Can You Score Well In This General Science Test?
Can you score well in this general science test? This question is perfect for helping you review all that you have learned in your classes and help you pass the final exams you are about to take. Do give it a shot and be sure to review tour notes once you are done. All the best and keep an eye out for others like it!
Questions and Answers
- 1.
Phosphoinositide-3-kinase forms inositol phospholipids that are phosphorylated in position 3. Normally, this phosphate group is removed by the lipid phosphatase PTEN. When a cell loses PTEN through somatic mutations, the most likely effect of this will be
- A.
Activation of the CREB protein
- B.
Enhanced activity of protein kinase C
- C.
Overactivity of MAP kinases
- D.
Elevated cAMP
- E.
Overactivity of protein kinase B (Akt)
Correct Answer
E. Overactivity of protein kinase B (Akt)Explanation
When a cell loses PTEN through somatic mutations, it will result in the overactivity of protein kinase B (Akt). PTEN is responsible for removing the phosphate group from inositol phospholipids, which helps regulate the activity of protein kinase B. Without PTEN, the phosphate group remains on the inositol phospholipids, leading to the overactivation of protein kinase B (Akt). This can have various effects on cell growth, survival, and metabolism, contributing to the development and progression of cancer.Rate this question:
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- 2.
Increased activity of the neurotransmitter dopamine at synapses in the mesolimbic dopamine system is known to cause euphoria. To get a good euphoriant effect, you should therefore design drugs that
- A.
Inhibit the sodium cotransporter for dopamine in the nerve terminal
- B.
Inhibit the transporter that pumps dopamine into synaptic vesicles
- C.
Inhibit tyrosine hydroxylase
- D.
Induce enhanced expression of catechol-O-methyl-transferase (COMT)
- E.
Block voltage-gated calcium channels in the nerve terminals
Correct Answer
A. Inhibit the sodium cotransporter for dopamine in the nerve terminalExplanation
Inhibiting the sodium cotransporter for dopamine in the nerve terminal would result in increased levels of dopamine in the synapses of the mesolimbic dopamine system. This increased activity of dopamine would lead to a euphoric effect.Rate this question:
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- 3.
The release of neurotransmitters from the nerve terminal depends on an increase in the cytoplasmic level of:
- A.
Sodium
- B.
Calcium
- C.
Magnesium
- D.
Cyclic AMP
- E.
IP3
Correct Answer
B. CalciumExplanation
The release of neurotransmitters from the nerve terminal is dependent on an increase in the cytoplasmic level of calcium. Calcium plays a crucial role in the process of neurotransmitter release by triggering the fusion of synaptic vesicles with the presynaptic membrane, leading to the release of neurotransmitters into the synaptic cleft. This influx of calcium ions into the cytoplasm is typically triggered by the depolarization of the presynaptic membrane, causing voltage-gated calcium channels to open and allowing calcium to enter the nerve terminal. Therefore, an increase in the cytoplasmic level of calcium is necessary for the release of neurotransmitters.Rate this question:
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- 4.
The synthesis of the neurotransmitter norepinephrine can be prevented by a drug that inhibits:
- A.
Choline-acetyl transferase
- B.
Tyrosinase
- C.
Monoamine oxidase
- D.
Catechol-O-methyl transferase
- E.
Tyrosine hydroxylase
Correct Answer
E. Tyrosine hydroxylaseExplanation
Tyrosine hydroxylase is an enzyme that is involved in the synthesis of norepinephrine, a neurotransmitter. Therefore, if a drug inhibits tyrosine hydroxylase, it would prevent the synthesis of norepinephrine. This is why the drug that inhibits tyrosine hydroxylase can prevent the synthesis of norepinephrine.Rate this question:
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- 5.
What is the difference between nicotinic and muscarinic acetylcholine receptors?
- A.
Nicotinic: has guanylate cyclase activity; muscarinic: ion channel
- B.
Nicotinic: ion channel; muscarinic: coupled to G-proteins
- C.
Nicotinic: coupled to G-proteins; muscarinic: has adenylate cyclase activity
- D.
Nicotinic: coupled to G-proteins; muscarinic: ion channel
- E.
Nicotinic: ion channel; muscarinic: has tyrosine-specific protein kinase activity
Correct Answer
B. Nicotinic: ion channel; muscarinic: coupled to G-proteinsExplanation
Nicotinic acetylcholine receptors are ion channels, meaning they allow the passage of ions across the cell membrane when activated by acetylcholine. On the other hand, muscarinic acetylcholine receptors are coupled to G-proteins. G-proteins are a type of protein that can transmit signals from the receptor to other molecules within the cell, resulting in a cellular response. Therefore, the difference between nicotinic and muscarinic acetylcholine receptors lies in their mode of action and signaling pathways.Rate this question:
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- 6.
Poor responsiveness to insulin is seen in many patients with type 2 diabetes. One possible way to increase the cellular response to insulin would be a drug that
- A.
Inhibits the Gq-protein, which is required for phospholipase C activation
- B.
Inhibits protein kinase B (Akt)
- C.
Inhibits the enzyme that hydrolyzes the phosphate group from position 3 of phosphorylated inositol lipids
- D.
Inhibits the phosphatidylinositol-3-kinase
- E.
Prevents the binding of calcium to calmodulin
Correct Answer
C. Inhibits the enzyme that hydrolyzes the pHospHate group from position 3 of pHospHorylated inositol lipidsExplanation
The enzyme that hydrolyzes the phosphate group from position 3 of phosphorylated inositol lipids plays a role in the insulin signaling pathway. By inhibiting this enzyme, the levels of phosphorylated inositol lipids would increase, leading to enhanced cellular response to insulin. This would be beneficial for patients with type 2 diabetes who have poor responsiveness to insulin.Rate this question:
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- 7.
You accidentally press your arm against a hot iron, and receive a second-degree burn on a patch of skin. Which of the following cell types is most likely to be killed by this burn?
- A.
Cells within the Meissner's corpuscle
- B.
Cells within the hair root matrix
- C.
Fat cells in the hypodermis
- D.
Keratinocytes of the stratum corneum
- E.
Smooth muscle cells of the arrector pilli muscle
Correct Answer
A. Cells within the Meissner's corpuscleExplanation
The Meissner's corpuscle is a sensory receptor located in the dermis of the skin that is responsible for detecting light touch and vibration. Second-degree burns can cause damage to the dermis, leading to the destruction of the Meissner's corpuscle cells. Therefore, the cells within the Meissner's corpuscle are most likely to be killed by this burn.Rate this question:
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- 8.
Barometric pressure at sea level is about 760 mm Hg. The world’s highest summit (Mt. Everest) is 29,035 ft. and the barometric pressure is 247 mm Hg. The inspired PO2 of a climber who was breathing 50% oxygen at the summit would be about:
- A.
65 mm Hg
- B.
40 mm Hg
- C.
100 mm Hg
- D.
150 mm Hg
Correct Answer
C. 100 mm HgExplanation
At sea level, the barometric pressure is 760 mm Hg. As altitude increases, the barometric pressure decreases. The barometric pressure at the summit of Mt. Everest is given as 247 mm Hg. The inspired PO2 (partial pressure of oxygen) can be calculated using the formula: PO2 = (barometric pressure - water vapor pressure) * fraction of inspired oxygen. Since the climber is breathing 50% oxygen, the fraction of inspired oxygen is 0.5. Assuming the water vapor pressure is negligible, the calculation would be: PO2 = (247 mm Hg - 0 mm Hg) * 0.5 = 123.5 mm Hg. Rounding it to the nearest whole number, the inspired PO2 would be about 100 mm Hg.Rate this question:
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- 9.
A patient is breathing air at sea level and has a respiratory exchange ratio of 0.8. The arterial blood values are: PO2 90 mm Hg PCO2 20 mm Hg pH 7.60 These values indicate that the:
- A.
Alveolar-arterial PO2 difference exceeds 20 mm Hg
- B.
Pulmonary vascular resistance is increased
- C.
Patient is hypoventilating
- D.
Patient is anemic.
Correct Answer
A. Alveolar-arterial PO2 difference exceeds 20 mm HgExplanation
The given arterial blood values indicate that the alveolar-arterial PO2 difference exceeds 20 mm Hg. This means that there is a significant difference between the partial pressure of oxygen in the alveoli and the arterial blood. This can be caused by factors such as ventilation-perfusion mismatch or diffusion impairment, which result in inadequate oxygenation of the blood.Rate this question:
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- 10.
A patient with severe arterial hypoxemia is given 100 percent O2 to breathe for 10 minutes, and arterial blood gases are then measured. An arterial PaO2 of 100 mm Hg following 10 minutes of breathing pure oxygen is due to:
- A.
Hypoventilation.
- B.
Ventilation/perfusion inequality
- C.
Diffusion abnormality
- D.
Normal value
- E.
Right-to-left shunting
Correct Answer
E. Right-to-left shuntingExplanation
The correct answer is right-to-left shunting. Right-to-left shunting refers to the mixing of deoxygenated blood with oxygenated blood in the systemic circulation, bypassing the lungs. This can occur due to certain heart defects or abnormalities, such as a ventricular septal defect or a patent foramen ovale. In this case, despite breathing 100 percent oxygen, the arterial PaO2 remains low because the oxygenated blood from the lungs is not effectively reaching the systemic circulation.Rate this question:
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- 11.
A patient exhales normally and then start to breathe into a spirometer containing 2-L of 2% helium. After several minutes, the helium concentration in the spirometer falls to 1%. The patient’s lung volume at FRC is approximately:
- A.
1L
- B.
2L
- C.
3L
- D.
4L
Correct Answer
B. 2LExplanation
The patient's lung volume at FRC is approximately 2L. This can be inferred from the fact that the helium concentration in the spirometer falls from 2% to 1% after several minutes of breathing into it. This decrease in helium concentration indicates that the patient's lungs have taken up some of the helium, resulting in a decrease in the volume of the spirometer. Since the starting volume of the spirometer is 2L and the helium concentration has decreased by half, it can be concluded that the patient's lung volume at FRC is approximately 2L.Rate this question:
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- 12.
Minute ventilation is controlled by numerous factors including blood gases, mechanical receptors, body temperature, and voluntary effort (input from cortex). Which of the following mechanisms produces the greatest minute ventilation?
- A.
Voluntary effort
- B.
Decreased PO2 plus an increased PCO2
- C.
Hypotension
- D.
Hyperthermia
- E.
Exercise
Correct Answer
A. Voluntary effortExplanation
Voluntary effort produces the greatest minute ventilation because it is under conscious control. When we voluntarily increase our breathing rate and depth, it leads to a higher minute ventilation. This allows for a greater exchange of gases in the lungs, ensuring an adequate supply of oxygen and removal of carbon dioxide. The other factors mentioned, such as decreased PO2 plus increased PCO2, hypotension, hyperthermia, and exercise, can also affect minute ventilation to some extent, but they are not directly controlled by conscious effort like voluntary breathing.Rate this question:
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- 13.
A patient is suffering from obstructive lung disease. His chest wall- and respiratory system compliances are measured as: Chest wall compliance (CCW) = 300 ml / cm H2O Respiratory system compliance (CRW) = 100 ml / cm H2O What is his Lung compliance (CL) in ml / cm H2O?
- A.
50
- B.
100
- C.
150
- D.
200
- E.
300
- F.
400
Correct Answer
C. 150Explanation
1/Crw = 1/Cl + 1/Ccw
1/Crw = 1Rate this question:
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- 14.
A gas mixture of several dry gases has a total pressure of 200 mm Hg and a total gas amount of 10 Mol. What is the partial pressure of one gas that has an amount 2 Mol?
- A.
20 mm Hg
- B.
30 mm Hg
- C.
40 mm Hg
- D.
50 mm Hg
- E.
60 mm Hg
Correct Answer
C. 40 mm HgExplanation
The partial pressure of a gas can be calculated using the formula: Partial Pressure = (Amount of Gas / Total Amount of Gas) x Total Pressure. In this case, the amount of the gas in question is 2 Mol and the total amount of gas is 10 Mol. The total pressure is given as 200 mm Hg. Plugging these values into the formula, we get: Partial Pressure = (2 Mol / 10 Mol) x 200 mm Hg = 40 mm Hg. Therefore, the partial pressure of the gas is 40 mm Hg.Rate this question:
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- 15.
A 45 year old male had a history of smoking 20 cigarettes a day for the past 2 years. He now complains of shortness of breath and coughing. Which of the following techniques would best determine that he has emphysema?
- A.
Sputum culture
- B.
Pulmonary function tests
- C.
Calculating pack years
- D.
Chest x ray
- E.
Bronchoscopy
Correct Answer
D. Chest x rayExplanation
A chest x-ray would be the best technique to determine if the 45-year-old male has emphysema. Emphysema is a type of chronic obstructive pulmonary disease (COPD) that is characterized by the destruction of the air sacs in the lungs. A chest x-ray can show signs of emphysema such as hyperinflation of the lungs, flattened diaphragm, and bullae (large air spaces) in the lung tissue. This imaging technique can help confirm the diagnosis of emphysema and assess the severity of the condition.Rate this question:
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- 16.
A 52 year old man has been a smoker for 30 years. He smoked 20 cigarettes a day for 20 years and 40/day for 10 years. How many pack years has he smoked?
- A.
65
- B.
40
- C.
20
- D.
30
- E.
10
- F.
52
Correct Answer
B. 40Explanation
The answer is 40 because pack years is a measure of smoking intensity calculated by multiplying the number of packs of cigarettes smoked per day by the number of years smoked. In this case, the man smoked 20 cigarettes a day for 20 years, which is equivalent to 1 pack per day for 20 years (20 pack years), and then he smoked 40 cigarettes a day for 10 years, which is equivalent to 2 packs per day for 10 years (20 pack years). Adding these two numbers together, he has smoked a total of 40 pack years.Rate this question:
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- 17.
The AMSA group of students are studying the vital capacity of individuals who work in a local quarry. Several persons were identified as having abnormal respiratory function. This screening exercise would be described as which kind of prevention?
- A.
Primary prevention
- B.
Secondary prevention
- C.
Tertiary prevention
- D.
Quaternary prevention
- E.
Adverse health prevention
Correct Answer
B. Secondary preventionExplanation
This screening exercise would be described as secondary prevention because it aims to identify and intervene early in individuals who already have abnormal respiratory function. Secondary prevention focuses on detecting and treating diseases or conditions in their early stages to prevent further progression or complications. In this case, the screening is being done to identify individuals with abnormal respiratory function and potentially provide them with appropriate interventions or treatments to prevent further respiratory issues.Rate this question:
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Nov 16, 2023Quiz Edited by
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Nov 11, 2011Quiz Created by
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