Exam 2

Question 1

In a DNA double helix:

A. purines pair with purines
B. the 2 DNA strands run parallel
C. the 2 DNA strands are identical
D. the 2 DNA strands run antiparallel
E. thymine pairs with cytosine

Answer 1

D. the 2 DNA strands run antiparallel

Question 2

The Central Dogma specifies that the information in:

A. DNA is copied directly into a protein.
B. DNA is copied into an RNA molecule during transcription, and the information in that RNA molecule is used to make a protein during translation.
C. RNA is copied into a DNA molecule during transcription, and the information in that DNA molecule is used to make a protein during translation.
D. DNA is copied into an RNA molecule during translation, and the information in that RNA molecule is used to make a protein during transcription.
E. Protein is copied into an RNA molecule during translation, and the information in that RNA molecule is used to make DNA during transcription.

Answer 2

B. DNA is copied into an RNA molecule during transcription, and the information in that RNA molecule is used to make a protein during translation.

Question 3

Which of the following chemical groups is NOT used to construct a DNA molecule?

A. nitrogen-containing base
B. five-carbon sugar
C. all of the above
D. phosphate
E. six-carbon sugar

Answer 3

E. six-carbon sugar

Question 4

The DNA from 2 different species can often be distinguished by a difference in the

A. ratio of A + T to G + C
B. none of the above
C. ratio of A + G to C + T
D. presence of bases other than A, G, C, and T
E. ratio of sugar to phosphate

Answer 4

A. ratio of A + T to G + C

Question 5

If 30% of the nitrogenous bases in a sample of DNA (double-stranded) from Drew, your PAL tutor, are cytosine, what percentage should be thymine?

Answer 5

20%

Question 6

For a science fair project, Aleena and Barrett decided to repeat teh Hershey and Chase experiment, with modifications. They decided to radioactively label the nitrogen of the DNA, rather than the phosphate. They reasoned that each nucleotide has only one phosphate and 2:5 nitrogen atoms. Thus, labeling the nitrogen atoms would provide a stronger signal than labeling the phosphates. Why won’t this experiment work?

A. Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins.
B. Radioactive nitrogen has a half-life of 100,000 years, and the material would be too dangerous for too long.
C. Although there are more nitrogen’s in a nucleotide, labeled phosphates actually have 16 extra neutrons; therefore, they are a more radioactive.
D. The science fair is to blame.
E. There is no radioactive isotope of nitrogen.

Answer 6

A. Amino acids (and thus proteins) also have nitrogen atoms; thus, the radioactivity would not distinguish between DNA and proteins.

Question 7

A world-renowned pathologist, Claire Ward, studies 2 strains of Streptococcus pneumonia, one that causes a lethal infection when injected into mice, and a second that is harmless. She observes that pathogenic bacteria that have been killed by heating can no longer cause an infection. But when these heat-killed bacteria are mixed with live, harmless bacteria, this mixture is capable of infecting and killing a mouse. what would Claire conclude from her experiment?

A. The infectious strain cannot be killed by heating.
B. The mice had lost their immunity to infection with S. pneumoniae.
C. The heat-killed pathogenic bacteria “transformed” the harmless strain into a lethal one.
D. I am Claire. I made this question and forgot.
E. The harmless strain somehow revived the heat-killed pathogenic bacteria.

Answer 7

C. The heat-killed pathogenic bacteria “transformed” the harmless strain into a lethal one.

Question 8

The human genome is a diploid genome. However, when germ-line cells produce gametes, these specialized cells are haploid. What is the total number of chromosomes found in each of the gametes (egg or sperm) in your body?

A. 46
B. 23
C. 22
D. 23 + X or Y
E. 44

Answer 8

B. 23

Question 9

Which statement is true about the association of histone proteins & DNA?

A. Each histone protein has a deep groove into which a DNA double helix tightly fits.
B. Histone proteins have a high proportion of positively charged amino acids, which bind tightly to the negatively charged DNA backbone.
C. Histone proteins from hydrogen bonds with the nucleotide bases of DNA.
D. Histone proteins have a high proportion of negatively charged amino acids, which bind tightly to the positively charged DNA backbone.
E. Histone proteins insert themselves into the major groove of DNA.

Answer 9

B. Histone proteins have a high proportion of positively charged amino acids, which bind tightly to the negatively charged DNA backbone.

Question 10

Which of the following statements is NOT true?

A. A cell will temporarily decondense its chromatin to give proteins rapid. localized access to specific DNA sequences.
B. When a cell divides, its chromatin structures will typically be inherited by its daughter cells.
C. A cell will temporarily decondense it chromatin to silence genes during differentiation.
D. A cell can permanently condense and silence an entire chromosome during development.
E. A cell will temporarily decondense its chromatin to allow access to specific DNA sequences for replication, repair or gene expression.

Answer 10

C. A cell will temporarily decondense its chromatin to silence genes during development

Question 11

You prepare bacterial extracts by lysing the (circular bacterial) cells and removing insoluble debris via centrifugation. These extracts provide the proteins required for DNA replication. Your DNA template is a small, double-stranded circular piece of DNA (a plasmid) with a single origin of replication and a single replication termination site. The termination site is on the opposite side of the plasmid from the origin. What part of the DNA replication process would most directly be affected if a strain of bacteria lacking helicase were used to make the cell extracts?

A. Okazaki fragment synthesis
B. termination of DNA synthesis
C. leading-strand elongation
D. lagging-strand completion
E. initiation of DNA synthesis

Answer 11

E. initiation of DNA synthesis

Question 12

! Which statement is FALSE?

A. DNA helicase produces transient nicks in the DNA backbone to relieve the tension built up by the unwinding of DNA ahead of the DNA to poison erase.
B. Single-strand DNA -binding protein binds to single-stranded DNA exposed by DNA helicase, preventing base pairs from re-forming before the lagging strand can be replicated.
C. DNA ligase uses the energy of AT hydrolysis to join Okazaki fragments made on the lagging-strand template.
D. Sliding clamp keeps DNA polymerase attached to the template, allowing the enzyme to move along without falling off as it synthesizes new DNA.
E. Primate synthesizes RNA primers along the lagging-strand template.

Answer 12

A. DNA helicase produces transient nicks in the DNA backbone to relieve the tension built up by the unwinding of DNA ahead of the DNA to poison erase

NOT — Sliding clamp keeps DNA polymerase attached to the template, allowing the enzyme to move along without falling off as it synthesizes new DNA — whic is TRUE

Question 13

Which of the following statements is NOT an accurate statement about thymine dimers?

A. Thymine dimers are covalent links between thymidines on opposite DNA strands.
B. Thymine dimers cannot be repaired.
C. Thymine dimers can cause the DNAr explication machinery to stall.
D. Prolonged exposure to sunlight causes thymine dimers to form.
E. Repair proteins recognize thymine dimers as a distortion in the DNA backbone.

Answer 13

A. Thymine dimers are covalent links between thymidines on opposite DNA strands.

Question 14

Which statement is TRUE?

A. homologous recombination cannot occur in prokaryotic cells, because they are haploid, and therefore ahve no extra copy of the chromosome to use as a template for repair.
B. The DNA template used to repair the broken strand is the homologous chromosome inherited from the other parent.
C. After damaged DNA has been repaired, nicks in the phosphate backbone are maintained as a way to identify the strand that was repaired.
D. Nonhomologous end joining is a mechanism that ensures that DNA double-strand breaks are repaired with a high degree of fidelity to the original DNA sequence.
E. all of the above

Answer 14

E. all of teh above

Question 15

Beki Gjyzeli and Lilliana Burton win No-Bell prize in 2021 for their discovery for the molecular mechanisms of DNA repair. Their discovery shows that double-strand DNA breaks caused by ionizing radiation often miss nucleotides at the broken ends. According to their discovery, what type of repair would likely be used, and what would be the result of repairing this type of damage?

A. Nonhomologous end joining would e used and would lead to properly repaired DNA.
B. If duplicated intact chromosomes are not available, nonhomologous end joining would be used to joint the DNA, but errors would still remain.
C. Direct repair would be used and would lead to properly repaired DNA.
D. DNA Mismatch repair would be used and would lead to properly repaired DNA.
E. Direct repair would be used to join the DNA, but errors would still remain.

Answer 15

B. If duplicated intact chromosomes are not available, nonhomologous end joining would be used to join gate DNA, but errors would still remain.

Question 16

The energy for DNA polymerization is provided by the hydrolysis of which of the following?

A. incoming nucleoside triphosphates (releasing PPi)
B. the 3’ end of the growing stand (releasing H2O)
C. GTP (releasing Pi)
D. ATP (releasing Pi)
E. pyrophsophate (releasing Pi)

Answer 16

A. incoming nucleoside triphosphates (releasing PPi)

Question 17

Which statement is true?

A. DNA synthesis in the 3’-to-5’ direction is chemically possible
B. In living systems, DNA synthesis in the 3’-to-5’ direction does not allow proofreading.
C. If DNA synthesis would occur in the 3’-to-5’ direction in living systems, there is no favorable hydrolysis reaction to drive the addition of new nucleotides once a mispaired nucleotide is removed.
D. The synthesis of DNA in living systems occurs in the 5’-to-3’ direction.
E. all of the above

Answer 17

E. all of the above

Question 18

Which of the following statements from our classmates is true?

A. Regan: DNA synthesis begins near promoter regions called replication origins. G and C nucleotides are rich in these regions.
B. Katie: ribosomal RNAs are synthesized and combined with proteins to form ribosomes in the Er & Golgi.
C. Addie: Primase requires a proofreading function that ensures that there are no errors in the RNA primers used for DNA replication.
D. Kaci: The attachment of single-strand DNA-binding proteins to the lagging-strand template powers thea cation of helicase the replication fork, where it opens up the double helix.
E. Autumn: Primase is needed to initiate DNA replication on both the leading strand and the lagging strand.

Answer 18

E. Autumn: Primase is needed to initiate DNA replication on both the leading strand and the lagging strand.

Question 19

Danielle isolated a bacterial strain with a mutation that removes the transcription termination signal from the squid operon. Which of the following statements describes the most likely effect of this mutation on Abd transcription?

A. Sigma factor will not dissociate from RNA polymerase when the squid operon is being transcribed in the mutant strain.
B. The squid RNA will not be produced in the mutant strain.
C. RNA polymerase will move in a backward fashion at the squid operon in the mutating strain.
D. none of the above
E. The squid RNA from the mutant strain will be longer than normal.

Answer 19

E. The squid RNA from the mutant strain will be longer than normal

Question 20

***Which of the following might decrease the transcription of only one specific gene in a bacterial cell?

A. a mutation that introduced extensive sequence changes into the DNA that precedes the gene’s transcription start site
B. a decrease in the amount of TBP
C. a decrease in the amount of RNA polymerase
D. a mutation that introduced a stop codon into the DNA that precedes the gene’s coding sequence
E. a decrease in the amount of sigma factor

Answer 20

A. a mutation that introduced extensive sequence changes into the DNA that precedes the gene’s transcription start site

Question 21

Transcription in bacteria differs from transcription in a eukaryotic cell because

A. RNA polymerase must be phosphorylation at its C-terminal tail for transcription to proceed.
B. RNA polymerase (along with its sigma subunit) requires the general transcription factors to assemble at the promoter before polymerase can begin transcription
C. the sigma subunit must associate with the appropriate type of RNA polymerase to produce mRNAs
D. RNA polymerase (along with its sigma subunit) can initiate transcription on its own
E. a decrease in the amount of sigma factor.

Answer 21

D. RNA polymerase (along with its sigma subunit) can initiate transcription on its own

Question 22

! Choose the statement that is NOT true regarding the function of bacterial RNA polymerase

A. Bacterial RNA polymerase does not require a primer
B. Bacterial RNA polymerase does not require a helicase
C. Bacterial RNA polymerase makes more errors than DNA polymerase
D. Bacterial RNA polymerase can transcribe along the DNA in either direction
E. Bacterial RNA polymerase uses ribonucleoside triphosphates as substrates

Answer 22

D. Bacterial RNA polymerase can transcribe along the DNA in either direction

NOT Bacterial RNA polymerase makes more errors than DNA polymerase — which is true

Question 23

What is true of bacterial mRNAs?

A. They are transcribed & translated simultaneously.
B. They contain extensive non coding sequences.
C. They are processed in the same way as eukaryotic mRNAs.
D. They must be folded into the correct three-dimensional shape before they can be translated.
E. They are translated after they are exported from the nucleus.

Answer 23

A. They are transcribed & translated simultaneously.

Question 24

Which of the following statements best describes the significance of the TATA box in the promoters of eukaryotes?

A. It is the recognition site for the binding of general transcription factors.
B. It is the recognition site for ribosomal binding during translation.
C. It is the recognition site for ribosomal binding during transcription.
D. It sets the reading frame of the mRNA during translation
E. none of the above

Answer 24

A. It is the recognition site for the binding of general transcription factors.

Question 25

Which of the following does not occur before a eukaryotic mRNA is exported from the nucleus?

A. The mRNA is polyadenylated at its 3’ end.
B. None of the above
C. A guanine nucleotide with a methyl group is added to the 5’ end of the mRNA.
D. RNA polymerase dissociates.
E. The ribosome binds to the mRNA.

Answer 25

E. The ribosome binds to the mRNA

Question 26

Which of the following is false about mRNA processing?

A. A long string of adenosine nucleotides is added to the 3’ end.
B. A 7-methyl-guanosine cap is added to the 5’ end.
C. All of the above are true about mRNA processing.
D. It is generally only seen in eukaryotes.
E. Spliceosomes are used to remove non-protein-coding regions (introns)

Answer 26

C. All of the above are true about mRNA processing

Question 27

In an experimental situation, Katie inserts an mRNA molecule into a eukaryotic cell after she has removed its 5’ cap and poly-A tail. Which of the following processes would you expect her to find to have occurred?

A. The molecule is digested by enzymes because it is not protected at the 5’ end.
B. The cell adds a new poly-A tail to the mRNA.
C. The mRNA attaches to a ribosome and is translated, but more slowly.
D. The mRNA is quickly converted into a ribosomal subunit.
E. Not fair. Katie would know but I don’t. I want to do the experiment for myself.

Answer 27

A. The molecule is digested by enzymes because it is not protected at the 5’ end.

Question 28

! Within the ribosome, the formation of peptide bonds it catalyzed by what component?

A. mRNA
B. a peptide we in the small ribosomal subunit
C. aminoacyl-tRNA synthetase
D. an RNA molecule in the large ribosomal subunit
E. tRNA

Answer 28

D. an RNA molecule in the large ribosomal subunit

NOT aminoacyl-tRNA synthetase — enzyme that attaches the appropriate amino acid onto its corresponding tRNA

Question 29

Translation takes place in a series of four steps. Which of these best describes this four-step cycle during elongation?

A. An aminoacyl-tRNA binds to the vacant E site on the ribosome; a peptide bond forms; the large subunit of the ribosome translocates, moving the bound tRNAs to the A and P sites; the small subunit of the ribosome translocates, ejecting the tRNA from the A site.
B. An aminoacyl-tRNA binds to the vacant A site on the ribosome; the large subunit of the ribosome translocates, moving the bound tRNA to the P site; a peptide bond forms; the small subunit of the ribosome translocates, ejecting the tRNA from the E site.
C. An aminoacyl-tRNA binds to the vacant A site on the ribosome; a peptide bond forms; the large subunit of the ribosome translocates, moving the bound tRNAs to the E and P sites; the small subunit of the ribosome translocates, ejecting the tRNA from the E site.
D. The large subunit of the ribosome translocates, moving the bound tRNAs to the E and P sites; an aminoacyl-tRNA binds to the vacant A site on the ribosome; a peptide bond forms; the small subunit of the ribosome translocates, ejecting the tRNA from the E site.
E. An aminoacyl-tRNA binds to the vacant P site on the ribosome; a peptide bond forms; the large subunit of the ribosome translocates, moving the bound tRNAs to the E and P sites; the small subunit of the ribosome translocates, ejecting the tRNA from the E site.

Answer 29

C. An aminoacyl-tRNA binds to the vacant A site on the ribosome; a peptide bond forms; the large subunit of the ribosome translocates, moving the bound tRNAs to the E and P sites; the small subunit of the ribosome translocates, ejecting the tRNA from the E site.

Question 30

This is a piece of Dae-Sung’s DNA. Which of the following RNA molecules could be transcribed from this piece of DNA?

5’-ATAGGCATTCGATCCGGATAGCAT-3’
3’-GATCCGTAAGCTAGGCCTATCGTA-3’

A. 5’-ATAGGCATTCGATCCGGATAGCAT-3’
B. 5’-GATCCGTAAGCTAGGCCTATCGTA-3’
C. 5’-UAUCCGUAAGCUAGGCCUAUGCUA-3’
D. 5’-UACGAUAGGCCUAGCUUACGGAUA-3’
E. 5’-CUAGGCAUUCGAUCCGGAUAGCAU-3’

Answer 30

E. 5’-CUAGGCAUUCGAUCCGGAUAGCAU-3’

Question 31

The greatest similarities among codons that specify the same amino acid occur ____.

A. in the first and third nucleotides of the triplet
B. in the last two nucleotides of the triplet
C. in the first and third nucleotides of the triplet
D. in the first two nucleotides of the triplet
E. in the middle nucleotide of the triplet

Answer 31

D. in the first two nucleotides of the triplet

Question 32

You attempt deletion mapping of a part of the promoter region of a particular gene. You remove a short sequence of nucleotides. Once the altered DNA is transferred into cells, the cells are able to transcribe the transferred DNA in a normal fashion. What do you conclude?

A. The sequence that was deleted is an important determinant of the ability to transcribe the gene.
B. The sequence that was removed is not an essential part of the promoter.
C. Bio329 is hard…
D. The sequence that was removed is an essential part of the promoter.
E. The deleted sequence binds tightly to RNA polymerase.

Answer 32

B. The sequence that was removed is not an essential part of the promoter.

Question 33

Which of the following statements about the proteasome is FALSE?

A. The protein stopper that surround the central cylinder of proteasome use the energy from ATP hydrolysis to move proteins into the proteasome inner chamber.
B. Ubiquitin is a small protein that is covalently attached to proteins to mark them for delivery to the proteasome.
C. Misfolded proteins are delivered to the proteasome, where they are sequestered from the cytoplasm and can attempt to refold.
D. None of the above.
E. Proteases reside in the central cylinder of a proteasome.

Answer 33

C. Misfolded proteins are delivered to the proteasome, where they are sequestered from the cytoplasm and can attempt to refold.

Question 34

Muscle cells differ from nerve cells mainly because they?

A. use different genetic codes.
B. contain different genes.
C. express different genes.
D. have unique ribosomes.
E. express same but independent genes.

Answer 34

C. express different genes.

Question 35

! Which of the following is true regarding gene expression?

A. A change in genotype always results in a changed phenotype.
B. Mistakes in transcription are corrected by RNA polymerase.
C. The ribosome binding site lies at the 3’ end of mRNA.
D. A second round of transcription can begin before the preceding transcript is completed.
E. Only one gene can be present within any RNA sequence.

Answer 35

D. A second round of transcription can begin before the preceding transcript is completed.

Question 36

Which of the following statements about transcriptional regulators is FALSE?

A. The DNA-binding motifs of transcription regulators usually bind in the major groove of the DNA helix.
B. Transcription regulators interact only with the sugar phosphate backbone on the outside of the double helix to determine where to bind on the DNA helix.
C. The binding of transcription regulators generally does not disrupt the hydrogen bonds that hold the double helix together.
D. Transcription regulators will form hydrogen bonds, ionic bonds, and hydrophobic interactions with DNA.
E. none of the above

Answer 36

B. Transcription regulators interact only with the sugar phosphate backbone on the outside of the double helix to determine where to bind on the DNA helix.

Question 37

! The tryptophan operator…

A. is an allosteric protein.
B. inducible operon.
C. is required for production of the mRNA encoded by the tryptophan operon.
D. is important for the production of the tryptophan repressor.
E. binds to the tryptophan repressor when the repressor is bound to tryptophan.

Answer 37

E. binds to the tryptophan repressor when the repressor is bound to tryptophan.

NOT is required for production of the mRNA encoded by the tryptophan operon.

Question 38

Which of the following statements about the Lac operon is FALSE?

A. The CAP activator can only bind DNA when it is bound to cAMP.
B. The lactose operon is inducible.
C. Even when the CAP activator is bound to DNA, if lactose is not present, the Lac operon will not be transcribed.
D. The Lac operon only produces RNA when lactose is present and glucose is absent.
E. The Lac repressor binds when lactose is present in the cell.

Answer 38

E. The Lac repressor binds when lactose is present in the cell.

Question 39

What is the role of general transcription factors (GTFs)?

A. GTFs are cis-acting regulatory sequences.
B. GTFs bind to enhancers or silencers and regulate transcription.
C. GTFs regulate the length of the mRNA.
D. GTFs bind to the core promoter and allow transcriptional initiation.
E. GTFs are part of the RNA polymerase II holoenzyme, and control transcription initiation.

Answer 39

D. GTFs bind to the core promoter and allow transcriptional initiation.

Question 40

Which of the following statements about nucleosomes is TRUE?

A. Nucleosomes activate transcription when bound to the promoter.
B. Histone acetyltransferases affect transcription by both altering chromatin structure to allow accessibility to the DNA and by adding acetyl groups to histones that can bind proteins that promote transcription.
C. none of the above
D. Although RNA polymerase can access DNA packed within nucleosomes, the general transcription factors and transcriptional regulators cannot.
E. Histone deacetylases remove lysines from histone tails.

Answer 40

B. Histone acetyltransferases affect transcription by both altering chromatin structure to allow accessibility to the DNA and by adding acetyl groups to histones that can bind proteins that promote transcription.

Question 41

! Which of the following statements about iPS cells is FALSE?

A. Stimulation by extracellular signal molecules causes iPS cells to differentiate.
B. iPS cells can be made by adding a combination of transcription regulators to a fibroblast.
C. A cell that is dedifferentiated to become an iPS cell will undergo changes to its gene expression profile.
D. iPS cells made from mouse cells can differentiate into almost any human cell type.
E. iPS was developed by the Novel Laureate Sir Gordon, who once failed in BIO329 with Dr. Hwangbo.

Answer 41

D. iPS cells made from mouse cells can differentiate into almost any human cell type

iPS cells can only be differentiated into different cell types of the SAME ORGANISM

Question 42

! Which of the following statements about DNA methylation in eukaryotes is FALSE?

A. Immediately after DNA replication, each daughter helix contains one methylated DNA strand, which corresponds to the newly synthesized strand.
B. Appropriate inheritance of DNA methylation patterns involves maintenance methyltransferase.
C. DNA methylation involves a covalent modification of cytosine bases.
D. Methylation of DNA attracts proteins that block gene expression.
E. all of the above

Answer 42

A. Immediately after DNA replication, each daughter helix contains one methylated DNA strand, which corresponds to the newly synthesized strand.

Question 43

Which of the following statements is TRUE?

A. all of the above
B. Transcription regulators recognize nucleotide sequences by fitting tightly to the surface features of the DNA double helix, allowing multiple weak interaction such as hydrogen bonds to form, yet resulting in a tight interaction.
C. Although all of the steps involved in expressing a gene can in principle be regulated, transcription initiation is the most important stage of control for most genes.
D. Even though control of eukaryotic gene expression is combinatorial, the effect of a single transcription regulator can still be decisive in switching any particular gene on or off.
E. Examples of epigenetic inheritance include the inheritance of patterns of chromosome condensation, the inheritance of methylation patterns in DNA, the inheritance of a regulatory protein that activates its own transcription.

Answer 43

A. all of the above

Question 44

! Which form of post-transcriptional control is being widely exploited by scientists to “knock down” genes of interest?

A. CRISPR/CAS9
B. promoter blocking
C. RNA interference
D. DNA methylation
E. histone deacetylation

Answer 44

C. RNA interference

NOT promoter blocking

Question 45

(EC) Calvin genetically alters the genome of an E. coli bacterium, introducing the gene that codes for telomerases. how will this affect the E. coli bacterium?

A. The bacterium will live longer.
B. Calvin's question is not good.
C. The bacterium will live shorter.
D. The bacterium will be unaffected.
E. The bacterium will have loner telomere and longer lifespan.

Answer 45

D. The bacterium will be unaffected

Question 46

(EC) If GTP supplies in the cell decreased, which of the following processes would be directly affected?

A. The degradation of unwanted/waste mRNA molecules.
B. The ability of aminoacyl-tRNA synthetases to charge tRNA molecules.
C. The folding process of tRNA molecules into the cloverleaf shape.
D. The formation of peptide bonds between amino acids within the ribosome during translation.
E. I don’t like the question. lol

Answer 46

D. The formation of peptide bonds between amino acids within the ribosome during translation