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Chapter 5, Origin and Evolution of Life, hsc, biology, science, maharashtra bore, latest edition, full solution,

Chapter 5: Origin and Evolution of Life

Multiple choice question.

1.Who proposed that the first form of life could have come from per- existing nonliving organic molecules?

OPTIONS

  • Alfred Wallace
  • Oparin and Haldane
  • Charles Darwin
  • Louis Pasteur

2.The sequence of origin of life may be-

OPTIONS

  • Organic materials- inorganic materials – Eobiont- colloidal aggregates- cell.
  • Inorganic materials – organic materials – colloidal aggregates – Eobiont- cell
  • Organic materials- inorganic materials – colloidal aggregates – cell
  • Inorganic materials- organic materials – Eobiont- colloidal aggregates – cell

3.In Hardy – Weinberg equation, the frequency of homozygous recessive individual is represented by:

OPTIONS

  • p2
  • pq
  • q2
  • 2pq

4.Select the analogous organs.

OPTIONS

  • Forelimbs of whale and bat
  • Flippers of dolphins and penguin
  • Thorn and tendrils of Bougainvillea and Cucurbita.
  • Vertebrates hearts or brains.

5.Archaeopteryx is known as missing link because it is a fossil and share characters of both

OPTIONS

  • Fishes and amphibians
  • Annelida and arthropoda
  • Birds and reptiles
  • Chordates and nonchordates

6.Identify the WRONG statement regarding evolution.

OPTIONS

  • Darwin’s variations are small and directional
  • Mutations are random and nondirectional
  • Adaptive radiations leads to divergent evolution
  • Mutations are non – random and directional

7.Gene frequency in a population remain constant due to –

OPTIONS

  • Mutation
  • Migration
  • Random mating
  • Non- random mating

8.Which of the following characteristic is not shown by the ape?

OPTIONS

  • Prognathous face
  • Tail is present
  • Chin is absent
  • Forelimbs are longer than hind limbs

9.______ can be considered as a connecting link between ape and man.

OPTIONS

  • Australopithecus
  • Homo habilis
  • Homo erectus
  • Neanderthal man

10.The cranial capacity of Neanderthal man was

OPTIONS

  • 600 cc
  • 940 cc
  • 1400 cc
  • 1600 cc

Very short answer question.

1.Define the Gene pool

SOLUTION

The total genetic information encoded in the sum total of genes in a Mendelian population is called gene pool.

2.Define the Gene frequency

SOLUTION

The proportion of an allele in the gene pool, to the total number of alleles at a given locus, is called gene frequency.

3.Define Organic evolution.

SOLUTION

Organic evolution can be defined as slow, gradual, continuous and irreversible changes through which the present-day complex forms of the life developed (or evolved) from their simple pre-existing forms.

4.Define Population.

SOLUTION

According to this theory all individuals of the same species constitute a population.

5.Define Speciation.

SOLUTION

The process of formation of a new species from the pre-existing species is called speciation.

6.What is adaptive radiation?

SOLUTION

The process of evolution which results in the transformation of original species to many different varieties is called adaptive radiation.

7.If variation occurs in a population by chance alone and not by natural selection and brings a change in frequencies of an allele. What is it called? 

SOLUTION

If the variation in a population occurs by chance alone and not by natural selection and brings about a change in frequencies of an allele, it is called genetic drift.

8.State the Hardy – Weinberg equilibrium.

SOLUTION

The Hardy-Weinberg equilibrium law states that ‘at the equilibrium point, both the gene (allele) frequency and genotypic frequency remain constant from generation to generation’.

9.What are homologous organs?

SOLUTION

Homologous organs are those organs, which are structurally similar but perform different functions.

10.What is vestigeal organ?

SOLUTION

Vestigeal organs or rudimentary organs are imperfectly developed and non-functional, degenerate structures that were functional in some related and other animals or in ancestors.

11.What is the scientific name of the modern man?

SOLUTION

The scientific name of the modern man is Homo sapiens.

12.What is coacervate?

SOLUTION

Coacervates are colloidal aggregations of hydrophobic proteins and lipids (lipoid bubbles).

13.Which period is known as “age of Reptilia”?

SOLUTION

Jurassic period is known as age of Reptilia.

14.Name the ancestor of human which is described as a man with an ape brain.

SOLUTION

Australopithecus is the ancestor of humans which is described as a man with an ape brain.


Short answer question.

1.Write a note on Genetic drift.

SOLUTION

1. Any alternation in allete frequency in the natural population by chance, is called genetic drift. e.g. Elimination of a particular allele from a population due to events like accidental death prior to mating of an organism.

2. The concept of genetic drift was first given Sewall Wright, and is hence also called as the Sewall Wright effect.

3. Genetic drifts are random or directionless.

4. The effect of genetic drift is more significant in small population than in large population.

5. Due to genetic drift, some alleles of a population are lost or reduced by chance and some others may be increased.

6. Sometimes, a few individuals become isolated from the large population and they produce a new population in a new geographical areas. The allele frequency of the new population becomes different. The original drifted population (i.e. colonizing ancestor/ pioneer) becomes ‘founders’ and the effect is called the founder effect.

7. A bottleneck effect is seen when much of a population is killed due to a natural disaster and only a few remaining individuals are left to begin a new population.

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2.Enlist the different factors that are responsible for changing gene frequency.

SOLUTION

The four major factors that are responsible for changing gene frequency are as follows:

1. Gene flow (Migration):

Gene flow is the movement of genes into or out of a population. Gene movement may be in the form of migration of organism, or gametes (dispersal of pollens) or segments of DNA (transformation). Thus, gene flow alters gene frequency causing evolutionary changes.

2. Genetic drift:

Any random fluctuation (alteration) in allele frequency, occurring in the natural population by pure chance, is called genetic drift. For example, when the size of a population is severely reduced due to natural disasters like earthquakes, floods, fires, etc. cause the elimination of particular alleles from a population. Smaller populations have greater chances for genetic drift. Thus, genetic drift will result in the change in the gene frequency and has the potential to bring about evolutionary change.

3. Natural selection:

Natural selection is the process by which better adapted organisms grow and produce more offsprings in the population. It brings about evolutionary changes by favouring differential reproduction of genes that bring about changes in gene frequency from one generation to the next generation.

iv. Mutations: www.asterclasses.com

Sudden permanent heritable changes are called mutations. Mutation can occur in the gene, in the chromosome and in the chromosome number. Mutation that occurs within the single gene, is called point mutation or gene mutation or in a larger segment of genes by chromosomal aberrations. Both point mutations and chromosomal aberrations can alter gene frequency. Mutation leads to the change in the phenotype of the organism, causing variation.

3.Draw a graph to show that natural selection leads to disruptive change.

SOLUTION

4.Give the significance of fossils.

SOLUTION

1. Fossils are the dead remains of plants and animals that lived in past in various geological layers.

2. The study of fossils provides the most convincing and direct evidence of evolution.

3. Study of fossils tells us that life forms were not the same millions of years ago (mya).

4. The geological time’s scale is based on fossil records.

5. From the fossil records we can trace the complete evolutionary history of animals.

6. Study of fossils is an important aspect of evolution since it can be used in paleontology and anthropology for determining age of the fossils and deducing information about their ancestors.

5.Write the objections to Mutation theory of Hugo de Vries.

SOLUTION

Objections to Mutation Theory are as follows:

1. The large and discontinuous variations observed by Hugo de Vries were actually due to chromosomal aberrations. Gene mutations usually bring about only minor changes.

2. Rate of mutation is very slow as compared to the requirement of evolution.

3. Chromosomal aberrations have little significance in evolution as they are quite unstable.

6.What is disrruptive selection?

SOLUTION

Disruptive Natural selection is a selection in which more number of individuals acquire peripheral character value at both ends of the distribution curve.

7.Give an example of disrruptive selection

SOLUTION

1. In this selection, nature selects extreme phenotypes and eliminate intermediates.

2. This results in the formation of two peaks in the distribution of traits.

3. This kind of selection is rare.

4. It ensures the effect on the entire gene pool of a population, considering all mating types or systems.

e.g. Disruptive selection was observed in the different beak sizes of African seed cracker finches. The birds have different sizes of the beak and they feed on seeds. The available seeds were of two kinds i.e., small and large-sized seeds. Large beak sized birds feed on large seeds while small beak sized birds feed on small seeds and their number was increased. Intermediate beak sized birds were unable to feed on either type of seeds so their population decreased gradually and then were eliminated by natural selection.


1.Match the following:

Column- IColumn- II
1. August Weismanna. Mutation theory
2. Hugo de vriesb. Germplasm theory
3. Charl Darwinc. Theory of acquired characters
4. Lamarkd. Theory of natural selection

SOLUTIONS

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Column- IColumn- II
1. August Weismannb. Germplasm theory
2. Hugo de vriesa. Mutation theory
3. Charl Darwind. Theory of natural selection
4. Lamarkc. Theory of acquired characters
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2.What is adaptive radiation? Explain with suitable example.

SOLUTION

1. The process of evolution which results in the transformation of original species to many different varieties is called adaptive radiation.

2. Darwin’s Finches is one of the best examples of adaptive radiation. During his visit to Galapagos Islands, Charles Darwin also noticed a variety of small birds. These birds are now called Darwin’s finches. Darwin concluded that the American mainland species of the bird was the original one from which they migrated to the different islands of Galapagos. These birds adapted to the different environmental conditions of these islands. From original seed-eating features, many other forms with altered beaks evolved into insectivorous features.

3. Another example of adaptive radiation is Australian Marsupials. In Australia, there are many marsupial mammals who evolved from a common ancestor.

3.By taking industrial melanism as one example. Explain the concept of natural selection.

SOLUTION

1. Natural selection encourages those genes or traits that assure the highest degree of adaptive efficiency between the population and its environment.

2. Industrial melanism is one of the best examples of natural selection.

3. In Great Britain, before industrialization (1845) grey white-winged moths (Biston betularia) were more in number than black-winged moth (Biston carbonara).

4. These moths are nocturnal and during the day time they rest on a tree trunk.

5. White-winged moths were camouflaged (hide in the background) well with the lichen-covered trees that helped them to escape from the predatory birds.

6. However, the black-winged moth resting on lichen-covered tree trunks were easy victims for the predatory birds and their number was reduced.

7. During the industrial revolution, large number of industries came up in Great Britain.

8. The industries released black sooty smoke that covered and killed the lichens growing on a tree and turn the tree black due to pollution.

9. This change became an advantage to the black-winged moths that camouflaged well with the black tree trunks and their number increased

10. The white-winged moths however became victims to predatory birds due to which their number reduced. Thus, natural selection has resulted in the establishment of a phenotypic trait in the changing environmental conditions.

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4.Describe the Urey and Miller experiment.

SOLUTION

Stanley L. Miller and his teacher Harold C. Urey provided the first experimental evidence in support of chemical evolution theory of Oparin.

1. Apparatus and Procedure:

They designed a glass-apparatus called spark-discharge apparatus. The apparatus was first sterilized and evacuated. Methane, ammonia, and hydrogen gases were pumped in the proportion of 1:2:2 into the glass chamber. A tube carrying water vapour was also connected to the chamber. The lightning effect was mimicked by the action of electric discharge in the chamber.

The process of evaporation and precipitation was also stimulated by the use of heating mantle and condenser respectively.

The mixture of CH4, NH3, H2 was exposed continuously to electric discharge for several days causing the gases to interact, after which these were condensed.

2. Observation:

It was observed that the liquid collected in the U-tube turned brown.

3. Results and Conclusions:

Chemical analysis of this liquid reported the presence of simple organic compounds. (urea, amino acids, lactic acid, etc.). This experiment strongly supports that the simple molecules present in the Earth’s early atmosphere combined to form the organic building blocks of life.

5.What is Isolation?

SOLUTION

It is the separation of the population of a particular species into smaller units which prevents interbreeding between them. A barrier that prevents gene flow or exchange of genes between isolated populations, is called isolating mechanism.

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5.Describe the different types of reproductive isolations.

SOLUTION

A number of isolating mechanisms are operated in nature and therefore divergence and speciation may occur.

The isolating mechanisms are of two types namely, geographical isolation and reproductive isolation.

1. Geographical Isolation:

It is also called as physical isolation. It occurs when an original population is divided into two or more groups by geographical barriers such as rivers, oceans, mountains, glaciers, etc. These barriers prevent interbreeding between isolated groups.

The separated groups are exposed to different kinds of environmental factors and they acquired new traits by mutations. The separated populations develop distinct gene pool and they do not interbreed. Thus, new species have been formed by geographical isolation. e.g. Darwin’s Finches.

2. Reproductive Isolation:

Reproductive isolation occurs due to change in genetic material, gene pool and structure of genital organs. It prevents interbreeding between populations. Types of Isolating Mechanisms:

a. Pre-mating or pre-zygotic isolating mechanism:

This mechanism prevents fertilization and zygote formation.

i. Habitat isolation or (Ecological isolation): Members of a population living in the same geographic region but occupying separate habitats in such a way that potential mate do not meet.

ii. Seasonal or temporal isolation: Members of a population living in the same geographic region but are sexually mature at different years or different times of the year.

iii. Ethological isolation: Due to specific mating behavior the members of the population do not mate.

iv. Mechanical Isolation: Members of two populations have a difference in the structure of reproductive organs.

2. Post-mating or Post-zygotic barriers:

i. Gamete mortality: Gametes have a limited life span. Due to one or the other reasons, if the union of the two gametes does not occur in the given time, it results in gamete mortality.

ii. Zygote mortality: Here, egg is fertilized but the zygote dies due to one or the other reasons.

iii. Hybrid sterility: Hybrids develop to maturity but become sterile due to the failure of proper gametogenesis (meiosis).

e.g. Mule is an inter-generic hybrid that is sterile.

6.What is Genetic variations?

SOLUTION

Genetic variations are caused due to various aspects of mutation, recombination, and migration. The change in gene and gene frequencies is known as genetic variation. 

7.Explain the different factors responsible for genetic variations.

SOLUTION

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Genetic variations are caused by the following factors:

a. Gene Mutation:

Sudden permanent heritable change is called a mutation. Mutation can occur in the gene, in the chromosome, and in the chromosome number. The mutation that occurs within the single gene is called point mutation or gene mutation.

Mutation leads to the change in the phenotype of the organism, causing variation.

b. Genetic recombination:

In sexually reproducing organisms, during gamete formation, the exchange of genetic material occurs between non-sister chromatids of homologous chromosomes. This is called crossing over. It produces new genetic combinations that result in variations. Fertilization between opposite mating gametes leads to various recombinations resulting in the phenotypic variations causing a change in the frequencies of alleles.

c. Gene flow:

Gene flow is the movement of genes into or out of a population. Gene movement may be in the form of migration of organism, or gametes (dispersal of pollens) or segments of DNA (transformation). Gene flow also alters gene frequency causing evolutionary changes.

d. Genetic drift:

Any random fluctuation (alteration) in allele frequency, occurring in the natural population by pure chance, is called genetic drift.

e.g. When the size of a population is severely reduced due to natural disasters like earthquakes, floods, fires, etc., it causes the elimination of particular alleles from a population. Smaller populations have greater chances for genetic drift. It will result in a change in the gene frequency. Genetic drift is also an important factor for evolutionary change.

e. Chromosomal aberrations:

The structural and morphological change in chromosome due to rearrangement is called chromosomal aberrations. It changes the arrangement of the genes (order or sequence) that results in the variation.

Chromosomal aberrations occur due to the following reasons:

1. Deletion: Loss of genes from the chromosome.

2. Duplication: Genes are repeated or doubled in number on the chromosome.

3. Inversion: A particular segment of the chromosome is broken and gets reattached to the same chromosome in an inverted position due to 180° twists. There is no loss or gain of the gene complement of the chromosome. www.asterclasses.com

4. Translocation: Transfer (transposition) of a part of a chromosome or a set of genes to a non-homologous chromosome is called translocation. It is affected naturally by the transposons present in the cell.

Chromosomal aberrations-


Long answer question.

1.Complete the chart.

EraDominating groupof animal
1. Cenozoic____________
2. _________Reptiles
3. Palaeozic____________
4. _________Invertebrates

SOLUTION

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EraDominating groupof animal
1. CenozoicMammals
2. MesozoicReptiles
3. PalaeozicAmphibians
4. PalaeozoicInvertebrates

          COMPLETED


Chapter 4, Molecular Basis of Inheritance, hsc, biology, science, maharashtra bore, latest edition, full solution,

Multiple Choice Question:

1.Griffith worked on ____________.

OPTIONS

  • Bacteriophage
  • Drosophila
  • Frog eggs
  • Streptococci

2.The molecular knives of DNA are _______.

OPTIONS

  • Ligases
  • Polymerases
  • Endonucleases
  • Transcriptase

3.Translation occurs in the _______.

OPTIONS

  • nucleus
  • cytoplasm
  • nucleolus
  • lysosomes

4.The enzyme required for transcription is _______.

OPTIONS

  • DNA polymerase
  • RNA polymerase
  • Restriction enzyme
  • RNAase

5.Transcription is the transfer of genetic information from _______.

OPTIONS

  • DNA to RNA
  • tRNA to mRNA
  • DNA to mRNA
  • mRNA to tRNA

6.Which of the following is NOT part of protein synthesis?

OPTIONS

  • Replication
  • Translation
  • Transcription
  • All of these

7.In the RNA molecule, which nitrogen base is found in place of thymine?

OPTIONS

  • Guanine
  • Cytosine
  • Thymine
  • Uracil

8.How many codons are needed to specify three amino acids?

OPTIONS

  • 3
  • 6
  • 9
  • 12

9.Which out of the following is not an example of an inducible operon?

OPTIONS

  • Lactose operon
  • Histidine operon
  • Arabinose operon
  • Tryptophan operon

10.Place the following event of translation in the correct sequence

i. Binding of met-tRNA to the start codon.

ii. Covalent bonding between two amino acids.

iii. Binding of second tRNA.

iv. Joining of small and large ribosome subunits.

OPTIONS

  • iii, iv, i, ii
  • i, iv, iii, ii
  • iv, iii, ii, i
  • ii, iii, iv, i

Very Short Answer Question:

1.What is the function of an RNA primer during DNA synthesis?

SOLUTION

RNA primers provide the starting point for DNA polymerase to initiate synthesizing a new DNA strand.

2.Why the genetic code is considered as commaless?

SOLUTION

Genetic code is commaless: There is no gap or punctuation mark between successive/ consecutive codons.

3.What is genome?

SOLUTION

The term genome refers to the total genetic constitution of an organism.

OR

It is a complete copy of genetic information (DNA) or one complete set of chromosomes (monoploid or haploid) of an organism.

4.Which enzyme does remove supercoils from replicating DNA?

SOLUTION

Super helix relaxing enzyme removes supercoils from replicating DNA.

5.Why are Okazaki fragments formed on lagging strand only?

SOLUTION

  1. The two strands in DNA are antiparallel i.e. one strand runs in 5’ → 3’ direction whereas the other runs in 3’ → 5’.
  2. The DNA polymerase synthesizes a new DNA strand in 5’ → 3’ direction only.
  3. Leading template is synthesized continuously and lagging template is synthesized discontinuously.
  4. Due to 5’ → 3’ polymerizing activity of DNA polymerase Okazaki fragments are formed only on lagging strand only.

6.When does DNA replication takes place?

SOLUTION

DNA replication occurs in the S-phase of the interphase of the cell cycle, prior to cell division.

7.Define term- codon and codogen.

SOLUTION

Codon:

A sequence of three adjacent nucleotides in mRNA that codes for one amino acid are known as a codon.

Codogen:

It is the smallest possible sequence (triplet) of nucleotides present on the DNA strand which can specify one particular amino acid.

8.What is degeneracy of genetic code?

SOLUTION

Usually, the single amino acid is encoded by a single codon. However, some amino acids are encoded by more than one codon. e.g. Cysteine has two codons, while isoleucine has three codons. This is called the degeneracy of the code. Degeneracy of the code is explained by the Wobble hypothesis. Here, the first two bases in different codons are identical but the third one varies.

9.Which are the nucleosomal ‘core’ histones?

SOLUTION

The nucleosome core is made up of two molecules of each of four types of histone proteins viz. H2A, H2B, H3 and H4.

Short Answer Question:

1.Write a short note on DNA packaging in the eukaryotic cell.

SOLUTION

  1. The organization of DNA is much more complex in eukaryotes.
  2. Histone proteins are rich in lysine and arginine residues which are basic in nature and are positively charged.
  3. These histones organize themselves to make a unit of 8 molecules known as histone octamer.
  4. The negatively charged helical DNA is wrapped around the positively charged histone octamer, forming a structure known as a nucleosome.
  5. The nucleosome core is made up of two molecules of each of four types of histone proteins viz. H2A, H2B, H3 and H4. H1 protein binds the DNA thread where it enters (arrives) and leaves the nucleosome.
  6. One nucleosome approximately contains 200 base pair long DNA helix wound around it.
  7. About 146 base pair long segment of DNA remains present in each nucleosome.
  8. Nucleosomes are the repeating units of chromatin, which are threadlike, stained (coloured) bodies present in nucleus. These look like ‘beads-on-string’, when observed under an electron microscope.
  9. DNA helix of 200 bp wraps around the histone octamer by 1¾ turns.
  10. Six such nucleosomes get coiled and then form solenoid that looks like coiled telephone wire.
  11. The chromatin is packed to form a solenoid structure of 30 nm diameter (300Å) and further supercoiling tends to form a looped structure called chromatin fiber, which further coils and condenses at the metaphase stage to form the chromosomes.
  12. The packaging of chromatin at higher levels, needs an additional set of proteins that are called Non-Histone Chromosomal proteins (NHC).

2.Enlist the characteristics of genetic code.

SOLUTION

Genetic code of DNA has certain following characteristics:

1. Genetic code is a triplet code:

The sequence of three consecutive bases constitutes a codon, which specifies one particular amino acid. The base sequence in a codon is always in 5’ → 3’ direction. In every living organism, genetic code is a triplet code.

2. Genetic code has distinct polarity:

Genetic code shows definite polarity i.e. direction. It is always read in 5’ → 3’ direction and not in 3’ → 5’ direction. Otherwise the message will change e.g. 5’ AUG 3’

3. Genetic code is non-overlapping:

Code is non-overlapping i.e. each single base is a part of only one codon. Adjacent codons do not overlap.

4. Genetic code is commaless:

There is no gap or punctuation mark between successive/ consecutive codons.

5. Genetic code has degeneracy:

Usually, the single amino acid is encoded by a single codon. However, some amino acids are encoded by more than one codon. e.g. Cysteine has two codons, while isoleucine has three codons. This is called the degeneracy of the code. Degeneracy of the code is explained by the Wobble hypothesis. Here, the first two bases in different codons are identical but the third one varies.

6. Genetic code is universal:

In most of the living organisms, the specific codon specifies the same amino acid. e.g. Codon AUG always specifies amino acid methionine.

7. Genetic code is non-ambiguous:

The specific amino acid is encoded by a particular codon. Alternatively, two different amino acids will never be encoded by the same codon.

8. Initiation codon and termination codon:

AUG is always an initiation codon in any and every mRNA. AUG codes for amino acid methionine. Out of 64 codons, three codons viz. UAA, UAG, and UGA are termination codons that terminate/ stop the process of elongation of a polypeptide chain, as they do not code for any amino acid.

9. Codon and anticodon:

A codon is a part of DNA e.g. AUG is codon. It is always represented as 5’ AUG 3’. Anticodon is a part of tRNA. It is always represented as 3’UAC 5’.

3.Write a note on applications of DNA fingerprinting.

SOLUTION

  1. In forensic science, DNA fingerprinting is used to solve problems of rape and some complicated murder cases.
  2. DNA fingerprinting is used to find out the biological father or mother or both, of the child, in case of disputed parentage.
  3. DNA fingerprinting is used in the pedigree analysis in cats, dogs, horses and humans.

4.Explain the role of lactose in ‘Lac Operon’.

SOLUTION

  1. A few molecules of lactose enter into the cell by an enzyme permease.
  2. A small amount of this enzyme is present even when the operon is switched off.
  3. A few molecules of lactose, act as inducer and bind to the repressor.
  4. This repressor – inducer complex fails to join with the operator gene, which is then turned on.
  5. Structural genes produce all enzymes. Thus, lactose acts as an inducer of its own breakdown.
  6. When the inducer level falls, the operator is blocked again by the repressor. So structural genes are repressed/inactivated again. This is negative feedback.

Short Answer Question:

1.Write a note on Human genome project (HGP).

SOLUTION

The human genome project was initiated in 1990 under the International administration of the Human Genome Organization (HUGO).

This project was coordinated by the US Department of Energy and the National Institute of health. Additional contributors included universities across the United States and international partners in the United Kingdom, France, Germany, Japan and China.

The Human Genome Project was completed in 2003.

Following are the main aims of the human genome project:

1. Mapping the entire human genome at the level of nucleotide sequences.

2. To store the information collected from the project in databases.

3. To develop tools and techniques for analysis of the data.

4. Transfer of the related technologies to the private sectors, such as industries.

5. Taking care of the legal, ethical and social issues which may arise from the project

2.Describe the structure of ‘Operon’.

SOLUTION

The concept of the operon was first proposed by Jacob and Monod. A unit of genetic material that functions in a coordinated manner by means of a regulator, an operator, a promoter, and one or more structural genes that are transcribed together is called an operon. The clusters of genes with related functions are called operons.

Components of operon:

1. Regulator gene:

i. This gene controls the operator gene in cooperation with an inducer present in the cytoplasm.

ii. The regulator gene precedes the promoter gene. It may not be present immediately adjacent to the operator gene.

iii. The regulator gene produces a protein called repressor protein.

iv. The repressor binds with the operator gene and represses (stops) its action. Therefore, it is called regulator protein.

2. Promoter gene:

i. This gene precedes the operator gene. It is present adjacent to the operator gene.

ii. RNA polymerase enzyme binds to the promoter gene.

iii. The promoter gene base sequence determines which strand of DNA acts a template.

iv. When the operator gene is turned on, the enzyme moves over the operator gene and transcription of structural genes starts.

3. Operator gene:

i. This gene lies adjacent to the structural genes and controls their functioning.

ii. When the operator gene is turned on by an inducer, the structural genes produce mRNA.

iii. The operator gene is turned off by a product of the repressor gene.

4. Structural gene:

i. When lactose is added to the E. coli culture, the structural genes produce mRNA which in turn produces polypeptides, on the ribosomes.

ii. The polypeptides formed, act as enzymes to metabolize lactose in the cell.

iii. There are 3 structural genes in the sequence lacZ, lacY and lacA.

iv. Enzymes produced by these genes are β-galactosidase, permease, and transacetylase respectively.

Short Answer Question:

1.In the figure below A, B and C are three types of ______.

SOLUTION

In the given figure A (Messenger RNA), B (Ribosomal RNA) and C (Transfer RNA) are three types of Ribonucleic acids (RNA).

2.Identify the labeled structures on the following diagram of translation.

  • Part A is the _______
  • Part B is the _______
  • Part A is the _______

SOLUTION

  • Part A is the Anticodon present on the anticodon loop of tRNA.
  • Part B is the Amino acid
  • Part A is the Large subunit of ribosome

3.Match the entries in column I with those of column II and choose the correct answer.

Column IColumn II
A. Alkali treatmenti. Separation of DNA fragments on gel slab
B. Southern blottingii. Split DNA fragments into single strands
C. Electrophoresisiii. DNA transferred to nitrocellulose sheet
D. PCRiv. X-ray photography
E. Autoradiographyv. Produce fragments of different sizes
F. DNA treated with RENvi. DNA amplification

SOLUTION

Column IColumn II
A. Alkali treatmentii. Split DNA fragments into single strands
B. Southern blottingiii. DNA transferred to nitrocellulose sheet
C. Electrophoresisi. Separation of DNA fragments on gel slab
D. PCRvi. DNA amplification
E. Autoradiographyiv. X-ray photography
F. DNA treated with RENv. Produce fragments of different sizes

Long Answer Question:

1.Explain the process of DNA replication.

SOLUTION

The process by which DNA duplicates to form identical copies is known as replication.

Semi-conservative method of replication:

1. After replication, each daughter DNA molecule has one old and other new strands.

2. As parental DNA is partly conserved in each daughter’s DNA, the process of replication is called semi-conservative.

3. The model of semi-conservative replication was proposed by Watson and Crick.

4. The semi-conservative model of DNA replication using the heavy isotope of nitrogen N15 and E. coli was experimentally proved by Meselson and Stahl (1958).

Mechanism of replication is as follows:

a. Activation of Nucleotides:

i. The four types of nucleotides of DNA i.e. dAMP, dGMP, dCMP and dTMP are present in the nucleoplasm.

ii. They are activated by ATP in presence of an enzyme phosphorylase.

iii. This results in the formation of deoxyribonucleotide triphosphates i.e. dATP, dGTP, dCTP and dTTP. This process is known as Phosphorylation.

b. Point of Origin or Initiation point:

i. Replication begins at a specific point ‘O’ origin and terminates at point ‘T’.

ii. Origin is flanked by ‘T’ sites. The unit of DNA in which replication occurs is called replicon.

iii. In prokaryotes, there is only one replicon however in eukaryotes, there are several replicons in tandem.

iv. At the point ‘O’, enzyme endonuclease nicks one of the strands of DNA, temporarily.

v. The nick occurs in the sugar-phosphate backbone or the phosphodiester bond.

c. Unwinding of DNA molecule:

i. Enzyme DNA helicase breaks weak hydrogen bonds in the vicinity of ‘O’.

ii. The strands of DNA separate and unwind. This unwinding is bidirectional and continues as ‘Y’ shaped replication fork.

iii. Each separated strand acts as a template.

iv. The two separated strands are prevented from recoiling (rejoining) by SSBP (Single-strand binding proteins).

v. SSB proteins remain attached to both the separated strands for facilitating the synthesis of new polynucleotide strands.

d. Replicating fork:

i. The point formed due to the unwinding and separation of two strands appears like a Y-shaped fork, called replicating/ replication fork.

ii. The unwinding of strands imposes strain which is relieved by the super-helix relaxing enzyme.

e. Synthesis of new strands:

i. Each separated strand acts as a mould or template for the synthesis of a new complementary strand.

ii. It requires a small RNA molecule, called RNA primer.

iii. RNA primer attaches to the 3’ end of the template strand and attracts complementary nucleotides from the surrounding nucleoplasm.

iv. These nucleotides bind to the complementary nucleotides on the template strand by forming hydrogen bonds (i.e. A=T or T=A; G = C or C = G).

v. The newly bound consecutive nucleotides get interconnected by phosphodiester bonds, forming a polynucleotide strand.

vi. The synthesis of a new complementary strand is catalyzed by enzyme DNA polymerase. 7. The new complementary strand is always formed in 5’→ 3’ direction.

f. Leading and Lagging strand:

i. The template strand with free 3’ end is called a leading template and with free 5’ end is called a lagging template.

ii. The process of replication always starts at the C-3 end of the template strand and proceeds towards C-5 end.

iii. As both the strands of the parental DNA are antiparallel, new strands are always formed in 5’ → 3’ direction.

iv. One of the newly synthesized strands which develop continuously towards the replicating fork is called the leading strand.

v. Another new strand develops discontinuously away from the replicating fork and is called the lagging strand.

vi. Maturation of Okazaki fragments: DNA synthesis on the lagging template takes place in the form of small fragments called as Okazaki fragments (named after scientist Okazaki).

vii. Okazaki fragments are joined by the enzyme DNA ligase.

viii. RNA primers are removed by DNA polymerase and replaced by DNA sequence with the help of DNA polymerase-I in prokaryotes and DNA polymerase-α in eukaryotes.

ix. Finally, DNA gyrase (topoisomerase) enzyme forms a double helix to form daughter DNA molecules.

g. Formation of daughter DNA molecules:

Asterclasses

i. At the end of the replication, two daughter DNA molecules are formed.

ii. In each daughter’s DNA, one strand is parental and the other one is totally newly synthesized.

iii. Thus, 50% is contributed by mother DNA. Hence, it is described as semiconservative replication.

2.Describe the process of transcription in protein synthesis.

SOLUTION

The process of copying of genetic information from one (template) strand of DNA into a single-stranded RNA transcript is called transcription.

The process of transcription is as follows:

  1. For transcription, promoter, structural gene, and terminator (together called transcription unit) are required.
  2. The DNA strand used for the synthesis of RNA is called antisense or template strand which is oriented in 3′ → 5′ direction, while the other strand not involved in RNA synthesis is called the coding strand. It is oriented in 5′ → 3′ direction.
  3. A small DNA sequence which provides a binding site for RNA polymerase is called promoter which is present towards 5′ end/upstream, while a small DNA sequence which terminates the transcription process called terminator is present towards 3′ end/downstream.
  4. The process of transcription, in both prokaryotes and eukaryotes, involves three stages viz. Initiation, Elongation, and Termination.
  5. During initiation, RNA polymerase binds to the promoter and moves along the DNA and causes local unwinding of DNA duplex into two chains in the region of the gene.
  6. Exposed ATCG bases project into the nucleoplasm.
  7. Only one strand functions as template (antisense strand) and the other strand is complementary which is actually a coding strand (sense strand).
  8. During elongation, the ribonucleoside triphosphates join bases of the DNA template chain.
  9. As transcription proceeds, the hybrid DNA-RNA molecule dissociates and makes mRNA molecules free.
  10. As transcription proceeds, the hybrid DNA-RNA molecule dissociates and makes mRNA molecules free.
  11.  

3.Describe the process of translation in protein synthesis.

SOLUTION

Definition:

The translation is the mechanism in which codons of mRNA are translated and specific amino acids in a sequence form a polypeptide on ribosomes.

The process of translation requires amino acids, mRNA, tRNA, ribosomes, ATP, Mg++ ions, enzymes, elongation, translocation and release factors.

  1. About 20 different types of amino acids available in the cytoplasm are known to form proteins.
  2. DNA controls the synthesis of proteins having amino acids in a specific sequence. This control is possible through the transcription of mRNA. Genetic code is specific for particular amino acids.
  3. RNAs serve as intermediate molecules between DNA and protein.
  4. Ribosomes serve as a site for protein synthesis. Each ribosome consists of large and small subunits. These subunits occur separately in the cytoplasm. Only during protein synthesis, in presence of Mg++ ions, these two subunits get associated together.

Mechanism of translation (Synthesis of polypeptide chain):

It involves three steps initiation, elongation and termination:

a. Initiation:

1. Activation of amino acids is essential before translation initiates.

2. The amino acid is activated by utilizing energy from ATP molecule. This amino acid binds with the amino acid binding site of tRNA and forms of tRNA- amino acid complex.

3. A small subunit of ribosome attaches to the mRNA at 5’ end.

4. The initiator codon, AUG is present on mRNA which initiates the process of protein synthesis.

5. Initiator charged tRNA (with activated amino acid methionine) binds with the initiation codon (AUG) by its anticodon (UAC) through hydrogen bonds.

6. It carries activated amino acid methionine (in eukaryotes) or formyl methionine (in prokaryotes).

7. It occupies the P site of the ribosome and the A- the site is vacant.

8. Now the large subunit of ribosome joins with the smaller subunit that requires Mg++ ions.

b. Elongations:

During this process, activated amino acids are added one by one to first amino acid (methionine). Addition of Amino acid occurs in 3 Step cycle –

1. Codon recognition- Amino acyl tRNA molecule enters the ribosome at A-site. Anticodon binds with the codon by hydrogen bonds.

2. Amino acid on the first initiator tRNA at P-site and amino acid on tRNA at A-site join by peptide bond. Here enzyme Ribozyme acts as a catalyst. At this time first tRNA at ‘P’ site is kicked off.

3. Translocation-

The tRNA at A-site carrying a dipeptide at A-site moves to the P site. This process is called translocation. In translocation, both the subunits of ribosome move along in relation to tRNA and mRNA. Hence, tRNA carrying dipeptide now gets positioned at ‘P’ site of the ribosome, making ‘A’ site vacant. At this site, then next charged tRNA molecule carrying amino acid will be received. During this process, the first uncharged tRNA is discharged from E-site. This process of arrival of tRNA- amino acid complex, the formation of the peptide bond, ribosomal translocation, and removal of the previous tRNA, are repeated.

c. Termination and release of polypeptide:

1. Towards the 3’ end of mRNA, there is a stop codon (UAA/ UAG/ UGA). It is exposed at the A-site.

2. It is not read and joined by the anticodon of any tRNA.

3. The release factor binds to the stop codon, thereby terminating the translation process.

4. The polypeptide is now released in the cytoplasm.

5. Two subunits of ribosome dissociate and last tRNA is set free in the cytoplasm.

6. mRNA also has some additional sequences that are not translated and are referred as untranslated regions (UTR).

7. The UTRs are present at both 5’-end (before start codon) and at 3’- end (after stop codon). They are required for an efficient translation process.

8. Finally, mRNA is also released in the cytoplasm. It gets denatured by nucleases immediately. Hence mRNA is short-lived.

4.Describe the ‘Lac-operon’.

SOLUTION

1. Lactose or lac operon of E. coli is an inducible operon. The operon is switched on when a chemical inducer- lactose is present in the medium.

2. Jacob and Monad proposed the classical model of Lac operon.

3. The Lac operon consists of the promoter site (P), regulatory site (i), and operator site (O).

4. It also has three structural genes, namely z, y and each producing an enzyme.

5. The following three enzymes are required for the metabolism of lactose in the cell.

Name of geneEnzyme producedFunction
lac zβ-galactosidaseLactose–β-galactosidase———–Glucose+GalactoseLactose→β-galactosidaseGlucose+Galactose
lac yPermeaseEntry of lactose in the cell
lac aTransacetylaseTransfers acetyl group from Acetyl CoA to β-galactosidase

6. If glucose is not available for cells, they will require another source of energy such as lactose.

7. If lactose is not available, the repressor protein produced by repressor gene will attach to the operator and block RNA polymerase.

8. Lactose acts as an inducer. If lactose is available, it will prevent the repressor from binding the operator, by forming an inducer-repressor complex and allow RNA polymerase to transcribe mRNA.

9. RNA polymerase will attach to the promoter and will begin transcribing mRNA.

10. RNA polymerase first transcribes the lac z gene which is responsible for synthesizing β-galactosidase.

11. RNA polymerase moves on to the next gene, lac y that synthesizes the enzyme permease.

12. RNA polymerase finally moves to the lac a gene that is responsible for synthesizing transacetylase.

13. β-galactosidase, permease and transacetylase are enzymes in the metabolic pathway used to get energy from lactose.

14. After lactose is used up and levels decrease, the repressor will attach to the operator blocking the production of β-galactosidase, permease and transacetylase, so that lactose levels increase.

Long Answer Question:

The DNA molecule is double-stranded and the RNA molecule is single-stranded.

OPTIONS

  • True
  • False

SOLUTION

True.

  1. The DNA is responsible for storing genetic information and also preserving it for the next generation of cells. Thus, it needs to be stable and resistant to enzymatic or oxidative alteration.
  2. The double-stranded DNA molecule is so designed that the part which stores the genetic information; the nitrogenous base pairs are stacked inward.
  3. Phosphate groups 
  4. (PO4-)
  5.  keep the base pairs safe inside form the backbone. DNA is further wrapped up around histone proteins into chromosomes which keep it condensed. This would not be possible if it were only single-stranded.
  6. RNA is not meant to last long and acts as a template carrying information which is copied from the DNA. Using protein-synthesizing machinery, it forms a protein. After which is disintegrates and digested.

Long Answer Question:

The process of translation occurs at the ribosome.

OPTIONS

  • True
  • False

SOLUTION

True.

Ribosomes are sites for protein synthesis. Ribosome is responsible for holding mRNA incorrect position. Therefore, the process of translation occurs at the ribosomes.

Long Answer Question:

The job of mRNA is to pick up amino acids and transport them to the ribosomes.

OPTIONS

  • True
  • False

SOLUTION

False.

The job of tRNA is to pick up amino acids and transport them to the ribosomes. mRNA carries message in the form of code from DNA.

Transcription must occur before translation may occur.

OPTIONS

  • True
  • False

SOLUTION

True.

Transcription is a process of formation of mRNA whereas translation is process of protein synthesis. For protein synthesis mRNA is required to act as a template. Therefore, transcription must occur before translation may occur.

6.Guess (i) the possible location of DNA on the collected evidence from a crime scene and (ii) the possible sources of DNA.

EvidencePossible location of DNA on the evidenceSources of DNA
e.g. Eyeglassese.g. Ear piecese.g. Sweat, Skin
Bottle, Can, GlassSides, mouthpiece__________
__________HandleSweat, skin, blood
Used cigaretteCigarette butt____________
Bite mark____________Saliva
__________Surface areaHair, semen, sweet, urine

SOLUTION

EvidencePossible location of DNA on the evidenceSources of DNA
e.g. Eyeglassese.g. Ear piecese.g. Sweat, Skin
Bottle, Can, GlassSides, mouthpieceSaliva, sweat
Door, baseball bat, a similar weaponHandleSweat, skin, blood
Used cigaretteCigarette buttSaliva
Bite markPersons skinor clothingSaliva
Blanket, pillow, bedsheet, dirty laundrySurface areaHair, semen, sweat, urine

COMPLETED


Chapter 3, Inheritance and Variation, hsc, biology, science, maharashtra bore, latest edition, full solution,

Multiple choice question.

1.Phenotypic ratio of incomplete dominance in Mirabilis jalapa.

OPTIONS

  • 2 : 1 : 1
  • 1 : 2 : 1
  • 3 : 1
  • 2 : 2

2.In a dihybrid cross, F2 generation offsprings show four different phenotypes while the genotypes are _______.

OPTIONS

  • six
  • nine
  • eight
  • sixteen

3.A cross between an individual with an unknown genotype for a trait with recessive plant for that trait is _______.

OPTIONS

  • back cross
  • reciprocal cross
  • monohybrid cross
  • test cross

4.When phenotypic and genotypic ratios are the same, then it is an example of _______.

OPTIONS

  • incomplete dominance
  • complete dominance
  • multiple alleles
  • cytoplasmic inheritance

5.If the centromere is situated near the end of the chromosome, the chromosome is called ______.

OPTIONS

  • metacentric
  • acrocentric
  • sub-metacentric
  • telocentric

6.Chromosomal theory of inheritance was proposed by _______.

OPTIONS

  • Sutton and Boveri
  • Watson and Crick
  • Miller and Urey
  • Oparin and Halden

7.If the genes are located in a chromosome as p-q-r-s-t, which of the following gene pairs will have the least probability of being inherited together?

OPTIONS

  • p and q
  • r and s
  • s and t
  • p and s

8.Find the mismatch pair:

OPTIONS

  • Down’s syndrome = 44 + XY
  • Turner’s syndrome = 44 + XO
  • Klinefelter syndrome = 44 + XXY
  • Super female = 44 + XXX

9.A colour-blind man marries a woman, who is homozygous for normal colour vision, the probability of their son being colour blind is –

OPTIONS

  • 0%
  • 25%
  • 50%
  • 100%

Very Short Answer Question.

1.Explain the statement of Test cross is back cross but back cross is not necessarily a test cross.

SOLUTION

  1. In back cross F1 generation can be crossed with either dominant or recessive parent.
  2. But in test cross, F1 generation is crossed with a recessive parent only.
  3. Thus, in the back cross, if F1 generation is crossed with a recessive parent it will be a test cross, but if F1 generation is crossed with a dominant parent it will not be a test cross. Therefore, the test cross is a back cross but the back cross is not necessarily a test cross.

2.Explain the statement of Law of dominance is not universal.

SOLUTION

  1. According to law of dominance, when two homozygous individuals with one or more sets of contrasting characters are crossed, the alleles (characters) that appear in F1 are dominant and those which do not appear in F1 are recessive.
  2. In many cases, the dominance is not complete or absent. This can be explained by two deviations of Mendel’s law of dominance: Incomplete dominance and codominance. Thus, law of dominance is significant and true, but it is not universally applicable.

3.Define dihybrid cross

SOLUTION

A cross between parents differing in two heritable traits is called a dihybrid cross.

OR

A cross between two pure (homozygous) parents in which the inheritance pattern of two pairs of contrasting characters is considered simultaneously is called a dihybrid cross.

1.Define the following:

1.Homozygous

SOLUTION 1

Homozygous – Diploid condition where both the alleles are identical is called homozygous.

SOLUTION 2

Homozygous (pure): An individual having identical alleles for a particular character is homozygous for that character. It is pure or true breeding for that trait. e.g. TT, tt.

2.Heterozygous

SOLUTION 1

Heterozygous – Diploid condition where both the alleles are different is called heterozygous.

SOLUTION 2

Heterozygous: An individual possessing contrasting (dissimilar) alleles for a particular trait is called heterozygous. It is a hybrid and does not breed true for that trait. e.g. Tt

3.Test cross.

SOLUTION

Test cross: The cross between F1 hybrid and its homozygous recessive parent is called a test cross.


Very Short Answer Question.

1.What is allosome?

SOLUTION

The chromosomes which are responsible for the determination of sex are known as Allosomes (sex chromosomes).

2.What is crossing over?

SOLUTION

  1. Crossing over is a process that produces new combinations (recombinations) of genes by interchanging and exchanging of corresponding segments between non-sister chromatids of homologous chromosomes.
  2. It occurs during pachytene of prophase I of meiosis.
  3. The mechanism of crossing over consists of four sequential steps such as synapsis, tetrad formation, crossing over, and terminalisation.
  4. The phenomenon of crossing over is universal and it is necessary for the natural selection because it increases the chances of variation.

3.Give one example of the autosomal recessive disorder.

SOLUTION

Autosomal recessive traits Phenyl ketonuria (PKU), Cystic fibrosis, and Sickle cell anaemia.

4.What are X-linked genes?

SOLUTION

The genes which are present on the non-homologous region of X-chromosome are known as X-linked genes.

5.What are holandric traits?

SOLUTION

The traits that are controlled by genes present only on the Y chromosome are known as holandric traits.

6.Give an example of a chromosomal disorder caused due to nondisjunction of autosomes.

SOLUTION

Down syndrome is an example of a chromosomal disorder caused due to non-disjunction of autosomes.

7.Give one example of complete sex linkage?

SOLUTION

Examples of X-linked traits are hemophilia, red-green colour blindness, myopia (near sightedness), and for Y-linked are hypertrichosis, ichthyosis, etc.


Short Answer Question.

1.Enlist seven traits of pea plant selected/ studied by Mendel.

SOLUTION

Following are the seven traits of pea plant selected/studied by Mendel:

CharacterContrasting form / traits
 DominantRecessive
1. Height of stemTallDwarf
2. Colour of flowerPurpleWhite
3. Position of flowerAxialTerminal
4. Pod shapeInflatedConstricted
5. Pod colourGreenYellow
6. Seed shapeRoundWrinkled
7. Seed colour (cotyledon)YellowGreen

2.Why law of segregation is also called the law of purity of gametes?

SOLUTION

  1. A diploid organism contains two factors for each trait in its diploid cells and the factors segregate during the formation of gametes.
  2. The two alleles (contrasting characters) do not mix, alter or dilute each other and the gametes formed are ‘pure’ for the characters which they carry.
  3. A gamete may carry either a dominant or recessive factor but not both.

Hence, this law is also called the law of purity of gametes.

3.Write a note on pleiotropy.

SOLUTION

  1. When a single gene controls two (or more) different traits it is called pleiotropic gene and the phenomenon is called pleiotropy or pleiotropism.
  2. The phenotypic ratio is 1:2 instead of 3:1 because of the death of recessive homozygote. The disease, sickle-cell anaemia, is caused by a gene HbS.
  3. Normal or healthy gene HbA is dominant. The carriers (heterozygotes HbA /HbS) show signs of mild anaemia as their RBCs become sickle shaped i.e. half-moon-shaped only under abnormally low O2 concentration.
  4. The homozygotes with recessive gene HbS die of fatal anaemia.
  5. Thus, the gene for sickle-cell anaemia is lethal in homozygous condition and produces sickle cell trait in the heterozygous carrier. Two different expressions are produced by a single gene.

4.What are the reasons for Mendel’s success?

SOLUTION

Following are the reasons for Mendel’s success:

  1. His experiments were carefully planned and involved a large sample.
  2. He carefully recorded the number of plants of each type and expressed his results as ratios.
  3. In the pea plant, contrasting characters can be easily recognized.
  4. The seven different characters in the pea plant were controlled by a single factor each.
  5. The factors are located on separate chromosomes and these factors are transmitted from generation to generation.

5.“Father is responsible for determination of sex of child and not the mother”. Justify.

Asterclasses.com

SOLUTION

  1. A human male has 44 autosomes + XY sex chromosomes, whereas a female has 44 autosomes + XX sex chromosomes.
  2. During gamete formation in male, the diploid germ cells in testis undergo spermatogenesis to produce two types of haploid sperms, 50% sperms contain 22 autosomes and X chromosome while 50% sperms contain 22 autosomes and Y chromosome.
  3. In females, the diploid germ cells in ovaries undergo oogenesis to produce only one type of egg. All eggs contain 22 autosomes and the X chromosome.
  4. Thus human male is heterogametic and female is homogametic.
  5. If a sperm containing X chromosome fertilizes the egg (ovum), the diploid zygote formed after fertilization grows into a female child.
  6. If a sperm containing Y chromosome fertilizes the egg, then diploid zygote formed after fertilization grows into a male child.
  7. The sex of a child depends on the type of sperm fertilizing the egg and hence the father is responsible for the determination of sex of the child and not the mother

6.What is a linkage? How many linkage groups do occur in human being?

SOLUTION

The tendency of two or more genes presents on the same chromosomes to be inherited together is known as linkage. The haploid number of chromosomes in humans is 23 therefore there are 23 linkage groups in humans.

7.Write note on –PKU.

SOLUTION

  1. Phenylketonuria is an inborn metabolic disorder caused due to deficiency of phenylalanine hydroxylase enzyme.
  2. Phenylketonuria is caused due to recessive autosomal genes.
  3. When recessive genes are present in homozygous condition, phenylalanine hydroxylase enzyme is not produced.
  4. This enzyme is essential for the conversion of amino acid phenylalanine into tyrosine.
  5. Due to the absence of this enzyme, phenylalanine is not converted into tyrosine.
  6. Hence, phenylalanine and its derivatives are accumulated in blood and cerebrospinal fluid (CSF).
  7. It affects development of the brain and causes mental retardation.
  8. Excess phenylalanine is excreted in urine, hence this disease is called phenylketonuria.

8.Compare X chromosome and Y chromosome.

SOLUTION

X ChromosomeY Chromosome
1. It is metacentric, hence appears X shaped.1. It is acrocentric, hence appears Y shaped
2. It is longer than ‘Y’ chromosomes.2. It is shorter than ‘X’ chromosomes.
3. It contains a large amount of euchromatin and a small amount of heterochromatin.3. It contains large amount of heterochromatin and small amount of euchromatin.
4. It is found in both males and females.4. It is found only in males.
5. Non-homologous part of X chromosome shows more genes than Y chromosome.5. Non-homologous part of Y chromosome contains few genes as compared to X chromosome.
6. X – linked genes are present on the X chromosome.6. Y-linked genes (Holandric genes) are present on Y chromosome.
7. Genes present on X chromosome show crisscross inheritance.7. Genes present on Y chromosome show straight inheritance.

9.Explain the chromosomal theory of inheritance.

SOLUTION

The chromosomal theory of inheritance was proposed by Sutton and Boveri.

Following are the postulates of chromosomal theory of inheritance:

  1. Chromosomes are found in pairs in somatic or diploid cells.
  2. During gamete formation, homologous chromosomes pair, segregate and assort independently at meiosis. Due to this, each gamete contains only one chromosome of a pair.
  3. Hereditary characters are carried by chromosomes which are present in the nucleus of these gametes.
  4. Gametes (sperm and egg) contain all the hereditary characters. They form the link between parents and offsprings.
  5. The union of sperm and egg during fertilization restores the diploid number of chromosomes.

10.Observe the given pedigree chart and answer the following question.

Identify whether the trait is sex-linked or autosomal.

SOLUTION

The trait represented in given pedigree is sex linked trait.

11.Observe the given pedigree chart and answer the following question.

Give an example of a trait in human beings which shows such a pattern of inheritance.

SOLUTION

Haemophilia, colour blindness are examples of sex-linked traits in humans.


1.Match the column-I with column-II and re-write the matching pairs.

Column-IColumn-II
1. 21 trisomya. Turner’s syndrome
2. X-monosomyb. Klinefelter’s syndrome
3. Holandric traitsc. Down’s syndrome
4. Feminized maled. Hypertrichosis

SOLUTION

Column-IColumn-II
1. 21 trisomyc. Down’s syndrome
2. X-monosomya. Turner’s syndrome
3. Holandric traitsd. Hypertrichosis
4. Feminized maleb. Klinefelter’s syndrome

Very Short Answer Questions.

1.Define dihybrid cross

SOLUTION

A cross between parents differing in two heritable traits is called a dihybrid cross.

OR

A cross between two pure (homozygous) parents in which the inheritance pattern of two pairs of contrasting characters is considered simultaneously is called a dihybrid cross.

2.Explain a dihybrid cross with suitable example and checker board method.

SOLUTION

The phenotypic ratio of different types of offsprings (with different combinations) obtained in F2 generation of dihybrid cross is called the dihybrid ratio. It is 9 : 3 : 3 : 1.

For example, when we cross a yellow round seed pea plant with a green wrinkled seed pea plant, we get 9 yellow round, 3 yellow wrinkled, 3 green round and 1 green wrinkled plants in the F2 generation.

F2 generation → 

YRYryRyr
 
YRYYRR Yellow roundYYRr Yellow roundYyRR Yellow roundYyRr Yellow round
YrYYRr Yellow roundYYrr Yellow wrinkledYyRr Yellow roundYyrr Yellow wrinkled
yRYyRR Yellow roundYyRr Yellow roundyyRR Green roundyyRr Green round
yrYyRr Yellow roundYyrr Yellow wrinkledyyRr Green roundyyrr Green wrinkled

Phenotypic ratio: Yellow round = 9 ;Yellow wrinkled = 3; Green round = 3; Green wrinkled = 1

Dihybrid ratio → 9 : 3 : 3 : 1

Genotypic ratio → 

YYRR YYRr YyRR YyRr YYrr Yyrr yyRR yyRr yyrr

  1   :   2    :   2   :   4   :   1  :  2  :   1   :  2  :   1

3.Explain with suitable example an independent assotrment.

SOLUTION

Law of independent assortment:

The law states that, when a hybrid possessing two (or more) pairs of contrasting factors (alleles) forms gametes, the factors in each pair segregate independently of the other pair.

OR

The law of independent assortment states that, when two parents differing from each other in two or more pairs of contrasting characters are crossed, then the inheritance of one pair of character is independent of the other pair of character

  1. This law is based on a dihybrid cross.
  2. It describes how different genes or alleles present on separate chromosomes independently separate from each other, during the formation of gametes. These alleles are then randomly united in fertilization.
  3. In dihybrid cross, F2 phenotypic ratio 9:3:3:1 indicates that the two pairs of characters behave independent of each other. It can be concluded that the two characters under consideration are assorted independently giving rise to different combinations.
Asterclasses.com

F2 generation → 

TRTrtRtr
  ↓
TRTTRR Tall redTTRrTall redTtRRTall redTtRrTall red
TrTTRrTall redTTrrTall whiteTtRrTall redTtrrTall white
tRTtRRTall redTtRrTall redttRR Dwarf redttRr Dwarf red
trTtRrTall redTtrrTall whitettRr Dwarf redttrr Dwarf white

Result: Tall red = 9; Tall white = 3; Dwarf red = 3; Dwarf white = 1

Phenotypic ratio → 9 : 3 : 3 : 1

Genotypic ratio → 

1    :     2     :  2   :   4  :    1    :   2  :   1   :   2  :   1

TTRR  TTRr  TtRR  TtRr  ttRR  ttRr  TTrr  Ttrr   ttrr

From the above results, it is obvious that the inheritance of character of tallness is not linked with the red colour of the flower. Similarly, the character of dwarfness is not linked with the white colour of the flower. This is due to the fact that in the above cross, the two pairs of characters segregate independently. In other words, there is an independent assortment of characters during inheritance.

4.Define test cross and explain its significance.

SOLUTION

Definition:

The cross between F1 hybrid and its homozygous recessive parent is called a test cross.

Significance of test cross:

  1. It helps to determine whether individuals exhibiting dominant character are genotypically homozygous or heterozygous.
  2. It has wide application in plant breeding experiments.
  3. In rapid crop improvement programmes, test cross is used to introduce a useful recessive trait in the hybrids.

4.What is parthenogenesis?

SOLUTION

Parthenogenesis:

The process by which a female produces offspring from unfertilized eggs is known as parthenogenesis.

5.Explain the haplo-diploid method of sex determination in the honey bee.

SOLUTION

Haplo-diploid method of sex determination in the honey bee:

Asterclasses.com
  1. In honey bees, the chromosomal mechanism of sex determination is of haplo-diploid type.
  2. In this type, the sex of the individual is determined by the number of the set of chromosomes received.
  3. Females are diploid (2n=32) and males are haploid (n=16).
  4. The female produces haploid eggs (n=16) by meiosis and male produces haploid sperms (n=16) by mitosis.
  5. If the egg is fertilized by sperm, the zygote develops into a diploid female (2n=32) (queen and worker) and an unfertilized egg develops into a haploid male (n=16) (Drone) by way of parthenogenesis.
  6. The diploid female gets differentiated into either worker or queen bee depending on the food they consume during their development.
  7. Diploid larvae which get royal jelly as food develops into queen (fertile female) and other develops into workers (sterile females).

6.In the answer for inheritance of X-linked genes, Madhav had shown carrier male. His answer was marked incorrect. Madhav was wondering why his marks were cut. Explain the reason.

SOLUTION

A male has X and Y chromosomes. The X-linked genes do not have their alleles on Y chromosome. Therefore, for males to suffer from disease only one copy of a defective gene is sufficient. In the inheritance of X-linked genes, females may be carriers because they have two X chromosomes and may carry one normal and another defective gene. This is not possible in the case of males due to the presence of a single X chromosome.

7.With the help of a neat labelled diagram, describe the structure of chromosome.

SOLUTION

Structure of chromosome:

i. Chromosomes are highly condensed and therefore are clearly visible in metaphase stage of cell division.

ii. A typical chromosome consists of two chromatids joined together at centromere also known as primary constriction.

iii. Primary constriction consists of a disk shape plate called kinetochore. During cell division, spindle fibres get attached to the kinetochore.

iv. Apart from primary constriction, some few chromosomes possess additional one or two constrictions called secondary constriction.

v. At secondary constriction I (nucleolar organizer), the nucleolus becomes organized during interphase.

vi. A satellite body (SAT body) is attached at secondary constriction II, in very few chromosomes.

vii. Each chromatid in turn contains a long, unbranched, slender, highly coiled double-stranded DNA thread, called chromonema, extending through the length of the chromatid.

viii. The ends of the chromosome (i.e. chromatids) are known as telomeres.

Structure of Chromosome-

8.What is cris-cross inheritance?

SOLUTION

The inheritance of characters from the father to his grandson through his daughter is called criss-cross inheritance. 

Long answer type question.

9.Explain cris-cross inheritance with suitable example.

SOLUTION

Criss-cross inheritance can be explained with the help of two examples: colour blindness and haemophilia.

i. Colour blindness:

a. A person suffering from colour blindness cannot differentiate between red and green colours. Both these colours appear grey to the colour blind person.

b. It is caused due to recessive X-linked genes (Xc) which prevent the formation of colour sensitive cells in the retina that are necessary for distinguishing red and green colours.

c. Dominant X linked gene (XC) is necessary for the formation of colour sensitive cells in the retina of eye.

d. The homozygous recessive females (XcXc) and hemizygous recessive male (XcY) are unable to distinguish between red and green colours. The frequency of colour blind women is much less than colour blind men.

e. When a normal man marries a carrier woman, half of their sons may be colour blind, while the remaining half will have normal vision. All their daughters will have normal vision and half of them will be carriers for the disease.

f. If a colour blind male (XcY) marries a female with normal vision (XCXC), then all the offsprings will have normal vision. The sons will have normal vision but daughters will be carriers for the disease. The carriers have normal vision.

ii. Haemophilia (Bleeder’s disease):

a. Haemophilia is X-linked recessive disorder in which blood fails to clot or coagulates very slowly.

b. The genes for normal clotting are dominant over the recessive genes for haemophilia.

c. The person having the recessive gene for haemophilia is deficient in clotting factors (VIII or IX) in blood.

d. Even minor injuries cause continuous bleeding, hence haemophilia is also called as bleeder’s disease.

e. The recessive gene for haemophilia is located on a nonhomologous region of X chromosome.

f. As there is no corresponding allele on Y chromosome to suppress its expression, so men suffer from this disease.

g. Women suffer only when both X chromosomes have recessive genes (alleles).

1. If a haemophilic male (XhY) marries a female with the normal clotting of blood (XX), then all the offsprings will show normal clotting of blood. The sons will have normal clotting of blood, but daughters will be carriers for the disease. The carriers have normal clotting of blood.

2. When carrier woman (XHXh) marries a normal man (XHY), then all the daughters will have normal clotting of blood but half of them will be carriers for the disease. Half the sons will be haemophilic while the remaining will have normal clotting of blood.

asterclasses

Long answer type question.

10.Describe the different types of chromosomes.

SOLUTION

Following are the four types of chromosomes with respect to the position of the centromere: Acrocentric (j shaped), Telocentric (i shaped), Sub-metacentric (L shaped) and Metacentric (V-shaped).

Type of chromosomeName of chromosomePosition of centromere
MetacentricMiddle of the chromosome
Sub-metacentricSome distance away from the centre of chromosomes
AcrocentricNear one end of the chromosome
TelocentricAt one end

                              COMPLETED


Chapter 2, Reproduction in Lower and Higher Animals, hsc, biology, science, maharashtra bore, latest edition, full solution,

Multiple choice question.

1.The number of nuclei present in a zygote is _______.

OPTIONS

  • two
  • one
  • four
  • eight

2.Which of these is the male reproductive organ in humans?

OPTIONS

  • sperm
  • seminal fluid
  • testes
  • ovary

3.Attachment of embryo to the wall of the uterus is known as _______.

OPTIONS

  • fertilization
  • gestation
  • cleavage
  • implantation

4.Rupturing of follicles and discharge of ova is known as _______.

OPTIONS

  • capacitation
  • gestation
  • ovulation
  • copulation

5.In human female, the fertilized egg gets implanted in uterus _______.

OPTIONS

  • After about 7 days of fertilization
  • After about 30 days of fertilization
  • After about two months of fertilization
  • After about 3 weeks of fertilization

6.Test tube baby technique is called _______.

OPTIONS

  • in vivo fertilization
  • in situ fertilization
  • in vitro fertilization
  • artificial insemination

7.The given figure shows human sperm. Various parts of it are labelled as A, B, C, and D. Which labelled part represents acrosome?

OPTIONS

  • B
  • C
  • D
  • A

8.Presence of beard in boys is a _______.

OPTIONS

  • primary sex organ
  • secondary sexual character
  • secondary sex organ
  • primary sexual character

Answer in one sentence.

1.What is the difference between a foetus and an embryo?

SOLUTION

The embryo is the developing organism from fertilization to the end of the eighth week of development. It develops into the foetus.

The foetus is the developing organism from the beginning of the third month to birth.

2.Outline the path of sperm upto the urethra.

SOLUTION

Seminiferous tubules → Rete testis → Vasa efferentia → Epididymis → Vas deferens → Ejaculatory duct → Urethra

3.Which glands contribute fluids to the semen?

SOLUTION

The seminal vesicle, prostate gland and Cowper’s / Bulbourethral gland contribute fluids to the semen.

4.Name the endocrine glands involved in maintaining the sex characteristics of males.

SOLUTION

Testes (gonads)

5.Where does fertilization and implantation occur?

SOLUTION

Fertilization usually takes place in the ampulla of the fallopian/uterine tube, while implantation occurs in the endometrium of the uterus.

6.Enlist the external genital organs in the female.

SOLUTION

The external genital organs of female include parts external to the vagina, collectively called vulva (covering or wrapping), or pudendum. They include the following parts:

  1. Vestibule:
    It is a median vertical depression of the vulva enclosing the urethral and vaginal opening.
  2. Labia minora:
    These are another pair of thin folds inner to the labia majora with which they merge posteriorly to form the fourchette (frenulum), while towards the anterior end they converge into a hood-like covering around the clitoris.
  3. Clitoris:
    A small conical and sensitive projection lying at the anterior end of labia minora. It has a pair of erectile tissue i.e. corpora cavernosa which is homologous to the penis.
  4. Labia majora:
    These are a pair of fleshy folds of skin forming the boundary of the vulva. They are homologous to the scrotum. They surround and protect the other parts of the external genitalia and enclose the urethral and vaginal openings in the vestibule.
  5. Mons pubis:
    It is a fleshy elevation above the labia majora. The Mons pubis and outer part of labia majora show pubic hair.

7.Give two differences between blastula and gastrula.

SOLUTION

BlastulaGastrula
1. It is formed during the early mitotic division through the process of blastulation.1. It is formed at a later stage through the process of gastrulation.
2. Blastula is formed from morula.2. Gastrula is formed blastula.
3. It is formed through rapid mitotic divisions.3. It is formed by slower mitotic divisions.
4. The cells do not move during the formation of the blastula.4. The cells move through morphogenetic movement during the formation of the gastrula.
5. Germinal layers absent5. The formation of three germinal layers occurs.
6. Cells are undifferentiated.6. Cells undergo differentiation

8.What is the difference between embryo and zygote?

SOLUTION

EmbryoZygote
1. It is multicellular.1. It is unicellular.
2. It follows the stage of zygote formation. 2. It is the first stage of development after fertilization.
3. Embryogenesis results in the formation of the embryo.3. Fertilization results in zygote formation.
4. It is formed in the uterus.4. It is formed in the fallopian/ uterine tube.

Fill in the blank:

1.The primary sex organ in human male is _______.

SOLUTION

The primary sex organ in human male is the testis.

2.The _______ is also called the womb.

SOLUTION

The uterus is also called the womb.

3.Sperm fertilizes ovum in the _______of fallopian tube.

SOLUTION

Sperm fertilizes ovum in the ampulla of fallopian tube.

4.The disc-like structure which helps in the transfer of substances to and from the fetus’s body is called _______.

SOLUTION

The disc-like structure which helps in the transfer of substances to and from the fetus’s body is called placenta.

5.Gonorrhoea is caused by _______ bacteria.

SOLUTION

Gonorrhoea is caused by Neisseria gonorrhoeae bacteria.

6.The hormone produced by the testis is _______.

SOLUTION

The hormone produced by the testis is testosterone.


Short answer question.

1.Write a note on budding in Hydra.

SOLUTION

  1. Asexual reproduction in Hydra takes place through budding.
  2. Budding normally occurs in favorable conditions.
  3. In Hydra, a small outgrowth is produced towards the basal end of the body.
  4. It develops as a bud which grows and forms tentacles.
  5. This bud eventually develops (get transformed) into a new individual.
  6. The young Hydra gets detached from the parent and becomes an independent new organism.

Budding in Hydra-

2.Explain the different methods of reproduction occurring in sponges.

SOLUTION

  • Asexual reproduction in sponges:
  1. Sponges reproduce asexually via. gemmule formation. Gemmule is an internal bud formed only in sponges to overcome unfavorable conditions.
  2. The structure of the gemmule includes the micropyle, spicule, inner layer, archaeocytes, and outer layer.
  3. It possesses an asexually produced mass or aggregation of dormant cells known as archaeocytes. These cells are capable of developing into a new organism (totipotent).
  4. The archaeocytes get coated by a thick resistant layer of secretion by amoebocytes.
  5. Monaxon spicules (developed by growth along a single axis) are secreted by scleroblasts in between the inner and outer membrane.
  6. On return of favorable conditions of water and temperature, the gemmules hatch and develop into a new individual.
  7. There is a minute opening called micropyle through which the cells (new individuals) come out during favourable conditions.
  • Sexual reproduction in sponges:
  1. Poriferans that reproduce by the sexual method are hermaphrodites and produce sperms and eggs at different times.
  2. Sperms disseminated into the water column, are subsequently captured by female sponges of the same species.
  3. Inside the female, the sperm is transported to eggs by means of archaeocytes.
  4. Fertilization occurs and zygotes develop into ciliated larvae.
  5. Once these larvae are in the water column, they settle and develop into juvenile sponges.
  6.  

3.Write a note on IVF.

SOLUTION

IVF (In-vitro Fertilization):

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. The zygote or early embryos (with up to 8 blastomeres) could be then transferred into the fallopian tube for further development.

4.Comment on any two mechanical contraceptive methods.

SOLUTION

Mechanical means / Barrier methods:

In this method, the ovum and sperm are prevented from physically meeting with the help of barriers. These mechanical barriers are of three types.:

1. Condom:

It is a thin rubber sheath that is used to cover the penis of the male during copulation.

It prevents the entry of ejaculated semen into the female reproductive tract. It can thus prevent conception. It is a simple and effective method and has no side effects.

Condoms should be properly discarded after every use.

Condom is also a safeguard against STDs and AIDS.

e.g. “Nirodh” is the most widely used contraceptive by males. It is easily available and is given free by the government.

2. Diaphragm, cervical caps and vaults:

These devices used by the female are made up of rubber. They prevent conception by blocking the entry of sperms through the cervix. The device is inserted into the female reproductive tract to cover the cervix during copulation.

3. Intra-uterine devices (IUDs):

These clinical devices are plastic or metal objects. A doctor or trained nurse places the IUDs into the uterus. These devices include Lippes loop, copper releasing IUDs (Cu-T, Cu7, multiload 375), and hormone-releasing IUDs (LNG-20, progestasert).

i. Lippes loop:

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 act as a contraceptive.

ii. Copper releasing IUDs:

Suppress sperm motility and the fertilizing capacity of sperms.

iii. Hormone releasing IUDs:

Make the uterus unsuitable for implantation and the cervix hostile to the sperms. It delays pregnancy for a longer period.

Drawbacks: Spontaneous expulsion, occasional haemorrhage, and chances of infection are the drawbacks of IUDs.

5.Write a note on tubectomy.

SOLUTION

  1. Tubectomy is a permanent birth control method in women.
  2. It is performed by removing a small part of the fallopian tube or tying it up through a small incision in the abdomen or through the vagina.
  3. These techniques are highly effective but their reversibility is highly poor.
  4. Sometimes it becomes necessary to use these methods either to prevent pregnancy or to delay or space pregnancy due to personal reasons.
  5. This method blocks gamete transport and prevents pregnancy.

6.Give the name of causal organism of syphilis and write on its symptoms.

SOLUTION

  1. Causative agent:
    Treponema pallidum (Bacteria)
  2. Symptoms:
    Primary lesion called chancre at the site of infection. The chancre is formed on the external genitalia, skin rashes, and mild fever, inflamed joints, loss of hair. Paralysis, Degenerative changes occur in the heart and brain.

7.What is colostrum?

SOLUTION

  1. Colostrum is sticky and yellow fluid secreted by the mammary glands soon after childbirth.
  2. It contains proteins, lactose, and mother’s antibodies e.g. IgA.
  3. The fat content in colostrum is low.
  4. The antibodies present in it helps in developing resistance for the new born baby at a time when its own immune response is not fully developed.

Answer the following question.

1.Describe the phases of the menstrual cycle and their hormonal control.

SOLUTION

The menstrual cycle involves a series of cyclic changes in the ovary and the female reproductive tract, mainly in the uterus.

It is divided into four phases:

i. Menstrual phase

ii. Proliferative phase

iii. Ovulatory phase

iv. Secretory phase or luteal phase

i. Menstrual phase:

The beginning of each cycle is taken as the first day when menses or loss of blood takes place. During this phase, about 45-100ml of blood is lost.

This phase lasts for approximately five days (average 3-7 days).

The blood in the menstrual discharge does not clot due to the presence of fibrinolysin.

Menstrual phase occurs when an ovulated egg does not get fertilized and it is thereby shed out along with the menstruum. This process is also referred to as the ‘funeral of unfertilized egg’.

Changes in the uterus:

The endometrium of the uterus breaks down under the effect of prostaglandins released due to decreased levels of progesterone and estrogen.

During menses, the blood, tissue fluid, mucus, endometrial lining, and the unfertilized oocyte is discharged through the vagina. Also, the endometrial lining becomes very thin i.e. about 1 mm.

Changes in the ovary:

During these five days, many primordial follicles develop into primary and few of them into secondary follicles under the effect of FSH.

ii. Proliferative phase / Follicular phase / Post menstrual phase:

This phase is the duration between the end of menstruation and the release of ovum (ovulation). The duration of this phase is more variable than other phases. Generally, it extends from 5th to 13th day of the menstrual cycle.

Changes in the ovary:

Generally, out of 6 to 12 secondary follicles that proceed to develop, only one develops into a Graafian follicle (mature follicle). while the rest of the follicles degenerate (atresia). The stimulation for proliferation of new follicles is influenced by GnRH which stimulates release of FSH. The developing secondary follicles secrete the hormone estrogen.

Changes in the uterus:

Endometrium begins to regenerate under the effect of gradually increasing the number of estrogens. Regeneration also involves the formation of endothelial cells, endometrial or uterine glands, and network of blood vessels. The thickness of the endometrium reaches 3-5 mm.

iii. Ovulatory phase:

It is the shortest phase of menstrual cycle.

Changes in the ovary:

It involves rupturing of the mature Graafian follicle and release of an ovum (secondary oocyte) into the pelvic cavity; usually on 14th day of the menstrual cycle. Rapid secretion of LH by a positive feedback mechanism causes the mature follicle to rupture. Ovulation may be accompanied by mild or severe pains in lower abdomen.

iv. Secretory phase / Luteal phase:

It is the phase between ovulation and the beginning of the next menses. This phase is the longest phase. It lasts for 14 days i.e., from 15th to 28th day of the cycle.

Changes in the ovary:

After release of secondary oocyte, remaining tissue of Graafian follicle transforms into a corpus luteum under the effect of LH. The corpus luteum releases progesterone, small amount of estrogen, and inhibin.

The ovulated egg may get fertilized within 24 hours. In the absence of fertilization: the Corpus luteum can survive for only two weeks and then degenerates into a white scar called corpus albicans. In case of fertilization: The embryo is implanted, there is a secretion of human chorionic gonadotropin (hCG), which extends the life of corpus luteum and stimulates its secretory activity. The presence of hCG in maternal blood and urine is an indicator of pregnancy. In absence of fertilization, the next menstrual cycle begins.

Changes in the uterus:

Under the influence of progesterone and estrogen, the endometrial glands grow, become coiled, and start uterine secretions. Endometrium becomes more vascularized and thickens up to 8-10 mm. Inhibin stops secretion of FSH. These changes are necessary for fertilization and subsequent implantation.

2.Explain the steps of parturition.

SOLUTION

Parturition is the process of giving birth to a baby. The physical activities involved in parturition like uterine and abdominal contractions, dilation of the cervix, and passage of baby are collectively called labour. Labour is accompanied by a localised sensation of discomfort or agony called labor pains. Parturition involves the following three steps:

1. Dilation stage:

Uterine contractions begin from the top, forcing the baby towards the cervix. Contractions are accompanied by pain caused by compression of blood vessels. Oxytocin induces uterine contractions which become stronger and stronger due to stimulatory reflex. As the baby is pushed down in the uterus, its head comes to lie against the cervix. The cervix gets dilated and the vagina also shows similar dilation. This stage of labour can normally last up to few hours. It ends in the rupturing of amniotic sac of the foetus.

2. Expulsion stage:

During this stage, the uterine and abdominal contractions become stronger. In normal delivery, the fetus passes out through the cervix and vagina with head in the forward direction. It takes around 20 to 60 minutes. The umbilical cord is tied and cut off close to the baby’s navel.

3. After birth:

After the delivery of the baby, the placenta separates from the uterus and is expelled out as “after birth”, due to severe contractions of the uterus. This process happens within 10 to 45 minutes of delivery.

3.Explain the histological structure of testis.

SOLUTION

Histology of Testis:

1. Externally, the testis is covered by three layers. These are:

a. Tunica vaginalis: It is the outermost incomplete peritoneal covering made up of connective tissue and epithelium.

b. Tunica albuginea: It is the middle layer formed by collagenous connective tissue.

c. Tunica vasculosa/vascularis: It is the innermost layers. It is a thin and membranous layer.

2. Each testis is divided into about 200-300 testicular lobules by fibres from tunica albuginea. Each lobule has 1 to 4 highly coiled seminiferous tubules.

3. Each seminiferous tubule is internally lined by a single layer of cuboidal germinal epithelial cells (spermatogonia) and few large pyramidal cells called Sertoli or sustentacular cells.

4. The germinal epithelial cells undergo gametogenesis to form spermatozoa.

5. Sertoli cells provide nutrition to the developing sperms.

6. Various stages of spermatogenesis can be seen in the seminiferous tubules. The innermost spermatogonial cell (2n), primary spermatocyte (2n), secondary spermatocyte (n), spermatids (n) and sperms (n).

7. Between seminiferous tubules, few groups of interstitial cells (Cells of Leydig) are present

8. After puberty, interstitial cells produce a type of androgen i.e. testosterone.

4.Describe the structure of blastula.

SOLUTION

Blastulation is the process of formation of the hollow and multicellular blastocyst. The process of blastulation can be summarized as follows:

  1. The embryo (blastocyst) that enters the uterus remains floating in uterine cavity for 2-4 days after its entry i.e. till the end of 7th day after fertilization.
  2. The outer layer of cells seen in the morula now forms the layer called the trophoblast.
  3. Cells from the trophoblast begin to absorb the glycogen rich uterine milk
  4. The blastocyst doubles in size from 0.15 mm to 0.30 mm.
  5. With more fluid entering inside the blastocyst cavity is formed.
  6. These outer cells become flat and are called trophoblast cells (since they help only in absorbing nutrition for the developing embryo).
  7. The larger inner cells form inner cell mass or embryoblast (the embryo proper develops from the embryoblasts).
  8. These remain attached to the trophoblasts on only one side.
  9. The trophoblast cells in contact with the embryonal knob are called cells of Rauber.
  10. At this stage, the blastocyst shows polarity i.e. the side with inner cell mass is called the embryonal end and the side opposite to it is the abembryonic end.
  11. By the end of the 7th day the blastocyst is fully formed and ready for implantation and gastrulation.
  12. The function of zona pellucida is to prevent the implantation of the embryo at an abnormal site. It does not expose the sticky and phagocytic trophoblast cells till it reaches the implantation site i.e. within the uterus, after which the zona pellucida ruptures.

5.Explain the histological structure of ovary in human.

SOLUTION

Histological structure of ovary:

Each ovary is a compact structure differentiated into a central part called medulla and the outer part called the cortex. The cortex is covered externally by a layer of germinal epithelium. The stroma of loose connective tissue of the medulla has blood vessels, lymph vessels, and nerve fibers.

The outer cortex is more compact and granular.

It shows large number of tiny masses of cells called ovarian follicles. These are collectively formed from the immature ova originating from cells of the dorsal endoderm of the yolk sac. The cells migrate to the gonadal ridge during embryonic development and divide mitotically.

Now these cells are called oogonia.

As the oogonia continue to grow in size they are surrounded by a layer of granulosa cells. This assembly forms the rudiments of the ovarian follicles. The process of oogenesis starts much before the birth of the female baby and by the end of twelve weeks the ovary is fully formed. The ovary has more than two million primordial follicles in it. The cells of the germinal epithelium give rise to groups of oogonia projecting into the cortex in the form of cords called egg tubes of Pfluger. Each cord at its end has a round mass of oogonial cells called egg nests, from which the primordial ovarian follicles develop. Each primordial follicle has, at its center a large primary oocyte (2n) surrounded by a single layer of flat follicular cells. The primary oocyte starts with its meiotic division but gets arrested it at meiosis I.

Of the two million primordial follicles embedded in the fetal ovary only about one million remains at birth and only about 40,000 remain at the time of puberty.

T.S. of Ovary-

The histological structure of the ovary shows the different stages of development of the oocyte in the ovary. These changes are cyclic and occur during each menstrual cycle. This development involves maturation of the primordial follicles into primary, secondary and Graafian follicles.

Each primary follicle has multi-layered cuboidal follicular cells. The stroma cells add theca over the follicle, which then changes into a secondary follicle.

There is the growth of the oocyte and the granulosa cells increase in number. They start producing the hormone estrogen.

The secondary follicle grows into the Graafian follicle by the addition of more follicular cells.

As this process of maturation of follicles takes place, they begin to move towards the surface of the ovary. The Graafian follicle presses against the thin wall of the ovary giving it a blistered appearance.

The egg is released from the Graafian follicle during ovulation and the remaining part of the follicle changes into a temporary endocrine gland called corpus luteum.

If fertilization does not take place the corpus luteum degenerates into a white scar called corpus albicans.

6.Describe the various methods of birth control to avoid pregnancy.

SOLUTION

Contraceptive methods are of two main types i.e. temporary and permanent.

1. Temporary methods:

i. Natural method/ Safe period / Rhythm method:

In the natural method, the principle of avoiding chances of fertilization is used. A week before and a week after menstrual bleeding is considered a safe period for sexual intercourse.

This method is based on the fact that ovulation occurs on the 14th day of the menstrual cycle. Drawback: High rate of failure.

ii. Coitus Interruptus or withdrawal:

In this method, the male partner withdraws his penis from the vagina just before ejaculation, so as to avoid insemination.

Drawback: Pre-ejaculation fluid may contain sperms and this can cause fertilization.

iii. Lactational amenorrhea (absence of menstruation):

This method is based on the fact that ovulation does not occur during the period of intense lactation following parturition. Therefore, as long as the mother breastfeeds the child fully, chances of conception are almost negligible.

Drawbacks: High chances of failure.

iv. Chemical means (spermicides):

In this method, chemicals like foam, tablets, jellies, and creams are used by the female partner. Before sexual intercourse, if these chemicals are introduced into the vagina, they adhere to the mucous membrane, immobilize and kill the sperms.

Drawback: It may cause allergic reaction. This method also has chances of failure.

v. Mechanical means / Barrier methods:

In this method, the ovum and sperm are prevented from physically meeting with the help of barriers.

These mechanical barriers are of three types.:

1. Condom:

It is a thin rubber sheath that is used to cover the penis of the male during copulation. It prevents the entry of ejaculated semen into the female reproductive tract. It can thus prevent conception. It is a simple and effective method and has no side effects. Condoms should be properly discarded after every use. A condom is also a safeguard against STDs and AIDS.

e.g.“Nirodh” is the most widely used contraceptive by males. It is easily available and is given free by the government.

2. Diaphragm, cervical caps and vaults:

These devices used by the female are made up of rubber. They prevent conception by blocking the entry of sperms through the cervix. The device is inserted into the female reproductive tract to cover the cervix during copulation.

3. Intra-uterine devices (IUDs):

These clinical devices are plastic or metal objects. A doctor or trained nurse places the IUDs into the uterus. These devices include the Lippes loop, copper releasing IUDs (Cu-T, Cu7, multiload 375), and hormone-releasing IUDs (LNG-20, progestasert).

  • Lippes loop:
    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 act as a contraceptive.
  • Copper releasing IUDs:
    Suppress sperm motility and the fertilizing capacity of sperms.
  • Hormone releasing IUDs:
    Make the uterus unsuitable for implantation and cervix hostile to the sperms. It delays pregnancy for a longer period.
    Drawbacks: Spontaneous expulsion, occasional haemorrhage and chances of infection are the drawbacks of IUDs.

vi. Physiological (Oral) Devices:

Physiological devices are used in the form of tablets/ pills. It is an oral contraceptive, used by the female which contains progesterone and estrogen. These hormones inhibit ovulation; hence no eggs are released from the ovary of the female using this pill and thus conception cannot occur.

They also alter the quality of cervical mucus to prevent the entry of sperms. The pill “Saheli” is an oral contraceptive for females which is non-steroidal. Saheli is to be taken once in a week. These pills are sponsored by the Government. Saheli is now a part of the National Family Programme as an oral contraceptive pill in India.

Drawback: Oral contraceptive pills have side effects such as nausea, weight gain, tenderness of breast, and slight blood loss between menstrual periods.

vii. Other contraceptives:

The birth control implant is a contraceptive used by the female. e.g. implanon, explanon, etc. It is a tiny, thin rod about the size of a matchstick. It is implanted under the skin of the upper arm and contains progesterone and estrogen. Their mode of action is similar to that of pills. They prevent pregnancy for 3-4 years.

2. Permanent Methods:

The permanent birth control method in men is called vasectomy and in women it is called tubectomy. These are surgical methods, also called sterilization. In vasectomy a small part of the vas deferens is tied and cut. In tubectomy, a small part of the fallopian tube is tied and cut. This blocks gamete transport and prevent pregnancy.

7.What are the goals of RCH programme.

SOLUTION

The goals of the Reproductive and Child Healthcare (RCH) programme are as follows:

i. To create awareness among people about various aspects related to reproduction.

ii. To provide facilities to people in order to understand and build up reproductive health.

iii. To provide support for building up a reproductively healthy society.

iv. To bring about a change mainly in three critical health indicators i.e. reducing total infertility rate, infant mortality rate, and maternal mortality rate.

8.Which hormones are involved in parturition?

SOLUTION

Parturition is controlled by a complex neuroendocrine mechanism.

1. Signals arise from the fully formed foetus and placenta cause mild uterine contractions.

2. This is accompanied by rise in estrogen- progesterone ratio, increase in oxytocin receptors in uterine muscles.

3. Increase in hormone ratio causes vigorous contractions of myometrium of uterus at the end of pregnancy.

4. The fully developed foetus gives signals for the uterine contractions by secreting Adrenocorticotropic Hormone (ACTH) from pituitary and corticosteroid from adrenal gland.

5. This, in turn, triggers the release of oxytocin from mother’s pituitary gland, which acts on the uterine muscles of the mother and causes vigorous uterine contractions leading to the expulsion of the baby from the uterus.

9.Which is the function of male accessory glands?

SOLUTION

Male accessory glands secrete substances that protect the gametes and facilitate their movement.

1. Seminal vesicles:

These are a pair of small fibromuscular pouches present on the posterior side of the urinary bladder. They secrete a seminal fluid (alkaline) containing citric acid, fructose, fibrinogen and prostaglandins. About 60% of the total volume of semen is made up of seminal fluid.

Fructose provides energy to sperms for swimming, while fibrinogen helps in coagulation of semen after ejaculation for quick propulsion into the vagina. The prostaglandins stimulate reverse peristalsis in the vagina and uterus aiding the faster movement of sperms towards the egg in the female body.

2. Prostate gland:

The prostate gland consists of 20 to 30 lobes and is located under the urinary bladder. It surrounds the urethra. It releases a milky white, alkaline fluid called prostatic fluid into the urethra. Prostatic fluid forms about 30% of the total volume of semen. It contains citric acid, acid phosphatase, and various other enzymes.

The acid phosphatase protects the sperm from the acidic environment of the vagina.

3. Cowper’s glands:

Cowper’s glands are also known as bulbourethral glands. These are pea-sized and lie on either side of membranous urethra. They secrete a viscous, alkaline, mucous like fluid which acts as a lubricant during copulation.

Semen: It is the viscous, alkaline and milky fluid (pH 7.2 to 7.7) ejaculated by the male reproductive system.

Generally, 2.5 to 4.0 ml of semen is given out during a single ejaculation and it contains about 400 million sperms.

Semen contains secretion of the epididymis and the accessory glands for nourishing (fructose), neutralizing acidity (Ca++, bicarbonates), activation for movement (prostaglandins).

10.What is capacitation? Give its importance.

SOLUTION

Capacitation generally requires 5-6 hours during which the acrosome membrane becomes thin, Ca++ enters the sperm, and sperm tails begin to show rapid whiplash movements.

As a result of capacitation, sperms become extra active and begin to start moving upwards from the vagina to the uterus and to the oviducts. Prostaglandins activate the sperms. The vestibular secretions of the female also enhance sperm’s motility. The sperms swim at an average speed of 1.5 to 3.0 mm/min and reach the ampulla. The contraction of uterus and fallopian tubes stimulated by oxytocin of females also aids in the movement of sperm. After capacitation the sperms may reach the ampulla within 5 minutes.

Answer the following question.

1.Explain the following parts of male reproductive system along with labelled diagram showing these parts- Testis, vasa deferentia, epididymis, seminal vesicle, prostate gland and penis.

SOLUTION

asterclasses

The male reproductive system consists of the following:

i. Primary sex organs (Gonads): Testes

ii. Accessory organs: Accessory ducts and accessory glands

iii. External genitalia: Scrotum and Penis

i. Primary sex organs: Testes

Testes are a pair of primary sex organs which are mesodermal in origin. They are located outside the abdomen in a pouch called scrotum, i.e. extra-abdominal in position. Testes develop in the abdominal cavity (early foetal life) and later descend into the scrotal sac through a passage called inguinal canal. They are suspended in the scrotal sac by the spermatic cord. Testes are connected to the wall of scrotum by a short fibromuscular band called gubernaculum.

They are oval in shape, about 4 to 5 cm long, 2 to 3 cm wide, and 3 cm thick.

The outermost covering of the testis is formed of a dense fibrous membrane called tunica albuginea.

ii. Accessory sex organs:

It includes accessory ducts, accessory glands.

a. Accessory ducts:

1. Rete testis:

The seminiferous tubules of the testis at the posterior surface form a network of tubules called rete testis. The rete testis opens into vasa efferentia. 2. Vasa efferentia:

Vasa efferentia are 12-20 fine tubules arising from the rete testis and join to the epididymis. They carry sperms from the testis and open into the epididymis.

3. Epididymis:

It is a long and highly coiled tube which is differentiated into an upper caput-, middle corpus- and lower cauda epididymis. The sperms undergo maturation in the epididymis.

4. Vasa deferens:

The vas deferens travels up to the abdominal cavity and loops over the ureter to open into the urethra. Vas deferens join the seminal vesicle to form ejaculatory duct.

5. Ejaculatory ducts:

The ejaculatory duct passes through the prostate gland and opens into the urethra.

6. Urethra:

The urethra provides a common passage for the urine and semen and hence is also called urinogenital duct. In males the urethra is long and extends through the penis. It opens to the outside by an opening called the urethral meatus or urethral orifice.

b. Accessory glands:

1. Seminal vesicles:

These are a pair of small fibromuscular pouches present on the posterior side of the urinary bladder. They secrete a seminal fluid (alkaline) containing citric acid, fructose, fibrinogen, and prostaglandins. About 60% of the total volume of semen is made up of seminal fluid. Fructose provides energy to sperms for swimming, while fibrinogen helps in coagulation of semen after ejaculation for quick propulsion into the vagina. The prostaglandins stimulate reverse peristalsis in vagina and uterus aiding the faster movement of sperms towards the egg in the female body.

2. Prostate gland:

Prostate gland consists of 20 to 30 lobes and is located under the urinary bladder. It surrounds the urethra. It releases a milky white, alkaline fluid called prostatic fluid into the urethra. Prostatic fluid forms about 30% of the total volume of semen. It contains citric acid, acid phosphatase and various other enzymes.

The acid phosphatase protects the sperm from the acidic environment of the vagina.

3. Cowper’s glands:

Cowper’s glands are also known as bulbourethral glands. These are pea-sized and lie on either side of membranous urethra. They secrete a viscous, alkaline, mucous like fluid which acts as a lubricant during copulation.

Semen: It is the viscous, alkaline and milky fluid (pH 7.2 to 7.7) ejaculated by the male reproductive system. Generally, 2.5 to 4.0 ml of semen is given out during a single ejaculation and it contains about 400 million sperms. Semen contains secretion of the epididymis and the accessory glands for nourishing (fructose), neutralizing acidity (Ca++, bicarbonates), activation for movement (prostaglandins).

iii. External genitalia:

a. Penis:

The penis is the male copulatory organ. It is cylindrical and muscular with three bundles of erectile tissue: a pair of postero-lateral tissue called corpora cavernosa and a median corpus spongiosum. The swollen tip of the penis is called glans penis. It is covered by a loose fold of skin called foreskin or prepuce.

b. Scrotum:

It is a loose pouch of pigmented skin lying behind the penis and is divided into a right and left scrotal sac by a septum of tunica dartos made of smooth muscle fibres. The foetal testes are guided into and retained in the scrotum by a short fibro muscular band called gubernaculum. The testes remain suspended in scrotum by a spermatic chord.

The failure of the testis to descend into the scrotum is called cryptorchidism which also results in sterility.

The cremaster and dartos muscles of scrotum help in drawing testes close or away from the body. This helps in maintaining the temperature of the testis 2-3 0C lower than the normal body temperature, necessary for spermatogenesis.

Long answer question.

2.Describe female reproductive system of human.

SOLUTION

The human female reproductive system consists of:

i. Internal genitalia:

It includes ovaries, oviducts, uterus, vagina.

Asterclasses

a. Ovary:

It is the primary female sex organ. It is a solid, oval, or almond-shaped organ. It is 3 cm in length, 1.5 cm in breadth, and 1 cm thick. It is located in the upper lateral part of the pelvis near the kidneys. Each ovary is held in position by ligaments by attaching it to the uterus and the abdominal wall. The largest of these is the broad ligament formed by a fold of peritoneum. It holds the ovary, oviduct and the uterus to the dorsal body wall. The ovarian ligament attaches the ovary to the uterus.

Functions:

Its main function is production of egg or ovum and the female reproductive hormones.

The ovary produces five hormones viz. estrogen, progesterone, relaxin, activin and inhibin.

b. Oviduct / Fallopian tube / Uterine tube:

These are a pair of muscular ducts lying horizontally over the peritoneal cavity. The proximal part of the tube lies close to the ovary and distally it opens into the uterus. Each tube is 10 to 12 cm in length. It is internally lined by ciliated epithelium. It can be divided into three regions:

1. Infundibulum:

The proximal funnel like the part with an opening called ostium surrounded by many finger-like processes called fimbriae (of these at least one is long and connected to the ovary). The cilia and the movement of fimbriae help in transporting the ovulated egg to the ostium.

2. Ampulla:

It is the middle, long and straight part of the oviduct. Fertilization of the ovum takes place in this region.

3. Isthmus / Cornua:

The distal narrow part of the duct opening into the uterus.

Functions:

Fallopian tubes carry the released egg from the ovary to the uterus. Ampulla provides the site for fertilization of the ovum.

c. Uterus:

It is commonly also called as the womb.

It is a hollow, muscular, pear-shaped organ, located above and behind the urinary bladder. It is about 7.5 cm long, 5 cm broad and 2.5 cm thick. Internally the uterine wall can be distinguished into three layers: The outermost perimetrium, middle thick muscular myometrium, made up of thick layer of smooth muscles. Vigorous contractions of these muscles cause labour during parturition (childbirth). The innermost layer called endometrium or mucosal membrane is made up of stratified epithelium. The thickness of this layer regularly undergoes changes during the menstrual cycle. It is richly supplied with blood vessels and uterine glands. These provide nourishment to the developing foetus.

The uterus can be divided into three regions:

1. Fundus: It is the upper dome shaped part. Normally implantation of the embryo occurs in the fundus.

2. Body: It is the broad part of the uterus which gradually tapers downwards.

3. Cervix: It is the narrow neck about 2.5 cm in length. It extends into the vagina. Its passage has two openings: an internal os/ orifice towards the body, and an external os/ orifice towards the vagina.

Functions:

Uterus receives the ovum. It provides site for implantation, gestation and parturition. It forms placenta for the development of foetus.

d. Vagina:

It is a tubular, female copulatory organ, 7 to 9 cm in length.

It lies between the cervix and the vestibule.

The vaginal wall has an inner mucosal lining, the middle muscular layer and an outer adventitia layer.

The mucosal epithelium is stratified and non-keratinized and stores glycogen.

There are no glands but the cervical secretion of mucus is received in the vagina.

The opening of the vagina into the vestibule is called vaginal orifice. The vaginal orifice is partially covered by the hymen.

Functions:

The vagina acts as a passage for menstrual flow as well as a birth canal during parturition.

ii. External genitalia (Vulva):

The external genital organs of female include parts external to the vagina, collectively called vulva (covering or wrapping), or pudendum. They include the following parts:

a. Vestibule:

It is a median vertical depression of vulva enclosing the urethral and vaginal opening.

b. Labia minora:

These are another pair of thin folds inner to the labia majora with which they merge posteriorly to form the fourchette (frenulum), while towards anterior end they converge into a hood-like covering around the clitoris.

c. Clitoris:

A small conical and sensitive projection lying at the anterior end of labia minora. It has a pair of erectile tissue i.e. corpora cavernosa which is homologous to the penis.

d. Labia majora:

These are a pair of fleshy folds of skin forming the boundary of the vulva. They are homologous to the scrotum. They surround and protect the other parts of the external genitalia and enclose the urethral and vaginal openings in the vestibule.

e. Mons pubis:

It is a fleshy elevation above the labia majora. The Mons pubis and outer part of labia majora show pubic hair.

iii. Accessory glands:

a. Vestibular glands / Bartholin’s glands:

It is a pair of glands homologous to the Bulbourethral or Cowper’s glands of the male. They open into the vestibule and release a lubricating fluid.

b. Mammary glands:

These are accessory organs of the female reproductive system for production and release of milk after parturition. The development of the mammary glands occur at puberty under the influence of estrogen and progesterone. Lactotropic hormone (LTH) or prolactin helps in the development of lactiferous tubules during pregnancy. The mammary glands are a pair of rounded structures present in the subcutaneous tissue of the anterior thorax in the pectoral region (from 2nd to 6th rib). These are modified sweat glands. Each mammary gland contains fatty connective tissue and numerous lactiferous ducts. The glandular tissue of each breast is divided into 15-20 irregularly shaped mammary lobes, each with alveolar glands and lactiferous duct.

Alveolar glands secrete milk which is stored in the lumen of alveoli. The alveoli open into mammary tubules. The tubules of each lobe join to form a mammary duct.

Many mammary ducts join to form a wider mammary ampulla, which is connected to lactiferous duct. These converge towards the nipple located near the tip of the breast.

It is surrounded by a dark brown coloured and circular area of the skin called areola.

3.Describe the process of fertilization.

SOLUTION

Fertilization is the process which involves the fusion of the haploid male and female gametes resulting in the formation of a diploid zygote (2n). The process of fertilization is internal and it usually takes place in the ampulla of the fallopian/uterine tube. The fertilized egg or zygote further develops into an embryo within the uterus.

The mechanism of fertilization is as follows:

i. Movement of sperm towards egg:

The ejaculated semen is made up of sperms and some other secretions. This coagulated semen undergoes liquefication and sperms become active. Once the sperms reach the vagina around 50% sperms are demobilized/broken/destroyed and the remaining sperms undergo capacitation.

Capacitation:

Capacitation generally requires 5-6 hours during which the acrosome membrane becomes thin, Ca++ enters the sperm, and sperm tails begin to show rapid whiplash movements. As a result of capacitation, sperms become extra active and begin to start moving upwards from the vagina to the uterus and to the oviducts. Prostaglandins activate the sperms. The vestibular secretions of the female also enhance sperms’ motility. The sperms swim at an average speed of 1.5 to 3.0 mm/min and reach the ampulla. The contraction of the uterus and fallopian tubes stimulated by oxytocin of females also aids in the movement of sperm. After capacitation, the sperms may reach the ampulla within 5 minutes.

ii. Entry of sperm into the egg:

Out of 200 to 400 million sperms, only few hundred manage to reach the ampulla, out of which only a single sperm fertilizes the ovum. After the sperm reaches the egg/ovum, its acrosome releases lysins: hyaluronidase and corona penetrating enzymes. These enzymes separate and dissolve the cells of corona radiata, so that the sperm head can pass through the zona pellucida of the egg. The zona pellucida has fertilizin receptor proteins (ZP3, ZP2). The fertilizin binds to specific acid protein- anti-fertilizin of sperm and brings about the attraction of sperms to the egg to enhance fertilization.

Acrosome reaction:

As the sperm head touches the zona pellucida in the animal pole region, its acrosome covering ruptures to release lytic enzymes, acrosin or zona lysin which act on the zona pellucida at the point of contact. This causes egg reaction during which a small fertilization cone/cone of reception is formed on the egg membrane. The sperm head comes in contact with this cone. It results in the production of a weak wave of depolarization. The plasma membrane of both cells dissolves at the point of contact.

The sperm nucleus and the centrioles enter the egg, while other parts remain outside.

As soon as the sperm head touches the vitelline membrane, a cortical reaction gets activated changing the vitelline membrane into a fertilization membrane by deactivating the sperm receptors of zona pellucida.

A distinct perivitelline space is created around the fertilization membrane.

This prevents any further entry of other sperms into the egg i.e. polyspermy is avoided.

iii. Activation of ovum:

The ovum before fertilization was at metaphase II stage. After the contact of sperm head to the vitelline membrane of egg, it gets activated to resume and complete meiosis II. After meiosis II, the second polar body is formed. The germinal vesicle organizes into female pronucleus also known as the true ovum or egg.

The fusion of egg and sperm:

The coverings of male and female pronuclei degenerate, allowing the chromosomal pairing. This results in the formation of a synkaryon by the process called syngamy or karyogamy. The zygote is thus formed. The proximal centriole received from the sperm helps in the formation of the synkaryon spindle and cleavage of the cell into two blastomeres.

The zygote is thus formed.

The proximal centriole received from the sperm helps in the formation of the synkaryon spindle and cleavage of the cell into two blastomeres.

4.Explain the process by which zygote divides and re-divides to form the morula.

SOLUTION

Asterclasses

The zygote formed as a result of syngamy is activated to divide.

1. Cleavage:

Cleavage is the process of early mitotic division of the zygote into a hollow multicellular blastula. It does not involve the growth of the daughter cells. The cells formed by cleavage are called blastomeres. Since there is no growth phase between the cleavages, the size of blastomeres will be reduced with every successive cleavage.

As the size reduces, the metabolic rate increases. Subsequent cleavages are thus faster than earlier ones. This requires rapid replication of DNA and high consumption of oxygen.

2. Process of cleavage:

In human, cleavage is holoblastic i.e. the whole zygote gets divided. The cleavage planes may be longitudinal or meridional and equatorial or horizontal. It is radial and indeterminate i.e. fate of each blastomere is not predetermined.

The 1st cleavage in the zygote is meridional and occurs at about 30 hours after fertilization.

It divides longitudinally into two blastomeres, one slightly larger than the other.

The 2nd cleavage is also longitudinal but at the right angle to the 1st one and occurs after 30 hours of 1st cleavage.

The 3rd cleavage is horizontal. After 3rd cleavage, the embryo is in the 8-cell stage.

While the cleavages occur, the young embryo is gradually being pushed towards the uterus.

By the end of 4th day after fertilization, the embryo is a solid ball of 16-32 cells and externally looking like mulberry. This stage is thus called a morula.

3. Morula:

The morula shows cells of two types:

a. smaller, clearer cells towards the outer side

b. inner cell mass of larger cells.

Cells are compactly arranged. Till the formation of morula, the zona pellucida is retained around the embryo and thus, there is no change in the overall size from zygote to morula. The morula reaches the isthmus and gains entry into the uterus by the end of day 4.


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