Chapter 3 - Gene Expression (Transcription), Synthesis of Proteins (Translation), and Regulation of Gene Expression Flashcards
Which one of the following is true for a double-stranded DNA molecule?
[A]=[T] / [U]=[A] / [G]=[C] / [C]=[T] / Overall charge
Base pairs per one turn of the helix

(A) Yes / No / Yes / No / Negative / 10
(B) Yes / Yes / No / Yes / Positive / 10
(C) Yes / No / Yes / No / Negative / 12
(D) No / Yes / Yes / Yes / Positive / 12
(E) No / No / No / No / Negative / 10
(F) No / Yes / Yes / Yes / Positive / 12
the answer is A. On a molar basis, DNA contains equal amounts of adenine and thymine and of guanine and cytosine. Uracil is not found in DNA. There are 10 base pairs per turn of the helix, and the overall charge on the molecule is negative, due to the phosphates in the back- bone (each phosphodiester bond contains one negative charge).
A bacterial mutant grows normally at 32°C but at 42°C accumulates short segments of newly synthesized DNA. Which one of the following enzymes is most
likely to be defective at the nonpermissive temperature (the higher temperature) in this mutant?
(A) DNA primase
(B) DNA polymerase
(c) An exonuclease
(d) An unwinding enzyme (helicase)
(E) DNA ligase
the answer is E. The short segments of the newly synthesized DNA that accumulate at 42°C are Okazaki fragments. They are usually joined together by DNA ligase, which most likely exhibits reduced activity at 42°C in this mutant. If the ligase is not functioning, Okazaki fragments
would not be joined during replication, so the cells would contain short fragments of the DNA. Endonucleases and exonucleases cleave DNA strands in the middle and at the ends, respec- tively. They do not join fragments together, nor does DNA polymerase. Unwinding enzymes “unzip” the parental strands, and if these were defective, DNA synthesis most likely would not occur at the nonpermissive temperature, and short DNA fragments would not accumulate.
An RNA produced from a fragment of DNA has the sequence of AAUUGGCU. The sequence of the nontemplate strand in the DNA that gave rise to this sequence is which one of the following? (A) AGCCAATT (B) AAUUGGCU (c) AATTGGCT (d) TTAACCGA (E) UUAACCGA
the answer is c. The nontemplate strand in the DNA is the same as the coding strand, and will have the same sequence of the RNA that is produced, except that T is in place of U. Both the nontemplate strand and RNA produced will be a complementary sequence to the template strand. As all sequences are written in the 5′ to 3′ direction, unless otherwise specified, an RNA sequence of AAUUGGCU would correspond to a DNA sequence, on the nontemplate strand, of AATTGGCT. The template strand would be the complement of the nontemplate strand, or AGCCAATT (written 5′ to 3′). Note that this sequence is also the complement of
the RNA that has been produced. The base U is not found in DNA (so answer choices B and E cannot be correct).
Which one of the following changes in the coding region of an mRNA (caused by a point mutation) would result in translation of a protein identical to the normal protein? (A) UCA → UAA (B) UCA → CCA (c) UCA → UCU (d) UCA → ACA (E) UCA → GCA
the answer is c. UCA is a codon for serine. Of the answer choices given, only the codon UCU is also a codon for serine, which would result in a silent mutation (serine would be placed in the protein even though the DNA had been mutated from TGA to TGT). Conversion of UCA
to UAA will generate a termination codon, and a truncated protein would be produced. The conversion of UCA to CCA would replace the serine with a proline in the amino acid. Conver- sion of UCA to ACA results in a threonine being placed in the protein in place of serine, and generation of GCA from UCA would result in alanine being incorporated into the protein in place of the serine. Only the change of UCA to UCU results in the exact same amino acid sequence being produced by the protein.
Proteins destined for secretion from eukaryotic cells have which of the following in common?

An N-terminal methionine in the mature protein is: / A signal peptide located at: / Synthesized on which type of ribosome? / Embedded within the ER membrane?
(A) Very likely / Carboxy terminus / Rough / Yes
(B) Very likely / Amino terminus / Cytoplasmic / No
(C) Very likely / Carboxy terminus / Rough / Yes
(D) Unlikely / Amino terminus / Rough / No
(E) Unlikely / Carboxy terminus / Cytoplasmic / Yes
(F) Unlikely / Amino terminus / Cytoplasmic / No
the answer is d. Proteins destined for secretion contain a signal sequence at the N-terminal end that causes the ribosomes on which they are being synthesized to bind to the SRP, which transfers the mRNA–ribosome complex to the RER. As they are being produced, they enter the cisternae of the RER, where the signal sequence, including the initial methionine, is removed. It is thus unlikely that the mature protein will contain an N-terminal methionine. Carbohydrate groups can be attached in the RER or the Golgi. Secretory vesicles bud from the Golgi, and the proteins are secreted from the cell by the process of exocytosis. If the proteins have a hydrophobic sequence that embeds in the membrane, they remain attached to the membrane and are not secreted, and become membrane-bound proteins.
Inducible bacterial operons exhibit which of the following properties?

Inducer binds to the repressor and: / Inducer effect on RNA polymerase binding to the promoter / Repressor produced by:
(A) Activates the repressor / Enhances / The polycistronic message
(B) Activates the repressor / Enhances / A separate gene
(C) Activates the repressor / Inhibits / The polycistronic message
(D) Inhibits the repressor / Inhibits / A separate gene
(E) Inhibits the repressor / No effect / The polycistronic message
(F) Inhibits the repressor / No effect / A separate gene
the answer is F. In induction, a regulatory gene produces an active repressor, which is inactivated by binding to the inducer. The inducer prevents binding of the repressor to the operator rather than stimulating the binding of RNA polymerase. The structural genes are coordinately expressed. Transcription yields a single, polycistronic mRNA, which is translated to produce a number of different proteins. The regulatory gene is not a part of the polycistronic message.
Processes that, in part, can lead to the activation of gene expression in eukaryotes can be best described as which one of the following?

Methylation of the gene / Formation of polycistronic messages / Histone acetylation levels
(A) Increased / Yes / Decreased (B) Decreased / Yes / Increased (C) Increased / Yes / Increased (D) Decreased / No / Decreased (E) Increased / No / Decreased (F) Decreased / No / Increased
the answer is F. A gene that is methylated is less readily transcribed than the one that is not methylated. Polycistronic mRNAs are only produced in prokaryotes, not in eukaryotic cells. Histone acetylation will be increased in regions of the chromatin that are being transcribed (euchromatin). Acetylation of histones reduces the positive charges on the proteins, thereby weakening their interaction with the negatively charged phosphates on the DNA.
Gene transcription rates and mRNA levels were determined for an enzyme that is induced by glucocorticoids. Compared with untreated levels, glucocorticoid treatment caused a 10-fold increase in the gene transcription
rate and a 20-fold increase in both mRNA levels and enzyme activity. These data indicate that a primary effect of glucocorti- coid treatment is to decrease which one of the following?
(A) The activity of RNA polymerase II
(B) The rate of mRNA translation
(c) The ability of nucleases to act on mRNA
(d) The rate of binding of ribosomes to mRNA
(E) The rate of transcription initiation by RNA polymerase II
the answer is c. If the rate of degradation of the mRNA is not altered by glucocorticoids, then the increase in mRNA levels should reflect the increase in transcription rate. Because the increase in mRNA level is greater than the increase in transcription rate, the glucocorticoids must also be increasing the mRNA stability (i.e., decreasing the rate of degradation by nucleases). The activity of RNA polymerase II is increased (transcription is increased, so transcription initiation is also increased), and the rate of translation (the binding of ribosomes to mRNA) is increased (the enzyme activity is increased), owing to the increased amount of mRNA available.
Which region (A to D) of the DNA strands shown could serve as the template for transcription of the region of an mRNA that contains the initial codon for translation of a protein 300 amino acids in length? Region A: GATGC Region B: TCATT Region C: AATGA Region D: GCATC
(A) A (B) B (c) C (d) D (E) None of the indicated areas would suffice.
the answer is B. Although D contains the sequence 3′-TAC-5′, which produces a start codon (5′-AUG-3′) in the mRNA, there is a sequence (3′-ATT-5′ in the DNA) that would produce a stop codon (5′-UAA-3′) in the mRNA in frame with this start codon. Sequence B, read 3′ to 5′ (from right to left), would produce a start codon in the mRNA transcribed from it. There are no stop codons in this sequence, so it could produce a protein 300 amino acids in length. Sequences A and C do not contain triplets corresponding to the start codon in mRNA.
A temperature-sensitive cell line would show early senescence when grown at the nonpermissive temperature, and examination of the chromosomes demonstrated many 3′ overhangs at the ends of the DNA fragments. The defective enzyme, at the nonpermissive temperature, is which one of the following? (A) Telomerase (B) DNA ligase (c) DNA polymerase (d) A repair DNA polymerase (E) A helicase
the answer is A. Telomerase is defective at the nonpermissive temperature. Owing to the dual requirements of DNA polymerases that they synthesize DNA in the 5′ to 3′ direction, and their need for a template, replicating the ends of linear chromosomes leads to a 3′ overhang after the replication is complete. The overhang is created when the DNA–RNA primer is removed from the template strand, and there is no primer for the DNA polymerase to extend to fill in the gap. Telomerase solves this problem by carrying its own RNA template, and extending the 3′ over- hang. After the gap is filled in as best it can, telomere-binding proteins cap the end of the chromosome to protect it from degradation. Lack of DNA ligase would result in a large number of gaps in the phosphodiester backbone (particularly on the lagging strand), but would not affect the telomeres. Lack of DNA polymerase activity would lead to an overall inhibition of DNA replication, and not just affect the telomeres. Lack of helicase activity would also affect the global DNA replication, not just the replication at the telomeres. Lack of repair polymerase would in- crease the amount of damage in the DNA, but would not specifically target the telomeres.
A family, while on a picnic, picked some wild mushrooms to add to their picnic salad. Shortly thereafter, all the members of the family became ill, with the youngest child showing the most severe symptoms. The family is suffering these effects owing to a primary inability to accomplish which one of the following in their cells and tissues? (A) Synthesize proteins (B) Synthesize lipids (c) Synthesize DNA (d) Synthesize carbohydrates (E) Repair damage in DNA
the answer is A. The poison in poisonous mushrooms is α-amanitin, an inhibitor of eukaryotic RNA polymerases, primarily RNA polymerase II. As the family ate the mushrooms contain- ing the poison, RNA polymerase II stopped functioning, and mRNA was no longer produced. This led to a lack of protein synthesis. There is no direct effect on the synthesis of lipids, carbohydrate, or DNA, other than replacing the required enzymes due to protein turnover. However, the net effect of α-amanitin poisoning would be to stop protein synthesis, which may then lead to a cessation of lipid or DNA synthesis. α-Amanitin has no direct effect on DNA repair.
A newborn has found to be very photophobic, and his skin burns even with minimal exposure to sunlight, eventually forming skin blisters. Neither parent exhibits this trait, although both are prone to burning when in the sun for a short period of time. As the child grows, he is found to be at average height and weight for his age, and is progressing normally along the developmental guidelines. He is, however, kept inside at all times, and is carefully wrapped if he has to leave the house. Fibroblasts isolated from this child are grown in culture, and in an experiment, exposed to UV light. An analysis of the fibroblast DNA will demonstrate which one of the following?
(A) A preponderance of apurinic sites and apyrimidinic sites
(B) An increase in sister chromatid exchange rate
(c) A preponderance of abnormal base pairs in the DNA
(d) Loss of telomeres within the DNA
(E) An increase in cross-linked bases within
the strands of DNA
the answer is E. The child has XP, a defect in nucleotide excision repair such that thymine dimers, created by exposure to UV light, cannot be removed from the DNA. XP will not affect the repair of apurinic or apyrimidinic sites (sites missing just the base from DNA, which requires the AP endonuclease for repair). An increase in sister chromatid exchange rates is a finding in Bloom’s syndrome, which is a defect in a helicase required for both DNA and RNA syntheses. Patients with Bloom’s syndrome are small for their age, unlike those with XP who follow normal developmental milestones. XP does not result in unusual base pairs in DNA, rather the formation of thymine dimers between the adjacent T residues in one strand of DNA. These T residues are still complementary to the A residues in the other strand. XP does not affect the ability of telomerase to extend the ends of the linear chromosomes in the cell.
A 15-year-old boy was diagnosed with skin cancer. He had always been sensitive to sun- light, and had remained indoors for most of his life. An analysis of his DNA, from isolated fibroblasts, indicated an increased level of thymine dimers when the cells were exposed to UV light. The boy developed a skin tumor owing to an increased mutation rate, which was caused by which one of the following?
(A) A lack of DNA primase activity
(B) Decreased recombination during mitosis
(c) Increased recombination during mitosis
(d) Loss of base excision repair activity
(E) Loss of nucleotide excision repair activity
the answer is E. The damage to the DNA caused by UV light (pyrimidine dimers) can be repaired by the nucleotide excision repair pathway. In some cases, the missing enzyme is a repair endonuclease. The boy has XP, as determined by the increase in thymine dimers in his DNA after exposure to UV light. Since the dimers cannot be repaired, the DNA polymerase will “guess” when replication occurs across the dimers, increasing the mutation rate of the cells. Eventually, a mutation occurs in a gene that regulates cell proliferation, and a cancer results. An increase or decrease in mutation rate is not related to the rate of recombination during mitosis, nor to a lack of DNA primase activity (which would lead to reduced DNA synthesis, not inaccurate DNA synthesis). Base excision repair is normal in patients with XP.
A 40-year-old male is well controlled on warfarin for a factor V leiden deficiency and recurrent deep vein thrombosis. He presents today with a community-acquired pneumonia, and is placed on erythromycin. Three days later, he develops bleeding and his INR is 8.0 (indicating an increased time for blood clotting to occur, where INR is international normalized ratio). Which of the following best explains why this bleeding occurred?
(A) The erythromycin inhibited cytochrome P450
(B) The erythromycin stimulated cytochrome P450
(c) The causative agent of the pneumonia inhibited vitamin K utilization
(d) The causative agent of the pneumonia stimulated vitamin K utilization
(E) The erythromycin inhibited mitochondrial translation
(F) The erythromycin inhibited mitochondrial transcription
the answer is A. Warfarin is metabolized by a specific subset of induced p450 isozymes. The p450 system is used by cells to modify the xenobiotic (in this case the warfarin) such that it can be more easily excreted. Erythromycin, along with other macrolide antibiotics, inhibits the p450 oxidizing system, which in this case would lead to a higher blood level of warfa-
rin and, therefore, the balance of clotting and bleeding is shifted toward excessive bleeding.
A stimulation of p450 production by erythromycin would lead to a lower level of warfarin (due to increased metabolism and loss of warfarin by p450) and the potential of excessive clotting. This effect of p450 is a common drug-drug interaction. The causative agents of community- acquired pneumonia do not affect vitamin K absorption in the small intestine, or distribution throughout the body. Erythromycin does not affect mitochondrial transcription, although it may affect mitochondrial translation. Inhibition of mitochondrial protein synthesis, however, will not alter the inhibition of cytochrome p450 activity, and the increased levels of warfarin present, which may lead to increased bleeding.
Your diabetic patient is using the short-acting insulin lispro to control his blood glucose levels. Lispro is a synthetic insulin formed by reversing the lysine and proline residues on the C-terminal end of the B-chain. This allows for more rapid absorption of insulin from the injection site. The engineering of this drug is an example of which of the following technologies? (A) Polymorphism (B) DNA fingerprinting (c) Site-directed mutagenesis (d) Repressor binding to a promoter (E) PCR
the answer is c. This is a prime example of recombinant DNA technology to manufacture a very useful treatment for human diseases, by using site-directed mutagenesis. As the amino acid and DNA sequences of mature insulin is known, the engineering of lispro required that
a proline codon be converted to a lysine codon, and the adjacent lysine codon converted to a proline codon. A polymorphism refers to the differences in DNA sequences amongst individu- als in a population at a particular location within the genome. A polymorphism would not result in the synthesis of lispro insulin. DNA fingerprinting is used to identify unknown DNA samples by comparing the polymorphisms present in the sample DNA to a particular indi- vidual’s DNA. Other than identical twins, everyone’s DNA is different, and can be distinguished by fingerprinting using genetic polymorphisms. Repressor binding to a promoter is part of gene regulation in prokaryotypes, and would not be useful in generating the genetic changes necessary to produce lispro insulin. The PCR can amplify a particular DNA segment between two known segments of DNA, but the technique does not lead to the alteration of amino acid sequence between lispro and normal insulin.
