NEET Previous Year Questions(2016-21): Molecular Basis of Inheritance Notes | Study Biology Class 12 - NEET

Q.1. Complete the flow chart on central dogma    (2021)

(a) (a) - Replication; (b) - Transcription;
     (c) - Translation; (d) -  Protein

(b) (a) - Transduction; (b) - Translation;
     (c) - Replication;  (d) - Protein

(c) (a) - Replication; (b) - Transcription;
     (c) - Transduction; (d) - Protein

(d) (a) - Transcription; (b) - Replication;
     (c) - Transcription; (d) - Transduction
Ans. (a)
Very soon, Francis Crick proposed the Central dogma in molecular biology, which states that the genetic information flows from DNA->RNA->Protein.

Q.2. What is the role of RNA polymerase III in the process of transcription in eukaryotes?     (2021)
(a) Transcribes precursor of mRNA
(b) Transcribes only snRNAs
(c) Transcribes rRNAs (28S, 18S, and 5.8S)
(d) Transcribes tRNA, 5s rRNA and snRNA
Ans. (d)
The RNA polymerase I transcribes rRNAs (28S, 18S, and 5.8S), whereas the RNA polymerase III is responsible for transcription of tRNA, 5srRNA, and snRNAs (small nuclear RNAs).

Q.3. DNA fingerprinting involves identifying differences in some specific regions in DNA sequence, called as    (2021)
(a) Single nucleotides
(b) Polymorphic DNA
(c) Satellite DNA
(d) Repetitive DNA
Ans. (d)
DNA fingerprinting involves identifying differences in some specific regions in DNA sequence called as repetitive DNA, because in these sequences, a small stretch of DNA is repeated many times.

Q.4. Identify the correct statement.    (2021)
(a) The coding strand in a transcription unit is copied to an mRNA.
(b) Split gene arrangement is characteristic of prokaryotes.
(c) In capping, methylguanosine triphosphate is added to the 3' end of hnRNA.
(d) RNA polymerase binds with the Rho factor to terminate the process of transcription in bacteria.
Ans. (d)

The RNA polymerase is only capable of catalysing the process of elongation.

It associates transiently with initiation-factor (σ) and termination-factor (ρ) to initiate and terminate the transcription, respectively.

Q.5. If Adenine makes 30% of the DNA molecule, what will be the percentage of Thymine, Guanine and Cytosine in it?     (2021)
(a) T:30 ; G:20 ; C:20
(b) T:20 ; G:25 ; C:25
(c) T:20 ; G:30 ; C:20
(d) T:20 ; G:20 ; C:30
Ans. (a)

• However, this proposition was also based on the 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.
  Adenine forms two hydrogen bonds with Thymine from opposite strand and vice-versa.
• Similarly, Guanine is bonded with Cytosine with three H-bonds.

Q.6. Which of the following RNAs is not required for the synthesis of protein?    (2021)
(a) rRNA
(b) siRNA
(c) mRNA
(d) tRNA
Ans. (b)

• In bacteria, there are three major types of RNAs: mRNA (messenger RNA), tRNA (transfer RNA), and rRNA (ribosomal RNA).
• All three RNAs are needed to synthesise a protein in a cell.

Q.7. Which is the "Only enzyme" that has "Capability to catalyze Initiation, Elongation, and Termination in the process of transcription in prokaryotes?    (2021)
(a) DNA Ligase
(b) DNase
(c) DNA dependent DNA polymerase
(d) DNA dependent RNA polymerase
Ans. (d)
There is single DNA-dependent RNA polymerase that catalyses transcription of all types of RNA in bacteria.

Q.8. Which one of the following statements about Histones is wrong?    (2021)
(a) Histones are rich in amino acids - Lysine and Arginine.
(b) Histones carry a positive charge in the side chain.
(c) Histones are organized to form a unit of 8 molecules.
(d) The pH of histones is slightly acidic.
Ans. (d)

• There is a set of positively charged, basic proteins called histones.
• Histones are rich in the basic amino acid residues lysine and arginine.
• Histones are organised to form a unit of eight molecules called histone octamer.
• The negatively charged DNA is wrapped around the positively charged histone octamer to form a structure called nucleosome (Figure Given.

Q.9. Statement I: The codon 'AUG codes for methionine and phenylalanine.
Statement II: AAA' and 'AAG are both codons that code for the amino acid lysine.
In the light of the above statements, choose the correct answer from the options given below.     (2021)
(a) Statement I is correct but Statement II is false
(b) Statement I is incorrect but Statement II is true
(c) Both Statement I and Statement II are true
(d) Both Statement I and Statement II are false
Ans. (b)

The Codons for the Various Amino Acids

Q.10. The first phase of translation is:    (2020)
(a) Aminoacylation of tRNA
(b) Recognition of an anti-codon
(c) Binding of mRNA to ribosome
(d) Recognition of DNA molecule
Ans. (a)
In translation the first phase is activation of amino acids in the presence of ATP. 

The activated amino acids are then linked to their cognate tRNAs, a process commonly called as charging of tRNA or aminoacylation of tRNAs.

Q.11. Name the enzyme that facilitates opening of DNA helix during transcription.     (2020)
(a) DNA polymerase
(b) RNA polymerase
(c) DNA ligase
(d) DNA helicase
Ans. (b)
RNA polymerase holoenzyme binds to the promoter, unwinds DNA (open complex) and form phosphodiester links between the initiating nucleotides. DNA polymerase, DNA ligase & DNA helicase are involved in the process of replication and not transcription.

Q.12. 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 dp, then the length of the DNA is approximately.     (2020)
(a) 2.2 meters
(b) 2.7 meters
(c) 2.0 meters
(d) 2.5 meters
Ans. (a)
The diploid content of human genome is 6.6 x 10⁹ base pairs. The distance between two consecutive base pairs is 0.34 nm (0.34 x 10^-9 m), so the length of DNA double helix in a typical mammalian cell is around 6.6 x 10^-9 bp x 0.34 x 10^-9m/bp = 2.2 metres.

Q.13. Match the following genes of the Lac operon with their respective products.    (2019)

Select the correct option.

Ans: (d)
In lac operon
I gene → Repressor
z gene → β-galactosidase
y gene → Permease
a gene → Transacetylase

Q.14. Purines found both in DNA and RNA are    (2019)
(a) Cytosine and thymine
(b) Adenine and thymine
(c) Adenine and guanine
(d) Guanine and cytosine
Ans: (c)
Purines are the same in both DNA and RNA. They are adenine and guanine. Thymine, cytosine and uracil are pyrimidine bases. Cytosine is found in both DNA and RNA. Thymine is present in DNA and RNA contains uracil in place of thymine.

Q.15. Under which of the following conditions there will be no change in the reading frame of following mRNA?    (2019)
5' AACAGCGGUGCUAUU 3'
(a) Deletion of GGU from 7^th, 8^th and 9^th positions
(b) Insertion of G at 5^th  position
(c) Deletion of G from 5^th position
(d) Insertion of A and G at 4^th and 5^th position respectively
Ans: (a)
Insertion or deletion of three or its multiple bases insert or delete one or multiple codon hence one or multiple amino acids, and reading frame remains unaltered from that point onwards.

Q.16. Expressed Sequence Tags (ESTs) refers to    (2019)
(a) Novel DNA sequences
(b) Genes expressed as RNA
(c) Polypeptide expression
(d) DNA polymorphism
Ans: (b)
Expressed Sequence Tags (ESTs) are genes that are expressed as RNA. It is used in sequencing of human genome.

Q.17. Which of the following features of genetic code does allow bacteria to produce human insulin by recombinant DNA technology?    (2019)
(a) Genetic code is specific.
(b) Genetic code is not ambiguous.
(c) Genetic code is redundant.
(d) Genetic code is nearly universal.
Ans: (d)
As genetic code is nearly universal means almost all organism (from a virus, bacteria to a tree or human being) will have amino acids coded by same kind of codons as given in checkerboard. So, this properties is utilised to produce human insulin using bacteria.

Q.18. What will be the sequence of mRNA produced by the following stretch of DNA?    (2019)
3'ATGCATGCATGCATG5'TEMPLATE 
STRAND
5' TACGTACGTACGTAC3' CODING 
STRAND
(a) 3'AUGCAUGCAUGCAUG5'
(b) 5'UACGUACGUACGUAC 3'
(c) 3' UACGUACGUACGUAC 5'
(d) 5' AUGCAUGCAUGCAUG 3'
Ans: (b)


Q.19. Match the following RNA polymerase with their transcribed products :

Select the correct option from the following :    (2019)
(a) I-i, 2-iii, 3-ii
(b) 1-i, 2-ii, 3-iii
(c) 1 -ii, 2-iii, 3-i
(d) 1 -iii, 2-ii, 3-i
Ans: (c)

Q.20. Select the correct statement.    (2018)
(a) Franklin Stahl coined the term ‘‘ linkage ’’.
(b) Punnett square was developed by a British scientist.
(c) Spliceosomes take part in translation.
(d) Transduction was discovered by S. Altman.
Ans: (b)
Franklin Stahl along with Matthew Meselson proved semiconservative mode of replication in DNA. Punnett square was developed by a British geneticist, Reginald C. Punnett. Spliceosome formation is part of post-transcriptional change in eukaryotes. Transduction was discovered by Joshua Lederberg and Norton Zinder in the bacterium Salmonella.

Q.21. The experimental proof for semi-conservative replication of DNA was first shown in a    (2018)
(a) Fungus
(b) Bacterium
(c) Plant
(d) Virus
Ans: (b)
Semiconservative replication of DNA was proved by the work of Matthew Meselson and Franklin Stahl (1958) using bacterium Escherichia coli.

Q.22. 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
Ans: (d)
Alec Jeffreys (1984) invented the DNA finger-printing technique. Alfred Hershey and Martha Chase proved that DNA is genetic material using T₂ bacteriophages in an experiment. Matthew Meselson and F. Stahl proved semiconservative mode of replication in bacterium E. coli.

Q.23. Select the correct match.    (2018)
(a) Ribozyme - Nucleic acid
(b) F₂ x Recessive parent - Dihybrid cross
(c) T.H. Morgan - Transduction
(d) G. Mendel - Transformation
Ans: (a)
Ribozymes (ribonucleic acid enzymes) are RNA molecules that are capable of catalysing specific biochemical reactions similar to the action of protein enzymes.

Q.24. Many ribosomes may associate with a single wiRNA to form multiple copies of a polypeptide simultaneously. Such strings of ribosomes are termed as    (2018)
(a) Polysome
(b) Polyhedral bodies
(c) Plastidome
(d) Nucleosome
Ans: (a)
The association of many ribosomes with single wRNA leads to formation of polyribosomes or polysomes. It occurs during translation process of protein synthesis.

Q.25. 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
Ans: (a)
Coding strand and mRNA have same nucleotide sequence except, ‘T ’ - Thymine is replaced by 'U' - Uracil in mRNA.

Q.26. All of the following are part of an operon except    (2018)
(a) An operator
(b) Structural genes
(c) An enhancer
(d) A promoter
Ans: (c)
Operon concept is for prokaryotes that consist of operator gene, promoter gene, regulator gene and structural gene. Structural, operator and regulator genes are also present in eukaryotic gene expression along with enhancer gene but enhancer gene is present only in eukaryotic gene expression. It changes the rate of transcription of structural genes.

Q.27. The final proof for DNA as the genetic material came from the experiments of    (2017)
(a) Hershey and Chase
(b) Avery, MacLeod and McCarty
(c) Hargobind Khorana
(d) Griffith
Ans: (a)
Hershey and Chase gave unequivocal proof which ended the debate between protein and DNA as genetic material.

Q.28. If there are 999 bases in an RNA that code 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)
(a) 11
(b) 33
(c) 333
(d) 1
Ans: (b)
1 codon consists of 3 bases, Therefore, a deletion on 901 position will affect 33 codons.

Q.29. During DNA replication, Okazaki fragments are used to elongate    (2017)
(a) The lagging strand towards replication fork
(b) The leading strand away from replication fork
(c) The lagging strand away from the replication fork
(d) The leading strand towards replication fork
Ans: (c)
Lagging strand is a replicated strand 0f DNA which is formed in short segments called Okazaki fragments. Its growth is discontinuous. The direction of growth of the lagging strand is 3' → 5’ though in each Okazaki fragment it is 5’ → 3’.

Q.30. Which of the following RNAs should be most abundant in animal cell?    (2017)
(a) tRNA
(b) mRNA
(c) miRNA
(d) rRNA
Ans: (d)
rRNA (ribosomal RNA) is the most abundant of all types of RNA (70-88%). Hence, it will be present in highest amount. Percentage of tRNA and mRNA is 15% and 2-5% respectively. miRNA (micro RNA) are 21-22 bp long RNA that bring degeneration of mRNA.

Q.31. Spliceosomes are not found in cells of    (2017)
(a) Fungi
(b) Animals
(c) Bacteria
(d) Plants
Ans: (c)
Spliceosomes help in removal of introns. They will not occur in prokaryotes because prokaryotes do not have introns and thus, processing does not require splicing of mRNA.

Q.32. The association of histone H₁ with a nucleosome indicates that    (2017)
(a) DNA replication is occurring
(b) The DNA is condensed into a chromatin fibre
(c) The DNA double helix is exposed
(d) Transcription is occurring
Ans: (b)
Histones help in packaging of DNA. In eukaryotes, DNA packaging is carried out with the help of positively charged basic proteins called histones. Histones are of five types - H₁ H₂A, H₂B, H₃ and H4. H₁ is attached over the linker DNA. Histone contains a large proportion of the positively charged (basic) amino acids, lysine and arginine in their structure. DNA is negatively charged due to the phosphate groups on its backbone. The result of these opposite charges is strong attraction and therefore, high binding affinity between histones and DNA.

Q.33. Taylor conducted the experiments to prove semiconservative mode of chromosome replication on    (2016)
(a) Vinca rosea
(b) Vicia faba
(c) Drosophila melanogaster
(d) E. coli
Ans: (b)
Taylor et al. (1957) conducted experiment on Vicia faba (broad bean) to prove semiconservative replication of DNA. He fed dividing cells of root tips of Vicia faba with radioactive ³H containing thymine instead of normal thymine and found that all the chromosomes became radioactive. Labelled thymine was then replaced with normal one. Next generation came to have radioactivity in one of the two chromatids of each chromosome while in subsequent generation radioactivity was present in 50% of the chromosomes. This is possible only if out of the two strands of a chromosome, one is formed afresh while the other is conserved at each replication.

Q.34. The equivalent of a structural gene is    (2016)
(a) Muton
(b) Cistron
(c) Operon
(d) Recon
Ans: (b)
Cislron (or gene) is a length ot DNA that contains the information for coding a specific Polypeptide chain or a functional RNA molecule transfer RNA or ribosomal RNA). Hence, cistron is a unit of function. Currently such a gene is called structural gene.

Q.35. Which of the following rRNAs acts as structural RNA as well as ribozyme in bacteria?    (2016)  
(a) 5S rRNA
(b) 18S rRNA
(c) 23S rRNA
(d) 5.8S rRNA
Ans: (c)
23S rRNA acts as structural RNA as well as ribozyme in bacteria.

Q.36. A molecule that can act as a genetic material must fulfill the traits given below, except    (2016)
(a) It should be able to express itself in the form of ‘Mendelian characters’
(b) It should be able to generate its replica
(c) It should be unstable structurally and chemically
(d) It should provide the scope for slow changes that are required for evolution.
Ans: (c)
Genetic material should be structurally and chemically stable otherwise its expression will change and lead to loss of several metabolic functions, etc.

Q.37. DNA-dependent RNA polymerase catalyses transcription on one strand of the DNA which is called the    (2016)
(a) Template strand
(b) Coding strand
(c) Alpha strand
(d) Antistrand
Ans: (a)
The strand of DNA on which RNA polymerase binds to catalyse transcription is called template strand. It is also known as master or antisense strand. It lias the polarity of 3' → 5'.

Q.38. Which one of the following is the starter codon?    (2016)
(a) UAA
(b) UAG
(c) AUG
(d) UGA
Ans: (c)
Polypeptide synthesis is signalled by two initiator codons or start codons i.c., AUG (methionine codon) and rarely by GUG (valine codon).

Q.39. Which of the following is required as inducer (s) for the expression of Lac operon?    (2016)
(a) Lactose
(b) Lactose and Galactose
(c) Glucose
(d) Galactose
Ans: (a)
In Lac operon, lactose is an inducer. It binds with suppressor and inactivates it. It allows RNA polymers access to the promoter and transcription proceeds.

Q.40. A complex of ribosomes attached to a single strand of RNA is known as    (2016)
(a) Polypeptide
(b) Okazaki fragment
(c) Polysome
(d) Polymer
Ans: (c)

• Large clusters of 10 to 100 ribosomes, actively translating protein synthesis are called polysomes. The ribosomes in a polysome are connected by a single molecule of mRNA that is being translated simultaneously by many closely spaced ribosomes.
• Polymers are large molecules composed of many monomer units. Polypeptide is large linear polymer of amino acids bonded together by peptide linkage.
• Okazaki fragments are small nucleic acid fragments formed by discontinuous replication of lagging strand during DNA replication and are later joined together by DNA ligase enzyme.