25% is the approximate production efficiency of this animal
Explanation:
Given ,
A ladybug eats 200 J of plant material
100 J is eliminated as feces
75 J are used in cellular respiration,
Now,
The approximate production efficiency of this animal :
200 - (100 +75) = 25 %
Answer:
1. Part A: No
2. Part B: Yes
3: Part C : Yes
4: Part D : No
Explanation:
1) Part A: Facilitated diffusion of glucose into a muscle cell:
No; sodium ion co - transport is required for active transport of glucose but not for facilitated diffusion of glucose
2) Part B: Active transport of dietary phenylalanine across the intestinal mucosa:
Yes; co - transport of sodium ions drives the inward movement of amino acids and can only occur if sodium ions are actively pumped back out again.
3) Part C: Uptake of potassium ions by red blood cells:
Yes; uptake of potassium ions can occur only via a pump that couples the inward pumping of potassium ions to the outward pumping of sodium ions.
4) Part D: Active uptake of lactose by the bacteria in your intestine
No; active uptake of sugars and amino acids in bacteria is driven by a proton gradient.
It results in four genetically different cells
<span>Centrosomes are made of from arrangement of two barrel-shaped clusters of microtubules, called “centrioles,” and a complex of proteins that help additional microtubules to form.</span>
