If something is not clear to you, please ask me during the exam. That is why I stay in the room. Good luck.

1. What is the relationship between methylation of eukaryotic DNA and transcription of that DNA? 5 points

DNA that is methylated cannot be transcribed. Non-methylated DNA may be transcribed (but non-methylation does not mean the DNA will be transcribed)

2. In the life cycle of T7 phage, how does transcription switch from the early genes to the late genes? 4 points

The early promoter is recognized by the E. coli RNA polymerase. One of the T7 early gene products is a T7 specific RNA polymerase. This T7 RNA polymerase will recognize the "late" promoter and transcribe the late genes.

3. The figure shows a restriction map and a gel electrophoresis pattern for a gene you are studying. The arrow in the figure indicates the region of the gene that is transcribed. If you were to do a Southern blot on this cloned DNA and probe the Southern with the transcribed RNA, which bands on the gel would hybridize with the probe. Sizes are indicated along the left edge of the gel. Indicate your answer by circling the bands. Your circles can include only one band each - any circle with more than one band in it will not be counted. 7 points

When a given location on a DNA is digested with a restriction enzyme, it will be cut into fragments. Because of small differences in the sequences for different "alleles", different individuals will have different fragment sizes. Fragments of different sizes that are formed from different alleles by a restriction enzyme digestion are referred to as Restriction Fragment Length Polymorphisms.

5. Let's talk about the Lac operon (as promised). For each of the following conditions, describe how regulation of lac operon gene expression will be altered (i.e. - how will it respond to the presence or absence of lactose). If no changes in regulation would be seen state so. 3 points each

a. a mutation in the repressor gene such that the repressor does not bind lactose

If the repressor does not bind lactose, the repressor will bind to and remain bound to the operator region - the operon will always be repressed.

b. a mutation in the repressor gene such that once lactose is bound to the repressor, the lactose is never released

This will maintain the repressor in an inactive state (the lactose is permanently bound) so the operon will never be repressed - it is always on.

c. a mutation in the operator region such that lac repressor no longer recognizes the operator

If repressor cannot bind to the operator, then the operon can never be repressed - it is always on.

d. a cell that is partially diploid and contains: 1) a mutated lac repressor gene on the bacterial DNA such that the repressor does not recognize lactose, and 2) a wild type lac repressor gene on a plasmid

A repressor can diffuse throughout the cell and bind to ANY operator region on ANY DNA. Therefore, the mutant repressor will bind to both the plasmid and the E. coli operators and turn those operons off. Since this repressor cannot recognize lactose, it will be bound permanently to the operators and those operons will always be off. The wild type repressor will have no effect (it will have nothing to bind to).

e. a cell that is partially diploid and contains: 1) a mutated lac operator region on the bacterial DNA such that it cannot be recognized by the lac repressor, and 2) a wild type lac operon on a plasmid

The lac operon in the bacterial DNA will always be on since it cannot be recognized by any repressor. The wild type operon on the plasmid will be regulated normally since it has a functional operator region.

6. Explain the process of chromosome walking. 5 points

Chromosome walking enables one to start off with a particular cloned piece of genomic DNA and then isolate neighboring pieces of DNA. This is accomplished by taking a small restriction fragment from one end of the cloned DNA (let's say the "left" end) and using it as a probe to screen the library. Some of the clones that are selected by this process will contain additional DNA sequence to the "left" of the original DNA clone. This process can be repeated by screening with a fragment from the "left" end of the new clone to isolate additional DNA - further to the "left".

7. Lambda phage can infect by either of two methods. The lytic infection resembles that for T4 phage, producing many progeny phage. The lysogenic infection results in a prophage that can later be "triggered" to remove itself from the host DNA and start a lytic infection cycle. Explain how a lysogenic infection might lead to gene transfer between different E. coli cells (transduction). 8 points

In the lysogenic infection, the lambda DNA is inserted into the host E. coli genome, where it remains dormant. However, physiological conditions can trigger this prophage to remove itself from the E. coli DNA and then initiate a lytic infection. It sometimes happens that when the prophage DNA is removed from the E. coli DNA some extra E. coli DNA is removed along with the phage DNA. This extra E. coli DNA will then be packaged into a phage. Any new cell that is infected by this particular phage will therefore have an additional copy of this particular fragment of E. coli DNA from the original infected cell - thus transferring the DNA from one individual to another.

8. Explain the experiment done in frogs which illustrated that no information is lost during development and that tissue differences are the result of differences in gene expression, not in gene content (i.e. totipotency of eukaryotic nuclei). 8 points

Nuclei were gently removed from intestinal cells and injected into an egg cell that had its own nucleus inactivated. After triggering development, a number of the eggs containing the intestinal cell nuclei developed normally into mature individuals. This indicates that the intestinal cell nuclei contain all the information necessary to guide development of the entire organism - no information has been lost from the nuclei during development of the intestine.

9. Expression libraries can be used to synthesize products from those DNA segments cloned into each vector. Explain how you might go about screening an expression library of human liver mRNA sequences can be screened for transcription factor genes that bind to a particular DNA sequence. 6 points

We are interested in a transcription factor (TF) that interacts with a particular DNA (let's called it DNAx). We can take advantage of this specific interaction (TF-DNAx) to screen the expression vector library. Basically, we are looking for bacterial cells that are expressing the TF we are interested in (they contain the TF gene which is being transcribed and translated). This can be done easily. First, grow the library on Petri dishes, transfer colonies to a filter, lyse the cells, and immobilize the proteins onto the filter. Next make DNAx radioactive and let it interact with the filter (note: this is not hybridization - you are looking for an interaction between the TF and DNAx). Wherever the DNA sticks to the filter, the TF must be present. The cells that are making this TF must contain a cDNA for the TF gene.

10. RNA polymerase II in eukaryotes does not have a "traditional" transcriptional stop signal, yet discrete sized pre-mRNAs are produced. How is this accomplished? 6 points

The RNA polymerase II continues transcription beyond the point that ultimately becomes the end of the mRNA. The RNA polymerase eventually falls of the DNA at an ill-defined location. The long RNA transcript then gets cleaved at a specific location near the 3' end to produce the discrete sized pre-mRNA.

11. Explain why polycistronic eukaryotic mRNAs can not work. 4 points

The initiation of protein synthesis from eukaryotic mRNAs requires that the ribosomes recognize the cap structure at the 5' end of the mRNA. Once the ribosome has bound to the cap it will start protein synthesis at the nearby AUG codon. Translation from a downstream coding region, therefore, is not possible because the ribosome does not have any cap structure to use for starting translation.

12. Enhancers play an important role in eukaryotic gene regulation.

a. What is an enhancer? 3 points

An enhancer is a region of DNA that contains binding sites for transcription factors (TFs) but is not part of a promoter. Binding of TFs to an enhancer can stimulate nearby promoters to much higher levels of transcription initiation than is possible without the enhancer (and its associated TFs).

b. How can an enhancer act at a distance? 3 points

It is believed the TFs which are bound to the enhancer can interact directly with TFs bound to the promoter by allowing the DNA to bend and loop out between these two locations. Thus, the distance between the enhancer and the promoter is not important because the TFs do not have to "reach" (diffuse) along the DNA to find each other - the DNA simply loops out to allow the TFs to interact.

13. The figure at the right shows the results of an experiment you perform to determine the location of a promoter. The horizontal line indicates a gene, whose transcript starts at position 1. A series of constructs containing different deletions are created using genetic engineering techniques. The blank (missing) segments of the line indicate the missing segments of the DNA (for example, construct `b' is missing nucleotides from -400 to -200). You put each of these DNAs into a cell and determine if transcription occurs from this gene. What is the range of positions that contains the promoter for this gene? Explain your answer. 6 points

Construct `a' works, so the DNA from -400 to -300 does not contain any part of the promoter. Likewise, construct `f' works so the DNA from -100 to 1 is not part of the promoter. Constructs `b' through `e' do not work. This indicates that the DNA from -300 to -100 is needed for initiation of transcription. The promoter must therefore be contained in the region between -300 and -100.

14. How can RNA splicing be used to produce different mRNAs from the same DNA which code for different proteins? 6 points

This can happen through the mechanism of alternative splicing. Let's say a gene consists of four exons: A--B--C--D. The pre-mRNA for this gene therefore will contain A--B--C--D. In some tissues RNA splicing will remove "exon" Balong with the introns to produce a final mRNA of ACD. In other tissues, "exon" Cmight be removed to produce a final mRNA of ABD. These two mRNAs code for different proteins.

15. Explain how the - and -globin gene clusters are thought to have arisen from a single ancestral globin gene. 8 points

An ancestral globin gene became duplicated and translocated to produce two copies of the globin gene, each on a different chromosome. One (or both) of these copies accumulated some mutations so that they produced slightly different proteins. One of these might have been the ancestor of the -globin cluster and the other the ancestor of the -globin cluster. Additional gene duplications occurred (on each of the chromosomes) to produce a cluster of related sequences on each chromosome. Each copy of the gene could also undergo some base changes. This process led to two clusters of genes, each having been derived from a common ancestor (one of the two original genes that arose from the original duplication event).