100,000 years ago is your answer
Answer:
The order must be K2→K1, since the permanently active K1 allele (K1a) is able to propagate the signal onward even when its upstream activator K2 is inactive (K2i). The reverse order would have resulted in a failure to signal (K1a→K2i), since the permanently active K1a kinase would be attempting to activate a dead K2i kinase.
Explanation:
- You characterize a double mutant cell that contains K2 with type I mutation and K1 with type II
mutation.
- You observe that the response is seen even when no extracellular signal is provided.
- In the normal pathway, i f K1 activat es K2, we expect t his combinat ion of two m utants to show no response with or without ext racell ular signal. This is because no matt er how active K1 i s, it would be unable to act ivate a mutant K2 that i s an activit y defi cient. If we reverse the order, K2 activating K1, the above observati on is valid. Therefore, in the normal signaling pathway, K2 activates K1.
The choices can be found elsewhere and as follows:
A. mature leaves
B. shoot apical meristem
C. cell elongation zone
<span>D. axillary buds
</span>
I think the correct answers are option B and D. It would be at the shoot apical meristem and the axillary buds that <span> a vascular plant would you expect to find totipotent cells. Hope this answers the question.</span>
Answer:
The gynoecium of the flower(s) forms all or part of the fruit, which occurs from the maturation of one or more blooms. One or more ovules contain the egg cell of the megagametophyte, which is found inside the ovary/ovaries. These ovules will form seeds after double fertilisation.
Explanation:
