Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 4 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

Class 12: Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 4 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

The document Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 4 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12 is a part of the Class 12 Course Sample Papers for Class 12 Medical and Non-Medical.
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Class-XII
Biology
Time: 90 Minutes
M.M: 40 Marks

General Instructions:

  1. The Question Paper contains four sections.
  2. Section A has 24 questions. Attempt any 20 questions.
  3. Section B has 24 questions. Attempt any 20 questions.
  4. Section C has 12 questions. Attempt any 10 questions.
  5. All questions carry equal marks.
  6. There is no negative marking.
Section - A

Q.1: Genetics deals with
(a) Relation between living things and environment
(b) Inheritance and variations
(c) Cell structure and functions
(d) Inheritance of characters

Correct Answer is Option (b)
Genetics deals with genes, genetic variation, gene mutation, and heredity; with a heavy focus on “trait inheritance'.


Q.2: Identify the four necessary ducts in a human male reproductive system.

Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 4 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

(a) 1- Vas deferens, 2-Rete testis, 3- Vasa efferentia, 4- Epididymis
(b) 1-, Vasa efferentia 2- Epididymis, 3- Vasa deferens, 4- Rete testis
(c) 1- Vas deferens, 2- Epididymis, 3- Vasa efferentia, 4- Rete testis
(d) 1- Vas deferens, 2- Epididymis, 3- Rete testis, 4- Vasa efferentia

Correct Answer is Option (c)
1- Vas deferens, 2- Epididymis, 3- Vasa efferentia, 4- Rete testis.


Q.3: Which one of the following pairs is correctly matched ?
(a) Okazaki fragments – Splicing
(b) RNA Polymerase – RNA Primer
(c) Central Dogma –Unidirectional
(d) AUG, ACG – Start

Correct Answer is Option (c)
Central Dogma was proposed by Francis Crick (1958). It states that the genetic information flows unidirectionally from DNA → RNA → Protein.


Q.4: From the statements given below choose the option that are true for a typical female gametophyte of a flowering plant:
(i) It is 8-nucleate and 7-celled at maturity
(ii) It is free-nuclear during the development
(iii) It is situated inside the integument but outside the nucellus
(iv) It has an egg apparatus situated at the chalazal end
(a) i and iv,
(b)  ii and iii
(c) i and ii
(d) ii and iv

Correct Answer is Option (c)
Statement (i) and (ii) are correct regarding female gametophyte of flowering plant. The female gametophyte or embryo sac is located inside the nucellus, enclosed within the integuments. In a majority of flowering plants, one of the megaspore is functional while the other three degenerates. Three repeated mitotic divisions of the functional megaspore results in the formation of seven-celled or eight nucleate embryo sac. Six of the eight nuclei are organised at the two poles. Three cells grouped at micropylar end forms egg apparatus and 3 at the chalazal end form antipodal cells. The large central cell at the centre has two polar nuclei. The meiotic divisions in the formation of embryo sac are strictly free nuclear, that is nuclear divisions are not followed immediately by cell-wall formation. Gametophyte is situated at micropylar end not at chalazal end.


Q.5: Functional unit of gene that specifies one polypeptide is
(a) Muton
(b) Codon
(c) Cistron
(d) Recon

Correct Answer is Option (c)
The term cistron is used to emphasize that genes exhibit a specific behaviour in a cis-trans test; distinct positions within a genome are cistronic.


Q.6: Mature Graafian follicle is generally present in the ovary of a healthy human female around stet.
(a) 5–8 days of menstrual cycle.
(b) 11–17 days of menstrual cycle.
(c) 18–23 days of menstrual cycle.
(d) 24–28 days of menstrual cycle.

Correct Answer is Option (b)
Mature Graafian follicle is the follicular stage present in the ovary. It is formed after the completion of first mitotic division but before ovulation. It therefore contains a 2N diploid oocyte. Graafian follicle is characterised by a large follicular antrum and releases one or more ova into the Fallopian tube and leaving behind the corpus luteum. It is generally present in the ovary of a healthy human female around 11–17 days of menstrual cycle.


Q.7: In operon model, RNA polymerase binds to
(a) Promoter gene
(b) Regulatory gene
(c) Operator gene
(d) Structural gene

Correct Answer is Option (a)
The genes present in the operon function together in the same or related metabolic pathway. There is an operator region for each operon. If there is no lactose (inducer), lac operon remains switched off. In the absence of inducer, repressor gene is active. The regulator gene synthesizes mRNA to produce the repressor protein, this protein binds to the operator genes and blocks RNA polymerase movement. So the structural genes are not expressed.


Q.8: During microsporogenesis, meiosis occurs in:
(a) Endothecium
(b) Microspore mother cells
(c) Microspore tetrads
(d) Pollen grains.

Correct Answer is Option (b)
During microsporogenesis, meiosis occurs in microspore mother cells. As the anther develops, the microspore mother cells of the sporogenous tissue undergo meiotic divisions to form microspore tetrads. The microspore tetrad after dehydration is separated into pollen grains.


Q.9: Pleiotropic gene is characterized by
(a) Multiple genotype
(b) Single genotype
(c) Multiple phenotype
(d) Single phenotype

Correct Answer is Option (a)
Pleiotropy is the phenomenon in which one gene controls many traits. For example, the gene in pea plants that controls the round and wrinkled texture of seeds also influences the phenotypic expression of starch grain size. Therefore, Pleiotropic gene is characterized by multiple phenotype.


Q.10: Reverse transcriptase is
(a) RNA dependent RNA polymerase
(b) RNA dependent DNA polymerase
(c) DNA dependent RNA polymerase
(d) DNA dependent DNA polymerase

Correct Answer is Option (b)
Reverse transcriptase, also known as RNA-dependent DNA polymerase, is a DNA polymerase enzyme that transcribes singlestranded RNA into DNA.


Q.11: Match between the following representing parts of the sperm and their functions and choose the correct option.
(a) A-ii, B-iv, C-i, D-iii
(b) A-iv, B-iii, C-i, D-ii
(c) A-iv, B-i, C-ii, D-iii
(d) A-ii, B-i, C-iii, D-iv

Correct Answer is Option (b)
A. Sperm head contains nucleus with densely coiled chromatin fibres surrounded by acrosome. Nucleus transfers the genetic material to next generation.
B. Middle piece of sperm contains large number of mitochondria to provide energy needed for the movement.
C. Acrosome present in the cap of sperm contains hydrolytic enzymes which help sperm to penetrate the egg. These enzymes break down the outer membrane of the ovum, called the zona pellucida, allowing the haploid nucleus in the sperm cell to join with the haploid nucleus in the ovum.
D. Tail of sperm helps in propelling or swims the sperm cell forwards to meet the egg.


Q.12: An individual with two identical alleles is
(a) Hybrid
(b) Dominant
(c) Homozygous
(d) Heterozygous

Correct Answer is Option (c)
Homozygous means having two of the same allele in the genotype.


Q.13: In prokaryotes, functionally related genes are sometimes position adjacent to each other in the genome and can under the control of the same regulatory machinery. What are these called?
(a) Promoters
(b) Operators
(c) Repressors
(d) Operons

Correct Answer is Option (d)
Prokaryotic organisms often have functionally related genes joined together on the chromosome under the direction of a single promoter called operons. Operons have additional sequences, called operators that can be bound by either repressor or activator proteins, which will repress or activate transcription of the operon.


Q.14: Which one is the not recessive in humans
(a) Albinism
(b) Rh factor
(c) Colour Blindness
(d) Haemophilia

Correct Answer is Option (b)
The Rh-factor is dominant whereas albinism, colour blindness and haemophilia are recessive in humans.


Q.15: The total number of nuclei involved in double fertilisation in angiosperm are
(a) two
(B) three
(c) four

(d) five

Correct Answer is Option (d)
Double fertilisation is the process in angiosperms. It involves fusion of one male gamete (haploid) with egg (haploid) to form zygote (diploid) that gives rise to embryo accompanied with fusion of other male gamete (haploid) with two polar nuclei (secondary nucleus) to form primary endosperm nucleus (PEN) that gives rise to a nutritive tissue called endosperm.


Q.16: When two organisms differing in atleast one set of characters are crossed, the Offspring of cross is called :
(a) Variant
(b) Monoploid
(c) Hybrid
(d) Mutant

Correct Answer is Option (c)
A hybrid is the offspring resulting from combining the qualities of two organisms of different breeds, varieties, species or genera through sexual reproduction.


Q.17: The figure given below is of DNA replication
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 4 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
Identify the enzyme ‘E’
(a) DNA polymerase
(b) DNA ligase
(c) RNA polymerase
(d) Transacetylase

Correct Answer is Option (b)
The DNA ligase enzyme joins or seals the discontinuous fragments of DNA. It helps in joining the DNA strands together by catalysing the formation of phosphodiester bond. It also plays an important role in repairing the single strand break in DNA duplex. It also plays an important role in joining the discontinuously synthesized fragments of  lagging strand (okazaki fragments) of DNA.


Q.18:
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 4 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
The cross shown above is an example of
(a) Co-dominance
(b) Monohybrid cross
(c) Incomplete Dominance
(d) Both (B) and (C)

Correct Answer is Option (c)
Incomplete dominance is a form of intermediate inheritance in which one allele for a specific trait is not completely expressed over its paired allele. This results in a third phenotype in which the expressed physical trait is a combination of the phenotypes of both allele.  Examples of incomplete dominance can be given by inheritance of flower colour in the dog flower (Snapdragon or Antirrhinum sp.)  In a cross between true-breeding redflowered (RR) and true breeding whiteflowered plants (rr), the F1 progenies were pink (Rr).   When the F1 was self-pollinated the F2 resulted in the following ratio 1 (RR) Red: 2 (Rr) Pink : 1 (rr) White.   Here, the genotype ratios were exactly as we would expect in any Mendelian monohybrid cross, but the phenotype ratios had changed from the 3 : 1, dominant : recessive ratio.  Since, R was not completely dominant over r, this made it possible to distinguish Rr as pink from RR (red) and rr (white).


Q.19: In a fertilised embryo sac, the haploid, diploid and triploid structures are:
(a) Synergid, zygote and primary endosperm nucleus
(b) Synergid, antipodal and polar nuclei
(c) Antipodal, synergid and primary endosperm nucleus
(d) Synergid, polar nuclei and zygote.

Correct Answer is Option (a)
In a fertilised embryo sac, the haploid, diploid and triploid structures are synergids, zygote and primary endosperm nucleus respectively.


Q.20: At what stage of life is oogenesis initiated in a human female ?
(a) At puberty
(b) During menarch
(c) During menopause
(d) During embryonic development

Correct Answer is Option (d)
During embryonic development the process of formation of haploid ova from diploid germinal cell is called oogenesis and the process occurred in the ovary. Oogenesis begins during embryonic development but is completed only at puberty of the secondary oocyte with the sperm.


Q. 21.
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 4 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

Identify A and B and choose the correct statement from the following:
(a) B flower always exhibit autogamy
(b) A flower always exhibit geitonogamy
(c) B flower exhibit both autogamy and geitonogamy
(d) A flower never exhibit autogamy

Correct Answer is Option (d)

A is chasmogamous flower  and B is cleistogamous flower. The pollination that occurs in opened flowers is called chasmogamy. It is of two types, that is, self-pollination (autogamy) and cross-pollination. Cross-pollination is of two types, that is, geitonogamy and xenogamy. So, we can say that chasmogamous flowers exhibit both autogamy (self-pollination) and allogamy (cross pollination). While, in cleistogamous flower the anthers and stigma lies close to each other within the closed flowers. When anthers dehisce in the flower buds, pollen grains come in contact with the stigma for effective pollination. Thus, these flowers are invariably autogamous as there is no chance of cross-pollen landing on the stigma.


Q.22: Which among the following has 23 chromosomes?
(a) Spermatogonia
(b) Zygote
(c) Secondary oocyte
(d) Oogonia

Correct Answer is Option (c)
Secondary oocyte (n = 23). Primary oocyte completes first meiotic division to form secondary oocyte (23 chromosomes) and polar body (23 chromosomes), whereas spermatogonia, zygote and oogonia have 46 chromosomes, hence diploid.


Q.23: Identify the figure given above and from among the sets of terms given below, choose the set that has those which are associated with the figure.
(a) Stigma, ovule, embryo sac, placenta
(b) Thalamus, pistil, style, ovule
(c) Ovule, ovary, embryo sac, tapetum
(d) Ovule, stamen, ovary, embryo sac

Correct Answer is Option (a)
The given figure is of gynoecium. Gynoecium indicates the female reproductive part of the flower which consists of pistil. Each pistil has three parts, that is, stigma, style and ovary. Inside the ovarian cavity, the placenta is located. Arising from the placenta there are the megasporangia, commonly called ovules. The functional megaspore undergoing the meiotic division develops into the female gametophyte or embryo sac. Thalamus, tapetum and stamen are not a part of gynoecium. Thalamus is the part of flower which form the base on which all the floral whorls rest up on. Tapetum is the inner most nutritive layer or microsporangium and stamens are male reproductive part (androecium) of plant.


Q.24: The phenomenon observed in some plants wherein parts of the sexual apparatus is used for forming embryos without fertilisation is called:
(a) Parthenocarpy
(b) Apomixis
(c) Vegetative propagation
(d) Sexual reproduction.

Correct Answer is Option (b)
Apomixis refers to the formation of seeds without fertilization. The embryos are genetically identical to the parental plant.

Section - B

Direction: Question No. 25 to 28 consist of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below:

Q. 25. Assertion (A) : Sponges are chemical methods of contraception.
Reason (R) : Diaphragms and cervical caps are disposable barrier method of contraception.
(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) A is true but R is false
(d) A is False but R is true

Correct Answer is Option (b)
Diaphragms, cervical caps and vaults are made of rubber, inserted into the female reproductive tract to cover the cervix before coitus. They prevent fertilization by blocking the entry of sperms through the cervix. These barriers are reusable.


Q.26: Assertion (A) : In RNA uracil is present at the place of thymine.

Reason (R) : 5-methyl uracil is chemical name of thymine.
(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) A is true but R is false
(d) A is False but R is true

Correct Answer is Option (b)
Since the half-life of the RNA molecules is shorter uracil would suffice to achieve the function of RNA. On the other hand, DNA remains same until cell dies/divides. The functions of thymine and uracil are the same.


Q. 27: Assertion (A): To achieve zero population growth rate, the replacement level should be slightly higher than two.
Reason (R) : Replacement level means the number of children that can replace parent equally
(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) A is true but R is false
(d) A is False but R is true

Correct Answer is Option (a)
Replacement level means the number of children that can replace parent equally. To achieve zero population growth rate, the replacement level should be slightly higher than two.


Q. 28: Assertion (A) : Haemophilia is an autosomal disorder.
Reason (R) : A haemophilic father can never pass the gene for haemophilia to his son.
(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) A is true but R is false
(d) A is False but R is true

Correct Answer is Option (d)
Haemophilia is a sexlinked recessive disorder in which X-chromosome has the haemophilic gene. A haemophilic father can never pass the gene for haemophilia to his son.


Q.29: During an experiment teacher showed a dicotyledonous plant that bears flower, she told the student that it never produces fruits and seeds. The most probable cause for the above situation is:
(a) Plant is dioecious and bear only pistillate flower
(b) Plant is dioecious and bear both pistillate and staminate flower
(c) Plant is monoecious and bear only staminate flower
(d) Plant is dioecious and bear only staminate flower

Correct Answer is Option (d)
Plant is dioecious and bears only staminate flowers. A dicotyledonous plant with flowers but no fruits and seeds could be a staminate (male) flower, with the plant being dioecious, that is, bearing male and female flowers on two separate plants. Here, the plant bearing female or pistillate flowers will give rise to fruits and seeds. These plants undergo cross pollination, majorly through Anemophily.


Q.30: In males the immature male germ cells undergo division to produce sperms by the process of spermatogenesis. Choose the correct one with reference to above.
(a) Spermatogonia have 46 chromosomes and always undergo meiotic cell division.
(b) Primary spermatocytes divide by mitotic cell division.
(c) Secondary spermatocytes have 23 chromosomes and undergo second meiotic division.
(d) Spermatozoa are transformed into spermatids

Correct Answer is Option (c)
Spermatogonia have 46 chromosomes and always undergo mitotic cell division. (B) Primary spermatocytes divide by meiotic cell division. (D) Spermatids are transformed into spermatozoa.


Q.31: Shravan is one year old his mother wants to have space between her children. The best contraceptive method she should use is
(a) Oral contraceptives
(b) Copper-T
(c) Tubectomy
(d) Diaphragm

Correct Answer is Option (b)
Copper-T is a contraceptive method under intra uterine contraceptive device which prevents the implantation and reduce the motility of sperm. It is implanted inside the uterus and works up to five years. So, it is the best method for spacing between two children.


Q.32: DNA is a polymer of nucleotides which are linked to each other by 3’-5’phosphodiester bond. To prevent polymerisation of nucleotides, which of the following modifications would you choose?
(a) Replace purine with pyrimidines.
(b) Remove/Replace 3’ OH group in deoxy ribose.
(c) Remove/Replace 2’ OH group with some other group in deoxyribose.
(d) Both (B) and (C).

Correct Answer is Option (b)
 If 3’ OH group is removed/ replaced in deoxyribose, there will be no formation of phosphodiester bonds this will prevent polymerisation of nucleotides.


Q.33: A dicotyledonous plant bears flowers but never produces fruits and seeds. The most probable cause for the above situation is:
(a) Plant is dioecious and bears only pistillate flowers
(b) Plant is dioecious and bears both pistillate and staminate flowers
(c) Plant is monoecious
(d) Plant is dioecious and bears only staminate flowers

Correct Answer is Option (b)
In dioecious plants, the unisexual male flower is staminate, that is, bearing stamens only, while the female is pistillate or bearing pistils only. For the production of fruits and seeds fertilization must take place, which is possible only in the presence of both male and female flowers. When the plant is dioecious, it will give rise to the following situations :
(i)  If the plant is dioecious and bears only pistillate flowers, fertilization can take place with the help of pollinators.
(ii) If the plant is dioecious and bears only staminate flowers, fertilization cannot take place, because female gamete is non-motile which can’t reach the male gamete in order to fuse with it. When the plant is monoecious, that is, carrying both stamen and pistil together, it may lead to self fertilization and production of seed.


Q.34: Regulatory proteins are the accessory proteins that interact with RNA polymerase and affect its role in transcription.   Which of the following statements is correct about regulatory protein?
(a) They only increase expression.
(b) They only decrease expression.
(c) They interact with RNA polymerase, but do not affect the expression.
(d) They can act both as activators and as repressors.

Correct Answer is Option (d)
Regulatory proteins, the accessory proteins that interact with RNA polymerase and affect its role in transcription. It controls the functions of structural genes and are called regulatory genes. Promoters, terminators, operators and repressor are some important regulatory genes. They can act both as activators and as repressors. both as activators and as repressors.


Q.35: In a flower, if the megaspore mother cell forms megaspores without undergoing meiosis and if one of the megaspores develops into an embryo sac, its nuclei would be:
(a) Haploid
(b) Diploid
(c) A few haploid and a few diploid
(d) With varying ploidy.

Correct Answer is Option (b)
In some species, the diploid egg cell is formed without reduction division and develops into an embryo without fertilization. It is an a sexual reproduction which occurs in the absence of pollinators or in extreme environments. In some species like citrus plants, nucellar cells surrounding the embryo sac start dividing and develop into embryos. It occurs in the megaspore mother cell without undergoing meiosis and produces diploid embryo sac through mitotic divisions. It helps in the preservation of desirable characters for indefinite period. Thus, it can be concluded that apomictic species produce diploid cells. Haploid cells will be formed during sexual reproduction when cell will undergo meiosis.


Q.36. If the sequence of nitrogen bases of the coding strand of DNA in a transcription unit is : 5'-ATGAATG-3', the sequence of bases in its RNA transcript would be
(a) 5'-AUGAAUG-3'
(b) 5'-UACUUAC-3'
(c) 5'-CAUUCAU-3'
(d) 5'-GUAAGUA-3'

Correct Answer is Option (a)
5'-ATGAATG-3' (coding strand) 5'-TACTTAC-3'  (complementary strand) 5'-AUGAAUG-3' (RNA)


Q.37: A woman has an X-linked condition on one of her X-chromosomes. This chromosome can be inherited by
(a) Only daughters
(b) Both sons and daughters
(c) Only grandchildren
(d) Only sons

Correct Answer is Option (b)
Here is this case, the women is a carrier. Both the son and daughter will inherit the X-chromosome but only the son will be diseased.


Q.38: (i) Promotes the supply of nutrients and oxygen to the embryo.
(ii) Facilitates the elimination of excretory wastes and carbon dioxide produced by the embryo.
(iii) Aids in the transportation of substances to and from the embryo.

The function discussed above is performed by:
(a) Ovary
(b) Placenta
(c) Fallopian tube
(d) Oviduct

Correct Answer is Option (b)
The placenta supply nutrients and oxygen to the embryo from mother’s blood and also withdraw excretory wastes and CO2 from the embryo. Placenta is connected to the embryo through the umbilical cord.


Q.39: Raman is planning for an artificial hybridization programme involving dioecious plants, which of the following steps would not be relevant for this programme:

(a) Bagging of female flower
(b) Dusting of pollen on stigma
(c) Emasculation
(d) Collection of pollen

Correct Answer is Option (c)
If the flower is dioecious then only bagging, collection of desired pollen grains and a dusting of pollen grains on stigma are done. The technique of emasculation is not carried out. Therefore the correct answer is option C, that is emasculation.


Q.40: A cross between two tall plants resulted in offspring having few dwarf plants. What would be the genotypes of both the parents?
(a) TT and Tt
(b) Tt and Tt
(c) TT and TT
(d) Tt and tt

Correct Answer is Option (b)
The genotypes of both the parents are Tt and Tt. Refer the given cross between true breeding tall plants and true breeding dwarf plants. When true breeding plants were crossed to each other, this is called a parental cross and offspring comprise the first filial or F1 - generation. When the members of the F1 - generation were crossed, this produced the F2-generation or second filial generation. A cross between true breeding tall and dwarf plants of the parent generation yield phenotypically tall plants. The cross between TT and Tt is called back cross, which results into two homozygous and two heterozygous dominant gametes. The cross between Tt and tt is called test cross which results into 1 : 1 ratio of gametes.


Q.41: During reproduction, the chromosome number (2n) reduces to half (n) in the gametes and again resume the original number (2n) in the offspring. What is the process through which these events take place?
(a) Gametogenesis
(b) Fertilisation
(c) Spermatogenesis
(d) Both (A) and (C)

Correct Answer is Option (b)
Halving of chromosomal number takes place during gametogenesis and regaining the 2n number occurs as a result of fertilisation.


Q.42: F2 generation in a Mendelian cross showed that both genotypic and phenotypic ratios are same as 1 : 2 : 1. It represent a case of :
(a) Co-dominance
(b) Dihybrid cross
(c) Monohybrid cross with incomplete dominance
(d) Monohybrid cross with complete dominance

Correct Answer is Option (c)
Monohybrid cross with incomplete dominance shows both genotypic and phenotypic ratio as same (1 : 2 : 1).


Q.43: Person having genotype IA IB would show the blood group as AB. This is because of
(a) Pleiotropy.
(b) Co-dominance.
(c) Segregation.
(d) Incomplete dominance.

Correct Answer is Option (b)
ABO blood grouping in humans is an example of co-dominance. ABO blood groups are controlled by gene I. Gene I has three alleles IA, IB, and Ii. When IA, and IB are present together, both express equally and produce the surface antigens A and B, whereas I is the recessive allele and does not produce any antigen. Pleiotropy referred the genetic effect of a single gene on multiple phenotypic traits. Incomplete dominance does not completely dominate another allele. Segregation is the separation of allele during the process of gametogenesis. This is the basis of reappearance of recessive character in F2 generation.


Q.44: In case of a couple where the male is having a very low sperm count which technique will be suitable for fertilization?
(a) Intrauterine transfer
(b) Gamete intracytoplasmic fallopian transfer
(c) Artificial Insemination
(d) Intracytoplasmic sperm injection

Correct Answer is Option (b)
Artificial insemination (AI) is a technique in which the semen collected from the husband or a healthy donor is artificially introduced either into the vagina or the uterus.


Q.45: While an mRNA strand is being translated in the ribosome subunit, the triplets in sequence were UAC and UAG. UAC codes for which amino acid?
(a) Phenylalanine
(b) Leucine
(c) Methionine
(d) Tyrosine

Correct Answer is Option (b)
UAC codes for tyrosine whereas UAG acts as terminator codon thus leads to the termination of polypeptide chain.


Q.46: In male, luteinizing hormones(LH) is called ICSH (Interstitial cells stimulating hormone) which stimulates the leydig cells of the testes to secrete testosterone hormone but in female, luteinzing hormone
(a) Promotes ovulation
(b) Controls the formation of corpus luteum after ovulation.
(c) Causes uterine contraction.
(d) Both (A) and (B)

Correct Answer is Option (d)
In female, luteinzing hormone promotes ovulation and controls the formation of corpus luteum after ovulation.


Q.47: In a certain taxon of insects some have 17 chromosomes and the others have 18 chromosomes. The 17 and 18 chromosome-bearing organisms are
(a) Males and females, respectively.
(b) Females and males, respectively.
(c) All males.
(d) All females.

Correct Answer is Option (a)
In certain taxon of insects, 17 and 18 chromosome bearing organisms are males and females respectively. Because, insects have XO type of sex determination method. In certain insects, such as cockroach, and some roundworms lacks Y-chromosome, so that the male has only one sex chromosome, that is, ‘X’ besides autosomes. This condition in the male is designated as XO (where O means absence of one sex chromosome) and in the female it is XX.


Q.48: The reproductive cycle in the human female and related primates is called the ___________. The first menstruation begins at puberty and is called __________. In human females, menstruation is repeated at an average interval of about _________ days.
(a) Menopause, menarch, 28/29
(b) Menstrual cycle, menarche, 28/29
(c) Menstrual cycle, menopause, 28/29
(d) Menstrual cycle, menarche, 20/25

Correct Answer is Option (b)
The first menstruation begins at puberty (at the age of 10-12 years) and is called menarche. In human females, menstruation is repeated at an average interval of about 28/29 days and the cycle of events starting from one menstruation till the next one is called the menstrual cycle.

Section - C

Case-I 

Direction : Section-C consists of one case followed by 6 questions linked to this case (Q.No.49 to 54).
Read the following text and answer the following questions on the basis of the same:
Down syndrome (sometimes called Down’s syndrome) is a condition in which a child is born with an extra copy of their 21st chromosome hence its other name, trisomy 21.  The affected individual mental retarded, short statured with small round, head, furrowed tongue and partially open mouth, Physical, psychomotor and mental development is retarded.

Q.49: The number of chromosomes a child with Down syndrome has is
(a) 45 

(b) 46 

(c) 47 

(d) 48

Correct Answer is Option (c)
The affected person inherited with one extra copy of 21st chromosome that forms trisomy condition.


Q.50: Down syndrome is
(a) Sex-linked
(b) Chromosomal
(c) dominant
(d) recessive

Correct Answer is Option (d)
Down syndrome is an autosomal recessive disorder which can be inherited through normal parents in the child.


Q.51: One of this trait is seen in a person with Down syndrome
(a) Upward slant eye
(b) Baldness
(c) Small stature
(d) Long neck

Correct Answer is Option (c)
The person affected with Down syndrome has symptoms like mental retarded, short statured with small round, head, furrowed tongue etc.


Q.52: Down Syndrome is an extra copy which chromosome
(a) 22nd chromosome
(b) 21st chromosome
(c) 45th chromosome
(d) 47th chromosome

Correct Answer is Option (b)
Down Syndrome is due to extra copy of 21st chromosome forming trisomy condition.


Q.53: Down syndrome is caused due to
(a) Bacterial infection
(b) A chromosomal abnormality lack of oxygen supply to the brain during birth
(c) Viral infection
(d) A chromosomal abnormality

Correct Answer is Option (d)
Down syndrome is due to autosomal chromosome abnormality.


Q.54: This disorder was first described by Langdon Down in
(a) 1856
(b) 1865
(c) 1866
(d) 1876

Correct Answer is Option (c)
Down syndrome was first described by Langdon Down in 1866.


Q.55: The graph given below shows the variation in the levels of ovarian hormones during various phases of menstrual cycle. Identify ‘A’ and ‘B’
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 4 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
(a) A–Estrogen ,  B–Progesterone
(b) A–Progesterone B–Estrogen
(c) A-Oxytocin,  B- Progestrone
(d) A – Estrogen,  B- Oxytocin

Correct Answer is Option (a)
A- estrogen and B – progesterone. The formation of Graafian follicle (releases estrogen) which is followed by the formation of corpus luteum (releases progesterone).

The role of estrogen leads to changes in the ovary and uterus/regeneration of endometrium through proliferation. The role of  progesterone is,maintenance of endometrium for implantation of the fertilized ovum and maintenance of other events of pregnancy.


Q.56: Given is a section of a Maize grain. Identify ‘A’ and state its function
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 4 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
(a) Coleorhiza, protecting the radicle.
(b) Coleoptile, protecting the shoot plumule
(c) Plumule, produces food for the growing embryonic plant
(d) Radicle, suck up water for the growing embryonic plant

Correct Answer is Option (b)
In AAC, one ‘A’ is replaced by ‘T’  hence, it is single base substitution that is, a Point mutation.


Q.57: Mention the polarity of DNA strands a-b & c-d shown in the replicating fork given below :
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 4 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

(a) Polarity of DNA strand a–b is 5'→3' and c–d polarity is 3'→5'
(b) Polarity of DNA strand a–b is 3'→5' and c–d polarity is 5'→3'.
(c) Polarity of DNA strand a–b is 3'→5' and c–d polarity is 3'→5'
(d) Polarity of DNA strand a–b is 5'→3' and c–d polarity is 5'→3'

Correct Answer is Option (b)
In the given replicating strand the polarity of DNA strand a–b is 3'→5'and  c–d has polarity 5'→3' .


Q.58: The below diagram shows human male reproductive system (one side only). Identify ‘X’ and ‘Z’ and  the accessory gland ‘Y’
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 4 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
(a) X- Epididymis, Y- Seminal vesicle, Z- Testicular lobules
(b) X- Testicular lobules, Y- Prostrate , Z- Epididymis
(c) X — Testicular lobules, Y — Seminal vesicle, Z — Epididymis
(d) X – Vasa efferentia, Y- Prostrate, Z- Epididymis

Correct Answer is Option (c)
X is Testicular lobules. Each testis contains about 250 compartments called testicular lobules. Each testicular lobules contain one to three highly coiled seminiferous tubules, in which sperms are produced. Y is Seminal vesicle. It secretes seminal plasma. Z is Epididymis and its function is to store sperms.


Q.59: Identify strands ‘A’ and ‘B’ in the diagram of transcription unit given below
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 4 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
(a) A-Template strand and B- Coding strand
(b) A - Coding strand and B- Coding strand
(c) Both A and B are Template strands
(d) Both A and B are Coding strands

Correct Answer is Option (a)
On the basis of polarity with respect to promoter, A is Template strand as it has polarity 3’-5’ and B is Coding strand as it has polarity 5’-3’ .


Q.60: The lac operon is represented in the diagram given below. The lac operon has 3 structural genes : z, y and a. These 3 genes code for three different enzymes (i). (ii) and (ii) respectively. Identify (i), (ii) and (iii)
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 4 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
(a) (i) Permease  (iii) Transacetylase (iii) ) β-galactosidase
(b) (i) β-galactosidase (ii) Permease  (iii) Transacetylase.
(c) (i) Transacetylase (ii) β-galactosidase(iii) Permease
(d) (i) β-galactosidase(ii) Transacetylase (iii) Permease

Correct Answer is Option (b)
3 structural genes:
(i) z gene : Codes for β-galactosidase (hydrolyze lactose to galactose and glucose).
(ii) y gene : Codes for permease (increase permeability of the cell to lactose).
(iii) a gene : Codes for a transacetylase.

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