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Some inherited diseases of the liver, including Wilson's Disease, are primarily or entirely genetically determined. Wilson's Disease results when a defect in a copper transporter in the small intestine occurs, leading to copper level disregulation in both the hepatocytes and the systemic circulatory system. Mutations have primarily been found in the copper transporter that helps load copper onto a transport protein, apoceruloplasmin, which normally creates serum-soluble ceruloplasmin with the addition of copper. Given this defect, serum studies of an individual with Wilson's Disease would likely show what kind of change in serum ceruloplasmin compared with a normal individual?
The question informs us that the mutational defect in the gene involves the enzyme's ability to load copper onto apoceruloplasmin. Healthy individuals are able to load copper to apoceruloplasmin, creating serum-soluble ceruloplasmin. With this process disrupted in an individual with Wilson's Disease, we would expect that less ceruloplasmin would be produced because copper could not be transported. We would expect to see reduced serum levels of the complete protein, and high levels of copper building up in hepatocytes and circulatory serum.
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An error occurs during DNA replication, resulting in the insertion of a base pair. Replication of the original 3' DNA strand (template) produces a mutant 3' strand (mutant), as diagrammed below:
Template
3' - AUGGCCATTTTTATA - 5'
Mutant
3'- AUGCGCCATTTTTATA - 5'
Of the answers below, which best describes the type of mutation depicted above?
The addition of a number of nucleotides that is not a multiple of three shifts the reading frame of the codons in the gene. One base pair was inserted early in the strand, thus shifting the codon reading frame +1 to the right.
Missense and nonsense mutations imply base pair substitutions, which did not occur in the diagram. Similarly, nothing was duplicated, and repeat expansion would require multiple repetitions of a short DNA sequence.
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In a eukaryotic cell, a molecule of pre-mRNA is found to have four exons and three introns. Which of the following are possible combinations of the exons, if the order in which they are written is the order in which they will be translated?
I. Exon 1, Exon 2, Exon 3, Exon 4
II. Exon 1, Exon 3, Exon 4
III. Exon 4, Exon 1, Exon 2, Exon 3
This question is asking about alternative splicing. Alternative splicing is a means by which several different proteins can arise from the same pre-mRNA due to the order in which the exons are organized. This typically takes the form of exon skipping. Therefore, both 1 and 2 are potential mature mRNAs that could arise from this pre-mRNA. Option 3 is not an acceptable transcript, however, because alternative splicing maintains the integrity of the genomic order of the exons (i.e. exon 4 will not come before exon 1, 2, or 3).
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__________ are parts of __________ molecules that do not contain information about a protein's primary structure.
After transcription, the resulting RNA molecule must undergo post-transcriptional modification before it becomes mature mRNA. Before these modifications, it is known as heteronuclear RNA (htRNA) or pre-mRNA.
Introns are portions of pre-mRNA molecules that are spliced prior to translation. Unlike exons, introns do not contain information about the structure of the protein. Only after intron splicing is the molecule considered mRNA.
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The primary transcript is much longer than the mRNA that will eventually be translated. This can be explained by which of the following?
Immediately following transcription, the primary transcript will undergo a variety of changes before being translated. One of the largest changes is that a spliceosome complex will remove introns from the primary transcript. Introns are not involved in protein creation, and their removal makes the transcript much shorter. The final mRNA transcript consists of a string of exons, a 5' cap, and a 3' poly-A tail.
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In most cases, introns are spliced out of mature messenger RNA (mRNA) and are not a part of the final translated protein product of a gene. Even though they are not included in the final protein, why are introns important?
These are all reasons that introns are important, despite the fact that they are not included in final proteins. Introns can allow for alternative splicing of exons, in which exons are placed in different orders to create different proteins from one gene. In the gene Dscam in Drosophila, alternative splicing allows for around 38,000 different proteins from one gene sequence. Some introns become non-coding RNAs that control expression of genes. Lastly, it has recently been shown that introns are involved in some special functions like mRNA export - in which mRNA's are moved between the nucleus and other cellular compartments.
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In eukaryotes, which of the following is true about introns and exons?
The primary RNA contains introns and exons because it has not been processed yet, and therefore the introns have not been spliced out. Mature mRNA contains only exons, which are the coding sequences that ultimately get translated. Intron regions are non-coding and are not included in mature transcripts. Note that post-translational modifications such as splicing only occurs in eukaryotes.
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If a gene produces a pre-RNA that is 1200 basepairs long and has the following intron-exon structure:
Exon 1 - 200 bp
Intron 1 - 100 bp
Exon 2 - 50 bp
Intron 2 - 150 bp
Exon 3 - 700 bp
How many basepairs long would we expect the mRNA to be?
This question requires you to know that preRNA contains both intronic and exonic regions, but the intronsget spliced out to produce the mRNA. Therefore, you had to subtract the total intron basepairs (250) from the total length of the preRNA (1200), which gives an mRNA length of 950 basepairs.
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Which of the following post-transcriptional modifications occur in the nucleus?
I. 5' capping
II. Splicing
III. Polyadenylation
All of the post-transcriptional modifications listed occur in the nucleus. Each is important in the process of turning pre-mRNA into mature mRNA that can successfully exit the nucleus and enter into translation. These modifications allow for the appropriate recognition by ribosomes and serve to enhance the stability of the mRNA molecule.
The 5' guanosine cap is added to one end of the RNA strand, and a poly-A tail is added to the other. These modifications serve to help with ribosome recognition and prevent degradation. Spicing involves the removal of non-coding introns from the RNA transcript, allowing for translation of the proper sequence.
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Which of the following post-translation modifications cannot be classified as lipidation?
Ubiquitination is the only option in which the modification to the protein does not include the binding of a lipid group to a protein. Rather, it is the addition of another peptide to the existing protein.
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The process of polyadenylation results in the addition of a poly-A tail to mRNA after transcription. The poly-A tail consists of approximately 150-200 adenine bases at the 3' end of the mRNA. Which of the following best describes the purpose of polyadenylation in mRNA processing?
Polyadenylation results in a long chain of adenosine monophosphate residues being added to the 3' end of a pre-mRNA as transcription is terminating. The poly-A tail provides stability to the mRNA molecule as it is transported through the cell to its ultimate location. Without this modification, the shorter mRNA would be degraded by enzymes within the cytoplasm. The other functions listed as answers are in no part dependent on the poly-A tail.
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Which of the following post-translational modifications has most typically been associated with protein degradation via activation of the cell's proteasome system?
Conjugation of ubiquitin molecules to a protein activates the ubiquitin proteasome system, which is required by cells to break down proteins into their component amino acid residues to be reallocated in protein building as necessary. While the other modifications may contribute to some degradation pathway, ubiquitin is classically considered a marker of protein destruction via the proteasome system.
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__________ is the addition of a long tail of adenine bases to an mRNA during mRNA processing, and this step is crucial for the stability of the mRNA as it gets exported to other parts of the cell.
Polyadenylation is the process in which a "Poly-a tail," or a tail of adenine bases, is added to the 3' end of an mRNA. The other processes listed do not carry out this function.
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Which of the following accurately describes the promoter?
The promoter is a specific segment of DNA that signals the starting point of transcription. RNA polymerase attaches to the promoter and proceeds to create the mRNA primary transcript.
DNA polymerase binds to the RNA primer to begin DNA replication. Ribosomes bind to the 5' cap on eukaryotic mRNA.
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The lac operon is typically found in prokaryotes in order to utilize lactose in the event that glucose is absent. How does the presence of lactose affect the lac operon?
The lac operon is set up in a way so that the lac repressor is able to be transcribed, regardless of glucose and lactose levels. The lac repressor will then attach to the operator, which inhibits transcription. If lactose is present, it will bind to the lac repressor, and make it detach from the operator.
This process allows the operon to be transcribed in the event that glucose is absent. If glucose is absent, but lactose is not present, then the repressor will remain in place and transcription will not take place.
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Which conditions would result in the largest levels of lac operon transcription?
The important thing to remember about the lac operon is that it is transcribed when glucose is absent from the cell, but lactose is present and can be utilized. As a result, the operon's transcription would be high if there are both high levels of lactose available, and very little amounts of glucose.
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Which of the following are commonly found in a eukaryotic RNA-polymerase-II-dependent promoter?
I. Shine-Delgarno sequence
II. TATA element
III. Ribosomal binding site
Of the three choices, the only element commonly found in a eukaryotic promoter is a TATA element. This is the site where TBP (TATA binding protein) binds and begins to recruit other transcriptional machinery.
The Shine-Delgarno sequence is commonly found on prokaryotic mRNA and serves as a ribosomal binding site. Because promoters are regions of DNA, both option I and II do not really apply.
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In eukaryotes, promoter sequences are regulatory elements found upstream of the transcription start site. Promoter sequences are required for transcription factors and RNA polymerase to recognize and bind to the DNA strand, thus promoting transcription of the genes on that strand and production of mRNA.
mRNA is ultimately translated into proteins, i.e. gene products. Consider a mutation in the promoter sequence that increases the affinity of RNA polymerase for the DNA strand. Compared to a sequence where the promoter sequence is wild-type, which of the outcomes below is most likely for this mutated promoter region?
The binding of RNA polymerase and transcription factors is tightly modulated by promoter elements. If affinity was increased compared to a wild-type sequence, we would expect that RNA polymerase would bind more easily to the sequence and produce more mRNA. Nothing about the nature of this mRNA is altered (since the coding sequences are unchanged); there is simply more of it, which would mean overexpression of the protein for which it codes.
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What is the region of DNA where transcription factors and RNA polymerase bind, and that is also responsible for regulating transcription?
The correct answer is promoter. The promoter is directly upstream of the start of transcription for a given gene. It is the site of transcription factor and RNA polymerase binding, and interacts with distant enhancers to regulate transcription.
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Eukaryotic transcription requires many proteins interacting in a coordinated fashion to drive the process of converting DNA to RNA. RNA polymerase, the enzyme that initiates transcription, needs a number of factors and components to being transcribing a gene. Which of the following answers is not one of these factors?
DNA polymerase is a crucial factor required for replication of DNA, but is not a component utilized in the process of transcription. The core promoter sequence, activators, and transcription factors are all needed in order for RNA polymerase to begin the process.
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