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.
60 years is three half life for the substance, so 40*1/2^3=5
Answer:
acetyl-CoA concentration increases
acetone concentration increases
oxaloacetate concentration decreases
Explanation:
Mark me brainly please
Answer:
Answer:
The correct answer would be 0.589
The Hardy-Weinberg equation is p + q = 1
In addition, the frequencies of p and q in a Hardy-Weinberg population remain the same generations over generations.
The value of q⁸ is given as 0.169
So, the value of q = ⇒ 0.411
Thus, the value of p = 1 - 0.411 ⇒ 0.589.
So, the expected value of p would be 0.589
The answer is c gametophyte
