Seek a detailed explanation of the secondary growth of E. coli

Updated on healthy 2024-06-08
6 answers
  1. Anonymous users2024-02-11

    When two metabolic substrates are injected into microorganisms for culture, if the enzyme system required for the metabolism of one substrate must be induced to be formed, the growth process of the microorganism is divided into two stages, and this phenomenon is called the secondary growth phenomenon, which is reported and named. [1]

    For example, with sucrose.

    and dextrin culture Bacillus subtilis.

    or with glucose.

    When E. coli is cultured with galactose, the enzyme required for the metabolism of another substrate is not induced when the substrate that is easily metabolized is used, and the induction of the enzyme begins to occur when the previous substrate is exhausted.

    Abbreviated as secondary growth. The microorganisms are grown in a culture medium containing two sugars, forming a biphasic growth curve.

    That is, the phenomenon of a degree of growth pause in the process of growth.

    This is due to the fact that the first of the two sugars is catalyzed by structural enzymes, and the second is catalyzed by inducible enzymes. Microorganisms are grown by using sugars catalyzed by structural enzymes and inhibiting the production of inducible enzymes using the second sugar. Only when the first sugar is used up can the second sugar induce the production of an inducible enzyme to utilize the second sugar.

    See Encyclopedia for details.

  2. Anonymous users2024-02-10

    When bacteria grow on a medium containing glucose and lactose, the bacteria first use glucose, and only after the glucose is used up does it start to use lactose. From the perspective of the growth curve, after an upward period, there is a pause period, at which time the curve is flat, and then there is a second upward period. This means that bacteria have to have an "adaptation process" before they can use lactose.

  3. Anonymous users2024-02-09

    Escherichia coli is commonly known as Escherichia coli, classified in the Enterobacteriaceae family, belonging to the genus Escherichia, with strong anabolic ability and good growth on ordinary medium containing inorganic salts, amine salts, and glucose.

    The optimum growth temperature is 37 and can still grow under 42-44 conditions, and the growth temperature range is 15-46.

    Growth on common nutrient agar exhibits 3 colony morphologies:

    1) Smooth type: The colony has a neat edge, shiny, moist, smooth, gray surface, and is easily dispersed in normal saline.

    2) Rough type: the colony is flat, dry, and the edges are not neat, and it is easy to self-coagulate in normal saline.

    3) Mucilage type: often strains containing capsule.

    This bacterium is facultatively anaerobic, grows well under aerobic conditions, the optimal growth pH is, the pH of the medium used, if the pH value is lower or higher, the growth is slow.

  4. Anonymous users2024-02-08

    Escherichia coli are bacteria and belong to the group of prokaryotes.

    The metabolic type is heterotrophic facultative anaerobic.

    Eosin Melin was added to the culture medium, and the colonies were dark purple, which could identify the presence of E. coli.

    The relationship between the human body and E. coli: in the case of non-pathogenic, it can be considered as a mutually beneficial symbiosis; In the case of pathogenesis, it can be considered parasitic.

  5. Anonymous users2024-02-07

    facultative anaerobic; Strong metabolic ability, good growth on ordinary medium containing inorganic salts, amine salts, glucose;

    The optimum growth temperature is 37 and can still grow under 42-44 conditions, and the growth temperature range is 15-46.

  6. Anonymous users2024-02-06

    a. The growth hormone produced by fermentation belongs to the secondary metabolite of Escherichia coli in peizhong, and a is wrong;

    b. Human growth hormone can be introduced into E. coli through genes and middle block engineering technology, and it can be stably inherited, b is correct;

    c. The content of purine and pyrimidine in the plasmid marker gene of E. coli is equal, and C is wrong;

    D. Growth hormone genes need RNA polymerase during transcription, while genes do not need helicase during transcription

    Therefore, b

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