CBSE Class 12 Biology Question Paper (Outside Delhi 2021) Term-I with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Biology with Solutions and CBSE Class 12 Biology Question Paper (Outside Delhi 2021) Term-I to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Biology Question Paper (Outside Delhi 2021) Term-I with Solutions

Time Allowed: 90 Minutes
Maximum Marks: 35

General Instructions:

1. The questions paper contains three sections: Section A, B and C.
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

This section consists of 24 questions. Attempt any 20 questions from this section. The first 20 questions attempted would be evaluated.

Question 1.
Enclosed within the integuments of a typical anatropous ovule is a diploid mass of cellular tissue known as :
(a) Megaspore mother cell
(b) Nucellus
(c) Synergids
(d) Embryo sac
Answer:
(b) Nucellus

Enclosed within the integuments is a mass of cells called the nucellus. Cells of the nuccllus have abundant reserve food materials. In the nucellus, the embryo sac or female gametophyte is present.

Note
At the time of fertilization, the nucellus consist of a bulky tissue ventral to the embryo sac and a rather thin layer elsewhere.

Question 2.
Researchers the world over are trying to transfer apomietic genes to hybrid varieties as hybrid characters in the progeny:
(a) do not segregate
(b) segregate
(c) develop genetic variations
(d) will remain unexpressed
Answer:
(a) do not segregate

Production of hybrid seeds is costly and hence the cost of hybrid seeds becomes too expensive for the farmers. If these hybrids are made into apomicts, there is no segregation of characters in the hybrid progeny. Then the farmers can keep on using the hybrid seeds to raise new crop year after year and he does not have to buy hybrid seeds every year.

Question 3.
The aquatic plant having long and ribbon like pollen grains is:
(a) Vallisneria
(b) Hydrilla
(c) Eicchornia
(d) Zoster a
Answer:
(d) Zoster a

Zostera is a submerged marine sea grass that releases long, ribbon-like pollen grains underwater. The pollen grains are carried passively by water and ultimately reach female flowers.

Question 4.
In a typical dicotyledonous embryo, the portion of embryonal axis above the level of cotyledons is :
(a) Plumule
(b) Coleoptile
(c) Epicotyle
(d) Hypocotyle
Answer:
(c) Epicotyle

The portion of embryonal axis above the level of cotyledons is the epicotyl, which terminates with the plumule or stem tip.

Question 5.
To overcome incompatible pollinations so as to get desired hybrids, a plant breeder must have the knowledge of _____
(a) pollen – nucellar interaction
(b) pollen – egg cell interaction
(c) pollen – pistil interaction
(d) pollen – embryo sac interaction
Answer:
(c) pollen – pistil interaction

To overcome incompatible pollination so as to get a desirable hybrid a plant breeder must have the knowledge of pollen pistil interaction.

Note
Pollination does not guarantee the transfer of right type of pollen C (comptabile pollen) of some species as the stigma.

Question 6.
Pollen grains retain viability for months in plants belonging to different families given below:
(i) Solanaceae
(ii) Leguminosae
(iii) Gramineae
(iv) Rosaceae
(v) Liliaceae
The correct option is :
(a) (i), (ii) and (v)
(b) (i), (ii) and (iv)
(c) (ii), (iv) and (v)
(d) (i), (iii) and (v)
Answer:
(b) (i), (ii) and (iv)

Question 7.
Given below is a diagramatic view of the human male reproductive system :

Identify the correct labelling for W, X, Y and Z and choose the correct option from the table below:

Answer:
(c)

Question 8.
During human embryonic development, the heart in the embryo is formed after:
(a) 15 days of pregnancy
(b) 30 days of pregnancy
(c) 45 days of pregnancy
(d) 60 days of pregnancy
Answer:
(b) 30 days of pregnancy
In human being after one month of pregnancy, the embryo’s heart is formed.

Note
The first sign of growing foetus may be noticed by listening to the heart sound carefully through stethoscope.

Question 9.
The uterus opens into the vagina through a narrow:
(a) Ampulla
(b) Isthmus
(c) Cervix
(d) Infundibulum
Answer:
(c) Cervix

The uterus opens into vagina through a narrow cervix and the cavity of the cervix is called cervical canal.

Note
The uterus is single and it is also known as womb.

Question 10.
In the tranverse section of a young anther shown below, identify the correct sequence of wall layers from outside to inside:

Answer:
(d)

Question 11.
Floral reward/s provided by insect pollinated flowers to sustain animal visit is/are:
(a) nectar and fragrance
(b) nectar and pollen grains
(c) pollen grains and fragrance
(d) fragrance and bright colour
Answer:
(b) nectar and pollen grains

To sustain animal visits, the flower have to provide rewards to the animals. Nectar and pollen grains are usual floral rewards.

Question 12.
The cause of Klinefelter’s syndrome in humans is:
(a) Absence of Y-chromosome
(b) Absence of X-chromosome
(c) Extra copy of an autosome
(d) Extra copy of an X-chromosome
Answer:
(d) Extra copy of an X-chromosome

Klinefelter’s Syndrome is a genetic disorder that is also caused due to the presence of an additional copy of the X chromosome resulting in a karyotype of 47, XXY.

Question 13.
Select the incorrect pair :
(a) Polygenic inheritance : Haemophilia
(b) Linkage: Drosophila
(c) Incomplete dominance : Antirrhinum
(d) Pleiotropy: Phenylketonuria
Answer:
(a) Polygenic inheritance : Haemophilia

Such trait that controlled by two or more than two genes are thus called polygenic inheritance. Human skin colour is one of the example of this.

Question 14.
According to Mendel, the nature of the unit factors that control the expression of traits were :
(a) Stable
(b) Blending
(c) Stable and discrete
(d) Discrete
Answer:
(c) Stable and discrete

According to Mendel, the natures of unit factor that control the expression of trait were stable and discrete.

Question 15.
Which of the following animals exhibit male heterogamety ?
(i) Fruit fly
(ii) Fowl
(iii) Human
(iv) Honey bee
(a) (i) and (iii)
(b) (ii)and(iv)
(c) (ii) and (iii)
(d) (i)and(iv)
Answer:
(a) (i) and (iii)

Males produce two different types of gametes, (a) either with or without X chromosome or (b) some gametes with X-chromosome and some with Y-chromosome. Such types of sex determination mechanisms are designated to be an example of male heterogamety. e.g.- Human beings, Drosophila

Question 16.
The probability of all possible genotypes of offsprings in a genetic cross can be obtained with the help of:
(a) Test cross
(b) Back cross
(c) Punnett square
(d) Linkage cross
Answer:
(c) Punnett square

Reginald Punnett, an English geneticist, established one of the simplest methods for calculating the mathematical chance of inheriting a given feature. Punnett square became popular as a result of his technique. It’s the simplest graphical method for determining all conceivable genotype combinations in children.

Note
Test-cross involves mating of an unknown genotypic individual with a known homozygous recessive.

Question 17.
The number of different types of gametes that would be produced from a parent with genotype AABBCc is:
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

Three genes A, B, and C control skin colour in humans with the dominant forms A, B and C responsible for dark skin colour and the recessive forms ‘c’ for light skin colour. Hence, The type of gametes produced by the parent with genotype AABBCc is 4.

Question 18.
Select the important goals of HGP from the given options :
(i) Store the information for data analysis
(ii) Cloning and amplification of human DNA
(iii) Identify all the genes present in human DNA
(iv) Use of DNA information to trace human history
(a) (i) and (ii)
(b) (ii) and (iii)
(c) (i) and (iii)
(d) (ii) and (iv)
Answer:
(c) (i) and (iii)

Some of the important goals of HGP were as follows:
(i) Identify all the approximately 20,000-25,000 genes in human DNA;
(ii) Determine the sequences of the 3 billion chemical base pairs that make up human DNA;
(iii) Store this information in databases;
(iv) Improve tools for data analysis;
(v) Transfer related technologies to other sectors, such as industries;
(vi) Address the ethical, legal, and social issues (ELSI) that may arise from the project.

Question 19.
A codon is a ‘triplet of bases’ was suggested by :
(a) Marshall Nirenberg
(b) Har Gobind Khorana
(c) Georgee Gamow
(d) Francis Crick
Answer:
(c) Georgee Gamow

George Gamow suggested that the genetic code should be made up of a combination of three nucleotides. He proposed that If 20 amino acids are to be coded by 4 bases, then the code should be made up of three nucleotides.

Question 20.
The correct feature of Double-helical structure of DNA as given by Waston and Crick is :
(a) Right-handed helix, pitch is 3.4 nm
(b) Left-handed helix, pitch is 3.8 nm
(c) Right-handed helix, pitch is 3.8 nm
(d) Left-handed helix, pitch is 3.4 nm
Answer:
(a) Right-handed helix, pitch is 3.4 nm

The two chains are coiled in a right-handed fashion. The pitch of the helix is 3.4 nm and there is roughly 10 bp in each turn.

Note
The structure of DNA is reffered to as a double helix as it resembles a twisted staircase.

Question 21.
Charging of tRNA during translation is necessary for:
(a) Binding of anticodons of tRNA to the respective codons of mRNA
(b) Peptide bond formation between two amino acids
(c) Movement of ribosomes from codon to codon
(d) Binding of ribosomes to the mRNA
Answer:
(b) Peptide bond formation between two amino acids

Charging of tRNA during translation process is called amino-acylation of tRNA. When two such charged tRNAs are brought close enough the formation of peptide bond between the corresponding amino acids would be favoured energetically. The presence of a catalyst would enhance the rate of peptide bond formation.

Question 22.
If E. coli were allowed to grow in the culture medium for 80 minutes by Matthew Meselson and Franklin Stabi in their experiments, the proportion of light and hybrid density DNA molecule would have been:
(a) 87.5% of light density DNA and 12.5% of hybrid density DNA
(b) 75.0% of light density DNA and 25% of hybrid density DNA
(c) 50% of light density DNA and 50% of hybrid density DNA.
(d) 12.5% of light density DNA and 87.5% of hybrid density DNA.
Answer:
(a) 87.5% of light density DNA and 12.5% of hybrid density DNA

E. coli divides in 20 minutes. So, after 80 minutes there will be 4 generations. In the first generation all the strands will be hybrid (as the heavy isotope will be incorporated in the newly synthesised strand of DNA) i.e the two DNA will be of intermediate nature. In the second generation, 50% the DNA will be light and 50% will be hybrid. In the 3rd generation, 25% will be hybrid and 75% will be light and in the fourth generation 12.5% will be hybrid and 87.5% will be light strand.

Question 23.
A diagrammatic illustration of the process of transcription by RNA polymerase-II in eukaryote is given below. Choose the most appropriate statement with respect to the fete of the precursor of mRNA transcribed that will be :

(a) Translation will take place once the precursor of mRNA leaves the nucleus.
(b) Translation on mRNA will not take place once the precursor of mRNA leaves the nucleus.
(c) Translation will take place in the nucleus.
(d) The precursor of mRNA has to be processed further in next step before being translated.
Answer:
(a) Translation will take place once the precursor of mRNA leaves the nucleus.

The fully processed hnRNA, now called mRNA, that is transported out of the nucleus for translation.

Question 24.
Identify the correct pair of codon with its corresponding pair of amino acid:
(a) UAA: Leucine
(b) UGA : Serine
(c) AUG : Histidine
(d) UUU : Phenylalanine
Answer:
(d) UUU : Phenylalanine

The code is nearly universal: for example, from bacteria to human UUU would code for Phenylalanine (phe).

Note
Codon is triplet of bases in the DNA coding far one amino-acid.

Section B

This section consists of 24 questions. Attempt any 20 questions from this section. The first 20 questions attempted should be evaluated. Direction: Question Nos. 25 to 28 consists of two statements – Assertion (A) and Reason (R). Answer these questions selecting the appropriate option given below:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.

Question 25.
Assertion (A): Very often persons suffering from Sexually Transmitted Diseases (STD) do not go for timely detection and proper treatment.
Reason (R): Absence or less significant symptoms in the early stages of STDs and the social stigma attached to the disease.
Answer:
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Infected person may often be asymptomatic and hence, may remain undetected for long. Absence or less significant symptoms in the early stages of infection and the social stigma attached to the STIs, deter the infected persons from going for timely detection and proper treatment.

Question 26.
Assertion (A) : Vasectomy is a sterilisation procedure advised for females as a terminal method.
Reason (R): In vasectomy, a small part of the vas deferens is removed or tied by blocking gamete transport therefore preventing conception.
Answer:
(d) Assertion (A) is false, but Reason (R) is true.

Sterilisation procedure in the male is called ‘vasectomy’ and that in the female is called ‘tubectomy’. The statement in reason is correct.

Question 27.
Assertion (A): Interstitial spaces outside the seminiferous tubule have blood vessels and sertoli cells.
Reason (R): Sertoli cells provide nutrition to the germ cells.
Answer:
(d) Assertion (A) is false, but Reason (R) is true.

The regions outside the seminiferous tubules called interstitial spaces, contain small blood vessels and interstitial cells or Leydig cells. The statement in reason is correct.

Question 28.
Assertion (A): Accumulation of phenylalanine in the brain results in metal retardation in Phenylketonuria.
Reason (R) : The affected person lacks phenylalanine which is therefore not converted to tyrosine.
Answer:
(c) Assertion (A) is true, but Reason (R) is false.

The affected individual lacks an enzyme that converts the amino acid phenylalanine into tyrosine. As a result of this phenylalanine is accumulated and converted into phenylpyruvic acid and other derivatives. Accumulation of these in brain results in mental retardation.

Note
Untreated phenylketouria can lead to brain damage, intellectual disabilities, behavioural symptoms or seizures.

Question 29.
Choose the correct option for the features of functional mammary gland of all female mammals from the statements below:
(i) Glandular tissue with variable amount of fat.
(ii) Mammarylobes, 30-40 in number called alveoli.
(iii) Mammary ducts joining to form mammary tubules.
(iv) Mammary ampulla connected to lactiferous duct.
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (i) and (iv)
(d) (ii) and (iv)
Answer:
(c) (i) and (iv)
The mammary glands are paired structures (breasts) that contain glandular tissue and a variable amount of fat. The glandular tissue of each breast is divided into 15-20 mammary lobes containing clusters of cells called alveoli. The cells of alveoli secrete milk, which is stored in the cavities (lumens) of alveoli. The alveoli open into mammary tubules. The tubules of each lobe join to form a mammary duct. Several mammary ducts join to form a wider mammary ampulla which is connected to the lactiferous duct.

Question 30.
Which condition of gynoecium (pistil) is shown the figures (i) and (ii) ?

(a) (i) multicarpellary apocarpous, (ii) multicarpellary syncarpous
(b) (i) multicarpellary syncarpous, (ii) multicarpellary apocarpous
(c) (i) bicarpellary apocarpous, (ii) bicarpellary syncarpous
(d) (i) bicarpellary syncarpous, (ii) bicarpellary apocarpous
Answer:
(b) (i) multicarpellary syncarpous, (ii) multicarpellary apocarpous

The gynoecium may consist more than one pistil is called multicarpellary. When there are more than one, the pistils maybe fused together (syncarpous) or maybe free (apocarpous).

Question 31.
An IUD recommended to promote the cervix hostility to the sperms is
(a) CuT
(b) Multiload-375
(c) LNG-20
(d) Cu7
Answer:
(c) LNG-20

The hormone releasing IUDs (Progestasert, LNG-20), make the uterus unsuitable for implantation and the cervix hostile to the sperms.

Note
An IUD is small plastic T-shape device used far birth control.

Question 32.
Identify the disease which is not a sexually transmitted disease :
(a) Gonnorhoea
(b) Syphilis
(c) Amoebiasis
(d) Chalamydiasis
Answer:
(c) Amoebiasis

Amoebiasis caused by protozoan parasites Entamoeba histolytica in large intestine of human. The main source of infection is contaminated drinking water and food.

Question 33.
The nature of meiotic division during oogenesis in a human female is:
(a) equal cell division
(b) suspended cell division
(c) continuous cell division
(d) rapid cell division
Answer:
(b) suspended cell division

In oogenesis, diploid oogonium go through mitosis until one develops into a primary oocyte, which will begin the first meiotic division, but then arrest; it will finish this division as it develops in the follicle, giving rise to a haploid secondary oocyte and a smaller polar body.

Question 34.
Choose the correct labellings for the parts X, Y arid Z in the given figure of the stages in embryo development in a dicot:

(a) X is suspensor, Y is radicle and Z is cotyledon
(b) X is radicle, Y is cotyledon and Z is suspensor
(c) X is cotyledon, Y is suspensor and Z is radicle
(d) X is zygote, Y is radicle and Z is cotyledon
Answer:
(c) X is cotyledon, Y is suspensor and Z is radicle

Question 35.
Which of the following outbreeding devices are used by majority of flowering plants to prevent inbreeding depression ?
(i) Pollen release and stigma receptivity are not synchronised.
(ii) Different positions of anther and stigma.
(iii) Production of different types of pollen grains.
(iv) Formation of unisexual flowers along with bisexual flowers.
(v) Preventing self-pollen from fertilising the ovules by inhibiting pollen germination.
(a) (i), (ii) and (v)
(b) (ii), (iii) and (v)
(c) (i), (iii) and (v)
(d) (iii), (iv) and (v)
Answer:
(a) (i), (ii) and (v)

Flowering plants Continued self-pollination result in inbreeding depression. For this it developed many devices to discourage self-pollination and encourage cross-pollination. In some species, pollen release and stigma receptivity are not synchronised.

In some other species, the anther and stigma are placed at different positions so that the pollen cannot come in contact with the stigma of the same flower.
The third device to prevent inbreeding is self-incompatibility.

Note
Majority of flowering plants prohermaphrodite flowers and pollen grain are likely to come in contact with stigma of same flower.

Question 36.
Histone proteins that help in forming the nucleosomes in the nucleus are rich in basic amino acids such as :
(a) Arginine and tyrosine
(b) Lysine and histidine
(c) Arginine and lysine
(d) Histidine and tryptophan
Answer:
(c) Arginine and lysine

Histones are rich in the basic amino acid residues lysine and arginine. Both the amino acid residues carry positive charges in their side chains.

Note
As in this question, only one trait is considered, it is an example of mono-hybrid cross.

Question 37.
In Pisum sativum, the flower position may be axial (allele A) or terminal (allele a). What would be the percentage of the offspring with respect to axial flower position, if a cross is made between parents Aa x aa ?
(a) 25%
(b) 50%
(c) 75%
(d) 100%
Answer:
(b) 50%

when cross is occur between axial (Aa) and terminal (aa) then,

as shown in punnett square the phenotype ratio of Axial flower is 50 % and terminal flower is 50 %.

Question 38.
In humans rolling of tongue is an autosomal dominant trait (R). In a family both the parents have the trait of rolling tongue but their daughter does not show the trait, whereas the sons have the trait of rolling of tongue.
The genotypes of the family would be :

Answer:
(b)

Mother and father have autosomal dominant traits with Rr and Rr respectively. After crossing between these two traits, the son is born with this autosomal dominant trait RR whereas the daughter is born with the recessive trait rr.

Question 39.
Study the pedigree analysis of human given below and identify the type of inheritance along with an example:

(a) Sex-linked recessive, Haemophilia
(b) Sex-linked dominant, Vitamin D resistant rickets
(c) Autosomal recessive, Sickle-cell anaemia
(d) Autosomal dominant, Myotonic Dystrophy
Answer:
(d) Autosomal dominant, Myotonic Dystrophy
Representative pedigree analysis is the example of Autosomal dominant trait Myotonic dystrophy.

Question 40.
Possibility of the blood groups of the children in a family where the father is heterozygous for blood group ‘A’ and the mother is heterozygous for blood group ‘B’, would be:
(a) Blood groups ‘A’, ‘B’
(b) Blood groups ‘A’, ‘B’, ‘O’
(c) Blood group ‘ AB ’, ‘ O ’
(d) Blood groups ‘A’, ‘B’, ‘AB‘, ‘O’
Answer:
(d) Blood groups ‘A’, ‘B’, ‘AB‘, ‘O’

If the Father is with heterozygous A blood group (I ^A I ^O ) and mother with heterozygous B blood group (I ^B I ^O ) then the progeny of such parents are,

Phenotype blood group of I ^A I ^B – AB blood group
I ^B I ^O – B blood group
I ^A I ^O – A blood group
I ^O I ^O – O blood group
Therefore, progeny is with blood group A,B,AB, and O.

Question 41.
The correct statement with respect to Thalassemia in humans is:
(a) α-Thalassemia is controlled by a single gene HBB.
(b) The gene for α-Thalassemia is located on chromosome-16.
(c) β-Thalassemia is controlled by two closely linked genesHBA-1 andHBA-2.
(d) In β-Thalassemia the production of a-globin chain is affected.
Answer:
(b) The gene for α-Thalassemia is located on chromosome-16.

a Thalassemia is controlled by a single gene HBB on chromosome 11 of each parent and occurs due to mutation of one or both the genes.
a Thalassemia, production of a globin chain is affected, a Thalassemia is controlled by two closely linked genes HBA1 and-HBA2 on chromosome 16 of each parent.

Question 42.
A region of coding strand of DNA has the following nucleotide sequence :
5′- TACGCCG – 3′
The sequence of bases on mRNA transcribed by this would be:
(a) 5 ‘-UACGCCG – 3’
(b) 3 ‘-UACGCCG-3′
(c) 5′-ATGCGGC-3′
(d) 3′-ATGCGGC – 3’
Answer:
(a) 5 ‘-UACGCCG – 3′

5’-TACGCCG-3’ Coding strand
3’ – ATGCGGC – 5 ’ Template strand
Therefore mRNA strand transcribed by template strand is 5’- UACGCCG -3’

Question 43.
A DNA molecule is 160 base pairs long. If it has 20% adenine, how many cytosine bases are present in this DNA molecule ?
(a) 48
(b) 64
(c) 96
(d) 192
Answer:
(a) 48

As per Chargaff rule ratio of Adenine is equal to Thymine and Guanine is equal to Cytosine.
Therefore, if Adenine is of20 % then Thymine is also 20% whereas Guanine is of 30% then cytosine is also of 30%. Hence, if DNA molecule with 160 bp then amount of cytosine is 30% of 160 = 48

Question 44.
A template strand in a bacterial DNA has the given base sequence:
5’AGGTTTAACG – 3′
What would be the RNA sequence transcribed from this template strand ?
(a) 5′-CGUUAAACCU – 3′
(b) 5′-AGGUUUUUCG – 3′
(c) 5′-TCCAAATTGC – 3′
(d) 5′-AGGTTTAACG-3′
Answer:
(a) 5′-CGUUAAACCU – 3’

5 ’- AGGTTTAACG -3’ Template strand
As template strand code from 3’ to 5’
Then the strand will become 3’- GCAATTTGGA-5’
Therefore the mRNA strand for this template is-5’- CGUUAAACCU -3’

Note
The RNA sequence is complementary to DNA sequence. RNA have “U” in place of “T”.

Question 45.
In the presence of allolactose, the lac repressor in the operon of E.coli:
(a) binds to the operator
(b) cannot bind to the operator
(c) binds to the promoter
(d) binds to the regulator.
Answer:
(b) cannot bind to the operator

In the presence of allolactose, a binary complex is formed between allolactose and the repressor that makes binding of the repressor to the operator region impossible.

Question 46.
Taylor and colleagues performed experiments on ____ using radioactive ___ to prove that
the DNA is chromosomes replicate semi-conservastively.
(Select the correct option for the blanks)
(a) Vicia faba, Uridine
(b) E. coli, Uridine
(c) Vicia faba, Thymidine
(d) E. coli, Thymidine
Answer:
(c) Vicia faba, Thymidine

The experiments involving use of radioactive thymidine to detect distribution of newly synthesised DNA in the chromosomes was performed on Vicia faba (faba beans) by Taylor and colleagues in 1958. The experiments proved that the DNA in chromosomes also replicate semiconservatively.

Question 47.
The reactive hydroxyl group in the nucleotide ofRNA is:
(a) 5’OH
(b) 4’OH
(c) 3’OH
(d) 2’OH
Answer:
(d) 2’OH

The 2′-OH hydroxyl protons are responsible for differences in conformation, hydration, and thermodynamic stability of RNA and DNA oligonucleotides. Additionally, the 2′-OH group plays a central role in RNA.

Note
The OH group present at 2’ of sugar molecule is the difference between DNA and RNA.

Question 48.
Given below are the pairs of contrasting traits in Pisum Sativum as studied by Mendel. Identify the incorrect pair of traits:

Answer:
(c) The dominant pod colour of Pisum sativum is green colour while recessive pod colour is yellow in colour.

Section B

This section consists of one case followed by 6 questions. Besides this 6 more questions are given. Attempt any 10 questions from this section. The first 10 questions attempted would be evaluated.

Case Study: (Qs. 49-54)

A group of medical students carried out a detailed study on the impact of various factors on the different hormones during the menstrual cycle in a human female. They collected the data with different factors. Given below is the graph plotted from the data collected showing the morning temperature and concentration of hormones FSH, LH, estrogen and progesterone during normal menstrual cycle in a woman.
Temperature Graph

Question 49.
The early morning recording of temperature in the graph during actual and during ovulation respectively are :
(a) low, high
(b) high, low
(c) low, low
(d) high, high
Answer:
(a) low, high
As per graph during menstrual estrogen i s low while during ovulation it reach to its higher peak.

Note
The reproductive cycle in female primates is called menstrual cycle.

Question 50.
The time of ovulation is of importance in cases of:
(i) couples having difficulty in conception.
(ii) to know the safe period for prevention of pregnancy.
(iii) to inhibit the process of ovulation.
(iv) to stimulate ovarian follicular development.
(a) (i) and (iv)
(b) (ii) and (iv)
(c) (i) and (ii)
(d) (iii) and (iv)
Answer:
(c) (i) and (ii)

Contraceptive are used to inhibit the ovulation process. Progesterone is important to stimulate ovarian follicular development. Hence only statements i and ii are important for ovulation.

Question 51.
The increase in the level of progesterone is maximum under the influence of LH during:
(a) Secretory phase
(b) Follicular phase
(c) Menstruation
(d) Proliferative phase
Answer:
(a) Secretory phase

The ovulation (ovulatory phase) is followed by the luteal phase during which the remaining parts of the Graafian follicle transform into the corpus luteum. The corpus luteum secretes large amounts of progesterone which is essential for the maintenance of the endometrium. Hence progesterone level is highest during secretory phase.

Question 52.
Which of the following hormone/hormones is’are showing rapid surge leading to changes in Graafian follicle just before ovulation ?
(a) LH
(b) FSH
(c) FSH and Estrogen
(d) FSH and LH
Answer:
(d) FSH and LH

Both LH and FSH attain a peak level in the middle of the cycle (about 14 ^th day). Rapid secretion of LH leading to its maximum level during the mid-cycle called LH surge induces rupture of Graafian follicle and thereby the release of an ovum (ovulation).

Question 53.
The human corpus luteum starts regressing ___ days after ovulation. (Identify the correct choice for the blank).
(a) 10 – 11
(b) 14 – 15
(c) 16 – 17
(d) 18 – 20
Answer:
(c) 16 – 17

In human ovulation followed by 14 ^th to 15 ^th day of menstrual cycle. Then corpus luteum starts regressing at 16 ^th -17 ^th days after ovulation.

Question 54.
As per the data plotted in the graph, in which period of the menstrual cycle is the chance of fertilisation very high in human female ?
(a) 3 ^rd – 9 ^th days
(b) 10 ^th – 17 ^th days
(c) 18 ^th – 23 ^rd days
(d) 23 ^rd – 28 ^th days
Answer:
(b) 10 ^th – 17 ^th days

The chances of fertilisation is highest during ovulation period hence, 10 ^th -17 ^th period of the menstrual cycle is the chance of fertilisation.

Note
The chances of fertilization is maximum during the period of ovulation.

Question 55.
A plant breeder crossed a pure breed tall plant having white flowers with a pure breed dwarf plant having blue flowers. He obtained 2002 F ₁ progeny and found that they are all tall having blue flowers. Upon selling these F ₁ plants he obtained a progeny of 2160 plants. Approximately how many of these are likely to be short having blue flowers ?
(a) 1215
(b) 405
(c) 540
(d) 135
Answer:
(b) 405

The cross between pure breed of tall plant with write colour flower (TTbb) with pure breed with short plant and blue flower (ttBB). Then the F2 progeny will be :
Tall plant with blue flower: 9
Tall plant with white flower: 3
Dwarf plant with blue flower: 3
Dwarf plant with white flower: 1
Therefore, the probability of short plant with blue flower is 3/16of2160 = 405

Question 56.
Given below is a Karyotype of a human foetus obtained for screening to find any probable genetic disorder :

Based on the Karyotype, the chromosomal disorder detected in unborn foetus and the consequent symptoms the child may suffer from are:
(a) Turner’s syndrome: Sterile ovaries, short stature
(b) Down’s syndrome: Gynaecomastia, overall masculine stature
(c) Turner’s syndrone: Small round head, flat back of head
(d) Down’s syndrome : Furrowed tongue, short stature.
Answer:
(a) Turner’s syndrome: Sterile ovaries, short stature
Turner’s Syndrome is a disorder is caused due to the absence of one of the X chromosomes, i.e., 45 with X0, Such females are sterile as ovaries arc rudimentary.

Note
Symptoms of Turner syndrome include short stature, heart defects and certain learning disabilities.

Question 57.
In the dihybrid cross that was conducted by Morgan involving mating between parental generation for genes yellow bodied, white eyed female Drosophila and wild type male Drosophila, upto F ₂ generation is given below :

Study the result obtained of the F ₂ progeny. Select the correct option from the given choices for the F ₂ progeny.
(a) Parental type, 1.3%: Strength of linkage high
(b) Recombinant types, 1.3%: Strength of linkage low
(c) Parental type 98.7%: Strength of linkage high
(d) Recombinant types, 98.7%: Strength of linkage low.
Answer:
(c) Parental type 98.7%: Strength of linkage high

Question 58
Study the given diagrammatic representation of Griffih’s experiment to demonstrate transformation in bacteria:

Select the option which is incorrectly representing the experiment:
(a) (i) and (iii)
(b) (ii) and (iii)
(c) (iii) and (iv)
(d) (ii) and (iv)
Answer:
(c) (iii) and (iv)

Griffith was able to kill bacteria by heating them. He observed that heat-killed S strain bacteria injected into mice did not kill them. When he

Note
In 1928, Fredericok Griffith, in a series of experiments with iffeptococcus pneumoniae.

Question 59.
Which one of the following diagram correctly represents DNA replication in eukaryotes?

Answer:
(d)

Question 60.
In the given figure of translation machinery of eukaryotes, select the correct labellings for (i), (ii), (iii) and (iv) :

(a) (i) Codon, (ii) Anticodon, (iii) tRNA, (iv) 3′ end of mRNA
(b) (i) Anticodon, (ii) Codon, (iii) 3’end of mRNA, (iv) 5’end of mRNA
(c) (i) Polypeptide chain, (ii) Large subunit of ribosome, (iii) 5′ end of mRNA, (iv) tRNA.
(d) (i) Ribozyme, (ii) Polypeptide chain, (iii) tRNA, (iv) 5′ end of tRNA.
Answer:
(c)

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