Homework Assignment #2
1. (3 pts)You wish to make a restriction map of a 3.0-kb BamHI restriction fragment. You digest three samples of the fragment with EcoRI, HpaII, and a mixture of EcoRI and HpaII. You then separate the fragments by gel electrophoresis and visualize the DNA bands by staining with ethidium bromide, as seen in the figure below. From these results, prepare a restriction map that shows the relative positions of the EcoRI and HpaII recognition sites and the distances in kilobases (kb) between them.
.
EcoRI EcoRI
Answer:
HpaII
2. (3 pts)Which of the enzymes listed below produce blunt ends? (Answer: D produces blunt ends) Isoschizomers are enzymes that have the same recognition sequence but do not necessarily cleave at the same sites; and isocaudamers are enzymes that produce identical sticky ends.
List from Table 1 (on page 2) those restriction endonucleases that are isoschizomers of
A. Enzyme A - D
B. Enzyme H - none
C. Enzyme F - E
Next list from Table 1 those restriction endonucleases that are isocaudamers of :
D. Enzyme A – F, G
E. Enzyme B – E, J
F. Enzyme C - none
Table 1, Y is for any pyrimidine R is for any purine ¯ cleavage site
enzyme a: T¯ ATA / enzyme f: G¯ ATCenzyme b: G¯ GATCC / enzyme g: RG¯ ATCY
enzyme c: T¯ CGA / enzyme h: CTGCA¯ G
enzyme d: TA¯ TA / enzyme i: C¯ TCGAG
enzyme e: ¯ GATC / enzyme j: A¯ GATCT
3. (2 pts) Some naturally occurring polynucleotide sequences are palindromic; that is, they are self-complementary about an axis of symmetry. Such a sequence is
…CAATGTCCAACTCTGGACATTG
GTTACAGGTTGAGACCTGTAAC
Show how this structure might form a double hairpin, or cruciform, conformation. Indicate the center of symmetry in the sequence and the bounds of the cruciform.
A cruciform is formed as indicated below; the green lines indicate the bounds, the red spot is the center of the symmetry.
4. (2 pts) You have previously found two actin genes in your coffee cup fungus. Now, you continue your study of actin genes in this fungus with several types of experiments.
a. First, you probe a Southern blot of the the fungus DNA with a labeled probe complementary to a highly conserved region of the acting gene. You find two bands that hybridize strongly with your probe and one band that hybridizes weakly. Propose an explanation for this hybridization pattern.
Answer: You’ve ran a gel and on this gel you separated various DNA fragments. You then electrotransfer the DNA onto a nitrocellulose membrane where it is then fixed. You then add a radiolabelled DNA probe to the membrane – if your DNA probe is complementary to anything on the membrane, then you will be able to visualize this on an autoradiograph (film). So two bands hybridize strongly and you conclude that your probe is strongly bound and therefore this DNA is present in the fungus. Further, you see weak hybridization in a third band. You conclude that there are 3 isoforms of the actin gene each of which hybridizes to the conserved region of the actin gene probe but to a varied degree; 2 of the genes are well conserved actin genes while the other is far less conserved.
b. You isolate clones for each of these bands. Two correspond to the ACT1 and ACT2 genes you have already identified. The third you name ACT3. Now you prepare labeled probes specific for each individual actin gene (i.e., they will not cross-hybridize with either of the other actin genes) and use these to probe a Northern blot containing fungus mRNA. Th ACT1 probe hybridizes strongly with a 2.5 kilobase mRNA, the ACT2 probe hybridizes weakly with a 2.5 kilobase mRNA, while the ACT3 probe does not hybridize with any mRNA. Propose an explanation for this hybridization pattern. ACT1 mRNA has been highly expressed in the fungus and is present on your gel. ACT2 mRNA has been weakly expressed and is present in the fungus and is present on your gel. ACT3 mRNA is not being expressed (at least not at the timepoint at which you harvested your fungal cells).
c. You now determine the nucleotide sequence of ACT3 and compare all three actin genes. ACT1 and ACT2 are identical at 88% of the nucleotides, while ACT3 is identical to ACT1 at 43% of the nucleotides. What can you conclude about these actin genes? All 3 actin genes are in the same gene family. While ACT1 and ACT2 show a high sequence similarity, the sequence of ACT3 has significantly diverged from this conserved sequence. Over the course of evolution, the ACT3 gene has taken on mutations or perhaps rearranged and thus it functions as a pseudogene which is transcrptionally inactive and highly divergent in sequence.
Problem Set #3
1. (2 pts) Promoters for protein-coding genes in eukaryotic cells contain a basal promoter element that is recognized by RNA polymerase II and a collection of basal transcription factors (e.g., TFIID, TFIIB). However, the basal activity of the promoter by itself is very low and is invariably influenced by other specific transcription factors that bind either to the adjacent upstream promoter region (-120 to -30) or to more distant enhancer sites. Analysis of these regulatory regions indicates that they generally contain many binding sites for different transcription factors that interact with the DNA in a sequence-specific fashion.
When a single factory binding site that makes up part of such a complex regulatory region is analyzed by itself in transfection assays, the activity of the factor binding site is generally low. However, it has been found that linking together multiple copies of the same binding site often results in synergistic promoter activation. The following data indicate the responses for a 20 base pair DNA binding motif for a transcription factory known as FE. This element was cloned upstream of a basal promoter and reporter gene and then introduced into cells in culture for 48 hours:
Number of copies Reporter gene activity
0 20
1 30
2 135
3 125
4 460
What could account for the pattern of activation observed in this experiment? Answer: One DNA binding element is not sufficient for the transcription factor FE to bind, but requires two binding sites. This is indicated because there is no increase over basal transcription levels when one site is inserted, but is when there are two. This is further indicated by transcription only increasing when you have pairs of DNA binding motifs ( compare 2 versus 3 – no increase, but there is an increase with 4). This would suggest that the FE factor is a dimer and requires two of its site present for the entire dimer to be stably bound. This situation is much like the two half sites for binding of steroid receptor proteins.
In another construct with four copies of the binding site, 30 nucleotides are inserted between each pair of binding sites. This construct is inactive. Why? Answer: The FE dimer has specific spatial requirements for the distance between the two sites. If the two sites are spaced too far apart, the dimer cannot bind to both sites at the same time and thus would not be able to stably bind to DNA.
2. (3 pts) Which transcription factors contain TBP? Why are they called positioning factors? Propose a model for how all three RNA polymerases can interact with TBP.
Answer: SL-1, TFIIIB, and TFIID all contain the TATA binding protein (TBP). Each of these transcription factors determine where the start site of transcription will be and thus must help either directly or indirectly position RNA polymerase over the start site. SL-1 and TFIIIB are both known to directly contact RNA polymerases I and III, respectively, and are themselves bound near the start site of transcription. The different TAFs associated with SL-1 or TFIIIB help the different RNA polymerases to differentiate between the different transcription factors, although each contains TBP. This would mean that for SL-1 and TFIIIB that one or more of the specific TAFs interact directly with RNA polymerase I and III and thus help recruit it to DNA. This is not likely to be the case for TFIID since RNA polymerase II can bind without its TAFs being present and it is the other general transcription factors such as TFIIB and IIF that bind directly or indirectly to TBP that provide the protein bridge between TBP and RNA polymerase II.
3. (2 pts) RNA polymerase III internal promoters are more than 50 nucleotides downstream of the initiation site. How is RNA polymerase III positioned for correct initiation?
Answer: The transcription factor TFIIIC and TFIIIA bind to these internal promoter elements and by themselves do not bind to RNA polymerase III. Instead these factors bind to TFIIIB and cause it to bind to a region just upstream of what will be the start site of transcription. TFIIIB does not bind to DNA by itself thus there is no promoter element for TFIIIB, but its position on DNA is determined by where TFIIIC binds. TFIIIB alone can bind to RNA polymerase III and cause it to initiate transcription a certain distance downstream of where TFIIIB is bound.
4. (3pts) Studies of the mouse albumin gene show that an element 10 kb upstream of the gene is required for high expression.
(a) What kind of element is this? Enhancer
(b) This element contains three binding sites for the transcription factor HNF3. HNF3 binds to this element, both as naked DNA and as chromatin. Shown below is a gel mobility shift assay using substrates of naked DNA (lanes 1-4) or the same DNA on to which chromatin was assembled (lanes 5-8). The same amount of labeled DNA is used in each lane. Each lane shows the substrate incubated with a different amount of purified HNF3.
What does this result reveal about HNF3? Answer: HNF3 binds both naked DNA and chromatin. Which substrate does it prefer (compare the binding between naked DNA and nucleosome - does it take the same amount of HNF3 to bind both substrates completely?
(c) How would you determine if HNF3 causes nucleosomes to be specifically positioned?
Answer: Perform an indirect end-labeling experiment as follows:
1. Isolate chromatin and add purified HNF3.
2. Digest chromatin with micrococcal nuclease (light digestion, titrate enzyme).
3. Remove proteins.
4. Cut with a few restriction enzymes upstream and downstream of the HNF3 sites.
5. Run on gel and Southern blot.
6. Probe for albumin upstream region.
If nucleosomes are phased, you will see specific bands. If they are not phased, you will see a smear.
(d) Further experiments show that HNF3 binds DNA best when its binding site is positioned exactly in the middle of the nucleosome core. Albumin transcription is increased when the transcription factor GATA4 is present. The DNA element also contains a binding site for GATA4, positioned 42 nucleotides away from an HNF3 site. If the distance between the GATA4 and HNF3 sites is changed to 38 nucleotides or 47 nucleotides, GATA4 does not stimulate transcription. Propose an explanation (a diagram may be helpful).
Answer: Adding or deleting five nucleotides between the HNF3 and GATA4 sites rotates the GATA4 site by one half turn of the helix. Because the HNF3 site is in the middle of the nucleosome-wrapped DNA, the GATA4 sites must also be within the same nucleosome (but on the opposite side, take a look at the nucleosome crystal structure). If you’ve changed the spacing from 42 bp to 38 or 47 bp, this is a change of 5 bp. Thus, rotation of the site would place it inside (facing the histones) and thereby unavailable for binding GATA4. (Take a look at your figure of B-DNA and select a nucleotide. Now go 10 base pairs up or downstream - which face of the helix is this nucleotide located. Likewise, select a nucleotide and go 5 bp up or downstream - which face of the helix is this nucleotide located? We’re talking about rotational positioning in this case (you will discuss this on Tuesday).