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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That property is called h<span>ydrogen bonding. </span>
5 chromosomes. Meiosis results in 4 haploid cells, meaning they have "n" pairs of chromosomes. The diploid cell that the 10 chromosomes originated in has "2n" pairs. This leaves us with a simple equation to solve the problem, 2n=10, so n must equal 5 chromosomes.
C , because Each strand of DNA acts as a template for synthesis of a new, complementary strand. Replication produces two identical DNA double helices, each with one new and one old strand.
I think the correct answer from the choices listed above is option A. The process that is being described above is called succession. Ecological succession is the process of change in species structure in a community as time goes by. Hope this answers the question.
