 Solution

In a bacterial transcription complex, RNA polymerase binds a promoter and locally separates the DNA strands into a transcription bubble. The template enters the active site in the $3^{\prime }$ to $5^{\prime }$ direction, while the complementary RNA grows $5^{\prime }$ to $3^{\prime }$. The displaced non-template strand passes around the bubble, downstream DNA re-forms a duplex, and the nascent RNA exits through a separate channel. A short RNA–DNA hybrid is maintained in an elongating complex; the first few nucleotides of an initiating complex do not yet make a full elongation-length hybrid.

The core enzyme has subunit composition ${\alpha }_2\beta {\beta }^{\prime }\omega$; association with a sigma factor forms a promoter-selective holoenzyme. Sigma recognizes promoter features and assists steps leading from a closed bound complex to the open complex. Different sigma factors redirect transcription towards different promoter classes. The familiar $-10$ and $-35$ elements are examples for a major bacterial sigma family, not a universal sequence rule for all sigma factors. Promoter escape changes these contacts, but complete sigma dissociation at one exact transcript length should not be assumed for every complex.

For the supplied non-template sequence, assuming its first base is the first transcribed position, the complementary template runs $3^{\prime }$-TCAGTAAGC-$5^{\prime }$. The RNA has the coding-strand order with uracil replacing thymine. Therefore the first five nucleotides areThe top-left panel of the original schematic in part (d) shows the strand polarities and polymerase organisation; it is not a copied experimental illustration.