Answer:
b to a
Explanation:
since cells in membranes move from high to low concentration (diffusion). b has more molecules, so it will go into a.
Answer:
male lions have females that they can mate with to relive all that tension in their balls that has been building up
Explanation:
#ballsdeepnosleep
Answer:
d. all endothelial cells would be venous
Explanation:
The Notch signaling pathway is a cell signaling system which consist basically of 4 NOTCH genes ( NOTCH1, NOTCH2, NOTCH3, and NOTCH4) each of them having the information to build a particular cell membrane receptor protein.
It has been demonstrated in zebrafish embryos that Notch signaling are highly determinant factors during vascular development, and a key function of these genes was specifically to regulate the differentiation of arterial fate in endothelial cells.
Notch signaling-deficient embryos showed a loss of expression of arterial markers such as ephrinB2 from arterial vessels with an accompanying expansion of venous markers into normally arterial domains. In contrast, embryos in which Notch signaling had been ectopically activated, meaning an activation out of "proper" place (where not expected, e.g. venous endothelial), exhibited the contrary phenotype, this is a suppression of vein-specific markers with ectopic expression of arterial markers in venous vessels.
To summarize, when Notch is supressed (in this case inhibited with a specific Notch inhibitor) arterial fate of endothelium is not followed, and venous fate is stablished. Thus, endothelial cells (those expected to form venous tissues but also arterial ones) would specified as venous.
This is a very confusing question since it sounds more like a statement. sorry really wish I could help :(
Answer:
Replicated chromosomes at metaphase I = 66
Sister chromatids at metaphase I = 66 x 2 = 132
Sister chromatids at prophase II = 66
Chromosomes in each sperm cells = 33
Explanation:
Metaphase I of meiosis I would have 66 replicated chromosomes in the testicular cells of the bird. Each of the replicated chromosomes would have two sister chromatids. So, a total of 66 replicated chromosomes would have 66 x 2 = 132 sister chromatids.
Due to segregation of homologous chromosomes towards opposite poles in anaphase I, each daughter cell formed by the end of meiosis I would have 33 replicated chromosomes. So, each of the daughter cells would have a total 33 x 2 = 66 sister chromatids at prophase II.
Since meiosis II maintains the chromosome number, each sperm cell formed by the end of meiosis II would have 33 chromosomes.
