Category: 2nd PUC

You can Download Chapter 12 Biotechnology and its Applications Questions and Answers, 1st PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 12 Biotechnology and its Applications

2nd PUC Biology Biotechnology and its Applications Ncert Text Book Questions and Answers

Question 1.
Crystals of Bt toxin produced by some bacteria do not kill the bacteria themselves because
(a) bacteria are resistant to the toxin
(b) toxin is immature;
(c) toxin is inactive;
(d) bacteria encloses toxin in a special sac.
Answer:
(c) Toxin is inactive. This is because it exists as protoxin, inactive form in bacteria.

Question 2.
What are transgenic bacteria? Illustrate using any one example. (CBSE – 2006)
Answer:
Transgenic bacteria are microbes carrying clones of foreign genes. It is also known as genetically modified bacteria.
These bacteria are being employed for many functions .
(a) Two DNA sequence (coding for A and B chains of human insulin) were introduced into the plasmid of bacteria E. Coli. This transgenic bacteria produced insulin chain- Used as biochemical factories

(b) Microbes have been genetically changed to help in cleaning the polluted environment, eg:- Pseudomonas putida for cleaning oil spills pseudomonas species for removing heavy metal pollutants. Aceto bacter aerogans for decomposition of DDT and Flavobacterium for decomposition of 2,4-D.

Question 3.
Compare and contrast the advantages and disadvantages of production of genetically modified crops.
Answer:
Advantages:-

• Crop plants can be made to grow fast and produce high yield through genetic modification.
• G.M crops increased the efficiency of mineral usage by plants.
• G.M crops helped to reduce post harvest losses.
• It enhances the nutritional value of food, eg: Vitamin A enriched rice.
• Transgenic plants can produce pharmaceuticals like insulin, interferons blood clotting factors, hormones etc.
• Resistance to viral diseases has been introduced in Tobacco, potato, tomato and rice. Cry gene from bacillus has been introduced in Bt. cotton and Bt. Corn. Nematode infection has been overcome in some plants through RNA interference.
• Plants have been modified genetically to produced biodiesel and other commercial products.

Disadvantages :-

• G.M. crops may, cause problem in human health by supplying allergens and transferring antibiotic resistance markers.
  (b)These crops may came damage to the natural environment.
• Weedicide genes are being introduced in to crop plants some of these crop may become super weeds
• Trans genes can be transferred from one plant to another plant, microbes and even animals. It shall disturb the genetic set of organisms and cause ecological imbalance.
• Cry gene being introduced in crop plants can pass into wild vegetation. Insects feeding on pollen and other parts will be killed resulting in destruction of pollination and disseminators.

Question 4.
What are Cry proteins? Name an organism that produce it. How has man exploited this protein to his benefit?
Answer:
Cry proteins are a group of toxic protein which are highly poisonous to deficient types of insects. It is produced by a soil bacterium Bacillus thuringiensis. The genes controlling their formation are called crygenes eg:- Cry I Ab, Cry I Ac, Cry II Ab, The bacterium produces protein in the crystal form of protoxin. Two cry genes have been incorporated in cotton (Bt cotton) while one has been introduced in corn (Bt corn) As a result Bt Cotton was disease resistance to boll worm and Bt corn was resistance to corn borer.

Question 5.
What is gene therapy? Illustrate using the example of adenosine deaminase (ADA) deficiency,
Answer:
It is the therapeutic treatment of defective heredity by the introduction healthy and functional gene which also silence the defective genes of an individual.
OR
The replacement of a nonfunctional or defective gene with a normal functional gene is called gene therapy. A person with defect in the gene for the enzyme adenosine deaminase (ADA) suffers with SC1D (Severe Combined Immuno Deficiency) The enzyme ADA is crucial for the immune system. Ideally gene therapy should be applied to the zygotes so that the progeny of defective individual also gets rid of effect. It is however, generally applied to somatic cells where the defect occurs.

At first step towards gene therapy, lymphocytes from the blood of the patient are grown in a culture outside the body. A functional ADA cDNA (using a retrovirus vector) is then introduced into these lymphocytes which are subsequently returned to the patient. Since these cells are not immortal, the patient requires periodic infusion of such genetically engineered lymphocytes.

Question 6.
Digrammatically represent the experimental steps in cloning and expressing an human gene (say the gene for growth hormone) into a bacterium like E. coli ?
Answer:

Question 7.
Can you suggest a method to remove oil (hydrocarbon) from seeds based on your understanding of rDNA technology and chemistry of oil?
Answer:
Oil is synthesized by the condensation of 3 fatty acid with a molecule of glycerol. Fatty acids are formed through an enzyme complex called fatty acid synthetase. The oil from the seeds can be removed by preventing the synthesis of either glycerol or the enzyme lipase which catalyses the synthesis of oil. It can be achieved by knocking out the genes coding for the enzyme lipase or the enzyme fatty acid synthetase.

Question 8.
Find out from internet what is golden rice.
Answer:
Golden rice is genetically engineered rice with high vitamin A content.

Question 9.
Does our blood have proteases and nucleases?
Answer:
No, our blood does not contain enzymes proteases and nucleases. If these two enzymes were there in the blood, it causes the degeneration of blood cells and lining cells of blood cells.

Question 10.
Consult internet and find out how to make orally active protein pharmaceutical. What is the major problem to be encountered?
Answer:
Orally active protein can be made through DNA technology for instance, hirudin a blood clotting protein has been produced in the seeds of Brassica napus. The transgenic cow named Rosie contain human protein a lactalbumin. How ever such proteins may cause allergic reaction in some people.

2nd PUC Biology Biotechnology and its Applications  Additional Questions and Answers

2nd PUC Biology Biotechnology and its Applications One Mark Questions

Question 1.
Why is the gene encoding Cry protein inserted into crop plant ?
Answer:
When the gene encoding cry protein inserted into a crop plant, the plant becomes resistant to insert pests.

Question 2.
What is the significance of the process of RNA inference (RNAi) in eukaryotic organism. (CBSE 2008)
Answer:
It is a method of cellular defense in eukaryotic organisms, which involves silencing of m.RNA.

Question 3.
State the principle on which ELISA works?
Answer:
The principle is antigen-antibody interaction.

Question 4.
What was the speciality of the milk produced by the transgenic cow, Rosie ? (AI 2008)
Answer:
The milk contained human protein alpha lactalbumin (2.4gm/liter) which is nutritionally important for human babies.

Question 5.
What is biopiracy?
Answer:
The use of bioresources by multinational companies and other organisation with out proper autherisation from the countries and people concerned.

Question 6.
What GEAC
Answer:
Genetic engineering Approval committee.

Question 7.
What is the role of the organisation GEACf
Answer:
It checks (Delhi 2008)

• The validity of G.M crops
• The safety of introducing genetically modified organisms for public services.

Question 8.
A multinational company outside India tried to sell new verities of turmeric without proper patent rights. What is such an act referred to as?
Answer:
Biopiracy

Question 9.
What is Hirudin ?
Answer:
It is a protein which prevent blood dotting.

2nd PUC Biology Biotechnology and its Applications Two Marks Questions

Question 1.
In case of Bt Cotton how does the toxic insecticide protein produced by bacterium kill the insect pest but not the cells of Bacillus thuringiensis where the toxic protein is generated?
Answer:
The toxin produced by Bacillus thuringiensis is an endotoxin called cry protein. It is crystalline and non toxic when formed being in protoxin stage. As it reaches the gut of insects, the cry protein is converted into toxic ‘ and soluble state. It attaches the receptors present on the epithelial cells of the gut produces pores and kills the cells resulting in death of the insects.

Question 2.
What are 4 main objectives of genetically modified crop plants ? (CBSE 2008)
Answer:

• Higher nutritional value eg: Vitamin. A in golden rice
• Abiotic stresses- Increased tolerance to drought etc.
• Post Harvest losses- Prevention of over-ripening and other post harvest losses.
• Insect and Pest resistance. eg: Bt-cotton.

Question 3.
Define transgenic organisms (CBSE 2006)
Answer:
They are organisms which have been modified genetically through introduction of genes of another organism artificially by the technique of genetic engineering instead of conventional hybridisation.

Question 4.
How is early detection of diseases possible using molecular diagnostics?
Answer:

• Low concentration of viral or bacterial DNA in the host cell/body can be detected much before the symptoms of the disease appear i.e. early detection of disease is possible
• Clones of genes can be used to as probe to detect the presence of mutual alleles in cancer suspected patients.

Question 5.
Name any two biological products that are produced in transgenic animals and mention their uses.
Answer:

• a-1- antitrypsin is used to treat emphysema.
• a – lactalbumin is produced in the milk of the transgenic cow, Rosie, this is nutritionally more balanced product for human babies than normal cow milk.

Question 6.
Give reason for each of the following.
(i) Insulin cannot be administered orally to diabetic patients.
(ii) Biopiracy affects India the most.
Answer:
(i) Insulin in a polypeptide and can be digested by peptidase in the alimentary canal. Hence it cannot be administered orally.

(ii) India is rich in biodiversity and traditional knowledge related to the use of bio-resources, than many other countries, hence it is affected the most with biopiracy.

Question 7.
What is the cause of adenosine deaminase deficiency in a person?
Answer:
It is due to the deletion of the gene coding for the enzyme adenosine deaminase, this enzyme is crucial for the functioning of the immune system.

Question 8.
Mention the uses of cloned genes in, molecular diagnostics.
Answer:

• Cloned genes are used as probe to detect the presence of its complementary DNA strand; the muted gene will not hybridise with the probe and hence will not appear in the photographic film.
• When cloned genes express themselves and produce recombinant proteins, they help in developing sensitive diagnostic techniques.

Question 9.
Why does Bt toxin not kill the bacillus ? How does it kill the insect larvae ?
Answer:

• When Bt toxin is ingested by an insect, it is converted into its active form when exposed to the alkaline pH in the gut.
• The activated toxin binds to the surface of the epithelial cells of the midgut create pores.
• Water enters the cells and causes their swelling and lysis.

Question 10.
Write the advantages of recombinant therapeutics?
Answer:
The recombinant therapeutics do not induce any unwanted immunological response like the similar products of non human origin such therapeutics are highly effective.

Question 11.
How is ELISA used to detect the pathogens in the body?
Answer:

• Pathogens are detected by the presence of antigens, which may be a protein or glycoprotein.
• Pathogens can be also be detected by the presence of antibodies synthesized against the pathogens.

2nd PUC Biology Biotechnology and its Applications Three Marks Questions

Question 1.
(a) Why is traditional knowledge related to bio-resources exploited ? Give 2 reasons
(b) Name 2 patents on Indian bio-resources that have been revoked.
Answer:
(a) Traditional knowledge is exploited to

• Develop modem applications of the resources
• Save time, effort, expenditure in the process of commercialisation of bio-resources.

(b) The two Indian patents include

• Pesticide property of neem
• Healing and antiseptic property of turmeric.

Question 2.
Mention three reasons for the success of green revolution in India.
Answer:

• Use of improved verities of crops
• Employing better management practice
• Use of agrochemical such as fertilizers and pesticides.

Question 3.
Enumerate the fields of application of biotechnology.
Answer:
The applications of biotechnology include

• Molecular diagnostics
• Bioremediation
• Waste treatment
• Energy production
• Therapeutics.

Question 4.
(a) Write the disadvantages of human use of insulin from other animal sources
(b) What are the concerns about transgenic insulin?
Answer:
(a) Disadvantages

• It causes allergy
• It causes other types of reaction

(b) Concerns about transgenic insulin

• The removal of C-peptide during maturation
• Assembling of the a and p polypeptides together to form mature form of insulin.

Question 5.
Name any 6 plants where Bt toxin producing genes have been included
Answer:

• Cotton
• Tomato
• Rice
• Potato
• Soybean
• Com.

2nd PUC Biology Biotechnology and its Applications Five Marks Questions

Question 1.
Two of the steps involved in producing nematode resistant tobacco plants base of on the process of RNA I are mentioned below. Write the missing steps in the proper sequence.
Answer:

• Nematode specific genes
• Production of both sense and antisense RNAs
• Double strand RNA
• Silencing specific mRNAs of nematode
• Death of nematodes
• Protection of transgenic plants from nematodes.

Question 2.
Describe by giving one example for each why transgenic animals are produced.
Answer:

• Transgenic animals are designed to allow the study of how genes are regulated and how they affect the normal functions of the body and its development
  eq: Information is obtained as to how insulin has a role as growth factor
• They are designed to increase our understanding as to how genes could control the development of disease, they serve as models of human disease.
• They produce useful biological compounds created by introducing a portion of the DNA that code for the products.
  eg: a-1 antitrypsin is produced for treating emphysema
• They are being developed to test the safety vaccines, eg: Polio vaccines has been tested on mice
• Transgenic animals with more sensitivity to toxic substance are being developed to the test the toxicity of drugs.

You can Download Chapter 11 Biotechnology: Principles and Processes Questions and Answers, 1st PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 11 Biotechnology: Principles and Processes

2nd PUC Biology Biotechnology: Principles and Processes NCERT Text Book Questions and Answers

Question 1.
Can you list 10 recombinant proteins which are used in medical practice? Find out where they are used as therapeutics (use the internet).
Answer:

• Human Insulin (Humulin) – Treatment of Diabetes Type 1
• Human Growth Hormone (HGH) – Replacement of deficient hormone in short stature persons.
• Calcitonin – Treatment of rickets.
• Chorionic Gonadotropin – Treatment of infertility.
• Erythropoietin – stimulates erythrocyte formation in anaemics.
• Tissue Plasminogen Activator – Dissolves blood clots after stroke and heart attack.
• Blood Clotting factors VIII and IX – Replacement of clotting factors missing in haemophitra A or B patients.
• Platelet Growth Factor – Stimulation of wound healing.
• Interferon (α,β,γ) – Treatment of viral infection and cancer.
• Interleukins – Enhancing activity of immune system.
• Vaccines – Preventing diseases like herpes, hepatitis B, etc.,

Question 2.
Make a chart (with diagrammatic representation) showing a restriction enzyme, the substrate . DNA on which it acts, the site at which it cuts DNA and the product it produces.
Answer:

Question 3.
From what you have learnt, can you tell whether enzymes are bigger or DNA is bigger in molecular size? How did you know?
Answer:
DNA molecules are biggest biomolecules containing number of genes. A gene operates through synthesis of polypeptide. An enzyme is formed of one or few polypeptides. It is always smaller than DNA which contains hundreds of genes.

Question 4.
What would be the molar concentration of human DNA in a human cell? Consult your teacher.
Answer:
Calculate molar concentration by finding weight of DNA in a cell and weight of whole cell.

Question 5.
Do eukaryotic cells have restriction endonucleases? Justify your answer.
Answer:
No. Eukaryotic cells do not have restriction endonucleases. Restriction endonucleases are present in bacteria. They protect bacteria from viral attack by disintegrating viral DNA without harming bacterial genome which has methylation of sensitive sites.

Question 6.
Besides better aeration and mixing properties, what other advantages do stirred tank bioreactors have over shake flasks?
Answer:
Foam control system, a temperature control system, pH control system and sampling port to take small volume of culture periodically are some other advantages of stirred tank bioreactors.

Question 7.
Collect 4 examples of palindromic DNA sequences by consulting your teacher. Better try to create a palindromic sequence by following base-pair rules.
Answer:

Question 8.
Can you recall meiosis and indicate at what stage a recombinant DNA is made?
Answer:
Pachytene of meioses I

Question 9.
Can you think and answer how a reporter enzyme can be used to monitor the transformation of host cells by foreign DNA in addition to a selectable marker? Ans: A selectable marker helps to identify transformed host cells and non transformed cells will be eliminated and only transformed cells will grow. Whereas reporter gene is the one whose phenotypic expression can be monitored and thus it reports about activity or change in advance of the effect of modification, in addition to eliminating non transformed cells by selectable markers. HLDescribe briefly the followings:
(a) Origin of replication
(b) Bioreactors
(c) Downstream processing
Answer:
(a) Origin of Replication (Ori):- It is DNA sequence which is specialised to initiate replication Bacterial chromosomes and plasmids possess a single origin of replication Eukaryote chromosomes have a number of origin of replication. Replication proceeds bidirectional from the site of origin of replication. The sequence also possesses nearby replication control which determines the number of copies it would form. Therefore the selected plasmid should have an origin of replication that supports high copy number.

(b) Bio Reactor:
Bio reactor used in biotechnology is generally 100-1000 litre cylindrical metal. container with a curved base to facilitate mixing of contents. The culture medium containing all nutrients, salts vitamins, hormones etc. is added along with inoculum of transformed cells with recombinant DNA. A stirrer helps in mixing and optimum availability of nutrients to culture cells. Supply of oxygen is maintained if the cells function better under aerobic conditions. Foam is kept under control. Gadgets are attached for knowing temperature and pH of the contents. Corrections are made when required. There is a sampling port where small volume of culture can be withdrawn to know the growth of cells and concentration of extractable product.

(c) Downstream processing:
It is the recovery of product from fully grown genetically modified cells, its purification and preservation. It is carried out after the sampling report indicates the completion of biosynthetic phase and presence of optimum product in the cells. After leaving a part of cellular mass of inoculum, the rest is crushed and chemically treated to separate the product. The separated product is purified and then formulated with suitable preservatives. Clinical traits are carried out to know its used and any immediate or long term adverse effect. Every batch of the product has to pass through strict quality control testing. Of course, the procedure and vigour of downstream processing and quality control varies from product to product.

Question 11.
Explain briefly
(a) PCR
(b) Restriction enzymes and DNA
(c) Chitinase
Answer:
(a) PCR:
PCR stands for Polymerase chain reaction. In this reaction multiple copies of the gene (DNA) of interest is synthesized in vitro using two sets of primers (small chemically synthesized oligonucleotides that are complementary to the region of DNA) and the enzyme DNA polymerase. The enzyme extends the primers using the nucleotides provided in the reaction and the genomic DNA as template. Then the process of replication of DNA is repeated many times i.e. one billion copies are made.

(b) Restriction enzyme:
The enzymes which are responsible for ‘ restricting the growth of bacteriophage in Escherichia coli are called restriction enzymes. One of this enzyme is added to the methyl groups of DNA, while the other cut DNA. This enzyme is known as restriction endonuclease enzyme. The first restriction endonuclease is Hind II. Now more than 900 restriction enzymes have been isolated from over 230 strains of bacteria.

(c) Chitinase:
It is an enzyme obtained from fungus Trichoderma which is specialised to digest chitin of fungal cell walls. The enzyme is used both as an effective fungicide as well as dissolving fungal cell walls to obtain their protoplasts for separation of DNA is genetic engineering.

Question 12.
Discuss with your teacher and find out how to distinguish between
(a) Plasmid DNA and Chromosomal DNA
(b) RNA and DNA
(c) Exonuclease and Endonuclease
Answer:

(a) Plasmid DNA and Chromosomal DNA

Plasmid DNA	Chromosomal DNA
(i) It is extra nuclear DNA	It is nuclear DNA
(ii) It caries nonvital genes	It possesses vital genes
(iii) A bacterial cell may carry one to several plasmid DNAs.	A bacterial cell carries only one chromosomed DNA

(b) RNA and DNA

RNA	DNA
(i) It is ribo nucleic acid	Deqxy ribo nucleic acid
(ii) It is single strandred	Double standard
(iii) RNA produced from DNA template	Parental DNA acts as DNA template
(iv) Nitrogen base Uracil in present instead of thymine, pairs with adenine.	Thymine pairs with adenine

(c) Exonuclease and Endonuclease

Exonuclease	Endonuclease
(i) It breaks DNA from ends	It cuts DNA from inside
(ii) The separated fragments are small nucleotides	These separated fragments  are generally large sized
(iii) The separated fragments cannot be used in genetic engineering	The desirable separated fragments are used in genetic engeneering

2nd PUC Biology Biotechnology: Principles and Processes Additional Questions and Answers

2nd PUC Biology Biotechnology: Principles and Processes One Mark Question

Question 1.
Expand
(i) PCR
(ii) Bt
Answer:
(i) PCR – Polymerase chain reaction
(ii) – Bacillus thurigenesis.

Question 2.
What is biotechnology?
Answer:
It is a branch of science that deals with techniques of using live organisms or enzymes from organisms, to produce products and processes useful to humans.

Question 3.
What is a plasmid? (Delhi 2008)
Answer:
A plasmid is an autonomously replicating circular extra chromosomal DNA found in bacterial cell.

Question 4.
Name the first restriction endonuclease discovered.
Answer:
Hind II is the first restriction endonuclease.

Question 5.
Name the technique used for separating DNA fragments in the latoratory
Answer:
Electrophoresis. (Dehli 2005)

Question 6.
What acts as “molecular scissors” in biotechnology.
Answer:
Restriction enzyme “endonuclease” acts as molecular scissors in biotechnology.

Question 7.
Write the function of restriction enzyme in bacterial cell?
Answer:
Responsible for restricting the growth of bacteriophage.

Question 8.
Name the substance used as the medium in gel electrophoresis.
Answer:
Agarose is the substance used in gel electrophoresis.

Question 9.
Why are plasmids and bacteriophages commonly used as cloning vectors?
Answer:
Plasmids and bacteriophages have the ability to replicate within bacterial cells independently of the chromosomal DNA.

Question 10.
Name the enzymes used to digest the cell wall of bacteria and fungi for genetic engineering.
Answer:

• Bacteria -Lysozyme
• Fungi – Chitinase

Question 11.
What is downstream processing?
Answer:
Downstream processing is the recovery of product from the fully grown genetically modified cells, its purification and preservation

Question 12.
Name any 4 products of Recombinant technology.
Answer:
Human insulin (Humulin). Human growth hormone (HGH) Chorionic gonadotropin Blood dotting factors VIII and IX Erythropoeitin Platelet growth factor Interferon (α,β,γ) (any 4)

Question 13.
What is gene therapy?
Answer:
It is the replacement of a defective gene by normal healthy and functional gene. This method helps to overcome the effect of various disorders like sickle cell anaemia, alkaptonuria, SCID, colour blindness etc.

Question 14.
Why does a cloning vector requires a selectable marker?
Answer:
Because it helps in identifying and selecting the recombinants and eliminating the non-recombinants.

Question 15.
What is amplification.
Answer:
It is the process of making multiple copies of gene/DNA segments of interest.

Question 16.
What is recombinant protein?
Answer:
It is biochemical compound or useful protein produced inside the heterologous host cell by recombinant biotechnology method.

Question 17.
What is bioreactor or fermenter?
Answer:
It is a container in which biochemical process is carried out by using living cells and their growth medium.

Question 18.
What meant by bioconversion?
Answer:
It is the process by which raw materials are biologically converted into specific products using microbes, plant or animal cells and or their enzymes.

Question 19.
Why are antibiotic resistance genes used as selectable markers for E.Coli?
Answer:
Since E.Coli doesn’t have any of antibiotic resistance genes, antibiotic resistance genes are used from outside as selectable marker.

2nd PUC Biology Biotechnology: Principles and Processes Two Marks Questions

Question 1.
Name the scientists who constructed recombinant DNA. Name the bacterium from which they isolated the gene.
Answer:
Stanley Cohen and Herbert Boyer were the first to construct a recombinant DNA. They isolated the gene from the bacterium. Salmonella typhimurium.

Question 2.
Few gaps have been left in the following table showing certain terms and their meanings, fill up the gaps.
Term Meanings
(i) ………….. Non-coding sequence in eukaryotic DNA
(ii) ………….. Technique used in solving paternity disputes
(iii) Restriction endonuclease …………..
(iv) Plasmids …………..
(v) Transgenics …………..
(vi) Nucleotide sequences with single base deficiencies …………….

Question 3.
Name the particular technique in biotechnology whose steps are shown in figure use the figure to summaries the technique in three steps.
Answer:

• Template stand
• DNA fingerprinting
• Extranuclear DNAs
• Organisms having genes of other organisms obtained through genetic engineering.
• Single nucleotide polymorphism. (SNPs)

Question 3.
Name the particular technique in biotechnology whose steps are shown in figure use the figure to summaries the technique in three steps.
Answer:

(a) Recombinant technology
(b)

• Cutting and isolation of human gene
• Incorporation of human gene into plasmid to produce recombinant DNA or plasmid
• Incorporation of recombinant plasmids into bacterium to obtain gene product

Question 4.
Refer the diagram and answer the following
(i) From what T₁– plasmid is obtained?
(ii) Name the enzyme which is involved in step I
(iii) What happens in step II
(iv) The plant produced is called hybrid or transgenic
(v) Will the plant produced have other genes along with desired genes? Yes or No explain.
Answer:

(i) Agrobacterium tumefaciens
(ii) Restriction endonuclease
(iii) Incorporation of genes in T₁ plasmid in the region of T-DNA.
(iv) Transgenic
(v) Yes. Selectable marker gene which is often an antibiotic resistance gene

Question 5.

Study the linking of DNA fragments shown above
(i) Name “a” DNA and “b” DNA
(ii) Name the restriction enzymes that recognizes this palindrome
(iii) Name the enzyme that can link these two DNA fragments (CBSE 2008)
Answer:
(i) (a) – vector DNA
(b) – foreign DNA

(ii) EcoRI
(iii) DNA ligase

Question 6.
Explain the importance of
(a) Ori
(b) amp R and
(c) rop in E. Coli vector shown below (CBSE 2008)
Answer:

• Ori – Origin of replication
• amp R – ampicillin antibiotic resistance gene
• rop – gene that produces proteins involved in the replication of plasmid.

Question 7.
An interesting property of restriction enzymes is molecular cutting and pasting Restriction enzymes typically recognize a symmetrical

Notice that the top strand is the same as the bottom strand but reads backward. When the enzyme cut the strand between G and A, it leaves overhanging chains

(A) What is the symmetrical sequence of DNA known as?
(B) What is the significance of these over hanging chains?
(C) Name the restriction enzyme that cuts the strand between G and A.
Answer:
(A) Palindromic sequence
(B) sticky ends
(C) Eco RI

Question 8.
Name any two cloning vectors. Describe the features required to facilitate cloning into a vector.
Answer:
a.

• Plasmid
• Viruses

b.

• Presence of origin of replication (Ori)
• Selectable marker for identifying recombinant from non-recombinant
• Single site recognition site for cloning

Question 9.
What are cloning vectors? What functions do these vectors perform?
Answer:
Cloning vectors are those organisms on their DNAs which can multiply independently of the host DNA and increase their copy number along with the alien DNA attached to them. Functions are:

• They help in linking the foreign/alien DNA with that of the host
• They also help in the selection of recombinants from non-recombinants

Question 10.
Name the organism from where the thermostable DNA polymerase is isolated. Write the importance of this enzyme is genetic engineering.
Answer:
Thermits acquaticus. This enzyme can remain active even during the high temperature induced denaturation of the double stranded DNA occurs.

Question 11.
How do “Ori” and “Cloning site” facilitate cloning into a vector? (AI 2008)
Answer:
“Ori” is a specific sequence of DNA, where replication occurs; the alien DNA must be linked to it, if it has to replicate. The restriction site where a foreign DNA is linked is called “cloning site”.

Question 12.
List 4 steps to isolate DNA from a bacterial cell. (AI 2008, 2009)
Answer:

• The bacterial cell is treated with lysozyme to break open the cell.
• The RNAs associated with the DNA are removed by treatment with ribonuclease (RNAases)
• The proteins are removed by treatment with proteases
• The purified DNA is precipitated with chilled ethanol.

2nd PUC Biology Biotechnology: Principles and Processes Three Marks Question

Question 1.
What are restriction enzymes? Describe the naming of the restriction enzyme.
Answer:
Restriction enzymes are those enzyme which are present in bacterial cells as a defence mechanism to restrict the growth of bacteriophage, by cutting the DNA at specific sites.

• The first letter of the name comes from the genus of the prokaryote.
• The second and third letters come from the name of the species of the cell from where it is obtained.
• The fourth letter comes from the strain of the prokaryote.
• The Roman numbers following these 4 letters indicate the order in which enzymes were isolated from the strain of the bacterise.
  e.g. E Coli RI is isolated from Escherichia coli strain RY13. The letter ‘R” stands for the strain.

Question 2.
Represent diagrammatically the E Coli cloning vector pBR 322 showing the restriction site.
Answer:

Question 3.
Draw a labelled diagram of a sparged stirred tank bioreactor.
Answer:

Question 4.
Read the following base sequence of a certain DNA strand and answer the questions that follow.

(i) What is called “Palindromic sequence” in a bNA?
(ii) Write the Palindromic nucleotide sequence shown in the DNA strand given and mention the enzyme that will recognise such a sequence.
(iii) State the significance of enzymes that identify palindromic nucleotide sequences.
(AI – 2008)
Answer:
(i) A palindromic sequence of DNA is a sequence of base pairs that reads the same on the two strands, when orientation of reading is kept the same, i.e. in the 5′ → 3′ direction.
(ii) 5′ GAA TT C – 3′
3′ CTTA AG – 5′
This sequence is restricted by the restriction enzyme Eco RI
(iii) The enzyme that identifies the palindromic nucletide sequence cut the strands between the same 2 bases, more often producing sticky ends; hence they are useful in the formation of recombinant DNA.

2nd PUC Biology Biotechnology: Principles and Processes Five Marks Question

Question 1.
Describe in detail the components of a simple stirred tank bioreactor along with a labelled diagram.
Answer:
A stiered tank bioreactor is usually a cylindrical vessel with a curved base to facilitate the mixing of the content. The stirrer facilitates even mixing and oxygen availability through out the bioreactor.

The bioreactor has an agitator system, an oxygen delivery system along with a foam control system, a temperature control system, pH control system and sampling ports to remove small volumes of culture periodically.

It provides optimum growth conditions of pH, temperatures, substrate, oxygen etc. for achieving the desired products.

Question 2.
In a bacterial culture, some of the colonies produced blue colour in the presence of a chromogenic substrate and some did not due to the presence or absence of an insert (r DNA) in the coding sequence of (3 – galactosidase).
(a) Mention the mechanism and the steps involved in the above experiment.
(b) How is it advantageous over simultaneous plating on two plates having different antibiotics.
Answer:
(a) The mechanism involved is called insertional inactivation, the phenomenon in which the enzyme becomes inactivated when a recombinant DNA is inserted with in the coding sequence of that enzyme.
Steps in process:

• A recombinant DNA is inserted into the DNA sequence coding for the enzyme galacto sidase, it results in the inactivation of the enzyme.
• In case of the recombinants, when the plasmid has an insert, there is no blue colour produced in the presence of a chromogenic substrate in the medium.
• In the case of nonrecombinants/non- tranforments i.e. when the plasmid has no insert, a blue colour is produced in the presence of a chromogenic substrate in the medium.

(b) Advantage:
Simultaneous plating method is quite cumbersome as it requires two plates where as this method is simple and easy.

Question 3.
(a) If the restriction enzyme has to cut a DNA, the DNA must be in pure form, i.e. free from the associated RNA and proteins. How is it achieved.
(b) Represent only diagrammatically the steps in the recombinant DNA (r DNA) technology.
Answer:
(a) The RNAs are removed by using enzymes called ribonucleases (RNases) The proteins are removed by using enzyme proteases.
(b)

Question 4.
(a) Show only diagrammatically the three steps in the polymerase chain reaction,
(b) How is repeated amplification achieved using this method?
Answer:

(b) Repeated amplification is achieved by the use of a thermostable DNA polymerase, it is isolated from the bacterium Thermus aquaticus. It remains active during the high temperature used for denaturation of the double-stranded DNA.

Question 5.
Make a diagrammatic representation of showing a restriction enzyme, the substrate DNA on u which it acts the site at which it cuts DNA and the product it produces.
Answer:

You can Download Chapter 10 Microbes in Human Welfare Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 10 Microbes in Human Welfare

2nd PUC Biology Microbes in Human Welfare NCERT Text Book Questions and Answers

Question 1.
Bacteria cannot be seen with the naked eyes, but these can be seen with the help of a microscope. If you have to carry a sample from your home to your biology laboratory to demonstrate the presence of microbes under a microscope, which sample would you carry and why?
Answer:
Curd – Contains Lactic Acid Bacteria Lactobacillus acidophilus).

Question 2.
Give examples to prove that microbes release gases during metabolism.
Answer:
Dough of dosa and idli is fermented by bacteria while dough of bread is fermented by yeast. The puffed-up appearance is due to CO₂ gas production.

Question 3.
In which food would you find lactic acid bacteria? Mention some of their useful applications.
Answer:
Curd
Uses:-

• Lactic acid bacteria (LAB) convert milk into curd by producing lactose. Converts sugar lactose into acid which converts casein in milk to curd.
• Removes lactose, produce Vitamin B₁₂ and checks growth of putrefying bacteria as well as harmful microbes.

Question 4.
Name some traditional Indian foods made of wheat, rice and Bengal gram (or their products) which involve use of microbes.
Answer:

• Bhatura, Bread, Jalebi (wheat).
• Dosa, Idli (Rice)
• Karhi (Bengal gram).

Question 5.
In which way have microbes played a major role in controlling diseases caused by harmful bacteria?
Answer:
Microbes have been a source of antibiotics. Antibiotics have been used successfully against pathogenic bacteria. E.g.: streptomycin (str. griseus), erythromycin
(str. erythreus),Bacitracin (Bacillus licheni – formis)

Question 6.
Name any two species of fungus, which are used in the production of the antibiotics.
Answer:

• Penicillin – Penicillin chrysogenum
• Griseofulvin – Penicillin griseofulvin
• Fumagillin – Aspergillus puniyatus
• Cephalosporin – Cephalosporium acremonium.

Question 7.
What is sewage? In which way can sewage be harmful to us?
Answer:
Sewage/municipal waste water is human excreta and other organic wastes containing large number of pathogenic microbes and harmful chemicals. It is harmful to us because

• Source of various water
• Causes eutrophication of water
• Produces scum and sludge, bad taste, foul smell and turbidity to water to which they are dumped.

Question 8.
What is the key difference between primary and secondary sewage treatment?
Answer:
Primary treatment is physical and removes grit and large prices of organic matter while secondary treatment is Biological causing digestion of organic matter by microbes.

Question 9.
Do you think microbes can also be used as source of energy? If yes, how?
Answer:
No, But their fermentation products Biogas, Alcohol, etc, are used as source of energy.

Question 10.
Microbes can be used to decrease the use of chemical fertilisers and pesticides. Explain how this can be accomplished.
Answer:
Chemical fertilizers are used to increase availability of minerals while pesticides are economically harmfull and toxic to human being, cattles. This disturbs biological environment. Microbes used as fertilizers (bio fertilizes) and pesticide (bio pesticide) are not harmful to ecology and humans. Organic farming uses them and provides increased production.

Question 11.
Three water samples namely river water, untreated sewage water and secondary effluent discharged from a sewage treatment plant were subjected to BOD test. The samples were labelled A, B and C; but the laboratory attendant did not note which was which. The BOD values of the three samples A, B and C were recorded as 20mg/L, 8mg/L and 400mg/ L, respectively. Which sample of the water is most polluted? Can you assign the correct label to each assuming the river water is relatively clean?
Answer:
B least BOD, so its river water A discharged from sewage treatment plant C highest BOD, untreated sewage water.

Question 12.
Find out the name of the microbes from which Cyclosporin A (an immuno-suppressive drug) and Statins (blood cholesterol lowering agents) are obtained.
Answer:

• Cyclosporin A from Trichoderma polysporum
• Statins from yeast Monascus purpureus.

Question 13.
Find out the role of microbes in the following and discuss it with your teacher.
(a) Single cell protein (SCP)
(b) Soil
Answer:
(i) Single Cell protein – Used as food and feed with all essential Amino Acids, low fat. E.g.: Spirullina, yeast, Mushroom.
(ii) Soil –

• Humification
• Mineralisation – release minerals during degradation
• Biofertilizers
• Denitrification – Discuss about all with teacher.

Question 14.
Arrange the following in the decreasing order (most important first) of their importance, for the welfare of human society. Give reasons for your answer. Biogas, Citric acid, Penicillin and Curd.
Answer:
Penicillin > Biogas > Curd> Citric Acid.

• Penicillin an antibiotic cures bacterial diseases.
• Biogas is source of energy and pollution free fuel.
• Curd easily digestible vitamin rich milk preparation.
• Citric Acid used as preservatives.

Question 15.
How do biofertilisers enrich the fertility of the soil?
Answer:
Biofertilisers are micro-organisms which increases soil fertility and enhances nutrients to crop plants.

• Nitrogen fixing bacteria and Cyano bacteria
  Converts free N₂ from atmosphere to salts of N₂.
• Phosphate bacteria – Secrete phosphatase that dissolves insoluble phosphate from soil for absorption by plants.
• Mycorrhiza – occurs in forest plants solubilizes and absorbs nutrients from organic matter.

2nd PUC Biology Microbes in Human Welfare Additional Questions And Answers

2nd PUC Biology Microbes in Human Welfare One Mark Questions

Question 1.
What is LAB?
Answer:
The microorganisms which commonly grow in milk and converted into curd are called Lactic Acid Bacteria or LAB.

Question 2.
What changes are brought about in milk by LAB?
Answer:
LAB produces acids that coagulate and partially digest the milk proteins.

Question 3.
Who discovered pencillin?
Answer:
Alexander Fleming.

Question 4.
What are prions?
Answer:
Prions are proteinaceous infectious agents

Question 5.
What are fermentors?
Answer:
Fermentors are very large vessel used for growing microbes for the production of microbial products on an industrial scale.

Question 6.
What are antibiotics?
Answer:
The chemical substances which are produced by certain microbes which can kill or retard the growth of other microbes.

Question 7.
What is BOD?
Answer:
The amount of oxygen that would be consumed for the oxidation or decomposition of the biodegradable organic waste present in the water is called BOD or biological oxygen demand.

Question 8.
Name the organism used for the preparation of Swiss cheese?
Answer:
Propioni bacterium sharmani.

Question 9.
Name the enzyme used “clot buster” is to remove blood clots from blood vessels and which organism produces this enzyme.
Answer:
Streptokinase is know as clot buster. It is produced by Streptococcus bacteria.

Question 10.
What is the scientific name of brewer’s or baker’s yeast. (AI2009)
Answer:
Saccharomyces cerevisiae.

Question 11.
What is STP stands for?
Answer:
Sewage Treatment plant.

Question 12.
Name the free-living nitrogen-fixing bacteria in the soil.
Answer:
Azospirillum and Azotobacter.

Question 13.
Which of the following is a Cyanobacteria that can fix atmospheric nitrogen? Spirulina, Azospirillum, Sonalika.
Answer:
Azospirillum.

Question 14.
Milk starts to coagulate when Lactic Acid Bacteria is added to warm milk as a starter. Mention any other two benefits LAB provides. (AI CBSE – 2009)
Answer:
Increase vitamin B₁₂ and also check disease-causing microbes in stomach.

Question 15.
Name the group of organisms and the substrate they act on to produce biogas.
(CBSE 2009)
Answer:
Methanogens, Cellulose present in cattle excreta (cow dung)

Question 16.
What are floes?
Answer:
The masses of bacteria associated with fungal filaments forming the mesh like structures are the floes.

Question 17.
What is biological control of Pests?
Answer:
It is the method of controlling pest in agriculture, that relies on natural predation rather than, the introduced elements.

Question 18.
Name the fungus that is being developed as a biocontrol.
Answer:
Trichoderma.

Question 19.
What is the key differences between primary and secondary sewage treatment?
Answer:
Primary treatment of sewage is a physical process, while the secondary treatment is a biological process.

Question 20.
Name a genus of a fungus that forms mycorrhiza.
Answer:
Glomus is the fungus that forms mycorrhiza.

Question 21.
List 2 advantages that a mycorrhizal association provides to the plant. (AI2008)
Answer:

• The fungal partner in mycorrhiza absorbs
• Phosphorus from soil and passes it on to the plant.
• The plants having mycorrhizal association show resistance to root borne pathogens and tolerance to salinity

2nd PUC Biology Microbes in Human Welfare Two Marks Questions

Question 1
(a) Name an eco-friendly hloherhicide which interferes with amino add synthesis and resistance to which has been obtained through transgenic culture.
(b) Name the first biopesticide. (CBSE 2004)
Answer:
Basta or phosphinothricin (ppt)
(a) It obtained from a product of Streptomyces species. Crop plants have been made resistant to it by transferring bar gene into them (e.g. wheat)
(b) Divine and college.

Question 2.
Match the columns

Answer:

Question 3.
Name the blank spaces ‘a’, ‘b’,’c’ and ‘d’ give in the following label.

Type of microbe	Scientific Name	Commercial or Product
Bacterium	a	Lactic acid
Fungus	b	Cyclosporin A stains
C	Monascus Purpur
Fungus	Penicillin Nolaltisin	d

Answer:

• a – Laetobaellhts
• b – Triehodmna polysptmtm
• c – Yeast (fungus)
• d – Penicillin

Question 4.
Name the blank spaces a, b, c, and d given in the following label. (CBSE – 2008)

Type of microbe	Scientific Name	Commercial or Product
Bacterium	a	Clot buster
b	Aspergillus Trichoderma	Enzyme citric acid
Fungus	Polysporum	c
Bacterium	d	Enzyme citric acid

Answer:
a – Streptococcus
b – Fungus
c – Cyclosporin A
d – Clostridium butylicum.

Question 5.
Fill in the blanks from Cephalosporin, Cyclosporin, Cycas, Soybean, Nepiatode, Fungus, Rhizobium. (CBSE 2008)
(a) Potent immunosuppressant drug is ………….. which is obtained from a …………..
(b) Roots pines and ………….. are associated will Amanita and …………..respectively.
Answer:
(a) Cyclosporin A, Fungus
(b) Soybean, Rhizobium.

Question 6.
(a) What is the source of statins and how they reduce the level of cholesterol in our body.
(b) Write the technical terms for VAM. Pickup examples of endomycorrhiza and ectomycorrhiza from sclerocystis, Laccaria, Gigaspora, glomas, Hebeloma, Psilocybes . (CBSE – 2006)
Answer:
(a) Stain which reduces the cholesterol level in our body is synthesized by the activity of yeast Monascus purpureus egs of statins are Lovastalin, simvastatin pravastatin, fluvastatin.

(b) VAM – It is vesicular arbuscular mycorrhiza where the hyphae send vesicular and branched haustoria into root cortical cells for obtaining nourishment in endomychorrhiza
eg:
Ectomychorrhiza – Laccaria, Hebeloma, Psilocybes
Endomuchorrhiza – Sclerocystis, Gigaspora, Glomos.

Question 7.
Name the blank spaces a, b, c and d from the table given below (CBSE 2007)

Type of Microbes	Scientific Name	Product	Medical Application
Fungus	a	Cyclosporin	b
c	Monascus	statin	d
	purpureus

Answer:
(a) Trichoderma polysporum
(b) Organs transplant patients
(c) Yeast (fungus)
(d) lowering blood cholesterol level.

Question 8.
Name the organism that causes large holes in “Swiss cheese”. How are these cause? (CBSE, Delhi 2009)
Answer:
Propioni bacterium sharmani. The large holes are due to the large quantity of carbon dioxide produced.

Question 9.
What are harmful effects of using chemical pesticides?
Answer:

• The chemical pesticides are harmful to many organisms (may be useful) other than the pests for which they used.
• They enter the food chain and cause diseases/disorders in various organism
• They are also highly toxic to man
• They cause pollution of our soil, water and air.

Question 10.
How do bacteria function as biofertilizers? Name 2 free-living soil bacteria that are biofertilizres.
Answer:

• Bacteria can fix atmospheric nitrogen into those nitrogen compounds which are used by the plants as their nutrients.
• Soil bacteria that are biofertilizers includes – Azotobacter and azospirillum.

2nd PUC Biology Microbes in Human Welfare Three Marks Questions

Question 1.
What are methanogens? Where are they generally found. Give examples.
Answer:
Methanogens are a group of anaerobic bacteria, which produce large quantities of methane from cellulosic materials.
E.g.: Methano bacterium Methanogens are found in

• The sewage wager
• Marshy place
• The rumen of cattle.

Question 2.
Describe the functions of anaerobic sludge in a sewage treatment plant.
Answer:

• A portion of the activated sludge from anaerobic sludge digester is pumped back to the aeration tank to serve as inoculum.
• The anaerobic bacteria digest the bacteria and fungi of the floes.
• The anaerobic bacteria in the sludge release CO₂, methane etc. during decomposition; these gases from the biogas is used as a source of energy, as it is inflammable.

Question 3.
What are mycorrhizae? How do they serve as biofertilizers or how are they useful to plants.
Answer:
Mycorrhizae are the symbiotic association between certain fungi and roots of higher plants.

• The fungus absorbs phosphorous from the soil and passes it to the plants
• Plants with mycorrhiza show resistance to root borne pathogens
• They show increased tolerance to salinity and drought
• There is an overall increase in plant growth and development.

Question 4.
Write about “Organic farming”.
Answer:
Organic farming is a holistic approach that seeks to develop an understandings of the webs of interaction among the myriads of organism that form the flora and fauna of the field.

• The organic matter works to create a system where the insects are not eradicated, but kept at managable levels by a complex system of checks and balance with in a leaving and vibrant ecosystem.
• According to former the eradication of the creatures, called pests, is not only possible but also undesirable because many beneficial predatory and parasitic insects cannot survive without them. Such method reduces the use the chemical pesticides and there by the pollution.

2nd PUC Biology Microbes in Human Welfare Five Marks Questions

Question 1.
Write a short note Cryprotein and its agricultural use. (CBSE 2006)
Answer:
Cryprotein is a potentially toxic chemical that is produced in pro toxin crystalline endotoxin state by bacterium Bacillus thurigenes. It can be extracted from the bacterium. The protein is also present in spores of the bacterium. The commercial product is called sporein, depel, biostol and thurigenesis. It is sprayed over the crop. As the protein reaches the midgut of the insect.it is converted into the toxic state by alkaline pH and digestive enzyme. The toxic binds to specific receptors present on epithetial cells, creates pores and ruptures the cells causing death of the insects.

There are several types of Cryproteins specific to different types of insects. The gene for cry protein called cry has been incorporated in some crop plants to provide resistance to insect pests. Such genetically modified crop plants are called Bt (after Bacillus thurigenesis) crops, example Bt cotton.

You can Download Chapter 9 Strategies for Enhancement in Food Production Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 9 Strategies for Enhancement in Food Production

2nd PUC Biology Strategies for Enhancement in Food Production NCERT Text Book Questions and Answers

Question 1.
Explain in brief the role of animal husbandry in human welfare.
Answer:
Animal husbandry provides livelihood for large number of persons with milk, egg, meat, etc., production.

• Milk-Only animal protein for vegetarians. It is a complete food. Milk is obtained from cow, buffalo, goat, sheep, camel and yak.
• Egg- Chick and duck are major sources. It is also a complete food.
• Meat – Protein rich diet obtained from goat, sheep, chicken, fish etc.,
• Honey – Sweet syrup from honey bees used for sweetening
• Fibres – wool, silk etc., for clothes etc.,
• Hides – For hides and leather from animals skin
• Work animals – To carry men and materials e.g.: Efuffalo, bullock, yak.
• Employment – For many persons by rearing and feeding camel etc.
• Waste production – Homs, feathers etc., all used for producing useful products.

Question 2.
If your family owned a dairy farm, what measures would you undertake to improve the quality and quantity of milk production?
Answer:
Measures should be taken under following headings:-

• Selection of Breed – Milk production depends mainly on breed. High yielding, disease resistant, acclimatised to climate condition of the area.
• Cattle shed – It should be spacious, roofed but airy with sloping floors draining wine and dung removal with permanent water supply.
• Feed – Daily, regular, fresh fodder and water in adequate quantity has to be provided.
• Grooming – To be regularly brushed, massaged and cleaned. Regular grooming keeps cattle healthy.
• Sanitation – Hygiene of cattle, handlers, transportation, mechanisation etc., to bee kept, Sanitation of cattle shed.
• Health care – Regular checkup, vaccination is mandatory

Question 3.
What is meant by the term ‘breed’? What are the objectives of animal breeding?
Answer:
Breed – Homogeneous group of animals within a species, subspecies or a variety which are related by descent and are similar in general appearance, size, configuration and other features. E.g.: Sahiwal, Brown swiss.
Objectives of Animal Breeding:

• Increase yield like that of milk, egg, meet and wool.
• Better quality of animal products
• Higher growth rate
• Resistance to diseases
• Longer productive life
• Better assimilation efficiency
• Higher reproductive rate.

Question 4.
Name the methods employed in animal breeding. According to you which of the methods is best? Why?
Answer:

• Inbreeding
• Out breeding

Question 5.
What is apiculture? How is it important in our lives?
Answer:
Apiculture (Bee-keeping) → Rearing, care and management of honey bees. Importance:-
(a) Honey

• Natural aromatic sweet syrup
• Natural sweetener, laxative, blood purifier.
• Immediate source of energy

(b) Bee wax – Used in cosmetics, creams, ointment
(c) Bee venom – Venom from sting used in Rheumatoid arthritis
(d) Propolis – Derived from plants for antiseptic and antibiotics
(e) Pollination – Honey Bees act as major pollinators

Question 6.
Discuss the role of fishery in enhancement of food production.
Answer:
Fish and other edible aquatic animals are important source of food rich is proteins, minerals and vitamins and is a source of oil. It provides with good financial gain . Enhanced food availability is called Blue Revolution.

Question 7.
Briefly describe various steps involved in plant breeding.
Answer:

• Collection of variability – variability is recorded
• Evaluation and selection of parents – desirable traits are searched and parents are selected. They are selfed to obtain homozyosity.
• Cross hybridisation – single to multiple crosses are made between parents of desirable traits to produce a single variety.
  They are protected from contamination
• Selection – seeds of desired characters are selected. Selection is made at every generation.
  Back cross done if desired characters of one parent is not incorporated.
• Testing, release and commercialisation – Tested by ICAR and released after giving a variety name.

Question 8.
Explain what is meant by biofortification.
Answer:
Crop breeding programme aimed at increasing quality of crop like high vitamin content, more minerals, complete proteins and heathier fats, eg: lysine and tryptophan rich maize, Vitamin A enriched carrot.

Question 9.
Which part of the plant is best suited for making virus-free plants and why?
Answer:
Shoot tip culture – Apical meristem is always free of virus. So we can obtain virus free plants.

Question 10.
What is the major advantage of producing plants by micro-propagation?
Answer:
A very large number of identical, virus free plants of desired characters (e.g. tolerance, mutation, resistance) can be raised in very short duration.

Question 11.
Find out what the various components of the medium used for propagation of an explant in vitro are?
Answer:
Medium should provide carbon source such as sucrose, inorganic salts, vitamins and growth regulators like auxins, cytokinins etc. It evens give required moisture.

Question 12.
Name any five hybrid varieties of crop plants which have been developed in India.
Answer:

• Hybrid Maize
• Hybrid Jowar
• Hybrid wheat
• Hybrid Bajra
• Hybrid Garden Pea.

2nd PUC Biology Strategies for Enhancement in Food Production Additional Questions and Answers

2nd PUC Biology Strategies for Enhancement in Food Production One Mark Questions

Question 1.
Define animal husbandry.
Answer:
It is the science of systematic breeding and raising of domesticated animals as per human requirement.

Question 2.
Define livestock.
Answer:
Domesticated animals which are raised for use or profit are collectively called livestock.

Question 3.
Define farm management.
Answer:
It is the controlled and scientific handling of farm animals in their rearing, grooming, breeding and caring so as to maximise their yield.

Question 4.
Mention the two major practices included in animal husbandry.
Answer:
The two major practices

• Management of the animals (caring, feeding, housing etc.)
• Animal breeding.

Question 5.
Define inbreeding depression. (CBSE 2004)
Answer:
Inbreeding depression refers to the reduced fertility and productivity associated with continued breeding.

Question 6.
What is plant breeding? (CBSE 2005)
Answer:
It is the manipulation of plant species so as to create new improved varieties that are better suited for cultivation, give better yield and are resistant .to damage from pest and pathogens.

Question 7.
Define mutation.
Answer:
Mutations are sudden, stable, inheritable discontinuous variation which develop in organism due to changes in their genetic system especially base sequence in genes. They produce alleles which were not present in parental type.

Question 8.
What is callus?
Answer:
Callus is an unorganised undifferentiated mass of actively dividing cells.

Question 9.
Define micro-propagation.
Answer:
The method of producing very large number of plants through tissue culture is known as micro-propagation.

Question 10.
What is embryo rescue? (Embryo culture)
Answer:
Embryos which normally don’t survive inside seeds can be grown in tissue culture to form new plants. This process is called embryo rescue.

Question 11.
What is SCP?
Answer:
Single Cell Protein.

Question 12.
Define apiculture?
Answer:
The maintenance of the hives of honey bees for the production of honey.

Question 13.
What is green revolution?
Answer:
Green revolution is the movement launched Which was responsible for the increased food production not merely to meet the national requirement of food production but also to export it.

Question 14.
Name a semidwarf verities of wheat which is known for high yielding and disease resistant.
Answer:
Kalyan Sona.

Question 15.
Name the varieties of rice from which semidwarf varieties have been developed in India.
Answer:

• 1R – 8 (from Philippines)
• TN – 1 (from Taiwan)

Question 16.
Name 2 better yielding semi dwarf varieties of rice developed in India.
Answer:
Jaya, Ratna, are the two varieties of rice

Question 17.
Name the organism commercially used for the production of single cell protein. (Delhi 2009)
Answer:
Spirullina.

Question 18.
Write the economic value of spirullina.
Answer:

• It is used as food rich in protein, carbohydrate vitamins and minerals.
• Since its culturing uses waste materials like molasses, waste water from potato processing factories, it reduces environmental pollution.

Question 19.
What is soma clone?
Answer:
These are genetically identical plants. developed through tissue culture from one plant.

Question 20.
What is somatic hybridisation.
Answer:
It is process in which a plant is raised from the product of fusion of isolated protoplasts of two different varieties or species of plants in vitro.

Question 21.
Define biofortification.
Answer:
Biofortification refers to breeding of crops to produce varieties with higher levels of nutrients like vitamins, minerals, high protein content or healthier fats.

Question 22.
Name a microorganism which is expected to produce large quantities of proteins (is in tonnes)
Answer:
Methylophilus methylo trophus

Question 23.
List any two economically important products for humans obtained from Apis India (CBSE 2008)
Answer:
Honey and beewax.

Question 24.
What is MOET? Write its objectives.
Answer:
Multiple Ovulation Embryo Transfer Technology. It used for the successful production of high yielding hybrids.

Question 25.
Write the economic value of Saccharum officinarum.
Answer:
This sugar cane has thicker stems and higher sugar content and grows well in South India. It is used for hybridisation with Saccharum barberi to improve the variety.

Question 26.
Define totipotency.
Answer:
The capacity to generate a whole plant from any cell or explant is called totipotency.

2nd PUC Biology Strategies for Enhancement in Food Production Two Marks Questions

Question 1.
What is meant by the term breed. What are the objectives of animal breeding.
Answer:
The term “breed” refers to a group of animals which are related by descent and are similar in most of their features.
The objectives of animal breeding are

• Increasing the quantity of yield.
• Improving the quality of yield.

Question 2.
The steps in a programme are
(a) collection of germ plasm
(b) cross breeding the selected parents
(c) selecting superior recombinant progeny
(d) testing, releasing and marketing new cultivars
(i) What is programme related to?
(ii) Name two special qualities as basis of selection of the progeny.
(iii) What is the outcome of programme.
(iv) What is the popular term given to this outcome? Also name the Indian scientist, Who is credited with taking out this programme.
Answer:
(i) Plant breeding
(ii) Better yield and disease resistance
(iii) Superior improved variety
(iv) Green revolution, M.S. Swaminathan.

Question 3.
Find out what are the various components of the medium used for propagation of explant in vctro.
Answer:
Medium should provide carbon sources such as sucrose, inorganic salts, vitamins, amino
acids and growth regulators like auxins cytokinins etc.

Question 4.
Name any 5 hybrid varieties of Crop plants which have been developed in India.
Answer:

• Hybrid maize
• Hybrid wheat
• Hybrid Jowar
• Hybrid bajra
• Hybrid Green pea.

Question 5.
What is commonly used to produce virus resistant plants.
Answer:
Hybridisation with disease resistant variety, wild relative or related species.

Question 6.
Name and define the two major methods of animal breeding.
Answer:
Inbreeding and out breeding are the two major methods of animal breeding.

• Inbreeding:- Inbreeding refers to the mating of more closely related individuals of the same breed.
• Outbreeding:- Outbreeding refers to the mating of the individuals of different breeds or the individuals of same breed having no common ancestors.

Question 7.
Why is bee keeping preferred?
Answer:
Beekeeping is preferred because

• There is an increased demand for honey and also beewax
• It can be practiced in any area where there are sufficient bee pastures
• Several species of honey bees can be reared

Question 8.
Write the differences between somaclones and somatic hybrids.
Answer:

Somaclones	Somatic hybrids
They are clones produced through tissue culture They are used in rapid multiplication of a desired variety	They are hybrids formed through protoplast tissue Hybrids develop even in those cases where their development is not possible through sexual means.

2nd PUC Biology Strategies for Enhancement in Food Production Five Marks Questions

Question 1.
Enumerate the points that have to be considered for successful bee keeping.
Answer:
The following are the points for successful beekeeing

• Knowledge of nature and habits of bee
• Selection of suitable location and keeping the beehives.
• Catching and hiving of swarms
• Management of beehives during deficient seasons
• Handling and collection of honey and bee wax.

Question 2
(a) Write the scientific names of sugarcane grown in North India and South India respectively Mention their characteristic features.
(b) Mention the characteristic features of hybrid variety produced by crossing these two varieties.
Answer:
(a) Sugar cane grown in North India is Saccharum barberi. It has poor yield and poor sugar content. Sugar cane grown in South India is Saccharum officinarum. It has thicker stems and high sugar content but cannot grow well in north Indian condition.

(b) They were successfully crossed to yield hybrid varieties having the following desirable qualities.

• High yield
• Thick stem
• High sugar content
• Ability to grow in North Indian sugar cane growing region.

Question 3.
Expand MOET. Explain the procedure of this technology in cattle improvement. (Delhi CBSE 2008)
Answer:
MOET – Multiple ovulation Embryo transfer technology .It is a method to improve the chances of successful production of hybrids. A cow is administered hormone with FSH like activity to induce follicular maturation and superovulation, i.e. production of 6-8 ova instead of one per cycle

• The cow is either mated with the selected superior bull or artificially inseminated
• The fertilize eggs are recovered non surgically at the 8-32 celled stage
• They are often transferred to the uterus of the surrogate mother and allowed IP develop till birth
• The genetic mother is now available for another cycle of superior ovulation.

Question 4.
IARI has released several varieties of crop plants that are biofortified. Give 3 examples of crops and there biofortification.
Answer:

• Vitamin A enriched carrots, spinach, pumpkin etc.
• Vitamin C enriched tomato, mustard, bitter guard.
• Calcium enriched spinach
• Proteins enriched broad bean, french beans and garden peas etc.

Question 5.
Biofortification can solve the problem of hidden hunger. Explain.
Answer:
Biofortification is the breeding of crops to produce varieties with higher levels of nutrients like vitamins, minerals, high protein content and healthier fats.

• It is the most practical means to improve public health.
• Many hybrid varieties of crops have been produced that are enriched with specific nutrients
  e.g.: Iron – riched spinach, vitamin C riched tomato, bitter gourd, mustard etc.,
•  By eating the normal food made from these nutrient – enriched varieties of crops there can be an end to deficiency diseases.

Question 6.
Name and explain the method employed these days for breeding for disease resistance give an example.
Answer:
Mutation breeding is used Mutation breeding is the process in which mutations are induced by the use of chemicals or radiation and selecting and using the plants that show the desirable characters, e.g.: By this process, mung bean is resistant to yellow mosaic virus and powdery mildew has been produced.

Question 7.
(a) Differentiate between Inbreeding and Out breeding.
(b) What is germ plasm collection?
Answer:
(a)

Inbreeding	Out breedings
Inbreeding refers to the mating of more closely related indi -viduals with in the same breed for 4-6 generations.	Out breeding refers to the mating of unrelated animals of same breed or deferent breeds of different species.

(b) Germ plasm collection refers to the entire collection of plants and seeds having all different alleles of all the genes in a crop and its closed relatives.

Question 8.
Write the steps in the production of new variety of plants.
Answer:

• Collection of variability or germ plasm collection
• Evaluation and selection of parents
• Cross hybridisation among selected parents
• Selection of superior in fields
• Release and commercialisation of new varieties.

Question 9.
Define totipotency of a cell. List the requirements of the objective to produce somaclones of a tomato plant in commercial scale.
Answer:
Totipotency is defined as the capacity of any plant cell to generate a whole plant. Requirements are

• An explant, i.e. any part/tissues of a tomato plant
• Nutrient medium containg a carbon source like sucrose, inorganic salts, aminoacids, vitamins and growth regulators like auxins and cytokinins.
• Sterile conditions.

Question 10.
What is artificial insemination in animal breeding? What are its advantages.
Answer:
Artificial insemination is the process in which the semen collected from a superior male animals is injected into the reproductive tract of selected female, using surgical instruments under aseptic condition.
Advantages:

• Semen cart be used immediately or stored or frozen and used at a later date, when the female animal is in the right phase of reproduction.
• Semen can be transported in the frozen form to where the selected/Superior female animal is present.
• Semen from a single selected male cow be used for insemination of a number of females.
• It helps us to overcome several problems associated with normal mating.

 

You can Download Chapter 8 Human Health and Disease Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

2nd PUC Biology Question Bank Chapter 8 Human Health and Disease

2nd PUC Biology Human Health and Disease NCERT Text Book Questions and Answers

Question 1.
What are the various public health measures, which you would suggest as safeguard against infectious diseases?
Answer:
Public Health Measures:-

• Garbage – regularly disposed at proper locations. Containers to be properly cleaned should be reused or recycled when applicable.
• Drainage – To be covered to prevent flies and mosquito. Stag nation of water near household to be removed. Slums to be cleaned.
• Water reservoirs- To be covered and periodically cleaned and disinfected.
• Sewage disposal – Properly treated before find disposal.
• Public places – To be kept clean and fogging to be done periodically.
• Pollution – Antipollution laws to be enforced strictly.
• Vaccination and medical care Periodic medical examination and vaccination to be taken properly.
• Drinking water- To be free from contamination.
• Facilities- Public toilets and bathrooms to be provided.

Question 2.
In which way has the study of biology helped us to control infectious diseases?
Answer:

• Helps to properly diagnose the problem and treat them
• Life cycle, host, reservoir, vectors, agents all can be studied and identified
• Mechanism of transmission of pathogens to humans are known
• Protective measures against catching the diseases are suggested.
• Vaccines and medicines against pathogen are worked out.

Question 3.
How does the transmission of each of the following diseases take place?
(a) Amoebiasis
(b) Malaria
(c) Ascariasis
(d) Pneumonia
Answer:
(a) Amoebiases – Family disposal of faecal matter causing contamination of drinking water and food. Houseflies and air current transport cysts Pet animals.
(b) Malaria – Female Anopheles inject sporozoites along with saliva
(c) Ascariasis – Food particles, dirty hands, taps, soaps etc., by faecal
(d) Pneumonia – Droplets and aerosols of infected person.

Question 4.
What measure would you take to prevent water-borne diseases?
Answer:

• All water resources, pools, tanks to be cleaned regularly and prevent this contamination. Only properly treated water should be added to it.
• Use only purified and disinfected water for drinking.
• Prevent passage of sewage and garbage into water bodies

Question 5.
Discuss with your teacher what does ‘a suitable gene’ means, in the context of DNA vaccines.
Answer:
In DNA vaccines suitable gene means gene controlling formation of immunogenic protein. Gene integrated with vectors immunize a person.

Question 6.
Name the primary and secondary lymphoid organs.
Answer:

• Primary lymphoid organs – Organs where B & T – lymphocytes are formed,mature and require antigen specific receptors – Bone marrow and thymus.
• Secondary lymphoid organs – place or residence of nature lymphocytes – lymph nodes, spleen, tonsils, Peyer’s patches and mucosal surfaces.

Question 7.
The following are some well-known abbreviations, which have been used in this chapter. Expand each one to its full form:
(a) MALT
(b) CMI
(c) AIDS
(d) NACO
(e) HIV
Answer:
(a) MALT – Mucosal associated lymphoid tissue.
(b) CMI – Cell Mediated Immunity
(c) AIDS – Acquired Immuno Deficiency Syndrome
(d) NACO – National AIDS Control Organisation
(e) HIV – Human immunodeficiency virus.

Question 8.
Differentiate the following and give examples of each:
(a) Innate and acquired immunity
(b) Active and passive immunity
Answer:
(a) Innate and acquired immunity

Innate	Innate
Present from birth	Develops during life time
Remains throughout life	Short or long life
Contact with antigen not necessary	Contact with antigen necessary
Is inheritable	Can be inherited for brief period to neonates
Prevents from disease contraction	Protects from pathogen and other members of some species.

(b) Active and passive immunity

Active	Passive
Due to contact with antigen	Obtain antibodies from outside
Not immediate	Develops immediately
Lasts for sufficiently long time	Lasts for few days
Antibodies are produced by body Side effects are few	Antibodies obtained from outside.

Question 9.
Draw a well-labelled diagram of an antibody molecule.
Answer:

Question 10.
What are the various routes by which transmission of human immunodeficiency virus takes place?
Answer:

• Sexual contact with infected person.
• Through blood transfusion.
• By sharing infected needles in case of intra venous drug abusers.
• From mother to child through placenta.

Question 11.
What is the mechanism by which the AIDS virus causes deficiency of immune system of the infected person?
Answer:
Macrophages helper T-cells and some nerve cells possess T-4, OV CD-4 antigen receptor sites over their surface. HIV gets attached to these site and passes into cells. It multiplies in cells. Virus particles bud out of the cells which die afterward.

The newly released – virus particles attack new cells especially the helper T-cells. Soon the number of helper T- cells decreases. Since helper T-cells are essential for functioning of immune system their reduced number results in deficient functioning of immune system of the infected person.

Question 12.
How is a cancerous cell different from a normal cell?
Answer:
Normal cells remain adhered to one another. They have a definite life span. As some old cells die, they are replaced by new cells which arise by cell division and differentiation. Cancerous cells:- Whenever there is break down of any regulatory mechanism, a cell develops the ability to undergo uncontrolled repeated division forms a clone at cells.

There is no adherence. The cells slip past one another forming a mass of undifferentiated cells called neoplasm or tumour. Tumour results in pressing of surrounding normal cells and tissues causing discomfort and disruption of their functioning.

Question 13.
Explain what is meant by metastasis.
Answer:
The rapid growth of cancerous tumour causes overcrowding and disruption of normal cells. It extends to neighbouring tissues. In the last stage, bits of tumour tissue breakoff and are carried by the circulating blood or lymphs to other parts of the body, where they invade new tissues and start new tumors called secondary tumors. This property is called metastasis. It is fated due to increasing interference with the body’s life processes.

Question 14.
List the harmful effects caused by alcohol/ drug abuse.
Answer:

• Addicts show reckless behaviour vandalism and violence. Self-confidence lost
• Neural and neuromuscular junctions are effected.
• Addiction disturbs peristalsis and secretion of digestive enzymes.
• Disturbs digestive and nervous systems leads to frequent nausea and vomiting.
• Effects on cardiovascular system.
• Damage to liver causes cirrhosis.
• Insufficiency develops in reproductive system.

Question 15.
Do you think that friends can influence one to take alcohol/drugs? If yes, how may one protect himself/herself from such an influence?
Answer:
Yes. This can be avoided by

• Choosing good peer group.
• Discussing ways and means to counteract the presence if any with family elders and teacher/counsellors
• Telling the programme of outing to family.
• Keeping contact with family while outside the home.

Question 16.
Why is that once a person starts taking alcohol or drugs, it is difficult to get rid of this habit? Discuss it with your teacher.
Answer:
A discusses to be done.

Question 17.
In your view what motivates youngsters to take to alcohol or drugs and how can this be avoided?
Answer:

• Choosing good peer group.
• Discussing ways and means to counteract the presence if any with family elders and teacher/counsellors
• Telling the programme of outing to family.
• Keeping contact with family while outside the home.

2nd PUC Biology Human Health and Disease Additional Questions and Answers

2nd PUC Biology Human Health and Disease One Mark Questions

Question 1.
What do you mean by “Health’?
Answer:
Health is the state of complete physical, mental and social well being.

Question 2.
Name 2 non infectious diseases which are the major cause of mortality?
Answer:
Hepatitis-B, AIDS.

Question 3.
What are pathogens?
Answer:
The organisms which cause diseases in man are called pathogens.

Question 4.
Define diseases.
Ans :
Disease is the condition when the functions of one or more organs/systems of the body are adversely affected by various symptoms.

Question 5.
“All pathogens are parasites” Why?
Answer:
Because they cause damage to the host body by living inside the body of host.

Question 6.
Name 2 protozoan disease.
Answer:
Amoebiasis, malaria, Kalaazar

Question 7.
Name the vectors of malaria and dengue fever?
Answer:

• Malaria fever – Female Anopheles mosquito
• Dengue fever- Female aedes mosquito

Question 8.
Give the scientific name of malarial parasite.
Answer:
Malarial parasite – Protozoan – Plasmodium falciparum

Question 9.
Patients suffering from filariasis suffer, swelling in the lower limbs. Why?
Answer:
The filarial worm live in the lymphatic vessels of the lower limbs, where they cause inflammation, that causes swelling of lower, limbs.

Question 10.
Expand PMNL.
Answer:
Polymorpho- nuclear leucocytes.

Question 11.
Name the common fungal diseases infects on man.
Answer:
Ring worms.

Question 12.
Name the vector responsible for chikungunya. (Delhi 2008)
Answer:
Aedes mosquito.

Question 13.
Write the role of macrophages in providing immunity to humans. (Delhi 2008)
Answer:
Macrophages are phagocytes and through phagocytes they destroy the microbes.

Question 14.
Write the full form of LSD. (CBSE 1994)
Answer:
D-Lysergic acid diethyl amide

Question 15.
What are Psychotropic drugs. (CBSE 1995)
Answer:
These are mood altering drugs which selectively affect behaviour perception and mental activity of a person using them.

Question 16.
Write the full name of pathogen that causes AIDS? (CBSE 1996)
Answer:
Human immunodeficiency virus – HIV

Question 17.
Define the term vaccine.
Answer:
Vaccine is a suspension of killed or alternated pathogenic macro organism or antigenic preparations made out of it which on administration provides immunity towards the pathogen.

Question 18.
What is AIDS (HS&B 98)
Answer:
Acquired Immuno Deficiency Syndrome.

Question 19.
Define Addiction. (HS&B 2001)
Answer:
Addiction is the state of giving up to a drug, alcohol or tobacco due to becoming physically, physiologically and psychologically attached to its certain effects like euphoria and temporary feeling of well being.

Question 20.
What is immunity. (HS&B 2003)
Answer:
Immunity is a natural or acquired ‘resistance of an individual to the development of pathological condition even after having received infective dose of virulent pathogen its taxis or an allergies.

Question 21.
Name 2 major group an of cells required in attaining specific immunity. (CBSE 2005)
Answer:
B-lymphocytes, T-lymphocytes.

Question 22.
Name the disease in humans where the myelin sheath of nerve cells is attacked by antibodies of itself. (CBSE 2005)
Answer:
Multiple sclerosis.

Question 23.
What are interferon?
Answer:
Interferon are the proteins produced by virus infected cells in our body.

Question 24.
Why is antibody-mediated immunity called humoral immunity?
Answer:
Antibodies are the glycol proteins that circulate in the body fluids (humors) hence it is called humoral immunity.

Question25.
Give 2 examples of passive immunity.
Answer:

• The antibodies IgA in colostrum
• Antibodies received by foetus through placenta.

Question 26.
How does colostrum provide initial protection against diseases to newborn infants? Give reason.
Answer:
Colostrum contains antibodies like IgA, which provide initial protection against diseases.

Question 27.
While 2 features of acquired immunity.
Answer:

• It is pathogen specific
• It is characterised by memory.

Question 28.
Name 2 types of cells in which the HIV multiplies after gaining entry into the human body.
Answer:
Macrophages and Helps T cells.

Question 29.
Why is the bone marrow considered as the main lymphoid tissue of our body.
Answer:
Because in bone marrow along with blood cells lymphocytes are produced.

Question 30.
What is MALT?
Answer:
MALT refers to the lymphoid tissue located with in the lining of the major tracts (respiratory, digestive, urogenital tract)

Question 31.
Expand ELISA.
Answer:
Enzyme linked Immuno-sorbent assay.

Question 32.
Why does an AIDS patient suffer from many other infectious disease?
Answer:
Due to the reduction in number of helper T-cells, the immune system becomes weak. Consequently, the patient suffers from infectious disease.

Question 33.
When is a tumour is referred to as Malignant?
Answer:
Malignant tumor is a mass of proliferating neoplastic cells, which spread to distant sites through body fluids to develop secondary tumors.

Question 34.
What are oncogenic virus?
Answer:
The viruses which cause cancer are called oncogenic virus.

Question 35.
How does tobacco smoking cause oxygen deficiency in the body.
Answer:
Smoking increases the carbon monoxide content of the blood that refuses.

Question 36.
What type of cells provide Innate immunity?
Answer:
Neutrophils.

Question 37.
What is drug abuse.
Answer:
When drugs are taken for a purpose other than their normal clinical use and in amounts concentrations or frequency that impairs one’s physical, physiological and psychological functions, it constitute drug abuse.

Question 38.
Write any 2 uses of morphine.
Answer:

• Used as selective
• Used as painkiller especially in patients who have undergone surgery.

Question 39.
Name the phylum and disease caused by Wucheria Bancroft (CBSE 2005)
Answer:

• Phylum – Nemathelminthes
• disease:- Elephantiasis (filaria)

2nd PUC Biology Human Health and Disease Short Answer Type Questions

Question 1.
Mention any 2 measures for prevention and control of alcohol and drug abuse among adolescents.
Answer:

• Communication: An adolescent should be able to disuses problems in studies, games, friends etc. with parents who should also know all the friends of the child.
• Avoid undue pressure: The child should not be asked to perform beyond thresholdjimits whether in studies, sports or extra curricular activities.

Question 2.
In which parts of the body of the hosts do the following events in the life cycle of plasmodiurfi take place? Name both the body parts and the host.
(a) Fertilization
(b) Development of gametocytes
(c) Release of Sporozoites
(d) Asexual reproductions.
Answer:
(a) Fertilization: Inside stomach/intestine of mosquito host.
(b) Development of Gamete cites:- In blood of human host.
(c) Release of sporozoites – In to the blood of human host
(d) Asexual Reproductions – Inside liver cells and-RBC’s of human host.

Question 3.
Write the specific symptoms of typhoid.
Name its causative agent. (AI – 2009)
Answer:
Symptoms of typhoid:

• Sustained high fever
• Loss of appetite
• Stomach pain
• Headache and weakness
• Intestinal perforations severe cases may cause death causative agent of typhoid is Salmonella typhi.

Question 4.
Write the scientific names of two helminthiasis that are pathogene to man. Name the disease caused by them.
Answer:

• Ascaris lubricoides causes Ascariasis
• Wuchereria bancroffi causes filariasis or elephantiasis.

Question 5.
Write the symptoms of pneumonia.
Answer:

• Fluid-filled alveoli
• Cough and head ache
• Fever
• The nails turns gray to bluish in severe cause.

Question 6.
Name and explain the type of barrier of innate immunity where some cells release inteifetofts when infected.
Answer:
Cytokinin barriers

• Cells infected by virus release interferons
• Interferons provide protection to non infected cells from infection.

Question 7.
How do saliva and tear help to prevent bacterial infection?
Answer:
Saliva and tear consists of lysozymes Lysosomes are enzyme which can digest the bacterial cell wall which helps to kill bacteria.

Question 8.
Name and explain the type of barrier of innate immune system, which involves.
Answer:
Cellular barrier
This barrier involves certain types of

• Polymorphs nuclear leucocytes
• Monocytes
• Natural killed cells
• Macrophages.

Question 9.
Differentiate between Primary response and secondary response.
Answer:

Primary response	Secondary response
The immune response that occurs at the first encounter of the body with antigen is called -primary immune response	The immune response that occurs at the second and subsequent, encounters of the body with the antigen is called secondary immune response
It declines rapidly	It is severe and lasts for long time

Question 10.
Differentiate between T lymphocyte and B- lymphocyte.
Answer:

T lymphocyte	B- lymphocyte.
These are lymphocyte which undergo maturation in thymus	Maturation undergo in bone marrow
They constitute cell mediated immune and don’t produce antibodies	They constitute the humoral immune response and produce antibodies
They react to organ transplants	Don’t respond to organ transplants.

Question 11.
How do B-lymphocyte cells direct humoral immunity?
Answer:
The B-lymphocyte cells multiply in number in response to the attack by antigen. They start producing characteristics proteins called antibodies. These antibodies circulate in the body fluids and act against the pathogen/ antigens.

Question 12.
Describe the cell mediate immunity.
Answer:
Cell mediate immune response is effected by T-lymphocyte. These are two types of T- lymphocytes.

• Cytotoxic or killer T- cells, which kill the antigens in a variety of mechanisms.
• Helper T-cells slimulate the B-lymphocyte to produce antibodies. This immunity is responsible for rejection of transplants.

Question 13.
Elst specific symptoms of amoebiases write its causative agent. (CBSE 2009)
Answer:
Symptoms of amoebiasis

• The patient passes out blood and mucus with the stool.
• Severe griping pain in the abdominal, fever nausea, exhaustion and nervousness.
  In chronic case, intestinal wall punctures. The causative agent is Entamoeba histolytica.

Question 14.
Explain metastasis. Why is it fatal?
Answer:
The rapid growth of cancerous tumour causes overcrowding and disruption of normal cells. It extends to neighbouring tissues. In the last stage, bits of tumour tissue breakoff and are carried by the circulating blood or lymphs to other parts of the body, where they invade new tissues and start new tumors called secondary tumors. This property is called metastasis. It is fated due to increasing interference with the body’s life processes.

Question 15.
What is vaccination? How does it help in producing immunity?
Answer:
The process of introducing a preparation of antigenic proteins of the pathogens or weakened or killed pathogen into the body.

• The vaccines induce quick multiplication of B and T lymphocytes, some of them are stored as memory cells.
• The B-lymphocytes quickly produce antibodies which neutralise the antigen during infection.

Question 16.
What are lymph nodes? What is their function in our immune system?
Answer:
Lymph nodes are small solid structures located at different points along the lymphatic system. They serve to tray the microbes and other antigens. These antigens trapped in the lymph nodes are responsible for the activation of lymphocytes to cause the immune system.

Question 17
(a) Explain the property that prevents normal cells from becoming cancerous
(b) All normal cells have inherent characteristics of becoming cancerous. Explain. (CBSE2009)
Answer:
(a) Contact inhibition
(b) All normal cells have several genes called cellular oncogenes or proto oncogenes.
These genes are activated under certain conditions can lead to oncogenic transformation of cells, i.e. cancer can result.

Question 18.
Write the importance of molecular biology in preventing cancer.
Answer:
Techniques of molecular biology can be applied to detect genes in individuals with inherited susceptibility to certain cancers. Identification of such genes, which predispose individuals to certain cancers may be helpful in prevention of cancer, such individuals are advised to avoid exposure to particular carcinogens to which they may be susceptible.

Question 19.
Why do sports persons always caught by authorities for cocaine addiction? (CBSE2008)
Answer:
Sports persons become addicted to cocaine because.

• Cocaine has a potent stimulating action on central nervous system.
• It produces a sense of euphoria and increased energy for performance.

Question 20.
Name one plant and the additive drug extracted from its latex. How does this drug affect the human body? (CBSE 2009)
Answer:
Morphine is obtained from the plant Papaver somniferum (poppy plant).
It is an opioid that binds to specific receptors in the central nervous system and gastro intestinal tract and slows down the body function.

Question 21.
Give four reason to justify the intake of cannabinoids by sports person (A.I. 2008)
Answer:

• Cannabinoids affect the cardio vascular system.
• They cause addiction and when the regular dose is not taken, there are unpleasant one term with drawl symptom, which may be life threatening.
• The person shows reckless behaviour vandatism and violence
• Excess does can lead to coma and death due to heart failure cerebral hemorrhage etc.

2nd PUC Biology Human Health and Disease Short Answer Type Questions

Question 1.
Mention 3 major categories of factors that affect the health of a person.
Answer:
Health is affected by
(i) genetic disorders:- the defects, deficiencies that are inherited from parents due to defective genes.
(ii) Infections caused by pathogens.
(iii) Life styles which includes

• Food and water (eating habits)
• Rest and exercise to the body
• Habits like drugs, alcohol etc.

Question 2.
How do normal cells get transformed into cancerous neoplastic cells? Mention the difference between viral oncogenes and cellular oncogenes (Foreign 2008)
Answer:
The transformation of normal cells into cancerous neoplastie cells is induced by physical clinical and biological agents collectively called careinogenes, they lose the property of contact inhibition.
Differences between viral oncogenes and cellular oncogenes are

• Viral oncogenes are the genes present in I the oncogenic virus which affect oncogenic transformation of the cells they infect.
• Cellular oncogenes are the genes presents in normal cells and code for growth factors, when activated under certain condition can cause oncogenic transformation of the cell.

Question 3.
What are the side effects of the use of anabolic steroids by females?
Answer:
The side effects in females includes

• Increased aggressiveness
• Abnormal menstrual cycle
• Excessive hair growth on the face and body
• Enlargement of elitoris
• Deepening of voice
• Mood swings and depression.

Question 4.
Give the side effects of the use of anabatic steroids in males.
Answer:
The side effects in males includes

• Reduction in the size of testicles
• Decreased sperm productions
• Breast enlargement
• Premature baldness
• Enlargement of prostrate gland
• Development of acne.

Question 5
(a) List any two situations when a medical doctor would recommend injection of performed antibodies into the body of a patient. Name this kind of immunisation
(b) Name the kind of immunity attained when instead of antibody weakend antigens are introduced into the body. (Delhi 2008)
Answer:
(a) The two conditions include the following

• When a person is infected by a deadly microbe, for which an immediate immune response is required.
• Snakebite
  This kind of immunisation is called passive imminisation

(b) Active immunity.

Question 6.
Name the stage of plasmodium that a female Anopheles mosquito injects into a healthy person during its bite. Represent schematically the stages of the pathogen developed in the human body, till it is picked up by a mosquito.
Answer:

Question 7.
(a) How are primary lymphoid organs different from secondary lymphoid organs. Give an example of each.
(b) Name any 4 types of cells involved in the cellular barriers of innate immunity.
Answer:
(a)

• Primary lymphoid organs: These are the lymphoid organs where immature lymphocytes differentiate into antigen sensitive lymphocytes.
  E.g. Bonemarrow, thymus.
• Secondary lymphoid organs: These are the lymphoid organs where the interaction occurs between lymphocytes and antigen and proliferation occurs.
  E.g. Spleen, tonsil.

(b) The cell types are

• Poly morpho nuclear leucocytes (PMNL)
• Monocyte
• Natural killer cells
• Macrophages

2nd PUC Biology Human Health and Disease Five Marks Questions

Question 1.
Justify the following statements
(i) Smoking leads to deficiency of oxygen in the body tissue.
(ii) Cancer patients are administered – interferon
(iii) Chewing of tobacco causes an increases in blood pressure.
Answer:
(i) Smoking increases the carbon monoxide (CO) content in blood and reduces the concentration of haem-bound oxygen hence there is deficiency of oxygen in the body tissues.

(ii) Tumor cells avoid detection and destruction by our immune system. Interferon like biological modifiers activate the immune system and help in destroying the tumor cells.

(iii) By chewing tobacco the nicotine reaches the blood and stimulates the adrenal glands to release adrenaline and noradrenaline, both these hormones increases the blood pressure.

Question 2.
(i) Draw a labelled schematic diagram of an antibody molecules
(ii) Mention any 4 warning signals of drug abuse in adolescents.
Answer:
(i)

(ii) Warning signals of drug abuse:
Aggressive and rebellious behaviour Loss of appetite
Lack of interest in personal hygiene Drop in academic performance Change in sleeping and eating habits .
(i) Define innate immunity Mention any 2 ways in which it is accomplished in our body.
(ii) How does spleen functions as an organ of our immune system.
Ans:
(i) Innate immunity is defined as the non¬specific type of defence with which an individual is bom and in always available to protect the body. It is accomplished by providing different types of barriers to the entry of antigens / pathogens
(a) Physical barriers: Skin of the body is the greater barrier which prevents the entry of microbes. Mucus coating on the epithelium of major tracts help in trapping the microbes that have entered the body.
(b) Physiological barrier:

• Saliva in the mouth and tear contain an anti bacterial substance.
• HC1 scented by the stomach (C) Cytokine barriers:
  Interferon are the proteins which are produced by the body cells in response to viral infection and they protect the non-infected cell from viral infection.
• Spleen acts as filter of blood by trapping the blood borne microbes. It contain phagocytes which destroy the microbes through phagocytosis.

Question 4.
(a) Differentiate between active immunity and passive immunity.
(b) Describe the different methods of treatment of cancer.
Answer:

Active immunity	Passive immunity
When antibodies are produced by our β lymphocytes in response to the antigen, it is called active immunity.	When ready made antibodies are injected into the body for defence, the immunity produced is called passive immunity.
It lasts for longer periods	It lasts for shorter period.

(b) Different methods for the treatment of cancer are

• Surgery: The tumour cells are surgically removed to reduce the growth of cancerous cells.
• Radiation therapy: The tumour cells are radiated lethally taking care of the surrounding normal cells.
• Chemotherapy: Certain drugs are specific for particular tumors, but have side effects like anaemia, hair loss etc.

Question 5.
(a) Represent schematically the life cycle of HIV or Replication of Retro virus.
(b) What are the various routes by which transmission of human immuno deficiency virus takes place?
Answer:
(a)

(b) The various ways of the transmission of ratio virus includes

• Sexual contact with infected persons
• By the transfusion of blood
• By the use of infected needles and syringes
• From infected mother to child through placenta.

Question 6.
Explain the different methods of detecting cancer.
Answer:
Cancer can be detected by

• Biopsy and histopathological studies of the tissue
• Using antibodies against cancer-specific antigens
• Blood and bone marrow test to check the count of WBC and RBC
• Use of C T scans (Computerised tomography) MRI (magnetic resonance imaging) and X-rays to detect the cancer of internal parts.
• Applying principles of molecular biology to detect the genes in individuals with inherited susceptibility to cancer.

 

You can Download Chapter 4 Reproductive Health Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 7 Evolution

2nd PUC Biology Evolution  NCERT Text Book Questions and Answers

Question 1.
Explain antibiotic resistance observed in bacteria in light of Darwinian selection theory.
Answer:
Penicillin when discovered was used as antibiotic against all bacteria. Soon many of these became resistant. This is because alleles of resistance which are already present in bacteria are of no importance in absence of antibiotics. Adjustment to change in environment due to genetic variation is adaptation.

Question 2.
Find out from newspapers and popular science articles any new fossil discoveries or controversies about evolution.
Answer:

• Fossil of small terrestrial dinosour with feathers covering limbs and body. (Archaeopteryx lithdgraphia)
• Mesohippus – intermediate horse size of goat with 3 toes on each foot and molar teeth had serration.

Question 3.
Attempt giving a clear definition of the term species.
Answer:
Species is a morphologically distinct and reproductively isolated one or more natural populations of individuals which resemble one another closely and interbreed freely amongst themselves.

Question 4.
Try to trace the various components of human evolution (hint: brain size and function, skeletal structure, dietary preference, etc.)
Answer:

Question 5.
Find out through internet and popular science articles whether animals other than man has self-consciousness.
Answer:
Please do survey in internet.

Question 6.
List 10 modern-day animals and using the internet resources link it to a corresponding
ancient fossil. Name both.
Answer:
Please do survey in internet.

Question 7.
Practise drawing various animals and plants.
Answer:
Please practice drawing.

Question 8.
Describe one example of adaptive radiation.
Answer:
Adaptive radiation – Formation of different species from a common ancestor with new species adapting to different geological niches.
Example: Darwin’s finches is Galapagos island have wolves from mainland finches. They underwent changes is shape, size of beaks, food habit, feathers.

Question 9.
Can we call human evolution as adaptive radiation?
Answer:
No, because parent species of homosapiens have evolved by progressive evolution
(Homo habilis —Homoerectus lineage)

Question 10.
Using various resources such as your school Library or the internet and discussions with your teacher, trace the evolutionary stages of any .one animal say horse.
Answer:
E.g.: Horse – See text book diagram

2nd PUC Biology Evolution Additional Questions and Answers

2nd PUC Biology Evolution One Mark Questions 

Question 1.
What is abiogenesis?
Answer:
It is the process of the appearance of first form of life slowly from non living molecules.

Question 2.
What are fossils?
Answer:
Fossils are the remnants or traces of ancient organism that lived in the past.
Or
Fossils are the organnic relies of the past which serve as evidences in tracing biological evolution.

Question 3.
What is evolutionary biology?
Answer:
The science which deals with the study of history of life forms on earth.

Question 4.
Define analogous organs. (CBSE 93,94,2007)
Answer:
The organs which performs the same function but differ in their origin and structures are called analogous organs.

Question 5.
Define homologons organs.
Answer:
The organs which have similar embryonic origin but they perform the different functions.

Question 6.
What is paleontology?
Answer:
It is the branch of science that deals with fossils.

Question 7.
Define homology.
Answer:
Homology refers to the similarities in the fundamental anatomy and embryology of organs of different groups of plants and animals.

Question 8.
Name any two vertebrate body parts that are homologous to human forelimbs.
(CBSE 2008)
Answer:
Wings of birds, flippers of whale forelimbs of cheetah, wings of bats.

Question 9.
What is analogy?
Answer:
It refers to structurally different organs evolving for the same function in different group of plants and animals.

Question 10.
Mention the type of evolution that has brought similarity as seen in potato tuber and sweet potato.
Answer:
Convergent evolution.

Question 11.
Why are the wings of a butterfly and of a bat called analogous? (CBSE 1996, 2009)
Answer:
They are of different embryonic origin but similar function.

Question 12.
Thorns of Bougainvillea and tendrils of cucurbit are analogous or homologous.
Or
Wings of bird and forelimbs of horse are homologons or analogous
Or
Flippers of penguin and dolphin are analogous or Homologies.
What type of evolution has brought similarity in these cases? (CBSE 1992, Delhi 2009, Foreign 2009)
Answer:

• All these are homologous.
• Divergent type evolution has brought these similarity.

Question 13.
Give 2 examples of evolution due to anthropogenic activities.
Answer:
DDT resistance in mosquitoes and Antibiotic resistance in microbes.

Question 14.
What kind of evidence is afforded by Darwin’s finches in support of orgaine evolution? (CBSE 1991)
Answer:
Adaptive radiation i.e. All these finches evolved from common ancestor but they diverge in various directions.

Question 15.
Name the fossil animal which serves as a connecting link believers reptiles and birds. (CBSE 1995)
Answer:
Archeopteryx – fossil bird.

Question 16.
Define ontogenetic law.
Answer:
Ontogeny recapitulates phytogeny.

Question 17.
Name the placental mammal corresponding to the Austrian “Spotted cuscus” and Tasmanian “tiger cat”. Which have evolved as a result of convergent evolution (CBSE 2008)
Answer:
Lemur and Bobcat.

Question 18.
As per Hugo deVries, what is the cause of speciation? (Delhi 2008, CBSE 1995)
Answer:
Mutation. (Single step large mutation)

Question 19.
What is meant by genelic equilibrium.
Answer:
When allele frequencies in a population are stable, the allele frequency of a population remains constant. It is called as genetic equilibrium, i.e. the sum total of all the allelic frequency is one.

Question 20.
Define evolution as per Hardy Weinberg.
Answer:
As per Hardy Weinberg, change of frequency of alleles in a population would be considered as evolution i.e. distubance in the genetic equilibrium.

Question 21.
What is meant by gene pool?
Answer:
The sum total of all the genes pooled by the members of a population.

Question 22.
What is meant by gene flow?
Answer:
Changes in the gene pool of population when there is continuous migration of organisms between them i.e. it refers to the addition or loss of genes.

Question 23.
Name the immediate ancestor of lycopods .
Answer:
Zosterophyllum.

Question 24.
Define natural selection?
Answer:
Natural selection is the process in which heritable variations that enable better survival, are enabled to produce and leave behind a greater number of progeny.

Question 25.
Darwin’s theory is known as “Theory of Natural selection”. How is Lamark theory know as? (CBSE 1995)
Answer:
Theory of inheritance of acquired characters.

Question 26.
Define the term “reproductive isolation”.
Answer:
If the population of 2 different species whether they are isolated or not, cannot be enter bread to produce offspring.

Question 27.
Define genetic drift.
Answer:
Random changes in the allelic frequencies of a population, occurring only by chance events, constitute genetic drift.

Question 28.
Name the group of animals that evolved into amphibian.
Answer:
Lobe-fins evolved into amphibian.

Question 29.
Mention the key concepts about the mechanism of biological evolution/ speciation according to
(a) Devries and
(b) Darwin (Delhi 2008)
Answer:
(a) De Vries – Mutation
(b) Darwin – Natural selection and branching decent.

2nd PUC Biology Evolution Two Marks Questions

Question 1.
Describe the theory of abiogenesis.
Answer:
According to this theory, the first form of life arose slowly through evolutionary forces from non living molecules.

• The reducing atmosphere and high temperature favoured the formation of diverse organic compounds from inorganic molecules.
• Then the first noncellular forms of life i.e. self duplicating molecules like RNA evolved.

Question 2.
Write the names of two dinosaurs that lived early in the geological history and two that lived later.
Answer:

• Lived early in the geological history:- Brachyosaurus and Stegosaurus
• Lived later:- Triceratops and Tyrannosaurus.

Question 3.
Amongst pea tendrils, opuntia spines, lemon thorns and cucurbit tendrils. Which ones are homologous structures? Why do you call them homologous? (CBSE 1999)
Answer:

• Pea tendrils and opuntia spines are homologous.
• Lemon thorm and cucurbit tendrils are homologous.

Reason:

• Pea tendrils and Opuntia spines are modified leaves (same orign) but perform different functions.
• Lemon thorns and cucurbit tendrils are modified stems. Both arise from axillary position. Both perform different functions.

Question 4.
Name one organ analogous to the wings of bird. Why are they both analogous. Can you include wings of bat also with them under the same category? Give reason. (CBSE 1997)
Answer:
Wings of insects and birds are analogous because both perform the sante function but have the dissimilar structured and origin. The wings of insects are modified outgrowth of the body without having bones, whereas wings of birds are modified fore limbs.
The wings of bat cannot be put under the same category. This became the wings of bat and birds are homologous organ.

Question 5.
Given below are the names of 2 pairs of limbs. Categorize them into homologous and analogous organs, give reason.
(i) Human arm and fore leg of cow
(ii) Bat’s wing and Grass hopper’s wing. (CBSE 1999)
Answer:
(i) Human arm and fore leg of cow are homologous organs because they are built upon the same fundamental plan (pentadactyl pattern) but they perform different functions as grasping in man and locomotion in cow.

(ii) Bat’s wing and grass hopper’s wing are analogous organs because they perform the same function bat differ in their origin and structure. Wings of bats are modified fore limbs where as grasshoppers wings are modified outgrowth of the body wall.

Question 6.
Give 2 examples each of analogy and homology in plants. (Hots)
Answer:
Analogy:

• Tubers of sweet potato and potato
• Tendrils of pea and cucurbits.

Homology:

• Tendrils of cucurbit and lemon thorn
• Tendrils of Pea and Spines of opuntia

Question 7.
How do Darwin’s finches illustrate adaptive radiation? (AI2008)
Answer:
Darwin finches are varieties of small black birds found in the Galapagos islands. They all must have evolved in the island itself. From the original seed eating birds, many other forms evolved, with altered beak enabling them to become insectivorous and vulgarian habits. This process of evolution of different species in the same geographical area starting with one spices and radiated to other areas is called adaptive radiation.

Question 8.
Define adaptive radiation with 2 eggs. (Hots)
Answer:
Adaptive radiation is defined as the process of evolution of different species in a given geographical area starting from single species and radiated to other habitats.
Example:

• Australian marsupials, each different from the other, have evolved from an ancestral stock.
• Darwin’s finches – from the original seed eating stock insectivores and vegetarian birds have evolved.

Question 9.
State Hardy – Wein berg principle of genetic equilibrium knowing that genetic drift disturbs this equilibrium. Mention what does this disturbance in genetic equilibrium lead to?
Answer:
Hardy Weinberg principle states that allele frequencies in a population are stable and remain constant from generation to generation.

• Genetic drift refers to a change in allele frequencies of a population acquiring by chance.
• Such a change in allele frequency may be so different that the population becomes a different species.
• The original drifted population is called founder and the effect is called founder effect.

2nd PUC Biology Evolution Three Marks Questions

Question 1.
Whose theory was put to test by Miller Urey and what was the theory? How did their experiment proved the abiotic origin of life on earth?
Answer:
Oparin and Haldane proposed that the first life form could have come from the „ nonliving organic molecules like RNA, protein etc.

• The orgaine molecules must have been produced by chemical evolution, i.e. formation of divers organic molecules from inorganic constituents.
• The condition, of the earth that favoured chemical origin were

(i) very high temperature
(ii) volcanic storms
(iii) Reducing atmosphere that contained methane, ammonia and water vapour.

• Energy must have been produced by U-V radiation and lightning.
• Analysis of the products of their experiment showed the presence of amino acids which help to form proteins.

Question 2.
Give a brief account as how evolution has taken pleace from the time the non cellular aggregate of giant molecules turned into cells.
Answer:
The first formed cells were anaerobic heterotrophs. But slowly some of these cells developed coloured proteins, that could release oxygen, in a process that could have been similar to the light reactions of photosynthesis. As oxygen started coming into the atmospheres is as the atmosphere started becoming an oxidising one, new formes of life could not arise from nonliving organic molecules.

The organisms started becoming aerobic and autotrophic. The single celled organism slowly became multicellular form, where some became autotrophic while many remained heterotrophic. Plants like bryophytes were the first to invade land, followed by reptiles among animals.

Question 3.
Differentiate between natural selection and artificial selection.
Answer:

Natural selection	Artificial selection
(a) It is the process occurring in nature over a number of generations to increase the number of fit individuals in a population. (b) The characters/ adaptationsare advantageous to the organism.	(a)  It is the process practised by man over a number of generations, to select organisms with better qualities. (b) The characters are advantageous human.

Question 4.
(i) The study of fossils support evolution of organism. Discuss?
(ii) Describe convergent evolution with an example each from animals and plants.
Answer:
(i) By the careful analyses of the distribution of fossils in the different strata of rocks gives the time of the history of the earth the study showed that life form varied over time, some were simple and those in the superfluid layers resembled modern organism. Some fossils indicate connecting links i.e. they share features of two groups of organisms indicating the evolution of one group into the other.

(ii) Convergent evolution refers to the evolutionary process of selection of similar adaptive features in different groups of organisms in a similar habitat towards the same function.

• It is the similar habilat that has resulted in selection of similar structures.
• Analogous structures that are not anatomically similar but perform the same functions result from convergent evolution. Examples: Wings of butterfly and wings of birds.

Question 5.
How is genetic drift differ from gene migration?
Answer:

Genetic drift	Gene migration
Random changes in the allele frequencies of a population occurring only by chance events constitute genetic drift.	It refers to the change in allele frequencies of a given population when individuals ‘ migrate into the population or leave the population.

Question 6
(a) Name the largest dinosaurs and mention any two characteristic features.
(b) How did Darwin explain the existence of different varieties of finches on Galapagos Islands?
Answer:
(a) Tyrannosaurus rex was the largest dinosaur it was near about 20 feet in height. It had huge fearsome dagger-like teeth

(b) Darwin explained that all the verities evolved on the island itself from the original seed-eating birds, many other forms with altered beaks arose some others became insectivorous while remained the vegetarian flinches. Such process of evolution of different species in a given geographied area, starting from a point and literally radiating to other habitats, is called adaptive radiation.

Question 7.
Explain Landmark’s theory of evolution with example.
Answer:
According to Lamarck, evolution of life forms had occurred, drives by use and disease of organs. This theory as known as theory of inheritance of acquired characters. He gave the examples of giraffes, who in an attempt to forage leaves on tall trees had to stretch their neck. As they passed this acquired character to the next generations giraffe slowly over the years, came to acquire tons necks.

Question 8.
How does the shift in Hardy-Weinberg equilibrium leads to founder effect. (Delhi 2008)
Answer:
A shift in Hardy – Weinberg equilibrium is caused by factors like gene migration, gametic drift, mutation etc. Gene migration refers to a change in the gene frequencies of the populations, when a section of a population migrates to another place and population.

If gene migration occurs a number of times there would be a gene flow. i.e. new genes/ alleles are added to the new population while they are lost from the original population.

Genetic drift refers to change in allele/ gene frequency at random or that occurs due to some chance event. Some times, the change in allele frequency is so different in the new sample of population that they become different species. The original drifted population becomes the founder and the effect produced is called founder effect.

You can Download Chapter 5 Principles of Inheritance and Variation Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 5 Principles of Inheritance and Variation

2nd PUC Biology Principles of Inheritance and Variation NCERT Text Book Questions and Answers

Question 1.
Mention the advantages of selecting pea plant for experiment by Mendel.
Answer:
Mendel selected pea plant because:-

• It possesses a large number of varieties.
• Pure varieties of this plant were available.
• Plant is small and flower is large enough to hand manually.
• Flowers are bisexual and produces fertile hybrids.
• Each plant produces a large number of seeds.
• Plants are self pollinated and can be cross pollinated.
• Plant is annual with growth period of few months.
• Chances of contamination are very less.

Question 2.
Differentiate between:-
(a) Dominance and Recessive.
Answer:

Dominance	Recessive
1. Able to express even in presence of contrasting allele. 2. Does not require another similar allele to produce its phenotype. 3. Produces complete polypeptide, protein or enzyme. 4.  It is usually wild type allele.	1. Cannot express in presence of contrasting allele. 2. Can produce its phenotype only along with similar allele. 3. Forms incomplete products. 4.  Usually a mutant allele.

(b) Homozygous and Heterozygous.
Answer:

Homozygous	Heterozygous
1. Possesses similar alleles. 2.  2 types – Homozygous dominant and recessive. 3. Individual is pure for the trait. 4. On self breeding, similar type of offspring are formed. 5. Only one type of gamete is formed	1. Possesses different alleles. 2.  Is of one type 3. Individual is seldom pure for trait. 4. On breeding, 3 types of offspring are formed. 5. 2 types of gametes produced.

(c) Monohybrid and Dihybrid
Answer:

Monohybrid	Dihybrid
1.   Cross made between individual having contrasting traits in order to study the inheritance of a pair of alleles. 2.  Phenotypic monohy brid ratio in F2 generation is 3:1. 3. Genotypic monohybrid ratio in F2 generation is 2:1.	1. Cross made between individuals having contrasting traits inorder to study the inheritance of 2 pair of alleles. 2. Phenotypic dihybrid ratio is 9:3:3:1 3. Genotypic dihybrid ratio is 1:2:1:2:4:2:1:2:1

Question 3.
A diploid organism is heterozygous for 4 loci, how many types of gametes can be produced?
Answer:
Number of gametes = 2ⁿ
n = Number of loci, n = 4
So, 2⁴ = 2×2×2×2 = 16 types of gametes.

Question 4.
Explain the law of Dominance using a monohybrid cross.
Answer:
Law of Dominance:

• Characters are controlled by discrete units called factors.
• Factors occurs in pairs.
• In a dissimilar pair of factors one member of the pair dominates (dominant) the other (recessive) supresses.

In a monohybrid cross:-

Question 5.
Define and design a test cross.
Answer:
Test cross: An organism showing dominant phenotype (genotype is to be determined.) from F₂ is crossed with recessive parent instead of self-crossing. Progenies of such a cross can be analysed to predict genotype of test organism. Eg: Tall and short plant taken into experiment. From F₂ generation, TT or Tt could be tall  I plant. Recessive parent will be tt. So,

So, if test plant was TT, resulting plant will only I tall. If test plant was Tt, resulting plant will be tall and short.

Question 6.
Using a punnett square, workout distribution of phenotypic features in the F₁ generation after cross between homozygous female and heterozygous male for single locus.
Answer:

Question 7.
When cross is made between hybrid tall plant with yellow seeds (Tt Yy) and tall plant with green seeds (Tt yy), What proportion phenotype in offspring could be expected to be
(a) Tall and Green
(b) Dwarf and Green
Answer:

Question 8.
Two heterozygous plants are crossed. If 2 loci are linked what would the distribution of phenotypic features in F₁ generation in dihybrid cross.
Answer:

Question 9.
Mention the contributions of TH Morgan in Genetics.
Answer:
TH Morgan is a Genetist who got Nobel Prize.

• He found fruit fly (Drosophila Melanogaster) to be a experimental material as it was easy to rear and multiply it.
• Established presence of genes over the chromosomes.
• Principle of linkage and crossing over.
• Discovered sex linkage and crossing over.
• He observed mutations.
• Developed technique of chromosome mapping,
• Wrote book “The theory of Gene”.

Question 10.
What is pedigree analysis ? Suggest how such analysis can be useful.
Answer:
Pedigree analysis: Study of transmission of particular traits graphically over the present and the last few generation for finding out possibility of their occurrence in future generations. So, analysis of traits in a several generations of a family is called Pedigree Analysis.
Importance:

• To know possibility of a recessive allele which may create disorder.
• Can indicate origin of a trait in the ancestors.
• Analysis is used for genetic counselling.
• Extensively used in medical research.

Question 11.
A child has blood group O. If the father has blood group A and mother B, work out genotypes of parent and the possible genotypes of other offsprings.
Answer:
Blood group O can appear with 1° 1° only (2 recessive allele).

Question 12.
How is sex determined in human beings?
Answer:
In humans there are 23 pairs of chromosomes. 22 pairs of these chromosomes do not take part in sex determination called autosomes. The 23rd pair determines the sex of individual called allosome or sex chromosome. If it is XX then female, if XY then male. Presence of Y¹ makes a person male. Human females produce only 1 type of gamete 22 + X. In male it could be 22 + X or 22+ Y.

Question 13.
Explain the following terms with example
(a) Co-dominance
(b) Incomplete dominance
Answer:
(a) Co-dominance: Phenomenon of 2 different alleles of same gene lacking dominance – recessive relationship and expressing their effect simultaneously in heterozygote.
Eg: 1^A and 1^B alleles in AB blood group of humans.

(b) Incomplete dominance: Phenomenon of neither of 2 alleles of a gene being dominant over each other so that when both of them are present together, a new phenotype is formed which is somewhat intermediate between both of their independent expression. Eg: Flower colour is Snapdragon and 4’0 clock plant.

Question 14.
What is point mutation ? Give example.
Answer:
Point mutation: Gene mutation that occurs due to change in single base pair of DNA. It results in incorporation of different DNA. Eg: Valine instead of glutamic acid in [3 -peptide in sickle cell anaemia.]

Question 15.
Who had proposed chromosomal theory of inheritance ?
Answer:
Sutton and Boveri (1902)

Question 16.
Mention 2 autosomal genetic disorder with their symptoms.
Ans:
(a) Sickle Cell Anaemia :- Autosomal Recessive (chr. 11)
Symptoms:- Anaemia due to RBC destruction.

(b) Phenylketonuria :- Autosomal Recessive (Chr.12)
Symptoms:- Brain fails to develop in infancy, mental retardation.

(c) Cystic Fibrosis :- Autosomal Recessive (Chr.7)
Symptoms:- Mucus clogging in lungs, liver and pancreas.

2nd PUC Biology Principles of Inheritance and Variation Additional Questions and Answers

2nd PUC Biology Principles of Inheritance and Variation One Mark Questions

Question 1.
Give the meaning of the term allele.
Answer:
It represents a pair of genes of 2 alter-natives of the same character and occupies the same position (locus) on the homologous chromosome. OR The various forms of genes are called alleles.

Question 2.
What is pleiotrophy? Give one example.
Answer:
A phenomenon where one gene controls more than one phenotypic characters of an organism. Eg: In Pisum sativum, a gene that regulates colour of flowers also regulates the colour of seed coat.

Question 3.
What is Genetics?
Answer:
The branch of biology that deals with the inheritance and variation of characters from parents to offspring.

Question 4.
What is variation in genetics?
Answer:
The degree by which the progeny differs from the parents in a character.

Question 5.
What term did Mendel use for what we now call the genes? [Hots]
Answer:
Factors or determination.

Question 6.
What is a true breeding line?
Answer:
It is one that has undergone continuous self pollination and shows the stable trait inheritance and expression of it for several generation.

Question 7.
Define Phenotype.
Answer:
The observable or external characteristic of an organism, constitute its phenotype.

Question 8.
Define genotype.
Answer:
The genetic constitution of an organism is called genotype.

Question 9.
What is monohybrid cross?
Answer:
The cross made between individuals of a species, considering the inheritance of the contrasting pair of a single character.

Question 10.
What is dihybrid cross?
Answer:
The cross made between individual of a species considering the inheritance of the contrasting pairs of 2 characters.

Question 11.
Name the plant in which Mendel performed his experiment.
Answer:
Pea plant – Pisum sativum.

Question 12.
What is the importance of pedigree analysis?
Answer:
It reveals the ancestral history of an individual and its possible genotype for a trait.

Question 13.
Mendel observed 2 kinds of ratios, i.e., 3:1 and 1:2:1 in F₂ generation in his experiments on garden pea. Name these two kinds of ratios respectively.
Answer:

• 3:1 – Phenotypic ratio
• 1:2:1 – Genotypic ratio.

Question 14.
What are.codominant alleles ? Give one Example.
Answer:
The genes of an allele morphes pair are not related as dominant or recessive but both are equally potent to express themselves in F₁ hybrid. Eg: Blood group AB where alleles
I^A and I^B are codominant.

Question 15.
What is epistasis?
Answer:
It is the interaction between two different genes (non-allelic genes) where one gene masks the effect of another gene.

Question 16.
When a tall pea plant was self pollinated, one fourth of the progeny were dwarf. Give the genotype of all the parent + the dwarf progenies.
Answer:

• Parent: Tt Dwarf
• progenies : tt

Question 17.
‘Give the scientific name and the common name of plant in which incomplete dominance was first discovered.
Answer:
Four O’clock plant – Mirabilis jalapa.

Question 18.
Define the term heterozygous.
Answer:
It is an individual who possesses two different alleles of a character on its homologous chromosome.

Question 19.
What is the phenotypic and genotypic ratio of incomplete dominance?
Answer:
The phenotypic and genotypic ratios are same i.e. 1AA : 2Aa : laa

Question 20.
Define the law of independent assortment of genes.
Answer:
The two genes of each character assort independently of the genes of other characters at the time of gamete formation and gets randomly rearranged in the offspring.

Question 21.
What are the glycoprotein found on the RBC’s of a person with blood group AB?
Answer:
Glycoprotein A and B are found.

Question 22.
Why is no glycoprotein found on the RBC’s of a person with ‘O’ blood groups?
Answer:
Glycoprotein found on the RBCs are coded by the dominant alleles, (I^A and I^B); since the person with the blood group “O” is homozygous recessive, no glycoprotein is found on RBCs.

Question 23.
What are complementary genes ?
Answer:
These are the two independent pairs of genes which interact to produce a trait together
but each dominant genes alone does not show its effect.

Question 24.
What is gene pool?
Answer:
It is the genotypes of all the individuals in a population.

Question 25.
Name the pigments that controls skin colouration in man. [Hots]
Answer:
Melanin.

Question 26.
Who first observed ‘X’ chromosome ?
Answer:
Henking. [Hots]

Question 27.
Why ‘X’ chromosome is called as sex chromosome ?
Answer:
It involves in the sex determination of an individual.

Question 28.
Why is Drosophila (male) referred to as heterogametic ?
Answer:
A male drosophila produce 2 types of gametes i.e., X and Y. Therefore they are heterogametic.

Question 29.
Define heterogamety. Give an example.
Answer:
It is a phenomenon in which organisms produce 2 types of (more than one type) gametes. Eg: Human male, male drosophilla / female fowl.

Question 30.
Define mutation.
Answer:
It is defined as sudden heritable change in the base sequence of DNA or structure of chromosome or a change in the number of chromosomes, that changes the phenotype of an organism.

Question 31.
Give the term for the factors which causes mutation.
Answer:
Mutagenes.

Question 32.
Mention 2 sex-linked Mendelian disorders.
Answer:
Haemophilia and colour blindness

Question 33.
Name some diseases which can be avoided in the progeny through pedigree analysis of parents.
Answer:
Colour blindness, Tuberculosis, Turner’s syndrome.

Question 34.
Give the reason for Down’s syndrome.
Answer:
Trisomy of 21st chromosome.

Question 35.
Define trisomic condition.
Answer:
When a particular chromosome is present in 3 copies in a diploid cell, the condition is called trisomic condition.

36.
Define monosomic condition.
Answer:
When a particular chromosome is present in a single copy in a diploid cell, the condition is called monosomic condition.

2nd PUC Biology Principles of Inheritance and Variation Two Marks Questions

Question 1.
Define test cross. What is its significance?
Answer:
It is the cross between an individual with dominant phenotype with an individual homozygous recessive for the trait. It is used to determine the genotype of an individual for any character trait.

Question 2.
What are multiple alleles ? Give an example.
Answer:
When a gene exist in more than 2 allelic form, the alleles are called multiple alleles.
Eg: The gene ‘I’ controls the human blood group. It exist in 3 different alleleic form i.e., I^A I^B and i.

Question 3.
Write T.H.Morgan’s contribution for genetics.
Answer:

• Morgan conducted experiments in Drosophilla melanogaster and discovered.sex linkage.
• He also found that linked genes may show the phenomenon called linkage where the recombinants in a test than 50%. cross progeny are less

Question 4.
Differentiate between monohybrid and dihybrid.
Answer:

Monohybrid	Dihybrid
It is a cross, where 2 forms of single trait are hybridised.	It is a cross where 2 forms of 2 different traits are hybridised.

Question 5.
Differentiate between genotypes and phenotype.
Answer:

Genotype	Phenotype
(a) It is the total genetic constitution of an individual.	(a) It is the external appearnce of an individual.
(b) It is the expression of genome or more specifically the alleles present at one locus.	(b) It is the expression of genotype produced under the influence of an environment.

Question 6.
Differentiate between test cross and reciprocal cross.
Answer:

Test cross	Reciprocal cross
It is a cross between, F1 hybrid and recessive parent. It confirms the purity of F1 hybrid whether it is homozygous or heterozygous.	It is the second cross involving the same strains carried by sexes opposite to those in the first cross. It is able to distinguish between nuclear chromosomal and sex linked inheritance.

Question 7.
Is not possible study the inheritance of traits in humans in the same way as in peas. Give 2 main reason for it and the alternative method used for such study.[Hots]
Answer:
Mendel’s laws are not applicable for human beings because:

• In human the generation time is too long and produces a small progeny. It creates difficulty in statistical computation of any generation of individuals.
• Human cannot be crossed at will. Pedigree analysis is the another method used to study the family instances and the transmission of particular trait generation after generation.

Question 8.
What is back cross?
Answer:
When an intercross done between 2 genetically different parents, a hybrid is produced which may be homozygous or heterozygous. To determine the purity of parents and to test genotypes of F₁ hybrid, a cross is made between F₁, hybrid and of the parent. Such cross is known as backcross.

Question 9.
How would you find the genotype of an organism exhibiting a dominant phenotypic trait?[Delhi 2008]
Answer:
First a test cross will be done between the dominant phenotypic individual (F₁ hybrid) with recessive phenotypic individual (parent). If the individuals are homozygous dominant, all the individuals in the progeny

If the individuals are heterozygous, the progeny will show dominant phenotype and recessive phenotype in the ratio 1:1.

Question 10.
What is pedigree analysis ? How such an analysis can be useful ?
Answer:
In human genetics controlled crosses are not possible. Pedigree analysis provide a strong tool to study the family history inheritance of a particular trait.It is uitlised to trace the inheritance of any particular abnormality or disease.

Question 11.
What is meant by chromosomal abberation? In which type ofvcells it commonly occurs?
Answer:
Alternation in chromosome is due to the deletion / duplication / addition of a segment of DNA is called chromosomal abberation. Commonly occurs in cancerous cells.

Question 12.
What is point mutation ? Give an example.
Answer:
The mutation that arises due to change in a single base pair of DNA, is known as point mutation. Eg: sickle cell anaemia.

Question 13.
Classify the following into chromosomal and Mendelian disorder.
(a) Cystic fibrosis
(b) Turner’s syndrome
(c) Klinefelter’s syndrome
(d) Haemophilin
Answer:
(a) Cystic fibrosis – Mendelian disorder
(b) Turner’s syndrome – Chromosomal disorder
(c) Klinefelter’s syndrome – Chromosomal disorder
(d) Haemophilia – Mendelian disorder

Question 14.
Give the chromosomal constitution and the resulting sex in each of the following syndrome.
(a) Turner’s syndrome
(b) Klinefelter’s syndrome
Answer:
Turner’s syndrome – 22 pairs of autosomes and one X chromosome.
(44 + XO = 45 chromosome)
The resulting individual is female. Klinefelter’s syndrome – 22 pairs of autosomes and XXY = [44 + XX Y = 47 chromosomes].
The resulting individual is male with more female character.

Question 15.
What is haemophilia due to? What happens in this disorder?
Answer:
Haemophilia is due to a defective recessive allele located on the ‘X’ chromosome. In this disorder, a single protein which is involved in clotting of blood is affected. As a result, the individual bleeds even from a simple cut, that may become fatal.

Question 16.
The human male never passes on the gene for haemophilia to his son. Why?
OR
Why do sons of haemophilia father never suffer from this trait?
Answer:
The genes for haemophilia is present on the ‘X’ chromosome only. A male has only one ‘X’ chromosome. Which he receives from his mother. He receives ‘Y’ chromosome from father. The human male passes the ‘X’ chromosomes to his daughters not to the male progeny.

Question 17.
Differentiate between haemophilia and sickle cell anaemia.
Answer:

Haemophilia	Sickle cell anaemia
1. It is due to recessive defective allele present on X – chromosome. 2.  It is a sex linked ‘disorder. 3.  A protein necessary for clotting of blood is deficient.	1. It is due to point mutation i.e., a single base pair change leading to a change in an aminoaied. 2. It is a autosomal disoder. 3. The defective haemoglobin leads to a change in the shape of RBCs, which become sickle celled.

Question 18.
Differentiate between Turners and Klinefelter’s syndrome.
Answer:

Turner’s Syndrome,	Klinefilter’s Syndrome
1. The individual is female.	1. The individual is male.
2. The individual has one X chromosome less i.e., she has 45 chromosomes.	2. The individual has an extra X chromosome i.e., 47 chromosome.
3.The individual has a short stature with an under developed feminine character.	3. The individual has a tall stature with a feminised character.

Question 19.
Describe Mendel’s observation on the hybridization experiments on garden pea.
Answer:

• The F hybrids always showed one of parental forms of the trait and there was no blending of traits.
• Both the parental form of traits appeared without any change / blending in the F₂ generation.
• The form of the trait appeared in the F₁ generation is called dominant character and it appeared in the F₂ generation about 3 times in frequency as that of its alternate form.

Question 20.
Why Mendel selected pea plants for his experiments?
Answer:
Mendel selected pea plants on the basis of following characters:

• Pea plant exhibited number of contrasting characters.
• It normally undergoes self pollination but can be cross pollinated manually.
• It is an annual plant and yield’s result in a year’s time, so it can be observed for many generation.
• Many varieties were available with observable alternative forms for a trait.
• Pure varieties of pea were available which always bred true.
• Large number of seeds are produced per plant which can be easily handled and cultivated.

Question 21.
List any 7 traits in garden pea which Mendel studied in his breeding
Answer:

Question 22.
What are the reasons for success of Mendel’s experiment on pea plant.
Answer:

• Mendel studied one or two characters at a time for his breeding experiment
• He selected only those traits for his experiments which did not exhibit linkage or incomplete dominance.
• He performed the reciprocal crosses to confirm the validity of the correct ratio.
• He performed his experiment to F₂ + F₃ generation.
• He selected true breeding varieties of pea plant for cross pollination.
• He performed emasculation process to prevent self pollination.
• He maintained complete records of all his ‘ experiments and used statistical method and law of probability for analysing his results.
• He took great care in finding and choosing the true breeding plants genetically.

Question 23.
A pea plant with Purple flowers was crossed with a plant with white flowers producing 40 plants with only purple flowers. On selfing these plants produced 470 plants with purple flower 162 with white flowers. What genetic mechanism account for this results ?
Answer:

F₂ generation 470 purple : 162 white. In F₁ generation only purple flowers producing plants appeared. This means purple colour is dominant. Which doesn’t allow the white colour to express itself.

In F₂ generation, purple and white coloured flowers were produced in the ratio 3:1. Here the parental character of white again reappeared in about one fourth of the progeny. This occurs due to segregation of genes in the gamete formation. This represents the law of segregation and monohybrid ratio.

Question 24.
In human beings, blue eye colour is recessive to brown eye colour. A brown eyed man has a blue eyed mother
(a) What is the genotype of the man and his mother?
(b) What are the possible genotype of his father?
(c) If a man marries a blue eyed woman, what are the possible genotypes of their offsprings ?
Answer:
As per the given condition, the brown eye colour is dominant over the blue eye colour.

(a) The mother’s genetype must be ‘bb’ as she is recessive for blue coloured eye.-The man is brown-eyed (dominant). Its possible genotype must be ‘Bb’ as he is procuring one of the recessive gene from his mother.

(b) As the genotype of the man is ‘Bb’ so the possible genotypes of his father may be BB or Bb

Question 25.
How do you relate dominance, co-dominance and incomplete dominance in the inheritance of character.
Answer:

• Dominance: It is the phenomenon in which one of the alleles of a gene expresses itself in the hybrid / heterozygous condition. While the other is suppressed (recessive) in the presence of the other.
• Codominarice: It is the phenomenon in which 2 alleles of a gene are equally dominant and express themselves in the presence of the other
• Incomplete dominance: It is the phenomenon in which neither of two alleles of a gene is completely dominant over the other and the hybrid is intermediate between the two parents.

Question 26.
Describe the XO- type of sex determination. [HOTS]
Answer:
XO – type of sex determination occurs in certain insects like grasshoppers. The males have only one X – chromosome and hence they have one chromosome less than that females. All the ova contains autosomes and one X chromosome. 50% of the sperm contains one X chromosome besides the autosomes, while 50% of sperm don’t have X – chromosomes. Sex of the individual is determined by the type of sperm fertilizing ovum. Egg fertilized by sperm having an X- chromosome developed into female. Egg fertilized by sperm having no chromosome develop into male.

Question 27.
Describe the different types of mutation.
Answer:

• Mutation is a phenomenon arising from alteration of DNA sequences or structure or number of chromosomes.
• Loss (deletion) or gain (insertion / duplication) of a segment of DNA results in the alteration in chromosome structure, which is called as chromosomal aberration.
• When mutation arises due to a change in single base pair of DNA, it is known as point mutation. Eg: Sickle cell anaemia.Deletion or addition of base pairs of DNA causes frame shift mutation.

2nd PUC Biology Principles of Inheritance and Variation Five Marks Questions

Question 1.
What is incomplete dominance ? Describe with one example.
Answer:
When 2 parents are intercrossed with each other, the hybrid that produced doesn’t resemble either of the parents, but mid way between 2 parents. The phenomenon of incomplete dominance occurs in Four O’ clock plant (Mirabilis Jalopa) and Snapdragon (Antirrhinum Majus). A cross between red flowered (RR) and a white flowered (rr) plant yields the hybrid flowered (Rr) one which is of pink colour in F, generation.

The pink hybrids on crossing with each other give the usual Mendelian ratio 1 : 2 : 1 (one red : 2 pink : 1 white). In this the gene “R is not completely dominant over the allele ‘r’. The heterozygous Rr synthesises only half pigment, so they are pink in colour.

Question 2.
In Snapdragons, tall (DD) is dominant over dwarf (dd) and Red (RR) are incompletely dominant over white (rr), the hybrid being pink. A pure tall white is crossed to a pure dwarf red and F₂ are self fertilized. Give the expected genotype and phenotype in F₁ and F₂ generations.
Answer:

Question 3.
Describe sex determination in certain Birds.
Answer:

• In birds, sex determination is of ZW type, both males and females have same number of chromosomes.
• The males have autosomes plus a pair of Z- chromosomes.
• The females along with autosomes ZW chromosomes.
• The males are homogametic and produce sperms carrying one Z – chromosome along with autosomes.
• The females are heterogametic and produces 50% of ova carrying one Z – chromosome and the remaining 50% carrying one W – chromosomes.
• Sex of the individual is determined by the type of ovum fertilized.
• When an ovum containing Z chromosome is fertilized; the zygote (ZZ) develops into male.
• When an ovum containing W chromosome is fertilized; the zygote (ZW) develops into female.

Question 4.
Study the Pedigree analysis chart and answer the question given below: [CBSE 2008]
Answer:

(a) Is the trait recessive or dominant?
(b) Is the trait sex-linked or autosomal?
(c) Give the genotypes of the parents in Generation I and of their 3rd and 4th child in Generation II.
Answer:
(a) The trait is recessive
(b) The trait is autosomal
(c) Parent Aa and Aa
3rd child is II generation – aa 4th child is II generation – Aa

Question 5.
Study the given pedigree chart show the inheritance pattern of a human trait and answer questions given below.
(a) Give the genotype of parent in 1st generation and of son and daughter shown in 2nd generation.
(b) give the genotype of daughters shown in 3rd generation.
(c) Is the trait sex linked or autosomal. Justify your answer.
Answer:
(a) Genotypes of parents in Generation 1. Male – Aa Female – Aa Son (Generation II) – Aa Daughter (Generation II) – aa.
(b) Genotype of the daughter in Generation III – Aa
(c) It is an autosomal trait and not sex-linked because if it is a sex-linked, the daughter in generation II cannot have it.

Question 6.
A girl baby has been reported to suffer from haemophilia. How is it possible ? Explain with the help of a cross.
Answer:
(a) Haemophilia is due to the presence of a recessive defective allele on X – chromosome.
(b) A female with XX – chromosome must be homozygous for the disease to appear.
(c) She must receive one of the alleles from her haemophilie father (X^hY) and the other from her mother who is also an haemophilie or at least a carrier i.e. the heterozygous for the disease XX^h.

Question 7.
How are Pedigree charts prepared ?
Answer:

• Squares denotes males.
• Circles denotes females.
• Mating is shown by horizontal line connecting a male symbol with a female symbol.
• Off springs symbols are rearranged from left to right in the order of their birth and connected by a horizontal line below the parents and this line is connected to the parents marriage line by a vertical line.
• A solid or blackened symbol represent the individual with the trait under study.
• An open / clear symbol represents the absence of trait under study.

  

 

You can Download Chapter 4 Reproductive Health Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 4 Reproductive Health

2nd PUC Biology Reproductive Health NCERT Text Book Questions and Answers

Question 1.
What do you think is significance of reproductive health in a society?
Answer:
Reproductive health means total well being in all aspects of reproduction i.e., physical, emotional, behavioural, social.

• Awareness is provided to both males and females to lead healthy and satisfying reproductive life.
• To make individuals aware of fertility regulating methods.
• Protect against STD’s.
• Planning children and family planning.
• Proper hygiene of genitalia and treat for reproductive diseases.

Question 2.
Suggest aspects of reproductive health which need to be given special attention in present scenario.
Answer:

• Sex education
• Family welfare information
• Reproductive health clinics.
• Prompting elders to give support and suggestions to children.

Question 3.
Is sex education necessary in schools? Why?
Answer:
Yes, it is necessary in schools because:

• It will provide correct information about sex and reproductive organs.
• Changes during adolescence are fully explained.
• Changes in behaviour are predicted.
• Horns of early life and marriage are explained.
• Cleanliness of genitalia are explained and misconception and myths are removed if any.
• Information is provided about STD’s
• Family planning and hygienic sexual practices are explained.

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

• Reduced MMR and IMR (Maternal Mortality Rate and Infant Mortality Rate)
• Reduced birth rate.
• Decline in STD’s in India.
• Couple protection by family planning has increased.

Question 5.
What are the suggested reasons for population explosion?
Answer:

• Improved medical facilities
• Decline in death rate, IMR, MMR
• Slower decline in birth rate.
• Longer life span.
• Lack of 100% family planning and education among village.

Question 6.
Is the use of contraceptives justified ? Give reasons.
Answer:

• It slows down growth of population
• Helps for proper spacing of children.
• Helps to prevent STDs and prevent its spreading.
• Helps couples to lead a healthy reproductive life.

Question 7.
Removal of gonads cannot be considered as contraceptive options ? Why ?
Answer:
Contraception is meant for preventing conceptions. But removal of gonads can lead to non-secretion of sex hormones. Virilism can appear in ladies and gents may have soft contour.

Question 8.
Amniocentesis for sex determination is banned in our country. Is this ban necessary ? Comment.
Answer:
Amniocentesis is a method for sex determination of the foetus. Due to ethical and spiritual reasons, in India female foetus is not accepted and foetus is destroyed by some families. So, sex – ratio has declined in India. In order to maintain sex – ratio and to prevent social problems in future generation, this test has been banned in India.

Question 9.
Suggest some methods to assist infertile couples.
Answer:
They are some Assisted Reproductive Technologies (ART) to assist infertile couples.
Some ARTs are:

• Test tube baby / IVF-ET (Invitro fertilization and Embryo transfer) : Ova and sperms are collected and fused (induced) to form zygote in laboratory conditions and transferred into fallopian tube (ZIFT – Zygote intra fallopian transfer) or Uterus (IUT – Intra uterine transfer). Here male / female / both may be donor.
• GIFT – (Gamete intra fallopian transfer). Gamete is from donor.
• ICSI – (Intra Cytoplasmic sperm injection). Sperm is directly injected into the ovum.
• AI (Artificial Insemination). Semen from husband ( AIH) or donor (AID) is injected directly into vagina or uterus (IUI – Intra uterine insemination)

Question 10.
What are the measures one has to take to prevent from STD’s ?
Answer:

• Avoid sex with unknown / multiple partners.
• Always use condoms.
• Avoid early or late marriages.
• Early detection and complete cure in case of doubt.

Question 11.
State True / False with explanation.
(a) Abortion could happen spontaneously too.
Answer:
True

• Due to lack of hormonal support.
• Due to congenital malformation of uterus.

(b) Infertility is defined as inability to produce viable offsprings and is always due to abnormality / defect in female partners.
Answer:
False. Infertility may be due to male or female partners.

c) Complete location could help as a natural method of contraception.
Answer:
True. Effective for a period of 6 months due to absence of menstruation.

(d) Creating awareness about sex related aspects is an effective method to improve reproductive health of the people.
Answer:
True. It can reduce STD’s and help practice hygienic sexual practices.

Question 12.
Correct the following statements:
(a) Surgical methods of contraception prevent gamete formation.
(b) All STD’s are completely curable.
(c) Oral pills are very popular contraceptive among rural women.
(d) In E.T techniques, embryos are always transferred into uterus.
Answer:
(a) Gamete formation → Gamete fusion.
(b) Completely curable → not curable
(c) Rural women → the woman (urban women piostly)
(d) Embryos → more than eight celled (zygote or early embryos).

2nd PUC Biology Reproductive Health Additional Questions and Answers

2nd PUC Biology Reproductive Health One Mark Questions 

Question 1.
What do you understand by the term “Reproductive health”?
Answer:
The term reproductive health refers to healthy reproductive organs and then normal functioning.

Question 2.
What is reproductive health according to the World Health Organisation ?
Answer:
According to the World Health Organisation reproductive health means a total well being in physical, emotional, behavioural and social aspects of reproduction.

Question 3.
Expand MMR and IMR.
Answer:

• MMR – Maternal Mortality Rate.
• IMR – Infant Mortality Rate.

Question 4.
Write one reason for the ban on amniocentesis.
Answer:
Amniocentesis is used to find out the sex of the foetus and it leads to female foeticide
i. e., killing of female foetus.

Question 5.
Mention any 2 probable reasons for rapid rise of population in our country from about 350 millions at the time of independence to about 1 billion by the year 2000.
Answer:
The reasons are

• Decline in death rate
• Decline in maternal mortality rate
• Decline in IMR.

Question 6.
Which of the following represents an increase or decrease in population
Answer:

• Decrease in population : Mortality and emigration.
• Increase hi population: Immigration and Natality.

Question 7.
Give 2 important measures taken by the government to tackle the problem of population explosion.
Answer:

• Statutory rising of marriageable age of the female to 18 years and that of the male 21 years.
• Incentives given to the couples with micro family.

Question 8.
Expand MTP and STD.
Answer:

• MTP – Medical termination of pregnancy.
• STD – Sexually transmitted disease.

Question 9.
Name 2 viral STDs that are incurable.
Answer:
Genital herpes, AIDS.

Question 10.
What are Assisted reproduction technologies. [ART]? [Delhi 2008]
Answer:
ART are the special techniques to help the infertile couples to produce children.

Question 11.
What are the different ways in which progesterone or progesterone – estrogen combination can be taken for contraception?
Answer:

• Oral pills
• Injections
• Implants.

Question 12.
Write the other 2 names given for STDs.
Answer:
Venereal diseases (VDs) and Reproductive tract infections (RTI).

Question 13.
Name 2 STDs that can be transmitted by sharing injection needles or surgical instruments.
Answer:
Hepatitis – B, AIDS.

Question 14.
What is In vitro fertilization ?
Answer:
It involves fertilization of ovum outside the body followed by the transfer of embryo inside the uterus.

Question 15.
Expand

• IVF
• ZIFT
• IUT
• GIFT
• ICSI
• AI
• IUI
• ART

Answer:

• IVF – In Vitro Fertilization
• ZIFT – Zygote intra fallopian transfer
• IUT – Intra Uterine Transfer
• GIFT – Gamete intra fallopian transfer.
• ICSI – Intra cytoplasmic Sperm Injection.
• AI – Artificial insemination.
• IUI – Intra uterine Insemination
• ART – Assisted Reproductive Technologies.

Question 16.
After a successful invitro fertilization, the fertile egg begins to divide. Where is this egg transferred before it reaches the 8 cells stage and What is this technique named ?
Answer:
It will be (egg) transferred into the fallopian tube; the technique is called zygote intra fallopian transfer (ZIFT).

2nd PUC Biology Reproductive Health Two Marks Questions

Question 1.
Amniocentesis for sex determination is banned in our country. Is this ban necessary ? Comment. [CBSE – 2006]
Answer:
Since the amniocentesis is misused for female foeticide, the ban is necessary. But by banning this, its advantage of finding out any chromosomal disorders and / or metabolic disorders of the foetus is lost. So, it should be legalized with some strict conditions to avoid its misuse.

Question 2.
Why is hormone releasing IUD considered a good contraceptive to space children ? [CBSE – 2008]
Answer:
Hormone releasing IUDs

• Make the uterus unsuitable for implantation.
• Make the cervix hostile to sperms.
• Increase the phagocytosis of sperm within the uterus.

Question 3.
How does Cu T act as an effective contraceptive for human females.
Answer:
Cu T is an intra – uterine device (IUD) and functions as follows:
The Cu ions released suppressed sperm motility and the fertilizing capacity of sperms.
IUDs increase phago cytosis of sperms with in the uterus.

Question 4.
How do pills act as contraceptives in human females ?
OR
Name the hormonal composition of the oral contraceptive used by human females. Explain how does it act as a contraceptive.
Answer:

• Pills contain progesterone or
• progesterone-estrogen combination.
• They inhibit ovulation
• They alter the quality of cervical mucus and retard the entry of sperms into cervix.

Question 5.
Explain any 2 methods of Assisted reproductive Technology (ART) that has helped couples to bear children.
[CBSE – 2008]
Answer:

• GIFT is method of transferring the ovum, collected from a donor, into the uterus of a female who cannot produce ova.
• ICSI is the process in which sperms are directly injected into the ovum under laboratory condition is and the embryo is transferred into the uterus / fallopian tube.
• In the so called test tube baby programme, the ova of the female and sperms of the male are made to fuse under laboratory conditions and the embryo is transferred into the uterus / fallopian tube of the female

Question 6.
Family planning techniques are not adopted by all in our country. Why?
Answer:
Because

• Religious belief
• People are not fully aware of the method.
• Emotional and social factors
• Fear of some of the ill effects.

Question 7.
How.do surgical procedures prevent conception in humans ? Mention the way it is achieved in human females.
Answer:
Surgical procedures block the transport of gametes and achieve contraception.
The sterilization procedure in human males is called vasectomy. In this method a small part of vas deferens is removed and then tied up through a small incision on the scrotum. Hence the continuity of the path of sperm is lost.

2nd PUC Biology Reproductive Health Three/Five Marks Questions

Question 1.
Give any six reproduction related issues.
Answer:

• Pregnancy
• Child birth / Parturition
• STDs
• Abortions
• Menstrual problem
• Infertility
• conceptive methods.

Question 2.
How lUDs prevent pregnancy?
Answer:

• IUDs increase phagocytosis of sperms within the uterus.
• The Cu ions released by IUDs suppress the term motility and their fertilizing capacity.
• The hormone releasing IUDs make the uterus unsuitable for implantation and cervix hostile to the sperms.

Question 3.
Represent diagrammatically the sterilization method, vasectomy in male reproductive system and tubectomy in female reproductive system.
Answer:

Question 4.
(a) What does gamete intra fallopian transfer (GIFT) represent ?
(b) How do Cu-T and Cu 7 acts as contraceptive devices ?
Answer:
(a) GIFT – It is the introduction of 2 unfertilized oocytes and 2-5 lac motile sperm into fallopian tube of a woman desires to have a child through laparoscope. The egg may be of her’s or of a donor. The sperm may be her husband’s or of a donor. Fertilization occurs inside the fallopian tube and the development of foetus takes place through natural process.

(b) Cu-T, Cu 7 are intra uterine, contraceptive devices having ionized copper. The copper defuses into uterus. It brings about the release of toxic cytokinins. They inhibit the sperm motility and therefore  fertilization of ovum. Another categories of IUDs are hormonal in nature. (Eg: LNG – 20).

You can Download Chapter 3 Human Reproduction Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 3 Human Reproduction

2nd PUC Biology Human Reproduction NCERT Text Book Questions and Answers

1. Fill in the blanks:

(a) Humans reproduce ………
Answer:
sexually

(b) Humans are ……………..
Answer:
viviparous

(c) Fertilization is …………….. in humans.
Answer:
internal

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

(e) Zygote is ………..
Answer:
diploid

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

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

(h) The fusion of male and female gamete is called ……….
Answer:
fertilization

(i) Fertilization takes place in …………
Answer:
ampullary isthmus in fallopian tube.

(j) Zygote divides to form which is implanted in uterus.
Answer:
Blastocyst

(k) The structure which provides vascular connection between foetus and uterus is
Answer:
umbilical cord.

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

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

Question 4.
Write two major functions each of testis and ovary.
Answer:
Testis :

• Formation of sperms
• Secretion of hormone testosterone.

Ovary :

• Formation of ova
• Secretion of hormone oestrogen and progesterone.

Question 5.
Describe the structure of a seminiferous tubule.
Answer:
Testis has about 250 compartments (Testicular lobules) which contains 1-3 highly coiled seminiferous tubules. Each tubule is lined by 2 types of cells. Male germ cells (immature) (spermatogonic) and Sertoli cells.

Male germ cells form sperms by meiosis and Sertoli cells provide nutrition to these germ cells. Male germ cells and Sertoli cells together form germinal epithelium. Seminiferous tubule is covered outside by basement membrane.

Question 6.
What is spermatogenesis ? Describe the process of spermatogenesis.
Answer:
Spermatogenesis is the production of sperms (n) by immature male germ cells (2n) at puberty inside the testis.

Hormonal role in spermatogenesis:
Spermatogenesis starts due to increase in secretion of GnRH (Gonadotropin Releasing Hormone) by hypothalamus. GnRH acts on Anterior pituitary gland and stimulates secretion of 2 gonadotropins – LH (Luteinizing Hormone) or ICSH (Interstitial Cell Stimulating Hormone) and FSH (Follicle Stimulating Hormone).

LH acts on Leydig cells for secreting testosterone and other androgens inturn stimulates process of spermatogenesis. FSH acts on sertoli cells which secretes some factors useful in spermiogenesis. Sertoli cells secrete inhebin that suppresses FSH synthesis.

 

It is production of haploid spermatozoa from diploid spermatogonia inside testis at puberty. At puberty, Spermatogonium undergoes mitosis forms 2 spermatogonia A and B. Both A and B are diploid with 46 chromosomes each.
A: Function as mother spermatogonia.
B : Grow in size to function as primary spermatocytes.
They then undergo meiosis to form 2 haploid cells called secondary spermatocytes (23 chromosomes). They then undergo 2nd meiotic division forming 4 haploid spermatids. These spermatids are transferred into spermatozoa (sperms) by spermiogenesis. After spermiogeneses, sperm heads become embedded into Sertoli cells and are released from seminiferous tubules by process called spermiation.

Question 7.
Name hormones involved in regulation of spermatogenesis.
Answer:
[Explained in Answer to Question Number 67-Hormones are: GnRH, FSH, LH / ICSH, testosterone, Inhibin

Question 8.
Define spermiogenesis and spermiation.
Answer:

• Spermiogenesis: Development and differentiation of a spermatozoa (n) from a spermatid (n).
• Spermiation: The detachment of fully mature spermatozoa from Sertoli cells is called spermiation.

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

Question 10.
What are the major components of seminal plasma?
Answer:
Fructose, fibrinogen, profibrinolysin, Calcium bicarbonate, prostaglandins, mucus, (seminal plasma + sperms = semen)

Question 11.
What are the major functions of male accessory ducts and glands?
Answer:
Male accessory ducts :
(i) Intra testicular Genital Duct system (till vasa efferentia) – Cilia lining them fails in passage of sperms.
(ii) Extra testicular / Excretory Genital Duct system:

• Epididymis – Store sperms, ejects sperms during ejaculation, destroys older sperms.
• Vasa deferentia – conduction of sperms
• Ejaculatory ducts – conduct sperms and secretion of seminal vesicles are also conducted
• Urinogenital duct – conduct sperms.

Male accessory glands:

• Seminal resides – produce seminal plasma (60 – 70 %)
• Prostate Gland – seminal plasma (20 – 30%)
• Bulbourethral gland – 5% seminal plasma

Question 12.
What is oogenesis ? Give a brief action of oogenesis.
Answer:
Process of formation, development and maturation of haploid ovum or female gamete from diploid germinal cell of the ovary.

Cells of the germinal epithelium of ovary undergoes repeated mitotic divisions to form diploid Oogonia or gamete mother cells. They are formed in the fetal ovary in large number by mitotic division form primary oocyte.

The primary oocyte enlarges and matures by taking food from the surrounding follicle cells. The mature primary oocyte undergoes its first meiotic division. It is an unequal division resulting in the formation of a large haploid secondary oocyte and a tiny first polar body or polocyte. The secondary oocyte remains bulk of the nutrient ride cytoplasm of the primary oocyte. The secondary oocyte undergoes second meiotic division which doesn’t proceed beyond metaphase until a sperm enters it. Ovulation occurs at this stage and the secondary oocyte is transferred to the fallopian tube.

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

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

Question 15.
Draw diagramatic representation of various events during menstrual cycle.
Answer:

Question 16.
Name the functions of the following.
(a) Corpus luteum
(b) Endometrium
(c) Acrosome
(d) Sperm tail
(e) Fimbriae
Answer:
(a) Corpus luteum : Secretion of mainly progesterone and small quantity of estrogen. Some androgens are also formed by theca cells.

(b) Endometrium : Nourishment and implantation of blastoeyst and later foetus if fertilization has occurred. Otherwise cyclic changes of growth and degeneration.

(c) Acrosome : Contains sperm lysin for separating cells of corona radiata and piercing through zona pellucida.

(d) Sperm Tail: Vibratile part that helps in swimming of sperm in the genital tract of female for reaching the ovum.

(e) Fimbriae : The finger like projection occurs at the edges of infundibulum which helps in the collection of ovum after ovulation.

Question 17.
Identify True / False statements. Correct each false statement to make it true,
(a) Androgens are produced by Sertoli cells.
(b) Spermatozoa get nutrition from Sertoli cells.
(c) Leydig cells are found in ovary
(d) Leydig cells synthesize androgens
(e) Oogenesis occurs in corpus luteum.
(f) Menstrual cycle ceases during pregnancy.
(g) Presence or absence of hymen is not a reliable indicator of virginity or sexual experience.
Answer:
(a) False: Spermatogenesis substances androgen binding protein and intubin are produced by Sertoli cells.
(b) False: Spermatozoa differentiate from spermatids with the help of spermatogenic factors produced by Sertoli cells.
(c) False: Leydig cells are found in testis.
(d) True.
(e) False: Oogenesis occurs inside the ovary.
(f) True.
(g) True.

Question 18.
What is menstrual cycle? Which hormone regulates menstrual cycle?
Answer:
A series of cyclic changes found in the reproductive tract of human female during her reproductive life that recur at intervals of about 28 days and is characterised by menstruation in the first 3-4 days.
Hormone :

• GnRH (Gonado tropin Releasing Hormone)
• FSH (Follicle Stimulating Hormone)
• LH (Luteinizing Hormone)
• Estrogen
• Progesterone

Question 19.
What is parturition ? Which hormones are involved in the induction of parturition ?
Answer:

• The process of delivering of full developed foetus or baby at the end of pregnancy period through vigorous contraction of uterus is called parturition.
• Estrogen (amount of estrogen is more than progesterone) and oxytocin are the hormones involved in the induction of parturition.

Question 20.
In our society women are blame for giving birth to daughters. Can you explain why this is not correct ?
Answer:
Sex chromosome pattern in female is xx, i.e., both the gametes are with ‘X’ chromosome. In males, sex chromosome is X Y i.e., one gamete with ‘X’ chromosome and other gamete with ‘Y’. So 50% sperms carry ‘X’ and 50% carry ‘Y’. Female child is produce when the sperm with ‘X’ chromosome fertilizes egg with ‘X’ chromosome. A male child is produce when sperm with ‘Y’ chromosome fertilizes egg with ‘X’ chromosome. Therefore sex of a baby depends on father not on mother.

Question 21.
(a) How many egg are released by a human ovary in a month ?
(b) How many eggs do you think would have been released if the mother gave birth to identical twins ?
(c) Would your answer change if the twins born were fraternal ?
Answer:
(a) One
(b) One
(c) Yes, fraternal twins are born due to the fertilization of 2 or more eggs.

Question 22.
How many eggs do you think were released by the ovary of a female dog, which gave birth to 6 puppies?
Answer:
Six eggs were released by the ovary of a female dog.

2nd PUC Biology Human Reproduction Additional Questions and Answers

2nd PUC Biology Human Reproduction One Mark Questions

Question 1.
Where are sperm produced in the testis?
Answer:
In the seminiferous tubules of testis.

Question 2.
What is the role of placenta?
Answer:
Provide nutrition to the developing embryo.

Question 3.
What is the function of amniotic fluid?
Answer:
Protects foetus from shock.

Question 4.
Name the germ layer from which gonad develops.
Answer:
Mesoderm

Question 5.
Name the sperm lysin ? Which organelle secretes it ?
Answer:
Hyaluronidase; Acrosome.

Question 6.
Define gametogenesis.
Answer:
The process of formation of male and female gamete in the gonads is called gametogenesis.

Question 7.
The spermatogonia of an animal contains 32 chromosomes. What will be the number of chromosomes in its
(a) secondary spermatocyte
(b) spermatids respectively
Answer:
(a) Secondary spermatocyte – 16 chromosome
(b) spermatids -16 chromosome

Question 8.
Name the structure formed from Graafian follicle after ovule?
Answer:
Corpus luteum

Question 9.
Name the hormone secreted by corpusluteum.
Answer:
Progesterone.

Question 10.
Define spermiogenesis ? Where does it occur ?
Answer:
The process of the transformation of spermatids into spermatozoa is called spermiogenesis. It occurs in the seminiferous tubules of testis.

Question 11.
Why middle piece of the sperm called power house of the sperm ?
Answer:
Middle piece contains numerous mitochondria which produce energy for sperm movement.

Question 12.
Define Oogenesis.
Answer:
The process of formation of mature female gametes.

Question 13.
Name the fluid filled space in the tertiary follicle.
Answer:
Antrum

Question 14.
What is the significance of secondary oocyte retaining the bulk of nutrient rich cytoplasm of the primary oocyte ?
Answer:
These reserve food materials nourish the embryo till implantation.

Question 15.
When do the levels of FSH and LH reach the maximum in the menstrual cycle ?
Answer:
The peak level of FSH and LH is reached in the middle of (14th day) menstrual cycle.

Question 16.
Define implantation?
Answer:
Implantation is the process in which the mammalian embryo (blastocyst) becomes attached to the endometriosis of the uterus.

Question 17.
What are the stem cells in human embryo?
Answer:
Stem cells are those cells in the inner mass of the blastocyst, which have the potency to give rise to all tissues and organs.

Question 18.
What is colostrum ? How does it provide initial protection against diseases to new born infants. Give reason.
Answer:
The milk produced during initial few days of lactation is called colostrum. It consists of several antibodies like IgA etc. which are essential for the development of resistance in new born babies.

2nd PUC Biology Human Reproduction Two Marks Questions

Question 1.
Name 2 types of cells present in the inner lining of seminiferous tubules. What are their functions?
Answer:
Two types of cells in the inner lining of seminiferous tubule are

• Spermatogonia : Spermatogonia produces male gamete called spermatozoa.
• Sertoli cells : Sertoli cells provide nutrition to the developing spermatozoans.

Question 2.
Where are Leydig cells present? What is their role in reproduction?
Answer:

• Ley dig cells are located in the interstitial space (space between the seminiferous tubule) in the testis.
• They secrete testicular hormones called androgens, mainly testosterone, this hormone regulates spermatogenesis.

Question 3.
Differentiate between vasa efferentia and vas deferens.
Answer:

• Vasa efferentia are the ducts that leave the testis to open into epididymis. These are located inside the testis and are extra abdominal and do not receive the ducts of any glands.
• Vas deferens is the duct continues from epididymis. This ascends into the abdominal cavity. It receives the ducts of seminal vesicle.

Question 4.
Where are fimbriae present in a human female reproductive system ? Give their function.
Answer:
Fimbriae are present in the free edges of the infundibulum of the fallopian tube. Then help in the easy capture of ova during ovulation.

Question 5.
Differentiate between endometrium and myometrium.
Answer:

• Endometrium is the innermost glandular layer that lines the uterine cavity. It undergoes cylindrical changes during menstrual cycle. Implantation occurs in this layer.
• Myometrium is the middle thick layer of smooth muscles of the uterine wall. It doesn’t undergo any changes during menstrual cycle. It is responsible for the uterine movement.

Question 6.
Give the differences between spermatogenesis and spermiogenesis.
Answer:
Spermatogenesis:
It is the process of the formation of mature spermatozoa in the testis. It involves meiotic and mitotic division. It is controlled by hormones like leutenising hormone (LH) and androgen (testesterone).

Spermiogenesis:
It is the process of the transformation of spermatids into spermatozoa. It doesn’t involve any cell division. It is controlled by LH which stimulates the Sertoli cells to secrete the factors needed for spermiogenesis.

Question 7.
Mention the sites of action of the hormones – GnRH and FSH, during spermatogenesis in human males. Give one function of each of the hormones.
Answer:
GnRH acts on anterior pituitary and FSH acts on Sertoli cells of seminiferous tubules.
GnRH stimulates the release of two gonadotropin – Follicle stimulating hormone and leutenising hormone (LH) from the anterior pituitary. FSH stimulates the Sertoli cells to secrete some factors necessary for spermiogenesis.

Question 8.
Why does the coitus not leads to pregnancy all the time?
Answer:
Pregnancy occurs when the sperm and ovum reaches the ampullary isthmus junction of the fallopian tube at the same time. During every coitus both are not reaching at this position together.

Question 9.
Given below is the diagram of the sectional view of human ovum just after ovulation.

Mention the site of fertilization in the fallopian tube of human female where the ovum and sperm meet.
Answer:
Ampullary – Isthmus Junction.

Question 10.
How it blastula / blastocyte differ from Morula ?

Blastocyte	Morula
(a) It is a hollow sphere of 32 or more cells formed by the rearrangement of blastomeres. (b) Zona pellucida disintegrates with the enlargement of blastocoel	(a) It is a solid sphere of 8-16 cells blastomeres by cleavage of zygote. (b)  Zona pellucida is intact.

Question 11.
What are chroionic villi ? What is their fate?
Answer:
The finger like projections of the trophoblast produced after implantation are called chronic villi. Chroionic villi and uterine tissue become interdigitated with each other and jointly form the placenta.

Question 12.
Name the hormones secreted by human placenta.
Answer:
The hormones secreted by human placenta are;

• Human chorionic gonadotropin
• Progesterone
• Human placental lactogen
• Estrogen

Question 13.
Write the functions of placenta in humans.
Answer:
(a) It helps to supply oxygen and nutrients to the foetus.
(b) It helps in the removal of CO₂ and other waste product formed by the foetus.
(c) Acts as endocrine gland by secreting hormone like human placental lactogen, human chorionic gonadotropin, estrogen and progesteron which are necessary to maintain pregnancy.

2nd PUC Biology Human Reproduction Three/Five Marks Questions

Question 1.
Describe the accessary ducts of human male reproductive system.
Answer:
The accessory ducts include rete testis, vasaefferentia, epididymis and vas deferens.
The seminiferous tubules end as short, straight tubules into rete testis. From the rete testis, 10-20 fine tubules called vasa efferentia leave the testis and open into epididymis. Epididymis is a single convoluted tubule that is located along the posterior surface of the testis.

The epididymis leads into vas deferens that ascends into the abdomen and loops over the urinary bladder. It receives the ducts of seminal vesicle to form ejaculatory duct, that runs through the prostrate and opens into the urethra, just after its origin from the urinary bladder. The urethra receives the ducts of prostrate and bulbo urethral glands and runs through the penis to its external opening called urethral meatus.

Question 2.
Name the cells found
(a) inside the seminiferous tubules
(b) outside the seminiferous tubules in a human testis. Mention the function of each of them.
Answer:
(a) Inside the seminiferous tubules are

• Spermatogonial cells
• Sertoli cells

(b) Outside the seminiferous tubules are Ley dig cells.
Functions:

• Spermatogonial cells form spermatozoa.
• Sertoli cells provide nutrition to the germ cells and help in spermiogenesis.
• Ley dig, cells secrete the male sex hormone, testesterone.

Question 3.
Name the accessory glands of human male reproductive system and mention their functions.
Answer:
The male accessory glands include

1. a pair of seminal vesicle
2. a prostrate gland
3. a pair of bulbo urethral gland.

(1) a pair of seminal vesicle:
The secretion from these glands constitute the seminal plasma, which is rich in calcium fructose and certain enzymes.

(2) a prostrate gland:
Seminal plasma provides the fluid medium for the-sperm to swim in the female reproduction tract, towards the ovum.

(3) a pair of bulbo urethral gland:
It provides nourishment to the sperm. The secretions from bulbourethral glands help in the lubrication of the penis.

Question 4.
Describe the structure of mammary glands of a human female with labelled diagram.
Answer:
A mammary gland consists of glandular tissue and variable quantity of fat. The glandular tissue is divided into 15 – 20 mammary lobes and each lobe contains clusters of cells called alveoli, which opens into mammary tubules. The mammary tubules of each be join to form a mammary duct. Several mammary ducts join to form a wider mammary ampulla, which is connected to lactiferous duct through which milk comes out.

Question 5.
(a) Explain the role of ovarian hormones in inducing changes in the uterus during menstrual cycle.
(b) What triggers release of oxytocin at the time of parturition ?
Answer:
(a) Estrogen influences the uterus in the follicular phase. The endometrin is regenerated through protection. Progesterone influences the uterus in the luteal phase, the endometrin becomes further thickened and vascular for implantation
(b) Foetal ejection triggers the release of oxytocin.

Question 6.
Describe the events that take place during in fertilization in human being.
Answer:
Fertilization refers to the fusion of a sperm and ovum in humans occurs in the ampullary isthmic junction of the fallopian tube.When a sperm comes in contact with the zona pellucida of the ovum, it induces changes in the membrane that. blocks the entry of other sperms. The secretions of acrosome help the sperm to digest the zona pellucida and plasma membrane of the ovum and enter into its cytoplasm.

The entry of sperm induces the completion of secondary meiotic division of the secondary oocyte resulting in the formation of a second polar body and a large ootid. The haploid nucleus of the ootid and that of the sperm fuse to form a diploid zygote.

Question 7.
Describe the major steps in the development of a fertilized egg upto complete differentiation into blastocyst ready for implantation.
Answer:

• The mitotic division called cleavage divisions start in the zygote as it moves through the isthmus of fallopian tube towards the uterus.
• The divisions result is 2,4,8,16 daughter cells, called blastomeres; the embryo with 8-16 blastomeres is a solid spherical structure and is called as morula.
• The morula continues to divide and the blastomeres rearrange themselves as it moves further into the uterus.
• As a result a hollow spherical structure, called blatocyst is formed.
• The blastocyst has an outer layer of cells, called trophoblast and an inner group of cells, called inner cell mass attached at one end of the trophoblast.
• The trophoblast layer becomes attached to the endometrium and the inner cell mass gets ready to form the embryo proper.

Question 8.
Draw a labelled diagram showing a human foetus developing within the uterus.
Answer:

Question 9.
What is Oogenesis ? Give a brief account of Oogenesis.
Answer:
Oogenesis is the process of formation of mature female gametes or ova. These cells start division and enter prophase I of meiosis and remain suspended at that stage; these are called primary oocyte.

Each primary oocyte is surrounded by a layer of granulosa cells and become the primary follicle. When the primary follicle becomes surrounded by more layers of granulosa cells, it is called secondary follicle.
The secondary follicle transforms into tertiary follicle. With the development of a fluid filled cavity (antrum) around the primary oocyte.

The granulosa cell become organised into an outer layer, called theca external and an internal interna. At this stage, the primary oocyte completes meiosis I and form a larger haploid secondary oocyte and a tiny polar body.

A tertiary follicle grow further and changes into mature follicle or agrarian follicle. The secondary oocyte secretes a new membrane called zona pellucida around it. At this stage the follicle ruptures to release the secondary oocyte, which moves into the fallopian tube. The secondary oocyte completes meiosis II only when a sperm enters its cytoplasm, it form a larger cell and Ootid and a smaller cell, the second polar body.

You can Download Chapter 2 Sexual Reproduction in Flowering Plants Questions and Answers, 2nd PUC Biology Question Bank with Answers, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Question Bank Chapter 2 Sexual Reproduction in Flowering Plants

2nd PUC Biology Sexual Reproduction in Flowering Plants NCERT Text Book Questions and Answers

Question 1.
Name the parts of an angiosperm flower in which development of male and female gametophyte takes place.
Answer:
Inside the anther, the cells of microsporangia develop as male gamete. Inside the ovary megasporangial cells develop as female gametes.

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

Microsporogenesis	Megasporogenesis
The process of formation of microspores from pollen mother cell through meiosis is called microsporo­genesis. Pollen grains are formed at the end of this event. Occurs inside the pollensac of the anther.	1. The process of formation of megaspores from megaspore mother cells through meiosis is called megaspore -genesis. 2.  Egg is, produced inside the embryo sac at the end of this event.

Question 3.
Arrange the following terms in the correct developmental sequence. Pollen grain, sporogenous tissue, microspore tetrad, pollen mother cell, male gamete.
Answer:
Sporogenous tissue → Pollen mother cell → microspore tetrad → pollen grain → male gamete.

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

The various parts of an ovule are:

1. Funicles: It is a stalk-like structure which represents the point of attachment of the ovule to the placenta of the ovary.

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

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

4. Micropyle: It is a narrow pore formed by the projection of integuments. It makes the point where the pollen tubes enters the ovule at the time of fertilization.

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

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

Question 5.
What is meant by monosporic development of female gametophyte?
Answer:
The female gametophyte or the embryo sac develops, from a single functional megaspore. This is known as monosporic development of the female gametophyte. In most flowering plants, a single megaspore mother cell present at the micropylar pole of the nucellus region of the ovule undergoes meiosis to produce four haploid megaspores. Later out of these 4 megaspores, only one functional megaspore develops into female gametophyte, while the remaining 3 degenerates.

Question 6.
With a neat labelled diagram, explain the 7 celled, 8 nucleate nature of the female gametophyte.
Answer:

The female gametophyte (embryo sac) develops from a single functional megaspore. Thus, megaspore undergoes three successive mitotic divisions to form 8 nucleate embryo sac. The first mitotic division in the megaspore forms 2 nuclei. One nucleus moves towards the micropylar end while the other nucleus moves towards the chalazal end. Then these nuclei divide at their respective ends and redivide to form 8 nucleate stages.

As a result there are 4 nuclei each at both the ends i.e., at the micropylar and the chalazal end in the embryo sac. At the micropylar end, out of 4 nuclei only 3 differentiate into 2 synergids and one egg cell. Together they are known as egg apparatus. Similarly, at the chalazal end 3 out of 4 nuclei differentiates as antipodal cells. The remaining 2 cells (of the micropylar and chalazal end) move towards the centre and are known as the polar nuclei, which are situated in the centre of embryo sac. Hence, at maturity, the female gametophyte appears as a 7 celled structure, though is has 8 nucleate.

Question 7.
What are chasmogamous flowers? Can cross-pollination occur in cleistogamous
flowers ? Give a reason for your answer.
Answer:
There are 2 types of flowers present in plants namely oxalis and viola chasmogamous and cleistogamous flowers. Chasmogamous flowers have exposed anthers and stigma similar to the flowers of other species.

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

Question 8.
Mention two strategies evolved to prevent self pollination in flowers.
Answer:
Self pollination involves the transfer of pollen from the stamen to the pistil of the same flower. Two strategies have evolved to prevent self pollination in flowers are as follows.

(a) In certain plants, the stigma of the flower has the capability to prevent the germination of pollen grains and hence, prevent the growth of the pollen tube. It is a genetic mechanism to prevent self pollination called self incompatibility. This may be between the individuals of the same species or between individuals of different species. Thus, incompatibility prevents breeding.

(b) In some plants, the gynoecium matures before the androeicum or vice – versa. This phenomenon is known as protogyny or protandry respectively. This prevents the pollen from coming in contact with the stigma of the same flower.

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

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

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

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

Question 11.
What is triple fusion? Where and how does it take place? Name the nuclei involved in triple fusion.
Answer:
Triple fusion is the fusion of the male gamete with 2 polar nuclei inside the embryo sac of the angiosperm. When pollen grains fall on the stigma, they germinate and give rise to the pollen tube that passes through the style and enters into the ovule.

After this, the pollen tube enters between one of the synergids and releases 2 male gametes there. Out of these 2 male gametes, one gamete fuses with nucleus of the egg cell and forms zygote (syngamy). The other male gametes fuses with 2 polar nuclei in the centre form a triploid primary endosperm nucleus. Since this process involves the fusion of 3 haploid nuclei, it is known as triple fusion. It results in the formation of endosperm.

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

Question 13.
Differentiate between
(a) Hypocotyl and epicotyl
(b) Coleoptile and coleorrhiza
(c) Integument and testa
(d) Perisperm and pericarp
Answer:
(a) Hypocotyl and epicotyl

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

(b) Coleoptile and coleorhiza

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

(c) Integument and testa

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

(d) Perisperm and pericarp

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

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

Question 15.
What is meant by emasculation ? When and why does a plant breeder employ this technique?
Answer:
Emasculation is the process of removing anthers from bisexual flowers without affecting the female reproductive part (pistil). Emasculation is performed by plant breeders in bisexual flowers to obtain the desired variety of a plant by crossing a particular plant with the desired pollen grain. After removing the anthers, the flowers are covered with a bag before they open. This ensures that the flower is pollinated by pollen grains obtained from desirable varieties only. Later, the mature, viable and stored pollen grains are dusted on the bagged stigma by breeders to allow artificial pollination to take place and obtain the desired plant variety.

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

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

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

2nd PUC Biology Sexual Reproduction in Flowering Plants Additional Questions and Answers

2nd PUC Biology Sexual Reproduction in Flowering Plants One Mark Questions

Question 1.
Name the protective substance present on the pollen envelop to tide over adverse condition. [CBSE 98]
Answer:
Sporopollenin.

Question 2.
Name some water pollinated plats.
Answer:

• Fresh water – vallesneria
• Hydrolla Aquatic – Zostera

Question 3.
What technical term is applied to fruits formed without fertilization ?
Answer:
Parthenocarpy

Question 4.
What is a false fruit ? Give an example.
Answer:
A fruit which is formed by any floral parts of the flower other than ovary, eg. apple, pear, cashew nut, etc. [CBSE – 95]

Question 5.
List the general characteristics of the pollen grains of wind pollinated plants.
Answer:
The pollen grains are smooth, dry, light, non-sticky and may be winged. [CBSE – 2K]

Question 6.
Which nuclei fuse to give endosperm ?
Answer:
Polar nuclei.

Question 7.
Name the stage of the occurrence of more than one embryo in a seed.
Answer:
Polyembryony.

Question 8.
The term for the early stages of embryo development.
Answer:
Embryogeny

Question 9.
State the function of a suspensor.
Answer:
Pushes the developing embryo into the endosperm for absorption of nutrients.

Question 10.
Define parthenocarpy.
Answer:
These are seedless fruits which are formed without pollination and fertilization.

Question 11.
What is fertilization ?
Answer:
It is the fusion of one male gamete with egg cell and second gamete with polar nuclei in angiosperm. [CBSE – 99]

Question 12.
State the difference between the endosperm of gymnosperms and angiosperms.
Answer:
Endosperm of gymnosperms is haploid gametophyte but in angiosperms it is triploid as it is formed after double fertilization.

Question 13.
What is epicotyl ?
Answer:
Portion of an embryonic axis between the plumule and cotyledon.

Question 14.
Define the term geitonogamy.
Answer:
A condition where pollen from one flower deposited on the stigma of another flower borne on the same plant.

Question 15.
What is the fate of secondary nucleus after fertilization.
Answer:
It forms the endosperms.

Question 16.
What is allogamy ?
Answer:
It is the transfer of pollen from one flower to the stigma of another flower on a separate plant of the same species.

Question 17.
What develops into a microspore mother cell in a flower ?
Answer:
Sporogenous cells.

Question 18.
What is coleorrhiza ? [CBSE – 97]
Answer:
It is the protective cap over the radicle of maize.

Question 19.
What is scutellum ?
Answer:
It is the single cotyledon of maize.

Question 20.
What is funiculus ?
Ans:
Stalk of the ovule by which it is attached to the placenta.

2nd PUC Biology Sexual Reproduction in Flowering Plants Two Marks Questions

Question 1.
Differentiate between parthenocarpy and parthenogenesis.
Answer:
Parthenocarpy is .the formation of fruit without fertilization where as parthenogenesis is the formation of embryo from unfertilized egg.

Question 2.
Draw a labelled diagram of mature pollen grain. [CBSE – 90]
Answer:

Question 3.
What is seed dormancy. Give any 2 advantages. [CBSE – 94]
Answer:
Seed dormancy is the condition of the seed when it fails to germinate even though the environmental conditions are favourable for active growth.
Two advantages of dormancy:

• It help the seed to disseminate in time and space in order to achieve maximum cooperative environment for the survival of species.
• To ensure the successful seed germination under most favourable condition.

Question 4.
What are false fruits? Give example.
Answer:
The fruits which develop from parts other than ripened ovary are called false fruits.
Eg: fruits of apple and pear develop from fleshy thalamus. [CBSE – 95]

Question 5.
Write the significance of double fertilization.
Answer:
Double fertilization leads to the development of triploid endosperm which provides nourishment to the developing healthy seed and this triploid endosperm compensates for the extreme reduction of female gametophyte in angiosperms.

Question 6.
Differentiate between non-albuminous and albuminous seeds with examples.
Answer:

• Non albuminous seeds have no residual endosperm as it is completely consumed during embryo development. Eg: Pea, ground nut.
• Albuminous seeds retain, a part of endosperm as it is not completely used up during embryo development. Eg: Wheat, maize, barley, castor.

Question 7.
Write the difference between coleoptile and coleorrhiza.
Answer:

• In the monocot seed, the region of embryonic axis below the cotyledon is the radicle covered with protective sheath is coleorrhiza.
• Above the point of attachment of the cotyledon, the embryonic axis becomes the plumule which is enclosed by a leaf like covering called coleoptile.

2nd PUC Biology Sexual Reproduction in Flowering Plants Five Marks Questions

Question 1.
Describe the process of development of dicotyledonous embryo.
Ans:
The process of development of mature embryo from diploid zygote is called embryogenesis. After fertilization the zygote of the ovule divides transversely into two cells – a small apical cell and a large basal cell. The basal cell lying towards the micropyle divides in one direction into a row of cells called a suspensor. It pushes the developing embryo into the endosperm for absorption of nutrients.

The apical cell located towards the antipodal end of the zygote undergoes two vertical divisions and one transverse division to form the embryonal mass. The cells of the embryonal mass divide repeatedly and the various parts of the embryo are formed. The anterior cells of the embryonal mass form _ plumule and two cotyledon. The main part of the radicle and the hypocotyl are formed from the posterior
embryonal mass cells.

Question 2.
Trace the events that would take place in a flower from the time the pollen grain of the same species falls on the stigma up to the completion of fertilization.
Answer:

The pollen grain develops on the stigma stimulated by the secretion of the stigma. The intine grows out as a protrubrance through one of the germ pore. This out growth continues to grow as pollen tube. The nucleus of the vegetative cell moves into it followed by the generative cell. The generative cell divides into two male gamete and moves to the tip of the pollen tube.

The pollen tube secretes enzyme to digest the tissues of the style and enters the ovule through micropylar end and discharges the two male gametes into one of the synergfds of the embryo sac. One of the male gametes fuses with the ovum to form a zygote. This process is called syngamy. The other male gamete fuses with the secondary nucleus to form primary endosperm nucleus.

This process is called triple fusion. Since there are two fusions (syngamy and triple fusion) inside the ovule during fertilization, it is known as double fertilization.

Question 3.
Draw a labelled diagram of a T.S of a dehisced anther.
Answer:

Question 4.
Describe the structure of a typical embryo sac found in flowering plant. Why is it generally referred to as monosporic ?
Answer:
A typical embryo sac is 8 nucleate and 7 celled. Six of the eight nuclei become surrounded by cell wall and are organised into cells. Two of the nuclei, called polar nuclei remain in the centre of the large central cell. Three cells are grouped together at the micropylar end forming the egg apparatus. Of these, one is the female gamete and two are synergids; the synergids have special cellular thickening called filiform apparatus at the micropylar tip.

Three cells are grouped at the clealazal end; they are called antipodals. The large cell in the centre of the embryo sac is the central cell. Later 2 polar nuclei in the centre cell fused to form a depolid secondary nucleus or endosperm nucleus. Thus the embryo sac of flowering plants is 8 nucleate 7 celled at maturity. This type of embryo sac is called monosporic because it is formed from only one of the 4 megaspores.