Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 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)- 2 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)- 2 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: Which of the following pairs is wrongly matched.
(a) Starch synthesis in pea : Multiple alleles
(b) ABO blood grouping : Co-dominance
(c) Flower colour in Snapdragon : Incomplete dominance
(d) T.H. Morgan : Linkage

Correct Answer is Option (a)
Starch synthesis in pea seed is controlled by one gene which has two alleles (B and b). Pleiotropy is shown by B gene.


Q.2: The factors that play a vital role on type of reproduction adopted by an organism are:
(a) Morphology of the organism
(b) Habitat of the organism
(c) Physiology of the organism
(d) All of the above

Correct Answer is Option (a)
There are various types of reproduction. The type of reproduction adopted by an organism depends on the organism’s habitat, physiology, morphology and genetic makeup.


Q.3: The first genetic material could be
(a) Protein
(b) Carbohydrates
(c) DNA
(d) RNA

Correct Answer is Option (d)
RNA was the first genetic material. There is now enough evidence to suggest that essential life processes (e.g., metabolism, translation and splicing), evolved from RNA. RNA is used to act as a genetic material as well as a catalyst (there are some important biochemical reactions in living systems that are catalysed by RNA catalysts and not by protein enzymes). But, RNA being a catalyst was reactive and hence unstable. Therefore, DNA has evolved from RNA with chemical modifications that make it more stable. DNA being double-stranded and having complementary strand, further resists changes by evolving a process of repair.


Q.4: Which of the following hormone is released by placenta?
(a) FSH
(b) HCG
(c) Relaxin
(d) LH

Correct Answer is Option (b)
Human chorionic gonadotropin hormone (HCG). This hormone is only produced during pregnancy, almost exclusively in the placenta.


Q.5: With regard to mature mRNA in eukaryotes
(a) Exons and introns do not appear in the mature RNA.
(b) Exons appear, but introns do not appear in the mature RNA.
(c) Introns appear but exons do not appear in the mature RNA.
(d) Both exons and introns appear in the mature RNA.

Correct Answer is Option (b)
In eukaryotes, the mono-cistronic structural genes have interrupted coding sequences, that is, the genes in eukaryotes are split. The coding sequences or expressed sequences are defined as exons. These sequences (exons) appear in mature or processed RNA, thus exons are interrupted by introns or intervening sequences which do not appear in mature or processed RNA.


Q.6: All genes located on the same chromosome
(a) Form different groups depending upon their relative distance.
(b) Form one linkage group.
(c) Will not from any linkage groups.
(d) Form interactive groups that affect the phenotype.

Correct Answer is Option (b)
All the genes, present on a particular chromosome form a linkage group. The number of linkage group of a species correspond to the total number of different chromosomes of that species. It is not simply the number of chromosomes in haploid set. For example, in human male there are 22 pairs of autosomes and X and Y sex chromosomes, that is, 24 linkage groups and in  female = 22 pairs autosomes + 2X-chromosomes, that is, 23 linkage groups.


Q.7: One of the following is true with respect to AUG.
(a) It codes for methionine only.
(b) It is also an initiation codon.
(c) It codes for methionine in both prokaryotes and eukaryotes.
(d) All of these

Correct Answer is Option (d)
Polypeptide synthesis is signalled by two initiation codons - commonly AUG or methionine codon and rarely GUG or valine codon. AUG serves two main functions. It signals the start of translation and codes for the incorporation of the methionine into the growing polypeptide chain. AUG codes for methionine in both prokaryotes and eukaryotes. 


Q.8: In triple fusion:

(a) Two polar nuclei and one male gamete are involved

(b) One polar nuclei and one male gamete are involved
(c) One polar nuclei and two male gamete are involved
(d) Two polar nuclei and two male gamete are involved

Correct Answer is Option (a)
Fusion of male gamete with diploid secondary nucleus to form triploid primary endosperm nucleus is known as triple fusion. In triple fusion two polar nuclei and one male gamete are involved.


Q.9: Distance between the genes and percentage of recombination shows
(a) A direct relationship.
(b) An inverse relationship.
(c) A parallel relationship.
(d) No relationship.

Correct Answer is Option (a)
Crossing over separates genes away from each other. So, the physical distance between the two genes and percentage of recombination shows a direct relationship. More the distance between two genes, more is the frequency of crossing over between them and hence more is the percentage of recombination.


Q.10: The promoter site and the terminator site for transcription are located at
(a) 3’ (downstream) end and 5’ (upstream) end, respectively of the transcription unit.
(b) 5’ (upstream) end and 3’ (downstream) end, respectively of the transcription unit.
(c) The 5’ (upstream) end.
(d) The 3’ (downstream) end.

Correct Answer is Option (b)
The promoter site and the terminator site for transcription are located at 5' (upstream) end and 3' (downstream) end, respectively of the transcription unit. The promoter is the binding site for RNA polymerase for initiation of transcription.


Q.11: Mesonephric duct is also known as _______.
(a) Wolffian duct
(b) Ejaculatory duct
(c) Major sublingual duct
(d) Cystic duct

Correct Answer is Option (a)
The Wolffian duct also known as the mesonephric duct is one of the paired embryogenic tubules that drain the primitive kidney (mesonephros) to the cloaca.


Q.12: Apple is a false fruit because:
(a) Ovary develops into the fruit
(b) Thalamus develops into fruit
(c) Ovules develops into the fruit
(d) Fertilisation does not takes place

Correct Answer is Option (b)

The fruits are formed from the ovary of a flower after fertilisation. Such fruits that develop from the ovary are called true fruits. On the other hand, fruits formed from any part of the flower other than ovary are called false fruits. Apple is called a false fruit because it develops from the thalamus and not from the ovary.


Q.13: How many spermatozoa does one secondary spermatocyte produce?
(a) Two diploids
(b) One haploid
(c) Three haploids
(d) Four haploids

Correct Answer is Option (b)
The secondary spermatocytes undergo meiotic division – II to generate four haploid spermatids which through the process of spermiogenesis are transformed into spermatozoa.


Q.14: Which one of the following is not a part of transcription unit in DNA ?
(a) The inducer
(b) A terminator
(c) A promoter
(d) The structural gene

Correct Answer is Option (a)
The segment of DNA that takes part in transcription is called transcription unit. It has three components (i) a promoter (ii) the structural gene and (iii) a terminator.


Q.15: Find the correct match:
(1) Plants that have both male and female sex organs on flowers-monoecious
(2) Plants having either male or female sex organ on flowers- dioecious
(3) Part of embryonal axis below the cotyledon- hypocotyl
(4) Part of embryonal axis above the cotyledon- epicotyl

(a) 1 and 3
(b) 2 and 4
(c) All of them are correct
(d) None of these

Correct Answer is Option (c)
(1)  Plants that have both male and female sex organs on flowers monoecious.
(2)  Plants having either male or female sex organ on flowers- dioecious.


Q.16: The RNA polymerase holoenzyme transcribes
(a) The promoter, structural gene and the terminator region.
(b) The promoter and the terminator gene.
(c) The structural gene and the terminator regions.
(d) The structural gene only.

Correct Answer is Option (c)
The RNA polymerase holoenzyme transcribes the structural gene and the terminator regions. RNA polymerase consists of a number of sub-units, including a sigma factor (transcription factor) that catalyses the process of transcription. It recognises the start signals or promoter region on DNA which then along with RNA polymerase binds to promoter to initiate the transcription. In eukaryotes there are three RNA polymerases : I, II and III. The process includes a proofreading mechanism.


Q.17: Which of the following will not result in variations among siblings?
(a) Independent assortment of genes
(b) Crossing over
(c) Linkage
(d) Mutation

Correct Answer is Option (c)
Linkage may be defined as the relationship between genes on the same chromosome that causes them to be inherited together, therefore it will not result in variations among siblings. In linkage there is a tendency to maintain the parental gene recombination except for occasional crossovers.  Independent assortment of genes means that allele pair separate during the formation of gametes independently; it means that traits are transmitted to offspring independently of one another. Crossing over is the exchange of genetic material between homologous chromosomes. It is one of the final phases of genetic recombination. Mutation is the sudden inheritable change in genetic material of an organism which transfers to next generation.


Q.18: Testosterone is a hormone released by:
(a) Testes
(b) Urethra
(c) Penis
(d) Prostrate

Correct Answer is Option (a)
Testosterone is the male sex hormone that is released by testes. Testosterone hormone levels are important to normal male sexual development and functions.


Q.19: The Government has banned amniocentesis to check on:
(a) Population explosion
(b) Sexually transmitted diseases
(c) The incidences of female foeticides
(d) Intense lactation

Correct Answer is Option (c)
The Government has banned amniocentesis to check on the incidences of female foeticides.


Q.20: A human female with Turner's syndrome.
(a) Has 45 chromosome with XO.
(b) Has one additional X chromosome
(c) Exhibits male characters.
(d) Is able to produce children with normal husband

Correct Answer is Option (a)
It is a disorder caused due to the absence of one of the X-chromosomes, i.e., 45 with XO.


Q.21: Identify the terms that are not technically correct names for a floral whorl:
(i) Androecium
(ii)  Petal
(iii) Corolla
(iv) Sepal
(a) (i) and (ii)
(b) (ii) and(iii)
(c) (ii) and (iv)
(d) All of the above

Correct Answer is Option (c)
The technically correct terms for the floral whorls are (from outermost to innermost) calyx, corolla, androecium and gynoecium. They are made up of sepals, petals, stamens and carpels respectively.


Q.22: Match the following and select the correct option:
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

(a) a: ii, b: i, c: iii
(b) a: i, b: ii, c: iii
(c) a: iii, b: i, c: ii
(d) a: ii, b: iii, c: i

Correct Answer is Option (a)
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12


Q.23: MTP is:
(a) Medical Termination of Pregnancy
(b) Medicinal Termination of Pregnancy
(c) Medicinal Termination of Progesterone
(d) Medical Termination of productivity

Correct Answer is Option (a)
MTP stands for Medical Termination of Pregnancy which is a procedure of terminating pregnancy using medicines.


Q.24: The uterus is also called
(a) Fimbriae
(b) Isthmus
(c) Ampulla
(d) Womb

Correct Answer is Option (d)
The uterus is where a foetus (unborn baby) develops and grows. Also called womb.

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) : Uterine tube, fallopian tube, and oviduct are terms used to refer to the different organs.
Reason (R) : These transport polar bodies formed during oogenesis from the ovaries to the uterus.
(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)
The uterine tubes, also known as oviducts or fallopian tubes, are the female structures that transport the ova from the ovary to the uterus each month. In the presence of sperm and fertilization, the uterine tubes transport the fertilized egg to the uterus for implantation. 


Q.26: Assertion (A) : Family planning is an action plan to attain reproductive health among people.
Reason (R) : Improved programs covering reproduction related areas were propagated by RCH to create awareness among people.
(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)
India was amongst the first countries in the world to initiate action plans and programs at a national level to attain total reproductive health as a social goal. These programs called 'family planning' were initiated in 1951 and were periodically assessed over the past decades.  Improved programs covering wider reproduction-related areas are currently in operation under the popular name 'Reproductive and Child Healthcare (RCH) programs. Creating awareness among people about various reproduction related aspects and providing facilities and support for building up a reproductively healthy society are the major tasks under these programs.


Q.27: Assertion (A) : Crossing of F1 hybrid with the recessive parent is known as test cross.
Reason (R) : Test cross helps to determine the unknown genotype by crossing it with the recessive parent.
(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)
On crossing F1 hybrid with parent plant then the recessive genotypic characters of F1 hybrid will inherit in the offspring as dominant genotypic characters.


Q.28: Assertion (A): The frequency of red-green colour blindness is many times higher in females than that in males.
Reason (R) : In females if only one X-chromosome of female possess allele for colour blind character she becomes the colour blind.
(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)
Colour blindness is X-linked sex inheritance. The frequency of red-green colour blindness is many times higher in males than that in the females because males have only one X chromosomes therefore they develop colour blindness when their sole X- chromosome has the allele for it. Thus, males are more prone to colour blindness while females are carriers. For becoming colour blind, the female must have the allele for it in both her X-chromosomes and if only one X-chromosome of female possess allele for colour blind character she becomes the carrier for this characteristic.


Q.29: The process of transfer of pollen grains from the anther to the stigma of a pistil is known as pollination. Identify the types of pollination occurring in plants A and B.
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

(a) X –- Xenogamy, Y- Autogamy, Z- Geitonogamy
(b) X- Autogamy, Y- Geitonogamy, Z- Xenogamy
(c) X- Xenogamy, Y – Geitonogamy. Z – Autogamy
(d) X—Geitonogamy, Y - Autogamy, Z- Xenogamy

Correct Answer is Option (a)
X-Xenogamy is when the pollen grains are transferred from anther to the stigma of a different plant. It is cross pollination, it brings about genetically different types of pollen grains to the stigma. Y- Autogamy is when the pollen grains are transferred from the anther to the stigma of the same flower, it is self pollination. Z- Geitonogamy is when the pollen grains are transferred from the anther to the stigma of another flower of the same plant. It involves pollination with the help of a pollinating agent. It is structurally cross-pollination but genetically self-pollination.


Q.30: A male has an overall masculine development, Gynaecomastia, and is sterile. What is the genetic disorder this human suffering from?
(a) Turner's syndrome
(b) Klinefelter's syndrome
(c) Down's syndrome
(d) Edward syndrome

Correct Answer is Option (b)
Klinefelter's syndrome is the genetic disorder which is caused due to the presence of an additional copy of X-chromosome resulting in karyotype of 47, XXY chromosome but only the son will be affected by the disease.


Q.31: Transcription unit is represented in the diagram given below

Identify X and Y and their functions.
(a) X – RNA Polymerase;transcribes the information in DNA into RNA molecules.
Y - Rho factor; terminates translation and release polypeptide from ribosome.
(b) X – Rho factor; terminates translation and release polypeptide from ribosome.
Y- RNA Polymerase; transcribe the information in DNA into RNA molecules.
(c) X – Promotor site; transcribes the information in DNA into RNA molecules.
Y- Sigma factor; determines the specificity of promoter DNA binding
(d) X – RNA Polymerase; transcribes the information in DNA into RNA molecules.
Y - Rho factor; determines the specificity of promoter DNA binding

Correct Answer is Option (a)
X- RNA Polymerase creates an mRNA copy of template DNA. The mRNA is then pushed into the cytoplasm of the cell where it is ready by ribosomes. Y- The Rho factor is a protein that acts in bacterial cells to mediate termination of transcription at distinct sites. Sigma factors are subunits of all bacterial RNA polymerases. They are responsible for determining the specificity of promoter DNA binding and control how efficiently RNA synthesis (transcription) is initiated. A promoter is a region of DNA where transcription of a gene is initiated.


Q.32: (1) They possess small male flowers that are not clearly visible.
(2) A large number of pollens are released in water that is caught by large, feathery stigma of female flowers.
(3) This pollen keeps floating on the water surface until they are caught by female flowers. Above mentioned characteristics are of :

(a) Wind pollinated flowers
(b) Insect pollinated flower
(c) Water pollinated flower
(d) None of these

Correct Answer is Option (c)
Water pollinated flowers

  • Possess small male flowers that are not clearly visible.
  • In them, large number of pollens are released in water that is caught by large, feathery stigma of female flowers.
  • The pollen keeps floating on the water surface until they are caught by female flowers.


Q.33: Which membrane of ovum is rich in glycoprotein?
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
(a) (i)
(b) (ii)
(c) (iii)
(d) (iv)

Correct Answer is Option (a)
Ovum has four membranes namely:
(a) Plasma membrane (Oolemma) : Innermost layer.
(b) Vitelline membrane : Attached to the plasma membrane.
(c)  Zona pellucida : Transparent noncellular, thick, glycoprotein rich layer found outer to the vitelline membrane.
(d)  Corona radiata : Outer layer formed of follicle cells. These cells are held together by a mucopolysaccharide called hyaluronic acid.


Q.34: Statistical data has shown that 8% of the human males are colour-blind whereas only 0.4% of females are colour-blind. The reasons for this is:
(a) Affected X chromosome has much high affinity to Y chromosome as compared to unaffected X chromosome.
(b) Males have only one X chromosome gene for colour blindness, if present in any one parent will always be expressed
(c) Males have only one Y chromosome gene for colour blindness, if present in any one parent will always be expressed
(d) Females need only one affected X chromosomes to be expressed.

Correct Answer is Option (b)
The colour blindness is found in about 8% of the males and only 0.4% of the females. The greater prevalence of the colour blindness in males is due to the presence of only one X chromosome and the hemizygous (X and Y) expression of the allele for colour blindness i.e. if gene for colour blindness is present on X-chromosome of male then it will always express while in case of females the incidence of disease of colour blindness is possible only in homozygous condition (XCXC) i.e. if both the X-chromosomes carry the allele (XCXC) for colour blindness. Occurrence of allele for colour blindness on one of the X chromosome makes the female a carrier XCX.


Q.35: A small stretch of DNA strand that codes for a polypeptide is shown below:
3`… … … …CAT ATA GAT GAA AC… … … 5`

How many amino acids will be translated from the above strand
(a) 20 amino acids
(b) 10 amino acids
(c) 5 amino acids
(d) 4 amino acids

Correct Answer is Option (d)
4 amino acids will be translated from the strand  3`… … … …CAT ATA GAT GAA AC… … … 5`


Q.36: Vinay is suffering from Gonorrhoea, a sexually transmitted disease caused by an bacterium Neisseria gonorrhoeae.The infection has incubation period of 2-5 days.Which mode of treatment would be useful for Vijay?
(a) Cryosurgery
(b) Use of Podophyllum preparation
(c) Use of antibiotic ampicillin
(d) It cannot be treated

Correct Answer is Option (c)
Gonorrhoea can be cured through use of appropriate antibiotics like penicillin and ampicillin.


Q.37: In humans, attached earlobes are a dominant feature over free earlobes while hypertrichosis of the ear (hair on pinna) is a holandric (Y-linked) feature. A man with attached earlobes and extensive hair on pinna married a woman having free earlobes. Their progenies will be
(i) All sons will have free earlobe and haiery pinna
(ii) 50%sons with attached earlobes and hairy pinna and 50% sons with free earlobes and hairy pinna.
(iii) All daughters with free earlobes
(iv) 50% daughters  will be have attached earlobes and 50% daughters will have free earlobe.
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
 

Correct Answer is Option (c)
A cross between a male having attached earlobes and female having a normal one, among the sons, one will have attached earlobe and another will have free ear lobe.

However, all sons will have hairy pinna  because it is Y-linked feature. Both the daughters will have free earlobes.


Q.38: Study the table given below and identify (i), (ii), (iii) and (iv)
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

(a) (i) AAA, (ii) UUU, (iii) GUG (iv) CAC
(b) (i) TTT, (ii) CAC, (iii) AAA, (iv) GUG
(c) (i) UUU, (ii) GUG, (iii) AAA, (iv) CAC
(d) (i) UUU, (ii) CAC, (iii) AAA, (iv) GUG

Correct Answer is Option (c)
For AAA (Phe) codon in mRNA and anticodon in tRNA will be UUU  and GUG respectively. For CAC (Val) codon in mRNA and anticodon in tRNA will be AAA  and CAC respectively.


Q.39: The graph given below shows the variation in the levels of ovarian hormones, A–Estrogen and  B– Progesterone during various phases of menstrual cycle.
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

Reason out : why A peaks before B ?
(a) Corpus luteum forms before that releases estrogen whereas, Graffian follicle develops later that forms progesterone.
(b) Corpus luteum forms before that releases progestrone whereas, Graffian follicle develops later that forms estrogen.
(c) Graafian follicle forms before that releases progestrone whereas, corpus luteum develops later that forms estrogen.
(d) Graafian follicle forms before that releases estrogen whereas, corpus luteum develops later that forms progesterone.

Correct Answer is Option (d)
A is the hormone estrogen. B is the hormone progesterone. Estrogen is produced by the Graafian follicle and Progesterone is produced by corpus luteum. Graafian follicle forms before that releases estrogen whereas, corpus luteum develops later that forms progesterone. A (estrogen) regenerates endometrium lining of the uterus whereas, B (progesterone) maintains the endometrium for implantation of the fertilized ovum and maintains pregnancy. 


Q.40: ____(i)______is the process of formation and maturation of the ovum.It takes place in ___(ii)_____. It is initiated in embryonic stage when millions of ____(iv)___ are formed within each ovary.They multiply to form primary____(v)_____.
(a) (i) Spermatogenesis, (ii) seminiferous tubules (iii) spermatogonia (iv) spermatids
(b) (i) Oogenesis, (ii) Corpus luteum, (iii) oocytes , (iv) oogonia
(c) (i) Spermatogenesis, (ii) seminiferous tubules (iii) spermatids (iv) spermatogonia
(d) (i) Oogenesis, (ii) Graafian follicles (iii) oogonia, (iv) oocytes

Correct Answer is Option (d)
Oogenesis is the process of formation and maturation of the ovum. It takes place in Graafian follicles . It is initiated in embryonic stage when millions of oogonia are formed within each ovary. They multiply to form primary oocytes.


Q.41: In human female reproductive system diagram, students are asked to label anarrow, hollow muscular organ located in front of the rectum and behind the urinary bladder, that opens into vagina. Which organ are they asked to label?
(a) Urethra
(b) Cervix
(c) Clitoris
(d) Vulva

Correct Answer is Option (b)
Lower end of uterus is called cervix that opens into vagina. A hollow muscular organ located in front of the rectum and behind the urinary bladder.


Q.42: The below pedigree chart represents a cross between a normal couple resulted in a son who was haemophilic and a normal daughter. In course of time, when the daughter was married to a normal man to their surprise, the grandson was also haemophilic.
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
What conclusion you draw from the is the pattern of inheritance in the above pedigree chart? inheritance pattern of this disease
(a) Sex -linked dominant
(b) Sex -linked recessive
(c) Autosomal dominant
(d) Autosomal recessive

Correct Answer is Option (b)
It is sex-linked recessive inheritance. Genotype of daughter is XXh,  Genotype of her husband is XY. XXh females are carriers of haemophilia. They are not suffering from haemophilia but can pass on the gene Xh to offspring. XhY males suffer from haemophilia. They seldom reach reproductive age. XX females are normal. XY males are normal too. XhXh females die in embryonic stage.


Q.43: Condoms are one of the most popular contraceptives because of the following reasons.
(a) These are effective barriers for in semination.
(b) They do not interfere with coitalact.
(c) These help in reducing the risk of STDs.
(d) All of the above

Correct Answer is Option (d)
Condoms can both prevent pregnancy by stopping sperm from meeting an egg. They also protect against sexually-transmitted infections (STIs). Condoms act as a barrier method of contraception. They are made up of very thin latex (rubber) and are designed to prevent pregnancy by stopping sperm from meeting an egg. They do not interfere with coital act.


Q.44: While studying a virus, the following proportions of nitrogenous bases were found to be present in it: adenine 33 %, guanine 47%, cytosine 33% and uracil 27%. Choose the correct statements regarding the virus.
(1) The genetic material of the virus is single stranded
(2) The genetic material of the virus is RNA
(3) The genetic material of the virus is double stranded (4) Base pairing rules in the virus is adenine: cytosine
(a) (1) and (3)
(b) (1) and (2)
(c) (3) and (4)
(d) (2) and (4)

Correct Answer is Option (b)
RNA is the genetic material because uracil is present in all virus. The percentage of guanine and cytosine are not equal so the genetic material is single stranded.


Q.45: The vas deferens receives duct from the seminal vesicle and opens into urethra as
(a) epididymis
(b) ejaculatory duct
(c) efferent ductules
(d) ureter

Correct Answer is Option (b)
The vas deferens receives duct from the seminal vesicle and opens into urethra as ejaculatory duct. These ejaculatory ducts which open into the urethra about halfway through the prostate gland function to mix the sperm stored in the ampulla with fluids secreted by the seminal vesicles and to transport these substances to the prostate.


Q.46: An infertile couple is advised to adopt test tube baby programme. The test-tube baby programme involves mainly two principal procedures. Choose the two technologies involved in this programme.
(i) IVF
(ii) IUDs
(iii ET
(iv) MTP
(a) (i) and (ii)
(b) (ii) and (iv)
(c) (iv) and (i)
(d) (iii) and (i)

Correct Answer is Option (d)
(i) In-vitro fertilization (IVF) : It is the fertilization outside the body in the conditions almost similar to those which exist in the body. The ova and sperms from the donor parents are taken and fused to form the zygote in the laboratory outside the body of female. Then the zygote is transferred into the fallopian tube.
(ii)  Intra Uterine Devices (IUDs) are ideal contraceptives for the females who want to delay pregnancy or spacing in children.
(iii) Embryo transfer (ET) : The embryo is transferred into the reproductive tract. The zygote is allowed to  divide so as to form about 8-blastomeres. The zygote or the embryo in early stage of development is transferred into the fallopian tube by Zygote Intra Fallopian Transfer (ZIFT) technique. If the embryo is in a bit later stage having about 8 blastomeres it is transferred into uterus by intra uterine transfer (IUT) of infertile female partner or into that of surrogate mother for giving birth to the baby.
(iv) Medical Termination of Pregnancy(MTP) : Intentional or voluntary termination of pregnancy before full term with help of an expert doctor.


Q.47: Where are promoters typically found in DNA?
(a) In the middle of the coding region of a gene
(b) Upstream of the coding region of a gene
(c) Downstream of the coding region of a gene
(d) mRNA region

Correct Answer is Option (b)
Promoters are the sites where transcription factors and RNA polymerase bind to initiate transcription. It makes sense that the promoter would be found upstream of a gene (i.e. before a gene).


Q.48: Aman has a disorder is located on the Y chromosome. This chromosome can be inherited by
(a) Only daughters
(b) Both sons and daughters
(c) Only grandchildren
(d) Only sons

Correct Answer is Option (d)
Because only males have a Y chromosome, in Y-linked inheritance, a variant can only be passed from father to son.

Section - C

Case-I
Direction : Section-C consists of one case followed by 6 questions linked to this case (Q.No.49 to 54).
To answer the questions, study the graphs below for Subject 1 and 2 showing different levels of certain hormones.
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12Q.49: The peak observed in Subject 1 and 2 is due to:
(a) estrogen
(b) progesterone
(c) luteinizing hormone
(d) follicle stimulating hormone

Correct Answer is Option (c)
The peak observed in Subject 1 and 2 is due to luteinizing hormone.


Q.50: Subject 2 has higher level of hormone B, which is:
(a) estrogen
(b) progesterone
(c) luteinizing hormone
(d) follicle stimulating hormone

Correct Answer is Option (b)
Subject 2 has higher level of hormone B, which is progesterone.


Q.51: If the peak of Hormone A does not appear in the study for Subject 1, which of the following statement is true ?
(a) Peak of Hormone B will be observed at a higher point in the graph
(b) Peak of Hormone B will be observed at a point lower than what is given in the graph
(c) There will be no observed data for Hormone B
(d) The graph for Hormone B will be a sharp rise followed by a plateau

Correct Answer is Option (c)
If the peak of Hormone A does not appear in the study for Subject 1, there will be no observed data for Hormone B.


Q.52: Which structure in the ovary will remain functional in subject 2 ?
(a) Corpus Luteum
(b) Tertiary follicle
(c) Graafian follicle
(d) Primary follicle

Correct Answer is Option (a)
Corpus Luteum will remain functional in subject 2.


Q.53: For subject 2 it is observed that the peak for hormone B has reached the plateau stage. After approximately how much time will the curve for hormone B descend ?
(a) 28 days
(b) 42 days
(c) 180 days
(d) 280 days

Correct Answer is Option (d)
Curve for hormone B descend after approximately 280 days.  


Q.54: Which of the following statements is true about the subjects ?
(a) Subject 1 is pregnant
(b) Subject 2 is pregnant
(c) Both subject 1 and 2 are pregnant
(d) Both subject 1 and 2 are not pregnant

Correct Answer is Option (b)
Subject 2 is pregnant.


Q.55: The gene that controls the ABO blood group system in human beings has three alleles - IA, IB and i. A child has blood group O. His father has blood group A and mother has blood group B. Genotypes of other off springs can be:
i. IBIB

ii. IA

iii. IB

iv. IAIB

v. ii
(a) i, ii, iii, v
(b) ii, iii ,iv, v
(c) iii, iv, v
(d) iv, iii, i

Correct Answer is Option (b)
The genotypes of other offsprings can be IAi, IBi, IAIB and ii.


Q.56: Placed below is a karyotype of a human being.
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
On the basis of this karyotype, which of the following conclusions can be drawn:(a) Normal human female
(b) Person is suffering from Colour blindness
(c) Affected individual is a female with Down’s syndrome
(d) Affected individual is a female with Turner ’s syndrome

Correct Answer is Option (c)
Affected individual is a female with Down's syndrome that have extra copy of 21st chromosome called Trisomy 21.


Q.57: Given below is a dihybrid cross performed on Drosophila.
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

Which of the following conclusions can be drawn on the basis of this cross ? When yellow bodied (y), white eyed (w) Drosophila females were hybridized with brown bodied (y+), red eyed males (w+) and F1 progenies were intercrossed, F2 generation would have shown the following ratio:
(a) 1:2:1 because of linkage of genes
(b) 9:3:3:1 because of recombination of genes
(c) Deviation from 9:3:3:1 ratio because of segregation of genes
(d) Deviation from 9:3:3:1 ratio because of linkage of genes

Correct Answer is Option (d)
When two genes are located on the same chromosome, the proportion of parental gene combinations was much higher than the non-parental type. Morgan attributed this due to the physical association or linkage of the two genes and coined the term linkage.


Q.58: Which cellular process is shown below ?
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
(a) DNA Replication(b) Translation - Initiation(c) Translation - Elongation
(d) Translation – Termination

Correct Answer is Option (c)
This is the elongation phase of translation.


Q.59: Origin of replication of DNA in E. coli is shown below, Identify the labelled parts (i), (ii), (iii) and (iv)
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12
(a) (i)- discontinuous synthesis , (ii)- continuous synthesis (iii) 3’ end (iv) 5’end(b) (i)- continuous synthesis , (ii)- discontinuous synthesis (iii) 5’ end (iv) 3’end
(c) (i)- discontinuous synthesis, (ii)- continuous synthesis (iii) 5’ end (iv) 3’end
(d) (i)- continuous synthesis , (ii)- discontinuous synthesis (iii) 3’ end (iv) 5’end

Correct Answer is Option (d)
During DNA replication , the DNA polymerase forms one new strand (leading strand) in a continuous strand in the 5’→3’ direction (Continuous synthesis). The other new strand formed is in small stretches called okazaki fragments in 5’→3’ direction (Discontinuous synthesis). The okazaki fragments are then joined together to form a new strand by an enzyme, DNA ligase. This new strand is called lagging strand.


Q.60: Transcription unit is represented in the diagram given below. 
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

Identify site (i), factor (ii) and Enzyme (iii) responsible for carrying out the process.
(a) (i) Promoter Site, (ii) Rho factor (iii) RNA polymerase
(b) (i) Terminator Site, (ii) Sigma factor (iii) RNA polymerase
(c) (i) Promoter Site, (ii) Sigma factor (iii) RNA polymerase
(d) (i) Promoter Site, (ii) Sigma factor (iii) DNA polymerase

Correct Answer is Option (c)
The label (i) is promoter site, (ii) is sigma factor and label (iii) is RNA polymerase.
Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 2 Notes | Study Sample Papers for Class 12 Medical and Non-Medical - Class 12

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