Answer:
In stem cell transplants, stem cells replace cells damaged by chemotherapy or disease or serve as a way for the donor's immune system to fight some types of cancer and blood-related diseases, such as leukemia, lymphoma, neuroblastoma and multiple myeloma. These transplants use adult stem cells or umbilical cord blood.Explanation:
TRUE OR FALSE?
All of the living things in a pond; algae, pond weeds, insects, birds, crustaceans, fish, and other organisms that reside and interact there make up a biological community.
True or False?
Answer:
shcdcg xsetr
Explanation:
xsdthfcq
Only red algae can grow in relatively deep regions of the ocean because red algae are: Select one: a. capable of absorbing carbon dioxide even from the carbon-dioxide-depleted deep waters and using the carbon dioxide to build sugars during photosynthesis. b. specialized for absorbing shorter wavelengths of light that penetrate deeper into the water for photosynthesis. c. able to perform chemosynthesis rather than photosynthesis. d. capable of absorbing ultraviolet light for photosynthesis, rather than visible light.
Answer:
b. specialized for absorbing shorter wavelengths of light that penetrate deeper into the water for photosynthesis.
Explanation:
There are two lights which are responsible for the photosynthesis in plants and algae i. e. red and blue light. These two lights are absorbed by the plants and algae in the process of photosynthesis while most of the light colors are reflected. In water red light remains in low depth while blue light penetrate deep into the water due to its shorter wavelength. So that's why red algae can grow due to the presence of blue light and photosynthesis occurs.
Correct Question and options:
Only red algae can grow in relatively deep regions of the ocean because red algae are:
a. capable of absorbing carbon dioxide even from the carbon dioxide deep waters and using the carbon dioxide to build sugars during photosynthesis.
b. specialized for absorbing longer wavelengths of light that penetrate deeper into the water for photosynthesis.
c. able to perform chemosynthesis, rather than photosynthesis.
d. capable of absorbing ultraviolet light for photosynthesis, rather than visible light.
Answer:
The correct answer is b. "specialized for absorbing longer wavelengths of light that penetrate deeper into the water for photosynthesis".
Explanation:
Red algae, also known as Rodophytes, are immobile organisms that inhabit the sea, but might also be found in freshwater and some of them can even be terrestrial. Due to its pigments, this algae is capable of absorbing sunlight in great depths, even much more than other types of algae. Rodophytes are photosynthetic organisms, which means that when they absorb sunlight they can transform it into energy.
Red algae´s coloration is due to the rhodoplasts that contain chlorophyll a, in addition to other pigments such as phycoerythrin and phycocyanin, that are phycobiliproteins. These last ones mask the chlorophyll and make a place to the characteristic algae´s red color.
Only red algae can grow in relatively deep-sea regions because they are specialized in photosynthesizing by absorbing longer light wavelengths that penetrate deeper into the water. Phycoerythrin is the one that absorbs the green wavelengths that reach deep regions and emits the red wavelengths of light. These proteins capture light energy and transfer it to chlorophylls during photosynthesis.
Based on the article, which tissue do you think is affected by polio?
Answer:
Nerve
Explanation:
Answer: NERVE
Explanation:
Scientific advances in the study of evolutionary developmental biology (evo-devo) have demonstrated that another factor has had as much of an impact as specific alterations in the genes on the macroevolutionary changes in body plans that have been observed through time. This factor is ___________.
Answer:
The phenomena of (evo-devo) is the part of biology related to how variations in embryonic growth throughout single peers are related to evolutionary changes that occur between generations.One of the scientist who worked on this phenomena named charles darwin argued for the importance of growth (embryology) in understanding evolution. Nevertheless, afterward the detection in 1900 of Mendel's research on genetics, any relationship between development and evolution was not considered important for understanding evolutionary processes or as a black box that was difficult to see. Research in the past two decades has opened that black box, revealing how evo-devo studies highlight the mechanisms that link genes (the genotype) to structures (the phenotype). This is vitally important because genes do not form structures. Development processes create structures using gene-provided road maps, but they also use many other signals - physical forces such as mechanical stimulation, ambient temperature, and interaction with chemicals produced by other species - often species in completely different kingdoms. As in communications amongst microorganisms and squid or between leaves and larvae. Genes not only do not form structures (the phenotype), but new properties and mechanisms emerge during embryonic development: genes are differentially regulated in different cells and locations; Similar cell aggregations provide the cellular resources (modules) from which tissues and organs arise; modules and differentiated cell populations interact to establish development along particular tracks; and organisms interact with their environment and create their niche in that environment. Such interactions are often called "epigenetic," meaning that they direct gene activity using mechanisms that are not encoded in the DNA of the genes. This article reviews the origins of evo - devo, how the field has changed in the last 30 years, assesses the recognition of the importance for the development and evolution of mechanisms that are not encoded in DNA, and assesses what the future holds. could bring for evo– devo. Though difficult to know, past communicates us that we could expect more of the same; expansion of evo - devo in other areas of biology (ecology, physiology, behavior); absorption of evo-devo by evolution or a unification of biology in which evo-devo plays an important role.
1. You are conducting some in vitro splicing reactions with three different RNAs (A, B, and C) that each have a single intron. The exons in each case are identical in size (although the sequences could be different). One RNA has a group I intron, another has a group II intron, and the third has a spliceosomal intron. You incubate each RNA in conditions that allow self-splicing [(-) lanes below] as well as with a nuclear extract/splicing extract. After incubation, you observe the products by denaturing PAGE and autoradiography (your RNAs were radioloabeled). Which RNA has which intron
Answer:
Explanation:
1. There are six lanes in total out of which three are being showing self-splicing activity and the remaining three are being provided with splicing enzymes. As can be seen in case of RNA A, in the absence of a nuclear extract or when no splicing enzymes are added, there would be no splicing which can be reflected in lane 1 showing formation of only a single band which indicates that splicing does not occur on its own, although when the nuclear extract containing the splicing enzyme, there will be separation of the exons and the introns leading to the formation of two bands. So, the first two lanes specific for RNA A will be having a Spliceosomal Intron.
2. Now the task remains to identify which out of the remaining RNA's will be either Group I or Group II introns. There is no data regarding the addition of Guanosine nucleotide which is usually required in case of Group 1 splicing. Now, as we know that the splicing mechanism in case of Group II and spliceosomal intron is similar being carried out in a way in which an Adenine which is located in the branch site shows binding to the 5'- splice site which shows a similar conformational change for both the groups of introns. So, the Ribosome 2 which will act as a ribozyme and specifies lane 3 and 4 should be Group II intron and thus the last two lanes namely 5 and 6 will specify Group I intron.
De novo purine synthesis occurs by the stepwise addition of atoms or groups of atoms to 5‑phosphoribosyl 1‑pyrophosphate (PRPP). The atoms of the purine rings are supplied by glutamate, glycine, glutamine, aspartate, and N 10 ‑formyltetrahydrofolate (THF). Inosine monophosphate (IMP), the product of the pathway, is a purine nucleotide that can be converted to either AMP or GMP. The structure of the base is labeled according to the numbering convention for purines. Identify the direct source of each atom in the purine ring of IMP.
Answer:
Explanation:
Like we all know, the purine ring of IMP is made up of a nine membered compound . they are heterocyclic aromatic organic compound that consist of a pyrimidine ring fused to an imidazole ring. there are four nitrogen atoms and five carbon atom.
The ring is imidazole ring is made up of N1, C2, N3, C4, C5, C6 with the pyrimidine sharing C4 and C5 with the imidazole ring and also made up of N7, C8, N9.
the direct source of N1 is from aspartate
the direct source of C2 and C8 is from N 10 ‑formyltetrahydrofolate (THF)
N3 and N9 is derived from the amide group of Glutamine
C4, C5 and N7 is derivd from Glycine
C6 is derived from bicarbonate
1. Discuss why you think the unit exercise only had you create two bones for the model’s hand rather than five? 2. How would rigging (and animation) be more difficult if you didn’t include guide bones, pole targets, and other rigging elements not directly part of the "skeleton" of the mesh? 3. Describe the ways your model originally looked "off" in either the deformation/skinning or in the actual animation/keyframing process. Also, explain specific ways that you fixed these deficiencies. 4. Imagine you were creating a robot instead of a human character. What design changes might you make to the model in terms of rigging and skinning to make the character more "robotic?" 5. How would you tackle the challenge of animating a "boneless" model such as, for example, a traditional RPG slime monster — or, alternatively, something like a snake or a detailed tail animation?
Answer:
1. Discuss why you think the unit exercise only caused you to create two bones for the model's hand instead of five.
R = why the prosthesis can be adapted according to the needs of each of the users.
2. How would the rigging (and animation) be more difficult if it did not include guide bones, pole targets, and other rigging elements that are not directly part of the mesh "skeleton"?
R = the bones of the fingers that are still present help us create the measurements of the missing fingers since in this way we create a more harmonic and sesthetic hand, and when these fingers are not present it is more difficult to create in the animation since it can be noticed a little deformed since the requests cannot be correct and the fingers can be seen in very different sizes from one to the other.
3. Describe the ways your model originally looked "off" in warp / skinning or in the animation / keyframe process
A = I would not know how to answer that because I do not have any key program to carry out the prosthesis but I can guide you a little bit about a prosthesis program which you can see in an animation program
The digital model of the affected hand was obtained using modeling techniques in Blender, Rhinoceros and Mesh to model conversion software in Solid Edge. As a result, a 3D digital model of the affected hand was obtained on which the mechanisms of the prosthesis to be implemented can be worked on and adjusted.
After developing the model of the prosthesis volume, measurements were taken and adjusted to the plaster model that was previously obtained. Figure 13 shows how the prosthesis model was fitted to the plaster hand model. Subsequently, with the measurements obtained from this model, the prosthesis mechanisms are designed for the three missing digits in the user's hand.
4. Imagine that you are creating a robot instead of a human character. What design changes could you make to the model in terms of rigging and skinning to make the character more "robotic"?
R = the physical aspect, and that the movements are more rigid and not so fluid to give the robotic touch, since currently robots move very fluidly and can make faster expressions and make it look more metallic and not treat it to do with such a humanoid aspect.
5. How would you face the challenge of animating a "boned" model, such as a traditional RPG slime monster, or alternatively a snake or detailed tail animation?
R = you have to make an animation character that looks gelatinous, as through the program you have to create a mass that looks translucent so that it can simulate that the character does not have bones, because if you make it a solid object we would be losing the effect that the character is gelatinous that is supported without bones.
Which are the major structures of the integumentary system? Select three options.
bones
hair
joints
muscles
nails
skin
tendons
Answer:
Correct options: Hair, Skin, and Nails.
Explanation:
The abilify to taste phenylthiocarbamide (prc) is determined by a dominant gene (T). Individuals who can taste pTC are called tasters. A man who is a taster and has a non-taster mother marries a woman who is a taster. She has five siblings, three ofwhom are non-tasters. what are the chances that the children of this marriage will be nonasters?
Answer:
The chances of producing children that will be non-tasters is 1/4.
Explanation:
The ability to taste is determined by T which can either be TT or Tt. Non tasters can only be inherited in the recessive condition (tt).
A man who is a taster with a non taster mother is heterozygous for the trait (Tt). He marries a woman who is a taster (she is a TT or Tt since she has siblings that are non tasters). To have a higher percentage of non-tasters in her (the woman) family, it means, one of the parent is heterozygous for the trait and the other is homozygous recessive for the trait. Thus, the taster woman will be an heterozygote.
Now the taster man that is heterozygous for the trait marries this heterozygote woman, the chances of producing children that will be nontasters is 1/4.
Answer:
1/4 or 25%
Explanation:
A man who is a taster and has a non taster mother is heterozygous and will have the genotype Tt.
A woman who is a taster will have the genotype Tt or TT. However, the woman has 5 siblings out of which 3 are non-tasters. It means that her parent had 6 children out of which 3 are tasters and 3 are non-tasters (1:1).
A 1:1 result is usually from a cross involving one homozygous non-taster and one heterozygous taster. Hence;
Tt x tt = Tt, Tt, tt and tt
It thus means that the taster woman is heterozygous with genotype Tt.
A marriage between Tt and Tt:
Tt x Tt = TT, 2Tt and tt
Recall that the trait is is a dominant one, hence TT and Tt are tasters while tt is a non-taster.
The chances of the children being a non-taster therefore is 1/4 or 25%.
You are snorkeling out in the ocean, when you come out across this sight. You cannot decide if the smaller fish is a parasite or a mutualist of the bigger fish. You decide to both do more observations on these two fish species in the wild, and to bring them back into the lab for experiments. Describe the observations you would make & an experiment you could do to test if this is a mutualism or a parasitism.
Answer:
The analysis comprise in watch the conduct of each gathering of fishes. Subjectively, we can mention objective facts of the swimming examples of every creature, on the off chance that they assault one another or if the pass on. Be that as it may, this investigation will be founded on quantitative outcomes. We will observe during multi month of the quantity of fishes in each gathering that still alive. Obviously, we will give the standard states of nourishment, oxygen and light to each gathering.
We need to determinate if the two species can get by their own without the nearness of the other one. Our hypothesis would be: If both kind of fishes have a mutual relationship, the two gatherings (1 and 2) will lessen the quantity of people continuously. On the off chance that the relationship is parasitic, one of the two gatherings will have less number of people toward the month's end.
In which of the two types of succession do you expect to take longer to get to Climax Community?
•Primary succession
•Secondary succession
Answer:
Secondary
Explanation:
Answer:
It is secondary succession.i had that question. hope it help and good luck ;)
What does Vivian Lee research? *
Answer:
she was an actor
Explanation:
She starred in Gone With the Wind when she was 25. she was a actress
When a Chinese hamster with white spots is crossed with another hamster that has no spots, approximately 1/2 of the offspring have white spots and ½ have no spots. When two hamsters with white spots are crossed, 2/3 of the offspring possess white spots and 1/3 have no spots.What is the genetic basis of white spotting in Chinese hamsters?Is it possible to produce Chinese hamsters that breed true for white spotting? Why or why not?
Answer:
A) One of the genotype i.e true breed white is lethal
B) No
Explanation:
A) Two hamsters with white spots are crossed, 2/3 of the offspring possess white spots and 1/3 has no spots.
This means that the two white hamsters would be carrier for the trait of no spot.
The ratio for offspring signifies that one of the offspring dies because of lethal genotype.
B) No, because the true breed for white spotting is lethal.
However, heterozygous white spotting can be detected.
The genetic basis of white spotting in Chinese hamsters is a simple Mendelian pattern of inheritance involving a single gene with two alleles. It is not possible to produce Chinese hamsters that breed true for white spotting unless both parents are homozygous for the allele.
Explanation:The genetic basis of white spotting in Chinese hamsters is a simple Mendelian pattern of inheritance involving a single gene with two alleles. The allele for white spotting (W) is dominant over the allele for no spots (w). When a Chinese hamster with white spots (Ww) is crossed with a hamster without spots (ww), approximately half of the offspring will have white spots (Ww) and half will have no spots (ww). When two hamsters with white spots (Ww) are crossed, two-thirds of the offspring will have white spots (Ww) and one-third will have no spots (ww).
The possibility of producing Chinese hamsters that breed true for white spotting depends on the genotype of the parents. If both parents are homozygous for the allele for white spotting (WW), then all of their offspring will also have white spots and they will breed true. However, if one or both of the parents are heterozygous for white spotting (Ww), there is a chance that some of their offspring will not have white spots. Therefore, it is not possible to produce Chinese hamsters that breed true for white spotting unless both parents are homozygous for the allele.
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Whether a G alpha subunit is active or inactive depends on which guanine nucleotide is bound to it. Binding of GDP or GTP results in the protein switching between two conformational states. Which of the following answers best describes the structural changes that occur in a G alpha subunit due to guanine nucleotide binding? Choose one: A. Phosphorylation of GDP to GTP by the G alpha subunit moves the switch II helix region from binding to G beta/gamma to binding effectors like adenylate cyclase. B. The helical region of G alpha, called switch II, which interacts with G beta/gamma in the inactive state, is brought into the interior of the G alpha protein, reducing contact with G beta/gamma. This permits G alpha interaction with effector proteins, like adenylate cyclase, since the switch Il region is now buried. C. Epinephrine directly binds and activates G alpha to allow the subunit to bind to an effector protein. D. Dissociation of GDP for GTP with the G alpha subunit structurally shifts the switch II helix region, allowing for the association of G alpha with its effector proteins, such as adenylate cyclase.
Answer:
Option D
Explanation:
Whether a G alpha subunit is active or inactive depends on which guanine nucleotide is bound to it. Binding of GDP or GTP results in the protein switching between two conformational states.
Dissociation of GDP for GTP with the G alpha subunit structurally shifts the switch II helix region, allowing for the association of G alpha with its effector proteins, such as adenylate cyclase,describes the structural changes that occur in a G alpha subunit due to guanine nucleotide binding
Hexaploid wheat was produced synthetically by He and co-workers. They mated the diploid species, Aegilops tauschii, and the tetraploid species, T. turgidum. Which of the following is an accurate statement about the relative contribution of each parent to the genome of the hexaploid offspring?
1) Aegilops tauschii contributed four chromosomes by failing to complete meiosis after chromosome replication, and T. turgidum contributed two chromosomes. [<-- this one is not correct]
2) Aegilops tauschii contributed two chromosomes, and T. turgidum contributed four chromosomes.
3) The hexaploid number appeared following mitosis with no subsequent cell division.
4) Each parent contributed equally to the genome of the offspring
Answer:
2.) Aegilops tauschii contributed two sets of chromosomes, and T. turgidum contributed four sets of chromosomes.
Explanation:
The mating of diploid and tetraploid species is one breeding method which has proven majorly satisfactory with the largest and most prominent mass producers of seedlings for uniform high-quality progeny, has been the mating of the tetraploid of good form with the free and prolific diploid.
The result yields a good progeny. This is due to the fact that the tetraploid parent influences twice as much as the diploid parent do. It's traits, form and color are more pronounced. This also implies that the diploid contributes two chromosomes and the tetraploid contributes four chromosomes. When hexaploid wheat was produced synthetically by He and co-workers, they mated diploid Aegilops tauschii, and the tetraploid species, T. turgidum. This implies that Aegilops tauschii contributed two chromosomes and T. turgidum contributes four chromosomes.
Aegilops tauschii, the diploid species, contributed two chromosomes, while the tetraploid species T. turgidum contributed four chromosomes to create a hexaploid wheat.
Explanation:The relative contribution of each parent to the genome of the hexaploid offspring involves Aegilops tauschii, the diploid species contributing two chromosomes, and T. turgidum, the tetraploid species, contributing four chromosomes. Thus, the correct statement would be 'Aegilops tauschii contributed two chromosomes, and T. turgidum contributed four chromosomes'. The hexaploid wheat produced synthetically by He and co-workers is a result of the genetic material from the diploid species Aegilops tauschii and the tetraploid species T. turgidum.
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Which ion has the greatest influence on the resting membrane potential of most neurons?
Potassium has the greatest influence on the resting membrane potential of most neurons due to the membrane's higher permeability to potassium ions through open non-gated potassium channels during the resting state.
The ion with the greatest influence on the resting membrane potential of most neurons is Potassium (K+). During the resting state, non-gated (leak) potassium channels are open, allowing potassium ions to move across the neuron's membrane. This movement of potassium is critical in establishing the resting membrane potential because the membrane is much more permeable to potassium than to sodium due to the higher number of open potassium channels. Essentially, these channels enable potassium to diffuse out of the neuron, influencing the membrane potential significantly. Although sodium (Na+) and chloride (Cl-) channels are also present, they are fewer in number, meaning that while these ions do play a role, their influence on the resting membrane potential is not as great as that of potassium. The resting membrane potential of a neuron is closer to the equilibrium potential of potassium, typically between -60 to -70 mV. Furthermore, the sodium-potassium pump helps to maintain the gradients of these ions, but the passive permeability largely dictates the resting membrane potential.
* The complete question is:
Which ion has the greatest influence on the resting membrane potential of most neurons?
Multiple Choice
Potassium
Sodium
Which best describes a roll of animals in the water cycle
Answer:
Animals contribute water mainly through breathing, perspiration and urination. ... When droplets of sweat evaporate from the surface of an animal's skin, they take a bit of the animal's body heat with them. They also turn into water vapor and enter the water cycle, just like water evaporating from plant leaves.
Explanation:
Why do some water masses in subsurface oceans have little or no oxygen? a. Large carnivores deplete oxygen in subsurface oceans via their high rates of oxygen metabolism. b. Rates of photosynthesis are low in overlying waters, limiting the availability of oxygen. c. Waters underneath the surface oceans are cold and thus have a limited capacity to carry oxygen in solution. d. Rates of photosynthesis are high in overlying waters, supporting high rates of respiration in waters that deplete the oxygen there.
Answer:
The Correct option is "D"
Explanation:
By and large there is accessibility of plenitude daylight on a superficial level which gives different tiny fish and diatoms a space to develop, which additionally accommodates thick algal sprout. These algal sprout are obtrusive to such an extent that their thickness may exhaust the oxygen level and furthermore can slaughter the microscopic fishes.
During the cross-bridge cycle in muscle cells, myosin motors hydrolyze ATP as fuel to create a pulling force on actin fibers. Please describe how the different states of ATP hydrolysis correspond to different interactions between myosin and actin to drive this cycle.
Answer and Explanation:
In rest, attraction strengths between myosin and actin filaments are inhibited by the tropomyosin. When the muscle fiber membrane depolarizes, the action potential caused by this depolarization enters the t-tubules depolarizing the inner portion of the muscle fiber. This activates calcium channels in the T tubules membrane and releases calcium into the sarcolemma. At this point, tropomyosin is obstructing binding sites for myosin on the thin filament. When calcium binds to the troponin C, the troponin T alters the tropomyosin by moving it and then unblocks the binding sites. Myosin heads bind to the uncovered actin-binding sites forming cross-bridges, and while doing it ATP is transformed into ADP and inorganic phosphate which is released. Myofilaments slide impulsed by chemical energy collected in myosin heads, producing a power stroke. The power stroke initiates when the myosin cross-bridge binds to actin. As they slide, ADP molecules are released. A new ATP links to myosin heads and breaks the bindings to the actin filament. Then ATP splits into ADP and phosphate, and the energy produced is accumulated in the myosin heads, which starts a new binding cycle to actin. Z-bands are then pulled toward each other, thus shortening the sarcomere and the I-band, and producing muscle fiber contraction.
It is good practice to autoclave all cultures before disposal, even if the culture is of a non-pathogenic organism. Autoclaving not only kills the cells in the culture, but denatures and destroys proteins and nucleic acids. Based on what you have learned in this laboratory exercise, why is it especially important to autoclave genetically engineered organisms before disposal
Answer:
Most of the GMO are hazardous therefore inactivation of the GMO is very important , so that they are completely or partially destroyed before disposal. This is needed to provide high level of protection for the surrounding environment inhabited by man and other organisms to be deposited..
In all cases at least 99-100% inactivation must be achieved.
Explanation:
Which of the following plants would most likely be found in the desert?
A.
a tree that has smooth bark so that water runs off of it more easily
B.
a vine that climbs on top of other plants in order to gain access to sunlight
C.
a plant that has a reduced number of stomata in its leaves
D.
a plant with a shallow root system to capture the soil's top layer nutrients
A large asteroid impact occurs, kicking up dust that blocks the sun and prevents plants from photosynthesizing.
What would most likely happen as a result of the asteroid impact? Check all that apply.
More earthquakes will occur across the planet.
Some plants will evolve to use dust as a source of food.
Many plants will die without sunlight for photosynthesis.
Some animals will adapt over time to survive in new conditions.
Many species will eventually die off because they lack a food source.
Answer:
- many plants will die w/out sunlight
- some animals will adapt over time
- many species will eventually die off
Answer:
3,4 and 5
Explanation:)
What are some plants that grow in mountains
Answer:
Trees & Shrubs
Explanation:
Which of the following statements about actin is correct?
a. If the actual amount of available ATP-bound G-actin is below the critical concentration at both ends of a filament, then the actin filament will depolymerize.
b. Actin monomers are called G-actin because they have a nucleotide binding site that accommodates a molecule of GTP.
c. The plus end of an actin monomer has a net positive charge.
d. Actin generally polymerizes more rapidly at the pointed end.
e. Polymerization of an actin filament will only occur at the plus end.
Answer:
Option C
Explanation:
The G actin are Actin monomers have a nucleotide binding site which accommodates a molecule of ATP where ATP is hydrolyzed after polymerization. The plus end of an actin monomer has a net positive charge. Polymerization of an actin filament occur at the both ends but elongation is faster to about 4-5 times at the plus end than at the minus end. Actin generally polymerizes more rapidly at the barbed end than the pointed end.
Answer:
The correct answer is option a. "If the actual amount of available ATP-bound G-actin is below the critical concentration at both ends of a filament, then the actin filament will depolymerize".
Explanation:
Actin is the monomeric subunit of most filaments in cells, constituting the the cytoskeleton that brings support and maintains the cells' structure. Actin filament depolymerization is a process that allows for the creation of turnovers in the cytoskeleton and ensures that enough actin subunits are available to produce the filaments. One signal that starts actin filament depolymerization occurs when the actual amount of available ATP-bound G-actin is below the critical concentration at both ends of a filament. This occurs in the opposite direction as well, because the ability of a solution of G-actin to polymerize is given when the available ATP-bound G-actin is above the critical concentration.
You isolate chromosomes from the salivary gland of an organism and prepare a glass slide of the chromosome preparation. When you look at the chromosomes on the slide, they can be seen to be extremely large and exhibit a distinct banding pattern. What name is used to describe these chromosomes?
Answer: the name used is polytene chromosomes.
Explanation:
Polytene chromosomes are produced when repeated rounds of deoxyribonucleic acid (DNA) replication without cell division forms a giant chromosome, they have thousand of DNA strands and provides high level of function in the salivary glands.
At interphase, polytene chromosomes are seen to have distinct thick and thin banding patterns, these bands are of 2 types, the dark band (dark stained,
contains more DNA and less RNA) and the interband (light stained, more RNA and less DNA). The bands enlarge and forms a swelling called puff in certain times, the puffing (which is the formation of puff) is caused by the uncoiling of individual chromomeres in a band. The puffs indicate the site of active genes where mRNA synthesis takes place. These distinct banding patterns are used to study the function of genes in transcription because they permit high level of gene expression.
Nijmegen breakage syndrome is a human disease that is characterized by immunodeficiency, cancer predisposition, and microcephaly. It is a disorder that is caused by the inheritance of two mutant copies of a gene called NBS1, whose product is involved in DNA repair. Knowing this, what does this tell you about the relationship between DNA repair processes and the immune system?
A. DNA repair processes must be critical for the proper functioning of the immune system, as when they are nonfunctional, the immune system does not function properly.
B. The NBS1 gene must have multiple alleles, one that affects DNA repair, and one that affects the immune system.
C. Nijmegen breakage syndrome is caused by an epistasis relationship between a gene affecting DNA repair and a gene affecting the immune system.
D. Nothing. Nijmegen breakage syndrome is the result of polygenic inheritance, and therefore the immunodeficiency seen in these patients is due to the inheritance of another gene.
Answer:
A. DNA repair processes must be critical for the proper functioning of the immune system, as when they are nonfunctional, the immune system does not function properly.
Explanation:
The Nijmegen breakage syndrome (NBS) affects the immune system and this disorder is produced by mutations in the NBS1 gene involved in DNA repair, thereby evidencing a causal relationship between the immune system and the mechanism of DNA repair
A series of phenomena will occur in Mio’s body to compensate for the high-salt diet. Predict how Mio’s body would compensate for a high-salt diet. (Answers may be used once, more than once or not at all.)
1. With the final result, a healthy individual 19s blood pressure will________ .
2. Next, water will________ the principal cells via the aquaporin-3 channels.
3. As a result, blood volume will__________ .
4. The net movement of water from the filtrate into the principal cells will then___________ .
5. Insertion of aquaporin-2 channels in the apical membrane of the principal cells in distal convoluted tubule and collecting duct will__________ their permeability to water.
6. Hence, an increased volume of water will__________ the peritubular fluid and capillary.
a. Diffuse out of
b. Stay the same
c. Exit
d. Increase
e. Decrease
f. Enter
g. Diffuse into
1. With the final result, a healthy individual 19s blood pressure will increase.
High salt diet :2. Next, water will diffuse out of the principal cells via the aquaporin-3 channels.
3. As a result, blood volume will increase.
4. The net movement of water from the filtrate into the principal cells will then increase .
5. Insertion of aquaporin-2 channels in the apical membrane of the principal cells in distal convoluted tubule and collecting duct will increase their permeability to water.
6. Hence, an increased volume of water will enter the peritubular fluid and capillary.
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Mio's body will compensate for a high-salt diet by mechanisms that lead to an increase in blood pressure, water reabsorption, and blood volume, highlighting the body's ability to maintain fluid balance and blood pressure.
A series of compensatory phenomena will take place in Mio’s body to adjust for a high-salt diet, influencing various physiological parameters. Here’s how Mio’s body is predicted to react:
With the final result, a healthy individual’s blood pressure will increase.Next, water will enter the principal cells via the aquaporin-3 channels.As a result, blood volume will increase.The net movement of water from the filtrate into the principal cells will then increase.Insertion of aquaporin-2 channels in the apical membrane of the principal cells in distal convoluted tubule and collecting duct will increase their permeability to water.Hence, an increased volume of water will enter the peritubular fluid and capillary.These events highlight the body’s capability to regulate fluid balance and blood pressure in response to dietary changes, such as a high intake of salt. The increase in blood volume and pressure is a direct response to the osmotic effect of the higher salt concentration, initiating a series of adaptative mechanisms to restore homeostasis.
DNA can be found in all cells EXCEPT which of the following?
Answer:
mature red blood cells/erythrocytes
Explanation:
where does photosynthesis occur in the biosphere
Answer:
In producers
Explanation:
How can the atmosphere be considered part of the hydrosphere?
it is a source of water
it blocks Ultraviolet rays from the sun
it contains oxygen necessary for life on earth *******
it traps pollutants that would otherwise harm the earth
7th grade science
Answer:
It is a source of water
Explanation:
water vapor could be considered part of the atmosphere as well as part of the hydrosphere: Water vapor is considered part of the hydrosphere because it's water.
The atmosphere should be considered part of the hydrosphere because it is the source of water.
What is water vapor?It refers to the gaseous phase of water. It should be considered as the one state of the water that lies within the hydrosphere. It could be generated from the evaporation or boiling the liquid water. Moreover, the water vapor should be the part of both the hydrosphere and atmosphere. Also it should be the part of hydrosphere since it is the water.
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