1 mol of H2SO4 contains 6.022 x 10^23 molecules
The probability of obtaining a yellow-fruited plant from the cross is ¼ or 25%.
<h3>WHAT IS A HETEROZYGOUS CROSS</h3>
A heterozygous cross is a cross between two individuals that are heterozygous for two genes.
According to this question, white (W) fruit color is dominant over yellow (w). If two white-fruited individuals with genotypes that are unknown are crossed i.e. Ww × Ww.
The offsprings of this cross are as follows:
Therefore, the probability of obtaining a yellow-fruited plant from the cross is ¼ or 25%.
Learn more about heterozygous cross at: brainly.com/question/14109187
Answer:
The correct option is B) adaptation
Explanation:
The features and properties of an organism that make it favourable to live in an environment are termed as the adaptations of that organism. Over time, evolution occurs and organisms with better adaptations survive as they are better suited to live in a particular environment. The light-coloured moths were adapted to live on Earth when the colour of the trees had not gone dark. With time, as the colour of trees got darker, the moth had to change colour to be better adapted to live on Earth.
Answer:
Constitutive activation is the alteration of a protein or signaling pathway such that it is functional or engaged even in the absence of an upstream activating event. For example, RasD is constitutively active because it cannot bind GAP and therefore remains in the GTP-bound, active state even when cells are not stimulated by growth factor to activate a receptor tyrosine kinase.
Constitutively active Ras is cancer promoting because cells will proliferate in the absence of growth factors, and thus normal regulatory mechanisms for cell proliferation are bypassed.
(a) A mutation that resulted in Smad3 binding Smad4, entering the nucleus, and activating transcription independent of phosphorylation by the TGFβ receptor would render Smad3 constitutively active.
(b) A mutation that made MAPK active as a kinase and able to enter the nucleus without being phosphorylated by MEK would render MAPK constitutively active.
(c) A mutation that prevented NF-KB from binding to IK-B or that allowed NF-KB to enter the nucleus and regulate transcription even when bound to IK-B would render NF-KB constitutively active.
Answer:
pretty sure its D!
Explanation:
Because Camouflage isn't related to comparative anatomy. If this is incorrect, sorry!
