it vii - neet 2020 preparation | neet exam | nta neet€¦ · (a) translation begins when mrna...
TRANSCRIPT
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Un
it V
II-
Gen
eti
cs &
Evo
luti
on
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01
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Recognise the figure and find out the
correct matching
(A) a-Ser, b-Tyr, f-tRNA, c-mRNA,
d-codon, e-anticodon
(B) a-Ser, b-Tyr, c-tRNA, f-mRNA,
e- codon, d-anticodon
(C) b-Ser, a-Tyr, c-tRNA, f-mRNA,
e- codon, d-anticodon
(D) b-Ser, a-Tyr, f-tRNA, c-mRNA,
e-codon, d-anticodon.
c
d
e f
U C A A U G
A G U U A C
a b
5’ 3’
I
V
IVIII
II
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02
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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tRNA has the function of
(A) Transcription
(B) Adapter for attaching amino acids over mRNA
template
(C) Transferring information to mRNA
(D) Carry genetic code to cytoplasm.
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03
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Aminoacylation of tRNA is essential for
(A) Replication of RNA
(B) Formation of peptide bond
(C) Splicing
(D) Initiation of transcription
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04
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Select correct statement about protein synthesis
(A) Translation begins when mRNA attaches to small
subunit of ribosome
(B) Peptidase catalyses formation of peptide bond
(C) UTRs are present between start and stop codons
(D) At the end of translation, release factor binds to
initiation codon.
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05
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
www.gangwarinstitute.com Contact No. : 8400582582
Which codon does not have tRNA
(A) Start codon
(B) Stop codon
(C) AUG
(D) GGG.
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06
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Amino acid binding site of tRNA is
(A) 5' end
(B) Anticodon loop
(C) DHU loop
(D) -CCA3' end.
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07
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Which is the site of translation of mRNA
(A) Nucleus
(B) Nucleolus
(C) Golgi body
(D) Ribosomes
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08
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Sequence of amino acids in a polypeptide is determined
by
(A) rRNA
(B) mRNA
(C) tRNA
(D) snRNA.
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09
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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What does the following signify ?
Amino acid + ATPAminoacyl AMP + PPi
(A) Chain termination
(B) Activation of amino acids
(C) Elongation of chain
(D) None of the above.
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10
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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mRNA directs building of protein through the sequence
of
(A) Introns
(B) Exons
(C) Anticodons
(D) Codons.
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11
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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The following figure shows the process of
(A) Transcription
(B) Translation
(C) Replication
(D) Central dogma. C U C U U G G G U C G UC UA U C U AA UA CC
5’ 3’
C G T C A A
tRNA tRNA
UU
A
Asn
A
GG
tRNA tRNA
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12
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Anticodon present on tRNA complementary to initiator
codon AUG of mRNA is
(A) CAU
(B) GUA
(C) TAC
(D) UAC
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13
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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A segment of DNA, AGCTTCGAA, has lost first base.
Assuming that translation starts from the next base the
effect would be
(A) No change in polypeptide form
(B) Only first amino acid will be different
(C) A complete change in type & sequence of amino acids
(D) Last amino acid of polypeptide will be different
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14
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Coding segment of DNA contains CAC. Which amino acid
is picked up during translation
(A) Leucine
(B) Glutamic acid
(C) Valine
(D) Methionine
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15
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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In bacteria formation of peptide bond during translation
is effected by
(A) Peroxisomes
(B) Lysosomes
(C) Ribozyme
(D) Nucleosome
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16
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Which of the following has no tRNA
(A) UAU
(B) UAA
(C) UGU
(D) UGG.
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17
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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A sequence of three bases on tRNA which binds to
mRNA codon is
(A) NODOC
(B) Nonsense codon
(C) Anticodon
(D) Both A and C.
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18
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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What is wrong about protein synthesis
(A) Only one strand of DNA acts as template for mRNA
synthesis
(B) DNA polymerase mediates mRNA synthesis
(C) mRNA passes into cytoplasm and gets attached to
30S ribosomal subunit
(D) tRNA bring amino acids one by one.
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19
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Amino acid sequence in protein synthesis is determined
by sequence of
(A) rRNA
(B) t RNA
(C) m RNA
(D) c DNA
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20
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Portion of gene which is transcribed but not translated is
(A) Exon
(B) Intron
(C) Cistron
(D) Codon.
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21
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Which one is correct matching RNA
(A) Anticodon-Site of tRNA having complementary bases
to a codon of
(B) Transcription- Synthesis of protein
(C) Ribosomal RNA-Carries amino acid to site of protein
synthesis
(D) Translation-Process of mRNA carrying information
from nucleus to ribosome.
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22
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Polyribosomes are formed when ribosomes are connected
by
(A) rRNA
(B) mRNA
(C) tRNA
(D) All of the above.
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23
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Enzyme catalysing synthesis of RNA over DNA template
is
(A) DNA polymerase
(B) RNA polymerase
(C) Reverse transcriptase
(D) Endonuclease.
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24
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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The processes by which DNA forms mRNA and mRNA
forms protein are respectively
(A) Translation and transcription
(B) Transcription and replication
(C) Transcription and translation
(D) Replication and translation.
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25
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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RNA that picks up specific amino acid from amino acid
pool of cytoplasm to carry it to ribosome during protein
synthesis is
(A) tRNA
(B) mRNA
(C) rRNA
(D) hnRNA.
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26
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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tRNA takes part in
(A) Transfer of genetic code to cytoplasm
(B) Carry amino acids to ribosomes
(C) Collection of RNA in ribosome
(D) Copy the genetic code from DNA in nucleus.
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27
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Lac operon is turned on when allolactose binds to
(A) Operator gene
(B) mRNA
(C) Repressor protein
(D) Promoter site.
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28
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Environmental agent triggering transcription from an
operon is
(A) Inducer
(B) Regulator
(C) Repressor
(D) Controlling factor.
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29
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Genes regulate growth and differentiation through
(A) Transcription and translation
(B) Transduction and translation
(C) Transformation
(D) Translation.
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30
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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In operon model, regulator gene functions as
(A) Repressor
(B) Regulator
(C) Inhibitor
(D) All of the above.
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31
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Operon contains
(A) Operator and regulator genes
(B) Operator and structural genes
(C) Operator and regulator genes, and repressor
(D) Operator gene, regulator gene, repressor, structural
genes and promoter gene.
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32
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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In Escherichia coli, the product of i gene combines with
(A) Operator gene to switch off structural genes
(B) Inducer gene to switch off structural genes
(C) Operator gene to switch on structural genes
(D) Regulator gene to switch off structural genes.
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33
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Operon is
(A) Sequence of three nitrogen bases determining single
amino acid
(B) A set of closely placed genes regulating a metabolic
pathway in prokaryotes
(C) Segment of DNA specifying a polypeptide
(D) Gene responsible for switching on and switching off
of other genes.
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34
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Match the columns and choose the correct option
Column I Column II
1. Taylor & colleagues a. Lac operon
2. Hershey and Chase b. DNA replicates semi-conservatively
3. Griffith c. Transforming principle
4. Jacob and Monod d. DNA is genetic material
e. Transcription
(A) 1-b, 2-e, 3-a, 4-c (B) 1-c, 2-d, 3-b, 4-a
(C) 1-b, 2-d, 3-c, 4-a (D) 1-a, 2-e, 3-d, 4-b
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35
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Lac operon concept was proposed by
(A) Jacob and Monod in 1961
(B) Jacob and Monod in 1969
(C) Watson and Crick in 1953
(D) Watson and Crick in 1963.
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36
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Escherichia coli growing on medium having glucose is
transferred to lactose containing medium. Which change
occurs
(A) Lac operon is induced
(B) Lac operon is suppressed
(C) All operons are induced
(D) The bacterium stops dividing.
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37
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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In lac-operon, repressor protein binds to
(A) Regulator gene
(B) Operator gene
(C) Promoter gene
(D) Structural gene.
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38
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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A gene of operon which forms the repressor protein is
(A) Operator
(B) Promoter
(C) Regulator
(D) Structural
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39
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Operon model of gene regulation and organisation of
procaryotes was proposed by
(A) Messelson and Franklin
(B) Wilkins and Franklin
(C) Beadle and Tatum
(D) Jacob and Monod.
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40
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Sequence of structural genes of lac operon is
(A) y, z, a
(B) z, y, a
(C) a, y, z
(D) a, z, y.
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41
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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What is correct gene expression pathway
(A) Gene – mRNA
(B) Transcription– gene – translation – mRNA – protein
(C) Gene – transcription – mRNA – translation – protein
(D) Gene – translation – mRNA – transcription – protein
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42
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Lac operon ‘inducer lactose’ serves as substrate for
enzyme
(A) Transacetylase
(B) Permease
(C) b-galactosidase
(D) Endonuclease
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43
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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In lac operon model, lactose functions as
(A) Inducer that binds to operator gene
(B) Repressor that binds to operator gene
(C) Inducer that binds to repressor protein
(D) Corepressor that binds to repressor protein.
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44
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Choose the wrong statement regarding the observation
drawn from Human Genome Project is
(A) Repetitive sequences are stretches of RNA
(B) Less than 2% of the genome codes for protein
(C) Chromosome 'Y' has the fewest number of genes
(D) SNPs help in tracing human history.
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45
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Coenorhabditis elegans is
(A) Free-living
(B) Non-pathogenic
(C) Nematode
(D) All of the above.
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46
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
www.gangwarinstitute.com Contact No. : 8400582582
Many non-humans model organisms have also been
sequenced along with the human genome, these are
(A) Bacteria and yeast
(B) Plants (rice and Arabidopsis)
(C) Fruitfly and Coenohabditis (nematode)
(D) All of the above
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47
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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The application of genetic engineering is in the field of
(A) Forensic science
(B) Genetic biodiversity
(C) Evolutionary biology
(D) All of the above.
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48
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
www.gangwarinstitute.com Contact No. : 8400582582
Most number of genes are located on chromosome
(A) 1
(B) 6
(C) X
(D) Y.
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49
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Automated–DNA sequencers is based or method developed
by
(A) Erwin Chargaff
(B) Maurice Wilkins
(C) Francis Crick
(D) Frederick Sanger.
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50
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Match the columns I and II, and choose the correct combination
from the options given
Column I Column II
a. Hershey and Chase experiment 1. 1928
b. Taylor's experiment 2. 1952
c. Meselson and Stahl experiment 3. 1958
d. Transforming Principle 4. 1990
e. Human Genome Project launching 5. 1953
(A) a-1, b-2, c-3, d-5, e-4 (B) a-2, b-3, c-5, d-1, e-4
(C) a-2, b-3, c-4, d-1, e-5 (D) a-2, b-3, c-3, d-1, e-4.
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51
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Single base DNA differences are called
(A) VNTR
(B) SCP
(C) SNPs
(D) Expressed sequence tags.
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52
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Haploid content of human DNA is
(A) 1.65 × 109 bp
(B) 3.1 – 3.3 × 109 bp
(C) 4.6 × 106 bp
(D) 3.3 × 109 kbp.
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53
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
www.gangwarinstitute.com Contact No. : 8400582582
Find the wrongly matched pair
(A) Hargobind Khorana-Synthesised RNA molecules chemically
(B) George Gamow-Codon in triplet
(C) Meselson and Stahl-Regulation of gene expression
(D) Alec Jeffreys-DNA fingerprinting.
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54
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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If an inheritable mutation is observed in a population
at high frequency, it is referred as
(A) DNA polymorphism
(B) Expressed sequence tag
(C) Sequence annonation
(D) Linkage.
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55
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
www.gangwarinstitute.com Contact No. : 8400582582
The VNTR belongs to a class of satellite DNA reffered
to as
(A) Repetitive DNA
(B) Mega-Satellite
(C) Mini-Satellite
(D) Micro-Satellite.
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56
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
www.gangwarinstitute.com Contact No. : 8400582582
Matching DNA sequence of criminal with suspect is
known as
(A) DNA fingerprinting
(B) DNA amplification
(C) Gene mapping
(D) DNA resolution
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57
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
www.gangwarinstitute.com Contact No. : 8400582582
One of the following is major requirement for DNA
fingerprinting
(A) Electron microscopy
(B) Electrophoresis
(C) ELISA
(D) HPLC.
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58
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
www.gangwarinstitute.com Contact No. : 8400582582
Which is an example of highly repetitive DNA
(A) Alu elements
(B) Histone gene cluster
(C) Dispersed DNA
(D) DNA minisatellite.
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59
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
www.gangwarinstitute.com Contact No. : 8400582582
Blood stain component to be used for DNA profiling
technique is
(A) Serum
(B) Leucocytes
(C) Platelets
(D) Erythrocytes.
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60
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
www.gangwarinstitute.com Contact No. : 8400582582
Which is useful in solving parental disputes
(A) Hybridoma technology
(B) DNA Fingerprinting
(C) Western blotting
(D) ELISA.
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61
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Regulation of gene expression in eukaryotes could be
exerted at
a. Transcriptional level (formation of primary transcript)
b Processing level (regulation of splicing)
c. Transport of mRNA from nucleus to the cytoplasm
d. Translational level
(A) a, b and c (B) b, c and d
(C) a, c and d (D) a, b, c and d
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62
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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In prokaryotes, the predominant site (or primary step)
of control of gene expression is at
(A) Control of rate of transcriptional initiation.
(B) Control of rate of translational initiation.
(C) Control of rate of initiation of replication
(D) Processing level.
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63
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Which of the following events would occur in 'Lac-operon of E. coli
when growth medium has high concentration of lactose
(A) Structural genes fail to produce polycistronic mRNA
(B) Repressor protein binds to RNA polymerase and prevents q
translation
(C) Repressor protein attaches to the promoter sequence and
represses the operator
(D) Inducer molecule binds to repressor protein and RNA
polymerase binds to promoter sequence.
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64
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Escherichia coli with mutated z gene of lac operon cannot grow in
medium containing only lactose as the source of energy because
(A) Lac operon is constitutively active in these cells
(B) They cannot synthesise functional beta- galactosidase
(C) In the presence of glucose, Escherichia coli cannot utilise lactose
(D) The bactetium cannot transport lactose from the medium into
the cell.
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65
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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In lac operon genes a, i, y and z code for
(A) Repressor protein, permease, b-galactosidase, transacetylase
(B) Trsnsacetylase, repressor protein, permase, b-galactosidase
(C) Trsnsacetylsse, permease, b-galactosidase, repressor protein
(D) Permease , transacetylase, repressor protin, b-galactosidase.
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66
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Sequencing the whole set of genome that contained all
the coding and non coding sequence and later assigning
different region in the sequence with functions are
referred to as
(A) Expressed Tags (B) Sequence Annotation
(C) ESTs (D) SNPs.
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67
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Repetitive DNA are seperated from bulk genomic DNA
as different peak during density gradient centrifugation.
The bulk DNA forms a major peaks and the other small
peaks are referred to as
(A) Repetitive DNA (B) SNPs (Snips)
(C) Taq-DNA (D) Satellite DNA
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68
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Largest gene in humans is
(A) Dystrophin
(B) Insulin gene
(C) Oncogene
(D) Huntingtin.
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69
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
www.gangwarinstitute.com Contact No. : 8400582582
Select the correct sequences of steps in DNA fingerprinting involving
Southern blot hybridisation using radiolabelled VNTR as proble
I. Hybridisation using labelled VNTR probe.
II. Isolation of DNA
III. Transferring (blotting) of separated DNA fragments to synthetic
membranes, such as nitrocellulose or nylon.
IV. Detection of hybridisation DNA fragments by autoradiography.
V. Separation of DNA fragments by electrophoresis.
VI. Digestion of DNA by restriction endonucleases
(A) I, V, VI, II, III and IV (B) II, VI, V, III, I and IV
(C) V, I, VI, III, IV and II (D) II, I, V, VI, IV and III.
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70
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Which is incorrect about DNA polymorphism
(A) The two alleles of a chromosome contain different copy
numbers of VNTR
(B) Different chromosomes contains different copy number of
VNTR
(C) Polymorphisms are inheritable from parents to children
(D) DNA from different tissue from an individual shows the
different degree of polymorphism.
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71
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Transfer of DNA bands from agarose gel to nitrocellulose
or nylon membrane is
(A) Southern transfer
(B) Western transfer
(C) Northern transfer
(D) Eastern transfer
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72
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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Which cannot be used for DNA fingerprinting in humans
(A) Leucocytes
(B) Erythrocytes/RBCs
(C) Hair bulbs
(D) Sperms.
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73
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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DNA fingerprinting technique was developed by
(A) Jeffreys et al
(B) Schleiden and Schwann
(C) Boysen and Jensen
(D) Edwards and Steptoe
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74
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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The probe used initially by Alec Jaffrey during development
of DNA fingerprinting was
(A) Ribozyme
(B) SAT chromosomes
(C) VNTR
(D) rDNA.
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75
Molecular Basis of Inheritance Unit VII- Genetics & Evolution
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DNA fingerprinting in forensic science was pioneered
in India by
(A) Lalji Singh
(B) H.G. Khorana
(C) J.C. Bose
(D) M.S. Swaminathan.
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