Genetics Quizzes for Midterm 3

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.

(Multi) In a diploid yeast cell:

Alpha-specific genes are constitutively down-regulated, Alpha2 and MCM1 down-regulate a-specific genes, a1 and alpha1 down-regulate haploid-specific genes.

(Multiselect) The GAL4 protein has which of the following functional domains?

An activation domain, A DNA-binding domain.

Which of the following is true of chromatin in prokaryotes?

Bacterial chromosomes are not organized into chromatin.

(Multiselect) GAL3 promotes GAL gene expression by:

Binding galactose, Binding ATP, And subsequently binding to the GAL Upstream Activation Sequence (UAS).

The transcription factors Tup-1-Mig-1 control GAL4 activity by:

Binding to silencer sites to repress transcription by deacylating local histones which down regulate local genes.

(Multiselect) For phage lambda to enter the lysogenic cycle:

Cl must occupy the R1 operator, Cl must occupy the R2 operator.

(Multiselect) GAL4 binds to UAS sequences and activates:

GAL1, GAL2, GAL7, and GAL10.

(One or two words, use style O(C) to answer) In an E. coli merozygote, an O(C) mutation is epistatic to:

I(S), Super repressor.

(Multiselect) For locus switching of the MATa locus (to MATalpha):

MATa is cut by HO, HMLalpha is up-regulated by MCM1 at the RE locus.

(Multiselect) A partial diploid of genotype I(-) P(-) O(+) Z(-) Y(+)/I(+) P(-) O(c) Z(+) Y(-) will show:

No production of beta-galactosidase, No production of permease.

(One of two words) What kind of transcription factor is lambda cro?

Repressor, A repressor.

What is the order of events in infection by lambda phage?

The phage enters E. coli, The phage P(R) and P(L) promoters are switched on, Cro and c(I) complete for the O(R1), O(R2), and O(R3) operators, Either the phage P(RM) or P(RE) promoter is switched on, Either the phage becomes lysogenic or lytic.

For the enhanceosome, binding sequences are on both strands of DNA. Why?

The proteins of the enhanceosome bind the opposite strand 5 bp out of phase with those on the normal strands so that are all on the same side of the double helix.

In eukaryotes, transcriptional gene control is mediated by:

Trans-acting factors binding to cis-acting elements.

If the GC content of a DNA molecule is 48%, what is the percentage of G bases in this molecule? a. 24% b. 26% c. 48% d. 52%

a. 24%

Which of the following sequences includes a clear 8 base pair palindrome? a. 5' -- CCGATCGATCCC -- 3' b. 5' -- CTGGGGTTTTGG -- 3' c. 5' -- CCGGGGAAAATT -- 3' d. 5' -- CTGATCCTAGCT -- 3' e. 5' -- AAAACCAACCAA -- 3'

a. 5' -- CCGATCGATCCC -- 3'

A normal 8-base restriction site starts with CGAT. What are the next four base pairs? a. ATCG b. TAGC c. GCTA d. CGAT

a. ATCG

Which of the following is true for replication in prokaryotes but not eukaryotes? a. It occurs in the cytoplasm. b. It requires leading and lagging strands. c. It requires helicase. d. It occurs in the nucleus.

a. It occurs in the cytoplasm.

Promoters are to transcription as _________ are to replication. a. Origins. b. Leading strands. c. Histones. d. Antiparallel strands. e. Okazaki fragments.

a. Origins.

The polymerase chain reaction requires ssDNA primers that: a. Anneal to the same strand of template DNA, though at distant sites. b. Anneal to opposite strands of template DNA at distant sites, with their 3' ends directed toward each other. c. Anneal to opposite strands of template DNA at distant sites, with their 5' ends directed toward each other. d. Anneal to each other to prime DNA polymerization. e. Hybridize only to DNA within the open reading frame of a selected gene.

b. Anneal to opposite strands of template DNA at distant sites, with their 3' ends directed toward each other.

ATG TCT CAG TAT GGG TAT GAG AAC GAC ATT CAC ATA ATT ... Met Ser Gln Tyr Gly Tyr Glu Asn Asp Ile His Ile Ile ... A mutant is found that has the following protein sequence: Met Ser Pro Val Trp Val Stop. What is the likely mutation? a. Between T4 and A5, insert G. b. Between C13 and A14, insert C. c. G9 -> del. d. Between G20 and G31 -> del. e. T24 -> A. f. T22 -> G. g. Digital copy: ACA TAA ATG TCT CAG TAT GGG TAT GAG AAC GAC ATT CAC ATA ATT.

b. Between C13 and A14, insert C.

The conserved adenine nucleotide in introns serves as the : a. Site for intron recognition for the spliceosome. 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. Branch point for the formation of the intronic "lariot".

When comparing the carious key models of DNA replication, the model that included the synthesis of a brand new double stranded DNA molecule from an original molecule was named: a. Dispersive replication. b. Conservative replication. c. Semi-conservative replication. d. Liberal replication. e. None of the above.

b. Conservative replication.

GGAAAACCGCTTTACGCACCAGAGGgtatgcttttgttcta...atgttccacagGTTATCAGAATAATTTGAAG Above is the partial genomic DNA sequence of two eons (UPPERCASE) and the intervening intron (lowercase). Most of the internal intron sequence is not shown. To help you get going, the amino end of the protein sequence is in frame with the 5' end of the first exon. The initial di-nucleotide consensus sequence for the beginning of the intron has its T mutated to an A. So the splice machinery finds the NEXT consensus sequence and uses it for splicing. What is the protein sequence of the new protein from the mutant sequence? You will need a codon table to translate the sequence. a. G K P L Y A P E G M L L L S E F E b. G K P L Y A P E G M L L L S E c. G K P L Y A P E G M L L S E d. G K P L Y A P E V H R L F e. G K P L Y A P E G M L L F f. G K P L Y A P E g. G K P L Y A P E G M L L F C S T G Y Q N N L K h. G K P L Y A P E G M L L S E F E i. Digital copy: GGA AAA CCG CTT TAC GCA CCA GAG Ggt atg ctt ttg ttc tan nnn nnn nnn atg ttc cac agG TTA TCA GAA TAA TTT GAA G

b. G K P L Y A P E G M L L L S E

Which of the following is true of mRNA but not pre-mRNA? a. It contains exons and introns. b. It contains exons but not introns. c. It is found in the nucleus but not in the cytoplasm. d. Both B and C are true for mRNA. e. Both A and C are true for mRNA.

b. It contains exons but not introns.

What is the function of DNA polymerase III? (Choose best answer): a. It is synthesizing the lagging strand of DNA. b. It is synthesizing the leading strand and the majority of the lagging strand of DNA. c. It is synthesizing the leading strand of DNA. d. It is synthesizing the lagging strand and the majority of the leading strand of DNA. e. It is synthesizing both the leading strand and the lagging strand of DNA. f. It is closing up nicks in the lagging strand of DNA. g. It is repairing mistakes in newly synthesized DNA. h. It is transcribing the lagging strand of DNA. i. It is removing primers, filling in missing nucleotides and repairing mistakes in the lagging strand of DNA.

b. It is synthesizing the leading strand and the majority of the lagging strand of DNA.

(Problem 31) 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. Yes, because the 5 prime cap on the prokaryotic mRNAs is recognized by eIF4a, b, and c. d. Yes, although the size may be affected due to intronic splicing.

b. No, translation would not occur because the eukaryotic initiation factors cannot recognize uncapped prokaryotic mRNAs.

RNA synthesis is always 5' to 3' because: a. The unwinding of the double-stranded DNA can only move one direction. b. Nucleotides can only be added to an available 3'-OH group on the transcript terminus. c. Nitrogenous bases cannot pair up in the 3' to 5' direction. d. The structure of ATP restricts 3' to 5' polymerization into RNA. e. RNA synthesis can move in the 3' to 5' direction.

b. Nucleotides can only be added to an available 3'-OH group on the transcript terminus.

Using the rules for Tm that we learned in class, how can you quickly adjust the selection to get the correct Tm? a. Shift the selection 3 bp to the left. b. Shift the selection 2 bp to the left. c. Shift the selection 1 bp to the left. d. Leave it alone; it is fine! e. Shift the selection 1 bp to the right. f. Shift the selection 2 bp to the right. g. Shift the selection 3 bp to the right.

b. Shift the selection 2 bp to the left.

Oswald Avery and colleagues strengthened scientific support that DNA was the transforming factor by replicating the Griffith experiment with some important differences in experimental design. The key difference was: a. Purifying S-strain DNA and RNA from other biomolecules. b. Systematically eliminating the impact of classes of S-strain biomolecules using enzymatic digestion before mixing with R-strain live cells. c. Injecting S-strain DNA directly into R-strain cells and observing the change in phenotype. d. Avery used Escherichia coli rather than Streptococcus as it is easier to transform. e. All of the above.

b. Systematically eliminating the impact of classes of S-strain biomolecules using enzymatic digestion before mixing with R-strain live cells.

Original: TTGACAT---16 bp---TATAAT Mutant: TATAAT---16 bp---TTGACAT A researcher was mutating prokaryotic cells by rearranging segments of DNA, and she made the above mutation. What does this original sequence represent? a. TATA box. b. The -35 and -10 consensus sequence. c. The rho factor. d. The sigma factor. e. The TBP factor.

b. The -35 and -10 consensus sequence.

Alfred Hershey and Martha Chase examined transformation using bacteriophage (bacterial DNA virus) and bacterial cells. If phage are labeled with radioactive sulfur and allowed to infect bacterial cells, the radioactive sulfur will be localized to: a. The inside of infected cells (in phage DNA). b. The outside of infected cells (in phage ghosts). c. In newly synthesized phage viruses in the host bacterial cell. d. No radioactivity will remain after infection. e. The sulfur will be metabolically consumed, and therefore the radioactivity will be destroyed.

b. The outside of infected cells (in phage ghosts).

(Problem 12) Is there a protein in prokaryotes analogous to the TATA binding protein? a. No. b. Yes, the sigma binding protein. c. Yes, the general transcription factors. d. yes, the ribosomal subunits. e. B and C are both correct. f. B, C, and D are all correct.

b. Yes, the sigma binding protein.

Assume that a certain bacterial chromosome has one origin of replication. Under some conditions of rapid cell division, replication could start from the origin before the preceding replication cycle is complete. How many replicons would be present under these conditions? a. 1. b. 2. c. 3. d. 4. e. 6. f. 10.

c. 3.

Zebrafish has 2 billion bp in its diploid genome. During gastrulation, the DNA is, on average, replicated in 32 min. Assuming that all replication forks move at a rate of 10,000 bp/min, how many replicons (replication units) are present during S phase in the nucleus? a. 781. b. 1563. c. 3125. d. 6250. e. 12,500. f. 25,000. g. 50,000. h. 100,000.

c. 3125.

(Problem 15) In the PCR process, if we assume that each cycle takes ten minutes, how many fold amplification would be accomplished in one hour? a. 12-fold. b. 32-fold. c. 64-fold. d. 512-fold. e. 4096-fold.

c. 64-fold.

During the last S-phase before meiosis in Neurospora, a single mutation occurred in a single nucleotide that resulted in a mismatched A to C base pair, with the A being the mutant nucleotide; C is the wild-type version. This mutation was not repaired before meiosis. In the final 8 asco spores, how many spores have the base pair G-C (wild type), the base pair A-C (mismatch), and the base pair A-T (mutant)? Remember, there is an extra S-phase after meiosis in Neurospora. a. 8, 0, 0. b. 7, 1, 0. c. 7, 0, 1. d. 6, 1, 1. e. 6, 0, 2. f. 5, 2, 1. g. 4, 0, 4. h. 4, 2, 2.

c. 7, 0, 1.

Figure out the mutation. You will need the codon table to answer this question. Wild type genomic sequence of a PORTION of a gene and the wild type sequence of a PORTION of the protein gene product. The protein sequence only matches partially. TAT AAA GGG CGA CCC CCT GGT AAT GGG ACT TTG AGG Tyr Lys Gly Arg Pro Pro Ala Pro Arg Gln Tyr Trp The five amino acids at the amino end of the WT protein match but not the six at the carboxyl end. You do not have the DNA sequence for the carboxyl end of the protein, but that shouldn't be a problem. A mutant is found that has the following protein sequence in that part: Tyr Lys Gly Arg Pro Leu Leu Leu Gly Asn Gly. what is the likely mutation? a. Between A12 and C13 insert T. b. Between C13 and C14 insert T. c. A15 -> deletion. d. C13 and C14 -> deletion. e. G8 -> T. f. G20 -> A. g. T21 -> deletion. h. Digital copy: TAT AAA GGG CGA CCA CCT GGT AAT GGG ACT TTG AGG.

c. A15 -> deletion.

If DNA with a coding strand ATG CAT GCA (written 5' to 3') were transcribed to RNA, the RNA would read (all written 5' to 3'): a. UGC AUG CAU b. TAC GTA CGT c. AUG CAU GCA d. ACG UAC GUA e. TGC ATG CAT f. UAC GUA CGU

c. AUG CAU GCA

Why can bacteria have poly-cistronic genes? a. Because they need multiple cistron organelles so they segregate evenly during cell division. b. Because they have many exons that are joined together before translation. c. Because ribosomes can be loaded at multiple Shine Delgano/AUG sequences. d. Because ribosomes are loaded at the single CAP site. e. Because the stability conferred by a poly-A tail is very limited. f. Because ribosomes limited by the number and placement of AUG sites . g. Because transcripts are limited by the number of promoter sites . h. Because ribosomes are loaded at multiple CAP sequences. i. Because ribosomes do not exit the mRNA at termination codons. j. Because the Polly-aannaa principle (Watson and Crick, 1960) states that transcription is optimistic. (the best answer, you think?)

c. Because ribosomes can be loaded at multiple Shine Delgano/AUG sequences.

Oswald Avery and colleagues strengthened scientific support that DNA was the transforming factor by replicating the Griffith experiment with some important differences in experimental design. They found that the enzyme __________ was effective at destroying the transforming capacity of S-strain biomolecules. a. Protease (protein destruction). b. RNase (RNA destruction). c. DNase (DNA destruction). d. Polysaccharide-destroying enzymes. e. Lipase (lipid destruction).

c. DNase (DNA destruction).

If a mutation that inactivated telomerase occurred in a cell (telomerase activity in the cell = zero), what do you expect the outcome to be? a. The position of the centromere would shift at each replication cycle eventually leading to mutations in the genetic information and cell death. b. The number of random repeats would increase with each replication cycle, eventually leading to a large number of unstable chromosomes and cell death. c. The telomeres would shorten at each replication cycle, eventually leading to loss of essential genetic information and cell death. d. DNA primate would be unable to bind to the 5' end of the template strand, eventually causing a reduction in chromosome size and cell death.

c. The telomeres would shorten at each replication cycle, eventually leading to loss of essential genetic information and cell death.

This is a rather interesting modification of a common part of an mRNA. I think I may have made a mistake.What is this structure and what is the mistake? a. This is the section of the mRNA that was just sliced and I forgot to change the triphosphate to a monophosphate. b. This is the 5' cap and I accidentally added a methyl group to that guanosine. c. This is the 5' cap and I attached the 3' end of the capping ribose to the triphosphate. d. This is the end of the splice lariat and I forgot to make the 5' to 2' inter-exon linkage. e. This is the 5' cap and, just kidding, there is no mistake!

c. This is the 5' cap and I attached the 3' end of the capping ribose to the triphosphate.

A purified DNA preparation of a circular plasmid is digested to completion with the restriction endonucleases shown above, singly and in combination. The diagram above shows the appearance of the original molecules and the digestion products, after separation by electrophoresis in an agarose gel. What is the map of the single band cut in lane #2? Use X and N for the restriction sites and the numbers between them for the size in kb. a. X 2 X 4 N 6 X b. X 2 N 1 N 3 X c. X 1 N 2 N 3 X d. N 2 N 3 X 1 N e. N 1 X 2 X 3 N f. X 1 N 3 N 2 X

c. X 1 N 2 N 3 X

Of the answers below, what is the smallest nucleotide sequence that could code for a protein with 300 amino acids? (tricky, tricky) a. 300 b. 600 c. 900 d. 1000 e. 2000 f. 3000

d. 1000

The intermediate temperature cycle (72 degree C) in the polymerase chain reaction enables: a. Hybridization between the template DNA and the PCR primers. b. Denaturation of the template DNA to enable primer access to matching sequence. c. Activation of the primers to being the amplification procedure. d. Activity of the thermostable DNA polymerase enzyme. e. A reduction of contamination in the amplification reaction.

d. Activity of the thermostable DNA polymerase enzyme.

GGT ATG(Met) GGG(Gly) ACT(Thr) TTG(Leu) AGG(Arg) ATG(Met) ATA(Ile) AGG(Arg) CGT(Arg) AAA(Lys) TAA ATA T A mutant is found that has the following protein sequence: Met Gly Thr Leu Arg Gly. What is the likely mutation? a. Between T14 and G15, insert T. b. G7 -> del. c. T5 -> del. d. Between T14 and G15, insert AA. e. A31 -> T. f. C11 -> del. g. Digital copy: GGT ATG GGG ACT TTG AGG ATG ATA AGG CGT AAA TAA ATA T.

d. Between T14 and G15, insert AA.

A purified DNA preparation of a certain plasmid is digested to completion with NheI restriction endonuclease. In separate reactions, the same preparation was digested to completion with XbaI and with a mixture of XbaI and NheI, respectively. The diagram above shows the appearance of the original molecules and the digestion products in the three digestion mixtures, after separation by electrophoresis in an agarose gel. Was the original plasmid DNA circular or linear and What was the original DNA molecule's length? a. Linear and 4 kb. b. Linear and 8 kb. c. Linear and 1.7 kb. d. Circular and 4 kb. e. Circular and 8 kb. f. Circular and 1.7 kb.

d. Circular and 4 kb.

Which of the following is true of mRNA but not tRNA? a. It is a non-coding RNA. b. It is transcribed by RNA polymerase from a gene. c. It is found in the cytoplasm of eukaryotes. d. It is a coding RNA. e. It contains uracil but not thymine. f. It is found in the nucleus.

d. It is a coding RNA.

E. coli is widely used in laboratories to produce proteins from other organisms. You have isolated a yeast gene that encodes a metabolic enzyme and want to produce this enzyme in E. coli. Would the yeast promoter work in E. coli? a. Yes, as long as you include the yeast promoter in the bacterial plasmid. b. Yes, but the yeast promoter will not be as efficient in E. coli. c. Yes, because the sigma factor would interact with the yeast TATA box at -30. d. No, because the sigma factor would not recognize the yeast TATA box at -30. e. Both A and C are correct. f. Both A and D are correct. g. Both B and C are correct.

d. No, because the sigma factor would not recognize the yeast TATA box at -30.

If Escherichia coli, grown for a period of time in 15N, is transferred to 14N for two generations of DNA replication, the resulting DNA should have 14N added to all new DNA. If semi-conservative replication is occurring, the 14N-containing new DNA will compose of: a. One strand of all bacterial chromosomes. b. Both strands of DNA in half of all bacterial chromosomes. c. Both strands of DNA in three quarters of the bacterial chromosomes. d. One strand of DNA in hand of all bacterial chromosomes and both strands in the rest. e. Regions dispersed throughout all bacterial DNA. f. None of the new DNA, it will only be found in the old DNA. g. None of the above.

d. One strand of DNA in half of all bacterial chromosomes and both strands in the rest.

The most common elements in living organisms are carbon, hydrogen, oxygen, nitrogen, phosphorus, and sulfur. Which of these is not found in protein? a. Hydrogen. b. Oxygen. c. Nitrogen. d. Phosphorus. e. Sulfur.

d. Phosphorus.

During S-phase, geminin is a protein that, among other things: a. Promotes Cdt1 binding to the ORC during early S-phase. b. Binds to the double helix to initiate replication during S-phase. c. Promotes translation of cell cycle factors during G1 phase. d. Targets Cdt1 for degradation during S-phase. e. Targets the Anaphase Promoting Complex for degradation during S-phase. f. Inhibits the function of the Anaphase Promoting Complex during G1 phase. g. Promotes CDC6 binding to the ORC during late G1 phase. h. Inhibits transcription during S-phase.

d. Targets Cdt1 for degradation during S-phase.

If a polyribonucleotide contains equal amounts of randomly positioned adenine, guanine and uracil bases, what proportion of its triplets will encode a stop codon? (Use your rules of probability!) a. 1/3 ratio. b. 1/4 ratio. c. 3/8 ratio. d. 2/9 ratio. e. 1/9 ratio. f. 1/18 ratio. g. 1/27 ratio.

e. 1/9 ratio.

The sense strand of DNA is: a. A synonym for the coding strand and an antonym for the contemplate strand or the nonsense strand. b. A synonym for the nonsense strand and an antonym for the nontemplate strand or the sense strand. c. A synonym for the template strand and an antonym for the noncoding strand or the template strand. d. A synonym for the noncoding strand and an antonym for the coding strand or the template strand. e. A synonym for the nontemplate strand and an antonym for the nonsense strand or the template strand. f. A synonym for the sense strand and an antonym for the coding strand or the template strand.

e. A synonym for the nontemplate strand and an antonym for the nonsense strand or the template strand.

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.) (remember wobble) a. CTA b. GAU c. ATC d. CUA e. AUU f. GUC

e. AUU

Why is telomerase not found in prokaryotes? a. Because bacteria only have one chromosome. b. Because bacteria do not have a nucleus. c. Because bacteria have their chromosome attached to the cell membrane already. d. Because bacterial chromosomes have only one origin of replication. e. Because bacterial chromosomes are circular. f. Because bacteria often have circular plasmids.

e. Because bacterial chromosomes are circular.

ATA GTG CGC TGT AGA CTC TAT GCG TCT GAG TCT ACG Ile Val Arg Lys Arg Leu Tyr Ala Ser Glu Ser Thr A mutant is found that has the following protein sequence: Ile Val Arg Cys Arg Leu Ala Met Arg Leu Ser Leu. What is the likely mutation? a. A13 -> T. b. A15 -> T. c. C16, T17, C18 -> del. d. T17, C18 -> del. e. Between T17 and C18 -> insert GG. f. Digital copy: ATA GTG CGC TGT AGA CTC TAT GCG TCT GAG TCT ACG.

e. Between T17 and C18 -> insert GG.

(Same protein sequence as previous slide for the initial protein). A mutant is found that has the following protein sequence in that part: Tye Lys Val Arg Pro Pro Ala Pro Arg Gln Trp Tyr. What is the likely mutation? a. Between A12 and C13 insert T. b. Between C13 and C14 insert T. c. A15 -> deletion. d. C13 and C14 -> deletion. e. G8 -> T. f. G20 -> A. g. T21 -> deletion. h. Digital copy: TAT AAA GGG CGA CCA CCT GGT AAT GGG ACT TTG AGG.

e. G8 -> T.

Which of the following is true of replication but not transcription? a. It requires promoters for initiation of the process. b. It requires a TBP to initiate. c. It requires a polymerase to catalyze the process. d. It requires DNA. e. It requires helicase to open the double helix.

e. It requires helicase to open the double helix.

How does the dual FUCCI transgender work to indicate the cell cycle? Remember, the cells are red in G1 and blue in S/G2/M. Which statement is most correct? a. The red protein is transcribed and translated by geminin. The blue protein is transcribed and translated by cdt1. b. The red protein is transcribed and translated by cdt1. The blue protein is transcribed and translated by geminin. c. The blue protein is degraded by geminin. The red protein is degraded by cdt1. d. The blue protein is degraded by cdt1. The red protein is degraded by geminin. e. The blue protein is targeted to the proteasome by APC-cdh1. The red protein is targeted to the proteasome by geminin. f. The blue protein is targeted to the proteasome by geminin. The red protein is targeted to the proteasome by cdt1.

e. The blue protein is targeted to the proteasome by APC-cdh1. The red protein is targeted to the proteasome by geminin.

1. Because Chargaff's rules, the bases are represented proportionally. If the bases face outward, complementarity of the two DNA strands would not be necessary. 2. DNA would not migrate to the positively charged electrode on a gel because the phosphates would be shielded from the water. 3. The oxygen-rich phosphates of the backbone would repel each other due to their electronegativity. The molecule would therefore be very unstable. 4. Ribosomes would confuse the DNA with single stranded mRNA. While James Watson and Francis Crick were trying to determine the structure of DNA, Linus Pauling published his thoughts about the molecular arrangement of DNA. He placed the phosphates within the interior of the helix, with the bases facing outward. Which reasons shown above are why this arrangement is unsatisfactory in light of what we now know. a. 1 b. 2 c. 3 d. 4 e. 1 and 2. f. 1 and 3. g. 2 and 4. h. 3 and 4.

f. 1 and 3.

Which of the following is 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. e. A and C. f. C and D. g. B and C. h. None of the above.

f. C and D.

The Dicer complex: a. Targets mRNA for destruction using double stranded miRNA. b. Separates DS miRNA into single stranded miRNA. c. Removes the template RNA strand from miRNA, making it a single strand. d. Cuts miRNA/mRNA hetero-duplexes, destroying mRNA. e. Degrades RNA having a short poly-A tail. f. Cuts up double stranded RNA into small 19 or 20 nucleotide pieces. g. Presents RNA to the Dicer complex for degradation. h. Only processes miRNA not other types of RNA, such as siRNAs or tRNAs. i. Cuts up RNA that is complementary to short strands of RNA.

f. Cuts up double stranded RNA into small 19 or 20 nucleotide pieces.

Topoisomerase and helicase have distinct functions that include which of the following? a. Topoisomerase is responsible for separating the strands of the double helix. b. Helicase relieves supercoiling that occurs in front of the replication fork. c. Topoisomerase adds RNA primers for making the Okazaki fragments. d. Helicase keeps the single stranded regions of DNA from re-annealing. e. The enzymes have nearly identical activities but are located at different sites during DNA replication. f. Helicase separates the strands of the double helix. g. Actually, that is wrong. These enzymes are nearly identical but have been given distinct names.

f. Helicase separates the strands of the double helix.

If a really short miRNA (a rs-miRNA, which only exists for this problem) had the sequence: AAA AAA CCC GGC CCG GCC AAA AAA AAA GGC CGG GGG AAA AAA, figure out the double stranded part of the rs-miRNA, and then figure out which RNA below is targeted: a. AGU AAG UAA GUA AGU CCC GGC CCG GCC AGU AAG UAA GUA AGU. b. AGU AAG UAA GUA AGU AAA AAA UAA GUA AGU AAG UAA GUA AGU. c. AGU AAG UAA GUA AGU GGC CGG GCC GGG UAA GUA AGU AAG UAA. d. AGU AAG UAA GUA AGU UAA GUA AGU AAG UAA GUA AGU GUA AGU. e. All of the above. f. A and B. g. A and C. h. B and D.

g. A and C.

(Same protein sequence as previous slide for the initial protein). A mutant is found that has the following protein sequence in that part: Tyr Lys Gly Arg Pro Pro Gly Met Gly Leu Stop. What is the likely mutation? a. Between A12 and C13 insert T. b. Between C13 and C14 insert T. c. A15 -> deletion. d. C13 and C14 -> deletion. e. G8 -> T. f. G20 -> A. g. T21 -> deletion. h. Digital copy: TAT AAA GGG CGA CCA CCT GGT AAT GGG ACT TTG AGG.

g. T21 -> deletion.

Forward Primer: GTG ATG CGG GTT TGT GCA GAC Please write the primer in its own 5 prime to 3 prime sequence, so we can send it out to be synthesized. a. CTC CAC GAG ACA GTC CGT TG b. GTG CTC CAC GAG ACA GTC CG c. ATC GCT GTG CTC CAC GAG ACA d. GAG GTG CTC TGT CAG GCA AC e. A CAC GAG GTG CTC TGT CAG GC f. TAG CGA CAC GAG GTG CTC TGT g. TCA ACG GAC TGT CTC GTG GAG h. CG GAC TGT CTC GTG GAG CAC AG i. TGT CTC GTG GAG CAC AGC GAT

g. TCA ACG GAC TGT CTC GTG GAG

During S-phase, cyclinA-CDK is responsible for: a. Binding CDT1 and CDC6 to the ORC. b. Attaching Helicase and MCM to the pre-replication complex. c. Assuring that each origin is only fired once during S-phase. d. The process of degrading Geminin. e. The process of inactivating the APC. f. The degradation of CDT1. g. The degradation of CDC6 and the phosphorylation of the ORC. h. Making a large BOOM.

g. The degradation of CDC6 and the phosphorylation of the ORC.

1. The structure must be accurately copied before cell division. 2. The structure must contain complex information (genetic code). 3. The structure must be composed of rare/unusual elements. 4. The structure must allow or tolerate small changes that facilitate non-lethal variation. 5. The structure must be enzymatically active, able to mediate cellular metabolism. From the list above, which is or are NOT a key structural property displayed by the heritable (transforming) material? a. none. b. 1 c. 2 d. 3 e. 4 f. 5 g. 1 and 2. h. 3 and 5. i. 2 and 5. j. 3 and 4.

h. 3 and 5.

RNA polymerase can transcribe in different directions on a chromosome. This is enabled because: a. Genes and their associated promoters can be oriented in either direction along a chromosome. b. RNA polymerase can transcribe from either strand of DNA, synthesizing RNA in the 5' to 3' direction. c. The chemistry of RNA synthesis is identical regardless of which strand is transcribed. d. The 3' to 5' nonsense strand of a gene is always the template strand. e. A and B. f. B and D. g. C and D. h. A, B, C, and D.

h. A, B, C, and D.

AGC AAT CAC CAG ATT ATG ATA TGA CCG TGT ACA TGT Six frame ORFs. What is or are the open reading frames that go through the above sequence? The labeling of the the reading frames is 0, +1 and +2 in the 5prime to 3prime direction, and RC-0, RC+1, and RC+2 for the reverse-complement reading frames. Do not consider the AUG in your analysis, just pure open reading frame. Just be clear, the RC+1 starts on the first C in the R-C, and the RC+2 starts on the second A in the R-C. a. 0 b. 1 c. 2 d. RC-0 e. RC+1 f. RC+2 g. 0, RC-0 h. +1, +2 i. +2, RC-0 j. +2, RC+1 k. RC-0, RC+2 l. +2, RC+1 m. +1, RC+2 n. Digital copy: AGC AAT CAC GAG ATT ATG ATA TGA CCG TGT ACA TGT

i. +2, RC-0

The RNA-induced silencing complex, or RISC: 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. f. Degrades RNA having a short poly-A tail. g. Presents RNA to the Dicer complex for degradation. h. Only processes miRNA not other types of RNA, such as siRNAs or tRNAs. i. Cuts up RNA that is complementary to short strands of RNA.

i. Cuts up RNA that is complementary to short strands of RNA.

What are the minimum sequences necessary for translation in bacteria? a. Cap, start codon, stop codon. b. Enhancers, silencers, operator, promoter, polyadenylation signal. c. 5 prime end of RNA, GU-splice site, branch point-A, AG-splice site, polyadenylation signal. d. Repressors, activators, cap, start codon, stop codon. e. Promoters, GU-splice site, branch point-A, AG-splice site, polyadenylation signal. f. Enhancers, silencers, promoters, polyadenylation signal. g. Repressors, activators. h. Attenuators, Shine-Dalgarno sequence, start codon, stop codon. i. Activator binding site, promoter, operator, termination site. j. Shine-Dalgarno sequence, start codon, stop codon. k. Attenuator sequence. l. Start codon, stop codon.

j. Shine-Dalgarno sequence, start codon, stop codon.