complementary bases present on the template DNA«. This process of opening of helix-and elongation of polynucleotide chain continues until the enzyme reaches the terminator region. As RNA polymerase reaches the terminator region, the newly synthesised mRNA transcripted along with enzyme is released. Another factor called terminator factor is required for the termination of the transcription.

(b) Polymorphism: It is a form of genetic variation in which distinct nucleotide sequence can exist at a particular site in a DNA molecule. This heritable mutation is observed at a high frequency in a population. It arises due to mutation either in somatic cell or in the germ cells. The germ cell mutation can be transmitted from parents to their offsprings. This results in accumulation of various mutations in a population, leading to variation and polymorphism in the population. This plays a very important role in the process of evolution and tracing human history.

(c) Translation: It is the process of polymerising amino acid to form a polypeptide chain. The triplet sequence of base pairs in mRNA defines the order and sequence of amino acids in a polypeptide chain.
The process of translation involves following three steps:

1. Initiation
2. Elongation
3. Termination

During the initiation of the translation, tRNA gets charged when the amino acid binds to it using ATP. The start (initiation) codon (AUG) present on mRNA is recognised only by the charged tRNA. The ribosome acts as an actual site for the process of translation and contains two separate sites in a large subunit for the attachment of subsequent amino acids. The small subunit of ribosome binds to mRNA at the initiation codon (AUG) followed by the large subunit. Then, it initiates the process of translation. During the elongation process, the ribosome moves one codon downstream along with mRNA so as to leave the space for binding of another charged tRNA. The amino acid brought by tRNA gets linked with the previous amino acid through a peptide bond and this process continues resulting in the formation of a polypeptide chain. When the ribosome reaches one or more STOP codon (VAA, UAG, and UGA), the process of translation gets terminated. The polypeptide chain is released and the ribosomes get detached from mRNA.

(d) Bioinformatics: It is the application of computational and statistical techniques to the field of molecular biology. It solves the practical problems arising from the management and analysis of biological data. The field of bioinformatics developed after the completion of human genome project (HGP). This is because enormous amount of data has been generated during the process of HGP that has to be managed and stored for easy access and interpretation for future use by various scientists. Hence, bioinformatics involves the creation of biological databases that store the vast information of biology.

It develops certain tools for easy and efficient access to the information and its utilisation. Bioinformatics has developed new algorithms and statistical methods to find out the relationship between the data, to predict protein structure and their functions, and to cluster the protein sequences into their related families.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 5 Principles of Inheritance and Variation Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 5 Principles of Inheritance and Variation

PSEB 12th Class Biology Guide Principles of Inheritance and Variation Textbook Questions and Answers

Question 1.
Mention the advantages of selecting pea plant for experiment by Mendel.
Answer:
Mendel selected pea plants to carry out his study on the inheritance of characters from parents to offspring. He selected a pea plant because of the following features:

• Peas have many visible contrasting characters such as tall/dwarf plants, round/wrinkled seeds, green/yellow pod, purple/white flowers, etc.
• Peas have bisexual flowers and therefore undergo self pollination easily. Thus, pea plants produce offsprings with same traits generation after generation.
• In pea plants, cross pollination can be easily achieved by emasculation in which the stamen of the flower is removed without affecting the pistil.
• Pea plants have a short life span and produce many seeds in one generation.

Question 2.
Differentiate between the following:
Dominance and Recessive Homozygous and Heterozygous
(a) Dominance and Recessive
(b) Homozygous and Heterozygous
(c) Monohybrid and Dihybrid.
Answer:
(a) Dominance and Recessive

Dominance	Recessive
1. A dominant trait expresses itself in the presence or absence of a recessive trait.	A recessive trait is able to express itself only in the absence of a dominant trait.
2. For example, tall plant, round seed, violet flower, etc. are dominant traits in a pea plant.	For example, dwarf plant, wrinkled seed, white flower, etc. are recessive traits in a pea plant.

(b) Homozygous and Heterozygous

Homozygous	Heterozygous
1. It contains two similar alleles for a particular trait.	It contains two different alleles for a particular trait.
2. Genotype for homozygous possess either dominant or recessive, but never both the alleles. For example, RR or rr	Genotype for heterozygous possess both dominant and recessive alleles. For example, Rr
3. It produces only one type of gamete.	It produces two different types of gametes.

(c) Monohybrid and Dihybrid

Monohybrid	Dihybrid
1. Monohybrid involves cross between parents’, which differs in only one pair of contrasting characters.	Dihybrid involves cross between parents, which differs in two pairs of contrasting characters.
2. For example, the cross between tall and dwarf pea plant is a monohybrid cross.	For example, the cross between pea plants having yellow wrinkled seed with those having green round seeds is a dihybrid cross.

Question 3.
A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
Answer:
Locus is a fixed position on a chromosome, which is occupied by a single or more genes. Heterozygous organisms contain different alleles for an allelic pair. Hence, a diploid organism, which is heterozygous at four loci, will have four different contrasting characters at four different loci.

For example, if an organism is heterozygous at four loci with four characters, say Aa, Bb, Cc, Dd, then during meiosis, it will segregate to form 8 separate gametes.

If the genes are not linked, then the diploid organism will produce 16 different gametes. However, if the genes are linked, the gametes will reduce their number as the genes might be linked and the linked genes will be inherited together during the process of meiosis.

Question 4.
Explain the Law of Dominance using a monohybrid cross.
Answer:
Mendel’s law of dominance states that a dominant allele expresses itself in a monohybrid cross and suppresses the expression of recessive allele. However, this recessive allele for a character is not lost and remains hidden or masked in the progenies of F₁ generation and reappears in the next generation.

For example, when pea plants with round seeds (RR) are crossed with plants with wrinkled seeds (rr), all seeds in F₁ generation were found to be round (Rr). When these round seeds were self fertilised, both the round and wrinkled seeds appeared in F₂ generation in 3 : 1 ratio. Hence, in F₂ generation, the dominant character (round seeds) appeared and the recessive character (wrinkled seeds) got suppressed, which reappeared in F₂ generation.

Question 5.
Define and design a test-cross.
Answer:
Test cross is a cross between an organism with unknown genotype and a recessive parent. It is used to determine whether the individual is homozygous or heterozygous for a trait.
If the progenies produced by a test cross show 50% dominant trait and 50% recessive trait, then the unknown individual is heterozygous for a trait. On the other hand, if the progeny produced shows dominant trait, then the unknown individual is homozygous for a trait.

Question 6.
Using a Punnett Square, workout the distribution of phenotypic features in the first filial generation after a cross between a homozygous female and a heterozygous male for a single locus.
Answer:
In guinea pigs, heterozygous male with black coat colour (Bb) is crossed with the homozygous female having white coat colour (bb). The male will produce two types of gametes, B and b, while the female will produce only one kind of gamete, b. The genotypic and phenotypic ratio in the progenies of Fx generation will be same i.e., 1:1.

Question 7.
When a cross is made between tall plant with yellow seeds (TtYy) and tall plant with green seed (Ttyy), what proportions of phenotype in the offspring could be expected to he
(a) tall and green.
(b) dwarf and green.
Answer:
A cross between tall plant with yellow seeds and tall plant with green seeds will produce
(a) three tall and green plants
(b) one dwarf and green plant

Question 8.
Two heterozygous parents are crossed. If the two loci are linked what would be the distribution of phenotypic features in F₁ generation for a dihybrid cross?
Answer:
When two individual heterozygous for two loci (Yy Rr) are crossed and the two loci are linked, the distribution of the phenotypic feature of F₁ generation will be in the ratio of 3:1 $\frac{3}{4}$ of the individuals will show
both the dominant traits and $\frac{1}{4}$ of the individuals will show both the
recessive traits. It is because the genes for both the traits are present on the same chro

Question 9.
Briefly mention the contribution of T.H. Morgan in genetics.
Answer:
Morgan’s work is based on fruit flies (Drosophila melanogaster). He formulated the chromosomal theory of linkage. He defined linkage as the co-existence of two or more genes in the same chromosome and performed dihybrid crosses in Drosophila to show that linked genes are inherited together and are located on X-chromosome. His experiments have also proved that tightly linked genes show very low recombination while loosely linked genes show higher recombination.

Question 10.
What is pedigree analysis? Suggest how such an analysis, can be useful.
Answer:
Pedigree analysis is a record of occurrence of a trait in several generations of a family. It is based on the fact that certain characteristic features are heritable in a family, for example, eye colour, skin colour, hair form and colour, and other facial characteristics. Along with these features, there are other genetic disorders such as Mendelian disorders that are inherited in a family, generation after generation. Hence, by using pedigree analysis for the study of specific traits or disorders, generation after generation, it is possible to trace the pattern of inheritance. In this analysis, the inheritance of a trait is represented as a tree, called family tree. Genetic counselors use pedigree chart for analysis of various traits and diseases in a family and predict their inheritance patterns. It is useful in preventing haemophilia, sickle cell anaemia, and other genetic disorders in the future generations.

Question 11.
How is sex determined in human beings?
Answer:
Human beings exhibit male heterogamy. In humans, males (XY) produce two different types of gametes, X and Y. The human female (XX) produces only one type of gametes containing X chromosomes. The sex of the baby is determined by the type of male gamete that fuses with the female gamete. If the fertilising sperm contains X chromosome, then the baby produced will be a girl and if the fertilising sperm contains Y chromosome, then the baby produced will be a boy. Hence, it is a matter of chance that determines the sex of a baby. There is an equal probability of the fertilising sperm being an X or Y chromosome. Thus, it is the genetic make up of the sperm that determines the sex of the baby.

Question 12.
A child has hlood group O. If the father has hlood group A and mother blood group B, work out the genotypes of the parents and the possible genotypes of the other offsprings.
Answer:
The blood group characteristic in humans is controlled by three set of alleles, namely, I^A,I^B and i. The alleles, I^A and I^B, are equally dominant whereas allele, i, is recessive to the other alleles. The individuals with genotype, I^A I^A and I^A i, have blood group A whereas the individuals with genotype, I^B I^B and I^B i, have blood group B. The persons with genotype I^A I^B have blood group AB while those with blood group O have genotype ii.
Hence, if the father has blood group A and mother has blood group B, then the possible genotype of the parents will be
Father
I^AI ^{or A}I^Ai

Mother
I^BI^B or I^Bi
A cross between homozygous parents will produce progeny with AB blood group.

A cross between heterozygous parents will produce progenies with AB blood group (I^A I^B) and O blood group (ii).

Question 13.
Explain the following terms with example
(a) Co-dominance
(b) Incomplete dominance
Answer:
(a) Co-dominance: Co-dominance is the phenomenon in which both the alleles of a contrasting character are expressed in heterozygous condition. Both the alleles of a gene are equally dominant. ABO blood group in human beings is an example of co-dominance. The blood group character is controlled by three sets of alleles, namely, I^A, I^B, and i. The alleles, I^A and I^B, are equally dominant and are said to be co-dominant as they are expressed in AB blood group. Both these alleles do not interfere with the expression of each other and produce their respective antigens. Hence, AB blood group is an example of co-dominance.

(b) Incomplete Dominance: Incomplete dominance is a phenomenon in which one allele shows incomplete dominance over the other member of the allelic pair for a character. For example, a monohybrid cross between the plants having red flowers and white flowers in Antirrhinum species will result in all pink flower plants in Fj generation. The progeny obtained in Fx generation does not resemble either of the parents and exhibits intermediate characteristics. This is because the dominant allele, R, is partially dominant over the other allele, r. Therefore, the recessive allele, r, also gets expressed in the generation resulting in the production of intermediate pink flowering progenies with Rr genotype.

What is point mutation? Give one example.
Answer:
Point mutation is a change in a single base pair of DNA by substitution, deletion, or insertion of a single nitrogenous base. An example of point mutation is sickle cell anaemia. It involves mutation in a single base pair in the beta-globin chain of haemoglobin pigment of the blood. Glutamic acid in short arm of chromosome II gets replaced with valine at the sixth position.

Question 15.
Who had proposed the chromosomal theory of the inheritance?
Answer:
Sutton and Boveri proposed the chromosomal theory of inheritance in 1902. They linked die inheritance of traits to the chromosomes.

Question 16.
Mention any two autosomal genetic disorders with their symptoms.
Answer:
Two autosomal genetic disorders are as follows:
1. Sickle Cell Anaemia: It is an autosomal linked recessive disorder, which is caused by point mutation in the beta-globin chain of haemoglobin pigment of the blood. The disease is characterised by sickle shaped red blood cells, which are formed due to the mutant haemoglobin molecule. The disease is controlled by Hb^A and Hb^S allele. The homozygous individuals with genotype, Hb^SHb^S, show the symptoms of this disease while the heterozygous individuals with genotype, Hb^A Hb^S, are not affected. However, they act as carriers of the disease.

Symptoms: Rapid heart rate, breathlessness, delayed growth and puberty, jaundice, weakness, fever, excessive thirst, chest pain, and decreased fertility are the major symptoms of sickle cell anaemia disease.

(b) Down’s Syndrome: It is an autosomal disorder that is caused by the trisomy of chromosome 21.
Symptoms : The individual is short statured with round head, open mouth, protruding tongue, short neck, slanting eyes, and broad short hands. The individual also shows retarded mental and physical growth.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 4 Reproductive Health Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 4 Reproductive Healths

PSEB 12th Class Biology Guide Reproductive Health Textbook Questions and Answers

Question 1.
What do you think is the significance of reproductive health in a society?
Answer:
Reproductive health is the total well being in all aspects of reproduction. It includes physical, emotional, behavioural, and social well being. Sexually transmitted diseases such as AIDS, gonorrhoea, etc. are transferred from one individual to another through sexual contact. It can also lead to unwanted pregnancies.

Hence, it is necessary to create awareness among people, especially the youth, regarding various reproduction related aspects as the young individuals are the future of the country and they are most susceptible of acquiring sexually transmitted diseases. Creating awareness about the available birth control methods, sexually transmitted diseases and their preventive measures, and gender equality will help in bringing up a socially conscious healthy family. Spreading. awareness regarding uncontrolled population growth and social evils among young individuals will help in building up a reproductively healthy society.

Question 2.
Suggest the aspects of reproductive health which need to be given special attention in the present scenario.
Answer:
Reproductive health is the total well being in all aspects of reproduction. The aspects which have to be given special attention in the present scenario are as follows:
1. Counselling and creating awareness among people, especially the youth, about various aspects of reproductive health, such as sexually transmitted diseases, available contraceptive methods, case of pregnant mothers, adolescence, etc.

2. Providing support and facilities such as medical assistance to people during pregnancy, STDs, abortions, contraceptives, infertility, etc. for building a reproductively healthy society.

Question 3.
Is sex education necessary in schools? Why?
Answer:
Yes, introduction of sex education in schools is necessary. It would provide right information to young individuals at the right time about various aspects of reproductive health such as reproductive organs, puberty, and adolescence related changes, safe sexual practices, sexually transmitted diseases, etc.

The young individual or adolescents are more susceptible in acquiring various sexually transmitted diseases. Hence, providing information to them at the right time would help them to lead a reproductively healthy life and also protect them from the myths and misconceptions about various sex related issues.

Question 4.
Do you think that reproductive health in our country has improved in the past 50 years? If yes, mention some such areas of improvement.
Answer:
Yes, the reproductive health has tremendously improved in India in the last 50 years. The areas of improvement are as follows:

1. Massive child immunisation programme, which has lead to a decrease in the infant mortality rate.
2. Maternal and infant mortality rate, which has been decreased drastically due to better post natal care.
3. Family planning, which has motivated people to have smaller families.
4. Use of contraceptive, which has resulted in a decrease in the rate of sexually transmitted diseases and unwanted pregnancies.

Question 5.
What are the suggested reasons for population explosion?
Answer:
The human population is increasing day by day, leading to population explosion. It is because of the following two major reasons:
(a) Decreased death rate
(b) Increased birth rate and longevity
The death rate has decreased in the past 50 years. The factor leading to decreased death rate and increased birth rate are control of diseases, awareness and spread of education, improvement in medical facilities, ensured food supply in emergency situation, etc. All this has resulted in an increase in the longevity of an individual.

Question 6.
Is the use of contraceptives justified? Give reasons.
Answer:
Yes, the use of contraceptives is absolutely justified. The human population is increasing tremendously. Therefore, to regulate the population growth by regulating reproduction has become a necessary demand in the present times. Various contraceptive devices have been devised to reduce unwanted pregnancies, which help in bringing down the increased birth rate and hence, in checking population explosion.

Question 7.
Removal of gonads cannot be considered as a contraceptive option. Why?
Answer:
Contraceptive devices are used to prevent unwanted pregnancy and to prevent the spreading of STDs. There are many methods, such as natural, barrier, oral, and surgical methods, that prevent unwanted pregnancy. However, the complete removal of gonads cannot be a contraceptive option because it will lead to infertility and unavailability of certain hormones that are required for normal functioning of accessory reproductive parts. Therefore, only those contraceptive methods can be used that prevent the chances of fertilisation rather than making the person infertile forever.

Question 8.
Amniocentesis for sex determination is banned in our country. Is this ban necessary? Comment.
Answer:
Amniocentesis is a pre-natal diagnostic technique that is used to determine the sex and metabolic disorders of the developing foetus in the mother’s uterus through the observation of the chromosomal patterns. This method was developed so as to determine any kind of genetic disorder present in the foetus. However, unfortunately, this technique is being misused to detect the sex of the child before birth and the female foetus is then aborted. Thus, to prevent the increasing female foeticides, it is necessary to ban the usage of amniocentesis technique for determining the sex of a child.

Question 9.
Suggest some methods to assist infertile couples to have children.
Answer:
Infertility is the inability of a couple to produce a baby even after unprotected intercourse. It might be due to abnormalities present in either male or female, or might be even both the partners. The techniques used to assist infertile couples to have children are as follows:
(a) Test Tube Babies: This involves in-vitro fertilisation where the sperms meet the egg outside the body of a female. The zygote, hence produced, is then transferred in the uterus or fallopian tube of a normal female. The babies produced from this method are known as test tube babies.

(b) Gamete Intra Fallopian Transfer (GIFT): It is a technique that involves the transfer of gamete (ovum) from a donor into the fallopian tube of the recipient female who is unable to produce eggs, but has the ability to conceive and can provide right conditions for the development of an embryo.

(c) Intra Cytoplasmic Sperm Injection (ICSI): It is a method of injecting sperm directly into the ovum to form an embryo in laboratory.

(d) Artificial Insemination: Artificial insemination is a method of transferring semen (sperm) from a healthy male donor into the vagina or uterus of the recipient female. It is employed when the male partner is not able to inseminate the female or has low sperm counts.

Question 10.
What are the measures one has to take to prevent from contracting STDs?
Answer:
Sexually transmitted diseases (STDs) get transferred from one individual to the other through sexual contact. Adolescents and young adults are at the greatest risk of acquiring these sexually transmitted diseases. Hence, creating awareness among the adolescents regarding its after-effects can prevent them from contracting STDs. The use of contraceptives, such as condoms, etc. while intercourse, can prevent the transfer of these diseases. Also, sex with unknown partners or multiple partners should be avoided as they may have such diseases. Specialists should be consulted immediately in case of doubt so as to assure early detection and cure of the disease.

Question 11.
State True/False with explanation.
(a) Abortions could happen spontaneously too. (True/False)
(b) Infertility is defined as the inability to produce a viable offspring and is always due to abnormalities/defects in the female partner. (True/False)
(c) Complete lactation could help as a natural method of contraception. (True/False)
(d) Creating awareness about sex related aspects is an effective method to improve reproductive health of the people. (True/False)
Answer:
(a) False
Abortion is term given for medical termination of pregnancy.

(b) False
Infertility is defined as the inability of the couple to produce baby even after unprotected coitus. It might occur due to abnormalities/defects in either male or female or both.

(c) False
Complete lactation or lactational amenorrhea is a natural method of contraception. Flowever, it is limited till lactation period, which continues till six months after parturition.

(d) True.

Question 12.
Correct the following statements:
(a) Surgical methods of contraception prevent gamete formation.
(b) All sexually transmitted diseases are completely curable.
(c) Oral pills are very popular contraceptives among the rural women.
(d) In E. T. techniques, embryos are always transferred into the uterus.
Answer:
(a) Surgical methods of contraception prevent the flow of gamete during intercourse.
(b) Some of the sexually transmitted diseases are curable if they are detected early and treated properly. AIDS is still an incurable disease.
(c) Oral pills are very popular contraceptives among urban women.
(d) In embryo transfer technique, 8 celled embryos are transferred into the fallopian tube while more than 8 celled embryos are transferred into the uterus.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 3 Human Reproduction Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 3 Human Reproduction

PSEB 12th Class Biology Guide Human Reproduction Textbook Questions and Answers

Question 1.
Fill in the blanks:
(a) Humans reproduce …………………. .(asexually/sexually)
Answer:
sexually.

(b) Humans are …………………… .(oviparous/viviparous/ovoviviparous)
Answer:
viviparous.

(c) Fertilisation is ………………………. in humans, (external/internal)
Answer:
internal

(d) Male and female gametes are …………………. .(diploid/haploid)
Answer:
haploid.

(e) Zygote is ………………….. .(diploid/haploid)
Answer:
diploid.

(f) The process of release of ovum from a mature follicle is called ……………………. .
Answer:
ovulation.

(g) Ovulation is induced by a hormone called …………………… .
Answer:
luteinising hormone.

(h) The fusion of male and female gametes is called ……………………… .
Answer:
fertilisation.

(i) Fertilisation takes place in ………………… .
Answer:
fallopian tube (ampullary-isthmic junction).

(j) Zygote divides to form ……………….., which is implanted in uterus.
Answer:
blastocyst

(k) The structure which provides vascular connection between foetus and uterus is called …………………… .
Answer:
placenta.

Question 2.
Draw a labelled diagram of male reproductive system.
Answer:

Question 3.
Draw a labelled diagram of female reproductive system.
Answer:

Question 4.
Write two major functions each of testis and ovary.
Answer:
Functions of the testis
(a) They produce male gametes called spermatozoa by the process of spermatogenesis.
(b) The leydig cells of the seminiferous tubules secrete the male sex hormone called testosterone. Testosterone aids the development of secondary sex characteristics in males.

Functions of the ovary
(a) They produce female gametes called ova by the process of oogenesis.
(b) The growing Graafian follicles secrete the female sex hormone called estrogen. Estrogen aids the development of secondary sex characteristics in females.

Question 5.
Describe the structure of a seminiferous tubule.
Answer:
The production of sperms in the testes takes place in a highly coiled structure called the seminiferous tubules. These tubules are located in the testicular lobules. Each seminiferous tubule is lined by germinal epithelium. It is lined on its inner side by two types of cells namely spermatogonia aid sertoli cells respectively. Spermatogonia are male germ cells which produce primary spermatocytes by meiotic divisions. Primary spermatocytes undergo further meiotic division to form secondary spermatocytes and finally, spermatids. Spermatids later metamorphoses into male gametes called spermatozoa. Sertoli cells are known as nurse cells of the testes as they provide nourishment to the germ cells. There are large polygonal cells known as interstitial cells or leydig cells just adjacent to seminiferous tubules. These cells secrete the male hormone called testosterone.

Question 6.
What is spermatogenesis? Briefly describe the process of spermatogenesis.
Answer:
Spermatogenesis is the process of the production of sperms from the immature germ cells in males. It takes place in seminiferous tubules present inside the testes. During spermatogenesis, a diploid spermatogonium (male germ cell) increases its size to form a diploid primary spermatocyte. This diploid primary spermatocyte undergoes first meiotic division (meiosis I), which is a reductional division to form two equal haploid secondary spermatocytes. Each secondary spermatocyte then undergoes second meiotic division (meiosis II) to form two equal haploid spermatids. Hence, a diploid spermatogonium produces four haploid spermatids. These spermatids are transformed into spermatozoa (sperm) by the process called spermiogenesis.

Question 7.
Name the hormones involved in regulation of spermatogenesis.
Answer:
Follicle-stimulating hormones (FSH) and luteinising hormones (LH) are secreted by gonadotropin releasing hormones from the hypothalamus. These hormones are involved in the regulation of the process of spermatogenesis. FSH acts on sertoli cells, whereas LH acts on leydig cells of the testis and stimulates the process of spermatogenesis.

Question 8.
Define spermiogenesis and spermiation.
Answer:
Spermiogenesis : It is the process of transforming spermatids into matured spermatozoa or sperms.
Spermiation : It is ‘the process when mature spermatozoa are released from the sertoli cells into the lumen of seminiferous tubules.

Question 9.
Draw a labelled diagram of sperm.
Answer:

Question 10.
What are the major components of seminal plasma?
Answer:
Semen (produced in males) is composed of sperms and seminal plasma. The major components of the seminal plasma in the male reproductive system are mucus, spermatozoa, and various secretions of accessory glands. The seminal plasma is rich in fructose, calcium, ascorbic acid, and certain enzymes. It provides nourishment and protection to sperms.

Question 11.
What are the major functions of male accessory ducts and glands?
Answer:
The male accessory ducts are vasa efferentia, epididymis, vas deferens, and rete testis. They play an important role in the transport and temporary storage of sperms. On the contrary, male accessory glands are seminal vesicles, prostate glands, and bulbourethral glands. These glands secrete fluids that lubricate the reproductive system and sperms. The sperms get dispersed in the fluid which makes their transportation into the female body easier. The fluid is rich in fructose, ascorbic acid, and certain enzymes. They also provide nutrients and activate the sperm.

Question 12.
What is oogenesis? Give a brief account of oogenesis.
Answer:
Oogenesis is the process of the formation of a mature ovum from the oogonia in females. It takes place in the ovaries. During oogenesis, a diploid oogonium or egg mother cell increases in size and gets transformed into a diploid primary oocyte. This diploid primary oocyte undergoes first meiotic division i.e., meiosis I or reductional division to form two unequal haploid cells. The smaller cell is known as the first polar body, while the larger cell is known as the secondary oocyte. This secondary oocyte undergoes second meiotic division i.e., meiosis II or equational division and gives rise to a second polar body and an ovum. Hence, in the process of oogenesis, a diploid oogonium produces a single haploid ovum while two or three polar bodies are produced.

Question 13.
Draw a labelled diagram of a section through ovary.
Answer:

Question 14.
Draw a labelled diagram of a Graafian follicle.
Answer:

Question 15.
Name the functions of the following:
(a) Corpus luteum
(b) Endometrium
(c) Acrosome
(d) Sperm tail
(e) Fimbriae
Answer:
(a) Corpus Luteum: It is formed from the ruptured Graafian follicle. It secretes progesterone hormone during the luteal phase of the menstrual cycle. A high level of progesterone inhibits the secretions of FSH and LH, thereby preventing ovulation. It also allows the endometrium of the uterus to proliferate and to prepare itself for implantation.

(b) Endometrium: It is the innermost lining of the uterus. It is rich in glands and undergoes cyclic changes during various phases of the menstrual cycle to prepare itself for the implantation of the embryo.

(c) Acrosome: It is a cap-like structure present in the anterior part of the head of the sperm. It contains hyaluronidase enzyme, which hydrolyses the outer membrane of the egg, thereby helping the sperm to penetrate the egg at the time of fertilisation.

(d) Sperm Tail: It is the longest region of the sperm that facilitates the movement of the sperm inside the female reproductive tract.

(e) Fimbriae: They are finger-like projections at the ovarian end of the fallopian tube. They help in the collection of the ovum (after ovulation), which is facilitated by the beating of the cilia.

Question 16.
Identify True/False statements. Correct each false statement to make it true.
(a) Androgens are produced by Sertoli cells. (True/False)
(b) Spermatozoa get nutrition from Sertoli cells. (True/False)
(c) Leydig cells are found in ovary. (True/False)
(d) Leydig cells synthesise androgens. (True/False)
(e) Oogenesis takes place in corpus luteum. (True/False)
(f) Menstrual cycle ceases during pregnancy. (True/False)
(g) Presence or absence of hymen is not a reliable indicator of virginity or sexual experience. (True/False)
Answer:
(a) Androgens are produced by Sertoli cells.
False Correct : Leydig cells.

(b) Spermatozoa get nutrition from Sertoli cells.
True

(c) Leydig cells are found in ovary.
False Correct : spermatogonia.

(d) Leydig cells synthesise androgens.
True

(e) Oogenesis takes place in corpus luteum.
False Correct : ovaries

(f) Menstrual cycle ceases during pregnancy.
True

(g) Presence or absence of hymen is not a reliable indicator of virginity or sexual experience.
True

Question 17.
What is menstrual cycle? Which hormones regulate menstrual cycle? ,
Answer:
The menstrual cycle is a series of cyclic physiologic changes that take place inside the female reproductive tract in primates. The whole cycle takes around 28 days to complete. The end of the cycle is accompanied by the breakdown of uterine endothelium, which gets released in the form of blood and mucus through the vagina. This is known as menses.

The follicle stimulating hormone (FSH), luteinising hormone (LH), L estrogen, and progesterone are the various hormones that regulate the menstrual cycle. The level of FSH and LH secreted from the anterior pituitary gland increases during the follicular phase. FSH secreted under the influence of RH (releasing hormone) from the hypothalamus , stimulates the conversion of a primary follicle into a graafian follicle.

The level of LH increases gradually leading to the growth of follicle and f secretion of estrogen. Estrogen inhibits the secretion of FSH and stimulates the secretion of luteinising hormone. It also causes the thickening of the uterine endometrium. The increased level of LH causes the rupturing of the graafian follicle and release the ovum into the fallopian tube. The ruptured graafian follicle changes to corpus luteum and starts secreting progesterone hormone during the luteal phase.

Progesterone hormone helps in the maintenance and preparation of endometrium for the implantation of the embryo. High levels of progesterone hormone in the blood decrease the secretion of LH and FSH, therefore inhibiting further ovulation.

Question 18.
What is parturition? Which hormones are involved in induction of parturition?
Answer:
Parturition is the process of giving birth to a baby as the development of the foetus gets completed in the mother’s womb. The hormones involved in this process are oxytocin and relaxin. Oxytocin leads to the contraction of smooth muscles of myometrium of the uterus, which directs the full term foetus towards the birth canal. On the other hand, relaxin hormone causes relaxation of the pelvic ligaments and prepares the uterus for child birth.

Question 19.
In our society the women are often blamed for giving birth to [ daughters. Can you explain why this is not correct?
Answer:
All human beings have 23 pairs of chromosomes. Human males have 22 pairs of autosomes and contain one or two types of sex chromosome. They are either X or Y. On the contrary, human females have 22 pairs of autosomes and contain only the X sex chromosome. The sex of an individual is determined by the type of the male gamete (X or Y), which fuses with the X chromosome of the female. If the fertilising sperm is X, then the baby will be a girl and if it is Y, then the baby will be a boy.
Hence, it is incorrect to blame a woman for the gender of the child.

Question 20.
How many eggs are released by a human ovary in a month? How many eggs do you think would have been released if the mother gave birth to identical twins? Would your answer change if the twins bom were fraternal?
Answer:
An ovary releases an egg every month. When two babies are produced in succession, they are called twins. Generally, twins are produced from a single egg by the separation of early blastomeres resulting from the first zygotic cleavage. As a result, the young ones formed will have the same genetic make-up and are thus, called identical twins.

If the twins born are fraternal, then they would have developed from two separate eggs. This happens when two eggs’ (one from each ovary) are released at the same time and get fertilised by two separate sperms. Hence, the young ones developed will have separate genes and are therefore, called non-identical or fraternal twins.

Question 21.
How many eggs do you think were released by the ovary of a female dog which gave birth to 6 puppies?
Answer:
Dogs and rodents are polyovulatory species. In these species, more than one ovum is released from the ovary at the time of ovulation. Hence, six eggs were released by the ovary of a female dog to produce six puppies.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 15 Biodiversity and Conservation Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 15 Biodiversity and Conservation

PSEB 12th Class Biology Guide Biodiversity and Conservation Textbook Questions and Answers

Question 1.
Name the three important components of biodiversity.
Answer:
Biodiversity is the variety of living forms present in various ecosystems. It includes variability among life forms from all sources including land, air, and water. Three important components of biodiversity are as follows :

• Genetic diversity
• Species diversity
• Ecological diversity

Question 2.
How do ecologists estimate the total number of species present in the world?
Answer:
The diversity of living organisms present on the Earth is very vast. According to an estimate by researchers, it is about seven million. the total number of species present in the world is calculated by statistical comparison between species richness of a well-studied group of insects of temperate and tropical regions. Then, these ratios are extrapolated with other groups of plants and animals to calculate the total species richness present on the Earth.

Question 3.
Give three hypotheses for explaining why tropics show greatest levels of species richness.
Answer:
There are three different hypotheses proposed by scientists for explaining species richness in the tropics.

1. Tropical latitudes receive more solar energy than temperate regions, which leads to high productivity and high species diversity.
2. Tropical regions have less seasonal variations and have a more or less constant environment. This promotes the niche specialization and thus, high species richness.
3. Temperate regions were subjected to glaciations during the ice age, while tropical regions remained undisturbed which led to an increase in the species diversity in this region.

Question 4.
What is the significance of the slope of regression in a species-area relationship?
Answer:
The slope of regression (z) has a great significance in order to find a species-area relationship. It has been found that in smaller areas (where the species-area relationship is analyzed), the value of slopes of regression is similar regardless of the taxonomic- group or the region. However, when a similar analysis is done in larger areas, then the slope of regression is much steeper.

Question 5.
What are the major causes of species losses in a geographical region?
Answer:
Biodiversity is the variety of living forms present in various ecosystems. It includes variability among life forms from all sources including land, air, and water. Biodiversity around the world is declining at a very fast pace. The following are the major causes for the loss of biodiversity around the world :

(i) Habitat Loss and Fragmentation: Habitats of various organisms are altered or destroyed by uncontrolled and unsustainable human activities such as deforestation, slash and burn agriculture, mining, and urbanisation. This results in the breaking up of the habitat into small pieces, which effects the movement of migratory animals and also, decreases the genetic exchange between populations leading to a declination of species.

(ii) Over-exploitation: Due to over-hunting and over-exploitation of various plants and animals by humans, many species have become endangered or extinct (such as the tiger and the passenger pigeon).

(iii) Alien Species Invasions: Accidental or intentional introduction of non-native species into a habitat has also led to the declination or extinction of indigenous species. For example, the Nile perch introduced in Lake Victoria in Kenya led to the extinction of more than two hundred species of native fish in the lake.

(iv) Co-extinction: In a native habitat, one species is connected to the other in an intricate network. The extinction of one species causes the extinction of other species, which is associated with it in an obligatory way. For example, the extinction of the host will cause the extinction of its parasites.

Question 6.
How is biodiversity important for ecosystem functioning?
Answer:
An ecosystem with high species diversity is much more stable than an ecosystem with low species diversity. Also, high biodiversity makes the ecosystem more stable in productivity and more resistant towards disturbances such as alien species invasions and floods.

If an ecosystem is rich in biodiversity, then the ecological balance would not get affected. As we all know, various trophic levels are connected through food chains. If anyone organism or all organisms of any one trophic level is killed, then it will disrupt the entire food chain. For example, in a food chain, if all plants are killed, then all deer’s will die due to the lack of food.

If all deer’s are dead, soon the tigers will also die. Therefore, it can be concluded that if an ecosystem is rich in species, then there will be other food alternatives at each trophic level which would not allow any organism to die due to the absence of their food resource. Hence, biodiversity plays an important role in maintaining the health and ecological balance of an ecosystem.

Question 7.
What are sacred groves? What is their role in conservation?
Answer:
Sacred groves are tracts of forest which are regenerated around places of worship. Sacred groves are found in Rajasthan, Western Ghats of Karnataka, and Maharashtra, Meghalaya, and Madhya Pradesh. Sacred grows help in the protection of many rare, threatened, and endemic species of plants and animals found in an area. The process of deforestation is strictly prohibited in this region by tribals. Hence, the sacred grove biodiversity is a rich area.

Question 8.
Among the ecosystem services are control of floods and soil erosion. How is this achieved by the biotic components of the ecosystem?
Answer:
The biotic components of an ecosystem include the living organisms such as plants and animals. Plants play a very important role in controlling floods and soil erosion. The roots of plants hold the soil particles together, thereby preventing the top layer of the soil to get eroded by wind or running water. The roots also make the soil porous, thereby allowing groundwater infiltration and preventing floods. Hence, plants are able to prevent soil erosion and natural calamities Fucii as floods and droughts. They also increase the fertility of soil and biodiversity.

Question 9.
The species diversity of plants (22 percent) is much less than that of animals (72 percent). What could be the explanations to how animals achieved greater diversification?
Answer:
72 percent of species recorded on the Earth are animals and only 22 percent species are plants. There is quite a large difference in their percentage This is because animals have adapted themselves to ensure their survival in changing environments in comparison to plants. For example, insects and other animals have developed a complex nervous system to control and coordinate their body structure. Also, repeated body/ segments with paired appendages and external cuticles have made insects versatile and have given them the ability to survive in vain JUS habitats as compared to other life forms.

Question 10.
Can you think of a situation where we deliberately want to make a species extinct? How would you justify it?
Answer:
Yes, there are various kinds of parasites and disease-causing microbes that we deliberately want to eradicate from the Earth. Since these micro-organisms are harmful to human beings, scientists are working hard to fight against them.

Scientists have been able to eliminate smallpox virus from the world through the use of vaccinations. This shows than humans deliberately want to make these species extinct. Several other eradication programs such as polio and Hepatitis B vaccinations are aimed to eliminate these disease-causing microbes.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 2 Sexual Reproduction in Flowering Plants Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 2 Sexual Reproduction in Flowering Plants

PSEB 12th Class Biology Guide Sexual Reproduction in Flowering Plants Textbook Questions and Answers

Question 1.
Name the parts of an angiosperm flower in which development of male and female gametophyte take place.
Answer:
The male gametophyte or the pollen grain develops inside the pollen chamber of the anther, whereas the female gametophyte (also known as the embryo sac) develops inside the nucellus of the ovule from the functional megaspore.

Question 2.
Differentiate between microsporogenesis and megasporogenesis. Which type of cell division occurs during these events? Name the structures formed at the end of these two events.
Answer:

Microsporogenesis	Megasporogenesis
1. It is the process of the formation of microspore tetrads from a microspore mother cell through meiosis.	It is the process of the formation of the four megaspores from a megaspore mother cell in the region of the nucellus through meiosis.
2. It occurs inside the pollen sac of the anther.	It occurs inside the ovule.

(b) Both events (microsporogenesis and megasporogenesis) involve the process of meiosis or reduction division which results in the formation of haploid gametes from the microspore and megaspore mother cells.

(c) Microsporogenesis results in the formation of haploid microspores from a diploid microspore mother cell. On the other hand, megasporogenesis results in the formation of haploid megaspores from a diploid megaspore mother cell.

Question 3.
Arrange the following terms in the correct developmental sequence:
Pollen grain, sporogenous tissue, microspore tetrad, pollen mother cell, male gametes.
Answer:
The correct developmental sequence is as follows:
Sporogenous tissue, pollen mother cell, microspore tetrad, pollen grain, male gametes.
During the development of microsporangium, each cell of the sporogenous tissue acts as a pollen mother cell and gives rise to a microspore tetrad, containing four haploid microspores by the process of meiosis (microsporogenesis). As the anther matures, these microspores dissociate and develop into pollen grains. The pollen grains mature and give rise to male gametes.

Question 4.
With a neat, labelled diagram, describe the parts of a typical ’ angiosperm ovule.
Answer:
An ovule is a female megasporangium where the formation of megaspores takes place.

The various parts of a typical angiospermic ovule are as follows :
1. Funiculus: It is a stalk-like structure which represents the point of attachment of the ovule to the placenta of the ovary.

2. Hilum: It is the point where the body of the ovule is attached to the funiculus.

3. Integuments: They are the outer layers surrounding the ovule that provide protection to the developing embryo.

4. Micropyle: It is a narrow pore formed by the projection of integuments. It marks the point where the pollen tube enters the ovule at the time of fertilisation.

5. Nucellus: It is a mass of the parenchymatous tissue surrounded by the integuments from the outside. The nucellus provides nutrition to the developing embryo. The embryo sac is located inside the nucellus.

6. Chalaza: It is the based swollen part of the nucellus from where the integuments originate.

Question 5.
What is meant by monosporic development of female gametophyte?
Answer:
The female gametophyte or the embryo sac develops from a single functional megaspore. This is known as monosporic development of the female gametophyte. In most flowering plants, a single megaspore mother cell present at the micropylar pole of the nucellus region of the ovule undergoes meiosis to produce four haploid megaspores. Later, out of these four megaspores, only one functional megaspoxe develops into the female gametophyte, while the remaining three degenerate.

Question 6.
With a neat diagram explain the 7-celled, 8-nucleate nature of the female gametophyte.
Answer:
Structure of the mature embryo sac

The female gametophyte (embryo sac) develops from a single functional megaspore. This megaspore undergoes three successive mitotic divisions to form eight nucleate embryo sacs.
The first mitotic division in the megaspore forms two nuclei. One nucleus moves towards the micropylar end while the other nucleus moves towards the chalazal end. Then, these nuclei divide at their respective ends and re-divide to form eight nucleate stages. As a result, there are four nuclei each at both the ends i.e., at the micropylar and the chalazal end in the embryo sac. At the micropylar end, out of the four nuclei only three differentiate into two synergids and one egg cell. Together they are known as the egg apparatus.

Similarly, at the chalazal end, three out of four nuclei differentiates as antipodal cells. The remaining two cells (of the micropylar and the chalazal end) move towards the centre and are known as the polar nuclei, which are situated in a large central cell. Hence, at maturity, the female gametophyte appears as a 7-celled structure, though it has 8-nucleate.

Question 7.
What are chasmogamous flowers? Can cross-pollination occur in cleistogamous flowers? Give reasons for your answer.
Answer:
There are two types of flowers present in plants namely Oxalis and Viola – chasmogamous and cleistogamous flowers. Chasmogamous flowers have exposed anthers and stigmata similar to the flowers of other species.

Cross-pollination cannot occur in cleistogamous flowers. This is because cleistogamous flowers never open at all. Also, the anther and the stigma lie close to each other in these flowers. Hence, only self-pollination is possible in these flowers.

Question 8.
Mention two strategies evolved to prevent self-pollination in flowers.
Answer:
Self-pollination involves the transfer of pollen from the stamen to the pistil of the same flower. Two strategies that have evolved to prevent self-pollination in flowers are given on next page:
1. Self-incompatibility: In certain plants, the stigma of the flower has the capability to prevent the germination of pollen grains and hence, prevent the growth of the pollen tube. It is a genetic mechanism to prevent self-pollination called self-incompatibility. Incompatibility may be between individuals of the same species or between individuals of different species. Thus, incompatibility prevents breeding.

2. Protandry: In some plants, the gynoecium matures before the androecium or vice-versa. This phenomenon is known as protogyny or protandry respectively. This prevents the pollen from coming in contact with the stigma of the same flower.

Question 9.
What is self-incompatibility? Why does self-pollination not lead to seed formation in self-incompatible species?
Answer:
Self-incompatibility is a genetic mechanism in angiosperms that prevents self-pollination. It develops genetic incompatibility between ‘ individuals of the same species or between individuals of different
species.

The plants which exhibit this phenomenon have the ability to prevent germination of pollen grains and thus, prevent the growth of the pollen tube on the stigma of the flower. This prevents the fusion of the gametes along with the development of the embryo. As a result, no seed formation takes place.

Question 10.
What is bagging technique? How is it useful in a plant breeding programme?
Answer:
Various artificial hybridisation techniques (under various crop improvement programmes) involve the removal of the anther from bisexual flowers without affecting the female reproductive part (pistil) through the process of emasculation. Then, these emasculated flowers are wrapped in bags to prevent pollination by unwanted pollen grains. This process is called bagging.

This technique is an important part of the plant breeding programme as
it ensures that pollen grains of only desirable plants are used for fertilisation of the stigma to develop the desired plant variety.

Question 11.
What is triple fusion? Where and how does it take place? Name the nuclei involved in triple fusion.
Answer:
Triple fusion is the fusion of the male gamete with two polar nuclei inside the embryo sac of the angiosperm. This process of fusion takes place inside the embryo sac.

When pollen grains fall on the stigma, they germinate and give rise to the pollen tube that passes through the style and enters into the ovule. After this, the pollen tube enters one of synergids and releases two male gametes there.

Out of the two male gametes, one gamete fuses with the nucleus of the egg cell and forms the zygote (syngamy). The other male gamete fuses with the two polar nuclei present in the central cell to form a triploid primary endosperm nucleus. Since this process involves the fusion of three haploid nuclei, it is known as triple fusion. It results in the formation of the endosperm.
One male gamete nucleus and two polar nuclei are involved in this process.

Question 12.
Why do you think the zygote is dormant for sometime in a fertilised ovule?
Answer:
The zygote is formed by the fusion of the male gamete with the nucleus of the egg cell. The zygote remains dormant for some time and waits for the endosperm to form, which develops from the primary endosperm cell resulting from triple fusion. The endosperm provides food for the growing embryo and after the formation of the endosperm, further development of the embryo from the zygote starts.

Question 13.
Differentiate between:
(a) hypocotyl and epicotyl;
(b) coleoptile and coleorrhiza;
(c) integument and testa;
(d) perisperm and pericarp.
Answer:
(a)

Hypocotyl	Epicotyl
1. The portion of the embryonal axis which lies below the cotyledon in a dicot embryo is known as the hypocotyl.	The portion of the embryonal axis which lies above the cotyledon in a dicot embryo is known as the epicotyl.
2. It terminates with the radicle.	It terminates with the plumule.

(b)

Coleoptile	Coleorrhiza
It is a conical protective sheath that encloses the plumule in a monocot seed.	It is an undifferentiated sheath that encloses the radicle and the root cap in a monocot seed.

(c)

Integument	Testa
It is the outermost covering of an ovule. It provides protection to it.	It is the outermost covering of a seed. It provides protection to the young embryo.

(d) Perisperm

Perisperm	Pericarp
It is the residual nucellus which persists. It is present in some seeds such as beet and black pepper.	It is the ripened wall of a fruit, which develops from the wall of an ovary.

Question 14.
Why is apple called a false fruit? Which part(s) of the flower forms the fruit?
Answer:
Fruits derived from the ovary and other accessory floral parts are called false fruits. On the contrary, true fruits are those fruits which develop from the ovary, but do not consist of the thalamus or any other floral part. In an apple, the fleshy receptacle forms the main edible part. Hence, it is a false fruit.

Question 15.
What is meant by emasculation? When and why does a plant breeder employ this technique?
Answer:
Emasculation is the process of removing anthers from bisexual flowers without affecting the female reproductive part (pistil), which is used in various plant hybridisation techniques.

Emasculation is performed by plant breeders in bisexual flowers to obtain the desired variety of a plant by crossing a particular plant with the desired pollen grain. To remove the anthers, the flowers are f covered with a bag before they open. This ensures that the flower is pollinated by pollen grains obtained from desirable varieties only. Later, the mature, viable, and stored pollen grains are dusted on the
bagged stigma by breeders to allow artificial pollination to take place and obtain the desired plant variety.

Question 16.
If one can induce parthenocarpy through the application of growth substances, which fruits would you select to induce parthenocarpy and why?
Answer:
Parthenocarpy is the process of developing fruits without involving the process of fertilisation or seed formation. Therefore, the seedless varieties of economically important fruits such as orange, lemon, water melon etc. are produced using this technique. This technique involves inducing fruit formation by the application of plant growth hormones such as auxins.

Question 17.
Explain the role of tapetum in the formation of pollen-grain wall.
Answer:
Tapetum is the innermost layer of the microsporangium. It provides nourishment to the developing pollen grains. During microsporogenesis, the cells of tapetum produce various enzymes, hormones, amino acids, and other nutritious material required for the development of pollen grains. It also produces the exine layer of the pollen grains, which is composed of the sporopollenin.

Question 18.
What is apomixis and what is its importance?
Answer:
Apomixis is the mechanism of seed production without involving the process of meiosis and syngamy. It plays an important role in hybrid seed production. The method of producing hybrid seeds by cultivation is very expensive for farmers. Also, by sowing hybrid seeds, it is difficult to maintain hybrid characters as characters segregate during meiosis. Apomixis prevents the loss of specific characters in the hybrid. Also, it is a cost-effective method for producing seeds.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 16 Environmental Issues Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 16 Environmental Issues

PSEB 12th Class Biology Guide Environmental Issues Textbook Questions and Answers

Question 1.
What are the various constituents of domestic sewage? Discuss the effects of sewage discharge on a river.
Answer:
Domestic sewage is the waste originating from the kitchen, toilet, laundry, and other sources. It contains impurities such as suspended solid (sand, salt, clay), colloidal, material (faecal matter, bacteria, plastic and cloth fibre), dissolved materials (nitrate, phosphate, calcium, sodium, ammonia), and disease-causing microbes. When organic wastes from the sewage enter the water bodies, it serves as a food source for micro-organisms such as algae and bacteria. As a result, the population of these micro-organisms in the water body increases.

Here, they utilise most of the dissolved oxygen for their metabolism. This results in an increase in the levels of Biological oxygen demand (BOD) in river water and results in the death of aquatic organisms. Also, the nutrients in the water lead to the growth of planktonic algal, causing algal bloom. This causes deterioration of water quality and fish mortality.

Question 2.
List all the wastes that you generate, at home, school or during your trips to other places. Could you very easily reduce the generation of these wastes? Which would be difficult or rather impossible to reduce?
Answer:
Wastes generated at home include plastic bags, paper napkins, toiletries, kitchen wastes (such as peelings of vegetables and fruits, tea leaves), domestic sewage, glass, etc.

Wastes generated at schools include waste paper, plastics, vegetable and fruit peels, food wrappings, sewage etc.
Wastes generated at trips or picnics include plastic, paper, vegetable and fruit peels, disposable cups, plates, spoons etc.

Yes, wastes can be easily reduced by the judicious use of the above materials. Wastage of paper can be minimised by writing on both sides of the paper and by using recycled paper. Plastic and glass waste can also be reduced by recycling and re-using.

Also, substituting plastic bags with biodegradable jute bags can reduce wastes generated at home, school, or during trips. Domestic sewage can be reduced by optimising the use of water while bathing, cooking, and other household activities. Non-biodegradable wastes such as plastic, metal, broken glass, etc. are difficult to decompose because micro-organisms do riot have the ability to decompose them.

Question 3.
Discuss the causes and effects of global warming. What measures need to be taken to control global warming?
Answer:
Global warming is defined as an increase in the average temperature of the Earth’s surface. Causes of Global Warming: Global warming occurs as a result of the increased concentration of greenhouse gases in the atmosphere. Greenhouse gases include carbon dioxide, methane, and water vapour. These gases trap solar radiations released back by the Earth. This helps in keeping our planet warm and thus, helps in human survival. However, an increase in the amount of greenhouse gases can lead to an excessive increase in the Earth’s temperature, leading to global warming. Global warming is a result of industrialisation, burning of fossil fuels, and deforestation.

Effects of Global Warming: It has been observed that in the past three decades, the average temperature of the Earth has increased by 0.6°C. As a result, the natural water cycle has been disturbed resulting in changes in the pattern of rainfall. It also changes the amount of rainwater. Also, it results in the melting of Polar ice caps and mountain glaciers, which has caused a rise in the sea level, leading to the inundation of coastal regions.

Control Measures for Preventing Global Warming:

• Reducing the use of fossil fuels
• Use of bio-fuels
• Improving energy efficiency
• Use of renewable source of energy such as CNG etc.
• Reforestation.
• Recycling of materials

Question 4.
Match the items given in column A and B:

Column A	Column B
(a) Catalytic converter	(i) Particulate matter
(b) Electrostatic precipitator	(ii) Carbon monoxide and nitrogen oxides
(c) Earmuffs	(iii) High noise level
(d) Landfills	(iv) Solid wastes

Answer:

Column A	Column B
(a) Catalytic converter	(ii) Carbon monoxide and nitrogen oxides
(b) Electrostatic precipitator	(i) Particulate matter
(c) Earmuffs	(iii) High noise level
(d) Landfills	(iv) Solid wastes

Question 5.
Write critical notes on the following:
(a) Eutrophication
(b) Biological magnification
(c) Groundwater depletion and ways for its replenishment
Answer:
(a) Eutrophication:
It is the natural ageing process of a lake caused due to nutrient enrichment. It is brought down by the runoff of nutrients such as animal wastes, fertilisers, and sewage from land which leads to an increased fertility of the lake. As a result, it causes a tremendous increase in the primary productivity of the ecosystem. This leads to an increased growth of algae, resulting, into, algal blooms. Later, the decomposition of these algae depletes the supply of oxygen, leading to the death of other aquatic animal life.

(b) Biological Magnification: Unknowingly some harmful chemicals enter our bodies through the food chain. We use several pesticides and other chemicals to protect our crops from diseases and pests. These chemicals are either washed down into the soil or into the water bodies. From the soil, these are absorbed by the plants along with water and minerals, and from the water bodies, these are taken up by aquatic plants and animals. This is one of the ways in which they enter the food chain. As these chemicals are not degradable, these get accumulated progressively at each trophic level. As human beings occupy the topmost level in any food chain, the maximum concentration of these chemicals get accumulated in our bodies. This phenomenon is known as biological magnification.

(c) Ground Water Depletion and Ways for its Replenishment: The level of groundwater has decreased in the recent years. The source of water supply is rapidly diminishing each year because of an increase in the population and water pollution. To meet the demand of water, water is withdrawn from water bodies such as ponds, rivers etc. As a result, the source of groundwater is depleting.

This is because the amount of groundwater being drawn for human use is more than the amount replaced by rainfall. Lack of vegetation cover also results in very small amounts of water seeping through the ground. An increase in water pollution is another factor that has reduced the availability of groundwater.

Measures for Replenishing Ground Water:

• Preventing over-exploitation of groundwater
• Optimising water use and reducing water demand
• Rainwater harvesting
• Preventing deforestation and plantation of more trees.

Question 6.
Why does ozone hole form over Antarctica? How will enhanced ultraviolet radiation affect us?
Answer:
The ozone hole is more prominent over the region of Antarctica. It is formed due to an increased concentration of chlorine in the atmosphere. Chlorine is mainly released from chlorofluorocarbons (CFC’s) widely used as refrigerants. The CFC’s magnate from the troposphere to the stratosphere, where they release chlorine atoms by the action of UV rays on them.

The release of Chlorine atoms causes the conversion of ozone into molecular oxygen. One atom of chlorine can destroy around 10,000 molecules of ozone and causes ozone depletion. The formation of the ozone hole will result in an increased concentration of UV – B radiations on the Earth’s surface. UV -B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV -B cause corneal cataracts in human beings.

Question 7.
Discuss the role of women and communities in protection and conservation of forests.
Answer:
Women and communities have played a major role in environmental conservation movements.
(i) Case Study of the Bishnoi Community: The Bishnoi community in Rajasthan strictly believes in the concept of living peacefully with nature. In 1731, the king of Jodhpur ordered his ministers to arrange wood for the construction of his new palace. For this purpose, the minister and 1116 workers went to Bishnoi village.

There, a Bishnoi woman called Amrita Devi along with her daughter and hundreds of other Bishnois showed the courage to step forward and stop them from cutting trees. They embraced the trees and lost their lives at the hands of soldiers of the king. This resistance by the people of the village forced the king to give up the idea of cutting trees.

(ii) Chipko Movement: The Chipko movement was started in 1974 in the Garhwal region of the Himalayas. In this movement, the women from the village stopped the contractors from cutting forest trees by embracing them.

Question 8.
What measures, as an individual, would you take to reduce environmental pollution?
Answer:
The following initiatives can be taken to prevent environmental pollution:
Measures for Preventing Air Pollution

• Planting more trees
• Use of clean and renewable energy sources such as CNG and bio-fuels
• Reducing the use of fossil fuels
• Use of catalytic converters in automobiles

Measures for Preventing Water Pollution:

• Optimising the use of water
• Using kitchen wastewater in gardening and other household purposes

Measures for Controlling Noise Pollution:

• Avoid burning crackers on Diwali
• Plantation of more trees

Measures for Decreasing Solid Waste Generation:

• Segregation of waste
• Recycling and reuse of plastic and paper
• Composting of biodegradable kitchen waste
• Reducing the use of plastics.

Question 9.
Discuss briefly the following:
(a) Radioactive wastes
(b) Defunct ships and e-wastes
(c) Municipal solid Wastes
Answer:
(a) Radioactive Wastes: Radioactive wastes are generated during the process of generating nuclear energy from radioactive materials. Nuclear waste is rich in radioactive materials that generate large quantities of ionising radiations such as gamma rays. These rays cause mutation in organisms, which often results in skin cancer. At high dosage, these rays can be lethal.

Safe disposal of radioactive wastes is a big challenge. It is recommended that nuclear wastes should be stored after pre-treatment in suitable shielded containers, which should then be buried in rocks.

(b) Defunct Ships and E-wastes: Defunct ships are dead ships that are no longer in use. Such ships are broken down for scrap metal in countries such as India and Pakistan. These ships are a source of various toxicants such as asbestos, lead, mercury etc. Thus, they contribute to solid* wastes that are hazardous to health.

E-waste or electronic wastes generally include electronic goods such as computers etc. Such wastes are rich in metals such as copper, iron, silicon, gold etc. These metals are highly toxic and pose serious health hazards. People of developing countries are involved in the recycling process of these metals and therefore, get exposed to toxic substances present in these wastes.

(c) Municipal Solid Wastes: Municipal solid wastes are generated from schools, offices, homes, and stores. It is generally rich in glass, metal, paper waste, food, rubber, leather, and textiles. The open dumps of municipal wastes serve as a breeding ground for flies, mosquitoes, and other disease-causing microbes. Hence, it is necessary to dispose of municipal solid waste properly to prevent the spreading of diseases. Sanitary landfills and incineration are the methods for the safe disposal of solid wastes.

Question 10.
What initiatives were taken for reducing vehicular air pollution in Delhi? Has air quality improved in Delhi?
Answer:
Delhi has been categorised as the fourth most polluted city of the world in a list of 41 cities. Burning of fossil fuels has added to the pollution of air in Delhi.
Various steps have been taken to improve the quality of air in Delhi.

(a) Introduction of CNG (Compressed Natural Gas): By the order of the supreme court of India, CNG-powered vehicles were introduced at the end of year 2006 to reduce the levels of pollution in Delhi. CNG is a clean fuel that produces very little unburnt particles.
(b) Phasing out of old vehicles
(c) Use of unleaded petrol
(d) Use of low-sulphur petrol and diesel
(e) Use of catalytic converters
(f) Application of stringent pollution-level norms for vehicles
(g) Implementation of Bharat stage I, which is equivalent to euro II norms in vehicles of major Indian cities.

The introduction of CNG-powered vehicles has improved Delhi’s air quality, which has lead to a substantial fall in the level of CO₂ and SO₂. However, the problem of suspended particulate matter (SPM) and respiratory suspended particulate matter,(RSPM) still persists.

Question 11.
Discuss briefly the following:
(a) Greenhouse gases
(b) Catalytic converter
(c) Ultraviolet B
Answer:
(a) Greenhouse Gases: The greenhouse effect refers to an overall increase in the average temperature of the Earth due to the presence of greenhouse gases. Greenhouse gases mainly consist of carbon dioxide, methane, and water vapour. When solar radiations reach the Earth, some of these radiations are absorbed.

These absorbed radiations are Released back into the atmosphere. These radiations are trapped by the greenhouse gases present in the atmosphere. This helps in keeping our planet warm and thus, helps in human survival.
However, an increase in the amount of greenhouse gases can lead to an excessive increase in the Earth’s temperature, thereby causing global warming.

(b) Catalytic Converter: Catalytic converters are devices fitted in automobiles to reduce automobile or vehicle pollution. These devices contain expensive metals such as platinum, palladium, and rhodium that act as catalysts.
As the vehicular discharge -passes through the catalytic converter, the unburnt hydrocarbons present in it get converted into carbon dioxide and water. Carbon monoxide and nitric oxide released by catalytic converters are converted into carbon dioxide and nitrogen gas respectively.

(c) Ultraviolet B: Ultraviolet-B is an electromagnetic radiation which has a shorter wavelength than visible light.
It is a harmful radiation that comes from sunlight and penetrates through the ozone hole onto the Earth’s surface.
It induces many health hazards in humans. UV -B damages DNA and activates the process of skin ageing. It also causes skin darkening and skin cancer. High levels of UV -B cause corneal cataracts in human beings.

Punjab State Board PSEB 12th Class Biology Book Solutions Chapter 1 Reproduction in Organisms Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Biology Chapter 1 Reproduction in Organisms

PSEB 12th Class Biology Guide Reproduction in Organisms Textbook Questions and Answers

Question 1.
Why is reproduction essential for organisms?
Answer:
Reproduction is a fundamental feature of all living organisms. It is a biological process through which living organisms produce offsprings (young ones) similar to them. Reproduction ensures the continuance of various species on the Earth. In the absence of reproduction, the species will not be able to exist for a long time and may soon get extinct.

Question 2.
Which is a better mode of reproduction sexual or asexual? Why?
Answer:
Sexual reproduction is a better mode of reproduction. It allows the formation of new variants by the combination of the DNA from two different individuals, typically one of each sex. It involves the fusion of the male and the female gamete to produce variants, which are not identical to their parents and to themselves. This variation allows the individual to adapt constantly changing and challenging environment. Also, it leads to the evolution of better suited organisms which ensures greater survival of a species. On the contrary, asexual reproduction allows very little or no variation at all. As a result, the individuals produced are exact copies of their parents and themselves.

Question 3.
Why is the offspring formed by asexual reproduction referred to as clone?
Answer:
A clone is a group of morphologically and genetically identical individuals. In the process of asexual reproduction, only one parent is involved and there is no fusion of the male and the female gamete. As a result, the offsprings so produced are morphologically and genetically similar to their parents and are thus, called clones.

Question 4.
Offspring formed due to sexual reproduction have better chances of survival. Why? Is this statement always true?
Answer:
Sexual reproduction involves the fusion of the male and the female gamete. This fusion allows the formation of new variants by the combination of the DNA from two (usually) different members of the species. The variations allow the individuals to adapt under varied environmental conditions for better chances of survival.

However, it is not always necessary that the offspring produced due to sexual reproduction has better chances of survival. Under some circumstances, asexual reproduction is more advantageous for certain organisms. For example, some individuals who do not move from one place to another and are well settled in their environment. Also, asexual reproduction is a fast and’ a quick mode of reproduction which does not consume much time and energy as compared to sexual reproduction.

Question 5.
How does the progeny formed from asexual reproduction differ from those formed by sexual reproduction?
Answer:

Progeny formed from asexual reproduction	Progeny formed from sexual reproduction
1. Asexual reproduction does not involve the fusion of the male and the female gamete. Organisms undergoing this kind ofreproduction produce offsprings that are morphologically and genetically identical to them.	Sexual reproduction involves the fusion of the male and the female gamete of two individuals, typically one of each sex. Organisms undergoing this kind of reproduction produce offsprings that are not identical to them.
2. Offsprings thus produced do not show variations and are called clones.	Offsprings thus produced show variations from each other and their parents.

Question 6.
Distinguish between asexual and sexual reproduction. Why is
vegetative reproduction also considered as a type of asexual reproduction?
Answer:

Sexual reproduction	Asexual reproduction
1. It involves the fusion of the male and the female gamete.	It does not involves the fusion of the male and the female gamete.
2. It requires two (usually) different individuals.	It requires only one individual.
3. The individuals produced are not identical to their parents and show variations from each other and also, from their parents.	The individuals produced are identical to the parent and are hence, called clones.
4. Most animals reproduce sexually. Both sexual and asexual modes of reproduction are found in plants.	Asexual modes of reproduction are common in organisms having simple organisations such as algae and fungi.
5. It is a slow process.	It is a fast process.

Vegetative reproduction is a process in which new plants are obtained without the production of seeds or spores. It involves the propagation of plants through certain vegetative parts such as the rhizome, sucker, tuber, bulb, etc. It does not involve the fusion of the male and the female gamete and requires only one parent. Hence, vegetative reproduction is considered as a type of asexual reproduction.

Question 7.
What is vegetative propagation? Give two suitable examples.
Answer:
Vegetative propagation is a mode of asexual reproduction in which new plants are obtained from the vegetative parts of plants. It does not involve the production of seeds or spores for the propagation of new plants. Vegetative parts of plants such as runners, rhizomes, suckers, tubers, etc. can be used as propagules for raising new plants.

Examples of vegetative reproduction are given below:
1. Eyes of potato: The surface of a potato has several buds called eyes. Each of these buds when buried in soil develops into a new plant, which is identical to the parent plant.

2. Leaf buds of Bryophyllum: The leaves of Bryophyllum plants bear several adventitious buds on their margins. These leaf buds have the ability to grow and develop into tiny plants when the leaves get detached from the plant and come in contact with moist soil.

Question 8.
Define:
(a) Juvenile phase,
(b) Reproductive phase,
(c) Senescent phase.
Answer:
(a) Juvenile Phase: All organisms have to reach a certain stage of growth and maturity in their life, before they can be reproduce sexually. This phase of growth is called the juvenile phase or vegetative phase in plants.

(b) Reproductive Phase: When the juvenile phase is over the organisms enter the period of reproductive phase or sexual maturity. It is indicated by showing various morphological and physiological changes such as development of secondary sexual characters in animals and by flowering in plants. This is the actual period of the life span of any organism when it is capable of producing offsprings. This phase is of variable duration in different organisms.

(c) Senescent Phase : This is the final and third stage of growth cycle. It can be considered as the end of reproductive phase. It is accompanied by reduction in functional capacity and increase in cellular break down and metabolic failures. It ultimately leads to death.

Question 9.
Higher organisms have resorted to sexual reproduction in spite of its complexity. Why?
Answer:
Although sexual reproduction involves more time and energy, higher organisms have resorted to sexual reproduction in spite of its complexity. This is because this mode of reproduction helps introduce new variations in progenies through the combination of the DNA from two (usually) different individuals. These variations allow the individual to cope with various environmental conditions and thus, make the organism better suited for the environment. Variations also lead to the evolution of better organisms and therefore, provide better chances of survival. On the other hand, asexual reproduction does not provide genetic differences in the individuals produced.

Question 10.
Explain why meiosis and gametogenesis are always interlinked?
Answer:
Meiosis is a process of reductional division in which the amount of genetic material is reduced. Gametogenesis is the process of the formation of gametes. Gametes produced by organisms are haploids (containing only one set of chromosomes), while the body of an organism is diploid. Therefore, for producing haploid gametes (gametogenesis), the germ cells of an organism undergo meiosis. During the process, the meiocytes of an organism undergo two successive nuclear and cell divisions with a single cycle of DNA replication to form the haploid gametes.

Question 11.
Identify each part in a flowering plant and write whether it is haploid (n) or diploid (2n).
(a) Ovary …………………………
(b) Anther ……………………..
(c) Egg ………………………….
(d) Pollen ……………………..
(e) Male gamete ……………….
(f) Zygote ………………..
Answer:
(a) diploid (2n)
(b) diploid (2n)
(c) haploid (n)
(d) haploid (n)
(e) haploid (n)
(f) diploid (2n)

Question 12.
Define external fertilisation. Mention its disadvantages.
Answer:
External fertilisation is the process in which the fusion of the male and the female gamete takes place outside the female body in an external medium, generally water. Fish, frog, starfish are some organisms that exhibit external fertilisation.

Disadvantages of external fertilisation
In external fertilisation, eggs have less chances of fertilisation. This can lead to the wastage of a large number of eggs produced during the process.
Further, there is an absence of proper parental care to the offspring, which results in a low rate of survival in the progenies.

Question 13.
Differentiate between a zoospore and a zygote.
Answer:

Zoospore	Zygote
1. A zoospore is a motile asexual spore that utilises the flagella for movement.	A zygote is a noh-motile diploid cell formed as a result of fertilisation.
2. It is an asexual reproductive structure.	It is formed as a result of sexual reproduction.
3. Zoospores are formed in simple plants like, algae or fungi.	Zygote is formed in complex organism.

Question 14.
Differentiate between gametogenesis from embryogenesis.
Answer:

Gametogenesis	Embryogenesis
1. It is the process of the formation of haploid male and female gametes from diploid meiocytes (gamete mother cell) through the process of meiosis.	It is the process of the development of the embryo from the repeated mitotic divisions of the diploid zygote.
2. Gametes may be either homogametes or heterogametes.	Animals may be either oviparous or viviparous.

Question 15.
Describe the post-fertilisation changes in a flower.
Answer:
As a result of double fertilisation in flowering plants, zygote (2n) and the primary endosperm nucleus (3n) is produced. The calyx, corolla, stamens and style wither away. The calyx may persist or even show growth in certain cases. The post fertilisation changes which take place are

• Endosperm formation
• Embryo formation
• Seed formation and
• Fruit formation.

The primary endosperm nucleus becomes active and forms a nutritive vegetative tissue. The endosperm at the expense of food present in the nucellus, Endosperm may be completely used up by the developing embryo (non-endospermic seeds e.g., pea) or may persist in the seed (endospermic seed e.g., castor). The zygote, waits for sometime till the formation of endosperm and then develops into embryo, by withdrawing nutrition from the endosperm. Ultimately the ovules are transformed into seeds and the ovary becomes a fruit. The formation of fruit helps in the nourishment and protection to the developing seeds and later helps in seed dispersal. Under favourable conditions the seeds germinate to form new plants.

Question 16.
What is a bisexual flower? Collect five bisexual flowers from your neighbourhood and with the help of your teacher find out their common and scientific names.
Answer:
A flower that contains both the male and female reproductive structure (stamen and pistil) is called a bisexual flower.
Examples of plants bearing bisexual flowers are:

1. Water lily {Nymphaea odorata)
2. Rose (Rosa multiflora)
3. Hibiscus (Hibiscus Rosa-sinensis)
4. Mustard (Brassica nigra)
5. Petunia (Petunia hybrida)

Question 17.
Examine a few flowers of any cucurbit plant and try to identify the staminate and pistillate flowers. Do you know any other plant that hears unisexual flowers?
Answer:
Cucurbit plant bears unisexual flowers as these flowers have either the stamen or the pistil. The staminate flowers bear bright, yellow coloured petals along with stamens that represent the male reproductive structure. On the other hand, the pistillate flowers bear only the pistil that represents the female reproductive structure.
Other examples of plants that bear unisexual flowers are corn, papaya, cucumber, etc.

Question 18.
Why are offspring of oviparous animals at a greater risk as compared to offspring of viviparous animals?
Answer:
Oviparous animals lay eggs outside their body. As a result, the eggs of these animals are under continuous threat from various environmental factors. On the other hand, in viviparous animals, the development of the egg takes place inside the body of the female. Hence, the offspring of an egg-laying or oviparous animal is at greater risk as compared to the offspring of a viviparous animal, which gives birth to its young ones.

Punjab State Board PSEB 12th Class Chemistry Book Solutions Chapter 9 Coordination Compounds Textbook Exercise Questions and Answers.

PSEB Solutions for Class 12 Chemistry Chapter 9 Coordination Compounds

PSEB 12th Class Chemistry Guide Coordination Compounds InText Questions and Answers

Question 1.
Explain the bonding in coordination compounds in terms of Werner’s postulates.
Answer:
(i) The primary valencies are satisfied by negative ions and equal-to the oxidation state of the metal.

(ii) The secondary valencies can be satisfied by neutral or negative ions. It is equal to the coordination number of the central metal atom and is fixed.

(iii) The ions bound to the central metal ion to secondary linkages have definite spatial arrangements and give geometry to the complex. While primary valency is non-directional.

Question 2.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1 : 1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1 : 4 molar ratio does not give the test of Cu²⁺ ion. Explain why?
Answer:
FeSO⁴ solution mixed with (NH⁴)²SO⁴ solution in 1 : 1 molar ratio forms double salt, FeSO⁴∙(NH⁴)²SO⁴∙6H²O which ionises in the solution to give Fe2+ ions. Hence, it gives the test of Fe²⁺ ions.

CuSO₄ solution mixed with aqueous ammonia in 1 : 4 molar ratio forms a complex, with the formula [Cu(NH₃)₄]SO₄. The complex ion, [CU(NH₃)]²⁺ does not ionise to give Cu²⁺ ions. Hence, it does not give the test of Cu²⁺ ion.

Question 3.
Explain with two examples each of the following: Coordination entity, ligand, coordination number, coordination polyhedron, homoleptic and heteroleptic.
Answer:
Coordination entity: A coordination entity constitutes usually a central metal atom or ion, to which a fixed number of other atoms or ions or groups are attached by coordinate bonds. A coordination entity may be neutral, positively or negatively charged. For examples : [Ni(CO)₄], [CoCl₃(NH₃)₃], [Co(NH₃)₆]^3+.

Ligand : A ligand is an ion or a small molecule having at least one lone pair of electrons and capable of forming a coordinate bond with central atom or ion in the coordination entity. For example: Cl^–, OH^–, CN^–, CO, NH₃, H₂O etc.

Coordination number : The coordination number of the central atom or ion is determined by the number of a bonds between the ligands and the central atom or ion. n bonds are not consider for the determination of coordination number. The a bonding electrons may be indicated by a pair of dots (:). For example, [Co(:NH₃)₆]³⁺ and [Fe(:CN)₆]^3-.

Coordination polyhedron : The spatial arrangement of the ligands which are directly attached to the central atom or ion called coordination polyhedron.
For example: [Co(NH₃)₆]³⁺ is octahedral, [Ni(CO)₄] is tetrahedral and [PtCl₄ ]^2- is square planar.

Homoleptic and heteroleptic : Complexes in which a metal is bound to only one type of donor groups are known as homoleptic.
For example : [Co(NH₃)₆]³⁺, [PtCl₆]^2- .
Complexes in which a metal is bound to more than one kind of donor groups are known as heteroleptic. ‘
For example : [Co(NH₃)₄Cl₂]⁺, [PdI₂(ONO)₂ (H₂O)₂],

Question 4.
What is meant by unidentate, didentate and ambidentate ligands? Give two examples for each.
Answer:
A molecule or an ion which has only one donor atom to form one coordinate bond with the central metal atoms is called unidentate ligand, e.g., Cl^– and NH₃.

A molecule or an ion which contains two donor atoms and hence forms two coordinate bonds with the central metal atoms is called a didentate ligand, e.g., NH₂—CH₂—CH₂—NH₂ and ^–OOC — COO^–.
A molecule or an ion which contains two donor atoms but only one of them forms a coordinate bond at a time with the central metal atom is called ambidentate ligand, e.g., CN^–or NC^– and or : ONO^–.

Question 5.
Specify the oxidation numbers of the metals in the following coordination entities:
(i) [Co(H₂O)(CN)(en)₂]²⁺
(ii) [CoBr₂(en)₂]⁺
(iii) [PtCl₂]^2-
(iv) K₃Fe(CN)₆]
(v) [Cr(NH₃)₂Cl₃]
Answer:
(i) x + (-1) + (0) + (0) = + 2 so x = +3 (III)
(ii) x + 2(-1) + 0 = +1 so x = +3 (III)
(iii) x + 4(-1) = -2 so x = +2(11)
(iv) x + 6(-1) = -3 so x = +3 (III)
(v) x + 3(-1) + 0 = 0 so x = +3 (III)

Question 6.
Using IUPAC norms write the formulas for the following:
(i) Tetrahydroxozincate(II)
(ii) Potassium tetrachloridopalladate(II)
(iii) Diamminedichloridoplatinum(II)
(iv) Potassium tetracyanonickelate(II)
(v) Pentaamminenitrito-O-cobalt(III)
(vi) Hexaamminecobalt(III) sulphate
(vii) Potassium tri(oxalato)chromate(III)
(viii) Hexaammineplatinum(IV)
(ix) Tetrabromidocuprate(Il)
(x) Pentaamminenitrito-N-cobalt(lll)
Answer:
(i) [Zn(OH)₄]^2-
(ii) K₂[PdCl₄]
(iii) pt(NH₃)₂Cl₂]
(iv) K₂[Ni(CN)₄]
(v) [Co(ONO) (NH₃)₅]²⁺
(vi) [CO(NH₃)₆]₂ (SO₄)₃
(vii) K₃[Cr(C₂O₄)₃]
(viii) [Pt(NH₃)₆]⁴⁺
(ix) [Cu(Br)₄]^2-
(x) [Co (NO₂) (NH₃)₅]²⁺

Question 7.
Using IUPAC norms write the systematic names of the following:
(i) [CO(NH₃)₆]Cl₃
(ii) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
(iii) [Ti(H₂O)₆]³⁺
(iv) [CO(NH₃)₄Cl(NO₂)]CI
(v) [Mn(H₂O)₆]²⁺
(vi) [NiCl₄]^2-
(vii) [Ni(NH₃)₆]Cl₂
(viii) [Co(en)₃]³⁺
(ix) [Ni(CO)₄]
Answer:
(i) Hexaamminecobalt(III) chloride
(ii) Diamminechlorido(methylamine) platinum(II) chloride
(iii) Hexaquatitanium(III) ion
(iv) Tetraamminechloridonitrito-N-Cobalt(III) chloride
(v) Hexaquamanganese(II) ion
(vi) Tetrachloridonickelate(II) ion
(vii) Hexamminenickel(II) chloride
(viii) Tris(ethane-1, 2-diamine) cobalt(III) ion
(ix) Tetracarbonylnickel(O)

Question 8.
List various types of isomerism possible for coordination compounds giving an example of each.
Answer:
Two principal types of isomerism are known among coordination compounds :
(A) Sterioisomerism,
(B) Structural isomerism.
Each of which can be further sub-divided as :
(A) Stereoisomerism
(i) Geometrical isomerism : It arises in heteroleptic complexes due to different possible geometric arrangements of the ligands.
Example: Pt[(NH₃)₂Cl₂]

(ii) Optical isomerism : It is common in octahedral complexes involving didentate ligands.
Example : [Pt Cl₂(en) ₂]²⁺

Optical isomers (d and l) of cis-[PtCl₂(en)₂]²⁺

(B) Structural isomerism
(i) Linkage isomerism.
Example: [Co(NH₃)₅ (NO₂)]Cl₂
(ii) Coordination isomerism.
Example: [Co(NH₃)₆] [Cr(CN)₆]
(iii) Ionisation isomerism.
Example: [Co(NH₃)₅SO₄]Br and [CO(NH₃)₅ Br]SO₄
(iv) Solvate isomerism.
Example : [Cr(H₂O)₆] Cl₃ (violet) its solvate isomer
[Cr(H₂O)₅Cl]Cl₂. H₂O (grey-green)

Question 9.
How many geometrical isomers are possible in the following coordination entities? ’
(i) [Cr(C₂O₄)₃]^3-
(ii) [Co(NH₃)₃Cl₃]
Answer:
(i) [Cr(C₂O₄)₃]^3-,
No geometric isomer is possible as it is a bidentate ligand.
(ii) [CO(NH₃)₃Cl₃] .
Two geometrical isomers are possible.

Question 10.
Draw the structures of optical isomers of:
(i) [Cr(C₂O₄)₃]^3-
(ii) [PtCl₂(en)₂]²⁺
(iii) [Cr(NH₃)₂ Cl₂ (en)]⁺
Answer:
(i) [Cr(C₂O₄)₃]^3-

Question 11.
Draw all the isomers (geometrical and optical) of:
(i) [CoCl₂ (en)₂]⁺
(ii) [Co(NH₃)Cl(en)₂]²⁺
(iii) [Co(NH₃)₂Cl₂(en)]⁺
Answer:
(i) [CoCl₂ (en)₂]⁺

Question 12.
Write all the geometrical isomers of [Pt(NH₃)(Br)(Cl)(py)] and how many of these will exhibit optical isomers?
Answer:
Three isomers are possible as follows :

Isomers of this type do not show any optical isomerism. Optical isomerism rarely occurs in square planar or tetrahedral complexes and that too when they contain unsymmetrical chelating ligand.

Question 13.
Aqueous copper sulphate solution (blue in colour) gives :
(i) a green precipitate with aqueous potassium fluoride, and
(ii) a bright green solution with aqueous potassium chloride. Explain these experimental results.
Answer:
Aqueous copper sulphate exists as [Cu(H₂O)₄]SO₄. It is a labile complex. The blue colour of the solution is due to [Cu(H₂O)₄]²⁺ ions,

(i) When KF is added, the weak H₂O ligands are replaced by F^– ligands forming [CuF₄]^2- ions, which is a green precipitate.

(ii) When KCl is added, Cl^– ligands replace the weak H₂O ligands forming [CuCl₄]^2- ion, which has bright green colour.

Question 14.
What is the coordination entity formed when excess of aqueous KCN is added to an aqueous solution of copper sulphate? Why is it that no precipitate of copper sulphide is obtained when H₂S (g) is passed through this solution?
Answer:
K₂[Cu(CN)₄] is formed when excess of aqueous KCN is added to an aqueous solution of CuSO₄.

As CN^–ions are strong ligands the complex is very stable. It is not replaced by S^2- ions when H₂S gas is passed through the solution and thus no precipitate of CuS is obtained.

Question 15.
Discuss the nature of bonding in the following coordination entities on the basis of valence bond theory:
(i) [Fe(CN)₆]^4-
(ii) [FeFe₆]^3-
(iii) [Co(C₂O₄)₃]^3-
(iv) [CoF₆]^3-
Answer:
(i) [Fe(CN)₆]^4-
In the above coordination complex, iron exists in the +2 oxidation state.
Fe = [Ar] 3d⁶ 4s²
Outer configuration of Fe²⁺ = 3d⁶ 4s⁰
Orbitals of Fe²⁺ ion:

As CN^– is a strong field ligand, it causes the pairing of the unpaired 3d electrons.

Since, there are six ligands around the central metal ion, the most feasible hybridisation is d²sp³. d²sp³ hybridised orbitals of Fe²⁺ are :

6 electron pairs from CN ions occupy the six hybrid d²sp³ orbitals.
Then,

Hence, the geometry of the complex is octahedral and the complex is diamagnetic (as there are no unpaired electrons).

(ii) [FeF₆]^3-
In this complex, the oxidation state of Fe is + 3.
Fe³⁺ = 3d⁵ 4s⁰
Orbitals of Fe³⁺ ion:

There are 6F^– ions. Thus, it will undergo d²sp³ or sp³d² hybridisation. As F^– is a weak field ligand, it does not cause the pairing of the electrons in the 3d orbital. Hence, the most feasible hybridisation is sp³d². sp³d² hybridised orbitals of Fe are:

Hence, the geometry of the complex is found to be octahedral.

(iii) [Co(C₂O₄)₃]^3-
Cobalt exists in the + 3 oxidation state in the given complex.
Outer configuration of Co = 3d₇ 4s₂
Co₃₊ = 3d₆4s₀
Orbitals of Co₃₊ ion:

Oxalate is a weak field ligand. Therefore, it cannot cause the pairing of the 3d electrons. As there are 6 ligands, hybridisation has to be either sp³d² or d²sp³ hybridisation. sp³d² hybridisation of Co³⁺.

The 6 electron pairs from the 3 oxalate ions (oxalate anion is a bidentate ligand) occupy these sp³d² orbitals.

Hence, the geometry of the complex is found to be octahedral.

(iv) [CoF₂]^3-
Cobalt exists in the + 3 oxidation state.
Orbitals of Co³⁺ ion:

Again, fluoride ion is a weak field ligand. It cannot cause the pairing of the 3d electrons. As a result, the Co³⁺ ion will undergo sp³d² hybridisation.
sp³d² hybridised orbitals of Co³⁺ ion are :

Hence, the geometry of complex is octahedral, 6 electron pants.

Question 16.
Draw figure to show the splitting of d-orbitals in an octahedral crystal field.
Answer:

Question 17.
What is spectrochemical series? Explain the difference between a weak field ligand and a strong field ligand.
Answer:
The arrangement of ligands in order of their increasing field strengths, i.e., increasing crystal field splitting energy (CFSE) values is called spectrochemical series.

The ligands with a small value of CFSE (△₀) are called weak field ligands whereas those with a large value of CFSE are called strong field ligands.

Question 18.
What is crystal field splitting energy? How does the magnitude of △₀ decide the actual configuration of d-orbitals in a coordi-nation entity?
Answer:
When ligands approach a transition metal ion, the d-orbitals split into two sets, one with lower energy and the other with higher energy. The difference of energy between the two sets of orbitals is called crystal field splitting energy (△₀) in case of octahedral field.

If △₀ < P, (pairing energy), the 4th electron enters one of the e_g orbitals giving the configuration $t_{2 g}^{3} e_{g}^{1}$, thereby forming high spin complexes.

Such ligands for which A 0 < P are called weak field ligands.
If △₀ > P, the 4th electron pairs up in one of the t2g orbitals giving the configuration $t_{2 g}^{4} e_{g}^{0}$, thus forming low spin complexes. Such ligands for which △₀ > P are called strong field ligands.

Question 19.
[Cr(NH₃)₆]³⁺ is paramagnetic while [Ni(CN)₄]^2- is diamagnetic. Explain why?
Answer:
Cr is in the +3 oxidation state i.e., d³ configuration. Also, NH₃ is a weak field ligand that does not cause the pairing of the electrons in the orbital.

Therefore, it undergoes d²sp³ hybridisation and the electrons in the 3d orbitals remain unpaired. Hence, it is paramagnetic in nature.
In [Ni(CN)₄]^2-, Ni exists in the + 2 oxidation state i. e., d⁸ configuration.

CN^– is a strong field ligand. It causes the pairing of the 3d electrons. Then, Ni²⁺ undergoes dsp² hybridisation.

As there are no unpaired electrons, it is diamagnetic.

Question 20.
A solution of [Ni(H₂O)₆]²⁺ is green but a solution of [Ni(CN)₄]^2- is colourless. Explain.
Answer:
In [Ni(H₂O)6]²⁺, $\mathrm{H}_{2} \ddot{\mathrm{O}}$ is a weak field ligand. Therefore, there are unpaired electrons in Ni²⁺. In this complex, the d electrons from the lower energy level can be excited to the higher energy level i. e., the possibility of d-d transition is present. Hence, [Ni(H₂O)₆]²⁺ is coloured.

In [Ni(CN)₄]²⁺, the electrons are all paired as CN^– is a strong field ligand. Therefore, d-d transition is not possible in [Ni(CN)₄]^2-. Hence, it is colourless.

Question 21.
[Fe(CN)₆]^4- and [Fe(H₂O)₆]²⁺ are of different colours in dilute solutions. Why?
Answer:
In both the complex compounds, Fe is in +2 oxidation state with configuration 3d⁶, i.e., it has four unpaired electrons. In the presence of weak H₂O ligands, the unpaired electrons do not pair up. But in the presence of strong ligand CN^– they get paired up. Then no unpaired electron is left. Due to this, difference in the number of unpaired electrons, both complex ions have different colours.

Question 22.
Discuss the nature of bonding in metal carbonyls.
Answer:
The metal carbon in metal carbonyls possesses both CT and π character. The ligand to metal is CT bond and metal to ligand is π bond. The effect of CT bond strengthens the rcbond and vice-versa. This is called synergic effect. This unique synergic provides stability to metal carbonyls.

Question 23.
Give the oxidation state, d-orbital occupation and coordination number of the central metal ion in the following complexes:
(i) K₃[CO(C₂O₄)₃]
(ii) cis-[Cr(en)₂Cl₂]Cl
(iii) (NH₄)₂[CoF₄]
(iv) [Mn(H₂O)₆]S0₄
Solution:
(i) K₃[CO(C₂O₄)₃]
The central metal ion is Co.
The oxidation state can be given as :
(+1) × 3 + × + (- 2) × 3 = 0
x – 6 = -3 ⇒ x = + 3
The d orbital occupation for Co³⁺ is $t_{2 g}^{6} e g^{0}$.
(as $\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$ is strong field ligand)
Coordination number of Co = 3 × denticity of C₂O₄
= 3 × 2 (as $\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$ is a bidentate ligand) = 6

(ii) cis-[Cr(en)₂Cl₂]Cl
The central metal ion is Cr.
The oxidation state can be given as:
x + 2(0) + 2(-1) + (-1) = 0
x – 2 – 1 = 0
x = + 3
The d orbital occupation for Cr³⁺ is $t_{2 g}^{3}$.
Coordination number of Cr
= 2 × denticity of en + 2
= 2 × 2 + 2 = 6

(iii) (NH₄)₂[CoF₄]
The central metal ion is Co.
The oxidation state can be given as:
(+1) × 2 + × + (-1) × 4 = 0
x – 4 = -2
x = + 2
The d orbital occupation for Co²⁺ is d⁷ or $t_{2 g}^{5} e_{g}^{2}$. (as F^– is a weak ligand)
Coordination number of Co = 4

(iv) [Mn(H₂O)₆]S0₄
The central metal ion is Mn.
The oxidation state can be given as:
x + (0) × 6 + (- 2) = 0
x = + 2
The d orbital occupation for Mn is d⁵ or [latext_{2 g}^{3} e_{g}^{2}][/latex].
Coordination number of Mn = 6

Question 24.
Write down the IUPAC name for each of the following complexes and indicate the oxidation state, electronic configuration and coordination number. Also give stereochemistry and magnetic moment of the complex:
(i) K[Cr(H₂O)₂(C₂O₄)₂] 3H₂O
(ii) [Co(NH₃)₅Cl]Cl₂
(iii) CrCl₃(py)₃
(iv) Cs[FeCl₄]
(v) K₄[Mn(CN)₆]
Answer:
(i) K[Cr(H₂O)₂ (C₂O₄)2 ] ∙ 3H₂O
IUPAC name : Potassium diaquadioxalatochromate (III) hydrate.
Oxidation state of chromium
+1 + x + (0) × 2 + (- 2) × 2 + 3(0) = 0
+ 1 + x – 4 = 0
x = + 3
Electronic configuration of Cr⁺³= 3d³ = ($t_{2 g}^{3} e_{g}^{0}$)
Coordination number = 6
Shape : Octahedral


Magnetic moment (μ) = $\sqrt{n(n+2)}$
= $\sqrt{3(3+2)}$
= $\sqrt{15}$ = 3.87 BM

(ii) [Co(NH₃)₅Cl]Cl₂
IUPAC name : Pentaammine chlorido cobalt(III) chloride
Oxidation state of Co
x + (0)5 + (-1) + (-1) × 2 = 0
x – 3 =0
x = + 3
Coordination number = 6
Shape: Octahedral.
Electronic configuration of Co³⁺ = 3d⁶ = $t_{2 g}^{6} e_{g}^{0}$
The complex does not exhibit geometrical as well as optical isomerism.
Magnetic Moment (μ) = $\sqrt{n(n+2)}$BM = $\sqrt{0(0+2)}$ BM = 0 BM

(iii) CrCl₃(py)₃
IUPAC name : Trichlorido tripyridine chromium (III) Oxidation state of Cr
x + (-1) × 3 + (0)3 = 0
x = + 3
Electronic configuration of Cr = 3d³ = ($t_{2 g}^{3} e_{g}^{0}$)
Coordination number = 6
Shape : Octahedral
Stereochemistry

Both isomers are optically active. Therefore, a total of 4 isomers exist.
Magnetic moment (μ) = $\sqrt{n(n+2)}$ = $\sqrt{3(3+2)}$
= $\sqrt{15}$ = 3.87 BM

(iv) Cs[FeCl₄]
IUPAC name : Caesium tetrachlorido ferrate (III)
Oxidation state of Fe
+ 1 + x + (-1) × 4= 0
x – 3 = 0
x = + 3
Electronic configuration of Fe = 3d⁵($t_{2 g}^{3} e_{g}^{2}$)
Coordination number = 4
Shape : Tetrahedral
The complex does not exhibit geometrical or optical isomerism, (stereo isomerism).
Magnetic moment (μ) = $\sqrt{n(n+2)}$
= $\sqrt{5(5+2)}$
= $\sqrt{35}$ = 5.92 BM

(v) K₄[Mn(CN)₆]
IUPAC name : Potassium hexacyanomanganate(II)
Oxidation state of Mn
(+1) × 4 + x + (-1) × 6 = 0
x – 2 = 0
x = + 2
Electronic configuration of Mn = 3d⁵ ($t_{2 g}^{5} e_{g}^{0}$)
Coordination number = 6
Shape : Octahedral.
The complex does not exhibit stereo isomerism.
Magnetic moment (μ) = $\sqrt{n(n+2)}$
= $\sqrt{1(1+2)}$
= $\sqrt{3}$
= 1.732 BM

Question 25.
What is meant by stability of a coordination compound in solution? State the factors which govern the stability of complexes.
Answer:
The stability of a coordination compound in solution refers to the degree of association between the two species involved in the state of equilibrium. The stability of the coordination compound is measured in term of magnitude of stability or formation of equilibrium constant.
M + 4L → ML₄
K = $\frac{\left[\mathrm{ML}_{4}\right]}{[\mathrm{M}][\mathrm{L}]^{4}}$
Larger the stability constant, the higher is the proportion of ML₄ that exists in solution.

Factors on which stability of the complex depends are as follows :

1. Charge on the central metal ion : Greater the charge on the central metal ion, greater is the stability of the complex.
2. Nature of the metal ion : Groups 3 to 6 and inner transition element form stable complexes when donor atoms of the ligands are N, O and F. The element after group 6 of the transition metals which have d-orbitals (e.g., Rh, Pd, Ag, Au, Hg, etc.) form stable complexes when the donor atoms of the ligands are heavier members of N, O and F family.
3. Basic nature of the ligand : Greater the basic strength of the ligand, greater is the stability of the complex.
4. Chelate effect: Presence of chelate rings in the complex increases its stability. It is called chelate effect. It is maximum for the 5- and 6- membered rings.
5. Effect of multidentate cyclic ligands : If the ligands happen to be multidentate and cyclic without any steric effect, the stability of the complex is further increased.

Question 26.
What is meant by chelate effect? Give an example.
Answer:
When a didentate or a polydentate ligand contains donor atoms positioned in such a way that when they coordinate with the central metal ion, a five or a six-membered ring is formed, the effect is called chelate effect. Example, [PtCl₂(en)].

Question 27.
Discuss briefly giving an example in each case the role of coordination compounds in:
(i) biological systems
(ii) medicinal chemistry
(iii) analytical chemistry
(iv) extraction/metallurgy of metals
Answer:
(i) Role of coordination compounds in biological systems :

• Haemoglobin, the oxygen carrier in blood, is a complex of Fe²⁺ with porphyrin.
• The pigment chlorophyll in plants, responsible for photosynthesis, is a complex of Mg²⁺ with porphyrin.
• Vitamin B₁₂ (cyanocobalamine) the antipemicious anaemia factor, is a complex of cobalt.

(ii) Role of coordination compounds in medicinal chemistry :

• The platinum complex cis-[Pt(NH₃)₂Cl₂] (cis-platin) is used in the treatment of cancer.
• EDTA complex of calcium is used in the treatment of lead poisoning. Ca-EDTA is a weak complex; when it is administered, calcium in the complex is replaced by the lead present in the body and is eliminated in the urine.
• The excess of copper and iron present in animal system are removed by the chelating ligands D-penicillamine and desferroxime B via the formation of complexes.

(iii) Role of coordination compounds in analytical chemistry :
Complex formation is frequently encountered in qualitative and quantitative chemical analysis.
(a) Qualitative analysis
I. Detection of Cu²⁺ is based on the formation of a blue tetraammine copper (II) ion.

II. Ni²⁺ is detected by the formation of a red complex with dimethyl glyoxime (DMG).

III. The separation of Ag⁺ and Hg²⁺ in group I is based on the fact that while AgCl dissolves in NH₃, forming a soluble complex, Hg₂Cl₂ forms an insoluble black substance.

(b) Quantitative analysis : Gravimetric estimation of Ni²⁺ is carried out by precipitating Ni²⁺ as red nickel dimethyl glyoxime complex in the presence of ammonia.

EDTA is used in the complexometric determination of several metal ions such as Ca²⁺, Zn²⁺, Fe²⁺, Co²⁺, Ni²⁺ etc.

(iv) Role of coordination compounds in extraction/metallurgy of metals : Extraction of various metals from their ore involves complex formation. For example, silver and gold are extracted from their ore by forming cyanide complex.

Purification of some metals can be achieved through complex formation. For example in Mond process, impure nickel is converted into [Ni(CO)₄] which is decomposed to yield pure nickel.

Question 28.
How many ions are produced from the complex Co(NH₃)₆ Cl₂ in solution?
(i) 6
(ii) 4
(iii) 3
(iv) 2
Answer:
The correct option is (iii)
Coordination number of cobalt = 6. It ionises in the solution as

Hence, 3 ion are produced.

Question 29.
Amongst the following ions, which one has the highest magnetic moment value?
(i)[Cr(H₂O)₆]³⁺
(ii)[Fe(H₂O)₆]²⁺
(iii) [Zn(H₂O)₆]²⁺
Answer:
The oxidation state are: Cr (III), Fe (II) and Zn (II).
Electronic configuration of Cr³⁺ = 3d³, unpaired electrons = 3
Electronic configuration of Fe²⁺ = 3d⁶, unpaired electrons = 4
Electronic configuration of Z²⁺ = 3d¹⁰, unpaired electrons = 0
As μ = $\sqrt{n(n+2)}$, therefore, (ii) has the highest magnetic moment.

Question 30.
The oxidation number of cobalt in K[Co(CO)₄] is
(i) +1
(ii) +3
(iii) -1
(iv) -3
Solution:
Oxidation number of Co : K[Co(CO)₄]
x+ (4 × 0) = -1; x = -1
Thus, correct answer is (iii).

Question 31.
Amongst the following, the most stable complex is
(i) [Fe(H₂O)₆]³⁺
(ii) [Fe(NH₃)₆]³⁺
(iii) [Fe(C₂O₄)₃]^3-
(iv) [FeCl₆]^3-
Answer:
In all these complexes, Fe is in +3 oxidation state. However, the complex (iii) is a chelate because three $\mathrm{C}_{2} \mathrm{O}_{4}^{2-}$ ions acts as the chelating ligands. Thus, the most stable complex is [Fe(C₂O₄)₃]^3-. Thus, correct answer is (iii).

Question 32.
What will be the correct order for the wavelengths of absorption in the visible region of the following:
[Ni(NO₂)₆]^4-, [Ni(NH₃)₆]²⁺, [Ni(H₂O)₆]²⁺
Answer:
As metal ion is fixed, the increasing CFSE values of the ligands from the spectrochemical series are in the order :
H₂O < NH₃ < $\mathrm{NO}_{2}^{-}$
Hence, the energies absorbed for excitation will be in the order :
[Ni(H₂O)₆]²⁺ < [Ni(NH₃)₆]²⁺ < [Ni(NO₂)₆]^4-
As E = $\frac{h c}{\lambda}$, therefore, the wavelengths absorbed will be in the opposite order,
[Ni(NO₂)₆]^4- < [Ni(NH₃)₆]²⁺ < [Ni(H₂O)₆]²⁺

Chemistry Guide for Class 12 PSEB Coordination Compounds Textbook Questions and Answers

Question 1.
Write the formulas for the following coordination compounds :
(i) Tetraamminediaquacobalt (III) chloride
(ii) Potassiumtetracyanidonickelate(II)
(iii) Tris(ethane-l,2-diammine)chromium(III) chloride
(iv) Amminebromidochloridonitrito-N-platinate(II)
(v) Dichloridobis(ethane-l,2-diammine) platinum (IV) nitrate
(vi) Iron(III)hexacyanidoferrate(II).
Answer:
(i) [Co(NH₃)₄(H₂O)₂]Cl₃
(ii) K₂[Ni(CN)₄
(iii) (Cr(en)₃]Cl₃
(iv) [Pt(NH₃)BrCl(NO₂)]^–
(v) [PtCl₂(en)₂] (NO₃)₂
(vi) Fe₄[Fe(CN)₆]₃

Question 2.
Write the IUPAC names of the following coordination compounds:
(i) [CO(NH₃)₆]Cl₃
(ii) [CO(NH₃)₅Cl]Cl₂
(iii) K₃[Fe(CN)₆]
(iv) K₃[Fe(C₂O₄)₃]
(v) K₂[PdCl₄]
(vi) [Pt(NH₃)₂Cl(NH₂CH₃)]Cl
Answer:
(i) Hexaamminecobalt(III)chloride
(ii) Pentaamminechloridocobalt(III)chloride
(iii) Potassiumhexacyanoferrate(III)
(iv) Potassiumtrioxalatoferrate (III)
(v) Potassiumtetrachloridopalladate (II)
(vi) Diamminechloridomethylamine platinum(II) chloride.

Question 3.
Indicate the types of isomerism exhibited by the following complexes and draw the structures of these isomers :
(i) K[Cr(H₂O)₂](C₂O₄)₂]
(ii) [Co(en)₃]Cl₃
(iii) [CO(NH₃)₅(NO₂)](NO₃)₂
(iv) [Pt(NH₃)(H₂O)Cl₂]
Answer:
(i) (a) Both geometrical isomer (cis and traits):

(b) Cis-isomer of this compound can exist as pair of optical is :

(ii) Complex will exist as optical isomers:

The compound will show ionisation as well as linkage isomerism.

(iii) Ionisation isomer :
[Co(NH₃)₅(NO₂)](NO₃)₂,
[Co(NH₃)₅ (NO)₃] (NO₂) (NO₃)
Linkage isomers :
[Co(NH₃)₅ (NO₂)](NO₃)₂;
[CO(NH₃)₅ (ONO)](NO₃)₂

(iv) Geometrical isomerism (cis and trans) :

Question 4.
Give evidence that [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅SO₄]Cl are ionisation isomers.
Answer:
When they are dissolved in water, they give different ions in the solution which can be tested by adding AgNO₃ solution and BaCl₂ solution. If Cl^– dons are the counter ions, a white precipitate will be obtained with AgNO₃ solution. If $\mathrm{SO}_{4}^{2-}$ ions are the counter ions, a white precipitate will be obtained with BaCl₂ solution.

Question 5.
Explain on the basis of valence bond theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and the [Ni(Cl)₂₄]^2- ion with tetrahedral geometry is paramagnetic.
Answer:
Nickel in [Ni(CN)₄]^2- is in the +2 oxidation state. The formation of [[Ni(CN)₄]^2- may be explained through hybridisation as follows :
Ni atom in the ground state

Since no unpaired electrons is present, the square planar complex is diamagnetic. In [Ni(CN)₄]^2-, Cl^– is a weak field ligand. It is, therefore, unable to pair up the unpaired electrons of the 3d orbital. Hence, the hybridisation involved is sp3 and the shape is tetrahedral. Since all the electrons are unpaired, it is paramagnetic

Question 6.
[Ni(CN)₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic though both are tetrahedral. Why?
Answer:
In [Ni(CO)₄] Ni is in zero oxidation state whereas in [NiCl₄]^2-, it is in
+ 2 oxidation state. In the presence of strong ligand, CO ligand, the unpaired d electrons of Ni pair up but Cl^– being a weak ligand is unable to pair up the unpaired electrons.

Question 7.
[Fe(H₂O)₆]³⁺ is strongly paramagnetic whereas [Fe(CN)₆]^3- is weakly paramagnetic. Explain.
Answer:
In presence of CN^– (a strong ligand), the 3d⁵ electrons pair up leaving only one unpaired electron. The hybridisation is d²sp³ forming an inner orbital complex. In the presence of H₂O (a weak ligand), 3d electrons do not pair up. The hybridisation is sp³d² forming an outer orbital complex containing five unpaired electrons. Hence, it is strongly paramagnetic.

Question 8.
Explain [Co(NH₃)₆]³⁺ is an inner orbital complex whereas [Ni(NH₃)₆]²⁺ is an outer orbital complex.
Answer:
In [CO(NH₃)₆]³⁺, CO is in +3 oxidation state and has d⁶ electrons. In the presence of NH₃, the 3d electrons pair up leaving two d-orbitals empty to be involved in d²sp³ hybridisation forming inner orbital complex. In [Ni(NH₃)₆]²⁺, Ni is in +2 oxidation state and has d⁸ configuration. The hybridisation involved is sp³d², forming the outer orbital complex.

Question 9.
Predict the number of unpaired electrons in the square planar [Pt(CN)₄]^2- ion.
Answer:
₇₈Pt lies in group 10 with the configuration 5d⁹6s¹. Thus Pt²⁺ has the configuration :

For square planar shape, the hybridisation is dsp². Hence, the unpaired electrons in 5d orbital pair up to make one d orbital empty for dsp² hybridisation.
Thus there is no unpaired electron.

Question 10.
The hexaquomanganese(II) ion contains five impaired electrons, while the hexacyano ion contains only one unpaired electron. Explain using crystal field theory.
Answer:
Mn in the + 2 oxidation state has the configuration 3d⁵. In the presence of H₂O a weak ligand, the distribution of these five electrons is $t_{2 g}^{3} e_{g}^{2}$
i.e., all the electrons remain unpaired

However, in the presence of CN^– the distribution of these electrons is , i.e., two t_2g orbitals contain paired electrons while the third t_2g orbital contains one unpaired electron.

Question 11.
Calculate the overall complex dissociation equilibrium constant for the $\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}$ ion, given that β₄ for this complex is 2. 1 × 10¹³.
Solution:
The overall complex dissociation equilibrium constant is the reciprocal of the overall stability constant, β₄.
∴ $\frac{1}{\beta_{4}}$ = $\frac{1}{2.1 \times 10^{13}}$
∴ = 4.7 × 10^-14

Punjab State Board PSEB 12th Class English Book Solutions English Grammar Translation Exercise Questions and Answers, Notes.

PSEB 12th Class English Grammar Translation

Some Common Proverbs

Exercise 1 (Solved)

Translate the following sentences into English:

Answer:
1. Every dog has his day.
2. He is completely a yes-man.
3. He has his own axe to grind.
4. No new taxes have been imposed upon us.
5. People (or subjects) have revolted against the king.

Exercise 2 (Solved)

Translate the following sentences into English:

Answer:
1. His poverty is an open secret.
2. He is at the root of this mischief.
3. My prosperity. is an eye-sore to him.
4. Ram and Sham are sworn enemies.
5. Where there is a will, there is a way.

Exercise 1 (For Practice)

Translate the following sentences into Hindi/Punjabi:
1. He works hard from dawn to dusk.
2. Most of our politicians are very corrupt.
3. Selfishness is the order of the day.
4. You must reduce your flab¹.
5. Spoken English is gaining importance these days.
6. Spoken English does not mean repeating in the parrot-fashion some crammed sentences.
7. Those who speak English fluently get easily noticed by others.
8. He is the only bread-earner² of the family.
9. Shaleenta is a very good Kathak dancer.
10. He is working as an executive³ in a bank.
11. My colleague in my college is a very good singer.
12. I feel pleased to meet you.
13. What is the nature of your job?
Hints:
1. ਸੀਟਾਧਾ, ਮੋਟਾਪਾ 2. ਹੇਜ਼ੀ ਸਾਰੇ ਭਾਗ, ਰੋਜ਼ੀ ਕਮਾਉਣ ਵਾਲਾ, 3. ਜਿਵੇਕ, ਨਿਦੇਸ਼ਕ ।

Note : Some of the sentences are very easy. It should not be difficult for many of you to translate them in simple Hindi and Punjabi. Hereafter we shall give a few more exercises with meanings in Hindi and Punjabi for your convenience.

Exercise 2 (For Practice)

Translate the following sentences into Hindi/Punjabi:
1. Gandhiji is known for his successful efforts¹ to liberate² India.
2. The lady is very mean in money matters³.
3. She did not allow her husband to spend much on himself.
4. This story tells us about the greed of well-to-do⁴ people.
5. Guru Gobind Singh was a great son of India who led India to sublime⁵ heights of glory⁶.
6. Why is this enormous⁷ increase in population taking place ?
7. Drug-addiction⁸ is a major social evil of modern times.
8. Nations become great by self-confidence⁹.
9. Mankind has undoubtedly progressed since medieval¹⁰ times.
10. Vidyasagar was a very generous¹¹ and charitable¹² man.
Hints:

Exercise 3 (For Practice)

Translate the following sentences into Hindi/Punjabi:
1. Everyone knows the events¹ of Buddha’s life.
2. Newton lived to a ripe age² and earned a great renown³.
3. Books are a wonderful blessing⁴.
4. Ishwara Chandra realised that he must go to England in oder to complete his scientific education.
5. Guru Teg Bahadur was the youngest son of Guru Hargobind Sahib.
6. Science is an angel⁵ in peace but a devil⁶ in war.
7. Wars of today are more dreadful⁷ than those of yesterday.
8. He stumbled⁸ at every step.
9. Silence prevailed⁹ all round the hospital.
10. It has been raining continuously¹⁰ for three hours.
11. It has been drizzling¹¹ for an hour.
Hints:

Exercise 4 (For Practice)

Translate the following sentences into Hindi/Punjabi:
1. This story is a typical example¹ of the storywriter’s work as a story.
2. He failed to find a man before whom he could express the feelings of his heart².
3. He wants to share his grief³ with somebody.
4. This story points out⁴ the truth that poverty dehumanises⁵ a man.
5. I am bereft⁶ of money at this time.
6. I want to know your opinion about etiquette⁷.
7. The teacher punished the students for being noisy in the class.
8. Some shopkeepers fleeces⁸ their customers.
9. A spendthrift lacks balance and perspective⁹.
10. He was reluctant¹⁰ to take the examination.
Hints:

Exercise 5 (For Practice)

Translate the following sentences into Hindi/Punjabi :
1. I have lost the novel¹ that you gave me.
2. She dressed in an elegant² manner.
3. We eat so that we may live.
4. God’s will be done³.
5. Prevention is better than cure⁴.
6. A bird in the hand is worth two in the bush⁵.
7. He is on the wrong side of sixty⁶.
8. We should not look down upon⁷ the poor.
9. I am at a loss to know⁸ what to do.
Hints:

Exercise 6 (For Practice)

Translate the following sentences into Hindi/Punjabi:
1. Trees give shade for the benefit of others.
2. Machines have become the slaves¹ of modern man.
3. A child should be trained to love the reading of books.
4. Politicians worship the rising sun².
5. Our body is like a machine.
6. Newton was a very kind-hearted³ man.
7. He never lost his temper⁴.
8. One should never read anybody’s letter without permission.
9. Children explode crackers on Diwali.
10. The life of modern cities is tiresome⁵.
Hints: