1. 
2. Help
3. Log In

MCB 229 Exam 2 Answer Key

Q.	Comments
1. E	Changing sigma factors allows entire blocks of genes to be turned on as a group. This is a drastic step, as it displaces recognition of previous genes, so it is only used for major life-style changes, such as survival in the heat shock response.
2. B	AUG is a codon (the "intiation codon") found on m-RNA and recognized by a ribosome in the context of the translation process. Codons have no functional significance in DNA or RNA synthesis.
3. C	CAP is an "activator protein", When bound to DNA, it activates genes for catabolism, such as the lac operon and others. But it only activates such genes in the absence of glucose, not in region A. Answer D is true, not false -- since lactose is present from time zero, lactose will complex with lac repressor and cause the lac repressor to cease binding to DNA. The failure to synthesize lac genes in region A is not due to lac repression at all, but to catabolite repression, a separate system.
4. D	This describes anaerobic respiration, which has nothing to do with chemolithotrophic growth.
5. D	If lac repressor doesn't function, then lac genes cannot be repressed -- they will act like constitutive genes (as long as glucose is not around to cause catabolite repression, a different level of control).
6. D	RNA polymerase is required for transcription, not translation. These are two separate processes. Other components listed are all part of the translation machinery
7. C	MS2 is an RNA virus; T4 and Lambda are both DNA viruses.
8. A	Positive control by definition requires some type of activation in order to proceed. The default is "off".
9. B	There are two major groups of photosynthetic bacteria: aerobic cyanobacteria, which carry out oxygenic photosynthesis, and the anaerobic green and purple bacteria, which carry out anoxygenic photosynthesis.
10. B	RNA polymerase is required for transcription, not replication. Other enzymes listed are all involved in replication.
11. B	Since "-" RNA cannot function as m-RNA, such viruses are initially harmless to a cell -- they can't make a single protein! The first thing they must do is to synthesize a "+" RNA strand in order to make viral proteins -- only then do they have weapons to attack the cell.
12. D	Example discussed in class and on reading quiz.
13. B	Definition.
14. B	Definition.
15. D	CAP is activated by cyclic AMP, not ATP. Once activated, it binds to DNA.
16. F	Mutagens and UV lead to induced mutations. Spontaneous mutations arise without external causes.
17. T	Anaerobic respiration and dissimilatory reduction are synoymous.
18. F	Oxygenic phototrophs ultimately use H2O, not H2S or H2, as a source of reducing power.
19. F	300 amino acids requires at least 900 base pairs of DNA, not 100.
20. T	Yes, transcription & translation occur simultaneously in prokaryotes.
21. F	m-RNA is actually the least abundant form of RNA, only about 5% of the cell content. 80% is ribosomal RNA and another 15% is t-RNA.
22. F	Anoxygenic phototrophs use cyclic photophosphorylation.
23. F	The molecule shown must be unwinding from R to L. Strand "a" has the 5'-end on the right, and must be synthesizing DNA from R to L, opposite the unwinding direction. So this is a lagging strand.
24. F	The statement defines auxotrophs, not prototrophs.
25. T	This exactly what the Ames test measures.
26.	CO2, ATP, and NADPH
27.	Long open reading frames (ORFs) are used by computers to predict the existence of genes.
28.	Nitrogen (N2) fixation
29.	Ammonia (NH3)
30.	Nitrification (nitrite oxidation by chemolithotrophs, nitrifying activity, Nitrobacter).
31.	Denitrification or Dissimilatory nitrate reduction (anaerobic respiration).
32.	Allows bacteria to use Hydrogen (H2) as an energy source = chemolithotroph.
33.	Allows organisms to "fix" CO2 into organic form (first step of the Calvin-Benson cycle).
34.	Joins or "charges" amino acids to t-RNA in preparation for translation (polypeptide synthesis)
35.	Allows RNA viruses to create complementary RNA (e.g. "+" RNA used as template for "-" RNA, vice versa).
36.	Comparing DNA sequences between different living organisms (comparing genes for 16S RNA sequence homology is the most common method).
37.	There are several critical assumptions, such as: (1) mutations are assumed to occur at the same rate over time, (2) "molecular clocks" are reliable, (3) mutational separation takes the shortest path (called the "parsimony" assumption), others.
38.	Repressible gene (used to regulate anabolic processes, different from inducible gene, typical of catabolic processes ), Note that this is not feedback inhibition, a term referring to modulating activity of allosteric proteins.
39.	A. C. = 0.6 (Out of 5 traits, 3 are shared. 3/5 = 0.6)
40.	5'-AUAGAACCAUAA-3' (Note that the "sense" strand is not the template strand! This is the one major error in your text, which I took pains to point out.)
41.	Conversion of a virus into a prophage (viral DNA inserts into host DNA, viral lytic genes are repressed)
42.	3 "start" codons (one for each of the 3 polypeptides).
43.	Lac repressor will not be bound to lac operon (it will be floating in the cytoplasm).
44.	Stromatolites are produced by cyanobacterial mats when they deposit minerals. They produce large fossils that allow evolutionary biologists to record the presence of cyanobacteria (and therefore O2 production) over time.
45.	Compare the 2 DNAs: there are 4 mutations that separate them, so the "mutational distance" is 4