Download Problem Set II Answer Key

 Names: Cell Biology; Zool 3200 Problem Set Exam 2 2/18/16 Given the DNA sequence below, which of the following RNA molecules could be transcribed? (2 points)
5′-ATAGGCATTCGATCCGGATAGCAT-3′
3′-TATCCGTAAGCTAGGCCTATCGTA-5′
a.) 5′‐UAUCCGUAAGCUAGGCCUAUGCUA‐3′ b.) 5′‐AUAGGCAUUCGAUCCGGAUAGCAU‐3′ c.) 5′‐UACGAUAGGCCUAGCUUACGGAUA‐3′ d.) 5’‐AUGCUAUCCGGAUCGAAUGCCUAU‐3’ e.) none of the above Indicate with an arrow (e.g., or ) the approximate transcription start site. Be sure to place the arrow on the strand that will be READ. (2 points) ‐35
‐10
You are interested in examining the regulation of the operon that encodes genes involved in the catabolism of trehalose, a disaccharide formed of two glucose units. Trehalose is not preferentially metabolized by bacteria that prefer instead to use glucose if it is available. The gene for Tre‐ase, an enzyme important in metabolizing trehalose into glucose, is contained within this bacterial operon. It is known that two DNA binding proteins, TreA and TreB, are important for binding to the promoter of the Tre‐ase gene and are involved in regulating the transcription of the Tre‐ase gene. TreA binds to the “A” site in the promoter region, and TreB binds to the “B” site. You make mutations in the TreA and TreB genes to create cells lacking these genes (and therefore lacking these proteins), observe what happens to transcription from the Tre operon, and obtain the results in the table on the back of this page. 1 + glucose only + trehalose only + glucose and trehalose Normal cells Off On On‐low level Tre A deficient cells Off On‐low level On‐low level Tre B deficient cells On‐low level On On‐low level Cells deficient in both Tre A and Tre B On‐low level On‐low level On‐low level What is/are the roles for both TreA and for TreB in controlling Tre‐ase expression? Explain how you were able to discern this. From these data above, what do you predict will happen to Tre‐ase transcription (compared with that in normal cells) in the presence of trehalose if you were to create a version of the TreA protein that will constitutively (i.e., always automatically) bind to the “A” element (the site to which A binds) in the Tre‐ase promoter? Explain. (6 points) Tre A is an activator and Tre B is a repressor. This is able to be determined because in the absence of trehalose (glucose only), the operon should be off with the repressor bound. When tre B is absent, the operon is on, so tre B must be a repressor. A is an activator because in the presence of trehalose only (no glucose), we would expect high expression of the operon due to activator activity. But in the absence of tre A, transcription is only at a low level. Thus, tre A is an activator. In the presence of trehalose, the repressor would be off so if tre A were always activating transcription (even in the presence of glucose), transcription of this operon would always be high. From the sequencing of the human genome, we believe that there are approximately 30,000 protein‐coding genes in the genome. Of these, an estimated 1500 to 3000 are transcription factors. If every gene, including these transcription factors and the other 90‐95% of genes, has a tissue‐specific and signal‐dependent transcription pattern, how can such a small number of transcriptional regulatory proteins generate a much larger set of transcriptional patterns? (4 points) Each transcription factor works in combination with others. With only three transcription factors, eight different combinations can be generated (23) so theoretically 8 different cell types can be produced. So with 1500‐3000 (5‐10% of our number of genes), we theoretically have the ability to generate 21500‐22000 different cell types with the different combinations of transcription factors. This is way more cell types than are necessary to comprise the human body! 2