Molecular Basis of Inheritance (last 10 yeas)

Question 1

1. If the length of a DNA molecule is 1.1 metres, what will be the approximate number of base pairs? (2022)

a. 6.6 × 106 bp
b. 3.3 × 109 bp
c. 6.6 × 109 bp
d. 3.3 × 106 bp

Answer 1

b. 3.3 × 109 bp

NCERT
DNA is a long polymer of deoxyribonucleotides.

The length of DNA is usually defined as number of nucleotides
(or a pair of nucleotide referred to as base pairs) present in it.

This also is the characteristic of an organism.

For example, a bacteriophage known as φ ×174 has 5386 nucleotides,
Bacteriophage lambda has 48502 base pairs (bp), Escherichia coli has 4.6 × 106 bp, and

haploid content of human DNA is 3.3 × 109 bp.

.

Question 2

2. Read the following statements and choose the set of correct statements (2022)

A. Euchromatin is loosely packed chromatin
B. Heterochromatin is transcriptionally active
C. Histone octomer is wrapped by negatively charged DNA in nucleosome
D. Histones are rich in lysine and arginine
E. A typical nucleosome contains 400 bp of DNA helix

Choose the correct answer from the options given below.
a. A, C and E only
b. B, D and E only
c. A, C and D only
d. B and E only

Answer 2

c. A, C and D only

NCERT
A. Euchromatin is loosely packed chromatin
C. Histone octomer is wrapped by negatively charged DNA in nucleosome
D. Histones are rich in lysine and arginine

NCERT
A. Non-histone Chromosomal (NHC) proteins. In a typical nucleus, some
region of chromatin are loosely packed (and stains light) and are referred to as euchromatin

B. Euchromatin is said to be transcriptionally active chromatin, whereas heterochromatin is inactive

C. The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome

D. Histones are rich in the basic amino acid residues lysine and arginine.

E. A typical nucleosome contains 200 bp of DNA helix

Question 3

3. Complete the flow chart on central dogma. (2021)

Insert the image

a. (A)-Translation;(B)-Replication; (C)-Transcription;(D)- Transduction

b. (A)-Replication;(B)-Transcription; (C)-Translation; (D)-Protein

c. (A)-Transduction;(B)-Translation; (C)-Replication; (D)-Protein

d. (A)-Replication;(B)-Transcription (C)-Transduction;(D)-Protein

Answer 3

b. (A)-Replication;(B)-Transcription; (C)-Translation; (D)-Protein

NCERT
Very soon, Francis Crick proposed the Central dogma in molecular biology, which states that the genetic information flows from DNA arrow RNA arrow Protein.

Add the image

.

Question 4

4. If Adenine makes 30% of the DNA molecule, what will be the percentage of Thymine, Guanine and Cytosine in it? (2021)

a. T : 20; G : 20; C : 30
b. T : 30; G : 20; C : 20
c. T : 20; G : 25; C : 25
d. T : 20; G : 30; C : 20

Answer 4

b. T : 30; G : 20; C : 20

NCERT
Observation of Erwin Chargaff that for a double stranded DNA, the ratios between Adenine and Thymine and Guanine and Cytosine are constant and equals one.

Question 5

5. Which one of the following statement about histones is wrong? (2021)

a. The pH of histones is slightly acidic.
b. Histones are rich in amino acids - Lysine and Arginine.
c. Histones carry positive charge in the side chain.
d. Histones are organized to form a unit of 8 molecules.

Answer 5

a. The pH of histones is slightly acidic.

NCERT
a. In eukaryotes, this organization is much more complex. There is a set of positively charged, basic
proteins called histones.

b. Histones are rich in the basic amino acid residues lysine and arginine.

c. A protein acquires charge depending upon the abundance of amino acids
residues with charged side chains. Both the amino acid residues carry positive charges
in their side chains.

d. Histones are organized to form a unit of eight molecules called histone octamer

Question 6

6. Which of the following statements is correct? (2020)

a. Adenine pairs with thymine through one H-bond
b. Adenine pairs with thymine through three H-bonds.
c. Adenine does not pair with thymine.
d. Adenine pairs with thymine through two H-bonds.

Answer 6

d. Adenine pairs with thymine through two H-bonds.

NCERT
Adenine forms two hydrogen bonds with Thymine from opposite strand and vice-versa.
Similarly, Guanine is bonded with Cytosine with three H-bonds.

The rule is that Adenine pairs with Thymine through two H-bonds, and Guanine with
Cytosine through three H-bonds. This makes one strand complementary to the other

Question 7

7. If the distance between two consecutive base pairs is 0.34 nm and the total number of base pairs of a DNA double helix in a typical mammalian cell is 6.6 × 109 bp, then the length of the DNA is approximately: (2020)

a. 2.5 meters
b. 2.2 meters
c. 2.7 meters
d. 2.0 meters

Answer 7

b. 2.2 meters

NCERT
Taken the distance between two consecutive base pairs as 0.34 nm (0.34×10–9 m), if the length of DNA double helix in a typical mammalian cell is calculated

(simply by multiplying the total number of bp with distance between two consecutive bp, that is,
6.6 × 109 bp × 0.34 × 10-9m/bp), it comes out to be approximately 2.2 metres

Question 8

8. In the polynucleotide chain of DNA, a nitrogenous base is linked to the –OH of: (2020-Covid)

a. 3’C pentose sugar
b. 5’C pentose sugar
c. 1’C pentose sugar
d. 2’C pentose sugar

Answer 8

c. 1’C pentose sugar

NCERT
A nitrogenous base is linked to the OH of 1’ C pentose sugar through a N-glycosidic
linkage to form a nucleoside, such as adenosine or deoxyadenosine,
guanosine or deoxyguanosine, cytidine or deoxycytidine and uridine or deoxythymidine.

Question 9

9. E. Coli has only 4.6 × 106 base pairs and completes the process of replication within 18 minutes; then the average rate of polymerization is approximately- (2020-Covid)

a. 3000 base pairs/second
b. 4000 base pairs/second
c. 1000 base pairs/second
d. 2000 base pairs/second

Answer 9

d. 2000 base pairs/second

In living cells, such as E. coli, the process of replication requires a set of catalysts (enzymes).

The main enzyme is referred to as DNA-dependent DNA polymerase, since it uses a DNA template to catalyze the polymerization of deoxynucleotides.

These enzymes are highly efficient enzymes as they have to catalyze polymerization of a large number of nucleotides in a very short time.

E. coli that has only 4.6 ×106 bp (compare it with human whose diploid content is 6.6 × 109 bp), completes the process of replication within 18 minutes; that means the average rate of polymerization has to be approximately 2000 bp per second.

Question 10

10. Purines found both in DNA and RNA are (2019)

a. Adenine and thymine
b. Adenine and guanine
c. Guanine and cytosine
d. Cytosine and thymine

Answer 10

b. Adenine and guanine

There are two types of nitrogenous bases – Purines (Adenine and Guanine), and Pyrimidines (Cytosine, Uracil and Thymine).

Cytosine is common for both DNA and RNA and Thymine is present in
DNA.

Question 11

11. The association of histone H1 with a nucleosome indicates: (2017-Delhi)

a. Transcription is occurring
b. DNA replication is occurring
c. The DNA is condensed into a chromatin fibre
d. The DNA double helix is exposed

Answer 11

c. The DNA is condensed into a chromatin fibre

NCERT (add the image)
The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome (Figure 5.4 a).

A typical nucleosome contains 200 bp of DNA helix.

Nucleosomes constitute the repeating unit of a structure in nucleus called chromatin, threadlike stained (coloured) bodies seen in nucleus.

The nucleosomes in chromatin are seen as ‘beads-on-string’ structure when viewed under
electron microscope (EM) (Figure 5.4 b).

Theoretically, how many such beads (nucleosomes) do you imagine are present in a mammalian cell?

The beads-on-string structure in chromatin is packaged to form chromatin fibers that are further coiled and condensed at metaphase stage of cell division to form chromosomes

Question 12

12. DNA fragments are: (2017-Delhi)

a. Positively charged
b. Negatively charged
c. Neutral
d. Either positively or negatively charged depending on their size

Answer 12

b. Negatively charged

NCERT
DNA (being negatively charged) is held with some proteins (that have positive
charges) in a region termed as ‘nucleoid’. The DNA in nucleoid is organised in large loops held by
proteins.

The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome

Question 13

13. Identify the correct order of organization of genetic material from largest to smallest: (2015 Re)

a. Genome, chromosome, nucleotide, gene
b. Genome, chromosome, gene, nucleotide
c. Chromosome, genome, nucleotide, gene
d. Chromosome, gene, genome, nucleotide

Answer 13

b. Genome, chromosome, gene, nucleotide

(see this question)

Question 14

14. The diagram shows an important concept in the genetic implication of DNA Fill in the blanks A to C: (2013)

insert an image

a. A-translation, B-extension, C-Rosalind Franklin
b. A-transcription, B-replication, C-James Watson
c. A-translation, B-transcription, C-Erwin Chargaff
d. A-transcription, B-translation, C-Francis Crick

Answer 14

d. A-transcription, B-translation, C-Francis Crick

NCERT
Very soon, Francis Crick proposed the Central dogma in molecular biology, which states that the genetic information flows from DNA arrow RNA arrow Protein.

Question 15

15. Ten E.coli with 15N- dsDNA are incubated in medium containing 14N nucleotide. After 60 minutes, how many E.coli cells will have DNA totally free from 15N? (2022)

a. 80 cells
b. 20 cells
c. 40 cells
d. 60 cells

Answer 15

d. 60 cells

(see in NCERT)

Question 16

16. The term ‘Nuclein’ for the genetic material was used by: (2020-Covid)

a. Meischer
b. Chargaff
c. Mendel
d. Franklin

Answer 16

a. Meischer

NCERT
DNA as an acidic substance present in nucleus was first identified by
Friedrich Meischer in 1869. He named it as ‘Nuclein’.

Even though the discovery of nuclein by Meischer and the proposition for principles of inheritance by Mendel were almost at the same time, but that the DNA acts as a genetic material took long to be discovered and proven.

Question 17

17. The experimental proof for semi-conservative replication of DNA was first shown in a: (2018)

a. Fungus
b. Bacterium
c. Plant
d. Virus

Answer 17

b. Bacterium

NCERT
It is now proven that DNA replicates semiconservatively. It was shown first in Escherichia coli and subsequently in higher organisms, such as plants

Question 18

18. Select the correct match (2018)

a. Ribozyme Nucleic acid
b. F2 × Recessive parent Dihybrid cross
c. T.H. Morgan Transduction
d. G. Mendel Transformation

Answer 18

a. Ribozyme Nucleic acid

NCERT
A. The ribosome also acts as a catalyst (23S rRNA in bacteria
is the enzyme- ribozyme) for the formation of peptide bond.

One of the rRNA acts as a catalyst for peptide bond formation, which is an example of RNA enzyme (ribozyme).

(couldn’t find the other options)

Question 19

19. The final proof for DNA as the genetic material came from the experiments of (2017-Delhi)

a. Griffith
b. Hershey and Chase
c. Avery, Mcleod and McCarty
d. Hargobind Khorana

Answer 19

b. Hershey and Chase

Question 20

20. A molecule that can act as a genetic material must fulfill the traits given below, except: (2016 - II)

a. It should be unstable structurally and chemically

b. It should provide the scope for slow changes that are required for evolution

c. It should be able to express itself in the form of ‘Mendelian characters’

d. It should be able to generate its replica

Answer 20

a. It should be unstable structurally and chemically

Question 21

21. Taylor conducted the experiment to prove semi- conservative mode of chromosome replication on: (2016 - II)

a. Drosophila melanogaster
b. E. coli
c. Vinca rosea
d. Vicia faba

Answer 21

d. Vicia faba

Question 22

22. Which of the following rRNA acts as structural RNA as well as ribozyme in bacteria? (2016 - II)

a. 23 S rRNA
b. 5.8 S rRNA
c. 5 S rRNA
d. 18 S rRNA

Answer 22

a. 23 S rRNA

Question 23

23. In sea urchin DNA, which is double stranded, 17% of the bases were shown to be cytosine. The percentages of the other three bases expected to be present in this DNA are: (2015)

a. G = 17%, A = 33%, T = 33%
b. G = 8.5 %, A = 50 %, T = 24.5 %
c. G = 34%, A = 24.5%, T = 24.5%
d. G = 17%, A = 16.5%, T = 32.5%

Answer 23

a. G = 17%, A = 33%, T = 33%

Question 24

24. Which one of the following is not applicable to RNA? (2015 Re)

a. 5′ phosphoryl and 3′ hydroxyl ends
b. Heterocyclic nitrogenous bases
c. Chargaff’s rule
d. Complementary base pairing

Answer 24

c. Chargaff’s rule

Question 25

25. Transformation was discovered by: (2014)

a. Watson and Crick
b. Messelson and Stahl
c. Hershey and Chase
d. Griffith

Answer 25

d. Griffith

Question 26

26. During DNA replication, Okazaki fragments are used to elongate (2017-Delhi)

a. The leading strand towards replication fork
b. The lagging strand towards replication fork
c. The leading strand away from replication fork
d. The lagging strand away from the replication fork

Answer 26

d. The lagging strand away from the replication fork

Question 27

27. Select the correct option: (2014)

Direction of RNA synthesis Direction of reading of the template DNA strand

a. 3′ → 5′ 3′ → 5′

b. 5′ → 3′ 3′ → 5′

c. 3′ → 5′ 5′ → 3′

d. 5′ → 3′ 5′ → 3′

Answer 27

b. 5′ → 3′ 3′ → 5′

Question 28

28. What is the role of RNA ploymerase III in the process of transcription in eukaryotes? (2021)

a. Transcribes tRNA, 5s rRNA and sn RNA
b. Transcribes precursor of mRNA
c. Transcribes only snRNAs
d. Transcribes rRNAs (28S, 18S and 5.8S)

Answer 28

a. Transcribes tRNA, 5s rRNA and sn RNA

Question 29

29. Identify the correct statement. (2021)

a. RNA polymerase binds with Rho factor to terminate the process of transcription in bacteria.

b. The coding strand in transcription unit is copied to an mRNA.

c. Split gene arrangement is characteristic of prokaryotes.

d. In capping, methyl guanosine triphosphate is added to the 3′ end of hnRNA of transcription in bacteria.

Answer 29

a. RNA polymerase binds with Rho factor to terminate the process of transcription in bacteria.

Question 30

30. Which is the “Only enzyme” that has “Capability” to catalyse Initiation, Elongation and Termination in the process of transcription in prokaryotes? (2021)

a. DNA dependent RNA polymerase
b. DNA Ligase
c. DNase
d. DNA dependent DNA polymerase

Answer 30

a. DNA dependent RNA polymerase

Question 31

31. Name the enzyme that facilitates opening of DNA helix during transcription. (2020)

a. DNA helicase
b. DNA polymerase
c. RNA polymerase
d. DNA ligase

Answer 31

c. RNA polymerase

Question 32

32. AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequence of the transcribed mRNA? (2018)

a. AGGUAUCGCAU
b. UGGTUTCGCAT
c. ACCUAUGCGAU
d. UCCAUAGCGUA

Answer 32

a. AGGUAUCGCAU

Question 33

33. Spliceosomes are not found in cells of: (2017-Delhi)

a. Plants
b. Fungi
c. Animals
d. Bacteria

Answer 33

d. Bacteria

Question 34

34. Which of the following RNAs should be most abundant in animal cell? (2017-Delhi)

a. r-RNA
b. t-RNA
c. m-RNA
d. mi-RNA

Answer 34

a. r-RNA

Question 35

35. DNA-dependent RNA polymerase catalyses transcription on one strand of the DNA which is called the: (2016 - II)

a. Alpha strand
b. Antistrand
c. Template strand
d. Coding strand

Answer 35

c. Template strand

Question 36

36. Statement I: The codon ‘AUG’ codes for methionine and
    phenylalanine.

Statement II: ‘AAA’ and ‘AAG’ both codons code for the
amino acid lysine.

In the light of the above statements, choose the correct answer from the options given below. (2021)

a. Both statement I and statement II are false
b. Statement I is correct but statement II is false
c. Statement I is incorrect but statement II is true
d. Both statement I and statement II are true

Answer 36

c. Statement I is incorrect but statement II is true

Question 37

37. Under which of the following conditions will there be no change in the reading frame of following mRNA? (2019)

5′AACAGCGGUGCUAUU3′

a. Insertion of G at 5th position
b. Deletion of G from 5th position
c. Insertion of A and G at 4th and 5th positions respectively
d. Deletion of GGU from 7th, 8th and 9th positions

Answer 37

d. Deletion of GGU from 7th, 8th and 9th positions

Question 38

38. Which of the following features of genetic code does allow bacteria to produce human insulin by recombinant DNA technology? (2019)

a. Genetic code is not ambiguous
b. Genetic code is redundant
c. Genetic code is nearly universal
d. Genetic code is specific

Answer 38

c. Genetic code is nearly universal

Question 39

39. If there are 999 bases in an RNA that codes for a protein with 333 amino acids, and the base at position 901 is deleted such that the length of the RNA becomes 998 bases, how many
    codons will be altered? (2017-Delhi)

a. 1
b. 11
c. 33
d. 333

Answer 39

c. 33

Question 40

40. Which one of the following is the starter codon? (2016 - I)

a. AUG
b. UGA
c. UAA
d. UAG

Answer 40

a. AUG

Question 41

41. The process of translation of mRNA to proteins begins as soon as: (2022)

a. The tRNA is activated and the larger subunit of ribosome encounters mRNA

b. The small subunit of ribosome encounters mRNA

c. The larger subunit of ribosome encounters mRNA

d. Both the subunits join together to bind with mRNA

Answer 41

b. The small subunit of ribosome encounters mRNA

Question 42

42. Which of the following RNAs is not required for the synthesis of protein? (2021)

a. tRNA
b. rRNA
c. siRNA
d. mRNA

Answer 42

c. siRNA

Question 43

43. The first phase of translation is: (2020)

a. Recognition of DNA molecule
b. Aminoacylation of tRNA
c. Recognition of an anti-codon
d. Binding of mRNA to ribosome

Answer 43

b. Aminoacylation of tRNA

Question 44

44. Which one of the following is wrongly matched? (2014)

a. Operon-Structural genes, operator and promoter
b. Transcription-Writing information from DNA to tRNA
c. Translation-Using information in mRNA to make protein
d. Repressor protein-Binds to operator to stop enzyme synthesis

Answer 44

b. Transcription-Writing information from DNA to tRNA

Question 45

45. In an E.coil strain i gene gets mutated and its product can not bind the inducer molecule. If growth medium is provided with lactose, what will be the outcome? (2022)

a. RNA polymerase will bind the promoter region
b. Only z gene will get transcribed
c. z, y, a genes will be transcribed
d. z, y, a genes will not be translated

Answer 45

d. z, y, a genes will not be translated

Question 46

46. Match the following genes of the Lac operon with their respective products : (2019)

A. i gene i. β-galactosidase

B. z gene ii. Permease

C. a gene iii. Repressor

D. y gene iv. Transacetylase

Select the correct option.

(A) (B) (C) (D)
a. (i) (iii) (ii) (iv)
b. (iii) (i) (ii) (iv)
c. (iii) (i) (iv) (ii)
d. (iii) (iv) (i) (ii)

Answer 46

c. (iii) (i) (iv) (ii)

Question 47

47. Select the correct match: (2018)

a. Alec Jeffreys – Streptococcus pneumoniae
b. Alfred Hershey and Martha Chase – TMV
c. Matthew Meselson and F. Stahl – Pisum sativum
d. Francois Jacob and Jacques Monod – Lac operon

Answer 47

d. Francois Jacob and Jacques Monod – Lac operon

Question 48

48. All of the following are part of an operon except: (2018)

a. An operator
b. Structural genes
c. An enhancer
d. A promoter

Answer 48

c. An enhancer

Question 49

49. The equivalent of a structural gene is: (2016 - II)

a. Operon
b. Recon
c. Muton
d. Cistron

Answer 49

d. Cistron

Question 50

50. Gene regulation governing lactose operon of E. coli that involves the lac I gene product is: (2015)

a. Negative and repressible because repressor protein prevents transcription

b. Feedback inhibition because excess of β-galactosidase can switch off transcription

c. Positive and inducible because it can be induced lactose

d. Negative and inducible because repressor protein prevents transcription

Answer 50

d. Negative and inducible because repressor protein prevents transcription

Question 51

51. Which enzyme/s will be produced in a cell in which there is a nonsense mutation in the lac Y gene? (2013)

a. Lactose permease and transacetylase
b. β-galactosidase
c. Lactose permease
d. Transacetylase

Answer 51

b. β-galactosidase

Question 52

52. DNA polymorphism forms the basis of: (2022)

a. Translation
b. Genetic mapping
c. DNA finger printing
d. Both genetic mapping and DNA finger printing

Answer 52

d. Both genetic mapping and DNA finger printing

Question 53

53. If a geneticist uses the blind approach for sequencing the whole genome of an organism, followed by assignment of function to different segments, the methodology adopted by
    him is called as: (2022)

a. Bioinformatics
b. Sequence annotation
c. Gene mapping
d. Expressed sequence tags

Answer 53

b. Sequence annotation

Question 54

54. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence, called as: (2021)

a. Repetitive DNA
b. Single nucleotides
c. Polymorphic DNA
d. Satellite DNA

Answer 54

a. Repetitive DNA

Question 55

55. Which is the basis of genetic mapping of human genome as well as DNA finger printing?
    (2020-Covid)

a. Single nucleotide polymorphism
b. Polymorphism in hnRNA sequence
c. Polymorphism in RNA sequence
d. Polymorphism in DNA sequence

Answer 55

d. Polymorphism in DNA sequence

Question 56

56. Expressed Sequence Tags (ESTs) refers to : (2019)

a. Genes expressed as RNA
b. Polypeptide expression
c. DNA polymorphism
d. Novel DNA sequences

Answer 56

a. Genes expressed as RNA

Question 57

57. Which of the following is not required for any of the techniques of DNA fingerprinting available at present? (2016 - I)

a. Polymerase chain reaction
b. Zinc finger analysis
c. Restriction enzymes
d. DNA-DNA hybridisation

Answer 57

b. Zinc finger analysis

Question 58

58. Which of the following is required as inducer(s) for the expression of Lac operon? (2016 - I)

a. Glucose
b. Galactose
c. Lactose
d. Lactose and Galactose

Answer 58

c. Lactose

Question 59

59. Satellite DNA is important because it: (2015 Re)

a. Shows high degree of polymorphism in population and also the same degree of polymorphism in an individual, which are heritable form parents to children.

b. Does not code for proteins and is same in all members of the population

c. Codes for enzymes needed for DNA replication

d. Codes for proteins needed in cell cycle.

Answer 59

a. Shows high degree of polymorphism in population and also the same degree of polymorphism in an individual, which are heritable form parents to children.

Question 60

60. Commonly used vectors for human genome sequencing are: (2014)

a. T/A Cloning Vectors
b. T-DNA
c. BAC and YAC
d. Expression Vectors

Answer 60

c. BAC and YAC

Question 61

61. An analysis of chromosomal DNA using the southern hybridization technique does not use: (2014)

a. PCR
b. Electrophoresis
c. Blotting
d. Autoradiography

Answer 61

a. PCR