Download Biochemistry - Second Examination with Answers | CHEM 642 and more Exams Biochemistry in PDF only on Docsity!
Chem 642-2011, Second exam Name:_______________
1. Describe three ways in which eukaryotic mRNA is processed prior to being exported from the nucleus.
mRNA is processed in three major ways prior to being exported from the nucleus: 5' capping of
the transcript, splicing, and 3' polyadenylation. In 5' capping, a modified guanine base is added to
the 5' end of the transcript. Splicing involves the removal of introns from the transcript to
generate the mature mRNA. Finally, 3' polyadenylation of the message involves cleavage of the
RNA and the addition of numerous adenine residues at the 3' end.
2. One of the many surprises to come from the various genome projects is that the number of genes an organism has is not necessarily a good predictor of its complexity. For example, humans are clearly much more complex than nematode worms, yet only have about 50% more genes. How could splicing help resolve this paradox?
Alternative splicing can dramatically increase the number of distinct mRNAs that are encoded by a single gene. Accordingly, in the case of humans, because human genes appear to undergo alternative splicing more frequently than do genes in other organisms, it is possible that humans actually produce a substantially greater number of proteins than other organisms, despite the apparent similarity in the number of genes.
**_3. Describe a spliceosome-mediated splicing reaction, including reactants, intermediates and the products. Draw the structure of adenosine present at the branch site of a lariat, and indicate the bonds that connect it to its immediate 5' and 3' neighbors as well as to the guanine at the 5' end of the intron.
- (5%) Self-splicing introns use two distinct strategies to accomplish splicing. Group I and Group II. Describe the main difference between the two, what are the structures of the excised_**
introns in both cases? Which mechanism most closely resembles pre-mRNA splicing catalyzed by the spliceosome?
5. How do prokaryotes and eukaryotes differ with respect to the number of open reading frames that are contained in a typical mRNA? Why does this difference make sense in view of the translation initiation strategies used by each type of organism?
Eukaryotic mRNAs almost always contain a single open reading frame (i.e., they are
monocistronic), whereas prokaryotic mRNAs frequently include multiple open-reading frames
(i.e., they are polycistronic).
This difference is consistent with the different translation initiation mechanisms used by
the two types of organisms. For example, in prokaryotes, translation is initiated when the small
ribosomal subunit binds directly to a sequence just upstream of the start codon, called the
ribosome binding site. Because the ribosome can bind to this site even when it is located at an
internal position within an mRNA, the translation machinery can translate multiple open-reading
frames within a single message with equal efficiency. In eukaryotes, in contrast, the ribosome
first binds to the cap at the 5' end of the mRNA, and then scans along the mRNA until it finds a
5'-AUG-3', which it then uses as a start codon. For this reason, the 5' most AUG in an mRNA is
generally the only codon at which translation can begin, meaning that eukaryotic mRNAs can
effectively only include a single open-reading frame. The ability of prokaryotes to synthesize
multicistronic mRNAs means that related sets of proteins can be regulated in concert through the
regulation of a single promoter. In eukaryotes, this particular type of regulation is impossible, as
individual promoters only control the expression of single genes.
6. Draw a charged tRNA, detailing how the amino acid is linked to the relevant nucleotide of the tRNA. Outline the steps by which aminoacyl tRNA synthetases charge tRNAs.
c. A site containing aminoacyl tRNA, P site containing peptidyl-tRNA, E site
containing deacylated tRNA;
d. A site of large subunit empty, A site of small subunit containing anticodon end of
peptidyl-tRNA; P site of large subunit containing protein-linked end of peptidyl-tRNA, P site of
small subunit containing anticodon end of deacylated-tRNA; E site of large subunit containing
acceptor stem end of deacylated tRNA, E site of small subunit containing anticodon end of
deacylated tRNA;
e. A site containing puromycin, P site containing peptidyl-tRNA, E site containing
deacylated tRNA;
f. A site containing release factor, P site containing peptidyl tRNA, E site
containing deacylated tRNA;
g. A site empty, P, E sites containing deacylated tRNA.
8. Outline the cycle of GTP binding and hydrolysis experienced by the elongation factors EF- Tu and EF-G, describing how the properties of each factor change depending on which form of guanine nucleotide is bound.
Both EF-Tu and EF-G use the cycle of GTP binding and hydrolysis to regulate their
respective functions during translation. In its GTP-bound form, for example, EF-Tu associates
with aminoacyl tRNA and escorts it to the ribosome. When the aminoacyl-tRNA enters the A
site of the ribosome, and the correct codon-anticodon match is present, then the ribosome
activates the GTPase present within EF-Tu, causing the factor to hydrolyze the bound GTP. In its
GDP bound form, however, EF-Tu has much less affinity for aminoacyl tRNA, and so it releases
the tRNA and dissociates from the ribosome.
EF-G's properties also change depending on the form of guanine nucleotide that is has
bound. For example, EF-G only has affinity for the ribosome when bound to GTP, and binds to
the A site within the large subunit after the peptidyl transferase reaction. Once it binds to the site,
however, the ribosome stimulates the GTPase activity of the factor, provoking a conformational
change in the protein that allows it to contact the small ribosomal subunit and shift the position
of the tRNA. Following translocation, the altered ribosome has much less affinity for EF-G (now
in its GDP bound form), causing the factor to dissociate from the ribosome.
9. Which nucleotides are capable of wobble base pairing? Why is it called "wobble" base pairing, and why does it only occur at the 5' end of the anticodon?
In general, the wobble base pairs are those that can adopt a geometry similar to the usual
G:C and A:U base pairs. Purine:purine or pyrimidine:pyrimidine base pairs are impossible, for
example, because the distance between the sugars would either be too long or too short,
respectively. Wobble base pairing is called "wobble" because it is based on the ability of the
nucleotide at the 5' position of the anticodon to change its position in order to accommodate the
unconventional base pairing. It apparently only occurs at the 5' end of the anticodon because the
5' nucleotide is at the end of a string of nucleotides (that includes the rest of the anticodon as well
as the two adjacent 3' nucleotides) that are aligned and thus stabilized by stacking interactions.
Being at the end of the series of five nucleotides, the 5' nucleotide may have more freedom in its
movement than those nucleotides that are internal to the series, such as the two other anticodon
nucleotides.
10. You have the enzyme polynucleotide phosphorylase, an unlimited supply of adenosine and uracil diphosphate, and a cell extract capable of supporting translation. Describe how these materials could be used to start unraveling the genetic code. What information about codon identity could be obtained using these resources? What limitations would you encounter, and how might you get around these limitations to learn more about the genetic code?
You could therefore begin by incubating the enzyme with high concentrations of either A
or U alone, in order to synthesize poly-A or poly-U RNA. This would tell you that the codon
AAA codes for lysine (that is, the poly-A RNA would direct the production of a polylysine), and
that UUU codes for phenylalanine. You could then incubate the enzyme with high concentrations
of both of the nucleotides, allowing you to create RNA containing both A and U. The A and U in
the RNA would be randomly ordered, meaning that all possible codons that can be formed by the
two nucleotides should be present (AAA, AAU, AUA, AUU, UAA, UAU, UUA, and UUU).
Although it would still preclude a definitive identification, you could learn more about
which codons code for which amino acids by altering the relative concentrations of the two
nucleotides used to prepare the RNA. For example, if you added an excess of A relative to U,
you would produce RNAs containing an excess of the codons AAA, AAU, AUA, and UAA
relative to the other four codons. This should lead to the production of polypeptides containing a
similar excess of the amino acids lysine, asparagine, and isoleucine (these polypeptides would
13. MerR is an activator that works by an allosteric mechanism rather than by recruitment. Describe the MerR system briefly.
MerR regulates the expression of genes involved in mercury metabolism, inhibiting their expression in the absence of mercury and inducing expression in its presence. At promoters under MerR control, the –10 and –35 elements are not optimally aligned, preventing the effective binding of RNA polymerase to the promoter. In the absence of mercury, MerR binds to the promoter and locks it in this unfavorable configuration. When mercury is present, however, it binds the DNA-bound MerR, triggering a conformational change in that protein, causing the DNA to twist and bulge, and restoring the optimal alignment of the promoter elements. Thisallosteric change in the DNA allows polymerase to bind and initiate transcription.
14. Which of the following mutational changes would you predict to be most deleterious to gene function? Explain your answer. 1. Insertion of a single nucleotide near the end of the coding sequence. 2. Removal of a single nucleotide near the beginning of the coding sequence. 3. Deletion of three consecutive nucleotides in the middle of the coding sequence. 4. Deletion of four consecutive nucleotides in the middle of the coding sequence. 5. Substitution of one nucleotide for another in the middle of the coding sequence.
2 & 4 are most harmful
15 (10%). In principle, a eukaryotic cell can regulate gene expression at any step in the pathway from DNA to the active protein.
A. Place the types of control listed below at appropriate places on the diagram in the above Figure.
- mRNA degradation control
- protein activity control
- protein stability control
- RNA processing control
- nuclear export and localization control
- transcription control
- translational control
2 6->4->5->
1 , 3
B. Which of the types of control listed above are unlikely to be used in bacteria?
4, 5
16 Which of the following statements is incorrect about riboswitches?
a. They regulate gene expression at the level of transcriptional or translation through changes in RNA secondary structure. b. They are regulatory elements found in the 3'-untranslated regions of the genes that they control. c. Each riboswitch is comprised of two components: the aptamer and the expression platform. d. Conformational changes in the secondary RNA structure is induced upon binding of the small-molecule ligands such as metabolites or uncharged tRNA to the aptamer. e. Conformational changes in the secondary structure results in either termination of transcription or inhibition of translation.
17 MicroRNAs (miRNAs) and small interfering RNAs (siRNAs) are small RNAs that play an important role in eukaryotic gene silencing. What are their similarities and differences?
The various small RNAs involved in gene silencing in eukaryotes have different names depending on their origin. miRNAs is a group of RNAs encoded as precursors by genes that have specific regulatory function. These longer transcripts (between 1-10 kb in length) are known as primary (pri-) miRNAs that carry a hairpin-shaped secondary structure. In animal cells, pri- miRNAs require two cleavages, by the ribonucleases Drosha and Dicer, to generate mature and functional single strand (ss) miRNAs. siRNAs refers to similar RNAs that are made synthetically or in vivo from double strand (ds) RNA precursors, and they require cleavage by Dicer only to become the mature and functional product.
