Previous Year Questions (2016-19) - Molecular Basis of Inheritance Notes | EduRev

Biology 28 Years NEET/AIPMT Question Papers of Class 12

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NEET : Previous Year Questions (2016-19) - Molecular Basis of Inheritance Notes | EduRev

The document Previous Year Questions (2016-19) - Molecular Basis of Inheritance Notes | EduRev is a part of the NEET Course Biology 28 Years NEET/AIPMT Question Papers of Class 12.
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Q.1. 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) (iii) (iv) (i) (ii)
 (b) (i) (iii) (ii) (iv)
 (c) (iii) (i) (ii) (iv)
 (d) (iii) (i) (iv) (ii)

Ans: (d)

Q.2. 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)

Q.3. Under which of the following conditions there will be no change in the reading frame of following mRNA?    (2019)
(a) Deletion of GGU from 7th, 8th and 9th positions
(b) Insertion of G at 5th  position
(c) Deletion of G from 5th position
(d) Insertion of A and G at 4th and 5th 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.4. 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.5. 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.
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.6. What will be the sequence of mRNA produced by the following stretch of DNA?    (2019)
Ans: (b)

Q.7. Match the following RNA polymerase with their transcribed products :

 (1) RNA polymerase I (i) tRNA
 (2) RNA polymerase II (ii) rRNA
 (3) RNA polymerase III (iii) hnRNA 

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.8. 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: 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.9. 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.10. Select the correct match.    (2018)
(a) Alec Jeffreys - Streptococcus pneumoniae
(b) Alfred Hershey and - TMV Martha Chase
(c) Matthew Meselson - Pisurn sativum and F. Stahl
(d) Francois Jacob and - Lac operon Jacques Monod

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.11. 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.12. 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.13. AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequence of the transcribed mRNA?    (2018)

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

Q.14. 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.15. The final proof for DNA as the genetic material came from the experiments of    (2017)
Hershey and Chase
Avery, MacLeod and McCarty
Hargobind Khorana
Ans: (a)

Q.16. 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.17. 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.18. 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.19. 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.20. 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.21. Taylor conducted the experiments to prove semiconservative mode of chromosome replication on    (2016)
Vinca rosea
Vicia faba
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.22. 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.23. Which ofthe followingrRNAs 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.24. 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 he structurally and chemically stable otherwise its expression will change and lead to loss of several metabolic functions, etc.

Q.25. 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.26. 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.27. 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.28. A complex of ribosomes attached to a single strand of RNA is known as    (2016)
Okazaki fragment

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