2nd PUC Biology Model Question Paper 1 with Answers

Students can Download 2nd PUC Biology Model Question Paper 1 with Answers, Karnataka 2nd PUC Biology Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Biology Model Question Paper 1 with Answers

Time: 3.15 Hours
Max Marks: 70

General Instructions:

  1. This question paper consists of four parts A, B, C and D. Part D consists of two parts, Section – I and Section – II.
  2. All the parts are Compulsory.
  3. Draw diagrams wherever necessary. Unlabelled diagrams or illustrations do not attract any marks.

Part – A

I. Answer the following questions in One Word or One Sentence each : ( 10 × 1 = 10 )

Question 1.
Define parthenogenesis.
Answer:
Development of the egg into a individual without fertilization.

Question 2.
What are meiocytes?
Answer:
Gamete mother cells / cells which undergo meiosis to form gametes.

Question 3.
How does repressor protein prevent the transcription of structural genes?
Answer:
Repressor protein bends to the operator region of the operon and prevents RNA polymerase from transcribing the operon.

Question 4.
Electrophoresis is an essential procedure that needs to be employed in genetic fingerprinting why?
Answer:
Required to separate DNA fragments obtained restriction enzyme digestion.

KSEEB Solutions

Question 5.
What are biofertilisers?
Answer:
Biofertilizers are organisms that enrich the nutrient quality of soil.

Question 6.
Restriction enzymes are considered as a type of endonucleases. Why?
Answer:
Restriction endonucleases make cuts at specific positions within the DNA.

Question 7.
State Gause’s competitive exclusion principle.
Answer:
Two closely related species competing for the same resources cannot co – exist indefinitely and the competitively inferior one will be eventually eliminated.

Question 8.
Why is pyramid of energy always upright?
Answer:
When energy flows from a particular tropic level to the next trophic level, some energy is always lost as heat at each step.

Question 9.
What are ‘biodiversity hotspots’?
Answer:
‘Biodiversity hotspots’ are regions with very high levels of species richness and high degree of endemism.

KSEEB Solutions

Question 10.
Ozone layer in the stratosphere becomes thinner due to the release of CFCs. Give a scientific reason for this.
Answer:
UV rays at on CFCs and release Cl atoms which act as catalysts, degrade, ozone, and release molecular oxygen.

Part – B

II. Answer any FIVE of the following questions in 3 – 5 sentences each, wherever applicable : ( 5 × 2 = 10 )

Question 11.
‘Unless foetal ejection reflex is produced, normal parturition does not occur’. Substanate the statement.
Answer:

  • Stimulation of pituitary to release oxytocin followed by stronger contractions of uterine muscles due to oxytocin.
  • Further secretion of oxytocin resulting in stronger and stronger contractions that causes the expulsion of the foetus the uterus.

Question 12.
What is infertility? Mention one Assited Reproductive Technology (ART).
Answer:
Infertility:
Inability to produce children (conceive) in spite of unprotected sex.
assisted reproductive technology. EVF – ET / ZIFT / GIFT / ICSI / Artificial insemination.

KSEEB Solutions

Question 13.
What are multiple alleles? Why the alleles IA and IB for blood group are considered codominant?
Answer:
Multiple alleles:

  • Alleles which occur in more than two alternate forms.
  • When IA and IB are present together in an individual, both A and B types of sugar polymers are produced / Both the alleles, express when they are together.

Question 14.
What is the karyotype in Turner’s syndrome? Mention two symptoms of the syndrome.
Answer:
Turner’s syndrome:

  • Karyotype is 45 chromsomes with XO / 44A – XO

Symptoms:

  • Lack of female secondary sexual characters.
  • sterility due to rudimentary ovaries
  • Short stature.

Question 15.
Mention the function of RNA polymerase I and RNA polymerase II in eukaryotes.
Answer:

  • RNA polymerase I – Catalyses the synthesis of rRNA (285, 385 & 5.85)
  • RNA polymerase II – catalyses the synthesis of reterogeneous nuclear RNA / hn RNA / precursor mRNA.

KSEEB Solutions

Question 16.
Explain two salient features of genetic code.
Answer:

  1. Genetic code is triplet: Each codon consists of sequence of three nitrogen bases.
  2. Genetic code is universal: A particular codon codes for the same amino acid in all organisms.
  3. Genetic code is non overlapping: The successive triplet codons are read in order without overlapping and they do not share any base.
  4. Genetic code is degenerate: A single amino acid is coded by more than one codon.
    Eg: valine is coded by 4 different codons GUA, GUC, GUU and GUG
  5. Genetic code is commaless: Codons are without punctuation and written in linear form. There is no signal to indicate the end of one codon or beginning of the next codon.
  6. Genetic code is non-ambiguous:
    Each codon specifies a particular amino acid in all organisms.
    Eg: AUG codes for methionine
  7. Initiator codons: Protein synthesis is always initiated by particular codons called initiator codons.
    Eg: AUG in eukaryotes, GUG in prokaryotes
  8. Terminator codons: Three codons that act as stop signals to terminate protein synthesis are called terminator codons or nonsense codons.
    E.gf.: UAA (Ochre), UGA (Amber) and UGA (Opal).

Question 17.
Differentiate homologous and analogous organs.
Answer:
Homologous:
Organs or structures that have the same origin and basic anatomical structure but perform different functions.

Analogous organs :
Structures that are different in their basic anatomical structure and origin but appear but similar and perform similar functions.

KSEEB Solutions

Question 18.
Define gene therapy. Name a genetic disorder that is being treated using the technique of gene therapy
Answer:
Gene therapy :
Gene therapy is the replacement of a defective mutant allele with a functional gene into an individual cell and tissues to treat genetic hereditary diseases.
OR
It is the insertion of normal functional genes into the individual or embryo with genetic defect to take over the function of and compensate for the non – functional gene.

Part – C

III. Answer any FIVE of the following Questions in 40-80 words each, wherever applicable. ( 5 × 3 = 15 )

Question 19.
Mention three differences between asexual and sexual reproduction.
Answer:
Asexual reproduction

  1. It is uniparental.
  2. Single parent is involved.
  3. No formation of male and female gametes.
  4. No fusion of gametes.
  5. Produces offspring that are identical to their parents.
  6. Offspring do not show variations.
  7. Offspring are not suitable for natural selection.
  8. Offspring have no better chances, of survival.

Sexual reproduction

  1. It is biparental.
  2. Two parents are involved.
  3. Male and female gametes are formed.
  4. Involves fusion of male and female gametes.
  5. Offspring are generally different from their parents.
  6. Offspring show variations due to genetic recombination.
  7. Offspring are suitable for natural selection since they show variations.
  8. Offspring have better chances of survival.

KSEEB Solutions

Question 20.
Explain the structure of a mature female gametophyte in flowering plants.
Answer:

  • Egg apparatus at the microphylar end, three antipodals at the cralazal end and a central cell.
  • Egg apparatus consists of one egg cell and two synergids which have allular thickenings called filiform apparatus at the microplayer tip.
  • Central cell with two polar nuclei.

Question 21.
How do intra – uterine devices prevent conception in humans?

  • Increase phagacytosis of sperms within the uterus.
  • Copper ions released by them decrease motility and fertilizing capacity of sperms.
  • Hormone releasing IUDS prevent implantation.
  • Hormone releasing IUDS also make the cervix hostile for sperms.

Question 22.
State Hardy – weinberg law. Write the role of any two factors that affect Hardy – weinberg equilibrium.
Answer:
Allele frequencies in a population are stable and are constant from generation to generation.
OR
Allele frequency or gene frequency in a population (gene pool) remains constant from generation to generation unless there are factors to upset it.

Role of factors that affect Hardy – weinberg equilibrium.
Gene flow: It is the movement of genes or alleles from one population to another when there is a migration of a section of population to another place and population this changes gene frequencies.

Genetic drift: It is the random ‘ change in the gene frequency by chance (in small populations). The change in gene frequency in the new population may be so different that the members become a different species (founder’s effect).

Mutation: It is the sudden hertitable change in genetic material which results in new genotypes and phenotypes, thus altering the gene frequency.

Genetic recombination: It occurs during gametogenesis (meiosis) and results in variations. This would result in gene frequency over a period of time.

Natural selection: Variations that occur as a result of mutations or recombinations or due to gene flow or genetic drift inhance reproductive success or fitness survivors would – leave more progeny and over a period of time the gene frequency of the population change.

KSEEB Solutions

Question 23.
What are carcinogens? Mention any two groups of carcinogens with an example for each.
Answer:
Carcinogens:
Agents that cause cancer are called carcinogens.

Groups of carcenogens:
Physical agents : X – rays, gamma rays, UV rays
Chemical agents: Chemical present in tobacco smoke.
Biological agents: Oncogenic viruses.

Question 24.
Differentiate out – crossing, cross breeding and inter specific hybridization.
Answer:
Out – crossing : It is the mating of the animals of the same breed but which have no common ancustors on either side of the pedigree up to 4 – 6 generations.

Cross breeding : It is the mating of superior males of one breed with superior females of another breed both with desirable characters).

Inter – specific hybridization : It is the mating of male and females of two different but related species.

Question 25.
Write a note an Bt toxin.
Answer:
Bt toxin:

  • Bt toxin is an insecticidal protein formed in the form of crystals during a particular phase of the growth of Bacillus thuringensis.
  • It exists are inactive protoxin in bacteria and is converted into an active form of toxin in insect due to the alkaline pH of the gut.
  • Active toxin binds to the surface of epithelial cells of the midgut and creates pores which cause cell swelling and lysis and eventually the death of the insect.

KSEEB Solutions

Question 26.
Explain any three major causes of biodiversity loss.
Answer:
1. Habitat loss and fragmentation:
Many large habitats like forests have been destroyed by Man either for cultivation or for conversion to grasslands and broken up into small fragments which leads to the decline in the population of living organsims.

2. Over – Exploitation:
Over – Exploitation of natural resources for the purpose of food, shelter and other purposes and harvesting of many mareine fish populations by humans leads to the elimination of many species.

3. Allien species invasions:
Unintentional or deliberate introduction of allien species may cause the decline or extinction of indigenous species as they may turn invasive.

4. Co – extinctions:
When a species becomes extinct, the plant and animal species associated with it in an obligatory way also becomes estinct.

Part – D

Section – I

IV. Answer any FOUR of the following questions on 200 – 250 words each, wherever applicable. ( 4 × 5 = 20 )

Question 27.
What is autogamy? Explain the devices that the plants have developed to prevent this.
Answer:
Autogamy:
Self pollination in which there is transfer of pollen grains from the anters to the stigma at the same flower.

Devices in flowering plants to prevent autogamy:
1. Release of pollen grains before the stigma becomes receptive or stigma becomes receptive much before the release of pollen grains.
2. Placement of anthers and stigma at different positions so that pollen cannot come in contact with the stigma of the same flower.
3. Self incompatibility – Genetic mechanism which inhibits the germination of pollen grain or growth of the pollen tube inside the style, thus preventing the fertilization process.
4. Unsexually : Production to unisexual flowers (male and female flowers on different plants or on the same plant).

KSEEB Solutions

Question 28.
Explain the different steps involved in translation.
Answer:
Translation in evkaryotes :

  • Charning of tRNA (amninoacylation of tRNA) : activation of animo acids in the presence of ATP and their linking to specific tRNA.
  • Binding of the ribosome (small sub unit) to mRNA at the initiator codon (AUG)
  • Binding of the initaitor tRNA carrying to amino acid (methionine) to the initiator codon to initiate protein synthesis.
  • Movement of ribosome from codon to codon along the mRNA and adding of amino acids (linked to tRNA) one by one – elongation of polypeptide chain.
  • Binding of the release factor to the stop codon located at the 3’ end of mRNA, terminating transulation and releasing the polypeptide from ribosome.

Question 29.
Write the schematic representation of the life cycle of HIV.
Answer:
2nd PUC Biology Model Question Paper 1 with Answers 1

KSEEB Solutions

Question 30.
Explain briefly the steps involved in recombinant DNA technology.
Answer:

1. Isolation of the genetic material
The gene of interest is isolated from the nucleus of the donor organism. This involves digestion of cell wall or cell membrane of the donor organism by enzymatic action to obtain pure form of DNA.

2. Cutting of DNA at specific locations

  • The pure DNA obtained is cut into several fragments using restriction endonucleases.
  • The fragments generated by restriction endonucleases are separated according to their size by gel electrophoresis.

3. Ampiflication of gene of interest using PCR

  • Polymerase chain reaction (PCR) is used to generate billions of copies of the desired DNA in vitro. This is called amplification.
  • The desired gene is inserted into a suitable vector (Plasmid).
  • The DNA ligases join the
    complementary ends of plasmid DNA with that of desired gene. The hybrid DNA is called recombinant DNA.

4. Insertion of recombinant DNA into host cell or organism

  • The recombinant DNA is transformed into a suitable bacterial host (E.coli) by transformation for the expression of the desired gene.
  • The cells into which rDNA is inserted are called transformed cells.
  • The transformed bacterial cells can be identified using suitable selectable markers.

5. Obtaining the foreign gene product

  • The transformed bacterial cells (recombinants) are grown in bioreactor having suitable nutrient medium.
  • Bioreactors provide optimum conditions for achieving the desired product by providing optimum growth conditions such as temperature, pH, suitable salts, vitamins, oxygen etc.
  • The desired protein is extracted from bacterial cells by separation and purification. These processes constitute downstream processing.

Question 31.
Briefly describe the different stages involved in decomposition.
Answer:
1. Fragmentation:
It is the break down detritus into smaller particles, by detritivores (earthworm).

2. Leaching:
It is the process in which water soluble inorganic nutrients go down into the soil horizon and get precipitated as unavailable salts.

3. Catabolism:
It is the process in which leads to bacterial and fungar enzymes degrade detritus into simpler inorganic substances.

4. Humification:
It is the process which leads to the accumulation of a dark coloured amorphous substance called humuns that is highly resistant to microbial action and undergoes very slow decomposition.

5. Mineralisation :
It is the degradation of humus by some microbes and the release of inorganic nutrients.

KSEEB Solutions

Question 32.
Explain five effects of water pollution.
Answer:
1. Sewage discharge, pollutes the water and makes it unfit for human use, decomposers in the aquatic ecosystem, decompose these pollutants and makes the water foul smelling.

2. The organic substances present in sewage are decomposed by decomposers aerobically as a result the oxygen content of the water decreases. This leads to biochemical oxygen demand (BOD). It kills large number of aquatic organisms.

3. Agricultural runoff Or excess wash from agricultural fields, carry pesticides and fertilizers to water bodies like lakes. The nutrient enrichment of lakes promotes the enormous growth of algae. This is called algal bloom; it blocks the free flow of water and O2 content and eventual death of aquatic organisms.

4. Industrial effluents discharged into ponds or lakes contain toxic substances, like mercury, DDT, etc. It is neither excreted nor metabolized, gradually they tend to accumulate in organisms through food chain. This is called biomagnification.

Biomagnifieation of DDT disturbs calcium metabolism in birds, which lay eggs with thin egg shell, it results in premature breaking and decline in bird population.

5. Oil slicks formed as a result of leakage of oils from ships carrying huge quantities of crude oil via sea routes, spread to several kilometers. It does not allow oxygen to penetrate into water and causes mass death of aquatic organisms.

6. The hot water discharged from thermal power stations or nuclear power stations into water bodies, increases the temperature; this results in evaporation of oxygen from the water surface leading to death of aquatic animals like fishes.

7. Sewage water contains many undesirable pathogenic microorganisms, when this water is discharged into nearby water bodies without proper treatment may cause outbreak of serious diseases like Jaundice, cholera, dysentery, typhoid, etc.

8. Industrial effluents released into surrounding land or into pits, percolate along with its chemicals into soil and contaminate ground water.

Section – II

Answer any THREE of the following questions in 200 – 250 words each, wherever applicable. ( 3 × 5 = 15 )

Question 33.
Draw the diagram of the sectional view of the female reproductive system in humans.
Answer:
2nd PUC Biology Model Question Paper 1 with Answers 2

KSEEB Solutions

Question 34.
Explain Mendel’s experiment that describes the inheritance of one gene.
Answer:

  • Crossing of true breeding tall pea plant with a true breeding dwarf plant. All the offspring (hybrid) in the F1 the tall.
  • Appearance of tall character in the F1 indicates that tallness is dominant over dwarf character.
  • Selfing of F, tall plants – production of both tall and dwarf plants in the ratio 3 : 1 in the F2.

The reappearance of dwarf character in the F2 generation indicates that alleles for tallness and dwarf character have segregated during gamate formation.
2nd PUC Biology Model Question Paper 1 with Answers 3
Based on the result of this monohybrid cross, Mendel put forward the law of segregation, i.e., factors or alleles for a pair of contrasting characters do not blend, but segregate or separate during gamete formation such that a gamete received only one of the factors (purity of gametes).

Question 35.
Draw a labelled diagram of a biogas plant.
Answer:
2nd PUC Biology Model Question Paper 1 with Answers 4

Question 36.
a) Explain the application of tissue culture in any three fields.    (3)
Answer:
Applications of tissues are :
1. Micropropagation:
It is used for the propagation / production of a large number of plants in short durations. Micropropagation produces somaclones (plants genetically identical to the original / parent plant).

2. Virus – free plants :
Meristems of plants are grown by tissue culture to get virus – free plants. By this method, healthly plants can be recovered from diseased plants.

3. Somatic hybridization:
Somatic hybrids obtained from somatic hybridization (fusion of naked protoplasts isolated from two different varieties og plants) can be further grown to form new plants by tissue culture.

b) Why continued inbreeding should be avoided in plants?    (1)
Answer:
Continued inbreeding results in inbreeding depression which decreases fertility and productivity of plants.

c) What is germplasm collection?     (1)
Answer:
Germplasm collection : It is the centre collection of plants or seeds having diverse alleles for all the genes in a given crop.

KSEEB Solutions

Question 37.
Explain Predation?
Answer:

  • Predation is a type of interaction in which a larger organism called predator of higher trophic level seeds on another called prey of lower trophic level.
  • In this type of interaction, one of the partner’s (predator) is benefitted while the prey is harmed.

Role played by predators :

  • They act as links for energy transfer across trophical levels.
  • They keep prey populations under control.

Defences developed by animals preys against predators :

  • Some show cryptic colouration to avoid being detected easily by the predator.
  • Some are poisonous and some are highly distantefal and therefore are avoided by the predators.

Defences developed by plant preys against predators :

  • Thorns (Acacia, cactus) are morphological means of defence.
  • Many plants produce and store chemicals like nicotine, caffeine, quinine, strychnine, opium etc., that make the herbivore sick when they are eaten, inhibit feeding or digestion, disrupt its reproduction of even kill it.

Leave a Comment