Q 1 :    

Given below are two statements:                                                                                                                  [2023]
Statement I: In prokaryotes, the positively charged DNA is held with some negatively charged proteins in a region called nucleoid.
Statement II: In eukaryotes, the negatively charged DNA is wrapped around the positively charged histone octamer to form nucleosome.

In the light of the above statements, choose the correct answer from the options given below:

  • Statement I is correct but Statement II is false.

     

  • Statement I incorrect but Statement II is true.

     

  • Both Statement I and Statement II are true.

     

  • Both Statement I and Statement II are false.

     

(2)

In prokaryotes, DNA being negatively charged is held with positively charged proteins in region termed as nucleoid.

 



Q 2 :    

Read the following statements and choose correct statements:                                [2022]
(A) Euchromatin is loosely packed chromatin.
(B) Heterochromatin is transcriptionally active.
(C) Histone octamer 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 below:

  • (B), (D), (E) only

     

  • (A), (C), (D) only

     

  • (B), (E) only

     

  • (A), (C), (E) only

     

(2)

In a typical nucleus, some region of chromatin are loosely packed (and stains light) and are referred to as euchromatin. Euchromatin is transcriptionally active chromatin whereas heterochromatin is inactive. Histone octamer is wrapped by negatively charged DNA to form nucleosome. Histones are rich in basic amino acid residues lysine and arginine. A typical nucleosome contains 200 bp of DNA helix.

 



Q 3 :    

If the length of a DNA molecule is 1.11 metres, what will be the approximate number of base pairs?         [2022]

  • 3.3×109 bp

     

  • 6.6×10 bp

     

  • 3.3×106 bp

     

  • 6.6×106 bp

     

(1)

Length of DNA double helix = Total number of bp × Distance between two consecutive bp.

So, total number of bp = Length of DNA double helixDistance between two consecutive bp

                                   = 1.11 metres0.34×10-9 m/bp=3.26×109bp

Therefore, approximate number of base pair will be 3.3×109 bp.

 



Q 4 :    

Complete the flow chart on central dogma.                                [2021]

[IMAGE 34]

  • (P)-Transduction; (Q)-Translation; (R)-Replication; (S)-Protein

     

  • (P)-Replication; (Q)-Transcription; (R)-Transduction; (S)-Protein

     

  • (P)-Translation; (Q)-Replication; (R)-Transcription; (S)-Transduction

     

  • (P)-Replication; (Q)-Transcription; (R)-Translation; (S)-Protein

     

(4)

 



Q 5 :    

If adenine makes 30% of the DNA molecule, what will be the percentage of thymine, guanine and cytosine in it?                  [2021]

  • T : 20; G : 25; C : 25

     

  • T : 20; G : 30; C : 20

     

  • T : 20; G : 20; C : 30

     

  • T : 30; G : 20; C : 20

     

(4)

According to the Chargaff’s rule,
A = T and G = C
A + G = T + C
∴ If A = 30%, then T = 30%
⇒ A + T = 60%
G + C = 100 – 60 = 40%

G = C = 20%
T = 30%, G = 20%, C = 20%

 



Q 6 :    

Which one of the following statements about histones is wrong?                  [2021]

  • Histones carry positive charge in the side chain.

     

  • Histones are organized to form a unit of 8 molecules.

     

  • The pH of histones is slightly acidic.

     

  • Histones are rich in amino acids - Lysine and Arginine.

     

(3)

Histones are positively charged basic proteins, thus, pH is alkaline.

 



Q 7 :    

Which of the following statements is correct?                [2020]

  • Adenine pairs with thymine through two H-bonds.

     

  • Adenine pairs with thymine through one H-bond.

     

  • Adenine pairs with thymine through three H-bonds.

     

  • Adenine does not pair with thymine.

     

(1)

Adenine pairs with thymine by forming two hydrogen bonds, A = T.

 



Q 8 :    

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]

  • 2.0 meters

     

  • 2.5 meters

     

  • 2.2 meters

     

  • 2.7 meters

     

(3)

If the distance between two consecutive base pairs is 0.34 nm i.e.,0.3×10-9 m and the total number of base pairs of a DNA double helix in a typical mammalian cell is 6.6×109 bp, the length of DNA is calculated by multiplying the total number of base pairs with distance between two consecutive base pairs i.e., 6.6×109 bp × 0.34 nm = 2.2 m (approx.)

 



Q 9 :    

Purines found both in DNA and RNA are                     [2019]

  • cytosine and thymine

     

  • adenine and thymine

     

  • adenine and guanine

     

  • guanine and cytosine

     

(3)

 



Q 10 :    

The association of histone H1 with a nucleosome indicates that              [2017]

  • DNA replication is occurring

     

  • the DNA is condensed into a chromatin fibre

     

  • the DNA double helix is exposed

     

  • transcription is occurring

     

(2)

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. The association of H1 histone with nucleosome indicates that DNA remains in its condensed form. Nucleosome is the unit of compaction. H1 is attached over the linker DNA. The linker DNA, consisting of H1 histone, connects two adjacent nucleosomes. They together constitute chromatosome. It gives rise to a chromatin fibre after further condensation.