rains of E. coli are isolated. They may be mutated on the ara, mal, leu and pro genes, only these 4 genes. The 5 strains (E1 to E5) are tested on 5 different media and onies are observed the next day. Rich Media (Lb) ES E4 ● E1 E3 E2Ⓡ MM + Mal+Leu E5 E4 X E1 E3 X E5 X MM + Mal E4 E1 E3 X E2 E5 E4 . MM + Mal+Leu +Pro E1 E3 E5 E2 MM + Ara E4 X X E1 E3 E2 X X onies are indicated by a black dot, absence of a colony is indicated by a cross. 1) Determine the genotypes of the 5 bacterial strains (E1 – E5)
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- You have several different media onto which you inoculated eight strains of yeast (A-H). The media include a rich medium, an unsupplemented minimal medium, and minimal media each supplemented with one vitamin. Of the yeast strains, one is a prototroph and seven are auxotrophs for a vitamin. After overnight incubation, the following results were observed (tan patches represent growth): D plate 1 (A) B DE F GH plate 5 plate 4 plate 6 Which plate contains an unsupplemented minimal medium? [Select] Which plate contains a rich medium? [Select] plate 2 Which strain is a prototroph? [Select] Strain E is an auxotroph for niacin. Which plate reveals this specific auxotrophy? [ Select] plate 3 plate 7 One strain is an auxotroph for both choline and pantothenic acid. Which one is this most likely to be? [Select]Three different strains of Escherichia coli, respectively A, B, and C have been inoculated in three different media, namely I, II, and III, and the growth of each bacterial strain has been followed by spectrophotometry (Figure 1). Strain A Strain B Strain C 4.5 45 Medium I 4 3.5 25 2.5 3 E 2.5 Minimal medium + Glucose 25 2.5 O 15 1 0.5 as 05 100 200 300 400 500 600 100 200 400 600 200 Time (minj Strain A Strain B StrainB 4.5 4s 4s Medium II 3.5 15 25 2 3 Minimal 25 2 medium + Glucose 1 + Arginine 0.5 as as 100 200 300 400 500 600 100 300 400 s00 600 100 300 400 600 Tme Time (min) Strain A me StreinB Streine 4.5 45 45 Medium II 4 4. 3.5 3.5 25 2 3 Minimal medium + Glucose 5 2.5 25 25 8 15 15 + Lactose 1 0.5 05 05 100 200 300 400 500 600 300 200 400 10 200 300 630 Time (min) Tine imin Time min Figure 1. Growth curves of the three E. coli strains A, B and C in three media: I, minimal medium containing glucose; II, minimal medium containing glucose and arginine; and III, minimal medium…You are given two strains of E. coli. The Hfr strain is arg+ ala+ glu+ pro+ leu+ T s; the F− strain is arg − ala− glu− pro−leu− T r. All the markers are nutritional except T, which determines sensitivity or resistance to phage T1. The order of entry is as given, with arg+ entering the recipient first and T slast. You find that the F− strain dies when exposed to penicillin (pens), but the Hfr strain does not (penr). How would you locate the locus for pen on the bacterial chromosome with respect to arg, ala, glu, pro, and leu? Formulate your answer in logical, well-explained steps, and draw explicit diagrams where possible.
- A high cell density culture of recombinant E. coli was carried out according to the following strategy:-Step 1: single batch with exponential growth until 98% conversion of the substrate, starting from V0= 4.0 L, S0=50 g/L/ X0= 1.0 g/LStep 2: batch fed with exponential flow (SF-800 g/L, μ= 0.1 h-1) until reaching X= 50.9 g/L;Step 3: batch fed with constant flow (F= 0.1 L/h) for 4 hours (induction phase with IPTG)Note: consider that the quasi-steady state is reached in both fed-batch stages.Extra data: YX/S = 0.4 gx/gs; μmax= 0.25 h-1; Ks== 1.0 g/L a) What was the cell concentration reached at the end of step 1?b) For step 3, considering that the substrate concentration in the feed was 1/4 of that used in step 2, what was the concentration of cells reached at the end of step 3?C) In terms of cell productivity, which of the three phases of cultivation was the most productive?A graduate student was assaying LD50 (lethal dose 50%) of two temperature-sensitive Francisella tularensis strains in HeLa cells (human cell line). Both strains can infect humans and cause fatal tularemia if untreated, but it is difficult to obtain LD50 values in human subjects. The data below shows LD50 (lethal dose 50%) values of the strains in human cell culture. Can you predict the more virulent strain of the two human pathogens? Francisella tularensis strain A: LD50 @ 20∘C= 100; LD50 @ 37∘C= 1000 Francisella tularensis strain B: LD50 @ 20∘C= 1000 LD50 @ 37∘C= 100 Group of answer choices It is not possible to determine the virulence of the two strains as human pathogens from the provided data Strain A and strain B are equally virulent as human pathogens, as they average out in virulence. Strain A is more virulent than strain A as a human pathogen. Strain B is more virulent than strain A as a human pathogen.Consider the five E. coli merodiploid strains listed here. Strain #1 I+P+O+Z-Y+/I+P+OcZ+Y+ Strain #2 I+P+O+Z+Y+/I-P+OcZ+Y- Strain #3 I+P+O+Z-Y+/I-P+OcZ+Y- Strain #4 I-P-O+Z+Y-/I+P+OcZ-Y+ Strain #5: ISP+O+Z+Y+/I-P+O+Z+Y- Which of these strains will be unable to express lactose permease under any conditions? Select all correct answers. A.)Strain #3 B.)Strain #4 C.)Strain #2 D.)Strain #5 E.)Strain #1
- A graduate student was assaying LD50 (lethal dose 50%) of two temperature-sensitive Francisella tularensis strains in HeLa cells (human cell line). Both strains can infect humans and cause fatal tularemia if untreated, but it is difficult to obtain LD50 values in human subjects. The data below shows LD50 (lethal dose 50%) values of the strains in human cell culture. Can you predict the more virulent strain of the two human pathogens? Francisella tularensis strain A: LD50 @ 20°C= 100; LD50 @ 37°C= 1000 Francisella tularensis strain B: LD50 @ 20°C= 1000 LD 50 @ 37°C= 100 O It is not possible to determine the virulence of the two strains as human pathogens from the provided data Strain A and strain B are equally virulent as human pathogens, as they average out in virulence. O Strain A is more virulent than strain A as a human pathogen. O Strain B is more virulent than strain A as a human pathogen.Consider the five E. coli merodiploid strains listed here. Strain #1 I+P+O+Z-Y+/I+P+OcZ+Y+ Strain #2 I+P+O+Z+Y+/I-P+OcZ+Y- Strain #3 I+P+O+Z-Y+/I-P+OcZ+Y- Strain #4 I-P-O+Z+Y-/I+P+OcZ-Y+ Strain #5: ISP+O+Z+Y+/I-P+O+Z+Y- Which of these strains will be inducible for expression of b-galactosidase? Select all correct answers. A.) Strain #2 B.) Strain #3 C.) Strain #1 D.) Strain #4 E.) None of these F. )Strain #5Consider the five E. coli merodiploid strains listed here. Strain #1 I+P+O+Z-Y+/I+P+OcZ+Y+ Strain #2 I+P+O+Z+Y+/I-P+OcZ+Y- Strain #3 I+P+O+Z-Y+/I-P+OcZ+Y- Strain #4 I-P-O+Z+Y-/I+P+OcZ-Y+ Strain #5: ISP+O+Z+Y+/I-P+O+Z+Y- Which of these strains will express b-galactosidase constitutively? Select all correct answers. A.)Strain #5 B.)Strain #1 C.)Strain #4 D.)Strain #2 E.)Strain #3
- Fifteen bacterial colonies growing on a complete medium (that means that they have all of nutrients that they need supplied in the dish, and they don't actually need to synthesize these compounds to survive) are transferred to minimal medium. Twelve of the colonies grow on minimal medium. Three colonies do not grow on minimal medium. But, if these three colonies are put on a plate that has minimal medium supplemented with the amino acid serine (min + Ser), they all What does this suggest about the three bacterial colonies (pick all that apply)? grow. They lack the ability to synthesize their own serine. O They are probably wild-type. They probably have a mutation that causes them to lack a certain protein. They probably have a mutation that causes them to be unable to perform translation. O They probably have a mutation that causes them to be unable to perform transcription. O They are able to synthesize everything that they need to grow except for serine.A phenylgalactoside-type compound, which yields galactose when enzymatically cleaved, was added as the sole carbon source to petri dishes containing E. coli culture medium. E. coli can utilize galactose for growth. Each of the following bacterial strains was plated on this medium and scored for growth after 1-2 days. Growth Genotype lacz+ lacI+ lacz+ lacIs lacz+ lacl- + oc lacZ+ lacI+ + a) What do you conclude about this compound with respect to its ability to be cleaved by ß- galactosidase? Why? b) What do you conclude about this compound with respect to its ability to inactivate the Lac repressor? Why? TIn the Avery, McLeod, McCarty Experiment where supernatant from heat killed, virulent S Strain pneumonia solutions were added to non-virulent R Strain pneumonia cell cultures and allowed to grow in liquid media (i.e., broth). In tubes where Protease was added to the supernatant prior to cell culture, what was the observed effect when plating and growing the S. pneumonia cells to solid media? Selected answer will be automatically saved. For keyboard navigation, press up/down arrow keys to select an answer. a b C d e All RNA was degraded and Transformation of the R Strain to S Strain occurred. All Protein was degraded and Transformation of the R Strain to S Strain occurred. All DNA was degraded and Transformation of the R Strain to S Strain occurred. All RNA was degraded and no Transformation occurred indicating RNA is the molecule of Transformation inheritance None of the above are true