Exam 3 Review Answers
Supplemental Instruction
IowaStateUniversity / Leader: / Matt C.
Course: / Biol/Gen 313
Instructor: / Dr. Rodermel
Date: / 11/09/2017

Introduction: The chapters to be covered by this exam are 11, 12, and 14.

Multiple Choice

  1. You are given an mRNA transcript below. Translation is just beginning off of this transcript. What will be the anticodon of the first tRNA to bind?

5’ – AAUAUGUUUGGAUAA – 3’

  1. 5’ – AUU – 3’
  2. 5’ – UUA – 3’
  3. 5’ – AUG – 3’
  4. 5’ – CAU – 3’
  5. None of the above.

This question is worded less distinctly than it ought to be. The intention was for you to find the start codon (5’-AUG-3’) and then the anticodon for that (3’-UAC-5’; the antiparallel complement of the start codon). If you read the question as “just translate this sequence, starting at the beginning,” then A. is the correct answer. Most likely, the test will just ask you to provide the anticodon for a given codon if such a question comes up.

  1. Which translation accessory factor chaperones tRNAs to the 70S ribosomal complex? (Assume the presence of GTP cofactors if necessary.)
  2. IF-1
  3. IF-2
  4. EF-Ts
  5. EF-Tu
  6. Several accessory factors chaperone tRNAs to the 70S ribosomal complex.

IF-2 chaperones the initiator tRNA to the initiation complex about the 30S ribosomal subunit, but only EF-Tu brings tRNAs to the full 70S complex (this occurs during elongation). Both IF-2 and EF-Tu require GTP in order to bind to their tRNAs. IF-1 and EF-Ts do not interact with the tRNAs. Also, you know that this is a prokaryotic system since it uses the 30S rRNA.

  1. What recognizes the Shine-Dalgarno sequence in prokaryotes at the beginning of translation initiation?
  2. 30S ribosomal subunit.
  3. 50S ribosomal subunit.
  4. IF-1.
  5. IF-2-GTP.
  6. Multiple of the above are involved in recognizing the Shine-Dalgarno sequence.

Specifically, the 16S rRNA in the 30S subunit recognizes the Shine-Dalgarno sequence on the mRNA.

  1. Translate the following sequence off of the template DNA strand to protein. Note the direction of the strand. The codon table is provided.

3’ – TATACAGATTGCGACCCACTG – 5’

5’ – AUAUGUCUAACGCUGGGUGAC – 3’

  1. Met – Ser – Asn – Ala – Gly
  2. Ile – Cys – Leu – Thr – Leu – Gly – Asp –
  3. Met – Ile – Cys – Leu – Thr – Leu – Gly – Asp –
  4. Gln – Trp – Val – Ala – Ile – Lys – Ile
  5. No peptide would be translated by this sequence.

Transcribe the template DNA to mRNA and then translate the mRNA by finding the first start codon and ending at the first stop codon.

  1. Which process within translation does not involve using energy from energy-containing molecules?
  2. Initiation.
  3. Elongation.
  4. Termination.
  5. None of the above use energy.
  6. All of the above use energy.

Initiation, elongation, and termination in translation all use the same energy-bearing molecule: GTP. IF-2, EF-Tu, EF-G, and RF-3 are the factors that perform their functions via GTP hydrolysis.

  1. A certain mRNA strand, starting with the AUG codon, has 26 nucleotides. The mRNA does have a stop codon, but the peptide encoded is as long as possible given the mRNA size. How many amino acids long is the mRNA?

  1. 4
  2. 5
  3. 6
  4. 7
  5. None of the above.

Codons contain three nucleotides so divide 26 by three and you’re left with 8 full codons (remainder of 3). Since a stop codon has to be a part of these 8 and stop codons don’t code for an amino acid, there must be only 7 codons that code for amino acids.

  1. Eukaryotic initiation involves scanning for a certain mRNA marker. What is this marker?
  2. An AUG codon.
  3. The 5’ cap.
  4. The Shine-Dalgarno sequence.
  5. The Kozak sequence.
  6. Eukaryotic initiation doesn’t involve scanning.

There is an AUG start codon in the Kozak sequence, but just have the AUG start codon is insufficient to be recognized by the ribosome. Scanning is indeed done during eukaryotic initiation.

  1. In which direction does the ribosome translocate at the end of elongation?
  2. N’ to C’ along the polypeptide.
  3. C’ to N’ along the polypeptide.
  4. 5’ to 3’ along the mRNA.
  5. 3’ to 5’ along the mRNA.
  6. None of the above.

You translate mRNAs from 5’ to 3’ because that’s the direction that ribosomes move. The ribosome needs to move 5’ to 3’ to get to the next codon in line.

  1. What catalyzes the formation of the peptide bond during elongation?
  2. The charged tRNA in the A site.
  3. A protein in the larger ribosomal subunit.
  4. An elongation factor protein.
  5. An rRNA of the larger ribosomal subunit.
  6. Peptide bond formation is spontaneous.

Specifically, the 23S rRNA in the 50S subunit (the large one).

  1. You discover a new operon in charge of modifying the sugar xylose. When xylose enters the cell, transcription of this xyl operon begins. A regulator protein involved in this operon becomes bound to the operator after binding to xylose. What is the method of regulation described here?

  1. Positive inducible
  2. Positive repressible
  3. Negative inducible
  4. Negative repressible
  5. None of the above

Xylose is set up as the allosteric effector / signal molecule in the system. Since xylose causes transcription to begin when it enters the cell, xylose must be an inducer and the system is therefore inducible. From the perspective of the regulator, the regulator becomes active when xylose binds and we just established that xylose being present means that transcription goes. This means that the regulator must be causing transcription to go (since the regulator enacts the signal of the allosteric effector). Therefore, the regulator is an activator and the control method is positive.

  1. A series of mutations occurs in a strain of E. coli that you’re studying that results in dramatically overexpressed and super-activated cAMP phosphodiesterases. These enzymes break down any cAMP that the cell can manage to make. What is the likely effect of this on the lac operon?
  2. The lac operon will be constitutively expressed.
  3. The lac operon will produce about half of the usual transcript count.
  4. The lac operon will not be transcribed.
  5. The lac operon will be unaffected.
  6. We can’t tell the effect on the lac operon from this information.

This is just a long way of saying that the cell doesn’t make cAMP. Without cAMP, CAP doesn’t get activated and therefore it can’t activate transcription of the lac operon. Therefore, the lac operon doesn’t exceed basal levels of transcription.

  1. Which of the following methods of expression regulation results in the shortest time between initiating regulation and observing a change in expression?
  2. Protein modification
  3. Histone deacetylation
  4. RNA degradation mediated by miRNA
  5. Translational control
  6. Transcriptional regulation

Don’t worry about this question. Just know from it that eukaryotes use more levels of control and the reason for why they use more levels is so they can be more specific in the timing of gene activation and the degree of gene activation.

  1. What is the function of lacI?
  2. It is the response element that is acted on to repress the lac operon.
  3. It is a molecule that binds near the lac promoter to physically impede RNA polymerase.
  4. It cleaves lactose into two monosaccharides that can then be metabolized.
  5. It binds to cAMP and represses the lac operon.
  6. None of the above.

LacI is the lac repressor protein. Answer B describes the mode of action of this repressor protein.

  1. Which of the below is cis-acting?

  1. LacP+
  2. LacIS
  3. LacOC
  4. Two of these.
  5. All of these.

Specifically, A and C are cis-acting. Cis-acting refers to a control element that only affects the operon on its own chromosome. Trans-acting refers to a control element that could affect the operon on either chromosome. Promoters and operators are cis-acting since they’re just DNA elements. Regulators (like lacI or CAP) are trans-acting since they’re protein products that can float throughout the cell.

  1. Given the following monoploid operons, which would not synthesize lactose permease under any circumstance?
  2. LacI+, LacP+, LacO+, LacZ+, LacY+
  3. LacI-, LacP+, LacO+, LacZ-, LacY+
  4. LacI+, LacP-, LacOC, LacZ+, LacY+
  5. LacIS, LacP+, LacOC, LacZ+, LacY+
  6. None of the above

No promoter, no transcription. Answer A will produce the structural genes only when lactose (allo-lactose) is present. B and D will constitutively express their active structural genes (repressor knockout in B and constitutive operator in D).

  1. Given the following partial diploid chromosome of the lac operon, what is the mode of regulation?

LacI-, LacP-, LacOC, LacZ+, LacY+

LacIS, LacP+, LacO+, LacZ+, LacY+

  1. Inducible transcription.
  2. Repressible transcription.
  3. Constitutive transcription.
  4. No transcription.
  5. This can’t be determined.

The top strand has a promoter knockout (cis-acting) and the bottom strand can be repressed by the super-repressor (IS).

  1. A mutation occurs in the CAP operator of the lac operon that makes it unrecognizable. What is the most likely effect?

  1. Inducible transcription.
  2. Repressible transcription.
  3. Constitutive transcription.
  4. No transcription.
  5. This can’t be determined.

Just like the lac repressor, CAP must have an operator since it has to bind to affect the promoter. However, if the operator can’t be recognized by CAP, then CAP doesn’t bind. If CAP doesn’t bind, then activation doesn’t happen and transcription never exceeds basal levels (effectively off).

  1. While preparing a gel for electrophoresis, the scientist accidentally adds half as much agarose as was desired. What is the likely effect?
  2. The DNA will move more slowly.
  3. The DNA will move more quickly.
  4. A higher voltage will be required for effective operation.
  5. A different ladder will have to be used for accurate results.
  6. None of the above.

If you add less agarose, then the gel is less dense and easier to move through. Think Jello before versus after hardening (less versus more agarose).

  1. A linear segment of DNA is digested with Sma1 and run on an agarose gel. Five bands are observed in the digested sample lane, one of which matches the control lane containing undigested sample. Assume that any strand that was digested was digested fully. How many restriction sites are in the segment of DNA?

  1. 2
  2. 3
  3. 4
  4. 5
  5. None of these.

As an added challenge, one of the bands in the digested lane is the undigested sample, indicating that the digestion wasn’t 100% efficient. This leaves you with the knowledge that four bands were obtained from digesting the sample. With a linear sample, this indicates three cut sites. If the sample were circular, there would have to be four cut sites (draw a line with three cuts versus a circle with four and determine how many segments there are).

  1. What is the purpose of the probe in Southern blotting?
  2. Ensure that the running buffer is sufficiently alkaline.
  3. Run current through the wetted blotting stack to pull the DNA towards the positive pole.
  4. Provide a dry, wicking material to osmotically draw buffer up through the gel and membrane.
  5. Stain target RNA sequences.
  6. None of the above.

The probe is used to highlight the DNA of interest and depending on the type of probe, you could potentially call this staining. However, a Southern blot is particular to DNA. A Northern blot is used for RNA so D is still wrong. Me being a prick here.

  1. An alien organism makes proteins out of a set of 70 amino acids. Based on similarity to our ribosomes, it’s surmised that each codon must have 3 nucleotides in it. How many bases must the organism be using in its genetic code at a minimum?

  1. 3
  2. 4
  3. 5
  4. 6
  5. This can’t be known.

With 3 bases used, there are 27 possible codons (33). With 4 bases, 64 codons possible (43, what we have). Five bases is 125 codons (53). Six bases is 216 codons (63). 64 is too few, but 125 is sufficient, so five is the minimum number of bases.

  1. Highly methylated promoters are less transcribed than un-methylated promoters. What is the leading hypothesis for why this is the case?
  2. The methylated bases cannot be read.
  3. The methylated region increases recruitment of HDAC complexes.
  4. Methylation marks the region for removal by DNA damage repair systems.
  5. Methylation causes DNA to wind more tightly around the nucleosome.
  6. None of the above.

Methylation doesn’t do anything directly. Histone de-acetylation complexes (HDAC – sorry for the improper additional “complexes”) de-acetylate histones which make them bind DNA more tightly. If it’s more tightly bound, it’s transcribed less.