Biology 4, Exam#2, February 25, 1998 (Answer Key)
see grade distribution
1. Bacterial cells have the ability to change the set of genes that
they express in response to changes in the environment (e.g -
changing food sources). This is manifested through activation
and inactivation of operons.
a. Explain why it is an advantage to the organism to be able to "turn on" only those genes that are needed at the time. 4 points
It is an advantage because it is the most efficient use of limited resources that might be available to the cell. By only synthesizing products when they are needed, the cell does not expend energy synthesizing unneeded RNA and protein
b.Explain why it is necessary to regulate expression of the Lac operon genes according to both the level of lactose and the level of glucose in the cell. 4 points
The cell prefers to use glucose as its energy source because its metabolism is geared to efficiently handle glucose. Operons like the Lac operon allow the cell to utilize other energy sources, but those sources are only useful if the cell does not have any glucose to use. Therefore, in the presence of lactose and glucose, the lac operon is not active -- the cell prefers to use the glucose. In the presence of just lactose (no glucose), the cell will be able to express the Lac operon products and metabolize the lactose.
2.The products of the Trp operon are responsible for the synthesis of the amino acid tryptophan.
a. Explain how the level of expression of this operon is normally regulated in the cell (a diagram is OK) 5 points
The Trp operon functions through the action of a repressor. When there is no Trp around, the repressor is inactive and does NOT bind to the Trp operator -- so the operon is transcribed and its products synthesize tryptophan. Once the level of tryptophan has built up, the tryptophan binds to the inactive repressor to activate it. The active repressor then binds to the Trp operator region to turn off transcription of the operon.
b.What would be the effect on regulation of the operon if there were a mutation in the operator so that it could no longer be recognized by the repressor? 3 points
This would make it impossible for the repressor to bind to the DNA so the operon would always be turned on. Levels of tryptophan would continue to increase.
c.What would be the effect on regulation of the operon if there were a mutation in the repressor gene so that it could no longer bind tryptophan? 3 points
Since the repressor could not bind Trp, it could never become functional. Therefore the operon would always be on as in part b.
3.Explain how the shift from early to middle operons is made in a T4 phage infection? 5 points
The early operon is transcribed by E. coli RNA polymerase. One of the products of the early operon prevents the E. coli sigma factor from binding to the core RNA polymerase, therefore preventing initiation at the early promoter. A different product of the early operon "replaces" sigma and binds to the core RNA polymerase in sigma's stead. This new complex will recognize the middle promoter of the T4 genome and synthesize products from the middle operon genes.
4.Prokaryotic promoters are often comprised of a single sequence
to which a control molecule can bind (e.g. a repressor). Eukaryotic
promoters, on the other hand, are much more complex, reflecting
the complexity of eukaryotic multicellular organisms.
a. Explain the basic structure/organization of a typical eukaryotic promoter (you may use a diagram) 3 points
A typical eukaryotic promoter consists of a number of sequence elements. Each of these sequence elements is actually a binding site for a transcription factor. The promoter elements could each be different, but there may be more than one copy of a given element.
b.How does this structure facilitate the precise regulation of transcription of specific genes by transcription factors in different tissues or in the same tissue under different conditions? 7 points
Each promoter sequence element can be the target for one or more transcription factors, and each transcription factor can recognize and bind to one or more kind of promoter element. The level of transcription from a promoter is determined by the number and nature of the transcription factors that are bound to that promoter. Further, different transcription factors can have different binding affinities to any given promoter element. So, in any given tissue at any given time, it is the population of transcription factors and their ratios that determine the level of transcription from any given promoter. Regulation of expression of the gene is therefore determined by the transcription factors that are available to interact with the promoter.
5.What are the major RNA products synthesized by each of the eukaryotic RNA polymerases? 6 points
RNA polymerase 1: ribosomal RNAs
RNA polymerase II: mRNAs (or pre-mRNAs)
RNA polymerase III: tRNAs
6.The transcription of eukaryotic protein encoding genes results
in mRNA precursors (pre-mRNAs). The pre-mRNAs undergo a number
of processes and modifications before becoming mature cytoplasmic
a. What is thought to be one of the function of the cap structure placed at the 5' end of the RNA? 3 points
It provides a binding site for ribosomes to be able to initiate protein synthesis.
(also - it might be involved in transport out of the nucleus)
(other answers might be acceptable)
b.How is the 3' end of the pre-mRNA generated? 3 points
The RNA polymerase II reads through the end of the gene and continues on into the DNA past the last exon of the gene. The newly synthesized RNA is then cleaved at a specific location in the RNA sequence.
c.What is the structure that is placed on the 3' end of the pre-mRNA? 3 points
7.How does the extent of methylation affect the ability of DNA to be transcribed? 4 points
Heavily methylated DNA cannot be transcribed -- it is transcriptionally dormant. No methylation (or greatly reduced methylation) along a DNA means that the DNA canbe transcribed (but might not necessarily be turned on).
8.The figure at the right shows the result of a hybridization between a mRNA and the eukaryotic genomic DNA that coded for it. Label the DNA and the RNA (2 points) and indicate intron(s) in the DNA (4 points). Be sure to precisely indicate what you are labeling. Vague arrows will be marked wrong.
9.The b-globin cluster resides on chromosome 11 in humans. Explain how the b-globin cluster of e-, d-, g-, and b-globin genes might have evolved from a single original b-globin-like gene on chromosome 11. 6 points
The original globin gene might have undergone a gene duplication, creating an exact copy. Since there were then two copies of the gene, one of the copies could have accumulated some base changes over time because the "first" gene would still function as it did before. Eventually this could lead to the "second" gene gaining some new properties while still being similar to the first gene. This process of gene duplication and divergence was repeated a number of times during evolution to lead to the current b-globin gene cluster.
10.Lambda phage is called a transducing phage.
a. What is a transducing phage? 3 points
A transducing phage is one that can transfer some DNA of the host organism from one cell to another through a normal infection process.
b.Explain how transduction works? 3 points
The transducing phage DNA is integrated into the host genome and then, upon excision from the genome, picks up a piece of the adjacent host DNA along with the phage DNA. The excised DNA, including the piece of host DNA, is then packaged into a phage and can deliver the new DNA into the next cell it infects.
11.In your lab work, you made a mistake and accidentally mixed together two different plasmid preparations -- one with an ampicillin resistance gene (AmpR gene), and the other with a tetracycline resistance gene (TetR gene). Your boss is upset and wants you to fix the problem by providing her with a bacterial culture that only contains the AmpR plasmids in the bacteria. You approach the problem by performing a transfection experiment using the mixed plasmid preparation. This creates a flask of bacteria containing one or both of these plasmids in each bacterial cell. Describe the procedure that you could use on this culture to select (grow) only those bacteria which have picked up only the AmpR containing plasmid while killing all the other bacterial cells (i.e. - those that contain TetR, or TetR + AmpR, or no plasmid DNA). 8 points
12.Explain the process of chromosome walking. (a diagram is OK) 8 points
This process involves starting with a piece of cloned DNA [A] and trying to find those sequences that were adjacent to the cloned DNA when it was in the genome of the source organism. Using restriction enzymes isolate a piece of DNA representing the "right" end of [A]. Use this DNA as a probe to rescreen the library of random fragments for that genome. New clones that hybridize with the probe, must contain the probe sequence somewhere in the new clone's DNA. Select one of the new clones[B] that have the original probe sequence near their "left" end. The DNA to the right of the probe sequence in [B] must also be to the right of the original cloned DNA [A]. Now choose a sequence near the right end of [B] and use it to screen the library again. This will lead to another clone [C], whose sequence lies still further to the right of [B] and of [A]. This process is repeated until you have walked as far as you want along the genomic DNA. A similar series of steps can be taken to walk to the left of the original sequence [A].
13.The figure at the right shows a restriction map of a genomic DNA fragment that you have cloned. Two restriction enzymes were used, BsaJI and SfcI, to digest the DNA. Their cut sites are shown in the map at the top with the exact cutting positions indicated in parentheses next to each enzyme name. Fragment lengths are indicated by the numbers below the restriction map. You perform a digest with each enzyme and with both enzymes combined and then separate the fragments by gel electrophoresis. The results are shown in the gel at the right. Fragment sizes are indicated by the scale along the left side of the gel. You perform a DNA blot of this gel and hybridize it with the cDNA for this gene. Those bands that hybridize with the cDNA are indicated with an asterisk (*) on the gel band.
a. Indicate on the restriction map which fragments in the double digest contain sequences found in the cDNA by placing an asterisk above the fragment on the map. 4 points
b. Does this result tell you anything about the intron/exon structure of this gene? If so, what does it indicate? If not, why not? 5 points
Since the cDNA hybridization occurs in two distinct locations on the genomic DNA, but does not occur in to the fragment in between those locations (1546 fragment), there must be at least two exons. We cannot tell if there are any more exons, because they might reside on a different cloned DNA fragment outside the range shown in this particular cloned DNA.
14.Explain what an RFLP is. 4 points
A Restriction Fragment Length Polymorphism is a site on genomic DNA that will produce different restriction enzyme digest patterns in different individuals. These differences can be thought of as being due to different "alleles" of a piece of DNA, each of which have a slightly different sequence resulting in different restriction site locations.
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