<span>The correct answer for is C. experimentation</span>
Answer:
A. We can deduce that microorganisms such as Citrobacter have a classification from which it is possible to establish the most favorable environment for it, since being gram-negative, lactose positive, optional anaerobic, not endospore, it will prefer the environment as humid sites and that contain wastewater and soil as well as the intestine, it can therefore remain in some contaminated food and be transmitted with consumption
B. Given the aforementioned characteristics of Citrobacter, it does not infect all people and prefers those who have deficits in its defense mechanisms, thus it becomes a favorable site for colonization of newborns, who do not have adequate immunity, according to the case. Clinically, the neonate was able to acquire the infection in various ways, although the medical personnel who have manipulated it do not have positive tests, the microorganism can adhere to contaminated surfaces and it will be necessary to take samples of the mother's and neonate's feces, as well as the soil. and the water where the infected was, to subsequently carry out disinfection processes in the areas and avoid new infections.
Answer:
d.0.48
Explanation:
When a population is in Hardy Weinberg equilibrium the <u>genotypic </u>frequencies are:
freq (AA) = p²
freq (Aa) = 2pq
freq (aa) = q²
<em>p</em> is the frequency of the dominant <em>A</em> allele and <em>q</em> is the frequency of the recessive <em>a</em> allele.
In this population of 100 individuals, 84 martians have the dominant phenotype and 16 have the recessive phenotype.
Therefore:
q²=16/100
q² = 0.16
q=√0.16
q = 0.4
And p+q=1, so:
p = 1 - q
p = 1-0.4
p = 0.6
The frequency of heterozygotes is:
freq (Aa) = 2pq = 2 × 0.4 × 0.6
freq (Aa) = 0.48
The answer is B) Amoeba
hope this helps
