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.
Vitamin K, one of the fat-soluble vitamins, plays a role in both bone health and blood clotting. There are two primary forms of the vitamin: phylloquinones and menaquinones. Choose the correct statement about vitamin K.a. Menaquinones are the primary form of vitamin K in the diet.b. Vitamin K deficiency inhibits the blood’s ability to clot.c. Excessive use of broad-spectrum antibiotics may cause vitamin K toxicity.d. The major source of vitamin K in our diets is from fortified grains.
Answer:
b. Vitamin K deficiency inhibits the blood’s ability to clot.
Explanation:
The main form is vitamin K1 (phylloquinone); followed by vitamin K2 (menaquinone), formed from the bacterial action on the large intestine tract and a third compound, vitamin K3 (menadione), a synthetic fat-soluble molecule. These vitamins can be found in many foods: in green vegetables (lettuce, cauliflower and spinach), in tomatoes, Brazil nuts, cashews, potatoes, contained in soybean oil, egg yolk, milk and to a lesser extent in wheat and oats.
Vitamin K participates in blood clotting, its deficiency can make it difficult to stop bleeding. Also causing damage to the body when in excessive concentration, for example: dyspnea (shortness of breath), and chest pain in adults with high vitamin K1 disorder and hyperbilirubinemia in newborns whose mothers in gestation underwent treatment with based on vitamin K3.
Human skin cells typically have 46 chromosomes. A human skin cell in G2 phase has ___________ chromosomes, ____________ chromatids, and ______________ DNA molecules.
A. 92 92 46
B. 46 92 92
C. 23 46 46
D. 46 46 46
E. 23 23 23
Answer:
B. 46 92 92
Explanation:
The cell cycle consists of three states: Interphase, cell division and Resting.
The interphase is composed of the pases G1, S and G2 (in this order), through which the cell prepares itself for the cell divison, where a single mother cell generates two (by mitosis) or four daughter cells (by meiosis).
As part of the preparation, the cell duplicates during the interphase all its genetic material (specifically in the phase S), forming replicated chromosomes. That is to say, a chromosome with two chromatides (sister chromatides).
A human cell that normally has 46 chromosomes will have in the G2 phase the same number of chromosomes: 46, however, these chromosomes will be composed of two chromatids (instead of a single one as normally), and therefore the cell will have 92 chromatides.
Because a chromatide is made up of a strand of DNA, in the G2 phase the cell will have also 92 DNA molecules.
ANSWER THIS ASAP WILL MARK BRAINLIEST EXTREME POINTS ANSWER NUMBER 21 ONLY
Answer:
dendrites
Explanation:
The neuron consists of a cell body and extensions called dendrites. In some neurons, one of the dendrites is elongated to form an axon. There are terminal and receptor dendrites. Receptor dendrites make contact with terminal dendrites of neighbouring neurons or receptor organs while terminal dendrites make contact with receptor dendrites of another adjacent neurons or effector organs.
There is a gene in eukaryotes under the control of a transcriptional activator that ultimately regulates chromatin structure. This transcription factor requires dimerization to function. Given your understanding of how transcription factors work, list four different mutations that could disrupt the regulation of this gene and the functional consequences of these individual mutations. Consider both cis- and trans- acting elements and different protein domains.
Answer:
Mutation that prevents Transcription Factors Dimerisation
Mutation that prevents Transcription Factors from binding to Chromatin Remodeling proteins.
Mutation that prevents Transcription Factors from binding DNA
Mutation that prevents binding of co- activator
Explanation:
Several different mutations can affect Transcription Activators to be unable to promote transcription of its target gene:
1. Mutation that prevents Transcription Factors Dimerisation: In the given example, because the Transcription Factors requires dimerisation to function. A mutation that disrupts dimer formation will prevent the Transcription Factor from activating gene transcription.
2. Mutation that prevents Transcription Factors from binding to Chromatin Remodeling proteins: Since this transcription activator results in Chromatin remodeling, a mutation in the domain that binds to and recruits chromatin remodeling proteins will prevent the transcription factor from being able to promote chromatin remodeling.
3. Mutation that prevents Transcription Factors from binding DNA: A mutation that abrogates either the ability or the specificity of transcription Activator to bind to DNA will prevent the activator to effectively promote transcription of target gene.
4. Mutation that prevents binding of co- activator: Many activators promote transcription from a locus in the presence of a co-activator. A mutation that prevents the binding of the activator to the activator will prevent expression of the target gene.
"An experiment was performed to determine the role that ATP plays in kinesin movement along microtubules. Kinesin and microtubules were incubated together in a test tube, but instead of ATP, a non-hydrolyzable analog of ATP was added to the tube. What impact on kinesin function do you expect to observe in the presence of this ATP analog?"
Answer:
There will be no movement of kinesin motor protein along the microtubules .
Explanation:
Microtubules is an important component of cytoskeleton which play an important role in maintaining cell shape,cell motility,cell division regulation etc.
Whereas kinesin belongs to the family of motor proteins present in the eukaryotic cell.Kinesin plays an important role axonal transport.
The hydrolysis of ATP generates energy to power up the movement of kinesin along the microtubules.
According to the given question if kinesin and microtubules were incubated with an non hydrolyzable ATP analog instead of ATP,then the movement of kinesin along the microtubules will not occur due to lack of energy because the non-hydrolyzable analog of ATP cannot undergo hydrolysis to release energy.
Two related misconceptions about photosynthesis and cellular respiration are that only plants perform photosynthesis and that cellular respiration is performed in animals as the equivalent of photosynthesis. Explain why these statements are inaccurate. Be sure to provide an accurate description of the relationship between photosynthesis and cellular respiration.
Answer:
Photosynthesis and cellular respiration is not restricted to just plants and just animals.
Explanation:
Photosynthesis is the opposite reaction of cellular respiration. The equation for photosynthesis is:
6CO2 + 6H2O + (energy) ---> C6H12O6 + 6O2
If you reverse the reaction:
C6H12O6 +6O2 ---> 6CO2 + 6H2O + (energy)
You get the chemical equation for cellular respiration.
The relationship between photosynthesis and cellular respiration is that they are opposite reactions of one another.
One example of these reactions not being limited to their respected plants/or/animals is that photosynthesis is also performed by protists, which include algae and other single-celled organisms, that are technically not considered plants OR animals.
Another example is that plants also participate in the process of cellular respiration. Plants use photosynthesis to grow and produce glucose, and they use cellular respiration to break down the glucose formed from photosynthesis to release energy.
Answer:
Exactly the statement is incorrect because both organism needs respration for energy production thus it's not restricted to one organism,
Addition**
Respiration takes place at the mitochondria and both they have
Which of the following macromolecules enter the nucleus of a eukaryotic cell through pores in the nuclear membrane? a. rRNAb. ribosomal proteins c. mRNA d. phospholipids
From the following macromolecules which enter the nucleus of a eukaryotic cell through pores in the nuclear membrane is mRNA. The correct option is c.
Macromolecules are big molecules, such as proteins, that are usually formed through the polymerization of smaller subunits known as monomers.
The nuclear pore is a conduit in the nuclear envelope that is lined with proteins. The nuclear pore controls the movement of molecules between the nucleus and the cytoplasm. The nucleus in eukaryotic cells is isolated from the cytoplasm and surrounded by a nuclear envelope.
mRNA is generated by DNA during the transcription process. Following synthesis, the newly formed molecule goes from the nucleus to the cytoplasm. A nuclear pore allows it to pass through the nuclear membrane. It will then connect with a ribosome, which is just coming assembled from two subunits, one large and one little.
Thus, the correct option is c.
Learn more about mRNA here,
https://brainly.com/question/29316969
#SPJ3
C)mRNA is the macromolecule that enters the nucleus of a eukaryotic cell through the pores in the nuclear membrane.
The nuclear pores allow large molecules such as mRNA transcribed from genes in the DNA to leave the nucleus and enter the cytoplasm to participate in protein synthesis.
Additionally, the RNA components of ribosomes are synthesized in the nucleolus, and after being produced, ribosomes are exported to the cytoplasm.
An individual with untreated type II diabetes has 400mg/100ml plasma glucose. Which of the following statements about this individual is INCORRECT?
-This individual will exhibit glucosoria
-This person's rate of glucose filtration will be greater that the rate of glucose reabsorption
-This person's SGLT carriers have reached saturation
-This person's rate of glucose excretion will be greater than the rate of glucose filtration
Answer:
The INCORRECT statement is This person's rate of glucose excretion will be greater than the rate of glucose filtration
Explanation:
The INCORRECT statement is This person's rate of glucose excretion will be greater than the rate of glucose filtration
It is incorrect because it is the opposite.
The incorrect statement is that the individual's rate of glucose excretion will be greater than the rate of glucose filtration.
An individual with untreated type II diabetes and a plasma glucose level of 400mg/100ml will indeed exhibit glucosuria, which occurs when the blood glucose level exceeds the renal threshold for glucose at 180mg/100ml.
At this point, the kidney's capacity to reabsorb glucose is surpassed, leading to the excretion of glucose in the urine. The statement that this person's rate of glucose excretion will be greater than the rate of glucose filtration is incorrect.
Which of the following is TRUE for BOTH mitosis and meiosis?
A. The parent and daughter cells have the same policy.
B. The daughter cells are genetically identical to each other.
C. Homologous chromosomes are separated during a division.
D. Crossing over occurs.
E. Sister chromatids are eventually separated.
Answer:
E. Sister chromatids are eventually separated.
Explanation:
Sister chromatids are separated and move towards the opposite poles of the cell during anaphase of mitosis and anaphase-II of meiosis. The splitting of centromere and shortening of microtubules assist the separation of sister chromatids and their segregation during mitosis and meiosis.
Crossing over and separation of homologous chromosomes occur during prophase-I and anaphase-I of meiosis respectively. Crossing over adds genetic variations in the daughter cells. Mitosis lack crossing over and produce genetically identical daughter cells.
"In the brain, vision originates in the rods and cones in the retina. Separate regions of the brain decode basic information, like color, shapes, intensity of light, and there are other regions that decode information like position in space, and awareness of patterns. As you use your visual system, all of these regions are working simultaneously. This simultaneous awareness of all regions working at the same time is due to which processing pattern listed below?"
a. oscillative processingb. reflexive processingc. parallel processingd. serial processing
Answer:
The correct answer will be option-Parallel processing
Explanation:
Parallel processing is the ability of the brain in which the brain processes the incoming stimuli of the different quality at the same time or simultaneously.
During the vision, the brain processes different stimuli at the same time like the shape, intensity, color and their spatial arrangement by different parts of the brain individually and then analyzing and comparing the stimuli to the stored memories. This parallel processing helps the brain to identify the vision.
Thus, parallel processing is the correct answer.
Jerry is a professional football player who injured his knee while running during a game. Jerry has had surgery and his knee is fully healed, but he is afraid to put pressure on it because he doesn’t want to re-injure it. The trainer implements a plan in which he first differentially reinforces standing, then walking, then jogging, and finally running at full speed. This is an example of using shaping to:
Answer:
reinstate a previous behavior
Explanation:
The rehabilitation to stop the pain and inflammation, for the recovery of the mobility and the muscular tone and the realization of a program of sports rehabilitation are essential to achieve a good recovery.
The recovery process must be carried out in a precise and multidisciplinary manner, the load that the knee must support and tolerate at all times, as well as rest and recovery times.
Prolonged use of lithium to treat bipolar disorder can cause damage to the kidneys such that epithelial cells of the collecting duct cannot insert aquaporin channels in their membranes. Which one of the following symptoms would you least expect in a person with such kidney damage?A. polyuria (excessive urine production)B. polydipsia (frequent drinking)C. excessive thirstD. high levels of vasopressin secretionE. glucosuria (glucose in the urine)
Answer:
E. glucosuria (glucose in the urine)
Explanation:
Generally, glycosuria occurs in patients with kidney changes due to diseases such as Wilson's disease or cystinosis, can also be a hereditary problem, but is not expected in patients with kidney damage caused by prolonged lithium use.
Normally, the kidneys filter the blood, eliminating all substances that are not necessary for the body to function, while glucose is reabsorbed in the blood because of its importance in energy production, but people with renal glycosuria do not reabsorb glucose. , which causes it to be eliminated in the urine, occurring glucosuria.
Anthocyanin is a pigment that gives flowers and leaves purple colors. The M gene codes for a transcription factor (Myb) that promotes expression of an enzyme that produces anthocyanin. The W gene codes for a different enzyme (Chs) that allows anthocyanin to be deposited in plant leaves and flowers. The dominant phenotype is the production of functional Myb and Chs. Use this information to answer the following question. Plants that have the mm genotype do not show any purple color. What is the best explanation for why this is?
Answer:
Anthocyanin is not produced in the plant cells
Explanation:
Anthocyanin is not produced in plant cells with the genotype mm.
As you can see from the question above, anthocyanin is responsible for the purple color of the flowers. Anthocyanin is encoded by the M gene, which is a dominant gene. Because it is a dominant gene, we know that it will be expressed in plants with the Mm and MM genotype, but will not be encoded by plants with the mm genotype. With this we can conclude that plants that have the mm genotype do not have purple color, because anthocyanin is not produced in the plant cells of these plants, since they do not have the M gene.
Scientific evidence suggests that the entire human population descended from just several thousand African migrants about 50,000 years ago." What sort of evidence would you expect to find if this claim were true? List the types of evidence you would expect to find if all humans living today came from a small population in Africa 50,000 years ago.
Final answer:
If the claim that all humans descended from a small population in Africa 50,000 years ago is true, we would expect to find genetic evidence, genetic variance, and archeological evidence supporting this claim.
Explanation:
Scientific evidence suggests that the entire human population descended from just several thousand African migrants about 50,000 years ago. If this claim were true, we would expect to find several types of evidence:
Genetic Evidence: Researchers have found that all human genomes tested outside of Africa have close ties to the genomes of people in Africa. This indicates that there is a genetic link between humans from different parts of the world, supporting the idea of a common African ancestry.
Genetic Variance: Genetic studies have also shown that there is a wider genetic variance in Africa compared to the rest of the world. This suggests that the ancestral population in Africa was larger and more genetically diverse, supporting the idea that all humans trace their ancestry back to Africa.
Archeological Evidence: Archeological findings such as fossils and artifacts can provide clues about human migration patterns. If all humans originated from Africa, we would expect to find evidence of early human settlements and migration routes out of Africa to different parts of the world.
DNA is different from RNA in that
A. RNA is made up five bases, whereas DNA is made up of four.
B. in general, RNA molecules are longer than DNA molecules.
C. RNA contains an additional oxygen atom on the ribose sugar.
D. RNA cannot exist as a double helix. All of these statements are true.
Answer:
RNA contains an additional oxygen atom on the ribose sugar. (Ans. C)
Explanation:
RNA contains a sugar which is known as ribose, and which contains an extra group of hydroxyl (OH-) while DNA contains deoxyribose sugar.
RNA contains four nitrogenous bases cytosine, adenine, guanine, and uracil, while DNA contains cytosine, adenine, guanine, and thymine.
Double helix means two stranded structure like DNA, while RNA is single stranded.
DNA molecule is much longer polymer than RNA molecule.
Considering the nitrogen cycle, how do all organisms depend on bacteria to produce and maintain adequate nitrogen in the environment?
I NEED HELP ! Choose the correct answer..
A. Bacteria eat the roots of plants that emit nitrogen.
B. Animals eat plants that contain bacteria, which contain nitrogen.
C. Since plants cannot use atmospheric nitrogen directly, they rely on bacteria in the soil to incorporate nitrogen into nitrates.
D.Plants and animals work together to create the appropriate amount of decay in the soil for nitrogen to form.
Answer:
Considering the nitrogen cycle, all organisms depend on bacteria to produce and maintain adequate nitrogen in the environment by eating the plants that contain bacteria which contain nitrogen.
Option: (B)
Explanation:
Nitrogen is a main element in the nucleic acid of both RNA and DNA which is most important for all living creatures and biological molecules. When plants doesn’t get enough nitrogen it doesn’t produce amino acid, without amino acid plant cannot make special proteins. Amino acid is building block for DNA which tends to the generation of organism. Nitrogen fixation is a process where bacteria convert atmospheric nitrogen to usage form for plant and animals get nitrogen by eating those plants.When a plant encounters a pathogen, it initiates a series of complex signaling pathways that may result in being able to fend off future attacks by the same pathogen. This mechanism may be the closest process plants have to an immune system. Which of the following describes this process?
a) systemic acquired resistance
b) herbivore-induced plant volatility
c) reactive oxygen species generation
d) chromatin modification.
The correct answer is a) Systemic acquired resistance
Explanation:
In biology, systemic acquired resistance occurs as one organism develops resistance to a specific pathogen such as a virus or bacteria, this means, the organism's system recognizes previous pathogens it has been exposed to and in this way responds appropriately. The systemic acquired resistance occurs mainly in plants and due to this, it is believed this response is similar to one of the immune systems in animals. Therefore, the mechanism that is the closest process plants have to an immune system is Systemic Acquired Resistance or SAR.
A drug that prevents disassembly of the mitotic spindle would stall the cell cycle
A. in the middle of G1
B. just before DNA replication
C. just before chromosome condensation
D. just before or just after the separation of sister chromatids
E. at the G2 checkpoint
Answer:
D. just before or just after the separation of sister chromatids
Explanation:
The mitotic spindle is a structure of the eukaryotic cell cytoskeleton involved in mitosis. Their function is to separate the sister chromatids during cell division so that they are included in the daughter cells. It consists of longitudinally aligned microtubule bundles that together have an ellipsoid-like shape.
For this reason, we can state that if a drug that prevents disassembly of the mitotic spindle paralyzes the cell cycle immediately before or shortly after separation of the sister chromatids.
In normal mitochondria the rate of electron transfer is tightly coupled to the demand for ATP. When the rate of use of ATP is relatively low, the rate of electron transport is low. When demand for ATP increases, electron-transport rate increases. Under these conditions of tight coupling, the number of ATP molecules produced per gram of oxygen consumed when NADH is the electron donor-the P/O ratio is about 2.5-3.0.
Ingestion of uncouplers causes profuse sweating and increase in body temperature. Explain this phenomenon in molecular terms. What happens to the P/O ratio in the presence of uncouplers.
Stanozolol is an anabolic, androgenic (testosterone-like) steroid that acts as an agonist at androgen receptors. Unlike testosterone, it is metabolized to inactive form very slowly. What happens to a man who takes large doses of stanozolol to bulk up? (Please select all correct answers.)
Answer:
It will result /lead to side effect that could be fatal and life threatening since its inactivation time is longer thus the effect of the large dose will be more pronounced and long lasting.
Explanation:
An hormone agonist is a compound/molecule that can bind to the hormone specific receptor leading to the activation of the endocrine gland, the biosynthesis of secreted hormone and the eliciting the physiological effects of such hormone. Androgenic anabolic efficacy of steroids is usually goverened by their receptor binding andspecificity and their biotransformation from active to inactive form. Possible side effect include elevated blood pressure, voice deepening, decease libido, early myocardial infarction and many more
g As a second step in thenerd-proposal, Wouldbe Fiancé needs to obtain DNA fragments that are the right size to make up the letters spelling out "Will you marry me?" How would they do that? A)Use restriction enzymes to cut the DNA at specific sequence sites. B)Waitfor helicase to unzip the genes. C)Use DNA ligase to stick together the Okazaki fragments. D)Alter the reading frame of the mRNA.
Answer:
A)Use restriction enzymes to cut the DNA at specific sequence sites.
Explanation:
Restriction enzymes cut DNA in specific sites. There are high and low frequency endonucleases. The easiest to get fragments to make the words: WILL YOU MARRY ME? is to amplify PCR fragments of known size and sequence, therefore, use restriction enzymes to cut the fragments of the desired size. It will be very useful make an in silico digestion to predict all the fragment sizes to be obtained, and a simulation of the gel before loading it.
How does the enzyme telomerase meet the challenge of replicating the ends of linear chromosomes?
A) It adds a single cap structure that resists degradation by nucleases.
B) It causes specific double-strand DNA breaks that result in blunt ends on both strands.
C) It catalyzes the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity.
D) It adds numerous GC pairs, which resist hydrolysis and maintain chromosome integrity
Answer:
The correct answer is C) It catalyzes the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity.
Explanation:
While replicating, the ends of the chromosomes always lose a part of the ends, therefore, the telomerase catalyzes this growing process, somehow compensating the information that will be lost.
Final answer:
Telomerase catalyzes the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity.
Explanation:
The telomerase enzyme meets the challenge of replicating the ends of linear chromosomes by catalyzing the lengthening of telomeres, compensating for the shortening that could occur during replication without telomerase activity. Telomerase contains a catalytic part and an RNA template. It attaches to the end of the chromosome and adds complementary bases to the RNA template on the 3' end of the DNA strand. Once the lagging strand template is sufficiently elongated, DNA polymerase can add nucleotides complementary to the ends of the chromosomes, ensuring the replication of chromosome ends.
A redwood tree, Sequoia sempervirens is hexaploid with a total of 66 chromosomes. Considering this, how many chromosomes did the tree inherit from each of it's parents? How many chromosomes can be expected to be in each of its gametes? How many chromosomes are likely to be in each somatic cell of the tree's offspring?
Answer:inherited 33 chromosomes from each parent; gametes have 33; somatic cells of offspring 66.
Explanation: S. sempervirens is sn autohexaploid, as were its parents. It inherited 33 chromosomes from each parent, and that is the chromosome number in its gametes, which means its offspring’s somatic cells will have 66 chromosomes
To study the molecular mechanism of DNA replication, you incubated soluble E. coli extracts with a mixture of dATP, dTTP, dGTP, and dCTP, all of which are labeled with the -phosphate group. After a while, the incubation mixture was treated with trichloroacetic acid, which precipitate the DNA but not the nucleotide or very short oligonucleotides. The precipitate was collected and the extent of the radioactively label nucleotide incorporation into the DNA was determined.
a.) If any one of the four nucleotides is omitted from the incubation mixture, would you detect radioactivity in the precipitate? Explainb.) Would 32P be incorporated into the DNA if only dTTP is labeled? Explainc.) Would radioactivity be found in the precipitate if 32P labeled the β and γ rather than the α phosphate of the deoxyribonucleotides? Explain.
Answer:
Explanation:
A) Since, mixture of dATP, dTTP, dGTP, and dCTP all were labeled; radioactivity can be detected even if one of the four nucleotides is omitted from the incubation mixture. This is because all other three nucleotides contain labeled phosphate and hence, radioactivity can be detected.
B) Yes. If 32P is labeled only dTTP, it can be detected in DNA. Since, dTTP can pair with 'A' nucleotide in DNA.
C) Mostly radioactivity can be detected in DNA when the alpha or terminal phosphates are labeled. It is difficult to detect radioactivity when phosphate groups other alpha like beta or gamma phosphates are labeled. Even if the radioactivity is detected it can be detected in rare cases like 5`end labeling of DNA on gamma phosphate.
To understand DNA replication, omitting any nucleotide from the mix or changing the radiolabel's location alters the assimilation of radioactivity in precipitated DNA. Presence of all nucleotides is crucial for complete DNA synthesis, and only the α-phosphate group incorporation leads to radioactive DNA.
Explanation:To study the molecular mechanism of DNA replication in E. coli extracts using nucleotides labeled at the α-phosphate, several scenarios illustrate how the radioactivity incorporation would be affected by experimental modifications.
If any one of the four nucleotides (dATP, dTTP, dGTP, dCTP) is omitted from the incubation mixture, radioactivity in the precipitate would likely be reduced or absent because DNA synthesis requires the presence of all four nucleotides. Each is necessary for building the complementary strand, thus, omitting one would halt DNA polymerase activity, and no or significantly less labeled DNA would be precipitated.If only dTTP is labeled with 32P, 32P would be incorporated into the DNA where thymine bases are incorporated. As DNA synthesis proceeds, dTTP is incorporated opposite adenine (A) in the template strand, resulting in radioactive labeling of the DNA.Should 32P label the β and γ rather than the α phosphate of the nucleotides, radioactivity would not be found in the precipitate. This is because, during DNA synthesis, the α-phosphate group is incorporated into the DNA backbone, while the β and γ phosphates are cleaved off during the formation of the phosphodiester bond.An electron loses potential energy when it:
A. Shifts to a less electronegative atom
B. Shifts to a more electronegative atom
C. Increases its kinetic energy
D. Increases its activity as an oxidizing agent
E. Moves further away from the nucleus of the atom
Answer:
B that is electron Shifts to a more electronegative atom
Explanation:
Electron loses potential energy when Shifts to a more electronegative atom. In an reduction-oxidation reaction, when an electron comes in proximity of high electronegative atom, it tends to lose some energy in form of potential energy. Electrons have more energy potential when associated with less electronegative atoms (such as C or H) and less energy potential when associated with a more electronegative atom (such as O).
hence the correct answer is B that is electron Shifts to a more electronegative atom
An electron loses potential energy when it shifts to a more electronegative atom, as the increased attraction to the atom's nucleus lowers potential energy and stabilizes the system.
Explanation:An electron loses potential energy when it shifts to a more electronegative atom. Electronegativity refers to the ability of an atom to attract electrons. When an electron moves from a less to a more electronegative atom, the electron is drawn closer to the atom's nucleus, resulting in a decrease in potential energy. This is because a more electronegative atom exerts a stronger pull on the electron, stabilizing the system by lowering the electron's potential energy.
Moreover, within an isolated atom, an electron loses potential energy when it moves closer to the nucleus (n decreases), making the system more stable and the atom ionized to a lower energy state if it captures an electron. This entire process enables the cell to incrementally use energy through the movement of high-energy electrons, which is crucial for metabolic pathways and the extraction of energy from food.
Learn more about Potential Energy of Electrons here:https://brainly.com/question/13681312
#SPJ6
As described in Investigating Life 20.1, scientists found that the range of the alpine chipmunk has shifted about 500 meters up in elevation, leading to isolated subpopulations on individual mountaintops. According to the researchers, what has caused this shift?
A.harmful UV radiation due to the ozone holeB.habitat destruction due to acid precipitationC.introduction of nonnative speciesD.climate warming due to increased CO2 concentrations
Final answer:
The shift in the Alpine chipmunk's range is caused by climate warming due to increased CO₂ concentrations, pushing species to higher elevations.
Explanation:
According to researchers, the shift in the Alpine chipmunk's range upwards by about 500 meters, resulting in isolated subpopulations on individual mountaintops, is primarily due to climate warming due to increased CO₂ concentrations. This conclusion is based on the observation that, globally, many species are moving towards higher altitudes or latitudes in response to changing climate gradients. With the warming climate, species adapted to colder environments find their habitable zones moving uphill. As a result, these species, including the Alpine chipmunk, are forced to follow the shift in their preferred climate range, leading to higher elevations and, unfortunately, potential habitat isolation and range restriction.
In general terms, normally a cell in your body crosses a checkpoint in the cell cycle when
A. It receives a "go-ahead" signal directly from your brain
B. Concentrations of proteins in the cytoplasm reach the appropriate level
C. It receives a "go-ahead" signal from your external environment
D. Concentrations of DNA in the nucleus reach the appropriate level
E. A mutation in the DNA occurs
Answer:
B. Concentrations of proteins in the cytoplasm reach the appropriate level
Explanation:
We can say that the cell cycle checkpoints is an important cell cycle control. In this regard, it is important to note that within a cell, the passage from one phase to another is controlled by regulatory factors that are characteristically proteinic and act at the so-called cell cycle checkpoints.
A cell passes the first checkpoint when protein concentrations in the cytoplasm reach the appropriate level.
The death cap mushroom, Amanita phalloides, contains several dangerous substances, including -amanitin. This toxin is a potent inhibitor of eukaryotic RNA polymerase II. Death, usually from liver dysfunction, occurs no earlier than several days after mushroom ingestion. Please speculate why
Answer:
Poisoning caused by the mushrooms like by eating Amanita phalloides can lead to the death of the organism after a few days. The chemical α- amanitin causes serious problems to humans as it shows its effect by inhibiting the RNA polymerase II enzyme which acts during the formation of mRNA.
If mRNA will not be formed, therefore the process of protein synthesis will not take place as a result, all the cellular processes will be halted and the organism will die.
The organisms do not die frequently after consumption of the mushroom instead dies few days only when the α- amanitin starts showing its effect on the liver and kidney and the liver and the kidney fails to perform their work.
Since the liver and kidney fails after a few days by the chemical α- amanitin, therefore, the organism dies after a few days.
The death cap mushroom contains α-amanitin, which inhibits RNA polymerase II, leading to delayed symptoms of poisoning such as liver dysfunction.
The death cap mushroom, Amanita phalloides, is notorious for its toxicity due to substances like α-amanitin. This particular toxin is a potent inhibitor of RNA polymerase II, which plays a crucial role in the synthesis of messenger RNA (mRNA) in eukaryotic cells. When α-amanitin inhibits RNA polymerase II, it disrupts the transcription process, eventually leading to cell death. However, the symptoms of poisoning, including liver dysfunction, do not manifest immediately because it takes time for a significant number of cells to be affected to the point of causing organ failure. Moreover, the body's compensatory mechanisms can temporarily mask the deterioration in function.
Due to the phenomenon of dominance,
A) different genotypes can give different characters.
B) different genotypes can result in the same phenotype.
C) the same genotype can produce different phenotypes.
D) the same genotype can produce different characters.
E) the same phenotype can produce different genotypes.
Answer:
The correct answer will be option-B
Explanation:
Dominance is the phenomenon observed in the genes coding for a trait in which the one variant of a gene called allele masks the effect of a complementary allele at same locus coding for the same trait.
The concept of dominance was observed by Gregor Mendel during his experiments on Pea related to study the inheritance pattern.
Since in dominance, one allele masks the effect of complementary allele therefore the dominant allele shows its effect and different genotypes produce the same phenotype.
Thus, Option-B is the correct answer
You treat bacteria with an extract made from peanuts that had a fungus growing on it. You then plate 10s of these bacteria onto a plate that contains penicillin. As a control you leave out peanut extract. You obtain 2 colonies from the control and 3 colonies after treatment with the corn extract. You conclude:
A. The peanut extract is not mutagenic.
B. The peanut extract is mutagenic.
C. Penicillin is mutagenic.
D. None of the answers listed above (ABC) are correct.
Answer:
C. Penicillin is mutagenic.
Explanation:
Mutagens are agents that change genetic material. They multiply the frequency of mutations of an organism over their standard level. The peanut extract that was left as control shows this natural or standard level, but when tenths of these bacteria were put into a plate that contained penicillin, the bacteria colonies increased the frequency of mutations of these bacteria after a determined time.