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Genetic Recombination
Recombination is crucial to this process because it allows genes to be reassorted into diverse combinations. Genetic recombination is the process of combining genetic components from two different origins into a single unit. In prokaryotes, genetic recombination takes place by the unilateral transfer of deoxyribonucleic acid. It includes transduction, transformation, and conjugation. The genetic exchange occurring between homologous deoxyribonucleic acid sequences (DNA) from two different sources is termed general recombination. For this to happen, an identical sequence of the two recombining molecules is required. The process of genetic exchange which occurs in eukaryotes during sexual reproduction such as meiosis is an example of this type of genetic recombination.
Microbial Genetics
Genes are the functional units of heredity. They transfer characteristic information from parents to the offspring.
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- For the juice samples in exercise 3-4, you will be using the pour plate technique. In this exercise, you will add juice to a petri dish and pour agar over the sample and mix. Why is the media (potato dextrose agar, PDA) kept in a water bath until you are ready to use it, and why must you work fairly quickly? (One response answers both questions). The media is already solid and will form chunks The media has a dye added to it that must be kept at 45°C until ready to use O PDA is a media that allows growth of many types of fungi, and the heat from the water bath keeps the fungal spores from growing. The media is kept liquid in the water bath with heat, removing the tube from the heat source will cause the PDA to solidifyShown below is the growth of E. coli in nutrient agar plates after exposure to different doses of UV light. What conclusion/s can be generated from these data? Exposure time (s) 3 Low inoculum (105 bacteria/plate) High inoculum (107 bacteria/plate) 1Here is an image of a UV-irradiation (wavelength 254nm) experiment performed on a plate of E. coli after incubation. Which side of the plate do you think has a 'zero-minute' negative control? A or B or neither? Explain with proper reasoning. ICA Exoli Eoli ky
- As shown in this diagram, you perform a ten-fold serial dilution of a culture to determine the number of colony forming units (CFU) per mL it contains. You do a plate count with these growth results (no. of colonies for each dilution): 1:10 too many to count; 1:100 too many to count; 1:1,000 312; 1:10,000 38; 1:100,000 no growth. The number of CFU per mL in the original culture was: A. 38 B. 380,000 C. 38,000 D. None of the other four answers (Correct answer not given) E. 10,000Answer the following questions: 1. Why do we pass the smear in the open flame of the bunsen burner 3 times? What do we want to achieve by doing this? 2. Is the size of the bacterial smear on the slide very critical to the analysis? Why or why not? 3. Is the thickness of the bacterial smear on the slide very critical to the analysis? Why or why not? 4. Why do we pass the smear in a gentle stream of water after flooding it with a stain? What do we want to prevent by doing this? 5. Is the duration or time of stain application very critical in simple staining? Why or why not? NOTE: Please try to answer all of the questions asked, i promise to give you a good ratingsa. What is the total dilution of Tube #4? Express the answer in exponential format. b. You plated 1 mL of the Tube #4, After incubating, you counted 500 colonies on the plate. What is the concentration of Tube #0? include units. c. How could you change the experiment in part B to get a plate in the countable range? Be specific about any dilution factors and/or plated volumes you would change.
- As shown in this diagram, you perform a ten-fold serial dilution of a culture to determine the number of colony forming units (CFU) per mL it contains. You do a plate count with these growth results (no. of colonies for each dilution): 1:10 too many to count; 1:100 too many to count; 1:1,000 174; 1:10,000 23; 1:100,000 no growth. The number of CFU per mL in the original culture was: 1 ml 1 ml Original inoculum Dilutions 9 ml broth in each tube 1:10 1 ml 174,000 1:100 1 ml 1 ml 1 ml 1:1000 1 ml 1:10,000 1 ml None of the other four answers (Correct answer not given) 1,000 230,000 174 1 ml 1:100,000 1 mlHere is an image of a UV-irradiation (wavelength 254nm) experiment performed on a plate of E. coll after incubation. Which side of the plate do you think has a zero-minute' negative control? A or B or neither? Explain with proper reasoning.Using the date in the photo of the two syringes below, answer these six questions: QUESTIONS 1. Which syringe had the most floating disks? Why? 2. What is the gas being released by the leaf disks? 3. Which syringe is the control? 4. Which syringe is the experimental? 5. What is the experimental variable in this experimental design? 6. List at least two controls in this experimental design.
- Which of the the following results of the extracellular enzyme tests is FALSE? Question 3 options: A Liquid gelatin positive for gelatinase B Clear area around bacterial growth on a starch plate after iodine is added means amylase is present C Clear zone around bacterial growth on a milk plate means caseinase is present D Solid gelatin after is positive for gelatinase E No clear zone around bacterial growth on milk plate means absence of casienaseAfter incubating for 24-48 hours, you retrieve your 4 agar plates for the osmotic pressure exercise from the 37°C incubator. You observe the following results: • Sa = Staphylococcus aureus • Pv = Proteus vulgaris Sa For each organism, you should indicate in the text box below: 1. At what salt concentrations (osmotic pressures) you saw growth. 2. The classification term you would use in order to classify the organism based on osmotic pressure. Choose your term based upon where you observed growth; not optimal growth. Each organism has only one classification term associated in describing its growth pattern(s). You will need to use the osmotic pressure classification descriptions and terms outlined in your lab manual.volume of the quasi-steady-state culture was V0= 500 L, and the nutrient solution containing glucose was added at a constant flow rate of F = 50 L/h.Data: X0 (at the beginning of feeding) = 20 g/L, S0 = 300 g/L, max = 0.2 h-1, KS = 0.5 g/L and Y x/s= 0.3 g/g a) Determine the volume of the culture at t = 10hb) Determine the concentration of glucose at t = 10 hc) determine the concentration and total mass of cells at t = 10 hd) If product is associated with growth with α = 1.5 and P0 = 0.1 g/L, determine the concentration of product at t = 10h. (answer:P = 67,55 g/L)