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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.