Prasanna

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 4 Molecular Basis of Inheritance Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 4 Molecular Basis of Inheritance

Multiple Choice Questions

Question 1.
How many of the following characteristics are shown by the R-strain of Streptococcus pneumonia? Avirulent, Smooth, Pathogenic, Capsulated ………………..
(a) One
(b) TWo
(c) Three
(d) Four
Answer:
(a) One

Question 2.
Griffith obtained …………….. from the blood of the dead mice.
(a) dead S-strain bacteria
(b) live R-strain bacteria
(c) dead R-strain bacteria
(d) live S-strain bacteria
Answer:
(d) live S-strain bacteria

Question 3.
Oswald T. Avery, Colin M. MacLeod and Maclyn McCarty demonstrated that ………………..
(a) transformation of live S-strain bacteria into R-strain type was because of DNA of bacteria of S-strain.
(b) the transforming substance was either a protein or RNA.
(c) only DNA was able to transform harmless R-strain into virulent S-strain.
(d) when DNA isolated from S-strain bacteria, was digested with DNase, the transformation occurred.
Answer:
(c) only DNA was able to transform harmless R-strain into virulent S-strain

Question 4.
Which of the following was NOT observed in Hershey and Chase experiment?
(a) Viruses grown in the presence of radioactive sulphur, had radioactive protein but not radioactive DNA.
(b) Radioactive ‘P’ remained in suspension.
(c) Only radioactive ‘P’ was found inside the bacterial cells in the pellet.
(d) Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive proteins.
Answer:
(b) Radioactive ‘P’ remained in suspension.

Question 5.
Enzymes like ……………….. and DNA topoisomerase-I, play important role in maintaining super-coiled state in prokaryotic DNA.
(a) DNA ligase
(b) DNA gyrase
(c) RNA polymerase
(d) None of these
Answer:
(b) DNA gyrase

Question 6.
Histone octamer of nucleosome has two molecules, each of ……………….. proteins.
(a) H₂A, H₂B, H₃ and H₄
(b) H₂A, H₂B, H₃ and H₁
(c) H₂A, H₂B. H₃A and H₃B
(d) H₁A, H₂B, H₃A and H₄
Answer:
(a) H₂A, H₂B, H₃ and H₄

Question 7.
Select the CORRECT statement.
(a) Euchromatin is mainly located near centromere and telomeres.
(b) Heterochromatin replicates at faster rate than euchromatin.
(c) Heterochromatin has 2 to 3 times more DNA than in the euchromatin.
(d) Heterochromatin is lightly stained region of chromonema.
Answer:
(c) Heterochromatin has 2 to 3 times more DNA than in the euchromatin

Question 8.
A DNA molecule in which both strands have 14N is allowed to replicate in an environment containing ¹⁵N. What will be the exact number of DNA molecules that contain the ¹⁴N after three replications?
(a) One
(b) Two
(c) Four
(d) Eight
Answer:
(b) Two

Question 9.
As the base sequence present on one strand of DNA decides the base sequence of other 8 strand, this strand is considered as ………………..
(a) descending strand
(b) leading strand
(c) lagging strand
(d) complementary strand
Answer:
(d) complementary strand

Question 10.
In prokaryotes ……………….. recognizes the promoter sequence.
(a) alpha factor
(b) rho factor
(c) theta factor
(d) sigma factor
Answer:
(d) sigma factor

Question 11.
If the base sequence in DNA is 5′ AAAA 3′, then the base sequence in m-RNA is ………………..
(a) 5′ UUUU 3′
(b) 3′ UUUU 5′
(c) 5′ AAAA 3′
(d) 3′ TTTT 5′
Answer:
(c) 5′ AAAA 3′

Question 12.
During capping, methylated guanosine tri¬phosphate is added to 5′ end of ………………..
(a) m-RNA
(b) t-RNA
(c) hnRNA
(d) r-RNA
Answer:
(c) hnRNA

Question 13.
If each codon has two nucleotides, then there will be ……………….. codons, which can encode for only …………….. different types of amino acids.
(a) 16, 16
(b) 16, 20
(c) 20, 16
(d) 64, 64
Answer:
(a) 16, 16

Question 14.
What would happen if in a gene encoding a polypeptide of 50 amino acids, 25th codon (UAU) is mutated to UAA?
(a) A polypeptide of 24 amino acids is formed.
(b) Two polypeptides of 24 and 25 amino acids will be formed.
(c) A polypeptide of 49 amino acids is formed.
(d) A polypeptide of 25 amino acids is formed.
Answer:
(a) A polypeptide of 24 amino acids is formed.

Question 15.
A strand of DNA has following base sequence – 3′ AAAAGTGAATAGTGA 5′. On transcription it produces an m-RNA. Which of the following anticodon of t-RNA recognizes the third codon of this m-RNA?
(a) AAA
(b) CUG
(c) AAG
(d) CTG
Answer:
(c) AAG

Question 16.
Polynucleotide chain consisting of only CUA repeats will give polypeptide chain with only one amino acid ………………..
(a) tryptophan
(b) leucine
(c) serine
(d) methionine
Answer:
(b) leucine

Question 17.
Select the INCORRECT statement.
(a) Dr. Khorana prepared polyribo-nucleotides chains with known repeated sequences of two or three nucleotides by using synthetic DNA.
(b) M. Nirenberg and Matthaei synthesized artificial m-RNA which was a homopolymer of uracil ribonucleotides.
(c) Enzyme polynucleotide phosphorylase polymerizes RNA with defined sequences in a template-dependent manner.
(d) Evidence for triplet nature of geneticcode, was given by Crick (1961) using “frame-shift mutation”.
Answer:
(c) Enzyme polynucleotide phosphorylase polymerizes RNA with defined sequences in a template-dependent manner.

Question 18.
…………… is/are based on complementarity principle.
(a) Replication and translation
(b) Replication and transcription
(c) Translation
(d) Only replication
Answer:
(b) Replication and transcription

Question 19.
Cysteine has codons, while isoleucin has ……………….. codons.
(a) two, three
(b) three, two
(c) two, four
(d) four, two
Answer:
(a) two, three

Question 20.
Out of 64 codons, only 61 code for the 20 different amino acids. This is known as ……………….. of genetic code.
(a) non-ambiguity
(b) overlapping nature
(c) ambiguity
(d) degeneracy
Answer:
(d) degeneracy

Question 21.
Mutation that results in Sickle-cell anaemia is a ………………..
(a) deletion
(b) frame-shiftmutation
(c) point mutation
(d) insertion
Answer:
(c) point mutation

Question 22.
Initiator charged t-RNA occupies the ……………….. of ribosome first.
(a) A-site
(b) P-site
(c) E-site
(d) either A-site or P-site
Answer:
(b) P-site

Question 23.
It takes ……………….. for formation of peptide bond.
(a) 10 seconds
(b) 0.1 second
(c) less than 0.1 second
(d) 60 seconds
Answer:
(c) less than 0.1 second

Question 24.
Anticodon and codon bind by ………………..
(a) glycosidic bond
(b) hydrogen bond
(c) phosphodiester bond
(d) none of these
Answer:
(b) hydrogen bond

Question 25.
The UTRs are present at ………………..
(a) 5′-end, before start codon and at 3′-end, after stop codon of m-RNA
(b) 5′-end, before start codon and at 3′-end, after stop codon of t-RNA
(c) only at 3′-end, after stop codon of m -RNA
(d) only at 5′-end, before start codon of m-RNA
Answer:
(a) 5′-end, before start codon and at 3′-end, after stop codon of m-RNA

Question 26.
The action of structural genes is regulated by …………….. site with the help of a …………….. protein.
(a) operator, inducer
(b) operator, repressor
(c) regulator, repressor
(d) regulator, inducer
Answer:
(b) operator, repressor

Question 27.
Repressor protein is produced by the action of ………………..
(a) gene z
(b) gene y
(c) gene i
(d) gene o
Answer:
(c) gene i

Question 28.
Select the correct pair.
(a) Gene z – Transacetylase
(b) Gene y – Beta-galactocidase
(c) Gene a – Beta-galactoside permease
(d) Gene I – Repressor
Answer:
(d) gene I – repressor

Question 29.
Structural genomics involves ……………….. of genome.
(a) mapping
(b) sequencing
(c) analysis
(d) all of these
Answer:
(d) all of these

Question 30.
The technique of transferring DNA fragments separated on agarose gel to a synthetic nitrocellulose membrane is known as ………………..
(a) Southern blotting
(b) Autoradiography
(c) Southern hybridization
(d) None of these
Answer:
(a) Southern blotting

Question 31.
Sequence of various steps in DNA fingerprinting is ………………..
i. Southern blotting.
ii. Restriction digestion
iii. Agarose gel electrophoresis
iv. DNA isolation
v. Photography.
vi. Selection of DNA probe
vii. Hybridization
(a) iv, iii, ii, i, v, vi, vii
(b) iv. v, iii, i, vi, vii, ii
(c) iv, ii, iii, i, vi, vii, v
(d) ii, iii, iv, i, vi, vii,
Answer:
(c) iv, ii, iii, i, vi. vii, v

Match the Columns

Question 1.

Column A	Column B
(1) Frederick Griffith	(a) Test tube assay
(2) Avery, McCarty and MacLeod	(b) Streptococcus pneumoniae
(3) Alfred Hershey and Martha Chase	(c) E. coli
(4) Meselson and Stahl	(d) Bacteriophages

Answer:

Column A	Column B
(1) Frederick Griffith	(b) Streptococcus pneumoniae
(2) Avery, McCarty and MacLeod	(a) Test tube assay
(3) Alfred Hershey and Martha Chase	(d) Bacteriophages
(4) Meselson and Stahl	(c) E. coli

Classify the following to form Column B as per the category given in Column A

Question 1.
(i) UUU
(ii) CUA
(iii) UAA
(iv) AUG
(v) UAG
(vi) UGA

Column A	Column B
1. Initiator codon	————–
2. Stop codons	————-
3. Codon that codes for Phenyl alanin	————–
4. Codon that codes for leucine	————-

Answer:

Column A	Column B
1. Initiator codon	(iv) AUG
2. Stop codons	(iii) UAA, (v) UAG, (vi) UGA
3. Codon that codes for Phenyl alanin	(i) UUU
4. Codon that codes for leucine	(ii) CUA

Question 2.
(i) Severo Ochoa
(ii) F. Jacob and J. Monod
(iii) Temin and Baltimore
(iv) H. Winkler
(v) T. H. Roderick
(vi) R Kornberg

Column A	Column B
(1) Lac Operon	————–
(2) Central dogma in retroviruses	————-
(3) Coined the term Genome	————–
(4) Coined the term Genomics	————-
(5) Enzymatic synthesis of RNA	————-
(6) DNA is associated with histones and non-histones	————–

Answer:

Column A	Column B
(1) Lac Operon	(ii) E Jacob and J. Monod
(2) Central dogma in retroviruses	(iii) Temin and Baltimore
(3) Coined the term Genome	(iv) H. Winkler
(4) Coined the term Genomics	(v) T. H. Roderick
(5) Enzymatic synthesis of RNA	(i) Severo Ochoa
(6) DNA is associated with histones and non-histones	(vi) R. Kornberg

Very Short Answer Questions

Question 1.
What are the two types of bacteria used by F. Griffith and which one out of these is avirulent?
Answer:
S-type and R-type strains of Streptococcus penumoniae were used by F. Griffith and out of these R-type is avirulent.

Question 2.
Enlist the characteristics of S-strain pneumoniae.
Answer:
S-strain pneumoniae are virulent, smooth and encapsulated.

Question 3.
What is the bacteriophage?
Answer:
Bacteriophage is a virus that infects bacterium and injects its genetic material in the bacterium.

Question 4.
What is the length of DNA double helix molecule in a typical mammalian cell?
Answer:
The length of DNA double helix molecule in a typical mammalian cell is approximately 2.2 meters.

Question 5.
What is the approximate size of a typical nucleus ?
Answer:
Approximate size of a typical nucleus is 10^-6 m.

Question 6.
What is size of E. coli cell?
Answer:
The size of E. coli cell size is 2-3 µm.

Question 7.
What determines the charge on protein molecules?
Answer:
A protein acquires its charge depending upon the abundance of amino acid residues with charged side chains.

Question 8.
What is nucleosome core?
Answer:
Nucleosome core is a histone octamer.

Question 9.
Where is H₁ histone present?
Answer:
H₁ histone binds the DNA thread where it enters and leaves the nucleosome.

Question 10.
How are solenoid fibres formed?
Answer:
Six nucleosomes get coiled and then form solenoid that looks like coiled telephone wire of 30 nm diameter (300Å).

Question 11.
How is chromatin fibre formed?
Answer:
Supercoiling of solenoid fibre forms a looped structure called chromatin fibre.

Question 12.
What is NHC?
Answer:
NHC stands for Nonhistone Chromosomal proteins.

Question 13.
List as many different enzyme activities required during DNA synthesis as you can.
Answer:
Phosphorylase, Helicase, DNA polymerase, Primase, DNA ligase, Super helix relaxing enzyme, Topoisomerase (gyrase) are different enzymes required during DNA synthesis.

Question 14.
How many replicons are present in prokaryotes and eukaryotes respectively?
Answer:
Prokaryotes have one replicon. Several replicons in tandem are present in eukaryotes.

Question 15.
What is the function of SSBP?
Answer:
During replication of DNA SSBP proteins remain attached to both the separated strands and prevent them from coiling back.

Question 16.
During which phases of cell cycle, transcription occurs in the nucleus?
Answer:
Transcription occurs in the nucleus during G₁ and G₂ phases of cell cycle.

Question 17.
Which strand of transcription unit gets transcribed ?
Answer:
DNA strand having 3’ → 5’ polarity acts as template strand and it gets transcribed.

Question 18.
What is a cryptogram?
Answer:
Cryptogram is a genetic code consisting of triplet codons on m-RNA that code for a specific amino acids.

Question 19.
What is meant by the polarity of genetic code?
Answer:
Genetic code is always read in 5′ → 3’ direction. This is called polarity of genetic code.

Question 20.
Which mutation can result in changes in the reading frame?
Answer:
Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion.

Question 21.
Which mutation can result in insertion or deletion of amino acids, but reading frame remains unaltered?
Answer:
Insertion or deletion of three or multiples of three bases results in insertion or deletion of amino acids and reading frame remains unaltered from that point onwards.

Question 22.
Which molecule serves as an intermediate molecules between DNA and protein during proteins synthesis?
Answer:
RNA serves as an intermediate molecule between DNA and protein during proteins synthesis.

Question 23.
What is the function of a groove present between two subunits of ribosome in eukaryotes ?
Answer:
The groove present between two subunits of ribosomes in eukaryotes protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.

Question 24.
Enlist different steps of protein synthesis.
Answer:
Steps in protein synthesis are:

1. Transcription
2. Activation of amino acids and formation of charged t-RNAs,
3. Synthesis of polypeptide chain:
4. initiation
5. elongation and
6. termination of polypeptide chain.

Question 25.
What is translocation?
Answer:
During elongation of polypeptide chain, the ribosome moves along the m-RNA in stepwise manner from start codon to stop codon (5′ → 3′), 1 codon ahead each time t, his movement is called translocation and due to this t-RNA carrying a dipeptide at A-site of the ribosome moves to the p-site.

Question 26.
What is meant by inducible enzymes?
Answer:
Bacteria like E.coli adapt to their chemical environment by synthesizing certain enzymes depending upon the substrate present. Such adaptive enzyme is called inducible enzyme.

Question 27.
What is meant by induction and inducer?
Answer:
A set of genes are switched on when a new substrate is to be metabolized. This phenomenon is called induction and small molecule responsible for this is known as inducer.

Question 28.
What is the role of a repressor gene?
Answer:
The role of a repressor gene is to produce repressor protein. Repressor binds with operator gene and this prevents transcription of structural genes in the operon.

Question 29.
Which molecule does act as inducer molecule in lac operon?
Answer:
Allolactose acts as inducer molecule in lac operon.

Question 30.
In which condition, lac operon is switched off?
Answer:
If E.coli bacteria do not have lactose in the surrounding medium as a source of energy, lac operon is switched off.

Question 31.
What lac operon consists of?
Answer:
Lac operon consists of a regulator promoter, operator and three structural genes z, y and a.

Question 32.
Which gene acts as a regulatory gene in lac operon?
Answer:
Repressor protein is produced by the action of gene i (inhibitor). This gene acts as a regulator gene.

Question 33.
When was Human Genome Project started ? When was it completed ?
Answer:
The Human Genome Project was started in 1990 and was completed in 2003.

Question 34.
What is functional genomics?
Answer:
Functional genomics is a branch of genomics that involves the study of functions of all gene sequences and their expressions in organisms.

Question 35.
What is the advantage of sequencing of genomes of non-human organisms?
Answer:
Sequencing of genomes of non-human model organisms allows researchers to study gene functions in these organisms. Since human beings possess many genes which are like those of flies, roundworms and mice, comparative studies will lead to greater understanding of human evolution.

Question 36.
What are VNTRs?
Answer:
Variable Number of Tandem Repeats (VNTRs) are unusual sequences of 20-100 base pairs, which are repeated several times and are arranged tandency.

Question 37.
Do different organisms have the same DNA?
Answer:
Different organisms differ in their DNA sequence.

Question 38.
What is the amino acid sequence encoded by base sequence UCA, UUU, UCC, GGG, AGU of an m-RNA segment?
Answer:
The amino acid sequence: Ser-Phe – Ser – Gly- Ser

Give definitions of the following

Question 1.
Replicon
Answer:
The unit of DNA in which replication occurs is known as replicon.

Question 2.
Transcription
Answer:
Transcription is defined as the process of copying of genetic information from template strand of DNA into a complementary single stranded RNA transcript.

Question 3.
Gene
Answer:
Gene is defined as the DNA sequence coding for m-RNA/ t-RNA or r-RNA.

Question 4.
Cistron
Answer:
Cistron is defined as a segment of DNA coding for a polypeptide.

Question 5.
Monocistronic gene
Answer:
Gene is called monocistronic, when there is a single structural gene in one transcription unit.

Question 6.
Polycistronic gene
Answer:
Gene is called polycistronic, when there is a set of various structural genes in one transcription unit.

Question 7.
Interrupted or Split genes
Answer:
Interrupted or split genes are the structural genes in eukaryotes which have both exons and introns.

Question 8.
Exons
Answer:
Exons are the coding sequences or express sequences in DNA/hnRNA/ m-RNA.

Question 9.
Introns
Answer:
Introns are the non-coding sequences in DNA or hnRNA.

Question 10.
Anticodon
Answer:
Anticodon is a triplet of nucleotides present on the anticodon loop of t-RNA, which is complementary to codon on m-RNA.

Question 11.
Mutation
Answer:
Mutation is a sudden heritable change in the DNA sequence that results in the change of genotype.

Question 12.
Translation
Answer:
Translation is the process in which sequence of codons of m-RNA is decoded and accordingly amino acids are added in specific sequence to form a polypeptide on ribosomes.

Question 13.
Genomics
Genomics is the study of genomes through analysis, sequencing and mapping of genes along with the study of their functions.

Question 14.
Repressors
Answer:
Repressors are proteins which are able to bind the operator region of operon and prevent the RNA polymerase from transcribing the operon.

Name the following

Question 1.
Enzyme that cleaves DNA.
Answer:
DNase

Question 2.
Enzyme that cleaves proteins.
Answer:
Protease

Question 3.
Enzyme involved in activation of nucleotides.
Answer:
Phosphorylase

Question 4.
Enzyme involved in unwinding of DNA.
Answer:
Helicase

Question 5.
Enzyme involved in synthesis of DNA.
Answer:
DNA polymerase

Question 6.
Enzyme involved in joining of Okazaki fragments.
Answer:
DNA ligase

Question 7.
Enzyme involved in synthesis of RNA primer.
Answer:
Primase

Question 8.
Enzyme involved in removal of RNA primer.
Answer:
DNA polymerase

Question 9.
Enzyme involved in replacement of gaps in prokaryotes.
Answer:
DNA polymerase – I

Question 10.
Enzyme involved in replacement of gaps in eukaryotes.
Answer:
DNA polymerase α

Question 11.
Enzyme involved in formation of double helix in daughter DNA molecules.
Answer:
Topoisomerase

Question 12.
Enzyme involved in releasing strain created by unwinding of DNA.
Answer:
Super helix relaxing enzyme

Question 13.
Enzyme involved in synthesis of hnRNA, m-RNA.
Answer:
RNA polymerase – II

Question 14.
Enzyme involved in synthesis of t-RNA, snRNA.
Answer:
RNA polymerase – III

Question 15.
Enzyme involved in synthesis of r-RNA
Answer:
RNA polymerase – I

Question 16.
Enzyme involved in polymerizing RNA in template independent manner.
Answer:
Polynucleotide phosphorylase

Question 17.
Enzyme involved in peptide bond synthesis.
Answer:
Ribozyme

Question 18.
Enzyme involved in Cutting DNA at specific sites.
Answer:
Restriction endonuclease

Question 19.
Name the initiator codon of protein synthesis.
Answer:
AUG is the initiator codon of protein synthesis.

Question 20.
Name three binding sites of ribosome.
Answer:
Three binding sites for t-RNA on ribosomes are P-site (peptidy t-RNA-site), A-site (aminoacyl – t-RNA-site) and E-site (exit site).

Question 21.
Name the different structural genes in sequence of lac operon.
Answer:
There are 3 structural genes in the sequence lac-Z, lac-Y and lac-A.

Question 22.
Name the organisms whose genomes have been sequenced?
Answer:
The genomes of several organisms such as bacteria e.g. E.coli, Caenorhabditis elegans (a free living non-pathogenic nematode), Saccharomyces cerevisiae (yeast), Drosophila (fruit fly), plants (rice and Arabidopsis), Mus musculus (mouse), etc. have been sequenced.

Give Significance/Functions of the following

Question 1.
DNA.
Answer:

1. DNA regulates and controls all the cellular activities.
2. It replicates and gets distributed equally to the daughter cells when the cell divides.
3. It is a carrier of genetic information.
4. Heterocatalytic function : DNA directs the synthesis of chemical molecules other than itself. E.g. Synthesis of RNA (transcription), synthesis of protein (Translation), etc.
5. Autocatalytic function : DNA directs the synthesis of DNA itself. E.g. Replication.
6. DNA is a master molecule of a cell that initiates, guides, regulates and controls the process of protein synthesis.

Question 2.
Proteins.
Answer:
Proteins serve as structural components, enzymes and hormones.

Distinguish between the following.

Question 1.
Euchromatin and Heterochromatin.
Answer:

Euchromatin	Heterochromatin.
1. Euchromatin is loosely packed region of the chromatain.	1. Heterochromatin is densely packed region of the chromatin.
2. Euchromatin stains lightly.	2. Heterochromatin stains darkly.
3. Euchromatin is transcriptionally active region of the chromatin.	3. Heterochromatin is transcriptionally inactive region of the chromatin.

Question 2.
DNA in prokaryotic and eukaryotic cells.
Answer:

DNA in prokaryotes	DNA in eukaryotic
1. It is present in the cytoplasm.	1. It is present in the nucleus.
2. It is not associated with histones.	2. It is associated with histones.
3. It is circular.	3. It is linear.
4. Genes do not contain introns.	4. Genes contain introns along with exons.
5. Genes are polycostronic.	5. Genes are monocistronic.

Question 3.
DNA and RNA
Answer:

DNA	RNA
1. DNA is deoxyribonucleic acid.	1. RNA is ribonucleic acid.
2. DNA is double stranded, helical molecule.	2. RNA is single stranded molecule.
3. In DNA, there is deoxyribose sugar.	3. In RNA, there is ribose sugar.
4. The pyrimidine nitrogen bases are cytosine and thymine.	4. The pyrimidine nitrogen bases are cytosine and uracil.
5. DNA is the genetic material in all types of organisms.	5. RNA is genetic material in few viruses only.
6. In eukaryotic cells, DNA is present in nucleus.	6. In eukaryotic cells, RNA is present in nucleus as well as cytoplasm.
7. The number of purine : pyrimidine ratio is always 1 : 1 in DNA molecule.	7. The number of purine : pyrimidine ratio may not be 1 : 1 in RNA molecule.
8. DNA sends the codon for the synthesis of proteins, but otherwise it does not participate in the protein synthesis.	8. RNA takes part in the protein synthesis through transcription and translation.

Question 4.
m-RNA, t-RNA and r-RNA.
Answer:

m-RNA	t-RNA	r-RNA.
1. m-RNA is a simple molecule which shows linear structure without any folds.	1. t-RNA is a single stranded molecule. There is regular pattern of folding shown by this molecule.	1. r-RNA is a single stranded RNA that is variously folded upon itself. In the folded regions it shows complementary base pairing.
2. It performs the function of transcription during protein synthesis.	2. It performs the function of transferring the amino acids during translation.	2. This RNA remains associated with the ribosomes permanently. It gives the binding site for m-RNA during the process of protein synthesis. It also orients the m-RNA molecule so as to read the message on codons properly.
3. Of the total cellular RNA, m-RNA forms 3-5%.	3. Of the total cellular RNA, t-RNA forms about 10-20%.	3. Of the total cellular RNA, r-RNA forms 80%
4. It has molecular weight of about 5,00,000.	4. t-RNA is the smallest RNA having only 73-93 nucleotides and has molecular weight of about 23,000 to 30,000 daltons.	4. Molecular weight is about 40,000 to 100,000 daltons.

Give Reasons

Question 1.
Nuclein was called as nucleic acid.
Answer:
Nuclein had acidic properties and it was isolated from nucleus. Hence, it was called as nucleic acid.

Question 2.
Initially proteins (and not DNA) were considered as genetic material.
Answer:

1. Proteins are large, complex molecules and store information required to govern cell metabolism. Hence it was assumed that variations found in species were caused by proteins.
2. On the other hand, DNA was considered as a small, simple molecule whose composition does not vary much among species.
3. Variations in the DNA molecules are different than the variation in shape, electrical charge and function shown by proteins.
4. Hence, initially proteins (and not DNA) were considered as genetic material.

Question 3.
On injecting a mixture of heat-killed S-bacteria and live R bacteria, the mice died.
Answer:

1. Griffith obtained live S-strain bacteria from the blood of the dead mice.
2. In a mixture of live R-bacteria and heat killed S-bacteria, live R-strain bacteria picked up something (transforming principle) from the heat-killed S bacterium and got changed into S-type.
3. Transforming principle allowed R-type bacteria to synthesize capsule and thus they became virulent.
4. Hence, on injecting a mixture of heat-killed S bacteria and live R bacteria, the mice died.

Question 4.
Viruses obtained by infecting bacteria having radioactive phosphorus contained radioactive DNA (labelled DNA), but not radioactive proteins.
Answer:
Viruses obtained by infecting bacteria having radioactive phosphorus, contained radioactive DNA (labelled DNA). but not radioactive proteins because DNA contains phosphorus (labelled DNA) but proteins do not.

Question 5.
Viruses obtained by infecting bacteria having radioactive sulphur contained radioactive protein but not radioactive DNA.
Answer:
Viruses obtained by infecting bacteria having radioactive sulphur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur and proteins contain sulphur.

Question 6.
In bacteria, m-RNA does not require any processing.
Answer:
In bacteria, m-RNA does not require any processing because it has no introns and it is synthesized in cytoplasm.

Question 7.
Eukaryotic DNA is condensed and supercoiled.
Answer:

1. In a typical mammalian cell, length of DNA double helix is approximately 2.2 metres.
2. The size of typical nucleus is approximately 10^-6 m
3. Such a long DNA molecule has to be fitted in small nuclear space.
4. Therefore, DNA is highly condensed, coiled and supercoiled so that it can be accommodated in the nucleus.

Question 8.
During translation, complementarity principle is not applicable.
Answer:
During translation, complementarity principle is not applicable as, genetic information is transferred from a polymer of nucleotides to a polymer of amino acids.

Question 9.
Protein synthesis is the most important and essential activity in the living cells.
Answer:

1. Proteins play a significant role in the metabolism of living cells.
2. The actual phenotypic expression of living cells is dependent on the biochemical reactions.
3. Each biochemical reaction needs a specific enzyme for its initiation and completion. All the enzymes are proteins.
4. In a cell there are many structural proteins too. Thousands of structural and catalytic proteins are constantly required within the cell at all times.
5. Many functional proteins like hormones are also important for metabolism. Thus, for the synthesis of all such proteins, protein synthesis has become the most important and essential activity of the living cell.

Question 10.
Only 20 amino acids are considered as standard.
Answer:

1. It was believed that there are total 20 amino acids in the living world. But 21st amino acid called selenocysteine was discovered later.
2. This amino acid is coded by UGA which is usually a termination codon.
3. In both prokaryotic and eukaryotic cells polypeptide chains contain 100-300 amino acids and they are formed by specific arrangement of 21 amino acids.
4. But formation of selenocysteine requires the availability of element selenium in the cells.
5. Therefore, only 20 amino acids are considered as standard.

Write Short Notes on the following

Question 1.
Friedrich Miescher’s nuclein
Answer:

1. Nuclein is an acidic substance, having high phosphorus content and it was isolated by Friedrich Miescher in 1869, from the nuclei of pus cells.
2. As Nuclein had acidic properties and it was isolated from nucleus, it was called as nucleic acid.
3. E Miescher started working with white blood cells (the major component of pus). He used a salt solution to wash the pus off the bandages. He lysed the cells by adding a weak alkaline solution and isolated nucleic acid from nuclei that precipitated out of the solution.
4. By the early 1900s, it was known that Miescher’s nuclein was a mixture of proteins and nucleic acids (DNA and RNA).

Question 2.
Packaging in Prokaryotes
Answer:

1. Size of cell in E. coli size is 2-3µ.
2. The nucleoid is small, circular, highly folded, naked DNA (1100 µm long in perimeter and contains about 4.6 million base pairs).
3. When the negatively charged DNA becomes circular, the size reduces to 350 µm in diameter.
4. Folding/looping (40-50 domains (loops) further reduce it to 30 µm in diameter.
5. RNA connectors assist in loop formation.
6. Further coiling and super coiling of each domain reduces the size to 2 µm in diameter.
7. This coiling (packaging) is assisted by positively charged HU (Histone like DNA binding proteins) proteins and enzymes like DNA gyrase and DNA topoisomerase-I, which maintain supercoiled state.

Question 3.
Experimental confirmation of semiconservative replication of DNA.
Answer:

(1) Matthew Meselson and Franklin Stahl (1958) used equilibrium – density – gradient – centrifugation technique to experimentally prove semiconservative DNA replication.

(2) They cultured bacteria E.coli in the medium containing ¹⁴N (light nitrogen). They obtained equilibrium density gradient band by using 6M CsCl₂. The position of this band is recorded.

(3) E. coli cells were then transferred to ¹⁵N medium (heavy isotopic nitrogen) and allowed to replicate for several generations. At equilibrium point density gradient band was obtained, by using 6M CsCl₂. The position of this band is recorded.

(4) The heavy DNA (¹⁵N) molecule can be distinguished from normal DNA by centrifugation in a 6M Cesium chloride (CsCl₂) density gradient. At the equilibrium point ¹⁵N DNA will form a band. In this both the strands of DNA are labelled with ¹⁵N.

(5) Such E. coli cells were then transferred to another medium containing ¹⁴N i.e. normal (light) nitrogen. After first generation, the density gradient band for ¹⁴N ¹⁵N was obtained and its position was recorded. After second generation, two density gradient bands were obtained – one at ¹⁴N ¹⁵N position and other at ¹⁴N position.

(6) The position of bands after two generations clearly proved that DNA replication is semiconservative.

Question 4.
Central Dogma of molecular biology.
Answer:
(1) Central dogma of molecular biology was postulated by EH.C. Crick in 1958.

(2) DNA gets transcribe to form m-RNA, m-RNA acts as a messenger and gets translated to form a polypeptide chain (protein) having specific amino acid sequence.

(3) This unidirectional flow of information from DNA to RNA and from RNA to proteins is referred as central dogma of molecular biology.

(4) Temin (1970) and Baltimore (1970) : Central dogma in retroviruses.

Question 5.
Processing of hnRNA.
Answer:

1. Eukaryotes have split genes.
2. Primary transcript or hnRNA is non-functional and contains both exons and introns.
3. Processing of hnRNA results in functional m-RNA.
4. The fully processed hnRNA is called m-RNA.
5. m-RNA comes out of the nucleus through nuclear pore for getting translated.

hnRNA undergoes capping, tailing and splicing.

1. Capping : Methylated guanosine tri¬phosphate is added to 5’ end of hnRNA.
2. Tailing : Polyadenylation take place at 3′ end.
3. Splicing : It is removal of introns.
4. DNA ligase joins exons in a definite sequence (order).

Question 6.
Cracking of genetic code.
Answer:

1. M. Nirenberg and Matthaei synthesized artificial poly-U m-RNA.
2. Using this synthetic poly-U m-RNA and cell free system of protein synthesis, a small polypeptide consisting of only amino acid phenylalanine was obtained.
3. It suggested that UUU codes for phenylalanine.
4. Dr. Har Gobind Khorana devised a technique for synthesis of artificial m-RNA with repeated sequences of known nucleotides.
5. He synthesized artificial RNA consisting of known repeated sequences of two or three nucleotides. E.g. CUC, UCU, CUC, UCU, by using synthetic DNA.
6. This resulted in formation of polypeptide chain consisting of alternate amino acids leucine and serine.
7. Synthetic RNA consisting of CUA, CUA, CUA, CUA repeats gave polypeptide chain with only one amino acid – leucine.
8. Severo Ochoa established that the enzyme (polynucleotide phosphorylase) also helps in polymerizing RNA of defined sequences in a template-independent manner (i.e. enzymatic synthesis of RNA).
9. Thus all the 64 codons in the dictionary of genetic code were deciphered.

Question 7.
Types of mutations.
Answer:

1. Chromosomal mutations : Loss (deletion) or gain (insertion/duplication) of a segment of DNA results in alteration in the chromosome.
2. Point mutations : They involve change in a single base pair of DNA. E.g. Mutation that results in Sickle-cell anaemia.
3. Deletion or insertion of base pairs of DNA : It causes frame-shift mutations or deletion mutation.
4. Insertion or deletion of one or two bases changes the reading frame from the point of insertion or deletion.
5. Insertion or deletion of three or multiples of three bases (insert or delete) results in insertion or deletion of amino acids and reading frame remains unaltered from that point onwards.

Question 8.
Transfer-RNA.
Answer:

(1) As t-RNA can read the codon and also can bind with the amino acid, t-RNA is considered as an adapter molecule.

(2) Clover leaf structure (2 dimensional) of t-RNA:

• t-RNA has four arms – DHU arm (has amino acyl binding loop), middle arm (has anticodon loop), Tif/C arm (has ribosome binding loop) and variable arm.
• It has G nucleotide at 5’ end.
• Amino acid acceptor end (3’ end) having unpaired CCA bases (i.e. amino acid binding site).

(3) For every amino acid, there is specific t-RNA.
(4) Initiator t-RNA is specific for methionine.
(5) There are no t-RNAs for stop codons.
(6) In the actual structure, the t-RNA molecule looks like inverted L (3 dimensional : structure)

Question 9.
UTRs.
Answer:

1. m-RNA has some additional sequences that are not translated. These sequences are referred as untranslated regions (UTR).
2. The UTRs are present at both 5′-end (before start codon) and at 3′-end (after stop codon).
3. They are required for efficient translation process.

Short Answer Questions

Question 1.
Why are Okazaki fragments formed on lagging strand only?
Answer:

1. The lagging template is the template strand with free 5’ end.
2. The replication always starts at C-3 end of template strand and proceeds towards C-5 end.
3. Both the strands of the parental DNA are antiparallel and new strands are always formed in 5′ → 3′ direction, i.e. DNA polymerase synthesizes new strand in only one direction i.e. 5′ → 3′ direction.
4. Hence, the lagging templates becomes available for replication only discontinuously in small patches.
5. The new lagging strand develops discontinuously away from the replicating fork in the form of small Okazaki fragments.
6. Hence, Okazaki fragments formed on lagging strand only.

Question 2.
Why t-RNA is called as adapter molecule?
Answer:
t-RNA can read the codon on m-RNA. It also can bind with the amino acid at 3’ end and transport it to m-RNA-ribosome complex during translation. It can bind with specific codon which is complementary to its anticodon. So t-RNA is considered as an adapter molecule.

Question 3.
Why DNA replication is called semiconservative replication?
Answer:

1. In each of the two daughter DNA molecules thus formed, one strand is parental and the other one is newly synthesized.
2. 50% is contributed by mother DNA.
3. Hence, DNA replication is described as semiconservative replication.

Question 4.
What are the functions of three types of RNAs in bacteria? Which enzyme is involved in transcription of DNA to form RNAs in bacteria? What is its function?
Answer:

1. In bacteria, m-RNA provides the encoded message for protein synthesis; t-RNA brings specific amino acid to the site of translation; r-RNA plays role in providing binding site to m-RNA and t-RNA.
2. There is single DNA dependent-RNA polymerase that catalyses transcription of all 3 types of RNA in bacteria.
3. RNA polymerase binds to promoter and initiates transcription (initiation) and synthesizes RNA.

Question 5.
Explain why it was suggested that codon is a sequence of three consecutive nucleotides on m-RNA.
Answer:

1. If each codon has only one nucleotide, then there will be 4¹ = 4 codons, which can encode for only four different types of amino acids.
2. If each codon has two nucleotides, then there will be 4² = 16 codons, which can encode for only 16 different types of amino acids.
3. If each codon has three nucleotides, then there will be 4³ = 64 codons, which are sufficient to specify 20 different types of amino acids.

Question 6.
Explain Genetic code is Non-overlapping.
Answer:

1. Each single base is a part of only one codon.
2. Adjacent codons do not overlap.
3. If it had been overlapping type, with 6 bases, there would be 4 amino acid molecules in a chain.
4. Experimental evidence favours non-overlapping nature of genetic code.

Question 7.
How degeneracy of the code is explained by Wobble hypothesis?
Answer:

1. In 1966, Crick proposed Wobble hypothesis.
2. According to this hypothesis, in codon- anticodon base pairing, the third base may not be complementary.
3. The third base of codon is called wobble base and this position is called wobble position.
4. This results in economy of t-RNA as anticodon of a t-RNA may bind with codon even when only first two bases are complementary.
5. For example, GUU, GUC, GUA and GUG codons code for amino acid Valine.
6. Degeneracy of genetic code means many codons can code for same amino acid.
7. Thus, the degeneracy of genetic code gets explained by Wobble hypothesis.

Question 8.
a. What is meant by universal genetic code? Give example.
b. Why genetic code is called Non- ambiguous?
Answer:
a. Universal genetic code means that the specific codon codes for same amino acid in all living organisms, e.g. codon AUG always specifies amino acid methionine.

b. Genetic code is called non-ambiguous because, each codon specifies a particular amino acid.

Question 9.
Give examples of termination codons. Why are they known as termination codons?
Answer:

1. UAA, UAG and UGA are known as termination codons.
2. They do not code for any amino acid.
3. They terminate or stop the process of elongation of polypeptide chain.
4. Hence, they are known as termination codons.

Question 10.
How many amino acids are required for protein synthesis? From where are they obtained?
Answer:

1. About 20 different types of amino acids are required for protein synthesis.
2. They are available in the cytoplasm.

Question 11.
How DNA regulates protein synthesis?
Answer:

1. DNA regulates protein synthesis by coding for the specific sequence of amino acids in a protein.
2. This control is possible through transcription of m-RNA.
3. Genetic code is specific for particular amino acid.

Question 12.
What is the role of ribosomes in protein synthesis?
Answer:

1. Ribosomes serve as site for protein synthesis.
2. A ribosome has one binding site for m-RNA and 3 binding sites for t-RNA. They are P-site (peptidylt-RNA-site), A-site (aminoacyl t-RNA-site) and E-site (exit site).
3. In Eukaryotes, a groove which is present between two subunits of ribosomes, protects the polypeptide chain from the action of cellular enzymes and also protects m-RNA from the action of nucleases.

Question 13.
Give examples of coordinated regulation or expression, of several sets of genes.
Answer:
Examples of coordinated regulation or expression of several sets of gene are:

1. If E.colt bacteria do not have lactose in the surrounding medium as a source of energy, then structural genes in Lac operon do not get transcribed and enzyme like β -galactosidase is not synthesized.
2. The development and differentiation of embryo into an adult organism.

Question 14.
Explain with example what is meant by positive control of gene regulation.
Answer:

1. A set of genes will be switched on when a substrate is to be metabolized.
2. This phenomenon is called induction and small molecule responsible for this, is known as inducer.
3. It is positive control of gene regulation.
4. For example, lactose acts as an inducer in Lac operon.

Question 15.
If operator gene is deleted due to mutation, how will E.coli metabolise lactose?
Answer:
If operator gene is deleted due to mutation, lac operon cannot be regulated. It will get transcribed continuously and enzymes required for lactose metabolism will get synthesized continuously.

Question 16.
Explain in brief the process of initiation during protein synthesis.
Answer:
Initiation of synthesis of polypeptide chain takes place as follows:

1. Small subunit of ribosome binds to the m-RNA at 5’ end.
2. Start codon is positioned properly at P-site.
3. Initiator t-RNA, (carrying methionine in eukaryotes or formyl methionine in prokaryotes) binds with initiation codon (AUG) of m-RNA by it’s anti-codon (UAC). Codon-anti-codon pairing involves formation of hydrogen bonds.
4. The large subunit and smaller subunit of ribosome bind in the presence of Mg⁺⁺.
5. Thus, initiator charged t-RNA occupies the P-site and A-site is vacant for next charged t-RNA.

Question 17.
What are the application of genomics?
Answer:
Applications of genomics are as follows:

1. Structural and functional genomics is used in the improvement of crop plant, human health and livestock.
2. The knowledge and understanding acquired by genomics research can be applied in medicine, biotechnology and social sciences.
3. It helps in the treatment of genetic disorders through gene therapy.
4. It helps in the development of transgenic crops having desirable characters.
5. Genetic markers have applications in forensic analysis.
6. Genomics can lead to introduction of new gene in microbes to produce enzymes, therapeutic proteins and biofuels.

Question 18.
Why is HGP important?
Answer:

1. HGP is associated with rapid development of Bioinformatics.
2. Knowledge gained about the functions of genes and proteins has a major impact in the fields like Medicine. Biotechnology and the Life sciences.
3. It has helped in identifying the genes that are associated with genetic characteristics.
4. The genetic basis of many hereditary diseases can be understood.
5. It has increased the understanding of gene structure and function in other species. As human beings have many of the genes same as those of flies, roundworms and mice, such studies will enhance understanding of human evolution.

Chart based / Table based Questions

Question 1.
Complete the following table:

Organism	Diploid chromosome number
Mouse	————-
Fruitfly	————
Roundworm	————-
Yeast	————–

Answer:

Organism	Diploid chromosome number
Mouse	40
Fruitfly	8
Roundworm	12
Yeast	32

Diagram Based Questions

Question 1.
a. Identify A and B in the following diagram.
b. Name the scientist who conducted this experiment.

Answer:
a. A : Smooth strain (III-S)
B : Rough Strain and Heat-killed Smooth Strain

b. F. Griffith conducted the experiment shown in the diagram.

Question 2.
a. Identify A and B in the given diagram.
b. What is the conclusion of the given experiment?

Answer:
a. A : Rough, nonvirulent R-strain B : Heat-killed virulent S-strain
b. When DNA of heat-killed S-strain bacteria is treated with DNase, mouse remains alive as transformation does not take place. This proves that DNA is the genetic material.

Question 3.
Identify A, B, C and name the scientists who carried out the experiment given in the diagram.

Answer:
Answer: A : Infection B : Blending C : Centrifugation Scientists who carried out experiment are Alfred Hershey and Martha Chase.

Question 4.
(1) Identify A, B and C.
(2) What are their sizes?

Answer:
(1) A : Circular, unfolded chromosome
b : Folded chromosome (40 to 50 loops)
c : Supercoiled, folded chromosome

(2) (A) 350 µ (B) 30 µ (C) 2 µ

Question 5.
Draw a labelled diagram of Nucleosome with H₁ histone.
Answer:

Question 6.
Identify A and B in the given diagram.

Answer:
A : Nucleosome
B : Linker DNA

Question 7.
(a) Identify A in the given diagram.
(b) What is the dimension denoted in ‘B’

Answer:
A : Solenoid B : 300 Å

Question 8.
Sketch and label process of formation of beads on strings/fibres from chain of nucleosomes.
Answer:

Question 9.
a. Identify A. Which enzyme joined them?
b. Identify B. What is its function ?

Answer:
a. A is Okazaki fragment. These fragments are joined by DNA ligase enzyme
b. B is RNA primer. It provides 3’ OH to which nucleotide gets attached by ester bond.

Question 10.
Draw a labelled diagram showing semi-conservative replication of DNA.
Answer:

Question 11.
Sketch and label, Meselson and Stahls experiment.
Answer:

Question 12.
Identify A, B and C in the following diagram.

Answer:
A : Transcription
B : Translation and
C : Reverse Transcription

Question 13.
a. Draw a labelled diagram of transcription unit.
b. What is the sequence of m-RNA and coding strand if sequence of template strand of DNA is

Answer:
a.
b.

Question 14.
a. The following diagram shows processing of ………….
b. What is capping?
c. Identify A, B and C.

Answer:
a. hnRNA
b. Capping is addition of methylated guanosine tri-phosphate at 5′ end of hnRNA.
c. A : Exon and B : Intron C : m-RNA

Question 15.
Draw a labelled diagram of t-RNA carrying Glutamic acid.
Answer:

Question 16.
Observe the diagrams (a), (b) and (c)

1. Which step of protein synthesis is shown in the following diagrams?
2. During initiation, which subunit of ribosome binds with m-RNA?
3. What are the three binding sites for t-RNA on ribosomes?
4. On which site of ribosome second and subsequent t-RNA arrives?
5. Which link is binding amino acids in diagram (b)?
6. Which chain is being released from ribosome in diagram (c) ?
   

Answer:

1. Translation
2. 30S or 40S
3. P site, A site and E site
4. A site
5. Peptide link
6. Polypeptide chain

Question 17.
Draw a labelled diagram – Working of lac operon.
Answer:

Question 18.
What is the process shown in the following diagram? Mention all the steps given in the diagram in a proper sequence.

Answer:
The process shown in the above diagram is DNA fingerprinting. The sequence of steps in the process are:

1. Isolation of DNA
2. Restriction digestion
3. Gel electrophoresis
4. Southern blotting
5. Selection of DNA probe
6. Hybridization
7. Photography,

Long Answer Questions

Question 1.
Describe Griffith’s transformation experiment.
OR
In the light of Griffith’s experiment, explain the action of two strains of streptococcus pneumoniae and give his conclusion.
Answer:

(1) In 1928, Frederick Griffith, carried out experiments with bacterium Streptococcus pneumoniae (which causes pneumonia in humans and other mammals).

(2) Griffith used two strains of Streptococcus pneumonia:

• S-type (Virulent, smooth, pathogenic and encapsulated).
• R-type (Non-virulent, rough, non- pathogenic and non-capsulated).

(3) Experiments carried out by E Griffith:

• Mice were injected with R-strain bacteria and they survived (no pneumonia).
• Mice injected with S-strain bacteria developed pneumonia and died.
• When heat-killed S-strain bacteria were injected in mice, the mice survived.
• On injecting a mixture of heat-killed S-bacteria and live R bacteria, the mice died.

(4) Griffith obtained live S-strain bacteria from the blood of the dead mice.

(5) He concluded that the live R-strain bacteria must have picked up something (which he called transforming principle) from the heat killed S bacterium, and got changed into S-type. Transforming principle allowed R-type to synthesize capsule and it became virulent.

(6) Thus, F. Griffith first demonstrated genetic transformations.

Question 2.
Describe Avery, McCarty and MacLeod’s experiments.
Answer:

• In 1944, Oswald T. Avery, Colin M. MacLeod and Maclyn McCarty proved that the DNA is a genetic material (transforming principle).
• They purified DNA, RNA, proteins (enzymes) and other materials from heat killed S-strain cells and mixed them with cells of R-strain bacteria separately to confirm which one could transform living R cells into S cells.
• Only DNA was able to transform harmless R-strain into virulent S-strain.
• They also demonstrated that proteases, RNases did not affect transformation. Thus it was proved that the transforming substance was neither a protein nor-RNA.
• When DNA was digested with DNase, there was no transformation.
• These experiments proved that the transformation of Live R-strain bacteria into S-strain type was because of DNA of bacteria of S-strain.
• Thus, they proved that the DNA is transforming principle.

Question 3.
Explain how Hershey – Chase experimentally proved that DNA is the genetic material.
Answer:

(1) Hershey and Chase worked with bacteriophages (viruses that infect bacteria and which are composed of DNA and protein coat).

(2) They cultured E. coli bacteria in medium containing radioactive phosphorus 32p. By infecting these bacteria with bacteriophages, Hershey and Chase could develop bacteriophages having DNA labelled with 32p. as DNA contains phosphorus and proteins do not.

(3) They also cultured E. coli bacteria in medium containing radioactive sulphur 35s. By infecting these bacteria with bacteriophages, they developed
bacteriophages whose protein coat was labelled with 35s, as proteins contain sulphur and DNA does not.
[Note : Viruses cannot be cultivated in medium.]

(4) Experiment involved three steps.

• Infection : Both the types of radioactive phages were allowed to infect E.coli bacteria grown on the medium containing normal ‘P‘ and ‘S’.
• Blending : Then bacterial cultures were agitated in blender to break contact between bacteria and parts of viruses that did not enter bacterial cells.
• Centrifugation : It was done to separate bacterial cells as a pellet. Parts of viruses which did not enter bacteria remained in the suspension.

(5) Observation:

• Radioactive ‘S’ remained in suspension.
• Only radioactive ‘P’ was found inside the bacterial cell in the pellet.

(6) Thus it was proved that DNA is the genetic material which enters bacterial cell and not protein.

Question 4.
Explain the formation of beads on string, solenoid fibre, chromatin fibre and chromosome.
Answer:

• Beads on string (11 nm in diameter) : Under an electron microscope, nucleosomes in thread like chromatin look like ’beads-on- string’.
• Solenoid fibre (30 nm in diameter) : Six such nucleosomes get coiled and then form solenoid that looks like coiled telephone wire.
• Chromatin fibre (200 nm in diameter) : Further supercoiling tends to form a looped structure called chromatin fibre.
• Chromosome (1400 nm in diameter) : Chromatin fibre further coils and condenses at metaphase stage to form the chromosomes. Each chromatid is 700 nm in diameter.
• Non-Histone Chromosomal Proteins (NHC) are the additional sets of proteins that contribute to the packaging of chromatin at a higher level.

Question 5.
Explain the components of a transcription unit with the help of a diagram?
Answer:

Transcription unit (Each transcribed segment of DNA) consists of the promoter, the structural gene and the terminator.
(1) The promoter:

• The promoter is located towards 5′ end of structural gene, i.e. upstream.
• It is a DNA sequence that provides binding site for enzyme RNA polymerase.
• In prokaryotes, sigma factor sub unit of the enzyme recognizes the promoter.

(2) Structural genes:

• Template strand (Antisense strand) : DNA strand having 3’→ 5’ polarity acts as template strand as DNA dependent- RNA polymerase catalyses polymerization in 5’ → 3’ direction.
• Sense strand : The other strand of DNA having 5’ → 3’ polarity is complementary to template strand. It is called as sense strand. The sequence of bases in this strand, is same as in RNA (where Thymine is replaced by Uracil). It is the actual coding strand.

(3) The terminator:

• The terminator is located at 3’ end of coding strand, i.e. downstream.
• It defines the end of the transcription process.

Question 6.
What have we learnt from the Human Genome Project?
Answer:
We have learnt following salient features of human genome from the Human Genome Project.

1. The human genome contains 3164.7 million nucleotide bases.
2. The average gene consists of 3000 bases.
3. Largest known human gene is dystrophin at 2.4 million bases.
4. Total number of genes is estimated to be 3000.
5. 99.9% nucleotide bases are exactly the same in all people.
6. The function of about 50% of the discovered genes are unknown.
7. Less than 2% of the genome codes for proteins.
8. Repeated sequences make up a very large portion of the human genome. They can shed light on chromosome structure, dynamics and evolution.
9. Chromosome 1 has most genes (2968) and the Y has the fewest (231).
10. Single nucleotide polymorphism have been identified at about 1.4 million locations. It is useful in finding chromosomal locations for disease-associated sequences and tracing human history.

Question 7.
Describe the steps involved in DNA fingerprinting ?
Answer:

Steps involved in DNA fingerprinting are as follows:
1. Isolation of DNA : The DNA can be isolated even from the small amount of tissue like blood, hair roots, skin, etc.

2. Restriction digestion:

• The isolated DNA is treated with restriction enzymes which cut the DNA at specific sites to form small fragments of variable lengths.
• Variations in the lengths of restriction fragments are known as Restriction Fragment Length Polymorphism (RFLP).

3. Gel electrophoresis:

• The DNA samples are loaded on agarose gel and electrophoresis is carried out.
• Negatively charged DNA fragments move to the positive pole.
• Separation of fragments depends on their length and it results in formation of bands.
• dsDNA is then denatured into ssDNA by alkali treatment.

4. Southern blotting : The separated DNA fragments are transferred to a nylon membrane or a nitrocellulose membrane.

5. Selection of DNA probe:

• DNA Probe is a known sequence of single- stranded DNA.
• It is obtained from organisms or prepared by cDNA preparation method.
• The DNA probe is labelled with radioactive isotopes.

6. Hybridization:

• In this process, probe is added to the nitrocellulose membrane containing DNA fragments.
• The single-stranded DNA probe pairs with the complementary base sequence of the DNA strand.
• As a result DNA-DNA hybrids are formed on the nitrocellulose membrane. Unbound single-stranded DNA probe fragments are washed off.

7. Photography : The nitrocellulose membrane is then kept in contact with X-ray film. DNA bands, due to radioactive probe, give photographic image on X-ray film. This is autoradiography.

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 3 Inheritance and Variation Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 3 Inheritance and Variation

Multiple Choice Questions

Question 1.
Which one of the following characters is recessive in the case of the pea plants?
(a) Axial flower
(b) Green pod
(c) Green seed
(d) Inflated pod
Answer:
(c) Green seed

Question 2.
Which of the following trait is dominant in Pisum sativum?
(a) White flowers
(b) Green seeds
(c) Yellow pods
(d) Inflated pods
Answer:
(d) Inflated pods

Question 3.
When phenotypic and genotypic ratio is the same, then it is an example of ……………….
(a) incomplete dominance
(b) cytoplasmic inheritance
(c) quantitative inheritance
(d) incomplete dominance and co-dominance
Answer:
(a) incomplete dominance

Question 4.
A pea plant with yellow and round seeds is crossed with another pea plant with green and wrinkled seeds produce 51 yellow round seeds and 49 yellow wrinkled seeds, the genotype of plant with yellow round seeds must be ……………….
(a) YYRr
(b) YyRr
(c) YyRR
(d) YYRR
Answer:
(a) YYRr

Question 5.
When a single gene produces two effects and one of it is lethal, then the ratio is ……………….
(a) 2 : 1
(b) 1 : 1
(c) 1 : 2 : 1
(d) 1 : 1 : 1 : 1
Answer:
(a) 2 : 1

Question 6.
When two genes control a single character and have cumulative effect, the ratio is ……………….
(a) 1 : 1 : 1 : 1
(b) 1 : 4 : 6 : 4 : 1
(c) 1 : 2 : 1
(d) 1 : 6 : 15 : 20 : 15 : 6 : 1
Answer:
(d) 1 : 6 : 15 : 20 : 15 : 6 : 1

Question 7.
Genes located on the same locus but show more than two different phenotypes are called ……………….
(a) polygenes
(b) multiple alleles
(c) co-dominants
(d) pleiotropic genes
Answer:
(b) multiple alleles

Question 8.
Genotype refers to the genetic composition of ……………….
(a) an organism
(b) an organ
(c) chromosomes
(d) germ cells
Answer:
(a) an organism

Question 9.
Individuals having identical alleles of a gene are known as ……………….
(a) homozygous
(b) heterozygous
(c) hybrids
(d) dominants
Answer:
(a) homozygous

Question 10.
If a heterozygous tall plant is crossed with a homozygous dwarf plant, the proportion of dwarf progeny will be ……………….
(a) 100 per cent
(b) 75 per cent
(c) 50 per cent
(d) 25 per cent
Answer:
(c) 50 percent

Question 11.
Inheritance of AB blood group is due to ……………….
(a) incomplete dominance
(b) polyploidy
(c) polygeny
(d) co-dominance
Answer:
(d) co-dominance

Question 12.
The recombination of characters in a dihybrid cross is related to ……………….
(a) law of dominance
(b) incomplete dominance
(c) co-dominance
(d) independent assortment
Answer:
(d) independent assortment

Question 13.
Which one of the following is a true pleiotropic gene?
(a) HbA
(b) Hbs
(c) HbD
(d) HbP
Answer:
(b) Hbs

Question 14.
For demonstrating the law of independent assortment, one should carry out ……………….
(a) back cross
(b) test cross
(c) dihybrid cross
(d) monohybrid cross
Answer:
(c) dihybrid cross

Question 15.
Which one of the following is an example of multiple alleles?
(a) Height in pea plant
(b) Hair colour in cattle
(c) Petal colour in four o’clock plant
(d) Wing-size in Drosophila
Answer:
(d) Wing-size in Drosophila

Question 16.
For the formation of 50 seeds, how many minimum meiotic divisions are necessary?
(a) 25
(b) 50
(c) 75
(d) 63
Answer:
(d) 63

Question 17.
A cross used to verify the unknown genotype of F₁ hybrid is a ………………. cross.
(a) test
(b) back
(c) dihybrid
(d) monohybrid
Answer:
(a) test

Question 18.
Appearance of new combinations in F₂ generation in a dihybrid cross proves the law of ……………….
(a) dominance
(b) segregation
(c) independent assortment
(d) purity of gametes
Answer:
(c) independent assortment

Question 19.
Genotype of human blood group ‘O’ will be ……………….
(a) I^AI^A
(b) I^AI^B
(c) ii
(d) I^Ai
Answer:
(c) ii

Question 20.
The genotype of human blood group B is ……………….
(a) I^AI^A
(b) I^Bi
(c) I^AI^B
(d) ii
Answer:
(b) I^Bi

Question 21.
……………… chromosome appears ‘V’-shaped during anaphase.
(a) Metacentric
(b) Acrocentric
(c) Telocentric
(d) Sub-Metacentric
Answer:
(a) Metacentric

Question 22.
The sister chromatids are held together by ……………….
(a) centrioles
(b) chromonemata
(c) chromomere
(d) centromere
Answer:
(d) centromere

Question 23.
Which of the following is not X-linked disorder ?
(a) Haemophilia
(b) Night-blindness
(c) Hypertrichosis
(d) Myopia
Answer:
(c) Hypertrichosis

Question 24.
Which of the following is also called bleeder’s disease ?
(a) Anaemia
(b) Thrombocytopenia
(c) Polycythemia
(d) Haemophilia
Answer:
(d) Haemophilia

Question 25.
The person with Turner’s syndrome has ……………….
(a) 45 autosomes and X sex chromosome
(b) 44 autosomes and XYY sex chromosome
(c) 45 autosomes and Y chromosome
(d) 44 autosomes and X chromosome
Answer:
(d) 44 autosomes and X chromosome

Question 26.
Which of the following is sex chromosomal disorder ?
(a) Colour blindness
(b) Turner’s syndrome
(c) Thalassemia
(d) Down’s syndrome
Answer:
(b) Turner’s syndrome

Question 27.
The word chroma means ……………….
(a) a part of nucleus
(b) a part of chromosome
(c) colour
(d) filamentous body
Answer:
(c) colour

Question 28.
Presence of whole sets of chromosomes is called ……………….
(a) aneuploidy
(b) euploidy
(c) ploidy
(d) chromatography
Answer:
(b) euploidy

Question 29.
The synonymous term for centromere is ……………….
(a) primary constriction
(b) secondary constriction
(c) telomere
(d) satellite
Answer:
(a) primary constriction

Question 30.
Small swellings on the surface of the chromosome are called ……………….
(a) centromeres
(b) chromonemata
(c) chromomeres
(d) telomeres
Answer:
(c) chromomeres

Question 31.
On what basis are the chromosomes usually classified?
(a) On the basis of their function
(b) On the basis of their length
(c) On the basis of the position of the centromere
(d) On the basis of their number
Answer:
(c) On the basis of the position of the centromere

Question 32.
Find the mismatched pair :
(a) Metacentric – V-shaped
(b) Sub-Metacentric – L-shaped
(c) Acrocentric – J-shaped
(d) Telocentric – S-shaped
Answer:
(d) Telocentric – S-shaped

Question 33.
Out of the following combinations which individual will have maximum genetically active DNA?
(a) 44 + XX
(b) 44 + XY
(c) 44 + XYY
(d) Down’s syndrome
Answer:
(a) 44 +XX

Question 34.
Crossing over occurs at the time of ……………….
(a) diplotene
(b) pachytene
(c) leptotene
(d) zygotene
Answer:
(b) pachytene

Question 35.
A mature woman has ………………. linkage groups.
(a) 44
(b) 22
(c) 46
(d) 23
Answer:
(d) 23

Question 36.
The pairing of homologous chromosomes is called ……………….
(a) crossing over
(b) terminalisation
(c) synapsis
(d) bivalent
Answer:
(c) synapsis

Question 37.
If only one ‘X’ chromosome is found in a female person, which of the following symptoms will she show?
(a) epicanthal skin fold
(b) webbing of neck
(c) small testis and absence of spermatogenesis
(d) presence of simian crease on the palm
Answer:
(b) webbing of neck

Question 38.
If centromere is situated in the middle of the chromosome, it is called ……………….
(a) metacentric
(b) acrocentric
(c) submetacentric
(d) telocentric
Answer:
(a) metacentric

Question 39.
In which of the following disorders the number of chromosomes present is (extra) 47?
(a) Turner’s syndrome
(b) Cushing’s syndrome
(c) Acquired immuno-deficiency syndrome
(d) Down’s syndrome
Answer:
(d) Down’s syndrome

Question 40.
Myopia is an example of ……………….
(a) complete sex linkage
(b) incomplete sex linkage
(c) recombination
(d) crossing over
Answer:
(a) complete sex linkage

Question 41.
Down’s syndrome is represented by ……………….
(a) n + 1
(b) 2n + 1
(c) 3n + 1
(d) n – 1
Answer:
(b) 2n + 1

Classify the following to form Column B as per the category given in Column A

Question 1.
Types of traits:
[Sickle-cell anaemia, Flower colour of Mirabelis jalapa, Coat colour of cattle, Human blood groups, Widow’s peak, Height in human beings.]

Column A	Column B
(1) Co-dominance	—————–
(2) Incomplete dominance	—————–
(3) Multiple allelism	—————-
(4) Pleiotropy	—————–
(5) Polygenes	—————-
(6) Autosomal dominance	——————

Answer:

Column A	Column B
(1) Co-dominance	Coat colour of cattle
(2) Incomplete dominance	Flower colour of Mirabelis jalapa
(3) Multiple allelism	Human blood groups
(4) Pleiotropy	Sickle-cell anaemia
(5) Polygenes	Height in human beings
(6) Autosomal dominance	Widow’s peak

Question 2.
Types of sex-linked genes:
[Haemophilia, Ichthyosis, Nephritis, Myopia, Hypertrichosis, Retinitis pigmentosa]

Column A	Column B
(1) Completely X-linked genes	—————–
(2) Completely Y-linked genes	—————–
(3) Incompletely sex-linked genes	—————-

Answer:

Column A	Column B
(1) Completely X-linked genes	Haemophilia, Myopia
(2) Completely Y-linked genes	Ichthyosis,Hypertrichosis
(3) Incompletely sex-linked genes	Nephritis, Retinitis pigmentosa

Question 3.
Genetic Disorders
[Turner’s syndrome, Sickle-cell anaemia, Thalassemia, Edward’s syndrome, Klinefelter’s syndrome, Down’s syndrome]

Column A	Column B
(A) Autosomal disorder	—————–
(B) Sex chromosomal disorder	—————–
(C) Mendelian disorder	—————-

Answer:

Column A	Column B
(A) Autosomal disorder	Edward’s syndrome, Down’s syndrome
(B) Sex chromosomal disorder	Turner’s syndrome, Klinefelter’s syndrome
(C) Mendelian disorder	Sickle-cell anemia, Thalassemia

Very Short Answer Questions

Question 1.
What is hybrid?
Answer:
Hybrid is a heterozygous individual produced from a cross involving two parents differing in one or more contrasting characters.

Question 2.
What are homologues?
Answer:
Homologues are homologous chromosomes which are morphologically similar to each other.

Question 3.
Which law of Mendelian genetics is universally applicable?
Answer:
The law of segregation of Mendelian genetics is universally applicable.

Question 4.
Which law of Mendelian genetics is not universally applicable?
Answer:
The law of independent assortment of Mendelian genetics is not universally applicable.

Question 5.
Give the alternative term for checker board.
Answer:
Punnett’s square is the alternative term for the checker board.

Question 6.
Give the genotypic dihybrid ratio.
Answer:
1 : 1 : 2 : 2 : 4 : 2 : 2 : 1 : 1 is the genotypic dihybrid ratio.

Question 7.
What is a lethal gene?
Answer:
The gene which causes the death of the bearer is called lethal gene.

Question 8.
A pea plant pure for yellow seed colour is crossed with a pea plant pure for green seed colour. In F₁ generation, all pea plants were with yellow seed. Which law of Mendel is applicable?
Answer:
Mendel’s law of dominance is applicable in this example.

Question 9.
Identify which one of the following is a test cross.

1. Tt × Tt
2. TT × tt
3. Tt × tt

Answer:
3. Tt × tt is a test cross.

Question 10.
What colouration do roans possess? Why?
Answer:
Roans possess the mixture of red and white colour side by side due to codominant alleles for red and white traits.

Question 11.
What are polygenes?
Answer:
When a character is controlled by two or more than two pairs of genes, the genes are called polygenes.

Question 12.
In which region of chromosomes does crossing over take place?
Answer:
Crossing over takes place in the homologous region of the chromosomes.

Question 13.
What are the four sequential steps of crossing over?
Answer:
There are four sequential steps such as synapsis, tetrad formation, crossing over and terminalisation.

Question 14.
Give one example of complete linkage.
Answer:
X chromosome of Drosophila males show complete linkage.

Question 15.
What is the number of linkage groups found in honey bee?
Answer:
The number of linkage group corresponds to the haploid number of chromosomes. Honey bee’s haploid chromosomes number is 16 and thus it has 16 linkage groups.

Question 16.
Name the term for genes located on non-homologous region of Y chromosomes.
Answer:
The genes located on non-homologous region of Y chromosomes are known as holandric genes or Y-linked genes.

Question 17.
What are linkage groups?
Answer:
The genes present on the same chromosome and inherited together are called linkage group.

Question 18.
How are RBCs changed due to sickle-cell anaemia ?
Answer:
RBCs undergo change in their shape and look like a sickle, resulting in reduced capacity to carry haemoglobin.

Question 19.
Which part of a chromosome is called nucleolar organizer?
Answer:
The secondary constriction present on the chromatid arms of a chromosome is called nucleolar organizer.

Question 20.
Why is Y chromosome genetically less active?
Answer:
Since Y-chromosome possesses small amount of euchromatin that contains DNA or genes, therefore it is genetically less active.

Question 21.
Why hypertrichosis is called holandric gene?
Answer:
Hypertrichosis is Y linked gene which can be seen only in males, therefore it is called holandric gene.

Question 22.
What is the genetic difference between total colour blindness and red-green colour blindness ?
Answer:
Total colour blindness is due to incomplete sex-linked genes while red-green colour blindness is due to complete sex linkage.

Question 23.
What happens if the gene for production of factor VIII and IX becomes recessive?
Answer:
The person having recessive gene for haemophilia is deficient in clotting factors (VIII or IX) in blood, such person’s blood does not clot and he thus becomes a patient of haemophilia.

Question 24.
What is the cause of Thalassemia?
Answer:
Thalassemia is caused due to deletion or mutation of gene which codes for alpha (α) and beta (β) globin chains, causing abnormal synthesis of haemoglobin. Thus it is a quantitative abnormality of polypeptide globin chain synthesis.

Question 25.
What is monosomy? Give one example of the same.
Answer:
Monosomy is lack of one chromosome from the usual chromosomal complement. Turner’s syndrome is the example of monosomy.

Give Definitions

Question 1.
Factor
Answer:
The unit of heredity which is responsible for the inheritance and expression of a character and which is responsible for the genetic character is called a factor.

Question 2.
Gene
Answer:
The specific segment of DNA or sequence of nucleotides which is responsible for the inheritance and expression of that character is called a gene.

Question 3.
Alleles or Allelomorphs
Answer:
The two or more alternative forms of a given gene which are present on the identical loci on the homologous chromosomes are called alleles of each other.

Question 4.
Phenotype
Answer:
The external appearance of an individual for any trait is called phenotype for that trait.

Question 5.
Genotype
Answer:
Genetic constitution of an organism with respect to a particular trait is called genotype.

Question 6.
Homologous Chromosomes?
Answer:
The morphologically, genetically and structurally essentially identical chromosomes present in a diploid cell are called homologous chromosomes.

Question 7.
Back cross
Answer:
The cross of Fx progeny with any of the parents, irrespective of being dominant or recessive, is called back cross.

Question 8.
Linkage
Answer:
Linkage is defined as the tendency of the genes to be inherited together because they are present in the same chromosome.

Question 9.
Non-disjunction
Answer:
Non-disjunction is the phenomenon in which chromosomes fail to separate at the time of cell division, resulting in abnormal chromosomal combinations.

Question 10.
Syndrome
Answer:
The appearance of different types of symptoms at the same time in an individual is called a syndrome.

Question 11.
Aneuploidy
Answer:
Addition or deletion of one or two chromosomes in a diploid chromosome set is called aneuploidy.

Distinguish Between

Question 1.
Homozygous and Heterozygous.
Answer:

Homozygous	Heterozygous
1. Individuals with similar gene pairs are called homozygous.	1. Individuals with different gene pairs are called heterozygous.
2. Homozygous individuals form only one type of gametes.	2. Heterozygous individuals form more than one type of gametes.
3. Individuals with similar gene pairs TT, tt, RR and rr are homozygous.	3. Individuals with dissimilar gene pairs Tt and Rr are heterozygous.
4. Homozygous are also called pure breed.	4. Heterozygous are referred to as hybrids.

Question 2.
Monohybrid cross and Dihybrid cross.
Answer:

Monohybrid cross	Dihybrid cross
1. Crosses involving a single pair of alleles are called monohybrid crosses.	1. Crosses involving two pairs of alleles are called dihybrid crosses.
2. Monohybrid crosses yield a phenotypic ratio of 3 : 1 in the F2 generation.	2. Dihybrid crosses yield a 9 : 3 : 3 : 1 ratio in F2 generation.
3. Monohybrid crosses yield 1 : 2 : 1 genotypic ratio in F2 generation.	3. Dihybrid crosses yield 1 : 1 : 2 : 2 : 4 : 2 : 2 : 1 : 1 genotypic ratio in F2 generation.
4. Application of the law of independent assortment is not applicable in monohybrid crosses.	4. Application of the law of independent assortment is applicable in dihybrid crosses.

Question 3.
Dominant characters and Recessive characters.
Answer:

Dominant characters	Recessive characters
1. The characters that are expressed in the F1 generation are called dominant characters.	1. The characters that are not expressed in the F1 generation are called recessive characters. They are prevented from expressing themselves, due to presence of dominant allele.
2. Dominant character is expressed either in homozygous or heterozygous combination.	2. Recessive characters are expressed only when they are in homozygous combination.
3. Dominant characters cannot be masked by recessive characters. E.g. Round seed and yellow seed are dominant characters in pea plant.	3. Recessive characters are masked by dominant characters. E.g. Wrinkled seed and green seed are recessive characters in pea plant.

Question 4.
Phenotype and Genotype.
Answer:

Phenotype	Genotype
1. Phenotype refers to the outward appearance of an individual such as shape, colour, sex, etc.	1. Genotype refers to the genetic composition of an individual.
2. Phenotype can be observed directly in an individual.	2. Genotype cannot be seen, but can be found out by modern techniques like DNA fingerprinting.
3. Individuals resembling each other may or may not have the same genotype.	3. Individuals possessing the same genotype usually have the same phenotypic expression.
4. The phenotypic ratio obtained in the F2 generation of a monohybrid cross is 3 : 1.	4. The genotypic ratio obtained in the F2 generation of a monohybrid cross is 1 : 2 : 1.

Question 5.
Incomplete dominance and Co-dominance.
Answer:

Incomplete dominance	Co-dominance
1. Incomplete dominance is seen when the phenotypes of the two parents blend together to create a new phenotype for their offspring.	1. Co-dominance is seen when the two parent phenotypes are expressed together in the offspring.
2. Both the genes of an allelomorphic pair express themselves partially in F1 hybrids.	2. Both the genes of an allelomorphic pair express themselves equally in F1 hybrids.
3. In incomplete dominance, a mixture of the alleles in the genotype is seen in the phenotype.	3. In co-dominance, both alleles in the genotype are seen in the phenotype.
4. The phenotypic effect of one allele is more prominent than the other.	4. The phenotypic effect of both the alleles is equally prominent.
5. Blending or intermixing of two alleles can be observed. A white flower and a red flower alleles mix and produce pink flowers. Example : Pink flowers in Mirabilis jalapa.	5. No intermixing or blending effect of two alleles is observed. The colours don’t mix but are seen in patches. Example : Roan colour in cattle.

Question 6.
Turner’s syndrome and Klinefelter’s syndrome.
Answer:

Turner’s syndrome	Klinefelter’s syndrome
1. Individual with Turner’s syndrome has total 45 chromosomes in each of her cell.	1. Individual with Klinefelter’s syndrome has total 47 chromosomes in each of his cell.
2. Turner’s syndrome is XO female, caused due to monosomy of X-chromosome.	2. Klinefelter’s syndrome is XXY male, caused due to trisomy of X chromosome.
3. The external phenotype is of female.	3. The external phenotype is of male.
4. The stature is short.	4. The stature is tall and thin.
5. Secondary sexual characteristics are not developed in Turner’s syndrome.	5. Secondary sexual characteristics are poorly developed in Klinefelter’s syndrome.

Give Reasons or Explain the Statements

Question 1.
Law of segregation is universally applicable.
Answer:

1. According to the law of segregation, the members of the allelic pair remain together without mixing with each other.
2. They segregate or separate when the gametes are formed.
3. Thus the gametes that are formed receive only one of the two factors.
4. Now it is known that the organisms are diploid and the gametes produced by them are haploid.
5. The law of segregation therefore is universally applicable.

Question 2.
Mendel selected garden pea for his breeding experiments.
Answer:
Mendel selected garden pea for his breeding experiments, because:

1. The pea plants were true breeding varieties.
2. The pea plants being annual, it was possible to cross and study many generations within a short period.
3. The pea plants had a number of distinguishable, contrasting characters such as tall habit and dwarf habit, round seed and wrinkled seed.
4. The pea plants were easy to handle for breeding experiments.

Question 3.
When Mendel crossed a tall pea plant with a dwarf pea plant the offspring obtained from this cross were all tall.
Answer:

1. The tall habit of the pea plant is dominant over the dwarf habit of the pea plant.
2. Hence, when Mendel crossed a tall pea plant with a dwarf pea plant, the offspring obtained from this cross were all tall.

Question 4.
A cross between a homozygous tall plant with a homozygous dwarf plant results in two types of tall plants in the F₂ generation.
Answer:

1. A cross between a homozygous tall (TT) and a homozygous dwarf (tt) gives rise to a heterozygous tall (Tt) plant in the F₁ generation.
2. When the F₁ plant is selfed (Tt × Tt), it gives rise to three tall plants of which two- are heterozygous (Tt) tall and one is homozygous (TT) tall.
3. Hence a cross between a homozygous tall plant with a homozygous dwarf plant results in two types of tall plants in the F₂ generation.

Question 5.
Possibility of female becoming a haemophilic is extremely rare.
Answer:

1. Haemophilia is caused due to X-linked recessive gene. Females have double X chromosomes.
2. Even if she has haemophilic gene on one of her X-chromosome, the dominant gene on other X-chromosome, suppresses its expression. Female therefore, does not become haemophilic.
3. If she inherits haemophilic gene on both of her X-chromosomes, this combination becomes lethal. Such embryo is aborted. If born, she dies soon. This makes the possibility of female becoming a haemophilic extremely rare.

Question 6.
Human female is referred to as carrier of colour blindness.
Answer:
Human female is referred to as carrier of colour blindness because of the following reasons:

1. Females possess double X-chromosomes in her gametes.
2. If one X-chromosome is carrying recessive gene for colour blindness, her other dominant X hides the expression of colour blindness and hence she does not become a patient.
3. But such female can carry the defective gene to her progeny. Thus she is called carrier of colour-blindness.
4. A female having one recessive gene on X-chromosome is a carrier female, while a female possessing both recessive genes on both the X-chromosomes will be colour blind which is very rare.

Write Short Notes

Question 1.
Linkage.
Answer:

1. Linkage is the tendency of genes to be inherited together because they are present in the same chromosome.
2. All the genes on a chromosome are linked to one another. In the linkage group some of the genes are included.
3. The number of linkage groups of a particular species corresponds to its haploid number of chromosomes present in the organism.
4. In human beings, there are 23 linkage groups which correspond to the pairs of chromosomes found in each cell.
5. Linkage groups can be separated only at the time of crossing over during meiosis. The linkage group can form a new combination of genes after crossing over.
6. Linkages Eire of two types, viz, complete linkage and incomplete linkage.
7. Morgan discovered linkage in animals while Bateson and Punnett discovered it in plants.

Question 2.
Multiple alleles.
Answer:

1. Multiple alleles are more than two alternative alleles of a gene in a population situated on the same locus on a chromosome or its homologue.
2. Multiple alleles arise by mutations of the wild type of gene. Series of multiple alleles are formed due to several mutations that take place in the wild type of allele. This series show alternative expression.
3. Different alleles in a series show dominant-recessive relation or may show co-dominance or incomplete dominance among themselves. Among all the wild type is the most dominant one over all other mutant alleles.
4. In Drosophila, a large number of multiple alleles are known. E.g. The size of wings from normal wings to vestigial wings is due to one allele (vg) in homozygous condition. The normal wing is dominant and wild type while vestigial wing is recessive type.
5. Human blood groups A, B, AB and O Eire also due to series of multiple alleles.

Question 3.
Autosomal inheritance.
Answer:

1. Transmission of body characters occurs due to autosomes. They are not concerned with sex determination or sex linkage.
2. All the body characters from parents are passed on to their offspring through autosomes. This is called autosomal inheritance.
3. Some autosomal characters are due to dominant genes while some other are due to recessive genes. E.g. Widow’s peak and Huntington’s disease is also autosomal dominant character, etc.
4. Phenyl ketonuria (PKU), Cystic fibrosis and Sickle-cell anaemia are autosomal recessive traits.

Question 4.
Widow’s peak.
Answer:

1. Widow’s peak is a prominent ‘V’ shaped hairline on forehead.
2. It is due to autosomal dominant gene.
3. Widow’s peak occurs in homozygous dominant (WW) and also heterozygous (Ww) individuals.
4. Individuals with homozygous recessive (ww) genotype do not have widow’s peak but have a straight hair line.
5. Both males and females have equal chance of inheritance.

Question 5.
Environmental sex determination
Answer:

1. Environmental sex determination is shown by lower organisms such as Bonellia viridis.
2. In this animal the environmental factors decide the sex of an offspring.
3. There is extreme sexual dimorphism in this worm. Female is about 10 cm long while male is tiny and parasitic in the reproductive parts of mature female.
4. If larva is reared in vicinity of mature female then it becomes a male. By settling on the proboscis of mature female, larva becomes parasitic, enters the female’s mouth and then takes permanent shelter in the female uterus. Such males then produce gametes and fertilize the eggs.
5. If larvae are drifted away from mature female or if they settle on the sea bottom, they develop into females. Thus determination of sex is due to environmental factors.

Question 6.
Y-linked or Holandric genes.
Answer:

1. Holandric means entirely of male sex. Y-linked genes are called holandric genes because they are located on non-homologous region of Y chromosome.
2. The Y-linked genes are inherited directly from male to male.
3. These genes are never seen in females due to lack of Y chromosome in them.
4. Hyper Mchosis and ichthyosis are examples of holandric genes.
5. Hypertrichosis means excessive development of hair on pinna of ear. This character is transmitted directly from father to son.
6. Ichthyosis person with rough skin.

Question 7.
X-body.
Answer:

1. German biologist, Henking in 1891, was studying spermatogenesis of the squash bug (Anasa tristis).
2. He noted that 50% of sperms receive the unpaired chromosomes while other 50% sperms do not receive it.
3. Henking gave a name to this structure as the X-body. He was unable to explain its role in sex determination.
4. Further investigations by other scientists led to conclusion that the ‘X-body’ of Henking was a chromosome and gave the name as X-Chromosome to X-body.

Question 8.
Thalassemia
Answer:
(1) Thalassemia is an autosomal-eeessive disorder. The synthesis of alpha ciiains are controlled by two genes, (HBA1 and HBA2) on chromosome 16. Beta chain synthesis is controlled by gene HBB located on chromosome 11. Two alpha chains and two beta chains together form four polypeptide chains that make heterotetrameric haemoglobin molecule. But when there is defective gene on either of chromosome 16 or 11, there is quantitative abnormality of polypeptide globin chain synthesis. This results into thalassemia.

(2) Depending upon which chain is affected, thalassemia is classified as, alpha (α) thalassemia and beta (β) thalassemia.

(3) The clinical symptoms of thalassemia are as follows:

• Pale yellow skin.
• Anaemia due to inability to synthesize haemoglobin.
• Slow growth and development.
• Variation in the shape and size of RBCs.

(4) Patients need regular blood transfusions to cope with the disorder.

Short Answer Questions

Question 1.
Write the statements of three laws of inheritance given by Mendel.
Answer:
(1) Statement of Law of Dominance : When two homozygous individuals with one or more sets of contrasting characters are crossed, the alleles that appear in F₁ are dominant and those which do not appear in F₁ are recessive.

(2) Statement of Law of Segregation or Law of purity of gametes : When F₁ hybrid forms gametes, the alleles segregate from each other and enter in different gametes. The gametes formed are pure because they carry only one either dominant allele or recessive allele each. Due to this the law is also called “Law of purity of gametes”.

(3) Statement of Law of Independent Assortment : When hybrid possessing two (or more) pairs of contrasting alleles forms gametes, these alleles in each pair segregate independently of the other pair.

Question 2.
Why are farmers and gardeners advised to buy new F₁ hybrid seeds every year?
Answer:

1. Farmers use hybrid seeds for agriculture or horticulture. Hybrid seeds are produced by crossing two unrelated parent plants.
2. Hybrid seed varieties give improved yields and crop vigour to the farmer.
3. Hybrids are made by crossing two highly inbred ‘parent’ plants. First generation hybrids, however, do not breed true to type, meaning that the seed they set may not grow into crops that are identical to the ‘parent’ plants.
4. This can result in variations in yield and quality therefore many farmers prefer to buy new hybrid seed each year to ensure consistency in their final product.

Question 3.
What are the main generalizations given after Mendel’s experiments on the pea plant?
Answer:
After the Mendel’s laws of inheritance and his experiments, following generalizations were made:

1. Single trait is shown due to single gene. Every single gene has two contrasting alleles.
2. Two alleles are always in interaction in which one is completely dominant while other is completely recessive.
3. Factors which were later called genes for different traits are always present on different chromosomes. These traits can assort independently of each other.

Question 4.
Mention the types of deviations from Mendel’s finding.
OR
Describe Neo-Mendelism in short.
Answer:
As the science of genetics progressed, many changes were seen from Mendel’s generalizations. These are called as Neo- Mendelism.
The deviations from Mendel’s findings can be categorised under following heads:

1. Intragenic interactions : These interactions : are seen between the alleles of same gene. e.g. incomplete dominance and co-dominance. They are also seen in multiple allele series of a gene.
2. Intergenic interactions : Intergenic interactions are between the alleles of different genes present on the same or different chromosomes, e.g. pleiotropy, polygenes, epistasis, supplementary and complementary genes, etc.

Question 5.
Why Drosophila is most suitable organism for genetics experiments?
Answer:
Drosophila is most suitable organism because of the following reasons:

1. Drosophila cam easily be cultured under laboratory conditions.
2. Life span of Drosophila is short for about two weeks.
3. Drosophila has high rate of reproduction and hence newer organisms can be obtained rapidly.

Question 6.
What are the causes of Down’s syndrome?
Answer:

1. Down’s syndrome is caused due to aneuploidy.
2. Aneuploidy is due to non-disjunction of chromosome at the time of gamete formation during meiosis. Due to non-disjunction, chromosomes fail to separate.
3. In addition to a homologous pair of 21st chromosome there is an extra 21st, therefore it is called trisomy (2n + 1) of 21st chromosome.

Question 7.
What are the characteristic symptoms of Down’s syndrome?
Answer:
Symptoms of Down’s syndrome:

1. Typical facial features.
2. An epicanthal skin fold, over the inner corner of eyes causing downward slanting eyes.
3. Typical flat face, rounded flat nose, mouth always open with protruding tongue.
4. Mental retardation.
5. Poor skeletal development.
6. Short stature, relatively small skull and arched palate.
7. Flat hand with simian crease that runs across the palm.

Question 8.
Write a brief account of Turner’s syndrome.
Answer:

1. Turner’s syndrome is a genetic disorder caused due to monosomy of X chromosome.
2. It was first described by H. H. Turner.
3. Turner’s syndrome is caused due to non-disjunction of sex chromosomes which takes place during gamete formation.
4. Chromosomal complement of Turner’s syndrome is 44 + XO, having a total of 45 chromosomes.

Symptoms of Turner’s syndrome are as follows:

1. Female phenotype.
2. Short stature with webbing of neck.
3. Low posterior hair line.
4. Secondary sexual characters fail to develop.
5. Mental retardation.

Question 9.
Write a brief account of Klinefelter’s syndrome.
Answer:

1. Klinefelter’s syndrome is a genetic disorder caused due to trisomy of X chromosome.
2. It was first described by Harry Klinefelter.
3. Klinefelter’s syndrome is caused due to non-disjunction of sex chromosomes which takes place during gamete formation.
4. Chromosomal complement of Klinefelter’s syndrome is 44+XXY, having a total of 47 chromosomes.

Symptoms of Klinefelter’s syndrome are as follows:

1. The Klinefelter’s syndrome individuals are tall, thin and eunuchoid.
2. They are sterile with poorly developed sexual characteristics.
3. Testes are underdeveloped and small. Spermatogenesis does not take place.
4. They have subnormal intelligence and show partial mental retardation.

Diagram Based Questions

Question 1.
Give the graphical representation of test cross and back cross.
Answer:
(1) Test cross
Answer:

The F₂ generation of test cross consists of 50% heterozygous tall plants and 50% homozygous dwarf plants.

(2) Back cross

F₁ crossed back with its dominant parent

Question 2.
Give a cross for incomplete dominance using a suitable example.
Answer:

Result:
Genotypic ratio – 1RR : 2 Rr : 11rr
Phenotypic ratio – 1Red : 2 Pink : 1 White

Question 3.
Give a cross of co-dominance using a suitable example.
Answer:
Coat colour in cattle
Red female RR × White male WW
P₁ generation : RR × WW
Gametes : R and W
F₁ generation all RW Roan coloured
P₂ generation RW × RW

	R	W
R	RR	RW
W	RW	WW

Genotypic ratio : 1 RR : 2 RW : 1 WW
Phenotypic ratio : 1 Red : 2 Roan : 1 White

Question 4.
Draw two crosses to show inheritance of colour blindness, (i) A cross between normal female and colour blind male, (ii) A cross of carrier woman with normal man.
Answer:
(i) A cross between normal female and colour-blind male.

(ii) A cross of carrier female with normal male.

Question 5.
Draw the following crosses to show inheritance of haemophilia : Normal female with haemophilic male. Show their progeny. If one of their carrier daughters marries a normal male what would be possible genotypes of this generation.
Answer:
(1) Haemophilic male crossed with normal female:

(2) Carrier female crossed with normal male :

Question 6.
Sketch and label structure of X and Y chromosomes.
Answer:

Question 7.
Give the graphical representation of pleiotropy to show inheritance of Sickle-cell anaemia.
Answer:

Long Answer Questions

Question 1.
There are 16 possible individuals in F₂ generation. Try to find out the phenotypes as well as the genotypic and phenotypic ratios.
Answer:
In the above dihybrid cross there are 4 phenotypes such as yellow round, yellow wrinkled, green round, green wrinkled.
There are 9 different genotypes as follows:
1 : RRYY / 2 : RRYy / 1 : RRyy / 2 : RrYY / 4 : RrYy / 2 : Rryy / 1 : rrYY /2 : rrYy / 1 : rryy

Question 2.
Explain with suitable diagram how test cross is used to find out genotype of dominant plant.
Answer:
Test cross is used to find out the exact genotype by crossing the F₁ individual with homozygous recessive one.
E.g. To find out the genotype of unknown violet flower obtained in F₁ generation, one can conduct two crosses as follows:
I. Unknown flower considering as RR (homozygous dominant)
RR × rr Homozygous dominant

	R	R
R	Rr	Rr
R	Rr	Rr

II. In such cross, all the flowers will be violet. II. Unknown flower considering as Rr (heterozygous)
Rr × rr Homozygous recessive with heterozygous

	R	R
R	Rr	Rr
R	Rr	Rr

In such cross half the flowers will be violet and half will be white.

Question 3.
Describe briefly Morgan’s Experiments showing linkage and crossing over. (Diagram is not needed)
Answer:
(1) Morgan used Drosophila melanogaster for his experiments.

(2) He carried out several dihybrid cross experiments to study sex-linked genes of Drosophila.

(3) Crosses between yellow-bodied, white-eyed female and brown-bodied, red-eyed males were done in P₁ generation. Brown-bodied and red-eyed forms were wild.

(4) Morgan intercrossed their F₁ progeny and noted that two genes did not segregate independently of each other and F₂ ratio deviated very significantly from Mendelian 9 : 3 : 3 : 1 ratio.

(5) When genes are grouped on the same chromosome, some genes are strongly linked. They show very few recombinations (1.3%).

(6) When genes are loosely linked, i.e. located away from each other on chromosome, they show more (higher) recombinations (37.2%).

(7) For example, the genes for yellow body and white eye were strongly linked and showed only 1.3 per cent recombination (in cross-I).

(8) White-bodied and miniature wings showed 37.2 per cent recombination (in cross-II). Cross I shows crossing over between genes y and w.

(9) Cross II shows crossing over between genes white (w) and miniature wing (m). Here dominant wild type alleles are represented with (+) sign.

(10) Parental combinations occur more due to linkage and new combinations less due to crossing over.

Question 4.
Describe the mechanism of sex determination in human beings with a suitable cross.
Answer:
1. Sex determination in human beings:

(1) In human beings, the sex is determined with the help of X and Y chromosomes. This chromosomal mechanism of sex determination is of XX-XY type.

(2) In male, the nucleus of each cell contains 46 chromosomes or 23 pairs of chromosomes. Of these 22 pairs are autosomes and one pair of sex chromosomes. Males are thus heteromorphic as they have two different types of sex chromosomes.

(3) Autosomes or somatic chromosomes are responsible for determination of other characters of the body, but not the sex.

(4) In female cells, there are 22 pairs of autosomes and one pair of X chromosomes. Females are thus homomorphic as they have similar sex chromosomes.

(5) Thus the genotypes of female and male are as follows:
Female : 46 chromosomes = 44 autosomes + XX sex chromosomes
Male : 46 chromosomes = 44 autosomes + XY sex chromosomes

(6) During gamete formation, the diploid germ cells in the testes and ovaries undergo meiosis to produce haploid gametes (sperms and eggs). The homologous chromosomes separate and enter into different gametes during this process.

(7) The human male produces two different types of sperms, one containing 22 autosomes and one X chromosome and the other containing 22 autosomes and one Y chromosome. Human males are therefore called heterogametic, i.e. they produce different types of gametes.

(8) The human female produces only one type of eggs containing 22 autosomes and one X chromosome and therefore she is homogametic.

(9) During fertilization, if X containing sperm fertilizes the egg having X chromosome, then a female child with XX chromosomes is conceived.

(10) If Y containing sperm fertilizes the egg having X chromosome then a male child with XY chromosomes is conceived.

(11) The sex of the child thus depends upon the type of sperm fertilizing the egg. The heterogametic parent determines the sex of the child and thus the father is responsible for the determination of the sex of the child and not the mother.

Question 5.
Explain the mechanism of sex determination in case of birds.
Answer:

1. Sex determination in birds is by ZW-ZZ mechanism.
2. In birds, males are homogametic while females are heterogametic.
3. Males produce all similar types of sperms, containing 8 autosomes and ‘Z’ sex chromosome.
4. Females produce two different types of eggs, one containing 8 autosomes and Z chromosome and the other containing 8 autosomes and W chromosome.
5. When Z bearing egg is fertilized by a sperm a male offspring is produced. If W bearing egg is fertilized then female offspring is produced.
   

[Note : In domestic fowl chromosome number is 18, with 16 autosomes and two sex chromosomes.]

Balbharti Maharashtra State Board 12th Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals Important Questions and Answers.

Maharashtra State Board 12th Biology Important Questions Chapter 2 Reproduction in Lower and Higher Animals

Multiple-choice questions

Question 1.
Gemmule formation takes place in ……………….
(a) Hydra
(b) Spongilla
(c) Planaria
(d) Human being
Answer:
(b) Spongilla

Question 2.
Which part of ovary in mammals acts as an endocrine gland after ovulation?
(a) stroma
(b) germinal epithelium
(c) vitelline membrane
(d) graafian follicle
Answer:
(d) graafian follicle

Question 3.
Cessation of menstrual cycle in female is called ……………….
(a) lactation
(b) ovulation
(c) menarche
(d) menopause
Answer:
(d) menopause

Question 4.
Capacitation of sperms occurs in ……………….
(a) vas deferens
(b) vas efferens
(c) vagina
(d) ejaculatory duct
Answer:
(c) vagina

Question 5.
How many sperms are formed from a secondary spermatocyte ?
(a) 4
(b) 8
(c) 2
(d) 1
Answer:
(c) 2

Question 6.
The middle piece of the sperm contains ……………….
(a) proximal centriole
(b) nucleus
(c) mitochondria
(d) distal centriole
Answer:
(c) mitochondria

Question 7.
About which day in a normal human menstrual cycle does rapid secretion of LH (popularly called LH surge) normally occurs ?
(a) 14th day
(b) 20th day
(c) 5th day
(d) 11th day
Answer:
(a) 14th day

Question 8.
Morphogenetic movements occur during ……………….
(a) blastulation
(b) gastrulation
(c) fertilization
(d) cleavage
Answer:
(b) gastrulation

Question 9.
The technique used to block the passage of sperm in male is ……………….
(a) tubectomy
(b) vasectomy
(c) coitus interruptus
(d) rhythm method
Answer:
(b) vasectomy

Question 10.
Planaria reproduces asexually through ……………….
(a) budding
(b) gemmule formation
(c) regeneration
(d) binary fission
Answer:
(c) regeneration

Question 11.
The role of Leydig cells is ……………….
(a) nourishment of sperms
(b) to give motility to sperms
(c) synthesis of testosterone
(d) to undergo spermatogenesis
Answer:
(c) synthesis of testosterone

Question 12.
Chancre are the primary lesions caused by ……………….
(a) Neisseria gonorrhoeae
(b) Treponema pallidum
(c) Plasmodium vivax
(d) Salmonella typhi
Answer:
(b) Treponema pallidum

Question 13.
Smooth muscles lining the wall of scrotum are called ……………….
(a) detrusor muscles
(b) dartos muscles
(c) gluteal muscles
(d) latissimus dorsi muscles
Answer:
(b) dartos muscles

Question 14.
The trophoblast cells in contact with embryonal knob are called ……………….
(a) inner mass cells
(b) blastomere
(c) amniogenic cells
(d) cells of Rauber
Answer:
(d) cells of Rauber

Question 15.
The external layer of collagenous connective tissue of human testis is ……………….
(a) tunica vasculosa
(b) tunica vaginalis
(c) tunica granulosa
(d) tunica albuginea
Answer:
(d) tunica albuginea

Question 16.
Which of the following is mesodermal in origin ?
(a) Retina
(b) Enamel of teeth
(c) Heart
(d) Liver
Answer:
(c) Heart

Question 17.
Pregnancy in second trimester is maintained by ……………….
(a) LH (luteinizing hormone)
(b) progesterone
(c) estrogen
(d) hCG (human Chorionic Gonadotropin)
Answer:
(b) progesterone

Question 18.
In human foetus, the heart begins to beat at developmental age of ……………….
(a) 4th week
(b) 3rd week
(c) 6th week
(d) 8th week
Answer:
(c) 6th week

Question 19.
……………… contribute about 60% of the total volume of the semen.
(a) Prostate gland
(b) Cowper’s glands
(c) Seminal vesicles
(d) Bartholin’s glands
Answer:
(c) Seminal vesicles

Question 20.
Which of the following is hormone releasing IUD?
(a) Lippes loop
(b) Cu 7
(c) LNG 20
(d) Multiload 375
Answer:
(c) LNG 20

Question 21.
Which of the following is incorrect regarding vasectomy?
(a) Vasa deferentia is cut and tied
(b) Irreversible sterility
(c) No sperm occurs in seminal fluid
(d) No sperm occurs in epididymis
Answer:
(d) No sperm occurs in epididymis

Question 22.
The test-tube baby programme employs which one of the following techniques?
(a) Gamete intra fallopian transfer (GIFT)
(b) Zygote intra fallopian transfer (ZIFT)
(c) Intra cytoplasmic sperm injection (ICSI)
(d) Intra uterine insemination (IUI)
Answer:
(b) Zygote intra fallopian transfer (ZIFT)

Question 23.
Medical Termination of Pregnancy (MTP) is considered safe up to how many weeks of pregnancy?
(a) 8 weeks
(b) 12 weeks
(c) 24 weeks
(d) 6 weeks
Answer:
(b) 12 weeks

Question 24.
‘Saheli? an oral contraceptive pill is to be taken ……………….
(a) daily
(b) weekly
(c) quarterly
(d) monthly
Answer:
(b) weekly

Question 25.
The role of copper releasing IUDs is to ……………….
(a) inhibit ovulation
(b) prevent fertilization
(c) inhibit implantation of blastocyst
(d) inhibit gametogenesis
Answer:
(b) prevent fertilization

Question 26.
The phenomenon of nuclear fusion of sperm and egg is known as ……………….
(a) karyogamy
(b) parthenogenesis
(c) vitellogenesis
(d) oogenesis
Answer:
(a) karyogamy

Question 27.
Acrosome of spermatozoa is formed from ……………….
(a) lysosomes
(b) Golgi bodies
(c) ribosomes
(d) mitochondria
Answer:
(b) Golgi bodies

Question 28.
Which of the following undergoes spermiogenesis ?
(a) Spermatids
(b) Spermatogonia
(c) Primary spermatocytes
(d) Secondary spermatocytes
Answer:
(a) Spermatids

Question 29.
In mammals, the estrogens are secreted by the graafian follicle from its ……………….
(a) theca externa
(b) theca interna
(c) membrane granulosa
(d) corona radiata
Answer:
(b) theca interna

Question 30.
Which hormone is essential for maintenance of the endometrium of uterus?
(a) FSH
(b) LH
(c) Progesterone
(d) Estrogen
Answer:
(c) Progesterone

Question 31.
Which of the following cells during gametogenesis is normally diploid?
(a) Spermatid
(b) Spermatogonia
(c) Second polar body
(d) Secondary oocyte
Answer:
(b) Spermatogonia

Question 32.
Fertilization takes place at ……………….
(a) cervix
(b) ampulla
(c) isthmus
(d) vagina
Answer:
(b) ampulla

Question 33.
In mammals, failure of testes to descend into scrotum is known as ……………….
(a) paedogenesis
(b) castration
(c) cryptorchidism
(d) impotency
Answer:
(c) cryptorchidism

Question 34.
Polar body is produced during the formation of ……………….
(a) sperm
(b) secondary oocyte
(c) oogonium
(d) spermatocytes
Answer:
(b) secondary oocyte

Question 35.
Menstrual flow occurs due to lack of ……………….
(a) vasopressin
(b) progesterone
(c) FSH
(d) oxytocin
Answer:
(b) progesterone

Question 36.
Approximately how many eggs are produced by a normal healthy human female up to the age of 25 years if the age of menarche is 12 years?
(a) 169
(b) 416
(c) 240
(d) 100
Answer:
(a) 169

Question 37.
In humans, at the end of the first meiotic division, the male germ cells differentiate into the ……………….
(a) spermatids
(b) spermatozoa
(c) primary spermatocytes
(d) secondary spermatocytes
Answer:
(d) Secondary spermotocytes

Question 38.
The part that carries sperms from testis to epididymis is ……………….
(a) rete testis
(b) vasa efferentia
(c) vasa differentia
(d) ejaculatory ducts
Answer:
(c) vasa differentia

Question 39.
Which period of menstrual cycle is called risky period of conception?
(a) 3rd to 7th day
(b) 7th to 13th day
(c) 10th to 17th day
(d) 15th to 25th day
Answer:
(c) 10th to 17th day

Question 40.
Which hormone confirms pregnancy?
(a) Progesterone
(b) Estrogen
(c) hCG
(d) LH
Ans
(c) hCG

Match the columns

Question 1.

Column I [Organs]	Column II [Functions]
(1) Epididymis	(a) Transport of sperms
(2) Sertoli cells	(b) Copulatory organ
(3) Vas deferens	(c) Nourishment to developing sperms
(4) Penis	(d) Maturation of sperms

Answer:

Column I [Organs]	Column II [Functions]
(1) Epididymis	(d) Maturation of sperms
(2) Sertoli cells	(c) Nourishment to developing sperms
(3) Vas deferens	(a) Transport of sperms
(4) Penis	(b) Copulatory organ

Question 2.

Column I [Organ/cells]	Column II [Hormones]
(1) Corpus luteum	(a) hCG
(2) Interstitial cells / Leydig’s cells	(b) Estrogen
(3) Placenta	(c) Progesterone
(4) Graafian follicle	(d) Testosterone

Answer:

Column I [Organ/cells]	Column II [Hormones]
(1) Corpus luteum	(c) Progesterone
(2) Interstitial cells / Leydig’s cells	(d) Testosterone
(3) Placenta	(a) hCG
(4) Graafian follicle	(b) Estrogen

Question 3.

Column I	Column II
(1) Graafian follicle	(a) Site of implantation
(2) Uterus	(b) Birth canal
(3) Fallopian tube	(c) Site of fertilization
(4) Vagina	(d) Release of secondary oocyte

Answer:

Column I	Column II
(1) Graafian follicle	(d) Release of secondary oocyte
(2) Uterus	(a) Site of implantation
(3) Fallopian tube	(c) Site of fertilization
(4) Vagina	(b) Birth canal

Question 4.

Column I [Phases]	Column II [Hormonal changes]
(1) Menstrual phase	(a) Rapid secretion of LH
(2) Proliferative phase	(b) Increased level of FSH and estrogen
(3) Ovulatory phase	(c) Increased level of progesterone
(4) Secretory phase	(d) Decrease in progesterone and estrogen

Answer:

Column I [Phases]	Column II [Hormonal changes]
(1) Menstrual phase	(d) Decrease in progesterone and estrogen
(2) Proliferative phase	(b) Increased level of FSH and estrogen
(3) Ovulatory phase	(a) Rapid secretion of LH
(4) Secretory phase	(c) Increased level of progesterone

Question 5.

Column I	Column II
(1) Acrosome	(a) Completion of IInd meiotic division of secondary oocyte
(2) Penetration of sperm into ovum	(b) Dissolution of zona pellucida
(3) Formation of fertilization membrane	(c) Secretion of Hyaluronidase
(4) Acrosin / Zona lysine	(d) Prevention of polyspermy

Answer:

Column I	Column II
(1) Acrosome	(c) Secretion of Hyaluronidase
(2) Penetration of sperm into ovum	(a) Completion of IInd meiotic division of secondary oocyte
(3) Formation of fertilization membrane	(d) Prevention of polyspermy
(4) Acrosin / Zona lysine	(b) Dissolution of zona pellucida

Question 6.

Column I	Column II
(1) Parturition	(a) Attachment of embryo to endometrium
(2) Gestation	(b) Release of egg from Graafian follicle
(3) Ovulation	(c) Delivery of baby from uterus
(4) Implantation	(d) Duration between pregnancy and birth

Answer:

Column I	Column II
(1) Parturition	(c) Delivery of baby from uterus
(2) Gestation	(d) Duration between pregnancy and birth
(3) Ovulation	(b) Release of egg from Graafian follicle
(4) Implantation	(a) Attachment of embryo to endometrium

Question 7.

Column I [Contraceptive method]	Column II [Mode of action]
(1) Pill	(a) Prevents sperms reaching cervix
(2) Condom	(b) Prevents implantation
(3) Vasectomy	(c) Prevents ovulation
(4) Copper T	(d) Semen contains no sperms

Answer:

Column I [Contraceptive method]	Column II [Mode of action]
(1) Pill	(c) Prevents ovulation
(2) Condom	(a) Prevents sperms reaching cervix
(3) Vasectomy	(d) Semen contains no sperms
(4) Copper T	(b) Prevents implantation

Question 8.

Column I	Column II
(1) Mechanical means	(a) Saheli
(2) Physiological device	(b) Jellies
(3) Chemical device	(c) Vasectomy
(4) Permanent method	(d) Diaphragm

Answer:

Column I	Column II
(1) Mechanical means	(d) Diaphragm
(2) Physiological device	(a) Saheli
(3) Chemical device	(b) Jellies
(4) Permanent method	(c) Vasectomy

Classify the following to form Column B as per the category given in Column A.

Question 1.
Classify the following contraceptives given below as per Column ‘A’ and complete Column ‘B’. Select from the given options:
(i) Foams
(ii) Lippe’s loop
(iii) Cervical caps
(iv) Multiload 375
(v) Diaphragms
(vi) Jellie

Column A	Column B
(1) Mechanical means	————–, ————
(2) Chemical means	————-, ————-
(3) Intra-uterine device	————-, ————

Answer:

Column A	Column B
(1) Mechanical means	Cervical caps, Diaphragms
(2) Chemical means	Foams, Jellies
(3) Intra-uterine device	Lippe’s loop, Multiload 375

Question 2.
Classify the following components of semen given below as per Column ‘A’ and complete the Column ‘B’. Select from the given options
(i) Acid phosphatase
(ii) Mucous like fluid
(iii) Prostaglandins
(iv) Citric acid
(v) Fructose
(vi) Fibrinogen

Column A	Column B
(1) Seminal fluid	————–, ————
(2) Prostatic fluid	————-, ————-
(3) Fluid from Cowper’s gland	————-, ————

Answer:

Column A	Column B
(1) Seminal fluid	Prostaglandins, Fructose, Fibrinogen
(2) Prostatic fluid	Acid phosphatase, Citric acid
(3) Fluid from Cowper’s gland	Mucous like fluid

Very short answer questions

Question 1.
How many sperms are present in single ejaculation?
Answer:
A single ejaculation contains about 400 millions of sperms.

Question 2.
What is gemmule? How is gemmule formed ?
Answer:
Gemmule is an internal bud formed by aggregation of archeocytes in sponges to overcome unfavourable season.

Question 3.
What is cryptorchidism?
Answer:
Failure of testis to descend into scrotum leading to sterility is called cryptorchidism.

Question 4.
What is the beginning of the menstrual cycle and cessation of menstrual cycle respectively called?
Answer:
The beginning of the menstrual cycle is called menarche while cessation of menstrual cycle is called menopause.

Question 5.
Which men have an increased risk of prostate cancer?
Answer:
Men who are over 50 years of age and have a daily high consumption of fat have an increased risk of prostate cancer.

Question 6.
What is capacitation with reference to sperm?
Answer:
Changes in a mammalian sperm which prepare it for fertilization of ovum is called capacitation.

Question 7.
Give any two examples each of seasonal breeders and continuous breeders among sexually reproducing animals.
Answer:
Example of seasonal breeders : Goat, Sheep and Donkey.
Example of continuous breeders : Humans, apes.

Question 8.
What does IUCD indicate?
Answer:
IUCD means Intra Uterine Contraceptive Device.

Question 9.
What is full form of IVF?
Answer:
IVF means In Vitro Fertilization.

Question 10.
From which germinal layers the nervous system is derived?
Answer:
The nervous system is derived from ectoderm.

Question 11.
A mother of a one-year-old child wanted to space her second child. Her doctor suggested ‘Copper-T’. Explain its contraceptive action.
Answer:
Copper ions released from ‘Copper-T’ suppress sperm motility and the fertilizing capacity of sperms.

Question 12.
Which options are available for infertile couples to have child?
Answer:
Infertile couples have many options to have a child such as fertility drugs, modern techniques such as IVE ZIFT, GIFT, ICSI, artificial insemination, IUI, using surrogate mother or taking the sperm from sperm bank.

Question 13.
How many primary spermatocytes and oocytes are required for the formation of 100 spermatozoa and ova?
Answer:
25 Primary spermatocytes and 100 primary oocytes will be required for the formation of 100 spermatozoa and ova respectively.

Question 14.
The entrance of fallopian tube of a lady is blocked. She wants motherhood. Which method will help her?
Answer:
The method of GIFT or Gamete Intra Fallopian Transfer is the method that will help the lady to have a child.

Question 15.
What is the role of birth control pills?
Answer:
Birth control pills are contraceptive pills that check the ovulation by inhibiting the secretion of FSH and LH.

Question 16.
In T.S. of ovary, can all stages of follicles be seen simultaneously?
Answer:
In T.S. of ovary, all the stages of follicles cannot be seen simultaneously. The stage of follicles develop alternately in the ovary as per timing of menstrual cycle under the influence of hormones of pituitary and ovaries.

Question 17.
What will be marriageable age for boy and girl as per the Indian law?
Answer:
As per the Hindu Marriage Act, minimum age for boy must be 21 years and for a girl 18 years, at the time of marriage.

Question 18.
What is MTP Act?
Answer:
MTP Act is for reducing the incidences of illegal abortions and maternal mortalities.

Question 19.
Which is the time period legally allowed by MTP ACT for terminating pregnancy?
Answer:
According to MTP Act, pregnancy may be terminated within first 12 weeks, more than 12 weeks but lesser than 20 weeks.

Give definitions of the following

Question 1.
Amphimixis
Answer:
It is the process which involves the production of offspring by the formation and fusion of gametes.

Question 2.
Gametogenesis
Answer:
The gametogenesis is the process of formation of gametes in sexually reproducing animals.

Question 3.
Spermiogenesis
Answer:
The process of transformation of non-motile and non¬functional spermatid into a functional and motile spermatozoa is called spermiogenesis.

Question 4.
Insemination
Answer:
The process of deposition of semen into the vagina of the female at the time of coitus or sexual intercourse is called insemination.

Question 5.
Cleavage
Answer:
The process of early mitotic division of the zygote to form a multicellular morula stage is called cleavage.

Question 6.
Implantation
Answer:
The process by which the blastocyst after its formation, gets implanted or embedded into the endometrium of the uterus is called implantation.

Question 7.
Gestation
Answer:
The condition of carrying one or more embryos in the uterus is called gestation.

Question 8.
Placenta
Answer:
A flattened, discoidal organ present in the uterus of pregnant mother and which acts as endocrine source and nutrition provider for growing foetus is called placenta.

Question 9.
Lactation
Answer:
The process of secretion of milk in the mammary glands and expelling it through nipples out to provide nourishment to the growing baby is called lactation.

Question 10.
Parturition
Answer:
Parturition is the process of giving birth to a baby.

Question 11.
Amniocentesis
Answer:
Amniocentesis is a process in which amniotic fluid containing foetal cells is collected using a hollow needle inserted into the uterus under ultrasound guidance.

Question 12.
Infertility
Answer:
Infertility is defined as the inability to conceive naturally after (one year of) regular unprotected intercourse.

Question 13.
IVF (In Vitro Fertilization)
Answer:
It is a process of fertilization where an egg is combined with sperm outside the body in a test tube or glass plate to form a zygote under simulated conditions in the laboratory.

Question 14.
Artificial Insemination (AI)
Answer:
It is the technique during which the sperms are collected from the male and artificially introduced into the cervix of female, for the purpose of achieving a pregnancy through in vivo fertilization (inside the body).

Question 15.
Adoption
Answer:
Adoption is a legal process by which a couple or a single parent gets legal rights, privileges and responsibilities that are associated to a biological child for the upbringing of the adopted child.

Give functions of the following

Question 1.
Corpus luteum.
Answer:
Corpus luteum is a secondary endocrine source that produces progesterone for maintaining pregnancy.

Question 2.
Scrotum.
Answer:
Scrotum protects the testis and also acts as thermoregulator.

Question 3.
Acrosome of sperm.
Answer:
Acrosome of the sperm releases hyaluronidase which digests the zona pellucida surrounding the ovum due to which sperm can fertilize the ovum.

Question 4.
Sertoli cells.
Answer:
Sertoli cells provide nourishment and surface to the sperm bundles during their development.

Question 5.
Interstitial cells / Leydig’s cells.
Answer:
Interstitial cells / Leydig’s cells secrete testosterone or androgen which is a male sex hormone.

Question 6.
Prostate gland.
Answer:
Prostate gland secretes prostatic fluid which forms 30% of semen, Citric acid and acid phosphatase present in this fluid protects the sperms from acidic environment of vagina.

Question 7.
Bulbourethral glands.
Answer:
Bulbourethral glands secrete alkaline, viscous mucus like fluid which provides lubrication during copulation.

Question 8.
Bartholin’s glands.
Answer:
Bartholin glands secrete lubricating mucus like fluid which is released in vestibule.

Question 9.
Uterus.
Answer:

1. Uterus receives ovum from fallopian tubes, develops placenta and provides site for implantation of embryo.
2. It provides protection and nourishment to the developing embryo.
3. It also provides path for sperms to ascend.
4. Due to contractions of uterus, baby is expelled out at the time of parturition.

Question 10.
Vagina
Answer:

1. Vagina acts as a copulatory passage.
2. It acts as a birth canal during parturition in normal delivery
3. It provides the passage for menstrual flow.

Name the following

Question 1.
The canal through which the testes descend into scrotum just before birth in human male child.
Answer:
Inguinal canal

Question 2.
The structure where sperms are matured.
Answer:
Epididymis

Question 3.
The part where the sperms are produced in the testes.
Answer:
Germinal epithelium of seminiferous tubules.

Question 4.
The gland in females homologous to Cowper’s gland.
Answer:
Bartholin’s glands or Vestibular glands.

Question 5.
Type of cleavage in human zygote
Answer:
Holoblastic, radial and indeterminate

Question 6.
The developmental stage of human being which gets implanted in the endometrium of uterus.
Answer:
Blastocyst

Question 7.
Name the primates who show presence of menstrual cycle.
Answer:
Human being and Apes like gorilla, chimpanzee, orangutan, etc.

Question 8.
Structures which help in transport of secondary oocyte through uterine tube.
Answer:
Ciliated epithelium

Question 9.
Hormones produced in women only during pregnancy.
Answer:
hCG, HPL (Human Placental Lactogen) and relaxin.

Question 10.
The oral contraceptive pill which is now. a part of the National Family Programme in India.
Answer:
Saheli

Question 11.
The scientific term for the animals giving birth to live young ones.
Answer:
Viviparous

Question 12.
The site of fertilization in woman.
Answer:
Ampulla of fallopian tube

Question 13.
The trophoblast cells lying over the embryonal knob.
Answer:
Cells of Rauber

Question 14.
The muscles which form the wall of scrotum.
Answer:

1. Dartos muscles
2. Cremaster muscles

Question 15.
Names of erectile tissues in penis.
Answer:

1. Corpora cavernosa
2. Corpus spongiosum

Question 16.
Any two copper releasing IUD.
Answer:

1. Copper-T, Cu 7
2. Multiload 375

Question 17.
Any two hormone-releasing IUDs.
Answer:

1. LNG-20
2. Progestaert

Question 18.
Two methods of birth control which have high chances of failure.
Answer:

1. Safe period
2. Lactational amenorrhea

Question 19.
Uterine walls.
Answer:

1. Perimetrium
2. Myometrium
3. Endometrium

Question 20.
Regions of the uterus.
Answer:

1. Fundus
2. Body
3. Cervix

Question 21.
Parts of fallopian tubes.
Answer:

1. Infundibulum
2. Ampulla
3. Isthmus

Question 22.
Layers of Graafian follicle which enclose antrum.
Answer:

1. Theca externa
2. Theca interna
3. Membrana granulosa

Question 23.
Stages of cells in spermatogenesis.
Answer:

1. Spermatogonia
2. Primary spermatocytes
3. Secondary spermatocytes
4. Spermatids
5. Sperms

Question 24.
Stages of cells in oogenesis.
Answer:

1. Oogonia
2. Primary oocytes
3. Secondary oocytes
4. Ootid
5. Ovum

Give significance of the following

Question 1.
Fertilization.
Answer:
Significance of fertilization:

1. Fertilization forms the zygote which eventually produces new offspring.
2. Fertilization restores diploid number of chromosomes in the zygote as two haploid gametes come together in a zygote.
3. During fertilization, centrioles are passed on to the ovum, due to this secondary oocyte can complete meiosis-II. The fertilization thus concludes the process of oogenesis.
4. By fertilization the genetic characters of two parents are mixed. This leads to variation and has significance in evolution.
5. Due to fertilization the sex of young one is determined.

Question 2.
Implantation.
Answer:
Gestation becomes possible due to implantation. Implantation protects the embryo and helps it to derive nourishment from the mother’s body through placenta.

Question 3.
Corpus luteum.
Answer:

1. Corpus luteum is the temporary source of female hormone, progesterone.
2. Corpus luteum is formed from empty Graafian follicle after the process of ovulation.
3. Due to progesterone secreted from corpus luteum, the endometrial wall of uterus undergoes repair and increase in thickness.
4. Progesterone is a gestational hormone and thus pregnancy is maintained if corpus luteum is well functional.

Question 4.
Fertilization membrane.
Answer:
Fertilization membrane prevents any further entry of other sperms into the egg, i.e. polyspermy is avoided.

Question 5.
Gastrulation.
Answer:

1. Due to the process of gastrulation, three germinal layers, viz. ectoderm, mesoderm and endoderm are formed.
2. Cells of embryonal knob become embryonic disc which develop into embryo due to gastrulation.
3. Gastrulation is necessary for the formation of amniotic cavity which is filled with amniotic fluid.

Question 6.
Trophoblast of blastocyst.
Answer:

1. Trophoblast cells help in absorbing nutrition for the developing embryo.
2. Trophoblast cells at the embryonal knob (cells of Rauber) help in implantation of blastocyst.
3. Synctiotrophoblast helps in implantation of fertilized ovum in the uterine endometrium.

Question 7.
hCG [human chorionic gonadotropin].
Answer:
hCG [human chorionic gonadotropin] is secreted in the pregnant female to extend the life of corpus luteum and stimulates its secretory activity. Presence of hCG in maternal blood and urine is an indication of pregnancy.

Question 8.
Colostrum.
Answer:

1. Colostrum is the first milk which is sticky and yellowish secreted by the mammary glands soon after the parturition.
2. Being high protein in its content, it nourishes the newly born child.
3. The antibodies present in it helps in developing resistance for the newborn baby at a time when its own immune response is not fully developed.

Distinguish between the following

Question 1.
Asexual reproduction and Sexual reproduction.
Answer:

Sexual reproduction	Asexual reproduction
1. Asexual reproduction requires single parent.	1. Sexual reproduction needs two different parents.
2. Meiosis does not take place in asexual reproduction. Only mitosis takes place.	2. Sexual reproduction involves meiosis and mitosis.
3. Gamete formation, fertilization and zygote formation does not take place.	3. Gamete formation, fertilization and zygote formation are important processes in sexual reproduction.
4. Progeny and parent Eire identical genetically.	4. Progeny and parents are genetically dissimilar.
5. Large number of progeny is developed by asexual reproduction. E.g. Spore formation, gemmule formation, budding, regeneration are the types of a sexual reproduction.	5. Limited number of progeny is developed by sexual reproduction. E.g. Sexual reproduction is only by a single method.

Question 2.
Primary sex organs and Secondary sex organs.
Answer:

Primary sex organs	Secondary / Accessory sex organs
1. Primary sex organs produce gametes.	1. Secondary sex organs do not produce gametes.
2. Primary sex organs secrete sex hormones.	2. Secondary sex organs do not secrete sex hormones.
3. Development of these organs is under the control of Gonadotropins released from Pituitary. E.g. Testes in male and Ovaries in females.	3. Development of these organs is under the control of estrogen and progesterone in females and testosterone in males. Eg. Prostate, seminal vesicles, vas deferens in males. Fallopian tubes, uterus and vagina in females.

Question 3.
Vasa efferentia and Vasa deferentia.
Answer:

Vasa efferentia	Vasa deferentia
1. Vasa efferentia arise from the rete testes and enter the epididymis.	1. Vasa deferentia arise from the epididymis and form ejaculatory duct after the union with seminal duct.
2. They are present in 15-20 number and are fine convoluted ductules.	2. They are thick and coiled ductules present in a single pair.
3. The spermatozoa are carried from rete testis to epididymis by vasa efferentia.	3. The spermatozoa are carried from epididymis to ejaculatory ducts by vasa deferentia.

Question 4.
Graafian follicle and Corpus luteum.
Answer:

Graafian follicle	Corpus luteum
1. Graafian follicle is produced by the maturation of the primary follicle.	1. Corpus luteum is produced by the cells of ruptured Graafian follicle.
2. It is formed in the ovary before ovulation.	2. It is formed in the ovary after ovulation.
3. It produces the hormone estrogen.	3. It produces the hormone progesterone.
4. It has secondary oocyte surrounded by follicle cells.	4. It has only follicle cells.

Question 5.
Menarche and Menopause.
Answer:

Menarche	Menopause
1. Menarche is the beginning of menstrual cycle.	1. Menopause is the stoppage of menstrual cycle.
2. Menarche is at the age of 10 to 14.	2. Menopause is at the age of 45 to 50.
3. Menarche begins with secretion of FSH and LH.	3. Menopause is caused due to decline of FSH and LH secretion.
4. Menarche is the beginning of the reproductive period.	4. Menopause is the end of the reproductive period.

Question 6.
Proliferative Phase and Secretory Phase.
Answer:

Proliferative Phase	Secretory Phase
1. Proliferative phase begins with the repair of endometrium.	1. Secretory phase begins with ovulation.
2. Time required for proliferative phase is 5th to 13th day of menstrual cycle.	2. Time required for secretory phase is 15th to 28th day of menstrual cycle.
3. Proliferative phase always ends with ovulation.	3. Secretory phase ends with menstruation if egg is not fertilized. It continues further if egg is fertilized.
4. Proliferative phase is in uterus which coincides . with follicular phase in ovary during which there is formation of Graafian follicle.	4. Secretory phase is in uterus which coincides with luteal phase in ovary during which there is formation of corpus luteum.
5. Proliferative phase is controlled by FSH from anterior pituitary.	5. Secretory phase is controlled by LH from anterior pituitary.
6. Hormone estrogen is secreted during this phase.	6. Hormone progesterone is secreted during this phase.
7. It causes the development of blood vessels and thickening of endometrium of uterus.	7. It causes further thickening and secretory activity of the glands of endometrium of uterus.

Question 7.
Spermatogenesis and Oogenesis.
Answer:

Spermatogenesis	Oogenesis
1. Spermatogenesis takes place in testis in mature and fertile males.	1. Oogenesis takes place in ovaries in mature and fertile females.
2. From one spermatogonium four haploid sperms are formed during spermatogenesis.	2. From one oogonium one haploid ovum and a polar body is formed during oogenesis.
3. Spermatid developed undergoes metamorphosis in the process of spermiogenesis.	3. There is no such process of metamorphosis in oogenesis.
4. Spermatid development takes place which later becomes a functional sperm.	4. Ootid development does not take place during oogenesis. It develops only after fertilization.
5. Spermatogonia, primary and secondary spermatocytes and spermatid are the stages of sperms formed during spermatogenesis.	5. Oogonia, primary and secondary oocytes are the stages formed during oogenesis. Ootid formation occurs only after fertilization.

Question 8.
Zona pellucida and Corona radiata.
Answer:

Zona pellucida	Corona radiata
1. Zona pellucida is inner, thin and transparent layer surrounding the secondary oocyte.	1. Corona radiata is the outer thick layer surrounding the secondary oocyte.
2. Zona pellucida is a non-cellular layer.	2. Corona radiata is a cellular layer.
3. Zona pellucida is secreted by the ovum itself.	3. Corona radiata is formed by follicular cells which are glued together by hyaluronic acid.
4. Zona pellucida is retained for more time after fertilization till the ovum gets implanted in the uterus.	4. Corona radiata is retained till the ovum gets fertilized.
5. Zona pellucida is digested by zona lysine or acrosin at the time of fertilization.	5. Corona radiata is digested by hyaluronidase enzyme at the time of fertilization.

Question 9.
Morula and Blastula.
Answer:

Morula	Blastula
1. Morula is the embryonic stage formed after the completion of cleavage.	1. Blastula is the embryonic stage formed after the completion of blastulation.
2. Morula is formed 4 to 6 days after the fertilization.	2. Blastula is formed 6 to 7 days after the fertilization.
3. Morula consists of 16 cells.	3. Blastula consists of more than 64 cells.
4. Morula is solid ball of cells.	4. Blastula is a hollow ball of cells.
5. Morula stage is passed in fallopian tube, once it reaches uterus, it starts developing into the next stage.	5. Blastula after reaching the uterus is implanted on the wall of uterus.
6. Morula does not have any distinction of its inner cell structure.	6. Blastula has a blastocoel, trophoblast and inner cell mass.

Question 10.
Blastula and Gastrula
OR
Give two differences between blastula and gastrula.
Answer:

Blastula	Gastrula
1. Blastula is formed from morula on 7th day after fertilization.	1. Gastrula is formed from blastula 15 days after fertilization.
2. Blastula has a blastocoel.	2. Gastrula has a gastrocoel or archenteron.
3. Blastula is produced by the process of blastulation.	3. Gastrula is produced by the process of gastrulation.
4. Blastula undergoes implantation followed by gastrulation.	4. Gastrula undergoes morphogenesis and then forms germs layers.
5. During blastula formation there is no movement of cells.	5. Gastrula formation results from the morphogenetic movement of cells.

Give reasons

Question 1.
Testes are located outside the body cavity in scrotal sacs.
Answer:

1. During early foetal life, the testes develop in the lumbar region of the abdominal cavity just below the kidney but during seventh month of development, they descend permanently into the respective scrotal sacs through a passage called inguinal canal.
2. For the development of the sperm, lesser temperature than the body temperature is required.
3. If the testes remain in the abdominal cavity, then the sperm production does not take place.
4. This may result in impotency. Therefore, testes are located outside the body cavity.

Question 2.
Urethra is also called urinogenital duct in males.
Answer:

1. Urinogenital duct means common duct for urine and the genital products.
2. In males, the penis lodges urethra throughout its entire length, through which urine as well as semen are given out of the body during urination or copulation.
3. Since the urethra carries both urine and semen, it is called urinogenital duct.

Question 3.
Proliferative phase is also called follicular phase.
Answer:

1. Proliferative phase means there is proliferation of endometrial cells in the uterus. Follicular means there is growth of ovarian follicles in the ovaries. Both these phases are simultaneous.
2. The follicular phase of ovaries is due to effect of FSH from adenohypophysis.
3. The ovaries follicles grow due to FSH and start secreting estrogen.
4. This estrogen from ovaries bring proliferative effect on the uterus.

Question 4.
Missing of menses is the first indication of pregnancy.
Answer:

1. Menstruation occurs if there is no fertilization of ovum.
2. The endometrium of uterus along with unfertilized egg is given out in the form of menstrual flow.
3. The sloughing off uterine endometrium takes place due to degeneration of corpus luteum.
4. In the absence of functional corpus luteum progesterone levels fall down. However, if the ovum is fertilized, the corpus luteum is maintained and it secretes progesterone which maintains the uterine endometrium. In such case, further growth of ovarian follicles and ovulation remains suspended and woman is said to be pregnant.
5. Endometrial wall of uterus now thickens and helps in the growth of placenta. Thus during pregnancy, menses will not take place.

Question 5.
Progesterone is called pregnancy hormone.
Answer:

1. Progesterone is secreted from corpus luteum which is formed from empty ovarian follicle after the ovulation.
2. Progesterone has the capacity to maintain pregnancy.
3. It acts on uterine endometrium and causes it to proliferate and develop in thickness.
4. Corpus luteum keeps on secreting progesterone till the placenta takes up the function of secreting the same.

Question 6.
Human female has restricted reproductive life.
Answer:

1. In human female, the reproductive period is about 30 – 33 years.
2. There is menarche at the age of about 13 and menopause at the age of 45-50.
3. During this span of 30 years, ovaries secrete sex hormones like estrogen and progesterone. After menopause this secretion is suspended.
4. Due to changes in hormonal level, human females cannot produce eggs later. Moreover, eggs in her ovaries are utilized by the age of 45.
5. Human female, therefore, has restricted reproductive period.

Question 7.
Zona pellucid is retained for sometime after fertilization.
Answer:

1. Fertilization of the ovum takes place in fallopian tube where it starts cleavages immediately.
2. Zona pellucida which remains on the surface of the ovum prevents the implantation of the blastocyst at an abnormal site such as fallopian tube.
3. Zona pellucida keeps the sticky and phagocytic trophoblast cells unexposed till the ovum reaches the uterine lumen.
4. Zona pellucida also protects the ovum. Therefore zona pellucida is retained for some time after fertilization.

Question 8.
The acrosome of sperm secretes an enzyme called hyaluronidase at the time of fertilization.
Answer:

1. The enzyme hyaluronidase secreted by acrosome of sperm dissolves the membranous covering of the ovum to facilitate the entry of sperm into the ovum.
2. It is a lytic enzyme causing lysis of egg membrane.
3. Owing to this, the acrosome of sperm secretes an enzyme called hyaluronidase at the time of fertilization.

Question 9.
The middle part of the human sperm is characterized by the presence of a number of mitochondria.
Answer:

1. Mitochondria provide energy required by sperms for their agile movement.
2. The agile movement of sperms helps them to reach the vicinity of the ovum at the time of fertilization.
3. Owing to this, the middle part of the human sperm is characterised by the presence of a number of mitochondria.

Question 10.
The size of morula remains almost same as that of ovum.
Answer:

1. The layer zone pellucida is retained around the embryo and thus, there is no change in the overall size from zygote to morula.
2. This layer is important because it prevents the implantation of the blastocyst at an abnormal site such as fallopian tube.
3. Though the number of blastomeres increase, the size of morula remains almost same as that of ovum till it reaches the uterus by the end of the day 4.

Question 11.
Placenta serves as the nutritive, respiratory and excretory organ of the embryo.
Answer:

1. Between the foetus and mother there is exchange of several materials. Food in the form of glucose, amino acids, simple proteins, lipids, mineral, salts, vitamins and hormones, antibodies, etc. is sent to foetus by maternal circulation.
2. Oxygen from mother’s blood is also given to the foetus.
3. The foetal metabolic wastes such as carbon dioxide, urea and water pass from foetus into the maternal blood.
4. This exchange takes place through the placenta.
5. In the placenta, foetal blood comes very close to maternal blood to permit these exchanges. Therefore placenta is said to serve as the nutritive, respiratory and excretory organ of the embryo.

Write short notes

Question 1.
Graafian follicle.
Answer:

1. Graafian follicle is a mature ovarian follicle.
2. There are following protective layers on the Graffian follicle : The outermost protective and fibrous covering, theca externa. Theca interna is the next layer which can secrete hormone estrogen.
3. Next to theca interna, there is membrana granulosa which forms discus proligerous and the corona radiata layer.
4. Graafian follicle contains an eccentric secondary oocyte. The oocyte is surrounded by a vitelline membrane which produces zona pellucida layer.
5. In the centre there is antrum which is filled with liquor folliculi fluid.

Question 2.
Mammary glands.
Answer:

1. Mammary glands are accessory organs of female reproductive system. These glands are essential for lactation after parturition.
2. They are modified sweat glands present in the subcutaneous tissue of the anterior thorax. They are in the pectoral region in the location between 2nd to 6th rib.
3. Each mammary gland consists of fatty connective tissue and many lactiferous ducts.
4. Each breast has glandular tissue which is divided into 15-20 irregularly shaped mammary lobes. Each lobe has an alveolar glands and lactiferous duct.
5. Milk is secreted by alveolar glands and it is stored in the lumen of alveoli. The alveoli open into mammary tubules and these in turn forms a mammary duct.
6. All the lactiferous ducts converge towards the nipple.
7. Nipple is surrounded by a dark brown coloured and circular area of the skin called areola.

Question 3.
Structure of sperm.
Answer:

1. Sperm is microscopic, elongated haploid motile male gamete produced by spermatogenesis.
2. It measures to about 0.055 mm or 60y in length.
3. The sperm consists of head, neck, middle piece and tail.

Head:

1. Head is the main part which is flat and oval and has a large nucleus and an acrosome.
2. Acrosome is formed from Golgi complex. It secretes enzyme hyaluronidase which helps in penetration of the egg during fertilization.
3. The acrosome and anterior half of nucleus is covered by a fibrillar sheath.

Neck : Neck is short region having two centrioles.

1. The proximal centriole plays a role in first cleavage of zygote.
2. The axial filament of the sperm is formed by the distal centriole.

Middle piece:

1. Middle piece acts as a power house for sperm.
2. It bears many spirally coiled mitochondria or Nebenkern around the axial filament.
3. The mitochondria supply energy for the sperm to swim in the female genital tract with a speed of about 1.5 to 3 mm per minute.
4. Posterior half of nucleus, neck and middle piece of sperm are covered by a sheath.

Tail:

1. The tail is formed of cytoplasm and is long, slender and tapering structure.
2. The axial filament is a fine thread-like structure that arises from the distal centriole and traverses the middle piece and tail.
3. Nine accessory fibres are present surrounding the two central longitudinal axial filaments.
4. Tail lashes and helps the spermatozoa to swim.

Question 4.
Structure of secondary oocyte.
Answer:

1. The unfertilized egg released through ovary at the time of ovulation is a secondary oocyte.
2. It is rounded, non-motile and haploid, non- cleidoic and microlecithal female gamete.
3. The size is approximately 0.1 mm (100 microns).
4. It has abundant cytoplasm called ooplasm which contains a large eccentric and prominent nucleus called germinal vesicle.
5. Centrioles are absent in secondary oocyte.
6. Various coverings seen around the oocyte are (i) vitelline membrane (ii) zona pellucida (iii) Corona radiata.
7. The cells are glued together with hyaluronic acid. Between vitelline membrane and zona pellucida, there is perivitelline space which lodges first polar body. This end is called 5 animal pole and the opposite is called vegetal pole.

Question 5.
Implantation
Answer:

1. Implantation is the process by which the blastocyst is embedded into the endometrium of uterus in the fundus region.
2. The implantation starts 7 days after fertilization and completed by the end of 10th day.
3. The trophoblast cells of blastocyst at the embryonal knob can stick to the uterine endometrium. The trophoblast layer then divides into inner cytotrophoblast and outer syncytiotrophoblast due to contact with endometrial cells.
4. Cytotrophoblast is the inner layer whose cells retain their cell boundaries.
5. Syncytiotrophoblast is the outer layer of cells without plasma membrane. The cells of syncytiotrophoblast appear multinucleate. This layer projects invasively into the endometrium and destroys endometrial cells by releasing lytic enzymes. Due to this blastocyst is buried deeply in the endometrium.

Question 6.
Fate of three germinal layers.
Answer:
Fate of germinal layers : The embryo after gastrulation develops the three germ layers, viz., ectoderm, mesoderm and endoderm. Later a process of histogenesis starts which leads to the development of different tissues and organs.
(i) Fate of ectoderm : Following tissues, structures and organs develop from the ectoderm : Epidermis of the skin, epidermal derivatives such as

1. hair and nails
2. sweat glands
3. conjunctiva
4. cornea
5. lens
6. retina
7. internal and external ear
8. enamel of teeth
9. nasal cavity
10. adrenal medulla
11. stomodaeum and proctodaeum
12. neurohypophysis and
13. entire nervous system.

(ii) Fate of mesoderm : The mesoderm forms the following derivatives:

1. All types of muscles
2. connective tissue
3. dermis of skin
4. adrenal cortex
5. kidney
6. circulatory system
7. heart
8. blood vessels
9. blood
10. lymphatic vessels
11. middle ear and
12. dentine of teeth.

(iii) Fate of endoderm : The following organs develop from the endoderm:

1. Epithelium of gut from pharynx to colon
2. glands of stomach and intestine
3. tongue and tonsils
4. lungs, trachea, bronchi, larynx, etc.
5. urinary bladder, vestibule and vagina
6. liver and pancreas
7. adenohypophysis
8. thymus, thyroid and parathyroid
9. eustachian tube
10. epithelium of urethra and associated glands.

Question 7.
Placenta.
Answer:

1. Placenta is a temporary organ derived from the tissues of the foetus as well as mother.
2. Human placenta is called chorionic placenta as it is made up of chorion which is an extra-embryonic membrane.
3. Blood vessels from the allantois vascularize the placenta. Branching villi emerge from the chorion and penetrate in the corresponding pits which are located in the uterine wall.
4. There are two parts of placenta, viz. foetal placenta and maternal placenta.
5. Foetal placenta is formed of chorionic villi.
6. Maternal placenta is formed of uterine wall which is in intimate contact with the chorionic villi.
7. Chorionic villi receive the blood from the embryo by umbilical artery. Umbilical vein returns the blood back to the embryo.
8. Human placenta is said to be haemochorial because a part of placenta is from foetus which has chorionic villi. The other highly vascularized part is from uterine wall of mother. Thus foetal and maternal placenta together is called haemocorial placenta.

Question 8.
Intratuterine devices (IUDs).
Answer:

1. IUDs are plastic or metal objects which act as contraceptive devices. They are placed into the uterus by a doctor or trained nurse.
2. E.g. Lippe’s loop, copper releasing IUDs (Cu-T, Cu-7, multiload 375) and hormone releasing IUDs (LNG-20, Progestaert).
3. Plastic double ‘S’ loop is called Lippe’s loop which stimulates accumulation of macrophages in the uterine cavity by attracting them. As phagocytosis increases the sperms are destroyed. Thus it acts as a contraceptive.
4. Copper releasing IUDs suppress sperm motility and the fertilizing capacity of sperms.
5. The hormone releasing IUDs make the : uterus unsuitable for implantation and ; cervix hostile to the sperms.
6. Their presence in the uterus acts as a minor irritant and thus makes the ovum to move quickly out of the body.
7. However, IUD can cause infection and occasional haemorrhage. It can cause discomfort for woman and may get spontaneously expelled out.

Question 9.
Physiological (Oral) Contraceptive « Devices.
Answer:

1. Physiological devices are in the form of oral contraceptive pills or birth control pills. 5 They are hormonal preparations and check ovulation by inhibiting the secretion of follicle stimulating hormone and luteinizing hormone.
2. Woman who is using pills does not release ovum at the time of ovulation and therefore conception does not occur.
3. Birth control pills have side effects such as nausea, breast tenderness, weight gain and ‘break through’ bleeding, i.e. slight bleeding between the menstrual periods. These health hazards are due to synthetic hormones.
4. These pills also alter the quality of cervical J mucus to prevent the entry of sperms.
5. The birth control pills contain progesterone and estrogen. Mala-D to be taken daily and Saheli to be taken weekly are two common birth control pills in India. These pills are non-steroidal.

Question 10.
Fate of trophoblast cells of blastocyst.
Answer:

1. Trophoblast cells do not form any part of the embryo proper.
2. They form ectoderm of the extra-embryonic membrane called chorion.
3. Chorion helps in supply of oxygen and nutrients to foetus from mother’s body. CO₂ and nitrogenous wastes are collected from foetus and passed in mother’s blood.
4. Thus, these cells have an important role in formation of placenta.

Question 11.
Medical Termination of Pregnancy (MTP).
Answer:

1. MTP or Medical Termination of Pregnancy is voluntary termination of pregnancy under medical supervision. It is an induced abortion.
2. Only during first trimester, MTP is safe for mother’s health.
3. Upon amniocentesis examination, if abnormality is detected, usually MTP is performed.
4. Government of India has legalized MTP There was MTP Act in 1971, which was later amended in 2017, to prevent its misuse, especially female foeticide should never be done through MTP
5. As per MTP Act, the procedure can be done only in first 12 weeks and never after 20 weeks of pregnancy.

Question 12.
Amniocentesis
Answer:

1. In amniocentesis, amniotic fluid containing foetal cells is collected using a hollow needle. This needle is inserted into the uterus of pregnant mother, under ultrasound guidance.
2. The chromosomes from the foetal cells are sujected to karyotyping. This helps to detect abnormalities in the developing foetus.
3. Amniocentesis is misused to determine the sex of the unborn child. This is illegal in India because it results into female foeticide.
4. Another risks involved in amniocentesis are miscarriage, needle injury to foetus, leaking amniotic fluid, infection, etc.
5. As per MTP Act (1971) the misuse of amniocentesis is curtailed.

Question 13.
ZIFT [Zygote Intra Fallopian Transfer].
Answer:

1. If there is a blockage in the fallopian tubes due to which fertilization is prevented, then ZIFT treatment is used.
2. The oocyte is removed form woman’s ovary. This oocyte is fertilized outside the body under sterile conditions with the known sperms. This forms zygote. This is In Vitro Fertilization (IVF).
3. Later the zygote is transferred in fallopian tube to achieve pregnancy.

Question 14.
GIFT [Gamete Intra Fallopian Transfer].
Answer:

1. When the oocyte is collected from donor and transferred into the fallopian tube of another female, the technique is called GIFT. This female provides suitable environment for further development.
2. When the entrance or upper segments of the fallopian tubes is blocked, this technique is used.
3. Ooocytes and sperms are directly injected into regions of the fallopian tubes. Here fertilization takes place forming a blastocyst. It later enters the uterus for implantation.
4. GIFT is successful in only 30 per cent cases.

Question 15.
Sterilization operations.
Answer:

1. Sterilization operations are the permanent means for the birth control. These can be performed on both the sexes. Usually these are performed after the couple does not desire another child.
2. These surgical interventions block the gamete transport and thus prevents the pregnancy.
3. Sterilization operation in males is called vasectomy while in females it is called tubectomy.
4. In vasectomy the vas deferens are tied and cut. In tubectomy fallopian tubes are ligated or cut.

Question 16.
Gonorrhoea.
Answer:
(1) Gonorrhoea is a sexually transmitted veneral disease caused by Diplococcus bacterium, Neisseria gonorrhoea.

(2) The incubation period is 2 to 14 days in males and 7 to 21 days in females.

(3) Infection sites are mucous membrane of urino-genital tract, rectum, throat and eye.

(4) Males show following symptoms : Partial blockage of urethra and reproductive ducts, pus from penis, pain and burning sensation : during urination, arthritis, etc.

(5) Symptoms in female include, pelvic inflammation of urinary tract, sterility, arthritis. The children born to affected mother suffer from gonococcal ophthalmia, In girl-child, there is occurrence of gonococcal vulvovaginitis before puberty.

(6) Preventive measures for gonorrhoea are as follows:

• Sexual hygiene
• Use of condom during coitus.
• Sex with unknown partner or multiple partners should be avoided.

(7) Gonorrhoea can be treated with Cefixime which is antibiotic.

Short-answer Questions

Question 1.
What are sexual dimorphic characters? Enlist these characters in human male and female.
Answer:
Sexual dimorphism is the phenomena in which the sexes of the individual can be identified externally. In human beings, even in infancy there is sexual dimorphism, by which one can identify the sex of the infant.

But when the male or female reaches puberty, then secondary sexual characters are developed due to sex hormones. These characters are called sexual dimorphic characters.
(i) Secondary sexual characters in males:

1. Presence of beard, Moustache.
2. Hair on the Chest, Axillary and Pubic Region.
3. Muscular body.
4. Enlarged larynx (Adam’s apple).

(ii) Secondary sexual characters in females:

1. Breast development.
2. Broadening of pelvis.
3. High pitched voice.

Question 2.
Describe the duct system that transports the sperms from seminiferous tubules to the exterior.
Answer:
(1) All the seminiferous tubules present in the testis show posterior network of tubules called rete testis. Vasa efferentia are the fine tubules which are 12-20 in number, are seen arising from rete testis. From testis to epididymis, the sperm transport is done by vasa efferentia.

(2) Epididymis has three parts, caput, corpus and cauda epididymis. In this long and highly coiled tube sperms undergo physiological maturation.

(3) Then from here sperms enter into vas deferens, which is a tube that arise from epididymis enters the abdominal cavity. On its course, later it joins the duct of seminal vesicle. Both together form the ejaculatory duct.

(4) Ejaculatory duct passes through the prostate gland and then opens into the urethra. Urethra is a common passage for urine and semen and hence it is also called urinogenital duct.

(5) Urethra passes through penis and opens to the outside by an opening called the urethral meatus or urethral orifice.

(6) Thus sperms are transported through vas deferens into urethra via ejaculatory duct and then to the outside through urethral orifice.

Question 3.
What is semen? Describe the composition of semen.
Answer:
(1) Semen is the viscous, alkaline and milky fluid having pH 7.2 to 7.7 ejaculated during sexual intercourse by male.

(2) A single ejaculation of semen i.e. 2.5 to 4 ml semen contains about 400 millions of sperms.

(3) Semen consists of sperms suspended in secretions of the epididymis and the accessory glands (seminal vesicles, prostate gland and Cowper’s gland). The semen nourishes the sperms by fructose, neutralizes acidity by Ca⁺⁺, ions and bicarbonates and also activates them for movement due to prostaglandins.

Question 4.
Describe in detail the external genitalia of human female reproductive system.
Answer:
The external genital organs of female are located external to the vagina. They have collective name, ‘vulva’ or pudendum. Following are the parts of vulva.
(1) Labia majora : Labia majora are homologous to scrotum of males. They are two large folds which form the boundary of the vulva. They are composed of skin, fibrous tissue and fat. These Eire prominent and longitudinal folds on right and left sides of the vestibule.

(2) Labia minora : Smaller and thinner lip-like folds located just medially are labia minora. Posteriorly the labia minora are fused together to form the fourchette.

(3) Mons veneris : Mons veneris is fleshy elevation above the labia majora.

(4) Clitoris : It is present at the anterior end of the labia minora. It shows the presence of erectile tissues.

(5) Vestibule : Vestibule is a median vertical depression of vulva enclosing vagina and urethral opening.

(6) Hymen : Hymen is a thin layer of mucous membrane which partially occludes the opening of the vagina.

(7) Vestibular glands:

1. Vestibular glands or Bartholin’s glands are homologous to the Cowper’s glands of the male.
2. These are paired glands situated on either side of the vaginal opening, secreting lubricating fluid.

Question 5.
How is puberty attained in females? Will a female normally remain reproductively capable even after age 50? If not then what makes her incapable?
Answer:
(1) Puberty is achieved due to gonadotropins such as FSH and LH secreted by the anterior pituitary. These hormones stimulate the ovaries. The ovaries in turn produce estrogen and progesterone, which brings about secondary sexual characters in female. Thus she attains the puberty. The beginning of menstrual cycle or menarche takes place due to these hormonal changes at about 10 to 14 years.

(2) But the women do not remain reproductively active after the age of 50 due to hormonal imbalance. This is called menopause or cessation of reproductive cycles. Absence of enough gonadotropins and unresponsive ovarian cells cause menopause at 45 to 50 years of age.

Question 6.
Why is menstruation painful in some women?
Answer:

1. The menstruation is painful in some women as the muscles in the uterus contract or tighten.
2. Women who experience painful periods can have higher levels of prostaglandins, a natural body chemical that causes contractions of the uterus and blood vessels.
3. Some women have a build-up of prostaglandins which means they experience stronger contractions and therefore due to spasmodic pain in some women menstruation is more painful.
4. Endometrial sloughing that takes at the time of menstruation also causes painful discomfort.

Question 7.
Why is it said that consumption of mother’s milk is safety for the newborn?
Answer:
Consumption of mother’s milk is safety for the newborn because of the following reasons:

1. Mother’s milk is the perfect food for babies in the first months of their lives. With the exception of vitamin D, it contains all the nutrients an infant needs.
2. Mother’s milk supplies antibodies [IgA] that protect the baby’s body organs from infections. Mother’s milk provides immunity and also help in maturation of the infant’s immune system which are lacking in ordinary milk. Natural acquired passive immunity is obtained only through mother’s milk.
3. Feeding of mother’s milk reduces the risk of overweight and obesity during childhood.
4. It also creates the bond between mother and child.

Question 8.
Which hygienic practices should be followed by the female during menstruation ?
Answer:
The following personal hygienic practices should be followed by the female during menstruation:

1. Keeping the pubic area clean.
2. Changing the sanitary napkin every 4-5 hours.
3. Reducing risk of infections by maintaining hygiene.
4. Proper disposal of soiled sanitary napkin.
5. Not to use damp and dirty clothes which can cause infections and bad odour. A sanitary napkin which is not changed in time can act as a perfect environment for rapid growth of infectious bacteria.

Question 9.
How can the goals of RCH be achieved?
Answer:
The goads of RCH can be achieved by the following ways:

1. Sex education in schools is introduced. Proper and scientific information about sexual organs and safe sexual acts should be given to students. They should be made aware of sexually transmitted diseases (STD, AIDS), and problems related to adolescence.
2. Audio-visual and the print media should be used by government and non-government organisations for creating awareness about reproductive health.
3. Younger generation should be educated about family planning measures, pre-natal and post-natal care of women and care of infant with knowledge about importance of breastfeeding.
4. Awareness should be spread about problems arising due to uncontrolled population growth, sex abuse and sex related crimes. Necessary steps to prevent these to be taken.
5. Statutory ban on amniocentesis for sex determination is practised. This should be known by all.
6. Details of child immunization programmes should be understood.
7. New parents should get the training for new born care so that infant and maternal mortality rate can be reduced.

Question 10.
How do addictions like smoking, alcoholism and drug abuse contribute in causing infertility in men?
Answer:

1. Tobacco, marijuana and other drugs, smoking may cause infertility in both men and women.
2. Nicotine blocks the production of sperm and decreases the size of testicles.
3. Alcoholism by men interferes with the synthesis of testosterone and has an impact on sperm count.
4. Use of cocaine or marijuana may temporarily reduce the number and quality of sperm.

Question 11.
Jayesh, a young married man of 26 years is suffering from T.B. for the last 2 years. He and his wife are desirous of a child but unable to have one, what could be the possible reason? Explain.
Answer:
Jayesh, though young, is suffering from TB for last 2 years. His wife is unable to conceive the child may be due to following reasons:

1. Tuberculosis disrupts sexual and reproductive function in patients.
2. Moreover T.B. patients have to take not less than 4 anti-TB drugs simultaneously for a long time.
3. These are basically a very high dose antibiotics which may hamper formation of sperms. In this way the anti-tuberculosis drugs may negatively influence on sexual function.
4. Pulmonary TB patient shows, deterioration of all parameters of copulatory act, from sexual desire to orgasm and thus the couple is unable to conceive.
5. Infertility is one of the most common symptoms of genital tuberculosis.

Question 12.
Neeta is 45 years old and the doctor advised her not to go for such a late pregnancy. She however wants to be the biological mother of a child without herself getting pregnant. Is this possible and how?
Answer:
(1) Neeta being 45 years old, she is approaching menopause. Therefore, she will be advised by the doctor to take the help of the modern remedial technique called surrogacy.

(2) In this technique the embryo is formed using intended father’s sperm and intended mother’s egg by In Vitro Fertilization (IVF) technique and then that embryo is implanted in a surrogate mother, sometimes called a gestational carrier.

(3) In surrogacy there is legal arrangement where the surrogate mother agrees to bear child for a couple. Remains pregnant with all the care and nourishment. Later she delivers a baby and hands it over to biological mother.

Chart based /Table based questions

Question 1.
Complete the following chart and rewrite

Female Reproductive Organs	Homology to Male Reproductive Organs
1. Labia majora	————–
2. ————-	Bulbourethral glands/ Cowper’s gland
3. Clitoris	—————

Answer:

Female Reproductive Organs	Homology to Male Reproductive Organs
1. Labia majora	Scrotum
2. Bartholin’s gland/ Vestibular gland	Bulbourethral glands/ Cowper’s gland
3. Clitoris	Penis

Question 2.

Hormones	Functions
1. Testosterone	————–
2. ————-	Stimulates contractions uterine during parturition
3. Progesterone	—————

Answer:

Hormones	Functions
1. Testosterone	Stimulates spermatogenesis
2. Oxytocin	Stimulates contractions uterine during parturition
3. Progesterone	Maintain endometrium of uterus during secretory phase and gestation.

Question 3.
Complete the following chart and rewrite

Hormones	Functions
1. Rapid regeneration of endometrium and maturation of Graafian follicle	————–
2. Secretion of endometrial glands and increased secretion of progesterone	————–
3. Breakdown of endometrium in absence of fertilization	————-

Answer:

Hormones	Functions
1. Rapid regeneration of endometrium and maturation of Graafian follicle	Proliferative phase / Follicular phase
2. Secretion of endometrial glands and increased secretion of progesterone	Secretory phase / Luteal phase
3. Breakdown of endometrium in absence of fertilization	Menstrual phase

Diagram based questions

Question 1.
Sketch and label Human male reproductive system.
Answer:

Question 2.
Label the given male reproductive system you have studied.

Answer:

1. Seminal vesicle
2. Ejaculatory duct
3. Cowper’s glands
4. Urethra
5. Epididymis
6. Testis
7. Urinary bladder
8. Prostate gland
9. Vas deferens
10. Penis

Question 3.
Sketch and label human female reproductive system.
Answer:

Question 4.
Give labels to given diagram of female reproductive system.

Answer:

1. Fallopian tube
2. Fundus of Uterus
3. Ampulla of fallopian tube
4. Ovarian ligament
5. Uterus
6. Ovary
7. Infundibulum with fimbriae
8. Endometrium of uterus
9. Cervix
10. Vagina

Question 5.
Sketch and label Seminiferous tubules as seen in T.S. of testis.
Answer:

Question 6.
Identify ‘A’ and ‘B’ in the diagram below and mention their functions.

Answer:
(1) A : Seminiferous tubule
Function : Seminiferous tubules produce sperms by spermatogenesis.

(2) B : Vas deferens
Function : Vas deferens carry sperm from epididymis to ejaculatory duct.

Question 7.
Sketch and label – T.S. of ovary.
Answer:

Question 8.
Sketch and label sectional view of mammary gland.
Answer:

Question 9.
Sketch and label – Graafian follicle.
Answer:

Question 10.
Sketch and label – Process spermatogenesis.
Answer:

Question 11.
Sketch and label process of oogenesis.
Answer:

Question 12.
Give the name and functions of ‘A’ and ‘B’ from the diagram given below

Answer:
(1) A is acrosome.
Function of acrosome : Acrosome produces lytic enzyme, hyalourinidase and thus helps in the penetration of the egg during fertilization.

(2) B is tail of the human sperm.
Function of tail : Tail lashes continuously and helps the movement of the sperm in the female genital tract.

Question 13.
The diagram represents a surgical sterilization method in males. Study the same and answer the questions that follow

1. Give the name of the surgical method represented in the diagram.
2. Which part is ligated or cut.
3. Name the corresponding surgical method conducted in females.
4. Name the part which is ligated in females and why?

Answer:

1. Vasectomy
2. Vas deferens
3. Tubectomy
4. Fallopian tubes are ligated so that the egg may not meet with the sperms.

Question 14.
Given below is the figure of an important structure developed during pregnancy.

1. Name the structure and its type.
2. Identify ‘A’. In which technique it is used.
3. Identify ‘B’ What is its function?

Answer:
(1) The given figure is placenta. The type of placenta in humans is haemochorial placenta.

(2) A is amnio tic fluid. Amnio tic fluid is withdrawn in amniocentesis technique. From this fluid foetal cells can be obtained, which are examined for any chromosomal abnormality by karyotyping.

(3) B is umbilical cord. This is the connection between placenta of mother and growing foetus. Through the umbilical cord, foetus gets nutrition and oxygen. Nitrogenous wastes and carbon dioxide is collected from foetus and brought into maternal circulation.

Question 15.
The diagram given below is that of a intra-uterine contraceptive device. Study the same and then answer the questions that follows:

1. Give the name of intra-uterine contraceptive device shown in the diagram.
2. What is its mode of action?

Answer:

1. The Intra-uterine contraceptive device shown in the diagram is Lippes loop.
2. It is a plastic double ‘S’ loop. It attracts the macrophages stimulating them to accumulate in the uterine cavity. Macrophages increase phagocytosis of sperms within the uterus and acts as a contraceptive.

Question 16.
Identify A in the given diagram. Write the function of the same.

Answer:
A in the above diagram is intrauterine device or IUD. It is a contraceptive device inserted in the uterus of woman. This is a hormone releasing IUD. It acts as a mechanical means of contraception and avoids pregnancy.

Long answer questions

Question 1.
With the help of diagrammatic representation, explain the process of spermatogenesis.
Answer:

(1) The process of spermatogenesis takes place in the male gonads or testis. The cells of germinal epithelium that line the seminiferous tubules undergo spermatogenesis.

(2) Primordial germ cells or germinal cells pass through three phases, viz. phase of multiplication, phase of growth and phase of maturation.

• Multiplication phase : Primordial germ cells undergo mitotic divisions to produce many diploid (2n) spermatogonia.
• Growth phase : Spermatogonium accumulates nutrients and grows in size, giving rise to primary spermatocyte (2n).
• Maturation phase : The primary spermatocyte undergoes first meiotic division or maturation division. Exchange of genetic material occurs between homologous chromosomes in each spermatocyte.

(3) The meiotic division gives rise to secondary spermatocyte which is haploid (n). At the end of first meiotic division two secondary spermatocytes are formed while at the end of second meiotic division four haploid spermatids are formed.

(4) Spermatids are non-motile. They undergo spermiogenesis and form motile spermatozoan (sperm).

(5) The changes taking place during spermiogeneis are as follows:

• Increase in length.
• Formation of proximal and distal centriole.
• Distal centriole forms the axial filament.
• Mitochondria become spirally coiled.
• Acrosome is formed from Golgi complex.

Question 2.
What is oogenesis? Describe it briefly.
Answer:

1. Oogenesis is the process of formation of the haploid female gamete, i.e. ovum.

2. The process of oogenesis takes place in the follicular cells inside the ovaries. The germinal epithelium cells undergo oogenesis.

3. They pass through three phases, viz. phase of multiplication, phase of growth and phase of maturation at the time of oogenesis.

• Multiplication phase : Germinal cells undergo mitotic divisions and produce large number of diploid (2n) oogonia. Oogonia are present in the ovaries of female even before she is born.
• Growth phase : During puberty changes, the FSH from pituitary makes one oogonium to develop at a time. The growth takes place as the follicle matures and larger primary oocyte (2n) is produced inside the Graafian follicle.
• Maturation phase : The primary oocyte undergoes first meiotic division. There are equal nuclear divisions during meiosis but the cytoplasm is unequally divided.

4. By the end of first meiotic division, larger haploid secondary oocyte and smaller haploid polar body are produced. Since the embryo develops from the egg, there is provision for more food in the secondary oocyte.

(5) The second meiotic division takes place in the secondary oocyte and polar body. But this division is arrested during metaphase.

(6) The secondary oocyte is released from the ovary in the process of ovulation. Remaining division takes place if and only if ovum is fertilized.

(7) The division is unequal and form functional female gamete or ovum at the time of fertilization.

Question 3.
What is gastrulation? What are the changes that are brought about by gastrulation?
Answer:
(1) Gastrulation : The process of formation of three germ layers by morphogenetic movements and rearrangements of the cells in blastula leading to the formation of gastrula is known as gastrulation.

(2) Cells on the free end of inner cell mass called hypoblasts (primitive endoderm) become flat, divide and grow towards the blastocoel to form endoderm.

(3) Endodermal cells grow within the blastocoel to form a Yolk sac.

(4) The remaining cell of the inner cell mass, in contact with cells of Rauber are called epiblasts (primary ectoderm) which further differentiate to form ectoderm.

(5) Cells of ectoderm divide and re-divide and move in such a way that they enclose the amniotic cavity. The floor of this cavity has the embryonal disc while roof is lined by amniogenic cells. Amnion is an extra embryonic membrane that surrounds and protects the embryo.

(6) Actual gastrulation occurs about days after fertilization.

(7) Trilaminar embryonic disc begins with the formation of primitive streak and a shallow groove on the surface called primitive groove. From the site of primitive streak, a third layer of cells called mesoderm extends between ectoderm and endoderm. Anterior end of the primitive groove communicates with yolk sac by an aperture called blastopore (future anus).

(8) The embryonal knob thus finally differentiate into three layers – ectoderm, mesoderm and endoderm.

Question 4.
Explain the major changes taking place during the three trimesters of pregnancy in woman.
Answer:
The pregnancy period of approximately nine months (280 days) is divided into three trimesters of three months each.
1. First Trimester : (From fertilization to 12th week)

• During first trimester there are radical changes in the body of mother as well as in the embryo.
• The embryo receives nutrients in the first 2-4 weeks directly from the endometrium.
• It is the main period of organogenesis and the development of body organs.
• By the end of eight weeks, the major structures found in the adult are formed in the embryo in a rudimentary form. It is now called foetus and is about 3 cm long.
• Arms, hands, fingers, feet, toes, CNS, excretory and circulatory system including heart are formed and begins to work.
• Progesterone level becomes high and menstrual cycle is suspended till the end of pregnancy.
• At the end of first trimester foetus is about 7-10 cm long.
• The maternal part of placenta grows, the uterus becomes larger. In this period, the mother experiences morning sickness, (nausea, vomiting, mood swings, etc.)

2. Second Trimester: (From 13th to 26th week)

• The foetus is very active and grows to about 30 cm.
• The uterus grows enough for the pregnancy to become obvious.
• Hormone levels stabilize as hCG declines, the corpus luteum deteriorates and the placenta completely takes over the production of progesterone which maintains the pregnancy.
• Head has hair, eyebrows and eyelashes appear, pinnae are distinct. Baby’s movement can be easily felt by the mother.
• The baby reaches half the size of a new born.

3. Third Trimester: (From 27th week till the parturition)

• Foetus grows to about 50 cm in length and about 3-4 kg in weight.
• As the foetus grows, the uterus expands around it, the mother’s abdominal organs become compressed and displaced, leading to frequent urination, digestive blockages and strain in the back muscles.
• At the end of third trimester the foetus becomes fully developed and ready for parturition.

Balbharti Maharashtra State Board 11th Biology Important Questions Chapter 15 Excretion and Osmoregulation Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 15 Excretion and Osmoregulation

Question 1.
What are metabolic waste products?
Answer:
Metabolism produces a variety of by-products, some of which need to be eliminated. Such by-products are called metabolic waste products.

Question 2.
Define excretion.
Answer:
The process of eliminating waste products from the body is called excretion.

Question 3.
Where are metabolic wastes produced?
Answer:
Metabolic wastes are produced inside body cells.

Question 4.
Enlist the various excretory products produced in the human body.
Answer:
The various excretory products produced by the human body are as follows:

1. Fluids such as water; gaseous wastes like CO₂; nitrogenous wastes like ammonia, urea and uric acid, creatinine; minerals; salts of sodium, potassium, calcium, etc. if present in body in excess are excreted through urine, faeces and sweat.
2. Pigments formed due to breakdown of haemoglobin like bilirubin (excreted through faeces) and urochrome (eliminated through urine).
3. The pigments present in consumed foodstuffs like beet root or excess of vitamins, hormones and drugs.
4. Volatile substances present in spices (eliminated through lungs).

Question 5.
Write a note on deamination.
Answer:

1. Deamination is the process of breakdown of excess amino acids.
2. It is an essential process, since the body of an organism is unable to store excess amino acids.
3. In this process, amino group is separated from the amino acid and ammonia is formed.
4. Toxic ammonia is either excreted or converted to less toxic forms like urea or uric acid before excretion.

Question 6.
Availability of water plays a key role in deciding the mode of excretion of an organism. Justify.
Answer:

1. Ammonia is the basic product of deamination process.
2. Ammonia is highly toxic and needs to be diluted immediately.
3. If there is no or limited access to water, need for conversion of ammonia becomes necessary.
   Hence, the availability of water plays a key role in deciding the mode of excretion of an organism.

Question 7.
What are the three main modes of excretion in animals?
Answer:
The three main modes of excretion in animals are as follows:
i. Ammonotelism
ii. Ureotelism
iii. Uricotelism

1. Ammonotelism:
   • Elimination of nitrogenous wastes in the form of ammonia is called as ammonotelism.
   • Ammonia is basic in nature and hence it can disturb the pH of the body, if not eliminated immediately.
   • Any change in pH would disturb all enzyme catalyzed reactions in the body and would also make the plasma membrane unstable.
   • Ammonia is readily soluble in water and needs large quantity of water to dilute and reduce its toxicity.
   • This is however an energy saving mechanism of excretion and hence all animals that have plenty of water available for dilution of ammonia, excrete nitrogenous wastes in the form of ammonia.
   • Animals that follow this mode of excretion are known as ammonotelic animals.
   • 1 gm ammonia needs about 300 – 500 ml of water for elimination.
   • Ammonotelic animals excrete ammonia through general body surface (skin), gills and kidneys.
     e.g. Ammonotelism is found in aquatic invertebrates, bony fishes, and aquatic / larval amphibians. Animals without excretory system (Protozoa) are also ammonotelic.
2. Ureotelism:
   • Elimination of nitrogenous wastes in the form of urea is called as ureotelism.
   • Urea is comparatively less toxic and less water-soluble than ammonia. Hence, it can be concentrated to some extent in body.
   • The body requires less water for elimination.
   • Since it is less toxic and less water soluble, ureotelism is suitable for animals that need to conserve water to some extent. Hence, ureotelism is common in terrestrial animals, as they have to conserve water.
   • It takes about 50 ml H2O for removal of 1 gm NH2 in form of urea.
   • Ureotelic animals generally convert ammonia to urea in the liver by operating ornithine / urea cycle in which 3 ATP molecules are used to produce one molecule of urea.
     e.g. Mammals, cartilaginous fishes (sharks and rays), many aquatic reptiles, most of the adult amphibians, etc. are ureotelic.
3. Uricotelism:
   • Elimination of nitrogenous wastes in the form of uric acid is called as uricotelism.
   • Uric acid is least toxic and hence, it can be retained in the body for some time in concentrated form.
   • It is least soluble in water. Hence there is minimum (about 5 – 10 ml for 1 gm) or no need of water for its elimination.
   • Those animals which need to conserve more water follow uricotelism. However, these animals need to spend more energy.
   • Ammonia is converted into uric acid by ‘inosinic acid pathway’ in the liver of birds, e.g. Birds, some insects, many reptiles, land snails, are uricotelic.

Question 8.
Fill in the blanks:
i. ________ is the basic product of deamination process.
ii. Aquatic amphibians excrete nitrogenous waste in the form of _______.
iii. Uricotelic organisms, convert ammonia to urea in the _______ by operating _____ cycle.
iv. Ammonia is converted into uric acid by ______ pathway in birds.
Answer:
i. Ammonia
ii. Ammonia
iii. liver, omithine/urea
iv. inosinic acid

Question 9.
Explain the following sentences,
i. Humans are ureotelic.
Answer:

• Urea is comparatively less toxic and less water – soluble than ammonia. Hence, it can be concentrated to some extent in the body.
• The body requires less water for elimination of urea.
  c. Due to these properties, ureotelism is suitable for animals which need to conserve water to some extent.
  Thus, humans are ureotelic.

ii. Sharks retain more urea in their blood.
Answer:

• Sharks retain more urea in their body fluid (blood) to make their blood isotonic to surrounding marine water (in order to maintain osmotic balance).
• This helps them to prevent possible loss of water by exosmosis.

Question 10.
Distinguish between Ureotelism and Uricotelism.
Answer:

No.	Ureotelism	Uricotelism
i.	It is the elimination of nitrogenous waste in the form of urea.	It is the elimination of nitrogenous waste in the form of uric acid.
ii.	Excretion of urea requires less (moderate ) amount of water.	Excretion of uric acid requires negligible amount of water.
iii.	Removal of 1 gm of urea requires 50 ml of water.	Removal of 1 gm of uric acid requires 5 – 10 ml of water, j
iv.	Urea is less toxic.	Uric acid is least toxic.
e.g.	It is generally seen in terrestrial animals. Mammals, cartilaginous fishes (sharks and rays), many aquatic reptiles, most adult amphibians, etc.	It is seen in birds, some insects, many reptiles, land snails, etc.

Question 11.
Distinguish between Ammonotelism and Uricotelism.
Answer:

No.	Ammonotelism	Uricotelism
i.	It is the elimination of nitrogenous waste in the form of ammonia.	It is the elimination of nitrogenous waste in the form of uric acid.
ii.	Excretion of ammonia requires plenty of water.	Excretion of uric acid requires negligible amount of water.
iii.	Removal of 1 gm of ammonia requires 300 – 500 ml of water.	Removal of 1 gm of uric acid requires 10ml of water.
iv.	Ammonia is very toxic.	Uric acid is less toxic.
e.g.	It is found in aquatic invertebrates, bony fishes and aquatic/ larval amphibians, etc.	It is seen in birds, some insects, many reptiles, land snails, etc.

Question 12.
Terrestrial animals are generally either ureotelic or uricotelic, not ammonotelic. Why?
Answer:

1. Ammonia is highly toxic to animals.
2. An animal requires large amount of water to dissolve and eliminate ammonia.
3. Terrestrial animals cannot lose such a large amount of water.
4. Ureotelic and uricotelic animals require less amount of water for removal of nitrogenous waste. Hence, to conserve water, ureotelism and uricotelism is adapted by terrestrial animals.

Question 13.
What is plasma creatinine? Why is it used as an index of kidney function?
Answer:

1. Plasma creatinine is produced from catabolism of creatinine phosphate during skeletal muscle contraction.
2. It provides a ready source of high energy phosphate.
3. Normally blood creatinine levels remain steady because the rate of production matches its excretion in urine.
4. Hence, plasma creatinine is used as an index of kidney function and its level above normal is an indication of poor renal function.
   [Note: Plasma creatinine is a waste product produced by muscles from the breakdown of a compound called ‘creatine phosphate.]

Question 14.
How can excretion play a role in homeostasis?
Answer:

1. Homeostasis is the maintenance of constant internal environment of the body.
2. Homeostasis is however dependent on osmoregulation, which is the process of controlling solute concentrations and water balance.
3. The composition of blood and therefore the internal environment is highly dependent on what the excretory organs retain in the body.
   Hence, excretion plays an important role in homeostasis.

Question 15.
How do different organisms carry out excretion?
Answer:
Different organisms carry out excretion in the following manner:

1. Unicellular organisms have contractile vacuoles which collect and discharge waste products outside the cell.
2. Excretion in sponges takes place by diffusion of waste material in water. This waste is discharged through the osculum.
3. True organs of excretion are found in those animals that show bilateral symmetry.
4. The most common type of excretory organ is a simple or branching tube that opens to the exterior, through pores called nephridiopores. This system is generally found in some annelids, Amphioxus, earthworms, etc.
5. In most of the insects, excretion takes place by a set of blind ended tubules called malpighian tubules.
6. Crustaceans have green glands as excretory organs.
7. Members of phylum Echinodermata do not have any specialised excretory organs. Waste materials directly diffuse into water or are excreted through tube feet.
8. The mammalian kidneys are a collection of functional units called nephrons, which are well designed to excrete metabolic waste.

Question 16.
What are nephridia? Explain the major types of nephridia in detail.
Answer:
Nephridia are simple or branching tubules used for excretion which open to the exterior through pores called nephridiopores.
Two major types of nephridia are as follows:
i. Protonephridia:
These are network of dead end tubes called flame cells. They are mostly found in animals that lack a true body cavity, e.g. Platyhelminthes, rotifers, some annelids and Amphioxus.

ii. Metanephridia:
These are unbranched coiled tubes that are connected to the body cavity through funnel like structures called nephrostomes. Body fluid enters the nephridium through nephrostome and gets discharged ‘ through nephridiopore. e.g. Earthworms.

Question 17.
Distinguish between the Ureter and Urethra.
Answer:

No.	Ureter	Urethra
i.	Ureters are two duct-like structures arising from the hilum of the kidney.	Urethra is a single tube-like structure arising from the urinary bladder.
ii.	Ureter carries urine from the kidney to the urinary bladder.	Urethra carries urine from the urinary bladder to the exterior of the body.
iii.	Ureters are paired structures.	Urethra is unpaired structure.

Question 18.
Write a short note on micturition.
Answer:

1. The process of release of urine from the urinary bladder is called micturition.
2. The average capacity of urinary bladder is 700 ml.
3. When urinary bladder is almost half filled, stretch receptors in urinary bladder transmit impulses to spinal cord, initiating a conscious desire to expel urine.
4. Micturition reflex center of spinal cord transmit impulses to the wall of urinary bladder and internal urethral sphincter.
5. Bladder muscles contract and muscles of internal urethral sphincter relax.
6. The external sphincter receives impulses from conscious centre of brain and relaxes.
7. This leads to elimination of urine from the bladder.

Question 19.
Explain the L.S of kidney with a neat and labelled diagram.
Answer:

1. Each kidney is covered by three layers of tissue, namely the outermost renal fascia, middle adipose capsule and innermost renal capsule.
   • The outermost layer, renal fascia is made up of a thin layer of fibrous connective tissue. It anchors the kidney to the abdominal wall as well as surrounding tissue.
   • The middle layer is a mass of fatty tissue called adipose capsule. It protects the kidneys by shock absorption.
   • The innermost layer, renal capsule is a smooth transparent fibrous membrane that is continuous with outer layer of ureters. It acts as a barrier against spread of infections in kidney.
2. The L.S. of kidney shows two distinct regions within the capsule. Histologically, kidney is divisible into two regions as renal cortex and renal medulla.
   • Renal cortex is the outer / peripheral, red coloured and granular region. It contains Malpighian bodies, convoluted tubules and blood vessels.
   • Medulla is inner region of kidney with pale red colour and striated appearance. Medulla mainly consists of Loops of Henle and collecting ducts. All these are arranged in conical manner to form renal pyramids.
   • Cortex extends in medulla as columns of Bertini / renal columns between pyramids. Narrow tip of pyramid is called as renal papilla. There are several pyramids.
   • Renal papilla open into the minor calyx. Minor calyces merge together to form major calyces and major calyces unite together to form renal pelvis.
   • Renal pelvis (renal sinus) is funnel-shaped area in the region of medulla of kidney. Renal pelvis
     

Question 20.
What is nephrology?
Answer:
Nephrology is branch of biology that deals with the structure, function and disorders of male and female urinary system.

Question 21.
Write a note on nephron.
Answer:

1. Nephrons are structural and functional units of kidney.
2. Each nephron consists of a 4 – 6 cm long, thin-walled tube called the renal tubule and a bunch of capillaries known as the glomerulus.
3. The wall of the renal tubule is made up of a single layer of epithelial cells.
4. Its proximal end is wide, blind, cup-like and is called as Bowman’s capsule, whereas the distal end is open.
5. The nephron is divisible into Bowman’s capsule, neck, proximal convoluted tubule (PCT), Loop of Henle (LoH), distal convoluted tubule (DCT) and collecting tubule (CT).
6. The glomerulus is present in the cup-like cavity of Bowman’s capsule and both are collectively known as renal corpuscle or Malpighian body.

Question 22.
With the help of a well labelled diagram, describe the structure of nephron.
Answer:
Nephron is the structural and functional unit of kidney.
Structure of nephron:
A nephron (uriniferous tubule) is a thin walled, coiled duct, lined by a single layer of epithelial cells. Each nephron is divided into two main parts:
i. Malpighian body
ii. Renal tubule

i. Malpighian body: Each Malpighian body is about 200pm in diameter and consists of a Bowman’s capsule and glomerulus.
a. Glomerulus:
Glomerulus is a bunch of fine blood capillaries located in the cavity of Bowman’s capsule.
A small terminal branch of the renal artery, called as afferent arteriole enters the cup cavity (Bowman capsule) and undergoes extensive fine branching to form network of several capillaries. This bunch is called as glomerulus.
The capillary wall is fenestrated (perforated).
All capillaries reunite and form an efferent arteriole that leaves the cup cavity.
The diameter of the afferent arteriole is greater than the efferent arteriole. This creates a high hydrostatic pressure essential for ultrafiltration, in the glomerulus.

b. Bowman’s capsule:
It is a cup-like structure having double walls composed of squamous epithelium.
The outer wall is called as parietal wall and the inner wall is called as visceral wall.
The parietal wall is thin consisting of simple squamous epithelium.
There is a space called as capsular space / urinary space in between two walls.
Visceral wall consists of special type of squamous cells called podocytes having a foot-like pedicel. These podocytes are in close contact with the walls of capillaries of glomerulus.
There are small slits called as filtration slits in between adjacent podocytes.

ii. Renal tubule:
a. Neck:
The Bowman’s capsule continues into the neck. The wall of neck is made up of ciliated epithelium. The lumen of the neck is called the urinary pole. The neck leads to proximal convoluted tubule.

b. Proximal Convoluted Tubule :
This is highly coiled part of nephron which is lined by cuboidal cells with brush border (microvilli) and surrounded by peritubular capillaries. Selective reabsorption occurs in PCT. Due to convolutions (coiling), filtrate flows slowly and remains in the PCT for longer duration, ensuring that maximum amount of useful molecules are reabsorbed.

c. Loop of Henle :
This is ‘U’ shaped tube consisting of descending and ascending limb.
The descending limb is thin walled and permeable to water and lined with simple squamous epithelium.
The ascending limb is thick walled and impermeable to water and is lined with simple cuboidal epithelium.
The LoH is surrounded by capillaries called vasa recta.
Its function is to operate counter current system – a mechanism for osmoregulation.
The ascending limb of Henle’s loop leads to DCT.

d. Distal convoluted tubule:
This is another coiled part of the nephron.
Its wall consists of simple cuboidal epithelium.
DCT performs tubular secretion / augmentation / active secretion in which, wastes are taken up from surrounding capillaries and secreted into passing urine.
DCT helps in water reabsorption and regulation of pH of body fluids.

e. Collecting tubule:
This is a short, straight part of the DCT which reabsorbs water and secretes protons.
The collecting tubule opens into the collecting duct.

Question 23.
What is the difference between Cortical nephrons and Juxtamedullary nephrons.
Answer:

	Cortical nephrons	Juxtamedullary nephrons
i.	They have a shorter loop of Henle.	They have a longer loop of Henle.
ii.	Loop of Henle of these nephrons extends very little into the medulla.	Loop of Henle of these nephrons run deep into the medulla.
iii.	Most nephrons are cortical nephrons.	Few nephrons are juxtamedullary nephrons.
iv.	Efferent arteriole forms peritubular capillary network around DCT, PCT and Henle’s loop of cortical nephrons	Efferent arteriole forms loop-shaped vasa recta around  Henle’s loop of juxtamedullary nephrons.

Question 24.
Sketch and label Malpighian body.
Answer:

Question 25.
What are podocytes? In which part of the nephron are they present?
Answer:
Podocytes are a special type of squamous cells that have a foot-like pedicel. They are present in the visceral wall of the Bowman’s capsule and are in close contact with the walls of capillaries of glomerulus

Question 26.
Write a short note on Juxta Glomerular Apparatus.
Answer:

1. Some smooth muscle cells of the wall of afferent arteriole are modified in such a way that their sarcoplasm is granular. These cells are called juxtaglomerular (JG) cells.
2. In each nephron, initial part of DCT makes contact with the afferent arteriole of same nephron.
3. Cells in the wall of DCT in this region are packed more densely than those in other region of DCT. This is called macula densa.
4. Macula densa and the JG cells together form Juxta Glomerular Apparatus (JGA).
5. The JGA plays an important role in blood pressure regulation within the kidney.

Question 27.
Explain the mechanism of urine formation in detail.
Answer:
Process of urine formation is completed in three steps, namely;
i. Ultrafiltration/ Glomerular filtration,
ii. Selective reabsorption,
iii. Tubular secretion / Augmentation

i. Ultrafiltration / Glomerular filtration :
Diameter of afferent arteriole is greater than the efferent arteriole. The diameter of capillaries is still smaller than both arterioles. Due to the difference in diameter, blood flows with greater pressure through the glomerulus. This is called as glomerular hydrostatic pressure (GHP) and normally, it is about 55 mmHg. GIIP is opposed by osmotic pressure of blood (normally, about 30 mm Hg) and capsular pressure (normally, about 15 mm Hg).

Hence net / effective filtration pressure (EFP) is 10 mm Hg.
EFP = Hydrostatic pressure in glomerulus – (Osmotic pressure of blood + Filtrate Hydrostatic pressure)
= 55 – (30 +15)
= 10 mm Hg

Under the effect of high pressure, the thin walls of the capillary become permeable to major components of blood (except blood cells and macromolecules like protein).
Thus, plasma except proteins oozes out through wall of capillaries.
About 600 ml blood passes through each kidney per minute.

The blood (plasma) flowing through kidney (glomeruli) is filtered as glomerular filtrate, at a rate of 125 ml / min. (180 L/d).

Glomerular filtrate / deproteinized plasma / primary urine is alkaline, contains urea, amino acids, glucose, pigments, and inorganic ions.

Glomerular filtrate passes through filtration slits into capsular space and then reaches the proximal convoluted tubule.

ii. Selective reabsorption :
Selective reabsorption occurs in proximal convoluted tubule (PCT). It is highly coiled so that glomerular filtrate passes through it very slowly. Columnar cells of PCT are provided with microvilli due to which absorptive area increases enormously.

This makes the process of reabsorption very effective.
These cells perform active (ATP mediated) and passive (simple diffusion) reabsorption.
Substances with considerable importance (high threshold) like – glucose, amino acids, vitamin C, Ca⁺⁺, K⁺, Na⁺, Cl^– are absorbed actively, against the concentration gradient. Low threshold substances like water, sulphates, nitrates, etc., are absorbed passively.
In this way, about 99% of glomerular filtrate is reabsorbed in PCT and DCT.

iii. Tubular secretion / Augmentation :
Finally filtrate reaches the distal convoluted tubule via loop of Henle. Peritubular capillaries surround DCT. Cells of distal convoluted tubule and collecting tubule actively absorb the wastes like creatinine and ions like K⁺, H⁺ from peritubular capillaries and secrete them into the lumen of DCT and CT, thereby augmenting the concentration of urine and changing its pH from alkaline to acidic.

Secretion of H⁺ ions in DCT and CT is an important homeostatic mechanism for pH regulation of blood. Tubular secretion is the only process of excretion in marine bony fishes and desert amphibians.

Question 28.
Marine bony fishes and desert amphibians rely on which process of excretion?
Answer:
Tubular secretion

Question 29.
Where does selective reabsorption of glomerular filtrate take place?
Answer:
Selective reabsorption of glomerular filtrate takes place in the proximal convoluted tubule (PCT).

Question 30.
What is glomerular filtration pressure or net effective filtration pressure?
Answer:
Glomerular filtration pressure (GFP)/ Effective filtration pressure (EFP) is the difference between the hydrostatic pressure (GHP) and the sum of osmotic pressure of blood and capsular pressure (CHP).
It can be represented as:
EFP = Glomerular hydrostatic pressure – (Osmotic pressure of blood + Filtrate Flydrostatic pressure)
= 55 – (30 + 15)
= 10 mmHg
[Note: Net filtration pressure = Glomerular blood hydrostatic pressure – (Capsular hydrostatic pressure + Blood colloid osmotic pressure)
Source: Tort or a, G., Derrickson, B. Principles of Anatomy and Physiology. 11^th Edition.]

Question 31.
Distinguish between Selective reabsorption and Tubular secretion.
Answer:

No.	Selective rcabsorption	Tubular secretion
i.	Selective reabsorption is concerned with the selective absorption of useful substances from the glomerular filtrate.	Tubular secretion is transfer of materials from peritubular capillaries to the renal tubular lumen.
ii.	Substances with considerable importance (high threshold) like – glucose, amino acids, vitamin C, Ca++, K+, Na+, Cl– are absorbed actively, against concentration gradient	In this process, substances like urea. amino acids, glucose, pigments, and inorganic ions arc removed from the blood and discharged along with the urine.
iii.	Selective reabsorption occurs in Proximal convoluted tubule, Henle’s loop, Distal convoluted tubule and collecting duct.	Tubular secretion occurs in Distal convoluted tubules only.

Question 32.
Explain the process of concentration of urine in deai1.
OR
Explain counter current mechanism in detail.
Answer:
Under the conditions like low water intake or high water loss due to sweating, humans can produce concentrated urine. This urine can be concentrated around four times i.e. 1200 mOsm/L, than the blood (300 mOsm/L). hence, a mechanism called countercurrent mechanism is operated in the human kidneys. The countercurrent mechanism operating in the Limbs of Henle’s loop of juxtamedullarv nephrons and vasa recta is as follows:
i. It involves the passage of fluid from descending to ascending limb of Henle’s loop.

ii. This mechanism is called countercurrent mechanism, since the flow of tubular fluid is in opposite direction through both limbs.

iii. In case of the vasa recta, blood flows from ascending to descending parts of itself.

iv. Wall of descending limb is thin and permeable to water, hence, water diffuses from tubular fluid into tissue fluid due to which, tubular fluid becomes concentrated.
v. The ascending limb is thick and impermeable to water. Its cells can reabsorb Na⁺ and Cl^– from tubular fluid and release into tissue fluid.

vi. Due to this, tissue fluid around descending limb becomes concentrated. This makes more water to move out from descending limb into tissue fluid by osmosis.

vii. Thus, as tubular fluid passes down through descending limb, its osmolarity (concentration) increases gradually due to water loss and on the other hand, progressively decreases due to Na⁺ and Cl^– secretion as it flows up through ascending limb.

viii. Whenever retention of water is necessary, the pituitary secretes ADH. ADH makes the cells in the wall of collecting ducts permeable to water.

ix. Due to this, water moves from tubular fluid into tissue fluid, making the urine concentrated.

x. Cells in the wall of deep medullar part of collecting ducts are permeable to urea. As concentrated urine flows through it, urea diffuses from urine into tissue fluid and from tissue fluid into the tubular fluid flowing through thin ascending limb of Henle’s loop.

xi. This urea cannot pass out from tubular fluid while flowing through thick segment of ascending limb, DCT and cortical portion of collecting duct due to impermeability for it in these regions.

xii. However, while flowing through collecting duct, water reabsorption is operated under the influence of ADII. Due to this, urea concentration increases in the tubular fluid and same urea again diffuses into tissue fluid in deep medullar region.

xiii. Thus, same urea is transferred between segments of renal tubule and tissue fluid of inner medulla. This is called urea recycling; operated for more and more water reabsorption from tubular fluid and thereby excreting small volumes of concentrated urine.

xiv. Osmotic gradient is essential in the renal medulla for water reabsorption by counter current multiplier system.

xv. This osmotic gradient is maintained by vasa recta by operating counter current exchange system.

xvi. Vasa recta also have descending and ascending limbs. Blood that enters the descending limb of the vasa recta has normal osmolarity of about 300 mOsm/L.

xvii. As it flows down in the region of renal medulla where tissue fluid becomes increasingly concentrated, Na⁺, Cl^– and urea molecules diffuse from tissue fluid into blood and water diffuse from blood into tissue fluid.

xviii. Due to this, blood becomes more concentrated which now flows through ascending part of vasa recta. This part runs through such region of medulla where tissue fluid is less concentrated.

xix. Due to this, Na⁺, Cl^– and urea molecules diffuse from blood to tissue fluid and water from tissue fluid to blood. This mechanism helps to maintain the osmotic gradient.

Question 33.
Camel excretes concentrated urine. Give reason.
Answer:
In order to reabsorb water to maximum capacity, loop of Henle is longer in desert mammals like camel. Hence, camel excretes concentrated urine.

Question 34.
Why is urine yellow in colour?
Answer:
Normal urine is pale yellow coloured transparent liquid, due to the pigment urochrome.

Question 35.
How is the composition of urine regulated?
Answer:
The composition of urine depends upon food and fluid consumed by an individual. There are two ways in which it the composition is regulated. They are as follows:

1. Regulating water reabsorption through ADH
2. Electrolyte reabsorption though RAAS
3. Atrial Natriuretic Peptide

i) Regulating water reabsorption through ADH:
Hypothalamus in the midbrain has special receptors called osmoreceptors which can detect change in osmolarity (measure of total number of dissolved particles per liter of solution) of blood.

If osmolarity of blood increases due to water loss from the body (after eating namkeen or due to sweating), osmoreceptors trigger release of Antidiuretic hormone (ADH) from neurohypophysis (posterior pituitary). ADH stimulates reabsorption of water from last part of DCT and entire collecting duct by increasing the permeability of cells.

This leads to reduction in urine volume and decrease in osmolarity of blood.

Once the osmolarity of blood comes to normal, activity of osmoreceptor cells decreases leading to decrease in ADH secretion. This is called negative feedback.

In case of hemorrhage or severe dehydration too, osmoreceptors stimulate ADH secretion. ADH is important in regulating water balance through kidneys.

In absence of ADH, diuresis (dilution of urine) takes place and person tends to excrete large amount of dilute urine. This condition called as diabetes insipidus.
[Note: Hypothalamus is a part of forebrain]

ii) Electrolyte reabsorption through RAAS:
Another regulatory mechanism is RAAS (Renin Angiotensin Aldosterone System) by Juxta Glomerular Apparatus (JGA).

Whenever blood supply (due to change in blood pressure or blood volume) to afferent arteriole decreases (e.g. low BP/dehydration), JGA cells release Renin. Renin converts angiotensinogen secreted by hepatocytes in liver to Angiotensin I. ‘Angiotensin converting enzyme’ further modifies Angiotensin I to Angiotensin II, the active form of hormone. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na and water, thereby increasing blood volume and pressure.

iii) Atrial natriuretic peptide (ANP):
A large increase in blood volume and pressure stimulates atrial wall to produce atrial natriuretic peptide (ANP). ANP inhibits Na+ and Cl reabsorption from collecting ducts inhibits release of renin, reduces aldosterone and ADH release too. This leads to a condition called Natriuresis (increased excretion of Na+ in urine) and diuresis.

Question 36.
What is renin? Give its function.
Answer:
Renin is an enzyme secreted by juxtaglomerular cells of afferent arteriole.
Function: It activates Angiotensinogen to Angiotensin-I.

Question 37.
What is the function of Angiotensin II?
Answer:
Functions of Angiotensin II:

1. It constricts arterioles in kidney thereby reducing blood flow and increasing blood pressure.
2. It stimulates PCT cells to enhance reabsorption of Na⁺, Cl^– and water.
3. It stimulates adrenal cortex to release another hormone called aldosterone that stimulates DCT and collecting ducts to reabsorb more Na and water, thereby increasing blood volume and pressure.

Question 38.
Which hormones and factors are involved in regulation of kidney function?
Answer:
Hormones like ADH, Renin, Angiotensin and Atrial Natriuretic Peptide (ANP) are involved in regulation of kidney function.

Question 39.
Can improper kidney function lead to brittle bones? Justify your answer.
Answer:
Yes, improper kidney function lead to brittle bones.

1. Kidneys participate in synthesis of calcitriol, the active form of Vitamin D which is needed for absorption of dietary calcium.
2. Deficiency of calcitriol can lead to brittle bones.

Question 40.
Do organs other than kidney participate in excretion? Explain.
Answer:
Yes, various organs other than the kidney participate in excretion. They are as follows:
i. Skin:
Skin acts as an accessory excretory organ. The skin of many organisms is thin and permeable. It helps in diffusion of waste products like ammonia.
Human skin however is thick and impermeable. It shows presence of two types of glands namely, sweat glands and sebaceous glands.
a. Sweat glands are distributed all over the skin. They are abundant in the palm and facial regions. These simple, unbranched, coiled, tubular glands open on the surface of the skin through an opening called sweat pore. Sweat is primarily produced for thermoregulation but it also excretes substances like water, NaCl, lactic acid and urea.
b. Sebaceous glands are present at the neck of hair follicles. They secrete oily substance called sebum. It forms a lubricating layer on skin making it softer. It protects skin from infection and injury.

ii. Lungs:
Lungs are the accessory excretory organs. They help in excretion of volatile substances like C02 and water vapour produced during cellular respiration. Along with CO₂, lungs also remove excess of H₂O in the form of vapours during expiration. They also excrete volatile substances present in spices and other food stuff.

Question 41.
Enlist the human excretory organs and their excretory products.
Answer:
Excretory OrgAnswer:

1. Lungs: Remove CO₂ and also water vapour to a considerable extent. Volatile substances present in spices and other food stuff are excreted through lungs
2. Kidneys: Remove nitrogenous waste products like ammonia, urea and uric acid, creatinine. They also remove excessive amount of water, salts and certain minerals.
3. Skin: Remove water, NaCl, lactic acid and urea by through of sweat.

Question 42.
What is albuminaria? What are its causes?
Answer:

• Albuminaria is the presence of excess albumin in the urine.
• Causes: Injury to the endothelial capsular membrane as a result of increased blood pressure, injury or irritation of kidney cells by substances such as toxins or heavy metals.

Question 43.
A patient report indicates presence of excessive quantities of ketone bodies in the urine. What does this indicate? How is it caused?
Answer:
Presence of excessive quantities of ketone bodies in the urine indicates that the patient is suffering from diabetes mellitus, starvation or too little carbohydrates in the diet.

Question 44.
Sheela is suffering from kidney infection. Presence of which type of cells in the urine can be indicative of this?
Answer:
Presence of leucocytes in the urine can be indicative of infection of kidney or other urinary organs.

Question 45.
Enlist the various disorders and diseases related to the excretory system.
Answer:
Some disorders and diseases related to the excretory system are as follows:

1. Kidney stones
2. Uremia
3. Nephritis
4. Renal Failure
5. Ketonuria
6. Albuminaria

Question 46.
Write a note on kidney stones with reference to types, symptoms and diagnosis.
Answer:
Kidney stones are also called renal calculi. They may be formed in any portion of urinary tract, from kidney tubules to external opening.
Types:
Depending on their composition, kidney stones are classified into the following types.

1. Calcium stones : These are usually calcium oxalate or calcium phosphate stones.
2. Struvite stones : These are formed in response to bacterial infection caused by urea – splitting bacteria. They grow in size quickly and become quite large.
3. Uric acid stones : These stones usually affect people drinking less water or consuming high protein diet.
4. Cystine stones : It is a genetic disorder that causes the kidney to excrete too much of certain amino acid.

Symptoms:
Intermittent pain below rib cage in back and side ways. Hazy, brownish/reddish/ pinkish urine. Frequent urge to pass urine. Pain during micturition.

Diagnosis:
Uric acid content of blood, colour of urine, kidney X-ray, sonography of kidney are different diagnostic tests prescribed depending on symptoms.

Question 47.
What is uremia?
Answer:
If the level of urea in blood rises above 0.05%, the condition is known as uremia. It may lead to kidney failure.

Question 48.
What is the normal content of urea in blood?
Answer:
The normal content of urea in blood is 0.01 to 0.03 %.

Question 49.
Write a note on nephritis.
Answer:

1. Nephritis is the inflammation of kidneys characterised by proteinuria.
2. It is caused due to increased permeability of glomerular capsular membrane, permitting large amounts of proteins to escape from blood to urine.
3. This leads to change in blood colloidal osmotic pressure, leading to movement of fluid from blood to interstitial spaces.
4. It is reflected as edema.

Question 50.
What is renal failure? Describe its types.
Answer:
Renal failure is the decrease or cessation of glomerular filtration and ¡s classified into two types.
i) Acute Renal failure (ARF):
ARF is sudden worsening of renal function that most commonly happens after severe bleeding. There is a decrease in urine output (oligouriaf scanty urine i.e., less than 400 mi/day or less than 0.5 ml/kg/li in children). Other causes of ARF may include acute obstruction of both ureters or nephrotoxic drugs. ARF can be detected biochemically by elevated serum creatinine levels.

ii) Chronic kidney disease (CKD) :
It is the progressive and generally irreversible decline in glomerular filtration rate (GFR). It may be caused due to chronic glomerulonephritis. It can be detected by reduced kidney size and possibility of anaemia.

Question 51.
When does a patient need to undergo haemodialysis? Explain the process in detail.
Answer:

1. When renal function of a person falls below 5 – 7 %, accumulation of harmful substances in blood begins. In such a condition, the person has to go for artificial means of filtration of blood i.e. haemodialysis.
2. In haemodialysis, a dialysis machine is used to filter blood. The blood is filtered outside the body using a dialysis unit.
3. In this procedure, the patients’ blood is removed; generally from the radial artery and passed through a cellophane tube that acts as a semipermeable membrane.
4. The tube is immersed in a fluid called dialysate which is isosmotic to normal blood plasma. Hence, only excess salts if present in plasma pass through the cellophane tube into the dialysate.
5. Waste substances being absent in the dialysate, move from blood into the dialyzing fluid.
6. Filtered blood is returned to vein.
7. In this process it is essential that anticoagulant like heparin is added to the blood while it passing through the tube and before resending it into the circulation, adequate amount of anti-heparin is mixed.
8. Also, the blood has to move slowly through the tube and hence the process is slow.

Question 52.
Write a note on peritoneal dialysis.
Answer:

1. In this method, the dialyzing fluid is introduced in abdominal cavity or peritoneal cavity.
2. The peritoneal membrane acts as semipermeable dialyzing membrane.
3. Toxic wastes and extra solutes pass into the fluid.
4. This fluid is drained out after a prescribed period of time.
5. Peritoneal dialysis can be repeated as per the need of the patient.
6. It can be carried out at home at work or while travelling. But it is not as efficient as haemodialysis.

Question 53.
What are the drawbacks of haemodialysis?
Answer:

• Kidneys are associated with secretion of erythropoietin, renin and calcitriol which is not possible using dialysis machine.
• During dialysis, the blood has to move slowly through the tube and hence the process is slow.

Question 54.
What is kidney transplant?
Answer:

1. It is the organ transplant of a healthy kidney into a patient with end – stage renal disease.
2. Kidney transplantation is classified as cadaveric (deceased donor) or living donor kidney transplant.
3. Living donor kidney transplant are further classified as genetically related (living-related) or non-related (living non-related) transplants.

Question 55.
Complete the diagram / chart with correct labels / information. Write the conceptual details regarding it.
i)

Answer:

Skin:
Skin acts as an accessory excretory organ. The skin of many organisms is thin and permeable. It helps in diffusion of waste products like ammonia.
Human skin however is thick and impermeable. It shows presence of two types of glands namely, sweat glands and sebaceous glands.
a. Sweat glands are distributed all over the skin.
They are abundant in the palm and facial regions. These simple, unbranched, coiled, tubular glands open on the surface of the skin through an opening called sweat pore. Sweat is primarily produced for thermoregulation but it also excretes substances like water, NaCI, lactic acid and urea.

b. Sebaceous glands are present at the neck of hair follicles.
They secrete oily substance called sebum. It forms a Lubricating layer on skin making it softer. It protects skin from infection and injury.

ii)

Answer:

1. In this method, the dialyzing fluid is introduced in abdominal cavity or peritoneal cavity.
2. The peritoneal membrane acts as semipermeable dialyzing membrane.
3. Toxic wastes and extra solutes pass into the fluid.
4. This tluid is drained out after a prescribed period of time.
5. Peritoneal dialysis can be repeated as per the need of the patient.
6. It can be carried out at home at work or while travelling. But it is not as efficient as haemodialysis.

Question 55.
Invertebrates, bony fishes, tadpoles, etc. are ammonotelic. Whereas birds, reptiles, land snails, etc. are uricotelic. What would be the probable reason for the difference in mode of excretion?
i. Ammonotelic organisms conserve water as conversion of ammonia to uric acid requires large amount of water.
ii. Elimination of ammonia requires large quantity of water, thus ammonotelism is seen in aquatic animals.
iii. Uricotelic organisms require moderate water for eliminating urea and formation of ammonia requires expenditure of energy. Thus, to conserve water and energy, these animals have uricotelism mode of excretion.
iv. Aquatic animals can retain ammonia and store it in the body for long time.
(A) i and iii
(B) only ii
(C) only iv
(D) ii and iii
Answer:
(B)

Question 56.
A lab technician was evaluating blood reports of some patients. She observed the following values of the report:

Sr. No.	Patient	Comments on urine/ blood report
i.	A	High levels of glucose in urine
ii.	B	Presence of excess quantities of ketone in urine
iii.	C	Presence of leucocytes in urine
iv.	D	0.06% urea in blood
V.	E	High level of proteins in blood

Read the given comments and discuss what disorder/disease the patient must be suffering from.

Answer:

Sr. No.	Patient	Comments on urine/ blood report	Disorder/ Disease
i.	A	High levels of glucose in urine	Glucosuria
ii.	B	Presence of excess quantities of ketone in urine	Ketonuria (Indicative of diabetes mellitus)
iii.	C	Presence of leucocytes in urine	Infection of kidney or urinary organs
iv.	D	0.06% urea in blood	Uremia
V.	E	High level of proteins in blood	Proteinuria/ Nephritis

Multiple Choice Questions

Question 1.
Uric acid is the main nitrogenous waste in
(A) birds
(B) cartilaginous fish
(C) mammals
(D) larval amphibians
Answer:
(A) birds

Question 2.
Protonephridia is the excretory organ of
(A) platyhelminthes
(B) coelenterates
(C) arthropods
(D) aschelminthes
Answer:
(A) platyhelminthes

Question 3.
The glomerulus receives blood through the
(A) vasa recta
(B) renal artery
(C) afferent arteriole
(D) efferent arteriole
Answer:
(C) afferent arteriole

Question 4.
More ADH results in
(A) reduced permeability of DCT
(B) dilute urine
(C) reduced blood pressure
(D) concentrated urine
Answer:
(D) concentrated urine

Competitive Corner

Question 1.
Match the items in Column-I with those in Column-II [NEET Odisha 2019]

	Column-I		Column-II
i.	Podocytes	a.	Crystallised oxalates
ii.	Protonephridia	b.	Annelids
iii.	Nephridia	c.	Amphioxus
iv.	Renal calculi	d.	Filtration slits

Select the correct option from the following:
(A) i-d, ii-b, iii-c, iv-a
(B) i-c, ii-d, iiii-b, iv-a
(C) i-c, ii-b, iii-d, iv-a
(D) i-d, ii-c, iii-b, iv-a
Answer:
(D) i-d, ii-c, iii-b, iv-a

Question 2.
Match the following parts of a nephron with their function: [NEET Odisha 2019]

i.	Descending limb of Henle’s loop	a.	Reabsorption of salts only
ii.	Proximal convoluted tubule	b.	Reabsorption of water only
iii.	Ascending limb of Henle’s loop	c.	Conditional reabsorption of sodium ions and water
iv.	Distal convoluted tubule	d.	Reabsorption of ions, water and organic nutrients

Select the correct option from the following:
(A) i-d, ii-a, iii-c, iv-b
(B) i-a, ii-c, iii-b, iv-d
(C) i-b, ii-d, iii-a, iv-c
(D) i-a, ii-d, iii-b, iv-c
Answer:
(C) i-b, ii-d, iii-a, iv-c

Question 3.
Which of the following factors is responsible for the formation of concentrated urine? [NEET(UG) 2019]
(A) Secretion of erythropoietin by Juxtaglomerular complex.
(B) Hydrostatic pressure during glomerular filtration.
(C) Low levels of antidiuretic hormone.
(D) Maintaining hyperosmolarity towards inner medullary interstitium in the kidneys.
Answer:
(D) Maintaining hyperosmolarity towards inner medullary interstitium in the kidneys.

Question 4.
Portions of renal cortex, which are projected into renal medulla, among the renal pyramids are called as _______. [MHT CET 2019]
(A) Columns of Bertini
(B) Columnae Camae
(C) Ampullae
(D) Ducts of Bellini
Answer:
(A) Columns of Bertini

Question 5.
Renal failure is typically detected by biochemical analysis of blood which shows _______. [MHT CET 2019]
(A) increased level of amino acids
(B) lower level of uric acid
(C) lower level of serum creatinine
(D) higher level of serum creatinine
Answer:
(D) higher level of serum creatinine

Question 6.
Which of the following statements is CORRECT with reference to nephron? [MHT CET 2019]
(A) ADH hormone increases permeability of PCT cells to reabsorb water.
(B) Efferent arteriole forms peritubular network all around tubule.
(C) Podocytes occur in ascending limb of loop of Henle.
(D) Descending limb of Henle’s loop is impenneable to water.
Answer:
(B) Efferent arteriole forms peritubular network all around tubule.

Question 7.
Match the items given in Column I with those in Column li and select the correct option given below. [NFET (UG) 2018]

	Column I (Function)		Column II (Part of excretory system)
i.	Ultrafiltration	a.	Henle’s loop
ii.	Concentration of urine	b.	Ureter
iii.	Transport of urine	c.	Urinary bladder
iv.	Storage of urine	d.	Malpighian corpuscle
v.		e.	Proximal convoluted tubule

(A) i-e, ii-d, iii-a, iv-b
(B) i-d, ii-a, iii-b, iv-c
(C) i-d, ii-e, iii-b, iv-c
(D) i-e, ii-d, iii-a, iv-c
Answer:
(B) i-d, ii-a, iii-b, iv-c

Question 8.
Match the items given in Column I with those in Column II and select the correct option given below. [NEET (UG) 2018]

	Column I		Column II
i.	Glycosuria	a.	Accumulation of uric acid in joints
ii.	Gout	b.	Mass of crystalised salts within the kidney
iii.	Renal calculi	c.	Inflammation of glomeruli
iv.	Glomerular nephritis	d.	Presence of glucose in urine

(A) i-b, ii-c, iii-a, iv-d
(B) i-a, ii-b, iii-c, iv-d
(C) i-c, ii-b, iii-d, iv-a
(D) i-d, ii-a, iii-b, iv-c
Answer:
(D) i-d, ii-a, iii-b, iv-c

Question 9.
In the given diagram of Malpighian body, blood is filtered from part labelled ________

(A) L
(B) M
(C) N
(D) O
Hint: L – Afferent arteriole
M – Efferent arteriole
N – Glomerulus
O – Proximal convoluted tubule
Blood filtration occurs in glomerulus. Afferent arteriole is the blood vessel leading to glomerulus. Efferent arteriole carries blood away from the glomerulus. Proximal convoluted tubule is involved in reabsorption of useful substances from the filtrate.
Answer:
(C) N

Question 10.
Majority of kidney stones consist crystals of [MHT CET 2018]
(A) calcium oxalate, sodium bicarbonate
(B) calcium oxalate, calcium phosphate
(C) calcium phosphate, sodium chloride
(D) calcium carbonate, copper sulphate
Answer:
(B) calcium oxalate, calcium phosphate

Question 11.
Which of the following group of animals is guanotelic? [MHT CEE 2018]
(A) Labeo, turtle, camel
(B) Lizard, snake, scorpion
(C) Penguin, spider, scorpion
(D) Spider, scorpion, snake
Answer:
(C) Penguin, spider, scorpion

Question 12.
Which of the following statements is CORRECT? [NEET (UG) 2017]
(A) The ascending limb of loop of Henle is impermeable to water.
(B) The descending limb of loop of Henle is impermeable to water.
(C) The ascending limb of loop of Henle is permeable to water.
(D) The descending limb of loop of Henle is permeable to electrolytes.
Hint: Descending limb of loop of Henle is permeable to water but impermeable to electrolytes. While ascending limb of loop of Henle is impermeable to water and permeable to electrolytes.
Answer:
(A) The ascending limb of loop of Henle is impermeable to water.

Question 13.
A decrease in blood pressure/volume will not cause the release of [NEET (UG) 2017]
(A) Renin
(B) Atrial Natriuretic Factor
(C) Aldosterone
(D) ADH
Hint: Decrease in blood pressure or volume stimulates the release of renin, aldosterone and ADH. ANF is released when blood pressure increases or blood volume increases. Release of ANF causes vasodilation and also inhibit renin angiotensin mechanism that decreases blood pressure and blood volume.
Answer:
(B) Atrial Natriuretic Factor

Question 14.
Formation of urea takes place in the _________. [MHT CET 2017]
(A) Heart
(B) Kidney
(C) Liver
(D) Lung
Answer:
(C) Liver

Question 15.
The yellow colour of normal urine is due to [MHT CET 2017]
(A) Bilirubin
(B) Biliverdin
(C) Urochrome
(D) Uric acid
Answer:

Question 16.
Uremia is indicated when the blood urea level rises above _______ [MHT CET 2017]
(A) 0.05%
(B) 0.04%
(C) 0.03%
(D) 0.02 %
Answer:
(A) 0.05%

Balbharti Maharashtra State Board 11th Biology Important Questions 16 Skeleton and Movement Important Questions and Answers.

Maharashtra State Board 11th Biology Important Questions Chapter 16 Skeleton and Movement

Question 1.
Name the three types of muscles which bring about movements in humans.
Answer:

1. Smooth / non-striated / visceral / involuntary muscles
2. Cardiac muscles
3. Skeletal / straited / voluntary muscles.

Question 2.
Name the type of muscles which bring about running and speaking.
Answer:
Skeletal muscles (Voluntary muscles)

Question 3.
Name the muscles which do not contract as per our will.
Answer:
Involuntary muscles (smooth muscles and cardiac muscles)

Question 4.
Which type of muscles show rhythmic contractions?
Answer:
Cardiac muscles

Question 5.
Which type of muscle is present in the diaphragm of the respiratory system?
Answer:
Skeletal muscle

Question 6.
State the functions of:

1. Smooth muscles
2. Cardiac muscles
3. Striated muscles

Answer:

1. Smooth muscles: They bring about involuntary movements like peristalsis in the alimentary canal, constriction and dilation of blood vessels.
2. Cardiac muscles: They bring about contraction and relaxation of the heart.
3. Striated muscles: They control voluntary movements of limbs, head, trunk, eyes, etc.

Question 7.
What is locomotion?
Answer:
The change in locus of whole body of living organism from one place to another place is called locomotion.

Question 8.
State the four basic types of locomotory movements seen in animals.
Answer:
The four basic types locomotory movements seen in animals are:

1. Amoeboid movement: It is performed by pseudopodia, e.g. leucocytes.
2. Ciliary movement: It is performed by cilia, e.g. ciliated epithelium. In Paramoecium, cilia help in passage of food through cytopharynx.
3. Whirling movement: It is performed by flagella, e.g. sperms.
4. Muscular movement: It is performed by muscles, with the help of bones and joints.

Question 9.
All locomotions are movements but all movements are not locomotion. Justify
Answer:
Locomotion occurs when body changes its position, however all movements may not result in locomotion. Thus, all locomotions are movements but all movements are not locomotion.

Question 10.
Kriti was diagnosed with knee tendon injury. She asked the doctor whether she will be able to walk due to the injury? If not then state the reason.
Answer:
Knee tendon injury affects the ability to walk. Kriti may not be able to walk freely as the tendons attached to the bones help in the movement of the parts of skeleton.

Question 11.
Explain the types of straited muscles.
Answer:
On the basis of movements striated muscles are of three types:

1. Agonists: These are considered as the prime movers. They bring about the initial movement of a part.
   e.g. biceps
2. Antagonists: They bring about the action opposite to that of prime movers. e.g. Triceps.
3. Svnergists They assist the prime movers. e.g. Brachialis assist biceps.

Question 12.
Describe the antagonistic muscles in detail.
Answer:
Following are the important antagonistic muscles:

1. Flexor and extensor: Flexor muscle on contraction results into bending or flexion of joint. e.g. Biceps. Extensor muscle on contraction results in straightening or extension of a joint. e.g. Triceps.
2. Abductor and adductor: Abductor muscle moves a body part away from the body axis. e.g. Deltoid muscle of shoulder moves the arm away from the body. Adductor muscle moves a body part towards the body axis.
   e.g. L.atissirnus dorsi of shoulder moves the arm near the body.
3. Pronator and supinator: Pronator turns the palm downwards and supinator turns the palm upward.
4. Levator and depressor: Levator raises a body part and the depressors lower the body part.
5. Protractor and retractor: Protractor moves forward, whereas the retractor moves backward.
6. Sphincters: Circular muscles present in the inner walls of anus, stomach. etc., for closure and opening.

Question 13.
Describe the structure of myosin and actin filaments, with the help of neat and labelled diagram.
Answer:
i. Myosin filament:

1. Each myosin filament is a polymerized protein. Many meromyosins (monomeric proteins) constitute one thick filament.
2. Myosin molecule consists of two heavy chains (heavy meromyosin / HMM) coiled around each other forming a double helix. One end of each of these chains is projected outwardly is known as cross bridge. This end folds to form a globular protein mass called myosin head.
3. Two light chains are associated with each head forming 4 light chains/light meromyosin / LMM.
4. Myosin head has a special ATPase activity. It can split ATP to produce energy.
5. Myosin contributes 55% of muscle proteins.
6. In sarcomere, myosin tails are arranged to point towards the centre of the sarcomere and the heads point to the sides of the myofilament band.
   

ii. Actin filament: It is a complex type of contractile protein. It is made up of three components:

1. F actin: It forms the backbone of actin filament. F actin is made up of two helical strands. Each strand is composed of polymerized G actin molecules. One ADP molecule is attached to G actin molecule.
2. Tropomyosin: The actin filament contains two additional protein strands that are polymers of tropomyosin molecules. Each strand is loosely attached to an F actin. In the resting stage, tropomyosin physically covers the active myosin-binding site of the actin strand.
3. Troponin: It is a complex of three globular proteins, is attached approx. 2/3^rd distance along each tropomyosin molecule. It has affinity for actin, tropomyosin and calcium ions. The troponin complex is believed to attach the tropomyosin to the actin. The strong affinity of troponin for calcium ions is believed to initiate the contraction process.
   

Question 14.
A muscle can only pull and not push the bone. Why?
Answer:
The fundamental characteristic of the muscle is contraction. Therefore, muscle can only pull and not push the bone.

Question 15.
Explain the physiology of muscle contraction.
Answer:
When the muscles are relaxed, the active sites remain covered with tropomyosin and troponin complex. Due to this, myosin cannot interact with active site of actin and thus contraction cannot occur.

1. When an impulse (action potential) comes to muscle through motor end plate, it spreads throughout the sarcolemma of the myofibril.
2. The transverse tubules of sarcoplasmic reticulum release large number of Ca⁺⁺ ions into sarcoplasm. These calcium ions interact with the troponin molecules and the interaction inactivates troponin-tropomyosin complex. This causes change in the structure of tropomyosin.
3. As a result, tropomyosin gets detached from the active site of actin (F actin) filament, exposing the active site for actin.
4. The myosin head cleaves the ATP to derive energy and gets attached to the uncovered active site of actin. This results into the formation of acto-myosin complex.
5. The myosin heads are now tilted backwards and pull the attached actin filament inwardly. This results in contraction of the muscle fibres.

Question 16.
Why do we shiver during winter?
Answer:

1. Humans are homeotherms as they can regulate their body temperature with respect to the surrounding temperature. During winter, when temperature falls, the thermoreceptors detect the change in temperature and send signals to the brain.
2. Shivering reflex i.e. rapid contraction of muscles is triggered by the brain to generate heat and raise the body temperature.

Question 17.
Write a short note on role of calcium ions in contraction and relaxation of muscles.
Answer:
Calcium ions play a major role in contraction and relaxation of muscles.

1. Calcium ions are released from the sarcoplasm during muscle contraction and stored in sarcoplasmic reticulum during muscle relaxation.
2. When a skeletal muscle is excited and an action potential travels along the T tubule, the concentration of calcium ions increases.
3. These calcium ions bind to troponin which in turn undergoes a conformational change that causes tropomyosin to move away from the myosin-binding sites on actin. Once these binding sites are free, myosin heads bind to them to form cross-bridges and the muscle fiber contracts.
4. The decrease in calcium ion concentration in the sarcoplasmic reticulum causes tropomyosin to slide back and block the myosin binding sites on actin. This causes the muscle to relax.

[Note: Students can scan the adjacent Q.R code to get conceptual clarity with the aid of a relevant video.]

Question 18.
Muscle contraction and relaxation are active processes. Give reason.
Answer:
Muscle contraction and relaxation are active processes as during both the processes energy is utilized by hydrolysis of ATP into ADP and inorganic phosphate by the enzyme ATPase.

Question 19.
Describe the properties of muscles observed on electrical stimulation.
Answer:
Properties of muscles observed on electrical excitation:

1. Single muscle twitch: It is a muscle contraction initiated by a single brief-stimulation. It occurs in 3 stages: a latent period of no contraction, a contraction period and a relaxation period.
2. Summation: If the muscle is stimulated before the end of the twitch, it generates greater tension i.e., summation or addition of effect takes place. Repeated stimuli will produce increasing strength of contraction (stair case phenomenon).
3. Tetanus: If stimulation is very rapid and frequent the muscle does not have time to relax. It remains in a
   state of contraction called tetanus.
4. Rcfraètory period: Immediately aller one stimulus, the muscle fibre cannot respond to another stimulus. This resting or refractory period is about 0.02 seconds.
5. Threshold stimulus: For a muscle fibre to contract, a certain minimum strength or intensity of stimulus is required. This is called threshold stimulus.
6. All or none principie: A stimulus below threshold will not result in contraction. A threshold stimulus will result in contraction. This contraction leads to maximum force. Higher stimulus will not increase force of contraction i.e. a muscle libre contracts either ftilly or not at all. This is ‘all or none’ principle. All types of muscle and nerve fibres obey this law.
7. Oxygen debt: During strenuous exercise, the oxygen supply of muscle rapidly becomes insufficient to maintain oxidative phosphorylation of respiratory substrate. At this stage, muscles contract anaerobically and accumulate lactic acid produced by anaerobic glycolysis. Lactic acid produces less AlP and is toxic. It causes tiredness, pain and muscle cramps. During recovery, oxygen consumption of the muscle far exceeds than that in the resting state. This extra oxygen consumed during recovery is called oxygen debt of the muscle.

Note: The duration of refractory period varies with the muscle involved. Cardiac muscles have a longer refractory period of about 250 msec whereas skeletal muscles have a short refractory period of about I msec]

Question 20.
What is difference between endoskeleton and exoskeleton?
Answer:
The supportive structures present inside the body form the endoskeleton and when the supportive structures are present on the outer surface of the body they form exoskeleton.

Question 21.
Name the tissues that form the structural framework of the body.
Answer:
Cartilage and bone

Question 22.
What type of bones are present in our body?
Answer:
Long bones, short bones, flat bones, irregular bones and sesamoid bones.

Question 23.
How do bones help in various ways?
Answer:

1. Bones form the framework of our body and thus provide shape to the body.
2. They protect vital organs thus help in the smooth functioning of body.
3. The joints between the bones help in movement and locomotion.
4. They provide firm surface for attachment of muscles.
5. They are reservoirs of calcium and form important site for hemopoiesis.

Question 24.
Explain the three types of lever found in human body.
Answer:
The three types of lever are as follows:

1. Class I lever: The joint between the first vertebra and occipital condyle of skull is an example of Class I lever. The force is directed towards the joints (fulcrum); contraction of back muscle provides force while the part of head that is raised acts as resistance.
   
2. Class II lever: Human body raised on toes is an example of Class II lever. Toe acts as fulcrum, contracting calf muscles provide the force while raised body acts as resistance.
   
3. Class III lever: Flexion of forearm at elbow exhibit lever of class III. Elbow joint acts as fulcrum and radius, ulna provides resistance. Contracting bicep muscles provides force for the movement.
   

Question 25.
Give an account of bones of human skull.
Answer:
Skull is made up of 22 bones. It is located at the superior end of vertebral column. The bones of skull are
joined by fixed or immovable joints except for jaw.
Skull consists of cranium or brain box and facial bones.

i. Cranium: It is made up of four median bones and two paired bones.

1. Frontal bone: It is median bone (unpaired) forming forehead, roof of orbit (eye socket) and the most anterior part of cranium. It is connected to two parietals, sphenoid and ethmoid bone.
2. Parietal bones: These paired bones form the roof of cranium and greater portion of sides of the cranium.
3. Temporal bones: These paired bones are situated laterally just above the ear on either side. Each temporal bone gives out zygomatic process that joins zygomatic bone to form zygomatic arch. Just at the base of zygomatic process is mandibular fossa, a depression for mandibles (lower jaw bone) that forms the only movable joint of the skull. This bone harbors the ear canal that directs sound waves into the ear. The processes of temporal bones provide points for attachment for various muscles of neck and tongue.
4. Occipital bone: It is a single bone present at the back of the head. It forms the posterior part and most of the base of cranium. The inferior part of this bone shows foramen magnum, the opening through which medulla oblongata connects with spinal cord. On the either sides of foramen magnum are two prominent protuberances called occipital condyles. These fit into the corresponding depressions present in 1st vertebra.
5. Sphenoid bone: Median bone present at the base of the skull that articulates with all other cranial bones and holds them together. This butterfly shaped bone has a saddle shaped region called sella turcica. In this hypophyseal fossa, the pituitary gland is lodged.
6. Ethmoid bone: This median bone is spongy in appearance. It is located anterior to sphenoid and posterior to nasal bones. It contributes to formation of nasal septum and is major supporting structure of nasal cavity.

ii. Facial Bones: Fourteen facial bones give a characteristic shape to the face. The growth of face stops of the age of 16.
Following bones comprise the facial bones:

1. Nasals: These are paired bones that form the bridge of nose.
2. Maxillae: These form the upper jaw bones. They are paired bones that join with all facial bones except mandible. Upper row of teeth are lodged maxillae.
3. Palatines: These are paired bones forming the roof of buccal cavity or floor of the nasal cavity.
4. Zygomatic bones: They are commonly called as cheek bones.
5. Lacrimal bones: These are the smallest amongst the facial bones.
   These bones form the medial wall of each orbit. They have lacrimal fossa that houses lacrimal sacs. These sacs gather tears and send them to nasal cavity.
6. Inferior nasal conchae: They form the part of lateral wall of nasal cavity. They help to swirl and filter air before it passes to lungs.
7. Vomer: The median, roughly triangular bone that forms the inferior portion of nasal septum.
8. Mandible: This median bone forms the lower jaw. It is the largest and strongest facial bone. It is the only movable bone of skull. It has curved horizontal body and two perpendicular branches i.e. rami. These help in attachment of muscles. It has lower row of teeth lodged in it.

Question 26.
Give an account of hyoid bone.
Answer:

1. It is a ‘U’ shaped bone that does not articulate with any other bone.
2. It is suspended from temporal bone by ligaments and muscles.
3. It is located between mandible and larynx.
4. It has horizontal body and paired projections called horns.
5. It provides site for attachment of some tongue muscles and muscles of neck and pharynx.
   

Question 27.
Sketch and label the anterior and ‘entraI view of skull.
Answer:

Question 28.
Mention the sutures in skull along with their location.
Answer:
Skull has many sutures (type of immovable joints) present, out of which four prominent ones are:

1. Coronal suture; Joins frontal bone with parietals.
2. Sagittal suture: Joins two parietal bones.
3. Lambdiodal suture: Joins two parietal bones with occipital bone.
4. Lateral/squamous sutures: Joins parietal and temporal bones on lateral side.

Question 29.
What are ear ossicles?
Answer:
The three tiny bones present in each middle ear namely malleus, incus and stapes, together are called ‘ear ossicles’

Question 30.
Draw a neat and labelled diagram of ear ossicles.
Answer:

Question 31.
Write a note on human vertebral column. Sketch and label the structure of the vertebral column.
Answer:

1. Human vertebral column or backbone is a part of axial skeleton.
2. It is made up of a chain of irregular bones called vertebrae.
   
3. Humans have 33 vertebrae in early years of growth, whereas adults have 26 vertebrae.
4. In adults, five sacral vertebrae lìjse to tòrm one sacrum and four coccygeal vertebrae fuse to form single coccyx.

Question 32.
Explain the cervical vertebrae in detail.
Answer:

1. Atlas vertebrae:
   
2. It is the 1^st cervical vertebra.
3. It is a ring like bone and consists of anterior and posterior arches.
4. It does not have centrum and spinous process.
5. The transverse processes and transverse foramina of atlas are large.
6. The vertebral foramen is large and divided into two parts by transverse ligament.
7. Spinal cord passes through anterior compartment.
8. Anterior zygopophyses are replaced by facets for attachment with occipital condyle of skull that forms ‘Yes joint’.

ii. Axis vertebrae:

1. It is the 2^nd cervical vertebrae.
2. The centrum of this vertebra gives out tooth-like odontoid process.
   
3. Odontoid process fits into the anterior portion of vertebral foramen of atlas vertebra thereby forming pivot joint, also known as ‘No joint’.

iii. Typical cervical vertebrae:

1. Vertebrae number 3 to 6 are called as typical cervical vertebrae.
2. They show short centrum and bifid spinous process.
3. The transverse processes of these vertebrae are reduced; each having large vertebrarterial canal at its base for the passage of vertebral artery.

iv. 7^th cervical vertebra (Vertebra prominens): It is the largest cervical vertebra where the neural spine is straight.

Question 33.
Write a note on thoracic vertebrae. Sketch and label thoracic vertebrae.
Answer:

1. There are 12 thoracic vertebrae found in the chest region.
2. The centrurn of thoracic vertebrae is heart shaped and all the processes are well developed.
3. Except for 11^th and 12^th vertebrae; transverse process of other thoracic vertebrae show facets for attachment with ribs.
   

Question 34.
Elaborate on lumbar vertebrae with help of a neat and labelled diagram.
Answer:

1. There are 5 lumbar vertebrae.
2. They are well developed vertebrae and exhibit all the characters of a typical vertebrae.
3. The centrum of the lumbar vertebrae is kidney shaped.

Question 35.
Give an account on:
i. Sacrum
ii. Coccyx
Answer:
i. Sacrum:

1. It is a triangular bone formed by the fusion of five sacral vertebrae.
2. It is located in pelvic cavity between two hip bones.
3. The anterior end of sacrum is broad and posterior end is narrow.
4. It consists of vertebral foramina formed by the fusion of vertebrae.
5. The reduced neural spines can be observed projecting from dorsal aspect of sacrum.
6. Function: It gives strength to pelvic girdle,

ii. Coccyx:

1. It is a triangular bone which is formed by fusion of four coccygeal vertebrae.
2. It is reduced and does not show vertebral foramina and spinous processes.
3. The transverse processes of coccygeal vertebrae are reduced.

Question 36.
Identify the given vertebrae and label it.

Answer:

Question 37.
Describe in brief the structure of thoracic cage.
Answer:
Thoracic cage consists of 12 pairs of ribs, breast bone or sternum.

i. Ribs:

1. A rib is a ‘C’ shaped bone. It is attached to respective thoracic vertebrae on dorsal side.
2. Twelve pairs of ribs are attached to twelve thoracic vertebrae. For attachment to the vertebrae the posterior ends of ribs have two protuberances namely the head and tubercle.
3. The head of rib attaches to facet formed by demifacets of adjacent thoracic vertebrae at the base of transverse processes.
   
4. The tip of transverse processes of these vertebrae also have facets for attachment of ribs where tubercles of ribs are attached.
5. Ribs provide space for attachment of intercostal muscles.
6. On ventral side, the ribs may or may not attach the sternum. Depending on their attachment, the ribs are classified into three types:
   1. True ribs: First seven pairs of ribs are attached directly to the sternum by means of their costal cartilages.
   2. False ribs: Costal cartilages of rib numbers 8, 9 and 10 are attached to rib number 7 on either side and not directly to the sternum. Hence, these are called false ribs.
   3. Floating ribs: Last two pairs of ribs have no ventral connection. Hence, they are called floating ribs.

ii. Sternum:

1. It is a flat, narrow bone, around 15 cms in length.
   
2. It is placed medially in anterior thoracic ] jugular notch wall (chest region).
3. It consists of three distinct parts – manubrium, body and xiphoid processes.
4. Manubrium shows two notches on anterio-lateral side for attachment with clavicle of each side. It also shows two notches on each of the lateral side for attachment of first two pairs of ribs.
5. Body of sternum is a flat bone that shows five notches on lateral aspect which are meant for direct or indirect attachment of ribs.
6. Xiphoid process is lowermost part of sternum which is initially cartilaginous and gets ossified in adults. It provides space for the attachment of diaphragm and abdominal muscles.
7. Ribs are attached to sternum by means of cartilaginous extensions called costal cartilages.

Question 38.
What is intercostal space?
Answer:
The space between the ribs is called as intercostal space.

Question 39.
Describe the bones of pectoral girdle.
Answer:

Pectoral girdle is called as shoulder girdle. It attaches forelimb skeleton with axial skeleton. There are two pectoral girdles, each consists of a shoulder blade or scapula and collar bone or clavicle.

i. Clavicle:

1. It is a ‘S’ shaped slender bone.
2. One end of clavicle is attached to acromion process of scapula. The other rounded end called sternal end attaches to manubrium of sternum.
3. This connects upper arm skeleton to axial skeleton.

ii. Scapula:

1. It is a large, flat, triangular bone that occupies posterior chest wall extending from 2nd to 7th ribs.
2. It is attached to axial skeleton by muscles and tendons.
3. Scapula bears a concave socket called glenoid cavity at its lateral angle.
4. Head of humerus (the upper arm bone) fits into the glenoid cavity.
5. A beak like coracoid process projects from lateral angle of scapula and acromion process arise from scapula. They can be easily felt as high point of shoulder. Both the processes are meant for attachment of muscles.

Question 40.
Describe the bones of forelimb.
Answer:
Forelimb consists of humerus, radius and ulna (together forming forearm bones), carpals (bones of wrist), metacarpals (bones of palm) and phalanges (bones of digits). It consists of 30 bones.

i. Humerus:

1. This is the bone of upper arm.
2. It has a hemispherical head at its proximal end. On either side of head of humerus are present a pair of projections termed greater and lesser tubercles.
   
3. Bicipital groove is a deep groove present between the tubercles where a tendon of biceps muscle is attached.
4. The shaft of humerus shows deltoid tuberosity. Distal end of humerus shows pulley like part called trochlea that articulates with ulna.

ii. Radius and Ulna:

1. Radius is located laterally on thumb side of the forearm.
2. The proximal end of radius has disc like head that articulates with humerus bone.
3. The shaft of radius widens distally to form styloid process.
4. Ulna is located medially on little finger side of forearm.
5. At the proximal end of ulna there is a prominent process called Olecranon process that forms elbow joint with humerus bone. On the lateral side, near the upper end of ulna is present the radial notch into which the side of head of radius is fixed.
   
   f. Radius and ulna articulate with each other at upper and lower extremities by superior and inferior radio-ulnar joints. In between the shaft of two bones, interosseous membrane is present.

iii. Carpals: These are bones of wrist, arranged in two rows of four each.

iv. Metacarpals: Five elongated metacarpals form the bones of palm. Their proximal ends join with carpals and distal ends form knuckles.

v. Phalanges: Phalanges form the bones of fingers and thumb. There are 14 phalanges in each hand (Four fingers have three phalanges each and thumb has two).

Question 41.
Explain briefly the pelvic girdle with a neat and labelled diagram.
Answer:

1. Pelvic girdle also known as hip girdle connects hind limb skeleton with axial skeleton.
2. It is made up of two hip bones called coxal bones. These coxal bones unite posteriorly with sacrum.
3. Coxal bone is a large irregularly shaped bone is made up of three parts – ilium, ischium and pubis.
4. At the point of fusion of ilium, ischium and pubis, a cavity called acetabulum is present that forms ball and socket joint with the thigh bone.
   
5. The two pubis bones are joined medially by cartilaginous joint called pubic symphysis. Pubis and ischium together form a ring of bone that encloses a space called obturator foramen.

Question 42.
Describe the bones of lower limb.
Answer:
The bones of lower limb are femur, patella, tibia and fibula, tarsals, metatarsals and phalanges.

i. Femur: The thigh bone is the longest bone in the body. The head is joined to shaft at an angle by a short neck. It forms ball and socket joint with acetabulum cavity of coxal bone. The lower one third region of shaft is triangular flattened area called popliteal surface. Distal end has two condyles that articulate with tibia and fibula.

ii. Patella: It is also called as knee cap. ft is a sesamoid bone (bone embedded in tendon). It is a flat rounded bone with a pointed lower end.

iii. Tibia and fibula: These are the two long bones of shank or lower leg. The two bones are connected to each other at the extremities. In between the two bones interosseous membrane is present.

1. Tibia: It is much thicker and stronger than fibula. Its broad and expanded upper end articulates with femur and the lower end articulates with talus (tarsal bone).
2. Fibula: It is a long slender bone on lateral side of tibia.
   

iv. Tarsals: These are the bones of ankle. Seven tarsals are arranged in three rows, two proximal, one intermediate and four distal.

v. Metatarsals: Five metatarsal bones support the foot. Proximally they attach with distal row of tarsals and distally the metatarsals articulate with phalanges.

vi. Phalanges: These are the bones of the toes. Except the big toe which has two phalanges, the other four toes have three phalanges each.

Question 43.
What is arthrology?
Answer:
The study of joints is called arthrology.

Question 44.
What makes synovial joint freely movable?
Answer:
Synovial joint is characterised by synovial cavity between the articulating bones which allows free movement at the joint. This makes the joint freely movable.

Question 45.
Complete the given table.

Answer:

Question 46.
State and explain the disorders related to muscles.
Answer:
i. Muscular dystrophy:

1. It is a gradual wasting disease affecting various groups of muscles.
2. It is a genetic disorder.
3. Usually the voluntary skeletal muscles are weakened whereas internal muscles such as diaphragm are not affected in the patient suffering from this disorder.
4. The Duchenne type of muscular dystrophy usually occurs in boys, affecting their lower limbs.
5. Limb girdle muscular dystrophy affects the muscles of shoulders or hips and it usually starts in adults between 20 – 30 years.
6. There is no cure for this disease.

ii. Myasthenia gravis:

1. It is a weakness of skeletal muscles.
2. It is caused by an abnormality at the neuromuscular junction that partially blocks muscle contraction.
3. It is an autoimmune disorder caused by excessive production of certain antibodies in the blood stream. These antibodies bind to acetylcholine receptors of neuromuscular junction. Thus, the transmission of nerve impulses to the muscle fibres is blocked. This causes progressive and extensive muscle weakness.
4. It may affect the eye and eyelid movements, facial expression and swallowing.
5. The degree of muscle weakness varies from local to general.
6. Symptoms: Ptosis (drooping or falling of upper eye lid), diplopia or double vision, difficulty in swallowing, chewing and speech.

Question 47.
Give an account on disorders related to bones.
Answer:

1. Arthritis: It is an inflammation of joints. It is a painful disorder of bones, ligaments, tendons, etc. In this disorder, joints become swollen, stiff and painful. It can lead to disability.
2. Arthritis is of three types:
   1. Osteoarthritis: In this disorder, the joint cartilage is degenerated. It is caused by various factors like aging, obesity, muscle weakness, etc. This is the most common type of arthritis that affects hands, knees and spine.
   2. Gouty arthritis (Gout): In this disorder, joint pain occurs due to the deposition of uric acid in joints. If uric acid is produced in excess or is not excreted, it accumulates in joints as sodium urate. The accumulated sodium urate degenerates the cartilage causing inflammation and pain. It generally affects the joints of feet.
   3. Rheumotoid arthritis: It is an autoimmune disorder where body’s immune system attacks its own tissues. In rheumatoid arthritis, synovial membrane swells up and starts secreting extra synovial fluid. This fluid exerts pressure on joint and makes it painful. Membrane may develop abnormal granulation tissue called pannus. Pannus may erode cartilage. Fibrous tissue gets ossified and may lead to stiffness in joints.
3. Osteoporosis:
   1. In this disorder, bones become porous and hence brittle. It is primarily age related disease and is more common in women than men.
   2. As age advances, bone resorption outpaces bone formation. Hence, the bones lose mass and become brittle. More calcium is lost in urine, sweat, etc., than it is gained through diet. Thus, prevention of disease is better than treatment by consuming adequate amount of calcium and exercise at young age.
   3. Osteoporosis may be caused due to decreasing estrogen secretion after menopause, deficiency of vitamin D, low calcium diet, decreased secretion of sex hormones and thyrocalcitonin.

Question 48.
Name the following.

Question 1.
The striated muscles that are known as prime movers
Answer:
Agonists

Question 2.
The antagonistic muscles that lower the body part
Answer:
Depressor

Question 3.
The contractile proteins of sarcomere
Answer:
Actin and myosin

Question 4.
Sliding filament theory is also known as
Answer:
Walk along theory or Rachet theory

Question 5.
The smallest facial bone
Answer:
Lacrimal bones

Question 6.
Number of bones in thoracic cage
Answer:
25

Question 7.
The first cervical vertebrae
Answer:
Atlas

Question 8.
The three bones of pelvic girdle
Answer:
Ilium, ischium and pubis

Question 9.
The bone which is known as the knee cap
Answer:
Patella

Question 10.
Age related disorder more common in woman than man.
Answer:
Osteoporosis

Question 49.
Fill in the blanks.

Question 1.
The actin filament contains two additional proteins strands that are polymers of ___ molecules.
Answer:
tropomyosin

Question 2.
___ of sarcoplasmic reticulum release large number of Ca⁺⁺ into sarcoplasm.
Answer:
Transverse (T) tubules

Question 3.
The joint between the first vertebra and occipital condyle of skull is an example of ____ class of lever.
Answer:
Class I

Question 4.
Endoskeleton of an adult human consists of ___ bones.
Answer:
206

Question 5.
___ are the bones that connect limbs to the axial skeleton.
Answer:
Girdles

Question 6.
___ bones form the roof of cranium.
Answer:
Parietal

Question 7.
Zveomatic bone is commonly known as ___ bone.
Answer:
Cheek

Question 8.
The largest and strongest facial bone is ___
Answer:
Mandible

Question 9.
There are ___ types of vertebrae.
Answer:
5

Question 10.
___ is also known as shoulder girdle.
Answer:
Pectoral girdle

Question 11.
___ are the bones of wrist.
Answer:
Carpals

Question 12.
___ is known as immovable or fixed joint.
Answer:
Synarthroses

Question 13.
Tibia and fibula are two long bones of ___.
Answer:
Shank or lower leg

Question 14.
___ is caused due to deposition of uric acid in joints.
Answer:
Gouty arthritis

Question 50.
State whether the following statements are True or False:

Question 1.
Saggital suture joins frontal bone with parietal bone.
Answer:
False. Saggital suture joins two parietal bones.

Question 2.
In human beings, there are 10 pairs of ribs.
Answer:
False. In human beings, there are 12 pairs of ribs.

Question 3.
Hyoid bone does not articulate with any other bone.
Answer:
True

Question 4.
Pivot joint is present between Atlas and Axis vertebrae.
Answer:
True

Question 5.
The two pubis bones are joined by cartilaginous joint called pubic symphysis.
Answer:
True

Question 6.
Rheumatoid arthritis is an autoimmune disorder.
Answer:
True

Question 50.
Identify the INCORRECTLY labelled part of pectoral girdle.

Answer:
Glenoid cavity is incorrectly labelled as obturator foramen.

Multiple Choice Questions

Question 1.
Locomotion in sperms takes place with the help of
(A) flagella
(B) cilia
(C) pseudopodia
(D) muscles
Answer:
(A) flagella

Question 2.
Levator muscles result into the action of
(A) lowering a body part
(B) tensing a body part
(C) relaxing a body part
(D) raising a body part
Answer:
(D) raising a body part

Question 3.
Which of these is not a skull bone?
(A) Frontal
(B) Scapula
(C) Occipital
(D) Temporal
Answer:
(B) Scapula

Question 4.
Immovable joint is in between the bones of
(A) frontal and parietal
(B) metacarpal and phalangeal
(C) femur and tibia
(D) radius and ulna
Answer:
(A) frontal and parietal

Question 5.
Total number of vertebrae in human beings is
(A) 22
(B) 33
(C) 24
(D) 12
Answer:
(B) 33

Question 6.
Which of the following is NOT a part of appendicular skeleton?
(A) Girdles
(B) Forelimb
(C) Hindlimb
(D) Vertebral column
Answer:
(D) Vertebral column

Question 7.
___ connects the upper arm to the axial skeleton.
(A) Clavicle
(B) Patella
(C) Tibia
(D) Femur
Answer:
(A) Clavicle

Question 8.
__ is a part of hind limb.
(A) Radius
(B) Ulna
(C) Humers
(D) Fibula
Answer:
(D) Fibula

Question 9.
Which of the following are toe bones?
(A) Tarsals
(B) Metatarsals
(C) Carpals
(D) Phalanges
Answer:
(D) Phalanges

Question 10.
Sutures on the skull are ___ joints.
(A) freely movable
(B) slightly movable
(C) fixed
(D) synovial
Answer:
(C) fixed

Question 11.
___ is an example of syndesmoses.
(A) Distal tibiofibular joint
(B) Rib-sternum junction
(C) Tooth and jaw bones
(D) Intervertebral disc
Answer:
(A) Distal tibiofibular joint

Question 12.
Elbow joint is
(A) ball and socket joint
(B) hinge joint
(C) suture joint
(D) gliding joint
Answer:
(B) hinge joint

Question 13.
Complete the analogy.
Hinge joint: Monoaxial movement:: ___ Biaxial movement
(A) Ball and socket joint
(B) Gliding joint
(C) Condyloid joint
(D) Both (A) and (B)
Answer:
(C) Condyloid joint

Question 14.
An autoimmune disorder in which an antibody reduces the efficiency of transmission between the motor neuron is called
(A) Myasthenia gravis
(B) Tetany
(C) Osteoarthritis
(D) Osteoporosis
Answer:
(A) Myasthenia gravis