Which of the following results in osteoporosis?
An increase in osteocyte activity while osteoclast activity continues at expected levels.
decline in osteoclast activity while osteoporosis activity reduces
An increase in osteocyte activity while osteoblast activity reduces.
A decline in osteoblast activity while osteoclast activity continues at expected levels.
Correct Answer : D
A decline in osteoblast activity while osteoclast activity continues at expected levels results in osteoporosis.
Osteoporosis is caused by an imbalance between the functioning of osteoclast and osteoblast cells.
Osteoblasts are responsible for forming new bone, while osteoclasts break down old bone.
If osteoblast activity declines while osteoclast activity continues at expected levels, this means that more bone is being broken down than is being formed, leading to a loss of bone density and an increased risk of osteoporosis.

Choice A is incorrect because an increase in osteocyte activity would not result in osteoporosis.
Osteocytes are mature bone cells that maintain the mineral concentration of the bone matrix.
Choice B is incorrect because a decline in osteoclast activity would not result in osteoporosis.
Osteoclasts break down old bone, so a decline in their activity would mean that less bone is being broken down.
Choice C is incorrect because an increase in osteocyte activity would not result in osteoporosis.
As mentioned earlier, osteocytes are mature bone cells that maintain the mineral concentration of the bone matrix.
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Related Questions
Correct Answer is C
Explanation
Plasma B cells.
Antibodies are produced by specialized white blood cells called B lymphocytes (or B cells).
When an antigen binds to the B-cell surface, it stimulates the B cell to divide and mature into a group of identical cells called a clone.
The mature B cells, called plasma cells, secrete millions of antibodies into the bloodstream and lymphatic system.

Choice A, Natural killer cells, is not the correct answer because natural killer cells are a type of white blood cell that play a major role in the host-rejection of both tumors and virally infected cells.
Choice B, Cytotoxic T-cells, is not the correct answer because cytotoxic T-cells are a type of white blood cell that kills cancer cells, cells that are infected (particularly with viruses), or cells that are damaged in other ways.
Choice D, Helper T-cells, is not the correct answer because helper T-cells are a type of white blood cell that play an important role in the immune system by helping other white blood cells fight infections.
Correct Answer is D
Explanation
The pleura is a double-layered serous membrane that covers each lung and lines the thoracic cage
The pleura is a vital part of the respiratory tract.
Its role is to cushion the lung and reduce any friction that may develop between the lung, rib cage, and chest cavity.
Each pleura (there are two) consists of a two-layered membrane that covers each lung.
The layers are separated by a small amount of viscous (thick) lubricant known as pleural fluid.
The pleura is comprised of two distinct layers: the visceral pleura and the parietal pleura.
The visceral pleura is the thin, slippery membrane that covers the surface of the lungs and dips into the areas separating the different lobes of the lungs (called the hilum).
Correct Answer is C
Explanation
Other scientists can validate or disprove the findings.
It is important for new scientific findings to be published so that other scientists can review the research and either validate or disprove the findings.
This process of peer-review helps to ensure the accuracy and reliability of scientific research.
Choice A.
Scientists will get paid if their findings are published is not correct because while some scientists may receive funding or grants for their research, the primary goal of publishing scientific findings is not for financial gain.
Choice B.
Publishing findings will help scientists become more biased is not correct because the goal of publishing scientific findings is to share information and promote transparency, not to promote bias.
Choice D.
This prevents other scientists from performing similar tests is not correct because publishing scientific findings allows other scientists to build upon the research and perform further tests to validate or disprove the findings.
Correct Answer is D
Explanation
Sodium bicarbonate neutralizes the acidity of chyme.
The pancreas secretes large amounts of sodium bicarbonate, which protects the duodenum by neutralizing the acid that comes from the stomach.
This compound helps neutralize stomach acid generated during the digestive process.

Choice A is incorrect because sodium bicarbonate is not a protease that digests carbohydrates.
Proteases are enzymes that break down proteins, while sodium bicarbonate is a chemical compound that helps neutralize stomach acid.
Choice B is incorrect because sodium bicarbonate does not stimulate the pyloric sphincter.
The pyloric sphincter is a ring of smooth muscle that separates the stomach from the duodenum and regulates the passage of partially digested food (chyme) into the small intestine.
Choice C is incorrect because sodium bicarbonate does not inhibit peristalsis.
Peristalsis is a series of wave-like muscle contractions that move food through the digestive tract.
Correct Answer is D
Explanation
The cytoskeleton of a cell is comprised of protein fibers that provide structural support and help maintain the shape of the cell.
These protein fibers include microfilaments, intermediate filaments, and microtubules.
Choice A.
Carbohydrates is not the correct answer because carbohydrates are a type of macromolecule that provides energy to cells and are not a component of the cytoskeleton.
Choice B.
Nucleic acids is not the correct answer because nucleic acids are macromolecules that store and transmit genetic information and are not a component of the cytoskeleton.
Choice C.
Lipids is not the correct answer because lipids are a type of macromolecule that makes up cell membranes and are not a component of the cytoskeleton.
Correct Answer is C
Explanation
Control.
A control group is a group in an experiment that does not receive the treatment or manipulation being tested and is used as a benchmark to measure how the other tested subjects do.
The control group is used to minimize the effects of all variables except the independent variable.
This allows researchers to determine if changes in the dependent variable are due to the manipulation of the independent variable or if they are due to some other factor.
Choice A.
Responding is not the correct answer because it refers to the dependent variable, which is the variable that is being measured in an experiment.
Choice B.
Manipulated is not the correct answer because it refers to the independent variable, which is the variable that is being manipulated in an experiment.
Choice D.
Variable is not the correct answer because it refers to any factor that can change in an experiment and can include both independent and dependent variables.
Correct Answer is B
Explanation
In DNA, the nitrogenous bases adenine (A) and thymine (T) pair together, while cytosine (C) and guanine (G) pair together.
Therefore, the complementary strand of the given DNA sequence 3' TCGATCGCA 5' would have the complementary nitrogenous bases as:
5’ AGCTAGCGT 3’
NOTE: The 5’ to 3’ direction of the complementary strand is opposite to that of the given strand.
Choice A.
3’ AGCTAGCGT 5’ is not correct because it is not complementary to the given strand.
Choice C.
5’ UCGAUCGCA 3’ is not correct because it contains uracil (U), which is a nitrogenous base found in RNA, not DNA.
Choice D.
3’ TCGUTCGCU 3’ is not correct because it also contains uracil (U), which is a nitrogenous base found in RNA, not DNA.
Correct Answer is B
Explanation
Electrophoresis is the most useful laboratory method for separating genomic DNA fragments by size.
Electrophoresis is a technique that uses an electric field to separate charged molecules, such as DNA fragments, based on their size and charge.
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Choice A is not correct because titration is a laboratory method used to determine the concentration of a solution.
Choice C is not correct because filtration is a laboratory method used to separate solids from liquids.
Choice D is not correct because spectrophotometry is a laboratory method used to measure the absorbance of light by a solution.
Correct Answer is B
Explanation
Calcium ions play a crucial role in initiating muscle contraction.
When a muscle cell is stimulated to contract by an action potential, calcium channels open in the sarcoplasmic membrane and release calcium into the sarcoplasm.
Some of this calcium attaches to troponin, which causes it to change shape.
This shape change exposes binding sites for myosin on the actin filaments.
Myosin’s binding to actin causes crossbridge formation, and contraction of the muscle begins.

The other ions mentioned in the question do not have this specific role in muscle contraction.
Potassium ions are important for maintaining the resting membrane potential of cells, but they do not bind to the troponin complex.
Phosphorus ions are important for energy metabolism, but they do not bind to the troponin complex.
Sodium ions are important for generating action potentials, but they do not bind to the troponin complex.
Correct Answer is D
Explanation
Proteins.
Proteins are made up of amino acids which are organic molecules that contain both an amine functional group (–NH2) and a carboxylic acid functional group (– COOH).
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Choice A, Lipids, is not the correct answer because lipids are a group of naturally occurring molecules that include fats, waxes, sterols, fat-soluble vitamins, monoglycerides, diglycerides, triglycerides, phospholipids, and others.
They do not contain both an amine and carboxyl group.
Choice B, Chitin, is not the correct answer because chitin is a long-chain polymer of N-acetylglucosamine, a derivative of glucose.
It does not contain both an amine and carboxyl group.
Choice C, Cellulose, is not the correct answer because cellulose is an organic compound with the formula (C6H10O5)n, a polysaccharide consisting of a linear chain of several hundred to many thousands of β(1→4) linked D-glucose units.
It does not contain both an amine and carboxyl group.
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