CBSE Sample Papers For Class 12 Biology Mock Paper 1

Q1.

If a genetic disease is transferred from a phenotypically normal but carrier female to only some of the male progeny, the disease is:

Opt:

a. one organism has given rise to another

b. they share a common ancestor

c. they perform the same function

​d. the have biochemical similarities

Ans:

they share a common ancestor

Q2.

​Identify the correct labeling for Q, R, S and T for the lac operon in E. Coli as given below:​

Choose the correct option from the given table:

Opt:

A

B

C

D

Ans:

D

Q3.

​Morgan hybridised Drosophila white eyed and yellow bodied female with red eyed and brown bodied male (wild type) and intercrossed their F₁, progeny. He observed that the two genes ​

Opt:

a. Did not segregate independent of each other.​

b. Did not segregate independent of each other.​

​c. May be located on two different chromosomes. ​

​d. Segregated independently of each other. ​

​Showed very high percentage of recombinants.​

Ans:

Did not segregate independent of each other.​

Q4.

A startup company is evaluating the potential of producing Single Cell Protein (SCP) from algae. Select which of the following organisms it should use for the production?

Opt:

a. Saccharomyces lipolytica and Candida utilis

b. Chaetomium cellulyticum and Fusarium graminearum

c. Spirulina maxima, Chlorella and Scenedesmus acutus

d. Hylophora cecropia, Phytophthora infestans and Drechslera maydis

Ans:

Spirulina maxima, Chlorella and Scenedesmus acutus

Q5.

The thymus of a newborn baby starts degenerating a few years after birth due to a genetic disorder. Predict its impact on the immune system.

Opt:

a. Production of B-lymphocytes will be affected.

b. Growth and maturation of B-lymphocytes will be affected.

c. Production of T-lymphocytes will be affected.

d. Growth and maturation of T-lymphocytes will be affected.

Ans:

Growth and maturation of T-lymphocytes will be affected.

Q6.

Match the following:

Opt:

a. (i) – 5, (ii) – 1, (iii) – 2, (iv)- 3, (v) – 4

b. (i) – 3, (ii) – 1, (iii) – 4, (iv)- 5, (v) – 2

c. (i) – 3, (ii) – 4, (iii) – 5, (iv)- 2, (v) – 1

d. (i) – 1, (ii) – 4, (iii) – 5, (iv)- 2, (v) – 3

Ans:

(i) – 3, (ii) – 4, (iii) – 5, (iv)- 2, (v) – 1

Q7.

Meghan took a test tube and mixed some sugar, yeast and water in it. After one hour, she observed the evolution of gas bubbles in the test tube. Identify the process due to which the gas bubbles evolved in the test tube.

Opt:

a. Reproduction

b. Growth

c. Excretion

d. Respiration

Ans:

Respiration

Q8.

Analyse the characteristic features visible in the given floral diagram and identify the plant family to which it belongs?

Opt:

Liliaceae

Solanaceae

Fabaceae

Compositae

Ans:

Liliaceae

Q9.

In IVF technique, the zygote is transferred into uterus at

Opt:

2 blastomere stage.

4 blastomere stage.

6 blastomere stage.

16 blastomere stage.

Ans:

16 blastomere stage.

Q10.

An American company got the patent rights for a variety of Basmati rice fraudulently.
What would be this act termed as?

Opt:

Anti-patency

Biopiracy

Fraudulently

Disclamation of trade mark

Ans:

Biopiracy

Q11.

Introduction of Nile Perch into lake Victoria (East Africa) caused the extinction of unique species of cichlid fish. This is termed as

Opt:

habitat loss.

fragmentation.

alien species invasion.

over exploitation.

Ans:

alien species invasion.

Q12.

​In prokaryotes like E. coli the DNA in the nucleoid region is organised as​

Opt:

​negatively charged DNA wrapped around histone.​

​densely packed chromatin with NHC proteins.

large loops held by the proteins.​

​many repeating units of nucleosomes.​

Ans:

large loops held by the proteins.​

Q13.

The given graph is showing a population growth curve.

Which of the following factors would cause region X on the graph to become steeper?

Opt:

Decrease in predation and decrease in food supply.

Increase in food supply and decrease in disease.

Increase in food supply and increase in predation.

Increase in predation and decrease in disease.

Ans:

Increase in food supply and decrease in disease.

Q14.

Choose the correct option from the following.
Assertion: Oxalis regnellii shows cleistogamy.
Reason: In cleistogamy, transfer of pollen grains occurs from anther to the stigma of a different plant.

Opt:

Both assertion and reason are true and reason is the correct explanation of assertion

Both assertion and reason are true but reason is not the correct explanation of assertion

Assertion is true and reason is false

Reason is true and assertion is false

Ans:

Assertion is true and reason is false

Q15.

Read the given statements carefully and select the correct option.

Assertion (A): Root nodules of leguminous plants harbor symbiotic bacteria like Rhizobium.

Reason (R): Rhizobium fixes atmospheric nitrogen into the soil as nitrogen compounds.

Opt:

Both A and R are true and R is the correct explanation of A

Both A and R are true but R is not the correct explanation of A

A is true but R is false

A is false but R is true

Ans:

Both A and R are true and R is the correct explanation of A

Q16.

Assertion: RNA produced during transcription in eukaryotic cells cannot be straight away used in photosynthesis.

Reason: RNA splicing phenomena help in the removal of exons.

Opt:

Both A and R are true and R is the correct explanation of A.

Both A and R are true and R is not the correct explanation of A.

A is true but R is false.

A is False but R is true.

Ans:

A is true but R is false.

Q17.

What is unidirectional flow of solar energy from plant to different organisms in an ecosystem?

Ans:

All organisms are dependent for their food on producers, either directly or indirectly, this shows the unidirectional flow of energy from the sun to producers and then to consumers.

Q18.

Sexual reproduction is the favoured mode of reproduction under unfavorable conditions. Explain.

Ans:

Sexual reproduction causes genetic variations and produce offsprings with different combination of characteristics, which help organism to adapt to unfavorable conditions and increase the chances of the survival of species in unfavorable conditions. Sexual reproduction also contributes to evolution of the species.

Q19.

Write names of the causative agents of the following:
AIDS, Leprosy, Malaria, Filariasis

Ans:

(a) AIDS: Human Immunodeficiency Virus (HIV)
(b) Leprosy: Mycobacterium leprae
(c) Malaria: Plasmodium sps.
(d) Filariasis: Wuchereria bancrofti

Q20.

What are the limitations of ecological pyramids ?

Ans:

The limitations of a ecological pyramid are:
– It does not take into account the same species belonging to two or more trophic level.
– It assumes a simple food chain, something that almost never exists in nature as it does not accommodate a food web.

Q21.

The species diversity of plants (22%) is much less than that of animals (72%). What could be the explanation to how animals achieved greater diversification?

OR

​Sexual reproduction is the favoured mode of reproduction under unfavorable conditions. Explain.

Ans:

The species diversity of plants is much less as compared to that of animals possibly due to the following reasons:

1. Animals could relocate to different places when the conditions became unsuitable in a particular area, unlike plants which remain at one place. This ability to move to a new geographical condition requires diversification. Plant species either remained in one place or became extinct.
2. Animals developed more complex body structures, like segmented body or complex nervous system, unlike plants which allowed them to diversify much more under changing environmental condition.

OR

Sexual reproduction causes genetic variations and produce offsprings with different combination of characteristics, which help organism to adapt to unfavorable conditions and increase the chances of the survival of species in unfavorable conditions. Sexual reproduction also contributes to evolution of the species.

Q22.

Two farmers cultivated two different types of crops namely, A and B. Crop ‘A’ reproduces by asexual means and crop ‘B’ reproduces by sexual means.
Both the farmers got their crops insured. The premium paid for crop A was more than the crop B.
What is the possible reason behind the difference in the premium?

Ans:

Genetic variation in asexual mode of reproduction is low as compared to the sexual mode of reproduction.
Plant breeding through sexual mode of reproduction will have more variants.
The emergence of any disease is more likely to devastate asexually reproducing plant species as compared to that of sexually reproducing species.
Hence, the premium of crop A was more than the crop B.

Q23.

Define the following.
i) Law of dominance
ii) Law of segregation
iii) Law of independent assortment

Ans:

i) Law of dominance: Only one parental character or the dominant character is expressed in the first filial generation while both the characters appear in the second filial generation.

ii) Law of segregation: The alleles normally occurring in pairs will segregate during the gamete formation; whereby one allele will be carried by each gamete.

These gametes will randomly unite with the gametes of the other sex during fertilisation.

iii) Law of independent assortment: When there are two sets of characters in a hybrid, the segregation of one pair of characters is independent of the other pair.

Q24.

Draw a well-labelled diagram of an antibody molecule.
Ans:

A well-labelled diagram of an antibody molecule is depicted below:

Q25.

What is the mechanism by which the AIDS virus causes deficiency of immune system of the infected person?

Ans:

AIDS is caused by the Human Immunodeficiency Virus (HIV), via sexual or blood-blood contact. After getting into the body of the person, the virus enters into macrophages where the RNA genome of the virus replicates to form viral DNA with the help of the enzyme reverse transcriptase. This viral DNA gets incorporated into the host cell’s DNA and directs the infected cells to produce virus particles. The macrophages continue to produce virus and in this way, they act as an HIV factory. HIV enters into helper T-lymphocytes (TH), replicates and produces progeny viruses. The progeny viruses released in the blood attack other helper T-lymphocytes. This repeated process leads to a progressive decrease in the number of helper T-lymphocytes in the body of the infected person, thereby decreasing the immunity of a person.

Q26.

Ans:

Q27.

Ans:

(a) Repetitive DNA and Satellite DNA

(b) mRNA and tRNA

c. Template strand and Coding strand

Q28.

Ans:

(a) Repetitive DNA and Satellite DNA

(b) mRNA and tRNA

c. Template strand and Coding strand

Q29.

Study the RNA segment given below which is completely translated into a polypeptide chain.

(a) Identify the bases a and b.

(b) What do the codon ‘aUG’ and ‘UAb’ code for?

(c) How is peptide bond formed between two amino acids in the ribosome?

OR

Ans:

​(a) ‘a’ is A (AUG) and ‘b’ is A/G (UAA/UAG).

(b) AUG codes for methionine.UAA/UAG is a stop codon /nonsense codon/does not code for any amino acid.

(c) Charged tRNAs are brought closer together on mRNA in the ribosome. Ribosome acts as a catalyst (ribozyme) in forming the peptide bond.

OR

Q30.

​Observe the diagram showing the structure of the antibody given below and answer the following questions:

a. Identify the part labeled as “(a)” and “(b)”.

b. What is the function of the part labeled as “(a)”?

c. Which bond is present between the part labeled as “(b)” and “(c)”?

d. How many light chains are present in the structure of antibody?

OR

d. Do both the arms of Y shaped antibody have the same antigen binding sites?​​

Ans:

​a. The part labeled as “(a)” is antigen binding site and “(b)” is light chain.

b. Antigen binding site (a) recognises and binds to a specific antigen in a lock and key fashion forming an antigen-antibody complex.

c. Disulphide bond is present between the part labeled as “(b)” and “(c)”.

d. Two identical light chains are present in the structure of antibody.

OR

d. Yes, both the arms of Y shaped antibody have the same antigen binding sites which recognise the same antigen.​

Q31.

​Enlist all the events that take place during a menstrual cycle.

OR

​“Successful gamete transfer and fusion of gametes is essential for the most critical event in sexual reproduction”.

i. Give the technical term for the fusion of gametes.

ii. What are the events that occur during gamete transfer in plants?

iii. What is formed as a result of fusion of gametes/syngamy?

iv. How will you categorise syngamy based on the fusion of gametes inside or outside the female body? v. What would happen if syngamy does not occur?

Ans:

The whole event of menstrual cycle is divided into following phases:-

i) Follicular phase (Proliferative phase)-a) Primary follicles in the ovary grow to become fully mature graffian follicles.b) Endometrium also develops simultaneously in the uterus.c) Secretion of gonadotropins increases gradually and stimulates follicular development and release of estrogen by growing follicles.

ii) Ovulatory phase – In the middle of the cycle (14^th day), the level of LH becomes quiet high and induces rupturing of graffian follicle and release of functional ovum into fallopian tube.

iii) Luteal phase – The left over part of graffian follicle is transformed into a yellow structure called corpus luteum, which secrete progesterone for maintaining endometrium for expected pregnancy.

iv) Menstrual phase – In the absence of fertilisation, corpus luteum degenerates, endometrium ruptures leading to menstruation.

OR

i. Fertilisation.

ii. In plants, pollen grains are the carriers of male gametes and ovules have the egg. These pollen grains have to reach the stigma before fertilisation. The process of pollination facilitates transfer of pollen grains to the stigma. Pollen grains germinate on stigma and the pollen tubes carrying the male gametes reach the ovule and release the male gametes.

iii. Fusion of gametes or syngamy results in the formation of a diploid zygote.

iv. If syngamy occurs outside the body of the organism, i.e., in the external medium (water), then it is called external fertilisation. If syngamy occurs inside the body of the organism, then it is called internal fertilisation.

v. In sexual reproduction, meiosis halves the diploid (2n) number of chromosome to the haploid number (n) of chromosome for forthcoming gametes. The two types of gametes (egg and sperm) fuse at fertilisation and the chromosome number is restored, giving the new individual two sets of chromosomes, one from each parent. If syngamy does not occur, two haploid gametes will not fuse and the chromosome number will not be restored.

Q32.

Decode the below-given sequence of amino acids and answer the following questions:

………GUCCAUCUAACACCUGAG………

a. What will happen if the codon ACA is replaced by ACG?

b. What will if the nucleotide A in the codon GAG is replaced by nucleotide U?

c. Are any of the two codons in the given sequence coding for the same amino acid? How can you identify this?

d. Can we isolate and mRNA from one organism and translate it by using the machinery of another organism? Give reason to support your answer.

e. In the mtDNA, what do the following codons code for:

i. AGA

ii. UGA

OR

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 addition of lactose in the medium?

Ans:

Decoded sequence:

………Val-His-Leu-Thr-Pro-Glu………

a. If the codon ACA is replaced by ACG, there will be no change in the amino acid sequence as both ACA and ACG code for the same amino acid, i.e., Thr (T).

b. If the nucleotide A in the codon GAG is replaced by nucleotide U, the codon will become GUG. Since GAG codes for amino acid Glu (E) and GUG codes for amino acid Val (V), the amino acid sequence will change leading to the occurrence of a fatal genetic disorder ‘phenylketonuria (PKU)’.

c. No two codons in the given sequence are coding for the same amino acid.

• When the first two nucleotides in two codons are identical and the last letter is U or C, those two codons will always code for the same amino acid.
• Similarly, when the first two nucleotides in two codons are identical and the last letter is A or G, those two codons will always code for the same amino acid.
• In some case, when the first two nucleotides in the codons are identical and the last letter is U, C, A or G, all those codons will always code for the same amino acid.

d. Yes, we can isolate and mRNA from one organism, translate it by using the machinery of another organism and synthesise the same protein as it would have been synthesised in the original organism because genetic code is universal and hence; all the organisms share the same genetic language, with few exceptions.

e.

i. AGA: Stop codon

ii. UGA: Trp (W)

OR

The lac operon consists of one regulatory gene (the i gene – the term i is derived from the word inhibitor) and three structural genes (z, y and a). The i gene codes for the repressor of the operon. The z gene codes for beta-galactosidase (β-gal), which is responsible for the hydrolysis of the sugar lactose into monomeric units, galactose and glucose. The y gene codes for permease, which increases cell permeability to β -galactosidases. The a gene encodes a transacetylase. Hence, all three genes of the operon are involved in lactose metabolism.

Lactose which is the substrate for the enzyme β –galactosidases, also acts as an inducer of the operon by switching on and off of the operon. When lactose is provided in the medium as a carbon source, it is transported into the cells by the action of a little amount of permease present in the cell. This lactose then induces the operon by inactivating the repressor. When lactose was absent, the repressor produced by the i gene remains bound to the operator region and prevents RNA polymerase from transcribing the operon. Once the repressor is inactivated in the presence of lactose, RNA polymerase initiates the transcription process.

However, after sometime, the lac operon shuts down because of catabolite repression of the lac operon. The level of lactose decreases and that of glucose rises. If lactose and glucose are present, the cell will use glucose before the lac operon is turned on. This type of control is termed catabolite repression. To prevent lactose metabolism, the second level of control of gene expression exists. The promoter of the lac operon has two binding sites. One site is where RNA polymerase binds. The second location is the binding site for a complex between the catabolite activator protein (CAP) and cyclic AMP (cAMP). The binding of the CAP-cAMP complex to the promoter site is required for transcription of the lac operon. The presence of this complex is closely associated with the presence of glucose in the cell. With the increase in the concentration of glucose, the amount of cAMP decreases. Thus, the amount of complex decreases. This decrease in the complex inactivates the promoter, and the lac operon is turned off. Because the CAP-cAMP complex is needed for transcription, the complex exerts positive control over the expression of the lac operon.

Q33.

Read the statement given below and answer the following questions:

‘X’ is used to cut DNA at specific loci.

i. What is ‘X’ in the given statement?

ii. How can we use ‘X’ to detect differences in DNA of different people?

OR

Expand PCR. Explain the steps involved in the technique. What is the PCR machine called?

Ans:

i. ‘X’ is a restriction enzyme in the given statement. Restriction enzymes are used to cut DNA at specific loci.

ii. We can use restriction enzymes to detect differences in DNA of different people as given below:

a. Restriction enzymes are sensitive to changes in DNA sequences.

b. They recognise only a particular site.

c. Change of a single base at a restriction site can cause it to be “lost”, i.e., no longer recognised by the enzyme

d. It results in varying lengths of DNA fragments in different people having differences in DNA.

OR

Polymerase Chain Reaction.The steps involved are: – A PCR starts with a denaturing step. The DNA sample is heated to 94-96⁰C to break the hydrogen bonds between the bases of the two strands.

– When the two strands separate out, the temperature is lowered to 50-56⁰C, which allows the primers (short DNA fragments that acts as starting point for replication) to anneal (base pair) with the denatured strands.

– The temperature is raised to 72⁰C at which the Taq polymerase (a thermostable polymerase) elongates the primer by adding nucleotides.

The PCR machine is called a thermocycler.

​