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 13 Organisms and Populations 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 13 Organisms and Populations

2nd PUC Biology Organisms and Populations NCERT Text Book Questions and Answers

Question 1.
How is diapause different from hibernation?
Answer:

Diapause	Hibernation
(i) Dormant stage in development of an organism	(i) State of inactivity in a mature organism.
(ii) Occurs in summers and winters	(ii) Only during winters
(iii) Reduction in free water	(iii) No such adaptations

Question 2.
If a marine fish is placed in a freshwater aquarium, will the fish be able to survive? Why or why not?
Answer:
It will not be able to survive for long because:-

• Water will enter the body by endosmosis
• As marine fishes drink lots of water, so excess water enters.
• It does not have salt absorbing mechanism as of freshwater fishes.

So, these fishes will die because of problem in osmolarity maintenance But due to hormonal mechanisms, migratory fishes are exceptions as they withstand the changes eg:- Hilsa, salmon

Question 3.
Define phenotypic adaptation. Give one example.
Answer:
Phenotypic adaptation:- It is non-genetic change in physical and physiological characteristics which develops in response to changes in environment and are useful to organisms in adjustment to prevailing Pvt. conditions.
eg:-

• Mountain sickness in normal people in high attitudes but tribes living in himalayas do not have mountain sickness
• Euphorbia hirta plant species.

Question 4.
Most living organisms cannot survive at temperature above 450C. How are some microbes able to live in habitats with temperatures exceeding 1000C?
Answer:
Temperature > 45°C causes precipitation of protoplasm and denaturation of enzymes. Therefore they die in most of organisms But some live in 1000°C (thermophiles) due to

• Branched-chain lipids that reduce fluidity of cell membranes.
• Inborn metabolism to produce heat-tolerant enzymes
• Reduced free water.

Question 5.
List the attributes that populations but not individuals possess.
Answer:

• Population density
• Birth rate
• Death rate
• Sex ratio
• Population growth
• Age distribution
• Population dispersal.

Question 6.
If a population growing exponentially double in size in 3 years, what is the intrinsic rate of increase (r) of the population?
Answer:

Question 7.
Name important defence mechanisms in plants against herbivory.
Answer:
Defence is of 2 types Morphological and chemical Defences.
Morphological defence –

• Thoms, spines, prickles, bristles
• Stinging hair, sticky hair, hairy coating
• Harbowing ants
• Resemblance to dreaded animals.

Chemical Defences

• Latex
• Alkaloids
• Tannins
• Silica
• Bitter taste, Offensive smell
• Irritating substance
• Poisonous cardiac glycosides.

Question 8.
An orchid plant is growing on the branch of mango tree. How do you describe this interaction between the orchid and the mango tree?
Answer:
Commensalism – orchid being space parasite benefitting from mango plant in obtaining space for residence but mango tree is not harmed.

Question 9.
What is the ecological principle behind the biological control method of managing with pest insects?
Answer:
The ecological principle behind control of pest insects is controlling growth of their population through natural enemies, predator, parasites.

Question 10.
Distinguish between the following:
(a) Hibernation and Aestivation
(b) Ectotherms and Endotherms
Answer:

Hibernation	Aestivation
(i) Condition of passing the winter in a resting or dormant condition.	(i) State of inactivity during hot and dry summer.
(ii) Animal rests in warm place	(ii) Animal rests in cool, shady, moist place.
(iii) Lasts for whole winter.	(iii) Lasts for hot dry day because nights are often cooler

(b)

Ectotherms/Cold Blooded	Entotherms/Wann Blooded
(i) Unable to regulated their body temperature.	(i) Able to regulate their body temperature.
(ii) Lives in areas with favourable body temperature as they change their body temperature with respect to environmental temperature	(ii) Live in all places as there is only little effect on their body temperature by environmental temperature.
(iii) They show hibernation, aestivallon and are less active.	(iii) More active and no such inactive states are expressed

Question 11.
Write a short note on
(a) Adaptations of desert plants and animals
Answer:
(a)
(1) Plants vegetation is sparse, grows during rains only. (Drought exoping)
(2) Annuals (Drought evaders)
(3) Succulents (Drought resisting) – stores water with mucilage and latex.
(4) Non succulent perennials (Drought enduring) – long roots and avoids transpiration by modification.
(5) Modifications

• leaves as spines and thorns
• Stems as water storage body with thick bark
• Long, deep roots with shoot system

Animals

• Lives in burrows
• Very little water requirement and can store water in modified body parts.
• Urine is concentrated and faeces is nearly solid with very little water and metabolic water is also used by them as source of water.
• Nasal counter- Current mechanism preventloss of water during breathing with small nostril size.
• Feeds on dry seeds and rehydrates and stores water when they reach a water body.
• Very small body surface are a so very less sweating so less water loss from body.

(b) Adaptations of plants to water scarcity
Answer:
same answer as above (drought/desert animals or plants).

(c) Behavioural adaptations in animals
Answer:
Snakes – in winter, the snakes will come out of their burrows and bask in the sun to warm their body for activity.

Desert lizards keep their body temperature fairly constant by their behaviour changes. When their body temperature drops below comfort zone, they would come out of burrow and bask in sun.

(d) Importance of light to plants
Answer:
In plants light influences photosynthesis growth and reproduction height influences flower induction, seed germination plant movement etc.

(e) Effect of temperature or water scarcity and the adaptations of animals.
Answer:
Animals living in arid areas are adapted to manage with scarcity of available water and reduce water loss to minimum.

Eg: Kangaroo rat feeds on dry seeds. Animals protect themselves from excessive cold by hibernation, thick for, constriction of peripheral blood vessel.

Question 12.
List the various abiotic environmental factors.
Answer:

• Atmospheric factors- Light, temperature, water, wind, humidity.
• Edaphic factors- Soil texture, soil water, soil air, soil pH soil micro organisms, minerals.
• Topographic factors – Slope, altitude, valley.

Question 13.
Give an example for:
(a) An endothermic animal
(b) An ectothermic animal
(c) An organism of benthic zone
Answer:
(a) An endothermic animal – Monkey (mammal)
(b) An endethsmml animal Frog (amphibian), snake (reptile)
(c) An organism of benthic zone – Angle Fish

Question 14.
Define population and community.
Answer:
Population Nearly/relatively permanent aggregation or grouping of interbreeding individuals of the same species present in a particular space or area at a given time who share the same gene pool.

Community (Biotic or Biological Community) Assemblage of different species present in the same area at the same time which show interactions and interdependence through trophic and spatial relationships.

Question 15.
Define the following terms and give one example for each:
(a) Commensalism
(b) Parasitism
(c) Camouflage
(d) Mutualism
(e) Interspecific competition
Answer:
(a) Commensalism:- Interaction where one species is benefitted and the other is neither benefitted nor harmed is called commensalism, eg: Clone Fish living in company of sea Anemones.

(b) Parasitism:- It is interspecific interaction in which small sized organism of one species obtains nourishment and spends a part or whole Of its life on or inside the body of large sized organism of other species, eg: Plasmodium in human, liver fluke in snail and fish.

(c) Camouflage:- Phenomenon of blending with surroundings with similar colour, marking, shape etc. to escape from predators and preys eg: Stick insect, leaf insect.

(d) Mutualism:- Interaction confers benefits on both the interacting species, eg: Lichen (fungus + Algae); Rhizobium and legumes.

(e) Interspecific competition :- struggle against, members of different species for obtaining some resource (food, water etc) eg: roots of plant species in same area of a forest
→ It is a potent force in organic evolution.

Question 16.
With the help of suitable diagram describe the logistic population growth curve.
Answer:

Logistic growth occurs because of natural selection. Here nature selects the filter (Survival of the fittest). Nature has resource to support maximum number of people (Carrying Capacity (K)) beyond which resources won’t support them due to their lock. This explain that log phase follows phase of acceleration. So we get sigmoid curve.

This growth is Verhulst-pearl Logistic Growth
\(D=\frac { dN }{ dt } =rN\left( \frac { K-N }{ K } \right) \)

N=Population density at time
K = Carrying capacity
r = Intrinsic rate of natural increase

Question 17.
Select the statement which explains best parasitism.
(a) One organism is benefited.
(b) Both the organisms are benefited.
(c) One organism is benefited, other is not affected.
(d) One organism is benefited, other is affected,
Answer:
(d) One organism is benefited, other is affected,

Question 18.
List any three important characteristics of a population and explain.
Answer:
Characteristics are: Population density, population growth, age distribution, birth rate, death rate, sex ratio etc.
(1) Population density:- Indicates size of population. It is number of individuals of a population found in a unit area or space at a given time. It is found out by total count
Random sampling.
\(D=\frac { N }{ S } \)

D – Population Density
N – total number of individuals of given area
S- unit of space or area

(2) Population growth:- It is increase in size of the population over a period of time.
N_t = N_o+ (natality +Immigration) – (Mortality +Emigration)
N_t= N_o +B + I-(D + E)
N_t = Population increase after time t;
N_o= Population at t = 0

(3) Age distribution :- Pre-reproductive, reproductive, post – reproductive.
it helps to provide information about the population – Expending, Declining or Stable population.

2nd PUC Biology Organisms and Populations Additional Questions and Answers

2nd PUC Biology Organisms and Populations One Mark Question

Question 1.
Define the term ecology.
Answer:
It is study of the relationships of living organisms with the biotic and abiotic components of the environment.

Question 2.
Name the factors essential for the formation ‘ of major biomes
Answer:

• The annual variation in the intensity and duration of temperature
• The annual variation in precipitation, account for the formation of major biomes.

Question 3.
Which one of the two stenothermals or eurythermals, shows wide range of distribution, on earth and why? (CBSE 2008)
Answer:
Eurythermals show a wide range of distribution as they can tolerate and thrive in a wide range of temperatures.

Question 4.
Between amphibians and birds, which will be stable to cope with global warming? (AI2008)
Answer:
Birds will be stable because they are eurythermals, which can tolerate wide range of temperatures.

Question 5.
Why many of the fresh water animals cannot live for long in sea water or vice versa. (Delhi 2008)
Answer:
This is because of the osmotic changes in the body.

Question 6.
Define homeostasis
Answer:
It is the self regulatory mechanism to maintain the internal environment constant despite changes in the external environment.

Question 7.
How do herbs and shrubs survive under the shadow of big canopied trees in the forest ? (AI 2008)
Answer:
They are adapted to photosynthesis optimally under very low light condition and hence survive under the shadow of big canopied trees.

Question 8.
Why is the polar region not a suitable habitat for tiny humming birds ? (AI 2008)
Answer:
The small birds have larger surface area to body volume ratio and tend to lose heat fast when it is cold outside; consequently they have to spend more energy to generate body heat.

Question 9.
Define the term migration
Answer:
Migration is the temporary, long-distance or short distance movements of animals from the stressful movement to a less stressful or comfortable environment and return to the original habitat when the stressful period is over.

Question 10.
Define the term hibernation.
Answer:
It is the process in which animals spend the winter in an inactive/dormant state,
eg: Frogs, polarbear.

Question 11.
When and why do some animals (frog, polarbear) go into hibernation (Delhi 2008)
Answer:
When the animals (like frog) are not able to migrate from unfavorable or stressful condition/habitat during very cold condition, they hibernate to avoid the stress by escaping in time.

Question 12.
Define the term aestivation?
Answer:
It is the process in which animals spend the dry, hot summer period in an inactive state,
eg: Snails.

Question 13.
When and why do some animals undergo into aestivation?
Answer:
When the snails are not able to migrate from stressful habitat in summers they undergo aestivation to avoid stress by escaping in time.

Question 14.
Define the term adaptation.
Answer:
The development of certain features in an organism in response to particular environment which may improve the chances of survivals are called adaptation.

Question 15.
What is Allen’s rule?
Answer:
Allen’s rule refers to the minimising of heat loss in animals by possessing shorter ears and limbs.

Question 16.
Why do people living in high altitude have a higher red blood cell count or total haemoglobin
Answer:
To compensate for the low atmospheric pressure and low oxygen content of air prevailing at high attitudes, people have higher, RBC count or total haemoglobin.

Question 17.
List any 2 physiological responses that help you to gradually get acclimatised to high attitudes when you go from the plains (Delhi 2008)
Answer:

• Increasing in breathing rate
• Decreasing in binding capacity of haemoglobin.
• Increase in RBC production.

Question 18.
Define Population density.
Answer:
The total number of individuals of a species present per unit area or volume at a given time.

Question 19.
Give an example where population estimation of an organism is done indirectly with out actually counting the organism. (AI2008)
Answer:

• The number of fish caught in a trap indicates the total population density in a lake
• The tiger census is based on the pug marks or faecal pellets.

Question 20.
When does a population growth curve become sigmoid?
Answer:
Sigmoid growth curve represents logistic growth where growth is slow initially and then becomes rapid the curve becomes steady due to resources become tending.

2nd PUC Biology Organisms and Populations Two Marks Questions

Question 1.
Why is thermoregulation more effectively achieved in larger animals than in smaller ones?
Answer:
Larger ones achieve thermoregulation more effectively than smaller ones because of less surface area/volume ratio in larger ones, hence lose body heat slowly.

Question 2.
Which one out of eurythermal or stenothermal species is likely to survive increased temperature?
Answer:
Eurythermal species as they can tolerate wide variations in environmental temperature; but stenothermal ones live with narrow range of temperature only.

Question 3.
Study 3 representative figure of age pyramids

(a). Mention names given to the 3 kinds of age profile (i), (ii), (iii).
Answer:
(i) Expanding population (Triangular)
(ii) Stable population (Bell shaped)
(iii) Declining population (Cube shaped)

(b). Which one of them is ideal for population and why?
Answer:
We take stable population pyramid as ideal because our aim is to stabilize the population.

(c). How do such age profile studies help policy makers get concerned about growing population?
Answer:
Age pyramid/profile can indicate

• Rate of growth of population
• Increase in elderly population and reproductive aged people.

These features help prepare for future planning on:

• Stress of family planning
• To help elderly and needful people to live comfortably.

Question 4.
(a) Write eg: for Verhulst Pearl logistic growth
Answer:
\(\frac { dN }{ dt } =rN\left( \frac { K-N }{ K } \right) \)
r=intrinsic rate of natural increase
N = Population density at time (t)
K = Carrying capacity.

(b) Draw graph for a population whose population density has reached the carrying capacity.
Answer:

(c) Why is logistic growth model considered more realistic one for most animal populations?
Answer:
Logistic growth model is considered realistic in animals because each area has a carrying capacity beyond which environmental resistance grows and restrict their growth. It decreases natality and increases mortality.

(d) Draw growth curve where resources are not limiting growth of population?
Answer:

Question 5.
In adjacent population growth curves

Answer:
Curve (a) – Non-limiting
curve (b) – Limiting and has reached carrying capacity.

(ii) In absence of predators which curve will be demonstrated?
Answer:
Curve (a) because of absence of controlling factor.

Question 6.
Mention any two significant roles predation plays in nature. (AI2008)
Answer:

• Predation is the natural way of energy transfer through trophic levels in a food chain.
• Predation keeps the prey population under control of ecosystem.
• Predation maintains spicies diversity.

Question 7.
List any two adaptive features evolved in parasites enabling them to live successfully on their hosts. (Delhi 2008)
Answer:
The adaptive features of parasites include

• Loss of unnecessary sense organs
• Presence of adhesive organs or suckers to cling on to the host.
• Loss of digestive system
• High reproductive capacity.

Question 8.
Name the four major biomes found in India.
Answer:

• Tropical rain forest
• Tropical deciduous forest
• Desert
• Seacoast

Question 9.
Write the influence of temperature in the life of organism.
Answer:

• The temperature affects the enzyme reaction in bur body.
• The thermal tolerance of organisms determines their geographical distribution.

Question 10.
Differentiate between Euryhaline and ‘ Stenohaline animals.
Answer:
Euryhaline animals Animals which can tolerate a wide range of salinity. Stenohaline animals Animals which can tolerate a narrow range of salinity.

Question 11.
What is the effect of light in the life of animals.
Answer:

• Intensity and duration of light influences the activities like foraging, migratory activities and reproduction.
• light influences the temperature and there by the metabolism of animals.

Question 12.
How do human being maintain constant body temperature?
Ans:

• In summer we sweat profusely; the evaporation of the sweat cools the body and brings the body temperature down.
• In winter, we shiver and that produces more ‘ heat, to raise the body temperature.

Question 13.
Mention 2 adaptations the mammals of colder region have, to minimize the loss of body heat.
Answer:

• Mammals of colder regions have shorter ears and limbs.
• Aquatic mammals of colder seas have thick layer of fat called blubber below the skin which acts as insulator reduces the loss of body heat.

Question 14.
Differentiate between Population size and Population Density.
Answer:

• Population size:- The number of individuals of a species, in a given area at a given time.
• Population density:- The number of individuals of a population per unit area at a given time.

Question 15.
Biomass is a more meaningful measure of population size. Explain.
Answer:

• If the population is huge, the total number is not as easily adaptable measure as counting takes a very long time or counting becomes practically impossible.
• For some ecological investigations, there is no need to know the absolute population size / density.

Question 16.
Define carrying capacity.
Answer:
Carrying capacity is the maximum number of individual of a population that can be sustained by a given habitat.

Question 17.
What is the role of predators in maintaining species diversity in an ecosystem?
Answer:
Predators help in maintaining species diversity in an ecosystem or community by reducing the intensity of competition among the competing prey species.

Question 18.
In what ways the prey species are adapted to escape from predators? Explain with example.
Answer:

• Some species of frogs and insects are cryptically coloured or camouflaged to avoid being detected easily by the predators.
• Some prey species are poisonous and are avoided by the predators.

Question 19.
(a) How is Cuscuta adapted to be a parasitic plant?
(b) Why do cattles avoid browsing on Calotropis plants? Explain. (CBSE foreign 2009)
Answer:
(a)

• Cuscuta has no chlorophyll and leaves.
• It develops haustoria to derive nutrition from the host it parasites.

(b) Calotropis plants possess a highly poisonous cardiac glycoside, which is fatal to animals.

Question 20.
Name two basic types of competition found amongst organisms. Which one of these is more intense and why?
Answer:
The two types of competitions are

• Interspecific competition and
• Intra specific competition

In this intra specific competition is more intense because it is an interaction among individuals of same species for similar requirements.

Question 21.
Explain interference competition with an example.
Answer:
In interference competition, even if the resource is abundant, the feeding efficiency of one species is reduced by the interfering and inhibitory presence of other species.
Eg: The Abingdon tortoise in Galapagos islands become extinct with in a decade due to the introduction of goat in that island.

Question 22.
Differentiate between commensalism and Amensatism.
Answer:

• Commensalism:- It is the interspecific interaction in which one species is benefited while the other is neither benefited nor affected.
• Amensatism:- It is an interspecific interaction in which one species is affected, while the other species is neither benefited or affected.

Question 23.
What is resource partitioning ? Give an example.
Answer:
It is the phenomenons in which two species competing for the same resource could avoid competition by choosing different patterns of foraging and different times of feeding. Eg:- Five closely related species of warblers living on the same tree have been shown to co exist and avoid competition by behavioural differences.

Question 24.
Give the interaction present in each.
(a) Cuckoo lays her eggs in the crow’s nest
(b) Orchid grows on a mango tree
(c) Cuscuta growing on a shoe flower plant
(d) Clown fish living among the tentacles of sea anemone
(e) Ticks live on the skin of dogs
(f) Mycorrhizae living on the roots of higher plants
(g) Sea anemone is often found in the shells of hermit crab (CBSE-Delhi 2008)
Answer:
(a) Brood parasitism
(b) Commensalism
(c) Commensalism
(d) Commensalism
(e) Parasitism
(f) Commensalism
(g) Commensalism.

Question 25.
Name and explain the interaction between fig trees and certain species of wasps.
Answer:
The interaction between fig trees and certain species of wasp is mutualism.

• The female wasp uses the ovary of flowers as a site of oviposition (egg laying) and the developing seeds as flood for the larvae.
• The flower is pollinated when the wasp is searching a place for oviposition.

Question 26.
What type of interaction is there between clown fish and sea anemone. Why?(AI 2008)
Answer:
The interaction between clown fish and sea anemone is commensalism. In this, one is benefitted and the other is neither benefitted nor harmed.
In this relation
(i) The sea anemone has stinging tentacles which keeps away the predators of clown fishes in this clown fish is getting benefit and the sea anemone is not getting any benefit or not harmed.

Question 27.
\(D=\frac { dN }{ dt } =rN\left( 1-\frac { N }{ K } \right) \)
(a) Which type of growth curve does it represents,
(b) What do the following notations represent
(i) \(\frac { dN }{ dt } \) (ii) r (iii) N (iv) K (CBSE 2005)
Answer:
(a) It represents logistic or Verhulst-Pearl logistic growth
(b) \(\frac { dN }{ dt } \) = rate of change in population size
r – intrinsic rate of natural increase
N – Number of ex histing individual
K – Carrying capacity.

Question 28.
Study the population growth curves shown in the diagam

(1) Identify curves ‘a’ and ‘b’
(2) Mentions the conditions responsible for the curves ‘a’ and ‘b’ respectively
(3) Give the equation for the curve ‘b'(CBSE 2006)
Answer:
(1) a – Exponential curve
b – logistic or sigmoid curve

(2)
a – Unlimited food and other resources
b – Limited food and other resources

(3)
\(\frac { dN }{ dt } =rN\left( \frac { K-N }{ K } \right) \)

Question 29.
Study the table given below in regard to population interactions and the questions given below.

[(+) plus – beneficial interaction; (-) minus – detrimental interaction; 0 (zero) – neutral interaction.
(a) Identify the interactions a to e
(b) Explain each one. (CBSE 2008 comptt)
Answer:
A – B
(a) Amensalism – Species A harmed species B unaffected
(b) Predation/parasitism – Species A benefited species B harmed
(c) Competition – Species A and B mutually inhibited
(d) Mutualism – Both A and B benefitted
(e) Commensalism – Species A is benefitted Species B is unaffected.

Question 30.
Is the statement in correctly matched ? If it is incorrect, rewrite the correct form Camoflague : An organism showing resemblance with external environment animals or plants for the purpose of protection
Answer:
Incorrect:- Camoflague is blending with the back ground or surroundings due to similar colours, shape or marketing so as to remain unnoticed by predators (for protection) and prey (advantage).

2nd PUC Biology Organisms and Populations Three Marks Questions

Question 1.
What are Conformers? Why they haven’t evolved mechanisms for temperature regulation or osmoregulation.
Answer:
Conformers are those organism, which change their body temperature, osmolarity etc. according to the external ‘environmental conditions.
(1) During the course of evolution, the costs and benefits of maintaining a constant internal environment are taken into consideration, when it is not advantageous, the organisms have not evolved mechanisms to maintain homeastasis.

(2) Heat loss or gain is a function of surface area and hence very small animals are not found in polar region i.e. they have become distributed according to the external conditions.

(3) Similarly aquatic animals, that are conformers can change the osmolarity of their body fluid according to the surroundings.

Question 2.
How do organisms manage with stressful conditions existing in their habitats for short duration ? Explain with an example.
Answer:
Under such conditions

• The organisms move away from the stressful habitat to hospital area and returned to that area when it becomes normal state. eg: Siberian birds migrates to Bharatpur sanctuary in Rajasthan.
• The organisms under go hiberation if they can’t migrate
  eg: Polarbear.
• In higher plants seeds and some other vegetative reproductive structure serve as means to tide over stress conditions.

Question 3.
Explain different types of interspecific interactions
Answer:
(1) Beneficial interaction:-

• The interspecific interaction which is beneficial to both the species is called mutualism.
• Beneficial to one, but neutral to other.
• Commensalism

(2) Detrimental interaction:-

• Competition is the type of interaction that is detrimental to both the species.
• Parasitism- Beneficial to one and detrimental to the other.
• Amensatism – Detrimental to one and neutral to the other.

(3) Neutral interaction:- Commensalism is a type in which interaction is detrimental one and beneficial to the other.

Question 4.
Differentiate between ectoparasite and endoparasite.
Answer:

Ecto parasit	Endo parasite
Those organisms which live on the surface of the other living organism and derive food from it eg: lice on human	Those organisms which live inside the body of other living organism and derive food from it eg: Plasmodium and liver fluke in man.

Question 5.
What is brood parasitism? Give an example what adaptation has evolved in this phenomenon
Answer:
Brood parasitism is a phenomenon in which one species (parasite) lays eggs in the nest of other species (host) of bird and lets the host incubate them, eg: Cuckoo lays eggs in crow’s nest Cuckoo’s eggs shows resemblance with the eggs of crow in colour size etc which failed to identify by the crow with their eggs.

Question 6.
Name and explain the kind of interaction in the following
(1) Algae and fungi in lichens
(2) Head louse and humans
(3) Hermit crab and sea anemone.
Answer:
(1) Algae and fungi in lichens:-

• This is an example of mutualism in which both are benefited
• Fungi provide protection and helps in absorbing water and minerals where as the algae prepare food for the fungus.

(2) Head louse and human: This is an example of ectoparasitism the parasite gets shelter as well as food from human’s head.

(3) Hermit crab and sea anemone:- This is an example of commensalism in which sea anemone gets benefit for moving to places rich in nutrients while the hermit crab neither benefitted nor harmed.

Question 7.

The above graph shows the organisms response to abiotic environmental factors. Name the categories of animals. Which show the pattern a, b, and c respectively and define each of them.
Answer:
(a) Conformers:
These are organisms which cannot maintain a constant internal environment but change the body temperature and osmolarity of their body fluids according to the external conditions.

(b) Regulators:
These organisms which maintain a relatively constant environment called homeostasis.

(c) Partial regulators:
These are organisms which have evolved the capacity to regulate the internal environment of the body to a limited range of external factors beyond which they become conformers.

Question 8.
Explain the following
(i) Predators are conducts of energy transfer
(ii) Predators are prudent in nature
Answer:
(i)

• A predator in a food chain feeds on the prey and obtains energy
• When the predator becomes a prey to another animal in the food chain, it transfers the energy obtained from ‘its’ prey.

(ii) If a predator is too efficient and overexploits the prey, the prey population becomes extinct. As a result predator will also become extinct due to the lack of food and hence predator are said to be prudent in nature.

Question 9.
The Mediterranean orched Ophrys employs a method called sexual deceit to get the flowers pollinated. Explain the phenomenon.
Answer:
In the flower of Ophrys, one petal resembles the female of a bee species in size, colour markings etc.

• The male bee perceives it as a female and pseudo copulates with it.
• During the process the pollengrains from the anthers become dusted on the body of bees. When this bee is attracted to another flower of this orchid species, the process is repeated and the pollengrains falls on the stigma hence pollination is achieved.

Question 10.
Mention any 4 adaptations seen is Xerophytic plants to cope up with water scarcity.
Answer:

• Xerophytes (Desert plants) have thick cuticle on their leaf surface.
• Leaves are absent or reduced to spines to prevent transpiration.
• Presence of sunken stomata to reduce water loss.
• The photo synthetic function is taken up by green flattened stems.

 

You can Download Chapter 6 Molecular Basis of Inheritance 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 6 Molecular Basis of Inheritance

2nd PUC Biology Molecular Basis of Inheritance Ncert Text Book Questions and Answers

Question 1.
Group the following as nitrogenous bases and nucleosides:
Adenine, Cytidine, Thymine, Guanosine, Uracil and Cytosine.
Answer:
Nitrogenous bases Adenine, Thymine, Uracil, Cytosine. NucleosidesCytidine, Guanosine.

Question 2.
If a double stranded DNA has 20 percent of cytosine, calculate the percent of adenine in the DNA.
Answer:

Question 3.
If the sequence of one strand of DNA is written as follows:
5ATGCATGCATGC ATGCATGCATGC ATGC-3′
Write down the sequence of complementary strand in 5’—>3’ direction.
Answer:
5′ – ATGCATGC ATGC ATGC ATGC ATGC ATGC – 3′ (In question)
complement 3′ – TACG TACG TACG TACG TACG TACG – 5′
complement in 5′ – 3′ direction 5′ – GCAT GCAT GCAT GCAT GCAT GCAT GC AT – 3′.

Question 4.
If coding strand in a transcription unit is written as follows:
5′-ATGCATGCATGCATGCATGCATGCATGC – 3′ Write down the sequence of m-RNA.
Answer:
m RNA carries same code as coding strand except T is replaced by U.
(Question) transcription strand 5′ – ATGC ATGC ATGC ATGC ATGC ATGC ATGC – 3′ m RNA → 5′ – AUGC AUGC AUGC AUGC AUGC AUGC AUGC – 3′.

Question 5.
Which property of DNA double helix led Watson and Crick to hypothesise semi-conservative mode of DNA replication? Explain.
Answer:
2 strands of DNA has complementary base pairs which run in opposite direction. Property was that during DNA replication one strand was conserved while the complementary is formed new.

Question 6.
Depending upon the chemical nature of the template (DNA or RNA) and the nature of nucleic acids synthesised from it (DNA or RNA), list the types of nucleic acid polymerases.
Answer:
DNA Template

• DNA polymerases for DNA replication.
• RNA polymerases for RNA synthesis or transcription

RNA Template

• RNA dependent RNA polymerases for synthesis of RNA is some RNA viruses.
• Reverse transcriptase to synthesize DNA over RNA template.

Question 7.
How did Hershey and Chase differentiate between DNA and protein in their experiment while proving that DNA is the genetic material?
Answer:
They raised 2 types of bacteriophages

• On radioactive phosphorous (³²P)
• On radioactive sulphur (³⁵S).

³⁵S gets into protein and ³²P into DNA When both bacteriophages infected bacteria differently and by shaking them, viral protein coat was separated

After raising these bacteria it was found that those infected with ³²P bacteriophage → radioactivity was found. But with ³⁵S → no radioactivity was found.

Question 8.
Differentiate between the followings:
(a) Repetitive DNA and Satellite DNA
Answer:

Repetitive DNA	Satellite DNA
(i) sequence of N2 bases present in more than 1 copy in a genome.	(i) part of DNA  having repeated short sequences of N2 bases
(ii) Repeated DNA sequence may or may not be present in text.	(ii) Repeated sequence occur in tandem
(iii) Variability may or may not be present	(ii) Variability occurs

(b) mRNA and tRNA
Answer:

m-RNA	t – RNA
(i) Large sized RNA with cap and tail. (ii) Carries codon information	(i)   Small sized with 3-4 loops and a limb (ii) Carries infor­mation for association of AA with anti Condon for incor­poration

(c) Template strand and Coding strand
Answer:

Template strand	Coding strand
(i) The strand of DNA which takes part in transcription.	(i) It does not takes part in transcription.

Question 9.
List two essential roles of ribosome during translation.
Answer:

• Provides sites for attachment of m-RNA and charged t – RNA for polypeptide synthesis.
• One of its r – RNA functions as peptidyl transferase for peptide bond formation.

Question 10.
In the medium where E. coli was growing, lactose was added, which induced the lac operon. Then, why does lac operon shut down some time after the addition of lactose in the medium?
Answer:
Due to decrease in lactose substrate concentration, lac operon shuts down as it is inducible operon.

Question 11.
Explain (in one or two lines) the function of the followings:

(a) Promoter
Answer:
It is a gene that lies near operator which functions as binding site structural genes if operator allows.

(b) tRNA
Answer:
Functions as adapter molecule that picks  up a particular amino – acid from cellular pool and takes same over A site, m – RNA for incorporation into polypeptide

(c) Exons
Answer:
Coding segments present in primary transcript which on splicing by snRNP’s get joined to form functional m-RNA.

Question 12.
Why is the Human Genome project called a mega project?
Answer:
Because it involves

• Sequencing of over 3 × 109 bp
• Identification of all genes and alleles in human genome and their function.
• Storage of data requires storage space equal to 3300 books 1000 pages, each page with 1000 letters.

Question 13.
What is DNA fingerprinting? Mention its applications.
Answer:
DNA fingerprinting is technique of determing similarity and dissimilarity of VNTR’s between 2 samples of DNA so as to bring out relationship if any VNTR’s are specific for each individual. They are derived from parents in 50: 50 ratio.
Applications

• Identification – of criminals
• Paternity and Maternity disputes.
• Migrations
• Human Lineage.

Question 14.
Briefly describe the following:
(a) Transcription
Answer:
Transcription – Formation of-RNA over template of DNA.

The snRNA formed has codon information similar to sense or coding strand of DNA with just U replacing T.

• The DNA strand which function as a template is template or anti-sense strand.
• Only one of the DNA strands is transcribed
• The main enzyme taking part in transcription is RNA polymerase.

(b) Polymorphism
Answer:
Polymorphism It is the occurrence of more than one form of genetic material
Types →

• Allelic polymorphism – multiple alleles in a gene. So it alters structure and function of protein formed.
• SNP or single nucleotide Polymorphism – Unique in every human being. Useful in locating specific alleles, disease associated sequences.
• RFLP or Restriction Fragment Length polymorphism – Different sized fragments are formed by cleavage with same enzyme most RFPL have no effect on phenotypes.

(c) Translation
Answer:
Translation (Biosynthesis of proteins) Coded genetic message brought by m – RNA from DNA is charged into polypeptide chain (proteins).

DNA transcription mRNA translation Proteins Materials required for translation are ribosomes, AA, t RNA’s, aninoacylt – RNA synthetase, m – RNA and same factors Step in Translation . Activation of Amino Acids → charging of t RNA→ Initiation → Elongation → Termination → Modification.

(d) Bioinformatics
Answer:
Science of handling, storing as databases, analysing, modelling and providing access to various aspects of biological information, especially molecules connected with genomies and proteomics
Applications

• Organisation of Biological Data
• Functional Genomics
• Proteomics, Pharmacogenomics
• Medical informatics, chemoinformatics
• Faster drey research.

2nd PUC Biology Molecular Basis of Inheritance Additional Questions and Answers

2nd PUC Biology Molecular Basis of Inheritance One Mark Questions

Question 1.
Expand the term DNA, RNA and tRNA.
Answer:
DNA – Deoxy ribonucleic acid RNA – Ribo nucleic acid tRNA – transfer – RNA

Question 2.
Define genetic code.
Answer:
Relationship between the sequence of nucleotides or bases on mRNA and the sequence of amino acids in the polypeptide.

Question 3.
What are Okazaki fragments?
Answer:
The small stretches of DNA formed due to the opposite running of DNA template at the time of replication is called Okazaki fragments.

Question 4.
What is codon?
Answer:
It is the triplet nitrogenous base sequence which code for one amino acid. It lies on the m- RNA.

Question 5.
What is anti codon?
Answer:
It in the triplet nitrogenous bases in the t-RNA complementary to an m-RNA codon. It identifies a particular codon on mRNA.

Question 6.
What is a translation?
Answer:
It is the process of the formation of protein directed by a m-RNA molecule.

Question 7.
Define Transcription?
Answer:
It is the process of the formation of m-RNA on a DNA template.

Question 8.
Name the enzyme that catalyses
(a) Replication of DNA and
(b) Formation of RNA. (CBSE 1995)
Answer:
(a) Topoisomerase.
(b) RNA polymerase.

Question 9.
What is the genetic material of tobacco mosaic virus (TMV)
Answer:
RNA.

Question 10.
Name the bond present between two adjacent nucleotides
Answer:
Phosphodiester bond.

Question 11.
Due to mistake during transcription, ATG forms UAG in m-RNA. What change would occur in polypeptide chain translated by this m-RNA? (Cbse 1996)
Answer:
UAG is a termination codon and so at point protein synthesis will get stopped.

Question 12.
Name three different non – sense codons.
Answer:
UAA, UAG and UGA.

Question 13.
If the base sequence of one strand of DNA is CAT TAG TAC GAC, what will be the base sequence
(a) Of complementary DNA strand, and
(b) Of its complementary RNA strand? (CBSE 1991)
Answer:
(a) GTA ATC ATG CTG
(b) GUA AUCAUGCUG.

Question 14.
Name the scientist who proposed one gene one enzyme hypothesis. (CBSE 1997)
Answer:
Beadle and Tatum.

Question 15.
Name the enzyme that joints the short pieces in the lagging strand during synthesis of DNA.
Answer:
DNA legase. (CBSE 1998)

Question 16.
In which direction 5′ – 3′ or 3′ -5′ are the new strands of DNA formed during replication ? (CBSE 1992, 2K)
Answer:
5′ – 3′ direction

Question 17.
Give the present-day representation of central dogma.
Answer:

Question 18.
State Chargaff s base complementary rule.
Answer:
The total molar amount of adenine in any specimen of DNA is always equal to that of thymine. In a given DNA A = T and G = C.

Question 19.
Define mutation
Answer:
It is an abrupt and distinct change in the structure of base pair. It is a discontinuous inheritable and sudden change in an organism.

Question 20.
What do you call the kind of mutation in which a single base is added to a base strand : (CBSE 2K)
Answer:
Frame shift mutation.

Question 21.
Sickle cell anaemia is caused due to abnormal haemoglobin. Which chain of haemoglobin is responsible for this disease?
Answer:
β – chain of haemoglobin

Question 22.
What do the triplets AUG and UGA respectively code for during proteins synthesis?
Answer:
AUG – Methionine UGA – Termination codon (Nonsense codon)

Question 23.
Name the technique used by Watson and Crick to propose the double-helical structure of DNA molecule?
Answer:

• X-ray crystallography
• X-ray diffraction method.

Question 24.
Who discovered nucleic acid DNA? What was it called them?
Answer:
Fredrich Meischer. It was called as nuclein.

Question 25.
(a) What is the length of the pitch of helix?
(b) What is the distance between 2 base pairs is a stand of DNA?
Answer:
(a) 3.4 nm
(b) 34 nm.

Question 26.
Why is the distance between the 2 nucleotide chains in a DNA maintained almost constant?
Answer:
The complementary base pairing between the 2 strands i.e. adenine and thymine (double bond) and Guanine and cytosine (triple bond) is responsible for the uniform distance between the 2 strands.

Question 27.
Name the process which occurs in virus where the formation of DNA occurs from RNA.
Answer:
Reverse transcription.

Question 28.
Name the components ‘a’ and ‘b’ in the nucleotides with a parent given below. (CBSE, Delhi 2008)
Answer:

Question 29.
What are histones?
Answer:
Histones are positively charges proteins found in association with DNA in eukaryotic cell.

Question 30.
Name 2 organisms where RNA as the genetic material.
Answer:
QB Bacteriophage, Tobacco mosaic virus.

Question 31.
Name the main enzyme involved in the replication of DNA.
Answer:
DNA dependent DNA polymerase enzyme.

Question 32.
Name the types of synthesis ‘a’ and ‘b’ occurring in the replication of DNA as shown below (CBSE 2008)
Answer:

• a – continuous synthesis
• b – discontinuous synthesis.

Question 33.
What is replication fork in DNA?
Answer:
During DNA replication the unwinding of DNA leads to the formation of ‘Y’ shaped structure to the 2 strands of DNA duplex. This is known as the replication fork.

Question 34.
Define a cistron
Answer:
A cistron is defined as the length of mRNA, that codes for a polypeptide.

Question 35.
What is meant by hnRNA?
Answer:
The hnRNA is the precursor of mRNA transcribed by RNA polymerase II in eukaryotic cells hn represents heteronuclear RNA.

Question 36.
When and at what end does the ‘tailing’ of hnRNA take place? (AI 2009)
Answer:
After splicing, tailing occurs at the 3′ end of hn RNA.

Question 37.
Why is hn RNA required to undergo splicing?
Answer:
Since hn RNA contain both the coding sequences (exon) and the non-coding sequences (introns), hn RNA has to undergo splicing for the removal of introns.

Question 38.
Who proposed the operon concept?
Answer:
Francois Jacob and Jacque Monod proposed the operon concept.

Question 39.
Name the induce of lac operon in E. Coli?
Answer:
Lactose.

Question 40.
Given below is a schematic representation of a lac operon in the absence of an inducer. Identify ‘a’ and ‘b’ in it
Answer:

a – Repressor
b – operator.

Question 41.
What term is given to a single base DNA differences?
Answer:
When the repressor binds to the operator, the operon is switched off and transcription is stopped.

Question 42.
Expand VNTR.
Answer:
Variable Number of Tandem Repeats.

Question 43.
Regulation of lac operon by repressor is referred to as negative why is it so?
Answer:
When the repressor binds to the operator, the operon is switched off and transcription is stopped.

Question 44.
Name 2 plants whose genome have been sequenced.
Answer:
Rice and Arabidopsis.

Question 45.
Define DNA polymorphism
Answer:
Inheritable mutations at high frequency in a population. It refers to the variation at genetic level.

Question 46.
Who discovered the techniques of DNA fingerprinting
Answer:
Alec Jeffreys.

Question 47.
What is a probe in DNA-finger printing?
Answer:
A probe is a short stretch of DNA, with the nucleotide sequence that is complementary to that of VNTR sequence.

Question 48.
Name the branch of science that HGP is closely associated with.
Answer:
Bioinformatics.

Question 49.
Given below is the sequence of steps of transcription in eukaryotic cell. Fill up the blank 1, 2, 3, 4 left in the sequence.
Answer:

Single nucleotide polymorphism (SNPS).
1 – RNA polymerase
2 – hn
3 – processed RNA
4 – tail.

Question 50.
Write any 3 unusual bases present in Yeast’s alanine tRNA with their sources.
(CBSE 2004)
Answer:
All tRNA have 2 unusual bases – dihydro uridine (derived from uracil) and pseudo uridine (from uraicl. The third common unusual base is hypoxanthine (from adenine).

2nd PUC Biology Molecular Basis of Inheritance Two Marks Questions

Question 1.
What are 5′ end, 3′ end of a polynucleotide chain?
Answer:

• The poly nucleotide chain has at one end a free phosphate at the 5′ end of the pentose sugar. Which is referred to as 5′ end.
• The other end of polynucleotide chain has a free. 3′ – OH group at the 3rd end of the pentose sugar which is referred to as 3′ end of polynucleotide chain.

Question 2.
Label the diagram 1, 2, 3, 4, 5, 6. (CBSE 2004)
Answer:

• 5′ end
• Ribosome binding site
• Start
• Stop signal
• Open reading frames (ORFs)
• 3′ end.

Question 3.
Name the components (parts) A and B of the transcription unit ‘given below.
Answer:

A – Promoter
B – Coding strand.

Question 4.
(a) Who first proposed semi-conservative mode of replication of DNA.
(b) Which organisms is used in this experiment?
(c) Name the techniques used
(d) What is the result of first, second and 3rd generation?
Answer:
(a) Watson and Crick (1953)

• Eschirisia Coli

(i) Use heavy isotope ¹⁵N instead of normal ¹⁴
(ii) Use of Cesium chloride based density gradient configuration with ethidium bromide as flarochrome.
(iii) First generation DNA is hybrid as with intermediate heaviness due to presence of both ¹⁴N and ¹⁵N strands. The parent generation is heaviest.
Second generation 50% light (¹⁴N/¹⁴N) and 50% intermediate heavy (¹⁴N/¹⁵N). Third generation75% light (¹⁴N/¹⁴N) and 25% of intermediate heaviness (¹⁴N/¹⁵N).          This is possible only if DNA is double stranded with semi conservative replication. (one parent strand and other new strand) UACGAG AGAUUUi

Question 5.

Study the messenger RNA segment given above which is complete to be translated into a polypeptide chain

• Write the codons and ‘a’ and ‘b’
• What do they code for?
• How is peptide bond formed between 2 amino acids in the ribosome (CBSE 2008)

Answer:
(i) a – AUG
b – UAA | UAG | UGA

(ii) AUG codes for methionine.
UAA/ UAG/ UGA is stop / nonsense codon.

(iii) Peptide bond is formed between -COOH group of P- site amino acid and NH₂ – group of A – site amino acid with the help of ribozine peptidyl transferase provided by ribosome.

Question 6.
State any one reason to explain why RNA virus mutate and evolve faster than other viruses.
Answer:
RNA is an unstable highly reactive molecule due to its single-stranded structure and exposure of its nitrogen bases. But DNA is stable, the molecule, as its, nitrogen bases are not exposed, became its double-helical nature.

Repressor binds to the operator region
(o)  and prevent RNA polymerase from transcribing the given.
Look at the figure above depicting lac operon in E-Coii.
(a) What could be a series of events when an inducer is present in the medium in which E. Coli is growing.
(b) Name the inducer.
Answer:
(a) When the inducer is present in the medium it is absorbed at first slowly into the bacterium. The inducer binds with the repressor attached to the operator gene the repressor leaves the operator gene and allows the RNA polymerase to pass from promotor to the structural genes for transcription or formation of a polycistronic mRNA.
(b) Lactose or galactoside.

Question 7.
Draw schematically a single polynucleotide strand (with at least three nucleotides)
Answer:

Question 8.
Differentiate between euchromatin and heterochromatin.
Answer:
Euchromation

• These are the regions where chromatin is loosely packed.
• Euchromation stains lighter
• This is transcriptionally more active.

Hetero chromatin

• These are the regions where chromatin is tightly packed
• Hetero chromatin stains darker
• This is transcriptionally less active or inert.

Question 9.
Write the functions of DNA polymerase.
Answer:

• DNA polymerase catalyses the polymerisation of nucleotides into a polynucleotide strand of DNA.
• It carries out proof reacting in prokaryotes; it moves back and removes any wrong bases added before polymerisation continues.

Question 10.
Compare the roles of the enzymes DNA polymerase and DNA ligase in the replication fork of DNA.
Answer:
DNA polymerases the nucleotides in the 5’ —> 3′ direction, as a continuous stretch on the template strand with 3′ —> 5′ polarity and short stretches on the template strand with 5′ —> 3′ polarity. —> DNA ligase joins the short stretches of DNA formed on the template strand with 5′-3′ polarity.

Question 11.
Write the differences between mono-cistronic and polycistronic mRNAs.
Answer:

Monocistromic mRNA	Polycistromic mRNA
(1) It is the mRNA that can code for only one polypeptide i.e. it has one cistron. (2) it is normally found in eukaryotic cell.	(1) It is the mRNA that can code for more than one polypeptide i.e. it has more than one cistron. (2) It is found in prokaryotic cell.

Question 12.
Differentiate between Exons and Introns.
Answer:
Exons (HOTS)

• They are the coding sequence of DNA/RNA transcript, that form parts of mRNA and code for different regions of the polypeptide.
• They are joined together during splicing to make the information continuous.

Introns

• Introns are non coding sequences of DNA/ RNA transcript that do not become part of mRNA.
• They are removed during splicing.

Question 13.
Genetic code is specific and nearly universal justify. (Delhi 2008)
Answer:
Since a codon codes only for a particular amino acid, genetic code is specific or unambiguous. A codon codes for one particular amino acid in all living, be it a bacterium or a human, hence it is universal.

Question 14.
Why is that transcription and translation can be coupled in prokaryotic cell but not in eukaryotic cells?
Answer:
(a) In prokaryotes, mRNA does not require any processing to become active

(b) Transcription and translation occur in the same compartment cytosol, as there is no well defined nuclear membrane. There for it can be coupled.

In eukaryotic m-RNA has to be processed (splicing) before it become active. Since RNA is synthesised inside the nucleus and translation occurs in the cytoplasm, coupling of transcription and translation is not possible.

Question 15.
Why is it essential that RNA binds to both an amino acid and an mRNA codon, during protein synthesis?
Answer:
(a) By binding an amino acid at its 3′ end the tRNA transports the amino acid to the site of protein synthesis.

(b) It binds to the codon on mRNA through hydrogen bonds with its anticodon as it is recognised by the codon of the amino acid it carries, it is to release the amino acid at the correct site for polypeptide elongation.

Question 16.
What is an operon? How does lactose act as an inducer in the lac operon? (Al 2008)
Answer:
All the genes controlling a metabolic pathway, collectively constitute an operon. Lactose binds to the repressor and inactivates it, and prevents it from binding to the operator. As a result, the RNA polymerase gets access to the promoter and transcription proceeds, ie., the operon is induced to function.

Question 17.
Explain VNTR as the basis of DNA finger printing.
Answer:

• Variable number of tandem repeats belong to the class of satellite DNA referred to as mini satellite.
• The number of repeats shows very high degree of polymorphism as a result the size of VNTR varies.
• After hybridisation of DNA sample with VNTR probe, an autoradiogram developed which gives many bands give the characteristic pattern of an individual DNA.

Question 18.
Give the applications of DNA fingerprinting?
Answer:

• To identify criminals in the forensic labs.
• To determine the biological parent in case of dispute.
• To verify whether an immigrant is really a close relative of the mentioned resident.
• To identify racial groups to rewrite the biological evolution.

Question 19.
How is nucleosome formed? Draw a diagram of the nucleosome.
Answer:

In eukaryotes, histones which are positively charged proteins, become organised as a unit of 8 molecules, called histone octamer. The negatively charged DNA is wraped around the positively charged histone octamer, to form the structure, called nucleosome.

Question 20.
What is meant by R – cells and S – cell with which Frederick Griffith carried out his experiments on Diplococcus pneumonial? What did he prove from these experiments?
Answer:
R – cells are those bacterial cells that form rough colonies, without a capsule and are non-virulent. S – cells are those cells, that form smooth colonies, with a capsule and are virulent. He proved that virulence of S – cells had some how been transferred into R – cells, which became transformed into S- cells, it is the genetic material that had effected transformation.

Question 21.
Draw the schematic representation to show continuous or discontinuous synthesis of DNA (replication of DNA) and label it.
Answer:

Question 22.
Explain frames shift mutation.
Answer:
(a) It is the type of mutation where insertion or deletion of one or two bases changes the reacting frame from the point of insertion or deletion.

(b) When three or multiples of three bases are added there is addition of one or more amino acids and the reacting of the frame remains unaltered after that. This provides the genetic proof that codons are triplets.

Question 23.
A t-RNA is charged with the amino acid phenylalanine.
(a) At what end of the tRNA is the amino acid attached ?
(b) Give the mRNA codon that codes for phenylalanine.
(c) Which enzyme is responsible for this attachment ?
Answer:
(a) Amino acid is attached to the 3′ end of t-RNA.
(b) UUC or UUU are the codes for phenylalanine.
(c) Amino Acyl tRNA synthetase enzyme is responsible for this attachment.

Question 24.
(a) Draw the schematic diagram of tRNA showing the following.
(i) Methionine attached to the amino acid accept site.
(ii) Correct base sequence at the anticodon loop
(b) Write the role of “untranslated regions” on mRNA segment play in protein synthesis?
Answer:
(a)

(b) They are needed for efficient translation

Question 25.
Draw a labelled diagram depiciting schematically the process of elongation in translation
Answer:

Question 26.
Describe the goals of Human genoine project.
Answer:
The major goals of Human genome project are

• Determine the sequence of the 3 billion base pairs present in Human DNA.
• Identify the genes in the human DNA.
• Store the information in data bases.
• Improve the tool for data analyses.
• Address ELSI (ethical, leagal, social issues) that arise from project.
• Transfer the technologies to other sectors.

2nd PUC Biology Molecular Basis of Inheritance Five Marks Questions

Question 1.
Describe the features of double helical model of DNA.
Answer:
The features of double helical DNA are

• It is made up of 2 polynucleotide of sugar phosphate and the nitrogen bases inside.
• The two chains have antiparallel polarity ie., one has 5′ —> 3’ polarity and the other with 3′ —> 5′ polarity.
• The bases of 2 strands are joined by double hydrogen bond between adenine and thymine and triple hydrogen bond between guanine cytosine.
• The distance between 2 base pairs is 0.34nm and the distance between each turn is 3.4nm. Each turn consists of 10 base pairs.
• The plane of one base pair stacks over the other in double helen this gives the stability of double helical structure.

Question 2.
(1) Represent diagramatically the Watson Crick model for semi conservation replication of DNA.
(2) Differentiate between continuous and discontinuous synthesis of DNA.
Answer:

(2)

Continuous	Discontinuous
(a) One stand of DNA is synthesised as a continuous strech in the 5′ —>3′ direction.	(a) Short stretches  are synthesised in the 5’—>3′ direction from replication fork
(b) The template strand of DNA strand is with 3’—>5′ polarity.	(b) The DNA strand with 5′ —> 3′ polarity is the template strand  for this.
(c) No need of enzyme ligase (for joining)	(c) DNA ligase  enzyme is required  for joining short stretches.
(d) There is no need for primers.	(d) There is need for primers.

Question 3.
Write short notes on different types of RNAs.
Answer:
The RNAs are of 3 types

• Messenger RNA (m – RNA)
• Transfer RNA (t – RNA)
• Ribosomal RNA (r – RNA)

m – RNA It provides the template for polypeptide synthesis, ie., it decides the sequence of amino acids in the polypeptide through the sequence of bases on it. t – RNA It has the shape of clover leaf in two dimensional structure It transports the amino acids to the site of protein synthesis it recognises the codon on m RNA. r – RNA It plays the structural and catalytic role during translation.

Question 4.
Define genetic code and write its salient features.
Answer:
Genetic code is the relationship between the sequence of nucleotides on mRNA and sequence of amino acids in the polypeptide.
Its features are

• The genetic code is universal ie., the codons code for one amino acid is same in all organism.
• Codons are triplet codons and there are 64 codons. 61 codons code for twenty different amino acids while the other three (UAA, UAG, UGA) are termination codons which do not code for any amino acids.
• Each codon codes for only one particular amino acid. Therefore it is unambiguous.
• Some amino acids are coded by more than one codon there for it is said to be degenerate.
• The codons are read in a continuous manner, without any punctuation ie., codons are comma less.
• AUG has dual functions of coding for metheonine as well as functioning as initiation codon.

Question 5.
(1) Represent schematically the process of transcription in eukaryotic cells.
(2) How does DNA polymerase function in the replication fork of DNA?
Answer:

(2) The DNA polymerase can catalase the polymerisation of nucleotides only in one direction is 5′ —> 3′ direction.
On the template stand with 3’—>5′ polarity the polymerisation occurs continuously and on the template strand with 5′ —> 3′ polarity, polymerisation occurs in short stretches (discontinuous synthesis).

Question 6.
(a) Explain with the help of schematic representation, the lac operon in E-Coli.
(b) Mention the role of lactose in this operon.
Answer:
(a) Jacob and Monod explained that lactose inducer the expression of genes leading to its catabolism.

(b) Role of lactose

• Lactose is a substrate for the enzyme (3- galactosidase.
• It function as the inducer and regulates the switching on and off of the operon.
• When lactose is present, it combines with the repressor protein which otherwise has a high affinity for the operator.
• This inactivates the repressor from binding to the operator and hence transcription continuous ie., the operon is switched on.

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 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 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.

 

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