Thickening of gill epithelia in rainbow trout, caused by chloride cell proliferation, could lead to an impairment of oxygen uptake under moderate to severe hypoxia (Thomas et al 1988; Bindon et al., 1994; Greco et al., 1995).
<h3>What results in an increase in AMS in interstitial lung disease?</h3>
The number of alveolar macrophages (AMs) can rise in interstitial lung disease. Precursor cells from the peripheral circulation may have been drawn in, and/or there may have been local lung growth, to create this.
<h3>What connection does sarcoidosis have between lymphocytes and proliferating cells?</h3>
Additionally, a strong association between the quantities of lymphocytes and proliferative cells in sarcoidosis and fibrosis was discovered in bronchoalveolar lavage (BAL). Eosinophil counts and proliferating cell counts were positively associated in fibrosis.
<h3>How do AMS patients and healthy controls differ in terms of propagating AMS?</h3>
With a substantial association between these two indices, there was a considerable increase in proliferating AMs in all patient groups when compared to healthy controls (4.2 versus 1.4% Feulgen, and 2.1 against 0.5% Ki67).
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Answer is B or the ball because other options don’t have any velocity and it makes the kinetic energy zero for them .the formula is k=1/2mv^2
Answer:
they form long chain-like molecules.
Explanation:
they combine using covalent bonds, and give off water molecules.
Hey There!!
The answer to this is: C) a factor that may affect an experiment or investigation. This is the best answer. A and B can be eliminated. D describes a controlled variable which is more specific than a scientific variable, thus C is the best answer. It is the best description of just a scientific variable. And, The second one is B: Robert Hooke was a famous scientist who experimented with springs (metal coils). He found that the farther you stretched the spring, the larger the force with which the spring pulled back on you.
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ItsNobody~ ☆
Answer:
1 out of 4
or
25%
Explanation:
First we draw a Punnett Square for this:
R r
R RR Rr
r Rr rr
So here we have the following genotypes possible for this cross:
RR = homozygous dominant
Rr = heterozygous dominant
rr = homozygous recessive
As you can see out of the four (4) possible outcomes, one (1) of them is rr. So the probability is 1 out of 4 chances or 25%.
