Answer:
Manufacturers could promote the recycling of cell phones.
Explanation:
Mining for the manufacture of cell phones and other electronic devices has caused at least six million refugees in the Democratic Republic of Congo in the last two decades. The high demand for minerals important for mobile technology causes humanitarian and environmental conflicts in the Democratic Republic of Congo. This is a serious issue that should be discussed among telephone companies to establish measures that will lower the human costs of depleting these ores.
One way to alleviate these problems would be through the recycling of disused electronic devices. Telephone companies could set up telephone collection points that are not in use and thus recover the ores needed to make new devices.
Minerals such as coltan, tungsten and cassiterite are widely used components in the manufacture of cell phones and electronic devices, many of which could be recovered and reused, but there is little awareness about cell phone recycling around the world.
Complex I, also called NADH dehydrogenase, is an entry point for electrons to the electron transport chain. Complex I transfers electrons from NADH to Q. The reduction of Q is linked to the active transport of protons out of the matrix. How many protons does complex I pump into the intermembrane space?
Final answer:
Complex I, or NADH dehydrogenase, of the electron transport chain pumps four protons into the intermembrane space for every two electrons it receives from NADH.
Explanation:
Complex I, also known as NADH dehydrogenase, is a crucial component of the electron transport chain in cellular respiration. Its role is to transfer electrons from NADH to coenzyme Q (ubiquinone), facilitating the conversion of ubiquinone to ubiquinol. Through the process of electron transfer, Complex I is able to pump protons from the mitochondrial matrix into the intermembrane space, thus contributing to the proton gradient used for ATP synthesis.
Importantly, Complex I is responsible for the translocation of four protons across the inner mitochondrial membrane for every two electrons it receives from NADH. This active transport of protons is crucial for the establishment and maintenance of the proton motive force that drives ATP synthesis via ATP synthase.
Think of the number of different parasites (species of bacteria, viruses, fungi, flatworm, apicomplexan, etc.) that can attack humans. Is it reasonable to expect that every other organism faces a similar array of parasites?No--most species should have a single, generalist parasite. (Generalists can exploit more than 1 species.)No--most species should have a single, specialist parasite.Yes, humans are a relatively young species so should have a relatively low number of parasites.Yes--every species represents a resource that may be exploited by an array of parasites
Answer:
The correct answer will be option-Yes--every species represents a resource that may be exploited by an array of parasites
Explanation:
Parasites are the organism which for survival depends on the resources of the host organism and utilize them to the extent that it could lead to the death of the host.
The interaction between the parasite and host proves harmful to the host but beneficial to the parasite.
A number of parasites exist for human species which can directly harm humans. Similarly, a large number of hosts exist for different species which belongs to another kingdom also like even the bacteria has a parasite called bacteriophage which utilizes the resources of the host.
This indicates that every species has some resources which can prove beneficial to another organism in the response of which they become host to a large number of parasites.
Thus, the selected option is the correct answer.
What is the advantage of having Complexes I, III, and IV associated with one another in the form of a respirasome?
a. It facilitates the energetically unfavorable electron transfer reactions carried out by Complexes I, III, and IV.
b. A respirasome composed of Complexes I, III, and IV allows the efficient transfer of electrons from to oxygen.
c. It enables the rapid transfer of electrons, thereby increasing the efficiency of proton pumping.
d. The respirasome allows and to interact directly with the inner mitochondrial membrane.
Answer:
The correct answer is c. It enables the rapid transfer of electrons, thereby increasing the efficiency of proton pumping.
Explanation:
Having three complexes will increase efficiency due to the fact that these complexes are connected, being more efficient in proton pumping.
Which of the following might explain the lack of cloud formation? Choose one or more:
A. The air package rises too fast for condensation to occur.
B. It is too hot for adiabatic cooling to take place.
C. The air package never reaches its dew point.
D. There are too few aerosols present in the air package.
Answer:
The following might explain the lack of cloud formation
The air package never reaches its dew point. There are too few aerosols present in the air package.Explanation:
For the formation of the cloud the air in the higher position should have much colder than that of the surface. The vapours travels till it reaches the point where it has lower temperature as well as the pressure so that the water will condense into the droplets which will result in the formation of the cloud. When those vapours condenses to the liquid or to the solid it will release some of it energy to the air. For the formation of the cloud three things are mandatory and they are moisture, cooling air and the condensation nuclei. Also, for the formation of clouds, suspended particles like aerosols are required for the water to condense. As there are few aerosols, water doesn't get to condense.
It has been observed that cancer cells take in much more glucose than normal cells. In fact, labeled glucose can be used to detect cancerous tumors through PET scans. Explain the changes in the metabolism of cancer cells that cause them to use so much glucose. Include the following terms in your answer: flux, regulation, ATP, and oxygen.
Answer:
Explanation:
Cancer cells are characterized by higher proliferation and replication of cells than normal cells. This causes a high demand for energy and nutrients. Cancer cells also thrive under hypoxic or low oxygen environment and they depend highly on glucose consumption for ATP production, that it, glycolysis is the main pathway for generating energy. Glucose transporters (GLUT's) are integral membrane proteins that mediate the flux of glucose in cells and have been implicated in cancer. For example GLUT2 has been found to be up-regulated in human carcinoma cells.
Your lab group is experimenting with the diffusion of molecules across a membrane. Dialysis tubing is used as a model cell membrane, because is it semi-permeable, allowing small molecules to cross the membrane, but NOT larger molecules, like starch. A starch solution is placed in the dialysis tubing and then the dialysis tubing is placed in an iodine and water solution. Iodine is used as an indicator for the presence of starch. Normally, iodine is a yellow-brown color, but in the presence of starch, it turns a deep purple. After observing the experimental setup for 20 minutes, the iodine solution remains brown, but the starch solution inside the dialysis tubing is purple and swollen. Based on what you learned from the lab, starch is MOST likely to cross the cell membrane at which of the points indicated on the diagram? A) Starch crosses the cell membrane at A. B) Starch crosses the cell membrane at B. C) Starch crosses the cell membrane at D. D) Starch crosses the cell membrane at E.
Answer:
Starch crosses the cell membrane at E.
Explanation:
i did the USATestprep
Answer:
C) Starch requires ATP energy for transport and therefore cannot cross the dialysis tubing into the iodine solution.
Explanation:
I did the USAtestprep
All the cells shown are part of the same organism. Which can you infer to be
true?
O
All of the cells came from the same cell.
O
All of the cells are able to produce food molecules.
O
All of the cells have the exact same function.
O
All of the cells have the exact same structures.
I need help!!! Please!!!!!
Answer:
hi there!
Explanation:
• Variation light and variation in temperatura
For me these are the problem, because they had to move some plant to the outdoors, because they don´t have enough space inside. So this changed of place could affect the plants.
Please match the Key Word to the correct definition.
1. h
2. b
3. c
4. d
5. g
6. a
7. f
8. e
List and describe four stages of food processing found in higher animals.
The visceral and parietal pleural membranes are normally held together by
A. loose connective tissue.
B. dense connective tissue.
C. decreased pressure in the pleural cavity.
D. surfactant in the pleural cavity.
E. elastic connective tissue.
Answer:
C) decreased pressure in the pleural cavity.
Explanation:
The pleura is a membrane that surrounds the lung and consists of two membranes, the parietal and the visceral, which are continuous in the hilum region of the lung. The parietal and visceral membranes delimit, for each lung, an independent cavity fully lined by the mesothelium. Typically, this pleural cavity is virtual, containing only a thin layer of lubricating fluid (pleural fluid), allowing smooth sliding of these two membranes during respiration, preventing friction between the visceral and parietal mesothelium.
The visceral and parietal pleural membranes are usually held together due to decreased pressure in the pleural cavity.
The visceral and parietal pleural membranes are held together by the decreased pressure in the pleural cavity, enabling smooth lung function with minimal friction.
Explanation:The visceral and parietal pleural membranes are normally held together by decreased pressure in the pleural cavity. The pleural cavity, which surrounds the lungs, contains a small amount of fluid that allows the visceral and parietal layers of the pleura to slide effortlessly over each other with minimal friction, aiding in lung movement during breathing. This fluid and the resulting suction-like effect keep these membranes close together.
Which statement is false? a. The number of double bonds in a fatty acid affects its melting point. b. Naturally occurring fatty acids usually have an even number of carbons. c. Wax esters contain long chain fatty acids and long chain alcohols. d. Triacylglycerols are polar due to their ester functional groups.
Answer:
d. Triacylglycerols are polar due to their ester functional groups.
Explanation:
Triacylglycerols are nonpolar due to the ester bond that forms between the glycerol polar hydroxyl (-OH) group and the fatty acid polar carboxylic acid (COOH) group. This leads to the release of a water molecule.
After eating lettuce contaminated with EHEC (E. coli) , what physical/chemical barriers to infection will the pathogen encounter en route to the intestine and once it reaches the intestine, but before it can attach to epithelial cells of the intestine?
Choose all the correct options below:a. Defensinsb. Cytolytic T-cellsc. Activated macrophagesd. Perforinse. Lysozymef. Mucing. Stomach acidh. IgGi. gut microbiotaj. tight junctions of gastrointestinal epithelial layers
Defensins, lysozymes, mucin, stomach acid, gut microbiota, tight junctions of gastrointenstinal epithelial cells are the physical/chemical barriers to infection will the pathogen encounter en route to the intestine and once it reaches the intestine after eating lettuce contaminated with EHEC
Explanation:
The enterohemorrhagic strain of Escherichia coli (EHEC) grows in the intestines, produces the highly potent Shiga toxin. E.coli is a foodborne pathogen.
Innate immunity acts against the EHEC pathogens through physical and chemical barriers with certain non-specific responses.
Physical barriers include skin and mucous membrane. Chemical barriers include stomach acid, lysozyme, defensin.
Tight junctions are the watertight regions lining the gastrointestinal mucosal surface with tightly packed epithelial cells which prevent the flow of E. coli bacteria between the host compartments.
The mucous membranes covered with tight junctions secrete mucin and lysozyme.
Mucin blocks the pathogen’s ability to attach to the cell surfaces and thereby inhibits its invasion.
Lysozymes are mucosal enzymes which act against the pathogens trying to adhere to the gut epithelium.
Defensins are peptides that attack the cytoplasmic membrane of the pathogens and kill them
Stomach acid (HCl) reduces the pH of the gut and maintains an acidic environment, and thereby inhibits the entry of foodborne pathogens
Gut microbes are commensal organisms that aid immunity by:
Colonization and creating competition for space and food Secreting toxins which can kill the pathogenic organismsWhen one allele is dominant to another, what is the most common biological mechanism for this dominance?
A. Dominant alleles often encode functional products, whereas recessive alleles often do not encode functional products
B. Dominant alleles subdue or repress recessive alleles
C. Cellular machinery recognizes which allele is dominant and only reads the
instructions from that one
D. A dominant allele on one chromosome causes the other, homologous
chromosome (containing the recessive allele) to be completely inactivated
Answer:
A. Dominant alleles often encode functional products, whereas recessive alleles often do not encode functional products
Explanation:
One gene can have it's variant forms called allele but in one individual only two allele form of a gene can be present. These two allele form can be the same(homozygous) or different(heterozygous). A dominant allele often encode a function protein of product while recessive allele often code for non- functional product.
Therefore when the dominant allele is present in the chromosome then function products are formed which can be seen in the phenotype and even a single dominant allele produce the functional product like in heterozygous condition so phenotype in heterozygous and homozygous dominant condition appears same. So the correct answer is A.
During the second set of photosynthetic reactions, called the Calvin cycle, all BUT ONE of the following occurs. A) Glucose is the end product of the Calvin cycle. B) By-products of the Calvin cycle include water vapor and oxygen. C) Carbon dioxide, a reactant of photo synthesis, is utilized during the Calvin cycle. D) ATP and NADPH, high energy molecules, are utilized to produce stored chemical energy.
Explanation:
False: B) By-products of the Calvin cycle include water vapor and oxygen
Photosynthesis is a chemical pathway that’s integral to producing energy in plants and other primary producers. Energy in the form of molecules of glucose is produced from light, water and carbon dioxide while oxygen (from the light reaction) is released.
Further Explanation:
This occurs in several complex steps, photosynthesis is a rate limited reaction, depends on several factors including carbon dioxide concentration, ambient temperature and light intensity; the energy is retrieved from photons, I.e. particles of light, and water is used as a reducing agent. This begins in the thykaloids, where pigment molecules like chlorophyll reside.
The chloroplast is a membrane-bound organelle found in plants. It contains several invaginations of a plasma membrane called the thylakoid membrane. This contains chlorophyll pigments, in stacks called granum, while the internal spaces of the organelle are called the lumen. Liquid surrounds the granum, forming the stroma.
Here, water is used as a reducing agent.
6CO2 + 6H20 + (energy) → C6H12O6 + 6O2
Carbon dioxide + water + energy= glucose + oxygen
Water supplies the chlorophyll in plant cell with replacement electrons for the ones removed from photosystem II. Additionally, water (H2O) split by light during photolysis into H+ and OH- acts as a source of oxygen along with functioning as a reducing agent; it reduces the molecule NADP to NADPH by providing H+ ions. NADP and NADPH are integral to the Calvin cycle where monosaccharides or sugars like glucose are produced after the modification of several molecules.
In the Calvin Cycle:
During the initial stage, three molecules of ribulose bisphosphate, RuBP are reduced and carbon is fixed from an inorganic CO2 to an organic form by the enzyme RuBisCO. With every molecule of CO2 fixed, 2 molecules of 3-phosphoglyceric acid are produced. 3-PGA is a 3 carbon molecule with a phosphate attached. Next, NADPH and ATP facilitate the addition of electrons to or reduction of 6 molecules of 3PGA to 6 molecules of G3P (glyeraldehyde 3-phosphate). The process uses 6 molecules of ATP and 6 of NADPH, after which the conversion of NADPH and ATP to ADP and NADP+ occur; these are returned to the light dependent reactions. A molecule of G3P, goes to the cytoplasm from this cycle, to form other complex molecules. Fixing sufficient carbon for this G3P export requires 3 revolutions of the cycle. Since two molecules of G3P are made per revolution, 3 turns make 6 G3Ps; 5 of these are used to regenerate RuBP for further fixation.Thus... no [tex]H_{2} O[/tex] or [tex]O_{2}[/tex] are produced...
Respiration in the mitochondria utilizes the oxygen waste for the production of ATP in the Krebs’s cycle via the oxidization of pyruvate( through the process of glycoysis). The electron transport chain, in which oxygen functions as the terminal electron acceptor occurs in both plants and animals.
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During the Calvin cycle, glucose is produced, carbon dioxide is utilized, and ATP and NADPH are used to create stored energy. However, water vapor and oxygen are not by-products of this part of photosynthesis.
Explanation:In the second set of photosynthetic reactions, the Calvin cycle, three important processes occur but one does not. A) Glucose is indeed the end product of the Calvin cycle. C) Carbon dioxide, a reactant of photosynthesis, is utilized during the Calvin cycle. D) ATP and NADPH, high energy molecules, are utilized to produce stored chemical energy. However, B) the by-products of the Calvin cycle are NOT water vapor and oxygen. These are the products of the light-dependent reactions of photosynthesis, not the Calvin cycle.
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The characters of "seed color" and "seed shape" showed independent assortment in Mendel's experiments because
A. the genes for these two characters were very close together on the same chromosome
B. these two characters were the product of the same gene
C. the genes for these two characters were on different chromosomes
D. the genes for both characters were sex-linked
E. either answer A or B could be correct
Answer:
C. the genes for these two characters were on different chromosomes
Explanation:
Mendel conducted several experiments with peas to demonstrate the mechanisms of heredity. He decided to use peas because they self-fertilize, have a short life cycle, are easily grown and produce large numbers of offspring. Peas, in addition to the advantages already mentioned, have simple characteristics to observe, such as the color and shape of the seed. Mendel analyzed seven of these characteristics that had no intermediates: seed texture, seed color, seed coat color, pod shape, pod color, flower position, and plant height.
Through his study, Mendel concluded that the characters "seed color" and "seed shape" showed an independent variety. This is because the genes related to these two traits are distant from each other in the chromosome and therefore have no dependence between them.
A woman reports a racing heart that began while she was playing softball 2 hours ago. Her ECG shows a rapid and regular heart rate, and she is given an intravenous bolus of a drug that activates G protein–coupled receptors. Which drug mechanism may lead to termination of the arrhythmia?
Answer: increase reflux of potassium
Explanation:
The woman possibly has supraventricular tachycardia which may precipitate by a premature atrial deplorization. Adenosine is administered to her which will bind and activate the G-protein coupled receptor that has adenosine binding site. This activation leads to the efflux of potassium and the hyperpolarization of supraventricular tissue.
The drug type used in this case should be beta blocker drug who has G protein as transduction unit where beta 1 or beta 2 adrenergic receptor is blocked.
Cellular respiration uses one molecules of glucose to produce approximately ___ molecules of ATP.
A. 36
B. 2
C. 4
D. 24
E. 12
Answer: 2 ATPs
Explanation:
In the overall reaction, glycolysis converts one six-carbon molecule of glucose into two three-carbon molecules of pyruvate.
The net products of this process are two molecules of ATP ( 4 ATP produced − 2 ATP used up).
Cellular respiration uses one molecules of glucose to produce approximately 36 molecules of ATP
What is cellular respirationDuring cellular respiration, one molecule of glucose is used to produce approximately 36 molecules of ATP (adenosine triphosphate). ATP is the energy currency of cells and is synthesized through the process of cellular respiration.
Although the exact number of ATP molecules produced can vary slightly depending on the specific conditions and efficiency of cellular respiration, the approximate yield of ATP from one glucose molecule is commonly estimated as 36 molecules.
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Which statement regarding ATP is false? Compared with other phosphate‑containing compounds in biochemistry, ATP has a higher free energy than nearly all of the others. Most enzymes that use ATP bind it in a form complexed with Mg2+. ATP concentrations in a cell are usually much higher than the equilibrium concentration for ATP hydrolysis. ATP can act as a group donor in adenylylation reactions.
Answer:
The correct answer will be option- ATP concentrations in a cell are usually much higher than the equilibrium concentration for ATP hydrolysis
Explanation:
Adenosine Triphosphate or ATP is known as the energy currency of the cell as it provides energy during the metabolic reactions.
The study on the structure of ATP revealed that it contains ribose sugar attached with three phosphates group bonded to each other via two phosphoanhydride bonds.
The metabolic reaction usually uses Magnesium bonded ATP as Magnesium covers the negative charge of the triphosphate which allows the ATP to bind hydrophobic clefts of the active site.
The ATP also transfers the adenyl group to the nucleophile which helps generate the pyrophosphate.
One of the genes that controls color vision is found on the X chromosome. The dominant allele results in normal color vision, and the recessive allele causes colorblindness. Ginny has normal color vision and is heterozygous, but Harry is colorblind. If Ginny and Harry have three children, what is the probability that at least one of them is a colorblind son?A. 0 B. 1/8 C. 9/64 D. 3/8 E. 27/64.
Answer:
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Rhonda tells Jack that a solution is a type of mixture. Jack disagrees and responds that he does not think that a solution is a mixture.
Which explanation below explains who is correct?
A. Rhonda is correct because a solution is a type of mixture. It is a mixture because the substances are mixed together, and each substance loses its primary physical properties. The substances can each be easily seen.
B. Jack is correct because a solution is not a type of mixture. It is not a mixture because one of the substances settles at the bottom of the solution. Each substance keeps it primary physical properties.
C. Rhonda is correct because the substances keep their primary physical properties, but one dissolved into the other. Each substance is not easily seen.
D. Jack is correct because a solution is not a type of mixture because the substances stay mixed together and cannot be separated. Each substance loses its its primary physical properties.
Answer:
Jack is correct because a solution is not a type of mixture because the substances stay mixed together and cannot be separated. Each substance loses its primary physical properties.
Option: (D)
Explanation:
A mixture is composed of 'two or more different substances', physically combined and it is in the form of a solution, suspension or a colloid. A mixture may be homogeneous (salt and water) or heterogeneous (sand and water). A homogeneous mixture of two or more compounds is termed as solution. It has a solute and a solvent. The 'solute' is the substance dissolved in the solvent. Both lose their primary physical properties.Even though she has been taking lithium for a year, you review some teaching about drug therapy with S.T. Which statement by S.T. reveals a need for further education? a. "I will call my doctor if I have severe vomiting or diarrhea." b. "I need to be careful because lithium is addictive."c. "I take the lithium tablets with meals." d. I will keep my appointments to have my drug levels checked.
Answer:
The answer is letter A.
Explanation:
Because vomiting and diarrhea are patological signs of lithium's intoxication. Which means ST is taking higher doses than she should.
In certain bacteria, an oval shape (O) is dominant over round (o) and thick cell walls (T) are dominant over thin (t). Show a cross between a heterozygous oval, thick cell walled bacteria with a round, thin cell walled bacteria. What are the phenotypes of the F1 and F2 offspring
Answer:
Phenotypes –
Oval Thick - OoTt
Oval Thin - Oott
Round Thick - ooTt
Round thin - oott
Explanation:
Given -
Allele "O" is dominant over allele "o"
and allele for thick cell wall "T" is dominant over thin celled wall "t"
Genotype of heterozygous oval, thick cell walled bacteria is OoTt
Genotype of round, thin cell walled bacteria is oott
The cross between the two parents will produce following offspring
OT Ot oT ot
ot OoTt Oott ooTt oott
ot OoTt Oott ooTt oott
ot OoTt Oott ooTt oott
ot OoTt Oott ooTt oott
Phenotypes –
Oval Thick - OoTt
Oval Thin - Oott
Round Thick - ooTt
Round thin - oott
Final answer:
The cross between a heterozygous oval, thick cell-walled bacteria and a round, thin cell-walled bacteria will result in all F1 offspring being oval and thick-walled. The F2 offspring will have a 3:1 phenotype ratio for both cell shape and wall thickness if complete dominance is assumed, or a 1:2:1 ratio for cell wall thickness if incomplete dominance occurs.
Explanation:
Genetics of Bacterial Shapes and Cell Walls
In regard to the genetics of bacterial characteristics, when crossing a heterozygous oval-shaped bacteria with thick cell walls (genotype OT) with a round, thin cell-walled bacteria (genotype oo tt), we must consider the dominance of the alleles. The dominant traits are oval shapes (O) over round (o), and thick cell walls (T) over thin (t). A Punnett square can be used to predict the genotypes of the F1 offspring, which would all be heterozygous oval-shaped with intermediate cell wall thickness (Oo Tt), assuming incomplete dominance for cell wall trait. The F1 phenotype would exhibit the dominant traits of oval shape and thick cell wall.
For the F2 generation, we would cross two F1 heterozygotes (Oo Tt x Oo Tt), which will produce offspring with varying combinations of these traits by following the principles of Mendelian genetics. The F2 phenotype ratio for oval and round would be 3:1 and for thick and thin cell walls also 3:1 if cell wall thickness shows complete dominance. If there is incomplete dominance for cell wall thickness, we would observe a 1:2:1 ratio of thick, intermediate, and thin cell walls.
You have added an irreversible inhibitor to a sample of enzyme and substrate. At this point, the reaction has stopped completely. What can you do to regain the activity of the enzyme? a. Removing the irreversible inhibitor should get the reaction working again.
b. Adding more substrate will increase the rate of reaction.
c. Adding more inhibitor should get the reaction up to speed again.
d. The enzyme is inactive at this point. New enzyme must be added to regain enzyme activity.
Answer:
The correct answer is d. The enzyme is inactive at this point. New enzyme must be added to regain enzyme activity.
Explanation:
This is because the inhibitor is irreversible, meaning that the enzyme is permanently modified, therefore, even if more substrate is added, the enzyme will no longer be able to work properly. If all of the inhibitor is consumed, adding more enzyme would regain the activity of the enzyme.
Irreversible inhibitor permanently inactivates enzymes. Removing the inhibitor, or adding more substrate or inhibitor will not recommence the enzyme’s activity. Only fresh, active enzyme will increase the rate of reaction.
Explanation:In biochemistry, an irreversible inhibitor permanently binds to an enzyme, rendering it incapable of catalyzing a reaction. As a consequence, the enzyme cannot regain its original function once it has encountered an irreversible inhibitor. Therefore, removing the irreversible inhibitor, adding more substrate or adding more inhibitor will not resurrect the enzyme’s activity.
The best answer is option D: The enzyme is inactive at this point. New enzyme must be added to regain enzyme activity. To increase the rate of reaction, fresh, active enzyme must be introduced to the system.
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What will happen to a red blood cell (RBC), which has an internal ion concentration of about 0.9 percent, if it is placed into a beaker of pure water?
Answer:
It wil swell and burst
Explanation:
"pure" water is only H2O so there are no solutes, because of osmosis, water will enter the red blood cell that does contain solutes until the cell and the surrounding environment (in this case the water) are iso-osmotic, this will never happen though because the water contains no solutes.
The cell cycle must be regulated in a multicellular organism in order to
A. maintain the proper cell number in tissues
B. activate cell division when repair of a tissue is needed
C. halt cell division when repair of a tissue is complete
D. prevent cancer
E. all of the above
Answer:
The correct answer is E. all of the above
Explanation:
The cell cycle has several checkpoints that helps in the regulation of the cell cycle. Cell cycle regulation is important because if the cell cycle is not regulated then the cell will constantly divide which will be a waste of energy and an abnormal increase in the number of cells can do damage to the tissue. So regulation is important to maintain a proper cell number in the tissue.
Cell division helps in repairing the damaged tissue, therefore, cell division must be stopped after tissue is repaired and regulation is also important to prevent cancer because cell division stops when the mutation occurs in the cell's genetic material. Therefore the correct answer is E.
What is the process of converting energy from a light - excited electron into ATP
Answer:
Photosynthesis
Explanation:
Photosynthesis is the process in which the energy will be converted from the light exited electron to phosphorylate the bond of ADP molecule to ATP.
The basic hereditary unit of information in biology is the
A. cell
B. gene
C. chromosome
D. genome
E. gamete
Answer: Option B
Explanation: A gene is the specific portion or sequence on the DNA (Deoxyribonucleic acid) that is able code for a functional protein. Thus, the entire DNA sequence in each cell consists of several genes and some non-coding portions that all makes up the complete hereditary information usually called genome.
Observe this linear chain
Cell -> Chromosomes -> DNA -> Gene.
The above illustrates that a gene is the basic carrier of hereditary information.
This will definitely help.
Which of the following indicates that a muscle fiber is relaxed?
ATP molecules are bound to the heads of the myosin filaments.
Myosin heads are attached to actin filaments, forming cross bridges.
Calcium ions are bound to troponin, causing the protein to change shape.
Z lines are pulled toward the center of the sarcomere.
ATP molecules are bound to the heads of the myosin filaments
In relaxed form of the sarcomere, ATP is bound to its site on the myosin heads. It is when the ATP is hydrolyzed to ADP and Pi that the mysoin forms a cross ridge with the actin fiber. Upon hydrolysis the ADP + Pi exits the site and another ATP binds. This again causes the myosin head to disengage the actin fiber and the cycle is repeated.
Explanation:
During contraction of muscle calcium ions are usually released from their sarcoplasmic reticulum where they bind to troponin enabling tropomyosin to be exposed such that myosin can bind. Contraction of sarcomere causes the z lines to move closer together.
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Match the appropriate labels to their respective targets.
a. Hormonal stimulus
b. Neural stimulus
c. Humoral stimulus
d. Humoral and/or hormonal stimulus
1. Aldosterone production
2. Parathyroid hormone production
3. Testosterone production
4. Epinephrine production
Answer:
Hormonal stimulus [tex]\rightarrow[/tex] testosterone production
Neural stimulus [tex]\rightarrow[/tex] epinephrine production
Hormonal stimulus [tex]\rightarrow[/tex] aldosterone production
Humoral stimulus [tex]\rightarrow[/tex] parathyroid hormone production
Explanation:
The correct matching for the given unmatched is given below
Hormonal stimulus [tex]\rightarrow[/tex] testosterone production
Neural stimulus [tex]\rightarrow[/tex] epinephrine production
Hormonal stimulus [tex]\rightarrow[/tex] aldosterone production
Humoral stimulus [tex]\rightarrow[/tex] parathyroid hormone production
Final answer:
Hormone release is regulated by humoral, hormonal, and neural stimuli. Aldosterone production is a result of a humoral stimulus, parathyroid hormone is regulated by hormonal stimulus, testosterone by both humoral and hormonal stimuli, and epinephrine by a neural stimulus.
Explanation:
The regulation of hormone release can be triggered by a variety of stimuli. Here, we match the appropriate labels to their respective targets representing the hormonal release mechanism:
Aldosterone production - c. Humoral stimulusParathyroid hormone production - a. Hormonal stimulusTestosterone production - d. Humoral and/or hormonal stimulusEpinephrine production - b. Neural stimulusEach of these targets are regulated by one of the three mechanisms:
Humoral stimuli are changes in ion or nutrient levels in the blood.Hormonal stimuli refer to the release of hormones in response to other hormones.Neural stimuli occur when a nerve impulse prompts the secretion of a hormone.