Supplemental Instruction
IowaStateUniversity / Leader: / Matt C.
Course: / Biol/Gen 313
Instructor: / Dr. Rodermel
Date: / 10/12/2017
Introduction: The chapters to be covered by this exam are 2,9, 10, and 11.
The exam is going to be a mixture of multiple choice and short answer, but these questions are tough enough that multiple choice will be a sufficient challenge. Work through my worksheets or problems out of the book for more short answer practice.
Multiple Choice
- A mutation knocks out expression of DNA gyrase in E. coli. What is the likely effect this would have?
- DNA replication would be strongly inhibited or stopped due to supercoiling.
- RNA transcription would be strongly inhibited or stopped due to supercoiling.
- DNA replication would be strongly inhibited or stopped due to the lack of ability to split hydrogen bonding between complementary bases.
- RNA transcription would be strongly inhibited or stopped due to the lack of ability to split hydrogen bonding between complementary bases.
- None of the above would happen.
- Telomerase activity helps solve which problem?
- Telomere dissociation.
- Base mismatch during DNA synthesis.
- Eukaryotic end-replication issues.
- Binding of the spindle apparatus during cell division.
- None of the above.
- A eukaryotic organism doesn’t add licensing factors during the G1 phase, but skips over the G1 checkpoint to begin DNA replication in S phase. What is likely to occur?
- The genome will be over-replicated.
- No DNA replication will occur.
- DNA replication will be completed, but with many more mutations than usual.
- Chromosomes will shorten following replication.
- None of the above.
- What is the primary function of sigma factors?
- To identify promoter regions in bacteria and assist in transcription initiation.
- To improve transcription elongation speed.
- To provide a proof-reading mechanism to the transcription holoenzyme.
- To put together the other components of the RNA polymerase complex.
- None of the above.
- Poly-adenylation accomplishes what?
- Lengthens telomere sequences.
- Creates a tail on the 3’ end of mRNAs.
- Protects the 5’ end of DNA from degradation.
- Modifies histone complexes.
- Multiple of the above are true.
- Below is an image of a DNA and mature mRNA hybrid. How many introns are contained in the DNA gene sequence?
- 2
- 3
- 4
- 5
- This cannot be determined.
- Several mutations occur in the terminator region of a bacterial gene following the inverted repeats that switches out A bases for T bases. What is the likely effect of this change?
- If the gene is rho-dependent, it won’t be transcribed.
- If the gene is rho-independent, it won’t be transcribed.
- The gene won’t be transcribed regardless of its rho-dependency.
- The gene will be transcribed, but it will continue past the terminator until it reaches the next terminator.
- Gene transcription will not be affected significantly.
- Which of the following is not an activity of DNA polymerase I?
- 5’-3’ DNA polymerase.
- 3’-5’ DNA polymerase.
- 5’-3’ exonuclease.
- 3’-5’ exonuclease.
- Multiple of the above are not activities of DNA polymerase I.
- All of the following are requirements for transcription EXCEPT:
- Priming with RNA primase.
- Sufficient NTPs.
- A DNA template.
- The RNA polymerase complex.
- All of these are requirements for transcription.
- Studies have shown that the knockout of telomerase in eukaryotic organisms leads to premature aging and death, however; telomerase isn’t synthesized at all in bacteria. What is the best explanation for this?
- Bacteria only have one chromosome.
- Bacteria have a circular chromosome.
- Bacteria only have one origin of replication.
- Bacteria don’t age regardless of telomere length.
- Bacteria actually do synthesize telomerase to deal with telomere shortening.
- Which of these proteins acts first during DNA replication at one origin?
- DNA primase.
- DNA helicase.
- DNA polymerase III.
- DNA ligase.
- Multiple of the above act at the same time at one origin.
- DNA polymerases are quite accurate with just their 5’-3’ polymerase activity, but also have a proofreading mechanism to improve that accuracy further. What is the name given to this mechanism?
- 3’-5’ polymerase activity.
- 2’-hydroxyl phosphodiester bond cleavage.
- 5’-3’ exonuclease activity.
- 3’-5’ exonuclease activity.
- None of these are involved in proofreading.
- For a certain gene, the sequence of the non-template strand is 5’-ATTGACCTG-3’. What is the mRNA sequence without processing?
- 5’-UTTGUCCTG-3’
- 5’-TAACTGGAC-3’
- 5’-AUUGACCUG-3’
- 5’-UAACUGGAC-3’
- You cannot know from this information.
Match the given terms with their representation marked below. You can reuse terms.
- Template strand.
- Coding strand.
- Promoter.
- Terminator.
- 5’- end.
- An RNA that normally associates with telomerase has sequence UAUCCCCUAUCCCC. A mutation occurs that converts the sequence to UAUCCCCUGGCCCC. What will likely happen as a result of the mutation?
- Proteins that normally associate with the telomere after synthesis won’t be able to recognize the new sequence.
- Telomerase will not be able to reattach after one round of synthesis.
- Telomerase won’t be able to bind at all. No new synthesis occurs.
- Telomerase will target the wrong DNA strand.
- None of these will happen. Telomerase activity would proceed as normal.
- A certain gene sequence in the DNA template has three exon regions with two introns in between. Alternative splicing occurs and connects exon 1 to exon 3. What happens to exon 2?
- Exon 2 will form its own mRNA and be transcribed.
- Exon 2 will attach to the end of exon 3.
- Exon 2 will be degraded with the two introns.
- Exon 2 will be degraded separately from the two introns since it doesn’t have a branch point site.
- None of the above since this splicing can’t occur.
- Consider the following amino acid. What best describes its sidechain?
- Polar; uncharged.
- Polar; charged.
- Nonpolar; uncharged.
- Nonpolar; charged.
- This isn’t an amino acid.
- Which polymerase synthesizes the majority of Okazaki fragments in eukaryotes?
- Polymerase I
- Polymerase III
- Polymerase α
- Polymerase δ
- Polymerase ε
- What type of enzyme is DNA gyrase in bacteria?
- Helicase
- Topoisomerase
- Exonuclease
- Endonuclease
- Polymerase
- A certain pre-mRNA has six exons. Exon 3 contains structural components vital for protein function and is never spliced out. How many mature mRNA products can be made via alternative splicing that retain exon 3?
- 5
- 6
- 8
- 16
- None of these.
- What is the eukaryotic analog for SSB proteins?
- ORC
- Topoisomerase
- RPA
- FEN1
- None of these.
- Which of the following is not a feature of mature mRNA in plants, but was a feature of the pre-mRNA?
- Introns
- Exons
- 5’-mG cap
- Promoter sequence
- None of the above
- Some people are born with a condition called Leber congenital amaurosis (LCA) that results in eventual blindness. What causes this disease according to the text?
- An inherited disorder in which cone cells don’t proliferate.
- Inhibited rod cell development as a result of mutagenic compounds passing the placental barrier before birth.
- An in utero bacterial infection arising from low availability of synthetic antibiotics in developing areas.
- Inheritance of a non-active gene involved in chemical modification of Vitamin A.
- None of the above.
- What is the biological function of restriction endonucleases?
- Assist in recombination.
- Destroy viruses.
- Re-sectioning of blunt ends in double-strand break repair.
- Degrade undesired RNA transcripts
- None of the above – restriction endonucleases are artificially engineered and don’t exist naturally in the biological world.
- Which of the following is a palindromic sequence?
- 5’-ATTCCG-3’
- 5’-ATTAAT-3’
- 5’-ATTTTA-3’
- 5’-ATTATT-3’
- Multiple of the above.
A short peptide is shown below. Provide the following numbers.
- Amino acids used to make the peptide.
- R-groups present.
- Peptide bonds present.
- Hydrophobic R-groups.
- Polar and uncharged R-groups.
- Charged R-groups.
- Which of the following is an example of a ribonucleoprotein?
- U1
- Telomerase
- RNA polymerase
- Poly-A polymerase
- Multiple of the above
- What can prokaryotes do that eukaryotes cannot?
- Couple transcription and translation.
- Splice mRNAs.
- Use promoter consensus sequences.
- Degrade viruses.
- Multiple of the above.