Question 1

How would a large chromosomal inversion affect the expression of a gene if the gene is located between the two break points but no breaks occur within the gene?
A)The inversion would probably have no effect on gene expression.
B)The gene would not be transcribed because it would be oriented in the wrong direction.
C)The gene would be transcribed in the 3' to 5' direction.
D)The gene would be transcribed normally but the mRNA would be translated in the 3' to 5' direction.

Accepted Answer

If no breaks occur within the gene, its sequence and regulatory elements should remain intact, and the gene should be able to be transcribed and translated normally. The orientation of the gene would also remain the same, so it would still be transcribed in the 5' to 3' direction. Therefore, the large chromosomal inversion would probably have no effect on gene expression.

Question 2

A scientist makes three artificial mRNA strands: 
(x)5' AAAUUUAAAUUUAAAUUUAAAUUUAAA 3'
(y)5' UUUCCCUUUCCCUUUCCCUUUCCCUUU 3'
(z)5' AUAUAUAUAUAUAUAUAUAUAUAUAU 3'
When he analyzes the polypeptides produced,he finds that:
X produces a polypeptide that is 50% phenylalanine and 50%lysine.
Y produces a polypeptide that is 50% phenylalanine and 50% proline.
Z produces a polypeptide that is 50% isoleucine and 50% tyrosine.
Based on these results only,the best conclusion to make is that
A)AUA codes for isoleucine
B)AAA codes for phenylalanine
C)AAA codes for lysine
D)AAA codes for lysine and AUA codes for isoleucine
E)AAA codes for phenylalanine and AUA codes for isoleucine

Accepted Answer

The mRNA strand X contains the sequence AAA, which results in lysine production, indicating that AAA codes for lysine.

Question 3

What is the likely consequence of a mutation that alters the branch point within an intron?
A)no effect,since introns are not expressed
B)failure to form a lariat
C)failure of snRNPs to recognize the 5' end of intron
D)no exon shuffling
E)failure of snRNAs to combine with protein and form snRNPs

Accepted Answer

The branch point within an intron is necessary for the formation of a lariat structure during splicing. Mutations that alter the branch point may cause splicing errors, leading to a failure to form a lariat and ultimately, aberrant mRNA processing.

Question 4

How does DNA polymerase differ from RNA polymerase?
A)Only RNA polymerase adds new nucleotides to the 3' end of a growing chain.
B)Only RNA polymerase requires a primer.
C)Only DNA polymerase uses a template DNA strand to direct synthesis of a new nucleotide strand.
D)Only DNA polymerase has a proofreading ability.

Accepted Answer

DNA polymerase has a proofreading ability, which allows it to check the accuracy of the newly synthesized DNA strand and correct any errors. RNA polymerase does not have this ability.

Question 5

The mutation responsible for Huntington's disease is a
A)missense mutation.
B)nonsense mutation.
C)frameshift mutation.
D)triplet repeat expansion mutation.

Accepted Answer

The mutation responsible for Huntington's disease is a triplet repeat expansion mutation, where a repeated sequence of three nucleotides (CAG) in the DNA is expanded, leading to a change in the protein produced by the gene. This results in the accumulation of abnormal proteins in the brain, leading to the symptoms of the disease. Missense, nonsense, and frameshift mutations involve different types of changes in the DNA sequence that can also lead to changes in protein structure and function, but are not specific to Huntington's disease.

Question 6

You are attempting to synthesize rRNA in a test tube using DNA isolated from mouse cells.In addition to the template DNA,ribonucleotides,and the necessary transcription factors,you should also add ________ to the test tube.
A)poly-A polymerase
B)RNA polymerase III
C)RNA polymerase II
D)RNA polymerase I

Accepted Answer

RNA polymerase I is responsible for synthesizing rRNA in eukaryotic cells. Therefore, adding RNA polymerase I to the test tube would be necessary for successful synthesis of rRNA. Poly-A polymerase adds poly-A tails to the 3' end of mRNA and RNA polymerase II and III are not involved in rRNA synthesis.

Question 7

Analysis of a bacterial gene indicates that 24% of the nucleotides in the coding strand contain the base A,18% contain the base T,26% contain the base G,and 32% contain the base C.What percent of the nucleotides in the mRNA transcribed from this gene will contain the base U? (Enter your answer as a number only without the percent sign,e.g.enter 5 not 5%.)

Accepted Answer

The answer of Analysis of a bacterial gene indicates that...

Question 8

In prokaryotes,the RNA polymerase holoenzyme consists of
A)the core polymerase plus two alpha subunits.
B)the core polymerase plus two beta subunits.
C)the core polymerase plus two alpha subunits,two beta subunits,and a sigma subunit.
D)the core polymerase plus a sigma subunit.
E)two alpha subunits,two beta subunits,and two sigma subunits.

Accepted Answer

The RNA polymerase holoenzyme in prokaryotes consists of the core polymerase plus a sigma subunit. The sigma subunit is necessary for the recognition of the promoter region on DNA and the initiation of transcription. Once transcription has begun, the sigma subunit is released, leaving behind the core polymerase to continue elongation.

Question 9

You are working to characterize a novel protein in mice.Analysis shows that high levels of the primary transcript that codes for this protein are found in tissue from the brain,muscle,liver,and pancreas.However,an antibody that recognizes the C-terminal portion of the protein indicates that the protein is present in brain,muscle,and liver,but not in the pancreas.What is the most likely explanation for this result?
A)The gene that codes for this protein is not transcribed in the pancreas.
B)The transcript is immediately degraded in the pancreas.
C)The transcript receives a 5' cap in the pancreas.
D)Alternative splicing in the pancreas creates a protein missing the domain that the antibody recognizes.
E)The transcript receives a 3' poly-A tail in the pancreas.

Accepted Answer

Alternative splicing is a mechanism that allows different transcripts to be produced from a single gene, by including or excluding different exons during RNA processing. In this case, it is likely that the transcript is spliced differently in the pancreas, leading to the production of a protein that lacks the C-terminal portion recognized by the antibody. This would explain why the transcript is present in the pancreas but the protein is not detected by the antibody.

Question 10

Within the transcription bubble,the 9 most recently added nucleotides in the newly synthesized RNA strand temporarily form a helix with the template DNA strand.How might transcription be affected if helix formation did not occur?
A)Rewinding the DNA molecule would be inhibited.
B)Unwinding the DNA molecule would be inhibited.
C)The position of the 5' end of the RNA would be unstable,inhibiting elongation.
D)The position of the 3' end of the RNA would be unstable,inhibiting elongation.
E)The position of the 5' end of the RNA would be unstable,stimulating elongation.

Accepted Answer

The helix formation between the newly synthesized RNA and the template DNA strand helps stabilize the 3' end of the RNA, ensuring proper alignment for elongation. Without this stabilization, the 3' end would be unstable, potentially inhibiting the elongation process.

Question 11

What is the best way to describe our current understanding of the one-gene/one-polypeptide hypothesis?
A)It continues to be an accurate description of how genes work.
B)It is oversimplified,especially in eukaryotes.
C)It has been replaced by the more accurate "one-gene/one-enzyme" hypothesis.
D)It was true for fungi like Neurospora,but not for other species.

Accepted Answer

Our current understanding has revealed that some genes can code for multiple polypeptides (through alternative splicing) and that some polypeptides are made up of multiple subunits coded for by different genes. Therefore, the one-gene/one-polypeptide hypothesis is oversimplified, especially in eukaryotes.

Question 12

Cells conserve energy and resources by making active proteins only when they are needed.If a protein is not needed,which of the following methods of control would be the most energy-efficient?
A)block transcription
B)degrade the mRNA after it is made
C)prevent translation of the mRNA
D)degrade the protein after it is made

Accepted Answer

Blocking transcription is the most energy-efficient method because it prevents the initial production of mRNA, thereby conserving resources that would be used in mRNA synthesis, translation, and potential degradation of unnecessary proteins.

Question 13

You are excited because your lab is studying the first microorganisms retrieved from Mars! Interestingly,their genetic material consists of a single-stranded nucleic acid with six different nitrogenous bases.Like organisms from Earth,segments of the nucleic acid contain the code for making proteins.If the genetic code in these organisms was read in groups of 2 bases instead of 3,what is the maximum number of amino acids that could be coded for assuming that only one codon functioned as a stop codon? (Enter your answer as a numeral not a word,e.g.enter 17 not seventeen.)

Accepted Answer

The answer of You are excited because your lab is...

Question 14

Two 6-base sequences are present in bacterial promoters: TATAAT (located 10 nt upstream from the start site)and TTGACA (located 35 nt upstream from the start site).What is the significance of the fact that these two base sequences are different?
A)Binding sites for both the holoenzyme and ATP are provided.
B)Both the location of the start site and the direction of transcription can be established.
C)Binding sites for both the core polymerase and holoenzyme are provided.
D)The transcription bubble can be properly formed.
E)It allows RNA polymerase to distinguish between the template strand and the coding strand of the DNA molecule.

Accepted Answer

The presence of distinct sequences at -10 (TATAAT) and -35 (TTGACA) positions in bacterial promoters allows RNA polymerase to recognize the start site of transcription and the direction in which transcription will proceed, ensuring accurate gene expression.

Question 15

If the genetic code were read in groups of two bases,how many different codons would be possible? (Enter your answer as a numeral not a word,e.g.enter 5 not "five".)

Accepted Answer

The answer of If the genetic code were read in...

Question 16

You are studying an individual with very low levels of insulin in her blood.Further analysis indicates that cells of her pancreas are producing normal levels of this protein,but it is accumulating in the cytoplasm rather than being secreted from the cells.Which hypothesis makes the most sense to explain this observation?
A)A small deletion has removed the nucleotides that code for the signal sequence at the amino terminus of the protein.
B)A missense mutation has caused premature termination during translation of this protein.
C)A chromosomal segment that includes the gene for insulin has been inverted.
D)A two-base deletion near the middle of the gene has altered the reading frame during translation of the protein.
E)A missense mutation has altered the ribosome-binding sequence at the 5' end of the mRNA.

Accepted Answer

The signal sequence is necessary for directing the protein to the endoplasmic reticulum (ER) where it can be properly processed and secreted. Without the signal sequence, the protein would not be able to enter the ER and would instead accumulate in the cytoplasm. This fits with the observation that the cells are producing normal levels of insulin, but it is not being secreted. The other choices do not provide a clear explanation for why the protein is not being secreted.

Question 17

A gene that codes for a protein was removed from a eukaryotic cell and inserted into a prokaryotic cell.Although the gene was successfully transcribed and translated,it produced a different protein than it produced in the eukaryotic cell.What is the most likely explanation?
A)There are slight differences in the genetic code for prokaryotes and eukaryotes.
B)Unlike eukaryotes,which have three different RNA polymerases,prokaryotes have a single RNA polymerase.
C)Eukaryotic genes often contain introns while prokaryotic genes do not.
D)Eukaryotic transcripts have a 5' cap while prokaryotic transcripts do not.

Accepted Answer

Prokaryotes do not have introns in their genes, whereas eukaryotes do. Therefore, the presence of introns in the eukaryotic gene may have resulted in a different protein being produced in the prokaryotic cell.

Question 18

What is required for formation of the transcription initiation complex in eukaryotes?
A)binding of a transcription factor to the TATA box,followed by recruitment of additional transcription factors and recruitment of RNA polymerase II
B)binding of a transcription factor to the transcription bubble,followed by recruitment of additional transcription factors and recruitment of RNA polymerase III
C)binding of the sigma subunit to the start site followed by recruitment of RNA polymerase II
D)binding of RNA polymerase II to the TATA box,followed by recruitment of transcription factors
E)binding of the sigma subunit to promoter elements at -35 and -10,followed by recruitment of the core polymerase

Accepted Answer

The formation of the transcription initiation complex in eukaryotes requires the binding of a transcription factor to the TATA box, followed by recruitment of additional transcription factors and recruitment of RNA polymerase II. This complex then forms the preinitiation complex, which is required for transcription to begin.

Question 19

A bacterial cell has a nonsense mutation that prevents it from producing a functional sigma subunit for RNA polymerase.Inability to synthesize a functional sigma subunit would have the most direct effect on
A)transcription initiation.
B)transcription elongation.
C)transcription termination.
D)translation initiation.
E)translation termination.

Accepted Answer

The sigma subunit of RNA polymerase is responsible for recognizing and binding to the promoter sequence on DNA to initiate transcription. Without a functional sigma subunit, the bacterial cell would not be able to properly initiate transcription, leading to a direct effect on transcription initiation. The other options (B, C, D, and E) are not directly influenced by the absence of a functional sigma subunit.

Quiz 15: Genes and How They Work

How would a large chromosomal inversion affect the expression of a gene if the gene is located between the two break points but no breaks occur within the gene?

What is the likely consequence of a mutation that alters the branch point within an intron?

How does DNA polymerase differ from RNA polymerase?

The mutation responsible for Huntington's disease is a

You are attempting to synthesize rRNA in a test tube using DNA isolated from mouse cells.In addition to the template DNA,ribonucleotides,and the necessary transcription factors,you should also add ________ to the test tube.

In prokaryotes,the RNA polymerase holoenzyme consists of

Within the transcription bubble,the 9 most recently added nucleotides in the newly synthesized RNA strand temporarily form a helix with the template DNA strand.How might transcription be affected if helix formation did not occur?

What is the best way to describe our current understanding of the one-gene/one-polypeptide hypothesis?

Cells conserve energy and resources by making active proteins only when they are needed.If a protein is not needed,which of the following methods of control would be the most energy-efficient?

Two 6-base sequences are present in bacterial promoters: TATAAT (located 10 nt upstream from the start site) and TTGACA (located 35 nt upstream from the start site) .What is the significance of the fact that these two base sequences are different?

If the genetic code were read in groups of two bases,how many different codons would be possible? (Enter your answer as a numeral not a word,e.g.enter 5 not "five".)

A gene that codes for a protein was removed from a eukaryotic cell and inserted into a prokaryotic cell.Although the gene was successfully transcribed and translated,it produced a different protein than it produced in the eukaryotic cell.What is the most likely explanation?

What is required for formation of the transcription initiation complex in eukaryotes?

A bacterial cell has a nonsense mutation that prevents it from producing a functional sigma subunit for RNA polymerase.Inability to synthesize a functional sigma subunit would have the most direct effect on