Ucsb mcdb 101 midterm | Biology homework help

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MCDB 101A Summer 2012 Midterm 2 Name____Key___________________________________


7/23/12 100 points Perm #__________________________________________


Circle the best answer for each of the multiple choice questions (3 points each)


1. A temperature sensitive (ts) mutation in 50 S ribosomal subunit protein prevents the binding of tRNAs to


the ribosome “E” at the non-permissive temperature. How many amino acids would be found in the longest


polypeptide chain that could be synthesized in this strain at the non-permissive temperature?


a. 0


b. 1


c. 2


d. 3


e. 4


2. In the presence of lactose, a merodiploid of genotype I-dP+O+ Z-Y+/I+ P-OCZ+Y- will produce


a. Both â galactosidase and permease


b. â galactosidase but no permease


c. Permease but no â galactosidase


d. Neither â galactosidase nor permease


e. Blue colonies on X-gal media


3. The presence of short nucleotide repeats or exposure of DNA to intercalating agents can produce


a. nonsense mutations


b. nucleotide insertions


c. nucleotide deletions


d. frameshift mutations


e. all of the above


4. When running a northern blot you must denature the RNA prior to its transfer to the blotting membrane.


What is the purpose of denaturing the RNA?


a. To prevent the RNA from forming double-stranded molecules


b. To prevent the RNA from forming intramolecular base pairs


c. To provide RNA with a negative charge


d. To fragment the RNA prior to gel electrophoresis


e. To release the RNA from the polyacrylamide gel matrix used in the electrophoresis


5. Polarity may increase the survival of bacteria in their natural environment by reducing


a. unnecessary transcription of monocistronic genes


b. unnecessary transcription of polycistronic genes


c. unnecessary translation of polycistronic genes


d. the mutation frequency of monocystronic genes


e. the mutation frequency of polycystronic genes


6. A mutation that changes one amino acid to another without altering protein function is known as a


a. silent mutation


b. neutral mutation


c. missense mutation


d. nonsense mutation


e. reversion mutation




7. Which of the following is true of the blue-white cloning assay?


a. It is based on á complementation


b. Insertion of a gene into a blue-white cloning vector is detected by the appearance of blue colonies


c. It requires recombination between the plasmid and the host genome


d. The genome copy of the β-galactosidase gene must contain a mutation in its C-terminal domain


e. Both a and d are correct


8. Which of the following is a potentially mutagenic effect of ionizing radiation?


a. Formation of oxygen radicals


b. single base pair insertions or deletions


c. Deamination


d. Formation of intrastrand covalent linkages


e. all of the above


9. (4 points; insert correct numbers in blanks) Wobble base pairing deceases the number of different tRNA


genes required to pair with all of the isoleucine (Ile) codons from _____3_______ to ______1_______.


10. (6 points) Can deaminination of cytosine lead to the reversion of a nonsense mutation, resulting in the


restoration of a sense codon at that location in the DNA molecule? Explain.


No (2 points). Cytosine deamination produces uracil, which will pair with adenine during replication.


This will cause a GC to AT transition. There are two stop codons that contain G: TGA and TAG. As a


result of deamination of the cytosine that is base paired with G, both of these codons would be


converted to another stop codon, TAA. As a result, cytosine deamination would not result in a


reversion (4 points for explanation).


11. (5 points) Explain why a small amount of histidine is added to growth media used in the Ames test.


Most mutations require replication (2 points). A limiting amount of histidine allows his- Salmonella


strains to undergo sufficient rounds of replication to generate mutations without producing colonies (3






12. (5 points) Following electrophoresis, a gel used in the Southern blotting procedure is soaked in a sodium


hydroxide (NaOH) solution. What effect does NaOH have on DNA and why is this step necessary?


NaOH denatures/separates the DNA strands (2 points). This allows the labeled probe to bind to its


complementary/target sequence (3 points).


13. You have isolated a mutant protein that is only half the size of its wild type counterpart. The gene that


codes for the mutant protein is the same length as the wild type gene and contains a single base change. The


mutation is located within a glycine (gly) codon.


a. (3 points) Give the mRNA sequence of the glycine codon before the mutation occurred.




b. (3 points) Give the RNA sequence of the mutant codon (mark the 5’ and 3’ ends of each sequence).




14. (4 points) For the genotype given below, is β-galactosidase production absent, inducible or constitutive?




15. Shown to the right is the imino form of the base __adenine______________.


This tautomer pairs with the ____amino________________ tautomer of the base


_________cytosine________________ (2 points for each correctly filled in blank)


16. (3 points) An E. coli cell with a nonsense suppressor tRNA could theoretically produce a fusion protein


consisting of the combined amino acid sequences encoded lacZ, lacY, and lacA. Give one reason why this


event would be unlikely to occur.


Any of the following will suffice:


1. Termination proteins are present in excess over any given tRNA


2. There are multiple genes for any given tRNA, and only one would likely contain a suppressor




3. More than one stop codon is normally present at the end of a protein coding sequence


4. There are three different stop codons


17. Regulation of the E. coli arabinose operon involves the binding of C protein to the ara I DNA sequence,


located just upstream of the transcription initiation site for the B, A, and D genes. The C protein binds to araI


only when arabinose is present in the media, and the binding of C to araI stimulates the transcription of the


polycistronic RNA that encodes the BAD proteins.


a. (3 points) Is this an example of positive or negative transcriptional regulation? positive


b. (4 points) Give the name of an assay that could be used to demonstrate that C protein binds to araI.


Gel retardation/shift assay




18. Ten individual tubes of media were inoculated with T1 sensitive E. coli. These were allowed to grow until


each reached a concentration of 1 X 107 cells/ml. One ml of a solution containing 1 X 106 T1 phage/ml was


combined with 0.1 ml of each bacterial culture and added to plates of solid media. The # of bacterial


colonies/plate is shown below. Show your work in the space provided.


Plate # 1 2 3 4 5 6 7 8 9 10


#colonies 22 23 20 25 21 20 26 22 24 21


a. (5 points) Use the information above (not the Poisson formula) to directly calculate the MOI.


MOI = #phage/#bacteria = _____1 X 106 phage/ml_____ = 1


1X 107 bacteria /ml X 0.1 ml


2 points for formula, 1 point for plugging in the correct data, and two points for correct answer


b. (5 points) Use the Poisson formula and the MOI you calculated above to determine the probability


of finding two colonies on any given plate.


Poisson formula: f(i) = em mi MOI = m = 1, i = 2, e-1 = 0.37




f(2) = 0.37 X 12 = 0.37


2! 2


f = 0.18


2 points for formula, 1 point for plugging in the correct data, and two points for correct answer


c. (6 points) Are the findings of this study consistent with the hypothesis that the T1 resistant colonies


were a consequence of random mutations? Why or why not?


No/inconsistent (2 points). Two findings argue against the hypothesis that T1 resistance resulted from a


random mutation (either explanation is sufficient for full credit). (1) The table shows that the number


of resistant mutants is almost the same in each of the 10 cultures. Random mutations occurring at


various times during the growth of each individual culture should produce a wide range of colony


numbers similar to that observed in the Luria and Delbruck experiment. (2) Random mutation events


should follow a Poisson distribution. Based on the the Poisson calculation above, nearly 20% of the


plates should contain 2 resistant colonies. However, none of the plates produced 2 colonies. (4 points


for explanation)




19. You have used molecular biology techniques to create a phage lambda construct that is identical to a wild


type ë phage except that the PR promoter has been replaced with the lac operon promoter-operator DNA


sequence. You infect an E. coli leu– auxotroph that is growing in rich lactose media with this phage.


a. (3 points) Which of the following types of media would be most favorable for lysogeny?


1. rich lactose media


2. minimal lactose media


3. rich glucose media


4. miminal glucose media


b. (6 points) Give a detailed explanation for your answer to part a. Briefly describe the basic function(s) of any


proteins you mention in your explanation.


Lysogeny would be favored by minimal media as this would inhibit production of integration host


factor/IHF proteins which degrade CII (2 points). CII is a transcriptional activator required for the


production of sufficient amounts of CI and integrase proteins needed respectively to inhibit the lytic


cycle and facilitate phage integration into the host genome (2 points). Lactose media is needed to


induce transcription of the mRNA that encodes CII from its lac operon promoter (2 points).


c. (5 points) Would lysogeny occur in the media you chose for part a in an E. coli strain with the genotype


I+P+O+ Z-Y+? Why or why not?


No (2 points). In the absence of â-galactosidase, lactose could not be converted to allolactose and


induction of CII from its lac operon promoter would not occur (3 points).


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