Answer:
B, starches
Explanation:
I’m not positive, but I think I remember this from 6th grade :)
Answer:starch
Explanation combines the glucose molecules a larger molecule , which is starch
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Reproduction in amoebas involves endocytosis, while reproduction in red algae involves spore formation.
The amoebas have to use endocytosis for photosynthesis while the red algae use pinocytosis during photosynthesis.
Endocytosis is the only way amoebas can bring food into their cells, while red algae can produce food by photosynthesis.
The correct answer is -- Endocytosis is the only way amoebas can bring food into their cells, while red algae can produce food by photosynthesis.
Amoeba is an unicellular organism. It is an heterotroph follows holozoic nutrition. It engulf its food with the help of pesudopodia. engulfing of solid food is called endocytosis, whereas red algae such as Polysiphonia is an autotroph which synthesis its own food by the process of photosynthesis. thus, in red algae endocytosis does not occur.
B) decomposer.
C) predator.
D) producer.
Answer:
the answer is b
Explanation:
b) phylogenetic tree II
c) phylogenetic tree III
d) phylogenetic tree IV
Answer:
Your question is incomplete, but I attached the exercise with the possible phylogenetic trees and all the information that your question missed. I hope it helps you.
The most parsimoniuous phylogenetic tree is c) phylogenetic tree III
Explanation:
The most parsimonious tree is the one that takes fewer base-change events, it's the simpler one. In this case it could be tree III or tree IV. But, if we look at the genetic evidence, group 2b and 3 are nearer than 2b and 2a; group 4 and 5 are also closely related. So it is possible that the group 2b, as well as groups 3,4 and 5, diverged from group 2a. So the answer is tree III.
The most parsimonious phylogenetic tree is the one that requires the least changes in the A, C, G, T genetic sequence coding for eye growth, thereby presenting the simplest explanation of evolutionary relationships among the alien species.
The question is asking which of the four possible phylogenetic trees represents the most parsimonious, or simplest, explanation of the shared genetic sequence for eye growth among the alien species. The most parsimonious tree would be one that requires the least amount of evolutionary change in terms of the genetic sequence. This would be determined by comparing each tree and the associated genetic sequences in the way they separate the six groups of aliens and particularly how they divide the single-tail three-eye group. This should provide the simplest explanation of evolutionary relationships based on the alien's genetic code of A, C, G and T.
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