POST-CLASS QUESTIONS for Week #7
(individual; open book; turn in July 08, 2008)
(Lose 20% if handed in July 09; lose everything if later!)
Do NOT turn this in! But turn in the answer sheet covering both Post-6 and Pre-7.

NAME

Several of the questions in this are engineered to push your minds. Can you go from what you know onwards just a little further?
Grades will be curved providing you correctly answer at least 4 of these:
This is a 14-question quiz. On your answer sheet, CROSS OUT one of the 15 questions you don't want counted.

1. You have irradiated an E. coli lac⁺ culture with UV light for 20 seconds and half of the cells are killed as demonstrated via plate-counts on MacConkey agar (contains lactose). On one of the plates there are two non-red colonies, M and N (can't use lactose to make acid), which you carefully transfer to two NA plates, to one of which you have added lactose. In a small test tube you permeablize some of the growth on each of those plates and add ONPG. No yellow appears for M (no lactase made in either case). M's mutation CANNOT be in which cistron?
(a) lacR not this because if R cannot be deactivated by lactose, gene is "off" under all conditions
(b) lacO not this because mutation might bind even deactivated R
(c) lacP a mutation here would not allow operon to be decoded - thus always "off"
(d) lacZ of course a mutation here would give the results shown
(e) lacY a mutation here would give yellow from colonies grown on lactose 8. E. coli is a "mixed-acid" fermenter. Two cultures are made in nutrient broth. Culture D is 0.01M in glucose (180 daltons), and Culture E is 0.01M in lactose (342 daltons). Both cultures are sparged with N₂ and acid production is monitored. (Notice! because of equal molarity, there is more weight of lactose present than glucose.)
(a) initially both drop the pH at the same rate
(b) both will be undergoing anaerobiosis
(c) at stationary phase D will be less acidic that E D had less weight of sugar with which to make acid.
(d) both will be fermenting
(e) all of the above
(f) none of the above
2. Now let's see if you can stretch your mind! N, from above, results in yellow only from the growth obtained from the NA+lactose plate. The mutation MUST be in which?
(a) a cistron Yes, of the lacY, which is a cistron!
(b) lacO this is a regulator region
(c) lacP a regulator region
(d) a regulator or control region 9. The compounds: lactose, ONPG, IPTG.
(a) all have glucose as one portion of their molecules (no, galactose is common to them all!)
(b) only one of them interacts with the lac-operon's repressor protein You have seen that both lactose and IPTG do it.
(c) one interacts with the protein coded by lacA
(d) lactase reacts with all three of them not with IPTG
(e) two are chemical analogs of the third one IPTG and ONPG are the analogs of lactose
(f) all of the above
(g) none of the above
3. Which of the following is NOT true?
(a) an abnormal enzyme can result from a mutation in a cistron true
(b) a constitutive mutant can result from a mutation in lacO true
(c) if lacZ is mutated, it is possible for lactase (β-galactosidase) to be produced whether or not lactose is present not true
(d) a lacA mutant is for all intents and purposes phenotypically lac^- true 10. "Arsenic" (actually arsenate, H₂AsO₄^-) is poisonous because it is
(a) an anion lots of things are anions and are necessary as "foods."
(b) similar in structure to phosphate, H₂PO₄^- Notice that As is directly under P on the periodic table - BIG HINT!
(c) instantly interfers with lactase production not ture
(d) causes phosphate-rich bones to disintegrate within hours no, death is due to slow deterioriation of metabolism
(e) all of the above
(f) none of the above
4. Jason has a culture of E. coli lac⁺ growing in nutrient broth (no sugars). He then dumps in some lactose and begins taking samples every minute and running the ONPG reaction on each sample. Which of the following curves is the most likely he would obtain?
Answer = "b". This is what you did in lab: remember the 10-minute delay before yellow was seen?
11. Cheryl has some anti-lactase (antibody against lactase). She isolates some E. coli lacZ mutant cells from NA+lactose and, after lysing them with lysozyme, does a slide-agglutination test on them using normal and lacO^- (off all the time) E. coli as positive and negative controls, respectively. She gets the results as shown.
(a) her lacZ mutant makes an inactive protein resembling lactase
(b) her lacZ mutant does not make an inactive protein resembling lactase
(c) her lacZ mutant uses lactose to make acid
(d) the lacO^- mutant's cytoplasm reacted with the anti-lactase
(e) the anti-lactase was inactive
(f) a and c only
(g) a and d only
not graded because a picture is missing.
5. You have a rapidly dividing culture of E. coli (g = 21 minutes). It's genome appears to the right. Consider that the bacterial DNA is also replicating. Which of the following are true regarding the concentrations of the various genes in the culture during this growth sprint? (FYI: his is the gene for making the amino acid histidine; leu for leucine; met for methionine. Also: brackets are chemists' notation for "concentration.")
(a) [met] = [leu] = [his]
(b) [met] > [leu] > [his]"Early genes - ones near oriC - are in higher "copy number" than late genes.
(c) [met] < [leu] < [his] 12. Radioisotopes are atomic analogs of their non-radioactive forms. The body rapidly takes up normal iodine (I¹²⁷), incorporates it into thyroxin, which lingers in the thyroid gland doing good. But in cases of hyperthyroidism, the thyroid can either be removed surgically or damaged by giving radioactive iodine (I¹³¹). Do not be confused by the question's saying nothing about iodide. Yes, iodide is in most salt, but if you make a solution of - say - KI, it will slowly turn orange indication the elemental iodine, I₂, is being formed.
(a) excessive I¹³¹ treatment leads to goiter-like metabolic symptoms kills thyroid so that body cannot make thyroxine, which is the same result as a thyroid with doesn't have the iodine to make thyroxine.
(b) in event of an "event" at a nuclear power station, downwind residents should immediately consume lots of I¹²⁷ true - e.g: as KI tablets
(c) if working in a lab that uses I¹³¹ in their experiments, always have open bottles of crystalline I¹²⁷ scattered about true
(d) common consumption of iodized salt (NaCl with 1 ppt I¹²⁷) has eliminated most goiter true
(e) all of the above
(f) none of the above
6. Mary Lou is confused: she was told that T4 virions are not inactivated by chloroform, CHCl₃, but that bacteria are killed and lysed by it - thus releasing all their intracellular contents for analysis - as also employed in the permeabilization for the ONPG test. She had a culture of E. coli containing a total of 2x10⁸ cfu's, and dumped in half that many T4 virions. She immediately began taking samples of the culture every minute and shaking each with CHCl₃, and then spreading 100 μL onto whole plates lawned with E. coli. In 6 hours plaques appeared in the lawns - each plaque was a "colony" of T4's that had "eaten" a hole in the growing bacterial lawn. What confused her was that she was sure she did everything identically from sample to sample, and yet her plates from 4 minutes to 10 minutes "didn't work" - just lawns without holes grew up. "Was that a bad batch of plates?" she wondered. Which graph best fits the data Mary Lou should have gotten? (Hint: be wary of the vertical axis!)
Answer = "c". Remember the "eclipse" phase when there are no virions present - either inside or outside the cells. Only virions can be transferred to the plates to make plaques.
13. (IMViC) You plate normal E. coli onto 2 different kinds of agar: NA+ONPG and NA+IPTG. The next day...
(a) a few colonies appear on the ONPG plate, but many on the IPTG plate These are lac^- on the ONPG plate, which don't hydrolyze ONPG and kill themselves with the ONP that is produced.
(b) those growing on the ONPG plate do not produce lactase, but those on the IPTG do (see previous comment)
(c) spontaneous mutations (our friends the cosmic rays!) allow those on the IPTG plate to grow not true: IPTG isn't hydrolyzed, no lethal product as with ONPG
(d)nothing grows on either plate because neither ONPG nor IPTG can be metabolized for energy you have other nutrients in NA
(e) both plates grow equally well - but form yellow colonies not true - see earlier comments
(f) only a and b are correct
(g) only a and c are correct
7. Wilbie returned from Syntech Corp. with two lac-transposition mutants, Q and R. These are normal in every way except that their lac-operons have been moved to other positions on the E. coli genome. These and normal E. coli ("N") were harvested at log-phase and 1 mL samples of exactly 0.1 optical density units of each were subjected to permeablization and ONPG. After 30 seconds with ONPG, the reactions were "stopped" by placing in ice-water, and then the amount of yellowness was quantified using the spectrophotometer. Q was twice as yellow as N, which was twice as yellow as R. Again this is a matter of "early" and "late" genes.
(a) Q's lac-operon is closer to oriC than is that of R
(b) R's lac-operon is closer to oriC than is that of Q
(c) Q's lac-operon is closer to oriC than is that of N
(d) R's lac-operon is closer to oriC than is that of N
(e) both a and c only
(f) both b and d only 14. The "T-phages" of E. coli
(a) consist of RNA, protein heads, and lipid attachment organs DNA phages, and protein attachment organs
(b) contain amounts of genetic material in their heads that are constant for their "species" "head-fulls," remember?
(c) are able to be grown axenically on NA viruses always require host cells; can never be grown axenically.
(d) cannot be grown anaerobically require whatever the host cells require
(e) all of the above
15. Syntech's Dr. Sørensen gave Wilbie a mutant strain of E. coli and told him to deliver it to his lac-operon friends in Bengston's lab over at the Kollege of Knowledge. Normally DNA-polymerase, the enzyme that replicates DNA, is coded by its own operon called the dnaE-operon. It normally has its own operator which responds to its own "signals." But Sørensen substituted dna-P and dna-O with lac-P and lac-O. Which one of the following is NOT true?
(a) these cells cannot divide in NA True: nothing turns on the DNA replicase operon.
(b) these cells cannot divide in NA+glucose same answer as above
(c) these cells soon die if grown in NA+0.3% lactose DNA-replicase gene is on all the time. Cell would be in a constant state of initiating DNA synthesis.
(d) these cells will all divide in unison 40 minutes after being fed on 0.01% lactose, and then stop dividing until another tiny dose of lactose is administered True!
(e) none of the above choices

	1. You have irradiated an E. coli lac⁺ culture with UV light for 20 seconds and half of the cells are killed as demonstrated via plate-counts on MacConkey agar (contains lactose). On one of the plates there are two non-red colonies, M and N (can't use lactose to make acid), which you carefully transfer to two NA plates, to one of which you have added lactose. In a small test tube you permeablize some of the growth on each of those plates and add ONPG. No yellow appears for M (no lactase made in either case). M's mutation CANNOT be in which cistron? (a) lacR not this because if R cannot be deactivated by lactose, gene is "off" under all conditions (b) lacO not this because mutation might bind even deactivated R (c) lacP a mutation here would not allow operon to be decoded - thus always "off" (d) lacZ of course a mutation here would give the results shown (e) lacY a mutation here would give yellow from colonies grown on lactose			8. E. coli is a "mixed-acid" fermenter. Two cultures are made in nutrient broth. Culture D is 0.01M in glucose (180 daltons), and Culture E is 0.01M in lactose (342 daltons). Both cultures are sparged with N₂ and acid production is monitored. (Notice! because of equal molarity, there is more weight of lactose present than glucose.) (a) initially both drop the pH at the same rate (b) both will be undergoing anaerobiosis (c) at stationary phase D will be less acidic that E D had less weight of sugar with which to make acid. (d) both will be fermenting (e) all of the above (f) none of the above
	2. Now let's see if you can stretch your mind! N, from above, results in yellow only from the growth obtained from the NA+lactose plate. The mutation MUST be in which? (a) a cistron Yes, of the lacY, which is a cistron! (b) lacO this is a regulator region (c) lacP a regulator region (d) a regulator or control region			9. The compounds: lactose, ONPG, IPTG. (a) all have glucose as one portion of their molecules (no, galactose is common to them all!) (b) only one of them interacts with the lac-operon's repressor protein You have seen that both lactose and IPTG do it. (c) one interacts with the protein coded by lacA (d) lactase reacts with all three of them not with IPTG (e) two are chemical analogs of the third one IPTG and ONPG are the analogs of lactose (f) all of the above (g) none of the above
	3. Which of the following is NOT true? (a) an abnormal enzyme can result from a mutation in a cistron true (b) a constitutive mutant can result from a mutation in lacO true (c) if lacZ is mutated, it is possible for lactase (β-galactosidase) to be produced whether or not lactose is present not true (d) a lacA mutant is for all intents and purposes phenotypically lac^- true			10. "Arsenic" (actually arsenate, H₂AsO₄^-) is poisonous because it is (a) an anion lots of things are anions and are necessary as "foods." (b) similar in structure to phosphate, H₂PO₄^- Notice that As is directly under P on the periodic table - BIG HINT! (c) instantly interfers with lactase production not ture (d) causes phosphate-rich bones to disintegrate within hours no, death is due to slow deterioriation of metabolism (e) all of the above (f) none of the above
	4. Jason has a culture of E. coli lac⁺ growing in nutrient broth (no sugars). He then dumps in some lactose and begins taking samples every minute and running the ONPG reaction on each sample. Which of the following curves is the most likely he would obtain? Answer = "b". This is what you did in lab: remember the 10-minute delay before yellow was seen?			11. Cheryl has some anti-lactase (antibody against lactase). She isolates some E. coli lacZ mutant cells from NA+lactose and, after lysing them with lysozyme, does a slide-agglutination test on them using normal and lacO^- (off all the time) E. coli as positive and negative controls, respectively. She gets the results as shown. (a) her lacZ mutant makes an inactive protein resembling lactase (b) her lacZ mutant does not make an inactive protein resembling lactase (c) her lacZ mutant uses lactose to make acid (d) the lacO^- mutant's cytoplasm reacted with the anti-lactase (e) the anti-lactase was inactive (f) a and c only (g) a and d only not graded because a picture is missing.
	5. You have a rapidly dividing culture of E. coli (g = 21 minutes). It's genome appears to the right. Consider that the bacterial DNA is also replicating. Which of the following are true regarding the concentrations of the various genes in the culture during this growth sprint? (FYI: his is the gene for making the amino acid histidine; leu for leucine; met for methionine. Also: brackets are chemists' notation for "concentration.") (a) [met] = [leu] = [his] (b) [met] > [leu] > [his]"Early genes - ones near oriC - are in higher "copy number" than late genes. (c) [met] < [leu] < [his]			12. Radioisotopes are atomic analogs of their non-radioactive forms. The body rapidly takes up normal iodine (I¹²⁷), incorporates it into thyroxin, which lingers in the thyroid gland doing good. But in cases of hyperthyroidism, the thyroid can either be removed surgically or damaged by giving radioactive iodine (I¹³¹). Do not be confused by the question's saying nothing about iodide. Yes, iodide is in most salt, but if you make a solution of - say - KI, it will slowly turn orange indication the elemental iodine, I₂, is being formed. (a) excessive I¹³¹ treatment leads to goiter-like metabolic symptoms kills thyroid so that body cannot make thyroxine, which is the same result as a thyroid with doesn't have the iodine to make thyroxine. (b) in event of an "event" at a nuclear power station, downwind residents should immediately consume lots of I¹²⁷ true - e.g: as KI tablets (c) if working in a lab that uses I¹³¹ in their experiments, always have open bottles of crystalline I¹²⁷ scattered about true (d) common consumption of iodized salt (NaCl with 1 ppt I¹²⁷) has eliminated most goiter true (e) all of the above (f) none of the above
	6. Mary Lou is confused: she was told that T4 virions are not inactivated by chloroform, CHCl₃, but that bacteria are killed and lysed by it - thus releasing all their intracellular contents for analysis - as also employed in the permeabilization for the ONPG test. She had a culture of E. coli containing a total of 2x10⁸ cfu's, and dumped in half that many T4 virions. She immediately began taking samples of the culture every minute and shaking each with CHCl₃, and then spreading 100 μL onto whole plates lawned with E. coli. In 6 hours plaques appeared in the lawns - each plaque was a "colony" of T4's that had "eaten" a hole in the growing bacterial lawn. What confused her was that she was sure she did everything identically from sample to sample, and yet her plates from 4 minutes to 10 minutes "didn't work" - just lawns without holes grew up. "Was that a bad batch of plates?" she wondered. Which graph best fits the data Mary Lou should have gotten? (Hint: be wary of the vertical axis!) Answer = "c". Remember the "eclipse" phase when there are no virions present - either inside or outside the cells. Only virions can be transferred to the plates to make plaques.			13. (IMViC) You plate normal E. coli onto 2 different kinds of agar: NA+ONPG and NA+IPTG. The next day... (a) a few colonies appear on the ONPG plate, but many on the IPTG plate These are lac^- on the ONPG plate, which don't hydrolyze ONPG and kill themselves with the ONP that is produced. (b) those growing on the ONPG plate do not produce lactase, but those on the IPTG do (see previous comment) (c) spontaneous mutations (our friends the cosmic rays!) allow those on the IPTG plate to grow not true: IPTG isn't hydrolyzed, no lethal product as with ONPG (d)nothing grows on either plate because neither ONPG nor IPTG can be metabolized for energy you have other nutrients in NA (e) both plates grow equally well - but form yellow colonies not true - see earlier comments (f) only a and b are correct (g) only a and c are correct
	7. Wilbie returned from Syntech Corp. with two lac-transposition mutants, Q and R. These are normal in every way except that their lac-operons have been moved to other positions on the E. coli genome. These and normal E. coli ("N") were harvested at log-phase and 1 mL samples of exactly 0.1 optical density units of each were subjected to permeablization and ONPG. After 30 seconds with ONPG, the reactions were "stopped" by placing in ice-water, and then the amount of yellowness was quantified using the spectrophotometer. Q was twice as yellow as N, which was twice as yellow as R. Again this is a matter of "early" and "late" genes. (a) Q's lac-operon is closer to oriC than is that of R (b) R's lac-operon is closer to oriC than is that of Q (c) Q's lac-operon is closer to oriC than is that of N (d) R's lac-operon is closer to oriC than is that of N (e) both a and c only (f) both b and d only			14. The "T-phages" of E. coli (a) consist of RNA, protein heads, and lipid attachment organs DNA phages, and protein attachment organs (b) contain amounts of genetic material in their heads that are constant for their "species" "head-fulls," remember? (c) are able to be grown axenically on NA viruses always require host cells; can never be grown axenically. (d) cannot be grown anaerobically require whatever the host cells require (e) all of the above
	15. Syntech's Dr. Sørensen gave Wilbie a mutant strain of E. coli and told him to deliver it to his lac-operon friends in Bengston's lab over at the Kollege of Knowledge. Normally DNA-polymerase, the enzyme that replicates DNA, is coded by its own operon called the dnaE-operon. It normally has its own operator which responds to its own "signals." But Sørensen substituted dna-P and dna-O with lac-P and lac-O. Which one of the following is NOT true? (a) these cells cannot divide in NA True: nothing turns on the DNA replicase operon. (b) these cells cannot divide in NA+glucose same answer as above (c) these cells soon die if grown in NA+0.3% lactose DNA-replicase gene is on all the time. Cell would be in a constant state of initiating DNA synthesis. (d) these cells will all divide in unison 40 minutes after being fed on 0.01% lactose, and then stop dividing until another tiny dose of lactose is administered True! (e) none of the above choices