The four groups of pathogenic E.coli are enteropathogenic, enterotoxigenic, verocytotoxigenic and enteroinvasive groups. These groups can best be isolated and recoved through luria broth.
<h3>What is Escherichia coli?</h3>
The pathogenic E. coli or Escherichia coli serotypes are grouped on the basis of their mechanism of causing symptoms in humans. The six groups of pathogenic E.coli are enteropathogenic, enterotoxigenic, verocytotoxigenic, enteroinvasive, enteroaggregative and diffusely adherent E. coli.
Luria broth (LB) is one of the most commonly used growth medium for E. coli. It promotes fast growth of the organism and also provides good plasmid yields, making it an excellent choice for most laboratory applications, especially the small-scale plasmid preps.
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Complete question:
A sample of butterflies contained 50% yellow-winged individuals and 50% black-winged individuals. In this species, wing color is determined by a single gene with two alleles, and the allele for black is dominant. Which of the following statements about the allele frequencies in the sample would most likely be true? Do not assume that this sample was obtained from a population in genetic equilibrium.
a)The frequency of the yellow allele is greater than that of the black allele.
b)The allele frequency of yellow is 3 times the allele frequency of black.
c)The allele frequency of yellow is twice the allele frequency of black.
d)The allele frequencies of black and yellow are equal.
e) The allele frequency of black is greater than the allele frequency of yellow.
2.Now assume that the population of butterflies sampled in question above is in Hardy-Weinberg equilibrium. Also assume that the sample is random and large enough that the allele frequencies in the sample equal the allele frequencies in the population. What is the frequency of the allele for a) yellow wings in the population? Assuming Hardy-Weinberg equilibrium in question 1, what would the correct answer be and why?
Answer:
- a)The frequency of the yellow allele is greater than that of the black allele.
- f(b) = q = 0.71
Explanation:
Due to technical problems, you will find the complete explanation in the attached files
The answer would be D, Schwann cells, because they are a part of the myelin sheath (which covers the neurons). c:
These are small eukaryotes
1. B.
2. C.
3. A.
4. B.
5. C.
6. C.
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