Biol101, Winter 2018 Name______________________________
Midterm 2
Page Total________________ 1
1. You discover a new eukaryote that has a single,
... [Show More] circular chromosome with multiple origins of replication. A fellow student
tells you that, when you sequence this organism’s genome, you will find that A) it does not have to contend with the end replication
problem, even though it uses RNA primers for DNA replication and, B) that it does not need telomerase.
Explain why your classmate is almost certainly correct. (12 pts)
First, with a circular chromosome, this organism has no DNA ends to worry about (6pts).
Second, because it has a circular chromosome, when RNA primers are removed, there will always be an adjacent DNA fragment
available to provide a primer and thus, there will be no end-replication problem (6pts).
2. You discover a mutation that abolishes the function of general transcription factor TFIID when cells are grown at 37 degrees.
That is, cells grow normally at 25 degrees, but within 5 minutes of shifting cells from 25 to 37 degrees, all TFIID is degraded and lost.
What effect will this mutation have on mRNA levels in the cell and why? (10 pts)
mRNA production will be abolished (5 pts). TFIID is a general transcription factor for RNA Pol II and is therefore required for
transcription of all RNA pol II transcribed genes (5pts).
3. Recall that the mediator is a very large, multi-protein complex. One of its subunits, Med14, directly interacts with the Gal4
transcription activator. You isolate a mutation in the Med14 gene that abolishes Gal4-Med14 binding (note that this mutation does not
affect the levels of Med14 or overall structure of Mediator). Cells that carry this mutation are healthy, but cannot grow if galactose is
their only carbon source. Why are these cells unable to use galactose? (6 pts) Be sure to explain your reasoning. (6 pts)
Because the activator cannot recruit mediator it it cannot bind it (6pts).
Mediator is a coactivator that is a bridge between transcription activators and Pol II, activators recruit mediator (bind to it), which
then recruits Pol II. A mutation that prevents an activator from binding mediator will prevent the activator from stimulating
transcription (6pts).
4. You are studying a new gene, RZA. When fused to a β-Gal reporter gene, you measure 100 units of activity. To map the cis
elements in this promoter, you construct 4 short deletion mutations (5 bp each) centered at 35 bp (deletion A), 50 bp (B), 75 bp (C)
and 120 bp (D) upstream of the transcription start site (see Figure 1 on the last page). You measure β-Gal activity in cells carrying
each of these, and measure values of 0, 100, 4 and 10 respectively for deletions A through D (see Figure 1).
a) Which cis element identified by these mutations (i.e., A-D) is most important for transcription of RZA? (6 pts) ____A_____
b) What type of proteins is most likely to bind to site A? (6 pts)__any one of:_General transcription factors/GTFs/TFIID___
5. The Gal4 protein is composed of 3 important and separable domains: a DNA binding domain, a transcription activation
domain and a dimerization domain. Under normal circumstances, mutations that inactivate Gal4’s dimerization domain abolish Gal4
function. However, when Gal4 protein is expressed at higher than normal levels, these dimerization domain mutations do not prevent
Gal4 from functioning. You then purify this mutant Gal4 protein (as well as WT Gal4 as a control) and use it in a gel mobility shift
assay with a probe that contains the Gal4 UAS (i.e., an enhancer with a Gal4 binding site; see Figure 2). This shows that the mutant
protein binds DNA less strongly. In addition, you see two DNA-protein complexes when working with low concentrations of the
mutant protein. What do these observations suggest about the function of the Gal4 dimerization domain and its role in DNA binding?
(12 pts) (Note, your answer should address the function(s) of Gal4 and its domains—you do not need to explain every experimental
result)
The Gal4 DNA dimerization domain is required for high affinity (i.e. strong DNA binding) (6pts). In the absence of dimerization,
Gal4 protein will bind as monomers (there are two monomer binding sites per Gal4 binding site), which have half the binding energy
as the dimer. Thus, at very high Gal4 protein concentration, the dimerization defective form of Gal4 can bind DNA, but a lower, more
normal concentrations this defective protein will not bind DNA (6pts) [Show Less]