Question 22.2: From the base sequence 5′ A–T–G–C–C–A 3′ in a DNA template s......

An RNA molecule cannot contain the base T. The base U is present instead. Therefore, U–A base pairing will occur instead of T–A base pairing. The other base-pairing combination, G–C, remains the same. The hnRNA product of the transcription process will therefore be
$\quad\quad$DNA template: 5$^{\prime}$ A–T–G–C–C–A 3$^{\prime}$
$\quad\quad\quad\quad\quad\quad\quad\quad\quad   \vdots    \vdots    \vdots    \vdots    \vdots    \vdots$
$\quad$hnRNA molecule: 3$^{\prime}$ U–A–C–G–G–U 5$^{\prime}$
Note that the direction of the hnRNA strand is antiparallel to that of the DNA template. This will always be the case during transcription.
It is standard procedure, when writing and reading base sequences for nucleic acids (both DNAs and RNAs), always to specify base sequence in the 5$^{\prime}$ → 3$^{\prime}$ direction unless otherwise directed. Thus
3$^{\prime}$ U–A–C–G–G–U 5$^{\prime}$      becomes      5$^{\prime}$ U–G–G–C–A–U 3$^{\prime}$