Class 12 Biology: CBSE Sample Question Paper- Term I (2021-22)- 1 | 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: Parthenogenesis is a process in which:
(a) An unfertilized egg develops into a new individual
(b) Development of mostly a female gamete takes place without fertilization
(c) Both (A) and (B)
(d) None of these

Correct Answer is Option (c)
Parthenogenesis is a process in which development of mostly a female gamete (unfertilized egg) into a new individual takes place without fertilization.

Q.2: The genotype of a plant showing the dominant phenotype can be determined by:
(a) Test cross
(b) Dihybird cross
(c) Pedigree analysis
(d) Back cross

Correct Answer is Option (a)
In a test cross an organism showing a dominant phenotype whose phenotype is to be determined (Whether it homozygous or heterozygous for that trait) is crossed with a recessive parent.

Q.3: In a DNA strand, shown below, the nucleotides are linked together by

(a) Glycosidic bonds
(b) Phosphodiester bonds
(c) Peptide bonds
(d) Hydrogen bonds

Correct Answer is Option (b)
In a DNA strand, the nucleotides are linked together by 3’– 5’ phosphodiester linkage (bonds) to form a dinucleotide. More nucleotides can be joined in such a manner to form a polynucleotide chain.

Q.4: A nucleoside differs from a nucleotide. It lacks the
(a) Base
(b) Sugar
(c) Phosphate group
(d) Hydroxyl group

Correct Answer is Option (c)
A nitrogenous base is attached to the pentose sugar by an N-glycosidic linkage to form a nucleoside, that is,   Nucleoside = Nitrogen base + Pentose sugar  When a phosphate group is attached to the 5’-OH of a nucleoside through phosphodiester linkage, a nucleotide is formed, that is,  Nucleotide = Nitrogen base + Pentose sugar + Phosphate (PO4).  So, a nucleoside differs from a nucleotide as it lacks the phosphate group.

Q.5: Some flowers which have single ovule in each ovary and are packed into inflorescence are usually pollinated by:
(a) Wasp
(b) Bear
(c) Water
(d) Wind

Correct Answer is Option (d)
Wind pollination or anemophily is favoured by flowers having a single ovule in each ovary, and numerous flowers packed in an inflorescence. Wind pollination is a non-directional pollination.

Q.6: Observe structures A and B given below. Both A and B belong to which class of sugars.

(a) Trioses
(b) Hexoses
(c) Pentoses
(d) Polysaccharides

Correct Answer is Option (c)
Ribose sugar and (B) Deoxyribose sugar both belong to the class pentoses as it contains '5' carbon atoms.

Q.7: Occasionally, a single gene may express more than one effect. The phenomenon is called
(a) Multiple allelism
(b) Mosaicism
(c) Pleiotropy
(d) Polygeny

Correct Answer is Option (c)
Pleiotropy is a phenomenon in which a single gene affects multiple effects. Sometimes, one trait will be very evident and others will be less evident, e.g., a gene for white eye in Drosophila also affect the shape of organs in male responsible for sperm storage as well as other structures. Similarly, sickle-cell anaemic individuals suffer from a number of problems, all of which are pleiotropic effects of the sickle-cell alleles. Multiple allelism is the state of having more than two alternative contrasting characters controlled by multiple alleles at a single genetic locus, for e.g. ABO blood group. Mosaicism describes the occurrence of cells that differ in their genetic component from other cells of the body. Polygeny refers to a single characteristic that is controlled by more than two genes. (it is also known as multifactorial inheritance).

Q.8: The net electric charge on DNA and histones is
(a) Both positive
(b) Both negative
(c) Negative and positive, respectively
(d) Zero

Correct Answer is Option (c)
DNA consists of a nitrogenous base, pentose sugar and a phosphate group. DNA has negative charge due to the presence of phosphate group. Histones are rich in the basic amino acids lysine and arginine, which carry positive charges in their side chains. Therefore, histones are positively charged.

Q.9: Mendel’s Law of independent assortment holds good for genes situated on the
(a) Non-homologous chromosomes
(b) Homologous chromosomes
(c) Extra nuclear genetic element
(d) Same chromosome

Correct Answer is Option (a)
Mendel’s Law of independent assortment holds good for genes situated on the nonhomologous chromosome. According to law of independent assortment when two or more characteristics are inherited, individual hereditary factors assort independently of one another during gamete production, giving different characters an equal opportunity of occurring together. It can be illustrated by the dihybrid cross (a cross between two truebreeding parents) that express different traits for two characteristics.  When the genes are on separate chromosome, the two alleles of one gene (A and a) will segregate into gametes independently of the two alleles of the other gene (B and b). Equal numbers of four different gametes will form AB, aB, Ab, ab. But if the two genes are on the same chromosome, then they will be linked and will segregate together during meiosis, producing only two kinds of gametes. Homologous chromosomes are essentially similar in size but not identical. Each carries the same genetic information in same order but the alleles for each trait may not be the same. Extra nuclear genetic elements (also called as plasmids) shows the pattern of maternal inheritance.
Homologous chromosomes are essentially similar in size but not identical. Each carries the same genetic information in same order but the alleles for each trait may not be the same. Extra nuclear genetic elements (also called as plasmids) shows the pattern of maternal inheritance.

Q.10: Which of the following are the functions of RNA?
(a) It is a carrier of genetic information from DNA to ribosomes synthesising polypeptides
(b) It carries amino acids to ribosomes.
(c) It is a constituent component of ribosomes.
(d) All of these

Correct Answer is Option (d)
Ribosomal RNA (rRNA), messenger RNA (mRNA) and transfer RNA (tRNA) are major classes of RNAs that are involved in gene expression.  rRNAs bind protein molecules and give rise to ribosomes. mRNA carries coded information for translation into polypeptide formation. rRNA is also called soluble or adaptor RNA and carries amino acids to mRNA during protein synthesis.

Q.11: Females with Turner’s syndrome have
(a) Less developed breasts
(b) Rudimentary ovaries
(c) Small sized uterus
(d) All of these

Correct Answer is Option (d)
Features of female with Turner’s syndrome; Ovaries are rudimentary, lack of other secondary sexual characters, dwarf, mentally retarded.

Q.12: Which of the following steps in transcription is catalysed by RNA polymerse?
 
(a) Initiation
(b) Elongation
(c) Termination
(d) All of the above

Correct Answer is Option (b)
The DNA-dependent RNA polymerase helps in transcription by catalysing the polymerisation in only one direction (i.e., 5' and 3').

Q.13: In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called:
(a) A-DNA
(b) B-DNA  
(c) cDNA
(d) rDNA

Correct Answer is Option (c)
In some viruses, like retroviruses (e.g., HIV), an enzyme called reverse transcriptase is used to generate complementary DNA (cDNA) from an RNA template. This process is termed reverse transcription.

Q.14: Which among the following has 23 chromosomes?
(a) Spermatogonia
(b) Zygote
(c) Secondary oocyte
(d) Oogonia

Correct Answer is Option (c)
Secondary oocyte has 23 chromosomes and is formed by meiosis-I of primary oocyte.

Q.15: Match the following:
(A) Stalk by which ovule is attached to placenta (i) Hilum
(B) Junction between ovule and funicle  (ii) Integument
(C) Protective envelopes around ovule  (iii) Micropyle
(D) A small opening at the tip of integuments. (iv)  Funicle
(a) a-i, b-iii, c-ii, d-iv
(b) a-ii b-i, c-ii, d-iv
(c) a-iv, b-i, c-ii, d-iii
(d) a-iii, b-iv, c-ii, d-i

Correct Answer is Option (c)
iv. Funicle - (a) Stalk by which ovule is attached  to placenta.
i.  Hilum - (b) Junction between ovule and funicle.
ii. Integument - (c) Protective envelopes around ovule.
iii. M icropyle - (d) A small opening at the tip of integuments.

Q.16: As Ovaries is to Ovulation : : Oviduct is to_______
(a) Fertilization
(b) Birth
(c) Pregnancy
(d) All of these

Correct Answer is Option (c)
Function of ovaries is ovulation whereas fertilization takes place in oviduct.

Q.17: Megaspore, the first cell of female gametophytic generation develops into:
(a) Pollens
(b) Embryo sac
(c) Ovule
(d) Anthers

Correct Answer is Option (b)
Megaspore is the first cell of female gametophytic generation in angiosperm. It undergoes three successive generations of free nuclear mitosis to form 8-nucleated and 7 -celled embryo sac.

Q.18: How many ovum(s) is/are released in one menstruation?
(a) 1
(b) 2
(c) 3
(d) 4

Correct Answer is Option (a)
All female born with all her eggs but when the egg develops when puberty starts. One egg develops and is released during each menstrual cycle.

Q.19: Select the correct option:
(a) The largest part of the rice grain is endosperm.
(b) The edible part of coconut is endosperm.
(c) Both a and b
(d) None of the above

Correct Answer is Option (c)
The largest part of the rice grain is endosperm. And the edible portions of the grain are bran, endosperm and embryo. The morphological nature of the edible part of coconut is endosperm.

Q.20: The figure given above shows the development of a fertilized human egg cell. In zygote, the first cleavage division occur at:

(a) Ovary
(b) Oviduct
(c) Fallopian tube
(d) Uterus

Correct Answer is Option (c)
Cleavage occurs within the fallopian tube and is holoblastic, dividing the zygote completely into blastomeres. The first cleavage divides the zygote longitudinally into two blastomeres wherein one is slightly larger than the other.

Q.21: WHO’s interpretation of reproductive health is:
(a) Physical well being
(b) Emotional well being
(c) Behavioural and social well being
(d) Total well-being in all respects of reproduction.

Correct Answer is Option (d)

WHO defines reproductive health as total well-being in all respects of reproduction including physical, emotional, behavioral and social.

Q.22: In XO type of sex determination
(a) Males produce two different types of gametes
(b) Females produce two different types of gametes
(c) Males produce gametes with Y chromosome.
(d) Females produce gametes with Y chromosome

Correct Answer is Option (a)
XO mechanism shows male heterogamety. Males produce two different types of gametes.

Q.23: In sickle cell anaemia glutamic acid is replaced by valine. Which one of the following triplets codes for valine?
(a) G G G
(b) A A G    
(c) G A A
(d) G U G

Correct Answer is Option (d)
Sickle-cell anaemia is a recessive autosomal gene disorder. This disease is controlled by a single pair of allele HbA and HbS.  It is caused due to inheritance of a defective allele coding for beta globulin. It results in the transformation of HbA into HbS in which glutamic acid (Glu) is replaced by valine (Val) at sixth position in each of two beta chains of haemoglobin. This substitution occurs due to the single base substitution of the beta globin gene from GAG (Glu) to GUG (Val). Whereas, the other codes GGG, AAG. GAA do not codes for valine.

Q.24: Control of gene expression takes place at the level of
(a) DNA-replication
(b) Transcription
(c) Translation
(d) None of the above

Correct Answer is Option (b)
Gene expression, which results in the formation of a polypeptide, can be regulated at several levels. In eukaryotes, the regulation occurs at transcriptional level (formation of primary transcript), processing level (regulation of splicing), and transport of mRNA from nucleus to the cytoplasm, and translational level. While in prokaryotes, control of the rate of transcriptional initiation is the predominant site for control of gene expression.

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) : The offspring of a cross made between the plants having two contrasting characters shows only one character without any blending.
Reason (R) : The factor controlling any character is discrete and independent.
(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)
According to law of segregation or law of purity of gametes, factors or alleles of a pair segregate from each other during gamete formation, such that a gamete receives only one of the two factors and do not show any blending.

Q.26: Assertion (A) : Cross of F1 individual with recessive homozygous parent is test cross.
Reason (R) : No recessive individual are obtained in the monohybrid test cross progeny.
(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)
In the monohybrid test cross progeny both heterozygous and recessive individuals are obtained in 1 : 1 ratio.

Q.27: Assertion (A) :  It is not necessary that breasts be large in order to nurse an infant.
Reason (R) : Breast size does not affect the ability of a woman to nurse breastfeed.
(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)
It is not necessary that breasts be large in order to nurse an infant. Breast size does not affect the ability of a woman to nurse breast feed.

Q.28: Assertion (A) : The chances of having a child with Down’s syndrome increases if the age of the mother is between 20 to 25.
Reason (R) : The chances of having a child with Downs syndrome increases with the age of the mother because age adversely affects meiotic chromosome behaviour.
(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)
Down’s syndrome increases if the age of the mother exceeds forty years because age adversely affects meiotic chromosome behaviour. Meiosis in the egg cells is not completed, until after fertilization. During this long gap (till meiosis is not completed) egg cells are arrested in prophase I and chromosomes are unpaired. The greater the time they remain unpaired greater the chance for unpairing and chromosome nondisjunction.

Q.29: A particular species of plant produces light, non-sticky pollen in large numbers and its stigmas are long and feathery. These modifications facilitate pollination by:
(a) Insects
(b) Water
(c) Wind
(d) Animals

Correct Answer is Option (c)
Plants use two abiotic (wind and water) and one biotic (animals) agent to achieve pollination. Majority of plants use biotic agents for pollination. Pollination by wind is more common amongst abiotic pollination. It requires the light and non-sticky pollen grains so that, they can be transported in wind currents. They often possess well exposed stamens (so that the pollens are easily dispersed into wind currents) and large often feathery stigma to easily trap air-borne pollen grains. Wind pollination is common in grasses. Pollination by water is called hydrophily which is quite rare in flowering plants but occurs in aquatic plants. Zoophily is pollination through the agency of animals. Entomophily (pollination by insects) is the most common type of zoophily which occurs through the agency of animals.

Q.30: 23-year-old Sahil has been diagnosed with an infection of reproductive tract caused by bacteria. He complains about burning sensation during urination, pus-containing discharge and pain around genitalia. This infection has incubation period of 2-5 days but can be cured. From which disease is Sahil suffering?
(a) Chlamydiasis
(b) Herpes
(c) Gonorrhoea
(d) Syphilis

Correct Answer is Option (c)
Sahil is suffering from Gonorrhoea, a sexually transmitted disease caused by an bacterium Neisseria gonorrhoeae.

Q.31: In a dihybrid cross carried out by T. H. Morgan in Drosophila the F2 ratio deviated from that of Mendel’s dihybrid F2 ratio. The reason was
(a) Mutation
(b) Homozygosity
(c) Linkage
(d) Polymorphism

Correct Answer is Option (c)
The F₂ ratio deviated from that of Mendel’s dihybrid F₂ ratio (9 : 3 : 3 : 1) in an experiment performed by Morgan on Drosophila because of Linkage. The genes were linked as they were located on the same chromosome and closely associated. Therefore, they failed to segregate at the time of gamete formation resulting in greater number of parental combinations and lesser number of new recombinations in F₂ generation, thereby deviating from the normal dihybrid Mendelian ratio.

Q.32: During a fire in an auditorium a large number of assembled guests got burnt beyond recognition. Suggest a modern technique that can help hand over the dead to their relatives.
(a) Lac operon
(b) DNA fingerprinting
(c) DNA replication
(d) Punnett Square graph

Correct Answer is Option (a)
DNA fingerprinting is the technique of determination of nucleotide sequence of certain areas of DNA, which are unique to each individual. 

Q.33: Identify the correct sequence of the names of the hormones responsible for ovarian changes during the menstrual cycle in the boxes provided.

(a) a: FSH; b: LH; c: Progesterone
(b) a: LH; b: FSH; c: Progesterone
(c) a: Progesterone; b: LH; c: FSH
(d) a: Progesterone; b: FSH; c: LH

Correct Answer is Option (a)

Q.34: While planning for an artificial hybridization programme involving dioecious plants, which of the following steps would not be relevant:
(A) Bagging of female flower
(B) Dusting of pollen on stigma
(C) Emasculation
(D) Collection of pollen

Correct Answer is Option (c)

Artificial hybridisation is one of the major methods of crop improvement programme. This cross will make sure that only the desired pollen grains are used for pollination and the stigma is protected from contamination (from unwanted pollen). This is achieved by emasculation and bagging techniques. If the female parent produces unisexual flowers;there is no need for emasculation (a process of removal of anther). The female flower buds are bagged before the flowers open. When the stigma becomes receptive, pollination is carried out using the desired pollen and the flower rebagged. This protects them from contamination by unwanted pollen grains. When the female parent bears bisexual flowers,removal of anthers from the flower bud before the anther dehisces is necessary.

Q.35: Study the figures given below and answer the question

Identify in which of the crosses the strength of linkage between the genes is higher
(a) Cross A
(b) Cross B
(c) Both have similar strength
(d) None of them.

Correct Answer is Option (a)
Strength of linkage between genes is higher in cross A than that of cross B because the two genes yw are located closely on the same chromosome. Whereas in case of cross B the genes w and m are located far apart on the same chromosome. Therefore, in the latter cross the chances of recombination are higher for crossing over because lesser the distance between genes greater the strength of linkage.

Q.36: One- gene one- enzyme concept is referred to as one __(i)____ one ___(ii)____ concept.
(a) (i) DNA (ii) polypeptide
(b) (i) DNA (ii)  nucleotide
(c) (i) gene (ii) polypeptide
(d) (i) gene (ii) nucleotide

Correct Answer is Option (c)
The theory one gene-one enzyme hypothesis was stated by the geneticist George Beadle in 1945 but later when it was realized that genes also encoded non-enzyme proteins and individual polypeptide chains it was modified to gene – one polypeptide.

Q.37: Mr. and Mrs. Vijay are unable to have a baby because Vijay 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 (c)
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.38: Human Genome Project is the first effort in identifying the sequence of nucleotides and mapping of all the genes in human genome. In Human Genome Project (HGP).? ‘YAC’ and ‘BAC’  are used as vector for cloning foreign DNA. What do ‘Y’ and ‘B’ stand for?
(a) Y = Yeast B = Bacteria
(b) Y = Replication Fork B = Base
(c) Y = Yeast B = Bacterial
(d) Y = Yeast B = Base

Correct Answer is Option (c)
YAC (Yeast Artificial Chromosomes) and BAC (Bacterial Artificial chromosomes) are cloning vectors. They are used in Human genome project for cloning or amplification of human DNA fragments.

Q.39: Dihybrid cross between two garden pea plant one homozygous tall with round seeds (TTRR) and the other dwarf with wrinkled seeds (ttrr) was carried. The genotype and phenotypes of the F1 progeny obtained from the cross will be
(a) Genotype: TTRR         Phenotype: Tall,Round
(b) Genotype: TtRr           Phenotype: Tall,Round
(c) Genotype: ttRr            Phenotype: Dwarf, Round
(d) Genotype: Ttrr            Phenotype: Tall, Wrinkled

Correct Answer is Option (d)

Q.40: Testes are located outside the abdominal cavity within a sac called scrotal sac or scrotum. The particular temperature is required for proper functioning of testes and for spermatogenesis. At what temperature scrotum keeps the testes?
(a) 20°C higher than the body temperature.
(b) 2°C lower than the body temperature.
(c) 4°C highhigher than the body temperature.
(d) 4°C high lower than the body temperature.

Correct Answer is Option (b)
Testes are located outside the abdominal cavity within a sac called scrotal sac or scrotum. Scrotum keeps the testes temperature at 2˚C lower than the body temperature.  The lower temperature is required for proper functioning of testes and for spermatogenesis.

Q.41: Acrosomal reaction of the sperm occurs due to
(a) Its contact with zona pellucida of the ova.
(b) Reactions within the uterine environment of the female.
(c) Reactions within the epididymal environment of the male.
(d) Androgens produced in the uterus.

Correct Answer is Option (a)
Acrosomal reaction of the sperm occurs due to its contact with zona pellucida of the ova. The reaction that occurs in acrosome of sperm is triggered by the release of fertilizin. The main purpose of the acrosomal reaction is to start the fusion of the oocyte membrane with the sperm cell membrane allowing the combination of genetic material contained in both gametes, leading to the fertilisation of the oocyte.

Q.42: In pea plants, the colour of the flower is either violet or white, whereas human skin colour shows many gradations. The reason is :
(a) Human skin colour is qualitative inheritance whereas the colour of the flower in pea is quantitative inheritance.
(b) Human skin colour is quantitative inheritance whereas the colour of the flower in pea is qualitative inheritance.
(c) Human skin colour is controlled by polygenic genes whereas the colour of the flower controlled by allelic complementary genes .
(d) Both (B) and (C).

Correct Answer is Option (d)
Human skin colour, an example of polygenic inheritance and it is controlled by at least three separate genes and inheritance of these genes is called quantitative inheritance. There can be total eight allelic combinations in the gametes of a person heterozygous for all the three genes, hence 64 combinations or gradations in colour are possible. In this type of inheritance, the dominant alleles have cumulative effect where in each dominant alleles expresses only a part of the trait and the trait in its full form is expressed only when all the dominant alleles are present. On the other hand, the colour of the flower in pea is controlled by allelic complementary genes, which independently show a complete effect. The inheritance is qualitative as here the presence of a single dominant allele expresses the trait in full form and the presence of two dominant alleles does not make any difference.

Q.43: The third stage of parturition is called “after-birth”. In this stage
(a) Excessive bleeding occurs
(b) Foetus is born and cervix and vagina contraction to normal condition happens .
(c) Foetus is born and contraction of uterine wall prevents excessive bleeding.
(d) Placenta is expelled out.

Correct Answer is Option (d)
Third process of parturition is the third and final stage, after the delivery until the placenta or after birth is expelled by powerful uterine contractions. Umbilical cord is cut close to the baby's navel. It lasts for 10-15 minutes after the birth of child.

Q.44: Given below are the sequence of nucleotides in a particular mRNA and amino acids coded by it:
UUU   AUG   UUC  GAG  UUA GUG UAA     
Phe –Met –Phe –Glu –Leu –Val
Which of the following  property cannot be correlated from the above given data.
(a) The codon is a triplet e.g. UUU, AUG etc.
(b) Mostly AUG work as an initiating codon
(c) One codon codes for only one amino acid and not other hence it is unambiguous and specific
(d) Code is commaless, continuous and does not have pauses

Correct Answer is Option (b)
Properties of genetic code that can be correlated are:
(i) The codon is a triplet e.g. UUU, AUG etc. They form triplets.
(ii) One codon codes for only one amino acid and not other hence it is unambiguous and specific.
(iii) AUG has a dual function as it codes for   methionine and also acts as the initiator codon.
(iv) UAA is the stop codon. It codes for valine.
(v) Code is commaless, continuous and does not have pauses.
(vi) The sequence of triple N-bases in mRNA corresponds to the sequence of amino-acids in a polypeptide. Property that cannot be correlated is  that mostly AUG work as an initiating codon.

Q.45: A haemophilic man marries a normal homozygous woman . What is the probability that their son will be haemophilic?
(a) 0%
(b) 50%
(c) 75%
(d) 100%

Correct Answer is Option (a)
Haemophilia is a sexlinked recessive disorder that has its been on the X chromosome. If a man is hemophilic and the female is normal homozygous female then the probability that their son will be haemophilic is 0% because they will receive one of their X chromosomes from the mother and as the mother is normal, the genes will not be transmitted.

Q.46: With reference to the below schematic representation of Oogenesis, how many chromatids are found during oogenesis in
(i) Primary oocyte and
(ii) First polar body in a human female

(a) (i) 92  (ii) 42
(b) (i) 42 (ii) 92
(c) (i) 92 (ii) 46
(d) (i) 46 (ii) 92

Correct Answer is Option (c)
Since replication has occurred by this stage
46 X 2 = 92 chromatids
Meiosis - I is completed by this time.
So, 92/2 =46 chromatids

Q.47: Although Mendel published his work on inheritance of characters in 1865 but for several reasons, it remained unrecognised till 1900. The reason for the delay in accepting his work could be:
(a) His could not be widely publicised due to poor communication.
(b) His concept of genes or factors as discrete units and which did not blend was not accepted due to continuous variation seen in nature.
(c) He used mathematics to explain biological phenomenon which was new and unacceptable.
(d) All of the above

Correct Answer is Option (d)
The communication was not easy in those days and his work could not be widely publicised. His concept of genes as stable and discrete units that controlled the expression of traits and of the pair of alleles which did not 'blend' with each other was not accepted by contemporaries as an explanation for the apparently continuous variation seen in nature. Mendel's approach of using statistical calculations to explain biological phenomena was totally new and unacceptable to many of the biologists of his time because they were beyond the comprehension of the biologists of the time. Though Mendel's work suggested that factors (genes) were discrete units, he could not provide any physical proof for the existence of factors and what they were made of.

Q.48: A man is suffering from sickle cell anaemia. It is found that his red blood cells have become elongated sickle shaped structures. He was told that the defect is caused by the substitution of Glutamic acid (Glu) by another amino acid at the sixth position of the beta globin chain of the haemoglobin molecule. Which amino acid has substituted Glutamic acid?
(a) Threonine (Thr)
(b) Leucine (Leu)
(c) Proline (Pro)
(d) Valine (Val)

Correct Answer is Option (d)
The defect is caused by the substitution of Glutamic acid (Glu) by Valine (Val) at the sixth position of the beta globin chain of the haemoglobin molecule. The substitution of amino acid in the globin protein results due to the single base substitution at the sixth codon of the beta globin gene from GAG to GUG. The mutant haemoglobin molecule undergoes polymerisation under low oxygen tension causing the change in the shape of the RBC from biconcave disc to elongated sickle like structure.

Case-I
Read the following text and answer the following questions on the basis of the same
The DNA replication is semi-conservative is proved by an experiment  conducted by Meselson and Stahl in 1958. To perform their experiment they use heavy nitrogen (15N) in E. coli. The process of replication in living cells requires a set of enzymes. The main enzyme is DNA dependent  DNA polymerase. The DNA-A dependent DNA polymerase catalyse polymerization only in one direction, that is 5→3'. In eukaryotes, the replication of DNA takes place at the S-phase of the cell cycle.

Q.49: Viruses grown in the presence of radioactive phosphorus contained radioactive ________ but not radioactive ________.
(a) DNA, protein
(b) Protein, DNA
(c) RNA, Nucleoside
(d) mRNA, Protein

Correct Answer is Option (a)
DNA is the genetic material came from the experiments of Hershey and Chase (1952). They worked with viruses that infect bacteria called bacteriophages. They worked to discover whether it was protein or DNA from the virus that entered the bacteria. They grew some viruses on a medium that contained radioactive phosphorus and some others on a medium that contained sulfur. Viruses grown in the presence of radioactive phosphorus contained radioactive DNA but not radioactive protein because DNA contains phosphorus but protein does not. Similarly, viruses grown on radioactive sulfur contained radioactive protein but not radioactive DNA because DNA does not contain sulphur.

Q.50: During DNA replication, the breaking of H-bonds is performed by:
(a) Topoisomerase
(b) Gyrase
(c) Helicases
(d) None

Correct Answer is Option (c)
DNA replication starts with unwinding of DNA duplexes which are held together by hydrogen bond. Helicases move along the double stranded DNA and separate the strands by breaking hydrogen bonds between base pairs.

Q.51: How many types of DNA polymerases are associated with eukaryotic cell ?
(a) Three
(b) Six
(c) Five
(d) One

Correct Answer is Option (c)
5 types of DNA polymerases are associated with eukaryotic cell.

Q.52: Which enzyme seals the Okazaki fragments together
(a) DNA Polymerase
(b) Ligase
(c) Helicase
(d) Primase

Correct Answer is Option (a)
DNA ligase joins the Okazaki fragments together into a single DNA molecule.

Q.53: DNA replication is:
(a) Semi-conservative, continuous
(b) Semi-continuous, conservative
(c) Semi-conservative, semi-discontinuous
(d) Conservative

Correct Answer is Option (c)
DNA replication is said to be semi-conservative because of the process of replication, where the resulting double helix is composed of both an old strand and a new strand. ... The two resulting double helices, which each contain one "old" strand and one "new" strand of DNA, are identical to the initial double helix. Due to this reason, replication occurs continuously on one strand and discontinuously on the other strand. This is known as the semi-discontinuous mode of replication. Every new DNA molecule that is formed has a new and an old strand of the DNA. Thus, during DNA replication, entirely new DNA copies are not generated.

Q.54: Teminism is bidirectional flow of information in which
(a) RNA is synthesized by DNA
(b) DNA is synthesized by DNA
(c) DNA is synthesized by RNA
(d) RNA is synthesized by RNA

Correct Answer is Option (c)
Teminism is popularly known as reverse transcription, i.e., DNA can be synthesized by RNA.

Q.55: A pedigree is shown below for a disease that is autosomal dominant. The genetic made up of the first generation is

(a) RR, Rr
(b) Rr, RR
(c) Rr, rr
(d) Rr, Rr

Correct Answer is Option (c)
A disease  is autosomal dominant hence, genes for the first generation will be Rr and rr.

Q.56: The below diagram shows human male reproductive system (one side only).

Name Z and state the function
(a) Z— testicular lobules; Function: Produce sperm
(b) Z — seminal vesicle; Function: Produce and store fluid that will eventually become semen.
(c) Z — epididymis; Function: Storage of sperms.
(d) Z — vasa efferentia; Function: Help the transportation of sperms

Correct Answer is Option (c)
The epididymis is a narrow, tightly-coiled tube connecting rear of the testicles to the deferent duct (ductus deferens or vas deferens). It stores sperm for maturation.

Q.57: The dihybrid cross carried on Drosophila melanogaster

Study the two figures and choose the correct statement:
(a) In cross A, genes are loosely linked and show very high recombination.
(b) In cross B, genes are loosely linked and show very low recombination.
(c) In cross A, genes are tightly linked and show very low recombination
(d) In cross B, genes are tightly linked and show very high recombination.

Correct Answer is Option (c)
In cross A genes are tightly linked, they showed very low recombination. In cross B genes were loosely linked they showed very high recombination. They deviated from 9:3:3:1 ratio because of segregation of genes.

Q.58: The following is the illustration of the sequence of ovarian events “a” to “i” in a human female :

What is the difference between “d” and “e” ?
(a) “d” is the developing corpus luteum and “e” is the tertiary follicle
(b) “d” is the developing tertiary follicle and “e” is the Graafian follicle.
(c) “d” is Luteinising hormone and “e’ is the corpus luteum
(d) “d” is developing Graafian follicle and “e” is the corpus luteum.

Correct Answer is Option (b)
‘a’, ‘b’, ’c’ and ‘d’ are developing tertiary follicle, ‘e’ is the Graafian follicle, ‘f ’ is ovulated follicle, ‘g’ is corpus luteum , ‘h’ and ‘i’ are regressing corpus luteum.

Q.59: The Cross shown above is a
(a) Back cross
(b) Test cross
(c) Both (A) and (B)
(d) None of the above

Correct Answer is Option (c)
The cross between heterozygous F₁ hybrid and the double recessive homozygous is known as the test cross. The cross between the F₁ progeny and any one of its parents is known as a back cross. The cross between the F₁ progeny and the recessive phenotypic parent is known as a test cross. Back crosses can only be referred to as a test cross if done with the recessive parent.

Q.60: Which of the following contains the actual genetic part of a sperm?
(a) Whole of it
(b) Tail
(c) Middle piece
(d) Head

Correct Answer is Option (d)
Head of the sperm is anterior, broad, flattened and almond shape. It consist of  two parts. The head portion is mainly a cell nucleus; it consists of genetic substances. and anterior small cap-like acrosome. The nucleus consists of condensed DNA and basic proteins.