genetics exam 3
Of the answers below, what is the smallest nucleotide sequence that could code for a protein with 300 amino acids? 300 600 900 1000 2000 3000
1000
A super repressor , I(S), results in no transcription. inducible transcription. transcription but no translation. no translation. constitutive transcription.
a
If the DNA template strand ATG CAT GCA (written 5' to 3') were transcribed to RNA, the RNA would read (all written 5' to 3'): UGC AUG CAU TAC GTA CGT ACG UAC GUA TGC ATG CAT UAC GUA CGU
a
In eukaryotes, transcriptional gene control is mediated by trans-acting factors binding to cis-acting elements. trans-acting factors failing to bind to cis-acting elements. repressor proteins binding to operator sites. metabolites binding to cis-acting elements. none of the above.
a
In the presence of the sugar galactose, the expression levels of GAL1, GAL2, GAL7, and GAL10 are ______________ as compared to the absence of galactose. 1000-fold higher 1000-fold lower 10-fold higher 10-fold lower the same
a
In vitro (in a test tube) translation of a synthetic RNA with the repeating sequence (AGA)20 would produce which of the following? Polypeptides containing arginine, polypeptides containing glutamic acid, and polypeptides containing lysine Polypeptides containing arginine Polypeptides containing a mixture of amino acids (mostly arginine, glutamic acid, and lysine) No translation product because the translation start site was not included on the RNA Polypeptides starting with methionine, followed by arginine additions
a
The A site, P site, and E sites each control ______________ (in order) during translation. binding incoming tRNAs (A), retention of the peptide chain during elongation (P), exit of deacylated tRNAs (E) activation of tRNAs (A), protease-based release of the protein product (P), exit of empty rRNAs (E) activation of ribosome activity (A), processing of amino acids (P), elongation of the peptide chain (E) amino acid bond transfer (A), protein processing (P), and elongation of the peptide chain (E) both A and C are correct.
a
The genetic code is said to be degenerate . This means that a single amino acid has multiple codons. a single codon has multiple amino acids. the code is universal. the code is different in prokaryotes and eukaryotes.
a
What type of library would be most valuable in the isolation of promoter sequences found in front of a mutant gene's transcriptional start site? A. Genomic DNA library B. cDNA library C. miRNA library D. Mutant genomic DNA library E. Repetitive DNA library
a
What type of library would be most valuable in the isolation of promoter sequences found in front of a gene's transcriptional start site? Genomic DNA library cDNA library miRNA library Mutant genomic DNA library Repetitive DNA library
a
Where does translation take place? A. in the cytoplasm of both prokaryotes and eukaryotes B. in the nucleus of eukaryotes and in the cytoplasm of prokaryotes C. in both the nucleus and cytoplasm of eukaryotes D. in the nucleus of both prokaryotes and eukaryotes E. None of the above are correct.
a
Where does translation take place? in the cytoplasm of both prokaryotes and eukaryotes in the nucleus of eukaryotes and in the cytoplasm of prokaryotes in both the nucleus and cytoplasm of eukaryotes None of the above are correct.
a
Which of the following is true of chromatin in prokaryotes? A. Bacterial chromosomes are not organized into chromatin. B. It is found in the nucleus. C. It is found in the cytoplasm. D. It is very similar to chromatin in eukaryotes. E. It uses more different types of histones than chromatin in eukaryotes.
a
Which of the following is true of tRNA but not mRNA? It is a non-coding RNA. It is transcribed by RNA polymerase from a gene. It is found in the cytoplasm of eukaryotes. It contains uracil but not thymine. It is found in the nucleus.
a
Which of the following is true of transcription but not replication? It requires promoters for initiation of the process. It requires a primer to initiate. It requires a polymerase to catalyze the process. It requires DNA. It requires helicase to open the double helix.
a
Which of the following sequences includes a clear 8 base pair palindrome? 5' -- CCGATCGATCCC -- 3' 5' -- CTGGGGTTTTGG -- 3' 5' -- CCGGGGAAAATT -- 3' 5' -- CTGATCCTAGCT -- 3' 5' -- AAAACCAACCAA -- 3'
a
The lac operon is controlled by A. the Lac repressor. B. attenuation. C. catabolite activator protein (CAP). A and C. A and B.
a and c
In addition to yeast cells, Gal4 has been shown to activate transcription of genes controlled by an UAS sequence in: insect cells. mouse cells. fly cells. zebrafish cells. all of the above. none of the above.
all
A key characteristic of bacterial RNAs that is not observed with eukaryotic RNAs is that: transcription generates RNAs that contain both introns and exons. transcription can occur in the same cellular region as translation. a prerequisite for translation initiation is RNA processing. the genetic code used by bacteria is different from other organisms. bacterial RNAs are generated as double stranded.
b
A merozygote of genotype I+ O+ Z+ / I+ Oc Z- will show A. inducible production of repressor. B. inducible production of �-galactosidase. C. constitutive production of �-galactosidase. D. no production of �-galactosidase. E. constitutive production of lactose.
b
A partial diploid of genotype I(+) O(+) Z(+) Y(-) / I(+) O(c) Z(-) Y(+) will show inducible production of beta-galactosidase and inducible production of permase. inducible production of beta-galactosidase and constitutive production of permase. inducible production of beta-galactosidase and no production of permase. constitutive production of beta-galactosidase and inducible production of permase. constitutive production of beta-galactosidase and constitutive production of permase. constitutive production of beta-galactosidase and no production of permase. Incorrect Response no production of beta-galactosidase and inducible production of permase. no production of beta-galactosidase and constitutive production of permase. no production of beta-galactosidase and no production of permase.
b
Degeneracy in the genetic code is best illustrated by: stop codons. threonine codons. tryptophan codons. methionine codons. operator sequences.
b
Following splicing, the conserved adenosine nucleotide within the intron displays an unusual array of bonds, including: a phosphate attached to the 4-carbon of the adenine nitrogenous base. phosphodiester bonds attached at 3 places on the ribose (2', 3', 5'). additional methyl (-CH3) groups added to the adenine. a DNA strand attached to the spliced RNA. attachment to an arginine amino acid.
b
Following splicing, the conserved adenosine nucleotide within the intron displays an unusual array of bonds, including: phosphate attached to the 4-carbon of the adenine nitrogenous base. phosphodiester bonds attached at 3 places on the ribose (2', 3', 5'). additional methyl (-CH3) groups added to the adenine. a DNA strand attached to the spliced RNA. attachment to an arginine amino acid.
b
Homologous recombination in yeast facilitates the segregation of genes during meiosis. the targeted replacement of a gene in a living yeast cell. the amplification of homologous yeast genes. the independent assortment of separate gene alleles. the expression of similar gene sequences via one promoter.
b
If the DNA template strand ATG CAT GCT (written 5' to 3') were transcribed to RNA, the RNA would read (all written 5' to 3'): A. AGC ATG CAT B. AUG CAU GCT C. UAC GUA CGA D. AGC AUG CAU E. UCG UAC GUA
b
In a eukaryotic translation system containing a cell extract from a eukaryotic cell, would a protein be produced by a bacterial mRNA? A. No, translation would not occur because of the lack of initiation factors. B. No, translation would not occur because the eukaryotic initiation factors cannot recognize ! !! uncapped prokaryotic mRNAs. C. No because eukaryotic initiation factors will not recognize a mRNA without a polyA tail. D. Yes, because the 5 prime cap on the prokaryotic mRNAs is recognized by eIF4a, b, and c. E. Yes, although the size may be affected due to intronic splicing.
b
In a eukaryotic translation system containing a cell extract from a eukaryotic cell, would a protein be produced by a bacterial mRNA? No, translation would not occur because of the lack of initiation factors. No, translation would not occur because the eukaryotic initiation factors cannot recognize uncapped prokaryotic mRNAs. Yes, because the 5 prime cap on the prokaryotic mRNAs is recognized by eIF4a, b, and c. Yes, although the size may be affected due to intronic splicing.
b
In the Sanger sequencing method, the use of dideoxy guanosine triphosphate stops nucleotide polymerization A. opposite A's in the template strand. B. opposite T's in the template strand. C. opposite G's in the template strand. D. opposite C's in the template strand. E. opposite any base selected randomly in the template strand.
b
In the presence of abundant tryptophan the attenuator stem-loop structure forms, allowing transcription to continue. the attenuator stem-loop structure forms, terminating transcription. the preemptor stem-loop structure forms, allowing transcription to continue. the preemptor stem-loop structure forms, terminating transcription. none of the above.
b
RNA synthesis is always 5' to 3' because: the unwinding of the double-stranded DNA can only move one direction. nucleotides can only be added to an available 3'-OH group on the transcript terminus. nitrogenous bases cannot pair up in the 3' to 5' direction. the structure of ATP restricts 3' to 5' polymerization into RNA. RNA synthesis can move in the 3_ to 5_ direction.
b
Recombinant DNA techniques almost always require the action of DNA polymerase and phosphatase. restriction enzymes and DNA ligase. RNA polymerase and RNA primase. reverse transcriptase. DNA helicase.
b
The "domain-swapping" experiment that grafts the GAL4 activation domain to the LexA DNA-binding domain generates a chimeric protein that will bind to the Gal4 site. bind to the LexA site. activate transcription of the Gal4 gene. activate transcription of the Gal7 gene. All of the above
b
The conserved adenine nucleotide in introns serves as the: A. site for intron recognition for the splicesome. B. branch point for the formation of the intronic "lariot." C. key point for ribosome assembly and initiation of translation. D. initial site of intronic RNA digestion and removal. E. molecular signal for RNA splicing.
b
The conserved adenine nucleotide in introns serves as the: site for intron recognition for the splicesome. branch point for the formation of the intronic 'lariot' key point for ribosome assembly and initiation of translation. initial site of intronic RNA digestion and removal. molecular signal for RNA splicing.
b
The polymerase chain reaction requires ssDNA primers that anneal to the same strand of template DNA, though at distant sites. anneal to opposite strands of template DNA at distant sites, with their 3' ends directed toward each other. anneal to opposite strands of template DNA at distant sites, with their 5' ends directed toward each other. anneal to each other to prime DNA polymerization. hybridize only to DNA within the open reading frame of a selected gene.
b
The splicesome includes both protein and a functional type of RNA known as: transfer RNA (tRNA). small nuclear RNAs (snRNAs). micro RNAs (miRNAs). small interfering RNAs (siRNAs). ribosomal RNAs (rRNAs).
b
Using the haploid genome size of 3.3 billion base pairs (for humans and chimps), what is the mass of nuclear DNA in a person composed of 20 trillion cells? The answer might surprise you. (I used moles of DNA per cell and Avogrado's number to solve, but there are other ways.) about 75 g about 150 g about 500 g about 1 kg about 7 or 8 kg about 15 kg
b
What is one of the ways transcription factors recognize binding sites on the Watson-Crick Helix? A. They recognize the sites by hydrogen bonding with the bases. B. They recognize the edges of the bases in the major groove. C. They bind to regions of high chromatin. D. They recognize the tertiary structure of the DNA. E. They recognize the phosphates on the outside of the DNA molecule.
b
What type of library would be most valuable in the isolation of a coding sequence for a mRNA? Genomic DNA library cDNA library miRNA library Mutant genomic DNA library Repetitive DNA library
b
Which of the following is an example of a cis-acting element? beta-galactosidase Operator site LacI repressor protein Lactose Permease
b
A bacterium of genotype I(+) O(c) Z(+) will show A. inducible production of repressor. B. inducible production of beta-galactosidase. C. constitutive production of beta-galactosidase. D. no production of beta-galactosidase. E. constitutive production of lactose.
c
A bacterium of genotype I+ Oc Z+ will show A. inducible production of repressor. B. inducible production of �-galactosidase. C. constitutive production of �-galactosidase. D. no production of �-galactosidase. E. constitutive production of lactose.
c
Degeneracy in the genetic code is poorly illustrated by: A. stop codons. B. threonine codons and arginine codons. C. tryptophan and methionine codons. D. alanine and proline codons. E. operator sequences.
c
GAL3 promotes GAL gene expression by binding galactose only and subsequently binding to Gal80. binding ATP only, and subsequently binding to Gal80. binding galactose and ATP, and subsequently binding to Gal80. binding galactose and ATP, and subsequently binding to Gal1, Gal7, and Gal10. Option
c
In the PCR process, if we assume that each cycle takes ten minutes, how many fold amplification would be accomplished in one hour? 12-fold 32-fold 64-fold 512-fold 4096-fold
c
The DICER complex: A. Targets mRNA for destruction using double stranded miRNA. B. Separates DS miRNA into single stranded miRNA C. Cuts double stranded miRNA into short pieces of double stranded RNA D. Removes the template RNA strand from miRNA, making it single stranded. E. Cuts miRNA/mRNA hetero-duplexes, destroying mRNA.
c
The GAL4 protein has which of the following functional domains? ! ! 1. A DNA-binding domain ! ! 2. An activation domain ! ! 3. A repressor domain A. 1 B. 3 C. 1 and 2 D. 1 and 3 E. 2 and 3
c
The cDNA for a eukaryotic gene B is about 900 nucleotide pairs long. A cDNA clone is used to isolate a genomic clone of gene B, and the gene is sequenced. From start to stop codon, the gene is found to be about 2700 nucleotide pairs long. The most probable reason for the discrepancy is that the mRNA broke during cDNA synthesis. the gene is present as a multiple duplication. the gene has about 1800 intronic nucleotide pairs. the genomic clone is not really gene B, just a related gene. there was a sequencing error.
c
The protein encoded by the cystic fibrosis gene is 1480 amino acids long, yet the gene spans 250 k. How is this difference possible? The gene spans two chromosomes. The protein is post-translationally modified to reduce its size to 1480 amino acids. The gene contains a large amount of intronic DNA, which is spliced out of the RNA transcript before it leaves the nucleus to be translated. Only a portion of the mRNA is translated. The gene encodes several proteins that are differentially expressed. One of them is the 1480-amino acid cystic fibrosis protein. The gene contains a large amount of exonic DNA, which is spliced out of the RNA transcript before it leaves the nucleus to be translated.
c
The yeast GAL4 enhancers are located _____ of the genes they regulate and are known as ______. downstream; downstream activation sequences (DAS) downstream; downstream enahancer sequences (DES) upstream; upstream activation sequences (UAS) upstream; upstream enhancer elements (UES) None of the above
c
Which of the following is NOT true of a cDNA clone? cDNAs are generally copies of mRNA from expressed genes. Reverse transcriptase activity is required for cDNA cloning. cDNA clones lack exons, as introns are spliced together before cloning. cDNAs are typically smaller than the chromosomal sequence for a particular gene. cDNAs can be fused to a promoter to enable expression of the encoded gene.
c
Which of the following is NOT true of a cDNA clone? A. cDNAs are generally copies of mRNA from expressed genes. B. Reverse transcriptase activity is required for cDNA cloning. C. cDNA clones lack exons, as introns are spliced together before cloning. D. cDNAs are typically smaller than the chromosomal sequence for a particular gene. E. cDNAs can be fused to a promoter to enable expression of the encoded gene.
c
Which of the following is an example of a trans-acting factor? beta-galactosidase Operator site LacI repressor protein Lactose Permease
c
Which of the following is or are one of the roles of the 5 prime cap? A. The cap helps the RNA polymerase find the promoter and initiate transcription. B. The cap plays a role in the removal of introns. C. The cap acts as a binding site for the ribosome. D. The cap protects the 5 prime end of the RNA from degradation. A and C C and D B and C None of the above
c and d
Analyzing mutations in the GAL80 and GAL3 indicate that both GAL80 and GAL3 inhibit GAL gene expression. both GAL80 and GAL3 promote GAL gene expression. GAL80 promotes GAL gene expression and GAL3 inhibits GAL gene expression. GAL80 inhibits GAL gene expression and GAL3 promotes GAL gene expression. GAL80 and GAL3 are not involved in GAL expression.
d
Enhancer is to activator as silencer is to A. activator B. operator C. UAS D. repressor E. promoter
d
Enhancer is to activator as silencer is to activator operator UAS repressor promoter
d
If a polyribonucleotide contains equal amounts of randomly positioned adenine and uracil bases, what proportion of its triplets will encode tyrosine? 1/2 proportion 1/4 proportion 1/6 proportion 1/8 proportion
d
If the GC content of a DNA molecule is 42%, what is the percentage of A bases in this molecule? A. 42% B. 21% C. 58% D. 29% E. None of the above.
d
In yeast, you have sequenced a piece of wild-type DNA and it clearly contains a gene, but you do not know what gene it is. Therefore, to investigate further, you would like to find out its mutant phenotype. How would you use the cloned wild-type gene to do so? Mutate the gene using a version of PCR that creates mutations in a gene sequence. Insert the mutant gene into the yeast, allowing the mutant gene to recombine and replace the normal gene. Now look for a mutant phenotype. Mutate the gene using a version of PCR that creates mutations in a gene sequence. Clone the mutant gene into bacteria and express the mutant protein using an in vitro method. Analyze the bacteria for a mutant phenotype. Using the wild-type gene, probe libraries of yeast DNA looking for clones that hybridize weakly to the wild-type gene. These are likely mutant copies of the gene. Isolate those yeast containing the putative mutant, grow them, and analyze for a mutant phenotype. Construct a recombinant expression vector containing the wild-type gene cloned into it and a selectable marker gene inserted into the wild-type gene. Transform yeast cells and grow under conditions that select for the marker gene. This procedure will knock out the gene of interest. Analyze for a mutant phenotype. Mutate the gene using CRISPRs that creates mutations in the sequence of the cloned gene. Transfect the mutant gene into zebrafish and express the mutant protein using an in vivo method. Analyze the zebrafish for a mutant phenotype.
d
The "wobble" base in a codon is found: A. at the 5' end of the mRNA codon's sense strand. B. at the 5' end of the mRNA codon's template strand. C. at the 3' end of the tRNA anticodon. D. at the 5' end of the tRNA template. E. within a tRNA hairpin loop structure.
d
The "wobble" base is less important than the other two nucleotides in a codon and is found: at the 5' end of the RNA codon's sense strand. at the 3' end of the mRNA codon's antisense strand. at the 3' end of the tRNA anticodon. at the 5' end of the tRNA anticodon. within a tRNA hairpin loop structure.
d
The attenuator region (AR) of the trp operon is located within the promoter region. between the promoter region and the operator site. within the operator site. between the operator site and the first structural gene. after the structural genes.
d
The intermediate temperature cycle (72 degree C) in the polymerase chain reaction enables hybridization between the template DNA and the PCR primers. denaturation of the template DNA to enable primer access to matching sequence. activation of the primers to being the amplification procedure. activity of the thermostable DNA polymerase enzyme. a reduction of contamination in the amplification reaction.
d
What would be the immediate consequence in the cloning process if someone forgot to add ligase? There would be no hybridization between the vectors and the donor DNA fragments. No complementary sticky ends would form in either the vectors or the donor DNAmolecules. Blunt-ended vectors would be able to form covalent bonds with donor DNA. The inserted DNA and the vector would be held together only by hydrogen bonds, so recombinant plasmids would lose the inserted DNA. Recombinant plasmids could not be separated from nonrecombinant plasmids.
d
Which of the following is an example of a trans-acting factor? A. beta-galactosidase B. Operator site C. LacI repressor protein D. Lactose E. Permease
d
A mutation in the DNA causes a 3bp insertion into exon sequences of gene A. What is the impact, if any, on the protein encoded by gene A? There is no impact, as exon sequences are spliced and not found in the mRNA. There will be one extra amino acid in the protein and all other amino acids will be the same. All of the amino acids after the 3bp insertion will be different. There will be one extra amino acid and 2 amino acids in the protein will be different. There will be one extra amino acid and 0, 1, or 2 amino acids in the protein may be different. There will problems splicing the introns and the protein will be nonfunctional.
e
Attenuation (as in the trp operon) would never work in eukaryotes because all RNA processing occurs in the nucleus. eukaryotic genes can not be polycistronic. All RNA processing in eukaryotes is done by RNA pol II. there is no room in the nucleus for protein folding. there are no ribosomes in the nucleus. eukaryotic RNA never has secondary structure
e
Figure out the mutation. You will need the codon table for this question. WT genomic sequence of a really small gene and matching complete amino acid sequence of its gene product. ! ! GGT ATG GGG ACT TTG AGG ATG ATA AGG CGT AAA TAA ATA T Met Gly Thr Leu Arg Met Ile Arg Arg Lys A mutant is found that has the following protein sequence: Met Gly Thr Leu Arg Met Ile Arg Arg. What is the likely mutation? A. insertion of another T in that cluster of Ts at positions 12 thru 14. B. deletion of G at 7. C. deletion of T at position 5. D. change of A to T at position 31. E. insertion of AA between T and G at positions 14 and 15.
e
Is there a protein in eukaryotes analogous to the sigma factor? A. No B. Yes, the TATA binding protein C. Yes, the general transcription factors D. Yes, the ribosomal subunits Answers B and C are both correct. Answers B, C, and D are all correct.
e
Which of the following mRNA codons would form a codon-anticodon base pairing interaction with the GAU tRNA anticodon? (All sequences written 5 prime to 3 prime.) CTA GAU ATC CUA AUC
e
A bacterium of genotype I(+) O(c) Z(+) Y(-) will show inducible production of beta-galactosidase and inducible production of permase. inducible production of beta-galactosidase and constitutive production of permase. Incorrect Response inducible production of beta-galactosidase and no production of permase. constitutive production of beta-galactosidase and inducible production of permase. constitutive production of beta-galactosidase and constitutive production of permase. constitutive production of beta-galactosidase and no production of permase. no production of beta-galactosidase and inducible production of permase. no production of beta-galactosidase and constitutive production of permase. no production of beta-galactosidase and no production of permase.
f
A partial diploid of genotype I(S) P(+) O(+) Z(-) Y(+) / I(+) P(+) O(C) Z(+) Y(-) will show inducible production of beta-galactosidase and inducible production of permase. inducible production of beta-galactosidase and constitutive production of permase. inducible production of beta-galactosidase and no production of permase. constitutive production of beta-galactosidase and inducible production of permase. constitutive production of beta-galactosidase and constitutive production of permase. constitutive production of beta-galactosidase and no production of permase. no production of beta-galactosidase and inducible production of permase. no production of beta-galactosidase and constitutive production of permase. Incorrect Response no production of beta-galactosidase and no production of permase.
f
A partial diploid of genotype I(-) P(-) O(+) Z(-) Y(+) / I(+) P(-) O(C) Z(+) Y(-) will show inducible production of beta-galactosidase and inducible production of permase. inducible production of beta-galactosidase and constitutive production of permase. inducible production of beta-galactosidase and no production of permase. constitutive production of beta-galactosidase and inducible production of permase. constitutive production of beta-galactosidase and constitutive production of permase. Incorrect Response constitutive production of beta-galactosidase and no production of permase. no production of beta-galactosidase and inducible production of permase. no production of beta-galactosidase and constitutive production of permase. no production of beta-galactosidase and no production of permase.
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