Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 3 | Sample Papers for Class 12 Medical and Non-Medical PDF Download

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.

Q.1: Gametes of parents have qualitatively different genetic composition therefore
(a) Offspring formed by sexual reproduction exhibit more variation
(b) Offspring formed by asexual reproduction exhibit more variation
(c) Genetic material can easily be mutated
(d) Greater amount of DNA is involved in sexual reproduction

Correct Answer is Option (b)
Gametes of parents have qualitatively different genetic composition. Therefore, offspring formed by sexual reproduction exhibit more variation. Gametes of parents have qualitatively different genetic composition. Therefore, offspring formed by sexual reproduction exhibit more variation.

Q.2: The fact that a purine base always paired through hydrogen bonds with a pyrimidine base leads to, in the DNA double helix
(a) The antiparallel nature
(b) The semi-conservative nature
(c) Uniform width throughout DNA
(d) Uniform length in all DNA

Correct Answer is Option (c)
The diameter of the strand is always constant due to a pairing of purine (adenine and guanine) and pyrimidine (cytosine and thymine). This specific bonding gives uniform width to the DNA.

Q.3: ZZ / ZW type of sex determination is seen in
(a) Platypus
(b) Snails
(c) Cockroach
(d) Peacock

Correct Answer is Option (d)
ZZ / ZW type of sex determination is seen in birds, reptiles and fish. Thus, peacock shows ZZ/ ZW sex determination type. In this type, female has heteromorphic (ZW) sex chromosomes and the male has homomorphic (ZZ) sex chromosomes.  In Platypus the sex determination is of XX-XY type in which both male and females has ten sex chromosome each. The male has XY, XY, XY, XY, XY and female has XX, XX, XX, XX, XX. In snails, the sex determination is environmentally induced, while in cockroaches it is of XX-XO types. In this type Y-chromosome is absent. In this the presence puffs unpaired X-chromosomes determines the masculine sex.

Q.4: To initiate translation, the mRNA first binds to
(a) The smaller ribosomal sub-unit
(b) The larger ribosomal sub-unit.
(c) The whole ribosome.
(d) No such specificity exists.

Correct Answer is Option (a)
The ribosome consists of structural RNAs and about 80 different proteins. In its inactive state, it exists as two subunits, a large subunit and a small subunit. When the smaller subunit encounters the mRNA, the process of translation of the mRNA to protein begins.

Q.5: The mother germ cells are transformed into a mature follicle through series of steps:
(a) Oogonia ->Primary oocyte -> Primary Follicle ->Secondary Follicle -> Tertiary Follicle ->Graafian Follicle
(b) Primary oocyte -> Oogonia -> Primary Follicle ->Secondary Follicle -> Tertiary Follicle ->Graafian Follicle
(c) Oogonia -> Primary Follicle ->Secondary Follicle -> Tertiary Follicle ->Graafian Follicle-> Primary oocyte
(d) Oogonia ->Primary oocyte -> Graafian Follicle-> Primary Follicle ->Secondary Follicle -> Tertiary Follicle

Correct Answer is Option (a)
In female oogenesis, the immature oogonia diploid germ cell undergoes mitotic division and give rise to primary oocyte and primary oocyte get surrounded by a layer of granulosa cells to develop into primary follicle. The primary follicles further grow and get surrounded by more layers of granulosa cells which develop into secondary follicles. The secondary follicle then transforms into a tertiary follicle. The primary oocyte and the tertiary follicle undergo meiotic division to become a secondary oocyte and a first polar body (haploid). The tertiary follicle further changes into the mature follicle or Graafian follicle.

Q.6: Which one of the following pairs of codons is correctly matched with their function or the signal for the particular amino acid?
(a) GUU, GCU – Alanine
(b) UAG, UGA – stop
(c) AUG, ACG – Start/methionine
(d) UUA, UCA – Leucine

Correct Answer is Option (b)
Three codons UAG, UAA and UGA are the stop or termination codons the end the process of translation.

Q.7: Objectives of RCHC are:
(1) Creating awareness about various reproduction related aspects e.g. STDs, birth control methods.
(2) Providing facilities and support for building up reproductive healthy society.
(3) Educating people about care of pregnant women, important of breast feeding.
(4) Creating awareness about sex abuse and sex related crimes.
(a) 1 and 4
(b) 2 and 3
(c) 1, 2 and 3
(d) 1,2,3,4

Correct Answer is Option (d)
RCHC refers to a popular programme called “Reproductive and child health care (RCHC) and the major tasks under these programme are:-Creating awareness about various reproduction related aspects e.g. STDs, birth control methods.

Provide Educating people about care of pregnant women, important of breast feeding.

Educating people about caring of pregnant women& importance of breast feeding for the new born. Creating awareness about sex abuse & sex related crimes facilities. It support for building up reproductive healthy society.

Q.8: A hybrid which expresses a character, is called
(a) Recessive
(b) Co-dominant
(c) Dominant
(d) Incomplete dominant

Correct Answer is Option (a)
Dominant factor or allele is one of a pair of alleles which can express itself whether present in homozygous or heterozygous state. e.g. T (tallness in pea), R (round seed in pea), A (axial flower in pea).

Q.9: The entire DNA in the haploid set of chromosomes of an organism is called a
(a) Gene
(b) Genome
(c) Codon
(d) Operon

Correct Answer is Option (b)
The entire DNA in the haploid set of chromosomes of an organism is called a Genome. In human genome, DNA is packed in 23 chromosomes. Human genome contains about 3 × 10⁹ bp.

Q.10: Given are the stages of post-fertilization in plants arrange them in order of their occurrence and select the correct option:
(a) Development of ovule into a seed
(b) Embryo improvement
(c) Development of ovary into a fruit
(d) Endosperm development
(a) c,d,b,a
(b) b,d,c,a
(c) c,a,b,d
(d) d,b,a,c

Correct Answer is Option (d)
In all flowering plants, the post-fertilization is a critical stage which occurs after the double fertilization and includes the series of steps : Endosperm development. Embryo improvement. Development of ovule into a seed. Development of ovary into a fruit.

Q.11: ___________ has dual function.
(a) UGA
(b) AUG
(c) UAG
(d) UAA

Correct Answer is Option (b)
AUG has dual functions. It codes for Methionine (met) and also acts as an initiator codon.

Q.12: Colour blindness is
(a) Z chromosome linked trait
(b) X-linked trait
(c) Y-linked trait
(d) Both X and Y linked

Correct Answer is Option (b)
Colour blindness is a recessive sex-linked trait. The gene which causes colour blindness is found only on the X chromosome. So, for a male to be colour blind the colour blindness  gene only has to appear on his X chromosome. For a female to be colour blind it must be present on both of her X chromosomes.

Q.13: Given below table provides a data, identify the correct option:
(a) 1 is correct
(b) 2 is correct
(c) 3 is correct
(d) All are correct

Correct Answer is Option (b)
The correct data is

Q.14: The process of fusion of one male gamete with egg to form diploid zygote and another male gamete with the polar nuclei or secondary nucleus to form primary endosperm nucleus is called:
(a) Parthenogenesis
(b) Double fertilization
(c) Triplody
(d) Dichogamy

Correct Answer is Option (b)

The process of fusion of one male gamete with egg to form diploid zygote and another male gamete with the polar nuclei or secondary nucleus to form primary endosperm nucleus is called double fertilization.

Q.15: Regulation of lac operon by repressor is called
(a) Neutral regulation
(b) Zero regulation
(c) Positive regulation
(d) Negative regulation

Correct Answer is Option (d)
Regulation of lac operon by repressor is called negative regulation. The lac repressor binds to the operator region and negatively controls (prevents) transcription.

Q.16: An example of an autosomal dominant disorder is
(a) Sickle cell anaemia
(b) Myotonic dystrophy
(c) Phenylketonuria
(d) Haemophilia

Correct Answer is Option (b)
Myotonic Dystrophy is an autosomal dominant disorder that is characterized by increasing contractility of muscles with decreasing relaxation. This leads to atrophy of muscles, particularly of the face and neck. Hypogonadism, balding and cardiac irregularities may also be caused due to this disorder.

Q.17: Mitochondria and 9 + 2 arrangement of microtubules will explain:
(a) Head part of the sperm
(b) Middle piece of the sperm
(c) Tail end of the sperm
(d) None of the above

Correct Answer is Option (b)
The middle piece of the sperm contains mitochondria that are arranged spirally around the axonema (axial filament). This axonema has the 9+2 arrangement of the microtubules.

Q.18: The “father of genetic” is
(a) Bateson
(b) Morgan
(c) Mendel
(d) Watson

Correct Answer is Option (c)
Gregor Johann Mendel is called the ‘father of genetic’ because through his work on pea plants, he discovered the fundamental laws of inheritance. He deduced that genes come in pairs and are inherited as distinct units, one from each parent.

Q.19: Embryo sac is related to ovule as _______ is related to an anther.
(a) Stamen
(b) Filament

(c) Pollen grain
(d) Androecium

Correct Answer is Option (c)
The pollen grains represent the male gametophytes. As the anthers mature and dehydrate, the microspores dissociate from each other and develop into pollen grains. So, embryo sac is to ovule as pollen grain is to an anther.

Q.20: Acrosome is filled with _________.
(a) Lipids
(b) Vasopressin
(c) Degradative enzymes
(d) Oxygenated blood

Correct Answer is Option (c)
In males, the acrosome contains degradative enzymes (including hyaluronidase and acrosin). 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.

Q.21: The outermost and innermost wall layers of microsporangium in an anther are respectively:
(a) Endothecium and tapetum
(b) Epidermis and endodermis
(c) Epidermis and middle layer
(d) Epidermis and tapetum

Correct Answer is Option (b)
The outermost and innermost wall layers of microsporangium in an anther are respectively, epidermis and tapetum. A typical microsporangium is generally surrounded by four-wall layers, that is, the epidermis, (outermost protective layer), endothecia, (middle fibrous layers) and the tapetum (innermost nutritive layer).

Q.22: A particular portion or region of the chromosomes representing a single gene is called
(a) Muton
(b) Recon
(c) Allele
(d) Locus

Correct Answer is Option (d)
A gene locus is a specific, fixed position on a chromosome where a particular gene or genetic marker is located.

Q.23: Signals for parturition originate from:
(a) Placenta only
(b) Fully developed foetus
(c) Both placenta as well as fully developed foetus
(d) Oxytocin released from maternal pituitary

Correct Answer is Option (c)
Signals for parturition originate from both placenta as well as the fully developed foetus. It is called foetal ejection reflex, i.e. mild uterine contractions.

Q.24: _________ synthesize and secrete testicular hormones called androgens.
(a) Ovum cells
(b) Leydig cells
(c) Sertoli cells
(d) None of the above

Correct Answer is Option (b)
Leydig cells synthesize and secrete testicular hormones called androgens.

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: Lactational amenorrhea is the natural method of contraception.
Reason : It increases the phagocytosis of sperm.

(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 (c)

Lactational amenorrhea is a natural method of contraception and  involves the prevention of conception by breast feeding the child.

Q.26: Assertion: Saheli, an oral contraceptive for females, contains a steroidal preparation.
Reason: It is a "once a week" pill with very few side effects.
(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)

Saheli is a non-steroidal preparation used as oral contraceptive pills. It is a ‘once a week’ pill with very few side effects and high contraceptive value.

Q.27: Assertion : Parturition is induced by a complex neuro endocrine mechanism. Reason : At the end of gestation period, the maternal pituitary releases prolactin which causes uterine contractions.
(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 (c)
Parturition is a neuroendocrine mechanism and is started by the signals from the developed foetus and the placenta. The signals originating from the foetus and placenta induce mild uterine contractions (fetal ejection reflex). This causes the release of oxytocin from maternal pituitary. Oxytocin causes stronger uterine muscle contractions which in turn stimulate further secretion of oxytocin. This process is continued leading to expulsion of the baby out of the uterus through the birth canal.

Q.28: Assertion: When the two genes in a dihybrid cross are situated on the same chromosome, the proportion of parental gene combinations is much higher than non-parental type.
Reason: Higher parental gene combinations can be attributed to crossing over between two genes.
(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)
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.29: Concentration of which of the following substances will decrease in the maternal blood  as it flows from embryo to placenta through the umbilical cord ?

(i) Oxygen
(ii) Amino acids
(iii) Carbon dioxide
(iv) Urea
(a) i and ii
(b) ii and iv
(c) iii and iv
(d) i and iv

Correct Answer is Option (a)
Placenta is connected to the embryo by an umbilical cord. It transports substances to and from the embryo.

Q.30: In a fertilized ovule, n, 2n and 3n conditions occur respectively in:
(a) antipodal, zygote and endosperm
(b) zygote, nucellus and endosperm
(c) endosperm, nucellus and zygote.
(d) antipodals, synergids and integuments

Correct Answer is Option (a)
In a fertilized ovule, antipodal cells are haploid (n), zygote is diploid  (2n)  while endosperm is triploid (3n).

Q.31: A botanist studying Viola (common pansy) noticed that one of the two flower types  withered and developed no further due to some unfavorable condition, but the other flower type on the same plant survived and it resulted in an assured seed set. Which of the following will be correct ?
(a) The flower type which survived is Cleistogamous and it always exhibits autogamy
(b) The flower type which survived is Chasmogamous and it always exhibits geitonogamy
(c) The flower type which survived is Cleistogamous and it exhibits both autogamy and geitonogamy.
(d) The flower type which survived is Chasmogamous and it never exhibits autogamy.

Correct Answer is Option (a)
The flower type which survived is Cleistogamous and it always exhibits autogamy. Cleistogamous flowers do not open at all. They have closed anthers and stigma, that lies close to each other. 

Q.32: During parturition, a pregnant woman is having prolonged labour pains and child birth has to be fastened. It is advisable to administer a hormone that can :
(A) increase the metabolic rate
(B) release glucose in the blood
(C) stimulate the ovary
(D) activate smooth muscles

Correct Answer is Option (d)
During parturition, a pregnant woman is having prolonged labour pains and child birth has to be fastened. It is advisable to administer a hormone that can activate smooth muscles of myometrium of the uterus, which directs the full term foetus towards the birth canal.

Q. 33. A female undergoing IVF treatment has blocked fallopian tubes. The technique by which the embryo with more than 8 blastomeres will be transferred into the female for further development is:
(a) ZIFT
(b) GIFT
(c) IUT
(d) AI

Correct Answer is Option (c)
IVF treatment technique is done to give birth test tube baby. The in vitro fertilisation of male and female gamete occurs outside the body and then it is cultured upto 8 blastomeres. After that it is transferred in the female body for further development.

Q.34: The mode of action of the copper ions in an IUD is to:
(a) increase the movement of sperms
(b) decrease the movement of the sperms
(c) make the uterus unsuitable for implantation
(d) make the cervix hostile to the sperms

Correct Answer is Option (b)
IUDs increase phagocytosis of sperms. The Cu ions suppress motility and fertilising capacity of sperms.

Q.35: To produce 400 seeds, the number of meiotic divisions required will be:
(a) 400 

(b) 200 

(c) 500 

(d) 800

Correct Answer is Option (c)
The meiotic division is involved in the formation male and the female gamete. The formation of the male gamete requires only one meiotic division which results in four male gametes. The formation of four female gametes, on the other hand requires four meiotic divisions. So, the five meiotic divisions result in four seeds. Thus 500 meiotic divisions result in 400 seeds.

Q.36: A cross is made between tall pea plants having green pods and dwarf pea plants having yellow pods. In the F2 generation, out of 80 plants how many are likely to be tall plants ?
(a) 15 

(b) 20 

(c) 45 

(d) 60

Correct Answer is Option (d)
Out of 9:3:3:1 = 16
9+3 will be tall.
Therefore, 12/16 x 80 = 60. 

Q.37: In Antirrhinum, RR is phenotypically red flowers, rr is white and Rr is pink. Select the correct phenotypic ratio in F1 generation when a cross is performed between RR X Rr :
(a) 1 red: 2 Pink: 1 white
(b) 2 Pink: 1 white
(c) 2 Red: 2 Pink
(d) All Pink

Correct Answer is Option (c)
The phenotypic ratio in F1 generation when a cross is performed between RR X Rr will be 2 red (RR) and 2 Pink (Rr).

Q.38: What would be the genotype of the parents if the offspring have the phenotypes in 1:1 proportion ?
(a) Aa X Aa
(b) AA X AA
(c) Aa X AA
(d) Aa x aa

Correct Answer is Option (d)
In the monohybrid cross, a test cross of a heterozygous individual resulted in a 1:1 ratio.

Q.39:

What is the pattern of inheritance in the above pedigree chart ?
(a) Autosomal dominant
(b) Autosomal recessive
(c) Sex -linked dominant
(d) Sex -linked recessive

Correct Answer is Option (b)
The pattern of inheritance in the above pedigree chart is autosomal recessive.

Q.40: A couple has two daughters. What is the probability that the third child will also be a female ?
(a) 25%
(b) 50%
(c) 75%
(d) 100%

Correct Answer is Option (b)
The probability that the third child will also be a female will be 50 %.

Q.41: Genotypic ratio of 1:2:1 is obtained in a cross between:
(a) AB × AB
(b) Ab × Ab
(c) Ab × ab
(d) ab × ab

Correct Answer is Option (b)
A cross of two F1 hybrids, that are heterozygous for a single trait and that displays incomplete dominance is predicted to give a 1:2:1 ratio among both the genotypes and phenotypes of the offspring.

Q.42: Total number of nucleotide sequences of DNA that codes for a hormone is 1530. The proportion of different bases in the sequence is found to be Adenine = 34%, Guanine = 19%, Cytosine = 23%, Thymine = 19%.
Applying Chargaff’s rule, what conclusion can be drawn?
(a) It is a double stranded circular DNA.
(b) It is a single stranded DNA.
(c) It is a double stranded linear DNA.
(d) It is a single stranded DNA coiled on Histones.

Correct Answer is Option (b)
It can be concluded that is a single stranded DNA.

Q.43: A stretch of an euchromatin has 200 nucleosomes. How many bp will there be in the stretch and what would be the length of the typical euchromatin?
(a) 20,000 bp and 13,000 × 10^–9 m
(b) 10,000 bp and 10,000 × 10^–9m
(c) 40,000 bp and 13,600 × 10^–9 m
(d) 40,000 bp and 13,900 × 10^–9 m

Correct Answer is Option (c)
A typical nucleosome contains about 200 base pairs (bp) of the DNA helix. So if a stretch of an euchromatin has 200 nucleosomes, there will be 200 × 200 = 40,000 base pairs in the stretch. Also, the length of the typical euchromatin will be 13,600 x10^-9 m

Q.44: Observe structures A and B given below. Which of the following statements are correct ?

(a) A is having 2'-OH group which makes it less reactive and structurally stable, whereas B is having 2'-H group which makes it more reactive and unstable.
(b) A is having 2'-OH group which makes it more reactive and structurally unstable, whereas B is having 2'-H group which makes it less reactive and structurally stable.
(c) A and B both have -OH groups which make it more reactive and structurally stable.
(d) A and B both are having -OH groups which make it less reactive and structurally stable

Correct Answer is Option (b)
Presence of 2’-OH, which is a highly reactive group, makes RNA labile and easily degradable than DNA.

Q.45: If Meselson and Stahl's experiment is continued for sixth generations in bacteria, the ratio of Heavy strands ¹⁵N/¹⁵N : Hybrid¹⁵N/¹⁴N : light ¹⁴N/¹⁴N containing DNA in the sixth generation would be:
(a) 1:1:1
(b) 0:1:7
(c) 0:1:15
(d) 0:1:31

Correct Answer is Option (d)
As per Meselson and Stahl's experiment, parent DNA was first isolated from E. coli grown in heavy 15N medium. It was then put in light 14N medium. Since, DNA shows semi conservative nature, hence, in sixth generation, the ratio of ¹⁵N/¹⁵N:¹⁵N/¹⁴N:¹⁴N/¹⁴N will be 0:1:31.

Q.46: Two important RNA processing events lead to specialized end sequences in most human mRNAs _____ (i) _____: at the 5’ end, and _____ (ii) _____ at the 3’ end. At the 5’ end the most distinctive specialized end nucleotide, _____ (iii) _____ is added and a sequence of about 200_____ (iv) is added to the 3’ end.
(a) (i) Initiator codon (ii) Promotor (iii) Terminator codon (iv) Release factors
(b) (i). Promotor (ii) Elongation (iii) Regulation (iv) Termination.
(c) (i) Capping (ii) Polyadenylation (iii) ^mG_ppp (iv) Poly(A).
(d) (i) Repressor (ii) Co repressor (iii) Operon (iv) sRelease factors

Correct Answer is Option (c)
Two important RNA processing events lead to specialized end sequences in most human mRNAs: Capping at the 5’ end, and polyadenylation at the 3’ end. At the 5’ end the most distinctive specialized end nucleotide, mGppp is added and a sequence of about 200 poly A is added to the 3’ end.

Q.47: What are minisatellites ?
(a) 10-40 bp sized small sequences within the genes
(b) Short coding repetitive region on the eukaryotic genome
(c) Short non-coding repetitive sequence forming large portion of eukaryotic genome
(d) Regions of coding strands of the DNA

Correct Answer is Option (c)
Minisatellites are short non-coding repetitive sequence forming large portion of eukaryotic genome.

Q.48: There was a mix-up at the hospital after a fire accident in the nursery division. Which of these children belong to the parents? 

(a) All of the children
(b) Children 2, 3 & 6
(c) Children 1 & 3
(d) Children 2 & 4

Correct Answer is Option (c)
Children from lane 1 and 3 belonged to the parents.

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 schematic representation of spermatogenesis given below:

Q.49: Which cell division occurs during multiplication phase?
(a) Mitosis
(b) Meiosis I
(c) Meiosis II
(d) Both (B) and (C)

Correct Answer is Option (a)

In multiplication phase, the primordial germ cells divide several times by mitosis to produce large number of spermatogonia.

Q.50: How many chromosomes are present in secondary spermatocyte and spermatid respectively?
(a) 46, 23
(b) 46, 46
(c) 23, 23
(d) 23, XY

Correct Answer is Option (c)
23 chromosomes are present in secondary spermatocyte and spermatid respectively.

Q.51: Transformation of spermatids into spermatozoa is known as
(a) Spermiation
(b) Spermatogenesis
(c) Spermateliosis
(d) Both B and C

Correct Answer is Option (d)
Transformation of spermatids (L) into spermatozoa (M) is known as spermiogenesis or spermateliosis.

Q.52: Select the correct option.
(a) Type A spermatogonia grows to larger primary spermatocyte.
(b) One spermatogonium forms two spermatids.
(c) Spermiation is the release of sperms from seminiferous tubules.
(d) Primary spermatocyte undergoes mitosis to form secondary spermatocytes.

Correct Answer is Option (c)
When spermatogonia develops into spermatids and completely mature into sperma that release of sperms from seminiferous tubules called spermiation.

Q.53: Which hormone acts on spermatogonia to stimulate sperm production?
(a) LH
(b) GnRH
(c) ABP
(d) FSH

Correct Answer is Option (c)
FSH is the follicle stimulating hormone that activates synthesis of spermatogonia to stimulate sperm production.

Q.54: Spermatogenesis occurs in
(a) Scrotum
(b) Testis
(c) Penis
(d) Seminal vesicle

Correct Answer is Option (b)
The process of formation spermatogenesis occurs in testis. In testis the immature male germ cells called spermatogonia produce sperms.

Q.55: In peas, tallness is dominant over dwarfness, and red colour of flowers is dominant over the white colour. When a tall plant bearing red flowers was pollinated by a dwarf plant bearing white flowers, the different phenotypic groups were obtained in the progeny in numbers mentioned against them.
Tall, Red = 138 Tall, White = 132 Dwarf, Red = 136 Dwarf, White = 128.
What will be the genotypes of the two parents?
(a) TtRr and ttrr
(b) TtRr and Ttrr
(c) ttRR and TTrr
(d) TtRR and ttrr

Correct Answer is Option (a)
In this case, tallness is dominant over dwarfness, and red colour of flowers is dominant over the white colour. The cross is done between a tall plant bearing red flowers and a dwarf plant bearing white flowers. Therefore the genotype of the two parents will be Tall and Red (Tt Rr) × Dwarf and White (ttrr).

Q.56: Mother and father of a person with ‘O’ blood group have ‘A’ and ‘B’ blood group respectively. What would be the genotype of both mother and father?
(a) Mother is homozygous for ‘A’ blood group and  father is heterozygous for ‘B’.
(b) Mother is heterozygous for ‘A’ blood group and father is homozygous for ‘B’.
(c) Both mother and father are heterozygous for ‘A’  and ‘B’ blood group, respectively.
(d) Both mother and father are homozygous for ‘A’  and ‘B’ blood group, respectively.

Correct Answer is Option (c)
If mother and father of a person with 'O' blood group have

'A' and 'B' blood group then both will be heterozygous for 'A' and 'B' blood group respectively. Therefore, both mother and father are heterozygous for ‘A’ and ‘B’ blood group, respectively. 

Q.57: Which type of cross is given below?

(a) Back cross
(b) Test cross
(c) Homozygous cross
(d) Both (A) and (B)

Correct Answer is Option (b)
It is a test cross because the progenies produced by a cross show 50% dominant trait and 50% recessive trait (1:1). A test cross is used to determine whether the individual is homozygous or heterozygous for a trait. 

Q.58: Nucleosome is represented in the diagram given below.Identify the labelled parts (i), (ii) and (iii)

(a) (i) H1 histone, (ii) DNA (iii)Histone octamer
(b) (i) Histone octamer (ii) DNA (iii) H1 histone
(c) (i) DNA (ii) H1 histone (iii) Histone octamer
(d) (i) DNA (ii) Histone octamer (iii) H1 histone

Correct Answer is Option (c)
In prokaryotes (e.g. E. coli), the DNA molecule is held with some positively charged non-histone basic proteins like negatively charged polyamines and form ‘nucleoid’. In eukaryotes, there is a set of positively charged basic proteins called histones. Histones proteins are rich in positively charged basic amino acid residues lysine and arginine. There are five types of histones proteins-H1, H2A, H2B, H3 and H4. Two molecules each of H2A, H2B, H3 and H4 organize to form a unit of eight molecules called as histone octamer. Negatively charged DNA is wrapped around positively charged histone octamer to form a structure called a nucleosome. Nucleosomes are connected with the help of linker DNA on which H1 Histone is present.

Q.59: Central Dogma is shown below, Identify the processes (i), (ii) and (iii)
(a) (i) Transcription (ii) Replication (iii) Translation
(b) (i) Replication (ii) Transcription (iii) Translation
(c) (i) Replication (ii) Translation (iii) Transcription
(d) (i) Translation (ii) Replication (iii) Transcription

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

Q.60: The structure of t-RNA is represented in the diagram given below. What is its role in the process of translation?
(a) Initiater t-RNA recognises start codon (AUG)
(b) t-RNA act as the adapter molecule that reads the genetic code.
(c) Initiater t-RNA recognises start codon (UUC)
(d) Both (A) and (B)

Correct Answer is Option (d)
t-RNA has an anticodon loop that has bases complementary to the code, and it has also an amino acid accepter end to which it binds to amino acid. Amino acid are activated in the presence of ATP, and linked to their cognate t-RNA, called as charging/ amino-acylation of t-RNA. Initiater t-RNA recognises start codon (AUG).

The t-RNA act as the adapter molecule that reads the genetic code.  Two such charged t-RNA are brought close enough to favour peptide bond formation.