exam3key - BIO 226R EXAM III — KEY Dr Earhart NAME UT...

Unformatted text preview: BIO 226R EXAM III — KEY April 16, 2007 Dr. Earhart NAME UT EIDI _ i. {it} points} Given: Two bacterial cultures, each growing with a doubling time ot‘t’ifl minutes. One culture is growing synchronously, the other asynchronously. {a} [4 pts] Diagram, using a semilog plot where the y axis is the log of cell no. as determined by plate counts, and the x axis is time in hours, the growth of both cultures through three generations. ‘5 nchronous ; ow_tl_t __ log cell no. time {generations} (b) Repeat the above growth experiment but instead of plotting the log of cell no. on the y axis, plot the log of turbidity of the culture as measured by a colorimeter or spectrophotometer. lug Synchronous & turbidity Asynchronous (Cl DNA synthesis is measured for three generations in the synchronous culture with the following result: relative rate of 2 DNA 1 synthesis o i .2 me" generations What does this data indicate regarding DNA synthesis during the cell division cycle? DNA synthesis in this culture occurs for ~ the first 413 min. of the fit} min. cell division cycle. Now-bom cells begin DNA synthesis immediately and 4t] min. later have completed DNA replication. There is then a ~20 min. period ofno DNA synthesis before the cell divides. 2. {it points} This question concerns the ribosomes of prokaryotic cells. {3} lb} 1|What is the composition and sedimentation coefficient [5 value} of a prokaryotic ribosome? /Tt§|s ~2i‘3 R, if} Protein Sfls 313s 233 rRNA loS IRMA 55 RNA m 2t} proteins ~32 proteins The ribosome undergoes a series of changes in its sedimentation coefficient during the synthesis of a protein. Explain. To initiate protein synthesis, mRNa associates with the His subunit {~3fls}. In the last stages of initiation. the 513s subunit joins and the approx. Tfls initiation complex is assembled. Once protein synthesis has started. the ribosome binding sites on the inRNa becomes available to bind more ribosomes and a polysome is value attiti] is formed. In the termination state. the process reverses itself as proteins are released and ribosomes leave as fills and 39s subunits. What catalytic activity is possessed by all ribosomes and what is unusual about this activity? The catalytic activity that synthesizes peptide bonds is an intrinsic activity of the ribosome. The nascent polypeptide chain in the P site is transferred them a tRNA molecule to the amino acid in the A site. The activity is unusual because it is catalyzed by a base on the EEIRNA; it is a tibozyme. 3. [8 points) What is a signal sequence, how can it be identified, and what role. if any. does it play in protein localization? The common signal sequence occurs at the N terminus of a protein. It is approximately 25aa long; the first 6-3 as have at least one basic {+ charged) aa (Fug, Lys}, the next ~ lSaa are hydrophobic. and the last {C terminus of signal peptide) aa sequence contains a leader pcptidase cleavage site {usually Ala X Ala]. All signal sequences play a role in protein localization. The signal sequencejust described would be present on the N terminus of exported (periplasm, outer membrane) proteins of E. coil and on some secreted proteins {type ll} as well. The signal sequence would be removed during passage of the protein through the cytoplasmic membrane. 4. {it points} Given: A mutant bacterium that produces a temperature-sensitive FtsZ protein. At non permissive temperatures, what would occur with respect to cell division? Why? Would this mutant give rise to colonies on agar plates if the plates were incubated at a non-permissive temperature? The cell would grow as filaments [snakes]. These would be no septa formation because FtsZ is the key structural protein of the system, assembling into long chains. The mutant is temperature sensitive which indicates it would be unable to form colonies at the non-permissive temperature. 5. {3 points} Describe the roles proteases may play in [a] regulation Some proteins, such as secondary sigma factorst normally have short half—lives because they are continually being degraded by proteases. The sigma factors accumulate when proteolysis is inhibited. {b} protein export Leader peptidases cleave signal sequences removing the N terminus ~ fine and in the process releasing this protein [exported or secreted) from the cytoplasmic membrane. if the leader peptidase is blocked. its substrate proteins remain bound to the cytoplasmic membrane. [c] dealing with hopelessly misfolded proteins Proteases destroy proteins that the cell is unable to refold, via chaperones. into active form. s. {3 points) F is aconjugative plasmid. {a} {bi Describe how it is able to move from one bacterium to another. The self-transmissablc plasmid F contains on genes, which enable it to move into F' (recipient) cells. Some Ire genes encode mating pair formation proteins, which include the genes for the synthesis of the sex pilus. This pilus can hind to a receptor on F' cells and then retract. pulling the F+ and Fl-cells together. Other fr'rr genes act to mobilize F+ DNA by acting to nick F at its oriT. covalently attaching the nickasc (relaxase then] and then moving to the F+-F' cell junction coupling protein. Rolling circle DNA synthesis drives a single strand of F DNA into the recipient. where a complementary stand is synthesized and the molecule circularized. How is the F plasmid able to move bacterial genes from one bacterium to another? To move bacterial DNA, the F factor must integrate into the bacterial chromosome. It usually does this by homologous recombination between identical insertion sequences on F and the bacterial chromosomes. Then, when mating pairs are formed and the oriT is nicked, some F factor DNA will be transferred to the F- cell but then bacterial DNA will be transferred. The orr'Tis not at one end of the integrated F factor — to transfer all of F one must first transfer some F. the entire bacterial chromosome and then the remainder of F. 'i'. (8 points] Distinguish between the roles and cellular locations of signal recognition particles and the Sec‘r’EG proteins. p. lfil Signal recognition particles are composed of RNA and proteins and located in the cytoplasm. They serve to assist the localization ofcytoplasmic membrane proteins. As the nascent cm pp is exiting the ribosome, its very hydrophobic N terminus is recognized by the SRP, which binds to the terminus, stops translation and delivers the entire mR. ribosome. nascent pp complex to the FtsY receptor in the cytoplasmic membrane, where protein synthesis resumes and the pp is inserted into the membrane [cotranslational insertion}. See't’EG proteins are integral cm proteins. They serve to permit the transmembrane passage of exported proteins and some proteins destined to be secreted. The Sec‘t’EG proteins form a translocon that has a hydrophilic channel permitting proteins {unthlded} to traverse the cm. Seek and leader peptidase are additional components of this path. The following questions are worth 3 points each. 3. Mutations (3} lb} {‘3} {£1} change the sequence of dcoxynucleotides ofa chromosome change the genotype of an organism might change the phenotype of an organism all of the above 9. l[C‘haperon es {:1} are found only in prokaryotes and arches (b) modulate the speed and path of protein folding {c} are composed of RNA and proteins [d] are membrane-associated It). The murein of gram positive cells {a} contains the same two sugars that are found in Gram negative cells {b} requires undecaprenylphosphate for its synthesis {c} has as its newest layer that closest to the cytoplasmic membrane {d} all of the above I 1. ln Gm+ conjugationi pheromones [a] are produced by the recipient cell [h] are encoded on the bacterial chromosome to] regulate the synthesis of a plasmid—home gene for aggregation substance {d} all of the above l2. The lactose operon encodes enzymes which {a} synthesize lactose [b] recognize and bind the lactose promoter to) split lactose into galactose and glucose [d] phosphorylate lactose l 3. Mutagenesis (a) is a way to inhibit growth of wild-type organisms [b]: alters the dcoxynucleon’de sequence of the chromosome {c} selects for growth of the mutant organism [d] is a technique for tbrrning partial diploids l4. Specialized and generalized transduction differ in that {a} generalized transducing bacteriophage can package any amount of DNA. {h} specialized transducing bacteriophage have the capability of integrating into the bacterial chromosome {c} the recipient bacterium after generalized transduction contains large amounts of bacteriophage DNA in plasmid form {d} high frequency transducing lysatcs can be prepared fiorn generalized transducing bacteriophage I5. Allosterie proteins (a) have their synthesis regulated by repressor proteins (b) insert into phospholipid bilayers spontaneously (e) are found only in fueling pathways {d} none of the above 1e. Minicells {a} result from polar, rather than mid-cell, septum “formation {b} contain the same amount of' DNA as a normal—sized cell {e} have defective divisomes {d} all of the above IT. The SOS response {a} is an example of negative regulation {b} 1would not occur in a cell lacking functional RceA proteins {e} brings about inhibition of cell division {d} all of the above 18. Type III [contact dependent) protein secretion {a} is See dependent {b} is important medically {e} requires only three proteins and ATP {d} involves an intermediate stage wherein the secreted protein exists independently in the periplasm. Ii}. Bacterial genera that are naturally transformable {a} take up only DNA from their own genera {b} are never competent {c} convert the transforming DNA. from double-stranded to single-stranded form during entry {d} all ot‘the above 2t}. Wild-type eoft' ean use lactose as its sole carbon and energy source. It" a mutation results in inactivation of the enzyme B-galactosidase (a) the mutant would be unable to grow on medium containing only lactose as a carbon and energy source [b the mutant would be unable to grow on any medium (c) the mutant could grow only if lactose were provided in the medium [d] none ct~ the above 21. Bacteria are haploid yet in fast-growing cells some genes may be present in two to four times the amount of other genes. Explain. pp. [TE-173 Gene dosage. Bacterial chromosomes replicate hidirectionally from a fixed origin. Therefore, a partially replicated chromoSome will have 2 times as many copies ofa gene near the origin thus near the terminus. 1‘Ir'ery rapidly growing cells may have 4 oristterm and therefore genes near the origin would be present in 4X the amount of genes near the terminus. ...
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