Which of the following cellular structures is characteristic of amoebas? a. microvilli b. flagella c. cilia d. pseudopodia

Answers

Answer 1
Answer:

Answer:

Answer is D. Pseudopodia.

Explanation:

Pseudopodia are described as a temporary extension of the cytoplasm. These are found in some certain unicellular protists like amoeba.

The pseudopodia is useful for movement and ingestion. They are useful in capturing or engulfing prey for a process known as phagocytosis.


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What is a well-known issue with cross-sectional methodology?

Answers

Answer:

This methodology enables to measure simultaneously both the result and exposure in the group under study

Explanation:

A cross-sectional method is a type of scientific design based on observational data. The cross-sectional studies involve the use of diverse groups that differ in the variable investigated but share other features (for example, individuals with the same educational background, ethnicity, economic status, etc). This class of studies is generally used in developmental psychology.

How many ATP and NADPH molecules are required in the C3 pathway to make one six-carbon sugar? If the synthesis of a molecule of ATP were to require four protons, would you expect that these relative requirements for ATP and NADPH could be met by noncyclic photophosphorylation in the absence of cyclic photophosphorylation?

Answers

Answer:

The correct answer is 18 ATP and 12 NADPH and noncyclic photophosphorylation can produce ATP molecule in the absence of cyclic photophosphorylation.

Explanation:

ATP and NADPH molecules are synthesized during light reaction of the photosynthesis which is utilized during the reactions of the dark phases. Dark reaction or Calvin cycle (C3 cycle) is the cyclic pathway of producing glucose triose phosphates (3C) from carbon dioxide and water. This reaction proceeds into 3 phases: carboxylation, reduction and regeneration.

First ATP and NADPH are utilized during the reduction step in the reduction of 3-phosphoglycerate to glucose-3-phosphate by the transfer of phosphate group from 6 ATP to 3-phosphoglycerate and 6 NADPH reduction as it donates an electron. Regeneration step also uses 3 ATP in conversion of G3P to RUBP molecules.  

A total of 9 ATP and 6 NADPH are utilized in producing 3C G3P molecule, So, to produce 6C sugar molecule 18 ATP AND 12 NADPH are used.

During chemiosmosis synthesis of ATP, 4 protons produce 1 molecule of ATP which could be able to generate 3 molecules of ATP for each pair of NADPH. so, noncyclic photophosphorylation or Z-scheme will be able to produce ATP in the absence of cyclic photophosphorylation.

Thus, 18 ATP and 12 NADPH and noncyclic photophosphorylation can produce ATP molecule in the absence of cyclic photophosphorylation are the correct answer.

The process of Meiosis results in what type of cells? (Choose all that apply)A. Sex cells
B. Somatic cells
C. Body cells
D. Gamtes

Answers

Answer: the answer is Gametes.

Answer:

sex cells , gamtes

Explanation:

A lens located directly above the stage. Many microscopes have more than one objective to provide a range ofviewing magnifications.
Cognitive lens
Orbital lens
Complex lens
Objective lens

Answers

Answer:

Objective lens

Explanation:

An experiment is designed to study the mechanism of sucrose uptake by plant cells. Cells are immersed in a sucrose solution, and the pH of the solution is monitored. Samples of the cells are taken at intervals and their sucrose concentration measured. The pH is observed to decrease until it reaches a steady, slightly acidic level, and then sucrose uptake begins. (a) Evaluate these results and propose a hypothesis to explain them. (b) Predict what would happen if an inhibitor of ATP regeneration by the cell were added to the beaker once the pH was at a steady level? Explain your thinking.

Answers

Answer:

a. Sucrose is present in the plant cells that provide them energy and helps in the metabolic process of the plants. The pH of the plant cells will increase and their cellular environment becomes basic in nature. The uptake of sucrose is pH specific and the acidic condition in the environment allows the uptake of sucrose. The decrease in pH concentration in the environment increases the pH inside the cells.

b. The inhibitor of ATP inhibits the production and functioning of the ATP molecule.  This effects the sucrose transport in the plant cells. As the sucrose movement requires the ATP and it is a active transport. The ATP inhibition decreases the sucrose uptake in the plant cells and the sucrose concentration decreases inside the plant cells.

Final answer:

Sucrose uptake in plant cells seems to require an acidic environment, brought about by the active transport of protons which requires ATP. An inhibitor of ATP regeneration would likely slow or stop this transport and, in turn, sucrose absorption.

Explanation:

The reported results suggest that the process of sucrose uptake in plant cells involves acidification of the surrounding medium prior to sucrose absorption. This can be explained by the proton-sucrose symport mechanism, in which protons (H+ ions) are actively pumped out of the cell in a process that requires ATP energy. When these protons combine with water (H2O) in the cell's environment, they form hydronium ions (H3O+), resulting in a lower pH or more acidic environment. Only after this acidic environment is established does sucrose uptake begin.

Based on this mechanism, introducing an inhibitor of ATP regeneration would be expected to decrease or halt this process, since ATP is required for the active transportation of protons. With less ATP, fewer protons will be pumped out, leading to a less acidic environment and, thus, lower sucrose uptake. This hypothesis is supported by how phosphofructokinase, a key enzyme in glycolysis (ATP production), is affected by low pH levels.

Learn more about Sucrose Uptake in Plant Cells here:

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Why can’t bears , tigers and lion live together

Answers

Because they would kill each other

Answer:

Bears, lions, and tigers all are predators that have similar hunting habits such as medium to large prey. Because of this, if they were living in the save environment, they would be competing for food and space. This would cause them to attack and kill off each other until one remained. That or the other two species would be driven out of the area.

Explanation: