Molecular Basis of Inheritance — NCERT Solutions

A nitrogenous base is the base part only, while a nucleoside is a base linked to a sugar. From the list:

• Bases: Adenine, Thymine, Uracil, Cytosine
• Nucleosides: Cytidine, Guanosine

In double-stranded DNA, A pairs with T and G pairs with C.

Given C = 20%, so G = 20%.

Thus, $C+G=40\%$.

Remaining percentage for A and T = $100\% - 40\% = 60\%$.

Since A = T,

$A = 60\%/2 = 30\%$.

The complementary strand follows base pairing: A–T and G–C.

Given strand:
$5'-ATGCATGCATGCATGCATGCATGC-3'$

So the complementary strand is antiparallel:
$3'-TACGTACGTACGTACGTACGTACG-5'$

Written in 5'→3' direction, it becomes:
$5'-CGTACGTACGTACGTACGTACGTA-3'$.

The question asks for the complementary strand in 5'-3' direction.

In a transcription unit, the mRNA sequence is complementary to the template strand and therefore same as the coding strand except that U replaces T.

Coding strand given:
$5'-ATGCATGCATGCATGCATGCATGC-3'$

So mRNA is:
$5'-AUGCAUGCAUGCAUGCAUGCAUGC-3'$

The property was complementary base pairing in the DNA double helix.

Watson and Crick proposed that the two strands are held together by specific hydrogen bonding between complementary bases: A pairs with T and G pairs with C. Because of this, if the two strands separate, each strand can serve as a template for the formation of a new complementary strand. Therefore, after replication, each daughter DNA molecule contains one parental strand and one newly synthesised strand, which is the semi-conservative mode of replication.

Based on the template and product:

• DNA → DNA : DNA-dependent DNA polymerase
• DNA → RNA : DNA-dependent RNA polymerase

These are the nucleic acid polymerases discussed in the chapter.

Hershey and Chase used radioactive labeling to distinguish DNA from protein in bacteriophages:

• Viruses grown in $^{32}P$ contained radioactive DNA, because DNA has phosphorus.
• Viruses grown in $^{35}S$ contained radioactive protein, because proteins contain sulfur but DNA does not.

They let these phages infect E. coli, then used a blender to remove the viral coats from the bacterial cells and a centrifuge to separate the bacterial pellet from the phage coats.

The bacteria infected with $^{32}P$-labeled phages were radioactive, showing that DNA entered the bacteria. The bacteria infected with $^{35}S$-labeled phages were not radioactive, showing that protein did not enter. Thus, DNA was identified as the genetic material.

Repetitive DNA means DNA sequences in which a small stretch of DNA is repeated many times. Satellite DNA is the bulk DNA separated from these repetitive sequences as distinct peaks during density gradient centrifugation; it includes highly repetitive sequences such as mini-satellites and micro-satellites.

mRNA is the messenger RNA that provides the template for protein synthesis. tRNA is the transfer RNA that brings amino acids to the ribosome and reads the genetic code through its anticodon; it acts as the adapter molecule.

The template strand is the DNA strand with 3'→5' polarity that is read by RNA polymerase to synthesise RNA. The coding strand has 5'→3' polarity and its sequence is the same as the RNA sequence except that thymine in DNA is replaced by uracil in RNA.

The ribosome has two essential roles during translation:

1. It binds the mRNA and positions the tRNAs correctly so amino acids can be added in order.
2. It catalyses peptide bond formation; in bacteria, 23S rRNA acts as the ribozyme for this reaction.