Exam 2 Review
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

  1. A mutation knocks out expression of DNA gyrase in E. coli. What is the likely effect this would have?
  2. DNA replication would be strongly inhibited or stopped due to supercoiling.
  3. RNA transcription would be strongly inhibited or stopped due to supercoiling.
  4. DNA replication would be strongly inhibited or stopped due to the lack of ability to split hydrogen bonding between complementary bases.
  5. RNA transcription would be strongly inhibited or stopped due to the lack of ability to split hydrogen bonding between complementary bases.
  6. None of the above would happen.

The function of gyrase is to relieve the supercoiling induced by the activity of DnaB – the bacterial helicase. Without gyrase, the replication bubble can’t open more.

  1. Telomerase activity helps solve which problem?
  2. Telomere dissociation.
  3. Base mismatch during DNA synthesis.
  4. Eukaryotic end-replication issues.
  5. Binding of the spindle apparatus during cell division.
  6. None of the above.

This is the issue where the lagging strand can’t get synthesized at the end of a linear chromosome, so a 3’ overhang is left. Without telomerase, the end gets degraded and the chromosome would shorten until it causes damage to the organism. Telomerase lengthens the end of the chromosomes – the telomeres.

  1. 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?
  2. The genome will be over-replicated.
  3. No DNA replication will occur.
  4. DNA replication will be completed, but with many more mutations than usual.
  5. Chromosomes will shorten following replication.
  6. None of the above.

ORC needs licensing factor to recognize origins of replication so without them, ORC can’t bind either and no replication occurs. Therefore, it’s a bit of a misnomer to say the cell has entered S phase.

  1. What is the primary function of sigma factors?
  2. To identify promoter regions in bacteria and assist in transcription initiation.
  3. To improve transcription elongation speed.
  4. To provide a proof-reading mechanism to the transcription holoenzyme.
  5. To put together the other components of the RNA polymerase complex.
  6. None of the above.

Sigma factor recognizes the -35 and -10 consensus sequences in bacteria (the promoter) and binds. From there, the RNA polymerase core complex binds to sigma and attaches to the DNA strand to begin synthesis. After that, sigma’s job is done and the core complex releases off to begin transcription.

  1. Poly-adenylation accomplishes what?
  2. Lengthens telomere sequences.
  3. Creates a tail on the 3’ end of mRNAs.
  4. Protects the 5’ end of DNA from degradation.
  5. Modifies histone complexes.
  6. Multiple of the above are true.

Specifically, the poly-A tail. This structure helps keep the mRNA from being degraded from the 3’ end.

  1. Below is an image of a DNA and mature mRNA hybrid. How many introns are contained in the DNA gene sequence?
  1. 2
  2. 3
  3. 4
  4. 5
  5. This cannot be determined.

Where the mRNA and the DNA are annealing is where their sequences match. These regions are the exons. The DNA has extra nucleotides where introns are since the introns are removed in mRNA. So the loop structures represent the introns that were removed in the mRNA.

  1. 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?
  2. If the gene is rho-dependent, it won’t be transcribed.
  3. If the gene is rho-independent, it won’t be transcribed.
  4. The gene won’t be transcribed regardless of its rho-dependency.
  5. The gene will be transcribed, but it will continue past the terminator until it reaches the next terminator.
  6. Gene transcription will not be affected significantly.

Rho-dependent termination just need the Rut site – upstream from the repeats – and the repeats themselves. Anything downstream from the repeats has no impact. Rho-independent termination needs the inverted repeats and then normally has a region of U-A pairing since the mechanism of Rho-independent termination relies on the weak U-A bonds. However, if some of the As in the DNA are switched to Ts, then there will still be equivalently weak A-T bonds between the mRNA transcript and DNA, allowing Rho-independent termination to proceed in the same manner.

  1. Which of the following is not an activity of DNA polymerase I?
  2. 5’-3’ DNA polymerase.
  3. 3’-5’ DNA polymerase.
  4. 5’-3’ exonuclease.
  5. 3’-5’ exonuclease.
  6. Multiple of the above are not activities of DNA polymerase I.

5’-3’ poly is the DNA-synthesizing activity, 5’-3’ exo removes RNA primers, and 3’-5’ exo is proofreading. 3’-5’ polymerization can’t happen.

  1. All of the following are requirements for transcription EXCEPT:
  2. Priming with RNA primase.
  3. Sufficient NTPs.
  4. A DNA template.
  5. The RNA polymerase complex.
  6. All of these are requirements for transcription.

DNA polymerases need primers, but RNA polymerases don’t. That’s how primase – an RNA polymerase itself – can make an RNA primer with the free 3’-OH. Transcription, then, doesn’t need a primer since the RNA polymerase can start by itself.

  1. 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?
  2. Bacteria only have one chromosome.
  3. Bacteria have a circular chromosome.
  4. Bacteria only have one origin of replication.
  5. Bacteria don’t age regardless of telomere length.
  6. Bacteria actually do synthesize telomerase to deal with telomere shortening.

No DNA ends; no end replication problem.

  1. Which of these proteins acts first during DNA replication at one origin?

  1. DNA primase.
  2. DNA helicase.
  3. DNA polymerase III.
  4. DNA ligase.
  5. Multiple of the above act at the same time at one origin.

The replication bubble has to be opened more in order for any of the other enzymes to do their jobs. In particular the order of these four enzymes will be helicase – primase – pol III – ligase.

  1. 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?
  2. 3’-5’ polymerase activity.
  3. 2’-hydroxyl phosphodiester bond cleavage.
  4. 5’-3’ exonuclease activity.
  5. 3’-5’ exonuclease activity.
  6. None of these are involved in proofreading.

As mentioned before, this is the proofreading activity since it regresses in the opposite direction of synthesis to fix mismatches that had just been made.

  1. For a certain gene, the sequence of the non-template strand is 5’-ATTGACCTG-3’. What is the mRNA sequence without processing?

  1. 5’-UTTGUCCTG-3’
  2. 5’-TAACTGGAC-3’
  3. 5’-AUUGACCUG-3’
  4. 5’-UAACUGGAC-3’
  5. You cannot know from this information.

The non-template strand is also called the coding strand since its sequence matches that of the RNA’s (switching Ts for Us). So for this question, just switch Ts for Us to get the RNA.

Match the given terms with their representation marked below. You can reuse terms.

  1. Template strand.C.
  2. Coding strand.B.
  3. Promoter.A.
  4. Terminator.E.
  5. 5’- end.G.

By convention, the upper leftmost end is 5’. Use this and the up-/downstream marker to orient yourself.

  1. An RNA that normally associates with telomerase has sequence 3’-UAUCCCCUAUCCCC-5’. A mutation occurs that converts the sequence to 3’-UAUCCCCUGGCCCC-5’. What will likely happen as a result of the mutation?
  2. Proteins that normally associate with the telomere after synthesis won’t be able to recognize the new sequence.
  3. Telomerase will not be able to reattach after one round of synthesis.
  4. Telomerase won’t be able to bind at all. No new synthesis occurs.
  5. Telomerase will target the wrong DNA strand.
  6. None of these will happen. Telomerase activity would proceed as normal.

Note: the 3’-5’ directionality of the RNA strand was added here and would’ve been mentioned as we got to it in the session. As is, I forgot to add that change to the review sheet I sent out. There is an argument for C. in the absence of this change.

Telomerase is a ribonucleoprotein whose catalytic activity is that of a reverse transcriptase. This means that telomerase has its own RNA and can use that RNA as a template to synthesize DNA. This functions to solve the end replication problem because the 3’-overhang of the telomere has a repeating pattern complementary to that of telomerase’s RNA (in this case it would be ATAGGGG). The first repeat of the telomerase RNA pairs with the final repeat of the DNA telomere. Then, telomerase adds onto the 3’-overhang, using the rest of its RNA as the template – so its creating more of the pattern repeats. Telomerase then detaches, moves down, and begins the process again to continue copying the telomere pattern. In this case, the RNA no longer has a repeating pattern. Telomerase can associate once since it will match up one time, but will not be able to attach again since the new DNA doesn’t match the first part of the sequence anymore.

  1. 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?
  2. Exon 2 will form its own mRNA and be transcribed.
  3. Exon 2 will attach to the end of exon 3.
  4. Exon 2 will be degraded with the two introns.
  5. Exon 2 will be degraded separately from the two introns since it doesn’t have a branch point site.
  6. None of the above since this splicing can’t occur.

Exon 2 is no longer an exon at this point, just another portion of the single, large intron that was removed and subsequently degraded.

  1. Consider the following amino acid. What best describes its sidechain?
  1. Polar; uncharged.
  2. Polar; charged.
  3. Nonpolar; uncharged.
  4. Nonpolar; charged.
  5. This isn’t an amino acid.

The R-group is bracketed. An easy way to distinguish R-groups (that happens to be functional in this class, but is loose with chemistry) is this: first, identify what is the R-group. That’s important to do and is based on knowing where the central (α) carbon is (identify the carbon attached to an amine group and a carboxyl group). Second: see if there is a formal charge (a + or a -) on the R-group. If there is, it’s polar and charged. Third: lacking a charge, check to see if there’s an -OH, an -SH, or an -NH2 on the R-group. If any of these are present, it’s polar and uncharged. Fourth: if neither of the other standards are met, it’s nonpolar.

In this example, we have the NH2 to indicate that the group is charged.

Note: charges on the other groups attached to the central carbon don’t matter here.

  1. Which polymerase synthesizes the majority of Okazaki fragments in eukaryotes?

  1. Polymerase I
  2. Polymerase III
  3. Polymerase α
  4. Polymerase δ
  5. Polymerase ε

Okazaki fragments make up the lagging strand and Pol δ is the lagging strand polymerase so it makes the Okazaki fragments. Note: the lagging strand is the strand formed from the ligation of many Okazaki fragments. Okazaki fragments are the several, smaller chunks of DNA synthesized discontinuously. Okazaki junctions are the gaps between adjacent fragments.

  1. What type of enzyme is DNA gyrase in bacteria?

  1. Helicase
  2. Topoisomerase
  3. Exonuclease
  4. Endonuclease
  5. Polymerase

Note: in eukaryotes, the analog for DNA gyrase is just a collection of topoisomerases.

  1. 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?

  1. 5
  2. 6
  3. 8
  4. 16
  5. None of these.

One method to solve this question is to write out all of the possible mature mRNAs with exon 3 in them, as such: 1,2,3,4,5,6 – 1,3,4,5,6 – 1,2,3,5,6 – 1,2,3,4,6 – 1,3,5,6 – 1,2,3,6 – 1,3,4,6 – 1,3,6. Alternatively, you can use a mathematical approach that goes as follows (explained generally first):

With splicing it must be true that the first and last exons are always retained following splicing since there is no 5’-splice site in front of the first exon nor a 3’-splice site behind the last exon. Those two are always present. However, every exon in between has introns to either side and therefore has a 5’-splice site somewhere in the 5’ direction from it and a 3’-splice site in the 3’ direction from it. So every exon in the middle may be removed by alternative splicing or may be left behind by alternative splicing. So for every middle exon, there are two options: stay or go. Since there are two options for every exon in the middle, you can multiply those numbers of options (2) by the number of exons in the middle. This is alternatively written as 2#Exons - 2. Convince yourself of this if you’d like. This means that – normally – if we have 6 exons, then we have 4 middle exons and thus, 24 = 16 different mature mRNA options. However, in this case, we’re forcibly retaining exon 3, so the only option that is left for that one is: stay. So we can change out math to 23·1 = 8 mature mRNAs.

Doing this higher level manipulation isn’t very important, but be very prepared to compute the number of possible mature mRNAs from a given number of exons or introns and explain how this happens.

  1. What is the eukaryotic analog for SSB proteins?

  1. ORC
  2. Topoisomerase
  3. RPA
  4. FEN1
  5. None of these.

SSBs are single-stranded binding proteins. The name in eukaryotes is RPAs, though they perform the same function.

  1. Which of the following is not a feature of mature mRNA in plants, but was a feature of the pre-mRNA?

  1. Introns
  2. Exons
  3. 5’-mG cap
  4. Promoter sequence
  5. None of the above

There are no introns in mature mRNA (plants are eukaryotes, too), but introns are present in pre-mRNA. Exons are present in both and the 5’-cap is only present in mature mRNA. The promoter sequence is never in mRNA.

  1. 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?
  2. An inherited disorder in which cone cells don’t proliferate.
  3. Inhibited rod cell development as a result of mutagenic compounds passing the placental barrier before birth.
  4. An in utero bacterial infection arising from low availability of synthetic antibiotics in developing areas.
  5. Inheritance of a non-active gene involved in chemical modification of Vitamin A.
  6. None of the above.

This is from the assigned reading of Chapter 14. Read it; it can be on the test. It’s also short. The problem in LCA is that the protein that helps convert Vitamin A to rhodopsin – the pigment that allows the receptor cells in your eyes to work – is nonfunctional so no rhodopsin is made.

Read the section.

  1. What is the biological function of restriction endonucleases?
  2. Assist in recombination.
  3. Destroy viruses.
  4. Re-sectioning of blunt ends in double-strand break repair.
  5. Degrade undesired RNA transcripts
  6. None of the above – restriction endonucleases are artificially engineered and don’t exist naturally in the biological world.

Restriction endonuclease – restriction enzyme. Know what these do! Ask me if you don’t understand how the book explains it.

  1. Which of the following is a palindromic sequence?
  2. 5’-ATTCCG-3’
  3. 5’-ATTAAT-3’
  4. 5’-ATTTTA-3’
  5. 5’-ATTATT-3’
  6. Multiple of the above.

A palindrome in DNA is a sequence that is the same as its complement. If you write the complement for B you get 3’-TAATTA-5’, but if you read it in the same direction, it reads 5’-ATTAAT-3’ just like the original. This is a palindrome in DNA. Don’t get it confused with an English palindrome since they are not the same.

A short peptide is shown below. Provide the following numbers.

  1. Amino acids used to make the peptide.4 – Blue boxes.
  2. R-groups present.Also 4 – Same as the number of amino acids. Brackets.
  3. Peptide bonds present.3 – Green circles.
  4. Hydrophobic R-groups.3 – Same as the number of nonpolar R-groups since nonpolar molecules are hydrophobic. These will normally be towards the inside of the protein. First three amino acid R-groups.
  5. Polar and uncharged R-groups.0 – Question 14 has an example of one, though.
  6. Charged R-groups.1 – Last amino acid R-group.

Know how to be able to do the reverse of this as well – be given two or more amino acids and make the resulting peptide.

  1. Which of the following is an example of a ribonucleoprotein?
  2. U1
  3. Telomerase
  4. RNA polymerase
  5. Poly-A polymerase
  6. Multiple of the above

U1 – a subunit of the spliceosome – and telomerase are ribonucleoproteins; proteins that contain RNAs.

  1. What can prokaryotes do that eukaryotes cannot?
  2. Couple transcription and translation.
  3. Splice mRNAs.
  4. Use promoter consensus sequences.
  5. Degrade viruses.
  6. Multiple of the above.

They can do this specifically because they don’t have a nucleus. Coupling means the two can happen at the same time.