Transcription

Bacterial genes

5'to3' strand: same sequence as RNA transcript except for having T instead of U 3'to5' strand: complementary to RNA transcript -35 to -10: recognized by sigma subunit of RNA polymerase polyadenine sequence: leads to an unstable RNA-DNA duplex inverted repeats: produce stem-loop structure in RNA transcript Bacterial transcription is a four-stage process. 1. Promoter recognition: RNA polymerase is a holoenzyme composed of a five-subunit core enzyme and a sigma (σ) subunit. Different types of σ subunits aid in the recognition of different forms of bacterial promoters. The bacterial promoter is located immediately upstream of the starting point of transcription (identified as the +1 nucleotide of the gene). The promoter includes two short sequences, the -10 and -35 consensus sequences, which are recognized by the σ subunit. 2. Chain initiation: The RNA polymerase holoenzyme first binds loosely to the promoter sequence and then binds tightly to it to form the closed promoter complex. An open promoter complex is formed once approximately 18 bp of DNA around the -10 consensus sequence are unwound. The holoenzyme then initiates RNA synthesis at the +1 nucleotide of the template strand. 3. Chain elongation: The RNA-coding region is the portion of the gene that is transcribed into RNA. RNA polymerase synthesizes RNA in the 5′ → 3′ direction as it moves along the template strand of DNA. The nucleotide sequence of the RNA transcript is complementary to that of the template strand and the same as that of the coding (nontemplate) strand, except that the transcript contains U instead of T. 4. Chain termination: Most bacterial genes have a pair of inverted repeats and a polyadenine sequence located downstream of the RNA-coding region. Transcription of the inverted repeats produces an RNA transcript that folds into a stem-loop structure. Transcription of the polyadenine sequence produces a poly-U sequence in the RNA transcript, which facilitates release of the transcript from the DNA.

Which three statements correctly describe the processing that takes place before a mature mRNA exits the nucleus?

A poly-A tail (50-250 adenine nucleotides) is added to the 3' end of the pre-mRNA, a cap consisting of a modified guanine nucleotide is added to the 5' end of the pre-mRNA, noncoding sequences calledintrons are spliced out by molecular complexes called spliceosomes. Once RNA polymerase II is bound to the promoter region of a gene, transcription of the template strand begins. As transcription proceeds, three key steps occur on the RNA transcript: Early in transcription, when the growing transcript is about 20 to 40 nucleotides long, a modified guanine nucleotide is added to the 5' end of the transcript, creating a 5' cap. Introns are spliced out of the RNA transcript by spliceosomes, and the exons are joined together, producing a continuous coding region. A poly-A tail (between 50 and 250 adenine nucleotides) is added to the 3' end of the RNA transcript. Only after all these steps have taken place is the mRNA complete and capable of exiting the nucleus. Once in the cytoplasm, the mRNA can participate in translation.

What is the order of events thought to occur during eukaryotic transcription involving RNA pol II?

TFIID binds to the TATA box TFIIB, TFIIF and RNA pol II bind TFIIE and TFIIH bind Synthesis of the pre-mRNA begins at the +1 nucleotide A 5' cap is added to the pre-mRNA Spliceosome complexes carry out intron splicing A poly-A tail is added to the pre-mRNA

Suppose that a portion of double-stranded DNA in the middle of a large gene is being transcribed by an RNA polymerase. As the polymerase moves through the sequence of six bases shown in the diagram below, what is the corresponding sequence of bases in the RNA that is produced?

The coding strand reads 3'CCGAGT5', so therefore the the sequence will read 5'UGAGCC3'. There are three principles to keep in mind when predicting the sequence of the mRNA produced by transcription of a particular DNA sequence. The RNA polymerase reads the sequence of DNA bases from only one of the two strands of DNA: the template strand. The RNA polymerase reads the code from the template strand in the 3' to 5' direction and thus produces the mRNA strand in the 5' to 3' direction. In RNA, the base uracil (U) replaces the DNA base thymine (T). Thus the base-pairing rules in transcription are A→U, T→A, C→G, and G→C, where the first base is the coding base in the template strand of the DNA and the second base is the base that is added to the growing mRNA strand.

Which of the following initially determines which DNA strand is the template strand, and therefore in which direction RNA polymerase II moves along the DNA?

the specific sequence of bases along the DNA strands. In eukaryotes, binding of RNA polymerase II to DNA involves several other proteins known as transcription factors. Many of these transcription factors bind to the DNA in the promoter region (shown below in green), located at the 3' end of the sequence on the template strand. Although some transcription factors bind to both strands of the DNA, others bind specifically to only one of the strands. Transcription factors do not bind randomly to the DNA. Information about where each transcription factor binds originates in the base sequence to which each transcription factor binds. The positioning of the transcription factors in the promoter region determines how the RNA polymerase II binds to the DNA and in which direction transcription will occur.