Exam IV Old Exams

Describe the events at the P and A sites of a ribosome during a round of elongation for translation in bacteria.

A peptide tRNA is bound to the codon within the P site. EF-TU (GTP) delivers the corresponding amino acyl tRNA (charged tRNA) to the site. GTP hydrolysis insures correct codon-anticoding interactions. Peptidyl transferase catalyzes the attack of the amino group of this charged tRNA onto the carbonyl C atom that is forming an ester bond and forms a new peptide bond, in effect transferring the peptide chain to the tRNA in the A site. EF-G (GTP) uses the energy of uncharged tRNA in the P site is shifted to the E site and the new peptide tRNA and its bound codon in the A site are shifted to the P site.

A number of antibiotics and toxins specifically target translation. For each antibiotic or toxin given below, the molecular target is indicated. What specific step during translation will be blocked/affected in each case. A. Chloramphenical B. Streptomycin C. Diptheria Toxin

A. Binds to and blocks 23S rRNA. Blocks peptide chain elongation by inhibition of peptidyl transferase on the large 50S subunit. B. Binds to and blocks the 16S rRNA. Prevents protein synthesis by interfering with the binding of the formyl-methionine-tRNA to the 30S subunit as the uncharged tRNA in the P site could not be shifted into the E site. C. Ribosylate eEF-2. This prevents translocation by preventing eEF2 from hydrolyzing GTP.

Describe the function and identify the position as well as the name of the sequence region needed for a. transcription initiation, b. transcription termination, c. translation initiation.

All sequences correspond to a non template strand of DNA. a. 35 and 10 sequences (TGGACAT,TATAAT) define promoters to position start of transcription at +1. b. An inverted repeat (to form a hairpin loop) followed by several T/s on the non template strand. c. Shine-Delgarno Sequence (GGAGGA) complimentary to 3' end of 16s rRNA to position small 30s ribosomal subunits.

The process of thermal denaturation of DNA is often called "melting" and is typically monitored using UV absorbance at 260 nm. Describe what happens to a DNA sample and UV absorbance as the temperature is raised from 25 C to 100 C and define what the "melting point" of a given sample of DNA describes. In doing so provide one explanation as to why the melting point of one sample of DNA would differ from that of another sample.

As the temperature is raised, the DNA will denature and transition from double strand duplex to single strands and as this happens the OD260nm will increase linearly. The melting point of the sample is the temperature at which 50% of the DNA duplexes have denatured. The GC% of the DNA is largely responsible for its Tm, as GC content rises so does the Tm due to the larger number of H bonds that have to be denatured in GC base pairs relative to AT compliments.

Discuss how eukaryotic TFIID and its function is both similar and dissimilar to a bacterial sigma factor and its function?

Both of these proteins confer promoter specificity to their respective polymerases. In bacteria sigma factor allows the holo-polymerase to recognize and bind the -10 and -35 sequences. TF2D, via its TBP subunit, binds to the TATA box and, along with other General Transcription Factors (GTFs or TFIIs), forms a binding site for polymerase II. Even though both of these proteins are promoter specificity factors, they function differently. Sigma factor is an integral subunit of RNA polymerase and thus allows the polymerase to bind and associate directly with the promoter DNA. In contrast, recognition of the promoter by TFIID is independent of polymerase, and only after it and other GTFs (TFIIs) bind is a binding site for polymerase generated.

Compare and contrast the functional roles of eukaryotic TFIID and a prokaryotic sigma factor.

Both of these proteins confer promoter specificity to their respective polymerases. In prokaryotes, there are multiple different sigma factor allows the RNA polymerase holopolymerase to recognize and bind the -10 and -35 sequences of particular sets of genes (sigma 70 works for most genes while sigma 35 works for heat shock genes). In eukaryotes, TFIID, via its TBP subunit, binds to the TATA box and along with other general transcription factors (GTFs or TFIIs) forms a binding site for RNA polymerase II. Even though both of these proteins are promoter specificity factors, they function differently. Sigma factor is an integral subunit of RNA polymerase and thus allows the polymerase to bind and associate directly with the promoter DNA. In contrast, recognition of the eukaryotic promoter by TFIID is independent of RNA polymerase, and only after it and other GTFs (TFIIs) bind is a binding site for polymerase generated.

Describe the role of adapter proteins in the activation of Ras by receptor tyrosine kinase AND the mechanism by which Ras is cycled from its active to inactive form.

Following dimerization and autophosphorylation of receptor tyrosine kinases, the adapter proteins GRB2 and Sos proteins couple the receptor to the inactive Ras-GDP complex. GRB2 has SH2 domains that recognize phosphotyrosines on the RTK and also SH3 protein-interaction domains that promote associations with Sos which also has those SH3 domains. Sos promotes dissociation of GDP from Ras, allowing GTP to bind. Sos acts as a guanine nucleotide exchange factor (GEF), which helps convert inactive Ras-GDP to active Ras-GDP. The Ras-GTP complex can then activate downstream effector molecules. Ras is cycled from its active to inactive form by the action of two proteins: guanine nucleotide exchange factor (GEF) and GTPase activating protein (GAP). When bound to GDP, Ras is in its inactive form. Binding of GEF to the Ras-GDP complex causes release of GDP and binding of GTP. Ras bound to GTP is in its active form. The active Ras-GTP complex possesses intrinsic low GTPase activity, which is elevated when GAP binds, causing conversion to inactive Ras-GDP.

Describe the general structure and function of the G-proteins that act together with seven trans-membrane G-protein couple receptors in mediating signal transduction. In your answer, please be sure to specifically address the basis for the large amount of diversity among members of this large family of proteins.

G-proteins of this type make up a large family of heterotrimeric proteins that are comprised of alpha, beta, and gamma subunits. These proteins are closely associated with the cytoplasmic face of the plasma membrane where they are able to interact with ligand activated seven transmembrane containing receptors. Prior to an interaction with the activated receptor, the G-protein is in an inactive form that has GDP bound to the G alpha subunit. As a result of interaction with the ligand activated receptor GDP is exchanged for GTP on the G alpha subunit causing it to disassociate from the G beta gamma complex. Once released, the G alpha GTP bound subunit is res to activate downstream effector proteins to produce the second messengers cAMP as well as IP3 and DAG. The broad diversity of G proteins is largely reflected in the G alpha subunits that are specific for interactions with different downstream effectors as well as show a restriction for particular signaling systems.

You have collected a sample of your own genomic DNA and would like to compare its melting point to a similarly prepared sample from a wild type strain of E coli bacteria. Would you expect these two DNA samples to differ in their melting point? Explain.

Given that DNA samples are derived from 2 divergent species (Homo sapiens/E coli) they are expected to have significant difference in their overall DNA content. The expectation is there should be significant change in GC content and therefore there should be a significant change in GC content and therefore there should be a significant difference in Tm. (FYI: Gc content in humans ~40% and E coli ~50.8%).

You have collected a sample of your own genomic DNA and would like to compare its melting point to a similarly prepared sample collected (with permission) from one of your classmates. Would you expect these two DNA samples to significantly differ in their melting point? Explain.

Given that both classmates are of the same species (Homo sapiens) they are not expected to have significant difference in their overall DNA content (humans and chimps share >98% of their DNA), the expectation is there should be no significant change in GC content between 2 humans and therefore no difference in Tm.

What would happen if the G alpha protein that interacts with the hormone glucagon receptor that normally acts via a cAMP 2nd messenger pathway were mutated so that it had lost its GTPase activity? What would happen if GTPase activity of a mutant G alpha subunit of the G protein were to have elevated GTPase activity?

If the G alpha subunit lacks GTPase activity, it will be unable to convert bound GTP to GDP and will, therefore, remain active. This means it will continue to induce the production of cAMP and the responses that cAMP triggers in the cell in question would remain active longer than under wild-type conditions, this could be a problem. If the G alpha subunit had elevated GTPase activity, it would not stay active very long. Therefore, cAMP levels would not rise as high as usual and the response to the hormone would be lessened or even non-existent.

Proteins A, B, and C are found in E. coli cells in the following relative levels: 1A: 3B: 5C. All three proteins have the same rate of degradation, and all three are encoded within a single polycistronic transcription unit (operon). Explain in details how you arrived at your rationale and how the three proteins can be present in amounts that differ from each other.

Inasmuch as these genes are encoded within a single operon, there can be no transcriptional issues and because this is in E. coli there is no differential splicing. Post translational control is ruled out since we are told that all 3 proteins have the same rate of degradation. This leaves translation control as the sole mechanism. Protein synthesis in bacteria requires the small ribosome subunit building to a start codon that is defined by a Shine-Dalgarno sequence just upstream of the start codon. The SD sequence is a consensus sequence that is complementary and antiparallel to the 3' end of the 16S rRNA of the small ribosome subunit. Since it is a consensus sequence, the SD for any given coding region (gene) may vary. The SD's for the genes referred to above would be present in the cell in equal concentrations polycistronic transcript since they are found on the same transcript. However, it one SD matched the consensus better than the others, there would be more initiation of translation for that coding region and thus more of that protein would be produced than the others. In short, the three coding regions each would have a different SD sequence and thus different efficiency of translation initiation.

Most eukaryotic mRNAs coding for protein have long stretches of adenine ribonucleotides (polyA) on the 3' end of the message. Does that mean that each gene from which these mRNAs are derived has a corresponding stretch of poly(dT) at the 5' end of the template from which the mRNA was formed? Please explain.

No, the poly A tail is not transcribed from the DNA of the gene, but instead it is added during mRNA processing by the enzyme poly A polymerase.

If a cell detects the presence of an unusual sugar, a typical response is to turn on a battery of genes that encode enzymes needed to metabolize the sugar. Please provide a single explanation as to how you expect 5 genes, all of which are needed to metabolize the sugar and which are needed in exactly the same amounts will be coordinately regulated in prokaryotes and eukaryotes.

Prokaryotes: The 5 genes are organized into a multi-cistronic transcription unit (an operon) that is coordinately controlled by a single promoter that is "turned on" by the cell in response to the need to digest this new sugar. Each transcript is translated into 5 independent proteins that together metabolize the sugar. Eukaryotes: The 5 genes are independent transcription units yet share common promoter or enhancer elements such that they bind identical sets of transcription factors (activators/repressors) that are expressed in response to the need to metabolize the new sugar and are therefore transcribed at the same time (albeit independently).

You have set up a cell free transcription system in a test tube. By mixing DNA, RNA polymerase, sigma factors, ribonucleotides and other essential substances, you are able to make mRNAs identical to those made under a variety of conditions in the cells from which these components were isolated. What would happen if you were ale to remove the sigma factor first added to the tube and replace it with another sigma factor?

Since sigma factors often can cause transcription from different genes with different promoters, the population of mRNAs made would likely be to change.

Describe what occurs when an in frame UGA codon (on the mRNA) enters the A site of a ribosome subunit during elongation of translation in prokaryotes.

Since there is no tRNA that has an anticodon that can bind to a UGA stop codon no amino acyl tRNA will be delivered to the A site. Instead release factor 2 (RF2) will recognize the UGA and work together with release factor 3-GTP which is hydrolyzed to RF3-GDP to drive the release from the ribosome. Extra credit: This alters the peptidyl transferase activity such that water can be used to hydrolyze the ester bond between the peptide and the tRNA that is sitting in the P site. This releases the now completed polypeptide chain.

Please describe the mechanistic basis for the lethality of mutations that delete one or more snoRNA genes in mammalian cells.

SnoRNA genes are vital because they give rise to non-coding snoRNAs (or snoRNPs) that are required to form the sequential processing of very large immature rRNA transcripts into their discrete rRNAs. This requires formation of complimentary duplexes at the junctions between rRNA sequences and spacer RNA that directs endonuclease cleavage at those double stranded sites. Without a critical snoRNA/RNP the cell would lack the necessary rRNAs to form ribosomes to carry out protein synthesis (translation) and die.

It makes sense that DNA, the genetic material of most organisms, should be sequestered in a specific area of the cell to protect it and prevent it from potential damage. What is an advantage of making the mRNA, the working copy of the genetic code, so much more unstable than DNA?

The instability of mRNA allows a rapid response to a changing environment by allowing the synthesis of new mRNAs while the older ones, which are no longer needed, can simply disappear because of their instability.

The mRNAs from two different transcription units (genes0 in E coli are produced at different rate with one produced twice as fast as the other (transcripts per minute) under the same conditions. Investigations show that the two genes have identical promoters, the two mRNAs are equally stable and there are no regulatory factors involved. Please provide a single plausible explanation as to how the rates of production of the two mRNAs can be different.

The two mRNAs are of different sizes, with one being 2x as long as the other. In that manner, 2x as many copies of the smaller transcript will be produced in the same amount of time.

There are a number of mechanisms a cell can use to regulate the amount of a particular protein in the cell. Very briefly describe 5 mechanisms by which a cell would use to ensure that it contains higher levels of protein A than protein B.

Transcriptional control: A is transcribed at a higher rate than B Alternative splicing: A and B are different splicing forms of the same gene, and splicing of the mRNA encoding A occurs more often than splicing of the mRNA encoding B. Differential rates of export from nucleus to cytoplasm such that more mRNA A is in cytoplasm than mRNA B. mRNA stability: mRNA encoding A is more stable. Control over level of protein stability: A Is more stable than B.

What differences, if any, occur if UAA is the codon in the A site?

UAA is a stop codon, so no charged tRNA can be delivered to the site in this case. Instead, a termination factor (TF) binds within the A site. This alters the peptide transferase activity so that water attacks the carbonyl C atom that is forming an ester bond to the tRNA in the P site. This results in hydrolysis of the ester bond and thus a release of the newly made peptide chain.

Provide a rationale for the observation that when comparing a prokaryotic to a eukaryotic cell, each with genomes containing 1000 protein coding genes, he eukaryotic cell produces a significantly more complex proteome (the entire set of proteins expressed at a given time) as compared to the prokaryotic cell?

Unlike prokaryotes where each gene encodes a single mRNA that leads to a single protein product, eukaryotic genes have introns that have to be spliced to generate messenger RNA, the splicing of introns brings with I the potential for alternative splicing which can significantly increase the number of potential messenger RNAs and resulting proteins that can be generated from a single gene.

What would the expected effect of the incorporation of 2', 3' Dideoxyadenosine as the terminal nucleotide of the tRNA?

When 2', 3' didexyadenosine is the terminal nucleotide at the 3' end of the tRNA it lacks a 3' OH group that is required for the synthetase to charge the tRNA. This would block the ability of the synthetase to complete its reaction and attach the tyrosine to the tRNA and the cell would rapidly run exhaust its tRNA stores and no longer be able to carry out protein synthesis- this would quickly kill the cell/organism.

Many antibiotics specifically target bacterial translation. For each antibiotic or toxin below, the molecular target is indicated. Please define the step during translation that will be blocked by each compound. a. Tetracycline b. Erythromycin

a. Binds to and blocks the A site of prokaryotic ribosomes. Prevent EF-Tu from delivering an amino acyl tRNA to the A site. b. Binds to blocks the E site of prokaryotic ribosomes.

Indicate/explain the following points regarding the process depicted in the diagram: a. Does it take place within a eukaryotic or prokaryotic cell? Why? b. Please identify which component in the diagram corresponds to a mature mRNA. c. Explain what bond J represents. d. Explain at which step in process depicted in the diagram removal of exons occurs. e. Indicate many amino acids will result from the translation of the mature mRNA.

a. Eukaryotic cells (presence of nucleus, mRNA splicing) b. Product G c. Bond J is a peptide bond between 2 amino acids. d. Exons are never removed. e. The mRNA will code for a protein containing 8 amino acids.

a. A hypothetical gene is known to have four exons, each 200 bp in length. At least how many introns must it have? Please explain. b. A mutation that maps 19 b upstream from the transcription start site eliminates all but basal levels of transcription of the gene. Please explain. c. The mRNA from this gene contains approximately 175 A residues at its 3' end. Would you expect to find a stretch of 175 T residues in Gene X? Please explain.

a. The gene must have at least 3 introns that interrupt its exons. there can be additional introns in either the 5' or 3' UTRs but the gene must have at least 3 introns. b. The distant mutation affects an enhancer (or can say that it prevents binding of an activator protein) that prevents any enhanced expression of the gene. c. No, the A residues are added by a Poly A polymerase.

a. Gene X has four exons, each 200 bp in length. Draw out the structure of gene X to show (and label) how many introns it has. b. The mRNA from this gene contains approximately 175 A residues at its 3' end. Would you expect to find a stretch of 175 T residues in Gene X? Please explain. c. Please explain how a mutation that maps 19 kb upstream from the transcription start site essentially eliminates transcription of the gene.

a. There are three introns: exon 1 intron 1 exon 2 intron 2 exon 3 intron 3 exon 4 b. No, the residues are added post-transcriptionally by a poly A polymerase. c. This mutation affects an enhancer or promoter element that maps upstream such that it prevents binding of an activator or other protein (transcription factor) that is required for activated transcription of gene X.

What are the substrates for tyrosyl-tRNA synthetase? What specific reaction does this enzyme catalyze?

tRNA, ATP, tyr(osine) The synthetase reaction is known as the charging of the tRNA requires the formation of an ester bond between the terminal 3' OH of the tRNA (typically an adenosine) and the carbonyl C atom of tyrosine.