In some viruses, DNA is synthesised by using RNA as template. Such a DNA is called
A-DNA
B-DNA
cDNA
$r$ DNA
If Meselson and Stahl's experiment is continued for four generations in bacteria, the ratio of $15_{\mathrm{N}} / 15_{\mathrm{N}}: 15_{\mathrm{N}} / 14_{\mathrm{N}}: 14_{\mathrm{N}} / 14_{\mathrm{N}}$ containing DNA in the fourth generation would be
$1: 1: 0$
$1: 4: 0$
$0: 1: 3$
$0: 1: 7$
If the sequence of nitrogen bases of the coding strand of DNA in a transcription unit is
$$ 5^{\prime} \text { - A T G A A T G - 3', } $$
the sequence of bases in its RNA transcript would be
$5^{\prime}-\mathrm{A} \cup \mathrm{GAA} \cup \mathrm{G}-3^{\prime}$
$5^{\prime}-\cup \mathrm{A} \subset \cup \cup \mathrm{A} \mathrm{C}-3^{\prime}$
$5^{\prime}-\mathrm{C} \mathrm{A} \cup \cup \mathrm{C} \mathrm{A} \cup-3^{\prime}$
$5^{\prime}-\mathrm{G} \cup \mathrm{A} A \mathrm{G} \cup \mathrm{A}-3^{\prime}$
The RNA polymerase holoenzyme transcribes
If the base sequence of a codon in $m R N A$ is $5^{\prime}-A \cup G-3^{\prime}$, the sequence of tRNA pairing with it must be
$5^{\prime}-\mathrm{UAC}-3^{\prime}$
$5^{\prime}-\mathrm{CAU}-3^{\prime}$
$5^{\prime}-\mathrm{AUG}-3^{\prime}$
$5^{\prime}-\mathrm{G} \cup \mathrm{A}-3^{\prime}$