which graph best represents disruptive selection

Answer:

Secondary

Explanation:

Answer:

It is secondary succession.i had that question. hope it help and good luck ;)

Answer:

e. meiosis → gametes → zygote → mitosis → embryo

Explanation:

In the vertebrates, the life cycle alternates between the diploid and haploid phase. The vertebrate body is diploid and the haploid phase is only observed during the gamete formation.

The sex organs of the vertebrates produce gametes from the gamete mother cell through meiosis, which reduces the chromosome number to half and produce four haploid gametes.

The male and female gametes undergo fertilization event and form a diploid zygote. The zygote develops into the embryo through mitosis as mitosis produces the cells with an equal number of chromosomes.

Thus, Option-E is correct.

Answer:

shcdcg xsetr

Explanation:

xsdthfcq

Answer:

Correct options: Hair, Skin, and Nails.

Explanation:

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.

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.

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.

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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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.

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:)

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

Answer:

In producers

Explanation:

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.

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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Answer:

Trees & Shrubs

Explanation:

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.

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

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.

Answer:

the answer is useing energy

Explanation:

the answer is useing energy bc a lion is a natural enemy of zebras. lions are very fast running animals so by running at extremely high speed,they preform hunting of zebra. by doing so,they utilize energy in getting their prey which is there meal. hope this helps!!:)

A lion eating a zebra exemplifies the characteristics of life, including nutrition, growth, response to stimuli, reproduction, homeostasis, and the role of evolution in adaptation to its environment.

The act of a lion eating a zebra exemplifies several key characteristics of life, primarily pertaining to the lion as an organism:

1. Nutrition: The lion obtains nourishment by consuming the zebra, demonstrating its ability to ingest and digest food.

2. Growth and Development: The lion's body will process the zebra's nutrients, contributing to its own growth and maintenance, a fundamental aspect of life.

3. Response to Stimuli: The lion's predatory behavior, hunting, and eating are responses to stimuli such as hunger and the presence of prey.

4. Reproduction: While not directly observed in this scenario, the lion's capacity to reproduce is a vital aspect of its life cycle.

5. Homeostasis: After eating, the lion must maintain internal balance (homeostasis) in terms of temperature, blood sugar, and other physiological factors.

6. Evolution: Over generations, the lion's predatory traits and behaviors may evolve in response to environmental pressures and changes in prey availability, highlighting the role of evolution in life.

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The complete question is:

A lion eating a zebra is an example of which characteristics of life what?

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.

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%.

Answer:

In an antibiogram, the larger the inhibition halo, the more sensitive bacteria will be to the drug that is exposed, that is why if the halo is 25 mm, it is considered quite wide, so these drugs such as chloramphenicol and amoxicillin affect the bacterial development in marine culture both bactericidal or bacteriostatic.

Both have a 25mm halo, so both would be SAME as specified.

Explication:

suppose that one of the two drugs has a smaller inhibition halo in the culture, that drug will be less effective in treating this bacterium, therefore more would allow its development, on the contrary, the larger halo would be the most effective drug (This is an example for you to understand that the greater the length of the inhibition halo, the greater the efficacy of the drug, and the lower the halo, the less efficacy)

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

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.

Answer:chase the engine

Because of the engine

Explanation:

Answer:The two systems work together in several ways. First, the respiratory system brings oxygen in the air to the alveoli in the lungs. The circulatory system then delivers the oxygen to the cells. Second, the circulatory system picks up carbon dioxide from the cells and carries it to the lungs, where it is released when we exhale.

Explanation:its the sample writing.

1. With the final result, a healthy individual 19s blood pressure will increase.

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:

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.

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:

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.

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.

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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Answer:

mature red blood cells/erythrocytes

Explanation: