101A_Su09_midterm2key - 1 MCDB 101A Summer 2009 2nd Midterm...

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Unformatted text preview: 1 MCDB 101A Summer 2009 2nd Midterm Name_________ ____Key_ __________________________ 7/20/09 (100 points total) Perm #__________________________________________ Multiple Choice: Circle the letter corresponding to the best answer (3 points for each correct answer) 1. Which is true of RecA protein? No answer was correct: 3 free points! a. It is required for excision of lambda from its host DNA in response to nutrient depletion b. RecA-dependent recombinational bypass requires the gene products UmuC and UmuD c. It is necessary for the integration of phage lambda into bacterial DNA d. It plays a role in the repair of thymidine dimers that result from UV exposure e. All the above 2. Very Short Patch repair can correct/repair a. mismatches resulting from the product of 5-methyl cytosine deamination b. DNA bases modified by the attachment of aflatoxin c. pyrimidine dimers d. small insertions or deletions resulting from polymerase errors e. both b and c 3. Which is true of a silent mutation? a. has no phenotype b. may prevent translation of downstream genes in an operon c. changes an amino acid to another of similar size and biochemical characteristics d. produces a phenotype that is identical to the pre-mutation phenotype e. may result from insertion of a single nucleotide into the protein coding sequence of a gene 4. Which is a correct statement about nonsense suppressor mutations? a. They result from large deletions b. They are sometimes found in the promoter region of protein genes c. They are located in the anticodon loop of tRNA genes d. They always convert a mutant codon back to the original codon e. They can be located within the same gene, but at a different location the site of the original mutation 5. Which of the following is used in PCR? a. a polymerase that can survive repeated cycles of freezing and thawing b. deoxy nucleotides: dATP, dUTP, dCTP, and dGTP c. a pair of RNA primers d. all of the above e. none of the above 6. Shown to the right is a base pair that resulted from tautomerization. The bases are a. enol cytosine and imino adenine b. keto guanine and enol adenine c. imino thymine and enol cytosine d. keto thymine and enol guanine e. enol thymine and amino xanthine 2 7. The membrane filter used in a Southern blotting assay is soaked in a NaOH solution for the purpose of a. accelerating the rate of binding of the labeled probe to the target DNA molecules b. removing supercoils from the DNA molecules c. denaturing the double stranded DNA into single strands d. enhancing the efficiency of transfer of DNA from the gel to the membrane filter e. digesting the DNA into fragments that will readily migrate through the gel 8. Shown below is a table of -galactosidase production by 3 E. coli strains grown in the presence (lactose+) and absence (lactose-) of lactose . One allele is missing from each genotype. Circle any alleles among the choices listed beneath the table for each genotype that would produce the indicated phenotype. choices listed beneath the table for each genotype that would produce the indicated phenotype....
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This note was uploaded on 10/22/2009 for the course MCDB 101A taught by Professor Thrower during the Summer '08 term at UCSB.

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101A_Su09_midterm2key - 1 MCDB 101A Summer 2009 2nd Midterm...

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