The bilaminar embryonic disc: Comes from inner layer of cells rather than outer layer. Develop into three germ layer during third week development. Contributes to formation of structure such as amnion & yolk sac.
Embryonic refers to the early stage of development in an organism, particularly in reference to the period when an embryo is formed. It encompasses the initial stages of growth and differentiation from a fertilized egg to the formation of the basic structures and organs of the developing organism. During embryonic development, cells undergo rapid division and specialization, forming the foundation for the subsequent stages of growth and maturation. This critical phase is characterized by high vulnerability to external influences and is essential for the establishment of proper structure and function in the developing organism.
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A decrease in the plasma volume occurring with an increased concentration of cells and larger molecules such as cholesterol is referred to as
The decrease in plasma volume occurring with an increased concentration of cells and larger molecules such as cholesterol is referred to as hemoconcentration.
Hemoconcentration is a condition where the proportion of red blood cells and other solid components in the blood becomes higher compared to the fluid component (plasma).
This can happen due to various reasons such as dehydration, excessive sweating, or certain medical conditions. In hemoconcentration, the volume of plasma decreases, causing an increase in the concentration of cells and larger molecules. This can lead to changes in blood viscosity and affect blood flow and overall circulation in the body.
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hypophysecotomy is done on an expiremental animal. Ten days later, the animal is sacrificied, and the adrenal glands are examined and found to be atrophic. Which of the following mitochondrial proteins is most likely decreased in the adrenal glands as a result of this process? Apaf-1: apoptotic protease activating factor 1 (could be decreased since its binded up in cytosol B. Bax PDA Φ Fr Suviuer A C. Bcl-2 Be clever ke tanti upaps 5 D. Caspase 8 De E. Caspase 9 EfFas idimmono infechur
The most likely mitochondrial protein to be decreased in the adrenal glands after hypophysectomy is Bcl-2.
Bcl-2 (B-cell lymphoma 2) is a protein involved in regulating apoptosis (cell death) and maintaining mitochondrial integrity.
It functions as an anti-apoptotic protein by preventing the release of cytochrome c from mitochondria, thus inhibiting the activation of caspases and subsequent apoptosis.
Hypophysectomy refers to the removal or destruction of the pituitary gland, which can lead to hormonal imbalances.
The pituitary gland secretes adrenocorticotropic hormone (ACTH), which stimulates the adrenal glands to produce cortisol. When the pituitary gland is removed or damaged, the production of ACTH decreases.
Cortisol is known to upregulate the expression of Bcl-2 in the adrenal glands. Therefore, after hypophysectomy, with decreased ACTH stimulation and subsequently reduced cortisol production, the levels of Bcl-2 are likely to decrease in the adrenal glands.
This decrease in Bcl-2 levels can contribute to mitochondrial dysfunction and result in atrophy of the adrenal glands. Therefore, Bcl-2 is the most likely mitochondrial protein to be decreased in this scenario.
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DNA methylation array analysis identifies breast cancer associated RPTOR, MGRN1 and RAPSN hypomethylation in peripheral blood DNA
The statement "DNA methylation array analysis identifies breast cancer-associated hypomethylation of RPTOR, MGRN1, and RAPSN in peripheral blood DNA" is false.
This finding suggests that these specific genes may be differentially methylated in individuals with breast cancer compared to healthy individuals. Hypomethylation refers to a decrease in DNA methylation, which can lead to altered gene expression patterns and potentially contribute to the development or progression of cancer.
The identification of these hypomethylated genes in peripheral blood DNA provides valuable insights into potential biomarkers for breast cancer detection or monitoring.
Further research is needed to fully understand the functional implications of these methylation changes and their role in breast cancer pathogenesis.
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Complete question:
DNA methylation array analysis identifies breast cancer-associated hypomethylation of RPTOR, MGRN1 and RAPSN in peripheral blood DNA. T/F
A patient in an emergency room complained to the doctor that she was not able to feel heat on her hand. The doctor knew that there were two nerve cells between the heat receptors in the hand and the heat-sensitive receptors in the brain. The arrangement of these receptors is shown in the following diagram. The doctor thought that the trouble might be in the synapse between A and B and made the following two hypotheses: A) No neurotransmitter is being released from nerve cell A. B) Nerve cell B does not have receptors for the neurotransmitter that is released from A. To test these hypotheses, the doctor designed an experiment to apply a neurotransmitter to the cell body of nerve cell A and observe any activity from nerve cell B. Evaluate whether or not this experiment would enable the doctor to support one hypothesis and reject the other. During World War I, physicians noted a phenomenon called "phantom pains'. Soldiers with amputated limbs complained of pain or itching in the missing limb. Using your knowledge of the nervous system explain why you think this phenomenon exists
In order to test these hypotheses, the doctor designed an experiment to apply a neurotransmitter to the cell body of nerve cell A and observe any activity from nerve cell B.
If the application of the neurotransmitter to the cell body of nerve cell A leads to nerve cell B getting active, this would imply that there are receptors for the neurotransmitter on nerve cell B. This would support hypothesis B and reject hypothesis A.However, if the application of the neurotransmitter to the cell body of nerve cell A does not lead to nerve cell B getting active, this would imply that there are no receptors for the neurotransmitter on nerve cell B. This would support hypothesis A and reject hypothesis B.A phenomenon called "phantom pains" occurs when a patient complains of pain or itching in a missing limb after amputation.
Phantom limb pain may be related to the brain's neuroplasticity, which refers to its ability to reorganize itself by forming new neural connections throughout life. When the sensory nerves in an amputated limb are severed, the area of the brain that previously received input from those nerves loses that input, and new connections may form between that area of the brain and the surrounding nerves that are still intact. When these new connections form, the brain may misinterpret the signals from the intact nerves as signals coming from the amputated limb, resulting in phantom limb pain.
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Which replicative step(s) do animal viruses and bacteriophages have in common? to be marked correct, you'll need to select all applicable statements, as there may be more than one correct answer.
Animal viruses and bacteriophages have replicative steps in common are adsorption, penetration, uncoating and synthesis. Replication of viruses and bacteriophages refers to the process by which viruses and bacteria multiply and produce new particles within host cell.
Adsorption is the initial stage of viral replication involves the binding of viruses to host cells. It occurs in a similar manner for both animal viruses and bacteriophages. Then comes penetration, once attached to the host cell, the virus can proceed to enter it. This represents the second stage of viral replication and is comparable for both animal viruses and bacteriophages. Next is uncoating, following entry into the host cell, the virus sheds its protective coat, releasing its genetic material. This step is common to both animal viruses and bacteriophages. During replication stage, the viral genetic material is utilized by the host cell to produce new viruses. This process is similar for both animal viruses and bacteriophages. Therefore, animal viruses and bacteriophages have adsorption, penetration, uncoating and synthesis in common.
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1. Give an example of a muscle whose name describes its: Action ONLY Shape ONLY Location ONLY Relative Size ONLY Attachment Points Number of Heads \& Location Location \& Relative Size Shape \& Location Action \& Relative Size Action, Location \& Relative Size 2. Define the following terms:
Examples of muscles: Flexor carpi radialis (Action: Flexing the wrist), Deltoid (Shape: Triangular), Brachioradialis (Location: Forearm), Gluteus maximus (Relative Size: Largest muscle in the buttocks).
Muscle names often provide valuable insights into their characteristics. The Flexor carpi radialis is named for its action of flexing the wrist, reflecting its role in hand and finger movements. The Deltoid muscle derives its name from its distinctive triangular shape, resembling the Greek letter delta. The Brachioradialis is aptly named, indicating its location in the forearm, where it connects the brachium (upper arm) to the radius bone.
The Gluteus maximus, as its name suggests, is the largest muscle in the buttocks, contributing to hip extension and thigh movement. These examples illustrate how muscle names convey information about their specific action, shape, location, and relative size, aiding in their identification and understanding within the human body.
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QUESTION 5 Which transport system can move an ion across the plasma membrane against its concentration gradient without using ATP? Oa. Primary active transport Ob. Secondary active transport Oc. Simple diffusion Od. Facilitated diffusion Oe. Facilitated diffusion via a carrier protein.
The transport system that can move an ion across the plasma membrane against its concentration gradient without using ATP is secondary active transport.
The transport system that can move an ion across the plasma membrane against its concentration gradient without using ATP is secondary active transport.
Primary active transport, such as the sodium-potassium pump, requires the direct expenditure of ATP to move ions against their concentration gradients. Simple diffusion and facilitated diffusion, including facilitated diffusion via a carrier protein, do not require ATP but can only move ions along their concentration gradient.
In secondary active transport, the movement of an ion against its concentration gradient is coupled with the movement of another molecule or ion down its concentration gradient. This coupling utilizes the energy stored in the electrochemical gradient of the second molecule to transport the ion against its concentration gradient. As a result, the transport of the ion is indirectly powered by the ATP-driven transport of the second molecule.
Therefore, secondary active transport is the transport system that can move an ion across the plasma membrane against its concentration gradient without using ATP.
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3. The cornea is normally translucent. However, here it is not. Why is this true?
4. What is unique about the cornea?
5. How is the cow eye similar to the human eye?
6. Name 4 muscles that control eye movements and describe how each moves the eye.
7. Which cranial nerve(s) control the contraction of these muscles?
8. What is the function of the fat capsule that surrounds the eye?
9. Describe the sclera.
10. Describe the appearance of the optic nerve as it exits the posterior of the cow eye.
The cornea is not translucent due to a condition called corneal opacity, which can be caused by various factors such as injury, infection, or disease.
The cornea is normally translucent because of its unique structure and composition. It is the clear, dome-shaped outermost layer of the eye that covers the iris, pupil, and anterior chamber. Its transparency allows light to enter the eye and helps to focus the incoming light onto the retina for vision.
However, in certain cases, the cornea can lose its transparency and become opaque, leading to a condition known as corneal opacity. This can occur due to a variety of reasons. One common cause is injury to the cornea, such as a deep cut or burn, which can disrupt the regular arrangement of its cells and result in scarring or clouding. Infections, such as severe cases of bacterial or viral keratitis, can also lead to corneal opacity by causing inflammation and damage to the corneal tissue.
Furthermore, certain diseases and conditions, such as corneal dystrophies, degenerations, or inherited disorders, can affect the cornea and lead to loss of transparency. These conditions may cause abnormal deposits, growths, or changes in the corneal structure, impairing its ability to transmit light effectively.
In summary, the cornea is normally translucent, allowing light to pass through and contribute to clear vision. However, corneal opacity can occur as a result of injury, infection, or various underlying conditions, causing the cornea to lose its transparency and impair vision.
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The cornea is normally translucent, but it can become opaque or cloudy due to various reasons such as injury, infection, inflammation, or certain diseases.
When the cornea is damaged or affected by these conditions, it can lose its transparency, resulting in an opaque appearance. The cornea is unique in several ways. It is the clear, dome-shaped outermost layer of the eye that covers the iris, pupil, and anterior chamber. It plays a crucial role in focusing light onto the retina and acts as a protective barrier against foreign objects, germs, and UV radiation. Unlike other parts of the eye, the cornea has no blood vessels and receives nutrients and oxygen from tears and the aqueous humor.
The cow eye is similar to the human eye in terms of general structure and basic functions. Both eyes have similar anatomical components, such as the cornea, iris, lens, retina, and optic nerve. They function similarly in terms of receiving light, focusing it onto the retina, and transmitting visual information to the brain.
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3. Describe the pathway of a molecule going through the following systems.
a. Respiratory System: Pathway of an oxygen molecule as it is breathed in, starting from the mouth and ending in the alveoli.
b. Circulatory System: Pathway of an oxygen molecule from the alveoli to the intestine capillary bed. Then continue the pathway with a carbon dioxide molecule from the intestine capillary bed back to the right atrium of the heart. Be sure to include the applicable blood vessels and heart valves.
c. Digestive System: Pathway of protein and its digestion products, starting from the mouth until absorbed into the bloodstream. Be sure to list the parts that are passed through and where the protein is digested- including the enzyme names.
a. Respiratory system enters the nasal cavity or oral cavity during inhalation. b. Circulatory System bloodstream from the alveoli, red blood cells. c . Digestive System the pathway of a protein molecule in the mouth
a. In the respiratory system, an oxygen molecule is breathed in through the mouth and travels down the respiratory tract. It enters the alveoli, where gas exchange takes place.
b. From the alveoli, the oxygen molecule diffuses into the bloodstream and enters the pulmonary capillaries. It is then carried by the pulmonary veins to the left side of the heart. From the left atrium, it is pumped into the left ventricle and then out of the heart through the aorta enzymes. The oxygen-rich blood travels through systemic arteries to reach various tissues, including the intestine. In the intestine, the oxygen molecule is delivered to the capillaries of the intestinal bed.
For the pathway of a carbon dioxide molecule, it is produced as a waste product in the tissues of the intestine. The carbon dioxide diffuses into the capillaries of the intestinal bed and is carried by systemic veins back to the right atrium of the heart. From the right atrium, it passes through the tricuspid valve into the right ventricle. Then, it is pumped out of the heart through the pulmonary artery and reaches the lungs. In the lungs, the carbon dioxide is expelled through gas exchange in the alveoli and exhaled.
c. In the digestive system, the pathway of a protein starts in the mouth where it is mechanically broken down by chewing. It then travels down the esophagus to the stomach, where it encounters gastric acid and the enzyme pepsin. In the stomach, the protein is further broken down into smaller peptide fragments. From the stomach, the partially digested protein enters the small intestine, where pancreatic enzymes, such as trypsin and chymotrypsin, continue the digestion process, breaking the peptide fragments into smaller peptides and amino acids. The final digestion and absorption of the protein occur in the small intestine, specifically in the lining of the small intestine called the villi. The small peptides and amino acids are absorbed into the bloodstream through the capillaries in the villi, and from there, they are transported to various tissues in the body for growth and repair.
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Human _______, creating a genetically identical human, is banned in many countries around the world, including the United States
Human cloning, the practice of creating a genetically identical human being through artificial means, is expressly prohibited in numerous countries worldwide, including the United States.
The act of human cloning involves duplicating the entire genetic makeup of an existing individual, resulting in the creation of an identical genetic copy. Due to ethical concerns, potential risks, and the lack of scientific consensus on its safety and long-term implications, many countries have implemented legal frameworks to ban human cloning.
These regulations aim to prevent the reproductive cloning of humans and emphasize the significance of protecting human dignity, individuality, and the integrity of the human genome.
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2 A. List the 13 steps of pulmonary circulation on left and then add each step and its corresponding number, correctly to the diagram illustrating pulmonary circulation on the right. (8 points). 2B. Name a congenital heart defect and discuss its significance in affecting pulmonary circulation above ( 2 points).
Surgical intervention is typically required to correct Tetralogy of Fallot, aiming to repair the defects and improve pulmonary circulation, allowing for better oxygenation and overall cardiac function.
A. List of the 13 steps of pulmonary circulation:
1. Deoxygenated blood enters the right atrium from the superior and inferior vena cava.
2. The right atrium contracts, forcing the blood through the tricuspid valve.
3. Blood flows into the right ventricle.
4. The right ventricle contracts, pushing the blood through the pulmonary valve.
5. Blood enters the pulmonary artery, which splits into left and right pulmonary arteries.
6. Pulmonary arteries carry deoxygenated blood to the lungs.
7. In the lungs, the blood moves through the pulmonary capillaries surrounding the alveoli.
8. Oxygen from the alveoli diffuses into the pulmonary capillaries, while carbon dioxide diffuses out of the capillaries into the alveoli.
9. Oxygenated blood returns to the heart via the pulmonary veins.
10. Pulmonary veins carry oxygenated blood from the lungs to the left atrium.
11. The left atrium contracts, pushing the blood through the mitral (bicuspid) valve.
12. Blood flows into the left ventricle.
13. The left ventricle contracts, forcing the oxygenated blood through the aortic valve and into the aorta.
B. Congenital heart defect affecting pulmonary circulation: Tetralogy of Fallot
Tetralogy of Fallot is a congenital heart defect that affects pulmonary circulation. It is a combination of four specific heart abnormalities, which include:
Ventricular septal defect (VSD): A hole in the wall (septum) that separates the right and left ventricles, allowing blood to flow from the right ventricle to the left ventricle.
Pulmonary stenosis: Narrowing of the pulmonary valve or the pulmonary artery, restricting blood flow from the right ventricle to the lungs.
The significance of Tetralogy of Fallot is that it causes a mixing of oxygenated and deoxygenated blood, leading to decreased oxygen levels in the systemic circulation. The ventricular septal defect allows blood from the right ventricle to flow into the left ventricle, resulting in systemic circulation receiving less oxygen-rich blood.
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Which of the following is NOT associated with Staphylococcus aureus?
a. Coagulase b. Catalase c. DNase d. Pyocyanin
e. a− toxin
The term that is not associated with Staphylococcus aureus is Pyocyanin. Pyocyanin is a pigment produced by Pseudomonas aeruginosa. Staphylococcus aureus is associated with the following terms Coagulase Catalase DNaseα-toxin Pyocyanin is not associated with aureus.
Pyocyanin is a pigment produced by Pseudomonas aeruginosa. Staphylococcus aureus is a bacterium that is responsible for several infections in humans, including skin infections, pneumonia, and endocarditis. It is also known for its ability to produce toxins that can cause severe illnesses. Some of the common toxins produced by this bacterium include α-toxin, β-toxin, and γ-toxin.
The bacterium is characterized by several enzymes, including coagulase, catalase, and DNase. Coagulase is an enzyme that helps the bacterium evade the host's immune system by converting fibrinogen to fibrin, which forms a clot around the bacteria. Catalase, on the other hand, is an enzyme that helps the bacterium detoxify hydrogen peroxide produced by the host's immune system. DNase is an enzyme that helps the bacterium evade the host's immune system by breaking down the host's DNA. DNase is secreted by the bacterium to break down the DNA released by the host's dying cells.
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5. Compare and contrast the characteristics of the four different tissue types. Recall basic anatomy Tissue types Epithelial tissue (layers and shapes) Serous membrane and mucous membrane Connective tissues (Loose or areolar; adipose; reticular; dense connective) Muscle tissue (skeletal, cardiac, smooth) Nerve tissue (neuron, neuroglia) Cell to cell connection Tight junction Adhering junction Gap junction NMJ Synapse Extracellular matrix Glycosaminoglycans (GAGs) Proteoglycans Adhesion molecules Cadherins Selectins Integrins Immunoglobulin superfamily
Epithelial tissue, connective tissue, muscle tissue, and nerve tissue differ in their composition, function, and cell-to-cell connections. Epithelial tissue forms protective layers with various shapes, while connective tissue provides support with an extracellular matrix. Muscle tissue enables contraction, and nerve tissue facilitates electrical signaling.
Explanation:
Epithelial tissue is characterized by closely packed cells that form protective layers. It can be classified into different layers, such as simple (single layer) or stratified (multiple layers), and shapes, including squamous (flat), cuboidal (cube-shaped), and columnar (column-shaped). It also forms serous membranes (lining body cavities) and mucous membranes (lining organs and passages).
Connective tissue, on the other hand, consists of cells dispersed within an abundant extracellular matrix. It includes loose or areolar connective tissue, which supports and surrounds organs; adipose tissue, responsible for fat storage; reticular tissue, which forms the framework in organs; and dense connective tissue, providing strength and support to various structures.
Muscle tissue is specialized for contraction and generating force. It includes skeletal muscle, responsible for voluntary movement; cardiac muscle, which contracts involuntarily to pump blood in the heart; and smooth muscle, found in the walls of organs and responsible for their involuntary movement.
Nerve tissue comprises neurons and supporting cells called neuroglia. Neurons transmit electrical signals, allowing communication throughout the body, while neuroglia provide support and insulation to neurons.
The cell-to-cell connections differ among the tissue types. Epithelial tissue utilizes tight junctions to form barriers, connective tissue relies on various types of adhesion molecules like cadherins, selectins, and integrins. Muscle tissue employs gap junctions for coordinated contractions, and nerve tissue relies on synapses for signal transmission.
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Question 10 i) Describe the composition of the glomerular filtrate. (2 marks) ii) What is a normal value for the glomerular filtration rate (GFR) in a healthy adult male? (1 mark) iii) What proportion of the renal plasma flow is usually filtered by the glomeruli? (1 mark) iv) Write the equation for calculating renal clearance defining all terms. (2 marks) v) Explain why the clearance of inulin can be used to measure GFR. (2 marks) vi) Which endogenous substance can be used instead of inulin to measure GFR? Where does this endogenous substance come from? (2 marks)
Answer: The glomerular filtration rate, or GFR, is a measure of how well your kidneys are cleaning your blood -- taking out waste and extra water whereas renal clearance tests laboratory tests that determine the ability of the kidney to remove certain substances from the blood.
Explanation: i) The glomerular filtrate is composed of water, electrolytes (such as sodium, potassium, calcium, and chloride ions), glucose, amino acids, waste products (such as urea and creatinine), and small molecules. It does not contain large molecules such as proteins or blood cells.
ii) The normal value for the glomerular filtration rate (GFR) in a healthy adult male is approximately 125 mL/min or 180 L/day.
iii) Typically, about 20% of the renal plasma flow is filtered by the glomeruli. This proportion is known as the filtration fraction.
iv) The equation for calculating renal clearance as follows:
Renal Clearance = (Urine Concentration of Substance × Urine Flow Rate) / Plasma Concentration of Substance
Where:
Urine Concentration of Substance refers to the concentration of the substance being measured in the urine.
Urine Flow Rate is the rate at which urine is produced.
Plasma Concentration of Substance is the concentration of the substance being measured in the blood plasma.
v) Inulin is a substance that is freely filtered by the glomeruli and is neither reabsorbed nor secreted by the renal tubules. Therefore, the clearance of inulin represents the glomerular filtration rate (GFR). Inulin is an ideal marker for GFR measurement because it meets the criteria of being freely filtered and not being reabsorbed or secreted by the kidneys.
vi) Another endogenous substance that can be used to measure GFR is creatinine. Creatinine is produced by the breakdown of creatine in muscle tissue and is continuously released into the bloodstream. It is filtered by the glomeruli and partially reabsorbed by the renal tubules. However, the amount of creatinine that is reabsorbed is relatively constant, allowing for a reasonably accurate estimation of GFR. Therefore, creatinine clearance is commonly used as an alternative to inulin clearance to measure GFR.
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What is the morphological difference between acid-fast organisms and non-acid-fast organisms (what chemical is found in the cell wall of acid-fast organisms
The morphological difference between acid-fast organisms and non-acid-fast organisms is the presence or absence of mycolic acid in their cell walls, respectively.
This difference in cell wall composition affects the staining properties of these organisms and requires different staining techniques for their visualization and identification.
The morphological difference between acid-fast organisms and non-acid-fast organisms lies in the composition of their cell walls. Acid-fast organisms have a unique chemical called mycolic acid in their cell walls, which makes them resistant to staining with traditional dyes. On the other hand, non-acid-fast organisms lack mycolic acid in their cell walls and are easily stained by conventional methods.
Explanation: Acid-fast organisms, such as Mycobacterium tuberculosis, have a waxy layer of mycolic acid in their cell walls. This mycolic acid layer makes their cell walls impermeable to many stains, including the commonly used Gram stain. As a result, acid-fast organisms cannot be easily visualized using traditional staining methods. Instead, a special staining technique called the acid-fast staining is used, which involves using a lipid-soluble stain and heat to penetrate the mycolic acid layer and stain the bacteria. This staining method helps in the identification and diagnosis of acid-fast organisms, particularly in the case of tuberculosis.
On the other hand, non-acid-fast organisms, such as Escherichia coli, lack mycolic acid in their cell walls. As a result, their cell walls are not impermeable to stains, and they can be easily stained using conventional staining methods, such as the Gram stain. These staining methods involve using a combination of crystal violet and iodine to form a complex with the peptidoglycan layer of the cell wall, followed by a decolorization step and counterstaining. This staining process helps in the identification and classification of non-acid-fast organisms based on their Gram stain characteristics.
In conclusion, the morphological difference between acid-fast organisms and non-acid-fast organisms is the presence or absence of mycolic acid in their cell walls, respectively. This difference in cell wall composition affects the staining properties of these organisms and requires different staining techniques for their visualization and identification.
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6. Create a SNP imputation problem that meets the following criteria. a. There should be 4 reference haplotypes that are 10 base pairs long b. The observed genotype should be 10 bp long with 3 unknown".." sites to be imputed c. Show the solution to your problem and highlight which haplotypes are sampled in the observed genotype
The SNP imputation problem can be created as follows:
a) Four reference haplotypes that are 10 base pairs long:
Hap 1: A T C G T A G T C GG A Hap 2: A T C G T A G T C AG A Hap 3: A T C G T A G T C TG A Hap 4: A T C G T A G T C CG Ab) The observed genotype should be 10 bp long with 3 unknown".." sites to be imputed. The observed genotype can be as follows: Observed: A T C G . . . T C G G A
c) Solution to the problem: The haplotypes that are sampled in the observed genotype are Hap 1, Hap 2, and Hap 3. The missing genotypes can be imputed based on the reference haplotypes and the observed genotype. The missing sites can be filled as follows: ATCGTACGTGGA: Hap 1ATCGTACGTCGA: Hap 2ATCGTACGTGTA: Hap 3
Therefore, the observed genotype can be fully reconstructed as follows: Observed: A T C G T A G T C G G A
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The graph below shows a typical time course of tubulin polymerization into microtubules (in vitro). Explain what is happening at each of the labeled parts of the curve by drawing a diagram showing the 'behavior' of tubulin heterodimers at each of the three phases. ( 4 points)
The graph below depicts a typical time course of tubulin polymerization into microtubules (in vitro). The labeled parts of the curve describe the behavior of tubulin heterodimers at each of the three phases. The three phases are the nucleation phase, the elongation phase, and the steady-state phase.Explanation of the graphThe nucleation phaseTubulin heterodimers initially form clusters or oligomers that are commonly referred to as the 'nucleation phase.'The nucleation phase requires that the concentration of tubulin heterodimers in the system is above the critical concentration (Cc) or the concentration needed to induce tubulin heterodimer self-assembly. It's difficult to nucleate microtubules below the Cc because of the need for a large enough concentration of tubulin heterodimers to meet the structural requirements for initiating the assembly process.Tubulin heterodimers aggregate to form nuclei or oligomers in the nucleation phase, which may be considered the 'birthplace' of microtubules. The critical nucleus (nucleus or oligomer) has a stable free energy and can continue to grow as long as new tubulin heterodimers can attach to it.The elongation phaseThe elongation phase begins when the concentration of tubulin heterodimers exceeds the Cc and proceeds when the tubulin heterodimers are added to the stable critical nucleus from the nucleation phase. Tubulin heterodimers can be added to either end of the microtubule, but they're usually added to the plus end, where polymerization is the fastest.The steady-state phaseThe steady-state phase begins when the concentration of tubulin heterodimers is insufficient to add new tubulin heterodimers to the microtubule's plus end. This occurs because the microtubule's plus end has a reduced binding affinity for tubulin heterodimers. At this point, the microtubule attains equilibrium, meaning that the loss of tubulin heterodimers through depolymerization is balanced by the gain of new tubulin heterodimers via polymerization.Furthermore, the graph below shows the behavior of tubulin heterodimers at each of the three phases: Figure 1:
The behavior of tubulin heterodimers at each of the three phases.About MicrotubulesMicrotubules are cell organelles, in the cytoplasm of all eukaryotic cells, in the form of long, hollow cylinders with an outer diameter of approximately 25 nm and an inner diameter of ± 12 nm. Their length varies from a few nanometers to several micrometers.
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The cranial nerve that has three major branches is the OC) vagus O E) glossopharyngeal. O A) abducens. OD) trigeminal. O B) facial.
The cranial nerve that has three major branches is the D) trigeminal.
The cranial nerve that has three major branches is the trigeminal nerve. The trigeminal nerve, also known as cranial nerve V, is the fifth cranial nerve and is responsible for providing sensory and motor innervation to the face.
The trigeminal nerve has three main branches, which are:
1. Ophthalmic branch (V1): This branch supplies sensory information from the forehead, scalp, upper eyelid, and nose.
2. Maxillary branch (V2): This branch provides sensory innervation to the lower eyelid, upper lip, cheek, and side of the nose.
3. Mandibular branch (V3): This branch is responsible for both sensory and motor functions. It provides sensory innervation to the lower lip, chin, lower teeth, gums, and part of the tongue. Additionally, it controls the muscles involved in chewing (mastication).
The trigeminal nerve is the largest cranial nerve and plays a vital role in various functions related to sensation and movement in the face. Its branches cover a wide range of areas, allowing for complex sensory input and motor control in the head and face region.
Thus, the correct option is D) trigeminal.
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gonadocorticoids are released by which part of the adrenal gland?
Gonadocorticoids are released by the zona reticularis of the adrenal gland.
The adrenal gland is composed of two main parts: the outer cortex and the inner medulla. The cortex is further divided into three layers: the zona glomerulosa, the zona fasciculata, and the zona reticularis. Each layer of the cortex produces different types of hormones. The zona reticularis specifically secretes gonadocorticoids, also known as sex hormones. These hormones include androgens (such as dehydroepiandrosterone, or DHEA) and some estrogenic compounds. While the zona reticularis is responsible for the production of gonadocorticoids, the other layers of the adrenal cortex produce different hormones, such as mineralocorticoids (aldosterone) and glucocorticoids (cortisol).
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If the diameter of the field of view at low power is 40mm, what would the diameter of the FOV be at high power? ▪ Use the formula in your lab book that I went over in the.lab video. ▪ Put your answer in this format: 22mm with NO space between the answer and.the units.
The diameter of the FOV at high power would be 10mm.
The formula for calculating the diameter of the field of view (FOV) at different magnifications is:
FOV2 = (Magnification1 / Magnification2) * FOV1
Given that the diameter of the FOV at low power is 40mm and we want to find the diameter at high power, we can substitute the values into the formula:
FOV2 = (Low Power Magnification / High Power Magnification) * 40mm
Since we are looking for the diameter of the FOV at high power, we can assume the low power magnification is 1x (as it is not specified) and the high power magnification is 4x (typical for high power). Plugging in these values:
FOV2 = (1x / 4x) * 40mm
FOV2 = (1/4) * 40mm
FOV2 = 10mm
Therefore, the diameter of the FOV at high power would be 10mm.
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You are artificially stimulating a neuron in a science experiment using a voltage source to produce action potentials in a SiNGLE ISOLATED NEURON, not. an entire nerve. You stimulate the neuron during the absolute refractory period, what happens? 1. Nothing, no action potentials can be generated during the absolute refractory period regardiess of the stirnulation. 2. You observe an action potential because a threshold voltage was used. 3. You see a small graded potental in the neuron but not an action potential. 4. Nothing. More voltage is needed to stimulate a neuron during the absolute refractory period.
In a science experiment where a voltage source is used to stimulate a single isolated neuron during the absolute refractory period, the expected outcome is that no action potentials can be generated regardless of the stimulation.
This is because the absolute refractory period is a brief period of time immediately following an action potential when the neuron is temporarily unable to generate another action potential, regardless of the strength of the stimulus.
During this period, the neuron's voltage-gated sodium channels are inactivated and unable to open, preventing the generation of action potentials.
Therefore, applying more voltage will not lead to the generation of action potentials during the absolute refractory period.
It is important to wait for the refractory period to end before attempting to stimulate the neuron again.
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Predict the acute effects of the following mutations/drugs on your ability to detect light (increase, decrease, or no effect). Explain your answer in a sentence or two. A) A Calcium chelator B) A GCAP inhibitor C) Defective RGS
The acute effect of calcium chelator on our ability to detect light is decreased. Calcium chelator binds to free Ca2+ ions, thus depleting them from intracellular stores.
The free Ca2+ ions play a vital role in the activation of the rod outer segment guanylate cyclase, leading to cGMP production. So, the depletion of Ca2+ ions results in the deactivation of the rod guanylate cyclase and a reduction in cGMP production. Therefore, the amount of cGMP-gated channels decreases, resulting in a decrease in the ability to detect light. The acute effect of GCAP inhibitor on our ability to detect light is decreased. GCAPs (guanylate cyclase activating proteins) are calcium-binding proteins that activate retinal guanylate cyclase (GC), resulting in the production of cGMP. Inhibiting GCAP activity will decrease the production of cGMP in response to light. Thus, the closure of cGMP-gated channels will not occur and a smaller current is produced. Therefore, the ability to detect light decreases. The acute effect of defective RGS on our ability to detect light is increased. RGS proteins (Regulator of G protein Signaling) inactivate the transduction cascade by enhancing the GTPase activity of the alpha-subunit of the G-protein. This reduces the duration and amplitude of the light response.
So, a defective RGS protein leads to a slower rate of the hydrolysis of GTP and a longer duration of the light response. Therefore, the ability to detect light is increased.
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lace the structures the sperm must pass through in the correct order: sperm cells penatrating secondary oocyte 1 2 3
The structures that a sperm passes through are va-gina, followed by cervix, followed by the uterus, fallopian tubes and finally the egg.
First is the va-gina. During se-xual intercourse, sperm is ejaculated into the va-gina. The cervix is the second stage is basically is the narrow opening at the lower end of the uterus. Sperm must pass through the cervix to enter the uterus.
The uterus, or womb, is where the fertilized egg implants and develops into a fetus. Sperm swim through the uterus in search of the fallopian tubes. The fallopian tubes are basically considered as the site of fertilization. If sperm encounters a secondary oocyte in the fallopian tube, fertilization can occur. If a sperm successfully penetrates the secondary oocyte, it fertilizes the egg, resulting in the formation of a zygote.
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1. Explain how and when fertilization takes place? 2. Describe how a zygote becomes and embryo?
1. Fertilization is the fusion of sperm and egg, usually occurring in the fallopian tubes, resulting in the formation of a zygote. 2). The zygote undergoes cleavage, forming a morula and then a blastocyst, which implants in the uterus and develops into an embryo through cellular differentiation and organ formation.
1. Fertilization is the process by which the sperm and egg fuse to form a zygote. It typically occurs in the fallopian tubes of the female reproductive system. When a mature egg is released from the ovary during ovulation, it is captured by the fallopian tube. If sexual intercourse has recently taken place, sperm cells are present in the female reproductive tract and can travel through the cervix, uterus, and into the fallopian tube.
The sperm cells swim through the fluid in the fallopian tube and can reach the egg. When a sperm cell successfully penetrates the egg's outer layer, it undergoes a series of changes, including the release of enzymes that help the sperm enter the egg's cytoplasm. Once a single sperm has fused with the egg, fertilization is considered complete, and the genetic material from both the sperm and egg combine to form a zygote.
2. After fertilization, the zygote begins a process of rapid cell division called cleavage. The zygote divides into two cells, then four cells, and so on, forming a solid ball of cells known as a morula. The morula then undergoes further division and rearrangement to form a hollow ball of cells called a blastocyst. The blastocyst consists of an outer layer of cells called the trophoblast, which will later develop into the placenta, and an inner cell mass, which will develop into the embryo.
The blastocyst implants itself into the lining of the uterus, and the trophoblast cells interact with the maternal tissues to establish the necessary structures for nutrient exchange and support. The inner cell mass undergoes further differentiation and specialization, forming the various embryonic tissues and organs. This process of cellular differentiation and organ development continues throughout pregnancy, leading to the formation of a fully developed embryo.
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Q5. Gene Ais twice the length of Gene B and is expressed three times higher. If the FPKM of gene B is 288, what is the FPKM of gene A? How many more reads are there of Gene A than Gene B?
The number of more reads of Gene A than Gene B is 3 * 288 = 864 reads. To calculate the FPKM (Fragments Per Kilobase of transcript per Million mapped reads) of gene A, we need additional information. FPKM is dependent on the length of the gene and the total number of mapped reads in the sample.
Assuming that the read counts are directly proportional to gene length, we can use the following formula:
FPKM_A = FPKM_B * (length_A / length_B) * (expression_A / expression_B)
Given that Gene A is twice the length of Gene B and expressed three times higher, we have:
length_A = 2 * length_B
expression_A = 3 * expression_B
Substituting these values into the formula, we get:
FPKM_A = 288 * (2 * length_B / length_B) * (3 * expression_B / expression_B)
= 288 * 2 * 3
= 1728
Therefore, the FPKM of Gene A is 1728.
To calculate the number of more reads of Gene A than Gene B, we need to compare their expression levels. Since Gene A is expressed three times higher than Gene B, there are three times more reads of Gene A.
So, the number of more reads of Gene A than Gene B is 3 * 288 = 864 reads.
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30. The rate of consumer biomass accumulation in a given area; analogous to NPP for producer a. Egested energy b. Respired energy c. assimilated energy d. Net secondary productivity. 31. Net primary productivity depends on autotrophs for energy. a. True b. False 32. Which of the following is false? a. Tropical rainforests are the most productive terrestrial ecosystems b. Arctic and alpine regions have low productivity c. Temperate open ocean waters tend to have a higher productivity than tropical waters d. The interface between terrestrial and aquatic ecosystems is highly productive e. Productivity of open oceans is generally quite high compared to coastal waters.
30. Net secondary productivity is the rate of consumer biomass accumulation in a given area, analogous to NPP for producers (d).
31. True, net primary productivity (NPP) depends on autotrophs for energy.
32. False,temperate open ocean waters tend to have a higher productivity than tropical waters.
30. The rate of consumer biomass accumulation in a given area, analogous to NPP for producers, is called net secondary productivity (d). Net secondary productivity represents the rate at which consumer biomass accumulates in a specific area.
31. True. Net primary productivity (NPP) depends on autotrophs for energy. Autotrophs are organisms that produce their own food through processes like photosynthesis. NPP measures the amount of energy captured by autotrophs and converted into biomass, which is available as food for other organisms in the ecosystem.
32. The false statement is: "Temperate open ocean waters tend to have a higher productivity than tropical waters." Here's an explanation:
Productivity refers to the rate at which energy and matter are converted into biomass within a given area. It is commonly measured in grams per square meter per year. The productivity of an ecosystem is influenced by factors such as sunlight, water, nutrients, temperature, and existing biomass.
In terms of terrestrial ecosystems, tropical rainforests are the most productive due to their high biodiversity and abundant sunlight. Arctic and alpine regions, on the other hand, have low productivity compared to other ecosystems due to harsh environmental conditions.
When it comes to aquatic ecosystems, temperate open ocean waters tend to have lower productivity compared to tropical waters. Coastal waters are generally more productive than open oceans. The interface between terrestrial and aquatic ecosystems, such as estuaries, can be highly productive due to the mixing of nutrients from both land and water sources.
In summary, the false statement is that temperate open ocean waters have higher productivity than tropical waters. The reality is that tropical waters, both terrestrial and aquatic, exhibit higher productivity levels compared to their temperate counterparts.
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Gastric acid commonly creats peptic ulcers in the _____? (select
all that apply)
-stomach
-duodenum
-illeum
-jejunum
Gastric acid commonly creates peptic ulcers in the stomach and duodenum.
Peptic ulcers are painful sores that occur in the stomach lining or the duodenum (the upper part of the small intestine). The majority of peptic ulcers are caused by the bacterium Helicobacter pylori, which is responsible for up to 90% of cases. In some instances, the long-term use of nonsteroidal anti-inflammatory drugs (NSAIDs) such as aspirin or ibuprofen can induce peptic ulcers. Peptic ulcers, as the name implies, are ulcers that develop in the stomach lining and the upper part of the small intestine known as the duodenum.
The duodenum is the area where stomach acid and digestive juices are introduced to the digestive system, and it is therefore more susceptible to peptic ulcer development.In conclusion, gastric acid commonly creates peptic ulcers in the stomach and duodenum.
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2(a) Discuss any FIVE (5) forms of energy with
relevant examples.
2(b) Explain the term Biotechnology? Provide any TWO (2)
differentiation between Green and Blue Biotechnology.
2(a) Forms of energy with examples:Energy is the capability to do work. There are different forms of energy with relevant examples. They are listed below. 1. Thermal Energy: Thermal energy is the energy produced by heat. It is the sum of kinetic energy in the molecules of an object.
Example: fire and geothermal energy 2. Electrical Energy: Electrical energy is a type of energy that is transmitted by electricity. Example: Batteries, lightning 3. Kinetic Energy: Kinetic energy is the energy possessed by a moving object. Example: Windmill, moving cars 4. Potential Energy: Potential energy is the energy stored in an object. Example: a stretched rubber band, water in a dam 5. Chemical Energy: Chemical energy is the energy released from the chemical reaction between substances. Example: Wood burning, gasoline combustion 2(b) Explanation of the term Biotechnology and differentiation between Green and Blue Biotechnology Biotechnology is a combination of biology and technology.
It is the use of living organisms or their products to modify or create useful products, improve existing products, and develop new treatments and cures for diseases.Biotechnology is divided into two main branches:Green Biotechnology and Blue Biotechnology. The following is an explanation of the differentiation between Green and Blue Biotechnology:Green Biotechnology:Green biotechnology refers to the biotechnology of plants. It is concerned with the use of biotechnology in plant breeding, genetic engineering, and agriculture. Example: Genetically modified crops, plant tissue culture.Blue Biotechnology:Blue biotechnology is a branch of biotechnology that focuses on marine and aquatic applications. Blue biotechnology is concerned with the use of marine organisms and products in industry, medicine, and agriculture. Example: Aquaculture, marine biotechnology.
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Identify a technique to control water or soil pollution at its source. describe any one pollutant that is converted into a harmless form by this technique. what is the pollutant converted into and how?
A technique to control water or soil pollution at its source is the use of bioremediation. Bioremediation is a process where microorganisms, such as bacteria or fungi, are used to break down pollutants into harmless forms.
One example of a pollutant that can be converted into a harmless form through bioremediation is petroleum hydrocarbons. These hydrocarbons, found in oil spills or contaminated soil, can be degraded by certain bacteria.
These bacteria use the hydrocarbons as a source of energy and convert them into simpler, non-toxic compounds like carbon dioxide and water through metabolic processes. This helps to eliminate the harmful effects of petroleum hydrocarbons on the environment.
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When the diaphragm contracts during inspiration a. 1. the lung volume decreases causing the air pressure in alveoli to increase
b. the lung volume increases causing the air pressure in alveoli to decrease c. 1. the lung volume decreases causing the air pressure in alveoli to decrease d. 1. The lung volume increases causing the air pressure in alveoli to increase
The correct option is a.1. the lung volume decreases causing the air pressure in alveoli to increase. The lung volume decreases, the air pressure in the alveoli decreases, and the air flows into the lungs.
The correct option is A.
During inspiration, the diaphragm, a thin dome-shaped muscle at the base of the thoracic cavity, contracts and moves downward. This causes an increase in the volume of the thoracic cavity. The lung volume decreases, the air pressure in the alveoli decreases, and the air flows into the lungs.
During expiration, the diaphragm relaxes and moves upward, causing a decrease in the volume of the thoracic cavity. The lung volume decreases, the air pressure in the alveoli increases, and the air flows out of the lungs. The pressure of the air within the lungs is determined by the volume of the lungs and the number of molecules present.
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