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Biochemistry 404 > Exam 3 > Flashcards

Exam 3 Flashcards

1
Q

The MutHLS repair system removes DNA around mismatched base pairs. How does the MutHLS repair system know which DNA strand to remove?

  • The Mut repair system indiscriminately removes either DNA strand
  • The Mut repair system interacts with DNA polymerase and thus knows which strand is the new one
  • The Mut repair system recognizes methylated DNA and removes the DNA strand that is unmethylated
  • The Mut repair system recognizes methylated DNA and removes the DNA strand that is methylated
  • The Mut repair system remoes teh strand of DNA that contains uracil
A

The Mut repair system recognizes methylated DNA and removes the DNA strand that is unmethylated

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2
Q

In Von Neumann’s analogy

  1. the genes represent hardware
  2. a viral protein coat is software
  3. the cell’s enzymes represent hardware
  4. the cell’s enzymes represent software
  5. the genes are the software
  6. A and D
  7. C and E
  8. H. none of these answers is correct
A

C and E

  • the cell’s enzymes represent hardware
  • the genes are the software
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3
Q

What is the order of development described by Dyson from earliest to latest?

  • pathogen > symbiont > parasite > integrated component
  • integrated component > symbiont > parasite > pathogen
  • pathogen > parasite > symbiont > integrated component
  • nucleic acid > clay > protein > membranes
  • none of the above
A

pathogen > parasite > symbiont > integrated component

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4
Q

Which has a lower error rate?

  • reproduction as used by Dyson
  • replication
  • metabolism
  • parasitic infection
  • the average student taking this exam
A

replication

(he sometimes calls it modern replication)

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5
Q

Where in a protein would you expect the mutation rate to be highest?

  • α helices
  • β strands
  • in the active site
  • in the loops
  • none of the above, as all regions would show the same rate
A

in the loops

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6
Q

Modification of DNA by methylases in bacteria is used

  1. To identify “self’ DNA and recognize “foreign” DNA
  2. To regulate the expression of certain genes
  3. To identify the original DNA strand in mismatch repair
  4. To mark DNA to be cut by restriction enzymes
  5. Two of the above
A

Two of the above

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7
Q
  1. The role of nick translation in DNA replication is
  2. to replace the RNA primer bases by the appropriate DNA bases
  3. to connect pieces of DNA to one another (ligation).
  4. to help unwind the double helix.
  5. to generate high energy phosphate bonds.
  6. none of the above.
A

to replace the RNA primer bases by the appropriate DNA bases

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8
Q

Splicing mRNA in eukaryotes

  1. Removes the introns
  2. Removes the exons
  3. Always involves folding of the mRNA
  4. Always involves proteins
  5. Two of the above
A

Two of the above

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9
Q

Which of the following is not true of translation initiation?

  1. Starts with a methionine-tRNA
  2. Needs a ribosome binding site on the mRNA
  3. Requires GTP
  4. The large subunit of the ribosome binds the mRNA first
  5. Two of the above are not true
A

The large subunit of the ribosome binds the mRNA first

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10
Q

When amino acids are attached to tRNAs,

  1. the N-terminal end is attached to the 2’ or 3’ hydroxyl of the ribose at the 3’ end of the tRNA.
  2. the N-terminal end is attached to the 5’ hydroxyl of the ribose at the 3’ end of the tRNA.
  3. the C-terminal end is attached to the 2’ or 3’ hydroxyl of the ribose at the 3’ end of the tRNA.
  4. the C-terminal end is attached to the 5’ hydroxyl of the ribose at the 3’ end of the tRNA.
  5. None of the above
A

the C-terminal end is attached to the 2’ or 3’ hydroxyl of the ribose at the 3’ end of the tRNA.

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11
Q

How are amino acids attached to tRNAs,

  1. the N-terminal end is attached to the 2’ or 3’ hydroxyl of the tRNA.
  2. the N-terminal end is attached to the 5’ hydroxyl of the tRNA.
  3. the C-terminal end is attached to the 2’ or 3’ hydroxyl of the tRNA.
  4. the C-terminal end is attached to the 5’ hydroxyl of the tRNA.
  5. None of the above
A

the C-terminal end is attached to the 2’ or 3’ hydroxyl of the tRNA.

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12
Q

When amino acids are attached to tRNAs,

  1. the N-terminal end is attached to the 2’ or 3’ hydroxyl of the ribose at the 3’ end of the tRNA.
  2. the N-terminal end is attached to the 5’ hydroxyl of the ribose at the 3’ end of the tRNA.
  3. the C-terminal end is attached to the 2’ or 3’ hydroxyl of the ribose at the 3’ end of the tRNA.
  4. the C-terminal end is attached to the 5’ hydroxyl of the ribose at the 3’ end of the tRNA.
  5. None of the above
A

the C-terminal end is attached to the 2’ or 3’ hydroxyl of the ribose at the 3’ end of the tRNA.

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13
Q

Proofreading during the elongation step of translation occurs

  • During insertion of the tRNA carrying the new amino acid into the ribosome
  • Just before peptide bond formation
  • Just after pepetide bond formation
  • During the translocation step to move the peptide from one site to another
  • None of the above; there is no proofreading during translation
A

Just before peptide bond formation

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14
Q

Reverse transcriptase

  • Is the enzyme responsible for degrading mRNA
  • Is an enzyme that works with RNA polymerase to proofread the new message
  • Makes an RNA copy of RNA
  • Makes a DNA copy of RNA
  • None of the above
A

Makes a DNA copy of RNA

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15
Q

What is the major role of DNA polymerase III?

  1. Repair of damaged DNA
  2. Fills in gaps between Okazaki fragments
  3. Synthesizes the majority of new DNA at replication forks
  4. The role of DNA polymerase III is not known
  5. None of the above
A

Synthesizes the majority of new DNA at replication forks

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16
Q

The lac operon is controlled by

  1. A repressor
  2. An activator
  3. Attenuation
  4. Both a repressor and activator
  5. Both attenuation and a repressor
A

Both a repressor and activator

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17
Q

Peptidyl transferase

  1. is needed for initiation of translation.
  2. is needed for elongation during translation
  3. is needed for termination of translation.
  4. is not needed for translation
  5. two of the above
A

is needed for elongation during translation

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18
Q

An iducible operon is one that is

  1. normally not transcribed unless a co-activator or an inducer is present that can interact with regulatory proteins to allow transcription
  2. normally transcribed unless an inhibitor or a corepressor is present that can interact with regulatory proteins to allow transcription
  3. only occurs with negatively controlled operons
  4. only occurs with positively controlled operons
A

normally not transcribed unless a co-activator or an inducer is present that can interact with regulatory proteins to allow transcription

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19
Q

Several different glycosylases can be involved in DNA repair. What do these enzymes do?

  1. Cleave the phosphate-sugar DNA backbone leaving a 5’ phosphate and a 3’ hydroxyl
  2. Cleave the phosphate-sugar DNA backbone leaving a 5’ hydroxyl and a 3’ phosphate
  3. Removes the sugar directly from the DNA backbone leaving an apyrimidinic site
  4. Removes the base attached to a sugar in the DNA backbone
  5. None of the above
A

Removes the base attached to a sugar in the DNA backbone

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20
Q

Which of the following contributes to the specificity with which regulatory proteins bind to particular DNA sequences?

  1. broad electrostatic attraction between DNA phosphates and positively charged protein side chains
  2. hydrogen bonding interactions between polar protein side chains and the parts of bases exposed to the solvent in the major groove
  3. hydrophobic interactions between the DNA riboses and hydrophobic amino acid side chains
  4. complementarity between the protein and the DNA sequence to which it binds.
  5. more than one of the above
A

more than one of the above

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21
Q

The binding reaction between the E. coli lac repressor and the lac operon can be represented by the dissociation reaction RO ⇔ R + 0. Kd, the dissociation equilibrium constant, for this reaction is 10-10 M. Kd for dissociation of the repressor from other E. coli DNA sequences is 10-4 M. This means that

  1. The repressor binds a million times more tightly to the operator than to other parts of the DNA.
  2. The repressor binds 10,000 times more tightly to the operator than to other parts of the DNA.
  3. The repressor binds a million times less tightly to the operator than to other parts of the DNA.
  4. The repressor binds 10,000 times less tightly to the operator than to other parts of the DNA.
  5. None of the above
A

The repressor binds a million times more tightly to the operator than to other parts of the DNA.

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22
Q

Binding of the lac repressor to the lac promoter in E. coli DNA

  1. stops RNA polymerase from elongating the operon’s mRNA
  2. occurs only when the lactose concentration is high
  3. prevents binding of RNA polymerase to the DNA
  4. causes the RNA polymerase to act as a hydrolase rather than as a polymerase
  5. E. more than one of the above.
A

prevents binding of RNA polymerase to the DNA

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23
Q

Leucine zippers

  1. are protein regions which prevent protein-protein interactions when these would interfere with gene regulation.
  2. are protein-protein interaction domains which interact with DNA polymerase
  3. are protein-protein interaction domains which help regulate RNA polymerase activity
  4. are a synthetic material used in certain types of clothing
  5. none of the above.
A

are protein-protein interaction domains which help regulate RNA polymerase activity

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24
Q

When the concentrations of both glucose and lactose are high,

  1. transcription of the lac operon is stimulated because lactose binding to the repressor causes it to dissociate from the operator.
  2. transcription of the lac operon is stimulated because glucose binds to the repressor, causing it to dissociate from the operator.
  3. the RNA polymerase will be stimulated because the elevated glucose concentration will produce an elevated cAMP concentration which, in turn, will bind to the catabolite gene activator protein (CAP), causing CAP to stimulate RNA polymerase transcription of the lac operon
  4. the RNA polymerase will not be stimulated by CAP, because the lowered cAMP concentration caused by elevated glucose will not allow much cAMP to bind to CAP, and that, in turn, will prevent CAP binding to the DNA
  5. None of the above
A
  • the RNA polymerase will not be stimulated by CAP, because the lowered cAMP concentration caused by elevated glucose will not allow much cAMP to bind to CAP, and that, in turn, will prevent CAP binding to the DNA
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25
Q

The presence of a high concentration of a signal molecule which binds to a regulatory protein leads to increased transcription of an operon and thence to increased protein synthesis. This implies that

  • the operon is under negative regulatory control
  • the operon is under positive regulatory control
  • the operon is under both positive and negative regulatory control
  • neither positive nor negative control is involved
  • cannot tell from data given.
A
  • the operon is under negative regulatory control
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26
Q

E. coli RNA polymerase contains no 3’-exonuclease site, and the error rate for mRNA synthesis is one error per 104 to 105 nucleotides. This is orders of magnitude higher than the error rate for DNA replication. Why do cells not die off from the production of defective proteins as a result?

  • There are other mechanisms to repair erroneous mRNA
  • Many mRNA copies are made of the same gene
  • Cells do die off because of such errors, but more cells replace them
  • mRNA molecules are usually degraded rapidly, allowing few defective proteins to be made
  • more than one of the above
A

more than one of the above

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27
Q

When a sigma factor is part of an RNA polymerase complex, the complex is called a holoenzyme, and it binds to the appropriate operator. In the absence of a sigma factor

  • the polymerase binds but doesn’t catalyze mRNA formation
  • the polymerase doesn’t bind at specific operators
  • the polymerase is called a halfenzyme
  • the polymerase is degraded rapidly
  • none of the above
A

the polymerase doesn’t bind at specific operators

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28
Q

In nucleic acid footprinting the binding to the nucleic acid of a protein

  • prevents nuclease hydrolysis of the nucleic acid region bound by the protein
  • promotes nuclease hydrolysis of the nucleic acid region bound by the protein
  • allows identification of promoter sequences
  • slows transcription
  • more than one of the above
A

more than one of the above

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29
Q

In which biochemical activity do we find “lariat” structures?

  • intron removal (splicing)
  • 5’-capping of mRNA
  • 3’-polyadenylation of mRNA
  • degradation of mRNA
  • none of these
A

intron removal (splicing)

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30
Q

Class I and Class II introns are described as self-splicing. This means that

  • they are joined together, and the exons are removed and degraded
  • their removal is catalyzed by the RNA transcript itself
  • they direct proteins, which catalyze the splicing
  • they form only ribosomal RNA
  • none of the above
A

their removal is catalyzed by the RNA transcript itself

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31
Q

Which of the following is facilitated by tethering the beginning of an mRNA molecule during transcription to the carboxy terminal domain (CTD) of RNA polymerase

  • A. protection against premature hydrolysis of the RNA
  • B. helping to bring the ends if introns close together to help in splicing
  • C. aiding in the export of the finished mRNA form the nucleus to the cytoplasm
  • A and B
  • B and C
A

A and B

  • A. protection against premature hydrolysis of the RNA
  • B. helping to bring the ends if introns close together to help in splicing
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32
Q

When an ancestral DNA sequence gives rise to two homologous sequences in curently living species, which of the following types of changes would be detectable by analyzing the DNA of the extant species?

  • single substitutions
  • convergent substitutions
  • back substitutions
  • parallel substitutions
  • none of these would be detectable
A

single substitutions

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33
Q

Why do the mutation rates for rats appear to be faster than for humans?

  • Rats evolve faster than humans
  • Humans evolve faster than rats
  • Correction for the faster generation time of rats shows that the rates of mutation are the same
  • Rat genes are less stable than human genes
  • None of the above
A

Correction for the faster generation time of rats shows that the rates of mutation are the same

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34
Q

Which of the following would have the lowest mutation rate, all other factos being equal?

  • Introns
  • DNA corresponding to peptides excised in post-translational processing of proteins
  • DNA corresponding to core secondary structures of proteins, e.g. alpha helices
  • DNA correspondin to surface loops in proteins
  • All the above would mutate at approximately teh same rate
A

DNA corresponding to core secondary structures of proteins, e.g. alpha helices

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35
Q

The immunological cross reactions of human and chimpanzee proteins are weaker than reactions involving humans and gorillas or chimpanzees and gorillas. Goodman, who performed these experiments in 1962 concluded from the results that

  • Chimpanzees are more closely related to gorillas than are humans
  • Humans are more closely related to chimpanzees that to gorillas
  • Gorillas, chimpanzees and humans are very closely related
  • Gorillas, chimpanzees and humans are very distantly related
  • none of the above
A

Humans are more closely related to chimpanzees that to gorillas

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36
Q

In arranging four species into a phylogenetic tree by maximum parsimony, which of the following types of site would be informative?

  • all the bases at the site are the same
  • all the bases at teh site are different
  • three bases are identical while the fourth is different
  • two bases are of one kind while the other two are also identical but different from the first two
  • none of these would be informative
A

two bases are of one kind while the other two are also identical but different from the first two

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37
Q

Which of the following is NOT involved in initiation of DNA replication in E. coli?

  • DnaB (helicase)
  • Primase
  • Single-stranded binding protein (SSB)
  • Gyrase
  • All of these are involved
A

All of these are involved

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38
Q

In John Von Neumann’s definition of hardware and software as described by Dyson, which of the following would not have hardware function?

  • tRNA
  • genomic RNA as in RNA viruses
  • ribosomal RNA
  • enzymes
  • all of these functions would have hardware functions
A

genomic RNA as in RNA viruses

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39
Q

Which of the following is NOT involved in elongation of DNA replication in E. coli?

  • DnaB (helicase)
  • Primase
  • Single-stranded binding protein (SSB)
  • Gyrase
  • All of these are involved
A

All of these are involved

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40
Q

In Dyson’s double origin hypothesis

  • cells originated first followed by enzymes and later by genes (Oparin)
  • genes originated first followed by enzymes and then by cells (Eigen et al.)
  • enzymes arose first followed by cells and then by genes (Cairns-Smith)
  • Spiegelman monsters arose first
  • none of these
A

cells originated first followed by enzymes and later by genes (Oparin)

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41
Q

At whic stage of DNA replication does nick translation play a major role?

  • initiation
  • elongation
  • termination
  • error correction
  • none of these
A

elongation

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42
Q

Why does Dyson prefer a double origin hypothesis?

  • Two unlikely events, origin of extact replication and origin of metabolism, are less likely to have occurred simultaneously than separately
  • Metabolism and simple non-genetic reproduction without exact replication would have been more tolerant of errors than is exact replication
  • The component of nucleic acids and the nucleic acis themselves are less stable than are amino acids and proteins
  • It is unfashionable
  • all of the above
A

all of the above

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43
Q

Which of the following DNA repair mechanisms involves cleavage of the ribose-phosphate polymer chain?

  • A. Mismatch repair
  • B. Base excision repair
  • C. Nucleotide excision repair
  • D. Direct repair
  • A and C
A

A and C

  • A. Mismatch repair
  • C. Nucleotide excision repair
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44
Q

DNA polymerase III, the main replicating enzyme, inserts approximately one incorrect base for every 104 to 105 nucleotides in E. coli. The overall error rate, however is one incorrect base in approximately 109 to 1010 base pairs.

Why are incorrect bases inserted at all?

  • Hydrogen bonding is a weak interaction. The difference in binding energy between correct and incorrect base pairing is therefore small, allowing errors to occur.
  • Tautomeric forms of the bases make non-Watson-Crick base pairings. These fit the polymerase active site almost as well as standard Watson-Crick pairing, allowing insertion of incorrect bases.
  • The speed at which DNA is replicated makes it impossible for all base pairings to be correct
  • When the energy charge of the cell is low, as happens occasionally, there is not enough energy to ensure correct base pairing.
  • None of the above

From above, the ratio of overall error rate to polymerase error is 109/104 = 1010/105 = 105. This means the overall rate is approximately 100,000 times smaller than the polymerase rate. What is the primary reason for this?

  • Cells with errors are usually not viable and are not therefore available to be counted
  • After the DNA is fully synthesized, housekeeping enzymes inspect it and mark errors for elimination
  • Replication forks usually disintegrate when errors occur, so that the apparatus can start from the beginning again to make correctly paired daughter DNA
  • The polymerase has a 3’ exonuclease activity, which detects and cuts off a mispaired base before the fork can move on
  • None of the above
A
  • Tautomeric forms of the bases make non-Watson-Crick base pairings. These fit the polymerase active site almost as well as standard Watson-Crick pairing, allowing insertion of incorrect bases.
  • The polymerase has a 3’ exonuclease activity, which detects and cuts off a mispaired base before the fork can move on
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45
Q

The Biebricher-Eigen-Luce experiment demonstrated that

  • a solution of nucleotide monomers would yield a nucleic acid polymer if given a template but no polymerase
  • a solution of nucleotide monomers would yield a nucleic acid polymer if given a polymerase but no template
  • a solution of nucleotide monomers would yield a nucleic acid polymer if given neither a polymerase nor a template
  • a solution of nucleotide monomers would yield a nucleic acid polymer only if given both a template and a polymerase
  • none of the above
A

a solution of nucleotide monomers would yield a nucleic acid polymer if given a polymerase but no template

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46
Q

If the eukaryotic analogs of the E. coli MutS and/or MutL proteins are less active than normal because of mutations, one observes an inherited susceptibility to cance. The probable mechanism for the susceptibility is

  • defective initiation of replication at certain points in the genome
  • defective elongation at certain base sequences
  • incorrect termination of replication
  • defective mismatch repair
  • defective base excision repair
A

defective mismatch repair

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47
Q

In Dyson’s view nucleic acid polymers

  • arose initially as biochemical accident which was toxic to the primitive cells containing them
  • became a molecular parasite c.v. Margulis’s work on mitochondria and chloroplasts
  • became a symbiont, providing help to the cell
  • eventually evolved into a central part of the modern genetic apparatus
  • all of the above
A

all of the above

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48
Q

Why are Okazaki fragments synthesized in a discontinuous manner? (This question is from last year’s exam.)

  1. The polymerase falls off the DNA and must restart, by which time the replication fork has moved on.
  2. The polymerase needs time to recharge with ATP, by which time the replication fork has moved on.
  3. Because both strands are replicated at the same time in the 5’ to 3’ direction, reading of the lagging strand in the 3’to 5’ direction must await opening of enough DNA for the polymerase to operate.
  4. The polymerase must await an AT-rich region as the replication fork moves in order to unwind the DNA.
  5. more than one of the above
A

Because both strands are replicated at the same time in the 5’ to 3’ direction, reading of the lagging strand in the 3’to 5’ direction must await opening of enough DNA for the polymerase to operate.

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49
Q

Which of the following is NOT involved in DNA mismatch repair in E. coli

  1. DNA glycosylase
  2. DNA helicase II
  3. DNA ligase
  4. DNA polymersase
  5. Exonuclease
A

DNA glycosylase

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50
Q

The first enzyme to take part in base excision repair is

  1. DNA ligase
  2. DNA glycosylase
  3. AP endonuclease
  4. DNA polymerase I
  5. none of the above
A

DNA glycosylase

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51
Q

The b subunit of DNA polymerase III is called the processivity factor. Its function is

  1. to unwind the double stranded helix.
  2. to stabilize single stranded sections of unwound DNA.
  3. to keep the polymerase from falling off the DNA too frequently.
  4. to connect the lagging strand fragments.
  5. none of the above.
A

to keep the polymerase from falling off the DNA too frequently.

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52
Q

One of the functions of DNA polymerase I (Komberg’s enzyme) is

  1. hydrolytic removal of bases which have been incorporated into the DNA in error.
  2. the polymerization of the leading strand in routine DNA synthesis.
  3. the synthesis of RNA primers.
  4. to back up polymerase III by taking over should polymerase III fall off the DNA
  5. none of the above
A

none of the above

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53
Q

Which of the following is characteristic of the polymerase III holoenzyme?

  1. It is twice the size of a half-enzyme.
  2. It is a stripped down protein with only some of the pol III catalytic activities.
  3. It contains ribonuclease H.
  4. It contains all the catalytic activities of pol III.
  5. none of the above.
A

It contains all the catalytic activities of pol III.

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54
Q

On the leading and lagging strands DNA polymerase III attaches the next base at the

  1. leading strand 3’-OH; lagging strand 3’-OH
  2. leading strand 3’-OH; lagging strand 5’-OH
  3. leading strand 5’-OH; lagging strand 3’-OH
  4. leading strand 5’-OH; lagging strand 5’-OH
  5. none of the above: it attaches at a phosphate of the growing strand
A

leading strand 3’-OH; lagging strand 3’-OH

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55
Q

In what order do the following enzymes act in the synthesis of Okazaki fragments?

  1. helicase, primase, polymerase III, polymerase I, ligase
  2. ligase, polymerase I, polymerase III, primase, helicase
  3. primase, polymerase I, helicase, polymerase III, ligase
  4. polymerase I, polymerase III, helicase, ligase, primase
  5. none of the above.
A

helicase, primase, polymerase III, polymerase I, ligase

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56
Q

What is the function of the 3’-exonuclease site of DNA polymerase III?

  1. When a mismatched base is added to the end of the growing chain, the polymerase slides back, and the exonuclease site removes the incorrect base.
  2. If synthesis of the leading strand outruns that of the lagging strand, the exonuclease removes enough leading strand bases to allow the lagging strand to catch up, whereupon synthesisresumes.
  3. to degrade invading viral DNA in the event the cell’s machinery starts to replicate the virus.
  4. to separate catenated chromosomes
  5. more than one of the above
A

When a mismatched base is added to the end of the growing chain, the polymerase slides back, and the exonuclease site removes the incorrect base.

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57
Q

What is the principal role of magnesium ions in the action of DNA dependent RNA polymerase?

  1. They help maintain the proper ionic strength.
  2. The help stabilize the polymerase.
  3. They prevent calcium from occupying the protein’s active site.
  4. They interact with and help stabilize and orient the phosphates of the incoming NTP.
  5. They have no role in RNA polymerization, as they are not involved
A

They interact with and help stabilize and orient the phosphates of the incoming NTP.

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58
Q

When DNA is being transcribed into RNA, it must unwind. This forces it to form positive supercoils, thereby putting conformational strain on it. This raises the DNA’s energy. When the DNA exits the polymerase, it rewinds, and the original negative supercoils that were present before polymerization began are restored. The enzyme which relieves the positive supercoil and controls the restoration of the negative supercoils is

  1. DNA dependent RNA polymerase.
  2. helicase
  3. topoisomerase
  4. Sigma factor
  5. the elongation factors
A

topoisomerase

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59
Q

When footprinting is used to find out where on a nucleic acid a protein binds, the nucleic acid sequences which interact with the protein are identified by the fact that

  1. they are rapidly degraded
  2. bands corresponding to them are seen on gel lanes whose samples were prepared with the protein present in the solution.
  3. bands corresponding to them are not seen on gellanes whose samples were prepared with the protein present in the solution but are seen in lanes whose solutions were prepared without protein present,
  4. they are not radioactively labeled.
  5. none of the above
A

bands corresponding to them are not seen on gellanes whose samples were prepared with the protein present in the solution but are seen in lanes whose solutions were prepared without protein present,

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60
Q

What is a promoter?

  1. an origin of replication
  2. a start site for binding of RNA polymerase
  3. a Shine-Dalgarno sequence
  4. a splice site
  5. someone who organizes boxing matches
A

a start site for binding of RNA polymerase

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61
Q

After binding to DNA by E. coli RNA polymerase, the correct order of events is

  1. closed complex formation, open complex formation, promoter clearance, start of RNA synthesis
  2. closed complex formation, open complex formation, start of RNA synthesis, promoter clearance
  3. open complex formation, closed complex formation, start of RNA synthesis, promoter clearance
  4. start of RNA synthesis, closed complex formation, open complex formation, promoter clearance
  5. start of RNA synthesis, open complex formation, closed complex formation, promoter clearance
A

closed complex formation, open complex formation, start of RNA synthesis, promoter clearance

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62
Q

Which of the following is NOT true of eukaryotic mRNA?

  1. Exons are used for polypeptide synthesis.
  2. Introns are complementary to their adjacent exons and will form hybrids with them.
  3. The mature mRNA is usually substantially shorter than the base sequences on the DNA which are complementary to the mature mRNA.
  4. The mRNA is originally synthesized in the nucleus but ends up in the cytoplasm.
  5. The splicing that yields a mature mRNA occurs at very specific sites in the primary transcript.
A

Introns are complementary to their adjacent exons and will form hybrids with them.

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63
Q

With respect to the endoplasmic reticulum (ER), what is the lumen?

  1. the places where ribosomes are attached
  2. the apparatus for secreting proteins to the rest of the cell
  3. the membranes separating the ER from the rest of the cell
  4. the interior aqueous space, which is surrounded by the membranes
  5. the part which produces i1/umination by means of the luciferin-luciferase reaction.
A

the interior aqueous space, which is surrounded by the membranes

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64
Q

What feature of EF-G helps it to “push” translocation of the ribosome along the mRNA?

  1. It binds weakly to the deacylated tRNA in the P site, gently pushing it toward the E site.
  2. It binds to the tRNA in the E site, pulling the message along by mass action.
  3. It’s shape resembles that of a charged tRNA in complex with EF-Tu, allowing it to bind in tlte A site, which forces the peptide into the adjacent P site
  4. It relieves conformational stress in the ribosome
  5. E. none of these
A

It’s shape resembles that of a charged tRNA in complex with EF-Tu, allowing it to bind in tlte A site, which forces the peptide into the adjacent P site

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65
Q

Protein degradation is a complex process, and many of the signals remain unknown. One known signal involves recognition of amino acids in a mature protein that lead to long half-lives such as Ala, Gly, Met, etc., and those leading to short half-lives, such as Arg, Asp, Leu, etc. These signal amino acids are located at

  1. a helix-turn-helix motif in the protein.
  2. a lysine-containing target sequence in the protein.
  3. a zinc finger structure in the protein.
  4. The carboxy terminus of the protein.
  5. the amino terminus of the protein.
A

the amino terminus of the protein.

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66
Q

Which of the following post-translational events would make a protein have a less negative charge?

  1. phosphorylation of a serine side chain
  2. phosphorylation of a tyrosine side chain
  3. carboxylation of a glulamate side chain
  4. formation ofa methyl ester on a glutamate side chain carboxyl group
  5. methylation of a lysine residue
A

formation ofa methyl ester on a glutamate side chain carboxyl group

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67
Q

In Crick’s wobble hypothesis

  1. the ribosome wobbles by one base on the m-RNA, shifting the reading frame of protein synthesis.
  2. the first (5’) base of m-RNA codons can make unusual hydrogen bonds with t-RNA anticodons explaining codon redundancy.
  3. a runner rounding third base doesn’t make it to home plate before being thrown out.
  4. the third (3’) base of m-RNA codons can make unusual hydrogen bonds with t-RNA anticodons explaining codon redundancy.
  5. mitochondrial codons replace chromosomal codons.
A

the third (3’) base of m-RNA codons can make unusual hydrogen bonds with t-RNA anticodons explaining codon redundancy.

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68
Q

The Spiegelman monster

  1. arose by spontaneous polymerization in the manner of the Eigen or Orgel experiments.
  2. eats cookies.
  3. arose by gradual loss of unneeded genes.
  4. defines the approximate minimum size of the replicative unit, given the modem set of enzymes and other materials.
  5. C and D
A

C and D

  • arose by gradual loss of unneeded genes
  • defines the approximate minimum size of the replicative unit, given the modem set of enzymes and other materials.
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69
Q

The error catastrophe

  1. leads to extinction when the loss of information through errors exceeds the gain of information forced on the system by natural selection.
  2. leads to extinction when the loss of information through errors is less than the gain of information forced on the system by natural selection.
  3. leads to extinction when environmental conditions change so that the fitness of the species is no longer high enough to sustain its continued existence.
  4. does not involve natural selection; it arises from neutral genetic drift.
  5. can lead students to do poorly on exams.
A

leads to extinction when the loss of information through errors exceeds the gain of information forced on the system by natural selection.

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70
Q

“Footprinting” or DNase protecfion is a technique used to identify:

  1. E. coli cells that contain a desired, cloned piece of DNA.
  2. the position of a particular gene of a chromosome
  3. the specific binding site of a repressor, polymerase, or other protein on the DNA
  4. the position of internally double-stranded regions in a single-stranded DNA molecule
  5. the shoe size of a criminal suspect prior to analysis of the suspect’s DNA. (Adapted from last year’s final exam.)
A

the specific binding site of a repressor, polymerase, or other protein on the DNA

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71
Q

A TATA box

  1. is part of the termination apparatus for m-RNA synthesis.
  2. contains a special methionine codon.
  3. is highly susceptible to failures of base excision repair
  4. is part of promoter regions, where it is involved in recognition by RNA polymerase
  5. is located after the first amino acid codon but before the second.
A

is part of promoter regions, where it is involved in recognition by RNA polymerase

72
Q

The term, ‘elongation bubble,’ refers to

  1. the region of DNA being rephcated.
  2. the region of a polypeptide being synthesized on a ribosome.
  3. the initiation complex of mRNA synthesis
  4. a section of unwound DNA containing RNA polymerase and a growing mRNA molecule
  5. none of the above.
A

a section of unwound DNA containing RNA polymerase and a growing mRNA molecule

73
Q

The binding of RNA polymerase to DNA is driven largely by electrostatic forces. This implies that

  1. the polymerase carries a substantial net negative charge.
  2. the polymerase carries a roughly even mixture of lysines, arginines, glutamic acids and aspartic acids.
  3. the polymerase exposes a large hydrophobic binding surface to the DNA
  4. a section of unwound DNA containing RNA polymerase and a growing mRNA molecue
  5. none of the above.

2015 options

  • A. the polymerase carries a substantial net negative charge.
  • B. the polymerase carries a roughly even mixture of lysines, arginines, glutamic acids and aspartic acids.
  • C. the polymerase exposes a large hydrophobic binding surface to the DNA.
  • D. the polymerase carries a substantial net positive charge.
  • E. none of the above.
A

a section of unwound DNA containing RNA polymerase and a growing mRNA molecue

Answer for 2015:

the polymerase carries a substantial net positive charge.

74
Q

Once σ (sigma) dissociates from the RNA polymerase-DNA complex,

  1. transcription terminales
  2. transcription continues, but slowly
  3. the elongation complex is stable and continues to the end of the cistron (gene or gene group)
  4. ribosomes are no longer able to bind to the growing m-RNA to start translation
  5. none of the above
A

the elongation complex is stable and continues to the end of the cistron (gene or gene group)

75
Q

When a DNA molecule is described as feplicating bidirectionally, it means that it has two

  1. chains.
  2. independently replicating segments
  3. origins of replication
  4. replication forks
  5. termination points
A

replication forks

76
Q

Which of the following statements aboat enzymes which interact with DNA is ture?

  • E. coli polymerase I is unusual in that it possesses only a 5’ to 3’ exonuclease activity.
  • Endonucleases degrade circular but not linear DNA molecules
  • Exonucleases degrade DNA at a free end
  • Many DNA polymerases have a proofreading 5’ to 3’ exonuclease activity
  • Primases syntehsize a short stretch of DNA to prime further synthesis
A

Exonucleases degrade DNA at a free end

77
Q

What is the role of the DAM methylase in E. roli?

  1. to label the template strand so that base errors in the newly synthesized strand can be detected
  2. to excise bases which are inserted erroneously by polymerases.
  3. to dispose of excess methyl groups.
  4. to stimulate helicase action
  5. the role is unknown; that’s why people curse it by calling it the DAMn methylase
A

to label the template strand so that base errors in the newly synthesized strand can be detected

78
Q

RNA polymerase

  1. binds tightly to a region of DNA thousands of base pairs away from the DNA to be satisfied
  2. can synthesize RNA de novo (without a primer)
  3. has a subunit called λ (lambda), which acts as a proofreading ribonuclease.
  4. separated DNA strands throughout a long region of DNA (up to thousands of base pairs) and then copies one of them.
  5. synthesizes RNA in the 3’ to 5’ direction.
A

can synthesize RNA de novo (without a primer)

79
Q

The σ (sigma) factor of E. coli RNA polymerase

  1. associates with the promoter before binding core enzyme
  2. combines with the core enzyme to confer specific binding to a promoter
  3. is inseparable from the core enzyme.
  4. is required for termination of an RNA chain.
  5. will catalyse RNA synthesis from both DNA strands in the absence of cote enzyme.
A

combines with the core enzyme to confer specific binding to a promoter

80
Q

Which of the following contributes to the specificity (accuracy) of amino acid incorporation into proteins?

  1. base pairing at the codon/anficodon interaction
  2. interactions between tRNA and aminoacyl-tRNA synthetase
  3. interactions between the amino acid and aminoacyl-tRNA synthetase
  4. activity of the hydrolytic site on the synthetase
  5. all of the above
A

all of the above

81
Q

which of the following is true?

  1. some tRNAs recognize more than one codon
  2. Some codons recognize more than one tRNA.
  3. Each amino acid can be charged onto one and only one tRNA.
  4. more than one of these is true
  5. none of these is ture
A

some tRNAs recognize more than one codon

82
Q

How do prokaryotes disfinguish between the codon AUG as the amino terminal amino acid during initiation of protein synthesis and AUG used later on within the polypeptide?

  1. They have two tRNAs, one for N-formyl met and one for met without an N-formyl group
  2. They have one tRNA for met, but N-forrnylation of the charged tRNA causes the 305 ribosome to recognize it as the start of the chain.
  3. If the AUG is near a promoter, it is used as the starting point.
  4. They don’t distinguish: met and fmet are introduced randomly and processed aftef translation is complete.
  5. none of the above
A

They have two tRNAs, one for N-formyl met and one for met without an N-formyl group

83
Q

In a eukaryotic polyribosome, which ribosome will have the longest growing polypepfide chain attached to it?

  1. the ribosome nearest the 3’ end of the mRNA
  2. the ribosome nearest the 5’ end of the mRNA.
  3. a ribosome near the middle of the mRNA, because synthesis is bidirectional.
  4. a ribosome with a chaperone complex adjacent to it.
  5. it depends on which protein is being made
A

the ribosome nearest the 3’ end of the mRNA

84
Q

You are working in a lab that focuses on the effects of estrogen on development. Your hypothesis is that treatment with estrogen results in changes in gene expression of the gene Summer (Sum) that functions to increase dopamine levels in your brain. Your thesis advisor thinks that it may involve the reduction of the inhibitor of Sum, (iSum), but you disagree and hypothesize that it is a change in chromatin remodeling.

  1. What technique(s) could you use to determine if there is a change in iSum RNA expression?
  • A knockout of iSum
  • Western analysis
  • siRNA
  • Restriction digests
  • Northern analysis
  1. While your advisor is on vacation, you decide to investigate changes in chromatin remodeling of Sum. Which of the following would be a technique you would use?
  • siRNA
  • Immunoporecipitation of histone proteins
  • RT-PCR analysis of Sum
  • Southern analysis of histone proteins
  • Western analysis of iSum
A
  1. Northern analysis
  2. Immunoporecipitation of histone proteins
85
Q

mRNA:

  • is the most abundant RNA in the cell
  • is not really important
  • can be separated from other RNA species by polyA isolation
  • can be used as a template for PCR
  • can be analyzed by Southern analysis
A

can be separated from other RNA species by polyA isolation

86
Q

Which of the following is a true statement about DNA:

  • A. All DNA in a eukaryotic organism is in heterochromatin form unless activated
  • B. Prokaryotic DNA is capped by telomere sequences
  • C. 1.5% of eukaryotic DNA is considered genes
  • D. All DNA in eukaryotic organisms is in euchromatin form unless activated
  • E. 1.5% of all eukaryotic DNA is coding.
A

E. 1.5% of all eukaryotic DNA is coding.

87
Q

Prokaryotic translation:

  • A. is exactly the same as eukaryotic transcription
  • B. requires a multiprotein complex that links the transcribed RNA to the translational machinery via the 5’ cap and poly A tail.
  • C. does not use AUG as the start codon
  • D. requires a shine-delgardo sequence to start translation.
  • E. is dependant on the sigma factor
A

D. requires a shine-delgardo sequence to start translation

88
Q

RNA polymerase II is distinct in that it

  • Is only used to generate tRNAs
  • Is used to make primers for DNA replication
  • it has only one subunit
  • Makes Okazaki fragments.
  • Is effective on a variety of different promoters
A

Is effective on a variety of different promoters

89
Q

In prokaryotes, RNA polymerase finds the start of transcription by:

  • A. Specific sequences in the promoter region
  • B. Identification of the cap structure
  • C. Sliding along the DNA and looking for unwound segments
  • D. Signaling the histone molecules to disconnect from the wound DNA
  • E. Sliding along the DNA to find a specific stem-loop structure
A

A. Specific sequences in the promoter region

90
Q

Which of the following statements is FALSE:

  • A. All amino acids have only one tRNA
  • B. Mitochondria have a slightly different “code” than that of the eukaryotic organism
  • C. The wobble hypothesis comes from the finding that there is greater freedom in pairing in the third position of the codon
  • D. Aminoacyl-tRNA synthetases function to add the amino acid onto the tRNA
  • E. Aminoacyl-tRNA synthetases can proofread
A

A. All amino acids have only one tRNA

91
Q

When a computer is looking for a potential open reading frame (ORF) it:

  • A. finds ATG and stop codon.
  • B. scans for translatable length higher than set threshold
  • C. repeats 3 times
  • D. A & B.
  • E. A, B & C
A

D. A & B.

  • A. finds ATG and stop codon.
  • B. scans for translatable length higher than set threshold
92
Q

The signal sequences that direct proteins to the nucleus are:

  • A. always at the amino terminus of the targeted protein.
  • B. cleaved after the protein arrives in the nucleus.
  • C. glycosyl moieties containing mannose 6-phosphate residues.
  • D. not located at the ends of the peptide sequence, but in the middle.
A

D. not located at the ends of the peptide sequence, but in the middle.

93
Q

Homologous recombination:

  • A. is only used during replication
  • B. Occurs when there is a double strand break in the DNA
  • C. is part of the process of antibody formation
  • D. is important for making a knockout mouse
  • E. B, C & D
A

E. B, C & D

  • B. Occurs when there is a double strand break in the DNA
  • C. is part of the process of antibody formation
  • D. is important for making a knockout mouse
94
Q

Given p(A)=0.5, p(C)=0.1, p(G)=0.2, and p(T)=0.2. Which string is most likely NON-RANDOM?

  • A. AACTG
  • B. AAAAA
  • C. ATATA
  • D. CCACG
  • E. CCCCC
A

E. CCCCC

(look for the ones with the lowest p value)

95
Q

What is the (3’-5’) DNA sequence encoded in this Sanger sequencing gel?

  • A. …ATGGTGACA…
  • B. …TACCACTGT…
  • C. …ACAGTGGTA…
  • D. …TGTCACCAT…
  • E. None of the above
A

B. …TACCACTGT…

(read top down, left to right, using complimentary nucleotides)

96
Q

Active transport through a biological membrane is:

  • A. driven by a difference of solute concentration.
  • B. driven by ATP.
  • C. ATP-producing.
  • D. required for all molecules
  • E. not specific with respect to the substrate
A

B. driven by ATP

97
Q

Which of these statements about the composition of biological membranes is false?

  • A. In a given eukaryotic cell type (e.g., a hepatocyte), all intracellular membranes have essentially the same complement of lipids and proteins.
  • B. The carbohydrate found in membranes is virtually all part of either glycolipids or glycoproteins.
  • C. The plasma membranes of the cells of vertebrate animals contain more cholesterol than the mitochondrial membranes.
  • D. The ratio of lipid to protein varies widely among cell types in a single organism.
  • E.. Triacylglycerols are not commonly found in membranes
A

A. In a given eukaryotic cell type (e.g., a hepatocyte), all intracellular membranes have essentially the same complement of lipids and proteins.

98
Q

In bacteria the elongation stage of protein synthesis does not involve:

  • A. aminoacyl-tRNAs.
  • B. EF-Tu.
  • C. GTP.
  • D. IF-2.
  • E. peptidyl transferase.
A

D. IF-2.

99
Q

Which of the following statements about tRNA molecules is false?

  • A. A, C, G, and U are the only bases present in the molecule.
  • B. Although composed of a single strand of RNA, each molecule contains several short, double-helical regions.
  • C. Any given tRNA will accept only one specific amino acid.
  • D. The amino acid attachment is always to an A nucleotide at the 3’ end of the molecule.
  • E. There is at least one tRNA for each of the 20 amino acids.
A

A. A, C, G, and U are the only bases present in the molecule.

100
Q

Which of the following is not true of sterols?

  • A. Cholesterol is a sterol that is commonly found in mammals.
  • B. They are commonly found in bacterial membranes.
  • C. They are more common in plasma membranes than in intracellular membranes (mitochondria, lysosomes, etc.).
  • D. They are precursors of steroid hormones.
  • E. They have a structure that includes four fused rings.
A

B. They are commonly found in bacterial membranes.

101
Q

Grb and Sos are:

  • A. sometimes covalently attached to lipid moieties.
  • B. sometimes covalently attached to carbohydrate moieties.
  • C. GDP/GTP exchange proteins that link signaling proteins
  • D. Tyrosine kinases
  • E. critical for the lac operon.
A

C. GDP/GTP exchange proteins that link signaling proteins

102
Q

An integral membrane protein:

  • A. Is linked to the membrane via covalent bonds
  • B. Can only be separated from the membrane by treatment with surfactants
  • C. Is the structure of traditional steroid receptors
  • D. Can be removed from the membrane by changes in pH.
  • E. Is linked to the membrane by glycosidic interactions.
A

B. Can only be separated from the membrane by treatment with surfactants

103
Q

In the “activation” of an amino acid for protein synthesis:

  • A. leucine can be attached to tRNAPhe, by the aminoacyl-tRNA synthetase specific for leucine.
  • B. methionine is first formylated, then attached to a specific tRNA.
  • C. the amino acid is attached to the 5’ end of the tRNA through a phosphodiester bond.
  • D. there is at least one specific activating enzyme and one specific tRNA for each amino acid.
  • E. two separate enzymes are required, one to form the aminoacyl adenylate, the other to attach the amino acid to the tRNA.
A

D. there is at least one specific activating enzyme and one specific tRNA for each amino acid.

104
Q
  1. The tryptophan operon of E. coli is repressed by tryptophan added to the growth medium. The tryptophan repressor probably:
  • A. binds to RNA polymerase when tryptophan is present.
  • B. binds to the trp operator in the absence of tryptophan.
  • C. binds to the trp intronic sequence in the presence of tryptophan.
  • D. is a DNA sequence.
  • E. is an attenuator
A

E. is an attenuator

105
Q

Differential RNA processing may result in:

  • A. a shift in the ratio of mRNA produced from two adjacent genes.
  • B. attachment of the poly(A) tail to the 5’ end of an mRNA.
  • C. inversion of certain exons in the final mRNA.
  • D. the production of the same protein from two different genes.
  • E. the production of two distinct proteins from a single gene
A

E. the production of two distinct proteins from a single gene

106
Q

Metabolic pathways that involve the degradation of large molecules to smaller ones are classified as ________.

  • A. anabolic
  • B. catabolic
  • C. amphibolic
  • D. intermediary
  • E. hermaneutic
A

B. catabolic

107
Q

In the experiment whose results are shown below, RNA polymerase is incubated with radiolabeled lac promoter DNA. After a period of time nucleases are added to the mixture, which is then run on a gel. The radioactivity is detected by autoradiography (exposure of the gel to x-ray film). In the left lane, marked −, no polymerase was added; the nucleases were added to the DNA sample without the polymerase. In the right lane, marked C, no polymerase was added either; instead of nucleases, however, the DNA was fragmented by a chemical reagent. The sample to which polymerase was added followed by hydrolases is in the center lane, marked +. One concludes from this experiment that

  • A. the DNA in the central lane was hydrolyzed, whereas the DNA in the left lane was not.
  • B. the two bare zones in the central lane arose because the polymerase prevented hydrolysis of those parts of the DNA by covering them.
  • C. RNA polymerase cannot bind to DNA promoters.
  • D. RNA polymerase transcribes promoters into mRNA, from which they are translated into protein.
  • E. none of the above
A

B. the two bare zones in the central lane arose because the polymerase prevented hydrolysis of those parts of the DNA by covering them.

108
Q

The division of E.coli cells can by synchronized so that all the cells divide at the same time. When cells are first grown on heavy nitrogen (15N) so that all their DNA contains that isotope and no other form of nitrogen, a single band is obtained in a density gradient centrifugation experiment. If the cells are transferred to a normal (14N) medium that contains none of the heavy isotope and synchronized, after one cell division, one band is obtained. After a second cell division, two bands are obtained with equal molar amounts of DNA in each. The same two bands are obtained after subsequent divisions but in different amounts after each division. What would be the proportions of the two bands after the fifth cell division?

  • A. 1 to 1
  • B. 1 to 2
  • C. 1 to 4
  • D. 1 to 8
  • E. none of these
A

D. 1 to 8

109
Q

The gene encoding the E. coli enzyme -galactosidase begins with the sequence ATGACCATGATTACG. What is the sequence of the RNA transcript specified by this part of the gene?

  • A. AUGACCAUGAUUACG
  • B. GCAUUAGUACCAGUA
  • C. UACUGGUACUAAUGC
  • D. CGUAAUCAUGGUCAU
  • E. none of these (adapted from the end of Chapter 26
A

A. AUGACCAUGAUUACG

(same sequence as DNA, except substitute T for U)

110
Q

If aminopterin, a potent inhibitor of dihydrofolate reductase is added to a culture of growing cells, which of the following major biosynthetic processes will be inhibited?

  • A. Carbohydrate synthesis
  • B. Lipid synthesis
  • C. DNA synthesis
  • D. RNA synthesis
  • E. Protein synthesis
A

C. DNA synthesis

111
Q

What happens at a Holliday junction?

  • A. Site specific recombination takes place.
  • B. Strand exchange takes place.
  • C. Vacationers meet.
  • D. Gene inversion can occur.
  • E. A, B and D
A

E. A, B and D

  • A. Site specific recombination takes place.
  • B. Strand exchange takes place
  • D. Gene inversion can occur
112
Q

Nuclear localization signals are short amino acid sequences found in eucaryotes and usually contain several Arg and Lys residues. Their presence leads to the protein being targeted to the nucleus. Why is such targeting necessary?

  • A. It is not necessary, as the proteins are already in the nucleus.
  • B. Proteins are synthesized in the cytoplasm and must be transported to the nucleus.
  • C. They come into action when protein synthesis in the nucleus is disrupted.
  • D. When cells divide, the nuclear membrane dissolves, releasing nuclear proteins, which must then be re-imported into the new nuclei.
  • E. more than one of the above
A

E. more than one of the above

113
Q

Dolichol phosphate has a large hydrophobic part with a phosphate at one end, and it spans the endoplasmic reticulum membrane. In one of the stranger features of biochemistry, its hydrophobic part spends its time in contact with the aqueous cytoplasm or aqueous ER lumen, flipping back and forth between these zones. In doing so, the phosphate must be pulled through the membrane. In what cellular function does it participate?

  • A. targeting of proteins to the Golgi apparatus
  • B. export of proteins from the cell
  • C. import of proteins into the cell
  • D. glycosylation of proteins
  • E. marking of proteins for degradation
A

D. glycosylation of proteins

114
Q

What happens in a proteasome?

  • A. Proteins are synthesized and folded while protected.
  • B. Ubiquitin is attached to proteins.
  • C. Ubiquitinated proteins are degraded.
  • D. The degradation products of protein hydrolysis are detoxified
  • E. Proteins are stored for later use.
A

C. Ubiquitinated proteins are degraded.

115
Q

Positive regulation of transcription initiation means that

  • A. transcription is “off” unless a protein activator is bound to the DNA.
  • B. transcription is “off” unless a small molecule ligand is bound to the activator.
  • C. is “off” unless a protein repressor is bound to the DNA.
  • D. transcription is “off” unless a small molecule ligand is bound to the repressor
  • E. more than one of the above.
A

A. transcription is “off” unless a protein activator is bound to the DNA.

116
Q

Why are protein domains that interact with DNA small?

  • A. Small domains are flexible, and flexibility enables the domain to adapt to the conformation of specific DNA regions.
  • B. Being small means there is relatively little energy cost to make them.
  • C. They must fit into the major groove, where there is not enough room for a large domain to interact with the DNA.
  • D. Being small means there are many places along the DNA with which they can interact.
  • E. They must fit into the minor groove, where there is not enough room for a large domain to interact with the DNA.
A

C. They must fit into the major groove, where there is not enough room for a large domain to interact with the DNA.

117
Q
A
118
Q

What is the primary role of the zinc ions in zinc finger motifs?

  • A. interaction with DNA phosphates
  • B. interaction with the edges of DNA bases
  • C. interactions with the π clouds of DNA bases
  • D. stabilization of the protein conformation by interacting with various side chains
  • E. none of the above.
A

D. stabilization of the protein conformation by interacting with various side chains

119
Q

Why would a low level of expression of the lac operon be useful even when lactose is not present in the medium?

  • A. It is not useful; the system is leaky, and some constitutive expression seems to be unavoidable.
  • B. Low level expression means that some lactose permease will be present in the cell membrane. Should it become necessary for the cell to use lactose, that permease would allow lactose to enter and turn on the operon.
  • C. Without some expression there would be no repressor made.
  • D. Glucose is occasionally converted to lactose accidentally. Having some -galactosidase present allows the lactose to be used.
  • E. none of the above
A

B. Low level expression means that some lactose permease will be present in the cell membrane. Should it become necessary for the cell to use lactose, that permease would allow lactose to enter and turn on the operon.

120
Q

In the synthesis of lagging strand DNA, polymerization of the Okazaki fragment near the replication fork stops when the RNA primer of the previous strand is reached. The RNA is then removed by which of the following processes?

  • A. kinetic proofreading
  • B. base excision repair
  • C. nick translation
  • D. ligation
  • E. none of the above
A

C. nick translation

121
Q

Free fatty acids in the bloodstream are:

  • A. Bound to hemoglobin
  • B. Present at levels that are independent of epinephrine.
  • C. Carried by the protein serum albumin.
  • D. Freely soluble in the aqueous phase of the blood.
  • E. Nonexistent; the blood does not contain free fatty acids.
A

C. Carried by the protein serum albumin.

122
Q

What is a signature sequence of amino acids or nucleotides?

  • A. It is a run of residues common to a set of related species but not to other groups of species.
  • B. It is a way to close a gap in a set of sequences.
  • C. It is a specific binding site for an enzyme that processes the protein or nucleic acid.
  • D. It is a set of residues which occurs only in gram positive and gram negative bacteria.
  • E. none of the above
A

A. It is a run of residues common to a set of related species but not to other groups of species.

123
Q

What sort of information is present in a rooted evolutionary tree but not in an unrooted tree?

  • A. A rooted tree shows the evolutionary relationships among species but not the sequence of evolutionary development.
  • B. A rooted tree shows both the evolutionary relationships among species and the sequence of evolutionary development.
  • C. A rooted tree offers several possible evolutionary routes from ancestral species to extant species whereas an unrooted tree shows only one possible route.
  • D. A rooted tree has nodes, whereas an unrooted one does not.
  • E. A rooted tree has leaves, whereas an unrooted one is dead and has no leaves
A

B. A rooted tree shows both the evolutionary relationships among species and the sequence of evolutionary development.

124
Q

As discussed in class, the numbers of rooted and unrooted trees are given by the formulas below. For a set of five OTUs the ratio of the number of possible rooted to unrooted trees is

  • A. 105
  • B. 15
  • C. 0.14
  • D. 7
  • E. none of these
A

D. 7

(N = 5, solve each formula, do the ratio)

125
Q

The components of a distance matrix are

  • A. the lengths of the “arms” in an unrooted tree.
  • B. composite OTUs made up of several species.
  • C. the time since species diverged from common ancestors.
  • D. the numbers of nucleotide or amino acid differences between each pair of species represented in the matrix.
  • E. more than one of the above
A

D. the numbers of nucleotide or amino acid differences between each pair of species represented in the matrix.

126
Q

Informative sites in studies of molecular evolution

  • A. are sites where the paleontological record can be reconciled with the nucleotide or protein sequence data.
  • B. are places in Wikipedia whose information can be trusted.
  • C. allow an investigator to favor one possible evolutionary tree over another.
  • D. allow an investigator to determine which species is the root of a rooted tree.
  • E. allow an investigator to group some species into a clade while excluding others.
A

C. allow an investigator to favor one possible evolutionary tree over another.

127
Q

The serological work of Goodman (p. 7 in the class notes) led him

  • A. to exclude humans from the same clade as other primates.
  • B. place humans and orangutans in the same clade and chimpanzees and gorillas in another clade.
  • C. to conclude that humans share no common ancestors at any level of evolution with the African apes.
  • D. to conclude that humans are closer in evolutionary terms to chimpanzees than to gorillas.
  • E. to place humans, chimpanzees and gorillas in single clade.
A

E. to place humans, chimpanzees and gorillas in single clade

128
Q

E. coli choose glucose over lactose as a carbon source when both sugars are present by

  • A. binding a CRP / c-AMP complex to the operator when repressor is not bound.
  • B. binding lac repressor.
  • C. binding neither repressor nor CRP, thereby allowing only a low level of lac operon transcription.
  • D. binding both repressor and CRP.
  • E. none of the abov
A

C. binding neither repressor nor CRP, thereby allowing only a low level of lac operon transcription.

129
Q

When tryptophan levels are high in E. coli

  • A. tryptophan binds to the trp repressor, causing it to fall off the DNA. Transcription of the trp operon then follows.
  • B. the trp repressor is not made.
  • C. synthesis of the leader peptide is disrupted.
  • D. tryptophan binds to the trp repressor thereby changing its conformation and enabling it to block the operon’s operator.
  • E. the cells excrete the excess amino acid.
A

D. tryptophan binds to the trp repressor thereby changing its conformation and enabling it to block the operon’s operator.

130
Q

What is the total number of possible RNA sequences of length 50?

  • A. 4*50
  • B. 450
  • C. 504
  • D. 6*504
  • E. 6*450
A

B. 450

131
Q

If there are 61 non-stop codons and all codons are uniformly distributed, what is the minimal length of a real ORF at alpha=0.01

  • A. log0.95 0.01
  • B. log0.01 0.95
  • C. log0.99 0.05
  • D. log0.95 0.99
  • E. none of the above
A

A. log0.95 0.01

132
Q

A predictor identifies people as blond (positive) or brunette (negative) on the basis of their genetic variation. It makes 100 predictions, identifies 5 brunettes as blond, 12 blondes as brunette, and gets the rest right. How many type I errors does it make?

  • A. 12
  • B. 5
  • C. 17
  • D. 7
  • E. 0.05
A

B. 5

  • type 1 error
    • false positive
    • occurs when a researcher incorrectly rejects a true null hypothesis
    • This means that your report that your findings are significant when in fact they have occurred by chance.
    • probability of making a type I error is represented by your alpha level (α), which is the p-value below which you reject the null hypothesis.
    • A p-value of 0.05 indicates that you are willing to accept a 5% chance that you are wrong when you reject the null hypothesis.
  • type 2 error
    • false negative
    • occurs when a researcher fails to reject a null hypothesis which is really false
    • researcher concludes there is not a significant effect, when actually there really is.
133
Q

What information about the genomes do we lose by representing it as a string?

  • A. Read accessibility
  • B. Codon frequency
  • C. ORF start/stop sites
  • D. All of the above
  • E. None of the above
A

Read accessibility

134
Q

E. coli cells are grown for many generations on 15N until all their nitrogenaous molecules are uniformly labeled, and their cell divisions are then synchronized so that all the cells divide at the same time. Once synchronized they are quickly diluted into a medium containing only 14 N. After three cell divisions what percent of the duplex, double stranded DNA molecules will contain 15N?

  • A. 100%
  • B. 50%
  • C. one third
  • D. 25%
  • E. none of these
A

D. 25%

135
Q

Which of the following is involved in initiation of DNA replication in E. coli?

  • A. DnaC
  • B. ATP
  • C. DnaB
  • D. Single strand binding protein
  • E. more than one of the above
A

E. more than one of the above

136
Q

Where does a large part of the driving free energy come from in the biosynthesis of both DNA and RNA from deoxynucleotide triphosphates and nucleotide triphosphates?

  • A. formation of a new phosphodiester bond
  • B. hydrolysis of the dNTP or NTP to dNDP or NDP
  • C. pyrophosphatase activity
  • D. dehydration.
  • E. none of the above
A

C. pyrophosphatase activity

137
Q

In the biosynthesis of mRNA in E. coli

  • A. The DNA template is transcribed in the 3’ to 5’ direction, while the RNA is built in the 5’ to 3’ direction.
  • B. The DNA template is transcribed in the 5’ to 3’ direction, while the RNA is built in the 3’ to 5’ direction.
  • C. The DNA template is transcribed in the 3’ to 5’ direction, while the RNA is also built in the 3’ to 5’ direction.
  • D. The DNA template is transcribed in the 5’ to 3’ direction, while the RNA is also built in the 5’ to 3’ direction.
  • E. It depends upon which DNA strand is being transcribed
A

A. The DNA template is transcribed in the 3’ to 5’ direction, while the RNA is built in the 5’ to 3’ direction.

138
Q

Why are upstream promoter regions of DNA rich in A and T bases?

  • A. They fit the RNA polymerase active site better than regions high in G and C.
  • B. They are easier to hydrolyze
  • C. They bind sigma factors more tightly.
  • D. They allow for easier strand separation so that polymerase activity can be initiated.
  • E. none of the above
A

D. They allow for easier strand separation so that polymerase activity can be initiated.

139
Q

What are there many different types of sigma subunits involved in E. coli RNA polymerase activity?

  • A. So much mRNA needs to be made that many sigma proteins are needed at the same time
  • B. Most arise from pseudogenes and are never used.
  • C. Each type is involved in transcription of one or more sets of proteins for specific purposes.
  • D. Many types are needed, as they are so short-lived that new ones are made all the time.
  • E. none of the above
A

C. Each type is involved in transcription of one or more sets of proteins for specific purposes.

140
Q

Which of the following amino acyl tRNA synthetases is most likely to have an editing function in addition to its ability to charge a tRNA with the amino acid?

  • A. tryptophan tRNA synthetase
  • B. arginine tRNA synthetase
  • C. glycine tRNA synthetase
  • D. leucine tRNA synthetase
  • E. all these are equally likely to have an editing function
A

D. leucine tRNA synthetase

141
Q

What is/are the role(s) of the catalytic site magnesium ions in the catalytic activity of DNA polymerase?

  • A. One aids in forming AT base pairs, while the other aids in forming GC base pairs.
  • B. They stabilize the interactions of aspartic acid side chains with the growing DNA.
  • C. One facilitates the attack of the 3’-hydroxyl of the growing DNA on the α–phosphate of the incoming NTP.
  • D. They facilitate departure of pyrophosphate from the incoming NTP.
  • E. more than one of the above
A

E. more than one of the above

142
Q

E. coli cells are grown for many generations on 15N until all their nitrogenous molecules are uniformly labeled, and their cell divisions are then synchronized so that all the cells divide at the same time. Once synchronized they are quickly diluted into a medium containing only 14 N. After four cell divisions what percent of the duplex, double stranded DNA molecules will contain 15N?

  • A. 100%
  • B. 50%
  • C. one third
  • D. 25%
  • E. none of these
A

E. none of these

143
Q

Which of these is NOT a standard bioinformatics question?

  • A. Where are the genes on the Mycoplasma genome?
  • B. How similar are human and mouse genomes?
  • C. What are the rates of mercury (de)methylation in periphyton biofilm?
  • D. What is the protein sequence encoded by the TP53 gene?
  • E. Where is the active site in a known Rab5 protein structure?
A

C. What are the rates of mercury (de)methylation in periphyton biofilm?

144
Q

Initiation of DNA replication in E. coli uses 14 ATP hydrolyses, eight from DnaA and six from DnaC. Why?

  • A. The energy of hydrolysis contributes to unwinding the DNA.
  • B. The energy of hydrolysis opens the DnaB (helicase) ring, allowing it to wrap around each strand.
  • C. The ATP hydrolysis stabilizes Pol III, allowing it to bind to the replication fork.
  • D. The hydrolysis activates the DAM methylase.
  • E. more than one of the above
A

E. more than one of the above

145
Q

If a replication fork moves from right to left in a diagram, in which direction is the lagging strand synthesized?

  • A. right to left
  • B. left to right
  • C. simultaneously in both directions
  • D. it depends on which proteins are present and functional
  • E. cannot tell without more data
A

B. left to right

146
Q

Magnesium ions are essential to the action of DNA polymerases. What is the primary source of their stability in the active site?

  • A. interactions with the phosphates of the growing DNA chain
  • B. interactions with the incoming nucleoside triphosphate
  • C. interactions with electronegative regions of the bases
  • D. interactions with negatively charged side chains of the protein
  • E. none of the above
A

D. interactions with negatively charged side chains of the protein

147
Q

In bacteria Okazaki fragments are typically about 1500 nucleotides long. The E. coli genome is 4.5 x 106 nucleotides long. In DNA polymerase a β-clamp is installed each time an Okazaki fragment is started as well as on the leading strand. The clamp raises the processivity so that the polymerase can remain on the DNA for 500,000 bases, even though for the fragments it falls off once the fragment is made. The clamp loader, a trimer, carries three ATP molecules, one per monomer, and the energy of hydrolysis of the ATP is used by the loader to install the clamp. How many ATP molecules are needed to install the clamps for one round of E. coli DNA replication?

  • A. 3000
  • B. 9027
  • C. 9
  • D. 4.5 x 106
  • E. 1500
A

B. 9027

148
Q

What is the role of nick translation in lagging strand synthesis?

  • A. It circumvents the need for DNA ligase activity.
  • B. It repairs proofreading errors.
  • C. It replaces the RNA primer with deoxynucleotides.
  • D. It seals the nick between Okazaki fragments.
  • E. none of the above
A

C. It replaces the RNA primer with deoxynucleotides.

149
Q

Deamination of cytosine yields uracil. If this is not repaired, what mutation would be introduced?

  • A. A GC base pair will be replaced by a CG base pair
  • B. A GC base pair will be replaced by an AT base pair
  • C. An additional AT base pair will be introduced at the mutation site.
  • D. A GC base pair will be deleted.
  • E. There will be no mutational effect at all
A

B. A GC base pair will be replaced by an AT base pair

150
Q

Promoters, where RNA transcription begins, are usually rich in AT base pairs. What is the effect of this?

  • A. The RNA polymerase binding site fits AT rich regions better than GC rich regions.
  • B. AT regions have lower mutational frequencies, which helps maintain promoter activity.
  • C. The DNA duplex must be unwound to make RNA from the template strand, and that is easier in AT rich regions than in GC regions.
  • D. It increases the overall transcription rate.
  • E. none of the above
A

C. The DNA duplex must be unwound to make RNA from the template strand, and that is easier in AT rich regions than in GC regions.

151
Q

In the discussion in class of RNA splicing, mention was made of the Pete Seeger song, “Green and Yeller,” also called “Henry My Son.” What did Henry eat that killed him?

  • A. eels
  • B. snakes
  • C. introns
  • D. exons
  • E. none of these
A

B. snakes

152
Q

If an oligopeptide had as its m-RNA sequence Codon Number in the order listed below, where N can represent any nucleotide and codon 14 is the only stop codon, in what sort of cellular environment would the oligopeptide be most likely to be found?

  1. NAN
  2. NAN
  3. NAN
  4. NAN
  5. NAN
  6. NUN
  7. NUN
  8. NUN
  9. NUN
  10. NUN
  11. NUN
  12. NUN
  13. NUN
  14. UAA
  • A. the cytoplasm
  • B. the plasma membrane
  • C. with its carboxy terminal region embedded in a membrane and its amino terminal region in the cytoplasm
  • D. only in the lumen of a eukaryotic endoplasmic reticulum
  • E. in a chloroplas
A

C. with its carboxy terminal region embedded in a membrane and its amino terminal region in the cytoplasm

153
Q

Which of the following sequences is/are likely to be part of an open reading frame?

Sequence 1: 5’ – CUU GUA AGU AAG UAU GUA AGU AAA – 3’

Sequence 2: 5’ – CUU UAA GUA AGU AAG UGA GUA AAA – 3’

Sequence 3: 5’ – CUC AAG UAA GUA AGU AAG UAG AAA – 3’

  • A. Sequence 1
  • B. Sequence 2
  • C. Sequence 3
  • D. Sequences 1 and 2 are equally likely to be parts of ORFs.
  • E. Sequences 1 and 3 are equally likely to be parts of ORFs
A

A. Sequence 1

154
Q

Which of the following is true of Crick’s wobble hypothesis?

  • A. The wobble means that some codons code for more than one amino acid.
  • B. Amino acids having several codons and which differ in the 1st or 2nd bases require different t- RNAs
  • C. First base of anticodon (5’ to 3’) determines how many codons the t-RNA will recognize. If A or C, only one, if U or G, two (less specific), and if I, then three, which is the maximum for any single t-RNA.
  • D. B and C
  • E. A and B
A

D. B and C

  • B. Amino acids having several codons and which differ in the 1st or 2nd bases require different t- RNAs
  • C. First base of anticodon (5’ to 3’) determines how many codons the t-RNA will recognize. If A or C, only one, if U or G, two (less specific), and if I, then three, which is the maximum for any single t-RNA.
155
Q

The Orgel experiment demonstrated that

  • A. a solution of nucleotide monomers would yield a nucleic acid polymer if given a template but no polymerase.
  • B. a solution of nucleotide monomers would yield a nucleic acid polymer if given a polymerase but no template.
  • C. a solution of nucleotide monomers would yield a nucleic acid polymer if given neither a polymerase nor a template.
  • D. a solution of nucleotide monomers would yield a nucleic acid polymer only if given both a template and a polymerase.
  • E. none of the above
A

A. a solution of nucleotide monomers would yield a nucleic acid polymer if given a template but no polymerase.

156
Q

Genome assembly REQUIRES

  • A. Knowledge of the reference genome sequence
  • B. Sanger sequencing
  • C. Pyrosequencing or Sanger sequencing
  • D. Shotgun sequencing
  • E. Matching overlapping read sequences
A

Matching overlapping read sequences

157
Q

If one estimate DNA sequence likelihoods, assuming that every position in the sequence is independent and the likelihoods P(A) = 0.18, P(C) = 0.22, P(G) = 0.27 and P(T) = 0.33, which is the most likely 4-mer?:

  • A. ATGC
  • B. ATCG
  • C. ATTC
  • D. AGTC
  • E. All are equally likely
A

C. ATTC

  • A. ATGC = 0.00352
  • B. ATCG = 0.00352
  • C. ATTC = 0.00431
  • D. AGTC = 0.00352
158
Q

Which of the following is not a type of post-translational modification?

  • A. Proteolysis
  • B. Protein folding
  • C. Glycosylation
  • D. Lipid addition
  • E. Methylation
A

B. Protein folding

159
Q

The first step in protein targeting is:

  • A. Transcription
  • B. Translocation to the Golgi apparatus
  • C. Translocation to the nucleus
  • D. Translocation to the ER
  • E. Transport to the lysosome
A

D. Translocation to the ER

160
Q

How many proteolytic cleavages produce the mature insulin?

  • A. 1
  • B. 2
  • C. 3
  • D. 4
  • E. 13
A

D. 4

161
Q

In the O-linked glycoprotein the carbohydrates are attached to:

  • A. Tyrosine
  • B. Asparagine
  • C. Serine
  • D. Tyrosine and Serine
  • E. Aspartic acid
A

C. Serine

162
Q

What amino-acid in the Ras proteins is modified in the process of farnesylation?

  • A. Methionine
  • B. Cystine
  • C. Threonine
  • D. Arginine
  • E. Lysine
A

B. Cystine

163
Q

Xeroderma pigmentosum (XP) is a rare skin disorder where a person is highly sensitive to sunlight and prone to developing skin cancers. XP is caused by mutations in proteins involved in one of DNA repair pathways. Which DNA repair pathway?

  • A. Nonhomologous end joining (NHEJ)
  • B. Error prone translesion synthesis (TLS)
  • C. Mismatch Repair (MMR)
  • D. Nucleotide Excision Repair (NER)
  • E. Base excision repair (BER)
A

D. Nucleotide Excision Repair (NER)

164
Q

In cryo-EM of unstained specimens, macromolecules are embedded in:

  • A. Water
  • B. Crystalline ice
  • C. Vitreous or amorphous ice
  • D. Uranyl acetate
  • E. Silicone
A

C. Vitreous or amorphous ice

165
Q

The “signal peptide” would be best described as:

  • A. A short sequence of amino acids located at the N-terminus of a protein.
  • B. A small protein that binds to another protein as it passes through a membrane.
  • C. A special receptor protein on the surface of membranes.
  • D. A short sequence of bases in the mRNA molecule that is not translated by the ribosome.
  • E. Six histidines located at the N-terminus of a protein.
A

A. A short sequence of amino acids located at the N-terminus of a protein.

166
Q

Clathrin is an example of:

  • A. A membrane receptor
  • B. A membrane transport protein
  • C. A vesicle coating protein
  • D. A phospholipid head group
  • E. None of the above answers is correc
A

C. A vesicle coating protein

167
Q

Which of the following processes take place in the Golgi apparatus?

  • A. Hydrolysis of proteins, lipids and carbohydrates
  • B. Synthesis of lipids
  • C. Synthesis of N-linked oligosaccharides on dolichol phosphate.
  • D. Covalent modifications of proteins by O-glycosylation.
  • E. All of the above answers are correct.
A

D. Covalent modifications of proteins by O-glycosylation.

168
Q

Which of the following statements is NOT true about the ER?

  • A. N-glycosylation of proteins occurs in the ER.
  • B. The appearance of rough ER is due to ribosomes bound to its surface.
  • C. The smooth ER is where proteins are packaged for transport to the Golgi apparatus.
  • D. In some places the ER membrane is attached to the polysaccharide cell wall.
  • E. Protein folding takes place in the ER.
A

D. In some places the ER membrane is attached to the polysaccharide cell wall.

169
Q

What are examples for biological information that is lost when DNA is represented as a string?

  • A. Read accessibility
  • B. Histone binding
  • C. Spatial proximity of DNA regions
  • D. DNA methylation
  • E. All of the above
A

E. All of the above

170
Q

In Sanger Sequencing, which of the following techniques is NOT applied:

  • A. The DNA is amplified.
  • B. A primer is annealed to the 5’ end of the template strand.
  • C. Template strand, DNA polymerase, dNTPs and ddNTPS are distributed into four containers one for each type of nucleotide.
  • D. Once the ddNTP binds to the DNA template ATP is released and the Luciferase converts ATP into a flash of light.
  • E. Gel electrophoresis is used to arrange each of the four containers’ DNA fragment contents by length, yielding the DNA sequence.
A

D. Once the ddNTP binds to the DNA template ATP is released and the Luciferase converts ATP into a flash of light.

171
Q

Proteolytic modifications of the polypeptide are an important process in the mechanism for protein sorting and transport

  • true
  • false
A

true

172
Q

The amino acid is the signal sequence in any polypeptide chain for

  • protein activity
  • glycosylation site
  • proteolytic site
  • sites for lipid addition
A
  • protein activity
  • proteolytic site
173
Q

Glycosylation is the addition of ______ to the protein.

  • carbohydrate
  • lipid
  • fat
  • minerals
A

carbohydrate

174
Q

In the N-linked glycoprotein teh carbohydrates are attached to which of the following amino-acids?

  • valine
  • threonine
  • asparagine
  • serine
A

asparagine

175
Q

In the O-linked glycoprotein the carbohydrates may be attached to how many amino acids?

  • 1
  • 2
  • 3
  • 4
A

2

176
Q

Initiatio of N-glycosylation occurs in the Golgi apparatus

  • true
  • false
A

false

177
Q

farnesylation adds farnesyl phyrophosphate groups to the _____ amino acid residues

  • methionine
  • cystine
  • threonine
  • arginine
A

cystine

Biochemistry 404 (13 decks)

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