Any two effects of gravitational force are as follows:
1. The Earth's gravitational force accelerates objects when they fall.
2. It constantly pulls, and the objects constantly speed up.
Answer:
The genotype for each of the parents must be
parent 1 : Gg
parent 2 : Gg
Explanation:
Please note that a dominant trait is a trait that is expressed phenotypically in a heterozygous state, while a recessive trait is a trait that can only be expressed in a homozygous state.
Now, since gray face (G) for Oompa Loompas is dominant, and orange face (g) is recessive, for an offspring to be orange faced, it means that the genotype of the offspring must be 'gg'. Also, since both parent contribute an allele in the pair of alleles in the offspring, both parents must have the recessive (g) in their genotype. Moreover, we are told that both parents are gray-faced, meaning that their genotypes were 'Gg' and 'Gg'. To confirm, let me do the cross
G g
G GG Gg
g Gg gg
from the cross above, we find out that out of 4 offspring, 3 were gray face (GG, Gg ) while one was orange face (gg).
<span>Trophic level—90% of energy consumed at trophic
level is used by the consumer for survival and reproduction. The remaining 10%
is transferred to the next trophic level. So the answer is 10,000 x 0.9
or 9,000 calories will be generally available
to primary consumer</span>
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.
