Molecular Basis Of Inheritance — DPP 2

20 NEET questions from Molecular Basis Of Inheritance, with answers and explanations. Attempt it as a timed test to get scoring, accuracy breakdown and mistake tracking.

Attempt this test free 20 questions · 30 minutes · +80 marks
1
Select the incorrect statement from the following.
  1. A In most of the eukaryotes, the structural gene in monocistronic
  2. B Exons are the expressed sequence of genes
  3. C Introns do not appear in processed RNA
  4. D A gene does not code for tRNA
Show answer & explanation

Correct answer: D

The DNA sequence coding for tRNA or rRNA molecules also define a gene.
2
In the technique of DNA fingerprinting digestion of DNA is followed by
  1. A Electrophoresis
  2. B Hybridisation
  3. C Denaturation
  4. D Southern blotting
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Correct answer: A

Gel electrophoresis will separate the DNA segment.
3
In eukaryotes, rRNAs are transcribed by
  1. A RNA polymerase I and III
  2. B RNA polymerase III only
  3. C RNA polymerase I and II
  4. D RNA polymerase I only
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Correct answer: A

RNA polymerase I-5.8S, 18S, and 28S rRNAs, RNA polymerase II - hnRNA, RNA polymerase III - tRNA, ScRNA, 5 S rRNA and SnRNA.
4
The non-human model organisms sequenced in Human Genome project were
  1. A A nematode and fruit fly
  2. B Wheat and rice
  3. C Fish and birds
  4. D Garden pea and fruit fly
Show answer & explanation

Correct answer: A

Caenorhabiditis elegna, Drosophila
5
Amongst the following, which one has least number of nucleotides in its genome?

  1. A $\phi \times 174$ bacteriophage
  2. B E. coli
  3. C Human
  4. D Lambda phage
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Correct answer: A

$\phi \times 174$ bacteriophage -5386 nucleotides, Bacteriophage lambda -48502 bp
E. coli $-4.6 \times 10^{6} \mathrm{bp}$
Human genome $-3.3 \times 10^{9}$ bp
6
If E. coli completes the process of DNA replication within 38 minutes then the rate of polymerisation of DNA would be approximately
  1. A $4.6 \times 10^{6} \mathrm{bp} / \mathrm{second}$
  2. B $500 \mathrm{bp} /$ second
  3. C $230 \mathrm{bp} /$ second
  4. D $2000 \mathrm{bp} / \mathrm{second}$
Show answer & explanation

Correct answer: D

DNA of E. coli has $4.6 \times 10^{6} \mathrm{bp}$. The rate of polymerisation in E. coli will be $\frac{4.6 \times 10^{6}}{38 \times 60} \mathrm{bp} / \mathrm{second} =2000 \mathrm{bp} /$ second
7
A hypothetical sequence of a transcription unit is represented below
$3^{\prime}$ C G GATATCCAA T-5' Template strand
$5^{\prime}$ G C C TA TA G G T TA-3' Coding strand
If we switch the position of promoter with terminator in the above given transcription unit and transcription occurs then the sequence of RNA transcribed will be
  1. A $5^{\prime}-\mathrm{C} \mathrm{G} \mathrm{G} \mathrm{A} \mathrm{U} \mathrm{A} \mathrm{U} \mathrm{C} \mathrm{C} \mathrm{A} \mathrm{A} \mathrm{U-3'}$
  2. B $5^{\prime}-\mathrm{G} \mathrm{C} \mathrm{C} \cup \mathrm{A} \cup \mathrm{A} \mathrm{G} \mathrm{G} \cup \cup \mathrm{A}-3^{\prime}$
  3. C $5^{\prime}-\mathrm{A} \cup \cup \mathrm{G} \mathrm{G} A \cup \mathrm{~A} \cup \mathrm{C} \mathrm{C} \mathrm{G}-3^{\prime}$
  4. D 5'-U A A C C U A U A G G C-3'
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Correct answer: D

If we switch the position of promoter with terminator in the transcription unit, the template strand becomes coding strand and the coding strand becomes template strand. By switching the positions of promoter with terminator, the template strand will be
Molecular Basis Of Inheritance – NEET-10028
8
A radioactive sulphur containing bacteriophage is allowed to attach to E. coli that contains radioactive phosphorus. The viruses produced in this bacteria by this infection will have
  1. A Radioactive protein capsule and nonradioactive genetic material
  2. B Radioactive protein capsule and radioactive genetic material
  3. C Non-radioactive protein capsule and radioactive genetic material
  4. D Non-radioactive protein capsule and nonradioactive genetic material
Show answer & explanation

Correct answer: C

Sulphur is present in proteins whereas phosphorus is present in genetic materials. Only genetic material of bacteriophage enters the bacteria and the protein coat of the virus is synthesised inside the bacteria will not contain radioactive sulphur. But the genetic material of bacteriophage formed inside the bacteria will have radioactive phosphorus.
9
Select the incorrect statement regarding the salient features of the double-helix structure of DNA

  1. A It is made of two polynucleotide chains
  2. B The two chains have anti-parallel polarity
  3. C Guanine is bonded with thymine by three H-bonds
  4. D The two chains are coiled in a right-handed fashion
Show answer & explanation

Correct answer: C

In dsDNA, purines of one strand are paired with pyrimidines of corresponding strand by formation of hydrogen bonds. Adenine forms two H-bonds with thymine and guanine is bonded with cytosine with three H-bonds.
10
Which of the following radioactive isotopes were utilised for labelling protein and DNA in transduction experiment respectively?
  1. A ${ }^{32} \mathrm{P},{ }^{35} \mathrm{P}$
  2. B ${ }^{35} \mathrm{~S},{ }^{35} \mathrm{P}$
  3. C ${ }^{35} \mathrm{~S},{ }^{32} \mathrm{P}$
  4. D ${ }^{32} \mathrm{~S},{ }^{35} \mathrm{P}$
Show answer & explanation

Correct answer: C

$\mathrm{S}^{35} \longrightarrow$ Protein coat $\quad \mathrm{P}^{32} \longrightarrow$ DNA
11
Which plant was used by Taylor to prove semiconservative replication at chromosomal level?
  1. A Haematoxylan
  2. B Vicia faba
  3. C Trillium
  4. D Ophioglossum
Show answer

Correct answer: B

12
Mark the correct match.
  1. A Catalytic RNA - 16 S rRNA and 23 S in bacteria rRNA- as ribozyme.
  2. B Val operon - Found in eukaryotes
  3. C Sanger - Determination of amino method acid sequences in proteins only
  4. D VNTR - Intron
Show answer & explanation

Correct answer: D

VNTR → Variable Number Tandem Repeat.
13
During polymerisation of deoxyribonucleosides triphosphates in bacteria which of the following enzymes is mainly required?
  1. A DNA dependent RNA polymerase
  2. B DNA dependent DNA polymerase
  3. C RNA dependent DNA polymerase
  4. D DNA gyrase
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Correct answer: B

DNA pol III → Read DNA segment and add nucleotide of DNA
14
In a transcription unit, the promoter is located towards
  1. A $5 ^ { \prime }$ end of the templates strand
  2. B $5 ^ { \prime }$ end of the coding strand
  3. C $3 ^ { \prime }$ end of the non-template strand
  4. D $3 ^ { \prime }$ end of the sense strand
Show answer & explanation

Correct answer: B

Promoter sequences are present towards $5 ^ { \prime }$ end of the structural gene, i.e., with respect to the polarity of coding strand.
15
The discontinuously synthesised fragments of DNA are joined by the enzyme [NCERT Pg. 90]
  1. A DNA ligase
  2. B DNA-dependent DNA polymerase
  3. C Phosphorylase
  4. D Topoisomerase
Show answer & explanation

Correct answer: A

On the template strand of DNA with polarity $5 ^ { \prime } \rightarrow 3 ^ { \prime }$ DNA synthesis is discontinuous in the form of Okazaki fragments. These fragments are joined by the enzyme DNA ligase.
16
What is incorrect for human chromosome 1?
  1. A It is one of the largest chromosome
  2. B Its sequence was completed in May 2007
  3. C It has maximum number of genes
  4. D It was the last chromosome to be sequenced
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Correct answer: B

Sequencing completed in the year 2006
17
What is incorrect for UTR?
  1. A Present in between the translational unit in mRNA
  2. B Not recognised by any tRNA
  3. C Required for efficient translation process
  4. D Provide stability to mRNA
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Correct answer: A

Present between Cap and Start point and stop codon and poly A tail.
18
If there is a flow of information from RNA to DNA then this process will be named as

  1. A Reverse transcription
  2. B Translation
  3. C Replication
  4. D Transformation
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Correct answer: A

Molecular Basis Of Inheritance – NEET-9984
19
Match the following columns and select the correct option
Molecular Basis Of Inheritance – NEET-10030
  1. A a(i), b(ii), c(iii), d(iv)
  2. B a(iv), b(iii), c(ii), d(i)
  3. C a(i), b(iii), c(ii), d(iv)
  4. D a(iv), b(ii), c(iii), d(i)
Show answer & explanation

Correct answer: C

Operator gene - Interacts with regulator molecule
Promoter gene - Provides attachment site for RNA polymerase
Structural gene - Transcribe mRNA for polypeptide synthesis
Regulator gene - Controls the activity of operator gene
20
In the experiment conducted by Griffith, when a mixture of heat killed S-strain and live R-strain of Pneumococcus bacteria was injected into a healthy mice, it developed pneumonia. This was due to
  1. A Reactivation of S-strain bacteria
  2. B Transformation of S-strain bacteria into R-strain bacteria
  3. C Synthesis of polysaccharide coat around heat killed S-strain bacteria
  4. D Transformed R-strain bacteria
Show answer & explanation

Correct answer: D

In the experiment conducted by Griffith, the R-strain bacteria had been transformed by the heat killed S-strain bacteria.

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