Week 8 Flashcards

1
Q

Why do proteins often bind to the major groove of DNA rather than the minor groove?
a) The major groove has more space for molecular interactions
b) The major groove contains more DNA bases
c) The minor groove is too rigid for protein binding
d) The major groove is negatively charged

A

The major groove has more space for molecular interactions and to figure out where protein synthesis starts

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2
Q

Which is responsible for making proteins such as enzymes, transporters, and structural proteins?
A. RRNA
B. MRNA
C. TRNA
D. NcRNA

A

Messenger RNA

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3
Q

How does DNA supercoiling affect transcription?
a) Negative supercoiling facilitates strand separation, aiding transcription
b) Positive supercoiling prevents RNA polymerase from binding
c) Supercoiling only affects replication, not transcription
d) Supercoiling makes RNA polymerase bind more strongly to DNA

A

Negative strands facilitate strand separation by reducing tension in order to carry ou transcription, translation and replication

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4
Q

What are the requirements to make supercoiled dna?

A
  • dna gyrase (a topoisomerase)
  • Energy from atp
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5
Q

What does it mean for DNA to be semi conservative?

A

During replication, the new dna strand has a copy of the old strand

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6
Q

If a mutation occurs in DnaA, what is the most likely effect on replication?
a) The origin of replication will not be recognized, preventing initiation
b) DNA strands will separate, but polymerization won’t begin
c) Helicase will compensate for the lack of DnaA function
d) Replication will start but proceed at a much slower rate

A

DnaA binds to oriC (the origin of replication)

A mutation in DnaA will cause the origin of replication to not be recognized

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7
Q

What is the role of DNA polymerase, and why does it require a primer?

A

Dna polymerase catalyzes synthesis of a complementary strand of DNA

It needs a primer because it can elongate a nucleic acid. It CANNOT form a nucleic acid out of nothing

dna helicase unwinds DNA for replication

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8
Q

Why is the 3’-OH end important for forming phosphodiester bonds during DNA replication?

a) It provides a reactive hydroxyl group for the incoming nucleotide’s phosphate to form a bond.
b) It ensures DNA polymerase recognizes the correct nucleotide for base pairing.
c) It prevents premature termination of DNA synthesis by stabilizing the DNA strand.
d) It allows DNA polymerase to proofread and correct errors in the newly synthesized strand.

A

The 3’-OH group of the growing DNA strand performs a nucleophilic attack on the 5’-phosphate of the incoming dNTP (deoxynucleotide triphosphate), allowing the formation of a phosphodiester bond and the release of pyrophosphate (PPi), which provides energy for polymerization. Without the 3’-OH, DNA polymerase cannot add nucleotides, which is why RNA primers (which have a 3’-OH) are necessary to start DNA synthesis

basically, the 3-OH’ end helps DNA polymerase add on more nucleotides

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9
Q

Why does lagging strand synthesis require more enzymes than the leading strand?
a) DNA polymerase III can only synthesize in one direction
b) The lagging strand template is more complex than the leading strand
c) DNA polymerase III works more efficiently on the leading strand
d) Okazaki fragments require extra steps to be joined together

A

Okazaki fragments require extra steps to be joined together
The lagging strand needs multiple primers and DNA ligase to seal the fragments.

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10
Q

How is DNAq subunit of DNA polymerase 3 proofreading related to bacterial survival?
a) It prevents excessive mutations that could be harmful
b) It ensures that replication happens faster
c) It removes excess nucleotides that slow down replication
d) It increases genetic diversity by altering base pairs

A

Prevents excessive mutations to be harmful

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11
Q

What enzyme is responzible for removing RNA primers? Are okazaki fragments filled in this space?

A

Dna polymerase 1
Okazaki fragments do not fill in the space, DNA ligase does that job

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12
Q

What is the mechanism of DNA ligase?

A

Forms a phosphodiester bond between the
- 3’ hydroxyl of the growing stranf
- 5’ phosphate of the okazaki fragment

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13
Q

Why is the Tus protein necessary for chromosome separation in bacterial replication?

A

Tus proteins prevent the replication forks from continuing past the termination sight, which ensures proper chromosome segregation

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14
Q

Why do topoisomerases liberate chromosomes after DNA replication?

a) To relieve supercoiling and untangle newly replicated chromosomes, ensuring proper segregation.

b) To initiate replication by unwinding the DNA at the origin of replication.

c) To break down RNA primers and replace them with DNA to complete replication.

d) To prevent mutations by proofreading newly synthesized DNA strands.

A

After bacterial DNA replication, the two daughter chromosomes can become physically intertwined (catenated).
Topoisomerases cut and rejoin DNA strands to untangle them, allowing proper chromosome segregation during cell division.

From lecture notes: topoisomerases liberate chromosomes by: cleaving DNA, thread strand through and religate DNA
*topoisomerases are responsible for making supercoiled DNA

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15
Q

Why do bacteria not require a 5’ cap and poly-A tail on their mRNA?
a) Transcription and translation occur simultaneously
b) Bacteria lack enzymes that recognize these modifications
c) The bacterial genome has a different nucleotide composition
d) mRNA stability is not necessary for bacterial gene expression

A

Because transcription and translation occur simultaneously

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16
Q

Why does trna spefically bring amino acids to the mRNA during translation? And how do they get attatched

A

Because they have an an amino acid attatchment site

Amino acids get attatched based on anticodon
- anticodon region binds to mrna to give trna its specifity

17
Q

Why is AMP important in the process of tRNA charging (aminoacylation)?

a) AMP activates the amino acid, allowing it to form a high-energy bond with tRNA for efficient translation.

b) AMP stabilizes the tRNA structure, preventing misfolding before it binds to the ribosome.

c) AMP provides energy directly for peptide bond formation during translation.

d) AMP modifies the anticodon loop of tRNA to ensure accurate codon recognition.

A

Amp activates the amino acid by using energy from phosphodiester hydrolysis to allow the amino acid to bind to the 3-OH end of RNA

18
Q

Why does trna need an amino acid

A

trna has a specific stop codon that codes for a specific amino acid

19
Q

What is the role of sigma factors in bacterial transcription, and what is a key characteristic of their function?

a) Sigma factors help RNA polymerase recognize promoter sequences, and different sigma factors allow bacteria to respond to environmental changes.

b) Sigma factors catalyze RNA synthesis and ensure that transcription occurs continuously without regulation.

c) Sigma factors bind directly to ribosomes to regulate translation and ensure accurate protein synthesis.

d) Sigma factors act as proofreading enzymes that remove incorrect nucleotides during mRNA synthesis.

A

Sigma factors have no catalytic activity but help core enzymes recognize the start of genes

Sigma factors help bind rna polymerase to a promotor (site where RNA polymerase inds and initiates transcription)

A

20
Q

Compare and contrast dna polymerase and rna polymerase

A

Dna poly; replicates DNA, requires a primer (starting point)

Rna poly: transcribes rna, does not need a primer (copies DNA into RNA)

21
Q

Which structure can only begin transcription?
A. Core enzymes
B. Sigma factors
C. Holoenzyme
D. Promoters

A

The holoenzyme (core enzymes, sigma factors)

22
Q

How does the Pribnow box facilitate transcription?
a) It acts as a recognition site for sigma factors
b) It signals the RNA polymerase where to terminate transcription
c) It codes for the ribosome binding site in mRNA
d) It slows down transcription to prevent errors

A

Primbow boxes in promoters contain a consensus sequence

The Pribnow box (a promoter sequence) helps RNA polymerase locate the transcription start site.

Sigma factors help rna locate the starting site of transcription

A

23
Q

What would happen if a bacterium had multiple sigma factors active at once?
a) Different genes would be transcribed simultaneously based on environmental conditions
b) RNA polymerase would not function due to competition
c) Transcription would be slower due to conflicting signals
d) Only the most abundant sigma factor would initiate transcription

24
Q

When does termination in transcription occur? And why?

A

When core RNA polymerase comes upon a terminator sequence, causing it to
dissociates from template DNA since

25
Q

What makes bacteria mRNA different than typical mRNA?

A
  • there is no 5’ cap/ long poly a tail
  • bacteria do not process their mRNA (do not modify their rna)
26
Q

What does it mean when bacteria dont process their RNA?

A
  • it means that they immediately use their mRNA for protein production with double checking for errors/ without modifications
27
Q

Why does wobble base pairing occur at the third codon position?
a) The third position is less critical for tRNA binding
b) It allows the cell to use fewer tRNAs to read all codons
c) It increases the speed of translation
d) It prevents mutations from affecting protein function

A

it eliminates the need for unique trna for each codon

(wobble base= codon with a different 3rd codon that still codes for the same protein eg: phenyalamine)

28
Q

How does codon usage bias affect bacterial translation?
a) Some codons are translated faster due to higher availability of matching tRNAs
b) Bacteria cannot translate certain codons, leading to gene silencing
c) Codon bias determines the location of transcription start sites
d) All codons are used equally, so bias has no effect

A

there are fewer trna’s in the trna pool for some codons, some specific trna’s are more abundant than others, so translation is slow sometimes

a

29
Q

If a mutation changed a stop codon (UAA) to UAG, what would likely happen?
a) Translation would still terminate, but efficiency may be reduced
b) The protein would be longer than normal
c) The ribosome would skip to the next stop codon
d) The protein would misfold due to an incorrect amino acid

A

Translation would still terminate, but efficiency may be reduced.

All three stop codons (UAA, UAG, UGA) function, but they have slightly different efficiencies in termination.

30
Q

How is DNA supercoiling related to transcription efficiency?
a) Negative supercoiling enhances transcription by loosening DNA
b) Supercoiling has no impact on transcription efficiency
c) Positive supercoiling increases transcription by stabilizing the RNA-DNA hybrid
d) Supercoiling makes it harder for RNA polymerase to bind

A

negative supercoiling enhances transcription by loosening DNA

negative supercoiling makes it easier for RNA to bind and transcribe genes

31
Q

How is sigma factor activity related to operon regulation?
a) Sigma factors determine which operons are transcribed based on environmental signals
b) Operons function independently of sigma factors
c) Sigma factors activate only individual genes, not operons
d) All sigma factors compete equally for RNA polymerase binding

A

sigma factors = help rnapolymerase bind to the promoter (in transcription)

operon: one mRNA for all genes

Sigma factors determine which operons are transcribed based on environmental signals.

Sigma factors HELP RNA polymerase bind to the promoter of an operon.

32
Q

How would a mutation in the Shine-Dalgarno sequence affect translation?
a) The ribosome may not recognize the mRNA, preventing translation
b) The protein would be synthesized but at a slower rate
c) The mRNA would degrade immediately
d) The start codon would shift to a new location

A

The ribosome may not recognize the mRNA, preventing translation.

The Shine-Dalgarno sequence helps align the ribosome with the start codon.

shine dragon is usually AGGAGG

33
Q

What would happen if a bacterium lost its rho factor?
a) Some mRNAs would not terminate properly, creating extra-long transcripts
b) Transcription would be faster but less accurate
c) RNA polymerase would start transcription at random points
d) The ribosome would attach to DNA instead of mRNA

A

Some mRNAs would not terminate properly, creating extra-long transcripts.

The rho factor helps terminate transcription in rho-dependent genes.

34
Q

Why might operons be advantageous for bacterial survival?
a) They allow coordinated expression of functionally related genes
b) They help bacteria synthesize proteins more slowly
c) They prevent mutations from occurring
d) They create more stable mRNA molecules

A

They allow coordinated expression of functionally related genes
* An operon is a group of genes that are regulated together and transcribed into a single mRNA molecule

Operons are advantageous for bacterial survival because they group functionally related genes together under the control of a single promoter. This allows bacteria to efficiently regulate the expression of multiple genes at once, ensuring that all the proteins needed for a specific function (e.g., lactose metabolism in the lac operon) are produced simultaneously and only when needed. This conserves energy and resources, helping bacteria adapt quickly to changing environments.

35
Q

If a bacterium has a mutation in DNA ligase, what will likely occur?

A

okazaki fragments wont be joined properly

36
Q

How does the presence of multiple ribosomes on an mRNA impact bacterial protein synthesis?
a) It allows multiple copies of a protein to be synthesized simultaneously
b) It ensures proteins fold correctly
c) It increases the accuracy of translation
d) It prevents degradation of mRNA

A

It allows multiple copies of a protein to be synthesized simultaneously

37
Q

what is the difference between the coding strand, template strand and mrna?

A

coding strand is what mrna is based on, and the template strand is used to make mrna

38
Q

what does no encode amino acids?

A

stop codons