Chapter 6 Pearson Flashcards
If you wanted to prevent the movement of a large protein into the cytoplasmic membrane, what would be the most effective approach?
Inactivate heat shock proteins in the cell
Inhibit the Tat translocase system
Remove the signal sequence from the protein
Modify the terminal amino acid of the protein
✅ Remove the signal sequence from the protein
The signal sequence is a short peptide found at the beginning of a protein that directs it to the appropriate cellular location. If the signal sequence is removed, the protein will not be recognized by the transport machinery, preventing it from reaching the cytoplasmic membrane.
The first amino acid added during bacterial translation is __________.
formylmethionine, although this may not be the first amino acid in the functional protein
methionine, although this may not be the first amino acid in the functional protein
formylmethionine, which is always the first amino acid in the functional protein
methionine, which is always the first amino acid in the functional protein
✅ formylmethionine, although this may not be the first amino acid in the functional protein
In bacteria, translation starts with N-formylmethionine (fMet), which is the first amino acid incorporated into the nascent polypeptide chain. However, this formyl group is often removed later, meaning fMet may not be present in the final functional protein.
The difference between a nucleoside and a nucleotide is that __________.
a nucleotide lacks a pentose sugar
a nucleoside lacks a phosphate group
a nucleoside lacks a nitrogenous base
a nucleotide lacks a phosphate group
✅ A nucleoside lacks a phosphate group
A nucleoside consists of a pentose sugar (ribose or deoxyribose) and a nitrogenous base, but no phosphate group.
A nucleotide is a nucleoside with one or more phosphate groups attached to the 5’ carbon of the sugar.
Why do gram-negative bacteria require more complex mechanisms for the transport of effectors?
They have a thicker peptidoglycan layer than Archaea.
They have a thicker peptidoglycan layer than gram-positive bacteria.
They have an outer membrane as well as a cytoplasmic membrane.\
They have a thicker plasma membrane than gram-positive bacteria.
✅ They have an outer membrane as well as a cytoplasmic membrane.
Gram-negative bacteria have two membranes—an outer membrane and a cytoplasmic (inner) membrane—which creates an additional barrier for transporting effectors (such as toxins or enzymes) outside the cell. This complexity requires specialized transport systems, like the Type III and Type VI secretion systems, to move molecules across both membranes.
If you find a cell that has operons and a TATA box, you can conclude that the cell is probably __________.
bacterial or archaeal
archaeal
bacterial
eukaryl
✅ archaeal
Explanation:
Operons are common in bacteria and archaea, but rare in eukaryotes.
The TATA box is a promoter element found in archaea and eukaryotes, but not typically in bacteria.
Since the cell has both operons and a TATA box, it is most likely archaeal, as archaea share transcriptional similarities with eukaryotes while still organizing genes into operons like bacteria.
Nucleotides in the nucleic acid backbone are connected by __________.
phosphodiester bonds
ether linkages
double bonds
hydrogen bonds
phosphodiester bonds
Transport into and across the membrane can be powered by __________.
guanosine triphosphate (GTP) and proton motive force only
adenosine triphosphate (ATP) only
adenosine triphosphate (ATP) and guanosine triphosphate (GTP) only
adenosine triphosphate (ATP), guanosine triphosphate (GTP), and proton motive force
✅ adenosine triphosphate (ATP), guanosine triphosphate (GTP), and proton motive force
Explanation:
ATP provides energy for active transport through ABC transporters and other ATP-dependent transport mechanisms.
GTP can power certain transport processes, especially those involving GTP-binding proteins.
Proton motive force (PMF), generated by the electron transport chain, drives transport through secondary active transporters like symporters and antiporters.
During bacterial protein synthesis, the incoming charged tRNAs after the first charged tRNA all initially bind to the __________.
A site, mostly on the 50S subunit
P site, mostly on the 30S subunit
A site mostly on the 30S subunit
P site, mostly on the 50S subunit
A site, mostly on the 50S subunit
During bacterial protein synthesis, the incoming charged tRNAs after the first charged tRNA all initially bind to the A site, mostly on the 50S subunit. The A site is the acceptor site. Each tRNA then moves to the P site (the peptide site) as translation proceeds.
An advantage of having operons is that __________.
groups of products needed simultaneously can be easily turned on and off together
genes needed for different environmental conditions can easily be turned on and off simultaneously
individual genes can be easily turned on and off independently
genes for varied biochemical pathways and conditions are easy to regulate simultaneously
✅ groups of products needed simultaneously can be easily turned on and off together
Explanation:
Operons allow bacteria and archaea to efficiently regulate genes that encode proteins needed for the same pathway or function. Since the genes are transcribed as a single mRNA, they can be turned on or off together in response to environmental changes, saving energy and resources.
To synthesize RNA, RNA polymerase adds nucleotides in a __________.
5’ to 3’ direction, opposite the direction of DNA synthesis
5’ to 3’ direction, as in DNA synthesis
3’ to 5’ direction, opposite the direction of DNA synthesis
3’ to 5’ direction, as in DNA synthesis
✅ 5’ to 3’ direction, as in DNA synthesis
Explanation:
RNA polymerase synthesizes RNA in the 5’ to 3’ direction, just like DNA polymerase does during DNA replication. It adds nucleotides to the 3’ hydroxyl (-OH) end of the growing RNA strand, using the DNA template strand (which is read in the 3’ to 5’ direction).
What would be the products of DNA replication if the process were conservative rather than semiconservative?
There would be two double-stranded daughter molecules that each have strands made up of alternating pieces of parental and new material.
There would be two double-stranded daughter molecules that each have one conserved parental strand and one new daughter strand.
There would be one entirely new double-stranded daughter molecule and one daughter molecule consisting of the two parental strands.
There would be two single-stranded daughter molecules that each contain a mix of new material and parental material.
✅ There would be one entirely new double-stranded daughter molecule and one daughter molecule consisting of the two parental strands.
Explanation:
In conservative DNA replication, the original (parental) DNA molecule remains intact, and an entirely new double-stranded DNA molecule is synthesized. This differs from the actual semiconservative model, where each daughter molecule consists of one parental strand and one newly synthesized strand.
Peptide bonds are broken down by hydrolysis, meaning that __________.
Water is a waste product formed when two amino acids separate
Water is added to the bond to separate two amino acids
Water is needed as a solvent but does not react with the amino acids as they separate
Water is a waste product formed when two amino acids join together to form a peptide bond
✅ Water is added to the bond to separate two amino acids
Explanation:
In hydrolysis, water is added to the peptide bond, breaking it apart and separating the two amino acids. This process is the reverse of condensation (or dehydration synthesis), where water is removed to form a peptide bond between amino acids.
When a new nucleotide is added to a DNA molecule, it is added __________.
by a bond between OH attached to the 5’ carbon of the terminal pentose of the existing chain and phosphate bound to the 3’ carbon of the pentose of the incoming nucleotide
by a bond between P of the phosphate group of the existing chain and OH bound to the 3’ carbon of the pentose of the incoming nucleotide
by a bond between OH attached to the 3’ carbon of the terminal pentose of the existing chain and phosphate bound to the 5’ carbon of the pentose of the incoming nucleotide
by a bond between P of the phosphate group of the existing chain and OH bound to the 5’ carbon of the pentose of the incoming nucleotide
✅ by a bond between OH attached to the 3’ carbon of the terminal pentose of the existing chain and phosphate bound to the 5’ carbon of the pentose of the incoming nucleotide
Explanation:
During DNA replication, a new nucleotide is added to the growing DNA strand by forming a phosphodiester bond between the 3’ hydroxyl group (OH) of the last nucleotide in the chain and the 5’ phosphate of the incoming nucleotide. This process extends the DNA strand in the 5’ to 3’ direction.
What is the most likely explanation if translation produces a polypeptide that differs in every amino acid from what is normally produced?
There was a shortage of amino acids, meaning that the correct ones were not available.
There have been mutations throughout the mRNA transcript.
There was insufficient guanosine triphosphate (GTP) for translation.
There was a shift in the reading frame.
✅ There was a shift in the reading frame.
Explanation:
A reading frame shift occurs when the mRNA is read incorrectly due to insertions or deletions of nucleotides that are not in multiples of three. This alters the grouping of codons, resulting in a completely different sequence of amino acids, which could explain why the polypeptide differs in every amino acid from the normal protein.
How would DNA replication be affected if helicase were not present?
OriC would be unable to function normally.
A new DNA strand could not be formed because the template strand would not be exposed.
Although new DNA strands could be formed, exonuclease activity would be inhibited.
RNA primers could be added, but no DNA strands would form.
A new DNA strand could not be formed because the template strand would not be exposed.
What is the most important reason that it is valuable for bacteria to have mRNAs that are short-lived?
This allows them to reduce the amount of energy needed to synthesize mRNAs.
They can quickly turn off production of a protein when the protein is not needed.
They usually only need proteins relatively briefly.
This allows them to recycle the components of the mRNAs to use them for other purposes.
✅ They can quickly turn off production of a protein when the protein is not needed.
Explanation:
Short-lived mRNAs in bacteria enable them to rapidly regulate gene expression. When a protein is no longer needed, the mRNA can be degraded quickly, preventing unnecessary protein synthesis and conserving energy and resources. This adaptability is crucial for bacteria to respond swiftly to changes in their environment.
How are prokaryotic cells able to rapidly carry out simultaneous transcription and translation when the same is not true in eukaryotic cells?
Prokaryotic cells have RNA that can act as both mRNA and rRNA.
Eukaryotic cells have RNA that can act as both mRNA and rRNA.
Prokaryotic cells have plasmids.
Eukaryotic cells have a nucleus.
✅ Eukaryotic cells have a nucleus.
Explanation:
In prokaryotic cells, transcription and translation occur in the cytoplasm, allowing both processes to happen simultaneously. In contrast, eukaryotic cells have a nucleus, where transcription occurs, and the RNA must then be processed and transported out to the cytoplasm before translation can begin. This separation of processes in eukaryotes limits the ability to perform them simultaneously.
How can DNA polymerase I and DNA polymerase III detect base pairing mismatches?
These enzymes repair mismatches that have been detected by primase.
Mismatches affect the shape of the DNA molecule, indicating where a base needs to be replaced.
As DNA polymerases add bases, they are unable to add incorrect bases and stall until the correct one is added.
These enzymes can identify correct and incorrect base sequences.
✅ Mismatches affect the shape of the DNA molecule, indicating where a base needs to be replaced.
Explanation:
DNA polymerases I and III are able to detect mismatched base pairs because incorrect pairings cause a distortion in the DNA helix. These enzymes can recognize and correct such distortions through their proofreading activity, using their exonuclease activity to remove the incorrect base and replace it with the correct one.
According to the wobble hypothesis, pairing can occur when __________.
the codon and anticodon differ at up to two bases
the codon and anticodon differ at the last base
the codon and anticodon differ at up to three bases
the codon and anticodon differ at the first base
✅ the codon and anticodon differ at the last base
Explanation:
The wobble hypothesis suggests that the third base of the codon (the 3’ end) can pair with a wider range of bases on the anticodon, allowing some flexibility (wobbling) in the base pairing. This flexibility primarily occurs at the third position, allowing for some variations in the codon-anticodon interaction without affecting protein synthesis.
While trying to classify a single-celled organism, you discover that the organism produces single mRNA molecules that encode multiple proteins and conclude that __________.
the cell is eukaryl
the cell is equally likely to be archaeal, bacterial, or eukaryl
the cell is most likely archaeal
the cell is archaeal or bacterial
✅ the cell is archaeal or bacterial
Explanation:
In bacteria and archaea, mRNA often encodes multiple proteins within a single transcript, forming operons. This is in contrast to eukaryotes, where each mRNA typically encodes only one protein. Therefore, the discovery of an mRNA encoding multiple proteins suggests the organism is more likely to be either bacterial or archaeal.
