CBSE Sample Papers for Class 12 Biology Set 1 with Solutions

Students must start practising the questions from CBSE Sample Papers for Class 12 Biology with Solutions Set 1 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Biology Set 1 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 70

General Instructions:

All questions are compulsory.
The question paper has five sections and 33 questions. ALL questions are compulsory.
Section-A has 16 questions of 1 mark each: Section-B has S questions of 2 marks each; Section-C has 7 questions of 3 marks each: Section-D has 2 case-based questions of 4 marks each; and Section-E has 3 questions of 5 marks each.
There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
Wherever necessary, neat and properly Labeled diagrams should be drawn.

Section – A (16 Marks)

Question 1.
An infertile couple was advised to undergo In vitro fertiLization by the doctor. Out of the options given below, select the correct
stage for transfer to the fallopian tube for successful results?
(a) Zygote only
(b) Zygote or early embryo upto 8 blastomeres
(c) Embryos with more than 8 blastomeres
(d) Btastocyst stage
Answer:
(b) Zygote or early embryo upto 8 blastom eres

Explanation:
The zygote or early embryo with up to 8 blastomeres is transferred into the Fallopian tube and the process is called Zygote lnçraFallopian Transfer or ZIFT. It is one of the successful technique of IVF.

Caution:
Morula is often confused with blastocyst as the stage that is implanted in the uterine wall. Morula is just the embryo with 8-16 blastomeres white blastocyst is formed after further divisions of the morula.

Question 2.
Given below are four contraceptive methods and their modes of action. Select the correct match:

Method	Mode of action
(A) Condom	(i) Ovum not able to reach Fallopian tube
(B) Vasectomy	(ii) Prevents ovulation
(C) Pill	(iii) P revents sperm reaching the cervix
(D) Tubectomy	(iv) Semen contains no sperms

(a) (A)-(i) (B)-(ii) (C)-(iii) (D)-(iv)
(b) (A)-(ii) (B)-(iii) (C)-(iii) (D)-(i)
(c) (A)-(iii) (B)-(iv) (C)-(ii) (D)-(i)
(d) (A)-(iv) (B)-(i) (C)-(iii) (D)-(ii)
Answer:
(c) (A)-(iii),(B)-(iv),(C)-(ii),(D)-(i)
Explanation:
(A) Condom is the contraceptive method which prevents sperm reaching the cervix.

(B) A man who has had o vasectomy stilt makes semen and is able to ejaculate. But the semen doesn’t contain sperm.

(C) Birth control methods include the combination oral contraceptive pill, also known as the birth control pill or simply “the pill ft works by preventing ovulation.

(D) In women. tubectomy, commonly referred to as tubal sterilisation, is a long-term technique of birth control Through a surgical procedure, the fallopian tubes are blocked, preventing the ovary’s released egg from reaching the uterus.

Related Theory:
in order to, increase the contraceptive efficiency of barrier methods (both mate and female condoms). certain spermicidal creams, jetties and foams are usually used along with these barriers

Question 3.
Which of the following amino acid residues will. constitute the histone core?
(a) Lysine and Arginine
(b) Asporagine and Arginine
(c) Gtutamine and Lysine
(d) Asparagine and Glutamine
Answer:
(a) Lysine and Arginine:

Explanation:
The fundamental building block of DNA packaging, known as a nucleosome, consists of a length of DNA wrapped in a cerrain order around an eight-histone protein core. Small alkaline proteins called histones contain a lot of the basic amino acids (arginine and lysine).

Question 4.
Evolutionary convergence is development of a:
(a) common set of functions in groups of different ancestry.
(b) dissimilar set of functions in closely related groups.
(c) common set of structures in closely related groups.
(d) dissimilar set of functions in unrelated groups.
Answer:
(a) common set of characters in groups of different ancestry.

Explanation:
Evolutionary convergence is the emergence of a shared set of traits in a population with different ancestries.

Question 5.
Apis mellifera are killer bees possessing toxic bee venom. Identify the treatment and the type of immunity developed from the given table to treat a person against the venom of this bee. [1]

Remedy	Immunity
(a) Inactivated proteins	Active
(b) Proteins of the venom	Passive
(c) Preformed antibodies	Passive
(d) Dead micro-organisms	Active

Answer:
(c) Preformed Antibodies, Passive

Explanation:
Artificial passive immunity is a condition when preformed antibodies are delivered directly into the body. It offers a rapid immunological, response. When a patient has been bitten by a snake, the injection contains antibodies that have been tested for snake venom, this is also called Passive immunisation.

Related Theory
Active Immunity: When an individual (host) is exposed to antigens in the form of living or dead microbes or other proteins it results in the production of anti bodies. This kind of immunity is slow and thus takes time to give its FÙU effective response, but its effect is long lasting. It has rio side effects

Question 6.
Interferons are most effective In making non-infected cells resistant against the spread of which of the following diseases in humans?
(a) asca riosis
(b) ringworm
(c) amoebiasis
(d) AIDS [1]
Answer:
(d) AIDS

Explanation:
Interferons are a group of signalling proteins made and released by host cells ¡n response to the presence of several viruses. AIDS is a chronic, potentially life-threatening condition caused by the human immunodeficiency virus (HIV).

Question 7.
Which of the following water samples in the table given below, wilt have a higher concentration of organic matter? [1]

Water Sample	Level of pollution	Value of BOO
(a)	High	High
(b)	Low	Low
(c)	Low	High
(d)	High	Low

Answer:
(a) High, High

Explanation:
Biochemical oxygen demand (BOD) is the most commonly used method for measuring the quantity of organic oxygen-demanding materials and is often used as an indicator of the degree of organic pollution of water,

Question 8.
The figure below shows the structure of a plasmid.

A foreign DNA was ligated at Barn HI. The transformants were then grown in a medium containing antibiotics tetracycline and ampicillin. Choose the correct observation for the growth of bacterial colonies from the given

Medium with Tetracycline	Medium with Ampicillin
(a) Growth	No growth
(b) No growth	Growth
(c) No growth	No Growth
(d) Growth	Growth

Answer:
(b) No growth, Growth

Explanation:
Insertional inactivation ¡s a technique used in recombinant DNA technology. In this procedure, a bacteria carrying recombinant ptasmids or a fragment of foreign DNA is made to insert into a restriction site inside a gene to resist antibiotics, hence causing the gene to turn non-functional or in an inactivated state.

Therefore, when transformants were grown in a medium containing tetracycline does not shows growth as it is inactivated when foreign DNA was Ligated at Barn HI.

Question 9.
Swathi was growing a bacterial colony in a culture flask under ideal laboratory conditions where the resources are replenished. Which of the following equations wilt represent the growth in this case?
(Where population size is N, birth rate ¡s b, death rate is d, unit time period is t, and carrying capacity is K).
(a) \(\frac {dN}{dt}\) = KN
(b) \(\frac {dN}{dt}\) = rN
(c) \(\frac {dN}{dt}\) = rN[k – \(\frac {N}{t}\)]
(d) \(\frac {dN}{dt}\) = rN[k + \(\frac {N}{t}\)]
Answer:
(b) \(\frac {dN}{dt}\) = rN

Explanation:
In this case, resources are replenished which means that resources are not Limiting the growth. plot is exponential and the equation which represent the growth in this case is:
(b) \(\frac {dN}{dt}\) = rN

Question 10.
Sea Anemone gets attached to the surface of the hermit crab. The kind of population interaction exhibited in this cose is:
(a) amensatism
(b) commensalism
(c) mutualism
(d) parasitism [1]
Answer:
(b) commensalism

Explanation:
Hermit crabs and sea anemones shows a commensalism relationship.

Caution:
it is important to remember in population interactions whether one species is benefitted or both and also whether it is detrimental to one or both.

Question 11.
Which of the following food chains is the major conduit for energy flow in terrestrial and aquatic ecosystems respectively?

Terrestrial Ecosystem	Aquatic Ecosystem
(a) Grazing	Grazing
(b) Detritus	Detritus
(c) Detritus	Grazing
(d) Grazing	Detritus

Answer:
(c) Detritus Grazing

Explanation:
Detritus food chains (DFC) are a key energy conduit in a terrestrial environment. An organism enters DFC as it acts or produces biological waste. In an aquatic ecosystem, the grazing food chain (GFC) is the major conduit for energy flow.

Question 12.
Which of the following is an example of ex-situ conservation?
(a) Sacred Groves
(b) National Park
(c) Biosphere Reserve
(d) Seed Bank
Answer:
(d) Seed Bank

Explanation:
Seed bank is an example of ex-situ conservation. Sacred groves, national parks and biosphere reserves are the examples of insitu conservation.

Questions 13 to 16 consist of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given
below:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is not the correct explanation of A.
(c) Astrue but Ris faLse.
(d) A is false but R is true.

Question 13.
Assertion (A): Apomictic embryos are genetically identical to the parent plant
Reason (R): Apomixis is the production of seeds without fertilization. 1
Answer:
(a) Both A and R are true and R ¡s the correct explanation of A.

Explanation:
Derivation of the egg from a diploid maternal cell without meiotic reduction, and its subsequent fertilization-independent development into an embryo means that the progeny derived from apomictic development are clonai and therefore genetically identical to the maternal parent.

Question 14.
Assertion (A): When white eyed, yellow bodied Drosophila females were hybridized with red eyed, brown-bodied males; and F₁progeny was intercrossed, F₂ ratio deviated from 9: 3 : 3: 1.
Reason (R): When two genes in a dihybrid are on the same chromosome, the proportion of parental gene combinations is much higher than the non-parental type. [1]
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation:
The mendelian dihybrid cross indicates that the parental combinations are predominate in the F₁ generation. Because the alleles segregate independently and the genes for one characteristic are not inherited along with the genes for another trait, non-parental combinations in the F₂ generation are higher than parental combinations.

According to Morgan’s experiment, when genes are connected, they are inherited together in both the F₁ and F₂ generations, increasing the percentage of eternal combination in both.

Assertion (A): Functional ADA cDNA genes must be inserted in the Lymphocytes at the early embryonic stage.
Reason (R): CeLLs in the embryonic stage are mortal, differentiated and easy to manipulate. [1]
Answer:
(c) AistruebutRistòlse.

Explanation:
Using a retroviral vector, a functional ADA cDNA is delivered to these cells. The patient receives these cells again. This procedure is repeated repeatedly. The defect might be permanently cured if a functional ADA gene is extracted from bone marrow cells and inserted at early embryonic stages.

Question 16.
Given below is the Age Pyramid of population in one of the states in India as per 2011 census. It depicts the mate population on the Left hand side, female population on the right hand side, newborns towards the base and gradually increasing age groups as we move from base to the top, with the oldest population at the top. Study this pyramid and comment upon the appropriateness of the assertion and the reason.

Assertion (A): ft is a stable population.
Reason (R): The pre-reproductive and reproductive individuals are almost in equal numbers and the post-reproductive individuals are relatively fewer. [1]
Answer:
(a) Both A and R are true and R is the correct explanation of A.

Explanation:
Population having the same number of pre-reproductive and post- reproductive age groups is stable.

Related Theory:
The other two types of age pyramids are:
(1) Expanding or growing: The age pyramid representing a growing population is formed when the population of pre-reproductive age group is highest followed by moderate number of reproductive individuals and fewer post-reproductive individuals.
(2) Deciining The age pyramid representing a declining population when the population of pre-reproductive age group is lower than that of reproductive age group.

Section-B (10 Marks)

Question 17.
In the figure given below, parts A and B show the Level of hormones which influence the menstrual cycle. Study the figure and answer the questions that follow:

(A) Name the organs which secrete the hormones represented in parts A and B.
(B) State the impact of the hormones in part B on the uterus of the human femaLe during 6 to 15 days of menstrual. cycle? [2]

Total Marks

Breakdown (As per CBSE Marking Scheme)

2m (SA-I)

(A) Write the name of the organs and B

(O.5m.+ O.5m)

(B) Write the impact of the

f hormone& (1m)

Answer:
(A) A-Pituitary gland; B: Ovary
(B) Endometnum of the uterus regenerates through proliferation.

Explanation:
(B) The follicular phase comes after the menstrual phase. In this stage, the ovary’s initial fóltictes develop into fully developed Graofiori follicles and the uterus’ endometrium regenerates through proliferation.

Question 18.
A true breeding pea plant, homozygous dominant for inflated green pods is crossed with another pea plant with constricted yellow pods (ffgg). With the help of Punnett square show the above cross and mention the results obtained phenotypically.y and genotypically in F₁ generation. [2]

Total Marks	Break down (As per CBSE Marking Scheme)
2m (SA-I)	Make the correct Punnett square. (1m) Write the genotype. (O.5m) Write the phenotype. (0.5m)

Answer:

Making the correct Punnett square
Phenotype – AlL Inflated green pods
Genotype – FfGg

Question 19.
During a field trip, one of your friends in the group suddenly became unwell, she started sneezing and had trouble in breathing. Name and explain the term associated with such sudden responses. What wouLd the doctor recommend for relief? [2]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Mention the name and define the term. (15m + O.5m) Explain about what doctor would recommend. (1m)

Answer:
(i) Allergy, the exaggerated response of the immune response to certain antigens present in the environment is catted alLergy.
(ii) Doctors would administer drugs Like antihistamines, adrenaline and steroids (anyone) to reduce the symptoms.

Question 20.
CTAAG
GAATTC
(A) What are such sequences catted? Name the enzyme used that recognizes such nucleotide sequences.
(B) What is their significance in biotechnology? [2]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	(A) Write the name of the term arid write the name of the enzyme. (O.5m + O.5m) (B) Write its significance in biotechnology. (1m)

Answer:
(A) Palindromic sequences endonucLease enzyme.
(B) Restriction enzymes can make complementary cut counterparts forming sticky ends for recombination DNA / rDNA technology/ to facilitate Ligation of vector and foreign DNA.

Question 21.
(A) Given below is a pyramid of biomass in an ecosystem where each bar represents the standing crop available in the trophic Level With the help of on example explain the conditions where this kind of pyramid is possible in nature?

(B) ‘Wilt the pyramid of energy be also of the same shape in this situation? Gave reason for your response.
OR
(A) Draw a pyramid of numbers where a Large number of insects are feeding on the Leaves of a tree. What is the shape of this pyramid?
(B) Wilt the pyramid of energy be also of the same shape ¡n this situation? Give reason for your response. [2]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	(A) Mention the conditions where this kind of pyramid is possible. (1m) (B) Give reason. (1m)

Answer:
(A) Inverted pyramids of biomass ore seen in aquatic conditions where a small standing crop of phytoplankton supports a large standing crop of Zooplankton/ fish/in terrestrial, ecosystem where a brge number of insects are feeding on the leaves of a tree.

(B) No, the pyràmid of energy is always upright, and can never be inverted because when energy flows from one trophic level to the next trophic revel some amount of energy is always lost as heat at each step.
OR

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	(A) Draw the pyramid and mention the shape of the pyramid. : (O.5m, O.5m) (B) Give roeson. (1m)

Inverted pyramid because a Large number of insects feed on one tree.

(B) No, the pyramid of energy is always upright, and can never be inverted because when energy flows from one trophic level to the next trophic level. some amount of energy is always lost as heat at each step.

Section – C (21 Marks)

Question 22.
Explain the functions of the following structures in the human male reproductive system.
(A) Scrotum
(B) Leydig cells
(C) Male accessory glands [3]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Explain the functions in each case. (1m +1m + 1m)

Answer:
(A) Scrotum: The testes are situated outside the abdominal cavity within a pouch-catted scrotum. The scrotum helps in maintaining the low temperature of the testes (2-2.5 degree celsius lower than the normal internal body temperature) necessary for spermatogenesis.

(B) Leydig cells: The regions outside the seminiferous tubules called interstitial spaces. contain small blood vessels and interstitial cells or Leydig cells. Leydig cells synthesize and secrete testicular hormones catted androgens.

(C) Mate accessory glands: The mate accessory glands include paired seminal vesicles, a prostate and paired bulbourethral glands. Secretions of these glands constitute the seminal plasma which is rich in fructose calcium and certain enzymes. The secretions of bulbourethral glands also help in the Lubrication of the penis.

Question 23.
State the agent(s) which help in pollinating in the following plants. Explain the adaptations in these plants to ensure pollination:
(A) Corn
(B) Water hyacinth
(C) VaWsneria [3]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Write the name of the agents in each case. (O.5m + O.Sm + O.Sm) Mention the adaptations n each case. (O.5m. O.5m + O.Sm)

Answer:
(A) Corn: Wind. Numerous flowers are packed in an inflorescence; the tassels seen in the corn cob are the stigma and style which wave in the wind to trap pollen grains.

(B) Water hyacinth: Insects or wind. In water hyacinth the flowers emerge above the Level of water and are pollinated by insects or wind as in most of the Land plants.

(C) VaUisneria: Water, In VaWsneria – the ‘female flower reaches the surface of water by the long stalk and the male flowers or pollen grains are released onto the surface of water. They are carried passiveLy by water currents some of them eventually reach the female flowers and the stigma.

Question 24.
(A) Identify the polarity of X to X’ in the diagram below and mention how many more amino acids are expected to be added to this polypeptide chain.

(B) Mention the codon and anticodon for alanine.
(C) Why are some untransLated sequences of bases seen in mRNA coding for a polypeptide? Where exactly are they present on mRNA? [3]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	(A) identify the polarity and mention the number of amino acids. (03m + 0.5m) (B) Write the codon and anticodon for alanine. (1m) (C) Give reason and position on mRNA. (0.5m + 0.5m)

Answer:
(A) X to X’ is 5’→3′ No more amino acids wilt be added
(B) GCA Anticodon is CGU
(C) The untranslated regions are required for an efficient translation process.
They are present before the initiation codon at the 5’ – end and after the stop/ termination codon, at the 3’ – end.

Question 25.
(A) How is Hardy-Weinberg’s expression “(p² + 2pq + q²) = 1” derived?
(B) List any two factors that can disturb the genetic equilibrium. [3]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	(A) Explain how Hardy Weinsberg1 expression derived. (1m) (B) List any two factors. (2m)

Answer:
(A) Sum Total of AIL the Allele Frequencies is

1. Let there be two alleles A and a in a population. The frequencies of alleles A and o are p’ and q’ respectively. The frequency of AA individuals in a population is p² and it can be explained that the probability that an allele A with a frequency of p would appear on both the chromosomes of a diploid individual. is simply the product of the probabilities. i.e p²;.

Similarly. the frequency of is q² and that of Aa is 2pq. p² + 2pq + q²) = 1, where p² represents the frequency of homozygous dominant genotype. 2pq represents the frequency of the heterozygous genotype and q² represents the frequency of the homozygous recessive.

(B) Factors that affect Hardy-Weinberg equilibrium:
(1) Gene migration or gene flow
(2) Genetic drift
(3) Mutation
(4) Genetic recombination
(5) Natural Selection

Question 26.
Highlight the structural importance of on antibody moLecuLe with a diagram. Name the four types of antibodies found to give a humoral immune response, mentioning the functions of two of them you have studied.
OR
(A) Explain the Ufe cycle of Plasmodium starting from its entry in the body of female Anopheles till the completion of its Life cycLe in humans.
(B) Explain the cause of periodic recurrence of chiLl and high fever during malarial attack in humans. [3]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Label the Light and heavy chain in diagram. (O.Sm + 05m) Write the name of Our antibodies. (1m) Write the function of any two antibodies. (0.Sm + 0.5m)

Answer:
An antibody molecule consists of four polypeptide chains, two are Long called heavy (H) chains white other two are short Called Light (L) chains. Both are arranged in the shape of Y. Hence, the antibody is represented as H₂ L₂

Diagram with LabeLs-Light chain, Heavy Chain
Types of Antibody-lgA. gM. lgE,lgG
IgA-Lactating Mother to protect their infant
IgE-To protect from allergen

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	(A) Label the diagram and mention the stages. (2.5m) (B) Write the cause. (O.5m)

(A) When a female Anopheles mosquito bites on infected person, the parasites enter the mosquito’s body as gametocytes. It Leads to fertilization and development in the gut of the mosquito and undergoes further development to form sporozoites that are stored in salivary glands until their transfer to human body. In the human body – the sporozoites reach the Liver and reproduce asexually, bursting the cells and releasing them into the RBCs as gametocyte & (Labeled diagram explaining the mentioned stages can also be considered)

(B) The rupture of RBCs reLeases o toxic substance catted haemozoin, which is responsible for the chill and high fever.

Question 27.
Carefully observe the given picture. A mixture of DNA with fragments ranging from 200 base pairs to 2500 base pairs was electrophoresed on agoras gel with the following arrangement.

(A) What result wilt be obtained on staining with ethidium bromide? Explain with reason.
(B) The above setup was modified and a band with 250 base pairs was obtained at X.

What change(s) were made to the previous design to obtain a band at X? Why did the band appear at the position X? [3]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Write the result obtained.(O.5m) and give the reason.(O.5m) Write the change made. (O.5m) and give reason. (O.5m)

Answer:
(A) No bands will be obtained os/Alt DNA will be seen in the well only. DNA fragments being negatively charged will, not move towards -ive end/ cathode. DNA being negative charged will remain stationed at the positive end / anodeend of the agar block.

(B) (1) Position of the positive terminat/ end/ anode and the negative terminal/ end/ cathode was interchonged.
(2)The fragment with least base pairs will get separated faster and move faster to the anode end.

Question 28.
(A) There was Loss of biodiversity in an ecosystem due to a new construction project in that area. What wouLd be its impact on the ecosystem? State any three.
(B) List any three major causes of Loss of biodiversity? [3]

Total Marks	Breakdown (As per CBSE Marking Scheme)
3m (SA-II)	Write any three impact. (O.5m. O.Sm + O.5m) Write any three causes. (O.Sm , O.5m + O.5m)

Answer:
Impacts of Loss of biodiversity on the ecosystem:
(A)
(1) Decline in plant production
(2) Lowered resistance to environmental perturbations such as drought
(3) Increased variability in certain ecosystems – processes such as plant productivity, water use, pest and disease cycles.

(B)
(1) Habitat loss and fragmentation
(2) Over-exploitation
(3) ALien invasive species
(4) Co-extinctions.

Section – D (8 Marks)

(Q. No. 29 and 30 are cese-based questions. Each question has subparts with internal choice in one subpart)

Question 29.
Study the Pedigree chart given below and answer the questions that follow

Symbols used in the given Pedigree Chart are as follows

(A) On the basis of the inheritance pattern exhibited in this pedigree chart, what conclusion can you draw about the pattern of inheritance? [1]
(B) If the female is homozygous for the affected trait ¡n this pedigree chart, then what percentage of her sons will be affected? [1]
(C) Give the genotype of offspring 1, 2, 3 and 4 In III generation. [1]
OR
(D) In this type of inheritance pattern, out of mate and femaLe children which one has Less probability of receiving the trait from the parents. Give a reason. [2]

Total Marks	Breakdown (As per CBSE Marking Scheme)
4m	(A) Mention whether it is X or Y-Linked. (O.5m) and mention the trait (O.5m) (B) Write the percentage of sons which will be affected. (1m) (C) Write the genotype of offspring 1,2.3 and 4. (O.5m + O.5m + O.5m + O.5m)

Answer:
(A) X- Linked, Recessive trait
(B) 100%
(C)XY OR 2. XX 3. XX 4.XX
OR

Total Marks	Breakdown (As per CBSE Marking Scheme)
4m	(C) Give any two reasons. (2m)

(C) The possibility of the female getting the trait is Less. The female will get the trait only if the mother is at Least a carrier and the father is affected.

Question 30.
The data below shows the concentration of nicotine smoked by a smoker taking 10 puffs/ minute.

(A) With reference to the above graph explain the concentration of nicotine in blood at 10 minutes. [1]
(B) How WILL this affect the concentration of carbon monoxide and homebound oxygen at 10 minutes? [1]
(C) How does cigarette smoking result in high blood pressure and increased in heart rate?
OR
(C) How does cigarette smoking result in Lung cancer and emphysema? [2]

Total Marks	Breakdown (As per CBSE Marking Scheme)
4m (CBQ)	(A) Mention the concentration of nicotine, (1m) (B) Write the effect of concentration of carbon monoxide. (1m) (C) Write any two effects of cigarette. (2m)

Answer:
(A) Concentration of nicotine is maximum at 10 minutes/ conc. of nicotine increases steadily in the blood to reach 45mg/cm³.
(B) The concentration of CO wilt increases resulting in reduced concentration of haemboundoxygen.
(C) Nicotine results in stimulating the adrenal gland which results in release of adrenaline ‘ noradrenaline in the blood resulting in increase of blood pressure and heart rate.
OR

Total Marks	Breakdown (As per CBSE Marking Scheme)
2m (CBQ)	(C) Give any two results. (2m)

(C) Chemical carcinogens present in tobacco smoke are the major cause of lung cancer. The cigarette smoke irritates the air passages of the lungs causing them to produce mucus which causes cough resulting in enlarging air spaces/ reduce surface area/loss their elasticity (any point can be mentioned) thus difficulty in breathing causing emphysema.

Section – E (15 Marks)

Question 31.
Trace the events from copulation to zygote Trace the development of a megaspore formation in a human female, mother cell to the formation of mature
OR
Trace the development of a megaspore mother cell to the formation of mature embryo sac in a flowering plant. [5]

Total
Marks

Breakdown

(As per CBSE Marking Scheme)

5m (LA)

I Write the events in ten points. (5m)

Answer:
(i) During copulation (coitus) semen is released by the penis into the vagina (insemination).

(ii) The motile sperms swim rapidly, pass through the cervix, enter into the uterus\ and finally reach the ampullary region of the Fallopian tube.

(iii) The ovum released by the ovary is also transported to the ampulLary region where fertilization takes place.

(iv) Fertilisation can only occur if the ovum and sperms are transported simultaneously to the ampullary region. This is the reason why not alt copulations Lead to fertilisation and pregnancy.

(v) The process of fusion of a sperm with an ovum is called fertilisation.

(vi) During fertilisation, a sperm comes in contact with the zona pellucid layer of the ovum and induces changes in the membrane that block the entry of additional sperms. Thus, it ensures that only one sperm can fertilise an ovum.

(vii) The secretions of the acrosome help the sperm enter into the cytoplasm of the ovum through the zona pellucid and the plasma membrane.

(viii) This induces the completion of the meiotic dMsion of the secondary oocyte.

(ix) The second meiotic division is also unequal and results in the fornication of a second polar body and o haploid ovum (ootid).

(x) Soon the haploid nucleus of the sperms and that of the ovum fuse together to form a diploid zygote.
OR

Total

Marks

Breakdown

(As per CBSE Marking Scheme)

5m (LA)

Write the development of’megaspore mother cell ir ten points, (5m)

Trace the development of a megaspore mother cell to the formation of mature embryo sac in a flowering plant The process of formation of megaspores from the megaspore mother cell, is called megasporogenesis.

(i) Ovules generally differentiate a single megaspore mother cell (MMC) in the micropylar region of the nucleolus. It is a large cell containing dense cytoplasm and o prominent nucleus. The MMC undergoes meiotic division to form megaspores.

(ii) n a majority of flowering plants, one of the megaspores is functional white the other three degenerate. Only the functional megaspore develops into the female gametophyte (embryo sac). This method of embryo sac formation from single megaspore is termed monosporic development.

(iii) The nucleus of the functional megaspore divides mitotically to form two nuclei which move to the opposite poles. forming the 2-nucleate embryo sac.

(iv) Two more sequentiaL mitotic nuclear divisions result in the formation of the 4-nucleate and later the 8-nucleate stages of the embryo sac.

(v) These mitotic divisions are strictly free nuclear, that is, nuclear divisions are not followed immediately by cell wall formation.

(vi) After the 8-nudeate stage, cell walls are laid down leading to the organisation of the typical female gametophyte or embryo soc

(vii) Six of the eight nuclei are surrounded by cell walls and organised into cells; the remaining two nuclei, called polar nudei are situated in the large central cell

(viii) Three cells are grouped together at the micropylar end and constitute the egg apparatus. The egg apparatus, in turn, consists of two synergids and one egg cell The synergids have special ceLlular thickenings at the micropylar tip caLled filiform apparatus

(ix) Three cells are at the chalazaL end and are called the antipodals.

(x) The large central cell, as mentioned earlier, has two polar nuclei. Which came to Lie below egg apparatus. Thus, a typical angiosperm embryo sac, at maturity, though 8-nucleate is 7-celled.

Question 32.
Observe the segment of mRNA given below.

(A) ExPlain and illustrate the steps involved to make fully processed hnRNA?
(B) Gene encoding RNA Polymerase I and III have been affected by mutation in a cell Explain its impact on the synthesis of polypeptide, stating reasons.
OR
Study the schematic representation of the genes involved in the Lac operon given below and answer the questions that follow:

(A) The active site of enzyme permease present in the cell membrane of a bacterium has been blocked by an inhibitor, how will it affect the Lac operon?
(B) The protein produced by the gene has become abnormaL due to unknown reasons. Explain its impact on Lactose metabolism stating the reason.
(C) If the nutrient medium for the bacteria contains only galactose; will operon be expressed? Justify your answer, [5]

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	(A) Illustrate the steps in six points. (3m) (B) State any two reasons. (2m)

Answer:
(A) The hnRNA undergoes processes called copping and tailing foLLowed by splicing. In capping, an unusual nucleotide is added to the SC-end of hnRNA methyl guanosine triphosphate. In taiLing. adenylate residues (about 200 – 300) are added at 3C-end in a template-independent manner. Now the hnRNA undergoes a process where the introns are removed and exons are joined to form mRNA called splicing.

(B) The process of translation will not happen. thus the polypeptide synthesis is stopped! hampered.

The reason for the above is:
RNA polymer I transcribe rRNAs which is the cellular factory for protein synthesis. RNA polymer III helps in the transcription of tRNA which ¡s the adaptor molecule/ that transfers amino acids to the site of protein synthesis.
OR

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	(A) Mention any two effects (2m) (B) Mention any two impacts. (2m) (C) State yes or no and give justification. (O.5m + O.5m)

(A) When the active site of enzyme permease present in the cell membrane of a bacterium has been blocked by an inhibitor, the Lactose is not transported into the cell As Lactose is the inducer, the lac operon will not be switched on.

(B) Since the repressor protein synthesized by the gene is abnormal it will not bind to the operator region of the operon, resulting in a continuous state of transcription process.

Question 33.
Oil spill is a major environmental issue. It has been found that different strains of Pseudomonas bacteria have genes to break down the four major groups of hydrocarbons in oil Trials are underway to use different biotechnological tools to incorporate these genes and create a genetically engineered strain of Pseudomonas – a ‘super-bug to break down the four major groups of hydrocarbons in oil Such bacteria might be sprayed onto surfaces polluted with oil to clean thin films of oil

(A) list two advantages of using bacteria for such biotechnological studies.
(B) For amplification of the gene of interest PCR was carried out. The PR was run with the help of polymerase which was functional only at a very Low temperature. How wilt this impact the efficiency of the PCR? Justify.
(C) If such bacteria are sprayed on water bodies with oil spills, how Wilt this have a positive or negative effect on the environment? Discuss.
OR
Insects in the Lepidopteran group toy eggs on maize crops. The Larvae on hatching feed on maize Leaf and tender cob. In order to arrest the spread of three such Lepidopteran pests, Bt maize crops were Introduced in an experimental field. A study was carried out to see which of the three species of Lepidopteran pests was most susceptible to Bt genes and its product The Lepidopteran pests were allowed to feed on the same Bt maize crops grown on 5 fields (A-E). The graph below shows the Leaf area damaged by these three pests after feeding on maize leaves for five days.

Insect gut pH was recorded as 10, 8 and 6 respectively for Species I, II and III respectively.
(A) Evaluate the efficacy of the Bt crop on the feeding habits of the three species of stem borer and suggest which species ¡s least susceptible to Bt toxin.
(B) Which species is most susceptible to Bt-maize, explain why?
(C) Using the given information, suggest why similar effect was not seen in the three insect species? [5]

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	(A) List any two advantages (1m) (B) Mention any one impact arid give justification. (1m + 1m) (C) Write the effect on the environment

Answer:
(A) You can easily grow a Large quantity of the bacteria/no ethical issues/have plasmids/ can easily transform.

(B) PCR will not amplify the gene. If the polymerase enzyme denatures at (ow temp. it will not be able to withstand high temperatures which is essential for separating/opening/unwinding/ denaturing DNA strand to open. Thus subsequent step of extending the primers using the nucleotides provided in the reaction and the genomic DNA as the template will not occur.

(C) Positive effect oil spills can be treated and the environment becomes better/ cleaner/ water becomes more potable! safe for aquatic forms/ safe for water birds [Ike seagulls. (Anyone) Negative effect: the bacteria can mutate! can harm other organisms/can conjugate with other non-virulent forms and make them superbugs with detrimental effects/ unpredictable! for a longer duration it may reduce the dissolved oxygen and Leading to mortality of aquatic organisms. (Anyone)
OR

Total Marks	Breakdown (As per CBSE Marking Scheme)
5m (LA)	(A) Evaluate the efficacy and name the species (0.5m + 0.5m) (B) Nome the species and state reason. (1m + 1m) (C) Give any two reasons (2m)

(A) Species III is least susceptible
(B) Bt toxin protoxins are converted into an active form in the gut which solubility the toxin crystals. The activated toxin binds to the surface of midgut epithelial cells and create pores that cause cell swelling and lysis and eventually cause death of the insect.
(C) insect species I and II have alkaline gut pH which solubilises the insecticidaL protein crystals of protoxin and makes it active. Species III has on acidic and the protoxin continues to remain in an inactive form doing no harm to insect species III.