MIC 205A Dr. Jan Decker
Exam 2
October 12 , 2005

Part 1 MULTIPLE CHOICE: Read each question and the answers carefully. Pick the one BEST answer and darken the corresponding letter on the scantron. Each answer is worth 3 points, for a Part one total of 90 points.

• Clostridium botulinum is a strict anaerobe.  That means that C. botulinum
a. cannot ferment glucose.
• b. dies in the presence of oxygen.
• c. forms endospores.
• d. uses oxygen as a terminal electron acceptor.

• DNA transformation does NOT require
• a. a sex pilus.
• b. DNA uptake by the recipient cell.
• c. naked DNA.
• d. the recipient cell to be competent.

• Catabolism is
• a. an energy-consuming reaction.
• b. the breakdown of complex molecules to simpler molecules.
• c. the synthesis of complex molecules from simpler molecules.
• d. the synthesis of proteins from amino acids.

• The end product of glycolysis is
• a. CO₂.
• b. glucose.
• c. glyceraldehyde-3-phosphate.
• d. pyruvic acid.

• Which of the following is NOT a mechanism of gene transfer in prokaryotic cells?
• a. conjugation
• b. spontaneous mutation
• c. transduction
• d. transformation

• Escherichia coli is transferred from culture media containing glucose to culture media containing lactose. If you did a growth curve beginning with the time the E. coli was put into lactose, the first phase of growth you would observe would be the
• a. death phase.
• b. exponential phase.
• c. lag phase.
• d. stationary phase.

• During translation, tRNA molecules
• a. are translated into protein.
• b. form a structural part of the ribosome.
• c. join amino acids together using peptide bonds.
• d. recognize the codon on the mRNA and bring the next amino acid into place.

• The operator site of an operon is the site at which
• a. RNA polymerase binds.
• b. the regulatory protein binds.
• c. transcription begins.
• d. transcription stops.

• During transcription,
• a. DNA polymerase makes a protein copy of DNA.
• b. the information in DNA is transcribed into RNA.
• c. the information in RNA is transcribed into DNA.
• d. the information in RNA is transcribed into protein.

• Which of the following statements about the lac operon is FALSE?
• a. The lac operon is inducible.
• b. The lac operon is never transcribed in the presence of lactose.
• c. The lac operon is transcribed when lactose is present and glucose is absent.
• d. The lac operon is not transcribed when high amounts of glucose are present

• Transduction is the mechanism by which
• a. competent cells take up naked DNA.
• b. DNA is transcribed into RNA and RNA is translated into protein.
• c. DNA is transferred from a donor cell to a recipient cell by a bacteriophage.
• d. plasmid DNA is transferred from donor to recipient through direct contact.

• DNA is always synthesized
• a. away from the replication fork.
• b. in the 3’ to 5’ direction.
• c. in the 5’ to 3’ direction.
• d. towards the replication fork.

• Pigs that can produce Factor VIII, a human blood clotting protein, are an example of
• a. biotechnology.
• b. genetic engineering.
• c. transgenic animals.
• d. All of the above.

•  In the last reaction in the TCA cycle, malate (4 carbons) + NAD⁺ are converted to oxaloacetate (4 carbons) + NADH + H⁺. This is an example of
• a. a decarboxylation reaction.
• b. an oxidation-reduction reaction.
• c. substrate level phosphorylation.
• d. sugar splitting.

• ATP is generated from reduced coenzymes like NADH by
• a. fermentation.
• b. glycolysis.
• c. the electron transport chain.
• d. the TCA cycle.

•  Non-competitive inhibitors of enzyme activity bind to
• a. the active site of an enzyme.
• b. the allosteric site of an enzyme.
• c. the promoter of an operon.
• d. the substrate of a reaction.

• In the lac operon, which of the following would happen if the operator site were mutated such that the LacI repressor could not bind to the operator?
• a. Lactose could not bind to the repressor.
• b. The lac operon would be constitutively transcribed regardless of whether lactose was present or not.
• c. The lac operon would be transcribed only in the absence of lactose.
• d. The lac operon would not be transcribed regardless of whether lactose was present or not.

• Which of the following processes produces the most energy in the form of ATP in prokaryotic cells?
• a. aerobic respiration
• b. anabolism
• c. anaerobic respiration
• d. fermentation

• A base change that generates a stop codon in place of that coding for an amino acid is called
• a. a frameshift mutation.
• b. a missense mutation.
• c. a nonsense mutation.
• d. a silent mutation.

• In DNA fingerprinting, DNA from several individuals is cut by restriction endonucleases and compared for identical
• a. chromosome numbers.
• b. gene sequence.
• c. migration in electrophoresis.
• d. recombinant DNA.

• Which of the following statements is NOT true? Induced mutations result from
• a. errors in protein translation.
• b. exposure to chemical mutagens.
• c. exposure to ionizing radiation.
• d. exposure to UV radiation.

• The site on an enzyme where the substrate binds is the
• a. active site.
• b. allosteric site.
• c. ATP-binding site.
• d. cofactor site.

• The polymerase chain reaction (PCR) is a technique used to
• a. copy DNA.
• b. cut and splice new DNA into recipient DNA.
• c. detect whether inserted DNA is correctly expressed.
• d. insert new DNA into a cell.

• In glycolysis, phosphoenolpyruvic acid (a phosphorylated three carbon compound) is converted to pyruvic acid (a three carbon compound) with the generation of ATP by what mechanism?
• a. oxidative phosphorylation
• b. photophosphorylation
• c. spontaneous phosphorylation
• d. substrate level phosphorylation

• Bacteria that produce only lactic acid from the fermentation of pyruvic acid are called
• a. alcohol fermenters.
• b. heterolactic bacteria.
• c. homolactic bacteria.
• d. mixed acid bacteria.

• Which of the following statements about mutations is TRUE?
• a. A mutation is a change in the base sequence of DNA.
• b. A mutation never results in a change in amino acid sequence of the translated protein.
• c. Mutations always cause a change the phenotype of the organism.
• d. Mutations always provide a selective advantage to the organism.

• Which of the following is NOT a mechanism of ATP generation?
• a. competitive inhibition
• b. oxidative phosphorylation
• c. photophosphorylation
• d. substrate level phosphorylation

• In the arg operon, arginine acts in repression of transcription as
• a. a corepressor.
• b. an inducer.
• c. a promoter.
• d. a substrate.

• In DNA replication, the enzyme primase
• a. joins fragments of the lagging strand together
• b. synthesizes Okazaki fragments during replication of the lagging strand
• c. synthesizes the leading strand
• d. synthesizes the RNA primers during replication of the lagging strand

• Which method below will tell you how many living cells you have in a bacterial culture?
• a. direct microscopic count.
• b. dry weight.
• c. plate count.
• d. turbidity.
  

Part 2 ESSAY: Answer 2 of the questions below in the space provided. Each correct answer is worth five points, for a Part 2 total of 10 points.

31. Describe the Ames Test and how it is used to detect chemical mutagens.

• -Uses a histidine biosynthesis mutant of Salmonella (histidine requiring) as the indicator strain
• -The strain is plated onto two plates. A control plate with liver enzymes, but no histidine and no test chemical. A test plate with liver enzymes and the test chemical, but no histidine. Liver enzymes are included because the test chemical may be converted by these enzymes in the body to a mutagenic substance.
• -The basis of the test is that mutagens will increase the back mutation of the his- strain to his+ (ie: will grow in the absence of histidine)
• -Colonies on the control plate are due to spontaneous mutation*
• -Any increase in the number of colonies on the test plate compared to the control plate indicates that the test chemical is a mutagen.
• -the more colonies on the test plate, the more mutagenic the test chemical is*
• -There is a big correlation between a chemical being mutagenic in the Ames test and being carcinogenic in animals*

32. Describe the process by which the lagging strand is synthesized during DNA replication.

• -lagging strand synthesis is discontinuous because it has to be synthesized in the 5’-3’ direction away from the replication fork
• -RNA primers are synthesized at various intervals along the lagging strand by Primase
• -Unlike DNA polymerase, primase does not require a free 3’ end to add nucleotides*
• -The free 3’ ends of the RNA primers are used by DNA polymerase III to synthesize DNA between the RNA primers in the 5’ to 3’ direction (Okazaki fragments)
• -DNA polymerase I removes the RNA primers and replaces them with DNA
• -DNA ligase then joins the gaps in the lagging strand to give a complete and contiguous lagging stran

33. Draw a bacterial growth curve, label the growth phases, and briefly explain what is happening during each phase.

• Lag phase: production of new enzymes needed for growth
• Log phase: maximum growth
• Stationary phase: growth = death; nutrients or waste products limiting growth
• Death phase: death > growth; lack of nutrients or accumulation of waste products killing cells