Microbiology Processes Study Guide Questions, Exams of Microbiology

Microbiology Processes Study Guide Questions

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Microbiology Processes Study Guide Questions
1.What are the 4 steps of Theta Replication: The double-stranded DNA
unwinds at the origin
Produces single-stranded DNA as template
Replication fork proceeds around in circle
Eventually 2 circular molecules are produced
2.What are the 4 steps of Rolling-Circle Replication: Replication
initiated by break in one of the nucleotide strands
DNA synthesis begins at 3' end of broken strand, inner strand used as a
template, 5' end of broken strand is displaced
Cleavage releases a single-stranded linear DNA and double-stranded
circular DNA The linear DNA may circularize and serve as a template for
synthesis of a comple- mentary strand
3.What are the 5 steps of Linear Chromosome Replication: Each
chromosome contains numerous origins of replication
At each origin DNA unwinds and produces replication bubble
DNA synthesis takes place on both strands at each end of the bubble as
replication forks move outward
Forks eventually run into eachother, DNA
segments fuse 2 identical linear molecules
produced
4.Initiator protein in prokaryotic DNA replication and what it binds to and
what it does to the DNA: DnaA, it binds to oriC , causes a short stretch of
DNA to unwind
5.What binds to helicase to initiate DNA replication in prokaryotes: A
DNA primase binds to helicase
6.Why are single-stranded binding proteins necessary for DNA replication,
which protein would be disrupted without a single-stranded binding
protein-
: So that the single-stranded DNA is stabilized and is not allowed to
form hairpins, it would disrupt helicase's activity
7.What is used to synthesize DNA on the leading and lagging strands
in prokaryotic DNA replication: DNA polymerase III
8.What is used to synthesize DNA on the leading and lagging strands in
eukaryotic DNA replication: DNA polymerase epsilon is used for leading
strand DNA polymerase delta is used for lagging strand
9.When is DNA polymerase III swapped out in prokaryotic DNA replication,
why and what is it swapped for?: When DNA polymerase III reaches the
end of the 5' primer on the lagging strand it must be swapped out for
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Microbiology Processes Study Guide Questions

  1. What are the 4 steps of Theta Replication: The double-stranded DNA unwinds at the origin Produces single-stranded DNA as template Replication fork proceeds around in circle Eventually 2 circular molecules are produced
  2. What are the 4 steps of Rolling-Circle Replication: Replication initiated by break in one of the nucleotide strands DNA synthesis begins at 3' end of broken strand, inner strand used as a template, 5' end of broken strand is displaced Cleavage releases a single-stranded linear DNA and double-stranded circular DNA The linear DNA may circularize and serve as a template for synthesis of a comple- mentary strand
  3. What are the 5 steps of Linear Chromosome Replication: Each chromosome contains numerous origins of replication At each origin DNA unwinds and produces replication bubble DNA synthesis takes place on both strands at each end of the bubble as replication forks move outward Forks eventually run into eachother, DNA segments fuse 2 identical linear molecules produced
  4. Initiator protein in prokaryotic DNA replication and what it binds to and what it does to the DNA: DnaA, it binds to oriC , causes a short stretch of DNA to unwind
  5. What binds to helicase to initiate DNA replication in prokaryotes: A DNA primase binds to helicase
  6. Why are single-stranded binding proteins necessary for DNA replication, which protein would be disrupted without a single-stranded binding protein- : So that the single-stranded DNA is stabilized and is not allowed to form hairpins, it would disrupt helicase's activity
  7. What is used to synthesize DNA on the leading and lagging strands in prokaryotic DNA replication: DNA polymerase III
  8. What is used to synthesize DNA on the leading and lagging strands in eukaryotic DNA replication: DNA polymerase epsilon is used for leading strand DNA polymerase delta is used for lagging strand
  9. When is DNA polymerase III swapped out in prokaryotic DNA replication, why and what is it swapped for?: When DNA polymerase III reaches the end of the 5' primer on the lagging strand it must be swapped out for

2 / 39 DNA polymerase I, this is because DNA polymerase I has 5' -> 3' exonuclease activity so it can remove the RNA primer and re-synthesize a short tract of DNA

  1. Does helicase bind to double or single stranded DNA in prokaryotes?: Sin- gle-stranded DNA
  2. Does helicase bind to double or single stranded DNA in eukaryotes?: Dou- ble-stranded DNA
  3. When does DNA replication occur in eukaryotes?: S phase
  4. Where does DNA replication begin in eukaryotes?: At autonomously replica- tion sequences (usually A-T rich)
  5. What does DNA polymerase alpha do? Is it in eukaryotes or prokaryotes- : DNA polymerase alpha has primase activity and generates the RNA primer in eukaryotes
  6. What does DNA polymerase epsilon do? Is it in eukaryotes or prokaryotes- : DNA polymerase epsilon performs DNA synthesis on the leading strand
  7. What does DNA polymerase delta do? Is it in eukaryotes or prokaryotes: - DNA polymerase delta performs DNA synthesis on the lagging strand
  8. What are nucleosomes made up of?: 2 of H2A, H2B, H3, H4 and 1 H
  9. What are 3 things nucleosomes do?: Affect stability Packaging and accessibility of DNA
  10. What needs to happen to nucleosomes in DNA replication?: They need to be removed from the parental DNA and properly reassembled on the newly synthesized DNA, when DNA is doubled histones also need to be doubled along with their modifications
  11. What protein plays a key role in the transfer of nucleosome/histone modi- fications? What factors help with their reassembly?: A chaperone protein and chromatin assembly factors
  12. What protein helps combat the un-replicated gap in DNA replication of eukaryotes? Does it require a template: Telomerases, they do not require a template
  13. The coding region of a gene is often a single continuous unit in prokary- otes or eukaryotes?: in prokaryotes
  14. How many RNA polymerases in prokaryotic transcription?: One
  15. What is the RNA polymerase called in prokaryotic transcription?: RNA polymerase holoenzyme
  16. What is not included in the RNA polymerase holoenzyme tetrameric core?- : Sigma factor

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  1. What does the omega subunit do?: Helps to stabilize the tetrameric core
  2. What does the sigma factor do?: It binds the RNA polymerase to the tetramer- ic core and assists in the correct initiation of transcription specifically at the promoter region
  3. How is transcription initiated in prokaryotes?: The sigma factor binds to the RNA polymerase forming the holoenzyme, the sigma factor recognizes the -35 and -10 consensus sequences in the promoter region, once bound the RNA polymerase is positioned over the transcription start site and has unwound the DNA to produce a single-stranded template
  4. Is there a primer needed for initiation of synthesis of RNA: No
  5. What happens after a rNTP complementary to the base at the start site serves as the first nucleotide in the RNA molecule up until elongation? (Prokaryotic): After the first nucleotide is added, TWO phosphate groups are cleaved from each subsequent rNTP creating an RNA nucleotide that is added to the 3' end of the growing RNA molecule, the sigma factor is released as the polymerase moves beyond the promoter
  6. What happens after the sigma factor is released in transcription, up until termination?: The RNA polymerase continues to move along the 3' to 5' DNA template strand with complementary nucleotides being added to the 3' end of the RNA molecule
  7. What activity does RNA polymerase have in transcription?: Both helix un- winding, rewinding, and polymerization activity
  8. What two types of termination are there in prokaryotes?: Rho- dependent and Rho-independent
  9. What are two sequence features of Rho-dependent termination that help termination?: DNA sequence of terminator site causes polymerization to pause DNA sequence upstream of terminator encodes a stretch of RNA that is C rich and devoid of secondary structure
  10. What happens in Rho-dependent termination?: Near the end of the gene, the polymerase encounters a G rich area, encoding a stretch of RNA that is C rich and it stalls. The Rho factor binds to the Rut site on the mRNA and moves towards the 3' end, and Rho catches up to the paused polymerase. The Rho factor has helicase activity, so it unwinds the RNA-DNA hybrid releasing the mRNA from the transcription bubble and brings transcription to an end
  11. What are two common features in Rho-independent termination?: Contains inverted repeats in the DNA

5 / 39 A string of 6-9 A's follows the inverted repeats in the DNA

  1. What happens in Rho-independent termination?: The inverted repeats are transcribed into the mRNA, then the string of U's encoded from the string of A's in

7 / 39 terminated when Rat1 reaches the transcription machinery.

  1. Why doesn't the Rat1 exonuclease chew up the protein coding mRNA in RNA polymerase II termination?: Because the protein coding mRNA is protected with a 5' cap.
  2. What interrupts the coding region of eukaryotic genes?: Introns

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  1. What are the 3 main processing steps in eukaryotic nuclear pre-mRNA: - Addition of a 7-methylguanosine cap on 5' end Addition of poly A tail on 3' end Removal of introns by splicing
  2. What are three primary regions of a mature mRNA and function?: 5' UTR allows for ribosomes to bind Protein coding region, comprised of the codons that specify amino acids, begins with start codon ends with stop codon 3' UTR affects stability of mRNA and regulates its translation
  3. What are the 5 steps of adding a 7-methylguanosine 5' cap: One of three phosphate groups at 5' end is removed on the mRNA A guanine NT with its phosphate group is added Methyl groups are added to position 7 of the base on terminal guanine NT And to the 2' position of the sugar in the second and sometimes third If the base on the second nucleotide is A that base may also be methylated
  4. How is the poly A tail added? What does it do?: The pre-mRNA is cleaved downstream of consensus sequence, addition of 50-250 As added to 3' end Affects the stability of the mRNA and facilitates attachment of the ribosome and export out of the nucleus
  5. The 3 conserved sequences for introns required for RNA splicing: A 5' splice site containing junction sequence GU A 3' splice site containing junction sequence AG A intron branch point with a conserved "A" residue located upstream of 3' splice site
  6. 2 Steps of RNA splicing (not including the snRNAs): The pre-mRNA is cut at the 5' splice site, this 5' end of intron attaches to the branch point forming a lariat (5' phosphate of G NT binds with the 2' OH group of the A NT at branch point) A cut is made at the 3' splice site and simultaneously the 3' end of exon 1 and exon2 become covalently attached and intron is released as lariat and degraded
  7. What is a spliceosome And how many snRNAs does it contain?: A spliceosome is an RNA/Protein structure Contains 5 snRNAs U1, U2, U4, U5, and U
  8. 6 Steps of RNA splicing including the snRNAs in this explanation:

10 / 39 Base pairing occurs between U6 and U2, and between U6 and the 5' splice site, this holds the spliceosome together The exons are joined together by two transesterification reactions carried out by U6, and the intron is released as a lariat

  1. 2 Alternative pathways of mRNA processing are: Alternative splicing Multiple 3' Cleavage sites
  2. Describe alternative splicing: Alternative splicing uses different combinations of exons For example, 3 exons and 2 introns Can remove two introns to yield an mRNA with all 3 exons and 2 introns OR Can remove 2 introns and the 2nd exon to yield a different mRNA and 1 large intron
  3. Describe Multiple 3' Cleavage Sites: There may be multiple 3' cleavage sites on the LAST exon, may or may not produce a different protein depending if its upstream or downstream of a stop codon For example there may be 2 cleavage sites on exon 2 and cleavage could take place at 3' site 1 OR 3' site 2 Producing mRNA products of different lengths after splicing
  4. What could a shorter 3' UTR resulting from Multiple 3' Cleavage sites do?: Could alter the regulation of the mRNA as the 3' UTR site is involved in stabilizing the RNA and regulating translation
  5. Describe RNA editing using guide RNAs and why is it useful?: The unedited mRNA pairs with the guide RNA The guide RNA serves as a template for the addition, deletion, or alteration of bases, the mature mRNA is now released. For example a bunch of uracil nucleotides could be added to the mRNA sequence. This permanently modifies the mRNA but NOT the DNA. It can be useful because it give the protein a different function post transcription while using the same DNA template.
  6. At what end would an amino acid be covalently attached to in tRNA, what nucleotides are added to the 3' end for all tRNAs?: The 3' end, and 5'- CCA-3'
  7. Describe tRNA modifications: A large precursor tRNA is cleaved to produce a individual tRNA molecule (removal of extra 5' and 3'

11 / 39 sequences) A long intron must be cleaved out, (NO spliceosome required) tRNA splices itself

13 / 39 Example: siRNA bind to mRNA and inhibit translation miRNA bind to mRNA and cause it to be degraded or inhibit translation

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  1. What do Crispr RNA do: Found in prokaryotes Encoded by DNA sequences found in prokaryotic genomes Works in association with Cas9 nuclease to cleave foreign DNA that may enter a host cell (prevents incorporation of foreign DNA into host genome)
  2. What does LncRNA do (long non-coding RNA molecules): Known to func- tion in eukaryotic cells Longer than miRNAs Regulate and control gene expression at the level of transcription or translation by binding mRNA or sequestering micro-RNAs that control gene expression OR Bind and recruit proteins involved in DNA modification
  3. Often times the base in the 3rd codon position can be changed and still specify the same amino acid, what does this feature of genetic code explain?: Degeneracy in the genetic code
  4. tRNAs can be post-transcriptionally modified at the nucleotide level, if the 1st nucleotide in the anticodon was Inosine, what can it base pair with?: C, U, or A
  5. What can the third position of a codon be if the anticodon's first position is G, U, or I: If it is G, then it can be U or C If it is U then it can be A or G If it is I then it can be C, U, or A
  6. What bonds are found between amino acids: Peptide bonds
  7. What are the two steps with charging a tRNA with a amino acid: First the amino acid reacts with ATP producing aminocyl-AMP and PPi In the second step the amino acid is transferred to the tRNA and AMP is released (tRNA is bound to the C-terminal of the aa)
  8. Describe initiation in translation of prokaryotes: IF-3 binds to the small subunit preventing the large subunit from binding, allowing the small subunit to attach to the mRNA at the Shine-Dalgarno sequence near the AUG start codon A tRNA charged with fMet forms a complex with IF-2 and GTP and binds to the initiation codon while IF-1 joins the small subunit, the IF-1 and IF-2 help position fMet-tRNA over the start codon All the initiation factors dissociate and GTP is hydrolyzed to GDP The large subunit then joins the small subunit creating the 70S initiation complex, having the first fMET positioned on the first codon which is

16 / 39 After the charged tRNA is placed into the A site, GTP is cleaved to GDP and the EF-Tu complex is released EF-Ts regenerates the EF-Tu-GTP complex A peptide bond forms between the amino acids in the P and A sites and the tRNA in the P site releases its amino acid (rRNA in large subunit catalyzes the formation of the peptide bond between two aa) The ribosome moves down the mRNA to the next codon (translocation) which require EF-G and GTP The tRNA that was in the P site is now in E site and move into the cytoplasm The tRNA that now occupied the A site is in P site and the A site is now open and ready to receive another tRNA

  1. Describe termination of translation of prokaryotes: The ribosome translo- cates to a stop codon which there is no tRNA with an anticodon for that to pair in the A site When a stop codon is encountered a release factor (RF-1 or RF-2) binds to A site ** RF-1 recognizes UAG and UAA, RF-2 recognizes UAA and UGA stop codons RF-1 attaches to the A site, and RF-3 forms a complex with GTP and binds to the ribosome, this alters the activity of the peptidyl transferase releasing the polypeptide from the tRNA in the P site leading to termination GTP associated with RF-3 is hydrolyzed to GDP and the tRNA, mRNA, and release factors are released from the ribosome
  2. Describe initiation of translation in eukaryotes: eIF-3 binds to small riboso- mal subunit, preventing large subunit from binding, the ribosomal subunit attaches to the 5' terminus of the mRNA This small ribosomal subunit scans for the first AUG, (it is the Kozak sequence which influences the efficiency of which AUG in the vicinity is used to start translation) eIF-2 and GTP form a complex with a charged amino acid Met and eIF-1 joins the small subunit (eIF-2 and eIF-1 help the position the tRNA over start codon) eIF-3 dissociates allowing all the eIF factors to dissociate and GTP is hydrolyzed to GDP And the large 60S subunit joins the 40S subunit The 3' poly A tail of the mRNA interacts with the 5' cap structure via cap binding protein complex to promote translation initiation to enhance the binding of the ribosome to the 5' end of the mRNA
  3. Describe elongation of translation in eukaryotes: The Met-tRNA

17 / 39 occupies the P site of the ribosome EF-Tu and GTP and a new charged tRNA form a complex which enters the A site of the ribosome After the charged tRNA is placed into the A site, GTP is cleaved to GDP and the

19 / 39 two base pairs which alters the reading frame of the gene distal to the site of the mutation Protein sequence changes dramatically AFTER the frameshift

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  1. What are tautomeric shifts, when do they happen: Tautomeric shifts are movement of H atoms from one position in a purine or pyrimidine base to another, it occurs during DNA replication when DNA is single-stranded, can cause some spontaneous mutations
  2. In what generation of progeny would a tautomeric shift cause a mutation in, and what kind of mutation: The second-generation progeny, and results in a transition mutation
  3. What bases have the rare enol form: Thymine and guanine
  4. What bases have the rare imino form: Adenine and cytosine
  5. When does expanding nucleotide repeats happen? Describe the process- : During DNA replication For example there may be a DNA molecule with eight copies of a CAG repeat, when the two strands separate and replicate a hairpin will form on the newly synthesized strand due to the DNA strand slipping back, causing a part of the template strand to be replicated twice and increasing the number of repeats on the newly synthesized strand, the two strands of the new DNA molecule separate and the strand with extra CAG copies serves as a template for replication, the resulting DNA molecule contains additional copies of the CAG repeat
  6. Forward mutation means: wild type -> mutant type
  7. Reverse mutation means: mutant type -> wild type
  8. Missense mutation means: amino acid -> different amino acid
  9. Nonsense mutation means: sense codon -> nonsense codon (stop codon)
  10. Silent mutation means: codon -> synonymous codon (doesn't change aa sequence)
  11. Neutral mutation is a: A missense mutation in which the amino acid is changed to one of a similar chemical type (ie. glycine to alanine so no observable affect on protein function)
  12. A loss of function mutation is the result of: Mutations that cause complete or partial loss of a normal protein function (ie. cystic fibrosis)
  13. A gain of function mutation is the result of: A mutation that causes the cell to produce a protein or gene product whose function is not normally present
  14. A conditional mutation is: Expressed only under certain conditions (ie. a temperature-sensitive mutation only observed at extreme temps)
  15. A lethal mutation causes: Premature death